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+\section{Dimension of Proper Subspace is Less Than its Superspace}
+Tags: Dimension of Proper Subspace
+
+\begin{theorem}
+Let $G$ be a [[Definition:Vector Space|vector space]] whose [[Definition:Dimension of Vector Space|dimension]] is $n$.
+Let $H$ be a [[Definition:Vector Subspace|subspace]] of $G$.
+Then $H$ is [[Definition:Finite Dimensional Vector Space|finite dimensional]], and $\map \dim H \le \map \dim G$.
+If $H$ is a [[Definition:Proper Vector Subspace|proper subspace]] of $G$, then $\map \dim H < \map \dim G$.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Vector Subspace|subspace]] of $G$.
+Every [[Definition:Linearly Independent Set|linearly independent subset]] of the [[Definition:Vector Space|vector space]] $H$ is a [[Definition:Linearly Independent Set|linearly independent subset]] of the [[Definition:Vector Space|vector space]] $G$.
+Therefore, it has no more than $n$ elements by [[Size of Linearly Independent Subset is at Most Size of Finite Generator]].
+So the set of all [[Definition:Natural Numbers|natural numbers]] $k$ such that $H$ has a [[Definition:Linearly Independent Set|linearly independent subset]] of $k$ [[Definition:Vector (Linear Algebra)|vectors]] has a largest element $m$, and $m \le n$.
+Now, let $B$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $H$ having $m$ [[Definition:Vector (Linear Algebra)|vectors]].
+If the subspace [[Definition:Generator of Vector Space|generated]] by $B$ were not $H$, then $H$ would contain a [[Definition:Linearly Independent Set|linearly independent subset]] of $m + 1$ [[Definition:Vector (Linear Algebra)|vectors]].
+This follows by [[Linearly Independent Subset also Independent in Generated Subspace]].
+This would contradict the definition of $m$.
+Hence $B$ is a [[Definition:Generator of Vector Space|generator]] for $H$ and is thus a [[Definition:Basis of Vector Space|basis]] for $H$.
+Thus $H$ is [[Definition:Finite Dimensional Vector Space|finite dimensional]] and $\map \dim H \le \map \dim G$.
+Now, if $\map \dim H = \map \dim G$, then a basis of $H$ is a [[Definition:Basis of Vector Space|basis]] of $G$ by [[Sufficient Conditions for Basis of Finite Dimensional Vector Space]], and therefore $H = G$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Results concerning Generators and Bases of Vector Spaces}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $E$ be a [[Definition:Vector Space|vector space]] of $n$ [[Definition:Dimension of Vector Space|dimensions]].
+Let $G$ be a [[Definition:Generator of Module|generator]] for $E$.
+Then $G$ has the following properties:
+\end{theorem}<|endoftext|>
+\section{Grassmann's Identity}
+Tags: Linear Algebra, Grassmann's Identity
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $M$ and $N$ be [[Definition:Finite Dimensional Vector Space|finite-dimensional]] [[Definition:Vector Subspace|subspaces]] of $G$.
+Then the [[Definition:Sum of Vector Subspaces|sum]] $M + N$ and [[Definition:Intersection|intersection]] $M \cap N$ are [[Definition:Finite Dimensional Vector Space|finite-dimensional]], and:
+:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$
+\end{theorem}
+
+\begin{proof}
+By the [[Second Isomorphism Theorem/Vector Spaces|second isomorphism theorem]]:
+:$\dfrac {M + N} M \equiv \dfrac N {M \cap N}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rank Plus Nullity Theorem}
+Tags: Linear Algebra, Linear Transformations, Named Theorems
+
+\begin{theorem}
+Let $G$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space]].
+Let $H$ be a [[Definition:Vector Space|vector space]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
+Let $\map \rho \phi$ and $\map \nu \phi$ be the [[Definition:Rank of Linear Transformation|rank]] and [[Definition:Nullity of Linear Transformation|nullity]] respectively of $\phi$.
+Then the [[Definition:Image of Mapping|image]] of $\phi$ is [[Definition:Finite Dimensional Vector Space|finite-dimensional]], and:
+:$\map \rho \phi + \map \nu \phi = n$
+By definition of [[Definition:Rank of Linear Transformation|rank]] and [[Definition:Nullity of Linear Transformation|nullity]], it can be seen that this is equivalent to the alternative way of stating this result:
+:$\map \dim {\Img \phi} + \map \dim {\map \ker \phi} = \map \dim G$
+{{wtd|and the theorem is applicable to matrices}}
+\end{theorem}
+
+\begin{proof}
+If $\phi = 0$ then the assertion is clear.
+Let $\phi$ be a non-zero [[Definition:Linear Transformation on Vector Space|linear transformation]].
+By [[Dimension of Proper Subspace is Less Than its Superspace]] and [[Generator of Vector Space Contains Basis]], there is an [[Definition:Ordered Basis|ordered basis]] $\sequence {a_n}$ of $G$ such that:
+:$\exists r \in \N_n: \set {a_k: r + 1 \le k \le n}$ is a basis of $\map \ker \phi$
+As a consequence:
+:$\map \nu \phi = n - r$
+and by [[Unique Linear Transformation Between Vector Spaces]]:
+:$\map \rho \phi = r$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation of Vector Space Monomorphism}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ and $H$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
+Then $\phi$ is a [[Definition:Vector Space Monomorphism|monomorphism]] {{iff}} for every [[Definition:Linearly Independent Sequence|linearly independent sequence]] $\sequence {a_n}$ of [[Definition:Vector (Linear Algebra)|vectors]] of $G$, $\sequence {\map \phi {a_n} }$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]] of [[Definition:Vector (Linear Algebra)|vectors]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\phi$ is a [[Definition:Vector Space Monomorphism|monomorphism]].
+Let $\sequence {a_n}$ be a [[Definition:Linearly Independent Sequence|linearly independent sequence]].
+Let:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k} = 0$
+Then:
+:$\displaystyle \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k} = 0$
+So by hypothesis:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = 0$
+Hence:
+:$\forall k \in \closedint 1 n: \lambda_k = 0$
+Suppose that for every [[Definition:Linearly Independent Sequence|linearly independent sequence]] $\sequence {a_n}$ of vectors of $G$, $\sequence {\map \phi {a_n} }$ is a linearly independent sequence of vectors of $H$.
+Let $\map \phi {a_1} = 0$.
+Then $a_1 = 0$, otherwise the sequence $\sequence {a_1}$ of one term would be linearly independent but $\sequence {\map \phi {a_1} }$ would not.
+Thus $\map \ker \phi = \set 0$ and by the [[Quotient Theorem for Group Epimorphisms]] $\phi$ is an [[Definition:Vector Space Isomorphism|isomorphism]] and therefore a [[Definition:Vector Space Monomorphism|monomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation of Vector Space Equivalent Statements}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Vector Space|vector spaces]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
+Then these statements are equivalent:
+: $(1): \quad \phi$ is an [[Definition:Vector Space Isomorphism|isomorphism]].
+: $(2): \quad \phi$ is a [[Definition:Vector Space Monomorphism|monomorphism]].
+: $(3): \quad \phi$ is an [[Definition:Vector Space Epimorphism|epimorphism]].
+: $(4): \quad \phi \left({B}\right)$ is a basis of $H$ for every [[Definition:Basis (Linear Algebra)|basis]] $B$ of $G$.
+: $(5): \quad \phi \left({B}\right)$ is a basis of $H$ for some [[Definition:Basis (Linear Algebra)|basis]] $B$ of $G$.
+\end{theorem}
+
+\begin{proof}
+* $(1)$ implies $(2)$ by definition.
+* $(2)$ implies $(4)$ by [[Linear Transformation of Vector Space Monomorphism]] and [[Results concerning Generators and Bases of Vector Spaces]].
+* $(4)$ implies $(5)$ by basic logic.
+* Suppose $\phi \left({B}\right)$ is a [[Definition:Basis (Linear Algebra)|basis]] of $H$.
+Then the [[Definition:Image of Mapping|image]] of $\phi$ is a [[Definition:Vector Subspace|subspace]] of $H$ [[Definition:Generator|generating]] $H$ and hence is $H$ itself.
+Thus $(5)$ implies $(3)$.
+* Finally, $(3)$ implies that $\phi$ is [[Definition:Injection|injective]].
+If $\phi$ is [[Definition:Surjection|surjective]], the [[Definition:Dimension (Linear Algebra)|dimension]] of its [[Definition:Kernel of Linear Transformation|kernel]] is $0$ by [[Rank Plus Nullity Theorem]].
+Hence $\phi$ is an [[Definition:Vector Space Isomorphism|isomorphism]] and therefore $(3)$ implies $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Results Concerning Annihilator of Vector Subspace}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]].
+Let $J: G \to G^{**}$ be the [[Definition:Evaluation Isomorphism|evaluation isomorphism]].
+Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
+Let $G^{**}$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G^*$.
+Let $M$ be an [[Definition:Dimension of Vector Space|$m$-dimensional]] [[Definition:Vector Subspace|subspace]] of $G$.
+Let $N$ be a [[Definition:Dimension of Vector Space|$p$-dimensional]] [[Definition:Vector Subspace|subspace]] of $G^*$.
+Let $M^\circ$ be the [[Definition:Annihilator on Algebraic Dual|annihilator]] of $M$.
+Then:
+:$(1): \quad M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$, and $M^{\circ \circ} = \map J M$
+:$(2): \map {\quad J^{-1} } {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$
+:$(3): \quad$ The [[Definition:Mapping|mapping]] $M \to M^\circ$ is a [[Definition:Bijection|bijection]] from the set of all $m$-dimensional subspaces of $G$ onto the set of all $\paren {n - m}$-dimensional subspaces of $G^*$
+:$(4): \quad$ Its inverse is the bijection $N \to \map {J^{-1} } {N^\circ}$.
+\end{theorem}
+
+\begin{proof}
+:$(1): \quad M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$, and $M^{\circ \circ} = \map J M$
+Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$ such that $\sequence {a_m}$ is an ordered basis of $M$.
+Let $\sequence {a'_n}$ be the [[Definition:Ordered Dual Basis|ordered dual basis]] of $G^*$.
+Let $\ds t' = \sum_{k \mathop = 1}^n \lambda_k a'_k \in M^\circ$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall j \in \closedint 1 m:
+      | l = \lambda_j
+      | r = \sum_{k \mathop = 1}^n \lambda_k a'_k \paren {a_j}
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{k \mathop = 1}^n \lambda_k a'_k} \paren {a_j}
+      | c = 
+}}
+{{eqn | r = t' \paren {a_j}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $t'$ is a [[Definition:Linear Combination|linear combination]] of $\set {a'_k: m + 1 \le k \le n}$.
+But $a'_k$ clearly belongs to $M^\circ$ for each $k \in \closedint {m + 1} n$.
+Therefore $M^\circ$ has dimension $n - m$.
+{{qed|lemma}}
+When we apply this result to $M^\circ$ instead of $M$, it is seen that the [[Definition:Annihilator on Algebraic Dual|annihilator]] $M^{\circ \circ}$ of $M^\circ$ has dimension $n - \paren {n - m} = m$.
+But clearly $\map J M \subseteq M^{\circ \circ}$.
+As $J$ is an [[Definition:Vector Space Isomorphism|isomorphism]], $\map J M$ has dimension $m$.
+So by [[Dimension of Proper Subspace is Less Than its Superspace]], $\map J M = M^{\circ \circ}$.
+As a consequence, $\map {J^{-1} } {M^{\circ \circ} } = M$.
+Hence the result: $M^{\circ \circ} = \map J M$
+{{qed|lemma}}
+:$(2) \quad \map {J^{-1} } {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$
+If $N$ is a $p$-dimensional subspace of $G$, then $N^\circ$ and hence also $\map {J^{-1} } {N^\circ}$ have dimension $n - p$ by what has just been proved.
+{{qed|lemma}}
+By definition: $\paren {\map {J^{-1} } {N^\circ} }^\circ = \set {z' \in G: \forall x \in G: \forall t' \in N: \map {t'} x = 0: \map {z'} x = 0}$
+Thus $N \subseteq \paren {\map {J^{-1} } {N^\circ} }^\circ$.
+But as $\paren {\map {J^{-1} } {N^\circ} }^\circ$ has dimension $n - \paren {n - p} = p$, it follows that $N = \paren {\map {J^{-1} } {N^\circ} }^\circ$ by [[Dimension of Proper Subspace is Less Than its Superspace]].
+{{explain|Where in the above assertion is $(3)$ proved?}}
+:$(4) \quad$ Its inverse is the bijection $N \to \map {J^{-1} } {N^\circ}$.
+The final assertion follows by the definition of an [[Definition:Inverse Mapping/Definition 2|inverse mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Subspace Dimension One Less}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $M$ be a [[Definition:Vector Subspace|subspace]] of the [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space $K^n$]].
+The following statements are equivalent:
+:$(1): \quad \map \dim M = n - 1$
+:$(2): \quad M$ is the [[Definition:Kernel of Linear Transformation|kernel]] of a nonzero [[Definition:Linear Form|linear form]]
+:$(3): \quad$ There exists a [[Definition:Sequence|sequence]] $\sequence {\alpha_n} $ of [[Definition:Scalar (Vector Space)|scalars]], not all of which are zero, such that:
+:::$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$
+Also, suppose the above hold.
+Let $\sequence {\beta_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Scalar (Vector Space)|scalars]] such that:
+:$M = \set {\tuple {\lambda_1, \ldots, \lambda_n} \in K^n: \beta_1 \lambda_1 + \cdots + \beta_n \lambda_n = 0}$
+Then there is a non-zero [[Definition:Scalar (Vector Space)|scalar]] $\gamma$ such that:
+:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
+\end{theorem}
+
+\begin{proof}
+Let $M^\circ$ be the [[Definition:Annihilator on Algebraic Dual|annihilator]] of $M$.
+Let $N = M^{\circ}$.
+By [[Results Concerning Annihilator of Vector Subspace]], $N$ is [[Definition:Dimension of Vector Space|one-dimensional]] and $M = \map {J^{-1} } {N^\circ}$.
+Let $\phi \in N: \phi \ne 0$.
+Then $N$ is the set of all [[Definition:Scalar Multiplication on Vector Space|scalar multiples]] of $\phi$.
+Because:
+:$\map {J^{-1} } {N^\circ} = \set {x \in K^n: \forall \psi \in N: \map \psi x = 0}$
+it follows that $\map {J^{-1} } {N^\circ}$ is simply the [[Definition:Kernel of Linear Transformation|kernel]] of $\phi$.
+Hence $(1)$ implies $(2)$.
+By [[Rank Plus Nullity Theorem]], $(2)$ also implies $(1)$.
+{{qed|lemma}}
+Suppose $\sequence {\alpha_n}$ is any sequence of [[Definition:Scalar (Vector Space)|scalars]].
+Let $\sequence {e'_n}$ be the [[Definition:Ordered Basis|ordered basis]] of $\paren {K^n}^*$ [[Definition:Ordered Dual Basis|dual to]] the [[Definition:Standard Ordered Basis|standard ordered basis]] of $K^n$.
+Let $\displaystyle \phi = \sum_{k \mathop = 1}^n \alpha_k e'_k$.
+Then, by simple calculation:
+:$\map \ker \phi = \set {\tuple {\lambda_1, \ldots, \lambda_n}: \alpha_1 \lambda_1 + \cdots + \alpha_n \lambda_n = 0}$
+{{explain|Prove the above.}}
+It follows that:
+:$\phi \ne 0 \iff \exists k \in \closedint 1 n: \alpha_k \ne 0$
+Thus $(2)$ and $(3)$ are equivalent.
+{{qed|lemma}}
+Suppose $M = \map \ker \psi$, where $\displaystyle \psi = \sum_{k \mathop = 1}^n \beta_k e'_k$.
+Then $\psi = M^\circ$.
+As $M^\circ$ is one-dimensional and since $\psi \ne 0$, it follows that:
+:$\exists \gamma \ne 0: \psi = \gamma \phi$
+Therefore:
+:$\forall k \in \closedint 1 n: \beta_k = \gamma \alpha_k$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rank and Nullity of Transpose}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector spaces]] over a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathcal L \left({G, H}\right)$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Let $u \in \mathcal L \left({G, H}\right)$.
+Let $u^t$ be the [[Definition:Transpose of Linear Transformation|transpose]] of $u$.
+Then:
+: $(1): \quad$ $u$ and $u^t$ have the same [[Definition:Rank of Linear Transformation|rank]] and [[Definition:Nullity of Linear Transformation|nullity]]
+: $(2): \quad$ $\ker \left({u^t}\right)$ is the [[Definition:Annihilator on Algebraic Dual|annihilator]] of the [[Definition:Image of Mapping|image]] of $u$
+: $(3): \quad$ The [[Definition:Image of Mapping|image]] of $u^t$ is the [[Definition:Annihilator on Algebraic Dual|annihilator]] of $\ker \left({u}\right)$.
+\end{theorem}
+
+\begin{proof}
+From the definitions of the [[Definition:Transpose of Linear Transformation|transpose]] $u^t$ and the [[Definition:Annihilator|annihilator]] $\left({u \left({G}\right)}\right)^\circ$, it follows that:
+:$u^t \left({y'}\right) = 0 \iff y' = \left({u \left({G}\right)}\right)^\circ$
+Thus:
+:$\ker \left({u^t}\right) = \left({u \left({G}\right)}\right)^\circ$.
+Let $x \in \ker \left({u}\right)$.
+Let $H^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $H$.
+Let $\left \langle {x, t'} \right \rangle$ be the [[Definition:Evaluation Linear Transformation|evaluation linear transformation]].
+Then:
+:$\forall y' \in H^*: \left \langle {x, u^t \left({y'}\right)} \right \rangle = \left \langle {u \left({x}\right), y'} \right \rangle = \left \langle {0, y'} \right \rangle = 0$
+So:
+:$u^t \left({H^*}\right) \subseteq \left({\ker \left({u}\right)}\right)^\circ$
+From [[Rank Plus Nullity Theorem]] and [[Results Concerning Annihilator of Vector Subspace]]:
+{{begin-eqn}}
+{{eqn | l = \dim \left({u^t \left({H^*}\right)}\right)
+      | r = n - \dim \left({\ker \left({u^t}\right)}\right)
+      | c = 
+}}
+{{eqn | r = n - \dim \left({\left({u \left({G}\right)}\right)^\circ}\right)
+      | c = 
+}}
+{{eqn | r = \dim \left({u \left({G}\right)}\right)
+      | c = 
+}}
+{{eqn | r = n - \dim \left({\ker \left({u}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \dim \left({\left({\ker \left({u}\right)}\right)^\circ}\right)
+      | c = 
+}}
+{{end-eqn}}
+So it follows that $u$ and $u^t$ have the same [[Definition:Rank of Linear Transformation|rank]] and [[Definition:Nullity of Linear Transformation|nullity]], and that:
+:$u^t \left({H^*}\right) = \left({\ker \left({u}\right)}\right)^\circ$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Linear Operator on the Plane}
+Tags: Linear Algebra, Analytic Geometry
+
+\begin{theorem}
+Let $\phi$ be a [[Definition:Linear Operator|linear operator]] on the [[Definition:Real Vector Space|real vector space]] of [[Definition:Dimension of Vector Space|two dimensions]] $\R^2$.
+Then $\phi$ is completely determined by an [[Definition:Ordered Tuple|ordered tuple]] of $4$ [[Definition:Real Number|real numbers]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi$ be a [[Definition:Linear Operator|linear operator]] on $\R^2$.
+Let $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ be the [[Definition:Real Number|real numbers]] which satisfy the equations:
+{{begin-eqn}}
+{{eqn | l = \phi \left({e_1}\right)
+      | r = \alpha_{11} e_1 + \alpha_{21} e_2
+      | c = 
+}}
+{{eqn | l = \phi \left({e_2}\right)
+      | r = \alpha_{12} e_1 + \alpha_{22} e_2
+      | c = 
+}}
+{{end-eqn}}
+where $\left({e_1, e_2}\right)$ is the [[Definition:Standard Ordered Basis|standard ordered basis]] of $\R^2$.
+Then, by linearity:
+{{begin-eqn}}
+{{eqn | l = \phi \left({\lambda_1, \lambda_2}\right)
+      | r = \phi \left({\lambda_1 e_1 + \lambda_2 e_2}\right)
+      | c = 
+}}
+{{eqn | r = \lambda_1 \phi \left({e_1}\right) + \lambda_2 \phi \left({e_2}\right)
+      | c = 
+}}
+{{eqn | r = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12} }\right) e_1 + \left({\lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) e_2
+      | c = 
+}}
+{{eqn | r = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right)
+      | c = 
+}}
+{{end-eqn}}
+Conversely, if $\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22} \in \R$ are ''any'' real numbers, then we can define the mapping $\phi$ as:
+: $\phi \left({\lambda_1, \lambda_2}\right) = \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22}}\right)$
+which is easily verified as being a [[Definition:Linear Operator|linear operator]] on $\R^2$:
+{{begin-eqn}}
+{{eqn | l = b \cdot \phi \left({\lambda_1, \lambda_2}\right) + c \cdot \phi \left({\lambda_3, \lambda_4}\right)
+      | r = b \left({\lambda_1 \alpha_{11} + \lambda_2 \alpha_{12}, \lambda_1 \alpha_{21} + \lambda_2 \alpha_{22} }\right) + c \left({\lambda_3 \alpha_{11} + \lambda_4 \alpha_{12}, \lambda_3 \alpha_{21} + \lambda_4 \alpha_{22} }\right)
+      | c = 
+}}
+{{eqn | r = \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} }\right) + \left({c \lambda_3 \alpha_{11} + c\lambda_4 \alpha_{12}, c\lambda_3 \alpha_{21} + c\lambda_4 \alpha_{22} }\right)
+      | c = 
+}}
+{{eqn | r = \left({b \lambda_1 \alpha_{11} + b \lambda_2 \alpha_{12} + c \lambda_3 \alpha_{11} + c \lambda_4 \alpha_{12}, b \lambda_1 \alpha_{21} + b \lambda_2 \alpha_{22} + c \lambda_3 \alpha_{21} + c \lambda_4 \alpha_{22} }\right)
+      | c = 
+}}
+{{eqn | r = \left({\left({b \lambda_1 + c \lambda_3}\right) \alpha_{11} + \left({b \lambda_2 + c \lambda_4}\right) \alpha_{12}, \left({b \lambda_1 + c \lambda_3}\right) \alpha_{21} + \left({c \lambda_2 + c \lambda_4}\right) \alpha_{22} }\right)
+      | c = 
+}}
+{{eqn | r = \left({b \lambda_1 + c \lambda_3, b \lambda_2 + c \lambda_4}\right)
+      | c = 
+}}
+{{end-eqn}}
+Thus, by [[Condition for Linear Transformation]], $\phi$ is a linear operator on $\R^2$.
+Thus each linear operator on $\R^2$ is completely determined by the [[Definition:Ordered Tuple|ordered tuple]]:
+:$\left({\alpha_{11}, \alpha_{12}, \alpha_{21}, \alpha_{22}}\right)$
+of [[Definition:Real Number|real numbers]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Similarity Mapping of Plane is Linear Operator}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\beta \in K$.
+Then the [[Definition:Mapping|mapping]]:
+:$s_\beta: G \to G$ defined by $\map {s_\beta} {\mathbf x} = \beta \mathbf x$
+is a [[Definition:Linear Operator|linear operator]] on $G$.
+If $\beta \ne 0$ then $s_\beta$ is an [[Definition:Vector Space Automorphism|automorphism]] of $G$, and $\paren {s_\beta}^{-1} = s_{\beta^{-1} }$
+The linear operators $s_\beta$, where $\beta \ne 0$, are called '''similarities of $G$'''.
+\end{theorem}
+
+\begin{proof}
+Since $\map \beta {\mathbf x + \mathbf y} = \beta \mathbf x + \beta \mathbf y$ and $\map \beta {\lambda \mathbf x} = \map \lambda {\beta \mathbf x}$, the fact of $s_\beta$ being a linear operator is immediately apparent.
+We have:
+:$\map {\paren {s_{\beta^{-1} } \circ s_\beta} } {\mathbf x} = \map {\beta^{-1} } {\beta \mathbf x} = \mathbf x = \map \beta {\beta^{-1} \mathbf x} = \map {\paren {s_\beta \circ s_{\beta^{-1} } } } {\mathbf x}$
+which proves the second bit.
+\end{proof}<|endoftext|>
+\section{Cantor-Dedekind Hypothesis}
+Tags: Analytic Geometry, Euclidean Geometry
+
+\begin{theorem}
+The [[Definition:Point|points]] on an [[Definition:Infinite Straight Line|infinite straight line]] are in [[Definition:Bijection|one-to-one correspondence]] with the set $\R$ of [[Definition:Real Number|real numbers]].
+Hence the [[Definition:Set|set]] of all [[Definition:Point|points]] on an [[Definition:Infinite Straight Line|infinite straight line]] and $\R$ are [[Definition:Set Equivalence|equinumerous]].
+\end{theorem}
+
+\begin{proof}
+=== Step 1 ===
+We will show that there exists a [[Definition:Mapping|mapping]] from the [[Definition:Infinite Straight Line|infinite straight line]] $L$ to the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$.
+Let us establish a [[Definition:Relation|relation]] $h$ between [[Definition:Point|points]] on $L$ and elements of $\R$.
+We allow the [[Axiom:Axiom of Choice|Axiom of Choice]] to set up a [[Definition:Choice Function|choice function]] to allow the points of $L$ to be selected systematically.
+Pick any [[Definition:Point|point]] on $L$, and label it $z$. This point can be referred to informally as the [[Definition:Origin|origin]].
+Map $z$ to [[Definition:Zero (Number)|zero]], that is, by allowing $\tuple {z, 0} \in h$. 
+Using the [[Definition:Choice Function|choice function]] pick any other point on $L$.
+From [[Axiom:Euclid's First Postulate|Euclid's first postulate]], we can draw a [[Definition:Line Segment|line segment]] between the two points.
+If the second point is to the right of the origin, let its [[Definition:Linear Measure|length]] be positive.
+If the second point is to the left of the origin, let its length be negative.
+The existence of the [[Definition:Choice Function|choice function]] allows that this process can be done for any point on $L$
+Hence:
+:$\forall p \in L: \exists x \in \R: \tuple {p, x} \in h$
+That is, $h$ is [[Definition:Left-Total Relation|left-total]].
+The method of construction of $h$ is such that every point on $L$ is assigned to exactly one element of $\R$.
+Thus it is seen that $h$ is [[Definition:Many-to-One Relation|many-to-one]].
+Hence, by definition, $h$ is a [[Definition:Mapping|mapping]].
+Since every point mapped to is associated to exactly one element of $L$, $h$ is [[Definition:Injection|injective]].
+{{qed|lemma}}
+=== Step 2 ===
+Now we need to demonstrate that there exists a [[Definition:Mapping|mapping]] from $\R$ to $L$.
+Let us establish a [[Definition:Relation|relation]] $h$ between elements of $\R$ and [[Definition:Point|points]] on $L$.
+We allow the [[Axiom:Axiom of Choice|Axiom of Choice]] to set up a [[Definition:Choice Function|choice function]] to allow the elements of $\R$ to be selected systematically.
+Pick any [[Definition:Point|point]] on $L$, and label it $z$.
+Map [[Definition:Zero (Number)|zero]] to $z$ that is, by allowing $\tuple {0, z} \in g$. 
+Using the [[Definition:Choice Function|choice function]] pick any other element $x$ of $\R$.
+Suppose $x$ is positive. 
+Then let the [[Definition:Image of Element under Mapping|image]] of $x$ be a point to the right of the origin. Define the magnitude of its [[Definition:Displacement|displacement]] be the [[Definition:Absolute Value|absolute value]] of $x$.
+Suppose $x$ is negative. 
+Then let the [[Definition:Image of Element under Mapping|image]] of $x$ be a point to the left of the origin. Define the magnitude of its [[Definition:Displacement|displacement]] be the [[Definition:Absolute Value|absolute value]] of $x$ multiplied by $-1$.
+The union of all such images, with the [[Definition:Zero Vector|zero vector]] from $z$ to $z$, is the [[Definition:Infinite Straight Line|infinite straight line]].
+By the method of construction, there exists a point on $L$ for all elements of $\R$, the relation $g$ is [[Definition:Left-Total Relation|left-total]].
+The method of construction of $g$ is also such that every element of $\R$ is assigned to exactly one point on $L$.
+Thus it is seen that $g$ is [[Definition:Many-to-One Relation|many-to-one]].
+Hence, by definition, $g$ is a [[Definition:Mapping|mapping]].
+Since every element of $L$ mapped to is associated to exactly one point on $\R$, $g$ is [[Definition:Injection|injective]].
+{{qed|lemma}}
+=== Step 3 ===
+Since:
+:$h: L \hookrightarrow \R$ is an [[Definition:Injection|injection]] 
+:$g: \R \hookrightarrow L$ is an [[Definition:Injection|injection]]
+by the [[Cantor-Bernstein-Schröder Theorem]], there exists a [[Definition:Bijection|one-to-one]] mapping between them.
+Hence by definition, $L$ and $\R$ are [[Definition:Set Equivalence|equinumerous]].
+{{qed}}{{proofread}}
+{{explain|From a careful study of the discussion in the [[Talk:Real Number Line|talk page]], I suspect that we may need to assume and invoke an axiom of {{AuthorRef|Alfred Tarski}}'s.}}
+{{AoC}}
+\end{proof}<|endoftext|>
+\section{Full Angle measures 2 Pi Radians}
+Tags: Euclidean Geometry
+
+\begin{theorem}
+One [[Definition:Full Angle|full angle]] is equal to $2 \pi$ [[Definition:Radian|radians]].
+:$2 \pi \approx 6 \cdotp 28318 \, 53071 \, 79586 \, 4769 \ldots$
+{{OEIS|A019692}}
+\end{theorem}
+
+\begin{proof}
+By definition, $1$ '''[[Definition:Radian|radian]]''' is the [[Definition:Angle|angle]] which sweeps out an [[Definition:Arc of Circle|arc]] on a [[Definition:Circle|circle]] whose [[Definition:Length (Linear Measure)|length]] is the [[Definition:Radius of Circle|radius]] $r$ of the [[Definition:Circle|circle]].
+From [[Perimeter of Circle]], the [[Definition:Length (Linear Measure)|length]] of the [[Definition:Circumference of Circle|circumference]] of a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $r$ is equal to $2 \pi r$.
+Therefore, $1$ [[Definition:Radian|radian]] sweeps out $\dfrac 1 {2 \pi}$ of a [[Definition:Circle|circle]].
+It follows that $2 \pi$ [[Definition:Radian|radians]] sweeps out the entire [[Definition:Circle|circle]], or one [[Definition:Full Angle|full angle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rotation of Plane about Origin is Linear Operator}
+Tags: Euclidean Geometry, Analytic Geometry, Geometric Rotations
+
+\begin{theorem}
+Let $r_\alpha$ be the [[Definition:Plane Rotation|plane rotation]] of [[Definition:The Plane|the plane]] about the [[Definition:Origin|origin]] through an [[Definition:Angle|angle]] of $\alpha$.
+That is, let $r_\alpha: \R^2 \to \R^2$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \R^2: \map {r_\alpha} x = \text { the point into which a rotation of } \alpha \text{ carries } x$
+Then $r_\alpha$ is a [[Definition:Linear Operator|linear operator]] determined by the [[Definition:Ordered Tuple|ordered sequence]]:
+: $\tuple {\cos \alpha  -\sin \alpha, \sin \alpha + \cos \alpha}$
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {\lambda_1, \lambda_2} = \tuple {\rho \cos \sigma, \rho \sin \sigma}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = r_\alpha \left({\lambda_1, \lambda_2}\right)
+      | r = \tuple {\rho \cos \alpha \cos \sigma - \rho \sin \alpha \sin \sigma, \rho \sin \alpha \cos \sigma + \rho \cos \alpha \sin \sigma}
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda_1 \cos \alpha - \lambda_2 \sin \alpha, \lambda_1 \sin \alpha + \lambda_2 \cos \alpha}
+      | c = 
+}}
+{{end-eqn}}
+The result follows from [[Linear Operator on the Plane]].
+{{ProofWanted|This definition requires to be approached from several conceptual directions.}}
+\end{proof}<|endoftext|>
+\section{Stretching and Contraction Mappings of Plane are Linear Operators}
+Tags: Linear Algebra, Euclidean Geometry, Analytic Geometry
+
+\begin{theorem}
+Let $s_\beta: \R^2 \to \R^2$ be a [[Similarity Mapping of Plane is Linear Operator|similarity]] of $\R^2$.
+Then $s_{-1}$ is the same as the [[Rotation of Plane about Origin is Linear Operator|rotation]] $r_{\pi}$ of [[Definition:The Plane|the plane]] about the [[Definition:Origin|origin]] one half turn.
+If $\beta \ge 1$, then $s_\beta$ is called a '''stretching''', and if $0 < \beta \le 1$, $s_\beta$ is called a '''contraction'''.
+If $\beta < 0$, then $s_\beta$ is a stretching or contraction followed by a [[Rotation of Plane about Origin is Linear Operator|rotation]] one half turn.
+It is also the same as a [[Rotation of Plane about Origin is Linear Operator|rotation]] one half turn followed by a stretching or contraction.
+\end{theorem}<|endoftext|>
+\section{Reflection of Plane in Line through Origin is Linear Operator}
+Tags: Linear Algebra, Euclidean Geometry, Analytic Geometry
+
+\begin{theorem}
+Let $M$ be a [[Definition:Infinite Line|straight line]] in [[Definition:The Plane|the plane]] passing through the [[Definition:Origin|origin]].
+Then the '''reflection''' $s_M$ of $\R^2$ in $M$ is the [[Rotation of Plane about Origin is Linear Operator|rotation]] of [[Definition:The Plane|the plane]] in [[Definition:Ordinary Space|space]] through one half turn about $M$ as an [[Definition:Axis|axis]].
+:$s_M \circ s_M = I_{\R^2}$
+and hence:
+:$s_M = s_M^{-1}$
+If $M$ is the [[Definition:X-Axis|$x$-axis]] then $\map {s_M} {\lambda_1, \lambda_2} = \tuple {\lambda_1, -\lambda_2}$.
+If $M$ is the [[Definition:Y-Axis|$y$-axis]] then $\map {s_M} {\lambda_1, \lambda_2} = \tuple {-\lambda_1, \lambda_2}$.
+In general, $s_M$ is a [[Definition:Linear Operator|linear operator]] for every [[Definition:Straight Line|straight line]] $M$ through the [[Definition:Origin|origin]].
+\end{theorem}<|endoftext|>
+\section{Projection of Straight Line on Another in Plane}
+Tags: Linear Algebra, Analytic Geometry
+
+\begin{theorem}
+Let $M$ and $N$ be distinct [[Definition:Straight Line|straight lines]] through [[Definition:The Plane|the plane]] through the [[Definition:Origin|origin]].
+Let $\operatorname{pr}_{M, N}$ be the [[Definition:Projection (Analytic Geometry)|projection on $M$ along $N$]].
+$M$ and $N$ are respectively the [[Definition:Codomain of Mapping|codomain]] and [[Definition:Kernel of Linear Transformation|kernel]] of $\operatorname{pr}_{M, N}$.
+{{explain|As the kernel is a concept defined in relation to a homomorphism, it needs to be clarified what homomorphism is being considered.}}
+:$\operatorname{pr}_{M, N} \left({x}\right) = x \iff x \in M$
+If $M$ is the [[Definition:X-Axis|$x$-axis]] and $N$ is the [[Definition:Y-Axis|$y$-axis]], then $\operatorname{pr}_{M, N} \left({\lambda_1, \lambda_2}\right) = \left({\lambda_1, 0}\right)$.
+If $M$ is the [[Definition:Y-Axis|$y$-axis]] and $N$ is the [[Definition:X-Axis|$x$-axis]], then $\operatorname{pr}_{M, N} \left({\lambda_1, \lambda_2}\right) = \left({0, \lambda_2}\right)$.
+Any such [[Definition:Projection (Analytic Geometry)|projection]] is a [[Definition:Linear Operator|linear operator]].
+\end{theorem}<|endoftext|>
+\section{Condition for Straight Lines in Plane to be Parallel}
+Tags: Analytic Geometry, Linear Algebra, Straight Lines
+
+\begin{theorem}
+Let $L: \alpha_1 x + \alpha_2 y = \beta$ be a [[Equation of Straight Line in Plane|straight line in $\R^2$]].
+Then the straight line $L'$ is [[Definition:Parallel Lines|parallel]] to $L$ iff there is a $\beta' \in \R^2$ such that:
+:$L' = \set {\tuple {x, y} \in \R^2: \alpha_1 x + \alpha_2 y = \beta'}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+When $L' = L$, the claim is trivial.
+Let $L' \ne L$ be [[Equation of Straight Line in Plane|described by]] the equation:
+:$\alpha'_1 x + \alpha'_2 y = \beta'$
+Without loss of generality, let $\alpha'_1 \ne 0$ (the case $\alpha'_2 \ne 0$ is similar).
+Then for $\tuple {x, y} \in L'$ to hold, one needs:
+{{begin-eqn}}
+{{eqn | l = \alpha'_1 x + \alpha'_2 y
+      | r = \beta'
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \frac {-\alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1}
+}}
+{{end-eqn}}
+For $L'$ to be [[Definition:Parallel Lines|parallel]] to $L$, it is required that then $\tuple {x, y} \notin L$, that is:
+{{begin-eqn}}
+{{eqn | l = \alpha_1 x + \alpha_2 y
+      | o = \ne
+      | r = \beta
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \alpha_1 \paren {\frac {- \alpha'_2} {\alpha'_1} y + \frac {\beta'} {\alpha'_1} } + \alpha_2 y
+      | o = \ne
+      | r = \beta
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y + \alpha_1 \frac {\beta'} {\alpha'_1}
+      | o = \ne
+      | r = \beta
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {\alpha_2 - \alpha_1 \frac {\alpha'_2} {\alpha'_1} } y
+      | o = \ne
+      | r = \beta - \alpha_1 \frac {\beta'} {\alpha'_1}
+}}
+{{end-eqn}}
+It follows that necessarily $\beta - \alpha_1 \frac {\beta'} {\alpha'_1} \ne 0$, or taking $y = 0$ would yield equality.
+The only remaining way to obtain the desired inequality for all $y$ is that:
+:$\alpha_2 - \alpha_1 \dfrac {\alpha'_2} {\alpha'_1} = 0$
+One observes that now $\alpha_1 = 0 \implies \alpha_2 = 0$.
+However, as $L: \alpha_1 x + \alpha_2 y = \beta$ is a [[Equation of Straight Line in Plane|straight line in $\R^2$]], it cannot be that $\alpha_1 = \alpha_2 = 0$.
+So $\alpha_1 \ne 0$, and one finds:
+:$\alpha'_2 = \dfrac {\alpha'_1} {\alpha_1} \alpha_2$
+Hence obtain:
+{{begin-eqn}}
+{{eqn | l = \alpha'_1 x + \alpha'_2 y
+      | r = \beta'
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \frac {\alpha'_1} {\alpha_1} \paren {\alpha_1 x + \alpha_2 y}
+      | r = \beta'
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \alpha_1 x + \alpha_2 y
+      | r = \beta' \frac {\alpha_1} {\alpha'_1}
+}}
+{{end-eqn}}
+That is, $L'$ is described by an equation of the required form.
+{{qed|lemma}}
+{{proofread}}
+=== Sufficient Condition ===
+Let $L' \ne L$ be a straight line given by the equation:
+:$\alpha_1 x + \alpha_2 y = \beta'$
+Suppose we have a point $\mathbf x = \tuple {x, y} \in L \cap L'$.
+Then, as $\mathbf x \in L$, it also satisfies:
+:$\alpha_1 x + \alpha_2 y = \beta$
+It follows that $\beta = \beta'$, so $L = L'$.
+This contradiction shows that $L \cap L' = \O$, that is, $L$ and $L'$ are [[Definition:Parallel Lines|parallel]].
+The remaining case is when $L' = L$. 
+By definition, $L$ is parallel to itself. 
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Plane}
+Tags: Linear Algebra, Solid Analytic Geometry
+
+\begin{theorem}
+A [[Definition:Plane|plane]] $P$ is the [[Definition:Set|set]] of all $\tuple {x, y, z} \in \R^3$, where:
+:$\alpha_1 x + \alpha_2 y + \alpha_3 z = \gamma$
+where $\alpha_1, \alpha_2, \alpha_3, \gamma \in \R$ are given, and not all of $\alpha_1, \alpha_2, \alpha_3$ are [[Definition:Zero (Number)|zero]].
+\end{theorem}<|endoftext|>
+\section{Condition for Planes to be Parallel}
+Tags: Linear Algebra, Solid Analytic Geometry
+
+\begin{theorem}
+Let $P: \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$ be a [[Equation of Plane|plane]] in $\R^3$.
+Then the plane $P'$ is [[Definition:Parallel Planes|parallel]] to $P$ iff there is a $\gamma' \in \R$ such that:
+:$P' = \left\{{ \left({x_1, x_2, x_3}\right) \in \R^3 : \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma' }\right\}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+{{ProofWanted}}
+=== Sufficient Condition ===
+Let $P' \ne P$ be a plane given by the equation:
+:$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma'$
+Suppose we have a point $\mathbf x = \left({x_1, x_2, x_3}\right) \in P \cap P'$.
+Then, as $\mathbf x \in P$, it also satisfies:
+:$\alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 = \gamma$
+It follows that $\gamma = \gamma'$, so $P = P'$.
+This contradiction shows that $P \cap P' = \varnothing$, i.e., $P$ and $P'$ are [[Definition:Parallel Planes|parallel]].
+The remaining case is when $P' = P$.
+By definition, $P$ is parallel to itself. 
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lines are Subspaces of Plane}
+Tags: Linear Algebra, Plane Analytic Geometry
+
+\begin{theorem}
+The [[Definition:Dimension of Vector Space|one-dimensional]] [[Definition:Vector Subspace|subspaces]] of $\R^2$ are precisely the [[Definition:Homogeneous (Analytic Geometry)|homogeneous lines]] of [[Definition:Plane Analytic Geometry|plane analytic geometry]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Vector Subspace Dimension One Less]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Planes are Subspaces of Space}
+Tags: Linear Algebra, Solid Analytic Geometry
+
+\begin{theorem}
+The [[Definition:Dimension of Vector Space|two-dimensional]] [[Definition:Vector Subspace|subspaces]] of $\R^3$ are precisely the [[Definition:Homogeneous (Analytic Geometry)|homogeneous planes]] of [[Definition:Solid Analytic Geometry|solid analytic geometry]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Vector Subspace Dimension One Less]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Space Semigroup under Hadamard Product}
+Tags: Hadamard Product
+
+\begin{theorem}
+Let $\map {\MM_S} {m, n}$ be the [[Definition:Matrix Space|matrix space]] over a [[Definition:Semigroup|semigroup]] $\struct {S, \cdot}$.
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\map {\MM_S} {m, n}, \circ}$, where $\circ$ is the [[Definition:Hadamard Product|Hadamard product]], is also a [[Definition:Semigroup|semigroup]].
+If $\struct {S, \cdot}$ is a [[Definition:Commutative Semigroup|commutative semigroup]] then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
+If $\struct {S, \cdot}$ is a [[Definition:Monoid|monoid]] then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
+\end{theorem}
+
+\begin{proof}
+$\struct {S, \cdot}$ is a [[Definition:Semigroup|semigroup]] and is therefore [[Definition:Closed Algebraic Structure|closed]] and [[Definition:Associative|associative]].
+As $\struct {S, \cdot}$ is [[Definition:Closed Algebraic Structure|closed]], then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from [[Closure of Hadamard Product]].
+As $\struct {S, \cdot}$ is [[Definition:Associative|associative]], then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from [[Associativity of Hadamard Product]].
+Thus if $\struct {S, \cdot}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
+If $\struct {S, \cdot}$ is [[Definition:Commutative Algebraic Structure|commutative]], then so is $\struct {\map {\MM_S} {m, n}, \circ}$ from [[Commutativity of Hadamard Product]].
+Thus if $\struct {S, \cdot}$ is a [[Definition:Commutative Semigroup|commutative semigroup]] then so is $\struct {\map {\MM_S} {m, n}, \circ}$.
+Let $\struct {S, \cdot}$ be a [[Definition:Monoid|monoid]], with [[Definition:Identity Element|identity]] $e$.
+Then from [[Zero Matrix is Identity for Hadamard Product]], $\struct {\map {\MM_S} {m, n}, \circ}$ also has an [[Definition:Identity Element|identity]] and is therefore also a [[Definition:Monoid|monoid]].
+{{Qed}}
+[[Category:Hadamard Product]]
+5bd25x7uey0z2g4kzye6z130c57p57z
+\end{proof}<|endoftext|>
+\section{Hadamard Product over Group forms Group}
+Tags: Hadamard Product, Examples of Groups
+
+\begin{theorem}
+Let $\struct {G, \cdot}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\map {\MM_G} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {G, \cdot}$.
+Then $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is [[Definition:Hadamard Product|Hadamard product]], is also a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+As $\struct {G, \cdot}$, being a [[Definition:Group|group]], is a [[Definition:Monoid|monoid]], it follows from [[Matrix Space Semigroup under Hadamard Product]] that $\struct {\map {\MM_G} {m, n}, \circ}$ is also a [[Definition:Monoid|monoid]].
+As $\struct {G, \cdot}$ is a [[Definition:Group|group]], it follows from [[Negative Matrix is Inverse for Hadamard Product]] that all [[Definition:Element|elements]] of $\struct {\map {\MM_G} {m, n}, \circ}$ have an [[Definition:Inverse Element|inverse]].
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Multiplication is Associative}
+Tags: Conventional Matrix Multiplication, Associativity
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+[[Definition:Matrix Product (Conventional)|Matrix multiplication (conventional)]] is [[Definition:Associative Operation|associative]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{p q}$ be [[Definition:Matrix|matrices]].
+From inspection of the subscripts, we can see that both $\paren {\mathbf A \mathbf B} \mathbf C$ and $\mathbf A \paren {\mathbf B \mathbf C}$ are defined:
+$\mathbf A$ has $n$ [[Definition:Column of Matrix|columns]] and $\mathbf B$ has $n$ [[Definition:Row of Matrix|rows]], while $\mathbf B$ has $p$ [[Definition:Column of Matrix|columns]] and $\mathbf C$ has $p$ [[Definition:Row of Matrix|rows]].
+Consider $\paren {\mathbf A \mathbf B} \mathbf C$.
+Let $\mathbf R = \sqbrk r_{m p} = \mathbf A \mathbf B, \mathbf S = \sqbrk s_{m q} = \mathbf A \paren {\mathbf B \mathbf C}$.
+Then:
+{{begin-eqn}}     
+{{eqn | l = s_{i j}
+      | r = \sum_{k \mathop = 1}^p r_{i k} \circ c_{k j}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | l = r_{i k}
+      | r = \sum_{l \mathop = 1}^n a_{i l} \circ b_{l k}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | ll= \leadsto
+      | l = s_{i j}
+      | r = \sum_{k \mathop = 1}^p \paren {\sum_{l \mathop = 1}^n a_{i l} \circ b_{l k} } \circ c_{k j}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^p \sum_{l \mathop = 1}^n \paren {a_{i l} \circ b_{l k} } \circ c_{k j}
+      | c = {{Ring-axiom|D}}
+}}
+{{end-eqn}}
+Now consider $\mathbf A \paren {\mathbf B \mathbf C}$.
+Let $\mathbf R = \sqbrk r_{n q} = \mathbf B \mathbf C, \mathbf S = \sqbrk s_{m q} = \mathbf A \paren {\mathbf B \mathbf C}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = s_{i j}
+      | r = \sum_{l \mathop = 1}^n a_{i l} \circ r_{l j}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | l = r_{l j}
+      | r = \sum_{k \mathop = 1}^p b_{l k} \circ c_{k j}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | ll= \leadsto
+      | l = s_{i j}
+      | r = \sum_{l \mathop = 1}^n a_{i l} \circ \paren {\sum_{k \mathop = 1}^p b_{l k} \circ c_{k j} }
+}}
+{{eqn | r = \sum_{l \mathop = 1}^n \sum_{k \mathop = 1}^p a_{i l} \circ \paren {b_{l k} \circ c_{k j} }
+      | c = {{Ring-axiom|D}}
+}}
+{{end-eqn}}
+Using {{Ring-axiom|M1}}:
+:$\displaystyle s_{i j} = \sum_{k \mathop = 1}^p \sum_{l \mathop = 1}^n \paren {a_{i l} \circ b_{l k} } \circ c_{k j} = \sum_{l \mathop = 1}^n \sum_{k \mathop = 1}^p a_{i l} \circ \paren {b_{l k} \circ c_{k j} } = s'_{i j}$
+It is concluded that:
+:$\paren {\mathbf A \mathbf B} \mathbf C = \mathbf A \paren {\mathbf B \mathbf C}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformations Isomorphic to Matrix Space}
+Tags: Linear Algebra, Matrix Algebra
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $F$, $G$ and $H$ be [[Definition:Free Module|free $R$-modules]] of [[Definition:Finite Dimensional Module|finite dimension]] $p,n,m>0$ respectively.
+Let $\left \langle {a_p} \right \rangle$, $\left \langle {b_n} \right \rangle$ and $\left \langle {c_m} \right \rangle$ be [[Definition:Ordered Basis|ordered bases]] 
+Let $\operatorname{Hom} \left({G, H}\right)$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Let $\mathcal M_R \left({m, n}\right)$ be the [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+Let $\left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$ be the [[Definition:Relative Matrix|matrix of $u$ relative to $\left \langle {b_n} \right \rangle$ and $\left \langle {c_m} \right \rangle$]].
+Let $M: \operatorname{Hom} \left({G, H}\right) \to \mathcal M_R \left({m, n}\right)$ be defined as:
+:$\forall u \in \operatorname{Hom} \left({G, H}\right): M \left({u}\right) = \left[{u; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right]$
+Then $M$ is an [[Definition:Module Isomorphism|isomorphism of modules]], and:
+:$\forall u \in \operatorname{Hom} \left({F, G}\right), v \in \operatorname{Hom} \left({G, H}\right): \left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$
+\end{theorem}
+
+\begin{proof}
+The proof that $M$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is straightforward.
+{{stub|The proof that $M$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].}}
+The relation:
+: $\left[{v \circ u; \left \langle {c_m} \right \rangle, \left \langle {a_p} \right \rangle}\right] = \left[{v; \left \langle {c_m} \right \rangle, \left \langle {b_n} \right \rangle}\right] \left[{u; \left \langle {b_n} \right \rangle, \left \langle {a_p} \right \rangle}\right]$
+follows from [[Relative Matrix of Composition of Linear Mappings]].
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Matrix Multiplication Distributes over Matrix Addition}
+Tags: Conventional Matrix Multiplication, Matrix Entrywise Addition, Distributive Operations
+
+\begin{theorem}
+[[Definition:Matrix Product (Conventional)|Matrix multiplication (conventional)]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Matrix Entrywise Addition|matrix entrywise addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}, \mathbf C = \sqbrk c_{n p}$ be [[Definition:Matrix|matrices]] over a [[Definition:Ring (Abstract Algebra)|ring]] $\struct {R, +, \circ}$.
+Consider $\mathbf A \paren {\mathbf B + \mathbf C}$.
+Let $\mathbf R = \sqbrk r_{n p} = \mathbf B + \mathbf C, \mathbf S = \sqbrk s_{m p} = \mathbf A \paren {\mathbf B + \mathbf C}$.
+Let $\mathbf G = \sqbrk g_{m p} = \mathbf A \mathbf B, \mathbf H = \sqbrk h_{m p} = \mathbf A \mathbf C$.
+Then:
+{{begin-eqn}}
+{{eqn | l = s_{i j}
+      | r = \sum_{k \mathop = 1}^n a_{i k} \circ r_{k j}
+      | c = 
+}}
+{{eqn | l = r_{k j}
+      | r = b_{k j} + c_{k j}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = s_{i j}
+      | r = \sum_{k \mathop = 1}^n a_{i k} \circ \paren {b_{k j} + c_{k j} }
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j} + \sum_{k \mathop = 1}^n a_{i k} \circ c_{k j}
+      | c = 
+}}
+{{eqn | r = g_{i j} + h_{i j}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\mathbf A \paren {\mathbf B + \mathbf C} = \paren {\mathbf A \mathbf B} + \paren {\mathbf A \mathbf C}$
+A similar construction shows that:
+:$\paren {\mathbf B + \mathbf C} \mathbf A = \paren {\mathbf B \mathbf A} + \paren {\mathbf C \mathbf A}$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Unity of Ring of Square Matrices}
+Tags: Rings of Square Matrices, Unit Matrices
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\struct {\map {\MM_R} n, +, \times}$ denote the [[Definition:Ring of Square Matrices|ring of square matrices of order $n$ over $R$]].
+The [[Definition:Unit Matrix|unit matrix]] over $R$:
+:$\mathbf I_n = \begin {pmatrix} 1_R & 0_R & 0_R & \cdots & 0_R \\ 0_R & 1_R & 0_R & \cdots & 0_R \\ 0_R & 0_R & 1_R & \cdots & 0_R \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0_R & 0_R & 0_R & \cdots & 1_R \end {pmatrix}$
+is the [[Definition:Identity Element|identity element]] of $\struct {\map {\MM_R} n, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+In [[Unit Matrix is Identity for Matrix Multiplication]], it is demonstrated that:
+:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$ 
+Hence the result, by definition of [[Definition:Identity Element|identity element]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Multiplication is Closed}
+Tags: Conventional Matrix Multiplication, Algebraic Closure
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\map {\MM_R} n$ be a [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
+Then [[Definition:Matrix Product (Conventional)|matrix multiplication (conventional)]] over $\map {\MM_R} n$ is [[Definition:Closure (Abstract Algebra)|closed]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Matrix Product (Conventional)|matrix multiplication]], the product of two [[Definition:Matrix|matrices]] is another [[Definition:Matrix|matrix]].
+The [[Definition:Order of Matrix|order]] of an $m \times n$ [[Definition:Matrix Product (Conventional)|multiplied]] by an $n \times p$ [[Definition:Matrix|matrix]] is $m \times p$.
+The [[Definition:Matrix Entry|entries]] of that [[Definition:Matrix Product (Conventional)|product]] [[Definition:Matrix|matrix]] are [[Definition:Element|elements]] of the [[Definition:Ring (Abstract Algebra)|ring]] over which the [[Definition:Matrix|matrix]] is formed.
+Thus an $n \times n$ [[Definition:Matrix|matrix]] over $R$ [[Definition:Matrix Product (Conventional)|multiplied]] by an $n \times n$ [[Definition:Matrix|matrix]] over $R$ gives another $n \times n$ [[Definition:Matrix|matrix]] over $R$.
+Hence the result.
+{{qed}}
+[[Category:Conventional Matrix Multiplication]]
+[[Category:Algebraic Closure]]
+9qwfqajxemqtzbxb0nsxfztyliq2lvf
+\end{proof}<|endoftext|>
+\section{Square Matrices over Real Numbers under Multiplication form Monoid}
+Tags: Matrix Algebra, Monoids
+
+\begin{theorem}
+Let $\map {\mathcal M_\R} n$ be a [[Definition:Matrix Space|$n \times n$ matrix space]] over the [[Definition:Real Number|set of real numbers $\R$]].
+Then the set of all $n \times n$ real matrices $\map {\mathcal M_\R} n$ under [[Definition:Matrix Product (Conventional)|matrix multiplication (conventional)]] forms a [[Definition:Monoid|monoid]].
+\end{theorem}
+
+\begin{proof}
+: [[Matrix Multiplication is Closed]].
+: [[Matrix Multiplication is Associative]].
+: The [[Unit Matrix is Unity of Ring of Square Matrices]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Square Matrices over Commutative Ring with Unity}
+Tags: Commutative Algebra, Rings of Square Matrices
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\struct {\map {\MM_R} n, +, \times}$ denote the [[Definition:Ring of Square Matrices|ring of square matrices of order $n$ over $R$]].
+Then $\struct {\map {\MM_R} n, +, \times}$ is a [[Definition:Ring with Unity|ring with unity]].
+However, for $n \ge 2$, $\struct {\map {\MM_R} n, +, \times}$ is not a [[Definition:Commutative Ring|commutative ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Square Matrices over Ring with Unity]] we have that $\struct {\map {\MM_R} n, +, \times}$ is a [[Definition:Ring with Unity|ring with unity]].
+However, [[Matrix Multiplication is not Commutative]].
+Hence $\struct {\map {\MM_R} n, +, \times}$ is not a [[Definition:Commutative Ring|commutative ring]] for $n \ge 2$.
+For $n = 1$ we have that:
+{{begin-eqn}}
+{{eqn | lo= \forall \mathbf A, \mathbf B \in \map {\MM_R} 1:
+      | l = \mathbf A \mathbf B
+      | r = a_{11} b_{11}
+      | c = where $\mathbf A = \begin {pmatrix} a_11 \end {pmatrix}$ and $\mathbf B = \begin {pmatrix} b_11 \end {pmatrix}$
+}}
+{{eqn | r = b_{11} a_{11}
+      | c = as $R$ is a [[Definition:Commutative Ring|commutative ring]]
+}}
+{{eqn | r = \mathbf {B A}
+      | c = 
+}}
+{{end-eqn}}
+Thus, for $n = 1$, $\struct {\map {\MM_R} n, +, \times}$ ''is'' a [[Definition:Commutative Ring|commutative ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Invertible Matrix corresponds to Automorphism}
+Tags: Matrix Algebra
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $G$ be an [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Module|$R$-module]].
+Let $\map {\mathcal M_R} n$ be the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
+Let $\map {\mathcal L_R} G$ be [[Definition:Set of All Linear Transformations|the set of all linear operators]] on $G$.
+Then the [[Definition:Invertible Matrix|invertible]] elements of the [[Definition:Ring of Square Matrices|ring of square matrices]] $\struct {\map {\mathcal M_R} n, +, \times}$ correspond directly to automorphisms of $\map {\mathcal L_R} G$.
+\end{theorem}<|endoftext|>
+\section{Change of Basis is Invertible}
+Tags: Matrix Algebra, Change of Basis
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $M$ be a [[Definition:Free Module|free $R$-module]] of [[Definition:Dimension (Linear Algebra)|finite dimension]] $n>0$.
+Let $\mathcal A$ and $\mathcal B$ be [[Definition:Ordered Basis|ordered bases]] of $M$.
+Let $\mathbf P$ be the [[Definition:Change of Basis Matrix|change of basis matrix]] from $\mathcal A$ to $\mathcal B$.
+Then $\mathbf P$ is [[Definition:Invertible Matrix|invertible]], and its [[Definition:Inverse Matrix|inverse]] $\mathbf P^{-1}$ is the [[Definition:Change of Basis Matrix|change of basis matrix]] from $\mathcal B$ to $\mathcal A$.
+\end{theorem}
+
+\begin{proof}
+From [[Product of Change of Basis Matrices]] and [[Change of Basis Matrix Between Equal Bases]]:
+* $\left[{I_M; \mathcal A, \mathcal B}\right] \left[{I_M; \mathcal B, \mathcal A}\right] = \left[{I_M; \mathcal A, \mathcal A}\right] = I_n$
+* $\left[{I_M; \mathcal B, \mathcal A}\right] \left[{I_M; \mathcal A, \mathcal B}\right] = \left[{I_M; \mathcal B , \mathcal B}\right] = I_n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Invertible Matrix Corresponds with Change of Basis}
+Tags: Linear Algebra, Matrix Algebra
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]] [[Definition:Ring with Unity|with unity]].
+Let $G$ be an [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Unitary Module|unitary $R$-module]].
+Let $\left \langle {a_n} \right \rangle$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
+Let $\mathbf P = \left[{\alpha}\right]_{n}$ be a [[Definition:Square Matrix|square matrix]] of order $n$ over $R$.
+Let $\displaystyle \forall j \in \left[{1 \,.\,.\, n}\right]: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$.
+Then $\left \langle {b_n} \right \rangle$ is an [[Definition:Ordered Basis|ordered basis]] of $G$ iff $\mathbf P$ is [[Definition:Invertible Matrix|invertible]].
+\end{theorem}
+
+\begin{proof}
+From [[Change of Basis is Invertible]], if $\left \langle {b_n} \right \rangle$ is an [[Definition:Ordered Basis|ordered basis]] of $G$ then $\mathbf P$ is [[Definition:Invertible Matrix|invertible]].
+Now let $\mathbf P$ be [[Definition:Invertible Matrix|invertible]].
+Then by [[Linear Transformations Isomorphic to Matrix Space/Corollary|the corollary to Linear Transformations Isomorphic to Matrix Space]], there is an [[Definition:Module Automorphism|automorphism]] $u$ of $G$ which satisfies $\mathbf P = \left[{u; \left \langle {a_n} \right \rangle}\right]$.
+Therefore, as $\forall j \in \left[{1 \,.\,.\, n}\right]: b_j = u \left({a_j}\right)$, it follows that $\left \langle {b_n} \right \rangle$ is also an ordered basis of $G$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Corresponding to Change of Basis under Linear Transformation}
+Tags: Linear Algebra, Change of Basis
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $G$ and $H$ be [[Definition:Free Module|free $R$-modules]] of [[Definition:Dimension (Linear Algebra)|finite dimensions]] $n,m>0$ respectively.
+Let $\left \langle {a_n} \right \rangle$ and $\left \langle {{a_n}'} \right \rangle$ be [[Definition:Ordered Basis|ordered bases]] of $G$.
+Let $\left \langle {b_m} \right \rangle$ and $\left \langle {{b_m}'} \right \rangle$ be [[Definition:Ordered Basis|ordered bases]] of $H$.
+Let $u: G \to H$ be a [[Definition:Linear Transformation|linear transformation]], and let $\left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$ be the [[Definition:Relative Matrix|matrix of $u$ relative to $\left \langle {a_n} \right \rangle$ and $\left \langle {b_m} \right \rangle$]].
+Let:
+: $\mathbf A = \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right]$
+: $\mathbf B = \left[{u; \left \langle {{b_m}'} \right \rangle, \left \langle {{a_n}'} \right \rangle}\right]$
+Then:
+:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
+where:
+: $\mathbf P$ is the [[Definition:Change of Basis Matrix|matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {{a_n}'} \right \rangle$]]
+: $\mathbf Q$ is the [[Definition:Change of Basis Matrix|matrix corresponding to the change of basis from $\left \langle {b_m} \right \rangle$ to $\left \langle {{b_m}'} \right \rangle$]].
+\end{theorem}
+
+\begin{proof}
+We have $u = I_H \circ u \circ I_G$
+and $\mathbf Q^{-1} = \left[{I_H; \left \langle {{b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right]$.
+Thus by [[Linear Transformations Isomorphic to Matrix Space]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf Q^{-1} \mathbf A \mathbf P
+      | r = \left[{I_H; \left \langle { {b_m}'} \right \rangle, \left \langle {b_m} \right \rangle}\right] \left[{u; \left \langle {b_m} \right \rangle, \left \langle {a_n} \right \rangle}\right] \left[{I_G; \left \langle {a_n} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]
+      | c = 
+}}
+{{eqn | r = \left[{I_H \circ u \circ I_G; \left \langle { {b_m}'} \right \rangle, \left \langle { {a_n}'} \right \rangle}\right]
+      | c = 
+}}
+{{eqn | r = \mathbf B
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Equivalence is Equivalence Relation}
+Tags: Matrix Algebra, Equivalence Relations
+
+\begin{theorem}
+[[Definition:Equivalent Matrices|Matrix equivalence]] is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+$\mathbf A = \mathbf{I_m}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all [[Definition:Matrix|$m \times n$ matrices]] $\mathbf A$.
+Thus [[Definition:Reflexive Relation|reflexivity]] holds.
+{{qed|lemma}}
+=== Symmetric ===
+Let $\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$.
+As $\mathbf P$ and $\mathbf Q$ are both [[Definition:Invertible Matrix|invertible]], we have:
+{{begin-eqn}}
+{{eqn | l=\mathbf Q \mathbf B \mathbf P^{-1}
+      | r=\mathbf Q \mathbf Q^{-1} \mathbf A \mathbf P \mathbf P^{-1}
+      | c=
+}}
+{{eqn | r=\mathbf{I_m} \mathbf A \mathbf{I_n}
+      | c=
+}}
+{{eqn | r=\mathbf A
+      | c=
+}}
+{{end-eqn}}
+Thus [[Definition:Symmetric Relation|symmetry]] holds.
+{{qed|lemma}}
+=== Transitive ===
+Let $\mathbf B = \mathbf Q_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf Q_2^{-1} \mathbf B \mathbf P_2$.
+Then $\mathbf C = \mathbf Q_2^{-1} \mathbf Q_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$.
+[[Definition:Transitive Relation|Transitivity]] follows from the definition of [[Definition:Invertible Matrix|invertible matrix]], that the product of two invertible matrices is itself invertible.
+{{qed|lemma}}
+Hence the result by definition of [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Similarity is Equivalence Relation}
+Tags: Matrix Algebra, Equivalence Relations, Matrix Similarity is Equivalence Relation
+
+\begin{theorem}
+[[Definition:Matrix Similarity|Matrix similarity]] is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Matrix Equivalence is Equivalence Relation]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+$\mathbf A = \mathbf{I_n}^{-1} \mathbf A \mathbf{I_n}$ trivially, for all order $n$ [[Definition:Square Matrix|square matrices]] $\mathbf A$.
+So [[Definition:Matrix Similarity|matrix similarity]] is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
+As $\mathbf P$ is [[Definition:Invertible Matrix|invertible]], we have:
+{{begin-eqn}}
+{{eqn | l = \mathbf P \mathbf B \mathbf P^{-1}
+      | r = \mathbf P \mathbf P^{-1} \mathbf A \mathbf P \mathbf P^{-1}
+      | c = 
+}}
+{{eqn | r = \mathbf{I_n} \mathbf A \mathbf{I_n}
+      | c = 
+}}
+{{eqn | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Matrix Similarity|matrix similarity]] is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+Let $\mathbf B = \mathbf P_1^{-1} \mathbf A \mathbf P_1$ and $\mathbf C = \mathbf P_2^{-1} \mathbf B \mathbf P_2$.
+Then $\mathbf C = \mathbf P_2^{-1} \mathbf P_1^{-1} \mathbf A \mathbf P_1 \mathbf P_2$.
+The result follows from the definition of [[Definition:Invertible Matrix|invertible matrix]], that the product of two invertible matrices is itself invertible.
+So [[Definition:Matrix Similarity|matrix similarity]] is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So, by definition, [[Definition:Matrix Similarity|matrix similarity]] is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Similar Matrices are Equivalent}
+Tags: Matrix Algebra
+
+\begin{theorem}
+If two [[Definition:Square Matrix|square matrices]] over a [[Definition:Ring with Unity|ring with unity]] $R$ are [[Definition:Matrix Similarity|similar]], then they are [[Definition:Matrix Equivalence|equivalent]].
+It follows directly that every [[Definition:Equivalence Class|equivalence class]] for the relation of similarity on $\mathcal M_R \left({n}\right)$ is contained in an equivalence class for the relation of matrix equivalence.
+Here, $\mathcal M_R \left({n}\right)$ denotes the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
+\end{theorem}
+
+\begin{proof}
+If $\mathbf A \sim \mathbf B$ then $\mathbf B = \mathbf P^{-1} \mathbf A \mathbf P$.
+Let $\mathbf Q = \mathbf P$.
+Then $\mathbf A$ are [[Definition:Matrix Equivalence|equivalent]] to $\mathbf B$, as:
+:$\mathbf B = \mathbf Q^{-1} \mathbf A \mathbf P$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalent Matrices have Equal Rank}
+Tags: Rank of Matrix
+
+\begin{theorem}
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|$m \times n$ matrices]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\map \phi {\mathbf A}$ denote the [[Definition:Rank of Matrix|rank]] of $\mathbf A$.
+Let $\mathbf A \equiv \mathbf B$ denote that $\mathbf A$ and $\mathbf B$ are [[Definition:Matrix Equivalence|matrix equivalent]].
+Then:
+:$\mathbf A \equiv \mathbf B$
+{{iff}}:
+:$\map \phi {\mathbf A} = \map \phi {\mathbf B}$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|$m \times n$ matrices]] over a [[Definition:Field (Abstract Algebra)|field]] $K$ such that $\mathbf A \equiv \mathbf B$.
+Let $S$ and $T$ be [[Definition:Vector Space|vector spaces]] of [[Definition:Dimension of Vector Space|dimensions]] $n$ and $m$ over $K$.
+Let $\mathbf A$ be the [[Definition:Relative Matrix|matrix of a linear transformation $u: S \to T$ relative]] to the [[Definition:Ordered Basis|ordered bases]] $\sequence {a_n}$ of $S$ and $\sequence {b_m}$ of $T$.
+Let $\psi: K^m \to T$ be the [[Definition:Vector Space Isomorphism|isomorphism]] defined as:
+:$\displaystyle \map \psi {\sequence {\lambda_m} } = \sum_{k \mathop = 1}^m \lambda_k b_k$
+Then $\psi$ takes the $j$th column of $\mathbf A$ into $\map u {a_j}$.
+Hence it takes the [[Definition:Vector Subspace|subspace]] of $K^m$ [[Definition:Generator of Module|generated]] by the [[Definition:Column Matrix|columns]] of $\mathbf A$ onto the [[Definition:Codomain of Mapping|codomain]] of $u$.
+Thus $\map \rho {\mathbf A} = \map \rho u$, and [[Definition:Matrix Equivalence|equivalent matrices]] over a [[Definition:Field (Abstract Algebra)|field]] have the same [[Definition:Rank of Matrix|rank]].
+Now let $\map {\mathcal L} {K^n, K^m}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $K^n$ to $K^m$.
+Let $u, v \in \map {\mathcal L} {K^n, K^m}$ such that $\mathbf A$ and $\mathbf B$ are respectively the [[Definition:Relative Matrix|matrices of $u$ and $v$ relative]] to the [[Definition:Standard Ordered Basis|standard ordered bases]] of $K^n$ and $K^m$.
+Let $r = \map \phi {\mathbf A}$.
+By [[Linear Transformation from Ordered Basis less Kernel]], there exist ordered bases $\sequence {a_n}, \sequence {a'_n}$ of $K^n$ such that:
+:$\sequence {\map u {a_r} }$ and $\sequence {\map v {a'_r} }$ are [[Definition:Ordered Basis|ordered bases]] of $\map u {K^n}$ and $\map v {K^n}$ respectively
+and such that:
+:$\set {a_k: k \in \closedint {r + 1} n}$ and $\set {a'_k: k \in \closedint {r + 1} n}$ are respectively bases of the [[Definition:Kernel of Linear Transformation|kernels]] of $u$ and $v$.
+Thus, by [[Results concerning Generators and Bases of Vector Spaces]] there exist ordered bases $\sequence {b_m}$ and $\sequence {b'_m}$ of $K^m$ such that $\forall k \in \closedint 1 r$:
+{{begin-eqn}}
+{{eqn | l = b_k
+      | r = \map u {a_k}
+      | c = 
+}}
+{{eqn | l = b'_k
+      | r = \map v {a'_k}
+      | c = 
+}}
+{{end-eqn}}
+Let $z$ be the [[Definition:Vector Space Automorphism|automorphism]] of $K^n$ which satisfies $\forall k \in \closedint 1 n: \map z {a'_k} = a_k$.
+Let $w$ be the [[Definition:Vector Space Automorphism|automorphism]] of $K^m$ which satisfies $\forall k \in \closedint 1 m: \map w {b'_k} = b_k$.
+Then:: $\map {\paren {w^{-1} \circ u \circ z} } {a'_k} = \begin{cases}
+\map {w^{-1} } {b_k} = \map v {a_k} & : k \in \closedint 1 r \\
+0 = \map v {a_k} & : k \in \closedint 1 {r + 1} n
+\end{cases}$
+So $w^{-1} \circ u \circ z = v$.
+Now let $\mathbf P$ be the matrix of $z$ relative to the [[Definition:Standard Ordered Basis|standard ordered bases]] of $K^n$, and let $\mathbf Q$ be the matrix of $w$ relative to the standard ordered basis of $K^m$.
+Then $\mathbf P$ and $\mathbf Q$ are [[Definition:Invertible Matrix|invertible]] and:
+:$\mathbf Q^{-1} \mathbf A \mathbf P = \mathbf B$
+and thus:
+:$\mathbf A \equiv \mathbf B$
+{{Qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Number of Matrix Equivalence Classes}
+Tags: Matrix Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathcal M_K \left({m, n}\right)$ be the [[Definition:Matrix Space|$m \times n$ matrix space]] over $K$.
+Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]] of [[Definition:Rank of Matrix|rank]] $r$ over $K$.
+Then:
+:$\mathbf A \equiv \begin{cases}
+\left[{0_K}\right]_{m n} & : r = 0 \\
+& \\
+\begin{bmatrix}
+  \mathbf I_r & \mathbf 0 \\
+  \mathbf 0 & \mathbf 0 
+\end{bmatrix} & : 0 < r < \min \left\{{n, m}\right\} \\
+& \\
+\begin{bmatrix}
+  \mathbf I_r & \mathbf 0 
+\end{bmatrix} & : r = m < n \\
+& \\
+\begin{bmatrix}
+  \mathbf I_r \\
+  \mathbf 0 
+\end{bmatrix} & : r = n < m \\
+& \\
+\mathbf I_r & : r = m = n
+\end{cases}$
+Thus there are exactly $\min \left\{{m, n}\right\} + 1$ [[Definition:Equivalence Class|equivalence classes]] for the relation of [[Definition:Matrix Equivalence|equivalence]] on $\mathcal M_K \left({m, n}\right)$, one of which contains only the [[Definition:Zero Matrix|zero matrix]].
+\end{theorem}
+
+\begin{proof}
+Follows from [[Equivalent Matrices have Equal Rank]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transpose of Matrix Product}
+Tags: Transposes of Matrices, Conventional Matrix Multiplication
+
+\begin{theorem}
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Matrix|matrices]] over a [[Definition:Commutative Ring|commutative ring]] such that $\mathbf A \mathbf B$ is [[Definition:Matrix Product (Conventional)|defined]].
+Then $\mathbf B^\intercal \mathbf A^\intercal$ is [[Definition:Matrix Product (Conventional)|defined]], and:
+:$\paren {\mathbf A \mathbf B}^\intercal = \mathbf B^\intercal \mathbf A^\intercal$
+where $\mathbf X^\intercal$ is the [[Definition:Transpose of Matrix|transpose of $\mathbf X$]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{n p}$
+Let $\mathbf A \mathbf B = \sqbrk c_{m p}$.
+Then from the definition of [[Definition:Matrix Product (Conventional)|matrix product]]:
+:$\displaystyle \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
+So, let $\paren {\mathbf A \mathbf B}^\intercal = \sqbrk r_{p m}$.
+The dimensions are correct, because $\mathbf A \mathbf B$ is an $m \times p$ matrix, thus making $\paren {\mathbf A \mathbf B}^\intercal$ a $p \times m$ matrix.
+Thus:
+:$\displaystyle \forall j \in \closedint 1 p, i \in \closedint 1 m: r_{j i} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
+Now, let $\mathbf B^\intercal \mathbf A^\intercal = \sqbrk s_{p m}$
+Again, the dimensions are correct because $\mathbf B^\intercal$ is a $p \times n$ matrix and $\mathbf A^\intercal$ is an $n \times m$ matrix.
+Thus:
+:$\displaystyle \forall j \in \closedint 1 p, i \in \closedint 1 m: s_{j i} = \sum_{k \mathop = 1}^n b_{k j} \circ a_{i k}$
+As the [[Definition:Underlying Structure of Matrix|underlying structure]] of $\mathbf A$ and $\mathbf B$ is a [[Definition:Commutative Ring|commutative ring]], then $a_{i k} \circ b_{k j} = b_{k j} \circ a_{i k}$.
+Note the order of the indices in the term in the summation sign on the {{RHS}} of the above.
+They are reverse what they would normally be because we are multiplying the transposes together.
+Thus it can be seen that $r_{j i} = s_{j i}$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rank is Dimension of Subspace}
+Tags: Matrix Algebra, Linear Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]] over $K$.
+Then the [[Definition:Rank of Matrix|rank]] of $\mathbf A$ is the [[Definition:Dimension of Vector Space|dimension]] of the [[Definition:Vector Subspace|subspace]] of $K^n$ [[Definition:Generator of Module|generated]] by the [[Definition:Row Matrix|rows]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Let $u: K^n \to K^m$ be the [[Definition:Linear Transformation|linear transformation]] such that $\mathbf A$ is the [[Definition:Relative Matrix|matrix of $u$ relative to]] the [[Definition:Standard Ordered Basis|standard ordered bases]] of $K^n$ and $K^m$.
+Let $\rho \left({\mathbf A}\right)$ be the [[Definition:Rank of Matrix|rank]] of $\mathbf A$.
+Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Similar notations on $u$ denote the [[Definition:Rank of Linear Transformation|rank]] and [[Definition:Transpose of Linear Transformation|transpose]] of $u$.
+We have $\rho \left({\mathbf A}\right) = \rho \left({u}\right)$ and $\rho \left({\mathbf A^\intercal}\right) = \rho \left({u^\intercal}\right)$, but $\rho \left({u^\intercal}\right) = \rho \left({u}\right)$ from [[Rank and Nullity of Transpose]].
+{{finish}}
+\end{proof}<|endoftext|>
+\section{General Linear Group is Group}
+Tags: General Linear Group
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\GL {n, K}$ be the [[Definition:General Linear Group|general linear group]] of [[Definition:Order of Square Matrix|order $n$]] over $K$.
+Then $\GL {n, K}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== Group Axiom $\text G 0$: Closure ===
+The [[Definition:Matrix Product (Conventional)|matrix product]] of two $n \times n$ matrices is another $n \times n$ matrix.
+The [[Definition:Matrix Product (Conventional)|matrix product]] of two [[Definition:Invertible Matrix|invertible matrices]] is another invertible matrix.
+Thus $\GL {n, K}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== Group Axiom $\text G 1$: Associativity ===
+[[Matrix Multiplication is Associative]].
+{{qed|lemma}}
+=== Group Axiom $\text G 2$: Identity ===
+From [[Unit Matrix is Unity of Ring of Square Matrices]], the [[Definition:Unit Matrix|unit matrix]] serves as the [[Definition:Identity Element|identity]] of $\GL {n, K}$.
+{{qed|lemma}}
+=== Group Axiom $\text G 3$: Inverses ===
+From the definition of [[Definition:Invertible Matrix|invertible matrix]], the [[Definition:Inverse Element|inverse]] of any [[Definition:Invertible Matrix|invertible matrix]] $\mathbf A$ is $\mathbf A^{-1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transpose of Row Matrix is Column Matrix}
+Tags: Transposes of Matrices
+
+\begin{theorem}
+Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a [[Definition:Row Matrix|row matrix]].
+Then $\mathbf x^\intercal$, the [[Definition:Transpose of Matrix|transpose]] of $\mathbf x$, is a [[Definition:Column Matrix|column matrix]]:
+:$\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
+\end{theorem}
+
+\begin{proof}
+Self-evident.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Transpose of Transpose of Matrix}
+Tags: Transposes of Matrices
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:Matrix|matrix]].
+Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Then:
+:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of the [[Definition:Transpose of Matrix|transpose of a matrix]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Solution to Simultaneous Linear Equations}
+Tags: Simultaneous Linear Equations
+
+\begin{theorem}
+Let $\displaystyle \forall i \in \closedint 1 m: \sum _{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$ be a system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]].
+where all of $\alpha_1, \ldots, a_n, x_1, \ldots x_n, \beta_i, \ldots, \beta_m$ are elements of a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Then $x = \tuple {x_1, x_2, \ldots, x_n}$ is a [[Definition:Solution to System of Simultaneous Equations|solution]] of this system {{iff}}:
+:$\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$
+where $\sqbrk a_{m n}$ is an [[Definition:Matrix|$m \times n$ matrix]].
+\end{theorem}
+
+\begin{proof}
+We can see the truth of this by writing them out in full.
+:$\displaystyle \sum_{j \mathop = 1}^n {\alpha_{i j} x_j} = \beta_i$
+can be written as:
+{{begin-eqn}}
+{{eqn | l = \alpha_{1 1} x_1 + \alpha_{1 2} x_2 + \ldots + \alpha_{1 n} x_n
+      | r = \beta_1
+      | c = 
+}}
+{{eqn | l = \alpha_{2 1} x_1 + \alpha_{2 2} x_2 + \ldots + \alpha_{2 n} x_n
+      | r = \beta_2
+      | c =
+}}
+{{eqn | o = \vdots
+}}
+{{eqn | l = \alpha_{m 1} x_1 + \alpha_{m 2} x_2 + \ldots + \alpha_{m n} x_n
+      | r = \beta_m
+      | c = 
+}}
+{{end-eqn}}
+while $\sqbrk \alpha_{m n} \sqbrk x_{n 1} = \sqbrk \beta_{m 1}$ can be written as:
+:$\begin {bmatrix}
+\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
+\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
+      \vdots &       \vdots & \ddots &       \vdots \\
+\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n}
+\end {bmatrix}
+\begin {bmatrix}
+x_1 \\ x_2 \\ \vdots \\ x_n
+\end {bmatrix}
+= \begin {bmatrix}
+\beta_1 \\ \beta_2 \\ \vdots \\ \beta_m
+\end {bmatrix}$
+So the question:
+:Find a solution to the following system of $m$ [[Definition:Simultaneous Linear Equations|simultaneous linear equations]] in $n$ variables
+is equivalent to:
+:Given the following element $\mathbf A \in \map {\MM_K} {m, n}$ and $\mathbf b \in \map {\MM_K} {m, 1}$, find the set of all $\mathbf x \in \map {\MM_K} {n, 1}$ such that $\mathbf A \mathbf x = \mathbf b$
+where $\map {\MM_K} {m, n}$ is the [[Definition:Matrix Space|$m \times n$ matrix space]] over $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Cyclic Group is Isomorphic to Integers}
+Tags: Infinite Cyclic Group, Additive Group of Integers, Examples of Group Isomorphisms
+
+\begin{theorem}
+Let $G$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Additive Group of Integers|additive group of integers]]: $G \cong \struct {\Z, +}$.
+\end{theorem}
+
+\begin{proof}
+From the definition of an [[Definition:Infinite Cyclic Group|infinite cyclic group]], we have:
+:$G = \gen a = \set {a^k: k \in \Z}$
+Let us define the [[Definition:Mapping|mapping]]:
+:$\phi: \Z \to G: \map \phi k = a^k$.
+We now show that $\phi$ is an [[Definition:Group Isomorphism|isomorphism]].
+From [[Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism]], $\phi$ is a [[Definition:Group Homomorphism|homomorphism]].
+Now we show that $\phi$ is a [[Definition:Surjection|surjection]].
+As $G$ is [[Definition:Cyclic Group|cyclic]], every [[Definition:Element|element]] of $G$ is a[[Definition:Power of Group Element|power]] of $a$ for some $a \in G$ such that $G = \gen a$.
+Thus:
+:$\forall x \in G: \exists k \in \Z: x = a^k$
+By the definition of $\phi$:
+:$\map \phi k = a^k = x$
+Thus $\phi$ is [[Definition:Surjection|surjective]].
+Now we show that $\phi$ is an [[Definition:Injection|injection]].
+This follows directly from [[Powers of Infinite Order Element]], where:
+:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
+Thus $\phi$ is an [[Definition:Injection|injective]], [[Definition:Surjection|surjective]] [[Definition:Group Homomorphism|homomorphism]], thus:
+: $G \cong \struct {\Z, +}$
+as required.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Infinite Cyclic Group is Infinite Cyclic Group}
+Tags: Subgroups, Cyclic Groups
+
+\begin{theorem}
+Let $G = \gen a$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]] [[Definition:Generator of Cyclic Group|generated]] by $a$, whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G, g \ne e: \exists k \in \Z, k \ne 0: g = a^k$.
+Let $H = \gen g$.
+Then $H \le G$ and $H \cong G$.
+Thus, all [[Definition:Non-Trivial Subgroup|non-trivial]] [[Definition:Subgroup|subgroups]] of an [[Definition:Infinite Cyclic Group|infinite cyclic group]] are themselves [[Definition:Infinite Cyclic Group|infinite cyclic groups]].
+A [[Definition:Subgroup|subgroup]] of $G = \gen a$ is denoted as follows:
+:$n G := \gen {a^n}$
+This notation is usually used in the [[Subgroups of Additive Group of Integers|context of $\struct {\Z, +}$]], where $n \Z$ is (informally) understood as '''the set of integer multiples of $n$'''.
+\end{theorem}
+
+\begin{proof}
+The fact that $H \le G$ follows from the definition of [[Definition:Generator of Subgroup|subgroup generator]].
+By [[Infinite Cyclic Group is Isomorphic to Integers]]:
+:$G \cong \struct {\Z, +}$
+Now we show that $H$ is of [[Definition:Infinite Group|infinite order]].
+Suppose $\exists h \in H, h \ne e: \exists r \in \Z, r > 0: h^r = e$.
+But:
+:$h \in H \implies \exists s \in \Z, s > 0: h = g^s$
+where $g = a^k$.
+Thus:
+:$e = h^r = \paren {g^s}^r = \paren {\paren {a^k}^s}^r = a^{k s r}$
+and thus $a$ is of [[Definition:Finite Order Element|finite order]].
+This would mean that $G$ was also of [[Definition:Finite Group|finite order]].
+So $H$ must be of [[Definition:Infinite Group|infinite order]].
+From [[Subgroup of Cyclic Group is Cyclic]], as $G$ is [[Definition:Cyclic Group|cyclic]], then $H$ must also be [[Definition:Cyclic Group|cyclic]].
+From [[Infinite Cyclic Group is Isomorphic to Integers]]:
+:$H \cong \struct {\Z, +}$
+Therefore, as $G \cong \struct {\Z, +}$:
+:$H \cong G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Infinite Cyclic Group by Subgroup}
+Tags: Cyclic Groups, Integers, Quotient Groups, Additive Groups of Integer Multiples, Additive Group of Integers
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of a Structure|order $n$]].
+Then:
+:$C_n \cong \dfrac {\struct {\Z, +} } {\struct {n \Z, +} } = \dfrac \Z {n \Z}$
+where:
+:$\Z$ is the [[Definition:Additive Group of Integers|additive group of integers]]
+:$n \Z$ is the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]]
+:$\Z / n \Z$ is the [[Definition:Quotient Group|quotient group]] of $\Z$ by $n \Z$.
+Thus, every [[Definition:Cyclic Group|cyclic group]] is [[Definition:Group Isomorphism|isomorphic]] to one of:
+:$\Z, \dfrac \Z \Z, \dfrac \Z {2 \Z}, \dfrac \Z {3 \Z}, \dfrac \Z {4 \Z}, \ldots$
+\end{theorem}
+
+\begin{proof}
+Let $C_n = \gen {a: a^n = e_{C_n} }$, that is, let $a$ be a [[Definition:Generator of Cyclic Group|generator]] of $C_n$.
+Let us define $\phi: \struct {\Z, +} \to C_n$ such that:
+:$\forall k \in \Z: \map \phi k = a^k$
+Then from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]]:
+:$\Img \phi = C_n = \struct {\Z, +} / \map \ker \phi$
+We now need to show that $\map \ker \phi = n \Z$.
+We have:
+:$\map \ker \phi = \set {k \in \Z: a^k = e_{C_n} }$
+Let $x \in \map \ker \phi$.
+Then $a^x = e_{C_n}$ and thus:
+:$n \divides x$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Operations on Coset Space of Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\powerset R$ be the [[Definition:Power Set|power set]] of $R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $X$ and $Y$ be [[Definition:Coset|cosets]] of $J$.
+Let $X +_\mathcal P Y$ be the [[Definition:Subset Product|sum]] of $X$ and $Y$, where $+_\mathcal P$ is the [[Definition:Subset Product|operation induced on $\powerset R$]] by $+$.
+Similarly, let $X \circ_\mathcal P Y$ be the [[Definition:Subset Product|product]] of $X$ and $Y$, where $\circ_\mathcal P$ is the [[Definition:Subset Product|operation induced on $\powerset R$]] by $\circ$.
+Then:
+* The [[Definition:Subset Product|sum]] $X +_\mathcal P Y$ in $\powerset R$ is also their [[Definition:Subset Product|sum]] in the [[Definition:Quotient Ring|quotient ring]] $R / J$.
+* The [[Definition:Subset Product|product]] $X \circ_\mathcal P Y$ in $\powerset R$ may be a [[Definition:Proper Subset|proper subset]] of their [[Definition:Subset Product|product]] in $R / J$.
+\end{theorem}
+
+\begin{proof}
+As $\struct {R, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]], it follows that $\struct {R, +}$ is an [[Definition:Abelian Group|abelian group]].
+Thus by [[Subgroup of Abelian Group is Normal]], all [[Definition:Subgroup|subgroups]] of $\struct {R, +, \circ}$ are [[Definition:Normal Subgroup|normal]].
+So from the definition of [[Definition:Quotient Group|quotient group]], it follows directly that $X +_\mathcal P Y$ in $\powerset R$ is also the sum in the [[Definition:Quotient Ring|quotient ring]] $R / J$.
+The set $\ideal 5$ of all integral multiples of $5$ is a [[Definition:Principal Ideal of Ring|principal ideal]] of the ring $\Z$.
+{{finish}}
+In the ring $\Z / \ideal 5$ we have:
+:$\ideal 5 \circ \ideal 5 = \ideal 5$
+However, in $\powerset \Z$, we have $\ideal 5 \circ_\mathcal P \ideal 5 = \ideal {25}$.
+{{finish|Plenty work needed here.}}
+\end{proof}<|endoftext|>
+\section{Property of Being an Ideal is not Transitive}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $J_1$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
+Let $J_2$ be an [[Definition:Ideal of Ring|ideal]] of $J_1$.
+Then $J_2$ need not necessarily be an [[Definition:Ideal of Ring|ideal]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $R = \Q \sqbrk X$ be the [[Definition:Polynomial Ring in One Variable|ring of polynomials in one variable]] $X$ over $\Q$.
+Let:
+:$J_1 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = 0}$
+and
+:$J_2 = \set {a_0 + a_1 X + \cdots + a_n X^n \in R : a_0 = a_1 = a_3 = 0}$
+First let us show that $J_1$ is an ideal of $R$.
+We establish the properties of the [[Test for Ideal|ideal test]] in order.
+$(1): \quad J_1 \ne \O$
+This follows from the fact that $X^2 \in J_1$.
+$(2): \quad \forall P, Q \in J_1: P + \paren {-Q} \in J_1$
+Let:
+:$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_1$
+:$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_1$
+Then by the definition of [[Definition:Addition of Polynomial Forms|addition of polynomials]]:
+:$\displaystyle P + \paren {-Q} = \sum_{i \mathop = 0}^{+\infty} c_i X^i$
+where:
+:$c_i = a_i - b_i$
+By assumption, $a_0 = a_1 = b_0 = b_1 = 0$.
+Therefore:
+:$c_0 = a_0 - b_0 = 0$
+:$c_1 = a_1 - b_1 = 0$
+Therefore $P + \left({-Q}\right) \in J_1$.
+$(3): \quad \forall P \in J_1, Q \in R: Q \cdot P \in J_1$
+Let
+:$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_1$
+:$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in R$
+By the definition of [[Definition:Multiplication of Polynomials|multiplication of polynomials]]:
+:$\displaystyle Q \cdot P = \sum_{i \mathop = 0}^{+\infty} c_i X^i$
+where:
+:$c_i = \sum_{j + k \mathop = i} a_j b_k$
+In particular, since $a_0 = 0$:
+:$c_0 = a_0 b_0 = 0$
+Since $a_0 = a_1 = 0$:
+:$c_ 1 = a_0 b_1 + a_1 b_0 = 0$
+Therefore $Q \cdot P \in J_1$.
+This shows that $J_1$ is an ideal of $R$.
+Next we show that $J$ is an ideal of $J_1$.
+Again, we verify the properties of the ideal test in turn.
+$(1): \quad J_2 \ne \O$
+This follows from the fact that $X^2 \in J_2$.
+$(2): \quad \forall P, Q \in J_2: P + \paren {-Q} \in J_2$
+Let:
+:$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_2$
+:$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_2$
+Then by the definition of [[Definition:Addition of Polynomial Forms|addition of polynomials]]:
+:$\displaystyle P + \paren {-Q} = \sum_{i \mathop = 0}^{+\infty} c_i X^i$
+where:
+:$c_i = a_i - b_i$
+By assumption:
+:$a_0 = a_1 = a_3 = b_0 = b_1 = b_3 = 0$
+Therefore:
+:$c_0 = a_0 - b_0 = 0$
+:$c_1 = a_1 - b_1 = 0$
+:$c_3 = a_3 - b_3 = 0$
+Therefore $P + \paren {-Q} \in J_2$.
+$(3): \quad \forall P \in J_2, Q \in J_1: Q \cdot P \in J_2$
+Let
+:$\displaystyle P = \sum_{i \mathop = 0}^{+\infty} a_i X^i \in J_2$
+:$\displaystyle Q = \sum_{i \mathop = 0}^{+\infty} b_i X^i \in J_1$
+By the definition of [[Definition:Multiplication of Polynomials|multiplication of polynomials]],
+:$\displaystyle Q \cdot P = \sum_{i \mathop = 0}^{+\infty} c_i X^i$
+where:
+:$c_i = \sum_{j + k \mathop = i} a_j b_k$
+Since $J_2 \subseteq J_1$, we have $P, Q \in J_1$.
+So we have already that $c_0 = c_1 = 0$.
+Moreover $a_0 = a_1 = b_0 = b_1 = 0$, so
+:$c_3 = a_0b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0 = 0$
+Therefore $Q \cdot P \in J_2$.
+This shows that $J_2$ is an ideal of $J_1$.
+Finally we wish to see that $J_2$ is not an ideal of $R$.
+We have that $X^2 \in J_2$ and $X \in R$.
+{{AimForCont}} $J_2$ were an ideal of $R$.
+This would imply that:
+:$X \cdot X^2 = X^3 \in J_2$
+But by the definition of $J_2$, we must have that the coefficient of $X^3$ is $0$.
+This is a contradiction, so $J_2$ is not an ideal of $R$.
+{{qed}}
+[[Category:Ideal Theory]]
+i65nuorbt880ya12pvev69hzk3c32li
+\end{proof}<|endoftext|>
+\section{Ideals Containing Ideal Form Lattice}
+Tags: Ideal Theory, Lattice Theory
+
+\begin{theorem}
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
+Let $\mathbb L_J$ be the set of all [[Definition:Ideal of Ring|ideal]] of $R$ which contain $J$.
+Then the [[Definition:Ordered Set|ordered set]] $\struct {\mathbb L_J, \subseteq}$ is a [[Definition:Lattice|lattice]].
+\end{theorem}
+
+\begin{proof}
+Let $b_1, b_2 \in \mathbb L_J$.
+Then from [[Set of Ideals forms Complete Lattice]]:
+:$(1): \quad b_1 + b_2 \in \mathbb L_J$ and is the [[Definition:Supremum of Set|supremum]] of $\set {b_1, b_2}$
+:$(2): \quad b_1 \cap b_2 \in \mathbb L_J$ and is the [[Definition:Infimum of Set|infimum]] of $\set {b_1, b_2}$
+Thus $\struct {\mathbb L_J, \subseteq}$ is a [[Definition:Lattice|lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ideals Containing Ideal Isomorphic to Quotient Ring}
+Tags: Ideal Theory, Quotient Rings
+
+\begin{theorem}
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of a [[Definition:Ring (Abstract Algebra)|ring]] $R$.
+Let $\mathbb L_J$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R$ which contain $J$.
+Let the [[Definition:Ordered Set|ordered set]] $\left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R / J$.
+Let the [[Definition:Mapping|mapping]] $\Phi_J: \left({\mathbb L_J, \subseteq}\right) \to \left({\mathbb L \left({R / J}\right), \subseteq}\right)$ be defined as:
+:$\forall a \in \mathbb L_J: \Phi_J \left({a}\right) = q_J \left({a}\right)$
+where $q_J: a \to a / J$ is the [[Definition:Quotient Epimorphism|quotient epimorphism]] from $a$ to $a / J$ from the definition of [[Definition:Quotient Ring|quotient ring]].
+Then $\Phi_J$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $b \in \mathbb L_J$.
+From the way $\mathbb L_J$ is defined:
+: $J \subseteq b$
+Thus by [[Preimage of Image of Subring under Ring Homomorphism]]:
+: $q_J^{-1} \left({q_J \left({b}\right)}\right) = b + J = b$
+Let $c$ be an ideal of $R / J$.
+Then, by [[Image of Preimage of Subring under Ring Epimorphism]]:
+: $q_J \left({q_J^{-1} \left({c}\right)}\right) = c$
+Thus by [[Bijection iff Left and Right Inverse]], $\Phi_J$ is a [[Definition:Bijection|bijection]].
+Hence:
+: $\forall c \in \mathbb L \left({R / J}\right): q_J^{-1} \left({\Phi_J}\right) c = q_J^{-1} \left({c}\right)$
+Now to show that $\Phi_J$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Let $b_1, b_2 \in \mathbb L_J$.
+Let $b_1 \subseteq b_2$.
+Then from [[Subset Maps to Subset]]:
+: $q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$
+Conversely, suppose $q_J \left({b_1}\right) \subseteq q_J \left({b_2}\right)$.
+By what we have just proved:
+: $b_1 = q_J^{-1} \left({q_J \left({b_1}\right)}\right) \subseteq  q_J^{-1} \left({q_J \left({b_2}\right)}\right) = b_2$
+Thus $\Phi_J$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Integers is Principal Ideal Domain}
+Tags: Integers, Principal Ideal Domains, Ring of Integers is Principal Ideal Domain
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ form a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\Z$.
+Then $J$ is a [[Definition:Subring|subring]] of $\Z$, and so $\left({J, +}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({\Z, +}\right)$.
+But by [[Integers under Addition form Infinite Cyclic Group]], the group $\left({\Z, +}\right)$ is [[Definition:Cyclic Group|cyclic]], generated by $1$.
+Thus by [[Subgroup of Cyclic Group is Cyclic]], $\left({J, +}\right)$ is cyclic, generated by some $m \in \Z$.
+Therefore from the definition of [[Definition:Principal Ideal of Ring|principal ideal]], $J = \left\{{k m: k \in \Z}\right\} = \left({m}\right)$, and is thus a [[Definition:Principal Ideal of Ring|principal ideal]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that [[Integers are Euclidean Domain]].
+Then we have that [[Euclidean Domain is Principal Ideal Domain]].
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $U$ be an arbitrary [[Definition:Ideal of Ring|ideal]] of $\Z$.
+Let $c$ be a non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] of $U$.
+Then both $c$ and $-c$ belong to $\ideal a$ and one of them is [[Definition:Positive Integer|positive]].
+Thus $U$ contains [[Definition:Strictly Positive Integer|strictly positive]] elements.
+Let $b$ be the [[Definition:Smallest Element|smallest]] [[Definition:Strictly Positive Integer|strictly positive]] element of $U$.
+By the [[Set of Integers Bounded Below by Integer has Smallest Element]], $b$ is guaranteed to exist.
+If $\ideal b$ denotes the [[Definition:Generator of Ideal|ideal generated by $b$]], then $\ideal b \subseteq U$ because $b\in U$ and $U$ is an ideal.
+Let $a \in U$.
+By the [[Division Theorem]]:
+:$\exists q, r \in \Z, 0 \le r < b: a = b q + r$
+As $a, b \in U$ it follows that so does $r = a - b q$.
+By definition of $b$ it follows that $r = 0$.
+Thus:
+:$a = b q \in \ideal b$
+and so:
+:$U \subseteq \ideal b$
+From the above:
+:$U = \ideal b$
+It follows by definition that $U$ is a [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$.
+Recall that $U$ was an arbitrary [[Definition:Ideal of Ring|ideal]] of $\Z$.
+Hence by definition $\Z$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principal Ideals of Integers}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $J$ be a [[Definition:Non-Null Ideal|non-zero]] [[Definition:Ideal of Ring|ideal]] of $\Z$.
+Then $J = \ideal b$ where $b$ is the smallest [[Definition:Strictly Positive|strictly positive]] [[Definition:Integer|integer]] belonging to $J$.
+\end{theorem}
+
+\begin{proof}
+It follows from [[Ring of Integers is Principal Ideal Domain]] that $J$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
+Let $c \in J, c \ne 0$.
+Then $-c \in J$ and by [[Natural Numbers are Non-Negative Integers]], exactly one of them is strictly positive.
+Thus $J$ ''does'' actually contain strictly positive elements, so that's a start.
+Let $b$ be the smallest strictly positive element of $J$.
+This exists because [[Natural Numbers are Non-Negative Integers]] and the [[Well-Ordering Principle]].
+By definition of a [[Definition:Principal Ideal of Ring|principal ideal]], we have $\ideal b \subseteq J$ as $b \in J$.
+We need to show that $J \subseteq \ideal b$.
+So, let $a \in J$.
+By the [[Division Theorem]], $\exists q, r: a = b q + r, 0 \le r < b$.
+As $a, b \in J$, then so does $r = a - b q$.
+So, by the definition of $b$, it follows that $r = 0$.
+Thus $a = b q \in \ideal b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers Set Equivalent to Ideals of Integers}
+Tags: Ideal Theory, Integers, Natural Numbers
+
+\begin{theorem}
+Let the [[Definition:Mapping|mapping]] $\psi: \N \to$ the set of all ideals of $\Z$ be defined as:
+:$\forall b \in \N: \psi \left({b}\right) = \left({b}\right)$
+where $\left({b}\right)$ is the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $b$.
+Then $\psi$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+First we show that $\psi$ is [[Definition:Injection|injective]].
+Suppose $0 < b < c$.
+Then $b \in \left({b}\right)$, but $b \notin \left({c}\right)$, because from [[Principal Ideals of Integers]], $c$ is the smallest [[Definition:Positive Integer|positive integer]] in $\left({c}\right)$.
+Thus $\left({b}\right) \ne \left({c}\right)$.
+It is also apparent that $b > 0 \implies \left({b}\right) \ne \left({0}\right)$ as $\left({0}\right) = \left\{{0}\right\}$.
+Thus $\psi$ is [[Definition:Injection|injective]].
+Surjectivity follows from [[Principal Ideals of Integers]]: every integer is the smallest strictly positive element of a [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Canonical Epimorphism from Integers by Principal Ideal}
+Tags: Modulo Arithmetic, Ideal Theory
+
+\begin{theorem}
+Let $m$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\left({m}\right)$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $m$.
+The [[Definition:Restriction of Mapping|restriction]] to $\N_m$ of the [[Quotient Mapping on Structure is Canonical Epimorphism|canonical epimorphism]] $q_m$ from the [[Definition:Ring (Abstract Algebra)|ring]] $\left({\Z, +, \times}\right)$ onto $\left({\Z, +, \times}\right) / \left({m}\right)$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from the [[Definition:Ring of Integers Modulo m|ring $\left({\N_m, +_m, \times_m}\right)$ of integers modulo $m$]] onto the [[Definition:Quotient Ring|quotient ring]] $\left({\Z, +, \times}\right) / \left({m}\right)$.
+In particular, $\left({\Z, +, \times}\right) / \left({m}\right)$ has $m$ elements.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \N_m$.
+By the [[Division Theorem]]:
+{{begin-eqn}}
+{{eqn | ll= \exists q, r \in \Z:
+      | l = x + y
+      | r = m q + r
+      | c = for $0 \le r < m$
+}}
+{{eqn | ll= \exists p, s \in \Z:
+      | l = x y
+      | r = m p + s
+      | c = for $0 \le s < m$
+}}
+{{end-eqn}}
+Then $x +_m y = r$ and $x \times_m y = s$, so:
+{{begin-eqn}}
+{{eqn | l = q_m \left({x +_m y}\right)
+      | r = q_m \left({r}\right)
+      | c = 
+}}
+{{eqn | r = q_m \left({m q}\right) + q_m \left({r}\right)
+      | c = 
+}}
+{{eqn | r = q_m \left({m q + r}\right)
+      | c = 
+}}
+{{eqn | r = q_m \left({x + y}\right)
+      | c = 
+}}
+{{eqn | r = q_m \left({x}\right) + q_m \left({y}\right)
+      | c = 
+}}
+{{end-eqn}}
+and similarly $q_m \left({x \times_m y}\right) = q_m \left({x y}\right) = q_m \left({x}\right) q_m \left({y}\right)$.
+So the restriction of $q_m$ to $\N_m$ is a homomorphism from $\left({\N_m, +_m, \times_m}\right)$ into $\left({\Z / \left({m}\right), +_{\left({m}\right)}, \times_{\left({m}\right)}}\right)$.
+Let $a \in \Z$.
+Then $\exists q, r \in \Z: a = q m + r: 0 \le r < m$, so $q_m \left({a}\right) = q_m \left({r}\right) \in q_m \left({\N_m}\right)$.
+Therefore $\Z / \left({m}\right) = q_m \left({\Z}\right) = q_m \left({\N_m}\right)$.
+Therefore the restriction of $q_m$ to $\N_m$ is [[Definition:Surjection|surjective]].
+If $0 < r < m$, then $r \notin \left({m}\right)$ and thus $q_m \left({r}\right) \ne 0$.
+Thus the [[Definition:Kernel of Ring Homomorphism|kernel]] of the restriction of $q_m$ to $\N_m$ contains only zero.
+Therefore by the [[Quotient Theorem for Group Epimorphisms]], the restriction of $q_m$ to $\N_m$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\N_m$ to $\Z / \left({m}\right)$.
+{{wtd|Reference is made throughout to $\left({\N_m, +_m, \times_m}\right)$. It needs to be shown that this is the same (at least up to isomorphism) as the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]] $\left({\Z, +_m, \times_m}\right)$.)}}
+\end{proof}<|endoftext|>
+\section{Integer Divisor is Equivalent to Subset of Ideal}
+Tags: Integers, Ideal Theory
+
+\begin{theorem}
+Let $\Z$ be the set of all [[Definition:Integer|integers]].
+Let $\Z_{>0}$ be the set of [[Definition:Strictly Positive Integer|strictly positive integers]].
+Let $m \in \Z_{>0}$ and let $n \in \Z$.
+Let $\ideal m$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $m$.
+Then:
+:$m \divides n \iff \ideal n \subseteq \ideal m$
+\end{theorem}
+
+\begin{proof}
+The [[Ring of Integers is Principal Ideal Domain|ring of integers is a principal ideal domain]].
+The result follows directly from [[Principal Ideals in Integral Domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principal Ideals in Integral Domain}
+Tags: Integral Domains, Principal Ideals, Factorization
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
+Let $U_D$ be the [[Definition:Group of Units of Ring|group of units]] of $D$.
+Let $\ideal x$ be the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $x$]].
+Let $x, y \in \struct {D, +, \circ}$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Principal Ideal Domain is Unique Factorization Domain}
+Tags: Integral Domains, Principal Ideal Domains, Unique Factorization Domains, Ideal Theory, Factorization
+
+\begin{theorem}
+Every [[Definition:Principal Ideal Domain|principal ideal domain]] is a [[Definition:Unique Factorization Domain|unique factorization domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Element of Principal Ideal Domain is Finite Product of Irreducible Elements]], each element which is neither $0$ nor a [[Definition:Unit of Ring|unit]] of a [[Definition:Principal Ideal Domain|principal ideal domain]] has a [[Definition:Factorization|factorization]] of [[Definition:Irreducible Element of Ring|irreducible elements]].
+{{proof wanted|Need to prove that the factorization is unique}}
+\end{proof}<|endoftext|>
+\section{Maximal Ideal iff Quotient Ring is Field}
+Tags: Quotient Rings, Maximal Ideal iff Quotient Ring is Field, Maximal Ideals of Rings, Field Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+The following are [[Definition:Logically Equivalent|equivalent]]:
+:$(1): \quad$ $J$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+:$(2): \quad$ The [[Definition:Quotient Ring|quotient ring]] $R / J$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+=== [[Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field|Maximal Ideal implies Quotient Ring is Field]] ===
+{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}}{{qed|lemma}}
+=== [[Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal|Quotient Ring is Field implies Ideal is Maximal]] ===
+{{:Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal}}{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\mathbb L_J$ be the set of all [[Definition:Ideal of Ring|ideals]] of $R$ which contain $J$.
+Let the [[Definition:Ordered Set|ordered set]] $\struct {\map {\mathbb L} {R / J}, \subseteq}$ be the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R / J$.
+Let the [[Definition:Mapping|mapping]] $\Phi_J: \struct {\mathbb L_J, \subseteq} \to \struct {\map {\mathbb L} {R / J}, \subseteq}$ be defined as:
+:$\forall a \in \mathbb L_J: \map {\Phi_J} a = \map {q_J} a$
+where $q_J: a \to a / J$ is the [[Definition:Quotient Ring Epimorphism|quotient epimorphism]] from $a$ to $a / J$.
+Then from [[Ideals Containing Ideal Isomorphic to Quotient Ring]], $\Phi_J$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+Now from [[Quotient Ring Defined by Ring Itself is Null Ring]], $\map {q_J} J$ is the [[Definition:Null Ideal|null ideal]] of $R / J$.
+At the same time, $\map {q_J} R$ is the entire [[Definition:Ring (Abstract Algebra)|ring]] $R / J$.
+Let $R / J$ not be the [[Definition:Null Ring|null ring]].
+Then by [[Epimorphism Preserves Rings]] and [[Epimorphism Preserves Commutativity]], $R / J$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+By definition, $J$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $R$ {{iff}} $\mathbb L_J = \set {J, R}$ and $J$ is a [[Definition:Proper Ideal of Ring|proper ideal]] of $R$.
+By [[Ideals of Field]], $R / J$ is a [[Definition:Field (Abstract Algebra)|field]] {{iff}}:
+:$\map {\mathbb L} {R / J} = \set {\map {q_J} J, \map {q_J} R}$
+and the [[Definition:Null Ideal|null ideal]] $\map {q_J} J$ is a [[Definition:Proper Ideal of Ring|proper ideal]] of $R / J$.
+As $\Phi_J: \mathbb L_J \to \map {\mathbb L} {R / J}$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], $J$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]] {{iff}} $J$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+We have by definition of [[Definition:Quotient Ring|quotient ring]] that $J$ is the [[Definition:Ring Zero|zero element]] of $R / J$.
+Let $A \in R / J$ be a non-[[Definition:Ring Zero|zero element]] of $R / J$.
+Let $x \in A$.
+Since $A \ne J$, we have that $x \notin J$.
+Let the [[Definition:Ideal of Ring|ideal]] $K = J + A$ of $R$ be formed.
+This contains all the [[Definition:Element|elements]] of the form $j + r a$, with $j \in J$ and $r \in R$.
+As $J$ is [[Definition:Maximal Ideal of Ring|maximal]] and $J \subsetneq K$, it follows that:
+:$K = R$
+and so:
+:$1_R \in K$
+That is:
+:$\exists j \in J, r \in R: j + r a = 1_R$
+Thus:
+:$\paren {r + J} \paren {a + J} = \paren {1 - u} J = 1_R + J$
+and so $\paren {r + J}$ is the [[Definition:Product Inverse|product inverse]] of $\paren {a + J}$.
+So every non-[[Definition:Ring Zero|zero element]] of $R / J$ has a [[Definition:Product Inverse|product inverse]].
+That is, $R / J$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed|lemma}}
+Let $R / J$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $K$ be an [[Definition:Ideal of Ring|ideal]] of $R$ such that:
+:$J \subsetneq K \subseteq R$
+Let $a \in K$ such that $a \notin J$.
+Then:
+:$J + \ideal a \subsetneq K$
+But as $a \notin J$, we have that $a + J$ is a non-[[Definition:Ring Zero|zero element]] of $R / J$.
+Thus as $R / J$ is a [[Definition:Field (Abstract Algebra)|field]], $a + J$ has a [[Definition:Product Inverse|product inverse]] $r + J$:
+:$\paren {r + J} \paren {a + J} = 1_R + J$
+So:
+:$\exists r \in R, j \in J: r a + \paren {-1_R} = j$
+That is:
+:$r a + \paren {-1_R} \in J$
+So:
+:$j + r a = 1_R$
+and from [[Ideal of Unit is Whole Ring/Corollary|Ideal of Unit is Whole Ring: Corollary]] this implies:
+:$J + \ideal a = R$
+So:
+:$K = R$
+and it follows by definition that $J$ is [[Definition:Maximal Ideal of Ring|maximal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Number iff Generates Principal Maximal Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\Z_{>0}$ be the set of [[Definition:Strictly Positive Integer|strictly positive integers]].
+Let $p \in \Z_{>0}$.
+Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal]] of $\Z$ generated by $p$. 
+Then $p$ is [[Definition:Prime Number|prime]] {{iff}} $\ideal p$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $\Z$.
+\end{theorem}
+
+\begin{proof}
+First, note that [[Ring of Integers is Principal Ideal Domain|$\Z$ is a principal ideal domain]], so all ideals are principal.
+Suppose $\ideal p$ is [[Definition:Prime Number|prime]].
+From [[Integer Divisor is Equivalent to Subset of Ideal]], $m \divides n \iff \ideal n \subseteq \ideal m$.
+But as $p$ is [[Definition:Prime Number|prime]], the only [[Definition:Divisor of Ring Element|divisors]] of $p$ are $1$ and $p$ itself.
+By [[Natural Numbers Set Equivalent to Ideals of Integers]], it follows that if $p$ is prime, then $\ideal p$ must be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+{{qed|lemma}}
+Conversely, let $p \in \Z_{>0}$ such that $\ideal p$ is [[Definition:Maximal Ideal of Ring|maximal]].
+Then if $\ideal p \subseteq \ideal q$ for some $q \in \Z_{>0}$, we have:
+:$\ideal q = \ideal p \implies q = p$
+or
+:$\ideal q = \ideal 1 \implies q = 1$
+Hence, as $p \in \ideal q$, we have:
+:$p \in q \iff q = 1 \text { or } q = p$
+and:
+:$p \in \ideal q \implies q \divides p$
+Hence $p$ is [[Definition:Prime Number|prime]], so $\ideal p$ is a [[Definition:Prime Ideal of Ring|prime ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Domain of Prime Order is Field}
+Tags: Integral Domains, Galois Fields
+
+\begin{theorem}
+Let $\left({\Z_p, +_p, \times_p}\right)$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]].
+The following statements are equivalent:
+: $(1): \quad p$ is a [[Definition:Prime Number|prime]].
+: $(2): \quad \left({\Z_p, +_p, \times_p}\right)$ is an [[Definition:Integral Domain|integral domain]].
+: $(3): \quad \left({\Z_p, +_p, \times_p}\right)$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+By [[Prime Number iff Generates Principal Maximal Ideal]] and [[Maximal Ideal iff Quotient Ring is Field]], $(1)$ implies $(3)$, and from [[Field is Integral Domain]], $(3)$ implies $(2)$.
+By the [[Definition:Integral Domain/Definition 2|definition of Integral Domain]], $\Z_p$ is an [[Definition:Integral Domain|integral domain]] iff $\left({\Z_p^*, \times_p}\right)$ is a [[Definition:Semigroup|semigroup]].
+Let $\left({p}\right)$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\left({\Z, +, \times}\right)$ generated by $p$]].
+From the [[Canonical Epimorphism from Integers by Principal Ideal]], $\left({\Z_p, +_p, \times_p}\right)$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $\left({\Z, +, \times}\right) / \left({p}\right)$.
+So, we can let $q_p \left({m}\right): \Z \to \Z_p$ be the [[Quotient Mapping on Structure is Canonical Epimorphism|quotient mapping]] from $\left({\Z, +, \times}\right)$ to $\left({\Z_p, +_p, \times_p}\right)$.
+Let $0_p$ denote the [[Definition:Ring Zero|zero]] of $\Z_p$.
+Suppose $p = m n$ where $1 < m < p, 1 < n < p$.
+Then in the [[Definition:Ring (Abstract Algebra)|ring]] $\Z_p$ we have $q_p \left({m}\right) \ne 0_p, q_p \left({n}\right) \ne 0_p$.
+But as $q_p$ is an [[Definition:Ring Epimorphism|epimorphism]] and therefore obeys the [[Definition:Morphism Property|morphism property]], $q_p \left({m}\right) \times_p q_p \left({n}\right) = q_p \left({m n}\right) = q_p \left({p}\right) = 0_p$.
+But by [[Definition:Ring Less Zero|definition]], $0_p \notin \Z_p^*$.
+Thus if $p = m n$, then $\left({\Z_p^*, \times_p}\right)$ is not a [[Definition:Semigroup|semigroup]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Integers and Zero}
+Tags: Quotient Rings, Integers
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ be the [[Integers form Integral Domain|integral domain of integers]].
+Let $\ideal 0$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\struct {\Z, +, \times}$ generated by $0$]].
+The [[Definition:Quotient Ring|quotient ring]] $\struct {\Z / \ideal 0, +, \times}$ is [[Definition:Ring Isomorphism|isomorphic]] to $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \ideal 0
+      | r = \set {\sum^n_{i \mathop = 1} r_i \times 0 \times s_i: n \in \N; r_i, s_i \in \Z}
+      | c = {{Defof|Principal Ideal of Ring}}
+}}
+{{eqn | r = \set {\sum^n_{i \mathop = 1} 0: n \in \N}
+      | c = $0$ is the [[Integer Multiplication has Zero|zero]] under [[Definition:Integer Multiplication|integer multiplication]]
+}}
+{{eqn | r = \set 0
+      | c = [[Integer Addition Identity is Zero]]
+}}
+{{end-eqn}}
+{{finish}}
+[[Category:Quotient Rings]]
+[[Category:Integers]]
+0pghfflaz3nxk2mqcj13qe7xh1y0fpw
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Integers and Principal Ideal from Unity}
+Tags: Ideal Theory, Quotient Rings, Integers
+
+\begin{theorem}
+Let $\left({\Z, +, \times}\right)$ be the [[Integers form Integral Domain|integral domain of integers]].
+Let $\left({1}\right)$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\left({\Z, +, \times}\right)$ generated by $1$]].
+The [[Definition:Quotient Ring|quotient ring]] $\left({\Z, +, \times}\right) / \left({1}\right)$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to the [[Definition:Null Ring|null ring]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Ideal Theory]]
+[[Category:Quotient Rings]]
+[[Category:Integers]]
+8vkfsw8gky0avcn8jnrex1w1lxqerkb
+\end{proof}<|endoftext|>
+\section{Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal}
+Tags: Principal Ideals, Principal Ideal Domains, Factorization, Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal, Maximal Ideals of Rings
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $\ideal p$ be the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $p$]].
+Then $p$ is [[Definition:Irreducible Element of Ring|irreducible]] {{iff}} $\ideal p$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $D$.
+\end{theorem}
+
+\begin{proof}
+=== [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication|Necessary Condition]] ===
+{{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication}}
+=== [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication|Sufficient Condition]] ===
+{{:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication}}
+[[Category:Principal Ideals]]
+[[Category:Principal Ideal Domains]]
+[[Category:Factorization]]
+[[Category:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]]
+[[Category:Maximal Ideals of Rings]]
+sqqhg4vygfe3be9qkn9ckj88k6xwfri
+\end{proof}<|endoftext|>
+\section{Subring Generated by Unity of Ring with Unity}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let the mapping $g: \Z \to R$ be defined as $\forall n \in \Z: g \left({n}\right) = n 1_R$, where $n 1_R$ the [[Definition:Power of Element|$n$th power of $1_R$]].
+Let $\left({x}\right)$ be the [[Definition:Principal Ideal of Ring|principal ideal of $\left({R, +, \circ}\right)$ generated by $x$]].
+Then $g$ is an [[Definition:Ring Epimorphism|epimorphism]] from $\Z$ onto the [[Definition:Subring|subring]] $S$ of $R$ [[Definition:Generated Subring|generated]] by $1_R$.
+If $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]], then $g$ is the only nonzero [[Definition:Ring Homomorphism|homomorphism]] from $\Z$ into $R$.
+The [[Definition:Kernel of Ring Homomorphism|kernel]] of $g$ is either:
+: $(1): \quad \left({0_R}\right)$, in which case $g$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\Z$ onto $S$
+or:
+: $(2): \quad \left({p}\right)$ for some [[Definition:Prime Number|prime]] $p$, in which case $S$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to the [[Integral Domain of Prime Order is Field|field $\Z_p$]].
+\end{theorem}
+
+\begin{proof}
+By the [[Index Laws for Monoids/Sum of Indices|Index Law for Sum of Indices]] and [[Powers of Ring Elements]], we have $\left({n 1_R}\right) \left({m 1_R}\right) = n \left({m 1_R}\right) = \left({n m}\right) 1_R$.
+Thus $g$ is an [[Definition:Ring Epimorphism|epimorphism]] from $\Z$ onto $S$.
+Assume that $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+By [[Kernel of Ring Epimorphism is Ideal]], the [[Definition:Kernel of Ring Homomorphism|kernel]] of $g$ is an ideal of $\Z$.
+By [[Ring of Integers is Principal Ideal Domain]], the [[Definition:Kernel of Ring Homomorphism|kernel]] of $g$ is $\left({p}\right)$ for some $p \in \Z_{>0}$.
+By [[Kernel of Ring Epimorphism is Ideal]] (don't think this is the correct reference - check it), $S$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $\Z_p$ and also has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+So from [[Integral Domain of Prime Order is Field]] either $p = 0$ or $p$ is [[Definition:Prime Number|prime]].
+Now we need to show that $g$ is unique.
+Let $h$ be  a non-zero [[Definition:Ring Homomorphism|(ring) homomorphism]] from $\Z$ into $R$.
+As $h \left({1}\right) = h \left({1^2}\right) = \left({h \left({1}\right)}\right)^2$, either $h \left({1}\right) = 1_R$ or $h \left({1}\right) = 0_R$ by [[Idempotent Elements of Ring with No Proper Zero Divisors]].
+But, by [[Homomorphism of Powers/Integers|Homomorphism of Powers: Integers]], $\forall n \in \Z: h \left({n}\right) = h \left({n 1}\right) = n h \left({1}\right)$
+So if $h \left({1}\right) = 0_R$, then $\forall n \in \Z: h \left({n}\right) = n 0_R = 0_R$.
+Hence $h$ would be a [[Definition:Zero Homomorphism|zero homomorphism]], which contradicts our stipulation that it is not.
+So $h \left({1}\right) = 1_R$, and thus $\forall n \in \Z: h \left({n}\right) = n 1 = g \left({n}\right)$.
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Null Ring iff Characteristic is One}
+Tags: Ring Theory
+
+\begin{theorem}
+The only [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Characteristic of Ring|characteristic]] is $1$ is the [[Definition:Null Ring|null ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Null Ring iff Zero and Unity Coincide]], $1_R \ne 0_R$ except when $R = \left\{{0_R}\right\}$.
+{{qed}}
+[[Category:Ring Theory]]
+6oqldbeeezzek7hrj6fbojap6vj42gf
+\end{proof}<|endoftext|>
+\section{Characteristic of Finite Ring with No Zero Divisors}
+Tags: Finite Rings, Rings with Unity, Characteristic of Finite Ring with No Zero Divisors
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Finite Ring|finite]] [[Definition:Ring with Unity|ring with unity]] with no [[Definition:Proper Zero Divisor|proper zero divisors]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $n \ne 0$ be the [[Definition:Characteristic of Ring|characteristic]] of $R$.
+Then:
+:$(1): \quad n$ must be a [[Definition:Prime Number|prime number]]
+:$(2): \quad n$ is the [[Definition:Order of Group Element|order]] of all non-zero elements in $\struct {R, +}$.
+It follows that $\struct {R, +} \cong C_n$, where $C_n$ is the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Subring Generated by Unity of Ring with Unity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose $\Char R = n$ where $n$ is [[Definition:Composite Number|composite]].
+Let $n = r s$, where $r, s \in \Z, r > 1, s > 1$.
+First note that:
+{{begin-eqn}}
+{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
+      | r = \paren {r s} \paren {1_R \circ 1_R}
+      | c = [[Powers of Ring Elements/General Result|Powers of Ring Elements]]
+}}
+{{eqn | r = \paren {r s} 1_R
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {r \cdot 1_R} \circ \paren {s \cdot 1_R}
+      | r = n \cdot 1_R
+      | c = 
+}}
+{{eqn | r = 0_R
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r \cdot 1_R = 0_R
+      | o = \lor
+      | r = s \cdot 1_R = 0_R
+      | c = 
+}}
+{{end-eqn}}
+as $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+But both $r$ and $s$ are less than $n$ which contradicting the minimality of $n$.
+So if $\Char R = n$ it follows that $n$ must be [[Definition:Prime Number|prime]].
+Now let $x \in R^*$.
+Then by [[Characteristic times Ring Element is Ring Zero]], $n \cdot x = 0_R$.
+It follows from [[Element to Power of Multiple of Order is Identity]] that:
+:$\order x \divides n$
+Since $n$ is [[Definition:Prime Number|prime]], either $\order x = 1$ or $\order x = n$.
+It cannot be $1$, from [[Null Ring iff Characteristic is One]], so the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Domain with Characteristic Zero}
+Tags: Integral Domains
+
+\begin{theorem}
+In an [[Definition:Integral Domain|integral domain]] with [[Definition:Characteristic of Ring|characteristic zero]], every non-zero element has [[Definition:Order of Group Element|infinite order]] under [[Definition:Ring Addition|ring addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]], whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$, such that $\Char D = 0$.
+Let $x \in D, x \ne 0_D$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \Z_{>0}:
+      | l = n \cdot x
+      | r = n \cdot \paren {x \circ 1_D}
+      | c = 
+}}
+{{eqn | r = \paren {n \circ 1_D} \cdot x
+      | c = [[Powers of Ring Elements]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n \cdot 1_D
+      | o = \ne
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n \cdot x
+      | o = \ne
+      | r = 0_D
+      | c = {{Defof|Integral Domain}}
+}}
+{{end-eqn}}
+That is, $x$ has [[Definition:Order of Group Element|infinite order]] in $\struct {D, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Field is Zero or Prime}
+Tags: Field Theory
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Then the [[Definition:Characteristic of Ring|characteristic]] of $F$ is either zero or a [[Definition:Prime Number|prime number]].
+\end{theorem}
+
+\begin{proof}
+From the definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Ring (Abstract Algebra)|ring]] with no [[Definition:Zero Divisor of Ring|zero divisors]].
+So by [[Characteristic of Finite Ring with No Zero Divisors]], if $\Char F \ne 0$ then it is [[Definition:Prime Number|prime]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Field of Characteristic Zero has Unique Prime Subfield}
+Tags: Field Theory, Subfields, Field of Characteristic Zero has Unique Prime Subfield
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]], whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$, with [[Definition:Characteristic of Ring|characteristic]] zero.
+Then there exists a unique $P \subseteq F$ such that:
+:$(1): \quad P$ is a [[Definition:Subfield|subfield]] of $F$
+:$(2): \quad P$ is [[Definition:Field Isomorphism|isomorphic]] to the [[Definition:Field of Rational Numbers|field of rational numbers]] $\struct {\Q, +, \times}$.
+That is, $P \cong \Q$ is a unique minimal [[Definition:Subfield|subfield]] of $F$, and all other [[Definition:Subfield|subfields]] of $F$ contain $P$.
+This field $P$ is called the [[Definition:Prime Subfield|prime subfield]] of $F$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+* [[Subring Generated by Unity of Ring with Unity]]
+* [[Quotient Theorem for Monomorphisms]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] such that $\Char F = 0$.
+Let $P$ be a [[Definition:Prime Subfield|prime subfield]] of $F$.
+From [[Field has Prime Subfield]], this has been shown to exist.
+As $P$ is a [[Definition:Subfield|subfield]] of $F$, we apply [[Zero and Unity of Subfield]] and see that the [[Definition:Unity of Field|unity]] of $P$ is $1_F$.
+As $P$ is [[Definition:Closed Algebraic Structure|closed]]:
+:$\forall m \in \Z: m \cdot 1_F \in P$ of which $0 \cdot 1_F = 0_F$ the only one that is [[Definition:Field Zero|zero]]
+:$\forall n \in \Z, n \ne 0: \paren {n \cdot 1_F}^{-1} \in P$
+So $P$ contains all elements of $F$ of the form:
+: $\paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}\^{-1}$
+where $m, n \in \Z, n \ne 0$.
+This can be expressed more clearly in [[Definition:Division Notation|division notation]] as:
+:$\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+Now let $P'$ consist of all the elements of $F$ of the form:
+:$\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+Let $\dfrac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \in P'$ and $\dfrac {m_2 \cdot 1_F} {n_2 \cdot 1_F} \in P'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}
+      | r = \frac {\paren {\paren {m_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} } + \paren {\paren {m_2 \cdot 1_F} \circ \paren {n_1 \cdot 1_F} } } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }
+      | c = [[Addition of Division Products]]
+}}
+{{eqn | r = \frac {\paren {\paren {m_1 n_2} \cdot \paren {1_F \circ 1_F} } + \paren {\paren {m_2 n_1} \cdot \paren {1_F \circ 1_F} } } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }
+      | c = [[Product of Integral Multiples]]
+}}
+{{eqn | r = \frac {\paren {\paren {m_1 n_2} \cdot 1_F} + \paren {\paren {m_2 n_1} \cdot 1_F} } {\paren {n_1 n_2} \cdot 1_F}
+      | c = as $1_F$ is the [[Definition:Unity of Field|unity]] of $F$
+}}
+{{eqn | r = \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = [[Integral Multiple Distributes over Ring Addition]]
+}}
+{{eqn | o = \in
+      | r = P'
+      | c = by definition of $P'$
+}}
+{{end-eqn}}
+So $P'$ is [[Definition:Closed Algebraic Structure|closed]] under $+$.
+Next:
+{{begin-eqn}}
+{{eqn | l = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \circ \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}
+      | r = \frac {\paren {m_1 \cdot 1_F} \circ \paren {m_2 \cdot 1_F} } {\paren {n_1 \cdot 1_F} \circ \paren {n_2 \cdot 1_F} }
+      | c = [[Product of Division Products]]
+}}
+{{eqn | r = \frac {\paren {m_1 m_2} \cdot \paren {1_F \circ 1_F} } {\paren {n_1 n_2} \cdot \paren {1_F \circ 1_F} }
+      | c = [[Product of Integral Multiples]]
+}}
+{{eqn | r = \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = as $1_F$ is the [[Definition:Unity of Field|unity]] of $F$
+}}
+{{eqn | o = \in
+      | r = P'
+      | c = by definition of $P'$
+}}
+{{end-eqn}}
+So $P'$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$.
+Next:
+{{begin-eqn}}
+{{eqn | l = -\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F}
+      | r = \frac {-\paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}
+      | c = [[Negative of Division Product]]
+}}
+{{eqn | r = \frac {-1 \cdot \paren {m_1 \cdot 1_F} } {n_1 \cdot 1_F}
+      | c = {{Defof|Integral Multiple}}
+}}
+{{eqn | r = \frac {-m_1 \cdot 1_F} {n_1 \cdot 1_F}
+      | c = [[Integral Multiple of Integral Multiple]]
+}}
+{{eqn | o = \in
+      | r = P'
+      | c = Definition of $P'$, as $-m_1 \in \Z$
+}}
+{{end-eqn}}
+So $P'$ is [[Definition:Closed Algebraic Structure|closed]] under taking inverses of $+$.
+Next, assuming that $m \ne 0$:
+{{begin-eqn}}
+{{eqn | l = \paren {\frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} }^{-1}
+      | r = \frac {n_1 \cdot 1_F} {m_1 \cdot 1_F}
+      | c = [[Inverse of Division Product]]
+}}
+{{eqn | o = \in
+      | r = P'
+      | c = Definition of $P'$
+}}
+{{end-eqn}}
+So $P' \setminus \set {0_F}$ is [[Definition:Closed Algebraic Structure|closed]] under taking inverses of $\circ$.
+Thus by [[Subfield Test]], $P'$ is a [[Definition:Subfield|subfield]] of $F$.
+It follows that $P = P'$, and so $P$ contains precisely the elements of the form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$.
+We can consistently define a [[Definition:Mapping|mapping]] $\phi: \Q \to P$ by:
+:$\forall m, n \in \Z: n \ne 0: \map \phi {\dfrac m n} = \dfrac {m \cdot 1_F} {n \cdot 1_F}$
+First we show that $\phi$ is well-defined.
+Suppose $\dfrac {m_1} {n_1} = \dfrac {m_2} {n_2}$.
+Then by multiplying both sides by $n_1 n_2$:
+:$m_1 n_2 = m_2 n_1$
+We need to show that:
+:$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$
+So:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\frac {m_1} {n_1} } + \paren {-\map \phi {\frac {m_2} {n_2} } }
+      | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \paren {-\frac {m_2 \cdot 1_F} {n_2 \cdot 1_F} }
+      | c = 
+}}
+{{eqn | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {-m_2 \cdot 1_F} {n_2 \cdot 1_F}
+      | c = from above: negative of element of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+}}
+{{eqn | r = \frac {\paren {m_1 n_2 - m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+}}
+{{eqn | r = \frac {0 \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = as $m_1 n_2 = m_2 n_1$
+}}
+{{eqn | r = 0_F
+      | c = 
+}}
+{{end-eqn}}
+That is:
+:$\map \phi {\dfrac {m_1} {n_1} } = \map \phi {\dfrac {m_2} {n_2} }$
+demonstrating that $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
+Next, we need to show that $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]].
+So:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\frac {m_1} {n_1} } + \map \phi {\frac {m_2} {n_2} }
+      | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} + \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {m_1 n_2 + m_2 n_1} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = from above: addition of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+}}
+{{eqn | r = \map \phi {\frac {m_1 n_2 + m_2 n_1} {n_1 n_2} }
+      | c = 
+}}
+{{eqn | r = \map \phi {\frac {m_1} {n_1} + \frac {m_2} {n_2} }
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\frac{m_1} {n_1} } \circ \map \phi {\frac {m_2} {n_2} }
+      | r = \frac {m_1 \cdot 1_F} {n_1 \cdot 1_F} \times \frac {m_2 \cdot 1_F} {n_2 \cdot 1_F}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {m_1 m_2} \cdot 1_F} {\paren {n_1 n_2} \cdot 1_F}
+      | c = from above: product of elements of form $\dfrac {m \cdot 1_F} {n \cdot 1_F}$
+}}
+{{eqn | r = \map \phi {\frac {m_1 m_2} {n_1 n_2} }
+      | c = 
+}}
+{{eqn | r = \map \phi {\frac {m_1} {n_1} \circ \frac {m_2} {n_2} }
+      | c = 
+}}
+{{end-eqn}}
+thus proving that $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+From [[Ring Homomorphism from Field is Monomorphism or Zero Homomorphism]], it follows that $\phi$ is a [[Definition:Ring Monomorphism|ring monomorphism]].
+It is also clear that $\phi$ is a [[Definition:Surjection|surjection]], as every element of $P$ is the image of some element of $\Q$.
+It follows that $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]].
+Now let $K$ be a [[Definition:Subfield|subfield]] of $F$.
+Let $P = \Img \phi$ as defined above.
+We know that $1_F \in K$.
+{{begin-eqn}}
+{{eqn | l = 1_F
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \forall k \in \Z: k \cdot 1_F
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \forall m, n \in \Z: n \ne 0: \paren {m \cdot 1_F} \circ \paren {n \cdot 1_F}^{-1}
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = P
+      | o = \subseteq
+      | r = K
+      | c = 
+}}
+{{end-eqn}}
+Thus $K$ contains a [[Definition:Subfield|subfield]] $P$ such that $P$ is [[Definition:Field Isomorphism|isomorphic]] to $\Q$.
+The [[Definition:Unique|uniqueness]] of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal [[Definition:Subfield|subfields]] of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Field of Prime Characteristic has Unique Prime Subfield}
+Tags: Prime Fields
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Characteristic of Ring|characteristic]] is $p$.
+Then there exists a [[Definition:Unique|unique]] $P \subseteq F$ such that:
+:$(1): \quad P$ is a [[Definition:Subfield|subfield]] of $F$
+:$(2): \quad P \cong \Z_p$.
+That is, $P \cong \Z_p$ is a unique minimal [[Definition:Subfield|subfield]] of $F$, and all other [[Definition:Subfield|subfields]] of $F$ contain $P$.
+This field $P$ is called the [[Definition:Prime Subfield|prime subfield]] of $F$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Field|unity]] is $1_F$ such that $\Char F = p$.
+Let $P$ be a [[Definition:Prime Subfield|prime subfield]] of $F$.
+From [[Field has Prime Subfield]], this has been shown to exist.
+We can consistently define a [[Definition:Mapping|mapping]] $\phi: \Z_p \to F$ by:
+:$\forall n \in \Z_p: \map \phi {\eqclass n p} = n \cdot 1_F$
+Suppose $a, b \in \eqclass n p$.
+Then:
+:$a = n + k_1 p, b = n + k_2 p$
+So:
+{{begin-eqn}}
+{{eqn | l = \map \phi a
+      | r = \map \phi {n + k_1 p}
+      | c = 
+}}
+{{eqn | r = \paren {n + k_1 p} \cdot 1_F
+      | c = 
+}}
+{{eqn | r = n \cdot 1_F + k_1 p \cdot 1_F
+      | c = 
+}}
+{{eqn | r = n \cdot 1_F
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $b$, showing that $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
+Let $C_a, C_b \in \Z_p$.
+Let $a \in C_a, b \in C_b$ such that $a = a' + k_a p, b = b' + k_b p$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {C_a} + \map \phi {C_b}
+      | r = \map \phi {a' + k_a p} + \map \phi {b' + k_b p}
+      | c = 
+}}
+{{eqn | r = \paren {a' + k_a p} \cdot 1_F + \paren {b' + k_b p} \cdot 1_F
+      | c = 
+}}
+{{eqn | r = \paren {a' + k_a p + b' + k_b p} \cdot 1_F
+      | c = [[Integral Multiple Distributes over Ring Addition]]
+}}
+{{eqn | r = \paren {a' + b'} \cdot 1_F + \paren {\paren {k_a p + k_b p} \cdot 1_F}
+      | c = 
+}}
+{{eqn | r = \paren {a' + b'} \cdot 1_F
+      | c = 
+}}
+{{eqn | r = \map \phi {C_a +_p C_b}
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\map \phi {C_a} \times \map \phi {C_b}$.
+So $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+From [[Ring Homomorphism from Field is Monomorphism or Zero Homomorphism]], it follows that $\phi$ is a [[Definition:Ring Monomorphism|ring monomorphism]].
+Thus it follows that $P = \Img \phi$ is a [[Definition:Subfield|subfield]] of $F$ such that $P \cong \Z_p$.
+Let $K$ be a [[Definition:Subfield|subfield]] of $F$.
+let $P = \Img \phi$ as defined above.
+We know that $1_F \in K$.
+It follows that $1_F \in K \implies P \subseteq K$.
+Thus $K$ contains a [[Definition:Subfield|subfield]] $P$ such that $P$ is [[Definition:Field Isomorphism|isomorphic]] to $\Z_p$.
+The [[Definition:Unique|uniqueness]] of $P$ follows from the fact that if $P_1$ and $P_2$ are both minimal [[Definition:Subfield|subfields]] of $F$, then $P_1 \subseteq P_2$ and $P_2 \subseteq P_1$, thus $P_1 = P_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of All Division Subrings is Prime Subfield}
+Tags: Subfields, Division Subrings
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $P$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $K$.
+Then $P$ is the [[Definition:Prime Subfield|prime subfield]] of $K$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection of Division Subrings is Division Subring]], the [[Definition:Set Intersection|intersection]] $P$ of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $K$ is a [[Definition:Division Ring|division ring]].
+Let $\map Z K$ be the [[Definition:Center of Ring|center]] of $K$.
+From [[Center of Ring is Commutative Subring]], $\map Z K$ is a [[Definition:Commutative Ring|commutative]] [[Definition:Subring|subring]] of $K$.
+Therefore $\map Z K$ is a [[Definition:Commutative Ring|commutative]] [[Definition:Division Ring|division ring]]
+Thus $\map Z K$ is a [[Definition:Subfield|subfield]] of $K$.
+But as $P$ is [[Definition:Subset|contained]] in $\map Z K$, it is itself [[Definition:Commutative Ring|commutative]].
+By its definition, $P$ [[Definition:Superset|contains]] no [[Definition:Proper Subfield|proper subfield]] and hence is a [[Definition:Prime Field|prime field]].
+Also, $P$ is [[Definition:Subset|contained]] in every other [[Definition:Subfield|subfield]] of $K$.
+Therefore $P$ is therefore the ''only'' [[Definition:Prime Subfield|prime subfield]] of $K$.
+{{qed}}
+[[Category:Subfields]]
+[[Category:Division Subrings]]
+daxy7sy883yl680few7ik2tj3tnrzis
+\end{proof}<|endoftext|>
+\section{Characteristic of Ordered Integral Domain is Zero}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+Let $\left({D, +, \circ}\right)$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $\operatorname{Char} \left({D}\right)$ be the [[Definition:Characteristic of Ring|characteristic of $D$]].
+Then $\operatorname{Char} \left({D}\right) = 0$.
+Let $g: \Z \to D$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \Z: g \left({n}\right) = n \cdot 1_D$
+where $n \cdot 1_D$ is defined as the [[Definition:Power of Element|$n$th power of $1_D$]].
+Then $g$ is the only [[Definition:Ring Monomorphism|monomorphism]] from the [[Definition:Ordered Ring|ordered ring]] $\Z$ onto the [[Definition:Ordered Ring|ordered ring]] $D$.
+\end{theorem}
+
+\begin{proof}
+* By [[Properties of Ordered Ring]] $(5)$:
+:$\forall n \in \Z_{>0}: n \cdot 1_D > 0$
+Thus $\operatorname{Char} \left({D}\right) \ne p$ for any $p > 0$.
+Hence $\operatorname{Char} \left({D}\right) = 0$ and so $g$ is a [[Definition:Ring Monomorphism|monomorphism]] from $\Z$ into $D$.
+Also, if $m < p$, then $p - m \in \Z_+$, so $p \cdot 1_D - m \cdot 1_D > 0_D$.
+Hence $g \left({m}\right) < g \left({p}\right)$ and thus by [[Monomorphism from Total Ordering]], $g$ is a monomorphism from $\Z$ into $D$.
+{{qed}}
+{{Proofread}}
+{{Improve|We have a new and more axiomatic definition of an [[Definition:Ordered Integral Domain|ordered integral domain]] which we may want to bring this into line with. Ultimately it boils down to the fact that an ordering on an integral domain is a lot more rigid than for a ring, such that totality is a consequence of the trichotomy law.}}
+\end{proof}<|endoftext|>
+\section{Monomorphism from Rational Numbers to Totally Ordered Field}
+Tags: Totally Ordered Fields, Ring Monomorphisms
+
+\begin{theorem}
+Let $\struct {F, +, \circ, \le}$ be a [[Definition:Totally Ordered Field|totally ordered field]].
+There is one and only one [[Definition:Ring Monomorphism|(ring) monomorphism]] from the [[Definition:Totally Ordered Field|totally ordered field]] $\Q$ onto $F$.
+Its [[Definition:Image of Mapping|image]] is the [[Definition:Prime Subfield|prime subfield]] of $F$.
+\end{theorem}
+
+\begin{proof}
+Follows from:
+: [[Characteristic of Ordered Integral Domain is Zero]]
+: [[Order Embedding between Quotient Fields is Unique]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Set of Polynomials over Integral Domain is Subring}
+Tags: Polynomial Theory, Subrings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
+Then $\forall x \in R$, the [[Definition:Set|set]] $D \sqbrk x$ of [[Definition:Polynomial in Ring Element|polynomials]] in $x$ over $D$ is a [[Definition:Subring|subring]] of $R$.
+\end{theorem}
+
+\begin{proof}
+By application of the [[Subring Test]]:
+As $D$ is an [[Definition:Integral Domain|integral domain]], it has a [[Definition:Unity of Ring|unity]] $1_D$ and so $x = 1_D x$.
+Hence $x \in D \sqbrk x$ and so $D \sqbrk x \ne \O$.
+Let $p, q \in D \sqbrk x$.
+Then let:
+:$\displaystyle p = \sum_{k \mathop = 0}^m a_k \circ x^k, q = \sum_{k \mathop = 0}^n b_k \circ x^k$
+Thus:
+:$\displaystyle -q = -\sum_{k \mathop = 0}^n b_k \circ x^k = \sum_{k \mathop = 0}^n \paren {-b_k} \circ x^k$
+and so:
+:$q \in D \sqbrk x$
+Thus as [[Polynomials Closed under Addition]], it follows that:
+: $p + \paren {-q} \in D \sqbrk x$
+Finally, from [[Polynomials Closed under Ring Product]], we have that $p \circ q \in D \sqbrk x$.
+All the criteria of the [[Subring Test]] are satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomials Closed under Ring Product}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\left({R, +, \circ}\right)$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\displaystyle f = \sum_{k \in Z} a_k \mathbf X^k$, $\displaystyle g =  \sum_{k \in Z} b_k \mathbf X^k$ be [[Definition:Polynomial|polynomials]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminates]] $\left\{{X_j: j \in J}\right\}$ over $R$, where $Z$ is the set of all [[Definition:Multiindex|multiindices]] indexed by $\left\{{X_j: j \in J}\right\}$.
+Define the product
+:$\displaystyle f \otimes g = \sum_{k \in Z} c_k \mathbf X^k$
+where
+:$\displaystyle c_k = \sum_{\substack{p + q = k \\ p, q \in Z}} a_p b_q$
+Then $f \otimes g$ is a polynomial.
+\end{theorem}
+
+\begin{proof}
+{{handwaving}}
+It is immediate that $f \otimes g$ is a map from the [[Definition:Free Commutative Monoid|free commutative monoid]] to $R$, so we need only prove that $f \otimes g$ is nonzero on finitely many $\mathbf X^k$, $k \in Z$.
+Suppose that for some $k \in Z$:
+:$\displaystyle \sum_{\substack{p + q = k \\ p, q \mathop \in Z}} a_p b_q \ne 0$
+Therefore if $c_k \ne 0$ there exist $p, q \in Z$ such that $p + q = k$ and $a_p$, $b_q$ are both nonzero.
+Since $f$ and $g$ are polynomials, the set $\left\{{p + q: a_p \ne 0, \ b_q \ne 0}\right\}$ is finite, and we are done.
+{{qed}}
+{{proofread}}
+[[Category:Polynomial Theory]]
+9e0x49comm43jlp1wrmcmg69fsco1f5
+\end{proof}<|endoftext|>
+\section{Unique Representation in Polynomial Forms}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
+Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental]] over $D$.
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $D$.
+Then each non-zero member of $D \left[{X}\right]$ can be expressed in just one way in the form:
+:$\ds f \in D \sqbrk X: f = \sum_{k \mathop = 0}^n {a_k \circ X^k}$
+\end{theorem}
+
+\begin{proof}
+Suppose $f \in D \sqbrk X \setminus \set {0_R}$ has more than one way of being expressed in the above form.
+Then you would be able to subtract one from the other and get a polynomial in $D \sqbrk X$ equal to zero.
+As $f$ is [[Definition:Transcendental over Integral Domain|transcendental]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Polynomial Forms is Integral Domain}
+Tags: Polynomial Rings, Integral Domains
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
+Let $\struct {D, +, \circ}$ be an [[Definition:Subdomain|integral subdomain]] of $R$.
+Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcendental over $D$]].
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $D$.
+Then $D \sqbrk X$ is an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+By [[Ring of Polynomial Forms is Commutative Ring with Unity]] we know that $D \sqbrk X$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let neither $\displaystyle \map f X = \sum_{k \mathop = 0}^n a_k x^k$ nor $\displaystyle \map g X = \sum_{k \mathop = 0}^m b_k X^k$ be the [[Definition:Null Polynomial over Ring|null polynomial]].
+Then their [[Definition:Leading Coefficient of Polynomial|leading coefficients]] $a_n$ and $b_m$ are non-[[Definition:Ring Zero|zero]].
+Therefore, as $D$ is an [[Definition:Integral Domain|integral domain]] and $a_n, b_m \in D$, so is their [[Definition:Ring Product|product]] $a_n b_m$.
+By the definition of [[Definition:Multiplication of Polynomials|polynomial multiplication]], it follows that $f g$ is not the [[Definition:Null Polynomial over Ring|null polynomial]].
+It follows that $D \sqbrk X$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+Hence $D \sqbrk X$ is an [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rings of Polynomials in Ring Elements are Isomorphic}
+Tags: Polynomial Theory, Ring Isomorphisms
+
+\begin{theorem}
+Let $R_1, R_2$ be [[Definition:Commutative Ring with Unity|commutative rings with unity]].
+Let $D$ be an [[Definition:Subdomain|integral subdomain]] of both $R_1$ and $R_2$.
+Let $X_1, X_2 \in R$ be [[Definition:Transcendental over Integral Domain|transcendental over $D$]].
+Let $D \sqbrk {X_1}, D \sqbrk {X_2}$ be the [[Definition:Ring of Polynomials in Ring Element|rings of polynomials]] in $X_1$ and $X_2$ over $D$.
+Then $D \sqbrk {X_1}$ is [[Definition:Ring Isomorphism|isomorphic]] to $D \sqbrk {X_2}$.
+\end{theorem}
+
+\begin{proof}
+First it is shown that the [[Definition:Mapping|mapping]] $\phi: D \sqbrk {X_1} \to D \sqbrk {X_2}$ given by:
+:$\displaystyle \map \phi {\sum_{k \mathop = 0}^n a_k \circ X_1^k} = \sum_{k \mathop = 0}^n a_k \circ X_2^k$
+is a [[Definition:Bijection|bijection]].
+Let $p, q \in \phi: D \sqbrk {X_1}$.
+Suppose $\map \phi p = \map \phi q$.
+Then the [[Definition:Polynomial Coefficient|coefficients]] of $\map \phi p$ and $\map \phi q$ are equal, and $p_1 = q_1$.
+Thus, by definition, $\phi$ is an [[Definition:Injection|injection]].
+By the same argument, the [[Definition:Mapping|mapping]] $\psi: D \sqbrk {X_2} \to D \sqbrk {X_1}$ defined as:
+:$\displaystyle \map \psi {\sum_{k \mathop = 0}^n a_k \circ X_2^k} = \sum_{k \mathop = 0}^n a_k \circ X_1^k$
+is similarly an [[Definition:Injection|injection]].
+Thus by [[Injection is Bijection iff Inverse is Injection]], $\phi$ is a [[Definition:Bijection|bijection]].
+It remains to show that $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]]. 
+Let $\displaystyle p = \sum_{k \mathop = 0}^n a_k \circ X_1^k, q = \sum_{k \mathop = 0}^n b_k \circ X_2^k \in D \sqbrk {X_1}$. 
+For convenience we set $a_k = 0$, $k > n$ and $b_k = 0$, $k > m$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \phi {p + q}
+      | r = \map \phi {\sum_{k \mathop = 0}^\infty \paren {a_k + b_k} X_1^k}
+      | c = {{Defof|Addition of Polynomials}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \left({a_k + b_k}\right) X_2^k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^m a_k X_2^k + \sum_{k \mathop = 0}^n b_k X_2^k
+      | c = 
+}}
+{{eqn | r = \map \phi p + \map \phi q
+      | c = 
+}}
+{{end-eqn}}
+Similarly for [[Definition:Multiplication of Polynomials|multiplication]]:
+{{begin-eqn}}
+{{eqn | l = \map \phi {p q}
+      | r = \map \phi {\sum_{k \mathop = 0}^{m n} \sum_{i + j = k} a_i b_j X_1^k}
+      | c = {{Defof|Multiplication of Polynomials}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{m n} \sum_{i + j \mathop = k} a_i b_j X_2^k
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{k \mathop = 0}^n a_k \circ X_1^k} \paren {\sum_{k \mathop = 0}^n b_k \circ X_2^k}
+      | c = 
+}}
+{{eqn | r = \map \phi p \, \map \phi q
+      | c = 
+}}
+{{end-eqn}}
+This completes the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Injection is Bijection iff Inverse is Injection}
+Tags: Injections, Bijections
+
+\begin{theorem}
+Let $\phi: S \to T$ be an [[Definition:Injection|injection]].
+Then $\phi$ is a [[Definition:Bijection|bijection]] {{iff}} its [[Definition:Inverse of Mapping|inverse]] $\phi^{-1}$ is also an [[Definition:Injection|injection]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\phi$ be a [[Definition:Bijection|bijection]].
+Then from [[Bijection iff Inverse is Bijection]], its inverse $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] and therefore by definition an [[Definition:Injection|injection]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\phi$ be an [[Definition:Injection|injection]] such that $\phi^{-1}$ is also an [[Definition:Injection|injection]].
+By [[Cardinality of Surjection]], and the [[Cantor-Bernstein-Schröder Theorem]], the result follows.
+{{qed}}
+[[Category:Injections]]
+[[Category:Bijections]]
+ieo6rvg42igne5veer4re8to6lotq4m
+\end{proof}<|endoftext|>
+\section{Epimorphism from Polynomial Forms to Polynomial Functions}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $D$ be an [[Definition:Integral Domain|integral domain]].
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
+Let $\map P D$ be the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] over $D$.
+The mapping $\kappa: D \sqbrk X \to \map P D$ given by:
+:$\displaystyle \map \kappa {\sum_{k \mathop = 0}^n {a_k \circ X^k} } = f$
+where $\displaystyle f = \sum_{k \mathop = 0}^n {a_k \circ x^k}, x \in D$
+is a [[Definition:Ring Epimorphism|ring epimorphism]].
+\end{theorem}<|endoftext|>
+\section{Division Theorem for Polynomial Forms over Field}
+Tags: Field Theory, Polynomial Theory, Division Theorem for Polynomial Forms over Field, Division Theorem
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $F$.
+Let $d$ be an [[Definition:Element|element]] of $F \sqbrk X$ of [[Definition:Degree of Polynomial over Field|degree]] $n \ge 1$.
+Then $\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
+:$(1): \quad r = 0_F$
+or:
+:$(2): \quad r \ne 0_F$ and $r$ has [[Definition:Degree of Polynomial over Field|degree]] that is less than $n$.
+\end{theorem}
+
+\begin{proof}
+From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
+So, if there is a [[Definition:Counterexample|counterexample]] to be found, it will have a [[Definition:Degree of Polynomial over Field|degree]].
+{{AimForCont}} there exists at least one [[Definition:Counterexample|counterexample]].
+By a version of the [[Well-Ordering Principle]], we can assign a number $m$ to the lowest [[Definition:Degree of Polynomial over Field|degree]] possessed by any [[Definition:Counterexample|counterexample]].
+So, let $f$ denote a [[Definition:Counterexample|counterexample]] which has that minimum [[Definition:Degree of Polynomial over Field|degree]] $m$.
+If $m < n$, the equation $f = 0_F \circ d + f$ would show that $f$ was not a [[Definition:Counterexample|counterexample]].
+Therefore $m \ge n$.
+Suppose $d \divides f$ in $F \sqbrk X$.
+Then:
+:$\exists q \in F \sqbrk X: f = q \circ d + 0_F$
+and $f$ would not be a [[Definition:Counterexample|counterexample]].
+So $d \nmid f$ in $F \sqbrk X$.
+So, suppose that:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \sum_{k \mathop = 0}^m {a_k \circ X^k}
+}}
+{{eqn | l = d
+      | r = \sum_{k \mathop = 0}^n {b_k \circ X^k}
+}}
+{{eqn | l = m
+      | o = \ge
+      | r = n
+}}
+{{end-eqn}}
+We can create the [[Definition:Polynomial in Ring Element|polynomial]] $\paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ which has the same [[Definition:Degree of Polynomial over Field|degree]] and [[Definition:Leading Coefficient of Polynomial|leading coefficient]] as $f$.
+Thus $f_1 = f - \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ is a [[Definition:Polynomial in Ring Element|polynomial]] of [[Definition:Degree of Polynomial over Field|degree]] less than $m$.
+Since $d \nmid f$, $f_1$ is a non-[[Definition:Null Polynomial over Ring|zero polynomial]].
+There is no [[Definition:Counterexample|counterexample]] of [[Definition:Degree of Polynomial over Field|degree]] less than $m$.
+Therefore:
+:$f_1 = q_1 \circ d + r$
+for some $q_1, r \in F \sqbrk X$, where either:
+:$r = 0_F$
+or:
+:$r$ is non-[[Definition:Null Polynomial over Ring|zero]] with [[Definition:Degree of Polynomial over Field|degree]] strictly less than $n$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = f_1 + \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d
+      | c = 
+}}
+{{eqn | r = \paren {q_1 + a_m \circ b_n^{-1} \circ X^{m - n} } \circ d + r
+      | c = 
+}}
+{{end-eqn}}
+Thus $f$ is not a [[Definition:Counterexample|counterexample]].
+From this [[Proof by Contradiction|contradiction]] follows the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose $\map \deg f < \map \deg d$.
+Then we take $\map q X = 0$ and $\map r X = \map a X$ and the result holds.
+Otherwise, $\map \deg f \ge \map \deg d$.
+Let:
+:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
+:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
+We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$:
+:$\map f X - \map d X \cdot \dfrac {a_m} {b_n} x^{m - n} = \map p X$
+where $\map p X$ is some polynomial whose [[Definition:Degree of Polynomial over Field|degree]] is less than that of $f$.
+If $\map p X$ still has [[Definition:Degree of Polynomial over Field|degree]] higher than that of $d$, we do the same thing again.
+Eventually we reach:
+:$\map f X - \map d X \cdot \paren {\dfrac {a_m} {b_n} x^{m - n} + \dotsb} = \map r X$
+where either $r = 0_F$ or $r$ has [[Definition:Degree of Polynomial over Field|degree]] that is less than $n$.
+This approach can be formalised using the [[Principle of Complete Induction]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
+=== Basis for the Induction ===
+$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.
+This is shown trivially to be true by taking $q = r = 0$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k$
+Then we need to show:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k + 1$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $f$ be such that $\map \deg f = n$.
+Let:
+:$f = a_0 + a_1 \circ x + a_2 \circ x^2 + \cdots + a_n \circ x^n$ where $a_n \ne 0$
+Let:
+:$d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ where $b_j \ne 0$
+If $n < l$ then take $q = 0, r = f$.
+If $n \ge l$, consider:
+:$c := f - a_n b_j^{-1} x^{n - j} d$
+This has been carefully arranged so that the [[Definition:Polynomial Coefficient|coefficient]] of $x^n$ in $c$ is zero.
+Thus $\map \deg c < n$.
+Therefore, by the [[Division Theorem for Polynomial Forms over Field/Proof 3#Induction Hypothesis|induction hypothesis]]:
+:$c = d q_0 + r$
+where $\map \deg r < \map \deg d$.
+Therefore:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = d \paren {q_0 + a_n b_j^{-1} x^{n - j} } + r
+      | c = 
+}}
+{{eqn | r = d q + r
+      | c = where $\map \deg r < \map \deg d$ and $q = q_0 + a_n b_j^{-1} x^{n - j}$
+}}
+{{end-eqn}}
+Thus the existence of $q$ and $r$ have been established.
+As for uniqueness, assume:
+:$d q + r = d q' + r'$
+with $\map \deg r < \map \deg d, \map \deg {r'} < \map \deg d$
+Then:
+:$d \paren {q - q'} = r' - r$
+By [[Degree of Sum of Polynomials]]:
+:$\map \deg {r' - r} \le \max \set {\map \deg {r'}, \map \deg r} < \map \deg d$
+and by [[Degree of Product of Polynomials over Integral Domain]]:
+:$\map \deg {d \paren {q - q'} } = \map \deg d + \map \deg {q - q'}$
+That is:
+:$\map \deg d < \map \deg d + \map \deg {q - q'}$
+and the only way for that to happen is for:
+:$\map \deg {q - q'} = -\infty$
+that is, for $q - q'$ to be the [[Definition:Null Polynomial|null polynomial]].
+That is, $q - q' = 0_F$ and by a similar argument $r' - r = 0_F$, demonstrating the uniqueness of $q$ and $r$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
+::$(1): \quad r = 0_F$
+:or:
+::$(2): \quad r \ne 0_F$ and $r$ has [[Definition:Degree of Polynomial over Field|degree]] that is less than $n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Forms over Field form Principal Ideal Domain}
+Tags: Field Theory, Polynomial Theory, Principal Ideal Domains, Polynomial Forms over Field form Principal Ideal Domain
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $X$ be [[Definition:Transcendental over Field|transcendental over $F$]].
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $F$.
+Then $F \sqbrk X$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+For any $d \in F \sqbrk X$, let $\ideal d$ denote the [[Definition:Principal Ideal of Ring|principal ideal of $F \sqbrk X$ generated by $d$]].
+Let $J$ be any [[Definition:Ideal of Ring|ideal]] of $F \sqbrk X$. What we need to prove is that $J$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
+Let us first [[Definition:Distinguish|distinguish]] the following two cases for $J$:
+:If $J = \set {0_F}$, then by [[Zero Element Generates Null Ideal]] $J = \ideal {0_F}$, and hence is a [[Definition:Principal Ideal of Ring|principal ideal]].
+:If $J = F \sqbrk X$, then by [[Ideal of Unit is Whole Ring/Corollary|Ideal of Unit is Whole Ring: Corollary]] $J = \ideal {1_F}$, and hence is a [[Definition:Principal Ideal of Ring|principal ideal]]. 
+Now suppose $J \ne \set {0_F}$ and $J \ne F \sqbrk X$.
+Then $J$ necessarily contains a non-[[Definition:Field Zero|zero]] [[Definition:Element|element]].
+By the [[Well-Ordering Principle]], we can introduce the lowest [[Definition:Degree of Polynomial over Field|degree]] of a non-[[Definition:Field Zero|zero]] [[Definition:Element|element]] of $J$.
+Denote this [[Definition:Degree of Polynomial over Field|degree]] by $n$.
+If $n = 0$, then $J$ contains a [[Definition:Polynomial over Ring in One Variable|polynomial]] of [[Definition:Degree of Polynomial over Field|degree]] $0$.
+This is a non-[[Definition:Field Zero|zero]] [[Definition:Element|element]] of $F$.
+As $F$ is a [[Definition:Field (Abstract Algebra)|field]], this is therefore a [[Definition:Unit of Ring|unit]] of $F$, and thus by [[Ideal of Unit is Whole Ring]], $J = F \sqbrk X$.
+Because the [[Definition:Degree of Polynomial over Field|degree]] of a non-[[Definition:Field Zero|zero]] [[Definition:Element|element]] is a [[Definition:Natural Numbers|natural number]], we conclude that $n \ge 1$.
+Now let $d$ be a [[Definition:Polynomial over Field|polynomial]] of degree $n$ in $J$, and let $f \in J$.
+By [[Division Theorem for Polynomial Forms over Field]], $f = q \circ d + r$ for some $q, r \in F \sqbrk X$ where either:
+:$r = 0_F$
+or:
+:$r$ is a [[Definition:Polynomial over Field|polynomial]] of [[Definition:Degree of Polynomial over Field|degree]] smaller than $n$.
+Because $J$ is an [[Definition:Ideal of Ring|ideal]] and $d \in J$, it follows that:
+:$q \circ d \in J$
+Since $f \in J$, we also conclude:
+:$r = f - q \circ d \in J$
+From the construction of $d$, it follows that we must have $r = 0_F$.
+Therefore:
+:$f = q \circ d$
+and thus:
+:$f \in \ideal d$.
+This reasoning shows that:
+:$J \subseteq \ideal d$
+From property $(3)$ of the [[Definition:Principal Ideal of Ring|principal ideal]] $\ideal d$, we conclude that:
+:$\ideal d \subseteq J$
+as $d \in J$.
+Hence $J = \ideal d$.
+These $2$ [[Definition:Distinguish|distinguished cases]] cover all of the possible [[Definition:Ideal of Ring|ideals]] of $F \sqbrk X$. 
+Hence $F \sqbrk X$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that [[Polynomial Forms over Field is Euclidean Domain]].
+We also have that [[Euclidean Domain is Principal Ideal Domain]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Consecutive Prime Number Gaps are Multiples of Six}
+Tags: Prime Numbers
+
+\begin{theorem}
+If you list the gaps between consecutive [[Definition:Prime Number|primes]] greater than $5$:
+:$2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, \ldots$
+you will notice that consecutive gaps that are equal are of the form $6 x$.
+This is ''always'' the case.
+{{OEIS|A001223}}
+\end{theorem}
+
+\begin{proof}
+Suppose there were two consecutive gaps between $3$ consecutive [[Definition:Prime Number|prime numbers]] that were equal, but not [[Definition:Divisor of Integer|divisible]] by $6$.
+Then the difference is $2 k$ where $k$ is not [[Definition:Divisor of Integer|divisible]] by $3$.
+Therefore the (supposed) [[Definition:Prime Number|prime numbers]] will be:
+:$p, p + 2 k, p + 4 k$
+But then $p + 4 k$ is [[Definition:Congruence Modulo Integer|congruent modulo $3$]] to $p + k$.
+That makes the three numbers [[Definition:Congruence Modulo Integer|congruent modulo $3$]] to $p, p + k, p + 2k$.
+One of those is [[Definition:Divisor of Integer|divisible]] by $3$ and so can not be [[Definition:Prime Number|prime]].
+So two consecutive gaps must be [[Definition:Divisor of Integer|divisible]] by $3$ and therefore (as they have to be [[Definition:Even Integer|even]]) by $6$.
+{{Qed}}
+[[Category:Prime Numbers]]
+3d5frank10lwfirx7e0rhzfct00dpq0
+\end{proof}<|endoftext|>
+\section{Standard Discrete Metric is Metric}
+Tags: Discrete Metrics
+
+\begin{theorem}
+The [[Definition:Standard Discrete Metric|standard discrete metric]] is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+Let $d: S \times S \to \R$ denote the [[Definition:Standard Discrete Metric|standard discrete metric]] on the [[Definition:Underlying Set of Metric Space|underlying set]] $S$ of some [[Definition:Metric Space|space]] $\left({S, d}\right)$.
+By definition:
+:$\forall x, y \in S: d \left({x, y}\right) = \begin{cases}
+0 & : x = y  \\ 
+1 & : x \ne y
+\end{cases}$
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d \left({x, x}\right)
+      | r = 0
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let $x = z$.
+{{begin-eqn}}
+{{eqn | l = d \left({x, z}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d \left({x, y}\right) + d \left({y, z}\right)
+      | o = \ge
+      | r = d \left({x, z}\right)
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{end-eqn}}
+Let $x \ne z$.
+Either $x \ne y$ or $y \ne z$, or both.
+So:
+{{begin-eqn}}
+{{eqn | l = d \left({x, y}\right) + d \left({y, z}\right)
+      | o = \ge
+      | r = 1
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{eqn | o = \ge
+      | r = d \left({x, z}\right)
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{end-eqn}}
+So in either case:
+: $d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$
+and [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+Let $x \ne y$.
+{{begin-eqn}}
+{{eqn | l = d \left({x, y}\right)
+      | r = 1
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{eqn | r = d \left({y, x}\right)
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of [[Definition:Standard Discrete Metric|Standard Discrete Metric]]
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Constant}
+Tags: Derivatives, Constant Mappings
+
+\begin{theorem}
+Let $\map {f_c} x$ be the [[Definition:Constant Mapping|constant function]] on $\R$, where $c \in \R$.
+Then:
+:$\map {f_c'} x = 0$
+\end{theorem}
+
+\begin{proof}
+The function $f_c: \R \to \R$ is defined as:
+:$\forall x \in \R: \map {f_c} x = c$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {f_c'} x
+      | r = \lim_{\delta x \mathop \to 0} \frac {\map {f_c} {x + \delta x} - \map {f_c} x} {\delta x}
+      | c = {{Defof|Differentiation}}
+}}
+{{eqn | r = \lim_{\delta x \mathop \to 0} \frac {c - c} {\delta x}
+      | c =
+}}
+{{eqn | r = \lim_{\delta x \mathop \to 0} \frac 0 {\delta x}
+      | c =
+}}
+{{eqn | r = 0
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Identity Function}
+Tags: Derivatives, Identity Mappings
+
+\begin{theorem}
+Let $X$ be either [[Definition:Set|set]] of either the [[Definition:Real Number|real numbers]] $\R$ or the [[Definition:Complex Number|complex numbers]] $\C$.
+Let $I_X: X \to X$ be the [[Definition:Identity Mapping|identity function]].
+Then:
+:$\map {I_X'} x = 1$
+where $\map {I_X'} x$ denotes the [[Definition:Derivative|derivative]] of $I_X$ {{WRT|Differentiation}} $x$.
+This can be presented for each of $\R$ and $\C$:
+\end{theorem}<|endoftext|>
+\section{Product Rule for Derivatives}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $\map f x, \map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]].
+Let $\map f x = \map j x \map k x$.
+Then:
+:$\map {f'} \xi = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi$
+It follows from the definition of [[Definition:Derivative on Interval|derivative]] that if $j$ and $k$ are both [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|interval]] $I$, then:
+:$\forall x \in I: \map {f'} x = \map j x \map {k'} x + \map {j'} x \map k x$
+Using [[Definition:Leibniz's Notation for Derivatives|Leibniz's notation for derivatives]], this can be written as:
+:$\map {\dfrac \d {\d x} } {y \, z} = y \dfrac {\d z} {\d x} + \dfrac {\d y} {\d x} z$
+where $y$ and $z$ represent functions of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {f'} \xi
+      | r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h
+      | c =
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j \xi \map k \xi} h
+      | c =
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\map j {\xi + h} \map k {\xi + h} - \map j {\xi + h} \map k \xi + \map j {\xi + h} \map k \xi - \map j \xi \map k \xi} h
+      | c =
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \paren {\map j {\xi + h} \frac {\map k {\xi + h} - \map k \xi} h + \frac {\map j {\xi + h} - \map j \xi} h \map k \xi}
+      | c =
+}}
+{{eqn | r = \map j \xi \map {k'} \xi + \map {j'} \xi \map k \xi
+      | c =
+}}
+{{end-eqn}}
+Note that $\map j {\xi + h} \to \map j \xi$ as $h \to 0$ because, from [[Differentiable Function is Continuous]], $j$ is continuous at $\xi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Composite Function}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f, g, h$ be [[Definition:Continuous Real Function|continuous real functions]] such that:
+:$\forall x \in \R: \map h x = \map {f \circ g} x = \map f {\map g x}$
+Then:
+:$\map {h'} x = \map {f'} {\map g x} \map {g'} x$
+where $h'$ denotes the [[Definition:Derivative of Real Function|derivative]] of $h$.
+Using the $D_x$ notation:
+:$\map {D_x} {\map f {\map g x} } = \map {D_{\map g x} } {\map f {\map g x} } \, \map {D_x} {\map g x}$
+This is often informally referred to as the '''chain rule (for differentiation)'''.
+\end{theorem}
+
+\begin{proof}
+Let $\map g x = y$, and let:
+{{begin-eqn}}
+{{eqn | l = \map g {x + \delta x}
+      | r = y + \delta y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \delta y
+      | r = \map g {x + \delta x} - \map g x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\delta y \to 0$ as $\delta x \to 0$
+and:
+:$(1): \quad \dfrac {\delta y} {\delta x} \to \map {g'} x$
+There are two cases to consider:
+=== Case 1 ===
+Suppose $\map {g'} x \ne 0$ and that $\delta x$ is small but non-zero.
+Then $\delta y \ne 0$ from $(1)$ above, and:
+{{begin-eqn}}
+{{eqn | l = \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x}
+      | r = \lim_{\delta x \mathop \to 0} \frac {\map f {\map g {x + \delta x} } - \map f {\map g x} } {\map g {x + \delta x} - \map g x} \frac {\map g {x + \delta x} - \map g x} {\delta x}
+      | c = 
+}}
+{{eqn | r = \lim_{\delta x \mathop \to 0} \frac {\map f {y + \delta y} - \map f y} {\delta y} \frac {\delta y} {\delta x}
+      | c = 
+}}
+{{eqn | r = \map {f'} y \, \map {g'} x
+      | c = 
+}}
+{{end-eqn}}
+hence the result.
+{{qed|lemma}}
+=== Case 2 ===
+Now suppose $\map {g'} x = 0$ and that $\delta x$ is small but non-zero.
+Again, there are two possibilities:
+=== Case 2a ===
+If $\delta y = 0$, then $\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = 0$.
+Hence the result.
+{{qed|lemma}}
+=== Case 2b ===
+If $\delta y \ne 0$, then:
+:$\dfrac {\map h {x + \delta x} - \map h x} {\delta x} = \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \dfrac {\delta y} {\delta x}$
+As $\delta y \to 0$:
+:$(1): \quad \dfrac {\map f {y + \delta y} - \map f y} {\delta y} \to \map {f'} y$
+:$(2): \quad \dfrac {\delta y} {\delta x} \to 0$
+Thus:
+:$\displaystyle \lim_{\delta x \mathop \to 0} \frac {\map h {x + \delta x} - \map h x} {\delta x} \to 0 = \map {f'} y \, \map {g'} x$
+Again, hence the result.
+{{qed|lemma}}
+All cases have been covered, so by [[Proof by Cases]], the result is complete.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Function}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $I = \closedint a b$ and $J = \closedint c d$ be [[Definition:Closed Real Interval|closed real intervals]].
+Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding [[Definition:Open Real Interval|open real intervals]].
+Let $f: I \to J$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on $I$ and [[Definition:Differentiable on Interval|differentiable]] on $I^o$ such that $J = f \sqbrk I$.
+Let either:
+:$\forall x \in I^o: D \map f x > 0$
+or:
+:$\forall x \in I^o: D \map f x < 0$
+Then:
+:$f^{-1}: J \to I$ exists and is [[Definition:Continuous on Interval|continuous]] on $J$
+:$f^{-1}$ is [[Definition:Differentiable on Interval|differentiable]] on $J^o$
+:$\forall y \in J^o: D \map {f^{-1} } y = \dfrac 1 {D \map f x}$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Monotone Function]], it follows that $f$ is either:
+:[[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$ (if $\forall x \in I^o: D \, \map f x > 0$)
+or:
+:[[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $I$ (if $\forall x \in I^o: D \, \map f x < 0$).
+Therefore from [[Inverse of Strictly Monotone Function]] it follows that $f^{-1}: J \to I$ exists.
+As $f$ is [[Definition:Continuous on Interval|continuous]], from [[Image of Interval by Continuous Function is Interval]] it follows that $J$ is an interval.
+By the [[Limit of Monotone Real Function/Increasing/Corollary|Corollary to Limit of Increasing Function]] and the [[Limit of Monotone Real Function/Decreasing/Corollary|Corollary to Limit of Decreasing Function]], $f^{-1}: J \to I$ is [[Definition:Continuous on Interval|continuous]].
+Next its [[Definition:Derivative|derivative]] is to be considered.
+Suppose $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+Let $y \in J^o$.
+Then $\map {f^{-1} } y \in I^o$.
+Let $k = \map {f^{-1} } {y + h} - \map {f^{-1} } y$.
+Thus:
+:$\map {f^{-1} } {y + h} = \map {f^{-1} } y + k = x + k$
+Thus:
+:$y + h = \map f {x + k}$
+and hence:
+:$h = \map f {x + k} - y = \map f {x + k} - \map f x$
+Since $f^{-1}$ is continuous on $J$, it follows that $k \to 0$ as $h \to 0$.
+Also, $f^{-1}$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] from [[Inverse of Strictly Monotone Function]] and so $k \ne 0$ unless $h = 0$.
+So by [[Limit of Composite Function]] we get:
+:$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h = \dfrac k {\map f {x + k} - \map f x}$
+Thus:
+:$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h \to \dfrac 1 {D \map f x}$
+as $h \to 0$.
+Suppose $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+Exactly the same argument applies.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Sum Never Smaller than Lower Sum}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed interval]] of the set $\R$ of [[Definition:Real Number|real numbers]].
+Let $P = \set {x_0, x_1, x_2, \ldots, x_{n - 1}, x_n}$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
+Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
+Let $f$ be [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$.
+Let $\map L P$ be the [[Definition:Lower Sum|lower sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$]].
+Let $\map U P$ be the [[Definition:Upper Sum|upper sum of $\map f x$ on $\closedint a b$ belonging to the subdivision $P$]].
+Then $\map L P \le \map U P$.
+\end{theorem}
+
+\begin{proof}
+For all $\nu \in 1, 2, \ldots, n$, let $\closedint {x_{\nu - 1} } {x_\nu}$ be a [[Definition:Closed Real Interval|closed subinterval]] of $\closedint a b$.
+As $f$ is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint a b$, it is [[Definition:Bounded Real-Valued Function|bounded]] on $\closedint {x_{\nu - 1} } {x_\nu}$.
+So, let $m_\nu$ be the [[Definition:Infimum of Real-Valued Function|infimum]] and $M_\nu$ be the [[Definition:Supremum of Real-Valued Function|supremum]] of $\map f x$ on the interval $\closedint {x_{\nu - 1} } {x_\nu}$.
+By definition, $m_\nu \le M_\nu$.
+So $m_{\nu} \paren {x_\nu - x_{\nu - 1} } \le M_{\nu} \paren {x_\nu - x_{\nu - 1} }$.
+It follows directly that $\displaystyle \sum_{\nu \mathop = 1}^n m_\nu \paren {x_\nu - x_{\nu - 1} } \le \sum_{\nu \mathop = 1}^n M_\nu \paren {x_\nu - x_{\nu - 1} }$.
+{{qed}}
+[[Category:Real Analysis]]
+9cugcptjnm98xa2lsslxmftu3hmjg09
+\end{proof}<|endoftext|>
+\section{Dedekind's Theorem}
+Tags: Real Analysis, Dedekind Cuts, Dedekind's Theorem
+
+\begin{theorem}
+Let $\tuple {L, R}$ be a [[Definition:Dedekind Cut|Dedekind cut]] of the set of [[Definition:Real Number|real numbers]] $\R$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Real Number|real number]] which is a [[Definition:Producer of Dedekind Cut|producer]] of $\tuple {L, R}$.
+\end{theorem}
+
+\begin{proof}
+Suppose $P$ and $Q$ are two [[Definition:Property|properties]] which are mutually exclusive.
+Suppose that one of either of $P$ and $Q$ are possessed by every $x \in \R$.
+Suppose that any number having $P$ is less than any which have $Q$.
+Let us call the numbers with $P$ the ''left hand set'' $L$, and the ones with $Q$ the ''right hand set'' $R$.
+There are two possibilities, as follows.
+* $L$ has a greatest element, or
+* $R$ has a least element.
+'''It is not possible that both of the above can happen.'''
+Because suppose $l$ is the greatest element of $L$ and $r$ is the least element of $R$.
+Then the number $\displaystyle \frac {l + r} 2$ is greater than $l$ and less than $r$, so it could not be in either class.
+'''However, one of the above ''must'' occur.'''
+Because, suppose the following.
+Let $L_1$ and $R_1$ be the [[Definition:Subset|subsets]] of $L$ and $R$ respectively consisting of only the [[Definition:Rational Number|rational numbers]] in $L$ and $R$.
+Then $L_1$ and $R_1$ form a [[Definition:Dedekind Cut|section]] of the set of [[Definition:Rational Number|rational numbers]] $\Q$.
+There are two cases to think about:
+'''Maybe $L_1$ has a greatest element $\alpha$.'''
+In this case, $\alpha$ must also be the greatest element of $L$.
+Because if not, then there's a greater one, which we can call $\beta$.
+There are always rational numbers between $\alpha$ and $\beta$ from [[Rational Numbers are Close Packed]].
+These are less than $\beta$ and thus belong to $L$ and (because they're rational) also to $L_1$.
+This is a contradiction, so if $\alpha$ is the greatest element of $L_1$, it's also the greatest element of $L$.
+'''On the other hand, $L_1$ may ''not'' have a greatest element.'''
+In this case, the section of the rational numbers formed by $L_1$ and $R_1$ is a real number $\alpha$.
+{{Explain|Justify the above assertion.}}
+It must belong to either $L$ or $R$.
+If it belongs to $L$ we can show, like we did before, that it is the greatest element of $L$.
+Similarly, if it belongs to $R$ we can show it is the least element of $R$.
+So in any case, either $L$ has a greatest element or $R$ has a least element.
+Thus, any section of the real numbers corresponds to a real number.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Proof of Uniqueness ===
+{{AimForCont}} both $\alpha$ and $\beta$ [[Definition:Producer of Dedekind Cut|produce]] $\tuple {L, R}$.
+By the [[Trichotomy Law for Real Numbers]] either $\beta < \alpha$ or $\alpha < \beta$.
+Suppose that $\beta < \alpha$.
+From [[Real Numbers are Close Packed]], there exists at least one [[Definition:Real Number|real number]] $c$ such that $\beta < c$ and $c < \alpha$.
+Because $c < \alpha$, it must be the case that $c \in L$.
+Because $\beta < c$, it must be the case that $c \in R$.
+That is:
+:$c \in L \cap R$
+But by the definition of [[Definition:Dedekind Cut|Dedekind Cut]], $\tuple {L, R}$ is a [[Definition:Set Partition|partition]] of $\R$.
+That is, $L$ and $R$ are [[Definition:Disjoint Sets|disjoint]].
+That is:
+:$L \cap R = \O$
+This is a [[Definition:Contradiction|contradiction]].
+Similarly, $\alpha < \beta$ also leads to a [[Definition:Contradiction|contradiction]].
+It follows that $\alpha$ is [[Definition:Unique|unique]].
+{{qed|lemma}}
+=== Proof of Existence ===
+Let $\alpha = \sup L$.
+Let $u_1 \in \R$ such that $u_1 < \alpha$.
+{{AimForCont}} $u_1 \in R$.
+Suppose there is no [[Definition:Real Number|real number]] $s \in L$ such that $u_1 < s \le \alpha$.
+Then $u_1$ would be an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] of $L$ which is less than $\sup {L}$.
+This is a [[Definition:Contradiction|contradiction]].
+So there exists at least one [[Definition:Real Number|real number]] $s \in L$ such that $u_1 < s \le \alpha$.
+But this is also a [[Definition:Contradiction|contradiction]] because all the [[Definition:Element|elements]] of $R$ are greater than all the [[Definition:Element|elements]] of $L$.
+It follows that $u_1 \in L$.
+Let $u_2 \in \R$ such that $\alpha < u_2$.
+{{AimForCont}} $u_2 \in L$.
+There exists $s' \in \R$ such that $\alpha < s' < u_2$.
+But $s' \notin L$ and thus $s' \in R$.
+This is a [[Definition:Contradiction|contradiction]] because all the [[Definition:Element|elements]] of $R$ are greater than all the [[Definition:Element|elements]] of $L$.
+It follows that $u_2 \in R$.
+Hence, $\alpha$ produces the [[Definition:Dedekind Cut|cut]] $\tuple {L, R}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Proof of Uniqueness ===
+{{AimForCont}} both $\alpha$ and $\beta$ [[Definition:Producer of Dedekind Cut|produce]] $\tuple {L, R}$.
+By the [[Trichotomy Law for Real Numbers]] either $\beta < \alpha$ or $\alpha < \beta$.
+Suppose that $\beta < \alpha$.
+From [[Real Numbers are Close Packed]], there exists at least one [[Definition:Real Number|real number]] $c$ such that $\beta < c$ and $c < \alpha$.
+Because $c < \alpha$, it must be the case that $c \in L$.
+Because $\beta < c$, it must be the case that $c \in R$.
+That is:
+:$c \in L \cap R$
+But by the definition of [[Definition:Dedekind Cut|Dedekind Cut]], $\tuple {L, R}$ is a [[Definition:Set Partition|partition]] of $\R$.
+That is, $L$ and $R$ are [[Definition:Disjoint Sets|disjoint]].
+That is:
+:$L \cap R = \O$
+This is a [[Definition:Contradiction|contradiction]].
+Similarly, $\alpha < \beta$ also leads to a [[Definition:Contradiction|contradiction]].
+It follows that $\alpha$ is [[Definition:Unique|unique]].
+{{qed|lemma}}
+=== Proof of Existence ===
+Let $\gamma$ be the [[Definition:Set|set]] of all [[Definition:Rational Number|rational numbers]] $p$ such that $p \in \alpha$ for some $\alpha \in L$.
+It is to be verified that $\gamma$ is a [[Definition:Cut (Analysis)|cut]].
+Because $L$ is not [[Definition:Empty Set|empty]], neither is $\gamma$.
+Suppose $\beta \in \R$ and $q \notin \beta$.
+Then because $\alpha < \beta$, we have that $q \notin \alpha$ for any $\alpha \in L$.
+Thus $q \notin \gamma$.
+Thus $\gamma$ satisfies criterion $(1)$ for being a [[Definition:Cut (Analysis)|cut]].
+Suppose $p \in \gamma$ and $q < p$.
+Then $p \in \alpha$ for some $\alpha \in L$.
+Hence $q \in \alpha$.
+Hence $q \in \gamma$.
+Thus $\gamma$ satisfies criterion $(2)$ for being a [[Definition:Cut (Analysis)|cut]].
+Suppose $p \in \gamma$.
+Then $p \in \alpha$ for some $\alpha \in L$.
+Hence there exists $q > p$ such that $q \in \alpha$.
+Hence $q \in \gamma$.
+Thus $\gamma$ satisfies criterion $(3)$ for being a [[Definition:Cut (Analysis)|cut]].
+Thus, by definition of the [[Definition:Real Number/Dedekind Cuts|real numbers]] by identifying them with [[Definition:Cut (Analysis)|cuts]], $\gamma$ is a [[Definition:Real Number|real number]].
+We have that:
+:$\alpha \le \gamma$
+for all $\alpha \in L$.
+{{AimForCont}} there exists $\beta \in R$ such that $\beta < \gamma$.
+Then there exists some [[Definition:Rational Number|rational number]] $p \in\ Q$ such that $p \in \gamma$ and $p \notin beta$.
+But if $p \in \gamma$ then $p \in \alpha$ for some $\alpha\ in L$.
+This implies that $\beta < \alpha$.
+But this [[Definition:Contradiction|contradicts]] the definition of [[Definition:Dedekind Cut/Definition 2|Dedekind cut]]:
+:$(3): \quad \forall x \in L: \forall y \in R: x < y$
+Thus $\gamma \le \beta$ for all $\beta \in R$.
+That is, $\gamma$ is a [[Definition:Producer of Dedekind Cut|producer]] of $\tuple {L, R}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Addition is Closed}
+Tags: Complex Addition, Complex Addition is Closed
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Complex Addition|addition]]:
+:$\forall z, w \in \C: z + w \in \C$
+\end{theorem}
+
+\begin{proof}
+From the informal definition of [[Definition:Complex Number/Definition 1|complex numbers]], we define the following:
+:$z = x_1 + i y_1$
+:$w = x_2 + i y_2$
+where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2 \in \R$.
+Then from the definition of [[Definition:Complex Addition|complex addition]]:
+:$z + w = \paren {x_1 + x_2} + i \paren {y_1 + y_2}$
+From [[Real Numbers under Addition form Abelian Group]], [[Definition:Real Addition|real addition]] is [[Definition:Closed Operation|closed]].
+So:
+:$\paren {x_1 + x_2} \in \R$ and $\paren {y_1 + y_2} \in \R$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the formal definition of [[Definition:Complex Number/Definition 2|complex numbers]], we have:
+:$z = \tuple {x_1, y_1}$
+:$w = \tuple {x_2, y_2}$
+where $x_1, x_2, y_1, y_2 \in \R$.
+Then from the definition of [[Definition:Complex Number/Definition 2/Addition|complex addition]]:
+:$z + w = \tuple {x_1 + x_2, y_1 + y_2}$
+From [[Real Numbers under Addition form Abelian Group]], [[Definition:Real Addition|real addition]] is [[Definition:Closed Operation|closed]].
+So:
+:$\paren {x_1 + x_2} \in \R$ and $\paren {y_1 + y_2} \in \R$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Addition is Associative}
+Tags: Complex Addition
+
+\begin{theorem}
+The operation of [[Definition:Complex Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Associative|associative]]:
+:$\forall z_1, z_2, z_3 \in \C: z_1 + \paren {z_2 + z_3} = \paren {z_1 + z_2} + z_3$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Number/Definition 2|complex numbers]], we define the following:
+{{begin-eqn}}
+{{eqn | l = z_1
+      | o = :=
+      | r = \tuple {x_1, y_1}
+}}
+{{eqn | l = z_2
+      | o = :=
+      | r = \tuple {x_2, y_2}
+}}
+{{eqn | l = z_3
+      | o = :=
+      | r = \tuple {x_3, y_3}
+}}
+{{end-eqn}}
+where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = z_1 + \paren {z_2 + z_3}
+      | r = \tuple {x_1, y_1} + \paren {\tuple {x_2, y_2} + \tuple {x_3, y_3} }
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{eqn | r = \tuple {x_1, y_1} + \paren {\tuple {x_2 + x_3, y_2 + y_3} }
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \tuple {x_1 + \paren {x_2 + x_3}, y_1 + \paren {y_2 + y_3} }
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \tuple {\paren {x_1 + x_2} + x_3, \paren {y_1 + y_2} + y_3}
+      | c = [[Real Addition is Associative]]
+}}
+{{eqn | r = \paren {\tuple {x_1 + x_2, y_1 + y_2} } + \tuple {x_3, y_3}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \paren {\tuple {x_1, y_1} + \tuple {x_2, y_2} } + \tuple {x_3, y_3}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \paren {z_1 + z_2} + z_3
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Addition is Commutative}
+Tags: Complex Addition
+
+\begin{theorem}
+The operation of [[Definition:Complex Addition|addition]] on the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] is [[Definition:Commutative Operation|commutative]]:
+:$\forall z, w \in \C: z + w = w + z$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Number/Definition 2|complex numbers]], we define the following:
+{{begin-eqn}}
+{{eqn | l = z
+      | o = :=
+      | r = \tuple {x_1, y_1}
+}}
+{{eqn | l = w
+      | o = :=
+      | r = \tuple {x_2, y_2}
+}}
+{{end-eqn}}
+where $x_1, x_2, y_1, y_2 \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z + w
+      | r = \tuple {x_1, y_1} + \tuple {x_2, y_2}
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{eqn | r = \tuple {x_1 + x_2, y_1 + y_2}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \tuple {x_2 + x_1, y_2 + y_1}
+      | c = [[Real Addition is Commutative]]
+}}
+{{eqn | r = \tuple {x_2, y_2} + \tuple {x_1, y_1}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = w + z
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers are Countably Infinite}
+Tags: Integers, Countable Sets
+
+\begin{theorem}
+The [[Definition:Set|set]] $\Z$ of [[Definition:Integer|integers]] is [[Definition:Countably Infinite|countably infinite]].
+\end{theorem}
+
+\begin{proof}
+Define the [[Definition:Inclusion Mapping|inclusion mapping]] $i: \N \to \Z$.
+From [[Inclusion Mapping is Injection]], $i: \N \to \Z$ is an [[Definition:Injection|injection]].
+Thus there exists an injection from $\N$ to $\Z$.
+Hence $\Z$ is [[Definition:Infinite|infinite]].
+Next, let us arrange $\Z$ in the following order:
+:$\Z = \set {0, 1, -1, 2, -2, 3, -3, \ldots}$
+Then we can directly see that we can define a [[Definition:Mapping|mapping]] $\phi: \Z \to \N$ as follows:
+:$\forall x \in \Z: \map \phi x = \begin{cases} 2 x - 1 & : x > 0 \\ -2 x & : x \le 0 \end{cases}$
+This is shown to be an [[Definition:Injection|injection]] as follows:
+Let $\map \phi x = \map \phi y$.
+Then one of the following applies:
+: $(1): \quad -2 x = -2 y$ in which case $x = y$
+: $(2): \quad 2 x - 1 = 2 y - 1$ in which case $2 x = 2 y$ and so $x = y$
+: $(3): \quad 2 x - 1 = -2 y$ in which case $y = -x + \frac 1 2$ and therefore $y \notin \Z$
+: $(4): \quad 2 y - 1 = -2 x$ in which case $x = -y + \frac 1 2$ and therefore $x \notin \Z$.
+So $2 x - 1 = -2 y$ and $2 y - 1 = -2 x$ can't happen and so $x = y$.
+Thus $\phi$ is [[Definition:Injection|injective]].
+The result follows from [[Domain of Injection to Countable Set is Countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Permutation of Determinant Indices}
+Tags: Determinants
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix of order $n$]] over a [[Definition:Field (Abstract Algebra)|field]].
+Let $\lambda: \N_{> 0} \to \N_{> 0}$ be any fixed [[Definition:Permutation on n Letters|permutation on $\N_{> 0}$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\struct {S_n, \circ}$ be the [[Definition:Symmetric Group on n Letters|symmetric group of $n$ letters]]. 
+Then:
+:$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$
+:$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$
+where:
+:the summation $\displaystyle \sum_{\mu \mathop \in S_n}$ goes over all the $n!$ [[Definition:Permutation on n Letters|permutations]] of $\set {1, 2, \ldots, n}$
+:$\map \sgn \mu$ is the [[Definition:Sign of Permutation|sign of the permutation]] $\mu$.
+\end{theorem}
+
+\begin{proof}
+First it is shown that:
+:$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} }$
+Let $\nu: \N_{> 0} \to \N_{> 0}$ be a [[Definition:Permutation|permutation]] on $\N_{> 0}$ such that $\lambda \circ \nu = \mu$.
+The product can be rearranged as:
+:$\displaystyle \prod_{k \mathop = 1}^n a_{\map \lambda k, \map \mu k} = a_{\map \lambda 1, \map \mu 1} a_{\map \lambda 2, \map \mu 2} \cdots a_{\map \lambda n, \map \mu n} = a_{1, \map \nu 1} a_{2, \map \nu 2} \cdots a_{n, \map \nu n} = \prod_{k \mathop = 1}^n a_{k, \map \nu k}$
+from {{Field-axiom|M2}}.
+By [[Parity Function is Homomorphism]]:
+:$\map \sgn \mu \map \sgn \lambda = \map \sgn \lambda \map \sgn \nu \map \sgn \lambda = \map {\sgn^2} \lambda \map \sgn \nu = \map \sgn \nu$
+and so:
+:$\displaystyle \map \det {\mathbf A} = \sum_{\nu \mathop \in S_n} \paren {\map \sgn \nu \prod_{k \mathop = 1}^n a_{k, \map \nu k} }$
+which is the usual definition for the [[Definition:Determinant of Matrix|determinant]].
+Next it is to be shown:
+:$\displaystyle \map \det {\mathbf A} = \sum_{\mu \mathop \in S_n} \paren {\map \sgn \mu \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \mu k, \map \lambda k} }$
+Let $\nu: \N_{> 0} \to \N_{> 0}$ be a [[Definition:Permutation|permutation]] on $\N_{> 0}$ such that $\mu \circ \nu = \lambda$.
+The result follows via a similar argument.
+{{Qed}}
+[[Category:Determinants]]
+ncgpkk0vowuqgjdvv37pev1rssrws1s
+\end{proof}<|endoftext|>
+\section{Determinant of Transpose}
+Tags: Determinants, Transposes of Matrices
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Then:
+:$\map \det {\mathbf A} = \map \det {\mathbf A^\intercal}$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \begin{bmatrix}
+a_{11} & a_{12} & \ldots & a_{1n} \\
+a_{21} & a_{22} & \cdots & a_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+a_{n1} & a_{n2} & \cdots & a_{nn} \\
+\end{bmatrix}$.
+Then $\mathbf A^\intercal = \begin{bmatrix}
+a_{11} & a_{21} & \ldots & a_{n1} \\
+a_{12} & a_{22} & \cdots & a_{n2} \\
+\vdots & \vdots & \ddots & \vdots \\
+a_{1n} & a_{2n} & \cdots & a_{nn} \\
+\end{bmatrix}$.
+Let $b_{r s} = a_{s r}$ for $1 \le r, s \le n$.
+We need to show that $\map \det {\sqbrk a_n} = \map \det {\sqbrk b_n}$.
+By the definition of [[Definition:Determinant of Matrix|determinant]] and [[Permutation of Determinant Indices]], we have:
+{{begin-eqn}}
+{{eqn | l = \map \det {\sqbrk b_n}
+      | r = \sum_\lambda \map {\sgn} \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}
+      | c = 
+}}
+{{eqn | r = \sum_\lambda \map {\sgn} \lambda a_{\map \lambda 1 1} a_{\map \lambda 2 2} \cdots a_{\map \lambda n n}
+      | c = 
+}}
+{{eqn | r = \map \det {\sqbrk a_n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant with Rows Transposed}
+Tags: Determinants, Determinant with Rows Transposed
+
+\begin{theorem}
+If two [[Definition:Row of Matrix|rows]] of a [[Definition:Matrix|matrix]] with [[Definition:Determinant of Matrix|determinant]] $D$ are [[Definition:Transposition|transposed]], its determinant becomes $-D$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $1 \le r < s \le n$.
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] that exchanging [[Definition:Row of Matrix|rows]] $r$ and $s$.
+Let $\mathbf B = \map e {\mathbf A}$.
+Let $\mathbf E$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e$.
+From [[Elementary Row Operations as Matrix Multiplications]]:
+:$\mathbf B = \mathbf E \mathbf A$
+From [[Determinant of Elementary Row Matrix/Exchange Rows|Determinant of Elementary Row Matrix: Exchange Rows]]:
+:$\map \det {\mathbf E} = -1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf B}
+      | r = \map \det {\mathbf E \mathbf A}
+      | c = [[Determinant of Matrix Product]]
+}}
+{{eqn | r = -\map \det {\mathbf A}
+      | c = as $\map \det {\mathbf E} = -1$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Matrix with Duplicate Rows has Zero Determinant}
+Tags: Matrix Algebra, Determinants, Square Matrix with Duplicate Rows has Zero Determinant
+
+\begin{theorem}
+If two [[Definition:Row of Matrix|rows]] of a [[Definition:Square Matrix|square matrix]] over a [[Definition:Commutative Ring|commutative ring]] $\struct {R, +, \circ}$ are the same, then its [[Definition:Determinant of Matrix|determinant]] is [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] over $n$, the [[Definition:Order of Square Matrix|order of the square matrix]].
+=== Basis for the Induction ===
+Let $n = 2$, which is the smallest [[Definition:Natural Number|natural number]] for which a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]] can have two identical [[Definition:Row of Matrix|rows]].
+Let $\mathbf A = \left[{a}\right]_2$ be a [[Definition:Square Matrix|square matrix]] over $R$ with two identical [[Definition:Row of Matrix|rows]].
+Then, by definition of [[Definition:Determinant of Matrix|determinant]]:
+:$\det \left({\mathbf A}\right) = a_{11}a_{22} - a_{12}a_{21} = a_{11}a_{22}-a_{22}a_{11} = 0$
+{{qed|lemma}}
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Assume for $n \in \N_{\ge 2}$ that any [[Definition:Square Matrix|square matrices of order $n$]] over $R$ with two identical [[Definition:Row of Matrix|rows]] has [[Definition:Determinant of Matrix|determinant]] equal to $0$.
+=== Induction Step ===
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix of order $n+1$]] over $R$ with two identical [[Definition:Row of Matrix|rows]].
+Let $i_1, i_2 \in \left\{ {1, \ldots, n+1}\right\}$ be the indices of the identical rows, and let $i_1 < i_2$.
+Let $i \in \left\{ {1, \ldots, n+1}\right\}$.
+Let $\mathbf A \left({i ; 1}\right)$ denote the [[Definition:Submatrix|submatrix]] obtained from $\mathbf A$ by removing row $i$ and [[Definition:Column of Matrix|column]] $1$.
+If $i \ne i_1$ and $i \ne i_2$, then $\mathbf A \left({i ; 1}\right)$ still contains two identical rows.
+By the [[Square Matrix with Duplicate Rows has Zero Determinant/Proof 1#Induction Hypothesis|induction hypothesis]]:
+: $\det \left({\mathbf A \left({i ; 1}\right) }\right) = 0$
+Now consider the [[Definition:Matrix|matrices]] $\mathbf A \left({i_1 ; 1}\right)$ and $\mathbf A \left({i_2 ; 1}\right)$.
+Let $r_j$ denote row $j$ of $\mathbf A \left({i_1 ; 1}\right)$.
+If we perform the following [[Definition:Sequence|sequence]] of $i_2 - i_1 -1$ [[Definition:Elementary Row Operation|elementary row operations]] on $\mathbf A \left({i_1 ; 1}\right)$:
+:$r_{i_1} \leftrightarrow r_{i_1 +1} \ ; \ r_{i_1 + 1} \leftrightarrow r_{i_1 +2}  \ ;  \ \ldots  \  ;  \ r_{i_2 - 1} \leftrightarrow r_{i_2}$
+we will transform $\mathbf A \left({i_1 ; 1}\right)$ into $\mathbf A \left({i_2 ; 1}\right)$.
+From [[Determinant with Rows Transposed]], it follows that $\det \left({\mathbf A \left({i_1 ; 1}\right) }\right) = \left({-1}\right)^{i_2 - i_1 - 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = \det \left({\mathbf A}\right)
+      | r = \sum_{k \mathop = 1 }^{n + 1} a_{k1} \left({-1}\right)^{k + 1} \det \left({\mathbf A \left({k ; 1}\right) }\right)
+      | c = [[Expansion Theorem for Determinants|expanding]] the determinant along column $1$
+}}
+{{eqn | r = a_{i_1 1}\left({-1}\right)^{i_1 + 1} \det \left({\mathbf A \left({i_1 ; 1}\right) }\right) + a_{i_2 1} \left({-1}\right)^{i_2 + 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right)
+}}
+{{eqn | r = a_{i_2 1} \left({-1}\right)^{i_1 + 1 + i_2 - i_1 - 1} \det \left({\mathbf A \left({i_2; 1}\right) }\right) + a_{i_2 1} \left({-1}\right)^{i_2 + 1} \det \left({\mathbf A \left({i_2 ; 1}\right) }\right)
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose that $\forall x \in R: x + x = 0 \implies x = 0$.
+From [[Determinant with Rows Transposed]], if you exchange two [[Definition:Row of Matrix|rows]] of a [[Definition:Square Matrix|square matrix]], the sign of its [[Definition:Determinant of Matrix|determinant]] changes.
+If you exchange two identical [[Definition:Row of Matrix|rows]] of a [[Definition:Square Matrix|square matrix]], then the sign of its [[Definition:Determinant of Matrix|determinant]] changes from $D$, say, to $-D$.
+But the [[Definition:Square Matrix|matrix]] stays the same.
+So $D = -D$ and so $D = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant with Row Multiplied by Constant}
+Tags: Determinants, Determinant with Row Multiplied by Constant
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf B$ be the [[Definition:Square Matrix|matrix]] resulting from one [[Definition:Row of Matrix|row]] of $\mathbf A$ having been multiplied by a [[Definition:Constant|constant]] $c$.
+Then:
+:$\map \det {\mathbf B} = c \map \det {\mathbf A}$
+That is, multiplying one [[Definition:Row of Matrix|row]] of a [[Definition:Square Matrix|square matrix]] by a [[Definition:Constant|constant]] multiplies its [[Definition:Determinant of Matrix|determinant]] by that [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] that [[Definition:Matrix Scalar Product|multiplies]] [[Definition:Row of Matrix|rows]] $i$ by the [[Definition:Scalar (Matrix Theory)|scalar]]$c$.
+Let $\mathbf B = \map e {\mathbf A}$.
+Let $\mathbf E$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e$.
+From [[Elementary Row Operations as Matrix Multiplications]]:
+:$\mathbf B = \mathbf E \mathbf A$
+From [[Determinant of Elementary Row Matrix/Exchange Rows|Determinant of Elementary Row Matrix: Exchange Rows]]:
+:$\map \det {\mathbf E} = c$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf B}
+      | r = \map \det {\mathbf E \mathbf A}
+      | c = [[Determinant of Matrix Product]]
+}}
+{{eqn | r = c \map \det {\mathbf A}
+      | c = as $\map \det {\mathbf E} = c$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant as Sum of Determinants}
+Tags: Determinants
+
+\begin{theorem}
+Let $\begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix}$ be a [[Definition:Determinant of Matrix|determinant]].
+Then $\begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix} = \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix} + \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix}$.
+Similarly:
+Then $\begin{vmatrix}
+  a_{11} & \cdots & a_{1s} + a'_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} + a'_{ns} & \cdots & a_{nn}
+\end{vmatrix} = \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix} + \begin{vmatrix}
+  a_{11} & \cdots & a'_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a'_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a'_{ns} & \cdots & a_{nn}
+\end{vmatrix}$.
+\end{theorem}
+
+\begin{proof}
+Let:
+: $B = \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} + a'_{r1} & \cdots & a_{rs} + a'_{rs} & \cdots & a_{rn} + a'_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix} = \begin{vmatrix}
+  b_{11} & \cdots & b_{1s} & \cdots & b_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  b_{r1} & \cdots & b_{rs} & \cdots & b_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  b_{n1} & \cdots & b_{ns} & \cdots & b_{nn}
+\end{vmatrix}$
+: $A_1 = \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{r1} & \cdots & a_{rs} & \cdots & a_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix}$
+: $A_2 = \begin{vmatrix}
+  a_{11} & \cdots & a_{1s} & \cdots & a_{1n} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a'_{r1} & \cdots & a'_{rs} & \cdots & a'_{rn} \\
+\vdots & \ddots & \vdots & \ddots & \vdots \\
+  a_{n1} & \cdots & a_{ns} & \cdots & a_{nn}
+\end{vmatrix}$
+Then: 
+{{begin-eqn}}
+{{eqn | l = B
+      | r = \sum_\lambda \left({\operatorname{sgn} \left({\lambda}\right) \prod_{k=1}^n b_{k \lambda \left({k}\right)} }\right)
+      | c = 
+}}
+{{eqn | r = \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots \left({a_{r \lambda \left({r}\right)} + a'_{r \lambda \left({r}\right)} }\right) \cdots a_{n \lambda \left({n}\right)}
+      | c = 
+}}
+{{eqn | r = \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots a_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)} + \sum_\lambda \operatorname{sgn} \left({\lambda}\right) a_{1 \lambda \left({1}\right)} \cdots a'_{r \lambda \left({r}\right)} \cdots a_{n \lambda \left({n}\right)}
+      | c = 
+}}
+{{eqn | r = A_1 + A_2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+The result for columns follows directly from [[Determinant of Transpose]].
+{{qed}}
+[[Category:Determinants]]
+444jdmhgv95gdwkmqke2otcqw0v698t
+\end{proof}<|endoftext|>
+\section{Multiple of Row Added to Row of Determinant}
+Tags: Determinants, Multiple of Row Added to Row of Determinant
+
+\begin{theorem}
+Let $\mathbf A = \begin {bmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{s 1} & a_{s 2} & \cdots & a_{s n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
+\end {bmatrix}$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ denote the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf B = \begin{bmatrix}
+a_{1 1}             & a_{1 2}             & \cdots &             a_{1 n} \\
+ \vdots             & \vdots              & \ddots &              \vdots \\
+a_{r 1} + k a_{s 1} & a_{r 2} + k a_{s 2} & \cdots & a_{r n} + k a_{s n} \\
+ \vdots             & \vdots              & \ddots &              \vdots \\
+a_{s 1}             & a_{s 2}             & \cdots &             a_{s n} \\
+ \vdots             & \vdots              & \ddots &              \vdots \\
+a_{n 1}             & a_{n 2}             & \cdots &             a_{n n} \\
+\end{bmatrix}$.
+Then $\map \det {\mathbf B} = \map \det {\mathbf A}$.
+That is, the value of a [[Definition:Determinant of Matrix|determinant]] remains unchanged if a [[Definition:Constant|constant]] multiple of any [[Definition:Row of Matrix|row]] is added to any other [[Definition:Row of Matrix|row]].
+\end{theorem}<|endoftext|>
+\section{Determinant of Matrix Product}
+Tags: Determinants, Conventional Matrix Multiplication, Determinant of Matrix Product
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be a [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf A \mathbf B$ be the [[Definition:Matrix Product (Conventional)|(conventional) matrix product]] of $\mathbf A$ and $\mathbf B$.
+Then:
+:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
+That is, the [[Definition:Determinant of Matrix|determinant]] of the [[Definition:Matrix Product (Conventional)|product]] is equal to the [[Definition:Multiplication|product]] of the [[Definition:Determinant of Matrix|determinants]].
+\end{theorem}
+
+\begin{proof}
+This proof assumes that $\mathbf A$ and $\mathbf B$ are $n \times n$-[[Definition:Matrix|matrices]] over a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] $\left({R, +, \circ}\right)$.
+Let $\mathbf C = \left[{c}\right]_n = \mathbf A \mathbf B$.
+From [[Square Matrix is Row Equivalent to Triangular Matrix]], it follows that $\mathbf A$ can be converted into a [[Definition:Upper Triangular Matrix|upper triangular matrix]] $\mathbf A'$ by a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\hat o_1, \ldots, \hat o_{m'}$.
+Let $\mathbf C'$ denote the matrix that results from using $\hat o_1, \ldots, \hat o_{m'}$ on $\mathbf C$.
+From [[Elementary Row Operations Commute with Matrix Multiplication]], it follows that $\mathbf C' = \mathbf A' \mathbf B$.
+[[Effect of Sequence of Elementary Row Operations on Determinant]] shows that there exists $\alpha \in R$ such that:
+:$\alpha \det \left({\mathbf A'}\right) = \det \left({\mathbf A}\right)$
+:$\alpha \det \left({\mathbf C'}\right) = \det \left({\mathbf C}\right)$
+Let $\mathbf B^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] of $B$.
+From [[Transpose of Matrix Product]], it follows that:
+: $\left({\mathbf C'}\right)^\intercal = \left({\mathbf A' \mathbf B}\right)^\intercal = \mathbf B^\intercal \left({\mathbf A'}\right)^\intercal$
+From [[Square Matrix is Row Equivalent to Triangular Matrix]], it follows that $\mathbf B^\intercal$ can be converted into a [[Definition:Triangular Matrix/Lower Triangular Matrix|lower triangular matrix]] $\left({\mathbf B^\intercal}\right)'$ by a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\hat p_1, \ldots, \hat p_{m''}$.
+Let $\mathbf C''$ denote the matrix that results from using $\hat p_1, \ldots, \hat p_{m''}$ on $\left({\mathbf C'}\right)^\intercal$.
+From [[Elementary Row Operations Commute with Matrix Multiplication]], it follows that:
+: $\mathbf C'' = \left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal$
+[[Effect of Sequence of Elementary Row Operations on Determinant]] shows that there exists $\beta \in R$ such that:
+:$\beta \det \left({\left({\mathbf B^\intercal}\right)'}\right) = \det \left({\mathbf B^\intercal}\right)$
+:$\beta \det \left({\mathbf C''}\right) = \det \left({ \left({\mathbf C'}\right)^\intercal }\right)$
+From [[Transpose of Upper Triangular Matrix is Lower Triangular]], it follows that $\left({\mathbf A'}\right)^\intercal$ is a lower triangular matrix.
+Then [[Product of Triangular Matrices]] shows that $\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal$ is a lower triangular matrix whose [[Definition:Diagonal Element|diagonal elements]] are the products of the diagonal elements of $\left({\mathbf B^\intercal}\right)'$ and $\left({\mathbf A'}\right)^\intercal$.
+From [[Determinant of Triangular Matrix]], we have that $\det \left({\left({\mathbf A'}\right)^\intercal}\right)$, $\det \left({\left({\mathbf B^\intercal}\right)' }\right)$, and $\det \left({\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal }\right)$ are equal to the product of their diagonal elements.
+Combinining these results shows that:
+:$\det \left({\left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal}\right) = \det \left({\left({\mathbf B^\intercal}\right)'}\right) \det \left({\left({\mathbf A'}\right)^\intercal }\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = \det \left({\mathbf C}\right)
+      | r = \alpha \det \left({\mathbf C'}\right)
+}}
+{{eqn | r = \alpha \det \left({ \left({\mathbf C'}\right)^\intercal}\right)
+      | c = [[Determinant of Transpose]]
+}}
+{{eqn | r = \alpha \beta \det \left({\mathbf C''}\right)
+}}
+{{eqn | r = \alpha \beta \det \left({ \left({\mathbf B^\intercal}\right)' \left({\mathbf A'}\right)^\intercal}\right)
+}}
+{{eqn | r = \alpha \beta \det \left({\left({\mathbf B^\intercal}\right)' }\right) \det \left({\left({\mathbf A'}\right)^\intercal}\right)
+}}
+{{eqn | r = \alpha \det \left({\left({\mathbf A'}\right)^\intercal}\right) \beta \det \left({\left({\mathbf B^\intercal}\right)' }\right) 
+      | c = [[Definition:Commutative Operation|Commutativity]] of [[Definition:Ring Product|Ring Product]] in $R$
+}}
+{{eqn | r = \alpha \det \left({\mathbf A'}\right) \det \left({\mathbf B^\intercal}\right)
+      | c = [[Determinant of Transpose]]
+}}
+{{eqn | r = \det \left({\mathbf A}\right) \det \left({\mathbf B}\right)
+      | c = [[Determinant of Transpose]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider two cases:
+:$(1): \quad \mathbf A$ is not [[Definition:Invertible Matrix|invertible]].
+:$(2): \quad \mathbf A$ is [[Definition:Invertible Matrix|invertible]].
+=== Proof of case $1$ ===
+Assume $\mathbf A$ is not [[Definition:Invertible Matrix|invertible]]. 
+Then:
+:$\map \det {\mathbf A} = 0$
+Also if $\mathbf A$ is not [[Definition:Invertible Matrix|invertible]] then neither is $\mathbf A \mathbf B$.
+Indeed, if $\mathbf A \mathbf B$ has an inverse $\mathbf C$, then $\mathbf A \mathbf B \mathbf C = \mathbf I$, whereby $\mathbf B \mathbf C$ is a right inverse of $\mathbf A$.
+It follows by [[Left or Right Inverse of Matrix is Inverse]] that in that case $\mathbf B \mathbf C$ is the inverse of $A$.
+It follows that:
+:$\map \det {\mathbf A \mathbf B} = 0$
+Thus:
+:$0 = 0 \cdot \map \det {\mathbf B}$
+:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \cdot \map \det {\mathbf B}$
+{{qed|lemma}}
+=== Proof of case $2$ ===
+Assume $\mathbf A$ is [[Definition:Invertible Matrix|invertible]].
+Then $\mathbf A$ is a product of [[[[Definition:Elementary Row Matrix|elementary row matrices]], $\mathbf E$.
+Let $\mathbf A = \mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1$.
+So:
+:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf E_1 \mathbf B}$
+It remains to be shown that for any [[Definition:Square Matrix|square matrix]] $\mathbf D$ of [[Definition:Order of Square Matrix|order]] $n$:
+:$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$
+Let $e_i \paren {\mathbf I} = \mathbf E_i$ for all $i \in \closedint 1 k$, then using [[Elementary Row Operations as Matrix Multiplications]] and [[Effect of Sequence of Elementary Row Operations on Determinant]] yields:
+:$\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E_k \mathbf E_{k - 1} \dotsm \mathbf {E_1} \mathbf D} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf D} } = \alpha \map \det {\mathbf D}$
+Using [[Elementary Row Operations as Matrix Multiplications]] and [[Effect of Sequence of Elementary Row Operations on Determinant]], '''and''' [[Unit Matrix is Unity of Ring of Square Matrices]]:
+ 
+:$\map \det {\mathbf E} = \map \det {\mathbf E_k \mathbf E_{k - 1} \cdots \mathbf {E_1} \mathbf I} = \map \det {e_k e_{k - 1} \cdots e_1 \paren {\mathbf I} } = \alpha \map \det {\mathbf I}$
+From [[Determinant of Unit Matrix]]:
+:$\map \det {\mathbf E} = \alpha$
+And so $\map \det {\mathbf E \mathbf D} = \map \det {\mathbf E} \cdot \map \det {\mathbf D}$
+{{qed|lemma}}
+Therefore:
+:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
+as required.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The [[Cauchy-Binet Formula]] gives:
+:$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$
+where:
+:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B$ is an [[Definition:Matrix|$n \times m$ matrix]].
+:For $1 \le j_1, j_2, \ldots, j_m \le n$:
+::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
+When $m = n$, the only set $j_1, j_2, \ldots, j_m$ that fulfils $1 \le j_1 < j_2 < \cdots < j_m \le n$ is $\left\{ {1, 2, \ldots, n}\right\}$.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Remember that $\det$ can be interpreted as an alternating multilinear map with respect to the columns.
+This property is sufficient to prove the theorem as follows.
+Let $\mathbf A, \mathbf B$ be two $n \times n$ matrices (with coefficients in a commutative field $\mathbb K$ like $\mathbb R$ or $\mathbb C$).
+Let us denote the vectors of the canonical basis of $\mathbb K^n$ by $\mathbf e_1, \ldots, \mathbf e_n$ (where $\mathbf e_i$ is a column with $1$ at $i$th row, zero elsewhere).
+Now, we are able to write the matrix $\mathbf B$ as a column block matrix :
+:$\mathbf B = \begin {pmatrix} \displaystyle \sum_{s_1 \mathop = 1}^n \mathbf B_{s_1, 1} \mathbf e_{s_1} & \cdots & \displaystyle \sum_{s_n \mathop = 1}^n \mathbf B_{s_n, n} \mathbf e_{s_n} \end {pmatrix}$
+We can rewrite the product $\mathbf A \mathbf B$ as a column-block matrix :
+:$\mathbf A \mathbf B = \begin {pmatrix} \displaystyle \sum_{s_1 \mathop = 1}^n \mathbf B_{s_1, 1} \mathbf A \mathbf e_{s_1} & \cdots & \displaystyle \sum_{s_n \mathop = 1}^n B_{s_n, n} \mathbf A \mathbf e_{s_n} \end {pmatrix} $
+Using linearity with respect to each columns, we get:
+:$\map \det {\mathbf A \mathbf B} = \displaystyle \sum_{1 \mathop \leqslant s_1, \ldots, s_n \mathop \leqslant n} \paren {\prod_{i \mathop = 1}^n \mathbf B_{s_i, i} } \det \begin {pmatrix} \mathbf A \mathbf e_{s_1} & \cdots & \mathbf A \mathbf e_{s_n} \end {pmatrix}$
+Now notice that $\det \begin {pmatrix} \mathbf A \mathbf e_{s_1} & \cdots & \mathbf A \mathbf e_{s_n} \end{pmatrix}$ is zero once two entries are the same (since $\det$ is an alternating map), it means that if for some $k \ne \ell$ we have $\mathbf A \mathbf e_{s_k} = \mathbf A \mathbf e_{s_\ell}$, then $\det \begin {pmatrix} \mathbf A \mathbf e_{s_1} & \cdots & \mathbf A \mathbf e_{s_n} \end {pmatrix} = 0$.
+Therefore the only nonzero summands are those one the $s_1, \ldots, s_n$ are all distinct.
+In other words, the "selector" $s$ represents some permutation of the numbers $1, \ldots, n$.
+As a result, the determinant of the product can now be expressed as a sum of precisely $n!$ terms using permutations:
+:$\map \det {\mathbf A \mathbf B} = \displaystyle \sum_{\sigma \in S_n} \paren {\prod_{i \mathop = 1}^n B_{\map \sigma i, i} } \det \begin {pmatrix} \mathbf A \mathbf e_{\map \sigma 1} & \cdots & \mathbf A \mathbf e_{\map \sigma n} \end {pmatrix}$
+where $S_n$ denotes the set of the permutations of numbers $1, \ldots, n$.
+However, the {{RHS}} determinant of the above equality corresponds to the determinant of permutated columns of $\mathbf A$.
+Whenever we transpose two columns, the determinant is modified by a factor $-1$.
+Indeed, let us apply some transposition $\tau_{i j}$ to a column-block matrix $\begin {pmatrix} \mathbf C_1 & \cdots & \mathbf C_n \end {pmatrix}$.
+By linearity it follows that for $i, j$ entries equal to $\mathbf C_i + \mathbf C_j$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_i + \mathbf C_j \cdots \mathbf C_j + \mathbf C_i \cdots \mathbf C_n \end {pmatrix}
+      | c = 
+}}
+{{eqn | r = \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_i \cdots \mathbf C_j \cdots \mathbf C_n \end {pmatrix} + \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_i \cdots \mathbf C_i \cdots \mathbf C_n \end {pmatrix}
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_j \cdots \mathbf C_j \cdots \mathbf C_n \end {pmatrix} + \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_j \cdots \mathbf C_i \cdots \mathbf C_n \end {pmatrix}
+      | c = 
+}}
+{{eqn | r = \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_i \cdots \mathbf C_j \cdots \mathbf C_n \end {pmatrix} + \det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_j \cdots \mathbf C_i \cdots \mathbf C_n \end {pmatrix}
+      | c = 
+}}
+{{end-eqn}}
+Hence, transpose two columns reverse determinant sign:
+:$\det \begin {pmatrix} \mathbf C_1 \cdots \mathbf C_n \end {pmatrix} = -\det \begin {pmatrix} \mathbf C_{\map {\tau_{i j} } 1} \cdots \mathbf C_{\map {\tau_{i j} } n} \end {pmatrix}$
+Since every permutation $\sigma \in S_n$ can be written as a product of transpositions, that is:
+:$\sigma = \tau_m \cdots \tau_1$
+for some transpositions $\tau_1, \ldots, \tau_m$, it follows that:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf C_1 \cdots \mathbf C_n}
+      | r = -\map \det {\mathbf C_{\map {\tau_1} 1} \cdots \mathbf C_{\map {\tau_1} n} }
+      | c = 
+}}
+{{eqn | r = \map \det {\mathbf C_{\map {\tau_2 \tau_1} 1} \cdots \mathbf C_{\map {\tau_2 \tau_1} n} }
+      | c = 
+}}
+{{eqn | r = \cdots
+      | c = 
+}}
+{{eqn | r = \paren {-1}^m \map \det {\mathbf C_{\map \sigma 1} \cdots \mathbf C_{\map \sigma n} }
+      | c = 
+}}
+{{end-eqn}}
+The number $\paren {-1}^m$ is the signature of the permutation $\sigma$ (see article [[Definition:Sign of Permutation|about the signature of permutations]]) and denoted by $\map \sgn \sigma$.
+It remains to apply several transpositions of columns to $\mathbf A$ to get for any permutation $\sigma$ the equality :
+:$\det \begin {pmatrix} \mathbf A \mathbf e_{\map \sigma 1} & \cdots & \mathbf A \mathbf e_{\map \sigma n} \end {pmatrix} = \map \sgn \sigma \det \begin {pmatrix} \mathbf A \mathbf e_1 & \cdots & \mathbf A \mathbf e_n \end {pmatrix} = \map \sgn \sigma \map \det {\mathbf A}$
+Since $\map \det {\mathbf A}$ is a constant quantity, we can go this factor out of the sum, then write:
+:$\displaystyle \map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \sum_{\sigma \mathop \in S_n} \map \sgn \sigma \prod_{i \mathop = 1}^n \mathbf B_{\map \sigma i, i}$
+But the above sum is exactly the definition of $\map \det {\mathbf B}$ using the Leibniz formula, and so:
+:$\map \det {\mathbf A \mathbf B} = \map \det {\mathbf A} \map \det {\mathbf B}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Expansion Theorem for Determinants}
+Tags: Determinants, Named Theorems
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $D = \map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$:
+:$\displaystyle \map \det {\mathbf A} := \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} } = \sum_\lambda \map \sgn \lambda a_{1 \map \lambda 1} a_{2 \map \lambda 2} \cdots a_{n \map \lambda n}$
+where:
+:the summation $\displaystyle \sum_\lambda$ goes over all the $n!$ [[Definition:Permutation on n Letters|permutations]] of $\set {1, 2, \ldots, n}$
+:$\map \sgn \lambda$ is the [[Definition:Sign of Permutation|sign of the permutation]] $\lambda$.
+Let $a_{p q}$ be an [[Definition:Element of Matrix|element]] of $\mathbf A$.
+Let $A_{p q}$ be the [[Definition:Cofactor of Element|cofactor]] of $a_{p q}$ in $D$.
+Then:
+:$(1): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
+:$(2): \quad \displaystyle \forall r \in \closedint 1 n: D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$
+Thus the value of a [[Definition:Determinant of Matrix|determinant]] can be found either by:
+:multiplying all the [[Definition:Element of Matrix|elements]] in a [[Definition:Row of Matrix|row]] by their [[Definition:Cofactor of Element|cofactor]]s and adding up the products
+or:
+:multiplying all the [[Definition:Element of Matrix|elements]] in a [[Definition:Column of Matrix|column]] by their [[Definition:Cofactor of Element|cofactor]]s and adding up the products.
+The identity:
+:$\displaystyle D = \sum_{k \mathop = 1}^n a_{r k} A_{r k}$
+is known as the '''expansion of $D$ in terms of [[Definition:Row of Matrix|row]] $r$''', while:
+:$\displaystyle D = \sum_{k \mathop = 1}^n a_{k r} A_{k r}$
+is known as the '''expansion of $D$ in terms of [[Definition:Column of Matrix|column]] $r$'''.
+\end{theorem}
+
+\begin{proof}
+Because of [[Determinant of Transpose]], it is necessary to prove only one of these identities.
+Let:
+:$D = \begin {vmatrix}
+a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
+\end {vmatrix}$
+First, note that from [[Determinant with Row Multiplied by Constant]], we have:
+:$\begin{vmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{r 1} &       0 & \cdots &       0 \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n}
+\end {vmatrix} = a_{r 1} \begin {vmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+      1 &       0 & \cdots &       0 \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n}
+\end {vmatrix}$
+and similarly:
+:$\begin {vmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+      0 & a_{r 2} & \cdots &       0 \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n}
+\end {vmatrix} = a_{r 2} \begin {vmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+      0 &       1 & \cdots &       0 \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n}
+\end{vmatrix}$
+and so on for the whole of [[Definition:Row of Matrix|row]] $r$.
+From [[Determinant as Sum of Determinants]]:
+:$\displaystyle \begin {vmatrix}
+a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{r 1} & \cdots & a_{r k} & \cdots & a_{r n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
+\end {vmatrix} = \sum_{k \mathop = 1}^n \paren {a_{r k} \begin {vmatrix}
+a_{1 1} & \cdots & a_{1 k} & \cdots & a_{1 n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+      0 & \cdots &       1 & \cdots &       0 \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & \cdots & a_{n k} & \cdots & a_{n n}
+\end {vmatrix} }$
+Consider the [[Definition:Determinant of Matrix|determinant]]:
+:$\begin{vmatrix}
+             a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 k} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
+              \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} k} & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
+                   0 & \cdots &                                  0 &                    1 &                                  0 & \cdots &                    0 \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} k} & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &                             \vdots &               \ddots &                            \vdots  & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n k} &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
+\end {vmatrix}$
+Exchange [[Definition:Row of Matrix|row]]s $r$ and $r - 1$, then (the new) [[Definition:Row of Matrix|row]] $r - 1$ with [[Definition:Row of Matrix|row]] $r - 2$, until finally [[Definition:Row of Matrix|row]] $r$ is at the top.
+[[Definition:Row of Matrix|Row]] 1 will be in [[Definition:Row of Matrix|row]] 2, [[Definition:Row of Matrix|row]] 2 in [[Definition:Row of Matrix|row]] 3, and so on.
+This is permuting the [[Definition:Row of Matrix|row]]s by a [[Definition:Cyclic Permutation|$k$-cycle]] of length $r$.
+Call that [[Definition:Cyclic Permutation|$k$-cycle]] $\rho$.
+Then from [[Parity of K-Cycle]]:
+:$\map \sgn \rho = \paren {-1}^{r - 1}$
+Thus:
+:$\begin {vmatrix}
+                   0 & \cdots &                    1 & \cdots &                    0 \\
+             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n}
+\end {vmatrix} = \paren {-1}^{r - 1} \begin {vmatrix}
+             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
+                   0 & \cdots &                    1 & \cdots &                    0 \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n k} & \cdots & a_{n n}
+\end {vmatrix}$
+The same argument can be applied to [[Definition:Column of Matrix|column]]s.
+Thus:
+:$\begin {vmatrix}
+                   1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\
+             a_{1 k} &              a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
+              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
+a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
+              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+             a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } &              \cdots & a_{n n}
+\end {vmatrix} = \paren {-1}^{k-1}\begin {vmatrix}
+                   0 & \cdots &                    1 & \cdots &                    0 \\
+             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n}
+\end {vmatrix}$
+and so:
+:$\begin{vmatrix}
+                   1 &                    0 & \cdots &                                  0 &                                  0 & \cdots &                    0 \\
+             a_{1 k} &              a_{1 1} & \cdots &               a_{1 \paren {k - 1}} &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
+              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} k} & a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
+a_{\paren {r + 1} k} & a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
+              \vdots &               \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+             a_{n k} &              a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
+\end {vmatrix} = \paren {-1}^{r + k} \begin {vmatrix}
+             a_{1 1} & \cdots &              a_{1 k} & \cdots &              a_{1 n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} k} & \cdots & a_{\paren {r - 1} n} \\
+                   0 & \cdots &                    1 & \cdots &                    0 \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} k} & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &               \vdots & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n k} & \cdots &              a_{n n}
+\end {vmatrix}$
+Then:
+:$\paren {-1}^{r + k} \begin {vmatrix}
+             a_{1 1} & \cdots &              a_{1 \paren {k - 1} } &              a_{1 \paren {k + 1} } & \cdots &              a_{1 n} \\
+              \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+a_{\paren {r - 1} 1} & \cdots & a_{\paren {r - 1} \paren {k - 1} } & a_{\paren {r - 1} \paren {k + 1} } & \cdots & a_{\paren {r - 1} n} \\
+a_{\paren {r + 1} 1} & \cdots & a_{\paren {r + 1} \paren {k - 1} } & a_{\paren {r + 1} \paren {k + 1} } & \cdots & a_{\paren {r + 1} n} \\
+              \vdots & \ddots &                             \vdots &                             \vdots & \ddots &               \vdots \\
+             a_{n 1} & \cdots &              a_{n \paren {k - 1} } &              a_{n \paren {k + 1} } & \cdots &              a_{n n}
+\end{vmatrix}$
+is $A_{r k}$, the [[Definition:Cofactor of Element|cofactor]] of $a_{r k}$ in $D$.
+But from [[Determinant with Unit Element in Otherwise Zero Row]], we have:
+:$\begin {vmatrix}
+      1 &       0 & \cdots &       0 \\
+b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+b_{n 1} & b_{n 2} & \cdots & b_{n n}
+\end {vmatrix} = \begin {vmatrix}
+b_{2 2} & \cdots & b_{2 n} \\
+ \vdots & \ddots &  \vdots \\
+b_{n 2} & \cdots & b_{n n}
+\end {vmatrix}$
+Assembling all the pieces derived above, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant with Unit Element in Otherwise Zero Row}
+Tags: Determinants
+
+\begin{theorem}
+Let $D$ be the [[Definition:Determinant of Matrix|determinant]]:
+:$D = \begin {vmatrix}
+      1 &       0 & \cdots &       0 \\
+b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+b_{n 1} & b_{n 2} & \cdots & b_{n n}
+\end {vmatrix}$
+Then:
+:$D = \begin {vmatrix}
+b_{2 2} & \cdots & b_{2 n} \\
+ \vdots & \ddots &  \vdots \\
+b_{n 2} & \cdots & b_{n n}
+\end {vmatrix}$
+\end{theorem}
+
+\begin{proof}
+We refer to the elements of:
+:$\begin {vmatrix}
+      1 &       0 & \cdots &       0 \\
+b_{2 1} & b_{2 2} & \cdots & b_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+b_{n 1} & b_{n 2} & \cdots & b_{n n}
+\end {vmatrix}$
+as $\begin {vmatrix} b_{i j} \end {vmatrix}$.
+Thus $b_{1 1} = 1, b_{1 2} = 0, \ldots, b_{1 n} = 0$.
+Then from the definition of [[Definition:Determinant of Matrix|determinant]]:
+{{begin-eqn}}
+{{eqn | l = D
+      | r = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n b_{k \map \lambda k} }
+      | c = 
+}}
+{{eqn | r = \sum_{\lambda} \map \sgn \lambda b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n}
+      | c = 
+}}
+{{end-eqn}}
+Now we note:
+{{begin-eqn}}
+{{eqn | l = \map \lambda 1 = 1
+      | o = \implies
+      | r = b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 1
+      | c = 
+}}
+{{eqn | l = \map \lambda 1 \ne 1
+      | o = \implies
+      | r = b_{1 \map \lambda 1} b_{2 \map \lambda 2} \cdots b_{n \map \lambda n} = 0
+      | c = 
+}}
+{{end-eqn}}
+So only those [[Definition:Permutation on n Letters|permutations]] on $\N^*_n$ such that $\map \lambda 1 = 1$ contribute towards the final [[Definition:Summation|summation]].
+Thus we have:
+:$\displaystyle D = \sum_\mu \map \sgn \mu b_{2 \map \mu 2} \cdots b_{n \map \mu n}$
+where $\mu$ is the collection of all [[Definition:Permutation on n Letters|permutations]] on $\N^*_n$ which [[Definition:Fixed Element of Permutation|fix]] $1$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Vandermonde Determinant}
+Tags: Determinants, Vandermonde Matrices, Vandermonde Determinant
+
+\begin{theorem}
+The '''Vandermonde determinant of order $n$''' is the [[Definition:Determinant of Matrix|determinant]] defined as follows:
+:$V_n = \begin {vmatrix}
+1 & x_1 & x_1^2 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\
+1 & x_2 & x_2^2 & \cdots & x_2^{n - 2} & x_2^{n - 1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+1 & x_n & x_n^2 & \cdots & x_n^{n - 2} & x_n^{n - 1}
+\end {vmatrix}$
+Its value is given by:
+:$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
+\end{theorem}
+
+\begin{proof}
+Let $V_n = \begin{vmatrix}
+  1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\
+  1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\
+  1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1}
+\end{vmatrix}$.
+By [[Multiple of Row Added to Row of Determinant]], we can subtract [[Definition:Row of Matrix|row]] 1 from each of the other rows and leave $V_n$ unchanged:
+:$V_n = \begin{vmatrix}
+  1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  0 & x_2 - x_1 & x_2^2 - x_1^2 & \cdots & x_2^{n-2} - x_1^{n-2} & x_2^{n-1} - x_1^{n-1} \\
+  0 & x_3 - x_1 & x_3^2 - x_1^2 & \cdots & x_3^{n-2} - x_1^{n-2} & x_3^{n-1} - x_1^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  0 & x_{n-1} - x_1 & x_{n-1}^2 - x_1^2 & \cdots & x_{n-1}^{n-2} - x_1^{n-2} & x_{n-1}^{n-1} - x_1^{n-1} \\
+  0 & x_n - x_1 & x_n^2 - x_1^2 & \cdots & x_n^{n-2} - x_1^{n-2} & x_n^{n-1} - x_1^{n-1}
+\end{vmatrix}$
+Similarly without changing the value of $V_n$, we can subtract, in order:
+:$x_1$ times [[Definition:Column of Matrix|column]] $n-1$ from [[Definition:Column of Matrix|column]] $n$
+:$x_1$ times [[Definition:Column of Matrix|column]] $n-2$ from [[Definition:Column of Matrix|column]] $n-1$
+and so on, till we subtract:
+: $x_1$ times [[Definition:Column of Matrix|column]] $1$ from [[Definition:Column of Matrix|column]] $2$.
+The first row will vanish all apart from the first element $a_{11} = 1$.
+On all the other rows, we get, with new $i$ and $j$:
+:$a_{ij} = \left({x_i^{j-1} - x_1^{j-1}}\right) - \left({x_1 x_i^{j-2} - x_1^{j-1}}\right) = \left({x_i - x_1}\right) x_i^{j-2}$:
+:$V_n = \begin{vmatrix}
+  1 & 0 & 0 & \cdots & 0 & 0 \\
+  0 & x_2 - x_1 & \left({x_2 - x_1}\right) x_2 & \cdots & \left({x_2 - x_1}\right) x_2^{n-3} & \left({x_2 - x_1}\right) x_2^{n-2} \\
+  0 & x_3 - x_1 & \left({x_3 - x_1}\right) x_3 & \cdots & \left({x_3 - x_1}\right) x_3^{n-3} & \left({x_3 - x_1}\right) x_3^{n-2} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  0 & x_{n-1} - x_1 & \left({x_{n-1} - x_1}\right) x_{n-1} & \cdots & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-3} & \left({x_{n-1} - x_1}\right) x_{n-1}^{n-2}\\
+  0 & x_n - x_1 & \left({x_n - x_1}\right) x_n & \cdots & \left({x_n - x_1}\right) x_n^{n-3} & \left({x_n - x_1}\right) x_n^{n-2}
+\end{vmatrix}$
+For all rows apart from the first, the $k$th row has the constant factor $\left({x_k - x_1}\right)$.
+So we can extract all these as factors, and from [[Determinant with Row Multiplied by Constant]], we get:
+:$\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) \begin{vmatrix}
+  1 & 0 & 0 & \cdots & 0 & 0 \\
+  0 & 1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\
+  0 & 1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  0 & 1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\
+  0 & 1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2}
+\end{vmatrix}$
+From [[Determinant with Unit Element in Otherwise Zero Row]], we can see that this directly gives us:
+:$\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) \begin{vmatrix}
+  1 & x_2 & \cdots & x_2^{n-3} & x_2^{n-2} \\
+  1 & x_3 & \cdots & x_3^{n-3} & x_3^{n-2} \\
+\vdots & \vdots & \ddots & \vdots & \vdots \\
+  1 & x_{n-1} & \cdots & x_{n-1}^{n-3} & x_{n-1}^{n-2}\\
+  1 & x_n & \cdots & x_n^{n-3} & x_n^{n-2}
+\end{vmatrix}$
+and it can be seen that:
+: $\displaystyle V_n = \prod_{k \mathop = 2}^n \left({x_k - x_1}\right) V_{n-1}$
+$V_2$, by the time we get to it (it will concern elements $x_{n-1}$ and $x_n$), can be calculated directly using the formula for calculating a [[Definition:Determinant of Order 2|Determinant of Order 2]]:
+: $V_2 = \begin{vmatrix}
+  1 & x_{n-1} \\
+  1 & x_n 
+\end{vmatrix} = x_n - x_{n-1}$
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+Let the [[Vandermonde Determinant]] be presented in the form as defined by [[Vandermonde Determinant#Mirsky|Mirsky]]:
+Let $V_n = \begin{vmatrix}
+  a_1^{n-1} & a_1^{n-2} & \cdots & a_1 & 1 \\
+  a_2^{n-1} & a_2^{n-2} & \cdots & a_2 & 1 \\
+\vdots & \vdots & \ddots & \vdots & \vdots \\
+  a_n^{n-1} & a_n^{n-2} & \cdots & a_n & 1 \\
+\end{vmatrix}$
+For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i - a_j}\right)$
+$P(1)$ is true, as this just says $\begin{vmatrix} 1 \end{vmatrix} = 1$.
+=== Basis for the Induction ===
+$P(2)$ holds, as it is the case:
+: $V_2 = \begin{vmatrix}
+  a_1 & 1 \\
+  a_2 & 1
+\end{vmatrix}$
+which evaluates to $V_2 = a_1 - a_2$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+: $\displaystyle V_k = \prod_{1 \mathop \le i \mathop < j \mathop \le k} \left({a_i - a_j}\right)$
+Then we need to show:
+: $\displaystyle V_{k+1} = \prod_{1 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Take the determinant:
+:$V_{k+1} = \begin{vmatrix}
+ x^k & x^{k-1} & \cdots & x^2 & x & 1 \\
+ a_2^k & a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\
+\vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
+ a_{k+1}^k & a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1
+\end{vmatrix}$
+Let the [[Expansion Theorem for Determinants]] be used to expand $V_n$ in terms of the first row
+It can be seen that it is a [[Definition:Real Polynomial Function|polynomial]] in $x$ whose [[Definition:Degree (Polynomial)|degree]] is no greater than $k$.
+Let that polynomial be denoted $f \left({x}\right)$.
+Let any $a_r$ be substituted for $x$ in the determinant.
+Then two of its rows will be the same.
+From [[Square Matrix with Duplicate Rows has Zero Determinant]], the value of such a determinant will be $0$.
+Such a substitution in the determinant is equivalent to substituting $a_r$ for $x$ in $f \left({x}\right)$.
+Thus it follows that:
+:$f \left({a_2}\right) = f \left({a_3}\right) = \ldots = f \left({a_{k+1}}\right) = 0$
+So $f \left({x}\right)$ is divisible by each of the factors $x - a_2, x - a_3, \ldots, x - a_{k+1}$.
+All these factors are distinct, otherwise the original determinant is zero.
+So:
+: $f \left({x}\right) = C \left({x - a_2}\right) \left({x - a_3}\right) \cdots \left({x - a_k}\right) \left({x - a_{k+1}}\right)$
+As the degree of $f \left({x}\right)$ is no greater than $k$, it follows that $C$ is independent of $x$.
+From the [[Expansion Theorem for Determinants]], the coefficient of $x^k$ is:
+: $\begin{vmatrix}
+a_2^{k-1} & \cdots & a_2^2 & a_2 & 1 \\
+\vdots & \ddots & \vdots & \vdots & \vdots \\
+a_{k+1}^{k-1} & \cdots & a_{k+1}^2 & a_{k+1} & 1
+\end{vmatrix}$.
+By the [[Vandermonde Determinant/Proof 2#Induction Hypothesis|induction hypothesis]], this is equal to:
+: $\displaystyle \prod_{2 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$
+This must be our value of $C$.
+So we have:
+: $\displaystyle f \left({x}\right) = \left({x - a_2}\right) \left({x - a_3}\right) \cdots \left({x - a_k}\right) \left({x - a_{k+1}}\right) \prod_{2 \mathop \le i \mathop < j \mathop \le k+1} \left({a_i - a_j}\right)$
+Substituting $a_1$ for $x$, we retrieve the proposition $P \left({k+1}\right)$.
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+: $\displaystyle V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i - a_j}\right)$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $V_n = \begin{vmatrix}
+  1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\
+  1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\
+  1 & x_n & x_n^2 & \cdots & x_n^{n-2} & x_n^{n-1}
+\end{vmatrix}$.
+Start by replacing number $x_n$ in $V_n$ with the unknown $x$.
+Thus $V_n$ is made into a function of $x$.
+:$P \left({x}\right) = \begin{vmatrix}
+  1 & x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  1 & x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\
+  1 & x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+  1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1} \\
+  1 & x & x^2 & \cdots & x^{n-2} & x^{n-1}
+\end{vmatrix}$.
+Let $x$ equal a value from the set $\set {x_1,\ldots,x_{n-1} }$.
+Then determinant $\map P {x}$ has [[Square Matrix with Duplicate Rows has Zero Determinant|equal rows]], giving:
+{{begin-eqn}}
+{{eqn | l = \map P {x}
+      | r = 0
+      | c = for $x = x_1, \ldots, x_{n-1}$ 
+}}
+{{end-eqn}}
+Perform row expansion by the last row.
+Then $P \left({x}\right)$ is seen to be a polynomial of degree $n-1$:
+:$P \left({x}\right) = \begin{vmatrix}
+  x_1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  x_2 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\
+  x_3 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+  x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1}
+\end{vmatrix} + 
+\begin{vmatrix}
+  1 & x_1^2 & \cdots & x_1^{n-2} & x_1^{n-1} \\
+  1 & x_2^2 & \cdots & x_2^{n-2} & x_2^{n-1} \\
+  1 & x_3^2 & \cdots & x_3^{n-2} & x_3^{n-1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+  1 & x_{n-1}^2 & \cdots & x_{n-1}^{n-2} & x_{n-1}^{n-1}
+\end{vmatrix}x \ \ + \ \ \cdots \ \ + \ \
+\begin{vmatrix}
+  1 & x_1 & x_1^2 & \cdots & x_1^{n-2} \\
+  1 & x_2 & x_2^2 & \cdots & x_2^{n-2} \\
+  1 & x_3 & x_3^2 & \cdots & x_3^{n-2} \\
+\vdots & \vdots & \vdots & \ddots & \vdots \\
+  1 & x_{n-1} & x_{n-1}^2 & \cdots & x_{n-1}^{n-2}
+\end{vmatrix}x^{n-1}$.
+<!---
+We can see that statement $P \left({x}\right) = 0$ holds true for all $x_1, x_2, \cdots x_{n-1}$ because if $x=x_1, x_2, \cdots x_{n-1}$ starting determinant would have two equal rows and by [[Square Matrix with Duplicate Rows has Zero Determinant]] would $V_n = 0$.
+{{explain|The meaning of the above statement needs to be made clearer, and the logical sense turned into a linear flow.}}
+Moved statement higher up so it reads linearly. Nov 6 2019 GBGustafson
+Also fixed one typo $x_{n-1} \to x^{n-1}$ and changed "returning" to "evaluating at".
+The cofactor expansion is missing checkerboard signs but the minors are correct. Not fixed.
+--->
+By the [[Polynomial Factor Theorem]]:
+:$P \left({x}\right) = C \left({x - x_1}\right) \left({x - x_2}\right) \dotsm \left({x - x_{n-1} }\right)$
+where $C$ is the leading coefficient (with $x^{n-1}$ power).
+Thus:
+:$P \left({x}\right) = V_{n-1} \left({x - x_1}\right) \left({x - x_2}\right) \dotsm \left({x - x_{n-1} }\right)$
+which by evaluating at $x = x_n$ gives: 
+:$V_n = V_{n-1} \left({x_n - x_1}\right) \left({x_n - x_2}\right) \dotsm \left({x_n - x_{n-1} }\right)$
+Repeating the process:
+{{begin-eqn}}
+{{eqn | l = V_n
+      | r = \prod_{1 \mathop \le i \mathop < n} \left({x_n - x_i}\right) V_{n-1}
+      | c = 
+}}
+{{eqn | r = \prod_{1 \mathop \le i \mathop < n} \left({x_n - x_i}\right) \prod_{1 \mathop \le i \mathop < n-1} \left({x_{n-1} - x_i}\right) V_{n-2}
+      | c = 
+}}
+{{eqn | r = \dotsm
+      | c = 
+}}
+{{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \left({x_j - x_i}\right)
+      | c = 
+}}
+{{end-eqn}}
+which establishes the solution.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+:$V_n = \begin{vmatrix}
+1      & x_1    & x_1^2  & \cdots & x_1^{n - 2} & x_1^{n - 1} \\
+1      & x_2    & x_2^2  & \cdots & x_2^{n - 2} & x_2^{n - 1} \\
+\vdots & \vdots & \vdots & \ddots & \vdots      & \vdots \\
+1      & x_n    & x_n^2  & \cdots & x_n^{n - 2} & x_n^{n - 1}
+\end{vmatrix}$
+Let $\map f x$ be '''any''' [[Definition:Monic Polynomial|monic polynomial]] of [[Definition:Degree of Polynomial|degree]] $n - 1$:
+:$\ds \map f x = x^{n - 1} + \sum_{i \mathop = 0}^{n - 2} a_i x^i$
+Apply [[Effect of Elementary Row Operations on Determinant|elementary column operations]] to $V_n$ repeatedly to show:
+:$V_n = W$
+where:
+:$ W = \begin{vmatrix}
+1      & x_1    & x_1^2  & \cdots & x_1^{n - 2} & \map f {x_1} \\
+1      & x_2    & x_2^2  & \cdots & x_2^{n - 2} & \map f {x_2} \\
+\vdots & \vdots & \vdots & \ddots & \vdots      & \vdots \\
+1      & x_n    & x_n^2  & \cdots & x_n^{n - 2} & \map f {x_n}
+\end{vmatrix}$
+Select a specific degree $n - 1$ monic polynomial:
+:$\ds \map f x = \prod_{k \mathop = 1}^{n - 1} \paren {x - x_k}$
+The selected polynomial is zero at all values $x_1, \ldots, x_{n - 1}$.
+Then the last column of $W$ is all zeros except the entry $\map f {x_n}$.
+Expand $\map \det W$ by cofactors along the last column to prove:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = V_n
+      | r = \map f {x_n} V_{n - 1}
+      | c = [[Expansion Theorem for Determinants]] for columns
+}}
+{{eqn | r = V_{n - 1} \prod_{k \mathop = 1}^{n - 1} \paren {x_n - x_k}
+}}
+{{end-eqn}}
+For $n \ge 2$, let $\map P n$ be the statement:
+:$\ds V_n = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$
+[[Definition:Mathematical Induction|Mathematical induction]] will be applied.
+=== Basis for the Induction ===
+By definition, determinant $V_1 = 1$.
+To prove $\map P 2$ is true, use equation $(1)$ with $n = 2$:
+:$V_2 = \paren {x_2 - x_1} V_1$
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+Let $\map P n$ is be assumed true.
+We are to prove that $\map P {n + 1}$ is true.
+As follows:
+{{begin-eqn}}
+{{eqn | l = V_{n + 1}
+      | r = V_n \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}
+      | c = from $(1)$, setting $n \to n + 1$
+}}
+{{eqn | r = \prod_{1 \mathop \le m \mathop < k \mathop \le n} \paren {x_k - x_m} \prod_{k \mathop = 1}^n \paren {x_{n + 1} - x_k}
+      | c = [[Vandermonde Determinant/Proof 4#Induction Hypothesis|Induction hypothesis]] with new indexing symbols: $i \to m, j \to k$
+}}
+{{eqn | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n + 1} \paren {x_j - x_i}
+      | c = simplifying
+}}
+{{end-eqn}}
+Thus $\map P {n + 1}$ has been shown to be true.
+The induction is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace's Expansion Theorem}
+Tags: Determinants, Laplace's Expansion Theorem
+
+\begin{theorem}
+Let $D$ be the [[Definition:Determinant of Matrix|determinant of order $n$]].
+Let $r_1, r_2, \ldots, r_k$ be [[Definition:Integer|integers]] such that:
+:$1 \le k < n$
+:$1 \le r_1 < r_2 < \cdots < r_k \le n$
+Let $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be an [[Definition:Minor of Determinant|order-$k$ minor of $D$]].
+Let $\map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$ be the [[Definition:Cofactor of Minor|cofactor]] of $\map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$.
+Then:
+:$\displaystyle D = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k} \, \map {\tilde D} {r_1, r_2, \ldots, r_k \mid u_1, u_2, \ldots, u_k}$
+A similar result applies for columns.
+\end{theorem}
+
+\begin{proof}
+Let us define $r_{k + 1}, r_{k + 2}, \ldots, r_n$ such that:
+:$1 \le r_{k + 1} < r_{k + 2} < \cdots < r_n \le n$
+:$\rho = \tuple {r_1, r_2, \ldots, r_n}$ is a [[Definition:Permutation on n Letters|permutation on $\N^*_n$]].
+Let $\sigma = \tuple {s_1, s_2, \ldots, s_n}$ be a [[Definition:Permutation on n Letters|permutation on $\N^*_n$]].
+Then by [[Permutation of Determinant Indices]] we have:
+{{begin-eqn}}
+{{eqn | l = D
+      | r = \sum_\sigma \map \sgn \rho \, \map \sgn \sigma \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}
+      | c = 
+}}
+{{eqn | r = \sum_\sigma \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + s_i} } \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = 1}^n a_{\map \rho j \, \map \sigma j}
+      | c = 
+}}
+{{end-eqn}}
+We can obtain all the [[Definition:Permutation on n Letters|permutations]] $\sigma$ exactly once by separating the numbers $1, \ldots, n$ in all possible ways into a set of $k$ and $n - k$ numbers.
+We let $\tuple {s_1, \ldots, s_k}$ vary over the first set and $\tuple {s_{k + 1}, \ldots, s_n}$ over the second set.
+So the summation over all $\sigma$ can be replaced by:
+:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
+:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
+:$\tuple {s_1, \ldots, s_k} = \map \sigma {u_1, \ldots, u_k}$
+:$\tuple {s_{k + 1}, \ldots, s_n} = \map \sigma {u_{k + 1}, \ldots, u_n}$
+Thus we get:
+{{begin-eqn}}
+{{eqn | l = D
+      | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \sum_{\map \sigma {u_1, \ldots, u_k} } \, \map \sgn {\map \rho {r_1, \ldots, r_k} } \, \map \sgn {\map \sigma {s_1, \ldots, s_k} } \prod_{j \mathop = 1}^k a_{\map \rho j \, \map \sigma j}
+      | c = 
+}}
+{{eqn | o = \times
+      | r = \sum_{\map \sigma {u_{k + 1}, \ldots, u_n} } \map \sgn {\map \rho {r_{k + 1}, \ldots, r_n} } \, \map \sgn {\map \sigma {s_{k + 1}, \ldots, s_n} } \prod_{j \mathop = k + 1}^n a_{\map \rho j \, \map \sigma j}
+      | c = 
+}}
+{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \begin {vmatrix} a_{r_1 u_1} & \cdots & a_{r_1 u_k} \\ \vdots & \ddots & \vdots \\ a_{r_k u_1} & \cdots & a_{r_k u_k} \end {vmatrix} \times \begin {vmatrix} a_{r_{k + 1} u_{k + 1} } & \cdots & a_{r_{k + 1} u_n} \\ \vdots & \ddots & \vdots \\ a_{r_n u_{k + 1} } & \cdots & a_{r_n u_n} \end {vmatrix}
+      | c = 
+}}
+{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \paren {-1}^{\sum_{i \mathop = 1}^k \paren {r_i + u_i} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map D {r_{k + 1}, \ldots, r_n \mid u_{k + 1}, \ldots, u_n}
+      | c = 
+}}
+{{eqn | r = \sum_{\map \sigma {u_1, \ldots, u_n} } \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \times \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k}
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le u_1 \mathop < \cdots \mathop < u_k \mathop \le n} \map D {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \, \map {\tilde D} {r_1, \ldots, r_k \mid u_1, \ldots, u_k} \sum_{u_{k + 1}, \ldots, u_n} 1
+      | c = 
+}}
+{{end-eqn}}
+That last inner sum extends over all integers which satisfy:
+:$\tuple {u_1, \ldots, u_n} = \map \sigma {1, \ldots, n}$
+:$u_1 < u_2 < \cdots < u_k, u_{k + 1} < u_{k + 2} < \cdots < u_n$
+But for each set of $u_1, \ldots, u_k$, then the integers $u_{k + 1}, \ldots, u_n$ are clearly uniquely determined.
+So that last inner sum equals 1 and the theorem is proved.
+{{Explain|I'm not too happy about this, it seems a bit handwavey and imprecise. I'm going to have to revisit it.}}
+The result for columns follows from [[Determinant of Transpose]]. 
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Equality of Polynomials}
+Tags: Polynomial Theory
+
+\begin{theorem}
+$f$ and $g$ are equal as polynomials {{iff}} $f$ and $g$ are equal as functions.
+Thus we can say $f = g$ without ambiguity as to what it means.
+{{explain|In the exposition, the term was "equal as forms", but it has now morphed into "equal as polynomials". Needs to be resolved.}}
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Proof missing. Also, I am not sure how general this result can be made. My suspicion is that if a comm. ring with $1$, $R$ has no idempotents save $0$ and $1$, then the result continue to hold, but not sure at the moment.}}
+[[Category:Polynomial Theory]]
+bswsxu6ck42asecntjn0fggs00z6dih
+\end{proof}<|endoftext|>
+\section{Value of Adjugate of Determinant}
+Tags: Determinants
+
+\begin{theorem}
+Let $D$ be the [[Definition:Determinant of Matrix|determinant]] of order $n$.
+Let $D^*$ be the [[Definition:Adjugate Matrix|adjugate]] of $D$.
+Then $D^* = D^{n-1}$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \begin{bmatrix}  a_{11} & a_{12} & \cdots & a_{1n} \\
+  a_{21} & a_{22} & \cdots & a_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+  a_{n1} & a_{n2} & \cdots & a_{nn}\end{bmatrix}$ and $\mathbf A^* = \begin{bmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
+  A_{21} & A_{22} & \cdots & A_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+  A_{n1} & A_{n2} & \cdots & A_{nn}\end{bmatrix}$.
+Thus $\left({\mathbf A^*}\right)^\intercal = \begin{bmatrix} A_{11} & A_{21} & \cdots & A_{n1} \\
+  A_{12} & A_{22} & \cdots & A_{n2} \\
+\vdots & \vdots & \ddots & \vdots \\
+  A_{1n} & A_{2n} & \cdots & A_{nn}\end{bmatrix}$ is the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A^*$.
+Let $c_{ij}$ be the typical element of $\mathbf A \left({\mathbf A^*}\right)^\intercal$.
+Then $\displaystyle c_{ij} = \sum_{k \mathop = 1}^n a_{ik} A_{jk}$ by definition of [[Definition:Matrix Product (Conventional)|matrix product]].
+Thus by the [[Expansion Theorem for Determinants/Corollary|corollary of the Expansion Theorem for Determinants]], $c_{ij} = \delta_{ij} D$.
+So $\det \left({\mathbf A \left({\mathbf A^*}\right)^\intercal}\right) = \begin{vmatrix} D & 0 & \cdots & 0 \\
+  0 & D & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+  0 & 0 & \cdots & D\end{vmatrix} = D^n$ by [[Determinant of Diagonal Matrix]].
+From [[Determinant of Matrix Product]], $\det \left({\mathbf A}\right) \det \left({\left({\mathbf A^*}\right)^\intercal}\right) = \det \left({\mathbf A \left({\mathbf A^*}\right)^\intercal}\right)$
+From [[Determinant of Transpose]]:
+: $\det \left({\left({\mathbf A^*}\right)^\intercal}\right) = \det \left({\mathbf A^*}\right)$
+Thus as $D = \det \left({\mathbf A}\right)$ and $D^* = \det \left({\mathbf A^*}\right)$ it follows that $DD^* = D^n$.
+Now if $D \ne 0$, the result follows.
+However, if $D = 0$ we need to show that $D^* = 0$.
+Let $D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
+  A_{21} & A_{22} & \cdots & A_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+  A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix}$.
+Suppose that at least one element of $\mathbf A$, say $a_{rs}$, is non-zero (otherwise the result follows immediately).
+By the [[Expansion Theorem for Determinants]] and its corollary, we can expand $D$ by row $r$, and get:
+:$\displaystyle D = 0 = \sum_{j \mathop = 1}^n A_{ij} t_j, \forall i = 1, 2, \ldots, n$
+for all $t_1 = a_{r1}, t_2 = a_{r2}, \ldots, t_n = a_{rn}$.
+But $t_s = a_{rs} \ne 0$.
+So, by '''(work in progress)''':
+:$D^* = \begin{vmatrix} A_{11} & A_{12} & \cdots & A_{1n} \\
+  A_{21} & A_{22} & \cdots & A_{2n} \\
+\vdots & \vdots & \ddots & \vdots \\
+  A_{n1} & A_{n2} & \cdots & A_{nn}\end{vmatrix} = 0$
+{{WIP|One result to document, I've got to work out how best to formulate it.}}
+[[Category:Determinants]]
+bkm55tefjheohcpkb9z8u2t6jx7dz5p
+\end{proof}<|endoftext|>
+\section{Determinant of Diagonal Matrix}
+Tags: Determinants, Diagonal Matrices
+
+\begin{theorem}
+Let $\mathbf A = \begin{bmatrix}
+a_{11} & 0 & \cdots & 0 \\
+0 & a_{22} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & a_{nn} \\
+\end{bmatrix}$ be a [[Definition:Diagonal Matrix|diagonal matrix]].
+Then the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$ is the product of the elements of $\mathbf A$.
+That is:
+:$\ds \map \det {\mathbf A} = \prod_{i \mathop = 1}^n a_{ii}$
+\end{theorem}
+
+\begin{proof}
+As a [[Definition:Diagonal Matrix|diagonal matrix]] is also a [[Definition:Triangular Matrix|triangular matrix]] (both upper and lower), the result follows directly from [[Determinant of Triangular Matrix]].
+{{qed}}
+[[Category:Determinants]]
+[[Category:Diagonal Matrices]]
+46j7gt7zea6di1bocalbw6n08ipn7hx
+\end{proof}<|endoftext|>
+\section{Row Equivalence is Equivalence Relation}
+Tags: Equivalence Relations, Row Operations, Row Equivalence
+
+\begin{theorem}
+[[Definition:Row Equivalence|Row equivalence]] is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary [[Definition:Matrix|matrices]] in a given [[Definition:Matrix Space|matrix space]] $\map \MM {m, n}$ for $m, n \in \Z{>0}$.
+We check in turn each of the conditions for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+Let $r_i$ denote an arbitrary [[Definition:Row of Matrix|row]] of $\mathbf A$.
+Let $e$ denote the [[Definition:Elementary Row Operation|elementary row operation]] $r_i \to 1 r_i$ applied to $\mathbf A$.
+Then trivially:
+:$\map e {\mathbf A} = \mathbf A$
+and so $\mathbf A$ is trivially [[Definition:Row Equivalence|row equivalent]] to itself.
+So [[Definition:Row Equivalence|row equivalence]] has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\mathbf A$ be [[Definition:Row Equivalence|row equivalent]] to $\mathbf B$.
+Let $\Gamma$ be the [[Definition:Row Operation|row operation]] that transforms $\mathbf A$ into $\mathbf B$.
+From [[Row Operation has Inverse]] there exists a [[Definition:Row Operation|row operation]] $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$.
+Thus $\mathbf B$ is [[Definition:Row Equivalence|row equivalent]] to $\mathbf A$.
+So [[Definition:Row Equivalence|row equivalence]] has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+Let $\mathbf A$ be [[Definition:Row Equivalence|row equivalent]] to $\mathbf B$, and let $\mathbf B$ be [[Definition:Row Equivalence|row equivalent]] to $\mathbf C$.
+Let $\Gamma_1$ be the [[Definition:Row Operation|row operation]] that transforms $\mathbf A$ into $\mathbf B$.
+Let $\Gamma_2$ be the [[Definition:Row Operation|row operation]] that transforms $\mathbf B$ into $\mathbf C$.
+From [[Sequence of Row Operations is Row Operation]], $\mathbf C$ is [[Definition:Row Equivalence|row equivalent]] to $\mathbf A$.
+So [[Definition:Row Equivalence|row equivalence]] has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+[[Definition:Row Equivalence|Row equivalence]] has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix is Row Equivalent to Reduced Echelon Matrix}
+Tags: Echelon Matrices
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_{m n}$ be a [[Definition:Matrix|matrix]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $F$.
+Then $A$ is [[Definition:Row Equivalence|row equivalent]] to a [[Definition:Reduced Echelon Matrix|reduced echelon matrix]] of [[Definition:Order of Matrix|order]] $m \times n$.
+\end{theorem}
+
+\begin{proof}
+Let the first [[Definition:Column of Matrix|column]] of $\mathbf A$ containing a non-[[Definition:Field Zero|zero]] [[Definition:Element of Matrix|element]] be [[Definition:Column of Matrix|column]] $j$.
+Let such a non-[[Definition:Field Zero|zero]] [[Definition:Element of Matrix|element]] be in [[Definition:Row of Matrix|row]] $i$.
+Take [[Definition:Element of Matrix|element]] $a_{i j} \ne 0$ and perform the [[Definition:Elementary Row Operation|elementary row operations]]:
+:$(1): \quad r_i \to \dfrac {r_i} {a_{i j}}$
+:$(2): \quad r_1 \leftrightarrow r_i$
+This gives a [[Definition:Matrix|matrix]] with $1$ in the $\tuple {1, j}$ position:
+:$\begin {bmatrix}
+     0 & \cdots &      0 &       1 & b_{1, j + 1} & \cdots & b_{1 n} \\
+     0 & \cdots &      0 & b_{2 j} & b_{2, j + 1} & \cdots & b_{2 n} \\
+\vdots & \ddots & \vdots &  \vdots &       \vdots & \ddots &  \vdots \\
+     0 & \cdots &      0 & b_{m j} & b_{m, j + 1} & \cdots & b_{m n} \\
+\end {bmatrix}$
+Now the [[Definition:Elementary Row Operation|elementary row operations]] $r_k \to r_k - b_{k j} r_1, k \in \set {2, 3, \ldots, m}$ gives the [[Definition:Matrix|matrix]]:
+:$\begin{bmatrix}
+0      & \cdots &      0 &      1 & c_{1, j + 1} & \cdots & c_{1 n} \\
+0      & \cdots &      0 &      0 & c_{2, j + 1} & \cdots & c_{2 n} \\
+\vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\
+0      & \cdots &      0 &      0 & c_{m, j + 1} & \cdots & c_{m n} \\
+\end{bmatrix}$
+If some [[Definition:Zero Row or Column|zero rows]] have appeared, do some further [[Definition:Elementary Row Operation|elementary row operations]], that is row interchanges, to put them at the bottom.
+We now repeat the process with the remaining however-many-there-are [[Definition:Row of Matrix|rows]]:
+:$\begin{bmatrix}
+\cdots &      0 &      1 & d_{1, j + 1} & \cdots & d_{1, k - 1} & d_{1 k} & d_{1, k + 1} & \cdots & d_{1 n} \\
+\cdots &      0 &      0 &            0 & \cdots &            0 &       1 & d_{2, k + 1} & \cdots & d_{2 n} \\
+\cdots &      0 &      0 &            0 & \cdots &            0 & d_{3 k} & d_{3, k + 1} & \cdots & d_{3 n} \\
+\ddots & \vdots & \vdots &       \vdots & \ddots &       \vdots &  \vdots &       \vdots & \ddots & \vdots  \\
+\cdots &      0 &      0 &            0 & \cdots &            0 & d_{n k} & d_{m, k + 1} & \cdots & d_{m n} \\ 
+\end{bmatrix}$
+Then we can get the [[Definition:Reduced Echelon Matrix|reduced echelon form]] by:
+:$r_i \to r_i - d_{i k} r_2, i \in \set {1, 3, 4, \ldots, m}$
+as follows:
+:$\begin{bmatrix}
+\cdots &      0 &      1 & {e_{1, j + 1 } } & \cdots & {e_{1, k - 1} } &      0 & {e_{1, k + 1} } & \cdots & {e_{1 n} } \\
+\cdots &      0 &      0 &                0 & \cdots &               0 &      1 & {e_{2, k + 1} } & \cdots & {e_{2 n} } \\
+\cdots &      0 &      0 &                0 & \cdots &               0 &      0 & {e_{3, k + 1} } & \cdots & {e_{3 n} } \\
+\ddots & \vdots & \vdots &           \vdots & \ddots &          \vdots & \vdots &          \vdots & \ddots & \vdots     \\
+\cdots &      0 &      0 &                0 & \cdots &               0 &      0 & {e_{m, k + 1} } & \cdots & {e_{m n} } \\
+\end{bmatrix}$
+Thus we progress, until the entire [[Definition:Matrix|matrix]] is in [[Definition:Reduced Echelon Matrix|reduced echelon]] form.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Square Matrix is Row Equivalent to Triangular Matrix}
+Tags: Square Matrices, Triangular Matrices, Square Matrix is Row Equivalent to Triangular Matrix
+
+\begin{theorem}
+Let $\mathbf A = \left[{a}\right]_n$ be a [[Definition:Square Matrix|square matrix of order $n$]] over a [[Definition:Commutative Ring|commutative ring]] $R$.
+Then $\mathbf A$ can be converted to an [[Definition:Upper Triangular Matrix|upper]] or [[Definition:Lower Triangular Matrix|lower triangular matrix]] by [[Definition:Elementary Row Operation|elementary row operations]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+We proceed by induction on $n$, the number of [[Definition:Row of Matrix|rows]] of $\mathbf A$.
+=== Basis for the Induction ===
+For $n = 1$, we have a matrix of just one [[Definition:Element of Matrix|element]], which is trivially [[Definition:Diagonal Matrix|diagonal]], hence both upper and lower triangular.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by [[Definition:Elementary Row Operation|elementary row operations]]. 
+If $R$ is a [[Definition:Field (Abstract Algebra)|field]], assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type 2.
+This forms our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+Let $\mathbf A = \left[{a}\right]_{n+1}$ be a [[Definition:Square Matrix|square matrix]] of order $n + 1$.
+When the first [[Definition:Column of Matrix|column]] of $\mathbf A$ contains only zeroes, it is upper triangularisable [[Definition:Iff|iff]] the [[Definition:Submatrix|submatrix]] $\mathbf A \left({1; 1}\right)$ is.
+Each [[Definition:Elementary Row Operation|elementary row operation]] used in triangularisation process of the submatrix $\mathbf A \left({1; 1}\right)$ will not change the zeros of the first column of $\mathbf A$.
+So when $\mathbf A \left({1; 1}\right)$ is upper triangularised, then $A$ will also be upper triangularised.
+From the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]], we conclude that $\mathbf A$ can be upper triangularised by elementary row operations.
+Now suppose that its first column contains a non-zero value.
+Suppose that $a_{11} = 0$.
+Let $j$ be the smallest row index such that $a_{j1} \ne 0$, and note that $j$ exists by assumption.
+Now apply the following operation of type 2:
+:$r_1 \to r_1 + r_j$
+As $a_{j1} \ne 0$, this enforces $a_{11} \ne 0$, and we continue as in the case below.
+Suppose $a_{11} \ne 0$. 
+We use the following operations for all $j \in \left\{{2, \ldots, n + 1}\right\}$:
+:$(1): \quad$ Put $c = a_{j1}$.
+:$(2): \quad r_j \to a_{11} r_j$
+:$(3): \quad r_j \to r_j - c r_1$
+This will put the first column to zero (except for the first element, $a_{11}$).
+It follows that $\mathbf A$ can be upper triangularised precisely when the [[Definition:Submatrix|submatrix]] $\mathbf A \left({1; 1}\right)$ can.
+Again, the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] renders $\mathbf A$ upper triangularisable by [[Definition:Elementary Row Operation|elementary row operations]].
+This completes the case distinction, and hence the result follows by [[Principle of Mathematical Induction|induction]].
+To put the matrix $\mathbf A$ into [[Definition:Lower Triangular Matrix|lower triangular form]], just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This proof assumes that $R$ is a [[Definition:Field (Abstract Algebra)|field]], which makes the triangulation process slightly quicker.
+By this assumptions, all [[Definition:Element of Matrix|elements]] of $\mathbf A$ have [[Definition:Inverse Element|multiplicative inverses]].
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+We proceed by induction on $n$, the number of [[Definition:Row of Matrix|rows]] of $\mathbf A$.
+=== Basis for the Induction ===
+For $n = 1$, we have a matrix of just one [[Definition:Element of Matrix|element]], which is trivially [[Definition:Diagonal Matrix|diagonal]], hence both upper and lower triangular.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n \in \N$, and assume all $n \times n$-matrices can be upper triangularised by [[Definition:Elementary Row Operation|elementary row operations]]. 
+If $R$ is a [[Definition:Field (Abstract Algebra)|field]], assume all $n \times n$-matrices can be upper triangularised by elementary row operations of type 2.
+This forms our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+Let $\mathbf A = \left[{a}\right]_{n + 1}$ be a [[Definition:Square Matrix|square matrix]] of order $n + 1$.
+When the first [[Definition:Column of Matrix|column]] of $\mathbf A$ contains only zeroes, it is upper triangularisable {{iff}} the [[Definition:Submatrix|submatrix]] $\mathbf A \left({1; 1}\right)$ is.
+Each [[Definition:Elementary Row Operation|elementary row operation]] used in triangularisation process of the submatrix $\mathbf A \left({1; 1}\right)$ will not change the zeros of the first column of $\mathbf A$.
+So when $\mathbf A \left({1; 1}\right)$ is upper triangularised, then $\mathbf A$ will also be upper triangularised.
+From the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]], we conclude that $\mathbf A$ can be upper triangularised by elementary row operations.
+Now suppose that its first column contains a non-zero value.
+Suppose that $a_{11} = 0$.
+Let $j$ be the smallest row index such that $a_{j1} \ne 0$, and note that $j$ exists by assumption.
+Now apply the following operation of type 2:
+:$r_1 \to r_1 + r_j$
+As $a_{j1} \ne 0$, this enforces $a_{11} \ne 0$, and we continue as in the case below.
+Suppose $a_{11} \ne 0$. 
+We use the following operations of type 2:
+:$\forall j \in \left\{{2, \ldots, n+1}\right\}: r_j \to r_j - \dfrac {a_{j1}} {a_{11}} r_1$
+This will put the first column to zero (except for the first element, $a_{11}$).
+It follows that $\mathbf A$ can be upper triangularised precisely when the [[Definition:Submatrix|submatrix]] $\mathbf A \left({1; 1}\right)$ can.
+Again, the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] renders $\mathbf A$ upper triangularisable by [[Definition:Elementary Row Operation|elementary row operations]].
+This completes the case distinction, and hence the result follows by [[Principle of Mathematical Induction|induction]].
+To put the matrix $\mathbf A$ into [[Definition:Lower Triangular Matrix|lower triangular form]], just do the same thing, but start with the last column and the last diagonal element $a_{n + 1 \; n + 1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Triangular Matrices}
+Tags: Triangular Matrices, Conventional Matrix Multiplication
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n, \mathbf B = \sqbrk b_n$ be [[Definition:Upper Triangular Matrix|upper triangular matrices]] of [[Definition:Order of Square Matrix|order]] $n$.
+Let $\mathbf C = \mathbf A \mathbf B$.
+Then
+:$(1): \quad$ the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf C$ are given by:
+::::$\forall j \in \closedint 1 n: c_{j j} = a_{j j} b_{j j}$
+:::That is, the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf C$ are those of the factor matrices multiplied together.
+:$(2): \quad$ The matrix $\mathbf C$ is itself [[Definition:Upper Triangular Matrix|upper triangular]].
+The same applies if both $\mathbf A$ and $\mathbf B$ are [[Definition:Lower Triangular Matrix|lower triangular matrices]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Matrix Product (Conventional)|matrix product]], we have:
+:$\displaystyle \forall i, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} b_{k j}$
+Now when $i = j$ (as on the [[Definition:Diagonal Element|main diagonal]]):
+:$\displaystyle c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j}$
+Now both $\mathbf A$ and $\mathbf B$ are [[Definition:Upper Triangular Matrix|upper triangular]].
+Thus:
+:if $k > j$, then $b_{k j} = 0$ and thus $a_{j k} b_{k j} = 0$
+:if $k < j$, then $a_{j k} = 0$ and thus $a_{j k} b_{k j} = 0$.
+So $a_{j k} b_{k j} \ne 0$ only when $j = k$.
+So:
+:$\displaystyle c_{j j} = \sum_{k \mathop = 1}^n a_{j k} b_{k j} = a_{j j} b_{j j}$
+Now if $i > j$, it follows that either $a_{i k}$ or $b_{k j}$ is [[Definition:Zero Element|zero]] for all $k$, and thus $c_{i j} = 0$.
+Thus $\mathbf C$ is [[Definition:Upper Triangular Matrix|upper triangular]].
+The same argument can be used for when $\mathbf A$ and $\mathbf B$ are both [[Definition:Lower Triangular Matrix|lower triangular matrices]].
+{{Qed}}
+[[Category:Triangular Matrices]]
+[[Category:Conventional Matrix Multiplication]]
+kfypsdf63kqm9gggh85dbj560bfaa6i
+\end{proof}<|endoftext|>
+\section{Effect of Elementary Row Operations on Determinant}
+Tags: Determinants, Elementary Row Operations
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+Let $\map \det {\mathbf A}$ denote the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Take the [[Definition:Elementary Row Operation|elementary row operations]]:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 1
+        | t = For some $\lambda$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Row of Matrix|row]] $i$ by $\lambda$
+        | m = r_i \to \lambda r_i
+}}
+{{axiom | n = \text {ERO} 2
+        | t = For some $\lambda$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Row of Matrix|row]] $j$ to [[Definition:Row of Matrix|row]] $i$
+        | m = r_i \to r_i + \lambda r_j
+}}
+{{axiom | n = \text {ERO} 3
+        | t = Exchange [[Definition:Row of Matrix|rows]] $i$ and $j$
+        | m = r_i \leftrightarrow r_j
+}}
+{{end-axiom}}
+Applying $\text {ERO} 1$ has the effect of multiplying $\map \det {\mathbf A}$ by $\lambda$.
+Applying $\text {ERO} 2$ has no effect on $\map \det {\mathbf A}$.
+Applying $\text {ERO} 3$ has the effect of multiplying $\map \det {\mathbf A}$ by $-1$.
+\end{theorem}
+
+\begin{proof}
+From [[Elementary Row Operations as Matrix Multiplications]], an [[Definition:Elementary Row Operation|elementary row operation]] on $\mathbf A$ is equivalent to [[Definition:Matrix Product (Conventional)|matrix multiplication]] by the [[Definition:Elementary Row Matrix|elementary row matrices]] corresponding to the [[Definition:Elementary Row Operation|elementary row operations]].
+From [[Determinant of Elementary Row Matrix]], the [[Definition:Determinant of Matrix|determinants]] of those [[Definition:Elementary Row Matrix|elementary row matrices]] are as follows:
+=== [[Determinant of Elementary Row Matrix/Scale Row|Scale Row]] ===
+{{:Determinant of Elementary Row Matrix/Scale Row}}
+=== [[Determinant of Elementary Row Matrix/Scale Row and Add|Add Scalar Product of Row to Another]] ===
+{{:Determinant of Elementary Row Matrix/Scale Row and Add}}
+=== [[Determinant of Elementary Row Matrix/Exchange Rows|Exchange Rows]] ===
+{{:Determinant of Elementary Row Matrix/Exchange Rows}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus in Terms of Conjugate}
+Tags: Complex Conjugates, Complex Modulus
+
+\begin{theorem}
+Let $z = a + i b$ be a [[Definition:Complex Number|complex number]].
+Let $\cmod z$ be the [[Definition:Complex Modulus|modulus]] of $z$.
+Let $\overline z$ be the [[Definition:Complex Conjugate|conjugate]] of $z$.
+Then:
+:$\cmod z^2 = z \overline z$
+\end{theorem}
+
+\begin{proof}
+Let $z = a + i b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z \overline z
+      | r = a^2 + b^2
+      | c = [[Product of Complex Number with Conjugate]]
+}}
+{{eqn | r = \cmod z^2
+      | c = {{Defof|Complex Modulus}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Positive Root of Positive Real Number}
+Tags: Real Numbers, Roots of Numbers, Existence and Uniqueness of Positive Root of Positive Real Number
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x \ge 0$.
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ne 0$.
+Then there always exists a [[Definition:Unique|unique]] $y \in \R: \paren {y \ge 0} \land \paren {y^n = x}$.
+Hence the justification for the terminology '''the positive [[Definition:Root (Analysis)|$n$th root]] of $x$''' and the notation $x^{1/n}$.
+\end{theorem}
+
+\begin{proof}
+The result follows from [[Existence of Positive Root of Positive Real Number]] and [[Uniqueness of Positive Root of Positive Real Number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Triangle Inequality}
+Tags: Named Theorems, Triangle Inequality, Inequalities
+
+\begin{theorem}
+=== [[Triangle Inequality/Geometry|Geometry]] ===
+{{:Triangle Inequality/Geometry}}
+=== [[Triangle Inequality/Real Numbers|Real Numbers]] ===
+{{:Triangle Inequality/Real Numbers}}
+=== [[Triangle Inequality/Complex Numbers|Complex Numbers]] ===
+{{:Triangle Inequality/Complex Numbers}}
+\end{theorem}<|endoftext|>
+\section{Even Power is Non-Negative}
+Tags: Powers
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $n \in \Z$ be an [[Definition:Even Integer|even integer]].
+Then $x^n \ge 0$.
+That is, all [[Definition:Even Power|even powers]] are [[Definition:Positive Real Number|positive]].
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z$ be an [[Definition:Even Integer|even integer]].
+Then $n = 2 k$ for some $k \in \Z$.
+Thus:
+:$\forall x \in \R: x^n = x^{2 k} = \paren {x^k}^2$
+But from [[Square of Real Number is Non-Negative]]:
+:$\forall x \in \R: \paren {x^k}^2 \ge 0$
+and so there is no [[Definition:Real Number|real number]] whose [[Definition:Square (Algebra)|square]] is [[Definition:Strictly Negative Real Number|negative]].
+The result follows from [[Solution to Quadratic Equation]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Sign of Odd Power}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $n \in \Z$ be an [[Definition:Odd Integer|odd integer]].
+Then:
+:$x^n = 0 \iff x = 0$
+:$x^n > 0 \iff x > 0$
+:$x^n < 0 \iff x < 0$
+That is, the sign of an [[Definition:Odd Power|odd power]] matches the number it is a [[Definition:Power (Algebra)|power]] of.
+\end{theorem}
+
+\begin{proof}
+If $n$ is an [[Definition:Odd Integer|odd integer]], then $n = 2 k + 1$ for some $k \in \N$.
+Thus $x^n = x \cdot x^{2 k}$.
+But $x^{2 k} \ge 0$ from [[Even Power is Non-Negative]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Absolute Values on Ordered Integral Domain}
+Tags: Absolute Value Function, Ordered Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is denoted by $0_D$.
+For all $a \in D$, let $\size a$ denote the [[Definition:Absolute Value on Ordered Integral Domain|absolute value]] of $a$.
+Then:
+:$\size a \times \size b = \size {a \times b}$
+\end{theorem}
+
+\begin{proof}
+Let $P$ be the [[Definition:Strict Positivity Property|(strict) positivity property]] on $D$.
+Let $<$ be the [[Definition:Strict Total Ordering|(strict) total ordering]] defined on $D$ as:
+:$a < b \iff a \le b \land a \ne b$
+Let $N$ be the [[Definition:Strict Negativity Property|strict negativity property]] on $D$.
+We consider all possibilities in turn.
+$(1): \quad a = 0_D$ or $b = 0_D$
+In this case, both the {{LHS}} $\size a \times \size b$ and the {{RHS}} are equal to [[Definition:Ring Zero|zero]].
+So:
+:$\size a \times \size b = \size {a \times b}$
+$(2): \quad \map P a, \map P b$
+First:
+{{begin-eqn}}
+{{eqn | l = \map P a, \map P b
+      | o = \leadsto
+      | r = \size a = a, \size b = b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{eqn | o = \leadsto
+      | r = \size a \times \size b = a \times b
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map P a, \map P b
+      | o = \leadsto
+      | r = \map P {a \times b}
+      | c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 2)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \size {a \times b} = a \times b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{end-eqn}}
+So:
+:$\size a \times \size b = \size {a \times b}$
+ 
+$(3): \quad \map P a, \map N b$
+First:
+{{begin-eqn}}
+{{eqn | l = \map P a, \map N b
+      | o = \leadsto
+      | r = \size a = a, \size b = -b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{eqn | o = \leadsto
+      | r = \size a \times \size b = -a \times b
+      | c = [[Product with Ring Negative]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map P a, \map N b
+      | o = \leadsto
+      | r = \map N {a \times b}
+      | c = [[Properties of Strict Negativity|Properties of Strict Negativity: $(5)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-a \times b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \size {a \times b} = -a \times b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{end-eqn}}
+So:
+:$\size a \times \size b = \size {a \times b}$
+Similarly $\map N a, \map P b$.
+$(4): \quad \map N a, \map N b$
+First:
+{{begin-eqn}}
+{{eqn | l = \map N a, \map N b
+      | o = \leadsto
+      | r = \size a = -a, \size b = -b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{eqn | o = \leadsto
+      | r = \size a \times \size b = a \times b
+      | c = [[Product of Ring Negatives]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map N a, \map N b
+      | o = \leadsto
+      | r = \map P {a \times b}
+      | c = [[Properties of Strict Negativity|Properties of Strict Negativity: $(4)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \map P {a \times b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \size {a \times b} = a \times b
+      | c = {{Defof|Absolute Value on Ordered Integral Domain}}
+}}
+{{end-eqn}}
+So:
+:$\size a \times \size b = \size {a \times b}$
+In all cases the result holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Absolute Value}
+Tags: Absolute Value Function, Inequalities, Negative of Absolute Value
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\size x$ denote the [[Definition:Absolute Value|absolute value]] of $x$.
+Then:
+:$-\size x \le x \le \size x$
+\end{theorem}
+
+\begin{proof}
+Either $x \ge 0$ or $x < 0$.
+:If $x \ge 0$, then:
+::$-\size x \le 0 \le x = \size x$
+:If $x < 0$, then:
+::$-\size x = x < 0 < \size x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Squares in Ordered Ring}
+Tags: Ordered Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $x, y \in \struct {R, +, \circ, \le}$ such that $0_R \le x, y$.
+Then:
+:$x \le y \implies x \circ x \le y \circ y$
+When $R$ is one of the standard [[Definition:Number|sets of numbers]], that is $\Z, \Q, \R$, then this translates into:
+:If $x, y$ are [[Definition:Positive|positive]] then $x \le y \implies x^2 \le y^2$.
+\end{theorem}
+
+\begin{proof}
+Assume $x \le y$.
+As $\le$ is [[Definition:Ordering Compatible with Ring Structure|compatible]] with the ring structure of $\struct {R, +, \circ, \le}$, we have:
+:$x \ge 0 \implies x \circ x \le x \circ y$
+:$y \ge 0 \implies x \circ y \le y \circ y$
+and thus as $\le$ is [[Definition:Ordering|transitive]], it follows that $x \circ x \le y \circ y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuum Property}
+Tags: Real Analysis, Named Theorems, Continuum Property
+
+\begin{theorem}
+Let $S \subset \R$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]] such that $S$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]].
+Then $S$ [[Definition:Supremum of Subset of Real Numbers|admits a supremum]] in $\R$.
+This is known as the '''least upper bound property''' of the [[Definition:Real Number|real numbers]].
+Similarly, let $S \subset \R$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]] such that $S$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]].
+Then $S$ [[Definition:Infimum of Subset of Real Numbers|admits an infimum]] in $\R$.
+This is sometimes called the '''greatest lower bound property''' of the [[Definition:Real Number|real numbers]].
+The two properties taken together are called the '''continuum property of $\R$'''.
+This can also be stated as:
+:The set $\R$ of [[Definition:Real Number|real numbers]] is [[Definition:Dedekind Complete|Dedekind complete]].
+\end{theorem}
+
+\begin{proof}
+Suppose that $S \subseteq \R_{\ge 0}$ has the [[Definition:Positive Real Number|positive real number]] $U$ as an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]].
+Then $\R_{\ge 0}$ can be represented as a [[Definition:Real Number Line|straight line]] $L$ whose sole [[Definition:Endpoint of Line|endpoint]] is the [[Definition:Point|point]] $O$. 
+Let $l_0 \in \R_{\ge 0}$ be the standard [[Definition:Unit of Measurement|unit]] of [[Definition:Length (Linear Measure)|length]].
+There exists a [[Definition:Unique|unique]] [[Definition:Point|point]] $X \in L$ such that $U \cdot l_0 = OX$.
+Furthermore, if $x \in S$, then:
+:$\map f x = x \cdot l_0$
+where $\cdot$ denotes [[Definition:Real Multiplication|(real) multiplication]].
+=== Segments of Finite Lines are Finite ===  
+No [[Definition:Line Segment|line segment]] of $OX$ is [[Definition:Infinite Line|infinite]]. 
+For suppose that the [[Definition:Line Segment|segment]] $s$ of $OX$ is [[Definition:Infinite Line|infinite]].
+Then $s$ is greater than every [[Definition:Line Segment|line segment]], including any [[Definition:Line Segment|line]] four times greater than $OX$.
+Therefore the less contains the greater: which is impossible.
+{{qed|lemma}}
+=== Existence of Second Endpoint of a Segment of $OX$ beginning at $O$ === 
+More precisely, every [[Definition:Line Segment|line segment]] $s$ of $OX$ having $O$ as one of its [[Definition:Endpoint of Line|endpoint]]s must have another [[Definition:Endpoint of Line|endpoint]] within $OX$.
+The second [[Definition:Endpoint of Line|endpoint]] of $s$ of $OX$ must exist. 
+For if the second [[Definition:Endpoint of Line|endpoint]] does not exist, then $s$ can be [[Definition:Production|continued]] to any [[Definition:Length (Linear Measure)|length]] however great and still remain a [[Definition:Line Segment|segment]] of $OX$.
+{{handwaving}}
+But then $s$ can be made over four times as great as $OX$, and still remain a [[Definition:Line Segment|segment]] of $OX$: which is impossible. 
+Therefore the second [[Definition:Endpoint of Line|endpoint]] exists. 
+{{qed|lemma}}
+=== Both Endpoints of a Segment of a Line lie Within the Line === 
+Every [[Definition:Point|point]] of a [[Definition:Line Segment|segment]] of a [[Definition:Line Segment|straight line]] $ab$ lies within $ab$.  
+This second [[Definition:Endpoint of Line|endpoint]] must be within $OX$.
+{{qed|lemma}}
+=== Formation of the Set $S^*$ Corresponding to the Set $S$ === 
+Therefore for every $x \in S$, there is a [[Definition:Unique|unique]] [[Definition:Point|point]] $\map w x$ such that:
+:$\map f x = O \cdot \map w x = x \cdot l_0$
+Thus let $S^*$ be the corresponding [[Definition:Set|set]] of all respective [[Definition:Line Segment|line segment]]s:
+:$\map f x = O \cdot \map w x$
+for all $x \in S$. 
+We have that $OX$ is greater than or equal to every [[Definition:Line Segment|line segment]] of $S^*$. 
+Therefore $OX$ contains every [[Definition:Line Segment|line segment]] of $S^*$.
+And for any two [[Definition:Line Segment|line segment]]s $P, Q$ of $L$ with an [[Definition:Endpoint of Line|endpoint]] at $O$, either:
+:$P, Q$ are identical
+:$P \subset Q$ but $Q \not \subset P$
+or:
+:$Q \subset P$ but $P \not \subset Q$. 
+Also:
+:$P \subset Q, Q \not \subset P \iff P > Q$
+The same can be proven for any other [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] $OY$ in $S^*$.
+=== Definition of $\Lambda$ === 
+Let $\Lambda$ be the [[Definition:Set Union|union]] of all [[Definition:Line Segment|line segment]]s of $S^*$. 
+=== Existence of $\Lambda$ ===
+Let $x \in S$.
+Then:
+:$\map f x = O \cdot \map w x$
+The [[Definition:Point|point]] $O$ is an [[Definition:Element|element]] of $O \cdot \map w x$. 
+Therefore there is a [[Definition:Point|point]] $p$ contained in at least one [[Definition:Line Segment|line segment]] of $S^*$. 
+But then there must exist an exhaustive and complete figure $F$ containing only all of those [[Definition:Point|point]]s $p$ contained in at least one [[Definition:Line Segment|line segment]] of $S^*$. 
+Set theory shows that this figure $F$ is precisely $\Lambda$.
+{{handwaving|The above needs to be clarified and explained with reference to set-theoretical proofs.}}
+{{qed|lemma}}
+=== Continuity of $\Lambda$ ===
+$\Lambda$ is everywhere continuous.
+{{explain|Define "continuous" in this context.}}
+For, given $p, q \in \Lambda$, such that $p$ and $q$ do not coincide, either $Op > Oq$ or $Op < Oq$. 
+{{WLOG}}, let $p$ be less [[Definition:Distance (Linear Measure)|distant]] from $O$ than $q$.
+Then:
+:$\exists x, h \in \R_{\ge 0}: x \in S \land x + h \in S \land p \in O \cdot \map w x \land q \in O \cdot \map w {x + h}$. 
+But:
+:$O \cdot \map w x \subset O \cdot \map w {x + h}$
+Therefore:
+:$p, q \in O \cdot \map w {x + h}$
+Therefore $O \cdot \map w {x + h}$ contains every [[Definition:Point|point]] in between $p, q \in \Lambda$.
+Thus suppose $r$ is between $p, q$. 
+Therefore:
+:$r \in O \cdot \map w {x + h}$
+But:
+:$O \cdot \map w {x + h} \in S^*$
+Also, $\Lambda$ contains all $p$ in at least one $O \cdot \map w x \in S^*$
+Therefore:
+:$r \in O \cdot \map w {x + h}$
+Therefore:
+:$\forall p, q \in \Lambda: \forall r: p < r < q: r \in \Lambda$
+Therefore $\Lambda$ is everywhere continuous. 
+{{qed|lemma}}
+=== $\Lambda$ is Finite ===
+$\Lambda \subseteq OX$.
+Let $p \in \Lambda$.
+Then:
+:$\exists y \in S: p \in O \cdot \map w y \in S^*$
+But it was proven that $OX$ contains every [[Definition:Line Segment|line segment]] of $S^*$. 
+Therefore:
+:$p \in O \cdot \map w y \subseteq OX$
+Therefore:
+:$p \in OX$
+Therefore:
+:$\Lambda \subseteq OX$
+ 
+Therefore $\Lambda$ has a second [[Definition:Endpoint of Line|endpoint]] $Z \in OX$, such that $Z$ is between $O, X$. 
+Therefore $\Lambda = OZ$.
+{{qed|lemma}}
+=== $\Lambda$ is an Upper Bound on $S^*$ ===
+$OZ$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$.
+ 
+For from set theory it is known that the [[Definition:Set Union|union]] of all the sets any given set $D$ contains every set of $D$.
+{{explain|Link to an appropriate result and express the above sentence in mathematical language.}}
+Therefore $OZ$ contains every [[Definition:Element|element]] of $S^*$. 
+But then no [[Definition:Element|element]] of $S^*$ can ever be greater than $OZ$. 
+Therefore $OZ$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$. 
+{{qed|lemma}}
+=== $\Lambda$ is the Supremum on $S^*$ ===
+$OZ$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] on $S^*$.
+For if $OY$ is any [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$, $\Lambda \subseteq OY$. 
+For if $p \in \Lambda$, there is some $y \in S$ such that $p \in O \cdot \map w y \in S^*$. 
+But it was remarked earlier that if $OY$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$, then $OY$ contains every [[Definition:Line Segment|line segment]] of $S^*$. 
+Therefore $p \in O \cdot \map w y \subseteq OY$.
+ 
+Therefore $p \in OY$. 
+Therefore $\Lambda \subseteq OY$. 
+Therefore $\Lambda \le OY$. 
+Therefore $OZ$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$ and less than or equal to every [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$.
+Therefore $OZ$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] on $S^*$.
+From [[Supremum is Unique]], $OZ$ is [[Definition:Unique|unique]].
+{{qed|lemma}}
+=== Definition of $z$ ===
+There is a [[Definition:Unique|unique]] $z \in \R_{\ge 0}$ such that:
+:$\map f z = O \cdot \map w z = OZ$
+=== $z$ is an Upper Bound on $S$ ===
+$z$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on all the [[Definition:Element|element]]s of $S$.
+For if not, then suppose there had been some $x \in S$ such that $x > z$. 
+$L$ is a representation of the [[Definition:Positive Real Number|positive]] [[Definition:Real Number Line|real number line]]. 
+But if $x \in S$, then $O \cdot \map w x \in S^*$. 
+Yet $O \cdot \map w x > OZ$, which is impossible because $OZ$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] on $S^*$. 
+Therefore $z$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on all the [[Definition:Element|element]]s of $S$.
+{{qed|lemma}}
+=== $z$ is the Supremum on $S$ ===
+$z$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] on all the [[Definition:Element|element]]s of $S$.
+ 
+For if $g \in \R_{\ge 0}$ and $g < z$, then $O \cdot \map w g < OZ$.
+ 
+Therefore $O \cdot \map w g$ is not an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S^*$.
+ 
+But on the contrary, there exists $\xi \in S$ such that $O \cdot \map w \xi \in S^*$ and $O \cdot \map w \xi > O \cdot \map w g$. 
+But then $\xi > g$ and $\xi \in S$. 
+Therefore no $g < z$ can be an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] on $S$. 
+Therefore $z$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] on all the [[Definition:Element|element]]s of $S$. 
+{{qed|lemma}}
+=== Conclusion ===
+Therefore the [[Definition:Set|set]] $S \subseteq \R_{\ge 0}$ with [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] $U$ has the [[Definition:Unique|unique]] [[Definition:Supremum of Subset of Real Numbers|supremum]] $z$. 
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be [[Definition:Bounded Above Subset of Real Numbers|bounded above]].
+Let $L$ be the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] defined as:
+:$\alpha \in L \iff \exists x \in S: \alpha < x$
+Let $R := \relcomp \R L$, where $\complement_\R$ denotes [[Definition:Relative Complement|complement in $\R$]].
+By construction of $L$, every [[Definition:Element|element]] of $L$ is less than some [[Definition:Element|element]] of $S$.
+Hence no [[Definition:Element|element]] of $L$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] of $S$.
+By construction of $R$, for every [[Definition:Element|element]] $x$ of $R$, there exists no [[Definition:Element|element]] of $S$ which is greater than $x$.
+Hence every [[Definition:Element|element]] of $R$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] of $S$.
+So, to prove the existence of $\sup S$, it is [[Definition:Sufficient Condition|sufficient]] to demonstrate that $R$ contains a [[Definition:Smallest Element|smallest number]].
+We verify that $L$ and $R$ fulfil the conditions for [[Dedekind's Theorem]] to hold.
+We confirm that $\tuple {L, R}$ is a [[Definition:Dedekind Cut|Dedekind cut]] of $\R$:
+:$(1): \quad \set {L, R}$ is a [[Definition:Partition|partition]] of $\R$
+:$(2): \quad L$ does not have a [[Definition:Greatest Element|greatest element]]
+:$(3): \quad \forall x \in L: \forall y \in R: x < y$
+By [[Union with Relative Complement]]:
+:$L \cup R = \R$
+By [[Set Difference and Intersection form Partition/Corollary 2|corollary $2$ to Set Difference and Intersection form Partition]], $\tuple {L, R}$ forms a [[Definition:Partition|partition]] of $\R$.
+So $(1)$ holds immediately.
+Let $\alpha \in L$.
+Then there exists $x \in S$ such that $\alpha < x$.
+Let $\alpha'$ be such that $\alpha < \alpha' < x$.
+Then $\alpha' \in L$
+So whatever $\alpha \in L$ is, it cannot be the [[Definition:Greatest Element|greatest element]] of $L$.
+Thus $(2)$ holds.
+Let $\alpha \in L$.
+Let $\beta \in R$.
+Then there exists $x \in S$ such that $\alpha < x$.
+By construction of $R$, $x \le \beta$.
+Thus $\alpha < \beta$ for all $\alpha \in L, \beta \in R$.
+Thus $(3)$ holds.
+By [[Dedekind's Theorem/Corollary|the corollary to Dedekind's Theorem]], either $L$ contains a [[Definition:Greatest Element|greatest element]] or $R$ contains a [[Definition:Smallest Element|smallest element]].
+We have shown that $L$ does not contain a [[Definition:Greatest Element|greatest element]].
+Hence $R$ contains a [[Definition:Smallest Element|smallest element]].
+Hence if $S$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]], it has a [[Definition:Supremum of Subset of Real Numbers|supremum]].
+Thus $\R$ is [[Definition:Dedekind Complete|Dedekind complete]] by definition.
+Now let $S$ be [[Definition:Bounded Below Subset of Real Numbers|bounded below]].
+By [[Dedekind Completeness is Self-Dual]], it follows that $S$ admits an [[Definition:Infimum of Subset of Real Numbers|infimum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Multiplication is Closed}
+Tags: Complex Multiplication, Algebraic Closure, Complex Multiplication is Closed
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Complex Multiplication|multiplication]]:
+:$\forall z, w \in \C: z \times w \in \C$
+\end{theorem}
+
+\begin{proof}
+From the informal definition of [[Definition:Complex Number#Informal Definition|complex numbers]], we define the following:
+: $z = x_1 + i y_1$
+: $w = x_2 + i y_2$
+where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2$.
+Then from the definition of [[Definition:Complex Multiplication|complex multiplication]]:
+: $z w = \left({x_1 x_2 - y_1 y_2}\right) + i \left({x_1 y_2 + x_2 y_1}\right)$
+From [[Real Numbers form Field]]:
+: $x_1 x_2 - y_1 y_2 \in \R$
+and:
+: $x_1 y_2 + x_2 y_1 \in \R$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the formal definition of [[Definition:Complex Number#Formal Definition|complex numbers]], we define the following:
+: $z = \left({x_1, y_1}\right)$
+: $w = \left({x_2, y_2}\right)$
+Then from the definition of [[Definition:Complex Multiplication|complex multiplication]]:
+: $z w = \left({x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}\right)$
+From [[Real Numbers form Field]]:
+: $x_1 x_2 - y_1 y_2 \in \R$
+and:
+: $x_1 y_2 + x_2 y_1 \in \R$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Multiplication is Associative}
+Tags: Complex Multiplication
+
+\begin{theorem}
+The operation of [[Definition:Complex Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Associative|associative]]:
+:$\forall z_1, z_2, z_3 \in \C: z_1 \paren {z_2 z_3} = \paren {z_1 z_2} z_3$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Number/Definition 2|complex numbers]], we define the following:
+{{begin-eqn}}
+{{eqn | l = z_1
+      | o = :=
+      | r = \tuple {x_1, y_1}
+}}
+{{eqn | l = z_2
+      | o = :=
+      | r = \tuple {x_2, y_2}
+}}
+{{eqn | l = z_3
+      | o = :=
+      | r = \tuple {x_3, y_3}
+}}
+{{end-eqn}}
+where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.
+Thus:
+{{begin-eqn}}
+{{eqn | r = z_1 \left({z_2 z_3}\right)
+      | o = 
+}}
+{{eqn | r = \tuple {x_1, y_1} \paren {\tuple {x_2, y_2} \tuple {x_3, y_3} }
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{eqn | r = \tuple {x_1, y_1} \tuple {x_2 x_3 - y_2 y_3, x_2 y_3 + y_2 x_3}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {x_1 \paren {x_2 x_3 - y_2 y_3} - y_1 \paren {x_2 y_3 + y_2 x_3}, y_1 \paren {x_2 x_3 - y_2 y_3} + x_1 \paren {x_2 y_3 + y_2 x_3} }
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {x_1 x_2 x_3 - x_1 y_2 y_3 - y_1 x_2 y_3 - y_1 y_2 x_3, y_1 x_2 x_3 - y_1 y_2 y_3 + x_1 x_2 y_3 + x_1 y_2 x_3}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \tuple {\paren {x_1 x_2 x_3 - y_1 y_2 x_3} - \paren {x_1 y_2 y_3 + y_1 x_2 y_3}, \paren {x_1 x_2 y_3 - y_1 y_2 y_3} + \paren {y_1 x_2 x_3 + x_1 y_2 x_3} }
+      | c = [[Real Multiplication is Commutative]]
+}}
+{{eqn | r = \tuple {\paren {x_1 x_2 - y_1 y_2} x_3 - \paren {x_1 y_2 + y_1 x_2} y_3, \paren {x_1 x_2 - y_1 y_2} y_3 + \paren {x_1 y_2 + y_1 x_2} x_3}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2} \tuple {x_3, y_3}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \paren {\tuple {x_1, y_1} \tuple {x_2, y_2} } \tuple {x_3, y_3}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \paren {z_1 z_2} z_3
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arithmetic Mean is Never Less than Harmonic Mean}
+Tags: Inequalities, Arithmetic Mean, Harmonic Mean
+
+\begin{theorem}
+Let $x_1, x_2, \ldots, x_n \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Let $A_n $ be the [[Definition:Arithmetic Mean|arithmetic mean]] of $x_1, x_2, \ldots, x_n$.
+Let $H_n$ be the [[Definition:Harmonic Mean|harmonic mean]] of $x_1, x_2, \ldots, x_n$.
+Then $A_n \ge H_n$.
+\end{theorem}
+
+\begin{proof}
+$A_n$ is defined as:
+:$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$
+$H_n$ is defined as:
+:$\displaystyle \frac 1 H_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {x_k} }$
+We have that:
+:$\forall k \in \closedint 1 n: x_k > 0$
+From [[Positive Real has Real Square Root]], we can express each $x_k$ as a [[Definition:Square (Algebra)|square]]:
+:$\forall k \in \closedint 1 n: x_k = y_k^2$
+without affecting the result.
+Thus we have:
+:$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2}$
+:$\displaystyle \frac 1 {H_n} = \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }$
+Multiplying $A_n$ by $\dfrac 1 {H_n}$:
+{{begin-eqn}}
+{{eqn | l = \frac {A_n} {H_n}
+      | r = \frac 1 n \paren {\sum_{k \mathop = 1}^n y_k^2} \frac 1 n \paren {\sum_{k \mathop = 1}^n \frac 1 {y_k^2} }
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n \frac {y_k} {y_k} }^2
+      | c = [[Cauchy's Inequality]]
+}}
+{{eqn | r = \frac 1 {n^2} \paren {\sum_{k \mathop = 1}^n 1}^2
+      | c = 
+}}
+{{eqn | r = \frac {n^2} {n^2} = 1
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\dfrac {A_n} {H_n} \ge 1$
+and so from [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$]]: [[Definition:Relation Compatible with Operation|compatible]] with [[Definition:Real Multiplication|multiplication]]:
+:$A_n \ge H_n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiple of Supremum}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S \subseteq \R: S \ne \varnothing$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers $\R$]].
+Let $S$ be [[Definition:Bounded Above Set|bounded above]].
+Let $z \in \R: z > 0$ be a [[Definition:Positive Real Number|positive real number]].
+Then:
+:$\displaystyle \map {\sup_{x \mathop \in S} } {z x} = z \map {\sup_{x \mathop \in S} } x$
+\end{theorem}
+
+\begin{proof}
+Let $B = \map \sup S$.
+Then by definition, $B$ is the [[Definition:Smallest Element|smallest]] [[Definition:Real Number|number]] such that $x \in S \implies x \le B$.
+Let $T = \set {z x: x \in S}$.
+Because $z > 0$, it follows that:
+:$\forall x \in S: z x \le z B$
+So $T$ is [[Definition:Bounded Above Set|bounded above]] by $z B$.
+By the [[Continuum Property]], $T$ has a [[Definition:Supremum of Set|supremum]] which we will call $C$.
+We need to show that $C = z B$.
+Since $z B$ is ''an'' [[Definition:Upper Bound of Set|upper bound]] for $T$, and $C$ is the ''smallest'' [[Definition:Upper Bound of Set|upper bound]] for $T$, it follows that $C \le z B$.
+Now as $z > 0$ and is a [[Definition:Real Number|real number]]:
+:$\exists z^{-1} \in \R: z^{-1} > 0$
+So we can reverse the roles of $S$ and $T$:
+:$S = \set {z^{-1} y: y \in T}$
+We know that $C$ is the [[Definition:Smallest Element|smallest]] [[Definition:Real Number|number]] such that:
+:$\forall y \in T: y \le C$
+So it follows that:
+:$\forall y \in T: z^{-1} y \le z^{-1} C$
+So $z^{-1} C$ is ''an'' [[Definition:Upper Bound of Set|upper bound]] for $S$.
+But $B$ is the ''[[Definition:Smallest Element|smallest]]'' [[Definition:Upper Bound of Set|upper bound]] for $S$.
+So:
+:$B \le z^{-1} C \implies z B \le C$
+So we have shown that:
+:$z B \le C$ and $C \le z B$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy's Mean Theorem}
+Tags: Arithmetic Mean, Geometric Mean, Inequalities, Cauchy's Mean Theorem
+
+\begin{theorem}
+Let $x_1, x_2, \ldots, x_n \in \R$ be [[Definition:Real Number|real numbers]] which are all [[Definition:Positive Real Number|positive]].
+Let $A_n$ be the [[Definition:Arithmetic Mean|arithmetic mean]] of $x_1, x_2, \ldots, x_n$.
+Let $G_n$ be the [[Definition:Geometric Mean|geometric mean]] of $x_1, x_2, \ldots, x_n$.
+Then:
+:$A_n \ge G_n$
+with equality holding {{iff}}:
+:$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
+That is, {{iff}} all [[Definition:Term of Sequence|terms]] are equal.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map f x = \ln x$
+for $x > 0$.
+With a view to apply [[Jensen's Inequality (Real Analysis)/Corollary|Jensen's Inequality: Real Analysis: Corollary]], we can show that $f$ is [[Definition:Concave Real Function|concave]] on $\openint 0 \infty$. 
+By [[Second Derivative of Concave Real Function is Non-Positive]], it is sufficient to show that $\map {f''} x \le 0$ for all $x > 0$. 
+We have, by [[Derivative of Natural Logarithm]]:
+:$\map {f'} x = \dfrac 1 x$
+We then have, by [[Derivative of Power]]:
+:$\map {f''} x = -\dfrac 1 {x^2}$ 
+As $x^2 > 0$ for all $x > 0$, we have:
+:$\dfrac 1 {x^2} > 0$
+Therefore: 
+:$-\dfrac 1 {x^2} = \map {f''} x < 0$
+so $f$ is indeed [[Definition:Concave Real Function|concave]] on $\openint 0 \infty$.
+As $x_1, x_2, \ldots, x_n$ are all [[Definition:Positive Real Number|positive]], they all lie in the interval $\openint 0 \infty$. 
+We therefore have, by [[Jensen's Inequality (Real Analysis)/Corollary|Jensen's Inequality: Real Analysis: Corollary]]: 
+:$\displaystyle \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} } \ge \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}$
+for [[Definition:Real Number|real]] $\lambda_1, \lambda_2, \ldots, \lambda_n \ge 0$, with at least one of which being non-zero. 
+As $n$ is a [[Definition:Positive Integer|positive integer]], we have: 
+:$\dfrac 1 n > 0$
+We can therefore set: 
+:$\lambda_i = \dfrac 1 n$
+for $1 \le i \le n$.
+This gives: 
+{{begin-eqn}}
+{{eqn	| l = \map \ln {\frac {\sum_{k \mathop = 1}^n \lambda_k x_k} {\sum_{k \mathop = 1}^n \lambda_k} }
+	| r = \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} {n \sum_{k \mathop = 1}^n \frac 1 n} }
+}}
+{{eqn	| r = \map \ln {\frac {\sum_{k \mathop = 1}^n x_k} n}
+}}
+{{eqn	| r = \map \ln {A_n}
+	| c = {{Defof|Arithmetic Mean}}
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn	| l = \frac {\sum_{k \mathop = 1}^n \lambda_k \map \ln {x_k} } {\sum_{k \mathop = 1}^n \lambda_k}
+	| r = \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } {n \sum_{k \mathop = 1}^n \frac 1 n}
+}}
+{{eqn	| r = \frac {\sum_{k \mathop = 1}^n \map \ln {x_k} } n
+}}
+{{eqn	| r = \frac 1 n \map \ln {\prod_{k \mathop = 1}^n x_k}
+	| c = [[Sum of Logarithms]]
+}}
+{{eqn	| r = \map \ln {\paren {\prod_{k \mathop = 1}^n x_k}^{1/n} }
+	| c = [[Logarithm of Power]]
+}}
+{{eqn	| r = \map \ln {G_n}
+	| c = {{Defof|Geometric Mean}}
+}}
+{{end-eqn}}
+We therefore have: 
+:$\map \ln {A_n} \ge \map \ln {G_n}$
+Note that for $x > 0$: 
+:$\dfrac 1 x = \map {f'} x > 0$
+Therefore, by [[Derivative of Monotone Function]], $f$ is [[Definition:Increasing Real Function|increasing]] on $\openint 0 \infty$. 
+We therefore have: 
+:$A_n \ge G_n$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The [[Definition:Arithmetic Mean|arithmetic mean]] of $x_1, x_2, \ldots, x_n$ is defined as:
+:$\displaystyle A_n = \frac 1 n \paren {\sum_{k \mathop = 1}^n x_k}$
+The [[Definition:Geometric Mean|geometric mean]] of $x_1, x_2, \ldots, x_n$ is defined as:
+:$\displaystyle G_n = \paren {\prod_{k \mathop = 1}^n x_k}^{1/n}$
+We prove the result by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:For all [[Definition:Positive Real Number|positive real numbers]] $x_1, x_2, \ldots, x_n: A_n \ge G_n$.
+$\map P 1$ is true, as this just says:
+:$\dfrac {x_1} 1 \ge x_1^{1/1}$
+which is trivially true.
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+:$\dfrac {x_1 + x_2} 2 \ge \sqrt {x_1 x_2}$
+As $x_1, x_2 > 0$ we can take their [[Definition:Square Root|square roots]] and do the following:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | r = \paren {\sqrt {x_1} - \sqrt {x_2} }^2
+      | c = 
+}}
+{{eqn | r = x_1 - 2\sqrt {x_1 x_2} + x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {x_1 x_2}
+      | o = \le
+      | r = \frac {x_1 + x_2} 2
+      | c = 
+}}
+{{end-eqn}}
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we show that:
+:$(1): \quad$ If $\map P {2^k}$ is true, where $k \ge 1$, then it logically follows that $\map P {2^{k + 1} }$ is true
+:$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true.
+The result will follow by [[Backwards Induction]].
+This is our first [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$A_{2^k} \ge G_{2^k}$
+Then we need to show:
+:$A_{2^{k + 1} } \ge G_{2^{k + 1} }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $m = 2^k$.
+Then $2^{k + 1} = 2 m$.
+Because $\map P m$ is true:
+:$\paren {x_1 x_2 \dotsm x_m}^{1/m} \le \dfrac 1 m \paren {x_1 + x_2 + \dotsb + x_m}$
+Also:
+:$\paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} \le \dfrac 1 m \paren {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} }$
+But we have $\map P 2$, so:
+:$\paren {\paren {x_1 x_2 \dotsm x_m}^{1/m} \paren {x_{m + 1} x_{m + 2} \dotsm x_{2 m} }^{1/m} }^{1/2} \le \dfrac 1 2 \paren {\dfrac {x_1 + x_2 + \cdots + x_m} m + \dfrac {x_{m + 1} + x_{m + 2} + \dotsb + x_{2 m} } m}$
+So:
+:$\paren {x_1 x_2 \dotsm x_{2 m} }^{1/2m} \le \dfrac {x_1 + x_2 + \dotsb + x_{2 m} } {2 m}$
+So $\map P {2 m} = \map P {2^{k + 1} }$ holds.
+So $\map P {2^n}$ holds for all $n$ by [[Principle of Mathematical Induction|induction]].
+Now suppose $\map P k$ holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1/k}
+      | o = \le
+      | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}
+      | o = \le
+      | r = \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k G_{k - 1}
+      | o = \le
+      | r = \paren {k - 1} A_{k - 1} + G_{k - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {k - 1} G_{k - 1}
+      | o = \le
+      | r = \paren {k - 1} A_{k - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = G_{k - 1}
+      | o = \le
+      | r = A_{k - 1}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k - 1}$ and the result follows by [[Backwards Induction]].
+Therefore $A_n \ge G_n$ for all $n$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Necessary Condition ===
+Let:
+:$\forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j = x$
+Then:
+{{begin-eqn}}
+{{eqn | l = A_n
+      | r = \dfrac 1 n \sum_{j \mathop = 1}^n x
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n n x
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = G_n
+      | r = \paren {\prod_{j \mathop = 1}^n x}^{\frac 1 n}
+      | c = 
+}}
+{{eqn | r = \paren {x^n}^{\frac 1 n}
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$A_n = G_n = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $A_n = G_n$.
+We prove the result by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
+$\map P 1$ is true, as this just says:
+{{begin-eqn}}
+{{eqn | l = A_1
+      | r = G_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 1 1 \sum_{j \mathop = 1}^1 x_j
+      | r = \paren {\prod_{j \mathop = 1}^1 x_j}^{1/1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1
+      | r = x_1
+      | c = 
+}}
+{{end-eqn}}
+which is trivially true.
+==== Basis for the Induction ====
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = A_2
+      | r = G_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {x_1 + x_2} 2
+      | r = \sqrt {x_1 x_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\dfrac {x_1 + x_2} 2}^2
+      | r = x_1 x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1^2 + 2 x_1 x_2 + x_2^2
+      | r = 4 x_1 x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1^2 - 2 x_1 x_2 + x_2^2
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x_1 - x_2}^2
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1
+      | r = x_2
+      | c = 
+}}
+{{end-eqn}}
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+==== Induction Hypothesis ====
+Now we show that:
+:$(1): \quad$ If $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {2 k}$ is true
+:$(2): \quad$ If $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k - 1}$ is true.
+The result will follow by [[Backwards Induction]].
+This is our first [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$A_k = G_k \implies \forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$
+Also, let:
+:$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
+and:
+:$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
+By the [[Cauchy's Mean Theorem/Proof of Equality Condition#Induction Hypothesis|induction hypothesis]]:
+:$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$
+We need to show:
+:$A_{2 k} = G_{2 k} \implies \forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$
+==== Induction Step ====
+This is our [[Definition:Induction Step|induction step]]:
+Suppose:
+:$A_{2 k} = G_{2 k}$
+Let:
+:$A_k' := \displaystyle \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j = y$
+and:
+:$G_k' := \displaystyle \paren {\prod_{j \mathop = k + 1}^{2 k} }^{1 / k} x_j = y$
+By the [[Cauchy's Mean Theorem/Proof of Equality Condition#Induction Hypothesis|induction hypothesis]]:
+:$A_k' = G_k' \implies \forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = y$
+Then:
+{{begin-eqn}}
+{{eqn | l = A_{2 k}
+      | r = \dfrac 1 {2 k} \sum_{j \mathop = 1}^{2 k} x_j
+      | c = {{Defof|Arithmetic Mean}}
+}}
+{{eqn | r = \dfrac 1 {2 k} \paren {\sum_{j \mathop = 1}^k x_j + \sum_{j \mathop = k + 1}^{2 k} x_j}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\dfrac 1 k \sum_{j \mathop = 1}^k x_j + \dfrac 1 k \sum_{j \mathop = k + 1}^{2 k} x_j}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {A_k + A_k'}
+      | c = {{Defof|Arithmetic Mean}}
+}}
+{{eqn | r = \dfrac 1 2 \paren {x + y}
+      | c = Definition of $A_k$ and $A_k'$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = G_{2 k}
+      | r = \paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / 2 k}
+      | c = {{Defof|Geometric Mean}}
+}}
+{{eqn | r = \paren {\paren {\prod_{j \mathop = 1}^{2 k} x_j}^{1 / k} }^{1 / 2}
+      | c = 
+}}
+{{eqn | r = \paren {\paren {\prod_{j \mathop = 1}^k x_j}^{1 / k} \times \paren {\prod_{j \mathop = k + 1}^{2 k} x_j}^{1 / k} }^{1 / 2}
+      | c = 
+}}
+{{eqn | r = \paren {G_k G_k'}^{1 / 2}
+      | c = {{Defof|Geometric Mean}}
+}}
+{{eqn | r = \paren {x y}^{1 / 2}
+      | c = Definition of $G_k$ and $G_k'$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = A_{2 k}
+      | r = G_{2 k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 1 2 \paren {x + y}
+      | r = \paren {x y}^{1 / 2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = [[Cauchy's Mean Theorem/Proof of Equality Condition#Basis for the Induction|Basis for the Induction]]
+}}
+{{end-eqn}}
+That is:
+:$\forall i, j \in \set {1, 2, \ldots, k}: x_i = x_j = x$
+:$\forall i, j \in \set {k + 1, k + 2, \ldots, 2 k}: x_i = x_j = x$
+Hence:
+:$\forall i, j \in \set {1, 2, \ldots, 2 k}: x_i = x_j = x$
+Now suppose $\map P k$ holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = G_{k - 1}
+      | r = A_{k - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {k - 1} G_{k - 1}
+      | r = \paren {k - 1} A_{k - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k G_{k - 1}
+      | r = \paren {k - 1} A_{k - 1} + G_{k - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {G_{k - 1}^{k - 1} G_{k - 1} }^{1/k}
+      | r = \dfrac {\paren {k - 1} A_{k - 1} + G_{k - 1} } k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}
+      | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}
+      | r = \dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k
+      | c = 
+}}
+{{end-eqn}}
+But:
+:$\paren {x_1 x_2 \dotsm x_{k - 1} G_{k - 1} }^{1 / k}$ is the [[Definition:Geometric Mean|geometric mean]] of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
+:$\dfrac {x_1 + x_2 + \dotsm + x_{k - 1} + G_{k - 1} } k$ is the [[Definition:Arithmetic Mean|arithmetic mean]] of $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$
+We have that $\set {x_1, x_2, \ldots, x_{k - 1}, G_{k - 1} }$ has $k$ [[Definition:Element|elements]].
+Hence by the [[Cauchy's Mean Theorem/Proof of Equality Condition#Induction Hypothesis|induction hypothesis]]:
+:$\forall i, j \in \set {1, 2, \ldots, k - 1}: x_i = x_j = G_{k - 1}$
+So $\map P k \implies \map P {k - 1}$ and the result follows by [[Backwards Induction]].
+Therefore $A_n \ge G_n$ for all $n$.
+:$\forall n \in \N: A_n = G_n \implies \forall i, j \in \set {1, 2, \ldots, n}: x_i = x_j$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance on Real Numbers is Metric}
+Tags: Real Analysis, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\map d {x, y}$ be the [[Definition:Distance between Real Numbers|distance]] between $x$ and $y$:
+:$\map d {x, y} = \size {x - y}$
+Then $\map d {x, y}$ is a [[Definition:Metric|metric]] on $\R$.
+Thus it follows that $\tuple {\R, d}$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+We check the [[Definition:Metric Space Axioms|metric space axioms]] in turn.
+=== Axiom $(\text M 1)$ ===
+The statement of this axiom is:
+:$(\text M 1): \forall x \in X: \size {x - x} = 0$
+This follows from the definition of [[Definition:Absolute Value|absolute value]].
+{{qed|lemma}}
+=== Axiom $(\text M 2)$ ===
+The statement of this axiom is:
+:$(\text M 2): \forall x, y, z \in X: \size {x - y} + \size {y - z} \ge \size {x - z}$
+We have:
+:$\paren {x - y} + \paren {y - z} = \paren {x - z}$
+The result follows from the [[Triangle Inequality for Real Numbers]].
+{{qed|lemma}}
+=== Axiom $(\text M 3)$ ===
+The statement of this axiom is:
+:$(\text M 3): \forall x, y \in X: \size {x - y} = \size {y - x}$
+As $x - y = -\paren {y - x}$, it follows from the definition of [[Definition:Absolute Value|absolute value]] that $\size {x - y} = \size {y - x}$.
+{{qed|lemma}}
+=== Axiom $(\text M 4)$ ===
+The statement of this axiom is:
+:$(\text M 4): \forall x, y \in X: x \ne y \implies \size {x - y} > 0$
+This follows from the definition of [[Definition:Absolute Value|absolute value]].
+{{qed|lemma}}
+Having verified all the axioms, we conclude $d$ is a [[Definition:Metric|metric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Sequences/Real}
+Tags: Combination Theorems for Sequences, Real Analysis
+
+\begin{theorem}
+Let $\sequence {x_n}$ and $\sequence {y_n}$ be [[Definition:Real Sequence|sequences in $\R$]].
+Let $\sequence {x_n}$ and $\sequence {y_n}$ be [[Definition:Convergent Real Sequence|convergent]] to the following [[Definition:Limit of Real Sequence|limits]]:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+:$\displaystyle \lim_{n \mathop \to \infty} y_n = m$
+Let $\lambda, \mu \in \R$.
+Then the following results hold:
+=== [[Combination Theorem for Sequences/Real/Sum Rule|Sum Rule]] ===
+{{:Combination Theorem for Sequences/Real/Sum Rule}}
+=== [[Combination Theorem for Sequences/Real/Difference Rule|Difference Rule]] ===
+{{:Combination Theorem for Sequences/Real/Difference Rule}}
+=== [[Combination Theorem for Sequences/Real/Multiple Rule|Multiple Rule]] ===
+{{:Combination Theorem for Sequences/Real/Multiple Rule}}
+=== [[Combination Theorem for Sequences/Real/Combined Sum Rule|Combined Sum Rule]] ===
+{{:Combination Theorem for Sequences/Real/Combined Sum Rule}}
+=== [[Combination Theorem for Sequences/Real/Product Rule|Product Rule]] ===
+{{:Combination Theorem for Sequences/Real/Product Rule}}
+=== [[Combination Theorem for Sequences/Real/Quotient Rule|Quotient Rule]] ===
+{{:Combination Theorem for Sequences/Real/Quotient Rule}}
+\end{theorem}<|endoftext|>
+\section{Convergent Sequence in Metric Space is Bounded}
+Tags: Limits of Sequences, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $M$]] which is [[Definition:Convergent Sequence (Metric Space)|convergent]], and so $x_n \to l$ as $n \to \infty$.
+Then $\sequence {x_n}$ is [[Definition:Bounded Sequence|bounded]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $M$]] which is [[Definition:Convergent Sequence (Metric Space)|convergent]], and so $x_n \to l$ as $n \to \infty$.
+From the definition, in order to prove [[Definition:Bounded Sequence|boundedness]], all we need to do is find $K \in \R$ such that $\forall n \in \N: \map d {x_n, l} \le K$.
+Since $\sequence {x_n}$ [[Definition:Convergent Sequence (Metric Space)|converges]], it is true that:
+:$\forall \epsilon > 0: \exists N: n > N \implies \map d {x_n, l} < \epsilon$
+In particular, this is true when $\epsilon = 1$, for example.
+That is:
+:$\exists N_1: \forall n > N_1: \map d {x_n, l} < 1$
+So now we set:
+:$K = \max \set {\map d {x_1, l}, \map d {x_2, l}, \ldots, \map d {x_{N_1}, l}, 1}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence Minus Limit}
+Tags: Limits of Sequences, Convergent Sequence Minus Limit
+
+\begin{theorem}
+Let $X$ be  one of the [[Definition:Standard Number Field|standard number fields]] $\Q, \R, \C$.
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Sequence|sequence in $X$]] which [[Definition:Convergent Sequence|converges]] to $l$.
+That is:
+: $\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+Then:
+: $\displaystyle \lim_{n \mathop \to \infty} \left|{x_n - l}\right| = 0$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+We need to show that there exists $N$ such that:
+:$\forall n > N: \size {\paren {\size {x_n - l} - 0} } < \epsilon$
+But:
+:$\size {\paren {\size {x_n - l} - 0} } = \size {x_n - l}$
+So what needs to be shown is just:
+:$x_n \to l$ as $n \to \infty$
+which is the definition of $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We note that all of $\Q, \R, \C$ can be considered as [[Definition:Euclidean Space|metric spaces]].
+Then under the [[Definition:Usual Metric|usual metric]]:
+: $d \left({x_n, l}\right) = \left|{x_n - l}\right|$.
+The result follows from the definition of [[Definition:Metric|metric]]: $d \left({x_n, l}\right) = 0 \iff x_n = l$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Convergence Theorem (Real Analysis)}
+Tags: Limits of Sequences, Named Theorems, Real Analysis, Monotone Convergence Theorem (Real Analysis)
+
+\begin{theorem}
+Every [[Definition:Bounded Real Sequence|bounded]] [[Definition:Monotone Real Sequence|monotone sequence]] is [[Definition:Convergent Real Sequence|convergent]].
+\end{theorem}<|endoftext|>
+\section{One Plus Reciprocal to the Nth}
+Tags: Limits of Sequences, Reciprocals
+
+\begin{theorem}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = \paren {1 + \dfrac 1 n}^n$.
+Then $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]] as $n$ [[Definition:Increase Without Bound|increases without bound]].
+\end{theorem}
+
+\begin{proof}
+First we show that $\sequence {x_n}$ is [[Definition:Increasing Real Sequence|increasing]].
+Let $a_1 = a_2 = \cdots = a_{n - 1} = 1 + \dfrac 1 {n - 1}$.
+Let $a_n = 1$.
+Let:
+:$A_n$ be the [[Definition:Arithmetic Mean|arithmetic mean]] of $a_1 \ldots a_n$
+:$G_n$ be the [[Definition:Geometric Mean|geometric mean]] of $a_1 \ldots a_n$
+Thus:
+:$A_n = \dfrac {\paren {n - 1} \paren {1 + \dfrac 1 {n - 1} } + 1} n = \dfrac {n + 1} n = 1 + \dfrac 1 n$
+:$G_n = \paren {1 + \dfrac 1 {n - 1} }^{\dfrac {n - 1} n}$
+By [[Cauchy's Mean Theorem]]:
+: $G_n \le A_n$
+Thus:
+:$\paren {1 + \dfrac 1 {n - 1} }^{\frac {n - 1} n} \le 1 + \dfrac 1 n$
+and so:
+:$x_{n - 1} = \paren {1 + \dfrac 1 {n - 1} }^{n - 1} \le \paren {1 + \dfrac 1 n}^n = x_n$
+Hence $\sequence {x_n}$ is [[Definition:Increasing Real Sequence|increasing]].
+Next, we show that $\sequence {x_n}$ is [[Definition:Bounded Above Real Sequence|bounded above]].
+Using the [[Binomial Theorem]]:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + \frac 1 n}^n
+      | r = 1 + n \paren {\frac 1 n} + \frac {n \paren {n - 1} } 2 \paren {\frac 1 n}^2 + \cdots + \paren  {\frac 1 n}^n
+}}
+{{eqn | r = 1 + 1 + \paren {1 - \frac 1 n} \frac 1 {2!} + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \frac 1 {3!} + \cdots + \paren {1 - \frac 1 n} \paren {1 - \frac 2 n} \cdots \paren {1 - \frac {n - 1} n} \frac 1 {n!}
+}}
+{{eqn | o = \le
+      | r = 1 + 1 + \frac 1 {2!} + \frac 1 {3!} + \cdots + \frac 1 {n!}
+}}
+{{eqn | o = \le
+      | r = 1 + 1 + \frac 1 2 + \frac 1 {2^2} + \cdots + \frac 1 {2^n}
+      | c = (because $2^{n - 1} \le n!$)
+}}
+{{eqn | r = 1 + \frac {1 - \paren {\frac 1 2}^n} {1 - \frac 1 2}
+}}
+{{eqn | r = 1 + 2 \paren {1 - \paren {\frac 1 2}^n}
+}}
+{{eqn | o = <
+      | r = 3
+}}
+{{end-eqn}}
+So $\sequence {x_n}$ is [[Definition:Bounded Above Real Sequence|bounded above]] by $3$.
+From the [[Monotone Convergence Theorem (Real Analysis)]], it follows that $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]].
+\end{proof}<|endoftext|>
+\section{Between two Real Numbers exists Rational Number}
+Tags: Real Analysis, Between two Real Numbers exists Rational Number
+
+\begin{theorem}
+Let $a, b \in \R$ be [[Definition:Real Number|real numbers]] such that $a < b$.
+Then:
+: $\exists r \in \Q: a < r < b$
+\end{theorem}
+
+\begin{proof}
+Suppose that $a \ge 0$.
+As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
+Thus:
+:$\dfrac 1 {b - a} \in \R$
+By the [[Archimedean Principle]]:
+:$\exists n \in \N: n > \dfrac 1 {b - a}$
+Let $M := \set {x \in \N: \dfrac x n > a}$.
+By the [[Well-Ordering Principle]], there exists $m \in \N$ such that $m$ is the [[Definition:Smallest Element|smallest element]] of $M$.
+That is:
+:$m > a n$
+and, by definition of [[Definition:Smallest Element|smallest element]]:
+:$m - 1 \le a n$
+As $n > \dfrac 1 {b - a}$, it follows from [[Ordering of Reciprocals]] that:
+:$\dfrac 1 n < b - a$
+Thus:
+{{begin-eqn}}
+{{eqn | l = m - 1
+      | o = \le
+      | r = a n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m
+      | o = \le
+      | r = a n + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac m n
+      | o = \le
+      | r = a + \frac 1 n
+      | c = 
+}}
+{{eqn | o = <
+      | r = a + \paren {b - a}
+      | c = 
+}}
+{{eqn | r = b
+      | c = 
+}}
+{{end-eqn}}
+Thus we have shown that $a < \dfrac m n < b$.
+That is:
+: $\exists r \in \Q: a < r < b$
+such that $r = \dfrac m n$.
+Now suppose $a < 0$.
+If $b > 0$ then $0 = r$ is a [[Definition:Rational Number|rational number]] such that $a < r < b$.
+Otherwise we have $a < b \le 0$.
+Then $0 \le -b < -a$ and there exists $r \in \Q$ such that:
+:$-b < r < -a$
+where $r$ can be found as above.
+That is:
+:$a < -r < b$
+All cases have been covered, and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+As $a < b$ it follows that $a \ne b$ and so $b - a \ne 0$.
+Thus:
+:$\dfrac 1 {b - a} \in \R$
+By the [[Archimedean Principle]]:
+:$\exists n \in \N: n > \dfrac 1 {b - a}$
+Let $M := \set {x \in \Z: x > a n}$.
+By [[Set of Integers Bounded Below has Smallest Element]], there exists $m \in \Z$ such that $m$ is the [[Definition:Smallest Element|smallest element]] of $M$.
+That is:
+:$m > a n$
+and, by definition of [[Definition:Smallest Element|smallest element]]:
+:$m - 1 \le a n$
+As $n > \dfrac 1 {b - a}$, it follows from [[Ordering of Reciprocals]] that:
+:$\dfrac 1 n < b - a$
+Thus:
+{{begin-eqn}}
+{{eqn | l = m - 1
+      | o = \le
+      | r = a n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m
+      | o = \le
+      | r = a n + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac m n
+      | o = \le
+      | r = a + \frac 1 n
+      | c = 
+}}
+{{eqn | o = <
+      | r = a + \paren {b - a}
+      | c = 
+}}
+{{eqn | r = b
+      | c = 
+}}
+{{end-eqn}}
+Thus we have shown that $a < \dfrac m n < b$.
+That is:
+:$\exists r \in \Q: a < r < b$
+such that $r = \dfrac m n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power over Factorial}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $x \in \R: x > 0$ be a positive [[Definition:Real Number|real number]].
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = \dfrac {x^n} {n!}$.
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence|converges]] to zero.
+\end{theorem}
+
+\begin{proof}
+We need to show that $x_n \to 0$ as $n \to \infty$.
+Let $N \in \N$ be the smallest [[Definition:Natural Numbers|natural number]] which satisfies $N > x$.
+(From the [[Archimedean Principle]], such an $N$ always exists.)
+First we show that:
+: $\forall n > N: \dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{n - N + 1}$
+Note that as $N > x$, $\dfrac x N < 1$.
+Also:
+:$m > n \implies \dfrac x m < \dfrac x n$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {x^n} {n!}
+      | r = \frac x 1 \frac x 2 \cdots \frac x {N - 1} \frac x N \frac x {N + 1} \cdots \frac x n
+      | c = 
+}}
+{{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \frac x N \frac x {N + 1} \cdots \frac x n
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{n - N + 1}
+      | c = as $\dfrac x {N + 1}, \dfrac x {N + 2}, \ldots, \dfrac x n < \dfrac x N$
+}}
+{{eqn | r = \frac {x^{N - 1} } {\paren {N - 1}!} \paren {\frac x N}^{1 - N} \paren {\frac x N}^n
+      | c = 
+}}
+{{end-eqn}}
+As $\dfrac x N < 1$, it follows from [[Sequence of Powers of Number less than One]] that $\paren {\dfrac x N}^n \to 0$ as $n \to \infty$.
+For a given $x$ and $N$, $\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N}$ is constant.
+Thus by the [[Multiple Rule for Real Sequences]]:
+:$\dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n \to 0$ as $n \to \infty$
+As (from above):
+:$\dfrac {x^n} {n!} \le \dfrac {x^{N - 1} } {\paren {N - 1}!} \paren {\dfrac x N}^{1 - N} \paren {\dfrac x N}^n$
+the result follows from the [[Squeeze Theorem for Real Sequences]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Reciprocals is Null Sequence}
+Tags: Limits of Sequences, Reciprocals, Sequence of Powers of Reciprocals is Null Sequence
+
+\begin{theorem}
+Let $r \in \Q_{>0}$ be a [[Definition:Strictly Positive Rational Number|strictly positive rational number]].
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as:
+: $x_n = \dfrac 1 {n^r}$
+Then $\sequence {x_n}$ is a [[Definition:Null Sequence (Analysis)|null sequence]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+We need to show that:
+:$\exists N \in \N: n > N \implies \size {\dfrac 1 {n^r} } < \epsilon$
+That is, that $n^r > 1 / \epsilon$.
+Let us choose $N = \ceiling {\paren {1 / \epsilon}^{1/r} }$.
+By [[Reciprocal of Strictly Positive Real Number is Strictly Positive]] and [[Power of Positive Real Number is Positive/Rational Number|power of positive real number is positive]], it follows that:
+:$\paren {\dfrac 1 \epsilon}^{1/r} \gt 0$
+Then by [[Positive Power Function on Non-negative Reals is Strictly Increasing]]:
+:$\forall n > N: n^r > N^r \ge 1 / \epsilon$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Number: Limit of Sequence implies Limit of Series}
+Tags: Euler's Number
+
+\begin{theorem}
+Let [[Definition:Euler's Number/Limit of Sequence|Euler's number $e$]] be defined as:
+:$\displaystyle e := \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$
+Then:
+:$\displaystyle e = \sum_{k \mathop \ge 0} \frac 1 {k!}$
+That is:
+:$\displaystyle e = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \frac 1 {4!} \cdots$
+\end{theorem}
+
+\begin{proof}
+We expand $\left({1 + \dfrac 1 n}\right)^n$ by the [[Binomial Theorem]], using that $\dfrac {n - k} n = 1 - \dfrac k n$:
+{{begin-eqn}}
+{{eqn | l = \left({1 + \frac 1 n}\right)^n
+      | r = 1 + n \left({\frac 1 n}\right) + \frac {n \left({n - 1}\right)} 2 \left({\frac 1 n}\right)^2 + \cdots + \left({\frac 1 n}\right)^n
+      | c = 
+}}
+{{eqn | r = \frac 1 {0!} + \frac 1 {1!} + \left({1 - \frac 1 n}\right) \frac 1 {2!} + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right)\frac 1 {3!} + \cdots + \left({1 - \frac 1 n}\right) \left({1 - \frac 2 n}\right) \cdots \left({1 - \frac {n - 1} n}\right) \frac 1 {n!}
+      | c = 
+}}
+{{end-eqn}}
+Take one of the terms in the above:
+:$x = \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {k - 1} n}\right) \dfrac 1 {k!}$
+From [[Sequence of Powers of Reciprocals is Null Sequence]], $\dfrac 1 n \to 0$ as $n \to \infty$.
+From the [[Combination Theorem for Sequences]]:
+:$\forall \lambda \in \R: \dfrac \lambda n \to 0$ as $n \to \infty$
+:$\forall \lambda \in \R: 1 - \dfrac \lambda n \to 1$ as $n \to \infty$
+:$x = \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {k - 1} n}\right) \dfrac 1 {k!} \to \dfrac 1 {k!}$ as $n \to \infty$
+Hence:
+:$\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+It will be shown that:
+:$\displaystyle \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = \sum_{k \mathop = 0}^\infty \frac 1 {k!}$
+Let $t_n := \left({1 + \dfrac 1 n}\right)^n$
+Then:
+:$t_n = \dfrac 1 {0!} + \dfrac 1 {1!} + \left({1 - \dfrac 1 n}\right) \dfrac 1 {2!} + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \dfrac 1 {3!} + \cdots + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {n-1} n}\right) \dfrac 1 {n!}$
+Now let:
+: $\displaystyle s_m := \sum_{k \mathop = 0}^m \frac 1 {k!}$
+We have that:
+:$\forall n: t_n \le s_n$
+Hence:
+:$\limsup \left({t_n}\right) \le e$
+Now, for all $m$, for $n \ge m$:
+:$t_n \ge \dfrac 1 {0!} + \dfrac 1 {1!} + \left({1 - \dfrac 1 n}\right) \dfrac 1 {2!} + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \dfrac 1 {3!} + \cdots + \left({1 - \dfrac 1 n}\right) \left({1 - \dfrac 2 n}\right) \cdots \left({1 - \dfrac {m-1} n}\right) \dfrac 1 {m!}$
+Hence, for all $m$, we have the right side as being a sequence in $n$, and then:
+:$\displaystyle \liminf \left({t_n}\right) \ge \sum_{k \mathop = 0}^m \frac 1 {m!}$
+Since this is true for all $m$:
+:$\liminf \left({t_n}\right) \ge e$
+So $\displaystyle \lim \left({t_n}\right)$ exists and is equal to $e$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Field Extensions is Multiplicative}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $E, K$ and $F$ be [[Definition:Field (Abstract Algebra)|fields]].
+Let $E / K$ and $K / F$ be [[Definition:Finite Field Extension|finite field extensions]].
+Then $E / F$ is a [[Definition:Finite Field Extension|finite field extension]], and:
+:$\index E F = \index E K \index K F$
+where $\index E F$ denotes the [[Definition:Degree of Field Extension|degree]] of $E / F$
+\end{theorem}
+
+\begin{proof}
+First, note that $E / F$ is a [[Definition:Field Extension|field extension]] as $F \subseteq K \subseteq E$.
+Suppose that $\index E K = m$ and $\index K F = n$.
+Let $\alpha = \set {a_1, \ldots, a_m}$ be a [[Definition:Basis (Linear Algebra)|basis]] of $E / K$, and $\beta = \set {b_1, \ldots, b_n}$ be a [[Definition:Basis of Vector Space|basis]] of $K / F$.
+We wish to prove that the [[Definition:Set|set]]:
+:$\gamma = \set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$
+is a [[Definition:Basis of Vector Space|basis]] of $E / F$.
+As $\alpha$ is a [[Definition:Basis of Vector Space|basis]] of $E / K$, we have, for all $c \in E$:
+:$\displaystyle c = \sum_{i \mathop = 1}^m c_i a_i$, for some $c_i \in K$.
+Define $\displaystyle b := \sum_{j \mathop = 1}^n b_i$ and $d_i := \dfrac {c_i} b$.
+Note $b \ne 0$ since $\beta$ is [[Definition:Linearly Independent Set|linearly independent]] over $F$, and $d_i \in K$ since $b, c_i \in K$.
+Now we have:
+:$\displaystyle c = \sum_{i \mathop = 1}^m \frac {c_i} b \cdot b \cdot a_i = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n d_i a_i b_j$
+Thus $\gamma$ is seen to be a [[Definition:Generator of Vector Space|spanning set]] of $E / F$.
+To show $\gamma$ is [[Definition:Linearly Independent Set|linearly independent]], suppose that for some $c_{i j} \in F$:
+:$\displaystyle \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j = 0$
+Then we have (as [[Definition:Field (Abstract Algebra)|fields]] are [[Definition:Commutative Ring|commutative rings]]):
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n c_{i j} a_i b_j
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall i: \sum_{j \mathop = 1}^n c_{i j} b_j
+      | r = 0
+      | c = by the [[Definition:Linearly Independent Set|linear independence]] of $\alpha$ over $K$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall i,j: c_{i j}
+      | r = 0
+      | c = by the [[Definition:Linearly Independent Set|linear independence]] of $\beta$ over $F$
+}}
+{{end-eqn}}
+Hence $\gamma$ is a [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Generator of Vector Space|spanning set]]; thus it is a [[Definition:Basis of Vector Space|basis]].
+Recalling the definition of $\gamma$ as $\set {a_i b_j: 1 \le i \le m, 1 \le j \le n}$, we have:
+:$\size {\gamma} = m n = \index E K \index K F$
+as was to be shown.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence in Metric Space has Unique Limit}
+Tags: Limits of Sequences, Metric Spaces, Convergent Sequence in Metric Space has Unique Limit
+
+\begin{theorem}
+Let $\left({X, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left \langle {x_n} \right \rangle$ be a [[Definition:Sequence|sequence]] in $\left({X, d}\right)$.
+Then $\left \langle {x_n} \right \rangle$ can have at most one [[Definition:Limit of Sequence (Metric Space)|limit]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\displaystyle \lim_{n \mathop \to \infty} x_n = l$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = m$.
+Let $\epsilon > 0$.
+Then, provided $n$ is [[Definition:Sufficiently Large|sufficiently large]]:
+{{begin-eqn}}
+{{eqn | l = \map d {l, m}
+      | o = \le
+      | r = \map d {l, x_n} + \map d {x_n, m}
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \epsilon + \epsilon
+      | c = {{Defof|Limit of Sequence (Metric Space)}}
+}}
+{{eqn | r = 2 \epsilon
+}}
+{{end-eqn}}
+So $0 \le \dfrac {\map d {l, m} } 2 < \epsilon$.
+This holds for ''any'' value of $\epsilon > 0$.
+Thus from [[Real Plus Epsilon]] it follows that $\dfrac {\map d {l, m} } 2 = 0$, that is, that $l = m$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that a [[Metric Space is Hausdorff]].
+The result then follows from [[Convergent Sequence in Hausdorff Space has Unique Limit]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower and Upper Bounds for Sequences}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $x_n \to l$ as $n \to \infty$.
+Then:
+:$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$
+:$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$
+\end{theorem}
+
+\begin{proof}
+$(1): \quad \forall n \in \N: x_n \ge a \implies l \ge a$:
+Let $\epsilon > 0$.
+Then:
+:$\exists N \in \N: n > N \implies \size {x_n - l} < \epsilon$
+So from [[Negative of Absolute Value]]:
+:$l - \epsilon < x_n < l + \epsilon$
+But $x_n \ge a$, so:
+:$a \le x_n < l + \epsilon$
+Thus, for ''any'' $\epsilon > 0$:
+:$a < l + \epsilon$
+From [[Real Plus Epsilon]] it follows that $a \le l$.
+{{qed|lemma}}
+$(2): \quad \forall n \in \N: x_n \le b \implies l \le b$:
+If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divergent Sequence may be Bounded}
+Tags: Limits of Sequences, Divergent Sequences, Divergent Sequence may be Bounded
+
+\begin{theorem}
+While every [[Convergent Sequence is Bounded]], it does not follow that every [[Definition:Bounded Sequence|bounded sequence]] is [[Definition:Convergent Sequence|convergent]].
+That is, there exist [[Definition:Bounded Sequence|bounded sequences]] which are [[Definition:Divergent Sequence|divergent]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] which forms the basis of [[Definition:Grandi's Series|Grandi's series]], defined as:
+:$x_n = \paren {-1}^n$
+It is clear that $\sequence {x_n}$ is [[Definition:Bounded Real Sequence|bounded]]: [[Definition:Bounded Above Real Sequence|above]] by $1$ and [[Definition:Bounded Below Real Sequence|below]] by $-1$.
+{{AimForCont}} $x_n \to l$ as $n \to \infty$.
+Let $\epsilon > 0$.
+Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$.
+But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$.
+It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$.
+From the [[Triangle Inequality for Real Numbers]], we have:
+{{begin-eqn}}
+{{eqn | l = 2
+      | r = \size {1 - \paren {-1} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {1 - l} + \size {l - \paren {-1} }
+      | c = 
+}}
+{{eqn | o = <
+      | r = 2 \epsilon
+      | c = 
+}}
+{{end-eqn}}
+This is a [[Definition:Contradiction|contradiction]] whenever $\epsilon < 1$.
+Thus $\sequence {x_n}$ has no [[Definition:Limit of Real Sequence|limit]] and, while definitely [[Definition:Bounded Real Sequence|bounded]], is unmistakably [[Definition:Divergent Sequence|divergent]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] which forms the basis of [[Definition:Grandi's Series|Grandi's series]],  defined as:
+:$x_n = \paren {-1}^n$
+It is clear that $\sequence {x_n}$ is [[Definition:Bounded Real Sequence|bounded]]: [[Definition:Bounded Above Real Sequence|above]] by $1$ and [[Definition:Bounded Below Real Sequence|below]] by $-1$.
+Note the following [[Definition:Subsequence|subsequences]] of $\sequence {x_n}$:
+:$(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the [[Definition:Integer Sequence|integer sequence]] defined as $n_r = 2 r$
+:$(2): \quad \sequence  {x_{n_s} }$ where $\sequence {n_s}$ is the [[Definition:Integer Sequence|integer sequence]] defined as $n_s = 2 s + 1$.
+We have that:
+:$\sequence {x_{n_r} }$ is the [[Definition:Real Sequence|sequence]] $1, 1, 1, 1, \ldots$
+:$\sequence {x_{n_s} }$ is the [[Definition:Real Sequence|sequence]] $-1, -1, -1, -1, \ldots$
+So $\sequence {x_n}$ has two [[Definition:Subsequence|subsequences]] with different [[Definition:Limit of Real Sequence|limits]].
+From [[Limit of Subsequence equals Limit of Real Sequence]], that means $\sequence {x_n}$ can not be [[Definition:Convergent Sequence|convergent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism}
+Tags: Division Rings, Monomorphisms, Ring Homomorphism from Division Ring is Monomorphism or Zero Homomorphism
+
+\begin{theorem}
+Let $\struct {R, +_R, \circ}$ and $\struct {S, +_S, *}$ be [[Definition:Ring (Abstract Algebra)|rings]] whose [[Definition:Ring Zero|zeros]] are $0_R$ and $0_S$ respectively.
+Let $\phi: R \to S$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+If $R$ is a [[Definition:Division Ring|division ring]], then either:
+: $(1): \quad \phi$ is a [[Definition:Ring Monomorphism|monomorphism]] (that is, $\phi$ is [[Definition:Injection|injective]])
+: $(2): \quad \phi$ is the [[Definition:Zero Homomorphism|zero homomorphism]] (that is, $\forall a \in R: \map \phi a = 0_S$).
+\end{theorem}
+
+\begin{proof}
+We have that:
+:The [[Kernel of Ring Epimorphism is Ideal|kernel of a homomorphism is an ideal of $R$]]
+:[[Ideals of Division Ring|the only ideals of a division ring are trivial]].
+So $\map \ker \phi = \set {0_R}$ or $R$.  
+If $\map \ker \phi = \set {0_R}$, then $\phi$ is [[Definition:Injection|injective]] by [[Kernel is Trivial iff Monomorphism]].
+If $\map \ker \phi = R$, $\phi$ is the [[Definition:Zero Homomorphism|zero homomorphism]] by definition.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Surjection by Restriction of Codomain]], we can restrict the [[Definition:Codomain of Mapping|codomain]] of $\phi$ and consider the mapping $\phi': R \to \operatorname {Im} \left({R}\right)$
+As $\phi'$ is now a [[Definition:Surjection|surjective]] [[Definition:Ring Homomorphism|homomorphism]], it is by definition an [[Definition:Ring Epimorphism|epimorphism]].
+Then we invoke the result that an [[Epimorphism from Division Ring to Ring|epimorphism from a division ring to a ring]] is either [[Definition:Null Ring|null]] or an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
+As an isomorphism is by definition [[Definition:Injection|injective]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Ring Epimorphism is Ideal}
+Tags: Ring Epimorphisms, Quotient Rings, Ideal Theory
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
+Then:
+:The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+:There is a unique [[Definition:Ring Isomorphism|ring isomorphism]] $g: R_1 / K \to R_2$ such that:
+::$g \circ q_K = \phi$
+:$\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]] {{iff}} $K = \set {0_{R_1} }$.
+\end{theorem}
+
+\begin{proof}
+=== Existence of Kernel ===
+By [[Kernel of Ring Homomorphism is Ideal]]:
+:The [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+{{qed|lemma}}
+=== Uniqueness of Quotient Mapping ===
+By [[Quotient Ring of Kernel of Ring Epimorphism]]:
+:there exists a unique [[Definition:Ring Isomorphism|ring isomorphism]] $g: R_1 / K \to R_2$ such that $g \circ q_K = \phi$
+:$\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]] {{iff}} $K = \set {0_{R_1} }$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unbounded Monotone Sequence Diverges to Infinity}
+Tags: Limits of Sequences, Unbounded Monotone Sequence Diverges to Infinity
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\sequence {x_n}$ be [[Definition:Monotone Real Sequence|monotone]], that is either [[Definition:Increasing Real Sequence|increasing]] or [[Definition:Decreasing Real Sequence|decreasing]].
+\end{theorem}<|endoftext|>
+\section{Reciprocal of Null Sequence}
+Tags: Limits of Sequences, Reciprocals, Reciprocal of Null Sequence
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\forall n \in \N: x_n > 0$.
+Then:
+:$x_n \to 0$ as $n \to \infty$ {{iff}} $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$
+\end{theorem}
+
+\begin{proof}
+Suppose $x_n \to 0$ as $n \to \infty$.
+Let $H > 0$.
+So $H^{-1} > 0$.
+Since $x_n \to 0$ as $n \to \infty$:
+:$\exists N: \forall n > N: \size {x_n} < H^{-1}$
+That is:
+:$\size {\dfrac 1 {x_n} } > H$.
+So:
+:$\exists N: \forall n > N: \size {\dfrac 1 {x_n} } > H$
+and thus:
+:$\sequence {\size {\dfrac 1 {x_n} } }$ [[Definition:Divergent Real Sequence to Positive Infinity|diverges to $+\infty$]].
+{{qed|lemma}}
+Suppose $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$.
+By reversing the argument above, we see that $x_n \to 0$ as $n \to \infty$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus of Limit}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $X$ be one of the [[Definition:Standard Number Field|standard number fields]] $\Q, \R, \C$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $X$]].
+Let $\sequence {x_n}$ be [[Definition:Convergent Sequence (Analysis)|convergent]] to the [[Definition:Limit of Sequence (Number Field)|limit]] $l$.
+That is, let $\displaystyle \lim_{n \mathop \to \infty} x_n = l$.
+Then
+:$\displaystyle \lim_{n \mathop \to \infty} \cmod {x_n} = \cmod l$
+where $\cmod {x_n}$ is the [[Definition:Modulus of Complex Number|modulus]] of $x_n$.
+\end{theorem}
+
+\begin{proof}
+By the [[Triangle Inequality]], we have:
+:$\cmod {\cmod {x_n} - \cmod l} \le \cmod {x_n - l}$
+Hence by the [[Squeeze Theorem]] and [[Convergent Sequence Minus Limit]], $\cmod {x_n} \to \cmod l$ as $n \to \infty$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Increasing Sequence of Natural Numbers}
+Tags: Sequences
+
+\begin{theorem}
+Let $\N_{>0}$ be the set of [[Definition:Natural Numbers|natural numbers]] without zero:
+: $\N_{>0} = \left\{{1, 2, 3, \ldots}\right\}$
+Let $\left \langle {n_r} \right \rangle$ be [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] in $\N_{>0}$.
+Then:
+: $\forall r \in \N_{>0}: n_r \ge r$
+\end{theorem}
+
+\begin{proof}
+This is to be proved by [[Principle of Mathematical Induction|induction]] on $r$.
+For all $r \in \N_{>0}$, let $P \left({r}\right)$ be the [[Definition:Proposition|proposition]] $n_r \ge r$.
+=== Basis for the Induction ===
+When $r = 1$, it follows that $n_1 \ge 1$ as $\N_{>0}$ is [[Definition:Bounded Below Set|bounded below]] by $1$.
+Thus $P(1)$ is true.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] is that $n_k \ge k$.
+Then we need to show that $n_{k+1} \ge k+1$.
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Suppose that $n_k \ge k$.
+Then as $\left \langle {n_r} \right \rangle$ is [[Definition:Strictly Increasing Sequence|strictly increasing]]:
+:$n_{k+1} > n_k \ge k$
+From [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]], we have:
+:$n_{k+1} > k \implies n_{k+1} \ge k+1$
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore $\forall r \in \N_{>0}: n_r \ge r$.
+{{qed}}
+{{improve|Better link instead of [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]}}
+[[Category:Sequences]]
+tl74r1b0764lp29mrj86082k7u4u23s
+\end{proof}<|endoftext|>
+\section{Limit of Subsequence equals Limit of Sequence}
+Tags: Convergence, Limits of Sequences
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $T$]].
+Let $l \in S$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+Let $\sequence {x_{n_r} }$ be a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Then:
+:$\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$
+That is, the [[Definition:Limit Point of Sequence|limit]] of a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in a [[Definition:Topological Space|topological space]] equals the [[Definition:Limit Point of Sequence|limit]] of any [[Definition:Subsequence|subsequence]] of it.
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau$ be an [[Definition:Open Set (Topology)|open set]] such that $l \in U$.
+By [[Definition:Convergent Sequence (Topology)|definition]] of convergence, we have:
+:$\exists N \in \N: \forall n > N: x_n \in U$.
+When $r > N$, we have $n_r > n_N > N$ by [[Strictly Increasing Sequence of Natural Numbers]].
+It follows that:
+:$\exists N \in \N: \forall r > N: x_{n_r} \in U$.
+Therefore, as $U$ was arbitrary, we have established $\displaystyle \lim_{r \mathop \to \infty} x_{n_r} = l$, by definition of [[Definition:Convergent Sequence (Topology)|convergence]].
+{{qed}}
+[[Category:Convergence]]
+[[Category:Limits of Sequences]]
+1c9kwdnmw5pp6isssb8tcvj4v1sitqm
+\end{proof}<|endoftext|>
+\section{Root of Number Greater than One}
+Tags: Analysis, Inequalities
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $n \in \N^*$ be a [[Definition:Natural Numbers|natural number]] such that $n > 0$.
+Then $x \ge 1 \implies x^{1/n} \ge 1$ where $x^{1/n}$ is the [[Definition:Root (Analysis)|$n$th root]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $y = x^{1/n}$.
+From the definition of the [[Definition:Root (Analysis)|$n$th root]] of $x$, it follows that $x = y^n$.
+We will show by [[Principle of Mathematical Induction|induction]] that $\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$.
+For all $n \in \N^*$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+: $y^n \ge 1 \implies y \ge 1$
+==== Basis for the Induction ====
+By [[Definition:Power (Algebra)|definition]], $y^1 = y$.
+Thus $P(1)$ is true, as this just says $y \ge 1 \implies y \ge 1$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+==== Induction Hypothesis ====
+* Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] is that:
+: $y^k \ge 1 \implies y \ge 1$
+Now we need to show that:
+: $y^{k+1} \ge 1 \implies y \ge 1$.
+==== Induction Step ====
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+By [[Definition:Power (Algebra)|definition]], $y^{k+1} = y \cdot y^k$.
+Suppose $y^k \ge 1$. From the [[#Induction Hypothesis|induction hypothesis]] it follows that $y > 1$.
+As $y \ge 1$ it follows that $y > 0$.
+Let $y^{k+1} = y \cdot y^k \ge 1$.
+Then $y \cdot y^k$
+By [[Real Number Ordering is Compatible with Multiplication]], $y \cdot y^k \ge y \times 1$ and hence the result.
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore $\forall n \in \N^*: y^n \ge 1 \implies y \ge 1$.
+As $y^n = x$ and $y = x^{1/n}$, the result follows.
+{{qed}}
+[[Category:Analysis]]
+[[Category:Inequalities]]
+jz0ugo79w1zoig24q41ykvss46x96vx
+\end{proof}<|endoftext|>
+\section{Limit of Root of Positive Real Number}
+Tags: Limits of Sequences, Limit of Root of Positive Real Number
+
+\begin{theorem}
+Let $x \in \R: x > 0$ be a [[Definition:Real Number|real number]].
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as:
+:$x_n = x^{1 / n}$
+Then $x_n \to 1$ as $n \to \infty$.
+\end{theorem}
+
+\begin{proof}
+Let us define $a_1 = a_2 = \cdots = a_{n-1} = 1$ and $a_n = x$.
+Let $G_n$ be the [[Definition:Geometric Mean|geometric mean]] of $a_1, \ldots, a_n$.
+Let $A_n$ be the [[Definition:Arithmetic Mean|arithmetic mean]] of $a_1, \ldots, a_n$.
+From their definitions:
+:$G_n = x^{1/n}$
+and:
+:$A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$
+From [[Arithmetic Mean Never Less than Geometric Mean]]:
+:$x^{1/n} \le 1 + \dfrac{x - 1} n$
+That is:
+:$x^{1/n} - 1 \le \dfrac{x - 1} n$
+There are two cases to consider: $x \ge 1$ and $0 < x < 1$.
+Let $x \ge 1$.
+From [[Root of Number Greater than One]], it follows that:
+:$x^{1/n} \ge 1$
+Thus:
+:$0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$
+But from [[Sequence of Powers of Reciprocals is Null Sequence]]:
+:$\dfrac 1 n \to 0$ as $n \to \infty$
+From the [[Combination Theorem for Sequences]]:
+:$\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$
+Thus by the [[Squeeze Theorem]]:
+:$x^{1/n} - 1 \to 0$ as $n \to \infty$
+Hence, again from the [[Combination Theorem for Sequences]]:
+:$x^{1/n} \to 1$ as $n \to \infty$
+Now let $0 < x < 1$.
+Then $x = \dfrac 1 y$ where $y > 1$.
+But from the above:
+:$y^{1/n} \to 1$ as $n \to \infty$
+Hence by the [[Combination Theorem for Sequences]]:
+:$x^{1/n} = \dfrac 1 {y^{1/n} } \to \dfrac 1 1 = 1$ as $n \to \infty$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for [[Limit of Root of Positive Real Number/Proof 1|proof 1]].
+From [[Root of Number Greater than One]]:
+:$x^{1/n} \ge 1$
+Hence $\sequence {x^{1/n} }$ is [[Definition:Bounded Below Real Sequence|bounded below]] by $1$.
+Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:
+{{begin-eqn}}
+{{eqn | l = \frac {x^{1/n} } {x^{\frac 1 {n + 1} } }
+      | r = x^{\frac 1 n - \frac 1 {n + 1} }
+      | c = 
+}}
+{{eqn | r = x^{\frac {n + 1 - n} {n \paren {n + 1} } }
+      | c = 
+}}
+{{eqn | r = x^{\frac 1 {n \paren {n + 1} } }
+      | c = 
+}}
+{{eqn | o = >
+      | r = 1
+      | c = [[Root of Number Greater than One]]
+}}
+{{end-eqn}}
+So:
+:$x^{1/n} > x^{\frac 1 {n + 1} }$
+and so $\sequence {x^{1 / n} }$ is [[Definition:Strictly Decreasing Real Sequence|strictly decreasing]].
+Hence from the [[Monotone Convergence Theorem (Real Analysis)]], it follows that $\sequence {x^{1 / n} }$ [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]] $l$ and that $l \ge 1$.
+Now, since we know that $\sequence {x^{1 / n} }$ is [[Definition:Convergent Real Sequence|convergent]], we can apply [[Limit of Subsequence equals Limit of Real Sequence]].
+That is, any [[Definition:Subsequence|subsequence]] of $\sequence {x^{1 / n} }$ must also [[Definition:Convergent Real Sequence|converge]] to $l$.
+So we take the [[Definition:Subsequence|subsequence]]:
+:$\sequence {x^{1 / {2 n} } }$
+From what has just been shown:
+:$x^{1 / {2 n} } \to l$ as $n \to \infty$
+Using the [[Combination Theorem for Sequences]], we have:
+:$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$
+But a [[Convergent Real Sequence has Unique Limit]], so $l^2 = l$ and so $l = 0$ or $l = 1$.
+But $l \ge 1$ and so $l = 1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hero's Method}
+Tags: Real Analysis, Square Roots, Hero's Method
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 0$.
+Let $x_1 \in \R$ be a [[Definition:Real Number|real number]] such that $x_1 > 0$.
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined recursively by:
+:$\forall n \in \N_{>0}: x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
+Then $x_n \to \sqrt a$ as $n \to \infty$.
+\end{theorem}
+
+\begin{proof}
+Consider $x_n - x_{n + 1}$.
+{{begin-eqn}}
+{{eqn | l = x_n - x_{n + 1}
+      | r = x_n - \frac {x_n + \dfrac a {x_n} } 2
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \paren {x_n^2 - a}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = for $n \ge 2$
+      | cc= [[Hero's Method/Lemma 2|Lemma 2]]
+}}
+{{end-eqn}}
+So, providing we ignore the first term (about which we can state nothing), the sequence $\sequence {x_n}$ is [[Definition:Decreasing Real Sequence|decreasing]] and [[Definition:Bounded Below Real Sequence|bounded below]] by $\sqrt a$.
+Thus by the [[Monotone Convergence Theorem (Real Analysis)]], $x_n \to l$ as $n \to \infty$, where $l \ge \sqrt a$.
+Now we want to find exactly what that value of $l$ actually is.
+By [[Limit of Subsequence equals Limit of Real Sequence]] we also have $x_{n + 1} \to l$ as $n \to \infty$.
+But $x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$.
+Because $l \ge \sqrt a$ it follows that $l \ne 0$.
+So by the [[Combination Theorem for Sequences]]:
+:$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2 \to \dfrac {l + \dfrac a l} 2$ as $n \to \infty$
+Since a [[Convergent Real Sequence has Unique Limit]], that means:
+:$l = \dfrac {l + \dfrac a l} 2$
+and so (after some straightforward algebra):
+:$l^2 = a$
+Thus:
+:$l = \pm \sqrt a$
+and as $l \ge +\sqrt a$ it follows that:
+:$l = +\sqrt a$
+Hence the result.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $a > 0$.
+We make no statement about $x_1$.
+We specify that:
+:$x_{n + 1} = \dfrac {x_n + \dfrac a {x_n} } 2$
+Now:
+{{begin-eqn}}
+{{eqn | l = x_{n + 1} - \sqrt a
+      | r = \frac {x_n + \dfrac a {x_n} } 2 - \sqrt a
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \paren {x_n^2 - 2 x_n \sqrt a + a}
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \paren {x_n - \sqrt a}^2
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \paren {\dfrac {\paren {x_{n - 1} - \sqrt a}^2} {2 x_{n - 1} } }^2
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \paren {x_{n - 1} - \sqrt a}^4
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \frac 1 {\paren {2 x_{n - 2} }^4} \paren {x_{n - 2} - \sqrt a}^8
+      | c = 
+}}
+{{eqn | r = \frac 1 {2 x_n} \frac 1 {\paren {2 x_{n - 1} }^2} \cdots \frac 1 {\paren {2 x_1}^{2 n - 1} } \paren {x_1 - \sqrt a}^{2 n}
+      | c = 
+}}
+{{end-eqn}}
+If we now assume that $x_1 \ge \sqrt a$, then it follows from [[Hero's Method/Lemma 2|Lemma 2]] that $x_n \ge \sqrt a$.
+So:
+{{begin-eqn}}
+{{eqn | l = \size {x_{n + 1} - \sqrt a}
+      | o = \le
+      | r = \paren {\frac 1 {2 \sqrt a} }^{1 + 2 + 2^2 + \cdots + 2^{n - 1} } \paren {x_1 - \sqrt a}^{2 n}
+      | c = 
+}}
+{{eqn | r = \paren {\frac 1 {2 \sqrt a} }^{\dfrac {2^n - 1} {2 - 1} } \paren {x_1 - \sqrt a}^{2 n}
+      | c = 
+}}
+{{eqn | r = 2 \sqrt a \paren {\frac {x_1 - \sqrt a} {2 \sqrt a} }^{2 n}
+      | c = 
+}}
+{{end-eqn}}
+If $\size y < 1$, then $y^n \to 0$ as $n \to \infty$ from [[Sequence of Powers of Number less than One]].
+So, by [[Limit of Subsequence equals Limit of Real Sequence]]:
+:$y^{2^n} \to 0$ as $n \to \infty$
+Thus we see that:
+:$x_n \to \sqrt a$ as $n \to \infty$
+provided that:
+:$\dfrac {x_1 - \sqrt a} {2 \sqrt a} < 1$
+that is, that:
+:$\sqrt a \le x_1 < 3 \sqrt a$
+We assumed above that $x_1 \ge \sqrt a$.
+Now we have shown that $x_n \to \sqrt a$ as $n \to \infty$ provided that $\sqrt a \le x_1 < 3 \sqrt a$.
+However, we have already shown that $x_n \to \sqrt a$ as long as $x_1 \ge 0$.
+The advantage to this analysis is that this gives us an opportunity to determine how close $x_n$ gets to $\sqrt a$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Integer to Reciprocal Power}
+Tags: Limits of Sequences, Reciprocals, Limit of Integer to Reciprocal Power
+
+\begin{theorem}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|real sequence]] defined as $x_n = n^{1/n}$, using [[Definition:Real Exponential Function|exponentiation]].
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence|converges]] with a [[Definition:Limit of Sequence (Number Field)|limit]] of $1$.
+\end{theorem}
+
+\begin{proof}
+From [[Number to Reciprocal Power is Decreasing]] we have that the [[Definition:Real Sequence|real sequence]] $\sequence {n^{1/n} }$ is [[Definition:Decreasing Real Sequence|decreasing]] for $n \ge 3$.
+Now, as $n^{1 / n} > 0$ for all [[Definition:Positive Integer|positive]] $n$, it follows that $\sequence {n^{1 / n} }$ is [[Definition:Bounded Below Real Sequence|bounded below]] (by $0$, for a start).
+Thus the [[Definition:Subsequence|subsequence]] of $\sequence {n^{1 / n} }$ consisting of all the [[Definition:Term of Sequence|terms]] of $\sequence {n^{1 / n} }$ where $n \ge 3$ is [[Definition:Convergent Real Sequence|convergent]] by the [[Monotone Convergence Theorem (Real Analysis)]].
+Now we need to demonstrate that this [[Definition:Limit of Real Sequence|limit]] is in fact $1$.
+Let $n^{1 / n} \to l$ as $n \to \infty$.
+Having established this, we can investigate the [[Definition:Subsequence|subsequence]] $\sequence {\paren {2 n}^{1 / {2 n} } }$.
+By [[Limit of Subsequence equals Limit of Real Sequence]], this will converge to $l$ also.
+From [[Limit of Root of Positive Real Number]], we have that $2^{1 / {2 n} } \to 1$ as $n \to \infty$.
+So $n^{1 / {2 n} } \to l$ as $n \to \infty$ by the [[Combination Theorem for Sequences]].
+Thus:
+:$n^{1 / n} = n^{1 / {2 n} } \cdot n^{1 / {2 n} } \to l \cdot l = l^2$
+as $n \to \infty$.
+So $l^2 = l$, and as $l \ge 1$ the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have the definition of the [[Definition:Power to Real Number|power to a real number]]:
+:$\displaystyle n^{1/n} = \map \exp {\frac 1 n \ln n}$
+From [[Powers Drown Logarithms]], we have that:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$
+Hence:
+:$\displaystyle \lim_{n \mathop \to \infty} n^{1/n} = \exp 0 = 1$
+and the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference Between Adjacent Square Roots Converges}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = \sqrt {n + 1} - \sqrt n$.
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence|converges]] to a zero [[Definition:Limit of a Sequence (Number Field)|limit]].
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | r = \sqrt {n + 1} - \sqrt n
+      | c = 
+}}
+{{eqn | r = \frac {\paren {\sqrt {n + 1} - \sqrt n} \paren {\sqrt {n + 1} + \sqrt n} } {\sqrt {n + 1} + \sqrt n}
+      | c = multiplying top and bottom by $\sqrt {n + 1} - \sqrt n$
+}}
+{{eqn | r = \frac {n + 1 - n} {\sqrt {n + 1} + \sqrt n}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \frac 1 {\sqrt {n + 1} + \sqrt n}
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 {\sqrt n}
+      | c = as $\sqrt {n + 1} + \sqrt n > \sqrt n$
+}}
+{{end-eqn}}
+But from [[Sequence of Powers of Reciprocals is Null Sequence]], $\dfrac 1 {\sqrt n} \to 0$ as $n \to \infty$.
+The result follows by the [[Squeeze Theorem]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Peak Point Lemma}
+Tags: Limits of Sequences, Named Theorems, Peak Point Lemma
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]] which is [[Definition:Infinite Sequence|infinite]].
+Then $\sequence {x_n}$ has an [[Definition:Infinite Sequence|infinite]] [[Definition:Subsequence|subsequence]] which is [[Definition:Monotone Sequence|monotone]].
+\end{theorem}
+
+\begin{proof}
+There are $2$ cases to consider.
+First, suppose that every [[Definition:Set|set]] $\set {x_n: n > N}$ has a [[Definition:Greatest Element|maximum]].
+If this is the case, we can find a [[Definition:Real Sequence|sequence]] $n_r \in \N$ such that:
+:$\displaystyle x_{n_1} = \max \set {x_n: n > 1}$
+:$\displaystyle x_{n_2} = \max \set {x_n: n > n_1}$
+:$\displaystyle x_{n_3} = \max \set {x_n: n > n_2}$
+and so on.
+From the method of construction, $n_1 < n_2 < n_3 < \cdots$, so at each stage we are taking the [[Definition:Greatest Element|maximum]] of a [[Definition:Subset|subset]] of the previous [[Definition:Set|set]].
+At each stage, the previous [[Definition:Greatest Element|maximum]] has already been taken as the previous [[Definition:Term of Sequence|term]] in the [[Definition:Real Sequence|sequence]].
+Thus, $\sequence {x_{n_r} }$ is a [[Definition:Decreasing Real Sequence|decreasing]] [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+{{qed|lemma}}
+Second, suppose it is ''not'' true that ''every'' set $\set {x_n: n > N}$ has a [[Definition:Greatest Element|maximum]].
+Then there will exist some $N_1$ such that $\set {x_n: n > N_1}$ has no [[Definition:Greatest Element|maximum]].
+So it follows that, given some $x_m$ with $m > N_1$, we can find an $x_n$ following that $x_m$ such that $x_n > x_m$.
+(Otherwise, the biggest of $x_{N_1 + 1}, \ldots, x_m$ would be a [[Definition:Greatest Element|maximum]] for $\set {x_n: n > N_1}$.)
+So, we define $x_{n_1} = x_{N_1 + 1}$.
+Then $x_{n_2}$ can be the first [[Definition:Term of Sequence|term]] after $x_{n_1}$ such that $x_{n_2} > x_{n_1}$.
+Then $x_{n_3}$ can be the first [[Definition:Term of Sequence|term]] after $x_{n_2}$ such that $x_{n_3} > x_{n_2}$.
+And so on.
+Thus we get an [[Definition:Increasing Real Sequence|increasing]] [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+{{qed|lemma}}
+There exist only these two possibilities.
+From each one we get a [[Definition:Subsequence|subsequence]] that is [[Definition:Monotone Real Sequence|monotone]]:
+:one is [[Definition:Decreasing Real Sequence|decreasing]]
+:one is [[Definition:Increasing Real Sequence|increasing]].
+We can of course choose instead to investigate whether every set $\set {x_n: n > N}$ has a [[Definition:Smallest Element|minimum]].
+The same conclusion will be reached.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+A coastal town in Spain has an infinite row of hotels along a road leading down to the beach.
+The tourist bureau which rates these hotels has a special designation for any hotel having a view of the sea.
+Any hotel which is at least as tall as the rest of the hotels on the road to the sea receives this view-rating.
+Starting with the hotel farthest from the sea, let $x_1, x_2, \ldots$ denote the heights of the hotels as we travel along the road to the beach.
+One possibility is that an infinite number of hotels are view-rated.
+In this case the heights of these hotels form a [[Definition:Decreasing Sequence|decreasing (i.e. nonincreasing)]] [[Definition:Subsequence|subsequence]] of the original sequence of heights.
+The other possibility is that only a finite number of hotels are view-rated.
+In this case we can obtain an [[Definition:Increasing Sequence|increasing]] [[Definition:Subsequence|subsequence]] as follows.
+Walk along the road to the beach past all the view-rated hotels.
+Let $n_1$ be the index of the next hotel.
+Since it does not have a view of the sea, there is a taller hotel nearer the shore which blocks the view.
+Let $n_2$ be the index of that taller hotel.
+Since it is not view-rated, it does not have a view of the sea.
+Thus there must be an even taller hotel nearer the shore which blocks the view.
+In this way we generate a list of indices $n_1 < n_2 < n_3 \cdots$ of taller and taller hotels.
+This gives an [[Definition:Increasing Sequence|increasing]] [[Definition:Subsequence|subsequence]] $x_{n_1}, x_{n_2}, x_{n_3}, \ldots$ of the original sequence.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bolzano-Weierstrass Theorem}
+Tags: Limits of Sequences, Real Analysis, Bolzano-Weierstrass Theorem
+
+\begin{theorem}
+Every [[Definition:Bounded Real Sequence|bounded sequence]] of [[Definition:Real Number|real numbers]] has a [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequence]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ be a [[Definition:Bounded Real Sequence|bounded sequence in $\R$]].
+By the [[Peak Point Lemma]], $\sequence {x_n}$ has a [[Definition:Monotone Real Sequence|monotone]] [[Definition:Subsequence|subsequence]] $\sequence {x_{n_r} }$.
+Since $\sequence {x_n}$ is [[Definition:Bounded Sequence|bounded]], so is $\sequence {x_{n_r} }$.
+Hence, by the [[Monotone Convergence Theorem (Real Analysis)]], the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\left \langle {x_n} \right \rangle_{n \in \N}$ be a [[Definition:Bounded Real Sequence|bounded]] [[Definition:Real Sequence|sequence in $\R$]].
+By definition there are real numbers $c, C \in \R$ such that $c < x_n < C$.
+Then at least one of the sets:
+:$\left\{{x_n : c < x_n < \dfrac{c + C} 2 }\right\}, \left\{{x_n : \dfrac{c + C} 2 < x_n < C }\right\}, \left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$
+contains infinitely many elements.
+If the set $\left\{{x_n : x_n = \dfrac{c + C} 2 }\right\}$ is infinite there's nothing to prove.
+If this is not the case, choose the first element from the infinite set, say $x_{k_1}$.
+Repeat this process for $\left \langle {x_n} \right \rangle_{n > k_1}$.
+As a result we obtain subsequence $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$.
+By construction $\left \langle {x_{k_n}} \right \rangle_{n \in \N}$ is a Cauchy sequence and therefore converges.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Maximum and Minimum of Bounded Sequence}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Bounded Real Sequence|bounded sequence in $\R$]] (which may or may not be [[Definition:Convergent Real Sequence|convergent]]).
+Let $L$ be the [[Definition:Set|set]] of all [[Definition:Real Number|real numbers]] which are the [[Definition:Limit of Real Sequence|limit]] of some [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Then $L$ has both a [[Definition:Greatest Element|maximum]] and a [[Definition:Smallest Element|minimum]].
+\end{theorem}
+
+\begin{proof}
+From the [[Bolzano-Weierstrass Theorem]]:
+:$L \ne \O$
+From [[Lower and Upper Bounds for Sequences]], $L$ is a [[Definition:Bounded Subset of Real Numbers|bounded subset of $\R$]].
+Thus $L$ does have a [[Definition:Supremum of Set|supremum]] and [[Definition:Infimum of Set|infimum]] in $\R$.
+The object of this proof is to confirm that:
+:$\overline l := \map \sup L \in L$
+and:
+:$\underline l := \map \inf L \in L$
+that is, that these points do actually belong to $L$.
+First we show that $\overline l \in L$.
+To do this, we show that:
+:$\exists \sequence {x_{n_r} }: x_{n_r} \to \overline l$ as $n \to \infty$
+where $\sequence {x_{n_r} }$ is a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Let $\epsilon > 0$.
+Then $\dfrac \epsilon 2 > 0$.
+Since $\overline l = \map \sup L$, and therefore by [[Definition:Supremum of Set|definition]] the ''smallest'' [[Definition:Upper Bound of Set|upper bound]] of $L$, $\overline l - \dfrac \epsilon 2$ is ''not'' an [[Definition:Upper Bound of Set|upper bound]] of $L$.
+Hence:
+:$\exists l \in L: \overline l \ge l > \overline l - \dfrac \epsilon 2$
+Therefore:
+:$\sequence {l - \overline l} < \dfrac \epsilon 2$
+Now because $l \in L$, we can find $\sequence {x_{m_r} }$, a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$, such that $x_{m_r} \to l$ as $n \to \infty$.
+So:
+:$\exists R: \forall r > R: \size {x_{m_r} - \overline l} < \dfrac \epsilon 2$
+So, for any $r > R$:
+{{begin-eqn}}
+{{eqn | l = \size {x_{m_r} - \overline l}
+      | r = \size {x_{m_r} - l + l - \overline l}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {x_{m_r} - l} + \size {l - \overline l}
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \frac \epsilon 2 + \frac \epsilon 2 = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+Thus we have shown that:
+:$\forall r > R: \size {x_{m_r} - \overline l} < \epsilon$
+So far, what has been shown is that, given any $\epsilon > 0$, there exists an [[Definition:Infinite Set|infinite set]] of [[Definition:Term of Sequence|terms]] of $\sequence {x_n}$ which satisfy $\size {x_n - \overline l} < \epsilon$.
+Next it is shown how to construct a [[Definition:Subsequence|subsequence]]:
+:$\sequence {x_{n_r} }$ such that $x_{n_r} \to \overline l$
+as $n \to \infty$.
+Take $\epsilon = 1$ in the above.
+Then:
+:$\exists n_1: \size {x_{n_1} - \overline l} < 1$
+Now take $\epsilon = \dfrac 1 2$ in the above.
+Then:
+:$\exists n_2 > n_1: \size {x_{n_2} - \overline l} < \dfrac 1 2$
+In this way a [[Definition:Subsequence|subsequence]] is contructed:
+:$\sequence {x_{n_r} }$ satisfying $\size {x_{n_r} - \overline l} < \dfrac 1 r$
+But $\dfrac 1 r \to 0$ as $r \to \infty$ from the [[Sequence of Reciprocals is Null Sequence]].
+From the [[Squeeze Theorem for Real Sequences]], it follows that:
+:$\size {x_{n_r} - \overline l} \to 0$ as $r \to \infty$
+Thus $\overline l \in L$ as required.
+{{qed|lemma}}
+A similar argument shows that the [[Definition:Infimum of Set|infimum]] $\underline l$ of $L$ is also in $L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interval Defined by Absolute Value}
+Tags: Real Intervals, Absolute Value Function
+
+\begin{theorem}
+Let $\xi, \delta \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\delta > 0$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Difference of Two Powers}
+Tags: Algebra, Polynomial Theory, Difference of Two Powers
+
+\begin{theorem}
+Let $\mathbb F$ denote one of the [[Definition:Standard Number System|standard number systems]], that is $\Z$, $\Q$, $\R$ and $\C$.
+Let $n \in \N$ such that $n \ge 2$.
+Then for all $a, b \in \mathbb F$:
+{{begin-eqn}}
+{{eqn | l = a^n - b^n
+      | r = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Geometric Sequence]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^{n - 1} x^j
+      | r = \frac {x^n - 1} {x - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\dfrac a b}^n - 1
+      | r = \paren {\dfrac a b - 1} \sum_{j \mathop = 0}^{n - 1} \paren {\dfrac a b}^j
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\dfrac {a^n - b^n} {b^n} }
+      | r = \paren {\dfrac {a - b} b} \paren {\paren {\dfrac a b}^{n - 1} + \paren {\dfrac a b}^{n - 2} + \dotsb + \paren {\dfrac a b}^1 + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^n - b^n
+      | r = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + \dotsb + a b^{n - 2} + b^{n - 1} }
+      | c = multiplying both sides by $b^n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The proof will proceed by the [[Principle of Complete Finite Induction]] on $\Z_{>0}$.
+Let $S$ be the [[Definition:Set|set]] defined as:
+:$\displaystyle S := \set {n \in \Z_{>0}: a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j}$
+That is, $S$ is to be the [[Definition:Set|set]] of all $n$ such that:
+:$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+=== Basis for the Induction ===
+We have that:
+:$\displaystyle a^1 - b^1 = \paren {a - b} \sum_{j \mathop = 0}^0 a^{1 - 0 - 1} b^j$
+So $1 \in S$.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It is to be shown that if $r \in S$ for all $r$ such that $1 \le r \le k$, then it follows that $k + 1 \in S$.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\forall r \in \Z_{>0}: 1 \le r \le k: \displaystyle a^r - b^r = \paren {a - b} \sum_{j \mathop = 0}^{r - 1} a^{r - j - 1} b^j$
+It is to be demonstrated that it follows that:
+:$\displaystyle a^{k + 1} - b^{k + 1} = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = a^{k + 1} - b^{k + 1}
+      | r = \paren {a + b} \paren {a^k - b^k} - a b \paren {a^{k - 1} - b^{k - 1} }
+      | c = 
+}}
+{{eqn | r = \paren {a + b} \paren {a - b} \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^j - a b \paren {a - b} \sum_{j \mathop = 0}^{k - 2} a^{k - j - 2} b^j
+      | c = [[Difference of Two Powers/Proof 4#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {a - b} \paren {\paren {a + b} \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^j - a b \sum_{j \mathop = 0}^{k - 2} a^{k - j - 2} b^j}
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{k - 1} a^{k - j} b^j + \sum_{j \mathop = 0}^{k - 1} a^{k - j - 1} b^{j + 1} - \sum_{j \mathop = 0}^{k - 2} a^{k - j - 1} b^{j + 1} }
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{k - 1} a^{k - j} b^j + a^0 b^k}
+}}
+{{eqn | r = \paren {a - b} \sum_{j \mathop = 0}^k a^{k - j} b^j
+}}
+{{end-eqn}}
+So $\forall r \in S: 0 \le r \le k: r \in S \implies k + 1 \in S$ and the result follows by the [[Principle of Complete Finite Induction]]:
+:$\forall n \in \Z_{>0}: \displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Factor Theorem}
+Tags: Polynomial Theory, Named Theorems
+
+\begin{theorem}
+Let $P \paren x$ be a [[Definition:Polynomial over Field|polynomial]] in $x$ over a [[Definition:Field (Abstract Algebra)|field]] $K$ of [[Definition:Degree of Polynomial|degree]] $n$.
+Then:
+:$\xi \in K: P \paren \xi = 0 \iff \map P x = \paren {x - \xi} \map Q x$
+where $Q$ is a [[Definition:Polynomial over Field|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n - 1$.
+Hence, if $\xi_1, \xi_2, \ldots, \xi_n \in K$ such that all are different, and $\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$, then:
+:$\displaystyle \map P x = k \prod_{j \mathop = 1}^n \paren {x - \xi_j}$
+where $k \in K$.
+\end{theorem}
+
+\begin{proof}
+Let $P = \paren {x - \xi} Q$.
+Then:
+:$\map P \xi = \map Q \xi \cdot 0 = 0$
+Conversely, let $\map P \xi = 0$.
+By the [[Division Theorem for Polynomial Forms over Field]], there exist polynomials $Q$ and $R$ such that:
+:$P = \map Q {x - \xi} + R$
+and:
+:$\map \deg R < \map \deg {x - \xi} = 1$
+Evaluating at $\xi$ we have:
+:$0 = \map P \xi = \map R \xi$
+But:
+:$\deg R = 0$
+so:
+:$R \in K$
+In particular:
+:$R = 0$
+Thus:
+:$P = \map Q {x - \xi}$
+as required.
+The fact that $\map \deg Q = n - 1$ follows from:
+:[[Ring of Polynomial Forms is Integral Domain]]
+and:
+:[[Degree of Product of Polynomials over Integral Domain]].
+We can then apply this result to:
+:$\map P {\xi_1} = \map P {\xi_2} = \dotsb = \map P {\xi_n} = 0$
+We can progressively work through:
+:$\map P x = \paren {x - \xi_1} \map {Q_{n - 1} } x$ 
+where $\map {Q_{n - 1} } x$ is a polynomial of order $n - 1$.
+Then, substituting $\xi_2$ for $x$:
+:$0 = \map P {\xi_2} = \paren {\xi_2 - \xi_1} \map {Q_{n - 1} } x$
+Since $\xi_2 \ne \xi_1$:
+:$\map {Q_{n - 1} } {\xi_2} = 0$
+and we can apply the above result again:
+:$\map {Q_{n - 1} } x = \paren {x - \xi_2} \map {Q_{n - 2} } x$
+Thus:
+:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \map {Q_{n - 2} } x$
+and we then move on to consider $\xi_3$.
+Eventually we reach:
+:$\map P x = \paren {x - \xi_1} \paren {x - \xi_2} \dotsm \paren {x - \xi_n} \map {Q_0} x$
+$\map {Q_0} x$ is a polynomial of [[Definition:Polynomial of Degree Zero|zero degree]], that is a [[Definition:Constant Polynomial|constant polynomial]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Telescoping Series/Example 1}
+Tags: Telescoping Series
+
+\begin{theorem}
+Let $\left \langle {b_n} \right \rangle$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\left \langle {a_n} \right \rangle$ be a [[Definition:Real Sequence|sequence]] whose [[Definition:Term of Sequence|terms]] are defined as:
+:$a_k = b_k - b_{k + 1}$
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n a_k = b_1 - b_{n + 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \displaystyle \sum_{k \mathop = 1}^n a_k
+      | r = \sum_{k \mathop = 1}^n \left({b_k - b_{k + 1} }\right)
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 1}^n b_{k + 1}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n b_k - \sum_{k \mathop = 2}^{n + 1} b_k
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = b_1 + \sum_{k \mathop = 2}^n b_k - \sum_{k \mathop = 2}^n b_k - b_{n + 1}
+      | c = 
+}}
+{{eqn | r = b_1 - b_{n + 1}
+      | c = 
+}}
+{{end-eqn}}
+If $\left \langle {b_k} \right \rangle$ [[Definition:Convergent Sequence|converges]] to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.
+Thus:
+: $\displaystyle \lim_{n \mathop \to \infty} s_n = b_1 - 0 = b_1$
+So:
+: $\displaystyle \sum_{k \mathop = 1}^\infty a_k = b_1$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Terms in Convergent Series Converge to Zero}
+Tags: Series, Convergent Series
+
+\begin{theorem}
+Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] in any of the [[Definition:Standard Number Field|standard number fields]] [[Definition:Rational Number|$\Q$]], [[Definition:Real Number|$\R$]], or [[Definition:Complex Number|$\C$]].
+Suppose that the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]] in any of the [[Definition:Standard Number Field|standard number fields]] [[Definition:Rational Number|$\Q$]], [[Definition:Real Number|$\R$]], or [[Definition:Complex Number|$\C$]].
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = 0$
+{{expand|Expand (on a different page) to Banach spaces}}
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle s = \sum_{n \mathop = 1}^\infty a_n$.
+Then $\displaystyle s_N = \sum_{n \mathop = 1}^N a_n \to s$ as $N \to \infty$.
+Also, $s_{N - 1} \to s$ as $N \to \infty$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a_N
+      | r = \paren {a_1 + a_2 + \cdots + a_{N - 1} + a_N} - \paren {a_1 + a_2 + \cdots + a_{N - 1} }
+      | c = 
+}}
+{{eqn | r = s_N - s_{N - 1}
+      | c = 
+}}
+{{eqn | o = \to
+      | r = s - s = 0 \text{ as } N \to \infty
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Convergent Series}
+Tags: Series
+
+\begin{theorem}
+Let $\sequence {a_n}_{n \mathop \ge 1}$ and $\sequence {b_n}_{n \mathop \ge 1}$ be [[Definition:Sequence|sequences]] of [[Definition:Real Number|real numbers]].
+Let the two [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ [[Definition:Convergent Series|converge]] to $\alpha$ and $\beta$ respectively.
+Let $\lambda, \mu \in \R$ be [[Definition:Real Number|real numbers]].
+Then the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty \paren {\lambda a_n + \mu b_n}$ [[Definition:Convergent Series|converges]] to $\lambda \alpha + \mu \beta$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 1}^N \paren {\lambda a_n + \mu b_n}
+      | r = \lambda \sum_{n \mathop = 1}^N a_n + \mu \sum_{n \mathop = 1}^N b_n
+      | c = [[Linear Combination of Indexed Summations]]
+}}
+{{eqn | o = \to
+      | r = \lambda \alpha + \mu \beta \text{ as } N \to \infty
+      | c = [[Combination Theorem for Sequences]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Abel's Theorem}
+Tags: Series
+
+\begin{theorem}
+Let $\displaystyle \sum_{k \mathop = 0}^\infty a_k$ be a [[Definition:Convergent Series|convergent]] [[Definition:Series|series in $\R$]].
+Then:
+:$\displaystyle \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$
+where $\displaystyle \lim_{x \mathop \to 1^-}$ denotes the [[Definition:Limit from Left|limit from the left]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+Let $\displaystyle \sum_{k \mathop = 0}^\infty a_k$ [[Definition:Convergent Series|converge]] to $s$.
+Then its [[Definition:Sequence of Partial Sums|sequence of partial sums]] $\sequence {s_N}$, where $\displaystyle s_N = \sum_{n \mathop = 1}^N a_n$, is a [[Convergent Sequence is Cauchy Sequence|Cauchy sequence]].
+So:
+: $\displaystyle \exists N: \forall k, m: k \ge m \ge N: \size {\sum_{l \mathop = m}^k a_l} < \frac \epsilon 3$
+From [[Abel's Lemma: Formulation 2]], we have:
+:$\displaystyle \sum_{k \mathop = m}^n u_k v_k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k u_l} \paren {v_k - v_{k+1}} } + v_n \sum_{k \mathop = m}^n u_k$
+We apply this, with $u_k = a_k$ and $v_k = x^k$:
+:$\displaystyle \sum_{k \mathop = m}^n a_k x^k = \sum_{k \mathop = m}^{n-1} \paren {\paren {\sum_{l \mathop = m}^k a_l} \paren {x^k - x^{k+1}} } + x^n \sum_{k \mathop = m}^n a_k$
+So it follows that $\forall n \ge m \ge N$ and $\forall 0 < x < 1$, we have:
+{{begin-eqn}}
+{{eqn | l = \size {\sum_{k \mathop = m}^n a_k x^k}
+      | o = <
+      | r = \paren {1 - x} \sum_{k \mathop = m}^{n-1} \frac \epsilon 3 x^k + \frac \epsilon 3 x^n
+      | c = replacing instances of $\displaystyle \sum_{l \mathop = m}^k a_l$ with $\dfrac \epsilon 3$
+}}
+{{eqn | o = <
+      | r = \frac \epsilon 3 \paren {1 - x} \frac {1 - x^n} {1 - x} + \frac \epsilon 3 x^n
+      | c = [[Sum of Geometric Progression]]
+}}
+{{eqn | r = \frac \epsilon 3
+      | c = 
+}}
+{{end-eqn}}
+So we conclude that:
+: $\displaystyle \size {\sum_{k \mathop = N}^\infty a_k x^k} \le \frac \epsilon 3$
+Next, note that from the above, we have $\forall x: 0 < x < 1$:
+:$\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} \le \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} + \frac \epsilon 3 + \frac \epsilon 3$
+But for finite $n$, we have that $1 - x^n \to 0$ as $x \to 1^-$.
+Thus:
+:$\displaystyle \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} \to 0$ as $x \to 1^-$
+So:
+: $\displaystyle \exists \delta > 0: \forall x: 1 - \delta < x < 1: \sum_{k \mathop = 0}^{N-1} \size {a_n} \paren {1 - x^n} < \frac \epsilon 3$
+So, for any given $\epsilon > 0$, we can find a $\delta > 0$ such that, for any $x$ such that $1 - \delta < x < 1$, it follows that:
+: $\displaystyle \size {\sum_{k \mathop = 0}^\infty a_k x^k - \sum_{k \mathop = 0}^\infty a_k} < \epsilon$
+That coincides with the definition for the [[Definition:Limit from Left|limit from the left]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Alternating Series Test}
+Tags: Convergence Tests, Named Theorems
+
+\begin{theorem}
+Let $\sequence {a_n}_{N \mathop \ge 0}$ be a [[Definition:Decreasing Sequence|decreasing sequence]] of [[Definition:Positive Real Number|positive]] [[Definition:Term of Sequence|terms]] in $\R$ which [[Definition:Convergent Sequence|converges]] with a [[Definition:Limit of Sequence (Number Field)|limit]] of zero.
+That is, let $\forall n \in \N: a_n \ge 0, a_{n + 1} \le a_n, a_n \to 0$ as $n \to \infty$
+Then the series:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n = a_1 - a_2 + a_3 - a_4 + \dotsb$ 
+[[Definition:Convergent Series|converges]].
+\end{theorem}
+
+\begin{proof}
+First we show that for each $n > m$, we have $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$.
+This will be achieved by means of the [[Second Principle of Mathematical Induction]].
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$
+=== Basis for the Induction ===
+$\map P 1$ holds, as $a_{m + 1} \ge 0$ by definition and $a_{m + 1} \le a_{m + 1}$.
+$\map P 2$ also holds, as $a_{m + 2} \le a_{m + 1}$ and so:
+:$0 \le a_{m + 1} - a_{m + 2} \le a_{m + 1}$
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+In order to simplify the algebra, let:
+:$b_k := a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_k$
+The $\pm$ signifies the fact that $a_k$ is [[Definition:Positive Real Number|positive]] for $k$ [[Definition:Odd Integer|odd]] and [[Definition:Negative Real Number|negative]] for $k$ [[Definition:Even Integer|even]].
+Suppose that $\forall j \le k: \map P j$ holds:
+:$0 \le b_j \le a_{m + 1}$
+This is the [[Definition:Induction Hypothesis|induction hypothesis]].
+We now show that $\forall k: \map P k \implies \map P {k + 1}$.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+Suppose $k$ is [[Definition:Odd Integer|odd]].
+By the [[Alternating Series Test#Induction Hypothesis|induction hypothesis]]:
+:$0 \le b_k \le a_{m + 1}$
+Because $\map P k$ holds:
+:$0 \le b_{k - 1} + a_k \le a_{m + 1}$
+But as $a_k \ge a_{k + 1}$:
+:$a_k - a_{k + 1} \ge 0$
+and so:
+:$0 \le b_{k - 1} + \paren {a_k - a_{k + 1} } = b_{k + 1}$
+But as $b_k \le a_{m + 1}$:
+:$b_k - a_{k + 1} = b_{k + 1} \le a_{m + 1}$
+So:
+:$0 \le b_{k + 1} \le a_{m + 1}$
+or:
+:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb - a_{k + 1} \le a_{m + 1}$
+Thus for [[Definition:Odd Integer|odd]] $k$ it follows that $\map P k \implies \map P {k + 1}$.
+Now suppose $k$ is [[Definition:Even Integer|even]].
+By the [[Alternating Series Test#Induction Hypothesis|induction hypothesis]]:
+:$0 \le b_k \le a_{m + 1}$
+Then:
+:$0 \le b_k + a_{k + 1} = b_{k + 1}$
+Because $\map P k$ holds:
+:$0 \le b_{k - 1} - a_k \le a_{m + 1}$
+But as $a_k \ge a_{k + 1}$:
+:$a_k - a_{k + 1} \ge 0$
+and so:
+:$b_{k + 1} = b_{k - 1} - a_k + a_{k + 1} =  b_{k - 1} - \paren {a_k - a_{k + 1} } = b_{k + 1} \le a_{m + 1}$
+So:
+:$0 \le b_{k + 1} \le a_{m + 1}$
+or:
+:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \cdots + a_{k + 1} \le a_{m + 1}$
+Thus for [[Definition:Even Integer|even]] $k$ it follows that $\map P k \implies \map P {k + 1}$.
+So for both [[Definition:Even Integer|even]] and [[Definition:Odd Integer|odd]] $k$ it follows that $\map P k \implies \map P {k + 1}$ and the result follows by the [[Second Principle of Mathematical Induction]].
+Therefore for each $n > m$, we have:
+:$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$
+Now, let $\sequence {s_n}$ be the [[Definition:Series|sequence of partial sums of the series:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n$]]
+Let $\epsilon > 0$.
+Since $a_n \to 0$ as $n \to \infty$:
+:$\exists N: \forall n > N: a_n < \epsilon$
+But $\forall n > m > N$, we have:
+{{begin-eqn}}
+{{eqn | l = \sequence {s_n - s_m}
+      | r = \size {\paren {a_1 - a_2 + a_3 - \dotsb \pm a_n} - \paren {a_1 - a_2 + a_3 - \dotsb \pm a_m} }
+      | c = 
+}}
+{{eqn | r = \size {\paren {a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsc \pm a_n} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = a_{m + 1}
+      | c = from the above
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = as $m + 1 > N$
+}}
+{{end-eqn}}
+Thus we have shown that $\sequence {s_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+The result follows from [[Convergent Sequence is Cauchy Sequence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Identity}
+Tags: Euler's Number, Pi, Complex Analysis
+
+\begin{theorem}
+:$e^{i \pi} + 1 = 0$
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Euler's Formula]] $e^{i z} = \cos z + i \sin z$, by plugging in $z = \pi$:
+:$e^{i \pi} + 1 = \cos \pi + i \sin \pi + 1 = -1 + i \times 0 + 1 = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intermediate Value Theorem}
+Tags: Named Theorems, Analysis
+
+\begin{theorem}
+Let $f: S \to \R$ be a [[Definition:Real Function|real function]] on some [[Definition:Subset|subset]] $S$ of $\R$.
+Let $I \subseteq S$ be a [[Definition:Real Interval|real interval]].
+Let $f: I \to \R$ be [[Definition:Continuous on Interval|continuous]] on $I$.
+Let $a, b \in I$.
+Let $k \in \R$ lie between $\map f a$ and $\map f b$.
+That is, either:
+:$\map f a < k < \map f b$
+or:
+:$\map f b < k < \map f a$
+Then $\exists c \in \openint a b$ such that $\map f c = k$.
+\end{theorem}
+
+\begin{proof}
+This theorem is a restatement of [[Image of Interval by Continuous Function is Interval]].
+From [[Image of Interval by Continuous Function is Interval]], the image of $\openint a b$ under $f$ is also a [[Definition:Real Interval|real interval]] (but not necessarily [[Definition:Open Real Interval|open]]).
+Thus if $k$ lies between $\map f a$ and $\map f b$, it must be the case that:
+:$k \in \Img {\openint a b}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tail of Convergent Series tends to Zero}
+Tags: Series
+
+\begin{theorem}
+Let $\sequence {a_n}_{n \mathop \ge 1}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Numbers|real numbers]].
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a [[Definition:Convergent Series|convergent series]].
+Let $N \in \N_{\ge 1}$ be a [[Definition:Natural Numbers|natural number]].
+Let $\displaystyle \sum_{n \mathop = N}^\infty a_n$ be the [[Definition:Tail of Series|tail]] of the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.
+Then:
+:$\displaystyle \sum_{n \mathop = N}^\infty a_n$ is [[Definition:Convergent Series|convergent]]
+:$\displaystyle \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
+That is, the [[Definition:Tail of Series|tail]] of a [[Definition:Convergent Series|convergent series]] tends to zero.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {s_n}$ be the [[Definition:Sequence of Partial Sums|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.
+Let $\sequence {s'_n}$ be the [[Definition:Sequence of Partial Sums|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = N}^\infty a_n$.
+It will be shown that $\sequence {s'_n}$ fulfils the [[Definition:Cauchy Criterion for Sequences|Cauchy criterion]].
+That is:
+:$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \size {s'_n - s'_m} < \epsilon$
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+As $\sequence {s_n}$ is [[Definition:Convergent Series|convergent]], it conforms to the [[Definition:Cauchy Criterion for Sequences|Cauchy criterion]] by [[Convergent Sequence is Cauchy Sequence]].
+Thus:
+:$\exists N: \forall m, n > N: \size {s_n - s_m} < \epsilon$
+Now:
+{{begin-eqn}}
+{{eqn | l = s_n
+      | r = \sum_{k \mathop = 1}^n a_k
+      | c =  
+}}
+{{eqn | r = \sum_{k \mathop = 1}^{N - 1} a_k + \sum_{k \mathop = N}^n a_k 
+      | c = [[Indexed Summation over Adjacent Intervals]]
+}}
+{{eqn | r = s_{N - 1} + s'_n
+}}
+{{end-eqn}}
+and similarly:
+:$s_m = s_{N - 1} + s'_m$
+Thus:
+:$s'_n = s_n - s_{N - 1}$
+and:
+:$s'_m = s_m - s_{N - 1}$
+So:
+{{begin-eqn}}
+{{eqn | l = \size {s_n - s_m}
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {s_n - s_{N - 1} - s_m + s_{N - 1} }
+      | o = <
+      | r = \epsilon
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\paren {s_n - s_{N - 1} } - \paren {s_m - s_{N - 1} } }
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\size {s_n - s_{N - 1} } - \size {s_m - s_{N - 1} } }
+      | o = <
+      | r = \epsilon
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | ll= \leadsto
+      | l = \size {s'_n - s'_m}
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+So $\displaystyle \sum_{n \mathop = N}^\infty a_n$ fulfils the [[Definition:Cauchy Criterion for Sequences|Cauchy criterion]].
+By [[Convergent Sequence is Cauchy Sequence]] it follows that it is [[Definition:Convergent Series|convergent]].
+Now it is shown that $\displaystyle \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
+We have that $\sequence {s_n}$ is [[Definition:Convergent Series|convergent]],
+Let its [[Definition:Limit of Sequence (Number Field)|limit]] be $l$.
+Thus we have:
+:$\displaystyle l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$
+So:
+:$\displaystyle \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$
+But $s_{N - 1} \to l$ as $N - 1 \to \infty$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Comparison Test}
+Tags: Series, Convergence Tests, Named Theorems
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be a [[Definition:Convergent Series|convergent series]] of [[Definition:Positive Real Number|positive real numbers]].
+Let $\sequence {a_n}$ be a [[Definition:Real Sequence|sequence $\R$]] or [[Definition:Complex Sequence|sequence in $\C$]].
+Let $\forall n \in \N_{>0}: \cmod {a_n} \le b_n$.
+Then the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+As $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ [[Definition:Convergent Series|converges]], its [[Tail of Convergent Series tends to Zero|tail tends to zero]].
+So:
+:$\displaystyle \exists N: \forall n > N: \sum_{k \mathop = n + 1}^\infty b_k < \epsilon$
+Let $\sequence {a_n}$ be the [[Definition:Series|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$.
+Then $\forall n > m > N$:
+{{begin-eqn}}
+{{eqn | l = \cmod {a_n - a_m}
+      | r = \cmod {\paren {a_1 + a_2 + \cdots + a_n} - \paren {a_1 + a_2 + \cdots + a_m} }
+      | c = 
+}}
+{{eqn | r = \cmod {a_{m + 1} + a_{m + 2} + \cdots + a_n}
+      | c = [[Indexed Summation over Adjacent Intervals]]
+}}
+{{eqn | o = \le
+      | r = \cmod {a_{m + 1} } + \cmod {a_{m + 2} } + \cdots + \cmod {a_n}
+      | c = [[Triangle Inequality for Indexed Summations]]
+}}
+{{eqn | o = \le
+      | r = b_{m + 1} + b_{m + 2} + \cdots + b_n
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \sum_{k \mathop = n + 1}^\infty b_k
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+So $\sequence {a_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+The result follows from [[Real Number Line is Complete Metric Space]] or [[Complex Plane is Complete Metric Space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ratio Test}
+Tags: Convergence Tests, Named Theorems
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a [[Definition:Series|series]] of [[Definition:Real Number|real numbers]] in $\R$, or a [[Definition:Series|series]] of [[Definition:Complex Number|complex numbers]] in $\C$.
+Let the [[Definition:Sequence|sequence]] $\sequence {a_n}$ satisfy:
+:$\displaystyle \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = l$
+:If $l > 1 $, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Divergent Series|diverges]].
+:If $l < 1 $, then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+\end{theorem}
+
+\begin{proof}
+From the statement of the theorem, it is necessary that $\forall n: a_n \ne 0$; otherwise $\size {\dfrac {a_{n + 1} } {a_n} }$ is not defined.
+Here, $\size {\dfrac {a_{n + 1} } {a_n} }$ denotes either the [[Definition:Absolute Value|absolute value]] of $\dfrac {a_{n + 1} } {a_n}$, or the [[Definition:Complex Modulus|complex modulus]] of $\dfrac {a_{n + 1} } {a_n}$.
+=== Absolute Convergence ===
+Suppose $l < 1$.
+Let us take $\epsilon > 0$ such that $l + \epsilon < 1$.
+Then:
+:$\exists N: \forall n > N: \size {\dfrac {a_n} {a_{n - 1} } } < l + \epsilon$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \size {a_n}
+      | r = \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 2} } {a_{N + 1} } } \size {a_{N + 1} }
+      | c = 
+}}
+{{eqn | o = <
+      | r = \paren {l + \epsilon}^{n - N - 1} \size {a_{N + 1} }
+      | c = 
+}}
+{{end-eqn}}
+By [[Sum of Infinite Geometric Progression]], $\displaystyle \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ [[Definition:Convergent Series|converges]].
+So by the the corollary to the [[Comparison Test/Corollary|comparison test]], it follows that $\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}$ [[Definition:Absolutely Convergent Series|converges absolutely]] too.
+{{qed}}
+=== Divergence ===
+Suppose $l > 1$.
+Let us take $\epsilon > 0$ small enough that $l - \epsilon > 1$.
+Then, for a sufficiently large $N$, we have:
+{{begin-eqn}}
+{{eqn | l = \size {a_n}
+      | r = \size {\frac {a_n} {a_{n - 1} } } \size {\frac {a_{n - 1} } {a_{n - 2} } } \dotsm \size {\frac {a_{N + 2} } {a_{N + 1} } } \size {a_{N + 1} }
+      | c = 
+}}
+{{eqn | o = >
+      | r = \paren {l - \epsilon}^{n - N + 1} \size {a_{N + 1} }
+      | c = 
+}}
+{{end-eqn}}
+But $\paren {l - \epsilon}^{n - N + 1} \size {a_{N + 1} } \to \infty$ as $n \to \infty$.
+So $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Divergent Series|diverges]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Subsequence of Bounded Sequence}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\sequence {x_n}$ be [[Definition:Bounded Real Sequence|bounded]].
+Let $b \in \R$ be a [[Definition:Real Number|real number]].
+Suppose that $\forall N: \exists n > N: x_n \ge b$.
+Then $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]] $l \ge b$.
+\end{theorem}
+
+\begin{proof}
+Let us pick $N \in \N$.
+Then $\exists n_1 > N: x_{n_1} \ge b$.
+Again, $\exists n_2 > n_1: x_{n_2} \ge b$.
+And so on: for each $n_k$ we find, $\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$.
+In this way we can build a [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$ each of whose terms are $b$ or bigger.
+By the [[Bolzano-Weierstrass Theorem]], this [[Definition:Subsequence|subsequence]] itself contains a [[Definition:Subsequence|subsequence]] $\sequence {x_{n_r} }$ which is [[Definition:Convergent Real Sequence|convergent]].
+Now, suppose $x_{n_r} \to l$ as $r \to \infty$.
+Since $x_{n_r} \ge b$ it follows from [[Lower and Upper Bounds for Sequences]] that $l \ge b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Terms of Bounded Sequence Within Bounds}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\sequence {x_n}$ be [[Definition:Bounded Sequence|bounded]].
+Let the [[Definition:Limit Superior|limit superior]] of $\sequence {x_n}$ be $\overline l$.
+Let the [[Definition:Limit Inferior|limit inferior]] of $\sequence {x_n}$ be $\underline l$.
+Then:
+:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
+:$\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$
+\end{theorem}
+
+\begin{proof}
+=== Upper Bound ===
+First we show that:
+:$\forall \epsilon > 0: \exists N: \forall n > N: x_n < \overline l + \epsilon$
+{{AimForCont}} this [[Definition:Proposition|proposition]] were to be [[Definition:False|false]].
+That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \ge \overline l + \epsilon$.
+From [[Limit of Subsequence of Bounded Sequence]] it would follow that there exists a [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequence]] whose limit was $l \ge \overline l + \epsilon$.
+This would contradict the definition of $\overline l$.
+{{qed|lemma}}
+=== Lower Bound ===
+Next, in the same way, we show that:
+: $\forall \epsilon > 0: \exists N: \forall n > N: x_n > \underline l - \epsilon$
+{{AimForCont}} this [[Definition:Proposition|proposition]] were to be [[Definition:False|false]].
+That would mean that for some $\epsilon > 0$ it would be true that for each $N$ we would be able to find $n > N$ such that $x_n \le \underline l + \epsilon$.
+From [[Limit of Subsequence of Bounded Sequence]] it would follow that there exists a [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequence]] whose limit was $l \le \underline l + \epsilon$.
+This would contradict the definition of $\underline l$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergence of Limsup and Liminf}
+Tags: Limits of Sequences, Convergence
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let the [[Definition:Limit Superior|limit superior]] of $\sequence {x_n}$ be $\overline l$.
+Let the [[Definition:Limit Inferior|limit inferior]] of $\sequence {x_n}$ be $\underline l$.
+Then $\left \langle {x_n} \right \rangle$ [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]] $l$ {{iff}} $\overline l = \underline l = l$.
+Hence a [[Definition:Bounded Real Sequence|bounded real sequence]] [[Definition:Convergent Real Sequence|converges]] {{iff}} all its [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequences]] have the same [[Definition:Limit of Real Sequence|limit]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+First, suppose that $\overline l = \underline l = l$.
+Let $\epsilon > 0$.
+By [[Terms of Bounded Sequence Within Bounds]]:
+:$\exists N_1: \forall n > N_1: x_n < l + \epsilon$
+Similarly:
+:$\exists N_2: \forall n > N_2: x_n > l - \epsilon$
+So take $N = \max \set {N_1, N_2}$.
+If $n > N$, both the above inequalities hold at the same time.
+So $l - \epsilon < x_n < l + \epsilon$ and so by [[Negative of Absolute Value]]:
+:$\size {x_n - l} < \epsilon$
+Thus $x_n \to l$ as $n \to \infty$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Now suppose that $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]] $l$.
+Then by [[Limit of Subsequence equals Limit of Real Sequence]], all [[Definition:Subsequence|subsequences]] have a [[Definition:Limit of Real Sequence|limit]] of $l$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limsup and Liminf are Limits of Bounds}
+Tags: Limits of Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\sequence {x_n}$ be [[Definition:Bounded Real Sequence|bounded]].
+Let $\displaystyle \overline l = \limsup_{n \mathop \to \infty} x_n$ be the [[Definition:Limit Superior|limit superior]] and $\displaystyle \liminf_{n \mathop \to \infty} x_n$ the [[Definition:Limit Inferior|limit inferior]] of $\sequence {x_n}$.
+Then:
+:$\displaystyle \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
+:$\displaystyle \underline l = \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\inf_{k \mathop \ge n} x_k}$
+\end{theorem}
+
+\begin{proof}
+First we show that:
+:$\displaystyle \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \mathop \ge n} x_k}$
+Let $M_n = \displaystyle \sup_{k \mathop \ge n} x_k$.
+By [[Supremum of Subset]], the [[Definition:Real Sequence|sequence]] $\sequence {M_n}$ [[Definition:Decreasing Real Sequence|decreases]], for:
+:$m \ge n \implies \set {k \in \N: k \ge m} \subseteq \set {k \in \N: k \ge n}$
+A [[Definition:Lower Bound of Set|lower bound]] for $\sequence {x_n}$ is also a [[Definition:Lower Bound of Set|lower bound]] for $\sequence {M_n}$.
+So from the [[Monotone Convergence Theorem (Real Analysis)]] it follows that $\sequence {M_n}$ [[Definition:Convergent Sequence|converges]].
+Suppose $M_n \to M$ as $n \to \infty$.
+From the [[Bolzano-Weierstrass Theorem]] there exists a [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Let $L$ be the [[Definition:Set|set]] of all numbers which are the [[Definition:Limit of Real Sequence|limit]] of some [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$.
+Let $l \in L$ be arbitrary.
+Let $\sequence {x_{n_r} }$ be a [[Definition:Convergent Real Sequence|convergent]] [[Definition:Subsequence|subsequence]] of $\sequence {x_n}$ such that $x_{n_r} \to l$ as $r \to \infty$.
+Then $\forall n_r \ge n: x_{n_r} \le M_n$.
+Hence $l \le M_n$ by [[Lower and Upper Bounds for Sequences]] and hence (from the same theorem) $l \le M$.
+This is true for all $l \in L$, so $\overline l = \limsup_{n \mathop \to \infty} x_n \le M$.
+Now, from [[Terms of Bounded Sequence Within Bounds]], we have that:
+:$\forall \epsilon > 0: \exists n: \forall k \ge n: x_k < \overline l + \epsilon$
+Thus $\overline l + \epsilon$ is an [[Definition:Upper Bound of Set|upper bound]] for $\set {x_k: k \ge n}$.
+So $M \le M_n \le \overline l + \epsilon$.
+Thus from [[Real Plus Epsilon]], $M \le \overline l$.
+Thus we conclude that $M = \overline l$ and hence:
+:$\displaystyle \overline l = \limsup_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop \to \infty} } {\sup_{k \ge n} x_k}$
+{{handwaving|"similar argument"}}
+:$\displaystyle \liminf_{n \mathop \to \infty} x_n = \map {\lim_{n \mathop to \infty} } {\inf_{k \mathop \ge n} x_k}$
+can be proved using a similar argument.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nth Root Test}
+Tags: Convergence Tests, Named Theorems
+
+\begin{theorem}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be a [[Definition:Series|series]] of [[Definition:Real Number|real numbers]] $\R$ or [[Definition:Complex Number|complex numbers]] $\C$.
+Let the [[Definition:Sequence|sequence]] $\sequence {a_n}$ be such that the [[Definition:Limit Superior|limit superior]] $\displaystyle \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = l$.
+Then:
+:If $l > 1$, the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Divergent Series|diverges]].
+:If $l < 1$, the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Absolutely Convergent Series|converges absolutely]].
+\end{theorem}
+
+\begin{proof}
+=== Absolute Convergence ===
+Let $l < 1$.
+Then let us choose $\epsilon > 0$ such that $l + \epsilon < 1$.
+Consider the [[Definition:Real Number|real]] sequence $\sequence {b_n}$ defined by $\sequence {b_n} = \sequence {\size {a_n} }$.
+Here, $\size {a_n}$ denotes either the [[Definition:Absolute Value|absolute value]] of $a_n$, or the [[Definition:Complex Modulus|complex modulus]] of $a_n$.
+Then:
+:$\displaystyle l = \limsup_{n \mathop \to \infty} {b_n}^{1/n}$
+It follows from [[Terms of Bounded Sequence Within Bounds]] that for sufficiently large $n$,:
+:$b_n < \paren {l + \epsilon}^n$
+By [[Sum of Infinite Geometric Sequence]], the [[Definition:Infinite Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty \paren {l + \epsilon}^n$ [[Definition:Convergent Sequence|converges]].
+By the [[Comparison Test|comparison test]], $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ [[Definition:Convergent Sequence|converges]].
+Hence $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Absolutely Convergent Series|converges absolutely]] by the definition of absolute convergence.
+{{qed|lemma}}
+=== Divergence ===
+Let $l > 1$.
+Then we choose $\epsilon > 0$ such that $l - \epsilon > 1$.
+{{AimForCont}} that there exist an [[Definition:Upper Bound of Set|upper bound]] for the set:
+:$S := \set {n \in \N: \size {a_n}^{1/n} > l - \epsilon}$
+Then for all [[Definition:Sufficiently Large|sufficiently large]] $n$:
+:$\size {a_n}^{1/n} \le l - \epsilon$
+However, this implies that:
+:$\displaystyle \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} \le l - \epsilon$
+which is false by the definition of $l$.
+The set $S$, then, is not [[Definition:Bounded Ordered Set|bounded]].
+This means that there exist [[Definition:Arbitrarily Large|arbitrarily large]] $n$ such that:
+:$\size {a_n} > \paren {l - \epsilon}^n$
+Thus:
+:$\displaystyle \lim_{n \mathop \to \infty} \size {a_n} \ne 0$
+and so $\displaystyle \lim_{n \mathop \to \infty} a_n \ne 0$.
+Hence from [[Terms in Convergent Series Converge to Zero]], $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ must be [[Definition:Divergent Series|divergent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Series of Power over Factorial Converges}
+Tags: Series
+
+\begin{theorem}
+The [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ [[Definition:Convergent Series|converges]] for all [[Definition:Real Number|real]] values of $x$.
+\end{theorem}
+
+\begin{proof}
+If $x = 0$ the result is trivially true as:
+:$\forall n \ge 1: \dfrac {0^n} {n!} = 0$
+If $x \ne 0$ we have:
+:$\left|{\dfrac{\left({\dfrac {x^{n+1}} {(n+1)!}}\right)}{\left({\dfrac {x^n}{n!}}\right)}}\right| = \dfrac {\left|{x}\right|} {n+1} \to 0$
+as $n \to \infty$.
+This follows from the results:
+:[[Sequence of Powers of Reciprocals is Null Sequence]], where $\dfrac 1 n \to 0$ as $n \to \infty$
+:The [[Squeeze Theorem for Real Sequences]], as $\dfrac 1 {n + 1} < \dfrac 1 n$
+:The [[Multiple Rule for Real Sequences]], putting $\lambda = \left|{x}\right|$.
+Hence by the [[Ratio Test]]: $\displaystyle \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$ [[Definition:Convergent Series|converges]].
+{{qed}}
+Alternatively, the [[Comparison Test]] could be used but this is more cumbersome in this instance.
+Another alternative is to view this as an example of [[Radius of Convergence of Power Series over Factorial]] setting $\xi = 0$.
+\end{proof}<|endoftext|>
+\section{Area of Triangle in Terms of Side and Altitude}
+Tags: Areas of Triangles
+
+\begin{theorem}
+The [[Definition:Area|area]] of a [[Definition:Triangle (Geometry)|triangle]] $\triangle ABC$ is given by:
+:$\dfrac {c \cdot h_c} 2 = \dfrac {b \cdot h_b} 2 = \dfrac {a \cdot h_a} 2$
+where:
+:$a, b, c$ are the [[Definition:Side of Polygon|sides]]
+:$h_a, h_b, h_c$ are the [[Definition:Altitude of Triangle|altitudes]] from $A$, $B$ and $C$ respectively.
+\end{theorem}
+
+\begin{proof}
+:[[File:Area-of-Triangle.png|400px]]
+Construct a point $D$ so that $\Box ABDC$ is a [[Definition:Parallelogram|parallelogram]].
+From [[Halves of Parallelogram Are Congruent Triangles]]:
+:$\triangle ABC \cong \triangle DCB$
+hence their [[Definition:Area|areas]] are equal.
+The [[Area of Parallelogram]] is equal to the [[Definition:Real Mulitplication|product]] of one of its [[Definition:Base of Parallelogram|bases]] and the associated [[Definition:Altitude of Parallelogram|altitude]].
+Thus 
+{{begin-eqn}}
+{{eqn | l = \paren {ABCD}
+      | r = c \cdot h_c
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \paren {ABC}
+      | r = c \cdot h_c
+      | c = because [[Axiom:Area Axioms|congruent surfaces have equal areas]]
+}}
+{{eqn | l = \paren {ABC}
+      | r = \frac {c \cdot h_c} 2
+}}
+{{end-eqn}}
+where $\paren {XYZ}$ is the [[Definition:Area|area]] of the [[Definition:Plane Figure|plane figure]] $XYZ$.
+A similar argument can be used to show that the statement holds for the other [[Definition:Side of Polygon|sides]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Series is Convergent}
+Tags: Absolute Convergence, Banach Spaces, Absolutely Convergent Series is Convergent
+
+\begin{theorem}
+Let $V$ be a [[Definition:Banach Space|Banach space]] with [[Definition:Norm on Vector Space|norm]] $\norm {\, \cdot \,}$.
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $V$.
+Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]].
+\end{theorem}
+
+\begin{proof}
+That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Absolutely Convergent Series|absolutely convergent]] means that $\displaystyle \sum_{n \mathop = 1}^\infty \norm {a_n}$ [[Definition:Convergent Series|converges]] in $\R$.
+Hence the sequence of [[Definition:Partial Sum|partial sums]] is a [[Definition:Cauchy Sequence|Cauchy sequence]] by [[Convergent Sequence is Cauchy Sequence]].
+Now let $\epsilon > 0$.
+Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
+:$\displaystyle \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
+This $N$ exists because the sequence is [[Definition:Cauchy Sequence|Cauchy]].
+Now observe that, for $m \ge n \ge N$, one also has:
+{{begin-eqn}}
+{{eqn | l = \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}
+      | r = \norm {\sum_{k \mathop = n + 1}^m a_k}
+}}
+{{eqn | o = \le
+      | r = \sum_{k \mathop = n + 1}^m \norm {a_k}
+      | c = [[Definition:Norm on Vector Space|Triangle inequality]] for $\norm {\, \cdot \,}$
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+It follows that the [[Definition:Series|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Cauchy Sequence|Cauchy]].
+As $V$ is a [[Definition:Banach Space|Banach space]], this implies that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Comparison Test}
+Tags: Convergence Tests, Series
+
+\begin{theorem}
+Let $\left \langle {a_n} \right \rangle$ and $\left \langle {b_n} \right \rangle$ be [[Definition:Real Sequence|sequences in $\R$]].
+Let $\displaystyle \frac {a_n}{b_n} \to l$ as $n \to \infty$ where $l \in \R_{>0}$.
+Then the [[Definition:Series|series]] $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ are either both [[Definition:Convergent Series|convergent]] or both [[Definition:Divergent Series|divergent]].
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be [[Definition:Convergent Series|convergent]].
+Then by [[Terms in Convergent Series Converge to Zero]], $\left \langle {b_n} \right \rangle$ [[Definition:Convergent Sequence|converges]] to zero.
+A [[Convergent Sequence is Bounded]].
+So it follows that:
+: $\exists H: \forall n \in \N_{>0}: a_n \le H b_n$
+Thus, by the corollary to the [[Comparison Test/Corollary|Comparison Test]], $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]].
+Since $l > 0$, from [[Sequence Converges to Within Half Limit]]:
+: $\exists N: \forall n > N: a_n > \dfrac 1 2 l b_n$
+Hence the convergence of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ implies the convergence of $\displaystyle \sum_{n \mathop = 1}^\infty b_n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area of Parallelogram}
+Tags: Areas of Quadrilaterals, Parallelograms, Areas of Parallelograms
+
+\begin{theorem}
+The [[Definition:Area|area]] of a [[Definition:Parallelogram|parallelogram]] equals the product of one of its [[Definition:Base of Parallelogram|bases]] and the associated [[Definition:Altitude of Parallelogram|altitude]].
+\end{theorem}
+
+\begin{proof}
+There are three cases to be analysed: the [[Definition:Square (Geometry)|square]], the [[Definition:Rectangle|rectangle]] and the general [[Definition:Parallelogram|parallelogram]].
+=== [[Area of Parallelogram/Square|Square]] ===
+{{:Area of Parallelogram/Square}}
+=== [[Area of Parallelogram/Rectangle|Rectangle]] ===
+{{:Area of Parallelogram/Rectangle}}
+=== [[Area of Parallelogram/Parallelogram|Parallelogram]] ===
+{{:Area of Parallelogram/Parallelogram}}
+\end{proof}<|endoftext|>
+\section{Area of Triangle in Terms of Inradius and Exradii}
+Tags: Areas of Triangles
+
+\begin{theorem}
+The area of a $\triangle ABC$ is given by the formula:
+:$(ABC) = \rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right) = \rho s = \sqrt {\rho_a \rho_b \rho_c \rho}$
+where:
+:$s$ is the [[Definition:Semiperimeter|semiperimeter]]
+:$I$ is the [[Definition:Incenter of Triangle|incenter]]
+:$\rho$ is the [[Definition:Inradius of Triangle|inradius]]
+:$I_a, I_b, I_c$ are the [[Definition:Excenter of Triangle|excenters]]
+:$\rho_a, \rho_b, \rho_c$ are the [[Definition:Exradius of Triangle|exradii]] from $I_a, I_b, I_c$, respectively.
+\end{theorem}
+
+\begin{proof}
+=== Proof of the First Part ===
+First, we show that the area is equal to $\rho_a \left({s - a}\right) = \rho_b \left({s - b}\right) = \rho_c \left({s - c}\right)$.
+{{WLOG}}, we pick an [[Definition:Excircle of Triangle|excircle]] $I_a$.
+[[File:Area1.PNG]]
+{{begin-eqn}}
+{{eqn | l = (ABC)
+      | r = (ABI_a) + (ACI_a) - (CBI_a)
+      | c = (see figure above)
+}}
+{{eqn | r = \frac {c \rho_a} 2 + \frac {b \rho_a} 2 - \frac {a \rho_a} 2
+      | c = [[Area of Triangle in Terms of Side and Altitude]]
+}}
+{{eqn | r = \rho_a \frac {b + c + a} 2 - \rho_a a
+}}
+{{eqn | r = \rho_a s - \rho_a a
+}}
+{{eqn | l = (ABC)
+      | r = \rho_a \left({s - a}\right)
+}}
+{{end-eqn}}
+A similar argument can be used to show that the statement holds for the other [[Definition:Excircle of Triangle|excircles]].
+{{qed|lemma}}
+=== Proof of the Second Part ===
+Second, we show the area is equal to $(ABC) = \rho s$.
+We take the [[Definition:Incircle of Triangle|incircle]] with [[Definition:Incenter of Triangle|incenter]] at $I$ and [[Definition:Inradius of Triangle|inradius]] $\rho$:
+[[File:T3.PNG]]
+{{begin-eqn}}
+{{eqn | l = (ABC)
+      | r = (ABI) + (BCI) + (CAI)
+      | c = (see figure above)
+}}
+{{eqn | r = \frac {a \rho} 2 + \frac {b \rho} 2 + \frac {c \rho} 2
+}}
+{{eqn | r = \frac {\rho} 2 \left({a + b + c}\right)
+}}
+{{eqn | l = (ABC)
+      | r = \rho s
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Proof of the Third Part ===
+Finally, we show that the area is equal to $\sqrt{\rho_a \rho_b \rho_c \rho}$:
+{{begin-eqn}}
+{{eqn | l = (ABC)^4
+      | r = \rho_a \left({s - a}\right) \rho_b \left({s - b}\right) \rho_c \left({s - c}\right) \rho s
+}}
+{{eqn | l = (ABC)^4
+      | r = s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \rho_a \rho_b \rho_c \rho
+}}
+{{eqn | l = (ABC)^4
+      | r = (ABC)^2 \rho_a \rho_b \rho_c \rho
+      | c = [[Heron's Formula]]
+}}
+{{eqn | l = (ABC)^2
+      | r = \rho_a \rho_b \rho_c \rho
+}}
+{{eqn | l = (ABC)
+      | r = \sqrt{\rho_a \rho_b \rho_c \rho}
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Area of Triangle in Terms of Circumradius}
+Tags: Areas of Triangles
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] whose [[Definition:Side of Polygon|sides]] are of [[Definition:Length (Linear Measure)|lengths]] $a, b, c$.
+Then the [[Definition:Area|area]] $\AA$ of $\triangle ABC$ is given by:
+:$\AA = \dfrac {a b c} {4 R}$
+where $R$ is the [[Definition:Circumradius of Triangle|circumradius]] of $\triangle ABC$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CircumradiusLengthProof.png|400px]]
+Let $O$ be the [[Definition:Circumcenter of Triangle|circumcenter]] of $\triangle ABC$.
+Let $\AA$ be the [[Definition:Area|area]] of $\triangle ABC$.
+Let a [[Definition:Perpendicular|perpendicular]] be dropped from $C$ to $AB$ at $E$.
+Let $h := CE$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \AA
+      | r = \frac {c h} 2
+      | c = [[Area of Triangle in Terms of Side and Altitude]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = h
+      | r = \frac {2 \AA} c
+      | c = 
+}}
+{{end-eqn}}
+Let a [[Definition:Diameter of Circle|diameter]] $CD$ of the [[Definition:Circumcircle of Triangle|circumcircle]] be passed through $O$.
+By definition of [[Definition:Circumradius of Triangle|circumradius]], $CD = 2 R$.
+By [[Thales' Theorem]], $\angle CAD$ is a [[Definition:Right Angle|right angle]].
+By [[Angles on Equal Arcs are Equal]], $\angle ADC = \angle ABC$.
+It follows from [[Sum of Angles of Triangle equals Two Right Angles]] that $\angle ACD = \angle ECB$.
+Thus by [[Equiangular Triangles are Similar]] $\triangle DAC$ and $\triangle BEC$ are [[Definition:Similar Triangles|similar]].
+So:
+{{begin-eqn}}
+{{eqn | l = \frac {CA} {CD}
+      | r = \frac {CE} {CB}
+      | c = $\triangle DAC$ and $\triangle BEC$ are [[Definition:Similar Triangles|similar]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac b {2 R}
+      | r = \frac h a
+      | c = 
+}}
+{{eqn | r = \frac {2 \AA} {a c}
+      | c = substituting for $h$ from $(1)$ above
+}}
+{{eqn | ll= \leadsto
+      | l = \AA
+      | r = \frac {a b c} {4 R}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stewart's Theorem}
+Tags: Triangles
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] with [[Definition:Side of Polygon|sides]] $a, b, c$.
+Let $CP$ be a [[Definition:Cevian|cevian]] from $C$ to $P$.
+:[[File:Stewart's Theorem.png|400px]]
+Then:
+:$a^2 \cdot AP + b^2 \cdot PB = c \paren {CP^2 + AP \cdot PB}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = b^2
+      | r = AP^2 + CP^2 - 2 AP \cdot CP \cdot \map \cos {\angle APC}
+      | c = [[Law of Cosines]]
+}}
+{{eqn | n = 2
+      | l = a^2
+      | r = PB^2 + CP^2 - 2 CP \cdot PB \cdot \map \cos {\angle BPC}
+      | c = [[Law of Cosines]]
+}}
+{{eqn | r = PB^2 + CP^2 + 2 CP \cdot PB \cdot \map \cos {\angle APC}
+      | c = [[Cosine of Supplementary Angle]]
+}}
+{{eqn | n = 3
+      | ll= \leadsto \quad
+      | l = b^2 \cdot PB
+      | r = AP^2 \cdot PB + CP^2 \cdot PB - 2 PB \cdot AP \cdot CP \cdot \map \cos {\angle APC}
+      | c = $(1) \ \times PB$
+}}
+{{eqn | n = 4
+      | l = a^2 \cdot AP
+      | r = PB^2 \cdot AP + CP^2 \cdot AP + 2 AP \cdot CP \cdot PB \cdot \map \cos {\angle APC}
+      | c = $(2) \ \times AP$
+}}
+{{eqn | ll= \leadsto \quad
+      | l = a^2 \cdot AP + b^2 \cdot PB
+      | r = AP^2 \cdot PB + PB^2 \cdot AP + CP^2 \cdot PB + CP^2 \cdot AP
+      | c = $(3) \ + \ (4)$
+}}
+{{eqn | r = CP^2 \left({PB + AP}\right) + AP \cdot PB \paren {PB + AP}
+}}
+{{eqn | r = c \paren {CP^2 + AP \cdot PB}
+      | c = as $PB + AP = c$
+}}
+{{end-eqn}}
+{{qed}}
+{{namedfor|Matthew Stewart|cat = Stewart}}
+It is also known as [[Apollonius's Theorem]] after {{AuthorRef|Apollonius of Perga}}.
+[[Category:Triangles]]
+knuni2ihszsgiqh1ceg11nxve4zy050
+\end{proof}<|endoftext|>
+\section{Length of Median of Triangle}
+Tags: Medians of Triangles, Length of Median of Triangle
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let $CD$ be the [[Definition:Median of Triangle|median]] of $\triangle ABC$ which [[Definition:Bisection|bisects]] $AB$.
+:[[File:MedianOfTriangle.png|400px]]
+The [[Definition:Length (Linear Measure)|length]] $m_c$ of $CD$ is given by:
+:${m_c}^2 = \dfrac {a^2 + b^2} 2 - \dfrac {c^2} 4$
+\end{theorem}
+
+\begin{proof}
+Let $\triangle ABC$ be embedded in the [[Definition:Complex Plane|complex plane]].
+:[[File:Length-of-Triangle-Median-Complex.png|300px]]
+Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$ and $C = \tuple {x_3, y_3}$ be represented by the [[Definition:Complex Number as Vector|complex numbers]] $z_1$, $z_2$ and $z_3$ respectively.
+Then:
+{{begin-eqn}}
+{{eqn | l = AC
+      | r = z_3 - z_1
+      | c = 
+}}
+{{eqn | l = BC
+      | r = z_3 - z_2
+      | c = 
+}}
+{{eqn | l = AB
+      | r = z_2 - z_1
+      | c = 
+}}
+{{eqn | l = AD
+      | r = \dfrac {AB} 2
+      | c = 
+}}
+{{eqn | r = \paren {\dfrac {z_2 - z_1} 2}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = AC + CD
+      | r = AD
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = CD
+      | r = AD - AC
+      | c = 
+}}
+{{eqn | r = \paren {\dfrac {z_2 - z_1} 2} - \paren {z_3 - z_1}
+      | c = 
+}}
+{{eqn | r = \paren {\dfrac {z_1 - z_3} 2} + \paren {\dfrac {z_2 - z_3} 2}
+      | c = 
+}}
+{{end-eqn}}
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Length of Angle Bisector}
+Tags: Triangles, Length of Angle Bisector
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let $AD$ be the [[Definition:Angle Bisector|angle bisector]] of $\angle BAC$ in $\triangle ABC$.
+:[[File:LengthOfAngleBisector.png|300px]]
+Let $d$ be the [[Definition:Length of Line|length]] of $AD$.
+Then $d$ is given by:
+:$d^2 = \dfrac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}$
+where $a$, $b$, and $c$ are the [[Definition:Opposite (in Triangle)|sides opposite]] $A$, $B$ and $C$ respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {BD} {DC} 
+      | r = \frac c b
+      | c = [[Angle Bisector Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {BD} {DC} + 1
+      | r = \frac c b + 1
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {BD + DC} {DC}
+      | r = \frac {b + c} b
+}}
+{{eqn | ll= \leadsto
+      | l = \frac a {DC}
+      | r = \frac {b + c} b
+}}
+{{eqn | ll= \leadsto
+      | l = DC
+      | r = \frac {a b} {b + c}
+}}
+{{end-eqn}}
+Similarly, or by symmetry, we get:
+:$BD = \dfrac {a c} {b + c}$
+From [[Stewart's Theorem]], we have:
+:$b^2 \cdot BD + c^2 \cdot DC = d^2 \cdot a + BD \cdot DC \cdot a$
+Substituting the above expressions for $BD$ and $DC$:
+{{begin-eqn}}
+{{eqn | l = b^2 \dfrac {a c} {b + c} + c^2 \frac {a b} {b + c}
+      | r = d^2 \cdot a + \dfrac {a c} {b + c} \cdot \frac {a b} {b + c} \cdot a
+}}
+{{eqn | ll= \leadsto
+      | l = a b c \frac {b + c} {b + c}
+      | r = d^2 \cdot a + \frac{a^2 b c} {\paren {b + c}^2} \cdot a
+}}
+{{eqn | ll= \leadsto
+      | l = b c
+      | r = d^2 + \frac {a^2 b c} {\paren {b + c}^2}
+}}
+{{eqn | ll= \leadsto
+      | l = d^2
+      | r = b c \paren {1 - \frac {a^2} {\paren {b + c}^2} }
+}}
+{{eqn | ll= \leadsto
+      | l = d^2
+      | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:LengthOfAngleBisector2.png|400px]]
+From [[Length of Angle Bisector/Proof 1|Length of Angle Bisector: Proof 1]], we have:
+:$BD = \dfrac {a c} {b + c}$
+:$DC = \dfrac {a b} {b + c}$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \angle BAD
+      | o = \cong
+      | r = \angle FAC
+      | c = Definition of [[Definition:Angle Bisector|Angle Bisector]]
+}}
+{{eqn | l = \angle ABD
+      | o = \cong
+      | r = \angle AFC
+      | c = [[Angles in Same Segment of Circle are Equal]]
+}}
+{{end-eqn}}
+Then from [[Triangles with Two Equal Angles are Similar]] we have:
+:$\triangle ABD \sim \triangle AFC$
+So:
+{{begin-eqn}}
+{{eqn | l = \frac c d
+      | r = \frac {AF} b
+      | c = as $\triangle ABD$ and $\triangle AFC$ are [[Definition:Similar Triangles|similar]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac c d
+      | r = \frac {d + DF} b
+}}
+{{end-eqn}}
+Now we use the [[Intersecting Chord Theorem]], which gives us $BD \cdot DC = d \cdot DF$.
+{{begin-eqn}}
+{{eqn | l = \frac c d
+      | r = \frac {d + \frac {BD \cdot DC} d} b
+}}
+{{eqn | ll= \leadsto
+      | l = b c
+      | r = d^2 + BD \cdot DC
+}}
+{{eqn | ll= \leadsto
+      | l = d^2
+      | r = b c - BD \cdot DC
+}}
+{{eqn | ll= \leadsto
+      | l = d^2
+      | r = b c - \frac {a c} {b + c} \cdot \frac {a b} {b + c}
+}}
+{{eqn | ll= \leadsto
+      | l = d^2
+      | r = \frac {b c} {\paren {b + c}^2} \paren {\paren {b + c}^2 - a^2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Subset}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({U, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $S \subseteq U$.
+Let $T \subseteq S$.
+Let $S$ admit a [[Definition:Supremum of Set|supremum]] (in $U$).
+If $T$ also admits a [[Definition:Supremum of Set|supremum]] (in $U$), then $\sup \left({T}\right) \preceq\sup \left({S}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $B = \sup \left({S}\right)$.
+Then $B$ is an [[Definition:Upper Bound of Set|upper bound]] for $S$.
+As $T \subseteq S$, it follows by the definition of a [[Definition:Subset|subset]] that $x \in T \implies x \in S$.
+Because $x \in S \implies x \preceq B$ (as $B$ is an [[Definition:Upper Bound of Set|upper bound]] for $S$) it follows that $x \in T \implies x \preceq B$.
+So $B$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$.
+Therefore $B$ [[Definition:Ordering|succeeds]] the [[Definition:Supremum of Set|supremum]] of $T$ in $S$.
+Hence the result.
+{{qed}}
+[[Category:Order Theory]]
+h6yid7mhhlbjzlqg1wmrmg3ej6mowmg
+\end{proof}
+
+\begin{proof}
+The [[Definition:Real Number|number]] $\sup S$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $S$.
+Therefore, $\sup S$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $T$ as $T$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Accordingly, $T$ has a [[Definition:Supremum of Subset of Real Numbers|supremum]] by the [[Continuum Property]].
+The [[Definition:Real Number|number]] $\sup S$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $T$.
+Therefore, $\sup S$ is greater than or equal to $\sup T$ as $\sup T$ is the [[Definition:Supremum of Subset of Real Numbers|least upper bound]] of $T$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the [[Continuum Property]], $T$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]].
+It follows from [[Supremum of Subset]] that $\sup T \le \sup S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+$S$ is [[Definition:Bounded Above Set|bounded above]] as $S$ has a [[Definition:Supremum of Subset of Real Numbers|supremum]].
+Therefore, $T$ is [[Definition:Bounded Above Set|bounded above]] as $T$ is a [[Definition:Subset|subset]] of $S$.
+Accordingly, $T$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]] by the [[Continuum Property]] as $T$ is [[Definition:Non-Empty Set|non-empty]].
+We know that $\sup T$ and $\sup S$ exist.
+Therefore by [[Suprema of two Real Sets]]:
+:$\forall \epsilon \in \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon \iff \sup T \le \sup S$
+We have:
+{{begin-eqn}}
+{{eqn | l = \forall \epsilon
+      | o = \in
+      | r = \R_{>0}: 0 < \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon
+      | o = \in
+      | r = \R_{>0}: \forall t \in T: t < t + \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon
+      | o = \in
+      | r = \R_{>0}: \forall t \in T: t < s + \epsilon \land s = t
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon
+      | o = \in
+      | r = \R_{>0}: \forall t \in T: \exists s \in S: t < s + \epsilon
+      | c = as $T \subseteq S$
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = \le
+      | r = \sup S
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition $\sup S$ is an [[Definition:Upper Bound|upper bound]] for $S$.
+Thus:
+:$\forall x \in S: x \le \sup S$
+As $T \subseteq S$ we have by definition of [[Definition:Subset|subset]] that:
+:$\forall x \in T: x \in S$
+Hence:
+:$\forall x \in T: x \le \sup S$
+So by definition $\sup S$ is an [[Definition:Upper Bound|upper bound]] for $T$.
+So $\sup S$ is at least as big as the [[Definition:Smallest Element|smallest]] [[Definition:Upper Bound|upper bound]] for $T$
+Thus by definition of [[Definition:Supremum|supremum]]:
+:$\sup T \le \sup S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area of Triangle}
+Tags: Areas of Triangles
+
+\begin{theorem}
+This page gathers a variety of formulas for the [[Definition:Area|area]] of a [[Definition:Triangle (Geometry)|triangle]].
+\end{theorem}
+
+\begin{proof}
+:[[File:TriangleAreaTwoSidesAngle.png|420px]]
+{{begin-eqn}} 
+{{eqn | l = \map \Area {ABC}
+      | r = \frac 1 2 h c
+      | c = [[Area of Triangle in Terms of Side and Altitude]]
+}}
+{{eqn | r = \frac 1 2 h \paren {p + q}
+}}
+{{eqn | r = \frac 1 2 a b \paren {\frac p a \frac h b + \frac h a \frac q b}
+}}
+{{eqn | r = \frac 1 2 a b \paren {\sin \alpha \cos \beta + \cos \alpha \sin \beta}
+      | c = {{Defof|Sine of Angle}} and {{Defof|Cosine of Angle}}
+}}
+{{eqn | r = \frac 1 2 a b \, \map \sin {\alpha + \beta}
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \frac 1 2 a b \sin C
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:TriangleAreaTwoSidesAngle-2.png|400px]]
+By definition of [[Definition:Sine|sine]]:
+:$h = b \sin C$
+From [[Area of Triangle in Terms of Side and Altitude]]:
+:$\map \Area {ABC} = \dfrac {a h} 2$
+Substituting:
+:$\map \Area {ABC} = \dfrac {a b \sin C} 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersecting Chord Theorem}
+Tags: Circles, Named Theorems, Intersecting Chord Theorem
+
+\begin{theorem}
+Let $AC$ and $BD$ both be [[Definition:Chord of Circle|chords]] of the same [[Definition:Circle|circle]].
+Let $AC$ and $BD$ [[Definition:Intersection (Geometry)|intersect]] at $E$.
+Then $AE \cdot EC = DE \cdot EB$.
+{{EuclidSaid}}
+:''If in a [[Definition:Circle|circle]] two [[Definition:Straight Line|straight lines]] cut one another, the [[Definition:Containment of Rectangle|rectangle contained]] by the [[Definition:Line Segment|segments]] of the one is equal to the [[Definition:Containment of Rectangle|rectangle contained]] by the [[Definition:Line Segment|segments]] of the other.''
+{{EuclidPropRef|III|35}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-III-35.png|300px]]
+Let $AC$ and $BD$ be intersecting [[Definition:Chord of Circle|chords]] of [[Definition:Circle|circle]] $ABCD$.
+Let the point of intersection be $E$.
+If $E$ is the [[Definition:Center of Circle|center]] of $ABCD$ the solution is trivial, as $AE = EC = BE = ED$ and so $AE \cdot EC = BE \cdot ED$.
+Otherwise, let $F$ be the [[Definition:Center of Circle|center]] of $ABCD$.
+[[Perpendicular through Given Point|Let $FG$ be drawn perpendicular]] to $AC$, and [[Perpendicular through Given Point|$FH$ be drawn perpendicular]] to $BD$.
+From [[Conditions for Diameter to be Perpendicular Bisector]], $G$ [[Definition:Bisect|bisects]] $AC$ and $H$ [[Definition:Bisect|bisects]] $BD$.
+So $AG = GC$ and $BH = HD$.
+From [[Difference of Two Squares]] we have that $AE \cdot EC + EG^2 = GC^2$.
+Let us add $GF^2$ to these.
+So $AE \cdot EC + EG^2 + GF^2 = GC^2 + GF^2$.
+But from [[Pythagoras's Theorem]] we have that:
+: $GC^2 + GF^2 = CF^2$
+: $EG^2 + GF^2 = EF^2$
+So:
+: $AE \cdot EC + EF^2 = CF^2$
+Using the same construction, we have that:
+: $DE \cdot EB + EF^2 = BF^2$
+But $BF = CF$ as both are the [[Definition:Radius of Circle|radius]] of the [[Definition:Circle|circle]] $ABCD$.
+That gives us:
+: $AE \cdot EC + EF^2 = DE \cdot EB + EF^2$
+It follows that $AE \cdot EC = DE \cdot EB$
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Join $A$ with $B$ and $C$ with $D$, as shown in this diagram:
+:[[File:Euclid-III-35-2.png|310px]]
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \angle AEB
+      | o = \cong
+      | r = \angle DEC
+      | c = [[Two Straight Lines make Equal Opposite Angles]]
+}}
+{{eqn | l = \angle BAE
+      | o = \cong
+      | r = \angle CDE
+      | c = [[Angles in Same Segment of Circle are Equal]]
+}}
+{{end-eqn}}
+By [[Triangles with Two Equal Angles are Similar]] we have $\triangle AEB \sim \triangle DEC$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {AE} {EB}
+      | r = \frac {DE} {EC}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = AE \cdot EC
+      | r = DE \cdot EB
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Plus Constant}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers $\R$]].
+Let $S$ be [[Definition:Bounded Above Set|bounded above]].
+Let $\xi \in \R$.
+Then:
+:$\displaystyle \map {\sup_{x \mathop \in S} } {x + \xi} = \xi + \map {\sup_{x \mathop \in S} } x$
+where $\sup$ denotes [[Definition:Supremum of Set|supremum]].
+\end{theorem}
+
+\begin{proof}
+Let $B = \sup S$.
+Let $T = \set {x + \xi: x \in S}$.
+Since $\forall x \in S: x \le B$ it follows that:
+:$\forall x \in S: x + \xi \le B + \xi$
+Hence $\xi + B$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$.
+If $C$ is the [[Definition:Supremum of Set|supremum]] for $T$ then $C \le \xi + B$.
+On the other hand:
+:$\forall y \in T: y \le C$
+Therefore:
+:$\forall y \in T: y - \xi \le C - \xi$
+Since $S = \set {y - \xi: y \in T}$ it follows that $C - \xi$ is an [[Definition:Upper Bound of Set|upper bound]] for $S$ and so $B \le C - \xi$.
+So we have shown that $C \le \xi + B$ and $C \ge \xi + B$, hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Suprema and Infima of Combined Bounded Functions}
+Tags: Analysis
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]].
+Let $c$ be a constant.
+\end{theorem}
+
+\begin{proof}
+=== Proof for Bounded Above ===
+First we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + c} = c + \map {\sup_{x \mathop \in S} } {\map f x}$:
+Let $T = \set {\map f x: x \in S}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\sup_{x \mathop \in S} } {\map f x + c}
+      | r = \map {\sup_{y \mathop \in T} } {y + c}
+      | c = 
+}}
+{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
+      | c = [[Supremum Plus Constant]]
+}}
+{{eqn | r = c + \map {\sup_{x \mathop \in S} } {\map f x}
+      | c = 
+}}
+{{end-eqn}}
+Next we show that $\displaystyle \map {\sup_{x \mathop \in S} } {\map f x + \map g x} \le \map {\sup_{x \mathop \in S} } {\map f x} + \map {\sup_{x \mathop \in S} } {\map g x}$:
+Let:
+:$\displaystyle H = \map {\sup_{x \mathop \in S} } {\map f x}$
+:$\displaystyle K = \map {\sup_{x \mathop \in S} } {\map g x}$
+Then:
+:$\forall x \in S: \map f x + \map g x \le H + K$
+Hence $H + K$ is an [[Definition:Upper Bound of Real-Valued Function|upper bound]] for $\set {\map f x + \map g x: x \in S}$.
+The result follows.
+{{qed}}
+=== Proof for Bounded Below ===
+This follows exactly the same lines.
+First we show that:
+:$\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + c} = c + \map {\inf_{x \mathop \in S} } {\map f x}$
+Let $T = \set {\map f x: x \in S}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\inf_{x \mathop \in S} } {\map f x + c}
+      | r = \map {\inf_{y \mathop \in T} } {y + c}
+      | c = 
+}}
+{{eqn | r = c + \map {\sup_{y \mathop \in T} } y
+      | c = [[Infimum Plus Constant]]
+}}
+{{eqn | r = c + \map {\inf_{x \mathop \in S} } {\map f x}
+      | c = 
+}}
+{{end-eqn}}
+Next we show that:
+:$\displaystyle \map {\inf_{x \mathop \in S} } {\map f x + \map g x} \ge \map {\inf_{x \mathop \in S} } {\map f x} + \map {\inf_{x \mathop \in S} } {\map g x}$
+Let:
+:$\displaystyle H = \map {\inf_{x \mathop \in S} } {\map f x}$
+:$\displaystyle K = \map {\inf_{x \mathop \in S} } {\map g x}$
+Then:
+:$\forall x \in S: \map f x + g \map x \ge H + K$
+Hence $H + K$ is a [[Definition:Lower Bound of Real-Valued Function|lower bound]] for $\set {\map f x + \map g x: x \in S}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiple of Infimum}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $T \subseteq \R: T \ne \varnothing$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers $\R$]].
+Let $T$ be [[Definition:Bounded Below Set|bounded below]].
+Let $z \in \R: z > 0$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then:
+:$\displaystyle \map {\inf_{x \mathop \in T} } {z x} = z \, \map {\inf_{x \mathop \in T} } x$
+where $\inf$ denotes [[Definition:Infimum of Set|infimum]].
+\end{theorem}
+
+\begin{proof}
+From [[Negative of Infimum is Supremum of Negatives]]:
+:$\displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$
+Let $S = \set {x \in \R: -x \in T}$.
+From [[Negative of Infimum is Supremum of Negatives]], $S$ is [[Definition:Bounded Above Set|bounded above]].
+From [[Multiple of Supremum]]:
+:$\displaystyle \map {\sup_{x \mathop \in S} } {z x} = z \, \map {\sup_{x \mathop \in S} } x$
+Hence:
+:$\displaystyle \map {\inf_{x \mathop \in T} } {z x} = -\map {\sup_{x \mathop \in T} } {-z x} = -z \, \map {\sup_{x \mathop \in T} } {-x} = z \, \map {\inf_{x \mathop \in T} } x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Law of Sines}
+Tags: Triangles, Sine Function, Named Theorems, Law of Sines
+
+\begin{theorem}
+For any [[Definition:Triangle (Geometry)|triangle]] $\triangle ABC$:
+:$\dfrac a {\sin A} = \dfrac b {\sin B} = \dfrac c {\sin C} = 2 R$
+where:
+:$a$, $b$, and $c$ are the [[Definition:Opposite (in Triangle)|sides opposite]] $A$, $B$ and $C$ respectively
+:$R$ is the [[Definition:Circumradius of Triangle|circumradius]] of $\triangle ABC$.
+\end{theorem}
+
+\begin{proof}
+Construct the [[Definition:Altitude of Triangle|altitude]] from $B$.
+:[[File:Law Of Sines 1.png|200px]]
+It can be seen from the [[Definition:Sine of Angle|definition of sine]] that:
+: $\sin A = \dfrac h c$ and $\sin C = \dfrac h a$
+Thus:
+: $h = c \sin A$ and $h = a \sin C$
+This gives:
+: $c \sin A = a \sin C$
+So:
+: $\dfrac a {\sin A} = \dfrac c {\sin C}$
+Similarly, constructing the altitude from $A$ gives:
+:$\dfrac b {\sin B} = \dfrac c {\sin C}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Construct the [[Definition:Circumcircle of Triangle|circumcircle]] of $\triangle ABC$, let $O$ be the [[Definition:Circumcenter of Triangle|circumcenter]] and $R$ be the [[Definition:Circumradius of Triangle|circumradius]].
+Construct $\triangle AOB$ and let $E$ be the foot of the [[Definition:Altitude of Triangle|altitude]] of $\triangle AOB$ from $O$.
+:[[File:Law-of-sines.png|350px]]
+By the [[Inscribed Angle Theorem]]:
+:$\angle ACB = \dfrac {\angle AOB} 2$
+From the definition of the [[Definition:Circumcenter of Triangle|circumcenter]]:
+:$AO = BO$
+From the definition of [[Definition:Altitude of Triangle|altitude]] and the fact that [[Axiom:Euclid's Fourth Postulate|all right angles are congruent]]:
+:$\angle AEO = \angle BEO$
+Therefore from [[Pythagoras's Theorem]]:
+:$AE = BE$
+and then from [[Triangle Side-Side-Side Equality]]:
+:$\angle AOE = \angle BOE$
+Thus:
+:$\angle AOE = \dfrac {\angle AOB} 2$
+and so:
+:$\angle ACB = \angle AOE$
+Then by the [[Definition:Sine of Angle|definition of sine]]:
+:$\sin C = \map \sin {\angle AOE} = \dfrac {c / 2} R$
+and so:
+:$\dfrac c {\sin C} = 2 R$
+The same argument holds for all three [[Definition:Internal Angle|angles]] in the [[Definition:Triangle (Geometry)|triangle]], and so:
+:$\dfrac c {\sin C} = \dfrac b {\sin B} = \dfrac a {\sin A} = 2 R$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin b \sin c \cos A
+      | r = \cos a - \cos b \cos c
+      | c = [[Spherical Law of Cosines]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sin^2 b \sin^2 c \cos^2 A
+      | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sin^2 b \sin^2 c \paren {1 - \sin^2 A}
+      | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A
+      | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
+      | c = multiplying out
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A
+      | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A
+      | r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
+      | c = multiplying out
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \sin^2 b \sin^2 c \sin^2 A
+      | r = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c
+      | c = rearranging and simplifying
+}}
+{{end-eqn}}
+Let $X \in \R_{>0}$ such that:
+:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
+Then from $(1)$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}
+      | o = =
+      | r = \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = X^2
+      | r = \dfrac {\sin^2 A} {\sin^2 a}
+      | c = 
+}}
+{{end-eqn}}
+In a [[Definition:Spherical Triangle|spherical triangle]], all of the [[Definition:Side of Spherical Triangle|sides]] are less than $\pi$ [[Definition:Radian|radians]].
+The same applies to the [[Definition:Spherical Angle|angles]].
+From [[Shape of Sine Function]]:
+:$\sin \theta > 0$ for all $0 < \theta < \pi$
+Hence the [[Definition:Negative Square Root|negative root]] of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:
+:$X = \dfrac {\sin A} {\sin a}$
+Similarly, from applying the [[Spherical Law of Cosines]] to $\cos B$ and $\cos C$:
+{{begin-eqn}}
+{{eqn | l = \sin a \sin c \cos B
+      | r = \cos b - \cos a \cos c
+}}
+{{eqn | l = \sin a \sin b \cos C
+      | r = \cos c - \cos a \cos b
+}}
+{{end-eqn}}
+we arrive at the same point:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = \dfrac {\sin B} {\sin b}
+}}
+{{eqn | r = \dfrac {\sin A} {\sin a}
+}}
+{{end-eqn}}
+where:
+:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
+as before.
+Hence we have:
+:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:Spherical-Cosine-Formula-2.png|500px]]
+Let $A$, $B$ and $C$ be the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Spherical Triangle|spherical triangle]] on the surface of a [[Definition:Sphere (Geometry)|sphere]] $S$.
+By definition of a [[Definition:Spherical Triangle|spherical triangle]], $AB$, $BC$ and $AC$ are [[Definition:Arc of Circle|arcs]] of [[Definition:Great Circle|great circles]] on $S$.
+By definition of a [[Definition:Great Circle|great circle]], the [[Definition:Center of Circle|center]] of each of these [[Definition:Great Circle|great circles]] is $O$.
+Let $O$ be joined to each of $A$, $B$ and $C$.
+Let $P$ be an arbitrary [[Definition:Point|point]] on $OC$.
+Construct $PQ$ [[Definition:Perpendicular|perpendicular]] to $OA$ meeting $OA$ at $Q$.
+Construct $PR$ [[Definition:Perpendicular|perpendicular]] to $OB$ meeting $OB$ at $R$.
+In the [[Definition:Plane|plane]] $OAB$:
+:construct $QS$ [[Definition:Perpendicular|perpendicular]] to $OA$
+:construct $RS$ [[Definition:Perpendicular|perpendicular]] to $OB$
+where $S$ is the [[Definition:Point|point]] where $QS$ and $RS$ [[Definition:Intersection (Geometry)|intersect]].
+Let $OS$ and $PS$ be joined.
+Let [[Definition:Tangent Line|tangents]] be constructed at $A$ to the [[Definition:Arc of Circle|arcs]] of the [[Definition:Great Circle|great circles]] $AC$ and $AB$.
+These [[Definition:Tangent Line|tangents]] [[Definition:Containment of Angle|contain]] the [[Definition:Spherical Angle|spherical angle]] $A$.
+But by construction, $QS$ and $QP$ are [[Definition:Parallel Lines|parallel]] to these [[Definition:Tangent Line|tangents]]
+Hence $\angle PQS = \sphericalangle A$.
+Similarly, $\angle PRS = \sphericalangle B$.
+Also we have:
+{{begin-eqn}}
+{{eqn | l = \angle COB
+      | r = a
+}}
+{{eqn | l = \angle COA
+      | r = b
+}}
+{{eqn | l = \angle AOB
+      | r = c
+}}
+{{end-eqn}}
+It is to be proved that $PS$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $AOB$.
+By construction, $OQ$ is [[Definition:Perpendicular|perpendicular]] to both $PQ$ and $QS$.
+Thus $OQ$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $PQS$.
+Similarly, $OR$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $PRS$.
+Thus $PS$ is [[Definition:Perpendicular|perpendicular]] to both $OQ$ and $OR$.
+Thus $PS$ is [[Definition:Perpendicular|perpendicular]] to every [[Definition:Straight Line|line]] in the [[Definition:Plane|plane]] of $OQ$ and $OR$.
+That is, $PS$ is [[Definition:Line Perpendicular to Plane|perpendicular]] to the [[Definition:Plane|plane]] $OAB$.
+In particular, $PS$ is [[Definition:Perpendicular|perpendicular]] to $OS$, $SQ$ and $SR$
+It follows that $\triangle PQS$ and $\triangle PRS$ are [[Definition:Right Triangle|right triangles]].
+From the [[Definition:Right Triangle|right triangles]] $\triangle OQP$ and $\triangle ORP$, we have:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = PQ
+      | r = OP \sin b
+}}
+{{eqn | n = 2
+      | l = PR
+      | r = OP \sin a
+}}
+{{eqn | n = 3
+      | l = OQ
+      | r = OP \cos b
+}}
+{{eqn | n = 4
+      | l = OR
+      | r = OP \cos a
+}}
+{{end-eqn}}
+From the [[Definition:Right Triangle|right triangles]] $\triangle PQS$ and $\triangle PRS$, we have:
+{{begin-eqn}}
+{{eqn | l = PS
+      | r = PS \sin \angle PRS
+}}
+{{eqn | r = PQ \sin A
+}}
+{{eqn | l = PS
+      | r = PR \sin \angle PRS
+}}
+{{eqn | r = PR \sin B
+}}
+{{eqn | ll= \leadsto
+      | l = OP \sin b \sin A
+      | r = OP \sin a \sin B
+      | c = from $(1)$ and $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\sin a} {\sin A}
+      | r = \dfrac {\sin b} {\sin B}
+      | c = 
+}}
+{{end-eqn}}
+The result follows by applying this technique [[Definition:Mutatis Mutandis|mutatis mutandis]] to the other [[Definition:Spherical Angle|angles]] of $ABC$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Supremum is Infimum of Negatives}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] $\R$.
+Let $S$ be [[Definition:Bounded Above Set|bounded above]].
+Then:
+:$(1): \quad \set {x \in \R: -x \in S}$ is [[Definition:Bounded Below Set|bounded below]]
+:$(2): \quad \displaystyle -\sup_{x \mathop \in S} x = \map {\inf_{x \mathop \in S} } {-x}$
+where $\sup$ and $\inf$ denote the [[Definition:Supremum of Subset of Real Numbers|supremum]] and [[Definition:Infimum of Subset of Real Numbers|infimum]] respectively.
+\end{theorem}
+
+\begin{proof}
+Let $B = \sup S$.
+Let $T = \set {x \in \R: -x \in S}$.
+Since $\forall x \in S: x \le B$ it follows that $\forall x \in S: -x \ge -B$.
+Hence $-B$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$.
+Thus $\set {x \in \R: -x \in S}$ is [[Definition:Bounded Below Set|bounded below]].
+If $C$ is the [[Definition:Infimum of Subset of Real Numbers|infimum]] of $T$, it follows that $C \ge -B$.
+On the other hand:
+:$\forall y \in T: y \ge C$
+Therefore:
+:$\forall y \in T: -y \le -C$
+Since $S = \set {x \in \R: -x \in T}$ it follows that $-C$ is an [[Definition:Upper Bound of Set|upper bound]] for $S$.
+Therefore $-C \ge B$ and so $C \le -B$.
+So $C \le -B$ and $C \ge -B$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Infimum is Supremum of Negatives}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $T$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] $\R$.
+Let $T$ be [[Definition:Bounded Below Set|bounded below]].
+Then:
+: $(1): \quad \set {x \in \R: -x \in T}$ is [[Definition:Bounded Above Set|bounded above]]
+: $(2): \quad \displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x}$
+where $\sup$ and $\inf$ denote the [[Definition:Supremum of Subset of Real Numbers|supremum]] and [[Definition:Infimum of Subset of Real Numbers|infimum]] respectively.
+\end{theorem}
+
+\begin{proof}
+As $T$ is [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Below Set|bounded below]], it follows by the [[Continuum Property]] that $T$ admits an [[Definition:Infimum of Subset of Real Numbers|infimum]].
+Let $B = \inf T$.
+Let $S = \set {x \in \R: -x \in T}$.
+Since $\forall x \in T: x \ge B$ it follows that:
+:$\forall x \in T: -x \le -B$
+Hence $-B$ is an upper bound for $S$, and so $\set {x \in \R: -x \in T}$ is [[Definition:Bounded Above Set|bounded above]].
+If $C$ is the [[Definition:Supremum of Subset of Real Numbers|supremum]] of $S$, it follows that $C \le -B$.
+On the other hand:
+:$\forall y \in S: y \le C$
+Therefore:
+:$\forall y \in S: -y \ge -C$
+Since $T = \set {x \in \R: -x \in S}$ it follows that $-C$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] for $T$.
+Therefore $-C \le B$ and so $C \ge -B$.
+So $C \ge -B$ and $C \le -B$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum Plus Constant}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $T$ be a [[Definition:Subset|subset]] of the set of [[Definition:Real Number|real numbers]].
+Let $T$ be [[Definition:Bounded Below Set|bounded below]].
+Let $\xi \in \R$.
+Then:
+:$\displaystyle \map {\inf_{x \mathop \in T} } {x + \xi} = \xi + \map {\inf_{x \mathop \in T} } x$
+where $\inf$ denotes [[Definition:Infimum of Set|infimum]].
+\end{theorem}
+
+\begin{proof}
+From [[Negative of Infimum is Supremum of Negatives]], we have that:
+:$\displaystyle -\inf_{x \mathop \in T} x = \map {\sup_{x \mathop \in T} } {-x} \implies \inf_{x \mathop \in T} x = -\map {\sup_{x \mathop \in T} } {-x}$
+Let $S = \set {x \in \R: -x \in T}$.
+From [[Negative of Infimum is Supremum of Negatives]], $S$ is [[Definition:Bounded Above Set|bounded above]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {\sup_{x \mathop \in S} } {x + \xi}
+      | r = \xi + \map {\sup_{x \mathop \in S} } x
+      | c = [[Supremum Plus Constant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\inf_{x \mathop \in T} } {x + \xi}
+      | r = -\map {\sup_{x \mathop \in T} } {-x + \xi}
+      | c = 
+}}
+{{eqn | r = \xi - \map {\sup_{x \mathop \in T} } {-x}
+      | c = 
+}}
+{{eqn | r = \xi + \map {\inf_{x \mathop \in T} } x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Subset}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {U, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $S \subseteq U$.
+Let $T \subseteq S$.
+Let $\struct {S, \preceq}$ admit an [[Definition:Infimum of Set|infimum]] in $U$.
+If $T$ also admits an [[Definition:Infimum of Set|infimum]] in $U$, then $\map \inf S \preceq \map \inf T$.
+\end{theorem}
+
+\begin{proof}
+Let $B = \map \inf S$.
+Then $B$ is a [[Definition:Lower Bound of Set|lower bound]] for $S$.
+As $T \subseteq S$, it follows by the definition of a [[Definition:Subset|subset]] that $x \in T \implies x \in S$.
+Because $x \in S \implies B \preceq x$ (as $B$ is a [[Definition:Lower Bound of Set|lower bound]] for $S$) it follows that $x \in T \implies B \preceq x$.
+So $B$ is a [[Definition:Lower Bound of Set|lower bound]] for $T$.
+Therefore $B$ [[Definition:Ordering|precedes]] the [[Definition:Infimum of Set|infimum]] of $T$ in $U$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Equilateral Triangle}
+Tags: Equilateral Triangles
+
+\begin{theorem}
+On a given [[Definition:Line Segment|straight line segment]], it is possible to construct an [[Definition:Equilateral Triangle|equilateral triangle]].
+{{:Euclid:Proposition/I/1}}
+\end{theorem}
+
+\begin{proof}
+As $A$ is the [[Definition:Center of Circle|center]] of circle $BCD$, it follows from {{EuclidDefLink|I|15|Circle}} that $AC = AB$.
+As $B$ is the [[Definition:Center of Circle|center]] of circle $ACE$, it follows from {{EuclidDefLink|I|15|Circle}} that $BC = AB$.
+So, as $AC = AB$ and $BC = AB$, it follows from {{EuclidCommonNotionLink|1}} that $AC = BC$.
+Therefore $AB = AC = BC$.
+Therefore $\triangle ABC$ is [[Definition:Equilateral Triangle|equilateral]].
+{{qed}}
+{{Euclid Note|1|I}}
+\end{proof}<|endoftext|>
+\section{Construction of Equal Straight Line}
+Tags: Lines
+
+\begin{theorem}
+At a given [[Definition:Point|point]], it is possible to construct a [[Definition:Line Segment|straight line segment]] of [[Definition:Length (Linear Measure)|length]] equal to that of any given [[Definition:Line Segment|straight line segment]].
+The given [[Definition:Point|point]] will be an [[Definition:Endpoint of Line|endpoint]] of the constructed [[Definition:Line Segment|straight line segment]].
+{{:Euclid:Proposition/I/2}}
+\end{theorem}
+
+\begin{proof}
+As $B$ is the [[Definition:Center of Circle|center]] of circle $CGH$, it follows from {{EuclidDefLink|I|15|Circle}} that $BC = BG$.
+As $D$ is the [[Definition:Center of Circle|center]] of circle $GKL$, it follows from {{EuclidDefLink|I|15|Circle}} that $DL = DG$.
+As $\triangle ABD$ is an [[Definition:Equilateral Triangle|equilateral triangle]], it follows that $DA = DB$.
+Therefore, by [[Axiom:Euclid's Common Notions|Common Notion 3]], $AL = BG$.
+As $AL = BG$ and $BC = BG$, it follows from [[Axiom:Euclid's Common Notions|Common Notion 1]] that $AL = BC$.
+Therefore, at the given point $A$, the required [[Definition:Line Segment|straight line segment]] $AL$ has been placed equal in length to $BC$.
+{{qed}}
+{{Euclid Note|2|I}}
+\end{proof}<|endoftext|>
+\section{Construction of Equal Straight Lines from Unequal}
+Tags: Lines
+
+\begin{theorem}
+Given two unequal [[Definition:Line Segment|straight line segments]], it is possible to cut off from the greater a [[Definition:Line Segment|straight line segment]] equal to the lesser.
+{{:Euclid:Proposition/I/3}}
+\end{theorem}
+
+\begin{proof}
+As $A$ is the [[Definition:Center of Circle|center]] of circle $DEF$, it follows from {{EuclidDefLink|I|15|Circle}} that $AE = AD$.
+But $C$ is also equal to $AD$.
+So, as $C = AD$ and $AD = AE$, it follows from [[Axiom:Euclid's Common Notions|Common Notion 1]] that $AE = C$.
+Therefore, given the two [[Definition:Line Segment|straight line segments]] $AB$ and $C$, from the greater of these $AB$, a length $AE$ has been cut off equal to the lesser $C$.
+{{qed}}
+{{Euclid Note|3|I}}
+\end{proof}<|endoftext|>
+\section{Triangle Side-Angle-Side Equality}
+Tags: Triangles
+
+\begin{theorem}
+If $2$ [[Definition:Triangle (Geometry)|triangles]] have:
+: $2$ [[Definition:Side of Polygon|sides]] equal to $2$ [[Definition:Side of Polygon|sides]] respectively
+: the [[Definition:Angle|angles]] [[Definition:Containment of Angle|contained]] by the equal [[Definition:Straight Line Segment|straight lines]] equal
+they will also have:
+: their third [[Definition:Side of Polygon|sides]] equal
+: the remaining two [[Definition:Angle|angles]] equal to their respective remaining [[Definition:Angle|angles]], namely, those which the equal [[Definition:Side of Polygon|sides]] [[Definition:Subtend|subtend]].
+\end{theorem}
+
+\begin{proof}
+[[File:Euclid-I-4.png|500px]]
+Let $\triangle ABC$ and $\triangle DEF$ be $2$ [[Definition:Triangle (Geometry)|triangles]] having [[Definition:Side of Polygon|sides]] $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$.
+If $\triangle ABC$ is placed on $\triangle DEF$ such that:
+: the [[Definition:Point|point]] $A$ is placed on [[Definition:Point|point]] $D$, and 
+: the [[Definition:Straight Line Segment|line]] $AB$ is placed on [[Definition:Straight Line Segment|line]] $DE$
+then the [[Definition:Point|point]] $B$ will also coincide with [[Definition:Point|point]] $E$ because $AB = DE$.
+So, with $AB$ coinciding with $DE$, the [[Definition:Straight Line Segment|line]] $AC$ will coincide with the [[Definition:Straight Line Segment|line]] $DF$ because $\angle BAC = \angle EDF$.
+Hence the [[Definition:Point|point]] $C$ will also coincide with the [[Definition:Point|point]] $F$, because $AC = DF$.
+But $B$ also coincided with $E$.
+Hence the [[Definition:Straight Line Segment|line]] $BC$ will coincide with line $EF$.
+(Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two [[Definition:Straight Line Segment|straight lines]] will enclose a [[Definition:Region of Plane|region]] which is impossible.)
+Therefore $BC$ will coincide with $EF$ and be equal to it.
+Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$.
+The remaining [[Definition:Angle|angles]] on $\triangle ABC$ will coincide with the remaining [[Definition:Angle|angles]] on $\triangle DEF$ and be equal to them.
+{{qed}}
+{{Euclid Note|4|I}}
+\end{proof}<|endoftext|>
+\section{Isosceles Triangle has Two Equal Angles}
+Tags: Isosceles Triangles
+
+\begin{theorem}
+In [[Definition:Isosceles Triangle|isosceles triangles]], the [[Definition:Angle|angles]] at the [[Definition:Base of Isosceles Triangle|base]] are equal to each other.
+Also, if the equal [[Definition:Line Segment|straight lines]] are extended, the [[Definition:Angle|angles]] under the [[Definition:Base of Isosceles Triangle|base]] will also be equal to each other.
+{{:Euclid:Proposition/I/5}}
+\end{theorem}
+
+\begin{proof}
+[[File:Euclid-I-5.png|200px]]
+Let $\triangle ABC$ be an [[Definition:Isosceles Triangle|isosceles triangle]] whose side $AB$ equals side $AC$.
+We [[Axiom:Euclid's Second Postulate|extend the straight lines]] $AB$ and $AC$ to $D$ and $E$ respectively.
+Let $F$ be a point on $BD$.
+We [[Construction of Equal Straight Lines from Unequal|cut off from $AE$ a length $AG$]] equal to $AF$.
+We [[Axiom:Euclid's First Postulate|draw line segments]] $FC$ and $GB$.
+Since $AF = AG$ and $AB = AC$, the two sides $FA$ and $AC$ are equal to $GA$ and $AB$ respectively.
+They [[Definition:Containment of Angle|contain]] a common [[Definition:Angle|angle]], that is, $\angle FAG$.
+So by [[Triangle Side-Angle-Side Equality]], $\triangle AFC = \triangle AGB$.
+Thus $FC = GB$, $\angle ACF = \angle ABG$ and $\angle AFC = \angle AGB$.
+Since $AF = AG$ and $AB = AC$, then $BF = CG$.
+But $FC = GB$, so the two sides $BF, FC$ are equal to the two sides $CG, GB$ respectively.
+Then $\angle BFC = \angle CGB$ while $CB$ is common to both.
+Therefore by [[Triangle Side-Angle-Side Equality]], $\triangle BFC = \triangle CGB$.
+Therefore $\angle FBC = \angle GCB$ and $\angle BCF = \angle CBG$.
+So since $\angle ACF = \angle ABG$, and in these $\angle BCF = \angle CBG$, then $\angle ABC = \angle ACB$.
+But $\angle ABC$ and $\angle ACB$ are at the base of $\triangle ABC$.
+Also, we have already proved that $\angle FBC = \angle GCB$, and these are the angles under the base of $\triangle ABC$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Triangle with Two Equal Angles is Isosceles}
+Tags: Isosceles Triangles, Triangle with Two Equal Angles is Isosceles
+
+\begin{theorem}
+If a [[Definition:Triangle (Geometry)|triangle]] has two [[Definition:Angle|angles]] equal to each other, the [[Definition:Side of Polygon|sides]] which [[Definition:Subtend|subtend]] the equal [[Definition:Angle|angles]] will also be equal to one another.
+Hence, by definition, such a triangle will be [[Definition:Isosceles Triangle|isosceles]].
+{{:Euclid:Proposition/I/6}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-I-6.png|200px]]
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]] in which $\angle ABC = \angle ACB$.
+Suppose side $AB$ is not equal to side $AC$. Then one of them will be greater.
+{{WLOG}}, Suppose $AB > AC$.
+We [[Construction of Equal Straight Lines from Unequal|cut off from $AB$ a length $DB$]] equal to $AC$.
+We [[Axiom:Euclid's First Postulate|draw the line segment]] $CD$.
+Since $DB = AC$, and $BC$ is common, the two sides $DB, BC$ are equal to $AC, CB$ respectively.
+Also, $\angle DBC = \angle ACB$.
+So by [[Triangle Side-Angle-Side Equality]], $\triangle DBC = \triangle ACB$.
+But $\triangle DBC$ is smaller than $\triangle ACB$, which is absurd.
+Therefore, have $AB \le AC$. 
+A similar argument shows the converse, and hence $AB = AC$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\angle ABC$ and $\angle ACB$ be the [[Definition:Angle|angles]] that are the same.
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \angle ABC
+      | r = \angle ACB
+      | c = Given
+}}
+{{eqn | n = 2
+      | l = BC
+      | r = CB
+      | c = [[Equality is Reflexive]]
+}}
+{{eqn | n = 3
+      | l = \angle ACB
+      | r = \angle ABC
+      | c = Given
+}}
+{{eqn | n = 4
+      | l = \triangle ABC
+      | r = \triangle ACB
+      | c = [[Triangle Angle-Side-Angle Equality]] by $\left({1}\right)$, $\left({2}\right)$ and $\left({3}\right)$
+}}
+{{eqn | n = 5
+      | l = AB
+      | r = AC
+      | c = from $\left({4}\right)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two Lines Meet at Unique Point}
+Tags: Lines
+
+\begin{theorem}
+Let two [[Definition:Line Segment|straight line segments]] be constructed on a [[Definition:Line Segment|straight line segment]] from its [[Definition:Endpoint of Line|endpoints]] so that they meet at a [[Definition:Point|point]].
+Then there cannot be two other [[Definition:Line Segment|straight line segments]] equal to the former two respectively, constructed on the same [[Definition:Line Segment|straight line segment]] and on the same side of it, meeting at a different [[Definition:Point|point]].
+{{:Euclid:Proposition/I/7}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-I-7.png|400px]]
+Let $AC$ and $CB$ be constructed on $AB$ meeting at $C$.
+Let two other [[Definition:Line Segment|straight line segments]] $AD$ and $DB$ be constructed on $AB$, on the same side of it, meeting at $D$, such that $AC = AD$ and $CB = DB$.
+Suppose, with a view to obtaining a [[Definition:Contradiction|contradiction]], $C$ and $D$ are different points.
+Let $CD$ be joined.
+Since $AC = AD$ it [[Isosceles Triangle has Two Equal Angles|follows that $\angle ACD = \angle ADC$]].
+Therefore $\angle ACD$ is greater than $\angle DCB$ because [[Axiom:Euclid's Common Notions|the whole is greater than the part]].
+Therefore $\angle CDB$ is much greater than $\angle DCB$.
+Now since $CB = DB$, it follows that $\angle CDB = \angle DCB$.
+But it was proved much greater than it.
+[[Proof by Contradiction|From this contradiction]] it follows that $C$ and $D$ can not be different points.
+Hence the result.
+{{qed}}
+{{Euclid Note|7|I}}
+\end{proof}<|endoftext|>
+\section{Triangle Side-Side-Side Equality}
+Tags: Triangles
+
+\begin{theorem}
+Let two [[Definition:Triangle (Geometry)|triangles]] have all $3$ [[Definition:Side of Polygon|sides]] equal.
+Then they also have all $3$ [[Definition:Angle|angles]] equal.
+Thus two [[Definition:Triangle (Geometry)|triangles]] whose [[Definition:Side of Polygon|sides]] are all equal are themselves [[Definition:Congruence (Geometry)|congruent]].
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-I-8.png|500px]]
+Let $\triangle ABC$ and $\triangle DEF$ be two [[Definition:Triangle (Geometry)|triangles]] such that:
+: $AB = DE$
+: $AC = DF$
+: $BC = EF$
+Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the [[Definition:Side of Polygon|side]] $BC$ on $EF$.
+Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$.
+{{AimForCont}} $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$.
+Then they will fall as, for example, $EG$ and $GF$.
+Thus there will be two pairs of [[Definition:Line Segment|straight line segments]] constructed on the same [[Definition:Line Segment|line segment]], on the same side as it, meeting at different points.
+This [[Definition:Contradiction|contradicts]] the theorem [[Two Lines Meet at Unique Point]].
+Therefore $BA$ coincides with $ED$ and $AC$ coincides with $DF$.
+Therefore $\angle BAC$ coincides with $\angle EDF$ and is equal to it.
+The same argument can be applied to the other two [[Definition:Side of Polygon|sides]], and thus we show that all corresponding [[Definition:Angle|angles]] are equal.
+{{qed}}
+{{Euclid Note|8|I}}
+\end{proof}<|endoftext|>
+\section{Bisection of Angle}
+Tags: Angles
+
+\begin{theorem}
+It is possible to [[Definition:Bisection|bisect]] any given [[Definition:Rectilineal Angle|rectilineal angle]].
+{{:Euclid:Proposition/I/9}}
+\end{theorem}
+
+\begin{proof}
+We have:
+: $AD = AE$
+: $AF$ is common
+: $DF = EF$
+Thus [[Triangle Side-Side-Side Equality|triangles $\triangle ADF$ and $\triangle AEF$ are equal]].
+Thus $\angle DAF = \angle EAF$.
+Hence $\angle BAC$ has been [[Definition:Bisection|bisected]] by $AF$.
+{{qed}}
+{{Euclid Note|9|I|There are quicker and easier constructions of a bisection, but this particular one uses only results previously demonstrated.}}
+\end{proof}<|endoftext|>
+\section{Bisection of Straight Line}
+Tags: Lines
+
+\begin{theorem}
+It is possible to [[Definition:Bisection|bisect]] a [[Definition:Line Segment|straight line segment]].
+{{:Euclid:Proposition/I/10}}
+\end{theorem}
+
+\begin{proof}
+As $\triangle ABC$ is an [[Definition:Equilateral Triangle|equilateral triangle]], it follows that $AC = CB$.
+The two [[Definition:Triangle (Geometry)|triangles]] $\triangle ACD$ and $\triangle BCD$ have side $CD$ in common, and side $AC$ of $\triangle ACD$ equals side $BC$ of $\triangle BCD$.
+The angle $\angle ACD$ [[Definition:Subtend|subtended]] by lines $AC$ and $CD$ equals the angle $\angle BCD$ [[Definition:Subtend|subtended]] by lines $BC$ and $CD$, as $\angle ACB$ was [[Bisection of Angle|bisected]].
+So [[Triangle Side-Angle-Side Equality|triangles $\triangle ACD$ and $\triangle BCD$ are equal]].
+Therefore $AD = DB$.
+So $AB$ has been [[Definition:Bisection|bisected]] at the point $D$.
+{{qed}}
+{{Euclid Note|10|I}}
+\end{proof}<|endoftext|>
+\section{External Angle of Triangle Greater than Internal Opposite}
+Tags: Triangles
+
+\begin{theorem}
+The [[Definition:External Angle|external angle]] of a [[Definition:Triangle (Geometry)|triangle]] is greater than either of the [[Definition:Opposite (in Triangle)|opposite]] [[Definition:Internal Angle|internal angles]].
+{{:Euclid:Proposition/I/16}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-I-16.png|250px]]
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let the side $BC$ be [[Axiom:Euclid's Second Postulate|extended to $D$]].
+Let $AC$ be [[Bisection of Straight Line|bisected]] at $E$.
+Let $BE$ be [[Axiom:Euclid's First Postulate|joined]] and [[Axiom:Euclid's Second Postulate|extended to $F$]].
+Let $EF$ be made [[Construction of Equal Straight Lines from Unequal|equal to $BE$]].
+(Technically we really need to extend $BE$ to a point beyond $F$ and then crimp off a length $EF$.)
+Let $CF$ be [[Axiom:Euclid's First Postulate|joined]].
+Let $AC$ be [[Axiom:Euclid's Second Postulate|extended to $G$]].
+We have $\angle AEB = \angle CEF$ from [[Two Straight Lines make Equal Opposite Angles]].
+Since $AE = EC$ and $BE = EF$, from [[Triangle Side-Angle-Side Equality]] we have $\triangle ABE = \triangle CFE$.
+Thus $AB = CF$ and $\angle BAE = \angle ECF$.
+But $\angle ECD$ is greater than $\angle ECF$.
+Therefore $\angle ACD$ is greater than $\angle BAE$.
+Similarly, if $BC$ were bisected, $\angle BCG$, which is equal to $\angle ACD$ by [[Two Straight Lines make Equal Opposite Angles]], would be shown to be greater than $\angle ABC$ as well.
+Hence the result.
+{{qed}}
+{{Euclid Note|16|I}}
+\end{proof}<|endoftext|>
+\section{Limit iff Limits from Left and Right}
+Tags: Analysis, Limits of Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Open Real Interval|open interval]] $\openint a b$ except possibly at a point $c \in \openint a b$.
+Then:
+:$\map f x \to l$ as $x \to c$
+{{iff}}:
+:$\map f x \to l$ as $x \to c^-$
+and
+:$\map f x \to l$ as $x \to c^+$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\map f x \to l$ as $x \to c$.
+Then from the definition of the [[Definition:Limit of Real Function|limit of a function]]:
+:$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
+So for any given $\epsilon$, there exists a $\delta$ such that:
+:$0 < \size {x - c} < \delta$
+implies that:
+:$l - \epsilon < \map f x < l + \epsilon$
+Now:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 0 < \size {x - c} < \delta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = - \delta < -\paren {x - c} < 0
+      | c = 
+}}
+{{eqn | l = \lor
+      | o = 
+      | r = 0 < \paren {x - c} < \delta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = c - \delta < x < c
+      | c = 
+}}
+{{eqn | l = \lor
+      | o = 
+      | r = c < x < c + \delta
+      | c = 
+}}
+{{end-eqn}}
+That is: $\forall \epsilon > 0: \exists \delta > 0$:
+:$(1): \quad c - \delta < x < c \implies \norm {\map f x - l} < \epsilon$
+:$(2): \quad c < x < c + \delta \implies \norm {\map f x - l} < \epsilon$
+So given that particular value of $\epsilon$, we can find a value of $\delta$ such that the conditions for both:
+:$(1): \quad f$ tending to the limit $l$ as $x$ tends to $c$ [[Definition:Limit from Left|from the left]]
+and :
+:$(2): \quad f$ tending to the limit $l$ as $x$ tends to $c$ [[Definition:Limit from Right|from the right]].
+Thus:
+:$\displaystyle \lim_{x \mathop \to c} \map f x = l$
+implies that:
+:$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$
+and:
+:$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\map f x \to l$ as $x \to c^-$ and $\map f x \to l$ as $x \to c^+$.
+This means that:
+:$(1): \quad \forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \size {\map f x - l} < \epsilon$
+and :
+:$(2): \quad \forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \size {\map f x - l} < \epsilon$
+In the same manner as above, the conditions on $\delta$ give us that:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = c - \delta < x < c
+      | c = 
+}}
+{{eqn | l = \land
+      | o = 
+      | r = c < x < c + \delta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = 0 < \size {x - c} < \delta
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
+Thus:
+:$\displaystyle \lim_{x \mathop \to c^-} \map f x = l$
+and:
+:$\displaystyle \lim_{x \mathop \to c^+} \map f x = l$
+together imply that:
+:$\displaystyle \lim_{x \mathop \to c} \map f x = l$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Function by Convergent Sequences}
+Tags: Metric Spaces, Limits of Sequences
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $S \subseteq A_1$ be an [[Definition:Open Set (Metric Space)|open set]] of $M_1$.
+Let $f$ be a [[Definition:Mapping|mapping]] defined on $S$, except possibly at the point $c \in S$.
+Then $\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$ [[Definition:Iff|iff]]:
+: for each [[Definition:Sequence|sequence]] $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$
+it is true that:
+: $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that:
+: $\displaystyle \lim_{x \mathop \to c} f \left({x}\right) = l$
+Let $\epsilon > 0$.
+Then by the definition of the [[Definition:Limit of Function (Metric Space)|limit of a function]]:
+: $\exists \delta > 0: d_2 \left({f \left({x}\right), l}\right) < \epsilon$
+provided $0 < d_1 \left({x, c}\right) < \delta$.
+Now suppose that $\left \langle {x_n} \right \rangle$ is a [[Definition:Sequence|sequence]] of points of $S$ such that such that:
+: $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$
+Since $\delta > 0$, from the definition of the [[Definition:Limit of Sequence (Metric Space)|limit of a sequence]]:
+: $\exists N: \forall n > N: d_1 \left({x_n, c}\right) < \delta$
+But:
+: $\forall n \in \N_{>0}: x_n \ne c$
+That means:
+: $0 < d_1 \left({x_n, c}\right) < \delta$
+by the definition of a [[Definition:Metric|metric]].
+But that implies that:
+: $d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$
+That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
+: $\forall n > N: d_2 \left({f \left({x_n}\right), l}\right) < \epsilon$
+Thus:
+: $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose that for each [[Definition:Sequence|sequence]] $\left \langle {x_n} \right \rangle$ of points of $S$ such that $\displaystyle \forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$, it is true that:
+: $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$
+What we will try to do is assume that it is ''not'' true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, and try to find a contradiction.
+So, if it is not true that $\displaystyle \lim_{x \to c} f \left({x}\right) = l$, then:
+:$\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < d_1 \left({x, c}\right) < \delta \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon$
+where $\land$ denotes [[Definition:And|logical and]].
+For all $n \in \N_{>0}$, define:
+:$S_n = \left\{{x \in S: 0 < d_1 \left({x, c}\right) < \dfrac 1 n \land d_2 \left({f \left({x}\right), l}\right) \ge \epsilon}\right\}$
+[[Definition:By Hypothesis|By hypothesis]], $S_n$ is non-[[Definition:Empty Set|empty]] for all $n \in \N_{>0}$.
+Using the [[Axiom:Axiom of Countable Choice|axiom of countable choice]], there exists a [[Definition:Sequence|sequence]] $\left \langle {x_n} \right \rangle$ of points in $S$ such that $x_n \in S_n$ for all $n \in \N_{>0}$.
+Then:
+: $\forall n \in \N_{>0}: x_n \ne c$ and $\displaystyle \lim_{n \to \infty} x_n = c$
+but it is ''not'' true that:
+: $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$
+So there is our [[Definition:Contradiction|contradiction]], and so the result follows.
+{{qed}}
+{{ACC}}
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Limits of Functions}
+Tags: Limits of Functions, Named Theorems
+
+\begin{theorem}
+Let $X$ be one of the [[Definition:Standard Number Field|standard number fields]] $\Q, \R, \C$.
+Let $f$ and $g$  be [[Definition:Function|functions]] defined on an [[Definition:Open Set (Metric Space)|open subset]] $S \subseteq X$, except possibly at the point $c \in S$.
+Let $f$ and $g$ tend to the following [[Definition:Limit of a Function|limits]]:
+:$\displaystyle \lim_{x \to c} \ f \left({x}\right) = l$
+:$\displaystyle \lim_{x \to c} \ g \left({x}\right) = m$
+Let $\lambda, \mu \in X$ be arbitrary [[Definition:Number|numbers]] in $X$.
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{Real Polynomial Function is Continuous}
+Tags: Analysis, Continuous Functions, Polynomial Theory
+
+\begin{theorem}
+A [[Definition:Real Polynomial Function|(real) polynomial function]] is [[Definition:Continuous Real Function|continuous]] at every point.
+Thus a [[Definition:Real Polynomial Function|(real) polynomial function]] is [[Definition:Continuous Real Function on Interval|continuous]] on every [[Definition:Real Interval|interval]] of $\R$.
+\end{theorem}
+
+\begin{proof}
+From [[Linear Function is Continuous]], setting $\alpha = 1$ and $\beta = 0$, we have that:
+:$\displaystyle \lim_{x \mathop \to c} x = c$
+Repeated application of the [[Product Rule for Limits of Functions]] shows us that:
+:$\displaystyle \forall k \in \N: \lim_{x \mathop \to c} x^k = c^k$
+Now let $\map P x = a_n x^N + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0$.
+By repeated application of the [[Combined Sum Rule for Limits of Functions]], we find that:
+:$\displaystyle \lim_{x \mathop \to c} \map P x = \map P c$
+So whatever value we choose for $c$, we have that $\map P x$ is [[Definition:Continuous Real Function|continuous]] at $c$.
+From the definition of [[Definition:Continuous Real Function on Interval|continuity on an interval]], the second assertion follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Function is Continuous}
+Tags: Continuous Functions, Linear Function is Continuous
+
+\begin{theorem}
+Let $\alpha, \beta \in \R$ be [[Definition:Real Number|real numbers]].
+Let $f : \R \to \R$ be the [[Definition:Real Function|real function]] with:
+:$\map f x = \alpha x + \beta$
+for all $x \in \R$.
+Then $f$ is [[Definition:Continuous Real Function at Point|continuous]] at every real number $c \in \R$.
+\end{theorem}
+
+\begin{proof}
+Let $c \in \R$. 
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|real sequence]] [[Definition:Convergent Real Sequence|converging]] to $c$. 
+Then:
+{{begin-eqn}}
+{{eqn	| l = \lim_{n \mathop \to \infty} \map f {x_n}
+	| r = \lim_{n \mathop \to \infty} \paren {\alpha x_n + \beta}
+}}
+{{eqn	| r = \alpha c + \beta
+	| c = [[Combination Theorem for Sequences/Real/Combined Sum Rule|Combination Theorem for Sequences: Real: Combined Sum Rule]]
+}}
+{{eqn	| r = \map f c
+}}
+{{end-eqn}}
+We therefore have: 
+:for all [[Definition:Real Sequence|real sequences]] $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converging]] to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
+So by [[Sequential Continuity is Equivalent to Continuity in the Reals]] $f$ is [[Definition:Continuous Real Function|continuous]] at $c$. 
+As $c \in \R$ was arbitrary, $f$ is continuous on $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Rational Function is Continuous}
+Tags: Real Analysis, Continuous Functions
+
+\begin{theorem}
+A [[Definition:Real Function|real]] [[Definition:Rational Function|rational function]] is [[Definition:Continuous Real Function|continuous]] at every point at which it is defined.
+Thus a [[Definition:Real Function|real]] [[Definition:Rational Function|rational function]] is [[Definition:Continuous on Interval|continuous]] on every [[Definition:Real Interval|interval]] of $\R$ not containing a [[Definition:Root of Polynomial|root]] of the [[Definition:Denominator|denominator]] of the function.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map R x = \dfrac {\map P x} {\map Q x}$
+be a [[Definition:Real Function|real]] [[Definition:Rational Function|rational function]], defined at all points of $\R$ at which $\map Q x \ne 0$.
+Let $c \in \R$.
+From [[Real Polynomial Function is Continuous]]:
+:$\displaystyle \lim_{x \mathop \to c} \map P x = \map P c$
+and:
+:$\displaystyle \lim_{x \mathop \to c} \map Q x = \map Q c$
+Thus by [[Quotient Rule for Limits of Functions]]:
+:$\displaystyle \lim_{x \mathop \to c} \map R x = \lim_{x \mathop \to c} \frac {\map P x} {\map Q x} = \frac {\map P c} {\map Q c}$
+whenever $\map Q c \ne 0$.
+So whatever value we choose for $c$ such that $\map Q c \ne 0$, we have that $\map P x$ is [[Definition:Continuous Real Function|continuous]] at $c$.
+From the definition of [[Definition:Continuous on Interval|continuity on an interval]], the second assertion follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Composite Function}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]].
+Let:
+:$\displaystyle \lim_{y \mathop \to \eta} \map f y = l$
+:$\displaystyle \lim_{x \mathop \to \xi} \map g x = \eta$
+Then, if either:
+:'''Hypothesis 1:''' $f$ is [[Definition:Continuous Real Function at Point|continuous at $\eta$]] (that is $l = \map f \eta$)
+or:
+:'''Hypothesis 2:''' for some [[Definition:Open Real Interval|open interval]] $I$ containing $\xi$, it is true that $\map g x \ne \eta$ for any $x \in I$ except possibly $x = \xi$
+then:
+: $\displaystyle \lim_{x \mathop \to \xi} \map f {\map g x} = l$
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+Since $\displaystyle \lim_{y \mathop \to \eta} \map f y = l$, we can find $\Delta > 0$ such that:
+:$\size {\map f y - l} < \epsilon$ provided $0 < \size {y - \eta} < \Delta$
+Let $y = \map g x$.
+Then, provided that $0 < \size {\map g x - \eta} < \Delta$, we have:
+:$\size {\map f {\map g x} - l} < \epsilon$
+But $\displaystyle \lim_{x \mathop \to \xi} \map g x = \eta$ and $\Delta > 0$.
+Hence:
+:$\exists \delta > 0: \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$
+We now need to establish the reason for the conditions under which $0 < \size {x - \xi} < \delta \implies \size {\map f {\map g x} - l} < \epsilon$.
+As it stands, this is not generally the case, as follows.
+Consider the functions:
+:$\map g x = \eta, \map f y = \begin {cases}
+y_1 & : y = \eta \\
+y_2 & : y \ne \eta
+\end{cases}$
+Then:
+:$\displaystyle \lim_{y \mathop \to \eta} \map f y = y_2$ and $\lim_{x \mathop \to \xi} \map g x = \eta$
+But it is not true that $\displaystyle \lim_{x \mathop \to \xi} \map f {\map g x} = y_2$ because $\forall x: \map f {\map g x} = y_1$.
+Now, if '''Hypothesis 1:''' $f$ is [[Definition:Continuous Real Function at Point|continuous at $\eta$]], then $l = \map f \eta$ and so $\size {\map f y - l} < \epsilon$ even when $y = \eta$.
+So we can write: provided that $\size {\map g x - \eta} < \Delta$, we have:
+:$\size {\map f {\map g x} - l} < \epsilon$
+and the argument holds.
+Otherwise, let us assume '''Hypothesis 2:''' For some [[Definition:Open Real Interval|open interval]] $I$ containing $\xi$, it is true that $\map g x \ne \eta$ for any $x \in I$ except possibly $x = \xi$.
+Then we can be sure that $\map g x \ne \eta$ provided that $0 < \size {x - \xi} < \delta$ for sufficiently small $\delta > 0$.
+So we can write: $0 < \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$, and the argument holds.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Continuity on Interval}
+Tags: Real Intervals, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $\mathbb I$.
+Then $f$ is [[Definition:Continuous on Interval|continuous]] on $\mathbb I$ {{iff}}:
+:$\forall x \in \mathbb I: \forall \epsilon > 0: \exists \delta > 0: y \in \mathbb I \land \size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \mathbb I$ such that $x$ is not an [[Definition:Endpoint of Real Interval|end point]].
+Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ is the same as $\size {x - y} < \delta$ provided $\delta$ is small enough.
+The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x} \map f y = \map f x$, that is, that $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $x$.
+Now suppose $x \in \mathbb I$ and $x$ is a [[Definition:Endpoint of Real Interval|left hand end point]] of $\mathbb I$.
+Then the condition $y \in \mathbb I \land \size {x - y} < \delta$ reduces to $x \le y < x + \delta$ provided $\delta$ is small enough.
+The criterion given therefore becomes the same as the statement $\ds \lim_{y \mathop \to x^+} \map f y = \map f x$, that is, that $f$ is [[Definition:Right-Continuous at Point|continuous on the right]] at $x$.
+Similarly, if $x \in \mathbb I$ and $x$ is a [[Definition:Real Interval|left hand end point]] of $\mathbb I$, then the criterion reduces to the statement that $f$ is [[Definition:Left-Continuous at Point|continuous on the left]] at $x$.
+Thus the assertions are equivalent to the statement that $f$ is [[Definition:Continuous Real Function at Point|continuous]] at all points in $\mathbb I$, that is, that $f$ is [[Definition:Continuous on Interval|continuous]] on $\mathbb I$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Image of Sequence}
+Tags: Continuous Mappings on Metric Spaces, Limits of Sequences
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] which is [[Definition:Continuous Mapping (Metric Spaces)|continuous]] at $a \in A_1$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] of points in $A_1$ such that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = a$
+where $\displaystyle \lim_{n \mathop \to \infty} x_n$ is the [[Definition:Limit of Sequence (Metric Space)|limit of $x_n$]].
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
+That is:
+:$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = \map f {\lim_{n \mathop \to \infty} x_n}$
+That is, for a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]], the limit and function symbols [[Definition:Commute|commute]].
+\end{theorem}
+
+\begin{proof}
+From [[Limit of Function by Convergent Sequences]], we have:
+:$\displaystyle \lim_{x \mathop \to a} \map f x = \map f a$
+{{iff}}:
+:for each [[Definition:Sequence|sequence]] $\sequence {x_n}$ of points of $A_1$ such that:
+::$\forall n \in \N_{>0}: x_n \ne a$
+:and:
+::$\displaystyle \lim_{n \mathop \to \infty} x_n = a$
+:it is true that:
+::$\displaystyle \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
+The result follows directly from this and the definition of [[Definition:Continuous Mapping (Metric Spaces)|continuity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Interval by Continuous Function is Interval}
+Tags: Real Analysis, Continuous Functions
+
+\begin{theorem}
+Let $I$ be a [[Definition:Real Interval|real interval]].
+Let $f: I \to \R$ be a [[Definition:Continuous on Interval|continuous]] [[Definition:Real Function|real function]].
+Then the [[Definition:Image of Mapping|image]] of $f$ is a [[Definition:Real Interval|real interval]].
+\end{theorem}
+
+\begin{proof}
+Let $J$ be the [[Definition:Image of Mapping|image]] of $f$.
+By definition of [[Definition:Real Interval|real interval]], it suffices to show that:
+:$\forall y_1, y_2 \in J: \forall \lambda \in \R: y_1 \le \lambda \le y_2 \implies \lambda \in J$
+So suppose $y_1, y_2 \in J$, and suppose $\lambda \in \R$ is such that $y_1 \le \lambda \le y_2$.
+Consider these subsets of $I$:
+: $S = \left\{{x \in I: f \left({x}\right) \le \lambda}\right\}$
+: $T = \left\{{x \in I: f \left({x}\right) \ge \lambda}\right\}$
+As $y_1 \in S$ and $y_2 \in T$, it follows that $S$ and $T$ are both [[Definition:Non-Empty Set|non-empty]].
+Also, $I = S \cup T$.
+So from [[Interval Divided into Subsets]], a point in one subset is at zero [[Distance from Subset of Real Numbers|distance]] from the other.
+So, suppose that $s \in S$ is at zero [[Distance from Subset of Real Numbers|distance]] from $T$.
+From [[Limit of Sequence to Zero Distance Point]], we can find a [[Definition:Sequence|sequence]] $\left \langle {t_n} \right \rangle$ in $T$ such that $\displaystyle \lim_{n \to \infty} t_n = s$.
+Since $f$ is [[Definition:Continuous on Interval|continuous]] on $I$, it follows from [[Limit of Image of Sequence]] that $\displaystyle \lim_{n \to \infty} f \left({t_n}\right) = f \left({s}\right)$.
+But $\forall n \in \N_{> 0}: f \left({t_n}\right) \ge \lambda$.
+Therefore by [[Lower and Upper Bounds for Sequences]], $f \left({s}\right) \ge \lambda$.
+We already have that $f \left({s}\right) \le \lambda$.
+Therefore $f \left({s}\right) = \lambda$ and so $\lambda \in J$.
+A similar argument applies if a point of $T$ is at zero distance from $S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+As before, let $J$ be the [[Definition:Image of Mapping|image]] of $f$.
+By [[Subset of Real Numbers is Interval iff Connected]] we need to show that $J$ is connected (and hence an interval).
+Suppose not.
+Then there exists a [[Definition:Separation (Topology)|separation]] $S \mid T$ of $J$.
+Define $A = f^{-1}(S)$ and $B = f^{-1}(T)$. $A$ and $B$ are both non-empty.
+Because $f$ is continuous, by [[Continuous iff inverse image of any open set is open]] we must have $A$ and $B$ open.
+Now, $A \cap B = f^{-1}(S) \cap f^{-1}(T) = f^{-1}(S \cap T) = \varnothing$, because $S \mid T$ is a separation.
+Also, $A \cup B = f^{-1}(S) \cup f^{-1}(T) = f^{-1}(S \cup T) = f^{-1}(J) = I$ ($S \mid T$ is a separation of $J$).
+Hence $A \mid B$ is a separation of $I$. $I$ can certainly not be an interval (because it is not connected).
+This is a contradiction. Thus $J$ must be an interval.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interval Divided into Subsets}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $\mathbb I$ be a [[Definition:Real Interval|real interval]].
+Let $S$ and $T$ be [[Definition:Non-Empty Set|non-empty]] subsets of $\mathbb I$ such that $\mathbb I \subseteq S \cup T$.
+Then one of $S$ or $T$ contains an [[Definition:Element|element]] at zero [[Distance from Subset of Real Numbers|distance]] from the other.
+\end{theorem}
+
+\begin{proof}
+$\mathbb I \subseteq S \cup T \implies \forall x \in \mathbb I: x \in S \lor x \in T$ from the definition of [[Definition:Set Union|union]].
+That is, every [[Definition:Element|element]] of $\mathbb I$ belongs either to $S$ or to $T$.
+The [[Distance from Subset of Real Numbers|distance]] of an [[Definition:Element|element]] $c \in \R$ from a [[Definition:Subset|subset]] $S$ of $\R$ is given as:
+:$\displaystyle \map d {c, S} = \map {\inf_{x \mathop \in S} } {\size {c - x} }$
+Assume that $S$ and $T$ have no [[Definition:Element|element]] in common, otherwise the result is trivial.
+{{WLOG}}, suppose that $\exists s \in S, t \in T$ such that $s < t$.
+(If not, then $\exists s \in S, t \in T$ such that $s > t$, and the following argument may be amended appropriately.)
+Let $T_0 = \set {x: x \in T: x > s}$.
+As $t \in T_0$ it follows that $T_0 \ne \O$.
+Also, $T_0$ is [[Definition:Bounded Below Set|bounded below]] by $s$.
+Let $b = \map \inf {T_0}$.
+If $b \notin T$ then $b \in S$.
+But from [[Distance from Subset of Real Numbers]], it follows that $\map d {b, T_0} = 0$.
+Thus we have found an [[Definition:Element|element]] in $S$ which is zero [[Distance from Subset of Real Numbers|distance]] from $T$.
+Otherwise, $b \in T$.
+Then $b > s$, and the [[Definition:Open Real Interval|open interval]] $\openint s b$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Hence $b$ is an [[Definition:Element|element]] in $T$ which is zero [[Distance from Subset of Real Numbers|distance]] from $S$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Distance from Subset of Real Numbers}
+Tags: Distance Function
+
+\begin{theorem}
+Let $S$ be a [[Definition:Subset|subset]] of the set of [[Definition:Real Number|real numbers]] $\R$.
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\map d {x, S}$ be the [[Definition:Distance between Element and Subset of Real Numbers|distance]] between $x$ and $S$.
+Then:
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Distance between Real Numbers|distance]]:
+:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
+Thus:
+:$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
+Consider the set $T = \set {\size {x - y}: y \in S}$.
+This has $0$ as a [[Definition:Lower Bound of Set|lower bound]] as [[Absolute Value is Bounded Below by Zero]].
+So:
+:$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} } \ge 0$
+If $x \in S$ then:
+:$\size {x - x} = 0 \in T$
+and so:
+:$\displaystyle 0 \le \map {\inf_{y \mathop \in S} } {\map d {x, y} }$
+Thus:
+:$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\map d {x, y} } = 0$
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Recall from [[Real Number Line is Metric Space]] that the [[Definition:Real Number|set of real numbers]] $\R$ with the [[Definition:Distance between Element and Subset of Real Numbers|distance function]] $d$ is a [[Definition:Metric Space|metric space]].
+The result is then seen to be an example of [[Distance from Subset to Element]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+From the definition of [[Definition:Distance between Real Numbers|distance]]:
+:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
+Thus:
+:$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
+Let $\xi = \sup S$.
+Then:
+:$\forall y \in S: \size {\xi - y} = \xi - y$
+So we need to show that no $h > 0$ can be a [[Definition:Lower Bound of Set|lower bound]] for $T = \set {\size {\xi - y}: y \in S}$.
+{{AimForCont}} $\exists h > 0: \forall y \in S: \xi - y \ge h$.
+But then:
+:$\forall y \in S: y \le \xi - h$
+and hence $\xi - h$ is an [[Definition:Upper Bound of Set|upper bound]] for $T$ smaller than $\xi = \sup S$.
+But this [[Definition:Contradiction|contradicts]] the definition of [[Definition:Supremum of Set|supremum]], that is the ''smallest'' [[Definition:Upper Bound of Set|upper bound]].
+So there is no such $h > 0$.
+Hence by [[Proof by Contradiction]] it follows that:
+:$\map d {\xi, S} = 0$.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Recall from [[Real Number Line is Metric Space]] that the [[Definition:Real Number|set of real numbers]] $\R$ with the [[Definition:Distance between Element and Subset of Real Numbers|distance function]] $d$ is a [[Definition:Metric Space|metric space]].
+The result is then seen to be an example of [[Distance from Subset to Supremum]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+From the definition of [[Definition:Distance between Real Numbers|distance]]:
+:$\forall x, y \in \R: \map d {x, y} = \size {x - y}$
+Thus:
+:$\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} }$
+Let $\xi = \inf S$.
+Consider $\map d {-\xi, S'}$ where $S' = \set {-\xi: \xi \in S}$.
+By [[Negative of Infimum is Supremum of Negatives]]:
+:$\xi = \inf S \implies -\xi = \sup S'$
+Thus from the above, $\map d {-\xi, S'} = 0$ and hence the result.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Recall from [[Real Number Line is Metric Space]] that the [[Definition:Real Number|set of real numbers]] $\R$ with the [[Definition:Distance between Element and Subset of Real Numbers|distance function]] $d$ is a [[Definition:Metric Space|metric space]].
+The result is then seen to be an example of [[Distance from Subset to Infimum]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Sequence to Zero Distance Point}
+Tags: Limits of Sequences, Limit of Sequence to Zero Distance Point
+
+\begin{theorem}
+Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\R$.
+Let the [[Distance from Subset of Real Numbers|distance]] $\map d {\xi, S} = 0$ for some $\xi \in \R$.
+Then there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}$ in $S$ such that $\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$.
+\end{theorem}
+
+\begin{proof}
+First it is shown that:
+:$\forall n \in \N_{>0}: \exists x_n \in S: \size {\xi - x_n} < \dfrac 1 n$
+{{AimForCont}} that:
+:$\exists n \in \N_{>0}: \not \exists x \in S: \size {\xi - x} < \dfrac 1 n$
+Then $\dfrac 1 n$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of the set $T = \set {\size {\xi - x}: x \in S}$.
+This [[Proof by Contradiction|contradicts]] the assertion that $\map d {\xi, S} = 0$.
+We have from [[Sequence of Powers of Reciprocals is Null Sequence]] that:
+:$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
+So as $\size {\xi - x_n} < \dfrac 1 n$ it follows from the [[Squeeze Theorem for Real Sequences]] that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Closed Real Interval is Bounded}
+Tags: Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Closed Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Then $f$ is [[Definition:Bounded Mapping|bounded]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+Suppose $f$ is not [[Definition:Bounded Mapping|bounded]] on $\closedint a b$.
+Then from the [[Limit of Sequence to Zero Distance Point/Corollary|corollary to Limit of Sequence to Zero Distance Point]], there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
+Since $\closedint a b$ is a [[Definition:Closed Real Interval|closed interval]], from [[Convergent Subsequence in Closed Interval]], $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] $\sequence {x_n}$ which [[Definition:Convergent Sequence|converges]] to some $\xi \in \closedint a b$.
+Because $f$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\closedint a b$, it follows from [[Limit of Image of Sequence]] that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
+But this contradicts our supposition that there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to +\infty$ as $n \to \infty$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Subsequence in Closed Interval}
+Tags: Analysis, Convergence, Limits of Sequences
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Then every [[Definition:Real Sequence|sequence]] of points of $\closedint a b$ contains a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Real Sequence|converges]] to a point in $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $\closedint a b$.
+Since $\closedint a b$ is [[Definition:Bounded Subset of Real Numbers|bounded in $\R$]], it follows that $\sequence {x_n}$ is a [[Definition:Bounded Sequence|bounded sequence]].
+By the [[Bolzano-Weierstrass Theorem]], $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] $\sequence {x_{n_r} }$ which is [[Definition:Convergent Sequence|convergent]].
+Suppose $x_{n_r} \to l$ as $n \to \infty$.
+Since $a \le x_{n_r} \le b$, from [[Lower and Upper Bounds for Sequences]] it follows that $a \le l \le b$.
+So $\sequence {x_{n_r} }$ [[Definition:Convergent Sequence|converges]] to a point in $\closedint a b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max and Min of Function on Closed Real Interval}
+Tags: Real Intervals, Max and Min of Function on Closed Real Interval
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Then $f$ reaches a [[Definition:Maximum Value|maximum]] and a [[Definition:Minimum Value|minimum]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+From [[Image of Closed Real Interval is Bounded]], we have that $f$ is [[Definition:Bounded Mapping|bounded]] on $\closedint a b$.
+Let $d$ be the [[Definition:Supremum of Real-Valued Function|supremum]] of $f$ on $\closedint a b$.
+Consider a [[Definition:Sequence|sequence]] $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to d$ as $n \to \infty$.
+From [[Limit of Sequence to Zero Distance Point/Corollary|the corollary to Limit of Sequence to Zero Distance Point]], this can always be found.
+Now $\closedint a b$ is a [[Definition:Closed Real Interval|closed interval]]
+So from [[Convergent Subsequence in Closed Interval]], $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] $\sequence {x_{n_r} }$ which [[Definition:Convergent Sequence|converges]] to some $\xi \in \closedint a b$.
+Because $f$ is [[Definition:Continuous on Interval|continuous]] on $\closedint a b$, it follows from [[Limit of Image of Sequence]] that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.
+So $\map f \xi = d$ and thus the [[Definition:Supremum of Real-Valued Function|supremum]] $d$ is indeed a [[Definition:Maximum Value|maximum]].
+A similar argument shows that the [[Definition:Infimum of Real-Valued Function|infimum]] is a [[Definition:Minimum Value|minimum]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This is an instance of the [[Extreme Value Theorem]].
+$\closedint a b$ is a [[Definition:Compact Subset of Metric Space|compact subset]] of a [[Definition:Metric Space|metric space]] from [[Real Number Line is Metric Space]].
+$\R$ itself is a [[Definition:Normed Vector Space|normed vector space]].
+{{MissingLinks|Find the reference to $\R$ being a [[Definition:Normed Vector Space|normed vector space]].}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Image of Closed Interval is Closed Interval}
+Tags: Real Analysis, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+Then the [[Definition:Image of Subset under Mapping|image]] of $\left[{a \,.\,.\, b}\right]$ under $f$ is also a [[Definition:Closed Real Interval|closed interval]].
+\end{theorem}
+
+\begin{proof}
+Let $I = \left[{a \,.\,.\, b}\right]$.
+Let $J = f \left({I}\right)$.
+From [[Image of Interval by Continuous Function is Interval]], $J$ is an [[Definition:Real Interval|interval]].
+From [[Image of Closed Real Interval is Bounded]], $J$ is [[Definition:Bounded Mapping|bounded]].
+From [[Max and Min of Function on Closed Real Interval]], $J$ includes its [[Definition:Endpoint of Real Interval|end points]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Retraction Theorem}
+Tags: Algebraic Topology, Named Theorems
+
+\begin{theorem}
+Let $M$ be a [[Definition:Compact Topological Space|compact]] [[Definition:Topological Manifold|manifold]] with [[Definition:Boundary (Topology)|boundary]] $\partial M$.
+Then there is no [[Definition:Smooth Mapping|smooth mapping]] $f: M \to \partial M$ such that $\partial f: \partial M \to \partial M$ is the [[Definition:Identity Mapping|identity]].
+\end{theorem}
+
+\begin{proof}
+{{MissingLinks|one-dimensional, for a start}}
+Aiming for a [[Definition:Contradiction|contradiction]], suppose such a [[Definition:Smooth Mapping|smooth mapping]] exists.
+By the [[Morse-Sard Theorem]], there exists a [[Definition:Regular Value|regular value]] $x \in \partial M$.
+By the [[Preimage Theorem]]:
+:$\map {f^{-1} } x$ is a [[Definition:Submanifold|submanifold]] of $M$ with boundary.
+We have that the [[Definition:Codimension|codimension]] of $\map {f^{-1} } x$ in $M$ equals the [[Definition:Codimension|codimension]] of $x$ in $\partial M$, that is, $\map \dim M - 1$.
+Then $\map {f^{-1} } x$ is one dimensional and [[Definition:Compact Topological Space|compact]].
+Since $\partial f$ is the [[Definition:Identity Mapping|identity mapping]]:
+:$\map {\partial f^{-1} } x = \map {f^{-1} } x \cap \partial M = \set x$
+This contradicts the [[Classification of Compact One-Manifolds]].
+{{qed}}
+[[Category:Algebraic Topology]]
+[[Category:Named Theorems]]
+5bhlxb7envkzp480dmd6wk5jwgarz0w
+\end{proof}<|endoftext|>
+\section{Classification of Compact One-Manifolds}
+Tags: Topology
+
+\begin{theorem}
+Every [[Definition:Compact Topological Space|compact]] one-[[Definition:Dimension (Topology)|dimensional]] manifold is [[Definition:Diffeomorphism|diffeomorphic]] to either a [[Definition:Circle|circle]] or a [[Definition:Closed Interval|closed interval]].
+\end{theorem}
+
+\begin{proof}
+=== Lemma 1 ===
+Let $f$ be a [[Definition:Function|function]] on $\closedint a b$ that is [[Definition:Smooth Mapping|smooth]] and has a positive [[Definition:Derivative|derivative]] everywhere except one [[Definition:Interior (Topology)|interior]] point, $c$.
+Then there exists a globally smooth function $g$ that agrees with $f$ near $a$ and $b$ and has a positive derivative everywhere.
+==== Proof of Lemma 1 ====
+Let $r$ be a smooth nonnegative function that vanishes outside a compact subset of $\openint a b$, which equals $1$ near $c$, and which satisfies $\displaystyle \int_a^b r = 1$.
+Define:
+:$\displaystyle \map g x = \map f a + \int_a^x \paren {k \map r s + \map {f'} s \paren {1 - \map r s} } \rd s$
+where the constant $\displaystyle k = \map f b= - \map f a - \int_a^b \map {f'} s \paren {1 - \map r s} \d s$.
+{{explain|why this function g satisfies the given criteria}}
+{{qed|lemma}}
+Let $f$ be a [[Definition:Morse Function|Morse function]] on a one-manifold $X$.
+Let $S$ be the [[Definition:Set Union|union]] of the [[Definition:Critical Point (Topology)|critical points]] of $f$ and $\partial X$.
+As $S$ is finite, $X - S$ consists of a finite number of one-manifolds, $L_1, L_2, \cdots, L_n$.
+=== Lemma 2 ===
+$f$ maps each $L_i$ diffeomorphically onto an open interval in $\R$
+==== Proof of Lemma 2 ====
+Let $L$ be any of the $L_i$.
+Because $f$ is a local diffeomorphism and $L$ is [[Definition:Connected Set (Topology)|connected]], $f \sqbrk L$ is [[Definition:Open Real Interval|open]] and connected in $\R$.
+We also have $f \sqbrk L \in f \sqbrk X$, the latter of which is compact.
+Hence there are numbers $c$ and $d$ such that $f \sqbrk L = \openint c d$.
+It suffices to show $f$ is [[Definition:One-to-One Mapping|one to one]] on $L$, because then $f^{-1}: \openint c d \to L$ is defined and locally smooth.
+Let $p$ be any point of $L$.
+Set $q = \map f p$.
+It suffices to show that every other point $z \in L$ can be joined to $p$ by a curve $\gamma: \closedint q y \to L$ such that $f \circ \gamma$ is the identity and $\map \gamma y = z$.
+Since $\map f z = y \ne q = \map f p$, this result shows $f$ is one to one.
+So let $Q$ be the set of points $x$ that can be so joined.
+Since $f$ is a local diffeomorphism, $Q$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)]].
+Hence $Q = L$.
+{{qed|lemma}}
+=== Lemma 3 ===
+Let $L$ be a [[Definition:Subset|subset]] of $X$ diffeomorphic to an open interval in $\R$, where $\dim X = 1$.
+Then the [[Definition:Closure (Topology)|closure]] $\map \cl L$ contains at most two points not in $L$.
+==== Proof of Lemma 3 ====
+Let $g$ be a diffeomorphism $g: \openint a b \to L$ and let $p \in \map \cl L - L$.
+Let $J$ be a closed subset of $X$ diffeomorphic to $\closedint 0 1$ such that:
+:$1$ corresponds to $p$
+:$0$ corresponds to some $\map g t$ in $L$.
+Consider the set $\set {s \in \openint a t: \map g s \in J}$.
+This set is both open and closed in $\openint a b$.
+Hence $J$ contains either $g \sqbrk {\openint a t}$ or $g \sqbrk {\openint t b}$.
+{{qed|lemma}}
+{{finish}}
+=== Proof of Corollary ===
+Follows trivially.
+{{qed}}
+{{proofread}}
+[[Category:Topology]]
+r8ydxytmtr0gkww1oq5cd0c3xh3w2f3
+\end{proof}<|endoftext|>
+\section{Differentiable Function is Continuous}
+Tags: Differentiable Real Functions, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $x_0 \in I$ such that $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $x_0$.
+Then $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $x_0$.
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], $\map {f'} {x_0}$ exists.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map f x - \map f {x_0}
+      | r = \frac {\map f x - \map f {x_0} } {x - x_0} \cdot \paren {x - x_0}
+      | c = 
+}}
+{{eqn | o = \to
+      | r = \map {f'} {x_0} \cdot 0
+      | c = as $x \to x_0$
+}}
+{{end-eqn}}
+Thus:
+:$\map f x \to \map f {x_0}$ as $x \to x_0$
+or in other words:
+:$\displaystyle \lim_{x \mathop \to x_0} \map f x = \map f {x_0}$
+The result follows by definition of [[Definition:Continuous Real Function at Point|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Rule for Derivatives}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $\map j x, \map k x$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $\xi \in I$ be a point in $I$ at which both $j$ and $k$ are [[Definition:Differentiable Real Function at Point|differentiable]].
+Define the [[Definition:Real Function|real function]] $f$ on $I$ by:
+:$\displaystyle \map f x = \begin{cases}
+\dfrac {\map j x} {\map k x} & : \map k x \ne 0 \\
+0 & : \text{otherwise}
+\end{cases}$
+Then, if $\map k \xi \ne 0$, $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at $\xi$, and furthermore:
+:$\map {f'} \xi = \dfrac {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi} {\paren {\map k \xi}^2}$
+It follows from the definition of [[Definition:Derivative on Interval|derivative]] that if $j$ and $k$ are both [[Definition:Differentiable on Interval|differentiable]] on the interval $I$, then:
+:$\displaystyle \forall x \in I: \map k x \ne 0 \implies \map {f'} x = \frac {\map {j'} x \map k x - \map j x \map {k'} x} {\paren {\map k x}^2}$
+\end{theorem}
+
+\begin{proof}
+Let $\xi$ be such that $\map k \xi \ne 0$.
+From [[Differentiable Function is Continuous]], $k$ is [[Definition:Continuous Real Function at Point|continuous]] at $\xi$.
+It follows that there exists an $\epsilon > 0$, such that $\size h < \epsilon \implies \map k {\xi + h} \ne 0$.
+So let $\size h < \epsilon$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \frac {\map f {\xi + h} - \map f \xi} h
+      | r = \frac 1 h \paren {\frac {\map j {\xi + h} } {\map k {\xi + h} } - \frac {\map j \xi} {\map k \xi } }
+      | c = as $\map k \xi, \map k {\xi + h} \ne 0$
+}}
+{{eqn | r = \frac {\map j {\xi + h} \map k \xi - \map k {\xi + h} \map j \xi} {h \map k {\xi + h} \map k \xi}
+}}
+{{eqn | r = \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}
+      | c = algebraic manipulation
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h
+      | o = 
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map k {\xi + h} - \map k \xi} h}
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac 1 {\map k {\xi + h} \map k \xi} \cdot \lim_{h \mathop \to 0} \paren {\frac {\map j {\xi + h} - \map j \xi} h \map k \xi - \map j \xi \frac {\map  k {\xi + h} - \map k \xi} h}
+      | c = [[Product Rule for Limits of Functions]]
+}}
+{{end-eqn}}
+Thus by:
+:[[Definition:Continuous Real Function at Point|continuity]] of $k$ at $\xi$
+:[[Definition:Differentiable Real Function at Point|differentiability]] of $j$ and $k$ at $\xi$
+:[[Combined Sum Rule for Limits of Functions]]:
+it is concluded that:
+:$\displaystyle \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h = \frac 1 {\map k \xi^2} \paren {\map {j'} \xi \map k \xi - \map j \xi \map {k'} \xi}$
+From the definition of [[Definition:Differentiable Real Function at Point|differentiability]], $f$ is differentiable at $\xi$, with stated value.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage Theorem}
+Tags: Named Theorems, Topology
+
+\begin{theorem}
+Let $y$ be a [[Definition:Regular Value|regular value]] of a [[Definition:Smooth Mapping|smooth]] [[Definition:Submersion|submersion]] $f:X \to Y$.
+Then the [[Definition:Preimage of Element under Mapping|preimage]] $f^{-1}(y)$ is a smooth [[Definition:Submanifold|submanifold]] of $X$, with $\dim f^{-1}(y) = \dim X - \dim Y$.
+\end{theorem}
+
+\begin{proof}
+Let $k,l$ be [[Definition:Natural Numbers|natural numbers]] with $k \geq l$.
+By the [[Local Submersion Theorem]], there exists coordinates in some open sets of $x,y$ such that $f(x_1, x_2, \ldots, x_k)=(x_1, \ldots,x_l)$ and $y$ corresponds to $(0, \ldots, 0)$.
+Let $V$ be that [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
+Then $f^{-1}(y) \cap V$ is the set of points where $x_1=0, \ldots, x_l=0$.
+The [[Definition:Mapping|functions]] $x_{l+1}, \ldots, x_k$ therefore form a [[Definition:Coordinate System|coordinate system]] on the set $f^{-1}(y) \cap V$, which is a [[Definition:Relatively Open|relatively open]] subset of $f^{-1}(y)$.
+Together these functions then form a [[Definition:Diffeomorphism|diffeomorphism]] to a [[Definition:Euclidean Space|Euclidean space]].
+We also have, by the regular value properties of $y$, a [[Definition:Surjection|surjection]] of [[Definition:Tangent Space|tangent spaces]] from $x$ to $y$.
+This ensures [[Definition:Smooth Mapping|smoothness]] of the solution set $f^{-1}(y)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Backwards Induction}
+Tags: Number Theory, Named Theorems, Mathematical Induction, Proof Techniques
+
+\begin{theorem}
+Let $P$ be a [[Definition:Propositional Function|propositional function]] on the [[Definition:Natural Numbers|natural numbers]] $\N$.
+Suppose that:
+:$(1): \quad \forall n \in \N: \map P {2^n}$ holds.
+:$(2): \quad \map P n \implies \map P {n - 1}$.
+Then $\map P n$ holds for all $\forall n \in \N$.
+The proof technique based on this result is called '''backwards induction'''.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\exists k \in \N$ such that $\map P k$ is [[Definition:False|false]].
+From [[Power of Real Number greater than One is Unbounded Above]], $\set {2^n: n \in \N}$ is [[Definition:Bounded Above Set|unbounded above]]
+Therefore we can find:
+:$M = 2^N > k$
+Now let us create the set:
+:$S = \set {n \in \N: n < M, \map P n \text { is false} }$
+As $k < M$ and $\map P k$ is false, $S \ne \O$ and as $\forall x \in S: x < M$ it follows that $S$ is [[Definition:Bounded Above Set|bounded above]].
+So from [[Set of Integers Bounded Above by Integer has Greatest Element]], $S$ has a [[Definition:Greatest Element|greatest element]].
+However, $\map P n$ holds for $m < n \le M$ and hence $\map P {m + 1}$ holds.
+But $\map P {m + 1} \implies \map P m$ and so $\map P m$ holds after all.
+So there can be no such $m$ and therefore $S = \O$, hence there can be no such $k \in \N$ such that $\map P k$ is [[Definition:False|false]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inequalities Concerning Roots}
+Tags: Roots of Numbers, Inequalities
+
+\begin{theorem}
+Let $\closedint X Y$ be a [[Definition:Closed Real Interval|closed real interval]] such that $0 < X \le Y$.
+Let $x, y \in \closedint X Y$.
+Then:
+:$\forall n \in \N_{> 0}: X Y^{1/n} \size {x - y} \le n X Y \size {x^{1/n} - y^{1/n} } \le Y X^{1/n} \size {x - y}$
+\end{theorem}
+
+\begin{proof}
+From [[Difference of Two Powers]]:
+:$\displaystyle a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+Let $a > b > 0$.
+Then:
+:$\dfrac {a^n - b^n} {a - b} = a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1}$
+Also, note that:
+:$b^{n - 1} = b^{n - j - 1} b^j \le a^{n - j - 1} b^j \le a^{n - j -1} a^j \le a^{n - 1}$
+Hence if $a > b > 0$:
+:$n b^{n - 1} \le \dfrac {a^n - b^n} {a - b} \le n a^{n - 1}$
+Taking [[Definition:Reciprocal|reciprocals]]:
+:$\dfrac 1 {n b^{n - 1} } \ge \dfrac {a - b} {a^n - b^n} \ge \dfrac 1 {n a^{n - 1} }$
+Thus:
+:$\dfrac {a^n - b^n} {n a^{n - 1} } \le a - b \le \dfrac {a^n - b^n} {n b^{n - 1} } \implies \dfrac {a \paren {a^n - b^n} } {a^n} \le n \paren {a - b} \le \dfrac {b \paren {a^n - b^n} } {b^n}$
+Now suppose $x > y$.
+Then:
+:$x - y = \size {x - y}$
+and:
+:$x^{1/n} - y^{1/n} = \size {x^{1/n} - y^{1/n} }$
+Then put $a = x^{1/n}$ and $b = y^{1/n}$ in the above:
+:$\dfrac {x^{1/n} \size {x - y} } x \le n \size {x^{1/n} - y^{1/n} } \le \dfrac {y^{1/n} \size {x - y} } y$
+Multiplying through by $x y$:
+:$x^{1/n} y \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le x y^{1/n} \size {x - y}$
+Similarly, if $y > x$:
+:$y^{1/n} x \size {x - y} \le n x y \size {x^{1/n} - y^{1/n} } \le y x^{1/n} \size {x - y}$
+The result follows after some algebra.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Root Function}
+Tags: Roots of Numbers, Continuous Functions
+
+\begin{theorem}
+Let $n \in \N_{>0}$ be a non-zero [[Definition:Natural Number|natural number]].
+Let $f: \hointr 0 \infty \to \R$ be the [[Definition:Real Function|real function]] defined by $\map f x = x^{1/n}$.
+Then $f$ is [[Definition:Continuous Real Function at Point|continuous]] at each $\xi > 0$ and [[Definition:Right-Continuous at Point|continuous on the right]] at $\xi = 0$.
+\end{theorem}
+
+\begin{proof}
+First suppose that $\xi > 0$.
+Let $X, Y \in \R$ such that $0 < X < \xi < Y$.
+Let $x \in \R$ such that $X < x < Y$.
+From [[Inequalities Concerning Roots]]:
+:$X Y^{1/n} \ \size {x - \xi} \le n X Y \ \size {x^{1/n} - \xi^{1/n} } \le Y X^{1/n} \ \size {x - \xi}$
+Thus:
+:$\dfrac 1 {n Y} Y^{1/n} \ \size {x - \xi} \le \size {x^{1/n} - \xi^{1/n} } \le \dfrac 1 {n X} X^{1/n} \ \size {x - \xi}$
+The result follows by applying the [[Squeeze Theorem]].
+Now we need to show that $\map f x \to 0$ as $x \to 0^+$.
+We need to show that:
+:$\forall \epsilon > 0: \exists \delta > 0: x^{1/n} = \size {x^{1/n} - 0} < \epsilon$
+provided $0 < x < \delta$.
+Clearly, for any given $\epsilon$, we can choose $\delta = \epsilon^n$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weak Whitney Immersion Theorem}
+Tags: Topology
+
+\begin{theorem}
+Every $k$-dimensional [[Definition:Topological Manifold|manifold]] $X$ admits a [[Definition:One-to-One Relation|one-to-one]] [[Definition:Immersion|immersion]] in $\R^{2 k + 1}$.
+{{MissingLinks|$k$-dimensional}}
+\end{theorem}
+
+\begin{proof}
+Let $M > 2 k + 1$ be a [[Definition:Natural Numbers|natural number]] such that $f: X \to \R^M$ is an [[Definition:Injection|injective]] [[Definition:Immersion|immersion]].
+Define a map $h: X \times X \times \R \to \R^M$ by:
+:$\map h {x, y, t} = \map t {\map f x - \map f y}$
+Define a map $g: \map T X \to \R^M$ by:
+:$\map g {x, v} = \map {\d f_x} v$
+where $\map T X$ is the [[Definition:Tangent Bundle|tangent bundle]] of $X$.
+Since $M > 2 k + 1$, the [[Morse-Sard Theorem]] implies $\exists a \in \R^M$ such that $a$ is in the image of neither function.
+Let $\pi$ be the projection of $\R^M$ onto the orthogonal complement of $a, H$.  
+Suppose:
+:$\map {\paren {\pi \circ f} } x = \map {\paren {\pi \circ f} } y$
+for some $x, y$.
+Then:
+:$\map f x - \map f y = t a$
+for some scalar $t$.
+{{AimForCont}} $x \ne y$.
+Then as $f$ is [[Definition:Injection|injective]]:
+:$t \ne 0$
+But then $\map h {x, y, 1/t} = a$, contradicting the choice of $a$.
+By [[Proof by Contradiction]]:
+:$x = y$
+and so $\pi \circ f$ is [[Definition:Injection|injective]].
+Let $v$ be a nonzero vector in $\map {T_x} X$ (the [[Definition:Tangent Space|tangent space]] of $X$ at $x$) for which:
+:$\map {\d \paren {\pi \circ f}_x} v = 0$
+Since $\pi$ is linear:
+:$\map {\d \paren {\pi \circ f}_x} = \pi \circ \d f_x$
+Thus:
+:$\map {\pi \circ \d f_x} v = 0$
+so:
+:$\map {\d f_x} v = t a$
+for some scalar $t$.
+Because $f$ is an immersion, $t \ne 0$.
+Hence:
+:$\map g {x, 1/t} = a$
+contradicting the choice of $a$.
+Hence $\pi \circ f: X \to H$ is an immersion. 
+$H$ is obviously isomorphic to $\R^{M-1}$.
+Thus whenever $M > 2 k + 1$ and $X$ admits of a one-to-one immersion in $\R^M$, it follows that $X$ also admits of a one-to-one immersion in $\R^{M-1}$.
+\end{proof}<|endoftext|>
+\section{P-adic Norm is Norm}
+Tags: Examples of Norms, P-adic Norm is Norm
+
+\begin{theorem}
+The [[Definition:P-adic Norm|$p$-adic norm]] forms a [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Number|rational numbers]] $\Q$.
+\end{theorem}
+
+\begin{proof}
+Let $v_p$ be the [[Definition:P-adic Valuation|$p$-adic valuation]] on the [[Definition:Rational Number|rational numbers]].
+Recall that the [[Definition:P-adic Norm|$p$-adic norm]] is defined as:
+:$\forall q \in \Q: \norm q_p := \begin{cases}
+  0 & : q = 0 \\
+  p^{- \map {\nu_p} q} & : q \ne 0
+\end{cases}$
+We must show the following hold for all $x$, $y \in \Q$:
+{{begin-axiom}}
+{{axiom | n = \text N 1
+        | q = \forall x \in \Q
+        | ml= \norm x_p = 0
+        | mo= \iff
+        | mr= x = 0
+}}
+{{axiom | n = \text N 2
+        | q = \forall x, y \in \Q
+        | ml= \norm {x y}
+        | mo= =
+        | mr= \norm x_p \times \norm y_p
+}}
+{{axiom | n = \text N 3
+        | q = \forall x, y \in \Q
+        | ml= \norm {x + y}_p
+        | mo= \le
+        | mr= \norm x_p + \norm y_p
+}}
+{{end-axiom}}
+=== Norm Axiom $(\text N 1)$ ===
+By [[Power of Positive Real Number is Positive/Real Number|Power of Positive Real Number is Positive]]:
+:$\displaystyle \forall s \in \R: \frac 1 {p^s} > 0$
+By definition of the [[Definition:P-adic Norm|$p$-adic norm]] it follows that:
+:$\forall x \in \Q: \norm x_p = 0 \iff x = 0$
+Thus the [[Definition:P-adic Norm|$p$-adic norm]] fulfils [[Definition:Norm Axioms|axiom $(\text N 1)$]].
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+Let $x = 0$ or $y = 0$.
+Then $\norm x_p = 0$ or $\norm y_p = 0$ from [[Definition:Norm Axioms|axiom $(\text N 1)$]], and:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \norm {x y}_p
+      | r = 0
+      | c = [[Definition:Norm Axioms|Norm Axioms: Axiom $(\text N 1)$]]
+}}
+{{eqn | r = \norm x_p \times \norm y_p
+      | c = 
+}}
+{{end-eqn}}
+Let $x, y\in \Q_{\ne 0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x y}_p
+      | r = \frac 1 {p^{\map {\nu_p} {x y} } }
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | r = \frac 1 {p^{\map {\nu_p} x + \map {\nu_p} y} }
+      | c = [[P-adic Valuation is Valuation|$p$-adic Valuation is Valuation]]
+}}
+{{eqn | r = \frac 1 {p^{\map {\nu_p} x} p^{\map {\nu_p} y} }
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \norm x_p \times \norm y_p
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{end-eqn}}
+Thus the [[Definition:P-adic Norm|$p$-adic norm]] fulfils [[Definition:Norm Axioms|axiom $(\text N 2)$]].
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+Let $x = 0$ or $y = 0$, or $x + y = 0$, the result is trivial.
+Let $x = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \norm x_p
+      | r = 0
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | ll= \leadsto
+      | l = \norm x_p + \norm y_p
+      | r = \norm y_p
+      | c = 
+}}
+{{eqn | r = \norm {x + y}_p
+      | c = 
+}}
+{{end-eqn}}
+and so $\norm {x + y}_p \le \norm x_p + \norm y_p$
+The same argument holds for $y = 0$.
+Let $x + y = 0$.
+{{begin-eqn}}
+{{eqn | l = x + y
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \norm {x + y}_p
+      | r = 0
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | o = \le
+      | r = \norm x_p + \norm y_p
+      | c = as $\norm x_p \ge 0$ and $\norm y_p \ge 0$ from [[Definition:Norm Axioms|Norm Axioms: Axiom $(N1)$]]
+}}
+{{end-eqn}}
+Let $x, y, x + y \in \Q_{\ne 0}$.
+From [[P-adic Valuation is Valuation|$p$-adic Valuation is Valuation]]:
+:$\map {\nu_p} {x + y} \ge \min \set {\map {\nu_p} x, \map {\nu_p} y}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}_p
+      | r = p^{-\map {\nu_p} {x + y} }
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | o = \le
+      | r = \max \set {p^{-\map {\nu_p} x}, p^{-\map {\nu_p} y} }
+}}
+{{eqn | r = \max \set {\norm x_p, \norm y_p}
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | o = \le
+      | r = \norm x_p + \norm y_p
+}}
+{{end-eqn}}
+Thus the [[Definition:P-adic Norm|$p$-adic norm]] fulfils [[Definition:Norm Axioms|axiom $(\text N 3)$]].
+{{qed|lemma}}
+All [[Definition:Norm Axioms|norm axioms]] are seen to be satisfied.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall that the [[Definition:P-adic Norm|$p$-adic norm]] is defined as:
+:$\forall q \in \Q: \norm r_p := \begin{cases}
+  0 & : r = 0 \\
+  p^{- k} & : r \ne 0
+\end{cases}$
+where:
+:$\displaystyle r = p^k \frac m n$
+and:
+:$k, n \in \Z, m \in \Z_{\ne 0} : p \nmid m, n$
+where $\nmid$ stands for [[Symbols:Number Theory/Does Not Divide|"does not divide"]].
+We must show the following hold for all $r_1$, $r_2 \in \Q$:
+{{begin-axiom}}
+{{axiom | n = \text N 1
+        | q = \forall r \in \Q
+        | ml= \norm r_p = 0
+        | mo= \iff
+        | mr= x = 0
+}}
+{{axiom | n = \text N 2
+        | q = \forall r_1, r_2 \in \Q
+        | ml= \norm {r_1 r_2}
+        | mo= =
+        | mr= \norm {r_1}_p \times \norm {r_2}_p
+}}
+{{axiom | n = \text N 3
+        | q = \forall r_1, r_2 \in \Q
+        | ml= \norm {r_1 + r_2}_p
+        | mo= \le
+        | mr= \norm {r_1}_p + \norm {r_2}_p
+}}
+{{end-axiom}}
+=== Norm Axiom $(\text N 1)$ ===
+Let $r \in \Q : r \ne 0$.
+Let $k, m\in \Z, n \in \Z_{\ne 0} : p \nmid m, n$.
+Suppose $r = 0$.
+By [[Definition:P-adic Norm|definition]]:
+:$\norm {r}_p = 0$
+Suppose $\displaystyle r = p^k \frac m n \ne 0$
+By [[Definition:P-adic Norm|definition]]:
+:$\displaystyle \norm {r}_p = \frac 1 {p^k} > 0$
+Suppose $\norm r_p = 0$.
+By [[Definition:P-adic Norm|definition]]:
+:$r = 0$
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+Suppose $r_1 = 0$ or $r_2 = 0$.
+From [[Definition:Norm Axioms|axiom $(\text N 1)$]], $\norm {r_1}_p = 0$ or $\norm {r_2}_p = 0$.
+Suppose $r_1 \ne 0 \ne r_2$.
+Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid n_1, n_2, m_1, m_2$
+Let $\displaystyle r_1 = p^{k_1} \frac {m_1} {n_1}, r_2 = p^{k_2} \frac {m_2} {n_2}$
+Then:
+:$\displaystyle r_1 r_2 = p^{k_1 + k_2} \frac {m_1 m_2}{n_1 n_2}$
+We have that $p \nmid m_1$, $p \nmid m_2$.
+Since $p$ is [[Definition:Prime Number|prime]]:
+:$p \nmid m_1 m_2$.
+Similarly:
+:$p \nmid n_1 n_2$.
+Therefore:
+{{begin-eqn}}
+{{eqn| l = \norm {r_1 r_2}_p 
+     | r = p^{- \paren {k_1 + k_2} }
+}}
+{{eqn| r = p^{-k_1} p^{-k_2} 
+}}
+{{eqn| r = \norm {r_1}_p \norm {r_2}_p
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+Suppose one of the following is true:
+:$r_1 = 0$
+:$r_2 = 0$
+:$r_1 + r_2 = 0$
+Then the result is straightforward.
+Suppose $r_1 \ne 0$, $r_2 \ne 0$, $r_1 + r_2 \ne 0$.
+Let $\displaystyle r_1 = p^{k_1} \frac {m_1}{n_1}, r_2 = p^{k_2} \frac{m_2}{n_2}$ where:
+:$\displaystyle k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$
+Then:
+{{begin-eqn}}
+{{eqn | l = r_1 + r_2
+      | r = \frac {p^{k_1} m_1 n_2 + p^{k_2} m_2 n_1} {n_1 n_2}
+}}
+{{eqn | r = p^{\map \min {k_1, k_2} } \frac {p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2}{n_1 n_2}
+      | c = {{Defof|Min Operation}}
+}}
+{{eqn | r = p^{\map \min {k_1, k_2} } \frac {\tilde m}{n_1 n_2}
+      | c = $\displaystyle \tilde m := p^{k_1 \mathop - \map \min {k_1, k_2} } m_1 n_2 + p^{k_2 \mathop - \map \min {k_1, k_2} } n_1 m_2$
+}}
+{{end-eqn}}
+By [[Fundamental Theorem of Arithmetic]]:
+:$\exists ! \tilde k \in \Z_{\ge 0} : \exists m \in \Z : p \nmid m : \tilde m = p^{\tilde k} m$
+Obviously, $p \nmid n_1 n_2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {r_1 + r_2}_p
+      | r = \frac 1 {p^{\tilde k + \map \min {k_1, k_2} } }
+}}
+{{eqn | o = \le
+      | r = \frac 1 {p^{\map \min {k_1, k_2} } }
+}}
+{{eqn | r = \map \max {p^{-k_1}, p^{-k_2} }
+      | c = {{Defof|Max Operation}}
+}}
+{{eqn | r = \map \max {\norm {r_1}_p, \norm {r_1}_p}
+}}
+{{eqn | o = \le
+      | r = \map \max {\norm {r_1}_p, \norm {r_2}_p} + \map \min {\norm {r_1}_p, \norm {r_2}_p}
+}}
+{{eqn | r = \norm {r_1}_p + \norm {r_2}_p
+}}
+{{end-eqn}}
+{{qed|lemma}}
+All [[Definition:Norm Axioms|norm axioms]] are seen to be satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fundamental Theorem of Algebra}
+Tags: Polynomial Theory, Algebra, Analysis, Fundamental Theorems, Fundamental Theorem of Algebra
+
+\begin{theorem}
+Every non-[[Definition:Constant Polynomial|constant]] [[Definition:Polynomial|polynomial]] with [[Definition:Polynomial Coefficient|coefficients]] in $\C$ has a [[Definition:Root of Polynomial|root]] in $\C$.
+\end{theorem}
+
+\begin{proof}
+Let $p \left({z}\right)$ be a [[Definition:Polynomial|polynomial]] in $\C$:
+: $p \left({z}\right) = z^m + a_1 z^{m-1} + \cdots + a_m$
+where not all of $a_1, \ldots, a_m$ are zero.
+Define a [[Definition:Free Homotopy|homotopy]]:
+: $p_t \left({z}\right) = t p \left({z}\right) + \left({1-t}\right) z^m$
+Then:
+: $\displaystyle \frac{p_t \left({z}\right)}{z^m} = 1 + t \left({a_1 \frac 1 z + \cdots + a_m \frac 1 {z^m}}\right)$
+The terms in the parenthesis go to $0$ as $z \to \infty$.
+{{explain}}
+Therefore, there is an $r \in \R_{>0}$ such that:
+: $\forall z \in \C: \left|{z}\right| = r: \forall t \in \left[{0 \,.\,.\, 1}\right]: p_t \left({z}\right) \ne 0$
+Hence the homotopy:
+: $\displaystyle \frac{p_t}{ \left|{p_t}\right|}:S \to \Bbb S^1$
+is [[Definition:Well-Defined Mapping|well-defined]] for all $t$.
+This shows that for any [[Definition:Complex Number|complex]] polynomial $p \left({z}\right)$ of order $m$, there is a circle $S$ of sufficiently large radius in $\C$ such that $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ and $\dfrac{z^m}{\left|{z^m}\right|}$ are [[Definition:Free Homotopy|freely homotopic]] maps $S \to \Bbb S^1$. 
+Hence $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ must have the same degree of $\left({z / r}\right)^m$, which is $m$.  
+When $m > 0$, that is $p$ is non-[[Definition:Constant Polynomial|constant]], this result and the [[Extendability Theorem for Intersection Numbers]] imply $\dfrac{p \left({z}\right)}{\left|{p \left({z}\right)}\right|}$ does not extend to the disk $\operatorname{int} \left({S}\right)$, implying $p \left({z}\right) = 0$ for some $z \in \operatorname{int} \left({S}\right)$.
+{{explain|$\operatorname{int} \left({S}\right)$}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $p: \C \to \C$ be a [[Definition:Complex Number|complex]] [[Definition:Polynomial|polynomial]] with $p \left({z}\right) \ne 0$ for all $z \in \C$.
+Then $p$ extends to a continuous [[Definition:Complex Transformation|transformation]] of the [[Definition:Riemann Sphere|Riemann sphere]]:
+:$\hat \C = \C \cup \left\{{\infty}\right\}$
+This extension also has no zeroes.
+{{MissingLinks|Write and link to appropriate version of "continuous".}}
+{{explain|why this extension has no zeroes}}
+By [[Riemann Sphere is Compact]], there is some $\epsilon \in \R_{>0}$ such that:
+:$\forall z \in \C: \left|{p \left({z}\right)}\right| \ge \epsilon$
+Now consider the [[Definition:Holomorphic Function|holomorphic function]] $g: \C \to \C$ defined by:
+: $g \left({z}\right) := \dfrac 1 {p \left({z}\right)}$
+We have:
+: $\forall z \in \C: \left|{g \left({z}\right)}\right| \le \dfrac 1 \epsilon$
+By [[Liouville's Theorem (Complex Analysis)|Liouville's Theorem]], $g$ is [[Definition:Constant Mapping|constant]].
+Hence $p$ is also [[Definition:Constant Polynomial|constant]], as claimed.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $p: \C \to \C$ be a [[Definition:Complex Number|complex]], non-[[Definition:Constant Polynomial|constant]] [[Definition:Polynomial|polynomial]].
+{{AimForCont}} that $\map p z \ne 0$ for all $z \in \C$.
+Now consider the [[Definition:Closed Complex Contour Integral|closed contour integral]]:
+:$\displaystyle \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z$
+where $\gamma_R$ is a [[Definition:Circle|circle]] with [[Definition:Radius of Circle|radius]] $R$ around the [[Definition:Origin|origin]].
+By [[Derivative of Complex Polynomial]], the polynomial $z \cdot \map p z$ is [[Definition:Holomorphic Function|holomorphic]].
+Since $\map p z$ is assumed to have no [[Definition:Root of Function|zeros]], the only zero of $z \cdot \map p z$ is $0 \in \C$.
+Therefore by [[Reciprocal of Holomorphic Function]] $\dfrac 1 {z \cdot \map p z}$ is [[Definition:Holomorphic Function|holomorphic]] in $\C \setminus \set 0$.
+Hence the [[Cauchy-Goursat Theorem]] implies that the value of this integral is independent of $R > 0$.
+On the one hand, one can calculate the value of this integral in the [[Definition:Limit of Function|limit]] $R \to 0$ (or use the [[Residue Theorem]]), using the [[Definition:Parameterization (Curve)|parameterization]] $z = R e^{i \phi}$ of $\gamma_R$:
+{{begin-eqn}}
+{{eqn | l = \lim \limits_{R \mathop \to 0} \oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z
+      | r = \frac 1 {\map p 0} \lim \limits_{R \mathop \to 0} \int \limits_0^{2 \pi} \frac 1 {R e^{i \phi} } \, i R e^{i \phi} \rd \phi
+      | c = [[Real Polynomial Function is Continuous]] and [[Product Rule for Limits of Functions|Product Rule]]
+}}
+{{eqn | r = \lim_{R \mathop \to 0} \frac 1 {\map p 0} \int_0^{2 \pi} i \rd \phi
+}}
+{{eqn | r = \frac {2 \pi i} {\map p 0}
+      | c =
+}}
+{{end-eqn}}
+which is non-zero.
+On the other hand, we have the following [[Definition:Upper Bound of Mapping|upper bound]] for the [[Definition:Absolute Value|absolute value]] of the integral:
+{{begin-eqn}}
+{{eqn | l = \size {\oint \limits_{\gamma_R} \frac 1 {z \cdot \map p z} \rd z}
+      | o = \le
+      | r = 2 \pi R \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {z \cdot \map p z} } }
+      | c = [[Estimation Lemma]]
+}}
+{{eqn | r = 2 \pi \max \limits_{\size z \mathop = R} \paren {\frac 1 {\size {\map p z} } }
+}}
+{{end-eqn}}
+But this goes to zero for $R \to \infty$.
+We arrive at a contradiction.
+Hence the assumption that $\map p z \ne 0$ for all $z \in \C$ must be wrong.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extendability Theorem for Intersection Numbers}
+Tags: Topology
+
+\begin{theorem}
+Let $X = \partial W$ be a [[Definition:Smooth Manifold|smooth manifold]] which is the [[Definition:Boundary (Topology)|boundary]] of a smooth [[Definition:Compact Topological Space|compact]] manifold $W$.
+Let $Y$ be a smooth manifold, $Z$ be a closed smooth [[Definition:Submanifold|submanifold]] of $Y$, and $f: X \to Y$ a smooth [[Definition:Mapping|map]].
+Let there exist a smooth map $g: W \to Y$ such that $g \restriction_X = f$.
+Then:
+:$I \left({f, Z}\right) = 0$
+where $I \left({f, Z}\right)$ is the [[Definition:Intersection Number|intersection number]].
+{{explain|what $I \left({f, Z}\right)$ is the intersection number of: presumably the words will go something like "... the intersection number of $f$ with respect to $Z$", or something.}}
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Topology]]
+5suse11w8ty3k67myho9jncdwdrvohm
+\end{proof}<|endoftext|>
+\section{Relative Homotopy is Equivalence Relation}
+Tags: Homotopy Theory
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
+Let $K \subseteq X$ be a (possibly [[Definition:Empty Set|empty]]) [[Definition:Subset|subset]] of $X$.
+Let $\map \CC {X, Y}$ be the set of all [[Definition:Continuous Mapping (Topology)|continuous mappings]] from $X$ to $Y$.
+Define a [[Definition:Relation|relation]] $\sim$ on $\map \CC {X, Y}$ as:
+:$f \sim g$ {{iff}} $f$ and $g$ are [[Definition:Relative Homotopy|homotopic relative to $K$]].
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+We examine each condition for [[Definition:Equivalence Relation|equivalence]].
+=== Reflexivity ===
+For any function $f: X \to Y$, define $H: X \times \closedint 0 1 \to Y$ by $\map H {x, t} := \map f x$.
+This yields a [[Definition:Homotopy|homotopy]] between $f$ and itself.
+Also, trivially, if $x \in K$ and $t \in \closedint 0 1$, then:
+:$\map f x = H \left({x, t}\right)$
+so that $H$ is a [[Definition:Relative Homotopy|homotopy relative to $K$]].
+Thus $\sim$ is a [[Definition:Reflexive Relation|reflexive relation]].
+{{qed|lemma}}
+=== Symmetry ===
+Given a [[Definition:Relative Homotopy|$K$-relative homotopy]]:
+:$H: X \times \closedint 0 1 \to Y$ 
+from $\map f x = \map H {x, 0}$ to $\map g x = \map H {x, 1}$, the function:
+:$\map G {x, t} = \map H {x, 1 - t}$
+is a [[Definition:Relative Homotopy|$K$-relative homotopy]] from $g$ to $f$.
+Thus $\sim$ is a [[Definition:Symmetric Relation|symmetric relation]].
+{{qed|lemma}}
+=== Transitivity ===
+Suppose that $f \sim g$ and $g \sim h$.
+Let $F, G: X \times \closedint 0 1 \to Y$ be [[Definition:Relative Homotopy|$K$-relative homotopies]] between $f$ and $g$, $g$ and $h$, respectively.
+Define $H: X \times \closedint 0 1 \to Y$ by:
+:$\map H {x, t} := \begin {cases}
+\map F {x, 2 t} & : 0 \le t \le \dfrac 1 2 \\
+\map G {x, 2 t - 1} & : \dfrac 1 2 \le t \le 1
+\end{cases}$
+By [[Continuous Mapping on Finite Union of Closed Sets]], $H$ is a [[Definition:Relative Homotopy|$K$-relative homotopy]] between $f$ and $h$.
+Thus $\sim$ is a [[Definition:Transitive Relation|transitive relation]].
+{{qed|lemma}}
+Having verified all three conditions, it follows that $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+[[Category:Homotopy Theory]]
+k4bkjkcd1kcgumsnpf4mlv4xrsm2po4
+\end{proof}<|endoftext|>
+\section{Homotopy Group is Homeomorphism Invariant}
+Tags: Homotopy Theory, Algebraic Topology
+
+\begin{theorem}
+Let $X$ and $Y$ be two [[Definition:Topological Space|topological spaces]].
+Let $\phi: X \to Y$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+Let $x_0 \in X$, $y_0 \in Y$.
+Then for all $n \in \N$ the [[Definition:Induced Mapping of Homotopy Groups|induced mapping]]:
+:$\phi_* : \pi_n \left({X, x_0}\right) \to \pi_n \left({Y, y_0}\right):$
+::$\left[{\!\left[{\, c \,}\right]\!}\right] \mapsto \left[{\!\left[{\, \phi \circ c \,}\right]\!}\right]$
+is an [[Definition:Group Isomorphism|isomorphism]], where $\pi_n$ denotes the $n$th [[Definition:Homotopy Group|homotopy group]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi: X \to Y$ be a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+We must show that:
+:$(1): \quad$ If $c: \left[{0 \,.\,.\, 1}\right]^n \to X$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]], then $ \phi \circ c: \left[{0 \,.\,.\, 1}\right]^n \to Y$ is also [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
+:$(2): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are [[Definition:Free Homotopy|freely homotopic]], then $\phi \circ c, \phi \circ d: \left[{0 \,.\,.\, 1}\right]^n \to Y$ are also [[Definition:Free Homotopy|freely homotopic]]
+:$(3): \quad$ If $c, d: \left[{0 \,.\,.\, 1}\right]^n \to X$ are not [[Definition:Free Homotopy|freely homotopic]], there can be no [[Definition:Free Homotopy|free homotopy]] between $\phi \circ c$ and $\phi \circ d$
+:$(4): \quad$ The image of the [[Definition:Concatenation (Topology)|concatenation]] of two maps, $\phi \left({c * d}\right)$, is [[Definition:Free Homotopy|freely homotopic]] to the [[Definition:Concatenation (Topology)|concatenation]] of the images, $\phi \left({c}\right) * \phi \left({d}\right)$.
+{{finish}}
+[[Category:Homotopy Theory]]
+[[Category:Algebraic Topology]]
+k56fmuxw5p3gs43xabyg5nndpghuet7
+\end{proof}<|endoftext|>
+\section{Homotopy Group is Group}
+Tags: Homotopy Theory, Algebraic Topology, Examples of Groups
+
+\begin{theorem}
+The set of all [[Definition:Homotopy Class|homotopy classes]] of [[Definition:Continuous Mapping (Topology)|continuous mappings]]:
+:$c: \closedint 0 1^n \to X$
+satisfying:
+:$\map c {\partial \closedint 0 1^n} = x_0$
+in a [[Definition:Topological Space|space]] $X$ at a base point $x_0$, under the operation of [[Definition:Concatenation (Topology)|concatenation]] on class members, forms a [[Definition:Group|group]].
+{{finish|This operation needs to be shown well-defined; probably a nasty proof}}
+This group is called the $n$th [[Definition:Homotopy Group|homotopy group]].
+\end{theorem}
+
+\begin{proof}
+We examine each of the [[Definition:Group Axioms|group axioms]] separately.
+=== $\text G 0$: Closure ===
+The [[Definition:Concatenation (Topology)|concatenation]] of any two [[Definition:Mapping|mappings]]:
+:$c_1, c_2: \closedint 0 1^n \to X$
+from any two (not necessarily distinct) [[Definition:Equivalence Class|equivalence classes]] is another [[Definition:Mapping|mapping]]:
+:$c_3: \closedint 0 1^n \to X$
+by the definition of concatenation, which will have its own [[Definition:Equivalence Class|equivalence class]].
+=== $\text G 1$: Associativity === 
+Let $c_1, c_2, c_3$ be three functions $\closedint 0 1^n \to X$, selected from three (not necessarily different) [[Definition:Equivalence Class|equivalence classes]].
+The concatenation $\paren {c_1 * c_2} * c_3$ is, from the definition of concatenation:
+:$\map A {\hat v} = \begin {cases}
+\map {c_1} {4 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 4} \\
+\map {c_2} {4 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac 1 4} {\dfrac 1 2} \\
+\map {c_3} {2 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 1 2} 1
+\end{cases}$
+Likewise, the concatenation $c_1 * \paren {c_2 * c_3}$ is by definition:
+:$\map B {\hat v} = \begin {cases}
+\map {c_1} {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\
+\map {c_2} {4 v_1 - 2, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac 1 2} {\dfrac 3 4} \\
+\map {c_3} {4 v_1 - 3, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 3 4} 1
+\end {cases}$
+We construct a homotopy:
+:$\map H {\hat v, t} = \begin {cases}
+\map {c_1} {\dfrac {4 v_1} {1 + t}, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac {1 + t} 4} \\
+\map {c_2} {4 v_1 - t - 1, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac {1 + t} 4} {\dfrac {2 + t} 4} \\
+\map {c_3} {\dfrac {4 v_1 - \paren {2 + t} } {2 - t}, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac {2 + t} 4} 1
+\end {cases}$
+We observe it satisfies $\map H {\hat v, 0} = \map A {\hat v}$ and $\map H {\hat v, 1} = \map B {\hat v}$.
+Therefore $A$ and $B$ are in the same [[Definition:Equivalence Class|equivalence class]].
+=== $\text G 2$: Identity === 
+The identity is simply the function $I: \closedint 0 1^n \to X$ defined as $\map I {\hat v} = x_0$, where $\hat v \in \closedint 0 1^n$.
+Suppose we are given the function $c: \closedint 0 1^n \to X$ and its concatenation with $i$:
+:$\map {\paren {c*i} } {\hat v} = \begin {cases}
+\map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\
+\map I {\hat v} = x_0 & : v_1 \in \closedint {\dfrac 1 2} 1
+\end {cases}$
+We construct a homotopy:
+:$\map H {\hat v, t} = \begin {cases}
+\map c {\dfrac {2 v_1} {2 - t}, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {1 - \dfrac {1 - t} 2} \\
+\map I {\hat v} = x_0 & : v_1 \in \closedint {1 - \dfrac {1 - t} 2} 1
+\end{cases}$ 
+which satisfies $\map H {\hat v, 0} = \map c {\hat v}$ and $\map H {\hat v, 1} = \map {\paren {c*i} } {\hat v}$.
+This shows $c$ and $c*i$ are in the same [[Definition:Equivalence Class|equivalence class]].
+=== $\text G 3$: Inverses ===
+For any $c: \closedint 0 1^n \to X$, $\map {c^{-1} } {\hat v} = \map c {\tuple {1, 0, \ldots, 0} - \hat v}$.
+Then we can construct a homotopy:
+:$\map H {\hat v, t} = \begin {cases}
+\map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac {1 - t} 2} \\
+\map c {1 - t, v_2, \ldots, v_n} & : v_1 \in \openint {\dfrac {1 - t} 2} {\dfrac {1 + t} 2} \\
+\map {c^{-1} } {2 v_1 - 1, v_2, \ldots v_n} & : v_1 \in \closedint {\dfrac {1 + t} 2} 1
+\end {cases}$
+We observe it satisfies:
+:$\map H {\hat v, 0} = \begin {cases}
+\map c {2 v_1, v_2, \ldots, v_n} & : v_1 \in \closedint 0 {\dfrac 1 2} \\
+\map {c^{-1} } {2 v_1 - 1, v_2, \ldots, v_n} & : v_1 \in \closedint {\dfrac 1 2} 1
+\end {cases}$
+and:
+:$\map H {\hat v, 1} = x_0$
+Hence $I$ and $c * c^{-1}$ are in the same [[Definition:Equivalence Class|equivalence class]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative at Maximum or Minimum}
+Tags: Differential Calculus, Derivative at Maximum or Minimum
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let $f$ have a [[Definition:Local Minimum|local minimum]] or [[Definition:Local Maximum|local maximum]] at $\xi \in \openint a b$.
+Then:
+:$\map {f'} \xi = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Local Maximum|local maximum]]:
+:$(1): \quad \map f {\xi + \epsilon} - \map f \xi < 0$
+for [[Definition:Sufficiently Small|sufficiently small]] $\epsilon \in \R_{>0}$.
+Similarly, definition of [[Definition:Local Minimum|local minimum]]:
+:$(2): \quad \map f {\xi + \epsilon} - \map f \xi > 0$
+for [[Definition:Sufficiently Small|sufficiently small]] $\epsilon \in \R_{>0}$.
+Let it be assumed that $\map f {\xi + \epsilon}$ can be expanded, using [[Taylor's Theorem]], in [[Definition:Positive Integer|positive]] [[Definition:Integer Power|integer powers]] of $\epsilon$
+Then:
+:$(3): \quad \map f {\xi + \epsilon} - \map f \xi = \epsilon \map {f'} \xi + \dfrac {\epsilon^2 \map {f''} \xi} 2 + \map \OO {\epsilon^3}$
+where $\map \OO {\epsilon^3}$ denotes [[Definition:Big-O Notation|big-$\OO$ notation]].
+That is, the quantity $\dfrac {\map \OO {\epsilon^3} } {\epsilon^3}$ is [[Definition:Bounded Real-Valued Function|bounded]].
+From $(1)$ and $(2)$, at a [[Definition:Local Maximum|local maximum]] or [[Definition:Local Minimum|local minimum]], the [[Definition:Sign of Number|sign]] of $\map f {\xi + \epsilon} - \map f \xi$ is independent the [[Definition:Sign of Number|sign]] of $\epsilon$ itself.
+So from $(3)$ it follows that $\map {f'} = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Behaviour of Function Near Limit}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]].
+Let $f \left({x}\right) \to l$ as $x \to \xi$.
+Then:
+* If $l > 0$, then $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: f \left({x}\right) > 0$
+* If $l < 0$, then $\exists h > 0: \forall x: \xi - h < x < \xi + h, x \ne \xi: f \left({x}\right) < 0$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Limit of Real Function|limit of a function]]:
+:$\forall \epsilon > 0: \exists \delta > 0: 0 < \left|{x - \xi}\right| < \delta \implies \left|{f \left({x}\right) - l}\right| < \epsilon$
+Let $l > 0$.
+Since this is true for all $\epsilon > 0$, it is also true for $\epsilon = l$.
+So let the value of $\delta$, for the above to be true, labelled $h$.
+Then:
+: $0 < \left|{x - \xi}\right| < h \implies \left|{f \left({x}\right) - l}\right| < l$
+That is:
+: $\xi - h < x < \xi + h, x \ne \xi \implies 0 = l - l < f \left({x}\right) < l + l = 2 l$
+Hence:
+: $\forall x: \xi - h < x < \xi + h, x \ne \xi: 0 < f \left({x}\right)$
+Now let $l < 0$, and consider $\epsilon = -l$.
+A similar thread of reasoning leads us to:
+: $\xi - h < x < \xi + h, x \ne \xi \implies -2 l < f \left({x}\right) < 0$
+and hence the second result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fundamental Group is Independent of Base Point for Path-Connected Space}
+Tags: Algebraic Topology
+
+\begin{theorem}
+Let $X$ be a [[Definition:Path-Connected|path-connected]] [[Definition:Topological Space|space]].
+For $x \in X$ let $\pi_1(X,x)$ denote the [[Definition:Fundamental Group|fundamental group]].
+For $x, y \in X$, there is an [[Definition:Group Isomorphism|isomorphism]]:
+:$\phi: \pi_1 \left({X, x}\right) \to \pi_1 \left({X, y}\right)$
+\end{theorem}
+
+\begin{proof}
+Since $X$ is [[Definition:Path-Connected|path-connected]] there exists a [[Definition:Path (Topology)|path]] $f$ connecting $y$ and $x$.
+We define $\phi_f: \pi_1 \left({X, x}\right) \to \pi_1 \left({X, y}\right)$ by $\phi_f \left({ \left[{g}\right] }\right) =\left[{ f^{-1} g f }\right]$.
+$\phi$ is a [[Definition:Group Homomorphism|homomorphism of groups]], as is seen from:
+:$\displaystyle \phi_f \left({ \left[{g h}\right] }\right) = \left[{ f^{-1} g h f }\right] = \left[{ f^{-1} g f f^{-1} h f }\right] = \left[{ f^{-1} g f }\right] \left[{ f^{-1} h f }\right] = \phi_f \left({ \left[{g}\right] }\right) \phi_f \left({ \left[{h}\right] }\right)$ 
+Also, $f^{-1}$ is a path from $x$ to $y$. Then by the same argument as before, $\phi_{f^{-1}}: \pi_1 \left({X, y}\right) \to \pi_1 \left({X, x}\right)$ where $\phi_{f^{-1}} \left({ \left[{g}\right] }\right) = \left[{ f g f^{-1} }\right]$, is a [[Definition:Group Homomorphism|homomorphism of groups]].
+Trivially, $\phi_f \circ \phi_{f^{-1}} = I_{\pi_1 \left({X, y}\right)}$ and $\phi_{f^{-1}} \circ \phi_f=I_{\pi_1 \left({X, x}\right)}$.
+Then $\phi_f^{-1} = \phi_{f^{-1}}$, so $\phi_f$ is [[Definition:Bijection|bijective]] and thus an [[Definition:Group Isomorphism|isomorphism]].
+{{qed}}
+[[Category:Algebraic Topology]]
+9vnhgyxsw38bkybqgtelfb0suywyrkj
+\end{proof}<|endoftext|>
+\section{List of Fundamental Groups for 2-Manifolds}
+Tags: Algebraic Topology
+
+\begin{theorem}
+For the following two-manifolds, the [[Definition:Homotopy Group|fundamental group]] for [[Fundamental Group is Independent of Base Point for Path-Connected Space|any point in $X$]], written $\pi_1 \left({X}\right)$ is [[Definition:Group Isomorphism|isomorphic]] to the listed group:
+:$\pi_1 \left({\Bbb S^1 \times \left[{0\,.\,.\,1}\right]}\right) = \Z$
+:$\pi_1 \left({\Bbb S^2}\right) = \left\{{e}\right\}$, the [[Definition:Trivial Group|trivial group]].
+:$\pi_1 \left({\Bbb T^2}\right) = \pi_1 \left({\Bbb S^1 \times \Bbb S^1}\right) = \Z \times \Z$
+:$\pi_1 \left({\Bbb {RP}^2}\right) = \Z_2$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Algebraic Topology]]
+2jz4qe9gjealmhsnl068shic4scjq2e
+\end{proof}<|endoftext|>
+\section{Rolle's Theorem}
+Tags: Differential Calculus, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is:
+:[[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$
+and:
+:[[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let $\map f a = \map f b$.
+Then:
+:$\exists \xi \in \openint a b: \map {f'} \xi = 0$
+\end{theorem}
+
+\begin{proof}
+We have that $f$ is [[Definition:Continuous on Interval|continuous]] on $\closedint a b$.
+It follows from [[Continuous Image of Closed Interval is Closed Interval]] that $f$ attains:
+:a [[Definition:Maximum Value|maximum]] $M$ at some $\xi_1 \in \closedint a b$
+and:
+:a [[Definition:Minimum Value|minimum]] $m$ at some $\xi_2 \in \closedint a b$.
+Suppose $\xi_1$ and $\xi_2$ are both end points of $\closedint a b$.
+Because $\map f a = \map f b$ it follows that $m = M$ and so $f$ is [[Definition:Constant Mapping|constant]] on $\closedint a b$.
+Then, by [[Derivative of Constant]], $\map {f'} \xi = 0$ for all $\xi \in \openint a b$.
+Suppose $\xi_1$ is not an end point of $\closedint a b$.
+Then $\xi_1 \in \openint a b$ and $f$ has a [[Definition:Local Maximum|local maximum]] at $\xi_1$.
+Hence the result follows from [[Derivative at Maximum or Minimum]].
+Similarly, suppose $\xi_2$ is not an end point of $\closedint a b$.
+Then $\xi_2 \in \openint a b$ and $f$ has a [[Definition:Local Minimum|local minimum]] at $\xi_2$.
+Hence the result follows from [[Derivative at Maximum or Minimum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mean Value Theorem}
+Tags: Differential Calculus, Mean Value Theorem, Named Theorems
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Closed Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then:
+:$\exists \xi \in \openint a b: \map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$
+\end{theorem}
+
+\begin{proof}
+From [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+By the [[Extreme Value Theorem]], there exist $m, M \in \closedint a b$ such that:
+:$\displaystyle \map f m = \min_{x \mathop \in \closedint a b} \map f x$
+:$\displaystyle \map f M = \max_{x \mathop \in \closedint a b} \map f x$
+Then, from [[Upper and Lower Bounds of Integral]]:
+:$\displaystyle \map f m \paren {b - a} \le \int_a^b \map f x \rd x \le \map f M \paren {b - a}$ 
+Dividing all terms by $\paren {b - a}$ gives:
+:$\displaystyle \map f m \le \frac 1 {b - a}\int_a^b \map f x \rd x \le \map f M$
+By the [[Intermediate Value Theorem]], there exists some $k \in \openint a b$ such that:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {b - a} \int_a^b \map f x \rd x
+      | r = \map f k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int_a^b \map f x \rd x
+      | r = \map f k \paren {b - a}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+Let $F : \closedint a b \to \R$ be a [[Definition:Real Function|real function]] defined by: 
+:$\displaystyle \map F x = \int_a^x \map f x \rd x$
+We are assured that this function is [[Definition:Well-Defined|well-defined]], since $f$ is [[Definition:Darboux Integrable Function|integrable]] on $\closedint a b$.
+From [[Fundamental Theorem of Calculus/First Part|Fundamental Theorem of Calculus: First Part]], we have: 
+: $F$ is [[Definition:Continuous Real Function|continuous]] on $\closedint a b$
+: $F$ is [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$ with [[Definition:Derivative of Real Function|derivative]] $f$
+By the [[Mean Value Theorem]], there therefore exists $k \in \openint a b$ such that: 
+:$\map {F'} k = \dfrac {\map F b - \map F a} {b - a}$
+As $F$ is [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$ with [[Definition:Derivative of Real Function|derivative]] $f$: 
+:$\map {F'} k = \map f k$
+We therefore have: 
+{{begin-eqn}}
+{{eqn	| l = \map f k
+	| r = \frac {\map F b - \map F a} {b - a}
+}}
+{{eqn	| r = \frac 1 {b - a} \paren {\int_a^b \map f x \rd x - \int_a^a \map f x \rd x}
+}}
+{{eqn	| r = \frac 1 {b - a} \int_a^b \map f x \rd x
+	| c = [[Definite Integral on Zero Interval]]
+}}
+{{end-eqn}}
+giving: 
+:$\displaystyle \int_a^b \map f x \rd x = \paren {b - a} \map f k$
+as required.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $g : \closedint a b \to \R$ be a [[Definition:Real Function|real function]] with: 
+:$\map g x = x$
+for all $x \in \closedint a b$. 
+By [[Power Rule for Derivatives]], we have: 
+:$g$ is [[Definition:Differentiable Real Function|differentiable]] with $\map {g'} x = 1$ for all $x \in \closedint a b$.
+Note that in particular: 
+:$\map {g'} x \ne 0$ for all $x \in \openint a b$.
+Since $f$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] on $\closedint a b$ and [[Definition:Differentiable on Interval|differentiable]] on $\openint a b$, we can apply the [[Cauchy Mean Value Theorem]].
+We therefore have that there exists $\xi \in \openint a b$ such that: 
+:$\dfrac {\map {f'} \xi} {\map {g'} \xi } = \dfrac {\map f b - \map f a} {\map g b - \map g a}$
+Note that: 
+:$\map {g'} \xi = 1$
+and: 
+:$\map g b - \map g a = b - a$
+so this can be rewritten: 
+:$\map {f'} \xi = \dfrac {\map f b - \map f a} {b - a}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Derivative implies Constant Function}
+Tags: Differential Calculus, Constant Mappings
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Suppose that:
+:$\forall x \in \left({a \,.\,.\, b}\right): f' \left({x}\right) = 0$
+Then $f$ is [[Definition:Constant Mapping|constant]] on $\left[{a \,.\,.\, b}\right]$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in \left[{a \,.\,.\, b}\right]$.
+Then $f$ satisfies the conditions of the [[Mean Value Theorem]] on $\left[{a \,.\,.\, y}\right]$.
+Hence:
+:$\exists \xi \in \left({a \,.\,.\, y}\right): f' \left({\xi}\right) = \dfrac {f \left({y}\right) - f \left({a}\right)} {y - a}$
+But:
+:$f' \left({\xi}\right) = 0$
+which means:
+:$f \left({y}\right) - f \left({a}\right) = 0$
+and hence:
+:$f \left({y}\right) = f \left({a}\right)$
+As $y$ is any $y \in \left[{a \,.\,.\, b}\right]$, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem}
+Tags: Normed Division Rings, P-adic Number Theory, Ostrowski's Theorem
+
+\begin{theorem}
+Every [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$ is [[Definition:Equivalent Division Ring Norms|equivalent]] to either:
+:the [[Definition:P-adic Norm|$p$-adic norm $\norm {\, \cdot \,}_p$]] for some [[Definition:Prime Number|prime]] $p$
+or:
+:the [[Definition:Absolute Value|absolute value]], $\size {\, \cdot \,}$.
+\end{theorem}
+
+\begin{proof}
+Let $\norm {\, \cdot \,}$ be a [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+=== [[Ostrowski's Theorem/Archimedean Norm|Archimedean Norm Case]] ===
+Let $\norm {\, \cdot \,}$ be an [[Definition:Archimedean Division Ring Norm|Archimedean norm]].
+{{:Ostrowski's Theorem/Archimedean Norm}}{{qed|lemma}}
+=== [[Ostrowski's Theorem/Non-Archimedean Norm|Non-Archimedean Norm Case]] ===
+Let $\norm {\, \cdot \,}$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean Norm]].
+{{:Ostrowski's Theorem/Non-Archimedean Norm}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy Mean Value Theorem}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f$ and $g$ be a [[Definition:Real Function|real functions]] which are [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Suppose that:
+:$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$
+Then:
+:$\exists \xi \in \left({a \,.\,.\, b}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$
+\end{theorem}
+
+\begin{proof}
+Let $F$ be the [[Definition:Real Function|real function]] defined on $\left[{a \,.\,.\, b}\right]$ by $F \left({x}\right) = f \left({x}\right) + h g \left({x}\right)$, where $h \in \R$ is a constant.
+The plan is to choose the [[Definition:Constant|constant]] $h$ such that $F \left({a}\right) = F \left({b}\right)$ and so apply [[Rolle's Theorem]].
+We need to make:
+:$f \left({a}\right) + h g \left({a}\right) = f \left({b}\right) + h g \left({b}\right)$
+We have that:
+:$\forall x \in \left({a \,.\,.\, b}\right): g' \left({x}\right) \ne 0$
+So, by [[Rolle's Theorem]]:
+:$g \left({a}\right) \ne g \left({b}\right)$
+Thus we can let:
+:$h = - \dfrac {f \left({b}\right) - f \left({a}\right)} {g \left({b}\right) - g \left({a}\right)}$
+So, by [[Rolle's Theorem]]:
+:$\exists \xi \in \left({a \,.\,.\, b}\right): 0 = F' \left({\xi}\right) = f' \left({\xi}\right) + h g' \left({\xi}\right)$
+That is:
+: $h = - \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)}$
+Hence the result, from the definition of the [[Definition:Derivative|derivative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homology Group is Group}
+Tags: Topology, Algebraic Topology
+
+\begin{theorem}
+The [[Definition:Homology Group|$p^{th}$ singular homology group]] of a space $X$ is a group.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Topology]]
+[[Category:Algebraic Topology]]
+7difl1riz3sxyegrwjint110egiqvtx
+\end{proof}<|endoftext|>
+\section{Limit of Absolute Value}
+Tags: Limits of Functions, Absolute Value Function
+
+\begin{theorem}
+Let $x, \xi \in \R$ be [[Definition:Real Number|real numbers]].
+Then:
+:$\left\vert{x - \xi}\right\vert \to 0$ as $x \to \xi$
+where $\left\vert{x - \xi}\right\vert$ denotes the [[Definition:Absolute Value|Absolute Value]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+Let $\delta = \epsilon$.
+From the definition of a [[Definition:Limit of Real Function|limit of a function]], we need to show that $\left\vert{f \left({x}\right) - 0}\right\vert < \epsilon$ provided that $0 < \left\vert{x - \xi}\right\vert < \delta$, where $f \left({x}\right) = \left\vert{x - \xi}\right\vert$.
+Thus, provided $0 < \left\vert{x - \xi}\right\vert < \delta$, we have:
+{{begin-eqn}}
+{{eqn | l=\left\vert{x - \xi}\right\vert - 0
+      | r=\left\vert{x - \xi}\right\vert
+      | c=
+}}
+{{eqn | o=<
+      | r=\delta
+      | c=
+}}
+{{eqn | r=\epsilon
+      | c=
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Function in Interval}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let $\xi \in \openint a b$
+Suppose that, $\forall x \in \openint a b$, either:
+:$\xi \le \map f x \le x$
+or:
+:$x \le \map f x \le \xi$
+Then $\map f x \to \xi$ as $x \to \xi$.
+\end{theorem}
+
+\begin{proof}
+Note that $\size {\map f x - \xi} \le \size {\xi - x}$.
+From [[Limit of Absolute Value]] we have that $\size {x - \xi} \to 0$ as $x \to \xi$.
+The result follows from the [[Squeeze Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Monotone Function}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+\end{theorem}<|endoftext|>
+\section{Strictly Monotone Real Function is Bijective}
+Tags: Real Analysis, Monotone Mappings, Bijections
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on $I \subseteq \R$.
+Let $f$ be [[Definition:Strictly Monotone Real Function|strictly monotone]] on $I$.
+Let the [[Definition:Image of Mapping|image]] of $f$ be $J$. 
+Then $f: I \to J$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+From [[Strictly Monotone Mapping with Totally Ordered Domain is Injective]], $f$ is an [[Definition:Injection|injection]].
+From [[Surjection by Restriction of Codomain]], $f: I \to J$ is a [[Definition:Surjection|surjection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Monotone Function}
+Tags: Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on $I \subseteq \R$.
+Let $f$ be [[Definition:Strictly Monotone Real Function|strictly monotone]] on $I$.
+Let the [[Definition:Image of Mapping|image]] of $f$ be $J$. 
+Then $f$ always has an [[Definition:Inverse Mapping|inverse function]] $f^{-1}$ and:
+: if $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] then so is $f^{-1}$
+: if $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] then so is $f^{-1}$.
+\end{theorem}
+
+\begin{proof}
+The function $f$ is a [[Definition:Bijection|bijection]] from [[Strictly Monotone Real Function is Bijective]].
+Hence from [[Bijection iff Inverse is Bijection]], $f^{-1}$ always exists and is also a [[Definition:Bijection|bijection]].
+From the definition of [[Definition:Strictly Increasing Real Function|strictly increasing]]:
+:$x < y \iff \map f x < \map f y$
+Hence:
+:$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} < \map {f^{-1} } {\map f y}$
+and so:
+:$\map {f^{-1} } x < \map {f^{-1} } y \iff x < y$
+Similarly, from the definition of [[Definition:Strictly Decreasing Real Function|strictly decreasing]]:
+:$x < y \iff \map f x > \map f y$
+Hence:
+:$\map {f^{-1} } x < \map {f^{-1} } y \iff \map {f^{-1} } {\map f x} > \map {f^{-1} } {\map f y}$
+and so:
+:$\map {f^{-1} } x < \map {f^{-1} } y \iff x > y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rokhlin's Theorem (Intersection Forms)}
+Tags: Topology
+
+\begin{theorem}
+Let $M$ be a [[Definition:Smooth Manifold|smooth 4-manifold]].
+Then:
+:$\map {\omega_2} {\map T M} = 0 \implies \operatorname {sign} Q_M = 0 \pmod {16}$
+where:
+:$Q_M$ is the [[Definition:Intersection Form|intersection form]]
+:$\map T M$ is the [[Definition:Tangent Bundle|tangent bundle]]
+:$\omega_2$ is the second [[Definition:Stiefel-Whitney Class|Stiefel-Whitney class]].
+\end{theorem}
+
+\begin{proof}
+{{Explain|What does "sign" mean in this context?}}
+{{ProofWanted}}
+{{Namedfor|Vladimir Abramovich Rokhlin|cat = Rokhlin}}
+[[Category:Topology]]
+eyik4t5vysus44jyt17o6qq3g6yjoz0
+\end{proof}<|endoftext|>
+\section{Convex Real Function is Continuous}
+Tags: Convex Real Functions, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Convex Real Function|convex]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then $f$ is [[Definition:Continuous on Interval|continuous]] on $\left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Convex Real Function is Left-Hand and Right-Hand Differentiable]], we have:
+:$\displaystyle \lim_{h \to 0^-} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^-} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^-} h}\right) = 0$
+and similarly:
+:$\displaystyle \lim_{h \to 0^+} f \left({x + h}\right) - f \left({x}\right) = \left({\lim_{h \to 0^+} \frac {f \left({x + h}\right) - f \left({x}\right)} h}\right) \left({\lim_{h \to 0^+} h}\right) = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convex Real Function is Left-Hand and Right-Hand Differentiable}
+Tags: Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is either [[Definition:Convex Real Function|convex]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Then the [[Definition:Left-Hand Derivative|left-hand derivative]] $f'_- \left({x}\right)$ and [[Definition:Right-Hand Derivative|right-hand derivative]] $f'_+ \left({x}\right)$ both exist for all $x \in \left({a \,.\,.\, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Convex Real Function|convex]] on $\left({a \,.\,.\, b}\right)$.
+Then by definition of [[Definition:Convex Real Function|convexity]]:
+:$\forall x_1, x_2, x_3 \in \left({a \,.\,.\, b}\right): x_1 < x_2 < x_3: \dfrac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \dfrac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$
+Let $0 < h_1 < h_2$.
+Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then:
+:$\dfrac {f \left({x + h_1}\right) - f \left({x}\right)} {h_1} \le \dfrac {f \left({x + h_2}\right) - f \left({x}\right)} {h_2}$
+Hence the function $F \left({h}\right) = \dfrac {f \left({x + h}\right) - f \left({x}\right)} h$ [[Definition:Increasing Real Function|increases]] in some $\left({0 \,.\,.\, \delta}\right)$.
+Thus from [[Limit of Increasing Function]] it follows that both $\displaystyle \lim_{h \to 0^+} F \left({h}\right) = f'_+ \left({x}\right)$ and $\displaystyle \lim_{h \to 0^-} F \left({h}\right) = f'_-\left({x}\right)$ exist.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Convex iff Derivative is Increasing}
+Tags: Differential Calculus, Convex Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Then:
+:$f$ is [[Definition:Convex Real Function|convex]] on $\openint a b$
+{{iff}}:
+:its [[Definition:Derivative on Interval|derivative]] $f'$ is [[Definition:Increasing Real Function|increasing]] on $\openint a b$.
+Thus the intuitive result that a [[Definition:Convex Real Function|convex function]] "gets steeper".
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Convex Real Function|convex]] on $\openint a b$.
+Let $x_1, x_2, x_3, x_4 \in \openint a b$ such that:
+:$x_1 < x_2 < x_3 < x_4$
+By the definition of [[Definition:Convex Real Function|convex function]]:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} \le \dfrac {\map f {x_4} - \map f {x_3} } {x_4 - x_3}$
+Ignore the middle term and let $x_2 \to x_1^+$ and $x_3 \to x_4^-$.
+Thus:
+:$\map {f'} {x_1} \le \map {f'} {x_4}$
+Hence $f'$ is [[Definition:Increasing Real Function|increasing]] on $\openint a b$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f'$ be [[Definition:Increasing Real Function|increasing]] on $\openint a b$.
+Let $x_1, x_2, x_3 \in \openint a b$ such that $x_1 < x_2 < x_3$.
+By the [[Mean Value Theorem]]:
+:$\exists \xi: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} = \map {f'} \xi$
+:$\exists \eta: \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2} = \map {f'} \eta$
+where $x_1 < \xi < x_2 < \eta < x_3$.
+Since $f'$ is [[Definition:Increasing Real Function|increasing]]:
+:$\map {f'} \xi \le \map {f'} \eta$
+Thus:
+:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \le \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
+Hence $f$ is [[Definition:Convex Real Function|convex]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Strictly Increasing Convex Real Function is Concave}
+Tags: Convex Real Functions, Concave Real Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Convex Real Function|convex]] on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $J = f \left[{I}\right]$.
+If $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$, then $f^{-1}$ is [[Definition:Concave Real Function|concave]] on $J$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X = f \left({x}\right) \in J$
+:$Y = f \left({y}\right) \in J$.
+From the definition of [[Definition:Convex Real Function|convex]]:
+: $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$
+Let $f$ be [[Definition:Strictly Increasing Real Function|strictly increasing]] on $I$.
+From [[Inverse of Strictly Monotone Function]] it follows that, $f^{-1}$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $J$.
+Thus:
+: $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$
+Hence $f^{-1}$ is [[Definition:Concave Real Function|concave]] on $J$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper and Lower Bounds of Integral}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Closed Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $\displaystyle \int_a^b \map f x \rd x$ be the [[Definition:Definite Integral|definite integral]] of $\map f x$ over $\closedint a b$.
+Then:
+:$\displaystyle m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
+where:
+:$M$ is the [[Definition:Maximal Element|maximum]] of $f$
+:$m$ is the [[Definition:Minimal Element|minimum]] of $f$
+on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+This follows directly from the definition of [[Definition:Definite Integral|definite integral]]:
+From [[Continuous Image of Closed Interval is Closed Interval]] it follows that $m$ and $M$ both exist.
+The [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ is a [[Definition:Finite Subdivision|finite subdivision]] of itself.
+By definition, the [[Definition:Upper Sum|upper sum]] is $M \paren {b - a}$, and the [[Definition:Lower Sum|lower sum]] is $m \paren {b - a}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Wilson's Theorem}
+Tags: Prime Numbers, Factorials, Modulo Arithmetic, Wilson's Theorem
+
+\begin{theorem}
+A [[Definition:Strictly Positive Integer|(strictly) positive integer]] $p$ is a [[Definition:Prime Number|prime]] {{iff}}:
+:$\paren {p - 1}! \equiv -1 \pmod p$
+\end{theorem}
+
+\begin{proof}
+If $p = 2$ the result is obvious.
+Therefore we assume that $p$ is an [[Definition:Odd Prime|odd prime]].
+=== [[Wilson's Theorem/Necessary Condition|Necessary Condition]] ===
+{{:Wilson's Theorem/Necessary Condition}}
+=== [[Wilson's Theorem/Sufficient Condition|Sufficient Condition]] ===
+{{:Wilson's Theorem/Sufficient Condition}}
+\end{proof}<|endoftext|>
+\section{Integral of Constant}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $c$ be a [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Let $F$ be a [[Definition:Primitive (Calculus)|primitive]] of $f$ on $\closedint a b$.
+By [[Primitive of Constant Multiple of Function]], $H = c F$ is a [[Definition:Primitive (Calculus)|primitive]] of $c f$ on $\closedint a b$.
+Hence by the [[Fundamental Theorem of Calculus]]:
+{{begin-eqn}}
+{{eqn | l = \int_a^b c \map f x \rd x
+      | r = \bigintlimits {c \map F x} a b
+      | c = 
+}}
+{{eqn | r = c \bigintlimits {\map F x} a b
+      | c = 
+}}
+{{eqn | r = c \int_a^b \map f x \rd x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $F$ be a [[Definition:Primitive (Calculus)|primitive]] of $f$ on $\closedint a b$.
+By [[Linear Combination of Definite Integrals]]:
+:$\displaystyle \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
+for [[Definition:Real Function|real functions]] $f$ and $g$ which are [[Definition:Integrable Function|integrable]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, where $\lambda$ and $\mu$ be [[Definition:Real Number|real numbers]].
+The result follows by setting $\lambda = c$ and $\mu = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Integrals on Adjacent Intervals for Continuous Functions}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on any [[Definition:Closed Real Interval|closed interval]] $I$.
+Let $a, b, c \in I$.
+Then:
+:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$
+\end{theorem}
+
+\begin{proof}
+By [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Definite Integral|integrable]] on $I$.
+The result follows by application of [[Sum of Integrals on Adjacent Intervals for Integrable Functions]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitives which Differ by Constant}
+Tags: Primitives
+
+\begin{theorem}
+Let $F$ be a [[Definition:Primitive (Calculus)|primitive]] for a [[Definition:Real Function|real function]] $f$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $G$ be a [[Definition:Real Function|real function]] defined on $\closedint a b$.
+Then $G$ is a [[Definition:Primitive (Calculus)|primitive]] for $f$ on $\closedint a b$ {{iff}}:
+:$\exists c \in \R: \forall x \in \closedint a b: \map G x = \map F x + c$
+That is, {{iff}} $F$ and $G$ differ by a constant on the whole interval.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $G$ is a [[Definition:Primitive (Calculus)|primitive]] for $f$.
+Then $F - G$ is [[Definition:Continuous on Interval|continuous]] on $\closedint a b$, [[Definition:Differentiable on Interval|differentiable]] on $\openint a b$, and for any $x \in \openint a b$, we have:
+{{begin-eqn}}
+{{eqn | l = \map {D_x} {\map F x - \map G x}
+      | r = \map {D_x} {\map F x} - \map {D_x} {\map G x}
+      | c = [[Sum Rule for Derivatives]]
+}}
+{{eqn | r = \map f x - \map f x
+      | c = $F, G$ are a [[Definition:Primitive (Calculus)|primitives]] for $f$
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+From [[Zero Derivative implies Constant Function]] it follows that $F - G$ is constant on $\closedint a b$, hence the result.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose $\map G x = \map F x + c$.
+We compute:
+{{begin-eqn}}
+{{eqn | l = D_x \map G x
+      | r = \map {D_x} {\map F x + c}
+}}
+{{eqn | r = \map {D_x} {\map F x} + 0
+      | c = [[Sum Rule for Derivatives]] and [[Derivative of Constant]]
+}}
+{{eqn | r = \map f x
+      | c = $F$ is a [[Definition:Primitive (Calculus)|primitive]] for $f$
+}}
+{{end-eqn}}
+Hence $G$ is also a [[Definition:Primitive (Calculus)|primitive]] for $f$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral of Function plus Constant}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $c$ be a constant.
+Then:
+:$\ds \int_a^b \paren {\map f t + c} \rd t = \int_a^b \map f t \rd t + c \paren {b - a}$
+\end{theorem}
+
+\begin{proof}
+Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
+Let $\map {L^{\paren {f + c} } } P$ be the [[Definition:Lower Sum|lower sum]] of $\map f x + c$ on $\closedint a b$ belonging to $P$.
+Let:
+:$\ds m_k^{\paren {f + c} } = \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}$
+where $k \in \set {0, 1, \ldots, n}$.
+So:
+{{begin-eqn}}
+{{eqn | l = m_k^{\paren {f + c} }
+      | r = \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x + c}
+      | c = 
+}}
+{{eqn | r = c + \map {\inf_{x \mathop \in \closedint {x_{k - 1} } {x_k} } } {\map f x}
+      | c =
+}}
+{{eqn | r = c + m_k^{\paren f}
+      | c =
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map {L^{\paren {f + c} } } P
+      | r = \sum_{k \mathop = 1}^n {m_k^{\paren {f + c} } } \paren {x_k - x_{k - 1} }
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n m_k^{\paren f} \paren {x_k - x_{k - 1} } + c \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }
+      | c = 
+}}
+{{eqn | r = \map {L^{\paren f} } P + c \paren {b - a}
+      | c = as $\ds \sum_{k \mathop = 1}^n \paren {x_k - x_{k - 1} }$ [[Telescoping Series/Example 2|telescopes]]
+}}
+{{end-eqn}}
+So from the definition of [[Definition:Definite Integral|definite integral]], it follows that:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \paren {\map f t + c} \rd t
+      | r = \map {\sup_P} {\map {L^{\paren {f + c} } } P}
+      | c = 
+}}
+{{eqn | r = \map {\sup_P} {\map {L^{\paren f} } P + c \paren {b - a} }
+      | c = 
+}}
+{{eqn | r = \map {\sup_P} {\map {L^{\paren f} } P} + c \paren {b - a}
+      | c = 
+}}
+{{eqn | r = \int_a^b \map f t \rd t + c \paren {b - a}
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Real Function is Darboux Integrable}
+Tags: Integral Calculus, Continuous Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Then $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We have the result $f$ is [[Definition:Bounded Real-Valued Function|bounded]] by [[Continuous Real Function is Bounded]].
+By [[Condition for Darboux Integrability]], it suffices to show that for all $\epsilon > 0$, there exists a [[Definition:Subdivision (Real Analysis)|subdivision]] $P$ of $\closedint a b$ such that:
+:$\map U P – \map L P < \epsilon$
+where $\map U P$ and $\map L P$ denote the [[Definition:Upper Sum|upper sum]] and [[Definition:Lower Sum|lower sum]] of $\map f x$ on $\closedint a b$ belonging to the [[Definition:Subdivision (Real Analysis)|subdivision]] $P$.
+Let $\epsilon > 0$.
+We have the result [[Continuous Function on Closed Interval is Uniformly Continuous]].
+By the definition of [[Definition:Uniformly Continuous Real Function|uniform continuity]], there exists a $\delta > 0$ such that if $x, y \in \closedint a b$ are such that $\size {x – y} < \delta$, then:
+:$\size {\map f x – \map f y} < \dfrac \epsilon {b - a}$
+Let $P = \set {x_0, x_1, x_2, \ldots, x_n}$ be a [[Definition:Subdivision (Real Analysis)|subdivision]] of $\closedint a b$ such that:
+:$\displaystyle \max_{1 \mathop \le k \mathop \le n} \paren {x_k – x_{k - 1} } < \delta$
+For all [[Definition:Integer|integers]] $k$ satisfying $1 \le k \le n$, it follows from the [[Heine-Borel Theorem/Real Line|Heine-Borel theorem]] that $\closedint {x_{k - 1} } {x_k}$ is [[Definition:Compact Space|compact]].
+So we can apply [[Continuous Image of Compact Space is Compact/Corollary 3|Corollary 3 to Continuous Image of Compact Space is Compact]] to conclude that there exist $u_k, v_k \in \closedint {x_{k - 1} } {x_k}$ such that:
+{{begin-eqn}}
+{{eqn | l = \map f {u_k}
+      | r = \sup \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }
+}}
+{{eqn | l = \map f {v_k}
+      | r = \inf \set {\map f x: x \in \closedint {x_{k - 1} } {x_k} }
+}}
+{{end-eqn}}
+By assumption, $x_k – x_{k - 1} < \delta$, so:
+:$\size {u_k – v_k} < \delta$
+It follows from the definition of $\delta$ that:
+:$\displaystyle \map f {u_k} – \map f {v_k} < \frac \epsilon {b – a}$
+This gives:
+{{begin-eqn}}
+{{eqn | l = \map U P – \map L P
+      | r = \sum_{k \mathop = 1}^n \map f {u_k} \paren {x_k – x_{k - 1} } - \sum_{k \mathop = 1}^n \map f {v_k} \paren {x_k – x_{k - 1} }
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {\map f {u_k} – \map f {v_k} } \paren {x_k – x_{k - 1} }
+}}
+{{eqn | o = <
+      | r = \frac \epsilon {b – a} \sum_{k \mathop = 1}^n \paren {x_k – x_{k - 1} }
+}}
+{{eqn | r = \frac \epsilon {b – a} \paren {x_n – x_0}
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral on Zero Interval}
+Tags: Definite Integrals
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on the [[Definition:Closed Real Interval|closed interval]] $\Bbb I := \closedint a b$, where $a < b$.
+Then:
+:$\displaystyle \forall c \in \Bbb I: \int_c^c \map f t \rd t = 0$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Definite Integral|definite integral]].
+There is only one [[Definition:Finite Subdivision|finite subdivision]] of $\closedint c c$ and that is $\set c$.
+Both the [[Definition:Lower Sum|lower sum]] and [[Definition:Upper Sum|upper sum]] of $\map f x$ on $\closedint c c$ belonging to the [[Definition:Finite Subdivision|finite subdivision]] $\set c$ are equal to [[Definition:Zero (Number)|zero]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Derivatives}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f \left({x}\right), g \left({x}\right)$ be [[Definition:Real Function|real functions]] defined on the [[Definition:Open Real Interval|open interval]] $I$.
+Let $\xi \in I$ be a point in $I$ at which both $f$ and $g$ are [[Definition:Differentiable Real Function at Point|differentiable]].
+Then:
+:$D \left({\lambda f + \mu g}\right) = \lambda D f + \mu D g$
+at the point $\xi$.
+It follows from the definition of [[Definition:Derivative on Interval|derivative]] that if $f$ and $g$ are both [[Definition:Differentiable on Interval|differentiable]] on the interval $I$, then:
+:$\forall x \in I: D \left({\lambda f \left({x}\right) + \mu g \left({x}\right)}\right) = \lambda D f \left({x}\right) + \mu D g \left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \frac 1 h \left({\lambda f \left({\xi + h}\right) + \mu g \left({\xi + h}\right) - \lambda f \left({\xi}\right) - \mu g \left({\xi}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \lambda \left({\frac {f \left({\xi + h}\right) - f \left({\xi}\right)} h}\right) + \mu \left({\frac {g \left({\xi + h}\right) - g \left({\xi}\right)} h}\right)
+      | c = 
+}}
+{{eqn | o = \to
+      | r = \lambda D f \left({\xi}\right) + \mu D g \left({\xi}\right)
+      | c = as $h \to 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Integrals}
+Tags: Integral Calculus
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Integrable Function|integrable]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $\lambda$ and $\mu$ be [[Definition:Real Number|real numbers]].
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{Integration by Parts}
+Tags: Integral Calculus, Named Theorems, Proof Techniques, Integration by Parts
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function on Closed Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $f$ and $g$ have [[Definition:Primitive (Calculus)|primitives]] $F$ and $G$ respectively on $\closedint a b$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Integration by Substitution}
+Tags: Integral Calculus, Named Theorems, Proof Techniques, Integration by Substitution
+
+\begin{theorem}
+Let $\phi$ be a [[Definition:Real Function|real function]] which has a [[Definition:Derivative|derivative]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $I$ be an [[Definition:Open Real Interval|open interval]] which contains the [[Definition:Image of Subset under Mapping|image]] of $\closedint a b$ under $\phi$.
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on $I$.
+\end{theorem}<|endoftext|>
+\section{Relative Sizes of Definite Integrals}
+Tags: Integral Calculus, Inequalities
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Continuous Real Function on Closed Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$, where $a < b$.
+If:
+:$\forall t \in \closedint a b: \map f t \le \map g t$
+then:
+:$\displaystyle \int_a^b \map f t \rd t \le \int_a^b \map g t \rd t$
+Similarly, if:
+:$\forall t \in \closedint a b: \map f t < \map g t$
+then:
+:$\displaystyle \int_a^b \map f t \rd t < \int_a^b \map g t \rd t$
+\end{theorem}
+
+\begin{proof}
+Suppose that $\forall t \in \closedint a b: \map f t \le \map g t$.
+From the [[Fundamental Theorem of Calculus]], $g - f$ has a [[Definition:Primitive (Calculus)|primitive]] on $\closedint a b$.
+Let $H$ be such a [[Definition:Primitive (Calculus)|primitive]].
+Then:
+:$\forall t \in \closedint a b: D_t \map H t = \map g t - \map f t \ge 0$
+By [[Derivative of Monotone Function]], it follows that $H$ is [[Definition:Increasing Real Function|increasing]] on $\closedint a b$.
+Thus:
+:$\map H b \ge \map H a$
+Hence:
+:$\displaystyle \int_a^b \map g t - \map f t \rd t = \map H b - \map H a \ge 0$
+The proof for the second case is similar.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value of Definite Integral}
+Tags: Integral Calculus, Triangle Inequality
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Then:
+:$\displaystyle \size {\int_a^b \map f t \rd t} \le \int_a^b \size {\map f t} \rd t$
+\end{theorem}
+
+\begin{proof}
+From [[Negative of Absolute Value]], we have for all $a \in \closedint a b$:
+:$-\size {\map f t} \le \map f t \le \size {\map f t}$
+Thus from [[Relative Sizes of Definite Integrals]]:
+:$\displaystyle -\int_a^b \size {\map f t} \rd t \le \int_a^b \map f t \rd t \le \int_a^b \size {\map f t} \rd t$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Test}
+Tags: Integral Calculus, Convergence Tests, Series
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Half Open Interval|continuous]], [[Definition:Positive Real Function|positive]] and [[Definition:Decreasing Real Function|decreasing]] on the [[Definition:Half-Open Real Interval|interval]] $\hointr 1 {+\infty}$.
+Let the [[Definition:Sequence|sequence]] $\sequence {\Delta_n}$ be defined as:
+:$\displaystyle \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
+Then $\sequence {\Delta_n} $ is [[Definition:Decreasing Real Sequence|decreasing]] and [[Definition:Bounded Below Set|bounded below]] by zero.
+Hence it [[Definition:Convergent Sequence|converges]].
+\end{theorem}
+
+\begin{proof}
+From [[Upper and Lower Bounds of Integral]], we have that:
+:$\displaystyle m \paren {b - a} \le \int_a^b \map f x \rd x \le M \paren {b - a}$
+where:
+:$M$ is the [[Definition:Maximum Value|maximum]]
+and:
+:$m$ is the [[Definition:Minimum Value|minimum]]
+of $\map f x$ on $\closedint a b$.
+Since $f$ decreases, $M = \map f a$ and $m = \map f b$.
+Thus it follows that:
+:$\displaystyle \forall k \in \N_{>0}: \map f {k + 1} \le \int_k^{k + 1} \map f x \rd x \le \map f k$
+as $\paren {k + 1} - k = 1$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \Delta_{n + 1} - \Delta_n
+      | r = \paren {\sum_{k \mathop = 1}^{n + 1} \map f k - \int_1^{n + 1} \map f x \rd x} - \paren {\sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x}
+      | c = 
+}}
+{{eqn | r = \map f {n + 1} - \int_n^{n + 1} \map f x \rd x
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \map f {n + 1} - \map f {n + 1}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Thus $\sequence {\Delta_n}$ is [[Definition:Decreasing Real Function|decreasing]].
+{{qed|lemma}}
+Also:
+{{begin-eqn}}
+{{eqn | l = \Delta_n
+      | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \int_k^{k + 1} \map f x \rd x
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \sum_{k \mathop = 1}^n \map f k - \sum_{k \mathop = 1}^{n - 1} \map f k
+      | c = 
+}}
+{{eqn | r = \map f n
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Non-Measurable Subset of Real Numbers}
+Tags: Measure Theory
+
+\begin{theorem}
+There exists a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]] which is not [[Definition:Measurable Set|measurable]].
+\end{theorem}
+
+\begin{proof}
+We construct such a [[Definition:Set|set]].
+For $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the [[Definition:Modulo Addition|sum modulo 1]]:
+:$x +_1 y = \begin{cases} x + y & : x + y < 1 \\ x + y - 1 & : x + y \ge 1 \end{cases}$
+Let $E \subset \left[{0 \,.\,.\, 1}\right)$ be a [[Definition:Measurable Set|measurable set]].
+Let $E_1 = E \cap \left[{0 \,.\,.\, 1 - x}\right)$ and $E_2 = E \cap \left[{1 - x \,.\,.\, 1}\right)$.
+By [[Measure of Interval is Length]], these [[Definition:Disjoint Sets|disjoint]] [[Definition:Real Interval|intervals]] are [[Definition:Measurable Set|measurable]].
+By [[Measurable Sets form Algebra of Sets]], so are these [[Definition:Set Intersection|intersections]] $E_1$ and $E_2$.
+So:
+:$m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$
+We have:
+: $E_1 +_1 x = E_1 + x$
+By [[Lebesgue Measure is Translation-Invariant]]:
+: $m \left({E_1 +_1 x}\right) = m \left({E_1}\right)$
+Also:
+: $E_2 +_1 x = E_2 + x - 1$, and so $m \left({E_2 +_1 x}\right) = m \left({E_2}\right)$
+Then we have:
+: $m \left({E +_1 x}\right) = m \left({E_1 +_1 x}\right) + m \left({E_2 +_1 x}\right) = m \left({E_1}\right) + m \left({E_2}\right) = m \left({E}\right)$
+So, for each $x \in \left[{0 \,.\,.\, 1}\right)$, the set $E +_1 x$ is [[Definition:Measurable Set|measurable]] and:
+:$m \left({E + x}\right) = m \left({E}\right)$
+Taking, as before, $x, y \in \left[{0 \,.\,.\, 1}\right)$, define the [[Definition:Relation|relation]]:
+:$x \sim y \iff x - y \in \Q$
+where $\Q$ is the [[Definition:Rational Number|set of rational numbers]].
+By [[Difference is Rational is Equivalence Relation]], $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+As this is an [[Definition:Equivalence Relation|equivalence relation]] we can invoke the [[Fundamental Theorem on Equivalence Relations|fundamental theorem on equivalence relations]].
+Hence $\sim$ [[Definition:Partition (Set Theory)|partitions]] $\left[{0 \,.\,.\, 1}\right)$ into [[Definition:Equivalence Class|equivalence classes]].
+By the [[Axiom:Axiom of Choice|axiom of choice]], there is a [[Definition:Set|set]] $P$ which contains exactly one [[Definition:Element|element]] from each [[Definition:Equivalence Class|equivalence class]].
+Let $\left\{{r_i}\right\}_{i \mathop = 0}^\infty$ be an [[Definition:Enumeration|enumeration]] of the [[Definition:Rational Number|rational numbers]] in $\left[{0 \,.\,.\, 1}\right)$ with $r_0 = 0$.
+Let $P_i := P +_1 r_i$.
+Then $P_0 = P$.
+Let $x \in P_i \cap P_j$.
+Then:
+:$x = p_i + r_i = p_j + r_j$
+where $p_i, p_j$ are [[Definition:Element|elements]] of $P$.
+But then $p_i - p_j$ is a [[Definition:Rational Number|rational number]].
+Since $P$ has only one [[Definition:Element|element]] from each [[Definition:Equivalence Class|equivalence class]]:
+:$i = j$
+The $P_i$ are [[Definition:Pairwise Disjoint|pairwise disjoint]].
+Each real number $x \in \left[{0 \,.\,.\, 1}\right)$ is in ''some'' [[Definition:Equivalence Class|equivalence class]] and hence is [[Definition:Equivalence Relation|equivalent]] to an [[Definition:Element|element]] of $P$.
+But if $x$ differs from an [[Definition:Element|element]] in $P$ by the [[Definition:Rational Number|rational number]] $r_i$, then $x \in P_i$ and so:
+:$\displaystyle \bigcup P_i = \left[{0 \,.\,.\, 1}\right)$
+Since each $P_i$ is a [[Definition:Translation|translation modulo $1$]] of $P$, each $P_i$ will be [[Definition:Measurable Set|measurable]] if $P$ is, with measure $m \left({P_i}\right) = m \left({P}\right)$.
+But if this were the case, then:
+:$\displaystyle m \left[{0 \,.\,.\, 1}\right) = \sum_{i \mathop = 1}^\infty m \left({P_i}\right) = \sum_{i \mathop = 1}^\infty m \left({P}\right)$
+Therefore:
+: $m \left({P}\right) = 0$ implies $m \left[{0 \,.\,.\, 1}\right) = 0$
+and:
+: $m \left({P}\right) \ne 0$ implies $m \left[{0 \,.\,.\, 1}\right) = \infty$
+This contradicts [[Measure of Interval is Length]].
+So the [[Definition:Set|set]] $P$ is not [[Definition:Measurable Set|measurable]].
+{{Qed}}
+{{AoC||4}}
+{{BPI}}
+{{explain|While BPI has been invoked as being necessary for this theorem, no explanation has been added as to why, or how, or where it would be applied.}}
+\end{proof}<|endoftext|>
+\section{Properties of Natural Logarithm}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Harmonic Series is Divergent}
+Tags: Harmonic Series is Divergent, Real Analysis, Harmonic Series, Examples of Divergent Series
+
+\begin{theorem}
+The [[Definition:Harmonic Series|harmonic series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n$ 
+[[Definition:Divergent Series|diverges]].
+\end{theorem}
+
+\begin{proof}
+:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$ 
+where $\displaystyle s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$
+From [[Ordering of Reciprocals]]:
+: $\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$
+so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k + 1} }$.
+The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:
+:$s_k > \dfrac{2^k} {2^{k + 1} } = \dfrac 1 2$
+Hence the harmonic sum $H_{2^m}$ satisfies the following inequality:
+{{begin-eqn}}
+{{eqn | l = H_{2^m}
+      | r = \sum_{n \mathop = 1}^{2^m} \frac 1 n
+      | c = 
+}}
+{{eqn | o = >
+      | r = \sum_{n \mathop = 1}^{2^m - 1} \frac 1 n
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^m \left({s_k}\right)
+      | c = 
+}}
+{{eqn | o = >
+      | r = 1 + \sum_{a \mathop = 0}^m \frac 1 2
+      | c = 
+}}
+{{eqn | r = 1 + \frac m 2
+      | c = 
+}}
+{{end-eqn}}
+The {{RHS}} diverges, from the [[Nth Term Test|$n$th term test]].
+The result follows from the the [[Comparison Test for Divergence]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Observe that all the terms of the harmonic series are [[Definition:Strictly Positive|strictly positive]].
+From [[Reciprocal Sequence is Strictly Decreasing]], the terms are [[Definition:Decreasing Sequence|decreasing]].
+Hence the [[Cauchy Condensation Test]] can be applied, and we examine the convergence of:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 1}^\infty 2^n \frac 1 {2^n}
+      | r = \sum_{n \mathop = 1}^\infty 1
+}}
+{{end-eqn}}
+This diverges, from the [[Nth Term Test|$n$th term test]].
+Hence $\displaystyle \sum \frac 1 n$ also diverges.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that the [[Integral of Reciprocal is Divergent]].
+Hence from the [[Integral Test]], the [[Definition:Harmonic Series|harmonic series]] also [[Definition:Divergent Series|diverges]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+For all $N \in \N$:
+:$\dfrac 1 N + \dfrac 1 {N + 1} + \cdots + \dfrac 1 {2 N} > N \cdot \dfrac 1 {2 N} = \dfrac 1 2$
+Hence, by [[Cauchy's Convergence Criterion for Series]], the Harmonic series is divergent.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Assume that for $G \ge 4$ that $\displaystyle \sum_{n = G}^\infty \frac{1}{n} = L < \infty$. Namely, that for some number $G$, there is a tail of the harmonic series which converges. 
+Then from [[Definition:Series/Sequence of Partial Sums]]:
+$s_N := \sum_{n = G}^N \frac{1}{n}$ is the partial sum of the above series. Which yields the sequence $\{ s_N \}$ of partial sums.  
+And, from [[Definition:Convergent Series]] we have that $\displaystyle \sum_{n = G}^\infty \frac{1}{n}$ converges iff $\{ s_N\}$ converges. 
+From [[Constant Sequence Converges to Constant in Normed Division Ring]]: The constant sequence $\{ G \}$ has limit $G$.  Note: $\R$ is a normed division ring as it is a field. 
+By [[Combination Theorem for Sequences/Real/Product Rule]]: The product of the sequences $\{ G \}$ and $\{ s_N\}$ has limit $GL$. Namely, the sequence $\{Gs_N\}$ has limit $GL$, by the opening assumption.
+$GL = \displaystyle \sum_{n = G}^\infty \frac{G}{n} =  \underbrace {1}_{s_0} + \underbrace{ \frac{G}{G+1} + \ldots + \frac{G}{G+4}}_{s_1} + \underbrace { \frac{G}{G+5} + \ldots + \frac{G}{G+12} }_{s_2} + \underbrace { \frac{G}{G+13} + \ldots + \frac{G}{G+28} }_{s_3} + \ldots $ 
+Where $s_0 = 1, s_1 = \frac{G}{G+1} + \ldots + \frac{G}{G+4}$ and for $k \ge 2, s_k = \displaystyle \sum_{i =  2^k + 2^{k-1} + \ldots 2^2 + 1}^{2^{k+1} + 2^{k} + \ldots 2^2} \frac{G}{G+i} $.
+From the above, $s_0 = 1, s_1 \ge \frac{4G}{G+4} \ge 1$ by inspection. 
+And for $k \ge 2$ then $s_k \ge \frac{2^{k+1}G}{G+2^{k+2}}$ since $\frac{G}{G+2^{k+2}}$ is smaller than the smallest summand of $s_k$. If summed $2^{k+1}$ many times, $2^{k+1}$ being the number of summands in $s_k$, it yields a result less than $s_k$. Note: The smallest summand of $s_k$ is $\frac{G}{G+2^{k+1}+ \ldots + 2^2}$.
+Claim: 
+For $k \ge 1, G \ge 4$ We have $\frac{2^{k+1}G}{G+2^{k+2}} \ge 1$ 
+Proof: 
+$\frac{2^{k+1}G}{G+2^{k+2}} \ge 1 \iff k + 1 \ge \log_2\frac{G}{G-2}$. If $G = 4$ then the righthand side of the second inequality is $\log_2(2) = 1$. If $G > 4$, then $ 1 < \frac{G}{G-2} < 2$. Namely as $G \uparrow$ we have $\frac{G}{G-2} \to 1$ meaning $\log_2\frac{G}{G-2} \to 0$.
+{{qed|lemma}}
+Now,
+$\displaystyle GL = s_0 + s_1 + s_2 + s_3 + \ldots \ge 1 + \frac{4G}{G+4} + \frac{8G}{G+16 } + \frac{16G}{G+32} + \ldots  \ge 1 + 1 + 1 + 1 +  \ldots \to \infty > GL$
+Deriving a contradiction.  Hence , the series does not converge which implies the sequence $\{s_N\}$ does not converge. Therefore by [[ Tail of Convergent Sequence ]] : A sequence $a_n$ converges iff the sequence $a_{n+N}, N \in \N$ converges. We have the tail of the harmonic series diverges for any $G$ thus the harmonic series will diverge.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Reciprocals of Primes is Divergent}
+Tags: Analytic Number Theory, Sum of Reciprocals of Primes is Divergent
+
+\begin{theorem}
+Let $n \in \N: n \ge 1$.
+There exists a [[Definition:Strictly Positive Real Number|(strictly) positive real number]] $C \in \R_{>0}$ such that:
+:$(1): \quad \displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - C$
+where $\Bbb P$ is the [[Definition:Set|set]] of all [[Definition:Prime Number|prime numbers]].
+:$(2): \quad \displaystyle \lim_{n \mathop \to \infty} \paren {\map \ln {\ln n} - C} = +\infty$
+\end{theorem}
+
+\begin{proof}
+By [[Sum of Reciprocals of Primes is Divergent/Lemma|Sum of Reciprocals of Primes is Divergent: Lemma]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {\map \ln {\map \ln n} - \frac 1 2} = +\infty$
+{{qed|lemma}}
+It remains to be proved that:
+:$\displaystyle \sum_{\substack {p \mathop \in \Bbb P \\ p \mathop \le n} } \frac 1 p > \map \ln {\ln n} - \frac 1 2$
+Assume all sums and product over $p$ are over the set of [[Definition:Prime Number|prime numbers]].
+Let $n \ge 1$.
+{{begin-eqn}}
+{{eqn | l = \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}
+      | r = \prod_{p \mathop \le n} \sum_{k \mathop = 0}^\infty \frac 1 {p^k}
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | o = \ge
+      | r = \sum_{k \mathop = 1}^n \frac 1 k 
+      | c = [[Fundamental Theorem of Arithmetic]]
+}}
+{{eqn | o = >
+      | r = \int_1^n \frac 1 x \rd x
+      | c = [[Integral Test]]
+}}
+{{eqn | n = 1
+      | r = \ln n
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1}
+      | r = \sum_{p \mathop \le n} \map \ln {1 - \frac 1 p}^{-1}
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = \sum_{p \mathop \le n} \sum_{k \mathop = 1}^\infty \frac 1 {k p^k}
+      | c = [[Power Series Expansion for Logarithm of 1 + x|Power Series Expansion for $\map \ln {1 + x}$]]
+}}
+{{eqn | o = <
+      | r = \sum_{p \mathop \le n} \frac 1 p + \sum_{p \mathop \le n} \frac 1 {2 p^2} \paren {\sum_{k \mathop = 0}^\infty \frac 1 {p^k} }
+      | c = 
+}}
+{{eqn | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{p \mathop \le n} \frac 1 {p \paren {p - 1} }
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | o = <
+      | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2 \sum_{n \mathop = 2}^\infty \frac 1 {n \paren {n - 1} }
+      | c = 
+}}
+{{eqn | r = \sum_{p \mathop \le n} \frac 1 p + \frac 1 2
+      | c = {{Defof|Telescoping Series}}
+}}
+{{eqn | o = >
+      | l = \sum_{p \mathop \le n} \frac 1 p
+      | r = \ln \prod_{p \mathop \le n} \paren {1 - \frac 1 p}^{-1} - \frac 1 2
+      | c = 
+}}
+{{eqn | o = >
+      | r = \map \ln {\ln n} - \frac 1 2
+      | c = by $(1)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
+Let $p_n$ denote the $n$th [[Definition:Prime Number|prime number]].
+Consider the [[Definition:Product Notation (Algebra)|product]]:
+:$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
+By [[Sum of Geometric Sequence]], we have:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {1 - \frac 1 2}
+      | r = 1 + \frac 1 2 + \frac 1 {2^2} + \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 3}
+      | r = 1 + \frac 1 3 + \frac 1 {3^2} + \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 5}
+      | r = 1 + \frac 1 5 + \frac 1 {5^2} + \cdots
+      | c = 
+}}
+{{eqn | o = \cdots
+      | c = 
+}}
+{{eqn | l = \frac 1 {1 - \frac 1 {p_n} }
+      | r = 1 + \frac 1 {p_n} + \frac 1 {p_n^2} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Consider what happens when all these [[Definition:Series|series]] are multiplied together.
+A new [[Definition:Series|series]] will be generated whose terms consist of all possible products of one term selected from each of the [[Definition:Series|series]] on the {{RHS}}.
+This new [[Definition:Series|series]] will [[Definition:Convergent Series|converge]] in any order to the product of the terms on the {{LHS}}.
+By the [[Fundamental Theorem of Arithmetic]], every [[Definition:Integer|integer]] greater than $1$ is uniquely expressible as a product of powers of different [[Definition:Prime Number|primes]].
+Hence the product of these [[Definition:Series|series]] is the [[Definition:Series|series]] of [[Definition:Reciprocal|reciprocals]] of all [[Definition:Strictly Positive Integer|(strictly) positive integers]] whose [[Definition:Prime Factor|prime factors]] are no greater than $p_n$.
+In particular, all [[Definition:Strictly Positive Integer|(strictly) positive integers]] up to $p_n$ have this property.
+So:
+:$\displaystyle \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}$
+{{begin-eqn}}
+{{eqn | l = \prod_{k \mathop = 1}^n \frac 1 {1 - 1 / p_k}
+      | o = \ge
+      | r = \sum_{k \mathop = 1}^{p_n} \frac 1 k
+      | c = 
+}}
+{{eqn | o = >
+      | r = \int_1^{p_n + 1} \dfrac {\mathrm d x} x
+      | c = 
+}}
+{{eqn | r = \map \ln {p_n + 1}
+      | c = 
+}}
+{{eqn | r = \ln p_n
+      | c = 
+}}
+{{end-eqn}}
+It follows by taking [[Definition:Reciprocal|reciprocals]] that:
+:$\displaystyle \prod_{k \mathop = 1}^n \paren {1 - \frac 1 {p_k} } < \frac 1 {\ln p_n}$
+Taking [[Definition:Logarithm|logarithms]] of each side:
+:$(1): \quad \displaystyle \sum_{k \mathop = 1}^n \map \ln {1 - \frac 1 {p_k} } < - \ln \ln p_n$
+Next, note that the [[Definition:Straight Line|line]] $y = 2 x$ in the [[Definition:Cartesian Plane|cartesian plane]] lies below the curve $y = \map \ln {1 + x}$ on the [[Definition:Half-Open Real Interval|interval]] $\closedint {-\frac 1 2} 0$.
+Also note that all [[Definition:Prime Number|primes]] are greater than or equal to $2$.
+Thus it follows that:
+:$-\dfrac 2 {p_k} < \map \ln {1 - \dfrac 1 {p_k} }$
+Applying this to $(1)$ yields:
+:$\displaystyle -2 \sum_{k \mathop = 1}^n \dfrac 1 {p_k} < -\ln \ln p_n$
+and so:
+:$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {p_k} > \dfrac 1 2 \ln \ln p_n$
+But:
+:$\displaystyle \lim_{n \mathop \to \infty} \ln \ln p_n \to \infty$
+and so the [[Definition:Series|series]]:
+:$\displaystyle \sum_{p \mathop \in \Bbb P} \frac 1 p$
+is [[Definition:Divergent Series|divergent]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{AimForCont}} the contrary.
+If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
+:$\displaystyle \sum_{n \mathop = k \mathop + 1}^{\infty} \frac 1 {p_n} < \frac 1 2$
+Let:
+:$\displaystyle Q = \prod_{i \mathop = 1}^k {p_i}$
+and:
+:$\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q}$
+Let $\map S {r, j}$ be the sum of all of the terms from $\map S r$ for which $1 + i Q$ has exactly $j$ prime factors.
+Notice that $1 + i Q$ is coprime with every prime factor in $Q$.
+Thus every prime factor of $1 + i Q$ where $i = 1, \ldots, r$ falls into some finite sequence of consecutive primes:
+:$\map P r = \set {p_{k + 1}, p_{k + 2}, \ldots, p_{\map m r} }$
+{{explain|What exactly is $\map m r$?}}
+Notice again that each term of $\map S {r, j}$ occurs at least once in the expansion of:
+:$\displaystyle \paren {\sum_{n \mathop = k \mathop + 1}^{\map m r} \frac 1 {p_n} }^j < \paren {\sum_{n \mathop = k \mathop + 1}^\infty \frac 1 {p_n} }^j < \paren {\frac 1 2}^j$
+and also by [[Sum of Infinite Geometric Sequence]]:
+:$\displaystyle \map S r = \sum_{j \mathop = 1}^r \map S {r, j} < \sum_{j \mathop = 1}^r \paren {\frac 1 2}^j < 1$
+for every $r$.
+Finally notice that:
+:$\displaystyle \map S r = \sum_{i \mathop = 1}^r \frac 1 {1 + i Q} > \frac 1 {1 + Q} \sum_{n \mathop = 1}^r \frac 1 n$
+which implies that $\map S r$ diverges towards $+\infty$ by [[Harmonic Series is Divergent]], a contradiction.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{AimForCont}} the contrary.
+If the prime reciprocal series converges then there must exist some $k \in \N$ such that:
+:$\displaystyle \sum_{n \mathop = k + 1}^\infty \frac 1 {p_n} < \frac 1 2$
+Let:
+:$\displaystyle M_x = \set {n \in \N: 1 \le n \le x \text { and } n \text { is not divisible by any primes greater than } p_k}$.
+Notice that:
+:$\size {M_x} \le {2^k} \sqrt x$
+because if:
+:$n = m^2r$
+then
+:$m \le \sqrt x$
+and there are at most $2^k$ distinct non-square prime compositions using primes less than $p_k$.
+Now let:
+:$N_{i, x} = \set {n \in \N: 1 \le n \le x \text{ and } n \text { is divisible by } p_i}$
+Notice:
+:$\displaystyle \size {\set {1, \dots, x} \setminus M_x} \le \sum_{i \mathop = k + 1}^\infty \size {N_{i, x} } < \sum_{i \mathop = k + 1}^\infty \dfrac x {p_i} \implies \dfrac x 2 < \size {M_x}$
+Finally notice that whenever:
+:$x \ge 2^{2 k + 2}$
+then it cannot be the case that both:
+:$\dfrac x 2 < \size {M_x} \le {2^k} \sqrt x$
+\end{proof}<|endoftext|>
+\section{Measure of Interval is Length}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $I$ be a [[Definition:Real Interval|real interval]] whose [[Definition:Endpoint of Real Interval|endpoints]] are $a$ and $b$.
+Then $I$ is [[Definition:Measurable Set|Lebesgue measurable]], and the value of the [[Definition:Lebesgue Measure|measure]] is the [[Definition:Length of Real Interval|length of the interval]] $b - a$.
+\end{theorem}
+
+\begin{proof}
+{{link wanted|Insert here: Some stuff proving intervals are measurable}}
+Let $L \subset \R$ be a [[Definition:Real Interval|real interval]].
+Then $L$ has two [[Definition:Distinct Objects|distinct]] [[Definition:Endpoint of Real Interval|endpoints]]: $a$ and $b$.
+Let $\displaystyle \left\{{I_n}\right\}_{n \mathop = 1}^\infty$ be a set of [[Definition:Open Real Interval|open real intervals]] satisfying:
+:$\displaystyle L \subseteq \bigcup_{n \mathop = 1}^\infty I_n$
+=== Case 1: $L$ is open and finite ===
+The theorem follows from the definition of [[Definition:Lebesgue Measure|Lebesgue measure]].
+One can construct:
+: $I_n = \begin{cases} L & : n = 1 \\ \varnothing & : n \ne 1 \end{cases}$
+which yields the [[Definition:Summation|sum]]:
+:$\displaystyle \sum l \left({I_n}\right) = b - a$
+This sum could not be any less because then:
+:$\displaystyle a + \sum l \left({I_n}\right) < b$
+Hence:
+:$m \left({L}\right) = b - a$
+{{qed|lemma}}
+=== Case 2: $L$ is closed and finite ===
+The open interval:
+:$\left({a - \epsilon \,.\,.\, b + \epsilon}\right)$
+contains $\left[{a \,.\,.\, b}\right]$ for each positive $\epsilon$.
+So:
+:$m \left({L}\right) \le l \left({a - \epsilon \,.\,.\, b + \epsilon}\right) = b - a + 2 \epsilon$
+Since this is true for any positive $\epsilon$, we have:
+:$m \left({L}\right) \le b - a$
+Now it must be shown that:
+:$m \left({L}\right) \ge b - a$
+which will demonstrate:
+:$m \left({L}\right) = b - a$
+By the [[Heine-Borel Theorem]], any [[Definition:Set|set]] of [[Definition:Open Real Interval|open intervals]] [[Definition:Open Cover|covering]] $\left[{a \,.\,.\, b}\right]$ contains a [[Definition:Finite Subcover|finite subcover]].
+The sum of the [[Definition:Length of Real Interval|lengths]] of the [[Definition:Finite Subcover|finite subcover]] is no greater than the sum of the [[Definition:Length of Real Interval|lengths]] of the [[Definition:Infinite Set|infinite]] [[Definition:Open Cover|cover]].
+Therefore it will suffice to show that $\displaystyle \sum l \left({I_n}\right) \ge b - a$ only for [[Definition:Finite Cover|finite covers]].
+Since $a \in \bigcup I_n$, there must be a [[Definition:Set|set]] in $\left\{{I_n}\right\}$ containing $a$.
+Call this [[Definition:Set|set]] $\left({a_1, b_1}\right)$.
+Necessarily:
+:$a_1 < a < b_1$
+If $b_1 \le b$, then:
+:$b_1 \in \left[{a \,.\,.\, b}\right]$
+So there must exist a [[Definition:Set|set]] in $\left\{{I_n}\right\}$ containing $b_1$.
+Let this [[Definition:Set|set]] be $\left({a_2 \,.\,.\, b_2}\right)$.
+Continuing in this fashion, construct a series of intervals:
+: $\left({a_1 \,.\,.\, b_1}\right), \left({a_2 \,.\,.\,  b_2}\right), \ldots, \left({a_k \,.\,.\, b_k}\right)$
+Since $\left\{{I_n}\right\}$ is [[Definition:Finite Set|finite]], this process must terminate at some [[Definition:Open Real Interval|interval]] $\left({a_k \,.\,.\, b_k}\right)$.
+But this process can only terminate if $b \in \left({a_k \,.\,.\, b_k}\right)$.
+Hence:
+:$\displaystyle \sum l \left({I_n}\right) \ge \sum_{i \mathop = 1}^k \left({b_i - a_i}\right) = b_k - a_1 - \sum_{j \mathop = 1}^{k-1} \left({a_{j+1} - b_j}\right) > b_k - a_1$
+since $a_i > b_{i-1}$.
+But $b_k > b$ and $a_1 < a$, and so:
+:$b_k - a_1 > b - a$
+Hence
+:$\displaystyle \sum l \left({I_n}\right) \ge b - a$
+{{qed|lemma}}
+=== Case 3: $L$ is a finite interval ===
+Regardless of whether the [[Definition:Set|set]] in question is [[Definition:Open Set (Real Analysis)|open]], [[Definition:Closed Set (Real Analysis)|closed]], or possibly neither, given $\epsilon >0$, there is a [[Definition:Closed Real Interval|closed interval]] $J \subset L$ such that:
+:$l \left({J}\right) > l \left({L}\right) - \epsilon$
+Hence:
+:$l \left({L}\right) - \epsilon < l \left({J}\right) = m \left({J}\right) \le m \left({L}\right) \le m \left({c \left({L}\right)}\right) = l \left({c \left({L}\right)}\right) = l \left({L}\right)$
+where $c \left({L}\right)$ is the [[Definition:Closure (Topology)|closure]] of $L$.
+Thus for each positive $\epsilon$:
+: $l \left({L}\right) - \epsilon < m \left({L}\right) \le l \left({L}\right)$
+and so:
+: $m \left({L}\right) = l \left({L}\right) = b - a$
+{{qed|lemma}}
+=== Case 4: Infinite Intervals ===
+If $L$ is [[Definition:Infinite Set|infinite]], either $a$ or $b$ is $\pm \infty$.
+Given any [[Definition:Real Number|real number]] $\Delta$, there is a [[Definition:Closed Real Interval|closed interval]] $J \subset L$ with $l \left({J}\right) = \Delta$.
+Hence $m \left({L}\right) \ge \Delta$ for arbitrarily large $\Delta$, and so:
+: $m \left({L}\right) = \infty$
+{{qed}}
+[[Category:Measure Theory]]
+2gfv9sp1szzjqpwslrprr1ojfd79ba8
+\end{proof}<|endoftext|>
+\section{Measurable Sets form Algebra of Sets}
+Tags: Measure Theory, Set Systems
+
+\begin{theorem}
+Let $\mu^*$ be an [[Definition:Outer Measure|outer measure]] on a [[Definition:Set|set]] $X$.
+Then the set of [[Definition:Measurable Set#Measurable Sets of an Arbitrary Outer Measure|$\mu^*$-measurable sets]] is an [[Definition:Algebra of Sets|algebra of sets]].
+\end{theorem}
+
+\begin{proof}
+For a [[Definition:Subset|subset]] $S \subseteq X$, let $\complement \left({S}\right)$ denote the [[Definition:Relative Complement|relative complement]] of $S$ in $X$.
+We first prove the second property of an [[Definition:Algebra of Sets|algebra of sets]], as described on that page.
+Let $S$ be $\mu^*$-measurable. For any [[Definition:Subset|subset]] $A \subseteq X$:
+{{begin-eqn}}
+{{eqn | l = \mu^* \left({A}\right)
+      | r = \mu^* \left({A \cap S}\right) + \mu^* \left({A \cap \complement \left({S}\right)}\right)
+}}
+{{eqn | r = \mu^* \left({A \cap \complement \left({\complement \left({S}\right)}\right)}\right) + \mu^* \left({A \cap \complement \left({S}\right)}\right)
+      | c = [[Complement of Complement]]
+}}
+{{end-eqn}}
+as desired.
+Now we prove the first property.
+Suppose that $S_1$ and $S_2$ are $\mu^*$-measurable sets.  Let $A$ be any [[Definition:Subset|subset]] of $X$.  
+Since:
+{{begin-eqn}}
+{{eqn | l = A \cap \left({S_1 \cup S_2}\right)
+      | r = \left[{\left({A \cap \left({S_1 \cup S_2}\right)}\right) \cap S_1}\right] \cup \left[{\left({A \cap \left({S_1 \cup S_2}\right)}\right) \setminus S_1}\right]
+      | c = [[Set Difference Union Intersection]] and [[Union is Commutative]]
+}}
+{{eqn | r = \left[{A \cap \left({\left({S_1 \cup S_2}\right) \cap S_1}\right)}\right] \cup \left[{\left({\left({S_1 \cup S_2}\right) \cap A}\right) \setminus S_1}\right]
+      | c = [[Intersection is Associative]] and [[Intersection is Commutative]]
+}}
+{{eqn | r = \left[{A \cap \left({\left({S_1 \cup S_2}\right) \cap S_1}\right)}\right] \cup \left[{\left({\left({S_1 \cup S_2}\right) \setminus S_1}\right) \cap A}\right]
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = \left({A \cap S_1}\right) \cup \left({\left({S_2 \setminus S_1}\right) \cap A}\right)
+      | c = [[Intersection Absorbs Union]] and [[Set Difference with Union is Set Difference]]
+}}
+{{eqn | r = \left({A \cap S_1}\right) \cup \left({\left({S_2 \cap A}\right) \setminus S_1}\right)
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = \left({A \cap S_1}\right) \cup \left({\left({A \cap S_2}\right) \setminus S_1}\right)
+      | c = [[Intersection is Commutative]]
+}}
+{{eqn | r = \left({A \cap S_1}\right) \cup \left({\left({A \setminus S_1}\right) \cap S_2}\right)
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{end-eqn}}
+{{wtd|I'm pretty sure the above, or at least significant parts of the above, have/has already been posted up as proofs in their own right. If I remember, or rather, when I'm in the mood, I may go through and find them.}}
+we have, by the [[Definition:Subadditive Function (Measure Theory)|subadditivity]] of an [[Definition:Outer Measure|outer measure]]:
+:$\mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) \leq \mu^* \left({A \cap S_1}\right) + \mu^* \left({\left({A \setminus S_1}\right) \cap S_2}\right)$
+Thus:
+{{begin-eqn}}
+{{eqn | r = \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) + \mu^* \left({A \setminus \left({S_1 \cup S_2}\right)}\right)
+}}
+{{eqn | r = \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right) + \mu^* \left({\left({A \setminus S_1}\right) \setminus S_2}\right)
+      | c = [[Set Difference with Union]]
+}}
+{{eqn | o = \le
+      | r = \mu^* \left({A \cap S_1}\right) + \mu^* \left({\left({A \setminus S_1}\right) \cap S_2}\right) + \mu^* \left({\left({A \setminus S_1}\right) \setminus S_2}\right)
+      | c = by the above argument
+}}
+{{eqn | r = \mu^* \left({A \cap S_1}\right) + \mu^* \left({A \setminus S_1}\right)
+      | c = Definition of [[Definition:Measurable Set#Measurable Sets of an Arbitrary Outer Measure|Measurability]] of $S_2$
+}}
+{{eqn | r = \mu^* \left({A}\right)
+      | c = Definition of [[Definition:Measurable Set#Measurable Sets of an Arbitrary Outer Measure|Measurability]] of $S_1$
+}}
+{{end-eqn}}
+The result follows by the [[Definition:Subadditive Function (Measure Theory)|subadditivity]] of an [[Definition:Outer Measure|outer measure]].
+{{refactor}}
+Alternatively, one could use the equality
+{{begin-eqn}}
+{{eqn | l = \mu^* \left({A \cap \left({S_1 \cup S_2}\right)}\right)
+      | r = \mu^* \left({A \cap \left({S_1 \cup S_2}\right) \cap S_1}\right) + \mu^* \left({A \cap \left({S_1 \cup S_2}\right) \setminus S_1}\right)
+}}
+{{eqn | r = \mu^* \left({A \cap S_1}\right) + \mu^* \left({A \cap S_2 \setminus S_1}\right)
+}}
+{{end-eqn}}
+to prove the result directly without the use of [[Definition:Subadditive Function (Measure Theory)|subadditivity]].
+{{qed}}
+[[Category:Measure Theory]]
+[[Category:Set Systems]]
+qv27l4hn8xkf63xmse0u1ewxu2nzud2
+\end{proof}<|endoftext|>
+\section{Classification of Compact Three-Manifolds Supporting Zero-Curvature Geometry}
+Tags: Topology
+
+\begin{theorem}
+Every [[Definition:Closed Set (Topology)|closed]], [[Definition:Orientable|orientable]], [[Definition:Path-Connected|path connected]] $3$-[[Definition:Dimension (Topology)|dimensional]] [[Definition:Riemannian Manifold|Riemannian manifold]] which supports a [[Definition:Riemannian Metric|geometry]] of [[Definition:Zero Gaussian Curvature|zero]] [[Definition:Gaussian Curvature|curvature]] is [[Definition:Homeomorphism|homeomorphic]] to one of the following:
+* [[Definition:Torus (Topology)|Torus $\mathbb T^3$]]
+* [[Definition:Half-Twist Cube|Half-Twist Cube]]
+* [[Definition:Quarter-Twist Cube|Quarter-Twist Cube]]
+* [[Definition:Hantschze-Wendt Manifold|Hantschze-Wendt Manifold]]
+* [[Definition:One Sixth Twist Hexagonal Prism|$\frac 1 6$-Twist Hexagonal Prism]]
+* [[Definition:One Third Twist Hexagonal Prism|$\frac 1 3$-Twist Hexagonal Prism]]
+The [[Definition:Torus (Topology)|$3$-torus]] is described on the [[Definition:Torus (Topology)|torus]] page.
+The other manifolds can be described using [[Definition:Quotient Space (Topology)|quotient spaces]] on familiar prisms, with the [[Definition:Equivalence Relation|equivalence relations]] described below.
+<gallery perRow="5">
+File:Halftwistcube.JPG|The [[Definition:Half-Twist Cube|Half-Twist Cube]]
+File:Hantschzewendt.JPG|The [[Definition:Hantschze-Wendt Manifold|Hantschze-Wendt Manifold]]
+File:Quartertwistcube.JPG|The [[Definition:Quarter-Twist Cube|Quarter-Twist Cube]]
+File:Sixthtwisthexagon.JPG|The [[Definition:One Sixth Twist Hexagonal Prism|$\frac 1 6$-Twist Hexagonal Prism]]
+File:Thirdtwisthexagon.JPG|The [[Definition:One Third Twist Hexagonal Prism|$\frac 1 3$-Twist Hexagonal Prism]]
+</gallery>
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Topology]]
+2q5k22nfp0qiqg60awvo3t84dni22hq
+\end{proof}<|endoftext|>
+\section{Sum of Logarithms}
+Tags: Logarithms, Sum of Logarithms
+
+\begin{theorem}
+{{:Sum of Logarithms/General Logarithm}}
+\end{theorem}<|endoftext|>
+\section{Heine-Borel Theorem/Real Line}
+Tags: Real Analysis, Euclidean Space, Compact Spaces, Heine-Borel Theorem
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]] considered as a [[Definition:Euclidean Space|Euclidean space]].
+Let $C \subseteq \R$.
+Then $C$ is [[Definition:Closed Set (Metric Space)|closed]] and [[Definition:Bounded Metric Space|bounded]] in $\R$ {{iff}} $C$ is [[Definition:Compact Subspace|compact]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $C$ be [[Definition:Closed Set (Metric Space)|closed]] and [[Definition:Bounded Metric Space|bounded]] in $\R$.
+Then, by [[Closed Bounded Subset of Real Numbers is Compact]], $C$ is [[Definition:Compact Subspace|compact]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $C$ be [[Definition:Compact Subspace|compact]] in $\R$.
+Then, by [[Compact Subspace of Real Numbers is Closed and Bounded]], $C$ is [[Definition:Closed Set (Metric Space)|closed]] and [[Definition:Bounded Metric Space|bounded]] in $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Measurable Image}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $\mathfrak M$ be the set of [[Definition:Measurable Set|measurable sets]] of $\R$.
+For any [[Definition:Extended Real-Valued Function|extended real-valued function]] $f: \R \to \R \cup \left\{{-\infty \,.\,.\, +\infty}\right\}$ whose [[Definition:Domain of Mapping|domain]] is [[Definition:Measurable Set|measurable]], the following statements are equivalent:
+: $(1): \quad \forall \alpha \in \R: \left\{{x: f \left({x}\right) > \alpha}\right\} \in \mathfrak M$
+: $(2): \quad \forall \alpha \in \R: \left\{{x: f \left({x}\right) \ge \alpha}\right\} \in \mathfrak M$
+: $(3): \quad \forall \alpha \in \R: \left\{{x: f \left({x}\right) < \alpha}\right\} \in \mathfrak M$
+: $(4): \quad \forall \alpha \in \R: \left\{{x: f \left({x}\right) \le \alpha}\right\} \in \mathfrak M$
+These statements imply:
+: $(5): \quad \forall \alpha \in \R \cup \left\{{-\infty \,.\,.\, +\infty}\right\}: \left\{{x:f \left({x}\right) = \alpha}\right\} \in \mathfrak M$
+{{refactor}}
+{{proofread}}
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Domain of Mapping|domain]] of $f$ be $D$.
+We have that [[Measurable Sets form Algebra of Sets]].
+First we note that, from [[Properties of Algebras of Sets]], the [[Definition:Set Difference|difference]] of two [[Definition:Measurable Set|measurable sets]] is measurable.
+So:
+: $\left\{{x: f \left({x}\right) \le \alpha}\right\} = D - \left\{{x: f \left({x}\right) > \alpha}\right\}$
+and so $(1) \iff (4)$.
+Similarly, $(2) \iff (3)$.
+Next we note that, also from [[Properties of Algebras of Sets]], the [[Definition:Set Intersection|intersection]] of a [[Definition:Sequence|sequence]] of [[Definition:Measurable Set|measurable sets]] is measurable.
+Now:
+:$\displaystyle \left\{{x: f \left({x}\right) \ge \alpha}\right\} = \bigcap_{n=1}^\infty \left\{{x: f \left({x}\right) > \alpha -\dfrac 1 n}\right\}$
+because if:
+:$x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$
+that is:
+:$f \left({x}\right) \ge \alpha$
+and since:
+:$\forall n \in \N: n > 0: \alpha > \alpha - \dfrac 1 n$
+then:
+:$\forall n \in \N: n > 0: f \left({x}\right) > \alpha - \dfrac 1 n$
+That is:
+:$\forall n \in \N: n > 0: x \in  \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$
+Hence:
+:$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$
+Conversely, suppose:
+:$\displaystyle x \in \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$
+that is:
+:$\forall n \in \N: n > 0: x \in  \left\{{x: f \left({x}\right) > \alpha - \dfrac 1 n}\right\}$
+Claim $f \left({x}\right) \ge \alpha $.
+Otherwise $f \left({x}\right) < \alpha$, say for example $f \left({x}\right) = \alpha - \left\vert{\epsilon}\right\vert$.
+Choose $N = \left\lceil{\dfrac 1 {\left\vert{\epsilon}\right\vert} }\right\rceil + 1 \in \N$.
+Therefore:
+:$N > \left\lceil{\dfrac 1 {\left\vert{\epsilon}\right\vert} }\right\rceil \ge \dfrac 1 {\left\vert{\epsilon}\right\vert}$
+and so:
+:$\alpha - \dfrac 1 N > \alpha - \left\vert{\epsilon}\right\vert$
+By hypothesis:
+:$\forall N \in \N: f \left({x}\right) > \alpha - \dfrac 1 N$
+and therefore by the just previous:
+:$f \left({x}\right) > \alpha - \left\vert{\epsilon}\right\vert$
+But we had $f \left({x}\right) = \alpha - \left\vert{\epsilon}\right\vert$, a contradiction.
+Therefore:
+:$f \left({x}\right) \ge \alpha $
+that is:
+$x \in \left\{{x: f \left({x}\right) \ge \alpha}\right\}$
+which was to be shown.
+So $(1) \implies (2)$.
+Similarly:
+:$\displaystyle \left\{{x: f \left({x}\right) > \alpha}\right\} = \bigcup_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) \ge \alpha + \dfrac 1 n}\right\}$
+and so $(2) \implies (1)$.
+This shows that $(1) \iff (2) \iff (3) \iff (4)$.
+For the fifth statement, we have:
+: $\left\{{x: f \left({x}\right) = \alpha}\right\} = \left\{{x: f \left({x}\right) \ge \alpha}\right\} \cap \left\{{x: f \left({x}\right) \le \alpha}\right\}$
+and so $(3) \land (4) \implies (5)$ for $\alpha \in \R$.
+Since:
+:$\displaystyle \left\{{x: f \left({x}\right) = +\infty }\right\} = \bigcap_{n \mathop = 1}^\infty \left\{{x: f \left({x}\right) \ge n}\right\}$
+we have that $(2) \implies (5)$ for $\alpha = +\infty$.
+Similarly $(4) \implies (5)$ for $\alpha = - \infty$.
+{{qed}}
+[[Category:Measure Theory]]
+4x9u1du0b8e1x5fwhum72uy3los1pwm
+\end{proof}<|endoftext|>
+\section{Lebesgue Integral is Extension of Darboux Integral}
+Tags: Analysis, Integral Calculus, Measure Theory
+
+\begin{theorem}
+Let $f: \closedint a b \to \R$ be a [[Definition:Darboux Integrable Function|Darboux integrable function]].
+Then it is also [[Definition:Lebesgue Integrable Function|Lebesgue integrable]], and furthermore:
+:$\displaystyle R \int_a^b \map f x \rd x = \int_{\closedint a b} f \rd \lambda$
+where $\displaystyle R \int_a^b$ is the [[Definition:Darboux Integral|Darboux integral]] and $\displaystyle \int_{\closedint a b}$ is the [[Definition:Lebesgue Integral|Lebesgue integral]].
+\end{theorem}
+
+\begin{proof}
+Since every [[Definition:Step Function|step function]] is also a [[Definition:Simple Function|simple function]], we have
+:$\displaystyle \map L P \le \sup_{\phi \mathop \le f} \int_a^b \map \phi x \rd x \le \inf_{\psi \mathop \ge f} \int_a^b \map \psi x \rd x \le \map U P$
+where $\map L P$ and $\map U P$ are the [[Definition:Lower Sum|lower sum]] and [[Definition:Upper Sum|upper sum]] as defined in the definition of [[Definition:Definite Integral|definite integral]].
+Since $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]], the inequalities are all equalities and $f$ is measurable by [[Infimum and Supremum Equivalent for Measurable Function|basic properties of measurable functions]].
+{{questionable|all reasoning is behind red links, and it's incomplete}}
+{{ProofWanted}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Exponential Function}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential of $x$]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Euclidean Metric on Real Vector Space is Metric}
+Tags: Euclidean Metric, Euclidean Metric on Real Vector Space is Metric
+
+\begin{theorem}
+The [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on a [[Definition:Real Vector Space|real vector space]] $\R^n$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Euclidean Metric/Real Vector Space|Euclidean metric]] on $\R^n$ is a special case of the [[Definition:P-Product Metric/Real Vector Space|$p$-product metric]].
+The result follows from [[P-Product Metric on Real Vector Space is Metric|$p$-Product Metric on Real Vector Space is Metric]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider the [[Definition:Real Euclidean Space|Euclidean space]] $M = \struct {\R^n, d_2}$ where $d_2$ is the [[Definition:Distance Function|distance function]] given by:
+:$\displaystyle \map {d_2} {x, y} = \paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2}^{\frac 1 2}$
+where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
+=== Proof of $\text M 1$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, x}
+      | r = \paren {\sum_{i \mathop = 1}^n \paren {x_i - x_i}^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum_{i \mathop = 1}^n 0^2}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 1$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $\text M 2$ ===
+It is required to be shown:
+:$\map {d_2} {x, y} + \map {d_2} {y, z} \ge \map {d_2} {x, z}$
+for all $x, y, z \in \R^n$.
+Let:
+:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
+:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
+:$(3): \quad x_i - y_i = r_i$
+:$(4): \quad y_i - z_i = s_i$.
+Thus we need to show that:
+:$\displaystyle \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2} \ge \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, y} + \map {d_2} {y, z}
+      | r = \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2} + \paren {\sum \paren {y_i - z_i}^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum r_i^2}^{\frac 1 2} + \paren {\sum s_i^2}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \paren {\sum \paren {r_i + s_i}^2}^{\frac 1 2}
+      | c = [[Minkowski's Inequality for Sums/Index 2|Minkowski's Inequality for Sums: index $2$]]
+}}
+{{eqn | r = \paren {\sum \paren {x_i - y_i + y_i - z_i}^2}^{\frac 1 2}
+      | c = Definition of $r_i$ and $s_i$
+}}
+{{eqn | r = \paren {\sum \paren {x_i - z_i}^2}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | r = \map {d_2} {x, z}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 2$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $\text M 3$ ===
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {x, y}
+      | r = \paren {\sum \paren {x_i - y_i}^2}^{\frac 1 2}
+      | c = Definition of $d_2$
+}}
+{{eqn | r = \paren {\sum \paren {y_i - x_i}^2}^{\frac 1 2}
+      | c = 
+}}
+{{eqn | r = \map {d_2} {y, x}
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 3$]] holds for $d_2$.
+{{qed|lemma}}
+=== Proof of $\text M 4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists k \in \closedint 1 n: x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_k - y_k
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\sum \paren {x_k - y_k}^2}^{\frac 1 2}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {d_2} {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d_2$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 4$]] holds for $d_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Induces Topology}
+Tags: Topology, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Topology Induced by Metric|topology $\tau$ induced]] by the [[Definition:Metric|metric]] $d$ is a [[Definition:Topology|topology]] on $M$.
+\end{theorem}
+
+\begin{proof}
+We examine each of the criteria for being a [[Definition:Topology|topology]] separately.
+:$(1): \quad$ By [[Union of Open Sets of Metric Space is Open]], the [[Definition:Set Union|union]] of any collection of [[Definition:Open Set of Metric Space|open sets]] of a [[Definition:Metric Space|metric space]] is [[Definition:Open Set of Metric Space|open]].
+:$(2): \quad$ By [[Finite Intersection of Open Sets of Metric Space is Open]], a [[Definition:Finite Intersection|finite intersection]] of [[Definition:Open Set of Metric Space|open sets]] of a [[Definition:Metric Space|metric space]] is [[Definition:Open Set of Metric Space|open]].
+:$(3): \quad$ By [[Open Sets in Metric Space]], $\O \in \tau$ and $A \in \tau$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Algebras of Sets}
+Tags: Algebras of Sets
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\mathfrak A$ be an [[Definition:Algebra of Sets|algebra of sets]] on $X$.
+Then the following hold:
+:$(1): \quad$ The [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] in $\mathfrak A$ is in $\mathfrak A$.
+:$(2): \quad$ The [[Definition:Set Difference|difference]] of two [[Definition:Set|sets]] in $\mathfrak A$ is in $\mathfrak A$.
+:$(3): \quad$ $X \in \mathfrak A$.
+:$(4): \quad$ The [[Definition:Empty Set|empty set]] $\O$ is in $\mathfrak A$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$X$ be a [[Definition:Set|set]]
+:$\mathfrak A$ be an [[Definition:Algebra of Sets|algebra of sets]] on $X$
+:$A, B \in \mathfrak A$
+By the definition of [[Definition:Algebra of Sets|algebra of sets]], we have that:
+:$A \cup B \in \mathfrak A$
+:$\relcomp X A \in \mathfrak A$
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = A, B \in \mathfrak A
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \relcomp X A \cup \relcomp X B \in \mathfrak A
+      | c = {{Defof|Algebra of Sets}}
+}}
+{{eqn | o = \leadsto
+      | r = \relcomp X {A \cap B} \in \mathfrak A
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{eqn | o = \leadsto
+      | r = A \cap B \in \mathfrak A
+      | c = {{Defof|Algebra of Sets}}
+}}
+{{end-eqn}}
+and so we have that the [[Definition:Set Intersection|intersection]] of two [[Definition:Set|sets]] in $\mathfrak A$ is in $\mathfrak A$.
+Next:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = A, B \in \mathfrak A
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = A \cap \relcomp X B \in \mathfrak A
+      | c = from above
+}}
+{{eqn | o = \leadsto
+      | r = A \setminus B \in \mathfrak A
+      | c = [[Set Difference as Intersection with Relative Complement]]
+}}
+{{end-eqn}}
+and so we have that the [[Definition:Set Difference|difference]] of two [[Definition:Set|sets]] in $\mathfrak A$ is in $\mathfrak A$.
+We have that $\mathfrak A \ne \O$ and so $\exists A \subseteq X: A \in \mathfrak A$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \relcomp X A
+      | o = \in
+      | r = \mathfrak A
+      | c = {{Defof|Algebra of Sets}}
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp X A \cup A
+      | o = \in
+      | r = \mathfrak A
+      | c = {{Defof|Algebra of Sets}}
+}}
+{{eqn | ll= \leadsto
+      | l = X
+      | o = \in
+      | r = \mathfrak A
+      | c = [[Union with Relative Complement]]
+}}
+{{end-eqn}}
+Also, $\relcomp X A \cap A \in \mathfrak A$ from above.
+So by [[Intersection with Relative Complement is Empty]]:
+:$\O \in \mathfrak A$
+{{qed}}
+[[Category:Algebras of Sets]]
+hq8mr16m39bsgf2dmbi1np8ux46w1mp
+\end{proof}<|endoftext|>
+\section{Countable Sets Have Measure Zero}
+Tags: Analysis
+
+\begin{theorem}
+Let $S$ be a [[Definition:Countable|countable set]].
+{{explain|Is it assumed that $S \subseteq \R$?}}
+Then the [[Definition:Lebesgue Measure|measure]] of $S$ is $\map m S = 0$.
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \set {x_i}_{i \mathop = 1}^\infty$ be an enumeration of the [[Definition:Element|elements]] of $S$.
+For any [[Definition:Strictly Positive Real Number|(strictly) positive real number]] $\epsilon$, define:
+:$A_i = \paren {x_i - 2^{-i} \epsilon, x_i + 2^{-i} \epsilon}$
+{{explain|The notation of the above seems to be of an open real interval. This needs to be clarified.}}
+Then:
+:$\displaystyle S \subseteq \bigcup_{i \mathop = 1}^\infty A_i$
+and:
+:$\displaystyle \map m {\bigcup A_i} \le \sum_{i \mathop = 1}^\infty 2^{1 - i} \epsilon = 2 \epsilon$
+Since our choice of $\epsilon$ was arbitrary, for any [[Definition:Strictly Positive Real Number|positive real]] $z$ we can construct a set $T$ such that $S \subseteq T$ and $\map m T \le z$.
+Hence $X$ has [[Definition:Measure Zero|zero measure]].
+{{qed}}
+[[Category:Analysis]]
+l9se6p00ujy712vxsgsx3xbbo2ikxro
+\end{proof}<|endoftext|>
+\section{Mean Value of Convex Real Function}
+Tags: Convex Real Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let $f$ be [[Definition:Convex Real Function|convex]] on $\openint a b$.
+Then:
+:$\forall \xi \in \openint a b: \map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$
+\end{theorem}
+
+\begin{proof}
+By the [[Mean Value Theorem]]:
+:$\exists \eta \in \openint x \xi: \map {f'} \eta = \dfrac {\map f x - \map f \xi} {x - \xi}$
+From [[Real Function is Convex iff Derivative is Increasing]], the [[Definition:Derivative|derivative]] of $f$ is [[Definition:Increasing Real Function|increasing]].
+Thus:
+:$x > \xi \implies \map {f'} \eta \ge \map {f'} \xi$
+:$x < \xi \implies \map {f'} \eta \le \map {f'} \xi$
+Hence:
+:$\map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bound of Natural Logarithm}
+Tags: Natural Logarithms, Inequalities, Upper Bound of Natural Logarithm
+
+\begin{theorem}
+Let $\ln y$ be the [[Definition:Natural Logarithm|natural logarithm]] of $y$ where $y \in \R_{>0}$.
+Then:
+:$\ln y \le y - 1$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {f_n}$ denote the [[Definition:Sequence|sequence]] of mappings $f_n: \R_{>0} \to \R$ defined as:
+:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
+Fix $x \in \R_{>0}$.
+We first show that
+$\forall n \in \N : n \paren {\sqrt [n] x - 1} < x - 1 $
+=== Case 1: $0 < x < 1$ ===
+Suppose $0 < x < 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | m = x
+      | mo= <
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | m = \sqrt [n] x^{n - k}
+      | mo= <
+      | r = 1
+      | c = [[Power Function on Base between Zero and One is Strictly Decreasing/Rational Number]]
+      | cc= $\forall k \in \set {0, 1, \ldots, n - 1}$
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | m = \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}
+      | mo= <
+      | r = n
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = \frac 1 n
+      | mo= <
+      | r = \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = [[Ordering of Reciprocals]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = 1
+      | mo= <
+      | r = \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = multiplying both sides by $n$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = x - 1
+      | mo= >
+      | r = \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = multiplying both sides by $x - 1 < 0$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = y^n - 1
+      | mo= >
+      | r = \frac {\paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }
+      | c = substituting $y = \sqrt [n] x$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = y^n - 1
+      | mo= >
+      | r = n \paren {y - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = x - 1
+      | mo= >
+      | r = n \paren {\sqrt [n] x - 1}
+      | c = substituting $\sqrt [n] x = y$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+      
+=== Case 2: $x = 1$ ===
+Suppose $x = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \ln x
+      | r = \map \ln 1
+}}
+{{eqn | r = 0
+      | c = [[Natural Logarithm of 1 is 0/Proof 3]]
+}}
+{{eqn | r = 1 - 1
+}}
+{{eqn | r = x - 1
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case 3: $x > 1$ ===
+Suppose $x > 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [n] x^{n - k}
+      | o = >
+      | r = 1
+      | c = [[Power Function on Base Greater than One is Strictly Increasing/Rational Number]]
+      | cc = $\forall k \in \set { 0, 1, \ldots, n - 1}$
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{k \mathop = 0}^{n - 1} \sqrt [n] x^{n - k}
+      | o = >
+      | r = n
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 n
+      | o = > 
+      | r = \frac 1 {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = [[Ordering of Reciprocals]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = > 
+      | r = \frac n {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = Multiply both sides by $n$
+}}
+{{eqn | ll= \leadsto
+      | l = x - 1
+      | o = >
+      | r = \frac {n \paren {x - 1} } {1 + \sqrt [n] x + \cdots + \sqrt [n] x^{n - 1} }
+      | c = multiplying both sides by $x - 1 > 1$
+}}
+{{eqn | ll= \leadsto
+      | l = y^n - 1
+      | o = >
+      | r = \frac {n \paren {y^n - 1} } {1 + y + \cdots + y^{n - 1} }
+      | c = substituting $y = \sqrt [n] x$
+}}
+{{eqn | ll= \leadsto
+      | l = y^n - 1
+      | o = >
+      | r = n \paren {y - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | l = x - 1
+      | o = >
+      | r = n \paren {\sqrt [n] x - 1}
+      | c = Substitute $\sqrt [n] x = y$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Thus:
+:$\forall n \in \N: n \paren {\sqrt [n] x - 1} \le x - 1$
+by [[Proof by Cases]].
+Thus:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {\sqrt [n] x - 1 } \le x - 1$
+from [[Limit of Bounded Convergent Sequence is Bounded]].
+Hence the result, from the [[Definition:Real Natural Logarithm|definition of $\ln$]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Logarithm is Strictly Concave]]:
+:$\ln$ is [[Definition:Strictly Concave Real Function|(strictly) concave]].
+From [[Mean Value of Concave Real Function]]:
+:$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$
+From [[Derivative of Natural Logarithm]]:
+:$D \ln 1 = \dfrac 1 1 = 1$
+So:
+:$\ln y - \ln 1 \le \left({y - 1}\right)$
+But from [[Logarithm of 1 is 0]]:
+:$\ln 1 = 0$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of Sum/Real Numbers}
+Tags: Exponential of Sum
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential of $x$]].
+Then:
+:$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$
+\end{theorem}<|endoftext|>
+\section{Exponential of Product}
+Tags: Exponential Function, Exponential of Product
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential of $x$]].
+Then:
+:$\map \exp {x y} = \paren {\exp y}^x$
+\end{theorem}
+
+\begin{proof}
+Let $Y = \exp y$.
+From [[Exponential of Natural Logarithm]]:
+:$\map \ln {\exp y} = y$
+From [[Logarithms of Powers]], we have:
+:$\ln Y^x = x \ln Y = x \, \map \ln {\exp y} = x y$
+Thus:
+:$\map \exp {x y} = \map \exp {\ln Y^x} = Y^x = \paren {\exp y}^x$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\exp y}^x
+      | r = \map \exp {x \, \map \ln {\exp y} }
+      | c = {{Defof|Power to Real Number}}
+}}
+{{eqn | r = \map \exp {x y}
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This proof assumes the [[Exponential of Sum/Real Numbers|Exponential of Sum]] property.
+First, for $n \in \Z_{\ge 0}$:
+{{begin-eqn}}
+{{eqn | l = \map \exp {n y}
+      | r = \map \exp {\sum_{k \mathop = 1}^n y}
+}}
+{{eqn | r = \prod_{k \mathop = 1}^n \exp y
+      | c = [[Exponential of Sum/Real Numbers|Exponential of Sum]]
+}}
+{{eqn | n = 1
+      | r = \paren {\exp y}^n
+}}
+{{end-eqn}}
+That is:
+:$\forall n \in \Z_{\ge 0}: \map \exp {n y} = \paren {\exp y}^n$
+Now let $n \in \Z_{<0}$.
+It follows that $-n \in \Z_{>0}$, so:
+{{begin-eqn}}
+{{eqn | l = \exp n
+      | r = \map \exp {-\paren {-n y} }
+}}
+{{eqn | r = \frac 1 {\map \exp {-n y} }
+      | c = [[Reciprocal of Real Exponential]]
+}}
+{{eqn | r = \frac 1 {\paren {\exp y}^{-n} }
+      | c = from $(1)$
+}}
+{{eqn | r = \paren {\exp y}^n
+      | c = [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]
+}}
+{{end-eqn}}
+Thus:
+:$(2): \quad \forall m \in \Z: \map \exp {m y} = \paren {\exp y}^m$
+Next, for $n \in \Z_{>0}$:
+{{begin-eqn}}
+{{eqn | l = \exp y
+      | r = \map \exp {n \frac y n}
+}}
+{{eqn | r = \paren {\map \exp {\frac y n} }^n
+      | c = from $(1)$
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \sqrt [n] {\exp y}
+      | r = \map \exp {\frac y n}
+}}
+{{end-eqn}}
+So fix $r \in \Q$.
+Let $r = \dfrac m n$, where $m \in \Z$ is an [[Definition:Integer|integer]] and $n \in \Z_{>0}$ is a [[Definition:Strictly Positive Integer|strictly positive integer]].
+From the above:
+{{begin-eqn}}
+{{eqn | l = \map \exp {r y}
+      | r = \map \exp {\frac m n y}
+}}
+{{eqn | r = \sqrt [n] {\map \exp {m y} }
+      | c = from $(3$
+}}
+{{eqn | r = \sqrt [n] {\paren {\exp y}^m}
+      | c = from $(2)$
+}}
+{{eqn | r = \paren {\exp y}^r
+}}
+{{end-eqn}}
+Thus, from the definition of $\paren {\exp y}^x$ as the [[Definition:Power (Algebra)/Real Number/Definition 2|unique continuous extension of $r \mapsto \paren {\exp y}^r$ from $\Q$ to $\R$]]:
+:$\map \exp {x y} = \paren {\exp y}^x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws}
+Tags: Analysis, Powers, Exponent Combination Laws
+
+\begin{theorem}
+Let $a, b \in \R_{>0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $a^x$ be defined as [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Derivative of Exponential at Zero}
+Tags: Derivatives involving Exponential Function, Derivative of Exponential at Zero
+
+\begin{theorem}
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential]] of $x$ for [[Definition:Real Number|real]] $x$.
+Then:
+: $\displaystyle \lim_{x \mathop \to 0} \frac {\exp x - 1} x = 1$
+\end{theorem}
+
+\begin{proof}
+For all $x \in \R$:
+:$\exp 0 - 1 = 0$ from [[Exponential of Zero]]
+:$\map {D_x} {\exp x - 1} = \exp x$ from [[Sum Rule for Derivatives]]
+:$D_x x = 1$ from [[Derivative of Identity Function]].
+Its prerequisites having been verified, [[L'Hôpital's Rule/Corollary 1|Corollary 1 to L'Hôpital's Rule]] yields immediately:
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\exp x - 1} x = \lim_{x \mathop \to 0} \frac {\exp x} 1 = \exp 0 = 1$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Note that this proof does not presuppose [[Derivative of Exponential Function]].
+We use the definition of the exponential [[Definition:Exponential Function/Real/Limit of Sequence|as a limit of a sequence]]:
+{{begin-eqn}}
+{{eqn | l = \frac {\exp h - 1} h
+      | r = \frac {\lim_{n \mathop \to \infty} \paren {1 + \dfrac h n}^n - 1} h
+      | c = {{Defof|Exponential Function/Real|subdef = Limit of Sequence|Exponential Function}}
+}}
+{{eqn | r = \frac {\displaystyle \lim_{n \mathop \to \infty} \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {\displaystyle \sum_{k \mathop = 0}^n {n \choose k} \paren {\frac h n}^k - 1} h
+      | c = as $h$ is constant
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \paren { {n \choose 0} 1 - 1 + {n \choose 1} \paren {\frac h n} \frac 1 h + \sum_{k \mathop = 2}^n {n \choose k} \paren {\frac h n}^k \frac 1 h}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty}1 + \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 1} }{n^k}
+      | c = [[Powers of Group Elements]]
+}}
+{{eqn | r = 1 + h \lim_{n \mathop \to \infty} \sum_{k \mathop = 2}^n {n \choose k} \frac {h^{k - 2} } {n^k}
+      | c = [[Powers of Group Elements]]
+}}
+{{end-eqn}}
+The right [[Definition:Summand|summand]] converges to zero as $h \to 0$, and so:
+:$\displaystyle \lim_{h \mathop \to 0} \frac {\exp h - 1} h = 1$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {e^x - 1} x
+      | r = \frac {e^x - e^0} x
+      | c = [[Exponential of Zero]]
+}}
+{{eqn | o = \to
+      | r = \intlimits {\dfrac \d {\d x} e^x} {x \mathop = 0} {}
+      | c = {{Defof|Derivative of Real Function at Point}}, as $x \to 0$
+}}
+{{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | r = 1
+      | c = [[Exponential of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Classification of Groups of Order up to 15}
+Tags: Order of Groups
+
+\begin{theorem}
+Up to [[Definition:Group Isomorphism|isomorphism]], every [[Definition:Group|group]] of order $\order G \le 15$ is one of the below:
+{| class="sortable wikitable"
+|- bgcolor="#ececec"
+! Order !! Abelian !! Non-Abelian
+|-
+|1 || $\Z_1$ || 
+|-
+|2 || $\Z_2$ ||
+|-
+|3 || $\Z_3$ ||
+|-
+|4 || $\Z_4, \Z_2 \oplus \Z_2$ || 
+|-
+|5 || $\Z_5$ ||
+|-
+|6 || $\Z_6$ || $D_3 = S_3$
+|- 
+|7 || $\Z_7$ ||
+|-
+|8 || $\Z_8, \Z_4 \oplus \Z_2, \Z_2 \oplus \Z_2 \oplus \Z_2$ || $D_4, \Dic 2$
+|-
+|9 || $\Z_9, \Z_3 \oplus \Z_3$ || 
+|-
+|10 || $\Z_{10}$ || $D_5$
+|-
+|11 || $\Z_{11}$ ||
+|-
+|12 || $\Z_{12}, \Z_6 \oplus \Z_2$ || $D_6, A_4, \Dic 3$
+|-
+|13 || $\Z_{13}$ ||
+|-
+|14 || $\Z_{14}$ || $D_7$
+|-
+|15 || $\Z_{15}$ || 
+|}
+where:
+: $D_n$ is the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$
+: $S_n$ is the [[Definition:Symmetric Group|$n$th symmetric group]]
+: $A_n$ is the [[Definition:Alternating Group|alternating group]] on $n$ points
+: $\Dic 2$ is the [[Definition:Dicyclic Group|dicyclic group]] of [[Definition:Order of Structure|order]] $4 n$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Abelian Group|Abelian]] cases are the direct result of the [[Fundamental Theorem of Finite Abelian Groups]].
+The non-Abelian cases follow from seven separate theorems:
+:$(1): \quad$ [[Trivial Group is Cyclic Group]] - determines theorem for order $1$
+:$(2): \quad$ [[Prime Group is Cyclic]] - determines theorem for orders $2$, $3$, $5$, $7$, $11$, and $13$
+:$(3): \quad$ [[Group of Order Prime Squared is Abelian]] - determines theorem for orders $4$ and $9$
+:$(4): \quad$ [[Cyclic Groups of Order p q]] - determines theorem for order $15$
+:$(5): \quad$ [[Groups of Order Twice a Prime]] - determines theorem for orders $6$, $10$, $14$
+:$(6): \quad$ [[Groups of Order 8]] - determines theorem for order $8$
+:$(7): \quad$ [[Groups of Order 12]] - determines theorem for order $12$
+{{qed}}
+[[Category:Order of Groups]]
+pp24q0v3ow83txltn6grzbjuoqymggr
+\end{proof}<|endoftext|>
+\section{Existence of Euler-Mascheroni Constant}
+Tags: Euler-Mascheroni Constant
+
+\begin{theorem}
+The [[Definition:Real Sequence|real sequence]]:
+:$\displaystyle \sequence {\sum_{k \mathop = 1}^n \frac 1 k - \ln n}$
+[[Definition:Convergent Real Sequence|converges]] to a [[Definition:Limit of Real Sequence|limit]].
+This limit is known as the [[Definition:Euler-Mascheroni Constant|Euler-Mascheroni constant]].
+\end{theorem}
+
+\begin{proof}
+Let $f: \R \setminus \set 0 \to \R: \map f x = \dfrac 1 x$.
+[[Definition:Clearly|Clearly]] $f$ is [[Definition:Continuous Real Function|continuous]] and [[Definition:Positive Real Function|positive]] on $\hointr 1 {+\infty}$.
+From [[Reciprocal Sequence is Strictly Decreasing]], $f$ is [[Definition:Decreasing Real Function|decreasing]] on $\hointr 1 {+\infty}$.
+Therefore the conditions of the [[Integral Test]] hold.
+Thus the [[Definition:Sequence|sequence]] $\sequence {\Delta_n}$ defined as:
+:$\displaystyle \Delta_n = \sum_{k \mathop = 1}^n \map f k - \int_1^n \map f x \rd x$
+is [[Definition:Decreasing Sequence|decreasing]] and [[Definition:Bounded Below Sequence|bounded below]] by zero.
+But from the definition of the [[Definition:Natural Logarithm|natural logarithm]]:
+:$\displaystyle \int_1^n \frac {\d x} x = \ln n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Groups of Order p q}
+Tags: Cyclic Groups, Cyclic Groups of Order p q, Groups of Order p q
+
+\begin{theorem}
+Let $p, q$ be [[Definition:Prime Number|primes]] such that $p < q$ and $p$ does not [[Definition:Divisor of Integer|divide]] $q - 1$.
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p q$.
+Then $G$ is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Let $K$ be a [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] of $G$.  
+By the [[Fourth Sylow Theorem]], the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$ is of the form $1 + k p$ and [[Definition:Divisor of Integer|divides]] $p q$.
+We have that $1 + k p$ cannot [[Definition:Divisor of Integer|divide]] $p$.
+Then $1 + k p$ must divide $q$.
+But as $q$ is [[Definition:Prime Number|prime]], either:
+:$1 + k p = 1$
+or:
+:$1 + k p = q$
+But:
+{{begin-eqn}}
+{{eqn | l = 1 + k p
+      | r = q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k p
+      | r = q - 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | o = \divides
+      | r = q - 1
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Contradiction|contradicts]] our condition that $p$ does not [[Definition:Divisor of Integer|divide]] $q - 1$.
+Hence $1 + k p = 1$.
+Thus there is only one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Similarly, there is only one [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] of $G$.
+Thus, by [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $H$ and $K$ are [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Let $H = \gen x$ and $K = \gen y$.
+To show $G$ is [[Definition:Cyclic Group|cyclic]], it is sufficient to show that $x$ and $y$ [[Definition:Commute|commute]], because then:
+:$\order {x y} = \order x \order y = p q$
+where $\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$ in $G$.
+{{explain|Why does it follow that $\order {x y} {{=}} \order x \order y {{=}} p q$?}}
+Since $H$ and $K$ are [[Definition:Normal Subgroup|normal]]:
+:$x y x^{-1} y^{-1} = \paren {x y x^{-1} } y^{-1} \in K y^{-1} = K$
+and 
+:$x y x^{-1} y^{-1} = x \paren {y x ^{-1} y^{-1} } \in x H = H$
+Now suppose $a \in H \cap K$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \order a
+      | o = \divides
+      | r = p
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \order a
+      | o = \divides
+      | r = q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order a
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+where $e$ is the [[Definition:Identity Element|identity]] of $G$.
+Thus:
+:$x y x^{-1}y^{-1} \in K \cap H = e$
+Hence $x y = y x$ and the result follows.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Let $K$ be a [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] of $G$.  
+By the [[Fourth Sylow Theorem]], the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$ is of the form $1 + k p$ and [[Definition:Divisor of Integer|divides]] $p q$.
+We have that $1 + k p$ cannot [[Definition:Divisor of Integer|divide]] $p$.
+Then $1 + k p$ must divide $q$.
+But as $q$ is [[Definition:Prime Number|prime]], either:
+:$1 + k p = 1$
+or:
+:$1 + k p = q$
+But:
+{{begin-eqn}}
+{{eqn | l = 1 + k p
+      | r = q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k p
+      | r = q - 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | o = \divides
+      | r = q - 1
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Contradiction|contradicts]] our condition that $p$ does not [[Definition:Divisor of Integer|divide]] $q - 1$.
+Hence $1 + k p = 1$.
+Thus there is only one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Similarly, there is only one [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] of $G$.
+Let the [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$ be denoted $P$.
+Let the [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] of $G$ be denoted $Q$.
+We have that:
+:$P \cap Q = \set e$
+where $e$ is the [[Definition:Identity Element|identity element]] of $G$.
+Hence in $P \cup Q$ there are $q + p - 1$ [[Definition:Element|elements]].
+As $p q \ge 2 q > q + p - 1$, there exists a non- [[Definition:Identity Element|identity element]] in $G$ that is not in $H$ or $K$.
+Its [[Definition:Order of Group Element|order]] must be $p q$.
+Hence, by definition, $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Logarithm at One}
+Tags: Derivatives, Logarithms, Derivative of Logarithm at One
+
+\begin{theorem}
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$ for [[Definition:Real Number|real]] $x$ where $x > 0$.
+Then:
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$
+\end{theorem}
+
+\begin{proof}
+[[L'Hôpital's Rule]] gives:
+:$\displaystyle \lim_{x \mathop \to c} \frac {\map f x} {\map g x} = \lim_{x \mathop \to c} \frac {\map {f'} x} {\map {g'} x}$
+(provided the appropriate conditions are fulfilled).
+Here we have:
+{{begin-eqn}}
+{{eqn | l = \map \ln {1 + 0}
+      | r = 0
+}}
+{{eqn | l = \map {D_x} {\map \ln {1 + x} }
+      | r = \dfrac 1 {1 + x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | l = \map {D_x} x
+      | r = 1
+      | c = [[Derivative of Identity Function]]
+}}
+{{end-eqn}}
+Thus:
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = \lim_{x \mathop \to 0} \frac {\paren {1 + x}^{-1} } 1 = \frac 1 {1 + 0} = 1$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x
+      | r = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} - \ln 1} x
+      | c = subtract $\ln 1 = 0$ from the numerator, from [[Logarithm of 1 is 0]]
+}}
+{{eqn | r = \intlimits {\dfrac {\d} {\d x} \ln x} {x \mathop = 1} {}
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{eqn | r = \frac 1 1
+      | c = [[Derivative of Natural Logarithm Function]]
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Note that this proof does not presuppose [[Derivative of Natural Logarithm Function]].
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x
+      | r = \lim_{n \mathop \to \infty} \frac {\map \ln {1 + \frac 1 n} } {\frac 1 n}
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} n \, \map \ln {1 + \frac 1 n}
+      | c = 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map \ln {\paren {1 + \frac 1 n}^n}
+      | c = 
+}}
+{{eqn | r = \ln e
+      | c = {{Defof|Euler's Number|subdef = Limit of Sequence|Euler's Number as Limit of Sequence}}
+}}
+{{eqn | r = 1
+      | c = [[Natural Logarithm of e is 1]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dicyclic Group is Group}
+Tags: Dicyclic Groups
+
+\begin{theorem}
+The [[Definition:Dicyclic Group|dicyclic group]] $Q_n$ is a [[Definition:Abelian Group|non-abelian]] [[Definition:Group|group]] on two [[Definition:Generator of Group|generators]].
+\end{theorem}
+
+\begin{proof}
+From the definition of the group, all of these statements follow:
+* $y^4 = 1$
+* $y^2x^k = x^{k+n} = x^ky^2$
+* $j = \pm 1 \implies y^jx^k = x^{-k}y^j$
+* $y^ky^{-1} = x^{k-n}y^ny^{-1} = x^{k-n}y^2y^{-1} = x^{k-n}y$
+Thus, every element of $Q_n$ can be uniquely written as $x^k y^j$, where $0 \leq k < 2n$ and $j \in \left\{{0,1}\right\}$. 
+The multiplication rules are given by
+*$a^k a^m = a^{k+m}$
+*$a^k a^m x = a^{k+m}x$
+*$a^k x a^m = a^{k-m}x$
+*$a^k x a^m x = a^{k-m+n}$
+{{proof wanted}}
+[[Category:Dicyclic Groups]]
+gj4zrw8a74cx3kk44s2su7icw3c8vaq
+\end{proof}<|endoftext|>
+\section{Existence of Interval of Convergence of Power Series}
+Tags: Power Series, Radius of Convergence, Existence of Interval of Convergence of Power Series
+
+\begin{theorem}
+Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
+Let $\displaystyle \map S x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a [[Definition:Power Series|power series]] about $\xi$.
+Then the [[Definition:Interval of Convergence|interval of convergence]] of $\map S x$ is a [[Definition:Real Interval|real interval]] whose [[Definition:Midpoint of Interval|midpoint]] is $\xi$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\map S x$ [[Definition:Convergent Series|converges]] when $x = y$.
+We need to show that it converges for all $x$ which satisfy $\size {x - \xi} < \size {y - \xi}$.
+So, let $\map S x$ converge when $x = y$.
+Then from [[Terms in Convergent Series Converge to Zero]]:
+:$a_n \paren {y - \xi}^n \to 0$ as $n \to \infty$
+Hence, from [[Convergent Sequence is Bounded]]:
+:$\sequence {a_n \paren {y - \xi}^n}$ is [[Definition:Bounded Sequence|bounded]]
+Thus:
+:$\exists H \in \R: \forall n \in \N_{>0}: \size {a_n \paren {y - \xi}^n} \le H$
+Now suppose $\size {x - \xi} < \size {y - \xi}$.
+Then:
+:$\rho = \dfrac {\size {x - \xi} } {\size {y - \xi} } < 1$
+(Note that if $\size {x - \xi} < \size {y - \xi}$ then $\size {y - \xi} > 0$ and the above fraction always exists.)
+Hence:
+:$\forall n \in \N_{>0} \size {a_n \paren {x - \xi}^n} = \rho^n \size {a_n \paren {y - \xi}^n} \le H \rho^n$
+By [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \rho^n$ [[Definition:Convergent Series|converges]]
+Thus $\displaystyle \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ converges by the [[Comparison Test]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Series is Differentiable on Interval of Convergence}
+Tags: Power Series, Convergence, Differential Calculus
+
+\begin{theorem}
+Let $\xi \in \R$ be a [[Definition:Real Number|real number]].
+Let $\displaystyle \map f x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a [[Definition:Power Series|power series]] about $\xi$.
+Let $\map f x$ have an [[Definition:Interval of Convergence|interval of convergence]] $I$.
+Then $\map f x$ is [[Definition:Continuous on Interval|continuous]] on $I$, and [[Definition:Differentiable on Interval|differentiable]] on $I$ except possibly at its [[Definition:Endpoint of Real Interval|endpoints]].
+Also:
+:$\displaystyle \map {D_x} {\map f x} = \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1}$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Radius of Convergence|radius of convergence]] of $\map f x$ be $R$.
+Suppose $x \in I$ such that $x$ is not an [[Definition:Endpoint of Real Interval|endpoint]] of $I$.
+Then there exists $x_0 \in I$ such that $x$ lies between $x_0$ and $\xi$.
+Thus:
+:$\size {x - \xi} < \size {x_0 - \xi} < R$
+Consider the series:
+:$\displaystyle \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$ where $X_0 = x_0 - \xi$
+From [[Radius of Convergence from Limit of Sequence]], we have:
+:$\displaystyle \frac 1 R = \limsup_{n \mathop \to \infty} \size {a_n}^{1/n} = \limsup_{n \mathop \to \infty} \paren {\size {\frac {n \paren {n - 1} } 2} a_n}^{1/n}$
+Thus we may deduce that:
+:$\displaystyle \sum_{n \mathop = 2}^\infty \size {\frac {n \paren {n - 1} } 2 a_n X_0^{n - 2} }$
+converges.
+Put $\delta = \size {x - x_0}$.
+For values of $y$ such that $0 < \size {x - y} < \delta$, we consider:
+:$\displaystyle \Delta = \frac {\map f y - \map f x} {y - x} - \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1} = \sum_{n \mathop = 1}^\infty a_n \paren {\frac {Y^n - X^n} {Y - X} - n X^{n - 1} }$
+where $X = x - \xi, Y = y - \xi$.
+We want to show that $\Delta \to 0$ as $Y \to X$.
+From [[Difference of Two Powers]], we have:
+{{begin-eqn}}
+{{eqn | l = \frac {Y^n - X^n} {Y - X} - n X^{n-1}
+      | r = \paren {Y^{n - 1} + Y^{n - 2} X + \cdots + Y X^{n - 2} + X^{n - 1} } - n X^{n - 1}
+      | c = 
+}}
+{{eqn | r = \paren {Y^{n - 1} - X^{n - 1} } + X \paren {Y^{n - 2} - X^{n - 2} } + \cdots + x^{n - 2} \paren {Y - X}
+      | c = 
+}}
+{{end-eqn}}
+It follows that from all of these terms we can extract $\left({Y - X}\right)$ as a factor.
+So the {{RHS}} reduces to a product of $\paren {Y - X}$ with the sum of $n \paren {n - 1} / 2$ terms of the form $X^r Y^s$ where $r + s = n - 2$.
+The number $n \paren {n - 1} / 2$ arises from [[Closed Form for Triangular Numbers]]:
+:$1 + 2 + \cdots + \paren {n - 1} = \dfrac {n \paren {n - 1} } 2$
+Since:
+:$\size X = \size {x - \xi} < \size {x_0 - \xi} = \size {X_0}$
+:$\size Y = \size {y - \xi} < \size {x - \xi} + \delta = \size {X_0}$
+we have:
+:$\size {\dfrac {Y^n - X^n} {Y - X} - n X^{n - 1} } < n \paren {n - 1} / 2 \size {X_0}^{n - 2} \size {Y - X}$
+So:
+:$\displaystyle \size \Delta \le \size {Y - X} \sum_{n \mathop = 2}^\infty \frac {n \paren {n - 1} } 2 \size {a_n X_0^{n - 2} } \to 0$ as $Y \to X$
+So $f$ is [[Definition:Differentiable Real Function at Point|differentiable]] at all $x \in I$ which is not an [[Definition:Endpoint of Real Interval|endpoint]], and that:
+:$\displaystyle \map {D_x} {\map f x} = \sum_{n \mathop = 1}^\infty n a_n \paren {x - \xi}^{n - 1}$
+Now we need to investigate the question of [[Definition:Left-Continuous at Point|left hand continuity]] and [[Definition:Right-Continuous at Point|right hand continuity]] at the [[Definition:Endpoint of Real Interval|endpoints]] of $I$.
+[[Abel's Theorem]] tells us that if $\displaystyle \sum_{k \mathop = 0}^\infty a_k$ is [[Definition:Convergent Series|convergent]], then $\displaystyle \lim_{x \mathop \to 1^-} \paren {\sum_{k \mathop = 0}^\infty a_k x^k} = \sum_{k \mathop = 0}^\infty a_k$.
+{{proof wanted}}
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Exponential Function}
+Tags: Exponential Function, Examples of Power Series, Taylor Series
+
+\begin{theorem}
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential function]].
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall x \in \R:
+      | l = \exp x
+      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Higher Derivatives of Exponential Function]], we have:
+:$\forall n \in \N: \map {f^{\paren n} } {\exp x} = \exp x$
+Since $\exp 0 = 1$, the [[Definition:Taylor Series|Taylor series]] expansion for $\exp x$ about $0$ is given by:
+:$\displaystyle \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$
+From [[Radius of Convergence of Power Series over Factorial]], we know that this [[Definition:Power Series|power series]] expansion converges for all $x \in \R$.
+From [[Taylor's Theorem]], we know that
+:$\displaystyle \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$
+where $0 \le \eta \le x$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \size {\exp x - \paren {1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} } }
+      | r = \size {\frac {x^n} {n!} \map \exp \eta}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \frac {\size {x^n} } {n!} \map \exp {\size x}
+      | c = [[Exponential is Strictly Increasing]]
+}}
+{{eqn | o = \to
+      | r = 0
+      | rr= \text { as } n \to \infty
+      | c = [[Series of Power over Factorial Converges]]
+}}
+{{end-eqn}}
+So the partial sums of the power series converge to $\exp x$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Real Sine Function}
+Tags: Sine Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\sin x$ be the [[Definition:Real Sine Function|sine of $x$]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Properties of Real Cosine Function}
+Tags: Cosine Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\cos x$ be the [[Definition:Real Cosine Function|cosine of $x$]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Derivative of Cosine Function}
+Tags: Derivatives of Trigonometric Functions, Cosine Function, Derivative of Cosine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Real Cosine Function|cosine function]], we have:
+:$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cos x}
+      | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n 2 n \frac {x^{2 n - 1} } {\paren {2 n}!}
+      | c = [[Power Series is Differentiable on Interval of Convergence]]
+}}
+{{eqn | r = \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n - 1} } {\paren {2 n - 1}!}
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^{n + 1} \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = changing summation index
+}}
+{{eqn | r = -\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
+}}
+{{end-eqn}}
+The result follows from the definition of the [[Definition:Real Sine Function|sine function]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cos x}
+      | r = \lim_{h \mathop \to 0} \frac {\map \cos {x + h} - \cos x} h
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\cos x \cos h - \sin x \sin h - \cos x} h
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\cos x \cos h - \cos x} h + \lim_{h \mathop \to 0} \frac {-\sin x \sin h} h
+      | c = [[Sum Rule for Limits of Functions]]
+}}
+{{eqn | r = \cos x \lim_{h \mathop \to 0} \frac {\cos h - 1} h - \sin x \lim_{h \mathop \to 0} \frac {\sin h} h
+      | c = [[Multiple Rule for Limits of Functions]]
+}}
+{{eqn | r = \cos x \times 0 - \sin x \times 1
+      | c = [[Limit of (Cosine (X) - 1) over X]] and [[Limit of Sine of X over X]]
+}}
+{{eqn | r = -\sin x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac \d {\d x} \cos x
+      | r = \frac \d {\d x} \map \sin {\frac \pi 2 - x}
+      | c = [[Sine of Complement equals Cosine]]
+}}
+{{eqn | r = -\map \cos {\frac \pi 2 - x}
+      | c = [[Derivative of Sine Function]] and [[Chain Rule for Derivatives]] 
+}}
+{{eqn | r = -\sin x
+      | c = [[Cosine of Complement equals Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cos x}
+      | r = \lim_{h \mathop \to 0} \frac {\map \cos {x + h} - \cos x} h
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\map \cos {\paren {x + \frac h 2} + \frac h 2} - \map \cos {\paren {x + \frac h 2} - \frac h 2} } h
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {-2 \map \sin {x + \frac h 2} \map \sin {\frac h 2} } h
+      | c = [[Simpson's Formula for Sine by Sine]]
+}}
+{{eqn | r = -\lim_{h \mathop \to 0} \map \sin {x + \frac h 2} \lim_{h \mathop \to 0} \frac {\map \sin {\frac h 2} } {\frac h 2}
+      | c = [[Multiple Rule for Limits of Functions]] and [[Product Rule for Limits of Functions]]
+}}
+{{eqn | r = -\sin x \times 1
+      | c = Continuity of Sine and [[Limit of Sine of X over X]]
+}}
+{{eqn | r = -\sin x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \cos^2 x + \sin^2 x
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {D_x} 1
+      | r = \map {D_x} {\cos^2 x + \sin^2 x}
+      | c = differentiating both sides
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = 2 \map {D_x} {\cos x} \cos x  + 2 \sin x \cos x
+      | c = [[Product Rule for Derivatives]] or [[Chain Rule for Derivatives]] and [[Derivative of Sine Function]] 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {D_x} {\cos x}
+      | r = -\sin x
+      | c = provided $\cos x \ne 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Sine of X over X}
+Tags: Differential Calculus, Sine Function, Limits of Functions, Limit of Sine of X over X
+
+\begin{theorem}
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin x
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = {{Defof|Real Sine Function}}
+}}
+{{eqn | r = \left({-1}\right)^0 \frac{x^{2 \cdot 0 + 1} } { \left({2 \cdot 0 + 1}\right)!} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
+}}
+{{eqn | r = x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to 0} \frac {\sin x} x
+      | r = \lim_{x \mathop \to 0} \frac {x + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x
+}}
+{{eqn | r = \lim_{x \mathop \to 0} \frac x x + \lim_{x \mathop \to 0} \frac{\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} } x
+}}
+{{eqn | r = 1 + \lim_{x \mathop \to 0} \frac {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} } 1
+      | c = [[Power Series is Differentiable on Interval of Convergence]] and [[L'Hôpital's Rule]]
+}}
+{{eqn | r = 1 + \lim_{x \mathop \to 0} \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}
+}}
+{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {0^{2 n} } {\paren {2 n}!}
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sine of Zero is Zero]]:
+:$\sin 0 = 0$
+From [[Derivative of Sine Function]]:
+:$D_x \left({\sin x}\right) = \cos x$
+Then by [[Cosine of Zero is One]]:
+:$\cos 0 = 1$
+From [[Derivative of Identity Function]]:
+:$D_x \left({x}\right) = 1$
+Thus [[L'Hôpital's Rule]] applies and so:
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {D_x \left({\sin x}\right)} {D_x \left({x}\right)} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Squares of Sine and Cosine}
+Tags: Sine Function, Cosine Function, Sum of Squares of Sine and Cosine
+
+\begin{theorem}
+:$\cos^2 x + \sin^2 x = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \cos 0
+      | c = [[Cosine of Zero is One]]
+}}
+{{eqn | r = \map \cos {x - x}
+      | c = 
+}}
+{{eqn | r = \cos x \, \map \cos {-x} - \sin x \, \map \sin {-x}
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \cos x \cos x - \paren {-\sin x \sin x}
+      | c = [[Cosine Function is Even]] and [[Sine Function is Odd]]
+}}
+{{eqn | r = \cos^2 x + \sin^2 x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+=== Notes ===
+Note that we need to start from the algebraic definitions of [[Definition:Complex Sine Function|sine]] and [[Definition:Complex Cosine Function|cosine]]:
+:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \frac {x^3} {3!} + \frac {x^5} {5!} - \cdots$
+:$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} = 1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \cdots$
+and then use the proofs of the [[Cosine of Sum]] that derive directly from these.
+Otherwise these proofs are circular.
+\end{proof}
+
+\begin{proof}
+From the trigonometric definitions of [[Definition:Sine of Angle|sine]] and [[Definition:Cosine of Angle|cosine]]:
+{{begin-eqn}}
+{{eqn | l = \sin x
+      | r = \frac{\text{opposite} } {\text{hypotenuse} }
+}}
+{{eqn | l = \cos x
+      | r = \frac{\text{adjacent} } {\text{hypotenuse} }
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \sin^2 x + \cos^2 x
+      | r = \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}
+      | c [[Definition:Square|squaring]] both sides and adding
+}}
+{{eqn | r = \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}
+      | c = [[Pythagoras's Theorem]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $P = \tuple {x, y}$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of a [[Definition:Unit Circle|unit circle]] whose [[Definition:Center of Circle|center]] is at the [[Definition:Origin|origin]] of a [[Definition:Cartesian Plane|cartesian plane]].
+From [[Sine of Angle in Cartesian Plane]] and [[Cosine of Angle in Cartesian Plane]]:
+:$P = \tuple {\cos \theta, \sin \theta}$
+The [[Definition:Graph of Relation|graph]] of the [[Definition:Unit Circle|unit circle]] is the [[Definition:Locus|locus]] of:
+:$x^2 + y^2 = 1$
+as given by [[Equation of Circle]].
+Substituting $x = \cos \theta$ and $y = \sin \theta$ yields:
+:$\cos^2 \theta + \sin^2 \theta = 1$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^2 x + \sin^2 x
+      | r = \left({\cos x + i \, \sin x}\right) \, \left({\cos x - i \, \sin x}\right)
+      | c = factoring over the [[Definition:Complex Number|complex numbers]]
+}}
+{{eqn | r = \left({\cos x + i \, \sin x}\right) \, \left({\cos \left({-x}\right) + i \, \sin \left({-x}\right)}\right)
+      | c = [[Cosine Function is Even]] and [[Sine Function is Odd]]
+}}
+{{eqn | r = e^{ix} \, e^{-ix}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos^2 x + \sin^2 x
+      | r = \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \sin^2 x
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \paren {\frac {e^{i x} + e^{-i x} } 2}^2 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | r = \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \paren {\frac {e^{i x} - e^{-i x} } {2 i} }^2
+      | c = [[Square of Sum]]
+}}
+{{eqn | r = \frac {\paren {e^{i x} }^2 + 2 e^{-i x} e^{i x} + \paren {e^{-i x} }^2} 4 + \frac {\paren {e^{i x} }^2 - e^{-i x} e^{i x} + \paren {e^{-i x} }^2} {-4} 
+      | c = [[Square of Difference]] and $i^2 = -1$
+}}
+{{eqn | r = \frac {e^{2 i x} + 2 + e^{-2 i x} } 4 + \frac {e^{2 i x} - 2 + e^{-2 i x} } {-4} 
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \frac {e^{2 i x} + 2 + e^{-2 i x} - e^{2 i x} + 2 - e^{-2 i x} } 4
+      | c = simplifying
+}}
+{{eqn | r = \frac 4 4
+      | c = simplifying
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Sum}
+Tags: Cosine Function, Cosine of Sum, Compound Angle Formulas
+
+\begin{theorem}
+:$\map \cos {a + b} = \cos a \cos b - \sin a \sin b$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {a + b} + i \, \map \sin {a + b}
+      | r = e^{i \paren {a + b} }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = e^{i a} e^{i b}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \paren {\cos a + i \sin a} \paren {\cos b + i \sin b}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \paren {\cos a \cos b - \sin a \sin b} + i \paren {\sin a \cos b + \cos a \sin b}
+      | c = [[Complex Numbers form Field]]
+}}
+{{end-eqn}}
+The result follows by equating the [[Definition:Real Part|real parts]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Recall the analytic definitions of [[Definition:Real Sine Function|sine]] and [[Definition:Real Cosine Function|cosine]]:
+:$\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n + 1}} {\left({2 n + 1}\right)!}$
+:$\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \left({-1}\right)^n \frac {x^{2 n}} {\left({2 n}\right)!}$
+Let:
+{{begin-eqn}}
+{{eqn | l = g \left({a}\right)
+      | r = \sin \left({a + b}\right) - \sin a \cos b - \cos a \sin b
+}}
+{{eqn | l = h \left({a}\right)
+      | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b
+}}
+{{end-eqn}}
+Let us [[Definition:Differentiation With Respect To|differentiate these with respect to]] $a$, keeping $b$ constant.
+Then from [[Derivative of Sine Function]] and [[Derivative of Cosine Function]], we have:
+{{begin-eqn}}
+{{eqn | l = g' \left({a}\right)
+      | r = \cos \left({a + b}\right) - \cos a \cos b + \sin a \sin b = h \left({a}\right)
+}}
+{{eqn | l = h' \left({a}\right)
+      | r = - \sin \left({a + b}\right) + \sin a \cos b + \cos a \sin b = - g \left({a}\right)
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = D_a \left({\left({g \left({a}\right)}\right)^2 + \left({h \left({a}\right)}\right)^2}\right)
+      | r = 2 g \left({a}\right) g' \left({a}\right) + 2 h \left({a}\right) h' \left({a}\right)
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Thus from [[Derivative of Constant]]:
+: $\forall a \in \R: g \left({a}\right)^2 + h \left({a}\right)^2 = c$
+In particular, it is true for $a = 0$, and so:
+:$g \left({0}\right)^2 + h \left({0}\right)^2 = 0$
+So:
+:$g \left({a}\right)^2 + h \left({a}\right)^2 = 0$
+But from [[Square of Real Number is Non-Negative]]:
+:$g \left({a}\right)^2 \ge 0$
+and:
+:$h \left({a}\right)^2 \ge 0$
+So it follows that:
+:$g \left({a}\right) = 0$
+and:
+:$h \left({a}\right) = 0$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+:[[File:Tri1.PNG]]
+$AB$, $AC$, $AE$, and $AD$ are [[Definition:Radius of Circle|radii]] of the [[Definition:Circle|circle]] centered at $A$.
+Let $\angle BAC = a$ and $\angle DAC = \angle BAE = b$.
+By [[Axiom:Euclid's First Postulate|Euclid's First Postulate]], we can construct [[Definition:Line Segment|line segments]] $BD$ and $CE$.
+By [[Axiom:Euclid's Common Notions|Euclid's second common notion]], $\angle DAB = \angle CAE$.
+Thus by [[Triangle Side-Angle-Side Equality]], $\triangle DAB \cong \triangle CAE$.
+Therefore, $DB = CE$.
+ 
+We now assign [[Definition:Cartesian Coordinate System|Cartesian coordinates]] to the points $B$, $C$, $D$, and $E$:
+{{begin-eqn}}
+{{eqn | l = B
+      | r = \tuple {1, 0}
+      | c = 
+}}
+{{eqn | l = C
+      | r = \tuple {\cos a, \sin a}
+      | c = 
+}}
+{{eqn | l = D
+      | r = \tuple {\map \cos {a + b}, \map \sin {a + b} }
+      | c = 
+}}
+{{eqn | l = E
+      | r = \tuple {\cos b, -\sin b}
+      | c = [[Cosine Function is Even]] and [[Sine Function is Odd]]
+}}
+{{end-eqn}}
+We use the definition of the [[Definition:Distance Function|distance function]] on the [[Definition:Euclidean Space|Euclidean space]] $\struct {\R^2, d}$ as defined by the [[Definition:Euclidean Metric on Real Number Plane|Euclidean metric]]:
+:$\forall x, y \in \R^2: \map d {x, y} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
+where $x = \tuple {x_1, y_1}, y = \tuple {x_2, y_2}$.
+Thus:
+:$DB \cong CE \iff \map d {D, B} = \map d {C, E}$
+So, plugging in the coordinates of $B, C, D, E$, we get:
+{{begin-eqn}}
+{{eqn | l = \paren {\map \cos {a + b} } - 1)^2 + \map {\sin^2} {a + b}
+      | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2
+}}
+{{eqn | ll= \leadsto
+      | l = \cos^2 \left({a + b}\right) + \sin^2 \left({a + b}\right)
+      | o = 
+      | c = multiplying out {{LHS}}
+}}
+{{eqn | l = {} - 2 \, \map \cos {a + b} + 1
+      | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 1 - 2 \, \map \cos {a + b} + 1
+      | r = \paren {\cos a - \cos b}^2 + \paren {\sin a + \sin b}^2
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = 2 - 2 \, \map \cos {a + b}
+      | r = \cos^2 a - 2 \cos a \cos b + \cos^2 b
+      | c = multiplying out {{RHS}}
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \sin^2 a + 2 \sin a \sin b + \sin^2 b
+}}
+{{eqn | ll= \leadsto
+      | l = 2 - 2 \, \map \cos {a + b}
+      | r = 2 - 2 \cos a \cos b + 2 \sin a \sin b
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \cos {a + b}
+      | r = \cos a \cos b - \sin a \sin b
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}} 
+{{eqn | l = \cos a \cos b - \sin a \sin b
+      | r = \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2} - \sin a \sin b
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \paren {\frac {e^{i a} + e^{-i a} } 2} \paren {\frac {e^{i b} + e^{-i b} } 2}
+          - \paren {\frac {e^{i a} - e^{-i a} } {2 i} } \paren {\frac {e^{i b} - e^{-i b} } {2 i} }
+      | c = [[Sine Exponential Formulation]]
+}}
+{{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } {4 i^2}
+}}
+{{eqn | r = \frac {e^{i a} e^{i b} + e^{i a} e^{-i b} + e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac {e^{i a} e^{i b} - e^{i a} e^{-i b} - e^{-i a} e^{i b} + e^{-i a} e^{-i b} } 4
+      | c = as $i^2 = -1$
+}}
+{{eqn | r = \frac {e^{i a} e^{i b} + e^{-ia} e^{-ib} } 2
+}}
+{{eqn | r = \frac {e^{i \paren {a + b} } + e^{-i \paren {a + b} } } 2
+      | c = [[Exponential of Sum]] 
+}}
+{{eqn | r = \map \cos {a + b}
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Sine and Cosine}
+Tags: Sine Function, Cosine Function, Definition Equivalences
+
+\begin{theorem}
+The definitions for [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]] are equivalent.
+That is:
+: $\displaystyle \sin x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n + 1} } {\paren {2 n + 1}!} \iff \sin x = \frac {\text{Opposite}} {\text{Hypotenuse}}$
+: $\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!} \iff \cos x = \frac {\text{Adjacent}} {\text{Hypotenuse}}$
+\end{theorem}
+
+\begin{proof}
+Let $\map s x: \R \to \R$, $\map c x: \R \to \R$ be two [[Definition:Real Function|functions]] that satisfy:
+:$(1): \quad \map {s'} x = \map c x$
+:$(2): \quad \map {c'} x = -\map s x$
+:$(3): \quad \map s 0 = 0$
+:$(4): \quad \map c 0 = 1$
+:$(5): \quad \forall x: \map {s^2} x + \map {c^2} x = 1$
+where $s'$ denotes the [[Definition:Derivative|derivative]] {{WRT|Differentiation}} $x$.
+Let $\map f x: \R \to \R$, $\map g x: \R \to \R$ also be two functions that satisfy:
+:$(1): \quad \map {f'} x = \map g x$
+:$(2): \quad \map {g'} x = -\map f x$
+:$(3): \quad \map f 0 = 0$
+:$(4): \quad \map g 0 = 1$
+:$(5): \quad \forall x: \map {f^2} x + \map {g^2} x = 1$
+It will be shown that:
+:$\map f x = \map s x$
+and:
+:$\map g x = \map c x$
+Define:
+:$\map h x = \paren {\map c x - \map g x}^2 + \paren {\map s x - \map f x}^2$
+Notice that:
+:$\paren {\forall x: \map h x = 0} \iff \paren {\forall x: \map c x = \map g x, \map s x = \map f x}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map h x
+      | r = \map {c^2} x - 2 \map c x \map g x + \map {g^2} x + \map {s^2} x - 2 \map s x \map f x + \map {f^2} x
+      | c =
+}}
+{{eqn | r = 2 - 2 \paren {\map c x \map g x + \map s x \map f x}
+      | c = Property $(5)$
+}}
+{{end-eqn}}
+By taking $\map {h'} x$:
+{{begin-eqn}}
+{{eqn | l = \map {h'} x
+      | r = -2 \paren {\map c x \paren {-\map f x} + \map g x \paren {-\map s x} + \map s x \map g x + \map c x \map f x}
+      | c = Properties $(1)$ and $(2)$ and [[Product Rule]]
+}}
+{{eqn | r = 0
+      | c =  
+}}
+{{end-eqn}}
+By [[Zero Derivative implies Constant Function]], $\map h x$ is a constant function:
+:$\map h x = k$
+Also:
+{{begin-eqn}}
+{{eqn | l = \map h 0
+      | r = \paren {1 - 1}^2 + \paren {0 - 0}^2
+      | c = Properties $(3)$ and $(4)$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Since $\map h x$ is constant, then:
+: $\forall x: \map h x = 0$
+Then:
+:$\map c x = \map g x$
+and:
+:$\map s x = \map f x$
+By:
+:[[Derivative of Sine Function]]
+:[[Derivative of Cosine Function]]
+:[[Sine of Zero is Zero]]
+:[[Cosine of Zero is One]]
+:[[Sum of Squares of Sine and Cosine]]
+both definitions satisfy all these properties.
+Therefore they must be the same.
+{{qed}}
+[[Category:Sine Function]]
+[[Category:Cosine Function]]
+[[Category:Definition Equivalences]]
+da4v7o00c8r9abiknxjy95m7utal6f4
+\end{proof}<|endoftext|>
+\section{Real Cosine Function is Bounded}
+Tags: Cosine Function
+
+\begin{theorem}
+: $\size {\cos x} \le 1$
+\end{theorem}
+
+\begin{proof}
+From the algebraic definition of the [[Definition:Real Cosine Function|real cosine function]]:
+: $\displaystyle \cos x = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n}!}$
+it follows that $\cos x$ is a [[Definition:Real Function|real function]].
+Thus $\cos^2 x \ge 0$.
+From [[Sum of Squares of Sine and Cosine]], we have that $\cos^2 x + \sin^2 x = 1$.
+Thus it follows that:
+: $\cos^2 x = 1 - \sin^2 x \le 1$
+From [[Ordering of Squares in Reals]] and the definition of [[Definition:Absolute Value|absolute value]], we have that:
+:$x^2 \le 1 \iff \size x \le 1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundedness of Sine X over X}
+Tags: Sine Function
+
+\begin{theorem}
+Let $x \in \R$.
+Then:
+:$\size {\dfrac {\sin x} x} \le 1$
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Sine Function]], we have:
+: $D_x \paren {\sin x} = \cos x$
+So by the [[Mean Value Theorem]], there exists $\xi \in \R$ between $0$ and $x$ such that:
+:$\dfrac {\sin x - \sin 0} {x - 0} = \cos \xi$
+From [[Real Cosine Function is Bounded]] we have that:
+: $\size {\cos \xi} \le 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine and Cosine are Periodic on Reals}
+Tags: Sine Function, Cosine Function, Periodic Functions
+
+\begin{theorem}
+The [[Definition:Real Sine Function|sine]] and [[Definition:Real Cosine Function|cosine]] functions are [[Definition:Periodic Function|periodic]] on the set of [[Definition:Real Number|real numbers]] $\R$:
+:$(1): \quad \map \cos {x + 2 \pi} = \cos x$
+:$(2): \quad \map \sin {x + 2 \pi} = \sin x$
+:[[File:SineCos.png|800px]]
+\end{theorem}
+
+\begin{proof}
+From [[Cosine of Zero is One]] we have that $\cos 0 = 1$.
+From [[Cosine Function is Even]] we have that $\cos x = \map \cos {-x}$.
+As the [[Cosine Function is Continuous]], it follows that:
+:$\exists \xi > 0: \forall x \in \openint {-\xi} \xi: \cos x > 0$
+{{AimForCont}} $\cos x$ were [[Definition:Positive Real Function|positive]] everywhere on $\R$.
+From [[Derivative of Cosine Function]]:
+:$\map {D_{xx} } {\cos x} = \map {D_x} {-\sin x} = -\cos x$
+Thus $-\cos x$ would always be [[Definition:Negative Real Function|negative]].
+Thus from [[Second Derivative of Concave Real Function is Non-Positive]], $\cos x$ would be [[Definition:Concave Real Function|concave]] everywhere on $\R$.
+But from [[Real Cosine Function is Bounded]], $\cos x$ is [[Definition:Bounded Real-Valued Function|bounded on $\R$]].
+By [[Differentiable Bounded Concave Real Function is Constant]], $\cos x$ would then be a [[Definition:Constant Function|constant function]].
+This [[Definition:Contradiction|contradicts]] the fact that $\cos x$ is not a [[Definition:Constant Function|constant function]].
+Thus by [[Proof by Contradiction]] $\cos x$ can not be [[Definition:Positive Real Function|positive]] everywhere on $\R$.
+Therefore, there must exist a smallest [[Definition:Positive Real Number|positive]] $\eta \in \R_{>0}$ such that $\cos \eta = 0$.
+By definition, $\cos \eta = \map \cos {-\eta} = 0$ and $\cos x > 0$ for $-\eta < x < \eta$.
+Now we show that $\sin \eta = 1$.
+From [[Sum of Squares of Sine and Cosine]]:
+:$\cos^2 x + \sin^2 x = 1$
+Hence as $\cos \eta = 0$ it follows that $\sin^2 \eta = 1$.
+So either $\sin \eta = 1$ or $\sin \eta = -1$.
+But $\map {D_x} {\sin x} = \cos x$.
+On the [[Definition:Closed Real Interval|interval]] $\closedint {-\eta} \eta$, it has been shown that $\cos x > 0$.
+Thus by [[Derivative of Monotone Function]], $\sin x$ is [[Definition:Increasing Real Function|increasing]] on $\closedint {-\eta} \eta$.
+Since $\sin 0 = 0$ it follows that $\sin \eta > 0$.
+So it must be that $\sin \eta = 1$.
+Now we apply [[Sine of Sum]] and [[Cosine of Sum]]:
+:$\map \sin {x + \eta} = \sin x \cos \eta + \cos x \sin \eta = \cos x$
+:$\map \cos {x + \eta} = \cos x \cos \eta - \sin x \sin \eta = -\sin x$
+Hence it follows, after some algebra, that:
+:$\map \sin {x + 4 \eta} = \sin x$
+:$\map \cos {x + 4 \eta} = \cos x$
+Thus $\sin$ and $\cos$ are [[Definition:Periodic Function|periodic]] on $\R$ with [[Definition:Period of Function|period]] $4 \eta$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiable Bounded Convex Real Function is Constant}
+Tags: Convex Real Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is:
+:$(1): \quad$ [[Definition:Differentiable on Interval|Differentiable]] on $\R$
+:$(2): \quad$ [[Definition:Bounded Real-Valued Function|Bounded]] on $\R$
+:$(3): \quad$ [[Definition:Convex Real Function|Convex]] on $\R$.
+Then $f$ is [[Definition:Constant Mapping|constant]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Differentiable on Interval|differentiable]] and [[Definition:Bounded Real-Valued Function|bounded]] on $\R$.
+Let $f$ be [[Definition:Convex Real Function|convex]] on $\R$.
+Let $\xi \in \R$.
+Aiming for a [[Proof by Contradiction|contradiction]], suppose $f' \left({\xi}\right) > 0$.
+Then by [[Mean Value of Convex Real Function]] it follows that:
+: $f \left({x}\right) \ge f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to +\infty$
+and therefore is not [[Definition:Bounded Real-Valued Function|bounded]].
+Similarly, suppose $f' \left({\xi}\right) < 0$.
+Then by [[Mean Value of Convex Real Function]] it follows that:
+: $f \left({x}\right) \ge f \left({\xi}\right) + f' \left({\xi}\right) \left({x - \xi}\right) \to + \infty$ as $x \to -\infty$
+and therefore is likewise not [[Definition:Bounded Real-Valued Function|bounded]].
+Hence $f' \left({\xi}\right) = 0$.
+From [[Zero Derivative implies Constant Function]], it follows that $f$ is [[Definition:Constant Mapping|constant]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Banach-Tarski Paradox}
+Tags: Veridical Paradoxes, Antinomies, Topology
+
+\begin{theorem}
+The [[Definition:Disk|unit ball]] $\mathbb D^3 \subset \R^3$ is [[Definition:Equidecomposable|equidecomposable]] to the union of two unit balls.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbb D^3$ be centered at the origin, and $D^3$ be some other unit ball in $\R^3$ such that $\mathbb D^3 \cap D^3 = \O$.
+Let $\mathbb S^2 = \partial \mathbb D^3$.
+By the [[Hausdorff Paradox]], there exists a [[Definition:Decomposition (Topology)|decomposition]] of $ \mathbb S^2$ into four sets $A, B, C, D$ such that $A, B, C$ and $B \cup C$ are [[Definition:Congruence (Topology)|congruent]], and $D$ is [[Definition:Countable|countable]].
+For $r \in \R_{>0}$, define a function $r^*: \R^3 \to \R^3$ as $\map {r^*} {\mathbf x} = r \mathbf x$, and define the [[Definition:Set|sets]]:
+{{begin-eqn}}
+{{eqn | l = W
+      | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} A
+}}
+{{eqn | l = X
+      | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} B
+}}
+{{eqn | l = Y
+      | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} C
+}}
+{{eqn | l = Z
+      | r = \bigcup_{0 \mathop < r \mathop \le 1} \map {r^*} D
+}}
+{{end-eqn}}
+Let $T = W \cup Z \cup \set {\mathbf 0}$.
+$W$ and $X \cup Y$ are clearly congruent by the congruency of $A$ with $B \cup C$, hence $W$ and $X \cup Y$ are [[Definition:Equidecomposable|equidecomposable]].
+Since $X$ and $Y$ are congruent, and $W$ and $X$ are congruent, $X \cup Y$ and $W \cup X$ are [[Definition:Equidecomposable|equidecomposable]].  
+$W$ and $X \cup Y$ as well as $X$ and $W$ are congruent, so $W \cup X$ and $W \cup X \cup Y$ are [[Definition:Equidecomposable|equidecomposable]].  
+Hence $W$ and $W \cup X \cup Y$ are [[Definition:Equidecomposable|equidecomposable]], by [[Equidecomposability is Equivalence Relation]].
+So $T$ and $\mathbb D^3$ are [[Definition:Equidecomposable|equidecomposable]], from [[Equidecomposability Unaffected by Union]].
+Similarly we find $X$, $Y$, and $W \cup X \cup Y$ are [[Definition:Equidecomposable|equidecomposable]].
+Since $D$ is only [[Definition:Countable|countable]], but $\map {\operatorname {SO} } 3$ is not, we have:
+:$\exists \phi \in \map {\operatorname {SO} } 3: \map \phi D \subset A \cup B \cup C$
+so that $I = \map \phi D \subset W \cup X \cup Y$.
+Since $X$ and $W \cup X \cup Y$ are [[Definition:Equidecomposable|equidecomposable]], by [[Subsets of Equidecomposable Subsets are Equidecomposable]], $\exists H \subseteq X$ such that $H$ and $I$ are [[Definition:Equidecomposable|equidecomposable]].
+Finally, let $p \in X - H$ be a point and define $S = Y \cup H \cup \set p$.
+Since:
+:$Y$ and $W \cup X \cup Y$
+:$H$ and $Z$
+:$\set 0$ and $\set p$
+are all [[Definition:Equidecomposable|equidecomposable]] in pairs, $S$ and $\mathbb B^3$ are [[Definition:Equidecomposable|equidecomposable]] by [[Equidecomposability Unaffected by Union]].
+Since $D^3$ and $\mathbb D^3$ are congruent, $D^3$ and $S$ are [[Definition:Equidecomposable|equidecomposable]], from [[Equidecomposability is Equivalence Relation]].  
+By [[Equidecomposability Unaffected by Union]], $T \cup S$ and $\mathbb D^3 \cup D^3$ are [[Definition:Equidecomposable|equidecomposable]].
+Hence $T \cup S \subseteq \mathbb D^3 \subset \mathbb D^3 \cup D^3$ are [[Definition:Equidecomposable|equidecomposable]] and so, by the [[Equidecomposable Nested Sets|chain property of equidecomposability]], $\mathbb D^3$ and $\mathbb D^3 \cup D^3$ are [[Definition:Equidecomposable|equidecomposable]].
+{{qed}}
+{{AoC|Hausdorff Paradox}}
+\end{proof}<|endoftext|>
+\section{Equidecomposability is Equivalence Relation}
+Tags: Topology
+
+\begin{theorem}
+The property of being [[Definition:Equidecomposable|equidecomposable]] is an [[Definition:Equivalence Relation|equivalence relation]] on the [[Definition:Power Set|power set]] $\mathcal P \left({\R^n}\right)$.
+\end{theorem}
+
+\begin{proof}
+=== Relexivity ===
+A set is necessarily [[Definition:Equidecomposable|equidecomposable]] with itself; the same [[Definition:Decomposition (Topology)|decomposition]] and set of [[Definition:Isometry (Metric Spaces)|isometries]] suffice for $A$ as for $A$.
+=== Symmetry ===
+There is no order to the relation of being [[Definition:Equidecomposable|equidecomposable]]; symmetry follows.
+=== Transitivity ===
+Suppose $A, B, C \subset \R^n$ are sets such that $A, B$ are [[Definition:Equidecomposable|equidecomposable]] and $B, C$ are [[Definition:Equidecomposable|equidecomposable]].
+Let $X_1, \dots, X_m$ be a [[Definition:Decomposition (Topology)|decomposition]] of $A, B$ together with [[Definition:Isometry (Metric Spaces)|isometries]] $\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m:\R^n \to \R^n$ such that:
+:$\displaystyle A = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)$ 
+and 
+:$\displaystyle B = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)$
+Further let $Y_1, \dots, Y_p$ together with $\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p$ be sets and [[Definition:Isometry (Metric Spaces)|isometries]] such that:
+:$\displaystyle B = \bigcup_{i \mathop = 1}^p \xi_i \left({Y_i}\right)$ 
+and:
+:$\displaystyle C = \bigcup_{i \mathop = 1}^p \tau_i \left({Y_i}\right)$
+Consider the sets
+:$Z_{i,j} = \nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)$
+where $1 \le i \le m$ and $1 \le j \le p$.
+We have:
+:$\displaystyle \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \left({\mu_i \circ \nu_i^{-1}}\right) \left({Z_{i,j}}\right) = \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \left({\mu_i \circ \nu_i^{-1}}\right) \left({\nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)}\right) = \bigcup_{i \mathop = 1}^m \left({\mu_i \circ \nu_i^{-1} \circ \nu_i}\right) \left({X_i}\right) = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right) = A$
+:$\displaystyle \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \left({\tau_j \circ \xi_j^{-1}}\right) \left({Z_{i,j}}\right) = \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \left({\tau_j \circ \xi_j^{-1}}\right) \left({\nu_i \left({X_i}\right) \cap \xi_j \left({Y_j}\right)}\right) = \bigcup_{j \mathop = 1}^p \left({\tau_j \circ \xi_j^{-1} \circ \xi_j}\right) \left({Y_j}\right) = \bigcup_{j \mathop = 1}^p \tau_j \left({Y_j}\right) = C$
+so $Z_{i, j}$ together with the [[Definition:Isometry (Metric Spaces)|isometries]] $\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1}$ is a [[Definition:Decomposition (Topology)|decomposition]] of $A$ and $C$.
+Hence these two are [[Definition:Equidecomposable|equidecomposable]].
+{{qed}}
+[[Category:Topology]]
+7ic97xbu5p8wuwze6h1ejtag6kjdbly
+\end{proof}<|endoftext|>
+\section{Equidecomposability Unaffected by Union}
+Tags: Topology
+
+\begin{theorem}
+{{explain|I guess the sets should be disjoint, in the sense that $S_i \cap S_j$ is empty for $i \neq j$. Similar for $T$.}}
+Let $\left\{{S_1, \ldots, S_m}\right\}, \left\{{T_1, \ldots, T_m }\right\}$ be [[Definition:Set of Sets|sets of sets]] in $\R^n$ such that:
+: for each $k \in \left\{{1, \dots, m}\right\}, S_k$ and $T_k$ are [[Definition:Equidecomposable|equidecomposable]]. 
+Then the set $\displaystyle S = \bigcup_{i \mathop = 1}^m S_i$ is [[Definition:Equidecomposable|equidecomposable]] with $\displaystyle T = \bigcup_{i \mathop = 1}^m T_i$.
+\end{theorem}
+
+\begin{proof}
+We have for each $k \in \left\{{1, \dots, m}\right\}$ a [[Definition:Decomposition (Topology)|decomposition]] $\left\{{A_{k, 1}, \cdots, A_{k, l_k}}\right\}$ and set of [[Definition:Isometry (Metric Spaces)|isometries]] $\phi_{i, j}: \R^n \to \R^n$ such that:
+:$\displaystyle S_k = \bigcup_{a \mathop = 1}^{l_k} \phi_{k, a} \left({A_{k, a} }\right)$
+and similarly for $T_k$ and some [[Definition:Isometry (Metric Spaces)|isometries]] $\theta_{i, j}: \R^n \to \R^n$:
+:$\displaystyle T_k = \bigcup_{a \mathop = 1}^{l_k} \theta_{k, a} \left({A_{k, a} }\right)$
+Thus:
+:$\displaystyle S = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \phi_{k, i} \left({A_{k, i} }\right)$
+and:
+:$\displaystyle T = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \theta_{k, i} \left({A_{k, i} }\right)$.
+This yields equivalent [[Definition:Decomposition (Topology)|decompositions]] of $S$ and $T$.
+{{qed}}
+[[Category:Topology]]
+khsd67uh7ztk2vsb1ngqax6ihrywa2e
+\end{proof}<|endoftext|>
+\section{Subsets of Equidecomposable Subsets are Equidecomposable}
+Tags: Topology
+
+\begin{theorem}
+Let $A, B \subseteq \R^n$ be [[Definition:Equidecomposable|equidecomposable]].
+Let $S \subseteq A$.
+Then there exists $T \subseteq B$ such that $S$ and $T$ are [[Definition:Equidecomposable|equidecomposable]].
+\end{theorem}
+
+\begin{proof}
+Let $X_1, \dots, X_m$ be a [[Definition:Decomposition (Topology)|decomposition]] of $A, B$ together with [[Definition:Isometry (Metric Spaces)|isometries]] $\mu_1, \ldots, \mu_m, \nu_1, \ldots, \nu_m: \R^n \to \R^n$ such that:
+:$\displaystyle A = \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)$ 
+and 
+:$\displaystyle B = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)$
+Define:
+:$Y_i = \mu_i^{-1} \left({S \cap \mu_i \left({X_i}\right)}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = \bigcup_{i \mathop = 1}^m \mu_i \left({Y_i}\right)
+      | r = \bigcup_{i \mathop = 1}^m \left({S \cap \mu_i \left({X_i}\right)}\right)
+      | c = 
+}}
+{{eqn | r = S \cap \bigcup_{i \mathop = 1}^m \mu_i \left({X_i}\right)
+      | c = 
+}}
+{{eqn | r = S \cap A
+      | c = 
+}}
+{{eqn | r = S
+      | c = 
+}}
+{{end-eqn}}
+and so $\left\{{Y_i}\right\}_{i \mathop = 1}^m$ forms a [[Definition:Decomposition (Topology)|decomposition]] of $S$.
+But for each $i$:
+:$\left({S \cap \mu_i \left({X_i}\right)}\right) \subseteq \mu_i \left({X_i}\right)$
+and so:
+{{begin-eqn}}
+{{eqn | l = Y_i
+      | r = \mu_i^{-1} \left({S \cap \mu_i \left({X_i}\right)}\right)
+      | c = 
+}}
+{{eqn | o = \subseteq
+      | r = \mu_i^{-1} \left({\mu_i \left({X_i}\right)}\right)
+      | c = 
+}}
+{{eqn | r = X_i
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\nu_i \left({Y_i}\right) \subseteq \nu_i \left({X_i}\right)$
+and so:
+{{begin-eqn}}
+{{eqn | l = \bigcup_{i \mathop = 1}^m \nu_i \left({Y_i}\right)
+      | o = \subseteq
+      | r = \bigcup_{i \mathop = 1}^m \nu_i \left({X_i}\right)
+      | c = 
+}}
+{{eqn | r = B
+      | c = 
+}}
+{{end-eqn}}
+Define:
+: $\displaystyle \bigcup_{i \mathop = 1}^m \nu_i \left({Y_i}\right) = T$
+Hence the result.
+{{qed}}
+[[Category:Topology]]
+d4nf9og3qpnwksv2n4mhypdck2ylcb4
+\end{proof}<|endoftext|>
+\section{Shape of Cosine Function}
+Tags: Cosine Function
+
+\begin{theorem}
+The [[Definition:Cosine|cosine]] function is:
+:$(1): \quad$ [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on the [[Definition:Closed Real Interval|interval]] $\closedint 0 \pi$
+:$(2): \quad$ [[Definition:Strictly Increasing Real Function|strictly increasing]] on the [[Definition:Closed Real Interval|interval]] $\closedint \pi {2 \pi}$
+:$(3): \quad$ [[Definition:Concave Real Function|concave]] on the [[Definition:Closed Real Interval|interval]] $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
+:$(4): \quad$ [[Definition:Convex Real Function|convex]] on the [[Definition:Closed Real Interval|interval]] $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
+\end{theorem}
+
+\begin{proof}
+From the discussion of [[Sine and Cosine are Periodic on Reals]], we know that:
+:$\cos x \ge 0$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
+and:
+:$\cos x > 0$ on the [[Definition:Open Real Interval|open interval]] $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
+From the [[Sine and Cosine are Periodic on Reals|same discussion]], we have that:
+:$\map \sin {x + \dfrac \pi 2} = \cos x$
+So immediately we have that $\sin x \ge 0$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint 0 \pi$, $\sin x > 0$ on the [[Definition:Open Real Interval|open interval]] $\openint 0 \pi$.
+But $\map {D_x} {\cos x} = -\sin x$ from [[Derivative of Cosine Function]].
+Thus from [[Derivative of Monotone Function]], $\cos x$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\closedint 0 \pi$.
+From [[Derivative of Sine Function]] it follows that:
+:$\map {D_{xx} } {\cos x} = -\cos x$
+On $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$ where $\cos x \ge 0$, therefore, $\map {D_{xx} } {\cos x} \le 0$.
+From [[Second Derivative of Concave Real Function is Non-Positive]] it follows that $\cos x$ is [[Definition:Concave Real Function|concave]] on $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$.
+The rest of the result follows similarly.
+=== [[Shape of Cosine Function/Graph|Graph of Cosine Function]] ===
+{{:Shape of Cosine Function/Graph}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent Function is Periodic on Reals}
+Tags: Tangent Function
+
+\begin{theorem}
+The [[Definition:Real Tangent Function|tangent function]] is [[Definition:Periodic Function|periodic]] on the set of [[Definition:Real Number|real numbers]] $\R$ with [[Definition:Period of Function|period $\pi$]].
+This can be written:
+:$\tan x = \tan \left({x \bmod \pi}\right)$
+where $x \bmod \pi$ denotes the [[Definition:Modulo Operation|modulo operation]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tan \left({x + \pi}\right)
+      | r = \frac {\sin \left({x + \pi}\right)} {\cos \left({x + \pi}\right)}
+      | c = {{Defof|Real Tangent Function}}
+}}
+{{eqn | r = \frac {-\sin x} {-\cos x}
+      | c = [[Sine and Cosine are Periodic on Reals]]
+}}
+{{eqn | r = \tan x
+      | c=
+}}
+{{end-eqn}}
+From [[Derivative of Tangent Function]], we have that:
+: $D_x \left({\tan x}\right) = \dfrac 1 {\cos^2 x}$
+provided $\cos x \ne 0$.
+From [[Shape of Cosine Function]], we have that $\cos > 0$ on the [[Definition:Open Real Interval|interval]] $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$.
+From [[Derivative of Monotone Function]], $\tan x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on that interval, and hence can not have a period of ''less'' than $\pi$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Tangent Function}
+Tags: Derivative of Tangent Function, Tangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tan x} = \sec^2 x = \dfrac 1 {\cos^2 x}$
+when $\cos x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\tan x}
+      | r = \lim_{h \mathop \to 0} \frac {\map \tan {x + h} - \tan x} h
+      | c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h} {1 - \tan x \tan h} - \tan x} h
+      | c = [[Tangent of Sum]]
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\frac {\tan x + \tan h - \tan x + \tan^2 x \tan h} {1 - \tan x \tan h} } h
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\tan h + \tan^2 x \tan h} {h \paren {1 - \tan x \tan h} } 
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {1 + \tan^2 x} {1 - \tan x \tan h} \cdot \lim_{h \mathop \to 0} \frac {\tan h} h
+      | c = [[Product Rule for Limits of Functions]]
+}}
+{{eqn | r = \frac {1 + \tan^2 x} {1 - \tan x \tan 0} \cdot 1
+      | c = [[Limit of Tan X over X]]
+}}
+{{eqn | r = 1 + \tan^2 x
+      | c = [[Tangent of Zero]]
+}}
+{{eqn | r = \sec^2 x
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = \frac 1 {\cos^2 x}
+      | c = [[Secant is Reciprocal of Cosine]] ($\cos x \ne 0$)
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the definition of the [[Definition:Tangent Function|tangent]] function:
+:$\tan x = \dfrac {\sin x} {\cos x}$
+From [[Derivative of Sine Function]]:
+:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
+From [[Derivative of Cosine Function]]:
+:$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\tan x}
+      | r = \frac {\cos x \cos x - \sin x \paren {-\sin x} } {\cos^2 x}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\cos^2 x + \sin^2 x} {\cos^2 x}
+      | c = 
+}}
+{{eqn | r = \frac 1 {\cos^2 x}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+This is valid only when $\cos x \ne 0$.
+The result follows from the [[Secant is Reciprocal of Cosine]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling's Formula}
+Tags: Asymptotics, Special Functions, Stirling's Formula, Factorials
+
+\begin{theorem}
+The [[Definition:Factorial|factorial]] function can be approximated by the formula:
+:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
+where $\sim$ denotes [[Definition:Asymptotically Equal Sequences|asymptotically equal]].
+\end{theorem}
+
+\begin{proof}
+Let $a_n = \dfrac {n!} {\sqrt {2 n} \paren {\frac n e}^n}$.
+=== Part 1 ===
+It will be shown that:
+:$\lim_{n \mathop \to \infty} a_n = a$
+for some [[Definition:Constant|constant]] $a$.
+This will imply that:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {n!} {a \sqrt{2 n} \paren {\frac n e}^n} = 1$
+By applying [[Power Series Expansion for Logarithm of 1 + x|Power Series Expansion for $\map \ln {1 + x}$]]:
+{{begin-eqn}}
+{{eqn | l = \map \ln {\frac {n + 1} n}
+      | r = \map \ln {1 + \frac 1 {2 n + 1} } - \map \ln {1 - \frac 1 {2 n + 1} }
+      | c = 
+}}
+{{eqn | r = 2 \sum_{k \mathop = 0}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k + 1}
+      | c = 
+}}
+{{end-eqn}}
+Let $b_n = \ln a_n$.
+{{begin-eqn}}
+{{eqn | l = b_n - b_{n + 1}
+      | r = \paren {n + \frac 1 2} \map \ln {\frac {n + 1} n} - 1
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}
+      | c = 
+}}
+{{eqn | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | l = b_n
+      | o = >
+      | r = b_{n + 1}
+      | c = 
+}}
+{{end-eqn}}
+so the [[Definition:Sequence|sequence]] $\sequence {b_n}$ is [[Definition:Strictly Decreasing Real Sequence|(strictly) decreasing]].
+From $(1)$:
+{{begin-eqn}}
+{{eqn | l = b_n - b_{n + 1}
+      | r = \sum_{k \mathop = 1}^\infty \frac 1 {2 k + 1} \paren {\frac 1 {2 n + 1} }^{2 k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b_n - b_{n + 1}
+      | o = <
+      | r = \sum_{k \mathop = 1}^\infty \paren {\frac 1 {2 n + 1} }^{2 k}
+      | c = as $\dfrac 1 {2 k + 1} < 1$
+}}
+{{eqn | r = \frac 1 {4 n \paren {n + 1} }
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = b_1 - b_n
+      | r = \sum_{m \mathop = 1}^{n - 1} b_m - b_{m + 1}
+      | c = {{Defof|Telescoping Series}}
+}}
+{{eqn | o = <
+      | r = \frac 1 4 \sum_{m \mathop = 1}^{n - 1} \frac 1 {m \paren {m + 1} }
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 4 \sum_{m \mathop = 1}^\infty \frac 1 {m \paren {m + 1} }
+      | c = 
+}}
+{{eqn | r = \frac 1 4
+      | c = [[Sum of Sequence of Products of Consecutive Reciprocals]]
+}}
+{{end-eqn}}
+{{explain|The use of [[Definition:Telescoping Series|Telescoping Series]] is unclear}}
+and so: 
+:$b_n > b_1 - \dfrac 1 4 = \dfrac 3 4 - \dfrac {\ln 2} 2$
+Thus by definition $\sequence {b_n}$ is [[Definition:Bounded Below Real Sequence|bounded below]]. 
+By [[Monotone Convergence Theorem (Real Analysis)|Monotone Convergence Theorem]], it follows that $\sequence {b_n}$ is [[Definition:Convergent Sequence (Analysis)|convergent]].
+Let $b$ denote its [[Definition:Limit of Sequence|limit]]. 
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} a_n = e^{\lim_{n \mathop \to \infty} b_n} = e^b = a$
+as required.
+{{qed|lemma}}
+=== Part 2 ===
+By [[Wallis's Product]], we have:
+:$\displaystyle \prod_{n \mathop = 1}^\infty \frac {2 n} {2 n - 1} \cdot \frac {2 n} {2 n + 1} = \frac \pi 2$
+or equivalently:
+:$(2): \quad \displaystyle \lim_{n \mathop \to \infty} \frac {2^{4 n} \paren {n!}^4} {\paren {\paren {2 n}!}^2 \paren {2 n + 1} } = \frac \pi 2$
+In [[Stirling's Formula/Proof 1#Part 1|Part 1]] it was proved that:
+:$n! \sim a \sqrt {2 n} \paren {\dfrac n e}^n$
+Substituting for $n!$ in $(2)$ yields:
+{{begin-eqn}}
+{{eqn | l = \frac \pi 2
+      | r = \lim_{n \mathop \to \infty} \frac {2^{4 n} a^4 \paren {4 n^2} n^{4 n} e^{-4 n} } {a^2 \paren {4 n} 2^{4 n} n^{4 n} e^{-4 n} \paren {2 n + 1} }
+      | c = 
+}}
+{{eqn | r = a^2 \lim_{n \mathop \to \infty} \frac n {2 n + 1}
+      | c = as most of it cancels out
+}}
+{{eqn | r = \frac {a^2} 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = \sqrt \pi
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider the [[Definition:Sequence|sequence]] $\sequence {d_n}$ defined as:
+:$d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
+From [[Stirling's Formula/Proof 2/Lemma 2|Lemma 2]] it is seen that $\sequence {d_n}$ is a [[Definition:Decreasing Real Sequence|decreasing sequence]].
+From [[Stirling's Formula/Proof 2/Lemma 3|Lemma 3]] it is seen that the [[Definition:Sequence|sequence]]:
+:$\sequence {d_n - \dfrac 1 {12 n} }$
+is [[Definition:Increasing Real Sequence|increasing]].
+In particular:
+:$\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 = \dfrac 1 {12}$
+and so $\sequence {d_n}$ is [[Definition:Bounded Below Real Sequence|bounded below]].
+From the [[Monotone Convergence Theorem (Real Analysis)]], it follows that $\sequence {d_n}$ is [[Definition:Convergent Sequence (Analysis)|convergent]].
+Let $d_n \to d$ as $n \to \infty$.
+From [[Exponential Function is Continuous]], we have:
+:$\exp d_n \to \exp d$ as $n \to \infty$
+Let $C = \exp d$.
+Then:
+:$\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$
+It remains to be shown that $C = \sqrt {2 \pi}$.
+Let $I_n$ be defined as:
+:$\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
+Then from [[Stirling's Formula/Proof 2/Lemma 4|Lemma 4]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
+So:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \dfrac {n!} {n^n \sqrt n e^{-n} }
+      | r = \sqrt {2 \pi}
+      | c = [[Stirling's Formula/Proof 2/Lemma 5|Lemma 5]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \dfrac {n!} {\sqrt {2 \pi n} \paren {\frac n e}^n}
+      | r = 1
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Cotangent Function}
+Tags: Derivatives of Trigonometric Functions, Cotangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cot x} = -\csc^2 x = \dfrac {-1} {\sin^2 x}$
+where $\sin x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Cotangent|cotangent]] function:
+:$\cot x = \dfrac {\cos x} {\sin x}$
+From [[Derivative of Sine Function]]:
+:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
+From [[Derivative of Cosine Function]]:
+:$\map {\dfrac \d {\d x} } {\cos x}= -\sin x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\cot x}
+      | r = \frac {\sin x \paren {-\sin x} - \cos x \cos x} {\sin^2 x}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {-\paren {\sin^2 x + \cos^2 x} } {\sin^2 x}
+      | c = 
+}}
+{{eqn | r = \frac {-1} {\sin^2 x}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+This is valid only when $\sin x \ne 0$.
+The result follows from the definition of the [[Definition:Real Cosecant Function|real cosecant]] function.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Shape of Sine Function}
+Tags: Sine Function
+
+\begin{theorem}
+The [[Definition:Sine|sine]] function is:
+:$(1): \quad$ [[Definition:Strictly Increasing Real Function|strictly increasing]] on the [[Definition:Closed Real Interval|interval]] $\closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
+:$(2): \quad$ [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on the [[Definition:Closed Real Interval|interval]] $\closedint {\dfrac \pi 2} {\dfrac {3 \pi} 2}$
+:$(3): \quad$ [[Definition:Concave Real Function|concave]] on the [[Definition:Closed Real Interval|interval]] $\closedint 0 \pi$
+:$(4): \quad$ [[Definition:Convex Real Function|convex]] on the [[Definition:Closed Real Interval|interval]] $\closedint \pi {2 \pi}$
+\end{theorem}
+
+\begin{proof}
+From the discussion of [[Sine and Cosine are Periodic on Reals]], we have that:
+: $\sin \paren {x + \dfrac \pi 2} = \cos x$
+The result then follows directly from the [[Shape of Cosine Function]].
+=== [[Shape of Sine Function/Graph|Graph of Sine Function]] ===
+{{:Shape of Sine Function/Graph}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Secant Function}
+Tags: Derivatives of Trigonometric Functions, Secant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sec x} = \sec x \tan x$
+where $\cos x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Real Secant Function|secant function]]:
+:$\sec x = \dfrac 1 {\cos x} = \paren {\cos x}^{-1}$
+From [[Derivative of Cosine Function]]:
+:$\map {\dfrac \d {\d x} } {\cos x} = -\sin x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\sec x}
+      | r = \map {\dfrac \d {\d x} } {\paren {\cos x}^{-1} }
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Laws]]
+}}
+{{eqn | r = \paren {-\sin x} \paren {-\cos^{-2} x}
+      | c = [[Chain Rule for Derivatives]], [[Power Rule for Derivatives|Power Rule]]
+}}
+{{eqn | r = \frac 1 {\cos x} \frac {\sin x} {\cos x}
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Laws]]
+}}
+{{eqn | r = \sec x \tan x
+      | c = Definitions of [[Definition:Real Secant Function|secant]] and [[Definition:Real Tangent Function|tangent]]
+}}
+{{end-eqn}}
+This is valid only when $\cos x \ne 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Cosecant Function}
+Tags: Derivatives of Trigonometric Functions, Cosecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csc x} = -\csc x \cot x$
+where $\sin x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Real Cosecant Function|cosecant]] function:
+:$\csc x = \dfrac 1 {\sin x}$
+From [[Derivative of Sine Function]]:
+:$\map {\dfrac \d {\d x} } {\sin x} = \cos x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\csc x}
+      | r = \cos x \frac {-1} {\sin^2 x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {-1} {\sin x} \frac {\cos x} {\sin x}
+      | c = 
+}}
+{{eqn | r = -\csc x \cot x
+      | c = {{Defof|Real Cosecant Function}} and {{Defof|Real Cotangent Function}}
+}}
+{{end-eqn}}
+This is valid only when $\sin x \ne 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Shape of Tangent Function}
+Tags: Tangent Function
+
+\begin{theorem}
+The nature of the [[Definition:Real Tangent Function|tangent function]] on the set of [[Definition:Real Number|real numbers]] $\R$ is as follows:
+: $\tan x$ is [[Definition:Continuous on Interval|continuous]] and [[Definition:Strictly Increasing Real Function|strictly increasing]] on the [[Definition:Open Real Interval|interval]] $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$
+: $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
+: $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$
+: $\tan x$ is not defined on $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi$, at which points it is [[Definition:Discontinuous|discontinuous]]
+: $\forall n \in \Z: \tan \left({n \pi}\right) = 0$.
+\end{theorem}
+
+\begin{proof}
+* $\tan x$ is [[Definition:Continuous on Interval|continuous]] and [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$:
+Continuity follows from the [[Quotient Rule for Continuous Functions]]:
+:$(1): \quad$ Both $\sin x$ and $\cos x$ are [[Definition:Continuous on Interval|continuous]] on $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$ from [[Real Sine Function is Continuous]] and [[Cosine Function is Continuous]]
+:$(2): \quad$ $\cos x > 0$ on this interval.
+The fact of $\tan x$ being [[Definition:Strictly Increasing Real Function|strictly increasing]] on this interval has been demonstrated in the discussion on [[Tangent Function is Periodic on Reals]].
+* $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$:
+From [[Sine and Cosine are Periodic on Reals]], we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.
+We have that:
+: $(1): \quad \cos x \to 0$ as $x \to \dfrac \pi 2 ^-$
+: $(2): \quad \sin x \to 1$ as $x \to \dfrac \pi 2 ^-$
+[[Infinite Limit Theorem|Thus it follows that]]:
+* $\tan x = \dfrac {\sin x} {\cos x} \to + \infty$ as $x \to \dfrac \pi 2 ^-$.
+* $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$:
+From [[Sine and Cosine are Periodic on Reals]], we have that $\sin x < 0$ and $\cos x > 0$ on $\left({-\dfrac \pi 2 \,.\,.\, 0}\right)$.
+We have that:
+: $(1): \quad \cos x \to 0$ as $x \to -\dfrac \pi 2 ^+$
+: $(2): \quad \sin x \to -1$ as $x \to -\dfrac \pi 2 ^+$
+Thus it follows that $\tan x = \dfrac {\sin x} {\cos x} \to - \infty$ as $x \to -\dfrac \pi 2 ^+$.
+* $\tan x$ is not defined and [[Definition:Discontinuous|discontinuous]] at $x = \left({n + \dfrac 1 2}\right) \pi$:
+From the discussion of [[Sine and Cosine are Periodic on Reals]], it was established that $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi \implies \cos x = 0$.
+As division by zero is not defined, it follows that at these points $\tan x$ is not defined either.
+Now, from the above, we have:
+: $(1): \quad \tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
+: $(2): \quad \tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$
+As $\tan \left({x + \pi}\right) = \tan x$ from [[Tangent Function is Periodic on Reals]], it follows that $\tan x \to - \infty$ as $x \to \dfrac \pi 2 ^+$.
+Hence the [[Definition:Limit from Left|left hand limit]] and [[Definition:Limit from Right|right hand limit]] at $x = \dfrac \pi 2$ are not the same.
+From the [[Tangent Function is Periodic on Reals|periodic nature]] of $\tan x$, it follows that the same applies $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi$.
+The fact of its discontinuity at these points follows from the definition of [[Definition:Discontinuous|discontinuity]].
+* $\tan \left({n \pi}\right) = 0$:
+Follows directly from [[Sine and Cosine are Periodic on Reals]]: $\forall n \in \Z: \sin \left({n \pi}\right) = 0$.
+=== [[Shape of Tangent Function/Graph|Graph of Tangent Function]] ===
+{{:Shape of Tangent Function/Graph}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arcsine Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arcsine Function
+
+\begin{theorem}
+:$\dfrac {\map \d {\arcsin x} } {\d x} = \dfrac 1 {\sqrt {1 - x^2} }$
+\end{theorem}
+
+\begin{proof}
+Let $y = \arcsin x$ where $-1 < x < 1$.
+Then:
+:$x = \sin y$
+Then from [[Derivative of Sine Function]]:
+:$\dfrac {\d x} {\d y} = \cos y$
+Hence from [[Derivative of Inverse Function]]:
+:$\dfrac {\d y} {\d x} = \dfrac 1 {\cos y}$
+From [[Sum of Squares of Sine and Cosine]], we have:
+:$\cos^2 y + \sin^2 y = 1 \implies \cos y = \pm \sqrt {1 - \sin^2 y}$
+Now $\cos y \ge 0$ on the [[Definition:Image of Mapping|image]] of $\arcsin x$, that is:
+:$y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$
+Thus it follows that we need to take the positive root of $\sqrt {1 - \sin^2 y}$.
+So:
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = \frac 1 {\sqrt {1 - \sin^2 y} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {\sqrt {1 - x^2} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arccosine Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arccosine Function
+
+\begin{theorem}
+:$\map {D_x} {\arccos x} = \dfrac {-1} {\sqrt {1 - x^2}}$
+\end{theorem}
+
+\begin{proof}
+Let $y = \arccos x$ where $-1 < x < 1$.
+Then:
+:$x = \cos y$
+Then from [[Derivative of Cosine Function]]:
+:$\dfrac {\d x} {\d y} = -\sin y$
+Hence from [[Derivative of Inverse Function]]:
+:$\dfrac {\d y} {\d x} = \dfrac {-1} {\sin y}$
+From [[Sum of Squares of Sine and Cosine]], we have:
+:$\cos^2 y + \sin^2 y = 1 \implies \sin y = \pm \sqrt {1 - \cos^2 y}$
+Now $\sin y \ge 0$ on the range of $\arccos x$, that is, for $y \in \closedint 0 \pi$.
+Thus it follows that we need to take the positive root of $\sqrt {1 - \cos^2 y}$.
+So:
+:$\dfrac {\d y} {\d x} = \dfrac {-1} {\sqrt {1 - \cos^2 y} }$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Arcsine and Arccosine}
+Tags: Arcsine Function, Arccosine Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $-1 \le x \le 1$.
+Then:
+: $\arcsin x + \arccos x = \dfrac \pi 2$
+where $\arcsin$ and $\arccos$ denote [[Definition:Arcsine|arcsine]] and [[Definition:Arccosine|arccosine]] respectively.
+\end{theorem}
+
+\begin{proof}
+Let $y \in \R$ such that:
+: $\exists x \in \left[{-1 \,.\,.\, 1}\right]: x = \cos \left({y + \dfrac \pi 2}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \cos \left({y + \frac \pi 2}\right)
+      | c = 
+}}
+{{eqn | r = -\sin y
+      | c = [[Cosine of Angle plus Right Angle]]
+}}
+{{eqn | r = \sin \left({-y}\right)
+      | c = [[Sine Function is Odd]]
+}}
+{{end-eqn}}
+Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
+Then we can write $-y = \arcsin x$.
+But then $\cos \left({y + \dfrac \pi 2}\right) = x$.
+Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that $0 \le y + \dfrac \pi 2 \le \pi$.
+Hence $y + \dfrac \pi 2 = \arccos x$.
+That is, $\dfrac \pi 2 = \arccos x + \arcsin x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arctangent Function}
+Tags: Derivative of Arctangent Function, Arctangent Function, Derivatives of Inverse Trigonometric Functions
+
+\begin{theorem}
+:$\dfrac {\map \d {\arctan x} } {\d x} = \dfrac 1 {1 + x^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \frac {\map \d {\map \arctan x} } {\d x} 
+	| r = \lim_{h \mathop \to 0} \frac {\map \arctan {x + h} - \map \arctan x} h 
+	| c = {{Defof|Derivative of Real Function at Point}}
+}}
+{{eqn	| r = \lim_{h \mathop \to 0} \frac {\arctan {x + h} + \map \arctan {-x} } h	
+	| c = [[Arctangent Function is Odd]]
+}}
+{{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac {x + h - x} {1 + x \paren {x + h} } }
+	| c = [[Sum of Arctangents]]
+}}
+{{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \map \arctan {\frac h {1 + x^2 + h x} }
+}}
+{{eqn	| r = \lim_{h \mathop \to 0} \frac 1 h \paren {\frac h {1 + x^2 + h x} - \frac 1 3 \paren {\frac h {1 + x^2 + h x} }^3 + \frac 1 5 \paren {\frac h {1 + x^2 + h x} }^5 + \map \OO {h^7} }
+	| c = {{Defof|Real Arctangent}}
+}}
+{{eqn	| r = \lim_{h \mathop \to 0} \paren {\frac 1 {1 + x^2 + h x} - \frac {h^2} {3 \paren {1 + x^2 + h x}^3} + \frac {h^4} {5 \paren {1 + x^2 + h x}^5} + \map \OO {h^6} }
+}}
+{{eqn	| r = \frac 1 {1 + x^2 + 0 x} - \frac {0^2} {3 \paren {1 + x^2 + 0 x}^3} + \frac {0^4} {5 \paren {1 + x^2 + 0 x}^5}
+}}
+{{eqn	| r = \frac 1 {1 + x^2} 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arctan x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \tan y
+      | c = {{Defof|Real Arctangent}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = \sec^2 y
+      | c = [[Derivative of Tangent Function]]
+}}
+{{eqn | r = 1 + \tan^2 y
+      | c = [[Difference of Squares of Secant and Tangent]]
+}}
+{{eqn | r = 1 + x^2
+      | c = Definition of $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac 1 {1 + x^2}
+      | c = [[Derivative of Inverse Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers form Field}
+Tags: Complex Numbers, Examples of Fields
+
+\begin{theorem}
+Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\C, +, \times}$, where:
+:$\C$ is the set of all [[Definition:Complex Number|complex numbers]]
+:$+$ is the operation of [[Definition:Complex Addition|complex addition]]
+:$\times$ is the operation of [[Definition:Complex Multiplication|complex multiplication]]
+Then $\struct {\C, +, \times}$ forms a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+From [[Complex Numbers under Addition form Abelian Group]], we have that $\struct {\C, +}$ forms an [[Definition:Abelian Group|abelian group]].
+From [[Non-Zero Complex Numbers under Multiplication form Abelian Group]], we have that $\struct {\C_{\ne 0}, \times}$ forms an [[Definition:Abelian Group|abelian group]].
+Finally, we have that [[Complex Multiplication Distributes over Addition]].
+Thus all the criteria are fulfilled, and $\struct {\C, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Multiplication is Commutative}
+Tags: Complex Multiplication
+
+\begin{theorem}
+The operation of [[Definition:Complex Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Commutative Operation|commutative]]:
+:$\forall z_1, z_2 \in \C: z_1 z_2 = z_2 z_1$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Number/Definition 2|complex numbers]], we define the following:
+{{begin-eqn}}
+{{eqn | l = z
+      | o = :=
+      | r = \tuple {x_1, y_1}
+}}
+{{eqn | l = w
+      | o = :=
+      | r = \tuple {x_2, y_2}
+}}
+{{end-eqn}}
+where $x_1, x_2, y_1, y_2 \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z_1 z_2
+      | r = \tuple {x_1, y_1} \tuple {x_2, y_2}
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {x_2 x_1 - y_2 y_1, x_1 y_2 + x_2 y_1}
+      | c = [[Real Multiplication is Commutative]]
+}}
+{{eqn | r = \tuple {x_2 x_1 - y_2 y_1, x_2 y_1 + x_1 y_2}
+      | c = [[Real Addition is Commutative]]
+}}
+{{eqn | r = \tuple {x_2, y_2} \tuple {x_1, y_1}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = z_2 z_1
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Multiplication Distributes over Addition}
+Tags: Complex Multiplication, Complex Addition, Distributive Operations
+
+\begin{theorem}
+The operation of [[Definition:Complex Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Complex Number|complex numbers]] $\C$ is [[Definition:Distributive Operation|distributive]] over the operation of [[Definition:Complex Addition|addition]].
+:$\forall z_1, z_2, z_3 \in \C:$
+::$z_1 \paren {z_2 + z_3} = z_1 z_2 + z_1 z_3$
+::$\paren {z_2 + z_3} z_1 = z_2 z_1 + z_3 z_1$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Complex Number/Definition 2|complex numbers]], we define the following:
+{{begin-eqn}}
+{{eqn | l = z_1
+      | o = :=
+      | r = \tuple {x_1, y_1}
+}}
+{{eqn | l = z_2
+      | o = :=
+      | r = \tuple {x_2, y_2}
+}}
+{{eqn | l = z_3
+      | o = :=
+      | r = \tuple {x_3, y_3}
+}}
+{{end-eqn}}
+where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = z_1 \paren {z_2 + z_3}
+      | r = \tuple {x_1, y_1} \paren {\tuple {x_2, y_2} + \tuple {x_3, y_3} }
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{eqn | r = \tuple {x_1, y_1} \tuple {x_2 + x_3, y_2 + y_3}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \tuple {x_1 \paren {x_2 + x_3} - y_1 \paren {y_2 + y_3}, x_1 \paren {y_2 + y_3} + y_1 \paren {x_2 + x_3} }
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {x_1 x_2 + x_1 x_3 - y_1 y_2 - y_1 y_3, x_1 y_2 + x_1 y_3 + y_1 x_2 + y_1 x_3}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \tuple {\paren {x_1 x_2 - y_1 y_2}\ + \paren {x_1 x_3 - y_1 y_3}, \paren {x_1 y_2 + y_1 x_2} + \paren {x_1 y_3 + y_1 x_3} }
+      | c = [[Real Addition is Commutative]]
+}}
+{{eqn | r = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + y_1 x_2} + \tuple {x_1 x_3 - y_1 y_3, x_1 y_3 + y_1 x_3}
+      | c = {{Defof|Complex Number/Definition 2/Addition|Complex Addition}}
+}}
+{{eqn | r = \tuple {x_1, y_1} \tuple {x_2, y_2} + \tuple {x_1, y_1} \tuple {x_3, y_3}
+      | c = {{Defof|Complex Number/Definition 2/Multiplication|Complex Multiplication}}
+}}
+{{eqn | r = z_1 z_2 + z_1 z_3
+      | c = {{Defof|Complex Number|index = 2}}
+}}
+{{end-eqn}}
+The result $\paren {z_2 + z_3} z_1 = z_2 z_1 + z_3 z_1$ follows directly from the above, and the fact that [[Complex Multiplication is Commutative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Modulus of Product of Complex Numbers}
+Tags: Complex Modulus, Complex Multiplication, Complex Modulus of Product of Complex Numbers
+
+\begin{theorem}
+Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\cmod z$ be the [[Definition:Complex Modulus|modulus]] of $z$.
+Then:
+:$\cmod {z_1 z_2} = \cmod {z_1} \cdot \cmod {z_2}$
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
+{{begin-eqn}}
+{{eqn | l = \cmod {z_1 z_2}
+      | r = \sqrt {\paren {x_1 x_2 - y_1 y_2}^2 + \paren {x_1 y_2 + x_2 y_1}^2}
+      | c = {{Defof|Complex Modulus}}, {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = \sqrt {\paren {x_1^2 x_2^2 + y_1^2 y_2^2 - 2 x_1 x_2 y_1 y_2} + \paren {x_1^2 y_2^2 + x_2^2 y_1^2 + 2 x_1 x_2 y_1 y_2} }
+      | c = 
+}}
+{{eqn | r = \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \cmod {z_1} \cdot \cmod {z_2}
+      | r = \sqrt {x_1^2 + y_1^2}  \sqrt {x_2^2 + y_2^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {\paren {x_1^2 + y_1^2} \paren {x_2^2 + y_2^2} }
+      | c = 
+}}
+{{eqn | r = \sqrt {x_1^2 x_2^2 + y_1^2 y_2^2 + x_1^2 y_2^2 + x_2^2 y_1^2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of $z$. 
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod {z_1 z_2}
+      | r = \sqrt {\paren {z_1 z_2} \overline {\paren {z_1 z_2} } }
+      | c = [[Modulus in Terms of Conjugate]]
+}}
+{{eqn | r = \sqrt {z_1 \overline {z_1} z_2 \overline {z_2} }
+      | c = [[Product of Complex Conjugates]], [[Complex Multiplication is Commutative]]
+}}
+{{eqn | r = \sqrt {z_1 \overline {z_1} } \sqrt {z_2 \overline {z_2} }
+      | c = [[Exponent Combination Laws/Power of Product|Power of Product]]
+}}
+{{eqn | r = \cmod {z_1} \cmod {z_2}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+: $z_1 = r_1 \left({\cos \theta_1 + i \sin \theta_1}\right)$
+: $z_2 = r_2 \left({\cos \theta_2 + i \sin \theta_2}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = \left\vert{z_1 z_2}\right\vert
+      | r = \left\vert{r_1 \left({\cos \theta_1 + i \sin \theta_1}\right) r_2 \left({\cos \theta_2 + i \sin \theta_2}\right)}\right\vert
+      | c = {{Defof|Polar Form of Complex Number}}
+}}
+{{eqn | r = \left\vert{r_1 r_2 \left({\cos \left({\theta_1 + \theta_2}\right) + i \sin \left({\theta_1 + \theta_2}\right)}\right)}\right\vert
+      | c = [[Product of Complex Numbers in Polar Form]]
+}}
+{{eqn | r = r_1 r_2
+      | c = {{Defof|Polar Form of Complex Number}}
+}}
+{{eqn | r = \left\vert{z_1}\right\vert \left\vert{z_2}\right\vert
+      | c = {{Defof|Polar Form of Complex Number}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Complex Conjugates}
+Tags: Complex Conjugates, Complex Addition
+
+\begin{theorem}
+Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of the [[Definition:Complex Number|complex number]] $z$.
+Then:
+:$\overline {z_1 + z_2} = \overline {z_1} + \overline {z_2}$
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \overline {z_1 + z_2}
+      | r = \overline {\paren {x_1 + x_2} + i \paren {y_1 + y_2} }
+      | c = 
+}}
+{{eqn | r = \paren {x_1 + x_2} - i \paren {y_1 + y_2}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{eqn | r = \paren {x_1 - i y_1} + \paren {x_2 - i y_2}
+      | c = {{Defof|Complex Addition}}
+}}
+{{eqn | r = \overline {z_1} + \overline {z_2}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Complex Conjugates}
+Tags: Complex Conjugates, Complex Multiplication, Product of Complex Conjugates
+
+\begin{theorem}
+Let $z_1, z_2 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of the [[Definition:Complex Number|complex number]] $z$.
+Then:
+:$\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$, where $x_1, y_1, x_2, y_2 \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \overline {z_1 z_2}
+      | r = \overline {\paren {x_1 x_2 - y_1 y_2} + i \paren {x_2 y_1 + x_1 y_2} }
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = \paren {x_1 x_2 - y_1 y_2} - i \paren {x_2 y_1 + x_1 y_2}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{eqn | r = \paren {x_1 x_2 - \paren {-y_1} \paren {-y_2} } + i \paren {x_2 \paren {-y_1} + x_1 \paren {-y_2} }
+      | c = 
+}}
+{{eqn | r = \paren {x_1 - i y_1} \paren {x_2 - i y_2}
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = \overline {z_1} \cdot \overline {z_2}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Complex Number with Conjugate}
+Tags: Complex Conjugates
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\overline z$ be the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+Let $\map \Re z$ be the [[Definition:Real Part|real part]] of $z$.
+Then:
+:$z + \overline z = 2 \, \map \Re z$
+\end{theorem}
+
+\begin{proof}
+Let $z = x + i y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z + \overline z
+      | r = \paren {x + i y} + \paren {x - i y}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{eqn | r = 2 x
+}}
+{{eqn | r = 2 \, \map \Re z
+      | c = {{Defof|Real Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Complex Number with Conjugate}
+Tags: Complex Conjugates
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\overline z$ be the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+Let $\map \Im z$ be the [[Definition:Imaginary Part|imaginary part]] of $z$.
+Then
+:$z - \overline z = 2 i \, \map \Im z$
+\end{theorem}
+
+\begin{proof}
+Let $z = x + i y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z - \overline z
+      | r = \paren {x + i y} - \paren {x - i y}
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{eqn | r = x + i y - x + i y
+}}
+{{eqn | r = 2 i y
+}}
+{{eqn | r = 2 i \, \map \Im z
+      | c = {{Defof|Imaginary Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Number equals Conjugate iff Wholly Real}
+Tags: Complex Conjugates
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Let $\overline z$ be the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+Then $z = \overline z$ {{iff}} $z$ is [[Definition:Wholly Real|wholly real]].
+\end{theorem}
+
+\begin{proof}
+Let $z = x + i y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \overline z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + i y
+      | r = x - i y
+      | c = {{Defof|Complex Conjugate}}
+}}
+{{eqn | ll= \leadsto
+      | l = +y
+      | r = -y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence by definition, $z$ is [[Definition:Wholly Real|wholly real]].
+{{qed|lemma}}
+Now suppose $z$ is [[Definition:Wholly Real|wholly real]].
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x + 0 i
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{eqn | r = x - 0 i
+      | c = 
+}}
+{{eqn | r = \overline z
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers cannot be Ordered Compatibly with Ring Structure}
+Tags: Complex Numbers, Complex Numbers cannot be Ordered Compatibly with Ring Structure
+
+\begin{theorem}
+Let $\struct {\C, +, \times}$ be the [[Definition:Field of Complex Numbers|field of complex numbers]].
+There exists no [[Definition:Total Ordering|total ordering]] on $\struct {\C, +, \times}$ which is [[Definition:Ordering Compatible with Ring Structure|compatible with the structure]] of $\struct {\C, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exists a [[Definition:Endorelation|relation]] $\preceq$ on $\C$ which is [[Definition:Ordering Compatible with Ring Structure|ordering compatible with the ring structure of $\C$]].
+That is:
+:$(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both
+:$(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$
+By [[Totally Ordered Ring Zero Precedes Element or its Inverse]], $(1)$ can be replaced with:
+:$(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both.
+As $i \ne 0$, it follows that:
+:$0 \prec i$ or $0 \prec -i$
+Suppose $0 \prec i$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \prec
+      | r = i \times i
+      | c = from $(2)$
+}}
+{{eqn | r = -1
+      | c = {{Defof|Complex Number|index = 1}}
+}}
+{{end-eqn}}
+Otherwise, suppose $0 \prec \paren {-i}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \prec
+      | r = \paren {-i} \times \paren {-i}
+      | c = from $(2)$
+}}
+{{eqn | r = -1
+      | c = {{Defof|Complex Number|index = 1}}
+}}
+{{end-eqn}}
+Thus by [[Proof by Cases]]:
+:$0 \prec -1$
+Thus it follows that:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \prec
+      | r = \paren {-1} \times \paren {-1}
+      | c = from $(2)$
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Thus both:
+:$0 \prec -1$
+and:
+:$0 \prec 1$
+This [[Definition:Contradiction|contradicts]] hypothesis $(1')$:
+:$(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both
+Hence, by [[Proof by Contradiction]], there can be no such ordering.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{AimForCont}} such a [[Definition:Total Ordering|total ordering]] $\preceq$ exists.
+By the definition of a [[Definition:Total Ordering/Definition 1|total ordering]], $\preceq$ is [[Definition:Connected Relation|connected]].
+That is:
+:$0 \preceq i \lor i \preceq 0$
+Using [[Proof by Cases]], we will prove that:
+:$0 \preceq -1$
+;Case 1
+Assume that $0 \preceq i$.
+By definition of an [[Definition:Ordering Compatible with Ring Structure|ordering compatible with the ring structure of $\C$]]:
+:$\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$
+Substituting $x = i$ and $y = i$ gives:
+:$0 \preceq i \times i$
+Simplifying:
+:$0 \preceq -1$
+which is the result required.
+;Case 2
+Assume that $i \preceq 0$.
+By definition of [[Definition:Relation Compatible with Operation|compatibility with addition]]:
+:$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$
+Substituting $x = i$, $y = 0$, $z = -i$ gives:
+:$i + \paren {-i} \preceq 0 + \paren {-i}$
+Simplifying:
+:$0 \preceq -i$
+By definition of an [[Definition:Ordering Compatible with Ring Structure|ordering compatible with the ring structure of $\C$]]:
+:$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$
+Substituting $x = -i$ and $y = -i$ gives:
+:$0 \preceq \paren {-i} \times \paren {-i}$
+Simplifying:
+:$0 \preceq -1$
+This has been demonstrated to follow from both cases, and so by [[Proof by Cases]]:
+:$0 \preceq -1$
+{{qed|lemma}}
+By definition of an [[Definition:Ordering Compatible with Ring Structure|ordering compatible with the ring structure of $\C$]]:
+:$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$
+Substituting $x = -1$ and $y = -1$:
+:$0 \preceq \paren {-1} \times \paren {-1}$
+Simplifying:
+:$0 \preceq 1$
+By definition of [[Definition:Relation Compatible with Operation|compatibility with addition]]:
+:$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$
+Substituting $x = 0$, $y = 1$, $z = -1$ gives:
+:$0 + \paren {-1} \preceq 1 + \paren {-1}$
+Simplifying:
+:$-1 \preceq 0$
+From the definition of [[Definition:Ordering/Definition 1|ordering]]:
+:$\forall a, b \in \C: \paren {a \preceq b \land b \preceq a} \implies a = b$
+Substituting $a = -1$ and $b = 0$ gives:
+:$-1 = 0$
+which is a [[Definition:Contradiction|contradiction]].
+Hence, from [[Proof by Contradiction]], there can be no such ordering.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Complex Numbers form Integral Domain]], $\struct {\C, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+{{AimForCont}} that $\struct {\C, +, \times}$ can be [[Definition:Ordered Integral Domain|ordered]].
+Thus, by definition, it possesses a [[Definition:Strict Positivity Property|(strict) positivity property]] $P$.
+Then from [[Strict Positivity Property induces Total Ordering]], let $\le$ be the [[Definition:Total Ordering Induced by Strict Positivity Property|total ordering induced by $P$]].
+From [[Unity of Ordered Integral Domain is Strictly Positive]]:
+:$1$ is [[Definition:Strictly Positive|strictly positive]].
+Thus by [[Definition:Strict Positivity Property|strict positivity, axiom $3$]]:
+:$-1$ is not [[Definition:Strictly Positive|strictly positive]].
+Consider the [[Definition:Element|element]] $i \in \C$.
+By definition of [[Definition:Strict Positivity Property|strict positivity, axiom $3$]], either:
+:$i$ is [[Definition:Strictly Positive|strictly positive]]
+or:
+:$-i$ is [[Definition:Strictly Positive|strictly positive]].
+Suppose $i$ is [[Definition:Strictly Positive|strictly positive]].
+Then by [[Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive]]:
+:$i^2 = -1$ is [[Definition:Strictly Positive|strictly positive]].
+Similarly, suppose $-i$ is [[Definition:Strictly Positive|strictly positive]].
+Then by [[Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive]]:
+:$\paren {-i}^2 = -1$ is [[Definition:Strictly Positive|strictly positive]].
+In both cases we have that $-1$ is [[Definition:Strictly Positive|strictly positive]].
+But it has already been established that $-1$ is not [[Definition:Strictly Positive|strictly positive]].
+Hence, by [[Proof by Contradiction]], there can be no such [[Definition:Ordered Integral Domain|ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Plane is Metric Space}
+Tags: Complex Analysis, Examples of Metric Spaces
+
+\begin{theorem}
+Let $\C$ be the [[Definition:Set|set]] of all [[Definition:Complex Number|complex numbers]].
+Let $d: \C \times \C \to \R$ be the [[Definition:Mapping|function]] defined as:
+:$\map d {z_1, z_2} = \size {z_1 - z_2}$
+where $\size z$ is the [[Definition:Complex Modulus|modulus]] of $z$.
+Then $d$ is a [[Definition:Metric|metric]] on $\C$ and so $\struct {\C, d}$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
+From the definition of [[Definition:Complex Modulus|modulus]]:
+:$\size {z_1 - z_2} = \sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
+This is the [[Definition:Euclidean Metric on Real Number Plane|euclidean metric]] on the [[Definition:Real Number Plane|real number plane]]
+This is shown in [[Euclidean Metric on Real Vector Space is Metric]] to be a [[Definition:Metric|metric]].
+Thus the [[Definition:Complex Plane|complex plane]] is a [[Definition:Dimension (Geometry)|2-dimensional]] [[Definition:Euclidean Space|Euclidean space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Line is Metric Space}
+Tags: Real Numbers, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Real Number Line|real number line]].
+Let $d: \R \times \R \to \R$ be defined as:
+:$\map d {x_1, x_2} = \size {x_1 - x_2}$
+where $\size x$ is the [[Definition:Absolute Value|absolute value]] of $x$.
+Then $d$ is a [[Definition:Metric|metric]] on $\R$ and so $\struct {\R, d}$ is a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $\text M 1$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, x}
+      | r = \size {x - x}
+      | c = Definition of $d$
+}}
+{{eqn | r = 0
+      | c = {{Defof|Absolute Value}}
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 1$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 2$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, y} + \map d {y, z}
+      | r = \size {x - y} + \size {y - z}
+      | c = Definition of $d$
+}}
+{{eqn | o = \ge
+      | r = \size {\paren {x - y} + \paren {y - z} }
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \size {x - z}
+      | c = 
+}}
+{{eqn | r = \map d {x, z}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 2$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 3$ ===
+{{begin-eqn}}
+{{eqn | l = \map d {x, y}
+      | r = \size {x - y}
+      | c = Definition of $d$
+}}
+{{eqn | r = \size {y - x}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | r = \map d {y, x}
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 3$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $\text M 4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x - y}
+      | o = >
+      | r = 0
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {x, y}
+      | o = >
+      | r = 0
+      | c = Definition of $d$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $\text M 4$]] holds for $d$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Product Metric on Real Vector Space is Metric}
+Tags: P-Product Metrics
+
+\begin{theorem}
+Let $\R^n$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Real Vector Space|real vector space]].
+Let $p \in \R_{\ge 1}$.
+Let $d_p: \R^n \times \R^n \to \R$ be the [[Definition:P-Product Metric/Real Vector Space|$p$-product metric]] on $\R^n$:
+: $\displaystyle d_p \left({x, y}\right) :=  \left({\sum_{i \mathop = 1}^n \left \vert {x_i - y_i} \right \vert^p}\right)^{\frac 1 p}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \R^n$.
+Then $d_p$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, x}\right)
+      | r = \left({\sum_{i \mathop = 1}^n \left \vert {x_i - x_i} \right \vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+It is required to be shown:
+:$d \left({x, y}\right) + d \left({y, z}\right) \ge d \left({x, z}\right)$
+for all $x, y, z \in \R^n$.
+Let:
+:$(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$
+:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
+:$(3): \quad \left\vert{x_i - y_i}\right\vert = r_i$
+:$(4): \quad \left\vert{y_i - z_i}\right\vert = s_i$.
+Thus we need to show that:
+:$\displaystyle \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p} \ge \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}$
+We have:
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)
+      | r = \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p} + \left({\sum \left\vert{y_i - z_i}\right\vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}
+      | c = [[Minkowski's Inequality for Sums]]
+}}
+{{eqn | o = \ge
+      | r = \left({\sum \left({\left\vert{x_i - y_i}\right\vert + \left\vert{y_i - z_i}\right\vert}\right)^p}\right)^{\frac 1 p}
+      | c = Definition of $r_i$ and $s_i$
+}}
+{{eqn | r = \left({\sum \left\vert{x_i - z_i}\right\vert^p}\right)^{\frac 1 p}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = d_p \left({x, z}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right)
+      | r = \left({\sum \left\vert{x_i - y_i}\right\vert^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum \left\vert{y_i - x_i}\right\vert^p}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | r = d_p \left({y, x}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \exists k \in \left[{1 \,.\,.\, n}\right]: x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x_k - y_k
+      | o = >
+      | r = 0
+      | c = as $d_k$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{eqn | ll= \implies
+      | l = \left({\sum \left\vert{x_k - y_k}\right\vert^p}\right)^{\frac 1 p}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_p \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_p$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This is an instance of [[P-Product Metric is Metric|$p$-Product Metric is Metric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Taxicab Metric is Metric}
+Tags: Taxicab Metric, Taxicab Metric is Metric
+
+\begin{theorem}
+The [[Definition:Taxicab Metric|taxicab metric]] is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[P-Product Metric is Metric]], where in this case $p = 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero and One are the only Consecutive Perfect Squares}
+Tags: Square Numbers, Zero and One are the only Consecutive Perfect Squares
+
+\begin{theorem}
+If $n$ is a [[Definition:Square Number|perfect square]] other than $0$, then $n+1$ is not a perfect square.
+\end{theorem}<|endoftext|>
+\section{ProofWiki:Jokes}
+Tags: Jokes
+
+\begin{theorem}
+If I cannot open these cans of food, I will die.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Contradiction|Suppose not]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Union of Bounded Subsets}
+Tags: Boundedness
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Then the [[Definition:Set Union|union]] of any [[Definition:Finite Set|finite]] number of [[Definition:Bounded Metric Space|bounded subsets]] of $M$ is itself [[Definition:Bounded Metric Space|bounded]].
+\end{theorem}
+
+\begin{proof}
+It is sufficient to prove this for two subsets, as the general result follows by [[Principle of Mathematical Induction|induction]].
+Suppose $S_1$ and $S_2$ are [[Definition:Bounded Metric Space|bounded subsets]] of $M = \left({A, d}\right)$.
+Let $a_1, a_2 \in A$.
+Let $K_1, K_2 \in \R$ such that:
+: $(1): \quad \forall x \in S_1: d \left({x, a_1}\right) \le K_1$
+: $(2): \quad \forall x \in S_2: d \left({x, a_2}\right) \le K_2$
+{{WLOG}}, let $a = a_1$ and $K = \max \left\{{K_1, K_2 + d \left({a_1, a_2}\right)}\right\}$.
+Then $\forall x \in S_1 \cup S_2$:
+Either:
+: $x \in S_1$ and so $d \left({x, a}\right) \le K_1 \le K$
+or:
+: $x \in S_2$ and so $d \left({x, a}\right) = d \left({x, a_1}\right) \le d \left({x, a_2}\right) + d \left({a_2, a_1}\right) \le K_2 + d \left({a_2, a_1}\right) \le K$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Product Metric is Metric}
+Tags: P-Product Metrics
+
+\begin{theorem}
+Let $M_{1'} = \left({A_{1'}, d_{1'}}\right), M_{2'} = \left({A_{2'}, d_{2'}}\right), \ldots, M_{n'} = \left({A_{n'}, d_{n'}}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\displaystyle \mathcal A = \prod_{i \mathop = 1}^n A_{i'}$ be the [[Definition:Finite Cartesian Product|cartesian product]] of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
+Let $p \in \R_{\ge 1}$.
+Let $d_p: \mathcal A \times \mathcal A \to \R$ be the [[Definition:P-Product Metric/General Definition|$p$-product metric]] on $\mathcal A$:
+: $\displaystyle d_p \left({x, y}\right) := \left({\sum_{i \mathop = 1}^n \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p}$
+where $x = \left({x_1, x_2, \ldots, x_n}\right), y = \left({y_1, y_2, \ldots, y_n}\right) \in \mathcal A$.
+Then $d_p$ is a [[Definition:Metric|metric]].
+\end{theorem}
+
+\begin{proof}
+=== Proof of $M1$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, x}\right)
+      | r = \left({\sum_{i \mathop = 1}^n \left({d_{i'} \left({x_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum_{i \mathop = 1}^n 0^p}\right)^{\frac 1 p}
+      | c = as $d_{i'}$ fulfils [[Definition:Metric Space Axioms|axiom $M1$]]
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M1$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M2$ ===
+Let:
+:$(1): \quad z = \left({z_1, z_2, \ldots, z_n}\right)$
+:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
+:$(3): \quad d_{i'} \left({x_i, y_i}\right) = r_i$
+:$(4): \quad d_{i'} \left({y_i, z_i}\right) = s_i$.
+Thus we need to show that:
+:$\displaystyle \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p} \ge \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}$
+We have:
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right) + d_p \left({y, z}\right)
+      | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p} + \left({\sum \left({d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum r_i^p}\right)^{\frac 1 p} + \left({\sum s_i^p}\right)^{\frac 1 p}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \left({\sum \left({r_i + s_i}\right)^p}\right)^{\frac 1 p}
+      | c = [[Minkowski's Inequality for Sums]]
+}}
+{{eqn | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right) + d_{i'} \left({y_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = Definition of $r_i$ and $s_i$
+}}
+{{eqn | o = \ge
+      | r = \left({\sum \left({d_{i'} \left({x_i, z_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = as $d_{i'}$ fulfils [[Definition:Metric Space Axioms|axiom $M2$]]
+}}
+{{eqn | r = d_p \left({x, z}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M2$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M3$ ===
+{{begin-eqn}}
+{{eqn | l = d_p \left({x, y}\right)
+      | r = \left({\sum \left({d_{i'} \left({x_i, y_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = Definition of $d_p$
+}}
+{{eqn | r = \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | c = as $d_i$ fulfils [[Definition:Metric Space Axioms|axiom $M3$]]
+}}
+{{eqn | r = d_p \left({y, x}\right)
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M3$]] holds for $d_p$.
+{{qed|lemma}}
+=== Proof of $M4$ ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \ne
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \exists k \in \left[{1 \,.\,.\, n}\right]: x_k
+      | o = \ne
+      | r = y_k
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_k \left({x_k, y_k}\right)
+      | o = >
+      | r = 0
+      | c = as $d_k$ fulfils [[Definition:Metric Space Axioms|axiom $M4$]]
+}}
+{{eqn | ll= \implies
+      | l = \left({\sum \left({d_{i'} \left({y_i, x_i}\right)}\right)^p}\right)^{\frac 1 p}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_p \left({x, y}\right)
+      | o = >
+      | r = 0
+      | c = Definition of $d_p$
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $M4$]] holds for $d_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler Triangle Formula}
+Tags: Triangles
+
+\begin{theorem}
+Let $d$ be the [[Definition:Distance (Linear Measure)|distance]] between the [[Definition:Incenter of Triangle|incenter]] and the [[Definition:Circumcenter of Triangle|circumcenter]] of a [[Definition:Triangle (Geometry)|triangle]].
+Then:
+:$d^2 = R \left({R - 2 \rho}\right)$
+where:
+:$R$ is the [[Definition:Circumradius of Triangle|circumradius]]
+:$\rho$ is the [[Definition:Inradius of Triangle|inradius]].
+\end{theorem}
+
+\begin{proof}
+=== [[Euler Triangle Formula/Lemma|Lemma]] ===
+{{:Euler Triangle Formula/Lemma}}
+{{qed|lemma}}
+:[[File:Incenter Circumcenter Distance.png|400px]]
+Let the [[Definition:Incenter of Triangle|incenter]] of $\triangle ABC$ be $I$.
+Let the [[Definition:Circumcenter of Triangle|circumcenter]] of $\triangle ABC$ be $O$.
+Let $OI$ be [[Definition:Production|produced]] to the [[Definition:Circumcircle of Triangle|circumcircle]] at $G$ and $J$.
+Let $F$ be the point where the [[Definition:Incircle of Triangle|incircle]] of $\triangle ABC$ meets $BC$.
+We are given that:
+:the [[Definition:Distance (Linear Measure)|distance]] between the [[Definition:Incenter of Triangle|incenter]] and the [[Definition:Circumcenter of Triangle|circumcenter]] is $d$
+:the [[Definition:Inradius of Triangle|inradius]] is $\rho$
+:the [[Definition:Circumradius of Triangle|circumradius]] is $R$.
+Thus:
+:$OI = d$
+:$OG = OJ = R$
+Therefore:
+:$IJ = R + d$
+:$GI = R - d$
+By the [[Intersecting Chord Theorem]]:
+: $GI \cdot IJ = IP \cdot CI$
+By the [[Euler Triangle Formula/Lemma|lemma]]:
+:$IP = PB$
+and so:
+:$GI \cdot IJ = PB \cdot CI$
+Now using the Extension of [[Law of Sines]] in $\triangle CPB$:
+:$\dfrac {PB} {\sin \left({\angle PCB}\right)} = 2 R$
+and so:
+:$GI \cdot IJ = 2 R \sin \left({\angle PCB}\right) \cdot CI$
+By [[Axiom:Euclid's Common Notion 4|the $4$th of Euclid's common notions]]:
+:$\angle PCB = \angle ICF$
+and so:
+:$(1): \quad GI \cdot IJ = 2 R \sin \left({\angle ICF}\right) \cdot CI$
+We have that:
+:$IF = \rho$
+and by [[Radius at Right Angle to Tangent]]:
+:$\angle IFC$ is a [[Definition:Right Angle|right angle]].
+By the definition of [[Definition:Sine of Angle|sine]]:
+:$\sin \left({\angle ICF}\right) = \dfrac {\rho} {CI}$
+and so:
+:$\sin \left({\angle ICF}\right) \cdot CI = \rho$
+Substituting in $(1)$:
+{{begin-eqn}}
+{{eqn | l = GI \cdot IJ
+      | r = 2 R \rho
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left({R + d}\right) \left({R - d}\right)
+      | r = 2 R \rho
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = R^2 - d^2
+      | r = 2 R \rho
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | ll= \implies
+      | l = d^2
+      | r = R^2 - 2 R \rho
+      | c = 
+}}
+{{eqn | r = R \left({R - 2 \rho}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+{{namedfor|Leonhard Paul Euler|cat = Euler}}
+[[Category:Triangles]]
+3epb0l67d8pfil90dh4r9ab1iotbqma
+\end{proof}<|endoftext|>
+\section{Open Ball of Point Inside Open Ball}
+Tags: Open Balls
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $B_\epsilon \left({x}\right)$ be an [[Definition:Open Ball|open $\epsilon$-ball]] in $M = \left({A, d}\right)$.
+Let $y \in B_\epsilon \left({x}\right)$.
+Then:
+: $\exists \delta \in \R: B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$
+That is, for every point in an [[Definition:Open Ball|open $\epsilon$-ball]] in a [[Definition:Metric Space|metric space]], there exists an [[Definition:Open Ball|open $\delta$-ball]] of that point entirely contained within that [[Definition:Open Ball|open $\epsilon$-ball]].
+\end{theorem}
+
+\begin{proof}
+Let $\delta = \epsilon - d \left({x, y}\right)$.
+From the definition of [[Definition:Open Ball|open ball]], this is [[Definition:Strictly Positive|strictly positive]], since $y \in B_\epsilon \left({x}\right)$.
+If $z \in B_\delta \left({y}\right)$, then $d \left({y, z}\right) < \delta$.
+So:
+: $d \left({x, z}\right) \le d \left({x, y}\right) + d \left({y, z}\right) < d \left({x, y}\right) + \delta = \epsilon$
+Thus $z \in B_\epsilon \left({x}\right)$.
+So $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$.
+{{qed}}
+This diagram illustrates the proof in $M = \left({\R^2, d_2}\right)$:
+:[[File:NeighborhoodInNeighborhood.png|300px]]
+In $\R^2$, we can draw a disc radius $\delta$ whose center is $y$ and which lies entirely within the larger disc whose center is $x$ with radius $\epsilon$.
+\end{proof}<|endoftext|>
+\section{Open Sets in Metric Space}
+Tags: Metric Spaces, Open Sets
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $\O$ and $A$ are both [[Definition:Open Set (Metric Space)|open]] in $M$.
+\end{theorem}
+
+\begin{proof}
+We have the results:
+: [[Empty Set is Open in Metric Space]]
+: [[Metric Space is Open in Itself]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Continuity on Metric Spaces}
+Tags: Continuous Mappings on Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]] from $A_1$ to $A_2$.
+{{TFAE|def = Continuous on Metric Space|view = continuity|context = Metric Space|contextview = metric spaces}}
+\end{theorem}
+
+\begin{proof}
+=== Definition by Points implies Definition by Open Sets ===
+Suppose that $f$ is [[Definition:Continuous at Point of Metric Space|continuous at every point]] $x \in A_1$.
+Let $U \subseteq M_2$ be [[Definition:Open Set (Metric Space)|open]] in $M_2$.
+Let $x \in f^{-1} \left({U}\right)$.
+Since $U$ is [[Definition:Open Set (Metric Space)|open]] in $M_2$:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right) \subseteq U$
+where $B_\epsilon \left({f \left({x}\right); d_2}\right)$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] of $f \left({x}\right)$ in $M_2$.
+By the definition of [[Definition:Continuous Mapping (Metric Space)/Point/Definition 3|continuity at a point]]:
+: $\exists \delta \in \R_{>0}: f \left({B_\delta \left({x}\right); d_1}\right) \subseteq B_\epsilon \left({f \left({x}\right); d_2}\right)$
+So:
+: $f \left({B_\delta \left({x}\right)}\right) \subseteq U$
+and so:
+: $B_\delta \left({x}\right) \subseteq f^{-1} \left({U}\right)$
+Thus $f^{-1} \left({U}\right)$ is [[Definition:Open Set (Metric Space)|open]] in $M_1$.
+{{qed|lemma}}
+=== Definition by Open Sets implies Definition by Points ===
+Suppose $f$ is defined to be [[Definition:Continuous Mapping (Metric Space)/Space/Definition 2|continuous]] in the sense that:
+:for every $U \subseteq A_2$ which is [[Definition:Open Set (Metric Space)|open in $M_2$]], $f^{-1} \left({U}\right)$ is [[Definition:Open Set (Metric Space)|open in $M_1$]].
+Let $x \in A_1$.
+Then by [[Open Ball of Point Inside Open Ball]]:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right)$ is open in $M_2$
+So by hypothesis, $f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$ is open in $M_1$.
+As $f \left({x}\right) \in B_\epsilon \left({f \left({x}\right); d_2}\right)$, it follows that:
+: $x \in f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$
+So by hypothesis:
+: $\exists \delta \in \R_{>0}: B_\delta \left({x; d_1}\right) \subseteq f^{-1} \left({B_\epsilon \left({f \left({x}\right); d_2}\right)}\right)$
+Then:
+: $f \left({B_\delta \left({x}\right)}\right) \subseteq B_\epsilon \left({f \left({x}\right)}\right)$
+Thus by the [[Definition:Continuous Mapping (Metric Space)/Point/Definition 3|$\epsilon$-Ball definition]], $f$ is [[Definition:Continuous at Point of Metric Space|continuous at $x$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Intersection of Open Sets of Metric Space is Open}
+Tags: Open Sets, Set Intersection
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $U_1, U_2, \ldots, U_n$ be [[Definition:Open Set (Metric Space)|open]] in $M$.
+Then $\ds \bigcap_{i \mathop = 1}^n U_i$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+That is, a [[Definition:Finite Intersection|finite intersection]] of [[Definition:Open Set (Metric Space)|open subsets]] is [[Definition:Open Set (Metric Space)|open]].
+\end{theorem}
+
+\begin{proof}
+Let $\ds x \in \bigcap_{i \mathop = 1}^n U_i$.
+By the definition of [[Definition:Intersection of Family|intersection]], for each $i \in \closedint 1 n$, we have $x \in U_i$.
+Thus, since $U_i$ is [[Definition:Open Set (Metric Space)|open in $M$]]:
+:$\exists \epsilon_i > 0: \map {B_{\epsilon_i} } x \subseteq U_i$
+where $\map {B_{\epsilon_i} } x$ is the [[Definition:Open Ball|open $\epsilon_i$-ball]] of $x$.
+Let $\displaystyle \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.
+Then, by [[Open Ball contains Smaller Open Ball]]:
+:$\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i} } x$
+for all $i \in \closedint 1 n$.
+So:
+:$\ds \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Open Sets of Metric Space is Open}
+Tags: Open Sets, Set Union
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+The [[Definition:Set Union|union]] of a [[Definition:Set|set]] of [[Definition:Open Set (Metric Space)|open sets]] of $M$ is [[Definition:Open Set (Metric Space)|open in $M$]].
+\end{theorem}
+
+\begin{proof}
+Let $I$ be any [[Definition:Indexing Set|indexing set]].
+Let $U_i$ be [[Definition:Open Set (Metric Space)|open in $M$]] for all $i \in I$.
+Let $\ds x \in \bigcup_{i \mathop \in I} U_i$.
+Then by [[Definition:Union of Family|definition of set union]], $x \in U_k$ for some $k \in I$.
+Since $U_k$ is [[Definition:Open Set (Metric Space)|open in $M$]]:
+:$\ds \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k$
+where $\map {B_\epsilon} x$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$ in $M$.
+By [[Set is Subset of Union]]:
+:$\ds U_k \subseteq \bigcup_{i \mathop \in I} U_i$
+Hence by [[Subset Relation is Transitive]]:
+:$\ds \map {B_\epsilon} x \subseteq \bigcup_{i \mathop \in I} U_i$
+and the result follows by definition of [[Definition:Open Set (Metric Space)|open set in metric space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lipschitz Equivalent Metric Spaces are Homeomorphic}
+Tags: Lipschitz Equivalence, Homeomorphic Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1$ and $M_2$ be [[Definition:Lipschitz Equivalent Metric Spaces|Lipschitz equivalent]].
+Then $M_1$ and $M_2$ are [[Definition:Homeomorphic Metric Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+Let $M_1$ and $M_2$ be [[Definition:Lipschitz Equivalent Metric Spaces|Lipschitz equivalent]].
+Then, by definition, $\exists h, k \in \R_{>0}$ such that:
+: $\forall x, y \in A_1: h d_1 \left({x, y}\right) \le d_2 \left({f \left({x}\right), f \left({y}\right)}\right) \le k d_1 \left({x, y}\right)$
+From the definition of [[Definition:Open Ball|open $\epsilon$-ball]]:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = B_{h \epsilon} \left({f \left({x}\right); d_2}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_2 \left({f \left({x}\right), f \left({y}\right)}\right)
+      | o = <
+      | r = h \epsilon
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_1 \left({x, y}\right)
+      | o = \le
+      | r = \frac {d_2 \left({f \left({x}\right), f \left({y}\right)}\right)} h < \epsilon
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = y
+      | o = \in
+      | r = B_\epsilon \left({x; d_1}\right)
+      | c = 
+}}
+{{end-eqn}}
+... and:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = B_{\epsilon / k} \left({x; d_1}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_1 \left({x, y}\right)
+      | o = <
+      | r = \frac \epsilon k
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = d_2 \left({f \left({x}\right), f \left({y}\right)}\right)
+      | o = \le
+      | r = k {d_1 \left({x, y}\right)} < \epsilon
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = y
+      | o = \in
+      | r = B_\epsilon \left({f \left({x}\right); d_2}\right)
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+: $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq B_\epsilon \left({x; d_1}\right)$
+: $B_{\epsilon / k} \left({x; d_1}\right) \subseteq B_\epsilon \left({f \left({x}\right); d_2}\right)$
+{{qed|lemma}}
+Now, suppose $U$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+Let $x \in U$.
+Then:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({f \left({x}\right); d_2}\right) \subseteq U$.
+Hence:
+: $B_{\epsilon / k} \left({x; d_1}\right) \subseteq U$
+Thus $U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+Similarly, suppose $U$ is [[Definition:Open Set (Metric Space)|$d_1$-open]].
+Let $x \in U$.
+Then:
+: $\exists \epsilon \in \R_{>0}: B_\epsilon \left({x; d_1}\right) \subseteq U$
+Hence:
+: $B_{h \epsilon} \left({f \left({x}\right); d_2}\right) \subseteq U$
+Thus $U$ is [[Definition:Open Set (Metric Space)|$d_2$-open]].
+The result follows by definition of [[Definition:Homeomorphic Metric Spaces|homeomorphic metric spaces]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Product Metrics on Real Vector Space are Topologically Equivalent}
+Tags: P-Product Metrics
+
+\begin{theorem}
+For $n \in \N$, let $\R^n$ be an [[Definition:Euclidean Space|Euclidean space]].
+Let $p \in \R_{\ge 1}$.
+Let $d_p$ be the [[Definition:P-Product Metric/Real Vector Space|$p$-product metric]] on $\R^n$.
+Let $d_\infty$ be the [[Definition:Chebyshev Distance/Real Vector Space|Chebyshev distance]] on $\R^n$.
+Then $d_p$ and $d_\infty$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent]].
+\end{theorem}
+
+\begin{proof}
+Let $r, t \in \R_{\ge 1}$.
+{{WLOG}}, assume that $r \le t$.
+For all $x, y \in \R^n$, we are going to show that:
+:$\displaystyle \map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$
+Then we can demonstrate [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalence]] between all of these metrics, from which [[Lipschitz Equivalent Metrics are Topologically Equivalent|topological equivalence follows]].
+Let $d_r$ be the [[Definition:Metric|metric]] defined as:
+:$\displaystyle \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$
+=== [[Relation between P-Product Metric and Chebyshev Distance on Real Vector Space|Inequalities for Chebyshev Distance]] ===
+{{:Relation between P-Product Metric and Chebyshev Distance on Real Vector Space}}
+{{qed|lemma}}
+=== [[P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case|Inequality for General Case]] ===
+{{:P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case}}
+When we combine the inequalities, we have:
+{{improve|In fact, this inequality (for [[Definition:Real Number|real]] $r \ge 1$) can be established quite easily without the humongous calculation.}}
+:$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y} \ge n^{-1} \map {d_r} {x, y}$
+Therefore, $d_r$ and $d_\infty$ are [[Definition:Lipschitz Equivalent Metrics|Lipschitz equivalent]] for all $r \in \R_{\ge 1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Function of Constant Multiple}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Differentiable on Interval|differentiable]] on $\R$.
+Let $c \in \R$ be a constant.
+Then:
+:$\map {D_x} {\map f {c x} } = c \map {D_{c x} } {\map f {c x} }$
+\end{theorem}
+
+\begin{proof}
+First it is shown that $\map {D_x} {c x} = c$:
+{{begin-eqn}}
+{{eqn | l = \map {D_x} {c x}
+      | r = c \map {D_x} x + x \map {D_x} c
+      | c = [[Product Rule]]
+}}
+{{eqn | r = c + x \map {D_x} c
+      | c = [[Derivative of Identity Function]]
+}}
+{{eqn | r = c + 0
+      | c = [[Derivative of Constant]]
+}}
+{{eqn | r = c
+}}
+{{end-eqn}}
+Next:
+{{begin-eqn}}
+{{eqn | l = \map {D_x} {\map f {c x} }
+      | r = \map {D_x} {c x} \map {D_{c x} } {\map f {c x} }
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = c \map {D_{c x} } {\map f {c x} }
+      | c = from above
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Differential Calculus]]
+jt5vlqqn2wg1t6e7w8q2vhlfcae82ro
+\end{proof}<|endoftext|>
+\section{Cotangent Function is Periodic on Reals}
+Tags: Cotangent Function
+
+\begin{theorem}
+The [[Definition:Cotangent|cotangent function]] is [[Definition:Periodic Function|periodic]] on the set of [[Definition:Real Number|real numbers]] $\R$ with [[Definition:Period of Function|period $\pi$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cot \left({x + \pi}\right)
+      | r = \frac {\cos \left({x + \pi}\right)} {\sin \left({x + \pi}\right)}
+      | c = Definition of [[Definition:Cotangent|Cotangent Function]]
+}}
+{{eqn | r = \frac {-\cos x} {-\sin x}
+      | c = [[Cosine of Angle plus Straight Angle]], [[Sine of Angle plus Straight Angle]]
+}}
+{{eqn | r = \cot x
+      | c = 
+}}
+{{end-eqn}}
+Also, from [[Derivative of Cotangent Function]], we have that:
+: $D_x \left({\cot x}\right) = -\dfrac 1 {\sin^2 x}$
+provided $\sin x \ne 0$.
+From [[Shape of Sine Function]], we have that $\sin$ is [[Definition:Strictly Positive Real Function|strictly positive]] on the [[Definition:Open Real Interval|interval]] $\left({0 \,.\,.\, \pi}\right)$.
+From [[Derivative of Monotone Function]], $\cot x$ is [[Definition:Strictly Decreasing|strictly decreasing]] on that interval, and hence can not have a period of ''less'' than $\pi$.
+Hence the result.
+{{qed}}
+[[Category:Cotangent Function]]
+3lwf13v4mp3rfvrjzzlc3jbgwu3lgnj
+\end{proof}<|endoftext|>
+\section{Shape of Cotangent Function}
+Tags: Cotangent Function
+
+\begin{theorem}
+The nature of the [[Definition:Cotangent|cotangent]] function on the set of [[Definition:Real Number|real numbers]] $\R$ is as follows:
+: $\cot x$ is [[Definition:Continuous on Interval|continuous]] and [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on the [[Definition:Open Real Interval|interval]] $\left({0 \,.\,.\, \pi}\right)$
+: $\cot x \to + \infty$ as $x \to 0^+$
+: $\cot x \to - \infty$ as $x \to \pi^-$
+: $\cot x$ is not defined on $\forall n \in \Z: x = n \pi$, at which points it is [[Definition:Discontinuous|discontinuous]]
+: $\forall n \in \Z: \cot \left({n + \dfrac 1 2}\right) \pi = 0$
+\end{theorem}
+
+\begin{proof}
+* $\cot x$ is [[Definition:Continuous on Interval|continuous]] and [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $\left({0 \,.\,.\, \pi}\right)$:
+Continuity follows from the [[Quotient Rule for Continuous Functions]]:
+:$(1): \quad$ Both $\sin x$ and $\cos x$ are [[Definition:Continuous on Interval|continuous]] on $\left({0 \,.\,.\, \pi}\right)$ from [[Real Sine Function is Continuous]] and [[Cosine Function is Continuous]]
+:$(2): \quad \sin x > 0$ on this interval.
+The fact of $\cot x$ being [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on this interval has been demonstrated in the discussion on [[Cotangent Function is Periodic on Reals]].
+* $\cot x \to + \infty$ as $x \to 0^+$:
+From [[Sine and Cosine are Periodic on Reals]], we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.
+We have that:
+:$(1): \quad \cos x \to 1$ as $x \to 0^+$
+:$(2): \quad \sin x \to 0$ as $x \to 0^+$
+Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to + \infty$ as $x \to 0^+$.
+* $\tan x \to - \infty$ as $x \to \pi^-$:
+From [[Sine and Cosine are Periodic on Reals]], we have that $\sin x > 0$ and $\cos x < 0$ on $\left({\dfrac \pi 2 \,.\,.\, \pi}\right)$.
+We have that:
+:$(1): \quad \cos x \to -1$ as $x \to \pi^-$
+:$(2): \quad \sin x \to 0$ as $x \to \pi^-$
+Thus it follows that $\cot x = \dfrac {\cos x} {\sin x} \to - \infty$ as $x \to \pi^-$.
+* $\cot x$ is not defined and [[Definition:Discontinuous|discontinuous]] at $x = n \pi$:
+From the discussion of [[Sine and Cosine are Periodic on Reals]], it was established that $\forall n \in \Z: x = n \pi \implies \sin x = 0$.
+As division by zero is not defined, it follows that at these points $\cot x$ is not defined either.
+Now, from the above, we have:
+:$(1): \quad \cot x \to + \infty$ as $x \to 0^+$
+:$(2): \quad \cot x \to - \infty$ as $x \to \pi^-$
+As $\cot \left({x + \pi}\right) = \cot x$ from [[Cotangent Function is Periodic on Reals]], it follows that $\cot x \to + \infty$ as $x \to \pi^+$.
+Hence the [[Definition:Limit from Left|left hand limit]] and [[Definition:Limit from Right|right hand limit]] at $x = \pi$ are not the same.
+From the [[Cotangent Function is Periodic on Reals|periodic nature]] of $\cot x$, it follows that the same applies $\forall n \in \Z: x = n \pi$.
+The fact of its discontinuity at these points follows from the definition of [[Definition:Discontinuous|discontinuity]].
+* $\cot \left({n + \dfrac 1 2}\right) \pi = 0$:
+Follows directly from [[Sine and Cosine are Periodic on Reals]]: $\forall n \in \Z: \cos \left({n + \dfrac 1 2}\right) \pi = 0$.
+=== [[Shape of Cotangent Function/Graph|Graph of Cotangent Function]] ===
+{{:Shape of Cotangent Function/Graph}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Change of Base of Logarithm}
+Tags: Logarithms, Change of Base of Logarithm
+
+\begin{theorem}
+:$\log_b x = \dfrac {\log_a x} {\log_a b}$
+\end{theorem}
+
+\begin{proof}
+Let:
+: $y = \log_b x \iff b^y = x$
+: $z = \log_a x \iff a^z = x$
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \map {\log_a} {b^y}
+}}
+{{eqn | r = y \log_a b
+      | c = [[Logarithms of Powers]]
+}}
+{{eqn | r = \log_b x \log_a b
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Function to Power of Function}
+Tags: Differential Calculus, Exponential Function
+
+\begin{theorem}
+Let $\map u x, \map v x$ be [[Definition:Real Function|real functions]] which are [[Definition:Differentiable on Interval|differentiable]] on $\R$.
+Then:
+:$\map {\dfrac \d {\d x} } {u^v} = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {u^v}
+      | r = \map {\dfrac \d {\d x} } {\map \exp {v \ln u} }
+      | c = {{Defof|Power to Real Number}}
+}}
+{{eqn | r = \map \exp {v \ln u} \map {\dfrac \d {\d x} } {v \ln u}
+      | c = [[Chain Rule for Derivatives]] and [[Derivative of Exponential Function]]
+}}
+{{eqn | r = \map \exp {v \ln u} \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + v \map {\dfrac \d {\d x} } {\ln u} }
+      | c = [[Product Rule]]
+}}
+{{eqn | r = u^v \paren {\paren {\ln u} \map {\dfrac \d {\d x} } v + \frac v u \map {\dfrac \d {\d x} } u}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = v u^{v - 1} \map {\dfrac \d {\d x} } u + u^v \paren {\ln u} \map {\dfrac \d {\d x} } v
+      | c = gathering terms
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Liouville's Theorem (Complex Analysis)}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: \C \to \C$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Entire Function|entire]] [[Definition:Complex Function|function]].
+Then $f$ is [[Definition:Constant Mapping|constant]].
+\end{theorem}
+
+\begin{proof}
+By assumption, there is $M \ge 0$ such that $\cmod {\map f z} \le  M$ for all $z \in \C$. 
+For any $R \in \R: R > 0$, consider the function:
+:$\map {f_R} z := \map f {R z}$
+Using the [[Cauchy Integral Formula]], we see that:
+:$\displaystyle \cmod {\map {f_R'} z} = \frac 1 {2 \pi} \cmod {\int_{\map {C_1} z} \frac {\map f w} {\paren {w - z}^2} \rd w} \le \frac 1 {2 \pi} \int_{\map {C_1} z} M \rd w = M$
+where $\map {C_1} z$ denotes the [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $1$ around $z$.
+Hence:
+:$\displaystyle \cmod {\map {f'} z} = \cmod {\map {f_R'} z} / R \le M / R$
+Since $R$ was arbitrary, it follows that $\cmod {\map {f'} z} = 0$ for all $z \in \C$.
+Thus $f$ is [[Definition:Constant Mapping|constant]].
+{{qed}}
+{{MissingLinks}}
+\end{proof}<|endoftext|>
+\section{Riemann Removable Singularities Theorem}
+Tags: Complex Analysis, Continuous Functions
+
+\begin{theorem}
+Let $U \subset \C$ be a domain, let $z_0 \in U$, and let $f: U \setminus \left\{ {z_0}\right\} \to \C$ be [[Definition:Holomorphic Function|holomorphic]].
+Then the following are equivalent:
+:$(1): \quad f$ [[Definition:Extension of Mapping|extends]] to a [[Definition:Holomorphic Function|holomorphic function]] $f: U \to \C$.
+:$(2): \quad f$ [[Definition:Extension of Mapping|extends]] to a [[Definition:Continuous Complex Function|continuous]] function $f: U \to \C$.
+:$(3): \quad f$ is [[Definition:Bounded Mapping|bounded]] in a [[Definition:Neighborhood (Complex Analysis)|neighborhood]] of $z_0$.
+:$(4): \quad f \left({z}\right) = o \left({\dfrac 1 {\left|{z - z_0}\right|} }\right)$ as $z \to z_0$.
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, we may assume that:
+:$U = \Bbb D := \left\{{z \in \C: \left|{z}\right| < 1}\right\}$
+and that $z_0 = 0$. 
+Otherwise, restrict $f$ to a suitable disk centered at $z_0$, and [[Definition:Precomposition|precompose]] with a suitable [[Definition:Affine Mapping|affine map]].
+A [[Definition:Holomorphic Function|holomorphic function]] is continuous.
+A continuous function is locally bounded.
+So $(1) \implies (2) \implies (3)$.
+Also, $(3) \implies (4)$ by definition. 
+That $(2) \implies (1)$ follows from the proof of the [[Residue Theorem]].
+(For completeness, we sketch a self-contained argument below.)
+{{refactor|Extract that proof referred to above, and put it in its own page, if necessary.}}
+Now assume that $(4)$ holds.
+That is:
+:$z \cdot f \left({z}\right) \to 0$ as $z \to 0$.
+We need to show that (b) holds. 
+By assumption, the function:
+:$g \left({z}\right) := \begin{cases} z \cdot f \left({z}\right) & : z \ne 0 \\ 0 & : \text{otherwise} \end{cases}$
+is continuous in $\Bbb D$ and [[Definition:Holomorphic Function|holomorphic]] in $\Bbb D \setminus \left\{ {0}\right\}$.
+By applying the direction (b) $\implies$ (a) to this function, we see that $g$ is holomorphic in $\Bbb D$.
+We have:
+:$\displaystyle g' \left({0}\right) = \lim_{z \to 0} \frac {g \left({z}\right)} z = \lim_{z \to 0} f \left({z}\right)$
+so $f$ extends continuously to $\Bbb D$, as claimed.
+To prove $(2) \implies (1)$, we use [[Morera's Theorem]] and the [[Residue Theorem]].
+By [[Morera's Theorem]], we need to show that:
+:$\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$
+for every closed curve $C$ in $\Bbb D$.
+It follows from the Cauchy integral theorem that we only need to check that:
+:$\displaystyle \int_C f \left({z}\right) \, \mathrm d z = 0$
+for a simple closed loop surrounding $0$, and that this integral is independent of the loop $C$.
+Letting $C = C_\epsilon \left({0}\right)$ be the circle of radius $\epsilon$ around $0$, we see that:
+:$\displaystyle \left|{\int_{C_\epsilon} f \left({z}\right) \, \mathrm d z}\right| \le 2 \pi \epsilon \max_{z \in C_\epsilon} \left|{f \left({z}\right)}\right| \to 0$
+as $\epsilon\to 0$ (because $f$ is continuous in $0$ by assumption).
+This completes the proof.
+{{qed}}
+{{Namedfor|Georg Friedrich Bernhard Riemann|cat = Riemann}}
+[[Category:Complex Analysis]]
+[[Category:Continuous Functions]]
+7l2wp2sscbwqqjmoijs79kwjblfak5j
+\end{proof}<|endoftext|>
+\section{Gaussian Integral}
+Tags: Definite Integrals involving Exponential Function
+
+\begin{theorem}
+:$\displaystyle \int_{-\infty}^\infty e^{-x^2} \rd x = \sqrt \pi$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \int_{-\infty}^\infty e^{-x^2} \rd x
+	| r = 2 \int_0^\infty e^{-x^2} \rd x
+	| c = [[Definite Integral of Even Function]]
+}}
+{{eqn	| r = \frac {2 \sqrt \pi} 2
+	| c = [[Integral to Infinity of Exponential of -t^2]]
+}}
+{{eqn	| r = \sqrt \pi
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Definite Integrals involving Exponential Function]]
+kxvyhs9dfoak7xid7n2a1pf5blmlt52
+\end{proof}<|endoftext|>
+\section{Indiscrete Topology is not Metrizable}
+Tags: Metrizable Topologies, Indiscrete Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with more than one [[Definition:Element|element]].
+The [[Definition:Indiscrete Topology|indiscrete topology]] on $S$ is not [[Definition:Metrizable Topology|metrizable]].
+\end{theorem}
+
+\begin{proof}
+In order to be [[Definition:Metrizable Topology|metrizable]], there needs to be a [[Definition:Metric|metric]] $d$ on $S$, so that $\struct {S, d}$ is a [[Definition:Metric Space|metric space]].
+As $S$ has more than one [[Definition:Element|element]], $\exists x, y \in S: x \ne y$.
+Then:
+:$\epsilon = \map d {x, y} > 0$
+So the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball]] $\map {B_{\epsilon / 2} } x$ is [[Definition:Open Set of Metric Space|$d$-open]].
+Hence $\map {B_{\epsilon / 2} } x$ is in the [[Definition:Topology|topology]] which $d$ [[Metric Induces Topology|induces]].
+But $x \in \map {B_{\epsilon / 2} } x$ while $y \notin \map {B_{\epsilon / 2} } x$.
+Thus $\map {B_{\epsilon / 2} } x \ne \O$ and $\map {B_{\epsilon / 2} } x \ne S$.
+So the [[Metric Induces Topology|topology induced by $d$]] is not the [[Definition:Indiscrete Topology|indiscrete topology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite of Continuous Mappings is Continuous}
+Tags: Topology, Continuous Mappings, Composite Mappings
+
+\begin{theorem}
+Let $T_1, T_2, T_3$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ and $g: T_2 \to T_3$ be [[Definition:Continuous Mapping (Topology)|continuous mappings]].
+Then the [[Definition:Composition of Mappings|composite mapping]] $g \circ f: T_1 \to T_3$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in T_3$ be [[Definition:Open Set (Topology)|open in $T_3$]].
+As $g$ is [[Definition:Continuous Mapping (Topology)|continuous]], $g^{-1} \sqbrk U \in T_2$ is [[Definition:Open Set (Topology)|open in $T_2$]].
+As $f$ is [[Definition:Continuous Mapping (Topology)|continuous]], $f^{-1} \sqbrk {g^{-1} \sqbrk U} \in T_1$ is [[Definition:Open Set (Topology)|open in $T_1$]].
+By [[Inverse of Composite Bijection]], $f^{-1} \sqbrk {g^{-1} \sqbrk U} = \paren {g \circ f}^{-1} \sqbrk U$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Mapping is Continuous on Induced Topological Spaces}
+Tags: Topology, Metric Spaces, Continuous Mappings
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\tau_{d_1}$ and $\tau_{d_2}$ be the [[Definition:Topology Induced by Metric|topologies induced by the metrics]] $d_1$ and $d_2$.
+Let $T_1 = \left({A_1, \tau_{d_1}}\right)$ and $T_2 = \left({A_2, \tau_{d_2}}\right)$ be the resulting [[Definition:Topological Space|topological spaces]].
+Let $f: A_1 \to A_2$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Continuous Mapping (Metric Spaces)|$\left({d_1, d_2}\right)$-continuous]] {{iff}} $f$ is [[Definition:Continuous Mapping (Topological Spaces)|$\left({\tau_{d_1}, \tau_{d_2}}\right)$-continuous]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+* the [[Definition:Metric Space Continuity by Open Set|open set definition]] of continuity on a [[Definition:Metric Space|metric space]]
+* the definition of [[Definition:Continuous Mapping (Topology)|continuity]] on a [[Definition:Topological Space|topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Synthetic Basis and Analytic Basis are Compatible}
+Tags: Topological Bases
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$ {{iff}} $\tau$ is the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by the synthetic basis $\BB$]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
+We proceed to check the axioms for $\BB$ to be a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
+[[Definition:Open Set Axioms|Open set axiom]] $(\text O 3)$ states that $S \in \tau$.
+Therefore, by the definition of an [[Definition:Analytic Basis|analytic basis]]:
+:$\displaystyle \exists \SS \subseteq \BB: S = \bigcup \SS$
+By [[Equivalent Conditions for Cover by Collection of Subsets]], $\BB$ is a [[Definition:Cover of Set|cover]] for $S$.
+That is, axiom $(\text B 1)$ for a [[Definition:Synthetic Basis|synthetic basis]] is satisfied by $\BB$.
+Suppose that $A, B \in \BB$.
+By the definition of an [[Definition:Analytic Basis|analytic basis]], $\BB \subseteq \tau$.
+Therefore $A, B \in \tau$.
+[[Definition:Open Set Axioms|Open set axiom]] $(\text O 2)$ states that $A \cap B \in \tau$.
+Therefore, by the definition of an [[Definition:Analytic Basis|analytic basis]]:
+:$\displaystyle \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$
+That is, axiom $(\text B 2)$ for a [[Definition:Synthetic Basis|synthetic basis]] is satisfied by $\BB$.
+Therefore, $\BB$ is a [[Definition:Synthetic Basis|synthetic basis]] on $S$.
+Let $\tau'$ be the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by the synthetic basis $\BB$]].
+It follows from the definition of an [[Definition:Analytic Basis|analytic basis]] that $\tau \subseteq \tau'$.
+Since the [[Subset Relation is Transitive|subset relation is transitive]], we can apply [[Definition:Open Set Axioms|open set axiom]] $(\text O 1)$ for a [[Definition:Topology|topology]] to conclude that:
+:$\displaystyle \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$
+That is, $\tau' \subseteq \tau$.
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\tau = \tau'$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $\BB$ is a [[Definition:Synthetic Basis|synthetic basis]] on $S$, and that $\tau$ is the [[Definition:Topology Generated by Synthetic Basis|topology on $S$ generated by $\BB$]].
+Then:
+:$\displaystyle \BB = \set {\bigcup \set B: \set B \subseteq \BB} \subseteq \tau$
+By definition, $\BB$ is an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Picard's Existence Theorem}
+Tags: First Order ODEs, Picard's Existence Theorem
+
+\begin{theorem}
+Let $\map f {x, y} : \R^2 \to \R$ be [[Definition:Continuous Real Function|continuous]] in a [[Definition:Region of Plane|region]] $D \subseteq \R^2$.
+Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
+Let $\map f {x, y}$ satisfy in $D$ the [[Definition:Lipschitz Condition (Real Function)|Lipschitz condition]] in $y$:
+:$\size{\map f {x, y_1} - \map f {x, y_2} } \le A \size {y_1 - y_2}$
+where $A$ is independent of $x, y_1, y_2$.
+Let the [[Definition:Rectangle|rectangle]] $R$ be defined as $\set {\tuple {x, y} \in \R^2: \size {x - a} \le h, \size {y - b} \le k}$ such that $M h \le k$.
+Let $R \subseteq D$.
+Then $\forall x \in \R: \size {x - a} \le h$, the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$y' = \map f {x, y}$
+has [[Definition:Exactly One|one and only one]] solution $y = \map y x$ for which $b = \map y a$.
+\end{theorem}
+
+\begin{proof}
+Let us define the following series of [[Definition:Real Function|functions]]:
+{{begin-eqn}}
+{{eqn | l = \map {y_0} x
+      | r = b
+      | c = 
+}}
+{{eqn | l = \map {y_1} x
+      | r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = b + \int_a^x \map f {t, \map {y_1} t} \rd t
+      | c = 
+}}
+{{eqn | o = \ldots
+      | c = 
+}}
+{{eqn | l = \map {y_n} x
+      | r = b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t
+      | c = 
+}}
+{{end-eqn}}
+What we are going to do is prove that $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_n} t$ is the required solution.
+There are five main steps, as follows:
+=== The curve lies in the rectangle ===
+We will show that for $a - h \le x \le a + h$, the curve $y = \map {y_n} x$ lies in the rectangle $R$.
+That is, that $b - k < y < b + k$.
+Suppose $y = \map {y_{n - 1} } x$ lies in $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\map {y_n} x - b}
+      | r = \size {\int_a^x \map f {t, \map {y_{n - 1} } t} \rd t}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = M \size {x - a}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = M h
+      | c = 
+}}
+{{eqn | o = <
+      | r = k
+      | c = 
+}}
+{{end-eqn}}
+Clearly $y_0$ lies in $R$, and the argument holds for $y_1$.
+So by [[Principle of Mathematical Induction|induction]], $y = \map {y_n} x$ lies in $R$ for all $n \in \N$.
+=== Bounded Nature of Adjacent Differences ===
+We will show that:
+:$\displaystyle \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$
+This is also to be proved by [[Principle of Mathematical Induction|induction]].
+Suppose that this holds for $n-1$ in place of $n$.
+Let this be the [[Definition:Induction Hypothesis|induction hypothesis]].
+We have:
+:$\displaystyle \map {y_n} x - \map {y_{n - 1} } x = \int_a^x \paren {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \rd t$
+We also have that:
+:$\size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le A \size {\map {y_{n - 1} } t - \map {y_{n - 2} } t}$
+by the [[Definition:Lipschitz Condition (Real Function)|Lipschitz condition]].
+By the [[Definition:Induction Hypothesis|induction hypothesis]], it follows that:
+$\displaystyle \size {\map f {t, \map {y_{n - 1} } t} - \map f {t, \map {y_{n - 2} } t} } \le \frac {M A^{n - 1} \size {t - a}^{n - 1} } {\paren {n - 1}!}$
+So:
+{{begin-eqn}}
+{{eqn | l = \size {\map {y_n} x - \map {y_{n - 1} } x}
+      | o = \le
+      | r = \frac {M A^{n - 1} } {\paren {n - 1}!} \size {\int_a^x \size {t - a}^{n - 1} \rd t}
+      | c = 
+}}
+{{eqn | r = \frac {M A^{n - 1} } {n!} \size {x - a}^n
+      | c = 
+}}
+{{end-eqn}}
+For the [[Definition:Basis for the Induction|base case]], we use $n = 1$:
+:$\displaystyle \size {\map {y_1} x - b} \le \size {\int_a^x \map f {t, b} \rd t} \le M \size {x - a}$
+Thus by [[Principle of Mathematical Induction|induction]]:
+:$\displaystyle \size {\map {y_n} x - \map {y_{n - 1} } x} \le \frac {M A^{n - 1} } {n!} \size {x - a}^n$
+for all $n$.
+=== Uniform Convergence of Sequence ===
+Next we show that the [[Definition:Sequence|sequence]] $\sequence {\map {y_n} x}$ [[Definition:Uniform Convergence|converges uniformly]] to a [[Definition:Limit of Sequence (Number Field)|limit]] for $a - h \le x \le a + h$.
+From [[Picard's Existence Theorem#Bounded Nature of Adjacent Differences|Bounded Nature of Adjacent Differences]] above, we have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots
+      | c = 
+}}
+{{eqn | o = \le
+      | r = b + M h + \cdots + \frac {M A^{n - 1} h^n} {n!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+From [[Radius of Convergence of Power Series over Factorial]], it follows that $b + M h + \cdots + \dfrac {M A^{n - 1} h^n} {n!} + \cdots$ is [[Definition:Absolutely Convergent Series|absolutely convergent]] for all $h$.
+Hence, by the [[Weierstrass M-Test]]:
+:$b + \paren {\map {y_1} x - b} + \cdots + \paren {\map {y_n} x - \map {y_{n - 1} } x} + \cdots$
+[[Definition:Uniform Convergence|converges uniformly]] for $a - h \le x \le a + h$.
+Since its terms are [[Definition:Continuous Real Function|continuous functions]] of $x$, its sum $\displaystyle \lim_{n \mathop \to \infty} \map {y_n} x = \map y x$ is also continuous from [[Combination Theorem for Sequences]].
+=== Solution Satisfies Differential Equation ===
+We now show that $y = \map y x$ satisfies the [[Definition:Differential Equation|differential equation]] $y' = \map f {x, y}$.
+Since:
+:$\map {y_n} x$ [[Definition:Uniform Convergence|converges uniformly]] to $\map y x$ in the [[Definition:Open Real Interval|open interval]] $\openint {a - h} {a + h}$ from [[Picard's Existence Theorem#Uniform Convergence of Sequence|Uniform Convergence of Sequence]] above
+:$\size {\map f {x, y} - \map f {x, y_n} } \le A \size {y - y_n}$ from the [[Definition:Lipschitz Condition (Real Function)|Lipschitz condition]] in $y$
+it follows that $\map f {x, \map {y_n} x}$ [[Definition:Uniform Convergence|tends uniformly]] to $\map f {x, \map y x}$.
+Letting $n \to \infty$ in:
+:$\displaystyle \map {y_n} x = b + \int_a^x \map f {t, \map {y_{n - 1} } t} \rd t$
+we get:
+:$\displaystyle \map y x = b + \int_a^x \map f {t, \map y t} \rd t$
+The [[Definition:Integrand|integrand]] $\map f {t, \map y t}$ is a [[Definition:Continuous Real Function|continuous function]] of $t$.
+Therefore the integral has the [[Definition:Derivative|derivative]] $\map f {x, y}$.
+Also, we have that $\map y a = b$.
+=== Uniqueness of Solution ===
+We now show that the solution $y = \map y x$ that we have found is the ''only'' solution where $\map y a = b$.
+{{AimForCont}} there is another such solution, $y = \map Y x$, say.
+Let $\size {\map Y x - \map y x} \le B$ when $\size {x - a} \le h$. (Certainly we could take $B = 2 k$.)
+Then:
+:$\displaystyle \map Y x - \map y x = \int_a^x \paren {\map f {t, \map Y t} - \map f {t, \map y t} } \rd t$
+But:
+:$\size {\map f {t, \map Y t} - \map f {t, \map y t} } \le A \size {\map Y t - \map y t} \le A B$
+So:
+:$\size {\map Y t - \map y t} \le A B \size {x - a}$
+Repeating the argument, we can get successive estimates for the upper bound of $\size {\map Y x - \map y x}$ in $\openint {a - h} {a + h}$.
+This gives:
+:$\displaystyle \frac {A^2 B} {2!} \size {x - a}^2, \ldots, \frac {A^n B} {n!} \size {x - a}^n, \ldots$
+But this [[Definition:Sequence|sequence]] tends to $0$.
+So $\map Y x = \map y x$ in $\openint {a - h} {a + h}$.
+This [[Definition:Contradiction|contradicts]] the supposition that $\map Y x$ and $\map y x$ are different.
+Hence by [[Proof by Contradiction]] it follows that $\map y x$ is unique.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let us define the following [[Definition:Partial Sum|series]] of [[Definition:Real Function|functions]]:
+{{begin-eqn}}
+{{eqn | l = \map {y_0} x
+      | r = b
+      | c = 
+}}
+{{eqn | l = \map {y_1} x
+      | r = b + \int_a^x \map f {t, \map {y_0} t} \rd t
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = b + \int_a^x \map f {t, \map {y_1} t} \rd t
+      | c = 
+}}
+{{eqn | o = \ldots
+      | c = 
+}}
+{{eqn | l = \map {y_n} x
+      | r = b + \int_a^x \map f {t, \map {y_{n-1} } t} \rd t
+      | c = 
+}}
+{{end-eqn}}
+Denote this [[Definition:Sequence|sequence]] by $\sequence {y_k}_{k \mathop \in \N_0}$.
+What we are going to do is prove that $\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_n} x$ is the required solution.
+=== The curve lies in the rectangle ===
+We will show that for $a - h \le x \le a + h$, the [[Definition:Curve|curve]] $y = \map {y_n}x$ lies in the rectangle $R$.
+That is, that $b - k < y < b + k$.
+Suppose $y = \map {y_{n - 1} } x$ lies in $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\map {y_n} x - b}
+      | r = \size {\int_a^x \map f {t, \map {y_{n - 1} } t} \rd t}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = M \size {x - a}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = M h
+      | c = 
+}}
+{{eqn | o = <
+      | r = k
+      | c = 
+}}
+{{end-eqn}}
+Clearly $y_0$ lies in $R$, and the argument holds for $y_1$.
+So by [[Principle of Mathematical Induction|induction]], $y = \map {y_n} x$ lies in $R$ for all $n \in \N$.
+=== Existence ===
+The [[Definition:Sequence|sequence]] $\sequence {y_k}_{k \mathop \in N_0}$ can be expressed as a [[Definition:Telescoping Series|telescoping series]]:
+:$\displaystyle y_{n + 1} = y_0 + \sum_{k = 0}^n \paren {y_{k + 1} - y_k}$
+The theorem contains more [[Definition:Real Variable|variables]] $\paren {\set {x, y_1, y_2} }$ and [[Definition:Parameter|parameters]] $\paren {\set{h, k, M, A} }$ than [[Definition:Inequality|inequality]] constraints.
+Thus, more relations between them can be chosen without affecting the constraints.
+Choose $\displaystyle h = \frac A 2$.
+For $a \le x \le h$ we have:
+{{begin-eqn}}
+{{eqn | l = \size {\map {y_{n+1} } x - \map {y_n} x}
+      | r = \size {\int_a^x \map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} \rd t}
+}}
+{{eqn | o = \le
+      | r = \int_a^x \size {\map f {t, \map {y_n} t} - \map f {t, \map {y_{n - 1} } t} } \rd t
+      | c = [[Absolute Value of Definite Integral]]
+}}
+{{eqn | o = \le
+      | r = \int_a^x A \size {\map {y_n} t - \map {y_{n - 1} } t} \rd t
+      | c = {{Defof|Lipschitz Condition (Real Function)}}
+}}
+{{eqn | o = \le
+      | r = \int_a^x A \norm {y_n - y_{n - 1} }_\infty \rd t
+      | c = {{Defof|Supremum Norm}}
+}}
+{{eqn | r = A \norm {y_n - y_{n-1} }_\infty \paren {x - a}
+}}
+{{eqn | o = \le
+      | r = A \norm {y_n - y_{n-1} }_\infty h
+}}
+{{eqn | r = \frac 1 2 \norm {y_n - y_{n - 1} }_\infty
+}}
+{{end-eqn}}
+By taking [[Definition:Supremum Norm/Continuous on Closed Interval Real-Valued Function|supremum norm]] of both sides, we get:
+:$\displaystyle \norm {y_{n + 1} - y_n}_\infty \le \frac 1 2 \norm {y_n - y_{n - 1} }_\infty$
+By [[Principle of Mathematical Induction|induction]], the [[Definition:Inequality|inequality]] can be extended:
+:$(1): \quad \displaystyle \norm {y_{n + 1} - y_n} \le \frac 1 {2^n} \norm {y_1 - y_0}$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \sum_{n = 0}^\infty \norm {y_{n + 1} - y_n}
+      | o = \le
+      | r = \norm {y_1 - y_0} \sum_{n \mathop = 0}^\infty \frac 1 {2^n}
+}}
+{{eqn | o = <
+      | r = \infty
+      | c = [[Sum of Infinite Geometric Progression]]
+}}
+{{end-eqn}}
+Same argument applies to $-h \le x \le a$.
+Hence, $\sequence {y_k}_{k \mathop \in \N_0}$ converges in $\struct {\map {\CC^1} {\size{x - a} \le h}, \norm {\cdot}_\infty}$ to $y \in \map {\CC^1} {\size {x - a} \le h}$ [[Definition:Absolutely Convergent Series|absolutely]].
+Therefore, the [[Definition:Sequence|sequence]] is [[Definition:Convergent Sequence in Normed Vector Space|convergent]]:
+:$\displaystyle \map y x = \lim_{n \mathop \to \infty} \map {y_{n + 1} } x = x_0 + \lim_{n \mathop \to \infty} \int_a^x \map f {x, \map {y_n} x} \rd x$
+To find the [[Definition:Convergent Sequence in Normed Vector Space|limit]], consider the following [[Definition:Sequence|sequence]]:
+:$\displaystyle \map {g_n} x = \map f {x, \map {y_n} x}$
+The [[Definition:Sequence|sequence]] $\sequence {g_n}_{n \mathop \in \N_0}$ is a [[Definition:Sequence|sequence]] of [[Definition:Partial Sum|partial sums]] $\displaystyle g_0 + \sum_{k \mathop = 0}^n \paren {g_{k + 1} - g_k}$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \norm {\map {g_{k + 1} } x - \map {g_k} x} 
+      | r = \norm {\map f {x, \map {y_{k+1} } x} - \map f {x, \map {y_k} x} } 
+}}
+{{eqn | o = \le
+      | r = L \norm {\map {y_{k + 1} } x - \map {y_k} x} 
+      | c = assumption in theorem
+}}
+{{eqn | o = \le 
+      | r = L \norm {y_{k + 1} - y_k}_\infty 
+      | c = {{Defof|Supremum Norm}}
+}}
+{{eqn | o = \le
+      | r = \frac 1 {2^k} \norm {y_1 - y_0}_\infty
+      | c = from $(1)$
+}}
+{{end-eqn}}
+So $\sequence {g_n}_{n \mathop \in \N_0}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] to some $g$ in $\struct {\map \CC {\size {x - a} \le h}, \norm {\, \cdot \,}_\infty}$ [[Definition:Absolutely Convergent Series|absolutely]].
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map g x 
+      | r = \lim_{n \mathop \to \infty} \map {g_n} x 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map f {x, \map {y_n} x}
+}}
+{{eqn | r = \map f {x, \map y x}
+}}
+{{end-eqn}}
+On the other hand,a  [[Definition:Riemann Integral|Riemann integral]] is a [[Definition:Continuous Mapping|continuous mapping]].
+From [[Continuous Map Preserves Convergent Sequences]]:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \int_a^x \map f {t, \map {y_n} t} \rd t 
+      | r = \lim_{n \mathop \to \infty} \int_a^x \paren {\map {g_0} t + \sum_{k \mathop = 0}^{n - 1} \paren {\map {g_{k + 1} } t - \map {g_k} t} } \rd t 
+}}
+{{eqn | r = \int_a^x \map g t \rd t
+}}
+{{eqn | r = \int_a^x \map f {t, \map y t} \rd t
+}}
+{{end-eqn}}
+We conclude that:
+:$\displaystyle \map y x = y_0 + \int_a^x \map f {t, \map y t} \rd t$
+where:
+:$\map y a = y_0 + 0 = b$
+and, by [[Fundamental Theorem of Calculus]]:
+:$\map {y'} x = 0 + \map f {x, \map y x}$
+for all $x \in \R : \size {x - a} \le h$.
+=== Uniqueness ===
+{{AimForCont}} that the [[Definition:Solution of Differential Equation|solution]] to [[Definition:Initial Value Problem|IVP]] is not [[Definition:Unique|unique]].
+Then, for the same [[Definition:Initial Condition|initial conditions]] there exists a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $R$ where [[Definition:Solution of Differential Equation|solutions]] differ.
+Let $y_1, y_2$ be [[Definition:Solution of Differential Equation|solutions]] to [[Definition:Initial Value Problem|IVP]] for $x \in \R : \size {x - a} \le h$.
+Let $x_* := \max \set {x \in \R : \size {x - a} \le h : \map {y_1} t = \map {y_2} t, \forall t \le x }$
+Then:
+:$\displaystyle \map {y_1} x - \map {y_1} {x_*} = \int_{x_*}^x \map {y_1'} t \rd t = \int_{x_*}^x \map {f_1} {t, \map {y_1} t} \rd t$
+:$\displaystyle \map {y_2} x - \map {y_2} {x_*} = \int_{x_*}^x \map {y_2'} t \rd t = \int_{x_*}^x \map {f_2} {t, \map {y_2} t} \rd t$
+After taking the [[Definition:Real Subtraction|difference]]:
+:$\displaystyle \map {y_1} x - \map {y_2} x = \int_{x_*}^x \paren {\map {f_1} {t, \map {y_1} t} - \map {f_2} {t, \map {y_2} t}} \rd t$
+Let $N \in \R$ be such that:
+:$N > \max \set {1, \dfrac 1 A, \dfrac 1 {A \paren {a \mathop + h \mathop - x_*} } }$
+For all cases it holds that:
+:$x_* + \dfrac 1 {A N} < a + h$
+Let:
+:$\displaystyle B = \max_{t \mathop \in \closedint {x_*} {x_* \mathop + \frac 1 {A N} } } \size {\map {x_2} t - \map {x_1} t} \le 2 k$
+Then $\forall x \in \closedint {x_*} {x_* + \dfrac 1 {AN}}$ we have:
+{{begin-eqn}}
+{{eqn | l = \size {\map {x_2} x - \map {x_1} x}
+      | r = \size {\int_{x_*}^x  \paren {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t}
+}}
+{{eqn | o = \le
+      | r = \int_{x_*}^x  \size {\map f {\map {x_2} t, t} - \map f {\map {x_1} t, t} } \rd t
+      | c = [[Absolute Value of Definite Integral]]
+}}
+{{eqn | o = \le
+      | r = \int_{x_*}^x A \size {\map {x_2} t - \map {x_1} t} \rd t
+      | c = {{Defof|Lipschitz Condition (Real Function)}}
+}}
+{{eqn | o = \le
+      | r = \int_{x_*}^x A B \rd t
+}}
+{{eqn | r = A B \paren {x - x_*}
+}}
+{{eqn | o = \le
+      | r = A B \paren {x_* + \frac 1 {A N} - x_*}
+}}
+{{eqn | r = \frac {A B} {A N}
+}}
+{{eqn | r = \frac B N
+}}
+{{end-eqn}}
+Thus: 
+:$\forall t \in \closedint {x_*} {x_* + \dfrac 1 {A N} } : \size {\map {x_1} t - \map {x_2} t} \le \dfrac B N$
+and $B \le \dfrac B N$ or $N \le 1$. 
+This brings us to a [[Definition:Contradiction|contradiction]].
+Hence our assumption that the [[Definition:Solution of Differential Equation|solution]] to [[Definition:Initial Value Problem|IVP]] is not [[Definition:Unique|unique]] was false.
+Hence the result, by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity Test using Basis}
+Tags: Topological Bases, Continuous Mappings
+
+\begin{theorem}
+Let $\left({X_1, \tau_1}\right)$ and $\left({X_2, \tau_2}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $f: X_1 \to X_2$ be a [[Definition:Mapping|mapping]].
+Let $\mathcal B$ be an [[Definition:Analytic Basis|analytic basis]] for $\tau_2$.
+Suppose that:
+:$\forall B \in \mathcal B: f^{-1} \left({B}\right) \in \tau_1$
+where $f^{-1} \left({B}\right)$ denotes the [[Definition:Preimage of Subset under Mapping|preimage]] of $B$ under $f$.
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $U \in \tau_2$.
+By the definition of an [[Definition:Analytic Basis|analytic basis]], it follows that:
+:$\displaystyle \exists \mathcal A \subseteq \mathcal B: U = \bigcup \mathcal A$
+Hence:
+{{begin-eqn}}
+{{eqn | l = f^{-1} \left({U}\right)
+      | r = f^{-1} \left({\bigcup \mathcal A}\right)
+}}
+{{eqn | r = \bigcup_{B \mathop \in \mathcal A} f^{-1} \left({B}\right)
+      | c = [[Preimage of Union under Mapping/General Result|Preimage of Union under Mapping: General Result]]
+}}
+{{eqn | o = \in
+      | r = \tau_1
+      | c = [[Definition:By Hypothesis|by hypothesis]], and by the definition of a [[Definition:Topology|topology]]
+}}
+{{end-eqn}}
+The result follows from the definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sub-Basis for Real Number Line}
+Tags: Topological Bases, Real Numbers
+
+\begin{theorem}
+Let the [[Definition:Real Number Line|real number line]] $\R$ be considered as a [[Definition:Topology|topology]] under the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]].
+Then:
+:$\mathcal B := \left\{{\left({-\infty \,.\,.\, a}\right), \left({b \,.\,.\, \infty}\right): a, b \in \R}\right\}$ is a [[Definition:Sub-Basis|sub-basis]] for $\R$.
+\end{theorem}
+
+\begin{proof}
+Let $\left({c \,.\,.\, d}\right)$ be an [[Definition:Open Real Interval|open real interval]].
+Then by definition:
+:$\left({c \,.\,.\, d}\right) = \left({-\infty \,.\,.\, d}\right) \cap \left({c \,.\,.\, \infty}\right)$
+and so $\left({c \,.\,.\, d}\right)$ is the [[Definition:Set Intersection|intersection]] of two [[Definition:Element|elements]] of $\mathcal B$.
+From [[Open Sets in Real Number Line]], any [[Definition:Open Set (Real Analysis)|open set]] of $\R$ is the [[Definition:Set Union|union]] of [[Definition:Countable Set|countably many]] [[Definition:Open Real Interval|open real intervals]].
+The result follows by definition of [[Definition:Sub-Basis|sub-basis]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Synthetic Basis formed from Synthetic Sub-Basis}
+Tags: Topological Bases, Sub-Bases
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\SS$ be a [[Definition:Synthetic Sub-Basis|synthetic sub-basis]] on $X$.
+Define:
+:$\displaystyle \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
+Then $\BB$ is a [[Definition:Synthetic Basis|synthetic basis]] on $X$.
+\end{theorem}
+
+\begin{proof}
+We consider $X$ as the [[Definition:Universe (Set Theory)|universe]].
+Thus, in accordance with [[Intersection of Empty Set]], we take the convention that:
+:$\displaystyle \bigcap \O = X \in \BB$
+By [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]], it follows that:
+:$\displaystyle X \subseteq \bigcup \BB$
+That is, axiom '''B1''' for a [[Definition:Synthetic Basis|synthetic basis]] is satisfied.
+We have that $\BB \subseteq \powerset X$.
+Let $B_1, B_2 \in \BB$. Then [[Definition:Existential Quantifier|there exist]] [[Definition:Finite Set|finite]] $\FF_1, \FF_2 \subseteq \SS$ such that:
+:$\displaystyle B_1 = \bigcap \FF_1$
+:$\displaystyle B_2 = \bigcap \FF_2$
+It follows that:
+:$\displaystyle B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$
+{{explain|proof that $\displaystyle \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ is equal to $\displaystyle \bigcap \bigcup_{i \mathop \in I} \mathbb S_i$}}
+By [[Union is Smallest Superset]], $\FF_1 \cup \FF_2 \subseteq \SS$.
+We have that [[Union of Finite Sets is Finite|$\FF_1 \cup \FF_2$ is finite]].
+Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom '''B2''' for a [[Definition:Synthetic Basis|synthetic basis]] is satisfied.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Epimorphism Preserves Distributivity}
+Tags: Epimorphisms, Distributive Operations
+
+\begin{theorem}
+Let $\struct {R_1, +_1, \circ_1}$ and $\struct {R_2, +_2, \circ_2}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: R_1 \to R_2$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
+:If $\circ_1$ is [[Definition:Left Distributive Operation|left distributive]] over $+_1$, then $\circ_2$ is [[Definition:Left Distributive Operation|left distributive]] over $+_2$.
+:If $\circ_1$ is [[Definition:Right Distributive Operation|right distributive]] over $+_1$, then $\circ_2$ is [[Definition:Right Distributive Operation|right distributive]] over $+_2$.
+Consequently, if $\circ_1$ is [[Definition:Distributive Operation|distributive]] over $+_1$, then $\circ_2$ is [[Definition:Distributive Operation|distributive]] over $+_2$.
+That is, [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] preserves [[Definition:Distributive Operation|distributivity]].
+\end{theorem}
+
+\begin{proof}
+Throughout the following, we assume the [[Definition:Morphism Property|morphism property]] holds for $\phi$ for both operations.
+=== Left Distributivity ===
+Suppose $\circ_1$ is [[Definition:Left Distributive Operation|left distributive]] over $+_1$. Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi x \circ_2 \paren {\map \phi y +_2 \map \phi z}
+      | r = \map \phi x \circ_2 \map \phi {y +_1 z}
+      | c = 
+}}
+{{eqn | r = \map \phi {x \circ_1 \paren {y +_1 z} }
+      | c = 
+}}
+{{eqn | r = \map \phi {\paren {x \circ_1 y} +_1 \paren {x \circ_1 z} }
+      | c = as $\circ_1$ is [[Definition:Left Distributive Operation|left distributive]] over $+_1$
+}}
+{{eqn | r = \map \phi {x \circ_1 y} +_2 \phi \paren {x \circ_1 z}
+      | c = 
+}}
+{{eqn | r = \paren {\map \phi x \circ_2 \map \phi y} +_2 \paren {\map \phi x \circ_2 \map \phi z}
+      | c = 
+}}
+{{end-eqn}}
+So $\circ_2$ is [[Definition:Left Distributive Operation|left distributive]] over $+_2$.
+{{qed}}
+=== Right Distributivity ===
+Suppose $\circ_1$ is [[Definition:Right Distributive Operation|right distributive]] over $+_1$. Then:
+{{begin-eqn}}
+{{eqn | l = \paren {\map \phi x +_2 \map \phi y} \circ_2 \map \phi z
+      | r = \map \phi {x +_1 y} \circ_2 \map \phi z
+      | c = 
+}}
+{{eqn | r = \map \phi {\paren {x +_1 y} \circ_1 z}
+      | c = 
+}}
+{{eqn | r = \map \phi {\paren {x \circ_1 z} +_1 \paren {y \circ_1 z} }
+      | c = as $\circ_1$ is [[Definition:Right Distributive Operation|right distributive]] over $+_1$
+}}
+{{eqn | r = \map \phi {x \circ_1 z} +_2 \map \phi {y \circ_1 z}
+      | c = 
+}}
+{{eqn | r = \paren {\map \phi x \circ_2 \map \phi z} +_2 \paren {\map \phi y \circ_2 \map \phi z}
+      | c = 
+}}
+{{end-eqn}}
+So $\circ_2$ is [[Definition:Right Distributive Operation|right distributive]] over $+_2$.
+{{qed}}
+=== Distributivity ===
+If $\circ_1$ is [[Definition:Distributive Operation|distributive]] over $+_1$, then it is both [[Definition:Right Distributive Operation|right]] and [[Definition:Left Distributive Operation|left distributive]] over $+_1$.
+Hence from the above, $\circ_2$ is both [[Definition:Right Distributive Operation|right]] and [[Definition:Left Distributive Operation|left distributive]] over $+_2$.
+That is, $\circ_2$ is [[Definition:Distributive Operation|distributive]] over $+_2$.
+{{qed}}
+[[Category:Epimorphisms]]
+[[Category:Distributive Operations]]
+ovqap1zaxej3ypd2qv9glmydamx4bfb
+\end{proof}<|endoftext|>
+\section{Epimorphism Preserves Groups}
+Tags: Group Epimorphisms
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure|algebraic structures]].
+Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
+Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
+Then $\struct {T, *}$ is also a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+From [[Epimorphism Preserves Semigroups]], if $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]] then so is $\struct {T, *}$.
+From [[Epimorphism Preserves Identity]], if $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]] $e_S$, then $\map \phi {e_S}$ is the [[Definition:Identity Element|identity]] for $*$.
+From [[Epimorphism Preserves Inverses]], if $x^{-1}$ is an [[Definition:Inverse Element|inverse]] of $x$ for $\circ$, then $\map \phi {x^{-1} }$ is an [[Definition:Inverse Element|inverse]] of $\map \phi x$ for $*$.
+The result follows from the definition of [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity of Composite with Inclusion}
+Tags: Topological Subspaces, Continuous Mappings, Inclusion Mappings, Composite Mappings
+
+\begin{theorem}
+Let $T = \left({A, \tau}\right)$ and $T' = \left({A', \tau'}\right)$ be [[Definition:Topological Space|topological spaces]].
+Let $H \subseteq A$.
+Let $T_H = \left({H, \tau_H}\right)$ be a [[Definition:Topological Subspace|topological subspace]] of $T$.
+Let $i: H \to A$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
+Let $f: A \to A'$ and $g: A' \to H$ be [[Definition:Mapping|mappings]].
+Then the following apply:
+\end{theorem}<|endoftext|>
+\section{Totally Bounded Metric Space is Bounded}
+Tags: Bounded Metric Spaces, Totally Bounded Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Totally Bounded Metric Space|totally bounded metric space]].
+Then $M$ is [[Definition:Bounded Metric Space|bounded]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be [[Definition:Totally Bounded Metric Space|totally bounded]].
+Then there exist $n \in \N$ and points $x_0, \dots, x_n \in A$ such that:
+:$\displaystyle \inf_{0 \mathop \le i \mathop \le n} \map d {x_i, x} \le 1$
+for all $x \in A$.
+Let us set:
+:$a := x_0$
+:$\displaystyle D := \max_{0 \mathop \le i \mathop \le n} \map d {x_0, x_i}$
+:$K := D + 1$
+Now let $x \in A$ be arbitrary.
+Then by assumption there exists $i$ such that $\map d {x_i, x} \le 1$.
+Hence:
+:$\map d {a, x} \le \map d {a, x_i} + \map d {x_i, x} \le 1 + D = K$
+So $M$ is [[Definition:Bounded Metric Space|bounded]], as claimed.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Subsequence of Cauchy Sequence}
+Tags: Cauchy Sequences
+
+\begin{theorem}
+Let $\left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $\left\langle{x_n}\right\rangle_{n \in \N}$ be a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]] in $A$. 
+Let $x \in A$.
+Then $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$ {{iff}} it has a [[Definition:Subsequence|subsequence]] that [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+If $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$, it trivially follows that $\left\langle{x_n}\right\rangle$ is a [[Definition:Subsequence|subsequence]] of itself that [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $\left\langle{x_{n_k}}\right\rangle$ is a [[Definition:Subsequence|subsequence]] of $\left\langle{x_n}\right\rangle$ that [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$.
+Let $\epsilon$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By the definition of a [[Definition:Cauchy Sequence (Metric Space)|Cauchy sequence]], there exists a [[Definition:Real Number|real number]] $M$ such that:
+:$\displaystyle \forall i, j \in \N: i, j > M \implies d \left({x_i, x_j}\right) < \frac \epsilon 2$
+By the definition of [[Definition:Convergent Sequence (Metric Space)|convergence]], there exists a [[Definition:Real Number|real number]] $N$ such that:
+:$\displaystyle \forall k \in \N: k > N \implies d \left({x_{n_k}, x}\right) < \frac \epsilon 2$
+By the [[Archimedean Principle|Archimedean principle]], there exists a [[Definition:Natural Numbers|natural number]] $K > \max \left\{{M, N}\right\}$.
+Therefore, by the [[Definition:Triangle Inequality|triangle inequality]]:
+:$\displaystyle \forall m \in \N: m > K \implies d \left({x_m, x}\right) \le d \left({x_m, x_K}\right) + d \left({x_K, x}\right) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
+That is, $\left\langle{x_n}\right\rangle$ [[Definition:Convergent Sequence (Metric Space)|converges]] to $x$.
+{{qed}}
+[[Category:Cauchy Sequences]]
+htq125kdcv946ebxpiasn2vm0n3eovm
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets}
+Tags: Sigma-Algebras
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Let $\mathcal G \subseteq \powerset X$ be a collection of [[Definition:Subset|subsets]] of $X$.
+Then $\map \sigma {\mathcal G}$, the [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebra generated by $\mathcal G$]], exists and is unique.
+\end{theorem}
+
+\begin{proof}
+=== Existence ===
+By [[Power Set is Sigma-Algebra]], there is at least one [[Definition:Sigma-Algebra|$\sigma$-algebra]] containing $\mathcal G$.
+Next, let $\Bbb E$ be the collection of [[Definition:Sigma-Algebra|$\sigma$-algebras]] containing $\mathcal G$:
+:$\Bbb E := \set {\Sigma': \mathcal G \subseteq \Sigma', \text{$\Sigma'$ is a $\sigma$-algebra} }$
+By [[Intersection of Sigma-Algebras]], $\Sigma := \bigcap \Bbb E$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]].
+Also, by [[Set Intersection Preserves Subsets/Families of Sets|Set Intersection Preserves Subsets]]:
+:$\mathcal G \subseteq \Sigma$
+Now let $\Sigma'$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] containing $\mathcal G$.
+By construction of $\Sigma$, and [[Intersection is Subset/General Result|Intersection is Subset: General Result]]:
+:$\Sigma \subseteq \Sigma'$
+{{qed|lemma}}
+=== Uniqueness ===
+Suppose both $\Sigma_1$ and $\Sigma_2$ are [[Definition:Sigma-Algebra Generated by Collection of Subsets|$\sigma$-algebras generated by $\mathcal G$]].
+Then property $(2)$ for these [[Definition:Sigma-Algebra|$\sigma$-algebras]] implies both $\Sigma_1 \subseteq \Sigma_2$ and $\Sigma_2 \subseteq \Sigma_1$. 
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\Sigma_1 = \Sigma_2$
+{{qed}}
+{{expand|Add proof considering $\Sigma$ as a magma of sets}}
+\end{proof}<|endoftext|>
+\section{Basis for Topology on Cartesian Product}
+Tags: Product Spaces
+
+\begin{theorem}
+Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $A_1 \times A_2$ be the [[Definition:Cartesian Product|Cartesian product]] of $A_1$ and $A_2$.
+Let $\PP = \set{ U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$
+Then $\PP$ is a [[Definition:Synthetic Basis|synthetic basis]] on $A_1 \times A_2$.
+\end{theorem}
+
+\begin{proof}
+We need to show that conditions '''B1''' and '''B2''' in the definition for [[Definition:Synthetic Basis|basis]] hold for $\PP$.
+:'''B1''': Since $A_1 \in \tau_1$ and $A_2 \in \tau_2$, $A_1 \times A_2 \in \PP$.
+:'''B2''': Suppose $U_1, V_1 \in \tau_1$ and $U_2, V_2 \in \tau_2$.
+From [[Cartesian Product of Intersections]], $\paren{U_1 \times U_2} \cap \paren{V_1 \times V_2} = \paren{U_1 \cap V_1} \times \paren{U_2 \cap V_2}$.
+Since $\paren{U_1 \cap V_1} \in \tau_1$ and $\paren{U_2 \cap V_2} \in \tau_2$, we have $\paren{U_1 \cap V_1} \times \paren{U_2 \cap V_2} \in \PP$.
+So $\paren{U_1 \times U_2} \cap \paren{V_1 \times V_2}$ is the [[Definition:Set Union|union]] of (one) set in $\PP$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection from Product Topology is Continuous}
+Tags: Tychonoff Topology, Continuous Mappings, Projections, Projection from Product Topology is Open and Continuous
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $T = \struct {T_1 \times T_2, \tau}$ be the [[Definition:Product Space (Topology)|product space]] of $T_1$ and $T_2$, where $\tau$ is the [[Definition:Tychonoff Topology|Tychonoff topology]] on $S$.
+Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the [[Definition:Projection (Mapping Theory)|first and second projections]] from $T$ onto its [[Definition:Factor Space|factors]].
+Then both $\pr_1$ and $\pr_2$ are are [[Definition:Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+From [[Natural Basis of Tychonoff Topology/Finite Product|Natural Basis of Tychonoff Topology:Finite Product]], a [[Definition:Synthetic Basis|basis]] for $\tau$ is:
+:$\BB = \set {U \times V: U \in \tau_1, V \in \tau_2}$
+et $U$ be [[Definition:Open Set (Topology)|open]] in $T_1$.
+Then $\map {\pr_1^{-1} } U = U \times T_2$ is one of the [[Definition:Open Set (Topology)|open sets]] in the [[Definition:Basis (Topology)|basis]] in the definition of [[Definition:Product Space (Topology)|product topology]].
+Thus $\pr_1$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+The same argument can be applied to $\pr_2$.
+\end{proof}<|endoftext|>
+\section{Continuous Mapping to Topological Product}
+Tags: Product Spaces, Continuous Mappings
+
+\begin{theorem}
+Let $T = T_1 \times T_2$ be a [[Definition:Product Space (Topology)|product space]] of two [[Definition:Topological Space|topological spaces]] $T_1$ and $T_2$.
+Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the [[Definition:Projection (Mapping Theory)|first and second projections]] from $T$ onto its [[Definition:Factor Space|factors]].
+Let $T'$ be a [[Definition:Topological Space|topological space]].
+Let $f: T' \to T$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}} $\pr_1 \circ f$ and $\pr_2 \circ f$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $f$ be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Then by:
+:[[Projection from Product Topology is Continuous]]
+:[[Composite of Continuous Mappings is Continuous]]
+so are $\pr_1 \circ f$ and $\pr_2 \circ f$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $\pr_1 \circ f$ and $\pr_2 \circ f$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Let $U_1 \subseteq T_1$ and $U_2 \subseteq T_2$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \map {f^{-1} } {U_1 \times U_2}
+      | r = \map {f^{-1} } {\paren{U_1 \times T_2} \cap \paren{T_1 \times U_2} }
+      | c = [[Cartesian Product of Intersections]]
+}}
+{{eqn | r = \map {f^{-1} } {U_1 \times T_2} \cap \map {f^{-1} } {T_1 \times U_2}
+      | c = 
+}}
+{{eqn | r = \map {f^{-1} } {\map {\pr_1^{-1} } {U_1} } \cap \map {f^{-1} } {\map {\pr_2^{-1} } {U_2} }
+      | c = 
+}}
+{{eqn | r = \map {\paren{\pr_1 \circ f}^{-1} } {U_1} \cap \map {\paren{\pr_2 \circ f}^{-1} } {U_2}
+      | c = 
+}}
+{{end-eqn}}
+Let $U_1$ and $U_2$ be [[Definition:Open Set (Topology)|open]] in $T_1$ and $T_2$ respectively.
+Then by [[Definition:Everywhere Continuous Mapping (Topology)|continuity]] of $\pr_1 \circ f$ and $\pr_2 \circ f$:
+:$\map {\paren{\pr_1 \circ f}^{-1} } {U_1}$ and $\map {\paren{\pr_2 \circ f}^{-1} } {U_2}$ are open in $T'$.
+So $\map {f^{-1} } {U_1 \times U_2}$ is [[Definition:Open Set (Topology)|open]] in $T'$.
+From [[Natural Basis of Tychonoff Topology of Finite Product]], a [[Definition:Synthetic Basis|basis]] for the [[Definition:Product Space (Topology)|product space]] $T$ is:
+:$\BB = \set{U_1 \times U_2 : U_1 \text{ is open in } T_1, U_2 \text{ is open in } T_2}$
+Then we have shown that:
+:$\forall B \in \BB : \map {f^{-1}} B$ is [[Definition:Open Set (Topology)|open]] in $T'$
+ 
+It follows from [[Continuity Test using Basis]] that $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open and Closed Sets in Topological Space}
+Tags: Clopen Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then $S$ and $\O$ are both [[Definition:Clopen Set|both open and closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Closed Set (Topology)|closed set]], $U$ is [[Definition:Open Set (Topology)|open]] in $T$ {{iff}} $S \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Underlying Set of Topological Space is Clopen]], $S$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] in $T$.
+From [[Empty Set is Element of Topology]], $\O$ is [[Definition:Open Set (Topology)|open]] in $T$.
+From [[Empty Set is Closed in Topological Space]], we have that $\O$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Continuous Real Function is Continuous}
+Tags: Continuous Real Functions, Absolutely Continuous Functions
+
+\begin{theorem}
+Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
+Let $f : I \to \R$ be an [[Definition:Absolute Continuity|absolutely continuous]] [[Definition:Real Function|function]].
+Then $f$ is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+From [[Absolutely Continuous Real Function is Uniformly Continuous]]:
+:$f$ is [[Definition:Uniform Continuity/Real Numbers|uniformly continuous]].
+Therefore, from [[Uniformly Continuous Function is Continuous/Real Function|Uniformly Continuous Real Function is Continuous]]:
+:$f$ is [[Definition:Continuous Real Function|continuous]].
+{{qed}}
+[[Category:Continuous Real Functions]]
+[[Category:Absolutely Continuous Functions]]
+n2cjhawz8cn1ke0wlnevycp12gjkgkt
+\end{proof}<|endoftext|>
+\section{Closed Set in Topological Subspace}
+Tags: Closed Sets, Topological Subspaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $T' \subseteq T$ be a [[Definition:Topological Subspace|subspace]] of $T$.
+Then $V \subseteq T'$ is [[Definition:Closed Set (Topology)|closed]] in $T'$ {{iff}} $V = T' \cap W$ for some $W$ [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $V \subseteq T'$ is [[Definition:Closed Set (Topology)|closed]] in $T'$.
+Then $T' \setminus V$ is [[Definition:Open Set (Topology)|open]] in $T'$ by definition.
+So, by definition of [[Definition:Topological Subspace|subspace topology]], $T' \setminus V = T' \cap U$ for some $U$ [[Definition:Open Set (Topology)|open]] in $T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = V
+      | r = T' \setminus \left({T' \setminus V}\right)
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | r = T' \setminus \left({T' \cap U}\right)
+      | c = from above
+}}
+{{eqn | r = T' \setminus U
+      | c = [[Set Difference with Intersection is Difference]]
+}}
+{{eqn | r = T' \cap \left({T \setminus U}\right)
+      | c = 
+}}
+{{end-eqn}}
+Thus $T \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Conversely, suppose $V = T' \cap W$ where $W$ [[Definition:Closed Set (Topology)|closed]] in $T$.
+Then $T' \setminus V = T' \setminus \left({T' \cap W}\right) = T' \cap \left({T \setminus W}\right)$ which is [[Definition:Open Set (Topology)|open]] in $T'$.
+So $V$ is [[Definition:Closed Set (Topology)|closed]] in $T'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuity Defined from Closed Sets}
+Tags: Closed Sets, Continuous Mappings
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]].
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] {{iff}} for all $V$ [[Definition:Closed Set (Topology)|closed]] in $T_2$, $f^{-1} \sqbrk V$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+\end{theorem}
+
+\begin{proof}
+First we show the following.
+Let $W \in T_2$.
+We note that $f^{-1} \sqbrk T_2 = T_1$.
+Hence, from [[Preimage of Set Difference under Mapping]], we have:
+:$f^{-1} \sqbrk {T_2 \setminus W} = T_1 \setminus f^{-1} \sqbrk W$
+=== Necessary Condition ===
+Suppose the condition on [[Definition:Closed Set (Topology)|closed sets]] holds.
+Let $U$ be [[Definition:Open Set (Topology)|open]] in $T_2$.
+By [[Relative Complement of Relative Complement]], $T_2 \setminus \paren { T_2 \setminus U } = U$.
+Therefore $T_2 \setminus \paren { T_2 \setminus U }$ is [[Definition:Open Set (Topology)|open]] in $T_2$.
+Then $T_2 \setminus U$ is [[Definition:Closed Set (Topology)|closed]] in $T_2$.
+By hypothesis, $f^{-1} \sqbrk {T_2 \setminus U} = T_1 \setminus f^{-1} \sqbrk U$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+So $f^{-1} \sqbrk U$ is [[Definition:Open Set (Topology)|open]] in $T_1$.
+This is true for any $U \in T_2$.
+Hence $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now let $f$ be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Let $V$ be [[Definition:Closed Set (Topology)|closed]] in $T_2$.
+Then $T_2 \setminus V$ is [[Definition:Open Set (Topology)|open]] in $T_2$.
+As $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], $f^{-1} \sqbrk {T_2 \setminus V} = T_1 \setminus f^{-1} \sqbrk V$ is [[Definition:Open Set (Topology)|open]] in $T_1$.
+So $f^{-1} \sqbrk V$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers are Uncountable}
+Tags: Real Numbers, Uncountable Sets, Real Numbers are Uncountable
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Uncountable Set|uncountably infinite]].
+\end{theorem}
+
+\begin{proof}
+We prove the equivalent result that every [[Definition:Sequence|sequence]] $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$.
+Let $\sequence {x_k}_{k \mathop \in \N}$ be a [[Definition:Sequence|sequence]] of [[Definition:Distinct|distinct]] [[Definition:Real Number|real numbers]].
+Let a [[Definition:Sequence|sequence]] of [[Definition:Closed Real Interval|closed real intervals]] $\sequence {I_n}$ be defined as follows:
+Let:
+:$a_k = \min \set {x_k, x_{k + 1} }$
+:$b_k = \max \set {x_k, x_{k + 1} }$
+and:
+:$I_k = \closedint {a_k} {b_k}$
+Since the [[Definition:Term of Sequence|terms]] of $\sequence {x_k}_{k \mathop \in \N}$ are [[Definition:Distinct|distinct]], $a_k \ne b_k$.
+Thus $I_k$ is not a [[Definition:Singleton|singleton]].
+Let:
+:$I_{n - 1} = \closedint {a_{n - 1} } {b_{n - 1} }$
+It can be assumed that infinitely many of the $x_k$ lie [[Definition:Interior (Topology)|inside]] $I_{n - 1}$.
+Otherwise the proof is complete.
+Let $y$ and $z$ be the first two such [[Definition:Term of Sequence|terms]] of $\sequence {x_k}_{k \mathop \in \N}$.
+Let:
+:$a_n = \min \set {y, z}$
+:$b_n = \max \set {y, z}$
+:$I_n = \closedint {a_n} {b_n}$
+Thus we have sequences:
+:$\sequence {a_k}_{k \mathop \in \N}$
+:$\sequence {b_k}_{k \mathop \in \N}$
+with:
+:$ a_1 < a_2 < \cdots < b_2 < b_1$
+So $\sequence {a_k}_{k \mathop \in \N}$ and $\sequence {b_k}_{k \mathop \in \N}$ are [[Definition:Monotone Sequence|monotone]], and [[Definition:Bounded Above Sequence|bounded above]] and [[Definition:Bounded Below Sequence|bounded below]] respectively.
+Therefore by the [[Monotone Convergence Theorem (Real Analysis)]] both are [[Definition:Convergent Real Sequence|convergent]].
+Let:
+:$\displaystyle A = \lim_{k \mathop \to \infty} a_k$
+:$\displaystyle B = \lim_{k \mathop \to \infty} b_k$
+Clearly we have $A \le B$.
+So:
+:$\closedint A B \ne \O$
+Let $h \in \closedint A B$.
+Then:
+:$h \ne a_k, b_k$ for all $k$.
+We claim that $h \ne x_k$ for all $k$. 
+Suppose that $h = x_k$ for some $k$. 
+Then there are only finitely many points in the sequence before $h$ occurs.
+{{Explain|"the sequence" -- which one in particular? There are a few under discussion here.}}
+Therefore only finitely many of the $a_k$ precede $h$.
+Let $a_d$ be the last element of the sequence $\sequence {a_k}_{k \mathop \in \N}$ preceding $h$.
+We defined $a_{d + 1}$, $b_{d + 1}$ to be interior points of $I_d$, and also $h \in I_{d + 1}$ by the definition of $h$.
+{{explain|Where are those interior points so defined?}}
+Therefore $a_{d + 1}$ must precede $h$ in the sequence, for the sequence is monotone increasing.
+This is a contradiction.
+{{explain|Specify precisely what assumption this contradiction falsifies.}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, a [[Definition:Perfect Set|perfect set]] is a [[Definition:Set|set]] $X$ such that every point $x \in X$ is the [[Definition:Limit of Sequence|limit of a sequence]] of points of $X$ distinct from $x$.
+From [[Real Numbers form Perfect Set]], $\R$ is [[Definition:Perfect Set|perfect]].
+Therefore it is sufficient to show that a [[Definition:Perfect Set|perfect]] [[Definition:Subset|subset]] of $X \subseteq \R^k$ is [[Definition:Uncountable Set|uncountable]].
+We prove the equivalent result that every [[Definition:Sequence|sequence]] $\sequence {x_k}_{k \mathop \in \N} \subseteq X$ omits at least one point in $X$.
+Let $y_1 \in X$.
+Let $B_1:= \map {\BB_r} {y_1}$ be a [[Definition:Closed Ball|closed ball]] centred at $y_1$.
+Consider a [[Definition:Closed Ball|closed ball]]:
+:$B_{n - 1} \supseteq B_n := \map {\BB_{\map \delta n} } {y_n}$
+such that:
+:$\map \delta n \le \dfrac {\map \delta {n - 1} } 2$
+:$y_n \in X$
+:$x_n \notin B_n$
+Note that $\map \delta 1 = r$.
+We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every ball centred at a point of $X$ contains infinitely many points of $X$.
+Since $\map \delta n \le \dfrac r {2^{n - 1} }$, $y_n$ is [[Definition:Cauchy Sequence|Cauchy]].
+Therefore from [[Perfect Set is Closed]], we may let $\displaystyle Y = \lim_{n \mathop \to \infty} y_n \in X$.
+For $n \in \N$ we have:
+:$\set {y_m: m > n} \subseteq B_n$
+so $Y \in B_n$.
+But by construction:
+:$\forall n \in \N: x_n \notin B_n$
+Therefore:
+:$\forall n \in \N: Y \ne x_n$
+and the proof is complete.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Surjection from Natural Numbers iff Countable]], a set $A$ is [[Definition:Countable Set|countable]] {{iff}} there exists a [[Definition:Surjection|surjection]] $f: \N \to A$.
+Suppose there exists a [[Definition:Surjection|surjection]] $f: \N \to \R$.
+Then $\forall x \in \R: \exists n \in \N: f \left({n}\right) = x$ as $f$ is [[Definition:Surjection|surjective]].
+Let $d_{n, 0}$ be the [[Definition:Integer|integer]] before the [[Definition:Decimal Point|decimal point]] of $f \left({n}\right)$.
+Similarly, for all $m > 0$, let $d_{n, m}$ be the $m$th [[Definition:Digit|digit]] in the [[Definition:Decimal Expansion|decimal expansion]] of $f \left({n}\right)$.
+Let $e_0$ be an [[Definition:Integer|integer]] different from $d_{0, 0}$.
+Similarly, for all $m > 0$, let $e_m$ be an [[Definition:Integer|integer]] different from $d_{m, m}$.
+Specifically, we can define $e_0$ to be $d_{0, 0} + 1$, and:
+:$e_m = \begin{cases}
+1 & : d_{m, m} \ne 1 \\
+2 & : d_{m, m} = 1
+\end{cases}$
+Now consider the [[Definition:Real Number|real number]] $\displaystyle x = e_0 + \sum_{n \mathop = 1}^\infty \frac {e_n} {10^n}$.
+Its [[Definition:Decimal Expansion|decimal expansion]] is:
+:$x = \left[{e_0 . e_1 e_2 e_3 \ldots}\right]_{10}$
+Since $e_0 \ne d_{0, 0}$, $x \ne f \left({0}\right)$.
+Similarly, for each $n \in \N$ such that $n \ge 1$, we have that $e_n \ne d_{n, n}$ and so $x \ne f \left({n}\right)$.
+Thus $x$ is a [[Definition:Real Number|real number]] which is not in the set $\left\{{f \left({n}\right): n \in \N}\right\}$.
+Hence $f$ can not be [[Definition:Surjection|surjective]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Cantor's Theorem]] there is no [[Definition:Surjection|surjection]]:
+: $\N \twoheadrightarrow \mathcal P \left({\N}\right)$
+Additionally, we have [[Power Set of Natural Numbers is not Countable]].
+Therefore, if we can show that $\mathcal P \left({\N}\right)$ [[Definition:Injection|injects]] into $\R$ then there is no [[Definition:Injection|injection]] $\R \hookrightarrow \N$ and $\R$ is uncountable.
+To prove the theorem we construct an [[Definition:Injection|injection]] $f: \mathcal P \left({\N}\right) \to \R$.
+For a subset $S \subseteq \N$, let $\chi_S$ be the [[Definition:Characteristic Function of Set|characteristic function]] of $S$, and let $d_i = \chi_S \left({i}\right)$ for all $i \in \N$.
+By the definition of [[Definition:Characteristic Function of Set|characteristic function]], $\left\langle{d_i}\right\rangle_{i \in \N}$ is an [[Definition:Infinite Sequence|infinite sequence]] of $1$s and $0$s.
+There are two cases: $\left\langle{d_i}\right\rangle_{i \in \N}$ terminates in an [[Definition:Infinite Sequence|infinite sequence]] of $1$s, or it does not.
+Suppose $\left\langle{d_i}\right\rangle_{i \in \N}$ does ''not'' terminate in an [[Definition:Infinite Sequence|infinite sequence]] of $1$s.
+Then $f \left({S}\right)$ is the [[Definition:Binary Notation|binary]] [[Definition:Basis Expansion|expansion]] of the following number in $\left[{0 \,.\,.\, 1}\right)$:
+:$0.d_1 d_2 d_3 d_4 \ldots$
+Otherwise $\left\langle{d_i}\right\rangle_{i \in \N}$ ''does'' terminate in an [[Definition:Infinite Sequence|infinite sequence]] of $1$s.
+Then $f \left({S}\right)$ is the [[Definition:Integer|integer]] expressed in [[Definition:Binary Notation|binary]] as:
+:$1 d_1 d_2 d_3 \ldots d_k$
+where $d_k$ is the last member of the sequence not equal to $1$.
+In either case, every [[Definition:Subset|subset]] of $\N$, that is, element of $\mathcal P \left({\N}\right)$, is mapped to an element of $\R$.
+That $f$ is an [[Definition:Injection|injection]] follows from the uniqueness statement of [[Existence of Base-N Representation]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+It is sufficient to show that the [[Definition:Real Interval|real interval]] $I = \set {x \in \R: 0 < x \le 1}$ is [[Definition:Uncountable Set|uncountable]].
+Let $x \in I$.
+From [[Existence of Base-N Representation]], $x$ has a [[Definition:Unique|unique]] representation of the form:
+:$x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$ 
+where $\epsilon_k = 0, 1$ or $2$ and an [[Definition:Infinite Set|infinite]] number of $\epsilon_k$ are different from $0$.
+Let $S \subseteq I$ be [[Definition:Countably Infinite Set|countably infinite]].
+Let $S = \set {x_1, x_2, \ldots}$.
+Let $\epsilon_{k 1}, \epsilon_{k 2}, \epsilon_{k 3}, \ldots$ be the [[Definition:Ternary Notation|ternary digits]] of $x_k$.
+Let $\epsilon_k = 1 + 2 \epsilon_{k k} - \epsilon_{k k}^2$ so that:
+:$\epsilon_k = 1$ if $\epsilon_{k k} = 0$ or $\epsilon_{k k} = 2$
+:$\epsilon_k = 2$ if $\epsilon_{k k} = 1$
+Then:
+:$(1): \quad \forall k: \epsilon_k \ne 0$
+:$(2): \quad \forall k: \epsilon_k \ne \epsilon_{k k}$
+We have that $0 < x \le 1$ so $x \in I$.
+But the [[Definition:Real Number|real number]] $\displaystyle x = \sum \epsilon_k 3^{-k}$ is different from every $x_k \in I$.
+Thus we have found an [[Definition:Element|element]] of $I$ which is not an [[Definition:Element|element]] of $S$.
+Therefore $S$ is a [[Definition:Proper Subset|proper subset]] of $I$.
+It follows by definition that $I$ is [[Definition:Uncountable Set|uncountable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Define a [[Definition:Mapping|mapping]] $f: \powerset {\N_{>0} } \to \R$ thus:
+:$\map f S = 0.d_1 d_2 \ldots$, interpreted as a [[Definition:Ternary Notation|ternary expansion]] where $\sequence {d_n}$ is the characteristic function of $S$.
+That is:
+:$\displaystyle \map f S = \sum_{i \mathop \in S} 3^{-i}$
+By the [[Real Numbers are Uncountable/Proof 2 using Ternary Notation/Lemma|lemma]], $f$ is an [[Definition:Injection|injection]].
+{{AimForCont}} that $\R$ is [[Definition:Countable Set|countable]].
+Then there is an [[Definition:Injection|injection]] $g: \R \to \N$.
+By [[Composite of Injections is Injection]], $g \circ f: \powerset \N \to \N$ is an [[Definition:Injection|injection]].
+But this [[Definition:Contradiction|contradicts]] [[No Injection from Power Set to Set]].
+Hence, by [[Proof by Contradiction]], $\R$ is not [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Point being in Closure}
+Tags: Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $x \in T$.
+Then $x \in H^-$ {{iff}} every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $x$ contains a point in $H$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Closure (Topology)|closure]], we have that $H^-$ is the [[Definition:Set Union|union]] of $H$ and all the [[Definition:Limit Point of Set|limit points]] of $H$ in $T$.
+=== Necessary Condition ===
+Suppose $x \in H^-$.
+Then either:
+:$(1): \quad x \in H$, in which case every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $x$ trivially contains a point in $H$ (that is, $x$ itself);
+:$(2): \quad x$ is a [[Definition:Limit Point of Set|limit point]] of $H$ in $T$.
+If the latter is the case, then it follows directly from the definition of [[Definition:Limit Point of Set/Definition 1|limit point]] that every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $x$ contains a point in $H$ other than $x$.
+=== Sufficient Condition ===
+Suppose that every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $x$ contains a point in $H$.
+If $x \in H$, then $x$ is in the [[Definition:Set Union|union]] of $H$ and all the [[Definition:Limit Point of Set|limit points]] of $H$ in $T$.
+Hence by definition $x \in H^-$.
+If $x \notin H$ then $x$ must be a [[Definition:Limit Point of Set|limit point]] of $H$ by definition.
+So again, $x$ is in the [[Definition:Set Union|union]] of $H$ and all the [[Definition:Limit Point of Set|limit points]] of $H$ in $T$.
+Hence by definition $x \in H^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Closure of Subset is Subset of Topological Closure}
+Tags: Subsets, Set Closures, Topological Closure of Subset is Subset of Topological Closure
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq K$ and $K \subseteq S$.
+Then:
+:$\map \cl H \subseteq \map \cl K$
+where $\map \cl H$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Closure (Topology)|closure]]:
+:$\map \cl H$ is the [[Definition:Set Union|union]] of $H$ and its [[Definition:Limit Point of Set|limit points]].
+Let $x \in \map \cl H$.
+If $x \in H$ then $x \in K \implies x \in \map \cl K$.
+Otherwise $x$ is a [[Definition:Limit Point of Set|limit point]] of $H$.
+That is, every [[Definition:Open Set (Topology)|open set]] $U$ of $T$ such that $x \in U$ contains $y \in H$ such that $y \ne x$.
+But as $y \in H$ it follows that $y \in K$.
+So every [[Definition:Open Set (Topology)|open set]] $U$ of $T$ such that $x \in U$ contains $y \in K$ such that $y \ne x$.
+This is the definition for a [[Definition:Limit Point of Set|limit point]] of $K$.
+Thus $x \in \map \cl K$.
+{{ExtractTheorem|Extract the above paragraph into something like "Limit Point of Subset is Limit Point" (which proves that $H \subseteq K \implies H' \subseteq K'$). Then use [[Set Union Preserves Subsets]] to prove this result.}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Topological Closure is Closed]], $\map \cl K$ is [[Definition:Closed Set (Topology)|closed]].
+From [[Set is Subset of its Topological Closure]]:
+:$K \subseteq \map \cl K$
+By [[Subset Relation is Transitive]], it follows that:
+:$H \subseteq \map \cl K$
+Hence, by definition of [[Definition:Closure (Topology)/Definition 3|closure]] as the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Closed Set (Topology)|closed set]] that [[Definition:Subset|contains]] $H$:
+:$\map \cl H \subseteq \map \cl K$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Topological Closure equals Closure}
+Tags: Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Then:
+: $\left({H^-}\right)^- = H^-$
+where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+It follows directly from [[Set is Subset of its Topological Closure]] that:
+: $H^- \subseteq \left({H^-}\right)^-$
+{{qed|lemma}}
+Let $x \in \left({H^-}\right)^-$.
+Then from [[Condition for Point being in Closure]], any $U$ which is [[Definition:Open Set (Topology)|open]] in $T$ such that $x \in U$ contains some $y \in H^-$.
+If we consider $U$ as an [[Definition:Open Set (Topology)|open set]] containing $y$, it follows that:
+: $U \cap H \ne \varnothing$
+Hence $x \in H^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Closure is Closed}
+Tags: Set Closures, Closed Sets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Then the [[Definition:Closure (Topology)|closure]] $\map \cl H$ of $H$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+\end{theorem}
+
+\begin{proof}
+From [[Closure of Topological Closure equals Closure]]:
+:$\map \cl {\map \cl H} = \map \cl H$
+From [[Set is Closed iff Equals Topological Closure]], it follows that $\map \cl H$ is [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Closure of Topological Subspace}
+Tags: Set Closures
+
+\begin{theorem}
+{{TFAE|def = Closure (Topology)|view = Closure|context = Topology (Mathematical Branch)|contextview = Topology}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+\end{theorem}
+
+\begin{proof}
+=== Definition $1$ is equivalent to Definition $2$ ===
+This is proved in [[Set Closure as Intersection of Closed Sets]].
+{{qed|lemma}}
+=== Definition $2$ is equivalent to Definition $3$ ===
+This is proved in [[Set Closure is Smallest Closed Set in Topological Space]].
+{{qed|lemma}}
+=== Definition $1$ is equivalent to Definition $4$ ===
+By the definition of the [[Definition:Interior (Topology)|interior]], and [[Set is Subset of its Topological Closure]], it easily follows that
+:$H^\circ \subseteq H \subseteq H^-$
+Then:
+{{begin-eqn}}
+{{eqn | l = H \cup \partial H
+      | r = H \cup \paren {H^- \setminus H^\circ} 
+      | c = {{Defof|Boundary (Topology)}}
+}}
+{{eqn | r = H \cup \paren {H \setminus H^\circ} \cup \paren {H^- \setminus H}
+      | c = [[Union of Relative Complements of Nested Subsets]]
+}}
+{{eqn | r = H \cup \paren {H^- \setminus H}
+      | c = [[Set Difference is Subset]] and [[Union with Superset is Superset]]
+}}
+{{eqn | r = H^-
+      | c = [[Union with Relative Complement]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Definition $1$ is equivalent to Definition $5$ ===
+Every [[Definition:Isolated Point (Topology)|isolated point]] of $H$ is a point of $H$.
+So by [[Set Union Preserves Subsets]]:
+:$(1): \quad H^i \cup H' \subseteq H \cup H'$
+If $S \setminus \paren {H^i \cup H'} = \O$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.
+Otherwise, let $x \in S \setminus \paren {H^i \cup H'} \ne \O$.
+From [[De Morgan's Laws (Set Theory)/Set Difference/Difference with Union|De Morgan's Laws]]:
+:$S \setminus \paren {H^i \cup H'} = \paren {S \setminus H^i} \cap \paren {S \setminus H'}$
+Thus $x$ is a point of $S$ that is neither an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ nor a [[Definition:Limit Point of Set|limit point]] of $H$.
+Then there exists an [[Definition:Open Set (Topology)|open set]] $U$ that contains $x$ such that:
+:$U \cap H \ne \set x$
+and:
+:$H \cap \paren {U \setminus \set x} = \O$
+This implies that
+:$U \cap H = \O$
+By [[Condition for Point being in Closure]]:
+:$x \notin H \cup H'$
+It has been shown that:
+:$x \notin H^i \cup H' \implies x \notin H \cup H'$
+From [[Rule of Transposition]]:
+:$x \in H \cup H' \implies x \in H^i \cup H'$
+And so
+:$(2): \quad H \cup H' \subseteq H^i \cup H'$
+Combining $(1)$ and $(2)$:
+:$H \cup H' = H^i \cup H'$
+Hence the result.
+{{qed|lemma}}
+=== Definition $1$ is equivalent to Definition $6$ ===
+By one of the equivalent definitions of an [[Definition:Adherent Point|adherent point]]:
+:A point $x \in S$ is an '''adherent point of $H$''' {{iff}} every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ satisfies $H \cap U \ne \O$
+From [[Condition for Point being in Closure]]:
+:$x \in H^-$ {{iff}} every [[Definition:Open Set (Topology)|open set]] of $T$ which contains $x$ contains a point in $H$
+where:
+:$H^-$ is the [[Definition:Set Union|union]] of $H$ and all the [[Definition:Limit Point of Set|limit points]] of $H$ in $T$
+The equivalence follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Intersection is Subset of Intersection of Closures}
+Tags: Set Intersection, Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\forall i \in I: H_i \subseteq T$.
+Then:
+: $\displaystyle \left({\bigcap_I H_i}\right)^- \subseteq \bigcap_I H_i^-$
+where $H_i^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H_i$.
+\end{theorem}
+
+\begin{proof}
+Since $\displaystyle \bigcap_I H_i^-$ is an intersection of [[Definition:Closed Set (Topology)|closed sets]], it is closed, from [[Topology Defined by Closed Sets]].
+Also, it contains $\displaystyle \bigcap_I H_i$ and so by the main definition of [[Definition:Closure (Topology)|closure]] also contains $\displaystyle \left({\bigcap_I H_i}\right)^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Finite Union equals Union of Closures}
+Tags: Set Union, Set Closures
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $n \in \N$.
+Let:
+:$\forall i \in \set {1, 2, \ldots, n}: H_i \subseteq T$
+Then:
+:$\displaystyle \bigcup_{i \mathop = 1}^n \map \cl {H_i} = \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
+\end{theorem}
+
+\begin{proof}
+From [[Closure of Union contains Union of Closures]] we have that:
+:$\displaystyle \bigcup_{i \mathop = 1}^n \map \cl {H_i} \subseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
+We need now to show that: 
+:$\displaystyle \bigcup_{i \mathop = 1}^n \map \cl {H_i} \supseteq \map \cl {\bigcup_{i \mathop = 1}^n H_i}$
+Let $\displaystyle K = \bigcup_{i \mathop = 1}^n \map \cl {H_i}$ and $\displaystyle H = \bigcup_{i \mathop = 1}^n H_i$.
+From [[Topological Closure is Closed]], all of $\map \cl {H_i}$ are [[Definition:Closed Set (Topology)|closed]].
+So $K$ is the [[Definition:Set Union|union]] of a [[Definition:Finite Set|finite number]] of [[Definition:Closed Set (Topology)|closed sets]].
+So $K$ is itself closed, from [[Topology Defined by Closed Sets]].
+From [[Set is Subset of its Topological Closure]]:
+:$\forall i \in \closedint 1 n: H_i \subseteq \map \cl {H_i}$
+and so $H \subseteq K$.
+So from [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\map \cl H \subseteq \map \cl K$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact First-Countable Space is Sequentially Compact}
+Tags: Sequences, Compact Spaces, First-Countable Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Compact Space|compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
+Then every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] [[Definition:Subsequence|subsequence]]; that is, $T$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $S$.
+{{AimForCont}} that $\sequence {x_n}_{n \mathop \in \N}$ does not have a [[Definition:Convergent Sequence (Topology)|convergent]] [[Definition:Subsequence|subsequence]].
+By [[Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit]], it follows that $\sequence {x_n}_{n \mathop \in \N}$ does not have an [[Definition:Accumulation Point of Sequence|accumulation point]] in $S$.
+Thus, for each $x \in S$, we can select an [[Definition:Open Set (Topology)|open set]] $U_x$ such that $x \in U_x$ and $U_x$ only contains $x_n$ for [[Definition:Finite Set|finitely many]] $n \in \N$.
+The [[Definition:Set|set]] $\UU = \set {U_x : x \in X}$ is clearly an [[Definition:Open Cover|open cover]] for $S$.
+{{wtd|The [[Axiom:Axiom of Choice|axiom of choice]] is implicitly invoked above}}
+By the definition of a [[Definition:Compact Space|compact space]], there exists a finite subcover $\set {U_{x_1}, U_{x_2}, \ldots, U_{x_m} }$ of $\UU$ for $X$.
+Then $U_{x_1} \cup U_{x_2} \cup \cdots \cup U_{x_m}$ contains all of $S$ since it is a cover.
+However, each [[Definition:Open Set (Topology)|open set]] in this union only contains $x_n$ for a finite number of $n$, and this is a finite union of sets.
+So the union only contains $x_n$ for finitely many $n$.
+This implies that $x_n \in X$ for only [[Definition:Finite Set|finitely many]] $n \in \N$.
+This is a [[Definition:Contradiction|contradiction]], since $\sequence {x_n}$ is a sequence in $S$, and so $x_n \in S$ for all $n \in \N$.
+{{qed}}
+{{Improve|To prove this in [[Definition:Zermelo-Fraenkel Set Theory|ZF]], one could simply use [[Compact Space is Countably Compact]] and [[Countably Compact First-Countable Space is Sequentially Compact]].}}
+[[Category:Sequences]]
+[[Category:Compact Spaces]]
+[[Category:First-Countable Spaces]]
+5slyzzaj1dyro8axdjxiz96vvkwg8hy
+\end{proof}
+
+\begin{proof}
+Follows directly from:
+: [[Infinite Sequence in Countably Compact Space has Accumulation Point]]
+: [[Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Countably Compact Space|countably compact]] [[Definition:First-Countable Space|first-countable]] [[Definition:Topological Space|topological space]].
+By definition of [[Definition:Sequentially Compact Space|sequentially compact]], it is sufficient to show that every [[Definition:Infinite Sequence|infinite sequence]] in $S$ has a [[Definition:Convergent Sequence (Topology)|convergent]] [[Definition:Subsequence|subsequence]].
+Let $\sequence {s_n}$ be any [[Definition:Infinite Sequence|sequence]] in $S$.
+By [[Infinite Sequence in Countably Compact Space has Accumulation Point]], $\sequence {s_n}$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] $p \in S$.
+As $T$ is [[Definition:First-Countable Space|first-countable]], $p$ has a [[Definition:Countable Set|countable]] [[Definition:Local Basis|local basis]], say:
+:$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$
+Then a [[Definition:Subsequence|subsequence]] $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, [[Definition:Convergent Sequence (Topology)|converges]] to $p$.
+{{handwaving|This last statement is "by intimidation"; justification needed}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Open Ball in Metric Space}
+Tags: Open Balls
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\map {B_\epsilon} x$ be an [[Definition:Open Ball|open $\epsilon$-ball]] in $M$.
+Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the [[Definition:Closure (Metric Space)|closure]] of $\map {B_\epsilon} x$.
+Then $\map d {x, y} \le \epsilon$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\map d {x, y} > \epsilon$.
+Then $\map {B_{\map d {x, y} - \epsilon} } y$ is an [[Definition:Open Set (Metric Space)|open set]] containing $y$ and not meeting $\map {B_\epsilon} x$.
+Hence:
+:$y \notin \map \cl {\map {B_\epsilon} x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nowhere Dense iff Complement of Closure is Everywhere Dense}
+Tags: Denseness, Set Closures
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Then $H$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}} $S \setminus H^-$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$, where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let:
+: $H^\circ$ denote the [[Definition:Interior (Topology)|interior]] of any $H \subseteq S$
+: $H^-$ denote the [[Definition:Closure (Topology)|closure]] of any $H \subseteq S$.
+From the definition, $H$ is [[Definition:Nowhere Dense|nowhere dense]] in $T$ {{iff}} $\paren {H^-}^\circ = \varnothing$.
+From the definition of [[Definition:Interior (Topology)|interior]], it follows that $\paren {H^-}^\circ = \varnothing$ {{iff}} every [[Definition:Open Set (Topology)|open set]] of $T$ contains a point of $S \setminus \paren {H^-}$.
+Thus $S \setminus \paren {H^-}$ is [[Definition:Everywhere Dense|everywhere dense]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergence in Indiscrete Space}
+Tags: Indiscrete Topology, Convergence
+
+\begin{theorem}
+Let $\left({S, \left\{{S, \varnothing}\right\}}\right)$ be an [[Definition:Indiscrete Topology|indiscrete space]].
+Let $\left \langle {x_n} \right \rangle$ be any [[Definition:Sequence|sequence in $S$]].
+Then $\left \langle {x_n} \right \rangle$ [[Definition:Convergent Sequence (Topology)|converges]] to any point $x$ of $S$.
+\end{theorem}
+
+\begin{proof}
+For any [[Definition:Open Set (Topology)|open set]] $U \subseteq S$ such that $x \in U$, we must have $U = S$.
+Hence:
+: $\forall n \ge 1: x_n \in U$
+The result follows from the definition of a [[Definition:Convergent Sequence (Topology)|convergent sequence in a topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Totally Bounded Metric Space is Separable}
+Tags: Separable Spaces, Metric Spaces, Totally Bounded Metric Spaces
+
+\begin{theorem}
+A [[Definition:Totally Bounded Metric Space|totally bounded metric space]] is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \left({A, d}\right)$ be a [[Definition:Totally Bounded Metric Space|totally bounded metric space]].
+By the definition of [[Definition:Totally Bounded Metric Space|total boundedness]], we can use the [[Axiom:Axiom of Countable Choice|axiom of countable choice]] to construct a [[Definition:Sequence|sequence]] $\left\langle{F_n}\right\rangle_{n \ge 1}$ such that:
+:For all [[Definition:Natural Numbers|natural numbers]] $n \ge 1$, $F_n$ is a [[Definition:Finite Net|finite $\left({1/n}\right)$-net]] for $M$.
+Let $\displaystyle S = \bigcup_{n \mathop \ge 1} F_n$.
+From [[Countable Union of Countable Sets is Countable]], it follows that $S$ is [[Definition:Countable Set|countable]].
+It suffices to prove that $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $M$.
+Let $S^-$ denote the [[Definition:Closure (Topology)|closure]] of $S$.
+Let $x \in X$.
+Let $U \subseteq X$ be [[Definition:Open Set of Metric Space|open]] in $M$ such that $x \in U$.
+By definition, there exists a [[Definition:Strictly Positive Real Number|strictly positive real number]] $\epsilon$ such that $B_{\epsilon} \left({x}\right) \subseteq U$.
+That is, the [[Definition:Open Ball of Metric Space|open $\epsilon$-ball of $x$ in $M$]] is [[Definition:Subset|contained]] in $U$.
+By the [[Archimedean Principle]], there exists a [[Definition:Natural Number|natural number]] $n > \dfrac 1 \epsilon$.
+That is, $\dfrac 1 n < \epsilon$, and so $B_{1 / n} \left({x}\right) \subseteq B_{\epsilon} \left({x}\right)$.
+Since [[Subset Relation is Transitive|$\subseteq$ is a transitive relation]], we have $B_{1/n} \left({x}\right) \subseteq U$.
+By the definition of a [[Definition:Net (Metric Space)|net]], there exists a $y \in F_n$ such that $x \in B_{1/n} \left({y}\right)$.
+By [[Definition:Metric Space Axioms|axiom $\left({M3}\right)$ for a metric]], it follows from the definition of an [[Definition:Open Ball of Metric Space|open ball]] that $y \in B_{1/n} \left({x}\right)$.
+Since $y \in S \cap U$, it follows that $x$ is an [[Definition:Adherent Point|adherent point]] of $S$.
+By definition of [[Definition:Adherent Point|adherent point]], we have $x \in S^-$.
+That is, $X \subseteq S^-$, and so $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $M$.
+{{qed}}
+{{ACC}}
+[[Category:Separable Spaces]]
+[[Category:Metric Spaces]]
+[[Category:Totally Bounded Metric Spaces]]
+ejynvni81404dwnx6gyadutj5l949qs
+\end{proof}<|endoftext|>
+\section{Convergent Sequence in Hausdorff Space has Unique Limit}
+Tags: Limits of Sequences, Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $\sequence {x_n}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence in $T$]].
+Then $\sequence {x_n}$ has exactly one [[Definition:Limit Point of Sequence|limit]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Convergent Sequence (Topology)|convergent sequence]], we have that $\sequence {x_n}$ converges to at least one [[Definition:Limit Point of Sequence|limit]].
+{{AimForCont}} $\ds \lim_{n \mathop \to \infty} x_n = l$ and $\displaystyle \lim_{n \mathop \to \infty} x_n = m$ such that $l \ne m$.
+As $T$ is [[Definition:Hausdorff Space|Hausdorff]], $\exists U \in \tau: l \in U$ and $\exists V \in \tau: m \in V$ such that $U \cap V = \O$.
+Then, from the definition of [[Definition:Convergent Sequence (Topology)|convergent sequence]]:
+{{begin-eqn}}
+{{eqn | ll= \exists N_U \in \R:
+      | l = n > N_U
+      | o = \implies
+      | r = x_n \in U
+}}
+{{eqn | ll= \exists N_V \in \R:
+      | l = n > N_V
+      | o = \implies
+      | r = x_n \in V
+}}
+{{end-eqn}}
+Taking $N = \max \set {N_U, N_V}$ we then have:
+:$\exists N \in \R: n > N \implies x_n \in U, x_n \in V$
+But $U \cap V = \O$.
+From that [[Definition:Contradiction|contradiction]] we can see that there can be no such two distinct $l$ and $m$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Space is Hausdorff}
+Tags: Metric Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Then $M$ is a [[Definition:Hausdorff Space|Hausdorff space]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in A: x \ne y$.
+Then from [[Distinct Points in Metric Space have Disjoint Open Balls]], there exist [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] $\map {B_\epsilon} x$ and $\map {B_\epsilon} Y$ which are [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set of Metric Space|open sets]] containing $x$ and $y$ respectively.
+Hence the result by the definition of [[Definition:Hausdorff Space|Hausdorff space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Real Sequence has Unique Limit}
+Tags: Limits of Sequences, Real Numbers, Convergent Real Sequence has Unique Limit
+
+\begin{theorem}
+Let $\sequence {s_n}$ be a [[Definition:Real Sequence|real sequence]].
+Then $\sequence {s_n}$ can have at most one [[Definition:Limit of Real Sequence|limit]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $\sequence {s_n}$ [[Definition:Convergent Real Sequence|converges]] to $l$ and also to $m$.
+That is, suppose that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = l$
+and:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = m$
+{{WLOG}}, assume that $l \ne m$.
+Let:
+:$\epsilon = \dfrac {\size {l - m} } 2$
+As $l \ne m$, it follows that $\epsilon > 0$.
+As $\sequence {s_n} \to l$:
+:$\exists N_1 \in \N: \forall n \in \N: n > N_1: \size {s_n - l} < \epsilon$
+Similarly, since $\sequence {s_n} \to m$:
+:$\exists N_2 \in \N: \forall n \in \N: n > N_2: \size {s_n - m} < \epsilon$
+Now set $N = \max \set {N_1, N_2}$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \size {l - m}
+      | r = \size {l - s_N + s_N - m}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \size {l - s_N} + \size {s_N - m}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = <
+      | r = 2 \epsilon
+      | c = 
+}}
+{{eqn | r = \size {l - m}
+      | c = 
+}}
+{{end-eqn}}
+This constitutes a [[Definition:Contradiction|contradiction]].
+It follows from [[Proof by Contradiction]] that $l = m$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that the [[Real Number Line is Metric Space|real number line is a metric space]]
+The result then follows from [[Convergent Sequence in Metric Space has Unique Limit]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distinct Points in Metric Space have Disjoint Open Balls}
+Tags: Open Balls
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $x, y \in M: x \ne y$.
+Then there exist [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Ball|open $\epsilon$-balls]] $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ containing $x$ and $y$ respectively.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in A: x \ne y$.
+Then $d \left({x, y}\right) > 0$.
+Put $\epsilon = \dfrac {d \left({x, y}\right)} 2$.
+Let $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ be the [[Definition:Open Ball|open $\epsilon$-balls]] of $x$ and $y$ respectively.
+Suppose $B_\epsilon \left({x}\right)$ and $B_\epsilon \left({y}\right)$ are not [[Definition:Disjoint Sets|disjoint]].
+Then $\exists z \in M$ such that $z \in B_\epsilon \left({x}\right)$ and $z \in B_\epsilon \left({y}\right)$.
+Then $d \left({x, z}\right) < \epsilon$ and $d \left({z, y}\right) < \epsilon$.
+Hence $d \left({x, z}\right) + d \left({z, y}\right) < 2 \epsilon = d \left({x, y}\right)$.
+This contradicts the definition of a [[Definition:Metric|metric]], so there can be no such $z$.
+Hence the open balls must be [[Definition:Disjoint Sets|disjoint]].
+{{qed}}
+[[Category:Open Balls]]
+qfipa27r1dmttasxcnjn6xadd0fmje2
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Lindelöf}
+Tags: Lindelöf Spaces, Metric Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $M$ be a [[Definition:Metric Space|metric space]].
+Let $M$ be [[Definition:Sequentially Compact Space|sequentially compact]].
+Then $M$ is also a [[Definition:Lindelöf Space|Lindelöf space]].
+That is, from every [[Definition:Open Cover|open cover]] of $M$, it is possible to extract a [[Definition:Countable Subcover|countable subcover]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \left({X, d}\right)$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
+Take any [[Definition:Open Cover|open cover]] $C$ of $M$.
+We need to find a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $C$ which still [[Definition:Cover of Set|covers]] $X$.
+We have that a [[Sequentially Compact Metric Space is Second-Countable]].
+Thus, by definition, the [[Definition:Topology Induced by Metric|topology]] of $M$ has a [[Definition:Countable Basis|countable basis]].
+Let $\mathcal B$ be a [[Definition:Countable Basis|countable basis]] for the [[Definition:Topology Induced by Metric|topology induced by $d$]] of $M$.
+Let $x \in X$.
+As $C$ [[Definition:Cover of Set|covers]] $X$:
+:$\exists U_x \in C: x \in U_x$
+As $\mathcal B$ is a [[Definition:Basis (Topology)/Analytic Basis/Definition 2|basis]]:
+:$\exists B_x \in \mathcal B: x \in B_x \subseteq U_x$
+Thus:
+: $(1): \quad \forall x \in X: \exists B_x \in \mathcal B: x \in B_x$
+Consider the set $\Sigma := \left\{{B_x: x \in X}\right\}$.
+$\Sigma$ is a [[Definition:Subset|subset]] of $\mathcal B$.
+Hence $\Sigma$ is [[Definition:Countable Set|countable]].
+As $\Sigma$ contains every $x \in X$ from $(1)$, $\Sigma$ [[Definition:Cover of Set|covers]] $X$.
+By construction of $\Sigma$, every [[Definition:Open Set (Topology)|open set]] in $\Sigma$ is contained in some $U \in C$.
+For each [[Definition:Open Set (Topology)|open set]] $B \in \Sigma$, choose one $U_B \in C$ such that $B \subseteq U_B$.
+{{explain|Does this need AoC or ACC?}}
+Let $\mathcal U$ be the [[Definition:Set|set]] defined as:
+:$\mathcal U = \left\{ {U_B: B \in \Sigma}\right\}$
+As $\mathcal U$ does not have more sets than $\Sigma$, $\mathcal U$ is [[Definition:Countable Set|countable]].
+We have that:
+: $\forall B \in \Sigma: B \subseteq U_B$
+Thus it follows that $\mathcal U$ [[Definition:Cover of Set|covers]] $X$.
+As:
+:$\forall U_B \in \mathcal U: U_B \in C$
+$U$ is a [[Definition:Countable Subcover|countable subcover]] of $C$.
+Thus a [[Definition:Countable Subcover|countable subcover]] has been obtained from $C$.
+As $C$ is arbitrary, it follows that $M$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Space is Compact}
+Tags: Metric Spaces, Sequentially Compact Spaces, Compact Spaces, Sequentially Compact Metric Space is Compact
+
+\begin{theorem}
+A [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]] is [[Definition:Compact Metric Space|compact]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {A, d}$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
+From [[Sequentially Compact Metric Space is Lindelöf]], every [[Definition:Open Cover|open cover]] of $A$ has a [[Definition:Countable Subcover|countable subcover]].
+Let $C$ be an [[Definition:Open Cover|open cover]] $A$.
+Extract from it a [[Definition:Countable Subcover|countable subcover]] $\set {U_1, U_2, \ldots}$.
+{{AimForCont}} there exists no [[Definition:Finite Subcover|finite subcover]] of $C$.
+Then, for all $n \in \N_{\ge 1}$, the [[Definition:Set|set]] $\set {U_1, \ldots, U_n}$ does ''not'' [[Definition:Cover of Set|cover]] $A$.
+Hence it is possible to choose $x_n \in A$ such that:
+:$x_n \notin U_1 \cup \cdots \cup U_n$.
+Thus we construct an [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n}_{n \mathop \ge 1}$ of points of $A$.
+By assumption $A$ is [[Definition:Sequentially Compact Space|sequentially compact]].
+Thus $\sequence {x_n}_{n \mathop \ge 1}$ has a [[Definition:Subsequence|subsequence]] which [[Definition:Convergent Sequence (Metric Space)|converges]] to some $x \in A$.
+But because the $U_i: i \ge 1$ forms a [[Definition:Cover of Set|cover]] for $A$, there exists some $U_m$ such that $x \in U_m$.
+Hence by one of the characterizations of convergence, there is an [[Definition:Infinite Set|infinite number]] of [[Definition:Term of Sequence|terms]] in the [[Definition:Sequence|sequence]] $\sequence {x_i}$ which are contained in $U_m$.
+{{explain|Determine which of those characterizations of convergence.}}
+But from the method of construction of $\sequence {x_i}$, each $U_n$ can contain only points $x_i$ with $i < n$.
+That is, each $U_n$ can only contain a [[Definition:Finite Set|finite number]] of the [[Definition:Term of Sequence|terms]] of $\sequence {x_i}$.
+This is a [[Definition:Contradiction|contradiction]].
+Thus the supposition that there exists no [[Definition:Finite Subcover|finite subcover]] of $C$ was false.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $M = \left({A, d}\right)$ be a [[Definition:Sequentially Compact Space|sequentially compact]] [[Definition:Metric Space|metric space]].
+Let $\mathcal U$ be any [[Definition:Open Cover|open cover]] of $M$.
+By [[Lebesgue's Number Lemma]], there exists a [[Definition:Lebesgue Number|Lebesgue number]] for $\mathcal U$.
+By [[Sequentially Compact Metric Space is Totally Bounded]], there exists $\left\{{x_1, x_2, \ldots, x_n}\right\}$ which is a [[Definition:Finite Net|finite $\epsilon$-net]] for $M$, where $\epsilon$ is this same [[Definition:Lebesgue Number|Lebesgue number]].
+Let $B_\epsilon \left({x_i}\right)$ be the [[Definition:Open Ball|open $\epsilon$-ball]] of $x_i$.
+By definition of [[Definition:Lebesgue Number|Lebesgue number]], $B_\epsilon \left({x_i}\right)$ is contained in some $U_i \in \mathcal U$.
+Since:
+: $\displaystyle M \subseteq \bigcup_{i \mathop = 1}^n B_\epsilon \left({x_i}\right) \subseteq \bigcup_{i \mathop = 1}^n U_i$
+we have a [[Definition:Finite Subcover|finite subcover]] $\left\{{U_1, U_2, \ldots, U_n}\right\}$ of $\mathcal{U}$ for $M$.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows directly from:
+: [[Sequentially Compact Space is Countably Compact]]
+: [[Countably Compact Metric Space is Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Subspace of Metric Space is Bounded}
+Tags: Metric Subspaces, Compact Spaces
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $C$ be a [[Definition:Metric Subspace|subspace]] of $M$.
+If $C$ is [[Definition:Compact Space|compact]], then it is [[Definition:Bounded Metric Space|bounded]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in M$.
+Let $n \in \N_{>0}$.
+Let $B_n \left({a}\right)$ be the [[Definition:Open Ball|open $n$-ball]] of $a$.
+Then $\displaystyle C \subseteq \bigcup_{n \mathop = 1}^\infty B_n \left({a}\right)$ because $\forall x \in C: d \left({x, a}\right) < n$ for some $n \in \N$.
+Thus the collection $\left\{{B_n \left({a}\right): n \in \N}\right\}$ forms an [[Definition:Open Cover|open cover]] of $C$.
+Because $C$ is [[Definition:Compact Space|compact]], it has a [[Definition:Finite Subcover|finite subcover]], say: $\left\{{B_{n_1} \left({a}\right), B_{n_2} \left({a}\right), \ldots, B_{n_r} \left({a}\right)}\right\}$.
+Let $n = \max \left\{{n_1, n_2, \ldots, n_r}\right\}$.
+Then:
+: $\displaystyle C \subseteq \bigcup_{n \mathop = 1}^r B_{n_r} \left({a}\right) = B_n \left({a}\right)$
+The result follows by definition of [[Definition:Bounded Metric Space|bounded]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Subspace of Hausdorff Space is Closed}
+Tags: Hausdorff Spaces, Compact Spaces, Closed Sets, Topological Subspaces, Compact Subspace of Hausdorff Space is Closed
+
+\begin{theorem}
+Let $H = \struct {A, \tau}$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $C$ be a [[Definition:Compact Topological Subspace|compact subspace]] of $H$.
+Then $C$ is [[Definition:Closed Set (Topology)|closed]] in $H$.
+\end{theorem}
+
+\begin{proof}
+For a [[Definition:Subset|subset]] $S \subseteq A$, let $S^{\complement}$ denote the [[Definition:Relative Complement|relative complement]] of $S$ in $A$.
+Consider an arbitrary [[Definition:Element|point]] $x \in C^{\complement}$.
+Define the [[Definition:Set|set]]:
+:$\displaystyle \mathcal O = \left\{{V \in \tau: \exists U \in \tau: x \in U \subseteq V^{\complement}}\right\}$
+By [[Empty Intersection iff Subset of Complement]], we have that:
+:$U \subseteq V^{\complement} \iff U \cap V = \varnothing$
+Hence, by the definition of a [[Definition:Hausdorff Space|Hausdorff space]], it follows that $\mathcal O$ is an [[Definition:Open Cover|open cover]] for $C$.
+By the definition of a [[Definition:Compact Topological Subspace|compact subspace]], there exists a [[Definition:Finite Subcover|finite subcover]] $\mathcal F$ of $\mathcal O$ for $C$.
+By the [[Principle of Finite Choice]], [[Definition:Existential Quantifier|there exists]] an [[Definition:Indexed Family of Subsets|$\mathcal F$-indexed family]] $\left\langle{U_V}\right\rangle_{V \in \mathcal F}$ of [[Definition:Element|elements]] of $\tau$ such that:
+:$\forall V \in \mathcal F: x \in U_V \subseteq V^{\complement}$
+Define:
+:$\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U_V$
+By [[General Intersection Property of Topological Space]], it follows that $U \in \tau$.
+Clearly, $x \in U$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = U
+      | o = \subseteq
+      | r = \bigcap_{V \mathop \in \mathcal F} \left({ V^{\complement} }\right)
+      | c = [[Set Intersection Preserves Subsets/Families of Sets|Set Intersection Preserves Subsets]]
+}}
+{{eqn | r = \left({\bigcup \mathcal F}\right)^{\complement}
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/General Case/Complement of Union|De Morgan's laws]]
+}}
+{{eqn | o = \subseteq
+      | r = C^{\complement}
+      | c = {{Defof|Cover of Set}} and [[Relative Complement inverts Subsets]]
+}}
+{{end-eqn}}
+From [[Subset Relation is Transitive]], we have that $U \subseteq C^{\complement}$.
+Hence $C^{\complement}$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+From [[Set is Open iff Neighborhood of all its Points]], we have that $C^{\complement} \in \tau$.
+That is, $C$ is [[Definition:Closed Set (Topology)|closed]] in $H$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Subspace of Hausdorff Space is Hausdorff]], a [[Definition:Topological Subspace|subspace]] of a [[Definition:Hausdorff Space|Hausdorff space]] is itself [[Definition:Hausdorff Space|Hausdorff]].
+Let $a \in A \setminus C$.
+We are going to prove that there exists an [[Definition:Open Set (Topology)|open set]] $U_a$ such that $a \in U_a \subseteq A \setminus C$.
+For any single point $x \in C$, the [[Definition:Hausdorff Condition|Hausdorff condition]] ensures the existence of [[Definition:Disjoint Sets|disjoint]] [[Definition:Open Set (Topology)|open set]] $\map U x$ and $\map V x$ containing $a$ and $x$ respectively.
+Suppose there were only a [[Definition:Finite Set|finite]] number of points $x_1, x_2, \ldots, x_r$ in $C$.
+Then we could take $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$.
+Now suppose $C$ is not [[Definition:Finite Set|finite]].
+The set $\set {\map V x: x \in C}$ is an [[Definition:Open Cover|open cover]] for $C$.
+As $C$ is [[Definition:Compact Subspace|compact]], it has a [[Definition:Finite Subcover|finite subcover]], say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$.
+Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$.
+Then $U_a$ is [[Definition:Open Set (Topology)|open]] because it is a [[Definition:Finite Set|finite]] [[Definition:Set Intersection|intersection]] of [[Definition:Open Set (Topology)|open sets]].
+Also, $a \in U_a$ because $a \in \map U {x_i}$ for each $i = 1, 2, \ldots, r$.
+Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in \map U {x_i}$.
+Because $\displaystyle C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$:
+:$b \notin \map V {x_i}$, so $b \notin C$
+Thus:
+:$U_a \subseteq A \setminus C$.
+Then:
+:$\displaystyle A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$
+So $A \setminus C$ is [[Definition:Open Set (Topology)|open]].
+It follows that $C$ is [[Definition:Closed Set (Topology)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Image of Compact Space is Compact}
+Tags: Compact Spaces, Continuous Invariants
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+If $T_1$ is [[Definition:Compact Topological Space|compact]] then so is its [[Definition:Image of Mapping|image]] $f \sqbrk {T_1}$ under $f$.
+That is, [[Definition:Compact Topological Space|compactness]] is a [[Definition:Continuous Invariant|continuous invariant]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\UU$ is an [[Definition:Open Cover|open cover]] of $f \sqbrk {T_1}$ by sets [[Definition:Open Set (Topology)|open in $T_2$]].
+Because $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], it follows that $f^{-1} \sqbrk U$ is [[Definition:Open Set (Topology)|open in $T_1$]] for all $U \in \UU$.
+The set $\set {f^{-1} \sqbrk U: U \in \UU}$ is an [[Definition:Open Cover|open cover]] of $T_1$, because for any $x \in T_1$, it follows that $\map f x$ must be in some $U \in \UU$.
+Because $T_1$ is [[Definition:Compact Topological Space|compact]], it has a [[Definition:Finite Subcover|finite subcover]] $\set {f^{-1} \sqbrk {U_1}, f^{-1} \sqbrk {U_2}, \ldots, f^{-1} \sqbrk {U_r} }$.
+It follows that $\set {U_1, U_2, \ldots, U_r}$ is a [[Definition:Finite Subcover|finite subcover]] of $f \sqbrk {T_1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Subspace of Compact Space is Compact}
+Tags: Closed Sets, Compact Spaces
+
+\begin{theorem}
+A [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of a [[Definition:Compact Topological Space|compact space]] is [[Definition:Compact Topological Space|compact]].
+That is, the [[Definition:Property|property]] of being [[Definition:Compact Topological Space|compact]] is [[Definition:Weakly Hereditary Property|weakly hereditary]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be a [[Definition:Compact Topological Space|compact space]].
+Let $C$ be a [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of $T$.
+Let $\UU$ be an [[Definition:Open Cover|open cover]] of $C$.
+Since $C$ is closed, it follows by definition of [[Definition:Closed Set (Topology)|closed]] that $T \setminus C$ is [[Definition:Open Set (Topology)|open]] in $T$.
+So if we add $T \setminus C$ to $\UU$, we see that $\UU \cup \set {T \setminus C}$ is also an [[Definition:Open Cover|open cover]] of $T$.
+As $T$ is [[Definition:Compact Topological Space|compact]], there is a [[Definition:Finite Subcover|finite subcover]] of $\UU \cup \set {T \setminus C}$, say $\VV = \set {U_1, U_2, \ldots, U_r}$.
+This covers $C$ by the fact that it covers $T$.
+If $T \setminus C$ is an element of $\VV$, then it can be removed from $\VV$ and the rest of $\VV$ still covers $C$.
+Thus we have a [[Definition:Finite Subcover|finite subcover]] of $\UU$ which covers $C$, and hence $C$ is [[Definition:Compact Topological Space|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Real Interval is Closed Real Interval}
+Tags: Real Intervals, Set Closures, Closure of Real Interval is Closed Real Interval
+
+\begin{theorem}
+Let $I$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Real Interval|real interval]] such that one of these holds:
+: $I = \left({a \,.\,.\, b}\right)$
+: $I = \left[{a \,.\,.\, b}\right)$
+: $I = \left({a \,.\,.\, b}\right]$
+: $I = \left[{a \,.\,.\, b}\right]$
+Let $I^-$ denote the [[Definition:Closure (Topology)|closure]] of $I$.
+Then $I^-$ is the [[Definition:Closed Real Interval|closed real interval]] $\left[{a \,.\,.\, b}\right]$.
+\end{theorem}
+
+\begin{proof}
+Let $I$ be one of the intervals as specified in the exposition.
+Note that:
+:$(1): \quad$ By [[Condition for Point being in Closure]], $x \in I^-$ {{iff}} every [[Definition:Open Set (Real Analysis)|open set]] in $\R$ containing $x$ contains a point in $I$.
+:$(2): \quad$ From [[Union of Open Sets of Metric Space is Open]], every [[Definition:Open Set (Real Analysis)|open set]] in $\R$ is a [[Definition:Set Union|union]] of [[Definition:Open Real Interval|open intervals]].
+Thus we also have that $x \in I^-$ {{iff}} every [[Definition:Open Real Interval|open interval]] containing $x$ also contains a point in $I$.
+This equivalence will be made use of throughout.
+=== Lemma: $x \in \left[{a \,.\,.\, b}\right] \implies x \in I^-$ ===
+Let $x \in \left[{a \,.\,.\, b}\right]$.
+Let $\left({c \,.\,.\, d}\right)$ be an [[Definition:Open Real Interval|open interval]] in $\R$ such that $x \in \left({c \,.\,.\, d}\right)$.
+We must show that $\left({c \,.\,.\, d}\right)$ contains a point in $I$.
+One of the following three possibilities holds:
+: $a < x < b$
+: $x = a$
+: $x = b$
+==== Case: $a < x < b$ ====
+In this case, $x \in I$ and $x \in \left({c \,.\,.\, d}\right)$.
+Therefore $\left({c \,.\,.\, d}\right)$ contains a point in $I$.
+{{qed|lemma}}
+==== Case: $x = a$ ====
+If $I$ contains $a$, then this means $x \in I$, and the proof is complete.
+So, assume that $a \notin I$.
+Since $I$ is nonempty but does not contain $a$, we must have $a < b$.
+Let $r$ be the [[Definition:Minimal Element|minimum]] of $d$ and $b$, so that $r \le d$ and $r \le b$.
+Since $a = x < d$ by choice of $d$ and since $a < b$ by assumption, we must have $a < r$.
+Thus, by [[Real Numbers are Close Packed]], there exists some $s \in \R$ such that $a < s < r$.
+To summarize, we have $c < x = a < s < r$, where $r \le d$ and $r \le b$.
+This means that $s$ satisfies both $c < s < d$ and $a < s < b$.
+Hence, $s$ is a point in $\left({c \,.\,.\, d}\right)$ which is also in $I$.
+The existence of such a point is what we wanted to show.
+{{qed|lemma}}
+==== Case: $x = b$ ====
+This case is analogous to case when $x = a$.
+Here we instead let $l$ be the maximum of $c$ and $a$, and select an $s$ such that $l < s < x = b < d$, where $c \le l$ and $a \le l$.
+{{qed|lemma}}
+=== Lemma: $x \notin \left[{a \,.\,.\, b}\right] \implies x \notin I^-$  ===
+Suppose $x \notin \left[{a \,.\,.\, b}\right]$.
+We must find an open interval containing $x$ which does not contain a point in $I$.
+There are two possibilities:
+: $x < a$
+or:
+: $x > b$
+==== Case: $x < a$ ====
+By [[Real Numbers are Close Packed]], there exists $r \in \R$ such that $x < r < a$.
+Thus $\left({x-1 \,.\,.\, r}\right)$ is an open interval, all of whose elements are less than $a$, and hence not in $I$.
+==== Case: $x > b$ ====
+This is similarly to the case when $x < a$.
+Here instead we pick $r$ such that $b < r < x$, and consider the interval $\left({r \,.\,.\, x+1}\right)$.
+{{qed|lemma}}
+By the two lemmas proven above:
+: $\left[{a \,.\,.\, b}\right] = I^-$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+There are four cases to cover:
+:$(1): \quad$ Let $I = \left({a \,.\,.\, b}\right)$.
+From [[Closure of Open Real Interval is Closed Real Interval]]:
+:$I^- = \left[{a \,.\,.\, b}\right]$
+{{qed|lemma}}
+:$(2): \quad$ Let $I = \left[{a \,.\,.\, b}\right)$.
+From [[Closure of Half-Open Real Interval is Closed Real Interval]]:
+:$I^- = \left[{a \,.\,.\, b}\right]$
+{{qed|lemma}}
+:$(3): \quad$ Let $I = \left({a \,.\,.\, b}\right]$.
+From [[Closure of Half-Open Real Interval is Closed Real Interval]]:
+:$I^- = \left[{a \,.\,.\, b}\right]$
+{{qed|lemma}}
+:$(4): \quad$ Let $I = \left[{a \,.\,.\, b}\right]$.
+From [[Closed Real Interval is Closed in Real Number Line]]:
+:$I$ is [[Definition:Closed Set (Topology)|closed]] in $\R$.
+From [[Set is Closed iff Equals Topological Closure]]:
+:$I^- = \left[{a \,.\,.\, b}\right]$
+{{qed|lemma}}
+Thus all cases are covered.
+The result follows by [[Proof by Cases]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Product of Compact Spaces}
+Tags: Topology
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Let $T_1 \times T_2$ be the [[Definition:Product Space (Topology)|product space]] of $T_1$ and $T_2$.
+Then $T_1 \times T_2$ is [[Definition:Compact Topological Space|compact]] {{iff}} both $T_1$ and $T_2$ are [[Definition:Compact Topological Space|compact]].
+\end{theorem}
+
+\begin{proof}
+Let $T_1 \times T_2$ be [[Definition:Compact Topological Space|compact]].
+From [[Projection from Product Topology is Continuous]], both $T_1$ and $T_2$ are the [[Definition:Continuous Mapping (Topology)|continuous]] [[Definition:Image of Subset under Mapping|images]] of $T_1 \times T_2$ under the [[Definition:Projection (Mapping Theory)|projections]] $\pr_1$ and $\pr_2$ respectively.
+So both $T_1$ and $T_2$ are [[Definition:Compact Topological Space|compact]] by [[Continuous Image of Compact Space is Compact]].
+Now let $T_1$ and $T_2$ be [[Definition:Compact Topological Space|compact]].
+Let $\WW$ be an [[Definition:Open Cover|open cover]] for $T_1 \times T_2$.
+Let $A \subseteq T_1$ be defined as $\text{good}$ (for $\WW$) if $A \times T_2$ is [[Definition:Cover of Set|covered]] by a [[Definition:Finite Subset|finite subset]] of $\WW$.
+We want to prove that $T_1$ is $\text{good}$.
+We will do this in stages, because it's complicated.
+=== Step 1 ===
+Suppose $A_1, A_2, \ldots, A_r$ are all $\text{good}$.
+Then so is $\displaystyle A = \bigcup_{i \mathop = 1}^r A_i$.
+For any given $i=1, 2, \ldots, r$ we have that $A_i \times T_2$ is [[Definition:Cover of Set|covered]] by a [[Definition:Finite Subset|finite subset]] of $\WW$, say $\WW_i \subseteq \WW$.
+Hence $\displaystyle A \times T_2 = \bigcup_{i \mathop = 1}^r \paren{A_i \times T_2}$ is [[Definition:Cover of Set|covered]] by the [[Definition:Finite Subset|finite subset]] $\displaystyle \bigcup_{i \mathop = 1}^r \WW_i$ of $\WW$.
+=== Step 2 ===
+Now we show that $T_1$ is '''locally''' $\text{good}$, in the sense that $\forall x \in T_1$, there is an [[Definition:Open Set (Topology)|open set]] $\map U x$ such that $x \in \map U x$ and $\map U x$ is $\text{good}$.
+Consider a fixed $x \in T_1$.
+For each $y \in T_2$, we have that $\tuple{x, y} \in \map W y$ for some $\map W y \in \WW$, since $\WW$ [[Definition:Cover of Set|covers]] $T_1$ and $T_2$.
+By the definition of the [[Definition:Product Topology|product topology]], $\exists \map U y, \map V y$ [[Definition:Open Set (Topology)|open]] in $T_1$ and $T_2$ respectively such that $\tuple{x, y} \in \map U y \times \map V y \subseteq \map W y$.
+The set $\set{\map V y : y \in T_2}$ is an [[Definition:Open Cover|open cover]] for $T_2$.
+As $T_2$ is [[Definition:Compact Topological Space|compact]], there is a [[Definition:Finite Subcover|finite subcover]] of $\map V y$, say $\set{\map V {y_1}, \map V {y_2}, \ldots, \map V {y_r}}$.
+Let $\map U x = \map U {y_1} \cap \map U {y_2} \cap \cdots \cap \map U {y_r}$.
+For each $i = 1, 2, \ldots, r$, we have $\map U x \times \map V {y_i} \subseteq \map U {y_i} \times \map V {y_i} \subseteq \map W {y_i}$.
+Hence $\displaystyle \map U x \times T_2 = \map U x \times \bigcup_{i \mathop = 1}^r \map V {y_i} \subseteq \map W {y_i}$.
+So $\map U x$ is $\text{good}$.
+And we have that $x \in \map U x$ and $\map U x$ is [[Definition:Open Set (Topology)|open]] in $T_1$ as required.
+=== Step 3 ===
+Now we pass from local to global.
+For each $x \in T_1$, let $\map U x$ be a $\text{good}$ [[Definition:Open Set (Topology)|open set]] containing $x$, by [[Topological Product of Compact Spaces#Step 2|step 2]].
+Then $\set{\map U x: x \in T_1}$ is an [[Definition:Open Cover|open cover]] for $T_1$.
+Because $T_1$ is [[Definition:Compact Topological Space|compact]], $\map U x$ has a [[Definition:Finite Subcover|finite subcover]], say $\set{\map U {x_1}, \map U {x_2}, \ldots, \map U {x_m}}$.
+Since each $\map U {x_i}$ is $\text{good}$, then so is $\displaystyle \bigcup_{i \mathop = 1}^m \map U {x_i}$ by [[Topological Product of Compact Spaces#Step 1|step 1]].
+But $\displaystyle \bigcup_{i \mathop = 1}^m \map U {x_i} = T_1$, so $T_1$ is $\text{good}$, as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Heine-Cantor Theorem}
+Tags: Metric Spaces, Compact Spaces, Continuous Mappings, Uniformly Continuous Mappings, Heine-Cantor Theorem
+
+\begin{theorem}
+Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $M_1$ be [[Definition:Compact Metric Space|compact]].
+Let $f: A_1 \to A_2$ be a [[Definition:Continuous Mapping (Metric Spaces)|continuous mapping]].
+Then $f$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+For all $x \in A_1$, define:
+:$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
+Define:
+:$\mathcal C = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map \Delta x}$
+where $B_{\delta} \left({x}\right)$ denotes the [[Definition:Open Ball|open $\delta$-ball of $x$ in $M_1$]].
+From the definition of [[Definition:Continuous Mapping (Metric Spaces)|continuity]], it follows that $\mathcal C$ is a [[Definition:Cover of Set|cover]] for $A_1$.
+From [[Open Ball of Metric Space is Open Set]], it therefore follows that $\mathcal C$ is an [[Definition:Open Cover|open cover]] for $A_1$.
+By the definition of a [[Definition:Compact Metric Space|compact metric space]], there exists a [[Definition:Finite Subcover|finite subcover]] $\set {\map {B_{\delta_1} } {x_1}, \map {B_{\delta_2} } {x_2}, \ldots, \map {B_{\delta_n} } {x_n} }$ of $\mathcal C$ for $A_1$.
+Define:
+:$\delta = \min \set {\delta_1, \delta_2, \ldots, \delta_n}$
+Let $x, y \in A_1$ satisfy $\map {d_1} {x, y} < \delta$.
+By the definition of a [[Definition:Cover of Set|cover]], there exists a $k \in \set{1, 2, \ldots, n}$ such that $\map {d_1} {x, x_k} < \delta_k$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {d_1} {y, x_k}
+      | o = \le
+      | r = \map {d_1} {y, x} + \map {d_1} {x, x_k}
+      | c = [[Definition:Triangle Inequality|Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \delta + \delta_k
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $\paren {M3}$]]
+}}
+{{eqn | o = \le
+      | r = 2 \delta_k
+}}
+{{end-eqn}}
+By the definition of $\Delta \left({x_k}\right)$, it follows that:
+:$\map {d_2} {\map f x, \map f {x_k} } < \dfrac \epsilon 2$
+:$\map {d_2} {\map f y, \map f {x_k} } < \dfrac \epsilon 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map {d_2} {\map f x, \map f y}
+      | o = \le
+      | r = \map {d_2} {\map f x, \map f {x_k} } + \map {d_2} {\map f {x_k}, \map f y}
+      | c = [[Definition:Triangle Inequality|Triangle Inequality]]
+}}
+{{eqn | o = <
+      | r = \frac \epsilon 2 + \frac \epsilon 2
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $\paren{M3}$]]
+}}
+{{eqn|r = \epsilon
+}}
+{{end-eqn}}
+The result follows from the definition of [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniform continuity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the [[Definition:Product Topology|product topology]].
+Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the [[Definition:Mapping|mapping]] defined as:
+:$\map F {x, y} = \tuple {\map f x, \map f y}$
+By [[Projection from Product Topology is Continuous]], we have that the (first and second) [[Definition:Projection (Mapping Theory)|projections]] on $A_1 \times A_1$ are [[Definition:Continuous Mapping (Topology)|continuous]].
+By [[Continuity of Composite Mapping]] and [[Continuous Mapping to Topological Product]], it follows that $F$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+By [[Distance Function of Metric Space is Continuous]] and [[Continuity of Composite Mapping]], it follows that $d_2 \circ F: A_1 \times A_1 \to \R$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+By [[Continuity Defined from Closed Sets]], the [[Definition:Set|set]]:
+:$C = \paren {d_2 \circ F}^{-1} \sqbrk {\hointr \epsilon \to} = \set {\tuple {x, y} \in A_1 \times A_1: \map {d_2} {\map f x, \map f y} \ge \epsilon}$
+is [[Definition:Closed Set (Topology)|closed]] in $A_1 \times A_1$.
+By [[Topological Product of Compact Spaces]], $A_1 \times A_1$ is [[Definition:Compact Space|compact]].
+By [[Closed Subspace of Compact Space is Compact]], $C$ is [[Definition:Compact Space|compact]].
+By [[Distance Function of Metric Space is Continuous]] and [[Continuous Image of Compact Space is Compact]], $d_1 \sqbrk C$ is [[Definition:Compact Space|compact]].
+Therefore, $d_1 \sqbrk C$ has a [[Definition:Smallest Element|smallest element]] $\delta$.
+{{explain}}
+By [[Definition:Metric Space Axioms|metric space axioms $(\text M 1)$ and $(\text M 4)$]], we have that $\delta > 0$.
+By construction, it follows that:
+:$\forall x, y \in A_1: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map f x, \map f y} < \epsilon$
+Hence, $f$ is [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Function on Closed Interval is Uniformly Continuous}
+Tags: Real Analysis, Continuous Real Functions, Uniformly Continuous Real Functions
+
+\begin{theorem}
+Let $\closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $f: \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
+Then $f$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+We have that [[Real Number Line is Metric Space|$\R$ is a metric space]] under the [[Definition:Euclidean Metric on Real Number Line|usual (Euclidean) metric]].
+We also have from the [[Heine-Borel Theorem]] that $\closedint a b$ is [[Definition:Compact Topological Space|compact]].
+So the result [[Heine-Cantor Theorem]] applies.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Complement equals Cosine}
+Tags: Sine Function, Cosine Function, Sine of Complement equals Cosine
+
+\begin{theorem}
+:$\sin \left({\dfrac \pi 2 - \theta}\right) = \cos \theta$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin \left({\frac \pi 2 - \theta}\right)
+      | r = \sin \frac \pi 2 \cos \theta - \cos \frac \pi 2 \sin \theta
+      | c = [[Sine of Difference]]
+}}
+{{eqn | r = 1 \times \cos \theta - 0 \times \sin \theta
+      | c = [[Sine of Right Angle]] and [[Cosine of Right Angle]]
+}}
+{{eqn | r = \cos \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {\frac \pi 2 - \theta}
+      | r = -\map \sin {\theta - \frac \pi 2}
+      | c = [[Sine Function is Odd]]
+}}
+{{eqn | r = \map \cos {\theta - \frac \pi 2 + \frac \pi 2}
+      | c = [[Cosine of Angle plus Right Angle]]
+}}
+{{eqn | r = \cos \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \sin {\dfrac \pi 2 - \theta}
+      | r = \map \Im {e^{i \paren {\frac \pi 2 - \theta} } }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \map \Im {e^{i \frac \pi 2} e^{-i \theta} }
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = \map \Im {\paren {\cos \dfrac \pi 2+i \sin \dfrac \pi 2} e^{-i \theta} }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \map \Im {i e^{-i \theta} }
+      | c = [[Cosine of Right Angle]], [[Sine of Right Angle]]
+}}
+{{eqn | r = \map \Re {e^{-i \theta} }
+      | c = 
+}}
+{{eqn | r = \map \cos {-\theta}
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \cos \theta
+      | c = [[Cosine Function is Even]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Complement equals Sine}
+Tags: Sine Function, Cosine Function, Cosine of Complement equals Sine
+
+\begin{theorem}
+:$\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {\frac \pi 2 - \theta}
+      | r = \cos \frac \pi 2 \cos \theta + \sin \frac \pi 2 \sin \theta
+      | c = [[Cosine of Difference]]
+}}
+{{eqn | r = 0 \times \cos \theta + 1 \times \sin \theta
+      | c = [[Cosine of Right Angle]] and [[Sine of Right Angle]]
+}}
+{{eqn | r = \sin \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cos \left({\frac \pi 2 - \theta}\right)
+      | r = \cos \left({\theta - \frac \pi 2}\right)
+      | c = [[Cosine Function is Even]]
+}}
+{{eqn | r = \sin \left({\theta - \frac \pi 2 + \frac \pi 2}\right)
+      | c = [[Sine of Angle plus Right Angle]]
+}}
+{{eqn | r = \sin \theta
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {\dfrac \pi 2 - \theta}
+      | r = \frac 1 2 \paren {e^{i \paren {\frac \pi 2 - \theta} } + e^{-i \paren {\frac \pi 2 - \theta} } }
+      | c = [[Cosine Exponential Formulation]]
+}}
+{{eqn | r = \frac 1 2 \paren {e^{i \frac \pi 2} e^{-i \theta} + e^{-i \frac \pi 2} e^{i \theta} }
+      | c = [[Exponential of Sum/Complex Numbers|Exponential of Sum: Complex Numbers]]
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {\map \cos {\frac \pi 2} + i \, \map \sin {\frac \pi 2} } e^{-i \theta} + \paren {\map \cos {-\frac \pi 2} + i \, \map \sin {-\frac \pi 2} } e^{i \theta} }
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = \frac 1 2 \paren {i e^{-i \theta} - i e^{i \theta} }
+      | c = [[Cosine of Right Angle]], [[Sine of Right Angle]], [[Cosine Function is Even]], [[Sine Function is Odd]]
+}}
+{{eqn | r = \frac 1 {2 i} \paren {e^{i \theta} - e^{-i \theta} }
+      | c = {{Defof|Imaginary Unit}}
+}}
+{{eqn | r = \sin \theta
+      | c = [[Sine Exponential Formulation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Bijection from Compact to Hausdorff is Homeomorphism}
+Tags: Homeomorphisms, Compact Spaces, Hausdorff Spaces, Continuous Mappings
+
+\begin{theorem}
+Let $T_1$ be a [[Definition:Compact Topological Space|compact space]].
+Let $T_2$ be a [[Definition:Hausdorff Space|Hausdorff space]].
+Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Bijection|bijection]].
+Then $f$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $g = f^{-1}$.
+We need to show that $g: T_2 \to T_1$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+For any $V \subseteq T_1$, we have $g^{-1} \left({V}\right) = f  \left({V}\right)$.
+We are to show that if $V$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$, then $g^{-1} \left({V}\right)$ is [[Definition:Closed Set (Topology)|closed]] in $T_2$.
+Suppose $V$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+Since $T_1$ is [[Definition:Compact Topological Space|compact]], $V$ is [[Definition:Compact Topological Space|compact]] by [[Closed Subspace of Compact Space is Compact]].
+So $f \left({V}\right)$ is [[Definition:Compact Topological Space|compact]] from [[Continuous Image of Compact Space is Compact]].
+Since $T_2$ is [[Definition:Hausdorff Space|Hausdorff]], $f  \left({V}\right)$ [[Definition:Closed Set (Topology)|closed]] by [[Compact Subspace of Hausdorff Space is Closed]].
+But $f  \left({V}\right) = g^{-1} \left({V}\right)$, so $g^{-1} \left({V}\right)$ is [[Definition:Closed Set (Topology)|closed]].
+From [[Continuity Defined from Closed Sets]], it follows that $g$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Thus by definition, $f$ is a [[Definition:Homeomorphism (Topological Spaces)|homeomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space}
+Tags: Connected Spaces, Equivalence of Definitions of Connected Topological Space
+
+\begin{theorem}
+{{TFAE|def = Connected Topological Space}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets|$(1) \iff (2)$: No Separation iff No Union of Closed Sets]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Separation iff No Clopen Sets|$(1) \iff (4)$: No Separation iff No Clopen Sets]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Separation iff No Clopen Sets}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary|$(2) \implies (3)$: No Union of Closed Sets implies No Subsets with Empty Boundary]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Union of Closed Sets implies No Subsets with Empty Boundary}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets|$(3) \implies (4)$: No Subsets with Empty Boundary implies No Clopen Sets]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Subsets with Empty Boundary implies No Clopen Sets}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets|$(4) \implies (5)$: No Clopen Sets implies No Union of Separated Sets]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Clopen Sets implies No Union of Separated Sets}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space|$(5) \implies (6)$: No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Union of Separated Sets implies No Continuous Surjection to Discrete Two-Point Space}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation|$(6) \implies (1)$: No Continuous Surjection to Discrete Two-Point Space implies No Separation]] ===
+{{:Equivalence of Definitions of Connected Topological Space/No Continuous Surjection to Discrete Two-Point Space implies No Separation}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Indirect Proof}
+Tags: Proof Techniques
+
+\begin{theorem}
+Let $P$ be a [[Definition:Proposition|proposition]] whose [[Definition:Truth Value|truth value]] is to be [[Definition:Proof|proved]] (either [[Definition:True|true]] or [[Definition:False|false]]).
+There are two aspects to this:
+\end{theorem}
+
+\begin{proof}
+For proofs, see:
+* [[Reductio ad Absurdum/Sequent Form]]
+* [[Proof by Contradiction/Sequent Form]]
+\end{proof}
+
+\begin{proof}
+Let $x$ be an [[Definition:Even Integer|even integer]].
+Let $y = 2 n + 5$.
+Assume $y = x + 5$ is not an [[Definition:Odd Integer|odd integer]].
+Then:
+:$y = x + 5 = 2 n$
+where $n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 2 n - 5
+      | c = 
+}}
+{{eqn | r = \paren {2 n - 6} + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {n - 3} + 1
+      | c = 
+}}
+{{eqn | r = 2 r + 1
+      | c = where $r = n - 3 \in \Z$
+}}
+{{end-eqn}}
+Hence $x$ is [[Definition:Odd Integer|odd]].
+That is, it is [[Definition:False|false]] that $x$ is [[Definition:Even Integer|even]].
+It follows by the [[Rule of Transposition]] that if $x$ is [[Definition:Even Integer|even]], then $y$ is [[Definition:Odd Integer|odd]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Mapping of Separation}
+Tags: Topology, Continuous Mappings
+
+\begin{theorem}
+Let $T$ and $T'$ be [[Definition:Topological Space|topological spaces]].
+Let $A \mid B$ be a [[Definition:Separation (Topology)|separation]] of $T$.
+Let $f: T \to T'$ be a [[Definition:Mapping|mapping]] such that the [[Definition:Restriction of Mapping|restrictions]] $f \restriction_A$ and $f \restriction_B$ are both [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Then $f$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] on the whole of $T$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Continuity from Union of Restrictions]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Continuity from Union of Restrictions}
+Tags: Topology, Continuous Mappings
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Then $f: T_1 \to T_2$ is [[Definition:Continuous Mapping (Topology)|continuous]] if either:
+: $\displaystyle T_1 = \bigcup_{i \mathop = 1}^n V_i$ where:
+:: each $V_i$ is [[Definition:Closed Set (Topology)|closed]] in $T_1$, and 
+:: the [[Definition:Restriction of Mapping|restriction]] $f \restriction_{V_i}$ is [[Definition:Continuous Mapping (Topology)|continuous]] for each $V_i$
+or:
+: $\displaystyle T_1 = \bigcup_{i \mathop \in I} U_i$ where:
+:: $I$ is any (possibly [[Definition:Infinite|infinite]]) [[Definition:Indexing Set|indexing set]];
+:: each $U_i$ is [[Definition:Open Set (Topology)|open in $T_1$]], and
+:: the [[Definition:Restriction of Mapping|restriction]] $f \restriction_{U_i}$ is [[Definition:Continuous Mapping (Topology)|continuous]] for each $U_i$.
+\end{theorem}
+
+\begin{proof}
+=== First assertion ===
+Let $\displaystyle T_1 = \bigcup_{i \mathop  =1}^n V_i$, with the $V_i$ [[Definition:Closed Set (Topology)|closed]] in $T_1$.
+Assume that for each $i$, the [[Definition:Restriction of Mapping|restriction]] $f \restriction_{V_i}$ is continuous.
+Then $f$ satisfies the hypotheses of [[Continuous Mapping on Finite Union of Closed Sets]].
+Hence $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed}}
+=== Second assertion ===
+Let $\displaystyle T_1 = \bigcup_{i \mathop \in I} U_i$, with the $U_i$ [[Definition:Open Set (Topology)|open]] in $T_1$.
+Assume that for each $i \in I$, the [[Definition:Restriction of Mapping|restriction]] $f \restriction_{U_i}$ is continuous.
+Then $f$ satisfies the hypotheses of [[Continuous Mapping on Union of Open Sets]].
+Hence $f$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fermat's Last Theorem}
+Tags: Number Theory, Classic Problems, Fermat's Last Theorem
+
+\begin{theorem}
+:$\forall a, b, c, n \in \Z_{>0}, \; n > 2$, the equation $a^n + b^n = c^n$ has no solutions.
+\end{theorem}
+
+\begin{proof}
+The proof of this theorem is beyond the current scope of {{ProofWiki}}, and indeed, is beyond the understanding of many high level mathematicians.
+For the curious reader, the proof can be found [http://math.stanford.edu/~lekheng/flt/wiles.pdf here], in a paper published by {{AuthorRef|Andrew John Wiles|Andrew Wiles}}, entitled ''Modular elliptic curves and Fermat's Last Theorem'', in volume 141, issue 3, pages 443 through 551 of the ''Annals of Mathematics''.
+It is worth noting that Wiles' proof was indirect in that he proved a special case of the [[Taniyama-Shimura Conjecture]], which then along with the already proved [[Epsilon Conjecture]] implied that integral solutions of the theorem were impossible.
+{{ProofWanted|add a broad overview of Wiles proof}}
+\end{proof}<|endoftext|>
+\section{Subset of Real Numbers is Interval iff Connected}
+Tags: Real Intervals, Connected Spaces
+
+\begin{theorem}
+Let the [[Definition:Real Number Line|real number line]] $\R$ be considered as a [[Definition:Topological Space|topological space]].
+Let $S$ be a [[Definition:Topological Subspace|subspace]] of $\R$.
+Then $S$ is [[Definition:Connected (Topology)|connected]] {{iff}} $S$ is an [[Definition:Real Interval|interval]] of $\R$.
+That is, the only [[Definition:Topological Subspace|subspaces]] of $\R$ that are connected are [[Definition:Real Interval|intervals]].
+\end{theorem}
+
+\begin{proof}
+From [[Rule of Transposition]], we may replace the ''only if'' statement by its [[Definition:Contrapositive Statement|contrapositive]].
+Therefore, the following suffices:
+=== Sufficient Condition ===
+Suppose $S$ is an [[Definition:Real Interval|interval]] of $\R$.
+Suppose further that $A \mid B$ is a [[Definition:Separation (Topology)|separation]] of $S$.
+Let $a \in A, b \in B$.
+{{WLOG}}, suppose that $a < b$.
+Since $a, b \in S$, and $S$ is an [[Definition:Real Interval|interval]], $\closedint a b \subseteq S$.
+Let $A' = A \cap \closedint a b$ and $B' = B \cap \closedint a b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = A' \cup B'
+      | r = \paren {A \cap \closedint a b} \cup \paren {B \cap \closedint a b}
+}}
+{{eqn | r = \paren {A \cup B} \cap \closedint a b
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = A' \cup B'
+      | r = \closedint a b
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{end-eqn}}
+By the definition of a [[Definition:Separation (Topology)|separation]], both $A$ and $B$ are [[Definition:Closed Set (Topology)|closed]] in $S$.
+Hence by [[Closed Set in Topological Subspace]], $A'$ and $B'$ are also [[Definition:Closed Set (Topology)|closed]] in $\closedint a b$.
+From [[Closed Set in Topological Subspace/Corollary|Closed Set in Topological Subspace: Corollary]], $A'$ and $B'$ are [[Definition:Closed Set (Topology)|closed]] in $\R$.
+Now, since $B' \ne \O$, and $B$ is [[Definition:Bounded Below Set|bounded below]] (by, for example, $a$), by the [[Continuum Property]] $b' := \map \inf {B'}$ exists, and $b' \ge a$.
+We have that $B'$ is [[Definition:Closed Set (Topology)|closed]] in $\R$
+Hence from [[Closure of Real Interval is Closed Real Interval]]:
+:$b' \in B'$
+Since $a \in A'$ and $A \cap B = \O$, it follows that $b' > a$.
+Now let $A'' = A' \cap \closedint a {b'}$.
+Using the same argument as for $B'$, we have that $a'' = \map \sup {A''}$ exists, that $a'' \in A''$ and also $a'' < b'$.
+Now $\openint {a''} {b'} \cap A' = \O$ or $a''$ would not be an [[Definition:Upper Bound of Set|upper bound]] for $A''$.
+Similarly, $\openint {a''} {b'} \cap B' = \O$ or $b'$ would not be a [[Definition:Lower Bound of Set|lower bound]] for $B''$.
+Thus:
+:$\openint {a''} {b'} \cap \paren {A' \cup B'} = \O$
+But since $a < a'' < b' < b$, we also have:
+:$\openint {a''} {b'} \subseteq \closedint a b$, and
+:$\openint {a''} {b'}$ is [[Definition:Non-Empty Set|non-empty]].
+So, there is an element $z \in \openint {a''} {b'}$, and hence in $\closedint a b$, which is not in $A' \cup B'$.
+This contradicts $(1)$ above, which says that we have $A' \cup B' = \closedint a b$.
+From this [[Proof by Contradiction|contradiction]] it follows that there can be no such [[Definition:Separation (Topology)|separation]] $A \mid B$ on the [[Definition:Real Interval|interval]] $S$.
+Therefore, by definition, $S$ is [[Definition:Connected (Topology)|connected]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose $S$ is not an [[Definition:Real Interval|interval]] of $\R$.
+Then $\exists x, y \in S$ and $z \in \R \setminus S$ such that $x < z < y$.
+Consider the sets $S \cap \openint \gets z$ and $S \cap \openint z \to$.
+Then $S \cap \openint \gets z$ and $S \cap \openint z \to$ are [[Definition:Open Set (Topology)|open]] by definition of the [[Definition:Topological Subspace|subspace topology]] on $S$.
+Neither is [[Definition:Empty Set|empty]] because they contain $x$ and $y$ respectively.
+They are [[Definition:Disjoint Sets|disjoint]], and their [[Definition:Set Union|union]] is $S$, since $z \notin S$.
+Therefore $S \cap \openint \gets z \mid S \cap \openint z \to$ is a [[Definition:Separation (Topology)|separation]] of $S$.
+It follows by definition that $S$ is [[Definition:Connected (Topology)|disconnected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Image of Connected Space is Connected}
+Tags: Connected Spaces, Continuous Mappings, Continuous Image of Connected Space is Connected
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Let $S_1 \subseteq T_1$ be [[Definition:Connected Set (Topology)|connected]].
+Let $f: T_1 \to T_2$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+Then the [[Definition:Image of Mapping|image]] $f \left({S_1}\right)$ is [[Definition:Connected Set (Topology)|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $i: f \sqbrk {T_1} \to T_2$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
+Let $g: T_1 \to f \sqbrk {T_1}$ be the [[Surjection by Restriction of Codomain|surjective restriction]] of $f$.
+Then $f = i \circ g$.
+Hence, by [[Continuity of Composite with Inclusion/Inclusion on Mapping|Continuity of Composite with Inclusion: Inclusion on Mapping]], it follows that $g$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+We use a [[Proof by Contradiction]].
+Suppose that $A \mid B$ is a [[Definition:Partition (Topology)|partition]] of $f \sqbrk {T_1}$.
+Then it follows that $g^{-1} \sqbrk A \mid g^{-1} \sqbrk B$ is a [[Definition:Partition (Topology)|partition]] of $T_1$.
+{{explain|Add the justification for the above - may already exist as a result, otherwise add a proof something like [[Preimage of Partition is Partition]].}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose that $S_2 = f \left({S_1}\right)$ is not [[Definition:Connected Set (Topology)|connected]] in $T_2$.
+Then [[Definition:Connected (Topology)/Set/Definition 2|by definition]] there exist [[Definition:Open Set (Topology)|open sets]] $U_2$ and $V_2$ in $T_2$ such that:
+:$S_2 \subseteq U_2 \cup V_2$
+:$U_2 \cap V_2 \cap S_2 = \varnothing$
+:$U_2 \cap S_2 \ne \varnothing$
+:$V_2 \cap S_2 \ne \varnothing$
+[[Definition:By Hypothesis|By hypothesis]], $f: T_1 \to T_2$ is [[Definition:Continuous Mapping (Topology)|continuous]].
+Thus $U_1 = f^{-1} \left({U_2}\right)$ and $V_1 = f^{-1} \left({V_2}\right)$ are [[Definition:Open Set (Topology)|open]] in $T_1$.
+We have that:
+:$U_2 \cap S_2 \ne \varnothing$
+Therefore:
+:$\exists x \in S_1: f \left({x}\right) \in U_2$
+Then:
+:$x \in f^{-1} \left({U_2}\right) = U_1$
+and:
+:$x \in S_1$
+so:
+:$U_1 \cap S_1 \ne \varnothing$
+Similarly:
+:$V_1 \cap S_1 \ne \varnothing$
+Suppose there exists $x \in S_1$ such that $x \in U_1 \cap V_1 \cap S_1$.
+Then:
+:$f \left({x}\right) \in U_2 \cap V_2 \cap S_2$
+which is a [[Definition:Contradiction|contradiction]].
+It follows that:
+:$U_1 \cap V_1 \cap S_1 = \varnothing$
+Thus by definition $S_1$ is not [[Definition:Connected Set (Topology)|connected]] in $T_1$.
+The result follows by the [[Rule of Transposition]].
+\end{proof}<|endoftext|>
+\section{Union of Connected Sets with Non-Empty Intersections is Connected}
+Tags: Connected Spaces, Union of Connected Sets with Non-Empty Intersections is Connected
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\AA = \family {A_\alpha}_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Subset|subsets]] of $S$, all [[Definition:Connected Set (Topology)|connected]] in $T$.
+Let $\AA$ be such that no two of its [[Definition:Element|elements]] are [[Definition:Disjoint Sets|disjoint]]:
+:$\forall B, C \in \AA: B \cap C \ne \O$
+Then $\displaystyle \bigcup \AA$ is itself [[Definition:Connected Set (Topology)|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $A := \displaystyle \bigcup \AA$.
+Let $D = \set {0, 1}$, with the [[Definition:Discrete Space|discrete topology]].
+Let $f: A \to D$ be [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+To show that $A$ is [[Definition:Connected Set (Topology)|connected]], we need to show that $f$ is not a [[Definition:Surjection|surjection]].
+Since each $C \in \AA$ is [[Definition:Connected Set (Topology)|connected]] and the [[Definition:Restriction of Mapping|restriction]] $f \restriction_C$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]:
+:$\map f C = \set {\map \epsilon C}$
+where $\map \epsilon C = 0$ or $1$.
+But, for all $B, C \in \AA$:
+:$B \cap C \ne \O$
+Hence $\map \epsilon B = \map \epsilon C$. 
+Thus $f$ is [[Definition:Constant Mapping|constant]] on $A$ as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Product Space is Connected iff Factors are Connected/Basis for the Induction}
+Tags: Product Spaces, Connected Spaces
+
+\begin{theorem}
+Let $T_1$ and $T_2$ be [[Definition:Topological Space|topological spaces]].
+Then the [[Definition:Product Space (Topology)|product space]] $T_1 \times T_2$ is [[Definition:Connected Topological Space|connected]] {{iff}} $T_1$ and $T_2$ are [[Definition:Connected Topological Space|connected]].
+\end{theorem}
+
+\begin{proof}
+==== Necessary Condition ====
+Let $T_1 \times T_2$ be [[Definition:Connected Topological Space|connected]].
+By [[Projection from Product Topology is Continuous]], $T_1$ and $T_2$ are [[Definition:Everywhere Continuous Mapping (Topology)|continuous images]] under the [[Definition:Projection (Mapping Theory)|projections]] $\operatorname{pr}_1$ and $\operatorname{pr}_2$.
+Hence by [[Continuous Image of Connected Space is Connected]], $T_1$ and $T_2$ are [[Definition:Connected Topological Space|connected]].
+{{qed|lemma}}
+==== Sufficient Condition ====
+Suppose that $T_1$ and $T_2$ are [[Definition:Connected Topological Space|connected]].
+Let:
+: $I = T_2$
+: $\forall y \in T_2: C_y = T_1 \times \left\{{y}\right\}$
+: $B = \left\{{x_0}\right\} \times T_2$ for some fixed $x_0 \in T_1$.
+Each $C_y$ is [[Definition:Homeomorphic Topological Spaces|homeomorphic]] to $T_1$ by [[Topological Product with Singleton]].
+By [[Continuous Image of Connected Space is Connected/Corollary 1|Connectedness is a Topological Property]], each $C_y$ is therefore [[Definition:Connected Topological Space|connected]].
+By the same argument, $B$ is also [[Definition:Connected Topological Space|connected]].
+Also:
+: $C_y \cap B = \left\{{\left({x_0, y}\right)}\right\}$ and hence is [[Definition:Non-Empty Set|non-empty]]
+: $\displaystyle T_1 \times T_2 = B \cup \bigcup_{y \mathop \in T_2} C_y$.
+So by the [[Union of Connected Sets with Non-Empty Intersections is Connected/Corollary|corollary to Union of Connected Sets with Non-Empty Intersections is Connected]], it follows that $T_1 \times T_2$ is [[Definition:Connected Topological Space|connected]].
+\end{proof}<|endoftext|>
+\section{Set between Connected Set and Closure is Connected}
+Tags: Set Closures, Connected Sets, Set between Connected Set and Closure is Connected
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H$ be a [[Definition:Connected Set (Topology)|connected set]] of $T$.
+Let $H \subseteq K \subseteq H^-$, where $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of $H$. 
+Then $K$ is [[Definition:Connected Set (Topology)|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $D$ be the [[Definition:Discrete Topology|discrete space]] $\left\{{0, 1}\right\}$.
+Let $f: K \to D$ be an arbitrary [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+From [[Continuity of Composite with Inclusion]], the [[Definition:Restriction of Mapping|restriction]] $f \restriction_H$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+We have that:
+:$H$ is [[Definition:Connected Set (Topology)|connected]]
+:$f \restriction_H$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
+Thus by definition of [[Definition:Connected Set (Topology)|connected set]]:
+:$f \left({H}\right) = \left\{{0}\right\}$ or $f \left({H}\right) = \left\{{1}\right\}$ 
+{{WLOG}}, let $f \left({H}\right) = \left\{{0}\right\}$.
+{{AimForCont}} $\exists k \in K: f \left({k}\right) = 1$.
+By definition of [[Definition:Discrete Topology|discrete space]], $\left\{{1}\right\}$ is [[Definition:Open Set (Topology)|open]] in $D$.
+Hence by definition of [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]]:
+: $f^{-1} \left({\left\{{1}\right\}}\right)$ is [[Definition:Open Set (Topology)|open]] in $K$.
+Let $K$ be given the [[Definition:Topological Subspace|subspace topology]].
+Then for some $U$ [[Definition:Open Set (Topology)|open]] in $T$:
+: $f^{-1} \left({\left\{{1}\right\}}\right) = K \cap U$
+We have that:
+: $k \in f^{-1} \left({\left\{{1}\right\}}\right) \subseteq U$
+and:
+:$k \in H^-$
+By definition of [[Definition:Closure (Topology)|topology]]:
+: $\exists x \in H \cap U$
+As $x \in H$, we have that:
+:$f \left({x}\right) = 0$
+But because $x \in H \cap U \subseteq K \cap U = f^{-1} \left({\left\{{1}\right\}}\right)$:
+: $f \left({x}\right) = 1$
+This [[Definition:Contradiction|contradicts]] the definition of [[Definition:Mapping|mapping]].
+Thus by [[Proof by Contradiction]], $f: K \to D$ can not be a [[Definition:Surjection|surjection]].
+Thus $K$ is [[Definition:Connected Set (Topology)|connected]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $T_K = \struct {K, \tau_K}$ be the [[Definition:Topological Subspace|topological subspace]] of $T$ whose [[Definition:Underlying Set of Topological Space|underlying set]] is $K$.
+Let $\map {\cl_K} H$ denote the [[Definition:Closure (Topology)|closure]] of $H$ in $K$.
+From [[Closure of Subset in Subspace]]:
+:$\map {\cl_K} H = K \cap H^-$
+[[Definition:By Hypothesis|By hypothesis]]:
+:$K \subseteq H^-$
+and so by [[Intersection with Subset is Subset]]:
+:$\map {\cl_K} H = K$
+Let $D$ be the [[Definition:Discrete Topology|discrete space]] $\set {0, 1}$.
+Let $f: K \to D$ be any [[Definition:Everywhere Continuous Mapping (Topology)|continuous mapping]].
+From [[Continuity of Composite with Inclusion]], the [[Definition:Restriction of Mapping|restriction]] $f \restriction_H$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+SWe have that:
+:$H$ is [[Definition:Connected Set (Topology)|connected]]
+:$f \restriction_H$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]
+Thus by definition of [[Definition:Connected Set (Topology)|connected set]]:
+:$f \sqbrk H = \set 0$ or $f \sqbrk H = \set 1$ 
+{{WLOG}}, let $f \sqbrk H = \set 0$.
+From [[Continuity Defined by Closure]]:
+:$f \sqbrk {\map {\cl_K} H} \subseteq \map {\cl_K} {f \sqbrk H} = \set 0^-$
+where $\set 0^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\set 0$ in $D$.
+As $D$ is the [[Definition:Discrete Space|discrete space]], it follows from [[Set in Discrete Topology is Clopen]] that $\set 0$ is [[Definition:Closed Set (Topology)|closed]] in $D$.
+Thus by [[Set is Closed iff Equals Topological Closure]]:
+:$\set 0^- = \set 0$
+That is, $f \sqbrk K = \set 0$.
+Thus $K$ is [[Definition:Connected Set (Topology)|connected]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Path-Connected Space is Connected}
+Tags: Path-Connected Spaces, Connected Spaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]] which is [[Definition:Path-Connected Space|path-connected]].
+Then $T$ is [[Definition:Connected Topological Space|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $D$ be the [[Definition:Discrete Topology|discrete space]] $\set {0, 1}$.
+Let $T$ be [[Definition:Path-Connected Space|path-connected]].
+Let $f: T \to D$ be a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
+Let $x, y \in T: \map f x = 0, \map f y = 1$.
+Let $I \subset \R$ be the [[Definition:Closed Real Interval|closed real interval]] $\closedint 0 1$.
+Let $g: I \to T$ be a [[Definition:Path (Topology)|path]] from $x$ to $y$.
+Then by [[Continuity of Composite Mapping]] it follows that $f \circ g: I \to D$ is a [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] [[Definition:Surjection|surjection]].
+This contradicts the [[Definition:Connected Topological Space|connectedness]] of $I$ as proved in [[Subset of Real Numbers is Interval iff Connected]].
+{{explain|Why does that follow? Explain what the chain of steps is}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hausdorff Paradox}
+Tags: Topology, Veridical Paradoxes, Antinomies
+
+\begin{theorem}
+There is a disjoint [[Definition:Decomposition (Topology)|decomposition]] of the [[Definition:Sphere (Topology)|sphere]] $\mathbb S^2$ into four sets $A, B, C, D$ such that $A, B, C, B \cup C$ are all [[Definition:Congruence (Topology)|congruent]] and $D$ is [[Definition:Countable|countable]].
+\end{theorem}
+
+\begin{proof}
+Let $R \subset \mathbb{SO} \left({3}\right)$ be the [[Definition:Group|group]] [[Definition:Generator of Group|generated]] by the $\pi$ and $\dfrac {2 \pi} 3$ rotations around different axes.
+The elements:
+:$\psi = \begin{pmatrix} -\tfrac 1 2 & \tfrac{\sqrt 3} 2 & 0 \\ -\tfrac{\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}$
+:$\phi = \begin{pmatrix} -\cos \left({\vartheta}\right) & 0 & \sin \left({\vartheta}\right) \\ 0 & -1 & 0 \\ \sin \left({\vartheta}\right) & 0 & \cos \left({\vartheta}\right) \\ \end{pmatrix}$
+form a [[Definition:Basis (Topology)|basis]] for $R$ for some $\vartheta$.
+We have:
+:$\psi^3 = \phi^2 = \mathbf I_3$
+where $\mathbf I_3$ is the [[Definition:Identity Mapping|identity mapping]] in $\mathbb R^3$.
+Therefore $\forall r \in R: r \ne \mathbf I_3, \psi, \phi, \psi^2$ there exists some $n \in \N$ and some set of numbers $m_k \in \left\{{1, 2}\right\}, 1 \le k \le n$ such that $r$ can written as one of the following:
+:$\displaystyle \text{(a)}: r = \prod_{k \mathop = 1}^n \phi \psi^{m_k}$
+:$\displaystyle \text{(b)}: r = \psi^{m_1} \left({\prod_{k \mathop = 2}^n \phi \psi^{m_k}}\right) \phi$
+:$\displaystyle \text{(c)}: r = \left({\prod_{k \mathop = 1}^n \phi \psi^{m_k}}\right) \phi$
+:$\displaystyle \text{(d)}: r = \psi^{m_1} \left({\prod_{k \mathop = 2}^n \phi \psi^{m_k}}\right)$
+Now we fix $\vartheta$ such that $\mathbf I_3$ cannot be written in any of the ways $\text{(a)}$, $\text{(b)}$, $\text{(c)}$, $\text{(d)}$. 
+The action of $R$ on $\mathbb S^2$ will leave two points unchanged for each element of $R$ (the intersection of the axis of rotation and the sphere, to be exact).  
+Since $R$ is finitely generated, it is a [[Definition:Countable|countable]] group.
+Therefore the set of points of $\mathbb S^2$ which are unchanged by at least one element of $R$ is also countable.  
+We call this set $D \subset \mathbb S^2$, so that $R$ acts freely on $\mathbb S^2 - D$.
+By [[Set of Orbits forms Partition]], this [[Definition:Set Partition|partitions]] $\mathbb S^2 - D$ into [[Definition:Orbit (Group Theory)|orbits]].
+By the [[Axiom:Axiom of Choice|Axiom of Choice]], there is a set $X$ containing one element of each [[Definition:Orbit (Group Theory)|orbit]].  
+For any $r \in R$, let $X_r$ be the [[Definition:Group Action|action]] of $r$ on $X$.
+We have:
+:$\displaystyle \mathbb S^2 - D = \bigcup_{r \mathop \in R} X_r$
+Define the sets $A, B, C$ to be the smallest sets satisfying 
+:$X \subseteq A$
+:$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi} \subset B, A, C, \text{respectively.}$
+:$\text{If } X_r \subset A, B, C, \text{ then } X_{r\psi} \subset B, C, A, \text{respectively.}$
+:$\text{If } X_r \subset A, B, C, \text{ then } X_{r\phi^2} \subset C, A, B, \text{respectively.}$
+These sets are defined due to the uniqueness of the properties $\text{(a)}$ to $\text{(d)}$
+Also, $A, B, C, B \cup C$ are congruent since they are rotations of each other, namely:
+:$A_\psi = B, B_{\psi^2} = C, A_\phi = B \cup C$
+Hence we have constructed the sets $A, B, C, D$ of the theorem.
+{{qed}}
+{{AoC}}
+Whether you view this result as a [[Definition:Veridical Paradox|veridical paradox]] or an [[Definition:Antinomy|antinomy]] depends your acceptance or otherwise of the [[Axiom:Axiom of Choice|Axiom of Choice]].
+{{namedfor|Felix Hausdorff|cat = Hausdorff}}
+\end{proof}<|endoftext|>
+\section{Closed Topologist's Sine Curve is not Path-Connected}
+Tags: Path-Connected Sets, Reciprocals, Sine Function
+
+\begin{theorem}
+Let $G$ be the [[Definition:Graph of Mapping|graph]] of the [[Definition:Real Function|function]] $y = \sin \left({\dfrac 1 x}\right)$ for $x > 0$.
+Let $J$ be the [[Definition:Line Segment|line segment]] joining the points $\left({0, -1}\right)$ and $\left({0, 1}\right)$ in $\R^2$.
+Then while $G \cup J$ is [[Definition:Connected Set (Topology)|connected]], it is '''not''' [[Definition:Path-Connected Set (Topology)|path-connected]].
+\end{theorem}
+
+\begin{proof}
+From [[Closed Topologist's Sine Curve is Connected]], $G \cup J$ is [[Definition:Connected Set (Topology)|connected]].
+Let $I \subseteq \R$ be the [[Definition:Closed Real Interval|closed real interval]] $\left[{0 \,.\,.\, 1}\right]$.
+Let $A = \left({\dfrac 1 \pi, 0}\right) \in \R^2$.
+This proof is based on the fact that a [[Definition:Path (Topology)|continuous path]] $f: I \to G \cup J$ beginning at $A$ will never actually arrive at $0 = \left({0, 0}\right) \in \R^2$ because $I$ is [[Definition:Compact Subset of Real Numbers|compact]].
+Suppose $f: I \to G \cup J$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and that $f \left({0}\right) = A$.
+Let $f_1: I \to \R$ and $f_2: I \to \R$ denote ${\operatorname{pr}_1} \circ i \circ f$ and ${\operatorname{pr}_2} \circ i \circ f$, where $i$ is the [[Definition:Inclusion Mapping|inclusion mapping]] and $\operatorname{pr}_1, \operatorname{pr}_2$ the [[Definition:Projection (Mapping Theory)|first and second projections]] on the $x$ and $y$ axes.
+Thus $f_2$ describes the vertical movement of the graph, and $f_1$ the horizontal movement.
+Now $f_2$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]] and $I$ [[Definition:Compact Subset of Real Numbers|compact]].
+So by the [[Heine-Cantor Theorem]], $f_2$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]] on $I$.
+Let $\delta > 0$ be such that $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| < 2$ for any $t, t' \in I$ such that $\left|{t - t'}\right| < \delta$.
+Let $0 < t_0 < t_1 < \ldots < t_n = 1$ be such that $t_i - t_{i-1} < \delta$ for all $i = 1, 2, \ldots, n$.
+Now: as $t$ goes from $t_0$ to $t_1$, $f \left({t}\right)$ (which starts at $A$), can not reach a point $C$ where $y = 1$ without passing through a point $B$ where $y = -1$ on the way.
+But then $f \left({t}\right) = B$ and $f \left({t'}\right) = C$ for some $t, t'$ where $\left|{t - t'}\right| < \delta$.
+And since $B$ and $C$ are $2$ apart, $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ which contradicts the choice of $\delta$.
+Similarly, when going from $t_1$ to $t_2$, $t$ can similarly not get past more than one hump.
+So, as $t$ goes from $0$ to $1$, $f \left({t}\right)$ can not traverse more than $n$ humps.
+We formalize this discussion by [[Principle of  Mathematical Induction|induction]].
+We will show that:
+: $\forall i = 0, 1, \ldots, n: f_1 \left({t_i}\right) > \dfrac 2 {\left({2i + 3}\right) \pi}$
+Since $f \left({t_0}\right) = f \left({0}\right) = \left({\dfrac 1 \pi, 0}\right)$, we have $f_1 \left({t_0}\right) = \dfrac 1 \pi > \dfrac 2 {3 \pi}$.
+Suppose that $f_1 \left({t_i}\right) > \dfrac 2 {\left({2 i + 3}\right) \pi}$ for some $i \ge 0$.
+{{AimForCont}} also that $f_1 \left({t_{i+1}}\right) \ge \dfrac 2 {\left({2 i + 5}\right) \pi}$.
+Then since $f_1$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]], by the [[Definition:Intermediate Value Property|I.V.P.]] there exists $t, t' \in \left[{t_1 \,.\,.\, t_{i + 1} }\right]$ such that:
+: $f_1 \left({t_i}\right) = \dfrac 2 {\left({2 i + 3}\right) \pi}, f_1 \left({t_{i + 1} }\right) = \dfrac 2 {\left({2 i + 5}\right) \pi}$
+But the only point in $G \cup J$ whose first coordinate is $\dfrac 2 {\left({2 i + 3}\right) \pi}$ is $\left({\dfrac 2 {\left({2 i + 3}\right) \pi}, \sin \left({\dfrac {\left({2 i + 3}\right) \pi} 2}\right)}\right)$.
+So:
+: $f_2 \left({t}\right) = \sin \left({\dfrac {\left({2 i + 3}\right) \pi} 2}\right)$
+Similarly:
+: $f_2 \left({t'}\right) = \sin \left({\dfrac {\left({2 i + 5}\right) \pi} 2}\right)$
+Hence $\left|{f_2 \left({t}\right) - f_2 \left({t'}\right)}\right| = 2$ while $\left|{t - t'}\right| < \delta$.
+This contradicts the choice of $\delta$.
+This proves the induction.
+The result follows from the fact that:
+: $f_1 \left({1}\right) > \dfrac 2 {\left({2 n + 3}\right) \pi} > 0$, so $f_1 \left({1}\right) \ne 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Two Sides of Triangle Greater than Third Side}
+Tags: Triangles, Triangle Inequality
+
+\begin{theorem}
+Given a [[Definition:Triangle (Geometry)|triangle]] $ABC$, the sum of the [[Definition:Length of Line|lengths]] of any two [[Definition:Side of Polygon|sides]] of the triangle is greater than the [[Definition:Length of Line|length]] of the third [[Definition:Side of Polygon|side]].
+{{:Euclid:Proposition/I/20}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Triangle Inequality.png|350 px]]
+Let $ABC$ be a [[Definition:Triangle (Geometry)|triangle]]
+[[Axiom:Euclid's Second Postulate|We can extend]] $BA$ past $A$ into a [[Definition:Straight Line|straight line]].
+[[Construction of Equal Straight Lines from Unequal|There exists a point $D$]] such that $DA = CA$.
+Therefore, from [[Isosceles Triangle has Two Equal Angles]]:
+:$\angle ADC = \angle ACD$
+Thus by [[Axiom:Euclid's Common Notions|Euclid's fifth common notion]]:
+:$\angle BCD > \angle BDC$
+Since $\triangle DCB$ is a triangle having $\angle BCD$ greater than $\angle BDC$, [[Greater Angle of Triangle Subtended by Greater Side|this means that $BD > BC$]].
+But:
+:$BD = BA + AD$
+and:
+:$AD = AC$
+Thus:
+:$BA + AC > BC$
+A similar argument shows that $AC + BC > BA$ and $BA + BC > AC$.
+{{qed}}
+{{Euclid Note|20|I|It is a [[Definition:Euclidean Geometry|geometric]] interpretation of the [[Triangle Inequality|triangle inequality]].}}
+\end{proof}<|endoftext|>
+\section{Lines Through Endpoints of One Side of Triangle to Point Inside Triangle is Less than Sum of Other Sides}
+Tags: Triangles
+
+\begin{theorem}
+Given a [[Definition:Triangle (Geometry)|triangle]] and a [[Definition:Point|point]] inside it, the sum of the lengths of the [[Definition:Line Segment|line segments]] from the [[Definition:Endpoint of Line|endpoints]] of one [[Definition:Side of Polygon|side]] of the triangle to the point is less than the sum of the other two sides of the triangle.
+{{:Euclid:Proposition/I/21}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Point Inside Triangle.png|250px]]
+Given a triangle $ABC$ and a point $D$ inside it.
+We can [[Axiom:Euclid's First Postulate|construct]] lines connecting $A$ and $B$ to $D$, and then [[Axiom:Euclid's Second Postulate|extend]] the line $AD$ to a point $E$ on $BC$.
+In $\triangle ABE$, [[Sum of Two Sides of Triangle Greater than Third Side|$AB + AE>BE$]].
+Then, $AB + AC = AB + AE + EC > BE + EC$ by [[Axiom:Euclid's Common Notions|Euclid's second common notion]].
+Similarly, $CE + ED > CD$, so $CE + EB = CE + ED + DB > CD + DB$.
+Thus, $AB + AC > BE + EC > CD + DB$.
+{{qed}}
+{{Euclid Note|21|I}}
+\end{proof}<|endoftext|>
+\section{Construction of Triangle from Given Lengths}
+Tags: Triangles
+
+\begin{theorem}
+Given three [[Definition:Straight Line|straight lines]] such that the sum of the lengths of any two of the lines is greater than the length of the third line, it is possible to construct a [[Definition:Triangle (Geometry)|triangle]] having the lengths of these lines as its side lengths.
+{{:Euclid:Proposition/I/22}}
+\end{theorem}
+
+\begin{proof}
+Since $F$ is the center of the circle with radius $FD$, it follows from {{EuclidDefLink|I|15|Circle}} that $DF = KF$, so $a = KF$ by [[Axiom:Euclid's Common Notions|Euclid's first common notion]].
+Since $G$ is the center of the circle with radius $GH$, it follows from {{EuclidDefLink|I|15|Circle}} that $GH = GK$, so $c = GK$ by [[Axiom:Euclid's Common Notions|Euclid's first common notion]].
+$FG = b$ by construction.
+Therefore the lines $FK$, $FG$, and $GK$ are, respectively, equal to the lines $a$, $b$, and $c$, so $\triangle FGK$ is indeed the required triangle.
+{{qed}}
+{{Euclid Note|22|I}}
+Note that the condition required of the lengths of the segments is the equality shown in [[Sum of Two Sides of Triangle Greater than Third Side|Proposition $20$: Sum of Two Sides of Triangle Greater than Third Side]].  Thus, this is a [[Definition:Necessary Condition|necessary condition]] for the construction of a triangle.
+When {{AuthorRef|Euclid}} first wrote the proof of this proposition in {{BookLink|The Elements|Euclid}}, he neglected to prove that the two circles described in the construction actually do intersect, just as he did in [[Construction of Equilateral Triangle|Proposition $1$: Construction of Equilateral Triangle]].
+\end{proof}<|endoftext|>
+\section{Connected Open Subset of Euclidean Space is Path-Connected}
+Tags: Connected Spaces, Path-Connected Spaces, Euclidean Space
+
+\begin{theorem}
+Let $\R^n$ be a [[Definition:Euclidean Space|Euclidean $n$-space]].
+Let $U$ be a [[Definition:Connected (Topology)|connected]] [[Definition:Open Set (Topology)|open subset]] of $\R^n$.
+Then $U$ is [[Definition:Path-Connected|path-connected]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in U$.
+Let $H \subseteq U$ be the [[Definition:Subset|subset]] of points in $U$ which can be joined to $a$ by a [[Definition:Path (Topology)|path]] in $U$.
+Let $K = U \setminus H$.
+Let $x \in H$.
+Then:
+: $\exists \epsilon > 0: B_\epsilon \left({x}\right) \subseteq U$
+where $B_\epsilon \left({x}\right)$ is the [[Definition:Open Ball|open $\epsilon$-ball]] of $x$.
+Given any $y \in B_\epsilon \left({x}\right)$, there is a ([[Definition:Straight Line|straight line]]) [[Definition:Path (Topology)|path]] $g$ in $B_\epsilon \left({x}\right) \subseteq U$ connecting $x$ and $y$.
+But since $x \in H$, there is a [[Definition:Path (Topology)|path]] $f$ in $U$ joining $a$ to $x$.
+From [[Joining Paths makes Another Path]], traversing $f$, and then $g$ forms a [[Definition:Path (Topology)|path]] from $a$ to $y$.
+It follows that $y \in H$, and therefore $B_\epsilon \left({x}\right) \subseteq H$.
+Thus $H$ is [[Definition:Open Set (Topology)|open]].
+By a similar argument, $K$ is also shown to be [[Definition:Open Set (Topology)|open]]:
+If $x \in K$, then $B_\epsilon \left({x}\right) \subseteq U$ for some $\epsilon > 0$.
+If any point in $B_\epsilon \left({x}\right)$ can be joined to $a$ by a [[Definition:Path (Topology)|path]] in $U$, then so could $x$.
+It is clear that $H \cap K = \varnothing$ and $H \cup K = U$ by definition of [[Definition:Set Difference|set difference]].
+As, trivially, $a \in H$, we have $H \ne \varnothing$.
+Knowing that $U$ is [[Definition:Connected (Topology)|connected]], it follows that $K = \varnothing$, and $H = U$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Joining Paths makes Another Path}
+Tags: Topology, Path-Connected Sets
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $I \subseteq \R$ be the [[Definition:Closed Real Interval|closed real interval]] $\closedint 0 1$.
+Let $f, g: I \to T$ be [[Definition:Path (Topology)|paths]] in $T$ from $a$ to $b$ and from $b$ to $c$ respectively.
+Let $h: I \to T$ be the [[Definition:Mapping|mapping]] given by:
+$\map h x = \begin{cases}
+\map f {2x} & : x \in \closedint 0 {\dfrac 1 2} \\
+\map g {2x - 1} & : x \in \closedint {\dfrac 1 2} 1
+\end{cases}$
+Then $h$ is a [[Definition:Path (Topology)|path]] in $T$.
+\end{theorem}
+
+\begin{proof}
+First we see that $h$ is [[Definition:Well-Defined Mapping|well-defined]], because on $\closedint 0 {\dfrac 1 2} \cap \closedint {\dfrac 1 2} 1 = \set {\dfrac 1 2}$ we have $\map f 1 = b = \map g 0$.
+Now $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } } \mathop = f \circ k$ where $k: \closedint 0 {\dfrac 1 2} \to \closedint 0 1$ is given by $\map k x = 2 x$.
+So by [[Continuity of Composite Mapping]], $\mathbin h {\restriction_{\closedint 0 {\frac 1 2} } }$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Similarly, $\mathbin h {\restriction_{\closedint {\frac 1 2} 1} }$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+By [[Continuity from Union of Restrictions]], it follows that $h$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]].
+Finally, $\map h 0 = \map f 0 = a$ and $\map h 1 = \map g 1 = c$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hinge Theorem}
+Tags: Triangles, Named Theorems
+
+\begin{theorem}
+If two [[Definition:Triangle (Geometry)|triangles]] have two pairs of [[Definition:Side of Polygon|sides]] which are the same length, the triangle with the larger included angle also has the larger third side.
+{{:Euclid:Proposition/I/24}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Hinge Theorem.png|250px]]
+Let $\triangle ABC$ and $DEF$ be two [[Definition:Triangle (Geometry)|triangles]] in which $AB = DE$, $AC = DF$, and $\angle CAB > \angle FDE$.
+[[Construction of Equal Angle|Construct $\angle EDG$]] on $DE$ at [[Definition:Point|point]] $D$.
+[[Construction of Equal Straight Lines from Unequal|Place $G$ so that $DG = AC$]].
+[[Axiom:Euclid's First Postulate|Join]] $EG$ and $FG$.
+Since $AB = DE$, $\angle BAC = \angle EDG$, and $AC = DG$, by [[Triangle Side-Angle-Side Equality]]:
+:$BC = GE$
+By [[Axiom:Euclid's Common Notions|Euclid's first common notion]]:
+:$DG = AC = DF$
+Thus by [[Isosceles Triangle has Two Equal Angles]]:
+:$\angle DGF = \angle DFG$
+So by [[Axiom:Euclid's Common Notions|Euclid's fifth common notion]]:
+:$\angle EFG \, > \, \angle DFG = \angle DGF \, > \, \angle EGF$
+Since $\angle EFG > \angle EGF$, by [[Greater Angle of Triangle Subtended by Greater Side]]:
+:$EG > EF$
+Therefore, because $EG = BC$, $BC > EF$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Component}
+Tags: Connected Sets
+
+\begin{theorem}
+{{TFAE|def = Component (Topology)|view = Component|context = Topology (Mathematical Branch)|contextview = Topology}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $x \in T$.
+\end{theorem}
+
+\begin{proof}
+Let $\CC_x = \set {A \subseteq S : x \in A \land A \text{ is connected in } T}$
+Let $C = \bigcup \CC_x$
+=== [[Equivalence of Definitions of Component/Lemma 1|Lemma]] ===
+{{:Equivalence of Definitions of Component/Lemma 1}}{{qed|lemma}}
+Let $C'$ be the [[Definition:Equivalence Class|equivalence class containing $x$]] of the [[Definition:Equivalence Relation|equivalence relation]] $\sim$ defined by:
+:$y \sim z$ {{iff}} $y$ and $z$ are [[Definition:Connected Points (Topology)|connected]] in $T$.
+=== [[Equivalence of Definitions of Component/Equivalence Class equals Union of Connected Sets|Equivalence Class equals Union of Connected Sets]] ===
+It needs to be shown that $C = C'$.
+{{:Equivalence of Definitions of Component/Equivalence Class equals Union of Connected Sets}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Component/Union of Connected Sets is Maximal Connected Set|Union of Connected Sets is Maximal Connected Set]] ===
+{{:Equivalence of Definitions of Component/Union of Connected Sets is Maximal Connected Set}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Component/Maximal Connected Set is Union of Connected Sets|Maximal Connected Set is Union of Connected Sets]] ===
+{{:Equivalence of Definitions of Component/Maximal Connected Set is Union of Connected Sets}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Converse Hinge Theorem}
+Tags: Triangles, Named Theorems
+
+\begin{theorem}
+If two [[Definition:Triangle (Geometry)|triangles]] have two pairs of [[Definition:Side of Polygon|sides]] which are the same [[Definition:Length (Linear Measure)|length]], the [[Definition:Triangle (Geometry)|triangle]] in which the third [[Definition:Side of Polygon|side]] is longer also has the larger [[Definition:Angle|angle]] [[Definition:Containment of Angle|contained]] by the first two [[Definition:Side of Polygon|sides]].
+{{:Euclid:Proposition/I/25}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Converse Hinge Theorem.png|300px]]
+Let $\triangle ABC$ and $\triangle DEF$ be two [[Definition:Triangle (Geometry)|triangles]] in which:
+:$AB = DF$
+:$AC = DE$
+:$BC > EF$
+{{AimForCont}} that $\angle BAC \not > \angle EDF$.
+Then either:
+:$\angle BAC = \angle EDF$
+or:
+:$\angle BAC < \angle EDF$
+Let $\angle BAC = \angle EDF$.
+Then by [[Triangle Side-Angle-Side Equality]]:
+:$BC = EF$
+But we know this is not the case, so by [[Proof by Contradiction]]:
+:$\angle BAC \ne \angle EDF$
+Suppose $\angle BAC < \angle EDF$.
+Then by [[Greater Angle of Triangle Subtended by Greater Side]]:
+:$EF > BC$
+But we know this is not the case, so by [[Proof by Contradiction]]:
+:$\angle BAC \not < \angle EDF$
+Thus:
+:$\angle BAC > \angle EDF$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequentially Compact Metric Subspace is Sequentially Compact in Itself iff Closed}
+Tags: Metric Subspaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $M$ be a [[Definition:Metric Space|metric space]].
+Let $C \subseteq M$ be a [[Definition:Metric Subspace|subspace]] of $M$ which is [[Definition:Sequentially Compact Space|sequentially compact]] in $M$.
+Then $C$ is [[Definition:Sequentially Compact In Itself|sequentially compact in itself]] {{iff}} $C$ is [[Definition:Closed Set (Topology)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Closure of Subset of Metric Space by Convergent Sequence]].
+{{qed}}
+[[Category:Metric Subspaces]]
+[[Category:Sequentially Compact Spaces]]
+mqj2iylshn2871w02l98raps0bmr2uw
+\end{proof}<|endoftext|>
+\section{Invariance of Extremal Length under Conformal Mappings}
+Tags: Geometric Function Theory
+
+\begin{theorem}
+Let $X, Y$ be [[Definition:Riemann Surface|Riemann surfaces]] (usually, [[Definition:Subset|subsets]] of the [[Definition:Complex Plane|complex plane]]).
+Let $\phi: X \to Y$ be a [[Definition:Conformal Isomorphism|conformal isomorphism]] between $X$ and $Y$.
+Let $\Gamma$ be a [[Definition:Indexed Family|family]] of [[Definition:Rectifiable Curve|rectifiable curves]] (or, more generally, of [[Definition:Set Union|unions]] of [[Definition:Rectifiable Curve|rectifiable curves]]) in $X$.
+Let $\Gamma'$ be the [[Definition:Indexed Family|family]] of their [[Definition:Image of Mapping|images]] under $\phi$.
+Then $\Gamma$ and $\Gamma'$ have the same [[Definition:Extremal Length|extremal length]]:
+:$\map \lambda \Gamma = \map \lambda {\Gamma'}$
+\end{theorem}
+
+\begin{proof}
+Let $\rho'$ be a [[Definition:Conformal Metric|conformal metric]] on $Y$ in the sense of the definition of [[Definition:Extremal Length|extremal length]], given in local coordinates as:
+:$\map {\rho'} z \size {\d z}$
+Let $\rho$ be the [[Definition:Metric|metric]] on $X$ obtained as the [[Definition:Pull-Back|pull-back]] of this [[Definition:Metric|metric]] under $\phi$.
+That is, $\rho$ is given in local coordinates as:
+:$\map {\rho'} {\map \phi w} \cdot \size {\map {\dfrac {\d \phi} {\d w} } w} \cdot \size {\d w}$
+Then the area of $X$ with respect to $\rho$ and the area of $Y$ with respect to $\rho'$ are equal by definition:
+:$\map A {\rho'} = \map A \rho$
+Furthermore, if $\gamma \in \Gamma$ and $\gamma' := \map \phi \gamma$, then also:
+:$\map L {\gamma, \rho} = \map L {\gamma', \rho'}$
+and hence:
+:$\map L {\Gamma, \rho} = \map L {\Gamma', \rho'}$
+In summary, for any metric $\rho'$ on $Y$, there is a [[Definition:Metric|metric]] $\rho$ on $X$ such that:
+:$\dfrac {\map L {\Gamma, \rho} } {\map A \rho} = \dfrac {\map L {\Gamma', \rho'} } {\map A {\rho'} }$
+It thus follows from the definition of [[Definition:Extremal Length|extremal length]] that:
+:$\map \lambda \Gamma \ge \map \lambda {\Gamma'}$
+The opposite inequality follows by exchanging the roles of $X$ and $Y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Line is Complete Metric Space}
+Tags: Complete Metric Spaces, Real Number Line with Euclidean Metric
+
+\begin{theorem}
+The [[Definition:Real Number Line with Euclidean Metric|real number line $\R$ with the usual (Euclidean) metric]] forms a [[Definition:Complete Metric Space|complete metric space]].
+\end{theorem}
+
+\begin{proof}
+From [[Real Number Line is Metric Space]], the [[Definition:Distance Function|distance function]] defined as $\map d {x, y} = \size {x - y}$ is a [[Definition:Metric|metric]] on $\R$.
+It remains to be shown that the [[Definition:Metric Space|metric space]] $\struct {\R, d}$ is [[Definition:Complete Metric Space|complete]].
+By definition, this is done by demonstrating that every [[Definition:Cauchy Sequence|Cauchy sequence]] of [[Definition:Real Number|real numbers]] has a [[Definition:Limit of Sequence (Metric Space)|limit]].
+This is demonstrated in [[Cauchy Sequence Converges on Real Number Line]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Series Law for Extremal Length}
+Tags: Geometric Function Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Riemann Surface|Riemann surface]].
+Let $\Gamma_1$, $\Gamma_2$ and $\Gamma$ be families of [[Definition:Rectifiable Curve|rectifiable curves]] (or, more generally, families of unions of rectifiable curves) on $X$.
+Let every $\gamma \in \Gamma$ contain a $\gamma_1 \in \Gamma_1$ and a $\gamma_2 \in \Gamma_2$ such that $\gamma_1 \cap \gamma_2 = \varnothing$.
+Then the [[Definition:Extremal Length|extremal lengths]] of $\Gamma_1$, $\Gamma_2$ and $\Gamma$ satisfy:
+:$\lambda \left({\Gamma}\right) \ge \lambda \left({\Gamma_1}\right) + \lambda \left({\Gamma_2}\right)$
+\end{theorem}
+
+\begin{proof}
+Let $\rho_1 = \rho_1 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ and $\rho_2 = \rho_2 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ be conformal metrics as in the [[Definition:Extremal Length|definition of extremal length]].
+It can be assumed that these are [[Definition:Extremal Length#Normalizations|normalized]]:
+:$A \left({\rho_j}\right) = L \left({\Gamma_j, \rho_j}\right)$ for $j \in \left\{ {1, 2}\right\}$. 
+We define another metric $\rho = \rho \left({z}\right) \left\vert{\mathrm d z}\right\vert$ by:
+:$\rho \left({z}\right) := \max \left({\rho_1 \left({z}\right), \rho_2 \left({z}\right)}\right)$
+Note that [[Series Law for Extremal Length/Rho is Well Defined|this is a well-defined metric]].
+By definition, the area form $\rho^2 \left({z}\right) \left\vert{\mathrm d z}\right\vert$ satisfies:
+:$\rho^2 \left({z}\right) \left\vert{\mathrm d z}\right\vert^2 = \max \left({\rho_1 \left({z}\right)^2, \rho_2 \left({z}\right)^2}\right) \left\vert{\mathrm d z}\right\vert^2 \le \left({\rho_1 \left({z}\right)^2 + \rho_2 \left({z}\right)^2}\right) \left\vert{\mathrm d z}\right\vert^2$
+Hence:
+:$A \left({\rho}\right) \le A \left({\rho_1}\right) + A \left({\rho_2}\right) = L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$
+On the other hand, let $\gamma \in \Gamma$.
+Let $\gamma_1$, $\gamma_2$ be as in the assumption.
+Then:
+{{begin-eqn}}
+{{eqn | l = L \left({\gamma, \rho}\right)
+      | o = \ge
+      | r = L \left({\gamma_1, \rho}\right) + L  \left({\gamma_2, \rho}\right)
+      | c = as $\gamma_1$ and $\gamma_2$ are disjoint
+}}
+{{eqn | o = \ge
+      | r = L \left({\gamma_1, \rho_1}\right) + L \left({\gamma_2, \rho_2}\right)
+      | c = Definition of $\rho$
+}}
+{{eqn | o = \ge
+      | r = L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)
+      | c = Definition of $L \left({\Gamma_j, \rho_j}\right)$
+}}
+{{end-eqn}}
+Thus 
+:$ L \left({\Gamma, \rho}\right) \ge L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$
+Combining this with the inequality for the area:
+:$\dfrac {L \left({\Gamma, \rho}\right)^2} {A \left({\rho}\right)} \ge \dfrac {\left({L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)}\right)^2} {L \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)} = L  \left({\Gamma_1, \rho_1}\right) + L \left({\Gamma_2, \rho_2}\right)$
+Taking the supremum over all metrics $\rho_1$ and $\rho_2$ as above:
+:$ L \left({\Gamma}\right) \ge L \left({\Gamma_1}\right) + L \left({\Gamma_2}\right)$
+as claimed.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extremal Length of Union}
+Tags: Geometric Function Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Riemann Surface|Riemann surface]].
+Let $\Gamma_1$ and $\Gamma_2$ be families of [[Definition:Rectifiable Curve|rectifiable curves]] (or, more generally, families of unions of rectifiable curves) on $X$. 
+Then the [[Definition:Extremal Length|extremal length]] of their union satisfies:
+:$\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} \le \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$
+Suppose that additionally $\Gamma_1$ and $\Gamma_2$ are disjoint in the following sense: there exist disjoint Borel subsets:
+:$A_1, A_2 \subseteq X$ such that $\displaystyle \bigcup \Gamma_1 \subset A_1$ and $\displaystyle \bigcup \Gamma_2 \subset A_2$
+Then
+:$\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} = \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$
+\end{theorem}
+
+\begin{proof}
+Set $\Gamma := \Gamma_1\cup \Gamma_2$.
+Let $\rho_1$ and $\rho_2$ be conformal metrics as in the [[Definition:Extremal Length|definition of extremal length]], [[Definition:Extremal Length#Normalizations|normalized]] such that:
+:$ L \left({\Gamma_1, \rho_1}\right) = L \left({\Gamma_2, \rho_2}\right) = 1$
+We define a new metric by:
+: $\rho := \max \left({\rho_1, \rho_2}\right)$.
+{{explain|Prove that $\rho$ is a metric}}
+Then:
+: $L \left({\Gamma, \rho}\right) \ge 1$
+and:
+: $A \left({\rho}\right) \le A \left({\rho_1}\right) + A \left({\rho_2}\right)$
+{{explain|What is $A$?}}
+Hence:
+{{begin-eqn}}
+{{eqn | l=\frac 1 {\lambda \left({\Gamma}\right)}
+      | o=\le
+      | r=\frac {A \left({\rho}\right)} {L \left({\Gamma, \rho}\right)}
+      | c=
+}}
+{{eqn | o=\le
+      | r=A \left({\rho}\right)
+      | c=
+}}
+{{eqn | o=\le
+      | r=A \left({\rho_1}\right) + A \left({\rho_2}\right)
+      | c=
+}}
+{{eqn | r=\frac 1 {L \left({\Gamma_1, \rho_1}\right)} + \frac 1 {L \left({\Gamma_2, \rho_2}\right)}
+      | c=
+}}
+{{end-eqn}}
+Taking the infimum over all metrics $\rho_1$ and $\rho_2$, the claim follows.
+Now suppose that the disjointness assumption holds, and let $\rho$ again be a Borel-measurable conformal metric, normalized such that $L \left({\Gamma, \rho}\right)= 1$. 
+We can define $\rho_1$ to be the restriction of $\rho$ to $A_1$, and likewise $\rho_2$ to be the restriction of $\rho$ to $A_2$.
+By this we mean that, in local coordinates, $\rho_j$ is given by
+:$ \rho_j \left({z}\right) \ \left|{\mathrm d z}\right| = \begin{cases}
+\rho \left({z}\right) \ \left|{\mathrm d z}\right| & : z \in A_j \\
+0 \ \left|{\mathrm d z}\right| & : \text{otherwise}
+\end{cases}$
+{{explain|The above section from "By this we mean" needs considerably more explanation, as none of the concepts introduced here can be understood without reference to links from elsewhere.}}
+Then:
+: $A \left({\rho}\right) = A \left({\rho_1}\right) + A \left({\rho_2}\right)$
+and:
+: $L \left({\Gamma_1, \rho_1}\right), L \left({\Gamma_2, \rho_2}\right) \ge 1$
+{{explain|How do these two statements follow from what went before?}}
+Hence:
+{{begin-eqn}}
+{{eqn | l=A \left({\rho}\right)
+      | r=A \left({\rho_1}\right) + A \left({\rho_2}\right)
+      | c=
+}}
+{{eqn | o=\ge
+      | r=\frac {A \left({\rho_1}\right)} {L \left({\Gamma_1, \rho}\right)} + \frac {A \left({\rho_2}\right)} {L \left({\Gamma_2, \rho}\right)}
+      | c=
+}}
+{{eqn | o=\ge
+      | r=\frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}
+      | c=
+}}
+{{end-eqn}}
+Taking the infimum over all metrics $\rho$, we see that:
+:$\dfrac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$
+Together with the first part of the Proposition, this proves the claim.
+{{qed}}
+[[Category:Geometric Function Theory]]
+4c236pxipdbjkcuxt6h6zrtidkaqptr
+\end{proof}<|endoftext|>
+\section{Parallel Law for Extremal Length}
+Tags: Geometric Function Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Riemann Surface|Riemann surface]].
+Let $\Gamma_1, \Gamma_2$ be families of [[Definition:Rectifiable Curve|rectifiable curves]] (or, more generally, families of disjoint unions of rectifiable curves) on $X$.
+Let $\Gamma_1$ and $\Gamma_2$ be disjoint, in the sense that:
+:there exist disjoint Borel subsets $A_1, A_2 \subseteq X$ such that:
+::for any $\gamma_1 \in \Gamma_1$ and $\gamma_2 \in \Gamma_2$, we have $\gamma_1 \subseteq A_1$ and $\gamma_2 \subseteq A_2$.
+Let $\Gamma$ be a third curve family, with the property that every element $\Gamma_1$ and every element of $\Gamma_2$ contains some element of $\Gamma$.
+Then the [[Definition:Extremal Length|extremal length]] of $\Gamma$ satisfies:
+:$\dfrac 1 {\lambda \left({\Gamma}\right)} \ge \dfrac 1 {\lambda \left({\Gamma_1}\right)} + \dfrac 1 {\lambda \left({\Gamma_2}\right)}$
+\end{theorem}
+
+\begin{proof}
+The assumption means that every element of $\Gamma_1 \cup \Gamma_2$ contains some element of $\Gamma$.
+Hence:
+{{begin-eqn }}
+{{eqn | l =\frac 1 {\lambda \left({\Gamma}\right)}
+      | o = \ge
+      | r = \frac 1 {\lambda \left({\Gamma_1 \cup \Gamma_2}\right)}
+      | c = [[Comparison Principle for Extremal Length]]
+}}
+{{eqn | r = \frac 1 {\lambda \left({\Gamma_1}\right)} + \frac 1 {\lambda \left({\Gamma_2}\right)}
+      | c = [[Extremal Length of Union]]
+}}
+{{end-eqn}}
+{{qed}}
+{{MissingLinks|Some results and definitions need to be linked.}}
+\end{proof}<|endoftext|>
+\section{Comparison Principle for Extremal Length}
+Tags: Geometric Function Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Riemann Surface|Riemann surface]].
+Let $\Gamma_1$ and $\Gamma_2$ be families of [[Definition:Rectifiable Curve|rectifiable curves]] (or, more generally, families of [[Definition:Set Union|unions]] of [[Definition:Rectifiable Curve|rectifiable curves]]) on $X$.
+Let every [[Definition:Element|element]] of $\Gamma_1$ contain some [[Definition:Element|element]] of $\Gamma_2$.
+Then the [[Definition:Extremal Length|extremal lengths]] of $\Gamma_1$ and $\Gamma_2$ are related by:
+:$\lambda \left({\Gamma_1}\right) \ge \lambda \left({\Gamma_2}\right)$
+More precisely, for every conformal metric $\rho$ as in the [[Definition:Extremal Length|definition of extremal length]], we have:
+:$L \left({\Gamma_1, \rho}\right) \ge L \left({\Gamma_2, \rho}\right)$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = L \left({\Gamma_1, \rho}\right)
+      | r = \inf_{\gamma \mathop \in \Gamma_1} L \left({\gamma, \rho}\right)
+      | c = by definition
+}}
+{{eqn | o = \ge
+      | r = \inf_{\gamma \mathop \in \Gamma_2} L \left({\gamma, \rho}\right)
+      | c = since every curve of $\Gamma_1$ contains a curve of $\Gamma_2$
+}}
+{{eqn | r = L \left({\Gamma_2, \rho}\right)
+      | c = by definition
+}}
+{{end-eqn}}
+This proves the second claim.
+The second claim implies the first by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reverse Triangle Inequality}
+Tags: Triangle Inequality, Named Theorems, Inequalities
+
+\begin{theorem}
+Let $M = \struct {X, d}$ be a [[Definition:Metric Space|metric space]].
+Then:
+:$\forall x, y, z \in X: \size {\map d {x, z} - \map d {y, z} } \le \map d {x, y}$
+=== [[Reverse Triangle Inequality/Normed Division Ring|Normed Division Ring]] ===
+{{:Reverse Triangle Inequality/Normed Division Ring}}
+=== [[Reverse Triangle Inequality/Normed Vector Space|Normed Vector Space]] ===
+{{:Reverse Triangle Inequality/Normed Vector Space}}
+=== [[Reverse Triangle Inequality/Real and Complex Fields|Real and Complex Numbers]] ===
+{{:Reverse Triangle Inequality/Real and Complex Fields}}
+\end{theorem}
+
+\begin{proof}
+Let $M = \struct {X, d}$ be a [[Definition:Metric Space|metric space]].
+By the [[Definition:Triangle Inequality|triangle inequality]], we have:
+:$\forall x, y, z \in X: \map d {x, y} + \map d {y, z} \ge \map d {x, z}$
+By subtracting $\map d {y, z}$ from both sides:
+:$\map d {x, y} \ge \map d {x, z} - \map d {y, z}$
+Now we consider 2 cases. 
+;Case $1$: Suppose $\map d {x, z} - \map d {y, z} \ge 0$.
+Then:
+:$\map d {x, z} - \map d {y, z} = \size {\map d {x, z} - \map d {y, z} }$
+and so:
+:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
+;Case 2: Suppose $\map d {x, z} - \map d {y, z} < 0$.
+Applying the [[Definition:Triangle Inequality|triangle inequality]] again, we have:
+:$\forall x, y, z \in X: \map d {y, x} + \map d {x, z} \ge \map d {y, z}$
+Hence:
+:$\map d {x, y} \ge \map d {y, z} - \map d {x, z}$
+Since we assumed $\map d {x, z} - \map d {y, z} < 0$, we have that:
+:$\map d {y, z} - \map d {x, z} > 0$
+and so:
+:$\map d {y, z} - \map d {x, z} = \size {\map d {y, z} - \map d {x, z} }$
+Thus we obtain:
+:$\map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
+Since these cases are exhaustive, we have shown that:
+:$\forall x, y, z \in X: \map d {x, y} \ge \size {\map d {x, z} - \map d {y, z} }$
+{{qed}}
+[[Category:Triangle Inequality]]
+[[Category:Named Theorems]]
+[[Category:Inequalities]]
+qymhy24nk2xkx1f0gpt8zzppyct76rb
+\end{proof}<|endoftext|>
+\section{Inverse Completion of Integral Domain Exists}
+Tags: Integral Domains, Inverse Completions
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Then an [[Definition:Inverse Completion|inverse completion]] of $\struct {D, \circ}$ can be constructed.
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Integral Domain|definition of an integral domain]]:
+:All elements of $D^* = D \setminus \set {0_D}$ are [[Definition:Cancellable Element|cancellable]]
+:$\struct {D^*, \circ}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+So by the [[Inverse Completion Theorem]], there exists an [[Definition:Inverse Completion|inverse completion]] of $\struct {D, \circ}$.
+From [[Construction of Inverse Completion]], this is done as follows:
+Let $\ominus$ be the [[Definition:Congruence Relation|congruence relation]] defined on $D \times D^*$ by:
+:$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
+The fact that this is a [[Definition:Congruence Relation|congruence relation]] is proved in [[Equivalence Relation on Semigroup Product with Cancellable Elements]].
+Let $\struct {D \times D^*, \otimes}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {D, \circ}$ with $\struct {D^*, \circ}$, where $\otimes$ is the [[Definition:External Direct Product|operation on $D \times D^*$ induced by $\circ$]].
+Let the [[Definition:Quotient Structure|quotient structure]] defined by $\ominus$ be $\struct {\dfrac {D \times D^*} \ominus, \otimes_\ominus}$.
+where $\otimes_\ominus$ is the [[Definition:Operation Induced on Quotient Set|operation induced on $\dfrac {D \times D^*} \ominus$ by $\otimes$]].
+Let us use $D'$ to denote the [[Definition:Quotient Set|quotient set]] $\dfrac {D \times D^*} {\ominus}$.
+Let us use $\circ'$ to denote the operation $\otimes_\ominus$.
+Thus $\struct {D', \circ'}$ is the [[Definition:Inverse Completion|inverse completion]] of $\struct {D, \circ}$.
+An element of $D'$ is therefore an [[Definition:Equivalence Class|equivalence class]] of the [[Definition:Congruence Relation|congruence relation]] $\ominus$.
+As the [[Inverse Completion is Unique]] up to [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], it follows that we can ''define'' the structure $\struct {K, \circ}$ which is isomorphic to $\struct {D', \circ'}$.
+An element of $D'$ is therefore an [[Definition:Equivalence Class|equivalence class]] of the [[Definition:Congruence Relation|congruence relation]] $\ominus$.
+So an element of $K$ is the isomorphic image of an element $\eqclass {\paren {x, y} } \ominus$ of $\dfrac {D \times D^*} \ominus$.
+Hence every element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
+Alternatively, from the definition of [[Definition:Division Product|division product]], of the form $\dfrac x y$.
+Hence we can therefore interpret any element of $\struct {K, \circ}$ as equivalence classes of elements of the form $\dfrac x y$.
+{{qed}}
+[[Category:Integral Domains]]
+[[Category:Inverse Completions]]
+l7kiwb0myfc07a0q01ixcexn3b2jvxf
+\end{proof}<|endoftext|>
+\section{Zero of Inverse Completion of Integral Domain}
+Tags: Integral Domains, Inverse Completions
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+Let $\struct {K, \circ}$ be the [[Definition:Inverse Completion|inverse completion]] of $\struct {D, \circ}$ as defined in [[Inverse Completion of Integral Domain Exists]].
+Let $x \in K: x = \dfrac p q$ such that $p = 0_D$.
+Then $x$ is equal to the [[Definition:Ring Zero|zero]] of $K$.
+That is, ''any'' element of $K$ of the form $\dfrac {0_D} q$ acts as the [[Definition:Ring Zero|zero]] of $K$.
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \ominus$ as in the [[Inverse Completion of Integral Domain Exists]].
+That is, $\eqclass {\tuple {a, b} } \ominus$ is an [[Definition:Equivalence Class|equivalence class]] of elements of $D \times D^*$ under the [[Definition:Congruence Relation|congruence relation]] $\ominus$.
+$\ominus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $D \times D^*$ by $\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$.
+By the [[Inverse Completion of Integral Domain Exists|method of its construction]], $\dfrac p q \equiv \eqclass {\tuple {p, q} } \ominus$.
+From [[Equality of Division Products]], two elements $\dfrac a b, \dfrac c d$ of $K$ are equal {{iff}} $a \circ d = b \circ c$.
+This correlates with the fact that two elements $\eqclass {\tuple {a, b} } \ominus, \eqclass {\tuple {c, d} } \ominus$ of $K$ are equal iff $a \circ d = b \circ c$.
+Suppose $a = 0_D$.
+{{begin-eqn}}
+{{eqn | l = a
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0_D \circ d
+      | r = b \circ c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b \circ c
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | r = 0_D
+      | c = as $b \in D^*$, so $b \ne 0$
+}}
+{{end-eqn}}
+Hence:
+:$\eqclass {\tuple {0_D, b} } \ominus  = \eqclass {\tuple {0_D, d} } \ominus$
+Thus all elements of $K$ of the form $\eqclass {\tuple {0_D, k} } \ominus$ are equal, for all $k \in D^*$.
+To emphasise the irrelevance of the $k$, we will [[Definition:Abuse of Notation|abuse our notation]] and write:
+:$\eqclass {\tuple {0_D, k} } \ominus$
+as
+:$\eqclass {0_D} \ominus$
+Next, by [[Product of Division Products]], we have that $\ds \frac a b \circ \frac c d = \frac {a \circ b} {c \circ d}$.
+Again abusing our notation, we will write:
+:$\eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {c, d} } \ominus$
+to mean:
+:$\eqclass {\tuple {a \circ c, b \circ d} } \ominus$
+So:
+{{begin-eqn}}
+{{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus
+      | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {a, b} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {0_D \circ a, k \circ b} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {0_D, k \circ b} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {0_D} \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {a \circ 0_D, b \circ k} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\eqclass {0_D} \ominus \circ \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus = \eqclass {\tuple {a, b} } \ominus \circ \eqclass {0_D} \ominus$
+So $\eqclass {0_D} \ominus$ fulfils the role of a [[Ring Product with Zero|zero]] for $\tuple {K, \circ}$ as required.
+Also we have that:
+{{begin-eqn}}
+{{eqn | l = \eqclass {0_D} \ominus \circ \eqclass {0_D} \ominus
+      | r = \eqclass {\tuple {0_D, k} } \ominus \circ \eqclass {\tuple {0_D, k} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {0_D \circ 0_D, k \circ k} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {\tuple {0_D, k \circ k} } \ominus
+      | c = 
+}}
+{{eqn | r = \eqclass {0_D} \ominus
+      | c = 
+}}
+{{end-eqn}}
+So $\eqclass {0_D} \ominus$ is [[Definition:Idempotent Element|idempotent]].
+It follows that $\eqclass {0_D} \ominus$ can be identified with $0_D$ from the mapping $\psi$ as defined in [[Construction of Inverse Completion#Quotient Mapping to Image is Isomorphism|Construction of Inverse Completion]].
+{{qed}}
+[[Category:Integral Domains]]
+[[Category:Inverse Completions]]
+retg8czrb5r3plkykgpud4aram1uv1f
+\end{proof}<|endoftext|>
+\section{Inverse Completion Less Zero of Integral Domain is Closed}
+Tags: Integral Domains, Inverse Completions
+
+\begin{theorem}
+Let $\left({D, +, \circ}\right)$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $\left({K, \circ}\right)$ be the [[Definition:Inverse Completion|inverse completion]] of $\left({D, \circ}\right)$.
+Then $\left({K^*, \circ}\right)$ is [[Definition:Closed Algebraic Structure|closed]], where $K^* = K \setminus \left\{{0_K}\right\}$.
+\end{theorem}
+
+\begin{proof}
+Let $\left({K, \circ}\right)$ be the [[Definition:Inverse Completion|inverse completion]] of $\left({D, \circ}\right)$.
+We define $\left({K, \circ}\right)$ of $\left({D, \circ}\right)$ by [[Inverse Completion of Integral Domain Exists]].
+The structure of $\left({K, \circ}\right)$ is such that element of $\left({K, \circ}\right)$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
+From [[Zero of Inverse Completion of Integral Domain]], $0_K$ is all elements of $D \times D^*$ of the form $\dfrac {0_D} x$.
+Therefore the elements of $K^*$ are those of the form $\dfrac x \circ y^{-1}$ where $x, y \in D^*$.
+By [[Product of Division Products]], $\displaystyle \frac a b \circ \frac c d = \frac {a \circ c} {b \circ d}$.
+As $\left({D, +, \circ}\right)$ is an [[Definition:Integral Domain|integral domain]], none of its non-zero elements are zero divisors.
+Therefore $\forall x, y \in D^*: x \circ y \ne 0_D$.
+So:
+: $\displaystyle \forall \frac a b, \frac c d \in K^*: \frac {a \circ c} {b \circ d} \in K^*$
+Hence the result.
+{{qed}}
+[[Category:Integral Domains]]
+[[Category:Inverse Completions]]
+6ivccqlw1ptq3qv4ftk49lrdkgm451r
+\end{proof}<|endoftext|>
+\section{Composition of Mappings is Associative}
+Tags: Composite Mappings, Associativity
+
+\begin{theorem}
+The [[Definition:Composition of Mappings|composition of mappings]] is an [[Definition:Associative Operation|associative]] [[Definition:Binary Operation|binary operation]]:
+:$\paren {f_3 \circ f_2} \circ f_1 = f_3 \circ \paren {f_2 \circ f_1}$
+where $f_1, f_2, f_3$ are arbitrary [[Definition:Mapping|mappings]] which fulfil the conditions for the relevant [[Definition:Composition of Mappings|compositions]] to be defined.
+\end{theorem}
+
+\begin{proof}
+{{expand|A commutative diagram would be nice. Anyone out there good at xyplot?}}
+From the definition, we know that a [[Definition:Mapping|mapping is a relation]].
+First, note that from the definition of [[Definition:Composition of Relations|composition of relations]], the following must be the case before the above expression is even to be defined:
+:$(1): \quad \Dom {f_2} = \Cdm {f_1}$
+:$(2): \quad \Dom {f_3} = \Cdm {f_2}$
+where $\Cdm f$ denotes the [[Definition:Codomain of Mapping|codomain]] of the [[Definition:Mapping|mapping]] $f$.
+The two '''composite relations''' can be seen to have the same [[Definition:Domain of Relation|domain]], as follows:
+{{begin-eqn}}
+{{eqn | l = \Dom {\paren {f_3 \circ f_2} \circ f_1}
+      | r = \Dom {f_1}
+      | c = [[Domain of Composite Relation]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \Dom {f_3 \circ \paren {f_2 \circ f_1} }
+      | r = \Dom {f_2 \circ f_1}
+      | c = [[Domain of Composite Relation]]
+}}
+{{eqn | r = \Dom {f_1}
+      | c = [[Domain of Composite Relation]]
+}}
+{{end-eqn}}
+Also they have the same [[Definition:Codomain of Relation|codomain]], as is seen by:
+{{begin-eqn}}
+{{eqn | l = \Cdm {\paren {f_3 \circ f_2} \circ f_1}
+      | r = \Cdm {f_3 \circ f_2}
+      | c = [[Codomain of Composite Relation]]
+}}
+{{eqn | r = \Cdm {f_3}
+      | c = [[Codomain of Composite Relation]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \Cdm {f_3 \circ \paren {f_2 \circ f_1} }
+      | r = \Cdm {f_3}
+      | c = [[Codomain of Composite Relation]]
+}}
+{{end-eqn}}
+As a mapping is a relation, we can use that the [[Composition of Relations is Associative]]:
+:$\forall x \in \Dom {f_1}: \map {\paren {f_3 \circ f_2} \circ f_1} x = \map {f_3 \circ \paren {f_2 \circ f_1} } x$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Abelian iff Cross Cancellation Property}
+Tags: Abelian Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then the following are equivalent:
+:$(1): \quad G$ is [[Definition:Abelian Group|abelian]]
+:$(2): \quad G$ has the [[Definition:Cross Cancellation Property|cross cancellation property]]
+\end{theorem}
+
+\begin{proof}
+Let us suppress the operation of $G$ for brevity.
+=== $(2) \implies (1)$ ===
+Suppose that $G$ has the [[Definition:Cross Cancellation Property|cross cancellation property]].
+Then, for all $x, y \in G$:
+{{begin-eqn}}
+{{eqn | l=y \left({x y}\right)
+      | r=\left({y x}\right) y
+      | c=[[Definition:Associative|Associativity]]
+}}
+{{eqn | ll=\implies
+      | l=x y
+      | r=y x
+      | c=[[Definition:Cross Cancellation Property|Cross Cancellation Property]]
+}}
+{{end-eqn}}
+Thus, $G$ is [[Definition:Abelian Group|abelian]].
+{{qed|lemma}}
+=== $(1) \implies (2)$ ===
+Conversely, suppose $G$ is [[Definition:Abelian Group|abelian]].
+Let $a, b, c \in G$ be such that $a b = c a$.
+Since $G$ is [[Definition:Abelian Group|abelian]], $c a = a c$.
+We conclude that:
+:$a b = c a = a c$
+Thus, by [[Cancellation Laws|left cancellation]], $b = c$.
+{{qed}}
+[[Category:Abelian Groups]]
+taj0pft4of6zsk03msxwfbb1l3egaff
+\end{proof}<|endoftext|>
+\section{Group Abelian iff Middle Cancellation Property}
+Tags: Abelian Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then the following are equivalent:
+:$(1): \quad G$ is [[Definition:Abelian Group|abelian]]
+:$(2): \quad G$ satisfies the [[Definition:Middle Cancellation Property|middle cancellation property]]
+\end{theorem}
+
+\begin{proof}
+Let us suppress the operation of $G$ for brevity.
+=== $(2) \implies (1)$ ===
+Suppose that $G$ satisfies the [[Definition:Middle Cancellation Property|middle cancellation property]].
+Then, for all $g, h \in G$:
+{{begin-eqn}}
+{{eqn | l=e h
+      | r=h e
+      | c=Definition of [[Definition:Identity (Abstract Algebra)|identity]]
+}}
+{{eqn | ll=\implies
+      | l=g g^{-1} h
+      | r=hg^{-1}g
+      | c=Definition of [[Definition:Inverse (Abstract Algebra)|inverse]]
+}}
+{{eqn | ll=\implies
+      | l=g h
+      | r=h g
+      | c=Definition of [[Definition:Middle Cancellation Property|middle cancellation property]]
+}}
+{{end-eqn}}
+Thus $G$ is [[Definition:Abelian Group|abelian]].
+{{qed|lemma}}
+=== $(1) \implies (2)$ ===
+Conversely, suppose $G$ is [[Definition:Abelian Group|abelian]].
+Then, for all $a, b, c, d, x \in G$:
+{{begin-eqn}}
+{{eqn | l=a x b
+      | r=c x d
+}}
+{{eqn | ll=\implies
+      | l=a b x
+      | r=c d x
+      | c=Definition of [[Definition:Abelian Group|Abelian Group]]
+}}
+{{eqn | ll=\implies
+      | l=a b
+      | r=c d
+      | c=[[Cancellation Laws|Right Cancellation]]
+}}
+{{end-eqn}}
+Thus the [[Definition:Middle Cancellation Property|middle cancellation property]] holds in $G$.
+{{qed}}
+[[Category:Abelian Groups]]
+tj0sq9li4dt91qjulmy5oyraictdbln
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Ordered Field}
+Tags: Examples of Fields, Rational Numbers
+
+\begin{theorem}
+The [[Definition:Rational Number|set of rational numbers]] $\Q$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]]: $\struct {\Q, +, \times, \le}$.
+\end{theorem}
+
+\begin{proof}
+Recall that by [[Integers form Ordered Integral Domain]], $\struct {\Z, +, \times, \le}$ is an [[Definition:Ordered Integral Domain|ordered integral domain]]
+By [[Rational Numbers form Field]], $\struct {\Q, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+In the [[Definition:Rational Number/Formal Definition|formal definition of rational numbers]], $\struct {\Q, +, \times}$ is the [[Definition:Field of Quotients|field of quotients]] of $\struct {\Z, +, \times, \le}$
+By [[Total Ordering on Field of Quotients is Unique]], it follows that $\struct {\Q, +, \times}$ has a unique [[Definition:Total Ordering|total ordering]] on it that is [[Definition:Ordering Compatible with Ring Structure|compatible with its ring structure]].
+{{explain|Review the ordering / total ordering question}}
+Thus $\struct {\Q, +, \times, \le}$ is an [[Definition:Ordered Field|ordered field]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are Close Packed}
+Tags: Analysis
+
+\begin{theorem}
+Let $a, b \in \Q$ such that $a < b$.
+Then $\exists c \in \Q: a < c < b$.
+That is, the [[Definition:Rational Number|set of rational numbers]] is [[Definition:Close Packed|close packed]].
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Rational Number|definition of rational numbers]], we can express $a$ and $b$ as $a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$.
+Thus from [[Mediant is Between]]:
+:$\dfrac {p_1} {q_1} < \dfrac {p_1 + p_2} {q_1 + q_2} < \dfrac {p_2} {q_2}$
+From [[Rational Numbers form Field]]:
+:$\dfrac {p_1 + p_2} {q_1 + q_2} \in \Q$
+Hence $c = \dfrac {p_1 + p_2} {q_1 + q_2}$ is an element of $\Q$ between $a$ and $b$.
+{{qed}}
+[[Category:Analysis]]
+cvubwxgv3tvpgfvlnsrbxlucyikct61
+\end{proof}<|endoftext|>
+\section{Real Addition is Closed}
+Tags: Real Addition, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] $\R$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Addition|addition]]:
+:$\forall x, y \in \R: x + y \in \R$
+\end{theorem}
+
+\begin{proof}
+From the definition, the [[Definition:Real Number|real numbers]] are the set of all [[Definition:Equivalence Class|equivalence classes]] $\eqclass {\sequence {x_n} } {}$ of [[Definition:Cauchy Sequence|Cauchy sequences]] of [[Definition:Rational Number|rational numbers]].
+Let $x = \eqclass {\sequence {x_n} } {}, y = \eqclass {\sequence {y_n} } {}$, where $\eqclass {\sequence {x_n} } {}$ and $\eqclass {\sequence {y_n} } {}$ are such [[Definition:Equivalence Class|equivalence classes]].
+From the definition of [[Definition:Real Addition|real addition]], $x + y$ is defined as:
+:$\eqclass {\sequence {x_n} } {} + \eqclass {\sequence {y_n} } {} = \eqclass {\sequence {x_n + y_n} } {}$
+We have that $\forall i \in \N: x_i \in \Q, y_i \in \Q$, therefore $x_i + y_i \in \Q$.
+So it follows that $\eqclass {\sequence {x_n + y_n} } {} \in \R$.
+{{qed}}
+\end{proof}
\ No newline at end of file