diff --git "a/wiki/proofwiki/shard_39.txt" "b/wiki/proofwiki/shard_39.txt"
new file mode 100644--- /dev/null
+++ "b/wiki/proofwiki/shard_39.txt"
@@ -0,0 +1,7252 @@
+\section{Space of Bounded Sequences with Supremum Norm forms Banach Space}
+Tags: Functional Analysis, Banach Spaces
+
+\begin{theorem}
+Let $\ell^\infty$ be a [[Definition:Space of Bounded Sequences|space of bounded sequences]].
+Let $\norm {\, \cdot \,}_\infty$ be a [[Definition:Supremum Norm|supremum norm]].
+Then $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is a [[Definition:Banach Space|Banach space]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Banach Space|Banach space]] is a [[Definition:Normed Vector Space|normed vector space]], where a [[Definition:Cauchy Sequence|Cauchy sequence]] [[Definition:Convergent Sequence in Normed Vector Space|converges]] {{WRT}} the supplied [[Definition:Norm on Vector Space|norm]].
+To prove the theorem, we need to show that a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]].
+We take a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$.
+Then we consider the $k$th component and show, that a [[Definition:Real Cauchy Sequence|real Cauchy sequence]] $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ [[Definition:Convergent Real Sequence|converges]] in $\struct {\R, \size {\, \cdot \,}}$ with the [[Definition:Limit of Real Sequence|limit]] $x^{\paren k}$ and denote the entire set as $\mathbf x$.
+Finally, we show that $\sequence {\mathbf x_n}_{n \in \N}$, composed of components $x_n^{\paren k},$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] in $\struct {\ell^\infty, \norm {\,\cdot\,}_\infty}$ with the [[Definition:Limit of Sequence in Normed Vector Space|limit]] $\mathbf x$.
+Let $\sequence {\mathbf x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$.
+Denote the $k$th component of $\mathbf x_n$ by $x_n^{\paren k}$.
+=== $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$=== 
+Let $\epsilon >0$.
+Then:
+:$\displaystyle \exists N \in \N : \forall m,n \in \N : m,n > N : \norm {\mathbf x_n - \mathbf x_m}_\infty < \epsilon$
+For same $N, m, n$ consider $\size {x_n^{\paren k} - x_m^{\paren k} } $:
+{{begin-eqn}}
+{{eqn | l = \size {x_n^{\paren k} - x_m^{\paren k} } 
+      | o = \le
+      | r = \sup_{k \mathop \in \N} \size {x_n^{\paren k} - x_m^{\paren k} }
+}}
+{{eqn | r = \norm {\mathbf x_n - \mathbf x_m}_\infty
+      | c = {{defof|Supremum Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+Hence, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\struct {\R, \size {\, \cdot \,}}$.
+From [[Real Number Line is Complete Metric Space]], $\R$ is a [[Definition:Complete Metric Space|complete metric space]].
+Consequently, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ [[Definition:Convergent Sequence|converges]] in $\struct {\R, \size {\, \cdot \,}}$.
+{{qed|lemma}}
+Denote the [[Definition:Limit of Real Sequence|limit]] $\displaystyle \lim_{n \mathop \to \infty} \sequence {x_n^{\paren k}}_{n \mathop \in \N} = x^{\paren k}$.
+Denote $\sequence {x^{\paren k}}_{k \mathop \in \N} = \mathbf x$.
+=== $\mathbf x$ belongs to $\ell^\infty$ ===
+From previous lemma, $\sequence {x_n^{\paren k}}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\struct {\R, \size {\, \cdot \,}}$.
+Fix any $n > N$ and $k \in \N$.
+Then:
+:$\displaystyle \forall m > N : \size {x_n^{\paren k} - x_m^{\paren k}} < \epsilon$
+Take the limit $m \to \infty$:
+:$\size {x_n^{\paren k} - x^{\paren k}} < \epsilon$
+Since $k$ was arbitrary:
+{{begin-eqn}}
+{{eqn | ll = \forall k \in \N :
+      | l = \size {x_n^{\paren k} - x^{\paren k} }
+      | o = \le
+      | r =  \sup_{k \mathop \in \N} \size {x_n^{\paren k} - x^{\paren k} } 
+      | c = {{defof|Supremum of Real Sequence}}
+}}
+{{eqn | r = \norm {\mathbf x_n - \mathbf x}_\infty
+      | c = {{defof|Supremum Norm}} }}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{eqn | o = <
+      | r = \infty
+}}
+{{end-eqn}}
+By [[Definition:Space of Bounded Sequences|definition]], $\mathbf x_n - \mathbf x \in \ell^{\infty}$.
+By [[Definition:Assumption|assumption]], $\mathbf x_n \in \ell^\infty$.
+Hence, $\mathbf x \in \ell^\infty$.
+{{qed|lemma}}
+=== $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ to $\mathbf x$===
+From previous lemma we have that for some $\epsilon > 0$, $N \in \N$ and fixed $n > N$:
+:$\displaystyle \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$
+Repeat the same argument for all $\epsilon \in \R_{>0}$ and note that $n$ was arbitrary:
+:$\displaystyle \forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {\mathbf x_n - \mathbf x}_\infty < \epsilon$
+Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] in $\struct {\ell^\infty, \norm{\, \cdot \,}_\infty}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}}
+Then:
+:$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$
+where $\card{B_1}$ and $\card{B_2}$ denote the [[Definition:Cardinality|cardinality]] of the [[Definition:Set|sets]] $B_1$ and $B_2$ respectively.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}}:
+:$\exists B_1, B_2 \in \mathscr B : \card{B_1} \ne \card{B_2}$
+{{WLOG}} and from [[Max Equals an Operand]] let:
+:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$
+{{WLOG}} let:
+:$\card{B_1} > \card{B_2}$
+From [[Set Difference of Larger Set with Smaller is Not Empty]]:
+:$\exists x \in B_1 \setminus B_2$
+From [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom $(\text B 1)$]]:
+:$\exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B$
+Let $B_3 = \paren {B_1 \setminus \set x} \cup \set y$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \card{B_3 \cap B_2}
+      | r = \card{\paren{\paren {B_1 \setminus \set x} \cup \set y} \cap B_2}
+}}
+{{eqn | r = \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \paren {\set y \cap B_2} }
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \card{\paren{\paren {B_1 \setminus \set x} \cap B_2} \cup \set y }
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \card{\paren{\paren {B_1 \cap B_2 } \setminus \set x } \cup \set y }
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = \card{\paren {B_1 \cap B_2 } \cup \set y }
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{eqn | o = >
+      | r = \card{B_1 \cap B_2 }
+      | c = [[Cardinality of Proper Subset of Finite Set]]
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the choice of $B_1$ and $B_2$ as:
+:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$
+It follows that:
+:$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$
+{{qed}}
+[[Category:Matroid Theory]]
+t0zxd9t5pt6bppfs5fvirdnv4yao2uk
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 4 Iff Definition 5/Lemma}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+:$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}}:
+:$\exists B_1, B_2 \in \mathscr B : \card{B_1} \ne \card{B_2}$
+{{WLOG}} and from [[Max Equals an Operand]] let:
+:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$
+{{WLOG}} let:
+:$\card{B_1} > \card{B_2}$
+From [[Set Difference of Larger Set with Smaller is Not Empty]]:
+:$\exists x \in B_1 \setminus B_2$
+From [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom $(\text B 5)$]]:
+:$\exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B$
+Let $B_3 = \paren {B_2 \setminus \set y} \cup \set x$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \card{B_3 \cap B_1}
+      | r = \card{\paren{\paren {B_2 \setminus \set y} \cup \set x} \cap B_1}
+}}
+{{eqn | r = \card{\paren{\paren {B_2 \setminus \set y} \cap B_1} \cup \paren {\set x \cap B_1} }
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \card{\paren{\paren {B_2 \setminus \set y} \cap B_1} \cup \set x }
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \card{\paren{\paren {B_2 \cap B_1 } \setminus \set y } \cup \set x }
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = \card{\paren {B_2 \cap B_1 } \cup \set x }
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{eqn | o = >
+      | r = \card{B_1 \cap B_2 }
+      | c = [[Cardinality of Proper Subset of Finite Set]]
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the choice of $B_1$ and $B_2$ as:
+:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : C_1, C_2 \in \mathscr B : \card{C_1} \ne \card{C_2}}$
+It follows that:
+:$\forall B_1, B_2 \in \mathscr B : \card{B_1} = \card{B_2}$
+{{qed}}
+[[Category:Equivalence of Definitions of Matroid Base Axiom]]
+pyv08opjnvc6w5sxlt3xrcjg4lii8vj
+\end{proof}<|endoftext|>
+\section{Negative of Field Negative}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
+Let $a \in F$ and let $-a$ be the [[Definition:Field Negative|field negative]] of $a$.
+Then:
+:$-\paren {-a} = a$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {-a} + a
+      | r = a + \paren {-a}
+      | c = {{Field-axiom|A2}}
+}}
+{{eqn | r = 0_F
+      | c = {{Field-axiom|A4}}
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = -\paren {-a}
+      | c = {{Defof|Field Negative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Product with Zero}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
+Let $a \in F$.
+Then:
+:$a \times 0_F = 0_F$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | r = a \times 1_F
+      | c = {{Field-axiom|M3}}: $1_F$ is the [[Definition:Unity of Field|unity]] of $F$
+}}
+{{eqn | r = a \times \paren {0_F + 1_F}
+      | c = {{Field-axiom|A3}}
+}}
+{{eqn | r = a \times 0_F + a \times 1_F
+      | c = {{Field-axiom|D}}
+}}
+{{eqn | r = a \times 0_F + a
+      | c = {{Field-axiom|M3}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \times 0_F + a} + \paren {-a}
+      | r = a + \paren {-a}
+      | c = adding $-a$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = a \times 0_F + \paren {a + \paren {-a} }
+      | r = a + \paren {-a}
+      | c = {{Field-axiom|A1}}
+}}
+{{eqn | ll= \leadsto
+      | l = a \times 0_F + 0_F
+      | r = 0_F
+      | c = {{Field-axiom|A4}}
+}}
+{{eqn | ll= \leadsto
+      | l = a \times 0_F
+      | r = 0_F
+      | c = {{Field-axiom|A3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product with Field Negative}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ 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 $a, b \in F$.
+Then:
+:$-\paren {a \times b} = a \times \paren {-b} = \paren {-a} \times b$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a \times b + a \times \paren {-b}
+      | r = a \times \paren {b + \paren {-b} }
+      | c = {{Field-axiom|A1}}
+}}
+{{eqn | r = a \times 0_F
+      | c = {{Field-axiom|A4}}
+}}
+{{eqn | r = 0_F
+      | c = [[Field Product with Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {a \times b}
+      | r = a \times \paren {-b}
+      | c = {{Defof|Ring Negative}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \paren {-a} \times b + a \times b
+      | r = \paren {\paren {-a} + a} \times b
+      | c = {{Field-axiom|A1}}
+}}
+{{eqn | r = 0_F \times a
+      | c = {{Field-axiom|A4}}
+}}
+{{eqn | r = 0_F
+      | c = [[Field Product with Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {a \times b}
+      | r = \paren {-a} \times b
+      | c = {{Defof|Ring Negative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product with Field Negative/Corollary}
+Tags: Field Theory
+
+\begin{theorem}
+:$\paren {-1_F} \times a = \paren {-a}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a + \paren {-1_F} \times a
+      | r = 1_F \times a + \paren {-1_F} \times a
+      | c = {{Field-axiom|M3}}
+}}
+{{eqn | r = \paren {1_F + \paren {-1_F} } \times a
+      | c = {{Field-axiom|D}}
+}}
+{{eqn | r = 0_F \times a
+      | c = {{Field-axiom|A4}}
+}}
+{{eqn | r = 0_F
+      | c = [[Field Product with Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-a}
+      | r = \paren {-1_F} \times a
+      | c = {{Defof|Ring Negative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Division by Field Elements to be Unity}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ 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 $a, b \in F$.
+Then:
+:$\dfrac a b = 1_F$
+{{iff}}
+:$a = b$
+where $\dfrac a b$ denotes [[Definition:Division|division]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $a = b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac a b
+      | r = a \times b^{-1}
+      | c = {{Defof|Division}}
+}}
+{{eqn | r = b \times b^{-1}
+      | c = as $a = b$
+}}
+{{eqn | r = 1_F
+      | c = {{Field-axiom|M4}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\dfrac a b = 1_F$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \times b^{-1}
+      | r = 1_F
+      | c = {{Defof|Division}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \times b^{-1} } \times b
+      | r = 1_F \times b
+      | c = multiplying both sides by $b$
+}}
+{{eqn | ll= \leadsto
+      | l = a \times \paren {b \times b^{-1} }
+      | r = 1_F \times b
+      | c = {{Field-axiom|M1}}
+}}
+{{eqn | ll= \leadsto
+      | l = a \times 1_F
+      | r = 1_F \times b
+      | c = {{Field-axiom|M4}}
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b
+      | c = {{Field-axiom|M3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Product with Non-Zero Element yields Unique Solution}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ 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 $a, b, x \in F$ such that $b \ne 0_F$.
+Let:
+:$b \times x = a$
+Then:
+:$x = a b^{-1}$
+That is:
+:$x = \dfrac a b$
+where $\dfrac a b$ denotes [[Definition:Division|division]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = b \times x
+      | r = a
+      | c = with $b \ne 0_F$
+}}
+{{eqn | ll= \leadsto
+      | l = b^{-1} \times \paren {b \times x}
+      | r = b^{-1} \times a
+      | c = multiplying both sides by $b^{-1}$, which exists because $b \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {b^{-1} \times b} \times x
+      | r = b^{-1} \times a
+      | c = {{Field-axiom|M1}}
+}}
+{{eqn | ll= \leadsto
+      | l = 1_F \times x
+      | r = b^{-1} \times a
+      | c = {{Field-axiom|M4}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = b^{-1} \times a
+      | c = {{Field-axiom|M3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Unity Divided by Element equals Multiplicative Inverse}
+Tags: Field Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ 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 $a \in F$.
+Then:
+:$\dfrac {1_F} a = a^{-1}$
+where $\dfrac {1_F} a$ denotes [[Definition:Division|division]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {1_F} a
+      | r = 1_F \times a^{-1}
+      | c = {{Defof|Division}}
+}}
+{{eqn | r = a^{-1} \times 1_F
+      | c = {{Field-axiom|M2}}
+}}
+{{eqn | r = a^{-1}
+      | c = {{Field-axiom|M3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellation Law for Field Product}
+Tags: Field Theory, Cancellation Laws
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ 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 $a, b, c \in F$.
+Then:
+:$a \times b = a \times c \implies a = 0_F \text { or } b = c$
+\end{theorem}
+
+\begin{proof}
+Let $a \times b = a \times c$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \ne
+      | r = 0_F
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo = \exists a^{-1} \in F:
+      | l = a^{-1} \times a
+      | r = 1_F
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{-1} \times \paren {a \times b}
+      | r = a^{-1} \times \paren {a \times c}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^{-1} \times a} \times b
+      | r = \paren {a^{-1} \times a} \times c
+      | c = {{Field-axiom|M1}}
+}}
+{{eqn | ll= \leadsto
+      | l = 1_F \times b
+      | r = 1_F \times c
+      | c = {{Field-axiom|M4}}
+}}
+{{eqn | ll= \leadsto
+      | l = b
+      | r = c
+      | c = {{Field-axiom|M3}}
+}}
+{{end-eqn}}
+That is:
+:$a \ne 0 \implies b = c$
+{{qed|lemma}}
+Suppose $b \ne c$.
+Then by [[Rule of Transposition]]:
+:$\neg \paren {a \ne 0}$
+that is:
+:$a = 0_F$
+and we note that in this case:
+:$a \times b = 0_F = a \times c$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Difference of Larger Set with Smaller is Not Empty}
+Tags: Set Difference, Cardinality
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Finite Set|finite]] [[Definition:Set|sets]].
+Let $\card S > \card T$.
+Then:
+:$S \setminus T \ne \O$
+\end{theorem}
+
+\begin{proof}
+From [[Cardinality of Subset of Finite Set]]:
+:$S \not \subseteq T$
+From the [[Definition:Contrapositive|contrapositive statement]] of [[Set Difference with Superset is Empty Set]]:
+:$S \setminus T \ne \O$.
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Cardinality]]
+ot8xec777hfs32dsdmty8u4wqsoqvgy
+\end{proof}<|endoftext|>
+\section{Autocovariance Matrix for Stationary Process is Variance by Autocorrelation Matrix}
+Tags: Autocovariance Matrices, Autocorrelation Matrices
+
+\begin{theorem}
+Let $S$ be a [[Definition:Strictly Stationary Stochastic Process|strictly stationary stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] of $n$ [[Definition:Successive Values of Equispaced Time Series|successive values]] of $T$:
+:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
+Let $\boldsymbol \Gamma_n$ denote the '''[[Definition:Autocovariance Matrix|autocovariance matrix]]''' associated with $S$ for $\sequence {s_n}$.
+Let $\mathbf P_n$ denote the '''[[Definition:Autocorrelation Matrix|autocorrelation matrix]]''' associated with $S$ for $\sequence {s_n}$.
+Then:
+:$\boldsymbol \Gamma_n = \sigma_z^2 \mathbf P_n$
+where $\sigma_z^2$ denotes the [[Definition:Variance|variance]] of $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Autocorrelation of Strictly Stationary Stochastic Process]]:
+:$\rho_k = \dfrac {\gamma_k} {\gamma_0}$
+Then from [[Autocovariance at Zero Lag for Strictly Stationary Stochastic Process is Variance]]:
+:$\gamma_0 = \sigma_z^2$
+Hence:
+:$\gamma_k = \sigma_z^2 \rho_k$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Variance of Linear Function of Observations of Stationary Process}
+Tags: Stationary Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Strictly Stationary Stochastic Process|strictly stationary stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] of $n$ [[Definition:Successive Values of Equispaced Time Series|successive values]] of $T$:
+:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
+Let $L_t$ be a [[Definition:Linear Function|linear function]] of $\sequence {s_n}$:
+:$L_t = l_1 z_t + l_2 z_{t - 1} + \dotsb + l_n z_{t - n + 1}$
+Then the [[Definition:Variance|variance]] of $L_t$ is given by:
+:$\var {L_t} = \displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n l_i l_j \gamma {\size {j - i} }$
+where $\gamma_k$ is the [[Definition:Autocovariance|autocovariance]] of $S$ at [[Definition:Lag|lag]] $k$.
+\end{theorem}<|endoftext|>
+\section{Autocorrelation Matrix is Positive Definite}
+Tags: Autocorrelation Matrices
+
+\begin{theorem}
+Let $S$ be a [[Definition:Strictly Stationary Stochastic Process|strictly stationary stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] of $n$ [[Definition:Successive Values of Equispaced Time Series|successive values]] of $T$:
+:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
+Let $\mathbf P_n$ denote the '''[[Definition:Autocorrelation Matrix|autocorrelation matrix]]''' associated with $S$ for $\sequence {s_n}$.
+Then $\mathbf P_n$ is a [[Definition:Positive Definite Matrix|positive definite matrix]].
+\end{theorem}<|endoftext|>
+\section{Determinant of Autocorrelation Matrix is Strictly Positive}
+Tags: Autocorrelation Matrices, Determinant of Autocorrelation Matrix is Strictly Positive
+
+\begin{theorem}
+Let $S$ be a [[Definition:Strictly Stationary Stochastic Process|strictly stationary stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] of $n$ [[Definition:Successive Values of Equispaced Time Series|successive values]] of $T$:
+:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
+Let $\mathbf P_n$ denote the '''[[Definition:Autocorrelation Matrix|autocorrelation matrix]]''' associated with $S$ for $\sequence {s_n}$.
+The [[Definition:Determinant|determinant]] of $\mathbf P_n$ is [[Definition:Strictly Positive Real Number|strictly positive]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Autocorrelation Matrix is Positive Definite]].
+The result follows from [[Determinant of Positive Definite Matrix is Strictly Positive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Wholly Real Numbers is Wholly Real}
+Tags: Complex Addition
+
+\begin{theorem}
+Let $x = \tuple {a, 0}$ and $y = \tuple {b, 0}$ be [[Definition:Wholly Real|wholly real]] [[Definition:Complex Number|complex numbers]].
+Then $x + y$ is also [[Definition:Wholly Real|wholly real]].
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = x + y
+      | r = \tuple {a, 0} + \tuple {b, 0}
+      | c = 
+}}
+{{eqn | r = \tuple {a + b, 0 + 0}
+      | c = {{Defof|Complex Addition}}
+}}
+{{eqn | r = \tuple {a + b, 0}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Wholly Real Numbers is Wholly Real}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Let $x = \tuple {a, 0}$ and $y = \tuple {b, 0}$ be [[Definition:Wholly Real|wholly real]] [[Definition:Complex Number|complex numbers]].
+Then $x y$ is also [[Definition:Wholly Real|wholly real]].
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = \tuple {a, 0} \tuple {b, 0}
+      | c = 
+}}
+{{eqn | r = \tuple {a \times b - 0 \times 0, a \times 0 + 0 \times b}
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {a \times b, 0}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Imaginary Unit with Itself}
+Tags: Imaginary Parts, Complex Multiplication
+
+\begin{theorem}
+Let $\tuple {0, 1}$ denote the [[Definition:Imaginary Unit|imaginary unit]].
+Then:
+:$\tuple {0, 1} \times \tuple {0, 1} = \tuple {-1, 0}$
+where $\times$ denotes [[Definition:Complex Multiplication|complex multiplication]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tuple {0, 1} \times \tuple {0, 1}
+      | r = \tuple {0 \times 0 - 1 \times 1, 0 \times 1 + 0 \times 1}
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{eqn | r = \tuple {-1, 0}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Set Difference of Distinct Equal Cardinality Sets is Not Empty}
+Tags: Set Difference, Cardinality
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Distinct|distinct]] [[Definition:Finite Set|finite]] [[Definition:Set|sets]].
+Let $\card S = \card T$.
+Then:
+:$S \setminus T \ne \O$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Contrapositive|contrapositive statement]] of [[Cardinality of Proper Subset of Finite Set]]:
+:$S \not \subset T$
+Because $S \ne T$:
+:$S \not \subseteq T$
+From the [[Definition:Contrapositive|contrapositive statement]] of [[Set Difference with Superset is Empty Set]]:
+:$S \setminus T \ne \O$.
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Cardinality]]
+4weoeikehps5mmuvv65hf05sttmp3up
+\end{proof}<|endoftext|>
+\section{Element of Matroid Base and Circuit has a Substitute}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $B \subseteq S$ be a [[Definition:Base of Matroid|base]] of $M$.
+Let $C \subseteq S$ be a [[Definition:Circuit (Matroid)|circuit]] of $M$.
+Let $x \in B \cap C$.
+Then:
+:$\exists y \in C \setminus B : \paren{B \setminus \set x} \cup \set y$ is a [[Definition:Base of Matroid|base]] of $M$
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Circuit (Matroid)|circuit]] we have:
+=== [[Element of Matroid Base and Circuit has a Substitute/Lemma 1|Lemma 1]] ===
+{{:Element of Matroid Base and Circuit has a Substitute/Lemma 1}}{{qed|lemma}}
+From [[Independent Subset of Matroid is Augmented by Base]]:
+:$\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X$ is a [[Definition:Base of Matroid|base]] of $M$
+By definition of [[Definition:Independent Subset (Matroid)|independent]] and [[Definition:Dependent Subset (Matroid)|dependent subsets]] we have:
+=== [[Element of Matroid Base and Circuit has a Substitute/Lemma 2|Lemma 2]] ===
+{{:Element of Matroid Base and Circuit has a Substitute/Lemma 2}} {{qed|lemma}}
+ 
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 3')$]] we have:
+=== [[Element of Matroid Base and Circuit has a Substitute/Lemma 3|Lemma 3]] ===
+{{:Element of Matroid Base and Circuit has a Substitute/Lemma 3}} {{qed|lemma}}
+ 
+From [[Independent Subset is Base if Cardinality Equals Cardinality of Base]]:
+:$\paren{B \setminus \set x} \cup \set y$ is a [[Definition:Base of Matroid|base]] of $M$
+Because $x \notin \paren{ C \setminus \set x} \cup X$:
+:$y \ne x$
+By definition of [[Definition:Set Difference|set difference]]:
+:$y \notin B \setminus \set x$:
+By definition of [[Definition:Set Union|set union]]:
+:$y \notin \paren{B \setminus \set x } \cup \set x = B$
+By the definition of a [[Definition:Subset|subset]]:
+:$y \notin X$
+By definition of [[Definition:Set Union|set union]]:
+:$y \in C \setminus \set x \subseteq C$
+By definition of [[Definition:Set Difference|set difference]]:
+:$y \in C \setminus B$
+{{qed}}
+[[Category:Matroid Theory]]
+ksu25n4mqulynitqsuv00d79tt9mrx8
+\end{proof}<|endoftext|>
+\section{Set Difference with Proper Subset is Proper Subset}
+Tags: Set Difference, Subsets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $T \subsetneq S$ be a [[Definition:Proper Subset|proper subset]] of $S$.
+ 
+Let $S \setminus T$ denote the [[Definition:Set Difference|set difference]] between $S$ and $T$.
+Then:
+:$S \setminus T$ is a [[Definition:Proper Subset|proper subset]] of $S$
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference is Subset]]:
+:$S \setminus T \subseteq S$
+From [[Set Difference with Proper Subset]]:
+:$S \setminus T \ne \O$
+By definition of [[Definition:Proper Subset|proper subset]]:
+:$T \ne \O$
+From [[Intersection with Subset is Subset]]:
+:$S \cap T = T$
+Hence:
+:$S \cap T \ne \O$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Set Difference with Disjoint Set]]:
+:$S \setminus T \ne S$
+It follows that $S \setminus T$ is a [[Definition:Proper Subset|proper subset]] by definition.
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Subsets]]
+d3yuafomz3kochodppm5iwa12u8pl6n
+\end{proof}<|endoftext|>
+\section{Element of Matroid Base and Circuit has a Substitute/Lemma 1}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$C \setminus \set x$ is an [[Definition:Independent Subset (Matroid)|independent]] [[Definition:Proper Subset|proper subset]] of $C$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Circuit (Matroid)|circuit]]:
+:$C$ is a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq C$ and $\set x \subseteq B$
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$\set x \in \mathscr I$
+Because $C \notin \mathscr I$:
+:$\set x \ne C$
+Hence:
+:$\set x \subsetneq C$
+From [[Set Difference with Proper Subset is Proper Subset]]:
+:$C \setminus \set x$ is a [[Definition:Proper Subset|proper subset]] of $C$
+As $C$ is a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$C \setminus \set x \in \mathscr I$
+{{qed}}
+[[Category:Matroid Theory]]
+a6zgx051bkt13r45ikrt73rhyuvwvce
+\end{proof}<|endoftext|>
+\section{Subspace of Normed Vector Space with Induced Norm forms Normed Vector Space}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}_X}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $Y \subseteq X$ be a [[Definition:Vector Subspace|vector subspace]].
+Let $\norm {\, \cdot \,}_Y$ be the [[Definition:Induced Norm|induced norm]] on $Y$.
+Then $\struct {Y, \norm {\, \cdot \,}_Y}$ is a [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+=== Positive definiteness ===
+By definition of [[Definition:Induced Norm|induced norm]]:
+:$\forall y \in Y : \norm {y}_Y = \norm {y}_X > 0$
+Suppose $y \in Y : \norm {y}_Y = 0$.
+Since $\norm {\, \cdot \,}_Y$ is an [[Definition:Induced Norm|induced norm]] in $\struct {X, \norm {\, \cdot \,}_X}$:
+:$\norm {y}_X = 0$.
+Therefore, $y = \mathbf 0 \in X$.
+By definition of a [[Definition:Vector Subspace|vector subspace]], $Y$ has the [[Definition:Vector Space Axioms|additive identity]]:
+:$\mathbf 0 \in Y$.
+Thus:
+:$y = \mathbf 0 \in Y \subseteq X$.
+=== Positive homogeneity ===
+Let $y \in Y$.
+Let $\alpha \in \R$.
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha \cdot y}_Y 
+      | r = \norm {\alpha \cdot y}_X 
+      | c = {{defof|Induced Norm}}
+}}
+{{eqn | r = \size \alpha \norm {y}_X 
+      | c =  [[Definition:Norm on Vector Space|Norm axiom (N2)]]
+}}
+{{eqn | r = \size \alpha \norm {y}_Y
+      | c = {{defof|Induced Norm}}; $y \in Y$
+}}
+{{end-eqn}}
+=== Triangle inequality ===
+Let $y_1, y_2 \in Y$.
+By [[Definition:Vector Space Axioms|closure axiom]] of [[Definition:Vector Space|vector space]]: 
+:$y_1 + y_2 \in Y$.
+Furthermore:
+{{begin-eqn}}
+{{eqn | l = \norm {y_1 + y_2}_Y
+      | r = \norm {y_1 + y_2}_X 
+      | c = {{defof|Induced Norm}}
+}}
+{{eqn | o = \le
+      | r = \norm {y_1}_X + \norm {y_2}_X 
+      | c = [[Definition:Norm on Vector Space|Norm axiom (N2)]]
+}}
+{{eqn | r = \norm {y_1}_Y + \norm {y_2}_Y
+      | c = {{defof|Induced Norm}}; $y_1, y_2 \in Y$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{All Bases of Matroid have same Cardinality/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $X \subseteq S$ be any [[Definition:Independent Subset (Matroid)|independent subset]] of $M$.
+Then:
+:$\card X \le \card B$
+\end{theorem}
+
+\begin{proof}
+From [[Independent Subset is Contained in Maximal Independent Subset ]]:
+:$\exists B' \subseteq S : X \subseteq B'$ and $B'$ is a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $S$
+By definition of a [[Definition:Base of Matroid|base]]:
+:$B'$ is a [[Definition:Base of Matroid|base]] of $M$
+From [[Cardinality of Subset of Finite Set]]:
+:$\card X \le \card {B'}$
+From [[All Bases of Matroid have same Cardinality]]:
+:$\card{B'} = \card B$
+Hence:
+:$\card X \le \card B$
+{{qed}}
+[[Category:Matroid Theory]]
+tqan50fl5k15k8ca13rj1fevjxdijq7
+\end{proof}<|endoftext|>
+\section{Independent Set can be Augmented by Larger Independent Set/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $B \subseteq S$ be a [[Definition:Base of Matroid|base]] of $M$.
+Then:
+:$\exists Z \subseteq B \setminus X : \card{X \cup Z} = \card B  : X \cup Z$ is a [[Definition:Base of Matroid|base]] of $M$
+\end{theorem}
+
+\begin{proof}
+From [[Cardinality of Independent Set of Matroid is Smaller or Equal to Base]]:
+:$\card X \le \card B$
+=== Case 1: $\card X < \card B$ ===
+From [[Independent Set can be Augmented by Larger Independent Set]]:
+:$\exists Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$
+{{qed|lemma}}
+=== Case 2: $\card X = \card B$ ===
+Let $Z = \O$.
+Then:
+:$Z \subseteq B \setminus X : X \cup Z \in \mathscr I : \card {X \cup Z} = \card B$
+{{qed|lemma}}
+From [[Independent Subset is Base if Cardinality Equals Cardinality of Base]]:
+:$X \cup Z$ is a [[Definition:Base of Matroid|base]] of $M$
+{{qed}}
+[[Category:Matroid Theory]]
+tnbbb5wfpvvleyboe6ehakzvdof7wuh
+\end{proof}<|endoftext|>
+\section{Independent Subset is Base if Cardinality Equals Rank of Matroid/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $B \subseteq S$ be a [[Definition:Base of Matroid|base]] of $M$.
+Let $X \subseteq S$ be any [[Definition:Independent Subset (Matroid)|independent subset]] of $M$.
+Let $\card X = \card B$.
+Then:
+:$X$ is a [[Definition:Base of Matroid|base]] of $M$.
+\end{theorem}
+
+\begin{proof}
+From [[All Bases of Matroid have same Cardinality]]:
+:$\card B = \map \rho S$
+where $\rho$ denotes the [[Definition:Rank Function (Matroid)|rank function]] on $M$.
+Hence:
+:$\card X = \map \rho S$
+From [[Independent Subset is Base if Cardinality Equals Rank of Matroid]]:
+:$X$ is a [[Definition:Base of Matroid|base]] of $M$.
+{{qed}}
+[[Category:Matroid Theory]]
+rt8h42qguk3o0v1qbyp50oaql4nlpmz
+\end{proof}<|endoftext|>
+\section{Element of Matroid Base and Circuit has a Substitute/Lemma 2}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$x \notin \paren{ C \setminus \set x} \cup X$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}}:
+:$x \in \paren{ C \setminus \set x} \cup X$
+Then:
+:$\set x, C \setminus \set x \subseteq \paren{ C \setminus \set x} \cup X$
+From [[Union of Subsets is Subset]]:
+:$C \subseteq \paren{ C \setminus \set x} \cup X$
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$C \in \mathscr I$
+This [[Definition:Contradiction|contradicts]]:
+:$C \notin \mathscr I$
+It follows that:
+:$x \notin \paren{ C \setminus \set x} \cup X$
+{{qed}}
+[[Category:Matroid Theory]]
+jjgqizigeavgbqieg5kl8mtvsbhg3kh
+\end{proof}<|endoftext|>
+\section{Element of Matroid Base and Circuit has a Substitute/Lemma 3}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$\exists y \in \paren{\paren{C \setminus \set x} \cup X} \setminus \paren{B \setminus \set x} : \paren{B \setminus \set x} \cup \set y \in \mathscr I : \card{\paren{B \setminus \set x} \cup \set y} = \card {\paren{ C \setminus \set x} \cup X}$
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference is Subset]]:
+:$B \setminus \set x \subseteq B$
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$B \setminus \set x \in \mathscr I$
+We have
+{{begin-eqn}}
+{{eqn | l = \card {B \setminus \set x} + 1
+      | r = \paren{\card B - \card{\set x} } + 1
+      | c = [[Cardinality of Set Difference with Subset]]
+}}
+{{eqn | r = \paren{\card B - 1 } + 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | r = \card B
+}}
+{{eqn | r = \card {\paren{ C \setminus \set x} \cup X}
+      | c = [[All Bases of Matroid have same Cardinality]]
+}}
+{{end-eqn}}
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 3')$]]:
+:$\exists y \in \paren{\paren{C \setminus \set x} \cup X} \setminus \paren{B \setminus \set x} : \paren{B \setminus \set x} \cup \set y \in \mathscr I : \card{\paren{B \setminus \set x} \cup \set y} = \card {\paren{ C \setminus \set x} \cup X}$
+{{qed}}
+[[Category:Matroid Theory]]
+shmd8jquqhipde4n19gs7glgxqunhci
+\end{proof}<|endoftext|>
+\section{Matroid Unique Circuit Property}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $X \subseteq S$ be an [[Definition:Independent Subset (Matroid)|independent subset]] of $M$.
+Let $x \in S$ such that:
+:$X \cup \set x$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $M$.
+Then there exists a unique [[Definition:Circuit|circuit]] $C$ such that:
+:$x \in C \subseteq X \cup \set x$
+\end{theorem}
+
+\begin{proof}
+From [[Dependent Subset Contains a Circuit]]:
+:there exists a [[Definition:Circuit|circuit]] $C$ such that $C \subseteq X \cup \set x$
+From [[Dependent Subset of Independent Set Union Singleton Contains Singleton]]:
+:$x \in C$
+{{AimForCont}} $C'$ is [[Definition:Circuit|circuit]] of $M$ such that:
+:$C' \ne C$
+:$C' \subseteq X \cup \set x$
+From [[Dependent Subset of Independent Set Union Singleton Contains Singleton]]:
+:$x \in C'$
+By the definition of [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$C \not \subseteq C'$
+By the definition of a [[Definition:Subset|subset]]:
+:$\exists y \in C \setminus C'$
+By the definition of [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$C \setminus \set y \in \mathscr I$
+From [[Independent Subset is Contained in Maximal Independent Subset]]:
+:$\exists Y \in \mathscr I : C \setminus \set y \subseteq Y \subseteq X \cup \set x : Y$ is a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $X \cup \set x$
+By assumption:
+:$X$ is a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $X \cup \set x$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 3''')$]]:
+:$\card Y = \card X$
+As $x \in C'$ and $y \notin C'$ then:
+:$x \ne y$
+Hence:
+:$x \in C \setminus \set y$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$C \setminus \set y \cup  \set y = C \not \subseteq Y$ 
+Hence:
+:$y \notin Y$
+Hence:
+:$Y \subseteq \paren{X \cup \set x} \setminus \set y$
+From [[Cardinality of Subset of Finite Set]]:
+:$\card Y \le \card{\paren{X \cup \set x} \setminus \set y}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \card{\paren{X \cup \set x} \setminus \set y}
+      | r = \card{\paren{X \cup \set x} } - \card{\set y}
+      | c =
+}}
+{{eqn | r = \card X + \card{\set x} - \card{\set y}
+      | c =
+}}
+{{eqn | r = \card X +  1 - 1
+      | c =
+}}
+{{eqn | r = \card X
+      | c =
+}}
+{{eqn | r = \card Y
+      | c =
+}}
+{{end-eqn}}
+From [[Cardinality of Proper Subset of Finite Set]]:
+:$Y = \paren{X \cup \set x} \setminus \set y$
+Hence:
+:$\paren{X \cup \set x} \setminus \set y \in \mathscr I$
+As $y \notin C'$ and $C' \subseteq X \cup \set x$ then:
+:$C' \subseteq \paren{X \cup \set x} \setminus \set y$
+From [[Superset of Dependent Set is Dependent]]:
+:$\paren{X \cup \set x} \setminus \set y \notin \mathscr I$
+This [[Definition:Contradiction|contradicts]] the previous statement that 
+:$\paren{X \cup \set x} \setminus \set y \in \mathscr I$
+It follows that $C$ is the unique [[Definition:Circuit|circuit]] such that:
+:$C \subseteq X \cup \set x$
+{{qed}}
+[[Category:Matroid Theory]]
+glbbo8kkqqsz457argdnbvn0w14wu1w
+\end{proof}<|endoftext|>
+\section{Absolute Value of Negative}
+Tags: Absolute Value Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then:
+:$\size x = \size {-x}$
+where $\size x$ denotes the [[Definition:Absolute Value|absolute value]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $x \ge 0$.
+By definition of [[Definition:Absolute Value|absolute value]]:
+:$\size x = x$
+We have that:
+:$-x < 0$
+and so by definition of [[Definition:Absolute Value|absolute value]]:
+:$\size {-x} = -\paren {-x} = x$
+{{qed|lemma}}
+Now let $x < 0$.
+By definition of [[Definition:Absolute Value|absolute value]]:
+:$\size x = -x$
+We have that:
+:$-x > 0$
+and so:
+and so by definition of [[Definition:Absolute Value|absolute value]]:
+:$\size {-x} = -x$
+{{qed|lemma}}
+In both cases it is seen that:
+:$\size x = \size {-x}$
+{{qed}}
+[[Category:Absolute Value Function]]
+n2lpetttt3eo8db79q0jju2cmvn2wzh
+\end{proof}<|endoftext|>
+\section{Dependent Subset Contains a Circuit}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\mathscr C$ denote the set of all [[Definition:Circuit (Matroid)|circuits]] of $M$.
+Let $A$ be a [[Definition:Dependent Subset (Matroid)|dependent subset]].
+Then:
+:$\exists C \in \mathscr C : C \subseteq A$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Ordered Set|ordered set]] $\struct {\powerset S \setminus \mathscr I, \subseteq}$.
+From [[Element of Finite Ordered Set is Between Maximal and Minimal Elements]]:
+:$\exists C \in \mathscr I : C \subseteq A$ and $A$ is [[Definition:Minimal Element|minimal]] in $\struct {\powerset S \setminus \mathscr I, \subseteq}$.
+By definition of a [[Definition:Circuit (Matroid)|circuit]]:
+:$C \in \mathscr C$
+{{qed}}
+[[Category:Matroid Theory]]
+mj27wuuxqy537wwzcf4zri7aadjq5z8
+\end{proof}<|endoftext|>
+\section{Condition for Linear Operation on Complex Numbers to be of Finite Order}
+Tags: Roots of Unity
+
+\begin{theorem}
+Let $A$ be the [[Definition:Operation|operation]] on the [[Definition:Complex Number|complex numbers]] $\C$ defined as:
+:$\map A x = \alpha x + \beta$
+Then $A$ is of [[Definition:Finite Order|finite order]] greater than $1$ {{iff}} $\alpha$ is a [[Definition:Complex Root of Unity|root of unity]] other than $1$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Not completely sure I understand the context}}
+\end{proof}<|endoftext|>
+\section{Positive Rational Numbers under Division do not form Group}
+Tags: Rational Numbers, Rational Division
+
+\begin{theorem}
+Let $\struct {\Q, /}$ denote the [[Definition:Algebraic Structure|algebraic structure]] consisting of the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] $\Q$ under the [[Definition:Binary Operation|operation]] $/$ of [[Definition:Rational Division|division]].
+We have that $\struct {\Q, /}$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+In order to be a [[Definition:Group|group]], it is necessary that $\struct {\Q, /}$ be an [[Definition:Associative Algebraic Structure|associative structure]].
+But consider the [[Definition:Element|elements]] $2, 6, 12$ of $\Q$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {12 / 6} / 2
+      | r = 2 / 1
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+whereas:
+{{begin-eqn}}
+{{eqn | l = 12 / \paren {6 / 2}
+      | r = 12 / 3
+      | c = 
+}}
+{{eqn | r = 4
+      | c = 
+}}
+{{end-eqn}}
+That is:
+:$\paren {12 / 6} / 2 \ne 12 / \paren {6 / 2}$
+So it is not generally the case that for $a, b, c \in \Q$:
+:$\paren {a / b} / c = a / \paren {b / c}$
+and so $/$ is not [[Definition:Associative Operation|associative]] on $\Q$.
+Hence $\struct {\Q, /}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Sine of x over a/Corollary 1}
+Tags: Derivative of Inverse Hyperbolic Sine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\ln \size {x + \sqrt {x^2 + a^2} } } = \dfrac 1 {\sqrt {x^2 + a^2} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\sinh^{-1} } {\frac x a}
+      | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 + a^2} }
+      | c = {{Defof|Inverse Hyperbolic Sine|subdef = Real|index = 2}}
+}}
+{{eqn | r = \map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 + a^2} } }
+      | c = 
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } - \ln a
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 + a^2} } }
+      | r = \map {\dfrac \d {\d x} } {\map {\sinh^{-1} } {\frac x a} + \ln a }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {\sqrt {x^2 + a^2} } + \map {\dfrac \d {\d x} } {\ln a}
+      | c = [[Derivative of Inverse Hyperbolic Sine of x over a]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {x^2 + a^2} } + 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+We have that $\sqrt {x^2 + a^2} > x$ for all $x$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = x + \sqrt {x^2 + a^2}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x + \sqrt {x^2 + a^2} }
+      | r = x + \sqrt {x^2 + a^2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\size {x + \sqrt {x^2 + a^2} } }
+      | r = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 + a^2} } }
+      | c = 
+}}
+{{end-eqn}}
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Logarithm of x plus Root x squared plus a squared}
+Tags: Logarithms
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then:
+:$-\map \ln {x + \sqrt {x^2 + a^2} } = \map \ln {-x + \sqrt {x^2 + a^2} } - \map \ln {a^2}$
+\end{theorem}
+
+\begin{proof}
+We have that $\sqrt {x^2 + a^2} > x$ for all $x$.
+Thus:
+:$x + \sqrt {x^2 + a^2} > 0$
+and so $\map \ln {x + \sqrt {x^2 + a^2} }$ is defined for all $x$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = -\map \ln {x + \sqrt {x^2 + a^2} }
+      | r = \map \ln {\dfrac 1 {x + \sqrt {x^2 + a^2} } }
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{eqn | r = \map \ln {\dfrac {x - \sqrt {x^2 + a^2} } {\paren {x + \sqrt {x^2 + a^2} } \paren {x - \sqrt {x^2 + a^2} } } }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $x - \sqrt {x^2 + a^2}$
+}}
+{{eqn | r = \map \ln {\dfrac {x - \sqrt {x^2 + a^2} } {x^2 - \paren {x^2 + a^2} } }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \map \ln {\dfrac {-x + \sqrt {x^2 + a^2} } {a^2} }
+      | c = simplifying
+}}
+{{eqn | r = \map \ln {-x + \sqrt {x^2 + a^2} } - \map \ln {a^2}
+      | c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Sine of x over a/Corollary 2}
+Tags: Derivative of Inverse Hyperbolic Sine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\ln \size {x - \sqrt {x^2 + a^2} } } = -\dfrac 1 {\sqrt {x^2 + a^2} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = -\map {\sinh^{-1} } {\frac x a}
+      | r = \map {\sinh^{-1} } {-\frac x a}
+      | c = [[Inverse Hyperbolic Sine is Odd Function]]
+}}
+{{eqn | r = \map \ln {-\paren {\frac x a} + \sqrt {\paren {-\frac x a}^2 + a^2} }
+      | c = {{Defof|Inverse Hyperbolic Sine|subdef = Real|index = 2}}
+}}
+{{eqn | r = \map \ln {-\frac x a + \dfrac 1 a \sqrt {x^2 + a^2} }
+      | c = 
+}}
+{{eqn | r = \map \ln {-\dfrac 1 a \paren {x - \sqrt {x^2 + a^2} } }
+      | c = 
+}}
+{{eqn | r = \map \ln {-x + \sqrt {x^2 + a^2} } - \ln a
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } - \ln a
+      | c = as $\sqrt {x^2 + a^2} > -x$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\ln \size {x - \sqrt {x^2 + a^2} } }
+      | r = \map {\dfrac \d {\d x} } {-\map {\sinh^{-1} } {\frac x a} + \ln a}
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = -\dfrac 1 {\sqrt {x^2 + a^2} } + \map {\dfrac \d {\d x} } {\ln a}
+      | c = [[Derivative of Inverse Hyperbolic Sine of x over a]]
+}}
+{{eqn | r = -\dfrac 1 {\sqrt {x^2 + a^2} } + 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form/Corollary}
+Tags: Primitive of Reciprocal of Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {-\sqrt {x^2 + a^2} } = \ln \size {x - \sqrt {x^2 + a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }
+      | r = \map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ in Logarithm Form]]
+}}
+{{eqn | l = \int \frac {\d x} {-\sqrt {x^2 + a^2} }
+      | r = -\map \ln {x + \sqrt {x^2 + a^2} } + C
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } + C + \map \ln {a^2}
+      | c = [[Negative of Logarithm of x plus Root x squared plus a squared/Corollary|Negative of Logarithm of x plus Root x squared plus a squared: Corollary]]
+}}
+{{eqn | r = \ln \size {x - \sqrt {x^2 + a^2} } + C
+      | c = subsuming $\map \ln {a^2}$ into the constant
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosine of x over a/Corollary 1}
+Tags: Derivative of Inverse Hyperbolic Cosine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 - a^2} } } = \dfrac 1 {\sqrt {x^2 - a^2} }$
+for $x > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\cosh^{-1} } {\frac x a}
+      | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - a^2} }
+      | c = {{Defof|Inverse Hyperbolic Cosine|subdef = Real|index = 2}}
+}}
+{{eqn | r = \map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 - a^2} }
+      | c = 
+}}
+{{eqn | r = \map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 - a^2} } }
+      | c = 
+}}
+{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\map \ln {x + \sqrt {x^2 - a^2} } }
+      | r = \map {\dfrac \d {\d x} } {\map {\cosh^{-1} } {\frac x a} - \ln a }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {\sqrt {x^2 - a^2} } + \map {\dfrac \d {\d x} } {\ln a}
+      | c = [[Derivative of Inverse Hyperbolic Cosine of x over a]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {x^2 - a^2} } + 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
+When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
+When $x < -a$ we have that $x + \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined.
+Hence the restriction on the [[Definition:Domain of Real Function|domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosine of x over a/Corollary 2}
+Tags: Derivative of Inverse Hyperbolic Sine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\map \ln {x - \sqrt {x^2 - a^2} } } = -\dfrac 1 {\sqrt {x^2 - a^2} }$
+for $x > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = -\map {\cosh^{-1} } {\frac x a}
+      | r = -\map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - a^2} }
+      | c = {{Defof|Inverse Hyperbolic Cosine|subdef = Real|index = 2}}
+}}
+{{eqn | r = -\map \ln {\frac x a + \dfrac 1 a \sqrt {x^2 - a^2} }
+      | c = 
+}}
+{{eqn | r = -\map \ln {\dfrac 1 a \paren {x + \sqrt {x^2 - a^2} } }
+      | c = 
+}}
+{{eqn | r = -\map \ln {x + \sqrt {x^2 - a^2} } + \ln a
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \map \ln {x - \sqrt {x^2 - a^2} } + \ln a^2 + \ln a
+      | c = [[Negative of Logarithm of x plus Root x squared minus a squared]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\map \ln {x - \sqrt {x^2 - a^2} } }
+      | r = \map {\dfrac \d {\d x} } {-\map {\cosh^{-1} } {\frac x a} - \ln a^3}
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = -\dfrac 1 {\sqrt {x^2 - a^2} } - \map {\dfrac \d {\d x} } {\ln a^3}
+      | c = [[Derivative of Inverse Hyperbolic Cosine of x over a]]
+}}
+{{eqn | r = -\dfrac 1 {\sqrt {x^2 - a^2} } + 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+When $x = a$ we have that $\sqrt {x^2 - a^2} = 0$ and then $\dfrac 1 {\sqrt {x^2 - a^2} }$ is not defined.
+When $\size x < a$ we have that $x^2 - a^2 < 0$ and then $\sqrt {x^2 - a^2}$ is not defined.
+When $x < -a$ we have that $x - \sqrt {x^2 - a^2} < 0$ and so $\map \ln {x + \sqrt {x^2 - a^2} }$ is not defined.
+Hence the restriction on the [[Definition:Domain of Real Function|domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form/Corollary}
+Tags: Primitive of Reciprocal of Root of x squared minus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {-\sqrt {x^2 - a^2} } = \ln \size {x - \sqrt {x^2 - a^2} } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {\sqrt {x^2 - a^2} }
+      | r = \ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Reciprocal of Root of x squared minus a squared/Logarithm Form|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$ in Logarithm Form]]
+}}
+{{eqn | l = \int \frac {\d x} {-\sqrt {x^2 + a^2} }
+      | r = -\ln \size {x + \sqrt {x^2 - a^2} } + C
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \ln \size {x - \sqrt {x^2 - a^2} } + C + \map \ln {a^2}
+      | c = [[Negative of Logarithm of x plus Root x squared minus a squared|Negative of Logarithm of x plus Root x squared minus a squared]]
+}}
+{{eqn | r = \ln \size {x - \sqrt {x^2 - a^2} } + C
+      | c = subsuming $\map \ln {a^2}$ into the constant
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Tangent of x over a/Corollary}
+Tags: Derivative of Inverse Hyperbolic Tangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } } = \dfrac 1 {a^2 - x^2}$
+where $\size x < a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 a \map {\tanh^{-1} } {\frac x a}
+      | r = \dfrac 1 a \cdot \dfrac 1 2 \map \ln {\dfrac {1 + \frac x a} {1 - \frac x a} }
+      | c = {{Defof|Inverse Hyperbolic Tangent|subdef = Real|index = 2}}
+}}
+{{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] of argument by $a$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } }
+      | r = \map {\dfrac \d {\d x} } {\dfrac 1 a \map {\tanh^{-1} } {\frac x a} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 a \cdot \dfrac a {a^2 - x^2}
+      | c = [[Derivative of Inverse Hyperbolic Tangent of x over a]]
+}}
+{{eqn | r = \dfrac 1 {a^2 - x^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cotangent of x over a/Corollary}
+Tags: Derivative of Inverse Hyperbolic Tangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } } = \dfrac 1 {a^2 - x^2}$
+where $\size x > a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 a \map {\coth^{-1} } {\frac x a}
+      | r = \dfrac 1 a \cdot \dfrac 1 2 \map \ln {\dfrac {\frac x a + 1} {\frac x a - 1} }
+      | c = {{Defof|Inverse Hyperbolic Cotangent|subdef = Real|index = 2}}
+}}
+{{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] of argument by $a$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d x} } {\dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } }
+      | r = \map {\dfrac \d {\d x} } {\dfrac 1 a \map {\coth^{-1} } {\frac x a} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 a \cdot \dfrac {-a} {x^2 - a^2}
+      | c = [[Derivative of Inverse Hyperbolic Cotangent of x over a]]
+}}
+{{eqn | r = \dfrac 1 {a^2 - x^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Set Difference of Matroid Circuit with Element is Independent}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$C \setminus \set x$ is an [[Definition:Independent Subset (Matroid)|independent subset]] of $C$
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference is Subset]]:
+:$C \setminus \set x \subseteq C$
+Because $x \in C$ and $x \notin C \setminus \set x$:
+:$C \setminus \set x \ne C$
+Hence:
+:$C \setminus \set x \subsetneq C$
+By definition of [[Definition:Circuit (Matroid)|circuit]]:
+:$C$ is a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]
+Hence:
+:$C \setminus \set x \in \mathscr I$
+{{qed}}
+[[Category:Matroid Theory]]
+l0nw0nrmr3yf8agel9n9z7hwfdruvn2
+\end{proof}<|endoftext|>
+\section{Union of Matroid Base with Element of Complement is Dependent}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$B \cup \set x$ is a [[Definition:Dependent Subset (Matroid)|dependent]] [[Definition:Superset|superset]] of $B$
+\end{theorem}
+
+\begin{proof}
+From [[Set is Subset of Union]]:
+:$B \subseteq B \cup \set x$
+Because $x \in B \cup \set x$ and $x \notin B$:
+:$B \ne B \cup \set x$
+Hence:
+:$B \subsetneq B \cup \set x$
+By definition of [[Definition:Base of Matroid|base]]:
+:$B$ is a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]]
+Hence:
+:$B \cup \set x \notin \mathscr I$
+{{qed}}
+[[Category:Matroid Theory]]
+05pwsmu9mo9f7zw2mm4sdq0xb4x0i3q
+\end{proof}<|endoftext|>
+\section{Primitive of Power of a x + b/Proof 2}
+Tags: Primitive of Power of a x + b
+
+\begin{theorem}
+{{:Primitive of Power of a x + b}}
+where $n \ne 1$.
+\end{theorem}
+
+\begin{proof}
+Let $u = a x + b$.
+Then:
+:$\dfrac {\d u} {\d x} = a$
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \paren {a x + b}^n \rd x
+      | r = \int \dfrac {u^n} a \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = \dfrac 1 a \dfrac {u^{n + 1} } {n + 1}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C
+      | c = substituting back for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Function Form}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \dfrac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 a \tanh^{-1} \dfrac x a + C & : \size x < a \\
+& \\
+\dfrac 1 a \coth^{-1} \dfrac x a + C & : \size x > a \\
+& \\
+\text {undefined} & : x = a \end {cases}$
+\end{theorem}
+
+\begin{proof}
+First note that if $x = a$ then $a^2 - x^2 = 0$ and so $\dfrac 1 {a^2 - x^2}$ is undefined.
+=== [[Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form/Proof|Proof for $\size x < a$]] ===
+{{:Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form/Proof}}{{qed|lemma}}
+=== [[Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form/Proof|Proof for $\size x > a$]] ===
+{{:Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form/Proof}}{{qed}}
+[[Category:Primitive of Reciprocal of a squared minus x squared]]
+psko6uv1r93j0d6ykdjnb58lpixw3u0
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \frac 1 a \coth^{-1} \frac x a + C$
+where $\size x > a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x less than a}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+Let $\size x < a$.
+Then:
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C$
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1/size of x greater than a}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+Let $\size x > a$.
+Then:
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C$
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Logarithm Form}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real]] [[Definition:Constant|constant]].
+Let $x \in \R$ such that $\size x \ne a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of a squared minus x squared/Logarithm Form 2}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \dfrac 1 {2 a} \ln \size {\dfrac {a + x} {a - x} } + C$
+\end{theorem}
+
+\begin{proof}
+From the [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form 1|$1$st logarithm form]]:
+:$\displaystyle \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a\\
+& \\
+ \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\
+& \\
+\text {undefined} & : \size x = a \end {cases}$
+From [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form/Lemma|Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma]]:
+:$\map \ln {\dfrac {a + x} {a - x} }$ is defined {{iff}} $\size x < a$
+:$\map \ln {\dfrac {x + a} {x - a} }$ is defined {{iff}} $\size x > a$
+Let $\size x < a$.
+Then $\map \ln {\dfrac {a + x} {a - x} }$ is defined.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \dfrac {a + x} {a - x}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {a + x} {a - x}
+      | r = \size {\dfrac {a + x} {a - x} }
+      | c = 
+}}
+{{end-eqn}}
+So the result holds for $\size x < a$.
+Let $\size x > a$.
+Then $\map \ln {\dfrac {x + a} {x - a} }$ is defined.
+We have:
+We have that:
+{{begin-eqn}}
+{{eqn | l = \dfrac {x + a} {x - a}
+      | r = -\dfrac {a + x} {a - x}
+      | c = 
+}}
+{{eqn | o = <
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {x + a} {x - a}
+      | r = \size {\dfrac {a + x} {a - x} }
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Logarithm Form}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real]] [[Definition:Constant|constant]].
+Let $x \in \R$ such that $\size x \ne a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Inverse Hyperbolic Tangent Form}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \tanh^{-1} {\frac x a} + C$
+where $\size x < a$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1}
+Tags: Primitive of Reciprocal of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\
+& \\
+\dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\
+& \\
+\text {undefined} & : \size x = a \end {cases}$
+\end{theorem}
+
+\begin{proof}
+=== [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x less than a|Case where $\size x < a$]] ===
+{{:Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x less than a}}
+=== [[Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a|Case where $\size x > a$]] ===
+{{:Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x less than a}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+Let $\size x < a$.
+Then:
+:$\displaystyle \int \frac {\d x} {x^2 - a^2} = \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C$
+\end{theorem}
+
+\begin{proof}
+Let $\size x < a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 - a^2}
+      | r = -\frac 1 a \tanh^{-1} {\frac x a} + C
+      | c = [[Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Tangent Form|Primitive of $\dfrac 1 {x^2 - a^2}$: $\tanh^{-1}$ form]]
+}}
+{{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} } } + C
+      | c = [[Inverse Hyperbolic Tangent of x over a in Logarithm Form|$\tanh^{-1} \dfrac x a$ in Logarithm Form]]
+}}
+{{eqn | r = -\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a}
+Tags: Primitive of Reciprocal of x squared minus a squared
+
+\begin{theorem}
+Let $\size x > a$.
+Then:
+:$\ds \int \frac {\d x} {x^2 - a^2} = \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C$
+\end{theorem}
+
+\begin{proof}
+Let $\size x > a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x^2 - a^2}
+      | r = -\frac 1 a \coth^{-1} {\frac x a} + C
+      | c = [[Primitive of Reciprocal of a squared minus x squared/Inverse Hyperbolic Cotangent Form|Primitive of $\dfrac 1 {x^2 - a^2}$: $\coth^{-1}$ form]]
+}}
+{{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C
+      | c = [[Inverse Hyperbolic Cotangent of x over a in Logarithm Form|$\coth^{-1} \dfrac x a$  in Logarithm Form]]
+}}
+{{eqn | r = -\dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C
+      | c = [[Reciprocal of Logarithm]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set is Closed under Complement}
+Tags: Power Set, Relative Complement
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Then:
+:$\forall A \in \powerset S: \relcomp S A \in \powerset S$
+where $\relcomp S A$ denotes the [[Definition:Relative Complement|complement of $A$ relative to $S$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = A
+      | o = \in
+      | r = \powerset S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | ll= \leadsto
+      | l = S \setminus A
+      | o = \subseteq
+      | r = S
+      | c = [[Set Difference is Subset]]
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S A
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | ll= \leadsto
+      | l = \relcomp S A
+      | o = \in
+      | r = \powerset S
+      | c = {{Defof|Power Set}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetric Difference with Intersection forms Boolean Ring}
+Tags: Idempotent Rings, Set Intersection, Symmetric Difference, Power Set
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then $\struct {\powerset S, *, \cap}$ is a [[Definition:Boolean Ring|Boolean ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Symmetric Difference with Intersection forms Ring]]:
+:$\struct {\powerset S, *, \cap}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+From [[Intersection is Idempotent]], $\cap$ is an [[Definition:Idempotent Operation|idempotent operation]] on $S$.
+Hence the result by definition of [[Definition:Boolean Ring|Boolean ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallelism is Reflexive Relation}
+Tags: Parallel Lines, Examples of Reflexive Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is [[Definition:Parallel Lines|parallel]] to $l_2$.
+Then $\parallel$ is a [[Definition:Reflexive Relation|reflexive relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Parallel Lines|parallel lines]], the contemporary definition is for a [[Definition:Straight Line|straight line]] to be declared [[Definition:Parallel Lines|parallel]] to itself.
+Hence for a [[Definition:Straight Line|straight line]] $l$:
+:$l \parallel l$
+Thus $\parallel$ is seen to be [[Definition:Reflexive Relation|reflexive]].
+\end{proof}<|endoftext|>
+\section{Parallelism is Symmetric Relation}
+Tags: Parallel Lines, Examples of Symmetric Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is [[Definition:Parallel Lines|parallel]] to $l_2$.
+Then $\parallel$ is a [[Definition:Symmetric Relation|symmetric relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $l_1 \parallel l_2$.
+By definition of [[Definition:Parallel Lines|parallel lines]], $l_1$ does not meet $l_2$ when [[Definition:Production|produced]] indefinitely.
+Hence $l_2$ similarly does not meet $l_1$ when [[Definition:Production|produced]] indefinitely.
+That is:
+:$l_2 \parallel l_1$
+Thus $\parallel$ is seen to be [[Definition:Symmetric Relation|symmetric]].
+\end{proof}<|endoftext|>
+\section{Parallelism is Equivalence Relation/Transitivity}
+Tags: Parallel Lines, Examples of Transitive Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is [[Definition:Parallel Lines|parallel]] to $l_2$.
+Then $\parallel$ is a [[Definition:Transitive Relation|transitive relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Parallelism is Transitive Relation]]:
+:$l_1 \parallel l_2$ and $l_2 \parallel l_3$ implies $l_1 \parallel l_3$.
+Thus $\parallel$ is seen to be [[Definition:Transitive Relation|transitive]].
+\end{proof}<|endoftext|>
+\section{1-Seminorm on Continuous on Closed Interval Real-Valued Functions is Norm}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $\CC \closedint a b$ be the [[Definition:Space of Continuous on Closed Interval Real-Valued Functions|space of continuous on closed interval real-valued functions]].
+Let $x \in \CC \closedint a b$ be a [[Definition:Continuous Real-Valued Vector Function|continuous real valued function]].
+Let $\displaystyle \norm x_1 := \int_a^b \size {\map x t} \rd t$ be the [[Definition:P-Seminorm|1-seminorm]].
+Then $\norm {\, \cdot \,}_1$ is a [[Definition:Norm on Vector Space|norm]] on $\CC \closedint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Positive definiteness ===
+Let $x \in \CC \closedint a b$.
+Then $\forall t \in \closedint 0 1 : \size {\map x t} \ge 0 $.
+Hence:
+:$\displaystyle \int_a^b \size {\map x t} \rd t = \norm x_1 \ge 0$.
+Suppose $\forall t \in \closedint a b : \map x t = 0$.
+Then $\norm x_1 = 0$.
+Therefore:
+:$\paren {x = 0} \implies \paren {\norm x_1 = 0}$
+Let $x \in \CC \closedint a b : \norm x_1 = 0$.
+Suppose:
+:$\forall t \in \openint a b : \map x t = 0$.
+By [[Definition:Assumption|assumption]] of [[Definition:Continuous Real Function on Closed Interval|continuity]] of $x$:
+:$\forall t \in \closedint a b : \map x t = 0$.
+{{AimForCont}}:
+:$\exists t_0 \in \openint a b : \map x {t_0} \ne 0$.
+By [[Definition:Assumption|assumption]], $x$ is [[Definition:Continuous Real Function on Closed Interval|continuous]] at $t_0$.
+:$\forall \epsilon \in \R_{> 0} : \exists \delta \in \R_{> 0} : \size {t - t_0} < \delta \implies \size {\map x t - \map x {t_0} } < \epsilon$
+Furthermore:
+:$\exists \delta \in \R_{> 0} : \paren{ a < t_0 - \delta} \land \paren {t_0 + \delta < b}$
+Let $\displaystyle \epsilon = \frac {\size {\map x {t_0}}} 2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \size {\map x t}
+      | r = \size {\map x t + \map x {t_0} - \map x {t_0} }
+}}
+{{eqn | o = \ge
+      | r = \size {\map x {t_0} } - \size {\map x t - \map x {t_0} }
+      | c = [[Reverse Triangle Inequality/Normed Vector Space|Reverse triangle inequality]]
+}}
+{{eqn | o = >
+      | r = \size {\map x {t_0} } - \frac {\size {\map x {t_0} } } 2
+}}
+{{eqn | r = \frac {\size {\map x {t_0} } } 2 
+}}
+{{eqn | o = >
+      | r = 0
+      | c = $\map x {t_0} \ne 0$
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \norm x_1
+}}
+{{eqn | r = \int_a^b \size {\map x t} \rd t
+}}
+{{eqn | o = \ge
+      | r = \int_{t_0 - \delta}^{t_0 + \delta} \size {\map x t} \rd t
+      | c = $t_0 + \delta < b$, $a < t_0 - \delta$
+}}
+{{eqn | o = \ge
+      | r = \int_{t_0 - \delta}^{t_0 + \delta} \frac {\size {\map x {t_0} } } 2 \rd t
+}}
+{{eqn | r = 2 \delta \frac {\size {\map x {t_0} } } 2
+}}
+{{eqn | r = \delta \size {\map x {t_0} }
+}}
+{{eqn | o = >
+      | r = 0
+}}
+{{end-eqn}}
+Hence, we reached a [[Definition:Contradiction|contradiction]].
+Therefore:
+:$\paren {\norm x_1 = 0} \implies \paren {x = 0}$
+=== Positive homogeneity ===
+Let $x \in \CC \closedint a b$, $\alpha \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\alpha x}_1
+      | r = \int_a^b \size {\alpha \map x t} \rd t
+}}
+{{eqn | r =\size \alpha \int_a^b \size {\map x t} \rd t
+}}
+{{eqn | r =\size \alpha \norm x_1
+}}
+{{end-eqn}}
+=== Triangle inequality ===
+Let $x, y \in \CC \closedint a b$
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}_1
+      | r = \int_a^b \size {\map {\paren {x + y} } t} \rd t
+}}
+{{eqn | r = \int_a^b \size {\map x t + \map y t} \rd t
+      | c = {{defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | o = \le
+      | r = \int_a^b \paren {\size x + \size y} \rd t
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \int_a^b \size x \rd t + \int_a^b \size y \rd t
+}}
+{{eqn | r = \norm x_1 + \norm y_1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Approximation/Examples/22 over 7}
+Tags: Approximations to Pi
+
+\begin{theorem}
+$\dfrac {22} 7$ is a convenient [[Definition:Approximation|approximation]] to $\pi$:
+:$\dfrac {22} 7 = 3 \cdotp \dot 14285 \dot 7$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {22} 7
+      | r = 3 \cdotp \dot 14285 \dot 7
+      | c = 
+}}
+{{eqn | l = \pi
+      | o = \approx
+      | r = 3 \cdotp 14159265
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {22} 7 - \pi
+      | o = \approx
+      | r = 0 \cdotp 0012645
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\frac {22} 7 - \pi} \pi
+      | o = \approx
+      | r = 0 \cdotp 0004025
+      | c = 
+}}
+{{eqn | o = \approx
+      | r = 0 \cdotp 04025 \%
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Perpendicularity is Symmetric Relation}
+Tags: Perpendiculars, Examples of Symmetric Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is [[Definition:Perpendicular Lines|perpendicular]] to $l_2$.
+Then $\perp$ is a [[Definition:Symmetric Relation|symmetric relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $l_1 \perp l_2$.
+By definition of [[Definition:Perpendicular Lines|perpendicular lines]], $l_1$ meets $l_2$ at a [[Definition:Right Angle|right angle]].
+Hence $l_2$ similarly meets $l_1$ at a [[Definition:Right Angle|right angle]].
+That is:
+:$l_2 \perp l_1$
+Thus $\parallel$ is seen to be [[Definition:Symmetric Relation|symmetric]].
+\end{proof}<|endoftext|>
+\section{Perpendicularity is Antireflexive Relation}
+Tags: Perpendiculars, Examples of Reflexive Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is [[Definition:Perpendicular Lines|perpendicular]] to $l_2$.
+Then $\perp$ is an [[Definition:Antireflexive Relation|antireflexive relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Perpendicular Lines|perpendicular lines]], for $l_1$ to be [[Definition:Perpendicular Lines|perpendicular]] to itself would mean it would have to meet itself in a [[Definition:Right Angle|right angle]].
+This it does not do.
+So $l_1 \not \perp l_1$.
+Thus $\perp$ is seen to be [[Definition:Antireflexive Relation|antireflexive]].
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma}
+Tags: Matroid Satisfies Base Axiom
+
+\begin{theorem}
+:$\exists B_3 \in \mathscr B$:
+:::$V \subseteq B_3$
+:::$\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$
+\end{theorem}
+
+\begin{proof}
+Let $U, V \in \mathscr I$ such that:
+:$\card V < \card U$
+=== [[Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 1|Lemma 1]] ===
+{{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 1}}{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 2|Lemma 2]] ===
+{{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 2}}{{qed|lemma}}
+From [[Subset of Set Difference iff Disjoint Set]]:
+:$U \setminus V \subseteq B_1 \setminus B_2$
+We have:
+{{begin-eqn}}
+{{eqn | l = \card{B_2} 
+      | r = \card{B_1}
+      | c = [[Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality]]
+}}
+{{eqn | ll = \leadsto 
+      | l = \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V} 
+      | r = \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}
+}}
+{{eqn | ll = \leadsto 
+      | l = \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V} 
+      | r = \card{B_1 \setminus B_2}
+      | c = Subtract $\card{B_1 \cap B_2}$ from both sides of equation
+}}
+{{eqn | o = \ge 
+      | r = \card{U \setminus V}
+      | c = [[Cardinality of Subset of Finite Set]]
+}}
+{{eqn | o = > 
+      | r = \card{V \setminus U}
+}}
+{{eqn | o = \ge 
+      | r = \card{V \setminus B_1}
+      | c = As $U \subseteq B_1$
+}}
+{{eqn | ll = \leadsto 
+      | l = \card{\paren{B_2 \setminus B_1} \setminus V} 
+      | o = >
+      | r = 0
+      | c = Subtract $\card{V \setminus B_1}$ from both sides of equation
+}}
+{{end-eqn}}
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Cardinality of Empty Set]]:
+:$\paren{B_2 \setminus B_1} \setminus V \ne \O$
+Hence:
+$\exists y \in \paren{B_2 \setminus B_1} \setminus V$
+From $(\text B 1)$:
+:$\exists z \in B_1 \setminus B_2 : \paren{B_2 \setminus \set y} \cup \set z \in \mathscr B$
+Let:
+:$B_3 = \paren{B_2 \setminus \set y} \cup \set z$
+We have:
+:{{begin-eqn}}
+{{eqn | l = V
+      | o = \subseteq
+      | r = B_2 \setminus \set y
+      | c = As $y \notin V$ and $V \subseteq B_2$
+}}
+{{eqn | o = \subseteq
+      | r = \paren{B_2 \setminus \set y} \cup \set z
+      | c = [[Set is Subset of Union]]
+}}
+{{eqn | r = B_3
+}}
+{{end-eqn}}
+=== [[Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 3|Lemma 3]] ===
+{{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 3}}{{qed|lemma}}
+Hence:
+:$\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$
+{{qed}}
+[[Category:Matroid Satisfies Base Axiom]]
+2j3iv9ails9r00l223vn9augij2zms8
+\end{proof}<|endoftext|>
+\section{Perpendicularity is Antitransitive Relation}
+Tags: Perpendiculars, Examples of Transitive Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \perp l_2$ denote that $l_1$ is [[Definition:Perpendicular Lines|perpendicular]] to $l_2$.
+Then $\perp$ is an [[Definition:Antitransitive Relation|antitransitive relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $l_1 \perp l_2$ and $l_2 \perp l_3$.
+Then $l_1$ and $l_3$ are [[Definition:Parallel Lines|parallel]], and not [[Definition:Perpendicular Lines|perpendicular]].
+Thus $\perp$ is seen to be [[Definition:Antitransitive Relation|antitransitive]].
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 2}
+Tags: Matroid Satisfies Base Axiom
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \card {B_1}
+      | r = \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}
+}}
+{{eqn | l = \card {B_2}
+      | r = \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \card {B_1}
+      | r = \card{ \paren{B_1 \cap B_2} \cup \paren{B_1 \setminus B_2} }
+      | c = [[Set Difference Union Intersection]]
+}}
+{{eqn | r = \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}
+      | c = [[Set Difference and Intersection are Disjoint]] and [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]]
+}}
+{{end-eqn}}
+and
+{{begin-eqn}}
+{{eqn | l = \card {B_2}
+      | r = \card{ \paren{B_2 \cap B_1} \cup \paren{B_2 \setminus B_1} }
+      | c = [[Set Difference Union Intersection]]
+}}
+{{eqn | r = \card{B_2 \cap B_1} + \card{B_2 \setminus B_1}
+      | c = [[Set Difference and Intersection are Disjoint]] and [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]]
+}}
+{{eqn | r = \card{B_2 \cap B_1} + \card{\paren{\paren{B_2 \setminus B_1} \cap V} \cup \paren{\paren{B_2 \setminus B_1} \setminus V} }
+      | c = [[Set Difference Union Intersection]]
+}}
+{{eqn | r = \card{B_2 \cap B_1} + \card{\paren{B_2 \setminus B_1} \cap V} + \card{\paren{B_2 \setminus B_1} \setminus V}
+      | c = [[Set Difference and Intersection are Disjoint]] and [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]]
+}}
+{{eqn | r = \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Matroid Satisfies Base Axiom]]
+ntzh0sbv53wpqpq4txhx6u57p808oq0
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 3}
+Tags: Matroid Satisfies Base Axiom
+
+\begin{theorem}
+:$\card{B_1 \cap B_3} = \card{B_1 \cap B_2} + 1$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = B_3 \cap B_1
+      | r = \paren{\paren{B_2 \setminus \set y} \cup \set z} \cap B_1
+}}
+{{eqn | r = \paren{\paren{B_2 \setminus \set y} \cap B_1 } \cup \paren{ \set z \cap B_1}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \paren{\paren{B_2 \setminus \set y} \cap B_1 } \cup \set z
+      | c = As $z \in B_1$
+}}
+{{eqn | r = \paren{\paren{B_2 \cap B_1} \setminus \paren{\set y \cap B_1} } \cup \set z
+      | c = [[Set Intersection Distributes over Set Difference]]
+}}
+{{eqn | r = \paren{\paren{B_2 \cap B_1} \setminus \O } \cup \set z
+      | c = As $y \notin B_1$
+}}
+{{eqn | r = \paren{B_2 \cap B_1} \cup \set z
+      | c = [[Set Difference with Empty Set is Self]]
+}}
+{{eqn | ll = \leadsto
+      | l = \card{B_3 \cap B_1}
+      | r = \card{\paren{B_2 \cap B_1} \cup \set z}
+}}
+{{eqn | r = \card{\paren{B_2 \cap B_1} } +  \card {\set z}
+      | c = As $z \notin B_2$
+}}
+{{eqn | r = \card{\paren{B_2 \cap B_1} } +  1
+      | c = [[Cardinality of Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Matroid Satisfies Base Axiom]]
+nvp65ssgy60e8jk5mdmozb5psrgp2hg
+\end{proof}<|endoftext|>
+\section{Equivalence of Versions of Axiom of Choice/Formulation 2 implies Formulation 1}
+Tags: Axiom of Choice
+
+\begin{theorem}
+The following formulation of the [[Axiom:Axiom of Choice|Axiom of Choice]]:
+\end{theorem}
+
+\begin{proof}
+Suppose that [[Axiom:Axiom of Choice/Formulation 2|Formulation 2]] holds.
+That is, the [[Definition:Cartesian Product|Cartesian product]] of a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family of Sets|family]] of [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]] is [[Definition:Non-Empty Set|non-empty]].
+Let $\CC$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set of non-empty sets]].
+$\CC$ may be converted into an [[Definition:Indexed Set|indexed set]] by using $\CC$ itself as the [[Definition:Indexing Set|indexing set]] and using the [[Definition:Identity Mapping|identity mapping]] on $\CC$ to do the [[Definition:Indexing Set|indexing]].
+Then the [[Definition:Cartesian Product|Cartesian product]] of all the [[Definition:Set|sets]] in $\CC$ has at least one [[Definition:Element|element]].
+An [[Definition:Element|element]] of such a [[Definition:Cartesian Product|Cartesian product]] is a [[Definition:Mapping|mapping]], that is a [[Definition:Indexed Family of Sets|family]] whose [[Definition:Domain of Mapping|domain]] is the [[Definition:Indexing Set|indexing set]] which in this context is $\CC$.
+The [[Definition:Value of Element under Mapping|value]] of this [[Definition:Mapping|mapping]] at each [[Definition:Index (Indexing Set)|index]] is an [[Definition:Element|element]] of the [[Definition:Set|set]] which bears that [[Definition:Index (Indexing Set)|index]].
+So we have a [[Definition:Mapping|mapping]] $f$ whose [[Definition:Domain of Mapping|domain]] is $\CC$ such that:
+:$A \in \CC \implies \map f A \in A$
+Now let $\CC$ be the [[Definition:Set|set]] of all [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subsets]] of some [[Definition:Set|set]] $X$.
+Then the assertion means that there exists a [[Definition:Mapping|mapping]] $f$ whose [[Definition:Domain of Mapping|domain]] is $\powerset X \setminus \set \O$ such that:
+:$A \in \powerset X \setminus \set \O \implies \map f A \in A$
+That is, [[Axiom:Axiom of Choice/Formulation 1|Formulation 1]] holds.
+\end{proof}<|endoftext|>
+\section{Equivalence of Versions of Axiom of Choice/Formulation 1 implies Formulation 3}
+Tags: Axiom of Choice
+
+\begin{theorem}
+The following formulation of the [[Axiom:Axiom of Choice|Axiom of Choice]]:
+\end{theorem}
+
+\begin{proof}
+Let $\SS$ be the set:
+:$\SS = \set {s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$
+Let $c$ be a [[Definition:Choice Function|choice function]] on $\SS$ and consider the [[Definition:Image of Mapping|image set]] $c \sqbrk \SS$:
+:$c \sqbrk \SS = \set {\map c s: \O \notin s \land \forall t, u \in s: t = u \lor t \cap u = \O}$
+By the definition of [[Definition:Choice Function|choice function]]:
+:$\map c s \in s$
+By construction of $\SS$, for any $s \in \SS$:
+:$s \cap c \sqbrk \SS = \set {\map c s}$
+\end{proof}<|endoftext|>
+\section{Equivalence of Versions of Axiom of Choice/Formulation 3 implies Formulation 1}
+Tags: Axiom of Choice
+
+\begin{theorem}
+The following formulation of the [[Axiom:Axiom of Choice|Axiom of Choice]]:
+\end{theorem}
+
+\begin{proof}
+Let $\BB$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Non-Empty Set|non-empty sets]] [[Definition:Indexing Set|indexed]] by $\II$.
+Consider [[Definition:Set|sets]] of the following form:
+:$\CC = \set {\tuple {B_i, x}: i \in \II, B_i \in \BB, x \in B_i}$
+That is, it is the [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]] of which the first [[Definition:Coordinate/Element of Ordered Pair|coordinate]] is a [[Definition:Set|set]] $B_i \in \BB$ and the second [[Definition:Coordinate/Element of Ordered Pair|coordinate]] is an [[Definition:Element|element]] of $B_i$.
+This is, by construction, a [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]]:
+:$\CC \subseteq \displaystyle \BB \times \bigcup_{i \mathop \in \II} B_i$
+[[Definition:By Hypothesis|By hypothesis]] all [[Definition:Set|sets]] $B$ are [[Definition:Non-Empty Set|non-empty]].
+Thus there exists at least one $\tuple {B_i, x}$ [[Definition:Element|element]] in $\CC$ for each $B_i \in \CC$.
+By [[Equality of Ordered Pairs]], if $B_j \ne B_k$ in $\BB$, then $\tuple {B_i, x} \ne \tuple {B_j, x}$ for all such [[Definition:Ordered Pair|pairs]] in $\CC$.
+So any two sets $\set {\tuple {B_j, x}: x \in B_j}$ and $\set {\tuple {B_k, x} :x \in B_k}$ are [[Definition:Disjoint Sets|disjoint]], by the inequality of their first [[Definition:Coordinate/Element of Ordered Pair|ccoordinates]].
+Then $\CC$ is an [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Non-Empty Set|non-empty sets]] in $\BB \times \bigcup_{i \mathop \in \II} B_i$.
+Hence $\CC$ satisfies the hypotheses of the [[Axiom:Axiom of Choice/Formulation 3|third formulation of the Axiom of Choice]].
+Let $c$ be a [[Definition:Set|set]] satisfying:
+:$\forall r \in \CC: c \cap r = \set t$
+All [[Definition:Element|elements]] of $c$ are [[Definition:Ordered Pair|ordered pairs]]:
+:the first [[Definition:Coordinate/Element of Ordered Pair|coordinate]] of which is a [[Definition:Set|set]] in $B_i \in \BB$
+and:
+:the second [[Definition:Coordinate/Element of Ordered Pair|coordinate]] of which is an [[Definition:Element|element]] $x \in B_i$.
+Viewed as a [[Definition:Relation|relation]], each [[Definition:Set|set]] in $\BB$ bears $c$ to an [[Definition:Element|element]] in that [[Definition:Set|set]].
+Every [[Definition:Set|set]] in $\BB$ bears $c$ to exactly one [[Definition:Element|element]] inside that [[Definition:Set|set]].
+So $c$ is in fact a [[Definition:Mapping|mapping]] and satisfies the criteria of a [[Definition:Choice Function|choice function]] as expounded in [[Axiom:Axiom of Choice/Formulation 1|formulation $(1)$]].
+\end{proof}<|endoftext|>
+\section{C^k Norm is Norm}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $I = \closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\struct {\map {\CC^k} I, +, \, \cdot \,}_\R$ be the [[Space of Real-Valued k-times Differentiable on Closed Interval Functions with Pointwise Addition and Pointwise Scalar Multiplication forms Vector Space|vector space of real-valued functions, k-times differentiable on]] $I$.
+Let $x \in \map {\CC^k} I$ be a [[Definition:Real-Valued Function|real-valued function]] of [[Definition:Differentiability Class|differentiability class]] $k$.
+Let $\norm {\, \cdot \,}_{\map {C^k} I}$ be the [[Definition:C^k Norm|$C^k$ norm]] on $I$.
+Then $\norm {\, \cdot \,}_{\map {C^k} I}$ is a [[Definition:Norm on Vector Space|norm]] on $\struct {\map {\CC^k} I, +, \, \cdot \,}_\R$.
+\end{theorem}
+
+\begin{proof}
+=== Positive definiteness ===
+Let $x \in \map {\CC^k} I$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm x_{\map {C^k} I}
+      | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty
+}}
+{{eqn | o = \ge
+      | r = \sum_{i \mathop = 0}^k 0
+      | c = [[Supremum Norm is Norm]], [[Definition:Norm Axioms (Vector Space)|Norm Axiom $\paren {N1}:$ Positive Definiteness]]
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Suppose $\norm x_{\map {C^k} I} = 0$.
+We have that the [[Sum of Nonnegative Real Numbers is Zero iff Every Element is Zero|sum of non-negatives is zero if every element is zero]].
+Hence:
+:$\forall i \in \N : 0 \le i \le k : \norm {x^{\paren i}}_\infty = 0$
+Namely, $\norm x_\infty = 0$.
+By [[Supremum Norm is Norm]] and [[Definition:Norm Axioms (Vector Space)|Norm Axiom $\paren {N1}:$ Positive Definiteness]]:
+:$\forall t \in I : \map x t = 0$.
+=== Positive homogeneity ===
+Let $x \in \map {\CC^k} I$, $\alpha \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha x}_{\map {C^k} I}
+      | r = \sum_{i \mathop = 0}^k \norm {\paren {\alpha x}^{\paren i} }_\infty
+      | c = {{defof|C^k Norm}}
+}}
+{{eqn | r = \sum_{i \mathop = 0}^k \norm {\alpha x^{\paren i} }_\infty
+      | c = {{defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \size \alpha \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty
+      | c = [[Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function|Supremum norm on continuous real-valued functions is a norm: positive homogeneity]]
+}}
+{{eqn | r = \size \alpha \norm {x}_{\map {C^k} I}
+      | c = {{defof|C^k Norm}}
+}}
+{{end-eqn}}
+=== Triangle inequality ===
+Let $x, y \in \map {\CC^k} I$
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}_{\map {C^k} I}
+      | r = \sum_{i \mathop = 0}^k \norm {\paren {x + y}^{\paren i} }_\infty
+      | c = {{defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} + y^{\paren i} }_\infty
+      | c = {{defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 0}^k \norm {x^{\paren i} }_\infty + \sum_{i \mathop = 0}^k \norm {y^{\paren i} }_\infty
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \norm x_{\map {C^k} I} + \norm y_{\map {C^k} I}
+      | c = {{defof|C^k Norm}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Minimal Polynomial for Square Matrix over Field}
+Tags: Linear Algebra, Minimal Polynomials
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $n$ be a [[Definition:Natural Number|natural number]].
+Let $K^{n \times n}$ be the set of $n \times n$ [[Definition:Matrix|matrices]] over $K$. 
+Let $A \in K^{n \times n}$.
+Then the [[Definition:Minimal Polynomial|minimal polynomial]] of $A$ exists and has [[Definition:Degree of Polynomial|degree]] at most $n^2$.
+\end{theorem}
+
+\begin{proof}
+By [[Matrices over Field form Vector Space]]:
+:$K^{n \times n}$ forms a [[Definition:Vector Space|vector space]] under usual [[Definition:Matrix Addition|matrix addition]] and scalar multiplication.
+By [[Dimension of Vector Space of Matrices]]:
+:$K^{n \times n}$ has [[Definition:Dimension (Linear Algebra)|dimension]] $n^2$. 
+Consider the collection of vectors: 
+:$I, A, A^2, \ldots, A^{n^2}$
+Since this is a collection of $n^2 + 1$ vectors, and $K^{n \times n}$ has [[Definition:Dimension (Linear Algebra)|dimension]] $n^2$, we have by [[Size of Linearly Independent Subset is at Most Size of Finite Generator]]:
+:$I, A, A^2, \ldots, A^{n^2}$ are [[Definition:Linearly Dependent|linearly dependent]].
+That is, there exists $\alpha_0, \alpha_1, \ldots, \alpha_{n^2} \in K$ not all zero such that: 
+:$\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i A^i = 0$
+That is, the [[Definition:Polynomial|polynomial]]: 
+:$\displaystyle \sum_{i \mathop = 0}^{n^2} \alpha_i X^i \in K \sqbrk X$
+has $\map P A = 0$, and [[Definition:Degree of Polynomial|degree]] at most $n^2$. 
+Let: 
+:$\displaystyle S = \set {P \in K \sqbrk X \setminus \set 0 \mid \map P A = 0}$ 
+$S$ is certainly [[Definition:Non-Empty Set|non-empty]] since we have found such an element in the computation above. 
+Now consider the set: 
+:$\displaystyle D = \set {\deg P \mid P \in S}$
+Since $D$ is a [[Definition:Subset|subset]] of the [[Definition:Natural Number|natural numbers]], it contains a [[Definition:Smallest/Ordered Set|least element]] $N$ by the [[Well-Ordering Principle]].
+Since the polynomial we constructed has [[Definition:Degree of Polynomial|degree]] at most $n^2$, we have $N \le n^2$.
+Let $Q \in S$ be of [[Definition:Degree of Polynomial|degree]] $N$.
+Let $a_N$ be the coefficient of the $X^N$ term in $Q$. 
+Then $\mu = \dfrac 1 {a_N} Q$ is a [[Definition:Monic Polynomial|monic polynomial]] of minimum [[Definition:Degree of Polynomial|degree]] with $\map \mu A = 0$. 
+So $\mu$ is a [[Definition:Minimal Polynomial|minimal polynomial]] for $A$. 
+{{qed}}
+[[Category:Linear Algebra]]
+[[Category:Minimal Polynomials]]
+tb8lxiwzhe3qho6xa228up6rkyunipb
+\end{proof}<|endoftext|>
+\section{Homeomorphic Topologies on Same Set may not be Identical}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\tau_1$ and $\tau_2$ both be [[Definition:Topology|topologies]] on $S$ such that the [[Definition:Topological Space|topological spaces]] $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ are [[Definition:Homeomorphic Topological Spaces|homeomorphic]].
+Then it is not necessarily the case that $\struct {S, \tau_1} = \struct {S, \tau_1}$.
+\end{theorem}
+
+\begin{proof}
+Let $p, q \in S$ such that $p \ne q$.
+Let $\tau_p$ and $\tau_q$ be the [[Definition:Particular Point Topology|particular point topologies]] on $S$ by $p$ and $q$ respectively.
+From [[Homeomorphic Non-Comparable Particular Point Topologies]], $\struct {S, \tau_p}$ and $\struct {S, \tau_q}$ are [[Definition:Homeomorphic Topological Spaces|homeomorphic]].
+But $\struct {S, \tau_p} \ne \struct {S, \tau_q}$, as $\set p \in \tau_p$ but $\set p \notin \tau_q$, and similarly $\set q \in \tau_q$ but $\set q \notin \tau_p$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Topologies on Set with More than One Element may not be Homeomorphic}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which contains at least $2$ [[Definition:Element|elements]].
+Let $\tau_1$ and $\tau_2$ be [[Definition:Topology|topologies]] on $S$.
+Then it is not necessarily the case that $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ are [[Definition:Homeomorphic Topological Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+Let $\tau_1$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on $S$.
+Let $\tau_2$ be the [[Definition:Discrete Topology|discrete topology]] on $S$.
+Then $\struct {S, \tau_1}$ has $2$ [[Definition:Element|elements]]: $S$ and $\O$.
+Let $a, b \in S$ such that $a \ne b$.
+Then $\set a \in \tau_2$ and $\set b \in \tau 2$, as well as $S$ and $\O$.
+So there cannot be a [[Definition:Bijection|bijection]] between $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Clopen Sets in Indiscrete Topology}
+Tags: Indiscrete Topology, Clopen Sets
+
+\begin{theorem}
+The only [[Definition:Subset|subsets]] of $S$ which are both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $T$ are $S$ and $\O$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Indiscrete Topology|indiscrete topological space]], the only [[Definition:Open Set (Topology)|open sets]] in $\struct {S, \tau}$ are $S$ and $\O$.
+From [[Open and Closed Sets in Topological Space]], both $S$ and $\O$ are both [[Definition:Closed Set (Topology)|closed]] and [[Definition:Open Set (Topology)|open]] in $\struct {S, \tau}$ are $S$ and $\O$. 
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Connected Topological Space/No Separation iff No Clopen Sets}
+Tags: 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}
+=== Definition by No Clopen Sets implies Definition by Separation ===
+Let $T$ be [[Definition:Connected (Topology)/Topological Space/Definition 4|connected by having no clopen sets]].
+{{AimForCont}} $T$ admits a [[Definition:Separation (Topology)|separation]], $A \mid B$ say.
+Then both $A$ and $B$ are [[Definition:Clopen Set|clopen sets]] of $T$, neither of which is either $S$ or $\O$.
+From this [[Definition:Contradiction|contradiction]] it follows that $T$ can admit no [[Definition:Separation (Topology)|separation]]
+{{qed|lemma}}
+=== Definition by Separation implies Definition by No Clopen Sets ===
+Let $T$ be [[Definition:Connected (Topology)/Topological Space/Definition 4|connected by admitting no separation]].
+Suppose $\exists H \subseteq S$ which is [[Definition:Clopen Set|clopen]].
+Then $\relcomp S H$ is also [[Definition:Clopen Set|clopen]].
+Hence $H \mid \relcomp S H$ is a [[Definition:Separation (Topology)|separation]] of $T$.
+From this [[Definition:Contradiction|contradiction]] it follows that $T$ can have no [[Definition:Proper Subset|proper subsets]] which are [[Definition:Clopen Set|clopen]].
+\end{proof}<|endoftext|>
+\section{Dependent Subset of Independent Set Union Singleton Contains Singleton}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $X$ be an [[Definition:Independent Subset (Matroid)|independent subset]] of $M$.
+Let $x \in S$.
+Let $C$ be a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $M$ such that:
+:$C \subseteq X \cup \set x$.
+Then:
+:$x \in C$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Superset of Dependent Set is Dependent]]:
+:$C \not \subseteq X$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Set Difference with Superset is Empty Set]]:
+:$C \setminus X \ne \O$
+From [[Set Difference over Subset]]:
+:$C \setminus X \subseteq \paren{X \cup \set x} \setminus X = \set x$
+From [[Power Set of Singleton]]:
+:$C \setminus X = \set x$
+From [[Set Difference is Subset]]:
+:$\set x \subseteq C$
+From [[Singleton of Element is Subset]]:
+:$x \in C$
+{{qed}}
+[[Category:Matroid Theory]]
+h3wje0fo4f9qmhw0uc6wwo1cj132rz2
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Independent Superset of Dependent Set Minus Singleton Doesn't Contain Singleton}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $C$ be a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $M$.
+Let $x \in S$.
+Let $X$ be an [[Definition:Independent Subset (Matroid)|independent subset]] of $M$ such that:
+:$C \setminus \set x \subseteq X$.
+Then:
+:$x \notin X$
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Contrapositive Statement|contrapositive statement]]:
+:$x \in X \implies X$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]].
+Let $x \in X$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren{C \setminus \set x} \cup \set x
+      | o = \subseteq
+      | r =  X
+      | c = [[Union of Subsets is Subset]]
+}}
+{{eqn | ll = \leadsto
+      | l = C \cup \set x
+      | o = \subseteq
+      | r =  X
+      | c = [[Set Difference Union Second Set is Union]]
+}}
+{{eqn | ll = \leadsto
+      | l = C
+      | o = \subseteq
+      | r =  X
+      | c = [[Set is Subset of Union]]
+}}
+{{end-eqn}}
+From [[Superset of Dependent Set is Dependent]]:
+:$X$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]]
+{{qed}}
+[[Category:Matroid Theory]]
+ndartfsndj1oorn109xfk0oppyk3883
+\end{proof}<|endoftext|>
+\section{Matroid Unique Circuit Property/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $B$ be a [[Definition:Base of Matroid|base]] of $M$.
+Let $x \in S \setminus B$.
+Then there exists a unique [[Definition:Circuit|circuit]] $C$ such that:
+:$x \in C \subseteq B \cup \set x$
+That is, $C$ is the [[Definition:Fundamental Circuit (Matroid)|fundamental circuit]] of $x$ in $B$.
+\end{theorem}
+
+\begin{proof}
+From [[Union of Matroid Base with Element of Complement is Dependent]]:
+:$B \cup \set x$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+From [[Matroid Unique Circuit Property]] there exists a unique [[Definition:Circuit|circuit]] $C$ such that:
+:$x \in C \subseteq B \cup \set x$
+{{qed}}
+[[Category:Matroid Theory]]
+eh68f7pq174bzyyyftuozku64gonq7v
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Base Substitution From Fundamental Circuit}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $B$ be a [[Definition:Base of Matroid|base]] of $M$.
+Let $y \in S \setminus B$.
+Let $\map C {y,B}$ denote the [[Definition:Fundamental Circuit (Matroid)|fundamental circuit]] of $y$ in $B$.
+Let $x \in B$.
+Then:
+:$\paren{B \setminus \set x} \cup \set y$ is a [[Definition:Base of Matroid|base]] of $M$ {{iff}} $x \in \map C {y,B}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\paren{B \setminus \set x} \cup \set y$ be a [[Definition:Base of Matroid|base]] of $M$.
+By definition of the [[Definition:Fundamental Circuit (Matroid)|fundamental circuit]]:
+:$\map C {y,B} \subseteq B \cup \set y$
+and
+:$\map C {y,B}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+{{AimForCont}} $x \notin \map C {y,B}$.
+Then:
+:$\map C {y,B} \subseteq \paren{B \setminus \set x} \cup \set y$
+By [[Definition:Matroid Axioms|matroid axiom $( \text I 2)$]]:
+:$\map C {y,B}$ is [[Definition:Independent Subset (Matroid)|independent]]
+This [[Definition:Contradiction|contradicts]] the fact that:
+:$\map C {y,B}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+It follows that:
+:$x \in \map C {y,B}$ 
+{{qed|lemma}}
+=== Sufficient Condition ===
+We prove the [[Definition:Contrapositive Statement|contrapositive statement]].
+Let $\paren{B \setminus \set x} \cup \set y$ not be a [[Definition:Base of Matroid|base]] of $M$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \card{\paren{B \setminus \set x} \cup \set y}
+      | r = \card{B \setminus \set x} + \card{\set y}
+      | c = [[Cardinality of Set Union/Corollary|Corollary of Cardinality of Set Union]]
+}}
+{{eqn | r = \paren{\card B - \card{\set x} } + \card{\set y}
+      | c = [[Cardinality of Set Difference]]
+}}
+{{eqn | r = \paren{\card B - 1 } + 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | r = \card B
+}}
+{{end-eqn}}
+From [[Definition:Contrapositive Statement|contrapositive statement]] of [[Independent Subset is Base if Cardinality Equals Cardinality of Base]]:
+:$\paren{B \setminus \set x} \cup \set y$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $M$
+From [[Dependent Subset Contains a Circuit]] there exists a [[Definition:Circuit (Matroid)|circuit]] $C'$:
+:$C' \subseteq \paren{B \setminus \set x} \cup \set y \subseteq B \cup \set y$
+From [[Matroid Base Union External Element has Fundamental Circuit]]:
+:$C' = \map C {y,B}$
+Hence:
+:$x \notin \map C {y, B}$
+From [[Rule of Transposition]] we have:
+:$x \in \map C {y, B} \implies \paren{B \setminus \set x} \cup \set y$ is a [[Definition:Base of Matroid|base]] of $M$
+{{qed}}
+[[Category:Matroid Theory]]
+nhqdzwrkojorfkazedzjs5tvqlktjlg
+\end{proof}<|endoftext|>
+\section{Socrates is Mortal/Variant}
+Tags: Classic Problems, Propositional Logic
+
+\begin{theorem}
+:$(1): \quad$ ''If {{AuthorRef|Socrates}} is a man then {{AuthorRef|Socrates}} is mortal.''
+:$(2): \quad$ ''{{AuthorRef|Socrates}} is a man.''
+:$(3): \quad$ ''Therefore {{AuthorRef|Socrates}} is mortal.''
+\end{theorem}
+
+\begin{proof}
+Let $P$ denote the [[Definition:Simple Statement|simple statement]] ''{{AuthorRef|Socrates}} is a man.''.
+Let $Q$ denote the [[Definition:Simple Statement|simple statement]] ''{{AuthorRef|Socrates}} is mortal.''.
+The argument can then be expressed as:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = P
+      | o = \implies
+      | r = Q
+      | c = 
+}}
+{{eqn | n = 2
+      | l = P
+      | o = 
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \therefore
+      | l = Q
+      | o = 
+      | c = [[Modus Ponendo Ponens]]
+}}
+{{end-eqn}}
+That is: 
+:''{{AuthorRef|Socrates}} is mortal.''
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Forward Implication/Formulation 2}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+:$\paren {p \implies \paren {q \implies r} } \implies \paren {\paren {p \implies q} \implies \paren {p \implies r} }$
+\end{theorem}<|endoftext|>
+\section{Self-Distributive Law for Conditional/Reverse Implication/Formulation 2}
+Tags: Self-Distributive Law for Conditional
+
+\begin{theorem}
+:$\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$
+\end{theorem}<|endoftext|>
+\section{Odd Integers under Addition do not form Group}
+Tags: Odd Integers, Integer Addition
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Odd Integer|odd integers]]:
+:$S = \set {x \in \Z: \exists n \in \Z: x = 2 n + 1}$
+Let $\struct {S, +}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by $S$ under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Addition|addition]].
+Then $\struct {S, +}$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+It is to be demonstrated that $\struct {S, +}$ does not satisfy the [[Definition:Group Axioms|group axioms]].
+Let $a$ and $b$ be [[Definition:Odd Integer|odd integers]].
+Then  $a = 2 n + 1$ and $b = 2 m + 1$ for some $m, n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a + b
+      | r = 2 n + 1 + 2 m + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {n + m + 1}
+      | c = 
+}}
+{{end-eqn}}
+and it is seen that $a + b$ is [[Definition:Even Integer|even]].
+That is:
+:$a + b \notin S$
+Thus $\struct {S, +}$ does not fulfil {{GroupAxiom|0}}.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Digits in Number}
+Tags: Number Theory
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $b \in \Z_{>1}$ be an [[Definition:Integer|integer]] greater than $1$.
+Let $n$ be expressed in [[Definition:Number Base|base]] $b$.
+Then the number of [[Definition:Digit|digits]] $d$ in this expression for $n$ is:
+:$d = 1 + \floor {\log_b n}$
+where:
+:$\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]]
+:$\log_b$ denotes the [[Definition:General Logarithm|logarithm to base $b$]].
+\end{theorem}
+
+\begin{proof}
+Let $n$ have $d$ [[Definition:Digit|digits]] when expressed in [[Definition:Number Base|base]] $b$.
+Then $n$ is expressed as:
+:$n = \sqbrk {n_{d - 1} n_{d - 2} \dotsm d_1 d_0}$
+where:
+:$n = \displaystyle \sum_{k \mathop = 0}^{d - 1} n_k b^k$
+Thus:
+:$b^{d - 1} \le n < b^d$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = b^{d - 1}
+      | o = \le
+      | m = n
+      | mo= <
+      | r = b^d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = d - 1
+      | o = \le
+      | m = \log_b n
+      | mo= <
+      | r = d
+      | c = 
+}}
+{{end-eqn}}
+By [[Integer equals Floor iff Number between Integer and One More]]:
+:$d - 1 \le \log_b n < d \iff \floor {\log_b n} = d - 1$
+and the result follows.
+{{Qed}}
+[[Category:Number Theory]]
+l1hru3y24mg0nhjqmv5bmly8y48rz1o
+\end{proof}<|endoftext|>
+\section{Every Tenth Power of Two Minus Every Third Power of Ten is Divisible By Three}
+Tags: 
+
+\begin{theorem}
+Let $x \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
+Then $2^{10 x} - 10^{3 x}$ is [[Definition:Divisor of Integer|divisible]] by $3$.
+That is:
+:$2^{10 x} - 10^{3 x} \equiv 0 \pmod 3$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 2^{10 x}
+      | r = \paren {2^{10} }^x
+      | c = [[Exponent Combination Laws/Power of Power|Power of Power]]
+}}
+{{eqn | r = 1024^x
+      | c = as $2^{10} = 1024$
+}}
+{{eqn | r = \paren {1000 + 24}^x
+      | c = rewriting $1024$ as the sum of a power of $10$ and some integer
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n 1000^{x - k} \, 24^k
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | n = 1
+      | r = 1000^x + \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^k
+      | c = extracting first term from [[Definition:Summation|summation]]
+}}
+{{eqn | r = 1000^x + 24 \paren {\sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1} }
+      | c = extracting $24$ as a [[Definition:Divisor of Integer|divisor]]
+}}
+{{eqn | ll= \leadsto
+      | l = 2^{10 x} - 1000^x
+      | r = 24 i
+      | c = setting $i = \displaystyle \sum_{k \mathop = 1}^n 1000^{x - k} \, 24^{k - 1}$
+}}
+{{eqn | ll= \leadsto
+      | l = 2^{10 x} - 10^{3 x}
+      | r = 24 i
+      | c = rewriting $1000^x$ as a power of $10$
+}}
+{{eqn | r = 3 k
+      | c = setting $k = 8 i$
+}}
+{{eqn | ll= \leadsto
+      | l = 2^{10x} - 10^{3 x}
+      | o = \equiv
+      | r = 0 \pmod 3
+      | c = {{Defof|Congruence Modulo Integer}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space of Almost-Zero Sequences is Everywhere Dense in 2-Sequence Space}
+Tags: Normed Vector Spaces, Denseness
+
+\begin{theorem}
+Let $\struct {\ell^2, \norm {\, \cdot \,}_2}$ be the [[Definition:P-Sequence Space|2-sequence space]] equipped with [[Definition:Euclidean Norm|Euclidean norm]].
+Let $c_{00}$ be the [[Definition:Space of Almost-Zero Sequences|space of almost-zero sequences]].
+Then $c_{00}$ is [[Definition:Everywhere Dense/Normed Vector Space|everywhere dense]] in $\struct {\ell^2, \norm {\, \cdot \,}_2}$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N} \in \ell^2$.
+By [[Definition:P-Sequence Space|definition]] of $\ell^2$:
+:$\displaystyle \sum_{i \mathop = 0}^\infty \size {x_i}^2 < \infty$
+Let $\displaystyle s_n := \sum_{i \mathop = 0}^n \size {x_i}^2$ be a [[Definition:Sequence|sequence]] of [[Definition:Partial Sum|partial sums]] of $\displaystyle s = \sum_{i \mathop = 0}^\infty \size {x_i}^2$.
+We have that $s$ is a [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]]:
+:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {s_n - s} < \epsilon$
+Note that:
+{{begin-eqn}}
+{{eqn | l = \size {s_n - s}
+      | r = \size {\sum_{i \mathop = 0}^n \size {x_i}^2 - \sum_{i \mathop = 0}^\infty \size {x_i}^2}
+}}
+{{eqn | r = \size {\sum_{i \mathop = n \mathop + 1}^\infty \size {x_i}^2}
+}}
+{{end-eqn}}
+Let $\displaystyle N \in \N : \sum_{n \mathop = N + 1}^\infty \size {x_n}^2 < \epsilon^2$
+Let $\mathbf y := \tuple {x_1 \ldots, x_N, 0, \ldots}$.
+By [[Definition:Space of Almost-Zero Sequences|definition]], $\mathbf y \in c_{00}$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x - \mathbf y}_2^2
+      | r = \sum_{i \mathop = 0}^\infty \size {x_i - y_i}^2
+      | c = {{defof|Euclidean Norm}}
+}}
+{{eqn | r = \sum_{i \mathop = N + 1}^\infty \size {x_i}^2
+}}
+{{eqn | o = <
+      | r = \epsilon^2
+}}
+{{eqn | ll = \leadsto
+      | l = \norm {\mathbf x - \mathbf y}_2
+      | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+By [[Definition:Definition|definition]], $c_{00}$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] in $\ell^2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gödel's Incompleteness Theorems/Second}
+Tags: Gödel's Incompleteness Theorems
+
+\begin{theorem}
+Let $T$ be the [[Definition:Set|set]] of [[Definition:Theorem of Logic|theorems]] of some [[Definition:Recursive Set|recursive set]] of [[Definition:Sentence|sentences]] in the [[Definition:Language of Arithmetic|language of arithmetic]] such that $T$ contains [[Definition:Minimal Arithmetic|minimal arithmetic]].
+Let $\map {\mathrm {Cons} } T$ be the [[Definition:Propositional Function|propositional function]] which states that $T$ is [[Definition:Consistent (Logic)|consistent]].
+Then it is not possible to prove $\map {\mathrm {Cons} } T$ by means of formal statements within $T$ itself.
+\end{theorem}<|endoftext|>
+\section{Linear Function on Stationary Stochastic Model is Stationary}
+Tags: Stationary Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Strictly Stationary Stochastic Process|strictly stationary stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $\sequence {s_n}$ be a [[Definition:Sequence|sequence]] of $n$ [[Definition:Successive Values of Equispaced Time Series|successive values]] of $T$:
+:$\sequence {s_n} = \tuple {z_1, z_2, \dotsb, z_n}$
+Let $L_t$ be a [[Definition:Linear Function|linear function]] of $\sequence {s_n}$:
+:$L_t = l_1 z_t + l_2 z_{t - 1} + \dotsb + l_n z_{t - n + 1}$
+Then $L_t$ is itself [[Definition:Strictly Stationary Stochastic Process|stationary]].
+\end{theorem}
+
+\begin{proof}
+Follows by definition of [[Definition:Strictly Stationary Stochastic Process|stationarity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering of Integers is Reversed by Negation}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $x, y \in \Z$ such that $x > y$.
+Then:
+:$-x < -y$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Integer/Formal Definition|formal definition of integers]], $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$.
+We have:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = >
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = >
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = b + c
+      | o = <
+      | r = a + d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {b, a} {}
+      | o = <
+      | r = \eqclass {d, c} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\eqclass {a, b} {}
+      | o = <
+      | r = -\eqclass {c, d} {}
+      | c = [[Negative of Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = -x
+      | o = <
+      | r = -y
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Ordering on Integers is Well-Defined}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $\eqclass {a, b} {}$ denote an [[Definition:Integer|integer]], as defined by the [[Definition:Integer/Formal Definition|formal definition of integers]].
+Let:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | r = \eqclass {a', b'} {}
+      | c = 
+}}
+{{eqn | l = \eqclass {c, d} {}
+      | r = \eqclass {c', d'} {}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = \eqclass {a', b'} {}
+      | o = <
+      | r = \eqclass {c', d'} {}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+This is a direct application of the [[Extension Theorem for Total Orderings]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Gaussian Process}
+Tags: Definitions: Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let the [[Definition:Probability Distribution|probability distribution]] of $T$ be a [[Definition:Multivariate Distribution|multivariate]] [[Definition:Normal Distribution|normal distribution]].
+Then $S$ is called a '''Gaussian process'''.
+\end{theorem}<|endoftext|>
+\section{Characterization of Stationary Gaussian Process}
+Tags: Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Gaussian Process|Gaussian]] [[Definition:Stochastic Process|stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let the the [[Definition:Mean of Stochastic Process|mean]] of $S$ be fixed.
+Let the [[Definition:Autocovariance Matrix|autocovariance matrix]] of $S$ also be fixed.
+Then $S$ is [[Definition:Stationary Stochastic Process|stationary]].
+\end{theorem}
+
+\begin{proof}
+From [[Characterization of Multivariate Gaussian Distribution]], the [[Definition:Gaussian Distribution|Gaussian distribution]] is completely characterized by its [[Definition:Expectation of Continuous Random Variable|expectation]] and its [[Definition:Variance of Continuous Random Variable|variance]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sufficient Conditions for Weak Stationarity of Order 2}
+Tags: Stationary Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let the [[Definition:Mean of Stochastic Process|mean]] of $S$ be fixed.
+Let the [[Definition:Autocovariance Matrix|autocovariance matrix]] of $S$ be of the form:
+:$\boldsymbol \Gamma_n = \begin {pmatrix}
+\gamma_0 & \gamma_1 & \gamma_2 & \cdots & \gamma_{n - 1} \\
+\gamma_1 & \gamma_0 & \gamma_1 & \cdots & \gamma_{n - 2} \\
+\gamma_2 & \gamma_1 & \gamma_0 & \cdots & \gamma_{n - 3} \\
+\vdots   & \vdots   & \vdots   & \ddots & \vdots \\
+\gamma_{n - 1} & \gamma_{n - 2} & \gamma_{n - 3} & \cdots & \gamma_0
+\end {pmatrix} = \sigma_z^2 \mathbf P_n = \begin {pmatrix}
+1 & \rho_1 & \rho_2 & \cdots & \rho_{n - 1} \\
+\rho_1 & 1 & \rho_1 & \cdots & \rho_{n - 2} \\
+\rho_2 & \rho_1 & 1 & \cdots & \rho_{n - 3} \\
+\vdots   & \vdots   & \vdots   & \ddots & \vdots \\
+\rho_{n - 1} & \rho_{n - 2} & \rho_{n - 3} & \cdots & 1
+\end {pmatrix}$
+Then $S$ is [[Definition:Weakly Stationary Stochastic Process|weakly stationary of order $2$]].
+\end{theorem}
+
+\begin{proof}
+Follows from the definition of [[Definition:Weakly Stationary Stochastic Process|weak stationarity]].
+\end{proof}<|endoftext|>
+\section{Second Order Weakly Stationary Gaussian Stochastic Process is Strictly Stationary}
+Tags: Stationary Stochastic Processes
+
+\begin{theorem}
+Let $S$ be a [[Definition:Gaussian Process|Gaussian]] [[Definition:Stochastic Process|stochastic process]] giving rise to a [[Definition:Time Series|time series]] $T$.
+Let $S$ be [[Definition:Weakly Stationary Stochastic Process|weakly stationary of order $2$]].
+Then $S$ is [[Definition:Strictly Stationary Stochastic Process|strictly stationary of order $2$]].
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Gaussian Process|Gaussian process]], the [[Definition:Probability Distribution|probability distribution]] of $T$ be a [[Definition:Multivariate Distribution|multivariate]] [[Definition:Gaussian Distribution|Gaussian distribution]].
+By definition, a [[Definition:Gaussian Distribution|Gaussian distribution]] is characterized completely by its [[Definition:Expectation of Discrete Random Variable|expectation]] and its [[Definition:Variance of Discrete Random Variable|variance]].
+That is, its $1$st and $2$nd [[Definition:Moment of Discrete Random Variable|moments]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering on Integers is Transitive}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $\eqclass {a, b} {}$ denote an [[Definition:Integer|integer]], as defined by the [[Definition:Integer/Formal Definition|formal definition of integers]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = \le
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \eqclass {c, d} {}
+      | o = \le
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \eqclass {a, b} {}
+      | o = \le
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{end-eqn}}
+That is, [[Definition:Ordering on Integers|ordering]] on the [[Definition:Integer|integers]] is [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Integer/Formal Definition|formal definition of integers]], we have that $a, b, c, d, e, f$ are all [[Definition:Natural Number|natural numbers]].
+To eliminate confusion between [[Definition:Ordering on Integers|integer ordering]] and the [[Definition:Ordering on Natural Numbers|ordering on the natural numbers]], let $a \preccurlyeq b$ denote that the [[Definition:Natural Number|natural number]] $a$ is less than or equal to the [[Definition:Natural Number|natural number]] $b$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = \le
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = a + d
+      | o = \preccurlyeq
+      | r = b + c
+      | c = {{Defof|Ordering on Integers}}
+}}
+{{eqn | l = \eqclass {c, d} {}
+      | o = <
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = c + f
+      | o = \preccurlyeq
+      | r = d + e
+      | c = {{Defof|Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d + f
+      | o = \preccurlyeq
+      | r = b + c + f
+      | c = adding $f$ to both sides of $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = a + d + f
+      | o = \preccurlyeq
+      | r = b + d + e
+      | c = from $(2)$: $b + \paren {c + f} \preccurlyeq b + \paren {d + e}$
+}}
+{{eqn | ll= \leadsto
+      | l = a + f
+      | o = \preccurlyeq
+      | r = b + e
+      | c = subtracting $d$ from both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = \le
+      | r = \eqclass {e, f} {}
+      | c = {{Defof|Ordering on Integers}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Orderings on Integers]]
+6ks88hv97gg4h02vn1o4gocu699l1ci
+\end{proof}<|endoftext|>
+\section{Strict Ordering on Integers is Asymmetric}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $\eqclass {a, b} {}$ denote an [[Definition:Integer|integer]], as defined by the [[Definition:Integer/Formal Definition|formal definition of integers]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | lo= \implies
+      | l = \eqclass {c, d} {}
+      | o = \not <
+      | r = \eqclass {a, b} {}
+      | c = 
+}}
+{{end-eqn}}
+That is, [[Definition:Strict Ordering on Integers|strict ordering]] on the [[Definition:Integer|integers]] is [[Definition:Asymmetric Relation|asymmetric]].
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Integer/Formal Definition|formal definition of integers]], we have that $a, b, c, d, e, f$ are all [[Definition:Natural Number|natural numbers]].
+To eliminate confusion between [[Definition:Strict Ordering on Integers|integer ordering]] and the [[Definition:Ordering on Natural Numbers|ordering on the natural numbers]], let $a \prec b$ denote that the [[Definition:Natural Number|natural number]] $a$ is less than the [[Definition:Natural Number|natural number]] $b$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \prec
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = b + c
+      | o = \not \prec
+      | r = a + d
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {c, d} {}
+      | o = \not <
+      | r = \eqclass {a, b} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Ordering on Integers is Transitive}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $\eqclass {a, b} {}$ denote an [[Definition:Integer|integer]], as defined by the [[Definition:Integer/Formal Definition|formal definition of integers]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = \eqclass {c, d} {}
+      | o = <
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{end-eqn}}
+That is, [[Definition:Strict Ordering on Integers|strict ordering]] on the [[Definition:Integer|integers]] is [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Integer/Formal Definition|formal definition of integers]], we have that $a, b, c, d, e, f$ are all [[Definition:Natural Number|natural numbers]].
+To eliminate confusion between [[Definition:Strict Ordering on Integers|integer ordering]] and the [[Definition:Ordering on Natural Numbers|ordering on the natural numbers]], let $a \prec b$ denote that the [[Definition:Natural Number|natural number]] $a$ is less than the [[Definition:Natural Number|natural number]] $b$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = a + d
+      | o = \prec
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | l = \eqclass {c, d} {}
+      | o = <
+      | r = \eqclass {e, f} {}
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = c + f
+      | o = \prec
+      | r = d + e
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d + f
+      | o = \prec
+      | r = b + c + f
+      | c = adding $f$ to both sides of $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = a + d + f
+      | o = \prec
+      | r = b + d + e
+      | c = from $(2)$: $b + \paren {c + f} \prec b + \paren {d + e}$
+}}
+{{eqn | ll= \leadsto
+      | l = a + f
+      | o = \prec
+      | r = b + e
+      | c = subtracting $d$ from both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {e, f} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Ordering on Integers is Trichotomy}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $\eqclass {a, b} {}$ and $\eqclass {c, d} {}$ be [[Definition:Integer|integers]], as defined by the [[Definition:Integer/Formal Definition|formal definition of integers]].
+Then exactly one of the following holds:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = =
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = >
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{end-eqn}}
+That is, [[Definition:Strict Ordering on Integers|strict ordering]] is a [[Definition:Trichotomy|trichotomy]].
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Integer/Formal Definition|formal definition of integers]], we have that $a, b, c, d, e, f$ are all [[Definition:Natural Number|natural numbers]].
+To eliminate confusion between [[Definition:Strict Ordering on Integers|integer ordering]] and the [[Definition:Ordering on Natural Numbers|ordering on the natural numbers]], let $a \preccurlyeq b$ denote that the [[Definition:Natural Number|natural number]] $a$ is less than or equal to the [[Definition:Natural Number|natural number]] $b$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {c, d} {}
+      | o = \not <
+      | r = \eqclass {a, b} {}
+      | c = [[Strict Ordering on Integers is Asymmetric]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \prec
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \ne
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = \ne
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = >
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {c, d} {}
+      | o = \not >
+      | r = \eqclass {a, b} {}
+      | c = [[Strict Ordering on Integers is Asymmetric]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = >
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \succ
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \ne
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = \ne
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = =
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = =
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \not \prec
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | lo= \land
+      | l = a + d
+      | o = \not \succ
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = \not <
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | lo= \land
+      | l = \eqclass {a, b} {}
+      | o = \not >
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+This demonstrates that $<$, $=$ and $>$ are mutually exclusive.
+Now:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \prec
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = b + c
+      | o = \preccurlyeq
+      | r = a + d
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {c, d} {}
+      | o = \le
+      | r = \eqclass {a, b} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = >
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \succ
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = b + c
+      | o = \succcurlyeq
+      | r = a + d
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {c, d} {}
+      | o = \ge
+      | r = \eqclass {a, b} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {}
+      | o = \ne
+      | r = \eqclass {c, d} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \ne
+      | r = b + c
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + d
+      | o = \prec
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | lo= \lor
+      | l = a + d
+      | o = \succ
+      | r = b + c
+      | c = {{Defof|Ordering on Natural Numbers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | o = <
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{eqn | lo= \land
+      | l = \eqclass {a, b} {}
+      | o = \>
+      | r = \eqclass {c, d} {}
+      | c = {{Defof|Strict Ordering on Integers}}
+}}
+{{end-eqn}}
+demonstrating that either $<$, $=$ or $>$ must hold.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Integer}
+Tags: Integers
+
+\begin{theorem}
+Let $x \in \Z$ be an [[Definition:Integer|integer]].
+Let $x = \eqclass {a, b} {}$ be defined from the [[Definition:Integer/Formal Definition|formal definition of integers]], where $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Then:
+:$-x = \eqclass {b, a} {}$
+\end{theorem}
+
+\begin{proof}
+Follows from [[Inverses for Integer Addition]].
+{{finish|The whole area of construction of integers from the inverse completion needs to be reviewed}}
+[[Category:Integers]]
+c8490t2014ko917aj6xcdymx2ob1gqk
+\end{proof}<|endoftext|>
+\section{Symmetric Function Theorem}
+Tags: Polynomial Theory, Named Theorems
+
+\begin{theorem}
+Let $f$ be a [[Definition:Polynomial over Ring in Multiple Variables|polynomial in $n$ variables]].
+Let $f$ be of [[Definition:Degree of Polynomial|degree $r$]] in each of its $n$ variables.
+Then $f$ is equal to a [[Definition:Polynomial over Ring in Multiple Variables|polynomial]] of [[Definition:Total Degree of Polynomial|total degree $r$]] with [[Definition:Integer|integer]] [[Definition:Coefficient of Polynomial|coefficients]] in the [[Definition:Elementary Symmetric Function|elementary symmetric functions]]:
+:$ds \sum x_i \sim x_i x_j, \dotsc, \prod x_j$
+and the [[Definition:Coefficient of Polynomial|coefficients]] of $f$.
+\end{theorem}<|endoftext|>
+\section{Euler's Integral Theorem}
+Tags: Integral Calculus
+
+\begin{theorem}
+:$\ds H_n = \ln n + \gamma + \map \OO {\dfrac 1 n}$
+where:
+:$H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]]
+:$\gamma$ denotes the [[Definition:Euler-Mascheroni Constant|Euler-Mascheroni constant]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Leonhard Paul Euler|cat = Euler}}
+\end{proof}<|endoftext|>
+\section{Ordering on Positive Integers is Equivalent to Ordering on Natural Numbers}
+Tags: Orderings on Integers
+
+\begin{theorem}
+Let $u, v \in \Z_{>0}$ be [[Definition:Natural Number|natural numbers]].
+Consider the [[Definition:Mapping|mapping]] $\phi: \N_{>0} \to \Z_{>0}$ defined as:
+:$\forall u \in \N_{>0}: \map \phi u = u'$
+where $u' \in \Z$ denotes the [[Definition:Strictly Positive Integer|(strictly) positive integer]] $\eqclass {b + u, b} {}$.
+Let $u', v' \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|strictly positive integers]].
+Then:
+:$u > v \iff u' > v'$
+\end{theorem}
+
+\begin{proof}
+Let $u' = \eqclass {b + u, b} {}$.
+Let $v' = \eqclass {c + v, c} {}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = u'
+      | o = >
+      | r = v'
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \eqclass {b + u, b} {}
+      | o = >
+      | r = \eqclass {c + v, c} {}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = b + u + c
+      | o = >
+      | r = c + v + b
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = u
+      | o = >
+      | r = v
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Absolute Values of Integers}
+Tags: Absolute Value Function, Integers
+
+\begin{theorem}
+Let $a, b \in \Z$ be [[Definition:Integer|integers]].
+Let $\size a$ denote the [[Definition:Absolute Value|absolute value]] of $a$:
+:$\size a = \begin {cases} a & : a \ge 0 \\ -a : a < 0 \end {cases}$
+Then:
+:$\size a \times \size b = \size {a \times b}$
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Ordered Integral Domain]], $\Z$ is an [[Definition:Ordered Integral Domain|ordered integral domain]].
+The result follows from [[Product of Absolute Values on Ordered Integral Domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of Self-Mappings on Finite Set}
+Tags: Cardinality of Set of All Mappings, Finite Sets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let the [[Definition:Cardinality|cardinality]] of $S$ be $n$.
+The [[Definition:Cardinality|cardinality]] of the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself (that is, the total number of [[Definition:Self-Map|self-maps]] on $S$) is:
+:$\card {S^S} = n^n$
+\end{theorem}
+
+\begin{proof}
+This is a specific example of [[Cardinality of Set of All Mappings]] where $S = T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Displacement of Particle under Force}
+Tags: Classical Mechanics
+
+\begin{theorem}
+Let $P$ be a [[Definition:Particle|particle]] of constant [[Definition:Mass|mass]] $m$.
+Let the [[Definition:Position|position]] of $P$ at [[Definition:Time|time]] $t$ be specified by the [[Definition:Position Vector|position vector]] $\mathbf r$.
+Let a [[Definition:Force|force]] applied to $P$ be represented by the [[Definition:Vector|vector]] $\mathbf F$.
+Then the [[Definition:Motion|motion]] of $P$ can be given by the [[Definition:Differential Equation|differential equation]]:
+:$\mathbf F = m \dfrac {\d^2 \mathbf r} {\d t^2}$
+or using [[Definition:Newtonian Notation for Derivatives|Newtonian notation]]:
+:$\mathbf F = m \ddot {\mathbf r}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf F
+      | r = \map {\dfrac \d {\d t} } {m \mathbf v}
+      | c = [[Newton's Second Law of Motion]]
+}}
+{{eqn | r = \map {\dfrac \d {\d t} } {m \dfrac {\d \mathbf r} {\d t} }
+      | c = {{Defof|Velocity}}
+}}
+{{eqn | r = m \map {\dfrac \d {\d t} } {\dfrac {\d \mathbf r} {\d t} }
+      | c = [[Derivative of Constant Multiple]]
+}}
+{{eqn | r = m \dfrac {\d^2 \mathbf r} {\d t^2}
+      | c = {{Defof|Second Derivative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{1-Sequence Space is Separable}
+Tags: Separable Spaces
+
+\begin{theorem}
+[[Definition:P-Sequence Space|$\ell^1$ space]] is a [[Definition:Separable Space/Normed Vector Space|separable space]].
+\end{theorem}
+
+\begin{proof}
+Let $D$ be the [[Definition:Set|set]] of all [[Definition:Finite Set|finitely]] [[Definition:Support of Mapping to Algebraic Structure|supported]] [[Definition:Sequence|sequences]] with [[Definition:Rational Number|rational]] terms:
+:$\displaystyle D = \set {\sequence {q_i}_{i \mathop \in \N} : n \in \N : i \le n : q_i \in \Q}$
+We have that:
+:[[Rational Numbers are Countably Infinite]]
+:A [[Definition:Finite Set|finite set]] is [[Definition:Countable Set/Definition 3|countable]]
+:$D$ is a union of [[Definition:Finite Set|finite sets]] indexed by $n$, which is [[Definition:Countable|countable]]
+By [[Countable Union of Countable Sets is Countable]], $D$ is [[Definition:Countable Set|countable]].
+Let $\mathbf x := \tuple {x_n}_{n \mathop \in \N} \in \ell^1$
+By [[Definition:P-Sequence Space|definition]] of $\ell^1$:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \size {x_n} < \infty$
+Let $\displaystyle s_n := \sum_{i \mathop = 0}^n \size {x_i}$ be a [[Definition:Sequence|sequence]] of [[Definition:Partial Sum|partial sums]] of $\displaystyle s = \sum_{i \mathop = 0}^\infty \size {x_i}$.
+Then $s$ is a [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]]:
+:$\forall \epsilon' \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \size {s_n - s} < \epsilon'$
+Note that:
+{{begin-eqn}}
+{{eqn | l = \size {s_n - s}
+      | r = \size {\sum_{i \mathop = 0}^n \size {x_i} - \sum_{i \mathop = 0}^\infty \size {x_i} }
+}}
+{{eqn | r = \size {\sum_{i \mathop = n \mathop + 1}^\infty \size {x_i} }
+}}
+{{eqn | r = \sum_{i \mathop = n \mathop + 1}^\infty \size {x_i}
+      | c = {{defof|Absolute Value}}
+}}
+{{end-eqn}}
+Let $\displaystyle \epsilon' = \frac \epsilon 2$.
+Let $\displaystyle N \in \N : \sum_{n \mathop = N + 1}^\infty \size {x_n} < \frac \epsilon 2$.
+We have that [[Rationals are Everywhere Dense in Reals/Normed Vector Space|$\Q$ is dense in $\R$]].
+:$\forall x \in \R : \exists \epsilon'' \in \R_{\mathop > 0} : \exists q \in \Q : \size {x - q} < \epsilon''$
+Then:
+:$\displaystyle \forall i \in \N : 1 \le i \le N : q_i \in \Q : \sum_{n \mathop = 1}^N \size {x_n - q_n} < \frac \epsilon 2$
+where
+:$\displaystyle \frac \epsilon 2 = \sum_{i \mathop = 1}^N \epsilon_i''$
+Let $\mathbf x' := \tuple {q_1, \ldots, q_N, 0, \ldots} \in D$.
+Then:
+:$\displaystyle \norm {\mathbf x - \mathbf x '} = \sum_{n \mathop = 1}^N \size {x_n - q_n} + \sum_{n \mathop = N + 1}^\infty \size {x_n} < \epsilon$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Root of Function under Half its Derivative}
+Tags: Primitives
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Integrable Function|integrable]].
+Then:
+:$\ds \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x = \sqrt {\map f x} + C$
+where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+By [[Integration by Substitution]] (with appropriate renaming of variables):
+:$\ds \int \map g u \rd u = \int \map g {\map f x} \map {f'} x \rd x$
+Let $\map u x = \sqrt {\map f x}$
+{{begin-eqn}}
+{{eqn | l = \map u x
+      | r = \sqrt {\map f x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d u} {\d x}
+      | r = \dfrac {\map {f'} x} {2 \sqrt {\map f x} }
+      | c = [[Chain Rule for Derivatives]], [[Derivative of Power]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \frac {\map {f'} x} {2 \sqrt {\map f x} } \rd x
+      | r = \int \rd u
+      | c = [[Integration by Substitution]]: putting $\map u x = \sqrt {\map f x}$
+}}
+{{eqn | r = u + C
+      | c = [[Primitive of Constant]]
+}}
+{{eqn | r = \sqrt {\map f x} + C
+      | c = Definition of $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Secant Function/Arctangent of Half Hyperbolic Tangent Form}
+Tags: Primitive of Hyperbolic Secant Function, Hyperbolic Tangent Half-Angle Substitutions
+
+\begin{theorem}
+:$\ds \int \sech x \rd x =  2 \map \arctan {\tanh \dfrac x 2} + C$
+\end{theorem}
+
+\begin{proof}
+Let $u = \tanh \dfrac x 2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \int \sech x \rd x
+      | r = \int \frac {1 - u^2} {1 + u^2} \frac {2 \rd u} {1 - u^2}
+      | c = [[Hyperbolic Tangent Half-Angle Substitution]]
+}}
+{{eqn | r = \int \frac {2 \rd u} {1 + u^2}
+      | c = simplifying
+}}
+{{eqn | r = 2 \arctan u + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$: Arctangent Form]]
+}}
+{{eqn | r = 2 \map \arctan {\tanh \dfrac x 2} + C
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y' + 2 x y = 1}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$y' + 2 x y = 1$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = e^{-{x^2} } \ds \int_a^x e^{t^2} \rd t$
+where $a$ is an [[Definition:Arbitrary Constant|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+This is a [[Definition:Linear First Order ODE with Constant Coefficients|linear first order ODE with constant coefficents]] in the form:
+:$\dfrac {\d y} {\d x} + a y = \map Q x$
+where:
+:$a = 2 x$
+:$\map Q x = 1$
+From [[Solution to Linear First Order Ordinary Differential Equation]]:
+:$\ds y = e^{-\int P \rd x} \paren {\int Q e^{\int P \rd x} \rd x + C}$
+Thus 
+{{begin-eqn}}
+{{eqn | l = y
+      | r = e^{-\int 2 x \rd x} \int e^{\int 2 x \rd x} \rd x + C
+      | c = 
+}}
+{{eqn | r = e^{- {x^2} } \int e^{x^2} \rd x + C
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+{{finish|Get this into the form as given -- the technique will be provided later in the book, one expects}}
+Further work on this is not trivial, as $\ds \int e^{x^2} \rd x$ has no solution in elementary functions.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Electric Potential over Conducting Surface is Constant}
+Tags: Electrostatics
+
+\begin{theorem}
+Let $S$ be a [[Definition:Conducting Surface|conducting surface]].
+The [[Definition:Electric Potential|electric potential]] $V$ over $S$ is [[Definition:Constant|constant]].
+This can be expressed using the [[Definition:Laplacian|Laplacian]]:
+:$\nabla^2 V = 0$
+and is thus seen to satisfy [[Definition:Laplace's Equation|Laplace's equation]].
+\end{theorem}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+{{TFAE|def=Base Axiom (Matroid)|view = Matroid Base Axiom}}
+\end{theorem}
+
+\begin{proof}
+==== Necessary Condition ====
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}}
+From [[Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom]]:
+:there exists a [[Definition:Matroid|matroid]] $M = \struct{S, \mathscr I}$ such that $\mathscr B$ is the [[Definition:Set|set]] of [[Definition:Base of Matroid|bases]] of $M$.
+Let $B_1, B_2 \in \mathscr B$.
+Let $x \in B_1 \setminus B_2$.
+From [[Matroid Base Union External Element has Fundamental Circuit]]:
+:there exists a [[Definition:Fundamental Circuit (Matroid)|fundamental circuit]] $\map C {x, B_2}$ of $M$ such that $x \in \map C {x, B_2} \subseteq B_2 \cup \set x$
+By definition of [[Definition:Set Intersection|set intersection]]:
+:$x \in B_1 \cap \map C {x, B_2}$
+From [[Element of Matroid Base and Circuit has a Substitute]]:
+:$\exists y \in \map C {x, B_2} \setminus B_1 : \paren{B_1 \setminus \set x} \cup \set y \in \mathscr B$
+We have:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \map C {x, B_2} \setminus B_1
+}}
+{{eqn | o = \subseteq
+      | r = \map C {x, B_2} \setminus \set x
+      | c = [[Set Difference with Subset is Superset of Set Difference]]
+}}
+{{eqn | o = \subseteq
+      | r = \paren{B_2 \cup \set x} \setminus \set x
+      | c = [[Set Difference over Subset]]
+}}
+{{eqn | o = \subseteq
+      | r = B_2 \setminus \set x
+      | c = [[Set Difference with Union is Set Difference]]
+}}
+{{eqn | o = \subseteq
+      | r = B_2
+      | c = [[Set Difference is Subset]]
+}}
+{{end-eqn}}
+From [[Leigh.Samphier/Sandbox/Matroid Base Substitution From Fundamental Circuit]]:
+:$\paren{B_2 \setminus \set y} \cup \set x \in \mathscr B$
+It follows that $\mathscr B$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 4|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 4}}{{qed|lemma}}
+==== Sufficient Condition ====
+Follows immediately from [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 4|Definition 4]] and [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|Definition 1]].
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 4 Iff Definition 5}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+{{TFAE|def=Base Axiom (Matroid)|view = Matroid Base Axiom}}
+\end{theorem}
+
+\begin{proof}
+==== Necessary Condition ====
+Follows immediately from [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 4|Definition 4]] and [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 5|Definition 5]].
+{{qed|lemma}}
+==== Sufficient Condition ====
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 5|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 5}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 3 Iff Definition 7}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+{{TFAE|def=Base Axiom (Matroid)|view = Matroid Base Axiom}}
+\end{theorem}
+
+\begin{proof}
+==== Necessary Condition ====
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}}
+Let $B_1, B_2 \in \mathscr B$.
+From $(\text B 3)$:
+:$\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y } \cup \set {\map \pi y} \in \mathscr B$
+Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the [[Definition:Inverse Mapping|inverse mapping]] of $\pi$.
+From [[Inverse of Bijection is Bijection]], $\pi^{-1}$ is a [[Definition:Bijection|bijection]].
+From [[Inverse Element of Bijection]]:
+:$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$
+Hence:
+:$\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set{\map {\pi^{-1}} x} } \cup \set x  = \paren {B_1 \setminus \set y} \cup \set {\map \pi y} \in \mathscr B$
+It follows that $\mathscr B$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7}}{{qed|lemma}}
+==== Sufficient Condition ====
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 7}}
+Let $B_1, B_2 \in \mathscr B$.
+From $(\text B 7)$:
+:$\exists \text{ a bijection } \pi : B_2 \setminus B_1 \to B_1 \setminus B_2 : \forall y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$
+Let $\pi^{-1} : B_1 \setminus B_2 \to B_2 \setminus B_1$ denote the [[Definition:Inverse Mapping|inverse mapping]] of $\pi$.
+From [[Inverse of Bijection is Bijection]], $\pi^{-1}$ is a [[Definition:Bijection|bijection]].
+From [[Inverse Element of Bijection]]:
+:$\forall x \in B_1 \setminus B_2 : \map {\pi^{-1}} x = y \iff \map \pi y = x$
+Hence:
+:$\forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set{\map {\pi^{-1}} x} = \paren {B_1 \setminus \set {\map \pi y} } \cup \set y \in \mathscr B$
+It follows that $\mathscr B$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+{{TFAE|def=Base Axiom (Matroid)|view = Matroid Base Axiom}}
+\end{theorem}
+
+\begin{proof}
+==== Necessary Condition ====
+Let $\mathscr B$ satisfy the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}}
+$\dots$
+It follows that $\mathscr B$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3}}{{qed|lemma}}
+==== Sufficient Condition ====
+By choosing $y = \map \pi x$ in [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 3|Definition 3]], [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|Definition 1]] follows immediately.
+\end{proof}<|endoftext|>
+\section{Number of Permutations with Repetition}
+Tags: Combinatorics, Number of Permutations with Repetition
+
+\begin{theorem}
+Set $S$ be a [[Definition:Set|set]] of $n$ [[Definition:Element|elements]].
+Let $\sequence T_m$ be a [[Definition:Sequence|sequence]] of $m$ [[Definition:Term|terms]] of $S$.
+Then there are $n^m$ different instances of $\sequence T_m$.
+\end{theorem}
+
+\begin{proof}
+Let $N_m$ denote the [[Definition:Set|set]] $\set {1, 2, \ldots, m}$.
+Let $f: N_m \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall k \in N_m: \map f t = t_m$
+By definition, $f$ corresponds to one of the specific instances of $\sequence T_m$.
+Hence the number of different instances of $\sequence T_m$ is found from [[Cardinality of Set of All Mappings]]:
+:$\card S^{\card {N_m} }$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Reciprocal of x Squared plus a Squared/Corollary}
+Tags: Definite Integrals involving x squared plus a squared, Definite Integral to Infinity of Reciprocal of x Squared plus a Squared
+
+\begin{theorem}
+:$\ds \int_0^\infty \dfrac {\d x} {1 + x^2} = \frac \pi 2$
+for $a \ne 0$.
+\end{theorem}
+
+\begin{proof}
+From [[Definite Integral to Infinity of Reciprocal of x Squared plus a Squared]]:
+:$\ds \int_0^\infty \dfrac {\d x} {x^2 + a^2} = \frac \pi {2 a}$
+which holds for for $a \ne 0$. 
+The result follows by setting $a = 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Norm is Norm/P-Sequence Space}
+Tags: P-Norms
+
+\begin{theorem}
+The [[Definition:P-Norm|$p$-norm]] on the [[Definition:P-Sequence Space|$p$-sequence space]] is a [[Definition:Norm on Vector Space|norm]].
+\end{theorem}
+
+\begin{proof}
+=== Norm Axiom $(\text N 1)$ ===
+Let $x \in \ell^p$ with $1 \le p < \infty$.
+By definition of [[Definition:P-Norm|$p$-norm]]:
+:$\ds \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p}^{1/p}$
+The [[Definition:Complex Modulus|complex modulus]] of $x_n$ is [[Definition:Real Number|real]] and [[Complex Modulus is Non-Negative|non-negative]].
+We have the results:
+:[[Sum of Non-Negative Reals is Non-Negative]]
+:[[Power of Positive Real Number is Positive]]
+:[[Definition:Power of Zero|Zero Raised to Positive Power is Zero]]
+Hence, $\norm {\mathbf x}_p \ge 0$.
+Suppose $\norm {\mathbf x}_p = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x}_p
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 0}^\infty \size {x_n}^p
+      | r = 0
+      | c = raising to power $p > 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x_n}
+      | r = 0
+      | c = [[Sum of Non-Negatives vanishes iff Summands vanish]]
+}}
+{{eqn | ll= \leadsto
+      | l = x_n
+      | r = 0
+      | c = [[Complex Modulus equals Zero iff Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = \bf x
+      | r = \sequence 0_{n \mathop \in \N}
+}}
+{{end-eqn}}
+Thus [[Definition:Norm Axioms|norm axiom $(\text N 1)$]] is satisfied.
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+Suppose $\alpha \in \set {\R, \C}$.
+{{begin-eqn}}
+{{eqn| l = \norm {\alpha \cdot \mathbf x}_p
+     | r = \paren {\sum_{n \mathop = 0}^\infty \size {\alpha x_n}^p }^{\frac 1 p}
+}}
+{{eqn| r = \paren {\sum_{n \mathop = 0}^\infty \size \alpha^p \size {x_n}^p }^{\frac 1 p}
+}}
+{{eqn| r = \size {\alpha} \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p }^{\frac 1 p}
+}}
+{{eqn| r = \size {\alpha} \norm {\mathbf x}_p
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+From [[P-Norm is Norm]] we have that:
+:$\ds \paren {\sum_{n \mathop = 0}^d \size {x_n + y_n}^p}^{\frac 1 p} \le \paren {\sum_{n \mathop = 0}^d \size {x_n}^p }^{\frac 1 p} + \paren {\sum_{n \mathop = 0}^d \size {y_n}^p }^{\frac 1 p}$
+$\map f z = z^{\frac 1 p}$ is a [[Definition:Continuous Function|continuous function]] for $z \ge 0$ and $p > 0$.
+For $\mathbf x \in \ell^p$, changing $d$ is equivalent to changing $z$ in the interval $0 \le z < \infty$. 
+Take the [[Limit of Composite Function|composite limit]] $d \to \infty$.
+Then:
+:$\norm {\mathbf x + \mathbf  y}_p \le \norm {\mathbf x}_p + \norm {\mathbf y}_p$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Norm is Norm/Complex Numbers}
+Tags: P-Norms, Examples of Norms
+
+\begin{theorem}
+The [[Definition:P-Norm|$p$-norm]] on the [[Definition:Complex Number|complex numbers]] is a [[Definition:Norm on Vector Space|norm]].
+\end{theorem}
+
+\begin{proof}
+Let $K \in \C^d$, where $d \in \N_{> 0}$.
+=== Norm Axiom $(\text N 1)$ ===
+Suppose $\sequence {x_n}_{n \mathop \in {\N}, \, n \mathop \le d} \in K$.
+By definition of [[Definition:P-Norm|$p$-norm]]:
+:$\displaystyle \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1/p}$
+The [[Definition:Complex Modulus|complex modulus]] of $x_n$ is [[Definition:Real Number|real]] and [[Complex Modulus is Non-Negative|non-negative]].
+We have the results:
+:[[Sum of Non-Negative Reals is Non-Negative]]
+:[[Power of Positive Real Number is Positive]]
+:[[Definition:Power of Zero|Zero Raised to Positive Power is Zero]]
+Hence, $\norm {\mathbf x}_p \ge 0$.
+Suppose, $\norm {\mathbf x}_p = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x}_p
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 0}^d \size {x_n}^p
+      | r = 0
+      | c = raising to power $p > 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x_n}
+      | r = 0
+      | c = [[Sum of Non-Negatives vanishes iff Summands vanish]]
+}}
+{{eqn | ll= \leadsto
+      | l = x_n
+      | r = 0
+      | c = [[Complex Modulus equals Zero iff Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = \bf x
+      | r = \sequence 0_{n \mathop \in \N, \, n \mathop \le d}
+}}
+{{end-eqn}}
+Thus [[Definition:Norm Axioms|norm axiom $(\text N 1)$]] is satisfied.
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+Suppose, $\lambda \in K$.
+{{begin-eqn}}
+{{eqn | l = \norm {\lambda \mathbf x}_p
+      | r = \paren {\sum_{n \mathop = 0}^d \size {\lambda x_n}^p}^{1/p}
+}}
+{{eqn | r = \paren {\size{\lambda}^p \sum_{n \mathop = 0}^d \size {x_n}^p}^{1/p}
+}}
+{{eqn | r = \size {\lambda} \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1/p}
+}}
+{{eqn | r = \size {\lambda} \norm {\mathbf x}_p
+}}
+{{end-eqn}}
+Thus [[Definition:Norm Axioms|norm axiom $(\text N 2)$]] is satisfied.
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+If $\mathbf x = \sequence 0$ and $\mathbf y = \sequence 0$, then by $\paren {\text N 1}$ we have [[Definition:Equality|equality]].
+If $\mathbf x + \mathbf y = \sequence 0$ and both $\bf x$ and $\bf y$ nonvanishing, then by $\paren {\text N 1}$ we get a [[Definition:Strict Ordering|strict]] [[Definition:Inequality|inequality]].
+If $\mathbf x + \mathbf y \ne \sequence 0$, then consider [[Definition:P-Norm|p-norm]] [[Definition:Real Power|raised to the power]] of $p$:
+{{begin-eqn}}
+{{eqn | l = \norm {\bf x + \bf y}_p^p
+      | r = \sum_{n \mathop = 0}^d \size {x_n + y_n} \size {x_n + y_n}^{p \mathop - 1}
+}}
+{{eqn | o = \le
+      | r = \sum_{n \mathop = 0}^d \size {x_n} \size {x_n + y_n}^{p \mathop - 1} + \sum_{n \mathop = 0}^d \size {y_n} \size {x_n + y_n}^{p \mathop - 1}
+      | c = [[Triangle Inequality for Complex Numbers]]
+}}
+{{eqn | o = \le
+      | r = \sum_{n \mathop = 0}^d \size {x_n \paren{x_n + y_n}^{p \mathop - 1} } + \sum_{n \mathop = 0}^d \size {y_n \paren{ x_n + y_n}^{p \mathop - 1} }
+      | c = [[Modulus of Product]]
+}}
+{{eqn | o = \le
+      | r = \norm {\bf x}_p \norm {\paren{\mathbf x + \mathbf y}^{p \mathop - 1} }_q + \norm {\mathbf y}_p \norm {\paren{\mathbf x + \mathbf y}^{p \mathop - 1} }_q
+      | c = [[Hölder's Inequality for Sums]]: $\dfrac 1 p + \dfrac 1 q = 1$
+}}
+{{eqn | o = \le
+      | r = \norm {\bf x}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1} + \norm {\mathbf y}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1}
+      | c = [[Transformation of P-Norm|Transformation of $p$-Norm]]: $q \paren {p - 1} = p$
+}}
+{{eqn | ll= \leadsto
+      | l = \norm {\mathbf x + \mathbf y}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1}
+      | o = \le
+      | r = \norm {\mathbf x}_p \norm {\mathbf x + \mathbf y}_p^{p \mathop - 1} + \norm {\bf y}_p \norm {\bf x + \bf y}_p^{p \mathop - 1}
+}}
+{{eqn | ll= \leadsto
+      | l = \norm {\bf x + \bf y}_p
+      | o = \le
+      | r = \norm {\bf x}_p  + \norm {\bf y}_p 
+      | c = [[Definition:Real Division|Division]] by $\norm {\bf x + \bf y}_p^{p \mathop - 1}$
+}}
+{{end-eqn}}
+Thus [[Definition:Norm Axioms|norm axiom $(\text N 3)$]] is satisfied.
+{{qed|lemma}}
+All [[Definition:Norm Axioms|norm axioms]] are seen to be satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-Norm is Norm/Real Numbers}
+Tags: Examples of Norms, P-Norms
+
+\begin{theorem}
+The [[Definition:P-Norm|$p$-norm]] on the [[Definition:Real Number|real numbers]] is a [[Definition:Norm on Vector Space|norm]].
+\end{theorem}
+
+\begin{proof}
+We have that [[P-Norm is Norm/Complex Numbers|$p$-norm is a norm on complex numbers]].
+Since [[Definition:Real Numbers|real numbers]] are [[Definition:Complex Number/Wholly Real|wholy real complex numbers]], the same results holds.
+{{qed}}
+[[Category:Examples of Norms]]
+[[Category:P-Norms]]
+d0t8j9qm69yeccq7h5c3xktgx0b9b4j
+\end{proof}<|endoftext|>
+\section{P-Sequence Space with P-Norm forms Normed Vector Space}
+Tags: P-Norms, Examples of Normed Vector Spaces
+
+\begin{theorem}
+[[Definition:P-Sequence Space|$P$-Sequence Space]] with [[Definition:P-Norm|$p$-norm]] forms [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[P-Sequence Space with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences form Vector Space|$P$-sequence space is a vector space]]
+:[[P-Norm is Norm/P-Sequence Space|$P$-norm on the $p$-sequence space is a norm]]
+By definition, $\struct {\ell^p, \norm {\, \cdot \,}_p}$ is a [[Definition:Normed Vector Space|normed vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space}
+Tags: Examples of Vector Spaces, Functional Analysis
+
+\begin{theorem}
+Let $\ell^\infty$ be the [[Definition:Space of Bounded Sequences|space of bounded sequences]].
+Let $\struct {\C, +_\C, \times_\C}$ be the [[Definition:Field of Complex Numbers|field of complex numbers]].
+Let $\paren +$ be the [[Definition:Pointwise Addition on Ring of Sequences|pointwise addition on the ring of sequences]].
+Let $\paren {\, \cdot \,}$ be the [[Definition:Pointwise Multiplication on Ring of Sequences|pointwise multiplication on the ring of sequences]].
+Then $\struct {\ell^\infty, +, \, \cdot \,}_\C$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \ell^\infty$.
+Let $\lambda, \mu \in \C$.
+Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a [[Definition:Complex-Valued Function|complex-valued function]].
+Let us use [[Definition:Complex Number|real number]] [[Definition:Complex Addition|addition]] and [[Definition:Complex Multiplication|multiplication]].
+Define [[Definition:Pointwise Addition on Ring of Sequences|pointwise addition]] as:
+:$\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\C \sequence {b_n}_{n \mathop \in \N}$.
+Define [[Definition:Pointwise Scalar Multiplication on Ring of Sequences|pointwise scalar multiplication]] as:
+:$\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\C \sequence {a_n}_{n \mathop \in \N}$
+Let the [[Definition:Ring of Sequences/Additive Inverse|additive inverse]] be $\sequence {-a_n} := - \sequence {a_n}$.
+=== Closure Axiom ===
+By [[Definition:Assumption|assumption]], $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N} \in \ell^\infty$.
+By [[Definition:Space of Bounded Sequences|definition]]:
+:$\displaystyle \sup_{n \mathop \in \N} \size {a_n} < \infty$
+:$\displaystyle \sup_{n \mathop \in \N} \size {b_n} < \infty$
+Consider the [[Definition:Real Sequence|sequence]] $\sequence {a_n + b_n}$.
+Then: 
+{{begin-eqn}}
+{{eqn| l = \sup_{n \mathop \in \N} \size {a_n + b_n}
+     | o = \le
+     | r = \sup_{n \mathop \in \N} \size {\size {a_n} + \size {b_n} }
+}}
+{{eqn| o = \le
+     | r = \sup_{n \mathop \in \N} \size {\map \max {\size {a_n}, \size {b_n} } + \map \max {\size {a_n}, \size {b_n} } }
+     | c = {{defof|Max Operation}}
+}}
+{{eqn| r = \sup_{n \mathop \in \N} \size {2 \map \max {\size {a_n}, \size {b_n} } }
+}}
+{{eqn | o = \le
+      | r = 2 \paren {\sup_{n \mathop \in \N} \size {a_n} + \sup_{n \mathop \in \N} \size {b_n} }
+}}
+{{eqn | o = <
+      | r = \infty
+      | c = $\sequence {a_n}, \sequence {b_n} \in \ell^\infty$
+}}
+{{end-eqn}}
+Hence:
+:$\sequence {a_n + b_n} \in \ell^\infty$
+{{qed|lemma}}
+=== Commutativity Axiom ===
+By [[Pointwise Addition on Ring of Sequences is Commutative]], $\sequence {a_n} + \sequence {b_n} = \sequence {b_n} + \sequence {a_n}$
+{{qed|lemma}}
+=== Associativity Axiom ===
+By [[Pointwise Addition on Ring of Sequences is Associative]], $\paren {\sequence {a_n} + \sequence {b_n} } + \sequence {c_n} = \sequence {a_n} + \paren {\sequence {b_n} + \sequence {c_n} }$.
+{{qed|lemma}}
+=== Identity Axiom ===
+{{begin-eqn}}
+{{eqn | l = \sequence {0 + a_n}
+      | r = \sequence 0 +_\C \sequence {a_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \tuple {0, 0, 0, \dots} +_\C \sequence {a_n}
+      | c = Definition of $\sequence 0$
+}}
+{{eqn | r = \sequence {a_n}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Inverse Axiom ===
+{{begin-eqn}}
+{{eqn | l = \sequence {a_n + \paren {-a_n} }
+      | r = \sequence {a_n} +_\C \sequence {-a_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {a_n} +_\C \paren {-1} \times_\C \sequence {a_n} 
+      | c = Definition of $\sequence {-a_n}$
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Distributivity over Scalar Addition ===
+{{begin-eqn}}
+{{eqn | l = \sequence {\paren {\lambda +_\C \mu} \cdot a_n }
+      | r = \paren {\lambda +_\C \mu} \times_\C \sequence {a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\C \sequence {a_n} +_\C \mu \times_\C \sequence {a_n}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n} +_\C \sequence {\mu \cdot a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n + \mu \cdot a_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Distributivity over Vector Addition ===
+{{begin-eqn}}
+{{eqn | l = \lambda \times_\C \sequence {a_n + b_n}
+      | r = \lambda \times_\C \paren {\sequence {a_n} +_\C \sequence {b_n} }
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\C \sequence {a_n} +_\C \lambda \times_\C \sequence {b_n}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n} +_\C \sequence {\lambda \cdot b_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n + \mu \cdot b_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Associativity with Scalar Multiplication ===
+{{begin-eqn}}
+{{eqn | l = \sequence {\paren {\lambda \times_\C \mu} \cdot a_n}
+      | r = \paren {\lambda \times_\C \mu} \times_\C \sequence {a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\C \paren {\mu \times_\C \sequence {a_n} }
+      | c = [[Real Multiplication is Associative]]
+}}
+{{eqn | r = \lambda \times_\C \sequence {\mu \cdot a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {\lambda \cdot \paren {\mu \cdot a_n} }
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Identity for Scalar Multiplication ===
+{{begin-eqn}}
+{{eqn | l = \sequence {1 \cdot a_n}
+      | r = 1 \times_\C \sequence {a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {a_n}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Vector has no Direction}
+Tags: Zero Vectors
+
+\begin{theorem}
+A [[Definition:Zero Vector Quantity|zero vector]] has no [[Definition:Direction|direction]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf 0$ denote a [[Definition:Zero Vector Quantity|zero vector]].
+{{AimForCont}} $\mathbf 0$ has a [[Definition:Direction|direction]].
+Then $\mathbf 0$ can be [[Definition:Arrow Representation of Vector Quantity|represented as an arrow]] in a [[Definition:Real Vector Space|real vector space]] $\R^n$ with a [[Definition:Cartesian Coordinate System|Cartesian frame]].
+Let $\mathbf 0$ be so embedded.
+Thus it consists of a [[Definition:Line Segment|line segment]] between two [[Definition:Point|points]] with an [[Definition:Initial Point of Vector|initial point]] $A$ and a [[Definition:Terminal Point of Vector|terminal point]] $B$.
+The [[Definition:Initial Point of Vector|initial point]] and a [[Definition:Terminal Point of Vector|terminal point]] are [[Definition:Distinct Elements|distinct]] from each other.
+Let these [[Definition:Point|points]] be identified as:
+{{begin-eqn}}
+{{eqn | l = A
+      | r = \tuple {a_1, a_2, \ldots, a_n}
+}}
+{{eqn | l = B
+      | r = \tuple {b_1, b_2, \ldots, b_n}
+}}
+{{end-eqn}}
+Hence we have that the [[Definition:Vector Length|length]] of $\mathbf 0$ is defined as:
+:$\norm {\mathbf 0} = \ds \sqrt {\sum_{i \mathop = 1}^n \paren {a_i - b_i}^2} > 0$
+which means that at least one of $a_i - b_i$ is non-[[Definition:Zero (Number)|zero]].
+But this [[Definition:Contradiction|contradicts]] the definition of $\mathbf 0$ being the [[Definition:Zero Vector Quantity|zero vector]].
+It follows by [[Proof by Contradiction]] that our assumption that $\mathbf 0$ has a [[Definition:Direction|direction]] must be false.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Set is Closed under Countable Unions}
+Tags: Power Set, Set Union, Countable Sets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Then:
+:$\forall A_n \in \powerset S: n = 1, 2, \ldots: \ds \bigcup_{n \mathop = 1}^\infty A_n \in \powerset S$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {A_i}$ be a [[Definition:Countable Set|countably infinite]] [[Definition:Sequence|sequence]] of [[Definition:Set|sets]] in $\powerset S$.
+Consider an element of the [[Definition:Set Union|union]] of all the [[Definition:Set|sets]] in this $\sequence {A_i}$:
+:$\ds x \in \bigcup_{i \mathop \in \N} A_i$
+By definition of [[Definition:Set Union|union]]:
+:$\exists i \in \N: x \in A_i$
+But $A_i \in \powerset S$ and so by definition $A_i \subseteq S$.
+By definition of [[Definition:Subset|subset]], it follows that $x \in S$.
+Hence, again by definition of [[Definition:Subset|subset]]:
+:$\ds \bigcup_{i \mathop \in \N} A_i \subseteq S$
+Finally, by definition of [[Definition:Power Set|power set]]:
+:$\ds \bigcup_{i \mathop \in \N} A_i \in \powerset S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Elementary Events belonging to k Events is Event}
+Tags: Event Spaces
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]] with a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
+Let $A_1, A_2, \ldots, A_m$ be [[Definition:Event|events]] in the [[Definition:Event Space|event space]] $\Sigma$ of $\EE$.
+Let $S$ denote the [[Definition:Set|set]] of all [[Definition:Elementary Event|elementary events]] of $\EE$ which are [[Definition:Element|elements]] of exactly $k$ of the [[Definition:Event|events]] $A_1, A_2, \ldots, A_m$.
+Then $S$ is an [[Definition:Event|event]] of $\Sigma$.
+\end{theorem}
+
+\begin{proof}
+Let $r_1, r_2, \ldots r_k$ be a [[Definition:Set|set]] of $k$ [[Definition:Element|elements]] of $\set {1, 2, \ldots, m}$.
+Then:
+:$\paren {A_{r_1} \cap A_{r_2} \cap \cdots \cap A_{r_k} } \setminus \paren { A_{r_{k + 1} } \cup A_{r_{k + 2} } \cup \cdots \cup A_{r_m} }$
+contains exactly those [[Definition:Element|elements]] of $\Omega$ which are contained in exactly those [[Definition:Event|events]] $A_{r_1}, A_{r_2}, \ldots, \cap A_{r_k}$.
+By [[Elementary Properties of Event Space]], that last [[Definition:Set|set]] is an [[Definition:Event|event]] of $\EE$.
+By selecting all [[Definition:Combination|combinations]] of $\set {1, 2, \ldots, m}$, we can form the [[Definition:Subset|subsets]] of $\Sigma$ containing all those [[Definition:Elementary Event|elementary events]] of exactly $k$ [[Definition:Event|events]].
+Each one is an [[Definition:Element|element]] of $\Sigma$.
+We then form the [[Definition:Set Union|union]] $S_k$ of each of those [[Definition:Element|element]] of $\Sigma$.
+By [[Elementary Properties of Event Space]], $S_k$ is an [[Definition:Event|event]] of $\EE$.
+By construction, $S_k$ contains all and only the [[Definition:Elementary Event|elementary events]] of $\EE$ which are [[Definition:Element|elements]] of exactly $k$ of the [[Definition:Event|events]] $A_1, A_2, \ldots, A_m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Event Space of Experiment with Final Sample Space has Even Cardinality}
+Tags: Event Spaces
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]] with a [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
+Let $\Omega$ be a [[Definition:Finite Set|finite set]].
+Then the [[Definition:Event Space|event space]] $\Sigma$ consists of an [[Definition:Even Integer|even number]] of [[Definition:Subset|subsets]] of $\Omega$.
+\end{theorem}
+
+\begin{proof}
+Let $A \in \Sigma$ be one of the [[Definition:Event|events]] of $\EE$.
+We have by definition that $\Omega$ is itself an [[Definition:Event|events]] of $\EE$.
+Hence by [[Set Difference of Events is Event]], $\Omega \setminus A$ is also an [[Definition:Event|event]] of $\EE$.
+As $A$ is arbitrary, the same applies to all [[Definition:Event|events]] of $\EE$.
+Thus all [[Definition:Event|events]] of $\EE$ come in [[Definition:Doubleton|pairs]]: $A$ and $\Omega \setminus A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Probability of Union of Disjoint Events is Sum of Individual Probabilities}
+Tags: Disjoint Events
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]].
+Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Measure|probability measure]] on $\EE$.
+Then:
+:$\forall A, B \in \Sigma: A \cap B = \O \implies \map \Pr {A \cup B} = \map \Pr A + \map \Pr B$
+\end{theorem}
+
+\begin{proof}
+From the [[Axiom:Kolmogorov Axioms|Kolmogorov Axioms]]:
+:$\ds \map \Pr {\bigcup_{i \mathop \ge 1} A_i} = \sum_{i \mathop \ge 1} \map \Pr {A_i}$
+where $\set {A_1, A_2, \ldots}$ is a [[Definition:Countable Set|countable set]] of [[Definition:Pairwise Disjoint|pairwise disjoint]] [[Definition:Event|events]] of $\EE$.
+This applies directly to $\map \Pr {A \cup B}$ where $A \cap B = \O$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Uniform Distribution gives rise to Probability Measure}
+Tags: Discrete Uniform Distribution
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]].
+Let the [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$ be defined as:
+:$\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$
+:$\Sigma = \powerset \Omega$
+:$\forall A \in \Sigma: \map \Pr A = \dfrac 1 n \card A$
+where:
+:$\powerset \Omega$ denotes the [[Definition:Power Set|power set]] of $\Omega$
+:$\card A$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+Then $\Pr$ is a [[Definition:Probability Measure|probability measure]] on $\struct {\Omega, \Sigma}$.
+\end{theorem}
+
+\begin{proof}
+From [[Power Set of Sample Space is Event Space]] we have that $\Sigma$ is an [[Definition:Event Space|event space]].
+{{qed|lemma}}
+We check the axioms defining a [[Definition:Probability Measure|probability measure]]:
+{{begin-axiom}}
+{{axiom | n = \text I
+        | q = \forall A \in \Sigma
+        | ml= \map \Pr A
+        | mo= \ge
+        | mr= 0
+}}
+{{axiom | n = \text {II}
+        | ml= \map \Pr \Omega
+        | mo= =
+        | mr= 1
+}}
+{{axiom | n = \text {III}
+        | q = \forall A \in \Sigma
+        | ml= \map \Pr A
+        | mo= =
+        | mr= \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}
+        | rc= where $e$ denotes the [[Definition:Elementary Event|elementary events]] of $\EE$
+}}
+{{end-axiom}}
+Axiom $\text I$ is seen to be satisfied by the observation that the [[Definition:Cardinality|cardinality]] of a [[Definition:Set|set]] is never negative.
+Hence $\map \Pr A \ge 0$.
+{{qed|lemma}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \map \Pr \Omega
+      | r = \dfrac 1 n \card \Omega
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n \times n
+      | c = Definition of $\Omega$: it has been defined as having $n$ [[Definition:Element|elements]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Axiom $\text {II}$ is thus seen to be satisfied.
+{{qed|lemma}}
+Let $A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$ where $k = \card A$.
+Then by [[Union of Set of Singletons]]:
+:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \Pr A
+      | r = \dfrac 1 n \card A
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} } }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n \card {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n \paren {\underbrace {1 + 1 + \cdots + 1}_{\text {$k$ times} } }
+      | c = 
+}}
+{{eqn | r = \underbrace {\dfrac 1 n + \dfrac 1 n + \cdots + \dfrac 1 n}_{\text {$k$ times} }
+      | c = 
+}}
+{{eqn | r = \map \Pr {\set {\omega_{r_1} } } + \map \Pr {\set {\omega_{r_2} } } + \cdots + \map \Pr {\set {\omega_{r_k} } }
+      | c = 
+}}
+{{eqn | r = \sum_{\bigcup \set e \mathop = A} \map \Pr {\set e}
+      | c = 
+}}
+{{end-eqn}}
+Hence Axiom $\text {III}$ is thus seen to be satisfied.
+{{qed|lemma}}
+All axioms are seen to be satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Probability Measure on Finite Sample Space}
+Tags: 
+
+\begin{theorem}
+Let $\Omega = \set {\omega_1, \omega_2, \ldots, \omega_n}$ be a [[Definition:Finite Set|finite set]].
+Let $\Sigma$ be a [[Definition:Sigma-Algebra|$\sigma$-algebra]] on $\Omega$.
+Let $p_1, p_2, \ldots, p_n$ be [[Definition:Non-Negative Real Number|non-negative real numbers]] such that:
+:$p_1 + p_2 + \cdots + p_n = 1$
+Let $Q: \Sigma \to \R$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall A \in \Sigma: \map Q A = \ds \sum_{i: \omega_i \in A} p_i$
+Then $\struct {\Omega, \Sigma, Q}$ constitutes a [[Definition:Probability Space|probability space]].
+That is, $Q$ is a [[Definition:Probability Measure|probability measure]] on $\struct {\Omega, \Sigma}$.
+\end{theorem}
+
+\begin{proof}
+Recall the [[Axiom:Kolmogorov Axioms|Kolmogorov axioms]]:
+{{:Axiom:Kolmogorov Axioms}}
+First we determine that $\Pr$ as defined is actually a [[Definition:Probability Measure|probability measure]].
+By definition, we have that $\map \Pr A$ is the sum of some subset of $\set {p_1, p_2, \ldots, p_n}$.
+Thus $0 \le \map \Pr A \le 1$ and [[Axiom:Kolmogorov Axioms|Axiom $(1)$]] is fulfilled trivially by definition.
+Let $A \in \Sigma$ be such that:
+:$A = \set {\omega_{r_1}, \omega_{r_2}, \ldots, \omega_{r_k} }$
+We have that:
+:$A = \set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} }$
+From [[Simple Events are Mutually Exclusive]], $\set {\set {\omega_{r_1} }, \set {\omega_{r_2} }, \ldots, \set {\omega_{r_k} } }$ constitutes a [[Definition:Set|set]] of [[Definition:Pairwise Disjoint Events|pairwise disjoint events]].
+Hence:
+:$\map \Pr {\set {\omega_{r_1} } \cup \set {\omega_{r_2} } \cup \cdots \cup \set {\omega_{r_k} } } = \ds \sum_{i \mathop = 1}^k \map \Pr {\omega_{r_1} }$
+and it is seen that [[Axiom:Kolmogorov Axioms|axiom $(3)$]] is fulfilled.
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = \map \Pr \Omega
+      | r = \map \Pr {\bigcup_{i \mathop = 1}^n \omega_i}
+      | c = 
+}}
+{{eqn | r = \ds \sum_{i \mathop = 1}^n \map \Pr {\omega_i}
+      | c = 
+}}
+{{eqn | r = \ds \sum_{i \mathop = 1}^n p_i
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Hence [[Axiom:Kolmogorov Axioms|axiom $(2)$]] is satisfied.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Probability Measure on Single-Subset Event Space}
+Tags: Probability Theory
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]] whose [[Definition:Sample Space|sample space]] is $\Omega$.
+Let $\O \subsetneqq A \subsetneqq \Omega$.
+Let $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ be the [[Definition:Event Space|event space]] of $\EE$.
+Let $\Pr: \Sigma \to \R$ be a [[Definition:Probability Measure|probability measure]] on $\struct {\Omega, \Sigma}$.
+Then $\Pr$ has the form:
+{{begin-eqn}}
+{{eqn | n = Pr 1
+      | l = \map \Pr \O
+      | r = 0
+}}
+{{eqn | n = Pr 2
+      | l = \map \Pr A
+      | r = p
+}}
+{{eqn | n = Pr 3
+      | l = \map \Pr {\Omega \setminus A}
+      | r = 1 - p
+}}
+{{eqn | n = Pr 4
+      | l = \map \Pr \Omega
+      | r = 1
+}}
+{{end-eqn}}
+for some $p \in \R$ satisfying $0 \le p \le 1$.
+\end{theorem}
+
+\begin{proof}
+From [[Event Space from Single Subset of Sample Space]], we have that $\Sigma$ is an [[Definition:Event Space|event space]].
+Recall the [[Axiom:Kolmogorov Axioms|Kolmogorov axioms]]:
+{{:Axiom:Kolmogorov Axioms}}
+First we determine that $\Pr$ as defined is actually a [[Definition:Probability Measure|probability measure]].
+[[Axiom:Kolmogorov Axioms|Axiom $(1)$]] and [[Axiom:Kolmogorov Axioms|axiom $(2)$]] are fulfilled trivially by definition.
+Then we note that, apart from $\O$, $\set {A, \Omega \setminus A}$ are the only [[Definition:Pairwise Disjoint Events|pairwise disjoint events]] whose [[Definition:Set Union|union]] is $\Omega$.
+Hence by definition of $\Pr$ we see that [[Axiom:Kolmogorov Axioms|axiom $(3)$]] degenerates to:
+{{begin-eqn}}
+{{eqn | l = \map \Pr {A \cup \paren {\Omega \setminus A} }
+      | r = \map \Pr \Omega
+      | c = 
+}}
+{{eqn | r = 1
+      | c = $\text {Pr 4}$
+}}
+{{eqn | r = p + \paren {1 - p}
+      | c = Definition of $p$
+}}
+{{eqn | r = \map \Pr A + \map \Pr {\Omega \setminus A}
+      | c = $\text {Pr 2}$ and $\text {Pr 3}$
+}}
+{{end-eqn}}
+and thus [[Axiom:Kolmogorov Axioms|axiom $(3)$]] is fulfilled.
+Thus $\Pr$ is indeed a [[Definition:Probability Measure|probability measure]].
+{{qed|lemma}}
+Now it is established that a [[Definition:Probability Measure|probability measure]] on $\Sigma$ is of the form of $\Pr$ as defined.
+First we note that from [[Probability of Empty Event is Zero]]:
+:$\map \Pr \O  = 0$
+Hence $\text {Pr 1}$ is satisfied.
+From [[Axiom:Kolmogorov Axioms|Axiom $(2)$]] we have that $\map \Pr \Omega = 1$
+Hence $\text {Pr 4}$ is satisfied.
+By definition of [[Definition:Probability Measure|probability measure]], $\map \Pr A$ is between $0$ and $1$ inclusive.
+That is:
+:$\map \Pr A = p$
+for some $p \in \R$ such that $0 \le p \le 1$.
+This is exactly $\text {Pr 2}$, which is hence seen to be satisfied.
+Then we note that from [[Probability of Event not Occurring]]:
+:$\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$
+That is:
+:$\map \Pr {\Omega \setminus A} = 1 - p$
+Hence $\text {Pr 3}$ is satisfied.
+Thus it has been proved that a [[Definition:Probability Measure|probability measure]] on such an an [[Definition:Event Space|event space]] $\Sigma$ has to be of the form $\Pr$ as defined.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Probability of Set Difference of Events}
+Tags: Probability Theory
+
+\begin{theorem}
+Let $\EE$ be an [[Definition:Experiment|experiment]] with [[Definition:Probability Space|probability space]] $\struct {\Omega, \Sigma, \Pr}$.
+Let $A, B \in \Sigma$ be [[Definition:Event|events]] of $\EE$.
+Let $\map \Pr A$ denote the [[Definition:Probability|probability]] of [[Definition:Event|event]] $A$ [[Definition:Occurrence of Event|occurring]].
+Then:
+:$\map \pr {A \setminus B} = \map \Pr A - \map \Pr {A \cap B}$
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference and Intersection form Partition]]:
+:$A$ is the [[Definition:Set Union|union]] of the two [[Definition:Disjoint Sets|disjoint sets]] $A \setminus B$ and $A \cap B$
+So, by the definition of [[Definition:Probability Measure|probability measure]]:
+:$\map \Pr A = \map \Pr {A \setminus B} + \map \Pr {A \cap B}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Rank Axioms}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\rho : \powerset S \to \Z$ be a [[Definition:Mapping|mapping]] from the [[Definition:Power Set|power set]] of $S$ to the [[Definition:Integer|integers]].
+{{TFAE|def=Rank Axioms (Matroid)|view=Matroid Rank Axioms}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Element of Arc in Orthogonal Curvilinear Coordinates}
+Tags: Orthogonal Curvilinear Coordinates
+
+\begin{theorem}
+Let $\tuple {q_1, q_2, q_3}$ denote a set of [[Definition:Orthogonal Curvilinear Coordinates|orthogonal curvilinear coordinates]].
+Let the relation between those [[Definition:Orthogonal Curvilinear Coordinates|orthogonal curvilinear coordinates]] and [[Definition:Cartesian Coordinates|Cartesian coordinates]] be expressed as:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \map x {q_1, q_2, q_3}
+}}
+{{eqn | l = y
+      | r = \map y {q_1, q_2, q_3}
+}}
+{{eqn | l = z
+      | r = \map z {q_1, q_2, q_3}
+}}
+{{end-eqn}}
+where $\tuple {x, y, z}$ denotes the [[Definition:Cartesian Coordinates|Cartesian coordinates]].
+The [[Definition:Arc Length|length]] $\d l$ of a small [[Definition:Arc of Curve|arc]] is given by:
+:$\d l = {h_1}^2 {\d q_1}^2 + {h_2}^2 {\d q_2}^2 + {h_3}^2 {\d q_3}^2$
+where:
+:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Lots of work to be done yet in this area}}
+\end{proof}<|endoftext|>
+\section{Laplacian of Function in Orthogonal Curvilinear Coordinates}
+Tags: Orthogonal Curvilinear Coordinates
+
+\begin{theorem}
+Let $\map \psi {q_1, q_2, q_3}$ denote a [[Definition:Real-Valued Function|real-valued function]] embedded in an [[Definition:Orthogonal Curvilinear Coordinates|orthogonal curvilinear coordinate system]].
+Then the [[Definition:Laplacian|Laplacian]] of $\psi$ can be expressed as:
+:$\nabla^2 \psi = \dfrac 1 {h_1 h_2 h_3} \paren {\map {\dfrac \partial {\partial q_1} } {\dfrac {h_2 h_3} {h_1} \dfrac {\partial \psi} {\partial q_1} } + \map {\dfrac \partial {\partial q_2} } {\dfrac {h_3 h_1} {h_2} \dfrac {\partial \psi} {\partial q_2} } + \map {\dfrac \partial {\partial q_3} } {\dfrac {h_1 h_2} {h_3} \dfrac {\partial \psi} {\partial q_3} } }$
+where:
+:${h_i}^2 = \paren {\dfrac {\partial x} {\partial q_i} }^2 + \paren {\dfrac {\partial y} {\partial q_i} }^2 + \paren {\dfrac {\partial z} {\partial q_i} }^2$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|a coherent understanding of exactly what it means would be useful here}}
+\end{proof}<|endoftext|>
+\section{Pressure of Gas for Density and Temperature}
+Tags: Pneumatics
+
+\begin{theorem}
+The [[Definition:Pressure|pressure]], [[Definition:Density|density]] and [[Definition:Temperature|temperature]] of a [[Definition:Body|body]] of [[Definition:Gas|gas]] are linked by the [[Definition:Equation|equation]]:
+:$p = k \rho \paren {1 + \alpha t}$
+where:
+:$p$ denotes the [[Definition:Pressure|pressure]] of the [[Definition:Gas|gas]]
+:$\rho$ denotes the [[Definition:Density|density]] of the [[Definition:Gas|gas]]
+:$t$ denotes the[[Definition:Temperature|temperature]] of the [[Definition:Gas|gas]]
+:$k$ and $\alpha$ are [[Definition:Constant|constants]].
+\end{theorem}<|endoftext|>
+\section{Shortest Distance between Two Points is Straight Line}
+Tags: Euclidean Geometry
+
+\begin{theorem}
+The shortest [[Definition:Distance (Linear Measure)|distance]] between $2$ [[Definition:Point|points]] is a [[Definition:Straight Line|straight line]].
+\end{theorem}
+
+\begin{proof}
+Let $s$ be the [[Definition:Arc Length|length]] of a [[Definition:Curve|curve]] between $2$ [[Definition:Point|points]] $A$ and $B$.
+The problem becomes one of finding the [[Definition:Curve|curve]] for which $\ds \int_a^B \rd s$ is a [[Definition:Absolute Minimum|minimum]].
+{{ProofWanted|In due course as the work progresses}}
+Hence such a [[Definition:Curve|curve]] has the [[Definition:Equation of Geometric Figure|equation]]:
+:$y = m x + c$
+which defines a [[Definition:Straight Line|straight line]].
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Circuit Axioms}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr C$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+Then $\mathscr C$ is the [[Definition:Set|set]] of [[Definition:Circuit (Matroid)|circuits]] of a [[Definition:Matroid|matroid]] on $S$ {{iff}} $\mathscr C$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Circuit Axioms (Matroid)/Definition 1|circuit axioms]]:
+{{: Leigh.Samphier/Sandbox/Definition:Circuit Axioms (Matroid)/Definition 1}}
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+{{qed|lemma}}
+=== Sufficient Condition ===
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Circuit Axioms}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr C$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+{{TFAE|def=Circuit Axioms (Matroid)|view=Matroid Circuit Axioms}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 implies Definition 2 ===
+{{qed|lemma}}
+=== Definition 2 implies Definition 1 ===
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Rank of Matroid Circuit is One Less Than Cardinality}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $C \subseteq S$ be a [[Definition:Circuit (Matroid)|circuit]] of $M$.
+Let $\rho: \powerset S \to \Z$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Then:
+:$\map \rho C = \card C -1$
+\end{theorem}
+
+\begin{proof}
+{{qed}}
+[[Category:Matroid Theory]]
+m7n5ztrqaltfh8fguktgz8q1dkx9w65
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Bound for Cardinality of Matroid Circuit}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $C \subseteq S$ be a [[Definition:Circuit (Matroid)|circuit]] of $M$.
+Let $\rho: \powerset S \to \Z$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Then:
+:$\card C \le \map \rho S + 1$
+\end{theorem}
+
+\begin{proof}
+{{qed}}
+[[Category:Matroid Theory]]
+jh9v9x5cew5hy6sp49kofbr5je244sb
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid with No Circuits Has Single Base}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]] with no [[Definition:Circuit (Matroid)|circuits]].
+Then:
+:$S$ is the only [[Definition:Base of Matroid|base]] on $M$.
+\end{theorem}
+
+\begin{proof}
+{{qed}}
+[[Category:Matroid Theory]]
+lthi01aw1ge2mhcoktt9npj8kpad8lc
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Proper Subset of Matroid Circuit is Independent}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $C \subseteq S$ be a [[Definition:Circuit (Matroid)|circuit]] of $M$.
+Then:
+:every [[Definition:Proper Subset|proper subset]] $A$ of $C$ is [[Definition:Independent Subset (Matroid)|independent]]
+\end{theorem}
+
+\begin{proof}
+{{qed}}
+[[Category:Matroid Theory]]
+s7kev08igj7spgrmz3ljon505kpyzb3
+\end{proof}<|endoftext|>
+\section{Supremum Norm is Norm/Space of Bounded Sequences}
+Tags: 
+
+\begin{theorem}
+The [[Definition:Supremum Norm|supremum norm]] on the [[Definition:Space of Bounded Sequences|space of bounded sequences]] is a [[Definition:Norm on Vector Space|norm]].
+\end{theorem}
+
+\begin{proof}
+=== Norm Axiom $(\text N 1)$ ===
+Let $x \in \ell^\infty$.
+By definition of [[Definition:Supremum Norm|supremum norm]]:
+:$\ds \norm {\mathbf x}_\infty = \sup_{n \mathop \in \N} \size {x_n}$
+The [[Definition:Complex Modulus|complex modulus]] of $x_n$ is [[Definition:Real Number|real]] and [[Complex Modulus is Non-Negative|non-negative]].
+Hence, $\norm {\mathbf x}_\infty \ge 0$.
+Suppose $\norm {\mathbf x}_\infty = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x}_\infty
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = \sup_{n \mathop \in \N} \size {x_n}
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = x_n
+      | r = 0
+      | c = [[Complex Modulus equals Zero iff Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = \bf x
+      | r = \sequence 0_{n \mathop \in \N}
+}}
+{{end-eqn}}
+Thus [[Definition:Norm Axioms|norm axiom $(\text N 1)$]] is satisfied.
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+Suppose $\alpha \in \C$.
+{{begin-eqn}}
+{{eqn| l = \norm {\alpha \cdot \mathbf x}_\infty
+     | r = \sup_{n \mathop \in \N} \size {\alpha x_n}
+}}
+{{eqn| r = \size {\alpha} \sup_{n \mathop \in \N} \size {x_n}
+}}
+{{eqn| r = \size {\alpha} \norm {\mathbf x}_\infty
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+Let $\mathbf x = \sequence {x_n}_{n \mathop \in \N}, \mathbf y = \sequence {y_n}_{n \mathop \in \N} \in \ell^\infty$.
+By [[Triangle Inequality for Complex Numbers]]:
+:$\forall n \in \N : \size {x_n + y_n} \le \size {x_n} + \size {y_n}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x + \mathbf y}_\infty
+      | r = \sup_{n \mathop \in \N} \size {x_n + y_n}
+      | c = {{defof|Supremum Norm}}
+}}
+{{eqn | o = \le
+      | r = \sup_{n \mathop \in \N} \size {x_n} + \sup_{n \mathop \in \N} \size {y_n}
+}}
+{{eqn | r = \norm {\mathbf x}_\infty + \norm {\mathbf y}_\infty
+      | c = {{defof|Supremum Norm}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallelogram Law for Vector Subtraction}
+Tags: Parallelogram Law
+
+\begin{theorem}
+Let $\mathbf u$ and $\mathbf v$ be [[Definition:Vector (Linear Algebra)|vectors]].
+Consider a [[Definition:Parallelogram|parallelogram]], two of whose adjacent sides represent $\mathbf y$ and $\mathbf v$ (in [[Definition:Magnitude|magnitude]] and [[Definition:Direction|direction]]).
+:[[File:ParallelogramLaw-Difference.png|400px]]
+Then the [[Definition:Diagonal of Quadrilateral|diagonal]] of the [[Definition:Parallelogram|parallelogram]] connecting the [[Definition:Terminal Point of Vector|terminal points]] of $\mathbf u$ and $\mathbf v$ represents the [[Definition:Magnitude|magnitude]] and [[Definition:Direction|direction]] of $\mathbf u - \mathbf v$, the [[Definition:Vector Difference|difference]] of $\mathbf u$ and $\mathbf v$.
+\end{theorem}
+
+\begin{proof}
+We can construct a [[Definition:Parallelogram|parallelogram]] as follows:
+:[[File:Vector-difference-parallelogram.png|400px]]
+and the construction is apparent.
+\end{proof}<|endoftext|>
+\section{Space of Bounded Sequences with Supremum Norm forms Normed Vector Space}
+Tags: Examples of Normed Vector Spaces
+
+\begin{theorem}
+[[Definition:Space of Bounded Sequences|Space of bounded sequences]] with [[Definition:Supremum Norm|supremum norm]] forms a [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Space of Bounded Sequences with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space|Space of bounded sequences is a vector space]]
+:[[Supremum Norm is Norm/Space of Bounded Sequences|Supremum norm on the space of bounded sequences is a norm]]
+By definition, $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is a [[Definition:Normed Vector Space|normed vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riesz's Lemma}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $X$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $Y$ be a [[Definition:Proper Subset|proper]] [[Definition:Closed Linear Subspace|closed linear subspace]] of $X$. 
+Let $\alpha \in \openint 0 1$.
+Then there exists $x_\alpha \in X$ such that: 
+:$\norm {x_\alpha} = 1$
+with:
+:$\norm {x_\alpha - y} > \alpha$
+for all $y \in Y$.
+\end{theorem}
+
+\begin{proof}
+Since $Y < X$:
+:$X \setminus Y$ is [[Definition:Non-Empty Set|non-empty]]. 
+Since $Y$ is [[Definition:Closed Set|closed]]:
+:$X \setminus Y$ is [[Definition:Open Set|open]]. 
+Let $x \in X \setminus Y$, then there exists $\epsilon > 0$ such that: 
+:$\map {B_\epsilon} x \subset X \setminus Y$
+So, for all $y \in Y$, we must have: 
+:$\norm {x - y} \ge \epsilon$
+That is: 
+:$\inf \left\{\norm {x - y} \colon y \in Y\right\} \ge \epsilon$
+For brevity, let:
+:$d = \inf \left\{\norm {x - y} \colon y \in Y\right\}$
+Since $\alpha^{-1} > 1$, there exists $z \in Y$ with: 
+:$\norm {x - z} < d \alpha^{-1}$
+otherwise the [[Definition:Infimum of Real-Valued Function|infimum]] would be at least $d \alpha^{-1}$, a contradiction.
+Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$.
+So, we can set: 
+:$\displaystyle x_\alpha = \frac {x - z} {\norm {x - z} }$
+Clearly:
+:$\displaystyle \norm {x_\alpha} = \frac {\norm {x - z} } {\norm {x - z} } = 1$
+Now, for any $y \in Y$ we have: 
+{{begin-eqn}}
+{{eqn	| l = \norm {x_\alpha - y}
+	| r = \norm {\frac {x - z} {\norm {x - z} } - y}
+}}
+{{eqn	| r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }
+}}
+{{end-eqn}}
+We've already seen that: 
+:$\norm {x - z} < d \alpha^{-1}$
+Since $Y$ is closed under linear combination, we have:
+:$z + \norm {x - z} y \in Y$
+and so: 
+:$\norm {x - \paren {z + \norm {x - z} y} } \ge d$
+We conclude that: 
+{{begin-eqn}}
+{{eqn	| l = \norm {x_\alpha - y}
+	| r = \frac {\norm {x - \paren {z + \norm {x - z} y} } } {\norm {x - z} }
+}}
+{{eqn	| o = >
+	| r = \frac d {d \alpha^{-1} } 
+}}
+{{eqn	| r = \alpha
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Frigyes Riesz|cat = Riesz F}}
+[[Category:Functional Analysis]]
+2dtquazxgxfx8hgz6crolrqva0ijaoz
+\end{proof}<|endoftext|>
+\section{Normed Vector Space is Finite Dimensional iff Unit Sphere is Compact}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $X$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $S = \map {S_1} 0$ be the [[Definition:Sphere in Normed Vector Space|unit sphere]] centred at $0$ in $X$.
+Then $X$ is [[Definition:Finite Dimensional Vector Space|finite dimensional]] {{iff}} $S$ is [[Definition:Compact Subset of Normed Vector Space|compact]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Functional Analysis]]
+63952ycd1i0rp508lbkg3nkn2rejl4w
+\end{proof}<|endoftext|>
+\section{Finite Dimensional Subspace of Normed Vector Space is Closed}
+Tags: Functional Analysis
+
+\begin{theorem}
+Let $V$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $W$ be a [[Definition:Finite Dimensional Vector Space|finite dimensional]] [[Definition:Vector Subspace|subspace]] of $V$. 
+Then $W$ is [[Definition:Closed Set of Normed Vector Space|closed]].
+\end{theorem}
+
+\begin{proof}
+Suppose that $\dim W = n$.
+Let: 
+:$\set {e_1, e_2, \ldots, e_n}$
+be a [[Definition:Basis of Vector Space|basis]] for $W$.
+{{AimForCont}} that $W$ is not a [[Definition:Closed Set of Normed Vector Space|closed set]].
+Then there exists a [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]] $\sequence {w_k}$ in $W$ such that: 
+:$w_k \to w$
+where $w \in V \setminus W$.
+{{explain|What does the notation $w_k \to w$ actually mean? As it stands, we are asked to accept at the end of this that $0 \to 1$ is "impossible", but that cannot be recognised as such unless we know what the notation means.}}
+Note that:
+:$\set {e_1, e_2, \ldots, e_n}$
+is [[Definition:Linearly Independent Set|linearly independent]] in $W$, and hence $V$.
+Note that since $w \not \in W$, $w$ cannot be written as a [[Definition:Linear Combination|linear combination]] of elements of $\set {e_1, e_2, \ldots, e_n}$.
+So:
+:$\set {e_1, e_2, \ldots, e_n, w}$
+is [[Definition:Linearly Independent Set|linearly independent]] in $V$.
+So consider the [[Definition:Vector Subspace|subspace]]:
+:$W^* = \mathrm {span} \set {e_1, e_2, \ldots, e_n, w}$.
+Using the [[Definition:Sequence|sequence]] $\sequence {w_n}$ from before, write: 
+:$w_k = \tuple {w_k^{\paren 1}, w_k^{\paren 2}, \ldots, w_k^{\paren n}, 0} \in W^*$
+and:
+:$w = \tuple {0, 0, \ldots, 0, 1} \in W^*$
+We necessarily have: 
+:$w_k^{\paren j} \to 0$
+for each $1 \le j \le n$.
+However, we would also have: 
+:$0 \to 1$
+Clearly this is impossible, so we have derived a contradiction.
+So $W$ is necessarily [[Definition:Closed Set of Normed Vector Space|closed]].
+{{qed}}
+[[Category:Functional Analysis]]
+2opja2acq5674ufrwguwbahy9fh4u7e
+\end{proof}<|endoftext|>
+\section{Catalan's Conjecture}
+Tags: Number Theory
+
+\begin{theorem}
+The only [[Definition:Solution to Equation|solution]] to the [[Definition:Diophantine Equation|Diophantine equation]]:
+:$x^a - y^b = 1$
+for $a, b > 1$ and $x, y > 0$, is:
+:$x = 3, a = 2, y = 2, b = 3$
+\end{theorem}<|endoftext|>
+\section{Total Force on Charged Particle from 2 Charged Particles}
+Tags: Electric Charge
+
+\begin{theorem}
+Let $p_1$, $p_2$ and $p_3$ be [[Definition:Charged Particle|charged particles]].
+Let $q_1$, $q_2$ and $q_3$ be the [[Definition:Electric Charge|electric charges]] on $p_1$, $p_2$ and $p_3$ respectively.
+Let $\mathbf F_{ij}$ denote the [[Definition:Force|force]] exerted on $q_j$ by $q_i$.
+Let $\mathbf F_i$ denote the [[Definition:Force|force]] exerted on $q_i$ by the combined action of the other two [[Definition:Charged Particle|charged particles]].
+Then the [[Definition:Force|force]] $\mathbf F_1$ exerted on $q_1$ by the combined action of $q_2$ and $q_3$ is given by:
+{{begin-eqn}}
+{{eqn | l = \mathbf F_1
+      | r = \mathbf F_{21} + \mathbf F_{31}
+      | c = 
+}}
+{{eqn | r = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}
+      | c = 
+}}
+{{end-eqn}}
+where:
+:$\mathbf F_{21} + \mathbf F_{31}$ denotes the [[Definition:Vector Sum|vector sum]] of $\mathbf F_{21}$ and $\mathbf F_{31}$
+:$\mathbf r_{ij}$ denotes the [[Definition:Displacement|displacement]] from $p_i$ to $p_j$
+:$r_{ij}$ denotes the [[Definition:Distance (Linear Measure)|distance]] between $p_i$ and $p_j$
+:$\varepsilon_0$ denotes the [[Definition:Vacuum Permittivity|vacuum permittivity]].
+\end{theorem}
+
+\begin{proof}
+:[[File:Two-charges-on-another.png|500px]]
+By definition, the [[Definition:Force|force]] $\mathbf F_{21}$ and $\mathbf F_{31}$ are [[Definition:Vector Quantity|vector quantities]].
+Hence their [[Definition:Resultant|resultant]] can be found by using the [[Parallelogram Law]].
+The result follows from [[Coulomb's Law of Electrostatics]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Total Force on Charged Particle from Multiple Charged Particles}
+Tags: Electric Charge
+
+\begin{theorem}
+Let $p_1, p_2, \ldots, p_n$ be [[Definition:Charged Particle|charged particles]].
+Let $q_1, q_2, \ldots, q_n$ be the [[Definition:Electric Charge|electric charges]] on $p_1, p_2, \ldots, p_n$ respectively.
+For all $i$ in $\set {1, 2, \ldots, n}$ where $i \ne j$, let $\mathbf F_{i j}$ denote the [[Definition:Force|force]] exerted on $q_j$ by $q_i$.
+For all $i$ in $\set {1, 2, \ldots, n}$, let $\mathbf F_i$ denote the [[Definition:Force|force]] exerted on $q_i$ by the combined action of all the other [[Definition:Charged Particle|charged particles]].
+Then the [[Definition:Force|force]] $\mathbf F_i$ exerted on $q_i$ by the combined action of all the other [[Definition:Charged Particle|charged particles]] is given by:
+{{begin-eqn}}
+{{eqn | l = \mathbf F_i
+      | r = \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \mathbf F_{j i}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}
+      | c = 
+}}
+{{end-eqn}}
+where:
+:the [[Definition:Summation|summation]] denotes the [[Definition:Vector Sum|vector sum]] of $\mathbf F_{21}$ and $\mathbf F_{31}$
+:$\mathbf r_{ij}$ denotes the [[Definition:Displacement|displacement]] from $p_i$ to $p_j$
+:$r_{ij}$ denotes the [[Definition:Distance (Linear Measure)|distance]] between $p_i$ and $p_j$
+:$\varepsilon_0$ denotes the [[Definition:Vacuum Permittivity|vacuum permittivity]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
+$\map P 2$ is the case:
+$\mathbf F_1 = \dfrac 1 {4 \pi \varepsilon_0} \dfrac {q_1 q_2} {r_{2 1}^3} \mathbf r_{2 1}$
+which is [[Coulomb's Law of Electrostatics]].
+Thus $\map P 0$ is seen to hold.
+=== Basis for the Induction ===
+$\map P 3$ is the case:
+:$\mathbf F_1 = \dfrac {q_2 q_1} {4 \pi \varepsilon_0 r_{2 1}^3} \mathbf r_{2 1} + \dfrac {q_3 q_1} {4 \pi \varepsilon_0 r_{3 1}^3} \mathbf r_{3 1}$
+which is proved in [[Total Force on Charged Particle from 2 Charged Particles]].
+Thus $\map P 3$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k  \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
+from which it is to be shown that:
+:$\ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1}  \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf F_i
+      | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le k \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i} + \dfrac {q_i q_{k + 1} } {4 \pi \varepsilon_0 r_{\paren {k + 1} i}^3}  \mathbf r_{\paren {k + 1} i}
+      | c = [[Total Force on Charged Particle from Multiple Charged Particles#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le {k + 1} \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}
+      | c = {{Defof|Summation}}
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 2}: \ds \mathbf F_i = \dfrac 1 {4 \pi \varepsilon_0} \sum_{\substack {1 \mathop \le j \mathop \le n \\ i \mathop \ne j} } \dfrac {q_i q_j} {r_{j i}^3} \mathbf r_{j i}$
+</onlyinclude>
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Scalar Multiplication by Zero gives Zero Vector}
+Tags: Zero Vectors, Scalar Multiplication
+
+\begin{theorem}
+Let $\mathbf a$ be a [[Definition:Vector Quantity|vector quantity]].
+Let $0 \mathbf a$ denote the [[Definition:Scalar Multiplication on Vector Quantity|scalar product]] of $\mathbf a$ with $0$.
+Then:
+:$0 \mathbf a = \bszero$
+where $\bszero$ denotes the [[Definition:Zero Vector Quantity|zero vector]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Scalar Multiplication on Vector Quantity|scalar product]]:
+:$\size {0 \mathbf a} = 0 \size {\mathbf a}$
+where $\size {\mathbf a}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf a$.
+Thus:
+:$\size {0 \mathbf a} = 0$
+That is: $0 \mathbf a$ is a [[Definition:Vector Quantity|vector quantity]] whose [[Definition:Magnitude|magnitude]] is [[Definition:Zero (Number)|zero]].
+Hence, by definition, $0 \mathbf a$ is the [[Definition:Zero Vector|zero vector]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm forms Normed Vector Space}
+Tags: Examples of Normed Vector Spaces
+
+\begin{theorem}
+[[Definition:Space of Continuous Functions of Differentiability Class k|Space of Continuously Differentiable on Closed Interval Real-Valued Functions]] with [[Definition:C^k Norm|$C^1$ norm]] forms a [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+Let $I := \closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+We have that:
+:[[Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space]]
+:[[C^k Norm is Norm|$\map {C^1} I$ norm on the space of continuously differentiable on closed interval real-valued functions is a norm]]
+By definition, $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a [[Definition:Normed Vector Space|normed vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Endomorphisms is Ring with Unity}
+Tags: Rings of Endomorphisms
+
+\begin{theorem}
+Let $\struct {G, \oplus}$ be an [[Definition:Abelian Group|abelian group]].
+Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Group Endomorphism|group endomorphisms]] of $\struct {G, \oplus}$.
+Let $\struct {\mathbb G, \oplus, *}$ denote the '''[[Definition:Ring of Endomorphisms|ring of endomorphisms]]''' on $\struct {G, \oplus}$.
+Then $\struct {\mathbb G, \oplus, *}$ is a [[Definition:Ring with Unity|ring with unity]].
+\end{theorem}
+
+\begin{proof}
+By [[Structure Induced by Group Operation is Group]], $\struct {\mathbb G, \oplus}$ is an [[Definition:Abelian Group|abelian group]].
+By [[Set of Homomorphisms is Subgroup of All Mappings]], it follows that $\struct {\mathbb G, \oplus}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G^G, \oplus}$.
+Next, we establish that $*$ is [[Definition:Associative Operation|associative]].
+By definition, $\forall u, v \in \mathbb G: u * v = u \circ v$ where $u \circ v$ is defined as [[Definition:Composition of Mappings|composition of mappings]].
+[[Definition:Associative Operation|Associativity]] of $*$ follows directly from [[Composition of Mappings is Associative]].
+Next, we establish that $*$ is [[Definition:Distributive Operation|distributive]] over $\oplus$.
+Let $u, v, w \in \mathbb G$.
+Then:
+:$\paren {u \oplus v} * w = \paren {u \oplus v} \circ w$
+:$u * \paren {v \oplus w} = u \circ \paren {v \oplus w}$
+So let $x \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\paren {u \oplus v} * w} } x
+      | r = \map {\paren {\paren {u \oplus v} \circ w} } x
+      | c = 
+}}
+{{eqn | r = \map {\paren {u \oplus v} } {\map w x}
+      | c = 
+}}
+{{eqn | r = \map u {\map w x} \oplus \map v {\map w x}
+      | c = 
+}}
+{{eqn | r = \map {\paren {u \circ w} } x \oplus \map {\paren {v \circ w} } x
+      | c = 
+}}
+{{eqn | r = \map {\paren {u * w} } x \oplus \map {\paren {v * w} } x
+      | c = 
+}}
+{{end-eqn}}
+So $\paren {u \oplus v} * w = \paren {u * w} \oplus \paren {v * w}$.
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {u * \paren {v \oplus w} } } x
+      | r = \map {\paren {u \circ \paren {v \oplus w} } } x
+      | c = 
+}}
+{{eqn | r = \map u {\map {\paren {v \oplus w} } x}
+      | c = 
+}}
+{{eqn | r = \map u {\map v x \oplus \map w x}
+      | c = 
+}}
+{{eqn | r = \map u {\map v x} \oplus \map u {\map w x}
+      | c = $u$ has the [[Definition:Morphism Property|morphism property]]
+}}
+{{eqn | r = \map {\paren {u \circ v} } x \oplus \map {\paren {u \circ w} } x
+      | c = 
+}}
+{{eqn | r = \map {\paren {u * v} } x \oplus \map {\paren {u * w} } x
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$u * \paren {v \oplus w} = \paren {u * v} \oplus \paren {u * w}$
+Thus $*$ is [[Definition:Distributive Operation|distributive]] over $\oplus$.
+The [[Definition:Ring Axioms|ring axioms]] are satisfied, and $\struct {\mathbb G, \oplus, *}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+The [[Definition:Ring Zero|zero]] is easily checked to be the [[Definition:Mapping|mapping]] which takes everything to the [[Definition:Identity Element|identity]]:
+:$e: G \to \set {e_G}: \map e x = e_G$
+The [[Definition:Unity of Ring|unity]] is easily checked to be the [[Definition:Identity Mapping|identity mapping]], which is known to be an [[Identity Mapping is Automorphism|automorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Endomorphisms is not necessarily Commutative Ring}
+Tags: Rings of Endomorphisms
+
+\begin{theorem}
+Let $\struct {G, \oplus}$ be an [[Definition:Abelian Group|abelian group]].
+Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Group Endomorphism|group endomorphisms]] of $\struct {G, \oplus}$.
+Let $\struct {\mathbb G, \oplus, *}$ denote the '''[[Definition:Ring of Endomorphisms|ring of endomorphisms]]''' on $\struct {G, \oplus}$.
+Then $\struct {\mathbb G, \oplus, *}$ is not necessarily a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
+\end{theorem}
+
+\begin{proof}
+We have that $\struct {\mathbb G, \oplus, *}$ is a [[Definition:Ring with Unity|ring with unity]].
+It remains to show that the operation $*$ is not necessarily [[Definition:Commutative Operation|commutative]].
+{{ProofWanted}}
+[[Category:Rings of Endomorphisms]]
+5h75pwdez5hb75bqi74jytghgcmbg3v
+\end{proof}<|endoftext|>
+\section{Set of Endomorphisms of Non-Abelian Group is not Ring}
+Tags: Rings of Endomorphisms
+
+\begin{theorem}
+Let $\struct {G, \oplus}$ be a [[Definition:Group|group]] which is non-[[Definition:Abelian Group|abelian]].
+Let $\mathbb G$ be the [[Definition:Set|set]] of all [[Definition:Group Endomorphism|group endomorphisms]] of $\struct {G, \oplus}$.
+Let $*: \mathbb G \times \mathbb G \to \mathbb G$ be the [[Definition:Binary Operation|operation]] defined as:
+:$\forall u, v \in \mathbb G: u * v = u \circ v$
+where $u \circ v$ is defined as [[Definition:Composition of Mappings|composition of mappings]].
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\mathbb G, \oplus, *}$ is not a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+In order to be a [[Definition:Ring (Abstract Algebra)|ring]], it is [[Definition:Necessary Condition|necessary]] that the [[Definition:Ring Addition|additive operation]] $\oplus$ is [[Definition:Commutative Operation|commutative]].
+However, as $\struct {G, \oplus}$ is specifically defined as being non-[[Definition:Abelian Group|abelian]], [[Definition:A Fortiori|a fortiori]] $\oplus$ is not [[Definition:Commutative Operation|commutative]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Positive Integers does not form Ring}
+Tags: Integers, Examples of Rings
+
+\begin{theorem}
+Let $\Z_{\ge 0}$ denote the [[Definition:Set|set]] of [[Definition:Positive Integer|positive integers]].
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z_{\ge 0}, +, \times}$ does not form a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+For $\struct {\Z_{\ge 0}, +, \times}$ to be a [[Definition:Ring (Abstract Algebra)|ring]], it is [[Definition:Necessary Condition|necessary]] for the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z_{\ge 0}, +}$ to form a [[Definition:Group|group]].
+But from the [[Natural Numbers under Addition do not form Group/Corollary|corollary to Natural Numbers under Addition do not form Group]]:
+:$\struct {\Z_{\ge 0}, +}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm is Banach Space}
+Tags: Banach Spaces
+
+\begin{theorem}
+Let $I := \closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\map \CC I$ be the [[Definition:Space of Continuous on Closed Interval Real-Valued Functions|space of real-valued functions, continuous]] on $I$.
+Let $\map {\CC^1} I$ be the [[Definition:Space of Continuous Functions of Differentiability Class k|space of real-valued functions, continuously differentiable]] on $I$.
+Let $\norm {\, \cdot \,}_{1, \infty}$ be the [[Definition:C^k Norm|$\CC^1$ norm]].
+$\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ be the [[Space of Continuously Differentiable on Closed Interval Real-Valued Functions with C^1 Norm forms Normed Vector Space|normed space of real-valued functions, continuously differentiable on]] $I$.
+Then $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ is a [[Definition:Banach Space|Banach space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$:
+:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m}_{1, \infty} < \epsilon$
+=== $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ to $x$ ===
+$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n - x_m}_\infty
+      | o = \le
+      | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty
+}}
+{{eqn | r = \norm {x_n - x_m}_{1, \infty}
+      | c = {{defof|C^k Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+      | c = {{defof|Cauchy Sequence}}
+}}
+{{end-eqn}}
+Hence, $\sequence {x_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$.
+We have that [[Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm form Banach Space|$\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ is a Banach space]].
+Therefore, $\sequence {x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]].
+Denote this [[Definition:Limit of Sequence in Normed Vector Space|limit]] as $x$:
+:$\ds \lim_{n \mathop \to \infty} x_n = x$
+where $x \in \map \CC I$.
+{{qed|lemma}}
+=== $\sequence {x'_n}_{n \mathop \in \N}$ converges in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ to $y$ ===
+$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N$ we have that:
+{{begin-eqn}}
+{{eqn | l = \norm {x'_n - x'_m}_\infty
+      | o = \le
+      | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty
+}}
+{{eqn | r = \norm {x_n - x_m}_{1, \infty}
+      | c = {{defof|C^k Norm}}
+}}
+{{end-eqn}}
+Hence, $\sequence {x'_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$.
+We have that [[Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm form Banach Space|$\struct {\map \CC I, \norm {\, \cdot \,}_\infty}$ is a Banach space]].
+Therefore, $\sequence {x'_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]].
+Denote this [[Definition:Limit of Sequence in Normed Vector Space|limit]] as $y$:
+:$\ds \lim_{n \mathop \to \infty} x'_n = y$
+where $y \in \map \CC I$.
+{{qed|lemma}}
+=== $x$ is differentiable, and $y' = x$ ===
+Let $t \in I$.
+By [[Fundamental Theorem of Calculus|fundamental theorem of calculus]] $\paren \star$:
+:$\ds \map {x_n} t - \map {x_n} a = \int_a^t \map {x'_n} \tau \rd \tau$
+Consider the following difference:
+:$\ds \size {\map {x_n} t - \map {x_n} a - \int_a^t \map y \tau \rd \tau}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\map {x_n} t - \map {x_n} a - \int_a^t \map y \tau \rd \tau}
+      | r = \size {\int_a^t \paren {\map {x'_n} \tau - \map y \tau} \rd \tau}
+      | c = Substitution of $\paren \star$
+}}
+{{eqn | o = \le 
+      | r = \int_a^t \size {\map {x'_n} \tau - \map y \tau} \rd \tau
+}}
+{{eqn | o = \le 
+      | r = \int_a^t \sup_{T \mathop \in I} \size {\map {x'_n} T - \map y T} \rd \tau
+}}
+{{eqn | o = \le
+      | r = \norm {x'_n - y}_\infty \paren {t - a}
+      | c = {{defof|Supremum Norm}}
+}}
+{{end-eqn}}
+[[Definition:Absolute Value|Absolute value]] is a [[Definition:Continuous Real Function|continuous function]].
+Therefore, we can take the [[Limit of Composite Function|limit of the composite function]].
+Passing the limit, as $n$ goes to infinity, gives:
+{{begin-eqn}}
+{{eqn | l = \size {\map x t - \map x a - \int_a^t \map y \tau \rd \tau}
+      | o = \le
+      | r = \norm {y - y}_\infty \paren {t - a}
+      | c = $\ds \sequence {x'_n}_{n \mathop \in \N}$ converges to $y$
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+In other words:
+:$\ds \map x t = \map x a + \int_a^t \map y \tau \rd \tau$
+By [[Fundamental Theorem of Calculus|fundamental theorem of calculus]], $x$ is a [[Definition:Primitive of Real Function|primitive]] of $y$.
+Thus, $x' = y \in \map \CC I$.
+Hence, $x \in \map {\CC^1} I$.
+{{qed|lemma}}
+=== $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\map {\CC^1} I, \norm {\, \cdot \,}_{1, \infty} }$ to $x$ ===
+Let $\epsilon \in \R_{> 0}$.
+Let $N \in \N : \forall m, n \in \N : m, n > N \implies \norm { x_n - x_m}_{1, \infty} < \epsilon$.
+Then for all $t \in I$ we have that:
+{{begin-eqn}}
+{{eqn | l = \size {\map {x_n} t - \map {x_m} t} + \size {\map {x'_n} t - \map {x'_m} t}
+      | o = \le
+      | r = \sup_{t \mathop \in I} \size {\map {x_n} t - \map {x_m} t}+ \sup_{t \mathop \in I} \size {\map {x'_n} t - \map {x'_m} t}
+}}
+{{eqn | r = \norm {x_n - x_m}_\infty + \norm {x'_n - x'_m}_\infty
+      | c = {{defof|Supremum Norm}}
+}}
+{{eqn | r = \norm {x_n - x_m}_{1, \infty}
+      | c = {{defof|C^k Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+[[Definition:Absolute Value|Absolute value]] is a [[Definition:Continuous Real Function|continuous function]].
+Therefore, we can take the [[Limit of Composite Function|limit of the composite function]].
+Letting $m$ go to [[Definition:Infinity|infinity]], it follows that:
+:$\forall n > N : \size {\map {x_n} t - \map x t} + \size {\map {x'_n} t - \map {x'} t} < \epsilon$
+Since $t \in I$ was arbitrary:
+{{begin-eqn}}
+{{eqn | l = \size {\map {x_n} t - \map x t} + \size {\map {x'_n} t - \map {x'} t} 
+      | o = \le
+      | r = \sup_{t \mathop \in I} \size {\map {x_n} t - \map x t} + \sup_{t \mathop \in I} \size {\map {x'_n} t - \map {x'} t}
+}}
+{{eqn | r = \norm {x_n - x}_{1, \infty}
+      | c = {{defof|Supremum Norm}}
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers whose Denominators are not Divisible by 4 do not form Ring}
+Tags: Rational Numbers, Examples of Rings
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] defined as:
+:$S = \set {\dfrac m n : m, n \in \Z, m \perp n, 4 \nmid n}$
+That is, $S$ is defined as the [[Definition:Set|set]] of [[Definition:Rational Number|rational numbers]] such that, when expressed in [[Definition:Canonical Form of Rational Number|canonical form]], their [[Definition:Denominator|denominators]] are not [[Definition:Divisor of Integer|divisible]] by $4$.
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, +, \times}$ is not a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+For $\struct {S, +, \times}$ to be a [[Definition:Ring (Abstract Algebra)|ring]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is a [[Definition:Semigroup|semigroup]].
+For $\struct {S, \times}$ to be a [[Definition:Semigroup|semigroup]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+That is:
+:$\forall x, y \in S: x \times y \in S$
+Let $x = \dfrac 1 2$ and $y = \dfrac 3 2$.
+Both $x$ and $y$ are in $S$, as both are [[Definition:Rational Number|rational numbers]] expressed in [[Definition:Canonical Form of Rational Number|canonical form]] whose [[Definition:Denominator|denominators]] are not [[Definition:Divisor of Integer|divisible]] by $4$.
+But then:
+:$x \times y = \dfrac 1 2 \times \dfrac 3 2 = \dfrac 3 4$
+which is a [[Definition:Rational Number|rational numbers]] expressed in [[Definition:Canonical Form of Rational Number|canonical form]] whose [[Definition:Denominator|denominator]] ''is'' [[Definition:Divisor of Integer|divisible]] by $4$.
+Hence $x \times y \notin S$ and so $\struct {S, \times}$ is not [[Definition:Closed Algebraic Structure|closed]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order 2 Square Matrices with Zero Diagonals do not form Ring}
+Tags: Square Matrices, Examples of Rings
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] whose [[Definition:Diagonal Element|diagonal elements]] are [[Definition:Zero (Number)|zero]].
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, +, \times}$ is not a [[Definition:Ring (Abstract Algebra)|ring]].
+Note that $\times$ denotes [[Definition:Matrix Product (Conventional)|conventional matrix multiplication]].
+\end{theorem}
+
+\begin{proof}
+For $\struct {S, +, \times}$ to be a [[Definition:Ring (Abstract Algebra)|ring]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is a [[Definition:Semigroup|semigroup]].
+For $\struct {S, \times}$ to be a [[Definition:Semigroup|semigroup]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+That is:
+:$\forall x, y \in S: x \times y \in S$
+Let $x = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix}$ and $y = \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix}$.
+Both $x$ and $y$ are in $S$, as both are [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] whose [[Definition:Diagonal Element|diagonal elements]] are [[Definition:Zero (Number)|zero]].
+But then:
+:$x y = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix} \begin {pmatrix} 0 & 2 \\ 2 & 0 \end {pmatrix} = \begin {pmatrix} 2 & 0 \\ 0 & 2 \end {pmatrix}$
+which is a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $2$]] whose [[Definition:Diagonal Element|diagonal elements]] are ''not'' [[Definition:Zero (Number)|zero]].
+Hence $x y \notin S$ and so $\struct {S, \times}$ is not [[Definition:Closed Algebraic Structure|closed]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Order m times n Matrices does not form Ring}
+Tags: Matrices, Examples of Rings
+
+\begin{theorem}
+Let $m, n \in \N_{>0}$ be non-[[Definition:Zero (Number)|zero]] [[Definition:Natural Number|natural numbers]] such that $m > n$.
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Matrix|matrices]] of [[Definition:Order of Matrix|order $m \times n$]].
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, +, \times}$ is not a [[Definition:Ring (Abstract Algebra)|ring]].
+Note that $\times$ denotes [[Definition:Matrix Product (Conventional)|conventional matrix multiplication]].
+\end{theorem}
+
+\begin{proof}
+For $\struct {S, +, \times}$ to be a [[Definition:Ring (Abstract Algebra)|ring]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is a [[Definition:Semigroup|semigroup]].
+For $\struct {S, \times}$ to be a [[Definition:Semigroup|semigroup]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+That is:
+:$\forall x, y \in S: x \times y \in S$
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Element|elements]] of $S$.
+The [[Definition:Matrix Product (Conventional)|matrix multiplication operation]] is defined on $\mathbf A$ and $\mathbf B$ {{iff}} $\mathbf A$ is of [[Definition:Order of Matrix|order]] $m \times p$ and $\mathbf A$ is of [[Definition:Order of Matrix|order]] $p \times n$, for some $m, n, p \in \N_{>0}$.
+That is, the second [[Definition:Dimensions of Matrix|dimension]] of $\mathbf A$ must be equal to the first [[Definition:Dimensions of Matrix|dimension]] of $\mathbf B$.
+But in this case, the second [[Definition:Dimensions of Matrix|dimension]] of $\mathbf A$ is $n$, and the first [[Definition:Dimensions of Matrix|dimension]] of $\mathbf B$ is $m$.
+As we have been given that $m \ne n$, it follows that [[Definition:Matrix Product (Conventional)|matrix multiplication operation]] is not defined on $S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Order 3 Vectors under Cross Product does not form Ring}
+Tags: Vectors, Examples of Rings
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Vector|vectors]] in a [[Definition:Vector Space|vector space]] of [[Definition:Dimension (Linear Algebra)|dimension]] $3$.
+Let $\times$ denote the [[Definition:Vector Cross Product|cross product operation]].
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, +, \times}$ is not a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+For $\struct {S, +, \times}$ to be a [[Definition:Ring (Abstract Algebra)|ring]], it is a [[Definition:Necessary Condition|necessary condition]] that $\struct {S, \times}$ is a [[Definition:Semigroup|semigroup]].
+For $\struct {S, \times}$ to be a [[Definition:Semigroup|semigroup]], it is a [[Definition:Necessary Condition|necessary condition]] that $\times$ is [[Definition:Associative Operation|associative]] on $S$.
+However, from [[Vector Cross Product is not Associative]], this is not the case here.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Subtraction do not form Semigroup}
+Tags: Integer Subtraction, Examples of Semigroups
+
+\begin{theorem}
+Let $\struct {\Z, -}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by the set of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Subtraction|subtraction]].
+Then $\struct {\Z, -}$ is not a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+It is to be demonstrated that $\struct {\Z, -}$ does not satisfy the [[Definition:Semigroup Axioms|semigroup axioms]].
+We then have [[Subtraction on Numbers is Not Associative]].
+So, for example:
+:$3 - \paren {2 - 1} = 2 \ne \paren {3 - 2} - 1 = 0$
+Thus it has been demonstrated that $\struct {\Z, -}$ does not satisfy {{SemigroupAxiom|1}}.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Necessary Condition}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\rho : \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of a [[Definition:Matroid|matroid]] on $S$.
+Then $\rho$ satisfies the [[Definition:Rank Axioms (Matroid)/Definition 1|rank axioms]]:
+{{:Definition:Rank Axioms (Matroid)/Definition 1}}
+\end{theorem}
+
+\begin{proof}
+Let $\rho$ be the [[Definition:Rank Function (Matroid)|rank function]] of a [[Definition:Matroid|matroid]] $M = \struct{S, \mathscr I}$ on $S$
+==== $\rho$ satisfies $(\text R 1)$ ====
+$\rho$ satisfies $(\text R 1)$ follows immediately from [[Rank of Empty Set is Zero]].
+{{qed|lemma}}
+==== $\rho$ satisfies $(\text R 2)$ ====
+Let:
+:$X \subseteq S$
+:$y \in S$
+From [[Rank Function is Increasing]]:
+:$\map \rho X \le \map \rho {X \cup \set y}$
+Let $Y$ be a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $X \cup \set y$.
+From [[Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set]]
+:$\card Y = \map \rho {X \cup \set y}$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$Y \setminus \set y \in \mathscr I$
+We now consider two cases.
+===== Case 1: $y \in Y$ =====
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \rho X
+      | o = \ge 
+      | r = \card{Y \setminus \set y}
+      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
+}}
+{{eqn | r = \card Y - \card{Y \cap \set y}
+      | c = [[Cardinality of Set Difference]]
+}}
+{{eqn | r = \card Y - \card{\set y}
+      | c = [[Singleton of Element is Subset]]
+}}
+{{eqn | r = \card Y - 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn |ll = \leadsto
+      | l = \map \rho X + 1 
+      | o = \ge
+      | r = \card Y
+      | c = Add one to both sides
+}}
+{{end-eqn}}
+{{qed|lemma}}
+===== Case 2: $y \notin Y$ =====
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \rho X + 1
+      | o = > 
+      | r = \map \rho X
+}}
+{{eqn | o = \ge 
+      | r = \card{Y \setminus \set y}
+      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
+}}
+{{eqn | r = \card Y - \card{Y \cap \set y}
+      | c = [[Cardinality of Set Difference]]
+}}
+{{eqn | r = \card Y - \O
+      | c = [[Intersection With Singleton is Disjoint if Not Element]]
+}}
+{{eqn | r = \card Y - 0
+      | c = [[Cardinality of Empty Set]]
+}}
+{{eqn | r = \card Y
+}}
+{{end-eqn}}
+{{qed|lemma}}
+In either case:
+:$\map \rho X + 1 \ge \card Y = \map \rho {X \cup \set y}$
+{{qed|lemma}}
+==== $\rho$ satisfies $(\text R 3)$ ====
+Let:
+:$X \subseteq S$
+:$y, z \in S$
+===== Case 1: $y, z \notin X$ =====
+Let $A = X \cup \set{y, z}$.
+From [[Rank Function is Increasing]]:
+:$\map \rho A \ge \map \rho X$
+We now prove the [[Definition:Contrapositive Statement|contrapositive statement]]:
+:$\map \rho A > \map \rho X \implies$ either $\map \rho {X \cup \set y} > \map \rho X$ or $\map \rho {X \cup \set z} > \map \rho X$
+Let $\map \rho A > \map \rho X$.
+Let $Y$ be a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $X$.
+Let $Z$ be a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $A$.
+From [[Leigh.Samphier/Sandbox/Cardinality of Maximal Independent Subset Equals Rank of Set]]:
+:$\card Y = \map \rho X < \map \rho A = \card Z$
+From [[Independent Set can be Augmented by Larger Independent Set]] there exists [[Definition:Non-Empty Set|non-empty set]] $Z' \subseteq Z \setminus Y$:
+:$Y \cup Z' \in \mathscr I$
+:$\card{Y \cup Z'} = \card Z$
+{{AimForCont}}:
+:$y, z \notin Z'$
+We have:
+{{begin-eqn}}
+{{eqn | l = Z'
+      | r = Z' \cap Z
+}}
+{{eqn | o = \subseteq
+      | r = Z' \cap A
+}}
+{{eqn | r = Z' \cap \paren{X \cup \set {y, z} }
+}}
+{{eqn | r = \paren{Z' \cap  X} \cup \paren{Z' \cap \set {y, z} }
+}}
+{{eqn | r = \paren{Z' \cap X} \cup \O
+}}
+{{eqn | r = \paren{Z' \cap X}
+}}
+{{eqn | o = \subseteq
+      | r = X
+}}
+{{eqn | ll = \leadsto
+      | l = Y \cup Z'
+      | o = \subseteq
+      | r = X
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \card{Y \cup Z'}
+      | r = \card Z
+}}
+{{eqn | r = \map \rho A
+}}
+{{eqn | o = > 
+      | r = \map \rho X
+}}
+{{eqn | o = \ge 
+      | r = \card{Y \cup Z'}
+}}
+{{end-eqn}}
+This is a [[Definition:Contradiction|contradiction]].
+Hence:
+:either $y \in Z'$ or $z \in Z'$
+{{WLOG}}, assume $y \in Z'$.
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$Y \cup \set y \in \mathscr I$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \rho {X \cup \set y}
+      | o = \ge
+      | r = \card{Y \cup \set y}
+      | c = {{Defof|Rank Function (Matroid)}}
+}}
+{{eqn | r = \card Y + \card{\set y}
+      | c = [[Cardinality of Set Union/Corollary|Corollary of Cardinality of Set Union]]
+}}
+{{eqn | r = \card Y + 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | o = >
+      | r = \card Y
+}}
+{{eqn | r = \map \rho X
+      | c = {{Defof|Rank Function (Matroid)}}
+}}
+{{end-eqn}}
+This proves the [[Definition:Contrapositive Statement|contrapositive statement]].
+{{qed|lemma}}
+===== Case 2: Either $y \in X$ or $z \in X$ =====
+{{WLOG}}, assume $y \in X$.
+Let:
+:$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$.
+We have:
+:$X \cup \set{y, z} = X \cup \set z$
+Hence:
+:$\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set z} = \map \rho X$
+{{qed|lemma}}
+In either case:
+:$\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set{y, z}} = \map \rho X$
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\rho : \powerset S \to \Z$ be a [[Definition:Mapping|mapping]] from the [[Definition:Power Set|power set]] of $S$ to the [[Definition:Integer|integers]].
+Let $\rho$ satisfy the [[Definition:Rank Axioms (Matroid)/Definition 1|rank axioms]]:
+{{:Definition:Rank Axioms (Matroid)/Definition 1}}
+Then $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] of a [[Definition:Matroid|matroid]] on $S$.
+\end{theorem}
+
+\begin{proof}
+===== [[Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition/Lemma|Lemma]] =====
+{{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition/Lemma}}{{qed|lemma}} 
+Let:
+:$\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$
+It is to be shown that:
+:* $\quad \mathscr I$ satisfies the [[Definition:Matroid Axioms|matroid axioms]]
+and
+:* $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] of the [[Definition:Matroid|matroid]] $M = \struct{S, \mathscr I}$
+=== Matroid Axioms ===
+==== Matroid Axiom $(\text I 1)$ ====
+From [[Cardinality of Empty Set]]:
+:$\card \O = 0$
+By [[Definition:Rank Axioms (Matroid)/Definition 1|rank axiom $(\text R 1)$]]:
+:$\map \rho \O = 0$
+Hence:
+:$\map \rho \O = \card \O$
+So:
+:$\O \in \mathscr I$
+Hence:
+$M$ satisfies [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]].
+{{qed|lemma}}
+==== Matroid Axiom $(\text I 2)$ ====
+Let
+:$X \in \mathscr I$
+{{AimForCont}}
+:$\exists Y \subseteq X : Y \notin \mathscr I$
+Let:
+:$Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$
+From [[Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition/Lemma|lemma]]:
+:$\map \rho {Y_0} < \card {Y_0}$
+As $\map \rho X = \card X$:
+:$Y_0 \ne X$
+From [[Set Difference with Proper Subset is Proper Subset]]:
+:$X \setminus Y_0 \neq \O$
+Let $y \in X \setminus Y_0$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \card {Y_0} + 1
+      | r = \card {Y_0 \cup \set y}
+      | c = [[Cardinality of Set Union/Corollary|Corollary of Cardinality of Set Union]]
+}}
+{{eqn | r = \map \rho {Y_0 \cup \set y}
+      | c = As $\card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$
+}}
+{{eqn | o = \le
+      | r = \map \rho {Y_0} + 1
+      | c = [[Definition:Rank Axioms (Matroid)/Definition 1|Rank axiom $(\text R 2)$]]
+}}
+{{eqn | o = <
+      | r = \card {Y_0} + 1
+      | c = As $\map \rho {Y_0} < \card {Y_0}$
+}}
+{{end-eqn}}
+This is a [[Definition:Contradiction|contradiction]].
+So:
+:$\forall Y \subseteq X : Y \in \mathscr I$
+Hence:
+:$M$ satisfies [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]].
+{{qed|lemma}}
+==== Matroid Axiom $(\text I 3)$ ====
+Let
+:$U \in \mathscr I$
+:$V \subseteq S$
+:$\card U < \card V$
+We prove the [[Definition:Contrapositive Statement|contrapositive statement]]:
+:$\paren{\forall x \in V \setminus U : U \cup \set x \notin \mathscr I} \implies V \notin \mathscr I$
+Let:
+:$\forall x \in V \setminus U : U \cup \set x \notin \mathscr I$
+{{qed|lemma}}
+=== $\rho$ is Rank Function ===
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Independent Subset is Contained in Maximal Independent Subset/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $X$ be a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $A$.
+Then:
+:$\card X = \map \rho A$
+where $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] on $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Independent Subset is Contained in Maximal Independent Subset]]:
+:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A : \card Y = \map \rho A$
+By definition of a [[Definition:Maximal Set|maximal]] [[Definition:Independent Subset (Matroid)|independent Subset]] of $A$:
+:$X = Y$
+The result follows.
+{{qed}}
+[[Category:Matroid Theory]]
+4332psornx28hd65ro444fgft84ne0w
+\end{proof}<|endoftext|>
+\section{Motion of Body with Constant Mass}
+Tags: Dynamics
+
+\begin{theorem}
+Let $B$ be a [[Definition:Body|body]] with [[Definition:Constant|constant]] [[Definition:Mass|mass]] $m$ undergoing a [[Definition:Force|force]] $\mathbf F$.
+Then the equation of motion of $B$ is given by:
+:$\mathbf F = m \mathbf a$
+where $\mathbf a$ is the [[Definition:Acceleration|acceleration]] of $B$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf F
+      | r = \map {\dfrac \d {\d t} } {m \mathbf v}
+      | c = [[Newton's Second Law of Motion]]
+}}
+{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}
+      | c = 
+}}
+{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + 0
+      | c = [[Derivative of Constant]]
+}}
+{{eqn | r = m \mathbf a
+      | c = {{Defof|Acceleration}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Like Vector Quantities are Multiples of Each Other}
+Tags: Scalar Multiplication
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Like Vector Quantities|like vector quantities]].
+Then:
+:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
+where:
+:$\size {\mathbf a}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf a$
+:$\dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$ denotes the [[Definition:Scalar Multiplication on Vector Quantity|scalar product]] of $\mathbf b$ by $\dfrac {\size {\mathbf a} } {\size {\mathbf b} }$.
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Like Vector Quantities|like vector quantities]]:
+:$\mathbf a$ and $\mathbf b$ are '''like [[Definition:Vector Quantity|vector quantities]]''' {{iff}} they have the same [[Definition:Direction|direction]].
+By definition of [[Definition:Unit Vector|unit vector]]:
+:$\dfrac {\mathbf a} {\size {\mathbf a} } = \dfrac {\mathbf b} {\size {\mathbf b} }$
+as both are in the same [[Definition:Direction|direction]], and both have [[Definition:Vector Length|length]] $1$.
+By definition of [[Definition:Scalar Division on Vector Quantity|scalar division]]:
+:$\dfrac 1 {\size {\mathbf a} } \mathbf a = \dfrac 1 {\size {\mathbf b} } \mathbf b$
+Hence, multiplying by $\size {\mathbf a}$:
+:$\mathbf a = \dfrac {\size {\mathbf a} } {\size {\mathbf b} } \mathbf b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product Norm is Norm}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be [[Definition:Normed Vector Space|normed vector spaces]].
+Let $V = X \times Y$ be a [[Definition:Direct Product of Vector Spaces/Finite Case|direct product of vector spaces]] $X$ and $Y$ together with [[Definition:Operation Induced by Direct Product|induced component-wise operations]].
+Let $\norm {\tuple {x, y} }$ be the [[Definition:Direct Product Norm|direct product norm]].
+Then $\norm {\tuple {x, y} }$ is a [[Definition:Norm on Vector Space|norm]] on $V$.
+\end{theorem}
+
+\begin{proof}
+=== Positive Definiteness ===
+Let $\tuple {x , y} \in V$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\tuple {x, y} } 
+      | r = \map \max {\norm x, \norm y}
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | o = \ge
+      | r =  0
+      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N1$: Positive Definiteness
+}}
+{{end-eqn}}
+Suppose $\norm {\tuple {x, y} } = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0 
+      | o = \le
+      | r = \norm x 
+      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N1$: Positive Definiteness      
+}}
+{{eqn | o = \le 
+      | r = \map \max {\norm x, \norm y} 
+}}
+{{eqn | r = \norm {\tuple {x, y} }
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+Hence, $\norm x = 0$
+{{qed|lemma}}
+=== Positive Homogeneity ===
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha \tuple {x, y} }
+      | r = \norm {\tuple {\alpha x, \alpha y} }
+      | c = [[Definition:Operation Induced by Direct Product|Induced component-wise operations]]
+}}
+{{eqn | r = \map \max {\norm {\alpha x}, \norm {\alpha y} } 
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | r = \map \max {\size \alpha \norm x, \size \alpha \norm y} 
+      | c = [[Definition:Norm on Vector Space|Norm Axiom]] $N2$ : Positive Homogeneity
+}}
+{{eqn | r = \size \alpha \max \set {\norm x, \norm y} 
+}}
+{{eqn | r = \size \alpha \norm {\tuple {x ,y} }
+      | c = {{defof|Direct Product Norm}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Triangle Inequality ===
+{{begin-eqn}}
+{{eqn | l = \norm {x_1 + x_2} 
+      | o = \le 
+      | r = \norm {x_1} + \norm {x_2} 
+}}
+{{eqn | o = \le
+      | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} } 
+}}
+{{eqn | r = \norm {\tuple {x_1, y_1} }+ \norm {\tuple {x_2, y_2} }
+      | c = {{defof|Direct Product Norm}}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \norm {y_1 + y_2} 
+      | o = \le 
+      | r = \norm {y_1} + \norm {y_2} 
+}}
+{{eqn | o = \le 
+      | r = \map \max {\norm {x_1}, \norm {y_1} } + \map \max {\norm {x_2}, \norm {y_2} } 
+}}
+{{eqn | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }
+      | c = {{defof|Direct Product Norm}}
+}}
+{{end-eqn}}
+Together:
+:$\map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2}} \le \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\tuple {x_1, y_1} + \tuple {x_2, y_2} } 
+      | r = \norm {\tuple {x_1 + x_2, y_1 + y_2} } 
+      | c = [[Definition:Operation Induced by Direct Product|Induced component-wise operations]]
+}}
+{{eqn | r = \map \max {\norm {x_1 + x_2}, \norm {y_1 + y_2} } 
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | o = \le 
+      | r = \norm {\tuple {x_1, y_1} } + \norm {\tuple {x_2, y_2} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Quantity as Scalar Product of Unit Vector Quantity}
+Tags: Scalar Multiplication
+
+\begin{theorem}
+Let $\mathbf a$ be a [[Definition:Vector Quantity|vector quantity]].
+Then:
+:$\mathbf a = \size {\mathbf a} \mathbf {\hat a}$
+where:
+:$\size {\mathbf a}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf a$
+:$\mathbf {\hat a}$ denotes the [[Definition:Unit Vector|unit vector]] in the [[Definition:Direction|direction]] $\mathbf a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {\size {\mathbf a} } \mathbf a
+      | r = \mathbf {\hat a}
+      | c = {{Defof|Unit Vector}}
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\mathbf a} \times \dfrac 1 {\size {\mathbf a} } \mathbf a
+      | r = \size {\mathbf a} \mathbf {\hat a}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf a
+      | r = \size {\mathbf a} \mathbf {\hat a}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Scalar Product of Magnitude by Unit Vector Quantity}
+Tags: Scalar Multiplication
+
+\begin{theorem}
+Let $\mathbf a$ be a [[Definition:Vector Quantity|vector quantity]].
+Let $m$ be a [[Definition:Scalar Quantity|scalar quantity]].
+Then:
+:$m \mathbf a = m \paren {\size {\mathbf a} \hat {\mathbf a} } = \paren {m \size {\mathbf a} } \hat {\mathbf a}$
+where:
+:$\size {\mathbf a}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf a$
+:$\hat {\mathbf a}$ denotes the [[Definition:Unit Vector|unit vector]] in the [[Definition:Direction|direction]] $\mathbf a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf a
+      | r = \size {\mathbf a} \hat {\mathbf a}
+      | c = [[Vector Quantity as Scalar Product of Unit Vector Quantity]]
+}}
+{{eqn | ll= \leadsto
+      | l = m \mathbf a
+      | r = m \paren {\size {\mathbf a} \hat {\mathbf a} }
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{finish|hard to prove something trivial}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Like Unit Vectors are Equal}
+Tags: Vectors
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Like Vector Quantities|like vector quantities]].
+Then:
+:$\mathbf {\hat a} = \mathbf {\hat b}$
+where $\mathbf {\hat a}$ and $\mathbf {\hat b}$ denote the [[Definition:Unit Vector|unit vectors]] in the [[Definition:Direction|direction]] of $\mathbf a$ and $\mathbf b$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Like Vector Quantities|like vector quantities]], $\mathbf a$ and $\mathbf b$ have the same [[Definition:Direction|direction]].
+By definition of [[Definition:Unit Vector|unit vector]], $\mathbf {\hat a}$ and $\mathbf {\hat b}$ are both of [[Definition:Magnitude|magnitude]] $1$.
+Hence the result, by [[Equality of Vector Quantities]].
+{{qed}}
+[[Category:Vectors]]
+4dc6u0f04r1xrbj90fu5p7eaj30d0uk
+\end{proof}<|endoftext|>
+\section{Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors}
+Tags: Vectors
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Vector Quantity|vector quantity]] embedded in [[Definition:Ordinary Space|space]].
+Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be [[Definition:Non-Coplanar Vectors|non-coplanar]].
+Then $\mathbf r$ can be expressed [[Definition:Unique|uniquely]] as the [[Definition:Resultant|resultant]] of $3$ [[Definition:Vector Quantity|vector quantities]] which are each [[Definition:Parallel Lines|parallel]] to one of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
+\end{theorem}
+
+\begin{proof}
+[[File:Resultant-of-3-non-coplanar-vectors.png|400px]]
+Let $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ be [[Definition:Unit Vector|unit vectors]] in the [[Definition:Direction|directions]] of $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
+Let $O$ be a [[Definition:Point|point]] in [[Definition:Ordinary Space|space]].
+Take $\vec {OP} := \mathbf r$.
+With $OP$ as its [[Definition:Space Diagonal|space diagonal]], construct a [[Definition:Parallelepiped|parallelepiped]] with [[Definition:Edge of Polyhedron|edges]] $OA$, $OB$ and $OC$ [[Definition:Parallel Lines|parallel]] to $\mathbf {\hat a}$, $\mathbf {\hat b}$ and $\mathbf {\hat c}$ respectively.
+Only one such [[Definition:Parallelepiped|parallelepiped]] can be so constructed.
+Let $x$, $y$ and $z$ be the [[Definition:Length of Line|length]] of the [[Definition:Edge of Polyhedron|edges]] $OA$, $OB$ and $OC$ respectively.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf r
+      | r = \vec {OA} + \vec {AF} + \vec {FP}
+      | c = 
+}}
+{{eqn | r = \vec {OA} + \vec {OB} + \vec {OC}
+      | c = 
+}}
+{{eqn | r = x \mathbf {\hat a} + y \mathbf {\hat b} + z \mathbf {\hat c}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\mathbf r$ is the [[Definition:Resultant|resultant]] of the $3$ [[Definition:Vector Quantity|vector quantities]] $x \mathbf {\hat a}$, $y \mathbf {\hat b}$ and $z \mathbf {\hat c}$ which, by construction, are [[Definition:Parallel Lines|parallel]] to $\mathbf a$, $\mathbf b$ and $\mathbf c$ respectively.
+The fact that only one [[Definition:Parallelepiped|parallelepiped]] can be constructed in the above proves [[Definition:Unique|uniqueness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vectors are Equal iff Components are Equal}
+Tags: Vectors, Equality
+
+\begin{theorem}
+Two [[Definition:Vector Quantity|vector quantities]] are [[Definition:Equality|equal]] {{iff}} they have the same [[Definition:Component of Vector|components]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector Quantity|vector quantities]].
+Then by [[Vector Quantity can be Expressed as Sum of 3 Non-Coplanar Vectors]], $\mathbf a$ and $\mathbf b$ can be expressed uniquely as [[Definition:Component of Vector|components]].
+So if $\mathbf a$ and $\mathbf b$ then the [[Definition:Component of Vector|components]] of $\mathbf a$ are the same as the [[Definition:Component of Vector|components]] of $\mathbf b$
+Suppose $\mathbf a$ and $\mathbf b$ have the same [[Definition:Component of Vector|components]]: $\mathbf x$, $\mathbf y$ and $\mathbf z$.
+Then by definition:
+:$\mathbf a = \mathbf x + \mathbf y + \mathbf z$
+and also:
+:$\mathbf b = \mathbf x + \mathbf y + \mathbf z$
+and trivially:
+:$\mathbf a = \mathbf b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristics of Birkhoff-James Orthogonality}
+Tags: Normed Vector Spaces
+
+\begin{theorem}
+Let $\struct {V, \norm {\,\cdot\,} }$ be a [[Definition:Normed Linear Space|normed linear space]].
+Let $x, y \in V$.
+Then $x$ and $y$ are '''[[Definition:Birkhoff-James Orthogonality|Birkhoff-James orthogonal]]''' {{iff}} either:
+:$(1): \quad x = 0$
+or:
+:$(2): \quad$ there exists a [[Definition:Continuous Functional|continuous functional]] $ f$ on $\struct {V, \norm {\,\cdot\,} }$ such that:
+::::$\norm f = 1$
+::::$\map f x = \norm x$
+::::$\map f y = 0$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \perp_B y$.
+Let $V' \subset V$ be the subspace spanned by $x$ and $y$.
+Define $\overline f$ on $V'$ as:
+:$\map {\overline f} {a x + b y} = a \norm x$
+for $a$ and $b$ scalars.
+Clearly, $\overline f$ is linear and:
+:$\map {\overline f} x = \norm x$
+:$\map {\overline f} y = 0$
+Further:
+{{begin-eqn}}
+{{eqn | l = \norm {a x + b y}
+      | r = \size a \norm {x + \dfrac b a y}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \size a \norm x
+      | c = 
+}}
+{{eqn | r = \size {\map {\overline f} {a x + b y} }
+      | c = 
+}}
+{{end-eqn}}
+proving that $\overline f$ is a bounded functional of norm $1$.
+Now by [[Hahn-Banach Theorem]], $\overline f$ can be extended to a functional $f$ on $V$ such that $\norm f = \norm {\overline f} = 1$
+This proves the necessity.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let such a functional $f$ on $V$ exist.
+Then for any scalar $\lambda$:
+:$\norm {x + \lambda y} \ge \size {\map f {x + \lambda y} } = \norm x$
+establishing the sufficiency.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Components of Vector in terms of Direction Cosines}
+Tags: Direction Cosines
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Vector Quantity|vector quantity]] embedded in a [[Definition:Cartesian 3-Space|Cartesian $3$-space]].
+Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the [[Definition:Unit Vector|unit vectors]] in the [[Definition:Positive Direction|positive directions]] of the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the [[Definition:Direction Cosine|direction cosines]] of $\mathbf r$ with respect to the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Let $x$, $y$ and $z$ be the [[Definition:Component of Vector|components]] of $\mathbf r$ in the $\mathbf i$, $\mathbf j$ and $\mathbf k$ [[Definition:Direction|directions]] respectively.
+Let $r$ denote the [[Definition:Magnitude|magnitude]] of $\mathbf r$, that is:
+:$r := \size {\mathbf r}$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = r \cos \alpha
+}}
+{{eqn | l = y
+      | r = r \cos \beta
+}}
+{{eqn | l = z
+      | r = r \cos \gamma
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Vector-Components-in-3-Space.png|480px]]
+By definition, the [[Definition:Direction Cosine|direction cosines]] are the [[Definition:Cosine|cosines]] of the [[Definition:Angle|angles]] that $\mathbf r$ makes with the [[Definition:Coordinate Axis|coordinate axes]].
+By definition of the [[Definition:Component of Vector|components of $\mathbf r$]]:
+:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
+Thus:
+:$\mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Magnitude of Vector Quantity in terms of Components}
+Tags: Vectors
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Vector Quantity|vector quantity]] embedded in a [[Definition:Cartesian 3-Space|Cartesian $3$-space]].
+Let $\mathbf r$ be expressed in terms of its [[Definition:Component of Vector|components]]:
+:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
+where $\mathbf i$, $\mathbf j$ and $\mathbf k$ denote the [[Definition:Unit Vector|unit vectors]] in the [[Definition:Positive Direction|positive directions]] of the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Then:
+:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2}$
+where $\size {\mathbf r}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf r$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Initial Point of Vector|initial point]] of $\mathbf r$ be $\tuple {x_1, y_1, z_1}$.
+Let the [[Definition:Terminal Point of Vector|terminal point]] of $\mathbf r$ be $\tuple {x_2, y_2, z_2}$.
+Thus, by definition of the [[Definition:Component of Vector|components of $\mathbf r$]], the [[Definition:Magnitude|magnitude]] of $\mathbf r$ equals the [[Definition:Distance (Linear Measure)|distance]] between $\tuple {x_1, y_1, z_1}$ and $\tuple {x_2, y_2, z_2}$.
+The result follows from [[Distance Formula in 3 Dimensions]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Components of Zero Vector Quantity are Zero}
+Tags: Zero Vectors
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Vector Quantity|vector quantity]] embedded in a [[Definition:Cartesian 3-Space|Cartesian $3$-space]].
+Let $\mathbf r$ be expressed in terms of its [[Definition:Component of Vector|components]]:
+:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
+Let $\mathbf r$ be the [[Definition:Zero Vector Quantity|zero vector]].
+Then:
+:$x = y = z = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Zero Vector Quantity|zero vector]], the [[Definition:Magnitude|magnitude]] of $\mathbf r$ is equal to [[Definition:Zero (Number)|zero]].
+By [[Magnitude of Vector Quantity in terms of Components]]:
+:$\size {\mathbf r} = \sqrt {x^2 + y^2 + z^2} = 0$
+where $\size {\mathbf r}$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf r$.
+As each of $x$, $y$ and $z$ are [[Definition:Real Number|real numbers]], each of $x^2$, $y^2$ and $z^2$ is [[Definition:Non-Negative Real Number|non-negative]].
+so in order for $\sqrt {x^2 + y^2 + z^2} = 0$, it must follow that each of $x$, $y$ and $z$ is [[Definition:Zero (Number)|zero]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Vector in terms of Direction Cosines}
+Tags: Direction Cosines, Unit Vectors
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Vector Quantity|vector quantity]] embedded in a [[Definition:Cartesian 3-Space|Cartesian $3$-space]].
+Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the [[Definition:Unit Vector|unit vectors]] in the [[Definition:Positive Direction|positive directions]] of the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the [[Definition:Direction Cosine|direction cosines]] of $\mathbf r$ with respect to the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Let $\mathbf {\hat r}$ denote the [[Definition:Unit Vector|unit vector]] in the [[Definition:Direction|direction]] of $\mathbf r$.
+Then:
+:$\mathbf {\hat r} = \paren {\cos \alpha} \mathbf i + \paren {\cos \beta} \mathbf j + \paren {\cos \gamma} \mathbf k$
+\end{theorem}
+
+\begin{proof}
+From [[Components of Vector in terms of Direction Cosines]]:
+:$(1): \quad \mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
+where $r$ denotes the [[Definition:Magnitude|magnitude]] of $\mathbf r$, that is:
+:$r := \size {\mathbf r}$
+By definition of [[Definition:Unit Vector|unit vector]]:
+:$\mathbf {\hat r} = \dfrac {\mathbf r} r$
+The result follows by [[Definition:Scalar Multiplication on Vector Quantity|multiplication]] of both sides of $(1)$ by $\dfrac 1 r$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Vectors of Cartesian 3-Space form Basis}
+Tags: Unit Vectors
+
+\begin{theorem}
+Consider the [[Definition:Cartesian 3-Space|Cartesian $3$-space]] $C$.
+Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ denote the [[Definition:Unit Vector|unit vectors]] in the [[Definition:Positive Direction|positive directions]] of the [[Definition:X-Axis|$x$-axis]], [[Definition:Y-Axis|$y$-axis]] and [[Definition:Z-Axis|$z$-axis]] respectively.
+Then $\set {\mathbf i, \mathbf j, \mathbf k}$ forms a [[Definition:Basis of Vector Space|basis]] for $C$.
+\end{theorem}<|endoftext|>
+\section{Vectors from Sum and Difference}
+Tags: Vector Addition, Vector Subtraction
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector Quantity|vector quantities]].
+Let $\mathbf c = \mathbf a + \mathbf b$ and $\mathbf d = \mathbf a - \mathbf b$ be given.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf a
+      | r = \dfrac 1 2 \paren {\mathbf c + \mathbf d}
+}}
+{{eqn | l = \mathbf b
+      | r = \dfrac 1 2 \paren {\mathbf c - \mathbf d}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 \paren {\mathbf c + \mathbf d}
+      | r = \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} + \paren {\mathbf a - \mathbf b} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\mathbf a + \mathbf b + \mathbf a - \mathbf b}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {2 \mathbf a}
+      | c = 
+}}
+{{eqn | r = \mathbf a
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 \paren {\mathbf c - \mathbf d}
+      | r = \dfrac 1 2 \paren {\paren {\mathbf a + \mathbf b} - \paren {\mathbf a - \mathbf b} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\mathbf a + \mathbf b - \mathbf a + \mathbf b}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {2 \mathbf b}
+      | c = 
+}}
+{{eqn | r = \mathbf b
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Banach Spaces is Banach Space}
+Tags: Banach Spaces
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be [[Definition:Normed Vector Space|normed vector spaces]].
+Let $V = X \times Y$ be a [[Definition:Direct Product of Vector Spaces/Finite Case|direct product of vector spaces]] $X$ and $Y$ together with [[Definition:Operation Induced by Direct Product|induced component-wise operations]].
+Let $\norm {\tuple {x, y} }$ be the [[Definition:Direct Product Norm|direct product norm]].
+Suppose $X$ and $Y$ are [[Definition:Banach Space|Banach spaces]].
+Then $V$ is a [[Definition:Banach Space|Banach space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $V$:
+: $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} } < \epsilon$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n - x_m} 
+      | o = \le
+      | r = \map \max {\norm {x_n - x_m}, \norm {y_n - y_m} }
+}}
+{{eqn | r = \norm {\tuple {x_n - x_m, y_n - y_m} }
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | r =  \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} }
+      | c = [[Definition:Operation Induced by Direct Product|Induced component-wise operations]]
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+Hence, $\sequence {x_n}_{n \mathop \in \N}$ is [[Definition:Cauchy Sequence|Cauchy sequence]] in $X$.
+By assumption, $X$ is a [[Definition:Banach Space|Banach space]].
+Therefore, $\sequence {x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $x \in X$.
+By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $y \in Y$.
+Then there exists $\epsilon \in \R_{> 0}$ such that:
+:$\exists N_x \in \N : \forall n \in \N : n > N_x : \norm {x - x_n}_n < \epsilon$
+:$\exists N_y \in \N : \forall n \in \N : n > N_y : \norm {y - y_n}_n < \epsilon$
+Let $N := \map \max {N_x, N_y}$.
+Then:
+:$\forall n \in \N : n > N : \paren {\norm {x - x_n}_n < \epsilon} \land \paren{\norm {y - y_n}_n < \epsilon}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \norm {\tuple {x, y} - \tuple {x_n, y_n} } 
+      | r = \norm {\tuple {x - x_n, y - y_n} } 
+}}
+{{eqn | r = \map \max {\norm {x - x_n}, \norm {y - y_n} } 
+      | c = {{defof|Direct Product Norm}}
+}}
+{{eqn | o = < 
+      | r = \epsilon
+}}
+{{end-eqn}}
+Therefore, $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $\tuple {x, y}$.
+By definition, $V$ is a [[Definition:Banach Space|Banach space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cartesian Plane Rotated with respect to Another}
+Tags: Coordinate Systems, Geometric Rotations, Vectors
+
+\begin{theorem}
+Let $\mathbf r$ be a [[Definition:Position Vector|position vector]] embedded in a [[Definition:Cartesian Plane|Cartesian plane]] $\CC$ with [[Definition:Origin|origin]] $O$.
+Let $\CC$ be [[Definition:Plane Rotation|rotated]] [[Definition:Anticlockwise|anticlockwise]] through an [[Definition:Angle|angle]] $\varphi$ about the [[Definition:Axis of Rotation|axis of rotation]] $O$.
+Let $\CC'$ denote the [[Definition:Cartesian Plane|Cartesian plane]] in its new position.
+Let $\mathbf r$ be kept fixed during this [[Definition:Plane Rotation|rotation]].
+Let $\tuple {x, y}$ denote the [[Definition:Component of Vector|components]] of $\mathbf r$ with respect to $\CC$.
+Let $\tuple {x', y'}$ denote the [[Definition:Component of Vector|components]] of $\mathbf r$ with respect to $\CC'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x'
+      | r = x \cos \varphi + y \sin \varphi
+}}
+{{eqn | l = y'
+      | r = -x \sin \varphi + y \cos \varphi
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Rotation-of-cartesian-plane.png|400px]]
+Let $\mathbf r$ be represented by a [[Definition:Directed Line Segment|directed line segment]] whose [[Definition:Initial Point of Vector|initial point]] coincides with the [[Definition:Origin|origin]] $O$.
+Let the [[Definition:Terminal Point of Vector|terminal point]] of $\mathbf r$ be identified with the [[Definition:Point|point]] $P$.
+Let $\CC$ be rotated to $\CC'$ through an [[Definition:Angle|angle]] $\varphi$ as shown, keeping $P$ fixed.
+We have that:
+{{begin-eqn}}
+{{eqn | l = x'
+      | r = OA + BP
+      | c = by inspection
+}}
+{{eqn | r = x \cos \varphi + y \sin \varphi
+      | c = {{Defof|Cosine}}, {{Defof|Sine}}
+}}
+{{eqn | l = y'
+      | r = xB - xA
+      | c = by inspection
+}}
+{{eqn | r = y \cos \varphi - x \sin \varphi
+      | c = {{Defof|Cosine}}, {{Defof|Sine}}
+}}
+{{end-eqn}}
+hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition/Lemma}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$\forall Y \subseteq S: \map \rho Y \le \card Y$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}}:
+:$\exists Y \subseteq S : \map \rho Y > \card Y$
+Let:
+:$Y_0 \subseteq S : \card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$
+By [[Definition:Rank Axioms (Matroid)/Definition 1|rank axiom $(\text R 1)$]]:
+:$\map \rho \O = 0$
+From [[Cardinality of Empty Set]]:
+:$\card \O = 0$
+So:
+:$\map \rho \O = \card \O$
+Hence:
+:$Y_0 \neq \O$
+Let $y \in Y_0$.
+Let $Y = Y_0 \setminus \set y$.
+From [[Cardinality of Set Difference]]:
+:$\card Y = \card{Y_0} - 1$
+By choice of $Y_0$:
+:$\map \rho Y \le \card Y$
+By [[Definition:Rank Axioms (Matroid)/Definition 1|rank axiom $(\text R 2)$]]:
+:$\map \rho {Y_0} = \map \rho {Y \cup \set y} \le \map \rho Y + 1 \le \card Y + 1 = \card{Y_0}$
+This [[Definition:Contradiction|contradicts]] the choice of $Y_0$:
+:$\card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$
+It follows that:
+:$\forall Y \subseteq S: \map \rho Y \le \card Y$
+{{qed}}
+[[Category:Matroid Theory]]
+gd78tscazxtxbfdp6j5pb8sm0b03ofd
+\end{proof}<|endoftext|>
+\section{Dot Product of Perpendicular Vectors}
+Tags: Dot Product
+
+\begin{theorem}
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Vector Quantity|vector quantities]].
+Let $\mathbf a$ and $\mathbf b$ be [[Definition:Perpendicular|perpendicular]].
+Then:
+:$\mathbf a \cdot \mathbf b = 0$
+where $\cdot$ denotes [[Definition:Dot Product|dot product]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Dot Product|dot product]]:
+:$\mathbf a \cdot \mathbf b = \norm {\mathbf a} \norm {\mathbf b} \cos \theta$
+where $\theta$ is the [[Definition:Angle|angle]] between $\mathbf a$ and $\mathbf b$.
+When $\mathbf a$ and $\mathbf b$ be [[Definition:Perpendicular|perpendicular]], by definition $\theta = 90 \degrees$.
+The result follows by [[Cosine of Right Angle]], which gives that $\cos 90 \degrees = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dot Product of Orthogonal Basis Vectors}
+Tags: Dot Product
+
+\begin{theorem}
+Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be an [[Definition:Orthonormal Basis of Vector Space|orthonormal basis]] of a [[Definition:Vector Space|vector space]] $V$.
+Then:
+:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \delta_{i j}$
+where:
+:$\mathbf e_i \cdot \mathbf e_j$ denotes the [[Definition:Dot Product|dot product]] of $\mathbf e_i$ and $\mathbf e_j$
+:$\delta_{i j}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Orthonormal Basis of Vector Space|orthonormal basis]]:
+:$(1): \quad \tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is an [[Definition:Orthogonal Basis of Vector Space|orthogonal basis of $V$]]
+:$(2): \quad \norm {\mathbf e_1} = \norm {\mathbf e_2} = \cdots = \norm {\mathbf e_1} = 1$
+From $(1)$ we have by definition that $\mathbf e_i$ and $\mathbf e_j$ are [[Definition:Perpendicular|perpendicular]] whenever $i \ne j$
+Thus, from [[Dot Product of Perpendicular Vectors]], for all $i$ and $j$, when $i \ne j$:
+:$\mathbf e_i \cdot \mathbf e_j = 0$
+From [[Dot Product of Vector with Itself]], for all $i$ we have:
+:$\mathbf e_i \cdot \mathbf e_i = \norm {\mathbf e_i}^2$
+From $(2)$, we have that:
+:$\norm {\mathbf e_i} = 1$
+and so:
+:$\mathbf e_i \cdot \mathbf e_i = 1$
+Putting this together, we have:
+:$\forall i, j \in \set {1, 2, \ldots, n}: \mathbf e_i \cdot \mathbf e_j = \begin {cases} 1 & : i = j \\ 0 & : i \ne j \end {cases}$
+The result follows by definition of the [[Definition:Kronecker Delta|Kronecker delta]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivative wrt x of sin x y over cos (x + y)}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+:$\dfrac \partial {\partial x} \dfrac {\sin x y} {\map \cos {x + y} } = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac \partial {\partial x} \frac {\sin x y} {\map \cos {x + y} }
+      | r = \frac {\map \cos {x + y} y \cos x y - \sin x y \paren {-\map \sin {x + y} } } {\map {\cos^2} {x + y} }
+      | c = [[Quotient Rule for Derivatives]], treating $y$ as a [[Definition:Constant|constant]]
+}}
+{{eqn | r = \dfrac {y \map \cos {x + y} \cos x y + \map \sin {x + y} \sin x y} {\map {\cos^2} {x + y} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Partial Derivative wrt r of ln (r^2 + s)}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+:$\dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s} = \dfrac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial^2} {\partial r^2} \map \ln {r^2 + s}
+      | r = \map {\dfrac \partial {\partial r} } {\dfrac \partial {\partial r} \map \ln {r^2 + s} }
+      | c = {{Defof|Second Partial Derivative}}
+}}
+{{eqn | r = \dfrac \partial {\partial r} \frac 1 {r^2 + s} 2 r
+      | c = [[Derivative of Natural Logarithm]], [[Chain Rule for Derivatives]], treating $s$ as a [[Definition:Constant|constant]]
+}}
+{{eqn | r = \frac {2 \paren {r^2 + s} - 2 r \paren {2 r} } {\paren {r^2 + s}^2}
+      | c = [[Quotient Rule for Derivatives]], again treating $s$ as a [[Definition:Constant|constant]]
+}}
+{{eqn | r = \frac {2 \paren {s - r^2} } {\paren {r^2 + s}^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of tan^2 (x^2 - y^2)}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$\map f {x, y} = \map {\tan^2} {x^2 - y^2}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {x, y}
+      | r = 4 x \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}
+}}
+{{eqn | l = \map {f_2} {1, 2}
+      | r = 8 \tan 3 \sec^2 3
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {x, y}
+      | r = \dfrac \partial {\partial x} {\map {\tan^2} {x^2 - y^2} }
+      | c = {{Defof|Partial Derivative}}
+}}
+{{eqn | r = 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot 2 x
+      | c = [[Derivative of Square of Tangent]], [[Derivative of Square Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = 4 x \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \map {f_2} {x, y}
+      | r = \dfrac \partial {\partial y} {\map {\tan^2} {x^2 - y^2} }
+      | c = {{Defof|Partial Derivative}}
+}}
+{{eqn | r = 2 \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2} \cdot \paren {-2 y}
+      | c = [[Derivative of Square of Tangent]], [[Derivative of Square Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -4 y \map \tan {x^2 - y^2} \map {\sec^2} {x^2 - y^2}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f_2} {1, 2}
+      | r = -4 \paren 2 \map \tan {1^2 - 2^2} \map {\sec^2} {1^2 - 2^2}
+      | c = 
+}}
+{{eqn | r = -8 \map \tan {-3} \map {\sec^2} {-3}
+      | c = 
+}}
+{{eqn | r = -8 \paren {-\tan 3} \sec^2 3
+      | c = [[Tangent Function is Odd]], [[Secant Function is Even]]
+}}
+{{eqn | r = 8 \tan 3 \sec^2 3
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Square of Tangent}
+Tags: Tangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tan^2 x} = 2 \tan x \sec^2 x$
+when $\cos x \ne 0$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\tan^2 x}
+      | r = 2 \tan x \map {\dfrac \d {\d x} } {\tan x}
+      | c = [[Chain Rule for Derivatives]], [[Derivative of Square Function]]
+}}
+{{eqn | r = 2 \tan x \sec^2 x 
+      | c = [[Derivative of Tangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Tangent Function]]
+nr7j1qx0qa2pkgzchptcsv4cp87ohvw
+\end{proof}<|endoftext|>
+\section{Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)}
+Tags: Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2), Examples of Partial Derivatives
+
+\begin{theorem}
+Let $z^2 = x^2 - 2 x y - 1$.
+Then:
+:$\valueat {\dfrac {\partial z} {\partial x} } {x \mathop = 1, y \mathop = -2, z \mathop = -2} = -\dfrac 3 2$
+\end{theorem}
+
+\begin{proof}
+First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:
+{{begin-eqn}}
+{{eqn | l = x^2 - 2 x y - 1
+      | r = \paren 1^2 - 2 \paren 1 \paren {-2} - 1
+      | c = at $\tuple {1, -2, -2}$
+}}
+{{eqn | r = 1 + 4 - 1
+      | c = 
+}}
+{{eqn | r = 4
+      | c = 
+}}
+{{eqn | r = \paren {-2}^2
+      | c = 
+}}
+{{eqn | r = z^2
+      | c = at $\tuple {1, -2, -2}$
+}}
+{{end-eqn}}
+and it is seen that this is the case.
+{{qed|lemma}}
+=== [[Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method|Explicit Method]] ===
+{{:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method}}
+=== [[Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Implicit Method|Implicit Method]] ===
+{{:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Implicit Method}}
+[[Category:Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)]]
+[[Category:Examples of Partial Derivatives]]
+krx6ob9fe8q70qq2cpcss95dyambcgy
+\end{proof}<|endoftext|>
+\section{Partial Derivative/Examples/u - v + 2 w, 2 u + v + 2 w, u - v + w}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+{{begin-eqn}}
+{{eqn | l = u - v + 2 w
+      | r = x + 2 z
+}}
+{{eqn | l = 2 u + v - 2 w
+      | r = 2 x - 2 z
+}}
+{{eqn | l = u - v + w
+      | r = z - y
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = 0
+}}
+{{eqn | l = \dfrac {\partial v} {\partial y}
+      | r = 2
+}}
+{{eqn | l = \dfrac {\partial w} {\partial y}
+      | r = 1
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+[[Definition:Partial Derivative|Partial differentiation]] {{WRT|Differentiation}} $y$ gives:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial y} - \dfrac {\partial v} {\partial y} + 2 \dfrac {\partial w} {\partial y}
+      | r = 0
+}}
+{{eqn | l = 2 \dfrac {\partial u} {\partial y} + \dfrac {\partial v} {\partial y} - 2 \dfrac {\partial w} {\partial y}
+      | r = 0
+}}
+{{eqn | l = \dfrac {\partial u} {\partial y} - \dfrac {\partial v} {\partial y} + \dfrac {\partial w} {\partial y}
+      | r = -1
+}}
+{{end-eqn}}
+which can be expressed in [[Definition:Matrix|matrix form]] as:
+:$\begin {pmatrix} 1 & -1 & 2 \\ 2 & 1 & -2 \\ 1 & -1 & 1 \end {pmatrix} \begin {pmatrix} \dfrac {\partial u} {\partial y} \\ \dfrac {\partial v} {\partial y} \\ \dfrac {\partial w} {\partial y} \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ -1 \end {pmatrix}$
+Solving by [[Cramer's Rule]]:
+{{explain|Actually this is not [[Cramer's Rule]], this will need to be taken care of}}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 2 & 1 & -2 & 0 \\ 1 & -1 & 1 & -1 \end {array} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 0 & 3 & -6 & 0 \\ 0 & 0 & -1 & -1 \end {array} }
+      | c = $\text r 2 \to \text r 2 - 2 \text r 1$, $\text r 3 \to \text r 3 - \text r 1$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \paren {\begin {array} {ccc {{|}} c} 1 & -1 & 2 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 1 \end {array} }
+      | c = $\text r 2 \to \text r 2 / 3$, $\text r 3 \to -\text r 3$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \paren {\begin {array} {ccc {{|}} c} 1 & 0 & 0 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 1 \end {array} }
+      | c = $\text r 1 \to \text r 1 - \text r 2$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \paren {\begin {array} {ccc {{|}} c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 1 \end {array} }
+      | c = $\text r 2 \to \text r 2 + 2 \text r 3$
+}}
+{{end-eqn}}
+The solution can be read directly:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = 0
+}}
+{{eqn | l = \dfrac {\partial v} {\partial y}
+      | r = 2
+}}
+{{eqn | l = \dfrac {\partial w} {\partial y}
+      | r = 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of x tan^-1 (x^2 + y)}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$\map f {x, y} = x \map \arctan {x^2 + y}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {1, 0}
+      | r = \dfrac \pi 4 + 1
+}}
+{{eqn | l = \map {f_2} {x, y}
+      | r = \dfrac x {1 + \paren {x^2 + y}^2}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {x, y}
+      | r = \map {\dfrac \partial {\partial x} } {x \map \arctan {x^2 + y} }
+      | c = {{Defof|Partial Derivative}}
+}}
+{{eqn | r = \map \arctan {x^2 + y} \map {\dfrac \partial {\partial x} } x + x \map {\dfrac \partial {\partial x} } {\map \arctan {x^2 + y} }
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \map \arctan {x^2 + y} + x \cdot \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 2 x
+      | c = [[Product Rule for Derivatives]], [[Derivative of Arctangent Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \map \arctan {x^2 + y} + \dfrac {2 x^2} {1 + \paren {x^2 + y}^2}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f_1} {1, 0}
+      | r = \map \arctan {1^2 + 0} + \dfrac {2 \times 1^2} {1 + \paren {1^2 + 0}^2}
+      | c = substituting $x = 1$ and $y = 0$
+}}
+{{eqn | r = \map \arctan 1 + \dfrac 2 2
+      | c = simplifying
+}}
+{{eqn | r = \dfrac \pi 4 + 1
+      | c = [[Tangent of 45 Degrees|Tangent of $45 \degrees$]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \map {f_2} {x, y}
+      | r = \map {\dfrac \partial {\partial y} } {x \map \arctan {x^2 + y} }
+      | c = {{Defof|Partial Derivative}}
+}}
+{{eqn | r = x \dfrac 1 {1 + \paren {x^2 + y}^2} \cdot 1
+      | c = [[Derivative of Identity Function]], [[Derivative of Arctangent Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac x {1 + \paren {x^2 + y}^2}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of x ln y^2 + y e^z}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$\map f {x, y, z} = x \ln y^2 + y e^z$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {1, -1, 0}
+      | r = 0
+}}
+{{eqn | l = \map {f_2} {x, x y, y + z}
+      | r = \dfrac 2 y + e^{y + z}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {f_1} {x, y, z}
+      | r = \ln y^2
+      | c = [[Derivative of Constant Multiple]], [[Derivative of Identity Function]], holding $y$ and $z$ [[Definition:Constant|constant]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f_1} {1, -1, 0}
+      | r = \map \ln {\paren {-1}^2}
+      | c = substituting $\tuple {1, -1, 0}$ for $\tuple {x, y, z}$
+}}
+{{eqn | r = \ln 1
+      | c = simplifying
+}}
+{{eqn | r = 0
+      | c = [[Natural Logarithm of 1 is 0]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \map {f_2} {x, y, z}
+      | r = \map {\dfrac \partial {\partial y} } {x \ln y^2 + y e^z}
+      | c = {{Defof|Partial Derivative}}
+}}
+{{eqn | r = \dfrac x {y^2} \cdot 2 y + e^z
+      | c = [[Derivative of Natural Logarithm]], [[Power Rule for Derivatives]], [[Chain Rule for Derivatives]], holding $x$ and $z$ [[Definition:Constant|constant]]
+}}
+{{eqn | r = \dfrac {2 x} y + e^z
+      | c = simplification
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f_2} {x, x y, y + z}
+      | r = \dfrac {2 x} {x y} + e^{y + z}
+      | c = substituting $\tuple {x, x y, y + z}$ for $\tuple {x, y, z}$
+}}
+{{eqn | r = \dfrac 2 y + e^{y + z}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of x^y^z}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$u = x^{\paren {y^z} }$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x}
+      | r = y^z x^{\paren {y^z - 1} }
+}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = x^{y^z} z y^{z - 1} \ln x
+}}
+{{eqn | l = \dfrac {\partial u} {\partial z}
+      | r = x^{\paren {y^z} } y^z \ln x \ln y
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x}
+      | r = y^z x^{\paren {y^z - 1} }
+      | c = [[Power Rule for Derivatives]], keeping $y$ and $z$ [[Definition:Constant|constant]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^{\paren {y^z} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ln u
+      | r = \map \ln {x^{\paren {y^z} } }
+      | c = 
+}}
+{{eqn | n = 1
+      | r = y^z \ln x
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 u \dfrac {\partial u} {\partial y}
+      | r = z y^{z - 1} \ln x
+      | c = [[Derivative of Natural Logarithm]], [[Power Rule for Derivatives]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y}
+      | r = u z y^{z - 1} \ln x
+      | c = multiplying both sides by $u$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y}
+      | r = x^{\paren {y^z} } z y^{z - 1} \ln x
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \ln u
+      | r = y^z \ln x
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 u \dfrac {\partial u} {\partial z}
+      | r = y^z \ln x \ln y
+      | c = [[Derivative of Natural Logarithm]], [[Derivative of General Exponential Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial z}
+      | r = u y^z \ln x \ln y
+      | c = multiplying both sides by $u$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial z}
+      | r = x^{\paren {y^z} } y^z \ln x \ln y
+      | c = substituting for $u$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of x^u + u^y}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$u = x^u + u^y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x}
+      | r = \frac {u^2} {x \paren {1 - u \ln x - y} }
+}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = \frac {u \ln u} {1 - u \ln x - y}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = u
+      | r = x^u + u^y
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \ln u
+      | r = u \ln x + y \ln u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 u \dfrac {\partial u} {\partial x}
+      | r = \frac u x + \ln x \dfrac {\partial u} {\partial x} + \frac y u \dfrac {\partial u} {\partial x}
+      | c = [[Derivative of Natural Logarithm]], [[Chain Rule for Derivatives]] keeping $y$ constant
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial x} \paren {\frac 1 u - \ln x - \frac y u}
+      | r = \frac u x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial x} \paren {\frac {1 - u \ln x - y} u}
+      | r = \frac u x
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial x}
+      | r = \frac {u^2} {x \paren {1 - u \ln x - y} }
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \ln u
+      | r = u \ln x + y \ln u
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 u \dfrac {\partial u} {\partial y}
+      | r = \ln x \dfrac {\partial u} {\partial y} + \ln u + \frac y u \dfrac {\partial u} {\partial y}
+      | c = [[Derivative of Natural Logarithm]], [[Product Rule for Derivatives]], [[Chain Rule for Derivatives]] keeping $x$ constant
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y} \paren {\frac 1 u - \ln x - \frac y u}
+      | r = \ln u
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y} \paren {\frac {1 - u \ln x - y} u}
+      | r = \ln u
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y}
+      | r = \frac {u \ln u} {1 - u \ln x - y}
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Derivatives of u^2 + x^2 + y^2, u - v^3 + 3 x}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+{{begin-eqn}}
+{{eqn | l = u^2 + x^2 + y^2
+      | r = 3
+}}
+{{eqn | l = u - v^3 + 3 x
+      | r = 4
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x}
+      | r = -\dfrac x u
+}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = -\dfrac y u
+}}
+{{eqn | l = \dfrac {\partial v} {\partial x}
+      | r = \dfrac {x - 3 u} {3 u v^2}
+}}
+{{eqn | l = \dfrac {\partial v} {\partial y}
+      | r = \dfrac y {3 u v^2}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 2 u \dfrac {\partial u} {\partial x} + 2 x
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial x}
+      | r = -\dfrac x u
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = 2 u \dfrac {\partial u} {\partial y} + 2 y
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \dfrac {\partial u} {\partial y}
+      | r = -\dfrac y u
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x} + 3 v^2 \dfrac {\partial v} {\partial x} + 3
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac x u + 3 v^2 \dfrac {\partial v} {\partial x} + 3
+      | r = 0
+      | c = substituting for $\dfrac {\partial u} {\partial x}$ from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial v} {\partial x}
+      | r = \dfrac {x - 3 u} {3 u v^2}
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial y} + 3 v^2 \dfrac {\partial v} {\partial y}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac y u + 3 v^2 \dfrac {\partial v} {\partial x}
+      | r = 0
+      | c = substituting for $\dfrac {\partial u} {\partial y}$ from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial v} {\partial y}
+      | r = \dfrac y {3 u v^2}
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Third Partial Derivatives of x^y}
+Tags: Examples of Partial Derivatives
+
+\begin{theorem}
+Let:
+:$u = x^y$
+Then:
+:$\dfrac {\partial^3 u} {\partial x^2 \partial y} = \dfrac {\partial^3 u} {\partial x \partial y \partial x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial y}
+      | r = x^y \ln x
+      | c = [[Derivative of General Logarithm Function]] keeping $x$ constant
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial^2 u} {\partial x \partial y}
+      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial u} {\partial y} }
+      | c = {{Defof|Second Partial Derivative}}
+}}
+{{eqn | r = y x^{y - 1} \ln x + \dfrac 1 x x^y
+      | c = [[Power Rule for Derivatives]], [[Derivative of Natural Logarithm]] keeping $y$ constant
+}}
+{{eqn | r = x^{y - 1} \paren {y \ln x + 1}
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial^3 u} {\partial x^2 \partial y}
+      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial x \partial y} }
+      | c = {{Defof|Third Partial Derivative}}
+}}
+{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial x} } {y \ln x + 1} + \paren {y \ln x + 1} \map {\dfrac \partial {\partial x} } {x^{y - 1} }
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = x^{y - 1} \paren {\dfrac y x} + \paren {y \ln x + 1} \paren {y - 1} x^{y - 2}
+      | c = [[Derivative of Natural Logarithm]], [[Power Rule for Derivatives]] keeping $y$ constant
+}}
+{{eqn | r = x^{y - 2} \paren {y + \paren {y \ln x + 1} \paren {y - 1} }
+      | c = simplifying
+}}
+{{eqn | r = x^{y - 2} \paren {y + y^2 \ln x + y -y \ln x - 1}
+      | c = multiplying out
+}}
+{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial u} {\partial x}
+      | r = y x^{y - 1}
+      | c = [[Power Rule for Derivatives]] keeping $y$ constant
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial^2 u} {\partial y \partial x}
+      | r = \map {\dfrac \partial {\partial y} } {\dfrac {\partial u} {\partial x} }
+      | c = {{Defof|Second Partial Derivative}}
+}}
+{{eqn | r = x^{y - 1} \map {\dfrac \partial {\partial y} } y + y \map {\dfrac \partial {\partial y} } {x^{y - 1} }
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = x^{y - 1} + y x^{y - 1} \ln x
+      | c = [[Derivative of Identity Function]], [[Derivative of General Logarithm Function]] keeping $x$ constant
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\partial^3 u} {\partial x \partial y \partial x}
+      | r = \map {\dfrac \partial {\partial x} } {\dfrac {\partial^2 u} {\partial y \partial x} }
+      | c = {{Defof|Third Partial Derivative}}
+}}
+{{eqn | r = \map {\dfrac \partial {\partial x} } {x^{y - 1} + y x^{y - 1} \ln x}
+      | c = 
+}}
+{{eqn | r = \paren {y - 1} x^{y - 2} + y \paren {x^{y - 1} \map {\dfrac \partial {\partial x} } {\ln x} + \ln x \map {\dfrac \partial {\partial x} } {x^{y - 1} } }
+      | c = [[Power Rule for Derivatives]], [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \paren {y - 1} x^{y - 2} + y x^{y - 1} \paren {\dfrac 1 x} + y \ln x \paren {y - 1} x^{y - 2}
+      | c = [[Derivative of Natural Logarithm]], [[Power Rule for Derivatives]]
+}}
+{{eqn | r = x^{y - 2} \paren {\paren {y - 1} + y + y \ln x \paren {y - 1} }
+      | c = factoring
+}}
+{{eqn | r = x^{y - 2} \paren {2 y + y \paren {y - 1} \ln x - 1}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed|lemma}}
+The two expressions are equal.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Terms in Homogeneous Polynomial}
+Tags: Homogeneous Polynomials
+
+\begin{theorem}
+The number of [[Definition:Term of Polynomial|terms]] in a [[Definition:Homogeneous Polynomial|homogeneous polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$ in $m$ [[Definition:Indeterminate|indeterminates]] is given by:
+:$N = \dbinom {n + m - 1} n = \dfrac {\paren {n + m - 1}!}{n! \, \paren {m - 1}!}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|By induction, probably, but really need to rationalise the existing material on polynomials}}
+\end{proof}
\ No newline at end of file