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+\section{Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Necessary Condition}
+Tags: Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent
+
+\begin{theorem}
+Let $S \subseteq \R$.
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $S \to \R$.
+Let $\sequence {f_n}$ be [[Definition:Uniform Convergence|uniformly convergent]] on $S$.
+Then $\sequence {f_n}$ is [[Definition:Uniform Cauchy Criterion|uniformly Cauchy]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Fix some $\epsilon \in \R_{> 0}$.
+Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that: 
+:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 2$
+for all $x \in S$ and $n > N$. 
+Then if $x \in S$ and $n, m > N$, we have:
+{{begin-eqn}}
+{{eqn	| l = \size {\map {f_n} x - \map {f_m} x}
+	| r = \size {\map {f_n} x - \map f x - \paren {\map {f_m} x - \map f x} }
+}}
+{{eqn	| r = \size {\map {f_n} x - \map f x} + \size {\map {f_m} x - \map f x}
+	| c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn	| o = <
+	| r = \epsilon
+}}
+{{end-eqn}}
+Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is [[Definition:Uniform Cauchy Criterion|uniformly Cauchy]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball of Point Inside Open Ball/Normed Vector Space}
+Tags: Open Balls
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\map {B_\epsilon} x$ be an [[Definition:Open Ball in Normed Vector Space|open $\epsilon$-ball]] in $M = \struct{X, \norm {\, \cdot \,}}$.
+Let $y \in \map {B_\epsilon} x$.
+Then:
+: $\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
+That is, for every point in an [[Definition:Open Ball in Normed Vector Space|open $\epsilon$-ball]] in a [[Definition:Normed Vector Space|normed vector space]], there exists an [[Definition:Open Ball in Normed Vector Space|open $\delta$-ball]] of that point entirely contained within that [[Definition:Open Ball in Normed Vector Space|open $\epsilon$-ball]].
+\end{theorem}
+
+\begin{proof}
+Let $\delta = \epsilon - \norm {x - y}$.
+From the definition of [[Definition:Open Ball in Normed Vector Space|open ball]], this is [[Definition:Strictly Positive|strictly positive]], since $y \in \map {B_\epsilon} x$.
+If $z \in \map {B_\delta} y$, then $\norm {y - z} < \delta$.
+So:
+: $\norm {x - z} \le \norm {x - y} + \norm {y - z} < \norm {x - y} + \delta = \epsilon$
+Thus $z \in \map {B_\epsilon} x$.
+So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function}
+Tags: Uniform Convergence, Continuous Functions
+
+\begin{theorem}
+Let $S \subseteq \R$.
+Let $x \in S$.
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $S \to \R$ [[Definition:Uniform Convergence|converging uniformly]] to $f : S \to \R$.
+Let $f_n$ be [[Definition:Continuous Real Function|continuous]] at $x$ for all $n \in \N$.
+Then $f$ is [[Definition:Continuous Real Function|continuous]] at $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{> 0}$. 
+Since $f_n \to f$ [[Definition:Uniform Convergence|uniformly]], there exists some $N \in \N$ such that: 
+:$\size {\map {f_n} x - \map f x} < \dfrac \epsilon 3$
+for all $x \in S$ and $n \ge N$.
+Since $f_N$ is [[Definition:Continuous Real Function|continuous]] at $x$, there exists some $\delta > 0$ such that: 
+:for all $y$ with $\size {x - y} < \delta$, we have $\size {\map {f_N} x - \map {f_N} y} < \dfrac \epsilon 3$
+Then for $y$ with $\size {x - y} < \delta$ we have: 
+{{begin-eqn}}
+{{eqn	| l = \size {\map f x - \map f y}
+	| r = \size {\map f x - \map {f_N} x + \map {f_N} x - \map {f_N} y + \map {f_N} y - \map f y}
+}}
+{{eqn	| r = \size {\paren {\map f x - \map {f_N} x} + \paren {\map {f_N} x - \map {f_N} y} + \paren {\map {f_N} y - \map f y} }
+}}
+{{eqn	| o = \le
+	| r = \size {\map f x - \map {f_N} x} + \size {\map {f_N} x - \map {f_N} y} + \size {\map {f_N} y - \map f y}
+	| c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn	| o = <
+	| r = 3 \times \frac \epsilon 3
+}}
+{{eqn	| r = \epsilon
+}}
+{{end-eqn}}
+Since $\epsilon$ was arbitrary, $f$ is [[Definition:Continuous Real Function|continuous]] at $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball is Open Set/Normed Vector Space}
+Tags: Open Balls, Open Sets
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $x \in X$.
+Let $\epsilon \in \R_{>0}$.
+Let $\map {B_\epsilon} x$ be an [[Definition:Open Ball in Normed Vector Space|open $\epsilon$-ball]] of $x$ in $M$.
+Then $\map {B_\epsilon} x$ is an [[Definition:Open Set in Normed Vector Space|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in \map {B_\epsilon} x$.
+From [[Open Ball of Point Inside Open Ball in Normed Vector Space]], there exists $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_\epsilon} x$
+The result follows from the definition of [[Definition:Open Set in Normed Vector Space|open set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Intersection of Open Sets of Normed Vector Space is Open}
+Tags: Open Sets, Set Intersection
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $U_1, U_2, \ldots, U_n$ be [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+Then $\displaystyle \bigcap_{i \mathop = 1}^n U_i$ is [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+That is, a [[Definition:Finite Intersection|finite intersection]] of [[Definition:Open Set in Normed Vector Space|open subsets]] is [[Definition:Open Set in Normed Vector Space|open]].
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle x \in \bigcap_{i \mathop = 1}^n U_i$.
+For each $i \in \closedint 1 n$, we have $x \in U_i$.
+Thus:
+: $\exists \epsilon_i > 0: \map {B_{\epsilon_i}} x \subseteq U_i$
+where $\map {B_{\epsilon_i}} x$ is the [[Definition:Open Ball in Normed Vector Space|open $\epsilon_i$-ball]] of $x$.
+Let $\displaystyle \epsilon = \min_{i \mathop = 1}^n \set {\epsilon_i}$.
+So:
+: $\epsilon > 0$.
+Let $y \in \map {B_\epsilon} x$.
+Then $\norm {x - y} < \epsilon$
+Hence:
+:$\forall i \in \closedint 1 n : \norm {x - y} < \epsilon_i$
+In other words:
+: $\map {B_\epsilon} x \subseteq \map {B_{\epsilon_i}} x \subseteq U_i$
+for all $i \in \closedint 1 n$.
+So:
+: $\displaystyle \map {B_\epsilon} x \subseteq \bigcap_{i \mathop = 1}^n U_i$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Open Sets of Normed Vector Space is Open}
+Tags: Open Sets, Set Union
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+The [[Definition:Set Union|union]] of a [[Definition:Set|set]] of [[Definition:Open Set in Normed Vector Space|open sets]] of $M$ is [[Definition:Open Set in Normed Vector Space|open in $M$]].
+\end{theorem}
+
+\begin{proof}
+Let $I$ be any [[Definition:Indexing Set|indexing set]].
+Let $U_i$ be [[Definition:Open Set in Normed Vector Space|open in $M$]] for all $i \in I$.
+Let $\displaystyle x \in \bigcup_{i \mathop \in I} U_i$.
+Then $x \in U_k$ for some $k \in I$.
+Since $U_k$ is [[Definition:Open Set in Normed Vector Space|open in $M$]]:
+: $\displaystyle \exists \epsilon > 0: \map {B_\epsilon} x \subseteq U_k \subseteq \bigcup_{i \mathop \in I} U_i$
+where $\map {B_\epsilon} x$ is the [[Definition:Open Ball in Normed Vector Space|open $\epsilon$-ball]] of $x$ in $M$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions}
+Tags: Integral Calculus, Uniform Convergence
+
+\begin{theorem}
+Let $a, b \in \R$ with $a < b$. 
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Riemann Integrable Function|Riemann integrable]] [[Definition:Real Function|real functions]] $\closedint a b \to \R$ [[Definition:Uniform Convergence|converging uniformly]] to $f : \closedint a b \to \R$.
+Then $f$ is integrable and: 
+:$\displaystyle \int_a^b \map f x \rd x = \lim_{n \to \infty} \int_a^b \map {f_n} x \rd x$
+\end{theorem}
+
+\begin{proof}
+By [[Limit of Uniformly Convergent Sequence of Integrable Functions is Integrable]], $f$ is integrable.
+We have:
+{{begin-eqn}}
+{{eqn	| l = \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x}
+	| r = \size {\int_a^b \paren {\map f x - \map {f_n} x} \rd x}
+}}
+{{eqn	| o = \le
+	| r = \int_a^b \size {\map f x - \map {f_n} x} \rd x
+	| c = [[Triangle Inequality for Integrals]]
+}}
+{{eqn	| o = \le
+	| r = \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x}
+	| c = [[Upper and Lower Bounds of Integral]]
+}}
+{{end-eqn}}
+Let $\varepsilon \in \R_{> 0}$.
+Since $f_n \to f$ uniformly, we can find $N \in \N$ such that for $n > N$ we have:
+:$\displaystyle \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \frac \varepsilon {b - a}$
+by the definition of [[Definition:Uniform Convergence|uniform convergence]].
+So, for $n > N$ we have: 
+:$\displaystyle \paren {b - a} \sup_{x \in \closedint a b} \size {\map f x - \map {f_n} x} < \varepsilon$
+and hence:
+:$\displaystyle \size {\int_a^b \map f x \rd x - \int_a^b \map {f_n} x \rd x} < \varepsilon$ 
+Since $\varepsilon$ was arbitrary it follows that: 
+:$\displaystyle \lim_{n \to \infty} \int_a^b \map {f_n} x \rd x = \int_a^b \map f x \rd x$
+as required.
+{{qed}}
+[[Category:Integral Calculus]]
+[[Category:Uniform Convergence]]
+hpgnx1cl23avi8v6oqmw2otmsa8ocqx
+\end{proof}<|endoftext|>
+\section{Events One of Which equals Union}
+Tags: Unions of Events, Events One of Which equals Union
+
+\begin{theorem}
+Let the [[Definition:Probability Space|probability space]] of an [[Definition:Experiment|experiment]] $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
+Let $A, B \in \Sigma$ be [[Definition:Event|events]] of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
+Let $A$ and $B$ be such that:
+:$A \cup B = A$
+Then whenever $B$ [[Definition:Occurrence of Event|occurs]], it is always the case that $A$ [[Definition:Occurrence of Event|occurs]] as well.
+\end{theorem}
+
+\begin{proof}
+From [[Union with Superset is Superset]]:
+:$A \cup B = A \iff B \subseteq A$
+Let $B$ [[Definition:Occurrence of Event|occur]].
+Let $\omega$ be the [[Definition:Outcome|outcome]] of $\EE$.
+Let $\omega \in B$.
+That is, by definition of [[Definition:Occurrence of Event|occurrence of event]], $B$ [[Definition:Occurrence of Event|occurs]].
+Then by definition of [[Definition:Subset|subset]]:
+:$\omega \in A$
+Thus by definition of [[Definition:Occurrence of Event|occurrence of event]], $A$ [[Definition:Occurrence of Event|occurs]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Complement of Union/Corollary}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$T_1 \cup T_2 = \overline {\overline T_1 \cap \overline T_2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = T_1 \cup T_2
+      | r = \overline {\overline {T_1 \cup T_2} }
+      | c = [[Complement of Complement]]
+}}
+{{eqn | r = \overline {\overline T_1 \cap \overline T_2}
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Union|De Morgan's Laws: Complement of Union]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Closed Sets is Closed/Normed Vector Space}
+Tags: Closed Sets
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Then the [[Definition:Set Intersection|intersection]] of an arbitrary number of [[Definition:Closed Set in Normed Vector Space|closed sets]] of $M$ (either [[Definition:Finite|finitely]] or [[Definition:Infinite|infinitely]] many) is itself [[Definition:Closed Set in Normed Vector Space|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] (either [[Definition:Finite|finite]] or [[Definition:Infinite|infinite]]).
+Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the [[Definition:Set Intersection|intersection]] of a [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Closed Set in Normed Vector Space|closed sets]] of $M$ indexed by $I$.
+By definition of [[Definition:Closed Set in Normed Vector Space|closed set]], each of $X \setminus V_i$ are by definition [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+From [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's laws: Difference with Intersection]]:
+:$\displaystyle X \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$
+We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i}$ is the [[Definition:Union of Family|union]] of a [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Open Set in Normed Vector Space|open sets]] of $M$ indexed by $I$.
+By [[Union of Open Sets of Normed Vector Space is Open]], $\displaystyle \bigcup_{i \mathop \in I} \paren {X \setminus V_i} = X \setminus \bigcap_{i \mathop \in I} V_i$ is likewise [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+Then by definition of [[Definition:Closed Set in Normed Vector Space|closed set]], $\displaystyle \bigcap_{i \mathop \in I} V_i$ is [[Definition:Closed Set in Normed Vector Space|closed]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normed Vector Space is Closed in Itself}
+Tags: Normed Vector Spaces, Closed Sets
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Then $X$ is [[Definition:Closed Set in Normed Vector Space|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Open in Normed Vector Space]], $\O$ is [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+But:
+:$X = \relcomp X \O$
+where $\complement_X$ denotes the [[Definition:Relative Complement|set complement relative to $X$]].
+The result follows by definition of [[Definition:Closed Set in Normed Vector Space|closed set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Series of Functions is Uniformly Convergent iff Sequence of Partial Sums is Uniformly Cauchy}
+Tags: Uniform Convergence
+
+\begin{theorem}
+Let $S \subseteq \R$.
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $S \to \R$.
+Then the infinite series: 
+:$\displaystyle \sum_{n \mathop = 1}^\infty f_n$
+[[Definition:Uniform Convergence/Infinite Series|converges uniformly]] on $S$ {{iff}} for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:
+:$\displaystyle \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$
+for all $x \in S$ and $n > m > N$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {s_n}$ be a sequence of real functions $S \to \R$ with: 
+:$\displaystyle \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_k} x$
+for each $n \in \N$ and $x \in S$.
+By the [[Definition:Uniform Convergence/Infinite Series|definition of uniform convergence of an infinite series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty f_n$ is uniformly convergent {{iff}} $\sequence {s_n}$ is [[Definition:Uniform Convergence/Real Numbers|uniformly convergent]].
+By [[Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent]]:
+:$\sequence {s_n}$ is uniformly convergent {{iff}} it is [[Definition:Uniform Cauchy Criterion|uniformly Cauchy]].
+That is for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that for all $m, n > N$ we have:
+:$\size {\map {s_n} x - \map {s_m} x} < \varepsilon$ for all $x \in S$.
+Note that if $n = m$: 
+:$\size {\map {s_n} x - \map {s_m} x} = 0 < \varepsilon$
+{{WLOG}}, we can therefore take $n > m$.
+If $n > m$, then:
+{{begin-eqn}}
+{{eqn	| l = \size {\map {s_n} x - \map {s_m} x}
+	| r = \size {\sum_{k \mathop = 1}^n \map {f_k} x - \sum_{k \mathop = 1}^m \map {f_k} x}
+}}
+{{eqn	| r = \size {\sum_{k \mathop = m + 1}^n \map {f_k} x}
+}}
+{{end-eqn}}
+So $\sequence {s_n}$ is uniformly Cauchy {{iff}} for all $\varepsilon \in \R_{> 0}$ there exists $N \in \N$ such that:
+:$\displaystyle \size {\sum_{k \mathop = m + 1}^n \map {f_k} x} < \varepsilon$
+for all $n > m > N$ and $x \in S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection/Corollary}
+Tags: De Morgan's Laws
+
+\begin{theorem}
+:$T_1 \cap T_2 = \overline {\overline T_1 \cup \overline T_2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = T_1 \cap T_2
+      | r = \overline {\overline {T_1 \cap T_2} }
+      | c = [[Complement of Complement]]
+}}
+{{eqn | r = \overline {\overline T_1 \cup \overline T_2}
+      | c = [[De Morgan's Laws (Set Theory)/Set Complement/Complement of Intersection|De Morgan's Laws: Complement of Intersection]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{LCM of 3 Integers in terms of GCDs of Pairs of those Integers/Lemma}
+Tags: Greatest Common Divisor, LCM of 3 Integers in terms of GCDs of Pairs of those Integers
+
+\begin{theorem}
+Let $a, b, c \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|strictly positive integers]].
+Then:
+:$\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = \gcd \set {a, b, c}$
+\end{theorem}
+
+\begin{proof}
+Let $\gcd \set {a, b, c} = d_1$.
+From [[Definition:Greatest Common Divisor of Integers|definition]]:
+:$d_1 \divides a$, $d_1 \divides b$ and $d_1 \divides c$.
+By [[Common Divisor Divides GCD]]:
+:$d_1 \divides \gcd \set {a, b}$ and $d_1 \divides \gcd \set {a, c}$.
+By [[Common Divisor Divides GCD]] again:
+:$d_1 \divides \gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$.
+On the other hand, let $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } = d_2$.
+From [[Definition:Greatest Common Divisor of Integers|definition]]:
+:$d_2 \divides \gcd \set {a, b}$ and $d_2 \divides \gcd \set {a, c}$.
+From [[Definition:Greatest Common Divisor of Integers|definition]] again:
+:$d_2 \divides a$, $d_2 \divides b$ and $d_2 \divides c$.
+Hence $d_2 \divides \gcd \set {a, b, c}$.
+Since $\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} }$ and $\gcd \set {a, b, c}$ divide each other, by [[Absolute Value of Integer is not less than Divisors]] they must be equal.
+{{qed}}
+[[Category:Greatest Common Divisor]]
+[[Category:LCM of 3 Integers in terms of GCDs of Pairs of those Integers]]
+1breqfkqm4awy95mtgjcqzl0krs2783
+\end{proof}<|endoftext|>
+\section{Events One of Which equals Intersection}
+Tags: Intersections of Events, Events One of Which equals Intersection
+
+\begin{theorem}
+Let the [[Definition:Probability Space|probability space]] of an [[Definition:Experiment|experiment]] $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
+Let $A, B \in \Sigma$ be [[Definition:Event|events]] of $\EE$, so that $A \subseteq \Omega$ and $B \subseteq \Omega$.
+Let $A$ and $B$ be such that:
+:$A \cap B = A$
+Then whenever $A$ [[Definition:Occurrence of Event|occurs]], it is always the case that $B$ [[Definition:Occurrence of Event|occurs]] as well.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection with Subset is Subset]]:
+:$A \cap B = A \iff A \subseteq B$
+Let $A$ [[Definition:Occurrence of Event|occur]].
+Let $\omega$ be the [[Definition:Outcome|outcome]] of $\EE$.
+Let $\omega \in A$.
+That is, by definition of [[Definition:Occurrence of Event|occurrence of event]], $A$ [[Definition:Occurrence of Event|occurs]].
+Then by definition of [[Definition:Subset|subset]]:
+:$\omega \in B$
+Thus by definition of [[Definition:Occurrence of Event|occurrence of event]], $B$ [[Definition:Occurrence of Event|occurs]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Event with Complement is Certainty}
+Tags: Complementary Events, Set Union
+
+\begin{theorem}
+Let the [[Definition:Probability Space|probability space]] of an [[Definition:Experiment|experiment]] $\EE$ be $\struct {\Omega, \Sigma, \Pr}$.
+Let $A \in \Sigma$ be an [[Definition:Event|events]] of $\EE$, so that $A \subseteq \Omega$.
+Then:
+:$A \cup \overline A = \Omega$
+where $\overline A$ is the [[Definition:Complementary Event|complementary event]] to $A$.
+That is, $A \cup \overline A$ is a [[Definition:Certain Event|certainty]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$A \subseteq \Omega$
+and:
+:$\overline A = \relcomp \Omega A$
+From [[Union with Relative Complement]]:
+:$A \cup \overline A = \Omega$
+We then have from [[Definition:Kolmogorov Axioms|Kolmogorov axiom $(2)$]] that:
+:$\map \Pr \Omega = 1$
+The result follows by definition of [[Definition:Certain Event|certainty]].
+{{qed}}
+{{LEM|Union with Relative Complement}}
+[[Category:Complementary Events]]
+[[Category:Set Union]]
+mbw4g8v1z5vt8du09qkcxemct1o1k0x
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 2}
+Tags: Equivalence of Definitions of Adherent Point
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$.
+{{TFAE|def = Adherent Point|view = adherent point of $A$}}
+\end{theorem}
+
+\begin{proof}
+Let $A^-$ denote the [[Definition:Closure (Topology)|closure]] of $A$.
+It is required to be shown that $x \in A^-$ {{iff}}, for every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$, the [[Definition:Set Intersection|intersection]] $A \cap U$ is [[Definition:Non-Empty Set|non-empty]].
+For a [[Definition:Subset|subset]] $H \subseteq S$, let $H^{\complement}$ denote the [[Definition:Relative Complement|relative complement]] of $H$ in $S$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = A \cap U
+      | r = \O
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = A
+      | o = \subseteq
+      | r = U^{\complement}
+      | c = Note that $U^{\complement}$ is [[Definition:Closed Set (Topology)|closed]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = A^-
+      | o = \subseteq
+      | r = U^{\complement}
+      | c = [[Set Closure is Smallest Closed Set in Topological Space]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = U
+      | o = \subseteq
+      | r = \paren {A^-}^{\complement}
+      | c = [[Relative Complement inverts Subsets]] and [[Relative Complement of Relative Complement]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = A^- \cap U
+      | o = =
+      | r = \O
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+: $x \in U \iff x \notin A^-$
+The result follows from the [[Rule of Transposition]].
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Adherent Point/Definition 1 iff Definition 3}
+Tags: Equivalence of Definitions of Adherent Point
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subseteq S$.
+{{TFAE|def = Adherent Point|view = adherent point of $A$}}
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ satisfy:
+:$A \cap U \ne \O$
+Let $N$ be any [[Definition:Neighborhood|neighborhood]] of $x$.
+By definition of a [[Definition:Neighborhood|neighborhood]]:
+:$\exists V \in \tau : x \in V \subseteq N$
+From [[Set is Open iff Neighborhood of all its Points]], $V$ is an [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+Thus:
+:$A \cap V \ne \O$
+By the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Subsets of Disjoint Sets are Disjoint]]:
+:$A \cap N \ne \O$
+Since $N$ was arbitrary then every [[Definition:Neighborhood|neighborhood]] $N$ of $x$ satisfies:
+:$A \cap N \ne \O$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let every [[Definition:Neighborhood|neighborhood]] $N$ of $x$ satisfy:
+:$A \cap N \ne \O$
+By definition, every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ is a [[Definition:Neighborhood|neighborhood]] of $x$.
+So every [[Definition:Open Neighborhood|open neighborhood]] $U$ of $x$ satisfies:
+:$A \cap U \ne \O$
+\end{proof}<|endoftext|>
+\section{Compact Sets in Fortissimo Space}
+Tags: Compact Spaces, Fortissimo Space
+
+\begin{theorem}
+A set in [[Definition:Fortissimo Space|Fortissimo space]] is [[Definition:Compact Topological Subspace|compact]] {{iff}} it is [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Fortissimo Space|Fortissimo space]].
+=== Necessary Condition ===
+By [[Finite Topological Space is Compact]], all [[Definition:Finite Set|finite sets]] are [[Definition:Compact Topological Subspace|compact]] in $T$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+We prove the [[Definition:Contrapositive|contrapositive]].
+Let $A$ be an [[Definition:Infinite Set|infinite]] [[Definition:Subset|subset]] of $S$.
+Let $C$ be a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]] of $A$ that does not contain $p$.
+For each $x \in A$, $C \setminus \set x$ is a [[Definition:Countably Infinite Set|countably infinite set]].
+Hence $\relcomp S {C \setminus \set x} = \relcomp S C \cup \set x$ is [[Definition:Open Set (Topology)|open]] in $T$.
+Thus $\CC = \set {\relcomp S C \cup \set x: x \in A}$ is an [[Definition:Open Cover|open cover]] of $A$.
+However, each set in $\CC$ contains exactly $1$ element in $C$, so a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\CC$ can only contain a [[Definition:Finite Set|finite]] number of elements in $C$.
+Therefore $\CC$ has no [[Definition:Finite Subcover|finite subcover]].
+Hence $A$ is not [[Definition:Compact Topological Subspace|compact]].
+This shows that if a set in [[Definition:Fortissimo Space|Fortissimo space]] is [[Definition:Compact Topological Subspace|compact]], it must be [[Definition:Finite Set|finite]].
+{{qed}}
+[[Category:Compact Spaces]]
+[[Category:Fortissimo Space]]
+jclupkhtqcxchvfqyn4lf2jpzpo26qi
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Boundary}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+{{TFAE|def = Boundary (Topology)|view = boundary of $H$}}
+\end{theorem}
+
+\begin{proof}
+=== Definition $1$ is equivalent to Definition $3$ ===
+This is demonstrated in [[Boundary is Intersection of Closure with Closure of Complement]].
+{{qed|lemma}}
+=== Definition $2$ is equivalent to Definition $3$ ===
+Let $x \in S$.
+By definition of the [[Definition:Closure (Topology)/Definition 6|closure]]:
+:$x \in H^-$ {{iff}} every [[Definition:Neighborhood of Point|neighborhood]] $N$ of $x$ satisfies $H \cap N \ne \O$
+:$x \in \paren{\overline H}^-$ {{iff}} every [[Definition:Neighborhood of Point|neighborhood]] $N$ of $x$ satisfies $\overline H \cap N \ne \O$
+Therefore $x \in H^- \cap \paren{\overline H}^-$ {{iff}} every [[Definition:Neighborhood|neighborhood]] $N$ of $x$ satisfies:
+:$H \cap N \ne \O$
+and
+:$\overline H \cap N \ne \O$
+{{qed}}
+[[Category:Boundaries]]
+0oiw3cdykopt9xx4g9gsbubbas49fah
+\end{proof}<|endoftext|>
+\section{Uniformly Convergent Series of Continuous Functions Converges to Continuous Function}
+Tags: Uniform Convergence, Continuous Functions
+
+\begin{theorem}
+Let $S \subseteq \R$.
+Let $x \in S$.
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]].
+Let $f_n$ be [[Definition:Continuous Function|continuous]] at $x$ for all $n \in \N$.
+Let the infinite series: 
+:$\displaystyle \sum_{n \mathop = 1}^\infty f_n$
+be [[Definition:Uniform Convergence/Infinite Series|uniformly convergent]] to a real function $f : S \to \R$.
+Then $f$ is [[Definition:Continuous Function|continuous]] at $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {s_n}$ be sequence of real functions $S \to \R$ with: 
+:$\displaystyle \map {s_n} x = \sum_{k \mathop = 1}^n \map {f_n} x$
+for each $n \in \N$ and $x \in S$.
+By [[Combination Theorem for Continuous Functions/Sum Rule|Combination Theorem for Continuous Functions: Sum Rule]]:
+:$s_n$ is continuous at $x$ for all $n \in \N$.
+Since additionally $s_n \to f$ uniformly, we have by [[Uniformly Convergent Sequence of Continuous Functions Converges to Continuous Function]]: 
+:$f$ is continuous at $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniformly Convergent Series of Continuous Functions Converges to Continuous Function/Corollary}
+Tags: Uniform Convergence, Continuous Functions
+
+\begin{theorem}
+Let $S \subseteq \R$.
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]].
+Let $f_n$ be [[Definition:Continuous Real Function|continuous]] for all $n \in \N$.
+Let the [[Definition:Infinite Series|infinite series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty f_n$
+be [[Definition:Uniform Convergence/Infinite Series|uniformly convergent]] to a [[Definition:Real Function|real function]] $f : S \to \R$.
+Then $f$ is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then $f_n$ is [[Definition:Continuous Real Function|continuous]] at $x$ for all $n \in \N$.
+Since:
+:$\displaystyle \sum_{n \mathop = 1}^\infty f_n$
+[[Definition:Uniform Convergence/Infinite Series|converges uniformly]] to $f$, we have by [[Uniformly Convergent Series of Continuous Functions Converges to Continuous Function]]:
+:$f$ is [[Definition:Continuous Real Function|continuous]] at $x$.
+As $x \in S$ was arbitrary, we have that: 
+:$f$ is [[Definition:Continuous Real Function|continuous]].
+{{qed}}
+[[Category:Uniform Convergence]]
+[[Category:Continuous Functions]]
+j1ahir88958io84stzghg2mres6r4bv
+\end{proof}<|endoftext|>
+\section{Euclidean Space is Banach Space/Proof 1}
+Tags: Euclidean Space is Banach Space
+
+\begin{theorem}
+Let $m$ be a [[Definition:Positive Integer|positive integer]].
+Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Norm|Euclidean norm]], forms a [[Definition:Banach Space|Banach space]] over $\R$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Euclidean Space|Euclidean space]] $\R^m$ is a [[Definition:Vector Space|vector space]] over $\R$.
+That the [[Definition:Norm Axioms (Vector Space)|norm axioms]] are satisfied is proven in [[Euclidean Space is Normed Space]].
+Then we have [[Euclidean Space is Complete Metric Space]].
+The result follows by the definition of a [[Definition:Banach Space|Banach space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euclidean Space is Banach Space/Proof 2}
+Tags: Euclidean Space is Banach Space
+
+\begin{theorem}
+Let $m$ be a [[Definition:Positive Integer|positive integer]].
+Then the [[Definition:Euclidean Space|Euclidean space]] $\R^m$, along with the [[Definition:Euclidean Norm|Euclidean norm]], forms a [[Definition:Banach Space|Banach space]] over $\R$.
+\end{theorem}
+
+\begin{proof}
+By definition, [[Definition:Euclidean Norm|Euclidean norm]] is the same as [[Definition:P-Norm|$p$-norm]] with $p = 2$.
+Let $\sequence {\mathbf x_n}_{n \mathop \in \N} = \sequence {\tuple {x_n^{\paren 1}, x_n^{\paren 2}, \ldots, x_n^{\paren m}}}_{n \mathop \in \N} $ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\R^m$.
+Let $k \in \N_{> 0} : k \le m$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {x_n^{\paren k} - x_m^{\paren k} }
+      | r = \paren {\paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | o = \le
+      | r = \paren {\sum_{k \mathop = 0}^m \paren {x_n^{\paren k} - x_m^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | r = \norm { {\bf x}_n - {\bf x}_m}_2
+}}
+{{end-eqn}}
+Hence, $\sequence {x_n^{\paren k} }_{n \mathop \in \N}$ is a [[Definition:Real Cauchy Sequence|Cauchy sequence]] in $\R$.
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n^{\paren k} = L^{\paren k}$
+Let $\mathbf L = \tuple {L^{\paren 1}, \ldots, L^{\paren m}} \in \R^m$
+We have that for all $n > N$:
+{{begin-eqn}}
+{{eqn | l = \norm {\mathbf x_n - \mathbf L}
+      | r = \paren {\sum_{k \mathop = 1}^m \size {x^{\paren k}_n - x^{\paren k} }^2}^{\frac 1 2}
+}}
+{{eqn | o = \le
+      | r = \paren {\sum_{k \mathop = 1}^m \frac {\epsilon^2} m}^{\frac 1 2}
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Therefore, $\sequence {\mathbf x_n}_{n \mathop \in \N}$ converges to $\mathbf L$ in $\struct {\R^m, \norm {\, \cdot \,}_2}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Function is of Bounded Variation}
+Tags: Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f : \closedint a b \to \R$ be a [[Definition:Monotone (Order Theory)/Real Function|monotone function]].
+Then $f$ is of [[Definition:Bounded Variation|bounded variation]].
+\end{theorem}
+
+\begin{proof}
+We use the notation from the [[Definition:Bounded Variation|definition of bounded variation]].
+Let $P = \set {x_0, x_1, \ldots, x_{\size P - 1} }$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
+As $f$ is [[Definition:Monotone (Order Theory)/Real Function|monotone]], it is either [[Definition:Increasing/Real Function|increasing]] or [[Definition:Decreasing/Real Function|decreasing]].
+First consider the case of $f$ [[Definition:Increasing/Real Function|increasing]], then: 
+:$\map f {x_i} \ge \map f {x_{i - 1} }$
+for all $i \in \N$ with $i \le \size P - 1$.
+By the definition of [[Definition:Absolute Value|the absolute value]], we have: 
+:$\size {\map f {x_i} - \map f {x_{i - 1} } } = \map f {x_i} - \map f {x_{i - 1} }$
+Therefore:
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} P
+	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^{\size P - 1} \paren {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| r = \map f {x_{\size P - 1} } - \map f {x_0}
+	| c = [[Telescoping Series/Example 2|Telescoping Series: Example 2]]
+}}
+{{eqn	| r = \map f b - \map f a
+}}
+{{end-eqn}}
+for all [[Definition:Finite Subdivision|finite subdivisions]] $P$.
+Note that $\map f b - \map f a$ is independent of the subdivision $P$.
+Therefore if $f$ is [[Definition:Increasing/Real Function|increasing]], it is of [[Definition:Bounded Variation|bounded variation]].
+Now consider the case of $f$ [[Definition:Decreasing/Real Function|decreasing]].
+We instead have:
+:$\map f {x_i} \le \map f {x_{i - 1} }$
+for all $i \in \N$ with $i \le \size P - 1$.
+Hence, by the definition of [[Definition:Absolute Value|the absolute value]], we instead have:
+:$\size {\map f {x_i} - \map f {x_{i - 1} } } = -\paren {\map f {x_i} - \map f {x_{i - 1} } }$
+Therefore:
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} P
+	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| r = -\sum_{i \mathop = 1}^{\size P - 1} \paren {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| r = -\paren {\map f b - \map f a}
+	| c = as in the previous calculation
+}}
+{{eqn	| r = \map f a - \map f b
+}}
+{{end-eqn}}
+for all [[Definition:Finite Subdivision|finite subdivisions]] $P$.
+Note again that $\map f a - \map f b$ is independent of the subdivision $P$. 
+Therefore if $f$ is [[Definition:Decreasing/Real Function|decreasing]], it is of [[Definition:Bounded Variation|bounded variation]].
+Hence, if $f$ is either increasing or decreasing, it is of bounded variation.
+Hence if $f$ is [[Definition:Monotone (Order Theory)/Real Function|monotone]], it is of bounded variation.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiable Function with Bounded Derivative is of Bounded Variation}
+Tags: Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
+Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$, with [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Derivative|derivative]].
+Then $f$ is of [[Definition:Bounded Variation|bounded variation]].
+\end{theorem}
+
+\begin{proof}
+For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write: 
+:$P = \set {x_0, x_1, \ldots, x_n }$
+with:
+:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
+Since the [[Definition:Derivative|derivative]] of $f$ is bounded, there exists some $M \in \R$ such that: 
+:$\size {\map {f'} x} \le M$
+for all $x \in \openint a b$.
+By the [[Mean Value Theorem]], for each $i \in \N$ with $i \le n$, there exists $\xi_i \in \openint {x_{i - 1} } {x_i}$ such that: 
+:$\map {f'} {\xi_i} = \dfrac {\map f {x_i} - \map f {x_{i - 1} } } {x_i - x_{i - 1} }$
+Note that, from the boundedness of $f'$: 
+:$\size {\map {f'} {\xi_i} } \le M$
+We also have from the fact that $x_i > x_{i - 1}$: 
+:$\size {x_i - x_{i - 1} } = x_i - x_{i - 1}$
+So, for each $i$: 
+{{begin-eqn}}
+{{eqn	| l = \size {\map f {x_i} - \map f {x_{i - 1} } }
+	| r = \size {\map {f'} {\xi_i} } \paren {x_i - x_{i - 1} }
+}}
+{{eqn	| o = \le
+	| r = M \paren {x_i - x_{i - 1} }
+}}
+{{end-eqn}}
+We therefore have: 
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} P
+	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }
+	| c = using the notation from the definition of [[Definition:Bounded Variation|bounded variation]]
+}}
+{{eqn	| o = \le
+	| r = M \sum_{i \mathop = 1}^n \paren {x_i - x_{i - 1} }
+}}
+{{eqn	| r = M \paren {x_n - x_0}
+	| c = [[Telescoping Series/Example 2|Telescoping Series: Example 2]]
+}}
+{{eqn	| r = M \paren {b - a}
+}}
+{{end-eqn}}
+for all [[Definition:Finite Subdivision|finite subdivisions]] $P$.
+So $f$ is of [[Definition:Bounded Variation|bounded variation]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Total Variation is Non-Negative}
+Tags: Total Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f : \closedint a b \to \R$ be a [[Definition:Real Function|function]] of [[Definition:Bounded Variation|bounded variation]].
+Let $V_f$ be the [[Definition:Total Variation|total variation]] of $f$ on $\closedint a b$.
+Then: 
+:$V_f \ge 0$
+with equality {{iff}} $f$ is [[Definition:Constant Mapping|constant]].
+\end{theorem}
+
+\begin{proof}
+We use the notation from the [[Definition:Bounded Variation|definition of bounded variation]].
+Note that by the definition of [[Definition:Absolute Value|absolute value]], we have: 
+:$\size x \ge 0$
+for all $x \in \R$.
+Let $P$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
+Then: 
+:$\displaystyle \map {V_f} P = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } } \ge 0$
+So, by the definition of [[Definition:Supremum of Mapping/Real-Valued Function|supremum]], we have: 
+:$\displaystyle V_f = \sup_P \paren {\map {V_f} P} \ge 0$
+Let $f$ be [[Definition:Constant Mapping|constant]].
+Then: 
+:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 0$
+for all $x_{i - 1}, x_i \in \closedint a b$.
+So, for any finite subdivision $P$ we have: 
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} P
+	| r = \sum_{i \mathop = 1}^{\size P - 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^{\size P - 1} 0
+}}
+{{eqn	| r = 0
+}}
+{{end-eqn}}
+We then have: 
+{{begin-eqn}}
+{{eqn	| l = V_f
+	| r = \sup_P \paren {\map {V_f} P}
+	| c = {{Defof|Total Variation}}
+}}
+{{eqn	| r = \sup \set {0}
+}}
+{{eqn	| r = 0
+	| c = {{Defof|Supremum of Subset of Real Numbers}}
+}}
+{{end-eqn}}
+So if $f$ is [[Definition:Constant Mapping|constant]], then $f$ is of [[Definition:Bounded Variation|bounded variation]] with:
+:$V_f = 0$
+It remains to show that if $V_f = 0$, then $f$ is constant.
+It suffices to show that if $f$ is not constant then $V_f > 0$.
+Since $f$ is non-constant, we can pick $x \in \openint a b$ such that either: 
+:$\map f x \ne \map f a$
+or:
+:$\map f x \ne \map f b$
+So that either:
+:$\size {\map f x - \map f a} > 0$
+or:
+:$\size {\map f b - \map f x} > 0$
+Note that we then have: 
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} {\set {a, x, b} }
+	| r = \size {\map f x - \map f a} + \size {\map f b - \map f x}
+}}
+{{eqn	| o = >
+	| r = 0
+}}
+{{end-eqn}}
+We must then have: 
+:$V_f \ge \map {V_f} {\set {a, x, b} } > 0$
+Hence the claim.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Norm Equivalence is Equivalence}
+Tags: Equivalence Relations, Norm Theory, Vector Spaces
+
+\begin{theorem}
+Let $X$ be a [[Definition:Vector Space|vector space]].
+Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be [[Definition:Equivalence of Norms|equivalent norms]] on $X$.
+Denote this [[Definition:Relation|relation]] by $\sim$: 
+:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+=== Reflexivity ===
+Let $\norm {\, \cdot \,}$ be a [[Definition:Norm|norm]] on $X$.
+Then for all $x \in X$ we have that:
+:$\norm x = 1 \cdot \norm {x}$.
+Therefore:
+:$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$
+Hence:
+:$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.
+{{qed|lemma}}
+=== Symmetry ===
+Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
+Then:
+:$\exists m, M \in \R_{> 0} : m \le M : \forall x \in X : m \norm x_b \le \norm x_a \le M \norm x_b$
+Consider two [[Definition:Inequality|inequalities]], obtained by [[Definition:Real Division|division]] by $M$ and $m$:
+:$\dfrac m M \norm x_b \le \dfrac 1 M \norm x_a \le \norm x_b$
+:$\norm x_b \le \dfrac 1 m \norm x_a \le \dfrac M m \norm x_b$
+Notice, that: 
+:$\dfrac 1 M \norm x_a \le \norm x_b \le \dfrac 1 m \norm x_a$
+We have that $m \le M$ implies $\dfrac 1 m \ge \dfrac 1 M$.
+Define $C := \dfrac 1 m$ and $c := \dfrac 1 M$.
+Hence, $c \le C$ and:
+:$\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_a$
+{{qed|lemma}}
+=== Transitivity ===
+Suppose, $\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$ and $\norm {\, \cdot \,}_b \sim \norm {\, \cdot \,}_c$.
+ 
+Then:
+:$\exists m_{a b}, M_{a b} \in \R_{> 0} : m_{a b} \le M_{a b} : \forall x \in X : m_{a b} \norm x_b \le \norm x_a \le M_{a b} \norm x_b$ 
+:$\exists m_{b c}, M_{b c} \in \R_{> 0} : m_{b c} \le M_{b c} : \forall x \in X : m_{b c} \norm x_c \le \norm x_b \le M_{b c} \norm x_c$
+Generate $2$ more [[Definition:Inequality|inequalities]] by [[Definition:Real Multiplication|multiplying]] the second [[Definition:Inequality|inequality]] by $m_{a b}$ and $M_{a b}$:
+:$m_{a b} m_{b c} \norm x_c \le m_{a b} \norm x_b \le m_{a b} M_{b c} \norm x_c$
+:$M_{a b} m_{b c} \norm x_c \le M_{a b} \norm x_b \le M_{a b} M_{b c} \norm x_c$
+From above it follows that:
+{{begin-eqn}}
+{{eqn | l = m_{a b} m_{b c} \norm x_c
+      | o = \le 
+      | r = m_{a b} \norm x_b 
+}}
+{{eqn | o = \le 
+      | r = \norm x_a 
+}}
+{{eqn | o = \le 
+      | r = M_{a b} \norm x_b 
+}}
+{{eqn | o = \le 
+      | r = M_{a b} M_{b c} \norm x_c
+}}
+{{end-eqn}}
+Define $k := m_{a b} m_{b c}$ and $K := M_{a b} M_{b c}$.
+Then $k \le K$ and:
+:$k \norm x_c \le \norm x_a \le K \norm x_c$
+Therefore:
+:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_c$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Non-Negative Real Function with Zero Integral is Zero Function}
+Tags: Definite Integrals, Continuous Non-Negative Real Function with Zero Integral is Zero Function
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$. 
+Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]]. 
+Let: 
+:$\map f x \ge 0$
+for all $x \in \closedint a b$.
+Let: 
+:$\displaystyle \int_a^b \map f x \rd x = 0$
+Then $\map f x = 0$ for all $x \in \closedint a b$.
+\end{theorem}
+
+\begin{proof}
+From [[Definite Integral of Constant]], if $\map f x = 0$ for all $x \in \closedint a b$, then: 
+:$\displaystyle \int_a^b \map f x \rd x = 0$
+We want to show that if:
+:$\displaystyle \int_a^b \map f x \rd x = 0$
+then:
+:$\map f x = 0$ for all $x \in \closedint a b$.
+Since $\map f x \ge 0$, by [[Relative Sizes of Definite Integrals]]:
+:$\displaystyle \int_a^b \map f x \rd x \ge 0$
+It therefore suffices to show that if: 
+:$\map f x > 0$ for some $x \in \closedint a b$
+then:
+:$\displaystyle \int_a^b \map f x \rd x > 0$
+We split this into three cases: 
+:$x = a$
+:$a < x < b$
+:$x = b$
+since [[Definition:Continuous Real Function|continuity]] on [[Definition:Interval/Ordered Set/Endpoint|endpoints]] is defined slightly differently.
+Consider the case: 
+:$a < x < b$
+By [[Definition:Continuous Real Function/Point|continuity]] at $x$ we have that there exists $\delta > 0$ such that: 
+:for all $y \in \openint {x - \delta} {x + \delta}$ we have $\size {\map f y - \map f x} < \dfrac {\map f x} 2$
+In particular for $y \in \openint {x - \delta} {x + \delta}$ we have: 
+:$0 < \dfrac {\map f x} 2 < \map f y$
+Pick $\delta$ sufficiently small so that $\openint {x - \delta} {x + \delta} \subseteq \closedint a b$
+We then have:
+{{begin-eqn}}
+{{eqn	| l = \int_a^b \map f y \rd y
+	| r = \int_a^{x - \delta} \map f y \rd y + \int_{x - \delta}^{x + \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y
+	| c = [[Sum of Integrals on Adjacent Intervals for Continuous Functions]]
+}}
+{{eqn	| o = \ge
+	| r = \int_{x - \delta}^{x + \delta} \map f y \rd y
+	| c = [[Relative Sizes of Definite Integrals]] gives $\displaystyle \int_a^{x - \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y \ge 0$
+}}
+{{eqn	| o = >
+	| r = \delta \map f x
+	| c = [[Relative Sizes of Definite Integrals]], [[Definite Integral of Constant]]
+}}
+{{eqn	| o = >
+	| r = 0
+}}
+{{end-eqn}}
+In the case $x = a$, by the definition of [[Definition:Continuous Real Function/Right-Continuous/Point|right continuity]], there exists $\delta > 0$ such that: 
+:for all $x \in \openint a {a + \delta}$ we have $\size {\map f x - \map f a} < \dfrac {\map f a} 2$
+That is, there exists some $x \in \openint a b$ such that: 
+:$\map f x > \dfrac {\map f a} 2 > 0$
+So the former proof applies in this case. 
+The case $x = b$ follows similarly.
+In the case $x = b$, by the definition of [[Definition:Continuous Real Function/Left-Continuous/Point|left continuity]], there exists $\delta > 0$ such that: 
+:for all $x \in \openint {b - \delta} b$ we have $\size {\map f x - \map f b} < \dfrac {\map f b} 2$
+That is, there exists some $x \in \openint a b$ such that: 
+:$\map f x > \dfrac {\map f b} 2 > 0$
+So, again, the proof for the case $a < x < b$ applies.
+We have covered all three cases, so we are done.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
+Let $F : \closedint a b \to \R$ be a [[Definition:Real Function|real function]] defined by: 
+:$\displaystyle \map F x = \int_a^x \map f x \rd x$
+We are assured that this function is [[Definition:Well-Defined|well-defined]], since $f$ is [[Definition:Darboux Integrable Function|integrable]] on $\closedint a b$.
+From [[Fundamental Theorem of Calculus/First Part|Fundamental Theorem of Calculus: First Part]], we have: 
+:$F$ is [[Definition:Continuous Real Function|continuous]] on $\closedint a b$
+:$F$ is [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$ with [[Definition:Derivative of Real Function|derivative]] $f$
+Note that: 
+{{begin-eqn}}
+{{eqn	| l = \map {F'} x
+	| r = \map f x
+}}
+{{eqn	| o = \ge
+	| r = 0
+}}
+{{end-eqn}}
+for all $x \in \openint a b$. 
+We therefore have, by [[Real Function with Positive Derivative is Increasing]]: 
+:$F$ is [[Definition:Increasing/Real Function|increasing]] on $\closedint a b$.
+However, by hypothesis:
+{{begin-eqn}}
+{{eqn	| l = \map F b
+	| r = \int_a^b \map f x \rd x
+}}
+{{eqn	| r = 0
+}}
+{{eqn	| r = \int_a^a \map f x \rd x
+	| c = [[Definite Integral on Zero Interval]]
+}}
+{{eqn	| r = \map F a
+}}
+{{end-eqn}}
+So, it must be the case that: 
+:$\map F x = 0$ for all $x \in \closedint a b$.
+We therefore have, from [[Derivative of Constant]]: 
+:$\map {F'} x = \map f x = 0$ for all $x \in \closedint a b$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition}
+Tags: Sequential Continuity is Equivalent to Continuity in the Reals
+
+\begin{theorem}
+Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $c \in A$.
+Let $f : A \to \R$ be a [[Definition:Real Function|real function]].
+Then if $f$ is [[Definition:Continuous Real Function|continuous]] at $c$:
+:for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ in $A$ that [[Definition:Convergent Real Sequence|converges]] to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
+\end{theorem}
+
+\begin{proof}
+It suffices to show that if $f$ is [[Definition:Discontinuous Real Function|discontinuous]] at $c$:
+:there exists a [[Definition:Real Sequence|real sequence]] $\sequence {x_n}$ in $A$ and $x \in \R$ such that $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to $c$ but $\sequence {\map f {x_n} }$ does not converge to $\map f c$.
+As $f$ is discontinuous, there exists some $\varepsilon > 0$ such that for all $\delta > 0$:
+:there exists $x \in A$ with $\size {x - c} < \delta$ we have $\size {\map f x - \map f c} \ge \varepsilon$.
+Using this property, we can construct a sequence $\sequence {x_n}$ as follows:
+:for each $n \in \N$, pick $x_n \in A$ such that $\size {x_n - c} \le \dfrac 1 n$ and $\size {\map f {x_n} - \map f c} \ge \varepsilon$
+Note that since: 
+:$\displaystyle \lim_{n \to \infty} \frac 1 n = 0$
+We have by the [[Squeeze Theorem/Sequences/Real Numbers|squeeze theorem for sequences of real numbers]]: 
+:$\displaystyle \lim_{n \to \infty} \size {x_n - c} = 0$
+so $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to $c$.
+However $\sequence {\map f {x_n} }$ cannot [[Definition:Convergent Real Sequence|converge]] to $\map f c$ since:
+:$\size {\map f {x_n} - \map f c} \ge \varepsilon > 0$
+for all $n \in \N$.
+Therefore, our $\sequence {x_n}$ satisfies our original demand.
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition}
+Tags: Sequential Continuity is Equivalent to Continuity in the Reals
+
+\begin{theorem}
+Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $c \in A$.
+Let $f : A \to \R$ be a [[Definition:Real Function|real function]].
+Then if $f$ is [[Definition:Continuous Real Function|continuous]] at $c$:
+:for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ in $A$ that [[Definition:Convergent Real Sequence|converges]] to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
+\end{theorem}
+
+\begin{proof}
+Let $c \in \R$. 
+Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence]] in $A$ that [[Definition:Convergent Real Sequence|converges]] to $c$.
+Let $\varepsilon \in \R_{> 0}$. 
+Since $f$ is [[Definition:Continuous Real Function|continuous]] at $c$, there exists $\delta > 0$ such that: 
+:for all $x \in A$ with $\size {x - c} < \delta$, we have $\size {\map f x - \map f c} < \varepsilon$.
+Additionally, since $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to $c$, there exists $N \in \N$ such that: 
+:for all $n > N$ we have $\size {x_n - c} < \delta$.
+Therefore, since $x_n \in A$ for all $n \in \N$: 
+:for all $n > N$, we have $\size {\map f {x_n} - \map f c} < \varepsilon$.
+Since $\varepsilon$ was arbitrary, we have: 
+:$\sequence {\map f {x_n} }$ [[Definition:Convergent Real Sequence|converges]] to $\map f c$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Closure and Topological Closure of Subset are Equivalent}
+Tags: Set Closures
+
+\begin{theorem}
+Let $M = \struct{A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $T = \struct{A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
+Let $H \subseteq A$.
+Then:
+:the [[Definition:Closure (Metric Space)|metric closure]] of $H$ in $M$ equals the [[Definition:Closure (Topology)|topological closure]] of $H$ in $T$
+\end{theorem}
+
+\begin{proof}
+Let $H^i$ be the [[Definition:Set|set]] of [[Definition:Isolated Point (Metric Space)/Subset|isolated points]] of $H$ in $M$.
+From [[Isolated Point in Metric Space iff Isolated Point in Topological Space]]:
+:$H^i$ equals the [[Definition:Set|set]] of [[Definition:Isolated Point (Topology)|isolated points]] of $H$ in the [[Definition:Topological Space|topological space]] $T$.
+Let $H'$ be the [[Definition:Set|set]] of [[Definition:Limit Point (Metric Space)|limit points]] of $H$ in $M$.
+From [[Limit Point in Metric Space iff Limit Point in Topological Space]]:
+:$H'$ equals the [[Definition:Set|set]] of [[Definition:Limit Point (Topology)|limit points]] of $H$ in the [[Definition:Topological Space|topological space]] $T$.
+Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$ in the [[Definition:Metric Space|metric space]] $M$.
+By definition of the [[Definition:Closure (Metric Space)|closure]] of $H$ in the [[Definition:Metric Space|metric space]] $M$
+:$H^- = H' \cup H^i$
+Let $\map \cl H$ denote the closure of the [[Definition:Closure (Topology)/Definition 5|closure]] of $H$ in the [[Definition:Topological Space|topological space]] $T$.
+By definition of the [[Definition:Closure (Topology)/Definition 5|closure]] of $H$ in the [[Definition:Topological Space|topological space]] $T$
+:$\map \cl H = H' \cup H^i$
+Thus:
+:$H^- = \map \cl H$
+{{qed}}
+[[Category:Set Closures]]
+k5csayawx16jp5g37kf7ulvg56ln4bc
+\end{proof}<|endoftext|>
+\section{Set together with Omega-Accumulation Points is not necessarily Closed}
+Tags: Set Closures, Omega-Accumulation Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $\Omega$ denote the [[Definition:Set|set]] of [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] of $H$.
+Then it is not necessarily the case that $H \cup \Omega$ is a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $T = \struct {\R, \tau}$ denote the [[Definition:Right Order Topology on Real Numbers|right order topology on $\R$]].
+Let $H \subseteq \R$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\R$.
+Let $\Omega$ denote the [[Definition:Set|set]] of [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] of $H$.
+From [[Finite Set of Right Order Topology with Omega-Accumulation Points is not Closed]], $H \cup \Omega$ is not a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set together with Condensation Points is not necessarily Closed}
+Tags: Set Closures, Condensation Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Condensation Point|condensation points]] of $H$.
+Then it is not necessarily the case that $H \cup \CC$ is a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $T = \struct {\R, \tau}$ denote the [[Definition:Right Order Topology on Real Numbers|right order topology on $\R$]].
+Let $H \subseteq \R$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\R$.
+Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Condensation Point|condensation points]] of $H$.
+From [[Finite Set of Right Order Topology with Condensation Points is not Closed]], $H \cup \CC$ is not a [[Definition:Closed Set (Topology)|closed set]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis Test for Isolated Point}
+Tags: Isolated Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
+Let $H \subseteq S$.
+Then $x \in H$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ {{iff}}:
+:$\exists U \in \BB : U \cap H = \set x$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in H$ be an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
+By definition of an [[Definition:Isolated Point (Topology)|isolated point]]:
+:$\exists U \in \tau: U \cap H  = \set x$
+By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$\exists V \in \BB: x \in V \subseteq U$
+From [[Set Intersection Preserves Subsets]]:
+:$V \cap H \subseteq U \cap H  = \set x$
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq V \cap H$
+From [[Definition:Set Equality|set equality]]:
+:$V \cap H  = \set x$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $U \in \BB : U \cap H = \set x$.
+By definition of [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$U \in \tau$
+Then $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ by definition.
+{{qed}}
+[[Category:Isolated Points]]
+5ljnuyr7azebz0poirvgt64csf3uyra
+\end{proof}<|endoftext|>
+\section{Basis Test for Limit Point}
+Tags: Limit Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
+Let $H \subseteq S$.
+Then $x \in S$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ {{iff}}:
+:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in S$ be a [[Definition:Limit Point (Topology)|limit point]] of $H$.
+By definition of a [[Definition:Limit Point (Topology)|limit point]] of $H$:
+:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
+By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$\BB \subseteq \tau$
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x$ satisfy:
+:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
+Let $V$ be any [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$\exists U \in \BB : x \in U \subseteq V$
+Then:
+:$H \cap U \setminus \set x \ne \O$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Subsets of Disjoint Sets are Disjoint]]:
+:$H \cap V \setminus \set x \ne \O$
+Thus $x$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ by definition.
+{{qed}}
+[[Category:Limit Points]]
+s2rci5ndexlisnsdafo9zdx1chhymx6
+\end{proof}<|endoftext|>
+\section{Basis Test for Adherent Point}
+Tags: Adherent Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
+Let $H \subseteq S$.
+Then $x \in S$ is an [[Definition:Adherent Point|adherent point]] of $H$ {{iff}}:
+:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in S$ be an [[Definition:Adherent Point|adherent point]] of $H$.
+By definition of an [[Definition:Adherent Point|adherent point]] of $H$:
+:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
+By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$\BB \subseteq \tau$
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x$ satisfy:
+:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$
+Let $V$ be any [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
+:$\exists U \in \BB : x \in U \subseteq V$
+Then:
+:$H \cap U \ne \O$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Subsets of Disjoint Sets are Disjoint]]:
+:$H \cap V \ne \O$
+Thus $x$ is an [[Definition:Adherent Point|adherent point]] of $H$ by definition.
+{{qed}}
+[[Category:Adherent Points]]
+c29nij74r4twhq1997ysbb0fpd0t097
+\end{proof}<|endoftext|>
+\section{Set of Liouville Numbers is Uncountable}
+Tags: Transcendental Numbers
+
+\begin{theorem}
+The set of [[Definition:Liouville Number|Liouville numbers]] is [[Definition:Uncountable Set|uncountable]].
+\end{theorem}
+
+\begin{proof}
+By [[Liouville's Constant is Transcendental/Corollary|Corollary to Liouville's Constant is Transcendental]], all numbers of the form:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }
+      | r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots
+      | c = 
+}}
+{{end-eqn}}
+where
+:$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$
+are [[Definition:Liouville Number|Liouville numbers]].
+Therefore each [[Definition:Sequence|sequence]] in $\set {1, 2, \ldots, 9}$ defines a unique [[Definition:Liouville Number|Liouville number]].
+By [[Set of Infinite Sequences is Uncountable]], there are [[Definition:Uncountable Set|uncountable]] [[Definition:Sequence|sequences]] in $\set {1, 2, \ldots, 9}$.
+As the set of [[Definition:Liouville Number|Liouville numbers]] has an [[Definition:Uncountable Set|uncountable]] [[Definition:Subset|subset]], it is also [[Definition:Uncountable Set|uncountable]] by [[Sufficient Conditions for Uncountability]].
+{{qed}}
+[[Category:Transcendental Numbers]]
+e83747fu0fg9s0g9ssbgyh2by3acyt2
+\end{proof}<|endoftext|>
+\section{Isolated Point in Metric Space iff Isolated Point in Topological Space}
+Tags: Isolated Points
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $T = \struct {A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
+Let $H \subseteq A$.
+Let $x \in H$
+Then:
+:$x$ is an [[Definition:Isolated Point (Metric Space)|isolated point]] of $H$ in $M$ {{iff}} $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ in $T$
+\end{theorem}
+
+\begin{proof}
+From [[Open Balls form Local Basis for Point of Metric Space]], the [[Definition:Set|set]]:
+:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0} }$
+is a [[Definition:Local Basis|local basis]] of $x$.
+From [[Local Basis Test for Isolated Point]]:
+:$x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ in $T$ {{iff}} $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$
+By definition of an [[Definition:Isolated Point (Metric Space)|isolated point]] in $M$:
+:$x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ in $T$ {{iff}} $x$ is an [[Definition:Isolated Point (Metric Space)|isolated point]] of $H$ in $M$
+{{qed}}
+[[Category:Isolated Points]]
+dutqp017g6dzcek07ew4iqp1wdenny4
+\end{proof}<|endoftext|>
+\section{Limit Point in Metric Space iff Limit Point in Topological Space}
+Tags: Limit Points
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $T = \struct {A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
+Let $H \subseteq A$.
+Then:
+:$x \in H$ is a [[Definition:Limit Point (Metric Space)|limit point]] in $M$ {{iff}} $x$ is a [[Definition:Limit Point (Topology)|limit point]] in $T$
+\end{theorem}
+
+\begin{proof}
+From [[Open Balls form Local Basis for Point of Metric Space]], the [[Definition:Set|set]]:
+:$\BB_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
+is a [[Definition:Local Basis|local basis]] of $x$.
+From [[Local Basis Test for Limit Point]]:
+:$x$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ in $T$ {{iff}} $\forall \epsilon \in \R_{>0}: H \cap \map {B_\epsilon} x \setminus \set x \ne \O$
+By definition of a [[Definition:Limit Point (Metric Space)|limit point]] in $M$:
+:$x$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ in $T$ {{iff}} $x$ is an [[Definition:Limit Point (Metric Space)|limit point]] of $H$ in $M$
+{{qed}}
+[[Category:Limit Points]]
+onrvxa4fd0s35dgsj7tyyamqf3pozjl
+\end{proof}<|endoftext|>
+\section{Boundary of Boundary is not necessarily Equal to Boundary}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq T$.
+Let $\partial H$ denote the [[Definition:Boundary (Topology)|boundary]] of $H$.
+While it is true that:
+:$\map \partial {\partial H} \subseteq \partial H$
+it is not necessarily the case that:
+:$\map \partial {\partial H} = \partial H$
+\end{theorem}
+
+\begin{proof}
+From [[Boundary of Boundary is Contained in Boundary]], we have that:
+:$\map \partial {\partial H} \subseteq \partial H$
+It remains to be proved that the equality does not always hold.
+[[Proof by Counterexample]]:
+Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space]].
+Let $H \subseteq S$ such that $H \ne \O$ and $H \ne S$.
+From [[Boundary of Subset of Indiscrete Space]]:
+:$\partial H = S$
+From [[Boundary of Boundary of Subset of Indiscrete Space]]:
+:$\map \partial {\partial H} = \O$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Functions of Bounded Variation is of Bounded Variation}
+Tags: Total Variation, Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation|bounded variation]].
+Let $V_f$ and $V_g$ be the [[Definition:Total Variation|total variations]] of $f$ and $g$ respectively. 
+Then $f + g$ is of [[Definition:Bounded Variation|bounded variation]] with:
+:$V_{f + g} \le V_f + V_g$
+where $V_{f + g}$ denotes the [[Definition:Total Variation|total variation]] of $f + g$.
+\end{theorem}
+
+\begin{proof}
+For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write: 
+:$P = \set {x_0, x_1, \ldots, x_n }$
+with:
+:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
+Then:
+{{begin-eqn}}
+{{eqn	| l = \map {V_{f + g} } P
+	| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {x_i} - \map {\paren {f + g} } {x_{i - 1} } }
+	| c = using the notation from the definition of [[Definition:Bounded Variation|bounded variation]]
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^n \size {\paren {\map f {x_i} - \map f {x_{i - 1} } } + \paren {\map g {x_i} - \map g {x_{i - 1} } } }
+}}
+{{eqn	| o = \le
+	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } + \sum_{i \mathop = 1}^n \size {\map g {x_i} - \map g {x_{i - 1} } }
+	| c = [[Triangle Inequality]]
+}}
+{{eqn	| r = \map {V_f} P + \map {V_g} P
+}}
+{{end-eqn}}
+Note that since $f$ and $g$ are of [[Definition:Bounded Variation|bounded variation]], there exists $M, K \in \R$ such that: 
+:$\map {V_f} P \le M$
+:$\map {V_g} P \le K$
+for all [[Definition:Finite Subdivision|finite subdivisions]] $P$ of $\closedint a b$. 
+We therefore have: 
+:$\map {V_{f + g} } P \le M + K$
+for all finite subdivisions $P$.
+So $f + g$ is of [[Definition:Bounded Variation|bounded variation]].
+Note then that: 
+{{begin-eqn}}
+{{eqn	| l = V_{f + g}
+	| r = \sup_P \paren {\map {V_{f + g} } P}
+	| c = {{Defof|Total Variation}}
+}}
+{{eqn	| o = \le
+	| r = \sup_P \paren {\map {V_f} P} + \sup_P \paren {\map {V_g} P}
+}}
+{{eqn	| r = V_f + V_g
+	| c = {{Defof|Total Variation}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Sets in Vector Spaces with Equivalent Norms Coincide}
+Tags: Equivalence Relations, Norm Theory, Vector Spaces
+
+\begin{theorem}
+Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
+Let $U \subseteq X$ be an [[Definition:Open Set in Normed Vector Space|open set]] in $M_a$.
+Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are [[Definition:Equivalence of Norms|equivalent norms]], i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
+Then $U$ is also [[Definition:Open Set in Normed Vector Space|open]] in $M_b$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Equivalence of Norms|equivalent norms]]:
+:$\exists m,M \in \R_{> 0} : m \le M : \forall x \in X:  m \norm x_b \le \norm x_a \le M \norm x_b$
+Since $U$ is [[Definition:Open Set in Normed Vector Space|open]] in $M_a$:
+:$\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon_a}} x \subseteq U$
+where $\map {B_{\epsilon_a}} x$ stands for an [[Definition:Open Ball in Normed Vector Space|open ball]], defined as:
+:$\map {B_{\epsilon_a}} x := \set {\forall y \in X : \norm {x - y}_a < \epsilon_a}$
+Define an [[Definition:Open Ball in Normed Vector Space|open ball]] $\map {B_{\epsilon_b}} x$ as:
+:$\map {B_{\epsilon_b}} x := \set {\forall y \in X : \norm {x - y}_b < \epsilon_b}$
+Then we have that:
+:$m \norm {x - y}_b \le \norm {x - y}_a < \epsilon_a$
+So far, $\epsilon_b$ was unspecified.
+Define $\epsilon_b := \dfrac {\epsilon_a} m$.
+So:
+:$\forall x \in U : y \in \map {B_{\epsilon_a}} x \implies y \in \map {B_{\epsilon_b}} x$
+with $\epsilon_a = m \epsilon_b$.
+Hence:
+:$\forall x \in U : \exists \epsilon_b \in \R_{> 0} : \map {B_{\epsilon_b}} x \subseteq U$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Local Basis Test for Isolated Point}
+Tags: Isolated Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $x \in H$.
+Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
+Then $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ {{iff}}:
+:$\exists U \in \BB_x : U \cap H = \set x$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in H$ be an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
+By definition of an [[Definition:Isolated Point (Topology)|isolated point]]:
+:$\exists U \in \tau: U \cap H  = \set x$
+By definition of a [[Definition:Local Basis|local basis]] of $T$:
+:$\exists V \in \BB_x : x \in V \subseteq U$
+From [[Set Intersection Preserves Subsets]]:
+:$V \cap H \subseteq U \cap H  = \set x$
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq V \cap H$
+From [[Definition:Set Equality|set equality]]:
+:$V \cap H  = \set x$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $U \in \BB_x : U \cap H = \set x$.
+By definition of [[Definition:Local Basis|local basis]] of $T$:
+:$U \in \tau$
+Then $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ by definition.
+{{qed}}
+[[Category:Isolated Points]]
+60b92i4by6anligwiu78ea676ya5pwh
+\end{proof}<|endoftext|>
+\section{Local Basis Test for Limit Point}
+Tags: Limit Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $x \in S$.
+Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
+Then $x \in S$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ {{iff}}:
+:$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in S$ be a [[Definition:Limit Point (Topology)|limit point]] of $H$.
+By definition of a [[Definition:Limit Point (Topology)|limit point]] of $H$:
+:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
+By definition of a [[Definition:Local Basis|local basis]] of $T$:
+:$\BB_x \subseteq \tau$
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x$ satisfy:
+:$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$
+Let $V$ be any [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By definition of a [[Definition:Local Basis|local basis]] of $T$:
+:$\exists U \in \BB : x \in U \subseteq V$
+Then:
+:$H \cap U \setminus \set x \ne \O$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Subsets of Disjoint Sets are Disjoint]]:
+:$H \cap V \setminus \set x \ne \O$
+Thus $x$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ by definition.
+{{qed}}
+[[Category:Limit Points]]
+ed1ztna0nbks5iayet5eu6ds6u85jx0
+\end{proof}<|endoftext|>
+\section{Local Basis Test for Adherent Point}
+Tags: Adherent Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $H \subseteq S$.
+Let $x \in S$.
+Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
+Then $x \in S$ is an [[Definition:Adherent Point|adherent point]] of $H$ {{iff}}:
+:$\forall U \in \BB_x : H \cap U \ne \O$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in S$ be an [[Definition:Adherent Point|adherent point]] of $H$.
+By definition of an [[Definition:Adherent Point|adherent point]] of $H$:
+:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
+By definition of a [[Definition:Local Basis|local basis]] of $T$:
+:$\BB_x \subseteq \tau$
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x$ satisfy:
+:$\forall U \in \BB_x : H \cap U \ne \O$
+Let $V$ be any [[Definition:Open Neighborhood|open neighborhood]] of $x$.
+By definition of a [[Definition:Local Basis|local basis]] of $T$:
+:$\exists U \in \BB : x \in U \subseteq V$
+Then:
+:$H \cap U \ne \O$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Subsets of Disjoint Sets are Disjoint]]:
+:$H \cap V \ne \O$
+Thus $x$ is an [[Definition:Adherent Point|adherent point]] of $H$ by definition.
+{{qed}}
+[[Category:Adherent Points]]
+e266iczrqukz5c6eywtzap4udjt4qzq
+\end{proof}<|endoftext|>
+\section{Closure of Subset of Metric Space is Closed}
+Tags: Set Closures
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
+Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$.
+Then $H^-$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Let $\overline {\paren{H^-}}$ denote the [[Definition:Relative Complement|complement]] of $H^-$.
+Let $x \in \overline {\paren{H^-}}$.
+By definition of the [[Definition:Closure (Metric Space)|closure]] of $H$:
+:$x$ is not a [[Definition:Limit Point (Topology)|limit point]] of $H$.
+So:
+:$\exists \epsilon \in \R_{> 0} : \paren{\map {B_\epsilon} x \setminus \set x} \cap H = \O$
+From [[Intersection with Set Difference is Set Difference with Intersection]]:
+:$\paren{\map {B_\epsilon} x \cap H} \setminus \set x = \O$
+From [[Set Difference with Superset is Empty Set]]:
+:$\map {B_\epsilon} x \cap H \subseteq \set x$
+By definition of the [[Definition:Closure (Metric Space)|closure]] of $H$:
+:$x$ is not an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
+So:
+:$\map {B_\epsilon} x \cap H \ne \set x$
+Thus:
+:$\map {B_\epsilon} x \cap H = \O$
+From [[Empty Intersection iff Subset of Complement]]:
+:$\map {B_\epsilon} x \subseteq \overline {\paren{H^-}}$
+It follows that $\overline {\paren{H^-}}$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+Thus $H^- $ is [[Definition:Closed Set (Metric Space)|closed]] in $M$ by definition.
+{{qed}}
+[[Category:Set Closures]]
+gln0tdiitfl0gxjq2wpzk430zbd3zdy
+\end{proof}<|endoftext|>
+\section{Convergent Sequences in Vector Spaces with Equivalent Norms Coincide}
+Tags: Equivalence Relations, Norm Theory, Vector Spaces
+
+\begin{theorem}
+Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
+Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]] in $M_a$.
+Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are [[Definition:Equivalence of Norms|equivalent norms]], i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
+Then $\sequence {x_n}_{n \mathop \in \N}$ is also [[Definition:Convergent Sequence in Normed Vector Space|convergent]] in $M_b$.
+\end{theorem}
+
+\begin{proof}
+Let $L \in X$.
+Then:
+:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_a < \epsilon_a$
+By [[Definition:Equivalence of Norms|equivalence of norms]]:
+:$\exists M \in \R_{> 0} : \norm {x_n - L}_b \le M \norm {x_n - L}_a < M \epsilon_a$
+Let $\epsilon_b := M \epsilon_a$
+Then:
+:$\forall \epsilon_b \in \R_{> 0}: \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_b < \epsilon_b$
+Therefore, $\sequence{x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $L$ also in $M_b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Mapping is not necessarily Closed Mapping}
+Tags: Open Mappings, Closed Mappings
+
+\begin{theorem}
+Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
+Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]] which is not a [[Definition:Bijection|bijection]].
+Let $f$ be an [[Definition:Open Mapping|open mapping]].
+Then it is not necessarily the case that $f$ is also a [[Definition:Closed Mapping|closed mapping]].
+\end{theorem}
+
+\begin{proof}
+Note that if $f$ is a [[Definition:Bijection|bijection]], the result [[Bijection is Open iff Closed]] applies.
+It is to be shown that if $f$ is not a [[Definition:Bijection|bijection]], this is not necessarily the case.
+This is achieved by [[Proof by Counterexample]]:
+Let $\struct {\R^2, d}$ be the [[Definition:Real Number Plane with Euclidean Topology|real number plane with the usual (Euclidean) topology]].
+Let $\rho: \R^2 \to \R$ be the [[Definition:First Projection|first projection]] on $\R^2$ defined as:
+:$\forall \tuple{x, y} \in \R^2: \map \rho {x, y} = x$
+From [[Projection on Real Euclidean Plane is Open Mapping]], $\rho$ is an [[Definition:Open Mapping|open mapping]].
+From [[Projection on Real Euclidean Plane is not Closed Mapping], $\rho$ is not a [[Definition:Closed Mapping|closed mapping]].
+The result is apparent.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Power of x by Cosine of a x/Corollary}
+Tags: Primitives involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int x^m \cos a x \rd x = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map {\sin} {x + \dfrac {\pi} 2 \paren {k - 1} } }$
+where $m^{\underline {k - 1} }$ denotes the $k - 1$th [[Definition:Falling Factorial|falling factorial]] of $m$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \int x^m \cos a x \rd x
+      | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \cos a x \rd x
+      | c = [[Primitive of Power of x by Cosine of a x]]
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \int x^{m - 2} \cos a x \rd x
+      | r = \frac {x^{m - 2} \sin a x} a + \frac {\paren {m - 2} x^{m - 3} \cos a x} {a^2} - \frac {\paren {m - 2} \paren {m - 3} } {a^2} \int x^{m - 4} \cos a x \rd x
+      | c = setting $m$ equal to $m - 2$
+}}
+{{eqn | ll= \leadsto
+      | l = \int x^m \cos a x \rd x
+      | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \paren {\frac {x^{m-2} \sin a x} a + \frac {\paren {m-2} x^{m - 3} \cos a x} {a^2} - \frac {\paren {m - 2} \paren {m - 3} } {a^2} \int x^{m - 4} \cos a x \rd x}
+      | c = substituting $(2)$ into $(1)$ above
+}}
+{{eqn | r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {\paren m \paren {m - 1} x^{m - 2} \sin a x} {a^3} - \frac {\paren m \paren {m - 1} \paren {m - 2} x^{m - 3} \cos a x} {a^4} + \frac {\paren m \paren {m - 1} \paren {m - 2} \paren {m - 3} } {a^4} \int x^{m - 4} \cos a x \rd x
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map \sin {x + \frac {\pi} 2 \paren { k - 1 } } }
+      | c = {{Defof|Falling Factorial}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Primitives involving Cosine Function]]
+9jbue7rvy7yr9my9w6go3hxdfzxwvq8
+\end{proof}<|endoftext|>
+\section{Homeomorphism may Exist between Non-Comparable Topologies}
+Tags: Homeomorphisms
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be [[Definition:Topological Space|topological spaces]] defined on the [[Definition:Underlying Set of Topological Space|underlying set]] $S$.
+Let $\tau_1$ and $\tau_2$ be non-[[Definition:Comparable Topologies|comparable]].
+Then it may possibly be the case that $T_1$ and $T_2$ are [[Definition:Homeomorphic Topological Spaces|homeomorphic]].
+\end{theorem}
+
+\begin{proof}
+A counterexample is demonstrated in [[Homeomorphic Non-Comparable Particular Point Topologies]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Limit Point of Filter Basis}
+Tags: Limit Points of Filter Bases
+
+\begin{theorem}
+{{TFAE|def = Limit Point of Filter Basis}}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\FF$ be a [[Definition:Filter on Set|filter]] on the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ of $T$.
+Let $\BB$ be a [[Definition:Filter Basis|filter basis]] of $\FF$.
+\end{theorem}
+
+\begin{proof}
+Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
+Let $\BB$ be a [[Definition:Filter Basis|filter basis]] of $\FF$.
+=== $(1)$ implies $(2)$ ===
+Let $x \in S$ be a [[Definition:Limit Point/Filter Basis/Definition 1|limit Point of $\BB$ by definition $1$]].
+Then by definition $\FF$ [[Definition:Convergent Filter|converges on $x$]].
+By definition of [[Definition:Convergent Filter|convergent filter]]:
+:$\forall N_x \subseteq S: N_x \in \FF$
+where $N_x$ is a [[Definition:Neighborhood of Point|neighborhood]] of $x$.
+That is, every [[Definition:Neighborhood of Point|neighborhood]] of $x$ is an [[Definition:Element|element]] of $\FF$.
+By definition of [[Definition:Filter Basis|filter basis]]:
+:$\forall U \in \FF: \exists V \in \BB: V \subseteq U$
+But we have shown that $N_x \in \FF$.
+Therefore:
+:$\exists V \in \BB: V \subseteq N_x$
+That is, every [[Definition:Neighborhood of Point|neighborhood]] of $x$ contains a set of $\BB$.
+Thus $x \in S$ is a [[Definition:Limit Point/Filter Basis/Definition 2|limit Point of $\BB$ by definition $2$]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $x \in S$ be a [[Definition:Limit Point/Filter Basis/Definition 2|limit Point of $\BB$ by definition $2$]].
+Then by definition every [[Definition:Neighborhood of Point|neighborhood]] of $x$ contains a set of $\BB$.
+By definition of [[Definition:Filter Basis|filter basis]], $\FF := \set {V \subseteq X: \exists U \in \BB: U \subseteq V}$ is a [[Definition:Filter on Set|filter]] on $S$ {{iff}}:
+:$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$
+But then letting $V_1 = V_2 = V$, we have:
+:$\forall V \in \BB: \exists U \in \BB: U \subseteq V$
+and so $V \in \FF$.
+Thus every [[Definition:Neighborhood of Point|neighborhood]] of $x$ is an [[Definition:Element|element]] of $\FF$.
+Thus, by definition, $\FF$ [[Definition:Convergent Filter|converges on $x$]].
+Thus $x \in S$ is a [[Definition:Limit Point/Filter Basis/Definition 1|limit Point of $\BB$ by definition $1$]].
+{{qed}}
+[[Category:Limit Points of Filter Bases]]
+b686t3dvy3zu8xc8nkijwc3361g4f7g
+\end{proof}<|endoftext|>
+\section{Richert's Theorem}
+Tags: Number Theory
+
+\begin{theorem}
+Let $S = \set {s_1, s_2, \dots}$ be an [[Definition:Infinite Set|infinite set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]], with the property:
+:$s_n < s_{n + 1}$ for every $n \in \N$
+Suppose there exists some [[Definition:Integer|integers]] $N, k$ such that every [[Definition:Integer|integer]] $n$ with $N < n \le N + s_{k + 1}$:
+:$n$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_k}$
+:$s_{i + 1} \le 2 s_i$ for every $i \ge k$
+Then for any $n > N$, $n$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $S$.
+\end{theorem}
+
+\begin{proof}
+We prove this using [[First Principle of Mathematical Induction]].
+Let $\map P n$ be the proposition:
+:For every [[Definition:Integer|integer]] $m$ with $N < m \le N + s_{n + 1}$:
+::$m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_n}$.
+=== Basis for the induction ===
+From our assumption above, $\map P k$ is true.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Suppose for some $n > k$, $\map P {n - 1}$ is true.
+That is, for every [[Definition:Integer|integer]] $m$ with $N < m \le N + s_n$:
+:$m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]].
+We need to show that $\map P n$ is true.
+That is, for every [[Definition:Integer|integer]] $m$ with $N < m \le N + s_{n + 1}$:
+:$m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_n}$.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By the induction hypothesis, for every [[Definition:Integer|integer]] $m$ with $N < m \le N + s_n$:
+:$m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.
+Hence we only need to consider $N + s_n < m \le N + s_{n + 1}$.
+For $m$ in this range:
+:$N < m - s_n \le N + s_{n + 1} - s_n \le N + s_n$
+By the induction hypothesis, $m - s_n$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_{n - 1}}$.
+Hence $m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_{n - 1}, s_n}$.
+Hence $\map P n$ is true.
+By the [[First Principle of Mathematical Induction]], $\map P n$ is true for all $n \ge k$.
+{{qed|lemma}}
+Since $1 \le s_n < s_{n + 1}$ for every $n \in N$:
+:$s_n \ge n$ for every $n \in \N$
+Let $K > N$.
+Then:
+{{begin-eqn}}
+{{eqn | l = N + s_{K + k + 1}
+      | o = \ge
+      | r = N + K + k + 1
+}}
+{{eqn | o = >
+      | r = K
+}}
+{{eqn | o = >
+      | r = N
+}}
+{{end-eqn}}
+By the result above, since $K + k \ge k$:
+:$K$ can be expressed as the [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s_{K + k}}$.
+Thus every number greater than $N$ can be expressed as the [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $S$.
+{{qed}}
+{{Namedfor|Hans-Egon Richert}}
+[[Category:Number Theory]]
+hu4tfm247b0ro2ijpsr8eu5hdkbi57h
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Filter Basis}
+Tags: Filter Bases
+
+\begin{theorem}
+{{TFAE|def = Filter Basis}}
+Let $S$ be a [[Definition:Set|set]].
+Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Let $\BB$ be a [[Definition:Filter Basis/Definition 1|filter basis of $\FF$ by definition $1$]].
+Then by definition:
+:$\BB \subset \powerset S$ such that $\O \notin \BB$ and $\BB \ne \O$
+and $\FF := \set {V \subseteq S: \exists U \in \BB: U \subseteq V}$ is a [[Definition:Filter on Set|filter]] on $S$ {{iff}}:
+:$\forall V_1, V_2 \in \BB: \exists U \in \BB: U \subseteq V_1 \cap V_2$
+Let $U \in \FF$.
+Then by definition of $\FF$:
+:$\exists V \in \BB: V \subseteq U$
+Thus $\BB$ is a [[Definition:Filter Basis/Definition 2|filter basis of $\FF$ by definition $2$]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $\BB$ be a [[Definition:Filter Basis/Definition 2|filter basis of $\FF$ by definition $2$]].
+By definition, $\BB$ is a '''filter basis''' of $\FF$ {{iff}}:
+:$\forall U \in \FF: \exists V \in \BB: V \subseteq U$
+Let $V_1, V_2 \in \BB$.
+Then:
+:$V_1, V_2 \in \FF$
+By definition of [[Definition:Filter on Set|filter]]:
+:$V_1 \cap V_2 \in \FF$
+and so:
+:$\exists U \in \BB: U \subseteq V_1 \cap V_2$
+Thus $\BB$ is a [[Definition:Filter Basis/Definition 1|filter basis of $\FF$ by definition $1$]].
+{{qed}}
+[[Category:Filter Bases]]
+qvl6ivhbdzhxpjlz7byc879nsga1kb2
+\end{proof}<|endoftext|>
+\section{Number as Sum of Distinct Primes}
+Tags: Prime Numbers
+
+\begin{theorem}
+For $n \ne 1, 4, 6$, $n$ can be expressed as the [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {s_n}_{n \mathop \in N}$ be the set of [[Definition:Prime Number|primes]].
+Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$.
+By [[Bertrand-Chebyshev Theorem]]:
+:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
+We observe that every [[Definition:Integer|integer]] $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, \dots, s_5} = \set {2, 3, 5, 7, 11}$.
+Hence the result by [[Richert's Theorem]].
+{{qed|lemma}}
+Here is a demonstration of our claim:
+{{begin-eqn}}
+{{eqn | l = 1
+      | o = \text {is}
+      | r = \text {less than the smallest prime } 2
+}}
+{{eqn | l = 2
+      | r = 2
+}}
+{{eqn | l = 3
+      | r = 3
+}}
+{{eqn | l = 4
+      | o = \ne
+      | r = 2 + 3
+}}
+{{eqn | l = 5
+      | r = 5
+}}
+{{eqn | l = 6
+      | o = \ne
+      | r = 2 + 3 \text { or } 2 + 5
+}}
+{{eqn | l = 7
+      | r = 7
+}}
+{{eqn | l = 8
+      | r = 3 + 5
+}}
+{{eqn | l = 9
+      | r = 2 + 7
+}}
+{{eqn | l = 10
+      | r = 2 + 3 + 5
+}}
+{{eqn | l = 11
+      | r = 11
+}}
+{{eqn | l = 12
+      | r = 2 + 3 + 7
+}}
+{{eqn | l = 13
+      | r = 2 + 11
+}}
+{{eqn | l = 14
+      | r = 3 + 11
+}}
+{{eqn | l = 15
+      | r = 3 + 5 + 7
+}}
+{{eqn | l = 16
+      | r = 5 + 11
+}}
+{{eqn | l = 17
+      | r = 2 + 3 + 5 + 7
+}}
+{{eqn | l = 18
+      | r = 2 + 5 + 11
+}}
+{{eqn | l = 19
+      | r = 3 + 5 + 11
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Prime Numbers]]
+moh4pi0nyd1rhipoc6kqau1oqjzdqar
+\end{proof}<|endoftext|>
+\section{Difference of Functions of Bounded Variation is of Bounded Variation}
+Tags: Total Variation, Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation|bounded variation]].
+Let $V_f$ and $V_g$ be the [[Definition:Total Variation|total variations]] of $f$ and $g$ respectively. 
+Then $f - g$ is of [[Definition:Bounded Variation|bounded variation]] with:
+:$V_{f - g} \le V_f + V_g$
+where $V_{f - g}$ denotes the [[Definition:Total Variation|total variation]] of $f - g$.
+\end{theorem}
+
+\begin{proof}
+By [[Multiple of Function of Bounded Variation is of Bounded Variation]], we have that: 
+:$-g$ is of [[Definition:Bounded Variation|bounded variation]].
+So, by [[Sum of Functions of Bounded Variation is of Bounded Variation]], we have that: 
+:$f + \paren{-g} = f - g$ is of [[Definition:Bounded Variation|bounded variation]]
+with: 
+:$V_{f - g} \le V_f + V_{-g}$
+where $V_{-g}$ is the [[Definition:Total Variation|total variation]] of $-g$. 
+We have, by [[Multiple of Function of Bounded Variation is of Bounded Variation]]:
+{{begin-eqn}}
+{{eqn	| l = V_{-g}
+	| r = \size {-1} V_g
+}}
+{{eqn	| r = V_g
+}}
+{{end-eqn}}
+so: 
+:$V_{f - g} \le V_f + V_g$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Functions of Bounded Variation is of Bounded Variation}
+Tags: Total Variation, Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation|bounded variation]].
+Let $V_f$ and $V_g$ be the [[Definition:Total Variation|total variations]] of $f$ and $g$ respectively. 
+Then $f \times g$ is of [[Definition:Bounded Variation|bounded variation]] with:
+:$V_{f \times g} \le A V_f + B V_g$
+where:
+:$V_{f \times g}$ denotes the [[Definition:Total Variation|total variation]] of $f \times g$
+:$A, B$ are [[Definition:Non-Negative Real Number|non-negative real numbers]].
+\end{theorem}
+
+\begin{proof}
+For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write: 
+:$P = \set {x_0, x_1, \ldots, x_n }$
+with:
+:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
+By [[Function of Bounded Variation is Bounded]]: 
+:$f$ and $g$ are [[Definition:Bounded Mapping/Real-Valued|bounded]]. 
+So, there exists $A, B \in \R$ such that: 
+:$\size {\map f x} \le B$
+:$\size {\map g x} \le A$
+for all $x \in \closedint a b$.
+Then:
+{{begin-eqn}}
+{{eqn	| l = \map {V_{f \times g} } P
+	| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {x_i} - \map {\paren {f \times g} } {x_{i - 1} } }
+	| c = using the notation from the definition of [[Definition:Bounded Variation|bounded variation]]
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } }
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_i} + \map f {x_{i - 1} } \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } }
+}}
+{{eqn	| o = \le
+	| r = \sum_{i \mathop = 1}^n \size {\map f {x_i} \map g {x_i} - \map f {x_{i - 1} } \map g {x_i} } + \sum_{i \mathop = 1}^n \size {\map f {x_{i - 1} } \map g {x_i} - \map f {x_{i - 1} } \map g {x_{i - 1} } }
+	| c = [[Triangle Inequality]]
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^n \size {\map g {x_i} } \size {\map f {x_i} - \map f {x_{i - 1} } } + \sum_{i \mathop = 1}^n \size {\map f {x_{i - 1} } } \size {\map g {x_i} - \map g {x_{i - 1} } }
+}}
+{{eqn	| o = \le
+	| r = A \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } } + B \sum_{i \mathop = 1}^n \size {\map g {x_i} - \map g {x_{i - 1} } }
+	| c = since $\size {\map g {x_i} } \le A$ and $\size {\map f {x_{i - 1} } } \le B$
+}}
+{{eqn	| r = A \map {V_f} P + B \map {V_g} P
+}}
+{{end-eqn}}
+Since $f$ and $g$ are of [[Definition:Bounded Variation|bounded variation]], there exists $M, K \in \R$ such that: 
+:$\map {V_f} P \le M$
+:$\map {V_g} P \le K$
+for all [[Definition:Finite Subdivision|finite subdivisions]] $P$.
+We therefore have: 
+:$\map {V_{f \times g} } P \le A M + B K$
+so $f \times g$ is of [[Definition:Bounded Variation|bounded variation]].
+Further, we have: 
+{{begin-eqn}}
+{{eqn	| l = V_{f \times g}
+	| r = \sup_P \paren {\map {V_{f \times g} } P}
+	| c = {{Defof|Total Variation}}
+}}
+{{eqn	| o = \le
+	| r = \sup_P \paren {A \map {V_f} P} + \sup_P \paren {B \map {V_g} P}
+}}
+{{eqn	| r = A \sup_P \paren {\map {V_f} P} + B \sup_P \paren {\map {V_g} P}
+	| c = [[Multiple of Supremum]]
+}}
+{{eqn	| r = A V_f + B V_g
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide}
+Tags: Equivalence Relations, Norm Theory, Vector Spaces, Cauchy Sequences
+
+\begin{theorem}
+Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
+Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_a$.
+Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are [[Definition:Equivalence of Norms|equivalent norms]], i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
+Then $\sequence {x_n}_{n \mathop \in \N}$ is also a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_b$.
+\end{theorem}
+
+\begin{proof}
+We have that $\sequence {x_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_a$.
+Then:
+:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$
+By [[Definition:Equivalence of Norms|equivalence of norms]]:
+:$\exists M \in \R_{> 0} : \norm {x_n - x_m}_b \le M \norm {x_n - x_m}_a < M \epsilon_a$
+Let $\epsilon_b := M \epsilon_a$
+Then:
+:$\forall \epsilon_b \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n, m > N \implies \norm {x_n - x_m}_b < \epsilon_b$
+Therefore, $\sequence {x_n}_{n \mathop \in \N}$ is also a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiable Function of Bounded Variation may not have Bounded Derivative}
+Tags: Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]] of [[Definition:Bounded Variation|bounded variation]].
+Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$.
+Then $f'$ is not necessarily [[Definition:Bounded Real-Valued Function|bounded]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Take $a = 0$, $b = 1$. 
+Let $f : \closedint 0 1 \to \R$ have: 
+:$\map f x = \sqrt x$
+for all $x \in \closedint 0 1$. 
+Note that $f$ is [[Definition:Increasing Real Function|increasing]], so by [[Monotone Function is of Bounded Variation]]:
+:$f$ is of  [[Definition:Bounded Variation|bounded variation]].
+By [[Derivative of Power]], $f$ is [[Definition:Differentiable Real Function|differentiable]] on $\openint 0 1$ with [[Definition:Derivative of Real Function|derivative]]:
+:$\map {f'} x = \dfrac 1 {2 \sqrt x}$
+Note that this is [[Definition:Unbounded Real-Valued Function|unbounded]] as $x \to 0^+$.
+So $f$ does not have a [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Derivative of Real Function|derivative]], despite being of [[Definition:Bounded Variation|bounded variation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Urysohn Function does not guarantee Normal Space}
+Tags: Normal Spaces, Urysohn Functions
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
+Let $T$ have the property that:
+:For all [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ of $T$ such that $A \cap B = \O$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $A$ and $B$.
+Then it is not necessarily the case that $T$ is a [[Definition:Normal Space|normal space]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ have the specified property.
+By definition of a [[Definition:Normal Space|normal space]], for $T$ to be [[Definition:Normal Space|normal]], it has to be both [[Definition:T4 Space|$T_4$ space]] and a [[Definition:T1 Space|$T_1$ space]].
+From [[Urysohn's Lemma Converse]], $T$ is a [[Definition:T4 Space|$T_4$ space]].
+It remains to be shown that $T$ is not necessarily a [[Definition:T1 Space|$T_1$ space]].
+This is done by [[Proof by Counterexample]]:
+Let $S$ be a [[Definition:Set|set]] and let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$ which is specifically not the [[Definition:Partition of Singletons|(trivial) partition of singletons]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] whose [[Basis for Partition Topology|basis]] is $\PP$.
+From [[Partition Topology is T4|Partition Topology is $T_4$]], we have that $T$ is a [[Definition:T4 Space|$T_4$ space]].
+From [[Urysohn's Lemma]], for all $A, B \subseteq S$ be [[Definition:Closed Set (Topology)|closed sets]] of $T$ such that $A \cap B = \O$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $A$ and $B$.
+From [[Partition Topology is not T1|Partition Topology is not $T_1$]], $T$ is not a [[Definition:T1 Space|$T_1$ space]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T3 1/2 Space is not necessarily T2 Space}
+Tags: T3 1: 2 Spaces, Hausdorff Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a be a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Then it is not necessarily the case that $T$ is a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S$ be a [[Definition:Set|set]] and let $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$ which is specifically not the [[Definition:Partition of Singletons|(trivial) partition of singletons]].
+Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition space]] whose [[Basis for Partition Topology|basis]] is $\PP$.
+From [[Partition Topology is T3 1/2|Partition Topology is $T_{3 \frac 1 2}$]], we have that $T$ is a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+From [[Partition Topology is not Hausdorff]], $T$ is not a [[Definition:Hausdorff Space|$T_2$ (Hausdorff) space]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Compact Space which Satisfies No Separation Axioms}
+Tags: Separation Axioms, Compact Spaces
+
+\begin{theorem}
+There exists at least one example of a [[Definition:Compact Space|compact space]] for which none of the [[Definition:Tychonoff Separation Axioms|Tychonoff separation axioms]] are satisfied.
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {S, \tau}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
+Let $D = \struct {\set {0, 1}, \vartheta}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on two points.
+Let $T \times D$ be the [[Definition:Double Pointed Topology|double pointed topology]] on $T$.
+From [[Double Pointed Finite Complement Topology is Compact]], $T \times D$ is [[Definition:Compact Topological Space|compact]].
+From [[Double Pointed Finite Complement Topology fulfils no Separation Axioms]], $T \times D$ is not a [[Definition:Kolmogorov Space|$T_0$ space]], [[Definition:Fréchet Space (Topology)|$T_1$ space]], [[Definition:Hausdorff Space|$T_2$ space]], [[Definition:T3 Space|$T_3$ space]], [[Definition:T4 Space|$T_4$ space]] or [[Definition:T5 Space|$T_5$ space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Everywhere Dense iff Interior of Complement is Empty}
+Tags: Denseness, Set Interiors
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A \subset S$.
+Then $A$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}}:
+:$\paren {\relcomp S A}^\circ = \O$
+where $A^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Everywhere Dense|everywhere dense]], $A$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}}:
+:$A^- = S$
+where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$.
+That happens {{iff}}:
+{{begin-eqn}}
+{{eqn | l = \paren {\relcomp S A}^\circ
+      | r = \relcomp S {A^-}
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = \relcomp S S
+}}
+{{eqn | r = \O
+      | c = [[Relative Complement with Self is Empty Set]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Denseness]]
+[[Category:Set Interiors]]
+c5ojyec4e826itgus8uvohb8qc6tea1
+\end{proof}<|endoftext|>
+\section{Greatest Set is Unique}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
+Then the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$, if it exists, must be [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \TT$ both be [[Definition:Greatest Set by Set Inclusion|greatest sets]] of $\TT$.
+Since $A$ is the [[Definition:Greatest Set by Set Inclusion|greatest set]]:
+:$B \subseteq A$
+Since $B$ is the [[Definition:Greatest Set by Set Inclusion|greatest set]]:
+:$A \subseteq B$
+Hence, by definition of [[Definition:Set Equality|set equality]]:
+:$A = B$
+Therefore the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$ is [[Definition:Unique|unique]].
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+ks9f2fsa7hkz4doxp5oul9yulae66pn
+\end{proof}<|endoftext|>
+\section{Greatest Set may not Exist}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
+The [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$ may not exist.
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
+Then since $\set 0 \not \subseteq \set 1$:
+:$\set 1$ is not the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$.
+Similarly, since $\set 1 \not \subseteq \set 0$:
+:$\set 0$ is not the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$.
+Therefore $\TT$ has no [[Definition:Greatest Set by Set Inclusion|greatest set]].
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+kar4bgf3c1wsygg84qvh6b2ranls27q
+\end{proof}<|endoftext|>
+\section{Smallest Set is Unique}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
+Then the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$, if it exists, must be unique.
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \TT$ both be [[Definition:Smallest Set by Set Inclusion|smallest sets]] of $\TT$.
+Since $A$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]]:
+:$A \subseteq B$
+Since $B$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]]:
+:$B \subseteq A$
+Hence, by definition of [[Definition:Set Equality|set equality]]:
+:$A = B$
+Therefore the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$ is [[Definition:Unique|unique]].
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+m04postjr6kfx509nqofattbnx1958e
+\end{proof}<|endoftext|>
+\section{Smallest Set may not Exist}
+Tags: Set Theory, Order Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
+Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
+The [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$ may not exist.
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
+Then since $\set 0 \not \subseteq \set 1$:
+:$\set 0$ is not the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$.
+Similarly, since $\set 1 \not \subseteq \set 0$:
+:$\set 1$ is not the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$.
+Therefore $\TT$ has no [[Definition:Smallest Set by Set Inclusion|smallest set]].
+{{qed}}
+[[Category:Set Theory]]
+[[Category:Order Theory]]
+3ifxyjo0zvbdefzijhv3jh37ipfsiw5
+\end{proof}<|endoftext|>
+\section{Mapping is Surjection if its Direct Image Mapping is Surjection}
+Tags: Surjections, Direct Image Mappings
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $f^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$.
+Let $f^\to$ be a [[Definition:Surjection|surjection]].
+Then $f: S \to T$ is also a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+Let $y \in S$.
+Since $f^\to$ is a [[Definition:Surjection|surjection]]:
+:$\exists U \in \powerset S: \map {f^\to} U = \set y$
+From definition of [[Definition:Direct Image Mapping of Mapping|direct image mapping]]:
+:$\set y \ne \O \implies U \ne \O$
+Let $x \in U$.
+Then from definition of [[Definition:Direct Image Mapping of Mapping|direct image mapping]]:
+:$\map f x = y$
+Since $y$ is arbitrary, $f$ is a [[Definition:Surjection|surjection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy Sequence in Metric Space is not necessarily Convergent}
+Tags: Metric Spaces, Cauchy Sequences
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]] in $M$.
+Then it is not necessarily the case that $M$ is a [[Definition:Convergent Sequence in Metric Space|convergent sequence]] in $M$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq \R$ be the [[Definition:Set|set]] of all points on $\R$ defined as:
+:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
+Let $M = \struct {A, \tau_d}$ be the [[Definition:Integer Reciprocal Space|integer reciprocal space]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] in $A$ that [[Definition:Convergent Sequence in Metric Space|converges]] to the [[Definition:Limit of Sequence (Metric Space)|limit]] $l \in A$.
+From [[Integer Reciprocal Space contains Cauchy Sequence with no Limit Point]], $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]] in $M$ which does not [[Definition:Convergent Sequence in Metric Space|converge]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nested Sequences in Complete Metric Space not Tending to Zero may be Disjoint}
+Tags: Complete Metric Spaces, Nested Sequences
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
+Let $\family {S_k}_{k \mathop \in \N}$ be a [[Definition:Nested Sequence|nested sequence]] of [[Definition:Closed Ball|closed balls]] in $M$.
+Let the [[Definition:Radius of Closed Ball|radii]] of $\family {S_k}_{k \mathop \in \N}$ be [[Definition:Convergent Sequence in Metric Space|convergent]] in $M$, but not to [[Definition:Zero (Number)|zero]].
+Then it is not necessarily the case that their [[Definition:Intersection of Family|intersection]] $\displaystyle \bigcap S_k$ is [[Definition:Non-Empty Set|non-empty]].
+\end{theorem}
+
+\begin{proof}
+Let $M = \struct {A, d}$ be [[Definition:Sierpinski's Metric Space|Sierpinski's metric space]]:
+:$A = \set {x_i: i = 1, 2, 3, \ldots}$
+:$\map d {x_i, x_j} = 1 + \dfrac 1 {i + j}$
+Let $S_k = \set {y \in A: \map d {y, x_k} \le 1 + \dfrac 1 {2 n} }$.
+From [[Nested Sequence of Closed Balls in Sierpinski's Metric Space with Empty Intersection]]:
+:$\displaystyle \bigcap S_k = \varnothing$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Interiors is Subset of Interior of Union}
+Tags: Set Interiors, Set Union
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
+That is, let $\H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
+Then the [[Definition:Set Union|union]] of the [[Definition:Interior (Topology)|interiors]] of the [[Definition:Element|elements]] of $\H$ is a [[Definition:Subset|subset]] of the [[Definition:Interior (Topology)|interior]] of the [[Definition:Set Union|union]] of $\H$.
+:$\displaystyle \bigcup_{H \mathop \in \H} H^\circ \subseteq \paren {\bigcup_{H \mathop \in \H} H}^\circ $
+\end{theorem}
+
+\begin{proof}
+In the following, $H^-$ denotes the [[Definition:Closure (Topology)|closure]] of the set $H$.
+{{begin-eqn}}
+{{eqn | l = \paren {\bigcup_{H \mathop \in \mathbb H} H}^\circ
+      | r = T \setminus \paren {T \setminus \bigcup_{H \mathop \in \mathbb H} H}^-
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-}
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{end-eqn}}
+At this point we note that:
+:$(1): \quad \displaystyle \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^- \subseteq \bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-$
+from [[Closure of Intersection is Subset of Intersection of Closures]].
+Then we note that:
+:$\displaystyle T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-} \subseteq T \setminus \paren {\paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H} }^-} $
+from $(1)$ and [[Set Complement inverts Subsets]].
+Then we continue:
+{{begin-eqn}}
+{{eqn | l = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} \paren {T \setminus H}^-}
+      | r = T \setminus \paren {\bigcap_{H \mathop \in \mathbb H} T \setminus H^\circ}
+      | c = [[Complement of Interior equals Closure of Complement]]
+}}
+{{eqn | r = T \setminus \paren {T \setminus \paren {\bigcup_{H \mathop \in \mathbb H} H^\circ} }
+      | c = [[De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union|De Morgan's Laws: Difference with Union]]
+}}
+{{eqn | r = \bigcup_{H \mathop \in \mathbb H} H^\circ
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Ball is Closed/Normed Vector Space}
+Tags: Normed Vector Spaces, Closed Balls
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $x \in X$.
+Let $\epsilon \in \R_{> 0}$.
+Let $\map {B_\epsilon^-} x$ be the [[Definition:Closed Ball in Normed Vector Space|closed $\epsilon$-ball]] of $x$ in $M$.
+Then $\map {B_\epsilon^-} x$ is a [[Definition:Closed Set in Normed Vector Space|closed set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+We show that the [[Definition:Set Complement|complement]] $X \setminus \map {B_\epsilon^-} x$ is [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+Let $y \in X \setminus \map {B_\epsilon^-} x$.
+Then by definition of [[Definition:Closed Ball in Normed Vector Space|closed ball]]:
+:$\norm {x - y} > \epsilon$
+Put:
+:$\delta := \norm {x - y} - \epsilon > 0$
+Then:
+:$\norm {x - y} - \delta = \epsilon$
+Let $z \in \map {B_\delta} y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x - z}
+      | o = \ge 
+      | r = \norm {x - y} - \norm {y - z}
+      | c = [[Reverse Triangle Inequality]]
+}}
+{{eqn | o = >
+      | r = \norm {x - y} - \delta
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+and so:
+:$z \notin \map {B_\epsilon^-} x$
+Then:
+:$\map {B_\delta} y \subseteq X \setminus \map {B_\epsilon^-} x$
+so $X \setminus \map {B_\epsilon^-} x$ is [[Definition:Open Set in Normed Vector Space|open]] in $M$.
+Hence, by definition of [[Definition:Closed Set in Normed Vector Space|closed set]]:
+:$\map {B_\epsilon^-} x$ is [[Definition:Closed Set in Normed Vector Space|closed]] in $M$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Sphere is Closed/Normed Vector Space}
+Tags: Closed Sets, Normed Vector Spaces
+
+\begin{theorem}
+Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\Bbb S := \set {x \in X : \norm {x} = 1}$ be a [[Definition:Unit Sphere/Normed Vector Space|unit sphere]] in $M$.
+Then $\Bbb S$ is [[Definition:Closed Set in Normed Vector Space|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an [[Definition:Open Ball in Normed Vector Space|open ball]].
+Let $\map {B_1^-} 0 = \set {x \in X : \norm x \le 1}$ be a [[Definition:Closed Ball in Normed Vector Space|closed ball]].
+Then:
+:$\displaystyle X = \Bbb S \bigcup \relcomp X {\Bbb S}$
+where
+:$\displaystyle \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map {B_1^-} 0}$
+is the [[Definition:Relative Complement|relative complement]] of $\Bbb S$ in $X$.
+We have that [[Closed Ball is Closed in Normed Vector Space]].
+By [[Definition:Closed Set in Normed Vector Space|definition]], $X \setminus \map {B_1^-} 0$ is [[Definition:Open Set in Normed Vector Space|open]].
+Furthermore, [[Open Ball is Open Set in Normed Vector Space]].
+By [[Union of Open Sets of Normed Vector Space is Open]], $\relcomp X {\Bbb S}$ is [[Definition:Open Set in Normed Vector Space|open]].
+By definition, $\Bbb S$ is [[Definition:Closed Set in Normed Vector Space|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continued Fraction Expansion of Euler's Number/Proof 1/Lemma}
+Tags: Euler's Number, Continued Fractions
+
+\begin{theorem}
+:For $n \in \Z , n \ge 0$:
+{{begin-eqn}}
+{{eqn | l = A_n
+      | r = q_{3 n} e - p_{3 n}
+}}
+{{eqn | l = B_n
+      | r = p_{3 n + 1} - q_{3 n + 1} e
+}}
+{{eqn | l = C_n
+      | r = p_{3 n + 2} - q_{3 n + 2} e
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:
+{{begin-eqn}}
+{{eqn | l = A_0
+      | r = \int_0^1 e^x \rd x
+}}
+{{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {x \mathop = 1}
+      | c = [[Primitive of Exponential Function]]
+}}
+{{eqn | r = e - 1
+}}
+{{eqn | r = q_0 e - p_0
+}}
+{{eqn | l = B_0
+      | r = \int_0^1 x e^x \rd x
+}}
+{{eqn | r = \bigintlimits {x e^x - e^x} {x \mathop = 0} {x \mathop = 1}
+      | c = [[Primitive of x by Exponential of a x]]
+}}
+{{eqn | r = 1
+}}
+{{eqn | r = p_1 - q_1 e
+}}
+{{eqn | l = C_0
+      | r = \int_0^1 \paren {x - 1} e^x \rd x
+}}
+{{eqn | r = \int_0^1 x e^x \rd x - \int_0^1 e^x \rd x
+}}
+{{eqn | r = B_0 - A_0
+}}
+{{eqn | r = 1 - \paren {e - 1}
+}}
+{{eqn | r = 2 - e
+}}
+{{eqn | r = p_2 - q_2 e
+}}
+{{end-eqn}}
+The final step needed to validate the assertion, we must demonstrate that the following three [[Definition:Recursive Sequence|recurrence relations]] hold:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = A_n
+      | r = -B_{n - 1} - C_{n - 1}
+}}
+{{eqn | n = 2
+      | l = B_n
+      | r = -2 n A_n + C_{n - 1}
+}}
+{{eqn | n = 3
+      | l = C_n
+      | r = B_n - A_n
+}}
+{{end-eqn}}
+To prove the '''first''' relation, we note that the [[Definition:Derivative|derivative]] of the [[Definition:Integrand|integrand]] of $A_n$ is equal to the sum of the [[Definition:Integrand|integrand]] of $A_n$ with the [[Definition:Integrand|integrand]] of $B_{n - 1}$ and the [[Definition:Integrand|integrand]] of $C_{n - 1}$.
+By integrating both sides of the equation, we verify the first [[Definition:Recursive Sequence|recurrence relation]]:
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n } {n!} e^x}
+      | r = \frac {x^n \paren {x - 1 }^n} {n!} e^x + \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x + \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^n} {n!} e^x}
+      | r = \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x + \int_0^1 \frac {x^n \paren {x - 1}^{n - 1} } {\paren {n - 1}!} e^x \rd x + \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x
+      | c = Integrating both sides of the equation over the interval from $0$ to $1$
+}}
+{{eqn | ll= \leadsto
+      | l = \intlimits {\frac {x^n \paren {x - 1}^n} {n!} e^x} {x \mathop = 0} {x \mathop = 1}
+      | r = A_n + B_{n - 1} + C_{n - 1}
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = A_n + B_{n - 1} + C_{n - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_n
+      | r = -B_{n - 1} - C_{n - 1}
+      | c = rearranging
+}}
+{{end-eqn}}
+To prove the '''second''' relation, we note that the [[Definition:Derivative|derivative]] of the [[Definition:Integrand|integrand]] of $C_n$ is equal to the sum of the [[Definition:Integrand|integrand]] of $B_n$ with two times n times the [[Definition:Integrand|integrand]] of $A_{n}$ minus the [[Definition:Integrand|integrand]] of $C_{n - 1}$.
+By integrating both sides of the equation, we verify the second [[Definition:Recursive Sequence|recurrence relation]]:
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}
+      | r = \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x + 2 n \frac {x^n \paren {x - 1}^n} {n!} e^x - \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int_0^1 \map {\frac \d {\d x} } {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x}
+      | r = \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x + 2 n \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^{n - 1} \paren {x - 1}^n} {\paren {n - 1}!} e^x \rd x
+      | c = Integrating both sides of the equation over the interval from $0$ to $1$
+}}
+{{eqn | ll= \leadsto
+      | l = \intlimits {\frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x} {x \mathop = 0} {x \mathop = 1}
+      | r = B_n + 2 n A_{n} - C_{n - 1}
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = B_n + 2 n A_n - C_{n - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B_n
+      | r = -2 n A_n + C_{n - 1}
+      | c = rearranging
+}}
+{{end-eqn}}
+To prove the '''third''' relation, we have:
+{{begin-eqn}}
+{{eqn | l = C_n
+      | r = \int_0^1 \frac {x^n \paren {x - 1}^{n + 1} } {n!} e^x \rd x
+      | c = 
+}}
+{{eqn | r = \int_0^1 \frac {x^n \paren {x - 1 }^n} {n!} e^x \paren {x - 1} \rd x
+      | c = factoring out $\paren {x - 1}$
+}}
+{{eqn | r = \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \paren 1 \rd x
+      | c = separate integrals
+}}
+{{eqn | r = \int_0^1 \frac {x^{n + 1} \paren {x - 1}^n} {n!} e^x \rd x - \int_0^1 \frac {x^n \paren {x - 1}^n} {n!} e^x \rd x
+      | c =
+}}
+{{eqn | r = B_n - A_n
+      | c = 
+}}
+{{end-eqn}}
+From the '''first''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
+{{begin-eqn}}
+{{eqn | l = A_n
+      | r = -B_{n - 1} - C_{n - 1}
+}}
+{{eqn | r = -\paren {p_{3 n - 2} - q_{3 n -2 } e} - \paren {p_{3 n - 1} - q_{3 n - 1} e}     
+}}
+{{eqn | r = q_{3 n} e - p_{3 n}
+}}
+{{end-eqn}}
+From the '''second''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
+{{begin-eqn}}
+{{eqn | l = B_n
+      | r = -2 n A_n + C_{n - 1}
+}}
+{{eqn | r = -2 n \paren {q_{3 n} e - p_{3 n} } + \paren {p_{3 n - 1} - q_{3 n - 1} e}     
+}}
+{{eqn | r = p_{3 n + 1} - q_{3 n + 1} e
+}}
+{{end-eqn}}
+From the '''third''' relation, combined with the initial condition at $n = 0$ being satisfied, we have:
+{{begin-eqn}}
+{{eqn | l = C_n
+      | r = B_n - A_n
+}}
+{{eqn | r = \paren {p_{3 n + 1} - q_{3 n + 1} e} - \paren {q_{3 n} e - p_{3 n} }     
+}}
+{{eqn | r = p_{3 n + 2} - q_{3 n + 2 } e
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Particular Point Space/Subset}
+Tags: Limit Points in Particular Point Space
+
+\begin{theorem}
+Let $U \subseteq S$ such that $p \in U$.
+Let $x \in S$ such that $x \ne p$.
+Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}<|endoftext|>
+\section{Limit Points in Closed Extension Space/Subset}
+Tags: Limit Points in Closed Extension Space
+
+\begin{theorem}
+Let $U \subseteq S^*_p$ such that $p \in U$.
+Let $x \in S$.
+Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by [[Definition:Closed Extension Topology|definition]].
+So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_p$ such that $x \in U$ contains $p$.
+So by definition of the [[Definition:Limit Point of Set|limit point of a set]], $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence in Particular Point Space}
+Tags: Particular Point Topology, Convergent Sequences
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $\sequence {a_i}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in $T$.
+Except for a [[Definition:Finite Set|finite number]] of [[Definition:Index of Term of Sequence|indices]], the [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$ for which $a_i \ne p$ are all equal.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {a_i}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in $T$ whose [[Definition:Limit Point of Sequence|limit]] is $\alpha$.
+Then by definition every [[Definition:Open Set (Topology)|open set]] in $T$ containing $\alpha$ contains all but a [[Definition:Finite Set|finite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$.
+This includes the [[Definition:Open Set (Topology)|open set]] $\set {\alpha, p}$.
+Thus $\sequence {a_i}$ is such that, except for a [[Definition:Finite Set|finite number]] of [[Definition:Term of Sequence|terms]], all are equal either to $\alpha$ or $p$.
+Thus $\sequence {a_i}$ is such that, except for a [[Definition:Finite Set|finite number]] of [[Definition:Term of Sequence|terms]], all such that $a_i \ne p$, are equal to $\alpha$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Accumulation Points for Sequence in Particular Point Space}
+Tags: Particular Point Topology, Accumulation Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+Let $\sequence {a_i}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $T$.
+Let $\beta$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
+Then $\beta$ is such that an [[Definition:Infinite Set|infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$ are equal either to $\beta$ or to $p$.
+\end{theorem}
+
+\begin{proof}
+Let $\beta$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
+Then by definition:
+:all [[Definition:Open Set (Topology)|open sets]] of $T$ which contain $\beta$ also contain an [[Definition:Infinite Set|infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$.
+This condition applies to the [[Definition:Open Set (Topology)|open set]] $\set {\beta, p}$.
+So $\set {\beta, p}$ contains an [[Definition:Infinite Set|infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_i}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value of Absolutely Continuous Function is Absolutely Continuous}
+Tags: Absolutely Continuous Functions
+
+\begin{theorem}
+Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
+Let $f : I \to \R$ be an [[Definition:Absolute Continuity|absolutely continuous]] [[Definition:Real Function|function]].
+Then $\size f$ is [[Definition:Absolute Continuity|absolutely continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
+Since $f$ is absolutely continuous, there exists [[Definition:Real Number|real]] $\delta > 0$ such that for all collections of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
+By the [[Reverse Triangle Inequality/Real and Complex Fields|Reverse Triangle Inequality]], we have: 
+:$\size {\map f {b_i} - \map f {a_i} } \ge \size {\size {\map f {b_i} } - \size {\map f {a_i} } }$
+Therefore: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\size {\map f {b_i} } - \size {\map f {a_i} } } \le \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
+whenever:
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+Since $\epsilon$ was arbitrary:
+:$\size f$ is [[Definition:Absolute Continuity|absolutely continuous]].
+{{qed}}
+[[Category:Absolutely Continuous Functions]]
+fp6q7lzsazm506llf7tqahhf1e5yh6a
+\end{proof}<|endoftext|>
+\section{Limit Points in Particular Point Space/Subset/Proof 2}
+Tags: Limit Points in Particular Point Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+{{:Limit Points in Particular Point Space/Subset}}
+\end{theorem}
+
+\begin{proof}
+Follows directly from:
+:[[Particular Point Topology is Closed Extension Topology of Discrete Topology]]
+:[[Limit Points in Subset of Closed Extension Space]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Points in Particular Point Space/Subset/Proof 1}
+Tags: Limit Points in Particular Point Space
+
+\begin{theorem}
+Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
+{{:Limit Points in Particular Point Space/Subset}}
+\end{theorem}
+
+\begin{proof}
+Every [[Definition:Open Set (Topology)|open set]] of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by [[Definition:Particular Point Topology|definition]].
+So every [[Definition:Open Set (Topology)|open set]] $U \in \tau_p$ such that $x \in U$ contains $p$.
+So by definition of the [[Definition:Limit Point of Set|limit point of a set]], $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Absolutely Continuous Functions is Absolutely Continuous}
+Tags: Absolutely Continuous Functions
+
+\begin{theorem}
+Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
+Let $f, g : I \to \R$ be [[Definition:Absolute Continuity|absolutely continuous]] [[Definition:Real Function|functions]].
+Then $f + g$ is [[Definition:Absolute Continuity|absolutely continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
+Since $f$ is [[Definition:Absolute Continuity|absolutely continuous]], there exists [[Definition:Real Number|real]] $\delta_1 > 0$ such that for all [[Definition:Set|sets]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
+Similarly, since $g$ is [[Definition:Absolute Continuity|absolutely continuous]], there exists [[Definition:Real Number|real]] $\delta_2 > 0$ such that whenever: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
+Let: 
+:$\delta = \map \min {\delta_1, \delta_2}$
+Then, for all [[Definition:Set|sets]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon 2$
+and:
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon 2$
+We then have:
+{{begin-eqn}}
+{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {b_i} - \map {\paren {f + g} } {a_i} }
+	| r = \sum_{i \mathop = 1}^n \size {\paren {\map f {b_i} - \map f {a_i} } + \paren {\map g {b_i} - \map g {a_i} } }
+}}
+{{eqn	| o = \le
+	| r = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }
+	| c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn	| o = <
+	| r = \frac \epsilon 2 + \frac \epsilon 2
+}}
+{{eqn	| r = \epsilon
+}}
+{{end-eqn}}
+whenever: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+Since $\epsilon$ was arbitrary:
+:$f + g$ is [[Definition:Absolute Continuity|absolutely continuous]].
+{{qed}}
+[[Category:Absolutely Continuous Functions]]
+nqfoj663e9mkgdcghrotrjnsi82vvuw
+\end{proof}<|endoftext|>
+\section{Norms on Finite-Dimensional Real Vector Space are Equivalent}
+Tags: Normed Vector Spaces, Equivalence Relations, Norm Theory, Vector Spaces
+
+\begin{theorem}
+[[Definition:Norm on Vector Space|Norms]] on [[Definition:Finite Dimensional Vector Space|finite-dimensional]] [[Definition:Real Vector Space|real vector space]] are [[Definition:Equivalence of Norms|equivalent]].
+\end{theorem}
+
+\begin{proof}
+We will prove that all [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]] to $\norm {\, \cdot \,}_2$.
+By definition, two [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]] on $\R^d$ {{iff}}:
+:$\forall \mathbf x \in \R^d : \exists m, M \in \R_{> 0} : m \norm {\mathbf x}_a \le \norm {\mathbf x}_b \le M \norm {\mathbf x}_a$
+=== "Less or equal" condition ===
+Let $\set {\mathbf e_1 \dots \mathbf e_n}$ be a [[Definition:Standard Basis/Vector Space|standard basis]] in $\R^d$.
+We have that each $\mathbf x \in \R^d$ is [[Expression of Vector as Linear Combination from Basis is Unique|uniquely]] expressible as:
+:$\displaystyle \mathbf x = \sum_{i \mathop = 1}^d x_i \mathbf e_i$
+where $x_i$ is a [[Definition:Scalar (Vector Space)|scalar]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\sum_{i \mathop = 1}^d x_i \mathbf e_i}
+      | o = \le
+      | r = \sum_{i \mathop = 1}^d \norm {x_i \mathbf e_i}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm axiom]]: triangle inequality
+}}
+{{eqn | r = \sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm axiom]]: positive homogeneity
+}}
+{{eqn | r = \sqrt {\paren {\sum_{i \mathop = 1}^d \size {x_i} \norm {\mathbf e_i} }^2 }
+}}
+{{eqn | o = \le
+      | r = \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \sqrt {\sum_{j \mathop = 1}^d \size {x_j}^2 }
+      | c = [[Cauchy-Schwarz Inequality]]
+}}
+{{eqn | o = \le
+      | r = \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2} \norm {\mathbf x}_2
+      | c = {{Defof|P-Norm|$p$-norm}}
+}}
+{{eqn | r = M \norm {\mathbf x}_2
+      | c = Define $\displaystyle M := \sqrt{\sum_{i \mathop = 1}^d \norm {\mathbf e_i}^2}$}}
+{{end-eqn}}
+By definition of [[Definition:Norm on Vector Space|norm]], its [[Definition:Image|image]] is the [[Definition:Set|set]] of [[Definition:Non-Negative Reals|non-negative real numbers]]: $M \in \R_{\ge 0}$.
+Hence:
+:$\forall \mathbf x \in \R^d : \exists M \in \R_{\ge 0} : \norm {\mathbf x} \le M \norm {\mathbf x}_2$
+{{qed|lemma}}
+=== Existence of $m$ ===
+Let $K := \set {\mathbf y \in \R^d : \norm {\mathbf y}_2 = 1}$ be a [[Definition:Unit Sphere/Normed Vector Space|unit sphere]] in $\struct {\R^d, \norm {\, \cdot \,}_2}$.
+By [[Unit Sphere is Closed in Normed Vector Space]], $K$ is [[Definition:Closed Set of Normed Vector Space|closed]] in $\struct {\R^d, \norm{\, \cdot \,}_2}$.
+By [[Definition:Bounded Normed Vector Space|definition]], $K$ is [[Definition:Bounded Normed Vector Space|bounded]] in $\struct {\R^d, \norm{\, \cdot \,}_2}$.
+Hence, by the [[Heine-Borel Theorem/Normed Vector Space|Heine-Borel theorem]], $K$ is a [[Definition:Compact Normed Vector Space|compact set]].
+By [[Norm on Vector Space is Continuous Function]], the [[Definition:Mapping|map]] $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ is [[Definition:Continuous Mapping|continuous]] from $\struct {K, \norm {\, \cdot \,}_2}$ to $\struct {\R_{\ge 0}, \size {\, \cdot \,}}$:
+:$\forall \mathbf y_1, \mathbf y_2 \in K : \size {\norm {\mathbf y_1} - \norm {\mathbf y_2}} \le \norm {\mathbf y_1 - \mathbf y_2} \le M \norm {\mathbf y_1 - \mathbf y_2}_2$
+By [[Weierstrass Extreme Value Theorem|Weierstrass theorem]], $\norm {\, \cdot \,} : K \to \R_{\ge 0}$ attains a [[Definition:Minimum Value of Real Function|minimum value]] $m$ for some $\mathbf y \in K$.
+Suppose $m = 0$.
+Then $\norm {\mathbf y} = 0$.
+By the [[Definition:Norm Axioms (Vector Space)|norm axiom]] of positive definiteness, $\mathbf y = 0$.
+Then $\mathbf y \notin K$.
+Hence, $m \ne 0$.
+So $m > 0$.
+Furthermore:
+:$\forall \mathbf y \in \R^d : \norm {\mathbf y}_2 = 1 : \norm {\mathbf y} \ge m $
+{{qed|lemma}}
+=== "Greater or equal" condition ===
+Suppose $\mathbf x = 0$.
+Then we have [[Definition:Equality|equality]].
+Suppose $\mathbf x \ne 0$.
+Let $\displaystyle \mathbf y = \frac {\mathbf x} {\norm {\mathbf x}_2}$.
+We have that $\norm {\mathbf y}_2 = 1$ and $\mathbf y \in K$.
+Then:
+{{begin-eqn}}
+{{eqn | l = m 
+      | o = \le 
+      | r = \norm {\mathbf y} 
+}}
+{{eqn | r = \norm {\frac {\mathbf x} {\norm {\mathbf x}_2} } 
+}}
+{{eqn | r = \frac {\norm {\mathbf x} } {\norm {\mathbf x}_2}
+}}
+{{end-eqn}} 
+This implies that:
+:$m \norm {\mathbf x}_2 \le \norm {\mathbf x}$
+Therefore:
+:$\forall \mathbf x \in \R^d : \exists m, M : m \norm {\mathbf x}_2 \le \norm {\mathbf x} \le M \norm {\mathbf x}_2$
+By definition, all [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]] to $\norm{\, \cdot \,}_2$.
+By [[Norm Equivalence is Equivalence]], the [[Definition:Equivalence Relation|equivalence relation]] $\norm {\, \cdot \,} \sim \norm {\, \cdot \,}_2$ is [[Definition:Transitive Relation|transitive]].
+Thus, all [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Distribution Function}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]] and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.
+For $\lambda > 0$, let $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.
+Then:
+:$\displaystyle \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda = \int_X \size f^{p + r} \rd \mu$
+and in particular:
+:$\displaystyle \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda = \int_X \size f^p \rd \mu$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda
+      | r = \int_0^\infty \int_{E_\lambda} p \lambda^{p - 1} \size f^r \rd \mu \rd \lambda
+}}
+{{eqn | r = \int_X \int_0^{\size {\map f x} } p \lambda^{p - 1} \size f^r \rd \lambda \rd \mu
+      | c = by [[Tonelli's Theorem]]
+}}
+{{eqn | r = \int_X \size f^r \int_0^{\size {\map f x} } p \lambda^{p - 1} \rd \lambda \rd \mu
+}}
+{{eqn | r = \int_X \size f^r \size f^p \rd \mu
+      | c = by [[Integral of Power]]
+}}
+{{eqn | r = \int_X \size f^{p + r} \rd \mu
+}}
+{{end-eqn}}
+We have that for any measurable $A \in \Sigma$:
+:$\map \mu A = \displaystyle \int_A 1 \rd \mu$
+Therefore, for $\lambda > 0$:
+:$\map \mu {E_\lambda} = \displaystyle \int_{E_\lambda} 1 \rd \mu$
+which can also be written:
+:$\map \mu {E_\lambda} = \displaystyle \int_{E_\lambda} \size f^0 \rd \mu$
+Therefore, taking $r = 0$ in the above, we obtain:
+{{begin-eqn}}
+{{eqn | l = \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda
+      | r = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^0 \rd \mu \rd \lambda
+      | c = 
+}}
+{{eqn | r = \int_X \size f^p \rd \mu
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Measure Theory]]
+e0w1gsd8i9iaj9udgcw28lqo0zizs99
+\end{proof}<|endoftext|>
+\section{Limit Points in Open Extension Space/Subset}
+Tags: Limit Points in Open Extension Space
+
+\begin{theorem}
+Let $U \subseteq S^*_p$.
+Then $p$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+\end{theorem}
+
+\begin{proof}
+Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by [[Definition:Open Extension Topology|definition]].
+So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such [[Definition:Open Set (Topology)|open set]]) contains $x$.
+So by definition of the [[Definition:Limit Point of Set|limit point of a set]], $p$ is a [[Definition:Limit Point of Set|limit point]] of $U$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differentiable Function with Bounded Derivative is Absolutely Continuous}
+Tags: Absolutely Continuous Functions
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f: \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
+Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$, with [[Definition:Bounded Real-Valued Function|bounded]] [[Definition:Derivative|derivative]].
+Then $f$ is [[Definition:Absolute Continuity|absolutely continuous]].
+\end{theorem}
+
+\begin{proof}
+Since the [[Definition:Derivative|derivative]] of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that: 
+:$\size {\map {f'} x} \le M$
+for all $x \in \openint a b$.
+Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]].
+Note that for each $i \in \set {1, 2, \ldots, n}$:
+:$f$ is [[Definition:Continuous Real Function|continuous]] on $\closedint {a_i} {b_i}$ and [[Definition:Differentiable Real Function|differentiable]] on $\openint {a_i} {b_i}$.
+So, by the [[Mean Value Theorem]], for each $i$ there exists some $\xi_i \in \openint {a_i} {b_i}$ such that:
+:$\map {f'} {\xi_i} = \dfrac {\map f {b_i} - \map f {a_i} } {b_i - a_i}$
+We then have:
+{{begin-eqn}}
+{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} }
+	| r = \sum_{i \mathop = 1}^n \size {\map {f'} {\xi_i} } \paren {b_i - a_i}
+}}
+{{eqn	| o = \le
+	| r = M \sum_{i \mathop = 1}^n \paren {b_i - a_i}
+	| c = since $\xi_i \in \openint a b$, we have $\size {\map {f'} {\xi_i} } \le M$
+}}
+{{end-eqn}}
+Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]]. 
+Then for all collections of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:
+:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \epsilon M$
+we have:
+:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
+Since $\epsilon$ was arbitrary:
+:$f$ is [[Definition:Absolute Continuity|absolutely continuous]].
+{{qed}}
+[[Category:Absolutely Continuous Functions]]
+0fg9ljkpexbqj5lldwzjnp4rnf0t3it
+\end{proof}<|endoftext|>
+\section{Either-Or Topology is Compact}
+Tags: Compact Spaces, Either-Or Topology
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
+Then $T$ is a [[Definition:Compact Space|compact space]].
+\end{theorem}
+
+\begin{proof}
+Any [[Definition:Open Cover|open cover]] $\CC$ of $T$ must contain an [[Definition:Open Set (Topology)|open set]] of $T$ which contains $0$.
+So $\openint {-1} 1$ will always be [[Definition:Cover of Set|covered]] by one [[Definition:Set|set]] in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other [[Definition:Set|sets]].
+So $\CC$ has a [[Definition:Subcover|subcover]] containing at most three [[Definition:Set|sets]].
+Hence $T$ is a [[Definition:Compact Space|compact space]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compact Space is Lindelöf}
+Tags: Compact Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Every [[Definition:Compact Topological Space|compact space]] is [[Definition:Lindelöf Space|Lindelöf]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Compact Space is Sigma-Compact]]
+:[[Sigma-Compact Space is Lindelöf]]
+Hence the result.
+{{qed}}
+[[Category:Compact Spaces]]
+[[Category:Lindelöf Spaces]]
+rq8ho9t43qo43ua1sdqq3ndsddg55j6
+\end{proof}<|endoftext|>
+\section{Product of Absolutely Continuous Functions is Absolutely Continuous}
+Tags: Absolutely Continuous Functions
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f, g : \closedint a b \to \R$ be [[Definition:Absolute Continuity|absolutely continuous]] [[Definition:Real Function|functions]].
+Then $f \times g$ is [[Definition:Absolute Continuity|absolutely continuous]].
+\end{theorem}
+
+\begin{proof}
+From [[Absolutely Continuous Real Function is Continuous]]:
+:$f$ and $g$ are [[Definition:Continuous Real Function|continuous]].
+From [[Closed Real Interval is Compact in Metric Space]]:
+:$\closedint a b$ is [[Definition:Compact Subset of Real Numbers|compact]].
+Therefore, by [[Continuous Function on Compact Subspace of Euclidean Space is Bounded]]:
+:$f$ and $g$ are [[Definition:Bounded Mapping|bounded]].
+That is, there exists $M_f, M_g \in \R_{> 0}$ such that: 
+:$\size {\map f x} \le M_f$
+:$\size {\map g x} \le M_g$
+for all $x \in \closedint a b$.
+Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a [[Definition:Set|set]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]].
+Then:
+{{begin-eqn}}
+{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} }
+	| r = \sum_{i \mathop = 1}^n \size {\map f {b_i} \map g {b_i} - \map f {a_i} \map g{a_i} + \map f {a_i} \map g {b_i} - \map f {a_i} \map g {b_i} }
+}}
+{{eqn	| r = \sum_{i \mathop = 1}^n \size {\map g {b_i} \paren {\map f {b_i} - \map f {a_i} } + \map f {a_i} \paren {\map g {b_i} - \map g {a_i} } }
+}}
+{{eqn	| o = \le
+	| r = \sum_{i \mathop = 1}^n \size {\map g {b_i} } \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map f {a_i} } \size {\map g {b_i} - \map g {a_i} }
+	| c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn	| o = \le
+	| r = M_g \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + M_f \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }
+	| c = since $a_i, b_i \in \closedint a b$
+}}
+{{end-eqn}}
+Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
+Since $f$ is [[Definition:Absolute Continuity|absolutely continuous]], there exists $\delta_1 > 0$ such that for all [[Definition:Set|sets]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$
+Similarly, since $g$ is [[Definition:Absolute Continuity|absolutely continuous]], there exists $\delta_2 > 0$ such that whenever:
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$
+we have:
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$
+Let:
+:$\delta = \map \min {\delta_1, \delta_2}$
+Then, whenever:
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+We have both:
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \epsilon {2 M_g}$
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \epsilon {2 M_f}$
+and hence: 
+{{begin-eqn}}
+{{eqn	| l = \sum_{i \mathop = 1}^n \size {\map {\paren {f \times g} } {b_i} - \map {\paren {f \times g} } {a_i} }
+	| o = <
+	| r = M_g \times \frac \epsilon {2 M_g} + M_f \times \frac \epsilon {2 M_f}
+}}
+{{eqn	| r = \epsilon
+}}
+{{end-eqn}}
+Since $\epsilon$ was arbitrary, we have:
+:$f \times g$ is [[Definition:Absolute Continuity|absolutely continuous]].
+{{qed}}
+[[Category:Absolutely Continuous Functions]]
+1dzalwwopk9m4yyu8d8dex5ma42gck3
+\end{proof}<|endoftext|>
+\section{Accumulation Point of Sequence is not necessarily Limit Point}
+Tags: Accumulation Points, Limit Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] in $T$.
+Let $q \in S$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_n}$.
+Then it is not necessarily the case that $q$ is also a [[Definition:Limit Point of Sequence|limit point]] of $\sequence {a_n}$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
+{{begin-eqn}}
+{{eqn | l = \sequence {a_n}
+      | r = \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}
+      | c = 
+}}
+{{eqn | r = \sequence {1, 1, 1, 2, 1, 3, 1, 4, \dotsc}
+      | c = 
+}}
+{{end-eqn}}
+Then $\sequence {a_n}$ has [[Definition:Unique|exactly one]] [[Definition:Accumulation Point of Sequence|accumulation point]], that is $1$.
+However, $1$ is not a [[Definition:Limit Point of Sequence|limit point]] of $\sequence {a_n}$, as $\sequence {a_n}$ has no [[Definition:Limit Point of Sequence|limit point]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1}
+Tags: Sequences
+
+\begin{theorem}
+Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] defined as:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
+      | c = 
+}}
+{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
+      | c = 
+}}
+{{end-eqn}}
+Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $S$ denote the [[Definition:Set|set]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ considered as a [[Definition:Subset|subset]] of $\struct {\R, \tau_d}$.
+Then $0$ is a [[Definition:Limit Point of Set|limit point]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
+Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$.
+Hence the result by definition of [[Definition:Limit Point of Set|limit point]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Omega-Accumulation Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1}
+Tags: Sequences, Omega-Accumulation Points
+
+\begin{theorem}
+Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] defined as:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
+      | c = 
+}}
+{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
+      | c = 
+}}
+{{end-eqn}}
+Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $S$ denote the [[Definition:Set|set]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ considered as a [[Definition:Subset|subset]] of $\struct {\R, \tau_d}$.
+Then $0$ is an [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
+We have that:
+:$\forall n \in \N: n \ge m$
+have the property that $0 < \dfrac 1 n < \epsilon$.
+Hence there are a [[Definition:Countably Infinite Set|countably infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
+Hence the result by definition of [[Definition:Omega-Accumulation Point|$\omega$-accumulation point]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Accumulation Point of Sequence of Reciprocals and Reciprocals + 1}
+Tags: Sequences, Accumulation Points
+
+\begin{theorem}
+Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] in $\struct {\R, \tau}$ defined as:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
+      | c = 
+}}
+{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
+      | c = 
+}}
+{{end-eqn}}
+Then $0$ is an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_n}$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
+We have that:
+:$\forall n \in \N: n \ge m$
+have the property that $0 < \dfrac 1 n < \epsilon$.
+Hence there are a [[Definition:Countably Infinite Set|countably infinite number]] of [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ such that $a_n \in \openint {-\epsilon} \epsilon$.
+Hence the result by definition of [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_n}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1}
+Tags: Sequences, Limit Points
+
+\begin{theorem}
+Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] in $\struct {\R, \tau}$ defined as:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
+      | c = 
+}}
+{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac 1 3, \dotsb}
+      | c = 
+}}
+{{end-eqn}}
+Then $0$ is not a [[Definition:Limit Point of Sequence|limit point]] of $\sequence {a_n}$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Open Real Interval|open interval]] $\openint 1 1$ contains $0$, and also contains all [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ with [[Definition:Odd Integer|odd]] [[Definition:Index of Term of Sequence|indices]] greater than $1$.
+However, all [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ with [[Definition:Even Integer|even]] [[Definition:Index of Term of Sequence|indices]] are outside $\openint {-\dfrac 1 2} {\dfrac 1 2}$.
+Hence $0$ cannot be a [[Definition:Limit Point of Sequence|limit point]] of $\sequence {a_n}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lagrange's Theorem (Number Theory)}
+Tags: Number Theory, Polynomial Theory, Proofs by Induction
+
+\begin{theorem}
+Let $f$ be a [[Definition:Polynomial over Field|polynomial in one variable]] of [[Definition:Degree of Polynomial|degree]] $n$ over $\Z_p$ for some [[Definition:Prime Number|prime]] $p$.
+Then $f$ has at most $n$ [[Definition:Root of Polynomial|roots]] in $\Z_p$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] on $n$:
+=== Basis for the Induction ===
+When $n = 1$, we have:
+:$\map f x = a x + b$ for some $a, b \in \Z_p$ and $a \ne 0$
+Suppose $x_1, x_2 \in \Z_p$ are two [[Definition:Root of Polynomial|roots]] of $\map f x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a x_1 + b
+      | o = \equiv
+      | r = a x_2 + b
+      | rr = \equiv 0
+      | rrr = \pmod p
+}}
+{{eqn | ll = \leadsto
+      | l = a x_1
+      | o = \equiv
+      | r = a x_2
+      | rrr = \pmod p
+}}
+{{eqn | ll = \leadsto
+      | l = x_1
+      | o = \equiv
+      | r = x_2
+      | rrr = \pmod p
+      | c = since $a \perp p$
+}}
+{{end-eqn}}
+Hence these two [[Definition:Root of Polynomial|roots]] must be the same, implying that there is at most $1$ [[Definition:Root of Polynomial|root]].
+This is our [[Definition:Basis for the Induction|base case]].
+=== Induction Hypothesis ===
+This is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:Any [[Definition:Polynomial over Field|polynomial in one variable]] of [[Definition:Degree of Polynomial|degree]] $k$ has at most $k$ [[Definition:Root of Polynomial|roots]] in $\Z_p$.
+It is to be demonstrated that:
+:Any [[Definition:Polynomial over Field|polynomial in one variable]] of [[Definition:Degree of Polynomial|degree]] $k + 1$ has at most $k + 1$ [[Definition:Root of Polynomial|roots]] in $\Z_p$.
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Consider $n = k + 1$, and let $f$ be a [[Definition:Polynomial over Field|polynomial in one variable]] of [[Definition:Degree of Polynomial|degree]] $k + 1$.
+If $f$ does not have a [[Definition:Root of Polynomial|root]] in $\Z_p$, our claim is satisfied.
+Hence suppose $f$ does have a root $x_0$.
+From [[Ring of Integers Modulo Prime is Field]], $\Z_p$ is a [[Definition:Field (Abstract Algebra)|field]].
+Applying the [[Polynomial Factor Theorem]], since $\map f {x_0} = 0$:
+:$\map f x = \paren {x - x_0} \map Q x$
+where $Q$ is a [[Definition:Polynomial over Field|polynomial]] of [[Definition:Degree of Polynomial|degree]] $k$.
+By [[Euclid's Lemma for Prime Divisors]]:
+:$\map f x = 0 \iff x - x_0 = 0$ or $\map Q x = 0$
+By induction hypothesis, $Q$ has at most $k$ [[Definition:Root of Polynomial|roots]].
+Hence $f$ has at most $k + 1$ [[Definition:Root of Polynomial|roots]].
+By the [[Principle of Mathematical Induction]], the theorem is true for any $n$.
+{{qed}}
+{{Namedfor|Joseph Louis Lagrange}}
+[[Category:Number Theory]]
+[[Category:Polynomial Theory]]
+[[Category:Proofs by Induction]]
+6lpsyjudkseo2swix5o3u7a4nui5874
+\end{proof}<|endoftext|>
+\section{1 plus Power of 2 is not Perfect Power except 9}
+Tags: Number Theory
+
+\begin{theorem}
+The only solution to:
+:$1 + 2^n = a^b$
+is:
+:$\tuple {n, a, b} = \tuple {3, 3, 2}$
+for [[Definition:Positive Integer|positive integers]] $n, a, b$ with $b > 1$.
+\end{theorem}
+
+\begin{proof}
+It suffices to prove the result for [[Definition:Prime Number|prime]] values of $b$.
+For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a [[Definition:Perfect Power|perfect power]].
+For $n > 0$, $1 + 2^n$ is [[Definition:Odd Integer|odd]].
+Hence for the equation to hold $a$ must be [[Definition:Odd Integer|odd]] as well.
+Writing $a = 2 m + 1$ we have:
+{{begin-eqn}}
+{{eqn | l = 1 + 2^n
+      | r = \paren {2 m + 1}^b
+}}
+{{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i \paren 1^{b - i}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = \sum_{i \mathop = 0}^b \binom b i \paren {2 m}^i
+}}
+{{eqn | r = 1 + \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | l = 2^n
+      | r = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^i
+}}
+{{eqn | r = 2 m \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}
+      | c = $m \ne 0$
+}}
+{{end-eqn}}
+Since all [[Definition:Divisor of Integer|factors]] of $2^n$ are [[Definition:Integer Power|powers]] of $2$:
+:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is a [[Definition:Integer Power|power]] of $2$.
+But since each summand is [[Definition:Non-Negative Integer|non-negative]]:
+:$\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1} \ge 2$
+we must have $\ds \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}$ is [[Definition:Even Integer|even]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^b \binom b i \paren {2 m}^{i - 1}
+      | r = \binom b 1 + \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 1}
+}}
+{{eqn | r = b + 2 m \sum_{i \mathop = 2}^b \binom b i \paren {2 m}^{i - 2}
+}}
+{{eqn | o = \equiv
+      | r = b
+      | rr = \pmod 2
+}}
+{{end-eqn}}
+Therefore we must have $b = 2$, the only [[Definition:Even Integer|even]] [[Definition:Prime Number|prime]].
+In that case:
+{{begin-eqn}}
+{{eqn | l = 2^n
+      | r = \paren {2 m + 1}^2 - 1
+}}
+{{eqn | r = 4 m^2 + 4 m + 1 - 1
+}}
+{{eqn | r = 4 m \paren {m + 1}
+}}
+{{end-eqn}}
+So $m$ and $m + 1$ are [[Definition:Integer Power|powers]] of $2$.
+The only $m$ satisfying this is $1$, giving the solution:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = 2 m + 1
+      | c = 
+}}
+{{eqn | r = 3
+      | c = 
+}}
+{{eqn | l = 2^n
+      | r = 3^2 - 1
+      | c = 
+}}
+{{eqn | r = 8
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | r = 3
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fermat Number is not Perfect Power}
+Tags: Fermat Numbers
+
+\begin{theorem}
+There exist no [[Definition:Fermat Number|Fermat numbers]] which are [[Definition:Perfect Power|perfect powers]].
+\end{theorem}
+
+\begin{proof}
+Each [[Definition:Fermat Number|Fermat number]] is in the form of $1 + 2^n$ for some $n \in \Z$.
+This $n$ must also be a [[Definition:Integer Power|power]] of $2$.
+From [[1 plus Power of 2 is not Perfect Power except 9]] we have:
+:$1 + 2^n = a^b$
+has only one solution $\tuple {n, a, b} = \tuple {3, 3, 2}$.
+But $3$ is not a [[Definition:Integer Power|power]] of $2$.
+Hence no [[Definition:Fermat Number|Fermat numbers]] are [[Definition:Perfect Power|perfect powers]].
+{{qed}}
+[[Category:Fermat Numbers]]
+blocil9c4mkehf14is8l2emcyx5njxl
+\end{proof}<|endoftext|>
+\section{Uncountable Closed Ordinal Space is Countably Compact}
+Tags: Ordinal Spaces, Countably Compact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
+Then $\closedint 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Closed Ordinal Space is Compact]]
+:[[Compact Space is Countably Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Euler Numbers by Binomial Coefficients Vanishes}
+Tags: Euler Numbers, Binomial Coefficients, Sum of Euler Numbers by Binomial Coefficients Vanishes
+
+\begin{theorem}
+$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
+where $E_k$ denotes the $k$th [[Definition:Euler Numbers|Euler number]].
+== Corollary ==
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = E_{2 n}
+      | r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
+      | c = 
+}}
+{{eqn | r = -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }
+      | c = 
+}}
+{{end-eqn}}
+where $E_n$ denotes the $n$th [[Definition:Euler Numbers|Euler number]].
+\end{theorem}
+
+\begin{proof}
+Take the definition of [[Definition:Euler Numbers|Euler numbers]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}
+      | r = \frac {2 e^x} {e^{2 x} + 1}
+      | c = 
+}}
+{{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }
+      | c = Multiply by $1$
+}}
+{{eqn | r = \paren {\frac 2 {e^x + e^{-x} } }
+      | c = 
+}}
+{{end-eqn}}
+From the definition of the [[Definition:Exponential Function/Real/Sum of Series|exponential function]]:
+{{begin-eqn}}
+{{eqn | l = e^x
+      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots
+      | c = 
+}}
+{{eqn | l = e^{-x}
+      | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-x}^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots
+      | c = 
+}}
+{{eqn | l = \paren {\frac {e^x + e^{-x} } 2}
+      | r = \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }
+      | c = 
+}}
+{{eqn | r = 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots
+      | c = [[Definition:Odd Integer|odd]] terms cancel in the [[Definition:Sum|sum]].
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \paren {\frac 2 {e^x + e^{-x} } } \paren {\frac {e^x + e^{-x} } 2}
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }
+      | c = 
+}}
+{{end-eqn}}
+By [[Product of Absolutely Convergent Series]], we will let:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \frac {E_n x^n} {n!}
+      | c = 
+}}
+{{eqn | l = b_n
+      | r = \frac {x^{2 n} } {\paren {2 n}!} 
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
+      | r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}
+      | rr= = 1
+      | c = 
+}}
+{{eqn | l = c_n
+      | r = \sum_{k \mathop = 0}^n a_k b_{n - k}
+      | c = 
+}}
+{{eqn | l = c_0
+      | r = \frac {E_0 x^0} {0!} \frac {x^0 } {0!} 
+      | rr= = 1
+      | c = $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 1}^\infty c_n
+      | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n} - a_0 b_0
+      | rr= = 0
+      | c = [[Definition:Subtraction|Subtract]] $1$ from both sides
+}}
+{{end-eqn}}
+We now have:
+{{begin-eqn}}
+{{eqn | l = c_1
+      | r = \frac {E_0 x^0} {0!} \frac {x^2} {2!} + \frac {E_1 x^1} {1!} \frac {x^0} {0!}
+      | c = $= a_0 b_1 + a_1 b_0$
+}}
+{{eqn | l = c_2
+      | r = \frac {E_0 x^0} {0!} \frac {x^4} {4!} + \frac {E_1 x^1} {1!} \frac {x^2} {2!} + \frac {E_2 x^2} {2!} \frac {x^0} {0!}
+      | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$
+}}
+{{eqn | l = c_3
+      | r = \frac {E_0 x^0} {0!} \frac {x^6} {6!} + \frac {E_1 x^1} {1!} \frac {x^4} {4!} + \frac {E_2 x^2} {2!} \frac {x^2} {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}
+      | c = $= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$
+}}
+{{eqn | l = c_4
+      | r = \frac {E_0 x^0} {0!} \frac {x^8} {8!} + \frac {E_1 x^1} {1!} \frac {x^6} {6!} + \frac {E_2 x^2} {2!} \frac {x^4} {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0} {0!}
+      | c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$
+}}
+{{eqn | o = \cdots
+}}
+{{eqn | l = c_n
+      | r = \frac {E_0 x^0} {0!} \frac {x^{2 n} } {\paren {2 n}!} + \frac {E_1 x^1} {1!} \frac {x^{2 n - 2} } {\paren {2 n - 2}!} + \frac {E_2 x^2} {2!} \frac {x^{2 n - 4} } {\paren {2 n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0} {0!}
+      | c = 
+}}
+{{end-eqn}}
+Grouping terms with [[Definition:Even Integer|even]] [[Definition:Power (Algebra)|exponents]] produces:
+{{begin-eqn}}
+{{eqn | l = \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2
+      | r = 0
+      | c = $x^2$ term from $c_1$ and $c_2$
+}}
+{{eqn | l = \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4
+      | r = 0
+      | c = $x^4$ term from $c_2$, $c_3$ and $c_4$
+}}
+{{eqn | o = \cdots
+}}
+{{eqn | l = \paren {\frac 1 {0! \paren {2 n}!} } E_0 + \paren {\frac 1 {2! \paren {2 n - 2 }!} } E_2 + \paren {\frac 1 {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\frac 1 {\paren {2 n}! 0!} } E_{2 n}
+      | r = 0
+      | c = $x^{2 n}$ term from $c_n$, $c_{n + 1} \cdots c_{2 n}$
+}}
+{{end-eqn}}
+$\forall n \in \Z_{>0}$, multiplying the [[Definition:Coefficient of Polynomial|coefficients]] of $x^{2 n}$ through by $\paren {2 n}!$ gives:
+:$\paren {\dfrac {\paren {2 n}! } {0! \paren {2 n}!} } E_0 + \paren {\dfrac {\paren {2 n}! } {2! \paren {2 n - 2 }!} } E_2 + \paren {\dfrac {\paren {2 n}! } {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\dfrac {\paren {2 n}! } {\paren {2 n}! 0!} } E_{2 n} = 0$
+But those [[Definition:Coefficient of Polynomial|coefficients]] are the [[Definition:Binomial Coefficient|binomial coefficients]]:
+:$\dbinom {2 n} 0 E_0 + \dbinom {2 n} 2 E_2 + \dbinom {2 n} 4 E_4 + \dbinom {2 n} 6 E_6 + \cdots + \dbinom {2 n} {2 n} E_{2 n} = 0$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is Countably Compact}
+Tags: Ordinal Spaces, Countably Compact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
+\end{theorem}
+
+\begin{proof}
+Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
+From [[Uncountable Closed Ordinal Space is Countably Compact]], $\closedint 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
+So every [[Definition:Sequence|sequence]] in $\hointr 0 \Omega$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] in $\closedint 0 \Omega$.
+{{LinkWanted|[[Definition:Sequence|sequence]] in $\hointr 0 \Omega$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] in $\closedint 0 \Omega$}}
+But $\Omega$ cannot be an [[Definition:Accumulation Point of Sequence|accumulation point]] of any [[Definition:Sequence|sequence]] in $\closedint 0 \Omega$.
+{{LinkWanted|$\Omega$ cannot be an [[Definition:Accumulation Point of Sequence|accumulation point]] of any [[Definition:Sequence|sequence]] in $\closedint 0 \Omega$}}
+So every [[Definition:Sequence|sequence]] in $\hointr 0 \Omega$ has an [[Definition:Accumulation Point of Sequence|accumulation point]] in $\hointr 0 \Omega$.
+This means that $\hointr 0 \Omega$ is [[Definition:Countably Compact Space|countably compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is not Metacompact}
+Tags: Ordinal Spaces, Metacompact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is not a [[Definition:Metacompact Space|metacompact space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Metacompact Space|metacompact space]].
+From [[Open Ordinal Space is not Compact in Closed Ordinal Space]] we have that $\hointr 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
+From [[Metacompact Countably Compact Space is Compact]] it follows that $\hointr 0 \Omega$ is a [[Definition:Compact Space|compact space]].
+But from [[Open Ordinal Space is not Compact in Closed Ordinal Space]] this [[Definition:Contradiction|contradicts]] the fact that $\hointr 0 \Omega$ is not a [[Definition:Compact Space|compact space]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is not Paracompact}
+Tags: Ordinal Spaces, Paracompact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is not a [[Definition:Paracompact Space|paracompact space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Paracompact Space|paracompact space]].
+From [[Paracompact Space is Metacompact]], it follows that $\hointr 0 \Omega$ is a [[Definition:Metacompact Space|metacompact space]].
+But from [[Uncountable Open Ordinal Space is not Metacompact]] this [[Definition:Contradiction|contradicts]] the fact that $\hointr 0 \Omega$ is not a [[Definition:Metacompact Space|metacompact space]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is not Lindelöf}
+Tags: Ordinal Spaces, Paracompact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is not a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+From [[Ordinal Space is Completely Normal]], $\hointr 0 \Omega$ is a [[Definition:Completely Normal Space|completely normal]].
+From [[Sequence of Implications of Separation Axioms]], $\hointr 0 \Omega$ is a [[Definition:T3 Space|$T_3$ space]].
+From [[Lindelöf T3 Space is Paracompact|Lindelöf $T_3$ Space is Paracompact]], it follows that $\hointr 0 \Omega$ is a [[Definition:Paracompact Space|paracompact space]].
+But this [[Definition:Contradiction|contradicts]] the fact that from [[Uncountable Open Ordinal Space is not Paracompact]], $\hointr 0 \Omega$ is not a [[Definition:Paracompact Space|paracompact space]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is not Sigma-Compact}
+Tags: Ordinal Spaces, Sigma-Compact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+From [[Sigma-Compact Space is Lindelöf]], $\hointr 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+But this [[Definition:Contradiction|contradicts]] the fact that from [[Uncountable Open Ordinal Space is not Lindelöf]], $\hointr 0 \Omega$ is not a [[Definition:Lindelöf Space|Lindelöf space]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Closed Ordinal Space is Lindelöf}
+Tags: Ordinal Spaces, Lindelöf Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
+Then $\closedint 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Closed Ordinal Space is Compact]]
+:[[Compact Space is Lindelöf]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Closed Ordinal Space is Sigma-Compact}
+Tags: Ordinal Spaces, Sigma-Compact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
+Then $\closedint 0 \Omega$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:[[Closed Ordinal Space is Compact]]
+:[[Compact Space is Sigma-Compact|Compact Space is $\sigma$-Compact]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Open Ordinal Space is Sequentially Compact}
+Tags: Ordinal Spaces, Sequentially Compact Spaces
+
+\begin{theorem}
+Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
+Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
+Then $\hointr 0 \Omega$ is a [[Definition:Sequentially Compact Space|sequentially compact space]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Uncountable Open Ordinal Space is First-Countable]]
+:[[Uncountable Open Ordinal Space is Countably Compact]]
+The result follows from [[First-Countable Space is Sequentially Compact iff Countably Compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Euler Numbers by Binomial Coefficients Vanishes/Corollary}
+Tags: Euler Numbers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = E_{2 n}
+      | r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
+      | c = 
+}}
+{{eqn | r = -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }
+      | c = 
+}}
+{{end-eqn}}
+where $E_n$ denotes the $n$th [[Definition:Euler Numbers|Euler number]].
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Euler Numbers by Binomial Coefficients Vanishes]] we have:
+$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
+If:
+:$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
+then:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}
+      | r = 0
+}}
+{{eqn | l = E_{2 n}
+      | r = - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Euler Numbers]]
+2qn8jp12mmcxnp5wqbpnk8phl58hd20
+\end{proof}<|endoftext|>
+\section{Integer to Power of Multiple of Order/Corollary}
+Tags: Integer to Power of Multiple of Order, Number Theory
+
+\begin{theorem}
+Then $\map \phi n$ is a [[Definition:Multiple of Integer|multiple]] of $c$, where $\map \phi n$ is the [[Definition:Euler Phi Function|Euler phi function]] of $n$.
+\end{theorem}
+
+\begin{proof}
+From [[Euler's Theorem]], we have $a^{\map \phi n} \equiv 1 \pmod n$.
+Applying [[Integer to Power of Multiple of Order]] we see that $\map \phi n$ is a [[Definition:Multiple of Integer|multiple]] of $c$.
+{{qed}}
+[[Category:Integer to Power of Multiple of Order]]
+[[Category:Number Theory]]
+kwzul7e0s2q95w09jdl9y42jzr6ycl5
+\end{proof}<|endoftext|>
+\section{Divisor of Fermat Number/Euler's Result}
+Tags: Divisor of Fermat Number
+
+\begin{theorem}
+Then $m$ is in the form:
+:$k \, 2^{n + 1} + 1$
+where $k \in \Z_{>0}$ is an [[Definition:Integer|integer]].
+\end{theorem}
+
+\begin{proof}
+It is sufficient to prove the result for [[Definition:Prime Divisor|prime divisors]].
+The general argument for all [[Definition:Divisor of Integer|divisors]] follows from the argument:
+:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
+So the product of two [[Definition:Divisor of Integer|factors]] of the form preserves that form.
+Let $n \in \N$.
+Let $p$ be a [[Definition:Prime Divisor|prime divisor]] of $F_n = 2^{2^n} + 1$.
+Then we have:
+:$2^{2^n} \equiv -1 \pmod p$
+Squaring both sides:
+:$2^{2^{n + 1}} \equiv 1 \pmod p$
+From [[Integer to Power of Multiple of Order]], the [[Definition:Multiplicative Order of Integer|order]] of $2$ modulo $p$ divides $2^{n + 1}$ but not $2^n$.
+Therefore it must be $2^{n + 1}$.
+Hence:
+{{begin-eqn}}
+{{eqn | ll = \exists k \in \Z:
+      | l = \map \phi p
+      | r = k \, 2^{n + 1}
+      | c = [[Integer to Power of Multiple of Order/Corollary|Corollary to Integer to Power of Multiple of Order]]
+}}
+{{eqn | l = p - 1
+      | r = k \, 2^{n + 1}
+      | c = [[Euler Phi Function of Prime]]
+}}
+{{eqn | l = p
+      | r = k \, 2^{n + 1} + 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisor of Fermat Number/Refinement by Lucas}
+Tags: Divisor of Fermat Number
+
+\begin{theorem}
+Let $n \ge 2$.
+Then $m$ is in the form:
+:$k \, 2^{n + 2} + 1$
+\end{theorem}
+
+\begin{proof}
+It is sufficient to prove the result for [[Definition:Prime Divisor|prime divisors]].
+The general argument for all [[Definition:Divisor of Integer|divisors]] follows from the argument:
+:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
+So the product of two [[Definition:Divisor of Integer|factors]] of the form preserves that form.
+Let $p$ be a [[Definition:Prime Divisor|prime divisor]] of $F_n = 2^{2^n} + 1$.
+From [[Divisor of Fermat Number/Euler's Result|Euler's Result]]:
+:$\exists q \in \Z: p = q \, 2^{n + 1} + 1$
+Since $n \ge 2$, $q \, 2^{n + 1}$ is [[Definition:Divisor of Integer|divisible]] by $2^{2 + 1} = 8$.
+Hence:
+:$p \equiv 1 \pmod 8$
+By [[Second Supplement to Law of Quadratic Reciprocity]]:
+:$\paren {\dfrac 2 p} = 1$
+so $2$ is a [[Definition:Quadratic Residue|quadratic residue]] modulo $p$.
+Hence:
+:$\exists x \in \Z: x^2 = 2 \pmod p$
+We have shown $2^{2^n} \equiv -1 \pmod p$ and $2^{2^{n + 1} } \equiv 1 \pmod p$.
+By [[Congruence of Powers]]:
+:$x^{2^{n + 1} } \equiv 2^{2^n} \equiv -1 \pmod p$
+:$x^{2^{n + 2} } \equiv 2^{2^{n + 1}} \equiv 1 \pmod p$
+From [[Integer to Power of Multiple of Order]], the [[Definition:Multiplicative Order of Integer|order]] of $x$ modulo $p$ divides $2^{n + 2}$ but not $2^{n + 1}$.
+Therefore it must be $2^{n + 2}$.
+Hence:
+{{begin-eqn}}
+{{eqn | ll= \exists k \in \Z:
+      | l = \map \phi p
+      | r = k \, 2^{n + 2}
+      | c = [[Integer to Power of Multiple of Order/Corollary|Corollary to Integer to Power of Multiple of Order]]
+}}
+{{eqn | l = p - 1
+      | r = k \, 2^{n + 2}
+      | c = [[Euler Phi Function of Prime]]
+}}
+{{eqn | l = p
+      | r = k \, 2^{n + 2} + 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Double Pointed Real Number Line}
+Tags: Definitions: Examples of Topologies, Definitions: Double Pointed Topologies, Definitions: Real Number Line with Euclidean Topology
+
+\begin{theorem}
+Let $T_\R = \struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $T_D = \struct {D, \tau_D}$ be the [[Definition:Indiscrete Topology|indiscrete topology]] on the [[Definition:Doubleton|doubleton]] $D = \set {a, b}$.
+Let $T = T_\R \times T_D$ be the[[Definition:Product Space (Topology)|product space]] of $T_\R$ and $T_D$.
+$T$ is known as the '''[[Definition:Double Pointed Topology|double pointed]] [[Definition:Real Number Line with Euclidean Topology|real number line]]'''.
+\end{theorem}<|endoftext|>
+\section{Set Closure is Smallest Closed Set/Normed Vector Space}
+Tags: Closed Sets, Normed Vector Spaces, Set Closure is Smallest Closed Set
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $S$ be a [[Definition:Subset|subset]] of $X$:
+:$S \subseteq X$
+Let $S^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $S$.
+Then $S^-$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Closed Set in Normed Vector Space|closed set]] which contains $S$.
+\end{theorem}
+
+\begin{proof}
+Let $F$ be a [[Definition:Closed Set in Normed Vector Space|closed set]] in $X$.
+Suppose $S \subseteq F$.
+=== $S^-$ is contained in $F$===
+Let $L$ be a [[Definition:Limit Point/Normed Vector Space|limit point]] of $S$.
+Then there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}_{n \mathop \in \N}$ in $S \setminus \set L$ which [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $L$.
+In other words:
+:$\forall n \in \N : x_n \in S \setminus \set L$.
+Furthermore:
+:$S \setminus \set L \subseteq S \subseteq F$
+Since $F$ is [[Definition:Closed Set in Normed Vector Space|closed]], $L \in F$.
+So all [[Definition:Limit Point/Normed Vector Space|limit points]] of $S$ belong to $F$.
+Hence, $S^- \subseteq F$.
+{{qed|lemma}}
+=== $S^-$ is closed ===
+Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $S^-$.
+Let $\sequence {x_n}_{n \mathop \in \N}$ [[Definition:Convergent Sequence in Normed Vector Space|converge]] to $L$:
+:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {x_n - L} < \epsilon$
+We can have either $L \in S$ or $L \notin S$.
+Suppose $L \in S$.
+:By [[Definition:Closure/Normed Vector Space|definition]], $L \in S^-$.
+Suppose $L \notin S$.
+Define a new [[Definition:Sequence|sequence]] $x_n'$ using $x_n$ as follows:
+:$(1): \quad$ if $x_n \in S$, then $x_n' := x_n$;
+:$(2): \quad$ if $x_n \notin S$, then $x_n$ is a [[Definition:Limit Point/Normed Vector Space/Set|limit point]] of $S$.
+::Then there is [[Definition:Open Ball in Normed Vector Space|open ball]] $\ds \map {B_{\frac 1 n}} {x_n}$ which has an element of $S$.
+::Define $x_n'$ such that $x_n' \in S$ and $x_n' \in \map {B_{\frac 1 n}} {x_n}$
+Suppose $x_n \in S$.
+Then:
+:$\norm {x_n' - L} = \norm {x_n - L}$
+Suppose $x_n \notin S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n' - L}
+      | r = \norm {x_n' - x_n + x_n - L}
+}}
+{{eqn | o = \le
+      | r = \norm {x_n' - x_n} + \norm {x_n - L}
+      | c = [[Definition:Norm|Norm axiom]]: triangle inequality
+}}
+{{eqn | o = <
+      | r = \frac 1 n + \norm {x_n - L}
+}}
+{{eqn | o = <
+      | r = \frac 1 n + \epsilon
+}}
+{{end-eqn}}
+Thus, $\sequence {x_n'}_{n \mathop \in \N}$ is a sequence in $S \setminus \set L$ which [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $L$.
+So $L$ is a [[Definition:Limit Point/Normed Vector Space|limit point]] of $S$:
+:$L \in S^-$.
+By [[Definition:Closed Set/Normed Vector Space/Definition 2|definition]], $S^-$ is [[Definition:Closed Set in Normed Vector Space|closed]].
+{{qed|lemma}}
+=== $S^-$ is the smallest closed set containing $S$ ===
+{{AimForCont}} there exists a [[Definition:Closed Set in Normed Vector Space|closed set]] $Q$ [[Definition:Smaller Set|smaller]] than $S^-$ which [[Definition:Contain|contains]] $S$.
+$S^-$ differs from $S$ only by [[Definition:Limit Point/Normed Vector Space|limit points]] of $S$.
+If $Q$ is [[Definition:Smaller Set|smaller]] than $S^-$, it has to [[Definition:Contain|contain]] fewer [[Definition:Limit Point/Normed Vector Space|limit points]] of $S$ than $S^-$.
+Hence, $Q$ would not [[Definition:Contain|contain]] all its [[Definition:Limit Point/Normed Vector Space|limit points]].
+By definition, $Q$ would not be [[Definition:Closed Set/Normed Vector Space/Definition 2|closed]].
+This is a [[Definition:Contradiction|contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounded Real Function may not be of Bounded Variation}
+Tags: Bounded Variation
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
+Let $f : \closedint a b \to \R$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Real Function|function]].
+Then $f$ is not necessarily of [[Definition:Bounded Variation|bounded variation]].
+\end{theorem}
+
+\begin{proof}
+Let $a = 0$, $b = 1$.
+Define $f : \closedint 0 1 \to \R$ by: 
+:$\map f x = \begin{cases}1 & x \in \Q \\ 0 & x \not \in \Q\end{cases}$
+For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint 0 1$, write: 
+:$P = \set {x_0, x_1, \ldots, x_n }$
+with:
+:$0 = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = 1$
+For each such subdivision also write:
+:$\displaystyle \map {V_f} P = \sum_{i \mathop = 1}^n \size {\map f {x_i} - \map f {x_{i - 1} } }$
+It suffices to find a sequence of [[Definition:Finite Subdivision|finite subdivisions]] $\sequence {P_n}$ such that: 
+:$\displaystyle \lim_{n \mathop \to \infty} \map {V_f} {P_n} = \infty$
+by the definition of [[Definition:Bounded Variation|bounded variation]].
+For each $n \in \N$, let $P_n$ be a [[Definition:Finite Subdivision|finite subdivision]] of size $n + 2$.
+Let $\sequence {x_n}_{0 \mathop \le i \mathop \le n + 2}$ be the [[Definition:Real Sequence|sequence]] forming this subdivision.
+Note that from [[Between two Rational Numbers exists Irrational Number]]:
+:it is possible to select an [[Definition:Irrational Number|irrational number]] [[Definition:Strictly Between|strictly between]] two [[Definition:Rational Number|rational numbers]].
+Similarly from [[Between two Real Numbers exists Rational Number]]:
+:it is possible to select a [[Definition:Rational Number|rational number]] [[Definition:Strictly Between|strictly between]] two [[Definition:Irrational Number|irrational numbers]].
+We can therefore define $x_i$ for $1 \le i \le n + 1$ as follows:
+:if $i$ is odd, let $x_i$ be an [[Definition:Irrational Number|irrational number]] between $x_{i - 1}$ and $b$
+:if $i$ is even, let $x_i$ be a [[Definition:Rational Number|rational number]] between $x_{i - 1}$ and $b$.
+Let $x_0 = 0$ and $x_{n + 2} = 1$ to complete the definition.
+Then:
+{{begin-eqn}}
+{{eqn	| l = \map {V_f} {P_n}
+	| r = \sum_{i \mathop = 1}^{n + 2} \size {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{eqn	| o = \ge
+	| r = \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } }
+}}
+{{end-eqn}}
+We omit the term $i = n + 2$ for convenience.
+Note that:
+:if $x_i$ is [[Definition:Rational Number|rational]] then $x_{i + 1}$ is [[Definition:Irrational Number|irrational]].
+That is:
+:if $\map f {x_i} = 1$ then $\map f {x_{i + 1} } = 0$.
+Similarly:
+:if $x_i$ is [[Definition:Irrational Number|irrational]] then $x_{i + 1}$ is [[Definition:Rational Number|rational]].
+That is:
+:if $\map f {x_i} = 0$ then $\map f {x_{i + 1} } = 1$.
+So for all $1 \le i \le n + 1$, we have:
+:$\size {\map f {x_i} - \map f {x_{i - 1} } } = 1$
+Hence:
+:$\displaystyle \sum_{i \mathop = 1}^{n + 1} \size {\map f {x_i} - \map f {x_{i - 1} } } = n + 1$
+So:
+:$\displaystyle \map {V_f} {P_n} \ge n + 1$
+Giving:
+:$\displaystyle \lim_{n \mathop \to \infty} \map {V_f} {P_n} = \infty$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element is Loop iff Member of Closure of Empty Set}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x \in S$.
+Then:
+:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $x \in \map \sigma \O$
+where $\map \sigma \O$ denotes the [[Definition:Closure Operator (Matroid)|closure]] of the [[Definition:Empty Set|empty set]].
+\end{theorem}
+
+\begin{proof}
+From [[Element is Loop iff Rank is Zero]]:
+:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\map \rho {\set x} = 0$
+where $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Now:
+{{begin-eqn}}
+{{eqn | r = x \in \map \sigma \O
+      | o =
+}}
+{{eqn | ll= \leadstoandfrom
+      | r = x \sim \O
+      | o =
+      | c = {{Defof|Closure Operator (Matroid)|Closure Operator}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | r = \map \rho {\set x} = \map \rho \O
+      | o =
+      | c = {{Defof|Depends Relation (Matroid)|Depends Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | r = \map \rho {\set x} = 0
+      | o =
+      | c = [[Rank of Empty Set is Zero]]
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton is Dependent implies Rank is Zero/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\map \rho {\set x} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Loop (Matroid)|loop]]:
+:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\set x \notin \mathscr I$
+From [[Singleton is Dependent implies Rank is Zero]]:
+:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
+From [[Singleton is Independent implies Rank is One]]:
+:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
+It follows that:
+:$\set x \notin \mathscr I$ {{iff}} $\map \rho {\set x} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Dependent Set is Dependent/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $A \subseteq S$.
+Let $x \in A$.
+If $x$ is a [[Definition:Loop (Matroid)|loop]] then $A$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+\end{theorem}
+
+\begin{proof}
+Let $x$ be a [[Definition:Loop (Matroid)|loop]].
+By definition of a [[Definition:Loop (Matroid)|loop]]:
+:$\set x \notin \mathscr I$
+By definition of a [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$\set x$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]]
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq A$
+From [[Superset of Dependent Set is Dependent]]:
+:$A$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Subset contains Loop}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x$ be a [[Definition:Loop (Matroid)|loop]] of $M$.
+Let $A \subseteq S$.
+Then:
+:$x \in \map \sigma A$
+where $\map \sigma A$ denotes the [[Definition:Closure Operator (Matroid)|closure]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Closure Operator (Matroid)|closure]] of $A$:
+:$x \in \map \sigma A$ {{iff}} $x \sim A$
+where $\sim$ is the [[Definition:Depends Relation (Matroid)|depends relation]] on $M$.
+By definition of the [[Definition:Depends Relation (Matroid)|depends relation]]:
+:$x \sim A$ {{iff}} $\map \rho {A \cup \set x} = \map \rho A$
+where $\rho$ is the [[Definition:Rank Function (Matroid)|rank funtion]] on $M$.
+So it remains to show that:
+:$\map \rho {A \cup \set x} = \map \rho A$
+By definition of the [[Definition:Rank Function (Matroid)|rank function]]:
+:$\map \rho {A \cup \set x} = \max \set {\size X : X \subseteq A \cup \set x \land X \in \mathscr I}$
+From [[Max Equals an Operand]]:
+:$\exists X \in \mathscr I : X \subseteq A \cup \set x \land \size X = \map \rho {A \cup \set x}$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Set is Dependent if Contains Loop]]:
+:$x \notin X$
+Now:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = \paren{A \cup \set x} \cap X
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \paren{A \cap X} \cup \paren {\set x \cap X}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \paren{A \cap X} \cup \O
+      | c = [[Intersection With Singleton is Disjoint if Not Element]]
+}}
+{{eqn | r = \paren{A \cap X}
+      | c = [[Union with Empty Set]]
+}}
+{{eqn | ll = \leadsto
+      | l = X
+      | o = \subseteq
+      | r = A
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{end-eqn}}
+From [[Max yields Supremum of Operands]]:
+:$\size X \le \max \set {\size Y : Y \subseteq A \land Y \in \mathscr I} = \map \rho A$
+From [[Rank Function is Increasing]]:
+:$\map \rho A \le \map \rho {A \cup \set x} = \size X$
+Thus:
+:$\map \rho A = \size X = \map \rho {A \cup \set x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element is Loop iff Singleton is Circuit}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x \in S$.
+Then:
+:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\set x$ is a [[Definition:Circuit (Matroid)|circuit]]
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x$ be a [[Definition:Loop (Matroid)|loop]].
+By definition of a [[Definition:Loop (Matroid)|loop]]:
+:$\set x$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $S$
+Let $A \subseteq \set x$ be a [[Definition:Dependent Subset (Matroid)|dependent subset]].
+From [[Power Set of Singleton]]:
+:$\powerset {\set x} = \set{\O, \set x}$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]]:
+:$\O$ is an [[Definition:Independent Subset (Matroid)|independent subset]]
+Then:
+:$A = \set x$
+It follows that:
+:$\set x$ is a [[Definition:Minimal|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]] of $S$.
+Then $\set x$ be a [[Definition:Circuit (Matroid)|circuit]] by definition. 
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\set x$ be a [[Definition:Circuit (Matroid)|circuit]].
+By definition of a [[Definition:Circuit (Matroid)|circuit]]:
+:$\set x$ is a [[Definition:Minimal|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]] of $S$.
+In particular, $\set x$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]] of $S$.
+Then $\set x$ is a [[Definition:Loop (Matroid)|loop]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element is Member of Base iff Not Loop}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\mathscr B$ denote the set of all [[Definition:Base of Matroid|bases]] of $M$.
+Let $x \in S$.
+Then:
+:$\exists B \in \mathscr B: x \in B$ {{iff}} $x$ is not a [[Definition:Loop (Matroid)|loop]]
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $B \in \mathscr B$ such that $x \in B$.
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq B$
+By definition of a [[Definition:Base of Matroid|base]]:
+:$B \in \mathscr I$
+From [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$\set x \in \mathscr I$
+Then $\set x$ is not a [[Definition:Dependent Subset (Matroid)|dependent subset]] by definition.
+It follows that $x$ is not a [[Definition:Loop|loop]] by definition.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x$ not be a [[Definition:Loop|loop]].
+By definition of a [[Definition:Loop|loop]]:
+:$x$ is not a [[Definition:Dependent Subset (Matroid)|dependent subset]]
+By definition of a [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$x \in \mathscr I$
+From [[Independent Subset is Contained in Base]]:
+:$\exists B \in \mathscr B: \set x \subseteq B$
+By definition of a [[Definition:Subset|subset]]:
+:$x \in B$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distinct Elements are Parallel iff Pair forms Circuit}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x, y \in S : x \ne y$.
+Then:
+:$x$ and $y$ are [[Definition:Parallel (Matroid)|parallel]] {{iff}} $\set {x, y}$ is a [[Definition:Circuit (Matroid)|circuit]]
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x$ and $y$ be [[Definition:Parallel (Matroid)|parallel]].
+By definition of [[Definition:Parallel (Matroid)|parallel]]:
+:$\set x$ is [[Definition:Independent Subset (Matroid)|independent]] 
+:$\set y$ is [[Definition:Independent Subset (Matroid)|independent]] 
+:$\set {x, y}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+Let $A \subseteq \set {x, y}$ be [[Definition:Dependent Subset (Matroid)|dependent]].
+Thus:
+:$A \ne \set x, \set y$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]]:
+:$\O$ is [[Definition:Independent Subset (Matroid)|independent]]
+Thus:
+:$A \ne \O$
+From [[Power Set of Doubleton]]:
+:$\powerset {\set {x, y} } = \set {\O, \set x, \set y, \set {x, y} }$
+Thus:
+:$A = \set {x, y}$
+By definition of a [[Definition:Minimal Set|minimal set]]:
+:$\set {x, y}$ is a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]
+It follows that $\set {x, y}$ is a [[Definition:Circuit (Matroid)|circuit]] by definition.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\set {x, y}$ by a [[Definition:Circuit (Matroid)|circuit]].
+By definition of a [[Definition:Circuit (Matroid)|circuit]]:
+:$\set {x, y}$ is a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]
+By definition of a [[Definition:Subset|subset]]:
+:$\set x, \set y \subseteq \set {x, y}$
+By definition of a [[Definition:Minimal Set|minimal]] [[Definition:Dependent Subset (Matroid)|dependent subset]]: 
+:$\set x, \set y$ are [[Definition:Independent Subset (Matroid)|independent subsets]]
+It follows that $x$ and $y$ are [[Definition:Parallel (Matroid)|parallel]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallel Relationship is Transitive}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x, y, z \in S : x \ne y, x \ne z, y \ne z$.
+If $x$ is [[Definition:Parallel (Matroid)|parallel]] to $y$ and $y$ is [[Definition:Parallel (Matroid)|parallel]] to $z$ then $x$ is [[Definition:Parallel (Matroid)|parallel]] to $z$.
+\end{theorem}
+
+\begin{proof}
+Let $x$ be [[Definition:Parallel (Matroid)|parallel]] to $y$ and $y$ be [[Definition:Parallel (Matroid)|parallel]] to $z$.
+By definition of [[Definition:Parallel (Matroid)|parallel]]:
+:$\set x$, $\set y$, $\set z$ are [[Definition:Independent Subset (Matroid)|independent subsets]]
+:$\set {x, y}$, $\set {y, z}$ are [[Definition:Dependent Subset (Matroid)|dependent subsets]]
+To show that $x$ is [[Definition:Parallel (Matroid)|parallel]] to $z$ it remains to show that:
+:$\set {x, z}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+{{AimForCont}} $\set {x, z}$ is [[Definition:Independent Subset (Matroid)|independent]].
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 3)$]]:
+:$\exists w \in \set{x, z} \setminus \set y : \set{w, y} \in \mathscr I$
+By definition of the [[Definition:Doubleton|doubleton]]:
+:$\set{x, y} \in \mathscr I \lor \set{z, y} \in \mathscr I$
+This [[Definition:Contradiction|contradicts]] the assumption that $\set {x, y}$, $\set {y, z}$ are [[Definition:Dependent Subset (Matroid)|dependent subsets]].
+It follows that:
+:$\set {x, z}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distinct Matroid Elements are Parallel iff Each is in Closure of Other}
+Tags: Matroid Theory, Distinct Matroid Elements are Parallel iff Each is in Closure of Other
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\sigma: \powerset S \to \powerset S$ denote the [[Definition:Closure Operator (Matroid)|closure operator]] of $M$.
+Let $x, y \in S : x \ne y$.
+Then $x$ is [[Definition:Parallel (Matroid)|parallel]] to $y$ {{iff}}:
+:$(1)\quad x$ and $y$ are not [[Definition:Loop (Matroid)|loops]]
+:$(2)\quad x \in \map \sigma {\set y}$
+:$(3)\quad y \in \map \sigma {\set x}$
+\end{theorem}
+
+\begin{proof}
+=== [[Distinct Matroid Elements are Parallel iff Each is in Closure of Other/Lemma|Lemma]] ===
+{{:Distinct Matroid Elements are Parallel iff Each is in Closure of Other/Lemma}}{{qed|lemma}}
+=== Necessary Condition ===
+Let $x$ and $y$ be [[Definition:Parallel (Matroid)|parallel]].
+By definition of [[Definition:Parallel (Matroid)|parallel]]:
+:$\set x$ and $\set y$ are [[Definition:Independent Subset (Matroid)|independent]] 
+:$\set {x, y}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+By definition of a [[Definition:Loop|loop]]:
+:$x$ and $y$ are not [[Definition:Loop (Matroid)|loops]] 
+From Lemma:
+:$x \in \map \sigma {\set y}$
+:$y \in \map \sigma {\set x}$
+It has been shown that [[Definition:Condition|conditions]] $(1), (2)$ and $(3)$ above hold.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let [[Definition:Condition|conditions]] $(1), (2)$ and $(3)$ above hold.
+By definition of a [[Definition:Loop|loop]]:
+:$\set x$ and $\set y$ are [[Definition:Independent Subset (Matroid)|independent]] 
+From Lemma:
+:$\set {y, x} \notin \mathscr I$
+It follows that $x$ is [[Definition:Parallel (Matroid)|parallel]] to $y$ by definition. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Subset Contains Parallel Elements}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\sigma: \powerset S \to \powerset S$ denote the [[Definition:Closure Operator (Matroid)|closure operator]] of $M$.
+Let $A \subseteq S$.
+Let $x, y \in S$.
+If $x \in \map \sigma A$ and $y$ is [[Definition:Parallel (Matroid)|parallel]] to $x$ then:
+:$y \in \map \sigma A$
+\end{theorem}
+
+\begin{proof}
+Let $\rho: \powerset S \to \Z$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $x \in \map \sigma A$
+Let $y$ be [[Definition:Parallel (Matroid)|parallel]] to $x$.
+By the definitions of the [[Definition:Closure Operator (Matroid)|closure operator]] and [[Definition:Depends Relation (Matroid)|depends depends]]:
+:$\map \rho A = \map \rho {A \cup \set x}$
+and:
+:$y \in \map \sigma A$ {{iff}} $\map \rho A = \map \rho {A \cup \set y}$
+From [[Rank Function is Increasing]]:
+:$\map \rho {A \cup \set x} = \map \rho A \le \map \rho {A \cup \set y} \le \map \rho {A \cup \set {x,y}}$
+To show $\map \rho A = \map \rho {A \cup \set y}$ it is sufficient to show:
+:$\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$
+From [[Rank Function is Increasing]]:
+:$\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$
+By the definition of [[Definition:Parallel (Matroid)|parallel]]:
+:$\set x$ is [[Definition:Independent Subset (Matroid)|independent]]
+From [[Independent Subset is Contained in Maximal Independent Subset]]:
+:$\exists X \in \mathscr I : \set x \subseteq X \subseteq A \cup \set x : \size X = \map \rho {A \cup \set x}$
+From [[Subset Relation is Transitive]]:
+:$X \subseteq A \cup \set {x, y}$
+From [[Independent Subset is Contained in Maximal Independent Subset]]:
+:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A \cup \set {x,y} : \size Y = \map \rho {A \cup \set {x,y}}$
+By definition of a [[Definition:Subset|subset]]:
+:$x \in Y$
+By definition of [[Definition:Parallel (Matroid)|parallel]]:
+:$\set{x, y}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+From [[Superset of Dependent Set is Dependent]]:
+:$\set{x, y} \not \subseteq Y$
+Then:
+:$y \notin Y$
+So:
+:$Y \subseteq A \cup \set x$
+By definition of the [[Definition:Rank Function (Matroid)|rank function]]:
+:$\size Y \le \map \rho {A \cup \set x}$
+As $\size Y = \map \rho {A \cup \set {x,y}}$ then:
+:$\map \rho {A \cup \set {x,y}} \le \map \rho {A \cup \set x}$
+As $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$ then:
+:$\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set with Two Parallel Elements is Dependent}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $x, y \in S$.
+Let $x, y$ be [[Definition:Parallel (Matroid)|parallel]] [[Definition:Element|elements]].
+If $x, y \in A$ then $A$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in A$.
+ 
+From [[Doubleton of Elements is Subset]]:
+:$\set{x, y} \subseteq A$
+By the definition of [[Definition:Parallel (Matroid)|parallel]] [[Definition:Element|elements]]:
+:$\set {x, y}$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+From [[Superset of Dependent Set is Dependent]]:
+:$A$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Loop Belongs to Every Flat}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $x \in S$.
+If $x$ is a [[Definition:Loop (Matroid)|loop]] and $A$ is a [[Definition:Flat (Matroid)|flat subset]] then $x \in A$.
+\end{theorem}
+
+\begin{proof}
+Let $\rho: \powerset S \to \Z$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+We proceed by [[Proof by Contraposition]].
+That is, it is shown that:
+:if $x \notin A$ then either $x$ is not a [[Definition:Loop (Matroid)|loop]] or $A$ is not a [[Definition:Flat (Matroid)|flat subset]]
+Let $x \notin A$.
+Let $x$ be a [[Definition:Loop (Matroid)|loop]].
+By definition of a [[Definition:Loop (Matroid)|loop]]:
+:$\set x$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]].
+From [[Rank Function is Increasing]]:
+:$\map \rho A \le \map \rho {A \cup \set x}$
+Let $X \in \mathscr I$ such that $X \subseteq A \cup \set x$.
+From [[Superset of Dependent Set is Dependent]]:
+:$\set x \not \subseteq X$
+From [[Singleton of Element is Subset]]:
+:$x \notin X$
+So:
+:$X \subseteq A$
+By definition of the [[Definition:Rank Function (Matroid)|rank function]]:
+:$\size X \le \map \rho A$
+From [[Max yields Supremum of Parameters]]:
+:$\map \rho {A \cup \set x} = \max \set{\size X : X \in \mathscr I \land X \subseteq A \cup \set x} \le \map \rho A$
+Then:
+:$\map \rho A = \map \rho {A \cup \set x}$
+If follows that $A$ is not a [[Definition:Flat (Matroid)|flat subset]] by definition.
+It has been shown that:
+:if $x \notin A$ then either $x$ is not a [[Definition:Loop (Matroid)|loop]] or $A$ is not a [[Definition:Flat (Matroid)|flat subset]]
+The theorem holds by the [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parallel Elements Depend on Same Subsets}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $x, y \in S$.
+Let $x$ be [[Definition:Parallel (Matroid)|parallel]] to $y$.
+Then:
+:$x$ [[Definition:Depends Relation (Matroid)|depends]] on $A$ {{iff}} $y$ [[Definition:Depends Relation (Matroid)|depends]] on $A$
+\end{theorem}
+
+\begin{proof}
+This follows directly from [[Closure of Subset Contains Parallel Elements]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matroid Contains No Loops iff Empty Set is Flat}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Then:
+:$M$ contains no [[Definition:Loop (Matroid)|loops]] {{iff}} the [[Definition:Empty Set|empty set]] is [[Definition:Flat (Matroid)|flat]].
+\end{theorem}
+
+\begin{proof}
+Let $\rho: \powerset S \to \Z$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+=== Necessary Condition ===
+Let $M$ contain no [[Definition:Loop (Matroid)|loops]].
+By definition of a [[Definition:Loop (Matroid)|loop]]:
+:$\forall x \in S : \set x \in \mathscr I$
+Let $x \in S \setminus \O$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \rho {\O \cup \set x}
+      | r = \map \rho {\set x}
+      | c = [[Union with Empty Set]]
+}}
+{{eqn | r = \size {\set x}
+      | c = [[Rank of Independent Subset Equals Cardinality]]
+}}
+{{eqn | r = 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | r = 0 + 1
+}}
+{{eqn | r = \map \rho \O + 1
+      | c = [[Rank of Empty Set is Zero]]
+}}
+{{end-eqn}}
+It follows that $\O$ is a [[Definition:Flat (Matroid)|flat subset]] by definition.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\O$ be a [[Definition:Flat (Matroid)|flat subset]].
+Let $x \in S$.
+From [[Set Difference with Empty Set is Self]]
+:$x \in S \setminus \O$
+{{begin-eqn}}
+{{eqn | l = \map \rho {\set x}
+      | r = \map \rho {\O \cup \set x}
+      | c = [[Union with Empty Set]]
+}}
+{{eqn | r = \map \rho \O + 1
+      | c = {{Defof|Flat (Matroid)|Flat Subset}}
+}}
+{{eqn | r = 0 + 1
+      | c = [[Rank of Empty Set is Zero]]
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+From [[Element is Loop iff Rank is Zero]]:
+:$x$ is not a [[Definition:Loop (Matroid)|loop]]
+It follows that $M$ contains no [[Definition:Loop (Matroid)|loops]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Continuous Real Function is Uniformly Continuous}
+Tags: Uniformly Continuous Real Functions, Absolutely Continuous Functions
+
+\begin{theorem}
+Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
+Let $f : I \to \R$ be an [[Definition:Absolute Continuity|absolutely continuous]] [[Definition:Real Function|function]].
+Then $f$ is [[Definition:Uniform Continuity/Real Numbers|uniformly continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
+Since $f$ is [[Definition:Absolute Continuity|absolutely continuous]], there exists [[Definition:Real Number|real]] $\delta > 0$ such that for all collections of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
+we have: 
+:$\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
+Consider specifically the case $n = 1$.
+From the absolute continuity of $f$, we have that whenever $a \le x \le y \le b$ and:
+:$y - x < \delta$
+we have:
+:$\size {\map f x - \map f y} < \epsilon$
+Notice however that: 
+:$\size {\map f x - \map f y} = \size {\map f y - \map f x}$
+We therefore have:
+:$\size {\map f x - \map f y} < \epsilon$
+when:
+:$x - y < \delta$
+as well.
+So, in fact, for all $x, y \in \closedint a b$ with:
+:$\size {x - y} < \delta$
+we have:
+:$\size {\map f x - \map f y} < \epsilon$
+Since $\epsilon$ was arbitrary:
+:$f$ is [[Definition:Uniform Continuity/Real Numbers|uniformly continuous]].
+{{qed}}
+[[Category:Uniformly Continuous Real Functions]]
+[[Category:Absolutely Continuous Functions]]
+g3gmkhld8qrtscjl505md6oi47b9sqy
+\end{proof}<|endoftext|>
+\section{Closure of Subspace of Normed Vector Space is Subspace}
+Tags: Set Closures
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $Y \subseteq X$ be a [[Definition:Vector Subspace|subspace]] of $X$.
+Let $Y^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $Y$.
+Then $Y^- \subseteq X$ is also a [[Definition:Vector Subspace|subspace]] of $X$.
+\end{theorem}
+
+\begin{proof}
+Suppose $y \in Y^-$.
+Then there is a [[Definition:Sequence|sequence]] $\displaystyle \sequence {y_n}_{n \mathop \in \N} \in Y$ which [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $y$.
+Suppose $y \in Y$ and $y$ is a [[Definition:Limit Point (Normed Vector Space)|limit point]].
+Then one can define a [[Definition:Constant|constant]] [[Definition:Sequence|sequence]] $\sequence {y_n}_{n \mathop \in \N} = y$.
+=== Closed under restriction of vector addition ===
+Let $x, y \in Y^-$.
+Let $\sequence {x_n}_{n \mathop \in \N}, \sequence {y_n}_{n \mathop \in \N}$ be [[Definition:Sequence|sequences]] in $Y$.
+Suppose:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = x$
+:$\displaystyle \lim_{n \mathop \to \infty} y_n = y$
+Since $Y$ is a [[Definition:Vector Subspace|subspace]]:
+:$\forall n \in \N : x_n + y_n \in Y \subseteq Y^-$
+Furthermore:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {x_n + y_n} = x + y$
+$Y^-$ is [[Definition:Closed Set in Normed Vector Space|closed]] and contains its [[Definition:Limit Point (Normed Vector Space)|limit points]].
+Hence, $x + y \in Y^-$.
+{{qed|lemma}}
+=== Closed under restriction of scalar multiplication ===
+Let $\alpha \in \Bbb K$, where $\Bbb K$ is a [[Definition:Field (Abstract Algebra)|field]].
+Let $y \in Y^-$.
+Let $\sequence {y_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $Y$.
+Suppose:
+:$\displaystyle \lim_{n \mathop \to \infty} y_n = y$
+Since $Y$ is a [[Definition:Vector Subspace|subspace]]:
+:$\forall n \in \N : \alpha \cdot y_n \in Y \subseteq Y^-$
+Furthermore:
+:$\displaystyle \lim_{n \mathop \to \infty} \alpha \cdot y_n = \alpha \cdot y$.
+$Y^-$ is [[Definition:Closed Set in Normed Vector Space|closed]] and contains its [[Definition:Limit Point (Normed Vector Space)|limit points]].
+Hence:
+:$\alpha \cdot y \in Y^-$
+{{qed|lemma}}
+=== Nonemptiness ===
+$Y$ is a [[Definition:Vector Subspace|subspace]].
+Hence:
+:$0 \in Y \subseteq Y^-$
+Thus $Y^-$ contains at least one [[Definition:Element|element]] and is [[Definition:Non-Empty Set|non-empty]].
+{{qed|lemma}}
+Hence, $Y^-$ is [[Definition:Closed Set in Normed Vector Space|closed]] [[Definition:Vector Subspace|subspace]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Convex Subset in Normed Vector Space is Convex}
+Tags: Set Closures
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $C \subseteq X$ be a [[Definition:Convex Set (Vector Space)|convex]] [[Definition:Subset|subset]] of $X$.
+Let $C^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $C$.
+Then $C^- \subseteq X$ is also a [[Definition:Convex Set (Vector Space)|convex]] [[Definition:Subset|subset]] of $X$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in C^-$.
+Suppose $x, y$ are [[Definition:Limit Point (Normed Vector Space)|limit points]].
+Then there are [[Definition:Sequence|sequences]] $\sequence {x_n}_{n \mathop \in \N}, \sequence {x_n}_{n \mathop \in \N}$ in $C$, such that:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = x$
+:$\displaystyle \lim_{n \mathop \to \infty} x_y = y$
+Let $\alpha \in \closedint 0 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - \alpha} x + \alpha y
+      | r = \paren {1 - \alpha} \lim_{n \mathop \to \infty} x_n + \alpha \lim_{n \mathop \to \infty} y_n
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \paren {\paren {1 - \alpha} x_n + \alpha y_n}
+}}
+{{end-eqn}}
+Since $C$ is [[Definition:Convex Set (Vector Space)|convex]]:
+:$\displaystyle \forall n \in \N : \paren {1 - \alpha} x_n + \alpha y_n \in C \subseteq C^-$
+$C^-$ is [[Definition:Closed Set in Normed Vector Space|closed]] and contains its [[Definition:Limit Point (Normed Vector Space)|limit points]].
+Hence:
+:$\paren {1 - \alpha} x + \alpha y \in C^-$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Number of Distinct Prime Factors for Integer to have Abundancy Index Exceed Given Number}
+Tags: Abundancy
+
+\begin{theorem}
+Let $r \in \R$.
+Let $\mathbb P^-$ be the set of [[Definition:Prime Number|prime numbers]] with possibly [[Definition:Finitely Many|finitely many]] numbers removed.
+Define:
+:$M = \min \set {m \in \N: \displaystyle \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > r}$
+where $p_i$ is the $i$th element of $\mathbb P^-$, [[Definition:Ordering on Natural Numbers|ordered by size]].
+Then $M$ satisfies:
+:$(1): \quad$ Every number formed with fewer than $M$ [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] in $\mathbb P^-$ has [[Definition:Abundancy Index|abundancy index]] less than $r$
+:$(2): \quad$ There exists some number formed with $M$ [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] in $\mathbb P^-$ with [[Definition:Abundancy Index|abundancy index]] at least $r$
+So $M$ is the minimal number of [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] in $\mathbb P^-$ a number must have for it to have [[Definition:Abundancy Index|abundancy index]] at least $r$.
+For $r$ an [[Definition:Integer|integer]] greater than $1$:
+If $\mathbb P^-$ is taken to be the set of all [[Definition:Prime Number|prime numbers]], the values of $M$ are:
+:$2, 3, 4, 6, 9, 14, 22, 35, 55, 89, 142, \cdots$
+{{OEIS|A005579}}
+This theorem shows that this [[Definition:Sequence|sequence]] is a [[Definition:Subsequence|subsequence]] of the [[Definition:Sequence|sequence]] A256969 in the OEIS, only differing by an offset.
+If we require the numbers to be [[Definition:Odd Integer|odd]], we remove $2$ from $\mathbb P^-$.
+The sequence of values of $M$ are:
+:$3, 8, 21, 54, 141, 372, 995, 2697, 7397, 20502, \cdots$
+{{OEIS|A005580}}
+\end{theorem}
+
+\begin{proof}
+First we show that [[Definition:Abundancy Index|abundancy index]] is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+Let $n \in \N$ and let $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ be its [[Definition:Prime Factorization|prime factorization]].
+Then the [[Definition:Abundancy Index|abundancy index]] of $n$ is:
+{{begin-eqn}}
+{{eqn | l = \frac {\map \sigma n} n
+      | r = \frac {\map \sigma {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} } } {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} }
+}}
+{{eqn | r = \frac {\map \sigma {p_1^{a_1} } \map \sigma {p_2^{a_2} } \cdots \map \sigma {p_k^{a_k} } } {p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} }
+      | c = [[Sigma Function is Multiplicative]]
+}}
+{{eqn | r = \frac {\map \sigma {p_1^{a_1} } } {p_1^{a_1} } \cdot \frac {\map \sigma {p_2^{a_2} } } {p_2^{a_2} } \cdot \cdots \cdot \frac {\map \sigma {p_k^{a_k} } } {p_k^{a_k} }
+}}
+{{end-eqn}}
+so [[Definition:Abundancy Index|abundancy index]] is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+{{qed|lemma}}
+Next we show that $M$ exists.
+Note that [[Sum of Reciprocals of Primes is Divergent]].
+By [[Divergent Sequence with Finite Number of Terms Deleted is Divergent]]:
+:the sum of [[Definition:Reciprocal|reciprocals]] of all elements of $\mathbb P^-$ is also [[Definition:Divergent Series|divergent]].
+Observe that:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \frac {\ln \frac {p_n} {p_n - 1} } {\frac 1 {p_n - 1} }
+      | r = \lim_{\frac 1 {p_n - 1} \mathop \to 0} \frac {\map \ln {1 + \frac 1 {p_n - 1} } } {\frac 1 {p_n - 1} }
+}}
+{{eqn | r = 1
+      | c = [[Derivative of Logarithm at One]]
+}}
+{{end-eqn}}
+By [[Limit Comparison Test]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \ln \frac {p_n} {p_n - 1}$ is [[Definition:Divergent Series|divergent]] as well.
+By [[Logarithm of Divergent Product of Real Numbers/Infinity|Logarithm of Divergent Product of Real Numbers]]:
+:$\displaystyle \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1}$ [[Definition:Divergent Product|diverges]] to [[Definition:Infinity|infinity]].
+Hence:
+:$\exists N \in \N: \forall n \ge N: \displaystyle \prod_{n \mathop = 1}^\infty \frac {p_n} {p_n - 1} > r$
+Therefore $\displaystyle \set {m \in \N: \displaystyle \prod_{i \mathop = 1}^m \frac {p_i} {p_i - 1} > r} \ne \O$.
+By the [[Well-Ordering Principle]], $M$ exists.
+{{qed|lemma}}
+Finally, we prove our claims $(1)$ and $(2)$.
+Let $n$ be a number formed with fewer than $M$ [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] in $\mathbb P^-$.
+Let $n = q_1^{a_1} q_2^{a_2} \cdots q_k^{a_k}$ be its [[Definition:Prime Factorization|prime factorization]], where $q_i \in \mathbb P^-$ and $k < M$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\map \sigma n} n
+      | r = \prod_{i \mathop = 1}^k \frac {\map \sigma {q_i^{a_i} } } {q_i^{a_i} }
+}}
+{{eqn | r = \prod_{i \mathop = 1}^k \frac {q_i^{a_i + 1} - 1} {q_i^{a_i} \paren {q_i - 1} }
+      | c = [[Sigma Function of Power of Prime]]
+}}
+{{eqn | r = \prod_{i \mathop = 1}^k \frac {q_i - q_i^{-a_i} } {q_i - 1}
+}}
+{{eqn | o = <
+      | r = \prod_{i \mathop = 1}^k \frac {q_i} {q_i - 1}
+      | c = as $q_i^{-a_i} > 0$
+}}
+{{eqn | o = \le
+      | r = \prod_{i \mathop = 1}^{M - 1} \frac {p_i} {p_i - 1}
+}}
+{{eqn | o = \le
+      | r = r
+      | c = by minimality condition on $M$
+}}
+{{end-eqn}}
+This proves $(1)$.
+{{qed|lemma}}
+Now define $M \bar \sharp = \displaystyle \prod_{i \mathop = 1}^M p_i$ (an analog of the [[Definition:Primorial|primorial]] for $\mathbb P^-$).
+Consider the sequence of [[Definition:Abundancy Index|abundancy indices]] of $\paren {M \bar \sharp}^n$, where $n$ is a [[Definition:Strictly Positive Integer|strictly positive integer]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n}
+      | r = \prod_{i \mathop = 1}^M \frac {\map \sigma {p_i^n } } {p_i^n}
+}}
+{{eqn | r = \prod_{i \mathop = 1}^M \frac {p_i - p_i^{-n} } {p_i - 1}
+      | c = similar to above
+}}
+{{end-eqn}}
+This product is strictly increasing and tends to $\displaystyle \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ as $n \to \infty$, which is strictly greater than $r$.
+From the definition of [[Definition:Convergent Real Sequence|convergence to a limit]]:
+:$\displaystyle \exists N \in \N: \forall n \ge N: \size {\frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - r$
+Since $\displaystyle \frac {\map \sigma {\paren {M \bar \sharp}^n} } {\paren {M \bar \sharp}^n} < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}$ for all $n$:
+:$\displaystyle r < \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1} - \size {\frac {\map \sigma {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N} - \prod_{i \mathop = 1}^M \frac {p_i} {p_i - 1}} = \frac {\map \sigma {\paren {M \bar \sharp}^N} } {\paren {M \bar \sharp}^N}$
+Therefore $\paren {M \bar \sharp}^N$ is a number formed with $M$ [[Definition:Distinct|distinct]] [[Definition:Prime Factor|prime factors]] in $\mathbb P^-$ with [[Definition:Abundancy Index|abundancy index]] at least $r$.
+This proves $(2)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{1 plus Square is not Perfect Power}
+Tags: Number Theory
+
+\begin{theorem}
+The equation:
+:$x^p = y^2 + 1$
+has no solution in the [[Definition:Integer|integers]] for $x, y, p > 1$.
+\end{theorem}
+
+\begin{proof}
+Suppose $p$ is [[Definition:Even Integer|even]]. 
+Write $p = 2 k$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = y^2 - x^{2 k}
+}}
+{{eqn | r = \paren {y - x^k} \paren {y + x^k}
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+Since both $y - x^k$ and $y + x^k$ are [[Definition:Integer|integers]], they must be equal to $\pm 1$.
+Summing them up, we have $2 y$ is one of $-2, 0, 2$.
+Thus $y$ is one of $-1, 0, 1$, and we ignore these solutions due to our condition $y > 1$.
+Now suppose $p$ is [[Definition:Odd Integer|odd]].
+Suppose $y$ is [[Definition:Odd Integer|odd]].
+Then $x^p = y^2 + 1$ is [[Definition:Even Integer|even]].
+Hence $x$ is [[Definition:Even Integer|even]].
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \equiv
+      | r = x^p
+      | rr = \pmod 8
+      | c = as $p \ge 3$
+}}
+{{eqn | o = \equiv
+      | r = y^2 + 1
+      | rr = \pmod 8
+}}
+{{eqn | o = \equiv
+      | r = 1 + 1
+      | rr = \pmod 8
+      | c = [[Odd Square Modulo 8]]
+}}
+{{eqn | o = \equiv
+      | r = 2
+      | rr = \pmod 8
+}}
+{{end-eqn}}
+which is a [[Definition:Contradiction|contradiction]].
+Hence $y$ must be [[Definition:Even Integer|even]], and $x$ must be [[Definition:Odd Integer|odd]].
+From [[Gaussian Integers form Euclidean Domain]], we can define [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] on $\Z \sqbrk i$, and it admits [[Definition:Unique Factorization Domain|unique factorization]].
+We factorize $y^2 + 1$:
+:$x^p = y^2 + 1 = \paren {1 + i y} \paren {1 - i y}$
+The [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of $1 + i y$ and $1 - i y$ must [[Definition:Divisor of Ring Element|divide]] their sum and product.
+Their sum is $2$ while their product is $y^2 + 1$, which is [[Definition:Odd Integer|odd]].
+Therefore we see that $1 + i y$ and $1 - i y$ are [[Definition:Coprime/Euclidean Domain|coprime]].
+From unique factorization we must have that both $1 + i y$ and $1 - i y$ is a product of a [[Definition:Unit of Ring|unit]] and a $p$th power.
+By [[Units of Gaussian Integers]], the [[Definition:Unit of Ring|units]] are $\pm 1$ and $\pm i$.
+Hence
+:$\exists u \in \set {\pm 1, \pm i}: \exists \alpha \in \Z \sqbrk i: 1 + i y = u \alpha^p, 1 - i y = \bar u \bar \alpha^p$
+Since $p$ is [[Definition:Odd Integer|odd]]:
+:$1^p = 1$
+:$\paren {-1}^p = -1$
+:$i^p = \pm i$
+:$\paren {-i}^p = -i^p = \mp i$
+therefore there is some [[Definition:Unit of Ring|unit]] $u' \in \set {\pm 1, \pm i}$ such that $u'^p = u$.
+By writing $\beta = u' \alpha$:
+:$1 + i y = u'^p \alpha^p = \beta^p, 1 - i y = \bar \beta^p$
+Write $\beta = a + i b$, where $a, b \in \Z$.
+By [[Sum of Two Odd Powers]]:
+:$2 a = \beta + \bar \beta \divides \beta^p + \bar \beta^p = 2$
+this gives $a = \pm 1$.
+We also have:
+{{begin-eqn}}
+{{eqn | l = 1 + y^2
+      | r = \beta^p \bar \beta^p
+}}
+{{eqn | r = \paren {\beta \bar \beta}^p
+}}
+{{eqn | r = \paren {a^2 + b^2}^p
+      | c = [[Product of Complex Number with Conjugate]]
+}}
+{{eqn | r = \paren {1 + b^2}^p
+}}
+{{end-eqn}}
+since $1 + y^2$ is [[Definition:Odd Integer|odd]], $b$ must be [[Definition:Even Integer|even]].
+Hence:
+{{begin-eqn}}
+{{eqn | l = 1 + i y
+      | r = \beta^p
+}}
+{{eqn | r = \paren {a + i b}^p
+}}
+{{eqn | r = \sum_{k \mathop = 0}^p \binom p k a^{p - k} \paren {i b}^k
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | o = \equiv
+      | r = a^p + p a^{p - 1} i b 
+      | rr = \pmod 4
+      | c = $k \ge 2$ vanish as all terms containing $b^2$ is divisible by $4$
+}}
+{{end-eqn}}
+In particular, comparing real parts gives $1 \equiv a^p \pmod 4$.
+Since $p$ is [[Definition:Odd Integer|odd]], we have $a = 1$.
+Now we have:
+{{begin-eqn}}
+{{eqn | l = 1 + i y
+      | r = \paren {1 + i b}^p
+}}
+{{eqn | r = \sum_{k \mathop = 0}^p \binom p k 1^{p - k} \paren {i b}^k
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | ll = \leadsto
+      | l = 1
+      | r = \sum_{k \mathop = 0}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k
+      | c = Comparing Real Parts; only even $k$ remain
+}}
+{{eqn | r = 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \binom p {2 k} b^{2 k} \paren {-1}^k
+}}
+{{eqn | r = 1 - \binom p 2 b^2 + \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac {p \paren {p - 1} } {2 k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k} \paren {-1}^k
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = 1 - \binom p 2 b^2 + \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
+      | c = [[Binomial Coefficient with Two]]
+}}
+{{eqn | ll = \leadsto
+      | l = \binom p 2 b^2
+      | r = \binom p 2 b^2 \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
+}}
+{{eqn | ll = \leadsto
+      | l = 1
+      | r = \sum_{k \mathop = 2}^{\paren {p - 1} / 2} \paren {\frac 1 {k \paren {2 k - 1} } } \binom {p - 2} {2 k - 2} b^{2 k - 2} \paren {-1}^k
+}}
+{{end-eqn}}
+The summands on the right hand side may not be an [[Definition:Integer|integer]], but if we can show:
+:In [[Definition:Canonical Form of Rational Number|canonical form]], the [[Definition:Numerator|numerator]] of each summand is [[Definition:Even Integer|even]]
+then the equation is never satisfied.
+This is because the sum of all the terms will be a rational number with [[Definition:Even Integer|even]] [[Definition:Numerator|numerator]] and [[Definition:Odd Integer|odd]] [[Definition:Denominator|denominator]], which cannot equal to $1$.
+Since $2 k + 1$ is always [[Definition:Odd Integer|odd]] and $\paren {-1}^k \dbinom {p - 2} {2 k - 2}$ is always an [[Definition:Integer|integer]], we only need to check $\dfrac {b^{2 k - 2} } k$.
+Since $b$ is [[Definition:Even Integer|even]]:
+:$2^{2 k - 2} \divides b^{2 k - 2}$
+But we have:
+{{begin-eqn}}
+{{eqn | l = 2^{2 k - 2}
+      | o = \ge
+      | r = 2^k
+      | c = as $k \ge 2$
+}}
+{{eqn | o = >
+      | r = k
+      | c = [[N less than M to the N]]
+}}
+{{end-eqn}}
+Hence the largest [[Definition:Integer Power|power]] of $2$ that [[Definition:Divisor of Integer|divides]] $k$ is less than $2^{2 k - 2}$.
+Therefore the [[Definition:Numerator|numerator]] of $\dfrac {b^{2 k - 2} } k$ is [[Definition:Even Integer|even]].
+And thus all the equations above are never satisfied.
+So our original equation:
+:$x^p = y^2 + 1$
+has no solution in the [[Definition:Integer|integers]] for $x, y, p > 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{1 plus Perfect Power is not Prime Power except for 9}
+Tags: Number Theory
+
+\begin{theorem}
+The only solution to:
+:$x^m = y^n + 1$
+is:
+:$\tuple {x, m, y, n} = \tuple {3, 2, 2, 3}$
+for [[Definition:Positive Integer|positive integers]] $x, y, m, n > 1$, and $x$ is a [[Definition:Prime Number|prime number]].
+This is a special case of [[Catalan's Conjecture]].
+\end{theorem}
+
+\begin{proof}
+It suffices to show the result for [[Definition:Prime Number|prime]] values of $n$.
+The case $n = 2$ is covered in [[1 plus Square is not Perfect Power]].
+So we consider the cases where $n$ is an [[Definition:Odd Integer|odd]] [[Definition:Prime Number|prime]].
+{{begin-eqn}}
+{{eqn | l = x^m
+      | r = y^n + 1
+}}
+{{eqn | r = y^n - \paren {-1}^n
+      | c = as $n$ is [[Definition:Odd Integer|odd]]
+}}
+{{eqn | r = \paren {y - \paren {-1} } \sum_{j \mathop = 0}^{n - 1} y^{n - j - 1} \paren {-1}^j
+      | c = [[Difference of Two Powers]]
+}}
+{{eqn | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + R}
+      | c = [[Division Theorem for Polynomial Forms over Field]]
+}}
+{{end-eqn}}
+where $Q$ is a polynomial in one unknown and $R$ is a degree zero polynomial, so $R$ is a constant.
+We have:
+{{begin-eqn}}
+{{eqn | l = R
+      | r = \map Q {-1} \paren {-1 + 1} + R
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - j - 1} \paren {-1}^j
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^{n - 1}
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} 1
+      | c = as $n$ is [[Definition:Odd Integer|odd]]
+}}
+{{eqn | r = n
+}}
+{{end-eqn}}
+Hence we have $x^m = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}$.
+Since $x$ is a [[Definition:Prime Number|prime]], we have:
+:$x \divides y + 1$
+:$x \divides \map Q y \paren {y + 1} + n$
+Hence $x \divides n$.
+Since $x > 1$ and $n$ is a [[Definition:Prime Number|prime]], we must have $x = n$.
+Now we write $y + 1 = x^\alpha$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = x^m
+      | r = \paren {y + 1} \paren {\map Q y \paren {y + 1} + n}
+}}
+{{eqn | r = x^\alpha \paren {\map Q y x^\alpha + x}
+}}
+{{eqn | r = x^{\alpha + 1} \paren {\map Q y x^{\alpha - 1} + 1}
+}}
+{{end-eqn}}
+For $\alpha > 1$, $x \nmid \map Q y x^{\alpha - 1} + 1$.
+Hence $\alpha = 1$.
+This gives $y + 1 = x = n$.
+The equation now simplifies to:
+:$\paren {y + 1}^m = y^n + 1$
+Expanding:
+{{begin-eqn}}
+{{eqn | l = \paren {y + 1}^m
+      | r = \sum_{j \mathop = 0}^m \binom m j y^j 1^{m - j}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = 1 + \sum_{j \mathop = 1}^m \binom m j y^j
+}}
+{{eqn | r = y^n + 1
+}}
+{{eqn | ll = \leadsto
+      | l = \sum_{j \mathop = 1}^m \binom m j y^{j - 1}
+      | r = y^{n - 1}
+}}
+{{eqn | ll = \leadsto
+      | l = \binom m 1 y^0
+      | r = 0
+      | rr = \pmod y
+      | c = as $y > 1$
+}}
+{{eqn | ll = \leadsto
+      | l = m
+      | r = 0
+      | rr = \pmod y
+      | c = [[Binomial Coefficient with One]]
+}}
+{{end-eqn}}
+hence we must have $y \divides m$.
+By [[Absolute Value of Integer is not less than Divisors]], $y \le m$.
+Moreover, from $\displaystyle \sum_{j \mathop = 1}^m \binom m j y^{j - 1} = y^{n - 1}$ we also have:
+:$y^{n - 1} > \dbinom m m y^{m - 1} = y^{m - 1}$
+Therefore we also have $n > m$.
+This gives $y = n - 1 \ge m$.
+The two inequalities forces $y = m$.
+Now our original equation is further simplified to:
+{{begin-eqn}}
+{{eqn | l = \paren {y + 1}^y
+      | r = y^{y + 1} + 1
+}}
+{{eqn | ll = \leadsto
+      | l = \paren {1 + \frac 1 y}^y
+      | r = y + \frac 1 {y^y}
+      | c = Dividing both sides by $y^y$
+}}
+{{end-eqn}}
+From [[Real Sequence (1 + x over n)^n is Convergent]]:
+:$\paren {1 + \dfrac 1 y}^y$ is [[Definition:Increasing Real Function|increasing]] and has [[Definition:Limit of Sequence|limit]] $e$.
+Then we have for all $y \in \N$:
+:$y + \dfrac 1 {y^y} < e < 3$
+Since $\dfrac 1 {y^y} > 0$ and $y > 1$, we can only have $y = 2$.
+This gives the solution $3^2 = 2^3 + 1$, and there are no others.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field is Principal Ideal Domain}
+Tags: Principal Ideal Domains, Field Theory
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Then $F$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $I \subset F$ be a [[Definition:Non-Null Ideal|non-null]] [[Definition:Ideal of Ring|ideal]] of $F$.
+Let $a \in I$ be non-[[Definition:Field Zero|zero]].
+Since $F$ is a [[Definition:Field (Abstract Algebra)|field]], $a^{-1}$ exists.
+We have that $1 = a^{-1} \cdot a \in I$.
+Since $1 \in I$, for every [[Definition:Element|element]] $b \in F$:
+:$b = b \cdot 1 \in I$
+we have that $I = F = \ideal 1$ if $I \ne \set 0$.
+Thus the only [[Definition:Ideal of Ring|ideals]] of $F$ are $\ideal 0 = \set 0$ and $\ideal 1 = F$, which are both [[Definition:Principal Ideal Domain|principal ideals]].
+Hence $F$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+{{qed}}
+[[Category:Principal Ideal Domains]]
+[[Category:Field Theory]]
+82o7uz5ng4f9faonjp1mx046d66y0sm
+\end{proof}<|endoftext|>
+\section{Complex Vector Space is Vector Space}
+Tags: Examples of Vector Spaces, Complex Numbers
+
+\begin{theorem}
+Let $\C$ denote the set of [[Definition:Complex Number|complex numbers]].
+Then the [[Definition:Complex Vector Space|complex vector space $\C^n$]] is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+=== Construction of Complex Vector Space ===
+From the definition, a [[Definition:Vector Space|vector space]] is a [[Definition:Unitary Module|unitary module]] whose [[Definition:Scalar Ring of Unitary Module|scalar ring]] is a [[Definition:Field (Abstract Algebra)|field]].
+In order to call attention to the precise scope of the operators, let [[Definition:Complex Addition|complex addition]] and [[Definition:Complex Multiplication|complex multiplication]] be expressed as $+_\C$ and $\times_\C$ respectively.
+Then we can express the [[Definition:Field of Complex Numbers|field of complex numbers]] as $\struct {\C, +_\C, \times_\C}$.
+From [[Complex Numbers under Addition form Abelian Group]], $\struct {\C, +}$ is a [[Definition:Group|group]].
+Again, in order to call attention to the precise scope of the operator, let [[Definition:Complex Addition|complex addition]] be expressed on $\struct {\C, +}$ as $+_G$.
+That is, the [[Definition:Group|group]] under consideration is $\struct {\C, +_G}$.
+Consider the [[Definition:Cartesian Product|cartesian product]]:
+:$\displaystyle \C^n = \prod_{i \mathop = 1}^n \struct {\C, +_G} = \underbrace {\struct {\C, +_G} \times \cdots \times \struct {\C, +_G} }_{n \text{ copies} }$
+Let:
+:$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
+:$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$
+be arbitrary elements of $\C^n$.
+Let $\lambda$ be an arbitrary element of $\C$.
+Let $+$ be the [[Definition:Binary Operation|binary operation]] defined on $\C^n$ as:
+:$\mathbf a + \mathbf b = \tuple {a_1 +_G b_1, a_2 +_G b_2, \ldots, a_n +_G b_n}$
+Also let $\cdot$ be the [[Definition:Binary Operation|binary operation]] defined on $\C \times \C^n$ as:
+:$\lambda \cdot \mathbf a = \tuple {\lambda \times_\C a_1, \lambda \times_\C a_2, \ldots, \lambda \times_\C a_n}$
+In this context, $\lambda \times_\C a_j$ is defined as [[Definition:Complex Multiplication|complex multiplication]], as is appropriate (both $\lambda$ and $a_j$ are [[Definition:Complex Number|complex numbers]]).
+With this set of definitions, the structure $\struct {\C^n, +, \cdot}$ is a [[Definition:Vector Space|vector space]], as is shown in [[Complex Vector Space is Vector Space#Proof of Complex Vector Space|Proof of Complex Vector Space]] below.
+=== Proof of Complex Vector Space ===
+In order to show that $\struct {\C^n, +, \cdot}$ is a [[Definition:Vector Space|vector space]], we need to show that:
+$\forall \mathbf x, \mathbf y \in \C^n, \forall \lambda, \mu \in \C$:
+:$(1): \quad \lambda \cdot \paren {\mathbf x + \mathbf y} = \paren {\lambda \cdot \mathbf x} + \paren {\lambda \cdot \mathbf y}$
+:$(2): \quad \paren {\lambda +_\C \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$
+:$(3): \quad \paren {\lambda \times_\C \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$
+:$(4): \quad \forall \mathbf x \in \C^n: 1 \cdot \mathbf x = \mathbf x$.
+where $1$ in this context means $1 + 0 i$, as derived in [[Complex Multiplication Identity is One]].
+From [[External Direct Product of Groups is Group]], we have that $\struct {\C^n, +}$ is a [[Definition:Group|group]] in its own right.
+Let:
+:$\mathbf x = \tuple {x_1, x_2, \ldots, x_n}$
+:$\mathbf y = \tuple {y_1, y_2, \ldots, y_n}$
+Checking the criteria in turn:
+$(1): \quad \lambda \cdot \paren {\mathbf x + \mathbf y} = \paren {\lambda \cdot \mathbf x} + \paren {\lambda \cdot \mathbf y}$:
+{{begin-eqn}}
+{{eqn | l = \lambda \cdot \paren {\mathbf x + \mathbf y}
+      | r = \tuple {\lambda \times_\C \paren {x_1 +_G y_1}, \lambda \times_\C \paren {x_2 +_G y_2}, \ldots, \lambda \times_\C \paren {x_n +_G y_n} }
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \tuple {\paren {\lambda \times_\C x_1 +_G \lambda \times_\C y_1}, \paren {\lambda \times_\C x_2 +_G \lambda \times_\C y_2}, \ldots, \paren {\lambda \times_\C x_n +_G \lambda \times_\C y_n} }
+      | c = [[Complex Multiplication Distributes over Addition|$\times_\C$ distributes over $+_G$]]
+}}
+{{eqn | r = \tuple {\lambda \times_\C x_1, \lambda \times_\C x_2, \ldots, \lambda \times_\C x_n} + \tuple {\lambda \times_\C y_1, \lambda \times_\C y_2, \ldots, \lambda \times_\C y_n}
+      | c = Definition of $+$ over $\C^n$
+}}
+{{eqn | r = \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \lambda \cdot \tuple {y_1, y_2, \ldots, y_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \lambda \cdot \mathbf x + \lambda \cdot \mathbf y
+      | c = Definition of $\mathbf x$ and $\mathbf y$
+}}
+{{end-eqn}}
+So $(1)$ has been shown to hold.
+$(2): \quad \paren {\lambda +_\C \mu} \cdot x = \paren {\lambda \cdot \mathbf x} + \paren {\mu \cdot \mathbf x}$:
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda +_\C \mu} \cdot x
+      | r = \tuple {\paren {\lambda +_\C \mu} \times_\C x_1, \paren {\lambda +_\C \mu} \times_\C x_2, \ldots, \paren {\lambda +_\C \mu} \times_\C x_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \tuple {\paren {\lambda \times_\C x_1 +_G \mu \times_\C x_1}, \paren {\lambda \times_\C x_2 +_G \mu \times_\C x_2}, \ldots, \paren {\lambda \times_\C x_n +_G \mu \times_\C x_n} }
+      | c = [[Complex Multiplication Distributes over Addition|$\times_\C$ distributes over $+_\C$]], and $+_\C$ is the same operation as $+_G$
+}}
+{{eqn | r = \tuple {\lambda \times_\C x_1, \lambda \times_\C x_2, \ldots, \lambda \times_\C x_n} + \tuple {\mu \times_\C x_1, \mu \times_\C x_2, \ldots, \mu \times_\C x_n}
+      | c = Definition of $+$ over $\C^n$
+}}
+{{eqn | r = \lambda \cdot \tuple {x_1, x_2, \ldots, x_n} + \mu \cdot \tuple {x_1, x_2, \ldots, x_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \lambda \cdot \mathbf x + \mu \cdot \mathbf x
+      | c = Definition of $\mathbf x$ and $\mathbf y$
+}}
+{{end-eqn}}
+So $(2)$ has been shown to hold.
+$(3): \quad \paren {\lambda \times_\C \mu} \cdot x = \lambda \cdot \paren {\mu \cdot \mathbf x}$:
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda \times_\C \mu} \cdot x
+      | r = \tuple {\paren {\lambda \times_\C \mu} \times_\C x_1, \paren {\lambda \times_\C \mu} \times_\C x_2, \ldots, \paren {\lambda \times_\C \mu} \times_\C x_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \tuple {\lambda \times_\C \paren {\mu \times_\C x_1}, \lambda \times_\C \paren {\mu \times_\C x_2}, \ldots, \lambda \times_\C \paren {\mu \times_\C x_n} }
+      | c = [[Complex Multiplication is Associative]]
+}}
+{{eqn | r = \lambda \cdot \tuple {\mu \times_\C x_1, \mu \times_\C x_2, \ldots, \mu \times_\C x_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \lambda \cdot \paren {\mu \cdot \tuple {x_1, x_2, \ldots, x_n} }
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \lambda \cdot \paren {\mu \cdot \mathbf x}
+      | c = Definition of $\mathbf x$
+}}
+{{end-eqn}}
+So $(3)$ has been shown to hold.
+$(4): \quad \forall \mathbf x \in \C^n: 1 \cdot \mathbf x = \mathbf x$:
+{{begin-eqn}}
+{{eqn | l = 1 \cdot \mathbf x = \mathbf x
+      | r = \tuple {1 \times_\C x_1, 1 \times_\C x_2, \ldots, 1 \times_\C x_n}
+      | c = Definition of $\cdot$ over $\C \times \C^n$
+}}
+{{eqn | r = \tuple {x_1, x_2, \ldots, x_n}
+      | c = [[Complex Multiplication Identity is One]]
+}}
+{{eqn | r = \mathbf x
+      | c = Definition of $\mathbf x$
+}}
+{{end-eqn}}
+So $(4)$ has been shown to hold.
+So the [[Definition:Module on Cartesian Product|$\C$-module $\C^n$]] is a [[Definition:Vector Space|vector space]], as we were to prove.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space}
+Tags: Normed Vector Spaces, Denseness, Set Closures
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ is a [[Definition:Normed Vector Space|normed vector space]].
+Let $D \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Let $D^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $D$.
+Then $D$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] iff $D^- = X$.
+\end{theorem}
+
+\begin{proof}
+=== [[Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition|Necessary Condition]] ===
+{{:Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition}}{{qed|lemma}}
+=== [[Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition|Sufficient Condition]] ===
+{{:Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition}
+Tags: Normed Vector Spaces, Denseness, Set Closures
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ is a [[Definition:Normed Vector Space|normed vector space]].
+Let $D \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Let $D^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $D$.
+Then $D$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] iff $D^- = X$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in X \setminus D$.
+Suppose $D$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] in $X$.
+Then:
+:$\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$
+where $\displaystyle \map {B_{\frac 1 n}} x$ is an [[Definition:Open Ball in Normed Vector Space|open ball]].
+Let $\sequence {d_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $D$.
+Then:
+:$\forall n \in \N : \norm {x - d_n} < \frac 1 n$
+Hence, $x$ is a [[Definition:Limit Point (Normed Vector Space)|limit point]] of $D$.
+In other words, $x \in D^-$.
+We have just shown that:
+:$x \in X \setminus D \implies x \in D^-$
+Hence:
+:$X \setminus D \subseteq D^-$.
+By definition of [[Definition:Closure/Normed Vector Space|closure]]:
+:$D \subseteq D^-$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = D \cup \paren {X \setminus D}
+}}
+{{eqn | o = \subseteq
+      | r = D^-
+}}
+{{eqn | o = \subseteq
+      | r = X
+}}
+{{end-eqn}}
+Thus:
+:$X = D^-$.
+\end{proof}<|endoftext|>
+\section{Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition}
+Tags: Normed Vector Spaces, Denseness, Set Closures
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ is a [[Definition:Normed Vector Space|normed vector space]].
+Let $D \subseteq X$ be a [[Definition:Subset|subset]] of $X$.
+Let $D^-$ be the [[Definition:Closure/Normed Vector Space|closure]] of $D$.
+Then $D$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] iff $D^- = X$.
+\end{theorem}
+
+\begin{proof}
+Let $X = D^-$.
+We have to show, that for every $x \in X$ there is an [[Definition:Open Ball in Normed Vector Space|open ball]] with an [[Definition:Element|element]] from $D^-$.
+We have that $X = D \cup \paren {X \setminus D}$.
+Suppose $x \in X \setminus D$.
+Then $x \in D^- \setminus D$.
+Hence, $x$ is a [[Definition:Limit Point (Normed Vector Space)|limit point]] of $D$.
+Therefore, there is a [[Definition:Sequence|sequence]] $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which [[Definition:Convergent Sequence in Normed Vector Space|converges]] to $x$.
+Thus:
+:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$
+In other words:
+:$\displaystyle d_N \in D \implies d_N \in \map {B_\epsilon} x$
+Therefore:
+:$d_N \in D \cap \map {B_\epsilon} x$.
+Suppose $x \in D$.
+Let $\epsilon > 0$.
+Then $x \in \map {B_\epsilon} x \cap D$.
+From both parts and definition we conclude that $D^-$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] in $X$.
+\end{proof}<|endoftext|>
+\section{Modulus of Limit/Normed Vector Space}
+Tags: Limits of Sequences, Normed Vector Spaces
+
+\begin{theorem}
+Let $\struct {X, \norm { \, \cdot \, } }$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\sequence {x_n}$ be a [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]] in $R$ to the [[Definition:Limit of Sequence in Normed Vector Space|limit]] $x$.
+That is, let $\displaystyle \lim_{n \mathop \to \infty} x_n = x$.
+Then
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n} = \norm x$
+\end{theorem}
+
+\begin{proof}
+By the [[Reverse Triangle Inequality]], we have:
+:$\cmod {\norm {x_n} - \norm x} \le \norm {x_n - x}$
+Hence by the [[Squeeze Theorem]] and [[Convergent Sequence Minus Limit]], $\norm {x_n} \to \norm x$ as $n \to \infty$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Entrywise Addition forms Abelian Group}
+Tags: Matrix Entrywise Addition, Examples of Groups, Matrix Entrywise Addition forms Abelian Group
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {R, +, \circ}$.
+Then $\struct {\map {\MM_R} {m, n}, +}$, where $+$ is [[Definition:Matrix Entrywise Addition|matrix entrywise addition]], is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+We have by definition that [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is a specific instance of a [[Definition:Hadamard Product|Hadamard product]].
+By definition of a [[Definition:Ring (Abstract Algebra)|ring]], the [[Definition:Algebraic Structure|structure]] $\struct {R, +}$ is a [[Definition:Group|group]].
+As $\struct {R, +}$ is [[Definition:A Fortiori|a fortiori]] a [[Definition:Monoid|monoid]], it follows from [[Matrix Space Semigroup under Hadamard Product]] that $\struct {\map {\MM_R} {m, n}, +}$ is also a [[Definition:Monoid|monoid]].
+As $\struct {R, +}$ is a [[Definition:Group|group]], it follows from [[Negative Matrix is Inverse for Matrix Entrywise Addition]] that all [[Definition:Element|elements]] of $\struct {\map {\MM_R} {m, n}, +}$ have an [[Definition:Inverse Element|inverse element]].
+From [[Matrix Entrywise Addition is Commutative]] it follows that $\struct {\map {\MM_R} {m, n}, +}$ is an [[Definition:Abelian Group|Abelian group]].
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Negative Matrix/General Group}
+Tags: Definitions: Negative Matrices
+
+\begin{theorem}
+Let $\struct {G, \cdot}$ be a [[Definition:Group|group]].
+Let $\map {\MM_G} {m, n}$ denote the [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {G, \cdot}$.
+Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Element|element]] of $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is the [[Definition:Hadamard Product|Hadamard product]].
+Then the '''negative (matrix) of $\mathbf A = \sqbrk a_{m n}$''' is denoted and defined as:
+:$-\mathbf A := \sqbrk {a^{-1} }_{m n}$
+where $a^{-1}$ is the [[Definition:Inverse Element|inverse element]] of $a \in G$.
+\end{theorem}<|endoftext|>
+\section{Negative Matrix is Inverse for Matrix Entrywise Addition over Ring}
+Tags: Negative Matrices
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {R, +, \circ}$.
+Let $\mathbf A$ be an [[Definition:Element|element]] of $\map {\MM_R} {m, n}$.
+Let $-\mathbf A$ be the [[Definition:Negative Matrix|negative]] of $\mathbf A$.
+Then $-\mathbf A$ is the [[Definition:Inverse Element|inverse]] for the operation $+$, where $+$ is [[Definition:Matrix Entrywise Addition over Ring|matrix entrywise addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \paren {-\mathbf A}
+      | r = \sqbrk a_{m n} + \paren {-\sqbrk a_{m n} }
+      | c = Definition of $\mathbf A$
+}}
+{{eqn | r = \sqbrk a_{m n} + \sqbrk {-a}_{m n}
+      | c = {{Defof|Negative Matrix over Ring}}
+}}
+{{eqn | r = \sqbrk {a + \paren {-a} }_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \sqbrk {0_R}_{m n}
+      | c = {{Defof|Ring Negative}}
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A + \paren {-\mathbf A}
+      | r = \mathbf 0_R
+      | c = {{Defof|Zero Matrix over Ring}}
+}}
+{{end-eqn}}
+The result follows from [[Zero Matrix is Identity for Matrix Entrywise Addition over Ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Matrix is Identity for Matrix Entrywise Addition over Ring}
+Tags: Matrix Entrywise Addition, Zero Matrix, Zero Matrix is Identity for Matrix Entrywise Addition
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+Let $\mathbf 0_R = \sqbrk {0_R}_{m n}$ be the [[Definition:Zero Matrix over Ring|zero matrix]] of $\map {\MM_R} {m, n}$.
+Then $\mathbf 0_R$ is the [[Definition:Identity Element|identity element]] for [[Definition:Matrix Entrywise Addition over Ring|matrix entrywise addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \mathbf 0_R
+      | r = \sqbrk a_{m n} + \sqbrk {0_R}_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {a + 0_R}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = {{Ring-axiom|A3}} is $0_R$
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A + \mathbf 0_R
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix over Ring}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \mathbf 0_R + \mathbf A
+      | r = \sqbrk {0_R}_{m n} + \sqbrk a_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {0_R + a}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = {{Ring-axiom|A3}} is $0_R$
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf 0_R + \mathbf A
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix over Ring}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' with respect to [[Definition:Ring Addition|ring addition]].
+We have from {{Ring-axiom|A3}} that the [[Definition:Identity Element|identity element]] of [[Definition:Ring Addition|ring addition]] is the [[Definition:Ring Zero|ring zero]] $0_R$.
+The result then follows directly from [[Zero Matrix is Identity for Hadamard Product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Matrix Entrywise Addition over Ring}
+Tags: Matrix Entrywise Addition
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $S$ over an [[Definition:Algebraic Structure|algebraic structure]] $\struct {R, +, \circ}$.
+Let $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$.
+Let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+The operation of [[Definition:Matrix Entrywise Addition over Ring|matrix entrywise addition]] satisfies the following properties:
+:$+$ is [[Definition:Closure (Abstract Algebra)|closed]] on $\map {\MM_R} {m, n}$
+:$+$ is [[Definition:Associative|associative]] on $\map {\MM_R} {m, n}$
+:$+$ is [[Definition:Commutative Operation|commutative]] on $\map {\MM_R} {m, n}$.
+\end{theorem}
+
+\begin{proof}
+=== [[Matrix Entrywise Addition over Ring is Closed]] ===
+{{:Matrix Entrywise Addition over Ring is Closed}}
+=== [[Matrix Entrywise Addition over Ring is Associative]] ===
+{{:Matrix Entrywise Addition over Ring is Associative}}
+=== [[Matrix Entrywise Addition over Ring is Commutative]] ===
+{{:Matrix Entrywise Addition over Ring is Commutative}}
+[[Category:Matrix Entrywise Addition]]
+ofmyntkjczgnuup663ekqa4e6fcfv7v
+\end{proof}<|endoftext|>
+\section{Matrix Entrywise Addition is Associative}
+Tags: Matrix Entrywise Addition, Associativity, Matrix Entrywise Addition is Associative
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over one of the [[Definition:Standard Number System|standard number systems]].
+For $\mathbf A, \mathbf B \in \map \MM {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+The operation $+$ is [[Definition:Associative Operation|associative]] on $\map \MM {m, n}$.
+That is:
+:$\paren {\mathbf A + \mathbf B} + \mathbf C = \mathbf A + \paren {\mathbf B + \mathbf C}$
+for all $\mathbf A$, $\mathbf B$ and $\mathbf C$ in $\map \MM {m, n}$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be [[Definition:Matrix|matrices]] whose [[Definition:Order of Matrix|order]] is $m \times n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf A + \mathbf B} + \mathbf C
+      | r = \paren {\sqbrk a_{m n} + \sqbrk b_{m n} } + \sqbrk c_{m n}
+      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
+}}
+{{eqn | r = \sqbrk {a + b}_{m n} + \sqbrk c_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk {\paren {a + b} + c}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk {a + \paren {b + c} }_{m n}
+      | c = [[Associative Law of Addition]]
+}}
+{{eqn | r = \sqbrk a_{m n} + \sqbrk {b + c}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n} + \paren {\sqbrk b_{m n} + \sqbrk c_{m n} }
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \mathbf A + \paren {\mathbf B + \mathbf C}
+      | c = Definition of $\mathbf A$, $\mathbf B$ and $\mathbf C$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Entrywise Addition over Ring is Commutative}
+Tags: Matrix Entrywise Addition, Commutativity, Matrix Entrywise Addition is Commutative
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\map {\MM_R} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+For $\mathbf A, \mathbf B \in \map {\MM_R} {m, n}$, let $\mathbf A + \mathbf B$ be defined as the [[Definition:Matrix Entrywise Addition over Ring|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+The operation $+$ is [[Definition:Commutative Operation|commutative]] on $\map {\MM_R} {m, n}$.
+That is:
+:$\mathbf A + \mathbf B = \mathbf B + \mathbf A$
+for all $\mathbf A$ and $\mathbf B$ in $\map {\MM_R} {m, n}$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be [[Definition:Element|elements]] of the [[Definition:Matrix Space|$m \times n$ matrix space]] over $R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \mathbf B
+      | r = \sqbrk a_{m n} + \sqbrk b_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf B$
+}}
+{{eqn | r = \sqbrk {a + b}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \sqbrk {b + a}_{m n}
+      | c = {{Ring-axiom|A2}}
+}}
+{{eqn | r = \sqbrk b_{m n} + \sqbrk a_{m n}
+      | c = {{Defof|Matrix Entrywise Addition over Ring}}
+}}
+{{eqn | r = \mathbf B + \mathbf A
+      | c = Definition of $\mathbf A$ and $\mathbf B$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' of $\mathbf A$ and $\mathbf B$ with respect to [[Definition:Ring Addition|ring addition]].
+We have from {{Ring-axiom|A2}} that [[Definition:Ring Addition|ring addition]] is [[Definition:Commutative Operation|commutative]].
+The result then follows directly from [[Commutativity of Hadamard Product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Matrix is Identity for Matrix Entrywise Addition}
+Tags: Matrix Entrywise Addition, Zero Matrix, Zero Matrix is Identity for Matrix Entrywise Addition
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over one of the [[Definition:Standard Number System|standard number systems]].
+Let $\mathbf 0 = \sqbrk 0_{m n}$ be the [[Definition:Zero Matrix|zero matrix]] of $\map \MM {m, n}$.
+Then $\mathbf 0$ is the [[Definition:Identity Element|identity element]] for [[Definition:Matrix Entrywise Addition|matrix entrywise addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \mathbf 0_R
+      | r = \sqbrk a_{m n} + \sqbrk {0_R}_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {a + 0_R}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = {{Ring-axiom|A3}} is $0_R$
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A + \mathbf 0_R
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix over Ring}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \mathbf 0_R + \mathbf A
+      | r = \sqbrk {0_R}_{m n} + \sqbrk a_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {0_R + a}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = {{Ring-axiom|A3}} is $0_R$
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf 0_R + \mathbf A
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix over Ring}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' with respect to [[Definition:Ring Addition|ring addition]].
+We have from {{Ring-axiom|A3}} that the [[Definition:Identity Element|identity element]] of [[Definition:Ring Addition|ring addition]] is the [[Definition:Ring Zero|ring zero]] $0_R$.
+The result then follows directly from [[Zero Matrix is Identity for Hadamard Product]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From:
+:[[Integers form Ring]]
+:[[Rational Numbers form Ring]]
+:[[Real Numbers form Ring]]
+:[[Complex Numbers form Ring]]
+the [[Definition:Standard Number System|standard number systems]] $\Z$, $\Q$, $\R$ and $\C$ are [[Definition:Ring (Abstract Algebra)|rings]] whose [[Definition:Zero Element|zero]] is the [[Definition:Zero (Number)|number $0$ (zero)]].
+Hence we can apply [[Zero Matrix is Identity for Matrix Entrywise Addition over Ring]].
+{{qed|lemma}}
+The above cannot be applied to the [[Definition:Natural Numbers|natural numbers]] $\N$, as they do not form a [[Definition:Ring (Abstract Algebra)|ring]].
+However, from [[Natural Numbers under Addition form Commutative Monoid]], the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is [[Definition:Zero (Number)|$0$ (zero)]].
+By definition, [[Definition:Matrix Entrywise Addition|matrix entrywise addition]] is the '''[[Definition:Hadamard Product|Hadamard product]]''' with respect to [[Definition:Addition|addition of numbers]].
+The result follows from [[Zero Matrix is Identity for Hadamard Product]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \mathbf 0
+      | r = \sqbrk a_{m n} + \sqbrk 0_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {a + 0}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = [[Identity Element of Addition on Numbers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A + \mathbf 0
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \mathbf 0 + \mathbf A
+      | r = \sqbrk 0_{m n} + \sqbrk a_{m n}
+      | c = Definition of $\mathbf A$ and $\mathbf 0_R$
+}}
+{{eqn | r = \sqbrk {0 + a}_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk a_{m n}
+      | c = [[Identity Element of Addition on Numbers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf 0 + \mathbf A
+      | r = \mathbf A
+      | c = {{Defof|Zero Matrix}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Scalar Product is Associative}
+Tags: Matrix Scalar Product, Associativity, Matrix Scalar Product is Associative
+
+\begin{theorem}
+Let $\Bbb F$ denote one of the [[Definition:Standard Number System|standard number systems]].
+Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\Bbb F$.
+For $\mathbf A \in \map \MM {m, n}$ and $\lambda$ \in $\Bbb F$, let $\lambda \mathbf A$ be defined as the [[Definition:Matrix Scalar Product|matrix scalar product]] of $\lambda$ and $\mathbf A$.
+The [[Definition:Matrix Scalar Product|matrix scalar product]] is [[Definition:Associative Operation|associative]] on $\map \MM {m, n}$, in the following sense:
+For all $\mathbf A$ in $\map \MM {m, n}$ and $\lambda, \mu \in \Bbb F$:
+:$\lambda \paren {\mu \mathbf A} = \paren {\lambda \mu} \mathbf A$
+\end{theorem}<|endoftext|>
+\section{Matrix Scalar Product Distributes over Number Addition}
+Tags: Matrix Scalar Product, Addition, Distributive Operations, Matrix Scalar Product Distributes over Number Addition
+
+\begin{theorem}
+Let $\Bbb F$ denote one of the [[Definition:Standard Number System|standard number systems]].
+Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\Bbb F$.
+For $\mathbf A \in \map \MM {m, n}$ and $\lambda$ \in $\Bbb F$, let $\lambda \mathbf A$ be defined as the [[Definition:Matrix Scalar Product|matrix scalar product]] of $\lambda$ and $\mathbf A$.
+The [[Definition:Matrix Scalar Product|matrix scalar product]] is [[Definition:Associative Operation|associative]] on $\map \MM {m, n}$, in the following sense:
+For all $\mathbf A$ in $\map \MM {m, n}$ and $\lambda, \mu \in \Bbb F$:
+:$\paren {\lambda + \mu} \mathbf A = \lambda \mathbf A + \mu \mathbf A$
+\end{theorem}<|endoftext|>
+\section{Negative Matrix is Inverse for Matrix Entrywise Addition}
+Tags: Negative Matrices
+
+\begin{theorem}
+Let $\Bbb F$ denote one of the [[Definition:Standard Number System|standard number systems]].
+Let $\map \MM {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\Bbb F$.
+Let $\mathbf A$ be an [[Definition:Element|element]] of $\map \MM {m, n}$.
+Let $-\mathbf A$ be the [[Definition:Negative Matrix|negative]] of $\mathbf A$.
+Then $-\mathbf A$ is the [[Definition:Inverse Element|inverse]] for the operation $+$, where $+$ is [[Definition:Matrix Entrywise Addition|matrix entrywise addition]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map \MM {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A + \paren {-\mathbf A}
+      | r = \sqbrk a_{m n} + \paren {-\sqbrk a_{m n} }
+      | c = Definition of $\mathbf A$
+}}
+{{eqn | r = \sqbrk a_{m n} + \sqbrk {-a}_{m n}
+      | c = {{Defof|Negative Matrix}}
+}}
+{{eqn | r = \sqbrk {a + \paren {-a} }_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \sqbrk 0_{m n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A + \paren {-\mathbf A}
+      | r = \mathbf 0
+      | c = {{Defof|Zero Matrix}}
+}}
+{{end-eqn}}
+The result follows from [[Zero Matrix is Identity for Matrix Entrywise Addition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Matrix is Zero for Matrix Multiplication}
+Tags: Conventional Matrix Multiplication, Zero Matrix
+
+\begin{theorem}
+Let $\struct {R, +, \times}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\mathbf A$ be a [[Definition:Matrix|matrix]] over $R$ of [[Definition:Order of Matrix|order]] $m \times n$
+Let $\mathbf 0$ be a [[Definition:Zero Matrix|zero matrix]] whose [[Definition:Order of Matrix|order]] is such that either:
+:$\mathbf 0 \mathbf A$ is defined
+or:
+:$\mathbf A \mathbf 0$ is defined
+or both.
+Then:
+:$\mathbf 0 \mathbf A = \mathbf 0$
+or:
+:$\mathbf A \mathbf 0 = \mathbf 0$
+whenever they are defined.
+The [[Definition:Order of Matrix|order]] of $\mathbf 0$ will be according to the [[Definition:Order of Matrix|orders]] of the factor [[Definition:Matrix|matrices]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$ be [[Definition:Matrix|matrices]].
+Let $\mathbf 0 \mathbf A$ be defined.
+Then $\mathbf 0$ is of [[Definition:Order of Matrix|order]] $r \times m$ for $r \in \Z_{>0}$.
+Thus we have:
+{{begin-eqn}}     
+{{eqn | l = \mathbf 0 \mathbf A
+      | r = \mathbf C
+      | c = 
+}}
+{{eqn | l = \sqbrk 0_{r m} \sqbrk a_{m n}
+      | r = \sqbrk c_{r n}
+      | c = Definition of $\mathbf 0$ and $\mathbf A$
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \closedint 1 r, j \in \closedint 1 n:
+      | l = c_{i j}
+      | r = \sum_{k \mathop = 1}^m 0_{i k} \times a_{k j}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^m 0
+      | c = {{Defof|Zero Matrix}}
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf 0 \mathbf A
+      | r = \sqbrk 0_{r n}
+      | c = 
+}}
+{{end-eqn}}
+Hence $\mathbf 0 \mathbf A$ is the [[Definition:Zero Matrix|Zero Matrix]] of [[Definition:Order of Matrix|order]] $r \times n$.
+Let $\mathbf A \mathbf 0$ be defined.
+Then $\mathbf 0$ is of [[Definition:Order of Matrix|order]] $n \times s$ for $s \in \Z_{>0}$.
+Thus we have:
+{{begin-eqn}}     
+{{eqn | l = \mathbf A \mathbf 0
+      | r = \mathbf C
+      | c = 
+}}
+{{eqn | l = \sqbrk a_{m n} \sqbrk 0_{n s}
+      | r = \sqbrk c_{m s}
+      | c = Definition of $\mathbf A$ and $\mathbf 0$
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall i \in \closedint 1 m, j \in \closedint 1 s:
+      | l = c_{i j}
+      | r = \sum_{k \mathop = 1}^n a_{i k} \times 0_{k j}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n 0
+      | c = {{Defof|Zero Matrix}}
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf 0 \mathbf A
+      | r = \sqbrk 0_{m s}
+      | c = 
+}}
+{{end-eqn}}
+Hence $\mathbf A \mathbf 0$ is the [[Definition:Zero Matrix|Zero Matrix]] of [[Definition:Order of Matrix|order]] $m \times s$.
+{{qed|lemma}}
+If $\mathbf 0$ is of [[Definition:Order of Matrix|order]] $n \times m$,then both $\mathbf A \mathbf 0$ and $\mathbf 0 \mathbf A$ are defined, and:
+{{begin-eqn}}
+{{eqn | l = \mathbf A \mathbf 0
+      | r = \sqbrk 0_{m m}
+}}
+{{eqn | l = \mathbf 0 \mathbf A
+      | r = \sqbrk 0_{n n}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Identity for Matrix Multiplication}
+Tags: Conventional Matrix Multiplication, Unit Matrices, Unit Matrix is Identity for Matrix Multiplication
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map {\MM_R} n$ denote the [[Definition:Matrix Space|metric space]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $n$]] over $R$.
+Let $\mathbf I_n$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
+Then:
+:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
+That is, the [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ is the [[Definition:Identity Element|identity element]] for [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] over $\map {\MM_R} n$.
+\end{theorem}
+
+\begin{proof}
+=== [[Unit Matrix is Identity for Matrix Multiplication/Left|Lemma: Left Identity]] ===
+{{:Unit Matrix is Identity for Matrix Multiplication/Left}}{{qed|lemma}}
+=== [[Unit Matrix is Identity for Matrix Multiplication/Right|Lemma: Right Identity]] ===
+{{:Unit Matrix is Identity for Matrix Multiplication/Right}}{{qed|lemma}}
+Thus:
+$\mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
+Hence, by definition, $\mathbf I_n$ is an [[Definition:Identity Element|identity element]] for [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] over $\map {\MM_R} n$.
+That $\mathbf I_n$ is ''the'' [[Definition:Identity Element|identity element]] follows from [[Identity is Unique]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Identity for Matrix Multiplication/Left}
+Tags: Unit Matrix is Identity for Matrix Multiplication
+
+\begin{theorem}
+Let $\map {\MM_R} {m, n}$ denote the [[Definition:Matrix Space|$m \times n$ metric space]] over $R$.
+Let $I_m$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $m$.
+Then:
+:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf I_m \mathbf A  = \mathbf A$
+\end{theorem}
+
+\begin{proof}
+Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
+Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall i \in \closedint 1 m, j \in \closedint 1 n
+      | l = b_{i j}
+      | r = \sum_{k \mathop = 1}^m \delta_{i k} a_{k j}
+      | c = where $\delta_{i k}$ is the [[Definition:Kronecker Delta|Kronecker delta]]: $\delta_{i k} = 1_R$ when $i = k$ otherwise $0_R$
+}}
+{{eqn | r = a_{i j}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_m$ is shown to be a [[Definition:Left Identity|left identity]].
+{{qed}}
+[[Category:Unit Matrix is Identity for Matrix Multiplication]]
+csco6xzhtl62zukbf5e63iq3i03fg0u
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Identity for Matrix Multiplication/Right}
+Tags: Unit Matrix is Identity for Matrix Multiplication
+
+\begin{theorem}
+Let $\map {\MM_R} {m, n}$ denote the [[Definition:Matrix Space|$m \times n$ metric space]] over $R$.
+Let $I_n$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
+Then:
+:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf A \mathbf I_n  = \mathbf A$
+\end{theorem}
+
+\begin{proof}
+Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
+Let $\sqbrk b_{m n} = \sqbrk a_{m n} \mathbf I_n$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall i \in \closedint 1 m, j \in \closedint 1 n:
+      | l = b_{i j}
+      | r = \sum_{k \mathop = 1}^n a_{i k} \delta_{k j}
+      | c = where $\delta_{k j}$ is the [[Definition:Kronecker Delta|Kronecker delta]]: $\delta_{k j} = 1_R$ when $k = j$ otherwise $0_R$
+}}
+{{eqn | r = a_{i j}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a [[Definition:Right Identity|right identity]].
+{{qed}}
+[[Category:Unit Matrix is Identity for Matrix Multiplication]]
+8pmjc10owrj8op15gsw82jetlcmq0k4
+\end{proof}<|endoftext|>
+\section{Left and Right Inverses of Square Matrix over Field are Equal}
+Tags: Inverse Matrices
+
+\begin{theorem}
+Let $\Bbb F$ be a [[Definition:Field (Abstract Algebra)|field]], usually one of the [[Definition:Standard Number Field|standard number fields]] $\Q$, $\R$ or $\C$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map \MM n$ denote the [[Definition:Matrix Space|matrix space]] of [[Definition:Order of Square Matrix|order]] $n$ [[Definition:Square Matrix|square matrices]] over $\Bbb F$.
+Let $\mathbf B$ be a [[Definition:Left Inverse Matrix|left inverse matrix]] of $\mathbf A$.
+Then $\mathbf B$ is also a [[Definition:Right Inverse Matrix|right inverse matrix]] of $\mathbf A$.
+Similarly, let $\mathbf B$ be a [[Definition:Right Inverse Matrix|right inverse matrix]] of $\mathbf A$.
+Then $\mathbf B$ is also a [[Definition:Right Inverse Matrix|right inverse matrix]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\map \MM {m, n}, +, \circ}$, where:
+:$+$ denotes [[Definition:Matrix Entrywise Addition|matrix entrywise addition]]
+:$\circ$ denotes [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]].
+From [[Ring of Square Matrices over Field is Ring with Unity]], $\struct {\map \MM {m, n}, +, \circ}$ is a [[Definition:Ring with Unity|ring with unity]].
+Hence [[Definition:A Fortiori|a fortiori]] $\struct {\map \MM {m, n}, +, \circ}$ is a [[Definition:Monoid|monoid]].
+The result follows directly from [[Left Inverse and Right Inverse is Inverse]].
+{{qed}}
+{{mistake|That's not what it actually says. What the above link says is that ''if'' $\mathbf A$ has both a [[Definition:Right Inverse Matrix|right inverse matrix]] ''and'' a [[Definition:Left Inverse Matrix|left inverse matrix]], then those are equal and can be called an [[Definition:Inverse Matrix|inverse matrix]]. It does not say that if $\mathbf B$ is a [[Definition:Left Inverse Matrix|left inverse matrix]] then it is automatically a [[Definition:Right Inverse Matrix|right inverse matrix]]. I'll sort this out when I get to exercise $1.15$.}}
+\end{proof}<|endoftext|>
+\section{Inverse of Square Matrix over Field is Unique}
+Tags: Inverse Matrices
+
+\begin{theorem}
+Let $\Bbb F$ be a [[Definition:Field (Abstract Algebra)|field]], usually one of the [[Definition:Standard Number Field|standard number fields]] $\Q$, $\R$ or $\C$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map \MM n$ denote the [[Definition:Matrix Space|matrix space]] of [[Definition:Order of Square Matrix|order]] $n$ [[Definition:Square Matrix|square matrices]] over $\Bbb F$.
+Let $\mathbf B$ be an [[Definition:Inverse Matrix|inverse matrix]] of $\mathbf A$.
+Then $\mathbf B$ is [[Definition:Unique|the only]] [[Definition:Inverse Matrix|inverse matrix]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\map \MM {m, n}, +, \circ}$, where:
+:$+$ denotes [[Definition:Matrix Entrywise Addition|matrix entrywise addition]]
+:$\circ$ denotes [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]].
+From [[Ring of Square Matrices over Field is Ring with Unity]], $\struct {\map \MM {m, n}, +, \circ}$ is a [[Definition:Ring with Unity|ring with unity]].
+Hence [[Definition:A Fortiori|a fortiori]] $\struct {\map \MM {m, n}, +, \circ}$ is a [[Definition:Monoid|monoid]].
+The result follows directly from [[Inverse in Monoid is Unique]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rank of Empty Set is Zero}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\rho : \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Then:
+:$\map \rho \O = 0$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]]:
+:$\O$ is [[Definition:Independent Subset (Matroid)|independent]]
+From [[Rank of Independent Subset Equals Cardinality]]:
+:$\map \rho \O = \size \O$
+From [[Cardinality of Empty Set]]:
+:$\size \O = 0$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rank Function is Increasing}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\rho: \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $A, B \subseteq S$ be [[Definition:Subset|subsets]] of $S$ such that $A \subseteq B$.
+Then:
+:$\map \rho A \le \map \rho B$
+\end{theorem}
+
+\begin{proof}
+Now:
+{{begin-eqn}}
+{{eqn | l = \map \rho A
+      | r = \max \set {\size X : X \subseteq A \land X \in \mathscr I}
+      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
+}}
+{{eqn | r = \max \set {\size X : X \in \powerset A \land X \in \mathscr I}
+      | c = {{Defof|Power Set}} of $\O$
+}}
+{{eqn | r = \max \set {\size X : X \in \powerset A \cap \mathscr I}
+      | c = {{Defof|Set Intersection}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \map \rho B
+      | r = \max \set {\size X : X \in \powerset B \cap \mathscr I}
+}}
+{{end-eqn}}
+From [[Power Set of Subset]]:
+:$\powerset A \subseteq \powerset B$
+From [[Set Intersection Preserves Subsets]]:
+:$\powerset A \cap \mathscr I \subseteq \powerset B \cap \mathscr I$
+It follows that:
+:$\set {\size X : X \in \powerset A \cap \mathscr I} \subseteq \set {\size X : X \in \powerset B \cap \mathscr I}$
+From [[Max of Subfamily of Operands Less or Equal to Max]]:
+:$\max \set {\size X : X \in \powerset B \cap \mathscr I} \le \max \set {\size X : X \in \powerset B \cap \mathscr I}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounds for Rank of Subset}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\rho: \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $A \subseteq S$ be [[Definition:Subset|subset]] of $S$.
+Then:
+:$0 \le \map \rho A \le \size A$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Rank Function (Matroid)|rank function]]:
+{{begin-eqn}}
+{{eqn | l = \map \rho A
+      | r = \max \set {\size X : X \subseteq A \land X \in \mathscr I}
+}}
+{{end-eqn}}
+From [[Cardinality of Subset of Finite Set]]:
+:$\forall X \subseteq A : \size X \le \size A$
+In particular:
+:$\forall X \subseteq A : X \in \mathscr I$ then $\size X \le \size A$
+From [[Max yields Supremum of Operands]]:
+:$\max \set {\size X : X \subseteq A \land X \in \mathscr I} \le \size A$
+From [[Empty Set is Subset of All Sets]]:
+:$\O \subseteq A$
+From [[Cardinality of Empty Set]]:
+:$\size \O = 0$
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]]:
+:$\O \in \mathscr I$
+From [[Max yields Supremum of Operands]]:
+:$0 \le \max \set {\size X : X \subseteq A \land X \in \mathscr I}$
+It follows that:
+:$0 \le \map \rho A \le \size A$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid satisfies 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]].
+Then $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] of a [[Definition:Matroid|matroid]] on $S$ {{iff}} $\rho$ satisfies the [[Definition:Rank Axioms (Matroid)/Definition 1|rank axioms]]:
+{{:Definition:Rank Axioms (Matroid)/Definition 1}}
+\end{theorem}
+
+\begin{proof}
+=== [[Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Necessary Condition|Necessary Condition]] ===
+{{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Necessary Condition}}{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition|Sufficient Condition]] ===
+{{:Leigh.Samphier/Sandbox/Matroid satisfies Rank Axioms/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Transpose of Matrix is Transpose of Inverse}
+Tags: Transposes of Matrices, Inverse Matrices
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:Matrix|matrix]] over a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathbf A^\intercal$ denote the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Let $\mathbf A$ be an [[Definition:Invertible Matrix|invertible matrix]].
+Then $\mathbf A^\intercal$ is also [[Definition:Invertible Matrix|invertible]] and:
+:$\paren {\mathbf A^\intercal}^{-1} = \paren {\mathbf A^{-1} }^\intercal$
+where $\mathbf A^{-1}$ denotes the [[Definition:Inverse Matrix|inverse]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf A^{-1} }^\intercal \mathbf A^\intercal
+      | r = \paren {\mathbf A \mathbf A^{-1} }^\intercal
+      | c = [[Transpose of Matrix Product]]
+}}
+{{eqn | r = \mathbf I^\intercal
+      | c = {{Defof|Inverse Matrix}}: $\mathbf I$ denotes [[Definition:Unit Matrix|Unit Matrix]]
+}}
+{{eqn | r = \mathbf I
+      | c = {{Defof|Unit Matrix}}
+}}
+{{end-eqn}}
+Hence $\paren {\mathbf A^{-1} }^\intercal$ is an [[Definition:Inverse Matrix|inverse]] of $\mathbf A^\intercal$.
+From [[Inverse of Square Matrix over Field is Unique]]:
+:$\paren {\mathbf A^{-1} }^\intercal = \paren {\mathbf A^\intercal}^{-1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Max yields Supremum of Parameters/General Case}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $x_1, x_2, \dots ,x_n \in S$ for some $n \in \N_{>0}$.
+Then:
+:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
+\end{theorem}
+
+\begin{proof}
+We will prove the result by [[Principle of Mathematical Induction|induction]] on the [[Definition:Cardinality|number]] of [[Definition:Operand|operands]] $n$.
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$\max \set {x_1} = \sup \set {x_1}$
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_1} = x_1$
+From [[Supremum of Singleton]]:
+:$\sup \set {x_1} = x_1$
+So:
+:$\max \set {x_1} = \sup \set {x_1} = x_1$
+Thus $\map P 1$ 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]]:
+:$\max \set {x_1, x_2, \dots ,x_k} = \sup \set {x_1, x_2, \dots ,x_k}$
+from which it is to be shown that:
+:$\max \set {x_1, x_2, \dots ,x_k, x_{k+1}} = \sup \set {x_1, x_2, \dots ,x_k, x_{k+1}}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]].
+Now:
+{{begin-eqn}}
+{{eqn | l = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }
+      | r = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }
+      | c = {{Defof|Max Operation (General Case)}}
+}}
+{{eqn | r = \max \set {\sup \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }
+      | c = [[Max yields Supremum of Parameters/General Case#Induction Hypothesis|Induction hypothesis]]
+}}
+{{end-eqn}}
+As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], all elements of $S$ are [[Definition:Comparable|comparable]] by $\preceq$.
+Therefore there are two cases to consider:
+==== Case 1: $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$ ====
+Let $x_{k + 1} \preceq \sup \set {x_1, x_2, \dotsc, x_k}$.
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$
+By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k}$.
+That is:
+:$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
+Thus:
+:$\forall 1 \le i \le k + 1 : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
+So $\sup \set {x_1, x_2, \dots ,x_k}$ is an [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
+Let $y$ be any other [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
+Then:
+:$y$ is an [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k}$.
+By definition of the [[Definition:Supremum|supremum]]:
+:$\sup \set {x_1, x_2, \dotsc, x_k} \preceq y$
+It has been shown that the [[Definition:Supremum|supremum]] of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$ is:
+:$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k}$
+Thus it follows:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
+==== Case 2: $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$ ====
+Let $\sup \set {x_1, x_2, \dotsc, x_k} \preceq x_{k + 1}$.
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$
+By definition, $\sup \set {x_1, x_2, \dotsc, x_k}$ is an [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k}$.
+That is:
+:$\forall 1 \le i \le k : x_i \preceq \sup \set {x_1, x_2, \dotsc, x_k}$
+By definition of an [[Definition:Ordering|ordering]]:
+:$\preceq$ is [[Definition:Transitive Relation|transitive]].
+Thus:
+:$\forall 1 \le i \le k : x_i \preceq x_{k + 1}$
+By definition of an [[Definition:Ordering|ordering]]:
+:$\preceq$ is [[Definition:Reflexive Relation|reflexive]].
+Thus:
+:$x_{k + 1} \preceq x_{k + 1}$
+It follows that $x_{k + 1}$ is an [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
+Let $y$ be any other [[Definition:Upper Bound|upper bound]] of $\set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$.
+Then:
+:$x_{k + 1} \preceq y$.
+It has been shown that
+:$\sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = x_{k + 1}$.
+Thus it follows:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1}} = \sup \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
+In either case, the result holds.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
+{{qed}}
+[[Category:Max and Min Operations]]
+ldrb4v46u61m7u5znizcgonxjhj334k
+\end{proof}<|endoftext|>
+\section{Max of Subfamily of Operands Less or Equal to Max}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
+Let $\set{k_1, k_2, \dotsc, k_m} \subseteq \set{1, 2, \dotsc, n}$
+Then:
+:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} \preceq \max \set {x_1, x_2, \dotsc, x_n}$
+where:
+:$\max$ denotes the [[Definition:Max Operation|max operation]]
+\end{theorem}
+
+\begin{proof}
+From [[Max yields Supremum of Operands]]:
+:$\max \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m}} = \sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} }$
+and
+:$\max \set {x_1, x_2, \dotsc, x_n} = \sup \set {x_1, x_2, \dotsc, x_n}$
+Since $\set {k_1, k_2, \dotsc, k_m} \subseteq \set {1, 2, \dotsc, n}$ then:
+:$\set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \subseteq \set {x_1, x_2, \dotsc, x_n}$
+From [[Supremum of Subset]]:
+:$\sup \set {x_{k_1}, x_{k_2}, \dotsc, x_{k_m} } \preceq \sup \set {x_1, x_2, \dotsc, x_n}$
+{{qed}}
+[[Category:Max and Min Operations]]
+dlvc1y82jhfq4bosca2cw83mjb3m8al
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Row Operation}
+Tags: Elementary Row Operations, Elementary Matrices, Elementary Matrix corresponding to Elementary Row Operation
+
+\begin{theorem}
+Let $\mathbf I$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $m$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $e$ be an [[Definition:Elementary Row Operation|elementary row operation]] on $\mathbf I$.
+Let $\mathbf E$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] of [[Definition:Order of Square Matrix|order]] $m$ [[Definition:Unique|uniquely]] defined as:
+:$\mathbf E = e \paren {\mathbf I}$
+where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
+Let $r_k$ denote the $k$th [[Definition:Row of Matrix|row]] of $\mathbf I$ for $1 \le k \le m$.
+\end{theorem}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Row Operation/Scale Row}
+Tags: Elementary Matrix corresponding to Elementary Row Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 1
+        | t = For some $\lambda \in K_{\ne 0}$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Row of Matrix|row]] $k$ of $\mathbf I$ by $\lambda$
+        | m = r_k \to \lambda r_k
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $m$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $k$ of $\mathbf I$ are to be [[Definition:Ring Product|multiplied]] by $\lambda$.
+By definition of [[Definition:Unit Matrix|unit matrix]], all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $k$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{k k}$, which is $1$.
+Thus in $\mathbf E$:
+:$E_{k k} = \lambda \cdot 1 = \lambda$
+The [[Definition:Element of Matrix|elements]] in all the other [[Definition:Row of Matrix|rows]] of $\mathbf E$ are the same as the corresponding [[Definition:Element of Matrix|elements]] of $\mathbf I$.
+Hence the result.
+{{qed}}
+[[Category:Elementary Matrix corresponding to Elementary Row Operation]]
+6u2jxz2ubtu2iv2m2jq3xwayfy66vm4
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Row Operation/Scale Row and Add}
+Tags: Elementary Matrix corresponding to Elementary Row Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 2
+        | t = For some $\lambda \in K$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Row of Matrix|row]] $j$ to [[Definition:Row of Matrix|row]] $i$
+        | m = r_i \to r_i + \lambda r_j
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $m$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $i$ of $\mathbf I$ are to have the corresponding [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $j$ added to them after the latter have been [[Definition:Ring Product|multiplied]] by $\lambda$.
+By definition of [[Definition:Unit Matrix|unit matrix]]:
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $i$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{i i}$, which is $1$.
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $j$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{j j}$, which is $1$.
+Thus in $\mathbf E$:
+:where $a \ne i$, $E_{a b} = \delta_{a b}$
+:where $a = i$:
+::$E_{a b} = \delta_{a b}$ where $b \ne j$
+::$E_{a b} = \delta_{a b} + \lambda \cdot 1$ where $b = j$
+That is:
+:$E_{a b} = \delta_{a b}$ for all [[Definition:Element of Matrix|elements]] of $\mathbf E$
+except where $a = i$ and $b = j$, at which [[Definition:Element of Matrix|element]]:
+:$E_{a b} = \delta_{a b} + \lambda$
+That is:
+:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Row Operation/Exchange Rows}
+Tags: Elementary Matrix corresponding to Elementary Row Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Row Operation|elementary row operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 3
+        | t = Interchange [[Definition:Row of Matrix|rows]] $i$ and $j$
+        | m = r_i \leftrightarrow r_j
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $m$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $i$ of $\mathbf I$ are to be exchanged with the corresponding [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $j$.
+By definition of [[Definition:Unit Matrix|unit matrix]]:
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $i$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{i i}$, which is $1$.
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Row of Matrix|row]] $j$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{j j}$, which is $1$.
+Thus in $\mathbf E$:
+:where $a \ne i$ and $a \ne j$, $E_{a b} = \delta_{a b}$ (all [[Definition:Row of Matrix|rows]] except $i$ and $j$ are unchanged)
+:where $a = i$, $E_{a b} = \delta_{j b}$ (the contents of row $j$)
+:where $a = j$, $E_{a b} = \delta_{i b}$ (the contents of row $i$)
+That is:
+:$E_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$
+Hence the result.
+{{qed}}
+[[Category:Elementary Matrix corresponding to Elementary Row Operation]]
+swum7srrfe0166pzo0sd3i1yzcw3b9d
+\end{proof}<|endoftext|>
+\section{Power Set of Singleton}
+Tags: Power Set, Singletons
+
+\begin{theorem}
+Let $x$ be an [[Definition:Object|object]].
+Then the [[Definition:Power Set|power set]] of the [[Definition:Singleton|singleton]] $\set x$ is:
+:$\powerset {\set x} = \set {\O, \set x}$
+\end{theorem}
+
+\begin{proof}
+From [[Empty Set is Subset of All Sets]]:
+:$\O \in \powerset {\set x}$
+Let $A \in \powerset {\set x}$ such that $A \ne \O$
+That is:
+{{begin-eqn}}
+{{eqn | r = A \subseteq \set x \land A \ne \O
+      | o =
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | r = A \subseteq \set x \land \exists y : y \in A
+      | o =
+      | c = {{Defof|Empty Set}}
+}}
+{{eqn | ll= \leadsto
+      | r = A \subseteq \set x \land \exists y : y \in A \land y \in \set x
+      | o =
+      | c = {{Defof|Subset}}
+}}
+{{eqn | ll= \leadsto
+      | r = A \subseteq \set x \land \exists y : y \in A \land y = x
+      | o =
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | ll= \leadsto
+      | r = A \subseteq \set x \land x \in A
+      | o =
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | r = A \subseteq \set x \land \set x \subseteq A
+      | o =
+      | c = [[Singleton of Element is Subset]]
+}}
+{{eqn | ll= \leadsto
+      | r = A = \set x
+      | o =
+      | c = {{Defof|Set Equality}}
+}}
+{{end-eqn}}
+So a [[Definition:Subset|subset]] of $\set x$ is either $\O$ or $\set x$.
+{{qed}}
+[[Category:Power Set]]
+[[Category:Singletons]]
+0z1pqmxvgh8yira067g3l8s1prfj25s
+\end{proof}<|endoftext|>
+\section{Row Operation to Clear First Column of Matrix}
+Tags: Row Operations, Row Operation to Clear First Column of Matrix
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Matrix|$m \times n$ matrix]] over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Then there exists a [[Definition:Row Operation|row operation]] to convert $\mathbf A$ into another [[Definition:Matrix|$m \times n$ matrix]] $\mathbf B = \sqbrk b_{m n}$ with the following properties:
+:$(1): \quad$ Except possibly for [[Definition:Element of Matrix|element]] $b_{1 1}$, all the [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $1$ are $0$
+:$(2): \quad$ If $b_{1 1} \ne 0$, then $b_{1 1} = 1$.
+This process is referred to as '''clearing the first column'''.
+\end{theorem}
+
+\begin{proof}
+The following [[Definition:Algorithm|algorithm]] generates a [[Definition:Sequence|sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] which convert $\mathbf A$ to $\mathbf B$.
+Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the state of $\mathbf A$ after having processed the latest step.
+After each step, an implicit step can be included that requires that the form of $\mathbf A'$ is inspected to see if it is in the form $\mathbf B$, and if so, terminating the algorithm, but this is not essential.
+:$(1): \quad$ Are all [[Definition:Element of Matrix|elements]] in the first [[Definition:Column of Matrix|column]] of $\mathbf A$ equal to $0$?
+:::If so, there is nothing to do, and the required [[Definition:Row Operation|row operation]] is the [[Definition:Unit Matrix|unit matrix]] $\mathbf I_m$.
+:::Otherwise, move on to step $(2)$.
+:$(2): \quad$ Is [[Definition:Element of Matrix|element]] $a_{1 1}$ equal to $0$?
+:::If so:
+::::$\text (a): \quad$ find the smallest $k$ such that [[Definition:Row of Matrix|row]] $k$ of $\mathbf A$ such that $a_{k 1} \ne 0$
+::::$\text (b): \quad$ use the [[Definition:Elementary Row Operation|elementary row operation]] $r_1 \leftrightarrow r_k$ which will result $a'_{1 1} = a_{k 1}$ and $a'_{k 1} = 0$.
+:Move on to step $(3)$.
+:$(3): \quad$ Is [[Definition:Element of Matrix|element]] $a'_{1 1}$ equal to $1$?
+:::If so, use the [[Definition:Elementary Row Operation|elementary row operation]] $r_1 \to \lambda r_1$ where $\lambda = \dfrac 1 {a'_{1 1} }$, which will result $a'_{1 1} = 1$.
+:Move on to step $4$
+:$(4): \quad$ For each [[Definition:Row of Matrix|row]] $j$ from $2$ to $m$, do the following:
+:::Is $a_{j 1} \ne 0$?
+::::If so, use the [[Definition:Elementary Row Operation|elementary row operation]] $r_j \leftrightarrow r_j + \mu r_1$, where $\mu = -\dfrac {a'_{j 1} } {a'{1 1} }$, which will result in $a'_{j 1} = 0$.
+This will result in an [[Definition:Matrix|$m \times n$ matrix]] in the required form.
+Exercising the above [[Definition:Algorithm|algorithm]] will have generated a [[Definition:Sequence|sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $e_1, e_2, \ldots, e_t$.
+For each $e_k$ we create the [[Definition:Elementary Row Matrix|elementary row matrix]] $\mathbf E_k$.
+We then assemble the [[Definition:Matrix Product (Conventional)|matrix product]]:
+:$\mathbf R := \mathbf E_t \mathbf E_{t - 1} \mathbf E_{t - 2} \dotsm \mathbf E_2 \mathbf E_1$
+From [[Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices]], $\mathbf R$ is the resulting [[Definition:Matrix|$m \times m$ matrix]] corresponding to the [[Definition:Row Operation|row operation]] which is used to convert $\mathbf A$ to $\mathbf B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix is Row Equivalent to Echelon Matrix}
+Tags: Echelon Matrices, Matrix is Row Equivalent to Echelon Matrix
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_{m n}$ be a [[Definition:Matrix|matrix]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $F$.
+Then $A$ is [[Definition:Row Equivalence|row equivalent]] to an [[Definition:Echelon Matrix|echelon matrix]] of [[Definition:Order of Matrix|order]] $m \times n$.
+\end{theorem}
+
+\begin{proof}
+Using the operation [[Row Operation to Clear First Column of Matrix]], $\mathbf A$ is converted to $\mathbf B$, which will be in the form:
+:$\begin{bmatrix}
+0      & \cdots &      0 &      1 & b_{1, j + 1} & \cdots & b_{1 n} \\
+0      & \cdots &      0 &      0 & b_{2, j + 1} & \cdots & b_{2 n} \\
+\vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\
+0      & \cdots &      0 &      0 & b_{m, j + 1} & \cdots & b_{m n} \\
+\end{bmatrix}$
+If some [[Definition:Zero Row or Column|zero rows]] have appeared, do some further [[Definition:Elementary Row Operation|elementary row operations]], that is row interchanges, to put them at the bottom.
+We then address our attention to the [[Definition:Submatrix|submatrix]]:
+:$\begin{bmatrix}
+b_{2, j + 1} & b_{2, j + 2} & \cdots & b_{2 n} \\
+b_{3, j + 1} & b_{3, j + 2} & \cdots & b_{3 n} \\
+      \vdots &       \vdots & \ddots &  \vdots \\
+b_{m, j + 1} & b_{m, j + 2} & \cdots & b_{m n} \\
+\end{bmatrix}$
+and perform the same operation on that.
+This results in the [[Definition:Submatrix|submatrix]] being transformed into the form:
+:$\begin{bmatrix}
+     1 & c_{2, j + 2} & \cdots & c_{2 n} \\
+     0 & c_{3, j + 2} & \cdots & c_{3 n} \\
+\vdots &       \vdots & \ddots &  \vdots \\
+     0 & c_{m, j + 2} & \cdots & c_{m n} \\
+\end{bmatrix}$
+Again, we process the [[Definition:Submatrix|submatrix]]:
+:$\begin{bmatrix}
+c_{3, j + 2} & \cdots & c_{3 n} \\
+      \vdots & \ddots &  \vdots \\
+c_{m, j + 2} & \cdots & c_{m n} \\
+\end{bmatrix}$
+Thus we progress, until the entire [[Definition:Matrix|matrix]] is in [[Definition:Echelon Form|echelon form]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Singleton is Independent implies Rank is One/Corollary}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$\set x$ is an [[Definition:Independent Subset (Matroid)|independent subset]] {{iff}} $\map \rho {\set x} = 1$
+\end{theorem}
+
+\begin{proof}
+By definition of an [[Definition:Independent Subset (Matroid)|independent subset]]:
+:$x$ is an [[Definition:Independent Subset (Matroid)|independent subset]] {{iff}} $\set x \notin \mathscr I$
+From [[Singleton is Independent implies Rank is One]]:
+:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
+From [[Singleton is Dependent implies Rank is Zero]]:
+:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
+It follows that:
+:$\set x \in \mathscr I$ {{iff}} $\map \rho {\set x} = 1$
+{{qed}}
+[[Category:Matroid Theory]]
+rkwlr71vzp2a4bgrb7osuudni8p1ilv
+\end{proof}<|endoftext|>
+\section{Singleton is Independent implies Rank is One}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $x \in S$.
+Let $\set x$ be [[Definition:Independent Subset (Matroid)|independent]].
+Then:
+:$\map \rho {\set x} = 1$
+where $\rho$ denotes the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Rank of Independent Subset Equals Cardinality]]:
+:$\map \rho {\set x} = \size {\set x}$
+From [[Cardinality of Singleton]]:
+:$\size {\set x} = 1$
+The result follows.
+{{qed}}
+[[Category:Matroid Theory]]
+nmy3shzkoghepoempvcg6zhnep4tw1q
+\end{proof}<|endoftext|>
+\section{Singleton is Dependent implies Rank is Zero}
+Tags: Matroid Theory
+
+\begin{theorem}
+:$\map \rho {\set x} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:$\set x \notin \mathscr I$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \rho {\set x}
+      | r = \max \set{\size A : A \in \powerset {\set x} \land A \in \mathscr I} 
+      | c = {{Defof|Rank Function (Matroid)|Rank Function}}
+}}
+{{eqn | r = \max \set {\size A : A \in \set {\O, \set x} \land A \in \mathscr I} 
+      | c = [[Power Set of Singleton]]
+}}
+{{eqn | r = \max \set {\size \O}
+      | c = As $\set x \notin \mathscr I$ and [[Definition:Matroid Axioms|Matroid axiom $(\text I 1)$]] : $\O \in \mathscr I$
+}}
+{{eqn | r = \max \set 0
+      | c = [[Cardinality of Empty Set]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Max Operation}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Matroid Theory]]
+51xabv2osf8qf1m5mgl08foem8ue5b2
+\end{proof}<|endoftext|>
+\section{System of Simultaneous Equations may have No Solution}
+Tags: Simultaneous Equations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simultaneous Equations|system of simultaneous equations]].
+Then it is possible that $S$ may have a [[Definition:Solution Set to System of Simultaneous Equations|solution set]] which is [[Definition:Empty Set|empty]].
+\end{theorem}
+
+\begin{proof}
+Consider this [[Simultaneous Linear Equations/Examples/Arbitrary System 2|system of simultaneous linear equations]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = x_1 + x_2
+      | r = 2
+}}
+{{eqn | n = 2
+      | l = 2 x_1 + 2 x_2
+      | r = 3
+}}
+{{end-eqn}}
+From its [[Simultaneous Linear Equations/Examples/Arbitrary System 2|evaluation]] it is seen to have no [[Definition:Solution to System of Simultaneous Equations|solutions]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{System of Simultaneous Equations may have Unique Solution}
+Tags: Simultaneous Equations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simultaneous Equations|system of simultaneous equations]].
+Then it is possible that $S$ may have a [[Definition:Solution Set to System of Simultaneous Equations|solution set]] which is a [[Definition:Singleton|singleton]].
+\end{theorem}
+
+\begin{proof}
+Consider this [[Simultaneous Linear Equations/Examples/Arbitrary System 1|system of simultaneous linear equations]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = x_1 - 2 x_2 + x_3
+      | r = 1
+}}
+{{eqn | n = 2
+      | l = 2 x_1 - x_2 + x_3
+      | r = 2
+}}
+{{eqn | n = 3
+      | l = 4 x_1 + x_2 - x_3
+      | r = 1
+}}
+{{end-eqn}}
+From its [[Simultaneous Linear Equations/Examples/Arbitrary System 1|evaluation]] it has the following [[Definition:Unique|unique]] [[Definition:Solution to System of Simultaneous Equations|solution]]:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | r = -\dfrac 1 2
+}}
+{{eqn | l = x_2
+      | r = \dfrac 1 2
+}}
+{{eqn | l = x_3
+      | r = \dfrac 3 2
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{System of Simultaneous Equations may have Multiple Solutions}
+Tags: Simultaneous Equations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simultaneous Equations|system of simultaneous equations]].
+Then it is possible that $S$ may have a [[Definition:Solution Set to System of Simultaneous Equations|solution set]] which is a [[Definition:Singleton|singleton]].
+\end{theorem}
+
+\begin{proof}
+Consider this [[Simultaneous Linear Equations/Examples/Arbitrary System 1|system of simultaneous linear equations]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = x_1 - 2 x_2 + x_3
+      | r = 1
+}}
+{{eqn | n = 2
+      | l = 2 x_1 - x_2 + x_3
+      | r = 2
+}}
+{{end-eqn}}
+From its [[Simultaneous Linear Equations/Examples/Arbitrary System 2|evaluation]] it has the following [[Definition:Solution to System of Simultaneous Equations|solutions]]:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | r = 1 - \dfrac t 3
+}}
+{{eqn | l = x_2
+      | r = \dfrac t 3
+}}
+{{eqn | l = x_3
+      | r = t
+}}
+{{end-eqn}}
+where $t$ is any [[Definition:Number|number]].
+Hence the are as many [[Definition:Solution to System of Simultaneous Equations|solutions]] as the [[Definition:Cardinality|cardinality]] of the [[Definition:Domain of Variable|domain]] of $t$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach}
+Tags: Absolute Convergence, Banach Spaces, Absolutely Convergent Series is Convergent
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $X$.
+Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]] {{iff}} $X$ is a [[Definition:Banach Space|Banach space]].
+\end{theorem}
+
+\begin{proof}
+=== [[Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Necessary Condition|Necessary Condition]] ===
+{{:Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Necessary Condition}}{{qed|lemma}}
+=== [[Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition|Sufficient Condition]] ===
+{{:Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Necessary Condition}
+Tags: Absolute Convergence, Banach Spaces, Absolutely Convergent Series is Convergent
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $X$.
+Suppose $X$ is a [[Definition:Banach Space|Banach space]].
+Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]].
+\end{theorem}
+
+\begin{proof}
+That $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Absolutely Convergent Series|absolutely convergent]] means that $\displaystyle \sum_{n \mathop = 1}^\infty \norm {a_n}$ [[Definition:Convergent Series|converges]] in $\R$.
+Hence the sequence of [[Definition:Partial Sum|partial sums]] is a [[Definition:Cauchy Sequence|Cauchy sequence]] by [[Convergent Sequence in Normed Vector Space is Cauchy Sequence]].
+Now let $\epsilon > 0$.
+Let $N \in \N$ such that for all $m, n \in \N$, $m \ge n \ge N$ implies that:
+:$\displaystyle \sum_{k \mathop = n + 1}^m \norm {a_k} = \size {\sum_{k \mathop = 1}^m \norm {a_k} - \sum_{k \mathop = 1}^n \norm {a_k} } < \epsilon$
+This $N$ exists because the sequence is [[Definition:Cauchy Sequence|Cauchy]].
+Now observe that, for $m \ge n \ge N$, one also has:
+{{begin-eqn}}
+{{eqn | l = \norm {\sum_{k \mathop = 1}^m a_k - \sum_{k \mathop = 1}^n a_k}
+      | r = \norm {\sum_{k \mathop = n + 1}^m a_k}
+}}
+{{eqn | o = \le
+      | r = \sum_{k \mathop = n + 1}^m \norm {a_k}
+      | c = [[Definition:Norm on Vector Space|Triangle inequality]] for $\norm {\, \cdot \,}$
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+It follows that the [[Definition:Series|sequence of partial sums]] of $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Cauchy Sequence|Cauchy]].
+As $X$ is a [[Definition:Banach Space|Banach space]], this implies that $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ [[Definition:Convergent Series|converges]].
+\end{proof}<|endoftext|>
+\section{Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach/Sufficient Condition}
+Tags: Absolute Convergence, Banach Spaces, Absolutely Convergent Series is Convergent
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be an [[Definition:Absolutely Convergent Series|absolutely convergent series]] in $X$.
+Suppose $\ds \sum_{n \mathop = 1}^\infty a_n$ is [[Definition:Convergent Series|convergent]].
+Then $X$ 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 $X$.
+We have that:
+:$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \norm {x_n - x_m} < \epsilon$
+We will prove the existence of a [[Definition:Subsequence|subsequence]] $\sequence {x_{n_k} }_{k \mathop \in \N}$ such that:
+:$n > n_k \implies \norm {x_n - x_{n_k} } < \dfrac 1 {2^k}$
+=== Basis for the Induction ===
+Let $\epsilon = \dfrac 1 2$.
+Choose $n_1$ such that:
+:$n, m \ge n_1 \implies \norm {x_n - x_m} < \dfrac 1 2$
+In particular, when $m = n_1$:
+:$n \ge n_1 \implies \norm {x_n - x_{n_1} } < \dfrac 1 2$
+=== Induction Step ===
+Suppose $x_{n_1}, \dots x_{n_k}$ have been constructed.
+Let $\epsilon = \dfrac 1 {2^{k + 1} }$.
+Choose $n_{k + 1}$ such that $n_{k + 1} > n_k$ and:
+:$n, m \ge n_{k + 1} \implies \norm {x_n - x_m} < \dfrac 1 {2^{k + 1} }$
+In particular, when $m = n_{k + 1}$:
+:$n \ge n_{k + 1} \implies \norm {x_n - x_{n_{k + 1} } } < \dfrac 1 {2^{k + 1}}$
+{{qed|lemma}}
+Define:
+:$u_1 := x_{n_1}$
+:$u_{k + 1} := x_{n_{k + 1} } - x_{n_k}$
+Now we have a [[Definition:Sequence|sequence]] $\sequence {u_k}_{k \mathop \in \N}$.
+Consider the [[Definition:Series|series]] $\ds \sum_{k \mathop = 1}^\infty \norm {u_k}$:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^\infty \norm {u_k}
+      | r = \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {u_k}
+}}
+{{eqn | r = \norm {u_1} + \sum_{k \mathop = 2}^\infty \norm {x_{n_k} - x_{n_{k - 1} } }
+}}
+{{eqn | r = \norm {u_1} + \sum_{k \mathop = 1}^\infty \norm {x_{n_{k + 1} } - x_{n_k} }
+      | c = Relabeling: $k \to k + 1$
+}}
+{{eqn | o = \le
+      | r = \norm {x_{n_1} } + \sum_{k \mathop = 1}^\infty \frac 1 {2^k}
+}}
+{{eqn | o = \le
+      | r = \norm {x_{n_1} } + 1
+}}
+{{eqn | o = <
+      | r = \infty
+}}
+{{end-eqn}}
+Thus, $\ds \sum_{k \mathop = 1}^\infty u_k$ is [[Definition:Absolutely Convergent Series|absolutely convergent]].
+By assumption in the theorem, $\ds \sum_{k \mathop = 1}^\infty u_k$ is [[Definition:Convergent Series/Normed Vector Space/Definition 2|convergent]].
+In other words:
+:$\ds \lim_{k \mathop \to \infty} \sum_{j \mathop = 1}^k u_j = u$.
+On the other hand:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^k u_j
+      | r = x_{n_1} + \sum_{j \mathop = 2}^k \paren {x_{n_j} - x_{n_{j \mathop - 1} } }
+}}
+{{eqn | r = x_{n_k}
+      | c = [[Definition:Telescoping Series|Telescoping Series]]
+}}
+{{end-eqn}}
+Therefore:
+:$\ds \lim_{k \mathop \to \infty} x_{n_k} = u =: x$
+So $\sequence {x_{n_k} }_{k \mathop \in \N}$ [[Definition:Convergent Series/Normed Vector Space/Definition 2|converges]] in $X$.
+We have that $\sequence {x_{n_k} }_{k \mathop \in \N}$ is a [[Definition:Convergent Sequence in Normed Vector Space|convergent]] [[Definition:Subsequence|subsequence]] of a [[Definition:Cauchy Sequence|Cauchy sequence]]$\sequence {x_n}_{n \mathop \in \N}$.
+By [[Convergent Subsequence of Cauchy Sequence in Normed Vector Space]], $\sequence {x_n}_{n \mathop \in \N}$ is [[Definition:Convergent Sequence in Normed Vector Space|convergent]] with the same [[Definition:Limit of Sequence in Normed Vector Space|limit]] $x$.
+By definition, the underlying [[Definition:Normed Vector Space|space]] is [[Definition:Banach Space|Banach]].
+\end{proof}<|endoftext|>
+\section{Sine of Integer Multiple of Argument/Formulation 2}
+Tags: Sine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \sin n \theta
+      | r = \cos^n \theta \paren {\paren {\tan \theta} - \dbinom n 3 \paren {\tan \theta}^3 + \dbinom n 5 \paren {\tan \theta}^5 - \cdots}
+      | c = 
+}}
+{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By [[De Moivre's Formula]]:
+:$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$
+As $n \in \Z_{>0}$, we use the [[Binomial Theorem]] on the {{RHS}}, resulting in:
+:$\displaystyle \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$
+When $k$ is [[Definition:Odd Integer|odd]], the expression being summed is [[Definition:Imaginary Number|imaginary]].
+Equating the [[Definition:Imaginary Part|imaginary parts]] of both sides of the equation, replacing $k$ with $2 k + 1$ to make $k$ [[Definition:Odd Integer|odd]], gives:
+{{begin-eqn}}
+{{eqn | l = \sin n \theta
+      | r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\cos^{n - \paren {2 k + 1} } \theta} \paren {\sin^{2 k + 1} \theta}
+      | c = 
+}}
+{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k + 1} \paren {\tan^{2 k + 1} \theta}
+      | c = factor out $\cos^n \theta$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Sine of Integer Multiple of Argument]]
+ixx0qmzthb6kv3ovxu3cq4c13kqywd1
+\end{proof}<|endoftext|>
+\section{Trivial Solution to System of Homogeneous Simultaneous Linear Equations is Solution}
+Tags: Simultaneous Linear Equations
+
+\begin{theorem}
+Let $S$ be a '''system of [[Definition:Homogeneous Simultaneous Linear Equations|homogeneous simultaneous linear equations]]''':
+:$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
+Consider the [[Definition:Trivial Solution to Homogeneous Simultaneous Linear Equations|trivial solution]] to $A$:
+:$\tuple {x_1, x_2, \ldots, x_n}$
+such that:
+:$\forall j \in \set {1, 2, \ldots, n}: x_j = 0$
+Then the [[Definition:Trivial Solution to Homogeneous Simultaneous Linear Equations|trivial solution]] is indeed a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $S$.
+\end{theorem}
+
+\begin{proof}
+Let $i \in \set {1, 2, \ldots, m}$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^n \alpha_{i j} x_j
+      | r = \sum_{j \mathop = 1}^n \alpha_{i j} \times 0
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+This holds for all $i \in \set {1, 2, \ldots, m}$.
+Hence:
+:$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = 0$
+and the result follows.
+{{qed}}
+[[Category:Simultaneous Linear Equations]]
+6rej10oi787y9zs6gtvdti9fplbgdqm
+\end{proof}<|endoftext|>
+\section{Sine of Integer Multiple of Argument/Formulation 3}
+Tags: Sine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \sin n \theta
+      | r = \sin \theta \cos^{n - 1} \theta \paren {1 + 1 + \frac {\cos 2 \theta} {\cos^2 \theta} + \frac {\cos 3 \theta} {\cos^3 \theta} + \cdots + \frac {\cos \paren {n - 1} \theta} {\cos^{n - 1} \theta} }
+      | c = 
+}}
+{{eqn | r = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop \ge 0} \frac {\cos k \theta} {\cos^k \theta}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sin \theta
+      | r = \sin \theta
+      | c = 
+}}
+{{eqn | r = \sin \theta \cos^{1 - 1} \theta \paren 1
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 1$ is seen to hold.
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sin 2 \theta
+      | r = 2 \sin \theta \cos \theta
+      | c = [[Double Angle Formula for Sine]]
+}}
+{{eqn | r = \sin \theta \cos^{2 - 1} \theta \paren {1 + 1}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 2$ is also seen to hold.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \sin {n \theta} = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta } {\cos^k \theta}$
+from which we are to show:
+:$\displaystyle \map \sin {\paren {n + 1} \theta} = \sin \theta \cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+For the first part:
+{{begin-eqn}}
+{{eqn | l = \map \sin {\paren {n + 1} \theta} 
+      | r = \map \sin {n \theta + \theta}
+      | c =
+}}
+{{eqn | r = \sin n \theta \cos \theta + \cos n \theta \sin \theta
+      | c = [[Sine of Sum]]
+}}
+{{eqn | r = \paren {\sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} } \cos \theta + \cos n \theta \sin \theta
+      | c = 
+}}
+{{eqn | r = \sin \theta \paren {\cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta} \cos \theta + \cos n \theta}
+      | c = Factor out $\sin \theta$
+}}
+{{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}  + \cos n \theta}
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^{n - 1} \frac {\cos k \theta} {\cos^k \theta}  + \cos n \theta \frac {\cos^n \theta} {\cos^n \theta} }
+      | c = multiply by $1$
+}}
+{{eqn | r = \sin \theta \paren {\cos^n \theta \sum_{k \mathop = 0}^n \frac {\cos k \theta} {\cos^k \theta} }
+      | c = 
+}}
+{{end-eqn}}
+The result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{>0}: \sin n \theta = \sin \theta \cos^{n - 1} \theta \sum_{k \mathop = 0}^{n - 1 } \frac {\cos k \theta} {\cos^k \theta}$
+{{qed}}
+[[Category:Sine of Integer Multiple of Argument]]
+p9h2uid0r5dmsd684420g9d55w5otf2
+\end{proof}<|endoftext|>
+\section{Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations}
+Tags: Elementary Row Operations, Simultaneous Linear Equations, Elementary Row Operation on Augmented Matrix leads to Equivalent System of Simultaneous Linear Equations
+
+\begin{theorem}
+Let $S$ be a system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]]:
+:$\displaystyle \forall i \in \set {1, 2, \ldots, m}: \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
+Let $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ denote the [[Definition:Augmented Matrix of Simultaneous Linear Equations|augmented matrix]] of $S$.
+Let $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ be obtained from $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ by means of an [[Definition:Elementary Row Operation|elementary row operation]].
+Let $S'$ be the system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]] of which $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ is the [[Definition:Augmented Matrix of Simultaneous Linear Equations|augmented matrix]].
+Then $S$ and $S'$ are [[Definition:Equivalent Systems of Simultaneous Linear Equations|equivalent]].
+\end{theorem}
+
+\begin{proof}
+We have that an [[Definition:Elementary Row Operation|elementary row operation]] $e$ is used to transform $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ to $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$.
+Now, whatever $e$ is, $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ is the [[Definition:Augmented Matrix of Simultaneous Linear Equations|augmented matrix]] of a system of [[Definition:Simultaneous Linear Equations|simultaneous linear equations]] $S'$.
+We investigate each type of [[Definition:Elementary Row Operation|elementary row operation]] in turn.
+In the below, let:
+:$r_k$ denote [[Definition:Row of Matrix|row]] $k$ of $\mathbf A$
+:$r'_k$ denote [[Definition:Row of Matrix|row]] $k$ of $\mathbf A'$
+for arbitrary $k$ such that $1 \le k \le m$.
+By definition of [[Definition:Elementary Row Operation|elementary row operation]], only the [[Definition:Row of Matrix|row]] or [[Definition:Row of Matrix|rows]] directly operated on by $e$ is or are different between $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ and $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$.
+Hence it is understood that in the following, only those [[Definition:Equation|equations]] corresponding to those [[Definition:Row of Matrix|rows]] directly affected will be under consideration.
+=== $\text {ERO} 1$: Scalar Product of Row ===
+Let $e \begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \to \lambda r_k$
+where $\lambda \ne 0$.
+Then the [[Definition:Equation|equation]] in $S$:
+:$(1 \text a): \displaystyle \sum_{i \mathop = 1}^n \alpha_{k i} x_i = \beta_k$
+is replaced in $S'$ by:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n \lambda \alpha_{k i} x_i
+      | r = \lambda \beta_k
+      | c = 
+}}
+{{eqn | n = 2a
+      | ll= \leadsto
+      | l = \lambda \sum_{i \mathop = 1}^n \alpha_{k i} x_i
+      | r = \lambda \beta_k
+      | c = 
+}}
+{{end-eqn}}
+It is seen that $\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(1 \text a)$ {{iff}} $\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(2 \text a)$.
+{{qed|lemma}}
+=== $\text {ERO} 2$: Add Scalar Product of Row to Another ===
+Let $e \begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \to r_k + \lambda r_l$
+Then the [[Definition:Equation|equation]] in $S$:
+:$(1 \text b): \displaystyle \sum_{i \mathop = 1}^n \alpha_{k i} x_i = \beta_k$
+is replaced in $S'$ by:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n \paren {\alpha_{k i} + \lambda \alpha_{l i} } x_i
+      | r = \beta_k + \lambda \beta_l
+      | c = 
+}}
+{{eqn | n = 2b
+      | ll= \leadsto
+      | l = \sum_{i \mathop = 1}^n \alpha_{k i} x_i + \lambda \sum_{i \mathop = 1}^n \alpha_{l i} x_i
+      | r = \lambda \beta_k + \lambda \beta_l
+      | c = 
+}}
+{{end-eqn}}
+It is seen that $\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(1 \text b)$ {{iff}} $\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(2 \text b)$.
+{{qed|lemma}}
+=== $\text {ERO} 3$: Exchange Rows ===
+Let $e \begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \leftrightarrow r_l$
+Then the [[Definition:Equation|equations]] in $S$:
+{{begin-eqn}}
+{{eqn | n = 1c
+      | l = \sum_{i \mathop = 1}^n \alpha_{k i} x_i
+      | r = \beta_k
+      | c = 
+}}
+{{eqn | n = 2c
+      | l = \sum_{i \mathop = 1}^n \alpha_{l i} x_i
+      | r = \beta_l
+      | c = 
+}}
+{{end-eqn}}
+exist unchanged in $S'$, but are in different positions.
+It follows trivially that:
+:$\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(1 \text c)$ in $S$
+{{iff}}:
+:$\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(1 \text c)$ in $S'$
+and:
+:$\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(2 \text c)$ in $S$
+{{iff}}:
+:$\tuple {x_1, x_2, \ldots x_n}$ is a [[Definition:Solution to Simultaneous Linear Equations|solution]] to $(2 \text c)$ in $S'$.
+{{qed|lemma}}
+Thus in all cases, for each [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$ to $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$, $S$ is [[Definition:Equivalent Systems of Simultaneous Linear Equations|equivalent]] to $S'$.
+Finally we note that from [[Existence of Inverse Elementary Row Operation]], there exists an [[Definition:Elementary Row Operation|elementary row operation]] $e'$ which transforms $\begin {pmatrix} \mathbf A' & \mathbf b' \end {pmatrix}$ to $\begin {pmatrix} \mathbf A & \mathbf b \end {pmatrix}$.
+Hence, [[Definition:Mutatis Mutandis|mutatis mutandis]], the above argument can be used to demonstrate that $S'$ is [[Definition:Equivalent Systems of Simultaneous Linear Equations|equivalent]] to $S$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Row Operation}
+Tags: Elementary Row Operations, Existence of Inverse Elementary Row Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be an [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Row Operation|inverse]] of $e$.
+Then $e'$ is an [[Definition:Elementary Row Operation|elementary row operation]] which always exists and is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let us take each type of [[Definition:Elementary Row Operation|elementary row operation]] in turn.
+For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new [[Definition:Matrix|matrix]] $\mathbf A'' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
+In the below, let:
+:$r_k$ denote [[Definition:Row of Matrix|row]] $k$ of $\mathbf A$
+:$r'_k$ denote [[Definition:Row of Matrix|row]] $k$ of $\mathbf A'$
+:$r''_k$ denote [[Definition:Row of Matrix|row]] $k$ of $\mathbf A''$
+for arbitrary $k$ such that $1 \le k \le m$.
+By definition of [[Definition:Elementary Row Operation|elementary row operation]]:
+:only the [[Definition:Row of Matrix|row]] or [[Definition:Row of Matrix|rows]] directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
+and similarly:
+:only the [[Definition:Row of Matrix|row]] or [[Definition:Row of Matrix|rows]] directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A''$.
+Hence it is understood that in the following, only those [[Definition:Row of Matrix|rows]] directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$.
+=== [[Existence of Inverse Elementary Row Operation/Scalar Product of Row|$\text {ERO} 1$: Scalar Product of Row]] ===
+{{:Existence of Inverse Elementary Row Operation/Scalar Product of Row}}{{qed|lemma}}
+=== [[Existence of Inverse Elementary Row Operation/Add Scalar Product of Row to Another|$\text {ERO} 2$: Add Scalar Product of Row to Another]] ===
+{{:Existence of Inverse Elementary Row Operation/Add Scalar Product of Row to Another}}{{qed|lemma}}
+=== [[Existence of Inverse Elementary Row Operation/Exchange Rows|$\text {ERO} 3$: Exchange Rows]] ===
+{{:Existence of Inverse Elementary Row Operation/Exchange Rows}}{{qed|lemma}}
+Thus in all cases, for each [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A'$ to $\mathbf A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Dependent Set is Dependent}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A, B \subseteq S$ such that $A \subseteq B$
+If $A$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]] then $B$ is a [[Definition:Dependent Subset (Matroid)|dependent subset]].
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]]:
+:$A \notin \mathscr I \implies B \notin \mathscr I$
+By the definition of a [[Definition:Dependent Subset (Matroid)|dependent subset]]:
+:If $A$ is not an [[Definition:Dependent Subset (Matroid)|dependent subset]] then $B$ is not an [[Definition:Dependent Subset (Matroid)|dependent subset]].
+{{qed}}
+[[Category:Matroid Theory]]
+37pbg3iuv6mpq72x8ovqrpjny6b0uit
+\end{proof}<|endoftext|>
+\section{Powers of 16 Modulo 20}
+Tags: Powers of 16, Modulo Arithmetic, Powers of 16 Modulo 20
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then:
+:$16^n \equiv 16 \pmod {20}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$16^n \equiv 16 \pmod {20}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$16^1 \equiv 16 \pmod {20}$
+Thus $\map P 1$ 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]]:
+:$16^k \equiv 16 \pmod {20}$
+from which it is to be shown that:
+:$16^{k + 1} \equiv 16 \pmod {20}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+We have:
+{{begin-eqn}}
+{{eqn | l = 16^{k + 1}
+      | r = 16^k \times 16
+}}
+{{eqn | o = \equiv
+      | r = 16 \times 16
+      | rr= \pmod {20}
+      | c = [[Powers of 16 Modulo 20#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | o = \equiv
+      | r = 256
+      | rr= \pmod {20}
+}}
+{{eqn | o = \equiv
+      | r = 16
+      | rr= \pmod {20}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and thus it follows by the [[Principle of Mathematical Induction]] that:
+:$\forall n \in \Z_{> 0}: 16^n \equiv 16 \pmod {20}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 16
+      | o = \equiv
+      | r = 16
+      | rr= \pmod {20}
+}}
+{{eqn | ll= \leadsto
+      | l = 16
+      | o = \equiv
+      | r = 0
+      | rr= \pmod 4
+}}
+{{eqn | lo= \text {and}
+      | l = 16
+      | o = \equiv
+      | r = 1
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = 16^n
+      | o = \equiv
+      | r = 0
+      | rr= \pmod 4
+}}
+{{eqn | lo= \text {and}
+      | l = 16^n
+      | o = \equiv
+      | r = 1
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = 16^n
+      | o = \equiv
+      | r = 16
+      | rr= \pmod {20}
+      | c = [[Chinese Remainder Theorem]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sine of Integer Multiple of Argument/Formulation 1/Lemma}
+Tags: Sine of Integer Multiple of Argument
+
+\begin{theorem}
+:For $n \in \Z$:
+{{begin-eqn}}
+{{eqn | l = \map \cos {n \theta} \map \sin {\theta} 
+      | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta} 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {n \theta} \map \sin {\theta} 
+      | r = \map \cos {n \theta} \map \sin {\theta} 
+}}
+{{eqn | r = \paren {\map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta} \map \cos {\theta} } + \map \cos {n \theta} \map \sin {\theta} 
+      | c = add zero
+}}
+{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \paren {\map \sin {n \theta} \map \cos {\theta} - \map \cos {n \theta} \map \sin {\theta} }
+      | c = regroup
+}}
+{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {n \theta - \theta}
+      | c = [[Sine of Difference]]
+}}
+{{eqn | r = \map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1} \theta}
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Sine of Integer Multiple of Argument]]
+9qv5qxvp6c5iv9vkuw2i3uqmbjdxrh4
+\end{proof}<|endoftext|>
+\section{Conjugacy Class of Identity is only Conjugacy Class which is Subgroup}
+Tags: Conjugacy Classes
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $e$ denote the [[Definition:Identity Element|identity]] of $G$.
+Let $\conjclass g$ denote the [[Definition:Conjugacy Class|conjugacy class]] of the element $g$.
+Then conjugacy class of identity is the only conjugacy class which is a [[Definition:Subgroup|subgroup]] of $G$:
+:$\conjclass g < G \iff g = e$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Assume $g = e$.
+Then by [[Identity of Group is in Singleton Conjugacy Class]], $\conjclass e = \set e$, which is the [[Definition:Trivial Subgroup|trivial subgroup]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Assume $g \neq e$.
+Then by [[Conjugacy Classes are Disjoint]], $e \notin \conjclass g$.
+Since $\conjclass g$ does not contain the [[Definition:Identity Element|identity element]], it could not be a [[Definition:Subgroup|subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Empty Group Word is Reduced}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]]
+Let $\epsilon$ be the [[Definition:Empty Group Word|empty group word]] on $S$.
+Then $\epsilon$ is [[Definition:Reduced Group Word on Set|reduced]].
+\end{theorem}
+
+\begin{proof}
+{{improve|Rather than mentioning them in passing and expecting the reader to understand what is meant, we need $w_i$ etc. to be defined. I know from experience that you can't rely on people involved in this website to have adequate knowledge, understanding or deductive abilities to be able to work it out for themselves.}}
+By definition, a [[Definition:Group Word on Set|group word]] $w = w_1 \cdots w_i \cdots w_n$ is [[Definition:Reduced Group Word on Set|reduced]] {{iff}} $w_i \ne w_{i + 1}^{-1}$ for all $i \in \set {1, \ldots, n - 1}$, which is [[Definition:Vacuous Truth|vacuously true]] for $\epsilon$.
+{{qed}}
+[[Category:Abstract Algebra]]
+i739mtq52b4yqdd9s8x4u2bfbp5t96q
+\end{proof}<|endoftext|>
+\section{Scalar Multiplication Corresponds to Multiplication by 1x1 Matrix}
+Tags: Matrix Scalar Product, Conventional Matrix Multiplication
+
+\begin{theorem}
+Let $\map \MM 1$ denote the [[Definition:Matrix Space|matrix space]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order]] $1$.
+Let $\map \MM {1, n}$ denote the [[Definition:Matrix Space|matrix space]] of [[Definition:Order of Square Matrix|order]] $1 \times n$.
+Let $\mathbf A = \begin {pmatrix} a \end {pmatrix} \in \map \MM 1$ and $\mathbf B = \begin {pmatrix} b_1 & b_2 & \cdots & b_n \end{pmatrix} \in \map \MM {1, n}$.
+Let $\mathbf C = \mathbf A \mathbf B$ denote the [[Definition:Matrix Product (Conventional)|(conventional) matrix product]] of $\mathbf A$ with $\mathbf B$.
+Let $\mathbf D = a \mathbf B$ denote the [[Definition:Matrix Scalar Product|matrix scalar product]] of $a$ with $\mathbf B$.
+Then $\mathbf C = \mathbf D$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Matrix Product (Conventional)|(conventional) matrix product]], $\mathbf C$ is of [[Definition:Order of Matrix|order]] $1 \times n$.
+By definition of [[Definition:Matrix Scalar Product|matrix scalar product]], $\mathbf D$ is also of [[Definition:Order of Matrix|order]] $1 \times n$.
+Consider arbitrary [[Definition:Element of Matrix|elements]] $c_i \in \mathbf C$ and $d_i \in \mathbf D$ for some [[Definition:Index of Matrix Element|index]] $i$ where $1 \le i \le n$.
+We have:
+{{begin-eqn}}
+{{eqn | l = c_i
+      | r = \sum_{j \mathop = 1}^i a_{j j} b_j
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = a b_j
+      | c = Definition of $\mathbf A$
+}}
+{{end-eqn}}
+and: 
+{{begin-eqn}}
+{{eqn | l = d_i
+      | r = a b_j
+      | c = {{Defof|Matrix Scalar Product}}
+}}
+{{eqn | r = c_i
+      | c = from above
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Unit Ball is Convex Set}
+Tags: Vector Spaces, Closed Balls
+
+\begin{theorem}
+Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
+Let $\map {B_1^-} 0$ be a [[Definition:Closed Unit Ball|closed unit ball]] in $X$.
+Then $\map {B_1^-} 0$ is [[Definition:Convex Set (Vector Space)|convex]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \map {B_1} 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {\paren {1 - \alpha} x + \alpha y}
+      | o = \le
+      | r = \norm {\paren {1 - \alpha} x} + \norm {\alpha y}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm Axiom $(\text N 3)$: Triangle Inequality]]
+}}
+{{eqn | r = \size {1 - \alpha} \norm x + \size \alpha \norm y
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm Axiom $(\text N 2)$: Positive Homogeneity]]
+}}
+{{eqn | r = \paren {1 - \alpha} \norm x + \alpha \norm y
+      | c = {{Defof|Convex Set (Vector Space)}}: $0 \le \alpha \le 1$
+}}
+{{eqn | o = \le
+      | r = \paren {1 - \alpha} + \alpha
+      | c = $x, y \in \map {B_1^-} 0$
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+Therefore, $\paren {1 - \alpha}x + \alpha y \in \map {B_1^-} 0$.
+By definition, $\map {B_1^-} 0$ is [[Definition:Convex Set (Vector Space)|convex]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uncountable Sum as Series/Corollary}
+Tags: Uncountable Sum as Series
+
+\begin{theorem}
+Let $f: X \to \closedint 0 {+\infty}$ have [[Definition:Uncountable Set|uncountably infinite]] [[Definition:Support of Real-Valued Function|support]].
+Then:
+:$\displaystyle \sum_{x \mathop \in X} \map f x = +\infty$
+\end{theorem}<|endoftext|>
+\section{Identity Matrix from Upper Triangular Matrix}
+Tags: Triangular Matrices, Identity Matrix from Upper Triangular Matrix
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_{m n}$ be an [[Definition:Upper Triangular Matrix|upper triangular matrix]] of [[Definition:Order of Matrix|order]] $m \times n$ with no [[Definition:Zero (Number)|zero]] [[Definition:Diagonal Element|diagonal elements]].
+Let $k = \min \set {m, n}$.
+Then $\mathbf A$ can be transformed into a [[Definition:Matrix|matrix]] such that the first $k$ [[Definition:Row of Matrix|rows]] and [[Definition:Column of Matrix|columns]] form the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $k$.
+\end{theorem}
+
+\begin{proof}
+By definition of $k$:
+:if $\mathbf A$ has more [[Definition:Row of Matrix|rows]] than [[Definition:Column of Matrix|columns]], $k$ is the number of [[Definition:Column of Matrix|columns]] of $\mathbf A$.
+:if $\mathbf A$ has more [[Definition:Column of Matrix|columns]] than [[Definition:Row of Matrix|rows]], $k$ is the number of [[Definition:Row of Matrix|rows]] of $\mathbf A$.
+Thus let $\mathbf A'$ be the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $k$ consisting of the first $k$ [[Definition:Row of Matrix|rows]] and [[Definition:Column of Matrix|columns]] of $\mathbf A$:
+:$\mathbf A' = \begin {bmatrix}
+a_{1 1} & a_{1 2} & a_{1 3} & \cdots &     a_{1, k - 1} &      a_{1 k} \\
+      0 & a_{2 2} & a_{2 3} & \cdots &     a_{2, k - 1} &      a_{2 k} \\
+      0 &       0 & a_{3 3} & \cdots &     a_{3, k - 1} &      a_{3 k} \\
+ \vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\
+      0 &       0 &       0 & \cdots & a_{k - 1, k - 1} & a_{k - 1, k} \\
+      0 &       0 &       0 & \cdots &                0 &      a_{k k} \\
+\end {bmatrix}$
+$\mathbf A$ can be transformed into [[Definition:Echelon Form|echelon form]] $\mathbf B$ by using the [[Definition:Elementary Row Operation|elementary row operations]]:
+:$\forall j \in \set {1, 2, \ldots, k}: e_j := r_j \to \dfrac 1 {a_{j j} } r_j$
+Again, let $\mathbf B'$ be the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $k$ consisting of the first $k$ [[Definition:Row of Matrix|rows]] and [[Definition:Column of Matrix|columns]] of $\mathbf B$:
+:$\mathbf B' = \begin {bmatrix}
+     1 & b_{1 2} & b_{1 3} & \cdots &     b_{1, k - 1} &      b_{1 k} \\
+     0 &       1 & b_{2 3} & \cdots &     b_{2, k - 1} &      b_{2 k} \\
+     0 &       0 &       1 & \cdots &     b_{3, k - 1} &      b_{3 k} \\
+\vdots &  \vdots &  \vdots & \ddots &           \vdots &      \vdots  \\
+     0 &       0 &       0 & \cdots &                1 & b_{k - 1, k} \\
+     0 &       0 &       0 & \cdots &                0 &            1 \\
+\end {bmatrix}$
+$\mathbf B$ is then transformed into [[Definition:Reduced Echelon Form|reduced echelon form]] $\mathbf C$ by means of the [[Definition:Elementary Row Operation|elementary row operations]]:
+:$\forall j \in \set {1, 2, \ldots, k - 1}: e_{j k} := r_j \to r_j - b_{j k} r_k$
+:$\forall j \in \set {1, 2, \ldots, k - 2}: e_{j, k - 1} := r_j \to r_j - b_{j, k - 1} r_{k - 1}$
+and so on, until:
+:$e_{1 2} := r_1 \to r_1 - b_{1 2} r_2$
+Again, let $\mathbf C'$ be the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $k$ consisting of the first $k$ [[Definition:Row of Matrix|rows]] and [[Definition:Column of Matrix|columns]] of $\mathbf C$:
+:$\mathbf C' = \begin {bmatrix}
+     1 &      0 &      0 & \cdots &      0 &      0 \\
+     0 &      1 &      0 & \cdots &      0 &      0 \\
+     0 &      0 &      1 & \cdots &      0 &      0 \\
+\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+     0 &      0 &      0 & \cdots &      1 &      0 \\
+     0 &      0 &      0 & \cdots &      0 &      1 \\
+\end {bmatrix}$
+By inspection, $\mathbf C$ is seen to be the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $k$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simultaneous Linear Equations have Solution iff Ranks of Matrix of Coefficients and Augmented Matrix are Equal}
+Tags: Simultaneous Linear Equations, Rank of Matrix
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simultaneous Linear Equations|system of simultaneous linear equations]]:
+:$\displaystyle \forall i \in \set {1, 2, \ldots, m} : \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
+Let $S$ be expressed in [[Definition:Matrix Representation of Simultaneous Linear Equations|matrix form]] as:
+:$\mathbf A \mathbf x = \mathbf b$
+where:
+:$\mathbf A = \begin {pmatrix}
+\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
+\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
+\end {pmatrix}$,  $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
+Then $S$ has at least one [[Definition:Solution to System of Simultaneous Equations|solution]] {{iff}}:
+:$\map \rho {\mathbf A} = \map \rho {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$
+where:
+:$\map \rho {\mathbf A}$ denotes the [[Definition:Rank of Matrix|rank]] of $\mathbf A$
+:$\paren {\begin {array} {c|c} \mathbf A & \mathbf b \end {array} }$ denotes the [[Definition:Augmented Matrix of Simultaneous Linear Equations|augmented matrix]] of $S$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|tedious}}
+\end{proof}<|endoftext|>
+\section{Simultaneous Linear Equations has Unique Solution iff Rank of Matrix of Coefficients equals Number of Columns}
+Tags: Simultaneous Linear Equations, Rank of Matrix
+
+\begin{theorem}
+Let $S$ be a [[Definition:Simultaneous Linear Equations|system of $m$ simultaneous linear equations in $n$ variables]]:
+:$\displaystyle \forall i \in \set {1, 2, \ldots, m} : \sum_{j \mathop = 1}^n \alpha_{i j} x_j = \beta_i$
+Let $S$ be expressed in [[Definition:Matrix Representation of Simultaneous Linear Equations|matrix form]] as:
+:$\mathbf A \mathbf x = \mathbf b$
+where:
+:$\mathbf A = \begin {pmatrix}
+\alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
+\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \\
+\end {pmatrix}$,  $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}$, $\mathbf b = \begin {pmatrix} \beta_1 \\ \beta_2 \\ \vdots \\ \beta_m \end {pmatrix}$
+Then $S$ has exactly one [[Definition:Solution to System of Simultaneous Equations|solution]] {{iff}}:
+:$\map \rho {\mathbf A} = n$
+where $\map \rho {\mathbf A}$ denotes the [[Definition:Rank of Matrix|rank]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|tedious}}
+\end{proof}<|endoftext|>
+\section{Max Equals an Operand}
+Tags: Max and Min Operations
+
+\begin{theorem}
+Let $x_1, x_2, \dotsc, x_n \in S$ for some $n \in \N_{>0}$.
+Then:
+:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
+\end{theorem}
+
+\begin{proof}
+We will prove the result by [[Principle of Mathematical Induction|induction]] on the [[Definition:Cardinality|number]] of [[Definition:Operand|operands]] $n$.
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$\exists i \in \closedint 1 1 : x_i = \max \set {x_1}$
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_1} = x_1$
+Thus $\map P 1$ 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]]:
+:$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$
+from which it is to be shown that:
+:$\exists i \in \closedint 1 {k + 1} : x_i = \max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} }$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]].
+By the definition of [[Definition: Max Operation (General Case)|Max Operation]]:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {\max \set {x_1, x_2, \dotsc, x_k}, x_{k + 1} }$
+By the [[Max yields Supremum of Parameters/General Case#Induction Hypothesis|Induction hypothesis]]:
+:$\exists i \in \closedint 1 k : x_i = \max \set {x_1, x_2, \dotsc, x_k}$
+So:
+:$\max \set {x_1, x_2, \dotsc, x_k, x_{k + 1} } = \max \set {x_i, x_{k + 1} }$
+As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], all elements of $S$ are [[Definition:Comparable|comparable]] by $\preceq$.
+Therefore there are two cases to consider:
+==== Case 1: $x_{k+1} \preceq x_i$ ====
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_i, x_{k + 1} } = x_i$
+{{qed|lemma}}
+==== Case 2: $x_i \preceq x_{k + 1}$ ====
+By definition of the [[Definition:Max Operation|max operation]]:
+:$\max \set {x_i, x_{k + 1} } = x_{k + 1}$
+{{qed|lemma}}
+In either case, the result holds.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\exists i \in \closedint 1 n : x_i = \max \set {x_1, x_2, \dotsc, x_n}$
+{{qed}}
+[[Category:Max and Min Operations]]
+jsficvqcwn813hmtxanb6d7e6u4q5n0
+\end{proof}<|endoftext|>
+\section{Sine of Integer Multiple of Argument}
+Tags: Sine Function, Sine of Integer Multiple of Argument
+
+\begin{theorem}
+For $n \in \Z_{>0}$:
+\end{theorem}<|endoftext|>
+\section{Trace of Sum of Matrices is Sum of Traces}
+Tags: Traces of Matrices
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order]] $n$.
+let $\mathbf A + \mathbf B$ debote the [[Definition:Matrix Entrywise Addition|matrix entrywise sum]] of $\mathbf A$ and $\mathbf B$.
+Then:
+:$\map \tr {\mathbf A + \mathbf B} = \map \tr {\mathbf A} + \map \tr {\mathbf B}$
+where $\map \tr {\mathbf A}$ denotes the [[Definition:Trace of Matrix|trace]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \tr {\mathbf A} + \map \tr {\mathbf B}
+      | r = \sum_{k \mathop = 1}^n a_{kk} + \sum_{k \mathop = 1}^n b_{kk}
+      | c = {{Defof|Trace of Matrix}}
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {a_{kk} + b_{kk} }
+      | c = [[Sum of Summations equals Summation of Sum]]
+}}
+{{eqn | r = \map \tr {\mathbf A + \mathbf B}
+      | c = {{Defof|Matrix Entrywise Addition}}, {{Defof|Trace of Matrix}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Similar Matrices have same Traces}
+Tags: Traces of Matrices
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order]] $n$.
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Similar Matrices|similar]].
+Then:
+:$\map \tr {\mathbf A} = \map \tr {\mathbf B}$
+where $\map \tr {\mathbf A}$ denotes the [[Definition:Trace of Matrix|trace]] of $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Similar Matrices|similar matrices]]
+:$\exists \mathbf P: \mathbf P^{-1} \mathbf A \mathbf P = \mathbf B$
+where $\mathbf P$ is an [[Definition:Invertible Matrix|invertible matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
+Thus it remains to show that:
+:$\map \tr {\mathbf P^{-1} \mathbf A \mathbf P} = \map \tr {\mathbf A}$
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Cosine of Integer Multiple of Argument/Formulation 1}
+Tags: Cosine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \cos n \theta
+      | r = \dfrac 1 2 \paren {\paren {2 \cos \theta }^n - \dfrac n 1 \paren {2 \cos \theta }^{n - 2} + \dfrac n 2 \dbinom {n - 3} 1 \paren {2 \cos \theta }^{n - 4} - \dfrac n 3 \dbinom {n - 4} 2 \paren {2 \cos \theta }^{n - 6} + \cdots }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\paren {2 \cos \theta }^n + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac n k \dbinom {n - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n - 2 k } }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta }^n + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac n k \dbinom {n - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n - 2 k } }$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \cos \theta
+      | r = \cos \theta
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \paren {2 \cos \theta}^1
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 1$ is seen to hold.
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \cos 2 \theta
+      | r = 2 \cos^2 \theta - 1
+      | c = [[Double Angle Formulas/Cosine/Corollary 1]]
+}}
+{{eqn | r = \dfrac 1 2 \paren {\paren {2 \cos \theta }^2 - \dfrac 2 1 \dbinom {2 - \paren {0 + 1 } } {1 - 1} \paren {2 \cos \theta }^{2 - 2 } }
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 2$ is also seen to hold.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P n$ is true, where $n > 2$, then it logically follows that $\map P {n + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \cos {n \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta }^n + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac n k \dbinom {n - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n - 2 k } }$
+from which we are to show:
+:$\displaystyle \map \cos {\paren {n + 1} \theta} = \dfrac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac {n + 1 } k \dbinom {n + 1 - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n + 1 - 2 k } }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+To proceed, we will require the following Lemma:
+=== [[Sine of Integer Multiple of Argument/Formulation 1/Lemma|Lemma]] ===
+{{:Sine of Integer Multiple of Argument/Formulation 1/Lemma}}{{qed|lemma}}
+Dividing through by $\sin \theta$, we obtain:
+{{begin-eqn}}
+{{eqn | l = \map \cos {n \theta}
+      | r = \frac {\map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta} } {\map \sin {\theta} }
+}}
+{{end-eqn}}
+We now have:
+{{begin-eqn}}
+{{eqn | l = \map \cos {\paren {n + 1} \theta} 
+      | r = \map \cos {n \theta + \theta }
+      | c =
+}}
+{{eqn | r = \cos n \theta \cos \theta - \sin n \theta \sin \theta
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \paren {\frac {\map \sin {n \theta} \map \cos {\theta} - \map \sin {\paren {n - 1 } \theta} } {\map \sin {\theta} } } \cos \theta - \sin n \theta \sin \theta
+      | c = substitution from above
+}}
+{{eqn | r = \sin n \theta \frac {\paren {\cos^2 \theta - \sin^2 \theta } } {\sin \theta } - \map \sin {\paren {n - 1 } \theta} \frac {\cos \theta } {\sin \theta }
+      | c = group terms
+}}
+{{eqn | r = \frac {\map \sin {n \theta} } {\sin \theta } \paren {2 \cos^2 \theta - 1 } - \frac {\map \sin {\paren {n - 1 } \theta} } {\sin \theta } \cos \theta
+      | c = [[Double Angle Formulas/Cosine/Corollary 1]]
+}}
+{{eqn | r = \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } \paren {2 \cos^2 \theta - 1 } - \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n - 1 - \paren {2 k + 1} } } \cos \theta
+      | c = [[Sine of Integer Multiple of Argument/Formulation 1]]
+}}
+{{eqn | r = \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } \paren {\frac {\paren {2 \cos \theta }^2 } 2 - 1 } - \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - 1 - \paren {k + 1} } k \paren {2 \cos \theta}^{n - 1 - \paren {2 k + 1} } } \frac {\paren {2 \cos \theta } } 2
+      | c = multiply by $1$
+}}
+{{eqn | r = \frac 1 2 \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } } - \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } } - \frac 1 2 \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 2} } k \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } } - \paren {\sum_{k \mathop \ge 0} \paren {-1}^k \paren {\binom {n - \paren {k + 1} } k + \frac 1 2 \binom {n - \paren {k + 2} } k } \paren {2 \cos \theta}^{n - \paren {2 k + 1} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \binom {n - \paren {k + 1} } k \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } } + \paren {\sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - \paren {\paren {k - 1 } + 1} } {k - 1 } + \frac 1 2 \binom {n - \paren {\paren {k - 1 } + 2} } {k - 1 } } \paren {2 \cos \theta}^{n - \paren {2 \paren {k - 1 } + 1} } }
+      | c = start all sums at $k = 1$
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - \paren {k + 1} } k + 2 \binom {n - k } {k - 1} + \binom {n - \paren {k + 1} } {k - 1 } } \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } }
+      | c = grouping all terms in one sum
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n - k } k + \binom {n - k } {k - 1} + \binom {n - k } {k - 1 } } \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } }
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \paren {\binom {n + 1 - k } k + \binom {n - k } {k - 1 } } \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } }
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \paren {\paren {\frac {n + 1 - k } k} \binom {n - k } {k - 1 } + \binom {n - k } {k - 1 } } \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } }
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = \frac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1}^k \paren {\paren {\frac {n + 1 - k } k + 1 } \binom {n - k } {k - 1 } } \paren {2 \cos \theta}^{n + 1 - \paren {2 k } } }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac {n + 1 } k \binom {n - k } {k - 1} \paren {2 \cos \theta }^{n + 1 - 2 k } }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\paren {2 \cos \theta }^{n + 1 } + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac {n + 1 } k \binom {n + 1 - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n + 1 - 2 k } }
+      | c = adding and subtracting $1$
+}}
+{{end-eqn}}
+The result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{>0}: \cos n \theta = \dfrac 1 2 \paren {\paren {2 \cos \theta }^n + \sum_{k \mathop \ge 1} \paren {-1 }^k \dfrac n k \dbinom {n - \paren {k + 1 } } {k - 1} \paren {2 \cos \theta }^{n - 2 k } }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Integer Multiple of Argument/Formulation 2}
+Tags: Cosine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \cos n \theta
+      | r = \cos^n \theta \paren {1 - \dbinom n 2 \paren {\tan \theta}^2 + \dbinom n 4 \paren {\tan \theta}^4 - \cdots}
+      | c = 
+}}
+{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \theta}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By [[De Moivre's Formula]]:
+:$\cos n \theta + i \sin n \theta = \paren {\cos \theta + i \sin \theta}^n$
+As $n \in \Z_{>0}$, we use the [[Binomial Theorem]] on the {{RHS}}, resulting in:
+:$\displaystyle \cos n \theta + i \sin n \theta = \sum_{k \mathop \ge 0} \binom n k \paren {\cos^{n - k} \theta} \paren {i \sin \theta}^k$
+When $k$ is [[Definition:Even Integer|even]], the expression being summed is [[Definition:Real Number|real]].
+Equating the [[Definition:Real Part|real parts]] of both sides of the equation, replacing $k$ with $2 k$ to make $k$ [[Definition:Even Integer|even]], gives:
+{{begin-eqn}}
+{{eqn | l = \cos n \theta
+      | r = \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\cos^{n - \paren {2 k } } \theta} \paren {\sin^{2 k } \theta}
+      | c = 
+}}
+{{eqn | r = \cos^n \theta \sum_{k \mathop \ge 0} \paren {-1}^k \dbinom n {2 k } \paren {\tan^{2 k } \theta}
+      | c = factor out $\cos^n \theta$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Cosine of Integer Multiple of Argument]]
+bcldz9phe21keab8wzscyuh1xqpihc1
+\end{proof}<|endoftext|>
+\section{Polygamma Reflection Formula}
+Tags: Gamma Function, Polygamma Function
+
+\begin{theorem}
+:$\map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z} = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn   | l = \map \Gamma z \map \Gamma {1 - z}
+	| r = \dfrac \pi {\sin \pi z}
+        | c = [[Euler's Reflection Formula]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \map \ln {\map \Gamma z \map \Gamma {1 - z} }
+	| r = \map \ln {\dfrac \pi {\sin \pi z} }
+        | c = applying $\ln$ on both sides
+}}
+{{eqn   | ll= \leadsto
+	| l = \map \ln {\map \Gamma z} + \map \ln {\map \Gamma {1 - z} }
+	| r = \map \ln \pi - \map \ln {\sin \pi z}
+        | c = [[Sum of Logarithms]] and [[Difference of Logarithms]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \dfrac \d {\d z} \map \ln {\map \Gamma z} + \dfrac \d {\d z} \map \ln {\map \Gamma {1 - z} }
+	| r = \dfrac \d {\d z} \map \ln \pi - \dfrac \d {\d z} \map \ln {\sin \pi z}
+        | c = taking first [[Definition:Derivative|derivative]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \dfrac {\map {\Gamma'} z} {\map \Gamma z} - \dfrac {\map {\Gamma'} {1 - z} } {\map \Gamma {1 - z} }
+	| r = 0 - \pi \cot \pi z
+        | c = [[Derivative of Natural Logarithm Function]], [[Derivative of Sine Function]], [[Chain Rule]], [[Derivative of Constant]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \map \psi z - \map \psi {1 - z}
+	| r = -\pi \cot \pi z
+        | c = {{Defof|Digamma Function}}
+}}
+{{eqn   | ll= \leadsto
+	| l = \dfrac {\d^n} {\d z^n} \map \psi z - \dfrac {\d^n} {\d z^n} \map \psi {1 - z}
+	| r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
+        | c = taking $n$th [[Definition:Derivative|derivative]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \map {\psi_n} z - \paren {-1}^n \map {\psi_n} {1 - z}
+	| r = -\pi \dfrac {\d^n} {\d z^n} \cot \pi z
+        | c = {{Defof|Polygamma Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polygamma Function in terms of Hurwitz Zeta Function}
+Tags: Hurwitz Zeta Function, Polygamma Function
+
+\begin{theorem}
+:$\displaystyle \map {\psi_n} z = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn   | ll= \leadsto
+	| l = \map \psi z
+	| r = -\gamma + \sum_{k \mathop = 1}^\infty \paren {\dfrac 1 n - \dfrac 1 {z + k - 1} }
+        | c = [[Reciprocal times Derivative of Gamma Function|Series for digamma function]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \dfrac {\d^n} {\d z^n} \map \psi z
+	| r = -\dfrac {\d^n} {\d z^n} \gamma + \dfrac {\d^n} {\d z^n} \sum_{k \mathop = 1}^\infty \paren {\dfrac 1 k - \dfrac 1 {z + k - 1} }
+        | c = taking $n$th [[Definition:Derivative|derivative]]
+}}
+{{eqn   | ll= \leadsto
+	| l = \map {\psi_n} z
+	| r = - \dfrac {\d^n} {\d z^n} \sum_{k \mathop = 1}^\infty \dfrac 1 {z + k - 1}
+        | c = {{Defof|Polygamma Function}}, [[Derivative of Constant]]
+}}
+{{eqn   | ll= \leadsto
+	| l = 
+	| r = - \dfrac {\d^n} {\d z^n} \sum_{k \mathop = 0}^\infty \dfrac 1 {z + k}
+        | c = reindexing $k$ from $1$ to $0$
+}}
+{{eqn   | ll= \leadsto
+	| l = 
+	| r = - \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^{n} \map \Gamma {n + 1} } {\paren {z + k}^{n + 1} }
+        | c = $n$th derivative of reciprocal function
+}}
+{{eqn   | ll= \leadsto
+	| l = 
+	| r = \paren {-1}^{n + 1} \map \Gamma {n + 1} \map \zeta {n + 1, z}
+        | c = {{Defof|Hurwitz Zeta Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area of Parallelogram from Determinant}
+Tags: Areas of Parallelograms
+
+\begin{theorem}
+Let $OABC$ be a [[Definition:Parallelogram|parallelogram]] in the [[Definition:Cartesian Plane|Cartesian plane]] whose [[Definition:Vertex of Polygon|vertices]] are located at:
+{{begin-eqn}}
+{{eqn | l = O
+      | r = \tuple {0, 0}
+}}
+{{eqn | l = A
+      | r = \tuple {a, c}
+}}
+{{eqn | l = B
+      | r = \tuple {a + b, c + d}
+}}
+{{eqn | l = C
+      | r = \tuple {b, d}
+}}
+{{end-eqn}}
+The [[Definition:Area|area]] of $OABC$ is given by:
+:$\map \Area {OABC} = \begin {vmatrix} a & b \\ c & d \end {vmatrix}$
+where $\begin {vmatrix} a & b \\ c & d \end {vmatrix}$ denotes the [[Definition:Determinant of Order 2|determinant of order $2$]].
+\end{theorem}
+
+\begin{proof}
+Arrange for the [[Definition:Parallelogram|parallelogram]] to be situated entirely in the [[Definition:First Quadrant|first quadrant]].
+:[[File:Area-of-Parallelogram-determinant.png|500px]]
+First need we establish that $OABC$ is actually a [[Definition:Parallelogram|parallelogram]] in the first place.
+Indeed:
+{{begin-eqn}}
+{{eqn | l = \vec {AB}
+      | r = \tuple {a + b - a, c + d - c}
+      | c = 
+}}
+{{eqn | r = \tuple {b, d}
+      | c = 
+}}
+{{eqn | r = \vec {CB}
+      | c = 
+}}
+{{eqn | l = \vec {OA}
+      | r = \tuple {a + b - b, c + d - d}
+      | c = 
+}}
+{{eqn | r = \tuple {a, c}
+      | c = 
+}}
+{{eqn | r = \vec {OA}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$OA = CB$
+:$OC = AB$
+and it follows from [[Opposite Sides Equal implies Parallelogram]] that $OABC$ is indeed a [[Definition:Parallelogram|parallelogram]].
+Now we calculate the [[Definition:Area|area]] of $OABC$ as equal to:
+:the [[Definition:Area|area]] occupied by the large [[Definition:Rectangle|rectangle]] in the diagram above
+less:
+:the $4$ [[Definition:Triangle (Geometry)|triangles]]
+:the $2$ small [[Definition:Rectangle|rectangles]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map \Area {OABC}
+      | r = \paren {a + b} \paren {c + d}
+      | c = the large [[Definition:Rectangle|rectangle]]
+}}
+{{eqn | o =
+      | ro= -
+      | r = \paren {\dfrac {a c} 2} - \paren {\dfrac {\paren {a + b - b} \paren {c + d - d} } 2}
+      | c = the $2$ [[Definition:Triangle (Geometry)|triangles]] at top and bottom
+}}
+{{eqn | o =
+      | ro= -
+      | r = \paren {\dfrac {b d} 2} - \paren {\dfrac {\paren {a + b - a} \paren {c + d - c} } 2}
+      | c = the $2$ [[Definition:Triangle (Geometry)|triangles]] at left and right
+}}
+{{eqn | o =
+      | ro= -
+      | r = \paren {a + b - a} c - b \paren {c + d - c}
+      | c = the $2$ small [[Definition:Rectangle|rectangles]]
+}}
+{{eqn | r = a c + a d + b c + b d - \dfrac {a c} 2 - \dfrac {a c} 2 - \dfrac {b d} 2 - \dfrac {b d} 2 - 2 b c
+      | c = multiplying out and simplifying
+}}
+{{eqn | r = a c - b d
+      | c = simplifying
+}}
+{{eqn | r = \begin {vmatrix} a & b \\ c & d \end {vmatrix}
+      | c = {{Defof|Determinant of Order 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix is Invertible iff Rank equals Order}
+Tags: Inverse Matrices, Rank of Matrix
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $\mathbf A \in R^{n \times n}$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
+Then $\mathbf A$ is [[Definition:Invertible Matrix|invertible]] {{iff}} its [[Definition:Rank of Matrix|rank]] also equals $n$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|tedious}}
+\end{proof}<|endoftext|>
+\section{Determinant of Upper Triangular Matrix}
+Tags: Determinants, Triangular Matrices
+
+\begin{theorem}
+Let $\mathbf T_n$ be an [[Definition:Upper Triangular Matrix|upper triangular matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf T_n}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf T_n$.
+Then $\map \det {\mathbf T_n}$ is equal to the product of all the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf T_n$.
+That is:
+:$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf T_n$ be an [[Definition:Upper Triangular Matrix|upper triangular matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+We proceed by [[Principle of Mathematical Induction|induction]] on $n$, the number of [[Definition:Row of Matrix|rows]] of $\mathbf T_n$.
+=== Basis for the Induction ===
+For $n = 1$, the [[Definition:Determinant of Matrix|determinant]] is $a_{11}$, which is clearly also the [[Definition:Diagonal Element|diagonal element]].
+This forms the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Fix $n \in \N$.
+Then, let:
+:$\mathbf T_n = \begin {bmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+      0 & a_{2 2} & \cdots & a_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+      0 &       0 & \cdots & a_{n n} \\
+\end {bmatrix}$
+be an [[Definition:Upper Triangular Matrix|upper triangular matrix]].
+Assume that:
+:$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
+This forms our [[Definition:Induction Hypothesis|induction hypothesis]].
+=== Induction Step ===
+Let $\mathbf T_{n + 1}$ be an [[Definition:Upper Triangular Matrix|upper triangular matrix]] of [[Definition:Order of Square Matrix|order $n + 1$]].
+Then, by the [[Expansion Theorem for Determinants]] (expanding across the $n + 1$th [[Definition:Row of Matrix|row]]):
+:$\displaystyle D = \map \det {\mathbf T_{n + 1} } = \sum_{k \mathop = 1}^{n + 1} a_{n + 1, k} T_{n + 1, k}$
+Because $\mathbf T_{n + 1}$ is [[Definition:Upper Triangular Matrix|upper triangular]], $a_{n + 1, k} = 0$ when $k < n + 1$.
+Therefore:
+:$\map \det {\mathbf T_{n + 1} } = a_{n + 1 \, n + 1} T_{n + 1, n + 1}$
+By the definition of the [[Definition:Cofactor of Element|cofactor]]:
+:$T_{n + 1, n + 1} = \paren {-1}^{n + 1 + n + 1} D_{n + 1, n + 1} = D_{n n}$
+where $D_{n n}$ is the [[Definition:Order of Determinant|order $n$]] [[Definition:Determinant of Matrix|determinant]] obtained from $D$ by deleting [[Definition:Row of Matrix|row]] $n + 1$ and [[Definition:Column of Matrix|column]] $n + 1$. 
+But $D_{n n}$ is just the [[Definition:Determinant of Matrix|determinant]] of an [[Definition:Upper Triangular Matrix|upper triangular matrix]] $\mathbf T_n$.
+Therefore:
+:$\map \det {\mathbf T_{n + 1} } = a_{n + 1, n + 1} \map \det {\mathbf T_n}$
+and the result follows by [[Principle of Mathematical Induction|induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cosine of Integer Multiple of Argument/Formulation 3}
+Tags: Cosine of Integer Multiple of Argument
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \cos n \theta
+      | r = \cos \paren {n - 1} \theta \cos \theta + \paren {1 - \sec^2 \theta } \cos^n \theta \paren {1 + 1 + \frac {\cos 2 \theta} {\cos^2 \theta} + \frac {\cos 3 \theta} {\cos^3 \theta} + \cdots + \frac {\cos \paren {n - 2} \theta} {\cos^{n - 2} \theta} }
+      | c = 
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta + \paren {1 - \sec^2 \theta } \cos^n \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \cos {n \theta } 
+      | r = \map \cos {\paren {n - 1} \theta + \theta}
+      | c =
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta -  \sin \paren {n - 1} \theta \sin \theta
+      | c = [[Cosine of Sum]]
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\sin \theta \cos^{n - 2 } \theta \sum_{k \mathop = 0}^{n - 2 } \frac {\cos k \theta} {\cos^k \theta} } \sin \theta
+      | c = [[Sine of Integer Multiple of Argument/Formulation 3]]
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\sin^2 \theta \cos^{n - 2 } \theta \sum_{k \mathop = 0}^{n - 2 } \frac {\cos k \theta} {\cos^k \theta} }
+      | c = 
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta - \paren {\paren {1 - \cos^2 \theta } \cos^{n - 2 } \theta \sum_{k \mathop = 0}^{n - 2 } \frac {\cos k \theta} {\cos^k \theta} }
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta + \paren {\paren {\cos^2 \theta - 1 } \cos^{n - 2 } \theta \sum_{k \mathop = 0}^{n - 2 } \frac {\cos k \theta} {\cos^k \theta} }
+      | c = 
+}}
+{{eqn | r = \cos \paren {n - 1} \theta \cos \theta + \paren {1 - \sec^2 \theta } \cos^n \theta \sum_{k \mathop = 0}^{n - 2} \frac {\cos k \theta} {\cos^k \theta}
+      | c = [[Definition:Secant Function/Complex]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Cosine of Integer Multiple of Argument]]
+nk013c8qxqdfsoyzcmu83w1f7ktbpof
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Row Matrix/Scale Row}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_1$ be the [[Definition:Elementary Row Operation|elementary row operation]] $\text {ERO} 1$:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 1
+        | t = For some $\lambda \ne 0$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Row of Matrix|row]] $k$ by $\lambda$
+        | m = r_k \to \lambda r_k
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_1$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e_1$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_1$ is:
+:$\map \det {\mathbf E_1} = \lambda$
+\end{theorem}
+
+\begin{proof}
+By [[Elementary Matrix corresponding to Elementary Row Operation/Scale Row|Elementary Matrix corresponding to Elementary Row Operation: Scale Row]], the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e_1$ is of the form:
+:$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$
+where:
+:$E_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf E_1$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$
+:$\delta_{a b}$ is the [[Definition:Kronecker Delta|Kronecker delta]]:
+::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
+Thus when $a \ne b$, $E_{a b} = 0$.
+This means that $\mathbf E_1$ is a [[Definition:Diagonal Matrix|diagonal matrix]].
+{{begin-eqn}}
+{{eqn | l = \displaystyle \map \det {\mathbf E_1}
+      | r = \prod_i E_{i i}
+      | c = [[Determinant of Diagonal Matrix]]
+      | cc= where the [[Definition:Index Variable of Indexed Product|index variable]] $i$ ranges over the [[Definition:Order of Square Matrix|order]] of $\mathbf E_1$
+}}
+{{eqn | r = \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }
+      | c = 
+}}
+{{eqn | r = \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda
+      | c = 
+}}
+{{eqn | r = 1 \times \lambda
+      | c = 
+}}
+{{eqn | r = \lambda
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Row Matrix/Scale Row and Add}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_2$ be the [[Definition:Elementary Row Operation|elementary row operation]] $\text {ERO} 2$:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 2
+        | t = For some $\lambda$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Row of Matrix|row]] $j$ to [[Definition:Row of Matrix|row]] $i$
+        | m = r_i \to r_i + \lambda r_j
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_2$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e_2$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_2$ is:
+:$\map \det {\mathbf E_2} = 1$
+\end{theorem}
+
+\begin{proof}
+By [[Elementary Matrix corresponding to Elementary Row Operation/Scale Row and Add|Elementary Matrix corresponding to Elementary Row Operation: Scale Row and Add]], $\mathbf E_2$ is of the form:
+:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
+where:
+:$E_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf E$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$
+:$\delta_{a b}$ is the [[Definition:Kronecker Delta|Kronecker delta]]:
+::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
+Because $i \ne j$ it follows that:
+:if $a = i$ and $b = j$ then $a \ne b$
+Hence when $a = b$ we have that:
+:$\delta_{a i} \cdot \delta_{j b} = 0$
+Hence the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf E_2$ are all equal to $1$.
+We also have that $\delta_{a i} \cdot \delta_{j b} = 1$ {{iff}} $a = i$ and $b = j$.
+Hence, all [[Definition:Element of Matrix|elements]] of $\mathbf E_2$ apart from the [[Definition:Diagonal Element|diagonal elements]] and $a_{i j}$ are equal to $0$.
+Thus $\mathbf E_2$ is a [[Definition:Triangular Matrix|triangular matrix]] (either [[Definition:Upper Triangular Matrix|upper]] or [[Definition:Lower Triangular Matrix|lower]]).
+From [[Determinant of Triangular Matrix]], $\map \det {\mathbf E_2}$ is equal to the [[Definition:Multiplication|product]] of all the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf E_2$.
+But as we have seen, these are all equal to $1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Row Matrix/Exchange Rows}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_3$ be the [[Definition:Elementary Row Operation|elementary row operation]] $\text {ERO} 3$:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 3
+        | t = Exchange [[Definition:Row of Matrix|rows]] $i$ and $j$
+        | m = r_i \leftrightarrow r_j
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_3$ be the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e_3$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_3$ is:
+:$\map \det {\mathbf E_3} = -1$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf I$ denote the [[Definition:Unit Matrix|unit matrix]] of arbitrary [[Definition:Order of Square Matrix|order]] $n$.
+By [[Determinant of Unit Matrix]]:
+:$\map \det {\mathbf I} = 1$
+Let $\rho$ be the [[Definition:Permutation on n Letters|permutation]] on $\tuple {1, 2, \ldots, n}$ which [[Definition:Transposition|transposes]] $i$ and $j$.
+From [[Parity of K-Cycle]], $\map \sgn \rho = -1$.
+By definition we have that $\mathbf E_3$ is $\mathbf I$ with [[Definition:Row of Matrix|rows]] $i$ and $j$ [[Definition:Transposition|transposed]].
+By the definition of a [[Definition:Determinant of Matrix|determinant]]:
+:$\displaystyle \map \det {\mathbf I} = \sum_{\lambda} \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }$
+By [[Permutation of Determinant Indices]]:
+:$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }$
+We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf E_3}
+      | r = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \rho k \map \lambda k} }
+      | c = 
+}}
+{{eqn | r = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{k \map \lambda k} }
+      | c = 
+}}
+{{eqn | r = -\map \det {\mathbf I}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Column Operation}
+Tags: Elementary Column Operations, Elementary Matrices, Elementary Matrix corresponding to Elementary Column Operation
+
+\begin{theorem}
+Let $\mathbf I$ denote the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order]] $n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $e$ be an [[Definition:Elementary Column Operation|elementary column operation]] on $\mathbf I$.
+Let $\mathbf E$ be the [[Definition:Elementary Column Matrix|elementary column matrix]] of [[Definition:Order of Square Matrix|order]] $n$ [[Definition:Unique|uniquely]] defined as:
+:$\mathbf E = e \paren {\mathbf I}$
+where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
+Let $\kappa_k$ denote the $k$th [[Definition:Column of Matrix|column]] of $\mathbf I$ for $1 \le k \le n$.
+\end{theorem}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Column Operation/Scale Column}
+Tags: Elementary Matrix corresponding to Elementary Column Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 1
+        | t = For some $\lambda \in K_{\ne 0}$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Column of Matrix|column]] $k$ of $\mathbf I$ by $\lambda$
+        | m = \kappa_k \to \lambda \kappa_k
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $m$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $k$ of $\mathbf I$ are to be [[Definition:Ring Product|multiplied]] by $\lambda$.
+By definition of [[Definition:Unit Matrix|unit matrix]], all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $k$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{k k}$, which is $1$.
+Thus in $\mathbf E$:
+:$E_{k k} = \lambda \cdot 1 = \lambda$
+The [[Definition:Element of Matrix|elements]] in all the other [[Definition:Column of Matrix|columns]] of $\mathbf E$ are the same as the corresponding [[Definition:Element of Matrix|elements]] of $\mathbf I$.
+Hence the result.
+{{qed}}
+[[Category:Elementary Matrix corresponding to Elementary Column Operation]]
+jb6yvslnfj0bla5ztd3536s3d31e4o2
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Column Operation/Scale Column and Add}
+Tags: Elementary Matrix corresponding to Elementary Column Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 2
+        | t = For some $\lambda \in K$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Column of Matrix|column]] $j$ to [[Definition:Row of Matrix|row]] $i$
+        | m = \kappa_i \to \kappa_i + \lambda r_j
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $m$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $i$ of $\mathbf I$ are to have the corresponding [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $j$ added to them after the latter have been [[Definition:Ring Product|multiplied]] by $\lambda$.
+By definition of [[Definition:Unit Matrix|unit matrix]]:
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $i$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{i i}$, which is $1$.
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $j$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{j j}$, which is $1$.
+Thus in $\mathbf E$:
+:where $b \ne i$, $E_{a b} = \delta_{a b}$
+:where $b = i$:
+::$E_{a b} = \delta_{a b}$ where $a \ne j$
+::$E_{a b} = \delta_{a b} + \lambda \cdot 1$ where $a = j$
+That is:
+:$E_{a b} = \delta_{a b}$ for all [[Definition:Element of Matrix|elements]] of $\mathbf E$
+except where $b = i$ and $a = j$, at which [[Definition:Element of Matrix|element]]:
+:$E_{a b} = \delta_{a b} + \lambda$
+That is:
+:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Matrix corresponding to Elementary Column Operation/Exchange Columns}
+Tags: Elementary Matrix corresponding to Elementary Column Operation
+
+\begin{theorem}
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] acting on $\mathbf I$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 3
+        | t = Interchange [[Definition:Column of Matrix|columns]] $i$ and $j$
+        | m = \kappa_i \leftrightarrow \kappa_j
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Unit Matrix|unit matrix]]:
+:$I_{a b} = \delta_{a b}$
+where:
+:$I_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf I$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$.
+By definition, $\mathbf E$ is the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]] formed by applying $e$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$.
+That is, all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $i$ of $\mathbf I$ are to be exchanged with the corresponding [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $j$.
+By definition of [[Definition:Unit Matrix|unit matrix]]:
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $i$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{i i}$, which is $1$.
+:all [[Definition:Element of Matrix|elements]] of [[Definition:Column of Matrix|column]] $j$ are $0$ except for [[Definition:Element of Matrix|element]] $I_{j j}$, which is $1$.
+Thus in $\mathbf E$:
+:where $a \ne i$ and $a \ne j$, $E_{a b} = \delta_{a b}$ (all [[Definition:Column of Matrix|columns]] except $i$ and $j$ are unchanged)
+:where $a = i$, $E_{a b} = \delta_{a j}$ (the contents of [[Definition:Column of Matrix|column]] $j$)
+:where $a = j$, $E_{a b} = \delta_{a i}$ (the contents of [[Definition:Column of Matrix|column]] $i$)
+That is:
+:$E_{a b} = \begin {cases} \delta_{a b} & : \text {if $b \ne i$ and $b \ne j$} \\ \delta_{a j} & : \text {if $b = i$} \\ \delta_{a i} & : \text {if $b = j$} \end {cases}$
+Hence the result.
+{{qed}}
+[[Category:Elementary Matrix corresponding to Elementary Column Operation]]
+3d6p4m50lbveddqbkwxoje3483kcrsf
+\end{proof}<|endoftext|>
+\section{Column Equivalence is Equivalence Relation}
+Tags: Equivalence Relations, Column Operations
+
+\begin{theorem}
+[[Definition:Column Equivalence|Column equivalence]] is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+In the following, $\mathbf A$, $\mathbf B$ and $\mathbf C$ denote arbitrary [[Definition:Matrix|matrices]] in a given [[Definition:Matrix Space|matrix space]] $\map \MM {m, n}$ for $m, n \in \Z{>0}$.
+We check in turn each of the conditions for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+Let $\kappa_i$ denote an arbitrary [[Definition:Column of Matrix|column]] of $\mathbf A$.
+Let $e$ denote the [[Definition:Elementary Column Operation|elementary column operation]] $\kappa_i \to 1 \kappa_i$ applied to $\mathbf A$.
+Then trivially:
+:$\map e {\mathbf A} = \mathbf A$
+and so $\mathbf A$ is trivially [[Definition:Column Equivalence|column equivalent]] to itself.
+So [[Definition:Column Equivalence|column equivalence]] has been shown to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\mathbf A$ be [[Definition:Column Equivalence|column equivalent]] to $\mathbf B$.
+Let $\Gamma$ be the [[Definition:Column Operation|column operation]] that transforms $\mathbf A$ into $\mathbf B$.
+From [[Column Operation has Inverse]] there exists a [[Definition:Column Operation|column operation]] $\Gamma'$ which transforms $\mathbf B$ into $\mathbf A$.
+Thus $\mathbf B$ is [[Definition:Column Equivalence|column equivalent]] to $\mathbf A$.
+So [[Definition:Column Equivalence|column equivalence]] has been shown to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+Let $\mathbf A$ be [[Definition:Column Equivalence|column equivalent]] to $\mathbf B$, and let $\mathbf B$ be [[Definition:Column Equivalence|column equivalent]] to $\mathbf C$.
+Let $\Gamma_1$ be the [[Definition:Column Operation|column operation]] that transforms $\mathbf A$ into $\mathbf B$.
+Let $\Gamma_2$ be the [[Definition:Column Operation|column operation]] that transforms $\mathbf B$ into $\mathbf C$.
+From [[Sequence of Column Operations is Column Operation]], the application of $\mathbf C$ is [[Definition:Column Equivalence|column equivalent]] to $\mathbf A$.
+So [[Definition:Column Equivalence|column equivalence]] has been shown to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+[[Definition:Column Equivalence|Column equivalence]] has been shown to be [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Hence by definition it is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Row Operation has Inverse}
+Tags: Row Operations
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma$ be a [[Definition:Row Operation|row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Then there exists another [[Definition:Row Operation|row operation]] $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that compose $\Gamma$.
+Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Row Matrix|elementary row matrices]].
+From [[Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf R \mathbf A = \mathbf B$
+where $\mathbf R$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
+:$\mathbf R = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
+By [[Elementary Row Matrix is Invertible]], each of $\mathbf E_i$ is [[Definition:Invertible Matrix|invertible]].
+By [[Product of Matrices is Invertible iff Matrices are Invertible]], it follows that $\mathbf R$ is likewise [[Definition:Invertible Matrix|invertible]].
+Thus $\mathbf R$ has an [[Definition:Inverse Matrix|inverse]] $\mathbf R^{-1}$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \mathbf R \mathbf A
+      | r = \mathbf B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf R^{-1} \mathbf R \mathbf A
+      | r = \mathbf R^{-1} \mathbf B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A
+      | r = \mathbf R^{-1} \mathbf B
+      | c = 
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = \mathbf R^{-1}
+      | r = \paren {\mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1}^{-1}
+      | c = 
+}}
+{{eqn | r = {\mathbf E_1}^{-1} {\mathbf E_2}^{-1} \dotsb {\mathbf E_{k - 1} }^{-1} {\mathbf E_k}^{-1}
+      | c = [[Inverse of Matrix Product]]
+}}
+{{end-eqn}}
+From [[Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse]], each of ${\mathbf E_i}^{-1}$ is the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to the [[Existence of Inverse Elementary Row Operation|inverse]] $e'_i$ of the corresponding [[Definition:Elementary Row Operation|elementary row operation]] $e_i$.
+Let $\Gamma'$ be the [[Definition:Row Operation|row operation]] composed of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.
+Thus $\Gamma'$ is a [[Definition:Row Operation|row operation]] which transforms $\mathbf B$ into $\mathbf A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers with Absolute Value form Normed Vector Space}
+Tags: Examples of Normed Vector Spaces
+
+\begin{theorem}
+Let $\R$ be the [[Definition:Set|set]] of [[Definition:Real Numbers|real numbers]].
+Let $\size {\, \cdot \,}$ be the [[Definition:Absolute Value|absolute value]].
+Then $\struct {\R, \size {\, \cdot \,}}$ is a [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Real Numbers form Vector Space]]
+:[[Absolute Value is Norm]]
+By definition, $\struct {\R, \size {\, \cdot \,}}$ is a [[Definition:Normed Vector Space|normed vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Dimensional Real Vector Space with Euclidean Norm form Normed Vector Space}
+Tags: Examples of Normed Vector Spaces
+
+\begin{theorem}
+Let $\R^n$ be an [[Definition:Dimension of Vector Space|n-dimensional]] [[Definition:Real Vector Space|real vector space]].
+Let $\norm {\, \cdot \,}_2$ be the [[Definition:Euclidean Norm|Euclidean norm]].
+Then $\struct {\R^n, \norm {\, \cdot \,}_2}$ is a [[Definition:Normed Vector Space|normed vector space]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Real Vector Space is Vector Space]]
+:By [[Euclidean Space is Normed Space]], $\norm {\, \cdot \,}_2$ is a [[Definition:Norm on Vector Space|norm]] on $\R^n$
+By [[Definition:Definition|definition]], $\struct {\R^n, \norm {\, \cdot \,}_2}$ is a [[Definition:Normed Vector Space|normed vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices}
+Tags: Row Operations, Proofs by Induction
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma$ be a [[Definition:Row Operation|row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+:$\mathbf R \mathbf A = \mathbf B$
+where $\mathbf R$ is the [[Definition:Matrix Product (Conventional)|product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Matrix|elementary row matrices]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+By definition, $\Gamma$ is a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] on $\mathbf A$.
+Let $\sequence e_k$ denote a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\tuple {e_1, e_2, \ldots, e_k}$ applied on $\mathbf A$ in order: first $e_1$, then $e_2$, then $\ldots$, then $e_k$.
+Let $\Gamma_k$ be the [[Definition:Row Operation|row operation]] which consists of $\sequence e_k$.
+Let $\mathbf E_k$ denote the [[Definition:Elementary Row Matrix|elementary row matrix]] of [[Definition:Order of Square Matrix|order]] $m$ formed by applying $e_k$ to the [[Definition:Unit Matrix|unit matrix]] $I_m$.
+For all $r \in \Z_{>0}$, let $\map P r$ be the [[Definition:Proposition|proposition]]:
+:For all $\Gamma_r$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R_r$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+::$\mathbf R_r \mathbf A = \mathbf B_r$
+:where:
+::$\Gamma_r$ is a [[Definition:Row Operation|row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B_r \in \map \MM {m, n}$.
+::$\mathbf R_r$ is the [[Definition:Matrix Product (Conventional)|product]] of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Matrix|elementary row matrices]]:
+:::$\mathbf R_r = \mathbf E_r \mathbf E_{r - 1} \dotsb \mathbf E_2 \mathbf E_1$
+=== Basis for the Induction ===
+$\map P 1$ is the case where $\Gamma_1$ is a single-[[Definition:Term of Sequence|term]] [[Definition:Finite Sequence|sequence]] consisting of one [[Definition:Elementary Row Operation|elementary row operation]] $e_1$.
+Let $e_1$ be an [[Definition:Elementary Row Operation|elementary row operation]] operating on $\mathbf A$, which transforms $\mathbf A$ into $\mathbf B_1$.
+By definition, there exists [[Definition:Unique|exactly one]] [[Definition:Elementary Row Matrix|elementary row matrix]] $\mathbf E_1$ of [[Definition:Order of Square Matrix|order $m$]] such that $\mathbf E_1$ is the result of applying $e_1$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$ of [[Definition:Order of Square Matrix|order $m$]].
+From the [[Elementary Row Operations as Matrix Multiplications/Corollary|corollary to Elementary Row Operations as Matrix Multiplications]]:
+:$\mathbf E_1 \mathbf A = \mathbf B_1$
+By [[Elementary Row Matrix is Invertible]], $E_1$ is [[Definition:Invertible Matrix|invertible]].
+Thus $\map P 1$ 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]]:
+:For all $\Gamma_k$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R_k$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+::$\mathbf R_k \mathbf A = \mathbf B_k$
+from which it is to be shown that:
+:For all $\Gamma_{k + 1}$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R_{k + 1}$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+::$\mathbf R_{k + 1} \mathbf A = \mathbf B_{k + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By definition, $\Gamma_{k + 1}$ is a [[Definition:Row Operation|row operation]] consisting of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\tuple {e_1, e_2, \ldots, e_k, e_{k + 1} }$ applied on $\mathbf A$ in order.
+Thus $\Gamma_{k + 1}$ consists of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\tuple {e_1, e_2, \ldots, e_k}$ applied on $\mathbf A$ in order, followed by a further [[Definition:Elementary Row Operation|elementary row operation]] $e_{k + 1}$.
+By the [[Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices#Induction Hypothesis|induction hypothesis]], there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R_k$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+:$\mathbf R_k \mathbf A = \mathbf B_k$
+where $\mathbf B_k \in \map \MM {m, n}$ is the result of applying $\sequence e_k$ to $\mathbf A$ in order.
+Let $e_{k + 1}$ be applied to $\mathbf B_k$.
+By definition, there exists [[Definition:Unique|exactly one]] [[Definition:Elementary Row Matrix|elementary row matrix]] $\mathbf E_{k + 1}$ of [[Definition:Order of Square Matrix|order $m$]] such that $\mathbf E_{k + 1}$ is the result of applying $e_{k + 1}$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$ of [[Definition:Order of Square Matrix|order $m$]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf B_{k + 1}
+      | r = \mathbf E_{k + 1} \mathbf B_k
+      | c = [[Elementary Row Operations as Matrix Multiplications/Corollary|Corollary to Elementary Row Operations as Matrix Multiplications]]
+}}
+{{eqn | r = \mathbf E_{k + 1} \paren {\mathbf R_k \mathbf A}
+      | c = 
+}}
+{{eqn | r = \paren {\mathbf E_{k + 1} \mathbf R_k} \mathbf A
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{end-eqn}}
+By [[Product of Matrices is Invertible iff Matrices are Invertible]], $\mathbf E_{k + 1} \mathbf R_k$ is [[Definition:Invertible Matrix|invertible]].
+We have that $\mathbf R_k$ is the [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] resulting from the application of $\sequence e_k$ on $\mathbf I_m$.
+Thus $\mathbf E_{k + 1} \mathbf R_k$ is the [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] resulting from the application of $\sequence e_{k + 1}$ on $\mathbf I_m$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore, for every [[Definition:Row Operation|row operation]] $\Gamma$ which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+:$\mathbf R \mathbf A = \mathbf B$
+where:
+:$\mathbf R$ is the [[Definition:Matrix Product (Conventional)|product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Matrix|elementary row matrices]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Matrices is Invertible iff Matrices are Invertible}
+Tags: Inverse Matrices, Conventional Matrix Multiplication
+
+\begin{theorem}
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Square Matrix|square matrices of order $n$]].
+Let $\mathbf A \mathbf B$ denote the [[Definition:Matrix Product (Conventional)|matrix product]] of $\mathbf A$ and $\mathbf B$.
+Let $\mathbf I$ be the $n \times n$ [[Definition:Unit Matrix|unit matrix]].
+Let $\mathbf A$ and $\mathbf B$ be [[Definition:Invertible Matrix|invertible]].
+Then:
+:$\mathbf A \mathbf B$ is [[Definition:Invertible Matrix|invertible]]
+{{iff}}
+:both $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let both $\mathbf A$ and $\mathbf B$ be [[Definition:Invertible Matrix|invertible]].
+By [[Matrix is Invertible iff Determinant has Multiplicative Inverse]]:
+:$\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$
+where $\map \det {\mathbf A}$ denotes the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+By [[Determinant of Matrix Product]]:
+:$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$
+Thus as both $\map \det {\mathbf A} \ne 0$ and $\map \det {\mathbf B} \ne 0$, it follows that:
+:$\map \det {\mathbf A \mathbf B} = \ne 0$
+Hence by [[Matrix is Invertible iff Determinant has Multiplicative Inverse]], $\map \det {\mathbf A \mathbf B}$ is [[Definition:Invertible Matrix|invertible]].
+=== Sufficient Condition ===
+Let $\mathbf A \mathbf B$ be [[Definition:Invertible Matrix|invertible]].
+{{AimForCont}} it is not the case that both $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible]].
+Then by [[Matrix is Invertible iff Determinant has Multiplicative Inverse]], either:
+:$\map \det {\mathbf A} = 0$
+or:
+:$\map \det {\mathbf B} = 0$
+By [[Determinant of Matrix Product]]:
+:$\map \det {\mathbf A} \map \det {\mathbf B} = \map \det {\mathbf A \mathbf B}$
+and so:
+:$\map \det {\mathbf A \mathbf B} = 0$
+Hence by [[Matrix is Invertible iff Determinant has Multiplicative Inverse]], $\map \det {\mathbf A \mathbf B}$ is not [[Definition:Invertible Matrix|invertible]].
+This [[Definition:Contradiction|contradicts]] the assumption that $\mathbf A \mathbf B$ is [[Definition:Invertible Matrix|invertible]].
+Hence by [[Proof by Contradiction]] it follows that both $\mathbf A$ and $\mathbf B$ are [[Definition:Invertible Matrix|invertible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Number Plus or Minus Square Root}
+Tags: Square Roots, Square Root of Number Plus or Minus Square Root
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]] such that $a^2 - b > 0$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Elementary Row Matrix is Invertible}
+Tags: Elementary Matrices
+
+\begin{theorem}
+Let $\mathbf E$ be an [[Definition:Elementary Row Matrix|elementary row matrix]].
+Then $\mathbf E$ is [[Definition:Invertible Matrix|invertible]].
+\end{theorem}
+
+\begin{proof}
+From [[Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse]] it is demonstrated that:
+:if $\mathbf E$ is the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to an [[Definition:Elementary Row Operation|elementary row operation]] $e$
+then:
+:the [[Definition:Inverse of Elementary Row Operation|inverse]] of $e$ corresponds to an [[Definition:Elementary Row Matrix|elementary row matrix]] which is the [[Definition:Inverse Matrix|inverse]] of $\mathbf E$.
+So as $\mathbf E$ has an [[Definition:Inverse Matrix|inverse]], [[Definition:A Priori|a priori]] it is [[Definition:Invertible Matrix|invertible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Column Operation}
+Tags: Elementary Column Operations, Existence of Inverse Elementary Column Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be an [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Column Operation|inverse]] of $e$.
+Then $e'$ is an [[Definition:Elementary Column Operation|elementary column operation]] which always exists and is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let us take each type of [[Definition:Elementary Column Operation|elementary column operation]] in turn.
+For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new [[Definition:Matrix|matrix]] $\mathbf A'' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
+In the below, let:
+:$\kappa_k$ denote [[Definition:Column of Matrix|column]] $k$ of $\mathbf A$
+:$\kappa'_k$ denote [[Definition:Column of Matrix|column]] $k$ of $\mathbf A'$
+:$\kappa''_k$ denote [[Definition:Column of Matrix|column]] $k$ of $\mathbf A''$
+for arbitrary $k$ such that $1 \le k \le m$.
+By definition of [[Definition:Elementary Column Operation|elementary column operation]]:
+:only the [[Definition:Column of Matrix|column]] or [[Definition:Column of Matrix|columns]] directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
+and similarly:
+:only the [[Definition:Column of Matrix|column]] or [[Definition:Column of Matrix|columns]] directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A''$.
+Hence it is understood that in the following, only those [[Definition:Column of Matrix|columns]] directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$.
+=== [[Existence of Inverse Elementary Column Operation/Scalar Product of Column|$\text {ECO} 1$: Scalar Product of Column]] ===
+{{:Existence of Inverse Elementary Column Operation/Scalar Product of Column}}{{qed|lemma}}
+=== [[Existence of Inverse Elementary Column Operation/Add Scalar Product of Column to Another|$\text {ECO} 2$: Add Scalar Product of Column to Another]] ===
+{{:Existence of Inverse Elementary Column Operation/Add Scalar Product of Column to Another}}{{qed|lemma}}
+=== [[Existence of Inverse Elementary Column Operation/Exchange Columns|$\text {ECO} 3$: Exchange Columns]] ===
+{{:Existence of Inverse Elementary Column Operation/Exchange Columns}}{{qed|lemma}}
+Thus in all cases, for each [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A'$ to $\mathbf A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Column Operation has Inverse}
+Tags: Column Operations
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma$ be a [[Definition:Column Operation|column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Then there exists another [[Definition:Column Operation|column operation]] $\Gamma'$ which transforms $\mathbf B$ back to $\mathbf A$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] that compose $\Gamma$.
+Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Column Matrix|elementary column matrices]].
+From [[Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf A \mathbf K = \mathbf B$
+where $\mathbf K$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
+:$\mathbf K = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$
+By [[Elementary Column Matrix is Invertible]], each of $\mathbf E_i$ is [[Definition:Invertible Matrix|invertible]].
+By [[Product of Matrices is Invertible iff Matrices are Invertible]], it follows that $\mathbf K$ is likewise [[Definition:Invertible Matrix|invertible]].
+Thus $\mathbf K$ has an [[Definition:Inverse Matrix|inverse]] $\mathbf K^{-1}$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \mathbf A \mathbf K
+      | r = \mathbf B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A \mathbf K \mathbf K^{-1}
+      | r = \mathbf B \mathbf K^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A
+      | r = \mathbf B \mathbf K^{-1}
+      | c = 
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = \mathbf K^{-1}
+      | r = \paren {\mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k}^{-1}
+      | c = 
+}}
+{{eqn | r = {\mathbf E_k}^{-1} {\mathbf E_{k - 1} }^{-1} \dotsb {\mathbf E_2}^{-1} {\mathbf E_1}^{-1}
+      | c = [[Inverse of Matrix Product]]
+}}
+{{end-eqn}}
+From [[Elementary Column Matrix for Inverse of Elementary Column Operation is Inverse]], each of ${\mathbf E_i}^{-1}$ is the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to the [[Existence of Inverse Elementary Column Operation|inverse]] $e'_i$ of the corresponding [[Definition:Elementary Column Operation|elementary column operation]] $e_i$.
+Let $\Gamma'$ be the [[Definition:Column Operation|column operation]] composed of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] $\tuple {e'_k, e'_{k - 1}, \ldots, e'_2, e'_1}$.
+Thus $\Gamma'$ is a [[Definition:Column Operation|column operation]] which transforms $\mathbf B$ into $\mathbf A$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elementary Column Matrix is Invertible}
+Tags: Elementary Matrices
+
+\begin{theorem}
+Let $\mathbf E$ be an [[Definition:Elementary Column Matrix|elementary column matrix]].
+Then $\mathbf E$ is [[Definition:Invertible Matrix|invertible]].
+\end{theorem}
+
+\begin{proof}
+From [[Elementary Column Matrix for Inverse of Elementary Column Operation is Inverse]] it is demonstrated that:
+:if $\mathbf E$ is the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to an [[Definition:Elementary Column Operation|elementary column operation]] $e$
+then:
+:the [[Definition:Inverse of Elementary Column Operation|inverse]] of $e$ corresponds to an [[Definition:Elementary Column Matrix|elementary column matrix]] which is the [[Definition:Inverse Matrix|inverse]] of $\mathbf E$.
+So as $\mathbf E$ has an [[Definition:Inverse Matrix|inverse]], [[Definition:A Priori|a priori]] it is [[Definition:Invertible Matrix|invertible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices}
+Tags: Column Operations, Proofs by Induction
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma$ be a [[Definition:Column Operation|column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf K$ of [[Definition:Order of Square Matrix|order $n$]] such that:
+:$\mathbf A \mathbf K = \mathbf B$
+where $\mathbf K$ is the [[Definition:Matrix Product (Conventional)|product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Matrix|elementary column matrices]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+By definition, $\Gamma$ is a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] on $\mathbf A$.
+Let $\sequence e_k$ denote a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] $\tuple {e_1, e_2, \ldots, e_k}$ applied on $\mathbf A$ in order: first $e_1$, then $e_2$, then $\ldots$, then $e_k$.
+Let $\Gamma_k$ be the [[Definition:Column Operation|column operation]] which consists of $\sequence e_k$.
+Let $\mathbf E_k$ denote the [[Definition:Elementary Column Matrix|elementary column matrix]] of [[Definition:Order of Square Matrix|order]] $n$ formed by applying $e_k$ to the [[Definition:Unit Matrix|unit matrix]] $I_n$.
+For all $r \in \Z_{>0}$, let $\map P r$ be the [[Definition:Proposition|proposition]]:
+:For all $\Gamma_r$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf K_r$ of [[Definition:Order of Square Matrix|order $n$]] such that:
+::$\mathbf A \mathbf K_r = \mathbf B_r$
+:where:
+::$\Gamma_r$ is a [[Definition:Column Operation|column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B_r \in \map \MM {m, n}$.
+::$\mathbf K_r$ is the [[Definition:Matrix Product (Conventional)|product]] of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Matrix|elementary column matrices]]:
+:::$\mathbf K_r = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{r - 1} \mathbf E_r$
+=== Basis for the Induction ===
+$\map P 1$ is the case where $\Gamma_1$ is a single-[[Definition:Term of Sequence|term]] [[Definition:Finite Sequence|sequence]] consisting of one [[Definition:Elementary Column Operation|elementary column operation]] $e_1$.
+Let $e_1$ be an [[Definition:Elementary Column Operation|elementary column operation]] operating on $\mathbf A$, which transforms $\mathbf A$ into $\mathbf B_1$.
+By definition, there exists [[Definition:Unique|exactly one]] [[Definition:Elementary Column Matrix|elementary column matrix]] $\mathbf E_1$ of [[Definition:Order of Square Matrix|order $m$]] such that $\mathbf E_1$ is the result of applying $e_1$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$ of [[Definition:Order of Square Matrix|order $n$]].
+From the [[Elementary Column Operations as Matrix Multiplications/Corollary|corollary to Elementary Column Operations as Matrix Multiplications]]:
+:$\mathbf A \mathbf E_1 = \mathbf B_1$
+By [[Elementary Column Matrix is Invertible]], $E_1$ is [[Definition:Invertible Matrix|invertible]].
+Thus $\map P 1$ 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]]:
+:For all $\Gamma_k$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf K_k$ of [[Definition:Order of Square Matrix|order $n$]] such that:
+::$\mathbf A \mathbf K_k = \mathbf B_k$
+from which it is to be shown that:
+:For all $\Gamma_{k + 1}$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf K_{k + 1}$ of [[Definition:Order of Square Matrix|order $n$]] such that:
+::$\mathbf A \mathbf K_{k + 1} = \mathbf B_{k + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By definition, $\Gamma_{k + 1}$ is a [[Definition:Column Operation|column operation]] consisting of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] $\tuple {e_1, e_2, \ldots, e_k, e_{k + 1} }$ applied on $\mathbf A$ in order.
+Thus $\Gamma_{k + 1}$ consists of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] $\tuple {e_1, e_2, \ldots, e_k}$ applied on $\mathbf A$ in order, followed by a further [[Definition:Elementary Column Operation|elementary column operation]] $e_{k + 1}$.
+By the [[Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices#Induction Hypothesis|induction hypothesis]], there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf K_k$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+:$\mathbf A \mathbf K_k = \mathbf B_k$
+where $\mathbf B_k \in \map \MM {m, n}$ is the result of applying $\sequence e_k$ to $\mathbf A$ in order.
+Let $e_{k + 1}$ be applied to $\mathbf B_k$.
+By definition, there exists [[Definition:Unique|exactly one]] [[Definition:Elementary Column Matrix|elementary column matrix]] $\mathbf E_{k + 1}$ of [[Definition:Order of Square Matrix|order $m$]] such that $\mathbf E_{k + 1}$ is the result of applying $e_{k + 1}$ to the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$ of [[Definition:Order of Square Matrix|order $m$]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf B_{k + 1}
+      | r = \mathbf B_k \mathbf E_{k + 1}
+      | c = [[Elementary Column Operations as Matrix Multiplications/Corollary|Corollary to Elementary Column Operations as Matrix Multiplications]]
+}}
+{{eqn | r = \paren {\mathbf A \mathbf K_k} \mathbf E_{k + 1}
+      | c = 
+}}
+{{eqn | r = \mathbf A \paren {\mathbf K_k \mathbf E_{k + 1} }
+      | c = [[Matrix Multiplication is Associative]]
+}}
+{{end-eqn}}
+By [[Product of Matrices is Invertible iff Matrices are Invertible]], $\mathbf K_k \mathbf E_{k + 1}$ is [[Definition:Invertible Matrix|invertible]].
+We have that $\mathbf K_k$ is the [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] resulting from the application of $\sequence e_k$ on $\mathbf I_m$.
+Thus $\mathbf K_k \mathbf E_{k + 1}$ is the [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] resulting from the application of $\sequence e_{k + 1}$ on $\mathbf I_m$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore, for every [[Definition:Column Operation|column operation]] $\Gamma$ which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$, there exists a [[Definition:Unique|unique]] [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] $\mathbf R$ of [[Definition:Order of Square Matrix|order $m$]] such that:
+:$\mathbf A \mathbf K = \mathbf B$
+where:
+:$\mathbf K$ is the [[Definition:Matrix Product (Conventional)|product]] of a [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Matrix|elementary column matrices]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Row Operation/Scalar Product of Row}
+Tags: Existence of Inverse Elementary Row Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 1
+        | t = For some $\lambda \in K_{\ne 0}$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Row of Matrix|row]] $i$ by $\lambda$
+        | m = r_i \to \lambda r_i
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Row Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e' := r_i \to \dfrac 1 \lambda r_i$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \to \lambda r_k$
+where $\lambda \ne 0$.
+Then $r'_k$ is such that:
+:$\forall a'_{k i} \in r'_k: a'_{k i} = \lambda a_{k i}$
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := r_k \to \dfrac 1 \lambda r_k$
+Because it is stipulated in the definition of an [[Definition:Elementary Row Operation|elementary row operation]] that $\lambda \ne 0$, it follows by definition of a [[Definition:Field (Abstract Algebra)|field]] that $\dfrac 1 \lambda$ exists.
+Hence $e'$ is defined.
+So applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | lo= \forall a''_{k i} \in r''_k:
+      | l = a''_{k i}
+      | r = \dfrac 1 \lambda a'_{k i}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 \lambda \paren {\lambda a_{k i} }
+      | c = 
+}}
+{{eqn | r = a_{k i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall a''_{k i} \in r''_k:
+      | l = a''_{k i}
+      | r = a_{k i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r''_k
+      | r = r_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Row Operation|elementary row operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Row Operation/Add Scalar Product of Row to Another}
+Tags: Existence of Inverse Elementary Row Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 2
+        | t = For some $\lambda \in K$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Row of Matrix|row]] $k$ to [[Definition:Row of Matrix|row]] $l$
+        | m = r_k \to r_k + \lambda r_l
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Row Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e' := r'_k \to r'_k - \lambda r'_l$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \to r_k + \lambda r_l$
+Then $r'_k$ is such that:
+:$\forall a'_{k i} \in r'_k: a'_{k i} = a_{k i} + \lambda a_{l i}$
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := r'_k \to r'_k - \lambda r'_l$
+Applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | lo= \forall a''_{k i} \in r''_k:
+      | l = a''_{k i}
+      | r = a'_{k i} - \lambda a'_{l i}
+      | c = 
+}}
+{{eqn | r = \paren {a_{k i} + \lambda a_{l i} } - \lambda a'_{l i}
+      | c = 
+}}
+{{eqn | r = \paren {a_{k i} + \lambda a_{l i} } - \lambda a_{l i}
+      | c = as $\lambda a'_{l i} = \lambda a_{l i}$: [[Definition:Row of Matrix|row]] $l$ was not changed by $e$
+}}
+{{eqn | r = a_{k i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall a''_{k i} \in r''_k:
+      | l = a''_{k i}
+      | r = a_{k i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r''_{k i}
+      | r = r_{k i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Row Operation|elementary row operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Row Operation/Exchange Rows}
+Tags: Existence of Inverse Elementary Row Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 3
+        | t = Exchange [[Definition:Row of Matrix|rows]] $k$ and $l$
+        | m = r_k \leftrightarrow r_l
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Row Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e' := r_k \leftrightarrow r_l$
+That is:
+:$e' = e$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Row Operation|elementary row operation]]:
+:$e := r_k \leftrightarrow r_l$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = r'_k
+      | r = r_l
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = r'_l
+      | r = r_k
+      | c = 
+}}
+{{end-eqn}}
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Row Operation|elementary row operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := r'_k \leftrightarrow r'_l$
+Applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | l = r''_k
+      | r = r'_l
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = r''_l
+      | r = r'_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r''_k
+      | r = r_k
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = r''_l
+      | r = r_l
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Row Operation|elementary row operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Column Operation/Scalar Product of Column}
+Tags: Existence of Inverse Elementary Column Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 1
+        | t = For some $\lambda \in K_{\ne 0}$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Column of Matrix|column]] $k$ by $\lambda$
+        | m = \kappa_k \to \lambda \kappa_k
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Column Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e := \kappa_k \to \lambda \kappa_k$
+where $\lambda \ne 0$.
+Then $\kappa'_k$ is such that:
+:$\forall a'_{k i} \in \kappa'_k: a'_{k i} = \lambda a_{k i}$
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := \kappa_k \to \dfrac 1 \lambda \kappa_k$
+Because it is stipulated in the definition of an [[Definition:Elementary Column Operation|elementary column operation]] that $\lambda \ne 0$, it follows by definition of a [[Definition:Field (Abstract Algebra)|field]] that $\dfrac 1 \lambda$ exists.
+Hence $e'$ is defined.
+So applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | lo= \forall a''_{i k} \in \kappa''_k:
+      | l = a''_{i k}
+      | r = \dfrac 1 \lambda a'_{i k}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 \lambda \paren {\lambda a_{i k} }
+      | c = 
+}}
+{{eqn | r = a_{i k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall a''_{i k} \in \kappa''_k:
+      | l = a''_{i k}
+      | r = a_{i k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \kappa''_k
+      | r = \kappa_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Column Operation|elementary column operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Column Operation/Add Scalar Product of Column to Another}
+Tags: Existence of Inverse Elementary Column Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 2
+        | t = For some $\lambda \in K$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Column of Matrix|column]] $l$ to [[Definition:Column of Matrix|column]] $k$
+        | m = \kappa_k \to \kappa_k + \lambda \kappa_l
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Column Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e' := \kappa_k \to \kappa_k - \lambda \kappa_l$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e := \kappa_k \to \kappa_k + \lambda r_l$
+Then $\kappa'_k$ is such that:
+:$\forall a'_{i k} \in \kappa'_k: a'_{i k} = a_{i k} + \lambda a_{i l}$
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := \kappa'_k \to \kappa'_k - \lambda \kappa'_l$
+Applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | lo= \forall a''_{i k} \in \kappa''_k:
+      | l = a''_{i k}
+      | r = a'_{i k} - \lambda a'_{i l}
+      | c = 
+}}
+{{eqn | r = \paren {a_{i k} + \lambda a_{i l} } - \lambda a'_{i l}
+      | c = 
+}}
+{{eqn | r = \paren {a_{i k} + \lambda a_{i l} } - \lambda a_{i l}
+      | c = as $\lambda a'_{i l} = \lambda a_{i l}$: [[Definition:Column of Matrix|column]] $l$ was not changed by $e$
+}}
+{{eqn | r = a_{i k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall a''_{i k} \in \kappa''_k:
+      | l = a''_{i k}
+      | r = a_{i k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \kappa''_{i k}
+      | r = \kappa_{i k}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Column Operation|elementary column operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Existence of Inverse Elementary Column Operation/Exchange Columns}
+Tags: Existence of Inverse Elementary Column Operation
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf A' \in \map \MM {m, n}$.
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 3
+        | t = Interchange [[Definition:Column of Matrix|columns]] $k$ and $l$
+        | m = \kappa_k \leftrightarrow \kappa_l
+}}
+{{end-axiom}}
+Let $\map {e'} {\mathbf A'}$ be the [[Definition:Inverse of Elementary Column Operation|inverse]] of $e$.
+Then $e'$ is the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e' := \kappa_k \leftrightarrow \kappa_l$
+That is:
+:$e' = e$
+\end{theorem}
+
+\begin{proof}
+Let $\map e {\mathbf A}$ be the [[Definition:Elementary Column Operation|elementary column operation]]:
+:$e := \kappa_k \leftrightarrow \kappa_l$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = \kappa'_k
+      | r = \kappa_l
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = \kappa'_l
+      | r = \kappa_k
+      | c = 
+}}
+{{end-eqn}}
+Now let $\map {e'} {\mathbf A'}$ be the [[Definition:Elementary Column Operation|elementary column operation]] which transforms $\mathbf A'$ to $\mathbf A''$:
+:$e' := \kappa'_k \leftrightarrow \kappa'_l$
+Applying $e'$ to $\mathbf A'$ we get:
+{{begin-eqn}}
+{{eqn | l = \kappa''_k
+      | r = \kappa'_l
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = \kappa''_l
+      | r = \kappa'_k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \kappa''_k
+      | r = \kappa_k
+      | c = 
+}}
+{{eqn | lo= \text {and}
+      | l = \kappa''_l
+      | r = \kappa_l
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A''
+      | r = \mathbf A
+      | c = 
+}}
+{{end-eqn}}
+It is noted that for $e'$ to be an [[Definition:Elementary Column Operation|elementary column operation]], the only possibility is for it to be as defined.
+\end{proof}<|endoftext|>
+\section{Elementary Column Operations as Matrix Multiplications}
+Tags: Conventional Matrix Multiplication, Elementary Column Operations, Elementary Matrices
+
+\begin{theorem}
+Let $e$ be an [[Definition:Elementary Column Operation|elementary column operation]].
+Let $\mathbf E$ be the [[Definition:Elementary Column Matrix|elementary column matrix]] of [[Definition:Order of Square Matrix|order]] $n$ defined as:
+:$\mathbf E = e \paren {\mathbf I}$
+where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
+Then for every $m \times n$ [[Definition:Matrix|matrix]] $\mathbf A$:
+:$e \paren {\mathbf A} = \mathbf A \mathbf E$
+where $\mathbf A \mathbf E$ denotes the [[Definition:Matrix Product (Conventional)|conventional matrix product]].
+\end{theorem}
+
+\begin{proof}
+Let $s, t \in \closedint 1 m$ such that $s \ne t$.
+=== Case $1$ ===
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\kappa_s \to \lambda \kappa_s$:
+:$E_{k i} = \begin{cases}
+\delta_{k i} & : i \ne s \\
+\lambda \delta_{k i} & : i = s
+\end{cases}$
+where $\delta$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {A E}_{i j}
+      | r = \sum_{k \mathop = 1}^m A_{j k} E_{k i} 
+      | c = 
+}}
+{{eqn | r = \begin {cases} A_{j i} & : i \ne r \\ \lambda A_{j i} & : i = r \end{cases}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf {A E}
+      | r = e \paren {\mathbf A}
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case $2$ ===
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\kappa_s \to \kappa_s + \lambda \kappa_t$:
+:$E_{k i} = \begin {cases}
+\delta_{k i} & : i \ne s \\
+\delta_{k s} + \lambda \delta_{k t} & : i = s
+\end {cases}$
+where $\delta$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {A E}_{j i}
+      | r = \sum_{k \mathop = 1}^m  A_{j k} E_{k i}
+      | c = 
+}}
+{{eqn | r = \begin {cases} A_{j i} & : j \ne s \\ A_{i j} + \lambda A_{j t} & : i = s \end {cases}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf {A E}
+      | r = e \paren {\mathbf A}
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case $3$ ===
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\kappa_s \leftrightarrow \kappa_t$:
+By [[Exchange of Columns as Sequence of Other Elementary Column Operations]], this [[Definition:Elementary Column Operation|elementary column operation]] can be expressed as:
+:$\paren {e_1 e_2 e_3 e_4 \mathbf A} = e \paren {\mathbf A}$
+where the $e_i$ are [[Definition:Elementary Column Operation|elementary column operation]] of the other two types.
+For each $e_i$, let $\mathbf E_i = e_i \paren {\mathbf I}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = e \paren {\mathbf A}
+      | r = e_1 e_2 e_3 e_4 \paren {\mathbf A}
+      | c = Definition of $e$
+}}
+{{eqn | r = \mathbf A \mathbf E_4 \mathbf E_3 \mathbf E_2 \mathbf E_1
+      | c = Cases $1$ and $2$
+}}
+{{eqn | r = \mathbf A e_4 \paren {\mathbf I} \mathbf E_3 \mathbf E_2 \mathbf E_1
+}}
+{{eqn | r = \mathbf A e_3 e_4 \paren {\mathbf I} \mathbf E_2 \mathbf E_1
+}}
+{{eqn | r = \mathbf A e_2 e_3 e_4 \paren {\mathbf I} \mathbf E_1
+}}
+{{eqn | r = \mathbf A e_1 e_2 e_3 e_4 \paren {\mathbf I}
+}}
+{{eqn | r = \mathbf A e \paren {\mathbf I}
+}}
+{{eqn | r = \mathbf A \mathbf E
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Number Plus Square Root/Proof 1}
+Tags: Square Root of Number Plus or Minus Square Root
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]] such that $a^2 - b > 0$.
+Then:
+{{:Square Root of Number Plus Square Root}}
+\end{theorem}
+
+\begin{proof}
+We are given that $a^2 - b > 0$.
+Then:
+:$a > \sqrt b$
+and so $\displaystyle \sqrt {a + \sqrt b}$ is defined on the [[Definition:Real Number|real numbers]].
+Let $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
+[[Definition:Square Function|Squaring]] both sides gives:
+{{begin-eqn}}
+{{eqn | l = a + \sqrt b
+      | r = \paren {\sqrt x + \sqrt y}^2
+      | c = 
+}}
+{{eqn | r = x + y + 2 \sqrt {x y}
+      | c = 
+}}
+{{end-eqn}}
+Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
+{{explain|How do you know that the $a$ and $b$ which are $x + y$ and $2 \sqrt {x y}$ are the same $a$ and $b$ that you started with?
+$x$ and $y$ are free to choose. They were introduced by hand, and can be set to any value provided that they satisfy the constraint above. This is similar to the proof of [[Cardano's Formula]].
+In that case it needs to be explained. As it stands, it looks as though $x$ and $y$ are pulled out of thin air, with no actual indication that having picked them, they bear the relations given to $a$ and $b$ as presented. It's incredibly confusing for beginners, and others whose abilities and understanding are limited, like me.}}
+From $\sqrt b = 2 \sqrt {x y}$ we get:
+{{begin-eqn}}
+{{eqn | l = \sqrt b
+      | r = 2 \sqrt {x y}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = b
+      | r = 4 x y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x y
+      | r = \frac b 4
+      | c = 
+}}
+{{end-eqn}}
+By [[Viète's Formulas]], $x$ and $y$ are solutions to the [[Definition:Quadratic Equation|quadratic equation]]:
+:$z^2 - a z + \dfrac b 4 = 0$
+From [[Solution to Quadratic Equation]]:
+:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
+where $a^2 - b > 0$ (which is a given)
+{{explain|the notation $z_{1, 2}$}}
+{{WLOG}}:
+{{begin-eqn}}
+{{eqn | l = x = z_1
+      | r = \dfrac {a + \sqrt {a^2 - b} } 2
+}}
+{{eqn | l = y = z_2
+      | r = \dfrac {a - \sqrt {a^2 - b} } 2
+}}
+{{end-eqn}}
+Subsituting into $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:
+{{begin-eqn}}
+{{eqn | l = \sqrt {a + \sqrt b}
+      | r = \sqrt x + \sqrt y
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Number Plus Square Root/Proof 2}
+Tags: Square Root of Number Plus or Minus Square Root
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]] such that $a^2 - b > 0$.
+Then:
+{{:Square Root of Number Plus Square Root}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2
+      | r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | c = multiplying out
+}}
+{{eqn | r = a + \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }
+      | c = simplifying
+}}
+{{eqn | r = a + \sqrt {a^2 - \paren {a^2 - b} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = a + \sqrt b
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | r = \sqrt {a + \sqrt b}
+      | c = taking [[Definition:Square Root|square root]] of both sides
+}}
+{{end-eqn}}
+{{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Number Minus Square Root/Proof 2}
+Tags: Square Root of Number Plus or Minus Square Root
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]] such that $a^2 - b > 0$.
+Then:
+{{:Square Root of Number Minus Square Root}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2
+      | r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 - 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | c = multiplying out
+}}
+{{eqn | r = a - \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }
+      | c = simplifying
+}}
+{{eqn | r = a - \sqrt {a^2 - \paren {a^2 - b} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = a - \sqrt b
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | r = \sqrt {a - \sqrt b}
+      | c = taking [[Definition:Square Root|square root]] of both sides
+}}
+{{end-eqn}}
+{{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of Number Minus Square Root/Proof 1}
+Tags: Square Root of Number Plus or Minus Square Root
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]] such that $a^2 - b > 0$.
+Then:
+{{:Square Root of Number Plus Square Root}}
+\end{theorem}
+
+\begin{proof}
+We are given that $a^2 - b > 0$.
+Then:
+:$a > \sqrt b$
+and so $\displaystyle \sqrt {a - \sqrt b}$ is defined on the [[Definition:Real Number|real numbers]].
+Let $\displaystyle \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
+Observe that:
+:$0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$
+[[Definition:Square Function|Squaring]] both sides gives:
+{{begin-eqn}}
+{{eqn | l = a - \sqrt b
+      | r = \paren {\sqrt x - \sqrt y}^2
+      | c = 
+}}
+{{eqn | r = x + y - 2 \sqrt {x y}
+      | c = 
+}}
+{{end-eqn}}
+Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
+From $\sqrt b = 2 \sqrt {x y}$ we get:
+{{begin-eqn}}
+{{eqn | l = \sqrt b
+      | r = 2 \sqrt {x y}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = b
+      | r = 4 x y
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x y
+      | r = \frac b 4
+      | c = 
+}}
+{{end-eqn}}
+By [[Viète's Formulas]], $x$ and $y$ are solutions to the [[Definition:Quadratic Equation|quadratic equation]]:
+:$z^2 - a z + \dfrac b 4 = 0$
+From [[Solution to Quadratic Equation]]:
+:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
+where $a^2 - b > 0$ (which is a given)
+Because we have that $x > y$:
+{{begin-eqn}}
+{{eqn | l = x = z_1
+      | r = \dfrac {a + \sqrt {a^2 - b} } 2
+}}
+{{eqn | l = y = z_2
+      | r = \dfrac {a - \sqrt {a^2 - b} } 2
+}}
+{{end-eqn}}
+Subsituting into $\displaystyle \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$:
+{{begin-eqn}}
+{{eqn | l = \sqrt {a - \sqrt b}
+      | r = \sqrt x - \sqrt y
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exchange of Columns as Sequence of Other Elementary Column Operations}
+Tags: Elementary Column Operations
+
+\begin{theorem}
+Let $\mathbf A$ be an $m \times n$ [[Definition:Matrix|matrix]].
+Let $i, j \in \closedint 1 m: i \ne j$
+Let $\kappa_k$ denote the $k$th [[Definition:Column of Matrix|column]] of $\mathbf A$ for $1 \le k \le n$:
+:$\kappa_k = \begin {pmatrix} a_{1 k} \\ a_{2 k} \\ \vdots \\ a_{m k} \end {pmatrix}$
+Let $e$ be the [[Definition:Elementary Column Operation|elementary column operation]] acting on $\mathbf A$ as:
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 3
+        | t = Interchange [[Definition:Column of Matrix|columns]] $i$ and $j$
+        | m = \kappa_i \leftrightarrow \kappa_j
+}}
+{{end-axiom}}
+Then $e$ can be expressed as a [[Definition:Finite Sequence|finite sequence]] of exactly $4$ instances of the other two [[Definition:Elementary Column Operation|elementary column operations]].
+{{begin-axiom}}
+{{axiom | n = \text {ERO} 1
+        | t = For some $\lambda \in K_{\ne 0}$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Column of Matrix|column]] $i$ by $\lambda$
+        | m = \kappa_i \to \lambda \kappa_i
+}}
+{{axiom | n = \text {ERO} 2
+        | t = For some $\lambda \in K$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Column of Matrix|column]] $j$ to [[Definition:Column of Matrix|column]] $i$
+        | m = \kappa_i \to \kappa_i + \lambda \kappa_j
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+In the below:
+:$\kappa_i$ denotes the initial state of [[Definition:Column of Matrix|column]] $i$
+:$\kappa_j$ denotes the initial state of [[Definition:Column of Matrix|column]] $j$
+:$\kappa_i'$ denotes the state of [[Definition:Column of Matrix|column]] $i$ after having had the latest [[Definition:Elementary Column Operation|elementary column operation]] applied
+:$\kappa_j'$ denotes the state of [[Definition:Column of Matrix|column]] $j$ after having had the latest [[Definition:Elementary Column Operation|elementary column operation]] applied.
+$(1)$: Apply [[Definition:Elementary Column Operation|$\text {ECO} 2$]] to [[Definition:Column of Matrix|column]] $j$ for $\lambda = 1$:
+:$\kappa_j \to \kappa_j + \kappa_i$
+After this operation:
+{{begin-eqn}}
+{{eqn | l = \kappa_i'
+      | r = \kappa_i
+}}
+{{eqn | l = \kappa_j'
+      | r = \kappa_i + \kappa_j
+}}
+{{end-eqn}}
+{{qed|lemma}}
+$(2)$: Apply [[Definition:Elementary Column Operation|$\text {ECO} 2$]] to [[Definition:Column of Matrix|column]] $i$ for $\lambda = -1$:
+:$\kappa_i \to \kappa_i + \paren {-\kappa_j}$
+After this operation:
+{{begin-eqn}}
+{{eqn | l = \kappa_i'
+      | r = \kappa_i - \paren {\kappa_i + \kappa_j}
+}}
+{{eqn | r = -\kappa_j
+}}
+{{eqn | l = \kappa_j'
+      | r = \kappa_i + \kappa_j
+}}
+{{end-eqn}}
+{{qed|lemma}}
+$(3)$: Apply [[Definition:Elementary Column Operation|$\text {ECO} 2$]] to [[Definition:Column of Matrix|column]] $j$ for $\lambda = 1$:
+:$\kappa_j \to \kappa_j + \kappa_i$
+After this operation:
+{{begin-eqn}}
+{{eqn | l = \kappa_i'
+      | r = -\kappa_j
+}}
+{{eqn | l = \kappa_j'
+      | r = \kappa_i + \kappa_j - \kappa_j
+}}
+{{eqn | r = \kappa_i
+}}
+{{end-eqn}}
+{{qed|lemma}}
+$(4)$: Apply [[Definition:Elementary Column Operation|$\text {ECO} 1$]] to [[Definition:Column of Matrix|column]] $i$ for $\lambda = -1$:
+:$\kappa_i \to -\kappa_i$
+After this operation:
+{{begin-eqn}}
+{{eqn | l = \kappa_i'
+      | r = -\paren {-\kappa_j}
+}}
+{{eqn | r = \kappa_j
+}}
+{{eqn | l = \kappa_j'
+      | r = \kappa_i
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Thus, after all the $4$ [[Definition:Elementary Column Operation|elementary column operations]] have been applied, we have:
+{{begin-eqn}}
+{{eqn | l = \kappa_i'
+      | r = \kappa_j
+}}
+{{eqn | l = \kappa_j'
+      | r = \kappa_i
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Effect of Elementary Column Operations on Determinant}
+Tags: Determinants, Elementary Column Operations
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix of order $n$]].
+Let $\map \det {\mathbf A}$ denote the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Take the [[Definition:Elementary Column Operation|elementary column operations]]:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 1
+        | t = For some $\lambda$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Column of Matrix|column]] $i$ by $\lambda$
+        | m = \kappa_i \to \lambda \kappa_i
+}}
+{{axiom | n = \text {ECO} 2
+        | t = For some $\lambda$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Column of Matrix|column]] $j$ to [[Definition:Column of Matrix|column]] $i$
+        | m = \kappa_i \to \kappa_i + \lambda \kappa_j
+}}
+{{axiom | n = \text {ECO} 3
+        | t = Exchange [[Definition:Column of Matrix|columns]] $i$ and $j$
+        | m = \kappa_i \leftrightarrow \kappa_j
+}}
+{{end-axiom}}
+Applying $\text {ECO} 1$ has the effect of multiplying $\map \det {\mathbf A}$ by $\lambda$.
+Applying $\text {ECO} 2$ has no effect on $\map \det {\mathbf A}$.
+Applying $\text {ECO} 3$ has the effect of multiplying $\map \det {\mathbf A}$ by $-1$.
+\end{theorem}
+
+\begin{proof}
+From [[Elementary Column Operations as Matrix Multiplications]], an [[Definition:Elementary Column Operation|elementary column operation]] on $\mathbf A$ is equivalent to [[Definition:Matrix Product (Conventional)|matrix multiplication]] by the [[Definition:Elementary Column Matrix|elementary column matrices]] corresponding to the [[Definition:Elementary Column Operation|elementary column operations]].
+From [[Determinant of Elementary Column Matrix]], the [[Definition:Determinant of Matrix|determinants]] of those [[Definition:Elementary Column Matrix|elementary column matrices]] are as follows:
+=== [[Determinant of Elementary Column Matrix/Scale Column|Scale Column]] ===
+{{:Determinant of Elementary Column Matrix/Scale Column}}
+=== [[Determinant of Elementary Column Matrix/Scale Column and Add|Add Scalar Product of Column to Another]] ===
+{{:Determinant of Elementary Column Matrix/Scale Column and Add}}
+=== [[Determinant of Elementary Column Matrix/Exchange Columns|Exchange Columns]] ===
+{{:Determinant of Elementary Column Matrix/Exchange Columns}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Column Matrix}
+Tags: Determinants, Elementary Matrices, Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $\mathbf E$ be an [[Definition:Elementary Column Matrix|elementary column matrix]].
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E$ is as follows:
+\end{theorem}<|endoftext|>
+\section{Determinant of Elementary Column Matrix/Scale Column}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_1$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\text {ECO} 1$:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 1
+        | t = For some $\lambda \ne 0$, [[Definition:Matrix Scalar Product|multiply]] [[Definition:Column of Matrix|column]] $k$ by $\lambda$
+        | m = \kappa_k \to \lambda \kappa_k
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_1$ be the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to $e_1$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_1$ is:
+:$\map \det {\mathbf E_1} = \lambda$
+\end{theorem}
+
+\begin{proof}
+By [[Elementary Matrix corresponding to Elementary Column Operation/Scale Column|Elementary Matrix corresponding to Elementary Column Operation: Scale Column]], the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to $e_1$ is of the form:
+:$E_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end{cases}$
+where:
+:$E_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf E_1$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$
+:$\delta_{a b}$ is the [[Definition:Kronecker Delta|Kronecker delta]]:
+::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
+Thus when $a \ne b$, $E_{a b} = 0$.
+This means that $\mathbf E_1$ is a [[Definition:Diagonal Matrix|diagonal matrix]].
+{{begin-eqn}}
+{{eqn | l = \displaystyle \map \det {\mathbf E_1}
+      | r = \prod_i E_{i i}
+      | c = [[Determinant of Diagonal Matrix]]
+      | cc= where the [[Definition:Index Variable of Indexed Product|index variable]] $i$ ranges over the [[Definition:Order of Square Matrix|order]] of $\mathbf E_1$
+}}
+{{eqn | r = \prod_i \paren {\begin {cases} 1 & : i \ne k \\ \lambda & : a = k \end{cases} }
+      | c = 
+}}
+{{eqn | r = \prod_{i \mathop \ne k} 1 \times \prod_{i \mathop = k} \lambda
+      | c = 
+}}
+{{eqn | r = 1 \times \lambda
+      | c = 
+}}
+{{eqn | r = \lambda
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Column Matrix/Scale Column and Add}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_2$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\text {ECO} 2$:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 2
+        | t = For some $\lambda$, add $\lambda$ [[Definition:Matrix Scalar Product|times]] [[Definition:Column of Matrix|column]] $j$ to [[Definition:Column of Matrix|column]] $i$
+        | m = \kappa_i \to \kappa_i + \lambda \kappa_j
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_2$ be the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to $e_2$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_2$ is:
+:$\map \det {\mathbf E_2} = 1$
+\end{theorem}
+
+\begin{proof}
+By [[Elementary Matrix corresponding to Elementary Column Operation/Scale Column and Add|Elementary Matrix corresponding to Elementary Column Operation: Scale Column and Add]], $\mathbf E_2$ is of the form:
+:$E_{a b} = \delta_{a b} + \lambda \cdot \delta_{b i} \cdot \delta_{j a}$
+where:
+:$E_{a b}$ denotes the [[Definition:Element of Matrix|element]] of $\mathbf E$ whose [[Definition:Index of Matrix Element|indices]] are $\tuple {a, b}$
+:$\delta_{a b}$ is the [[Definition:Kronecker Delta|Kronecker delta]]:
+::$\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
+Because $i \ne j$ it follows that:
+:if $a = i$ and $b = j$ then $a \ne b$
+Hence when $a = b$ we have that:
+:$\delta_{b i} \cdot \delta_{j a} = 0$
+Hence the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf E_2$ are all equal to $1$.
+We also have that $\delta_{b i} \cdot \delta_{j a} = 1$ {{iff}} $a = i$ and $b = j$.
+Hence, all [[Definition:Element of Matrix|elements]] of $\mathbf E_2$ apart from the [[Definition:Diagonal Element|diagonal elements]] and $a_{i j}$ are equal to $0$.
+Thus $\mathbf E_2$ is a [[Definition:Triangular Matrix|triangular matrix]] (either [[Definition:Upper Triangular Matrix|upper]] or [[Definition:Lower Triangular Matrix|lower]]).
+From [[Determinant of Triangular Matrix]], $\map \det {\mathbf E_2}$ is equal to the [[Definition:Multiplication|product]] of all the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf E_2$.
+But as we have seen, these are all equal to $1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiple of Column Added to Column of Determinant}
+Tags: Determinants
+
+\begin{theorem}
+Let $\mathbf A = \begin {bmatrix}
+a_{1 1} & \cdots & a_{1 r} & \cdots & a_{1 s} & \cdots & a_{1 n} \\
+a_{2 1} & \cdots & a_{2 r} & \cdots & a_{2 s} & \cdots & a_{2 n} \\
+ \vdots & \ddots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & \cdots & a_{n r} & \cdots & a_{n s} & \cdots & a_{n n} \\
+\end {bmatrix}$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ denote the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf B = \begin{bmatrix}
+a_{1 1} & \cdots & a_{1 r} + \lambda a_{1 s} & \cdots & a_{1 s} & \cdots & a_{1 n} \\
+a_{2 1} & \cdots & a_{2 r} + \lambda a_{2 s} & \cdots & a_{2 s} & \cdots & a_{2 n} \\
+ \vdots & \ddots &                    \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & \cdots & a_{n r} + \lambda a_{n s} & \cdots & a_{n s} & \cdots & a_{n n} \\
+\end{bmatrix}$.
+Then $\map \det {\mathbf B} = \map \det {\mathbf A}$.
+That is, the value of a [[Definition:Determinant of Matrix|determinant]] remains unchanged if a [[Definition:Constant|constant]] multiple of any [[Definition:Column of Matrix|column]] is added to any other [[Definition:Column of Matrix|column]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\mathbf A^\intercal = \begin {bmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{s 1} & a_{s 2} & \cdots & a_{s n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
+\end {bmatrix}$
+where $\mathbf A^\intercal$ denotes the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Similarly, we have that:
+:$\mathbf B^\intercal = \begin{bmatrix}
+a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{r 1} + \lambda a_{s 1} & a_{r 2} + \lambda a_{s 2} & \cdots & a_{r n} + \lambda a_{s n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{s 1} & a_{s 2} & \cdots & a_{s n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
+\end {bmatrix}$
+From [[Multiple of Row Added to Row of Determinant]]:
+:$\map \det {\mathbf B^\intercal} = \map \det {\mathbf A^\intercal}$
+From from [[Determinant of Transpose]]:
+:$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
+:$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant with Column Multiplied by Constant}
+Tags: Determinants
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\mathbf B$ be the [[Definition:Square Matrix|matrix]] resulting from one [[Definition:Column of Matrix|column]] of $\mathbf A$ having been multiplied by a [[Definition:Constant|constant]] $c$.
+Then:
+:$\map \det {\mathbf B} = c \map \det {\mathbf A}$
+That is, multiplying one [[Definition:Column of Matrix|column]] of a [[Definition:Square Matrix|square matrix]] by a [[Definition:Constant|constant]] multiplies its [[Definition:Determinant of Matrix|determinant]] by that [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\mathbf A = \begin{bmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 r} & \cdots & a_{1 n} \\
+a_{2 1} & a_{2 2} & \cdots & a_{2 r} & \cdots & a_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\
+\end{bmatrix}$
+:$\mathbf B = \begin{bmatrix}
+b_{1 1} & b_{1 2} & \cdots & b_{1 r} & \cdots & b_{1 n} \\
+b_{2 1} & b_{2 2} & \cdots & b_{2 r} & \cdots & b_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+b_{n 1} & b_{n 2} & \cdots & b_{n r} & \cdots & b_{n n} \\
+\end{bmatrix} = \begin{bmatrix}
+a_{1 1} & a_{1 2} & \cdots & c a_{1 r} & \cdots & a_{1 n} \\
+a_{2 1} & a_{2 2} & \cdots & c a_{2 r} & \cdots & a_{1 n} \\
+ \vdots &  \vdots & \ddots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n r} & \cdots & a_{n n} \\
+\end{bmatrix}$
+We have that:
+:$\mathbf A^\intercal = \begin {bmatrix}
+a_{1 1} & a_{1 2} & \cdots & a_{1 n} \\
+a_{2 1} & a_{2 2} & \cdots & a_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{r 1} & a_{r 2} & \cdots & a_{r n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
+\end {bmatrix}$
+where $\mathbf A^\intercal$ denotes the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+Similarly, we have that:
+:$\mathbf B^\intercal = \begin{bmatrix}
+a_{1 1} & a_{1 2} & \ldots & a_{1 n} \\
+a_{2 1} & a_{2 2} & \ldots & a_{2 n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+c a_{r 1} & c a_{r 2} & \cdots & c a_{r n} \\
+ \vdots &  \vdots & \ddots &  \vdots \\
+a_{n 1} & a_{n 2} & \cdots & a_{n n} \\
+\end {bmatrix}$
+From [[Determinant with Row Multiplied by Constant]]:
+:$\map \det {\mathbf B^\intercal} = c \map \det {\mathbf A^\intercal}$
+From from [[Determinant of Transpose]]:
+:$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$
+:$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Elementary Column Matrix/Exchange Columns}
+Tags: Determinant of Elementary Matrix
+
+\begin{theorem}
+Let $e_3$ be the [[Definition:Elementary Column Operation|elementary column operation]] $\text {ECO} 3$:
+{{begin-axiom}}
+{{axiom | n = \text {ECO} 3
+        | t = Exchange [[Definition:Column of Matrix|columns]] $i$ and $j$
+        | m = \kappa_i \leftrightarrow \kappa_j
+}}
+{{end-axiom}}
+which is to operate on some arbitrary [[Definition:Matrix Space|matrix space]].
+Let $\mathbf E_3$ be the [[Definition:Elementary Column Matrix|elementary column matrix]] corresponding to $e_3$.
+The [[Definition:Determinant of Matrix|determinant]] of $\mathbf E_3$ is:
+:$\map \det {\mathbf E_3} = -1$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf I$ denote the [[Definition:Unit Matrix|unit matrix]] of arbitrary [[Definition:Order of Square Matrix|order]] $n$.
+By [[Determinant of Unit Matrix]]:
+:$\map \det {\mathbf I} = 1$
+Let $\rho$ be the [[Definition:Permutation on n Letters|permutation]] on $\tuple {1, 2, \ldots, n}$ which [[Definition:Transposition|transposes]] $i$ and $j$.
+From [[Parity of K-Cycle]], $\map \sgn \rho = -1$.
+By definition we have that $\mathbf E_3$ is $\mathbf I$ with [[Definition:Column of Matrix|columns]] $i$ and $j$ [[Definition:Transposition|transposed]].
+By the definition of a [[Definition:Determinant of Matrix|determinant]]:
+:$\displaystyle \map \det {\mathbf I} = \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }$
+By [[Permutation of Determinant Indices]]:
+:$\displaystyle \map \det {\mathbf E_3} = \sum_\lambda \paren {\map \sgn \rho \map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }$
+We can take $\map \sgn \rho = -1$ outside the summation because it is constant, and so we get:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf E_3}
+      | r = \map \sgn \rho \sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k \map \rho k} }
+      | c = 
+}}
+{{eqn | r = -\sum_\lambda \paren {\map \sgn \lambda \prod_{k \mathop = 1}^n a_{\map \lambda k k} }
+      | c = 
+}}
+{{eqn | r = -\map \det {\mathbf I}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Determinant of Rescaling Matrix/Corollary}
+Tags: Determinants
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\lambda$ be a [[Definition:Scalar (Matrix Theory)|scalar]].
+Let $\lambda \mathbf A$ denote the [[Definition:Matrix Scalar Product|scalar product]] of $\mathbf A$ by $\lambda$.
+Then:
+:$\map \det {\lambda \mathbf A} = \lambda^n \map \det {\mathbf A}$
+where $\det$ denotes [[Definition:Determinant of Matrix|determinant]].
+\end{theorem}
+
+\begin{proof}
+For $1 \le k \le n$, let $e_k$ be the [[Definition:Elementary Row Operation|elementary row operation]] that [[Definition:Matrix Scalar Product|multiplies]] [[Definition:Row of Matrix|row]] $k$ of $\mathbf A$ by $\lambda$.
+By definition of the [[Definition:Matrix Scalar Product|scalar product]], $\lambda \mathbf A$ is obtained by [[Definition:Matrix Scalar Product|multiplying]] every [[Definition:Row of Matrix|row]] of $\mathbf A$ by $\lambda$.
+That is the same as applying $e_k$ to $\mathbf A$ for each of $k \in \set {1, 2, \ldots, n}$.
+Let $\mathbf E_k$ denote the [[Definition:Elementary Row Matrix|elementary row matrix]] corresponding to $e_k$.
+By [[Determinant of Elementary Row Matrix/Scale Row|Determinant of Elementary Row Matrix: Scale Row]]:
+:$\map \det {\mathbf E_k} = \lambda$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \lambda \mathbf A
+      | r = \prod_{k \mathop = 1}^n \mathbf E_k \mathbf A
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \det {\lambda \mathbf A}
+      | r = \map \det {\prod_{k \mathop = 1}^n \mathbf E_k \mathbf A}
+      | c = 
+}}
+{{eqn | r = \paren {\prod_{k \mathop = 1}^n \map \det {\mathbf E_k} } \map \det {\mathbf A}
+      | c = [[Determinant of Matrix Product]]
+}}
+{{eqn | r = \paren {\prod_{k \mathop = 1}^n \lambda} \map \det {\mathbf A}
+      | c = [[Determinant of Elementary Row Matrix/Scale Row|Determinant of Elementary Row Matrix: Scale Row]]
+}}
+{{eqn | r = \lambda^n \map \det {\mathbf A}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Row Operations is Row Operation}
+Tags: Row Operations
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma_1$ be a [[Definition:Row Operation|row operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Let $\Gamma_2$ be a [[Definition:Row Operation|row operation]] which transforms $\mathbf B$ to another new [[Definition:Matrix|matrix]] $\mathbf C \in \map \MM {m, n}$.
+Then there exists another [[Definition:Row Operation|row operation]] $\Gamma$ which transforms $\mathbf A$ back to $\mathbf C$ such that $\Gamma$ consists of $\Gamma_1$ followed by $\Gamma_2$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that compose $\Gamma_1$.
+Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Row Matrix|elementary row matrices]].
+Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] that compose $\Gamma_2$.
+Let $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Row Matrix|elementary row matrices]].
+From [[Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf R_1 \mathbf A = \mathbf B$
+where $\mathbf R$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
+:$\mathbf R_1 = \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
+Also from [[Row Operation is Equivalent to Pre-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf R_2 \mathbf B = \mathbf C$
+where $\mathbf R_2$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$:
+:$\mathbf R_2 = \mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1$
+Hence we have:
+:$\mathbf R_2 \mathbf R_1 \mathbf A = \mathbf C$
+where $\mathbf R := \mathbf R_2 \mathbf R_1$ is the [[Definition:Matrix Product (Conventional)|product]]:
+:$\mathbf F_l \mathbf F_{l - 1} \dotsb \mathbf F_2 \mathbf F_1 \mathbf E_k \mathbf E_{k - 1} \dotsb \mathbf E_2 \mathbf E_1$
+Let $\Gamma$ be the [[Definition:Row Operation|row operation]] composed of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Row Operation|elementary row operations]] $\tuple {e_1, e_2, \ldots, e_{k - 1}, e_k, f_1, f_2, \ldots, f_{l - 1}, f_l}$.
+Thus $\Gamma$ is a [[Definition:Row Operation|row operation]] which transforms $\mathbf A$ into $\mathbf C$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Column Operations is Column Operation}
+Tags: Column Operations
+
+\begin{theorem}
+Let $\map \MM {m, n}$ be a [[Definition:Metric Space|metric space]] of [[Definition:Order of Matrix|order]] $m \times n$ over a [[Definition:Field (Abstract Algebra)|field]] $K$.
+Let $\mathbf A \in \map \MM {m, n}$ be a [[Definition:Matrix|matrix]]. 
+Let $\Gamma_1$ be a [[Definition:Column Operation|column operation]] which transforms $\mathbf A$ to a new [[Definition:Matrix|matrix]] $\mathbf B \in \map \MM {m, n}$.
+Let $\Gamma_2$ be a [[Definition:Column Operation|column operation]] which transforms $\mathbf B$ to another new [[Definition:Matrix|matrix]] $\mathbf C \in \map \MM {m, n}$.
+Then there exists another [[Definition:Column Operation|column operation]] $\Gamma$ which transforms $\mathbf A$ back to $\mathbf C$ such that $\Gamma$ consists of $\Gamma_1$ followed by $\Gamma_2$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {e_i}_{1 \mathop \le i \mathop \le k}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] that compose $\Gamma_1$.
+Let $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Column Matrix|elementary column matrices]].
+Let $\sequence {f_i}_{1 \mathop \le i \mathop \le l}$ be the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] that compose $\Gamma_2$.
+Let $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$ be the corresponding [[Definition:Finite Sequence|finite sequence]] of the [[Definition:Elementary Column Matrix|elementary column matrices]].
+From [[Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf A \mathbf R_1 = \mathbf B$
+where $\mathbf R$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf E_i}_{1 \mathop \le i \mathop \le k}$:
+:$\mathbf R_1 = \mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k$
+Also from [[Column Operation is Equivalent to Post-Multiplication by Product of Elementary Matrices]], we have:
+:$\mathbf B \mathbf R_2 = \mathbf C$
+where $\mathbf R_2$ is the [[Definition:Matrix Product (Conventional)|product]] of $\sequence {\mathbf F_i}_{1 \mathop \le i \mathop \le l}$:
+:$\mathbf R_2 = \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$
+Hence we have:
+:$\mathbf A \mathbf R_1 \mathbf R_2 = \mathbf C$
+where $\mathbf R := \mathbf R_1 \mathbf R_2$ is the [[Definition:Matrix Product (Conventional)|product]]:
+:$\mathbf E_1 \mathbf E_2 \dotsb \mathbf E_{k - 1} \mathbf E_k \mathbf F_1 \mathbf F_2 \dotsb \mathbf F_{l - 1} \mathbf F_l$
+Let $\Gamma$ be the [[Definition:Column Operation|column operation]] composed of the [[Definition:Finite Sequence|finite sequence]] of [[Definition:Elementary Column Operation|elementary column operations]] $\tuple {e_1, e_2, \ldots, e_{k - 1}, e_k, f_1, f_2, \ldots, f_{l - 1}, f_l}$.
+Thus $\Gamma$ is a [[Definition:Column Operation|column operation]] which transforms $\mathbf A$ into $\mathbf C$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Determinant}
+Tags: Determinants
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+{{TFAE|def = Determinant of Matrix|view = the determinant of $\mathbf A$}}
+\end{theorem}
+
+\begin{proof}
+This is proved in [[Expansion Theorem for Determinants]].
+{{qed}}
+[[Category:Determinants]]
+ejpg48vysa61ntozt8m9l4q41r80xl6
+\end{proof}<|endoftext|>
+\section{Intersection With Singleton is Disjoint if Not Element}
+Tags: Singletons, Disjoint Sets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\set x$ be the [[Definition:Singleton|singleton of $x$]].
+Then:
+:$x \notin S$ {{iff}} $\set x \cap S = \O$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o =
+      | r = \set x \cap S = \O
+      | c =
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : \lnot \paren{ y \in \set x \cap S}
+      | c = {{Defof|Empty Set}}
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : \lnot \paren{ y \in \set x \land y \in S}
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : \lnot \paren{ y \in \set x \land \lnot \lnot \paren{y \in S} }
+      | c = [[Double Negation Introduction]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : y \in \set x \implies \lnot \paren{y \in S}
+      | c = [[Implication Equivalent to Negation of Conjunction with Negative]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : y \in \set x \implies y \notin S
+      | c = {{Defof|Element|Not Element}}
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = \forall y : y = x \implies y \notin S
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = x \notin S
+      | c = {{Defof|Equality}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Singletons]]
+[[Category:Disjoint Sets]]
+osfxprp7vvy9sl3nsfye0u7x1a4su30
+\end{proof}<|endoftext|>
+\section{Determinant of Lower Triangular Matrix}
+Tags: Determinants, Triangular Matrices
+
+\begin{theorem}
+Let $\mathbf T_n$ be a [[Definition:Lower Triangular Matrix|lower triangular matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf T_n}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf T_n$.
+Then $\map \det {\mathbf T_n}$ is equal to the product of all the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf T_n$.
+That is:
+:$\displaystyle \map \det {\mathbf T_n} = \prod_{k \mathop = 1}^n a_{k k}$
+\end{theorem}
+
+\begin{proof}
+From [[Transpose of Upper Triangular Matrix is Lower Triangular]], the [[Definition:Transpose of Matrix|transpose]] $\mathbf T_n^\intercal$ of $\mathbf T_n$ is an [[Definition:Upper Triangular Matrix|upper triangular matrix]].
+From [[Determinant of Upper Triangular Matrix]], the [[Definition:Determinant of Matrix|determinant]] of $\mathbf T_n^\intercal$ is equal to the product of all the [[Definition:Diagonal Element|diagonal elements]] of $\mathbf T_n^\intercal$.
+From [[Determinant of Transpose]], the [[Definition:Determinant of Matrix|determinant]] of $\mathbf T_n^\intercal$ equals the [[Definition:Determinant of Matrix|determinant]] of $\mathbf T_n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Matrix with Adjugate equals Determinant by Unit Matrix}
+Tags: Adjugate Matrices, Determinants
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\adj {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Then:
+:$\paren {\adj {\mathbf A} } \mathbf A = \map \det {\mathbf A} \mathbf I = \mathbf A \paren {\adj {\mathbf A} }$
+where $\mathbf I$ denotes the [[Definition:Unit Matrix|unit matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+\end{theorem}<|endoftext|>
+\section{Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant}
+Tags: Adjugate Matrices, Determinants, Inverse of Matrix is Scalar Product of Adjugate by Reciprocal of Determinant
+
+\begin{theorem}
+Let $\mathbf A = \sqbrk a_n$ be an [[Definition:Invertible Matrix|invertible]] [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order $n$]].
+Let $\map \det {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Let $\adj {\mathbf A}$ be the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$.
+Then:
+:$\mathbf A^{-1} = \dfrac 1 {\map \det {\mathbf A} } \cdot \adj {\mathbf A}$
+where $\mathbf A^{-1}$ denotes the [[Definition:Inverse Matrix|inverse]] of $\mathbf A$
+\end{theorem}<|endoftext|>
+\section{Greedy Algorithm yields Maximal Set}
+Tags: Maximization Problem for Independence Systems
+
+\begin{theorem}
+Let $\struct{S,\mathscr F}$ be an [[Definition:Independence System|independence system]].
+Let $w : S \to \R_{\ge 0}$ be a [[Definition:Weight Function|weight function]].
+Then the [[Maximization Problem (Greedy Algorithm)|Greedy Algorithm]] selects a [[Definition:Maximal Set|maximal set]] $A_0$ in $\mathscr F$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Maximization Problem for Independence Systems]]
+omhyyhdrmhuqurhcydk3z4mmkhrgbcg
+\end{proof}<|endoftext|>
+\section{Greedy Algorithm may not yield Maximum Weight}
+Tags: Maximization Problem for Independence Systems
+
+\begin{theorem}
+Let $\struct{S,\mathscr F}$ be an [[Definition:Independence System|independence system]].
+Let $w : S \to \R_{\ge 0}$ be a [[Definition:Weight Function|weight function]].
+Then the [[Definition:Maximal Set|maximal set]] $A_0 \in \mathscr F$ selected by the [[Maximization Problem (Greedy Algorithm)|Greedy Algorithm]] may not have [[Definition:Maximum|maximum]] [[Definition:Extended Weight Function|weight]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Maximization Problem for Independence Systems]]
+r9h7nporqvvqh3bx6w73oitt97putpg
+\end{proof}<|endoftext|>
+\section{Greedy Algorithm guarantees Maximum Weight iff Matroid}
+Tags: Maximization Problem for Independence Systems
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr I$ be a [[Definition:Non-Empty|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$.
+Then $\mathscr I$ is the [[Definition:Set|set]] of [[Definition:Independent Subset (Matroid)|independent subsets]] of a [[Definition:Matroid|matroid]] on $S$ {{iff}}:
+:$(1) \quad \struct{S, \mathscr I}$ is an [[Definition:Independence System|independence system]]
+:$(2) \quad$ for all [[Definition:Non-Negative Reals|non-negative]] [[Definition:Weight Function|weight functions]] $w : S \to \R_{\ge 0}$, the [[Maximization Problem (Greedy Algorithm)|Greedy Algorithm]] selects $A_w \in \mathscr I$:
+:::::$\forall B \in \mathscr I: \map {w^+} {A_w} \ge \map {w^+} B$
+:where $w^+$ denotes the [[Definition:Extended Weight Function|extended weight function]] of $w$.
+\end{theorem}<|endoftext|>
+\section{Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space}
+Tags: Examples of Vector Spaces, Functional Analysis
+
+\begin{theorem}
+Let $I := \closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\map \CC I$ be a [[Definition:Space of Continuous on Closed Interval Real-Valued Functions|space of continuous on closed interval real-valued functions]].
+Let $\struct {\R, +_\R, \times_\R}$ be the [[Definition:Field of Real Numbers|field of real numbers]].
+Let $\paren +$ be the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise addition of real-valued functions]].
+Let $\paren {\, \cdot \,}$ be the [[Definition:Pointwise Scalar Multiplication of Real-Valued Functions|pointwise scalar multiplication of real-valued functions]].
+Then $\struct {\map \CC I, +, \, \cdot \,}_\R$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+Let $f, g, h \in \map \CC I$ such that:
+:$f, g, h : I \to \R$
+Let $\lambda, \mu \in \R$.
+Let $\map 0 x$ be a [[Definition:Real-Valued Function|real-valued function]] such that:
+:$\map 0 x : I \to 0$.
+Let us use [[Definition:Real Number|real number]] [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]].
+$\forall x \in I$ define [[Definition:Pointwise Addition of Real-Valued Functions|pointwise addition]] as:
+:$\map {\paren {f + g}} x := \map f x +_\R \map g x$.
+Define [[Definition:Pointwise Scalar Multiplication of Real-Valued Functions|pointwise scalar multiplication]] as:
+:$\map {\paren {\lambda \cdot f}} x := \lambda \times_\R \map f x$
+Let $\map {\paren {-f} } x := -\map f x$.
+=== Closure Axiom ===
+By [[Sum Rule for Continuous Functions]], $f + g \in \map \CC I$
+{{qed|lemma}}
+=== Commutativity Axiom ===
+By [[Pointwise Addition on Real-Valued Functions is Commutative]], $f + g = g + f$
+{{qed|lemma}}
+=== Associativity Axiom ===
+By [[Pointwise Addition is Associative]], $\paren {f + g} + h = f + \paren {g + h}$.
+{{qed|lemma}}
+=== Identity Axiom ===
+{{begin-eqn}}
+{{eqn | l = \map {\paren {0 + f} } x
+      | r = \map 0 x +_\R \map f x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = 0 +_\R \map f x
+      | c = Definition of $\map 0 x$
+}}
+{{eqn | r = \map f x
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Inverse Axiom ===
+{{begin-eqn}}
+{{eqn | l =  \map {\paren {f + \paren {-f} } } x 
+      | r = \map f x +_\R \map {\paren {-f} } x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = \map f x +_\R \paren {-1} \times_\R \map f x 
+      | c = Definition of $\map {\paren {-f} } x$
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Distributivity over Scalar Addition ===
+{{begin-eqn}}
+{{eqn | l = \map {\paren { \paren {\lambda +_\R \mu} f} } x
+      | r = \paren {\lambda +_\R \mu} \times_\R \map f x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \lambda \times_\R \map f x +_\R \mu \times_\R \map f x
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \map {\paren {\lambda \cdot f} } x +_\R \map {\paren {\mu\cdot f} } x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \map {\paren {\lambda \cdot f + \mu \cdot f} } x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Distributivity over Vector Addition ===
+{{begin-eqn}}
+{{eqn | l = \lambda \times_\R \map {\paren {f + g} } x
+      | r = \lambda \times_\R \paren {\map f x +_\R \map g x}
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | r = \lambda \times_R \map f x +_\R \lambda \times_\R \map g x
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \map {\paren{\lambda \cdot f} } x +_\R \map {\paren{\lambda \cdot g} } x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \map {\paren {\lambda \cdot f + \mu \cdot f} } x
+      | c = {{Defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Associativity with Scalar Multiplication ===
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\paren {\lambda \times_\R \mu} \cdot f} } x
+      | r = \paren {\lambda \times_\R \mu} \times_\R \map f x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \lambda \times_\R \paren {\mu \times_\R \map f x}
+      | c = [[Real Multiplication is Associative]]
+}}
+{{eqn | r = \lambda \times_\R \map {\paren {\mu \cdot f} } x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \map {\paren {\lambda \cdot \paren {\mu \cdot f} } } x
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Identity for Scalar Multiplication ===
+{{begin-eqn}}
+{{eqn | l = \map {\paren {1 \cdot f} } x 
+      | r = 1 \times_\R \map f x 
+      | c = {{Defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \map f x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Independent Subset is Contained in Base}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\mathscr B$ denote the set of all [[Definition:Base of Matroid|bases]] of $M$.
+Let $A \in \mathscr I$.
+Then:
+:$\exists B \in \mathscr B : A \subseteq B$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Ordered Set|ordered set]] $\struct {\mathscr I, \subseteq}$.
+From [[Element of Finite Ordered Set is Between Maximal and Minimal Elements]]:
+:$\exists B \in \mathscr I : A \subseteq B$ and $B$ is [[Definition:Maximal Element|maximal]] in $\struct {\mathscr I, \subseteq}$.
+By definition of a [[Definition:Base of Matroid|base]]:
+:$B \in \mathscr B$
+{{qed}}
+[[Category:Matroid Theory]]
+bct1wemdzy528q031kddmjegtcudbpp
+\end{proof}<|endoftext|>
+\section{Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements/Corollary}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct{S, \preceq}$ be a [[Definition:Finite|finite]] [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then there exists a [[Definition:Maximal Element|maximal element]] $M \in S$ and a [[Definition:Minimal Element|minimal element]] $m \in S$ such that:
+:$m \preceq x \preceq M$
+\end{theorem}
+
+\begin{proof}
+Let $T = \set{y : x \preceq y}$.
+By the [[Definition:Reflexive Relation|reflexivity]] of the [[Definition:Ordering|ordering]] $\preceq$:
+:$x \preceq x$
+So $x \in T$ and $T$ is [[Definition:Non-Empty Set|non-empty]].
+From [[Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements]]:
+:$\struct{T, \preceq}$ has a [[Definition:Maximal Element|maximal element]] $M \in T$
+We now show that $M$ is a [[Definition:Maximal Element|maximal element]] in $\struct{S, \preceq}$. 
+Let $y \in S$ such that:
+:$M \preceq y$
+By the [[Definition:Transitive Relation|transitiviy]] of the [[Definition:Ordering|ordering]] $\preceq$:
+:$x \prec y$
+So $y \in T$.
+By the definition of a [[Definition:Maximal Element|maximal element]]:
+:$y = M$
+Similarly for $T' = \set{y : y \preceq x}$:
+:$\struct{T', \preceq}$ has a [[Definition:Minimal Element|minimal element]] $m \in T'$
+and $m$ is a [[Definition:Minimal Element|minimal element]] in $\struct{S, \preceq}$
+{{qed}}
+[[Category:Order Theory]]
+qptimv3q5zr0p4ls4ge0z0tr9gluzdq
+\end{proof}<|endoftext|>
+\section{Equivalent Conditions for Element is Loop}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\sigma$ denote the [[Definition:Closure Operator (Matroid)|closure operator]] on $M$.
+Let $\rho$ denote the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $\mathscr B$ denote the set of all [[Definition:Base of Matroid|bases]] of $M$.
+Let $x \in S$.
+{{TFAE}}
+:$(1)\quad x$ is a [[Definition:Loop (Matroid)|loop]]
+:$(2)\quad x \in \map \sigma \O$ 
+:$(3)\quad \map \rho {\set x} = 0$
+:$(4)\quad \set x$ is a [[Definition:Circuit (Matroid)|circuit]]
+:$(5)\quad x$ is not an [[Definition:Element|element]] of any $B \in \mathscr B$
+\end{theorem}
+
+\begin{proof}
+=== Condition $(1)$ iff Condition $(2)$ ===
+Follows immediately from [[Element is Loop iff Member of Closure of Empty Set]].
+{{qed|lemma}} 
+=== Condition $(1)$ iff Condition $(3)$ ===
+Follows immediately from [[Element is Loop iff Rank is Zero]].
+{{qed|lemma}} 
+=== Condition $(1)$ iff Condition $(4)$ ===
+Follows immediately from [[Element is Loop iff Singleton is Circuit]].
+{{qed|lemma}} 
+=== Condition $(1)$ iff Condition $(5)$ ===
+Follows immediately from the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Element is Member of Base iff Not Loop]].
+{{qed}}
+[[Category:Matroid Theory]]
+84i77rbmvcca6a8jdcv7zwft3kdctww
+\end{proof}<|endoftext|>
+\section{Power Set of Doubleton}
+Tags: Power Set, Doubletons
+
+\begin{theorem}
+Let $x, y$ be [[Definition:Distinct|distinct]] [[Definition:Object|objects]].
+Then the [[Definition:Power Set|power set]] of the [[Definition:Doubleton|doubleton]] $\set {x, y}$ is:
+:$\powerset {\set {x, y}} = \big \{ \O, \set x, \set y, \set {x,y} \big \}$
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Subset|subset]]:
+:$\set x , \set y, \set {x, y} \subseteq \set{x, y}$
+Let $A \subseteq \set {x, y}$:
+:$A \ne \set x, \set y, \set {x,y}$
+From [[Definition:Set Equality|set equality]]:
+:$\set {x,y} \not \subseteq A$
+From [[Doubleton of Elements is Subset]]:
+:either $x \notin A$ or $y \notin A$.
+{{WLOG}} assume that $x \notin A$.
+From [[Intersection With Singleton is Disjoint if Not Element]]:
+:$A \cap \set x = \O$
+From [[Subset of Set Difference iff Disjoint Set]]:
+:$A \subseteq \set{x, y} \setminus \set x$
+From [[Set Difference of Doubleton and Singleton is Singleton]]:
+:$A \subseteq \set y$
+From [[Definition:Set Equality|set equality]]:
+:$\set y \not \subseteq A$
+From [[Singleton of Element is Subset]]:
+:$y \notin A$.
+From [[Intersection With Singleton is Disjoint if Not Element]]:
+:$A \cap \set y = \O$
+From [[Subset of Set Difference iff Disjoint Set]]:
+:$A \subseteq \set y \setminus \set y$
+From [[Set Difference with Self is Empty Set]]:
+:$A \subseteq \O$
+From [[Empty Set is Subset of All Sets]]:
+:$\O \subseteq A$
+From [[Definition:Set Equality|set equality]]:
+:$A = \O$
+It follows that:
+:$\powerset {\set {x, y}} = \big \{ \O, \set x, \set y, \set {x,y} \big \}$
+{{qed}}
+[[Category:Power Set]]
+[[Category:Doubletons]]
+mcqoyqldh5icrnt4mrvnznotubp8h9r
+\end{proof}<|endoftext|>
+\section{Doubleton of Elements is Subset}
+Tags: Subsets, Doubletons
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\set {x,y}$ be the [[Definition:Doubleton|doubleton]] of distinct $x$ and $y$.
+Then:
+:$x, y \in S \iff \set {x,y} \subseteq S$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x, y \in S$.
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq S$
+:$\set y \subseteq S$
+From [[Union of Subsets is Subset]]:
+:$\set x \cup \set y \subseteq S$
+From [[Union of Disjoint Singletons is Doubleton]]:
+: $\set x \cup \set y = \set {x, y}$
+Hence:
+:$\set {x,y} \subseteq S$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\set {x,y} \subseteq S$.
+From the definition of a [[Definition:Subset|subset]]:
+:$x \in \set {x,y} \implies x \in S$
+:$y \in \set {x,y} \implies y \in S$
+{{qed}}
+[[Category:Subsets]]
+[[Category:Doubletons]]
+snl24lnninfwoacfvoj1ighxevol7be
+\end{proof}<|endoftext|>
+\section{Sum of Unitary Divisors of Power of Prime}
+Tags: Prime Numbers, Sum of Unitary Divisors
+
+\begin{theorem}
+Let $n = p^k$ be the [[Definition:Power (Algebra)|power]] of a [[Definition:Prime Number|prime number]] $p$.
+Then the sum of all [[Definition:Positive Integer|positive]] [[Definition:Unitary Divisor|unitary divisors]] of $n$ is $1 + n$.
+\end{theorem}
+
+\begin{proof}
+Let $d \divides n$.
+By [[Divisors of Power of Prime]], $d = p^a$ for some [[Definition:Positive Integer|positive integer]] $a \le k$.
+We have $\dfrac n d = p^{k - a}$.
+Suppose $d$ is a [[Definition:Unitary Divisor|unitary divisor]] of $n$.
+Then $d$ and $\dfrac n d$ are [[Definition:Coprime Integers|coprime]].
+If both $a, k - a \ne 0$, $p^a$ and $p^{k - a}$ have a [[Definition:Common Divisor|common divisor]]: $p$.
+Hence either $a = 0$ or $k - a = 0$.
+This leads to $d = 1$ or $p^k$.
+Hence the sum of all [[Definition:Positive Integer|positive]] [[Definition:Unitary Divisor|unitary divisors]] of $n$ is:
+:$1 + p^k = 1 + n$
+{{qed}}
+[[Category:Prime Numbers]]
+[[Category:Sum of Unitary Divisors]]
+ebj0fv5t3kwe6rz3pfhvp8mk2uyf2lt
+\end{proof}<|endoftext|>
+\section{Sum of Unitary Divisors is Multiplicative}
+Tags: Sum of Unitary Divisors, Multiplicative Functions
+
+\begin{theorem}
+Let $\map {\sigma^*} n$ denote the sum of [[Definition:Unitary Divisor|unitary divisors]] of $n$.
+Then the function:
+:$\displaystyle \sigma^*: \Z_{>0} \to \Z_{>0}: \map {\sigma^*} n = \sum_{\substack d \mathop \divides n \\ d \mathop \perp \frac n d} d$
+is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b$ be [[Definition:Coprime Integers|coprime integers]].
+Because $a$ and $b$ have no [[Definition:Common Divisor of Integers|common divisor]], the [[Definition:Divisor of Integer|divisors]] of $a b$ are [[Definition:Integer|integers]] of the form $a_i b_j$, where $a_i$ is a [[Definition:Divisor of Integer|divisor]] of $a$ and $b_j$ is a [[Definition:Divisor of Integer|divisor]] of $b$.
+That is, any [[Definition:Divisor of Integer|divisor]] $d$ of $a b$ is in the form:
+:$d = a_i b_j$
+in a [[Definition:Unique|unique]] way, where $a_i \divides a$ and $b_j \divides b$.
+First we show that:
+:$d$ is an [[Definition:Unitary Divisor|unitary divisor]] of $a b$ {{iff}} $a_i, b_j$ are [[Definition:Unitary Divisor|unitary divisors]] of $a, b$ respectively
+In the forward implication we are given $d \perp \dfrac {a b} d$.
+By [[Divisor of One of Coprime Numbers is Coprime to Other]]:
+:$a_i, b_j \perp \dfrac {a b} d$
+By [[Divisor of One of Coprime Numbers is Coprime to Other]] again:
+:$a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$
+In the backward implication we are given $a_i \perp \dfrac a {a_i} \land b_j \perp \dfrac b {b_j}$.
+By [[Divisor of One of Coprime Numbers is Coprime to Other]]:
+:$a \perp b \implies \paren {a_i \perp b_j \land \dfrac a {a_i} \perp \dfrac b {b_j} }$
+By [[Product of Coprime Pairs is Coprime]]:
+:$d = a_i b_j \perp \dfrac a {a_i} \dfrac b {b_j} = \dfrac {a b} d$
+{{qed|lemma}}
+We can list the [[Definition:Unitary Divisor|unitary divisors]] of $a$ and $b$ as 
+:$1, a_1, a_2, \ldots, a$
+and:
+:$1, b_1, b_2, \ldots, b$
+and thus the sum of their [[Definition:Unitary Divisor|unitary divisors]] are:
+:$\displaystyle \map {\sigma^*} a = \sum_{i \mathop = 1}^r a_i$
+:$\displaystyle \map {\sigma^*} b = \sum_{j \mathop = 1}^s b_j$
+Consider all [[Definition:Unitary Divisor|unitary divisors]] of $a b$ with the same $a_i$.
+Their sum is:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^s a_i b_j
+      | r = a_i \sum_{j \mathop = 1}^s b_j
+      | c = 
+}}
+{{eqn | r = a_i \map {\sigma^*} b
+      | c = 
+}}
+{{end-eqn}}
+Summing over all $a_i$:
+{{begin-eqn}}
+{{eqn | l = \map {\sigma^*} {a b}
+      | r = \sum_{i \mathop = 1}^r \paren {a_i \map {\sigma^*} b}
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{i \mathop = 1}^r a_i} \map {\sigma^*} b
+      | c = 
+}}
+{{eqn | r = \map {\sigma^*} a \map {\sigma^*} b
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Sum of Unitary Divisors]]
+[[Category:Multiplicative Functions]]
+c6yun19lmtgq08xdz2qamz8jtidjz7i
+\end{proof}<|endoftext|>
+\section{Sum of Unitary Divisors of Integer}
+Tags: Sum of Unitary Divisors
+
+\begin{theorem}
+Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
+Let $\map {\sigma^*} n$ be the sum of all positive [[Definition:Unitary Divisor|unitary divisors]] of $n$.
+Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
+:$\displaystyle n = \prod_{1 \mathop \le i \mathop \le r} p_i^{k_i} =  p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
+Then:
+:$\displaystyle \map {\sigma^*} n = \prod_{1 \mathop \le i \mathop \le r} \paren {1 + p_i^{k_i}}$
+\end{theorem}
+
+\begin{proof}
+We have that the [[Sum of Unitary Divisors is Multiplicative]].
+From [[Value of Multiplicative Function is Product of Values of Prime Power Factors]], we have:
+:$\map {\sigma^*} n = \map {\sigma^*} {p_1^{k_1} } \map {\sigma^*} {p_2^{k_2} } \ldots \map {\sigma^*} {p_r^{k_r} }$
+From [[Sum of Unitary Divisors of Power of Prime]], we have:
+:$\displaystyle \map {\sigma^*} {p_i^{k_i} } = \frac {p_i^{k_i + 1} - 1} {p_i - 1}$
+Hence the result.
+{{qed}}
+[[Category:Sum of Unitary Divisors]]
+4ga8cqg48g3xpy1pbq6s76o1kdvqcya
+\end{proof}<|endoftext|>
+\section{Range of Infinite Sequence may be Finite}
+Tags: Sequences
+
+\begin{theorem}
+Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Infinite Sequence|infinite sequence]].
+Then it is possible for the [[Definition:Range of Sequence|range]] of $\sequence {x_n}$ to be [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Infinite Sequence|infinite sequence]] $\sequence {x_n}_{n \mathop \in \N}$ defined as:
+:$\forall n \in \N: x_n = \dfrac {1 + \paren {-1}^n} 2$
+Thus:
+:$\sequence {x_n}_{n \mathop \in \N} = 1, 0, 1, 0, \dotsc$
+Hence the [[Definition:Range of Sequence|range]] of $\sequence {x_n}$ is $\set {0, 1}$, which is [[Definition:Finite Set|finite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Squares Ending in n Occurrences of m-Digit Pattern}
+Tags: Number Theory, Recreational Mathematics, Squares Ending in n Occurrences of m-Digit Pattern
+
+\begin{theorem}
+Suppose there exists some [[Definition:Integer|integer]] $x$ such that $x^2$ ends in some $m$-[[Definition:Digit|digit]] pattern ending in an [[Definition:Odd Integer|odd number]] not equal to $5$ and is preceded by another [[Definition:Odd Integer|odd number]], i.e.:
+:$\exists x \in \Z: x^2 \equiv \sqbrk {1 a_1 a_2 \cdots a_m} \pmod {2 \times 10^m}$
+where $a_m$ is [[Definition:Odd Integer|odd]], $a_m \ne 5$ and $m \ge 1$.
+Then for any $n \ge 1$, there exists some [[Definition:Integer|integer]] with not more than $m n$-[[Definition:Digit|digits]] such that its [[Definition:Square Number|square]] ends in $n$ occurrences of the $m$-[[Definition:Digit|digit]] pattern.
+\end{theorem}
+
+\begin{proof}
+We prove that there exists a [[Definition:Integer Sequence|sequence]] $\sequence {b_n}$ with the properties:
+:$b_n < 10^{m n}$
+:$b_n^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{n \text { occurrences}} \pmod {2 \times 10^{m n}}$
+by [[Principle of Mathematical Induction|induction]]:
+=== Basis for the Induction ===
+For $n = 1$, we choose the number $b_1 = x \pmod {10^m}$ with $b_1 < 10^m$.
+Note that:
+{{begin-eqn}}
+{{eqn | l = b_1^2
+      | r = \paren {x - k \times 10^m}^2
+      | c = for some $k \in \Z$
+}}
+{{eqn | r = k^2 \times 10^{2 m} - 2 x k \times 10^m + x^2
+      | c = [[Square of Sum]]
+}}
+{{eqn | o = \equiv
+      | r = x^2
+      | rr = \pmod {2 \times 10^m}
+}}
+{{eqn | o = \equiv
+      | r = \sqbrk {1 a_1 a_2 \cdots a_m}
+      | rr = \pmod {2 \times 10^m}
+      | c = by assumption
+}}
+{{end-eqn}}
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+This is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:There exists some $b_r$ such that:
+::$b_r < 10^{m r}$
+::$b_r^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{r \text { occurrences}} \pmod {2 \times 10^{m r}}$
+Now we need to show true for $n = r + 1$:
+:There exists some $b_{r + 1}$ such that:
+::$b_{r + 1} < 10^{m \paren {r + 1} }$
+::$b_{r + 1}^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{r + 1 \text { occurrences}} \pmod {2 \times 10^{m \paren {r + 1}}}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $b < 10^m$ and $b_{r + 1} = b \times 10^{m r} + b_r$.
+Note that:
+{{begin-eqn}}
+{{eqn | l = b_{r + 1}^2
+      | r = \paren {b \times 10^{m r} + b_r}^2
+}}
+{{eqn | r = b^2 \times 10^{2 m r} + 2 b b_r \times 10^{m r} + b_r^2
+      | c = [[Square of Sum]]
+}}
+{{eqn | o = \equiv
+      | r = 2 b b_r \times 10^{m r} + \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m} } }_{r \text { occurrences} } + 2 k \times 10^{m r}
+      | rr = \pmod {2 \times 10^{m \paren {r + 1} } }
+      | c = [[Squares Ending in n Occurrences of m-Digit Pattern#Induction Hypothesis|Induction Hypothesis]]; for some $k \in \Z$
+}}
+{{end-eqn}}
+The rightmost $m r$ [[Definition:Digit|digits]] already satisfy the condition, so we consider the next $m + 1$ [[Definition:Digit|digits]].
+We want:
+:$2 b b_r + 1 + 2 k \equiv \sqbrk {1 a_1 \cdots a_m} \pmod {2 \times 10^m}$
+We first take Modulo $10^m$:
+:$2 b b_r + 1 + 2 k \equiv \sqbrk {a_1 \cdots a_m} \pmod {10^m}$
+So we need to solve:
+:$b b_r + t \times 10^m = \dfrac {\sqbrk {a_1 \cdots a_m} - 1} 2 - k$
+for [[Definition:Integer|integer]] solutions $b, t$.
+Since $a_m \ne 5$ and is [[Definition:Odd Integer|odd]]:
+:$2, 5 \nmid b_r$
+so $b_r$ and $10^m$ are [[Definition:Coprime Integers|coprime]].
+By [[Bezout's Lemma]], a solution for $b, t$ exists.
+We can also find a solution with $0 \le b < 10^m$.
+For this $b$, if;
+:$2 b b_r + 1 + 2 k \equiv \sqbrk {1 a_1 \cdots a_m} \pmod {2 \times 10^m}$
+then take $b' = b$, $b_{r + 1} = b' \times 10^{m r} + b_r$ and we are done.
+It may happen that:
+:$2 b b_r + 1 + 2 k \equiv \sqbrk {0 a_1 \cdots a_m} \pmod {2 \times 10^m}$
+In this case, take $b' = b \pm 5 \times 10^{m - 1}$, whichever is between $0$ and $10^m$.
+We have:
+{{begin-eqn}}
+{{eqn | l = 2b' b_r + 1 + 2 k
+      | r = 2 b_r \paren {b \pm 5 \times 10^{m - 1} } + 2 k + 1
+}}
+{{eqn | r = 2 b_r b + 2 k + 1 \pm 10^m 
+}}
+{{eqn | r = \equiv \sqbrk {0 a_1 \cdots a_m} \pm 10^m
+      | rr = \pmod {2 \times 10^m}
+}}
+{{eqn | r = \equiv \sqbrk {1 a_1 \cdots a_m}
+      | rr = \pmod {2 \times 10^m}
+}}
+{{end-eqn}}
+hence $b_{r + 1} = b' \times 10^{m r} + b_r$ satisfy our conditions as well.
+By the [[Principle of Mathematical Induction]], the sequence $\sequence {b_n}$ exists.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Largest Number not Expressible as Sum of Multiples of Coprime Integers}
+Tags: Largest Number not Expressible as Sum of Multiples of Coprime Integers, Integer Combinations
+
+\begin{theorem}
+Let $a, b$ be [[Definition:Coprime Integers|coprime integers]], each greater than $1$.
+Then the largest number not expressible as a sum of multiples of $a$ and $b$ is the number:
+:$a b - a - b = \paren {a - 1} \paren {b - 1} - 1$
+\end{theorem}
+
+\begin{proof}
+First we show that $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
+{{AimForCont}} $a b - a - b = s a + t b$ for some $s, t \in \N$.
+Note that $t b \le s a + t b < a b - b = \paren {a - 1} b$.
+This gives $t < a - 1$.
+We also have $\paren {a - t - 1} b = \paren {s + 1} a$.
+Hence $a \divides \paren {a - t - 1} b$.
+Since $a$ and $b$ are [[Definition:Coprime Integers|coprime]], by [[Euclid's Lemma]]:
+:$a \divides a - t - 1$
+As $a - t - 1 > 0$, by [[Absolute Value of Integer is not less than Divisors]]:
+:$a \le a - t - 1$
+which is a [[Definition:Contradiction|contradiction]].
+Hence $a b - a - b$ is not expressible as a sum of multiples of $a$ and $b$.
+{{qed|lemma}}
+Next we need to show that all numbers greater than $a b - a - b$ can be so expressed.
+{{WLOG}} assume that $a > b$ and we split the numbers into two cases:
+=== Case $1$: $x = a b - a - b + k$ for $1 \le k \le b$ ===
+For $k = b$, we have $a b - a = a \paren {b - 1}$.
+Notice that for $1 \le k < b$ and $0 \le s \le b - 2$:
+{{begin-eqn}}
+{{eqn | l = a b - a - b + k - s a
+      | o = \ge
+      | r = a b - a - b + k - \paren {b - 2} a
+}}
+{{eqn | r = a - b + k
+}}
+{{eqn | o = >
+      | r = 0
+}}
+{{end-eqn}}
+and we see that, by [[Absolute Value of Integer is not less than Divisors]]:
+:$a b - a - b + k - \paren {b - 1} a = k - b$
+cannot be a multiple of $b$.
+We claim that one of $a b - a - b + k - s a$ is [[Definition:Divisor of Integer|divisible]] by $b$, where $0 \le s \le b - 1$.
+Suppose not. Then we consider each of the [[Definition:Remainder|remainders]] when dividing by $b$.
+There are $b - 1$ [[Definition:Remainder|remainders]] excluding $0$.
+However we have a set of $b$ [[Definition:Integer|integers]].
+By [[Pigeonhole Principle]], two [[Definition:Integer|integers]] must share the same [[Definition:Remainder|remainder]].
+Suppose we have $a b - a - b + k - s_1 a$ and $a b - a - b + k - s_2 a$ both having [[Definition:Remainder|remainder]] $r$ with $s_1 \ne s_2$:
+:$\exists p, q \in \Z: a b - a - b + k - s_1 a = p b + r, a b - a - b + k - s_2 a = q b + r$
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {s_1 - s_2} a
+      | r = \size {a b - a - b + k - s_2 a - \paren {a b - a - b + k - s_1 a} }
+}}
+{{eqn | r = \size {q b + r - p b - r}
+}}
+{{eqn | r = \size {q - p} b
+}}
+{{end-eqn}}
+Hence $b \divides \size {s_1 - s_2} a$.
+By [[Euclid's Lemma]], $b \divides \size {s_1 - s_2}$.
+But $0 < \size {s_1 - s_2} \le b - 1$.
+This contradicts [[Absolute Value of Integer is not less than Divisors]].
+Therefore one of $a b - a - b + k - s a$ must be [[Definition:Divisor of Integer|divisible]] by $b$, where $0 \le s \le b - 1$.
+We have shown that $s \ne b - 1$.
+Hence for some $s$ with $0 \le s \le b - 2$, $a b - a - b + k - s a$ is a [[Definition:Positive Integer|positive]] multiple of $b$.
+This gives:
+:$a b - a - b + k = s a + t b$ for some $s, t \in \N$
+as required.
+{{qed|lemma}}
+=== Case $2$: $x > a b - a$ ===
+By [[Division Theorem]]:
+:$\exists q, r \in \Z: 0 \le r < b: x - \paren {a b - a - b + 1} = q b + r$
+We have:
+{{begin-eqn}}
+{{eqn | l = q b
+      | r = x - \paren {a b - a - b + 1} - r
+}}
+{{eqn | o = >
+      | r = b - 1 - r
+}}
+{{eqn | o = \ge
+      | r = 0
+}}
+{{end-eqn}}
+hence $q > 0$.
+Moreover:
+:$x - q b = a b - a - b + \paren {r + 1}$
+which falls into Case $1$.
+We have shown that:
+:$\exists s, t \in \N: a b - a - b + \paren {r + 1} = s a + t b$
+Hence:
+:$x = s a + t b + q b = s a + \paren {t + q} b$
+as required.
+{{qed}}
+[[Category:Largest Number not Expressible as Sum of Multiples of Coprime Integers]]
+[[Category:Integer Combinations]]
+4b3abrh2xjnj9ikb35rk8022bjsqny2
+\end{proof}<|endoftext|>
+\section{Space of Continuously Differentiable on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space}
+Tags: Examples of Vector Spaces, Functional Analysis
+
+\begin{theorem}
+Let $I := \closedint a b$ be a [[Definition:Closed Real Interval|closed real interval]].
+Let $\map \CC I$ be a [[Definition:Space of Continuous on Closed Interval Real-Valued Functions|space of continuous on closed interval real-valued functions]].
+Let $\map {\CC^1} I$ be a [[Definition:Space of Continuous Functions of Differentiability Class k|space of continuously differentiable functions]] on [[Definition:Closed Real Interval|closed interval]] $I$.
+Let $\struct {\R, +_\R, \times_\R}$ be the [[Definition:Field of Real Numbers|field of real numbers]].
+Let $\paren +$ be the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise addition of real-valued functions]].
+Let $\paren {\, \cdot \,}$ be the [[Definition:Pointwise Scalar Multiplication of Real-Valued Functions|pointwise scalar multiplication of real-valued functions]].
+Then $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+We [[Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space|have]] that $\struct {\map \CC I, +, \, \cdot \,}_\R$ is a [[Definition:Vector Space|vector space]].
+By [[Differentiable Function is Continuous]], $\map {\CC^1} I \subset \map \CC I$.
+Let $f, g \in \map {\CC^1} I$.
+Let $\alpha \in \R$.
+Let $\map 0 x$ be a [[Definition:Real-Valued Function|real-valued function]] such that:
+:$\map 0 x : I \to 0$
+[[Definition:Restriction of Operation|Restrict]] $\paren +$ to $\map {\CC^1} I \times \map {\CC^1} I$.
+[[Definition:Restriction of Operation|Restrict]] $\paren {\, \cdot \,}$ to $\R \times \map {\CC^1} I$.
+=== Closure under vector addition ===
+By [[Sum Rule for Derivatives]]:
+:$f + g \in \map {\CC^1} I$
+{{qed|lemma}}
+=== Closure under scalar multiplication ===
+By [[Derivative of Constant Multiple]]:
+:$\alpha \cdot f \in \map {\CC^1} I$
+{{qed|lemma}}
+=== Nonemptiness ===
+By [[Derivative of Constant]], a [[Definition:Constant Mapping|constant mapping]] is [[Definition:Differentiable Real-Valued Function|differentiable]].
+Hence, $\map 0 x \in \map {\CC^1} I$.
+{{qed|lemma}}
+We have that $\map {\CC^1} I$ is closed under [[Definition:Restriction of Operation|restrictions]] of $\paren +$ and $\paren {\, \cdot \,}$ to $\map {\CC^1} I$.
+Also, $\map {\CC^1} I$ is [[Definition:Non-Empty Set|non-empty]].
+By definition, $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ is a [[Definition:Vector Subspace|vector subspace]] of $\struct {\map \CC I, +, \, \cdot \,}_\R$.
+Since $\struct {\map {\CC^1} I, +, \, \cdot \,}_\R$ satisfies [[Definition:Vector Space Axioms|vector space axioms]] under given [[Definition:Restriction of Operation|restrictions]], it is a [[Definition:Vector Space|vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fermat Quotient of 2 wrt p is Square iff p is 3 or 7/Generalization}
+Tags: Fermat Quotients
+
+\begin{theorem}
+The [[Definition:Fermat Quotient|Fermat quotient]] of $2$ with respect to $p$:
+:$\map {q_p} 2 = \dfrac {2^{p - 1} - 1} p$
+is a [[Definition:Perfect Power|perfect power]] {{iff}} $p = 3$ or $p = 7$.
+\end{theorem}
+
+\begin{proof}
+To show that these are the only ones, we observe that since $p$ is an [[Definition:Odd Prime|odd prime]], write:
+:$p = 2 n + 1$ for $n \ge 1$.
+Let $\map {q_p} 2$ be a [[Definition:Perfect Power|perfect power]].
+Then $2^{p - 1} - 1 = p x^y$ for some [[Definition:Integer|integers]] $x, y$.
+Note that:
+:$2^{p - 1} - 1 = 2^{2 n} - 1 = \paren {2^n - 1} \paren {2^n + 1}$
+and we have:
+:$\gcd \set {2^n - 1, 2^n + 1} = \gcd \set {2^n - 1, 2} = 1$
+so $2^n - 1$ and $2^n + 1$ are [[Definition:Coprime Integers|coprime]].
+Hence there are $2$ cases:
+=== Case $1$: $p \divides 2^n - 1$ ===
+By [[Divisor of One of Coprime Numbers is Coprime to Other]]:
+:$\gcd \set {\dfrac {2^n - 1} p, 2^n + 1} = 1$
+Hence both the numbers are [[Definition:Perfect Power|perfect $y$th powers]].
+In particular we have:
+:$\exists k \in \Z: 2^n + 1 = k^y$
+by [[1 plus Power of 2 is not Perfect Power except 9]], the only solution to the equation above is:
+:$n = k = 3, y = 2$
+This gives $p = 2 n + 1 = 7$.
+{{qed|lemma}}
+=== Case $2$: $p \divides 2^n + 1$ ===
+By [[Divisor of One of Coprime Numbers is Coprime to Other]]:
+:$\gcd \set {\dfrac {2^n + 1} p, 2^n - 1} = 1$
+Hence both the numbers are [[Definition:Perfect Power|perfect $y$th powers]].
+In particular we have:
+:$\exists k \in \Z: 2^n - 1 = k^y$
+By [[1 plus Perfect Power is not Prime Power except for 9]], this equation has no solution for $n, k, y > 1$.
+Hence we must have $n = 1$.
+This gives $p = 2 n + 1 = 3$.
+{{qed|lemma}}
+In the main theorem we have already shown that $p = 3, 7$ gives [[Definition:Perfect Power|perfect power]] [[Definition:Fermat Quotient|Fermat quotients]].
+Hence the result.
+{{qed}}
+[[Category:Fermat Quotients]]
+rqzhs2gdzagzjs1p0zkge6ndnlx5vig
+\end{proof}<|endoftext|>
+\section{Taxicab Norm is Norm}
+Tags: Taxicab Norm
+
+\begin{theorem}
+The [[Definition:Taxicab Norm|taxicab norm]] is a [[Definition:Norm on Vector Space|norm]] on the [[Definition:Real Number|real]] and [[Definition:Complex Number|complex numbers]].
+\end{theorem}
+
+\begin{proof}
+By [[P-Norm is Norm]], $\norm {\, \cdot \,}_p$ is a [[Definition:Norm on Vector Space|norm]].
+By definition, the [[Definition:Taxicab Norm|taxicab norm]] is $\norm {\, \cdot \,}_1$.
+Therefore, the [[Definition:Taxicab Norm|taxicab norm]] is a [[Definition:Norm on Vector Space|norm]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplication by 2 over 3 in Egyptian Fractions}
+Tags: Egyptian Fractions, Multiplication by 2 over 3 in Egyptian Fractions
+
+\begin{theorem}
+Let $\dfrac 1 n$ be an [[Definition:Egyptian Fraction|Egyptian fraction]] not equal to $\dfrac 2 3$.
+In order to [[Definition:Rational Multiplication|multiply]] $\dfrac 1 n$ by $\dfrac 2 3$ and have it that $\dfrac 1 n \times \dfrac 2 3$ is also expressed in [[Definition:Egyptian Fraction|Egyptian form]], we have:
+:$\dfrac 1 n \times \dfrac 2 3 = \dfrac 1 {2 n} + \dfrac 1 {6 n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {2 n} + \dfrac 1 {6 n}
+      | r = \dfrac 3 {6 n} + \dfrac 1 {6 n}
+      | c = 
+}}
+{{eqn | r = \dfrac {3 + 1} {6 n}
+      | c = 
+}}
+{{eqn | r = \dfrac 2 {3 n}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 n \times \dfrac 2 3
+      | c = 
+}}
+{{end-eqn}}
+Note the case where we multiply $\dfrac 2 3$ by $\dfrac 2 3$ itself:
+{{begin-eqn}}
+{{eqn | l = \dfrac 2 3 \times \dfrac 2 3
+      | r = \dfrac 4 9
+      | c = 
+}}
+{{eqn | r = \dfrac 3 9 + \dfrac 1 9
+      | c = 
+}}
+{{eqn | r = \dfrac 1 3 + \dfrac 1 9
+      | c = which is in [[Definition:Egyptian Fraction|Egyptian form]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Proper Fraction can be Expressed as Finite Sum of Unit Fractions}
+Tags: Unit Fractions, Egyptian Fractions, Proper Fraction can be Expressed as Finite Sum of Unit Fractions
+
+\begin{theorem}
+Let $\dfrac p q$ denote a [[Definition:Proper Fraction|proper fraction]] expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
+Then it is always possible to express $\dfrac p q$ as the [[Definition:Integer Addition|sum]] of a [[Definition:Finite Set|finite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Unit Fraction|unit fractions]]:
+{{begin-eqn}}
+{{eqn | l = \dfrac p q
+      | r = \sum_{\substack {1 \mathop \le k \mathop \le m \\ n_j \mathop \le n_{j + 1} } } \dfrac 1 {n_k}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {n_1} + \dfrac 1 {n_2} + \dotsb + \dfrac 1 {n_m}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}<|endoftext|>
+\section{Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm}
+Tags: Fibonacci's Greedy Algorithm
+
+\begin{theorem}
+Let $\dfrac p q$ denote a [[Definition:Proper Fraction|proper fraction]] expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
+Let $\dfrac p q$ be expressed as the [[Definition:Integer Addition|sum]] of a [[Definition:Finite Set|finite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Unit Fraction|unit fractions]] using [[Fibonacci's Greedy Algorithm]].
+Then $\dfrac p q$ is expressed using no more than $p$ [[Definition:Unit Fraction|unit fractions]].
+\end{theorem}
+
+\begin{proof}
+Let $\dfrac {x_k} {y_k}$ and $\dfrac {x_{k + 1} } {y_{k + 1} }$ be consecutive stages of the calculation of the [[Definition:Unit Fraction|unit fractions]] accordingly:
+:$\dfrac {x_k} {y_k} - \dfrac 1 {\ceiling {y_n / x_n} } = \dfrac {x_{k + 1} } {y_{k + 1} }$
+By definition of [[Fibonacci's Greedy Algorithm]]:
+:$\dfrac {x_{k + 1} } {y_{k + 1} } = \dfrac {\paren {-y_k} \bmod {x_k} } {y_k \ceiling {y_k / x_k} }$
+It is established during the processing of [[Fibonacci's Greedy Algorithm]] that:
+:$\paren {-y_k} \bmod {x_k} < x_k$
+Hence successive [[Definition:Numerator|numerators]] decrease by at least $1$.
+Hence there can be no more [[Definition:Unit Fraction|unit fractions]] than there are [[Definition:Natural Number|natural numbers]] between $1$ and $p$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Smallest n for which 3 over n produces 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient}
+Tags: Fibonacci's Greedy Algorithm, 25
+
+\begin{theorem}
+Consider [[Definition:Proper Fraction|proper fractions]] of the form $\dfrac 3 n$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
+Let [[Fibonacci's Greedy Algorithm]] be used to generate a [[Definition:Sequence|sequence]] $S$ of [[Definition:Egyptian Fraction|Egyptian fractions]] for $\dfrac 3 n$.
+The smallest $n$ for which $S$ consists of $3$ [[Definition:Term of Sequence|terms]], where $2$ would be sufficient, is $25$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+{{begin-eqn}}
+{{eqn | l = \frac 3 {25}
+      | r = \frac 1 9 + \frac 2 {225}
+      | c = as $\ceiling {25 / 3} = \ceiling {8.333\ldots} = 9$
+}}
+{{eqn | r = \frac 1 9 + \frac 1 {113} + \frac 1 {25 \, 425}
+      | c = as $\ceiling {225 / 2} = \ceiling {112.5} = 113$
+}}
+{{end-eqn}}
+But then we have:
+{{begin-eqn}}
+{{eqn | l = \frac 3 {25}
+      | r = \frac 6 {50}
+      | c = 
+}}
+{{eqn | r = \frac 5 {50} + \frac 1 {50}
+      | c = 
+}}
+{{eqn | r = \frac 1 {10} + \frac 1 {50}
+      | c = 
+}}
+{{end-eqn}}
+By [[Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient]], we are to find the smallest $n$ such that:
+:$n \equiv 1 \pmod 6$
+:$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$
+The first few $n \ge 4$ which satisfies $n \equiv 1 \pmod 6$ are:
+:$7, 13, 19, 25$
+of which $7, 13, 19$ are [[Definition:Prime Number|primes]], so they do not have a [[Definition:Divisor of Integer|divisor]] of the form $d \equiv 2 \pmod 3$.
+We see that $5 \divides 25$ and $5 \equiv 2 \pmod 3$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum Norm is Norm/Continuous on Closed Interval Real-Valued Function}
+Tags: Examples of Norms
+
+\begin{theorem}
+Let $I = \closedint a b$ be a [[Definition:Closed Interval|closed interval]].
+Let $\struct {\map \CC I, +, \, \cdot \,}_\R$ be the [[Space of Continuous on Closed Interval Real-Valued Functions with Pointwise Addition and Pointwise Scalar Multiplication form Vector Space|vector space of real-valued functions, continuous on]] $I$.
+Let $\map x t \in \map \CC I$ be a [[Definition:Continuous Real Function on Subset|continuous real function]].
+Let $\size {\, \cdot \,}$ be the [[Definition:Absolute Value|absolute value]].
+Let $\norm {\, \cdot \,}_\infty$ be the [[Definition:Supremum Norm/Continuous on Closed Interval Real-Valued Function|supremum norm on real-valued functions, continuous on]] $I$.
+Then $\norm {\, \cdot \,}_\infty$ is a [[Definition:Norm on Vector Space|norm]] over $\struct {\map \CC I, +, \, \cdot \,}_\R$.
+\end{theorem}
+
+\begin{proof}
+=== Positive definiteness ===
+{{begin-eqn}}
+{{eqn | l = \norm {x}_\infty
+      | r = \sup_{t \mathop \in I} \size {\map x t}
+      | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
+}}
+{{eqn | r = \max_{t \mathop \in I} \size {\map x t}
+      | c = [[Weierstrass Extreme Value Theorem]]
+}}
+{{eqn | o = \ge
+      | r = \size {\map x t}
+      | c = {{defof|Max Operation}}
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = [[Complex Modulus is Non-Negative]]
+}}
+{{end-eqn}}
+Suppose $\norm {x}_\infty = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \norm {x}_\infty
+}}
+{{eqn | r = \sup_{t \mathop \in I} \size {\map x t}
+      | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
+}}
+{{eqn | r = \max_{t \mathop \in I} \size {\map x t}
+      | c = [[Weierstrass Extreme Value Theorem]]
+}}
+{{eqn | o = \ge
+      | r = \size {\map x t}
+      | c = {{defof|Max Operation}}
+}}
+{{eqn | o = \ge 
+      | r = 0
+      | c = [[Complex Modulus is Non-Negative]]
+}}
+{{eqn | ll = \leadsto
+      | l = \size {\map x t}
+      | r = 0
+}}
+{{eqn | ll = \leadsto
+      | l = \map x t
+      | r = 0
+      | c = [[Complex Modulus equals Zero iff Zero]]
+}}
+{{end-eqn}}
+Therefore:
+:$\displaystyle \forall t \in I : \map x t = 0$
+=== Positive homogeneity === 
+Let $\alpha \in \R$.
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha \cdot x}_\infty
+      | r = \sup_{t \mathop \in I} \size {\map {\paren {\alpha \cdot x} } t}
+      | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
+}}
+{{eqn | r = \max_{t \mathop \in I} \size {\map {\paren {\alpha \cdot x} } t}
+      | c = [[Weierstrass Extreme Value Theorem]]
+}}
+{{eqn | r = \max_{t \mathop \in I} \size {\alpha {\map x t} }
+      | c = {{defof|Pointwise Scalar Multiplication of Real-Valued Functions}}
+}}
+{{eqn | r = \max_{t \mathop \in I} \size \alpha \size {\map x t}
+      | c = [[Absolute Value of Product]]
+}}
+{{eqn | r = \size \alpha \max_{t \mathop \in I} \size {\map x t}
+}}
+{{eqn | r = \size \alpha \sup_{t \mathop \in I} \size {\map x t}
+      | c = [[Weierstrass Extreme Value Theorem]]
+}}
+{{eqn | r = \size \alpha \norm {x}_\infty
+      | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
+}}
+{{end-eqn}}
+=== Triangle inequality ===
+{{begin-eqn}}
+{{eqn | l = \size {\map {\paren {x_1 + x_2} } t}
+      | r = \size {\map {x_1} t + \map {x_2} t}
+      | c = {{defof|Pointwise Addition of Real-Valued Functions}}
+}}
+{{eqn | o = \le
+      | r = \size {\map {x_1} t} + \size {\map {x_2} t}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \max_{t \mathop \in I} \size {\map {x_1} t} + \max_{t \mathop \in I} \size {\map {x_2} t}
+      | c = {{defof|Max Operation}}
+}}
+{{eqn | r = \sup_{t \mathop \in I} \size {\map {x_1} t } + \sup_{t \mathop \in I} \size { \map {x_2} t }
+      | c = [[Weierstrass Extreme Value Theorem]]
+}}
+{{eqn | r = \norm {x_1}_\infty + \norm {x_2}_\infty
+      | c = {{defof|Supremum Norm on Space of Continuous on Closed Interval Real-Valued Functions}}
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Union with Disjoint Singleton is Dependent if Element Depends on Subset}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $x \in S : x \notin A$.
+If $x$ [[Definition:Depends Relation (Matroid)|depends]] on $A$ then $A \cup \set x$ is [[Definition:Dependent Subset (Matroid)|dependent]]
+\end{theorem}
+
+\begin{proof}
+We proceed by [[Proof by Contraposition]].
+Let $A \cup \set x$ be [[Definition:Independent Subset (Matroid)|independent]].
+By [[Definition:Matroid Axioms|matroid axiom $( \text I 2)$]]:
+:$A$ is [[Definition:Independent Subset (Matroid)|independent]]
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \rho {A \cup \set x} 
+      | r = \size {A \cup \set x}
+      | c = [[Rank of Independent Subset Equals Cardinality]]
+}}
+{{eqn | r = \size A + \size{\set x}
+      | c = [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]]
+}}
+{{eqn | r = \size A + 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | o = >
+      | r = \size A
+}}
+{{eqn | r = \map \rho A
+      | c = [[Rank of Independent Subset Equals Cardinality]]
+}}
+{{end-eqn}}
+Then $x$ does not [[Definition:Depends Relation (Matroid)|depend]] on $A$ by definition.
+The theorem holds by the [[Rule of Transposition]].
+{{qed}}
+[[Category:Matroid Theory]]
+1imatcs3l1aayw7qezhwfd1eluhtu15
+\end{proof}<|endoftext|>
+\section{Element Depends on Independent Set iff Union with Singleton is Dependent}
+Tags: Matroid Theory, Element Depends on Independent Set iff Union with Singleton is Dependent
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $X \in \mathscr I$.
+Let $x \in S : x \notin X$.
+Then:
+:$x \in \map \sigma X$ {{iff}} $X \cup \set x$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \in \map \sigma X$.
+By definition of the [[Definition:Closure Operator (Matroid)|closure]]:
+:$x$ [[Definition:Depends Relation (Matroid)|depends]] on $X$.
+From [[Union with Disjoint Singleton is Dependent if Element Depends on Subset]]:
+:$X \cup \set x$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $X \cup \set x$ be [[Definition:Dependent Subset (Matroid)|dependent]].
+==== [[Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma|Lemma]] ====
+{{:Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma}}{{qed|lemma}}
+From [[Max yields Supremum of Operands]]:
+:$\map \rho {X \cup \set x} = \max \set {\size A : A \subseteq X \cup \set x \land A \in \mathscr I} \le \size X$
+By assumption:
+:$X \in \mathscr I$
+From [[Max yields Supremum of Operands]]:
+:$\size X \le \map \rho {X \cup \set x}$
+Thus:
+:$\size X = \map \rho {X \cup \set x}$
+From [[Rank of Independent Subset Equals Cardinality]]:
+:$\map \rho X = \size X$
+Thus:
+:$\map \rho X = \map \rho {X \cup \set x}$
+It follows that $x$ [[Definition:Depends Relation (Matroid)|depends]] on $X$ by definition.
+So:
+:$x \in \map \sigma X$
+{{qed}}
+[[Category:Matroid Theory]]
+[[Category:Element Depends on Independent Set iff Union with Singleton is Dependent]]
+jzb8a7ulf7qe7hmju1wbvtzkkjojfjb
+\end{proof}<|endoftext|>
+\section{Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma}
+Tags: Element Depends on Independent Set iff Union with Singleton is Dependent
+
+\begin{theorem}
+Let $A \in \mathscr I$ such that $A \subseteq X \cup \set x$.
+Then:
+:$\size A \le \size X$
+\end{theorem}
+
+\begin{proof}
+==== Case 1: $x \in A$ ====
+Let $x \in A$.
+We have:
+{{begin-eqn}}
+{{eqn | l = A \setminus \set x
+      | o = \subseteq
+      | r = \paren {X \cup \set x} \setminus \set x
+      | c = [[Set Difference over Subset]]
+}}
+{{eqn | r = \paren {X \setminus \set x} \cup \paren {\set x \setminus \set x}
+      | c = [[Set Difference is Right Distributive over Union]]
+}}
+{{eqn | r = X \cup \paren {\set x \setminus \set x}
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{eqn | r = X \cup \O
+      | c = [[Set Difference with Superset is Empty Set]]
+}}
+{{eqn | r = X
+      | c = [[Union with Empty Set]]
+}}
+{{end-eqn}}
+{{AimForCont}}:
+:$A \setminus \set x = X$
+Then:
+{{begin-eqn}}
+{{eqn | l = X \cup \set x
+      | r = \paren {A \setminus \set x} \cup \set x
+      | c = 
+}}
+{{eqn | r = A
+      | c = [[Set Difference Union Second Set is Union]]
+}}
+{{end-eqn}}
+So:
+:$X \cup \set x$ is [[Definition:Independent Subset (Matroid)|independent]].
+This [[Definition:Contradiction|contradicts]]:
+:$X \cup \set x$ is [[Definition:Dependent Subset (Matroid)|dependent]].
+So:
+:$A \setminus \set x \subsetneq X$
+Then:
+{{begin-eqn}}
+{{eqn | l = \size X
+      | o = >
+      | r = \size {A \setminus \set x}
+      | c = [[Cardinality of Proper Subset of Finite Set]]
+}}
+{{eqn | r = \size A - \size {\set x}
+      | c = [[Cardinality of Set Difference with Subset]]
+}}
+{{eqn | r = \size A - 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{end-eqn}}
+So:
+:$\size A \le \size X$
+{{qed|lemma}}
+==== Case 2: $x \notin A$ ====
+Let $x \notin A$.
+Then:
+{{begin-eqn}}
+{{eqn | l = A
+      | r = \paren {X \cup \set x} \cap A
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | r = \paren {X \cap A} \cup \paren {\set x \cap A}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | r = \paren {X \cap A}  \cup \O
+      | c = [[Intersection With Singleton is Disjoint if Not Element]]
+}}
+{{eqn | r = X \cap A
+      | c = [[Union with Empty Set]]
+}}
+{{end-eqn}}
+From [[Intersection with Subset is Subset]]:
+:$A \subseteq X$
+From [[Cardinality of Subset of Finite Set]]:
+:$\size A \le \size X$
+{{qed|lemma}}
+In either case:
+:$\size A \le \size X$
+{{qed}}
+[[Category:Element Depends on Independent Set iff Union with Singleton is Dependent]]
+10jyue4m0c7k6n9v64pm847l618e8c9
+\end{proof}<|endoftext|>
+\section{Rank of Independent Subset Equals Cardinality}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\rho : \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $X \in \mathscr I$
+Then:
+:$\map \rho X = \size X$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \rho X
+      | r = \max \set {\size Y : Y \subseteq X \land X \in \mathscr I}
+      | c = {{Defof|Rank Function (Matroid)}}
+}}
+{{eqn | r = \max \set {\size Y : Y \in \powerset X \land X \in \mathscr I}
+      | c = {{Defof|Power Set}} of $X$
+}}
+{{eqn | r = \max \set {\size Y : Y \in \powerset X \cap \mathscr I}
+      | c = {{Defof|Intersection}}
+}}
+{{eqn | r = \max \set {\size Y : Y \in \powerset X}
+      | c = [[Definition:Matroid Axioms|Matroid axiom $(\text I 2)$]]
+}}
+{{eqn | r = \size X
+      | c = [[Cardinality of Proper Subset of Finite Set]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Matroid Theory]]
+k9pv9s0dp4n4bglwvc5wk3460w4qt6l
+\end{proof}<|endoftext|>
+\section{Generating Function for Lucas Numbers}
+Tags: Generating Functions, Lucas Numbers
+
+\begin{theorem}
+Let $\map G z$ be the [[Definition:Real Function|function]] defined as:
+:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
+Then $\map G z$ is a [[Definition:Generating Function|generating function]] for the [[Definition:Lucas Number|Lucas numbers]].
+\end{theorem}
+
+\begin{proof}
+Let the form of $\map G z$ be assumed as:
+{{begin-eqn}}
+{{eqn | l = \map G z
+      | r = \sum_{k \mathop \ge 0} L_k z^k
+      | c = 
+}}
+{{eqn | r = L_0 + L_1 z + L_2 z^2 + L_3 z^3 + L_4 z^4 + \cdots
+      | c = 
+}}
+{{eqn | r = 2 + z + 3 z^2 + 4 z^3 + 7 z^4 + \cdots
+      | c = 
+}}
+{{end-eqn}}
+where $L_n$ denotes the [[Definition:Lucas Number|$n$th Lucas number]].
+Then:
+{{begin-eqn}}
+{{eqn | l = z \map G z
+      | r = L_0 z + L_1 z^2 + L_2 z^3 + L_3 z^4 + L_4 z^5 + \cdots
+      | c = 
+}}
+{{eqn | l = z^2 \map G z
+      | r = L_0 z^2 + L_1 z^3 + L_2 z^4 + L_3 z^5 + L_4 z^6 + \cdots
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - z - z^2} \map G z
+      | r = L_0 + \paren {L_1 - L_0} z + \paren {L_2 - L_1 - L_0} z^2 + \paren {L_3 - L_2 - L_1} z^3 + \cdots
+      | c = 
+}}
+{{eqn | r = L_0 + \paren {L_1 - L_0} z
+      | c = {{Defof|Lucas Number}}: $L_n = L_{n - 1} + L_{n - 2}$
+}}
+{{eqn | r = 2 - z
+      | c = {{Defof|Lucas Number}}: $L_0 = 2, L_1 = 1$
+}}
+{{end-eqn}}
+Hence the result:
+:$\map G z = \dfrac {2 - z} {1 - z - z^2}$
+{{qed}}
+[[Category:Generating Functions]]
+[[Category:Lucas Numbers]]
+1w3pj53p6rjf7zepkedhxtc72mut7w4
+\end{proof}<|endoftext|>
+\section{492 Cubed is Sum of 3 Positive Cubes in 13 Ways}
+Tags: 492, Sums of Cubes
+
+\begin{theorem}
+The [[Definition:Cube (Algebra)|cube]] of $492$ can be expressed as the [[Definition:Integer Addition|sum]] of $3$ [[Definition:Positive Integer|positive]] [[Definition:Cube Number|cubes]] in $13$ different ways:
+{{begin-eqn}}
+{{eqn | l = 492^3
+      | r = 24^3 + 204^3 + 480^3
+}}
+{{eqn | r = 48^3 + 85^3 + 491^3
+}}
+{{eqn | r = 72^3 + 384^3 + 396^3
+}}
+{{eqn | r = 113^3 + 264^3 + 463^3
+}}
+{{eqn | r = 114^3 + 360^3 + 414^3
+}}
+{{eqn | r = 149^3 + 336^3 + 427^3
+}}
+{{eqn | r = 176^3 + 204^3 + 472^3
+}}
+{{eqn | r = 190^3 + 279^3 + 449^3
+}}
+{{eqn | r = 207^3 + 297^3 + 438^3
+}}
+{{eqn | r = 226^3 + 332^3 + 414^3
+}}
+{{eqn | r = 243^3 + 358^3 + 389^3
+}}
+{{eqn | r = 246^3 + 328^3 + 410^3
+}}
+{{eqn | r = 281^3 + 322^3 + 399^3
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Brute force.
+\end{proof}<|endoftext|>
+\section{Maximum Area of Isosceles Triangle}
+Tags: Isosceles Triangles
+
+\begin{theorem}
+Consider two [[Definition:Line Segment|line segments]] $A$ and $B$ of equal [[Definition:Length of Line|length]] $a$ which are required to be the [[Definition:Legs of Isosceles Triangle|legs]] of an [[Definition:Isosceles Triangle|isosceles triangle]] $T$.
+Then the [[Definition:Area|area]] of $T$ is greatest when the [[Definition:Apex of Isosceles Triangle|apex]] of $T$ is a [[Definition:Right Angle|right angle]].
+The [[Definition:Area|area]] of $T$ in this situation is equal to $\dfrac {a^2} 2$.
+\end{theorem}
+
+\begin{proof}
+:[[File:Maximum-size-isosceles-triangle.png|500px]]
+Let $\triangle OAB$ be the [[Definition:Isosceles Triangle|isosceles triangle]] $T$ formed by the [[Definition:Legs of Isosceles Triangle|legs]] $OA$ and $OB$.
+Thus the [[Definition:Apex of Isosceles Triangle|apex]] of $T$ is at $O$.
+Let $\theta$ be the [[Definition:Angle|angle]] $\angle AOB$.
+We see that by keeping $OA$ fixed, $B$ can range over the [[Definition:Semicircle|semicircle]] $AOB$.
+Thus $\theta$ can range from $0$ to $180 \degrees$, that is, $2$ [[Definition:Right Angle|right angles]].
+From [[Area of Triangle in Terms of Two Sides and Angle]], the [[Definition:Area|area]] $\AA$ of $T$ is:
+:$\AA = \dfrac 1 2 a^2 \sin \theta$
+This is a maximum when $\sin \theta = 1$, that is, when $\theta$ is a [[Definition:Right Angle|right angle]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inscribing Equilateral Triangle inside Square with a Coincident Vertex}
+Tags: Squares, Equilateral Triangles, Inscribing Equilateral Triangle inside Square with a Coincident Vertex
+
+\begin{theorem}
+Let $\Box ABCD$ be a [[Definition:Square (Geometry)|square]].
+It is required that $\triangle DGH$ be an [[Definition:Equilateral Triangle|equilateral triangle]] [[Definition:Inscribe/Polygon within Polygon|inscribed]] within $\Box ABCD$ such that [[Definition:Vertex of Polygon|vertex]] $D$ of $\triangle DGH$ coincides with [[Definition:Vertex of Polygon|vertex]] $D$ of $\Box ABCD$.
+\end{theorem}<|endoftext|>
+\section{Construction of Perpendicular using Rusty Compass}
+Tags: Lines, Angles, Rusty Compass Constructions
+
+\begin{theorem}
+Let $AB$ be a [[Definition:Line Segment|line segment]].
+Using a [[Definition:Straightedge|straightedge]] and [[Definition:Rusty Compass|rusty compass]], it is possible to construct a [[Definition:Straight Line|straight line]] at [[Definition:Right Angle|right angles]] to $AB$ from the [[Definition:Endpoint of Line|endpoint]] $A$, without extending $AB$ past $A$.
+\end{theorem}
+
+\begin{proof}
+As $DE = CD = DA$, the [[Definition:Point|points]] $A$, $C$ and $E$ all lie on a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $AC$.
+$CE$ is a [[Definition:Straight Line|straight line]] through the [[Definition:Center of Circle|centers]] of [[Definition:Circle|circle]] $ACE$ and so is a [[Definition:Diameter|diameter]] of [[Definition:Circle|circle]] $ACE$.
+Hence by [[Thales' Theorem]], $\angle CAE$ is a [[Definition:Right Angle|right angle]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division of Straight Line into Equal Parts using Rusty Compass}
+Tags: Lines, Rusty Compass Constructions
+
+\begin{theorem}
+Let $AB$ be a [[Definition:Line Segment|line segment]].
+Using a [[Definition:Straightedge|straightedge]] and [[Definition:Rusty Compass|rusty compass]], it is possible to divide $AB$ into as many equal parts as required.
+\end{theorem}<|endoftext|>
+\section{Construction of Regular Pentagon using Rusty Compass}
+Tags: Pentagons, Rusty Compass Constructions
+
+\begin{theorem}
+Using a [[Definition:Straightedge|straightedge]] and [[Definition:Rusty Compass|rusty compass]], it is possible to [[Definition:Polygon Inscribed in Circle|inscribe]] a [[Definition:Regular Pentagon|regular pentagon]] inside a [[Definition:Circle|circle]].
+\end{theorem}<|endoftext|>
+\section{Sum to Infinity of 2x^2n over n by 2n Choose n}
+Tags: Central Binomial Coefficients
+
+\begin{theorem}
+For $\cmod x < 1$:
+:$\displaystyle \frac {2 x \arcsin x} {\sqrt {1 - x^2} } = \sum_{n \mathop = 1}^\infty \frac {\paren {2 x}^{2 n} } {n \dbinom {2 n} n}$
+\end{theorem}
+
+\begin{proof}
+By [[Gregory Series]]:
+:$\displaystyle \arctan t = \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}$
+Let $t = \dfrac x {\sqrt {1 - x^2} }$.
+Let $y = \arcsin x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = t
+      | r = \frac {\sin y} {\sqrt {1 - \sin^2 y} }
+}}
+{{eqn | r = \frac {\sin y} {\cos y}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \tan y
+}}
+{{end-eqn}}
+Hence $\arctan t = \arcsin x$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac {2 x \arcsin x} {\sqrt {1 - x^2} }
+      | r = 2 t \arctan t
+}}
+{{eqn | r = 2 t \sum_{m \mathop = 0}^\infty \frac {\paren {-1}^m t^{2 m + 1} } {2 m + 1}
+      | c = [[Gregory Series]]
+}}
+{{eqn | r = 2 t \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m - 1} } {2 m - 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} t^{2 m} } {2 m - 1}
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {\paren {2 m - 1} \paren {1 - x^2}^m}
+}}
+{{eqn | r = 2 \sum_{m \mathop = 1}^\infty \frac {\paren {-1}^{m - 1} x^{2 m} } {2 m - 1} \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}
+      | c = [[Binomial Theorem for Negative Index and Negative Parameter]]
+}}
+{{end-eqn}}
+It remains to show the the coefficient of $x^{2 n}$ on the {{RHS}} is equal to $\dfrac {2^{2 n} } {n \dbinom {2 n} n}$, that is:
+:$\displaystyle 2 \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {r + n - r - 1} {r - 1} = \frac {2^{2 n} } {n \dbinom {2 n} n}$
+The {{LHS}} above is generated by picking, for each $m > 0$, the corresponding $k = n - m$ from the right sum $\displaystyle \sum_{k \mathop = 0}^\infty \dbinom {m + k - 1} {m - 1} x^{2 k}$.
+We have:
+{{begin-eqn}}
+{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 1}^n \frac {\paren {-1}^{r - 1} } {2 r - 1} \dbinom {n - 1} {r - 1}
+      | o =
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \sum_{r \mathop = 0}^{n - 1} \frac {\paren {-1}^r} {2 r + 1} \dbinom {n - 1} r
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \sum_{r \mathop = 0}^{n - 1} \paren {-1}^r \dbinom {n - 1} r y^{2 r} \d y
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \int_0^1 \paren {1 - y^2}^{n - 1} \d y
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \int_{\frac \pi 2}^0 \sin^{2 n - 2} \theta \, \frac {\d y} {\d \theta} \d \theta
+      | c = by substitution of $y = \cos \theta$
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \int_0^{\frac \pi 2} \sin^{2 n - 1} \theta \, \d \theta
+}}
+{{eqn | r = 2 n \dbinom {2 n} n \frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!}
+      | c = [[Definite Integral from 0 to Half Pi of Odd Power of Sine x]]
+}}
+{{eqn | r = 2 n \paren {\frac {\paren {2 n}!} {n! \, n!} } \paren {\frac {\paren {2^{n - 1} \paren {n - 1}!}^2} {\paren {2 n - 1}!} }
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = 2 n \paren {\frac {2 n} {n^2} } \paren {2^{2 n - 2} }
+}}
+{{eqn | r = 2^{2 n}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x}
+Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
+\end{theorem}
+
+\begin{proof}
+Fix $b \in \R$ and define: 
+:$\displaystyle \map I a = \int_0^\pi \map \ln {a + b \cos x} \rd x$
+for $a \ge \size b$.
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} a
+	| r = \frac \d {\d a} \int_0^\pi \map \ln {a + b \cos x} \rd x
+}}
+{{eqn	| r = \int_0^\pi \frac \partial {\partial a} \paren {\map \ln {a + b \cos x} } \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = \int_0^\pi \frac 1 {a + b \cos x} \rd x
+	| c = [[Derivative of Natural Logarithm]]
+}}
+{{eqn	| r = \frac 1 2 \int_0^{2 \pi} \frac 1 {a + b \cos x} \rd x
+	| c = [[Definite Integral of Even Function]]
+}}
+{{eqn	| r = \frac \pi {\sqrt {a^2 - b^2} }
+	| c = [[Definite Integral from 0 to 2 Pi of Reciprocal of a plus b Cosine x|Definite Integral from $0$ to $2 \pi$ of $\dfrac 1 {a + b \cos x}$]]
+}}
+{{end-eqn}}
+So, by [[Primitive of Root of x squared minus a squared/Logarithm Form|Primitive of $\sqrt {x^2 - a^2}$: Logarithm Form]]: 
+:$\displaystyle \map I a = \pi \map \ln {a + \sqrt {a^2 - b^2} } + C$
+for all $a \ge \size b$, for some $C \in \R$.
+We now split up depending on the sign of $b$. 
+If $b = 0$, then we have: 
+{{begin-eqn}}
+{{eqn	| l = \map I a
+	| r = \pi \map \ln {a + \sqrt {a^2} } + C
+}}
+{{eqn	| r = \pi \map \ln {a + \size a} + C
+	| c = {{Defof|Absolute Value/Definition 2|Absolute Value: Definition 2}}
+}}
+{{eqn	| r = \pi \map \ln {2 a} + C
+	| c = since $a \ge 0$
+}}
+{{end-eqn}}
+On the other hand: 
+{{begin-eqn}}
+{{eqn	| l = \map I a
+	| r = \int_0^\pi \ln a \rd x
+}}
+{{eqn	| r = \pi \ln a
+	| c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+so, by [[Sum of Logarithms]]:
+:$\pi \ln 2 + \pi \ln a + C = \pi \ln a$
+so:
+:$C = -\pi \ln 2$
+giving: 
+{{begin-eqn}}
+{{eqn	| l = \displaystyle \int_0^\pi \map \ln {a + b \cos x} \rd x 
+	| r = \pi \map \ln {a + \sqrt {a^2 - b^2} } - \pi \ln 2
+}}
+{{eqn	| r = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}
+	| c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+in the case $b = 0$.
+Suppose that $b > 0$, then: 
+{{begin-eqn}}
+{{eqn	| l = \map I b
+	| r = \pi \map \ln {b + \sqrt {b^2 - b^2} } + C
+}}
+{{eqn	| r = \pi \ln b + C
+}}
+{{end-eqn}}
+On the other hand:
+{{begin-eqn}}
+{{eqn	| l = \map I b
+	| r = \int_0^\pi \map \ln {b + b \cos x} \rd x
+}}
+{{eqn	| r = \int_0^\pi \ln b \rd x + \int_0^\pi \map \ln {1 + \cos x} \rd x
+	| c = [[Sum of Logarithms]]
+}}
+{{eqn	| r = \pi \ln b + \int_0^\pi \map \ln {2 \cos^2 \frac x 2} \rd x
+	| c = [[Primitive of Constant]], [[Double Angle Formulas/Cosine/Corollary 1|Double Angle Formulas: Cosine: Corollary 1]]
+}}
+{{eqn	| r = \pi \ln b + \int_0^\pi \ln 2 \rd x + 2 \int_0^\pi \map \ln {\cos \frac x 2} \rd x
+	| c = [[Sum of Logarithms]], [[Logarithm of Power]]
+}}
+{{eqn	| r = \pi \ln b + \pi \ln 2 + 4 \int_0^{\frac \pi 2} \map \ln {\cos u} \rd u
+	| c = [[Integration by Substitution|substituting]] $u = \dfrac x 2$
+}}
+{{eqn	| r = \pi \ln b + \pi \ln 2 - 2 \pi \ln 2
+	| c = [[Definite Integral from 0 to Half Pi of Logarithm of Cosine x|Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\cos x}$]]
+}}
+{{eqn	| r = \pi \ln b - \pi \ln 2
+}}
+{{end-eqn}}
+giving: 
+:$C = -\pi \ln 2$
+So we have: 
+:$\displaystyle \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
+in the case $b > 0$ too.
+In the case $b < 0$ we have: 
+{{begin-eqn}}
+{{eqn	| l = \map I {\size b}
+	| r = \pi \map \ln {\size b + \sqrt {\size b^2 - b^2} } + C
+}}
+{{eqn	| r = \pi \map \ln {\size b} + C
+}}
+{{end-eqn}}
+On the other hand:
+{{begin-eqn}}
+{{eqn	| l = \map I {\size b}
+	| r = \int_0^\pi \map \ln {\size b + b \cos x} \rd x
+}}
+{{eqn	| r = \int_0^\pi \map \ln {\size b} \rd x + \int_0^\pi \map \ln {1 + \frac b {\size b} \cos x} \rd x
+	| c = [[Sum of Logarithms]]
+}}
+{{eqn	| r = \pi \ln {\size b} + \int_0^\pi \map \ln {1 - \cos x} \rd x
+	| c = since $\dfrac b {\size b} = -1$ for $b < 0$ and using [[Primitive of Constant]]
+}}
+{{eqn	| r = \pi \ln {\size b} + \int_0^\pi \map \ln {2 \sin^2 \frac x 2} \rd x
+	| c = [[Double Angle Formulas/Cosine/Corollary 2|Double Angle Formulas: Cosine: Corollary 2]]
+}}
+{{eqn	| r = \pi \ln \size b + \int_0^\pi \ln 2 \rd x + \int_0^\pi \map \ln {\sin^2 \frac x 2} \rd x
+	| c = [[Sum of Logarithms]]
+}}
+{{eqn	| r = \pi \ln \size b + \pi \ln 2 + 2 \int_0^\pi \map \ln {\sin \frac x 2} \rd x
+	| c = [[Primitive of Constant]], [[Logarithm of Power]]
+}}
+{{eqn	| r = \pi \ln \size b + \pi \ln 2 + 4 \int_0^{\pi/2} \map \ln {\sin u} \rd u
+	| c = [[Integration by Substitution|substituting]] $u = \dfrac x 2$
+}}
+{{eqn	| r = \pi \ln \size b + \pi \ln 2 - 2 \pi \ln 2
+	| c = [[Definite Integral from 0 to Half Pi of Logarithm of Sine x|Definite Integral from $0$ to $\dfrac \pi 2$ of $\map \ln {\sin x}$]]
+}}
+{{eqn	| r = \pi \ln \size b - \pi \ln 2
+}}
+{{end-eqn}}
+We have in this case: 
+:$C = -\pi \ln 2$
+So we have: 
+:$\displaystyle \int_0^\pi \map \ln {a + b \cos x} \rd x = \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
+in the case $b < 0$ too.
+We've covered all possible values for $b$ so we're done. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccosine in terms of Arctangent}
+Tags: Arccosine Function, Arctangent Function
+
+\begin{theorem}
+:$\displaystyle \arccos x = 2 \map \arctan {\sqrt {\frac {1 - x} {1 + x} } }$
+\end{theorem}
+
+\begin{proof}
+Let: 
+:$\theta = \arccos x$
+Then:
+:$x = \cos \theta$
+and: 
+:$0 \le \theta < \pi$
+by the [[Definition:Inverse Cosine/Real/Arccosine|definition of arccosine]].
+Then: 
+{{begin-eqn}}
+{{eqn	| l = 2 \map \arctan {\sqrt {\frac {1 - x} {1 + x} } }
+	| r = 2 \map \arctan {\sqrt {\frac {1 - \cos \theta} {1 + \cos \theta} } }
+}}
+{{eqn	| r = 2 \map \arctan {\sqrt {\frac {2 \sin^2 \frac \theta 2} {2 \cos^2 \frac \theta 2} } }
+	| c = [[Double Angle Formulas/Cosine/Corollary 1|Double Angle Formulas: Cosine: Corollary 1]], [[Double Angle Formulas/Cosine/Corollary 2|Double Angle Formulas: Cosine: Corollary 2]]
+}}
+{{eqn	| r = 2 \map \arctan {\tan \frac \theta 2}
+	| c = for $0 \le \theta < \dfrac \pi 2$ we have $\sin \theta \ge 0$ and $\cos \theta > 0$
+}}
+{{eqn	| r = \theta
+	| c = {{Defof|Arctangent}}
+}}
+{{eqn	| r = \arccos x
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Arccosine Function]]
+[[Category:Arctangent Function]]
+9w15qwluzce1ugsmhchgoisx49c76w6
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Reciprocal of a plus b Cosine x}
+Tags: Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x = \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a}$
+\end{theorem}
+
+\begin{proof}
+Since $a > b > 0$, we have $a^2 > b^2$.
+So:
+{{begin-eqn}}
+{{eqn	| l = \int_0^{\pi/2} \frac 1 {a + b \cos x} \rd x
+	| r = \intlimits {\frac 2 {\sqrt {a^2 - b^2} } \map \arctan {\sqrt {\frac {a - b} {a + b} } \tan \frac x 2} } 0 1
+	| c = [[Primitive of Reciprocal of p plus q by Cosine of a x|Primitive of $\dfrac 1 {p + q \cos x}$]]
+}}
+{{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {a - b} {a + b} } } }
+}}
+{{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \paren {2 \map \arctan {\sqrt {\frac {1 - \frac b a} {1 + \frac b a} } } }
+}}
+{{eqn	| r = \frac 1 {\sqrt {a^2 - b^2} } \map \arccos {\frac b a}
+	| c = [[Arccosine in terms of Arctangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Exponential of -a x by Sine of b x over x}
+Tags: Definite Integrals involving Sine Function, Definite Integrals involving Exponential Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x = \map \arctan {\frac b a}$
+\end{theorem}
+
+\begin{proof}
+Take $a$ constant and define: 
+:$\displaystyle \map I b = \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} b
+	| r = \frac \d {\d b} \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x
+}}
+{{eqn	| r = \int_0^\infty \frac \partial {\partial b} \paren {\frac {e^{-a x} \sin b x} x} \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = \int_0^\infty e^{-a x} \cos b x \rd x
+	| c = [[Derivative of Cosine of a x|Derivative of $\cos a x$]]
+}}
+{{eqn	| r = \frac a {a^2 + b^2}
+	| c = [[Definite Integral to Infinity of Exponential of -a x by Cosine of b x|Definite Integral to Infinity of $e^{-a x} \cos b x$]]
+}}
+{{end-eqn}}
+so:
+{{begin-eqn}}
+{{eqn	| l = \map I b
+	| r = a \int \frac 1 {b^2 + a^2} \rd b
+}}
+{{eqn	| r = \frac a a \arctan \frac b a + C
+	| c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn	| r = \arctan \frac b a + C
+}}
+{{end-eqn}}
+for some constant $C \in \R$.
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map I 0
+	| r = \int_0^\infty \frac {e^{-a x} \sin 0} x \rd x
+}}
+{{eqn	| r = \int_0^\infty 0 \rd x
+}}
+{{eqn	| r = 0
+}}
+{{end-eqn}}
+on the other hand:
+{{begin-eqn}}
+{{eqn	| l = \map I b
+	| r = \arctan 0 + C
+}}
+{{eqn	| r = C
+}}
+{{end-eqn}}
+so:
+:$C = 0$
+So we have: 
+:$\displaystyle \int_0^\infty \frac {e^{-a x} \sin b x} x \rd x = \map \arctan {\frac b a}$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Sine of m x over Exponential of 2 Pi x minus One}
+Tags: Definite Integrals involving Sine Function, Definite Integrals involving Exponential Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x = \frac 1 4 \coth \frac m 2 - \frac 1 {2 m}$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{begin-eqn}}
+{{eqn	| l = \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x
+	| r = \int_0^\infty \frac {e^{-2 \pi x} \sin m x} {1 - e^{-2 \pi x} } \rd x
+}}
+{{eqn	| r = \int_0^\infty e^{-2 \pi x} \sin m x \paren {\sum_{k = 0}^\infty e^{-2 \pi k x} } \rd x
+	| c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn	| r = \sum_{k = 0}^\infty \paren {\int_0^\infty e^{-x \paren {2 \pi \paren {1 + k} } } \sin m x \rd x}
+}}
+{{eqn	| r = \sum_{k = 0}^\infty \frac m {\paren {2 \pi \paren {1 + k} }^2 + m^2}
+	| c = [[Definite Integral to Infinity of Exponential of -a x by Sine of b x|Definite Integral to Infinity of $e^{-a x} \sin b x$]]
+}}
+{{eqn	| r = \frac 1 {2 \pi} \sum_{k = 0}^\infty \frac {\frac m {2 \pi} } {\paren {k + 1}^2 + \paren {\frac m {2 \pi} }^2}
+	| c = dividing by $\paren {2 \pi}^2$
+}}
+{{eqn	| r = \frac 1 {2 \pi} \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2}
+	| c = shifting the index
+}}
+{{end-eqn}}
+By [[Mittag-Leffler Expansion for Hyperbolic Cotangent Function]] we have: 
+:$\displaystyle \sum_{k = 1}^\infty \frac z {k^2 + z^2} = \frac \pi 2 \map \coth {\pi z} - \frac 1 {2 z}$
+for all $z \in \C$ where $z$ is not an [[Definition:Integer|integer]] multiple of $i$. 
+Since all variables concerned in this instance all real-valued, we can apply this identity. 
+Setting $z = \dfrac m {2 \pi}$ in the above we obtain:
+{{begin-eqn}}
+{{eqn	| l = \sum_{k = 1}^\infty \frac {\frac m {2 \pi} } {k^2 + \paren {\frac m {2 \pi} }^2}
+	| r = \frac \pi 2 \map \coth {\pi \times \frac m {2 \pi} } - \frac 1 {2 \times \frac m {2 \pi} }
+}}
+{{eqn	| r = \frac \pi 2 \coth \frac m 2 - \frac \pi m
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \int_0^\infty \frac {\sin m x} {e^{2 \pi x} - 1} \rd x = \frac 1 {2 \pi} \paren {\frac \pi 2 \coth \frac m 2 - \frac \pi m} = \frac 1 4 \coth \frac m 2 - \frac 1 {2 m}$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Exponential of -a x^2 by Cosine of b x}
+Tags: Definite Integrals involving Exponential Function, Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$
+\end{theorem}
+
+\begin{proof}
+Fix $a$ and define: 
+:$\displaystyle \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x$
+for all $b \in \R$. 
+Then, we have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} b
+	| r = \frac \d {\d b} \paren {\int_0^\infty e^{-a x^2} \cos b x \rd x}
+}}
+{{eqn	| r = \int_0^\infty \frac \partial {\partial b} \paren {e^{-a x^2} \cos b x} \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = -\int_0^\infty \paren {x e^{-a x^2} } \sin b x \rd x
+	| c = [[Derivative of Cosine of a x|Derivative of $\cos a x$]]
+}}
+{{eqn	| r = -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x}
+	| c = [[Integration by Parts]]
+}}
+{{end-eqn}}
+Note that: 
+{{begin-eqn}}
+{{eqn	| l = \size {\frac 1 {2 a} e^{-a x^2} \sin b x}
+	| o = \le
+	| r = \frac 1 {2 a} e^{-a x^2}
+	| c = noting that $\size {\sin x} \le 1$
+}}
+{{eqn	| o = \to
+	| r = 0
+	| c = [[Exponential Tends to Zero and Infinity]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn	| l = -\paren {\intlimits {-\frac 1 {2 a} e^{-a x^2} \sin b x} 0 \infty - b \int_0^\infty \paren {-\frac 1 {2 a} e^{-a x^2} } \cos b x \rd x}
+	| r = -\frac b {2 a} \int_0^\infty e^{-a x^2} \cos b x \rd x
+}}
+{{eqn	| r = -\frac b {2 a} \map I b
+}}
+{{end-eqn}}
+We then have: 
+:$\displaystyle \frac {\map {I'} b} {\map I b} = -\frac b {2 a}$
+Integrating, by [[Primitive of Function under its Derivative]] and [[Primitive of Constant]]:
+:$\displaystyle \ln \size {\map I b} = -\frac {b^2} {4 a} + C$
+for some $C \in \R$.
+{{finish|This obviously only gives us an expression for $\ln \size {\map I b}$, we then need to determine that $\map I b > 0$ which seems nontrivial}}
+So: 
+:$\displaystyle \map I b = A \map \exp {-\frac {b^2} {4 a} }$
+for some $A \in \R$.
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map I 0
+	| r = \int_0^\infty e^{-a x^2} \rd x
+}}
+{{eqn	| r = \frac 1 2 \sqrt {\frac \pi a}
+	| c = [[Definite Integral to Infinity of Exponential of -a x^2|Definite Integral to Infinity of $e^{-a x^2}$]]
+}}
+{{end-eqn}}
+on the other hand we have: 
+{{begin-eqn}}
+{{eqn	| l = \map I 0
+	| r = A \map \exp 0
+}}
+{{eqn	| r = A
+	| c = [[Exponential of Zero]]
+}}
+{{end-eqn}}
+So we have: 
+$\displaystyle \map I b = \int_0^\infty e^{-a x^2} \cos b x \rd x = \frac 1 2 \sqrt {\frac \pi a} \map \exp {-\frac {b^2} {4 a} }$
+for all $b \in \R$ as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fourier Series/Logarithm of Sine of x over 0 to Pi}
+Tags: Examples of Fourier Series
+
+\begin{theorem}
+:$\displaystyle \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$
+\end{theorem}
+
+\begin{proof}
+We find the [[Definition:Half-Range Fourier Cosine Series|Half-Range Fourier Cosine Series]] over $\openint 0 {\dfrac \pi 2}$ for $\map \ln {\sin x}$.
+By definition:
+:$\displaystyle \map \ln {\sin x} \sim \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x$
+where for all $n \in \Z_{\ge 0}$:
+:$\displaystyle a_n = \frac 4 \pi \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x$
+By [[Definite Integral from 0 to Half Pi of Logarithm of Sine x]]:
+:$a_0 = \dfrac 4 \pi \paren {- \dfrac \pi 2 \ln 2} = -2 \ln 2$
+By [[Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx]]:
+:$a_n = \dfrac 4 \pi \paren {- \dfrac \pi {4 n}} = -\dfrac 1 n$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map \ln {\sin x}
+      | o = \sim
+      | r = \frac {a_0} 2 + \sum_{n \mathop = 1}^\infty a_n \cos 2 n x
+}}
+{{eqn | r = - \ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of Fourier Series]]
+90o3kbsvaey3is713tdddlykcy0litm
+\end{proof}<|endoftext|>
+\section{Subset of Set Difference iff Disjoint Set}
+Tags: Set Difference, Disjoint Sets
+
+\begin{theorem}
+Let $S, T$ be [[Definition:Set|sets]].
+Let $A \subseteq S$
+Then:
+:$A \cap T = \varnothing \iff A \subseteq S \setminus T$
+where:
+:$A \cap T$ denotes [[Definition:Set Intersection|set intersection]]
+:$\varnothing$ denotes the [[Definition:Empty Set|empty set]]
+:$S \setminus T$ denotes [[Definition:Set Difference|set difference]].
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = A \cap \paren {S \setminus T} 
+      | r = \paren{A \cap S} \setminus T
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = A \setminus T
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | ll = 
+      | o =
+      | r = A \subseteq \paren{ S \setminus T}
+      | c =
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = A = A \cap \paren{ S \setminus T}
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = A = A \setminus T
+      | c = As $A \cap \paren {S \setminus T} = A \setminus T$
+}}
+{{eqn | ll = \leadstoandfrom
+      | o =
+      | r = A \cap T = \O
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Disjoint Sets]]
+20ifr7cnqwnyfplovl2gvqlufalz85n
+\end{proof}<|endoftext|>
+\section{Set Difference of Doubleton and Singleton is Singleton}
+Tags: Singletons, Doubletons
+
+\begin{theorem}
+Let $x, y$ be [[Definition:Distinct|distinct]] [[Definition:Object|objects]].
+Then:
+:$\set{x, y} \setminus \set x = \set y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \set{x, y} \setminus \set x
+      | r = \big \{ z: z \in \set{x, y} \land z \notin \set x \big \}
+      | c = {{Defof|Set Difference}}
+}}
+{{eqn | r = \big \{ z: \paren{ z = x \lor z = y} \land z \notin \set x \big \}
+      | c = {{Defof|Doubleton}}
+}}
+{{eqn | r = \big \{ z: \paren{ z = x \lor z = y} \land z \ne x \big \}
+      | c = {{Defof|Singleton}}
+}}
+{{eqn | r = \big \{ z: \paren{z = x \land z \ne x} \lor \paren {z = y \land z \ne x}  \big \}
+      | c = [[Conjunction Distributes over Disjunction]]
+}}
+{{eqn | r = \big \{ z: \bot \lor \paren {z = y \land z \ne x} \big \}
+      | c = {{Defof|Contradiction}}
+}}
+{{eqn | r = \big \{ z: \paren {z = y \land z \ne x} \big \}
+      | c = [[Disjunction with Contradiction]]
+}}
+{{eqn | r = \big \{ z : z = y \big \}
+      | c = [[Rule of Simplification]]
+}}
+{{eqn | r = \set y
+      | c = {{Defof|Singleton}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Singletons]]
+[[Category:Doubletons]]
+4jshkpfs1zzd6deak4twch4o3ox2k2a
+\end{proof}<|endoftext|>
+\section{Egyptian Formula for Area of Quadrilateral}
+Tags: Area Formulas, Quadrilaterals
+
+\begin{theorem}
+Let $\Box ABCD$ be a [[Definition:Quadrilateral|quadrilateral]].
+Let the [[Definition:Side of Polygon|sides]] of $\Box ABCD$ be $a$, $b$, $c$ and $d$ such that $a$ is [[Definition:Opposite (in Polygon)|opposite]] $c$ and $b$ is [[Definition:Opposite (in Polygon)|opposite]] $d$.
+Then the [[Definition:Area|area]] of $\Box ABCD$ can be approximated by:
+:$\map \Area {\Box ABCD} \approx \dfrac {a + c} 2 \times \dfrac {b + d} 2$
+The closer $\Box ABCD$ is to a [[Definition:Rectangle|rectangle]], the better the approximation.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Need to consider how to approach this}}
+\end{proof}<|endoftext|>
+\section{1 plus Perfect Power is not Power of 2}
+Tags: Number Theory
+
+\begin{theorem}
+The equation:
+:$1 + a^n = 2^m$
+has no solutions in the [[Definition:Integer|integers]] for $n, m > 1$.
+This is an elementary special case of [[Catalan's Conjecture]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there is a solution.
+Then:
+{{begin-eqn}}
+{{eqn | l = a^n
+      | r = 2^m - 1
+}}
+{{eqn | o = \equiv
+      | r = -1
+      | rr = \pmod 4
+      | c = as $m > 1$
+}}
+{{end-eqn}}
+$a$ is immediately seen to be [[Definition:Odd Integer|odd]].
+By [[Square Modulo 4]], $n$ must also be [[Definition:Odd Integer|odd]].
+Now:
+{{begin-eqn}}
+{{eqn | l = 2^m
+      | r = a^n + 1
+}}
+{{eqn | r = \paren {a + 1} \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a^{n - k - 1}
+      | c = [[Sum of Two Odd Powers]]
+}}
+{{end-eqn}}
+The latter sum has $n$ powers of $a$, which sums to an [[Definition:Odd Integer|odd number]].
+The only [[Definition:Odd Integer|odd]] [[Definition:Divisor of Integer|divisor]] of $2^m$ is $1$.
+However, if the sum is $1$, we have:
+:$a^n + 1 = a + 1$
+giving $n = 1$, [[Definition:Contradiction|contradicting]] our constraint $n > 1$.
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+[[Category:Number Theory]]
+ex9a8rtweepg9xdh7umpt71f6nn8kuk
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Logarithm of Sine x by Cosine of 2nx}
+Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Sine Function
+
+\begin{theorem}
+For $n \in \N_{>0}$:
+:$\displaystyle \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x = -\frac \pi {4 n}$
+\end{theorem}
+
+\begin{proof}
+First we have:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x
+      | r = \lim_{x \mathop \to 0} \frac {\map \ln {\sin x} } {\csc 2 n x}
+      | c = {{Defof|Cosecant}}
+}}
+{{eqn | r = \lim_{x \mathop \to 0} \frac {\cot x} {- 2 n \cot 2 n x \csc 2 n x}
+      | c = [[L'Hôpital's Rule/Corollary 2]]
+}}
+{{eqn | r = \lim_{x \mathop \to 0} \frac {\cos x} {- 2 n \cos 2 n x} \frac {\sin^2 2 n x} {\sin x}
+      | c = {{Defof|Cosecant}}, {{Defof|Cotangent}}
+}}
+{{eqn | l = \lim_{x \mathop \to 0} \frac {\sin^2 2 n x} {\sin x}
+      | r = \lim_{x \mathop \to 0} \frac {4 n \sin 2 n x \cos 2 n x} {\cos x}
+      | c = [[L'Hôpital's Rule]]
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+hence $\displaystyle \lim_{x \mathop \to 0} \map \ln {\sin x} \sin 2 n x = 0$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int_0^{\pi/2} \map \ln {\sin x} \cos 2 n x \ \d x
+      | r = \frac 1 {2 n} \int_0^{\pi/2} \map \ln {\sin x} \ \map \d {\sin 2 n x}
+      | c = [[Primitive of Cosine Function/Corollary]]
+}}
+{{eqn | r = \frac 1 {2 n} \paren {\bigintlimits {\map \ln {\sin x} \sin 2 n x} 0 {\pi/2} - \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} } }
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin n x \ \map \d {\map \ln {\sin x} }
+      | c = From above
+}}
+{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \sin 2 n x \frac {\cos x} {\sin x} \ \d x
+      | c = [[Primitive of Cotangent Function]]
+}}
+{{eqn | r = -\frac 1 {2 n} \int_0^{\pi/2} \frac {\sin \paren {2 n + 1} x + \sin \paren {2 n - 1} x} {2 \sin x} \ \d x
+      | c = [[Simpson's Formulas/Sine by Cosine]]
+}}
+{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \frac {\sin \paren {\paren {2 n + 1} u/2} + \sin \paren {\paren {2 n - 1} u/2} } {2 \sin \paren {u/2} } \ \d u
+      | c = [[Integration by Substitution|Substituting]] $u = 2 x$
+}}
+{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} \paren {\frac 1 2 + \sum_{k \mathop = 1}^n \map \cos {k u} + \frac 1 2 + \sum_{k \mathop = 1}^{n - 1} \map \cos {k u} } \ \d u
+      | c = [[Sum of Cosines of Multiples of Angle]]
+}}
+{{eqn | r = -\frac 1 {4 n} \int_0^{\pi} 1 \ \d u
+      | c = All integrals involving $\cos k u$ evaluates to $0$
+}}
+{{eqn | r = -\frac \pi {4 n}
+      | c = [[Integral of Constant/Definite]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Definite Integrals involving Logarithm Function]]
+[[Category:Definite Integrals involving Sine Function]]
+swrdggg8dsb6vrtuf3lswb34vohkf7i
+\end{proof}<|endoftext|>
+\section{Distinct Matroid Elements are Parallel iff Each is in Closure of Other/Lemma}
+Tags: Distinct Matroid Elements are Parallel iff Each is in Closure of Other
+
+\begin{theorem}
+Let $a, b \in S$.
+Let $\set a$ and $\set b$ be [[Definition:Independent Subset (Matroid)|independent]].
+Then $\set {a, b}$ is [[Definition:Dependent Subset (Matroid)|dependent]] {{iff}}:
+:$a \in \map \sigma {\set b}$
+and
+:$b \in \map \sigma {\set a}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \set {a, b}
+      | o = \notin
+      | r = \mathscr I
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \set a \cup \set b
+      | o = \notin
+      | r = \mathscr I
+      | c = [[Union of Disjoint Singletons is Doubleton]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a
+      | o = \in
+      | r = \map \sigma {\set b}
+      | c = [[Element Depends on Independent Set iff Union with Singleton is Dependent]]
+}}
+{{eqn | lo= \land
+      | l = b
+      | o = \in
+      | r = \map \sigma {\set a}
+      | c = [[Element Depends on Independent Set iff Union with Singleton is Dependent]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Distinct Matroid Elements are Parallel iff Each is in Closure of Other]]
+a3szsfqgb984wsvqqb2yy8i472rnz2j
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Sine m x over x by x Squared plus a Squared}
+Tags: Definite Integrals involving Sine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$
+\end{theorem}
+
+\begin{proof}
+Fix $a$ and set: 
+:$\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x$
+for $m \ge 0$. 
+We have:
+{{begin-eqn}}
+{{eqn	| l = \map {I'} m
+	| r = \frac \d {\d m} \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x
+}}
+{{eqn	| r = \int_0^\infty \frac \partial {\partial m} \paren {\frac {\sin m x} {x \paren {x^2 + a^2} } } \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = \int_0^\infty \frac {\cos m x} {x^2 + a^2} \rd x
+	| c = [[Derivative of Sine of a x|Derivative of $\sin a x$]]
+}}
+{{eqn	| r = \frac \pi {2 a} e^{-m a}
+	| c = [[Definite Integral to Infinity of Cosine m x over x Squared plus a Squared|Definite Integral to Infinity of $\dfrac {\cos m x} {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+So by [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]:
+:$\displaystyle \map I m = -\frac \pi {2 a^2} e^{-m a} + C$
+for some constant $C \in \R$.
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map I 0
+	| r = \int_0^\infty \frac {\sin 0 x} {x^2 + a^2} \rd x
+}}
+{{eqn	| r = \int_0^\infty 0 \rd x
+	| c = [[Sine of Zero is Zero]]
+}}
+{{eqn	| r = 0
+}}
+{{end-eqn}}
+On the other hand: 
+{{begin-eqn}}
+{{eqn	| l = \map I 0
+	| r = -\frac \pi {2 a^2} e^0 + C
+}}
+{{eqn	| r = -\frac \pi {2 a^2} + C
+	| c = [[Exponential of Zero]]
+}}
+{{end-eqn}}
+So:
+:$C = \dfrac \pi {2 a^2}$
+giving: 
+:$\displaystyle \map I m = \int_0^\infty \frac {\sin m x} {x \paren {x^2 + a^2} } \rd x = \frac \pi {2 a^2} \paren {1 - e^{-m a} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Square of Logarithm of Sine x}
+Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Sine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
+\end{theorem}
+
+\begin{proof}
+From [[Fourier Series/Logarithm of Sine of x over 0 to Pi|Fourier Series of $\map \ln {\sin x}$ from $0$ to $\pi$]]:
+:$\displaystyle \map \ln {\sin x} = -\ln 2 - \sum_{n \mathop = 1}^\infty \frac {\cos 2 n x} n$
+Then, by [[Parseval's Theorem]]:
+{{begin-eqn}}
+{{eqn	| l = \frac 2 \pi \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x
+	| r = 2 \paren {\ln 2}^2 + \sum_{n = 1}^\infty \frac 1 {n^2}
+}}
+{{eqn	| r = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6
+	| c = [[Basel Problem]]
+}}
+{{end-eqn}}
+We then have: 
+{{begin-eqn}}
+{{eqn	| l = \int_0^\pi \paren {\map \ln {\sin x} }^2 \rd x
+	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_{\pi/2}^\pi \paren {\map \ln {\sin x} }^2 \rd x
+	| c = [[Sum of Integrals on Adjacent Intervals for Integrable Functions]]
+}}
+{{eqn	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x + \int_0^{\pi/2} \paren {\map \ln {\map \sin {\pi - x} } }^2 \rd x
+}}
+{{eqn	| r = 2 \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x
+	| c = [[Sine of Supplementary Angle]]
+}}
+{{end-eqn}}
+So we have: 
+:$\displaystyle \frac 4 \pi \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = 2 \paren {\ln 2}^2 + \frac {\pi^2} 6$
+multiplying by $\dfrac \pi 4$ we have: 
+:$\displaystyle \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Half Pi of Square of Logarithm of Cosine x}
+Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^{\pi/2} \paren {\map \ln {\cos x} }^2 \rd x = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \int_0^{\pi/2} \paren {\map \ln {\cos x} }^2 \rd x
+	| r = \int_0^{\pi/2} \paren {\map \ln {\map \cos {\frac \pi 2 - x} } }^2 \rd x
+}}
+{{eqn	| r = \int_0^{\pi/2} \paren {\map \ln {\sin x} }^2 \rd x
+	| c = [[Cosine of Complement equals Sine]]
+}}
+{{eqn	| r = \frac \pi 2 \paren {\ln 2}^2 + \frac {\pi^3} {24}
+	| c = [[Definite Integral from 0 to Half Pi of Square of Logarithm of Sine x|Definite Integral from $0$ to $\dfrac \pi 2$ of $\paren {\map \ln {\sin x} }^2$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to 2 Pi of Logarithm of a plus b Cosine x}
+Tags: Definite Integrals involving Logarithm Function, Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^{2 \pi} \map \ln {a + b \cos x} \rd x = 2 \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_0^{2 \pi} \map \ln {a + b \cos x} \rd x
+      | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_\pi^{2 \pi} \map \ln {a + b \cos x} \rd x
+      | c = [[Sum of Integrals on Adjacent Intervals for Integrable Functions]]
+}}
+{{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x - \int_\pi^0 \map \ln {a + b \map \cos {2 \pi - x} } \rd x
+      | c = [[Integration by Substitution|substituting]] $x \mapsto 2 \pi - x$
+}}
+{{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \map \cos {2 \pi - x} } \rd x
+      | c = [[Reversal of Limits of Definite Integral]]
+}}
+{{eqn | r = \int_0^\pi \map \ln {a + b \cos x} \rd x + \int_0^\pi \map \ln {a + b \cos x} \rd x
+      | c = [[Cosine of Conjugate Angle]]
+}}
+{{eqn | r = 2 \int_0^\pi \map \ln {a + b \cos x} \rd x
+}}
+{{eqn | r = 2 \pi \map \ln {\frac {a + \sqrt {a^2 - b^2} } 2}
+      | c = [[Definite Integral from 0 to Pi of Logarithm of a plus b Cosine x]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral from 0 to Pi of Sec x by Logarithm of One plus b Cosine x over One plus a Cosine x}
+Tags: Definite Integrals involving Cosine Function, Definite Integrals involving Logarithm Function
+
+\begin{theorem}
+:$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$
+\end{theorem}
+
+\begin{proof}
+Note that by [[Difference of Logarithms]]: 
+:$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \int_0^{\pi/2} \sec x \map \ln {1 + b \cos x} \rd x - \int_0^{\pi/2} \sec \map \ln {1 + a \cos x} \rd x$
+For each $\alpha \in \openint {-1} 1$, set: 
+:$\displaystyle \map I \alpha = \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x$
+Then: 
+:$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \map I b - \map I a$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} \alpha
+	| r = \frac \d {\d \alpha} \int_0^{\pi/2} \sec x \map \ln {1 + \alpha \cos x} \rd x
+}}
+{{eqn	| r = \int_0^{\pi/2} \frac \partial {\partial \alpha} \paren {\sec x \map \ln {1 + \alpha \cos x} } \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = \int_0^{\pi/2} \frac {\cos x \sec x} {1 + \alpha \cos x} \rd x
+	| c = [[Chain Rule]], [[Derivative of Natural Logarithm]]
+}}
+{{eqn	| r = \int_0^{\pi/2} \frac 1 {1 + \alpha \cos x} \rd x
+}}
+{{eqn	| r = \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} }
+	| c = [[Definite Integral from 0 to Half Pi of Reciprocal of a plus b Cosine x|Definite Integral from $0$ to $\dfrac \pi 2$ of $\dfrac 1 {a + b \cos x}$]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn	| l = \map I b - \map I a
+	| r = \int_a^b \map {I'} \alpha \rd \alpha
+	| c = [[Fundamental Theorem of Calculus/Second Part|Fundamental Theorem of Calculus: Second Part]]
+}}
+{{eqn	| r = \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha
+}}
+{{end-eqn}}
+From [[Derivative of Arccosine Function]], we have: 
+:$\displaystyle \frac \d {\d \alpha} \arccos \alpha = -\frac 1 {\sqrt {1 - \alpha^2} }$
+so:
+{{begin-eqn}}
+{{eqn	| l = \int_a^b \frac {\arccos \alpha} {\sqrt {1 - \alpha^2} } \rd \alpha
+	| r = \int_{\arccos a}^{\arccos b} x \frac {-\sqrt {1 - \alpha^2} } {\sqrt {1 - \alpha^2} } \rd x
+	| c = [[Integration by Substitution|substituting]] $x = \arccos \alpha$
+}}
+{{eqn	| r = -\int_{\arccos a}^{\arccos b} x \rd x
+}}
+{{eqn	| r = \intlimits {-\frac {x^2} 2} {\arccos a} {\arccos b}
+	| c = [[Primitive of Power]]
+}}
+{{eqn	| r = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}
+}}
+{{end-eqn}}
+so:
+:$\displaystyle \int_0^{\pi/2} \sec x \map \ln {\frac {1 + b \cos x} {1 + a \cos x} } \rd x = \frac 1 2 \paren {\paren {\arccos a}^2 - \paren {\arccos b}^2}$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x by Secant of p x}
+Tags: Definite Integrals involving Exponential Function, Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} }$
+\end{theorem}
+
+\begin{proof}
+Fix $p$ and set: 
+:$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \sec p x} \rd x$
+for all $\alpha \ge 0$.
+Then: 
+:$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x = \map I a - \map I b$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} \alpha
+	| r = \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \sec p x} \rd x
+}}
+{{eqn	| r = \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \sec p x} } \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = -\int_0^\infty e^{-\alpha x} \cos p x \rd x
+	| c = [[Derivative of Exponential of a x|Derivative of $e^{a x}$]], {{Defof|Secant Function}}
+}}
+{{eqn	| r = -\frac \alpha {\alpha^2 + p^2}
+	| c = [[Definite Integral to Infinity of Exponential of -a x by Cosine of b x|Definite Integral to Infinity of $e^{-a x} \cos b x$]]
+}}
+{{end-eqn}}
+so:
+{{begin-eqn}}
+{{eqn	| l = \map I \alpha
+	| r = -\int \frac \alpha {\alpha^2 + p^2} \rd \alpha
+}}
+{{eqn	| r = -\frac 1 2 \map \ln {\alpha^2 + p^2} + C
+	| c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+for all $\alpha \ge 0$, for some constant $C \in \R$.
+We then have: 
+{{begin-eqn}}
+{{eqn	| l = \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \sec p x} \rd x
+	| r = \map I a - \map I b
+}}
+{{eqn	| r = \paren {-\frac 1 2 \map \ln {a^2 + p^2} + C} - \paren {-\frac 1 2 \map \ln {b^2 + p^2} + C}
+}}
+{{eqn	| r = \frac 1 2 \paren {\map \ln {b^2 + p^2} - \map \ln {a^2 + p^2} }
+}}
+{{eqn	| r = \frac 1 2 \map \ln {\frac {b^2 + p^2} {a^2 + p^2} }
+	| c = [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Cosine p x minus Cosine q x over x Squared}
+Tags: Definite Integrals involving Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x = \frac {\pi \paren {\size q - \size p} } 2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \int_0^\infty \frac {\cos p x - \cos q x} {x^2} \rd x
+	| r = \int_0^\infty \frac {1 - \cos q x - \paren {1 - \cos p x} } {x^2} \rd x
+}}
+{{eqn	| r = \int_0^\infty \frac {1 - \cos q x} {x^2} \rd x - \int_0^\infty \frac {1 - \cos p x} {x^2} \rd x
+}}
+{{eqn	| r = \frac \pi 2 \size q - \frac \pi 2 \size p
+	| c = [[Integral to Infinity of One minus Cosine p x over x Squared|Integral to Infinity of $\dfrac {1 - \cos p x} {x^2}$]]
+}}
+{{eqn	| r = \frac {\pi \paren {\size q - \size p} } 2
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral to Infinity of Exponential of -a x minus Exponential of -b x over x by Cosecant of p x}
+Tags: Definite Integrals involving Exponential Function, Definite Integrals involving Sine Function
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \arctan \frac b p - \arctan \frac a p$
+\end{theorem}
+
+\begin{proof}
+Fix $p$ and set: 
+:$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x$
+for all $\alpha \ge 0$.
+Then: 
+:$\displaystyle \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x = \map I a - \map I b$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {I'} \alpha
+	| r = \frac \d {\d \alpha} \int_0^\infty \frac {e^{-\alpha x} } {x \csc p x} \rd x
+}}
+{{eqn	| r = \int_0^\infty \frac \partial {\partial \alpha} \paren {\frac {e^{-\alpha x} } {x \csc p x} } \rd x
+	| c = [[Definite Integral of Partial Derivative]]
+}}
+{{eqn	| r = -\int_0^\infty e^{-\alpha x} \sin p x \rd x
+	| c = [[Derivative of Exponential of a x|Derivative of $e^{a x}$]], {{Defof|Cosecant/Real Function|Cosecant}}
+}}
+{{eqn	| r = -\frac p {\alpha^2 + p^2}
+	| c = [[Definite Integral to Infinity of Exponential of -a x by Sine of b x|Definite Integral to Infinity of $e^{-a x} \sin b x$]]
+}}
+{{end-eqn}}
+so:
+{{begin-eqn}}
+{{eqn	| l = \map I \alpha
+	| r = -p \int \frac 1 {\alpha^2 + p^2} \rd \alpha
+}}
+{{eqn	| r = -\arctan \frac \alpha p + C
+	| c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+for all $\alpha \ge 0$ and constant $C \in \R$.
+We therefore have: 
+{{begin-eqn}}
+{{eqn	| l = \int_0^\infty \frac {e^{-a x} - e^{-b x} } {x \csc p x} \rd x
+	| r = \map I a - \map I b
+}}
+{{eqn	| r = \paren {-\arctan \frac a p + C} - \paren {-\arctan \frac b p + C}
+}}
+{{eqn	| r = \arctan \frac b p - \arctan \frac a p
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral of Periodic Function}
+Tags: Definite Integrals, Periodic Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Darboux Integrable Function|Darboux integrable]] [[Definition:Periodic Function|periodic function]] with [[Definition:Periodic Function/Period|period]] $L$.
+Let $\alpha \in \R$ and $n \in \Z$.
+Then:
+:$\displaystyle \int_\alpha^{\alpha + n L} \map f x \d x = n \int_0^L \map f x \d x$
+\end{theorem}
+
+\begin{proof}
+For $n \ge 0$:
+{{begin-eqn}}
+{{eqn | l = \int_\alpha^{\alpha + n L} \map f x \d x
+      | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f x \d x + \int_{n L}^{\alpha + n L} \map f x \d x
+      | c = [[Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary]]
+}}
+{{eqn | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_{k L}^{\paren {k + 1} L} \map f {x - k L} \d x + \int_{n L}^{\alpha + n L} \map f {x - n L} \d x
+      | c = [[General Periodicity Property]]
+}}
+{{eqn | r = \int_\alpha^0 \map f x \d x + \sum_{k \mathop = 0}^{n - 1} \int_0^L \map f x \d x + \int_0^\alpha \map f x \d x
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = n \int_0^L \map f x \d x
+      | c = [[Reversal of Limits of Definite Integral]]
+}}
+{{end-eqn}}
+For $n < 0$:
+{{begin-eqn}}
+{{eqn | l = \int_\alpha^{\alpha + n L} \map f x \d x
+      | r = -\int_{\alpha + n L}^\alpha \map f x \d x
+      | c = [[Reversal of Limits of Definite Integral]]
+}}
+{{eqn | r = -\int_{\alpha + n L}^{\alpha + n L + \paren {-n L} } \map f x \d x
+}}
+{{eqn | r = -\paren {-n \int_0^L \map f x \d x}
+      | c = by the above; $-n > 0$
+}}
+{{eqn | r = n \int_0^L \map f x \d x
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Definite Integrals]]
+[[Category:Periodic Functions]]
+bewj8ikxf8sp0alog7mr5srk702adv3
+\end{proof}<|endoftext|>
+\section{Independent Subset is Contained in Maximal Independent Subset}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $A \subseteq S$.
+Let $X \in \mathscr I$ such that $X \subseteq A$.
+Then:
+:$\exists Y \in \mathscr I : X \subseteq Y \subseteq A : \size Y = \map \rho A$
+where $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] on $M$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Rank Function (Matroid)|rank function]] on $M$:
+:$\size X \le \map \rho A$
+=== Case 1 : $\size X = \map \rho A$ ===
+Let $\size X = \map \rho A$.
+Let $Y = X$ and the result follows.
+{{qed|lemma}}
+ 
+=== Case 2 : $\size X < \map \rho A$ ===
+ 
+Let $\size X < \map \rho A$.
+By definition of the [[Definition:Rank Function (Matroid)|rank function]] on $M$:
+:$\map \rho A = \max \set{\size I : I \subseteq A \land I \in \mathscr I}$
+From [[Max Equals an Operand]]:
+:$\exists Z \in \mathscr I : Z \subseteq A$ and $\size Z = \map \rho A$
+From [[Independent Set can be Augmented by Larger Independent Set]]:
+:$\exists Y' \subseteq Z \setminus X: \size {X \cup Y'} = \size Z$ and $X \cup Y' \in \mathscr I$
+We have:
+{{begin-eqn}}
+{{eqn | l = Y'
+      | o = \subseteq
+      | r = Z \setminus X
+}}
+{{eqn | o = \subseteq
+      | r = Z
+      | c = [[Set Difference is Subset]]
+}}
+{{eqn | o = \subseteq
+      | r = A
+      | c = by choice of $Z$
+}}
+{{end-eqn}}
+From [[Union of Subsets is Subset]]:
+:$X \cup Y' \subseteq A$
+Let $Y = X \cup Y'$ and the result follows.
+{{qed|lemma}}
+In either case, the result follows.
+{{qed}}
+[[Category:Matroid Theory]]
+6ujcur7ahw1f3k9r25liuy0a1cduvu2
+\end{proof}<|endoftext|>
+\section{Automorphic Numbers in Base 10}
+Tags: Automorphic Numbers
+
+\begin{theorem}
+If leading zeroes are allowed, there are exactly $4$ $n$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic numbers]] in [[Definition:Number Base|base $10$]]:
+:$00 \dots 00$
+:$00 \dots 01$
+:$5^{2^{n - 1} } \pmod {10^n}$
+:$6^{5^{n - 1} } \pmod {10^n}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Definition:Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:There are exactly $4$ $n$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic numbers]] of the forms above.
+=== Basis for the Induction ===
+For $n = 1$:
+$0, 1, 5, 6$ are the only $1$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic numbers]], and we have:
+:$5^{2^0} = 5$
+:$6^{5^0} = 6$
+Thus $\map P 1$ 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.
+We assume that for some $k \ge 1$, there are exactly $4$ $k$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic numbers]] of the forms above.
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+We aim to construct $x$, a $\paren {k + 1}$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic number]].
+By [[Left-Truncated Automorphic Number is Automorphic]], after removing the leftmost [[Definition:Digit|digit]], what remains is a $k$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic number]].
+Write $x = 10^k a + b$, where $a$ is the leftmost [[Definition:Digit|digit]] of $x$.
+We have:
+{{begin-eqn}}
+{{eqn | l = x^2
+      | r = \paren {10^k a + b}^2
+}}
+{{eqn | r = 10^{2 k} a^2 + 2 \times 10^k a b + b^2
+      | c = [[Square of Sum]]
+}}
+{{eqn | r = 10^{2 k} a^2 + 2 \times 10^k a b + \paren {b + 10^k N}
+      | c = for some $N \in \N$; {{Defof|Automorphic Number}}
+}}
+{{eqn | o = \equiv
+      | r = 10^k \paren {2 a b + N} + b
+      | rr= \pmod {10^{k + 1} }
+}}
+{{eqn | o = \equiv
+      | r = x
+      | rr= \pmod {10^{k + 1} }
+      | c = {{Defof|Automorphic Number}}
+}}
+{{eqn | o = \equiv
+      | r = 10^k a + b
+      | rr= \pmod {10^{k + 1} }
+}}
+{{eqn | ll = \leadsto
+      | l = 2 a b + N
+      | o = \equiv
+      | r = a
+      | rr= \pmod {10}
+}}
+{{eqn | ll = \leadsto
+      | l = N
+      | o = \equiv
+      | r = a \paren {1 - 2 b}
+      | rr= \pmod {10}
+}}
+{{end-eqn}}
+By [[Left-Truncated Automorphic Number is Automorphic]], $b$ must end in $0, 1, 5, 6$.
+We see that:
+:$a \equiv N \pmod {10}$ for $b$ ending in $0$ or $5$;
+:$a \equiv -N \pmod {10}$ for $b$ ending in $1$ or $6$;
+So the choice of $x$ is completely determined by $b$.
+This shows that there are exactly $4$ $\paren {k + 1}$-[[Definition:Digit|digit]] [[Definition:Automorphic Number|automorphic numbers]].
+{{qed|lemma}}
+Now we show that the $4$ numbers are indeed of the forms given above.
+For any $n \in \Z_{>0}$, we have:
+{{begin-eqn}}
+{{eqn | l = 0^2 - 0
+      | o = \equiv
+      | r = 0
+      | rr= \pmod {10^n}
+}}
+{{eqn | l = 1^2 - 1
+      | o = \equiv
+      | r = 0
+      | rr= \pmod {10^n}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {5^{2^{n - 1} } }^2 - 5^{2^{n - 1} }
+      | r = 5^{2^{n - 1} } \paren {5^{2^n} - 1}
+}}
+{{eqn | r = 5^{2^{n - 1} } \paren {5 - 1} \paren {5 + 1} \paren {5^2 + 1} \dots \paren {5^{2^{n - 2} } + 1}
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+Since $2^{n - 1} \ge n$ and the each factor on the right is [[Definition:Divisor of Integer|divisible]] by $2$, the above product is [[Definition:Divisor of Integer|divisible]] by $5^n 2^n = 10^n$.
+{{begin-eqn}}
+{{eqn | l = \paren {6^{5^{n - 1} } }^2 - 6^{5^{n - 1} }
+      | r = 6^{5^{n - 1} } \paren {6^{5^{n - 1} } - 1}
+}}
+{{eqn | r = 6^{5^{n - 1} } \paren {6^{5^{n - 2} } - 1} \paren {6^{4 \times 5^{n - 2} } + 6^{3 \times 5^{n - 2} } + 6^{2 \times 5^{n - 2} } + 6^{5^{n - 2} } + 1}
+      | c = [[Difference of Two Odd Powers]]
+}}
+{{eqn | r = 6^{5^{n - 1} } \paren {6^{5^{n - 2} } - 1} \paren {5 N}
+      | c = for some $N \in \N$: [[Congruence of Powers]]
+}}
+{{eqn | o = :
+}}
+{{eqn | r = 2^{5^{n - 1} } 3^{5^{n - 1} } \paren {6^{5^0} - 1} \paren {5^{n - 1} C}
+      | c = for some $C \in \N$
+}}
+{{eqn | r = 2^{5^{n - 1} } 3^{5^{n - 1} } \paren {5^n C}
+}}
+{{end-eqn}}
+Since $5^{n - 1} \ge n$, the above product is [[Definition:Divisor of Integer|divisible]] by $2^n 5^n = 10^n$.
+Hence for each number above:
+:$x^2 \equiv x \pmod {10^n}$
+showing they are indeed [[Definition:Automorphic Number|automorphic]].
+So $\map P k \implies \map P {k + 1}$ and thus it follows by the [[Principle of Mathematical Induction]] that these are all the [[Definition:Automorphic Number|automorphic numbers]].
+{{qed}}
+{{expand|Add a proof via p-adic numbers}}
+[[Category:Automorphic Numbers]]
+jhzrvo7ggjkanrgbcquux60qq8dx0u8
+\end{proof}<|endoftext|>
+\section{Seventeen Horses/General Problem 1}
+Tags: Seventeen Horses, Unit Fractions
+
+\begin{theorem}
+A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $3$ heirs.
+They are to be distributed in the [[Definition:Ratio|ratio]] $\dfrac 1 a : \dfrac 1 b : \dfrac 1 c$.
+Let $\dfrac 1 a + \dfrac 1 b + \dfrac 1 c < 1$.
+Then there are $7$ possible values of $\tuple {n, a, b, c}$ such that the required shares are:
+:$\dfrac {n + 1} a, \dfrac {n + 1} b, \dfrac {n + 1} c$
+These values are:
+:$\tuple {7, 2, 4, 8}, \tuple {11, 2, 4, 6}, \tuple {11, 2, 3, 12}, \tuple {17, 2, 3, 9}, \tuple {19, 2, 4, 5}, \tuple {23, 2, 3, 8}, \tuple {41, 2, 3, 7}$
+leading to shares, respectively, of:
+:$\tuple {4, 2, 1}, \tuple {6, 3, 2}, \tuple {6, 4, 1}, \tuple {9, 6, 2}, \tuple {10, 5, 4}, \tuple {12, 8, 3}, \tuple {21, 14, 6}$
+\end{theorem}
+
+\begin{proof}
+It is taken as a condition that $a \ne b \ne c \ne a$.
+We have that:
+:$\dfrac 1 a + \dfrac 1 b + \dfrac 1 c + \dfrac 1 n = 1$
+and so we need to investigate the solutions to the above equations.
+From [[Sum of 4 Unit Fractions that equals 1]], we have that the only possible solutions are:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4
+      | r = 1
+}}
+{{end-eqn}}
+From these, we can eliminate the following, because it is not the case that $a \ne b \ne c \ne a$:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 5 + \dfrac 1 5 + \dfrac 1 {10}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 6 + \dfrac 1 6 + \dfrac 1 6
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 3 + \dfrac 1 6 + \dfrac 1 {6}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 3 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 6
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4 + \dfrac 1 4
+      | r = 1
+}}
+{{end-eqn}}
+Then we can see by inspection that the following are indeed solutions to the problem:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 7 + \dfrac 1 {42}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 8 + \dfrac 1 {24}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 9 + \dfrac 1 {18}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {12} + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 5 + \dfrac 1 {20}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 6 + \dfrac 1 {12}
+      | r = 1
+}}
+{{eqn | l = \dfrac 1 2 + \dfrac 1 4 + \dfrac 1 8 + \dfrac 1 8
+      | r = 1
+}}
+{{end-eqn}}
+The remaining tuple we have is:
+:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} + \dfrac 1 {15} = 1$
+But we note that:
+:$\dfrac 1 2 + \dfrac 1 3 + \dfrac 1 {10} = \dfrac {28} {30}$
+which is not in the correct form.
+Hence the $7$ possible solutions given.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Seventeen Horses/General Problem 2}
+Tags: Seventeen Horses
+
+\begin{theorem}
+A man dies, leaving $n$ indivisible and indistinguishable objects to be divided among $m$ heirs.
+They are to be distributed in the [[Definition:Ratio|ratio]] $\dfrac 1 {a_1} : \dfrac 1 {a_2} : \cdots : \dfrac 1 {a_m}$.
+Let $t = \dfrac q r = \displaystyle \sum_{k \mathop = 1}^m \dfrac 1 {a_k}$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
+Let $t \ne 1$.
+Then it is possible to achieve the required share by adding $s$ objects to the existing $n$ such that:
+:$s + q = r$
+when $q = n$.
+This still works whether $q$ is [[Definition:Positive Integer|positive]] or [[Definition:Negative Integer|negative]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|More clarity of exposition is required. Might not even be correct}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Products of 3 Consecutive Reciprocals}
+Tags: Sums of Sequences, Reciprocals, Sum of Sequence of Products of 3 Consecutive Reciprocals
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }$
+\end{theorem}
+
+\begin{proof}
+We observe that:
+{{begin-eqn}}
+{{eqn | l = \frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2}
+      | r = \frac {\paren {j + 1} \paren {j + 2} - 2 j \paren {j + 2} + j \paren {j + 1} } {j \paren {j + 1} \paren {j + 2} }
+}}
+{{eqn | r = \frac {j^2 + 3 j + 2 - 2 j^2 - 4 j + j^2 + j} {j \paren {j + 1} \paren {j + 2} }
+}}
+{{eqn | r = \frac 2 {j \paren {j + 1} \paren {j + 2} }
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }
+      | r = \frac 1 2 \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 2 {j + 1} + \frac 1 {j + 2} }
+}}
+{{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 1}^n \frac 2 {j + 1} + \sum_{j \mathop = 1}^n \frac 1 {j + 2} }
+}}
+{{eqn | r = \frac 1 2 \paren {\sum_{j \mathop = 1}^n \frac 1 j - \sum_{j \mathop = 2}^{n + 1} \frac 2 j + \sum_{j \mathop = 3}^{n + 2} \frac 1 j}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 + \sum_{j \mathop = 3}^n \frac 1 j - \frac 2 2 - \frac 2 {n + 1} - \sum_{j \mathop = 3}^n \frac 2 j + \frac 1 {n + 1} + \frac 1 {n + 2} + \sum_{j \mathop = 3}^n \frac 1 j}
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 1 + \frac 1 2 - \frac 2 2 - \frac 2 {n + 1} + \frac 1 {n + 1} + \frac 1 {n + 2} }
+}}
+{{eqn | r = \frac 1 2 \paren {\frac 1 2 - \frac 1 {n + 1} + \frac 1 {n + 2} }
+}}
+{{eqn | r = \frac {\paren {n + 1} \paren {n + 2} - 2 \paren {n + 2} + 2 \paren {n + 1} } {4 \paren {n + 1} \paren {n + 2} }
+}}
+{{eqn | r = \frac {n^2 + 3 n + 2 - 2 n - 4 + 2 n + 2} {4 \paren {n + 1} \paren {n + 2} }
+}}
+{{eqn | r = \frac {n^2 + 3 n} {4 \paren {n + 1} \paren {n + 2} }
+}}
+{{eqn | r = \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Sums of Sequences]]
+[[Category:Reciprocals]]
+[[Category:Sum of Sequence of Products of 3 Consecutive Reciprocals]]
+qkvjhla4ruv1a7xa74143x5lni4gbfy
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Products of 3 Consecutive Reciprocals/Corollary}
+Tags: Limits of Series, Reciprocals, Sum of Sequence of Products of 3 Consecutive Reciprocals
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} } = \frac 1 4$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \paren {j + 1} \paren {j + 2} }
+      | r = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} \paren {j + 2} }
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {n \paren {n + 3} } {4 \paren {n + 1} \paren {n + 2} }
+      | c = [[Sum of Sequence of Products of 3 Consecutive Reciprocals]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {1 + \frac 3 n} {4 \paren {1 + \frac 1 n} \paren {1 + \frac 2 n} }
+      | c = dividing [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $n^2$
+}}
+{{eqn | r = \frac 1 4
+      | c = [[Definition:Basic Null Sequence|Basic Null Sequences]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Limits of Series]]
+[[Category:Reciprocals]]
+[[Category:Sum of Sequence of Products of 3 Consecutive Reciprocals]]
+5rbergex7jbt15a4pgmcgfn3s1ft25m
+\end{proof}<|endoftext|>
+\section{Heronian Triangle whose Altitude and Sides are Consecutive Integers}
+Tags: Heronian Triangles
+
+\begin{theorem}
+There exists [[Definition:Unique|exactly one]] [[Definition:Heronian Triangle|Heronian triangle]] one of whose [[Definition:Altitude of Triangle|altitudes]] and its [[Definition:Side of Polygon|sides]] are all consecutive [[Definition:Integer|integers]].
+This is the [[Definition:Heronian Triangle|Heronian triangle]] whose [[Definition:Side of Polygon|sides]] are $\tuple {13, 14, 15}$ and which has an [[Definition:Altitude of Triangle|altitude]] $12$.
+\end{theorem}
+
+\begin{proof}
+We note that a [[Definition:Heronian Triangle|Heronian triangle]] whose [[Definition:Side of Polygon|sides]] are all consecutive [[Definition:Integer|integers]] is also known as a [[Definition:Fleenor-Heronian Triangle|Fleenor-Heronian triangle]].
+From [[Definition:Fleenor-Heronian Triangle/Sequence|Sequence of Fleenor-Heronian Triangles]], we have that the smallest such [[Definition:Fleenor-Heronian Triangle|triangles]] are as follows:
+:$\tuple {1, 2, 3}$, which has an [[Definition:Altitude of Triangle|altitude]] of $0$
+This is the [[Definition:Degenerate Case|degenerate case]] where the [[Definition:Heronian Triangle|Heronian triangle]] is a [[Definition:Straight Line|straight line]].
+While $0, 1, 2, 3$ is a [[Definition:Finite Sequence|sequence]] of $4$ consecutive [[Definition:Integer|integers]], this is not technically a [[Definition:Triangle (Geometry)|triangle]].
+:$\tuple {3, 4, 5}$ with [[Definition:Area|area]] $6$.
+It has [[Definition:Altitude of Triangle|altitudes]] $3$, $4$ and $\dfrac {12} 5$.
+:$\tuple {13, 14, 15}$
+This can be constructed by placing the $2$ [[Definition:Pythagorean Triangle|Pythagorean triangles]] $\tuple {5, 12, 13}$ and $\tuple {9, 12, 15}$ together along their common side $12$:
+:[[File:Heronian-Triangle-Consecutive-Altitude-Sides.png|500px]]
+Thus the [[Definition:Altitude of Triangle|altitude]] and [[Definition:Side of Polygon|sides]] are:
+:$\tuple {12, 13, 14, 15}$
+and this is the [[Definition:Heronian Triangle|Heronian triangle]] we seek.
+It has [[Definition:Area|area]] $84$.
+The next largest [[Definition:Fleenor-Heronian Triangle|Fleenor-Heronian triangle]] has sides $\tuple {51, 52, 53}$.
+Using [[Heron's Formula]], its [[Definition:Area|area]] is given by:
+:$\AA = \sqrt {78 \times 25 \times 26 \times 27} = 1170$
+Hence its [[Definition:Altitude of Triangle|altitudes]] are:
+:$45 \frac {45} {51}$, $45$, $44 \frac 8 {53}$
+For still larger triangles, the [[Definition:Altitude of Triangle|altitudes]] are never within $1$ unit of the sides:
+Consider the [[Definition:Triangle (Geometry)|triangle]] with sides $\tuple {a - 1, a, a + 1}$.
+Using [[Heron's Formula]], its [[Definition:Area|area]] is given by:
+{{begin-eqn}}
+{{eqn | l = \AA
+      | r = \sqrt {s \paren {s - a + 1} \paren {s - a} \paren {s - a - 1} }
+}}
+{{eqn | r = \sqrt {\frac 3 2 a \paren {\frac 1 2 a + 1} \paren {\frac 1 2 a} \paren {\frac 1 2 a - 1} }
+}}
+{{eqn | r = \frac a 4 \sqrt {3 \paren {a + 2} \paren {a - 2} }
+}}
+{{eqn | r = \frac a 4 \sqrt {3 a^2 - 12}
+}}
+{{end-eqn}}
+Its longest [[Definition:Altitude of Triangle|altitude]] is therefore:
+{{begin-eqn}}
+{{eqn | l = \frac {2 a} {4 \paren {a - 1} } \sqrt {3 a^2 - 12}
+      | o = <
+      | r = \frac {a^2 \sqrt 3} {2 \paren {a - 1} }
+}}
+{{end-eqn}}
+and we have:
+{{begin-eqn}}
+{{eqn | l = \frac {a^2 \sqrt 3} {2 \paren {a - 1} }
+      | o = <
+      | r = \paren {a - 1} - 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a^2 \sqrt 3
+      | o = <
+      | r = 2 \paren {a - 1}^2 - 2 \paren {a - 1}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 2 a^2 - 4 a + 2 - 2 a + 2 - \sqrt 3 a^2
+      | o = >
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {2 - \sqrt 3} a^2 - 6 a + 4
+      | o = >
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = >
+      | r = \frac {6 + \sqrt {6^2 - 4 \times 4 \paren {2 - \sqrt 3} } } {2 \paren {2 - \sqrt 3} }
+      | c = [[Quadratic Formula]]
+}}
+{{eqn | o = \approx
+      | r = 21.7
+}}
+{{end-eqn}}
+This shows that for $a \ge 22$, all [[Definition:Altitude of Triangle|altitudes]] of the [[Definition:Triangle (Geometry)|triangle]] is less than $a - 2$.
+Hence there are no more examples.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Heronian Triangle can be Scaled so Area equals Perimeter}
+Tags: Heronian Triangles
+
+\begin{theorem}
+Let $T_1$ be an [[Definition:Integer Heronian Triangle|integer Heronian triangle]] whose [[Definition:Side of Polygon|sides]] are $a$, $b$ and $c$.
+Then there exists a [[Definition:Rational Number|rational number]] $k$ such that the [[Definition:Heronian Triangle|Heronian triangle]] $T_2$ whose [[Definition:Side of Polygon|sides]] are $k a$, $k b$ and $k c$ such that the [[Definition:Perimeter|perimeter]] of $T$ is equal to the [[Definition:Area|area]] of $T$.
+\end{theorem}
+
+\begin{proof}
+For a given [[Definition:Triangle (Geometry)|triangle]] $T$:
+:let $\map \AA T$ denote the [[Definition:Area|area]] of $T$
+:let $\map P T$ denote the [[Definition:Perimeter|perimeter]] of $T$.
+We are given that $T_1$ is an [[Definition:Integer Heronian Triangle|integer Heronian triangle]] whose [[Definition:Side of Polygon|sides]] are $a$, $b$ and $c$.
+Let $\map P {T_1} = k \map \AA {T_1}$.
+Let $T_2$ have [[Definition:Side of Polygon|sides]] $k a$, $k b$ and $k c$.
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = \map P {T_2}
+      | r = k \map P {T_1}
+      | c = 
+}}
+{{eqn | l = \map A {T_2}
+      | r = k^2 \map A {T_1}
+      | c = 
+}}
+{{eqn | r = k \map P {T_1}
+      | c = 
+}}
+{{eqn | r = \map P {T_2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Heronian Triangles]]
+hfgq1h2x4tu21n5ln7trqjwr8qpy72w
+\end{proof}<|endoftext|>
+\section{3 Proper Integer Heronian Triangles whose Area and Perimeter are Equal}
+Tags: Heronian Triangles
+
+\begin{theorem}
+There are exactly $3$ [[Definition:Proper Heronian Triangle|proper]] [[Definition:Integer Heronian Triangle|integer Heronian triangles]] whose [[Definition:Area|area]] and [[Definition:Perimeter|perimeter]] are equal.
+These are the [[Definition:Triangle (Geometry)|triangles]] whose [[Definition:Side of Polygon|sides]] are:
+:$\tuple {6, 25, 29}$
+:$\tuple {7, 15, 20}$
+:$\tuple {9, 10, 17}$
+\end{theorem}
+
+\begin{proof}
+First, using [[Pythagoras's Theorem]], we establish that these [[Definition:Integer Heronian Triangle|integer Heronian triangles]] are indeed [[Definition:Proper Heronian Triangle|proper]]:
+{{begin-eqn}}
+{{eqn | l = 6^2 + 25^2
+      | r = 661
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = 29^2
+      | c = so not [[Definition:Right Triangle|right-angled]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 7^2 + 15^2
+      | r = 274
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = 20^2
+      | c = so not [[Definition:Right Triangle|right-angled]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 9^2 + 10^2
+      | r = 181
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = 17^2
+      | c = so not [[Definition:Right Triangle|right-angled]]
+}}
+{{end-eqn}}
+Now we show they have [[Definition:Area|area]] equal to [[Definition:Perimeter|perimeter]].
+We use [[Heron's Formula]] throughout:
+:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+where:
+:$\AA$ denotes the [[Definition:Area|area]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$a$, $b$ and $c$ denote the [[Definition:Length of Line|lengths]] of the [[Definition:Side of Polygon|sides]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$s = \dfrac {a + b + c} 2$ denotes the [[Definition:Semiperimeter|semiperimeter]] of the [[Definition:Triangle (Geometry)|triangle]].
+Thus we take the $3$ [[Definition:Triangle (Geometry)|triangles]] in turn:
+{{begin-eqn}}
+{{eqn | n = 6, 25, 29
+      | l = s
+      | r = \frac {6 + 25 + 29} 2
+      | c = 
+}}
+{{eqn | r = 30
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+      | r = \sqrt {30 \paren {30 - 6} \paren {30 - 25} \paren {30 - 29} }
+      | c = 
+}}
+{{eqn | r = \sqrt {30 \times 24 \times 5 \times 1}
+      | c = 
+}}
+{{eqn | r = \sqrt {3600}
+      | c = 
+}}
+{{eqn | r = 60
+      | c = 
+}}
+{{eqn | r = 6 + 25 + 29
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | n = 7, 15, 20
+      | l = s
+      | r = \frac {7 + 15 + 20} 2
+      | c = 
+}}
+{{eqn | r = 21
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+      | r = \sqrt {21 \paren {21 - 7} \paren {21 - 15} \paren {21 - 20} }
+      | c = 
+}}
+{{eqn | r = \sqrt {21 \times 14 \times 6 \times 1}
+      | c = 
+}}
+{{eqn | r = \sqrt {1764}
+      | c = 
+}}
+{{eqn | r = 42
+      | c = 
+}}
+{{eqn | r = 7 + 15 + 20
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | n = 9, 10, 17
+      | l = s
+      | r = \frac {9 + 10 + 17} 2
+      | c = 
+}}
+{{eqn | r = 18
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+      | r = \sqrt {18 \paren {18 - 9} \paren {18 - 10} \paren {18 - 17} }
+      | c = 
+}}
+{{eqn | r = \sqrt {18 \times 9 \times 8 \times 1}
+      | c = 
+}}
+{{eqn | r = \sqrt {1296}
+      | c = 
+}}
+{{eqn | r = 36
+      | c = 
+}}
+{{eqn | r = 9 + 10 + 17
+      | c = 
+}}
+{{end-eqn}}
+It remains to be demonstrated that these are indeed the only such [[Definition:Proper Heronian Triangle|proper]] [[Definition:Integer Heronian Triangle|integer Heronian triangles]] which match the criterion.
+Let $\tuple {a, b, c}$ be the sides of such a triangle.
+Using [[Heron's Formula]], we have:
+{{begin-eqn}}
+{{eqn | l = 2 s
+      | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+}}
+{{eqn | ll= \leadsto 
+      | l = 4 s^2
+      | r = s \paren {s - a} \paren {s - b} \paren {s - c}
+}}
+{{eqn | ll= \leadsto
+      | l = 4 s
+      | r = \paren {s - a} \paren {s - b} \paren {s - c}
+}}
+{{end-eqn}}
+Note that:
+:$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
+Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
+:$4 \paren {x + y + z} = x y z$
+By [[Semiperimeter of Integer Heronian Triangle is Composite]], $s$ is an [[Definition:Integer|integer]].
+Hence $s, x, y, z \in \N_{>0}$.
+By [[Triple with Product Quadruple the Sum]], our equation has solutions:
+:$\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$
+Using:
+:$a = s - x = x + y + z - x = y + z$
+:$b = s - y = x + z$
+:$c = s - z = x + y$
+the possible sets of side lengths are:
+:$\tuple {29, 25, 6}, \tuple {20, 15, 7}, \tuple {17, 10, 9}, \tuple {13, 12, 5}, \tuple {10, 8, 6}$
+of which the final $2$ are [[Definition:Pythagorean Triple|Pythagorean Triples]], so they are not [[Definition:Proper Heronian Triangle|proper Heronian triangles]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Heronian Triangle is Similar to Integer Heronian Triangle}
+Tags: Heronian Triangles
+
+\begin{theorem}
+Let $\triangle {ABC}$ be a [[Definition:Heronian Triangle|Heronian triangle]].
+Then there exists an [[Definition:Integer Heronian Triangle|integer Heronian triangle]] $\triangle {A'B'C'}$ such that $\triangle {ABC}$ and $\triangle {A'B'C'}$ are [[Definition:Similar Triangles|similar]].
+\end{theorem}
+
+\begin{proof}
+Let $\triangle {ABC}$ have [[Definition:Side of Polygon|sides]] whose [[Definition:Length of Line|lengths]] are $a$, $b$ and $c$.
+By definition of [[Definition:Heronian Triangle|Heronian triangle]], each of $a$, $b$ and $c$ are [[Definition:Rational Number|rational]].
+By definition of [[Definition:Rational Number|rational number]], we can express:
+:$a = \dfrac {p_a} {q_a}$, $b = \dfrac {p_b} {q_b}$ and $c = \dfrac {p_c} {q_c}$
+where each of $p_a, q_a, p_b, q_b, p_c, q_c$ are [[Definition:Integer|integers]].
+Now let:
+{{begin-eqn}}
+{{eqn | l = a'
+      | r = a q_a q_b q_c
+      | c = 
+}}
+{{eqn | l = b'
+      | r = b q_a q_b q_c
+      | c = 
+}}
+{{eqn | l = c'
+      | r = c q_a q_b q_c
+      | c = 
+}}
+{{end-eqn}}
+Let $\triangle {A'B'C'}$ be the [[Definition:Triangle (Geometry)|triangle]] whose [[Definition:Side of Polygon|sides]] have [[Definition:Length of Line|lengths]] $a'$, $b'$ and $c'$.
+By definition, $\triangle {ABC}$ and $\triangle {A'B'C'}$ are [[Definition:Similar Triangles|similar]].
+Each of $a'$, $b'$ and $c'$ are [[Definition:Integer|integers]].
+Consider the [[Definition:Area|area]] of [[Definition:Triangle (Geometry)|triangle]] $\triangle {A'B'C'}$
+Let the [[Definition:Area|area]] of $\triangle {ABC}$ be $A$.
+Then the [[Definition:Area|area]] $\triangle {A'B'C'}$ is $q_a q_b q_c A$, which is [[Definition:Rational Number|rational]].
+Hence $\triangle {A'B'C'}$ is an [[Definition:Integer Heronian Triangle|integer Heronian triangle]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Triple with Sum and Product Equal}
+Tags: 6
+
+\begin{theorem}
+For $a, b, c \in \Z$, $a \le b \le c$, the solutions to the equation:
+:$a + b + c = a b c$
+are:
+:$\tuple {1, 2, 3}$
+:$\tuple {-3, -2, -1}$
+and the trivial solution set:
+:$\set {\tuple {-z, 0, z}: z \in \N}$
+\end{theorem}
+
+\begin{proof}
+Suppose one of $a, b, c$ is [[Definition:Zero|zero]].
+Then $a b c = 0 = a + b + c$.
+The remaining two numbers [[Definition:Integer Addition|sum]] to $0$, giving the solution set:
+:$\set {\tuple {-z, 0, z}: z \in \N}$
+{{qed|lemma}}
+Suppose $a < 0$ and $0 < b \le c$.
+Then $a b c \le a < a + b + c$.
+Hence equality never happens.
+Similarly, for $a \le b < 0$ and $c > 0$:
+:$a b c \ge c > a + b + c$
+Hence equality never happens.
+{{qed|lemma}}
+Now it remains the case $0 < a \le b \le c$.
+Suppose $a \ge 2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a b c
+      | o = \ge
+      | r = 4 c
+}}
+{{eqn | o = \ge
+      | r = a + b + c + c
+}}
+{{eqn | o = >
+      | r = a + b + c
+}}
+{{end-eqn}}
+hence $a = 1$.
+Suppose $b \ge 3$.
+Then:
+{{begin-eqn}}
+{{eqn | l = b c
+      | o = \ge
+      | r = c + c + c
+}}
+{{eqn | o = \ge
+      | r = b + c + c
+}}
+{{eqn | o = >
+      | r = b + c
+}}
+{{end-eqn}}
+hence $b = 1$ or $b = 2$.
+For $b = 1$:
+:$c = 1 + 1 + c$
+which is a [[Definition:Contradiction|contradiction]].
+For $b = 2$:
+:$2 c = 1 + 2 + c$
+giving $c = 3$.
+Therefore $\tuple {1, 2, 3}$ is the only solution for [[Definition:Strictly Positive Integer|strictly positive integer]] values of $a, b, c$.
+Similarly, for $a \le b \le c < 0$, we have $0 < -c \le -b \le -a$.
+Also:
+{{begin-eqn}}
+{{eqn | l = \paren {-c} + \paren {-b} + \paren {-a}
+      | r = -\paren {a + b + c}
+}}
+{{eqn | r = -a b c
+}}
+{{eqn | r = \paren {-c} \paren {-b} \paren {-a}
+}}
+{{end-eqn}}
+and we have already show that $-c = 1, -b = 2, -a = 3$.
+Therefore we have $\tuple {a, b, c} = \tuple {-3, -2, -1}$.
+{{qed|lemma}}
+We have considered all possible signs of $a, b, c$.
+Hence the result.
+{{qed}}
+[[Category:6]]
+1qfmk0u0l45yr8r2onnev4hi91jlyuf
+\end{proof}<|endoftext|>
+\section{Triple with Product Quadruple the Sum}
+Tags: 
+
+\begin{theorem}
+Let $a, b, c \in \N$ such that $a \le b \le c$.
+Then the solutions to:
+:$a b c = 4 \paren {a + b + c}$
+are:
+:$\tuple {0, 0, 0}, \tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}, \tuple {2, 3, 10}, \tuple {2, 4, 6}$
+\end{theorem}
+
+\begin{proof}
+Suppose $a \ge 4$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a b c
+      | o = \ge
+      | r = 16 c
+      | c = as $4 \le a \le b$
+}}
+{{eqn | o = \ge
+      | r = 4 \paren {a + b + c + c}
+      | c = as $a \le b \le c$
+}}
+{{eqn | o = >
+      | r = 4 \paren {a + b + c}
+      | c = as $c > 0$
+}}
+{{end-eqn}}
+hence $0 \le a \le 3$.
+For $a = 0$, we have $4 \paren {b + c} = 0$.
+This forces $b = c = 0$, giving the trivial solution:
+:$\tuple {0, 0, 0}$
+{{qed|lemma}}
+For $a \ge 1$, note that:
+{{begin-eqn}}
+{{eqn | l = 4 \paren {a + b + c}
+      | r = a b c
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 4 \paren {a + b}
+      | r = \paren {a b - 4} c
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = c
+      | r = \frac {4 \paren {a + b} } {a b - 4}
+}}
+{{end-eqn}}
+In the second step, {{LHS}} is [[Definition:Strictly Positive Integer|strictly positive]].
+Hence $a b > 4$ and the final step is justified.
+For $a = 1$, we have $b c = 4 \paren {1 + b + c}$.
+Since $a b > 4$, $b > 4$.
+Suppose $b \ge 9$.
+Then:
+{{begin-eqn}}
+{{eqn | l = b c
+      | o = \ge
+      | r = 9 c
+}}
+{{eqn | o = \ge
+      | r = 4 \paren {b + c} + c
+      | c = as $b \le c$
+}}
+{{eqn | o = >
+      | r = 4 \paren {1 + b + c}
+      | c = as $c \ge b > 4$
+}}
+{{end-eqn}}
+hence $b \le 8$.
+We find the value of $c$ by substituting $a$ and $b$:
+:$b = 5: c = \dfrac {4 \paren {1 + 5} } {5 - 4} = 24$
+:$b = 6: c = \dfrac {4 \paren {1 + 6} } {6 - 4} = 14$
+:$b = 7: c = \dfrac {4 \paren {1 + 7} } {7 - 4} = \dfrac {32} 3$
+:$b = 8: c = \dfrac {4 \paren {1 + 8} } {8 - 4} = 9$
+and we see that $\tuple {1, 5, 24}, \tuple {1, 6, 14}, \tuple {1, 8, 9}$ are valid solutions.
+{{qed|lemma}}
+For $a = 2$, we have $2 b c = 4 \paren {2 + b + c}$.
+Since $a b > 4$, $b > 2$.
+Suppose $b \ge 5$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 2 b c
+      | o = \ge
+      | r = 10 c
+}}
+{{eqn | o = \ge
+      | r = 4 \paren {b + c} + 2 c
+      | c = as $b \le c$
+}}
+{{eqn | o = >
+      | r = 4 \paren {2 + b + c}
+      | c = as $c \ge b > 4$
+}}
+{{end-eqn}}
+hence $b \le 4$.
+We find the value of $c$ by substituting $a$ and $b$:
+:$b = 3: c = \dfrac {4 \paren {2 + 3} } {2 \times 3 - 4} = 10$
+:$b = 4: c = \dfrac {4 \paren {2 + 4} } {2 \times 4 - 4} = 6$
+and we see that $\tuple {2, 3, 10}, \tuple {2, 4, 6}$ are valid solutions.
+{{qed|lemma}}
+For $a = 3$, we have $3 b c = 4 \paren {3 + b + c}$.
+Suppose $b \ge 4$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 3 b c
+      | o = \ge
+      | r = 12 c
+}}
+{{eqn | o = \ge
+      | r = 4 \paren {b + c} + 4 c
+      | c = as $b \le c$
+}}
+{{eqn | o = >
+      | r = 4 \paren {3 + b + c}
+      | c = as $c \ge b > 1$
+}}
+{{end-eqn}}
+hence $b \le 3$.
+Since $3 = a \le b$, this forces $b = 3$
+We find the value of $c$ by substituting $a$ and $b$:
+:$c = \dfrac {4 \paren {3 + 3} } {3 \times 3 - 4} = \dfrac {24} 5$
+and we see that it is not a valid solution.
+{{qed|lemma}}
+We have considered all possible values of $a$.
+Hence the result.
+{{qed}}
+mgja9o3yt8d0kuze46853enzwrydbf2
+\end{proof}<|endoftext|>
+\section{Area of Integer Heronian Triangle is Multiple of 6}
+Tags: Heronian Triangles
+
+\begin{theorem}
+Let $\triangle {ABC}$ be an [[Definition:Integer Heronian Triangle|integer Heronian triangle]].
+Then the [[Definition:Area|area]] of $\triangle {ABC}$ is a [[Definition:Integer Multiple|multiple]] of $6$.
+\end{theorem}
+
+\begin{proof}
+[[Heron's Formula]] gives us that:
+:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+where:
+:$\AA$ denotes the [[Definition:Area|area]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$a$, $b$ and $c$ denote the [[Definition:Length of Line|lengths]] of the [[Definition:Side of Polygon|sides]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$s = \dfrac {a + b + c} 2$ denotes the [[Definition:Semiperimeter|semiperimeter]] of the [[Definition:Triangle (Geometry)|triangle]].
+We set out to eliminate $s$ and simplify as best possible:
+{{begin-eqn}}
+{{eqn | l = \AA
+      | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+      | c = 
+}}
+{{eqn | l = \AA^2
+      | r = \dfrac {a + b + c} 2 \paren {\dfrac {a + b + c} 2 - a} \paren {\dfrac {a + b + c} 2 - b} \paren {\dfrac {a + b + c} 2 - c}
+      | c = substituting for $s$ and [[Definition:Square Function|squaring]]
+}}
+{{eqn | ll= \leadsto
+      | l = 16 \AA^2
+      | r = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}
+      | c = multiplying through by $16$ and simplifying
+}}
+{{eqn | r = 2 a^2 b^2 + 2 b^2 c^2 + 2 c^2 a^2 - a^4 - b^4 - c^4
+      | c = multiplying out and simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {4 \AA}^2 + \paren {b^2 + c^2 - a^2}
+      | r = \paren {2 b c}^2
+      | c = factorising
+}}
+{{end-eqn}}
+This is now in the form $p^2 + q^2 = r^2$.
+From [[Solutions of Pythagorean Equation]], $\tuple {p, q, r}$ has the parametric solution:
+:$\tuple {m^2 - n^2, 2 m n, m^2 + n^2}$
+There are two steps to showing $6 \divides \AA$:
+=== Step $1$: $2 \divides \AA$ ===
+By [[Euclid's Lemma for Prime Divisors]], we just need to show:
+:$2 \divides \AA^2 = s \paren {s - a} \paren {s - b} \paren {s - b}$
+By [[Semiperimeter of Integer Heronian Triangle is Composite]], $s \in \N$.
+There are $3$ cases:
+==== Case $1$: There are sides with odd and even lengths ====
+Under this condition, one of $s - x$ will be [[Definition:Even Integer|even]].
+Then $\AA^2$ is also [[Definition:Even Integer|even]].
+{{qed|lemma}}
+==== Case $2$: All sides are of odd length ====
+Suppose all [[Definition:Side of Polygon|sides]] are of [[Definition:Odd Integer|odd]] [[Definition:Length of Line|length]].
+Then the [[Definition:Perimeter|perimeter]] is also [[Definition:Odd Integer|odd]].
+But then the [[Definition:Semiperimeter|semiperimeter]] cannot be an [[Definition:Integer|integer]].
+This is a [[Definition:Contradiction|contradiction]].
+{{qed|lemma}}
+==== Case $3$: All sides are of even length ====
+Suppose all [[Definition:Side of Polygon|sides]] are of [[Definition:Even Integer|even]] [[Definition:Length of Line|length]].
+If the [[Definition:Semiperimeter|semiperimeter]] is [[Definition:Even Integer|even]], the result follows.
+Therefore we consider the case where $s$ is [[Definition:Odd Integer|odd]].
+Then $x := s - a$, $y := s - b$ and $z := s - c$ are also [[Definition:Odd Integer|odd]].
+Note that $x + y + z = 3 s - a - b - c = s$.
+Hence $\AA^2 = x y z \paren {x + y + z}$.
+Note that each of $x, y, z$ must be equivalent to $\pm 1 \pmod 4$.
+If all of $x, y, z \equiv 1 \pmod 4$:
+:$x + y + z \equiv -1 \pmod 4$
+If two of $x, y, z \equiv 1 \pmod 4$:
+:$x + y + z \equiv 1 \pmod 4$
+If one of $x, y, z \equiv 1 \pmod 4$:
+:$x + y + z \equiv -1 \pmod 4$
+If none of $x, y, z \equiv 1 \pmod 4$:
+:$x + y + z \equiv 1 \pmod 4$
+In any of the above cases, we have:
+:$\AA^2 \equiv -1 \pmod 4$
+which is impossible by [[Square Modulo 4]].
+This is a [[Definition:Contradiction|contradiction]].
+{{qed|lemma}}
+In each valid case, we see that $2 \divides \AA^2$.
+{{qed|lemma}}
+=== Step $2$: $3 \divides \AA$ ===
+By [[Euclid's Lemma for Prime Divisors]], we just need to show:
+:$3 \divides 16 \AA^2 = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}$
+We split the problem into $4$ cases:
+==== Case $1$: None of $a, b, c$ are divisible by $3$ ====
+If $a \equiv b \equiv c \pmod 3$:
+:$a + b + c \equiv 3 a \equiv 0 \pmod 3$
+Hence:
+:$3 \divides \paren {a + b + c}$
+If two of the lengths have the same remainder when divided by $3$, say $a \equiv b \not \equiv c \pmod 3$:
+:$a \equiv b \equiv -c \pmod 3$
+Hence:
+:$a + b - c \equiv 3 a \equiv 0 \pmod 3$
+Thus:
+:$3 \divides \paren {a + b - c}$
+{{qed|lemma}}
+==== Case $2$: One of $a, b, c$ is divisible by $3$ ====
+{{WLOG}} suppose that number is $a$.
+If $b \equiv c \pmod 3$:
+:$a + b - c \equiv a \equiv 0 \pmod 3$
+Hence
+:$3 \divides \paren {a + b - c}$
+If $b \not \equiv c \pmod 3$:
+:$b \equiv -c \pmod 3$
+Hence:
+:$a + b + c \equiv -c + c \equiv 0 \pmod 3$
+Thus:
+:$3 \divides \paren {a + b + c}$
+{{qed|lemma}}
+==== Case $3$: Two of $a, b, c$ is divisible by $3$ ====
+{{WLOG}}, suppose $3 \divides a, b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 16 \AA^2
+      | r = \paren {a + b + c} \paren {-a + b + c} \paren {a - b + c} \paren {a + b - c}
+}}
+{{eqn | o = \equiv
+      | r = c \cdot c \cdot c \cdot \paren {-c}
+      | rr = \pmod 3
+}}
+{{eqn | o = \equiv
+      | r = -c^4
+      | rr = \pmod 3
+}}
+{{eqn | o = \equiv
+      | r = -1
+      | rr = \pmod 3
+      | c = [[Square Modulo 3]]
+}}
+{{end-eqn}}
+By [[Square Modulo 3]], $16 \AA^2 = \paren {4 \AA}^2$ is not a [[Definition:Square Number|square]].
+This is a [[Definition:Contradiction|contradiction]].
+Therefore this case is not valid.
+{{qed|lemma}}
+==== Case $3$:  $a, b, c$ are all divisible by $3$ ====
+We have:
+:$3 \divides \paren {a + b + c}$
+{{qed|lemma}}
+We see that in every valid case, $3 \divides 16 \AA^2$.
+Hence $3 \divides \AA$, and thus $6 \divides \AA$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Proper Integer Heronian Triangle whose Area is 24}
+Tags: Heronian Triangles
+
+\begin{theorem}
+There exists [[Definition:Unique|exactly one]] [[Definition:Proper Heronian Triangle|proper]] [[Definition:Integer Heronian Triangle|integer Heronian triangle]] whose [[Definition:Area|area]] equals $24$.
+That is, the [[Definition:Obtuse Triangle|obtuse triangle]] whose [[Definition:Side of Polygon|sides]] are of [[Definition:Length of Line|length]] $4$, $13$ and $15$.
+\end{theorem}
+
+\begin{proof}
+First we show that the $\tuple {4, 13, 15}$ [[Definition:Triangle (Geometry)|triangle]] is actually [[Definition:Heronian Triangle|Heronian]].
+[[Heron's Formula]] gives us that:
+:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$
+where:
+:$\AA$ denotes the [[Definition:Area|area]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$a$, $b$ and $c$ denote the [[Definition:Length of Line|lengths]] of the [[Definition:Side of Polygon|sides]] of the [[Definition:Triangle (Geometry)|triangle]]
+:$s = \dfrac {a + b + c} 2$ denotes the [[Definition:Semiperimeter|semiperimeter]] of the [[Definition:Triangle (Geometry)|triangle]].
+Hence:
+{{begin-eqn}}
+{{eqn | l = s
+      | r = \frac {4 + 13 + 15} 2
+      | c = 
+}}
+{{eqn | r = 16
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+      | r = \sqrt {16 \paren {16 - 4} \paren {16 - 13} \paren {16 - 15} }
+      | c = 
+}}
+{{eqn | r = \sqrt {16 \times 12 \times 3 \times 1}
+      | c = 
+}}
+{{eqn | r = \sqrt {576}
+      | c = 
+}}
+{{eqn | r = 24
+      | c = 
+}}
+{{end-eqn}}
+It can be constructed by taking a $9-12-15$ [[Definition:Pythagorean Triangle|Pythagorean triangle]] and removing a $5-12-13$ [[Definition:Pythagorean Triangle|Pythagorean triangle]] from it:
+:[[File:Heronian-Triangle-Area-24.png|500px]]
+Let $\tuple {a, b, c}$ be the sides of such a triangle.
+Using [[Heron's Formula]], we have:
+{{begin-eqn}}
+{{eqn | l = 24
+      | r = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }
+}}
+{{eqn | ll= \leadsto 
+      | l = 576
+      | r = s \paren {s - a} \paren {s - b} \paren {s - c}
+}}
+{{end-eqn}}
+Note that:
+:$\paren {s - a} + \paren {s - b} + \paren {s - c} = 3 s - a - b - c = s$
+Hence by substituting $x = s - a$, $y = s - b$, $z = s - c$:
+:$x y z \paren {x + y + z} = 576$
+By [[Semiperimeter of Integer Heronian Triangle is Composite]], $s$ is an [[Definition:Integer|integer]].
+Hence $s, x, y, z \in \N_{>0}$.
+{{WLOG}} suppose $x \le y \le z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 576
+      | r = x y z \paren {x + y + z}
+}}
+{{eqn | o = \ge
+      | r = x^3 \paren {3 x}
+}}
+{{eqn | r = 3 x^4
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \le
+      | r = \sqrt [4] {\frac {576} 3}
+}}
+{{eqn | o = \approx
+      | r = 3.72
+}}
+{{end-eqn}}
+so we need to check $1 \le x \le 3$.
+Note that for fixed $x$:
+{{begin-eqn}}
+{{eqn | l = 576
+      | r = x y z \paren {x + y + z}
+}}
+{{eqn | o = >
+      | r = x y^2 \paren {2 y}
+}}
+{{eqn | r = 2 x y^3
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \le
+      | r = \sqrt [3] {\frac {576} {2 x} }
+}}
+{{end-eqn}}
+For $x = 1$:
+:$1 \le y \le \sqrt [3] {\frac {576} 2} \approx 6.60$
+For $x = 2$:
+:$2 \le y \le \sqrt [3] {\frac {576} 4} \approx 5.24$
+For $x = 3$:
+:$3 \le y \le \sqrt [3] {\frac {576} 6} \approx 4.58$
+Finally, for fixed $x$, $y$:
+{{begin-eqn}}
+{{eqn | l = x y z \paren {x + y + z}
+      | r = 576
+}}
+{{eqn | ll= \leadsto
+      | l = z^2 + \paren {x + y} z - \frac {576} {x y}
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = \frac {- x - y + \sqrt {\paren {x + y}^2 + \frac {2304} {x y} } } 2
+      | c = [[Quadratic Formula]]
+}}
+{{end-eqn}}
+Since $z \in \Z$:
+:$\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} } = 2 z + x + y \in \Z$
+We check the value of $\sqrt {\paren {x + y}^2 + \dfrac {2304} {x y} }$ for the valid values of $x, y$.
+We do not need to check $y = 5$ since $5 \nmid 576$.
+We have:
+:$\tuple {1, 1}: \sqrt {\paren {1 + 1}^2 + \dfrac {2304} {1 \times 1} } = \sqrt {2308} \notin \Z$
+:$\tuple {1, 2}: \sqrt {\paren {1 + 2}^2 + \dfrac {2304} {1 \times 2} } = \sqrt {1161} \notin \Z$
+:$\tuple {1, 3}: \sqrt {\paren {1 + 3}^2 + \dfrac {2304} {1 \times 3} } = \sqrt {784} = 28$
+:$\tuple {1, 4}: \sqrt {\paren {1 + 4}^2 + \dfrac {2304} {1 \times 4} } = \sqrt {601} \notin \Z$
+:$\tuple {1, 6}: \sqrt {\paren {1 + 6}^2 + \dfrac {2304} {1 \times 6} } = \sqrt {433} \notin \Z$
+:$\tuple {2, 2}: \sqrt {\paren {2 + 2}^2 + \dfrac {2304} {2 \times 2} } = \sqrt {592} \notin \Z$
+:$\tuple {2, 3}: \sqrt {\paren {2 + 3}^2 + \dfrac {2304} {2 \times 3} } = \sqrt {409} \notin \Z$
+:$\tuple {2, 4}: \sqrt {\paren {2 + 4}^2 + \dfrac {2304} {2 \times 4} } = \sqrt {324} = 18$
+:$\tuple {3, 3}: \sqrt {\paren {3 + 3}^2 + \dfrac {2304} {3 \times 3} } = \sqrt {292} \notin \Z$
+:$\tuple {3, 4}: \sqrt {\paren {3 + 4}^2 + \dfrac {2304} {3 \times 4} } = \sqrt {241} \notin \Z$
+Now we calculate $z$ for $\tuple {x, y} = \tuple {1, 3}, \tuple {2, 4}$.
+:$\tuple {1, 3}: z = \dfrac {- 1 - 3 + 28} 2 = 12$
+:$\tuple {2, 4}: z = \dfrac {- 2 - 4 + 18} 2 = 6$
+Using:
+:$a = s - x = x + y + z - x = y + z$
+:$b = s - y = x + z$
+:$c = s - z = x + y$
+the possible sets of side lengths are:
+:$\tuple {15, 13, 4}, \tuple {10, 8, 6}$
+But note that the $\tuple {6, 8, 10}$ triangle is [[Definition:Right Triangle|right-angled]], and hence not a [[Definition:Proper Heronian Triangle|proper Heronian Triangle]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Semiperimeter of Integer Heronian Triangle is Composite}
+Tags: Heronian Triangles
+
+\begin{theorem}
+The [[Definition:Semiperimeter|semiperimeter]] of an [[Definition:Integer Heronian Triangle|integer Heronian triangle]] is always a [[Definition:Composite Number|composite number]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c$ be the [[Definition:Side of Polygon|side]] [[Definition:Length of Line|lengths]] of an [[Definition:Integer Heronian Triangle|integer Heronian triangle]].
+By [[Heron's Formula]], its [[Definition:Area|area]] is given by:
+:$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} } \in \N$
+where the [[Definition:Semiperimeter|semiperimeter]] $s$ is given by:
+:$s = \dfrac {a + b + c} 2$
+First we prove that $s$ is indeed an [[Definition:Integer|integer]].
+{{AimForCont}} not.
+Since $2 s = a + b + c \in \N$, $2 s$ must be [[Definition:Odd Integer|odd]].
+Hence $2 s - 2 a, 2 s - 2 b, 2 s - 2 c$ are [[Definition:Odd Integer|odd]] as well.
+Thus:
+{{begin-eqn}}
+{{eqn | l = 16 \AA^2
+      | r = 16 s \paren {s - a} \paren {s - b} \paren {s - c}
+}}
+{{eqn | r = 2 s \paren {2 s - 2 a} \paren {2 s - 2 b} \paren {2 s - 2 c}
+}}
+{{end-eqn}}
+Since $16 \AA^2$ is a product of [[Definition:Odd Integer|odd numbers]], it must be [[Definition:Odd Integer|odd]].
+But then $\AA^2$ is not an [[Definition:Integer|integer]], a [[Definition:Contradiction|contradiction]].
+Therefore $s \in \N$.
+{{qed|lemma}}
+Now we show that $s$ is [[Definition:Composite Number|composite number]].
+{{AimForCont}} not.
+Then $s$ is either $1$ or [[Definition:Prime Number|prime]].
+Since $a, b, c \ge 1$, $s \ge \dfrac 3 2 > 1$.
+Hence $s$ is [[Definition:Prime Number|prime]].
+Since:
+:$\AA^2 = s \paren {s - a} \paren {s - b} \paren {s - c}$
+We have $s \divides \AA^2$.
+By [[Prime Divides Power]], $s^2 \divides \AA^2$.
+Thus $s \divides \paren {s - a} \paren {s - b} \paren {s - c}$.
+By [[Euclid's Lemma for Prime Divisors/General Result|Euclid's Lemma]], $s$ divides some $s - x$.
+However by [[Absolute Value of Integer is not less than Divisors]]:
+:$s \le s - x$
+which is a [[Definition:Contradiction|contradiction]].
+Therefore $s$ is [[Definition:Composite Number|composite]].
+{{qed}}
+[[Category:Heronian Triangles]]
+9rjbrxvgau2ac4m16h5h5g1n9j9h3ym
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Necessary Condition}
+Tags: Matroid Satisfies Base Axiom
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\mathscr B$ be the set of [[Definition:Base of Matroid|bases]] of the [[Definition:Matroid|matroid]] on $M$.
+Then $\mathscr B$ satisfies the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}}
+\end{theorem}
+
+\begin{proof}
+Let $B_1, B_2 \in \mathscr B$.
+Let $x \in B_1 \setminus B_2$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \size {B_1 \setminus \set x}
+      | r = \size {B_1} - \size {\set x}
+      | c = [[Cardinality of Set Difference with Subset]]
+}}
+{{eqn | r = \size {B_2} - \size {\set x}
+      | c = [[All Bases of Matroid have same Cardinality]]
+}}
+{{eqn | r = \size {B_2} - 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | o = <
+      | r = \size {B_2}
+}}
+{{end-eqn}}
+By [[Definition:Matroid Axioms|matroid axiom $(\text I 3)$]]:
+:$\exists y \in B_2 \setminus \paren{B_1 \setminus \set x} : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$
+We have:
+{{begin-eqn}}
+{{eqn | l = B_2 \setminus \paren{B_1 \setminus \set x}
+      | r = \paren{B_2 \setminus B_1} \cup \paren{B_2 \cap \set x}
+      | c = [[Set Difference with Set Difference is Union of Set Difference with Intersection]]
+}}
+{{eqn | r = \paren{B_2 \setminus B_1} \cup \O
+      | c = [[Intersection With Singleton is Disjoint if Not Element]]
+}}
+{{eqn | r = B_2 \setminus B_1
+      | c = [[Union with Empty Set]]
+}}
+{{end-eqn}}
+Then:
+:$\exists y \in B_2 \setminus B_1 : \paren{ B_1 \setminus \set x} \cup \set y \in \mathscr I$
+We have:
+{{begin-eqn}}
+{{eqn | l = \size{\paren { B_1 \setminus \set x} \cup \set y}
+      | r = \size{B_1 \setminus \set x} + \size{\set y}
+      | c = [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]]
+}}
+{{eqn | r = \size{B_1} - \size{\set x} + \size{\set y}
+      | c = [[Cardinality of Set Difference with Subset]]
+}}
+{{eqn | r = \size{B_1} - 1 + 1
+      | c = [[Cardinality of Singleton]]
+}}
+{{eqn | r = \size{B_1}
+}}
+{{end-eqn}}
+From [[Independent Subset is Base if Cardinality Equals Cardinality of Base]]:
+:$\paren { B_1 \setminus \set x} \cup \set y \in \mathscr B$
+Since $x$, $B_1$ and $B_2$ were arbitrary then the result follows.
+\end{proof}<|endoftext|>
+\section{Independent Subset is Base if Cardinality Equals Rank of Matroid}
+Tags: Matroid Theory
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
+Let $\rho: \powerset S \to \Z$ be the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
+Let $B \in \mathscr I$ such that:
+:$\size B = \map \rho S$
+Then:
+:$B$ is a [[Definition:Base of Matroid|base]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Let $Z \in \mathscr I$ such that:
+:$B \subseteq Z$
+From [[Cardinality of Subset of Finite Set]]:
+:$\size B \le \size Z$
+By definition of the [[Definition:Rank Function (Matroid)|rank function]]:
+:$\size Z \le \map \rho S$
+Then:
+:$\size Z = \size B$
+From the [[Definition:Contrapositive Statement|contrapositive statement]] of [[Cardinality of Proper Subset of Finite Set]]:
+:$B = Z$
+It follows that $B$ is a [[Definition:Maximal|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] by definition.
+That is, $B$ is a [[Definition:Base of Matroid|base]] by definition.
+{{qed}}
+[[Category:Matroid Theory]]
+4qx1rhugi5cnbb38luslx8mvxpmnhlu
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition}
+Tags: Matroid Satisfies Base Axiom
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let $\mathscr B$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $S$ satisfying the [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|base axiom]]:
+{{:Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1}}
+Then $\mathscr B$ is the set of [[Definition:Base of Matroid|bases]] of a [[Definition:Matroid|matroid]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathscr I = \set{X \subseteq S : \exists B \in \mathscr B : X \subseteq B}$
+It is to be shown that:
+:* $\quad \mathscr I$ satisfies the [[Definition:Matroid Axioms|matroid axioms]]
+and
+:* $\quad \mathscr B$ is the [[Definition:Set|set]] of [[Definition:Base of Matroid|bases]] of the [[Definition:Matroid|matroid]] $M = \struct{S, \mathscr I}$
+=== Matroid Axioms ===
+==== Matroid Axiom $(\text I 1)$ ====
+We have $\mathscr B$ is [[Definition:Non-Empty Set|non-empty]].
+Let $B \in \mathscr B$.
+From [[Empty Set is Subset of All Sets]]:
+:$\O \subseteq B$
+By definition of $\mathscr I$:
+:$\O \in \mathscr I$
+It follows that $\mathscr I$ satisfies the [[Definition:Matroid Axioms|matroid axiom $(\text I 1)$]] by definition.
+{{qed|lemma}}
+==== Matroid Axiom $(\text I 2)$ ====
+Let $X \in \mathscr I$.
+Let $Y \subseteq X$.
+By definition of $\mathscr I$:
+:$\exists B \in \mathscr B : X \subseteq B$
+From [[Subset Relation is Transitive]]:
+:$Y \subseteq B$
+By definition of $\mathscr I$:
+:$Y \in \mathscr I$
+It follows that $\mathscr I$ satisfies the [[Definition:Matroid Axioms|matroid axiom $(\text I 2)$]] by definition.
+{{qed|lemma}}
+==== Matroid Axiom $(\text I 3)$ ====
+Let $U, V \in \mathscr I$ such that:
+:$\card V < \card U$
+By definition of $\mathscr I$:
+:$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2$
+From [[Max Equals an Operand]]:
+:$\exists B_1, B_2 \in \mathscr B : U \subseteq B_1, V \subseteq B_2 : \card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$
+{{AimForCont}}:
+:$B_2 \cap \paren{U \setminus V} = \O$
+===== [[Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma|Lemma]] =====
+{{:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma}}{{qed|lemma}}
+This [[Definition:Contradiction|contradicts]] the choice of $B_1, B_2$ such that:
+:$\card{B_1 \cap B_2} = \max \set{\card{C_1 \cap C_2} : U \subseteq C_1, V \subseteq C_2 \text{ and } C_1, C_2 \in \mathscr B}$
+Hence:
+:$B_2 \cap \paren{U \setminus V} \ne \O$
+Let $x \in B_2 \cap \paren{U \setminus V}$.
+From [[Union of Subsets is Subset]]:
+:$V \cup \set x \subseteq B_2$
+By definition of $\mathscr I$:
+:$V \cup \set x \in \mathscr I$
+It follows that $\mathscr I$ satisfies the [[Definition:Matroid Axioms|matroid axiom $(\text I 3)$]] by definition.
+{{qed|lemma}}
+This completes the proof that $M = \struct{S, I}$ forms a [[Definition:Matroid|matroid]]. 
+{{qed|lemma}}
+=== $\mathscr B$ is Set of Bases ===
+Let $B \in \mathscr B$.
+From [[Set is Subset of Itself]]:
+:$B \in \mathscr I$
+Let $U \in \mathscr I$ such that:
+:$B \subseteq U$
+By definition of $\mathscr I$:
+:$\exists B' \in \mathscr B : I \subseteq B'$
+From [[Subset Relation is Transitive]]:
+:$B \subseteq B'$
+From [[Leigh.Samphier/Sandbox/Matroid Base Axiom Implies Sets Have Same Cardinality]]:
+:$\card B = \card {B'}$
+From [[Cardinality of Proper Subset of Finite Set]]:
+:$B = B'$
+By definition of [[Definition:Set Equality|set equality]]:
+:$U = B$
+It has been shown that $B$ is a [[Definition:Maximal Set|maximal]] [[Definition:Subset|subset]] of the [[Definition:Ordered Set|ordered set]] $\struct{\mathscr I, \subseteq}$.
+It follows that $\mathscr B$ is the [[Definition:Set|set]] of [[Definition:Base of Matroid|bases]] of the [[Definition:Matroid|matroid]] $M = \struct{S, I}$ by definition.
+\end{proof}<|endoftext|>
+\section{Morley's Trisector Theorem}
+Tags: Triangles, Morley's Trisector Theorem
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let the [[Definition:Internal Angle|internal angles]] of $\triangle ABC$ be [[Definition:Trisection|trisected]].
+Let the [[Definition:Point|points]] where these [[Definition:Angle Trisector|angle trisectors]] first [[Definition:Intersection (Geometry)|intersect]] be $D$, $E$ and $F$.
+:[[File:Morleys-Theorem.png|500px]]
+Then $\triangle EDF$ is [[Definition:Equilateral Triangle|equilateral]].
+\end{theorem}<|endoftext|>
+\section{Mean Number of Elements Fixed by Self-Map}
+Tags: Combinatorics, Probability Theory, Mean Number of Elements Fixed by Self-Map
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $n$.
+Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself.
+Let $\map \mu n$ denote the [[Definition:Arithmetic Mean|arithmetic mean]] of the number of [[Definition:Fixed Point|fixed points]] of all the [[Definition:Mapping|mappings]] in $S^S$.
+Then:
+:$\map \mu n = 1$
+\end{theorem}
+
+\begin{proof}
+Let $f \in S^S$ be an arbitrary [[Definition:Mapping|mapping]] from $S$ to itself.
+Let $s \in S$ be an arbitrary [[Definition:Element|element]] of $S$.
+$s$ has an equal [[Definition:Probability|probability]] of being mapped to any [[Definition:Element|element]] of $S$.
+Hence the [[Definition:Probability|probability]] that $\map f s = s$ is equal to $\dfrac 1 n$.
+There are $n$ [[Definition:Element|elements]] of $S$.
+By the above argument, each one has a [[Definition:Probability|probability]] of $\dfrac 1 n$ that it is a [[Definition:Fixed Point|fixed point]].
+Thus the [[Definition:Expectation|expectation]] of the number of [[Definition:Fixed Point|fixed points]] is $n \times \dfrac 1 n = 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient}
+Tags: Fibonacci's Greedy Algorithm, Egyptian Fractions
+
+\begin{theorem}
+Consider [[Definition:Proper Fraction|proper fractions]] of the form $\dfrac 3 n$ expressed in [[Definition:Canonical Form of Rational Number|canonical form]].
+Let [[Fibonacci's Greedy Algorithm]] be used to generate a [[Definition:Sequence|sequence]] $S$ of [[Definition:Egyptian Fraction|Egyptian fractions]] for $\dfrac 3 n$.
+Then $S$ consists of $3$ [[Definition:Term of Sequence|terms]], where $2$ would be sufficient {{iff}} the following conditions hold:
+:$n \equiv 1 \pmod 6$
+:$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$
+\end{theorem}
+
+\begin{proof}
+By [[Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm]], $S$ consists of no more than $3$ [[Definition:Term of Sequence|terms]].
+Suppose $n$ has our desired property.
+Since $\dfrac 3 n$ is [[Definition:Proper Fraction|proper]], $n \ge 4$.
+Since $\dfrac 3 n$ is in [[Definition:Canonical Form of Rational Number|canonical form]], $3 \nmid n$.
+We also have that $S$ consists of at least $2$ [[Definition:Term of Sequence|terms]].
+Consider the case $n = 3 k - 1$.
+:$\dfrac 3 n = \dfrac 1 k + \dfrac 1 {k \paren {3 k - 1} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 1 3} = k$
+[[Fibonacci's Greedy Algorithm]] produces $2$ [[Definition:Term of Sequence|terms]] only.
+hence it must be the case that $n = 3 k - 2$.
+{{qed|lemma}}
+We have:
+:$\dfrac 3 n = \dfrac 1 k + \dfrac 2 {k \paren {3 k - 2} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 2 3} = k$
+If $k$ is [[Definition:Even Integer|even]], $\dfrac 1 {\paren {k / 2} \paren {3 k - 2} }$ is an [[Definition:Egyptian Fraction|Egyptian fraction]].
+Then [[Fibonacci's Greedy Algorithm]] would produce $2$ [[Definition:Term of Sequence|terms]] only.
+hence it must be the case that $k$ is [[Definition:Odd Integer|odd]].
+This happens {{iff}} $n$ is [[Definition:Odd Integer|odd]].
+{{qed|lemma}}
+We have shown that for [[Fibonacci's Greedy Algorithm]] to produce $3$ terms, $n$ must be [[Definition:Odd Integer|odd]] and $n = 3 k - 2$.
+By [[Chinese Remainder Theorem]], these conditions can be merged into:
+:$n \equiv 1 \pmod 6$
+We need find when [[Fibonacci's Greedy Algorithm]] gives minimum terms.
+Write:
+{{begin-eqn}}
+{{eqn | l = \frac 3 n
+      | r = \frac 1 x + \frac 1 y
+}}
+{{eqn | r = \frac {x + y} {x y}
+}}
+{{end-eqn}}
+Since $\dfrac 3 n$ is in [[Definition:Canonical Form of Rational Number|canonical form]], $x y \divides n$.
+By [[Divisor of Product]], we can find $p, q \in \N$ such that:
+:$p \divides x$, $q \divides y$
+:$p q = n$
+Rewrite:
+{{begin-eqn}}
+{{eqn | l = \frac 3 n
+      | r = \frac 1 {p a} + \frac 1 {q b}
+}}
+{{eqn | r = \frac {q b + p a} {p q a b}
+}}
+{{eqn | ll= \leadsto
+      | l = p a + q b
+      | r = 3 a b
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | r = \frac {b \paren {3 a - q} } a
+}}
+{{end-eqn}}
+If $\exists d: d \divides n$ and $d \equiv 2 \pmod 3$, set $q = 1$ and $a = \dfrac {d + 1} 3$.
+Then $p = n$ and $b = \dfrac {a n} {3 a - 1} = \dfrac {n \paren {d + 1} } {3 d}$, which is a solution:
+:$\dfrac 3 n = \dfrac 3 {n \paren {d + 1} } + \dfrac {3 d} {n \paren {d + 1} }$
+{{qed|lemma}}
+Now suppose $d$ does not exist.
+Then any [[Definition:Divisor of Integer|divisor]] of $n$ is equivalent to $1 \pmod 3$.
+Hence $q \equiv 1 \pmod 3$.
+Thus $3 a - q \equiv 2 \pmod 3$.
+Let $r = \gcd \set {3 a - q, a}$.
+We have:
+:$a \divides b \paren {3 a - q} = b r \paren {\dfrac {3 a - q} r}$
+Thus:
+:$\dfrac a r \divides b \paren {\dfrac {3 a - q} r}$
+By [[Integers Divided by GCD are Coprime]]:
+:$\dfrac a r \perp \dfrac {3 a - q} r$
+Finally, by [[Euclid's Lemma]]:
+:$\dfrac a r \divides b$
+Hence $\dfrac {3 a - q} r, \dfrac {b r} a \in \Z$, and $p = \paren {\dfrac {3 a - q} r} \paren {\dfrac {b r} a}$.
+Thus we also have:
+:$\dfrac {3 a - q} r \divides p \divides n$
+:$r = \gcd \set {q, a} \divides q \divides n$
+Hence:
+:$\dfrac {3 a - q} r \equiv 1 \pmod 3$
+:$r \equiv 1 \pmod 3$
+Taking their product:
+{{begin-eqn}}
+{{eqn | l = 3 a - q
+      | r = \frac {3 a - q} r \times r
+}}
+{{eqn | o = \equiv
+      | r = 1 \times 1
+      | rr = \pmod 3
+}}
+{{eqn | o = \equiv
+      | r = 1
+      | rr = \pmod 3
+}}
+{{end-eqn}}
+which is a [[Definition:Contradiction|contradiction]].
+Therefore $n$ cannot be expressed as the [[Definition:Integer Addition|sum]] of $2$ [[Definition:Egyptian Fraction|Egyptian fractions]].
+Hence the result.
+{{qed}}
+[[Category:Fibonacci's Greedy Algorithm]]
+[[Category:Egyptian Fractions]]
+lkbiny67byhfdimpr3o1qt6xwno8ec4
+\end{proof}<|endoftext|>
+\section{P-Sequence Space with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences form Vector Space}
+Tags: Examples of Vector Spaces, Functional Analysis
+
+\begin{theorem}
+Let $\ell^p$ be the [[Definition:P-Sequence Space|p-sequence space]].
+Let $\struct {\R, +_\R, \times_\R}$ be the [[Definition:Field of Real Numbers|field of real 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 $\struct {\ell^p, +, \, \cdot \,}_\R$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \ell^p$.
+Let $\lambda, \mu \in \R$.
+Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a [[Definition:Real-Valued Function|real-valued function]].
+Let us use [[Definition:Real Number|real number]] [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]].
+Define [[Definition:Pointwise Addition on Ring of Sequences|pointwise addition]] as:
+:$\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\R \sequence {b_n}_{n \mathop \in \N}$.
+Define [[Definition:Pointwise Scalar Multiplication on Ring of Sequences|pointwise scalar multiplication]] as:
+:$\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\R \sequence {a_n}_{n \mathop \in \N}$
+Let the [[Definition:Ring of Sequences/Additive Inverse|additive inverse]] be $\sequence {-a_n} := - \sequence {a_n}$.
+=== Closure Axiom ===
+By [[Definition:Assumption|assumption]], $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N} \in \ell^p$.
+By [[Definition:P-Sequence Space|definition]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \size {a_n}^p < \infty$
+:$\displaystyle \sum_{n \mathop = 1}^\infty \size {b_n}^p < \infty$
+Consider the [[Definition:Real Sequence|sequence]] $\sequence {a_n + b_n}$.
+Then: 
+{{begin-eqn}}
+{{eqn| l = \sum_{n \mathop = 1}^\infty \size {a_n + b_n}^p
+     | o = \le
+     | r = \sum_{n \mathop = 1}^\infty \paren {\size {a_n} + \size {b_n} }^p
+}}
+{{eqn| o = \le
+     | r = \sum_{n \mathop = 1}^\infty \paren {\map \max {\size {a_n}, \size {b_n} } + \map \max {\size {a_n}, \size {b_n} } }^p
+     | c = {{defof|Max Operation}}
+}}
+{{eqn| r = \sum_{n \mathop = 1}^\infty 2^p \paren {\map \max {\size {a_n}, \size {b_n} } }^p
+}}
+{{eqn | o = \le
+      | r = 2^p \sum_{n \mathop = 1}^\infty \paren {\size {a_n}^p + \size {b_n}^p}
+}}
+{{eqn | o = <
+      | r = \infty
+      | c = $\sequence {a_n}, \sequence {b_n} \in \ell^p$
+}}
+{{end-eqn}}
+Hence:
+:$\sequence {a_n + b_n} \in \ell^p$
+{{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 +_\R \sequence {a_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \tuple {0, 0, 0, \dots} +_\R \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} +_\R \sequence {-a_n}
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \sequence {a_n} +_\R \paren {-1} \times_\R \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 +_\R \mu} \cdot a_n }
+      | r = \paren {\lambda +_\R \mu} \times_\R \sequence {a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\R \sequence {a_n} +_\R \mu \times_\R \sequence {a_n}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n} +_\R \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_\R \sequence {a_n + b_n}
+      | r = \lambda \times_\R \paren {\sequence {a_n} +_\R \sequence {b_n} }
+      | c = {{Defof|Pointwise Addition on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\R \sequence {a_n} +_\R \lambda \times_\R \sequence {b_n}
+      | c = [[Real Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \sequence {\lambda \cdot a_n} +_\R \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_\R \mu} \cdot a_n}
+      | r = \paren {\lambda \times_\R \mu} \times_\R \sequence {a_n}
+      | c = {{Defof|Pointwise Scalar Multiplication on Ring of Sequences}}
+}}
+{{eqn | r = \lambda \times_\R \paren {\mu \times_\R \sequence {a_n} }
+      | c = [[Real Multiplication is Associative]]
+}}
+{{eqn | r = \lambda \times_\R \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_\R \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{Orthogonal Latin Squares of Order 6 do not Exist}
+Tags: Latin Squares
+
+\begin{theorem}
+Two [[Definition:Orthogonal Latin Squares|orthogonal Latin squares]] of [[Definition:Order of Latin Square|order]] $6$ do not exist.
+\end{theorem}<|endoftext|>
+\section{Divisor of Product}
+Tags: Coprime Integers
+
+\begin{theorem}
+Let $a, b, c \in \Z$ be [[Definition:Integer|integers]].
+Let the [[Definition:Symbol|symbol]] $\divides$ denote the [[Definition:Divisor of Integer|divisibility relation]].
+Let $a \divides b c$.
+Then there exist [[Definition:Integer|integers]] $r, s$ such that:
+:$a = r s$, where $r \divides b$ and $s \divides c$.
+\end{theorem}
+
+\begin{proof}
+Let $r = \gcd \set {a, b}$.
+By [[Integers Divided by GCD are Coprime]]:
+:$\exists s, t \in \Z: a = r s \land b = r t \land \gcd \set {s, t} = 1$
+So we have written $a = r s$ where $r$ [[Definition:Divisor of Integer|divides]] $b$.
+We now show that $s$ [[Definition:Divisor of Integer|divides]] $c$.
+Since $a$ [[Definition:Divisor of Integer|divides]] $b c$ there exists $k$ such that $b c = k a$.
+Substituting for $a$ and $b$:
+:$r t c = k r s$
+which gives:
+:$t c = k s$
+So $s$ [[Definition:Divisor of Integer|divides]] $t c$.
+But we have that:
+:$s \perp t$
+Hence by [[Euclid's Lemma]] $s \divides c$ as required.
+{{qed}}
+[[Category:Coprime Integers]]
+k9j3wdqnu5y7uqj2ft8uko6w8dii5n7
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid}
+Tags: Matroid Theory, Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+{{TFAE|def=Matroid}}
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Matroid/Definition 1 implies Definition 2|Definition 1 implies Definition 2]] ===
+{{:Equivalence of Definitions of Matroid/Definition 1 implies Definition 2}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Matroid/Definition 2 implies Definition 3|Definition 2 implies Definition 3]] ===
+{{:Equivalence of Definitions of Matroid/Definition 2 implies Definition 3}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Matroid/Definition 3 implies Definition 1|Definition 3 implies Definition 1]] ===
+{{:Equivalence of Definitions of Matroid/Definition 3 implies Definition 1}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Matroid/Definition 1 implies Definition 4|Definition 1 implies Definition 4]] ===
+{{:Equivalence of Definitions of Matroid/Definition 1 implies Definition 4}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Matroid/Definition 4 implies Definition 1|Definition 4 implies Definition 1]] ===
+{{:Equivalence of Definitions of Matroid/Definition 4 implies Definition 1}}{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm is Well Defined}
+Tags: 
+
+\begin{theorem}
+[[Definition:P-adic Norm|P-adic norm]] $\norm {\, \cdot \,}_p$ is [[Definition:Well-Defined|well defined]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\norm {\, \cdot \,}_p$ is not [[Definition:Well-Defined|well defined]].
+Then, given $r \in \Q$, for two [[Equivalent Representations of Rational Numbers|equivalent representations]] of $r$ $\norm {r}_p$ will yield two different results. 
+Let $k_1, k_2, m_1, m_2 \in \Z, n_1, n_2 \in \Z_{\ne 0} : p \nmid m_1, m_2, n_1, n_2$.
+Let $\displaystyle r = p^{k_1} \frac {m_1} {n_1} = p^{k_2} \frac {m_2} {n_2}$, with $k_1 \ne k_2$.
+Suppose $k_2 < k_1$.
+Then:
+:$p^{k_1 - k_2} m_1 n_2 = m_2 n_1$
+Therefore:
+:$p \divides m_2 n_1$
+Since $p$ is [[Definition:Prime Number|prime]], it cannot be expressed as a [[Definition:Product (Algebra)|product]] of selected [[Definition:Divisor of Integer|divisors]] of both $m_2$ and $n_1$.
+Hence $p \divides m_2$ or $p \divides n_1$.
+This is a [[Definition:Contradiction|contradiction]].
+Thus, $k_1 \le k_2$.
+Similarly, [[Definition:Assumption|assuming]] $k_1 < k_2$ leads to a [[Definition:Contradiction|contradiction]].
+Hence, $k_1 \ge k_2$.
+Since $k_1 \ne k_2$, both $k_1$ and $k_2$ have to be such that:
+:$k_1 < k_2$
+:$k_1 > k_2$
+are satisfied.
+No [[Definition:Integer|integers]] satisfy this.
+{{help|because [[Integers form Totally Ordered Ring]]. How to use this argument rigorously?}}
+Hence, we reached a [[Definition:Contradiction|contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid/Definition 1 implies Definition 2}
+Tags: Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+Let $M$ also satisfy:
+{{begin-axiom}}
+{{axiom | n = \text I 3
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+Then $M$ satisfies:
+{{begin-axiom}}
+{{axiom | n = \text I 3'
+        | q = \forall U, V \in \mathscr I
+        | mr= \size U = \size V + 1 \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+Since:
+:$\forall U, V \in \mathscr I : \size U = \size V + 1 \implies \size V < \size U$
+If follows that if $M$ satisfies condition $(\text I 3)$ then $M$ satisfies condition $(\text I 3')$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid/Definition 2 implies Definition 3}
+Tags: Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+Let $M$ also satisfy:
+{{begin-axiom}}
+{{axiom | n = \text I 3'
+        | q = \forall U, V \in \mathscr I
+        | mr= \size U = \size V + 1 \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+Then $M$ satisfies:
+{{begin-axiom}}
+{{axiom | n = \text I 3''
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists Z \subseteq U \setminus V : \paren{V \cup Z \in \mathscr I} \land \paren{ \size {V \cup Z} = \size U}
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+From [[Independent Set can be Augmented by Larger Independent Set]] it follows that if $M$ satisfies condition $(\text I 3')$ then $M$ satisfies condition $(\text I 3'')$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid/Definition 3 implies Definition 1}
+Tags: Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+Let $M$ also satisfy:
+{{begin-axiom}}
+{{axiom | n = \text I 3''
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}
+}}
+{{end-axiom}}
+Then $M$ satisfies:
+{{begin-axiom}}
+{{axiom | n = \text I 3
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+Let $M$ satisfy condition $(\text I 3'')$.
+Let $U, V \in \mathscr I$ such that $\size V < \size U$.
+By condition $(\text I 3'')$:
+:$\exists Z : \exists Z \subseteq U \setminus V : \paren {V \cup Z \in \mathscr I} \land \paren {\size {V \cup Z} = \size U}$
+Then:
+:$V \cup Z \ne V$
+From [[Union with Empty Set]]:
+:$Z \ne \O$
+Then:
+:$\exists x : x \in Z$
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq Z$
+From [[Set Union Preserves Subsets]]:
+:$V \cup \set x \subseteq V \cup Z$
+From [[Definition:Independence System Axioms|independence system axiom $(\text I 2)$]]:
+:$V \cup \set x \in \mathscr I$
+By definition of a [[Definition:Subset|subset]]:
+:$x \in U \setminus V$
+It follows that $M$ satisfies condition $(\text I 3)$.
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid/Definition 1 implies Definition 4}
+Tags: Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+Let $M$ also satisfy:
+{{begin-axiom}}
+{{axiom | n = \text I 3
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+Then $M$ satisfies:
+{{begin-axiom}}
+{{axiom | n = \text I 3'''
+        | q = \forall A \subseteq S
+        | mr= \text{ all maximal subsets } Y \subseteq A \text{ with } Y \in \mathscr I \text{ have the same cardinality}
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+Let $M$ satisfy condition $(\text I 3)$.
+Let $A \subseteq S$.
+Let $Y_1, Y_2$ be [[Definition:Maximal|maximal]] [[Definition:Independent Subset (Matroid)|independent subsets]] of $A$.
+{{WLOG}}, let:
+:$\size {Y_2} \le \size {Y_1}$
+{{AimForCont}}
+:$\size {Y_2} < \size {Y_1}$
+By condition $(\text I 3)$:
+:$\exists y \in Y_1 \setminus Y_2 : Y_2 \cup \set y \in \mathscr I$
+From [[Union of Subsets is Subset]]:
+:$Y_2 \cup \set y \subseteq A$
+This [[Definition:Contradiction|contradicts]] the [[Definition:Maximal|maximality]] of $Y_2$.
+Then:
+:$\size {Y_2} = \size {Y_1}$
+It follows that $M$ satisifies $(\text I 3''')$
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Matroid/Definition 4 implies Definition 1}
+Tags: Equivalence of Definitions of Matroid
+
+\begin{theorem}
+Let $M = \struct {S, \mathscr I}$ be an [[Definition:Independence System|independence system]].
+Let $M$ also satisfy:
+{{begin-axiom}}
+{{axiom | n = \text I 3'''
+        | q = \forall A \subseteq S
+        | mr= \text{ all maximal subsets } Y \subseteq A \text{ with } Y \in \mathscr I \text{ have the same cardinality}
+}}
+{{end-axiom}}
+Then $M$ satisfies:
+{{begin-axiom}}
+{{axiom | n = \text I 3
+        | q = \forall U, V \in \mathscr I
+        | mr= \size V < \size U \implies \exists x \in U \setminus V : V \cup \set x \in \mathscr I
+}}
+{{end-axiom}}
+\end{theorem}
+
+\begin{proof}
+Let $M$ satisfy condition $(\text I 3''')$.
+Let $U, V \in \mathscr I$ such that $\size V < \size U$.
+Let $W$ be a [[Definition:Maximal|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $U \cup V$ containing $U$.
+Then:
+:$\size U \le \size W$
+By condition $(\text I 3''')$:
+:$V$ is not a [[Definition:Maximal|maximal]] [[Definition:Independent Subset (Matroid)|independent subset]] of $U \cup V$
+Then:
+:$\exists x \in \paren {U \cup V} \setminus V$ such that $V \cup \set x \in \mathscr I$
+From [[Set Difference with Union is Set Difference]]:
+:$\exists x \in U \setminus V$ such that $V \cup \set x \in \mathscr I$
+It follows that $M$ satisifies $(\text I 3)$
+\end{proof}<|endoftext|>
+\section{Cardinality of Set Difference}
+Tags: Set Difference, Cardinality
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]] such that $T$ is [[Definition:Finite Set|finite]].
+Then:
+:$\card {S \setminus T} = \card S - \card {S \cap T}$
+where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection is Subset]]:
+:$S \cap T \subseteq S$
+:$S \cap T \subseteq T$
+From [[Subset of Finite Set is Finite]]:
+:$S \cap T$ is [[Definition:Finite Set|finite]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \card {S \setminus T}
+      | r = \card {S \setminus \paren {S \cap T} }
+      | c = [[Set Difference with Intersection is Difference]]
+}}
+{{eqn | r = \card S -  \card {S \cap T}
+      | c = [[Cardinality of Set Difference with Subset]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Cardinality]]
+ev0k49h3v3w0sf3xazascpr8dqudj59
+\end{proof}<|endoftext|>
+\section{Set Difference and Intersection are Disjoint}
+Tags: Set Difference, Set Intersection
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Then: 
+:$S \setminus T$ and $S \cap T$ are [[Definition:Disjoint Sets|disjoint]] 
+where $S \setminus T$ denotes [[Definition:Set Difference|set difference]] and $S \cap T$ denotes [[Definition:Set Intersection|set intersection]].
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference Intersection with Second Set is Empty Set]]:
+:$\paren {S \setminus T} \cap T = \O$
+and hence immediately from [[Intersection with Empty Set]]:
+:$\paren {S \setminus T} \cap \paren {S \cap T} = \O$
+So $S \setminus T$ and $S \cap T$ are [[Definition:Disjoint Sets|disjoint]].
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Set Intersection]]
+5bipag1bp4tpinkf1sdxe4njjr7mdn9
+\end{proof}<|endoftext|>
+\section{Straight Line has Zero Curvature}
+Tags: Straight Lines, Curvature
+
+\begin{theorem}
+A [[Definition:Straight Line|straight lines]] has zero [[Definition:Curvature|curvature]].
+\end{theorem}
+
+\begin{proof}
+From [[Equation of Straight Line in Plane/Slope-Intercept Form|Equation of Straight Line in Plane: Slope-Intercept Form]], a [[Definition:Straight Line|straight line]] has the [[Definition:Equation|equation]]:
+:$y = m x + c$
+[[Definition:Differentiation|Differentiating]] twice {{WRT|Differentiation}} $x$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = m
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d^2 y} {\d x^2}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+By definition, the [[Definition:Curvature/Cartesian Form|curvature]] of a [[Definition:Curve|curve]] is defined as:
+:$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$
+But we have that:
+:$y'' := \dfrac {\d^2 y} {\d x^2} = 0$
+and so, as in general $y' := \dfrac {\d y} {\d x} = m \ne 0$:
+:$\kappa = \dfrac 0 {\paren {1 + m^2}^{3/2} }$
+the [[Definition:Curvature|curvature]] is zero.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Differential Equation of Spheres in 3-Space}
+Tags: Partial Differentiation, Solid Analytic Geometry
+
+\begin{theorem}
+The set of [[Definition:Sphere (Geometry)|spheres]] in [[Definition:Real Cartesian Space|real Cartesian $3$-dimensional space]] can be described by the [[Definition:System of Differential Equations|system]] of [[Definition:Partial Differential Equation|partial differential equations]]:
+:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
+and if the [[Definition:Sphere (Geometry)|spheres]] are expected to be [[Definition:Real Number|real]]:
+:$z_{xx} z_{yy} > z_{xy}$
+\end{theorem}
+
+\begin{proof}
+From [[Equation of Sphere/Rectangular Coordinates|Equation of Sphere]], we have that the [[Definition:Equation|equation]] defining a general [[Definition:Sphere (Geometry)|sphere]] $S$ is:
+:$\paren {x - a}^2 + \paren {y - b}^2 + \paren {z - c}^2 = R^2$
+where $a$, $b$ and $c$ are [[Definition:Arbitrary Constant|arbitrary constants]].
+We use the technique of [[Elimination of Constants by Partial Differentiation]].
+Taking the [[Definition:Partial Derivative|partial first derivatives]] {{WRT|Differentiation}} $x$ and $y$ and simplifying, we get:
+{{begin-eqn}}
+{{eqn | l = \paren {x - a} + \paren {z - c} \dfrac {\partial z} {\partial x}
+      | r = 0
+}}
+{{eqn | l = \paren {y - b} + \paren {z - c} \dfrac {\partial z} {\partial y}
+      | r = b
+}}
+{{end-eqn}}
+$2$ [[Definition:Equation|equations]] are insufficient to dispose of $3$ [[Definition:Constant|constants]], so the process continues by taking the [[Definition:Partial Derivative|partial second derivatives]] {{WRT|Differentiation}} $x$ and $y$:
+{{begin-eqn}}
+{{eqn | l = 1 + \paren {\dfrac {\partial z} {\partial x} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial x^2}
+      | r = 0
+}}
+{{eqn | l = \dfrac {\partial z} {\partial x} \dfrac {\partial z} {\partial y} + \paren {z - c} \dfrac {\partial^2 z} {\partial x \partial y}
+      | r = 0
+}}
+{{eqn | l = 1 + \paren {\dfrac {\partial z} {\partial y} }^2 + \paren {z - c} \dfrac {\partial^2 z} {\partial y^2}
+      | r = 0
+}}
+{{end-eqn}}
+Eliminating $z - c$:
+:$\dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_x} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$
+Let $\lambda = \dfrac {1 + z_x^2} {z_{xx} } = \dfrac {z_x z_y} {z_{xy} } = \dfrac {1 + z_y^2} {z_{yy} }$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \lambda^2
+      | r = \dfrac {1 + z_x^2} {z_{xx} } \dfrac {1 + z_y^2} {z_{yy} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lambda^2 \paren {z_{xx} z_{yy} }
+      | r = 1 + z_x^2 + z_y^2 + z_x^2 z_y^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lambda^2 \paren {z_{xx} z_{yy} } - z_{xy} \dfrac {z_x z_y} {z_{xy} }
+      | r = 1 + z_x^2 + z_y^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lambda^2 \paren {z_{xx} z_{yy} - z_{xy} }
+      | r = 1 + z_x^2 + z_y^2
+      | c = 
+}}
+{{eqn | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z_{xx} z_{yy} - z_{xy}
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+and so:
+:$z_{xx} z_{yy} > z_{xy}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fermat's Right Triangle Theorem}
+Tags: Number Theory
+
+\begin{theorem}
+$x^4 + y^4 = z^2$ has no solutions in the [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+\end{theorem}
+
+\begin{proof}
+This proof using [[Method of Infinite Descent]] was created by {{AuthorRef|Pierre de Fermat}}.
+Suppose there is such a solution.
+Then there is one with $\gcd \set {x, y, z} = 1$.
+By [[Parity of Smaller Elements of Primitive Pythagorean Triple]] we can assume that $x^2$ is [[Definition:Even Integer|even]] and $y^2$ is [[Definition:Odd Integer|odd]].
+By [[Solutions of Pythagorean Equation/Primitive|Primitive Solutions of Pythagorean Equation]], we can write:
+:$x^2 = 2 m n$
+:$y^2 = m^2 - n^2$
+:$z = m^2 + n^2$
+where $m, n$ are [[Definition:Coprime Integers|coprime]] [[Definition:Positive Integer|positive integers]].
+Similarly we can write:
+:$n = 2 r s$
+:$y = r^2 - s^2$
+:$m = r^2 + s^2$
+where $r, s$ are [[Definition:Coprime Integers|coprime]] [[Definition:Positive Integer|positive integers]], since $y$ is [[Definition:Odd Integer|odd]], forcing $n$ to be [[Definition:Even Integer|even]].
+We have:
+:$\paren {\dfrac x 2}^2 = m \paren {\dfrac n 2}$
+Since $m$ and $\dfrac n 2$ are [[Definition:Coprime Integers|coprime]], they are both [[Definition:Square Number|squares]].
+Similarly we have:
+:$\dfrac n 2 = r s$
+Since $r$ and $s$ are [[Definition:Coprime Integers|coprime]], they are both [[Definition:Square Number|squares]].
+Therefore $m = r^2 + s^2$ becomes an equation of the form $u^4 + v^4 = w^2$.
+Moreover:
+:$z^2 > m^4 > m$
+and so we have found a smaller set of solutions.
+By [[Method of Infinite Descent]], no solutions can exist.
+{{qed}}
+{{Namedfor|Pierre de Fermat|cat = Fermat}}
+[[Category:Number Theory]]
+ewnxr39df8505pmfqljt4eegmfj4f5m
+\end{proof}<|endoftext|>
+\section{Taylor's Theorem/One Variable with Two Functions}
+Tags: Taylor's Theorem
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] satisfying following conditions:
+:$(1): \quad f$ is $n + 1$ times [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a x$
+:$(2): \quad f$ is of [[Definition:Differentiability Class|differentiability class]] $C^n$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint a x$
+:$(3): \quad g$ is $k + 1$ times [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a x$
+:$(4): \quad g$ is of [[Definition:Differentiability Class|differentiability class]] $C^k$ on the [[Definition:Closed Real Interval|closed interval]] $\closedint a x$
+:$(5): \quad \map {g^{\paren {k + 1}}} t \ne 0$ for any $t \in \openint a x$
+Then the following equation holds for some real number $\xi \in \openint a x$:
+:$\dfrac {\map {f^{\paren {n + 1} } } \xi /n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}$
+or equivalently:
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \map f a + \map {f'} a \paren {x - a} + \dfrac {\map {f''} a} {2!} \paren {x - a}^2 + \dotsb + \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n + R_n
+}}
+{{eqn | l = R_n
+      | r = \dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi / k!} \paren {x - \xi}^{n - k} \paren {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+We define $F$ and $G$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map F t
+      | r = \map f t + \map {f'} t \paren {x - t} + \dfrac {\map {f''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n
+}}
+{{eqn | l = \map G t
+      | r = \map g t + \map {g'} t \paren {x - t} + \dfrac {\map {g''} t} {2!} \paren {x - t}^2 + \dotsb + \dfrac {\map {g^{\paren k} } t} {k!} \paren {x - t}^k
+}}
+{{end-eqn}}
+Then $F$ and $G$ are [[Definition:Continuous on Interval|continuous]] on $\closedint a x$ and [[Definition:Differentiable on Interval|differentiable]] on $\openint a x$.
+Differentiating {{WRT|Differentiation}} $t$:
+{{begin-eqn}}
+{{eqn | l = \map {F'} t
+      | r = \map {f'} t + \paren {\map {f''} t \paren {x - t} - \map {f'} t} + \paren {\frac {\map {f^{\paren 3} } t} {2!} \paren {x - t}^2 - \frac {\map {f^{\paren 2} } t} {1!} \paren {x - t} } + \dotsb + \paren {\frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n - \frac {\map {f^{\paren n} } t} {\paren {n - 1}!} \paren {x - t}^{n - 1} }
+}}
+{{eqn | r = \frac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - t}^n
+}}
+{{eqn | l = \map {G'} t
+      | r = \frac {\map {f^{\paren {k + 1} } } t} {k!} \paren {x - t}^k
+}}
+{{end-eqn}}
+By condition $(5)$ of the statement of the theorem, it follows that $G$ does not vanish on $\openint a x$.
+By [[Cauchy Mean Value Theorem]], there exists a [[Definition:Real Number|real number]] $\xi \in \openint a x$ such that:
+:$\dfrac {\map {F'} \xi} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$
+That is:
+:$\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g''} a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Even Integers is Countably Infinite}
+Tags: Countable Sets, Odd Integers
+
+\begin{theorem}
+Let $\Bbb E$ be the [[Definition:Set|set]] of [[Definition:Even Integer|even integers]].
+Then $\Bbb E$ is [[Definition:Countably Infinite Set|countably infinite]].
+\end{theorem}
+
+\begin{proof}
+Let $f: \Bbb E \to \Z$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \Bbb E: \map f x = \dfrac x 2$
+$f$ is [[Definition:Well-Defined Mapping|well-defined]] as $x$ is [[Definition:Even Integer|even]] and so $\dfrac x 2 \in \Z$.
+Let $x, y \in \Bbb E$ such that $\map f x = \map f y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \map f y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac x 2
+      | r = \dfrac y 2
+      | c = Definition of $f$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Thus $f$ is [[Definition:Injection|injective]] by definition.
+Consider the [[Definition:Inverse of Mapping|inverse]] $f^{-1}$.
+By inspection:
+:$\forall x \in \Z: \map {f^{-1} } x = 2 x$
+$f^{-1}$ is [[Definition:Well-Defined Mapping|well-defined]], and $2 x$ is [[Definition:Even Integer|even]].
+Thus $f^{-1}$ is a [[Definition:Mapping|mapping]] from $\Z$ to $\Bbb E$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f^{-1} } x
+      | r = \map {f^{-1} } y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 x
+      | r = 2 y
+      | c = Definition of $f^{-1}$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Thus $f^{-1}$ is [[Definition:Injection|injective]] by definition.
+It follows by the [[Cantor-Bernstein-Schröder Theorem]] that there exists a [[Definition:Bijection|bijection]] between $\Z$ and $\Bbb E$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis Expansion of Rational Number}
+Tags: 
+
+\begin{theorem}
+Let $x$ be a [[Definition:Rational Number|rational number]].
+{{WIP|Time has caught up with me, I'll continue this later. I'm late for work.}}
+q2llusuyprk7567f7o3l3jnktxipxcq
+\end{theorem}<|endoftext|>
+\section{Integration by Parts/Definite Integral}
+Tags: Integration by Parts
+
+\begin{theorem}
+:$\displaystyle \int_a^b \map f t \map G t \rd t = \bigintlimits {\map F t \map G t} a b - \int_a^b \map F t \map g t \rd t$
+\end{theorem}
+
+\begin{proof}
+By [[Product Rule for Derivatives]]:
+:$\map D {F G} = f G + F g$
+Thus $F G$ is a [[Definition:Primitive (Calculus)|primitive]] of $f G + F g$ on $\closedint a b$.
+Hence, by the [[Fundamental Theorem of Calculus]]:
+:$\displaystyle \int_a^b \paren {\map f t \map G t + \map F t \map g t} \rd t = \bigintlimits {\map F t \map G t} a b$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integration by Parts/Primitive}
+Tags: Integration by Parts
+
+\begin{theorem}
+:$\ds \int \map f t \map G t \rd t = \map F t \map G t - \int \map F t \map g t \rd t$
+on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d t} } {\map F t \map G t}
+      | r = \map f t \map G t + \map F t \map g t
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \paren {\map f t \map G t + \map F t \map g t} \rd t
+      | r = \map F t \map G t
+      | c = [[Fundamental Theorem of Calculus]]: [[Definition:Integration|integrating]] both sides {{WRT|Integration}} $t$
+}}
+{{eqn | ll= \leadsto
+      | l = \int \map f t \map G t \rd t + \int \map F t \map g t \rd t
+      | r = \map F t \map G t
+      | c = [[Linear Combination of Primitives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \map f t \map G t \rd t
+      | r = \map F t \map G t - \int \map F t \map g t \rd t
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integration by Substitution/Primitive}
+Tags: Integration by Substitution
+
+\begin{theorem}
+The [[Definition:Primitive (Calculus)|primitive]] of $f$ can be evaluated by:
+:$\ds \int \map f x \rd x = \int \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$
+where $x = \map \phi u$.
+\end{theorem}<|endoftext|>
+\section{Integration by Substitution/Definite Integral}
+Tags: Integration by Substitution
+
+\begin{theorem}
+The [[Definition:Definite Integral|definite integral]] of $f$ from $a$ to $b$ can be evaluated by:
+:$\displaystyle \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$
+where $x = \map \phi u$.
+\end{theorem}
+
+\begin{proof}
+Let $F$ be an [[Definition:Primitive (Calculus)|antiderivative]] of $f$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d u} } {\map F x}
+      | r = \map {\frac \d {\d u} } {\map F {\map \phi u} }
+      | c = Definition of $\map \phi u$
+}}
+{{eqn | r = \dfrac \d {\d x} \map F {\map \phi u} \dfrac \d {\d u} \map \phi u
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = f {\map \phi u} \dfrac \d {\d u} \map \phi u
+      | c = as $\map F x = \ds \int \map f x \rd x$
+}}
+{{end-eqn}}
+Hence $\map F {\map \phi u}$ is an [[Definition:Primitive (Calculus)|antiderivative]] of $\map f {\map \phi u} \d {\d u} \map \phi u$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u
+      | r = \bigintlimits {\map F {\map \phi u} } a b
+      | c = [[Fundamental Theorem of Calculus/Second Part|Fundamental Theorem of Calculus: Second Part]]
+}}
+{{eqn | n = 1
+      | r = \map F {\map \phi b} - \map F {\map \phi a}
+      | c = 
+}}
+{{end-eqn}}
+However, also:
+{{begin-eqn}}
+{{eqn | l = \int_{\map \phi a}^{\map \phi b} \map f t \rd t
+      | r = \bigintlimits {\map F t} {\map \phi a} {\map \phi b}
+      | c = 
+}}
+{{eqn | r = \map F {\map \phi b} - \map F {\map \phi a}
+      | c = 
+}}
+{{eqn | r = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u
+      | c = from $(1)$
+}}
+{{end-eqn}}
+which was to be proved. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Formula for Norms on Non-zero Rationals/Lemma}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $z \in \Z_{\ne 0}$.
+Then the following [[Definition:Infinite Product|infinite product]] [[Definition:Convergent Real Sequence|converges]]:
+:$\size z \times \displaystyle\prod_{p \mathop \in \Bbb P}^{} \norm z_p = 1$
+\end{theorem}
+
+\begin{proof}
+=== Case 1 : $z \in \Z_{>0}$ ===
+Let $z \in \Z_{>0}$.
+From [[Fundamental Theorem of Arithmetic]], we can factor $z$ as a [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]]:
+:$z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k}$
+Then for every [[Definition:Prime Number|prime number $q$]]:
+:$\norm z_q = \begin{cases}
+p_i^{-b_i} & : \exists i \in \closedint 1 k :q = p_i \\
+1 & : \forall i \in \closedint 1 k : q \ne p_i \\
+\end {cases}$
+By definition of [[Definition:Absolute Value|absolute value]] on $\Q$:
+:$\size z = p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} $
+For $n \ge \max \set{p_1, p_2 \dots p_k}$:
+{{begin-eqn}}
+{{eqn | l = \size z \times \prod_{p \mathop \in \Bbb P \mathop : p \mathop \le n } \norm z_p
+      | r = \paren {p_1^{b_1} p_2^{b_2} \dots p_k^{b_k} } \times \paren {p_1^{-b_1} p_2^{-b_2} \dots p_k^{-b_k} } 
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \size z \times \prod_{p \mathop\in \Bbb P} \norm z_p
+      | r = \lim_{n \mathop \to \infty} \paren{\size z \times \prod_{p \mathop \in \Bbb P \mathop : p \mathop \le n} \norm z_p}
+      | c = {{Defof|Infinite Product}}
+}}
+{{eqn | r = 1
+      | c = [[Eventually Constant Sequence Converges to Constant]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case 2 : $z \in \Z_{<0}$ ===
+Let $z \in \Z_{<0}$.
+Hence
+: $-z \in \Z_{>0}$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \size z \times \prod_{p \mathop\in \Bbb P} \norm z_p
+      | r = \size {-z} \times \prod_{p \mathop \in \Bbb P} \norm {-z}_p
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]
+}}
+{{eqn | r = 1
+      | c = [[Product Formula for Norms on Non-zero Rationals/Lemma#Case 1|Case 1]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+In either case:
+:$\size z \times \displaystyle \prod_{p \mathop \in \Bbb P} \norm z_p = 1$
+{{qed}}
+[[Category:P-adic Number Theory]]
+dwzm2ub4btf424j902li9t47zqm25fb
+\end{proof}<|endoftext|>
+\section{P-adic Open Ball is Instance of Open Ball of a Norm}
+Tags: Definitions: Open Balls, Definitions: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $a \in \Q_p$.
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $B \subseteq \Q_p$.
+Then: 
+:$B$ is an [[Definition:Open Ball in P-adic Numbers|open ball in $p$-adic numbers]] with [[Definition:Radius of Open Ball in P-adic Numbers|radius]] $\epsilon$ and [[Definition:Center of Open Ball in P-adic Numbers|centre]] $a$
+{{iff}}:
+:$B$ is an [[Definition:Open Ball of Normed Division Ring|open ball]] of the [[Definition:Normed Division Ring|normed division ring]] $\struct {\Q_p, \norm {\,\cdot\,}_p}$ with [[Definition:Radius of Open Ball of Normed Division Ring|radius]] $\epsilon$ and [[Definition:Center of Open Ball of Normed Division Ring|centre]] $a$ .
+That is, the definition of an [[Definition:Open Ball in P-adic Numbers|open ball in $p$-adic numbers]] is a specific instance of the general definition of an [[Definition:Open Ball of Normed Division Ring|open ball in a normed division ring]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:P-adic Numbers|$p$-adic numbers]] are the [[Definition:Unique up to Isomorphism|unique (up to isometric isomorphism)]] [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean valued field]] that [[Definition:Completion (Normed Division Ring)|completes]] $\struct {\Q, \norm {\,\cdot\,}_p}$ and $\norm {\,\cdot\,}_p$ is a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]].
+The definition of an [[Definition:Open Ball in P-adic Numbers|open ball in $p$-adic numbers]] is identical to the definition of an [[Definition:Open Ball of Normed Division Ring|open ball of a normed division ring]] with respect to the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}_p$.
+{{qed}}
+[[Category:Definitions/Open Balls]] 
+[[Category:Definitions/P-adic Number Theory]]
+pet2kommg23356y64mnh4ttpfocrvep
+\end{proof}<|endoftext|>
+\section{P-adic Closed Ball is Instance of Closed Ball of a Norm}
+Tags: Definitions: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $a \in \Q_p$.
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $B \subseteq \Q_p$.
+Then: 
+:$B$ is a [[Definition:Closed Ball in P-adic Numbers|closed ball in $p$-adic numbers]] with [[Definition:Radius of Closed Ball in P-adic Numbers|radius]] $\epsilon$ and [[Definition:Center of Closed Ball in P-adic Numbers|centre]] $a$
+{{iff}}:
+:$B$ is a [[Definition:Closed Ball of Normed Division Ring|closed ball]] of the [[Definition:Normed Division Ring|normed division ring]] $\struct {\Q_p, \norm {\,\cdot\,}_p}$ with [[Definition:Radius of Closed Ball of Normed Division Ring|radius]] $\epsilon$ and [[Definition:Center of Closed Ball  of Normed Division Ring|centre]] $a$ 
+That is, the definition of a [[Definition:Closed Ball in P-adic Numbers|closed ball in $p$-adic numbers]] is a specific instance of the general definition of a [[Definition:Closed Ball of Normed Division Ring|closed ball in a normed division ring]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:P-adic Numbers|$p$-adic numbers]] are the [[Definition:Unique up to Isomorphism|unique (up to isometric isomorphism)]] [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean valued field]] that [[Definition:Completion (Normed Division Ring)|completes]] $\struct {\Q, \norm {\,\cdot\,}_p}$ and $\norm {\,\cdot\,}_p$ is a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]].
+The definition of a [[Definition:Closed Ball in P-adic Numbers|closed ball in $p$-adic numbers]] is identical to the definition of a [[Definition:Closed Ball of Normed Division Ring|closed ball of a normed division ring]] with respect to the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}_p$.
+{{qed}}
+[[Category:Definitions/P-adic Number Theory]]
+4vy5e5qipmoxnwxru1tx0wc988p7ujm
+\end{proof}<|endoftext|>
+\section{P-adic Sphere is Instance of Sphere of a Norm}
+Tags: Definitions: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $a \in \Q_p$.
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $S \subseteq \Q_p$.
+Then: 
+:$S$ is a [[Definition:Sphere in P-adic Numbers|sphere in $p$-adic numbers]] with [[Definition:Radius of Sphere in P-adic Numbers|radius]] $\epsilon$ and [[Definition:Center of Sphere in P-adic Numbers|centre]] $a$
+{{iff}}:
+:$S$ is a [[Definition:Sphere in Normed Division Ring|sphere]] of the [[Definition:Normed Division Ring|normed division ring]] $\struct {\Q_p, \norm {\,\cdot\,}_p}$ with [[Definition:Radius of Sphere in Normed Division Ring|radius]] $\epsilon$ and [[Definition:Center of Sphere in Normed Division Ring|centre]] $a$ 
+That is, the definition of a [[Definition:Sphere in P-adic Numbers|closed ball in $p$-adic numbers]] is a specific instance of the general definition of a [[Definition:Sphere in Normed Division Ring|sphere in a normed division ring]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:P-adic Numbers|$p$-adic numbers]] are the [[Definition:Unique up to Isomorphism|unique (up to isometric isomorphism)]] [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean valued field]] that [[Definition:Completion (Normed Division Ring)|completes]] $\struct {\Q, \norm {\,\cdot\,}_p}$ and $\norm {\,\cdot\,}_p$ is a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]].
+The definition of a [[Definition:Sphere in P-adic Numbers|sphere in $p$-adic numbers]] is identical to the definition of a [[Definition:Sphere in Normed Division Ring|sphere in a normed division ring]] with respect to the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}_p$.
+{{qed}}
+[[Category:Definitions/P-adic Number Theory]]
+p2xvi2tweh0f7d100wnsm1lgz788tbn
+\end{proof}<|endoftext|>
+\section{Sphere is Set Difference of Closed Ball with Open Ball/P-adic Numbers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\Q_p$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $a \in \Q_p$.
+Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $\map {{B_\epsilon}^-} a$ denote the [[Definition:Closed Ball in P-adic Numbers|$\epsilon$-closed ball of $a$]] in $\Q_p$.
+Let $\map {B_\epsilon} a$ denote the [[Definition:Open Ball in P-adic Numbers|$\epsilon$-open ball of $a$]] in $\Q_p$.
+Let $\map {S_\epsilon} a$ denote the [[Definition:Sphere  in P-adic Numbers|$\epsilon$-sphere of $a$]] in $\Q_p$.
+Then:
+:$\map {S_\epsilon} a = \map { {B_\epsilon}^-} a \setminus \map {B_\epsilon} a$
+\end{theorem}
+
+\begin{proof}
+The result follows directly from:
+:[[P-adic Closed Ball is Instance of Closed Ball of a Norm]]
+:[[P-adic Open Ball is Instance of Open Ball of a Norm]]
+:[[P-adic Sphere is Instance of Sphere of a Norm]]
+:[[Sphere is Set Difference of Closed and Open Ball in Normed Division Ring]]
+{{qed}}
+[[Category:P-adic Number Theory]]
+4ssdrfo0qp1k4n27x56x7rjf0q63hay
+\end{proof}<|endoftext|>
+\section{Set is Closed in Metric Space iff Closed in Induced Topological Space}
+Tags: Closed Sets, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced]] by the [[Definition:Metric|metric]] $d$.
+Let $F$ be a [[Definition:Subset|subset]] of $M$.
+Then:
+:$F$ is [[Definition:Closed Set (Metric Space)|closed in $M$]] {{iff}} $F$ is [[Definition:Closed Set (Topology)|closed in $\struct {A, \tau}$]]
+\end{theorem}
+
+\begin{proof}
+By definition of a [[Definition:Closed Set (Metric Space)|closed set in $M$]]:
+:$F$ is [[Definition:Closed Set (Metric Space)|closed set in $M$]] {{iff}} $A \setminus F$ is [[Definition:Open Set (Metric Space)|open in $M$]]
+By definition of the [[Definition:Topology Induced by Metric|topology $\tau$ induced]] by the [[Definition:Metric|metric]] $d$:
+:$A \setminus F$ is [[Definition:Open Set (Metric Space)|open in $M$]] {{iff}} $A \setminus F$ is [[Definition:Open Set (Topology)|open in $\struct {A, \tau}$]]
+By definition of a [[Definition:Closed Set (Topology)|closed set in $\struct{A, \tau}$]]:
+:$A \setminus F$ is [[Definition:Open Set (Topology)|open in $\struct {A, \tau}$]] {{iff}} $F$ is [[Definition:Closed Set (Topology)|closed set in $\struct {A, \tau}$]]
+The result follows.
+{{qed}}
+[[Category:Closed Sets]]
+[[Category:Metric Spaces]]
+l4ycvi1ahs62ph5bgwvds2evtt68snc
+\end{proof}<|endoftext|>
+\section{Primitive of Hyperbolic Cosecant Function/Inverse Hyperbolic Cotangent of Hyperbolic Cosine Form}
+Tags: Primitive of Hyperbolic Cosecant Function, Primitives involving Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \int \csch x \rd x = -\map {\coth^{-1} } {\cosh x} + C$
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Root of a squared minus x squared/Arccosine Form}
+Tags: Arccosine Function, Primitive of Reciprocal of Root of a squared minus x squared
+
+\begin{theorem}
+:$\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = -\arccos \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac 1 {\sqrt {a^2 - x^2} } \rd x
+      | r = \int \frac {\rd x} {\sqrt {a^2 \paren {1 - \frac {x^2} {a^2} } } } 
+      | c = factor $a^2$ out of the [[Definition:Radicand|radicand]]
+}}
+{{eqn | r = \int \frac {\rd x} {\sqrt{a^2} \sqrt {1 - \paren {\frac x a}^2} }
+      | c = 
+}}
+{{eqn | r = \frac 1 a \int \frac {\rd x} {\sqrt {1 - \paren {\frac x a}^2} }
+}}
+{{end-eqn}}
+[[Integration by Substitution|Substitute]]:
+:$\cos \theta = \dfrac x a \iff x = a \cos \theta$
+for $\theta \in \openint 0 \pi$.
+From [[Real Cosine Function is Bounded]] and [[Shape of Cosine Function]], this substitution is valid for all $x / a \in \openint {-1} 1$.
+[[Definition:By Hypothesis|By hypothesis]]:
+{{begin-eqn}}
+{{eqn | l = a^2
+      | o = > 
+      | r = x^2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \frac {x^2} {a^2}
+      | c = dividing both terms by $a^2$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \paren {\frac x a}^2
+      | c = [[Powers of Group Elements]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | o = >
+      | r = \size {\frac x a}
+      | c = taking the square root of both terms
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = -1
+      | o = <
+      | r = \paren {\frac x a} < 1
+      | c = [[Negative of Absolute Value]]
+}}
+{{end-eqn}}
+so this substitution will not change the domain of the integrand.
+Then:
+{{begin-eqn}}
+{{eqn | l = x 
+      | r = a \cos \theta
+      | c = from above
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | r = -a \sin \theta \frac {\rd \theta} {\rd x}
+      | c = differentiating {{WRT|Differentiation}} $x$, [[Derivative of Cosine Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | l = \frac 1 a \int \frac 1 {\sqrt {1 - \paren {\frac x a}^2 } } \rd x
+      | r = \frac 1 a \int \frac {-a \sin \theta} {\sqrt {1 - \cos^2 \theta} } \frac {\rd \theta} {\rd x} \rd x
+      | c = from above
+}}
+{{eqn | r = -\frac a a \int \frac {\sin \theta} {\sqrt {1 - \cos^2 \theta} } \rd \theta
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\int \frac {\sin \theta} {\sqrt {\sin^2 \theta} } \rd \theta
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = -\int \frac {\sin \theta} {\size {\sin \theta} } \rd \theta
+      | c = 
+}}
+{{end-eqn}}
+We have defined $\theta$ to be in the [[Definition:Open Real Interval|open interval]] $\openint 0 \pi$.
+From [[Sine and Cosine are Periodic on Reals]], $\sin \theta > 0$ for the entire interval. Therefore the [[Definition:Absolute Value|absolute value]] is unnecessary, and the integral simplifies to:
+{{begin-eqn}}
+{{eqn | l = -\int \rd \theta
+      | r = -\theta + C
+}}
+{{end-eqn}}
+As $\theta$ was stipulated to be in the [[Definition:Open Real Interval|open interval]] $\openint 0 \pi$:
+:$\cos \theta = \dfrac x a \iff \theta = \arccos \dfrac x a$
+The answer in terms of $x$, then, is:
+:$\displaystyle \int \frac 1 {\sqrt {a^2 - x^2} } \rd x = -\arccos \frac x a + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Negative of Logarithm of x plus Root x squared minus a squared}
+Tags: Logarithms
+
+\begin{theorem}
+Let $x \in \R: \size x > 1$.
+Let $x > 1$.
+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}
+First we note that if $x > 1$ then $x + \sqrt {x^2 - a^2} > 0$.
+Hence $\map \ln {x + \sqrt {x^2 - a^2} }$ is defined.
+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{Supremum Norm on Vector Space of Real Matrices is Norm}
+Tags: Examples of Norms
+
+\begin{theorem}
+[[Definition:Supremum Norm|Supremum Norm]] forms a [[Definition:Norm on Vector Space|norm]] on the [[Definition:Vector Space|vector space]] of [[Definition:Real Matrix|real matrices]].
+\end{theorem}
+
+\begin{proof}
+Let $M \in \R^{m \times n} : m, n \in \N_{\mathop > 0}$ be a [[Definition:Real Matrix|real matrix]].
+Denote the $\paren {i, j}$-th entry of $M$ by $a_{ij}$.
+Note that the [[Definition:Set|set]] of [[Definition:Element of Matrix|matrix elements]] of $M$ is a [[Definition:Finite Set|finite set]] of [[Definition:Real Number|real numbers]].
+We have that:
+:[[Real Numbers form Ordered Field]]
+:[[Non-Empty Finite Set has Greatest Element]]
+{{Help|Something of this sort}}
+Therefore, $M$ has the [[Definition:Greatest Element|greatest element]].
+=== Norm Axiom $(\text N 1)$ ===
+{{begin-eqn}}
+{{eqn | l = \norm M_\infty
+      | r = \max_{\begin {split} 
+                  1 \mathop \le i \mathop \le m \\
+                  1 \mathop \le j \mathop \le n 
+                  \end {split} } 
+            \size {a_{ij} }
+      | c = [[Greatest Element is Supremum]], {{defof|Max Operation}}
+}}
+{{eqn | o = \ge
+      | r = 0
+}}
+{{end-eqn}}
+[[Definition:Equality|Equality]] is obtained for $M$ being a [[Definition:Zero Matrix|zero matrix]].
+Suppose $\norm M_\infty = 0$.
+Then: 
+:$\displaystyle \forall i, j : 1 \le i \le m, 1 \le j \le n : \size {a_{ij}} = 0$
+In other words, $M$ is a [[Definition:Zero Matrix|zero matrix]].
+{{qed|lemma}}
+=== Norm Axiom $(\text N 2)$ ===
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha \cdot M}_\infty
+      | r = \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} }
+             \size {\alpha m_{ij} }
+      | c = [[Greatest Element is Supremum]]
+}}
+{{eqn | r = \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} } \size \alpha \size {a_{ij} }
+      | c = [[Absolute Value of Product]]
+}}
+{{eqn | r = \size \alpha \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} } \size {a_{ij} }
+}}
+{{eqn | r = \size \alpha \norm M_\infty
+      | c = [[Greatest Element is Supremum]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Norm Axiom $(\text N 3)$ ===
+Let $P, Q \in \R^{m \times n}$.
+Denote their $\paren {i, j}$-th [[Definition:Element of Matrix|matrix elements]] as $p_{ij}$ and $q_{ij}$ respectively.
+Fix $i,j \in \N : 1 \le i \le m, 1 \le j \le n$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \size {p_{ij} + q_{ij} }
+      | o = \le 
+      | r = \size {p_{ij} } + \size {q_{ij} }
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | o = \le
+      | r = \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} } \size {p_{ij} } + 
+            \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} } \size {q_{ij} }
+      | c = {{defof|Max Operation}}
+}}
+{{eqn | r = \norm P_\infty + \norm Q_\infty
+      | c = [[Greatest Element is Supremum]], {{defof|Supremum Norm}}
+}}
+{{end-eqn}}
+This holds for any $i,j$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {P + Q}_\infty
+      | r = \max_{\begin {split}
+                  & 1 \mathop \le i \mathop \le m\\
+                  & 1 \mathop \le j \mathop \le n
+                  \end {split} } \size {p_{ij} + q_{ij} }
+      | c = [[Greatest Element is Supremum]]
+}}
+{{eqn | o = \le
+      | r = \norm P_\infty + \norm Q_\infty
+}}
+{{end-eqn}}
+{{qed|lemma}}
+All [[Definition:Norm Axioms (Vector Space)|norm axioms]] are seen to be satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arccotangent Logarithmic Formulation}
+Tags: Arccotangent Function
+
+\begin{theorem}
+For any [[Definition:Real Number|real number]] $x$:
+:$\arccot x = \dfrac 1 2 i \, \map \ln {\dfrac {1 + i x} {1 - i x} }$
+where $\arccot x$ is the [[Definition:Arccotangent|arccotangent]] and $i^2 = -1$.
+\end{theorem}
+
+\begin{proof}
+{{Proofread}}
+Assume $y \in \R$, $ -\dfrac \pi 2 \le y \le \dfrac \pi 2 $.
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \arccot x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \cot y
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = i \frac {1 + e^{2 i y} } {1 - e^{2 i y} }
+      | c = [[Cotangent Exponential Formulation]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x
+      | r = \frac {e^{2 i y} + 1} {e^{2 i y} - 1}
+      | c = $ i^2 = -1 $
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x \paren {e^{2 i y} - 1}
+      | r = e^{2 i y} + 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = i x e^{2 i y} - i x
+      | r = e^{2 i y} + 1
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y} - i x e^{2 i y}
+      | r = 1 + i x
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = e^{2 i y}
+      | r = \frac {1 + i x} {1 - i x}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | r = \frac 1 2 i \map \ln {\frac {1 + i x} {1 - i x} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x squared plus a squared/Arccotangent Form}
+Tags: Primitive of Reciprocal of x squared plus a squared
+
+\begin{theorem}
+:$\ds \int \frac {\d x} {x^2 + a^2} = -\frac 1 a \arccot \frac x a + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \int \frac {\d x} {x^2 + a^2}
+	| r = \frac 1 a \int \frac {\d t} {t^2 + 1}
+	| c = [[Integration by Substitution|Substitution of $x \to a t$]]}}
+{{eqn	| r = \frac 1 a \int \frac {\d t} {\paren {1 + i t} \paren {1 - i t} }
+	| c = Factoring
+}}
+{{eqn	| r = \frac 1 {2 a} \paren {\int \frac {\d t} {1 + i t} + \int \frac {\d t} {1 - i t} }
+	| c = {{Defof|Partial Fractions Expansion}}
+}}
+{{eqn	| r = \frac 1 {2 a} \paren {i \map \ln {1 - i t} - i \map \ln {1 + i t} } + C
+	| c = [[Primitive of Reciprocal]]
+}}
+{{eqn	| r = \frac i {2 a} \map \ln {\frac {1 - i t} {1 + i t} } + C
+	| c = [[Sum of Logarithms]]
+}}
+{{eqn	| r = -\frac i {2 a} \map \ln {\frac {1 + i t} {1 - i t} } + C
+	| c = [[Logarithm of Reciprocal]]
+}}
+{{eqn	| r = -\frac 1 a \arccot \frac x a + C
+	| c = [[Arccotangent Logarithmic Formulation]] and substituting back $t \to \dfrac x a$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of x by Root of x squared plus a squared/Reciprocal Logarithm Form}
+Tags: Primitive of Reciprocal of x by Root of x squared plus a squared
+
+\begin{theorem}
+:$\displaystyle \int \frac {\d x} {x \sqrt {x^2 + a^2} } = \frac 1 a \map \ln {\frac x {a + \sqrt {x^2 + a^2} } } + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int \frac {\d x} {x \sqrt {x^2 + a^2} }
+      | r = -\frac 1 a \map \ln {\frac {a + \sqrt {x^2 + a^2} } x} + C
+      | c = [[Primitive of Reciprocal of x by Root of x squared plus a squared/Logarithm Form|Primitive of Reciprocal of $x \sqrt {x^2 + a^2}$: Logarithm form]]
+}}
+{{eqn | r = \frac 1 a \map \ln {\frac x {a + \sqrt {a^2 + x^2} } } + C
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Sine}
+Tags: Derivative of Hyperbolic Sine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sinh x} = \cosh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sinh x}
+      | r = \map {\frac \d {\d x} } {\dfrac {e^x - e ^{-x} } 2}
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac 1 2 \paren {\map {\frac \d {\d x} } {e^x} - \map {\frac \d {\d x} } {e^{-x} } }
+      | c = [[Linear Combination of Derivatives]]
+}}
+{{eqn | r = \frac 1 2 \paren {e^x - \paren {-e^{-x} } }
+      | c = [[Derivative of Exponential Function]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {e^x + e^{-x} } 2
+      | c = simplification
+}}
+{{eqn | r = \cosh x
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sinh x}
+      | r = \lim_{h \mathop \to 0} \frac {\map \sinh {x + h} - \sinh x} h
+      | c = {{Defof|Derivative}}
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cosh {\frac {x + h + x} 2} \map \sinh {\frac {x + h - x} 2} } h
+      | c = [[Hyperbolic Sine minus Hyperbolic Sine]]
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {2 \map \cosh {x + \frac h 2} \map \sinh {\frac h 2} } h
+      | c = 
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\map \cosh {x + \frac h 2} \map \sinh {\frac h 2} } {\frac h 2}
+      | c = 
+}}
+{{eqn | r = \lim_{2 d \mathop \to 0} \frac {\map \cosh {x + d} \map \sinh d} d
+      | c = where $d = \dfrac h 2$
+}}
+{{eqn | r = \lim_{d \mathop \to 0} \frac {\map \cosh {x + d} \map \sinh d} d
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {\map \sinh d} d
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {e^d - e^{-d} } {2 d}
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac {e^{2 d} - 1 } {2 d e^d}
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \frac {e^{2 d} - 1} {2 d}
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \lim_{d \mathop \to 0} \frac {e^{2 d} - 1} {2 d}
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d} \lim_{2 d \mathop \to 0} \frac {e^{2 d} - 1} {2 d}
+      | c = 
+}}
+{{eqn | r = \cosh x \lim_{d \mathop \to 0} \frac 1 {e^d}
+      | c = [[Derivative of Exponential at Zero]]
+}}
+{{eqn | r = \cosh x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn 	| l = \map {\frac \d {\d x} } {\sinh x}
+	| r = -i \map {\frac \d {\d x} } {\sin i x}
+	| c = [[Hyperbolic Sine in terms of Sine]] 
+}}
+{{eqn 	| r = \cos i x 
+	| c = [[Derivative of Sine Function]] 
+}}
+{{eqn 	| r = \cosh x 
+	| c = [[Hyperbolic Cosine in terms of Cosine]] 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Open Ball is Open Set/Metric Space}
+Tags: Open Balls, Open Sets
+
+\begin{theorem}
+Let $M = \left({A, d}\right)$ be a [[Definition:Metric Space|metric space]].
+Let $x \in A$.
+Let $\epsilon \in \R_{>0}$.
+Let $B_\epsilon \left({x}\right)$ be an [[Definition:Open Ball|open $\epsilon$-ball]] of $x$ in $M$.
+Then $B_\epsilon \left({x}\right)$ is an [[Definition:Open Set (Metric Space)|open set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in B_\epsilon \left({x}\right)$.
+From [[Open Ball of Point Inside Open Ball]], there exists $\delta \in \R_{>0}$ such that $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$
+The result follows from the definition of [[Definition:Open Set (Metric Space)|open set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Cosine}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Cosine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cosh x} = \sinh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\cosh x}
+      | r = \map {\dfrac \d {\d x} } {\dfrac {e^x + e ^{-x} } 2}
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \dfrac 1 2 \map {\dfrac \d {\d x} } {e^x + e^{-x} }
+      | c = [[Derivative of Constant Multiple]]
+}}
+{{eqn | r = \dfrac 1 2 \paren {e^x + \paren {-e^{-x} } }
+      | c = [[Derivative of Exponential Function]], [[Chain Rule for Derivatives]], [[Linear Combination of Derivatives]]
+}}
+{{eqn | r = \dfrac {e^x - e^{-x} } 2
+}}
+{{eqn | r = \sinh x
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Tangent}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tanh x} = \sech^2 x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\tanh x}
+      | r = \map {\dfrac \d {\d x} } {\dfrac {\sinh x} {\cosh x} }
+      | c = {{Defof|Hyperbolic Tangent|index = 2}}
+}}
+{{eqn | r = \dfrac {\paren {\dfrac \d {\d x}  \sinh x} \cosh x - \sinh x \paren {\dfrac \d {\d x}  \cosh x} } {\cosh^2 x}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {\cosh^2 x - \sinh x \paren {\dfrac \d {\d x} \cosh x} } {\cosh^2 x}
+      | c = [[Derivative of Hyperbolic Sine]]
+}}
+{{eqn | r = \dfrac {\cosh^2 x - \sinh^2 x} {\cosh^2 x}
+      | c = [[Derivative of Hyperbolic Cosine]]
+}}
+{{eqn | r = \dfrac 1 {\cosh^2 x}
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{eqn | r = \sech^2 x
+      | c = {{Defof|Hyperbolic Secant|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Cotangent}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\coth x} = -\csch^2 x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\coth x}
+      | r = \map {\dfrac \d {\d x} } {\frac {\cosh x} {\sinh x} }
+      | c = {{Defof|Hyperbolic Cotangent|index = 2}}
+}}
+{{eqn | r = \frac {\sinh x \dfrac \d {\d x} \cosh x - \cosh x \dfrac \d {\d x} \sinh x} {\sinh^2 x}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\sinh x \sinh x - \cosh x \dfrac \d {\d x} \cosh x} {\sinh^2 x}
+      | c = [[Derivative of Hyperbolic Cosine]]
+}}
+{{eqn | r = \frac {\sinh x \sinh x - \cosh x \cosh x} {\sinh^2 x}
+      | c = [[Derivative of Hyperbolic Sine]]
+}}
+{{eqn | r = \frac {-1} {\sinh^2 x}
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{eqn | r = -\csch^2 x
+      | c = {{Defof|Hyperbolic Cosecant|index = 2}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Secant}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Secant Function, Derivative of Hyperbolic Secant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sech x} = -\sech x \tanh x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \map {\frac \d {\d x} } {\sech x}
+	| r = 2 \map {\frac \d {\d x} } {\frac {e^x} {e^{2 x} + 1} }
+	| c = {{Defof|Hyperbolic Secant}}
+}}
+{{eqn	| r = \frac 2 {\paren {e^{2 x} + 1}^2} \paren {\map {\frac \d {\d x} } {e^x} \paren {e^{2 x} + 1} - e^x \map {\frac \d {\d x} } {e^{2 x} + 1} }
+	| c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn	| r = -\frac 2 {\paren {e^{2 x} + 1}^2} \paren {2 e^{2 x} \cdot e^x - e^x \cdot e^{2 x} - e^x}
+	| c = [[Derivative of Exponential Function]]
+}}
+{{eqn	| r = -\frac {2 \paren {e^{3 x} - e^x} } {\paren {e^{2 x} + 1}^2}
+}}
+{{eqn	| r = -\frac {2 e^x} {\paren {e^{2 x} + 1} } \cdot \frac {e^{2 x} - 1} {e^{2 x} + 1} 
+}}
+{{eqn	| r = -\sech x \tanh x
+	| c = {{Defof|Hyperbolic Secant}}, {{Defof|Hyperbolic Tangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sech x}
+      | r = \map {\frac \d {\d x} } {\frac 1 {\cosh x} }
+      | c = {{Defof|Hyperbolic Secant}}
+}}
+{{eqn | r = \map {\frac \d {\d x} } {\paren {\cosh x}^{-1} }
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Laws]]
+}}
+{{eqn | r = -\paren {\cosh x}^{-2} \sinh x
+      | c = [[Derivative of Hyperbolic Cosine]], [[Power Rule for Derivatives]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {-1} {\cosh x} \frac {\sinh x} {\cosh x}
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = -\sech z \tanh z
+      | c = {{Defof|Hyperbolic Secant}} and {{Defof|Hyperbolic Tangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Hyperbolic Cosecant}
+Tags: Derivatives of Hyperbolic Functions, Hyperbolic Cosecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csch x} = -\csch x \coth x$
+\end{theorem}
+
+\begin{proof}
+It is noted that at $x = 0$, $\csch x$ is undefined.
+Hence the restriction of the domain.
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d x} } {\csch x}
+      | r = \map {\dfrac \d {\d x} } {\frac 1 {\sinh x} }
+      | c = {{Defof|Hyperbolic Cosecant}}
+}}
+{{eqn | r = \map {\dfrac \d {\d x} } {\paren {\sinh z}^{-1} }
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Laws]]
+}}
+{{eqn | r = -\paren {\sinh x}^{-2} \cosh x
+      | c = [[Derivative of Hyperbolic Cosine]], [[Power Rule for Derivatives]], [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {-1} {\sinh x} \ \frac {\cosh x} {\sinh x}
+      | c = [[Exponent Combination Laws]]
+}}
+{{eqn | r = -\csch x \coth x
+      | c = {{Defof|Hyperbolic Cosecant}} and {{Defof|Hyperbolic Cotangent}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Sine Function}
+Tags: Derivative of Inverse Hyperbolic Sine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sinh^{-1} u} = \dfrac 1 {\sqrt {1 + u^2} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sinh^{-1} u}
+      | r = \map {\frac \d {\d u} } {\sinh^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {1 + u^2} } \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosine Function}
+Tags: Derivative of Inverse Hyperbolic Cosine
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cosh^{-1} u} = \dfrac 1 {\sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$ 
+where $u > 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cosh^{-1} u}
+      | r = \map {\frac \d {\d u} } {\cosh^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {u^2 - 1} } \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Tangent Function}
+Tags: Derivative of Inverse Hyperbolic Tangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tanh^{-1} u} = \dfrac 1 {1 - u^2} \dfrac {\d u} {\d x}$ 
+where $\size u < 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\tanh^{-1} u}
+      | r = \map {\frac \d {\d u} } {\tanh^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {1 - u^2} \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Tangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cotangent Function}
+Tags: Derivative of Inverse Hyperbolic Cotangent
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\coth^{-1} u} = \dfrac {-1} {u^2 - 1} \dfrac {\d u} {\d x}$ 
+where $\size u > 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\coth^{-1} u}
+      | r = \map {\frac \d {\d u} } {\coth^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {-1} {u^2 - 1} \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Secant Function}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Secant
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sech^{-1} u} = \dfrac {-1} {u \sqrt {1 - u^2} } \dfrac {\d u} {\d x}$ 
+where $0 < u < 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sech^{-1} u}
+      | r = \map {\frac \d {\d u} } {\sech^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {-1} {u \sqrt {1 - u^2} } \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Secant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Inverse Hyperbolic Cosecant Function}
+Tags: Derivatives of Inverse Hyperbolic Functions, Inverse Hyperbolic Cosecant
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csch^{-1} u} = \dfrac {-1} {\size u \sqrt {1 + u^2} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+[[File:Arccosech.png|600px|right]]
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\csch^{-1} u}
+      | r = \map {\frac \d {\d u} } {\csch^{-1} u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {-1} {\size u \sqrt {1 + u^2} } \frac {\d u} {\d x}
+      | c = [[Derivative of Inverse Hyperbolic Cosecant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Closed Sets is Closed/Topology}
+Tags: Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then the [[Definition:Set Intersection|intersection]] of an arbitrary number of [[Definition:Closed Set (Topology)|closed sets]] of $T$ (either [[Definition:Finite|finitely]] or [[Definition:Infinite|infinitely]] many) is itself [[Definition:Closed Set (Topology)|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $I$ be an [[Definition:Indexing Set|indexing set]] (either [[Definition:Finite|finite]] or [[Definition:Infinite|infinite]]).
+Let $\displaystyle \bigcap_{i \mathop \in I} V_i$ be the [[Definition:Set Intersection|intersection]] of a [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Closed Set (Topology)|closed sets]] of $T$ indexed by $I$.
+Then from [[De Morgan's Laws (Set Theory)/Set Difference/Family of Sets/Difference with Intersection|De Morgan's laws: Difference with Intersection]]:
+:$\displaystyle S \setminus \bigcap_{i \mathop \in I} V_i = \bigcup_{i \mathop \in I} \paren {S \setminus V_i}$
+By definition of [[Definition:Closed Set (Topology)|closed set]], each of $S \setminus V_i$ are by definition [[Definition:Open Set (Topology)|open]] in $T$.
+We have that $\displaystyle \bigcup_{i \mathop \in I} \paren {S \setminus V_i}$ is the [[Definition:Union of Family|union]] of a [[Definition:Indexed Family of Subsets|indexed family]] of [[Definition:Open Set (Topology)|open sets]] of $T$ indexed by $I$.
+Therefore, by definition of a [[Definition:Topology|topology]], $\displaystyle \bigcup_{i \mathop \in I} \paren {S \setminus V_i} = S \setminus \bigcap_{i \mathop \in I} V_i$ is likewise [[Definition:Open Set (Topology)|open]] in $T$.
+Then by definition of [[Definition:Closed Set (Topology)|closed set]], $\displaystyle \bigcap_{i \mathop \in I} V_i$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Union of Closed Sets is Closed/Topology}
+Tags: Closed Sets
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then the [[Definition:Set Union|union]] of [[Definition:Finite|finitely many]] [[Definition:Closed Set (Topology)|closed sets]] of $T$ is itself [[Definition:Closed Set (Topology)|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ be the [[Definition:Set Union|union]] of a [[Definition:Finite|finite]] number of [[Definition:Closed Set (Topology)|closed sets]] of $T$.
+Then from [[De Morgan's Laws (Set Theory)|De Morgan's laws]]:
+:$\displaystyle S \setminus \bigcup_{i \mathop = 1}^n V_i = \bigcap_{i \mathop = 1}^n \left({S \setminus V_i}\right)$
+By definition of [[Definition:Closed Set (Topology)|closed set]], each of the $S \setminus V_i$ is by definition [[Definition:Open Set (Topology)|open]] in $T$.
+We have that $\displaystyle \bigcap_{i \mathop = 1}^n \left({S \setminus V_i}\right)$ is the [[Definition:Set Intersection|intersection]] of a [[Definition:Finite|finite]] number of [[Definition:Open Set (Topology)|open sets]] of $T$.
+Therefore, by definition of a [[Definition:Topology|topology]], $\displaystyle \bigcap_{i \mathop = 1}^n \left({S \setminus V_i}\right) = S \setminus \bigcup_{i \mathop = 1}^n V_i$ is likewise [[Definition:Open Set (Topology)|open]] in $T$.
+Then by definition of [[Definition:Closed Set (Topology)|closed set]], $\displaystyle \bigcup_{i \mathop = 1}^n V_i$ is [[Definition:Closed Set (Topology)|closed]] in $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Ball is Closed/Metric Space}
+Tags: Metric Spaces, Closed Balls
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $x \in A$.
+Let $\epsilon \in \R_{>0}$.
+Let $\map {B_\epsilon^-} x$ be the [[Definition:Closed Ball|closed $\epsilon$-ball]] of $x$ in $M$.
+Then $\map {B_\epsilon^-} x$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $M$.
+\end{theorem}
+
+\begin{proof}
+We show that the [[Definition:Set Complement|complement]] $A \setminus B_\epsilon^- \left({x}\right)$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+Let $a \in A \setminus \map {B_\epsilon^-} x$.
+Then by definition of [[Definition:Closed Ball|closed ball]]:
+:$\map d {x, a} > \epsilon$
+Put:
+:$\delta := \map d {x, a} - \epsilon > 0$
+Then:
+:$\map d {x, a} - \delta = \epsilon$
+Let $b \in \map {B_\delta} a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {x, b}
+      | o = \ge 
+      | r = \map d {x, a} - \map d {a, b}
+      | c = [[Reverse Triangle Inequality]]
+}}
+{{eqn | o = >
+      | r = \map d {x, a} - \delta
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+and so:
+:$b \notin \map {B_\epsilon^-} x$
+Then:
+:$\map {B_\delta} a \subseteq A \setminus \map {B_\epsilon^-} x$
+so $A \setminus \map {B_\epsilon^-} x$ is [[Definition:Open Set (Metric Space)|open]] in $M$.
+Hence, by definition of [[Definition:Closed Set (Metric Space)|closed set]]:
+:$\map {B_\epsilon^-} x$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Closed Balls]]
+i0t1kn6zyg0qs0ijte71kxgpqtx78ez
+\end{proof}<|endoftext|>
+\section{Derivative of Sine of Function}
+Tags: Derivatives of Trigonometric Functions, Sine Function, Derivative of Sine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sin u} = \cos u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sin u}
+      | r = \map {\frac \d {\d u} } {\sin u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \cos u \frac {\d u} {\d x}
+      | c = [[Derivative of Sine Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Cosine of Function}
+Tags: Derivatives of Trigonometric Functions, Cosine Function, Derivative of Cosine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cos u} = -\sin u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cos u}
+      | r = \map {\frac \d {\d u} } {\cos u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\sin u \frac {\d u} {\d x}
+      | c = [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Tangent of Function}
+Tags: Derivatives of Trigonometric Functions, Tangent Function, Derivative of Tangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\tan u} = \sec^2 u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\tan u}
+      | r = \map {\frac \d {\d u} } {\tan u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \sec^2 u \frac {\d u} {\d x}
+      | c = [[Derivative of Tangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Cotangent of Function}
+Tags: Derivatives of Trigonometric Functions, Cotangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\cot u} = -\csc^2 u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\cot u}
+      | r = \map {\frac \d {\d u} } {\cot u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\csc^2 u \frac {\d u} {\d x}
+      | c = [[Derivative of Cotangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Secant of Function}
+Tags: Derivatives of Trigonometric Functions, Secant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\sec u} = \sec u \tan u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\sec u}
+      | r = \map {\frac \d {\d u} } {\sec u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \sec u \tan u \frac {\d u} {\d x}
+      | c = [[Derivative of Secant Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Cosecant of Function}
+Tags: Derivatives of Trigonometric Functions, Cosecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\csc u} = \csc u \cot u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\csc u}
+      | r = \map {\frac \d {\d u} } {\csc u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\csc u \cot u \frac {\d u} {\d x}
+      | c = [[Derivative of Cosecant Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of General Logarithm of Function}
+Tags: Derivatives, Logarithms
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\log_a u} = \dfrac {\log_a e} u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\log_a u}
+      | r = \map {\frac \d {\d u} } {\log_a u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {\log_a e} u \frac {\d u} {\d x}
+      | c = [[Derivative of General Logarithm Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Natural Logarithm of Function}
+Tags: Derivatives, Logarithms
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\ln u} = \dfrac 1 u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\ln u}
+      | r = \map {\frac \d {\d u} } {\ln u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 u \frac {\d u} {\d x}
+      | c = [[Derivative of Natural Logarithm Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Constant to Power of Function}
+Tags: Derivatives involving Exponential Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {a^u} = a^u \ln a \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {a^u}
+      | r = \map {\frac \d {\d u} } {a^u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = a^u \ln a \frac {\d u} {\d x}
+      | c = [[Derivative of Power of Constant]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Exponential of Function}
+Tags: Derivatives involving Exponential Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {e^u} = e^u \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {e^u}
+      | r = \map {\frac \d {\d u} } {e^u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = e^u \frac {\d u} {\d x}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arcsine of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arcsine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arcsin u} = \dfrac 1 {\sqrt {1 - u^2} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arcsin u}
+      | r = \map {\frac \d {\d u} } {\arcsin u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {1 - u^2} } \frac {\d u} {\d x}
+      | c = [[Derivative of Arcsine Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arccosine of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arccosine Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arccos u} = -\dfrac 1 {\sqrt {1 - u^2} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{:Graph of Arccosine Function}}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arccos u}
+      | r = \map {\frac \d {\d u} } {\arccos u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\dfrac 1 {\sqrt {1 - u^2} } \frac {\d u} {\d x}
+      | c = [[Derivative of Arccosine Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arctangent of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arctangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arctan u} = \dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arctan u}
+      | r = \map {\frac \d {\d u} } {\arctan u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {1 + u^2} \frac {\d u} {\d x}
+      | c = [[Derivative of Arctangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arccotangent of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arccotangent Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arccot u} = -\dfrac 1 {1 + u^2} \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arccot u}
+      | r = \map {\frac \d {\d u} } {\arccot u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\dfrac 1 {1 + u^2} \frac {\d u} {\d x}
+      | c = [[Derivative of Arccotangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arcsecant of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arcsecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arcsec u} = \dfrac 1 {\size u \sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arcsec u}
+      | r = \map {\frac \d {\d u} } {\arcsec u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac 1 {\size u \sqrt {u^2 - 1} } \frac {\d u} {\d x}
+      | c = [[Derivative of Arcsecant Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Arccosecant of Function}
+Tags: Derivatives of Inverse Trigonometric Functions, Arccosecant Function
+
+\begin{theorem}
+:$\map {\dfrac \d {\d x} } {\arccsc u} = -\dfrac 1 {\size u \sqrt {u^2 - 1} } \dfrac {\d u} {\d x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {\arccsc u}
+      | r = \map {\frac \d {\d u} } {\arccsc u} \frac {\d u} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\dfrac 1 {\size u \sqrt {u^2 - 1} } \frac {\d u} {\d x}
+      | c = [[Derivative of Arccosecant Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Even Function is Odd}
+Tags: Even Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Differentiable Real Function|differentiable real function]] such that $f$ is [[Definition:Even Function|even]].
+Then its [[Definition:Derivative|derivative]] $f'$ is an [[Definition:Odd Function|odd function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \map f {-x}
+      | c = {{Defof|Even Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac \d {\d x} \map f x
+      | r = \frac \d {\d x} \map f {-x}
+      | c = [[Definition:Differentiation|differentiating]] both sides {{WRT|Differentiation}} $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f'} x
+      | r = \map {f'} {-x} \times \paren {-1}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\map {f'} {-x}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Odd Function|odd function]].
+{{qed}}
+[[Category:Even Functions]]
+[[Category:Differential Calculus]]
+q12f604kwqnh0uz516rkh6s0hud65ih
+\end{proof}<|endoftext|>
+\section{Derivative of Odd Function is Even}
+Tags: Odd Functions, Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Differentiable Real Function|differentiable real function]] such that $f$ is [[Definition:Odd Function|odd]].
+Then its [[Definition:Derivative|derivative]] $f'$ is an [[Definition:Even Function|even function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = -\map f {-x}
+      | c = {{Defof|Odd Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac \d {\d x} \map f x
+      | r = -\frac \d {\d x} \map f {-x}
+      | c = [[Definition:Differentiation|differentiating]] both sides {{WRT|Differentiation}} $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {f'} x
+      | r = -\map {f'} {-x} \times \paren {-1}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \map {f'} {-x}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Even Function|even function]].
+{{qed}}
+[[Category:Odd Functions]]
+[[Category:Differential Calculus]]
+efatzudp5edp8236s4w6wm3n5kyic8k
+\end{proof}<|endoftext|>
+\section{Form of Prime Sierpiński Number of the First Kind}
+Tags: Sierpiński Numbers of the First Kind
+
+\begin{theorem}
+Suppose $S_n = n^n + 1$ is a [[Definition:Prime Number|prime]] [[Definition:Sierpiński Number of the First Kind|Sierpiński number of the first kind]].
+Then:
+:$n = 2^{2^k}$
+for some [[Definition:Integer|integer]] $k$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $n$ has an [[Definition:Odd Integer|odd]] [[Definition:Divisor of Integer|divisor]] $d$.
+By [[Sum of Two Odd Powers]]:
+:$\paren {n^{n/d} + 1} \divides \paren {\paren {n^{n/d}}^d + 1^d} = S_n$
+thus $S_n$ is [[Definition:Composite Number|composite]], which is a [[Definition:Contradiction|contradiction]].
+Hence $n$ has no [[Definition:Odd Integer|odd]] [[Definition:Divisor of Integer|divisors]].
+That is, $n$ is a [[Definition:Integer Power|power]] of $2$.
+Write $n = 2^m$.
+{{AimForCont}} that $m$ has an [[Definition:Odd Integer|odd]] [[Definition:Divisor of Integer|divisor]] $f$.
+By [[Sum of Two Odd Powers]]:
+:$\paren {2^{n m/f} + 1} \divides \paren {\paren {2^{n m/f}}^f + 1^f} = 2^{m n} + 1 = S_n$
+thus $S_n$ is [[Definition:Composite Number|composite]], which is a [[Definition:Contradiction|contradiction]].
+Hence $m$ has no [[Definition:Odd Integer|odd]] [[Definition:Divisor of Integer|divisors]].
+That is, $m$ is a [[Definition:Integer Power|power]] of $2$.
+Therefore we must have:
+:$n = 2^{2^k}$
+for some [[Definition:Integer|integer]] $k$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Motion of Body Falling through Air}
+Tags: Gravity, Examples of Differential Equations
+
+\begin{theorem}
+The motion of a [[Definition:Body|body]] $B$ falling through air can be described using the following [[Definition:Differential Equation|differential equation]]:
+:$m \dfrac {\d^2 y} {\d t^2} = m g - k \dfrac {d y} {d t}$
+where:
+:$m$ denotes [[Definition:Mass|mass]] of $B$
+:$y$ denotes the [[Definition:Height (Linear Measure)|height]] of $B$ from an arbitrary reference
+:$t$ denotes [[Definition:Time|time elapsed]] from an arbitrary reference
+:$g$ denotes the [[Acceleration Due to Gravity]] of $B$
+:$k$ denotes the coefficient of resistive [[Definition:Force|force]] exerted on $B$ by the air (assumed to be [[Definition:Proportional|proportional]] to the [[Definition:Speed|speed]] of $B$)
+\end{theorem}
+
+\begin{proof}
+From [[Newton's Second Law of Motion]], the [[Definition:Force|force]] on $B$ equals its [[Definition:Mass|mass]] multiplied by its [[Definition:Acceleration|acceleration]].
+Thus the [[Definition:Force|force]] $F$ on $B$ is given by:
+:$F = m \dfrac {\d^2 y} {\d t^2}$
+where it is assumed that the [[Definition:Acceleration|acceleration]] is in a downward direction.
+The [[Definition:Force|force]] on $B$ due to [[Definition:Gravity|gravity]] is $m g$.
+The [[Definition:Force|force]] on $B$ due to the air it is passing through is $k$ multiplied by the [[Definition:Speed|speed]] of $B$, in the opposite direction to its travel.
+That is::
+:$k \dfrac {d y} {d t}$
+Hence the required [[Definition:Differential Equation|differential equation]]:
+:$m \dfrac {\d^2 y} {\d t^2} = m g - k \dfrac {d y} {d t}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rationals are Everywhere Dense in Reals/Normed Vector Space}
+Tags: Real Analysis, Rational Number Space, Denseness
+
+\begin{theorem}
+Let $\struct {\R, \size {\, \cdot \,}}$ be the [[Real Numbers with Absolute Value form Normed Vector Space|normed vector space of real numbers]].
+Let $\Q$ be the [[Definition:Rational Number|set of rational numbers]].
+Then $\Q$ are [[Definition:Everywhere Dense/Normed Vector Space|everywhere dense]] in $\struct {\R, \size {\, \cdot \,}}$
+\end{theorem}
+
+\begin{proof}
+We have that [[Between two Real Numbers exists Rational Number]]:
+:$\forall a, b \in \R : a < b : \exists r \in \Q : a < r < b$
+Let $a := x$ with $x \in \R$.
+Let $\epsilon \in \R_{\mathop > 0} : r - a < \epsilon$.
+Let $b := x + \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x - \epsilon
+      | o = <
+      | r = x
+}}
+{{eqn | o = <
+      | r = r
+}}
+{{eqn | o = <
+      | r = x + \epsilon
+}}
+{{end-eqn}}
+Since this holds for all $x$, we have that:
+:$\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists r \in \Q : \size {x - r} < \epsilon$
+By [[Definition:Definition|definition]], $\Q$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] in $\R$.
+\end{proof}<|endoftext|>
+\section{Number of Parameters of Autoregressive Model}
+Tags: Autoregressive Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+Let $M$ be an '''[[Definition:Autoregressive Model|autoregessive model]]''' on $S$ of order $p$:
+:$\tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
+Then $M$ has $p + 2$ [[Definition:Parameter of Autoregressive Model|parameters]].
+</onlyinclude>
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Parameter of Autoregressive Model|parameters]] of $M$:
+{{:Definition:Parameter of Autoregressive Model}}
+Thus:
+:there are $p$ [[Definition:Parameter of Autoregressive Model|parameters]] of the form $\phi_j$
+:$1$ [[Definition:Parameter of Autoregressive Model|parameter]] $\mu$
+:$1$ [[Definition:Parameter of Autoregressive Model|parameter]] $\sigma_a^2$.
+That is: $p + 1 + 1 = p + 2$ [[Definition:Parameter of Autoregressive Model|parameters]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Autoregressive Model is Special Case of Linear Filter Model}
+Tags: Autoregressive Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+Let $M$ be an '''[[Definition:Autoregressive Model|autoregessive model]]''' on $S$ of order $p$:
+:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
+Then $M$ is a special case of a [[Definition:Linear Filter|linear filter model]].
+</onlyinclude>
+\end{theorem}
+
+\begin{proof}
+We can eliminate $\tilde z_{t - 1}$ from the {{RHS}} of $(1)$ by substituting:
+:$\tilde z_{t - 1} = \phi_1 \tilde z_{t - 2} + \phi_2 \tilde z_{t - 3} + \dotsb + \phi_p \tilde z_{t - p - 1} + a_{t - 1}$
+Similarly we can substitute for $\tilde z_{t - 2}$, and so on.
+Eventually we get an [[Definition:Infinite Series|infinite series]] in $a_{t - j}$.
+Hence:
+:$\map \phi B \tilde z_t = a_t$
+is equivalent to:
+:$\tilde z_t = \map \psi B a_t$
+such that:
+:$\map \psi B = \map {\phi^{-1} } B$
+Hence the result by definition of [[Definition:Linear Filter|linear filter model]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Irrationals are Everywhere Dense in Reals/Topology}
+Tags: Real Analysis, Real Number Line with Euclidean Topology, Irrational Number Space, Denseness
+
+\begin{theorem}
+Let $T = \struct {\R, \tau}$ denote the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $\R \setminus \Q$ be the [[Definition:Irrational Number|set of irrational numbers]].
+Then $\R \setminus \Q$ is [[Definition:Everywhere Dense|everywhere dense]] in $T$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R$.
+Let $U \subseteq \R$ be an [[Definition:Open Set (Topology)|open set]] of $T$ such that $x \in U$.
+From [[Basis for Euclidean Topology on Real Number Line]], there exists an [[Definition:Open Real Interval|open interval]] $V_0 = \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V_0$.
+From [[Between two Real Numbers exists Rational Number]]:
+:$\exists p \in \Q: p \in \openint {x - \epsilon} {x + \epsilon}$
+Thus, we can define the [[Definition:Open Real Interval|open interval]] $V_1 = \openint x p \subseteq V_0$.
+Similarly:
+:$\exists q \in \Q: q \in \openint x p$
+We can then define an [[Definition:Open Real Interval|open interval]] $V_2 = \openint q p \subseteq V_1$.
+We have $V_2 \subseteq V_1$, $V_1 \subseteq V_0$ and $V_0 \subseteq U$.
+By successively applying [[Subset Relation is Transitive]], it follows that $V_2 \subseteq U$.
+Note that $x \notin V_2$, since $x < q < p < x + \epsilon$.
+From [[Between two Rational Numbers exists Irrational Number]], there exists $y \in \R \setminus \Q: y \in \openint q p = V_2$.
+As $x \notin V_2$, it must be the case that $x \ne y$.
+Since $V_2 \subseteq U$, $U$ is an [[Definition:Open Set (Topology)|open set]] of $T$ containing $x$ which also contains an element of $\R \setminus \Q$ other than $x$.
+As $U$ is arbitrary, it follows that every [[Definition:Open Set (Topology)|open set]] of $T$ containing $x$ also contains an element of $\R \setminus \Q$ other than $x$.
+That is, $x$ is by definition a [[Definition:Limit Point of Set|limit point]] of $\R \setminus \Q$.
+As $x$ is arbitrary, it follows that all elements of $\R$ are [[Definition:Limit Point of Set|limit points]] of $\R \setminus \Q$.
+The result follows from the definition of [[Definition:Everywhere Dense|everywhere dense]].
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Real Number Line with Euclidean Topology]]
+[[Category:Irrational Number Space]]
+[[Category:Denseness]]
+09xf05ij99hlu19r9bah8fmexbn8quc
+\end{proof}<|endoftext|>
+\section{Irrationals are Everywhere Dense in Reals/Normed Vector Space}
+Tags: Real Analysis, Normed Vector Spaces, Irrational Number Space, Denseness
+
+\begin{theorem}
+Let $\struct {\R, \size {\, \cdot \,}}$ be the [[Real Numbers with Absolute Value form Normed Vector Space|normed vector space of real numbers]].
+Let $\R \setminus \Q$ be the [[Definition:Irrational Number|set of irrational numbers]].
+Then $\R \setminus \Q$ are [[Definition:Everywhere Dense/Normed Vector Space|everywhere dense]] in $\struct {\R, \size {\, \cdot \,}}$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R$.
+Let $\epsilon \in \R_{\mathop > 0}$
+Either $x \in \Q$ or $x \in \R \setminus \Q$.
+Suppose $x \in \R \setminus \Q$.
+Let $y := x$.
+Then:
+:$\size {x - y} < \epsilon$
+Suppose $x \in \Q$.
+Let $\displaystyle n \in \N : n > \frac {\sqrt 2} \epsilon$
+Let $\displaystyle y := x + \frac {\sqrt 2} n$
+Then $y \in \R \setminus \Q$.
+Furthermore:
+{{begin-eqn}}
+{{eqn | l = \size {x - y}
+      | r = \size {\frac {\sqrt 2} n}
+}}
+{{eqn | o = <
+     | r = \epsilon
+}}
+{{end-eqn}}
+In both cases $x$ was arbitrary.
+Hence:
+:$\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists y \in \R \setminus \Q : \size {x - y} < \epsilon$
+By [[Definition:Definition|definition]], $\R \setminus \Q$ is [[Definition:Everywhere Dense/Normed Vector Space|dense]] in $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Subtracted from Itself leaves Zero}
+Tags: Subtraction
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then:
+:$x - x = 0$
+where $x - x$ denotes the operation of [[Definition:Real Subtraction|real subtraction]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x - x
+      | r = x + \paren {-x}
+      | c = {{Defof|Real Subtraction}}
+}}
+{{eqn | r = 0
+      | c = [[Inverses for Real Addition]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Ordering is Compatible with Multiplication/Positive Factor/Corollary}
+Tags: Real Number Ordering is Compatible with Multiplication
+
+\begin{theorem}
+:$\forall a, b, c, d \in \R: 0 < a < b \land 0 < c < d \implies a c < b d$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a < b
+      | o = \implies
+      | r = a \times c < b \times c
+      | c = [[Real Number Ordering is Compatible with Multiplication/Positive Factor|Real Number Ordering is Compatible with Multiplication: Positive Factor]] as $c > 0$
+}}
+{{eqn | l = c < d
+      | o = \implies
+      | r = b \times c < b \times d
+      | c = [[Real Number Ordering is Compatible with Multiplication/Positive Factor|Real Number Ordering is Compatible with Multiplication: Positive Factor]] as $b > 0$
+}}
+{{end-eqn}}
+The result follows by [[Real Number Ordering is Transitive]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom}
+Tags: Matroid Theory, 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}
+==== [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma|Lemma]] ====
+{{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma}}{{qed|lemma}}
+=== Definition 1 iff Definition 2 ===
+[[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 1|Definition 1]] holds {{iff}} [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 2|Definition 2]] holds follows immediately from the [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma|lemma]].
+{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3|Definition 1 iff Definition 3]] ===
+{{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 3}}{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4|Definition 1 iff Definition 4]] ===
+{{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 1 Iff Definition 4}}{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 4 Iff Definition 5|Definition 4 iff Definition 5]] ===
+{{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 4 Iff Definition 5}}{{qed|lemma}}
+=== Definition 5 iff Definition 6 ===
+[[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 5|Definition 5]] holds {{iff}} [[Leigh.Samphier/Sandbox/Definition:Base Axiom (Matroid)/Definition 6|Definition 6]] holds follows immediately from the [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma|lemma]].
+{{qed|lemma}}
+=== [[Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 3 Iff Definition 7|Definition 3 iff Definition 7]] ===
+{{:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Definition 3 Iff Definition 7}}{{qed}}
+[[Category:Matroid Theory]]
+[[Category:Equivalence of Definitions of Matroid Base Axiom]]
+b2e3wn7jv8hbi7bkec9m0257n1rjulc
+\end{proof}<|endoftext|>
+\section{Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma}
+Tags: Equivalence of Definitions of Matroid Base Axiom
+
+\begin{theorem}
+Let $B_1, B_2 \subseteq S$.
+Let $x \in B_1 \setminus B_2$.
+Let $y \in B_2 \setminus B_1$.
+Then:
+:$\paren{B_1 \setminus \set x} \cup \set y = \paren{B_1 \cup \set y} \setminus \set x$
+\end{theorem}
+
+\begin{proof}
+From [[Singleton of Element is Subset]]:
+:$\set x \subseteq B_1 \setminus B_2$
+and
+:$\set y \subseteq B_2 \setminus B_1$
+From [[Set Difference is Disjoint with Reverse]]:
+:$\paren{B_1 \setminus B_2} \cap \paren{B_2 \setminus B_1} = \O$
+From [[Subsets of Disjoint Sets are Disjoint]]:
+:$\set x \cap \set y = \O$
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren{B_1 \cup \set y} \setminus \set x
+      | r = \paren {B_1 \setminus \set x} \cup \paren {\set y \setminus \set x}
+      | c = [[Set Difference is Right Distributive over Union]]
+}}
+{{eqn | r = \paren {B_1 \setminus \set x} \cup \set y
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Equivalence of Definitions of Matroid Base Axiom]]
+qzqxo0oq4xm2gd56eptqyn4fx03vi6k
+\end{proof}<|endoftext|>
+\section{Number of Parameters of Moving Average Model}
+Tags: Moving Average Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t$ be the [[Definition:Deviation from Mean|deviation]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+Let $M$ be an '''[[Definition:Moving Average Model|moving average model]]''' on $S$ of order $q$:
+:$\tilde z_t = a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+Then $M$ has $q + 2$ [[Definition:Parameter of Moving Average Model|parameters]].
+</onlyinclude>
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Parameter of Moving Average Model|parameters]] of $M$:
+{{:Definition:Parameter of Moving Average Model}}
+Thus:
+:there are $q$ [[Definition:Parameter of Moving Average Model|parameters]] of the form $\theta_j$
+:$1$ [[Definition:Parameter of Moving Average Model|parameter]] $\mu$
+:$1$ [[Definition:Parameter of Moving Average Model|parameter]] $\sigma_a^2$.
+That is: $q + 1 + 1 = q + 2$ [[Definition:Parameter of Moving Average Model|parameters]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Parameters of ARMA Model}
+Tags: ARMA Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+Let $M$ be an '''[[Definition:ARMA Model|ARMA model]]''' on $S$ of order $p$:
+:$\tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+Then $M$ has $p + q + 2$ [[Definition:Parameter of ARMA Model|parameters]].
+</onlyinclude>
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Parameter of ARMA Model|parameters]] of $M$:
+{{:Definition:Parameter of ARMA Model}}
+Thus:
+:there are $p$ [[Definition:Parameter of ARMA Model|parameters]] of the form $\phi_i$
+:there are $q$ [[Definition:Parameter of ARMA Model|parameters]] of the form $\theta_j$
+:$1$ [[Definition:Parameter of ARMA Model|parameter]] $\mu$
+:$1$ [[Definition:Parameter of ARMA Model|parameter]] $\sigma_a^2$.
+That is: $p + q + 1 + 1 = p + q + 2$ [[Definition:Parameter of ARMA Model|parameters]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Field by Annihilator/Characteristic Zero}
+Tags: Field Theory
+
+\begin{theorem}
+Suppose that:
+:$\map {\mathrm {Ann} } F = \set 0$
+That is, the [[Definition:Annihilator of Ring|annihilator]] of $F$ consists of the [[Definition:Field Zero|zero]] only.
+Then:
+:$\Char F = 0$
+That is, the [[Definition:Characteristic of Ring|characteristic]] of $F$ is zero.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Field Zero|zero]] of $F$ be $0_F$ and the [[Definition:Unity of Field|unity]] of $F$ be $1_F$.
+By definition of [[Definition:Characteristic of Ring|characteristic]], $\Char F = 0$ {{iff}}:
+:$\not \exists n \in \Z, n > 0: \forall r \in F: n \cdot r = 0_F$
+That is, there exists no $n \in \Z, n > 0$ such that $n \cdot r = 0_F$ for all $r \in F$.
+But note that $\forall r \in F: 0 \cdot r = 0_F$ by definition of [[Definition:Integral Multiple|integral multiple]].
+{{AimForCont}} there exists a non-[[Definition:Field Zero|zero]] [[Definition:Element|element]] of $\map {\mathrm {Ann} } F$.
+From [[Non-Trivial Annihilator Contains Positive Integer]], $\map {\mathrm {Ann} } F$ must contain a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+But this would [[Definition:Contradiction|contradict]] the statement that $\Char F = 0$.
+So it follows that:
+:$\map {\mathrm {Ann} } F = \set 0 \iff \Char F = 0$
+{{qed|lemma}}
+\end{proof}<|endoftext|>
+\section{Characteristic of Field by Annihilator/Prime Characteristic}
+Tags: Field Theory
+
+\begin{theorem}
+Suppose that:
+:$\exists n \in \map {\mathrm {Ann} } F: n \ne 0$
+That is, there exists (at least one) non-zero [[Definition:Integer|integer]] in the [[Definition:Annihilator of Ring|annihilator]] of $F$.
+If this is the case, then the [[Definition:Characteristic of Ring|characteristic]] of $F$ is non-zero:
+:$\Char F = p \ne 0$
+and the [[Definition:Annihilator of Ring|annihilator]] of $F$ consists of the [[Definition:Set of Integer Multiples|set of integer multiples]] of $p$:
+:$\map {\mathrm {Ann} } F = p \Z$
+where $p$ is a [[Definition:Prime Number|prime number]].
+\end{theorem}
+
+\begin{proof}
+Let $A := \map {\mathrm {Ann} } F$.
+We are told that:
+:$\exists n \in A: n \ne 0$
+Consider the set $A^+ \set {n \in A: n > 0}$.
+From [[Non-Trivial Annihilator Contains Positive Integer]] we have that $A^+ \ne \O$.
+As $A^+ \subseteq \N$ it follows from the [[Well-Ordering Principle|well-ordering principle]] that $A^+$ has a least value $p$, say.
+{{AimForCont}} $p$ is not a [[Definition:Prime Number|prime number]].
+Then $p$ can be expressed as $p = a b$ where $1 < a, b < p$.
+{{begin-eqn}}
+{{eqn | l = 0_R
+      | r = p \cdot 1_F
+      | c = {{Defof|Annihilator of Ring}}
+}}
+{{eqn | r = \paren {a b} \cdot 1_F
+      | c = 
+}}
+{{eqn | r = \paren {a \cdot 1_F} \times \paren {b \cdot 1_F}
+      | c = [[Product of Integral Multiples]]
+}}
+{{eqn | r = \paren {a \cdot 1_F} = 0_F \lor \paren {b \cdot 1_F} = 0_F
+      | c = [[Field has no Proper Zero Divisors]]
+}}
+{{end-eqn}}
+But then either $a \in A$ or $b \in A$, and so $p$ is not the minimal [[Definition:Strictly Positive Integer|positive]] [[Definition:Element|element]] of $A$ after all.
+So from this [[Definition:Contradiction|contradiction]] it follows that $p$ is necessarily [[Definition:Prime Number|prime]].
+Next let $n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {n p} \cdot 1_F
+      | r = n \cdot \paren {p \cdot 1_F}
+      | c = [[Integral Multiple of Integral Multiple]]
+}}
+{{eqn | r = n \cdot 0_F
+      | c = as $p$ is in the [[Definition:Annihilator of Ring|annihilator]] of $F$
+}}
+{{eqn | r = 0_F
+      | c = {{Defof|Integral Multiple}}
+}}
+{{end-eqn}}
+So all [[Definition:Integer Multiple|multiples]] of $p$ are in $A$.
+Finally, suppose $k \in A$.
+By the [[Division Theorem]] $k = q p + r$ where $0 \le r < p$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0_F
+      | r = k \cdot 0_F
+      | c = {{Defof|Integral Multiple}}
+}}
+{{eqn | r = \paren {q p + r} \cdot 1_F
+      | c = 
+}}
+{{eqn | r = \paren {q p} \cdot 1_F + r \cdot 1_F
+      | c = [[Integral Multiple Distributes over Ring Addition]]
+}}
+{{eqn | r = 0_F + r \cdot 1_F
+      | c = from above: $q p$ is a multiple of $p$
+}}
+{{eqn | r = r \cdot 1_F
+      | c = {{Defof|Field Zero}}
+}}
+{{end-eqn}}
+So $r \in A$ contradicting the stipulation that $p$ is the smallest [[Definition:Strictly Positive Integer|positive]] [[Definition:Element|element]] of $A$.
+Hence all and only [[Definition:Integer Multiple|multiples]] of $p$ are in $A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Necessary Condition for Autoregressive Process to be Stationary}
+Tags: Autoregressive Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+Let $M$ be an '''[[Definition:Autoregressive Model|autoregessive model]]''' on $S$ of order $p$:
+:$\map \phi B \tilde z_t = a_t$
+where $\map \phi B := 1 - \phi_1 B - \phi_2 B^2 - \dotsb - \phi_p B^p$ is the [[Definition:Autoregressive Operator|autoregressive operator]] of order $p$.
+Consider the [[Definition:Polynomial Equation|polynomial equation]] in $B$ of [[Definition:Degree of Polynomial|degree]] $p$:
+:$(1): \quad \map \phi B = 0$
+Let $\map R \phi \subseteq \C$ denote the [[Definition:Set|set]] of [[Definition:Root of Polynomial|roots]] of $(1)$, considered as a [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $p$.
+It is noted that the [[Definition:Element|elements]] of $\map R \phi$ may be [[Definition:Real Number|real]] or [[Definition:Complex Number|complex]].
+For $S$ modelled by $M$ to be a [[Definition:Stationary Stochastic Process|stationary process]], it is [[Definition:Necessary Condition|necessary]] that the [[Definition:Element|elements]] of $\map R \phi$ have a [[Definition:Complex Modulus|complex modulus]] greater than $1$:
+:$\forall z \in \map R \phi: \size z > 1$
+\end{theorem}<|endoftext|>
+\section{ARIMA Model subsumes ARMA Model}
+Tags: ARIMA Models, ARMA Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let $M$ be an [[Definition:ARMA Model|ARMA model]] for $S$.
+Then $M$ is also an implementation of an [[Definition:ARIMA Model|ARIMA model]].
+\end{theorem}
+
+\begin{proof}
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+By definition of [[Definition:ARMA Model|ARMA model]], $M$ is implemented as:
+:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+The general [[Definition:ARIMA Model|ARIMA model]] is implemented as:
+:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+where:
+:$w_t = \nabla^d z_t$
+Letting $d = 0$ we have:
+:$w_t = z_t$
+Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{ARIMA Model subsumes Autoregressive Model}
+Tags: ARIMA Models, Autoregressive Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let $M$ be an [[Definition:Autoregressive Model|autoregressive model]] for $S$.
+Then $M$ is also an implementation of an [[Definition:ARIMA Model|ARIMA model]].
+\end{theorem}
+
+\begin{proof}
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+By definition of [[Definition:Autoregressive Model|autoregressive model]], $M$ is implemented as:
+:$(1): \quad \tilde z_t = \phi_1 \tilde z_{t - 1} + \phi_2 \tilde z_{t - 2} + \dotsb + \phi_p \tilde z_{t - p} + a_t$
+The general [[Definition:ARIMA Model|ARIMA model]] is implemented as:
+:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+where:
+:$w_t = \nabla^d z_t$
+Let $\theta_j = 0$ for all $j$.
+Letting $d = 0$ we have:
+:$w_t = z_t$
+Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{ARIMA Model subsumes Moving Average Model}
+Tags: ARIMA Models, Moving Average Models
+
+\begin{theorem}
+Let $S$ be a [[Definition:Stochastic Process|stochastic process]] based on an [[Definition:Equispaced Time Series|equispaced time series]].
+Let $M$ be a [[Definition:Moving Average Model|moving average model]] for $S$.
+Then $M$ is also an implementation of an [[Definition:ARIMA Model|ARIMA model]].
+\end{theorem}
+
+\begin{proof}
+Let the values of $S$ at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$ be $z_t, z_{t - 1}, z_{t - 2}, \dotsc$
+Let $\tilde z_t, \tilde z_{t - 1}, \tilde z_{t - 2}, \dotsc$ be [[Definition:Deviation from Mean|deviations]] from a [[Definition:Constant Mean Level|constant mean level]] $\mu$:
+:$\tilde z_t = z_t - \mu$
+Let $a_t, a_{t - 1}, a_{t - 2}, \dotsc$ be a [[Definition:Sequence|sequence]] of [[Definition:Independent Shocks|independent shocks]] at [[Definition:Timestamp of Time Series Observation|timestamps]] $t, t - 1, t - 2, \dotsc$
+By definition of [[Definition:Moving Average Model|moving average model]], $M$ is implemented as:
+:$(1): \quad \tilde z_t = a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+The general [[Definition:ARIMA Model|ARIMA model]] is implemented as:
+:$w_t = \phi_1 w_{t - 1} + \phi_2 w_{t - 2} + \dotsb + \phi_p w_{t - p} + a_t - \theta_1 a_{t - 1} - \theta_2 a_{t - 2} - \dotsb - \theta_q a_{t - q}$
+where:
+:$w_t = \nabla^d z_t$
+Let $\phi_i = 0$ for all $i$.
+Letting $d = 0$ we have:
+:$w_t = z_t$
+Setting $w_t = z_t - \mu = \tilde z$, we recover $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero of Field is Unique}
+Tags: Field Theory, Zero of Field is Unique
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
+The [[Definition:Field Zero|zero]] of $F$ is unique.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Product|ring product]] less [[Definition:Ring Zero|zero]] is an [[Definition:Abelian Group|abelian group]].
+The result follows from [[Ring Zero is Unique]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $0_1$ and $0_2$ both be [[Definition:Element|elements]] of $F$ such that:
+:$\forall a \in F: a + 0_1 = a$
+:$\forall a \in F: a + 0_2 = a$
+Then:
+:$0_1 + 0_2 = 0_2$
+because $0_1$ is a [[Definition:Zero Element|zero element]]
+:$0_1 + 0_2 = 0_1$
+because $0_2$ is a [[Definition:Zero Element|zero element]]
+Hence:
+:$0_1 = 0_2$
+and the two [[Definition:Field Zero|zero elements]] are the same.
+{{qed}}
+\end{proof}
\ No newline at end of file