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+\section{Distance-Preserving Image Isometric to Domain for Metric Spaces}
+Tags: Definitions: Metric Spaces, Definitions: Isometries
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
+Then:
+:$\phi: M_1 \to \Img \phi$
+is an [[Definition:Isometry|isometry]].
+\end{theorem}
+
+\begin{proof}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]] from $M_1$ to $M_2$.
+Let $A = \Img \phi$ be the [[Definition:Image of Mapping|image]] of $\phi$.
+By [[Subspace of Metric Space is Metric Space]], $\struct {A, d_2}$ is a [[Definition:Metric Space|metric space]].
+As $\phi$ is a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]], by [[Distance-Preserving Mapping is Injection of Metric Spaces]] it is [[Definition:Injection|injective]].
+From [[Restriction of Injection is Injection]], $\phi: M_1 \to \Img \phi$ is an [[Definition:Injection|injection]].
+From [[Restriction of Mapping to Image is Surjection]], $\phi: M_1 \to \Img \phi$ is a [[Definition:Surjection|surjection]].
+Thus $\phi \to \Img \phi$ is by definition a [[Definition:Bijection|bijection]].
+Thus $\phi: M_1 \to \Img \phi$ is a [[Definition:Bijection|bijective]] [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
+Hence, by definition, $\phi: M_1 \to \Img \phi$ is an [[Definition:Isometry|isometry]].
+{{qed}}
+[[Category:Definitions/Metric Spaces]]
+[[Category:Definitions/Isometries]]
+mzp3gc1t8oxve33zhyembmh0h7gwq7z
+\end{proof}<|endoftext|>
+\section{Distance-Preserving Mapping is Injection of Metric Spaces}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be [[Definition:Metric Space|metric spaces]].
+Let $\phi: M_1 \to M_2$ be a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
+Then $\phi$ is an [[Definition:Injective|injection]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in A_1$ and suppose that $\map \phi a = \map \phi b$.
+Then by the definition of a [[Definition:Metric Space|metric space]]:
+:$\map {d_2} {\map \phi a, \map \phi b} = 0$
+By the definition of a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]] then:
+:$\map {d_1} {a, b} = 0$
+Thus by the definition of a [[Definition:Metric Space|metric space]]:
+:$a = b$
+Hence $\phi$ is [[Definition:Injection|injective]].
+{{qed}}
+[[Category:Metric Spaces]]
+psyxnk22y9f84t2gcazrzrfubf3j17p
+\end{proof}<|endoftext|>
+\section{Odd Power Function is Surjective}
+Tags: Powers, Surjections, Real Functions
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be an [[Definition:Odd Integer|odd]] [[Definition:Strictly Positive Integer|positive integer]].
+Let $f_n: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map {f_n} x = x^n$
+Then $f_n$ is a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Positive Root of Positive Real Number]] we have that:
+:$\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$
+From [[Power of Ring Negative]]:
+:$\paren {-x}^n = -\paren {x^n}$
+and so:
+:$\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$
+Thus:
+:$\forall x \in \R: \exists y \in \R: y^n = x$
+and so $f_n$ is a [[Definition:Surjection|surjection]].
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+[[Category:Surjections]]
+[[Category:Real Functions]]
+0ejl6jmukqy21eb92502g3vn2r5quxu
+\end{proof}<|endoftext|>
+\section{Residue at Multiple Pole}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f: \C \to \C$ be a [[Definition:Complex Function|function]] [[Definition:Meromorphic Function|meromorphic]] on some [[Definition:Region (Complex Analysis)|region]], $D$, containing $a$. 
+Let $f$ have a single [[Definition:Pole|pole]] in $D$, of order $N$, at $a$.
+Then the [[Definition:Residue (Complex Analysis)|residue]] of $f$ at $a$ is given by:
+:$\displaystyle \Res f a = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac { \d^{N - 1} } { \d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$
+\end{theorem}
+
+\begin{proof}
+By [[Existence of Laurent Series]], there exists a [[Definition:Laurent Series|Laurent series]]: 
+:$\displaystyle \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$ 
+convergent on $D \setminus \set a$. 
+As $f$ has a pole of order $N$ at $a$, we have $c_n = 0$ for $n < -N$. 
+So: 
+:$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = -N}^\infty c_n \paren {z - a}^{n + N}$ 
+Which can be rewritten: 
+:$\displaystyle \paren {z - a}^N \map f z = \sum_{n \mathop = 0}^\infty c_{n - N} \paren {z - a}^n$
+Note that this is a [[Definition:Taylor Series|Taylor series]] with centre $a$. 
+By the [[Definition:Residue (Complex Analysis)|definition of a residue]]: 
+:$\displaystyle \Res f a = c_{-1}$
+This corresponds to the $\paren {N - 1}$th in the Taylor series of $\paren {z - a}^N \map f z$ about $a$. 
+We therefore have by [[Taylor Series of Holomorphic Function]]: 
+:$\displaystyle c_{-1} = \frac 1 {\paren {N - 1}!} \lim_{z \mathop \to a} \frac { \d^{N - 1} } { \d z^{N - 1} } \paren {\paren {z - a}^N \map f z}$
+Hence the result. 
+{{qed}}
+[[Category:Complex Analysis]]
+hsgyngvd2qagmuo7wyg4mdhcjcyobuj
+\end{proof}<|endoftext|>
+\section{Bijection from Cartesian Product of Initial Segments to Initial Segment}
+Tags: Bijections, Natural Numbers
+
+\begin{theorem}
+Let $\N_k$ be used to denote the [[Definition:Set|set]] of the first $k$ [[Definition:Non-Zero Natural Number|non-zero natural numbers]]:
+:$\N_k := \set {1, 2, \ldots, k}$
+Then a [[Definition:Bijection|bijection]] can be established between $\N_k \times \N_l$ and $\N_{k l}$, where $\N_k \times \N_l$ denotes the [[Definition:Cartesian Product|Cartesian product]] of $\N_k$ and $\N_l$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi: \N_k \times \N_l \to \N_{k l}$ be defined as:
+:$\forall \tuple {m, n} \in \N_k \times \N_l: \map \phi {m, n} = \paren {m - 1} \times l + n$
+First it is confirmed that the [[Definition:Codomain of Mapping|codomain]] of $\phi$ is indeed $\N_{k l}$.
+{{finish|fiddly and tedious, can't think of an elegant way to prove it}}
+\end{proof}<|endoftext|>
+\section{Bijection between S x T and T x S}
+Tags: Cartesian Product
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$.
+Then there exists a [[Definition:Bijection|bijection]] from $S \times T$ to $T \times S$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi: S \times T \to T \times S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {s, t} \in S \times T: \map \phi {s, t} = \tuple {t, s}$
+Then $\phi$ is the [[Definition:Bijection|bijection]] required, as follows:
+The [[Definition:Domain of Mapping|domain]] of $\phi$ is $S \times T$.
+Let $\tuple {t, s} \in T \times S$.
+Then there exists $\tuple {s, t} \in S \times T$ such that $\map \phi {s, t} = \tuple {t, s}$.
+Thus $\phi$ is a [[Definition:Surjection|surjection]].
+Let $\map \phi {s_1, t_1} = \map \phi {s_2, t_2}$ for some $\tuple {s_1, t_1}$ and $\tuple {s_2, t_2}$ in $S \times T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {s_1, t_1}
+      | r = \map \phi {s_2, t_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {t_1, s_1}
+      | r = \tuple {t_2, s_2}
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {s_1, t_1}
+      | r = \tuple {s_2, t_2}
+      | c = {{Defof|Ordered Pair}}
+}}
+{{end-eqn}}
+and so $\phi$ is an [[Definition:Injection|injection]].
+Hence the result by definition of [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection between R x (S x T) and (R x S) x T}
+Tags: Cartesian Product
+
+\begin{theorem}
+Let $R$, $S$ and $T$ be [[Definition:Set|sets]].
+Let $S \times T$ be the [[Definition:Cartesian Product|Cartesian product]] of $S$ and $T$.
+Then there exists a [[Definition:Bijection|bijection]] from $R \times \paren {S \times T}$ to $\paren {R \times S} \times T$.
+Hence:
+:$\card {R \times \paren {S \times T} } = \card {\paren {R \times S} \times T}$
+\end{theorem}
+
+\begin{proof}
+Let $\phi: R \times \paren {S \times T} \to \paren {R \times S} \times T$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}: \map \phi {s, t} = \tuple {\tuple {r, s}, t}$
+Then $\phi$ is the [[Definition:Bijection|bijection]] required, as follows:
+The [[Definition:Domain of Mapping|domain]] of $\phi$ is $R \times \paren {S \times T}$.
+Let $\tuple {\tuple {r, s}, t} \in \paren {R \times S} \times T$.
+Then there exists $\tuple {r, \tuple {s, t} } \in R \times \paren {S \times T}$ such that $\map \phi {r, \tuple {s, t} } = \tuple {\tuple {r, s}, t}$.
+Thus $\phi$ is a [[Definition:Surjection|surjection]].
+Let $\map \phi {r_1, \tuple {s_1, t_1} } = \map \phi {r_2, \tuple {s_2, t_2} }$ for some $\tuple {r_1, \tuple {s_1, t_1} }$ and $\tuple {r_2, \tuple {s_2, t_2} }$ in $R \times \paren {S \times T}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {r_1, \tuple {s_1, t_1} }
+      | r = \map \phi {r_2, \tuple {s_2, t_2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {\tuple {r_1, s_1}, t_1}
+      | r = \tuple {\tuple {r_2, s_2}, t_2}
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {r_1, \tuple {s_1, t_1} }
+      | r = \tuple {r_2, \tuple {s_2, t_2} }
+      | c = {{Defof|Ordered Pair}}
+}}
+{{end-eqn}}
+and so $\phi$ is an [[Definition:Injection|injection]].
+Hence the result by definition of [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection between Power Set of Disjoint Union and Cartesian Product of Power Sets}
+Tags: Power Set, Set Union, Cartesian Product
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Disjoint Sets|disjoint sets]].
+Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$.
+Then there exists a [[Definition:Bijection|bijection]] between $\powerset {S \cup T}$ and $\paren {\powerset S} \times \paren {\powerset T}$.
+Hence:
+:$\powerset {S \cup T} \sim \paren {\powerset S} \times \paren {\powerset T}$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi: \paren {\powerset S} \times \paren {\powerset T} \to  \powerset {S \cup T}$ be defined as:
+:$\forall \tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}: \map \phi {A, B} = A \cup B$
+In order to show that $\phi$ is a [[Definition:Bijection|bijection]], it needs to be shown that $\phi$ has the following [[Definition:Property|properties]]:
+:$\phi$ is a [[Definition:Mapping|mapping]], that is:
+::$\phi$ is [[Definition:Left-Total Relation|left-total]]
+::$\phi$ is [[Definition:Many-to-One Relation|many-to-one]]
+:$\phi$ is a [[Definition:Surjection|surjection]]
+:$\phi$ is an [[Definition:Injection|injection]].
+Let $A \subseteq S$ and $B \subseteq T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = A
+      | o = \in
+      | r = \powerset S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | lo= \land
+      | l = B
+      | o = \in
+      | r = \powerset T
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {A, B}
+      | o = \in
+      | r = \paren {\powerset S} \times \paren {\powerset T}
+      | c = {{Defof|Cartesian Product}}
+}}
+{{end-eqn}}
+Thus:
+:$\forall A \subseteq S, B \subseteq T: \tuple {A, B} \in \Dom \phi$
+and so $\phi$ is [[Definition:Left-Total Relation|left-total]].
+{{qed|lemma}}
+Let $A_1, A_2 \subseteq S$ and $B_1, B_2 \subseteq T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {A_1, B_1}
+      | o = \ne
+      | r = \map \phi {A_2, B_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_1 \cup B_1
+      | o = \ne
+      | r = A_2 \cup B_2
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = A_1 \cup B_1
+      | o = \nsubseteq
+      | r = A_2 \cup B_2
+      | c = {{Defof|Set Equality}}
+}}
+{{eqn | lo= \lor
+      | l = A_2 \cup B_2
+      | o = \nsubseteq
+      | r = A_1 \cup B_1
+      | c = 
+}}
+{{end-eqn}}
+{{WLOG}}, suppose $A_1 \cup B_1 \nsubseteq A_2 \cup B_2$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \exists x \in A_1 \cup B_1:
+      | l = x
+      | o = \notin
+      | r = A_2 \cup B_2
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | ll= x \in A_1 \lor x \in B_1:
+      | l = x \notin A_2
+      | o = \land
+      | r = x \notin B_2
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = A_1 \ne A_2
+      | o = \lor
+      | r = A_1 \ne B_2
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = B_1 \ne A_2
+      | o = \lor
+      | r = B_1 \ne B_2
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = \tuple {A_1, B_1}
+      | o = \ne
+      | r = \tuple {A_2, B_2}
+      | c = 
+}}
+{{end-eqn}}
+That is:
+:$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2} \implies \tuple {A_1, B_1} \ne \tuple {A_2, B_2}$
+demonstrating that $\phi$ is [[Definition:Many-to-One Relation|many-to-one]].
+Thus $\phi$ has been shown to be a [[Definition:Mapping|mapping]].
+{{qed|lemma}}
+Let $A \cup B \in \powerset {S \cup T}$.
+Then:
+:$\tuple {A, B} \in \paren {\powerset S} \times \paren {\powerset T}$
+by definition of [[Definition:Power Set|power set]] and [[Definition:Cartesian Product|Cartesian product]].
+That is, $\phi$ is a [[Definition:Surjection|surjection]] by definition.
+{{qed|lemma}}
+It remains to be shown that $\phi$ is an [[Definition:Injection|injection]].
+Let $\tuple {A_1, B_1} \ne \tuple {A_2, B_2}$.
+Then either $A_1 \ne A_2$ or $B_1 \ne B_2$.
+That is, at least one of the following holds:
+{{begin-eqn}}
+{{eqn | l = A_1
+      | o = \nsubseteq
+      | r = A_2
+}}
+{{eqn | l = A_2
+      | o = \nsubseteq
+      | r = A_1
+}}
+{{eqn | l = B_1
+      | o = \nsubseteq
+      | r = B_2
+}}
+{{eqn | l = B_2
+      | o = \nsubseteq
+      | r = B_1
+}}
+{{end-eqn}}
+{{WLOG}}, suppose $A_1 \nsubseteq A_2$.
+Then:
+:$\exists x \in A_1: x \notin A_2$
+and so by [[Set is Subset of Union]]:
+:$x \in A_1 \cup B_1$
+But we have that $S$ and $T$ are [[Definition:Disjoint Sets|disjoint]].
+That is:
+:$x \in A_1 \implies x \notin B_2$
+Thus:
+:$\exists x \in A_1 \cup B_1: x \notin A_2 \cup B_2$
+and so:
+:$A_1 \cup B_1 \ne A_2 \cup B_2$
+That is:
+:$\map \phi {A_1, B_1} \ne \map \phi {A_2, B_2}$
+demonstrating that $\phi$ is an [[Definition:Injection|injection]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bijection between Power Set of nth Initial Section and Initial Section of nth Power of 2}
+Tags: Power Set, Integer Powers
+
+\begin{theorem}
+Let $\N_n$ be used to denote the [[Definition:Initial Segment of One-Based Natural Numbers|first $n$ non-zero natural numbers]]:
+:$\N_n = \set {1, 2, \ldots, n}$
+Then there exists a [[Definition:Bijection|bijection]] between the [[Definition:Power Set|power set]] of $\N_n$ and $\N_{2^n}$.
+\end{theorem}
+
+\begin{proof}
+Let $\phi: \powerset {\N_n} \to \N_{2^n}$ be defined as:
+:$\forall A \in \powerset {\N_n}: \map \phi A = \begin{cases} \displaystyle \sum_{k \mathop \in A} 2^{k - 1} & : A \ne \O \\ 2^k & : A = \O \end{cases}$
+Apart from $\O$, every $A \in \powerset {\N_n}$ consists of a [[Definition:Set|set]] of [[Definition:Integer|integers]] between $1$ and $n$.
+The expression $\displaystyle \sum_{k \mathop \in A} 2^{k - 1}$ is the [[Definition:Summation|summation]] of a [[Definition:Set|set]] of [[Definition:Integer Power|powers of $2$]] between $2^0$ and $2^{n - 1}$.
+Hence $A$ is seen to be [[Definition:Left-Total Relation|left-total]] and [[Definition:Many-to-One Relation|many-to-one]] and so is a [[Definition:Mapping|mapping]].
+By the [[Basis Representation Theorem]], every [[Definition:Integer|integer]] between $1$ and $2^n - 1$ can be expressed uniquely as the [[Definition:Sum (Addition)|sum]] of [[Definition:Integer Power|powers of $2$]] between $2^0$ and $2^{n - 1}$.
+The final [[Definition:Integer|integer]] $2^n$ is mapped to from $\O$.
+Thus it is seen that $\phi$ is both an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Group is Group}
+Tags: Trivial Group
+
+\begin{theorem}
+The [[Definition:Trivial Group|trivial group]] is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Let $G = \struct {\set e, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+=== G0: Closure ===
+For $G$ to be a [[Definition:Group|group]], it must be [[Definition:Closed Algebraic Structure|closed]].
+So it must be the case that:
+:$\forall e \in G: e \circ e = e$
+{{qed|lemma}}
+=== G1: Associativity ===
+$\circ$ is [[Definition:Associative|associative]]:
+:$e \circ \paren {e \circ e} = e = \paren {e \circ e} \circ e$
+trivially.
+{{qed|lemma}}
+=== G2: Identity Element ===
+$e$ is the [[Definition:Identity Element|identity]]:
+:$\forall e \in G: e \circ e = e$
+{{qed|lemma}}
+=== G3: Inverse Elements ===
+Every [[Definition:Element|element]] of $G$ (all one of them) has an [[Definition:Inverse Element|inverse]]:
+This follows from the fact that the [[Identity is Self-Inverse]], and the only [[Definition:Element|element]] of $G$ is indeed the [[Definition:Identity Element|identity]]:
+:$e \circ e = e \implies e^{-1} = e$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetry Group of Line Segment is Group}
+Tags: Symmetry Group of Line Segment
+
+\begin{theorem}
+The [[Definition:Symmetry Group of Line Segment|symmetry group of the line segment]] is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Let us refer to this group as $D_1$.
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== G0: Closure ===
+From the [[Definition:Cayley Table|Cayley table]] it is seen directly that $D_1$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== G1: Associativity ===
+[[Composition of Mappings is Associative]].
+{{qed|lemma}}
+=== G2: Identity ===
+The [[Definition:Identity Element|identity]] is $e$.
+{{qed|lemma}}
+=== G3: Inverses ===
+Each element is seen to be [[Definition:Self-Inverse Element|self-inverse]]:
+:$r^{-1} = r$
+{{qed|lemma}}
+No more need be done. $D_1$ is seen to be a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Dihedral Group}
+Tags: Dihedral Groups
+
+\begin{theorem}
+The [[Definition:Dihedral Group|dihedral group]] $D_n$ is of [[Definition:Order of Structure|order]] $2 n$.
+\end{theorem}
+
+\begin{proof}
+By definition, $D_n$ is the [[Definition:Symmetry Group|symmetry group]] of the [[Definition:Regular Polygon|regular polygon]] of $n$ [[Definition:Side of Polygon|sides]].
+:[[File:SymmetryGroupOddPolygon.png|500px]] [[File:SymmetryGroupEvenPolygon.png|500px]]
+Let $P$ be a [[Definition:Regular Polygon|regular $n$-gon]].
+By inspection, it is seen that:
+:$(1): \quad$ there are $n$ [[Definition:Symmetry Mapping|symmetries]] of the [[Definition:Vertex of Polygon|vertices]] of $P$ by [[Definition:Plane Rotation|rotation]]
+:$(2): \quad$ there are a further $n$ [[Definition:Symmetry Mapping|symmetries]] of the [[Definition:Vertex of Polygon|vertices]] of $P$ by [[Definition:Plane Rotation|rotation]] after [[Definition:Plane Reflection|reflected]] in any of the [[Definition:Axis of Symmetry|axes of symmetry]] of $P$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Klein Four-Group is Group}
+Tags: Klein Four-Group
+
+\begin{theorem}
+The [[Definition:Klein Four-Group|Klein $4$-group]] $K_4$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+From [[Klein Four-Group as Subgroup of S4|Klein Four-Group as Subgroup of $S_4$]] it is demonstrated that $K_4$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group|$4$th symmetric group]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dihedral Group is Non-Abelian}
+Tags: Dihedral Groups
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n > 2$.
+Let $D_n$ denote the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$.
+Then $D_n$ is not [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+From [[Group of Order less than 6 is Abelian]] we have that $D_1$ and $D_2$ are [[Definition:Abelian Group|abelian]], which is why the condition on $n$.
+From [[Group Presentation of Dihedral Group]] we have:
+:$\beta \alpha = \alpha^{n - 1} \beta$
+for some $\alpha, \beta \in D_n$ such that $\alpha \ne \beta$.
+We also have:
+:$\alpha^n = e$
+But if $D_n$ were [[Definition:Abelian Group|abelian]], that would mean:
+:$\beta \alpha = \beta \alpha^{n - 1}$
+and so by the [[Cancellation Laws]]:
+:$\alpha = \alpha^{n - 1}$
+and so:
+:$\alpha^2 = e$
+But by definition of the [[Definition:Dihedral Group|dihedral group]], this is not the case.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Entrywise Addition is Commutative}
+Tags: Matrix Entrywise Addition, Commutativity, Matrix Entrywise Addition is Commutative
+
+\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:Commutative Operation|commutative]] on $\map \MM {m, n}$.
+That is:
+:$\mathbf A + \mathbf B = \mathbf B + \mathbf A$
+for all $\mathbf A$ and $\mathbf B$ in $\map \MM {m, n}$.
+\end{theorem}
+
+\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]].
+Hence we can apply [[Matrix Entrywise Addition over Ring is Commutative]].
+{{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]].
+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 [[Commutativity of Hadamard Product]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n}$ and $\mathbf B = \sqbrk b_{m n}$ be [[Definition:Matrix|matrices]] whose [[Definition:Order of Matrix|order]] is $m \times n$.
+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}}
+}}
+{{eqn | r = \sqbrk {b + a}_{m n}
+      | c = [[Commutative Law of Addition]]
+}}
+{{eqn | r = \sqbrk b_{m n} + \sqbrk a_{m n}
+      | c = {{Defof|Matrix Entrywise Addition}}
+}}
+{{eqn | r = \mathbf B + \mathbf A
+      | c = Definition of $\mathbf A$ and $\mathbf B$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Special Linear Group is not Abelian}
+Tags: Special Linear Group
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_K$ and [[Definition:Unity of Field|unity]] is $1_K$.
+Let $\SL {n, K}$ be the [[Definition:Special Linear Group|special linear group of order $n$ over $K$]].
+Then $\SL {n, K}$ is not an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From [[Special Linear Group is Subgroup of General Linear Group]] we have that $\SL {n, K}$ is a [[Definition:Group|group]].
+From [[Matrix Multiplication is not Commutative]] it follows that $\SL {n, K}$ is not [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation Formula (Complex Analysis)}
+Tags: Complex Analysis
+
+\begin{theorem}
+:$\displaystyle \sum_{n \in \Z \setminus X} \map f n = - \sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$
+\end{theorem}
+
+\begin{proof}
+By [[Summation Formula (Complex Analysis)/Lemma|Summation Formula: Lemma]], there exists a constant $A$ such that: 
+:$\displaystyle \cmod {\cot \paren {\pi z} } < A$ 
+for all $z$ on $C_N$. 
+Let $X_N$ be the poles of $f$ contained in $C_N$. 
+From [[Poles of Cotangent Function]], $\cot \paren {\pi z}$ has poles at $z \in \Z$. 
+Let $A_N = \set {n \in \Z : -N \le n \le N}$
+We then have: 
+{{begin-eqn}}
+{{eqn	| l = \oint_{C_N} \pi \cot \paren {\pi z} \map f z \rd z
+	| r = 2 \pi i \sum_{z_0 \mathop \in X_N \cap A_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}
+	| c = [[Residue Theorem]] 
+}}
+{{eqn	| r = 2 \pi i \paren {\sum_{n \mathop \in A_N \setminus X_N} \Res {\pi \cot \paren {\pi z} \map f z} n + \sum_{z_0 \in X_N} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} }
+}}
+{{end-eqn}}
+We then have: 
+{{begin-eqn}}
+{{eqn	| l = \Res {\pi \cot \paren {\pi z} \map f z} n
+	| r = \lim_{z \mathop \to n} \paren {\paren {z - n} \pi \cot \paren {\pi z} \map f z}
+	| c = [[Residue at Simple Pole]]
+}}
+{{eqn	| r = \map f n \lim_{z_0 \mathop \to n} \paren {\frac {z - n} z + 2 \sum_{k \mathop = 1}^\infty \frac {z \paren {z - n} } {z^2 - k^2} }
+	| c = [[Mittag-Leffler Expansion for Cotangent Function]]
+}}
+{{eqn	| r = \map f n \cdot 2 \lim_{z \mathop \to n} \paren {\frac z {z + n} }
+}}
+{{eqn	| r = \map f n \frac {2 n} {2 n}
+}}
+{{eqn	| r = \map f n
+}}
+{{end-eqn}}
+We also have: 
+{{begin-eqn}}
+{{eqn	| l = \cmod {\oint_{C_N} \pi \cot \paren {\pi z} \map f z \rd z}
+	| o = \le
+	| r = \frac {\pi A M \paren {8 N + 4} } {N^k}
+	| c = [[Estimation Lemma]]
+}}
+{{eqn	| r = \frac {8 \pi A M} {N^{k - 1} } + \frac {4 \pi A M} {N^k}
+}}
+{{eqn	| o = \to
+	| r = 0
+	| c = as $N \to \infty$, as $k - 1 > 0$
+}}
+{{end-eqn}}
+So, taking $N \to \infty$: 
+:$\displaystyle 0 = 2 \pi i \paren {\sum_{n \mathop \in \Z \setminus X} \map f n + \sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0} }$
+which gives:
+:$\displaystyle \sum_{n \mathop = -\infty}^\infty \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \cot \paren {\pi z} \map f z} {z_0}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation Formula (Complex Analysis)/Lemma}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $C_N$ be the square with vertices $\left({N + \frac 1 2}\right) \left({\pm 1 \pm i}\right)$ for $N \in \N$. 
+Then there exists a constant $A$ independent of $N$ such that: 
+:$\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert < A$ 
+for all $z$ on $C_N$.
+\end{theorem}
+
+\begin{proof}
+Let $z = x + iy$ for real $x, y$. 
+=== Case 1: $y > \frac 1 2$ === 
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert
+	| r = \left\vert{\frac {e^{i \pi z} + e^{-i \pi z} } {e^{i \pi z} - e^{-i \pi z} } }\right\vert 
+	| c = [[Cotangent Exponential Formulation]]
+}}
+{{eqn	| r = \left\vert{\frac {e^{i \pi x - \pi y} + e^{-i \pi x + \pi y} } {e^{i \pi x - \pi y} - e^{-i \pi x + \pi y} } }\right\vert
+	| c = as $z = x + iy$
+}}
+{{eqn	| o = \le
+	| r = \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert} 
+	| c = [[Triangle Inequality]]
+}}
+{{eqn	| r = \frac {e^{-\pi y} + e^{\pi y} } {e^{\pi y} - e^{-\pi y} }
+}}
+{{eqn	| r = \frac {e^{-2 \pi y} + 1} {1 - e^{-2 \pi y} }
+	| c = multiplying through $\dfrac {e^{-\pi y} } {e^{-\pi y} }$
+}}
+{{eqn	| o = \le
+	| r = \frac {1 + e^{-\pi} } {1 - e^{-\pi} }
+}}
+{{eqn	| r = A_1
+}}
+{{end-eqn}}
+=== Case 2: $y < -\frac 1 2$ === 
+Similarly: 
+{{begin-eqn}}
+{{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert
+	| r = \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert} 
+}}
+{{eqn	| r = \frac {e^{-\pi y} + e^{\pi y} } {e^{-\pi y} - e^{\pi y} } 
+}}
+{{eqn	| r = \frac {1 + e^{2 \pi y} } {1 - e^{2 \pi y} }
+	| c = multiplying through $\dfrac {e^{\pi y} } {e^{\pi y} }$
+}}
+{{eqn	| o = \le
+	| r = \frac {1 + e^{-\pi} } {1 - e^{-\pi} }
+}}
+{{eqn	| r = A_1
+}}
+{{end-eqn}}
+=== Case 3: $-\frac 1 2 \le y \le \frac 1 2$ === 
+First consider $z = N + \frac 1 2 + iy$. 
+Then: 
+{{begin-eqn}}
+{{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert
+	| r = \left\vert{\cot \pi \left({N + \frac 1 2 + iy}\right)}\right\vert
+}}
+{{eqn	| r = \left\vert{\cot \left({\frac \pi 2 + i \pi y}\right)}\right\vert
+	| c = [[Cotangent Function is Periodic on Reals]]
+}}
+{{eqn	| r = \left\vert{\tanh \pi y}\right\vert
+	| c = [[Cotangent of Complement equals Tangent]], [[Hyperbolic Tangent in terms of Tangent]]
+}}
+{{eqn	| r = \left\vert{\tanh \frac \pi 2}\right\vert
+}}
+{{eqn	| r = A_2
+}}
+{{end-eqn}}
+Similarly in the case of $z = -N - \frac 1 2 + iy$, we have: 
+{{begin-eqn}}
+{{eqn	| l = \left\vert{\cot \left({\pi z}\right)}\right\vert
+	| r = \left\vert{\cot \pi \left({-N - \frac 1 2 + iy}\right)}\right\vert
+}}
+{{eqn	| r = \left\vert{\cot \left({-\frac 1 2 + iy}\right)}\right\vert
+	| c = [[Cotangent Function is Periodic on Reals]]
+}}
+{{eqn	| r = \left\vert{\tanh \pi y}\right\vert
+	| c = [[Cotangent of Complement equals Tangent]], [[Hyperbolic Tangent in terms of Tangent]]
+}}
+{{eqn	| r = \left\vert{\tanh \frac \pi 2}\right\vert
+}}
+{{eqn	| r = A_2
+}}
+{{end-eqn}}
+Picking $A = \max \left({A_1, A_2}\right)$ gives the desired bound. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Generating Elements of Dihedral Group}
+Tags: Dihedral Groups
+
+\begin{theorem}
+Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$.
+Let $D_n$ be defined by its [[Group Presentation of Dihedral Group|group presentation]]:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+Then for all $k \in \Z_{\ge 0}$:
+:$\beta \alpha^k = \alpha^{n - k} \beta$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $k \in \Z_{\ge 0}$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$\beta \alpha^k = \alpha^{n - k} \beta$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = \beta \alpha^0
+      | r = \beta e
+      | c = 
+}}
+{{eqn | r = e \beta
+      | c = 
+}}
+{{eqn | r = \alpha^n \beta
+      | c = 
+}}
+{{eqn | r = \alpha^{n - 0} \beta
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 0$ is seen to hold.
+=== Basis for the Induction ===
+We have:
+{{begin-eqn}}
+{{eqn | l = \beta \alpha \beta 
+      | r = \alpha^{-1}
+      | c = [[Group Presentation of Dihedral Group]]
+}}
+{{eqn | ll= \leadsto
+      | l = \beta \alpha \beta^2
+      | r = \alpha^{-1} \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \beta \alpha
+      | r = \alpha^{-1} \beta
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\beta \alpha^r = \alpha^{n - r} \beta$
+from which it is to be shown that:
+:$\beta \alpha^{r + 1} = \alpha^{n - r - 1} \beta$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \beta \alpha^{r + 1}
+      | r = \alpha^{n - r} \beta \alpha
+      | c = 
+}}
+{{eqn | r = \alpha^{n - r} \alpha^{-1} \beta
+      | c = [[Product of Generating Elements of Dihedral Group#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \alpha^{n - r - 1} \beta
+      | c = 
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall k \in \Z_{\ge 0}: \beta \alpha^k = \alpha^{n - k} \beta$
+{{qed}}
+[[Category:Dihedral Groups]]
+3x2tn8gwziutax0mwls4l9m97plr2oq
+\end{proof}<|endoftext|>
+\section{Center of Dihedral Group}
+Tags: Dihedral Groups, Centers of Groups
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$.
+Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+Let $\map Z {D_n}$ denote the [[Definition:Center of Group|center of $D_n$]].
+Then:
+:$\map Z {D_n} = \begin{cases} e & : n \text { odd} \\ \set {e, \alpha^{n / 2} } & : n \text { even} \end{cases}$
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Center of Group|center]] of $D_n$ is:
+:$\map Z {D_n} = \set {g \in D_n: g x = x g, \forall x \in D_n}$
+For $n \le 2$ we have that $\order {D_n} \le 4$ and so by [[Group of Order less than 6 is Abelian]] $D_n$ is [[Definition:Abelian Group|abelian]] for $n < 3$.
+Hence by [[Group equals Center iff Abelian]] $\map Z {D_n} = D_n$ for $n < 3$.
+So, let $n \ge 3$.
+By [[Group Presentation of Dihedral Group]]:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+From [[Product of Generating Elements of Dihedral Group]]:
+:$\beta \alpha^k = \alpha^{n - k} \beta$
+for all $k \in \Z_{\ge 0}$.
+We have that $D_n$ is [[Definition:Generator of Group|generated]] by $\alpha$ and $\beta$.
+Thus:
+:$x \in \map Z {D_n} \iff x \alpha = \alpha x \land x \beta = \beta x$
+Let $x \in \map Z {D_n}$.
+We have that $x$ can be expressed in the form:
+:$x = \alpha^i \beta^j$
+As $x \in \map Z {D_n}$, we have:
+{{begin-eqn}}
+{{eqn | l = x \alpha
+      | r = \alpha x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^i \beta^j \alpha
+      | r = \alpha^{i + 1} \beta^j
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \beta^j \alpha
+      | r = \alpha \beta^j
+      | c = 
+}}
+{{end-eqn}}
+But for $j = 1$ this means:
+{{begin-eqn}}
+{{eqn | l = \alpha \beta
+      | r = \beta \alpha
+      | c = 
+}}
+{{eqn | r = \alpha^{-1} \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha
+      | r = \alpha^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^2
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+But the [[Definition:Order of Group Element|order]] of $\alpha$ is $n$ and $n > 2$, and hence:
+:$\alpha^2 \ne e$
+So if $x \in \map Z {D_n}$ it follows that $x$ has to be in the form:
+:$x = \alpha^i$
+for some $i \in \Z_{\ge 0}$.
+Again, as $x \in \map Z {D_n}$, we have:
+{{begin-eqn}}
+{{eqn | l = x \beta
+      | r = \beta x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^i \beta
+      | r = \beta \alpha^i
+      | c = 
+}}
+{{eqn | r = \alpha^{n - i} \beta
+      | c = [[Product of Generating Elements of Dihedral Group]]
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^i
+      | r = \alpha^{n - i}
+      | c = 
+}}
+{{eqn | r = \alpha^n \alpha^{-i}
+      | c = 
+}}
+{{eqn | r = \alpha^{-i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^{2 i}
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = \divides
+      | r = 2 i
+      | c = {{Defof|Order of Group Element}}
+}}
+{{end-eqn}}
+So either $i = 0$ or $n = 2 i$, as $0 \le i \le n$.
+If $i = 0$ then $x = \alpha^0 = e$.
+If $2 i = n$ then $n$ is [[Definition:Even Integer|even]] and so:
+:$x = \alpha^{n / 2}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Additive Groups of Integer Multiples}
+Tags: Additive Groups of Integer Multiples
+
+\begin{theorem}
+Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\struct {m \Z, +}$ and $\struct {n \Z, +}$ be the corresponding [[Definition:Additive Group of Integer Multiples|additive groups of integer multiples]].
+Then:
+:$\struct {m \Z, +} \cap \struct {n \Z, +} = \struct {\lcm \set {m, n} \Z, +}$
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$m \Z = \set {x \in \Z: m \divides x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = m \Z \cap n \Z
+      | r = \set {x \in \Z: n \divides x} \cap \set {x \in \Z: m \divides x}
+      | c = 
+}}
+{{eqn | r = \set {x \in \Z: n \divides x \land m \divides x}
+      | c = 
+}}
+{{eqn | r = \set {x \in \Z: \lcm \set {m, n} \divides x}
+      | c = 
+}}
+{{eqn | r = \lcm \set {m, n} \Z
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroups of Additive Group of Integers Modulo m}
+Tags: Additive Groups of Integers Modulo m
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers|additive group of integers modulo $m$]].
+The [[Definition:Subgroup|subgroups]] of $\struct {\Z_m, +_m}$ are the [[Definition:Additive Group of Integers Modulo m|additive groups of integers modulo $k$]] where:
+:$k \divides m$
+\end{theorem}
+
+\begin{proof}
+From [[Integers Modulo m under Addition form Cyclic Group]], $\struct {\Z_m, +_m}$ is [[Definition:Cyclic Group|cyclic]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +_m}$
+From [[Subgroup of Cyclic Group is Cyclic]], $H$ is of the form $\struct {\Z_k, +_k}$ for some $k \in \Z$.
+From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], it follows that $k \divides m$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Additive Group of Integers Generated by Two Integers}
+Tags: Additive Groups of Integer Multiples
+
+\begin{theorem}
+Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\gen {m, n}$ be the [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ [[Definition:Generator of Subgroup|generated]] by $m$ and $n$.
+Then:
+:$\gen {m, n} = \struct {\gcd \set {m, n} \Z, +}$
+That is, the [[Definition:Additive Group of Integer Multiples|additive groups of integer multiples]] of $\gcd \set {m, n}$, where $\gcd \set {m, n}$ is the [[Definition:Greatest Common Divisor|greatest common divisor]] of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$\gen {m, n} = \set {x \in \Z: \gcd \set {m, n} \divides x}$
+{{Handwaving|Sorry, I would make the effort, but it's tedious.}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroups of Cartesian Product of Additive Group of Integers}
+Tags: Additive Group of Integers, Additive Groups of Integer Multiples
+
+\begin{theorem}
+Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $m, n \in \Z_{> 0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\struct {\Z \times \Z, +}$ denote the [[Definition:Cartesian Product|Cartesian product]] of $\struct {\Z, +}$ with itself.
+The [[Definition:Subgroup|subgroups]] of $\struct {\Z \times \Z, +}$ are not all of the form:
+:$\struct {m \Z, +} \times \struct {n \Z, +}$
+where $\struct {m \Z, +}$ denotes the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]] of $m$.
+\end{theorem}
+
+\begin{proof}
+Consider the map $\phi: \struct {m \Z, +} \times \struct {n \Z, +} \mapsto \struct {\Z, +} \times \struct {\Z, +}$ defined by:
+:$\forall c, d \in \Z: \map \phi {m c, n d} = \tuple {c, d}$
+which is a [[Definition:Group Isomorphism|group isomorphism]].
+{{explain|Prove the above statement}}
+Hence, $\struct {m \Z, +} \times \struct {n \Z, +}$ is a [[Definition:Free Abelian Group on Set|free abelian group]] of [[Definition:Rank (Free Abelian Group)|rank]] $2$.
+Therefore, any [[Definition:Subgroup|subgroup]] [[Definition:Generator of Subgroup|generated]] by a [[Definition:Singleton|singleton]], for example, $\set {\tuple {x, 0}: x \in \Z}$ is a [[Definition:Subgroup|subgroup]] not in the form $\struct {m \Z, +} \times \struct {n \Z, +}$.
+{{refactor|Implement "singly generated" as a definition page in its own right and then link to it, hence keep this page clean}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence Relation on Symmetric Group by Image of n is Congruence Modulo Subgroup}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]] $\set {1, \dots, n}$.
+Let $\sim$ be the [[Definition:Relation|relation on $S_n$]] defined as:
+:$\forall \pi, \tau \in S_n: \pi \sim \tau \iff \map \pi n = \map \tau n$
+Then $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] which is [[Definition:Congruence Modulo Subgroup|congruence modulo a subgroup]].
+{{explain|Work needed to be done to explain exactly what is happening here.}}
+\end{theorem}
+
+\begin{proof}
+We claim that $\sim$ is [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $S_{n - 1}$]], the [[Definition:Symmetric Group on n Letters|symmetric group on $n - 1$ letters]] $\set {1, \dots, n - 1}$.
+Notice that every element of $S_{n - 1}$ fixes $n$.
+For all $\pi, \tau \in S_n$ such that $\pi \sim \tau$:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\pi^{-1} \circ \tau} } n
+      | r = \map {\pi^{-1} } {\map \tau n}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\pi^{-1} } {\map \pi n}
+      | c = Definition of $\sim$
+}}
+{{eqn | r = \map {\paren {\pi^{-1} \circ \pi} } n
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = n
+      | c = {{Defof|Inverse Element}}
+}}
+{{end-eqn}}
+so $\pi^{-1} \circ \tau$ fixes $n$ as well.
+This shows that $\pi^{-1} \circ \tau \in S_{n - 1}$.
+By definition of [[Definition:Left Congruence Modulo Subgroup|Left Congruence Modulo Subgroup]]:
+:$\pi \equiv^l \tau \pmod {S_{n - 1} }$
+Now we show the converse.
+Suppose $\pi \equiv^l \tau \pmod {S_{n - 1} }$.
+Then $\pi^{-1} \circ \tau \in S_{n - 1}$.
+Hence $\map {\paren {\pi^{-1} \circ \tau} } n = n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \pi n
+      | r = \map \pi {\map {\paren {\pi^{-1} \circ \tau} } n}
+}}
+{{eqn | r = \map {\paren {\pi \circ \pi^{-1} \circ \tau} } n
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map \tau n
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | ll = \leadsto
+      | l = \pi
+      | o = \sim
+      | r = \tau
+}}
+{{end-eqn}}
+Therefore $\sim$ and [[Definition:Left Congruence Modulo Subgroup|left congruence modulo $S_{n - 1}$]] are equivalent.
+The fact that $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]] follows from [[Left Congruence Modulo Subgroup is Equivalence Relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Additive Group of Integers Modulo m}
+Tags: Additive Groups of Integers Modulo m
+
+\begin{theorem}
+Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+The [[Definition:Order of Group|order]] of $\struct {\Z_m, +_m}$ is $m$.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Order of Group|order]] of a [[Definition:Group|group]] is the [[Definition:Cardinality|cardinality]] of its [[Definition:Underlying Set|underlying set]].
+By definition, the [[Definition:Underlying Set|underlying set]] of $\struct {\Z_m, +_m}$ is the [[Definition:Set of Residue Classes|set of residue classes]] $\Z_m$:
+:$\Z_m = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
+From [[Cardinality of Set of Residue Classes]], $\Z_m$ has $m$ [[Definition:Element|elements]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Multiplicative Group of Reduced Residues}
+Tags: Multiplicative Groups of Reduced Residues
+
+\begin{theorem}
+Let $\struct {\Z'_m, \times_m}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]].
+The [[Definition:Order of Group|order]] of $\struct {\Z'_m, \times_m}$ is $\map \phi m$, where $\phi$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Order of Group|order]] of a [[Definition:Group|group]] is the [[Definition:Cardinality|cardinality]] of its [[Definition:Underlying Set|underlying set]].
+By definition, the [[Definition:Underlying Set|underlying set]] of $\struct {\Z'_m, \times_m}$ is the [[Definition:Reduced Residue System|reduced residue system ]] $\Z'_m$:
+:$\Z'_m = \set {\eqclass {a_1} m, \eqclass {a_2} m, \ldots, \eqclass {a_{\map \phi m} } m}$
+where:
+:$\forall k: a_k \perp m$
+From [[Cardinality of Reduced Residue System]], $\Z'_m$ has $\map \phi m$ [[Definition:Element|elements]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Circle Group is Uncountably Infinite}
+Tags: Circle Group, Uncountable Sets
+
+\begin{theorem}
+The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is an [[Definition:Uncountable Set|uncountably]] [[Definition:Infinite Group|infinite group]].
+\end{theorem}
+
+\begin{proof}
+From [[Quotient Group of Reals by Integers is Circle Group]], $\struct {K, \times}$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Quotient Group|quotient group]] of $\struct {\R, +}$ by $\struct {\Z, +}$.
+But $\dfrac {\struct {\R, +} } {\struct {\Z, +} }$ is the [[Definition:Half-Open Real Interval|half-open interval]] $\hointr 0 1$.
+A [[Definition:Real Interval|real interval]] is [[Definition:Uncountable Set|uncountable]] by (some result).
+Hence the result.
+{{qed}}
+{{MissingLinks|results in here}}
+\end{proof}<|endoftext|>
+\section{Subgroups of Symmetry Group of Regular Hexagon}
+Tags: Symmetry Group of Regular Hexagon
+
+\begin{theorem}
+Let $\mathcal H = ABCDEF$ be a [[Definition:Regular Hexagon|regular hexagon]].
+Let $D_6$ denote the [[Definition:Symmetry Group of Regular Hexagon|symmetry group of $\mathcal H$]].
+:[[File:SymmetryGroupRegularHexagon.png|520px]]
+Let $e$ denote the [[Definition:Identity Mapping|identity mapping]]
+Let $\alpha$ denote [[Definition:Plane Rotation|rotation]] of $\mathcal H$ [[Definition:Anticlockwise|anticlockwise]] through $\dfrac \pi 3$ radians ($60 \degrees$)
+Let $\beta$ denote [[Definition:Plane Reflection|reflection]] of $\mathcal H$ in the $AD$ [[Definition:Axis of Reflection|axis]].
+The [[Definition:Subset|subsets]] of $D_6$ which form its [[Definition:Subgroup|subgroups]] are as follows:
+;[[Definition:Order of Group|Order $1$]]:
+:$\set e$
+;[[Definition:Order of Group|Order $2$]]:
+:$\set {e, \alpha^3}$
+:$\set {e, \beta}$
+:$\set {e, \alpha \beta}$
+:$\set {e, \alpha^2 \beta}$
+:$\set {e, \alpha^3 \beta}$
+:$\set {e, \alpha^4 \beta}$
+:$\set {e, \alpha^5 \beta}$
+;[[Definition:Order of Group|Order $3$]]:
+:$\set {e, \alpha^2, \alpha^4}$
+;[[Definition:Order of Group|Order $4$]]:
+:$\set {e, \alpha^3, \beta, \alpha^3 \beta}$
+:$\set {e, \alpha^3, \alpha \beta, \alpha^4 \beta}$
+:$\set {e, \alpha^3, \alpha^2 \beta, \alpha^5 \beta}$
+;[[Definition:Order of Group|Order $6$]]:
+:$\set {e, \alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^5}$
+:$\set {e, \alpha^2, \alpha^4, \beta, \alpha^2 \beta, \alpha^4 \beta}$
+:$\set {e, \alpha^2, \alpha^4, \alpha \beta, \alpha^3 \beta, \alpha^5 \beta}$
+;[[Definition:Order of Group|Order $12$]]:
+:$D_6$ itself.
+\end{theorem}<|endoftext|>
+\section{Subgroup of Symmetric Group that Fixes n}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $H$ denote the [[Definition:Subgroup|subgroup]] of $S_n$ which consists of all $\pi \in S_n$ such that:
+:$\map \pi n = n$
+Then:
+:$H = S_{n - 1}$
+and the [[Definition:Index of Subgroup|index]] of $H$ in $S_n$ is given by:
+:$\index {S_n} H = n$
+\end{theorem}
+
+\begin{proof}
+We have that $S_{n - 1}$ is the [[Definition:Symmetric Group on n Letters|symmetric group on $n - 1$ letters]].
+Let $\pi \in S_{n - 1}$.
+Then $\pi$ is a [[Definition:Permutation on n Letters|permutation on $n - 1$ letters]].
+Hence $\pi$ is also a [[Definition:Permutation on n Letters|permutation on $n$ letters]] which [[Definition:Fixed Element of Permutation|fixes]] $n$.
+So $S_{n - 1} \subseteq H$.
+Now let $\pi \in H$.
+Then $\pi$ is a [[Definition:Permutation on n Letters|permutation on $n$ letters]] which [[Definition:Fixed Element of Permutation|fixes]] $n$.
+That is, $\pi$ is a [[Definition:Permutation on n Letters|permutation on $n - 1$ letters]].
+Thus $\pi \in S_{n - 1}$.
+So we have that $H = S_{n - 1}$.
+We also have that $S_{n - 1}$ is a [[Definition:Group|group]], and:
+:$\forall \rho \in S_{n - 1}: \rho \in S_n$
+So $S_{n - 1}$ is a [[Definition:Subset|subset]] of $S_n$ which is a [[Definition:Group|group]].
+Hence $S_{n - 1}$ is a [[Definition:Subgroup|subgroup]] of $S_n$ by definition.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \index {S_n} {S_{n - 1} }
+      | r = \dfrac {\order {S_n} } {\order {S_{n - 1} } }
+      | c = {{Defof|Index of Subgroup}}
+}}
+{{eqn | r = \dfrac {n!} {\paren {n - 1}!}
+      | c = [[Order of Symmetric Group]]
+}}
+{{eqn | r = \dfrac {n \paren {n - 1}!} {\paren {n - 1}!}
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Orders of Abelian Group Elements Divides LCM of Order of Product}
+Tags: Order of Group Elements, Abelian Groups
+
+\begin{theorem}
+Let $G$ be an [[Definition:Abelian Group|abelian group]].
+Let $a, b \in G$.
+Then:
+:$\order {a b} \divides \lcm \set {\order a, \order b}$
+where:
+:$\order a$ denotes the [[Definition:Order of Group Element|order]] of $a$
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]]
+:$\lcm$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]].
+\end{theorem}
+
+\begin{proof}
+Let $\order a = m, \order b = n$.
+Let $c = \lcm \set {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = c
+      | r = r m
+      | c = for some $r \in \Z$
+}}
+{{eqn | r = s n
+      | c = for some $s \in \Z$
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \paren {a b}^c
+      | r = a^c b^c
+      | c = [[Power of Product of Commuting Elements in Semigroup equals Product of Powers]]
+}}
+{{eqn | r = a^{r m} b^{s n}
+      | c = 
+}}
+{{eqn | r = \paren {a^m}^r \paren {b^n}^s
+      | c = 
+}}
+{{eqn | r = e^r e^s
+      | c = {{Defof|Order of Group Element}}
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {a b}
+      | o = \divides
+      | r = c
+      | c = [[Element to Power of Multiple of Order is Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elements of Abelian Group whose Order Divides n is Subgroup}
+Tags: Order of Group Elements, Subgroups, Abelian Groups
+
+\begin{theorem}
+Let $G$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] .
+Let $G_n$ be the [[Definition:Subset|subset]] of $G$ defined as:
+:$G_n = \set {x \in G: \order x \divides n}$
+where:
+:$\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then $G_n$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Identity is Only Group Element of Order 1]]:
+:$\order e = 1$
+and so from [[One Divides all Integers]]:
+:$\order e \divides n$
+Thus $G_n \ne \O$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = G_n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order x
+      | o = \divides
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {x^{-1} }
+      | o = \divides
+      | r = n
+      | c = [[Order of Group Element equals Order of Inverse]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1}
+      | o = \in
+      | r = G_n
+      | c = 
+}}
+{{end-eqn}}
+Let $a, b \in G_n$ such that $\order a = r, \order b = s$.
+Then from [[Product of Orders of Abelian Group Elements Divides LCM of Order of Product]]:
+:$\order {a b} = \divides \lcm \set {a, b}$
+But $r \divides n$ and $s \divides n$ by definition of $G_n$.
+Therefore, by definition of [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]:
+:$\order {a b} \divides n$
+Thus we have:
+:$G_n \ne \O$
+:$x \in G_n \implies x^{-1} \in G_n$
+:$a, b \in G_n \implies a b \in G_n$
+and the result follows by the [[Two-Step Subgroup Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 4}
+Tags: Groups of Order 4
+
+\begin{theorem}
+There exist exactly $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $4$, up to [[Definition:Group Isomorphism|isomorphism]]:
+:$C_4$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $4$
+:$K_4$, the [[Definition:Klein Four-Group|Klein $4$-group]].
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Cyclic Group of Order n]] we have that one such [[Definition:Group|group]] of [[Definition:Order of Group|order]] $4$ is the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $4$:
+This is exemplified by the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $4$]], whose [[Modulo Addition/Cayley Table/Modulo 4|Cayley table]] can be presented as:
+{{:Modulo Addition/Cayley Table/Modulo 4}}
+From [[Group whose Order equals Order of Element is Cyclic]], any [[Definition:Group|group]] with an [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $4$ is [[Definition:Cyclic Group|cyclic]].
+From [[Cyclic Groups of Same Order are Isomorphic]], no other [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $4$ which are not [[Definition:Isomorphic Groups|isomorphic]] to $C_4$ can have an [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $4$.
+{{qed|lemma}}
+From [[Order of Element Divides Order of Finite Group]], any other [[Definition:Group|group]] of [[Definition:Order of Group|order]] $4$ must have [[Definition:Element|elements]] of [[Definition:Order of Group|order]] $2$.
+We have the [[Definition:Klein Four-Group|Klein $4$-group]], whose [[Klein Four-Group/Cayley Table|Cayley table]] can be presented as:
+{{:Klein Four-Group/Cayley Table}}
+and is seen to have that property.
+{{qed|lemma}}
+We have that [[Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic]].
+{{qed|lemma}}
+It remains to be shown that the [[Definition:Klein Four-Group|Klein $4$-group]] is the only [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $4$ whose [[Definition:Element|elements]] are all of [[Definition:Order of Group|order]] $2$ (except the [[Definition:Identity Element|identity]]).
+Let the [[Definition:Cayley Table|Cayley table]] be populated as far as can be directly established:
+:$\begin{array}{c|cccc}
+  & e & a & b & c \\
+\hline
+e & e & a & b & c \\
+a & a & e &   &   \\
+b & b &   & e &   \\
+c & c &   &   & e \\
+\end{array}$
+Consider $a b$.
+As $a^2 = e$, $a b \ne e$.
+As $a e = a$, $a b \ne a$.
+As $e b = b$, $a b \ne b$.
+It follows that $a b = c$.
+Hence we have:
+:$\begin{array}{c|cccc}
+  & e & a & b & c \\
+\hline
+e & e & a & b & c \\
+a & a & e & c &   \\
+b & b &   & e &   \\
+c & c &   &   & e \\
+\end{array}$
+and the rest of the table is completed by following the result that [[Group has Latin Square Property]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 6}
+Tags: Groups of Order 6
+
+\begin{theorem}
+There exist exactly $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $6$, up to [[Definition:Group Isomorphism|isomorphism]]:
+:$C_6$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $6$
+:$S_3$, the [[Definition:Symmetric Group on n Letters|symmetric group on $3$ letters]].
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Cyclic Group of Order n]] we have that one such [[Definition:Group|group]] of [[Definition:Order of Group|order]] $6$ is $C_6$ the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $6$:
+This is exemplified by the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $6$]], whose [[Modulo Addition/Cayley Table/Modulo 6|Cayley table]] can be presented as:
+{{:Modulo Addition/Cayley Table/Modulo 6}}
+Then we have the [[Definition:Symmetric Group on n Letters|symmetric group on $3$ letters]].
+From [[Order of Symmetric Group]], this has [[Definition:Order of Group|order]] $6$.
+It can be exemplified by the [[Definition:Symmetry Group of Equilateral Triangle|symmetry group of the equilateral triangle]], whose [[Symmetry Group of Equilateral Triangle/Cayley Table|Cayley table]] can be presented as:
+{{:Symmetry Group of Equilateral Triangle/Cayley Table}}
+It remains to be shown that these are the only $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $6$.
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $6$ whose [[Definition:Identity Element|identity]] is $e$.
+By the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]].
+By the [[Third Sylow Theorem]], the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] is a [[Definition:Divisor of Integer|divisor]] of $6$.
+By the [[Fourth Sylow Theorem]], the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] is equivalent to $1 \pmod p$.
+Combining these results, this number is therefore $1$.
+Call this [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] $P$.
+By [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ is [[Definition:Normal Subgroup|normal]].
+From [[Prime Group is Cyclic]], $P = \gen x$ for some $x \in G$ for $x^3 = e$.
+By the [[First Sylow Theorem]], $G$ also has at least one [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of [[Definition:Order of Group|order]] $2$.
+Thus $G$ has an [[Definition:Element|element]] $y$ such that $y^2 = e$.
+We have that $P$ is [[Definition:Normal Subgroup|normal]].
+Therefore:
+:$y^{-1} x y \in P$
+Therefore one of the following applies:
+{{begin-eqn}}
+{{eqn | l = y^{-1} x y
+      | r = e
+}}
+{{eqn | l = y^{-1} x y
+      | r = x
+}}
+{{eqn | l = y^{-1} x y
+      | r = x^2
+}}
+{{end-eqn}}
+If $y^{-1} x y = e$ then it follows that $x = 1$, which is contrary to our deduction that $x^3 = e$.
+Hence there remain two possibilities.
+First suppose $y^{-1} x y = x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y^{-1} x y
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | r = y x
+      | c = 
+}}
+{{end-eqn}}
+Hence we can calculate the [[Definition:Power of Group Element|powers]] of $x y$ in turn:
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^2
+      | r = x \paren {y x} y
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = x \paren {x y} y
+      | c = as $x y = y x$ [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = x^2 y^2
+      | c = 
+}}
+{{eqn | n = 1
+      | r = x^2
+      | c = as $y^2 = e$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^3
+      | r = \paren {x y}^2 \paren {x y}
+      | c = 
+}}
+{{eqn | r = x^2 x y
+      | c = as $\paren {x y}^2 = x^2$ by $(1)$ above
+}}
+{{eqn | r = x^3 y
+      | c = 
+}}
+{{eqn | n = 2
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^4
+      | r = \paren {x y} \paren {x y}^3
+      | c = 
+}}
+{{eqn | r = x y y
+      | c = as $\paren {x y}^3 = y$ by $(2)$ above
+}}
+{{eqn | r = x y^2
+      | c = 
+}}
+{{eqn | n = 3
+      | r = x
+      | c = as $y^2 = e$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^5
+      | r = \paren {x y}^4 \paren {x y}
+      | c = 
+}}
+{{eqn | r = x x y
+      | c = as $\paren {x y}^4 = x$ by $(3)$ above
+}}
+{{eqn | n = 4
+      | r = x^2 y
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^6
+      | r = \paren {x y}^5 \paren {x y}
+      | c = 
+}}
+{{eqn | r = x^2 y x y
+      | c = as $\paren {x y}^5 = x^2 y$ by $(4)$ above
+}}
+{{eqn | r = x^2 y y x
+      | c = as $x y = y x$ [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = x^3
+      | c = as $y^2 = e$
+}}
+{{eqn | r = e
+      | c = as $x^3 = e$
+}}
+{{end-eqn}}
+Thus the [[Definition:Order of Group Element|order]] of $x y$ is $6$, and so $G$ is [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Group|order $6$]].
+Now suppose $y^{-1} x y = x^2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y^{-1} x y
+      | r = x^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | r = y x^{-1}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^2
+      | r = x y y x^{-1}
+      | c = 
+}}
+{{eqn | r = x x^{-1}
+      | c = as $y^2 = e$
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+It remains to investigate $x^2 y$:
+{{begin-eqn}}
+{{eqn | l = \paren {x^2 y}^2
+      | r = \paren {x y x^{-1} } \paren {x^2 y}
+      | c = as $x y = y x^{-1}$
+}}
+{{eqn | r = x y x y
+      | c = 
+}}
+{{eqn | r = e
+      | c = from above: $\paren {x y}^2 = e$
+}}
+{{end-eqn}}
+Thus we have $6$ [[Definition:Element|elements]]:
+:$e, x, x^2$ which form a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $3$
+:$y, x y, x^2 y$ all of which are [[Definition:Self-Inverse Element|self-inverse]].
+Thus in this case $G$ is the [[Definition:Symmetric Group on n Letters|symmetric group on $3$ letters]] $S_3$.
+The possibilities are exhausted.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Group whose Subsets form Nest is Cyclic P-Group}
+Tags: Cyclic Groups, P-Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $G$ be such that its [[Definition:Subgroup|subgroups]] form a [[Definition:Nest|nest]].
+Then $G$ is a [[Definition:Cyclic Group|cyclic]] [[Definition:P-Group|$p$-group]].
+\end{theorem}
+
+\begin{proof}
+Suppose $G$ is not a [[Definition:P-Group|$p$-group]].
+Then there exist two [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|primes]] $p_1, p_2$.
+By [[Cauchy's Group Theorem]], there exist [[Definition:Subgroup|subgroups]] $H, K$ such that:
+:$\order H = p_1$
+:$\order K = p_2$
+That is:
+:$H = \gen a$
+:$k = \gen b$
+for some $a, b \in G: a \ne b$, where:
+:$\order a = p_1$
+:$\order b = p_2$
+and so both $H \nsubseteq K$ and $K \nsubseteq H$
+Thus, by definition, the [[Definition:Subgroup|subgroups]] of $G$ do not form a [[Definition:Nest|nest]].
+It follows by the [[Rule of Transposition]] that $G$ is a [[Definition:P-Group|$p$-group]].
+It remains to be shown that $G$ is [[Definition:Cyclic Group|cyclic]].
+Suppose to the contrary that $G$ is not [[Definition:Cyclic Group|cyclic]].
+Then it is not the case that $G$ is [[Definition:Generator of Cyclic Group|generated]] by a single [[Definition:Element|element]] of $G$.
+Thus there exist $a, b \in G, a \ne b$ such that $H = \gen a$ and $K = \gen b$ are [[Definition:Subgroup|subgroups]] of $G$.
+But as $a \in H, b \in K, a \notin K, b \notin H$ it follows that $H \nsubseteq K$ and $K \nsubseteq H$.
+It follows by the [[Rule of Transposition]] that $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Circle Group Generated by Distinct Roots of Unity}
+Tags: Multiplicative Groups of Complex Roots of Unity, Circle Group
+
+\begin{theorem}
+Let $K$ be the [[Definition:Circle Group|circle group]].
+Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $d = \lcm \set {m, n}$ be the [[Definition:Least Common Multiple|least common multiple]] of $m$ and $n$.
+Let $\alpha$ be a [[Definition:Primitive Complex Root of Unity|primitive $n$th root of unity]].
+Let $\beta$ be a [[Definition:Primitive Complex Root of Unity|primitive $m$th root of unity]].
+Let $\gamma$ be a [[Definition:Primitive Complex Root of Unity|primitive $d$th root of unity]].
+Let $H = \gen {\alpha, \beta}$ be the [[Definition:Subgroup|subgroup]] of $K$ [[Definition:Generator of Subgroup|generated]] by $\alpha, \beta$.
+Then $H = \gen \gamma$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Complex Roots of Unity is Subgroup of Circle Group}
+Tags: Circle Group, Multiplicative Groups of Complex Roots of Unity
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 0$.
+Let $\struct {U_n, \times}$ denote the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $n$th roots of unity]].
+Let $\struct {K, \times}$ denote the [[Definition:Circle Group|circle group]].
+Then $\struct {U_n, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {K, \times}$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $n$th roots of unity]]:
+:$U_n := \set {z \in \C: z^n = 1}$
+By definition of the [[Definition:Circle Group|circle group]]:
+:$K = \set {z \in \C: \cmod z = 1}$
+By [[Modulus of Complex Root of Unity equals 1]]:
+:$\forall z \in U_n: \cmod z = 1$
+Thus:
+:$U_n \subseteq K$
+We further have that the [[Definition:Binary Operation|operation]] $\times$ on both $U_n$ and $K$ is [[Definition:Complex Multiplication|complex multiplication]].
+Finally, from [[Roots of Unity under Multiplication form Cyclic Group]], we have that $\struct {U_n, \times}$ is a [[Definition:Group|group]].
+The result follows by definition of [[Definition:Subgroup|subgroup]].
+{{qed}}
+[[Category:Circle Group]]
+[[Category:Multiplicative Groups of Complex Roots of Unity]]
+lfvp2f3yr0fc3tttjne1u8897x8k13i
+\end{proof}<|endoftext|>
+\section{Intersection of Multiplicative Groups of Complex Roots of Unity}
+Tags: Multiplicative Groups of Complex Roots of Unity, Circle Group
+
+\begin{theorem}
+Let $\struct {K, \times}$ denote the [[Definition:Circle Group|circle group]].
+Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $c = \lcm \set {m, n}$ be the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $m$ and $n$.
+Let $\struct {U_n, \times}$ denote the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $n$th roots of unity]].
+Let $\struct {U_m, \times}$ denote the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of complex $m$th roots of unity]].
+Let $H = U_m \cap U_n$.
+Then $H = U_c$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|This is (probably) a specialisation of a more general result on cyclic groups.}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Normal Subgroups is Normal}
+Tags: Group Direct Products, Normal Subgroups
+
+\begin{theorem}
+Let $G$ and $G'$ be [[Definition:Group|groups]].
+Let:
+:$H \lhd G$
+:$H' \lhd G'$
+where $\lhd$ denotes the relation of being a [[Definition:Normal Subgroup|normal subgroup]].
+Then:
+:$\paren {H \times H'} \lhd \paren {G \times G'}$
+where $H \times H'$ denotes the [[Definition:Group Direct Product|group direct product]] of $H$ and $H'$
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {x, x'} \in G \times G'$ and $\tuple {y, y'} \in H \times H'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \tuple {x, x'} \tuple {y, y'} \tuple {x, x'}^{-1}
+      | r = \tuple {x, x'} \tuple {y, y'} \tuple {x^{-1}, x'^{-1} }
+      | c = 
+}}
+{{eqn | r = \tuple {x y x^{-1}, x' y' x'^{-1} }
+      | c = 
+}}
+{{eqn | o = \in
+      | r = H \times H'
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{end-eqn}}
+Hence:
+:$\paren {H \times H'} \lhd {G \times G'}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer of Rotation in Dihedral Group}
+Tags: Dihedral Groups, Normalizers
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$.
+Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+Let $\map {N_{D_n} } {\set \alpha}$ denote the [[Definition:Normalizer|normalizer]] of the [[Definition:Singleton|singleton]] containing the [[Definition:Plane Rotation|rotation]] element $\alpha$.
+Then:
+:$\map {N_{D_n} } {\set \alpha} = \gen \alpha$
+where $\gen \alpha$ is the [[Definition:Generated Subgroup|subgroup generated]] by $\alpha$.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Normalizer|normalizer]] of $\set \alpha$ is:
+:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \set \alpha g^{-1} = \set \alpha}$
+That is:
+:$\map {N_{D_n} } {\set \alpha} := \set {g \in D_n: g \alpha = \alpha g}$
+First let $g = \alpha^k$ for some $k \in \Z$.
+Then:
+:$\alpha \alpha^k = \alpha^k \alpha$
+which includes $k = 0$, that is $e$.
+Thus:
+:$\forall k \in \Z: \alpha^k \in \map {N_{D_n} } {\set \alpha}$
+Now let $g = \alpha^k \beta$.
+Suppose $g \alpha = \alpha g$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \alpha^k \beta \alpha
+      | r = \alpha \alpha^k \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \beta \alpha
+      | r = \alpha \beta
+      | c = [[Cancellation Laws]]
+}}
+{{end-eqn}}
+But $\beta \alpha \ne \alpha \beta$ in $D_n$ except for $n < 3$.
+Hence the result:
+:$\map {N_{D_n} } {\set \alpha} = \set {e, \alpha, \alpha^2, \ldots, \alpha^{n - 1} }$
+Hence the result, by definition of [[Definition:Generator of Subgroup|generator of subgroup]].
+{{qed}}
+[[Category:Dihedral Groups]]
+[[Category:Normalizers]]
+l8t1v1hdfwid63js3d7czck2lhjh8dr
+\end{proof}<|endoftext|>
+\section{Normalizer of Reflection in Dihedral Group}
+Tags: Dihedral Groups, Normalizers
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 3$.
+Let $D_n$ be the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Structure|order]] $2 n$, given by:
+:$D_n = \gen {\alpha, \beta: \alpha^n = \beta^2 = e, \beta \alpha \beta = \alpha^{−1} }$
+Let $\map {N_{D_n} } {\set \beta}$ denote the [[Definition:Normalizer|normalizer]] of the [[Definition:Singleton|singleton]] containing the [[Definition:Plane Reflection|reflection]] element $\beta$.
+Then:
+:$\map {N_{D_n} } {\set \beta} = \begin{cases} \set {e, \beta} & : n \text { odd} \\ \set {e, \beta, \alpha^{n / 2}, \alpha^{n / 2} \beta} & : n \text { even} \end{cases}$
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Normalizer|normalizer]] of $\set \beta$ is:
+:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \set \beta g^{-1} = \set \beta}$
+That is:
+:$\map {N_{D_n} } {\set \beta} := \set {g \in D_n: g \beta = \beta g}$
+First let $g = \beta^k$ for $k \in \set {0, 1}$.
+Then:
+:$\beta \beta^k = \beta^k \beta$
+Thus:
+:$\forall k \in \set {0, 1}: \beta^k \in \map {N_{D_n} } {\set \beta}$
+Now let $g = \alpha^j \beta^k$ for $0 < j < n, k \in \set {0, 1}$.
+Suppose $g \alpha = \alpha g$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \alpha^j \beta^k \beta
+      | r = \beta \alpha^j \beta^k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^j \beta \beta^k
+      | r = \beta \alpha^j \beta^k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^j \beta
+      | r = \beta \alpha^j
+      | c = [[Cancellation Laws]]
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^j \beta
+      | r = \alpha^{n - j} \beta 
+      | c = [[Product of Generating Elements of Dihedral Group]]
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^j
+      | r = \alpha^{n - j}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha^2 j
+      | r = \alpha^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 j
+      | r = n
+      | c = 
+}}
+{{end-eqn}}
+Thus when $n$ is [[Definition:Odd Integer|odd]], there is no such $j$ such that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
+But when $n$ is [[Definition:Even Integer|even]] such that $n = 2 j$, we have that $\alpha^j \beta \in \map {N_{D_n} } {\set \beta}$.
+Hence the result.
+{{qed}}
+[[Category:Dihedral Groups]]
+[[Category:Normalizers]]
+or2apavxjj0cj0ux4v60y43iu4pql3h
+\end{proof}<|endoftext|>
+\section{Normalizer of Subgroup of Symmetric Group that Fixes n}
+Tags: Symmetric Groups, Normalizers
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $H$ denote the [[Definition:Subgroup|subgroup]] of $S_n$ which consists of all $\pi \in S_n$ such that:
+:$\map \pi n = n$
+The [[Definition:Normalizer|normalizer]] of $H$ is given by:
+:$\map {N_{S_n} } H = \map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
+\end{theorem}
+
+\begin{proof}
+We have from [[Subgroup of Symmetric Group that Fixes n]] that $N = S_{n - 1}$.
+By definition of [[Definition:Normalizer|normalizer]]:
+:$\map {N_{S_n} } {S_{n - 1} } := \set {\rho \in S_n: \rho S_{n - 1} \rho^{-1} = S_{n - 1} }$
+We have from [[Group is Normal in Itself]] that:
+:$\forall \rho \in S_{n - 1}: \rho S_{n - 1} \rho^{-1} \in S_{n - 1}$
+and so:
+:$S_{n - 1} \subseteq \map {N_{S_n} } {S_{n - 1} }$
+It remains to be shown that $\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$.
+This will be done by demonstrating that:
+: $S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
+where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+Let $\rho \in S_n$ such that $\rho \notin S_{n - 1}$.
+Thus:
+:$\map \rho n \ne n$
+and so:
+:$\exists a \in S_n, a \ne n: \map \rho a = n$
+for some $a \in S_{n - 1}$.
+Then:
+:$\map {\rho^{-1} } n = a$
+Let $\pi \in S_{n - 1}$ such that:
+: $\map \pi a = b$
+for some $b \ne a$.
+As $\rho$ is a [[Definition:Permutation on n Letters|permutation]], $\rho$ is by definition a [[Definition:Bijection|bijection]].
+Hence:
+:$\map \rho b \ne n$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {\rho \pi \rho^{-1} } n
+      | r = \map {\rho \pi} a
+      | c = as $\map {\rho^{-1} } n = a$
+}}
+{{eqn | r = \map \rho b
+      | c = as $\map \pi a = b$
+}}
+{{eqn | o = \ne
+      | r = n
+      | c = as $\map \rho b \ne n$
+}}
+{{end-eqn}}
+Thus $\rho \pi \rho^{-1}$ does not [[Definition:Fixed Element of Permutation|fix]] $n$.
+That is:
+:$\rho \pi \rho^{-1} \notin S_{n - 1}$
+Since this is the case for all arbitrary $\rho \in S_n \setminus S_{n - 1}$, it follows that:
+:$S_n \setminus S_{n - 1} \cap \map {N_{S_n} } {S_{n - 1} } = \O$
+So from [[Intersection with Complement is Empty iff Subset]]:
+:$\map {N_{S_n} } {S_{n - 1} } \subseteq S_{n - 1}$
+and so by definition of [[Definition:Set Equality|set equality]]:
+:$\map {N_{S_n} } {S_{n - 1} } = S_{n - 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Center of Quaternion Group}
+Tags: Quaternion Group, Centers of Groups
+
+\begin{theorem}
+Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the [[Definition:Quaternion Group|quaternion group]].
+Let $\map Z {\Dic 2}$ denote the [[Definition:Center of Group|center of $\Dic 2$]].
+Then:
+:$\map Z {\Dic 2} = \set {e, a^2}$
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Center of Group|center]] of $\Dic 2$ is:
+:$\map Z {\Dic 2} = \set {g \in \Dic 2: g x = x g, \forall x \in \Dic 2}$
+We are given that:
+:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
+We have that $\Dic 2$ is [[Definition:Generator of Group|generated]] by $\alpha$ and $\beta$.
+Thus:
+:$x \in \map Z {\Dic 2} \iff x a = a x \land x b = b x$
+Let $x \in \map Z {\Dic 2}$.
+We have that $x$ can be expressed in the form:
+:$x = a^i b^j$
+for $i \in \set {0, 1, 2, 3}$ and $j \in \set {0, 1}$.
+As $x \in \map Z {\Dic 2}$, we have:
+{{begin-eqn}}
+{{eqn | l = x a
+      | r = a x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^i b^j a
+      | r = a^{i + 1} b^j
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b^j a
+      | r = a b^j
+      | c = 
+}}
+{{end-eqn}}
+For $j = 1$ this means:
+{{begin-eqn}}
+{{eqn | l = a b
+      | r = b a
+      | c = 
+}}
+{{eqn | r = a^{-1} \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = a^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^2
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+But the [[Definition:Order of Group Element|order]] of $a$ is $4$, not $2$, and hence:
+:$a^2 \ne e$
+So if $x \in \map Z {\Dic 2}$ it follows that $x$ has to be in the form:
+:$x = a^i$
+for some $i \in \Z_{\ge 0}$.
+Again, as $x \in \map Z {\Dic 2}$, we have:
+{{begin-eqn}}
+{{eqn | l = x b
+      | r = b x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^i b
+      | r = b a^i
+      | c = 
+}}
+{{eqn | r = a^{-i} b
+      | c = [[Product of Generating Elements of Quaternion Group]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^i
+      | r = a^{-i}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{2 i}
+      | r = e
+      | c = 
+}}
+{{eqn | r = a^4
+      | c = {{Defof|Order of Group Element}}
+}}
+{{end-eqn}}
+So either $i = 0$ or $2 i = 4$, as $0 \le i \le 4$.
+If $i = 0$ then $x = a^0 = e$.
+If $2 i = 4$ then:
+:$x = a^{4 / 2} = a^2$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Generating Elements of Quaternion Group}
+Tags: Quaternion Group
+
+\begin{theorem}
+Let $Q = \Dic 2$ be the [[Definition:Quaternion Group|quaternion group]]:
+:$\Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$
+Then for all $k \in \Z_{\ge 0}$:
+:$b a^k = a^{-k} b$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $k \in \Z_{\ge 0}$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
+:$b a^k = a^{-k} b$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = b a^0
+      | r = b e
+      | c = 
+}}
+{{eqn | r = e b
+      | c = 
+}}
+{{eqn | r = a^{-0} b
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 0$ is seen to hold.
+=== Basis for the Induction ===
+We have:
+{{begin-eqn}}
+{{eqn | l = a b a
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b a^1
+      | r = a^{-1} b
+      | c = 
+}}
+{{end-eqn}}
+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 $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$b a^r = a^{-r} b$
+from which it is to be shown that:
+:$b a^{r + 1} = a^{-r - 1} b$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = b a^{r + 1}
+      | r = b a^r a
+      | c = 
+}}
+{{eqn | r = a^{-r} b a
+      | c = [[Product of Generating Elements of Quaternion Group#Induction Hypothesis|Product of Generating Elements of Quaternion Group]]
+}}
+{{eqn | r = a^{-r} \alpha^{-1} b
+      | c = [[Product of Generating Elements of Quaternion Group#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = a^{-r - 1} b
+      | c = 
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall k \in \Z_{\ge 0}: b a^k = a^{-k} b$
+{{qed}}
+[[Category:Quaternion Group]]
+4qzdhrhtoodtvkbxkhcfxt7ith3i7ic
+\end{proof}<|endoftext|>
+\section{Conjugacy Classes of Quaternion Group}
+Tags: Quaternion Group, Examples of Conjugacy Classes
+
+\begin{theorem}
+Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the [[Definition:Quaternion Group|quaternion group]].
+The [[Definition:Conjugacy Class|conjugacy classes]] of $\Dic 2$ are:
+:$\set e, \set {a^2}, \set {a, a^3}, \set {b, a^2 b}, \set {a b, a^3 b}$
+\end{theorem}
+
+\begin{proof}
+From [[Center of Quaternion Group]], we have:
+:$\map Z {\Dic 2} = \set {e, a^2}$
+Thus from [[Conjugacy Class of Element of Center is Singleton]], $\set e$ and $\set {a^2}$ are two of those [[Definition:Conjugacy Class|conjugacy classes]].
+By inspection of the [[Quaternion Group/Cayley Table|Cayley table]]:
+{{:Quaternion Group/Cayley Table}}
+we investigate the remaining $6$ elements in turn, starting with $a$:
+{{begin-eqn}}
+{{eqn | l = a a a^{-1}
+      | o = 
+      | rr= = a
+      | c = 
+}}
+{{eqn | l = a^2 a \paren {a^2}^{-1}
+      | r = a^2 a a^2 = a^5
+      | rr= = a
+      | c = 
+}}
+{{eqn | l = a^3 a \paren {a^3}^{-1}
+      | r = a^3 a a = a^5
+      | rr= = a
+      | c = 
+}}
+{{eqn | l = b a b^{-1}
+      | r = b a \paren {a^2 b} = b a^3 b
+      | rr= = a^3
+      | c = 
+}}
+{{eqn | l = \paren {a b} a \paren {a b}^{-1}
+      | r = \paren {a b} a \paren {a^3 b} = b a^3 b
+      | rr= = a^3
+      | c = 
+}}
+{{eqn | l = \paren {a^2 b} a \paren {a^2 b}^{-1}
+      | r = \paren {a^2 b} a b
+      | rr= = a^3
+      | c = 
+}}
+{{eqn | l = \paren {a^3 b} a \paren {a^3 b}^{-1}
+      | r = \paren {a^3 b} a \paren {a b} = \paren {a^2 b} \paren {a b}
+      | rr= = a^3
+      | c = 
+}}
+{{end-eqn}}
+So we have a [[Definition:Conjugacy Class|conjugacy class]]:
+:$\set {a, a^3}$
+Investigating the remaining $4$ elements in turn, starting with $b$:
+{{begin-eqn}}
+{{eqn | l = a b a^{-1}
+      | r = a b a^3
+      | rr= = a^2 b
+      | c = 
+}}
+{{eqn | l = a^2 b \paren {a^2}^{-1}
+      | r = a^2 b a^2
+      | rr= = b
+      | c = 
+}}
+{{eqn | l = a^3 b \paren {a^3}^{-1}
+      | r = a^3 b a
+      | rr= = a^2 b
+      | c = 
+}}
+{{eqn | l = b b b^{-1}
+      | o = 
+      | rr= = b
+      | c = 
+}}
+{{eqn | l = \paren {a b} b \paren {a b}^{-1}
+      | r = \paren {a b} b \paren {a^3 b} = \paren {a b} a^3
+      | rr= = a^2 b
+      | c = 
+}}
+{{eqn | l = \paren {a^2 b} b \paren {a^2 b}^{-1}
+      | r = \paren {a^2 b} b b = \paren {a^2 b} a^2
+      | rr= = b
+      | c = 
+}}
+{{eqn | l = \paren {a^3 b} b \paren {a^3 b}^{-1}
+      | r = \paren {a^3 b} b \paren {a b} = \paren {a^3 b} a
+      | rr= = a^2 b
+      | c = 
+}}
+{{end-eqn}}
+So we have a [[Definition:Conjugacy Class|conjugacy class]]:
+:$\set {b, a^2 b}$
+Investigating the remaining $2$ elements, starting with $a b$:
+{{begin-eqn}}
+{{eqn | l = a \paren {a b} a^{-1}
+      | r = a \paren {a b} a^3 = a^2 b a^3
+      | rr= = a^3 b
+      | c = 
+}}
+{{end-eqn}}
+We need go no further: the remaining elements $a b$ and $a^3 b$ are in the same [[Definition:Conjugacy Class|conjugacy class]]:
+:$\set {a b, a^3 b}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Action on Subsets is Group Action}
+Tags: Conjugacy Action
+
+\begin{theorem}
+Let $\powerset G$ be the set of all [[Definition:Subgroup|subgroups]] of $G$.
+For any $S \in \powerset G$ and for any $g \in G$, the [[Definition:Conjugacy Action on Subsets|conjugacy action]]:
+:$g * S := g \circ S \circ g^{-1}$
+is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+Clearly {{GroupActionAxiom|1}} is fulfilled as $e * S = S$.
+{{GroupActionAxiom|2}} is shown to be fulfilled thus:
+{{begin-eqn}}
+{{eqn | l = \paren {g_1 \circ g_2} * S
+      | r = \paren {g_1 \circ g_2} \circ S \circ \paren {g_1 \circ g_2}^{-1}
+      | c = 
+}}
+{{eqn | r = g_1 \circ \paren {g_2 \circ S \circ g_2^{-1} } \circ g_1^{-1}
+      | c = 
+}}
+{{eqn | r = g_1 * \paren {g_2 \circ S \circ g_2^{-1} }
+      | c = 
+}}
+{{eqn | r = g_1 * \paren {g_2 * S}
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Normed Vector Space Requires Multiplicative Norm on Division Ring}
+Tags: Normed Division Rings, Norm Theory
+
+\begin{theorem}
+Let $R$ be a [[Definition:Normed Division Ring|normed division ring]] with a [[Definition:Submultiplicative Norm on Ring|submultiplicative norm]] $\norm {\, \cdot \,}_R$.
+Let $V$ be a [[Definition:Vector Space|vector space]] that is not a [[Definition:Trivial Vector Space|trivial vector space]].
+Let $\norm {\, \cdot \,}: V \to \R_{\ge 0}$ be a mapping from $V$ to the [[Definition:Positive Real Number|positive real numbers]] satisfying the [[Definition:Norm Axioms (Vector Space)|vector space norm axioms]].
+Then $\norm {\, \cdot \,}_R$ is a [[Definition:Multiplicative Norm on Ring|multiplicative norm]].
+That is:
+:$\forall r, s \in R: \norm {r s}_R = \norm r_R \norm s_R$
+\end{theorem}
+
+\begin{proof}
+Since $V$ is not a [[Definition:Trivial Vector Space|trivial vector space]]:
+:$\exists \mathbf v \in V: \mathbf v \ne 0$
+By [[Definition:Norm Axioms (Vector Space)|Norm axiom (N1)]]:
+:$\norm {\mathbf v} > 0$
+Let $r, s \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm {r s}_R \norm {\mathbf v}
+      | r = \norm {\paren {r s} \mathbf v}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm axiom (N2)]]
+}}
+{{eqn | r = \norm {r \paren {s \mathbf v} }
+      | c = {{Vector-space-axiom|7}}
+}}
+{{eqn | r = \norm r_R \norm {s \mathbf v}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm axiom (N2)]]
+}}
+{{eqn | r = \norm r_R \norm s_R \norm {\mathbf v}
+      | c = [[Definition:Norm Axioms (Vector Space)|Norm axiom (N2)]]
+}}
+{{end-eqn}}
+By dividing both sides of the equation by $\norm {\mathbf v}$ then:
+:$\norm {r s}_R = \norm r_R \norm s_R$
+The result follows.
+{{qed}}
+[[Category:Normed Division Rings]]
+[[Category:Norm Theory]]
+n9j7rsaatesqxu9zjp0c8vke73w4pay
+\end{proof}<|endoftext|>
+\section{Ring with Multiplicative Norm has No Proper Zero Divisors}
+Tags: Norm Theory, Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let its [[Definition:Ring Zero|zero]] be denoted by $0_R$.
+Let $\norm {\,\cdot\,}$ be a [[Definition:Multiplicative Norm on Ring|multiplicative norm]] on $R$.
+Then $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]]. 
+That is:
+:$\forall x, y \in R^*: x \circ y \ne 0_R$
+where $R^*$ is [[Definition:Ring Less Zero|defined as $R \setminus \set {0_R}$]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}}:
+:$\exists x, y \in {R^*} : x \circ y = 0_R$
+By [[Definition:Multiplicative Norm on Ring|positive definiteness]]:
+:$x, y \ne 0_R \iff \norm x, \norm y \ne 0$
+Thus:
+:$\norm x \norm y \ne 0$
+But we also have:
+{{begin-eqn}}
+{{eqn | l = \norm x \norm y
+      | r = \norm {x \circ y}
+      | c = [[Definition:Multiplicative Norm on Ring|Multiplicativity]]
+}}
+{{eqn | r = \norm {0_R}
+      | c = by assumption
+}}
+{{eqn | r = 0
+      | c = [[Definition:Multiplicative Norm on Ring|Positive Definiteness]]
+}}
+{{end-eqn}}
+a contradiction.
+{{qed}}
+[[Category:Norm Theory]]
+[[Category:Ring Theory]]
+azlqrazgbpgo5vzoy5kkmcuepobfreg
+\end{proof}<|endoftext|>
+\section{Finite Ring with Multiplicative Norm is Field}
+Tags: Ring Theory, Field Theory, Norm Theory
+
+\begin{theorem}
+Let $R$ be a [[Definition:Finite Set|finite]] [[Definition:Ring (Abstract Algebra)|ring]] with a [[Definition:Multiplicative Norm on Ring|multiplicative norm]].
+Then $R$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring with Multiplicative Norm has No Proper Zero Divisors]], $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+From [[Finite Ring with No Proper Zero Divisors is Field]], $R$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+[[Category:Ring Theory]]
+[[Category:Field Theory]]
+[[Category:Norm Theory]]
+lwtk5hog282ltnhd0cbmwawzoi1ae1i
+\end{proof}<|endoftext|>
+\section{Composition of Isometries is Isometry}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let:
+:$\struct {X_1, d_1}$
+:$\struct {X_2, d_2}$
+:$\struct {X_3, d_3}$
+be [[Definition:Metric Space|metric spaces]].
+Let:
+:$\phi: \struct {X_1, d_1} \to \struct {X_2, d_2}$
+:$\psi: \struct {X_2, d_2} \to \struct {X_3, d_3}$
+be [[Definition:Isometry (Metric Spaces)|isometries]].
+Then the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ is also an [[Definition:Isometry (Metric Spaces)|isometry]].
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Isometry (Metric Spaces)|isometry]] is a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]] which is also a [[Definition:Bijection|bijection]].
+From [[Composition of Distance-Preserving Mappings is Distance-Preserving]], $\psi \circ \phi$ is a [[Definition:Distance-Preserving Mapping|distance-preserving mapping]].
+From [[Composite of Bijections is Bijection]], $\psi \circ \phi$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Metric Spaces]]
+m8cjqegtg9dd5z42hpcuok44xqc5ih0
+\end{proof}<|endoftext|>
+\section{Subgroup Action is Group Action}
+Tags: Subgroup Action
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\struct {H, \circ}$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Subgroup Action|subgroup action]] defined for all $h \in H, g \in G$ as:
+:$\forall h \in H, g \in G: h * g := h \circ g$
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+Let $g \in G$.
+First we note that since $G$ is [[Definition:Closed Algebraic Structure|closed]], and $h \circ g \in G$, it follows that $h * g \in G$.
+Next we note:
+:$e * g = e \circ g = g$
+and so {{GroupActionAxiom|2}} is satisfied.
+Now let $h_1, h_2 \in G$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {h_1 \circ h_2} * g
+      | r = \paren {h_1 \circ h_2} \circ g
+      | c = Definition of $*$
+}}
+{{eqn | r = h_1 \circ \paren {h_2 \circ g}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = h_1 * \paren {h_2 * g}
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+and so {{GroupActionAxiom|1}} is satisfied.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Orbit of Subgroup Action is Coset}
+Tags: Subgroup Action
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\struct {H, \circ}$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Subgroup Action|subgroup action]] defined for all $h \in H, g \in G$ as:
+:$\forall h \in H, g \in G: h * g := h \circ g$
+The [[Definition:Orbit (Group Theory)|orbit]] of $x \in G$ is the [[Definition:Right Coset|right coset]] by $x$ of $H$:
+:$\Orb x = H x$
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup Action is Group Action]] we have that $*$ is a [[Definition:Group Action|group action]].
+Let $x \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \Orb x
+      | r = \set {g \in G: \exists h \in H: g = h * x}
+      | c = {{Defof|Orbit (Group Theory)|Orbit}}
+}}
+{{eqn | r = \set {g \in G: \exists h \in H: g = h \circ x}
+      | c = Definition of $*$
+}}
+{{eqn | r = H x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Right Coset|right coset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Removable Singularity at Infinity implies Constant Function}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $f : \C \to \C$ be an [[Definition:Entire Function|entire function]].
+Let $f$ have an [[Definition:Removable Singularity|removable singularity]] at $\infty$. 
+Then $f$ is [[Definition:Constant Function|constant]].
+\end{theorem}
+
+\begin{proof}
+By [[Riemann Removable Singularities Theorem]], as $f$ has a [[Definition:Removable Singularity|removable singularity]] at $\infty$, $f$ must be [[Definition:Bounded Mapping/Complex-Valued|bounded]] in a [[Definition:Neighborhood of Infinity (Complex Analysis)|neighborhood of $\infty$]].
+That is, there exists a [[Definition:Real Number|real number]] $M > 0$ such that $\cmod {\map f z} \le M$ for all $z \in \set{z : \cmod z > r}$ for some real $r \ge 0$.
+However, by [[Continuous Function on Compact Space is Bounded]], $f$ is also bounded on $\set{z : \cmod z \le r}$. 
+As $\set{z : \cmod z > r} \cup \set{z : \cmod z \le r} = \C$, $f$ is therefore bounded on $\C$.
+Therefore by [[Liouville's Theorem (Complex Analysis)|Liouville's Theorem]], $f$ is constant.
+{{qed}}
+[[Category:Complex Analysis]]
+ft58yz7272i4zngzeagf3r0b1n94if3
+\end{proof}<|endoftext|>
+\section{Stabilizer of Subgroup Action is Identity}
+Tags: Subgroup Action, Stabilizers
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\struct {H, \circ}$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: H \times G \to G$ be the [[Definition:Subgroup Action|subgroup action]] defined for all $h \in H, g \in G$ as:
+:$\forall h \in H, g \in G: h * g := h \circ g$
+The [[Definition:Stabilizer|stabilizer]] of $x \in G$ is $\set e$:
+:$\Stab x = \set e$
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup Action is Group Action]] we have that $*$ is a [[Definition:Group Action|group action]].
+Let $x \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \Stab x
+      | r = \set {h \in H: h * x = x}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {h \in H: h \circ x = x}
+      | c = Definition of $*$
+}}
+{{eqn | r = \set {h \in H: h = x \circ x^{-1} }
+      | c = 
+}}
+{{eqn | r = \set {h \in H: h = e}
+      | c = 
+}}
+{{eqn | r = \set e
+      | c = 
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Right Coset|right coset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizers of Elements in Same Orbit are Conjugate Subgroups}
+Tags: Group Actions, Stabilizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group Action|group acting]] on a set $X$.
+Let:
+:$y, z \in \Orb x$
+where $\Orb x$ denotes the [[Definition:Orbit (Group Theory)|orbit]] of some $x \in X$.
+Then their [[Definition:Stabilizer|stabilizers]] $\Stab y$ and $\Stab z$ are [[Definition:Conjugate of Group Subset|conjugate]] [[Definition:Subgroup|subgroups]].
+\end{theorem}
+
+\begin{proof}
+From [[Stabilizer is Subgroup]] we have that both $\Stab y$ and $\Stab z$ are [[Definition:Subgroup|subgroups]] of $G$.
+From [[Definition:Orbit (Group Theory)|definition of orbits]]:
+:$\exists h_1, h_2 \in G: y = h_1 * x, z = h_2 * x$
+Then $y = h_1 * \paren {h_2^{-1} * z} = h_1 h_2^{-1} * z$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \Stab y
+      | r = \set {g \in G: g * y = y}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {g \in G: g * \paren {h_1 h_2^{-1} * z} = h_1 h_2^{-1} * z}
+}}
+{{eqn | r = \set {g \in G: h_1^{-1} h_2 * \paren {g h_1 h_2^{-1} * z} = z}
+}}
+{{eqn | r = \set {g \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z}
+}}
+{{eqn | r = \paren {h_1 h_2^{-1} } \set {\paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } \in G: \paren {h_1 h_2^{-1} }^{-1} g \paren {h_1 h_2^{-1} } * z = z} \paren {h_1 h_2^{-1} }^{-1}
+}}
+{{eqn | r = \paren {h_1 h_2^{-1} } \Stab z \paren {h_1 h_2^{-1} }^{-1}
+      | c = {{Defof|Stabilizer}}
+}}
+{{end-eqn}}
+This shows that $\Stab y$ and $\Stab z$ are [[Definition:Conjugate of Group Subset|conjugate]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizer of Subgroup Action on Left Coset Space}
+Tags: Stabilizers, Group Action on Coset Space
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let $K$ [[Definition:Group Action|act on]] the [[Definition:Left Coset Space|left coset space]] $G / H^l$ by:
+:$\forall \tuple {k, g H} \in K \times G / H^l: k * g H := \paren {k g} H$
+The [[Definition:Stabilizer|stabilizer]] of $g H$ is $K \cap H^g$, where $H^g$ denotes the [[Definition:Conjugate (Group Theory)/Subset|$G$-conjugate of $H$ by $g$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \Stab {g H}
+      | r = \set {k \in K: \paren {k g} H = g H}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {k \in K: g^{-1} \paren {k g} \in H}
+      | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | r = \set {k \in K: k \in g H g^{-1} }
+      | c = 
+}}
+{{eqn | r = K \cap H^g
+      | c = {{Defof|Conjugate (Group Theory)|subdef = Subset|Conjugate of Set}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Orbit of Subgroup Action on Left Coset Space}
+Tags: Group Action on Coset Space
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let $K$ [[Definition:Group Action|act on]] the [[Definition:Left Coset Space|left coset space]] $G / H^l$ by:
+:$\forall \tuple {k, g H} \in K \times G / H^l: k * g H := \paren {k g} H$
+The [[Definition:Length of Orbit|length]] of the [[Definition:Orbit (Group Theory)|orbit]] of $g H$ is $\index K {K \cap H^g}$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \card {\Orb {g H} }
+      | r = \index K {\Stab {g H} }
+      | c = [[Orbit-Stabilizer Theorem]]
+}}
+{{eqn | r = \index K {K \cap H^g}
+      | c = [[Stabilizer of Subgroup Action on Left Coset Space]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set Difference with Subset}
+Tags: Set Difference, Cardinality
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]] such that $T$ is [[Definition:Finite Set|finite]].
+Let $T \subseteq S$.
+Then:
+:$\card {S \setminus T} = \card S - \card T$
+where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Set Difference with Superset is Empty Set]]:
+:$T \subseteq S \iff T \setminus S = \O$
+From [[Set Difference and Intersection form Partition]]:
+:$S = \paren {S \setminus T} \cup T$
+Thus from [[Cardinality of Set Union]]:
+:$\card S = \card T + \card {S \setminus T} - \card {T \cap \paren {S \setminus T} }$
+But from [[Set Difference Intersection with Second Set is Empty Set]]:
+:$T \cap \paren {S \setminus T} = 0$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Function at Non-Positive Integers}
+Tags: Riemann Zeta Function
+
+\begin{theorem}
+Let $n \ge 0$ be a [[Definition:Integer|integer]]. 
+Then: 
+:$\map \zeta {-n} = \paren {-1}^n \dfrac {B_{n + 1} } {n + 1}$ 
+where:
+:$B_n$ is the [[Definition:Bernoulli Numbers|$n$th Bernoulli number]]
+:$\zeta$ is the [[Definition:Riemann Zeta Function|Riemann Zeta function]]
+\end{theorem}
+
+\begin{proof}
+By [[Hankel Representation of Riemann Zeta Function]]: 
+:$\displaystyle \map \zeta {-n} = \frac {i \map \Gamma {1 + n} } {2 \pi} \oint_C \frac 1 {z^{n + 1} \paren {e^z - 1} } \rd z$ 
+where $C$ is the [[Definition:Hankel Contour|Hankel contour]].
+Note that the integrand is [[Definition:Meromorphic Function|meromorphic]], with a pole at $z = 0$ lying inside the contour.
+So: 
+{{begin-eqn}}
+{{eqn	| l = \map \zeta {-n}
+	| r = \frac {i n!} {2 \pi} \oint_C \frac 1 {\paren {-z}^{n + 2} } \cdot \frac {-z} {e^z - 1} \rd z
+	| c = [[Gamma Function Extends Factorial]]
+}}
+{{eqn	| r = -\frac {i n!} {2 \pi} \cdot 2 \pi i \Res {\frac 1 {\paren {-z}^{n + 2} } \cdot \frac {-z} {e^z - 1} } 0
+	| c = [[Residue Theorem]] 
+}}
+{{eqn	| r = n! \Res {\paren {-1}^2 \paren {-1}^n \sum _{k \mathop = 0}^\infty \frac {B_k} {k!} z^{k - n - 2} } 0
+	| c = {{Defof|Bernoulli Numbers/Generating Function|Bernoulli Numbers}}, $i^2 = -1$
+}}
+{{eqn	| r = n! \paren {-1}^n \Res {\sum _{k \mathop = 0}^\infty \frac {B_k} {k!} z^{k - n - 2} } 0
+}}
+{{end-eqn}}
+By the definition of a [[Definition:Residue (Complex Analysis)|residue]], the residue at $0$ is given by the coefficient of the $\dfrac 1 z$ term.
+This is the term where $k = n + 1$, so:
+{{begin-eqn}}
+{{eqn	| l = \map \zeta {-n}
+	| r = n! \paren {-1}^n \frac {B_{n + 1} } {\paren {n + 1}!} 
+}}
+{{eqn	| r = \paren {-1}^n \frac {B_{n + 1} } {n + 1}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Riemann Zeta Function]]
+svniiy7i2jmd86hhra1u1ymikmr4he0
+\end{proof}<|endoftext|>
+\section{Remainder on Division is Least Positive Residue}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ge 0$ and $b \ne 0$.
+Let $r$ be the [[Definition:Remainder|remainder]] resulting from the operation of [[Definition:Integer Division|integer division]] of $a$ by $b$:
+$a = q b + r, 0 \le r < \size b$
+Then $r$ is equal to the [[Definition:Least Positive Residue|least positive residue]] of $a \pmod b$.
+\end{theorem}<|endoftext|>
+\section{Floor Function/Examples/Floor of Root 5}
+Tags: Examples of Floor Function
+
+\begin{theorem}
+:$\floor {\sqrt 5} = 2$
+\end{theorem}
+
+\begin{proof}
+The [[Square Root of 5|decimal expansion of $\sqrt 5$]] is:
+:$\sqrt 5 \approx 2 \cdotp 23606 \, 79774 \, 99789 \, 6964 \ldots$
+Thus:
+:$2 \le \sqrt 5 < 3$
+Hence $2$ is the [[Definition:Floor Function|floor]] of $\sqrt 5$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor Function/Examples/Floor of 3}
+Tags: Examples of Floor Function
+
+\begin{theorem}
+:$\floor 3 = 3$
+\end{theorem}
+
+\begin{proof}
+We have that $3$ is an [[Definition:Integer|integer]].
+Thus this is a specific example of [[Real Number is Integer iff equals Floor]]:
+$\floor x = x \iff x \in \Z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisibility by 12}
+Tags: Divisibility Tests, 12
+
+\begin{theorem}
+Let $N \in \N$ be expressed as:
+:$N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$
+Then $N$ is [[Definition:Divisor of Integer|divisible]] by $12$ {{iff}} $a_0 - 2 a_1 + 4 \paren {\displaystyle \sum_{r \mathop = 2}^n a_r}$ is  [[Definition:Divisor of Integer|divisible]] by $12$.
+\end{theorem}
+
+\begin{proof}
+We first prove that $100 \times 10^n = 4 \pmod {12}$, where $n \in \N$.
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $P \paren n$ be the [[Definition:Proposition|proposition]]:
+:$100 \times 10^n = 4 \pmod {12}$
+=== Basis for the Induction ===
+$P(0)$ says $100 = 4 \pmod {12}$, which is true as $100 = 8 \times 12 + 4$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \paren k$ is true, where $k \ge 0$, then it logically follows that $P \paren{k+1}$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$100 \times 10^k = 4 \pmod {12}$
+Then we need to show:
+:$100 \times 10^{k+1} = 4 \pmod {12}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = 100 \times 10^{k+1}
+      | r = 10 \times 100 \times 10^k \pmod {12}
+}}
+{{eqn | r = 10 \times 4 \pmod {12}
+      | c = [[Divisibility by 12#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 4 \pmod {12}
+      | c = $10 \times 4 = 40 = 3 \times 12 + 4$
+}}
+{{end-eqn}}
+So $P \paren k \implies P \paren{k+1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$100 \times 10^n = 4 \pmod {12}$
+And then:
+{{begin-eqn}}
+{{eqn | l = N
+      | r = a_0 + a_1 10 + \sum^n_{r \mathop = 2} a_r 10^r
+      | c = 
+}}
+{{eqn | r = a_0 + a_1 10 + \sum^n_{r \mathop = 2} a_r \paren {100 \times 10^{r-2} } \pmod {12}
+      | c = 
+}}
+{{eqn | r = a_0 + a_1 10 + 4 \sum^n_{r \mathop = 2} a_r \pmod {12}
+      | c = Applying earlier result
+}}
+{{eqn | r = a_0 - 2 a_1 + 4 \sum^n_{r \mathop = 2} a_r \pmod {12}
+      | c = $-2 = -1 \times 12 + 10$
+}}
+{{end-eqn}}
+Therefore:
+:$N = 0 \pmod {12} \iff a_0 - 2 a_1 + 4 \paren {\displaystyle \sum_{r \mathop = 2}^n a_r} = 0 \pmod {12}$
+The result follows from [[Congruent to Zero if Modulo is Divisor]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential on Real Numbers is Injection}
+Tags: Exponential Function, Injections
+
+\begin{theorem}
+Let $\exp: \R \to \R$ be the [[Definition:Real Exponential Function|exponential function]]:
+:$\map \exp x = e^x$
+Then $\exp$ is an [[Definition:Injection|injection]].
+\end{theorem}
+
+\begin{proof}
+<onlyinclude>
+From [[Exponential is Strictly Increasing]]:
+:$\exp$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\R$.
+From [[Strictly Monotone Mapping with Totally Ordered Domain is Injective]]:
+:$\exp$ is an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Surjective Restriction of Real Exponential Function}
+Tags: Exponential Function, Surjections
+
+\begin{theorem}
+Let $\exp: \R \to \R$ be the [[Definition:Real Exponential Function|exponential function]]:
+:$\map \exp x = e^x$
+Then the [[Definition:Restriction of Mapping|restriction]] of the [[Definition:Codomain|codomain]] of $\exp$ to the [[Definition:Strictly Positive Real Number|strictly positive real numbers]]:
+:$\exp: \R \to \R_{>0}$
+is a [[Definition:Surjective Restriction|surjective restriction]].
+Hence:
+:$\exp: \R \to \R_{>0}$
+is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+We have [[Exponential on Real Numbers is Injection]].
+Let $y \in \R_{> 0}$.
+Then $\exists x \in \R: x = \map \ln y$
+That is:
+:$\exp x = y$
+and so $\exp: \R \to \R_{>0}$ is a [[Definition:Surjection|surjection]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximum Rule for Real Sequences}
+Tags: Real Analysis, Sequences
+
+\begin{theorem}
+:$\displaystyle \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {l, m}$
+\end{theorem}
+
+\begin{proof}
+=== Case $1$: $l = m$ ===
+Let $l = m$.
+Then:
+:$\max \set {l, m} = l = m$
+Let $\epsilon > 0$ be given.
+By definition of the [[Definition:Limit of Real Sequence|limit of a real sequence]], we can find $N_1$ such that:
+:$\forall n > N_1: \size {x_n - l} < \epsilon$
+Similarly we can find $N_2$ such that:
+:$\forall n > N_2: \size {y_n - m} < \epsilon$
+Let $N = \max \set {N_1, N_2}$.
+Then if $n > N$, both the above inequalities will be true:
+:$n > N_1$
+:$n > N_2$
+Thus $\forall n > N$:
+:if $y_n < x_n$ then:
+::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {x_n - l} < \epsilon$
+:if $x_n \le y_n$ then:
+::$\size {\max \set {x_n, y_n} - \max \set {l, m} } = \size {y_n - m} < \epsilon$
+In either case:
+: $\size {\max \set {x_n, y_n} - \max \set {l, m} } < \epsilon$
+The result follows.
+{{qed|lemma}}
+=== Case 2: $l > m$ ===
+Let $l > m$.
+Then:
+:$l = \max \set {l, m}$
+Let $\delta = \dfrac {l - m} 2$.
+Then:
+:$\delta > 0$
+Let $\epsilon > 0$ be given.
+Let $\epsilon' = \min \set {\delta, \epsilon}$.
+Then:
+:$\epsilon' > 0$
+By definition of the [[Definition:Limit of Real Sequence|limit of a real sequence]], we can find $N_1$ such that:
+:$\forall n > N_1: \size {x_n - l} < \epsilon'$
+Similarly we can find $N_2$ such that:
+:$\forall n > N_2: \size {y_n - m} < \epsilon'$
+Let $N = \max \set {N_1, N_2}$.
+Then if $n > N$, both the above inequalities will be true:
+:$n > N_1$
+:$n > N_2$
+From [[Negative of Absolute Value/Corollary 3|Corollary 3 of Negative of Absolute Value]], we have that $\forall n > N$:
+:$\size {x_n - l} < \epsilon' \implies x_n > l - \epsilon'$
+:$\size {y_n - m} < \epsilon' \implies y_n < m + \epsilon'$
+As $\epsilon' \le \delta$:
+{{begin-eqn}}
+{{eqn | l = l - \epsilon'
+      | r = l - \delta
+      | o = \ge
+}}
+{{eqn | r = l - \dfrac {l - m} 2
+}}
+{{eqn | r = \dfrac {2 l - l + m} 2
+}}
+{{eqn | r = \dfrac {l + m} 2
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = m + \epsilon'
+      | r = m + \delta
+      | o = \le
+}}
+{{eqn | r = m + \dfrac {l - m} 2
+}}
+{{eqn | r = \dfrac {2 m + l - m} 2
+}}
+{{eqn | r = \dfrac {l + m} 2
+}}
+{{end-eqn}}
+So:
+:$x_n > l - \epsilon' \ge \dfrac {l + m} 2 \ge  m + \epsilon' > y_n$
+Hence:
+:$x_n = \max \set{x_n, y_n}$
+So $\forall n > N$:
+:$\size {\max \set {x_n, y_n} - \max \set {l, m} }  = \size {x_n - l} < \epsilon' \le \epsilon$
+The result follows.
+{{qed|lemma}}
+=== Case 3: $m > l$ ===
+Similar to Case 2 by interchanging $l$ with $m$ and $x_n$ with $y_n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimum Rule for Real Sequences}
+Tags: Real Analysis, Sequences
+
+\begin{theorem}
+:$\displaystyle \lim_{n \mathop \to \infty} \min \set {x_n, y_n} = \min \set {l, m}$
+\end{theorem}
+
+\begin{proof}
+By [[Sum Less Maximum is Minimum]]:
+:$\forall n \in \R: \min \set {x_n, y_n} = x_n + y_n - \max \set {x_n, y_n}$
+and
+:$\min \set {m, l} = m + l - \max \set {m, l}$
+By [[Maximum Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \max \set {x_n, y_n} = \max \set {m, l}$
+By the [[Multiple Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} - \max \set {x_n, y_n} = - \max \set {m, l}$
+By the [[Sum Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty}  x_n + y_n - \max \set {x_n, y_n} = m + l - \max \set {m, l}$
+Hence:
+:$\displaystyle \lim_{n \mathop \to \infty} \min \set {x_n, y_n} = \min \set {l, m}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Subring of Normed Division Ring}
+Tags: Division Rings, Norm Theory
+
+\begin{theorem}
+Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $S$ be a [[Definition:Division Subring|division subring]] of $R$.
+Then:
+:$\struct {S, \norm {\, \cdot \,}_S}$ is a [[Definition:Normed Division Subring|normed division subring]] of $\struct {R, \norm {\, \cdot \,} }$
+where $\norm {\, \cdot \,}_S$ is the [[Definition:Norm on Division Ring|norm]] $\norm{\,\cdot\,}$ [[Definition:Restriction/Mapping|restricted]] to $S$.
+\end{theorem}
+
+\begin{proof}
+=== (N1) : [[Definition:Positive Definite (Ring)|Positive Definiteness]] ===
+{{begin-eqn}}
+{{eqn | lo= \forall x \in S:
+      | l = \norm x_S
+      | r = 0
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm x
+      | r = 0
+      | c = Definition of $\norm x_S$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = 0
+      | c = [[Definition:Norm on Division Ring|Norm axiom (N1)]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== (N2) : [[Definition:Multiplicative Function|Multiplicativity]] ===
+{{begin-eqn}}
+{{eqn | lo= \forall x, y \in S:
+      | l = \norm {x y}_S
+      | r = \norm {x y}
+      | c = Definition of $\norm {\, \cdot \,}_S$
+}}
+{{eqn | r = \norm x \norm y
+      | c = [[Definition:Norm on Division Ring|Norm axiom (N2)]]
+}}
+{{eqn | r = \norm x_S \norm y_S
+      | c = Definition of $\norm {\, \cdot \,}_S$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== (N3) : Triangle Inequality ===
+{{begin-eqn}}
+{{eqn | lo= \forall x, y \in S:
+      | l = \norm {x + y}_S
+      | r = \norm {x + y}
+      | c = Definition of $\norm {\, \cdot \,}_S$
+}}
+{{eqn | o = \le
+      | r = \norm x + \norm y
+      | c = [[Definition:Norm on Division Ring|Norm axiom (N3)]]
+}}
+{{eqn | r = \norm x_S + \norm y_S
+      | c = Definition of $\norm {\, \cdot \,}_S$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Division Rings]]
+[[Category:Norm Theory]]
+ky2s7b1nsucwi1zc3srsir3ym7gws58
+\end{proof}<|endoftext|>
+\section{Normed Division Ring is Dense Subring of Completion}
+Tags: Normed Division Rings, Complete Metric Spaces, Completion of Normed Division Ring
+
+\begin{theorem}
+Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $\struct {R', \norm {\, \cdot \,}' }$ be a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] of $\struct {R, \norm {\, \cdot \,} }$
+Then:
+:$\struct {R, \norm {\, \cdot \,} }$ is [[Definition:Isometric Isomorphism|isometrically isomorphic]] to a [[Definition:Everywhere Dense|dense]] [[Definition:Normed Division Subring|normed division subring]] of $\struct {R', \norm {\, \cdot \,}' }$.
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] then:
+:$(1): \quad$ there exists a [[Definition:Distance-Preserving Mapping|distance-preserving]] [[Definition:Ring Monomorphism|ring monomorphism]] $\phi: R \to R'$.
+:$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a [[Definition:Complete Metric Space|complete metric space]].
+:$(3): \quad \phi \sqbrk R$ is a [[Definition:Everywhere Dense|dense]] [[Definition:Topological Subspace|subspace]] in $\struct {R', \norm {\, \cdot \,}' }$.
+By [[Ring Homomorphism Preserves Subrings/Corollary|image of a ring homomorphism is a subring]] then $\phi \sqbrk R$ is a [[Definition:Subring|subring]] of $R'$ and $\phi: R \to \phi \sqbrk R$ is an [[Definition:Ring Isomorphism|isomorphism]].
+By [[Epimorphism from Division Ring to Ring]] then $\phi \sqbrk R$ is a [[Definition:Division Subring|division subring]] of $R'$.
+By [[Division Subring of Normed Division Ring|Division Subring of Normed Division Ring]] then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is a [[Definition:Normed Division Subring|normed division subring]] of $\struct {R', \norm {\, \cdot \,}' }$.
+By [[Distance-Preserving Surjection is Isometry of Metric Spaces]] then $\phi: R \to \phi \sqbrk R$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+The result follows.
+{{qed}}
+[[Category:Normed Division Rings]]
+[[Category:Complete Metric Spaces]]
+[[Category:Completion of Normed Division Ring]]
+jig8wbrux01r8hlv2ykm1x9upahaez2
+\end{proof}<|endoftext|>
+\section{Inverse of Isometric Isomorphism}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be [[Definition:Normed Division Ring|normed division rings]].
+Let $\phi:R \to S$ be a mapping.
+Then $\phi:R \to S$ is an [[Definition:Isometric Isomorphism|isometric isomorphism]] {{iff}} $\phi^{-1}: S \to R$ is also an [[Definition:Isometric Isomorphism|isometric isomorphism]].
+\end{theorem}
+
+\begin{proof}
+By [[Inverse of Algebraic Structure Isomorphism is Isomorphism]] then:
+:$\phi: R \to S$ is an [[Definition:Ring Isomorphism|ring isomorphism]] {{iff}} $\phi^{-1}: S \to R$ is also an [[Definition:Ring Isomorphism|ring isomorphism]].
+Let $d_R$ and $d_S$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by the [[Definition:Norm on Division Ring|norms]] $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.
+By [[Inverse of Isometry of Metric Spaces is Isometry]] then:
+:$\phi: \struct {R, d_R} \to \struct {S, d_S}$ is an [[Definition:Isometry (Metric Spaces)|isometry]] {{iff}} $\phi^{-1}: \struct {S, d_S} \to \struct {R, d_R}$ is also an [[Definition:Isometry (Metric Spaces)|isometry]].
+The result follows.
+{{qed}}
+[[Category:Normed Division Rings]]
+py0czj6bpbe3o9u87vjpmzmaqi3wrxw
+\end{proof}<|endoftext|>
+\section{Isometric Isomorphism is Norm-Preserving}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be [[Definition:Normed Division Ring|normed division rings]].
+Let $\phi: R \to S$ be a [[Definition:Ring Isomorphism|ring isomorphism]].
+Then $\phi: R \to S$ is an [[Definition:Isometric Isomorphism|isometric isomorphism]] {{iff}} $\phi$ satisfies:
+:$\forall x \in R: \norm {\map \phi x}_S = \norm x_R $
+\end{theorem}
+
+\begin{proof}
+Let $d_R$ and $d_S$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by the [[Definition:Norm on Division Ring|norms]] $\norm {\,\cdot\,}_R$ and $\norm {\,\cdot\,}_S$ respectively.
+=== Necessary Condition ===
+Let $\phi: R \to S$ be an [[Definition:Isometric Isomorphism|isometric isomorphism]].
+Then for $x \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm {\map \phi x}_S
+      | r = \norm {\map \phi x - 0_S}_S
+      | c = $0_S$ is the [[Definition:Ring Zero|zero]] of $S$
+}}
+{{eqn | r = \norm {\map \phi x - \map \phi {0_R} }_S
+      | c = [[Ring Homomorphism Preserves Zero]]
+}}
+{{eqn | r = \map {d_S} {\map \phi x, \map \phi {0_R} }
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \map {d_R} {x, {0_R} }
+      | c = {{Defof|Isometry (Metric Spaces)}}
+}}
+{{eqn | r = \norm {x - {0_R} }_R
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \norm x_R
+      | c = $0_R$ is the [[Definition:Ring Zero|zero]] of $R$
+}}
+{{end-eqn}}
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\phi: R \to S$ satisfy:
+:$\forall x \in R: \norm {\map \phi x}_S = \norm x_R$
+Then for $x, y \in R$:
+{{begin-eqn}}
+{{eqn | l = \map {d_S} {\map \phi x, \map \phi y}
+      | r = \norm {\map \phi x - \map \phi y}_S
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \norm {\map \phi {x - y} }_S
+      | c = $\phi$ is a [[Definition:Ring Isomorphism|Ring Isomorphism]]
+}}
+{{eqn | r = \norm {x - y}_R
+      | c = by hypothesis
+}}
+{{eqn | r = \map {d_R} {x, y}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+[[Category:Normed Division Rings]]
+lry0tga0ndyp1ra3oqf88egnkmdshws
+\end{proof}<|endoftext|>
+\section{Taylor Series of Holomorphic Function}
+Tags: Complex Analysis
+
+\begin{theorem}
+:$\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \frac {\map {f^n} a} {n!} \paren {z - a}^n$
+\end{theorem}
+
+\begin{proof}
+In [[Holomorphic Function is Analytic]], it is shown that: 
+:$\displaystyle \map f z = \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t} \paren {z - a}^n$ 
+for all $z \in D$.
+From [[Cauchy's Integral Formula/General Result|Cauchy's Integral Formula: General Result]], we have: 
+:$\displaystyle \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t = \frac {\map {f^n} a} {n!}$
+Hence the result. 
+{{qed}}
+[[Category:Complex Analysis]]
+r6d3bo3tpqrs7hrsjjgxuobtbdq2eo7
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Square Function}
+Tags: Square Function
+
+\begin{theorem}
+Let $\F$ denote one of the [[Definition:Number|standard classes of numbers]]: $\N$, $\Z$, $\Q$, $\R$, $\C$.
+{{TFAE|def = Square Function}}
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Integer Power|$n$th power]] (for [[Definition:Positive Integer|positive]] $n$):
+:$x^n = \begin{cases}
+1 & : n = 0 \\
+x \times x^{n - 1} & : n > 0
+\end{cases}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = x^2
+      | r = x \times x^1
+      | c = 
+}}
+{{eqn | r = x \times x \times x^0
+      | c = 
+}}
+{{eqn | r = x \times x \times 1
+      | c = 
+}}
+{{eqn | r = x \times x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Square Function]]
+0unisqa74pxlosnkezxy36diu2jkfgx
+\end{proof}<|endoftext|>
+\section{Subring of Non-Archimedean Division Ring}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]] $\norm {\, \cdot \,}$.
+Let $\struct {S, \norm {\, \cdot \,}_S }$ be a [[Definition:Normed Division Subring|normed division subring]] of $R$.
+Then:
+:$\norm {\, \cdot \,}_S$ is a [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]].
+\end{theorem}
+
+\begin{proof}
+$\forall x, y \in S$:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}_S
+      | r = \norm {x + y}
+      | c = Definition of $\norm {\,\cdot\,}_S$
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm x, \norm y}
+      | c = [[Definition:Non-Archimedean Division Ring Norm|$(\text N 4)$: Ultrametric Inequality]]
+}}
+{{eqn | r = \max \set {\norm x_S, \norm y_S}
+      | c = Definition of $\norm {\, \cdot \,}_S$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Normed Division Rings]]
+mnmsno5i821o90pknrspo4twwhbguqt
+\end{proof}<|endoftext|>
+\section{Isometrically Isomorphic Non-Archimedean Division Rings}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}_R}$ and $\struct {S, \norm {\,\cdot\,}_S}$ be [[Definition:Normed Division Ring|normed division rings]].
+Let $\phi:R \to S$ be an [[Definition:Isometric Isomorphism|isometric isomorphism]].
+Then:
+:$\norm {\,\cdot\,}_R$ is a [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]] {{iff}} $\norm {\,\cdot\,}_S$ is a [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\norm {\,\cdot\,}_R$ be a [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]].
+Then for all $x,y \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}_S
+      | r = \norm {\map \phi {\map {\phi^{-1} } x} + \map \phi {\map {\phi^{-1} } y} }_S
+      | c = $\phi$ is a [[Definition:Bijection|bijection]]
+}}
+{{eqn | r = \norm {\map {\phi^{-1} } x  + \map {\phi^{-1} } y}_R
+      | c = $\phi$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {\map {\phi^{-1} } x}_R, \norm {\map {\phi^{-1} } y}_R}
+      | c = $\norm {\,\cdot\,}_R$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+}}
+{{eqn | r = \max \set {\norm {\map \phi {\map {\phi^{-1} } x} }_S, \norm {\map \phi {\map {\phi^{-1} } y} }_S}
+      | c = $\phi$ is an [[Definition:Isometry (Metric Spaces)|isometry]].
+}}
+{{eqn | r = \max \set {\norm x_S, \norm y_S}
+      | c = $\phi$ is a [[Definition:Bijection|bijection]].
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\norm {\, \cdot \,}_S$ be a [[Definition:Non-Archimedean Division Ring Norm|non-archimedean norm]].
+By [[Inverse of Isometric Isomorphism]], $\phi^{-1}: S \to R$ is an [[Definition:Isometric Isomorphism|isometric isomorphism]].
+By the '''[[Isometrically Isomorphic Non-Archimedean Division Rings#Necessary Condition|necessary condition]]''', $\norm {\, \cdot \,}_R$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+{{qed}}
+[[Category:Normed Division Rings]]
+m3vzfu8nxtrx8s5lgtieza28ol5q02r
+\end{proof}<|endoftext|>
+\section{Non-Archimedean Division Ring Iff Non-Archimedean Completion}
+Tags: Normed Division Rings, Non-Archimedean Norms
+
+\begin{theorem}
+Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $\struct {R', \norm {\, \cdot \,}' }$ be a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] of $\struct {R, \norm {\, \cdot \,} }$
+Then:
+:$\norm {\, \cdot \,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]] {{iff}} $\norm {\, \cdot \,}'$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Completion (Normed Division Ring)|normed division ring completion]] then:
+:$(1): \quad$ there exists a [[Definition:Distance-Preserving Mapping|distance-preserving]] [[Definition:Ring Monomorphism|ring monomorphism]] $\phi: R \to R'$.
+:$(2): \quad \struct {R', \norm {\, \cdot \,}' }$ is a [[Definition:Complete Metric Space|complete metric space]].
+:$(3): \quad \phi \sqbrk R$ is a [[Definition:Everywhere Dense|dense]] [[Definition:Topological Subspace|subspace]] in $\struct {R', \norm {\, \cdot \,}' }$.
+By [[Normed Division Ring is Dense Subring of Completion]], $\phi \sqbrk R$ is a [[Definition:Dense|dense]] [[Definition:Normed Division Subring|normed division subring]] of $R'$ and $\phi: R \to \map \phi R$ is an [[Definition:Isometric Isomorphism|isometric isomorphism]].
+By [[Isometrically Isomorphic Non-Archimedean Division Rings]], $\struct {R, \norm {\, \cdot \,} }$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]] {{iff}} $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is.
+So it remains to show that $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]] {{iff}} $\struct {R', \norm {\, \cdot \,}' }$ is.
+=== Necessary Condition ===
+Let $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ be [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+Let $x', y' \in R'$.
+By the definition of a [[Definition:Dense|dense subset]]:
+:$\map \cl {\phi \sqbrk R} = R'$
+By [[Closure of Subset of Metric Space by Convergent Sequence]] then:
+:there exists a [[Definition:Sequence|sequence]] $\sequence {x_n'} \subseteq \phi \sqbrk R$ that [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x'$, that is, $\displaystyle \lim_{n \mathop \to \infty} x_n' = x'$
+:there exists a [[Definition:Sequence|sequence]] $\sequence {y_n'} \subseteq \phi \sqbrk R$ that [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $y'$, that is, $\displaystyle \lim_{n \mathop \to \infty} y_n' = y'$
+By [[Sum Rule for Sequences in Normed Division Ring]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n' + y_n' = x' + y'$
+By [[Modulus of Limit/Normed Division Ring|modulus of limit]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n'}' = \norm {x'}'$
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {y_n'}' = \norm {y'}'$
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' = \norm {x' + y'}'$
+By [[Maximum Rule for Real Sequences]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$
+Since $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]] then:
+:$\forall n: \norm {x_n' + y_n'}' \le \max \set {\norm {x_n'}', \norm {y_n'}' }$
+By [[Inequality Rule for Real Sequences]] then:
+:$\displaystyle \norm {x' + y'}' = \lim_{n \mathop \to \infty} \norm {x_n' + y_n'}' \le \lim_{n \mathop \to \infty} \max \set {\norm {x_n'}', \norm {y_n'}' } = \max \set {\norm {x'}', \norm {y'}'}$
+The result follows.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\struct {R', \norm {\, \cdot \,}' }$ be [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+By [[Subring of Non-Archimedean Division Ring]] then $\struct {\phi \sqbrk R, \norm {\, \cdot \,}' }$ is [[Definition:Non-Archimedean Division Ring Norm|non-archimedean]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Integer Square Function}
+Tags: Square Function
+
+\begin{theorem}
+The [[Definition:Domain of Mapping|domain]] of the [[Definition:Integer Square Function|integer square function]] is the entire [[Definition:Integer|set of integers]] $\Z$.
+\end{theorem}
+
+\begin{proof}
+The operation of [[Definition:Integer Multiplication|integer multiplication]] is defined on all [[Definition:Integer|integers]].
+Thus:
+:$\forall x \in \Z: \exists y \in \Z: x^2 = y$
+Hence the result by definition of [[Definition:Domain of Mapping|domain]].
+{{qed}}
+[[Category:Square Function]]
+thaqmabzebfpu0jntuszq9f2o226uxb
+\end{proof}<|endoftext|>
+\section{Image of Integer Square Function}
+Tags: Square Function
+
+\begin{theorem}
+The [[Definition:Image of Mapping|image]] of the [[Definition:Integer Square Function|integer square function]] is the [[Definition:Set|set]] of [[Definition:Square Number|square numbers]].
+\end{theorem}
+
+\begin{proof}
+Follows directly by definition of the [[Definition:Integer Square Function|integer square function]].
+{{qed}}
+[[Category:Square Function]]
+crigqa9nrja5iymme5m6dh111j0ao06
+\end{proof}<|endoftext|>
+\section{Restriction of Real Square Mapping to Positive Reals is Bijection}
+Tags: Square Function
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Square Function|real square function]]:
+:$\forall x \in \R: \map f x = x^2$
+Let $g: \R_{\ge 0} \to R_{\ge 0} := f {\restriction_{\R_{\ge 0} \times R_{\ge 0} } }$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to the [[Definition:Positive Real Number|positive real numbers]] $\R_{\ge 0}$.
+Then $g$ is a [[Definition:Bijective Restriction|bijective restriction]] of $f$.
+\end{theorem}
+
+\begin{proof}
+From [[Order is Preserved on Positive Reals by Squaring]], $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $\R_{\ge 0}$.
+By definition, a [[Definition:Strictly Increasing Real Function|strictly increasing real function]] is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+The result follows from [[Strictly Monotone Real Function is Bijective]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Real Square Function on Positive Reals}
+Tags: Square Function
+
+\begin{theorem}
+Let $f: \R_{\ge 0} \to R_{\ge 0}$ be the [[Definition:Restriction of Mapping|restriction]] of the [[Definition:Real Square Function|real square function]] to the [[Definition:Positive Real Number|positive real numbers]] $\R_{\ge 0}$.
+The [[Definition:Inverse Mapping|inverse]] of $f$ is $f^{-1}: \R_{\ge 0} \times R_{\ge 0}$ defined as:
+:$\forall x \in \R_{\ge 0}: \map {f^{-1} } x = \sqrt x$
+where $\sqrt x$ is the [[Definition:Positive Square Root|positive square root]] of $x$.
+\end{theorem}
+
+\begin{proof}
+From [[Restriction of Real Square Mapping to Positive Reals is Bijection]], $f$ is a [[Definition:Bijection|bijection]].
+By definition of the [[Definition:Positive Square Root|positive square root]]:
+:$y = \sqrt x \iff x = y^2$
+for $x, y \in \R_{\ge 0}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Square Function is not Bijective}
+Tags: Square Function
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Square Function|real square function]]:
+:$\forall x \in \R: \map f x = x^2$
+Then $f$ is not a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+From [[Real Square Function is not Injective]], $f$ is not an [[Definition:Injection|injection]].
+From [[Real Square Function is not Surjective]], $f$ is not a [[Definition:Surjection|surjection]].
+The result follows by definition of [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Linear Function on Real Numbers}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $a, b \in \R$ be [[Definition:Real Number|real numbers]] such that $a \ne 0$.
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = a x + b$
+Then the [[Definition:Inverse Mapping|inverse]] of $f$ is given by:
+:$\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
+\end{theorem}
+
+\begin{proof}
+We have that [[Linear Function on Real Numbers is Bijection]].
+Let $y = \map f x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \map f x
+      | c = 
+}}
+{{eqn | r = a x + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \dfrac {y - b} a
+      | c = 
+}}
+{{end-eqn}}
+and so:
+:$\forall y \in \R: \map {f^{-1} } y = \dfrac {y - b} a$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Function on Real Numbers is Bijection}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $a, b \in \R$ be [[Definition:Real Number|real numbers]].
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = a x + b$
+Then $f$ is a [[Definition:Bijection|bijection]] {{iff}} $a \ne 0$.
+\end{theorem}
+
+\begin{proof}
+Let $a \ne 0$.
+Let $y = \map f x$.
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \map f x
+      | c = 
+}}
+{{eqn | r = a x + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \dfrac {y - b} a
+      | c = 
+}}
+{{end-eqn}}
+and so:
+:$\forall y \in \R: \exists x \in \R; y = \map f x$
+demonstrating that $f$ is [[Definition:Surjection|surjective]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {x_1}
+      | r = \map f {x_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a x_1 + b
+      | r = a x_2 + b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a x_1
+      | r = a x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1
+      | r = x_2
+      | c = as $a \ne 0$
+}}
+{{end-eqn}}
+demonstrating that $f$ is [[Definition:Injection|injective]].
+Thus $f$ is a [[Definition:Bijection|bijection]] by definition.
+{{qed|lemma}}
+Let $a = 0$.
+Then:
+:$\forall x \in \R: \map f x = b$
+Thus for example:
+:$\map f 1 = \map f 2$
+and $f$ is trivially not [[Definition:Injection|injective]].
+Also:
+:$\forall y \in \R: y \ne b \implies \nexists x \in \R: \map f x = y$
+and $f$ is equally trivially not [[Definition:Surjection|surjective]] either.
+Thus $f$ is not a [[Definition:Bijection|bijection]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Linear Real Functions}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\theta_{a, b}: \R \to \R$  be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
+Let $\theta_{c, d} \circ \theta_{a, b}$ denote the [[Definition:Composition of Mappings|composition]] of $\theta_{c, d}$ with $\theta_{a, b}$.
+Then:
+:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\theta_{c, d} \circ \theta_{a, b} } } x
+      | r = \map {\theta_{c, d} } {\map {\theta_{a, b} } x}
+      | c = 
+}}
+{{eqn | r = \map {\theta_{c, d} } {a x + b}
+      | c = 
+}}
+{{eqn | r = c \paren {a x + b} + d
+      | c = 
+}}
+{{eqn | r = \paren {a c} x + \paren {b c + d}
+      | c = 
+}}
+{{eqn | r =  \theta_{a c, b c + d}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Composition of Linear Real Functions to be Commutative}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $a, b, c, d \in \R$ be [[Definition:Real Number|real numbers]].
+Let $\theta_{a, b}: \R \to \R$  be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
+Let $\theta_{c, d} \circ \theta_{a, b}$ denote the [[Definition:Composition of Mappings|composition]] of $\theta_{c, d}$ with $\theta_{a, b}$.
+Then:
+:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$
+{{iff}}:
+:$b c + d = a d + b$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\theta_{c, d} \circ \theta_{a, b} } x
+      | r = \map {\theta_{a, b} \circ \theta_{c, d} } x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \theta_{a c, b c + d}
+      | r = \theta_{c a, a d + b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b c + d
+      | r = a d + b
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Right Inverse with Mapping is Idempotent}
+Tags: Composite Mappings
+
+\begin{theorem}
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $g: T \to S$ be a [[Definition:Right Inverse Mapping|right inverse mapping]] of $f$.
+Then:
+:$\paren {g \circ f} \circ \paren {g \circ f} = g \circ f$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {g \circ f} \circ \paren {g \circ f}
+      | r = g \circ \paren {f \circ g} \circ f
+      | c = [[Composition of Mappings is Associative]]
+}}
+{{eqn | r = g \circ I_T \circ f
+      | c = {{Defof|Right Inverse Mapping}}
+}}
+{{eqn | r = g \circ f
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Even Integers is Equivalent to Set of Integers}
+Tags: Set Equivalence, Integers
+
+\begin{theorem}
+Let $\Z$ denote the [[Definition:Set|set]] of [[Definition:Integer|integers]].
+Let $2 \Z$ denote the [[Definition:Set|set]] of [[Definition:Even Integer|even integers]].
+Then:
+:$2 \Z \sim \Z$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+\end{theorem}
+
+\begin{proof}
+To demonstrate [[Definition:Set Equivalence|set equivalence]], it is sufficient to construct a [[Definition:Bijection|bijection]] between the two [[Definition:Set|sets]].
+Let $f: \Z \to 2 \Z$ defined as:
+:$\forall x \in \Z: \map f x = 2 x$
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \map f y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 x
+      | r = 2 y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+demonstrating [[Definition:Injection|injectivity]].
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = 2 \Z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x \in \Z:
+      | l = y
+      | r = 2 x
+      | c = {{Defof|Even Integer}}
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \in
+      | r = f \sqbrk \Z
+      | c = 
+}}
+{{end-eqn}}
+demonstrating [[Definition:Surjection|surjectivity]].
+Hence by definition $f$ is a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sets of Permutations of Equivalent Sets are Equivalent}
+Tags: Permutation Theory
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Set|sets]] such that:
+:$A \sim B$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+Let $\map \Gamma A$ denote the [[Definition:Set|set]] of [[Definition:Permutation|permutations]] on $A$.
+Then:
+:$\map \Gamma A \sim \map \Gamma B$
+\end{theorem}<|endoftext|>
+\section{Composition of Permutations is not Commutative}
+Tags: Permutation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\map \Gamma S$ denote the [[Definition:Set|set]] of [[Definition:Permutation|permutations]] on $S$.
+Let $\pi, \rho$ be [[Definition:Element|elements]] of $\map \Gamma S$
+Then it is not necessarily the case that:
+:$\pi \circ \rho = \rho \circ \pi$
+where $\circ$ denotes [[Definition:Composition of Mappings|composition]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S := \set {1, 2, 3}$.
+Let:
+{{begin-eqn}}
+{{eqn | l = \pi
+      | o = :=
+      | r = \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}
+}}
+{{eqn | l = \rho
+      | o = :=
+      | r = \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2}
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \pi \circ \rho
+      | r = \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1} \circ \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} 
+}}
+{{eqn | r = \dbinom {1 \ 2 \ 3} {3 \ 2 \ 1}
+}}
+{{end-eqn}}
+while:
+{{begin-eqn}}
+{{eqn | l = \rho \circ \pi
+      | r = \dbinom {1 \ 2 \ 3} {1 \ 3 \ 2} \circ \dbinom {1 \ 2 \ 3} {2 \ 3 \ 1}
+}}
+{{eqn | r = \dbinom {1 \ 2 \ 3} {2 \ 1 \ 3}
+}}
+{{eqn | o = \ne
+      | r = \pi \circ \rho
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 2}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm{\,\cdot\,}$,
+Let $x, y \in R$ and $\norm x \ne \norm y$.
+Then:
+:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y}$
+\end{theorem}
+
+\begin{proof}
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] $\norm {\,\cdot\,}$.
+By [[Non-Archimedean Norm iff Non-Archimedean Metric]] then $d$ is a [[Definition:Non-Archimedean Metric|non-Archimedean metric]] and $\struct {R, d}$ is an [[Definition:Ultrametric Space|ultrametric space]].
+Let $x, y \in R$ and $\norm x \ne \norm y$.
+By the definition of the [[Definition:Non-Archimedean Metric|non-Archimedean metric]] $d$ then:
+:$\norm x = \norm {x - 0} = \map d {x, 0}$
+and similarly:
+:$\norm y = \map d {y, 0}$
+By assumption then:
+:$\map d {x, 0} \ne \map d {y, 0}$
+By [[Three Points in Ultrametric Space have Two Equal Distances]] then:
+:$\norm {x - y} = \map d {x, y} = \max \set {\map d {x, 0}, \map d {y, 0} } = \max \set {\norm x, \norm y}$
+By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]] then:
+:$\norm {y - x} = \norm {x - y} = \max \set {\norm x, \norm y}$
+By the definition of the [[Definition:Non-Archimedean Metric|non-Archimedean metric]] $d$ then:
+:$\norm y = \norm {0 - \paren {-y} } = \map d {0, -y} = \map d {-y, 0}$
+By assumption then:
+:$\map d {x, 0} \ne \map d {-y, 0}$
+By [[Three Points in Ultrametric Space have Two Equal Distances]] then:
+:$\map d {x, -y} = \max \set {\map d {x, 0}, \map d {-y, 0} } = \max \set {\norm x, \norm y}$
+By the definition of the [[Definition:Non-Archimedean Metric|non-Archimedean metric]] $d$ then:
+:$\map d {x, -y} = \norm {x - \paren {-y} } = \norm {x + y}$
+So $\norm {x + y} = \max \set {\norm x, \norm y}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:$y \in \map {B_r} x \implies \map {B_r} y = \map {B_r} x$
+\end{theorem}
+
+\begin{proof}
+Let $y \in \map {B_r} x$.
+Let $a \in \map {B_r} y$.
+By the definition of an [[Definition:Open Ball of Normed Division Ring|open ball]], then:
+:$\norm {a - y} < r$
+:$\norm {y - x} < r$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {a - x}
+      | r = \norm {a - y + y - x}
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {a - y}, \norm{y - x} } 
+      | c = {{Defof|Non-Archimedean Division Ring Norm}}
+}}
+{{eqn | o = <
+      | r = r
+      | c = 
+}}
+{{end-eqn}}
+By the definition of an [[Definition:Open Ball of Normed Division Ring|open ball]]: 
+:$a \in \map {B_r} x$
+Hence:
+:$\map {B_r} y \subseteq \map {B_r} x$
+By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]:
+:$\norm {x - y} < r$
+By the definition of an [[Definition:Open Ball of Normed Division Ring|open ball]]:
+:$x \in \map {B_r} y$
+Similarly it follows that:
+:$\map {B_r} x \subseteq \map {B_r} y$
+By [[Definition:Set Equality/Definition 2|set equality]]:
+:$\map {B_r} y = \map {B_r} x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Centers of Closed Balls}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:$y \in \map { {B_r}^-} x \implies \map { {B_r}^-} y = \map { {B_r}^-} x$
+\end{theorem}
+
+\begin{proof}
+Let $y \in \map { {B_r}^-} x$.
+Let $a \in \map { {B_r}^-} y$.
+By the definition of an [[Definition:Closed Ball of Normed Division Ring|closed ball]], then:
+:$\norm {a - y} \le r$
+:$\norm {y - x} \le r$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm {a - x}
+      | r = \norm {a - y + y - x}
+}}
+{{eqn | r = \max \set {\norm {a - y}, \norm {y - x} } 
+      | o = \le
+      | c = {{Defof|Non-Archimedean Division Ring Norm}}
+}}
+{{eqn | r = r
+      | o = \le
+      | c = 
+}}
+{{end-eqn}}
+By the definition of a [[Definition:Closed Ball of Normed Division Ring|closed ball]], then: 
+:$a \in \map { {B_r}^-} x$.
+Hence:
+:$\map { {B_r}^-} y \subseteq \map { {B_r}^-} x$
+By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]] then:
+:$\norm {x - y} \le r$
+By the definition of a [[Definition:Closed Ball of Normed Division Ring|closed ball]], then:
+:$x \in \map { {B_r}^-} y$
+Similarly it follows that:
+:$\map { {B_r}^-} x \subseteq \map { {B_r}^-} y$
+By [[Definition:Set Equality/Definition 2|set equality]]:
+:$\map { {B_r}^-} x = \map { {B_r}^-} y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:The [[Definition:Open Ball of Normed Division Ring|open $r$-ball of $x$]], $\map {B_r} x$, is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by $\norm {\,\cdot\,}$.
+\end{theorem}
+
+\begin{proof}
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] $\norm {\,\cdot\,}$.
+By the definition of an [[Definition:Open Ball of Normed Division Ring|open ball]] in $\norm {\,\cdot\,}$:
+:$\map {B_r} x$ is an [[Definition:Open Ball|open ball]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+By [[Open Ball of Metric Space is Open Set]] then $\map {B_r} x$ is [[Definition:Open Set of Metric Space|open]] in $\struct {R, d}$.
+So it remains to show that $\map {B_r} x$ is [[Definition:Closed Set of Metric Space|closed]] in $\struct {R, d}$.
+Let $\map \cl {\map {B_r} x}$ denote the [[Definition:Closure (Metric Space)|closure]] of $\map {B_r} x$.
+Let $y \in \map \cl {\map {B_r} x}$.
+By the definition of the [[Definition:Closure (Metric Space)|closure]] of $\map {B_r} x$ then:
+:$\forall s > 0: \map {B_s} y \cap \map {B_r} x \ne \O$
+In particular:
+:$\map {B_r} y \cap \map {B_r} x \ne \O$
+Let $z \in \map {B_r} y \cap \map {B_r} x$.
+By [[Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls|Centers of Open Balls]]:
+:$\map {B_r} y = B_r \paren{z} = \map {B_r} x$
+By the definition of an [[Definition:Open Ball|open ball]]:
+:$y \in \map {B_r} y = \map {B_r} x$.
+Hence:
+:$\map \cl {\map {B_r} x} \subseteq \map {B_r} x$
+By [[Subset of Metric Space is Subset of its Closure]] then:
+:$\map {B_r} x \subseteq \map \cl {\map {B_r} x}$
+So by definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$\map \cl {\map {B_r} x} = \map {B_r} x$
+By [[Set is Closed iff Equals Topological Closure]] then $\map {B_r} x$ is [[Definition:Closed Set of Metric Space|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Closed Balls are Clopen}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:The [[Definition:Closed Ball of Normed Division Ring|closed $r$-ball of $x$]], $\map { {B_r}^-} x$, is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] by $\norm {\,\cdot\,}$.
+\end{theorem}
+
+\begin{proof}
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] $\norm {\,\cdot\,}$.
+By the definition of a [[Definition:Closed Ball of Normed Division Ring|closed ball]] in $\norm {\,\cdot\,}$ then:
+:$\map { {B_r}^-} x$ is a [[Definition:Closed Ball|closed ball]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+By [[Closed Ball is Closed in Metric Space]] then $\map { {B_r}^-} c$ is [[Definition:Closed Set of Metric Space|closed]] in $d$.
+So it remains to show that $\map { {B_r}^-} x$ is [[Definition:Open Set of Metric Space|open]] in $d$.
+Let $y \in \map { {B_r}^-} x$.
+By [[Topological Properties of Non-Archimedean Division Rings/Centers of Closed Balls|Centers of Closed Balls]] then:
+:$\map { {B_r}^-} y = \map { {B_r}^-} x$
+By the definition of an [[Definition:Open Ball|open ball]] then:
+:$y \in \map {B_r} y \subseteq \map { {B_r}^-} y = \map { {B_r}^-} x$
+By the definition of an [[Definition:Open Set of Metric Space|open set]] in a [[Definition:Metric Space|metric space]], $\map { {B_r}^-} x$ is [[Definition:Open Set of Metric Space|open]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Intersection of Open Balls}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:$\map {B_r} x \cap \map {B_s} y \ne \O \iff \map {B_r} x \subseteq \map {B_s} y$ or $\map {B_s} y \subseteq \map {B_r} x$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $z \in \map {B_r} x \cap \map {B_s} y$.
+If $r \le s$ then:
+{{begin-eqn}}
+{{eqn| l = \map {B_r} x
+     | r = \map {B_r} z
+     | c = [[Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls|Every element in an open ball is the center]]
+}}
+{{eqn| o = \subseteq 
+     | r = \map {B_s} z
+     | c = as $r \le s$
+}}
+{{eqn| r = \map {B_s} y
+     | c = [[Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls|Every element in an open ball is the center]]
+}}
+{{end-eqn}}
+Similarly, if $s \le r$ then:
+{{begin-eqn}}
+{{eqn| l = \map {B_s} y 
+     | o = \subseteq 
+     | r = \map {B_r} x
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let 
+:$\map {B_r} x \subseteq \map {B_s} y$ 
+or 
+:$\map {B_s} y \subseteq \map {B_r} x$
+By the definition of an [[Definition:Open Ball|open ball]] then:
+:$x \in \map {B_r} x \ne \O$
+:$y \in \map {B_s} y \ne \O$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Intersection of Closed Balls}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:$\map { {B_r}^-} x \cap \map { {B_s}^-} y \ne \O \iff \map { {B_r}^-} x \subseteq \map { {B_s}^-} y$ or $\map { {B_s}^-} y \subseteq \map { {B_r}^-} x$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $z \in \map { {B_r}^-} x \cap \map { {B_s}^-} y$.
+If $r \le s$ then:
+{{begin-eqn}}
+{{eqn| l = \map { {B_r}^-} x 
+     | r = \map { {B_r}^-} z
+     | c = [[Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls|Every element in an open ball is the center]]
+}}
+{{eqn| o = \subseteq
+     | r = \map { {B_s}^-} z
+     | c = as $r \le s$
+}}
+{{eqn| r = \map { {B_s}^-} y
+     | c = [[Topological Properties of Non-Archimedean Division Rings/Centers of Open Balls|Every element in an open ball is the center]]
+}}
+{{end-eqn}}
+Similarly, if $s \le r$ then:
+{{begin-eqn}}
+{{eqn| l = \map { {B_s}^-} y 
+     | o = \subseteq
+     | r = \map { {B_r}^-} x
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let:
+:$\map { {B_r}^-} x \subseteq \map { {B_s}^-} y$ 
+or:
+:$\map { {B_s}^-} y \subseteq \map { {B_r}^-} x$
+By the definition of an [[Definition:Open Ball|open ball]] then:
+:$x \in \map { {B_r}^-} x \ne \O$
+:$y \in \map { {B_s}^-} y \ne \O$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Hyperbolic Cotangent Function}
+Tags: Mittag-Leffler Expansions, Hyperbolic Cotangent Function
+
+\begin{theorem}
+:$\displaystyle \pi \, \map \coth {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 + n^2}$
+where: 
+:$z \in \C$ is not an [[Definition:Integer|integer]] multiple of $i$
+:$\coth$ is the [[Definition:Hyperbolic Cotangent|hyperbolic cotangent function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \pi \, \map \coth {\pi z}
+	| r = \pi i \, \map \cot {\pi i z}
+	| c = [[Hyperbolic Cotangent in terms of Cotangent]]
+}}
+{{eqn	| r = i \paren {\frac 1 {i z} + 2 i \sum_{n \mathop = 1}^\infty \frac z {\paren {i z}^2 - n^2} }
+	| c = [[Mittag-Leffler Expansion for Cotangent Function]]
+}}
+{{eqn	| r = \frac 1 z - 2 \sum_{n \mathop = 1}^\infty \frac z {-z^2 - n^2}
+	| c = $i^2 = -1$
+}}
+{{eqn	| r = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \frac z {z^2 + n^2}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Secant Function}
+Tags: Mittag-Leffler Expansions, Secant Function
+
+\begin{theorem}
+:$\displaystyle \pi \, \map \sec {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1}^2 - 4 z^2}$
+where: 
+:$z \in \C$ is not a [[Definition:Half-Integer|half-integer]]
+:$\sec$ is the [[Definition:Secant Function|secant function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Pringsheim's Theorem}
+Tags: 
+
+\begin{theorem}
+Let $f$ be a [[Definition:Holomorphic Function|holomorphic function]] defined on a unit disc centered at the origin of the complex plane and is denoted by its [[Definition:Taylor Series|Taylor series]]:
+:$\map f z = \displaystyle \sum_{n \mathop = 0}^{\infty} c_n z^n$
+Let:
+:$(1): \quad \forall n \ge 0: c_n \ge 0$
+:$(2): \quad$ the [[Definition:Radius of Convergence of Complex Power Series|radius of convergence]] of the [[Definition:Taylor Series|Taylor series]] of function $f$ is $1$.
+Then $z = 1$ is an [[Definition:Isolated Singularity|isolated singularity]] of $f$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Alfred Pringsheim|cat = Pringsheim}}
+kxqv9jc0q8o2yx3gz6o85k0s5g5p8et
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Tangent Function}
+Tags: Mittag-Leffler Expansions, Tangent Function
+
+\begin{theorem}
+:$\displaystyle \pi \map \tan {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {\paren {2 n + 1}^2 - 4 z^2}$
+where: 
+:$z \in \C$ is not a [[Definition:Half-Integer|half-integer]]
+:$\tan$ is the [[Definition:Tangent Function|tangent function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Hyperbolic Tangent Function}
+Tags: Mittag-Leffler Expansions, Hyperbolic Tangent Function
+
+\begin{theorem}
+:$\displaystyle \pi \map \tanh {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {4 z^2 + \paren {2 n + 1}^2}$
+where:
+:$z \in \C$ is not a [[Definition:Half-Integer|half-integer]] multiple of $i$
+:$\tanh$ is the [[Definition:Hyperbolic Tangent|hyperbolic tangent function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Hyperbolic Secant Function}
+Tags: Mittag-Leffler Expansions, Hyperbolic Secant Function
+
+\begin{theorem}
+:$\displaystyle \pi \, \map \sech {\pi z} = 4 \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {2 n + 1} {\paren {2 n + 1}^2 + 4 z^2}$
+where: 
+:$z \in \C$ is not a [[Definition:Half-Integer|half-integer]] multiple of $i$ 
+:$\sech$ is the [[Definition:Hyperbolic Secant|hyperbolic secant function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Mittag-Leffler Expansion for Hyperbolic Cosecant Function}
+Tags: Mittag-Leffler Expansions, Hyperbolic Cosecant Function
+
+\begin{theorem}
+:$\ds \pi \map \csch {\pi z} = \frac 1 z + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac z {z^2 + n^2}$
+where:
+:$z \in \C$ is not an [[Definition:Integer|integer]] multiple of $i$
+:$\csch$ is the [[Definition:Hyperbolic Cosecant|hyperbolic cosecant function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Subtraction has no Identity Element}
+Tags: Numbers, Subtraction, Identity Elements
+
+\begin{theorem}
+The [[Definition:Binary Operation|operation]] of [[Definition:Subtraction|subtraction]] on [[Definition:Standard Number System|numbers]] of any kind has no [[Definition:Identity Element|identity]].
+\end{theorem}
+
+\begin{proof}
+Supposed there exists an [[Definition:Identity Element|identity]] $e$ in one of the [[Definition:Standard Number System|standard number systems]] $\F$.
+{{begin-eqn}}
+{{eqn | lo= \forall x \in \F:
+      | l = x
+      | r = x - e
+      | c = 
+}}
+{{eqn | r = e - x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + \paren {-e}
+      | r = e + \paren {-x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + x
+      | r = e + e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+That is:
+:$\forall x \in \F: x = e$
+from which it follows that $\F$ has no such $e$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Element of Multiplication on Numbers}
+Tags: Numbers
+
+\begin{theorem}
+On all the number systems:
+* [[Definition:Natural Numbers|natural numbers]] $\N$
+* [[Definition:Integer|integers]] $\Z$
+* [[Definition:Rational Number|rational numbers]] $\Q$
+* [[Definition:Real Number|real numbers]] $\R$
+* [[Definition:Complex Number|complex numbers]] $\C$
+the [[Definition:Zero Element|zero element]] of [[Definition:Multiplication|multiplication]] is [[Definition:Zero (Number)|zero ($0$)]].
+\end{theorem}
+
+\begin{proof}
+This is demonstrated by showing that:
+:$n \times 0 = 0 = 0 \times n$
+for all $n$ in all [[Definition:Standard Number System|standard number systems]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Addition on Numbers has no Zero Element}
+Tags: Numbers, Addition, Zero Elements
+
+\begin{theorem}
+On all the number systems:
+* [[Definition:Natural Numbers|natural numbers]] $\N$
+* [[Definition:Integer|integers]] $\Z$
+* [[Definition:Rational Number|rational numbers]] $\Q$
+* [[Definition:Real Number|real numbers]] $\R$
+* [[Definition:Complex Number|complex numbers]] $\C$
+there exists no [[Definition:Zero Element|zero element]] for [[Definition:Addition|addition]].
+\end{theorem}
+
+\begin{proof}
+Suppose $z$ is a [[Definition:Zero Element|zero element]] for [[Definition:Addition|addition]] in a [[Definition:Standard Number System|standard number system]] $\F$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \F
+      | l = n + z
+      | r = z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | r = 0
+      | c = subtracting $z$ from both sides
+}}
+{{end-eqn}}
+As $n$ is arbitrary, and therefore not always $0$, it follows there can be no such $z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $F$ and its [[Definition:Partial Derivative|partial derivatives]] $F_y, F_{y'}$ be [[Definition:Real Function|real functions]], defined on the [[Definition:Closed Interval|closed interval]] $I = \closedint a b$.
+Let $F, F_y, F_{y'} $ be [[Definition:Continuous on Interval|continuous]] at every [[Definition:Point|point]] $\tuple {x, y}$ for all [[Definition:Finite|finite]] $y'$.
+Suppose there exists a [[Definition:Constant Mapping|constant]] $k > 0$ such that:
+:$\map {F_y} {x, y, y'} > k$
+Suppose there exist [[Definition:Real Function|real functions]] $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ [[Definition:Bounded Real-Valued Function|bounded]] in every [[Definition:Region|bounded region]] of the [[Definition:The Plane|plane]] such that:
+:$\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
+Then one and only one [[Definition:Integral Curve|integral curve]] of [[Definition:Differential Equation/Ordinary|equation]] $y'' = \map F {x, y, y'}$ passes through any two [[Definition:Point|points]] $\tuple {a, A}$ and $\tuple {b, B}$ such that $a \ne b$.
+\end{theorem}
+
+\begin{proof}
+=== Lemma 1 (Uniqueness) ===
+{{AimForCont}} there are two [[Definition:Integral Curve|integral curves]] $y = \map {\phi_1} x$ and $y = \map {\phi_2} x$ such that:
+:$\map {y''} x = \map F {x, y, y'}$
+Define a [[Definition:Mapping|mapping]] $\delta: I \to S \subset \R$:
+:$\map \delta x = \map {\phi_2} x - \map {\phi_1} x$
+From definition it follows that:
+:$\map \delta a = \map \delta b = 0$
+Then the [[Definition:Second Derivative|second derivative]] of $\delta$ yields:
+{{begin-eqn}}
+{{eqn | l = \map {\delta''} x 
+      | r = \map {\phi_2''} x - \map {\phi_1''} x 
+      | c = Definition of $\delta$
+}}
+{{eqn | r = \map F {x, \phi_2, \phi_2'} - \map F {x, \phi_1, \phi_1'}
+      | c = as $y'' = \map F {x, y', y''}$
+}}
+{{eqn | r = \map F {x, \phi_1 + \delta, \phi_1' + \delta'} - \map F {x, \phi_1, \phi_1'}
+      | c = Definition of $\delta$
+}}
+{{eqn | n = 1
+      | r = \map \delta x F_y^* + \map {\delta'} x F_{y'}^*
+      | c = Multivariate [[Mean Value Theorem]] {{WRT}} $y, y'$
+}}
+{{end-eqn}}
+where:
+:$F_y^* = \map {F_y} {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$
+:$F_{y'}^* = \map {F_{y'} } {x, \phi_1 + \theta \delta, \phi_1' + \theta \delta'}$
+and $0 < \theta < 1$.
+Suppose:
+:$\forall x \in I: \map {\phi_2} x \ne \map {\phi_1} x$
+Then there are two possibilities for $\delta$:
+:$\map \delta x$ attains a [[Definition:Positive Real Number|positive]] [[Definition:Local Maximum|maximum]] within $\tuple {a, b}$
+:$\map \delta x$ attains a [[Definition:Negative Real Number|negative]] [[Definition:Local Minimum|minimum]] within $\tuple {a, b}$.
+Denote this [[Definition:Point|point]] by $\xi$.
+Suppose $\xi$ denotes a [[Definition:Local Maximum|maximum]].
+Then:
+:$\map {\delta''} \xi \le 0$
+:$\map \delta \xi > 0$
+:$\map {\delta'} \xi = 0$
+These, together with $(1)$, imply that $F_y^* \le 0$. 
+This is in [[Proof by Contradiction|contradiction]] with [[Definition:Assumption|assumption]].
+For the [[Definition:Local Minimum|minimum]] the [[Definition:Inequality|inequalities]] are reversed, but the last [[Definition:Equality|equality]] is the same.
+Therefore, it must be the case that:
+:$\map {\phi_1} x = \map {\phi_2} x$
+{{qed|lemma}}
+=== Lemma 2 ===
+Suppose
+:$\map {y''} x = \map F {x, y, y'}$
+for all $x \in \closedint a c$, where:
+:$\map y a = a_1$
+:$\map y c = c_1$
+Then the following [[Definition:Bound of Real-Valued Function|bound]] holds:
+$\size {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \le \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
+=== Proof ===
+As a consequence of $y'' = \map F {x, y, y'}$ we have:
+{{begin-eqn}}
+{{eqn | l = y''
+      | r = \map F {x, y, y'}
+}}
+{{eqn | r = \map F {x, y, y'} - \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, y'} + \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, y'}
+}}
+{{eqn | n = 2
+      | r = \sqbrk {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \map {F_y} {x, \psi, y'} + \map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a},y'}
+      | c = [[Mean Value Theorem]] {{WRT}} $y$
+}}
+{{end-eqn}}
+where:
+:$\psi = \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + \theta \sqbrk {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} }$
+and:
+:$0 < \theta < 1$
+{{explain|the meaning of the square brackets above}}
+Note that the term $\chi$, defined as:
+:$\chi = \map y x - \dfrac {a_1 \paren {c - a} + c_1 \paren {x - a} } {c - a}$ 
+vanishes at $x = a$ and $x = c$.
+Unless $\chi$ is identically zero, there exists a [[Definition:Point|point]] $\xi\in\openint a b$ such one of the following holds:
+:$\chi$ attains a [[Definition:Positive Real Number|positive]] [[Definition:Local Maximum|maximum]] at $\xi$
+:$\chi$ attains a [[Definition:Negative Real Number|negative]] [[Definition:Local Minimum|minimum]] at $\xi$. 
+In the first case:
+:$\map {y''} \xi \le 0$,
+:$\map {y'} \xi = \dfrac {c_1 - a} {c - a}$ 
+which implies:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \ge
+      | r = \map {y''} \xi
+}}
+{{eqn | r = \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } + \map {F_y} {x, \map \psi \xi, \map {y'} x} \sqbrk {\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} }
+      | c = equation $(2)$
+}}
+{{eqn | o = \ge
+      | r = \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } + k \sqbrk {\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} }
+      | c = Assumption of Theorem
+}}
+{{end-eqn}}
+{{explain|the meaning of the square brackets above}}
+Hence:
+:$\map y \xi - \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a} \le -\dfrac 1 k \map F {\xi, \dfrac {a_1 \paren {c - \xi} + c_1 \paren {\xi - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} }$
+The [[Definition:Negative Real Number|negative]] [[Definition:Local Minimum|minimum]] part is proven analogously.
+Hence, the following holds: 
+:$\size {\map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } \le \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
+{{qed|lemma}}
+=== Lemma 3 ===
+Suppose that for $x \in I$:
+:$\map {y''} x = \map F {x, y, y'}$
+where:
+:$\map y a = a_1$
+:$\map y c = c_1$
+Then:
+:$\forall x \in I: \exists M \in \R: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le M$
+=== Proof ===
+Let $\AA$ and $\BB$ be the [[Definition:Least Upper Bound|least upper bounds]] of $\map \alpha {x, y}$ and $\map \beta {x, y}$ respectively in the [[Definition:Rectangle|rectangle]] $a \le x \le c$, $\size y \le m + max \set {\size a_1, \size c_1}$
+where:
+:$m = \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
+Suppose that $\AA \ge 1$.
+Let $u, v$ be [[Definition:Real Function|real functions]] such that 
+{{begin-eqn}}
+{{eqn | n = 3
+      | l = \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m
+      | r = \dfrac {\Ln u} {2 \AA}
+}}
+{{eqn | n = 4
+      | l = -\map y x + \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m
+      | r = \dfrac {\Ln v} {2 \AA}
+}}
+{{end-eqn}}
+{{explain|The precise meaning of $\Ln$ in this context}}
+Due to Lemma 2, for $x \in I$ the {{LHS}}s of $(3)$ and $(4)$ are not [[Definition:Negative|negative]]. 
+Thus:
+:$\forall x \in I: u, v \ge 1$
+[[Definition:Differentiation|Differentiate]] [[Definition:Equation|equations]] $(3)$ and $(4)$ {{WRT|Differentiation}} $x$:
+{{begin-eqn}}
+{{eqn| n = 5
+     | l = \map {y'} x - \dfrac {c_1 - a_1} {c - a}
+     | r = \dfrac {u'} {2 \AA u} 
+}}
+{{eqn| n = 6
+     | l = -\map {y'} x + \dfrac {c_1 - a_1} {c - a}
+     | r = \dfrac {v'} {2 \AA v}
+}}
+{{end-eqn}}
+[[Definition:Differentiation|Differentiate]] again: 
+{{begin-eqn}}
+{{eqn| n = 7
+     | l = \map {y''} x
+     | r = \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}
+}} 
+{{eqn| n = 8
+     | l = -\map {y''} x
+     | r = \dfrac {v''} {2 \AA v} - \dfrac {v'^2} {2 \AA v^2}
+}}
+{{end-eqn}}
+By [[Definition:Assumption|assumption]]:
+{{begin-eqn}}
+{{eqn | l = \size {\map F {x, y, y'} }
+      | r = \size {\map {y''} x}
+}}
+{{eqn | o = \le
+      | r = \map \alpha {x, y} \map {y'^2} x + \map \beta {x, y}
+}}
+{{eqn | o = \le
+      | r = \AA \map {y'^2} x + \BB
+}}
+{{eqn | o = \le
+      | r = 2 \AA \map {y'^2} x + \BB
+}}
+{{eqn | r = 2 \AA {\paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2} + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2 - 2 \map {y'} x \dfrac {c_1 - a_1} {c - a} + \BB
+}}
+{{eqn | n = 9
+      | o = \le
+      | r = 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1
+}}
+{{end-eqn}} 
+where:
+:$\BB_1 = \BB + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2$
+Then:
+{{begin-eqn}}
+{{eqn | l = y''
+      | r = \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}
+      | c = Equation $(7)$
+}}
+{{eqn | o = \ge
+      | r = -2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 - \BB_1
+      | c = Inequality $(9)$
+}}
+{{eqn | r = -2 \AA \paren {\dfrac {u'} {2 \AA u} }^2 - \BB_1
+      | c = Equation $(5)$
+}}
+{{end-eqn}}
+[[Definition:Multiplication|Multiply]] the [[Definition:Inequality|inequality]] by $2 \AA u$ and simplify:
+{{begin-eqn}}
+{{eqn | l = u''
+      | o = \ge
+      | r = -2 \AA \BB_1 u
+}}
+{{end-eqn}}  
+Similarly: 
+{{begin-eqn}}
+{{eqn | l = y''
+      | r = - \dfrac {v''} {2 \AA v} + \dfrac {v'^2} {2 \AA v^2}
+      | c = Equation $(8)$
+}}
+{{eqn | o = \le
+      | r = 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1
+      | c = Inequality $(9)$
+}}
+{{eqn | r = 2 \AA \paren {\dfrac {v'} {2 \AA v} }^2 + \BB_1
+      | c = Equation $(6)$
+}}
+{{end-eqn}}
+[[Definition:Multiplication|Multiply]] the [[Definition:Inequality|inequality]] by $-2 \AA v$ and simplify:
+{{begin-eqn}}
+{{eqn | l = v''
+      | o = \ge
+      | r = -2 \AA \BB_1 v
+}}
+{{end-eqn}}
+Note that $\map u a = \map u c$ and $\map v a = \map v c$.
+From [[Intermediate Value Theorem]] it follows that
+:$\exists K \subset I: \forall x_0 \in K: \map {y'} {x_0} - \dfrac {c_1 - a_1} {c - a} = 0$
+[[Definition:Point|Points]] $x_0$ divide $I$ into [[Definition:Subinterval|subintervals]].
+Due to $(5)$ and $(6)$ both $\map {u'} x$ and $\map {v'} x$ maintain [[Definition:Sign of Number|sign]] in the [[Definition:Subinterval|subintervals]] and vanish at one or both [[Definition:Endpoint of Real Interval|endpoints]] of each [[Definition:Subinterval|subinterval]].
+Let $J$ be one of the [[Definition:Subinterval|subintervals]].
+Let [[Definition:Real Function|functions]] $\map {u'} x$, $\map {v'} x$ be zero at $\xi$, the right [[Definition:Endpoint of Real Interval|endpoint]]. 
+The quantity 
+:$\map {y'} x - \dfrac {c_1 - a_1} {c - a}$
+has to be either [[Definition:Positive|positive]] or [[Definition:Negative|negative]].
+Suppose it is [[Definition:Positive|positive]] in $J$. 
+From $(5)$, $u'$ is not [[Definition:Negative|negative]].
+[[Definition:Multiplication|Multiply]] both sides of $(10)$ by $u'$: 
+:$u'' u' \ge - 2 \AA \BB_1 u u'$
+[[Definition:Definite Integral|Integrating]] this from $x \in J$ to $\xi$, together with $\map {u'} \xi = 0$, yields:
+:$-\map {u'^2} x \ge - 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {u'^2} x
+      | o = \le
+      | r = 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}
+}}
+{{eqn | o = \le
+      | r = 2 \AA \BB_1 \map {u^2} \xi
+}}
+{{eqn | r = 2 \AA \BB_1 \exp \sqbrk {4 \AA \paren {m + \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } }
+      | c = Equation $(3)$
+}}
+{{eqn | o = \le
+      | r = 2 \AA \BB_1 e^{8 \AA m}
+      | c = Lemma 2
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\map {u'^2} x} {\map {u^2} x}
+      | o = \le
+      | r = 2 \AA \BB_1 e^{8 \AA m}
+      | c = $u \ge 1$
+}}
+{{eqn | ll= \leadsto
+      | l = 4 \AA^2 \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2
+      | o = \le
+      | r = 2 \AA \BB_1 e^{8 \AA m}
+      | c = Equation $(5)$
+}}
+{{end-eqn}}
+{{explain|square brackets in the above}}
+Hence:
+:$\forall x \in J: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le \sqrt {\dfrac {\BB_1} {2 \AA} } e^{4 \AA m}$
+Similar arguments for aforementioned quantity being [[Definition:Negative|negative]].
+{{qed|lemma}}
+Consider a [[Definition:Plane|plane]] with [[Definition:Coordinate Axis|axes]] denoted by $x$ and $y$:
+:[[File:BernsteinCurve.png|500px]]
+Put the [[Definition:Point|point]] $A \tuple {a, a_1}$.
+Through this [[Definition:Point|point]] draw an [[Definition:Arc of Curve|arc]] of the [[Definition:Integral Curve|integral curve]] such that $\map {y'} a = 0$.
+On this [[Definition:Arc of Curve|arc]] put another [[Definition:Point|point]] $D \tuple {d, d_1}$.
+For $x \ge d$ draw the [[Definition:Straight Line|straight line]] $y = d_1$.
+Put the [[Definition:Point|point]] $B \tuple {b, b_1}$.
+For $y \ge d_1$ draw the [[Definition:Straight Line|straight line]] $x = b_1$.
+Denote the [[Definition:Intersection (Geometry)|intersection]] of these two [[Definition:Straight Line|straight lines]] by $Q$.
+Then the broken [[Definition:Curve|curve]] $DQB$ connects [[Definition:Point|points]] $D$ and $B$.
+Choose any [[Definition:Point|point]] of $DQB$ and denote it by $P \tuple {\xi, \xi_1}$.
+Consider a [[Definition:Family|family]] of [[Definition:Integral Curve|integral curves]] $y = \map \phi {x, \alpha}$, passing through the [[Definition:Point|point]] $A$, where $\alpha = \map {y'} a$.
+For $\alpha = 0$ the [[Definition:Integral Curve|integral curve]] concides with $AD$.
+Suppose [[Definition:Point|point]] $P$ is sufficiently close to the [[Definition:Point|point]] $D$.
+By Lemma 1, there exists a unique [[Definition:Curve|curve]] $AP$.
+Then, $\alpha$ can be found uniquely from:
+:$d_1 = \map \phi {\xi, \alpha}$.
+Due to uniqueness and [[Definition:Continuous on Interval|continuity]], it follows that $\xi$ is a [[Definition:Monotone Real Function|monotonic function]] of $\alpha$.
+Hence, $\alpha$ is a [[Definition:Monotone Real Function|monotonic function]] of $\xi$.
+Put the [[Definition:Point|point]] $R$ in between of $D$ and $Q$.
+Suppose, that, except for $R$, any [[Definition:Point|point]] of $DR$ can be reached by the aforementioned procedure.
+When $\xi$ approaches the [[Definition:Abscissa|abscissa]] $r$ of $R$, $\alpha$ [[Definition:Monotone Real Function|monotonically]] approaches a [[Definition:Limit of Real Function|limit]].
+If it is different from $\pm \dfrac \pi 2$, [[Definition:Point|point]] $R$ is attained.
+By assumption, $R$ is not attained.
+Thus: 
+:$\displaystyle \lim_{\xi \mathop \to r} \alpha = \pm \dfrac \pi 2$
+In other words, as $P$ approaches $R$, the [[Definition:Derivative of Real Function at Point|derivative]] of $\map y x$ joining $A$ to $P$ will not be [[Definition:Bound of Real-Valued Function|bounded]] at $x = a$.
+This contradicts the [[Definition:Bound of Real-Valued Function|bounds]] from Lemma 2 and 3, and the fact that the [[Definition:Subtraction|difference]] of [[Definition:Abscissa|abscissas]] of $A$ and $P$ does not approach $0$.
+Therefore, $R$ can be reached.
+Similar argument can be repeated for the [[Definition:Line/Segment|line segment]] $QB$.
+{{qed}}
+{{Namedfor|Sergei Natanovich Bernstein|cat = Bernstein, Sergei}}
+\end{proof}<|endoftext|>
+\section{Real Numbers under Subtraction do not form Semigroup}
+Tags: Real Subtraction, Semigroups
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] under [[Definition:Real Subtraction|subtraction]] $\struct {\R, -}$ does not form a [[Definition:Semigroup|semigroup]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Subtraction on Numbers is Not Associative]].
+Hence $\struct {\R, -}$ is not a [[Definition:Semigroup|semigroup]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group has Latin Square Property/Additive Notation}
+Tags: Group has Latin Square Property
+
+\begin{theorem}
+Let $\struct {G, +}$ be a [[Definition:Group|group]].
+Then $G$ satisfies the [[Definition:Latin Square Property|Latin square property]].
+That is, for all $a, b \in G$, there exists a [[Definition:Unique|unique]] $g \in G$ such that $a + g = b$.
+Similarly, there exists a [[Definition:Unique|unique]] $h \in G$ such that $h + a = b$.
+\end{theorem}
+
+\begin{proof}
+From [[Group has Latin Square Property]], we have that:
+{{begin-eqn}}
+{{eqn | l = a + g
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g
+      | r = \paren {-a} + b
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = h + a
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = b + \paren {-a}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Count of Binary Operations on Set/Sequence}
+Tags: Count of Binary Operations on Set
+
+\begin{theorem}
+The [[Definition:Integer Sequence|sequence]] of $N$ for each $n$ begins:
+$\begin{array} {c|rr}
+n & n^2 & N = n^{\paren {n^2} } \\
+\hline
+1 & 1 & 1 \\
+2 & 4 & 16 \\
+3 & 9 & 19 \ 683 \\
+4 & 16 & 4 \ 294 \ 967 \ 296 \\
+\end{array}$
+There are still only $4$ [[Definition:Element|elements]] in a [[Definition:Set|set]], and already there are over $4$ thousand million different possible [[Definition:Algebraic Structure|algebraic structures]].
+\end{theorem}<|endoftext|>
+\section{Count of Commutative Binary Operations on Set/Sequence}
+Tags: Count of Commutative Binary Operations on Set
+
+\begin{theorem}
+The [[Definition:Integer Sequence|sequence]] of $N$ for each $n$ begins:
+$\begin{array} {c|cr}
+n & \dfrac {n \paren {n + 1} }2 & n^{\frac {n \paren {n + 1} } 2} \\
+\hline
+1 & 1 & 1 \\
+2 & 3 & 8 \\
+3 & 6 & 729 \\
+4 & 10 & 1 \ 048 \ 576 \\
+\end{array}$
+and so on.
+\end{theorem}<|endoftext|>
+\section{Mittag-Leffler’s Expansion Theorem}
+Tags: Mittag-Leffler Expansions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Meromorphic Function|meromorphic function]] with only [[Definition:Simple Pole|simple poles]] continuous, or with a [[Definition:Removable Singularity|removable singularity]], at $0$. 
+Let $X$ be the set of [[Definition:Pole|poles]] of $f$. 
+For $N \in \N$, let $C_N$ be a circle, centred at the origin, of radius $R_N$, where $R_N \to \infty$ as $N \to \infty$, such that $\partial C_N$ contains no poles of $f$ for any $N$. 
+Let $M > 0$ be a [[Definition:Real Number|real number]] independent of $N$ such that for all $z \in \partial C_N$, $\cmod {\map f z} < M$ , for all $N \in \N$.
+Then:
+:$\displaystyle \map f z = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {z - n} + \frac 1 n}$ 
+where:
+:$\Res f n$ is the [[Definition:Residue|residue]] of $f$ at $n$
+:$z$ is not a pole of $f$
+:$\displaystyle \map f 0 = \lim_{z \mathop \to 0} \map f z$ if $f$ has a removable singularity at $0$.
+\end{theorem}
+
+\begin{proof}
+Let $\zeta \in \C \setminus X$. 
+Then: 
+:$\displaystyle \frac {\map f z} {z - \zeta}$ 
+has simple poles for $z \in X \cup \set \zeta$. 
+Let $X_N$ be the set of poles contained within $C_N$. 
+Then: 
+{{begin-eqn}}
+{{eqn	| l = \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z - \zeta} \rd z
+	| r = \Res {\frac {\map f z} {z - \zeta} } \zeta + \sum_{n \mathop \in X_N} \Res {\frac {\map f z} {z - \zeta} } n
+	| c = [[Residue Theorem]]
+}}
+{{eqn	| r = \lim_{z \mathop \to \zeta} \paren {\frac {\paren {z - \zeta} \map f z} {z - \zeta} } + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \frac {\paren {z - n} \map f z} {z - \zeta} }
+	| c = [[Residue at Simple Pole]]
+}}
+{{eqn	| r = \lim_{z \mathop \to \zeta} \map f z + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \paren {\paren {z - n} \map f z } \cdot \lim_{z \mathop \to n} \frac 1 {z - \zeta} }
+	| c = [[Combination Theorem for Limits of Functions/Product Rule|Product Rule for Limits]] 
+}}
+{{eqn	| r = \map f \zeta + \sum_{n \mathop \in X_N} \frac {\Res f n} {n - \zeta}
+	| c = $f$ is continuous at $\zeta$, [[Residue at Simple Pole]]
+}}
+{{end-eqn}}
+Setting $\zeta = 0$, we obtain: 
+:$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$ 
+So: 
+:$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
+It remains to show that the integral on the left hand side vanishes as $N \to \infty$.
+We have:
+{{begin-eqn}}
+{{eqn	| l = \cmod {\frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z}
+	| r = \cmod {\frac \zeta {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z \paren {z - \zeta} } \rd z}
+}}
+{{eqn	| o = \le
+	| r = \frac {\cmod \zeta} {2 \pi} \cdot \frac M {R_N \paren {R_N - \cmod \zeta} } \cdot 2 \pi R_N
+	| c = [[Triangle Inequality for Contour Integrals]], [[Reverse Triangle Inequality/Real and Complex Fields|Reverse Triangle Inequality]], noting that $\cmod z$ for $z \in \partial C_N$  
+}}
+{{eqn	| o = \sim
+	| r = \frac 1 {R_N}
+}}
+{{eqn	| o = \to
+	| r = 0
+	| c = as $N \to \infty$, $R_N \to \infty$
+}}
+{{end-eqn}}
+Letting $N \to \infty$ gives: 
+:$\displaystyle 0 = \map f \zeta + \sum_{n \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$
+Giving:
+:$\displaystyle \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$ 
+{{qed}}
+{{Namedfor|Magnus Gustaf Mittag-Leffler|cat = Mittag-Leffler}}
+\end{proof}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+Let $\struct {R, +, *, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] $\norm {\,\cdot\,}$.
+Let $p \in \R_{\ge 1} \cup \set \infty$.
+Let $d_p$ be the [[Definition:P-Product Metric|$p$-product metric]] on $R \times R$.
+Let $R^* = R \setminus \set 0$
+Let $d^*$ be the [[Definition:Restriction|restriction]] of $d$ to $R^*$.
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous/Addition}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+:$+ : \struct {R \times R, d_p} \to \struct{R,d}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+\end{theorem}
+
+\begin{proof}
+By [[P-Product Metric Induces Product Topology|$p$-Product Metric Induces Product Topology]] and [[Continuous Mapping is Continuous on Induced Topological Spaces]], it suffices to consider the case $p = \infty$.
+Let $\tuple {x_0, y_0} \in R \times R$.
+Let $\epsilon > 0$ be given.
+Let $\tuple {x, y} \in R \times R$ such that:
+:$\map {d_\infty} {\tuple {x, y},\tuple{x_0, y_0} } < \dfrac \epsilon 2$
+By the definition of the [[Definition:P-Product Metric|product metric $d_\infty$]] then:
+:$\max \set {\map d {x, x_0}, \map d {y, y_0} } < \dfrac \epsilon 2$
+or equivalently:
+:$\map d {x, x_0} < \dfrac \epsilon 2$
+:$\map d {y, y_0} < \dfrac \epsilon 2$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {x_0 + y_0, x + y}
+      | r = \norm {\paren {x_0 + y_0} - \paren {x+y} }
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \norm {\paren {x_0 - x} + \paren {y_0 - y} }
+      | c = 
+}}
+{{eqn | r = \norm {x_0 - x } + \norm {y_0 - y}
+      | o = \le
+      | c = [[Definition:Norm Axioms|Norm Axiom (N3) (Triangle Inequality)]]
+}}
+{{eqn | r = \map d {x_0, x} + \map d {y_0, y}
+      | o = \le
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \dfrac \epsilon 2 + \dfrac \epsilon 2
+      | o = <
+      | c = 
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Since $\tuple {x_0, y_0}$ and $\epsilon$ were arbitrary, by the definition of [[Definition:Continuous Mapping (Metric Space)|continuity]] then the [[Definition:Mapping|mapping]]:
+:$+ : \struct {R \times R, d_\infty} \to \struct {R, d}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous/Negation}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+:$\eta: \struct {R, d} \to \struct {R, d}: \map \eta x = -x$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $x_0 \in R$.
+Let $\epsilon > 0$ be given.
+Let $x \in R$ such that:
+:$\map d {x, x_0} < \epsilon$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {-x, -x_0}
+      | r = \norm {-x - \paren {-x_0} }
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \norm {-x + x_0}
+}}
+{{eqn | r = \norm {x_0 - x}
+      | c = $+$ is [[Definition:Commutative Operation|Commutative]].
+}}
+{{eqn | r = \map d {x_0, x}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \map d {x, x_0}
+      | c = [[Definition:Metric Space Axioms|Metric Space Axiom $(\text M 2)$]]
+}}
+{{eqn | r = \epsilon
+      | o = <
+}}
+{{end-eqn}}
+Since $x_0$ and $\epsilon$ were arbitrary, by the definition of [[Definition:Continuous Mapping (Metric Space)|continuity]] then the [[Definition:Mapping|mapping]]:
+:$\eta: \struct {R, d} \to \struct {R, d} : \map \eta x = -x$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous/Multiplication}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+:$* : \struct {R \times R, d_p} \to \struct {R, d}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+\end{theorem}
+
+\begin{proof}
+By [[P-Product Metric Induces Product Topology|$p$-Product Metric Induces Product Topology]] and [[Continuous Mapping is Continuous on Induced Topological Spaces]], it suffices to consider the case $p = \infty$.
+Let $\tuple {x_0, y_0} \in R \times R$.
+Let $\epsilon > 0$ be given.
+Let $\delta = \min \set {\dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} }, 1}$
+Since $1 + \norm {y_0} + \norm {x_0} > 0$ then $\delta > 0$
+Let $\tuple {x,y} \in R \times R$ such that:
+:$\map {d_\infty} {\tuple {x, y}, \tuple{x_0, y_0} } < \delta$
+By the definition of the [[Definition:P-Product Metric|$p$-product metric $d_\infty$]]:
+:$\max \set {\map d {x, x_0}, \map d {y, y_0}} < \delta$
+or equivalently:
+:$\map d {x, x_0} < \delta$
+:$\map d {y, y_0} < \delta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm y
+      | r = \norm {y - y_0 + y_0}
+      | c =
+}}
+{{eqn | o = \le
+      | r = \norm {y - y_0}  + \norm {y_0}
+      | c = {{NormAxiom|3}}
+}}
+{{eqn | o = \le
+      | r = \map d {y, y_0}  + \norm {y_0}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | o = <
+      | r = \delta  + \norm {y_0}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = 1  + \norm {y_0}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map d {x y, x_0 y_0}
+      | r = \norm {x y - x_0 y_0}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \norm {x y - x_0 y + x_0 y - x_0 y_0}
+      | c =
+}}
+{{eqn | o = \le
+      | r = \norm {x y - x_0 y} + \norm {x_0 y - x_0 y_0}
+      | c = {{NormAxiom|3}}
+}}
+{{eqn | o = \le
+      | r = \norm {x - x_0} \norm y + \norm {x_0} \norm {y - y_0}
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = \le
+      | r = \map d {x, x_0} \norm y + \norm {x_0} \map d {y, y_0}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | o = <
+      | r = \delta \norm y + \norm {x_0} \delta
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \delta \paren {\norm y + \norm {x_0} }
+      | c =
+}}
+{{eqn | o = <
+      | r = \delta \paren {1 + \norm {y_0} + \norm {x_0} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \dfrac \epsilon {1 + \norm {y_0} + \norm {x_0} } \paren {1 + \norm {y_0} + \norm {x_0} }
+      | c = 
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+We have that $\tuple {x_0, y_0}$ and $\epsilon$ are arbitrary.
+Hence, by the definition of [[Definition:Continuous Mapping (Metric Space)|continuity]], the [[Definition:Mapping|mapping]]:
+:$* : \struct {R \times R, d_\infty} \to \struct {R, d}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous/Inversion}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+:$\iota : \struct {R^* ,d^*} \to \struct {R, d} : \map \iota x = x^{-1}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $x_0 \in R^*$.
+Let $\epsilon \gt 0$ be given.
+Let $\delta = \min \set {\dfrac {\norm {x_0} } 2, \dfrac {\norm {x_0}^2 \epsilon} 2 }$
+Let $x \in R^*$ such that:
+:$\map {d^*} {x, x_0} < \delta$
+By the definition of the [[Definition:Metric Subspace|subspace metric]] on $R^*$ and the definition of the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] on $R$:
+:$\map {d^*} {x, x_0} = \map d {x, x_0} = \norm {x - x_0} < \delta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x_0}
+      | o = \le
+      | r = \norm {x - x_0} + \norm x
+      | c = {{NormAxiom|3}}
+}}
+{{eqn | o =
+}}
+{{eqn | o = <
+      | r = \delta + \norm x
+      | c = by assumption: $\norm {x - x_0} < \delta$
+}}
+{{eqn | o =
+}}
+{{eqn | o = \le
+      | r = \dfrac {\norm {x_0} } 2 + \norm x
+      | c = by assumption: $\delta \le \dfrac {\norm {x_0} } 2$
+}}
+{{eqn | o =
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\norm {x_0} } 2
+      | o = <
+      | r = \norm x
+      | c = subtracting $\dfrac {\norm {x_0} } 2$ from both sides
+}}
+{{eqn | o =
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 2 {\norm {x_0} }
+      | o = >
+      | r = \dfrac 1 {\norm x}
+      | c = inverting both sides of the equation
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map d {x^{-1}, x_0^{-1} }
+      | r = \norm {x^{-1} - x_0^{-1} }
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | o =
+}}
+{{eqn | r = \dfrac 1 {\norm x} \paren {\norm x \norm {x^{-1} - x_0^{-1} } \norm {x_0} } \dfrac 1 {\norm {x_0} } 
+      | c = 
+}}
+{{eqn | o =
+}}
+{{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x \paren {x^{-1} - x_0^{-1} } x_0} }
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o =
+}}
+{{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x x^{-1} x_0 - x x_0^{-1} x_0} }
+      | c = [[Definition:Ring (Abstract Algebra)|Ring Axiom $(\text D)$: Product is Distributive over Addition]]
+}}
+{{eqn | o =
+}}
+{{eqn | r = \dfrac 1 {\norm x \norm {x_0} } \paren {\norm {x_0 - x} }
+      | c = {{Defof|Division Ring}}
+}}
+{{eqn | o =
+}}
+{{eqn | o = <
+      | r = \dfrac 2 {\norm {x_0}^2} \paren {\norm {x_0 - x} }
+      | c = from $\dfrac 1 {\norm x} < \dfrac 2 {\norm {x_0} }$ above
+}}
+{{eqn | o =
+}}
+{{eqn | o = <
+      | r = \dfrac {2 \delta} {\norm {x_0}^2}
+      | c = by assumption: $\norm {x - x_0} < \delta$
+}}
+{{eqn | o =
+}}
+{{eqn | o = \le
+      | r = \dfrac 2 {\norm {x_0}^2} \paren {\dfrac {\norm {x_0}^2 \epsilon} 2}
+      | c = by assumption: $\delta \le \dfrac {\norm {x_0}^2 \epsilon} 2$
+}}
+{{eqn | o =
+}}
+{{eqn | r = \epsilon
+      | c = cancelling terms
+}}
+{{end-eqn}}
+Since $x_0$ and $\epsilon$ were arbitrary, by the definition of [[Definition:Continuous Mapping (Metric Space)|continuity]] then the [[Definition:Mapping|mapping]]:
+:$\iota: \struct {R^*, d^*} \to \struct {R, d} : \map \iota x = x^{-1}$ 
+is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normed Division Ring Operations are Continuous/Corollary}
+Tags: Normed Division Rings, Topological Division Rings
+
+\begin{theorem}
+Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced by the metric]] $d$.
+Then:
+:$\struct {R, \tau}$ is a [[Definition:Topological Division Ring|topological division ring]].
+\end{theorem}
+
+\begin{proof}
+Let $d_\infty$ be the [[Definition:Chebyshev Distance|Chebyshev distance metric]] on $R \times R$.
+Let $\tau^\times$ be the [[Definition:Product Topology|product topology]] on $R \times R$.
+By [[P-Product Metric Induces Product Topology|$p$-Product Metric Induces Product Topology]], $\tau^\times$ is the [[Definition:Topology Induced by Metric|topology induced by the metric]] $d_\infty$.
+Let $R^* = R \setminus \set 0$.
+Let $d^*$ be the [[Definition:Restriction|restriction]] of $d$ to $R^*$.
+Let $\tau^*$ be the [[Definition:Subspace Topology|subspace topology]] on $R^*$.
+By [[Metric Subspace Induces Subspace Topology]] then $\tau^*$ is the [[Definition:Topology Induced by Metric|topology induced by the metric]] $d^*$
+By [[Normed Division Ring Operations are Continuous]] and [[Continuous Mapping is Continuous on Induced Topological Spaces]], the [[Definition:Mapping|mappings]]:
+::$\phi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \phi {x, y} = x + y$ 
+::$\theta : \struct {R ,\tau} \to \struct {R, \tau} : \map \theta x = -x$ 
+::$\psi : \struct {R \times R, \tau^\times} \to \struct {R, \tau} : \map \psi {x, y} = x y$ 
+::$\xi : \struct {R^* ,\tau^*} \to \struct {R, \tau} : \map \xi x = x^{-1}$ 
+are [[Definition:Continuous Mapping (Topology)|continuous]].
+By the definition of a [[Definition:Topological Division Ring|topological division ring]] then the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Count of Binary Operations with Identity/Sequence}
+Tags: Count of Binary Operations with Identity
+
+\begin{theorem}
+The [[Definition:Integer Sequence|sequence]] of $N$ for each $n$ begins:
+$\begin{array} {c|cr}
+n & \paren {n - 1}^2 + 1 & n^{\paren {n - 1}^2 + 1}\\
+\hline
+1 & 1 & 1 \\
+2 & 2 & 4 \\
+3 & 5 & 243 \\
+4 & 10 & 1 \ 048 \ 576 \\
+\end{array}$
+\end{theorem}<|endoftext|>
+\section{Structure with Element both Identity and Zero has One Element}
+Tags: Identity Elements, Zero Elements
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $z \in S$ such that $z$ is both an [[Definition:Identity Element|identity element]] and a [[Definition:Zero Element|zero element]].
+Then:
+:$S = \set z$
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then
+{{begin-eqn}}
+{{eqn | l = x
+      | r = x \circ z
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = z
+      | c = {{Defof|Zero Element}}
+}}
+{{end-eqn}}
+and so there is no other [[Definition:Element|element]] of $S$ but $z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group/Examples/Linear Functions}
+Tags: Examples of Groups
+
+\begin{theorem}
+Let $G$ be the [[Definition:Set|set]] of all [[Definition:Real Function|real functions]] $\theta_{a, b}: \R \to \R$ defined as:
+:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
+where $a, b \in \R$ such that $a \ne 0$.
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {G, \circ}$, where $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]], is a [[Definition:Group|group]].
+$\struct {G, \circ}$ is specifically non-[[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+We verify the [[Definition:Group Axioms|group axioms]], in the following order (for convenience):
+=== G0: Closure ===
+Let $\theta_{a, b}$ and $\theta_{c, d}$ be [[Definition:Element|elements]] of $G$.
+From [[Composition of Linear Real Functions]]:
+:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$
+As $a \ne 0$ and $c \ne 0$ it follows that $a c \ne 0$.
+Thus $\theta_{a c, b c + d} \in G$
+This proves [[Definition:Closed Algebraic Structure|closure]] of $\circ$.
+{{qed|lemma}}
+=== G1: Associativity ===
+We have that [[Composition of Mappings is Associative]].
+{{qed|lemma}}
+=== G2: Identity ===
+We assert that $\theta_{1, 0}$ is the identity with respect to $\circ$.
+{{begin-eqn}}
+{{eqn | l = \theta_{a, b} \circ \theta_{1, 0}
+      | r = \theta_{1 \times a, \, 0 \times a + b}
+      | c = [[Composition of Linear Real Functions]]
+}}
+{{eqn | r = \theta_{a, b}
+      | c = 
+}}
+{{eqn | r = \theta_{a \times 1, \, b \times 1 + 0}
+      | c = 
+}}
+{{eqn | r = \theta_{1, 0} \circ \theta_{a, b}
+      | c = [[Composition of Linear Real Functions]]
+}}
+{{end-eqn}}
+So indeed $\theta_{1, 0}$ is the [[Definition:Identity Element|identity]] for $\circ$.
+{{qed|lemma}}
+=== G3: Inverses ===
+For any element $\theta_{a, b} \in S$, we claim that $\theta_{1 / a, \, -b / a}$ is the [[Definition:Inverse Element|inverse]] for $\theta_{a, b}$.
+From [[Inverse of Linear Function on Real Numbers]]:
+:$\theta^{-1}_{a, b} = \theta_{1 / a, \, -b / a}$
+Verifying this:
+{{begin-eqn}}
+{{eqn | l = \theta_{1 / a, \, -b / a} \circ \theta_{a, b}
+      | r = \theta_{a \times 1 / a, \, b \times 1 / a + \paren {-b / a} }
+      | c = [[Composition of Linear Real Functions]]
+}}
+{{eqn | r = \theta_{1, 0}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \theta_{a, b} \circ \theta_{1 / a, \, -b / a}
+      | r = \theta_{\paren {1 / a} \times a, \, \paren {-b / a} \times a + b}
+      | c = [[Composition of Linear Real Functions]]
+}}
+{{eqn | r = \theta_{1, 0}
+      | c = 
+}}
+{{end-eqn}}
+We conclude that $\theta_{1 / a, \, -b / a}$ is indeed the [[Definition:Inverse Element|inverse]] for $\theta_{a, b}$.
+{{qed|lemma}}
+It follows that $\struct {G, \circ}$ is indeed a [[Definition:Group|group]].
+{{qed|lemma}}
+=== Non-Abelian ===
+Let $\theta_{a, b}$ and $\theta_{c, d}$ be [[Definition:Element|elements]] of $G$.
+From [[Condition for Composition of Linear Real Functions to be Commutative]], it is not true in general that $\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$.
+Thus $\struct {G, \circ}$ is non-[[Definition:Abelian Group|abelian]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation Formula for Alternating Series}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $C_N$ be the square with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$ for some [[Definition:Real Number|real]] $N \in \N$. 
+Let $f$ be a function [[Definition:Meromorphic Function|meromorphic]] on $C_N$.
+Let $\cmod {\map f z} < \dfrac M {\cmod z^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z \in \partial C_N$. 
+Let $X$ be the set of [[Definition:Pole|poles]] of $f$.
+Then: 
+:$\displaystyle \sum_{n \mathop \in \Z \mathop \setminus X} \paren {-1}^n \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \csc {\pi z} \map f z} {z_0}$
+If $X \cap \Z = \O$, this becomes: 
+:$\displaystyle \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f n = -\sum_{z_0 \mathop \in X} \Res {\pi \map \csc {\pi z} \map f z} {z_0}$
+\end{theorem}<|endoftext|>
+\section{Integer Multiples under Addition form Subgroup of Integers}
+Tags: Additive Groups of Integer Multiples, Additive Group of Integers
+
+\begin{theorem}
+Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
+Then $\struct {n \Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$.
+Hence $\struct {n \Z, +}$ can be justifiably referred to as the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]].
+\end{theorem}
+
+\begin{proof}
+Clearly $0 \in n \Z$ so $n \Z \ne \O$.
+Now suppose $x, y \in n \Z$.
+Then $\exists r, s \in \Z: x = n r, y = n s$.
+Also, $-y = - n s$.
+So $x - y = n \paren {r - s}$.
+As $r - s \in \Z$ it follows that $x - y \in n \Z$.
+So by the [[One-Step Subgroup Test]] it follows that $\struct {n \Z, +}$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup Generated by Subgroup}
+Tags: Generated Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then:
+:$H = \gen H$
+where $\gen H$ denotes the [[Definition:Generated Subgroup|subgroup generated]] by $H$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Generated Subgroup|generated subgroup]], $\gen H$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Subgroup|subgroup]] of $H$ containing $H$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group is Generated by Itself}
+Tags: Generated Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then:
+:$G = \gen G$
+where $\gen G$ denotes the [[Definition:Generator of Group|group generated]] by $G$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Generated Subgroup|generated subgroup]], $\gen G$ is the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Subgroup|subgroup]] of $G$ containing $G$.
+Hence the result by [[Group is Subgroup of Itself]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation Formula for Alternating Series over Half-Integers}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $C_N$ be the square with vertices $N \paren {\pm 1 \pm i}$ for [[Definition:Real Number|real]] $N \in \N$. 
+Let $f$ be a function [[Definition:Meromorphic Function|meromorphic]] on $C_N$.
+Let $\cmod {\map f z} < \dfrac M {\cmod z^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z \in \partial C_N$. 
+Let $X$ be the set of [[Definition:Pole|poles]] of $f$.
+Let $Y$ be the set of [[Definition:Pole|poles]] of $\map f {\dfrac {2 z + 1} 2}$. 
+Then: 
+:$\displaystyle \sum_{n \in \Z \setminus Y} \paren {-1}^n \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \sec \paren {\pi z} \map f z} {z_0}$
+If $Y \cap \Z = \O$, this becomes: 
+:$\displaystyle \sum_{n \mathop = -\infty}^\infty \paren {-1}^n \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \sec \paren {\pi z} \map f z} {z_0}$
+\end{theorem}<|endoftext|>
+\section{Summation Formula over Half-Integers}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $C_N$ be the square with vertices $N \paren {\pm 1 \pm i}$ for [[Definition:Real Number|real]] $N \in \N$. 
+Let $f$ be a function [[Definition:Meromorphic Function|meromorphic]] on $C_N$.
+Let $\cmod {\map f z} < \dfrac M {\cmod z^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z \in \partial C_N$. 
+Let $X$ be the set of [[Definition:Pole|poles]] of $f$.
+Let $Y$ be the set of [[Definition:Pole|poles]] of $\map f {\dfrac {2 z + 1} 2}$.
+Then: 
+:$\displaystyle \sum_{n \in \Z \setminus Y} \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$
+If $Y \cap \Z = \O$, this becomes: 
+:$\displaystyle \sum_{n \mathop = -\infty}^\infty \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$
+\end{theorem}<|endoftext|>
+\section{Heaviside Expansion Formula}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $P, Q$ be [[Definition:Polynomial/Complex Numbers|polynomials]] with coefficients in $\C$. 
+Let $\deg Q \ge \deg P + 1$.
+Let $\map Q z$ have a [[Definition:Order of Zero/Simple Zero|simple zero]] for $z \in X$. 
+Let $\map {\laptrans f} z = \dfrac {\map P z} {\map Q z}$. 
+Then:
+:$\displaystyle \map f t = \sum_{z \mathop \in X} e^{z t} \frac {\map P z} {\map {Q'} z}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}} 
+{{Namedfor|Oliver Heaviside|cat = Heaviside}}
+\end{proof}<|endoftext|>
+\section{Action of Inverse of Group Element}
+Tags: Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Set|sets]].
+Let $*: G \times S \to S$ be a [[Definition:Group Action|group action]].
+Then:
+:$g * a = b \iff g^{-1} * b = a$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = g * a
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^{-1} * \paren {g * a}
+      | r = g^{-1} * b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {g^{-1} \circ g} * a
+      | r = g^{-1} * b
+      | c = {{GroupActionAxiom|2}}
+}}
+{{eqn | ll= \leadsto
+      | l = e * a
+      | r = g^{-1} * b
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = g^{-1} * b
+      | c = {{GroupActionAxiom|1}}
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = g^{-1} * b
+      | r = a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g * \paren {g^{-1} * b}
+      | r = g * a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {g \circ g^{-1} } * b
+      | r = g * a
+      | c = {{GroupActionAxiom|2}}
+}}
+{{eqn | ll= \leadsto
+      | l = e * b
+      | r = g * a
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | ll= \leadsto
+      | l = b
+      | r = g * a
+      | c = {{GroupActionAxiom|1}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union Operation on Supersets of Subset is Closed}
+Tags: Set Union
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $T \subseteq S$ be a given [[Definition:Subset|subset]] of $S$.
+Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$
+Let $\mathscr S$ be the [[Definition:Subset|subset]] of $\powerset S$ defined as:
+:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\mathscr S, \cup}$ is [[Definition:Closed Algebraic Structure|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \mathscr S$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = A
+      | c = Definition of $\mathscr S$
+}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = B
+      | c = Definition of $\mathscr S$
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = T
+      | o = \subseteq
+      | r = A \cup B
+      | c = [[Set is Subset of Union]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = A
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | l = B
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | ll= \leadsto
+      | l = A \cup B
+      | o = \subseteq
+      | r = S
+      | c = [[Union is Smallest Superset]]
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = A \cup B
+      | o = \in
+      | r = \powerset S
+      | c = {{Defof|Power Set}}
+}}
+{{end-eqn}}
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = A \cup B
+      | c = from $(1)$
+}}
+{{eqn | l = A \cup B
+      | o = \in
+      | r = \powerset S
+      | c = from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = A \cup B
+      | o = \in
+      | r = \mathscr S
+      | c = Definition of $\mathscr S$
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection Operation on Supersets of Subset is Closed}
+Tags: Set Union
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $T \subseteq S$ be a given [[Definition:Subset|subset]] of $S$.
+Let $\powerset S$ denote the [[Definition:Power Set|power set]] of $S$
+Let $\mathscr S$ be the [[Definition:Subset|subset]] of $\powerset S$ defined as:
+:$\mathscr S = \set {Y \in \powerset S: T \subseteq Y}$
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\mathscr S, \cap}$ is [[Definition:Closed Algebraic Structure|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $A, B \in \mathscr S$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = A
+      | c = Definition of $\mathscr S$
+}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = B
+      | c = Definition of $\mathscr S$
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = T
+      | o = \subseteq
+      | r = A \cap B
+      | c = [[Intersection is Largest Subset]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = A
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | l = B
+      | o = \subseteq
+      | r = S
+      | c = {{Defof|Power Set}}
+}}
+{{eqn | ll= \leadsto
+      | l = A \cap B
+      | o = \subseteq
+      | r = S
+      | c = [[Intersection is Subset]]
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = A \cap B
+      | o = \in
+      | r = \powerset S
+      | c = {{Defof|Power Set}}
+}}
+{{end-eqn}}
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = T
+      | o = \subseteq
+      | r = A \cap B
+      | c = from $(1)$
+}}
+{{eqn | l = A \cap B
+      | o = \in
+      | r = \powerset S
+      | c = from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = A \cap B
+      | o = \in
+      | r = \mathscr S
+      | c = Definition of $\mathscr S$
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Magma with no Proper Submagma}
+Tags: Magmas
+
+\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]] $n$:
+:$\card S = n$
+Then there exists an [[Definition:Binary Operation|operation]] $\circ$ on $S$ such that:
+:$\struct {S, \circ}$ is a [[Definition:Magma|magma]]
+:$\struct {S, \circ}$ has no [[Definition:Submagma|submagma]] $\struct {T, \circ}$ such that $T$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subset]] of $S$.
+\end{theorem}
+
+\begin{proof}
+For $n = 1$ the result follows trivially: there are no [[Definition:Non-Empty Set|non-empty]] [[Definition:Proper Subset|proper subsets]] of a [[Definition:Singleton|singleton]].
+Let $S = \set {s_1, s_2, \ldots, s_n}$.
+Let $\circ$ be defined on $S$ such that:
+:$\forall s_a \in S: s_a \circ s_a = \begin{cases} s_{a + 1} & : a < n \\ s_1 & : a = n \end{cases}$
+For $a \ne b$ the operation $s_a \circ s_b$ can be arbitrary as long as $s_a \circ s_b \in S$.
+Let $T$ be such that $\O \subsetneq T \subsetneq S$.
+Then either:
+:$\exists s_k \in T: s_{k + 1} \notin T$
+or:
+:$s_n \in T$ but $s_1 \notin T$
+otherwise $T = S$.
+Thus either:
+:$s_k \circ s_k \notin T$
+or:
+:$s_n \circ s_n \notin T$
+In either case, $\circ$ is not [[Definition:Closed Operation|closed in $T$]].
+Thus by definition, $\struct {T, \circ}$ is not a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$.
+As $T$ is arbitrary, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup}
+Tags: Abelian Groups, Subset of Abelian Group Generated by Product of Element with Inverse Element is Subgroup
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]].
+Let $S \subset G$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $G$ such that $\struct {S, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+Let $H$ be the [[Definition:Set|set]] defined as:
+:$H := \set {x \circ y^{-1}: x, y \in S}$
+Then $\struct {H, \circ}$ is a [[Definition:Subgroup|subgroup]] of $\struct {G, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Then:
+:$x \circ x^{-1} \in H$
+and so $H \ne \O$.
+Now let $a, b \in H$.
+Then:
+:$a = x_a \circ y_a^{-1}$
+and:
+:$b = x_b \circ y_b^{-1}$
+for some $x_a, y_a, x_b, y_b \in S$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a \circ b^{-1}
+      | r = \paren {x_a \circ y_a^{-1} } \circ \paren {x_b \circ y_b^{-1} }^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {\paren {y_b^{-1} }^{-1} \circ x_b^{-1} }
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \paren {x_a \circ y_a^{-1} } \circ \paren {y_b \circ x_b^{-1} }
+      | c = [[Inverse of Group Inverse]]
+}}
+{{eqn | r = x_a \circ \paren {y_a^{-1} \circ y_b} \circ x_b^{-1}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = x_a \circ \paren {y_b\circ y_a^{-1} } \circ x_b^{-1}
+      | c = {{Defof|Abelian Group}}
+}}
+{{eqn | r = \paren {x_a \circ y_b} \circ \paren {y_a^{-1} \circ x_b^{-1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = \paren {x_a \circ y_b} \circ \paren {x_b \circ y_a}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{end-eqn}}
+As $\struct {S, \circ}$ is [[Definition:Closed Algebraic Structure|closed]], both $x_a \circ y_b \in S$ and $x_b \circ y_a \in S$.
+Thus $a \circ b$ is in the form $x \circ y^{-1}$ for $x, y \in S$.
+Thus $a \circ b \in H$
+Hence by the [[One-Step Subgroup Test]], $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Local Basis Test}
+Tags: Local Bases
+
+\begin{theorem}
+Let $\struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $x \in S$.
+Let $\BB$ be a [[Definition:Local Basis|local basis]] for $x$ in $\struct {S, \tau}$.
+Let $\CC$ be a [[Definition:Set|set]] of [[Definition:Open Neighborhood|open neighborhoods]] of $x$.
+Then:
+:$\CC$ is a [[Definition:Local Basis|local basis]] {{iff}}:
+::$\forall B \in \BB \implies \exists C \in \CC: C \subseteq B$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\CC$ be a [[Definition:Local Basis|local basis]].
+Let $B \in \BB$.
+Since $\BB$ is a [[Definition:Local Basis|local basis]], by the definition of a [[Definition:Local Basis|local basis]] then $B$ is [[Definition:Open Set|open]].
+Since $\CC$ is a [[Definition:Local Basis|local basis]], by the definition of a [[Definition:Local Basis|local basis]] then:
+:$\exists C \in \CC : C\subseteq B$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\CC$ satisfy:
+:$\forall B \in \BB \implies \exists C \in \CC: C\subseteq B$
+Let $U \in \tau$ and $x \in U$.
+By the definition of a [[Definition:Local Basis|local basis]] then:
+:$\exists B \in \BB : B\subseteq U$
+By the assumption then:
+:$\exists C \in \CC: C\subseteq B \subseteq U$
+By the definition of a [[Definition:Local Basis|local basis]] then $\CC$ is a [[Definition:Local Basis|local basis]].
+{{qed}}
+[[Category:Local Bases]]
+58qxh2muzay6vkkz4kn3r0hmnbx7er5
+\end{proof}<|endoftext|>
+\section{Non-Zero Integers under Multiplication are not Subgroup of Reals}
+Tags: Integer Multiplication, Real Multiplication
+
+\begin{theorem}
+Let $\struct {\Z_{\ne 0}, \times}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by the [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]].
+Let $\struct {\R_{\ne 0}, \times}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by the [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real numbers]] under [[Definition:Real Multiplication|multiplication]].
+Then, while $\struct {\Z_{\ne 0}, \times}$ is [[Definition:Closed Algebraic Structure|closed]], it is not a [[Definition:Subgroup|subgroup]] of $\struct {\R_{\ne 0}, \times}$.
+\end{theorem}
+
+\begin{proof}
+We have that [[Non-Zero Real Numbers under Multiplication form Group]].
+We also have that the [[Definition:Set|set]] of [[Definition:Zero (Number)|non-zero]] [[Definition:Integer|integers]] $\Z_{\ne 0}$ form a [[Definition:Subset|subset]] of $\R_{\ne 0}$.
+From [[Non-Zero Integers Closed under Multiplication]]:
+:$\forall a, b \in \Z_{\ne 0}: a \times b \in \Z_{\ne 0}$
+We have that:
+:$\forall x \in \Z_{\ne 0}: 1 \times x= x = x \times 1$
+and so $1$ is the [[Definition:Identity Element|identity]] of $\struct {\Z_{\ne 0}, \times}$.
+But for $x \in \Z_{\ne 0}: x \ne 1$, there exists no $y \in \Z_{\ne 0}: x \times y = 1$.
+Thus $\struct {\Z_{\ne 0}, \times}$ does not have [[Definition:Inverse Element|inverses]] for all $x \in \Z_{\ne 0}$.
+Thus, by definition, $\struct {\Z_{\ne 0}, \times}$ is not a [[Definition:Group|group]].
+It follows that $\struct {\Z_{\ne 0}, \times}$ is not a [[Definition:Subgroup|subgroup]] of $\struct {\R_{\ne 0}, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Elements of Group to be in Subgroup}
+Tags: Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $x, y \in G$ be such that $2$ [[Definition:Element|elements]] of $\set {x, y, x y}$ are [[Definition:Element|elements]] of $h$.
+Then ''all'' the [[Definition:Element|elements]] of $\set {x, y, x y}$ are in $H$.
+\end{theorem}
+
+\begin{proof}
+As $H$ is a [[Definition:Subgroup|subgroup]] of $G$, it is a [[Definition:Group|group]] in its own right.
+Thus the [[Definition:Group Axioms|group axioms]] all apply to $H$.
+Let $x, y \in H$.
+Then by {{GroupAxiom|0}}, $x y \in H$.
+Let $x, x y \in H$.
+As $x \in H$, it follows that $x^{-1} \in H$ by {{GroupAxiom|3}}.
+Thus by {{GroupAxiom|0}}, $x^{-1} \paren {x y} = y \in H$.
+Let $y, x y \in H$.
+As $y \in H$, it follows that $y^{-1} \in H$ by {{GroupAxiom|3}}.
+Thus by {{GroupAxiom|0}}, $\paren {x y} y^{-1} = x \in H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Subgroups/Corollary 1}
+Tags: Union of Subgroups
+
+\begin{theorem}
+Let $H \cup K$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then either $H \subseteq K$ or $K \subseteq H$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} neither $H \subseteq K$ nor $K \subseteq H$.
+Then from [[Union of Subgroups]] it follows that $H \cup K$ is not a [[Definition:Subgroup|subgroup]] of $G$.
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup Generated by Commuting Elements is Abelian}
+Tags: Abelian Groups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S \subseteq G$ such that:
+:$\forall x, y \in S: x \circ y = y \circ x$
+Then the [[Definition:Generated Subgroup|subgroup generated]] by $S$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $H = \gen S$ denote the [[Definition:Generated Subgroup|subgroup generated]] by $S$.
+Let $a, b \in H$.
+Then:
+:$a = s_1$
+:$b = s_2$
+for some [[Definition:Word (Group Theory)|words]] $s_1, s_2$ of the [[Definition:Set of Words|set of words]] $\map W S$ of $S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ b
+      | r = s_1 \circ s_2
+      | c = 
+}}
+{{eqn | r = s_2 \circ s_1
+      | c = as all [[Definition:Element|elements]] of $S$ [[Definition:Commuting Elements|commute]] with each other
+}}
+{{eqn | r = b \circ a
+      | c = 
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Generated Subgroup}
+Tags: Generated Subgroups
+
+\begin{theorem}
+{{TFAE|def = Generated Subgroup}}
+Let $G$ be a [[Definition:Group|group]].
+Let $S \subset G$ be a [[Definition:Subset|subset]].
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ is equivalent to $(2)$ ===
+Let $H$ be the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Subgroup|subgroup]] containing $S$.
+Let $\mathbb S$ be the set of [[Definition:Subgroup|subgroups]] containing $S$.
+To show the equivalence of the two definitions, we need to show that $H = \bigcap \mathbb S$.
+Since $H$ is a [[Definition:Subgroup|subgroup]] containing $S$:
+:$H \in \mathbb S$
+By [[Intersection is Subset/General Result|Intersection is Subset]]:
+:$\bigcap \mathbb S \subseteq H$
+On the other hand, by [[Intersection of Subgroups is Subgroup/General Result|Intersection of Subgroups is Subgroup]]:
+:$\bigcap \mathbb S$ is a [[Definition:Subgroup|subgroup]] containing $S$.
+Since $H$ be the [[Definition:Smallest Set by Set Inclusion|smallest]] [[Definition:Subgroup|subgroup]] containing $S$:
+:$H \subseteq \bigcap \mathbb S$
+By definition of [[Definition:Set Equality/Definition 2|set equality]]:
+:$H = \bigcap \mathbb S$
+Hence the result.
+{{qed|lemma}}
+=== $(1)$ is equivalent to $(3)$ ===
+This is shown in [[Set of Words Generates Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizer of Element after Group Action}
+Tags: Stabilizers
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Set|set]].
+Let $*_S: G \times S \to S$ be a [[Definition:Group Action|group actions]].
+Let $x \in S, a \in G$.
+Then:
+:$\Stab {a * x} = a^{-1} \circ \Stab x \circ a$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \Stab {a * x}
+      | r = \set {g \in G: g * \paren {a * x} = a * x}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {g \in G: \paren {g \circ a} * x = a * x}
+      | c = {{GroupActionAxiom|2}}
+}}
+{{eqn | r = \set {g \in G: a^{-1} * \paren {g \circ a} * x = a^{-1} * \paren {a * x} }
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \paren {a^{-1} \circ \paren {g \circ a} } * x = \paren {a^{-1} \circ a} * x}
+      | c = {{GroupActionAxiom|2}}
+}}
+{{eqn | r = \set {g \in G: \paren {a^{-1} \circ g \circ a} * x = x}
+      | c = {{GroupAxiom|1}} and {{GroupAxiom|3}}
+}}
+{{eqn | r = a^{-1} \circ \set {g \in G: g * x = x} \circ a
+      | c = {{Defof|Subset Product}}
+}}
+{{eqn | r = a^{-1} \circ \Stab x \circ a
+      | c = {{Defof|Stabilizer}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group/Subset}
+Tags: Group Action of Symmetric Group
+
+\begin{theorem}
+Let $r \in \N: 0 < r \le n$.
+Let $B_r$ denote the [[Definition:Set|set]] of all [[Definition:Subset|subsets]] of $\N_n$ of [[Definition:Cardinality|cardinality]] $r$:
+:$B_r  := \set {S \subseteq \N_n: \card S = r}$
+Let $*$ be the [[Definition:Mapping|mapping]] $*: S_n \times B_r \to B_r$ defined as:
+:$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$
+where $\pi \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image of $S$ under $\pi$]].
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Group Action Axioms|group action axioms]] are investigated in turn.
+Let $\pi, \rho \in S_n$.
+Let $S \in B_r$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \pi * \paren {\rho * S}
+      | r = \pi * \rho \sqbrk S
+      | c = Definition of $*$
+}}
+{{eqn | r = \pi \sqbrk {\rho \sqbrk S}
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {\pi \circ \rho} \sqbrk S
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \paren {\pi \circ \rho} * n
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|1}} holds.
+Then:
+{{begin-eqn}}
+{{eqn | l = I_{\N_n} * S
+      | r = I_{\N_n} \sqbrk S
+      | c = where $I_{\N_n}$ is the [[Definition:Identity Mapping|identity mapping]] on $\N_n$
+}}
+{{eqn | r = S
+      | c = {{Defof|Identity Mapping}}
+}}
+{{end-eqn}}
+demonstrating that {{GroupActionAxiom|2}} holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group on Subset is Transitive}
+Tags: Group Action of Symmetric Group, Transitive Group Actions
+
+\begin{theorem}
+Let $r \in \N: 0 < r \le n$.
+Let $B_r$ denote the [[Definition:Set|set]] of all [[Definition:Subset|subsets]] of $\N_n$ of [[Definition:Cardinality|cardinality]] $r$:
+:$B_r  := \set {S \subseteq \N_n: \card S = r}$
+Let $*$ be the [[Definition:Mapping|mapping]] $*: S_n \times B_r \to B_r$ defined as:
+:$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$
+where $\pi \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image of $S$ under $\pi$]].
+Then $*$ is a [[Definition:Transitive Group Action|transitive group action]].
+\end{theorem}
+
+\begin{proof}
+From [[Group Action of Symmetric Group on Subset]] it is established that $*$ is a [[Definition:Group Action|group action]].
+Let $U = \set {u_1, u_2, \ldots, u_r}$ and $V = \set {v_1, v_2, \ldots, v_r}$ be [[Definition:Element|elements]] of $B_r$.
+Then there exists a [[Definition:Permutation|permutation]] $\rho \in S_n$ such that:
+:$\map \rho {u_k} = v_k$
+for all $k \in \set {1, 2, \ldots, r}$.
+Thus:
+:$\rho \sqbrk U = V$
+Thus $B_r$ is the [[Definition:Orbit of Group Action|orbit]] of all $U \in B_r$.
+Hence the result by definition of [[Definition:Transitive Group Action|transitive group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Coset Product on Non-Normal Subgroup is not Well-Defined}
+Tags: Coset Product
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ which is not [[Definition:Normal Subgroup|normal]].
+Let $a, b \in G$.
+Then it is not necessarily the case that the [[Definition:Coset Product|coset product]]:
+:$\paren {a \circ H} \circ \paren {b \circ H} = \paren {a \circ b} \circ H$
+is [[Definition:Well-Defined Operation|well-defined]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S_3$ denote the [[Symmetric Group on 3 Letters]], whose [[Symmetric Group on 3 Letters/Cayley Table|Cayley table]] is given as:
+{{:Symmetric Group on 3 Letters/Cayley Table}}
+Consider the [[Symmetric Group on 3 Letters/Subgroups|subgroups]] of $S_3$:
+{{:Symmetric Group on 3 Letters/Subgroups}}
+Let $H = \set {e, \tuple {12} }$.
+From [[Symmetric Group on 3 Letters/Normal Subgroups|Normal Subgroups in Symmetric Group on 3 Letters]], $H$ is not [[Definition:Normal Subgroup|normal]].
+Let $A = \set {\tuple {123}, \tuple {23} }$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \tuple {123} H 
+      | r = \set {\tuple {123}, \tuple {23} }
+      | c = 
+}}
+{{eqn | l = \tuple {23} H 
+      | r = \set {\tuple {123}, \tuple {23} }
+      | c = 
+}}
+{{end-eqn}}
+and so $A$ is the [[Definition:Left Coset|left coset]] of $H$ by both $\tuple {123}$ and $\tuple {23}$.
+Let $B = \set {\tuple {132}, \tuple {13} }$.
+{{begin-eqn}}
+{{eqn | l = \tuple {132} H 
+      | r = \set {\tuple {132}, \tuple {13} }
+      | c = 
+}}
+{{eqn | l = \tuple {13} H 
+      | r = \set {\tuple {132}, \tuple {13} }
+      | c = 
+}}
+{{end-eqn}}
+and so $B$ is the [[Definition:Left Coset|left coset]] of $H$ by both $\tuple {132}$ and $\tuple {13}$.
+Now consider:
+{{begin-eqn}}
+{{eqn | l = A \circ B
+      | r = \tuple {123} H \tuple {132} H
+      | c = 
+}}
+{{eqn | r = \tuple {123} \tuple {132} H
+      | c = 
+}}
+{{eqn | r = e H
+      | c = 
+}}
+{{eqn | r = H
+      | c = 
+}}
+{{end-eqn}}
+But:
+{{begin-eqn}}
+{{eqn | l = A \circ B
+      | r = \tuple {23} H \tuple {13} H
+      | c = 
+}}
+{{eqn | r = \tuple {23} \tuple {13} H
+      | c = 
+}}
+{{eqn | r = \tuple {132} H
+      | c = 
+}}
+{{eqn | r = B
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = H
+      | c = 
+}}
+{{end-eqn}}
+So two different evaluations of the [[Definition:Coset Product|coset product]] give two different result.
+Hence by definition the [[Definition:Coset Product|coset product]] is not [[Definition:Well-Defined Operation|well-defined]] on $H$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Klein Four-Group is Normal in A4}
+Tags: Alternating Group on 4 Letters, Examples of Normal Subgroups
+
+\begin{theorem}
+Let $A_4$ denote the [[Alternating Group on 4 Letters|alternating group on $4$ letters]], whose [[Alternating Group on 4 Letters/Cayley Table|Cayley table]] is given as:
+{{:Alternating Group on 4 Letters/Cayley Table}}
+<onlyinclude>
+The [[Definition:Subset|subsets]] of $A_4$ which form [[Definition:Subgroup|subgroups]] of $A_4$ are as follows:
+Consider the [[Definition:Order of Group|order $4$]] [[Definition:Subgroup|subgroup]] $V$ of $A_4$, presented by [[Klein Four-Group/Cayley Table|Cayley table]]:
+:$\begin{array}{c|cccc}
+\circ & e & t & u & v \\
+\hline
+e & e & t & u & v \\
+t & t & e & v & u \\
+u & u & v & e & t \\
+v & v & u & t & e \\
+\end{array}$
+Then $V$ is [[Definition:Normal Subgroup|normal]] in $A_4$.
+Its [[Definition:Index of Subgroup|index]] is:
+:$\index {A_4} V = \dfrac {\order {A_4} } {\order V} = \dfrac {12} 4 = 3$
+The [[Definition:Left Coset|(left) cosets]] of $V$ are:
+:$V$
+:$A := a V$
+:$P := p V$
+and the [[Definition:Cayley Table|Cayley table]] of the [[Definition:Quotient Group|quotient group]] $A_4 / V$ is given by:
+:$\begin{array}{c|ccc}
+\circ & V & A & P \\
+\hline
+V & V & A & P \\
+A & A & P & V \\
+P & P & V & A \\
+\end{array}$
+Note that while $A_4 / V$ is [[Definition:Abelian Group|Abelian]], $A_4$ is not.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Straightforward but tedious, unless someone has a short cut better than testing all the products}}
+:$\index {A_4} V = 3$ follows from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]].
+\end{proof}<|endoftext|>
+\section{Coset of Trivial Subgroup is Singleton}
+Tags: Cosets, Trivial Group
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $E := \struct {\set e, \circ}$ denote the [[Definition:Trivial Subgroup|trivial subgroup]] of $\struct {G, \circ}$.
+Let $g \in G$.
+Then the [[Definition:Left Coset|left coset]] and [[Definition:Right Coset|right coset]] of $E$ by $g$ is $\set g$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = g \circ \set e
+      | r = \set {g \circ x: x \in \set e}
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | r = \set {g \circ e}
+      | c = 
+}}
+{{eqn | r = \set g
+      | c = 
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \set e \circ g
+      | r = \set {x \circ g: x \in \set e}
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | r = \set {e \circ g}
+      | c = 
+}}
+{{eqn | r = \set g
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Elements of Right Transversal is Left Transversal}
+Tags: Transversals (Group Theory)
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $S \subseteq G$ be a [[Definition:Right Transversal|right transversal]] for $H$ in $G$.
+Let $T$ be the [[Definition:Set|set]] defined as:
+:$T := \set {x^{-1}: x \in S}$
+where $x^{-1}$ is the [[Definition:Inverse Element|inverse]] of $x$ in $G$.
+Then $T$ is a [[Definition:Left Transversal|left transversal]] for $H$ in $G$.
+\end{theorem}
+
+\begin{proof}
+Let $g \in G$.
+We show that $g H$ contains exactly $1$ element of $T$.
+Since $S$ is a [[Definition:Right Transversal|right transversal]]:
+:$\exists ! x \in S: x \in H g^{-1}$
+By [[Right Cosets are Equal iff Element in Other Right Coset]]:
+:$H x = H g^{-1}$
+By [[Right Cosets are Equal iff Left Cosets by Inverse are Equal]]:
+:$x^{-1} H = g H$
+We have from definition $x^{-1} \in T$.
+The result follows from [[Definition:Unique|uniqueness]] of $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Subset of Group to be Right Transversal}
+Tags: Transversals (Group Theory)
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] in $G$ is $n$:
+:$\index G H = n$
+Let $S \subseteq G$ be a [[Definition:Subset|subset]] of $G$ of [[Definition:Cardinality|cardinality]] $n$.
+Then $S$ is a [[Definition:Right Transversal|right transversal]] for $H$ in $G$ {{iff}}:
+:$\forall x, y \in S: x \ne y \implies x y^{-1} \notin H$
+\end{theorem}
+
+\begin{proof}
+From {{Defof|Right Transversal}}, $S$ is a [[Definition:Right Transversal|right transversal]] for $H$ in $G$ {{iff}} every [[Definition:Right Coset|right coset]] of $H$ contains exactly one [[Definition:Element|element]] of $S$.
+Since there are $n$ [[Definition:Right Coset|right cosets]] of $H$  and $S$ has [[Definition:Cardinality|cardinality]] $n$, if $S$ is not a [[Definition:Right Transversal|right transversal]] for $H$ in $G$, at least one [[Definition:Right Coset|right coset]] of $H$ contain more than one [[Definition:Element|element]] of $S$.
+Thus the [[Definition:Contrapositive|contrapositive]] of the theorem is as follows:
+:At least one [[Definition:Right Coset|right coset]] of $H$ contain more than one [[Definition:Element|element]] of $S$ {{iff}}:
+::$\exists x, y \in S: x \ne y \land x y^{-1} \in H$
+This is a consequence of [[Elements in Same Right Coset iff Product with Inverse in Subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Coset Space is Transitive}
+Tags: Group Action on Coset Space, Transitive Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|action on the (left) coset space]]:
+:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
+Then $G$ is a [[Definition:Transitive Group Action|transitive group action]].
+\end{theorem}
+
+\begin{proof}
+It is established in [[Action of Group on Coset Space is Group Action]] that $*$ is a [[Definition:Group Action|group action]].
+It remains to be shown that:
+:$\forall g' H \in G / H: \Orb {g' H} = G / H$
+where $\Orb {g' H}$ denotes the [[Definition:Orbit (Group Theory)|orbit]] of $g' H \in G / H$ under $*$.
+Let $a H, b H \in G / H$ such that $a H \ne b H$.
+We have that:
+:$\exists x \in G: x a = b$
+and so:
+:$x * a H = \paren {x a} H = b H$
+and so:
+:$b H \in \Orb {a H}$
+As both $a$ and $b$ are arbitrary, the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizer of Coset under Group Action on Coset Space}
+Tags: Group Action on Coset Space, Stabilizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|action on the (left) coset space]]:
+:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
+Then the [[Definition:Stabilizer|stabilizer]] of $a H$ under $*$ is given by:
+:$\Stab {a H} = a H a^{-1}$
+\end{theorem}
+
+\begin{proof}
+It is established in [[Action of Group on Coset Space is Group Action]] that $*$ is a [[Definition:Group Action|group action]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \Stab {a H}
+      | r = \set {g \in G: g * a H = a H}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {g \in G: \paren {g a} H = a H}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: g H = a H a^{-1} }
+      | c = 
+}}
+{{eqn | r = a H a^{-1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Index of Subgroup equals Index of Conjugate}
+Tags: Subgroups, Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then:
+:$\index G H = \index G {a H a^{-1} }$
+where $\index G H$ denotes the [[Definition:Index of Subgroup|index]] of $H$ in $G$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|tired}}
+\end{proof}<|endoftext|>
+\section{Normality Relation is not Transitive}
+Tags: Normal Subgroups, Normality Relation is not Transitive
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $K$ be a [[Definition:Normal Subgroup|normal subgroup]] of $N$.
+Then it is not necessarily the case that $K$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $D_4$ denote the [[Definition:Dihedral Group D4|dihedral group $D_4$]].
+Let $D_4$ be presented in [[Dihedral Group D4/Matrix Representation/Formulation 1|matrix representation]]:
+{{:Dihedral Group D4/Matrix Representation/Formulation 1}}
+Its [[Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table|Cayley table]] is given as:
+{{:Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table}}
+Consider the [[Definition:Subgroup|subgroup]] $H$ whose [[Definition:Underlying Set of Structure|underlying set]] is:
+:$H = \set {\mathbf I, \mathbf A, -\mathbf I, -\mathbf A}$
+From [[Subgroup of Index 2 is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $D_4$.
+Consider the [[Definition:Subgroup|subgroup]] $H$ whose [[Definition:Underlying Set of Structure|underlying set]] is:
+:$K = \set {\mathbf I, \mathbf A} = \gen {\mathbf A}$
+From [[Subgroup of Index 2 is Normal]], $K$ is [[Definition:Normal Subgroup|normal]] in $H$.
+It remains to be demonstrated that $K$ is not [[Definition:Normal Subgroup|normal]] in $D_4$.
+From the [[Dihedral Group D4/Matrix Representation/Formulation 1/Cayley Table|Cayley table]]:
+:$\mathbf C \mathbf A \mathbf C^{-1} = \mathbf B \mathbf C = -\mathbf A \notin K$
+Hence $K$ is not [[Definition:Normal Subgroup/Definition 6|normal]] in $D_4$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stabilizer is Normal iff Stabilizer of Each Element of Orbit}
+Tags: Stabilizers, Normal Subgroups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Set|set]].
+Let $*: G \times S \to S$ be a [[Definition:Group Action|group action]].
+Let $x \in S$.
+Let $\Stab x$ denote the [[Definition:Stabilizer|stabilizer]] of $x$ under $*$.
+Let $\Orb x$ denote the [[Definition:Orbit (Group Theory)|orbit]] of $x$ under $*$.
+Then $\Stab x$ is [[Definition:Normal Subgroup|normal]] in $G$ {{iff}} $\Stab x$ is also the [[Definition:Stabilizer|stabilizer]] of every [[Definition:Element|element]] in $\Orb x$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\Stab x$ be [[Definition:Normal Subgroup|normal]] in $G$.
+{{AimForCont}} $y \in \Orb x$ such that $\Stab x \ne \Stab y$.
+Then:
+{{ProofWanted}}
+=== Sufficient Condition ===
+Let $\Stab x$ be the [[Definition:Stabilizer|stabilizer]] of every [[Definition:Element|element]] in $\Orb x$.
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Power of Coset Product is Coset of Power}
+Tags: Coset Product
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $a \in G$.
+Then:
+:$\forall n \in \Z: \paren {a \circ N} = \paren {a^n} \circ N$
+\end{theorem}
+
+\begin{proof}
+From [[Quotient Group is Group]], the operation:
+:$\forall a, b \in G: \paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
+is the [[Definition:Group Operation|group operation]] in the [[Definition:Quotient Group|quotient group]] $\struct {G / N, \circ}$.
+The result follows directly by definition of [[Definition:Power of Group Element|power of group element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Power of Element of Quotient Group to be Identity}
+Tags: Coset Product
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $a \in G$.
+Then:
+:$\paren {a N}^n$ is the [[Definition:Identity Element|identity]] of the [[Definition:Quotient Group|quotient group]] $G / N$
+{{iff}}:
+:$a^n \in N$
+\end{theorem}
+
+\begin{proof}
+Let $\paren {a N}^n$ be the [[Definition:Identity Element|identity]] of $G / N$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {a N}^n
+      | r = N
+      | c = [[Quotient Group is Group]]: {{GroupAxiom|2}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {a^n} N
+      | r = N
+      | c = [[Power of Coset Product is Coset of Power]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a^n
+      | o = \in
+      | r = N
+      | c = [[Coset Equals Subgroup iff Element in Subgroup]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Integers is Normal Subgroup of Reals}
+Tags: Additive Group of Integers, Additive Group of Real Numbers, Examples of Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Then $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\R, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Integers is Subgroup of Reals]], $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\R, +}$.
+As the [[Definition:Additive Group of Real Numbers|additive group of real numbers]] is [[Definition:Abelian Group|abelian]], from [[Subgroup of Abelian Group is Normal]] it follows that $\struct {\Z, +} \lhd \struct {\R, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Element of Quotient Group of Additive Group of Reals by Integers to be of Finite Order}
+Tags: Integers, Real Numbers, Quotient Groups
+
+\begin{theorem}
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\R / \Z$ denote the [[Definition:Quotient Group|quotient group]] of $\struct {\R, +}$ by $\struct {\Z, +}$.
+Let $x + \Z$ denote the [[Definition:Coset|coset]] of $\Z$ by $x \in \R$.
+Then $x + \Z$ is of [[Definition:Finite Order Element|finite order]] {{iff}} $x$ is [[Definition:Rational Number|rational]].
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Integers is Normal Subgroup of Reals]], we have that $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\R, +}$.
+Hence $\R / \Z$ is indeed a [[Definition:Quotient Group|quotient group]].
+By definition of [[Definition:Rational Number|rational number]], what is to be proved is:
+:$x + \Z$ is of [[Definition:Finite Order Element|finite order]]
+{{iff}}:
+:$x = \dfrac m n$
+for some $m \in \Z, n \in \Z_{> 0}$.
+Let $x + \Z$ be of [[Definition:Finite Order Element|finite order]] in $\R / \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \exists n \in \Z_{\ge 0}:
+      | l = \paren {x + \Z}^n
+      | r = \Z
+      | c = {{Defof|Quotient Group}}: {{GroupAxiom|2}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n x
+      | o = \in
+      | r = \Z
+      | c = [[Condition for Power of Element of Quotient Group to be Identity]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n x
+      | r = m
+      | c = for some $m \in \Z$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = \dfrac m n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Integers is Normal Subgroup of Rationals}
+Tags: Additive Group of Integers, Additive Group of Rational Numbers, Examples of Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
+Then $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\Q, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Integers is Subgroup of Rationals]], $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q, +}$.
+From [[Rational Numbers under Addition form Abelian Group]], $\struct {\Q, +}$ is an [[Definition:Abelian Group|abelian group]].
+From [[Subgroup of Abelian Group is Normal]] it follows that $\struct {\Z, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\Q, +}$.
+{{qed}}
+[[Category:Additive Group of Integers]]
+[[Category:Additive Group of Rational Numbers]]
+[[Category:Examples of Normal Subgroups]]
+7jc6hb49k1u71jvloy78brhxaj4hu5e
+\end{proof}<|endoftext|>
+\section{Mapping from Additive Group of Integers to Powers of Group Element is Homomorphism}
+Tags: Additive Group of Integers, Examples of Group Homomorphisms
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $g \in G$.
+Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\phi_g: \struct {\Z, +} \to \struct {G, \circ}$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall k \in \Z: \map {\phi_g} k = g^k$
+Then $\phi_g$ is a [[Definition:Group Homomorphism|(group) homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $k, l \in \Z$.
+{{begin-eqn}}
+{{eqn | l = \map {\phi_g} {k + l}
+      | r = a^{k + l}
+      | c = 
+}}
+{{eqn | r = a^k a^l
+      | c = 
+}}
+{{eqn | r = \map {\phi_g} k \circ \map {\phi_g} l
+      | c = 
+}}
+{{end-eqn}}
+thus proving that $\phi_g$ is a [[Definition:Group Homomorphism|homomorphism]] as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Automorphisms form Subgroup of Symmetric Group}
+Tags: Inner Automorphisms, Symmetric Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\struct {\map \Gamma G, \circ}$ be the [[Definition:Symmetric Group|symmetric group]] on $G$.
+Let $\Inn G$ denote the [[Definition:Inner Automorphism Group|inner automorphism group]] of $G$.
+Then:
+:$\Inn G \le \struct {\map \Gamma G, \circ}$
+where $\le$ denotes the relation of being a [[Definition:Subgroup|subgroup]].
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Inner Automorphism|inner automorphism]] is a [[Definition:Permutation|permutation]] on $G$ by definition.
+From [[Inner Automorphisms form Subgroup of Automorphism Group]]:
+:$\Inn G \le \Aut G$
+where $\Aut G$ denotes the set of [[Definition:Group Automorphism|automorphisms]] of $G$.
+From [[Automorphism Group is Subgroup of Symmetric Group]]:
+:$\Aut G \le \struct {\map \Gamma G, \circ}$
+Thus $\Inn G \le \struct {\map \Gamma G, \circ}$ as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order is Preserved by Group Isomorphism}
+Tags: Group Isomorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $\phi: G \to H$ be a [[Definition:Group Isomorphism|(group) isomorphism]].
+Then:
+:$\order G = \order H$
+where $\order {\, \cdot \,}$ denotes the [[Definition:Order of Group|order]] of a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Group Isomorphism|isomorphism]] is a [[Definition:Bijection|bijection]].
+By definition, the [[Definition:Order of Group|order]] of a [[Definition:Group|group]] is the [[Definition:Cardinality|cardinality]] of its [[Definition:Underlying Set of Structure|underlying set]].
+The result follows by definition of [[Definition:Set Equivalence|set equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Reals is Subgroup of Complex}
+Tags: Additive Group of Real Numbers, Additive Group of Complex Numbers
+
+\begin{theorem}
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
+Then $\struct {\R, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, +}$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
+As $x$ and $y$ are [[Definition:Wholly Real|wholly real]], we have that $x, y \in \R$.
+Then $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ which is also [[Definition:Wholly Real|wholly real]].
+Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also [[Definition:Wholly Real|wholly real]].
+Thus by the [[Two-Step Subgroup Test]], $\struct {\R, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, +}$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{C6 is not Isomorphic to S3}
+Tags: Examples of Cyclic Groups, Symmetric Group on 3 Letters
+
+\begin{theorem}
+Let $C_6$ denote the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $6$]].
+Let $S_3$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $3$ letters]].
+Then $C_6$ and $S_3$ are not [[Definition:Group Isomorphism|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+Note that both $C_6$ and $S_3$ are of [[Definition:Order of Group|order $6$]].
+From [[Cyclic Group is Abelian]], $C_6$ is [[Definition:Abelian Group|abelian]].
+From [[Symmetric Group is not Abelian]], $S_6$ is not [[Definition:Abelian Group|abelian]].
+From [[Isomorphism of Abelian Groups]], if two [[Definition:Group|groups]] are [[Definition:Group Isomorphism|isomorphic]], they are either both [[Definition:Abelian Group|abelian]] or both not [[Definition:Abelian Group|abelian]].
+Hence $C_6$ and $S_3$ are not [[Definition:Group Isomorphism|isomorphic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Alternating Group}
+Tags: Alternating Groups
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n > 1$.
+Let $A_n$ be the [[Definition:Alternating Group|alternating group on $n$ letters]].
+Then:
+:$\order {A_n} = \dfrac {n!} 2$
+where $\order {A_n}$ denotes the [[Definition:Order of Group|order]] of $A_n$.
+\end{theorem}
+
+\begin{proof}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+From [[Alternating Group is Normal Subgroup of Symmetric Group]]:
+:$\index {S_n} {A_n} = 2$
+where $\index {S_n} {A_n}$ denotes the [[Definition:Index of Subgroup|index]] of $A_n$ in $S_n$.
+From [[Order of Symmetric Group]]:
+:$\order {S_n} = n!$
+The result follows from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping to Power is Endomorphism iff Abelian}
+Tags: Abelian Groups, Group Endomorphisms
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Let $\phi: G \to G$ be defined as:
+:$\forall g \in G: \map \phi g = g^n$
+Then $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} $\phi$ is a [[Definition:Group Endomorphism|(group) endomorphism]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]].
+Let $a, b \in G$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {a \circ b}
+      | r = \paren {a \circ b}^n
+      | c = Definition of $\phi$
+}}
+{{eqn | r = a^n \circ b^n
+      | c = [[Power of Product of Commutative Elements in Group]]
+}}
+{{eqn | r = \map \phi a \circ \map \phi b
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.
+Thus $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]] from $G$ to $G$.
+So by definition, $\phi$ is a [[Definition:Group Endomorphism|group endomorphism]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\phi: G \to G$ as defined above be a [[Definition:Group Endomorphism|group endomorphism]].
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall a, b \in G:
+      | l = \map \phi {a \circ b}
+      | r = \map \phi a \circ \map \phi b
+      | c = {{Defof|Group Endomorphism}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall a, b \in G:
+      | l = \paren {a \circ b}^n
+      | r = a^n \circ b^n
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+From [[Power of Product of Commutative Elements in Group]] it follows that $G$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Groups of Integers and Integer Multiples are Isomorphic}
+Tags: Additive Groups of Integer Multiples, Additive Group of Integers, Infinite Cyclic Group
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\struct {n \Z, +}$ denote the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]].
+Let $\struct {\Z, +}$ denote the [[Definition:Additive Group of Integers|additive group of integers]].
+Then $\struct {n \Z, +}$ is [[Definition:Group Isomorphism|isomorphic]] to $\struct {\Z, +}$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:[[Infinite Cyclic Group is Isomorphic to Integers]].
+:[[Integer Multiples under Addition form Infinite Cyclic Group]].
+:[[Infinite Cyclic Group is Unique up to Isomorphism]]
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers}
+Tags: Multiplicative Group of Real Numbers, Additive Group of Real Numbers, Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers
+
+\begin{theorem}
+Let $\struct {\R, +}$ denote the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $\struct {\R_{\ne 0}, \times}$ denote the [[Definition:Multiplicative Group of Real Numbers|multiplicative group of real numbers]].
+Then $\struct {\R, +}$ is not [[Definition:Group Isomorphism|isomorphic]] to $\struct {\R_{\ne 0}, \times}$.
+\end{theorem}<|endoftext|>
+\section{Normal Subgroup is Kernel of Group Homomorphism}
+Tags: Normal Subgroups, Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then there exists a [[Definition:Group Homomorphism|group homomorphism]] of which $N$ is the [[Definition:Kernel of Group Homomorphism|kernel]].
+\end{theorem}
+
+\begin{proof}
+Let $G / N$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $N$.
+Let $q_N: G \to G / N$ be the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to $G / N$:
+:$\forall x \in G: \map {q_N} x = x N$
+Then from [[Quotient Group Epimorphism is Epimorphism]], $N$ is the [[Definition:Kernel of Group Homomorphism|kernel]] of $q_n$
+Thus $q_N$ is that [[Definition:Group Homomorphism|group homomorphism]] of which $N$ is the [[Definition:Kernel of Group Homomorphism|kernel]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphic Image of Cyclic Group is Cyclic Group}
+Tags: Cyclic Groups, Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $g$.
+Let $H$ be a [[Definition:Group|group]].
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|(group) homomorphism]].
+Let $\Img G$ denote the [[Definition:Homomorphic Image|homomorphic image]] of $G$ under $\phi$.
+Then $\Img G$ is a [[Definition:Cyclic Group|cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $\map \phi g$.
+That is:
+:$\phi \sqbrk {\gen g} = \gen {\map \phi g}$
+\end{theorem}
+
+\begin{proof}
+Let $y \in \Img G$.
+Then $\exists x \in G: y = \map \phi x$.
+As $G$ be a [[Definition:Cyclic Group|cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $g$, $x = g^n$ for some $n \in \Z$.
+Thus by [[Homomorphism of Power of Group Element]]:
+:$y = \paren {\map \phi g}^n$
+and so is a [[Definition:Power of Group Element|power]] of $\map \phi g$.
+As $y$ is arbitrary, it follows that all [[Definition:Element|elements]] of $\Img G$ are [[Definition:Power of Group Element|powers]] of $\map \phi g$.
+Hence the result, by definition of [[Definition:Cyclic Group|cyclic group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Group Element in Kernel of Homomorphism iff Power of Image is Identity}
+Tags: Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e_G$.
+Let $H$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e_H$.
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|(group) homomorphism]].
+Let $x^n \in \map \ker \phi$ for some [[Definition:Integer|integer]] $n$.
+Then:
+:$\paren {\map \phi x}^n = e_H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^n
+      | o = \in
+      | r = \map \ker \phi
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi {x^n}
+      | r = e_H
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {\map \phi x}^n
+      | r = e_H
+      | c = [[Homomorphism of Power of Group Element]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Homomorphism on Cyclic Group}
+Tags: Cyclic Groups, Group Homomorphisms
+
+\begin{theorem}
+Let $G = \gen g$ be a [[Definition:Cyclic Group|cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $g$.
+Let $H$ be a [[Definition:Group|group]].
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|(group) homomorphism]].
+Let $\map \ker \phi$ denote the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+Let $\Img G$ denote the [[Definition:Homomorphic Image|homomorphic image]] of $G$ under $\phi$.
+Then:
+:$\map \ker \phi = \gen {g^m}$
+where:
+:$m = 0$ if $\Img \phi$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]]
+:$m = \order {\Img \phi}$ if $\Img \phi$ is a [[Definition:Finite Cyclic Group|finite cyclic group]].
+\end{theorem}
+
+\begin{proof}
+From [[Kernel of Group Homomorphism is Subgroup]] and [[Subgroup of Cyclic Group is Cyclic]]:
+:$\exists m \in \N: \map \ker \phi = \gen {g^m}$
+From [[Homomorphic Image of Cyclic Group is Cyclic Group]]:
+:$\Img \phi$ is a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated by]] $\map \phi g$.
+=== Case $1$: $\Img \phi$ is infinite ===
+{{AimForCont}} $m \ne 0$.
+Then $g^m$ is not the [[Definition:Identity of Group|identity]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map \phi {g^m}
+      | r = \paren {\map \phi g}^m
+      | c = [[Homomorphism of Power of Group Element]]
+}}
+{{eqn | o = \ne
+      | r = e_H
+      | c = {{Defof|Infinite Cyclic Group}}
+}}
+{{end-eqn}}
+However this [[Definition:Contradiction|contradicts]] $g^m \in \gen {g^m} = \map \ker \phi$.
+Hence we must have $m = 0$.
+{{qed|lemma}}
+=== Case $2$: $\Img \phi$ is finite ===
+From [[Order of Cyclic Group equals Order of Generator]], the [[Definition:Order of Group Element|order]] of $\map \phi g$ is $\order {\Img \phi}$.
+Let $n \in \Z$.
+By [[Division Theorem]]:
+:$\exists q, r \in \Z: 0 \le r < \order {\Img \phi}: n = q \order {\Img \phi} + r$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \phi {g^n}
+      | r = \paren {\map \phi g}^n
+      | c = [[Homomorphism of Power of Group Element]]
+}}
+{{eqn | r = \paren {\map \phi g}^{q \order {\Img \phi} + r}
+}}
+{{eqn | r = \paren {\paren {\map \phi g}^{\order {\Img \phi} } }^q \paren {\map \phi g}^r
+}}
+{{eqn | r = {e_H}^q \paren {\map \phi g}^r
+      | c = as $\paren {\map \phi g}^{\order {\Img \phi} } = e_H$
+}}
+{{eqn | r = \paren {\map \phi g}^r
+      | c = [[Power of Identity is Identity]]
+}}
+{{end-eqn}}
+From definition of [[Definition:Order of Group Element|order of group element]]:
+:$0 < r < \order {\Img \phi} \implies \paren {\map \phi g}^r \ne e_H$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \phi {g^n} = e_H
+      | o = \iff
+      | r = r = 0
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = g^n \in \ker \phi
+      | o = \iff
+      | r = n = q \order {\Img \phi}
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = g^n \in \gen {g^m}
+      | o = \iff
+      | r = g^n \in \gen {g^\order {\Img \phi} }
+      | c = [[Group Generated by Singleton]]
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = \gen {g^m}
+      | r = \gen {g^\order {\Img \phi} }
+      | c = {{Defof|Set Equality}}
+}}
+{{end-eqn}}
+and we see that $m = \order {\Img \phi}$ satisfies the above relation.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping from Group Element to Inner Automorphism is Homomorphism}
+Tags: Inner Automorphisms, Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\kappa: G \to \Aut G$ be the [[Definition:Mapping|mapping]] from $G$ to the [[Definition:Automorphism Group of Group|automorphism group]] of $G$ defined as:
+:$\forall x \in G: \map \kappa x := \kappa_x$
+where $\kappa_x$ is the [[Definition:Inner Automorphism|inner automorphism]] on $x$:
+:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
+Then $\kappa$ is a [[Definition:Group Homomorphism|homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in G$.
+By definition of [[Definition:Automorphism Group of Group|automorphism group]], we have that:
+:$\map \kappa x \map \kappa y = \kappa_x \circ \kappa_y$
+where $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+Then $\forall g \in G$:
+{{begin-eqn}}
+{{eqn | l = \map {\kappa_x \circ \kappa_y} g
+      | r = \map {\kappa_x} {\map {\kappa_y} g}
+      | c = for all $g \in G$
+}}
+{{eqn | r = \map {\kappa_x} {y g y^{-1} }
+      | c = Definition of $\kappa_y$
+}}
+{{eqn | r = x \paren {y g y^{-1} } x^{-1}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = \paren {x y} g \paren {y^{-1} x^{-1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = \paren {x y} g \paren {x y}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \map {\kappa_{x y} } g
+      | c = Definition of $\kappa_{x y}$
+}}
+{{end-eqn}}
+And so:
+:$\forall g \in G: \map {\kappa_{x y} } g = \map {\kappa_x \circ \kappa_y} g$
+Thus by definition of $\kappa$:
+:$\map \kappa x \map \kappa y = \map \kappa{x y}$
+demonstrating that $\kappa$ is a [[Definition:Group Homomorphism|homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Mapping from Group Element to Inner Automorphism is Inner Automorphism Group}
+Tags: Inner Automorphisms, Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\kappa: G \to \Aut G$ be the [[Definition:Mapping|mapping]] from $G$ to the [[Definition:Automorphism Group of Group|automorphism group]] of $G$ defined as:
+:$\forall x \in G: \map \kappa x := \kappa_x$
+where $\kappa_x$ is the [[Definition:Inner Automorphism|inner automorphism]] on $x$:
+:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
+Then $\Img \kappa$ is the [[Definition:Inner Automorphism Group|inner automorphism group]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $\Inn G$ denote the [[Definition:Inner Automorphism Group|inner automorphism group]] of $G$.
+For all $x \in G$, $\map \kappa x = \kappa_x \in \Inn G$.
+Hence $\Img \kappa \subseteq \Inn G$. 
+Let $\phi \in \Inn G$. Then:
+:$\exists y \in G: \forall g \in G: \map \phi g = y g y^{-1}$
+Then $\map \kappa y = \phi$.
+Hence $\Inn G \subseteq \Img \kappa$. 
+Therefore $\Img \kappa = \Inn G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Monomorphic Image of Group Element}
+Tags: Group Monomorphisms
+
+\begin{theorem}
+Let $G$ and $H$be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
+Let $\phi: G \to H$ be a [[Definition:Group Monomorphism|monomorphism]].
+Let $g \in G$ be of [[Definition:Order of Group Element|finite order]].
+Then:
+:$\forall g \in G: \order {\map \phi g} = \order g$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Group Monomorphism|monomorphism]], $\phi$ is a [[Definition:Group Homomorphism|homomorphism]] which is also an [[Definition:Injection|injection]].
+From [[Order of Homomorphic Image of Group Element]]:
+:$\forall g \in G: \order {\map \phi g} \divides \order g$
+{{begin-eqn}}
+{{eqn | l = \map \phi {g^m}
+      | r = \paren {\map \phi g}^m
+      | c = [[Homomorphism of Power of Group Element]]
+}}
+{{eqn | r = e_H
+      | c = 
+}}
+{{eqn | r = \map \phi {e_G}
+      | c = [[Homomorphism to Group Preserves Identity]]
+}}
+{{end-eqn}}
+So $g^m = e$, as $\phi$ is [[Definition:Injection|injective]].
+From the definition of [[Definition:Order of Group Element|order of group element]], that means $n \le m$ since $n$ is the ''smallest'' such power.
+Thus $m = n$ and the result holds.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Image under Epimorphism of Center is Subset of Center}
+Tags: Centers of Groups, Group Epimorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $\theta: G \to H$ be an [[Definition:Group Epimorphism|epimorphism]].
+Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
+Then:
+:$\theta \sqbrk {\map Z G} \subseteq \map Z H$
+\end{theorem}
+
+\begin{proof}
+Let $y \in \theta \sqbrk {\map Z G}$.
+Let $t \in H$.
+We have that:
+:$y = \map \theta z$
+for some $z \in \map Z G$
+As $\theta$ is an [[Definition:Group Epimorphism|epimorphism]], it is by definition [[Definition:Surjection|surjective]].
+Then:
+:$t = \map \theta s$
+for some $s \in G$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y t
+      | r = \map \theta z \map \theta s
+      | c = 
+}}
+{{eqn | r = \map \theta {z s}
+      | c = {{Defof|Group Epimorphism}}
+}}
+{{eqn | r = \map \theta {s z}
+      | c = {{Defof|Center of Group}}
+}}
+{{eqn | r = \map \theta s \map \theta z
+      | c = {{Defof|Group Epimorphism}}
+}}
+{{eqn | r = t y
+      | c = {{Defof|Group Epimorphism}}
+}}
+{{end-eqn}}
+As $t$ is arbitrary, it follows that $y \in \map Z H$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Metric Subspace Induces Subspace Topology}
+Tags: Metric Spaces, Topology
+
+\begin{theorem}
+Let $M = \struct {A,d}$ be a [[Definition:Metric Space|metric space]].
+Let $H \subseteq A$.
+Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced by the metric $d$]].
+Let $\tau_H$ be the [[Definition:Subspace Topology|subspace topology]] induced by $\tau$ on $H$.
+Let $d_H$ be the [[Definition:Metric Subspace|subspace metric]] induced by $d$ on $H$.
+Let $\tau_{d_H}$ be the [[Definition:Topology Induced by Metric|topology induced by the metric $d_H$]].
+Then:
+:$\tau_{d_H} = \tau_H$
+\end{theorem}
+
+\begin{proof}
+Let $\BB$ be the set of [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] in $M$.
+Let $\BB_H$ be the set of [[Definition:Open Ball of Metric Space|open $\epsilon$-balls]] in $\struct {H, d_H}$.
+Let $U \in \tau_{d_H}$.
+By the definition of the [[Definition:Topology Induced by Metric|topology induced by the metric $d_H$]]:
+:$\exists \AA_H \subseteq \BB_H: U = \bigcup \AA_H$
+Let $\AA = \set {B': B' \in \BB, B' \cap H \in \AA_H}$.
+Let $V = \bigcup \set {B' : B' \in \AA}$.
+By the definition of the [[Definition:Topology Induced by Metric|topology induced by the metric $d$]]:
+:$V \in \tau$
+By the definition of the [[Definition:Metric Subspace|subspace metric]]:
+:$\forall B \in \BB_H: \exists B' \in \BB: B = B' \cap H$
+Hence:
+{{begin-eqn}}
+{{eqn | l = U
+      | r = \bigcup \set {B' \cap H : B' \in \AA}
+}}
+{{eqn | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn|r= V \cap H
+}}
+{{end-eqn}}
+By the definition of the [[Definition:Subspace Topology|subspace topology]] induced by $\tau$ on $H$:
+:$U \in \tau_H$
+Hence:
+:$\tau_{d_H} \subseteq \tau_H$.
+{{qed|lemma}}
+Let $U \in \tau_H$.
+By the definition of the [[Definition:Subspace Topology|subspace topology]] induced by $\tau$ on $H$:
+:$\exists V \in \tau : U = V \cap H$
+By the definition of the [[Definition:Topology Induced by Metric|topology induced by the metric $d$]]:
+:$\exists \AA \subseteq \BB: V = \bigcup \set {B' : B' \in \AA}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = U
+      | r = \paren {\bigcup \set {B' : B' \in \AA} } \cap H
+}}
+{{eqn | r = \bigcup \set {B' \cap H : B' \in \AA}
+      | c = [[Intersection Distributes over Union]]
+}}
+{{end-eqn}}
+By the definition of the [[Definition:Metric Subspace|subspace metric]]:
+:$\forall B' \in \BB: B' \cap H \in \BB_H$
+By the Definition of the [[Definition:Topology Induced by Metric|topology induced by the metric $d_H$]]:
+:$U \in \tau_{d_H}$
+Hence:
+:$\tau_H \subseteq \tau_{d_H}$
+By [[Equivalence of Definitions of Set Equality]]:
+:$\tau_H = \tau_{d_H}$
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Topology]]
+mmcdd43jmur3u3f18rwm2yx8tqqj6lg
+\end{proof}<|endoftext|>
+\section{Index of Intersection of Subgroups/Corollary}
+Tags: Index of Intersection of Subgroups
+
+\begin{theorem}
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $K$ be a [[Definition:Subgroup|subgroup]] of [[Definition:Finite Index|finite index]] of $G$.
+Then:
+:$\index H {H \cap K} \le \index G K$
+\end{theorem}
+
+\begin{proof}
+Note that $H \cap K$ is a [[Definition:Subgroup|subgroup]] of $H$.
+From [[Index of Intersection of Subgroups]], we have:
+:$\index G {H \cap K} \le \index G H \index G K$
+Setting $G = H$, we have:
+:$\index H {H \cap K} \le \index H H \index H K$
+{{finish|This does not get us where we want}}
+{{CircularStructure|This is [[Index in Subgroup]], which is used to prove the main theorem}}
+\end{proof}<|endoftext|>
+\section{Quotient of Cauchy Sequences is Metric Completion/Lemma 1}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\quad \mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ and $\sequence {y_n}$ be [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequences]] in $\mathcal C$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sequence {x_n} + \mathcal N = \sequence {y_n} + \mathcal N
+      | o = \leadstoandfrom
+      | r = \sequence {x_n} - \sequence {y_n} \in \mathcal N
+      | c = [[Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} x_n - y_n = 0_R
+      | o = \leadstoandfrom
+      | c = Definition of $\mathcal N$
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n} = 0
+      | o = \leadstoandfrom
+      | c = {{Defof|Convergent Sequence in Normed Division Ring}}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n - y_n} = 0
+      | o = \leadstoandfrom
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | r = \sequence {x_n} \sim \sequence {y_n}
+      | o = \leadstoandfrom
+      | c = Definition of [[Definition:Equivalence Relation|equivalence relation]] $\sim$
+}}
+{{end-eqn}}
+Hence:
+:$\sequence {x_n}$ and $\sequence {y_n}$ belong to the same [[Definition:Equivalence Class|equivalence class]] in $\mathcal C \,\big / \mathcal N$ {{iff}} $\sequence {x_n}$ and $\sequence {y_n}$ belong to the same [[Definition:Equivalence Class|equivalence class]] in $\tilde {\mathcal C}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Cauchy Sequences is Metric Completion/Lemma 2}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\quad d' = \tilde d$
+\end{theorem}
+
+\begin{proof}
+By [[Quotient of Cauchy Sequences is Metric Completion/Lemma 1|Lemma 1 of Quotient of Cauchy Sequences is Metric Completion]] we have that:
+:$\mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$
+Let $\eqclass {x_n} {}$ and $\eqclass {x_n} {}$ be [[Definition:Equivalence Class|equivalence classes]] in $\mathcal C \,\big / \mathcal N = \tilde {\mathcal C}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {d'} {\eqclass {x_n}{}, \eqclass {x_n}{} }
+      | r = \norm {\eqclass {x_n}{} - \eqclass {x_n}{} }
+      | c = Definition of $d'$
+}}
+{{eqn | r = \norm {\eqclass {x_n - y_n}{} }
+      | c = {{Defof|Quotient Ring}} Addition
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n - y_n}
+      | c = {{Defof|Norm on Division Ring}}: $\norm {\,\cdot\,}$ on $\mathcal C \,\big / \mathcal N$
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \map d {x_n, y_n}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}: by the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$ on $R$
+}}
+{{eqn | r = \map {\tilde d} {\eqclass {x_n}{}, \eqclass {x_n}{} }
+      | c = Definition of $\tilde d$
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generator for Quaternion Group}
+Tags: Quaternion Group
+
+\begin{theorem}
+The [[Definition:Quaternion Group|Quaternion Group]] can be [[Definition:Generator of Group|generated]] by the [[Definition:Square Matrix|matrices]]:
+:$\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+\qquad
+\mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
+where $i$ is the [[Definition:Imaginary Unit|imaginary unit]]:
+:$i^2 = -1$
+\end{theorem}
+
+\begin{proof}
+Note that:
+:$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
+is the [[Definition:Identity Element|identity]] for [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] of [[Definition:Order of Square Matrix|order $2$]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \mathbf a^2
+      | r = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \times 0 + 1 \times \paren {-1} & 0 \times 1 + 1 \times 0 \\ \paren {-1} \times 0 + 0 \times \paren {-1} & \paren {-1} \times 1 + 0 \times 0 \end{bmatrix}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \mathbf a^4
+      | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} \paren {-1} \times \paren {-1} + 0 \times 0 & \paren {-1} \times 0 + 0 \times \paren {-1} \\ 0 \times \paren {-1} + \paren {-1} \times 0 & 0 \times 0 + \paren {-1} \times \paren {-1} \end{bmatrix}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf I
+      | c = 
+}}
+{{end-eqn}}
+Next we have:
+{{begin-eqn}}
+{{eqn | l = \mathbf b^2
+      | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \times 0 + i \times i & 0 \times i + i \times 0 \\ \paren i \times 0 + 0 \times i & i \times i + 0 \times 0 \end{bmatrix}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{end-eqn}}
+and so:
+:$\mathbf b^2 = \mathbf a^2$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \mathbf a \mathbf b \mathbf a
+      | r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} \paren {-1} \times 0 + 0 \times i & \paren {-1} \times i + 0 \times 0 \\ \paren 0 \times 0 + \paren {-1} \times i & 0 \times i + \paren {-1} \times 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 0 \times \paren {-1} + \paren {-i} \times 0 & 0 \times 0 + \paren {-i} \times \paren {-1} \\ \paren {-i} \times \paren {-1} + 0 \times 0 & \paren {-i} \times 0 + 0 \times \paren {-1} \end{bmatrix}
+      | c = {{Defof|Matrix Product (Conventional)}}
+}}
+{{eqn | r = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \mathbf b
+      | c = 
+}}
+{{end-eqn}}
+Thus $\gen {\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} }$ fulfils the conditions for the [[Quaternion Group/Group Presentation|group presentation]] of $\Dic 2$:
+{{:Quaternion Group/Group Presentation}}
+Hence the result.
+{{qed}}
+[[Category:Quaternion Group]]
+pt2e4tktnyza63mtq1yg1ju5syuu1r7
+\end{proof}<|endoftext|>
+\section{Condition for Nu Function to be 1}
+Tags: Cyclic Groups, Nu Function, Condition for Nu Function to be 1
+
+\begin{theorem}
+Let:
+:$n = \displaystyle \prod_{i \mathop = 1}^s p_i^{m_i}$
+where $p_1, p_2, \ldots, p_s$ are [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|primes]].
+Then:
+:$(1): \quad m_1, m_2, \ldots, m_s = 1$, that is, $n$ is [[Definition:Square-Free|square-free]]
+:$(2): \quad \forall i, j \in \set {1, 2, \ldots, s}: p_i \not \equiv 1 \pmod {p_j}$
+{{iff}}:
+:every [[Definition:Group|group]] $G$ of [[Definition:Order of Group|order]] $n$ is [[Definition:Cyclic Group|cyclic]] and so $\map \nu n = 1$.
+\end{theorem}<|endoftext|>
+\section{Group of Order 15 is Cyclic Group}
+Tags: Groups of Order 15, Cyclic Groups, Group of Order 15 is Cyclic Group, Groups of Order p q
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $15$.
+Then $G$ is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+We have that $15 = 3 \times 5$.
+Thus:
+:$15$ is [[Definition:Square-Free|square-free]]
+:$5 \equiv 2 \pmod 3$
+:$3 \equiv 3 \pmod 5$
+The conditions are fulfilled for [[Condition for Nu Function to be 1]].
+Thus $\map \nu {15} = 1$ and so all [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $15$ are [[Definition:Cyclic Group|cyclic]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+From [[Number of Sylow p-Subgroups in Group of Order 15]]:
+:the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] is in the [[Definition:Set|set]] $\set {1, 4, 7, \ldots}$
+:the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] is in the [[Definition:Set|set]] $\set {1, 6, 11, \ldots}$.
+From the [[Fifth Sylow Theorem]]
+:the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] is a [[Definition:Divisor of Integer|divisor]] of $15$
+:the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] is a [[Definition:Divisor of Integer|divisor]] of $15$.
+Combining the above:
+:there is a [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] in $G$, which we will denote $P$
+:there is a [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] in $G$, which we will denote $Q$.
+Let $x \in G$ be of [[Definition:Order of Group Element|order $3$]].
+Then by [[Non-Trivial Group has Non-Trivial Cyclic Subgroup]], $\gen x$ has $3$ [[Definition:Element|elements]].
+Thus $\gen x$ is a [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] and so:
+:$\gen x = P$
+and so $x$ is one of the $2$ non-[[Definition:Identity Element|identity]] [[Definition:Element|elements]] of $P$.
+Similarly, let $y \in G$ be of [[Definition:Order of Group Element|order $5$]].
+Then by a similar argument:
+:$\gen y = Q$
+and so $y$ is one of the $4$ non-[[Definition:Identity Element|identity]] [[Definition:Element|elements]] of $Q$.
+So $G$ has:
+:$1$ [[Definition:Element|element]] of [[Definition:Order of Group Element|order $1$]] (that is, the [[Definition:Identity Element|identity]])
+:$2$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order $3$]]
+:$4$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order $5$]]
+which leaves $8$ [[Definition:Element|elements]] whose [[Definition:Order of Group Element|order]] is still to be determined.
+From [[Order of Element Divides Order of Finite Group]], they are all of [[Definition:Order of Group Element|order]] $1$, $3$, $5$ or $15$.
+As the [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $1$, $3$ and $5$ have been accounted for, they must all be of [[Definition:Order of Group Element|order]] $15$.
+So $G$ has $8$ [[Definition:Distinct Elements|distinct elements]] of [[Definition:Order of Group Element|order]] $15$.
+Hence $G$ must be [[Definition:Cyclic Group|cyclic]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+{{AimForCont}} $G$ is non-[[Definition:Abelian Group|abelian]].
+Let $n_3$ denote the number of [[Definition:Element|elements]] of $G$ of [[Definition:Order of Group Element|order]] $3$.
+From [[Number of Elements of Order p in Group of Order pq is Multiple of q]], $n_3$ is a [[Definition:Integer Multiple|multiple]] of $5$.
+From [[Number of Order p Elements in Group with m Order p Subgroups]], $n_3$ is a [[Definition:Integer Multiple|multiple]] of $2$.
+Therefore $n_3$ is a [[Definition:Integer Multiple|multiple]] of $10$.
+Let $n_5$ denote the number of [[Definition:Element|elements]] of $G$ of [[Definition:Order of Group Element|order]] $5$.
+From [[Number of Elements of Order p in Group of Order pq is Multiple of q]], $n_5$ is a [[Definition:Integer Multiple|multiple]] of $3$.
+From [[Number of Order p Elements in Group with m Order p Subgroups]], $n_5$ is a [[Definition:Integer Multiple|multiple]] of $4$.
+Therefore $n_5$ is a [[Definition:Integer Multiple|multiple]] of $12$.
+Together with the [[Definition:Identity Element|identity element]] which has [[Definition:Order of Group Element|order]] $1$, that makes $1 + 12 a + 10 b = 15$ for some [[Definition:Positive Integer|positive integers]] $a$ and $b$.
+This is impossible.
+Hence by [[Proof by Contradiction]] it follows that $G$ must be [[Definition:Abelian Group|abelian]].
+Since $15$ is a product of $2$ distinct [[Definition:Prime Number|primes]], by [[Abelian Group of Semiprime Order is Cyclic]], $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Abelian Groups}
+Tags: Abelian Groups
+
+\begin{theorem}
+Let $n \in \Z_{\ge 1}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let:
+:$n = \displaystyle \prod_{i \mathop = 1}^s p_i^{m_i}$
+where the $p_i$ are [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|primes]].
+Let $\map {\nu_a} n$ denote the number of [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Group|order]] $n$.
+Then:
+:$\map {\nu_a} n = \displaystyle \prod_{i \mathop = 1}^s \map {\nu_a} {p_i^{m_i} }$
+where:
+:$\map {\nu_a} {p_i^{m_i} }$ is the number of [[Definition:Integer Partition|integer partitions]] of $m_i$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|long and heavy proof which needs plenty work}}
+\end{proof}<|endoftext|>
+\section{Order of Quotient Group}
+Tags: Quotient Groups, Order of Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $G / N$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $N$.
+Then:
+:$\dfrac {\order G} {\order N} = \order {G / N}$
+where $\order G$ denotes the [[Definition:Order of Group|order]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\dfrac {\order G} {\order N} = \index G N$
+where $\index G N$ is the [[Definition:Index of Subgroup|index]] of $N$ in $G$.
+By definition of [[Definition:Index of Subgroup|index]]:
+:$\index G N = \order {G / N}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Composition Series/Composition Length}
+Tags: Definitions: Normal Series
+
+\begin{theorem}
+Let $\HH$ be a [[Definition:Composition Series|composition series]] for $G$.
+The '''composition length''' of $G$ is the [[Definition:Length of Normal Series|length]] of $\HH$.
+\end{theorem}<|endoftext|>
+\section{Definition:Composition Series/Composition Factor}
+Tags: Definitions: Normal Series
+
+\begin{theorem}
+Let $\HH = \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$ be a [[Definition:Composition Series|composition series]] for $G$.
+Each of the [[Definition:Quotient Group|quotient groups]]:
+:$G_1 / G_0, G_2 / G_1, \ldots, G_n / G_{n - 1}$
+are the '''composition factors''' of $G$.
+\end{theorem}<|endoftext|>
+\section{Non-Abelian Simple Finite Groups are Infinitely Many}
+Tags: Simple Groups
+
+\begin{theorem}
+There exist [[Definition:Infinite Set|infinitely many]] [[Definition:Group Type|types of group]] which are non-[[Definition:Abelian Group|abelian]] and [[Definition:Finite Group|finite]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Alternating Group is Simple]].
+\end{proof}<|endoftext|>
+\section{Centralizer of Self-Inverse Element of Non-Abelian Finite Simple Group is not That Group}
+Tags: Centralizers, Simple Groups, Self-Inverse Elements, Finite Groups
+
+\begin{theorem}
+Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Finite Group|finite]] [[Definition:Simple Group|simple group]].
+Let $t \in G$ be a [[Definition:Self-Inverse Element|self-inverse element]] of $G$.
+Then:
+:$\map {C_G} t \ne G$
+where $\map {C_G} t$ denotes the [[Definition:Centralizer|centralizer]] of $t$ in $G$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Finite Group|finite]] [[Definition:Simple Group|simple group]].
+Let $t \in G$ which is not the [[Definition:Identity Element|identity]].
+By definition of a [[Definition:Simple Group|simple group]] and [[Center of Group is Normal Subgroup]]:
+:either $\map Z G = G$ or $\map Z G$ is the [[Definition:Trivial Group|trivial group]].
+By definition of an [[Definition:Abelian Group/Definition 2|abelian group]]:
+:$\map Z G = G$ {{iff}} $G$ is abelian
+Hence we must have $\map Z G$ is the [[Definition:Trivial Group|trivial group]].
+Thus $t \notin \map Z G$.
+From definition of [[Definition:Center of Group|center]]:
+:$\exists x \in G: t x \ne x t$
+For this $x$, $x \notin \map {C_G} t$.
+Hence $\map {C_G} t \ne G$.
+{{Questionable|So trivial it's scary: something stronger was proved </br> Remark: the identity is a self-inverse element with $\map {C_G} e {{=}} G$ trivially}}
+\end{proof}<|endoftext|>
+\section{Upper Bound of Order of Non-Abelian Finite Simple Group}
+Tags: Centralizers, Simple Groups, Self-Inverse Elements, Finite Groups
+
+\begin{theorem}
+Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Finite Group|finite]] [[Definition:Simple Group|simple group]].
+Let $t \in G$ be a [[Definition:Self-Inverse Element|self-inverse element]] of $G$.
+Let $\map {C_G} t$ denote the [[Definition:Centralizer|centralizer]] of $t$ in $G$.
+Let $m = \order {\map {C_G} t}$ be the [[Definition:Order of Group|order]] of $\map {C_G} t$.
+Then:
+:$\order G \le \paren {\dfrac {m \paren {m + 1} } 2}!$
+\end{theorem}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 1}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\norm {\, \cdot \,}_1$ is [[Definition:Well-Defined Mapping|well-defined]].
+That is, 
+:$(1): \quad \forall \eqclass {x_n}{}: \lim_{n \mathop \to \infty} \norm{x_n}$ exists.
+:$(2): \quad \displaystyle \forall \eqclass {x_n}{}, \eqclass {y_n}{} \in \mathcal C \,\big / \mathcal N: \eqclass {x_n}{} = \eqclass {y_n}{} \implies \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$
+\end{theorem}
+
+\begin{proof}
+By [[Norm Sequence of Cauchy Sequence has Limit]] then:
+:for each $\eqclass {x_n}{}$ the $\displaystyle \lim_{n \mathop \to \infty} \norm{x_n}$ exists.
+Suppose $\eqclass {x_n}{} = \eqclass {y_n}{}$.
+By [[Left Cosets are Equal iff Product with Inverse in Subgroup|Left Cosets are Equal iff Difference in Subgroup]] then:
+:$\sequence {x_n} - \sequence {y_n} =  \sequence {x_n - y_n} \in \mathcal N$
+By [[Equivalent Cauchy Sequences have Equal Limits of Norm Sequences]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n}$
+Hence:
+:$\displaystyle \norm { \eqclass {x_n}{} }_1 = \lim_{n \mathop \to \infty} \norm{x_n} = \lim_{n \mathop \to \infty} \norm{y_n} =  \norm { \eqclass {x_n}{} }_1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 2}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\norm {\, \cdot \,}_1$ satisfies {{NormAxiom|1}}
+That is:
+:$\forall \eqclass {x_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} }_1 = 0 \iff \eqclass {x_n} {} = \eqclass {0_R} {} $
+\end{theorem}
+
+\begin{proof}
+By [[Quotient Ring of Cauchy Sequences is Division Ring]] the [[Definition:Ring Zero|zero]] of $\CC \,\big / \NN$ is $\eqclass {0_R} {}$.
+{{begin-eqn}}
+{{eqn | l = \norm {\eqclass {0_R} {} }_1 = 0
+      | o = \leadstoandfrom
+      | r = \lim_{n \mathop \to \infty} \norm {x_n} = 0
+      | c = Definition of $\norm {\,\cdot\,}_1$
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \sequence {x_n} \in \NN
+      | c = Definition of $\NN$
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \eqclass {x_n} {} = \eqclass {0_R} {}
+      | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 3}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\norm {\, \cdot \,}_1$ satisfies the {{NormAxiom|2}}.
+That is:
+:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} \eqclass {y_n} {} }_1 = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} }_1$
+\end{theorem}
+
+\begin{proof}
+Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
+{{begin-eqn}}
+{{eqn | l = \norm {\eqclass {x_n} {} \eqclass {y_n} {} } _1
+      | r = \norm {\eqclass {x_n y_n} {} }_1
+      | c = Multiplication on quotient ring
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n y_n}
+      | c = Definition of $\norm {\,\cdot\,}_1$
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \norm {y_n}
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} \times \lim_{n \to \infty} \norm {y_n }
+      | c = [[Product Rule for Real Sequences]]
+}}
+{{eqn | r = \norm {\eqclass {x_n} {} }_1 \times \norm {\eqclass {y_n} {} } _1
+      | c = Definition of $\norm {\,\cdot\,}_1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Normed Division Ring/Lemma 4}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+:$\norm {\, \cdot \,}_1$ satisfies the {{NormAxiom|3}}.
+That is:
+:$\forall \eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN: \norm {\eqclass {x_n} {} +  \eqclass {y_n} {} }_1 \le \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} }_1$
+\end{theorem}
+
+\begin{proof}
+Let $\eqclass {x_n} {}, \eqclass {y_n} {} \in \CC \,\big / \NN$
+{{begin-eqn}}
+{{eqn | l = \norm {\eqclass {x_n} {} + \eqclass {y_n} {} } _1
+      | r = \norm {\eqclass {x_n + y_n} {} }_1
+      | c = Addition on quotient ring
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n + y_n}
+      | c = Definition of $\norm {\,\cdot\,}_1$
+}}
+{{end-eqn}}
+By {{NormAxiom|3}}:
+:$\forall n: \norm {x_n + y_n} \le \norm {x_n} + \norm {y_n}$
+So:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \norm {x_n + y_n}
+      | o = \le
+      | r = \lim_{n \mathop \to \infty} \norm {x_n} + \norm {y_n}
+      | c = [[Inequality Rule for Real Sequences]] 
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \norm {x_n} + \lim_{n \mathop \to \infty} \norm {y_n}
+      | c = [[Sum Rule for Real Sequences]]
+}}
+{{eqn | r = \norm {\eqclass {x_n} {} }_1 + \norm {\eqclass {y_n} {} } _1
+      | c = Definition of $\norm {\,\cdot\,}_1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Invertible Elements of Monoid form Subgroup}
+Tags: Monoids, Subgroups
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity element]] is $e$.
+Let $U \subseteq S$ be the [[Definition:Subset|subset]] of $S$ consisting of the [[Definition:Invertible Element|invertible elements]] of $S$.
+Then $\struct {U, \circ}$ forms a [[Definition:Subgroup|subgroup]] of $S$.
+\end{theorem}
+
+\begin{proof}
+We have from [[Inverse of Identity Element is Itself]] that $e$ is [[Definition:Invertible Element|invertible]].
+Hence $e \in U$ and so $U \ne \O$.
+Let $x, y \in U$.
+As $x$ and $y$ are [[Definition:Invertible Element|invertible]], it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$.
+Both $x^{-1}$ and $y^{-1}$ also have [[Definition:Inverse Element|inverses]] $x$ and $y$ respectively, and so themselves are [[Definition:Invertible Element|invertible]].
+Hence both $x^{-1} \in U$ and $y^{-1} \in U$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \paren {x \circ y} \circ \paren {y^{-1} \circ x^{-1} }
+      | r = x \circ \paren {y \circ y^{-1} } \circ x^{-1}
+      | c = $\circ$ is [[Definition:Associative Operation|associative]] in $S$
+}}
+{{eqn | r = x \circ e \circ x^{-1}
+      | c = 
+}}
+{{eqn | r = x \circ x^{-1}
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$
+Hence $x \circ y$ is [[Definition:Invertible Element|invertible]].
+The result follows by the [[Two-Step Subgroup Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Combination Theorem for Sequences/Normed Division Ring/Inverse Rule/Lemma}
+Tags: Combination Theorem for Sequences in Normed Division Rings
+
+\begin{theorem}
+:$\displaystyle \lim_{n \mathop \to \infty} y_n^{-1} = l^{-1}$
+\end{theorem}
+
+\begin{proof}
+By [[Limit of Subsequence equals Limit of Sequence/Normed Division Ring|Limit of Subsequence equals Limit of Sequence]] then $\sequence {y_n}$ is [[Definition:Convergent Sequence in Normed Division Ring|convergent]] with:
+:$\displaystyle \lim_{n \mathop \to \infty} y_n = l$
+Let $\epsilon > 0$ be given.
+Let $\epsilon' = \dfrac {\epsilon  {\norm l}^2 } {2}$.
+Then:
+:$ \epsilon' > 0$
+As $\sequence {y_n} \to l$, as $n \to \infty$, we can find $N_1$ such that:
+:$\forall n > N_1: \norm {y_n - l} < \epsilon'$
+As $\sequence {y_n}$ converges to $l \ne 0$, by [[Sequence Converges to Within Half Limit/Normed Division Ring|Sequence Converges to Within Half Limit]]:
+:$\exists N_2 \in \N: \forall n > \N_2: \dfrac {\norm l} 2 < \norm {y_n}$
+or equivalently:
+:$\exists N_2 \in \N: \forall n > \N_2: 1 < \dfrac {2 \norm {y_n} } {\norm l}$
+Let $N = \max \set {N_1, N_2}$. 
+Then $\forall n > N$:
+:$(1): \quad \norm {y_n - l} < \epsilon'$
+:$(2): \quad 1 < \dfrac {2 \norm {y_n} } {\norm l}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm { {y_n }^{-1} - l^{-1} }
+      | o = <
+      | r = \dfrac {2 \norm {y_n} } {\norm l} \norm { {y_n}^{-1} - l^{-1} }
+      | c = from $(1)$
+}}
+{{eqn | r = \dfrac 2 {\norm l^2} \paren {\norm { {y_n } } \norm { {y_n}^{-1} - l^{-1} } \norm l}
+      | c = multiplying and dividing by $\norm l$
+}}
+{{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n \paren { {y_n}^{-1} - l^{-1} } l}
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | r = \dfrac 2 {\norm l^2} \norm {y_n y_n^{-1} l - y_n l^{-1} l}
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = \dfrac 2 {\norm l^2} \norm {l - y_n}
+      | c = [[Definition:Division Ring|Inverse Property of Division Ring]]
+}}
+{{eqn | o = <
+      | r = \dfrac 2 {\norm l^2} \epsilon'
+      | c = from $(2)$
+}}
+{{eqn | r = \dfrac {2} {\norm l^2} \paren { \dfrac {\epsilon \norm l^2 } 2}
+      | c = Definition of $\epsilon'$
+}}
+{{eqn | r = \epsilon
+      | c = cancelling terms
+}}
+{{end-eqn}}
+Hence:
+:$\sequence { {y_n}^{-1} }$ is [[Definition:Convergent Sequence in Normed Division Ring|convergent]] with $\displaystyle \lim_{n \mathop \to \infty} {y_n}^{-1} = l^{-1}$.
+{{qed}}
+[[Category:Combination Theorem for Sequences in Normed Division Rings]]
+s9m3b9wd0jhdnb1n3cmd1fpdiz6vbae
+\end{proof}<|endoftext|>
+\section{Division Ring Norm is Continuous on Induced Metric Space}
+Tags: Normed Division Rings, Norm Theory
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}}$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $d$ be the [[Definition:Metric Induced by Norm|metric induced]] by the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$.
+The [[Definition:Mapping|mapping]] $\norm {\,\cdot\,} : \struct {R, d} \to \R$ is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $x_0 \in R$.
+Let $\epsilon \in \R_{>0}$. 
+Let $x \in R: \norm {x - x_0} < \epsilon$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\norm x - \norm {x_0} }
+      | o = \le
+      | r = \norm {x - x_0}
+      | c = [[Reverse Triangle Inequality/Normed Division Ring|Reverse triangle inequality]]
+}}
+{{eqn | o = <
+      | r = \epsilon
+}}
+{{end-eqn}}
+By the definition of [[Definition:Metric Induced by Norm|metric induced by a norm]] and the definition of a [[Definition:Continuous Mapping (Metric Space)|continuous mapping]], $\norm {\,\cdot\,}$ is [[Definition:Continuous Mapping (Metric Space)|continuous]].
+{{qed}}
+[[Category:Normed Division Rings]]
+[[Category:Norm Theory]]
+81dpy5f0xjivpcxu1h0pvr5llskobwv
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Squares of Fibonacci Numbers}
+Tags: Fibonacci Numbers, Sums of Sequences, Proofs by Induction
+
+\begin{theorem}
+:$\forall n \ge 1: \displaystyle \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
+That is:
+:${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
+=== Basis for the Induction ===
+$\map P 1$ is the case ${F_1}^2 = 1 = F_3 - 1$, which holds from the definition of [[Definition:Fibonacci Numbers|Fibonacci numbers]].
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^1 {F_j}^2
+      | r = {F_1}^2
+      | c = 
+}}
+{{eqn | r = 1 \times 1
+      | c = 
+}}
+{{eqn | r = F_1 \times F_2
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $\map P 1$ holds.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{j \mathop = 1}^k {F_j}^2 = F_k F_{k + 1}$
+Then we need to show:
+:$\displaystyle \sum_{j \mathop = 1}^{k + 1} {F_j}^2 = F_{k + 1} F_{k + 2}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^{k + 1} {F_j}^2
+      | r = \sum_{j \mathop = 1}^k {F_j}^2 + {F_{k + 1} }^2
+      | c = 
+}}
+{{eqn | r = F_k F_{k + 1} + {F_{k + 1} }^2
+      | c = [[Sum of Sequence of Squares of Fibonacci Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {F_k + F_{k + 1} } F_{k + 1}
+      | c = 
+}}
+{{eqn | r = F_{k + 2} F_{k + 1}
+      | c = {{Defof|Fibonacci Number}}
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \ge 1: \displaystyle \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring}
+Tags: Normed Division Rings, Properties of Norm on Division Ring
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] with [[Definition:Ring Zero|zero]] $0_R$ and [[Definition:Unity of Ring|unity]] $1_R$.
+Let $\norm {\,\cdot\,}$ be a [[Definition:Norm on Division Ring|norm]] on $R$.
+Let $x, y \in R$.
+Then the following hold:
+\end{theorem}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Negative}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$\norm {-x} = \norm {x}$
+\end{theorem}
+
+\begin{proof}
+By [[Properties of Norm on Division Ring/Norm of Negative of Unity|Norm of Negative of Unity]] then:
+:$\norm{-1_R} = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm{-x}
+      | r = \norm{-1_R \circ x}
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = \norm{-1_R} \norm{x}
+      | c = [[Definition:Norm Axioms|Norm axiom (N2) (Multiplicativity)]]
+}}
+{{eqn | r = \norm{x}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative of Unity|Norm of Negative of Unity]]
+}}
+{{end-eqn}}
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Unity}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$\norm {1_R} = 1$.
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Norm Axioms|norm axiom (N2) (Multiplicativity)]] then:
+:$\forall x, y \in R: \norm {x \circ y} = \norm{x} \norm{y}$
+In particular:
+:$\norm{1_R} = \norm{1_R \circ 1_R} = \norm{1_R} \norm{1_R}$
+By the [[Definition:Norm Axioms|norm axiom (N1) (Positive defintiteness)]] then:
+:$\norm{1_R} \ne 0$
+So $\norm{1_R}$ has an inverse in $\R$.
+By multiplying by this inverse, then:
+:$ \norm{1_R} \norm{1_R} =\norm{1_R} \iff \norm{1_R} = 1$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Negative of Unity}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$\norm{-1_R} = 1$
+\end{theorem}
+
+\begin{proof}
+By [[Product of Ring Negatives]] then:
+:$-1_R \circ -1_R = 1_R \circ 1_R=1_R$.
+So:
+{{begin-eqn}}
+{{eqn | l = \norm{-1_R}^{2}
+      | r = \norm{-1_R} \norm{-1_R}
+}}
+{{eqn | r = \norm{-1_R \circ -1_R}
+      | c = [[Definition:Norm Axioms|Norm axiom (N2) (Multiplicativity)]]
+}}
+{{eqn | r = \norm{1_R}
+      | c = [[Product of Ring Negatives]]
+}}
+{{eqn | r = 1
+      | c = [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity of Division Ring]]
+}}
+{{end-eqn}}
+Thus:
+:$\norm{-1_R} = \pm 1$
+By the [[Definition:Norm Axioms|norm axiom (N1) (Positive defintiteness)]] then: 
+:$\norm{-1_R} \ge 0$
+Hence:
+:$\norm{-1_R} = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Repunit is Zuckerman Number}
+Tags: Repunits, Zuckerman Numbers
+
+\begin{theorem}
+Let $n$ be a [[Definition:Repunit|repunit]].
+Then $n$ is also a [[Definition:Zuckerman Number|Zuckerman number]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Digit|digits]] of a [[Definition:Repunit|repunit]] are by definition all $1$.
+Thus the [[Definition:Integer Multiplication|product]] of the [[Definition:Digit|digits]] of a [[Definition:Repunit|repunit]] is $1$.
+By [[One Divides all Integers]], $1$ is a [[Definition:Divisor|divisor]] of $n$.
+Hence the result, by definition of [[Definition:Zuckerman Number|Zuckerman number]].
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Difference}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$\norm {x - y} \le \norm x + \norm y$
+\end{theorem}
+
+\begin{proof}
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x - y}
+      | r = \norm {x + \paren {-y} }
+}}
+{{eqn | o = \le
+      | r = \norm x + \norm {-y}
+      | c = [[Definition:Norm Axioms|Norm axiom $(\text N 3)$: Triangle Inequality]]
+}}
+{{eqn | r = \norm x + \norm y
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]
+}}
+{{end-eqn}}
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Inverse}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$x \ne 0_R \implies \norm {x^{-1} } = \dfrac 1 {\norm x}$
+\end{theorem}
+
+\begin{proof}
+Let $x \ne 0_R$.
+By [[Definition:Norm Axioms|Norm axiom $(\text N 1)$: Positive Definiteness]]:
+:$\norm x \ne 0$
+So:
+{{begin-eqn}}
+{{eqn| l = \norm x \norm {x^{-1} }
+     | r = \norm {x \circ x^{-1} }
+     | c = [[Definition:Norm Axioms|Norm axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn| r = \norm {1_R}
+     | c = {{Defof|Product Inverse}}
+}}
+{{eqn| r = 1
+     | c = [[Norm of Unity of Division Ring]]
+}}
+{{eqn| ll= \leadsto
+     | l = \norm {x^{-1} }
+     | r = \dfrac 1 {\norm x}
+     | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Quotient}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$y \ne 0_R \implies \norm{xy^{-1}} = \norm{y^{-1}x} = \dfrac {\norm{x}}{\norm{y}}$
+\end{theorem}
+
+\begin{proof}
+Let $y \ne 0_R$.
+By [[Definition:Norm Axioms|Norm axiom (N1) (Positive Definiteness)]] then:
+:$\norm {y} \ne 0$
+So:
+{{begin-eqn}}
+{{eqn| l = \norm{x \circ y^{-1} }
+     | r = \norm{x} \norm{y^{-1} }
+     | c = [[Definition:Norm Axioms|Norm axiom (N2) (Multiplicativity)]]
+}}
+{{eqn| r = \dfrac {\norm{x} } {\norm{y} }
+     | c = [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]]
+}}
+{{end-eqn}}
+Similarly:
+:$\norm{y^{-1}x} = \dfrac {\norm{x}}{\norm{y}}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Power Equals Unity}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+:$\forall n \in \N_{\gt 0}: \norm {x^n} = 1 \implies \norm x = 1$
+\end{theorem}
+
+\begin{proof}
+Let $n \in \N_{\gt 0}$.
+Let $\norm {x^n} = 1$.
+By [[Definition:Norm Axioms|Norm axiom (N2) (Multiplicativity)]] then:
+:$\norm x^n = 1$
+Since $\norm x \ge 0$, by [[Positive Real Complex Root of Unity]] then:
+:$\norm x = 1$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Norm on Division Ring/Norm of Integer}
+Tags: Properties of Norm on Division Ring
+
+\begin{theorem}
+For all $n \in \N_{\gt 0}$, let $n \cdot 1_R$ denote the sum of $1_R$ with itself $n$-times. That is:
+:$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{n \, times}$
+Then:
+:$\norm {n \cdot 1_R} \le n$.
+\end{theorem}
+
+\begin{proof}
+Let $n \in \N_{\gt 0}$.
+Then:
+{{begin-eqn}}
+{{eqn| l = \norm {n \cdot 1_R}
+     | r = = \norm {1_R + 1_R + \dots + 1_R}
+     | o =
+}}
+{{eqn| o =
+}}
+{{eqn| o = 
+     | r = \le \underbrace {\norm {1_R} + \norm {1_R} + \dots + \norm {1_R} }_{n \, times}
+     | c = [[Definition:Norm Axioms|Norm axiom (N3) (Triangle Inequality)]]
+}}
+{{eqn| o =
+}}
+{{eqn| o =
+     | r = = \underbrace {1 + 1 + \dots + 1 }_{n \, times}
+     | c = [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity]]
+}}
+{{eqn| o =
+}}
+{{eqn| o =
+     | r = = n
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conversion from Hexadecimal to Binary}
+Tags: Hexadecimal Notation, Binary Notation, Conversion from Hexadecimal to Binary
+
+\begin{theorem}
+Let $n$ be a [[Definition:Positive Integer|(positive) integer]] expressed in [[Definition:Hexadecimal Notation|hexadecimal notation]] as:
+:$n = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H$
+Then $n$ can be expressed in [[Definition:Binary Notation|binary notation]] as:
+:$n = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2$
+where $\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2$ is the expression of the [[Definition:Hexadecimal Notation|hexadecimal digit]] $a_j$ in [[Definition:Binary Notation|binary notation]].
+That is, you take the [[Definition:Binary Notation|binary expression]] of each [[Definition:Hexadecimal Notation|hexadecimal digit]], padding them out with [[Definition:Zero Digit|zeroes]] to make them $4$ [[Definition:Bit|bits]] long, and simply [[Definition:Concatenation (Formal Systems)|concatenate]] them.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^r a_j 16^j
+      | c = {{Defof|Hexadecimal Notation}}
+}}
+{{end-eqn}}
+We have that:
+:$0 \le a_j < 16$
+and so:
+{{begin-eqn}}
+{{eqn | l = a_j
+      | r = \sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^3 b_{j k} 2^k
+      | c = {{Defof|Binary Notation}}
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \sqbrk {a_r a_{r - 1} \dotso a_1 a_0}_H
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^r a_j 16^j
+      | c = {{Defof|Hexadecimal Notation}}
+}}
+{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 16^j
+      | c = {{Defof|Binary Notation}}
+}}
+{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sum_{k \mathop = 0}^3 b_{j k} 2^k} 2^{4 j}
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^r \paren {\sqbrk {b_{j 3} b_{j 2} b_{j 1} b_{j 0} }_2} 2^{4 j}
+      | c = {{Defof|Binary Notation}}
+}}
+{{eqn | r = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} }_2 2^{4 r} + \sqbrk {b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} }_2 2^{4 {r - 1} } + \cdots + \sqbrk {b_{1 3} b_{1 2} b_{1 1} b_{1 0} }_2 2^4 + \sqbrk {b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2
+      | c = 
+}}
+{{eqn | r = \sqbrk {b_{r 3} b_{r 2} b_{r 1} b_{r 0} b_{\paren {r - 1} 3} b_{\paren {r - 1} 2} b_{\paren {r - 1} 1} b_{\paren {r - 1} 0} \dotso b_{1 3} b_{1 2} b_{1 1} b_{1 0} b_{0 3} b_{0 2} b_{0 1} b_{0 0} }_2
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Birthday Paradox/General/3}
+Tags: Birthday Paradox
+
+\begin{theorem}
+Let $n$ be a [[Definition:Set|set]] of people.
+Let the [[Definition:Probability|probability]] that at least $3$ of them have the same birthday be greater than $50 \%$.
+Then $n \ge 88$.
+\end{theorem}
+
+\begin{proof}
+Let $\map F {r, n}$ be the number of ways to distribute $r$ objects into $n$ cells such that there are no more than $2$ objects in each cell.
+Let there be $d$ cells which are each occupied by $2$ objects.
+These can be chosen in $\dbinom n d$ ways.
+There remain $s = r - 2 d$ objects which can then be distributed among $n - d$ cells in $\dbinom {n - d} s$ ways.
+In each such arrangement, the $r$ objects may be [[Definition:Permutation|permuted]] in:
+:$\dbinom r 2 \dbinom {r - 2} 2 \cdots \dbinom {r - 2 d + 2} 2 \paren {r - 2 d}! = \dfrac {r!} {2^d}$
+different ways.
+Hence:
+:$\map F {r, n} = \dbinom n d \dbinom {n - d} s \dfrac {r!} {2^d}$
+So the [[Definition:Probability|probability]] of exactly $d$ pairs and $s$ singletons, where $d - s \le n$, is given by:
+:$\dfrac {\map F {r, n} } {n^r}$
+If we assume a $365$-day year, we have that the [[Definition:Probability|probability]] that at least $3$ of them have the same birthday is given by:
+:$\map \Pr r = 1 - \displaystyle \sum_{d \mathop = 0}^{\floor {r / 2} } \dfrac {n! \, r!} {n^r 2^d d! \paren {r - 2 d}! \paren {n + d - r}!}$
+where $n = 365$.
+We require the smallest $r$ for which $\map \Pr r > \dfrac 1 2$.
+The result yields to calculation.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Convergent Sequence is Cauchy Sequence/Normed Division Ring}
+Tags: Convergent Sequence in Normed Division Ring is Cauchy Sequence
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}} $ be a [[Definition:Normed Division Ring|normed division ring]].
+Every [[Definition:Convergent Sequence in Normed Division Ring|convergent sequence]] in $R$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]].
+\end{theorem}<|endoftext|>
+\section{Sum of 3 Squares in 2 Distinct Ways}
+Tags: Square Numbers, 27
+
+\begin{theorem}
+$27$ is the smallest [[Definition:Positive Integer|positive integer]] which can be expressed as the [[Definition:Integer Addition|sum]] of $3$ [[Definition:Square Number|square numbers]] in $2$ [[Definition:Distinct|distinct]] ways:
+{{begin-eqn}}
+{{eqn | l = 27
+      | r = 3^2 + 3^2 + 3^2
+}}
+{{eqn | r = 5^2 + 1^2 + 1^2
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Can be performed by brute-force investigation.
+\end{proof}<|endoftext|>
+\section{Triangular Numbers which are Sum of Two Cubes}
+Tags: Cube Numbers, Triangular Numbers, 28
+
+\begin{theorem}
+The [[Definition:Integer Sequence|sequence]] of [[Definition:Triangular Number|triangular numbers]] which are the [[Definition:Integer Addition|sum]] of $2$ [[Definition:Cube Number|cubes]] begins:
+:$28, 91, 351, 2926, 8001, 46971, 58653, 93528, 97461, \dots$
+{{OEIS|A113958}}
+\end{theorem}
+
+\begin{proof}
+Can be demonstrated by brute force.
+For example:
+{{begin-eqn}}
+{{eqn | l = 28
+      | r = 1 + 27
+      | c = 
+}}
+{{eqn | r = 1^3 + 3^3
+      | c = 
+}}
+{{eqn | r = \dfrac {7 \paren {7 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 91
+      | r = 27 + 64
+      | c = 
+}}
+{{eqn | r = 3^3 + 4^3
+      | c = 
+}}
+{{eqn | r = \dfrac {13 \paren {13 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 351
+      | r = 125 + 216
+      | c = 
+}}
+{{eqn | r = 5^3 + 6^3
+      | c = 
+}}
+{{eqn | r = \dfrac {26 \paren {26 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 2976
+      | r = 729 + 2197
+      | c = 
+}}
+{{eqn | r = 5^3 + 6^3
+      | c = 
+}}
+{{eqn | r = \dfrac {76 \paren {76 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 8001
+      | r = 1 + 8000
+      | c = 
+}}
+{{eqn | r = 1^3 + 20^3
+      | c = 
+}}
+{{eqn | r = \dfrac {126 \paren {126 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 46 \, 971
+      | r = 4096 + 42 \, 875
+      | c = 
+}}
+{{eqn | r = 16^3 + 35^3
+      | c = 
+}}
+{{eqn | r = \dfrac {306 \paren {306 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 58 \, 653
+      | r = 8000 + 50 \, 653
+      | c = 
+}}
+{{eqn | r = 20^3 + 37^3
+      | c = 
+}}
+{{eqn | r = \dfrac {342 \paren {343 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 93 \, 528
+      | r = 42 \, 875 + 50 \, 653
+      | c = 
+}}
+{{eqn | r = 35^3 + 37^3
+      | c = 
+}}
+{{eqn | r = \dfrac {432 \paren {433 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 97 \, 461
+      | r = 125 + 97 \, 336
+      | c = 
+}}
+{{eqn | r = 5^3 + 46^3
+      | c = 
+}}
+{{eqn | r = \dfrac {441 \paren {442 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Factors of Perfect Number}
+Tags: Perfect Numbers
+
+\begin{theorem}
+Let $P$ be the [[Definition:Perfect Number|perfect number]] $2^{n - 1} \paren {2^n - 1}$.
+Then:
+:$\displaystyle \prod_{d \mathop \divides P} d = P^n$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Divisor of Integer|factors]] of $P$ are:
+:$1, 2, 4, \dots, 2^{n - 1}, 2^n - 1, 2 \paren {2^n - 1}, \dots, 2^{n - 1} \paren {2^n - 1}$
+Therefore their product is:
+{{begin-eqn}}
+{{eqn | l = \prod_{d \mathop \divides P} d
+      | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i} \paren {\prod_{i \mathop = 0}^{n - 1} 2^i \paren {2^n - 1} }
+}}
+{{eqn | r = \paren {\prod_{i \mathop = 0}^{n - 1} 2^i}^2 \paren {2^n - 1}^n
+}}
+{{eqn | r = \paren {2^\frac {n \paren {n - 1} } 2}^2 \paren {2^n - 1}^n
+}}
+{{eqn | r = \paren {2^{n - 1} }^n \paren {2^n - 1}^n
+}}
+{{eqn | r = P^n
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence is Bounded in Norm iff Bounded in Metric}
+Tags: Normed Division Rings, Convergence, Metric Spaces
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} } $ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] on $R$ be the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $R$]].
+Then:
+:$\sequence {x_n} $ is a [[Definition:Bounded Sequence in Normed Division Ring|bounded sequence]] in the [[Definition:Normed Division Ring|normed division ring]] $\struct {R, \norm {\,\cdot\,} }$ {{iff}} $\sequence {x_n} $ is a [[Definition:Bounded Sequence in Metric Space|bounded sequence]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+\end{theorem}
+
+\begin{proof}
+=== [[Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition|Necessary Condition]] ===
+{{:Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition}}{{qed|lemma}}
+=== [[Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition|Sufficent Condition]] ===
+{{:Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence is Bounded in Norm iff Bounded in Metric/Necessary Condition}
+Tags: Normed Division Rings, Boundedness, Metric Spaces
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}} $ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] on $R$ be the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $R$]].
+Let $\sequence {x_n} $ be a [[Definition:Bounded Sequence in Normed Division Ring|bounded sequence]] in the [[Definition:Normed Division Ring|normed division ring]] $\struct {R, \norm {\,\cdot\,}}$ 
+Then:
+:$\sequence {x_n} $ is a [[Definition:Bounded Sequence in Metric Space|bounded sequence]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n} $ be a [[Definition:Bounded Sequence in Normed Division Ring|bounded sequence]] in $\struct {R, \norm {\,\cdot\,} }$.
+Then:
+:$\exists K \in \R_{\gt 0} : \forall n : \norm {x_n} \le K$
+Then $\forall n, m \in \N$:
+{{begin-eqn}}
+{{eqn | l = \map d { x_n , x_m }
+      | r = \norm {x_n - x_m}
+      | c = {{Defof|Metric Induced by Norm on Division Ring}}
+}}
+{{eqn | o = \le
+      | r = \norm {x_n} + \norm {x_m}
+      | c = [[Properties of Norm on Division Ring/Norm of Difference|Norm of Difference]]
+}}
+{{eqn | o = \le
+      | r = K + K
+      | c = {{Defof|Bounded Sequence in Normed Division Ring}}
+}}
+{{eqn | r = 2 K
+}}
+{{end-eqn}}
+Hence the [[Definition:Sequence|sequence]] $\sequence {x_n} $ is [[Definition:Bounded Sequence in Metric Space|bounded]] by $2 K$ in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+\end{proof}<|endoftext|>
+\section{Sequence is Bounded in Norm iff Bounded in Metric/Sufficient Condition}
+Tags: Normed Division Rings, Boundedness, Metric Spaces
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}} $ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] on $R$ be the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence in $R$]].
+Let $\sequence {x_n} $ be a [[Definition:Bounded Sequence in Metric Space|bounded sequence]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$
+Then:
+:$\sequence {x_n} $ is a [[Definition:Bounded Sequence in Normed Division Ring|bounded sequence]] in the [[Definition:Normed Division Ring|normed division ring]] $\struct {R, \norm {\,\cdot\,} }$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n} $ be a [[Definition:Bounded Sequence in Metric Space|bounded sequence]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+Then:
+:$\exists K \in \R_{> 0} : \forall n, m : \map d {x_n , x_m} \le K$
+By the definition of the [[Definition:Metric Induced by Norm on Division Ring|metric induced by a norm]] this is equivalent to:
+:$\exists K \in \R_{> 0} : \forall n, m : \norm {x_n - x_m} \le K$
+Then $\forall n \in \N$:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n}
+      | r = \norm {x_n - x_1 + x_1}
+}}
+{{eqn | o = \le
+      | r = \norm {x_n - x_1} + \norm {x_1}
+      | c = [[Definition:Norm Axioms|Norm axiom (N3)]] (Triangle Inequality)
+}}
+{{eqn | o = \le
+      | r = K + \norm {x_1}
+      | c = Assumption of [[Definition:Bounded Sequence in Metric Space|bounded sequence]]
+}}
+{{end-eqn}}
+Hence the [[Definition:Sequence|sequence]] $\sequence {x_n}$ is [[Definition:Bounded Sequence in Normed Division Ring|bounded]] by $K + \norm {x_1}$ in the [[Definition:Normed Division Ring|normed division ring]] $\struct {R, \norm {\,\cdot\,} }$.
+\end{proof}<|endoftext|>
+\section{Cauchy Sequence is Bounded/Normed Division Ring/Proof 1}
+Tags: Cauchy Sequence in Normed Division Ring is Bounded
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Every [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence in $R$]] is [[Definition:Bounded Sequence in Normed Division Ring|bounded]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n} $ be a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence in $R$]].
+Then by definition:
+:$\forall \epsilon \in \R_{\gt 0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$
+Let $n_1$ satisfy:
+:$\forall n, m \ge n_1: \norm {x_n - x_m} < 1$
+Then $\forall n \ge n_1$:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n}
+      | r = \norm {x_n - x_{n_1} + x_{n_1} }
+}}
+{{eqn | o = \le
+      | r = \norm {x_n - x_{n_1} } + \norm {x_{n_1} }
+      | c = [[Definition:Norm Axioms|Norm axiom (N3)]] (Triangle Inequality).
+}}
+{{eqn | o = \le
+      | r = 1 + \norm {x_{n_1} }
+      | c = as $n, n_1 \ge n_1$
+}}
+{{end-eqn}}
+Let $K = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_{n_1 - 1}}, 1 + \norm {x_{n_1} } }$.
+Then:
+:$\forall n < n_1: \norm {x_n} \le K$
+:$\forall n \ge n_1: \norm {x_n} \le 1 + \norm {x_{n_1} } \le K$
+It follows by definition that $\sequence {x_n}$ is [[Definition:Bounded Sequence in Normed Division Ring|bounded]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Magic Constant of Magic Cube}
+Tags: Magic Cubes
+
+\begin{theorem}
+The [[Definition:Magic Constant|magic constant]] of a [[Definition:Magic Cube|magic cube]] of [[Definition:Order of Magic Square|order $n$]] is given by:
+:$C_n = \dfrac {n \paren {n^3 + 1} } 2$
+\end{theorem}
+
+\begin{proof}
+Let $M_n$ denote a [[Definition:Magic Cube|magic cube]] of [[Definition:Order of Magic Cube|order $n$]].
+By [[Sum of Terms of Magic Cube]], the total of all the entries in $M_n$ is given by:
+:$T_n = \dfrac {n^3 \paren {n^3 + 1}} 2$
+There are $n^2$ [[Definition:Row of Matrix|rows]] in $M_n$, each one with the same [[Definition:Magic Constant|magic constant]].
+Thus the [[Definition:Magic Constant|magic constant]] $C_n$ of the [[Definition:Magic Cube|magic cube]] $M_n$ is given by:
+{{begin-eqn}}
+{{eqn | l = C_n
+      | r = \dfrac {T_n} {n^2}
+      | c = 
+}}
+{{eqn | r = \dfrac {n^3 \paren {n^3 + 1} } {2 n^2}
+      | c = [[Sum of Terms of Magic Cube]]
+}}
+{{eqn | r = \dfrac {n \paren {n^3 + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+=== [[Magic Constant of Magic Cube/Sequence|Sequence]] ===
+{{:Magic Constant of Magic Cube/Sequence}}
+[[Category:Magic Cubes]]
+l460snwa7l9sgr0y71rc3rmnc1ggo1y
+\end{proof}<|endoftext|>
+\section{Sum of Terms of Magic Cube}
+Tags: Magic Cubes
+
+\begin{theorem}
+The total of all the entries in a [[Definition:Magic Cube|magic cube]] of [[Definition:Order of Magic Cube|order $n$]] is given by:
+:$T_n = \dfrac {n^3 \paren {n^3 + 1} } 2$
+\end{theorem}
+
+\begin{proof}
+Let $M_n$ denote a [[Definition:Magic Cube|magic cube]] of [[Definition:Order of Magic Cube|order $n$]].
+$M_n$ is by definition an arrangement of the first $n^3$ [[Definition:Strictly Positive Integer|(strictly) positive integers]] into an $n \times n \times n$ [[Definition:Cube (Geometry)|cubic]] [[Definition:Array|array]] containing the [[Definition:Positive Integer|positive integers]] from $1$ upwards.
+Thus there are $n^3$ entries in $M_n$, going from $1$ to $n^3$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = T_n
+      | r = \sum_{k \mathop = 1}^{n^3} k
+      | c = 
+}}
+{{eqn | r = \frac {n^3 \paren {n^3 + 1} } 2
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{end-eqn}}
+{{qed}}
+=== [[Sum of Terms of Magic Cube/Sequence|Sequence]] ===
+{{:Sum of Terms of Magic Cube/Sequence}}
+[[Category:Magic Cubes]]
+f5o21efpxqfe4sl7t92bh1tdmolatpu
+\end{proof}<|endoftext|>
+\section{Smallest Magic Cube is of Order 3}
+Tags: Magic Cubes
+
+\begin{theorem}
+Apart from the trivial [[Magic Cube/Examples/Order 1|order $1$ magic cube]]:
+{{:Magic Cube/Examples/Order 1}}
+the smallest [[Definition:Magic Cube|magic cube]] is the [[Magic Cube/Examples/Order 3|order $3$ magic cube]]:
+{{:Magic Cube/Examples/Order 3}}
+\end{theorem}
+
+\begin{proof}
+Suppose there were an [[Definition:Order of Magic Cube|order $2$]] [[Definition:Magic Cube|magic cube]].
+Take one row of this [[Definition:Magic Cube|magic cube]].
+From [[Magic Constant of Magic Cube]], the row and column total is $9$.
+Any row or column with a $1$ in it must therefore also have an $8$ in it.
+But there are:
+:one row
+:one column
+both of which have a $1$ in them.
+Therefore the $8$ would need to go in $2$ distinct cells.
+But $8$ appears in a [[Definition:Magic Cube|magic cube]] exactly once.
+Hence there can be no [[Definition:Order of Magic Cube|order $2$]] [[Definition:Magic Cube|magic cube]].
+{{qed}}
+[[Category:Magic Cubes]]
+ge077rmjlfvg5amhhg2wh5d3hlqzemj
+\end{proof}<|endoftext|>
+\section{Fourth Power as Summation of Groups of Consecutive Integers}
+Tags: Fourth Powers, Fourth Power as Summation of Groups of Consecutive Integers
+
+\begin{theorem}
+Take the [[Definition:Strictly Positive Integer|positive integers]] and group them in [[Definition:Set|sets]] such that the $m$th [[Definition:Set|set]] contains the next $m$ [[Definition:Strictly Positive Integer|positive integers]]:
+:$\set 1, \set {2, 3}, \set {4, 5, 6}, \set {7, 8, 9, 10}, \set {11, 12, 13, 14, 15}, \ldots$
+Remove all the [[Definition:Set|sets]] with an [[Definition:Even Integer|even number]] of [[Definition:Element|elements]].
+Then the [[Definition:Integer Addition|sum]] of all the [[Definition:Integer|integers]] in the first $n$ [[Definition:Set|sets]] remaining equals $n^4$.
+\end{theorem}
+
+\begin{proof}
+Let $S_m$ be the $m$th [[Definition:Set|set]] of $m$ consecutive [[Definition:Integer|integers]].
+Let $S_k$ be the $k$th [[Definition:Set|set]] of $m$ consecutive [[Definition:Integer|integers]] after the [[Definition:Set|sets]] with an [[Definition:Even Integer|even number]] of [[Definition:Element|elements]] have been removed.
+Then $S_k = S_m$ where $m = 2 k - 1$.
+By the method of construction:
+:the largest [[Definition:Integer|integer]] in $S_m$ is $T_m$, the $m$th [[Definition:Triangular Number|triangular number]]
+:there are $m$ [[Definition:Integer|integers]] in $S_m$.
+We also have that the middle [[Definition:Integer|integer]] in $S_m$ is $T_m - \dfrac {m - 1} 2$ (by inspection).
+Thus the [[Definition:Integer Addition|sum]] of the [[Definition:Element|elements]] of $S_k$ is:
+{{begin-eqn}}
+{{eqn | l = \sum S_k
+      | r = m \times \paren {T_m - \dfrac {m - 1} 2}
+      | c = 
+}}
+{{eqn | r = m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}
+      | c = 
+}}
+{{eqn | r = m \times \paren {\dfrac {m \paren {m + 1} } 2 - \dfrac {m - 1} 2}
+      | c = 
+}}
+{{eqn | r = m \times \paren {\dfrac {m^2 + 1} 2}
+      | c = 
+}}
+{{eqn | r = \dfrac {m^3 + m} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {2 k - 1}^3 + \paren {2 k - 1} } 2
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {8 k^3 - 12 k^2 + 6 k - 1} + \paren {2 k - 1} } 2
+      | c = 
+}}
+{{eqn | r = 4 k^3 - 6 k^2 + 4 k - 1
+      | c = 
+}}
+{{end-eqn}}
+We need to calculate the [[Definition:Integer Addition|sum]] of all $S_k$ from $1$ to $n$.
+Hence we have:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n S_k
+      | r = \sum_{k \mathop = 1}^n \paren {4 k^3 - 6 k^2 + 4 k - 1}
+      | c = 
+}}
+{{eqn | r = 4 \sum_{k \mathop = 1}^n k^3 - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1
+      | c = 
+}}
+{{eqn | r = 4 \paren {\frac {n^2 \paren {n + 1}^2} 4} - 6 \sum_{k \mathop = 1}^n k^2 + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1
+      | c = [[Sum of Sequence of Cubes]]
+}}
+{{eqn | r = n^2 \paren {n + 1}^2 - 6 \paren {\frac {n \paren {n + 1} \paren {2 n + 1} } 6} + 4 \sum_{k \mathop = 1}^n k - \sum_{k \mathop = 1}^n 1
+      | c = [[Sum of Sequence of Squares]]
+}}
+{{eqn | r = n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 4 \paren {\frac {n \paren {n + 1} } 2} - \sum_{k \mathop = 1}^n 1
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = n^2 \paren {n + 1}^2 - n \paren {n + 1} \paren {2 n + 1} + 2 n \paren {n + 1} - n
+      | c = simplification
+}}
+{{eqn | r = n^4 + 2 n^3 + n^2 - 2 n^3 - 3 n^2 - n + 2 n^2 + 2 n - n
+      | c = multiplying everything out
+}}
+{{eqn | r = n^4
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Power of 3 as Sum of Consecutive Positive Integers}
+Tags: Powers of 3, Even Power of 3 as Sum of Consecutive Positive Integers
+
+\begin{theorem}
+Take the [[Definition:Strictly Positive Integer|positive integers]] and group them in [[Definition:Set|sets]] such that the $n$th [[Definition:Set|set]] contains the next $3^n$ [[Definition:Strictly Positive Integer|positive integers]]:
+:$\set 1, \set {2, 3, 4}, \set {5, 6, \ldots, 13}, \set {14, 15, \cdots, 40}, \ldots$
+Let the $n$th such set be denoted $S_{n - 1}$, that is, letting $S_0 := \set 1$ be considered as the [[Definition:Zeroth|zeroth]].
+Then the [[Definition:Integer Addition|sum]] of all the [[Definition:Element|elements]] of $S_n$ is $3^{2 n}$.
+\end{theorem}
+
+\begin{proof}
+The total number of [[Definition:Element|elements]] in $S_0, S_1, \ldots, S_r$ is:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^r \card {S_j}
+      | r = \sum_{j \mathop = 0}^r 3^j
+      | c = 
+}}
+{{eqn | r = \dfrac {3^{r + 1} - 1} {3 - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = \dfrac {3^{r + 1} - 1} 2
+      | c = simplifying
+}}
+{{end-eqn}}
+Thus for any given $S_n$:
+:$\displaystyle \sum S_n = \sum k \sqbrk {\dfrac {3^n - 1} 2 < k \le \dfrac {3^{n + 1} - 1} 2}$
+using [[Definition:Iverson's Convention|Iverson's convention]].
+So $\displaystyle \sum S_n$ can be evaluated as the [[Definition:Integer Subtraction|difference]] between two [[Definition:Triangular Number|triangular numbers]]:
+{{begin-eqn}}
+{{eqn | l = \sum S_n
+      | r = \sum_{k \mathop = 1}^{\paren {3^{n + 1} - 1} / 2} k - \sum_{k \mathop = 1}^{\paren {3^n - 1} / 2} k
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1} 2 + 1} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1} 2 + 1} }
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = \dfrac 1 2 \paren {\dfrac {3^{n + 1} - 1} 2 \paren {\dfrac {3^{n + 1} - 1 + 2} 2} - \dfrac {3^n - 1} 2 \paren {\dfrac {3^n - 1 + 2} 2} }
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 8 \paren {\paren {3^{n + 1} - 1} \paren {3^{n + 1} + 1} - \paren {3^n - 1} \paren {3^n + 1} }
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 8 \paren {\paren {3^{2 n + 2} - 1} - \paren {3^{2 n} - 1} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \dfrac 1 8 \paren {3^{2 n + 2} - 3^{2 n} }
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 8 \paren {3^{2 n} \paren {3^2 - 1} }
+      | c = extracting $3^{2 n}$ as a factor
+}}
+{{eqn | r = 3^{2 n}
+      | c = Oh look, that second factor magically equals $8$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Smallest Number with 2^n Divisors}
+Tags: Tau Function, 120
+
+\begin{theorem}
+The smallest [[Definition:Positive Integer|positive integer]] with $2^n$ [[Definition:Divisor of Integer|divisors]] is found by multiplying together the first $n$ numbers in this [[Definition:Integer Sequence|sequence]]:
+:$2, 3, 4, 5, 7, 9, 11, 13, 16, 17, 19, \ldots$
+which consists of all the [[Definition:Prime Number|primes]] and [[Definition:Prime Power|powers of primes]].
+\end{theorem}<|endoftext|>
+\section{Convergent Subsequence of Cauchy Sequence/Normed Division Ring}
+Tags: Convergent Sequences in Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $\sequence{x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence (Normed Division Ring)|Cauchy sequence]] in $\struct {R, \norm {\,\cdot\,} }$. 
+Let $x \in R$.
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x$ {{iff}} $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] that [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x$.
+\end{theorem}
+
+\begin{proof}
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced]] on $R$ be the [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}$.
+By the definition of a [[Definition:Convergent Sequence in Normed Division Ring|convergent sequence in a normed division ring]] then:
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x$ in $\struct {R, \norm {\,\cdot\,} }$ {{iff}} $\sequence {x_n}$ [[Definition:Convergent Sequence in Metric Space|converges]] to $x$ in $\struct {R, d}$.
+By [[Convergent Subsequence of Cauchy Sequence|Convergent Subsequence of Cauchy Sequence in Metric Space]]:
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Metric Space|converges]] to $x$ in $\struct {R, d}$ {{iff}} $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] that [[Definition:Convergent Sequence in Metric Space|converges]] to $x$ In $\struct {R, d}$.
+By the definition of a [[Definition:Convergent Sequence in Normed Division Ring|convergent sequence in a normed division ring]]:
+:$\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] that [[Definition:Convergent Sequence in Metric Space|converges]] to $x$ In $\struct {R, d}$ {{iff}} $\sequence {x_n}$ has a [[Definition:Subsequence|subsequence]] that [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x$ in $\struct {R, \norm {\,\cdot\,} }$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal}
+Tags: Cauchy Sequences, Normed Division Rings, Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+Let $\struct {R, \norm {\, \cdot \,} }$ be a [[Definition:Normed Division Ring|normed division ring]].
+Let $\CC$ be the [[Definition:Ring of Cauchy Sequences|ring of Cauchy sequences over $R$]].
+Let $\NN$ be the [[Definition:Set|set]] of [[Definition:Null Sequence in Normed Division Ring|null sequences]].
+That is:
+:$\NN = \set {\sequence {x_n}: \displaystyle \lim_{n \mathop \to \infty} x_n = 0 }$
+Then $\NN$ is a [[Definition:Ideal of Ring|ring ideal]] of $\CC$ that is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] and a [[Definition:Maximal Right Ideal of Ring|maximal right ideal]].
+\end{theorem}
+
+\begin{proof}
+By [[Convergent Sequence is Cauchy Sequence/Normed Division Ring|every convergent sequence is a Cauchy sequence]] then $\NN \subseteq \CC$.
+The proof is completed in these steps:
+:$(1): \quad \NN$ is an [[Definition:Ideal of Ring|ideal]] of $\CC$.
+:$(2): \quad \NN$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]].
+:$(3): \quad \NN$ is a [[Definition:Maximal Right Ideal of Ring|maximal right ideal]].
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 1|Lemma 1]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 1}}{{qed|lemma}}
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 2|Lemma 2]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 2}}{{qed|lemma}}
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 3|Lemma 3]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 3}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 1}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\mathcal N$ is an [[Definition:Ideal of Ring|ideal]] of $\mathcal C$.
+\end{theorem}
+
+\begin{proof}
+The [[Test for Ideal]] is applied to prove the result.
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.1|Lemma 1.1]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.1|Lemma 1}}{{qed|lemma}}
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.2|Lemma 1.2]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.2}}{{qed|lemma}}
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.3|Lemma 1.3]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.3}}{{qed|lemma}}
+By [[Test for Ideal]] then the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 2}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\mathcal N$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]].
+\end{theorem}
+
+\begin{proof}
+By [[Null Sequences form Maximal Left and Right Ideal/Lemma 1|Lemma 1 of Null Sequences form Maximal Left and Right Ideal]]:
+:$\NN$ is an [[Definition:Ideal of Ring|ideal]] of $\CC$.
+Hence $\NN$ is a [[Definition:Left Ideal of Ring|left ideal]] of $\CC$.
+It remains to show that $\NN$ is [[Definition:Maximal Left Ideal of Ring|maximal]].
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1|Lemma 2.1]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1}}{{qed|lemma}}
+=== [[Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.2|Lemma 2.2]] ===
+{{:Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.2}}{{qed|lemma}}
+The result follows by definition of [[Definition:Maximal Left Ideal of Ring|maximal left ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 3}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\NN$ is a [[Definition:Maximal Right Ideal of Ring|maximal right ideal]].
+\end{theorem}
+
+\begin{proof}
+By [[Null Sequences form Maximal Left and Right Ideal/Lemma 1|Lemma 1 of Null Sequences form Maximal Left and Right Ideal]] then $\NN$ is an [[Definition:Ideal of Ring|ideal]] of $\CC$.
+Hence $\NN$ is a [[Definition:Right Ideal of Ring|right ideal]] of $\CC$.
+It remains to show that $\NN$ is [[Definition:Maximal Right Ideal of Ring|maximal]].
+By [[Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1|Lemma 2.1 of Null Sequences form Maximal Left and Right Ideal]] then $\NN \subsetneq \CC$.
+By [[Definition:Maximal Right Ideal of Ring|maximal right ideal]] it needs to be shown that:
+:There is no [[Definition:Right Ideal of Ring|right ideal]] $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$
+Let $\JJ$ be a [[Definition:Right Ideal of Ring|Right ideal]] of $\CC$ such that $\NN \subsetneq \JJ \subseteq \CC$.
+It will be shown that $\JJ$ = $\CC$, from which the result will follow.
+Let $\sequence {x_n} \in \JJ \setminus \NN$
+By [[Combination Theorem for Cauchy Sequences/Inverse Rule|Inverse Rule for Cauchy sequences]] then 
+:$\exists K \in \N: \sequence {\paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]].
+Let $\sequence {y_n}$ be the sequence defined by:
+:$y_n = \begin{cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end{cases}$
+By [[Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence]] then $\sequence {y_n} \in \CC$
+By the definition of a [[Definition:Right Ideal of Ring|right ideal]] the product $\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n} \in \JJ$
+By the definition of $\sequence {y_n}$ then:
+:$x_n y_n = \begin{cases} 0 & : n \le K \\ 1 & : n > K \end{cases}$
+Let $\mathcal 1 = \tuple {1, 1, 1, \dots}$ be the [[Definition:Unity of Ring|unity]] of $\CC$
+Then $\mathcal 1 - \sequence {x_n} \sequence {y_n}$ is the sequence $\sequence {w_n}$ defined by:
+:$w_n = \begin {cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$
+By [[Convergent Sequence with Finite Elements Prepended is Convergent Sequence]] then $\sequence {w_n}$ is convergent to 0.
+So $\sequence {w_n} \in \NN \subsetneq \JJ$
+Since \sequence {x_n} $\sequence {y_n}, \sequence {w_n} \in \JJ$ by the definition of a [[Definition:Ideal of Ring|ring ideal]] then:
+:$\sequence {w_n} + \sequence {x_n} \sequence {y_n} = \mathcal 1 \in \JJ$
+By the definition of a [[Definition:Right Ideal of Ring|right ideal]] then:
+:$\forall \sequence {a_n} \in \CC, \mathcal 1 \circ \sequence {a_n} = \sequence {a_n} \in \JJ$
+Hence $\JJ = \CC$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Decomposition of 8th Fermat Number}
+Tags: Fermat Numbers
+
+\begin{theorem}
+The [[Definition:Prime Decomposition|prime decomposition]] of the $8$th [[Definition:Fermat Number|Fermat number]] is given by:
+{{begin-eqn}}
+{{eqn | l = 2^{\paren {2^8} } + 1
+      | r = 115 \, 792 \, 089 \, 237 \, 316 \, 195 \, 423 \, 570 \, 985 \, 008 \, 687 \, 907 \, 853 \, 269 \, 984 \, 665 \, 640 \, 564 \, 039 \, 457 \, 584 \, 007 \, 913 \, 129 \, 639 \, 937
+      | c = 
+}}
+{{eqn | r = 1 \, 238 \, 926 \, 361 \, 552 \, 897
+      | c = 
+}}
+{{eqn | o =
+      | ro= \times
+      | r = 93 \, 461 \, 639 \, 715 \, 357 \, 977 \, 769 \, 163 \, 558 \, 199 \, 606 \, 896 \, 584 \, 051 \, 237 \, 541 \, 638 \, 188 \, 580 \, 280 \, 321
+      | c = 
+}}
+{{eqn | r = \paren {2 \times 157 \times 3 \, 853 \, 149 \, 761 \times 2^{10} + 1}
+      | c = 
+}}
+{{eqn | o =
+      | ro= \times
+      | r = \paren {2 \times 3 \times 5 \times 7 \times 13 \times 31 \, 618 \, 624 \, 099 \, 079 \times 1 \, 057 \, 372 \, 046 \, 781 \, 162 \, 536 \, 274 \, 034 \, 354 \, 686 \, 893 \, 329 \, 625 \, 329 \times 2^{10} + 1}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}<|endoftext|>
+\section{Product of Sequence of Fermat Numbers plus 2}
+Tags: Fermat Numbers
+
+\begin{theorem}
+Let $F_n$ denote the $n$th [[Definition:Fermat Number|Fermat number]].
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \Z_{>0}:
+      | l = F_n
+      | r = \prod_{j \mathop = 0}^{n - 1} F_j + 2
+      | c = 
+}}
+{{eqn | r = F_0 F_1 \dotsm F_{n - 1} + 2
+      | 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]]:
+:$F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = F_1
+      | r = 2^{\paren {2^1} } + 1
+      | c = {{Defof|Fermat Number}}
+}}
+{{eqn | r = 5
+      | c = 
+}}
+{{eqn | r = 3 + 2
+      | c = 
+}}
+{{eqn | r = \paren {2^{\paren {2^0} } + 1} + 2
+      | c = 
+}}
+{{eqn | r = F_0 + 2
+      | c = {{Defof|Fermat Number}}
+}}
+{{eqn | r = \displaystyle \prod_{j \mathop = 0}^{1 - 1} F_j + 2
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#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 [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$F_k = \displaystyle \prod_{j \mathop = 0}^{k - 1} F_j + 2$
+from which it is to be shown that:
+:$F_{k + 1} = \displaystyle \prod_{j \mathop = 0}^k F_j + 2$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = F_{k + 1}
+      | r = 2^{\paren {2^{k + 1} } } + 1
+      | c = {{Defof|Fermat Number}}
+}}
+{{eqn | r = 2^{\paren {2 \times 2^k} } + 1
+      | c = 
+}}
+{{eqn | r = 2^{\paren {2^k} } \times 2^{\paren {2^k} } + 1
+      | c = 
+}}
+{{eqn | r = \paren {F_k - 1} \times \paren {F_k - 1} + 1
+      | c = {{Defof|Fermat Number}}
+}}
+{{eqn | r = \paren {\paren {\prod_{j \mathop = 0}^{k - 1} F_j + 2} - 1} \paren {F_k - 1} + 1
+      | c = [[Product of Sequence of Fermat Numbers plus 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {\prod_{j \mathop = 0}^{k - 1} F_j} F_k + F_k - \prod_{j \mathop = 0}^{k - 1} F_j - 1 + 1
+      | c = multiplying out
+}}
+{{eqn | r = \prod_{j \mathop = 0}^k F_j + F_k - \paren {F_k - 2}
+      | c = [[Product of Sequence of Fermat Numbers plus 2#Induction Hypothesis|Induction Hypothesis]], and simplifying
+}}
+{{eqn | r = \prod_{j \mathop = 0}^k F_j + 2
+      | c = simplifying
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>0}: F_n = \displaystyle \prod_{j \mathop = 0}^{n - 1} F_j + 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{492 is Sum of 3 Cubes in 3 Ways}
+Tags: 492, Sums of Cubes
+
+\begin{theorem}
+$492$ can be expressed as the [[Definition:Integer Addition|sum]] of $3$ [[Definition:Cube Number|cubes]], either [[Definition:Positive Integer|positive]] or [[Definition:Negative Integer|negative]] in $3$ known ways.
+{{begin-eqn}}
+{{eqn | l = 492
+      | r = 50^3 + \paren {-19}^3 + \paren {-49}^3
+}}
+{{eqn | r = 123 \, 134^3 + 9179^3 + \paren {-123 \, 151}^3
+}}
+{{eqn | r = 1 \, 793 \, 337 \, 644^3 + \paren {-81 \, 3701 \, 167}^3 + \paren {-1 \, 735 \, 662 \, 109}^3
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Brute force.
+\end{proof}<|endoftext|>
+\section{Smallest n needing 6 Numbers less than n so that Product of Factorials is Square}
+Tags: 527, Factorials, Square Numbers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]].
+Then it is possible to choose at most $6$ [[Definition:Positive Integer|positive integers]] less than $n$ such that the [[Definition:Integer Multiplication|product]] of their [[Definition:Factorial|factorials]] is [[Definition:Square Number|square]].
+The smallest $n$ that actually requires $6$ numbers to be chosen is $527$.
+\end{theorem}
+
+\begin{proof}
+Obviously the product cannot be a [[Definition:Square Number|square]] if $n$ is a [[Definition:Prime Number|prime]].
+For $n$ [[Definition:Composite Number|composite]], we can write:
+:$n = a b$
+where $a, b \in \Z_{>1}$.
+Then:
+{{begin-eqn}}
+{{eqn | o =
+      | r = n! \paren {n - 1}! \paren {a!} \paren {a - 1}! \paren {b!} \paren {b - 1}!
+}}
+{{eqn | r = n a b \paren {\paren {n - 1}! \paren {a - 1}! \paren {b - 1}!}^2
+}}
+{{eqn | r = \paren {n! \paren {a - 1}! \paren {b - 1}!}^2
+}}
+{{end-eqn}}
+which is a [[Definition:Square Number|square]].
+Hence no more than $6$ [[Definition:Factorial|factorials]] is required.
+To show that $527$ is the smallest that actually requires $6$, observe that:
+{{tidy}}
+{{explain|It might be worth extracting some of the below statements into lemmata, for example: "If $n$ is itself [[Definition:Square Number|square]], then so is $n! \paren {n - 1}!$" and "... Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is [[Definition:Square Number|square]]" -- they're really easy to prove, even I can do them :-) but it takes more than a glance to recognise that they are true.}}
+If $n$ is itself [[Definition:Square Number|square]], then so is $n! \paren {n - 1}!$.
+If $n$ is not [[Definition:Square-Free Integer|square-free]], write $n = a^2 b$, where $b$ is [[Definition:Square-Free Integer|square-free]].
+Then $n! \paren {n - 1}! b! \paren {b - 1}!$ is [[Definition:Square Number|square]].
+If $n$ is [[Definition:Divisor of Integer|divisible]] by $2$, write $n = 2 m$.
+Then $\paren {2 m}! \paren {2 m - 1}! \paren {m!} \paren {m - 1}! \paren {2!}$ is [[Definition:Square Number|square]].
+If $n$ is [[Definition:Divisor of Integer|divisible]] by $3$, write $n = 3 m$.
+Then $\paren {3 m}! \paren {3 m - 1}! \paren {2 m}! \paren {2 m - 1}! \paren {3!}$ is [[Definition:Square Number|square]].
+If $n$ is [[Definition:Divisor of Integer|divisible]] by $5$, write $n = 5 m$.
+Then $\paren {5 m}! \paren {5 m - 1}! \paren {m!} \paren {m - 1}! \paren {6!}$ is [[Definition:Square Number|square]].
+If $n$ is [[Definition:Divisor of Integer|divisible]] by $7$, write $n = 7 m$.
+Then $\paren {7 m}! \paren {7 m - 1}! \paren {5 m}! \paren {5 m - 1}! \paren {7!}$ is [[Definition:Square Number|square]].
+If $n$ is [[Definition:Divisor of Integer|divisible]] by $11$, write $n = 11 m$.
+Then $\paren {11 m}! \paren {11 m - 1}! \paren {7 m}! \paren {7 m - 1}! \paren {11!}$ is [[Definition:Square Number|square]].
+The remaining numbers less than $527$ that are not of the above forms are:
+:$221, 247, 299, 323, 377, 391, 403, 437, 481, 493$
+Each of the following is a product of $5$ factorials which is square:
+:$221! \, 220! \, 18! \, 11! \, 7!$
+:$247! \, 246! \, 187! \, 186! \, 20!$
+:$299! \, 298! \, 27! \, 22!$
+:$323! \, 322! \, 20! \, 14! \, 6!$
+:$377! \, 376! \, 29! \, 23! \, 10!$
+:$391! \, 389! \, 24! \, 21! \, 17!$
+:$403! \, 402! \, 33! \, 30! \, 14!$
+:$437! \, 436! \, 51! \, 49! \, 28!$
+:$481! \, 479! \, 38! \, 33! \, 22!$
+:$493! \, 491! \, 205! \, 202! \, 7!$
+{{finish|The fact that $527$ has no such representation can be verified by a direct (but lengthy) computation.}}
+\end{proof}<|endoftext|>
+\section{Numbers whose Product with Reverse are Equal/Historical Note}
+Tags: Recreational Mathematics, 651, 156, 372, 273
+
+\begin{theorem}
+According to {{AuthorRef|David Wells}} in his {{BookLink|Curious and Interesting Numbers|David Wells}} of $1986$, this result appeared in a $1939$ issue of {{JournalLink|name = Scripta Mathematica}}, attributed to {{AuthorRef|A.A.K. Iyangar}}.
+However, it has not been possible to corroborate this.
+\end{theorem}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.1}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\mathcal N \ne \O$
+\end{theorem}
+
+\begin{proof}
+From [[Constant Sequence Converges to Constant in Normed Division Ring]], the [[Definition:Ring Zero|zero]] $\tuple {0, 0, 0, \dots}$ of $\mathcal C$ to [[Definition:Convergent Sequence in Normed Division Ring|converges]] $0 \in R$.
+Therefore $\tuple {0, 0, 0, \dots} \in \mathcal N$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.2}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\forall \sequence {x_n}, \sequence {y_n} \in \NN: \sequence {x_n} + \paren {-\sequence {y_n} } \in \NN$
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$ and $\displaystyle \lim_{n \mathop \to \infty} y_n = 0$.
+The [[Definition:Sequence|sequence]] $\sequence {x_n} + \paren {-\sequence {y_n} } = \sequence {x_n - y_n}$.
+By [[Combination Theorem for Sequences/Difference Rule|Difference Rule for Sequences]], $\displaystyle \lim_{n \mathop \to \infty} x_n - y_n = 0 - 0 = 0.$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 1/Lemma 1.3}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\quad \forall \sequence {x_n} \in \NN, \sequence {y_n} \in \CC: \sequence {x_n} \sequence {y_n} \in \NN, \sequence {y_n} \sequence {x_n} \in \NN$
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle \lim_{n \mathop \to \infty} x_n = 0$
+By the definition of the [[Definition:Ring of Cauchy Sequences|product on the ring of Cauchy sequences]] then:
+:$\sequence {x_n} \sequence {y_n} = \sequence {x_n y_n}$
+:$\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n}$
+By [[Product of Sequence Converges to Zero with Cauchy Sequence Converges to Zero|product of sequence converges to zero with Cauchy sequence]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n y_n = 0$
+:$\displaystyle \lim_{n \mathop \to \infty} y_n x_n = 0$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.1}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:$\NN \subsetneq \CC$.
+\end{theorem}
+
+\begin{proof}
+By [[Convergent Sequence is Cauchy Sequence/Normed Division Ring|every convergent sequence is a Cauchy sequence]] then $\NN \subseteq \CC$.
+From [[Constant Sequence Converges to Constant in Normed Division Ring]], the [[Definition:Unity of Ring|unity]] $\tuple {1, 1, 1, \dotsC}$ of $\CC$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $1 \in R$, and therefore $\tuple {1, 1, 1, \dotsc} \in \CC \setminus \NN$
+So $\NN \subsetneq \CC$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Lemma 2/Lemma 2.2}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+:There is no [[Definition:Left Ideal of Ring|left ideal]] $\JJ$ of $\CC$ such that $\NN \subsetneq \JJ \subsetneq \CC$
+\end{theorem}
+
+\begin{proof}
+Let $\JJ$ be a [[Definition:Left Ideal of Ring|left ideal]] of $\CC$ such that $\NN \subsetneq \JJ \subseteq \\CC$.
+It will be shown that $\JJ$ = $\CC$, from which the result will follow.
+Let $\sequence {x_n} \in \JJ \setminus \NN$
+By [[Combination Theorem for Cauchy Sequences/Inverse Rule|Inverse Rule for Cauchy sequences]] then 
+:$\exists K \in \N: \sequence { \paren {x_{K + n} }^{-1} }_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]].
+Let $\sequence {y_n}$ be the [[Definition:Sequence|sequence]] defined by:
+:$y_n = \begin {cases} 0 & : n \le K \\ \paren {x_n}^{-1} & : n > K \end {cases}$
+By [[Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence]] then $\sequence {y_n} \in \CC$
+By the definition of a [[Definition:Left Ideal of Ring|left ideal]] the product $\sequence {y_n} \sequence {x_n} = \sequence {y_n x_n} \in \JJ$
+By the definition of $\sequence {y_n}$ then:
+:$y_n x_n = \begin {cases} 0 & : n \le K \\ 1 & : n > K \end {cases}$
+Let $\mathcal {1} = \tuple {1, 1, 1, \dotsc}$ be the [[Definition:Unity of Ring|unity]] of $\CC$
+Then $\mathcal {1} - \sequence {y_n} \sequence {x_n}$ is the [[Definition:Sequence|sequence]] $\sequence {w_n}$ defined by
+:$w_n = \begin{cases} 1 & : n \le K \\ 0 & : n > K \end {cases}$
+By [[Convergent Sequence with Finite Elements Prepended is Convergent Sequence]] then $\sequence {w_n}$ is convergent to 0.
+So $\sequence {w_n} \in \NN \subsetneq \JJ$
+Since $\sequence {y_n} \sequence {x_n}, \sequence {w_n} \in \JJ$ by the definition of a [[Definition:Ideal of Ring|ring ideal]] then:
+:$\sequence {w_n} + \sequence {y_n} \sequence {x_n} = \mathcal {1} \in \JJ$
+By the definition of a [[Definition:Left Ideal of Ring|left ideal]] then:
+:$\forall \sequence {a_n} \in \CC, \sequence {a_n} \circ \mathcal {1} = \sequence {a_n} \in \JJ$
+Hence $\JJ = \CC$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference between Two Squares equal to Repdigit}
+Tags: Difference between Two Squares equal to Repunit
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = 6^2 - 5^2
+      | r = 11
+      | c = 
+}}
+{{eqn | l = 56^2 - 45^2
+      | r = 1111
+      | c = 
+}}
+{{eqn | l = 556^2 - 445^2
+      | r = 111 \, 111
+      | c = 
+}}
+{{eqn | o = :
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 7^2 - 4^2
+      | r = 33
+      | c = 
+}}
+{{eqn | l = 67^2 - 34^2
+      | r = 3333
+      | c = 
+}}
+{{eqn | l = 667^2 - 334^2
+      | r = 333 \, 333
+      | c = 
+}}
+{{eqn | o = :
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 8^2 - 3^2
+      | r = 55
+      | c = 
+}}
+{{eqn | l = 78^2 - 23^2
+      | r = 5555
+      | c = 
+}}
+{{eqn | l = 778^2 - 223^2
+      | r = 555 \, 555
+      | c = 
+}}
+{{eqn | o = :
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 9^2 - 2^2
+      | r = 77
+      | c = 
+}}
+{{eqn | l = 89^2 - 12^2
+      | r = 7777
+      | c = 
+}}
+{{eqn | l = 889^2 - 112^2
+      | r = 777 \, 777
+      | c = 
+}}
+{{eqn | o = :
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Let $a, b$ be [[Definition:Integer|integers]] with $1 \le b < a \le 8$ and $a + b = 9$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + \sum_{k \mathop = 0}^n a 10^k}^2 - \paren {1 + \sum_{k \mathop = 0}^n b 10^k}^2
+      | r = \paren {1 + \sum_{k \mathop = 0}^n a 10^k - 1 - \sum_{k \mathop = 0}^n b 10^k} \paren {1 + \sum_{k \mathop = 0}^n a 10^k + 1 + \sum_{k \mathop = 0}^n b 10^k}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {2 + \sum_{k \mathop = 0}^n 9 \times 10^k}
+      | c = $a + b = 9$
+}}
+{{eqn | r = \paren {\sum_{k \mathop = 0}^n \paren {a - b} 10^k} \paren {1 + 10^{n + 1} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = 0}^n \paren {a - b} 10^{k + n + 1}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \paren {a - b} 10^k + \sum_{k \mathop = n + 1}^{2 n + 1} \paren {a - b} 10^k
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{2 n + 1} \paren {a - b} 10^k
+}}
+{{end-eqn}}
+which is a [[Definition:Repdigit Number|repdigit number]].
+The examples above are instances with $\tuple {a, b} = \tuple {5, 4}, \tuple {6, 3}, \tuple {7, 2}, \tuple {8, 1}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Largest Number Not Expressible as Sum of Fewer than 8 Cubes}
+Tags: Sums of Cubes, 8042
+
+\begin{theorem}
+$8042$ is (probably) the largest [[Definition:Positive Integer|positive integer]] that cannot be expressed as the [[Definition:Integer Addition|sum]] of fewer than $8$ [[Definition:Cube Number|cubes]].
+\end{theorem}
+
+\begin{proof}
+{{Mistake}}
+$8042 = 1^3 + 4^3 + 4^3 + 10^3 + 10^3 + 10^3 + 17^3$, among many other expressions.
+However:
+$8042$ is conjectured to be the largest [[Definition:Positive Integer|positive integer]] that cannot be expressed as the [[Definition:Integer Addition|sum]] of fewer than $\bf 7$ [[Definition:Cube Number|cubes]].
+$\bf {454}$ is proven to be the largest [[Definition:Positive Integer|positive integer]] that cannot be expressed as the [[Definition:Integer Addition|sum]] of fewer than $8$ [[Definition:Cube Number|cubes]].
+\end{proof}<|endoftext|>
+\section{Infinite Number of Even Fermat Pseudoprimes}
+Tags: Fermat Pseudoprimes
+
+\begin{theorem}
+Despite their relative rarity, there exist an [[Definition:Infinite Set|infinite number]] of [[Definition:Even Integer|even]] [[Definition:Fermat Pseudoprime|Fermat pseudoprimes]].
+\end{theorem}
+
+\begin{proof}
+In the context of Wells, [[Definition:Fermat Pseudoprime|Fermat pseudoprime]] probably means [[Definition:Poulet Number|Fermat pseudoprimes to the base $2$]].
+Consider the equation:
+:$2^m - 2 \equiv 0 \pmod m$
+Any $m$ satisfying the above equation is a [[Definition:Fermat Pseudoprime|Fermat pseudoprime]].
+We show that for each even $m$ satisfying the above equation, there exists a [[Definition:Prime Number|prime]] $p$ such that $m p$ also satisfies the equation.
+Write $m = 2 n$.
+Then $2^{2 n - 1} - 1 \equiv 0 \pmod n$.
+By [[Zsigmondy's Theorem]], there exists a [[Definition:Prime Number|prime]] $p$ such that:
+:$p \divides 2^{2 n - 1} - 1$
+:$p \nmid 2^k - 1$ for all $k < 2 n - 1$
+By this we have $2^{2 n - 1} \equiv 1 \pmod p$.
+By [[Fermat's Little Theorem]] $2^{p - 1} \equiv 1 \pmod p$.
+Let $d = \gcd \set {2 n - 1, p - 1}$.
+We have $d \le 2 n - 1$ and $2^d \equiv 1 \pmod p$.
+This gives $d = 2 n - 1$ and thus $2 n - 1 \divides p - 1$.
+Thus there exists $x \in \N$ such that:
+:$p = \paren {2 n - 1} x + 1 > n$
+This shows that $p \perp n$.
+From $2 p n - 1 = 2 n - 1 + 2 n \paren {p - 1} = \paren {2 n - 1} \paren {1 + 2 n x}$ we have:
+:$2^{m p - 1} = 2^{\paren {2 n - 1} \paren {1 + 2 n x}} \equiv 1 \pmod {n, p}$
+By [[Chinese Remainder Theorem]]:
+:$2^{m p - 1} \equiv 1 \pmod {n p}$
+By [[Congruence by Product of Moduli]]:
+:$2^{m p} \equiv 2 \pmod {m p}$
+showing that $m p$ is also an [[Definition:Even Integer|even]] [[Definition:Fermat Pseudoprime|Fermat pseudoprime]].
+The first [[Definition:Even Integer|even]] [[Definition:Fermat Pseudoprime|Fermat pseudoprime]] is $161 \, 038$.
+We can use the above method repeatedly to construct infinitely many [[Definition:Even Integer|even]] [[Definition:Fermat Pseudoprime|Fermat pseudoprimes]], though it is impractical since these numbers grow exponentially.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of 5 Consecutive Non-Primable Numbers by Changing 1 Digit}
+Tags: Prime Numbers, Recreational Mathematics, Numbers that cannot be made Prime by changing 1 Digit
+
+\begin{theorem}
+The following [[Definition:Integer Sequence|sequence]] of $5$ consecutive [[Definition:Positive Integer|positive integers]] cannot be made into [[Definition:Prime Number|prime numbers]] by changing just one [[Definition:Digit|digit]]:
+:$872\,894, 872\,895, 872\,896, 872\,897, 872\,898$
+{{OEIS|A192545}}
+\end{theorem}
+
+\begin{proof}
+Numbers ending in $0$, $2$, $4$, $6$ and $8$ are not [[Definition:Prime Number|prime]] because by [[Divisibility by 2]] they are [[Definition:Divisor of Integer|divisible]] by $2$.
+Numbers ending in $0$ and $5$ are not [[Definition:Prime Number|prime]] because by [[Divisibility by 5]] they are [[Definition:Divisor of Integer|divisible]] by $5$.
+Hence each of $872\,894$, $872\,895$, $872\,896$ and $872\,898$ remain [[Definition:Composite Number|composite]] when you change any of their digits except the last one.
+So we inspect the [[Definition:Prime Factor|prime factors]] of the following, only bothering to check the numbers ending in $1$, $3$, $7$ and $9$:
+{{begin-eqn}}
+{{eqn | l = 872\,891
+      | r = 271 \times 3221
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,893
+      | r = 7 \times 124699
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,897
+      | r = 263 \times 3319
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,897
+      | r = 17 \times 51347
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+All we need to do now is to inspect $872\,897$.
+{{begin-eqn}}
+{{eqn | l = 072\,897
+      | r = 3 \times 11 \times 472
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 172\,897
+      | r = 41 \times 4217
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 272\,897
+      | r = 19 \times 53 \times 271
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 372\,897
+      | r = 3^3 \times 7 \times 1973
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 472\,897
+      | r = 37 \times 12781
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 572\,897
+      | r = 13 \times 127 \times 347
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 672\,897
+      | r = 3 \times 224299
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 772\,897
+      | r = 757 \times 1021
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 972\,897
+      | r = 3 \times 324299
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 802\,897
+      | r = 53 \times 15149
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 812\,897
+      | r = 733 \times 1109
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 822\,897
+      | r = 3^2 \times 91433
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 832\,897
+      | r = 13 \times 79 \times 811
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 842\,897
+      | r = 11 \times 19 \times 37 \times 109
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 852\,897
+      | r = 3 \times 107 \times 2657
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 862\,897
+      | r = 7 \times 131 \times 941
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 882\,897
+      | r = 3 \times 151 \times 1949
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 892\,897
+      | r = 607 \times 1471
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 870\,897
+      | r = 3 \times 61 \times 4759
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 871\,897
+      | r = 13 \times 47 \times 1427
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 873\,897
+      | r = 3 \times 291299
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 874\,897
+      | r = 23 \times 38039
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 875\,897
+      | r = 11 \times 79627
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 876\,897
+      | r = 3^2 \times 7 \times 31 \times 449
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 877\,897
+      | r = 17 \times 113 \times 457
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 878\,897
+      | r = 139 \times 6323
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 879\,897
+      | r = 3 \times 37 \times 7927
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 872\,097
+      | r = 3 \times 149 \times 1951
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,197
+      | r = 59 \times 14783
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,297
+      | r = 191 \times 4567
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,397
+      | r = 3^3 \times 79 \times 409
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,497
+      | r = 37 \times 23581
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,597
+      | r = 11 \times 23 \times 3449
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,697
+      | r = 3 \times 7 \times 29 \times 1433
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,797
+      | r = 17 \times 51341
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,997
+      | r = 3 \times 290999
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 872\,807
+      | r = 13 \times 67139
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,817
+      | r = 3 \times 11 \times 26449
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,827
+      | r = 23 \times 137 \times 277
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,837
+      | r = 7^2 \times 47 \times 379
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,847
+      | r = 3^2 \times 293 \times 331
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,857
+      | r = 43 \times 53 \times 383
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,867
+      | r = 31 \times 37 \times 761
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,877
+      | r = 3 \times 290959
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 872\,887
+      | r = 523 \times 1669
+      | c = Not [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+The result has been proven.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Smith Number from Prime Repunit}
+Tags: Smith Numbers, Repunits
+
+\begin{theorem}
+Let $R_n$ be a [[Definition:Repunit|repunit]] which is [[Definition:Prime Number|prime]] where $n \ge 3$.
+Then $3304 \times R_n$ is a [[Definition:Smith Number|Smith number]].
+$3304$ is not the only number this works for, but it is the smallest.
+\end{theorem}
+
+\begin{proof}
+Let $\map S n$ denote the [[Definition:Integer Addition|sum]] of the [[Definition:Digit|digits]] of a [[Definition:Positive Integer|positive integer]] $n$.
+Let $\map {S_p} n$ denote the [[Definition:Integer Addition|sum]] of the [[Definition:Digit|digits]] of the [[Definition:Prime Decomposition|prime decomposition]] of $n$.
+Then $\map S n = \map {S_p} n$ {{iff}} $n$ is a [[Definition:Smith Number|Smith number]].
+Let $n \ge 3$.
+We have that:
+:$3304 = 2 \times 2 \times 2 \times 7 \times 59$
+and so for a [[Definition:Prime Number|prime]] [[Definition:Repunit|repunit]] $R_n$:
+:$3304 \times R_n = 2 \times 2 \times 2 \times 7 \times 59 \times \underbrace {111 \ldots 11}_{n \text { ones} }$
+and so:
+:$\map {S_p} {3304 \times R_n} = 2 + 2 + 2 + 7 + 5 + 9 + n \times 1 = n + 27$
+Then:
+:$\map S {3304 \times R_n} = n + 27$
+because:
+{{explain|that because}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arithmetic Sequence of 16 Primes}
+Tags: Prime Numbers, Arithmetic Sequences
+
+\begin{theorem}
+The $16$ [[Definition:Integer|integers]] in [[Definition:Arithmetic Sequence|arithmetic sequence]] defined as:
+:$2\,236\,133\,941 + 223\,092\,870 n$
+are [[Definition:Prime Number|prime]] for $n = 0, 1, \ldots, 15$.
+\end{theorem}
+
+\begin{proof}
+First we note that:
+:$2\,236\,133\,941 - 223\,092\,870 = 2\,013\,041\,071 = 53 \times 89 \times 426\,763$
+and so this [[Definition:Arithmetic Sequence|arithmetic sequence]] of [[Definition:Prime Number|primes]] does not extend to $n < 0$.
+{{begin-eqn}}
+{{eqn | l = 2\,236\,133\,941 + 0 \times 223\,092\,870
+      | r = 2\,236\,133\,941
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 1 \times 223\,092\,870
+      | r = 2\,459\,226\,811
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 2 \times 223\,092\,870
+      | r = 2\,682\,319\,681
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 3 \times 223\,092\,870
+      | r = 2\,905\,412\,551
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 4 \times 223\,092\,870
+      | r = 3\,128\,505\,421
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 5 \times 223\,092\,870
+      | r = 3\,351\,598\,291
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 6 \times 223\,092\,870
+      | r = 3\,574\,691\,161
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 7 \times 223\,092\,870
+      | r = 3\,797\,784\,031
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 8 \times 223\,092\,870
+      | r = 4\,020\,876\,901
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 9 \times 223\,092\,870
+      | r = 4\,243\,969\,771
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 10 \times 223\,092\,870
+      | r = 4\,467\,062\,641
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 11 \times 223\,092\,870
+      | r = 4\,690\,155\,511
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 12 \times 223\,092\,870
+      | r = 4\,913\,248\,381
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 13 \times 223\,092\,870
+      | r = 5\,136\,341\,251
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 14 \times 223\,092\,870
+      | r = 5\,359\,434\,121
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 2\,236\,133\,941 + 15 \times 223\,092\,870
+      | r = 5\,582\,526\,991
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+But note that $2\,236\,133\,941 + 16 \times 223\,092\,870 = 5\,805\,619\,861 = 79 \times 73\,488\,859$ and so is not [[Definition:Prime Number|prime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Smallest 17 Primes in Arithmetic Sequence}
+Tags: Prime Numbers, Arithmetic Sequences
+
+\begin{theorem}
+The smallest $17$ [[Definition:Prime Number|primes]] in [[Definition:Arithmetic Sequence|arithmetic sequence]] are:
+:$3\,430\,751\,869 + 87\,297\,210 n$
+for $n = 0, 1, \ldots, 16$.
+\end{theorem}
+
+\begin{proof}
+First we note that:
+:$3\,430\,751\,869 - 87\,297\,210 = 3\,343\,454\,659 = 17\,203 \times 194\,353$
+and so this [[Definition:Arithmetic Sequence|arithmetic sequence]] of [[Definition:Prime Number|primes]] does not extend to $n < 0$.
+{{begin-eqn}}
+{{eqn | l = 3\,430\,751\,869 + 0 \times 87\,297\,210
+      | r = 3\,430\,751\,869
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 1 \times 87\,297\,210
+      | r = 3\,518\,049\,079
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 2 \times 87\,297\,210
+      | r = 3\,605\,346\,289
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 3 \times 87\,297\,210
+      | r = 3\,692\,643\,499
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 4 \times 87\,297\,210
+      | r = 3\,779\,940\,709
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 5 \times 87\,297\,210
+      | r = 3\,867\,237\,919
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 6 \times 87\,297\,210
+      | r = 3\,954\,535\,129
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 7 \times 87\,297\,210
+      | r = 4\,041\,832\,339
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 8 \times 87\,297\,210
+      | r = 4\,129\,129\,549
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 9 \times 87\,297\,210
+      | r = 4\,216\,426\,759
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 10 \times 87\,297\,210
+      | r = 4\,303\,723\,969
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 11 \times 87\,297\,210
+      | r = 4\,391\,021\,179
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 12 \times 87\,297\,210
+      | r = 4\,478\,318\,389
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 13 \times 87\,297\,210
+      | r = 4\,565\,615\,599
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 14 \times 87\,297\,210
+      | r = 4\,652\,912\,809
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 15 \times 87\,297\,210
+      | r = 4\,740\,210\,019
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 3\,430\,751\,869 + 16 \times 87\,297\,210
+      | r = 4\,827\,507\,229
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+But note that $3\,430\,751\,869 + 17 \times 87\,297\,210 = 4\,914\,804\,439 = 41 \times 97 \times 1 235807$ and so is not [[Definition:Prime Number|prime]].
+{{ProofWanted|It remains to be shown that there are no smaller such APs}}
+\end{proof}<|endoftext|>
+\section{Smallest 18 Primes in Arithmetic Sequence}
+Tags: Prime Numbers, Arithmetic Sequences
+
+\begin{theorem}
+The smallest $18$ [[Definition:Prime Number|primes]] in [[Definition:Arithmetic Sequence|arithmetic sequence]] are:
+:$107\,928\,278\,317 + 9\,922\,782\,870 n$
+for $n = 0, 1, \ldots, 16$.
+\end{theorem}
+
+\begin{proof}
+First we note that:
+:$107\,928\,278\,317 - 9\,922\,782\,870 = 98\,005\,495\,447 = 29 \times 149 \times 22\,681\,207$
+and so this [[Definition:Arithmetic Sequence|arithmetic sequence]] of [[Definition:Prime Number|primes]] does not extend to $n < 0$.
+{{begin-eqn}}
+{{eqn | l = 107\,928\,278\,317 + 0 \times 9\,922\,782\,870
+      | r = 107\,928\,278\,317
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 1 \times 9\,922\,782\,870
+      | r = 117\,851\,061\,187
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 2 \times 9\,922\,782\,870
+      | r = 127\,773\,844\,057
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 3 \times 9\,922\,782\,870
+      | r = 137\,696\,626\,927
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 4 \times 9\,922\,782\,870
+      | r = 147\,619\,409\,797
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 5 \times 9\,922\,782\,870
+      | r = 157\,542\,192\,667
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 6 \times 9\,922\,782\,870
+      | r = 167\,464\,975\,537
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 7 \times 9\,922\,782\,870
+      | r = 177\,387\,758\,407
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 8 \times 9\,922\,782\,870
+      | r = 187\,310\,541\,277
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 9 \times 9\,922\,782\,870
+      | r = 197\,233\,324\,147
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 10 \times 9\,922\,782\,870
+      | r = 207\,156\,107\,017
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 11 \times 9\,922\,782\,870
+      | r = 217\,078\,889\,887
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 12 \times 9\,922\,782\,870
+      | r = 227\,001\,672\,757
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 13 \times 9\,922\,782\,870
+      | r = 236\,924\,455\,627
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 14 \times 9\,922\,782\,870
+      | r = 246\,847\,238\,497
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 15 \times 9\,922\,782\,870
+      | r = 256\,770\,021\,367
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 16 \times 9\,922\,782\,870
+      | r = 266\,692\,804\,237
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{eqn | l = 107\,928\,278\,317 + 17 \times 9\,922\,782\,870
+      | r = 276\,615\,587\,107
+      | c = which is [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+But note that $107\,928\,278\,317 + 18 \times 9\,922\,782\,870 = 286\,538\,369\,977 = 23 \times 181 \times 68\,829\,779$ and so is not [[Definition:Prime Number|prime]].
+{{ProofWanted|It remains to be shown that there are no smaller such APs}}
+\end{proof}<|endoftext|>
+\section{Prime Gap of 654}
+Tags: Prime Gaps
+
+\begin{theorem}
+There exists a [[Definition:Prime Gap|prime gap]] of $654$ between $11\,000\,001\,446\,613\,353$ and $11\,000\,001\,446\,614\,007$.
+\end{theorem}
+
+\begin{proof}
+$11\,000\,001\,446\,613\,353$ is a [[Definition:Prime Number|prime number]].
+$11\,000\,001\,446\,614\,007$ is a [[Definition:Prime Number|prime number]].
+It can be checked that all numbers between these two are [[Definition:Composite Number|composite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pair of Large Twin Primes}
+Tags: Twin Primes: Examples
+
+\begin{theorem}
+The [[Definition:Integer|integers]] defined as:
+:$1\,159\,142\,985 \times 2^{2304} \pm 1$
+are a pair of [[Definition:Twin Primes|twin primes]] each with $703$ [[Definition:Digit|digits]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|It remains to be shown that they are prime.}}
+\end{proof}<|endoftext|>
+\section{Integers under Subtraction do not form Group}
+Tags: Integer Subtraction, Examples of Groups
+
+\begin{theorem}
+Let $\struct {\Z, -}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by the set of [[Definition:Integer|integers]] under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Subtraction|subtraction]].
+Then $\struct {\Z, -}$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+It is to be demonstrated that $\struct {\Z, -}$ does not satisfy the [[Definition:Group Axioms|group axioms]].
+First it is noted that [[Integer Subtraction is Closed]].
+Thus $\struct {\Z, -}$ fulfils {{GroupAxiom|0}}.
+However, we then have [[Subtraction on Numbers is Not Associative]].
+So, for example:
+:$3 - \paren {2 - 1} = 2 \ne \paren {3 - 2} - 1 = 0$
+Thus it has been demonstrated that $\struct {\Z, -}$ does not satisfy the [[Definition:Group Axioms|group axioms]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One/Necessary Condition}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One
+
+\begin{theorem}
+Let $x \in \R$ be such that $\size{x} < 1$.
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = x^n$.
+Then $\sequence {x_n}$ is a [[Definition:Real Null Sequence|null sequence]].
+\end{theorem}<|endoftext|>
+\section{Odd Integers under Multiplication do not form Group}
+Tags: Odd Integers, Integer Multiplication
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Odd Integer|odd integers]]:
+:$S = \set {x \in \Z: \exists n \in \Z: x = 2 n + 1}$
+Let $\struct {S, \times}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by $S$ under the [[Definition:Binary Operation|operation]] of [[Definition:Integer Multiplication|multiplication]].
+Then $\struct {S, \times}$ is not a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+It is to be demonstrated that $\struct {S, \times}$ does not satisfy the [[Definition:Group Axioms|group axioms]].
+First it is noted that [[Integer Multiplication is Closed]].
+Then from [[Odd Number multiplied by Odd Number is Odd]], $S$ is [[Definition:Closed Algebraic Structure|closed]] under $\times$.
+Thus $\struct {S, \times}$ fulfils {{GroupAxiom|0}}.
+From [[Integer Multiplication is Associative]], we have that $\times$ is [[Definition:Associative Operation|associative]] on $S$.
+Thus $\struct {S, \times}$ fulfils {{GroupAxiom|1}}.
+Then we have that:
+:$\forall x \in S: 1 \times x = x = x \times 1$
+and as $1 \in S$ it follows that $1$ is the [[Definition:Identity Element|identity element]] of $\struct {S, \times}$
+Thus $\struct {S, \times}$ fulfils {{GroupAxiom|2}}.
+Now consider $3 \in S$.
+There exists no $x \in S$ such that $3 \times x = 1$.
+Thus $x$ has no [[Definition:Inverse Element|inverse element]] in $S$.
+Thus $\struct {S, \times}$ does not fulfil {{GroupAxiom|3}}.
+Thus it has been demonstrated that $\struct {S, \times}$ does not satisfy the [[Definition:Group Axioms|group axioms]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One/Sufficient Condition}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One
+
+\begin{theorem}
+Let $x \in \R$.
+Let $\sequence {x_n}$ be the [[Definition:Sequence|sequence in $\R$]] defined as $x_n = x^n$.
+Let $\sequence {x_n}$ be a [[Definition:Real Null Sequence|null sequence]].
+Then $\size x < 1$.
+\end{theorem}
+
+\begin{proof}
+By [[Reciprocal of Null Sequence]]:
+:$\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to $0$ {{iff}} $\sequence {\dfrac 1 {x_n} }$ [[Definition:Unbounded Divergent Real Sequence|diverges to $\infty$]].
+By the definition of [[Definition:Unbounded Divergent Real Sequence|divergence to $\infty$]]:
+:$\exists N \in \N: \forall n \ge N: \size {\dfrac 1 {x_n} } > 1$
+In particular:
+:$\size {\dfrac 1 {x_N} } > 1$
+By [[Ordering of Reciprocals]]:
+:$\size {x_N} < 1$
+That is:
+:$\size {x_N} = \size {x^N} = \size x^N < 1$
+{{AimForCont}} $\size x \ge 1$.
+By [[Inequality of Product of Unequal Numbers]]:
+:$\size x^N \ge 1^N = 1$
+This is a [[Definition:Contradiction|contradiction]].
+So $\size x < 1$ as required.
+\end{proof}<|endoftext|>
+\section{Square Matrices with +1 or -1 Determinant under Multiplication forms Group}
+Tags: Examples of Groups, Matrix Groups
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order]] $n$ of [[Definition:Real Numbers|real numbers]] whose [[Definition:Determinant of Matrix|determinant]] is either $1$ or $-1$.
+Let $\struct {S, \times}$ denote the [[Definition:Algebraic Structure|algebraic structure]] formed by $S$ whose [[Definition:Binary Operation|operation]] is [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]].
+Then $\struct {S, \times}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $G \, 0$: Closure ===
+Let $\mathbf A, \mathbf B \in S$.
+By definition of [[Definition:Matrix Product (Conventional)|matrix product]], $\mathbf {A B}$ is a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
+From [[Determinant of Matrix Product]]:
+:$\det \mathbf A \det \mathbf B = \map \det {\mathbf {A B} }$
+Hence the [[Definition:Determinant of Matrix|determinant]] of $\mathbf {A B}$ is either $1$ or $-1$.
+Thus $\mathbf {A B} \in S$ and so $\struct {S, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $G \, 1$: Associativity ===
+We have that [[Matrix Multiplication is Associative]].
+Thus $\times$ is [[Definition:Associative|associative]] on $\struct {S, \times}$.
+{{qed|lemma}}
+=== $G \, 2$: Identity ===
+From [[Unit Matrix is Unity of Ring of Square Matrices]], the [[Definition:Unit Matrix|unit matrix]] $\mathbf I$ serves as the [[Definition:Identity Element|identity element]] of $\struct {S, \times}$.
+{{qed|lemma}}
+=== $G \, 3$: Inverses ===
+Because the [[Definition:Determinant of Matrix|determinants]] of the [[Definition:Element|elements]] of $S$ are $1$ or $-1$, they are by definition [[Definition:Invertible Matrix|invertible]].
+We have that $\mathbf I$ is the [[Definition:Identity Element|identity element]] of $\struct {\R, \circ}$.
+From the definition of [[Definition:Invertible Matrix|invertible matrix]], the [[Definition:Inverse Element|inverse]] of any [[Definition:Invertible Matrix|invertible matrix]] $\mathbf A$ is $\mathbf A^{-1}$.
+{{qed|lemma}}
+All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\struct {S, \times}$ is a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Sine Function is neither Injective nor Surjective}
+Tags: Sine Function, Examples of Injections, Examples of Surjections
+
+\begin{theorem}
+The [[Definition:Real Sine Function|real sine function]] is neither an [[Definition:Injection|injection]] nor a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+This is immediately apparent from the [[Graph of Sine Function|graph of the sine function]]:
+{{:Graph of Sine Function}}
+For example:
+:$\map \sin 0 = \map \sin \pi = 0$
+and so the [[Definition:Real Sine Function|real sine function]] is not an [[Definition:Injection|injection]].
+Then, for example:
+:$\nexists x \in \R: \map \sin x = 2$
+and so the [[Definition:Real Sine Function|real sine function]] is not a [[Definition:Surjection|surjection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One/Complex Numbers}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One
+
+\begin{theorem}
+Let $z \in \C$.
+Let $\sequence {z_n}$ be the [[Definition:Complex Sequence|sequence in $\C$]] defined as $z_n = z^n$.
+Then:
+:$\size z < 1$ {{iff}} $\sequence {z_n}$ is a [[Definition:Complex Null Sequence|null sequence]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Convergent Complex Sequence|convergence]]:
+:$\displaystyle \lim_{n \mathop \to \infty} z_n = 0 \iff \lim_{n \mathop \to \infty} \size {z_n} = 0$
+By [[Modulus of Product]]:
+:$\forall n \in \N: \size {z_n} = \size {z^n} = \size z^n$
+So:
+:$\displaystyle \lim_{n \mathop \to \infty} \size {z_n} = 0 \iff \lim_{n \mathop \to \infty} \size z^n = 0$
+Since $\size z \in \R_{\ge 0}$, by [[Sequence of Powers of Number less than One|Sequence of Powers of Real Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \size z^n = 0 \iff \size z < 1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One/Normed Division Ring}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One, Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}}$ be a [[Definition:Normed Division Ring|normed division ring]]
+Let $x \in \R$.
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = x^n$.
+Then:
+:$\norm x < 1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]].
+\end{theorem}
+
+\begin{proof}
+Let $0_R$ be the [[Definition:Ring Zero|zero]] of $R$.
+ 
+By the definition of [[Definition:Convergent Sequence in Normed Division Ring|convergence]]:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = 0_R \iff \lim_{n \mathop \to \infty} \norm {x_n} = 0$
+By [[Definition:Norm Axioms|norm axiom (N2) (Multiplicativity)]] then for each $n \in \N$:
+:$\norm {x_n} = \norm {x^n} = \norm x^n$.
+So:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n} = 0 \iff \lim_{n \mathop \to \infty} \norm x^n = 0$
+Since $\norm x \in \R_{\ge 0}$, by [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm x^n = 0 \iff \norm x < 1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One/Rational Numbers}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One
+
+\begin{theorem}
+Let $x \in \Q$.
+Let $\sequence {x_n}$ be the [[Definition:Rational Sequence|sequence in $\Q$]] defined as $x_n = x^n$.
+Then:
+:$\size x < 1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Rational Null Sequence|null sequence]].
+\end{theorem}
+
+\begin{proof}
+By the definition of [[Definition:Convergent Rational Sequence|convergence of a rational sequence]]:
+:$\sequence {x_n}$ is a [[Definition:Rational Null Sequence|null sequence in the rational numbers]] {{iff}} $\sequence {x_n}$ is a [[Definition:Real Null Sequence|null sequence in the real numbers]]
+By [[Sequence of Powers of Number less than One|Sequence of Powers of Real Number less than One]]:
+:$\sequence {x_n}$ is a [[Definition:Real Null Sequence|null sequence in the real numbers]] {{iff}} $\size x < 1$
+{{qed}}
+[[Category:Limits of Sequences]]
+[[Category:Sequence of Powers of Number less than One]]
+hzx5m2gf5fkreya5guv3y0jmzbwmq3f
+\end{proof}<|endoftext|>
+\section{Composite of Injection on Surjection is not necessarily Either}
+Tags: Injections, Surjections, Composite Mappings
+
+\begin{theorem}
+Let $f$ be an [[Definition:Injection|injection]].
+Let $g$ be a [[Definition:Surjection|surjection]].
+Let $f \circ g$ denote the [[Definition:Composition of Mappings|composition]] of $f$ with $g$.
+Then it is not necessarily the case that $f \circ g$ is either a [[Definition:Surjection|surjection]] or an [[Definition:Injection|injection]].
+\end{theorem}
+
+\begin{proof}
+Let $X, Y, Z$ be [[Definition:Set|sets]] defined as:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = \set {a, b, c}
+      | c = 
+}}
+{{eqn | l = Y
+      | r = \set {1, 2}
+      | c = 
+}}
+{{eqn | l = Z
+      | r = \set {z, y, z}
+      | c = 
+}}
+{{end-eqn}}
+Let $g: X \to Y$ be defined in [[Definition:Two-Row Notation|two-row notation]] as:
+:$\dbinom {a \ b \ c } {1 \ 2 \ 2}$
+which is seen by inspection to be a [[Definition:Surjection|surjection]].
+Let $f: Y \to Z$ be defined in [[Definition:Two-Row Notation|two-row notation]] as:
+:$\dbinom {1 \ 2} {x \ y}$
+which is seen by inspection to be an [[Definition:Injection|injection]].
+The [[Definition:Composition of Mappings|composition]] $f \circ g$ is seen to be:
+:$\dbinom {a \ b \ c} {x \ y \ y}$
+which is:
+:not an [[Definition:Injection|injection]] (both $b$ and $c$ map to $y$)
+:not a [[Definition:Surjection|surjection]] (nothing maps to $z$).
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 3 is Unique}
+Tags: Groups of Order 3
+
+\begin{theorem}
+There exists exactly $1$ [[Definition:Group|group]] of [[Definition:Order of Group|order]] $3$, up to [[Definition:Group Isomorphism|isomorphism]]:
+:$C_3$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $3$.
+\end{theorem}
+
+\begin{proof}
+From [[Existence of Cyclic Group of Order n]] we have that one such [[Definition:Group|group]] of [[Definition:Order of Group|order]] $3$ is the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $3$.
+This is exemplified by the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $3$]], whose [[Modulo Addition/Cayley Table/Modulo 3|Cayley table]] can be presented as:
+{{:Modulo Addition/Cayley Table/Modulo 3}}{{qed|lemma}}
+Consider an arbitrary [[Definition:Group|group]] $\struct {G, \circ}$ whose [[Definition:Identity Element|identity element]] is $e$.
+Let the [[Definition:Underlying Set of Structure|underlying set]] of $G$ be:
+:$G = \set {e, a, b}$
+where $a, b \in G$ are arbitrary.
+Since $e$ is the [[Definition:Identity Element|identity]], we can start off the [[Definition:Cayley Table|Cayley table]] for $G$ as:
+:$\begin{array}{c|ccc}
+\circ & e & a & b \\
+\hline
+e & e & a & b \\
+a & a &   &   \\
+b & b &   &   \\
+\end{array}$
+Consider the [[Definition:Element|element]] at $a \circ a$.
+We have that $a \circ a$ must be either $e$ or $b$, as from [[Group has Latin Square Property]] it cannot be $a$.
+If $a \circ a = e$, then that leaves only $b$ to complete the middle row.
+But $b$ already exists in the final column.
+So $a \circ a$ cannot be $e$, so must be $b$.
+Hence the [[Definition:Cayley Table|Cayley table]] for $G$ so far is:
+:$\begin{array}{c|ccc}
+\circ  & e & a & b \\
+\hline
+e & e & a & b \\
+a & a & b &   \\
+b & b &   &   \\
+\end{array}$
+The rest of the table is completed by following the result that [[Group has Latin Square Property]]:
+:$\begin{array}{c|ccc}
+\circ  & e & a & b \\
+\hline
+e & e & a & b \\
+a & a & b & e \\
+b & b & e & a \\
+\end{array}$
+and it is seen by inspection that $G$ is indeed the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Powers of Number less than One}
+Tags: Limits of Sequences, Sequence of Powers of Number less than One
+
+\begin{theorem}
+Let $x \in \R$.
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|sequence in $\R$]] defined as $x_n = x^n$.
+Then:
+:$\size x < 1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Real Null Sequence|null sequence]].
+\end{theorem}
+
+\begin{proof}
+=== [[Sequence of Powers of Number less than One/Necessary Condition|Necessary Condition]] ===
+{{refactor}}
+[For other proofs of the [[Sequence of Powers of Number less than One/Necessary Condition|Necessary Condition]] visit [[Sequence of Powers of Number less than One/Necessary Condition|here]].]
+{{:Sequence of Powers of Number less than One/Necessary Condition/Proof 1}}{{qed|lemma}}
+=== [[Sequence of Powers of Number less than One/Sufficient Condition|Sufficient Condition]] ===
+{{:Sequence of Powers of Number less than One/Sufficient Condition}}{{qed}}
+[[Category:Limits of Sequences]]
+[[Category:Sequence of Powers of Number less than One]]
+5faaam982gem8e9cdufdfzgmq3qwztx
+\end{proof}<|endoftext|>
+\section{Set of Rotations is Subgroup of Symmetry Group}
+Tags: Symmetry Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Symmetry Group|symmetry group]].
+Let $H$ be the [[Definition:Subset|subset]] of $G$ consisting of the [[Definition:Rotation (Geometry)|rotations]] in $G$ about a given [[Definition:Axis of Rotation|axis]].
+Then $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Needs a more formal definition of rotation. Surprised this hasn't already been covered properly.}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms}
+Tags: Normed Division Rings, Norm Theory, Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $R$ be a [[Definition:Division Ring|division ring]].
+Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be [[Definition:Norm on Division Ring|norms]] on $R$.
+Let $d_1$ and $d_2$ be the [[Definition:Metric Induced by Norm|metrics induced]] by the [[Definition:Norm on Division Ring|norms]] $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ respectively.
+{{TFAE|def = Equivalent Division Ring Norms}}
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent|Topologically Equivalent implies Convergently Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence Equivalent|Convergently Equivalent implies Null Sequence Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence Equivalent}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Null Sequence Equivalent implies Open Unit Ball Equivalent|Null Sequence Equivalent implies Open Unit Ball Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Null Sequence Equivalent implies Open Unit Ball Equivalent}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm|Open Unit Ball Equivalent implies Norm is Power of Other Norm]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent|Norm is Power of Other Norm implies Topologically Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent|Norm is Power of Other Norm implies Cauchy Sequence Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent}}{{qed|lemma}}
+=== [[Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent|Cauchy Sequence Equivalent implies Open Unit Ball Equivalent]] ===
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $d_1$ and $d_2$ be [[Definition:Topologically Equivalent Metrics|topologically equivalent metrics]].
+Then:
+:$d_1$ and $d_2$ are [[Definition:Equivalent Metrics|convergently equivalent metrics]].
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converge]] to $l$ in $\norm {\, \cdot \,}_1$.
+Let $\epsilon \in \R_{> 0}$ be given.
+Let $\map {B_\epsilon^2} i$ denote the [[Definition:Open Ball|open ball]] [[Definition:Center of Open Ball|centered]] on $l$ of [[Definition:Radius of Open Ball|radius]] $\epsilon$ in $\struct {R, \norm {\, \cdot \,}_2}$.
+By [[Open Ball of Metric Space is Open Set]] then $\map {B_\epsilon^2} l$ is [[Definition:Open Set of Metric Space|open set]] in $\struct {R, d_2}$.
+Since $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent metrics]] then $\map {B_\epsilon^2} l$ is [[Definition:Open Set of Metric Space|open set]] in $\struct {R, d_1}$.
+By the definition of an [[Definition:Open Set of Metric Space|open set in a metric space]] then:
+:$\exists \delta \in \R_{> 0}: \map {B_\delta^1} l \subseteq \map {B_\epsilon^2} l$
+Hence:
+:$\forall x \in R: \norm {x - l}_1 < \delta \implies \norm {x - l}_2 < \epsilon$
+Since $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_1$ then:
+:$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_1 < \delta$
+Hence:
+:$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_2 < \epsilon$
+Since $\sequence {x_n}$ and $\epsilon > 0$ were arbitrary then it has been shown that for all [[Definition:Sequence|sequences]] $\sequence {x_n}$ in $R$: 
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_2$. 
+By a similar argument it is shown that for all [[Definition:Sequence|sequences]] $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_1$. 
+The result follows. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm{\, \cdot \,}_1 \iff \sequence {x_n}$ is a [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $l$ in $\norm {\, \cdot \,}_2$
+Then for all sequences $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_2$
+\end{theorem}
+
+\begin{proof}
+Let $0_R$ be the [[Definition:Ring Zero|zero]] of $R$, then:
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $0_R$ in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $0_R$ in $\norm {\, \cdot \,}_2$
+Hence:
+:$\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm{\,\cdot\,}_1  \iff \sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm{\,\cdot\,}_2$ 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Null Sequence Equivalent implies Open Unit Ball Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_1  \iff \sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_2$ 
+Then $\forall x \in R$:
+:$\norm x_1 < 1 \iff \norm x_2 < 1$
+\end{theorem}
+
+\begin{proof}
+Let $x \in R$.
+Let $\sequence {x_n}$ be the [[Definition:Sequence|sequence]] defined by: $\forall n: x_n = x^n$.
+{{begin-eqn}}
+{{eqn | l = \norm x_1 < 1 \quad
+      | o = \leadstoandfrom
+      | c = $\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_1$
+      | cc= [[Sequence of Powers of Number less than One/Normed Division Ring|Sequence of Powers of Number less than One in Normed Division Ring]]
+}}
+{{eqn | o = \leadstoandfrom
+      | c = $\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_2$
+      | cc= Assumption
+}}
+{{eqn | o = \leadstoandfrom
+      | c = $\norm x_2 < 1$
+      | cc= [[Sequence of Powers of Number less than One/Normed Division Ring|Sequence of Powers of Number less than One in Normed Division Ring]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:$\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$ 
+Then:
+:$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
+\end{theorem}
+
+\begin{proof}
+==== Case 1 ====
+For all $x \in R: x \ne 0_R$, let $x$ satisfy $\norm x_1 \ge 1$.
+===== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1|Lemma 1]] =====
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1}}
+By assumption, for all $x \in R, x \ne 0_R$, then $\norm x_2 \ge 1$.
+Similarly $\norm {\, \cdot \,}_2$ is the [[Definition:Trivial Norm on Division Ring|trivial norm]].
+Hence $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are [[Definition:Equal|equal]].
+For $\alpha = 1$ the result follows.
+{{qed|lemma}}
+==== Case 2 ====
+Let $x_0 \in R$ such that $x_0 \ne 0_R$ and $\norm {x_0}_1 < 1$.
+By assumption then $\norm {x_0}_2 < 1$.
+Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.
+Then $\norm {x_0}_1 = \norm {x_0}_2^\alpha$.
+As $\norm {x_0}_1, \norm {x_0}_2 < 1$:
+:$\log \norm {x_0}_1 < 0$
+:$\log \norm {x_0}_2 < 0$
+So $\alpha > 0$.
+====== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2|Lemma 2]] ======
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Subgroups of Prime Power Order}
+Tags: Subgroups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p^a k$, where:
+:$a \in \Z_{>0}$ is a [[Definition:Strictly Positive Integer|(strictly) positive integer]]
+:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
+Let $P \le G$ be a [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order]] $p^a$.
+Let $Q \le G$ be a [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order]] $p^b$, where $0 < b \le a$.
+Let it be the case that $Q$ is not a [[Definition:Subgroup|subgroup]] of $P$.
+Then $P Q$ is not a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Subgroups is Subgroup]], $P \cap Q$ is a [[Definition:Subgroup|subgroup]] of $P$.
+Thus:
+:$\order {P \cap Q} = p^c$ for some $c \in \Z$ such that $0 \le c \le a$
+where $\order {P \cap Q}$ denotes the [[Definition:Order of Group|order]] of $P \cap Q$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \order {P Q}
+      | r = \frac {\order P \order Q} {\order {P \cap Q} }
+      | c = [[Order of Subgroup Product]]
+}}
+{{eqn | r = \frac {p^a p^b} {p^c}
+      | c = 
+}}
+{{eqn | r = p^{a + b - c}
+      | c = 
+}}
+{{end-eqn}}
+{{AimForCont}} $P Q$ is a [[Definition:Subgroup|subgroup]] of $G$.
+By [[Lagrange's Theorem (Group Theory)]]:
+:$\order {P Q} \divides \order G$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+We have that $\order {P Q}$ is a [[Definition:Integer Power|power]] of $p$.
+The highest [[Definition:Integer Power|power]] of $p$ which [[Definition:Divisor of Integer|divides]] $\order G$ is $p^a$.
+Thus $P Q$ could have [[Definition:Order of Group|order]] $p^a$ at most.
+Thus:
+:$a + b - c \le a$
+That is:
+:$b \le c$
+But $P \cap Q$ is a [[Definition:Subgroup|subgroup]] of $P$.
+Hence it must be the case that $b = c$.
+Thus:
+:$P \cap Q = Q$
+and so $Q$ is a [[Definition:Subgroup|subgroup]] of $P$.
+This [[Proof by Contradiction|contradicts]] the fact that $Q$ is not a [[Definition:Subgroup|subgroup]] of $P$.
+It must follow that $P Q$ is not a [[Definition:Subgroup|subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Independent Set is Contained in some Basis}
+Tags: Bases of Vector Spaces, Linearly Independent Set is Contained in some Basis
+
+\begin{theorem}
+There exists a [[Definition:Basis of Vector Space|basis]] $B$ for $E$ such that $H \subseteq B$.
+\end{theorem}
+
+\begin{proof}
+By hypothesis there is a [[Definition:Basis of Vector Space|basis]] $B$ of $E$ with $n$ elements.
+Then $H \cup B$ is a [[Definition:Generator of Module|generator]] for $E$.
+So by [[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]] there exists a [[Definition:Basis of Vector Space|basis]] $C$ of $E$ such that $H \subseteq C \subseteq H \cup B$.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $H = \set {\xi_1, \xi_2, \ldots, \xi r}$.
+Consider the [[Definition:Basis of Vector Space|basis]] $B = \set {\alpha_1, \alpha_2, \ldots, \alpha_n}$ of $E$.
+Consider the [[Definition:Set|set]] $G = H \cup B = \set {\xi_1, \xi_2, \ldots, \xi r, \alpha_1, \alpha_2, \ldots, \alpha_n}$.
+We have that $G$ is a [[Definition:Generator of Vector Space|generator]] of $E$.
+As $B$ is a [[Definition:Basis of Vector Space|basis]], it follows that each of $H$ is a [[Definition:Linear Combination|linear combination]] of $B$.
+Thus $G = H \cup B$ is [[Definition:Linearly Dependent Set|linearly dependent]].
+Thus one of the [[Definition:Element|elements]] $\alpha_i$ of $B$ is a [[Definition:Linear Combination|linear combination]] of the preceding [[Definition:Element|elements]] of $G$.
+Eliminate this one, and do the same thing with with $G \setminus \set {\alpha_i}$. 
+Eventually there will exist a [[Definition:Set|set]] which is a [[Definition:Basis of Vector Space|basis]] of $E$ containing all the [[Definition:Element|elements]] of $H$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Independent Set is Basis iff of Same Cardinality as Dimension}
+Tags: Bases of Vector Spaces, Dimension of Vector Space
+
+\begin{theorem}
+$H$ is a [[Definition:Basis of Vector Space|basis]] for $E$ {{iff}} it contains exactly $n$ [[Definition:Element|elements]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:By Hypothesis|By hypothesis]], let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $E$
+=== Necessary Condition ===
+Let $H$ be a [[Definition:Basis of Vector Space|basis]] for $E$.
+By definition of [[Definition:Dimension of Vector Space|dimension of vector space]], a [[Definition:Basis of Vector Space|basis]] for $E$ contains exactly $n$ [[Definition:Element|elements]].
+By [[Bases of Finitely Generated Vector Space have Equal Cardinality]], it follows that $H$ also contains exactly $n$ [[Definition:Element|elements]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $H$ contain exactly $n$ [[Definition:Element|elements]].
+By [[Sufficient Conditions for Basis of Finite Dimensional Vector Space]] $H$ is itself a [[Definition:Basis of Vector Space|basis]] for $E$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Generator of Vector Space is Basis iff Cardinality equals Dimension}
+Tags: Generators of Vector Spaces, Bases of Vector Spaces, Dimension of Vector Space
+
+\begin{theorem}
+:$G$ is a [[Definition:Basis of Vector Space|basis]] for $E$ {{iff}} $\card G = n$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $G$ be a [[Definition:Basis of Vector Space|basis]] for $E$.
+From [[Cardinality of Basis of Vector Space]], $\card G = n$.
+{{qed}}
+=== Sufficient Condition ===
+Let $\card G = n$.
+From [[Sufficient Conditions for Basis of Finite Dimensional Vector Space]], $G$ is a [[Definition:Basis of Vector Space|basis]] for $E$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{ISBN-10 is Error-Correcting Code}
+Tags: International Standard Book Numbers, Error-Correcting Codes, ISBN-10 is Error-Correcting Code
+
+\begin{theorem}
+[[Definition:ISBN-10|ISBN-$10$]] is an [[Definition:Error-Correcting Code|error-correcting code]] in the following sense:
+\end{theorem}<|endoftext|>
+\section{ISBN-10 is Error-Correcting Code/Transposition Error}
+Tags: ISBN-10 is Error-Correcting Code
+
+\begin{theorem}
+If any two of the first $9$ [[Definition:Digit|digits]] are transposed, the [[Definition:Check Digit|check digit]] will be wrong.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Ongoing}}
+[[Category:ISBN-10 is Error-Correcting Code]]
+mwbruqai92ntf6l904fwb5runnoin1b
+\end{proof}<|endoftext|>
+\section{ISBN-10 is Error-Correcting Code/Transmission Error}
+Tags: ISBN-10 is Error-Correcting Code
+
+\begin{theorem}
+If an error has been made in any one of the first $9$ [[Definition:Digit|digits]], the [[Definition:Check Digit|check digit]] will be wrong.
+\end{theorem}
+
+\begin{proof}
+Let $S$ denote an [[Definition:ISBN-10|ISBN-$10$]] whose $k$th [[Definition:Digit|digit]] is $d_k$.
+Let $d_S$ denote the [[Definition:Check Digit|check digit]] of $S$.
+Let $S'$ denote the [[Definition:ISBN-10|ISBN-$10$]] $S$ whose $n$th [[Definition:Digit|digit]] has been transmitted incorrectly, as $d'_n$.
+Let $d'_S$ denote the [[Definition:Check Digit|check digit]] calculated on $S'$ according to the algorithm via which calculated $d_S$ on $S$.
+It will be demonstrated that $d'_S \ne d_S$.
+By definition of [[Definition:ISBN-10|ISBN-$10$]]:
+:$d_S = \displaystyle \sum_{k \mathop = 1}^9 k d_k \pmod {11}$
+where $d_k$ is assigned the [[Definition:Symbol|symbol]] $X$ if it calculates to $10$.
+Then:
+:$d'_S = \displaystyle \sum_{k \mathop = 1}^9 k d_k - n d_n + n d'_n \pmod {11}$
+and so:
+:$d'_S - d_S = n \paren {d'_n - d_n} \pmod {11}$
+It follows that:
+:$d'_S - d_S \ne 0$
+and so the [[Definition:Check Digit|check digit]] on $d_S$ is the incorrect [[Definition:Check Digit|check digit]] for $S'$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Master Code}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $\map V {n, p}$ be a [[Definition:Master Code|master code]] of [[Definition:Length of Sequence|length]] $n$ modulo $p$.
+Then there are $p^n$ [[Definition:Element|elements]] of $\map V {n, p}$.
+\end{theorem}
+
+\begin{proof}
+For each [[Definition:Term of Sequence|term]] of a [[Definition:Finite Sequence|sequence]] in $\map V {n, p}$ there are $p$ possible values.
+There are $n$ such [[Definition:Term of Sequence|terms]].
+Hence there are $\underbrace {p \times p \times \cdots \times p}_{n \text { times} } = p^n$ different possible [[Definition:Finite Sequence|sequences]] in $\map V {n, p}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Master Code forms Vector Space}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $\map V {n, p}$ be a [[Definition:Master Code|master code]] of [[Definition:Length of Sequence|length]] $n$ modulo $p$.
+Then $\map V {n, p}$ forms a [[Definition:Vector Space|vector space]] over $\Z_p$ of [[Definition:Dimension of Vector Space|$n$ dimensions]].
+\end{theorem}
+
+\begin{proof}
+Recall the [[Definition:Vector Space Axioms|vector space axioms]]:
+{{:Definition:Vector Space Axioms}}
+First, the [[Definition:Set|set]] of [[Definition:Finite Sequence|sequences]] $\tuple {x_1, x_2, \ldots, x_n}$, for $x_1, x_2, \ldots, x_n \in \Z_p$, has to be shown to fulfil the [[Definition:Abelian Group Axioms|abelian group axioms]].
+This follows from:
+:[[Integers Modulo m under Addition form Cyclic Group]]
+and:
+:[[Cyclic Group is Abelian]].
+{{ProofWanted|needs to be finished off, but it's tedious and I lack the patience tonight.}}
+\end{proof}<|endoftext|>
+\section{Conditions Satisfied by Linear Code}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\Z_p$ be the [[Definition:Set of Residue Classes|set of residue classes modulo $p$]].
+Let $C := \tuple {n, k}$ be a [[Definition:Linear Code|linear code]] of a [[Definition:Master Code|master code]] $\map V {n, p}$.
+Then $C$ satisfies the following conditions:
+:$(C \, 1): \quad \forall \mathbf x, \mathbf y \in C: \mathbf x + \paren {-\mathbf y} \in C$
+:$(C \, 2): \quad \forall \mathbf x \in C, m \in \Z_p: m \times \mathbf x \in C$
+where $+$ and $\times$ are the operations of [[Definition:Addition of Codewords in Linear Code|codeword addition]] and [[Definition:Multiple of Codeword in Linear Code|codeword multiplication]] respectively.
+{{expand|Add a page defining the difference between codewords.}}
+\end{theorem}
+
+\begin{proof}
+From [[Master Code forms Vector Space]], $\map V {n, p}$ is a [[Definition:Vector Space|vector space]].
+By definition, $\tuple {n, k}$ is a [[Definition:Vector Subspace|subspace]] of $\map V {n, p}$.
+The result follows by the fact that a [[Definition:Vector Subspace|subspace]] is itself a [[Definition:Vector Space|vector space]].
+{{finish|I lose patience with the fine detail.}}
+\end{proof}<|endoftext|>
+\section{Distance between Linear Codewords is Distance Function}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $\map V {n, p}$ be a [[Definition:Master Code|master code]].
+Let $d: V \times V \to \Z$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall u, v \in V: \map d {u, v} =$ the [[Definition:Distance between Linear Codewords|distance]] between $u$ and $v$
+that is, the number of corresponding [[Definition:Term of Sequence|terms]] at which $u$ and $v$ are different.
+Then $d$ defines a [[Definition:Distance Function|distance function]] in the sense of a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+It is to be demonstrated that $d$ satisfies all the [[Definition:Metric Space Axioms|metric space axioms]].
+Let $u, v, w \in \map V {n, p}$ be arbitrary.
+=== Proof of $(\text M 1)$ ===
+By definition of [[Definition:Distance between Linear Codewords|distance]]:
+:$\map d {u, u} = 0$
+So [[Definition:Metric Space Axioms|axiom $(\text M 1)$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $(\text M 2)$ ===
+Consider $\map d {u, v} + \map d {v, w}$.
+Let $\map d {u, w} \ne 0$.
+Then at each [[Definition:Term of Sequence|term]] at which $u$ and $w$ are different, those corresponding terms in either $u$ and $v$ or $v$ and $w$ must be different.
+So every contribution to the value of $\map d {u, w}$ is present in either $\map d {u, v}$ or $\map d {v, w}$.
+It follows that $\map d {u, v} + \map d {v, w} \ge \map d {u, w}$.
+So [[Definition:Metric Space Axioms|axiom $(\text M 2)$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $(\text M 3)$ ===
+$\map d {u, v} = \map d {v, u}$ by definition of [[Definition:Distance between Linear Codewords|distance]].
+So [[Definition:Metric Space Axioms|axiom $(\text M 3)$]] holds for $d$.
+{{qed|lemma}}
+=== Proof of $(\text M 4)$ ===
+{{begin-eqn}}
+{{eqn | l = u
+      | o = \ne
+      | r = v
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map d {u, v}
+      | o = >
+      | r = 0
+      | c = {{Defof|Distance between Linear Codewords}}
+}}
+{{end-eqn}}
+So [[Definition:Metric Space Axioms|axiom $(\text M 4)$]] holds for $d$.
+{{qed|lemma}}
+Thus $d$ satisfies all the [[Definition:Metric Space Axioms|metric space axioms]] and so is a [[Definition:Metric|metric]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear $\tuple {n, k}$-code]] whose [[Definition:Master Code|master code]] is $\map V {n, p}$.
+Let $\map d C$ denote the [[Definition:Minimum Distance of Linear Code|minimum distance]] of $C$.
+Then:
+:$\map d C = \displaystyle \min_{u \mathop \in C} \map w u$
+where $\map w u$ denotes the [[Definition:Weight of Linear Codeword|weight of $u$]].
+\end{theorem}
+
+\begin{proof}
+Let $f := \displaystyle \min_{u \mathop \in C} \map w u$.
+Let $\mathbf 0$ denote the [[Definition:Codeword of Linear Code|codeword]] in $\map V {n, p}$ consisting of all [[Definition:Zero Digit|zeroes]].
+As $C$ is a [[Definition:Vector Subspace|subspace]] of $\map V {n, p}$, we have that $\mathbf 0 \in C$.
+Let $w$ be a [[Definition:Codeword of Linear Code|codeword]] with [[Definition:Weight of Linear Codeword|weight]] $f$.
+Then:
+:$\map d {w, \mathbf 0} = f$
+so $f \ge \map d C$.
+Let $u, v \in C$ such that $\map d {u, v} = \map d C$.
+We have that $C$ is a [[Definition:Linear Code|linear code]].
+Therefore:
+:$u - v \in C$
+where $u - v$ denotes the [[Definition:Difference between Linear Codewords|difference]] between $u$ and $v$.
+But $u - v$ has [[Definition:Weight of Linear Codeword|weight]] $\map d C$.
+Thus:
+:$\map d C \le f$
+and it follows that $\map d C = f$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Error Detection Capability of Linear Code}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear code]].
+Let $C$ have a [[Definition:Minimum Distance of Linear Code|minimum distance]] $d$.
+Then $C$ detects $d - 1$ or fewer [[Definition:Transmission Error|transmission errors]].
+\end{theorem}
+
+\begin{proof}
+Let $C$ be a [[Definition:Linear Code|linear code]] whose [[Definition:Master Code|master code]] is $V$.
+Let $c \in C$ be a [[Definition:Transmitted Codeword|transmitted codeword]].
+Let $v$ be the [[Definition:Received Word|received word]] from $c$.
+By definition, $v$ is an [[Definition:Element|element]] of $V$.
+Let $v$ have a [[Definition:Distance between Linear Codewords|distance]] $f$ from $c$, where $f \le d - 1$.
+Thus there have been $f$ [[Definition:Transmission Error|transmission errors]].
+As $d$ is the [[Definition:Minimum Distance of Linear Code|minimum distance]] it is clear that $v$ cannot be a [[Definition:Codeword of Linear Code|codeword]] of $C$.
+Hence it can be understood that $C$ has detected that $v$ has as many as $d - 1$ [[Definition:Transmission Error|transmission errors]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Error Correction Capability of Linear Code}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear code]].
+Let $C$ have a [[Definition:Minimum Distance of Linear Code|minimum distance]] $d$.
+Then $C$ corrects $e$ [[Definition:Transmission Error|transmission errors]] for all $e$ such that $2 e + 1 \le d$.
+\end{theorem}
+
+\begin{proof}
+Let $C$ be a [[Definition:Linear Code|linear code]] whose [[Definition:Master Code|master code]] is $V$.
+Let $c \in C$ be a [[Definition:Transmitted Codeword|transmitted codeword]].
+Let $v$ be the [[Definition:Received Word|received word]] from $c$.
+By definition, $v$ is an [[Definition:Element|element]] of $V$.
+Let $v$ have a [[Definition:Distance between Linear Codewords|distance]] $e$ from $c$, where $2 e + 1 \le d$.
+Thus there have been $e$ [[Definition:Transmission Error|transmission errors]].
+{{AimForCont}} $c_1$ is a [[Definition:Codeword of Linear Code|codeword]] of $C$, [[Definition:Distinct Objects|distinct]] from $c$, such that $\map d {v, c_1} \le e$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map d {c, c_1}
+      | o = \le
+      | r = \map d {c, v} + \map d {v, c_1}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = e + e
+      | c = 
+}}
+{{eqn | o = <
+      | r = d
+      | c = 
+}}
+{{end-eqn}}
+So $c_1$ has a [[Definition:Distance between Linear Codewords|distance]] from $c$ less than $d$.
+But $C$ has a [[Definition:Minimum Distance of Linear Code|minimum distance]] $d$.
+Thus $c_1$ cannot be a [[Definition:Codeword of Linear Code|codeword]] of $C$.
+From this [[Proof by Contradiction|contradiction]] it follows that there is no [[Definition:Codeword of Linear Code|codeword]] of $C$ closer to $v$ than $c$.
+Hence there is a [[Definition:Unique|unique]] [[Definition:Codeword of Linear Code|codeword]] of $C$ which has the smallest [[Definition:Distance between Linear Codewords|distance]] from $v$.
+Hence it can be understood that $C$ has corrected the [[Definition:Transmission Error|transmission errors]] of $v$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Generation of Linear Code from Standard Generator Matrix}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $G$ be a [[Definition:Standard Generator Matrix for Linear Code|(standard) generator matrix]] for a [[Definition:Linear Code|linear code]].
+The following methods can be used to generate a [[Definition:Linear Code|linear code]] from $G$:
+\end{theorem}<|endoftext|>
+\section{Generation of Linear Code from Standard Generator Matrix/Method 1}
+Tags: Linear Codes
+
+\begin{theorem}
+A [[Definition:Linear Code|linear code]] $C$ can be obtained from $G$ by:
+:considering the [[Definition:Row of Matrix|rows]] of $G$ as [[Definition:Codeword of Linear Code|codewords]]
+:forming all possible [[Definition:Linear Combination|linear combinations]] of those [[Definition:Codeword of Linear Code|codewords]], considering them as [[Definition:Vector (Linear Algebra)|vectors]] of a [[Definition:Vector Space|vector space]].
+\end{theorem}<|endoftext|>
+\section{Generation of Linear Code from Standard Generator Matrix/Method 2}
+Tags: Linear Codes
+
+\begin{theorem}
+A [[Definition:Linear Code|linear code]] $C$ can be obtained from $G$ by:
+:taking the [[Definition:Set|set]] $U$ of all [[Definition:Finite Sequence|sequences]] of [[Definition:Length of Sequence|length]] $k$ over $\Z_p$ and expressing them as $1 \times k$ [[Definition:Matrix|matrices]]
+:forming all possible [[Definition:Matrix Product (Conventional)|matrix products]] $u G$ for all $u \in U$.
+\end{theorem}<|endoftext|>
+\section{Golay Ternary Code has Minimum Distance 5}
+Tags: Golay Ternary Code
+
+\begin{theorem}
+The [[Definition:Golay Ternary Code|Golay ternary code]] has a [[Definition:Minimum Distance of Linear Code|minimum distance]] of $5$.
+\end{theorem}
+
+\begin{proof}
+Let $C$ denote the [[Definition:Golay Ternary Code|Golay ternary code]].
+By inspection of the [[Definition:Standard Generator Matrix for Linear Code|standard generator matrix]] $G$ of $C$:
+:$G := \begin{pmatrix}
+1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 \\
+0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 2 & 2 \\
+0 & 0 & 1 & 0 & 0 & 0 & 2 & 1 & 0 & 1 & 2 \\
+0 & 0 & 0 & 1 & 0 & 0 & 2 & 2 & 1 & 0 & 1 \\
+0 & 0 & 0 & 0 & 1 & 0 & 1 & 2 & 2 & 1 & 0 \\
+0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 
+\end{pmatrix}$
+it is seen that the smallest [[Definition:Weight of Linear Codeword|weight]] of all the [[Definition:Codeword of Linear Code|codewords]] of $C$ that can be found in $G$ is $5$.
+So it is immediately seen that the [[Definition:Minimum Distance of Linear Code|minimum distance]] of $C$ is at least $5$.
+It remains to be shown that the [[Definition:Minimum Distance of Linear Code|minimum distance]] of $C$ is no more than $5$.
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Golay Ternary Code Corrects 2 Errors}
+Tags: Golay Ternary Code
+
+\begin{theorem}
+The [[Definition:Golay Ternary Code|Golay ternary code]] corrects $2$ [[Definition:Transmission Error|transmission errors]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Golay Ternary Code has Minimum Distance 5]].
+The result follows from [[Error Correction Capability of Linear Code]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Decoding Received Word using Coset Decoding Table}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear code]].
+Let $v$ be a [[Definition:Received Word|received word]], which may have [[Definition:Transmission Error|transmission errors]].
+To find out the [[Definition:Transmitted Codeword|transmitted codeword]] $u$ corresponding to $v$:
+:$(1): \quad$ Find $v$ in the [[Definition:Coset Decoding Table|coset decoding table]] for $C$.
+:$(2): \quad$ The corresponding [[Definition:Transmitted Codeword|transmitted codeword]] $u$ will be found at the top of the [[Definition:Column of Array|column]] where $v$ can be found.
+\end{theorem}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+:$\norm {\, \cdot \,}_1$ is the [[Definition:Trivial Norm on Division Ring|trivial norm]].
+\end{theorem}
+
+\begin{proof}
+We prove the [[Definition:Contrapositive Statement|contrapositive]].
+Let $\norm {\, \cdot \,}_1$ be a [[Definition:Nontrivial Division Ring Norm|nontrivial norm]].
+Then:
+:$\exists y \in R: \norm y_1 \ne 0, \norm y_1 \ne 1$.
+By [[Real Numbers form Ordered Field]] either $\norm y_1 < 1$ or $\norm y_1 > 1$.
+Suppose $\norm y_1 > 1$.
+By [[Definition:Norm on Division Ring|Norm axiom $(\text N 1)$: Positive Definiteness]]:
+:$y \ne 0_R$
+By [[Norm of Inverse in Division Ring]]:
+:$\norm {y^{-1} }_1 = \dfrac 1 {\norm y_1} < 1$
+So either $\norm y_1 < 1$ or $\norm {y^{-1} }_1 < 1$
+So:
+:$\exists x \in R, x \ne 0_R:\norm x_1 < 1$
+The theorem now follows by the [[Rule of Transposition]]. 
+{{qed}}
+[[Category:Equivalence of Definitions of Equivalent Division Ring Norms]]
+m7xz77776pfypxq1omyk1sb2jg57gwo
+\end{proof}<|endoftext|>
+\section{Syndrome is Zero iff Vector is Codeword}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear $\tuple {n, k}$-code]] whose [[Definition:Master Code|master code]] is $\map V {n, p}$
+Let $G$ be a [[Definition:Standard Generator Matrix for Linear Code|(standard) generator matrix]] for $C$.
+Let $P$ be a [[Definition:Standard Parity Check Matrix|standard parity check matrix]] for $C$.
+Let $w \in \map V {n, p}$.
+Then the [[Definition:Syndrome|syndrome]] of $w$ is [[Definition:Zero Codeword|zero]] {{iff}} $w$ is a [[Definition:Codeword of Linear Code|codeword]] of $C$.
+\end{theorem}
+
+\begin{proof}
+Let $G = \paren {\begin{array} {c|c} \mathbf I & \mathbf A \end{array} }$.
+Let $c \in \map V {n, p}$.
+Then, by definition of $G$, $c$ is a [[Definition:Codeword of Linear Code|codeword]] of $C$ {{iff}} $c$ is of the form $u G$, where $u \in \map V {k, p}$.
+Thus $c \in C$ {{iff}}:
+{{begin-eqn}}
+{{eqn | l = c
+      | r = u G
+      | c = 
+}}
+{{eqn | r = u \paren {\begin{array} {c {{!}} c} \mathbf I & \mathbf A \end{array} }
+      | c = 
+}}
+{{eqn | r = \paren {\begin{array} {c {{!}} c} u & v \end{array} }
+      | c = 
+}}
+{{end-eqn}}
+where:
+:$v = u \mathbf A$
+:$\paren {\begin{array} {c|c} u & v \end{array} }$ denotes the $1 \times n$ [[Definition:Matrix|matrix]] formed from the $k$ elements of $u$ and the $n - k$ elements of $v$.
+Let $w \in \map V {n, p}$.
+$w$ can be expressed in the form:
+:$w = \paren {\begin{array} {c|c} u_1 & v_1 \end{array} }$
+where $u_1 \in \map V {k, p}$.
+The [[Definition:Syndrome|syndrome]] of $v$ is then calculated as:
+{{begin-eqn}}
+{{eqn | l = \map S v
+      | r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } w^\intercal
+      | c = 
+}}
+{{eqn | r = \paren {\begin{array} {c {{!}} c} -\mathbf A^\intercal & \mathbf I \end{array} } \paren {\begin{array} {c {{!}} c} u_1^\intercal & v_1^\intercal \end{array} }
+      | c = 
+}}
+{{eqn | r = -\mathbf A^\intercal u_1^\intercal + v_1^\intercal
+      | c = 
+}}
+{{end-eqn}}
+It follows that the [[Definition:Syndrome|syndrome]] of $w$ is [[Definition:Zero Codeword|zero]] {{iff}} $w$ is the [[Definition:Concatenation of Matrices|concatenation]] of $u_1$ and $v_1$, where:
+:$v_1^\intercal = \mathbf A^\intercal u_1^\intercal = \paren {u_1 \mathbf A}^\intercal$
+Thus the [[Definition:Syndrome|syndrome]] of $w$ is [[Definition:Zero Codeword|zero]] {{iff}} $w$ is a [[Definition:Codeword of Linear Code|codeword]] of $C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Vectors to have Same Syndrome}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear $\tuple {n, k}$-code]] whose [[Definition:Master Code|master code]] is $\map V {n, p}$
+Let $G$ be a [[Definition:Standard Generator Matrix for Linear Code|(standard) generator matrix]] for $C$.
+Let $P$ be a [[Definition:Standard Parity Check Matrix|standard parity check matrix]] for $C$.
+Let $u, v \in \map V {n, p}$.
+Then $u$ and $v$ have the same [[Definition:Syndrome|syndrome]] {{iff}} they are in the same [[Definition:Coset|coset]] of $C$.
+\end{theorem}
+
+\begin{proof}
+Let $u, v \in \map V {n, p}$.
+Let $\map S u$ denote the [[Definition:Syndrome|syndrome]] of $u$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map S u
+      | r = \map S v
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = P u^\intercal
+      | r = P v^\intercal
+      | c = {{Defof|Syndrome}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = P \paren {u^\intercal - v^\intercal}
+      | r = \mathbf 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = P \paren {u - v}^\intercal
+      | r = \mathbf 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = u - v
+      | o = \in
+      | r = C
+      | c = [[Syndrome is Zero iff Vector is Codeword]]
+}}
+{{end-eqn}}
+Hence the result from [[Elements in Same Coset iff Product with Inverse in Subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Syndrome Decoding}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear $\tuple {n, k}$-code]] whose [[Definition:Master Code|master code]] is $\map V {n, p}$
+To decode a given [[Definition:Vector (Linear Algebra)|vector]] $v$ of $\map V {n, p}$, the [[Definition:Syndrome|syndrome]] of $v$ can be used as follows.
+Create an [[Definition:Array|array]] $T$ of $2$ [[Definition:Column of Array|column]] consisting of the following:
+:The top [[Definition:Row of Array|row]] contains:
+::in [[Definition:Column of Array|column]] $1$: the [[Definition:Zero Codeword|zero]] of $C$ 
+::in [[Definition:Column of Array|column]] $2$: its [[Definition:Syndrome|syndrome]].
+:The $r$th [[Definition:Row of Array|row]] subsequent contains:
+::in [[Definition:Column of Array|column]] $1$: any [[Definition:Element|element]] of $\map V {n, p}$ of minimum [[Definition:Weight of Linear Codeword|weight]] which is not already included in the first $r - 1$ [[Definition:Row of Array|rows]]
+::in [[Definition:Column of Array|column]] $2$: its [[Definition:Syndrome|syndrome]].
+To decode a given [[Definition:Vector (Linear Algebra)|vector]] $v$ of $\map V {n, p}$:
+:Calculate its [[Definition:Syndrome|syndrome]]
+:Find it in [[Definition:Column of Array|column]] $2$ of $T$
+:See what is in [[Definition:Column of Array|column]] $1$ of $T$, and call it $u$, say
+:Subtract $u$ from $v$.
+\end{theorem}<|endoftext|>
+\section{Euler's Equation/Independent of x}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $y$ be a [[Definition:Mapping|mapping]].
+Let $J$ be a [[Definition:Real Functional|functional]] such that:
+:$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$
+Then the corresponding [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]] can be reduced to:
+:$F - y' F_{y'} = C$ 
+where $C$ is an [[Definition:Arbitrary Constant|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+Assume that:
+:$\ds J \sqbrk y = \int_a^b \map F {y, y'} \rd x$
+Then:
+{{begin-eqn}}
+{{eqn| l = \delta J
+     | r = 0
+}}
+{{eqn| ll= \leadsto
+     | l = F_y - \dfrac \d {\d x} F_{y'}
+     | r = F_y - \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} }
+}}
+{{eqn| r = F_y - y' F_{y'y} - y'' F_{y'y'}
+}}
+{{end-eqn}}
+Multiply this [[Definition:Differential Equation|differential equation]] by $y'$.
+This gives:
+{{begin-eqn}}
+{{eqn | l = F_y y' - {y'}^2 F_{y'y} - y'y'' F_{y'y'}
+      | r = \paren {F_{y} y' + F_{y'} y''} - F_{y'} y'' - y' \paren {y' F_{y'y} + y'' F_{y'y'} }
+}}
+{{eqn | r = \dfrac {\d F} {\d x} - y' \paren {\dfrac {\d y} {\d x} \dfrac {\partial F_{y'} } {\partial y} + \dfrac {\d y'} {\d x} \dfrac {\partial F_{y'} } {\partial y'} } - F_{y'} y''
+}}
+{{eqn | r = \dfrac {\d F} {\d x} - y' \dfrac {\d F} {\d x} - \dfrac {\d y'} {\d x} F_{y'}
+}}
+{{eqn | r = \dfrac \d {\d x} \paren {F - y' F_{y'} }
+}}
+{{eqn | r = 0
+}}
+{{end-eqn}}
+[[Definition:Integration|Integration]] yields the desired result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Equation/Independent of y}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $y$ be a [[Definition:Mapping|mapping]]
+Let $J$ be a [[Definition:Real Functional|functional]] such that
+:$\ds J \sqbrk y = \int_a^b \map F {x,y'} \rd x$
+Then the corresponding [[Definition:Euler's Equation for Vanishing Variation|Euler's equation]] can be reduced to:
+:$F_{y'} = C$ 
+where $C$ is an [[Definition:Arbitrary Constant|arbitrary constant]].
+\end{theorem}
+
+\begin{proof}
+Assume that:
+:$\ds J \sqbrk y = \int_a^b \map F {x,y'} \rd x$
+[[Definition:Euler's Equation for Vanishing Variation|Euler's equation]] for $J$ is:
+:$\dfrac \d {\d x} F_{y'} = 0$
+[[Definition:Integration|Integration]] yields:
+:$F_{y'} = C$ 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Equation/Independent of y'}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $y$ be a [[Definition:Mapping|mapping]].
+Let $J$ a [[Definition:Real Functional|functional]] be such that
+:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
+Then the corresponding [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]] can be reduced to:
+:$F_y = 0$ 
+Furthermore, this is an algebraic [[Definition:Equation|equation]].
+\end{theorem}
+
+\begin{proof}
+Assume that:
+:$\ds J \sqbrk y = \int_a^b \map F {x,y} \rd x$
+Then [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]] for $J$ is:
+:$F_y = 0$
+Since $F$ is independent of $y'$, the [[Definition:Equation|equation]] is algebraic.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Equation/Integrated wrt Length Element}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $y$ be a [[Definition:Real Function|real mapping]] belonging to $C^2$ [[Definition:Differentiability Class|differentiability class]].
+Assume that:
+:$\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \rd s$
+where
+:$\rd s = \sqrt {1 + y'^2} \rd x$
+Then [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]] can be reduced to:
+:$f_y - f_x y' - f_{y'} y' y'' - f \dfrac {y''} {\paren {1 + y'^2}^{\frac 3 2} } = 0$
+\end{theorem}
+
+\begin{proof}
+Substitution of $\rd s$ into $J$ results in the following [[Definition:Real Functional|functional]]:
+:$\ds J \sqbrk y = \int_a^b \map f {x, y, y'} \sqrt {1 + y'^2} \rd x$
+We can consider this as a [[Definition:Real Functional|functional]] with the following effective $F$:
+:$F = \map f {x, y, y'} \sqrt {1 + y'^2}$
+Find [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]]:
+{{begin-eqn}}
+{{eqn | l = F_y - \dfrac \d {\d x} F_{y'}
+      | r = f_y \sqrt {1 + y'^2} - \dfrac \d {\d x} \paren {\map f {x, y, y'} \frac {y'} {\sqrt {1 + y'^2} } }
+}}
+{{eqn | r = f_y \sqrt {1 + y'^2} - \dfrac \d {\d x} \map f {x, y, y'} \frac {y'} {\sqrt {1 + y'^2} } - \map f {x, y, y'} \dfrac \d {\d x} \frac {y'} {\sqrt {1 + y'^2} }
+}}
+{{eqn | r = f_y \sqrt {1 + y'^2} - f_x \frac {y'} {\sqrt {1 + y'^2} } - f_y \frac {y'^2} {\sqrt {1 + y'^2} } - f_{y'} \frac {y' y''} {\sqrt {1 + y'^2} } - f \frac {y'' \sqrt {1 + y'^2} - \frac {y'^2 y''} {\sqrt {1 + y'^2} } } {1 + y'^2}
+}}
+{{eqn | r = \frac 1 {\sqrt {1 + y'^2} } \paren {f_y - f_x y' - f_{y'} y' y'' - f \frac {y''} {\paren {1 + y'^2}^{\frac 3 2} } }
+}}
+{{end-eqn}}
+Due to [[Definition:Assumption|assumptions]] on $y$, the prefactor does not vanish.
+By [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]], the last [[Definition:Expression|expression]] vanishes.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Linear Code with Even Weight Codewords}
+Tags: Linear Codes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Linear Code|linear code]].
+Let $C^+$ be the [[Definition:Subset|subset]] of $C$ consisting of all the [[Definition:Codeword of Linear Code|codewords]] of $C$ which have [[Definition:Even Integer|even]] [[Definition:Weight of Linear Codeword|weight]].
+Then $C^+$ is a [[Definition:Subgroup|subgroup]] of $C$ such that either $C^+ = C$ or such that $\order {C^+} = \dfrac {\order C} 2$.
+\end{theorem}
+
+\begin{proof}
+Note that the [[Definition:Zero Codeword|zero codeword]] is in $C^+$ as it has a [[Definition:Weight of Linear Codeword|weight]] of $0$ which is [[Definition:Even Integer|even]].
+Let $c$ and $d$ be of [[Definition:Even Integer|even]] [[Definition:Weight of Linear Codeword|weight]], where $c$ and $d$ agree in $k$ ordinates.
+Let $\map w c$ denote the [[Definition:Weight of Linear Codeword|weight]] of $c$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map w {c + d}
+      | r = \map w c - k + \map w d - k
+      | c = 
+}}
+{{eqn | r = \map w c + \map w d - 2 k
+      | c = 
+}}
+{{end-eqn}}
+which is [[Definition:Even Integer|even]].
+Since the negative of a [[Definition:Vector (Linear Algebra)|vector]] $\mathbf v$ in $\Z_2$ equals $\mathbf v$, it follows that the [[Definition:Inverse Element|inverse]] of $c \in C$ is also in $C$.
+It follows from the [[Two-Step Subgroup Test]] that $C^+$ is a [[Definition:Subgroup|subgroup]] of $C$.
+Let $C \ne C^+$.
+Then $C$ contains a [[Definition:Codeword of Linear Code|codeword]] $c$ of [[Definition:Odd Integer|odd]] [[Definition:Weight of Linear Codeword|weight]].
+Let $C^-$ denote the [[Definition:Subset|subset]] of $C$ consisting of all the [[Definition:Codeword of Linear Code|codewords]] of $C$ which have [[Definition:Odd Integer|odd]] [[Definition:Weight of Linear Codeword|weight]].
+Adding $c$ to each [[Definition:Codeword of Linear Code|codeword]] of $C^+$ gives distinct [[Definition:Codeword of Linear Code|codewords]] of [[Definition:Odd Integer|odd]] [[Definition:Weight of Linear Codeword|weight]], so:
+:$\order {C^-} \ge \order {C^+}$
+Similarly, adding $c$ to each [[Definition:Codeword of Linear Code|codeword]] of $C^-$ gives distinct [[Definition:Codeword of Linear Code|codewords]] of [[Definition:Even Integer|even]] [[Definition:Weight of Linear Codeword|weight]], so:
+:$\order {C^-} \le \order {C^+}$
+As $C = C^+ \cup C^-$ it follows that:
+:$\order C = 2 \order {C^+}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Smooth Surface of Revolution}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $\map y x$ be a [[Definition:Real Function|real mapping]] in 2-dimensional [[Definition:Real Euclidean Space|real Euclidean space]].
+Let $y$ pass through the [[Definition:Point|points]] $\tuple {x_0, y_0}$ and $\tuple {x_1, y_1}$.
+Consider a [[Definition:Surface of Revolution|surface of revolution]] constructed by rotating $y$ around the $x$-axis.
+Suppose this [[Definition:Surface|surface]] is [[Definition:Smooth Real Function|smooth]] for any $x$ between $x_0$ and $x_1$.
+Then its [[Definition:Surface|surface]] [[Definition:Area|area]] is minimized by the following [[Definition:Curve|curve]], known as a catenoid:
+:$y = C \map \cosh {\dfrac {x + C_1} C}$
+Furthermore, its [[Definition:Area|area]] is:
+:$A = \paren {x_1 - x_0} C \pi + \dfrac {\pi C^2} 2 \paren {\map \sinh {\dfrac {2 \paren {x_1 + C_1} } C} - \map \sinh {\dfrac {2 \paren {x_0 + C_1} } C} }$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Area|area]] [[Definition:Real Functional|functional]] of the [[Definition:Surface of Revolution|surface of revolution]] is:
+:$\ds A \sqbrk y = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x$
+The [[Definition:Integrand|integrand]] does not depend on $x$.
+By [[Euler's Equation/Independent of x|Euler's Equation]]:
+:$F - y' F_{y'} = C$
+that is:
+:$y \sqrt {1 + y'^2} - \dfrac {y y'^2} {\sqrt {1 + y'^2} } = C$
+which is equivalent to:
+:$y = C \sqrt {1 + y'^2}$
+Solving for $y'$ yields:
+ 
+:$y' = \sqrt {\dfrac {y^2 - C^2} {C^2} }$
+Consider this as a [[Definition:Differential Equation|differential equation]] for $\map x y$:
+:$\dfrac {\d x} {\d y} = \dfrac C {\sqrt {y^2 - C^2} }$
+[[Definition:Integration|Integrate]] this once:
+:$x + C_1 = C \ln {\dfrac {y + \sqrt {y^2 - C^2} } C}$
+Solving for $y$ results in:
+:$y = C \cosh {\dfrac {x + C_1} C}$
+To find the [[Definition:Area|area]], substitute this [[Definition:Real Function|function]] into the [[Definition:Formula|formula]] for the [[Definition:Area|area]]:
+{{begin-eqn}}
+{{eqn| l = A \sqbrk y
+     | r = 2 \pi \int_{x_0}^{x_1} y \sqrt {1 + y'^2} \rd x
+}}
+{{eqn| r = 2 \pi \int_{x_0}^{x_1} C \, \map {\cosh^2} {\frac {x + C_1} C} \rd x
+}}
+{{eqn| r = \paren {x_1 - x_0} C \pi + \frac {\pi C^2} 2 \paren {\map \sinh {\frac {2 \paren {x_1 + C_1} } C} - \map \sinh {\frac {2 \paren {x_0 + C_1} } C} }
+}}
+{{end-eqn}}
+{{Finish|Connect this to the fact that not always this minimizes the area}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+:$\forall x \in R: \norm x_1 = \norm x_2^\alpha$
+\end{theorem}
+
+\begin{proof}
+Since $\norm {x_0}_1 < 1$ then $\norm {x_0}_2 < 1$ and:
+:$\log \norm {x_0}_1 < 0$
+:$\log \norm {x_0}_2 < 0$
+Hence $\alpha > 0$
+Since $\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$: 
+===== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.1|Lemma 1]] =====
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.1}}
+===== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2|Lemma 2]] =====
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2}}
+Let $x \in R, x \ne 0_R$.
+===== Case 1 =====
+Let $\norm x_1 = 1$.
+By [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2|Lemma 2]] then:
+:$\norm x_2 = 1$
+Hence:
+:$\norm x_1 = 1 = 1^\alpha = \norm x_2^\alpha$
+{{qed|lemma}} 
+===== Case 2 =====
+Let $\norm x_1 \ne 1$.
+By [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2|Lemma 2]] then:
+:$\norm x_2 \ne 1$
+Let $\beta = \dfrac {\log \norm x_1 } {\log \norm x_2 }$.
+Then $\norm x_1 = \norm x_2^\beta$.
+====== [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3|Lemma 3]] ======
+{{:Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3}}
+Hence $\norm x_1 = \norm x_2^\alpha$.
+</onlyinclude>
+{{qed}}
+[[Category:Equivalence of Definitions of Equivalent Division Ring Norms]]
+d424573c8eu49vw4soo5v51psipeox4
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.1}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Then:
+:$\forall y \in R: \norm y_1 > 1 \iff \norm y_2 > 1$
+\end{theorem}
+
+\begin{proof}
+For $y \in R$ then:
+{{begin-eqn}}
+{{eqn | l = \norm y_1 > 1
+      | r = \dfrac 1 {\norm y_1 } < 1
+      | o = \leadstoandfrom
+      | c = 
+}}
+{{eqn | r = \norm {y^{-1} }_1 < 1 
+      | o = \leadstoandfrom
+      | c = [[Norm of Inverse in Division Ring]]
+}}
+{{eqn | r = \norm {y^{-1} }_2 < 1 
+      | o = \leadstoandfrom
+      | c = by assumption
+}}
+{{eqn | r = \dfrac 1 {\norm y_2 } < 1
+      | o = \leadstoandfrom
+      | c = [[Norm of Inverse in Division Ring]]
+}}
+{{eqn | r = \norm y_2 > 1
+      | o = \leadstoandfrom
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Equivalence of Definitions of Equivalent Division Ring Norms]]
+7c3kw7jre137tryfpo9a96n979a9bxq
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Then:
+:$\forall y \in R:\norm y_1 = 1 \iff \norm y_2 = 1$
+\end{theorem}
+
+\begin{proof}
+By assumption:
+:$\forall y \in R:\norm y_1 \ge 1 \iff \norm y_2 \ge 1$
+By [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1|Lemma 1]]:
+:$\forall y \in R:\norm y_1 \le 1 \iff \norm y_2 \le 1$
+Hence $\forall y \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm y_1 = 1 
+      | o = \leadstoandfrom
+      | r = \norm y_1 \le 1, \norm y_1 \ge 1
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm y_2 \le 1, \norm y_2 \ge 1
+      | c = [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1|Lemma 1]] and by assumption
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm y_2 = 1
+}}
+{{end-eqn}} 
+{{qed}}
+[[Category:Equivalence of Definitions of Equivalent Division Ring Norms]]
+i3p50dq69v4rnr033jf1x68v04xmtjz
+\end{proof}<|endoftext|>
+\section{Symmetric Group on 3 Letters is Isomorphic to Dihedral Group D3}
+Tags: Symmetric Group on 3 Letters, Dihedral Group D3, Examples of Group Isomorphisms
+
+\begin{theorem}
+Let $S_3$ denote the [[Symmetric Group on 3 Letters]].
+Let $D_3$ denote the [[Definition:Dihedral Group D3|dihedral group $D_3$]].
+Then $S_3$ is [[Definition:Isomorphism|isomorphic]] to $D_3$.
+\end{theorem}
+
+\begin{proof}
+Consider $S_3$ as presented by its [[Symmetric Group on 3 Letters/Cayley Table|Cayley table]]:
+{{:Symmetric Group on 3 Letters/Cayley Table}}
+Consider $D_3$ as presented by its [[Group Presentation of Dihedral Group D3|group presentation]]:
+{{:Group Presentation of Dihedral Group D3}}
+and its [[Dihedral Group D3/Cayley Table|Cayley table]]:
+{{:Dihedral Group D3/Cayley Table}}
+Let $\phi: S_3 \to D_3$ be specified as:
+{{begin-eqn}}
+{{eqn | l = \map \phi {1 2 3}
+      | r = a
+      | c = 
+}}
+{{eqn | l = \map \phi {2 3}
+      | r = b
+      | c = 
+}}
+{{end-eqn}}
+Then by inspection, we see:
+{{begin-eqn}}
+{{eqn | l = \map \phi {1 3 2}
+      | r = a^2
+      | c = 
+}}
+{{eqn | l = \map \phi {1 3}
+      | r = a b
+      | c = 
+}}
+{{eqn | l = \map \phi {1 2}
+      | r = a^2 b
+      | c = 
+}}
+{{end-eqn}}
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism from Group of Cube Roots of Unity to Itself}
+Tags: Multiplicative Groups of Complex Roots of Unity, Cyclic Group of Order 3
+
+\begin{theorem}
+Let $\struct {U_3, \times}$ denote the [[Definition:Multiplicative Group of Complex Roots of Unity|multiplicative group of the complex cube roots of unity]].
+Here, $U_3 = \set {1, \omega, \omega^2}$ where $\omega = e^{2 i \pi / 3}$.
+Let $\phi: U_3 \to U_3$ be defined as:
+:$\forall z \in U_3: \map \phi z = \begin{cases} 1 & : z = 1 \\ \omega^2 & : z = \omega \\ \omega & : z = \omega^2 \end{cases}$
+Then $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
+\end{theorem}
+
+\begin{proof}
+It is noted that
+:$\paren {\omega^2}^2 = \omega$
+and so $\phi$ is the [[Definition:Square Function|square function]].
+By [[Roots of Unity under Multiplication form Cyclic Group]] and [[Cyclic Group is Abelian]], $U_3$ is [[Definition:Abelian Group|abelian]].
+Thus for all $a, b \in U_3$:
+{{begin-eqn}}
+{{eqn | l = \map \phi a \map \phi b
+      | r = a^2 b^2
+}}
+{{eqn | r = a b a b
+      | c = {{Defof|Abelian Group}}
+}}
+{{eqn | r = \paren {a b}^2
+}}
+{{eqn | r = \map \phi {a b}
+}}
+{{end-eqn}}
+showing that $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+:$\alpha = \beta$
+\end{theorem}
+
+\begin{proof}
+Because $x, y \in R \setminus 0_R$:
+:$\norm x_1 , \norm y_1, \norm x_2 , \norm y_2 > 0$.
+Because $\norm{x}_1 , \norm {y}_1 \ne 1$, by [[Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2|Lemma 2]]:
+:$\norm x_2 , \norm y_2 \ne 1$.
+Hence:
+:$\log \norm x_1 , \log \norm y_1, \log \norm x_2, \log \norm y_2 \ne 0$ 
+and $\alpha, \beta$ are well-defined.
+Let $r = \dfrac n m \in \Q$ be any [[Definition:Rational Number|rational number]] where $n, m \in \Z$ are [[Definition:Integer|integers]] and $m \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm y_1^n < \norm x_1^m
+      | o = \leadstoandfrom
+      | r = \dfrac {\norm y_1^n} {\norm x_1^m} < 1
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm y_1^n \norm {x^{-1} }_1^m
+      | c = [[Norm of Inverse in Division Ring]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y^n}_1 \norm {\paren {x^{-1} }^m}_1 < 1 
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y^n \paren {x^{-1} }^m}_1 < 1 
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y^n \paren {x^{-1} }^m}_2 < 1 
+      | c = By assumption
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y^n}_2 \norm {\paren {x^{-1} }^m}_2 < 1 
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm y_2^n \norm {x^{-1} }_2^m
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \dfrac {\norm y_2^n} {\norm x_2^m} < 1
+      | c = [[Norm of Inverse in Division Ring]]
+}}
+{{eqn | r = \norm y_2^n < \norm x_2^m
+      | o = \leadstoandfrom
+}}
+{{end-eqn}}
+By [[Logarithm is Strictly Increasing]]:
+:$\log \norm y_1^n < \log \norm x_1^m \iff \log \norm y_2^n < \log \norm x_2^m$
+By [[Sum of Logarithms]]:
+:$n \log \norm y_1 < m \log \norm x_1 \iff n \log \norm y_2 < m \log \norm x_2$
+Because $m, \log \norm x_1, \log \norm x_2 \ne 0$:
+:$r = \dfrac n m < \dfrac {\log \norm x_1} {\log \norm y_1} \iff r = \dfrac n m < \dfrac {\log \norm x_2} {\log \norm y_2}$
+By [[Between two Real Numbers exists Rational Number]]:
+:$\dfrac {\log \norm x_1} {\log \norm y_1} = \dfrac {\log \norm x_2} {\log \norm y_2}$
+Because $\log \norm x_2, \log \norm y_2 \ne 0$:
+:$\dfrac {\log \norm x_1} {\log \norm x_2} = \dfrac {\log \norm y_1} {\log \norm y_2}$
+That is:
+:$\alpha = \beta$
+{{qed}}
+[[Category:Equivalence of Definitions of Equivalent Division Ring Norms]]
+0w9fh9fzzjpsmeh7kgmw6d8fc585lm2
+\end{proof}<|endoftext|>
+\section{Product with Inverse on Homomorphic Image is Group Homomorphism}
+Tags: Group Homomorphisms, Group Direct Products, Product with Inverse on Homomorphic Image is Group Homomorphism
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be an [[Definition:Abelian Group|abelian group]].
+Let $\theta: G \to H$ be a [[Definition:Group Homomorphism|(group) homomorphism]].
+Let $\phi: G \times G \to H$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {g_1, g_2} \in G \times G: \map \phi {g_1, g_2} = \map \theta {g_1} \map \theta {g_2}^{-1}$
+Then $\phi$ is a [[Definition:Group Homomorphism|homomorphism]].
+\end{theorem}
+
+\begin{proof}
+First note that from [[Group Homomorphism Preserves Inverses]]:
+:$\map \theta {g_2}^{-1} = \paren {\map \theta {g_2} }^{-1} = \map \theta { {g_2}^{-1} }$
+and so $\map \theta {g_1} \map \theta {g_2}^{-1}$ is not [[Definition:Ambiguous|ambiguous]]:
+:$\map \theta {g_1} \map \theta {g_2}^{-1} = \map \theta {g_1} \paren {\map \theta {g_2} }^{-1} = \map \theta {g_1} \map \theta { {g_2}^{-1} }$
+From [[External Direct Product of Groups is Group]], $G \times G$ is a [[Definition:Group|group]].
+Let $a_1, a_2, b_1, b_2 \in G$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \phi {a_1 b_1, a_2 b_2}
+      | r = \map \theta {a_1 b_1} \map \theta {a_2 b_2}^{-1}
+      | c = 
+}}
+{{eqn | r = \map \theta {a_1 b_1} \paren {\map \theta {a_2 b_2} }^{-1}
+      | c = [[Group Homomorphism Preserves Inverses]]
+}}
+{{eqn | r = \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {a_2} \map \theta {b_2} }^{-1}
+      | c = {{Defof|Group Homomorphism}}
+}}
+{{eqn | r = \map \theta {a_1} \map \theta {b_1} \paren {\map \theta {b_2}^{-1} } \paren {\map \theta {a_2}^{-1} }
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \map \theta {a_1} \paren {\map \theta {a_2}^{-1} } \map \theta {b_1} \paren {\map \theta {b_2}^{-1} }
+      | c = {{Defof|Abelian Group}}: $H$ is [[Definition:Abelian Group|Abelian]]
+}}
+{{eqn | r = \map \phi {a_1, a_2} \map \phi {b_1, b_2}
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Group Homomorphism|homomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Max is Associative}
+Tags: Max and Min Operations
+
+\begin{theorem}
+The [[Definition:Max Operation|Max]] operation is [[Definition:Associative Operation|associative]]:
+: $\max \left({\max \left({x, y}\right), z}\right) = \max \left({x, \max \left({y, z}\right)}\right)$
+Thus we are justified in writing $\max \left({x, y, z}\right)$.
+\end{theorem}
+
+\begin{proof}
+To simplify our notation:
+: Let $\max \left({x, y}\right)$ be (temporarily) denoted $x \overline \wedge y$
+There are the following cases to consider:
+:$(1): \quad x \le y \le z$
+:$(2): \quad x \le z \le y$
+:$(3): \quad y \le x \le z$
+:$(4): \quad y \le z \le x$
+:$(5): \quad z \le x \le y$
+:$(6): \quad z \le y \le x$
+Taking each one in turn:
+:$(1): \quad$ Let $x \le y \le z$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge z
+      | rr= = z
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = y \overline \wedge z
+      | rr= = z
+}}
+{{end-eqn}}
+:$(2): \quad$ Let $x \le z \le y$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge y
+      | rr= = y
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = y \overline \wedge z
+      | rr= = y
+}}
+{{end-eqn}}
+:$(3): \quad$ Let $y \le x \le z$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge z
+      | rr= = z
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = x \overline \wedge z
+      | rr= = z
+}}
+{{end-eqn}}
+:$(4): \quad$ Let $y \le z \le x$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge z
+      | rr= = x
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = x \overline \wedge z
+      | rr= = x
+}}
+{{end-eqn}}
+:$(5): \quad$ Let $z \le x \le y$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge y
+      | rr= = y
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = y \overline \wedge z
+      | rr= = y
+}}
+{{end-eqn}}
+:$(6): \quad$ Let $z \le y \le x$. Then:
+{{begin-eqn}}
+{{eqn | l = x \overline \wedge \left({y \overline \wedge z}\right)
+      | r = x \overline \wedge y
+      | rr= = x
+}}
+{{eqn | l = \left({x \overline \wedge y}\right) \overline \wedge z
+      | r = x \overline \wedge z
+      | rr= = x
+}}
+{{end-eqn}}
+Thus in all cases it can be seen that the result holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Min is Associative}
+Tags: Max and Min Operations
+
+\begin{theorem}
+The [[Definition:Min Operation|Min]] operation is [[Definition:Associative Operation|associative]]:
+: $\min \left({\min \left({x, y}\right), z}\right) = \min \left({x, \min \left({y, z}\right)}\right)$
+Thus we are justified in writing $\min \left({x, y, z}\right)$.
+\end{theorem}
+
+\begin{proof}
+To simplify our notation:
+: Let $\min \left({x, y}\right)$ be (temporarily) denoted $x \underline \vee y$.
+There are the following cases to consider:
+:$(1): \quad x \le y \le z$
+:$(2): \quad x \le z \le y$
+:$(3): \quad y \le x \le z$
+:$(4): \quad y \le z \le x$
+:$(5): \quad z \le x \le y$
+:$(6): \quad z \le y \le x$
+Taking each one in turn:
+:$(1): \quad$ Let $x \le y \le z$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = x \underline \vee y
+      | rr= = x
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = x \underline \vee z
+      | rr= = x
+}}
+{{end-eqn}}
+:$(2): \quad$ Let $x \le z \le y$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = x \underline \vee z
+      | rr= = x
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = x \underline \vee z
+      | rr= = x
+}}
+{{end-eqn}}
+:$(3): \quad$ Let $y \le x \le z$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = x \underline \vee y
+      | rr= = y
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = y \underline \vee z
+      | rr= = y
+}}
+{{end-eqn}}
+:$(4): \quad$ Let $y \le z \le x$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = x \underline \vee y
+      | rr= = y
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = y \underline \vee z
+      | rr= = y
+}}
+{{end-eqn}}
+:$(5): \quad$ Let $z \le x \le y$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = x \underline \vee z
+      | rr= = z
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = x \underline \vee z
+      | rr= = z
+}}
+{{end-eqn}}
+:$(6): \quad$ Let $z \le y \le x$. Then:
+{{begin-eqn}}
+{{eqn | l = x \underline \vee \left({y \underline \vee z}\right)
+      | r = y \underline \vee z
+      | rr= = z
+}}
+{{eqn | l = \left({x \underline \vee y}\right) \underline \vee z
+      | r = y \underline \vee z
+      | rr= = z
+}}
+{{end-eqn}}
+Thus in all cases it can be seen that the result holds.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Theoretical Justification for Cycle Notation}
+Tags: Permutation Theory
+
+\begin{theorem}
+Let $\N_k$ be used to denote the [[Definition:Initial Segment of Natural Numbers|initial segment of natural numbers]]:
+:$\N_k = \closedint 1 k = \set {1, 2, 3, \ldots, k}$
+Let $\rho: \N_n \to \N_n$ be a [[Definition:Permutation on n Letters|permutation of $n$ letters]].
+Let $i \in \N_n$.
+Let $k$ be the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] for which $\map {\rho^k} i$ is in the [[Definition:Set|set]]:
+:$\set {i, \map \rho i, \map {\rho^2} i, \ldots, \map {\rho^{k - 1} } i}$
+Then:
+:$\map {\rho^k} i = i$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\map {\rho^k} i = \map {\rho^r} i$ for some $r > 0$.
+As $\rho$ has an [[Definition:Inverse Element|inverse]] in $S_n$:
+:$\map {\rho^{k - r} } i = i$
+This [[Definition:Contradiction|contradicts]] the definition of $k$, because $k - r < k$
+Thus:
+:$r = 0$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
+Then $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent metrics]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in R$ and $\epsilon \in \R_{\gt 0}$
+Then for $y \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm {y - x}_1 < \epsilon
+      | o = \leadstoandfrom
+      | r = \norm {y - x}_2^\alpha < \epsilon
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y - x}_2 < \epsilon^{1 / \alpha}
+}}
+{{end-eqn}}
+Hence:
+:$\map {B^1_\epsilon} x = \map {B^2_{\epsilon^{1 / \alpha} } } x$
+where:
+:$\map {B^1_\epsilon} x$ is the [[Definition:Open Ball|open ball]] in $d_1$ [[Definition:Center of Open Ball|centered]] on $x$ with [[Definition:Radius of Open Ball|radius]] $\epsilon$
+:$\map {B^2_{\epsilon^{1 / \alpha} } } x$ is the [[Definition:Open Ball|open ball]] in $d_2$ [[Definition:Center of Open Ball|centered]] on $x$ with [[Definition:Radius of Open Ball|radius]] $\epsilon^{1 / \alpha}$
+Since $x$ and $\epsilon$ were arbitrary then:
+:every [[Definition:Open Ball|$d_1$-open ball]] is a [[Definition:Open Ball|$d_2$-open ball]].
+Similarly, for $y \in R$:
+{{begin-eqn}}
+{{eqn | l = \norm {y - x}_2 < \epsilon
+      | o = \leadstoandfrom
+      | r = \norm {y - x}_2^\alpha < \epsilon^\alpha
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \norm {y - x}_1 < \epsilon^\alpha
+}}
+{{end-eqn}}
+So:
+:every [[Definition:Open Ball|$d_2$-open ball]] is a [[Definition:Open Ball|$d_1$-open ball]].
+By the definition of an [[Definition:Open Set of Metric Space|open set of a metric space]] it follows that $d_1$ and $d_2$ are [[Definition:Topologically Equivalent Metrics|topologically equivalent metrics]],
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $R$ be a [[Definition:Division Ring|division ring]].
+Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be [[Definition:Norm on Division Ring|norms]] on $R$.
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:for all sequences $\sequence {x_n}$ in $R$: $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$
+Then $\forall x \in R$:
+:$\norm x_1 < 1 \iff \norm x_2 < 1$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Contrapositive|contrapositive]] is proved.
+Let there exist $x \in R$ such that $\norm x_1 < 1$ and $\norm x_2 \ge 1$.
+Let $\sequence {x_n}$ be the [[Definition:Sequence|sequence]] defined by: $\forall n: x_n = x^n$.
+By [[Sequence of Powers of Number less than One/Normed Division Ring|Sequence of Powers of Number less than One in Normed Division Ring]] then $\sequence {x_n}$ is a [[Definition:Null Sequence in Normed Division Ring|null sequence]] in $\norm {\, \cdot \,}_1$.
+By [[Convergent Sequence is Cauchy Sequence/Normed Division Ring|convergent sequence in normed division ring is a Cauchy sequence]] then $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$.
+Let $0_R$ be the [[Definition:Ring Zero|zero]] of $R$ and $1_R$ be the [[Definition:Unit of Ring|unit]] of $R$.
+By [[Norm of Unity of Division Ring]] and the assumption $\norm x_1 < 1$:
+:$x \ne 1_R$
+Hence:
+:$x - 1_R \ne 0_R$
+By [[Definition:Norm on Division Ring|norm axiom $(\text N 1)$: Positive Definiteness]]:
+:$\norm {x - 1_R}_2 > 0$
+Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$.
+Then $\norm {x - 1_R}_2 > \epsilon$.
+We have that $\norm x_2 \ge 1$.
+Hence for all $n \in \N$:
+{{begin-eqn}}
+{{eqn | l = \norm {x_n}_2
+      | r = \norm {x^n}_2
+      | c = Definition of $x_n$
+}}
+{{eqn | r = \norm x_2^n
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = \ge
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+For all $n \in \N$:
+{{begin-eqn}}
+{{eqn | l = \norm {x_{n + 1} - x_n}_2
+      | r = \norm {x^{n + 1} - x^n}_2
+}}
+{{eqn | r = \norm {x^n x - x^n}_2
+}}
+{{eqn | r = \norm {x^n \paren {x - 1_R} }_2
+}}
+{{eqn | r = \norm {x^n}_2 \norm {x - 1_R}_2
+      | c = [[Definition:Norm on Division Ring|Norm Axiom $(\text N 2)$: Multiplicativity]]
+}}
+{{eqn | o = >
+      | r = \epsilon
+}}
+{{end-eqn}} 
+So $\sequence {x_n}$ is not a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$.
+The theorem now follows by the [[Rule of Transposition]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Order Elements may not be Conjugate}
+Tags: Order of Group Elements, Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]]
+Let $x, y \in G$ be [[Definition:Element|elements]] of $G$ such that:
+:$\order x = \order y$
+where $\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$.
+Then it is not necessarily the case that $x$ and $y$ are [[Definition:Conjugate of Group Element|conjugates]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Dihedral Group D4|dihedral group $D_4$]], whose [[Group Presentation of Dihedral Group D4|group presentation]] is:
+{{:Group Presentation of Dihedral Group D4}}
+We have that:
+:$\order {a^2} = 2$
+and:
+:$\order b = 2$
+but $a^2$ and $b$ are not [[Definition:Conjugate of Group Element|conjugate]] to each other.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group on Complex Vector Space}
+Tags: Examples of Group Actions, Group Action of Symmetric Group on Complex Vector Space, Symmetric Groups
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $V$ denote a [[Definition:Vector Space|vector space]] over the [[Definition:Complex Number|complex numbers $\C$]].
+Let $V$ have a [[Definition:Basis of Vector Space|basis]]:
+:$\mathcal B := \set {v_1, v_2, \ldots, v_n}$
+Let $*: S_n \times V \to V$ be a [[Definition:Group Action|group action]] of $S_n$ on $V$ defined as:
+:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
+where:
+: $v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+Let $\rho, \sigma \in S_n$.
+Let $v = \displaystyle \sum_{k \mathop = 1}^n \lambda_k v_k$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \rho * \paren {\sigma * v}
+      | r = \rho * \paren {\sigma * \sum_{k \mathop = 1}^n \lambda_k v_k}
+      | c = Definition of $v$
+}}
+{{eqn | r = \rho * \sum_{k \mathop = 1}^n \lambda_k v_{\map \sigma k}
+      | c = Definition of $*$
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho {\map \sigma k} }
+      | c = Definition of $*$
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map {\rho \mathop \circ \sigma} k}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \paren {\rho \circ \sigma} * \sum_{k \mathop = 1}^n \lambda_k v_k
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {\rho \circ \sigma} * v
+      | c = Definition of $v$
+}}
+{{end-eqn}}
+Hence $*$ fulfils {{GroupActionAxiom|1}}.
+Let $e$ denote the [[Definition:Identity Element|identity element]] of $S_n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * v
+      | r = e * \sum_{k \mathop = 1}^n \lambda_k v_k
+      | c = Definition of $v$
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map e k}
+      | c = Definition of $*$
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_k
+      | c = Definition of $e$
+}}
+{{eqn | r = v
+      | c = Definition of $v$
+}}
+{{end-eqn}}
+Hence $*$ fulfils {{GroupActionAxiom|2}}.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group on Complex Vector Space/Orbit}
+Tags: Group Action of Symmetric Group on Complex Vector Space
+
+\begin{theorem}
+The [[Definition:Orbit (Group Theory)|orbit]] of an [[Definition:Element|element]] $v \in V$ is:
+:$\Orb v = \displaystyle \set {w \in V: \exists \rho \in S_n: w = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho k} }$
+\end{theorem}
+
+\begin{proof}
+By definition:
+{{begin-eqn}}
+{{eqn | l = \Orb v
+      | r = \set {w \in V: \exists \rho \in S_n: w = \rho * v}
+      | c = {{Defof|Orbit (Group Theory)|Orbit}}
+}}
+{{eqn | r = \set {w \in V: \exists \rho \in S_n: w = \rho * \sum_{k \mathop = 1}^n \lambda_k v_k}
+      | c = Definition of $v$
+}}
+{{eqn | r = \set {w \in V: \exists \rho \in S_n: w = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho k} }
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action of Symmetric Group on Complex Vector Space/Stabilizer}
+Tags: Stabilizers, Group Action of Symmetric Group on Complex Vector Space
+
+\begin{theorem}
+The [[Definition:Stabilizer|stabilizer]] of an [[Definition:Element|element]] $v \in V$ is:
+:$\Stab v = \displaystyle \set {\rho \in S_n: \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_{\map \rho k} v_k}$
+\end{theorem}
+
+\begin{proof}
+By definition:
+{{begin-eqn}}
+{{eqn | l = \Stab v
+      | r = \set {\rho \in S_n: \rho * v = v}
+      | c = {{Defof|Stabilizer}}
+}}
+{{eqn | r = \set {\rho \in S_n: \rho * \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_k v_k}
+      | c = Definition of $v$
+}}
+{{eqn | r = \set {\rho \in S_n: \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_{\map \rho k} v_k}
+      | c = Definition of $*$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent}
+Tags: Equivalence of Definitions of Equivalent Division Ring Norms
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
+Then for all sequences $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$ {{iff}} $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$.
+Let $\epsilon > 0$ be given.
+Since $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] then:
+:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$
+Then:
+:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$
+Hence:
+:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$
+So $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$
+It follows that for all sequences $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$
+{{qed|lemma}}
+Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2$.
+Let $\epsilon > 0$ be given.
+Since $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] then:
+:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$
+Then:
+:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$
+Hence:
+:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$
+So $\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$
+It follows that for all sequences $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm {\, \cdot \,}_1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Classes of Symmetric Group}
+Tags: Symmetric Groups, Conjugacy Classes
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+The [[Definition:Conjugacy Class|conjugacy classes]] of $S_n$ are determined entirely by the [[Definition:Cycle Type|cycle type]].
+That is, the [[Definition:Conjugacy Class|conjugacy class]] $\conjclass x$ of an [[Definition:Element|element]] $x$ of $S_n$ consists of all the [[Definition:Element|elements]] of $S_n$ whose [[Definition:Cycle Type|cycle type]] is the same as the [[Definition:Cycle Type|cycle type]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\sigma \in S_n$ have [[Definition:Cycle Type|cycle type]] $\tuple {k_1, k_2, \ldots, k_n}$.
+Let $\rho$ be [[Definition:Conjugate of Group Element|conjugate]] to $\sigma$
+From [[Conjugate Permutations have Same Cycle Type]], $\rho$ has the same [[Definition:Cycle Type|cycle type]] $\tuple {k_1, k_2, \ldots, k_n}$ as $\sigma$.
+That is, all the [[Definition:Element|elements]] of the same [[Definition:Conjugacy Class|conjugacy class]] have the same [[Definition:Cycle Type|cycle type]].
+{{qed|lemma}}
+Let $\sigma, \rho \in S_n$ have the same [[Definition:Cycle Type|cycle type]] $\tuple {k_1, k_2, \ldots, k_n}$.
+It is to be demonstrated that $\sigma$ and $\rho$ are in the same [[Definition:Conjugacy Class|conjugacy class]].
+From [[Existence and Uniqueness of Cycle Decomposition]], $\sigma$ and $\rho$ can each be expressed uniquely as the [[Definition:Product Element|product]] of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]]:
+{{begin-eqn}}
+{{eqn | l = \sigma
+      | r = \alpha_1 \alpha_2 \dotsm \alpha_l
+}}
+{{eqn | l = \rho
+      | r = \beta_1 \beta_2 \dotsm \beta_l
+}}
+{{end-eqn}}
+where $\alpha_i$ and $\beta_i$ are [[Definition:Cyclic Permutation|$k_i$-cycles]].
+For each $i$, let the [[Definition:Cyclic Permutation|$k_i$-cycles]] $\alpha_i$ and $\beta_i$ be expressed as:
+{{begin-eqn}}
+{{eqn | l = \alpha_i
+      | r = \alpha_{i1} \alpha_{i2} \dotsm \alpha_{i k_l}
+}}
+{{eqn | l = \beta_i
+      | r = \beta_{i1} \beta_{i2} \dotsm \beta_{i k_l}
+}}
+{{end-eqn}}
+For all $i, j$ such that $1 \le i \le l$, $1 \le j \le k_i$, let:
+:$\tau := \map \tau {\alpha_{i j} } = \beta_{i j}$
+Such a $\tau$ is bound to exist in $S_n$, as the [[Definition:Underlying Set of Structure|underlying set]] of $S_n$ is the [[Definition:Set|set]] of all [[Definition:Permutation|permutations]] of $\set {1, 2, \ldots, n}$.
+Thus from [[Product of Conjugates equals Conjugate of Products]]:
+:$\tau \alpha_i \tau^{-1} = \beta_i$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \tau \sigma \tau^{-1}
+      | r = \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1}
+}}
+{{eqn | r = \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1}
+}}
+{{eqn | r = \beta_1 \beta_2 \dotsm \beta_l
+}}
+{{eqn | r = \rho
+}}
+{{end-eqn}}
+demonstrating that $\sigma$ and $\rho$ are [[Definition:Conjugate of Group Element|conjugate]].
+That is, $\sigma$ and $\rho$ are in the same [[Definition:Conjugacy Class|conjugacy class]].
+Hence the result.
+{{qed}}
+[[Category:Symmetric Groups]]
+[[Category:Conjugacy Classes]]
+lqejkraisnceer65vizzoa9d6n6znq1
+\end{proof}<|endoftext|>
+\section{Identity of Group is in Center}
+Tags: Centers of Groups, Identity Elements
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
+Then $e$ is in the [[Definition:Center of Group|center]] of $G$:
+:$e \in \map Z G$
+\end{theorem}
+
+\begin{proof}
+From [[Center is Intersection of Centralizers]]:
+:$\displaystyle \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$
+where $\map {C_G} g$ denotes the [[Definition:Centralizer of Group Element|centralizer]] of $g$.
+From [[Centralizer of Group Element is Subgroup]], each of $\map {C_G} g$ is a [[Definition:Subgroup|subgroup]] of $G$.
+From [[Identity of Subgroup]]:
+:$\forall g \in G: e \in \map {C_G} g$
+Hence by definition of [[Definition:Set Intersection|set intersection]]:
+:$e \in \displaystyle \bigcap_{g \mathop \in G} \map {C_G} g$
+whence the result.
+{{qed}}
+[[Category:Centers of Groups]]
+[[Category:Identity Elements]]
+kgz29nmgfr9rmlqsa8b3we40xlppoae
+\end{proof}<|endoftext|>
+\section{Identity of Group is in Singleton Conjugacy Class}
+Tags: Identity Elements, Conjugacy Classes
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
+Then $e$ is in its own [[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy class]]:
+:$\conjclass e = \set e$
+\end{theorem}
+
+\begin{proof}
+From [[Identity of Group is in Center]]:
+:$e \in \map Z G$
+where $\map Z G$ is the [[Definition:Center of Group|center]] of $G$.
+From [[Conjugacy Class of Element of Center is Singleton]]:
+:$\conjclass e = \set e$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Group with 2 Conjugacy Classes has 2 Elements}
+Tags: Conjugacy Classes
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $G$ have exactly $2$ [[Definition:Conjugacy Class|conjugacy classes]].
+Then $G$ has exactly $2$ [[Definition:Element|elements]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be of [[Definition:Order of Group|order]] $n$.
+Let $G$ have exactly $2$ [[Definition:Conjugacy Class|conjugacy classes]].
+Let $x \in G$ such that $x \ne e$.
+Let $\conjclass x$ denote the [[Definition:Conjugacy Class|conjugacy class]] of $x$.
+From [[Identity of Group is in Singleton Conjugacy Class]]:
+$\conjclass e = \set e$
+where $\conjclass e$ denotes the [[Definition:Conjugacy Class|conjugacy class]] of $e$
+The other [[Definition:Element|elements]] of $G$ are in $\conjclass x$.
+Thus:
+:$\card {\conjclass x} = n - 1$
+where $\card {\, \cdot \,}$ denotes the [[Definition:Cardinality|cardinality]] of a [[Definition:Set|set]].
+We have that [[Size of Conjugacy Class is Index of Normalizer]].
+From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Index of Subgroup|index]] of any [[Definition:Subgroup|subgroup]] of $G$ is a [[Definition:Divisor of Integer|divisor]] of $G$.
+Thus:
+:$\paren {n - 1} \divides n$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n = 2$.
+Hence the result by definition of [[Definition:Order of Group|order of group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 15 has Cyclic Subgroups of Order 3 and Order 5}
+Tags: Groups of Order 15
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $15$.
+Then $G$ has
+:a [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] of [[Definition:Order of Group|order]] $3$
+and:
+:a [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] of [[Definition:Order of Group|order]] $5$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $15$.
+We have that $15 = 3 \times 5$.
+Thus from the [[First Sylow Theorem]]:
+:$G$ has at least one [[Definition:Subgroup|subgroup]] $H_3$ of [[Definition:Order of Group|order]] $3$
+and:
+:$G$ has at least one [[Definition:Subgroup|subgroup]] $H_5$ of [[Definition:Order of Group|order]] $5$.
+From [[Prime Group is Cyclic]], all such [[Definition:Subgroup|subgroups]] of [[Definition:Order of Group|order]] $3$ and [[Definition:Order of Group|order]] $5$ are [[Definition:Cyclic Group|cyclic]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Number of Sylow p-Subgroups in Group of Order 15}
+Tags: Groups of Order 15, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $15$.
+Then:
+:the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] is in the [[Definition:Set|set]] $\set {1, 4, 7, \ldots}$
+:the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] is in the [[Definition:Set|set]] $\set {1, 6, 11, \ldots}$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $15$.
+From the [[Fourth Sylow Theorem]]:
+:the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] is equivalent to $1 \pmod p$
+We have that $15 = 3 \times 5$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Sylow p-Subgroups is Sylow p-Subgroup}
+Tags: Group Direct Products, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G_1$ and $G_2$ be [[Definition:Group|groups]].
+Let $H_1$ and $H_2$ be [[Definition:Subgroup|subgroups]] of $G_1$ and $G_2$ respectively.
+Let $H_1$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1$.
+Let $H_2$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_2$.
+Then $H_1 \times H_2$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1 \times G_2$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]]:
+:$\order {H_1} = p^r$, where $p^r$ is the highest [[Definition:Integer Power|power]] of $p$ which is a [[Definition:Divisor of Integer|divisor]] of $\order {G_1}$.
+:$\order {H_2} = p^s$, where $p^s$ is the highest [[Definition:Integer Power|power]] of $p$ which is a [[Definition:Divisor of Integer|divisor]] of $\order {G_2}$.
+We have that:
+:$\order {H_1 \times H_2} = p^{r + s}$
+We also have that $p^{r + s}$ is the highest [[Definition:Integer Power|power]] of $p$ which is a [[Definition:Divisor of Integer|divisor]] of $\order {G_1 \times G_2}$.
+Hence it follows by definition that $H_1 \times H_2$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1 \times G_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Unique Sylow p-Subgroups is Unique Sylow p-Subgroup}
+Tags: Group Direct Products, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G_1$ and $G_2$ be [[Definition:Group|groups]].
+Let $H_1$ and $H_2$ be [[Definition:Subgroup|subgroups]] of $G_1$ and $G_2$ respectively.
+Let $G_1$ be such that $H_1$ is the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1$.
+Let $G_2$ be such that $H_2$ is the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_2$.
+Then $H_1 \times H_2$ is the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1 \times G_2$.
+\end{theorem}
+
+\begin{proof}
+From [[Direct Product of Sylow p-Subgroups is Sylow p-Subgroup]], $H_1 \times H_2$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1 \times G_2$.
+By [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], each of $H_1$ and $H_2$ are [[Definition:Normal Subgroup|normal]] in $G_1$ and $G_2$ respectively.
+By [[Direct Product of Normal Subgroups is Normal]], $H_1 \times H_2$ is [[Definition:Normal Subgroup|normal]] in $G_1 \times G_2$.
+Again by [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $H_1 \times H_2$ is the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G_1 \times G_2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Sylow p-Subgroup with Subgroup not necessarily Sylow p-Subgroup}
+Tags: Sylow p-Subgroups, Set Intersection
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $P \cap H$ is not necessarily a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $H$.
+\end{theorem}
+
+\begin{proof}
+We note that from [[Intersection of Subgroups is Subgroup]] that $P \cap H$ is a [[Definition:Subgroup|subgroup]] of $G$ and also of $H$.
+Let $G$ be the [[Definition:Dihedral Group D3|dihedral group $D_3$]], given by its [[Group Presentation of Dihedral Group D3|group presentation]]:
+{{:Group Presentation of Dihedral Group D3}}
+By definition of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]], $\gen a$ is a [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] of $G$.
+However, $\gen b$ is also a [[Definition:Subgroup|subgroup]] of $G$, of [[Definition:Order of Group|order $2$]].
+But:
+:$\gen b \cap \gen a = e$
+and $e$ is not a [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sylow p-Subgroups of Group of Order 2p}
+Tags: Sylow p-Subgroups, Groups of Order 2 p, Sylow p-Subgroups of Group of Order 2p
+
+\begin{theorem}
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $2 p$.
+Then $G$ has [[Definition:Unique|exactly one]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]].
+This [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+Let $n_p$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_p \equiv 1 \pmod p$
+From the [[Fifth Sylow Theorem]]:
+:$n_p \divides 2 p$
+that is:
+:$n_p \in \set {1, 2, p, 2 p}$
+But $p$ and $2 p$ are [[Definition:Congruence Modulo Integer|congruent to $0$ modulo $p$]]
+So:
+:$n_p \notin \set {p, 2 p}$
+Also we have that $p > 2$.
+Hence:
+:$2 \not \equiv 1 \pmod p$
+and so it must be that $n_p = 1$.
+It follows from [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]] that this [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $n_p$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+From the [[First Sylow Theorem]], there exists at least $1$ [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Let $P$ be such a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+The [[Definition:Index of Subgroup|index]] of $P$ is $2$.
+From [[Subgroup of Index 2 is Normal]], $P$ is [[Definition:Normal Subgroup|normal]] in $G$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]] it follows that there is only $1$ [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This is a specific instance of [[Group of Order p q has Normal Sylow p-Subgroup|Group of Order $p q$ has Normal Sylow $p$-Subgroup]], where $q = 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 2p}
+Tags: Groups of Order 2 p, Cyclic Groups, Dihedral Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]].
+Let the [[Definition:Order of Group|order]] of $G$ be $2 p$.
+Then $G$ is either:
+:the [[Definition:Cyclic Group|cyclic group $C_{2 p}$]]
+or:
+:the [[Definition:Dihedral Group|dihedral group $D_p$]].
+\end{theorem}
+
+\begin{proof}
+When $p = 2$, the result follows from [[Groups of Order 4]].
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+From [[Sylow p-Subgroups of Group of Order 2p]], $G$ has [[Definition:Unique|exactly $1$]] [[Definition:Normal Subgroup|normal subgroup]] $P$ of [[Definition:Order of Group|order $p$]].
+$p$ is [[Definition:Prime Number|prime number]].
+So from [[Prime Group is Cyclic]], $P$ is a [[Definition:Cyclic Group|cyclic group]].
+Let $P = \gen x$ for some $x \in G$.
+By the [[First Sylow Theorem]] there exists at least one [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order $2$]].
+Hence:
+:$\exists y \in G: y^2 = e$
+It follows that the [[Definition:Element|elements]] of $G$ are known:
+:$G = \set {e, x, \ldots, x^{p - 1}, y, y x, y x^{p - 1} }$
+Then:
+{{begin-eqn}}
+{{eqn | l = y x y^{-1}
+      | o = \in
+      | r = P
+      | c = as $P$ is [[Definition:Normal Subgroup|normal]] in $G$.
+}}
+{{eqn | ll= \leadsto
+      | l = y x y^{-1}
+      | r = x^i
+      | c = for some $i \in \Z_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = y x y^{-1} x
+      | r = x^{i + 1}
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = y x y x = \paren {y x}^2
+      | r = x^{i + 1}
+      | c = $y^2 = e$, so $y = y^{-1}$
+}}
+{{end-eqn}}
+Thus:
+:the [[Definition:Even Integer|even]] [[Definition:Power of Group Element|powers]] of $y x$ are [[Definition:Power of Group Element|powers]] of $x$
+:the [[Definition:Even Integer|odd]] [[Definition:Power of Group Element|powers]] of $y x$ are of the form $y x^j$ for some $j \in \Z_{>0}$.
+By [[Order of Element Divides Order of Finite Group]]:
+:$\order {y x} \divides 2 p$
+where:
+:$\order {y x}$ denotes the [[Definition:Order of Group Element|order]] of $y x$
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+We have that:
+:$y x \ne e$
+and so by [[Identity is Only Group Element of Order 1]]:
+:$\order {y x} \ne 1$
+Thus:
+:$\order {y x} \in \set {2, p, 2 p}$
+{{qed|lemma}}
+Suppose $i \ne -1$ in $(1)$ above.
+Then:
+:$\paren {y x}^2 \ne e$
+and so:
+:$\order {y x} \ne 2$
+Because [[Definition:Even Integer|odd]] [[Definition:Power of Group Element|powers]] of $y x$ are of the form $y x^j$:
+:$\paren {y x}^p \ne e$
+and so:
+:$\order {y x} \ne p$
+It follows that:
+:$\order {y x} = 2 p$
+and from [[Group whose Order equals Order of Element is Cyclic]], $G$ is [[Definition:Cyclic Group|cyclic]].
+Thus, when $i \ne -1$:
+:$G = \gen {y x}$
+and so is [[Definition:Cyclic Group|cyclic]].
+Thus by [[Cyclic Group is Abelian]]:
+:$y x = x y$
+{{qed|lemma}}
+When $i = -1$ in $(1)$ above, we have that:
+{{begin-eqn}}
+{{eqn | l = y x y^{-1} x
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y x y^{-1}
+      | r = x^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y x
+      | r = x^{-1} y
+      | c = 
+}}
+{{end-eqn}}
+leading to the [[Definition:Group Presentation|group presentation]] of $G$:
+:$G = \gen {x, y: x^p = e = y^2, y x = x^{-1} y}$
+which is the [[Group Presentation of Dihedral Group]] $D_p$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order p q has Normal Sylow p-Subgroup}
+Tags: Sylow p-Subgroups, Normal Subgroups, Groups of Order p q
+
+\begin{theorem}
+Let $p$ and $q$ be [[Definition:Prime Number|prime numbers]] such that $p > q$.
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p q$.
+Then $G$ has [[Definition:Unique|exactly one]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]].
+This [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+Let $n_p$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] in $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_p \equiv 1 \pmod p$
+From the [[Fifth Sylow Theorem]]:
+:$n_p \divides p q$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+The [[Definition:Divisor of Integer|divisors]] of $p q$ are $1$, $p$, $q$ and $p q$.
+Of these:
+:$p$ and $p q$ are $\equiv 0 \pmod p$
+and as $p > q$:
+:$q \equiv q \pmod p$
+So the only possibility for $n_p$ is $1$.
+Let $P$ denote this [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]].
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms}
+Tags: Normed Division Rings, Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Unity of Ring|unity]] $1_R$.
+Then $\norm {\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] {{iff}}:
+:$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$
+where:
+:$n \cdot 1_R = \underbrace {1_R + 1_R + \dotsb + 1_R}_{\text {$n$ times} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition|Necessary Condition]] ===
+{{:Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition}}{{qed|lemma}}
+=== [[Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition|Sufficient Condition]] ===
+{{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 21}
+Tags: Groups of Order 21
+
+\begin{theorem}
+There exist exactly $2$ [[Definition:Group|groups]] of [[Definition:Order of Group|order]] $21$, up to [[Definition:Group Isomorphism|isomorphism]]:
+:$(1): \quad C_{21}$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $21$
+:$(2): \quad$ the [[Definition:Group|group]] whose [[Definition:Group Presentation|group presentation]] is:
+:::$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be of [[Definition:Order of Group|order]] $21$.
+From [[Group of Order p q has Normal Sylow p-Subgroup|Group of Order $p q$ has Normal Sylow $p$-Subgroup]], $G$ has [[Definition:Unique|exactly one]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]], which is [[Definition:Normal Subgroup|normal]].
+Let this [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]] of $G$ be denoted $P = \gen {x: x^7 = 1}$.
+From the [[First Sylow Theorem]], $G$ also has at least one [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]].
+Thus there exists $y \in G$ of [[Definition:Order of Group Element|order]] $3$.
+As $P$ is [[Definition:Normal Subgroup|normal]]:
+:$y x y^{-1} = x^i$
+for some $i \in \set {0, 1, \ldots, 6}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = y^3 x y^{-3}
+      | c = 
+}}
+{{eqn | r = y^2 \paren {y x y^{-1} } y^{-2}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = y^2 x^i y^2
+      | c = as $y x y^{-1} = x^i$
+}}
+{{eqn | r = \paren {y^2 x y^{-2} }^i
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {y \paren {y x y^{-1} } y^{-1} }^i
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = \paren {y x^i y^{-1} }^i
+      | c = as $y x y^{-1} = x^i$
+}}
+{{eqn | r = y \paren {x^i}^i y^{-1}
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = y \paren {x^{i^2} } y^{-1}
+      | c = [[Powers of Group Elements]]
+}}
+{{eqn | r = \paren {y x y^{-1} }^{i^2}
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^i}^{i^2}
+      | c = as $y x y^{-1} = x^i$
+}}
+{{eqn | r = x^{i^3}
+      | c = [[Powers of Group Elements]]
+}}
+{{end-eqn}}
+So $x^1 = x^{i^3}$ and so:
+:$i^3 \equiv 1 \pmod 7$
+and so:
+:$7 \divides \paren {i^3 - 1}$
+where $\divides$ indicates [[Definition:Divisor of Integer|divisibility]].
+Let us consider the $7$ possible values of $i$ in turn.
+{{begin-eqn}}
+{{eqn | ll= i = 0:
+      | l = 0^3 - 1
+      | r = -1
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 6
+      | rr= \pmod 7
+      | c = so $0$ is not a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 1:
+      | l = 1^3 - 1
+      | r = 0
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 0
+      | rr= \pmod 7
+      | c = so $1$ is a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 2:
+      | l = 2^3 - 1
+      | r = 7
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 0
+      | rr= \pmod 7
+      | c = so $2$ is a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 3:
+      | l = 3^3 - 1
+      | r = 26
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 5
+      | rr= \pmod 7
+      | c = so $3$ is not a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 4:
+      | l = 4^3 - 1
+      | r = 63
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 0
+      | rr= \pmod 7
+      | c = so $4$ is a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 5:
+      | l = 5^3 - 1
+      | r = 124
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 5
+      | rr= \pmod 7
+      | c = so $5$ is not a possible value of $i$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | ll= i = 6:
+      | l = 6^3 - 1
+      | r = 215
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = 5
+      | rr= \pmod 7
+      | c = so $6$ is not a possible value of $i$
+}}
+{{end-eqn}}
+Thus $i \bmod 7 \in \set {1, 2, 4}$.
+{{qed|lemma}}
+Suppose $i \equiv 1 \pmod 7$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y x y^{-1}
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y x
+      | r = x y
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^3
+      | r = x^3 y^3
+      | c = [[Power of Product of Commutative Elements in Group]]
+}}
+{{eqn | r = x^3
+      | c = as $y^3 = e$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \paren {x y}^7
+      | r = x^7 y^7
+      | c = [[Power of Product of Commutative Elements in Group]]
+}}
+{{eqn | r = y^7
+      | c = as $x^7 = e$
+}}
+{{eqn | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{21}
+      | r = y^3
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+It follows that:
+:$\order {x y} = 21$
+where $\order {x y}$ denotes the [[Definition:Order of Group Element|order]] of $x y$.
+Thus $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed|lemma}}
+Suppose that $i \equiv 2 \pmod 7$.
+Thus, let $y$ be an [[Definition:Element|element]] of [[Definition:Order of Group Element|order $3$]] for which $u x y^{-1} = x^2$.
+Then $z = y^2$ is an [[Definition:Element|element]] of [[Definition:Order of Group Element|order $3$]] for which $z x z^{-1} = x^4$.
+Thus the [[Definition:Group|group]] as defined here where $i = 2$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Group|group]] as defined here where $i = 4$.
+Thus, apart from $C_{21}$, the other [[Definition:Group|group]] of [[Definition:Order of Group|order]] $21$ has the [[Definition:Group Presentation|group presentation]]:
+:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2 }$
+{{qed}}
+=== [[Groups of Order 21/Matrix Representation of Non-Abelian Instance|Matrix Representation of Non-Abelian Instance]] ===
+{{:Groups of Order 21/Matrix Representation of Non-Abelian Instance}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 21/Matrix Representation of Non-Abelian Instance}
+Tags: Groups of Order 21
+
+\begin{theorem}
+Let $G$ be the [[Definition:Group|group]] of [[Definition:Order of Group|order]] $21$ whose [[Definition:Group Presentation|group presentation]] is:
+:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$
+Then $G$ can be instantiated by the following pair of [[Definition:Square Matrix|matrices]] over $\Z_7$:
+:$X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$
+\end{theorem}
+
+\begin{proof}
+We calculate the powers of $X$ and $Y$ in turn:
+{{begin-eqn}}
+{{eqn | l = X^2
+      | r = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = X^3
+      | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{end-eqn}}
+and so on to:
+{{begin-eqn}}
+{{eqn | l = X^7
+      | r = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
+      | c = as $7 \equiv 0 \pmod 7$
+}}
+{{end-eqn}}
+Thus we have:
+:$X^7 = \mathbf I$
+where $\mathbf I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the [[Definition:Identity Matrix|identity matrix]].
+Then:
+{{begin-eqn}}
+{{eqn | l = Y^2
+      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = Y^3
+      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{end-eqn}}
+Thus we have:
+:$Y^3 = \mathbf I$
+and:
+:$Y^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = Y X Y^{-1}
+      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 4 & 4 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}
+      | c = 
+}}
+{{eqn | r = X^2
+      | c = 
+}}
+{{end-eqn}}
+and the result is apparent.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Sylow p-Subgroups in Group of Order 12}
+Tags: Groups of Order 12, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be of [[Definition:Order of Group|order]] $12$.
+Then $G$ has either:
+:a [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]]
+or:
+:a [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]].
+\end{theorem}
+
+\begin{proof}
+Note that a [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of $G$ is of [[Definition:Order of Group|order $4$]].
+From [[Sylow Theorems/Examples/Sylow 3-Subgroups in Group of Order 12|Sylow $3$-Subgroups in Group of Order 12]], there are either $1$ or $4$ [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]].
+Suppose there is [[Definition:Unique|exactly $1$]] [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] $P$.
+Then from [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ is [[Definition:Normal Subgroup|normal]].
+{{qed|lemma}}
+Suppose there are $4$ [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] $P_1$, $P_2$, $P_3$ and $P_4$.
+Each [[Definition:Set Intersection|intersection]] $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the [[Definition:Trivial Subgroup|trivial subgroup]] of $G$:
+:$P_i \cap P_j = \set e$
+Thus $G$ contains:
+:The [[Definition:Identity Element|identity element]] $e$
+:$8$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$
+:$3$ more [[Definition:Element|elements]], which (along with $e$) must form the [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of [[Definition:Order of Group|order $4$]].
+This [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] $Q$ must be [[Definition:Unique|unique]].
+Hence by [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $Q$ is [[Definition:Normal Subgroup|normal]].
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order p^2 q has Normal Sylow p-Subgroup}
+Tags: Sylow p-Subgroups, Normal Subgroups, Groups of Order p^2 q
+
+\begin{theorem}
+Let $p$ and $q$ be [[Definition:Prime Number|prime numbers]] such that $p \ne q$.
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p^2 q$.
+Then $G$ has a [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]].
+\end{theorem}
+
+\begin{proof}
+Let $n_p$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] in $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_p \equiv 1 \pmod p$
+From the [[Fifth Sylow Theorem]]:
+:$n_p \divides p^2 q$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Thus $n_p \in \set {1, q}$.
+Suppose $p > q$.
+Then:
+:$q \not \equiv 1 \pmod p$
+and so $n_p \ne q$.
+Hence $n_p = 1$.
+From [[Sylow p-Subgroup is Unique iff Normal]] it follows that this [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed|lemma}}
+Suppose $q > p$.
+Then $n_p$ could equal $q$ if $q \equiv 1 \pmod p$.
+Let $n_q$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $q$-subgroups]] in $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_q \equiv 1 \pmod q$
+and so $n_q$ is not a [[Definition:Integer Multiple|multiple]] of $q$.
+From the [[Fifth Sylow Theorem]]:
+:$n_q \divides p^2 q$
+It follows that:
+:$n_q \in \set {1, p, p^2}$
+If $n_q = 1$ then it follows from [[Sylow p-Subgroup is Unique iff Normal]] that this [[Definition:Sylow p-Subgroup|Sylow $q$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed|lemma}}
+As $p \not \equiv 1 \pmod q$, it follows that $n_q \ne p$.
+The final possibility to be explored is when $n_q = p^2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = p^2
+      | o = \equiv
+      | r = 1
+      | rr= \pmod q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = q
+      | o = \divides
+      | r = p^2 - 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = q
+      | o = \divides
+      | r = \paren {p + 1} \paren {p - 1}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | ll= \leadsto
+      | l = q
+      | r = p + 1
+      | c = as $q > p$
+}}
+{{end-eqn}}
+But then $p$ and $q$ are consecutive [[Definition:Prime Number|prime numbers]], and so:
+:$p = 2, q = 3$
+This gives the specific case:
+:$\order G = 2^2 \times 3$
+and the result follows from [[Normal Sylow p-Subgroups in Group of Order 12]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 30 has Normal Cyclic Subgroup of Order 15}
+Tags: Groups of Order 30
+
+\begin{theorem}
+Let $G$ be of [[Definition:Order of Group|order]] $30$.
+Then $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $15$ which is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+By [[Group of Order 15 is Cyclic Group]], any [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order]] $15$ is [[Definition:Cyclic Group|cyclic]].
+It remains to be proved that a [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order]] $15$ exists, and that it is [[Definition:Normal Subgroup|normal]].
+Let $n_3$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] of $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_3 \equiv 1 \pmod 3$
+and from the [[Fifth Sylow Theorem]]:
+:$n_3 \divides 10$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_3 \in \set {1, 10}$.
+Let $n_5$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] of $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_5 \equiv 1 \pmod 5$
+and from the [[Fifth Sylow Theorem]]:
+:$n_5 \divides 30$
+It follows that $n_5 \in \set {1, 6}$.
+Suppose $n_3 = 1$.
+Let $P$ denote the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] of $G$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ is [[Definition:Normal Subgroup|normal]].
+Thus we may form the [[Definition:Quotient Group|quotient group]] $G / P$, which is of [[Definition:Order of Group|order $10$]].
+From [[Groups of Order 2p]], $G / P$ has a [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]], which we will denote $N / P$.
+By the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]], $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ with $15$ [[Definition:Element|elements]].
+As noted, $N$ is [[Definition:Cyclic Group|cyclic]].
+{{qed|lemma}}
+Suppose $n_3 = 10$.
+Let the [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]] of $G$ be denoted $P_1, P_2, \ldots, P_{10}$.
+Each [[Definition:Set Intersection|intersection]] $P_i \cap P_j$ for $i, j \in \set {1, 2, \ldots, 10}, i \ne j$ is the [[Definition:Trivial Subgroup|trivial subgroup]] of $G$:
+:$P_i \cap P_j = \set e$
+Thus $G$ contains:
+:The [[Definition:Identity Element|identity element]] $e$
+:$20$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $3$, of which $2$ each are in $P_1, P_2, \ldots, P_{10}$
+:$9$ more [[Definition:Element|elements]] of $G$ which are in [[Definition:Subgroup|subgroups]] of $G$ different from the [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]], and so have [[Definition:Order of Group Element|orders]] different from $3$.
+Suppose $n_5 = 6$.
+By a similar argument to the above, these $6$ [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] contribute $24$ [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $5$ to $G$.
+This is not possible because there are only $9$ [[Definition:Element|elements]] of $G$ so far unaccounted for.
+So if $G$ has $10$ [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]], it can have only $1$ [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]], which we will denote $Q$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $Q$ is [[Definition:Normal Subgroup|normal]].
+Thus we may form the [[Definition:Quotient Group|quotient group]] $G / Q$, which is of [[Definition:Order of Group|order $16$]].
+From [[Groups of Order 2p]], $G / Q$ has a [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]], which we will denote $N / Q$.
+Again by the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]], $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ with $15$ [[Definition:Element|elements]].
+As noted, $N$ is [[Definition:Cyclic Group|cyclic]].
+{{qed|lemma}}
+Both cases have been accounted for:
+:that $G$ has $1$ [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]]
+:that $G$ has $10$ [[Definition:Sylow p-Subgroup|Sylow $3$-subgroups]].
+Both cases result in $G$ having a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $15$ which is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 30/Lemma}
+Tags: Groups of Order 30
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $30$.
+Then $G$ is one of the following:
+:The [[Definition:Cyclic Group|cyclic group]] $C_{30}$
+:The [[Definition:Dihedral Group|dihedral group]] $D_{15}$
+:[[Definition:Group Isomorphism|Isomorphic]] to one of:
+::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
+::$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
+\end{theorem}
+
+\begin{proof}
+By [[Group of Order 30 has Normal Cyclic Subgroup of Order 15]], $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $15$ which is [[Definition:Cyclic Group|cyclic]].
+Let this [[Definition:Normal Subgroup|normal]] [[Definition:Cyclic Group|cyclic]] [[Definition:Order of Group|order]] $15$ [[Definition:Subgroup|subgroup]] be denoted $N$:
+:$N = \gen x$
+Let $y$ be the [[Definition:Generator of Subgroup|generator]] for any [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of $G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y x y^{-1}
+      | o = \in
+      | r = N
+      | c = as $N$ is [[Definition:Normal Subgroup|normal]]
+}}
+{{eqn | ll= \leadsto
+      | l = y x y^{-1}
+      | r = x^i
+      | c = for some $i \in \Z_{\ge 0}$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = y^2 x y^{-2}
+      | c = as $y^2 = e$
+}}
+{{eqn | r = y \paren {y x y^{-1} } y^{-1}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = y x^i y^{-1}
+      | c = as $y x y^{-1} = x^i$
+}}
+{{eqn | r = \paren {y x y^{-1} }^i
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^i}^i
+      | c = as $y x y^{-1} = x^i$
+}}
+{{eqn | r = x^{i^2}
+      | c = [[Powers of Group Elements]]
+}}
+{{end-eqn}}
+and so:
+:$i^2 - 1 \equiv 0 \pmod {15}$
+Investigating the powers of $i$, case by case, searching for those which satisfy this congruence, yields:
+:$i \in \set {1, 4, 11, 14}$
+The case $i \equiv 1 \pmod {15}$ leads to the [[Definition:Cyclic Group|cyclic group]] $C_{30}$.
+The case where $i \equiv {14} \equiv {-1} \pmod {15}$ leads to the [[Definition:Dihedral Group|dihedral group]] $D_{15}$.
+The other two cases lead to:
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Subgroup of Group of Order 24}
+Tags: Groups of Order 24
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $24$.
+Then $G$ has either:
+:a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $8$
+or:
+:a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $4$.
+\end{theorem}
+
+\begin{proof}
+We note that:
+:$24 = 3 \times 2^3$
+Hence a [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of $G$ is of [[Definition:Order of Group|order]] $8$.
+Let $n_2$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $2$-subgroups]] of $G$.
+By the [[Fourth Sylow Theorem]]:
+:$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is [[Definition:Odd Integer|odd]]
+and from the [[Fifth Sylow Theorem]]:
+:$n_2 \divides 24$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_3 \in \set {1, 3}$.
+Suppose $n_2 = 1$.
+Let $P$ denote the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] of $G$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ is [[Definition:Normal Subgroup|normal]].
+This is therefore the required [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $8$.
+{{qed|lemma}}
+Suppose $n_2 = 3$.
+Let $S_1$, $S_2$ and $S_3$ denote the $3$ [[Definition:Sylow p-Subgroup|Sylow $2$-subgroups]] of $G$.
+Each has [[Definition:Order of Group|order]] $8$, as noted earlier.
+Consider the [[Definition:Subset Product|product]] $S_1 S_2$.
+By [[Order of Subgroup Product]]:
+:$\order {S_1 S_2} = \dfrac {2^3 2^3} {2^r}$
+where:
+:$\order {S_1 S_2}$ denotes the [[Definition:Order of Group|order]] of $S_1 S_2$
+:$\order {S_1 \cap S_2} = 2^r$
+As $S_1 S_2 \subseteq G$ and $\order G = 24$:
+:$\order {S_1 S_2} \le 24$
+Hence:
+{{begin-eqn}}
+{{eqn | l = 2^3 2^3
+      | o = \le
+      | r = \order {S_1 S_2} \times 2^r
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 64
+      | o = \le
+      | r = 24 \times 2^r
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r
+      | o = \ge
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+As $S_1 \cap S_2$ is a [[Definition:Proper Subgroup|proper subgroup]] of $S_1$, it can have no more than $2^2 = 4$ [[Definition:Element|elements]].
+So if $G$ has $3$ [[Definition:Sylow p-Subgroup|Sylow $2$-subgroups]], the [[Definition:Set Intersection|intersection]] of any $2$ of them is of [[Definition:Order of Group|order]] $4$.
+Let $T = S_1 \cap S_2$, so that $\order T = 4$.
+We have that:
+:$\index T {S_1} = 2$
+where $\index T {S_1}$ denotes the [[Definition:Index of Subgroup|index]] of $T$ in $S_1$.
+From [[Subgroup of Index 2 is Normal]], $T$ is [[Definition:Normal Subgroup|normal]] in $S_1$.
+Similarly, $T$ is [[Definition:Normal Subgroup|normal]] in $S_2$.
+Thus $S_1$ and $S_2$ are both [[Definition:Subgroup|subgroups]] of the [[Definition:Normalizer|normalizer]] $\map {N_G} T$ of $T$ in $G$.
+Thus $H = \gen {S_1, S_2}$ is a [[Definition:Subgroup|subgroup]] of $\map {N_G} T$.
+Hence $T$ is a [[Definition:Normal Subgroup|normal subgroup]] of $H$.
+As $H$ is a [[Definition:Subgroup|subgroup]] of $\map {N_G} T$, it contains $S_1 S_2$ as a [[Definition:Subset|subset]].
+But $S_1 S_2$ contains $\dfrac {2^6} {2^2} = 16$ [[Definition:Element|elements]].
+The only [[Definition:Subgroup|subgroup]] of $G$ containing $16$ [[Definition:Element|elements]] has to be $G$ itself.
+So $H = G$ and so $\map {N_G} T = G$.
+That is, $T$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+This is therefore the required [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $4$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 35 is Cyclic Group}
+Tags: Groups of Order 35, Cyclic Groups, Group of Order 35 is Cyclic Group
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $35$.
+Then $G$ is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+We have that $35 = 5 \times 7$.
+Then we have that $5$ and $7$ are [[Definition:Prime Number|primes]] such that $5 < 7$ and $5$ does not [[Definition:Divisor of Integer|divide]] $7 - 1$.
+Thus [[Cyclic Groups of Order p q|Cyclic Groups of Order $p q$]] can be applied.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $G$ be of [[Definition:Order of Group|order $35$]].
+We have that $35 = 5 \times 7$ where both $5$ and $7$ are [[Definition:Prime Number|prime]].
+Hence from the [[First Sylow Theorem]], $G$ has:
+:at least one [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]]
+and:
+:at least one [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]]
+Let $n_5$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] of $G$.
+Let $n_7$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] of $G$.
+By the [[Fourth Sylow Theorem]]:
+:$n_5 \equiv 1 \pmod 5$
+and from the [[Fifth Sylow Theorem]]:
+:$n_5 \divides 35$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_5 = 1$.
+Let $P$ denote this [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]].
+By the [[Fourth Sylow Theorem]]:
+:$n_7 \equiv 1 \pmod 7$
+and from the [[Fifth Sylow Theorem]]:
+:$n_7 \divides 35$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_7 = 1$.
+Let $Q$ denote this [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], $P$ and $Q$ are [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Consider $P \cap Q$.
+By [[Intersection of Subgroups is Subgroup]]:
+:$P \cap Q$ is a [[Definition:Subgroup|subgroup]] of $P$
+:$P \cap Q$ is a [[Definition:Subgroup|subgroup]] of $Q$
+and:
+:$P \cap Q$ is a [[Definition:Subgroup|subgroup]] of $G$
+By [[Definition:Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\order {P \cap Q} \divides \order P = 5$
+:$\order {P \cap Q} \divides \order Q = 7$
+where:
+:$\order {P \cap Q}$ denotes the [[Definition:Order of Group|order]] of $P \cap Q$
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Thus:
+:$\order {P \cap Q} = 1$
+and so:
+:$P \cap Q = \set e$
+where $e$ is the [[Definition:Identity Element|identity]] of $G$.
+From [[Prime Group is Cyclic]], both $P$ and $Q$ are [[Definition:Cyclic Group|cyclic]].
+Let:
+:$P = \gen a$
+:$Q = \gen b$
+where $\gen a$ denotes the [[Definition:Generated Subgroup|subgroup of $G$ generated by $a$]].
+As $P$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$:
+:$b^{-1} a^{-1} b \in P$
+and so as $a \in P$, from {{GroupAxiom|0}}:
+:$b^{-1} a^{-1} b a \in P$
+As $Q$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$:
+:$a^{-1} b a \in Q$
+and so $b^{-1} \in Q$, from {{GroupAxiom|0}}:
+:$b^{-1} a^{-1} b a \in Q$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$b^{-1} a^{-1} b a \in P \cap Q$
+But $P \cap Q = \set e$.
+So:
+:$b^{-1} a^{-1} b a = e$
+and so from [[Product of Commuting Elements with Inverses]]
+:$a b = b a$
+From [[Power of Product of Commutative Elements in Group]]:
+:$\paren {a b}^7 = a^7 b^7 = a^7 \ne e$
+and:
+:$\paren {a b}^5 = a^5 b^5 = b^5 \ne e$
+By [[Order of Element Divides Order of Finite Group]]:
+:$\order {a b} \divides 35$
+where $\order {a b}$ denotes the [[Definition:Order of Group Element|order]] of $a b$.
+Thus $\order {a b} \in \set {1, 5, 7, 35}$
+As it has been established that $\order {a b}$ is not $1$, $5$ or $7$, it follows that:
+:$\order {a b} = 35$
+Hence from [[Group whose Order equals Order of Element is Cyclic]], $G$ is a [[Definition:Cyclic Group|cyclic group]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup}
+Tags: Groups of Order 105, Sylow p-Subgroups, Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $105$.
+Then $G$ has either:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]]
+or:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $105$ whose [[Definition:Identity Element|identity]] is $e$.
+We have that:
+:$105 = 3 \times 5 \times 7$
+From the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]], [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] and [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+Let:
+:$n_5$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] of $G$
+:$n_7$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] of $G$.
+$5$ and $7$ appear in $105$ with [[Definition:Multiplicity of Prime Factor|multiplicity $1$]].
+Hence any such [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] are [[Definition:Prime Number|prime groups]].
+From [[Sylow p-Subgroup is Unique iff Normal]]:
+:if $n_5 = 1$ then the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] is [[Definition:Normal Subgroup|normal]]
+and:
+:if $n_7 = 1$ then the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+It remains to be shown that either $n_5 = 1$ or $n_7 = 1$.
+By the [[Fourth Sylow Theorem]]:
+:$n_5 \equiv 1 \pmod 5$
+and from the [[Fifth Sylow Theorem]]:
+:$n_5 \divides 105$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_5 \in \set {1, 21}$.
+By the [[Fourth Sylow Theorem]]:
+:$n_7 \equiv 1 \pmod 7$
+and from the [[Fifth Sylow Theorem]]:
+:$n_7 \divides 105$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_7 \in {1, 15}$.
+If either $n_5 = 1$ or $n_7 = 1$ the proof is finished.
+Suppose $n_7 = 15$.
+As all of these [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] are [[Definition:Prime Group|prime]], the [[Definition:Set Intersection|intersection]] of any $2$ of them is $\set e$.
+Thus, these $15$ [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] contribute $6 \times 15 = 90$ [[Definition:Distinct Elements|distinct elements]] to $G$.
+This leaves $15$ [[Definition:Element|elements]] still to be accounted for.
+Suppose $n_5 = 21$.
+By a similar argument, these $21$ [[Definition:Sylow p-Subgroup|Sylow $5$-subgroups]] contribute $4 \times 21 = 84$ [[Definition:Distinct Elements|distinct elements]] to $G$.
+None of these [[Definition:Element|elements]], apart from $e$, can also be [[Definition:Element|elements]] of one of the [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]].
+But there are only $15$ [[Definition:Element|elements]] available.
+So it is not possible for both $n_7 = 15$ and $n_5 = 21$.
+So either $n_7 = 1$ or $n_5 = 1$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order 105 has Normal Cyclic Subgroup of Index 3}
+Tags: Groups of Order 105, Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $105$.
+Then $G$ has a [[Definition:Normal Subgroup|normal]] [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] $N$ such that:
+:$\index G N = 3$
+where $\index G N$ denotes the [[Definition:Index of Subgroup|index]] of $N$ in $G$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $105$ whose [[Definition:Identity Element|identity]] is $e$.
+From [[Group of Order 105 has Normal Sylow 5-Subgroup or Normal Sylow 7-Subgroup]], $G$ has either:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]]
+or:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+Suppose $G$ has [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]], which we will denote $P$.
+Then $G / P$ is the [[Definition:Quotient Group|quotient group]] of $G$ by $P$ and is of [[Definition:Order of Group|order]] $21$.
+From [[Group of Order p q has Normal Sylow p-Subgroup|Group of Order $p q$ has Normal Sylow $p$-Subgroup]], $G / P$ has [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]] which we denote $N / P$.
+From the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]], $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ of [[Definition:Order of Group|order]] $7 \times 5 = 35$.
+It follows from [[Group of Order 35 is Cyclic Group]] that $N$ is [[Definition:Cyclic Group|cyclic]].
+{{qed|lemma}}
+Now suppose that $G$ has [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]], which we will denote $Q$.
+Then $G / Q$ is the [[Definition:Quotient Group|quotient group]] of $G$ by $Q$ and is of [[Definition:Order of Group|order]] $15$.
+From [[Group of Order p q has Normal Sylow p-Subgroup|Group of Order $p q$ has Normal Sylow $p$-Subgroup]], $G / Q$ has [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] which we denote $M / Q$.
+From the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]], $m$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ of [[Definition:Order of Group|order]] $5 \times 7 = 35$.
+It follows from [[Group of Order 35 is Cyclic Group]] that $M$ is [[Definition:Cyclic Group|cyclic]].
+{{qed|lemma}}
+Both cases have been covered, and the existence of a [[Definition:Normal Subgroup|normal]] [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] of [[Definition:Order of Group|order]] $35$ has been proved.
+The result follows by definition of the [[Definition:Index of Subgroup|index of a subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Reflexive}
+Tags: Diagonal Relation, Reflexive Relations
+
+\begin{theorem}
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is a [[Definition:Reflexive Relation|reflexive relation]] in $S$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo= \forall x \in S:
+      | l = \tuple {x, x}
+      | o = \in
+      | r = \Delta_S
+      | c = {{Defof|Diagonal Relation}}
+}}
+{{end-eqn}}
+So $\Delta_S$ is [[Definition:Reflexive Relation|reflexive]].
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Symmetric}
+Tags: Diagonal Relation, Symmetric Relations
+
+\begin{theorem}
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is a [[Definition:Symmetric Relation|symmetric relation]] in $S$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo= \forall x, y \in S:
+      | l = \tuple {x, y}
+      | o = \in
+      | r = \Delta_S
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = x
+      | r = y
+      | c = {{Defof|Diagonal Relation}}
+}}
+{{eqn | lll=\leadsto
+      | l = y
+      | r = x
+      | c = [[Equality is Symmetric]]
+}}
+{{eqn | lll=\leadsto
+      | l = \tuple {y, x}
+      | o = \in
+      | r = \Delta_S
+      | c = {{Defof|Diagonal Relation}}
+}}
+{{end-eqn}}
+So $\Delta_S$ is [[Definition:Symmetric Relation|symmetric]].
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Transitive}
+Tags: Diagonal Relation, Transitive Relations
+
+\begin{theorem}
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is a [[Definition:Transitive Relation|transitive relation]] in $S$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo= \forall x, y, z \in S:
+      | l = \tuple {x, y}
+      | o = \in
+      | r = \Delta_S \land \tuple {y, z} \in \Delta_S
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = x
+      | r = y \land y = z
+      | c = {{Defof|Diagonal Relation}}
+}}
+{{eqn | lll=\leadsto
+      | l = x
+      | r = z
+      | c = [[Equality is Transitive]]
+}}
+{{eqn | lll=\leadsto
+      | l = \tuple {x, z}
+      | o = \in
+      | r = \Delta_S
+      | c = {{Defof|Diagonal Relation}}
+}}
+{{end-eqn}}
+So $\Delta_S$ is [[Definition:Transitive Relation|transitive]].
+\end{proof}<|endoftext|>
+\section{Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup}
+Tags: Groups of Order 56, Sylow p-Subgroups, Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $56$.
+Then $G$ has either:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]]
+or:
+:[[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $56$ whose [[Definition:Identity Element|identity]] is $e$.
+We have that:
+:$56 = 2^3 \times 7$
+From the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] and [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]].
+Let:
+:$n_2$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $2$-subgroups]] of $G$
+:$n_7$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] of $G$.
+From [[Sylow p-Subgroup is Unique iff Normal]]:
+:if $n_2 = 1$ then the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] is [[Definition:Normal Subgroup|normal]]
+and:
+:if $n_7 = 1$ then the [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+It remains to be shown that either $n_2 = 1$ or $n_7 = 1$.
+By the [[Fourth Sylow Theorem]]:
+:$n_2 \equiv 1 \pmod 2$ (that is, $n_2$ is [[Definition:Odd Integer|odd]])
+and from the [[Fifth Sylow Theorem]]:
+:$n_2 \divides 56$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_2 \in \set {1, 7}$.
+By the [[Fourth Sylow Theorem]]:
+:$n_7 \equiv 1 \pmod 7$
+and from the [[Fifth Sylow Theorem]]:
+:$n_7 \divides 56$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_7 \in {1, 8}$.
+If either $n_2 = 1$ or $n_7 = 1$ the proof is finished.
+Suppose $n_7 = 8$.
+As all of these [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] are [[Definition:Prime Group|prime]], the [[Definition:Set Intersection|intersection]] of any $2$ of them is $\set e$.
+Thus, these $8$ [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] contribute $6 \times 8 = 48$ [[Definition:Distinct Elements|distinct elements]] to $G$.
+This leaves $7$ [[Definition:Element|elements]] still to be accounted for.
+A [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]] is of [[Definition:Order of Group|order]] $2^3 = 8$.
+Thus all $7$ of these remaining [[Definition:Element|elements]] must all be in that one [[Definition:Sylow p-Subgroup|Sylow $2$-subgroup]]
+So if $n_7 \ne 1$, then $n_2 = 1$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Direct Product is not necessarily Direct Product of Subgroups}
+Tags: Group Direct Products
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $G \times H$ denote the [[Definition:Group Direct Product|direct product]] of $G$ and $H$.
+Let $K$ be a [[Definition:Subgroup|subgroup]] of $G \times H$.
+Then it is not necessarily the case that $K$ is of the form:
+:$G' \times H'$
+where:
+:$G'$ is a [[Definition:Subgroup|subgroup]] of $G$
+:$H'$ is a [[Definition:Subgroup|subgroup]] of $H$.
+\end{theorem}
+
+\begin{proof}
+Let $G = H = C_2$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $2$]].
+Let $G = \gen x$ and $H = \gen y$, so that:
+:$G = \set {e_G, x}$
+:$H = \set {e_H, y}$
+where $e_G$ and $e_H$ are the [[Definition:Identity Element|identity elements]] of $G$ and $H$ respectively.
+Consider the [[Definition:Element|element]] $\tuple {x, y} \in G \times H$.
+We have that:
+:$\gen {\tuple {x, y} } =\set {\tuple {e_G, e_H}, \tuple {x, y} }$
+but this is not the [[Definition:Group Direct Product|direct product]] of a [[Definition:Subgroup|subgroup]] of $G$ with a [[Definition:Subgroup|subgroup]] of $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 30}
+Tags: Groups of Order 30
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $30$.
+Then $G$ is one of the following:
+:The [[Definition:Cyclic Group|cyclic group]] $C_{30}$
+:The [[Definition:Dihedral Group|dihedral group]] $D_{15}$
+:The [[Definition:Group Direct Product|group direct product]] $C_5 \times D_3$
+:The [[Definition:Group Direct Product|group direct product]] $C_3 \times D_5$
+\end{theorem}
+
+\begin{proof}
+First we introduce a [[Definition:Lemma|lemma]]:
+=== [[Groups of Order 30/Lemma|Lemma]] ===
+{{:Groups of Order 30/Lemma}}{{qed|lemma}}
+From the [[Groups of Order 30/Lemma|lemma]], it remains to be shown that the [[Definition:Group Presentation|group presentations]]:
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
+give the [[Definition:Group|groups]] $C_5 \times D_3$ and $C_3 \times D_5$.
+=== [[Groups of Order 30/C 5 x D 3|Direct Product $C_5 \times D_3$]] ===
+{{:Groups of Order 30/C 5 x D 3}}
+=== [[Groups of Order 30/C 3 x D 5|Direct Product $C_3 \times D_5$]] ===
+{{:Groups of Order 30/C 3 x D 5}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Unity of Ring|unity]] $1_R$.
+Then:
+:$\norm {\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] $\implies \forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$.
+where:
+$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} }$
+\end{theorem}
+
+\begin{proof}
+Let $\norm {\,\cdot\,}$ be [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+Then by the definition of a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]], for $n \in \N$:
+{{begin-eqn}}
+{{eqn | lo= \forall n \in \N_{>0}:
+      | l = \norm {n \cdot 1_R}
+      | r = \norm {1_R + \dots + 1_R}
+      | c = ($n$ summands)
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {1_R}, \ldots, \norm {1_R} }
+      | c = 
+}}
+{{eqn | r = 1
+      | c = because $\norm {1_R} = 1$
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Unity of Ring|unity]] $1_R$.
+Then:
+:$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1 \implies \norm {\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm]]
+where:
+:$n \cdot 1_R = \underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} }$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$
+Let $x, y \in R$.
+Let $y = 0_R$ where $0_R$ is the [[Definition:Ring Zero|zero]] of $R$.
+Then $\norm {x + y} = \norm x = \max \set {\norm x, 0} = \max \set {\norm x, \norm y}$
+==== [[Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 1|Lemma 1]] ====
+{{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 1}}
+{{qed|lemma}}
+Hence to complete the proof it is sufficient to prove:
+:$\forall x \in R: \norm {x + 1_R} \le \max \set {\norm x, 1}$
+For $n \in \N$:
+{{begin-eqn}}
+{{eqn | l = \norm {x + 1_R}^n
+      | r = \norm {\sum_{i \mathop = 0}^n \binom n i \cdot x^i}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 0}^n \norm {\binom n i \cdot x^i}
+      | c = {{NormAxiom|3}}
+}}
+{{eqn | r = \sum_{i \mathop = 0}^n \norm {\binom n i \cdot 1_R} \norm x^i
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 0}^n \norm x^i
+      | c = $\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$
+}}
+{{end-eqn}}
+==== [[Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2|Lemma 2]] ====
+{{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2}}
+{{qed|lemma}}
+Hence
+{{begin-eqn}}
+{{eqn | l = \norm {x + 1_R}^n
+      | o = \le
+      | r = \sum_{i \mathop = 0}^n \norm x^i
+      | c = continuing from above
+}}
+{{eqn | o = \le
+      | r = \sum_{i \mathop = 0}^n \max \set {\norm x^n , 1}
+      | c = [[Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2|Lemma 2]]
+}}
+{{eqn | r = \paren {n + 1} \max \set {\norm x^n , 1}
+}}
+{{end-eqn}}
+Taking $n$th roots yields:
+:$\norm {x + 1_R} \le \paren {n + 1}^{1/n} \max \set {\norm x, 1}$
+==== [[Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3|Lemma 3]] ====
+{{:Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3}}{{qed|lemma}}
+By the [[Multiple Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} \max \set {\norm x, 1} = \max \set {\norm x, 1}$
+By [[Inequality Rule for Real Sequences]]:
+:$\norm {x + 1_R} \le \max \set {\norm x, 1}$
+The result follows.
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 1}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $y \ne 0_R$ where $0_R$ is the [[Definition:Ring Zero|zero]] of $R$.
+Then:
+:$\norm {x + y} \le \max \set {\norm x, \norm y} \iff \norm {x y^{-1} + 1_R} \le \max \set {\norm {x y^{-1} }, 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | o = \le
+      | r = \max \set {\norm x, \norm y}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {x + y} \norm {y^{-1} }
+      | o = \le
+      | r = \max \set {\norm x \norm {y^{-1} }, \norm y \norm {y^{-1} } }
+      | c = Multiply through by $\norm{y^{-1} }$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {\paren {x + y} y^{-1} }
+      | o = \le
+      | r = \max \set {\norm {x y^{-1} }, \norm {y y^{-1} } }
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {\paren {x y^{-1} + y y^{-1} } }
+      | o = \le
+      | r = \max \set {\norm {x y^{-1} }, \norm {y y^{-1} } }
+      | c = [[Definition:Ring (Abstract Algebra)|Ring axiom $\text D$: Product is Distributive over Addition]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {\paren {x y^{-1} + 1_R } }
+      | o = \le
+      | r = \max \set {\norm {x y^{-1} }, \norm {1_R } }
+      | c = [[Definition:Division Ring|Product with Inverse is Unit]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \norm {\paren {x y^{-1} + 1_R } }
+      | o = \le
+      | r = \max \set {\norm {x y^{-1} }, 1 }
+      | c = [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 2}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Then for all $i$, $0 \le i \le n$:
+:$\norm x^i \le \max \set {\norm x^n , 1}$
+\end{theorem}
+
+\begin{proof}
+If $\norm x > 1$ then for all $i$, $0 \le i \le n$:
+:$\norm x^i \le \norm {x}^n \le \max \set {\norm x^n, 1}$
+If $\norm x \le 1$ then for all $i$, $0 \le i \le n$:
+:$\norm x^i \le 1 \le \max \set {\norm x^n, 1}$
+In either case for all $i$, $0 \le i \le n$:
+:$\norm x^i \le \max \set {\norm x^n , 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Epimorphism preserves Central Subgroups}
+Tags: Central Subgroups, Group Epimorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $\theta: G \to H$ be an [[Definition:Group Epimorphism|epimorphism]].
+Let $Z \le G$ be a [[Definition:Central Subgroup|central subgroup]] of $G$.
+Then $\theta \sqbrk Z$ is a [[Definition:Central Subgroup|central subgroup]] of $H$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Central Subgroup|central subgroup]]:
+:$Z \subseteq \map Z G$
+where $\map Z G$ denotes the [[Definition:Center of Group|center]] of $G$.
+From [[Image under Epimorphism of Center is Subset of Center]]:
+:$\theta \sqbrk {\map Z G} \subseteq \map Z H$
+From [[Image of Subset under Mapping is Subset of Image]] it follows that:
+:$\theta \sqbrk Z \subseteq \map Z H$
+The result follows.
+{{qed}}
+[[Category:Central Subgroups]]
+[[Category:Group Epimorphisms]]
+9fdj8xamuhphxaj2xl5xszjyb657b1r
+\end{proof}<|endoftext|>
+\section{Direct Product of Central Subgroups}
+Tags: Central Subgroups, Group Direct Products
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $Z$ and $W$ be [[Definition:Central Subgroup|central subgroups]] of $G$ and $H$ respectively.
+Then $Z \times W$ is a [[Definition:Central Subgroup|central subgroup]] of $G \times H$.
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {z, w} \in Z \times W$.
+Let $\tuple {x, y} \in G \times H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \tuple {x, y} \tuple {z, w}
+      | r = \tuple {x z, y w}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{eqn | r = \tuple {z x, w y}
+      | c = {{Defof|Central Subgroup}}
+}}
+{{eqn | r = \tuple {z, w} \tuple {x, y}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{end-eqn}}
+$\tuple {x, y}$ is arbitrary.
+Thus $\tuple {z, w}$ [[Definition:Commuting Elements|commutes]] with all [[Definition:Element|elements]] of $G \times H$.
+Hence $\tuple {z, w}$ is in the [[Definition:Center of Group|center]] of $G \times H$
+That is, $Z \times W$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Center of Group|center]] $G \times H$.
+The result follows by definition of [[Definition:Central Subgroup|central subgroup]].
+{{qed}}
+[[Category:Central Subgroups]]
+[[Category:Group Direct Products]]
+t3bozwduh2dv43876c7vq6v7lgk7520
+\end{proof}<|endoftext|>
+\section{Groups of Order 30/C 5 x D 3}
+Tags: Groups of Order 30
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $30$.
+Let $G$ have the [[Definition:Group Presentation|group presentation]]:
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Group Direct Product|group direct product]] of the [[Definition:Cyclic Group|cyclic group]] $C_5$ and the [[Definition:Dihedral Group|dihedral group $D_3$]]:
+:$G \cong C_5 \times D_3$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be defined by its [[Definition:Group Presentation|group presentation]]:
+:$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^{11} }$
+Let $z$ denote $x^3$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y z y^{-1}
+      | r = y x^3 y^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {y x y^{-1} }^3
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^{11} }^3
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{33}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{15} x^{15} x^3
+      | c = [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | r = e \cdot e \cdot x^3
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^3
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = z
+      | c = Definition of $z$
+}}
+{{eqn | ll= \leadsto
+      | l = y z
+      | r = z y
+      | c = [[Definition:Product Element|Product]] of both sides with $y$
+}}
+{{end-eqn}}
+So $z$ [[Definition:Commuting Elements|commutes]] with $y$.
+As $z$ is a [[Definition:Power of Group Element|power]] of $x$, $z$ also [[Definition:Commuting Elements|commutes]] with $x$.
+Hence by definition of [[Definition:Center of Group|center]]:
+:$z \in \map Z G$
+It follows that $\gen z$ is a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order $5$]].
+Let $K$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x^5$ and $y$.
+Note that:
+{{begin-eqn}}
+{{eqn | l = y x^5 y^{-1}
+      | r = \paren {y x y^{-1} }^5
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^{11} }^5
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{55}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{15} x^{15} x^{15} x^{10}
+      | c = [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | r = e \cdot e \cdot e \cdot x^{10}
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{10}
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = x^{-5}
+      | c = as $x^{15} = e$
+}}
+{{end-eqn}}
+Hence the [[Definition:Generator of Subgroup|generator]] of $K$ satisfies:
+:$\paren{x^5}^3 = e = y^2$
+and:
+:$y x^5 y^{-1} = x^{-5}$
+Let $w := x^5$.
+Then $K$ is [[Definition:Generator of Subgroup|generated]] by $w$ and $y$ where:
+:$w^3 = 1 = y^2$
+and:
+:$w y = y w^2 = y w^{-1}$
+and it is seen that $K$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Dihedral Group D3|dihedral group $D_3$]].
+It is now to be shown that $G$ is an [[Definition:Internal Group Direct Product|internal group direct product]] of $K$ and $\gen z$.
+We have that $K \cap \gen z = \set e$.
+From the [[Internal Direct Product Theorem]], we need to prove only that $K$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+We have that:
+:$\order K = 6$
+where $\order K$ denotes the [[Definition:Order of Group|order]] of $K$.
+We also have that $K$ is a [[Definition:Subgroup|subgroup]] of its [[Definition:Normalizer|normalizer]] $\map {N_G} K$.
+Hence by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$6 \divides \order {\map {N_G} K}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Again by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\order {\map {N_G} K} \divides 30$
+We have:
+{{begin-eqn}}
+{{eqn | l = x w x^{-1}
+      | r = x x^5 x^{-1}
+      | c = Definition of $w$
+}}
+{{eqn | r = x^5
+      | c = 
+}}
+{{eqn | r = w
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $w$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x y x^{-1}
+      | r = x y^{-1} x^{-1}
+      | c = as $y^2 = e$
+}}
+{{eqn | r = y x^{11} x^{-1}
+      | c = from [[Definition:Group Presentation|Group Presentation]]: $y x y^{-1} = x^{11}$
+}}
+{{eqn | r = y x^{10}
+      | c = 
+}}
+{{eqn | r = y w^2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = K
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $y$.
+Thus $x \in \map {N_G} K$ and so $\order {\map {N_G} K} > 6$.
+As $6 \divides \order {\map {N_G} K}$ and $\order {\map {N_G} K} \divides 30$, it follows that:
+:$\map {N_G} K = G$
+and so $K$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Thus:
+: $K$ and $\gen z$ are [[Definition:Normal Subgroup|normal]] in $G$
+: $K \cap \gen z = \set e$
+: $K \gen z = G$
+and it therefore follows from the  [[Internal Direct Product Theorem]] that:
+:$G = C_5 \times D_3$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Groups of Order 30/C 3 x D 5}
+Tags: Groups of Order 30
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $30$.
+Let $G$ have the [[Definition:Group Presentation|group presentation]]:
+:$\gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Group Direct Product|group direct product]] of the [[Definition:Cyclic Group|cyclic group]] $C_3$ and the [[Definition:Dihedral Group|dihedral group $D_5$]]:
+:$G \cong C_3 \times D_5$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be defined by its [[Definition:Group Presentation|group presentation]]:
+:$G = \gen {x, y: x^{15} = e = y^2, y x y^{-1} = x^4}$
+Let $z$ denote $x^5$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y z y^{-1}
+      | r = y x^5 y^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {y x y^{-1} }^5
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^4}^5
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{20}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{15} x^5
+      | c = [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | r = e \cdot x^5
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^5
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = z
+      | c = Definition of $z$
+}}
+{{eqn | ll= \leadsto
+      | l = y z
+      | r = z y
+      | c = [[Definition:Product Element|Product]] of both sides with $y$
+}}
+{{end-eqn}}
+So $z$ [[Definition:Commuting Elements|commutes]] with $y$.
+As $z$ is a [[Definition:Power of Group Element|power]] of $x$, $z$ also [[Definition:Commuting Elements|commutes]] with $x$.
+Hence by definition of [[Definition:Center of Group|center]]:
+:$z \in \map Z G$
+It follows that $\gen z$ is a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order $3$]].
+Let $N$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x^3$ and $y$.
+Note that:
+{{begin-eqn}}
+{{eqn | l = y x^3 y^{-1}
+      | r = \paren {y x y^{-1} }^3
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^4}^5
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{12}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{-3}
+      | c = as $x^{15} = e$
+}}
+{{end-eqn}}
+Hence the [[Definition:Generator of Subgroup|generator]] of $N$ satisfies:
+:$\paren{x^3}^5 = e = y^2$
+and:
+:$y x^3 y^{-1} = x^{-3}$
+Let $w := x^3$.
+Then $N$ is [[Definition:Generator of Subgroup|generated]] by $w$ and $y$ where:
+:$w^5 = 1 = y^2$
+and:
+:$w y = y w^4 = y w^{-1}$
+and it is seen that $N$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Dihedral Group|dihedral group $D_5$]].
+It is now to be shown that $G$ is an [[Definition:Internal Group Direct Product|internal group direct product]] of $N$ and $\gen z$.
+We have that $N \cap \gen z = \set e$.
+From the [[Internal Direct Product Theorem]], we need to prove only that $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+We have that:
+:$\order N = 10$
+where $\order N$ denotes the [[Definition:Order of Group|order]] of $N$.
+We also have that $N$ is a [[Definition:Subgroup|subgroup]] of its [[Definition:Normalizer|normalizer]] $\map {N_G} N$.
+Hence by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$10 \divides \order {\map {N_G} N}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Again by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\order {\map {N_G} N} \divides 30$
+We have:
+{{begin-eqn}}
+{{eqn | l = x w x^{-1}
+      | r = x x^3 x^{-1}
+      | c = Definition of $w$
+}}
+{{eqn | r = x^3
+      | c = 
+}}
+{{eqn | r = w
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $w$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x y x^{-1}
+      | r = x y^{-1} x^{-1}
+      | c = as $y^2 = e$
+}}
+{{eqn | r = y x^4 x^{-1}
+      | c = from [[Definition:Group Presentation|Group Presentation]]: $y x y^{-1} = x^4$
+}}
+{{eqn | r = y x^3
+      | c = 
+}}
+{{eqn | r = y w
+      | c = 
+}}
+{{eqn | o = \in
+      | r = N
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $y$.
+Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 10$.
+As $10 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 30$, it follows that:
+:$\map {N_G} N = G$
+and so $N$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Thus:
+: $N$ and $\gen z$ are [[Definition:Normal Subgroup|normal]] in $G$
+: $N \cap \gen z = \set e$
+: $N \gen z = G$
+and it therefore follows from the  [[Internal Direct Product Theorem]] that:
+:$G = C_3 \times D_5$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dihedral Group D6 is Internal Direct Product of C2 with D3}
+Tags: Dihedral Group D6, Examples of Internal Group Direct Products
+
+\begin{theorem}
+The [[Definition:Dihedral Group D6|dihedral group $D_6$]] is an [[Definition:Internal Group Direct Product|internal direct product]] of the [[Definition:Cyclic Group|cyclic group]] $C_2$ of [[Definition:Order of Group|order $2$]] and the [[Definition:Dihedral Group D3|dihedral group $D_3$]]:
+:$D_6 = C_2 \times D_3$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be defined by its [[Definition:Group Presentation|group presentation]]:
+:$G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^{-1} }$
+or:
+:$G = \gen {x, y: x^6 = e = y^2, y x y^{-1} = x^5}$
+Let $z$ denote $x^3$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y z y^{-1}
+      | r = y x^3 y^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {y x y^{-1} }^3
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^5}^3
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{15}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{12} x^3
+      | c = [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | r = e \cdot x^3
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^3
+      | c = {{GroupAxiom|2}}
+}}
+{{eqn | r = z
+      | c = Definition of $z$
+}}
+{{eqn | ll= \leadsto
+      | l = y z
+      | r = z y
+      | c = [[Definition:Product Element|Product]] of both sides with $y$
+}}
+{{end-eqn}}
+So $z$ [[Definition:Commuting Elements|commutes]] with $y$.
+As $z$ is a [[Definition:Power of Group Element|power]] of $x$, $z$ also [[Definition:Commuting Elements|commutes]] with $x$.
+Hence by definition of [[Definition:Center of Group|center]]:
+:$z \in \map Z G$
+It follows that $\gen z$ is a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order $2$]].
+Let $N$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x^2$ and $y$.
+Note that:
+{{begin-eqn}}
+{{eqn | l = y x^2 y^{-1}
+      | r = \paren {y x y^{-1} }^2
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = \paren {x^5}^2
+      | c = from [[Definition:Group Presentation|Group Presentation]]
+}}
+{{eqn | r = x^{10}
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = x^{-2}
+      | c = as $x^{10} = e$
+}}
+{{end-eqn}}
+Hence the [[Definition:Generator of Subgroup|generator]] of $N$ satisfies:
+:$\paren{x^2}^3 = e = y^2$
+and:
+:$y x^2 y^{-1} = x^{-2}$
+Let $w := x^3$.
+Then $N$ is [[Definition:Generator of Subgroup|generated]] by $w$ and $y$ where:
+:$w^3 = 1 = y^2$
+and:
+:$w y = y w^2 = y w^{-1}$
+and it is seen that $N$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Dihedral Group|dihedral group $D_3$]].
+It is now to be shown that $G$ is an [[Definition:Internal Group Direct Product|internal group direct product]] of $N$ and $\gen z$.
+We have that $N \cap \gen z = \set e$.
+From the [[Internal Direct Product Theorem]], we need to prove only that $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+We have that:
+:$\order N = 6$
+where $\order N$ denotes the [[Definition:Order of Group|order]] of $N$.
+We also have that $N$ is a [[Definition:Subgroup|subgroup]] of its [[Definition:Normalizer|normalizer]] $\map {N_G} N$.
+Hence by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$6 \divides \order {\map {N_G} N}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Again by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\order {\map {N_G} N} \divides 12$
+We have:
+{{begin-eqn}}
+{{eqn | l = x w x^{-1}
+      | r = x x^2 x^{-1}
+      | c = Definition of $w$
+}}
+{{eqn | r = x^2
+      | c = 
+}}
+{{eqn | r = w
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $w$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x y x^{-1}
+      | r = x y^{-1} x^{-1}
+      | c = as $y^2 = e$
+}}
+{{eqn | r = y x^5 x^{-1}
+      | c = from [[Definition:Group Presentation|Group Presentation]]: $y x y^{-1} = x^5$
+}}
+{{eqn | r = y x^2
+      | c = 
+}}
+{{eqn | r = y w
+      | c = 
+}}
+{{eqn | o = \in
+      | r = N
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $x$ is [[Definition:Conjugate of Group Element|conjugate]] to $y$.
+Thus $x \in \map {N_G} N$ and so $\order {\map {N_G} N} > 6$.
+As $6 \divides \order {\map {N_G} N}$ and $\order {\map {N_G} N} \divides 12$, it follows that:
+:$\map {N_G} N = G$
+and so $N$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Thus:
+: $N$ and $\gen z$ are [[Definition:Normal Subgroup|normal]] in $G$
+: $N \cap \gen z = \set e$
+: $N \gen z = G$
+and it therefore follows from the  [[Internal Direct Product Theorem]] that:
+:$G = C_2 \times D_3$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Integers defining Abelian Group}
+Tags: Abelian Groups, Sequence of Integers defining Abelian Group
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $C_n$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+Then $C_n$ is of the form:
+:$C_{n_1} \times C_{n_2} \times \cdots \times C_{n_r}$
+such that:
+:$n = \displaystyle \prod_{k \mathop = 1}^r n_k$
+:$\forall k \in \set {2, 3, \ldots, r}: n_k \divides n_{k - 1}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|This is probably just a statement of [[Fundamental Theorem of Finite Abelian Groups]], which needs to be studied to see what it actually means}}
+\end{proof}<|endoftext|>
+\section{Parallelism is Equivalence Relation}
+Tags: Parallel Lines, Examples of Equivalence Relations
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Straight Line|straight lines]] in [[Definition:The Plane|the plane]].
+For $l_1, l_2 \in S$, let $l_1 \parallel l_2$ denote that $l_1$ is [[Definition:Parallel Lines|parallel]] to $l_2$.
+Then $\parallel$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== [[Parallelism is Equivalence Relation/Reflexivity|Reflexivity]] ===
+{{:Parallelism is Equivalence Relation/Reflexivity}}{{qed|lemma}}
+=== [[Parallelism is Equivalence Relation/Symmetry|Symmetry]] ===
+{{:Parallelism is Equivalence Relation/Symmetry}}{{qed|lemma}}
+=== [[Parallelism is Equivalence Relation/Transitivity|Transitivity]] ===
+{{:Parallelism is Equivalence Relation/Transitivity}}{{qed|lemma}}
+$\parallel$ 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{Connected Equivalence Relation is Trivial}
+Tags: Connected Relations, Equivalence Relations, Trivial Relation, Connected Equivalence Relation is Trivial
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\mathcal R$ be a [[Definition:Endorelation|relation on $S$]] which is both [[Definition:Connected Relation|connected]] and an [[Definition:Equivalence Relation|equivalence relation]].
+Then $\mathcal R$ is the [[Definition:Trivial Relation|trivial relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Equivalence Relation|equivalence relation]], $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence relation]] {{iff}}:
+:$\Delta_S \cup \mathcal R^{-1} \cup \mathcal R \circ \mathcal R \subseteq \mathcal R$
+From [[Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation]]:
+:$\Delta_S \cup \mathcal R^{-1} \cup \mathcal R = S \times S$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Sufficient Condition/Lemma 3}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $\sequence {x_n}$ be the [[Definition:Real Sequence|real sequence]] defined as $x_n = \paren {n + 1}^{1/n}$, using [[Definition:Real Exponential Function|exponentiation]].
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence|converges]] with a [[Definition:Limit of Sequence (Number Field)|limit]] of $1$.
+\end{theorem}
+
+\begin{proof}
+We have the definition of the [[Definition:Power to Real Number|power to a real number]]:
+:$\paren {n + 1}^{1/n} = \map \exp {\dfrac 1 n \, \map \ln {n + 1} }$
+For $n >= 1$ then $n + 1 \le 2 n$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \frac 1 n \, \map \ln {n + 1}
+      | o = \le
+      | r = \frac 1 n \, \map \ln {2 n}
+      | c = [[Logarithm is Strictly Increasing]]
+}}
+{{eqn | o =
+}}
+{{eqn | r = \frac 1 n \paren {\ln 2 + \ln n}
+      | c = [[Logarithm on Positive Real Numbers is Group Isomorphism]]
+}}
+{{eqn | o =}}
+{{eqn | r = \frac {\ln 2} n + \frac 1 n \ln n
+      | c = 
+}}
+{{end-eqn}}
+By [[Powers Drown Logarithms]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$
+By [[Sequence of Reciprocals is Null Sequence]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac 1 n = 0$
+By [[Combined Sum Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {\frac {\ln 2} n + \frac 1 n \ln n} = \ln 2 \cdot 0 + 0 = 0$
+By the [[Squeeze Theorem for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = 0$
+Hence:
+:$\displaystyle \lim_{n \mathop \to \infty} \paren {n + 1}^{1/n} = \exp 0 = 1$
+and the result follows.
+{{qed}}
+[[Category:Characterisation of Non-Archimedean Division Ring Norms]]
+4gtwgr743fe4lo5o5j9ncxmmwnpxg30
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Corollary 1}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+$\norm {\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] {{iff}}:
+:$\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0}} = 1$.
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\norm {\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] {{iff}}:
+::$\sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} \le 1$
+By [[Properties of Norm on Division Ring/Norm of Unity|norm of unity]] then:
+:$\norm {1_R} = 1$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Corollary 3}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+$\norm {\,\cdot\,}$ is [[Definition:Archimedean Division Ring Norm|Archimedean]] {{iff}}:
+:$\sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0} } = +\infty$
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms]]:
+:$\norm{\,\cdot\,}$ is [[Definition:Archimedean Division Ring Norm|Archimedean]] $\iff \sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} \gt 1$
+By [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 2|Corollary 2]]:
+:$\sup \set {\norm{n \cdot 1_R}: n \in \N_{\gt 0} } \gt 1 \iff \sup \set {\norm {n \cdot 1_R}: n \in \N_{\gt 0}} = +\infty$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Corollary 2}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0} } = C < +\infty$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $C > 1$.
+By [[Characterizing Property of Supremum of Subset of Real Numbers]]:
+:$\exists m \in \N_{> 0}: \norm {m \cdot 1_R} > 1$
+Let   
+:$x = m \cdot 1_R$ 
+:$y = x^{-1}$
+By [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]]:
+:$\norm y < 1$
+By [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm y^n = 0$
+By [[Reciprocal of Null Sequence]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac 1 {\norm y^n} = +\infty$
+For all $n \in \N_{> 0}$:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {\norm y^n}
+      | r = \norm {y^{-1} }^n
+      | c = [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]]
+}}
+{{eqn | r = \norm x^n
+      | c = Definition of $y$
+}}
+{{eqn | r = \norm {x^n}
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | r = \norm {\paren {m \cdot 1_R}^n}
+      | c = Definition of $x$
+}}
+{{eqn | r = \norm {m^n \cdot 1_R}
+}}
+{{end-eqn}}
+So:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {m^n \cdot 1_R} = +\infty$
+Hence:
+:$\sup \set {\norm{n \cdot 1_R}: n \in \N_{> 0} } = +\infty$
+This [[Definition:Contradiction|contradicts]] the assumption that $C < +\infty$.
+{{qed|lemma}}
+It follows that $C \le 1$.
+Then:
+:$\forall n \in \N_{>0}: \norm{n \cdot 1_R} \le 1$
+By [[Characterisation of Non-Archimedean Division Ring Norms]], $\norm{\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+By [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 1|Corollary 1]]:
+:$\sup \set {\norm {n \cdot 1_R}: n \in \N_{> 0} } = 1$
+So $C = 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Corollary 4}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+Let $R$ have [[Definition:Characteristic of Ring|characteristic]] $p > 0$.
+Then $\norm {\,\cdot\,}$ is a [[Definition:Non-Archimedean Division Ring Norm|non_Archimedean norm]] on $R$.
+\end{theorem}
+
+\begin{proof}
+Because $R$ has [[Definition:Characteristic of Ring|characteristic]] $p > 0$, the [[Definition:Set|set]]:
+:$\set {n \cdot 1_k: n \in \Z}$
+has [[Definition:Cardinality|cardinality]] $p - 1$.
+Therefore:
+:$\sup \set {\norm {n \cdot 1_R}: n \in \Z} = \max \set {\norm {1 \cdot 1_R}, \norm {2 \cdot 1_R}, \cdots, \norm {\paren {p - 1} \cdot 1_R} } < +\infty$
+By [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 2|Corollary 2]]:
+:$\norm{\,\cdot\,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] and $C = 1$.
+{{qed}}
+[[Category:Characterisation of Non-Archimedean Division Ring Norms]]
+qpz1930i9qr4qtkaa39un5u70jdvup9
+\end{proof}<|endoftext|>
+\section{Norms Equivalent to Absolute Value on Rational Numbers}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\alpha \in \R_{\gt 0}$.
+Let $\norm{\,\cdot\,}:\Q \to \R$ be the [[Definition:Mapping|mapping]] defined by:
+:$\forall x \in \Q: \norm{x} = \size {x}^\alpha$
+where $\size {x}$ is the [[Definition:Absolute Value|absolute value]] of $x$ in $\Q$.
+Then:
+:$\norm{\,\cdot\,}$ is a [[Definition:Norm/Division Ring|norm]] on $\Q$ {{iff}} $\,\,\alpha \le 1$
+\end{theorem}
+
+\begin{proof}
+=== [[Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition|Necessary Condition]] ===
+{{:Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition}}{{qed|lemma}}
+=== [[Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition|Sufficient Condition]] ===
+{{:Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Reflexive Relation on Set of Cardinality 2 is Transitive}
+Tags: Reflexive Relations, Transitive Relations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is equal to $2$:
+:$\card S = 2$
+Let $\odot \subseteq S \times S$ be a [[Definition:Reflexive Relation|reflexive relation]] on $S$.
+Then $\odot$ is also [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, let $S = \set {a, b}$.
+Let $\odot$ be [[Definition:Reflexive Relation|reflexive]].
+By definition of [[Definition:Reflexive Relation|reflexive relation]]:
+:$\Delta_S \subseteq \odot$
+where $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal relation]]:
+:$\Delta_S = \set {\tuple {x, x}: x \in S}$
+That is:
+:$\set {\tuple {a, a}, \tuple {b, b} } \subseteq \odot$
+Suppose $\set {\tuple {a, a}, \tuple {b, b} } = \odot$.
+Then by [[Diagonal Relation is Equivalence]], $\odot$ is [[Definition:Transitive Relation|transitive]].
+Suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \notin \odot$.
+Then:
+:$x \odot x, x \odot y \implies x \odot y$
+and:
+:$x \odot y, y \odot y \implies x \odot y$
+Now suppose there exists $\tuple {x, y} \in \odot$ such that $\tuple x \ne y$ and $\tuple {y, x} \in \odot$.
+Then we have:
+:$x \odot y, y \odot x \implies x \odot x$
+and:
+:$y \odot x, x \odot y \implies y \odot y$
+which hold because $\odot$ is [[Definition:Reflexive Relation|reflexive]]
+Hence in all cases, a [[Definition:Reflexive Relation|reflexive relation on $S$]] is also [[Definition:Transitive Relation|transitive]].
+{{qed}}
+[[Category:Reflexive Relations]]
+[[Category:Transitive Relations]]
+lhc0zlz6rinxrqyqa85wu1x5pkupht9
+\end{proof}<|endoftext|>
+\section{Relation on Set of Cardinality 2 cannot be Non-Symmetric and Non-Transitive}
+Tags: Symmetric Relations, Transitive Relations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is equal to $2$:
+:$\card S = 2$
+Let $\odot \subseteq S \times S$ be a [[Definition:Endorelation|relation on $S$]].
+Then it is not possible for $\odot$ to be not [[Definition:Symmetric Relation|symmetric]] and also not [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, let $S = \set {a, b}$.
+Let $\odot$ not be [[Definition:Symmetric Relation|symmetric]].
+{{AimForCont}} $\odot$ is not [[Definition:Transitive Relation|transitive]].
+As $\odot$ is not [[Definition:Symmetric Relation|symmetric]]:
+:$\exists \tuple {x, y} \in \odot: \tuple {y, x} \notin \odot$
+Thus there are two possibilities:
+:$\exists \tuple {a, b} \in \odot: \tuple {b, a} \notin \odot$
+:$\exists \tuple {b, a} \in \odot: \tuple {a, b} \notin \odot$
+{{WLOG}}, let $\tuple {a, b} \in \odot$.
+As $\odot$ is not [[Definition:Transitive Relation|transitive]], either:
+:$\exists \tuple {x, a} \in \odot: \tuple {x, b} \notin \odot$
+or:
+:$\exists \tuple {b, y} \in \odot: \tuple {a, y} \notin \odot$
+In the first case, $x \ne b$ as that would make $\odot$ [[Definition:Symmetric Relation|symmetric]].
+In the second case, $y \ne a$ as that would also make $\odot$ [[Definition:Symmetric Relation|symmetric]].
+Hence $x = a$ and $y = b$, and we have either:
+:$\exists \tuple {a, a} \in \odot: \tuple {a, b} \notin \odot$
+or:
+:$\exists \tuple {b, b} \in \odot: \tuple {a, b} \notin \odot$
+But it has been established that $\tuple {a, b} \in \odot$.
+From this [[Definition:Contradiction|contradiction]] it follows that $\odot$ is [[Definition:Transitive Relation|transitive]].
+Hence it is not possible for $\odot$ to be neither [[Definition:Symmetric Relation|symmetric]] nor [[Definition:Transitive Relation|transitive]].
+{{qed}}
+[[Category:Symmetric Relations]]
+[[Category:Transitive Relations]]
+8tm3k4p73d1idz37803ov16ps7v5yv1
+\end{proof}<|endoftext|>
+\section{Relations with Combinations of Reflexivity, Symmetry and Transitivity Properties}
+Tags: Reflexive Relations, Symmetric Relations, Transitive Relations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] which has at least $3$ [[Definition:Element|elements]].
+Then it is possible to set up a [[Definition:Endorelation|relation $\circledcirc$ on $S$]] which has any combination of the $3$ properties:
+:[[Definition:Reflexive Relation|Reflexivity]]
+:[[Definition:Symmetric Relation|Symmetry]]
+:[[Definition:Transitive Relation|Transitivity]]
+but this is not possible for a [[Definition:Set|set]] which has fewer than $3$ [[Definition:Element|elements]].
+\end{theorem}
+
+\begin{proof}
+In the following:
+:$S_n$ denotes the [[Definition:Set|set]] $S_n = \set {s_1, s_2, \ldots, s_n}$ of [[Definition:Cardinality|cardinality]] $n$, where $n \in \Z_{\ge 0}$ is a [[Definition:Non-Negative Integer|non-negative integer]].
+:$\circledcirc$ denotes an arbitrary [[Definition:Endorelation|relation on $S_n$]].
+Let:
+:$\map R \circledcirc$ denote that $\circledcirc$ is [[Definition:Reflexive Relation|reflexive]]
+:$\map S \circledcirc$ denote that $\circledcirc$ is [[Definition:Symmetric Relation|symmetric]]
+:$\map T \circledcirc$ denote that $\circledcirc$ is [[Definition:Transitive Relation|transitive]]
+=== Set of Cardinality $0$ ===
+When $S = S_0 = \O$, the result [[Relation on Empty Set is Equivalence]] applies.
+Hence $\circledcirc$ is always [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]]:
+:$\map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+{{qed|lemma}}
+=== Set of Cardinality $1$ ===
+Let $S = S_1 = \set {s_1}$.
+From [[Relation on Set of Cardinality 1 is Symmetric and Transitive]], $\circledcirc$ [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Thus when $n = 1$, it is not possible for a [[Definition:Endorelation|relation on $S_n$]] to be either [[Definition:Non-symmetric Relation|non-symmetric]] or [[Definition:Non-transitive Relation|non-transitive]]
+Hence either:
+:$\neg \map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+when $\circledcirc$ is the [[Definition:Null Relation|null relation]], or:
+:$\map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+when $\circledcirc$ is the [[Definition:Trivial Relation|trivial relation]].
+{{qed|lemma}}
+=== Set of Cardinality $2$ ===
+Let $S = S_2 = \set {s_1, s_2}$.
+From [[Reflexive Relation on Set of Cardinality 2 is Transitive]], it is not possible for $\circledcirc$ to be both [[Definition:Reflexive Relation|reflexive]] and [[Definition:Non-transitive Relation|non-transitive]].
+From [[Relation on Set of Cardinality 2 cannot be Non-Symmetric and Non-Transitive]], it is not possible for $\circledcirc$ to be neither [[Definition:Symmetric Relation|symmetric]] nor [[Definition:Transitive Relation|transitive]].
+Thus the following combinations of $R$, $S$ and $T$ are not possible with $S_2$:
+:$\neg \map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$
+:$\map R \circledcirc, \neg \map S \circledcirc, \neg \map T \circledcirc$
+:$\map R \circledcirc, \map S \circledcirc, \neg \map T \circledcirc$
+However, we note the following examples of the possible combinations of properties for [[Definition:Endorelation|relations on $S_2$]]:
+$(1): \quad \circledcirc := \O$, which is the [[Definition:Null Relation|null relation]] on $S_2$.
+From [[Null Relation is Antireflexive, Symmetric and Transitive]], $\circledcirc$ is always [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]], and so:
+:$\neg \map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+$(2):$
+:$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2}, \tuple {s_2, s_1}, \tuple {s_2, s_2} } = S \times S$
+that is, the [[Definition:Trivial Relation|trivial relation]].
+From [[Trivial Relation is Equivalence]]:.
+:$\map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+$(3): \quad \circledcirc := \set {\tuple {s_1, s_2} }$:
+By inspection:
+:$\neg \map R \circledcirc$
+:$\map S \circledcirc$
+:$\neg \map T \circledcirc$
+$(4):$
+:$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_1, s_2} }$
+By inspection:
+:$\neg \map R \circledcirc$
+:$\neg \map S \circledcirc$
+:$\map T \circledcirc$
+$(5):$
+:$\circledcirc := \set {\tuple {s_1, s_1}, \tuple {s_2, s_2}, \tuple {s_1, s_2} }$
+By inspection:
+:$\map R \circledcirc$
+:$\neg \map S \circledcirc$
+:$\map T \circledcirc$
+{{qed|lemma}}
+=== Set of Cardinality $\ge 3$ ===
+Let $S_n = \set {s_1, s_2, \ldots, s_n}$.
+In the following, the smallest examples of [[Definition:Endorelation|relations on $S_n$]] will be given to exemplify each combination of $R$, $S$ and $T$,
+$(1): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$
+We have:
+:$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$
+:$\neg \map S \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$
+:$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$.
+$(2): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_3}, \tuple {s_1, s_3} }$
+We have:
+:$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$
+:$\neg \map S \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$
+We note that there exists exactly one [[Definition:Ordered Pair|pair]] of [[Definition:Element|elements]] of $\circledcirc$ of the form $\tuple {\tuple {x, y}, \tuple {y, z} }$:
+:$\tuple {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$
+and in this case we also have that:
+:$\tuple {\tuple {s_1, s_3} }$
+Hence:
+:$\map T \circledcirc$
+$(3): \quad \circledcirc = \set {\tuple {s_1, s_2}, \tuple {s_2, s_1} }$
+We have:
+:$\neg \map R \circledcirc$ because for example $\tuple {s_1, s_1} \notin \circledcirc$
+:$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_1} \in \circledcirc$ but $\tuple {s_1, s_1} \notin \circledcirc$.
+We have:
+:$\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_1} \in \circledcirc$
+and so in all cases:
+:$\tuple {x, y} \in \circledcirc \implies \tuple {y, x} \in \circledcirc$
+Hence:
+:$\map S \circledcirc$
+$(4): \quad \circledcirc = \O$, that is, the [[Definition:Null Relation|null relation]].
+From [[Null Relation is Antireflexive, Symmetric and Transitive]]:
+:$\neg \map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+It remains to provide examples of $\circledcirc$ for which $\circledcirc$ is [[Definition:Reflexive Relation|reflexive]].
+By definition, $\circledcirc$ is [[Definition:Reflexive Relation/Definition 2|reflexive]] {{iff}} it is a [[Definition:Superset|superset]] of the [[Definition:Diagonal Relation|diagonal relation]]: 
+:$\Delta_S \subseteq \circledcirc$
+So in the remaining examples it is taken as read that $\circledcirc$ is [[Definition:Reflexive Relation|reflexive]]
+$(5): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2}, \tuple {s_2, s_3} }$
+:$\map R \circledcirc$ as discussed
+:$\neg \map S \circledcirc$ because $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$
+:$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$.
+$(6): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2} }$
+:$\map R \circledcirc$ as discussed
+:$\neg \map S \circledcirc$ because $\tuple {s_1, s_2} \in \circledcirc$ but $\tuple {s_2, s_1} \notin \circledcirc$
+Apart from the trivial $\tuple {\tuple {x, x}, \tuple {x, x} }$ pairs there exist exactly two [[Definition:Ordered Pair|pair]] of [[Definition:Element|elements]] of $\circledcirc$ of the form $\tuple {\tuple {x, y}, \tuple {y, z} }$:
+:$\tuple {\tuple {s_1, s_1}, \tuple {s_1, s_2} } \implies \tuple {s_1, s_2}$
+:$\tuple {\tuple {s_1, s_2}, \tuple {s_2, s_2} } \implies \tuple {s_1, s_2}$
+and so:
+:$\map T \circledcirc$
+$(7): \quad \circledcirc = \Delta_{S_n} \cup \set {\tuple {s_1, s_2}, \tuple {s_2, s_3}, \tuple {s_3, s_2}, \tuple {s_2, s_1} }$
+:$\map R \circledcirc$ as discussed
+:$\map S \circledcirc$ by inspection
+:$\neg \map T \circledcirc$ because for example $\tuple {s_1, s_2} \in \circledcirc$ and $\tuple {s_2, s_3} \in \circledcirc$ but $\tuple {s_1, s_3} \notin \circledcirc$.
+$(8): \quad \circledcirc = \Delta_{S_n}$
+From [[Diagonal Relation is Equivalence]]:
+:$\map R \circledcirc$
+:$\map S \circledcirc$
+:$\map T \circledcirc$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation on Set of Cardinality 1 is Symmetric and Transitive}
+Tags: Symmetric Relations, Transitive Relations
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] whose [[Definition:Cardinality|cardinality]] is equal to $1$:
+:$\card S = 1$
+Let $\odot \subseteq S \times S$ be a [[Definition:Endorelation|relation on $S$]].
+Then $\odot$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, let $S = \set a$.
+There are $2$ [[Definition:Endorelation|relations on $S$]]:
+$(1): \quad \odot := \O$, which is the [[Definition:Null Relation|null relation]] on $S$.
+From [[Null Relation is Antireflexive, Symmetric and Transitive]], $\odot$ is [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Thus in this case $\odot$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+$(2): \quad \odot := \set {\tuple {a, a} }$, which is the [[Definition:Trivial Relation|trivial relation]] on $S$.
+From [[Trivial Relation is Equivalence]], $\odot$ is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+Thus in this case also, $\odot$ is both [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
+{{qed}}
+[[Category:Symmetric Relations]]
+[[Category:Transitive Relations]]
+3mj1jd4zecex4gsih4bo4ufrwlqndfh
+\end{proof}<|endoftext|>
+\section{Congruence of Triangles is Equivalence Relation}
+Tags: Examples of Equivalence Relations, Triangles
+
+\begin{theorem}
+Let $S$ denote the [[Definition:Set|set]] of all [[Definition:Triangle (Geometry)|triangles]] in [[Definition:The Plane|the plane]].
+Let $\triangle A \cong \triangle B$ denote the [[Definition:Relation|relation]] that $\triangle A$ is [[Definition:Congruence (Geometry)|congruent]] to $\triangle B$.
+Then $\cong$ is an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+Let $\triangle A$ be a [[Definition:Triangle (Geometry)|triangle]].
+By definition, by [[Triangle Side-Side-Side Equality]], $\triangle A$ is trivially [[Definition:Congruence (Geometry)|congruent]] to itself.
+Thus $\cong$ is seen to be [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+Let $\triangle A \cong \triangle B$.
+Then:
+:all the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Side of Polygon|sides]] of $\triangle B$
+:all the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle B$.
+It follows directly that:
+:all the [[Definition:Side of Polygon|sides]] of $\triangle B$ are equal to the [[Definition:Side of Polygon|sides]] of $\triangle A$
+:all the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle B$ are equal to the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle A$.
+That is:
+: $\triangle B \cong \triangle A$
+Thus $\cong$ is seen to be [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let:
+:$\triangle A \cong \triangle B$
+:$\triangle B \cong \triangle C$
+Then:
+:all the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Side of Polygon|sides]] of $\triangle B$
+:all the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle B$.
+and:
+:all the [[Definition:Side of Polygon|sides]] of $\triangle B$ are equal to the [[Definition:Side of Polygon|sides]] of $\triangle C$
+:all the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle B$ are equal to the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle C$.
+From [[Equality is Equivalence Relation]], it follows that:
+:all the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Side of Polygon|sides]] of $\triangle C$
+:all the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle A$ are equal to the [[Definition:Angle|angles]] contained by the [[Definition:Side of Polygon|sides]] of $\triangle C$.
+That is:
+: $\triangle A \cong \triangle C$
+Thus $\cong$ is seen to be [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+$\cong$ 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{Cardinality of Mapping}
+Tags: Mapping Theory, Cardinality
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]] whose [[Definition:Cardinality|cardinality]] is $n$:
+:$\card S = n$
+Let $T$ be a [[Definition:Non-Empty|non-empty set]]
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then:
+:$\card f = n$
+\end{theorem}
+
+\begin{proof}
+First suppose $T = \O$ there are no [[Definition:Element|elements]] in $f$
+From [[Null Relation is Mapping iff Domain is Empty Set]], there are no [[Definition:Element|elements]] in $f$.
+Hence in this case $\card f = 0$, whatever $\card S$ may be,
+By definition of [[Definition:Mapping|mapping]], $f$ is a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]] $\tuple {s, t}$ where $s \in S$ and $t \in T$, such that:
+:$(1): \quad \forall s \in S: \exists \tuple {s, t} \in f$
+and:
+:$(2): \quad \forall x \in S: \tuple {s, t_1} \in f \land \tuple {s, t_2} \in f \implies t_1 = t_2$
+From $(1)$ it follows that $\card f \ge n$.
+{{AimForCont}} that $\card f > n$.
+Then by the [[Pigeonhole Principle]]:
+:$\exists x \in f: \exists y_1, y_2 \in T, y_1 \ne y_2: \tuple {x, y_1} \in f \land \tuple {x, y_2} \in f$
+and so $f$ is not a [[Definition:Mapping|mapping]].
+From this [[Definition:Contradiction|contradiction]] it follows that $\card f = n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Integer Plus 5 is Odd}
+Tags: Even Integers, Odd Integers, Even Integer Plus 5 is Odd
+
+\begin{theorem}
+Let $x \in \Z$ be an [[Definition:Even Integer|even integer]].
+Then $x + 5$ is [[Definition:Odd Integer|odd]].
+\end{theorem}
+
+\begin{proof}
+Let $x$ be an [[Definition:Even Integer|even integer]].
+Let $y = 2 n + 5$.
+Assume $y = x + 5$ is not an [[Definition:Odd Integer|odd integer]].
+Then:
+:$y = x + 5 = 2 n$
+where $n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 2 n - 5
+      | c = 
+}}
+{{eqn | r = \paren {2 n - 6} + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {n - 3} + 1
+      | c = 
+}}
+{{eqn | r = 2 r + 1
+      | c = where $r = n - 3 \in \Z$
+}}
+{{end-eqn}}
+Hence $x$ is [[Definition:Odd Integer|odd]].
+That is, it is [[Definition:False|false]] that $x$ is [[Definition:Even Integer|even]].
+It follows by the [[Rule of Transposition]] that if $x$ is [[Definition:Even Integer|even]], then $y$ is [[Definition:Odd Integer|odd]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x$ be an [[Definition:Even Integer|even integer]].
+Then by definition:
+:$x = 2 n$
+for some [[Definition:Integer|integer]] $n$.
+{{AimForCont}} $y = x + 5 = 2 m$ for some [[Definition:Integer|integer]] $m$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 2 m - 5
+      | c = 
+}}
+{{eqn | r = \paren {2 m - 6} + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {m - 3} + 1
+      | c = 
+}}
+{{eqn | r = 2 r + 1
+      | c = where $r = m - 3 \in \Z$
+}}
+{{end-eqn}}
+Hence $x$ is [[Definition:Odd Integer|odd]].
+But this [[Definition:Contradiction|contradicts]] our [[Definition:Premise|premise]] that $x$ is [[Definition:Even Integer|even]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Integer Plus 5 is Odd/Direct Proof}
+Tags: Even Integer Plus 5 is Odd
+
+\begin{theorem}
+{{:Even Integer Plus 5 is Odd}}
+\end{theorem}<|endoftext|>
+\section{Even Integer Plus 5 is Odd/Indirect Proof}
+Tags: Even Integer Plus 5 is Odd
+
+\begin{theorem}
+{{:Even Integer Plus 5 is Odd}}
+\end{theorem}
+
+\begin{proof}
+Let $x$ be an [[Definition:Even Integer|even integer]].
+Let $y = 2 n + 5$.
+Assume $y = x + 5$ is not an [[Definition:Odd Integer|odd integer]].
+Then:
+:$y = x + 5 = 2 n$
+where $n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 2 n - 5
+      | c = 
+}}
+{{eqn | r = \paren {2 n - 6} + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {n - 3} + 1
+      | c = 
+}}
+{{eqn | r = 2 r + 1
+      | c = where $r = n - 3 \in \Z$
+}}
+{{end-eqn}}
+Hence $x$ is [[Definition:Odd Integer|odd]].
+That is, it is [[Definition:False|false]] that $x$ is [[Definition:Even Integer|even]].
+It follows by the [[Rule of Transposition]] that if $x$ is [[Definition:Even Integer|even]], then $y$ is [[Definition:Odd Integer|odd]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Integer Plus 5 is Odd/Proof by Contradiction}
+Tags: Even Integer Plus 5 is Odd
+
+\begin{theorem}
+{{:Even Integer Plus 5 is Odd}}
+\end{theorem}
+
+\begin{proof}
+Let $x$ be an [[Definition:Even Integer|even integer]].
+Then by definition:
+:$x = 2 n$
+for some [[Definition:Integer|integer]] $n$.
+{{AimForCont}} $y = x + 5 = 2 m$ for some [[Definition:Integer|integer]] $m$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 2 m - 5
+      | c = 
+}}
+{{eqn | r = \paren {2 m - 6} + 1
+      | c = 
+}}
+{{eqn | r = 2 \paren {m - 3} + 1
+      | c = 
+}}
+{{eqn | r = 2 r + 1
+      | c = where $r = m - 3 \in \Z$
+}}
+{{end-eqn}}
+Hence $x$ is [[Definition:Odd Integer|odd]].
+But this [[Definition:Contradiction|contradicts]] our [[Definition:Premise|premise]] that $x$ is [[Definition:Even Integer|even]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Cubes/Historical Note}
+Tags: Historical Notes
+
+\begin{theorem}
+The result '''[[Sum of Sequence of Cubes]]''' was documented by {{AuthorRef|Aryabhata the Elder}} in his work ''Āryabhaṭīya'' of $499$ CE.
+\end{theorem}<|endoftext|>
+\section{Sum of Sequence of Odd Squares/Formulation 2}
+Tags: Sum of Sequence of Odd Squares
+
+\begin{theorem}
+:$\displaystyle \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$
+\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]]:
+:$\displaystyle \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^1 \paren {2 i - 1}^2
+      | r = \paren {2 \times 1 - 1}^2
+      | c = 
+}}
+{{eqn | r = 1^2
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {4 \times 1 - 1}^3} 3
+      | c = 
+}}
+{{end-eqn}}
+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]]:
+:$\displaystyle \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 = \frac {4 k^3 - k} 3$
+from which it is to be shown that:
+:$\displaystyle \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2 = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^{k + 1} \paren {2 i - 1}^2
+      | r = \sum_{i \mathop = 1}^k \paren {2 i - 1}^2 + \paren {2 \paren {k + 1} - 1}^2
+      | c = 
+}}
+{{eqn | r = \frac {4 k^3 - k} 3 + \paren {2 k + 1}^2
+      | c = [[Sum of Sequence of Odd Squares/Formulation 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \frac {4 k^3 - k + 12 k^2 + 12 k + 3} 3
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1} \paren {4 k^2 + 8 k + 3} } 3
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1} \paren {4 \paren {k + 1}^2  - 1} } 3
+      | c = 
+}}
+{{eqn | r = \frac {4 \paren {k + 1}^3 - \paren {k + 1} } 3
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{> 0}: \sum_{i \mathop = 1}^n \paren {2 i - 1}^2 = \frac {4 n^3 - n} 3$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers}
+Tags: Sums of Sequences
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | lo= \forall n \in \Z_{>0}:
+      | l = \sum_{j \mathop = 1}^n j \paren {j + 2}
+      | r = 1 \times 3 + 2 \times 4 + 3 \times 5 + \dotsb + n \paren {n + 2}
+      | c = 
+}}
+{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 7} } 6
+      | 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]]:
+:$\displaystyle \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^1 j \paren {j + 2}
+      | r = 1 \times 3
+      | c = 
+}}
+{{eqn | r = 3
+      | c = 
+}}
+{{eqn | r = \frac {1 \paren {1 + 1} \paren {2 \times 1 + 7} } 6
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#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 [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$
+from which it is to be shown that:
+:$\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2}
+      | r = \sum_{j \mathop = 1}^k j \paren {j + 2} + \paren {k + 1} \paren {k + 3}
+      | c = 
+}}
+{{eqn | r = \frac {k \paren {k + 1} \paren {2 k + 7} } 6 + \paren {k + 1} \paren {k + 3}
+      | c = [[Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {k + 1} \paren {\frac {k \paren {2 k + 7} + 6 \paren {k + 3} } 6}
+      | c = 
+}}
+{{eqn | r = \paren {k + 1} \paren {\frac {2 k^2 + 13 k + 18} 6 }
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 k + 9} } 6
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$
+{{qed}}
+</onlyinclude>
+\end{proof}<|endoftext|>
+\section{Arcsin as an Integral}
+Tags: 
+
+\begin{theorem}
+:$\displaystyle \map \arcsin x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$
+\end{theorem}
+
+\begin{proof}
+=== [[Arcsin as an Integral/Lemma 1|Lemma 1]] ===
+{{:Arcsin as an Integral/Lemma 1}}{{qed|lemma}}
+=== [[Arcsin as an Integral/Lemma 2|Lemma 2]] ===
+{{:Arcsin as an Integral/Lemma 2}}{{qed|lemma}}
+:$\displaystyle \map \arcsin x = \map {\arcsin_A} x = \map {\arcsin_G} x = \int_0^\Theta \frac {\d x} {\sqrt {1 - x^2} }$
+{{qed}}
+hd79ptfc8vvlm8f8hye4haku88a0b3m
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Products of Consecutive Fibonacci Numbers}
+Tags: Sums of Sequences, Fibonacci Numbers
+
+\begin{theorem}
+=== [[Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers]] ===
+{{:Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers}}
+=== [[Sum of Even Sequence of Products of Consecutive Fibonacci Numbers]] ===
+{{:Sum of Even Sequence of Products of Consecutive Fibonacci Numbers}}
+\end{theorem}<|endoftext|>
+\section{Lucas Number as Element of Recursive Sequence}
+Tags: Lucas Numbers
+
+\begin{theorem}
+Let $L_k$ be the $k$th [[Definition:Lucas Number/Definition 2|Lucas number]], defined as the sum of two [[Definition:Fibonacci Number|Fibonacci numbers]]:
+:$L_n = F_{n - 1} + F_{n + 1}$
+Then $L_n$ can be defined as the $n$th [[Definition:Element|element]] of the [[Definition:Recursive Sequence|recursive sequence]]:
+:$L_n = \begin{cases}
+2 & : n = 0 \\
+1 & : n = 1 \\
+L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Second Principle of Mathematical Induction|induction]]:
+Let $L_n$ be the [[Definition:Lucas Number/Definition 2|Lucas number]] defined as the sum of two [[Definition:Fibonacci Number|Fibonacci numbers]]:
+:$L_n = F_{n - 1} + F_{n + 1}$
+For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$L_n = \begin{cases}
+2 & : n = 0 \\
+1 & : n = 1 \\
+L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
+=== Basis for the Induction ===
+We have that:
+:$L_0 = F_{-1} + F_1 = 1 + 1 = 2$
+:$L_1 = F_0 + F_2 = 0 + 1 = 1$
+:$L_2 = F_1 + F_3 = 1 + 2 = 3$
+Thus $\map P 0$, $\map P 1$ and $\map P 2$ hold.
+$\map P 3$ is the case:
+: $L_3 = F_2 + F_4 = 1 + 3 = 4$
+So $\map P 3$, as $L_3 = L_1 + L_2$.
+This is our [[Second Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Let us make the supposition that, for some $k \in \N: k \ge 1$, the proposition $\map P j$ holds for all $j \in \N: 1 \le j \le k$.
+We shall show that it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Second Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\forall 1 \le j \le k: L_j = L_{j - 1} + L_{j - 2}$
+Then we need to show:
+:$L_{k + 1} = L_k + L_{k - 1}$
+=== Induction Step ===
+This is our [[Second Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = L_{k + 1}
+      | r = F_k + F_{k + 2}
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = F_{k - 2} + F_{k - 1} + F_k + F_{k + 1}
+      | c = {{Defof|Fibonacci Number}}
+}}
+{{eqn | r = \paren {F_{k - 2} + F_k} + \paren {F_{k - 1} + F_{k + 1} }
+      | c = 
+}}
+{{eqn | r = L_{k - 1} + L_k
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{end-eqn}}
+Hence $L_n = L_{n - 2} + L_{n - 1}$ follows by the [[Second Principle of Mathematical Induction]].
+That is: $\sequence {L_n}$ is the [[Definition:Sequence|sequence]] defined as:
+:$L_n = \begin{cases}
+2 & : n = 0 \\
+1 & : n = 1 \\
+L_{n - 1} + L_{n - 2} & : \text{otherwise} \end{cases}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of nth Lucas and Fibonacci Numbers}
+Tags: Lucas Numbers, Fibonacci Numbers
+
+\begin{theorem}
+Let $L_k$ be the $k$th [[Definition:Lucas Number|Lucas number]].
+Let $F_k$ be the $k$th [[Definition:Fibonacci Number|Fibonacci number]].
+Then:
+:$\forall n \in \N_{>0}: F_n L_n = F_{2 n}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Lucas Number/Definition 2|Lucas numbers]]:
+:$L_n = F_{n - 1} + F_{n + 1}$
+Hence:
+:$F_n L_n = F_n \paren {F_{n - 1} + F_{n + 1} }$
+From [[Fibonacci Number in terms of Smaller Fibonacci Numbers]]:
+:$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
+The result follows by setting $m = n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Representation of Integers in Balanced Ternary}
+Tags: Balanced Ternary Representation
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+$n$ can be represented [[Definition:Unique|uniquely]] in [[Definition:Balanced Ternary Representation|balanced ternary]]:
+:$\displaystyle n = \sum_{j \mathop = 0}^m r_j 3^j$
+:$\sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}$
+such that:
+where:
+:$m \in \Z_{>0}$ is a [[Definition:Strictly Positive Integer|strictly positive integer]] such that $3^m < \size {2 n} < 3^{m + 1}$
+:all the $r_j$ are such that $r_j \in \set {\underline 1, 0, 1}$, where $\underline 1 := -1$.
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z$.
+Let $m \in \Z_{\ge 0}$ be such that:
+:$3^m + 1 \le \size {2 n} \le 3^{m + 1} - 1$
+where $\size {2 n}$ denotes the [[Definition:Absolute Value|absolute value]] of $2 n$.
+As $2 n$ is [[Definition:Even Integer|even]], this is always possible, because $3^r$ is always an [[Definition:Odd Integer|odd integer]] for [[Definition:Non-Negative Integer|non-negative]] $r$.
+Let $d = \dfrac {3^{m + 1} - 1} 2$.
+Let $k = n + d$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \size {2 n}
+      | o = \le
+      | r = 3^{m + 1} - 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size n
+      | o = \le
+      | r = d
+      | c = Definition of $d$
+}}
+{{eqn | ll= \leadsto
+      | l = -d
+      | o = \le
+      | r = n \le d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | r = n + d \le 3^{m + 1} - 1
+      | c = 
+}}
+{{end-eqn}}
+Let $k = n + d \in \Z$ be represented in [[Definition:Ternary Notation|ternary notation]]:
+:$k = \displaystyle \sum_{j \mathop = 0}^m s_j 3^j$
+where $s_j \in \set {0, 1, 2}$.
+By the [[Basis Representation Theorem]], this expression for $k$ is [[Definition:Unique|unique]].
+Now we have:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = \dfrac {3^{m + 1} - 1} {3 - 1}
+      | c = by definition
+}}
+{{eqn | r = \sum_{j \mathop = 0}^m 3^j
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{end-eqn}}
+Hence we see:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = k - d
+      | c = by definition
+}}
+{{eqn | r = \sum_{j \mathop = 0}^m s_j 3^j - \sum_{j \mathop = 0}^m 3^j
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^m \paren {s_j - 1} 3^j
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^m r_j 3^j
+      | c = where $r_j \in \set {-1, 0, 1}$
+}}
+{{end-eqn}}
+Hence $n$ has a representation in [[Definition:Balanced Ternary Representation|balanced ternary]].
+The representation for $k$ in [[Definition:Ternary Notation|ternary notation]] is [[Definition:Unique|unique]], as established.
+Hence the representation in [[Definition:Balanced Ternary Representation|balanced ternary]] for $n$ is also [[Definition:Unique|unique]].
+{{qed}}
+{{Proofread|Not completely sure that uniqueness has been properly proved.}}
+\end{proof}<|endoftext|>
+\section{Absolute Value is Norm}
+Tags: Examples of Norms, Absolute Value Function
+
+\begin{theorem}
+The [[Definition:Absolute Value|absolute value]] is a [[Definition:Norm on Division Ring|norm]] on the [[Definition:Real Number|set of real numbers]] $\R$.
+\end{theorem}
+
+\begin{proof}
+By [[Complex Modulus is Norm]] then the [[Definition:Complex Modulus|complex modulus]] satisfies the [[Definition:Norm on Division Ring|norm axioms]] on the [[Definition:Complex Number|set of complex numbers]] $\C$.
+Since the [[Definition:Real Number|real numbers]] $\R$ is a [[Definition:Subset|subset]] of the [[Definition:Complex Number|complex numbers]] $\C$ then the [[Definition:Complex Modulus|complex modulus]] satisfies the [[Definition:Norm on Division Ring|norm axioms]] on the [[Definition:Real Number|real numbers]] $\R$.
+By [[Complex Modulus of Real Number equals Absolute Value]] then the [[Definition:Absolute Value|absolute value]] satisfies the [[Definition:Norm on Division Ring|norm axioms]] on [[Definition:Real Number|set of real numbers]] $\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bounds for Integer Expressed in Base k}
+Tags: Number Bases
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Let $k \in \Z$ such that $k \ge 2$.
+Let $n$ be expressed in [[Definition:Number Base|base $k$ notation]]:
+:$n = \displaystyle \sum_{j \mathop = 1}^s a_j k^j$
+where each of the $a_j$ are such that $a_j \in \set {0, 1, \ldots, k - 1}$.
+Then:
+:$0 \le n < k^{s + 1}$
+\end{theorem}
+
+\begin{proof}
+As none of the [[Definition:Coefficient of Polynomial|coefficients]] $a_j$ in $\displaystyle \sum_{j \mathop = 1}^s a_j k^j$ is [[Definition:Strictly Negative Integer|(strictly) negative]], the [[Definition:Summation|summation]] itself likewise cannot be [[Definition:Strictly Negative Integer|negative]]
+Thus:
+:$0 \le n$
+The equality is satisfied when $a_j = 0$ for all $j$.
+We then have:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \sum_{j \mathop = 1}^s a_j k^j
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \paren {k - 1} \sum_{j \mathop = 1}^s k^j
+      | c = as $a_j \le k - 1$ for all $j$
+}}
+{{eqn | r = \paren {k - 1} \dfrac {k^{s + 1} - 1} {k - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = k^{s + 1} - 1
+      | c = 
+}}
+{{eqn | o = <
+      | r = k^{s + 1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Different Representations to Number Base represent Different Integers}
+Tags: Number Bases
+
+\begin{theorem}
+Let $k \in \Z$ such that $k \ge 2$.
+Let $a$ and $b$ be representations of [[Definition:Integer|integers]] in [[Definition:Number Base|base $k$ notation]]:
+:$a = \displaystyle \sum_{j \mathop = 0}^r a_j k^j$
+:$b = \displaystyle \sum_{j \mathop = 0}^s b_j k^j$
+such that either:
+:$r \ne s$
+or:
+:$\exists j \in \set {0, 1, \ldots, r}: a_j \ne b_j$
+Then $a$ and $b$ represent different [[Definition:Integer|integers]].
+\end{theorem}
+
+\begin{proof}
+First suppose that $r \ne s$.
+{{WLOG}}, suppose $r > s$.
+Then from [[Bounds for Integer Expressed in Base k]]:
+{{begin-eqn}}
+{{eqn | l = a_r k^r
+      | o = >
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = >
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = \ne
+      | r = b
+      | c = 
+}}
+{{end-eqn}}
+Otherwise $r = s$.
+Let $l$ be the largest such that $a_l \ne b_l$.
+{{WLOG}}, suppose $a_l > b_l$.
+For all $j > l$, let $a_1 = a - a_j k_j$ and $b_1 = b - b_j k_j$.
+As $a_j = b_j$ in this range, $a_j k_j = b_j k_j$ and so the same amount is being subtracted from both.
+So consider $\paren {a_l - b_l} k_l$.
+From [[Bounds for Integer Expressed in Base k]]:
+:$\paren {a_l - b_l} k_l > \displaystyle \sum_{j \mathop = 0}^{l - 1} a_j k^j$
+and:
+:$\paren {a_l - b_l} k_l > \displaystyle \sum_{j \mathop = 0}^{l - 1} b_j k^j$
+and so $a_1 > b_1$.
+Hence:
+:$a_1 + a_j k_j > b_1 + b_j k_j$
+Hence $a \ne b$ and the result.
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Existence of q for which j - qk is Positive}
+Tags: Number Theory
+
+\begin{theorem}
+Let $j, k \in \Z$ be [[Definition:Integer|integers]] such that $k > 0$.
+Then there exist $q \in \Z$ such that $j - q k > 0$.
+\end{theorem}
+
+\begin{proof}
+Let $q = -\size j - 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = j - q k
+      | r = j - \paren {-\size j - 1} k
+      | c = 
+}}
+{{eqn | r = j + \size j + k
+      | c = 
+}}
+{{end-eqn}}
+We have that:
+:$\forall j \le 0: j + \size j = 0$
+and:
+:$\forall j > 0: j + \size j = 2 j$
+So:
+:$j - q k \ge k$
+and as $k > 0$ the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function on Base between Zero and One is Strictly Decreasing/Real Number}
+Tags: Power Function on Base between Zero and One is Strictly Decreasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $0 \lt a \lt 1$.
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map f x = a^x$
+where $a^x$ denotes [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$ be such that $x < y$.
+Since $0 < a < 1$, we have that:
+:$\dfrac 1 a > 1$
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = \paren {\dfrac 1 a}^x 
+      | o = <
+      | r = \paren {\dfrac 1 a}^y
+      | c = [[Real Power Function on Base Greater than One is Strictly Increasing]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \dfrac 1 {a^x}
+      | o = <
+      | r = \dfrac 1 {a^y}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a^x
+      | o = >
+      | r = a^y
+      | c = [[Reciprocal Function is Strictly Decreasing]]
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+[[Category:Power Function on Base between Zero and One is Strictly Decreasing]]
+5b3907fc75dupddxgephkodb55l5y9t
+\end{proof}<|endoftext|>
+\section{Integral Ideal is Ideal of Ring}
+Tags: Integral Ideals, Ideal Theory
+
+\begin{theorem}
+Let $J$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Integer|set of integers]] $\Z$.
+Then:
+:$J$ is an [[Definition:Integral Ideal|integral ideal]]
+{{iff}}:
+:$J$ is an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Let $J \subseteq \Z$ fulfil the conditions of an [[Definition:Integral Ideal|integral ideal]]:
+:$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$
+:$(2): \quad n \in J, r \in \Z \implies r n \in J$
+First note that $J$ is [[Definition:Non-Empty Set|non-empty]] by definition.
+Then from $(1)$ we have in particular:
+:$n, m \in J \implies m - n \in J$
+Thus by the [[One-Step Subgroup Test]]:
+:$\struct {J, +}$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Additive Group of Integers|additive group of integers $\struct {\Z, +}$]].
+Then from $(2)$ we have:
+:$n \in J, r \in \Z \implies r n \in J$
+and by [[Integer Multiplication is Commutative]] it follows that:
+:$n \in J, r \in \Z \implies n r \in J$
+Thus it is seen that the conditions are fulfilled for $J$ to be an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$.
+{{qed|lemma}}
+Suppose $J \subseteq \Z$ such that:
+:$(3): \quad \struct {J, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$
+:$(4): \quad \forall n \in J: \forall r \in R: n r \in J \land r n \in J$
+From $(4)$, $(2)$ immediately follows.
+From $(3)$ it follows from the [[One-Step Subgroup Test]] that:
+:$\forall n, m \in J: m - n \in J$
+which is a restatement of $(1)$.
+Thus the conditions are fulfilled for $J$ to be an [[Definition:Integral Ideal|integral ideal]].
+{{qed}}
+[[Category:Integral Ideals]]
+[[Category:Ideal Theory]]
+1gbbe9q46i0bmjvtwjlszsur1l5oy0a
+\end{proof}<|endoftext|>
+\section{Set of Integer Multiples is Integral Ideal}
+Tags: Integral Ideals, Sets of Integer Multiples
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]].
+Let $m \Z$ denote the [[Definition:Set of Integer Multiples|set of integer multiples]] of $m$.
+Then $m \Z$ is an [[Definition:Integral Ideal|integral ideal]].
+\end{theorem}
+
+\begin{proof}
+First note that $m \times 0 \in m \Z$ whatever $m$ may be.
+Thus $m \Z \ne \O$.
+Let $a, b \in m \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a + b
+      | r = m j + m k
+      | c = for some $j, k \in \Z$ by definition of $m \Z$
+}}
+{{eqn | r = m \paren {j + k}
+      | c = 
+}}
+{{eqn | o = \in
+      | r = m \Z
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = a - b
+      | r = m j - m k
+      | c = for some $j, k \in \Z$ by definition of $m \Z$
+}}
+{{eqn | r = m \paren {j - k}
+      | c = 
+}}
+{{eqn | o = \in
+      | r = m \Z
+      | c = 
+}}
+{{end-eqn}}
+Let $r \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = r a
+      | r = r \paren {m j}
+      | c = for some $j \in \Z$ by definition of $m \Z$
+}}
+{{eqn | r = m \paren {r j}
+      | c = 
+}}
+{{eqn | o = \in
+      | r = m \Z
+      | c = 
+}}
+{{end-eqn}}
+Thus the conditions for $m \Z$ to be an [[Definition:Integral Ideal|integral ideal]] are fulfilled.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimal Smooth Surface Spanned by Contour}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $\map z {x, y}: \R^2 \to \R$ be a [[Definition:Real-Valued Function|real-valued function]].
+Let $\Gamma$ be a [[Definition:Closed Contour|closed contour]] in $3$-[[Definition:Dimension of Vector Space|dimensional]] [[Definition:Real Euclidean Space|Euclidean space]].
+Suppose this [[Definition:Surface|surface]] is [[Definition:Smooth Real Function|smooth]] for every $x$ and $y$.
+Then it has to satisfy the following [[Necessary Condition for Integral Functional to have Extremum/Two Variables|Euler's equation]]:
+:$r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$
+where:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = z_x
+}}
+{{eqn | l = q
+      | r = z_y
+}}
+{{eqn | l = r
+      | r = z_{xx}
+}}
+{{eqn | l = s
+      | r = z_{xy}
+}}
+{{eqn | l = t
+      | r = z_{yy}
+}}
+{{end-eqn}}
+with subscript denoting respective [[Definition:Partial Derivative|partial derivatives]].
+In other words, its [[Definition:Mean Curvature|mean curvature]] has to vanish.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Surface|surface]] [[Definition:Area|area]] for a [[Definition:Smooth Real Function|smooth]] [[Definition:Surface|surface]] embedded in $3$-[[Definition:Dimension of Vector Space|dimensional]] [[Definition:Real Euclidean Space|Euclidean space]] is given by:
+:$\displaystyle A \sqbrk z = \iint_\Gamma \sqrt {1 + z_x^2 + z_y^2} \rd x \rd y$
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \dfrac \d {\d x} \frac \partial {\partial z_x} \sqrt {1 + z_x^2 + z_y^2}
+      | r = \dfrac \d {\d x} \frac {z_x} {\sqrt {1 + z_x^2 + z_y^2} }
+}}
+{{eqn | r = \frac {z_{xx} + z_y^2 z_{xx} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} }
+}}
+{{eqn | r = \frac {r \paren {1 + q^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} }
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \dfrac \d {\d y} \frac \partial {\partial z_y} \sqrt {1 + z_x^2 + z_y^2}
+      | r = \dfrac \d {\d y} \frac {z_y} {\sqrt {1 + z_x^2 + z_y^2} }
+}}
+{{eqn | r = \frac {z_{yy} + z_x^2 z_{yy} - z_x z_y z_{xy} } {\paren {1 + z_x^2 + z_y^2}^{\frac 3 2} }
+}}
+{{eqn | r = \frac {t \paren {1 + p^2} - p q s } {\paren {1 + p^2 + q^2}^{\frac 3 2} }
+}}
+{{end-eqn}}
+By [[Definition:Euler's Equation for Vanishing Variation|Euler's equation]]:
+:$\dfrac {r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} } {\paren {1 + p^2 + q^2}^{\frac 3 2} } = 0$
+Due to the [[Definition:Smooth Real Function|smoothness]] of the [[Definition:Surface|surface]], $1 + p^2 + q^2$ is [[Definition:Bounded Real-Valued Function|bounded]].
+Hence, the following equation is sufficient:
+:$r \paren {1 + q^2} - 2 s p q + t \paren {1 + p^2} = 0$
+Introduce the following change of variables:
+{{begin-eqn}}
+{{eqn | l = E
+      | r = 1 + p^2
+}}
+{{eqn | l = F
+      | r = p q
+}}
+{{eqn | l = G
+      | r = 1 + q^2
+}}
+{{eqn | l = e
+      | r = \dfrac r {\sqrt {1 + p^2 + q ^2} }
+}}
+{{eqn | l = f
+      | r = \dfrac s {\sqrt {1 + p^2 + q^2} }
+}}
+{{eqn | l = g
+      | r = \dfrac t {\sqrt {1 + p^2 + q^2} }
+}}
+{{end-eqn}}
+Then [[Definition:Euler's Equation for Vanishing Variation|Euler's equation]] can be rewritten as:
+:$\dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} } = 0$
+By definition, [[Definition:Mean Curvature|mean curvature]] is:
+:$M = \dfrac {E g - 2 F f + G e} {2 \paren {E G - F^2} }$
+Hence:
+:$M = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Ideal is Set of Integer Multiples}
+Tags: Integral Ideals, Sets of Integer Multiples
+
+\begin{theorem}
+Let $J$ be an [[Definition:Integral Ideal|integral ideal]].
+Then $J$ is in the form of a [[Definition:Set of Integer Multiples|set of integer multiples]] $m \Z$ for some $m \in \Z$.
+\end{theorem}
+
+\begin{proof}
+By definition, $J$ satisfies the following conditions:
+:$(1): \quad n, m \in J \implies m + n \in J, m - n \in J$
+:$(2): \quad n \in J, r \in \Z \implies r n \in J$
+First note that the [[Definition:Null Ideal|null ideal]] $\set 0$ is an [[Definition:Integral Ideal|integral ideal]].
+This is of the form $0 \Z$.
+Let $J \ne \set 0$.
+Then $\exists a \in J: a \ne 0$.
+As $0 \in \Z$ it follows from $(2)$ that:
+:$0 \times a = 0 \in J$
+Thus from $(1)$ we have that:
+:$0 - a \in J$
+so both $a \in J$ and $-a \in J$
+Thus there exists $a \in J: a > 0$.
+Consider the set $S = \set {x \in J: x > 0}$.
+$S$ is [[Definition:Bounded Below Set|bounded below]] by $0$, for example.
+$S$ is not [[Definition:Empty Set|empty]], because it has been established that $a \in S$.
+By [[Set of Integers Bounded Below has Smallest Element]], $S$ has a [[Definition:Smallest Element|smallest element]], $m$, say.
+Consider $y \in m \Z$.
+By definition of $m \Z$, $y = r m$ for some $r \in \Z$.
+Thus, by $(2)$, $y \in J$.
+So by definition of [[Definition:Subset|subset]], $m \Z \subseteq J$.
+It remains to be demonstrated that $J \subseteq m \Z$.
+{{AimForCont}} $k \in J: k \notin m \Z$.
+As $k \in J$, it follows from $(1)$ that $0 - k = -k \in J$ also.
+Either $k > 0$ or $-k > 0$.
+{{WLOG}}, let $k > 0$.
+If $k < m$, then that [[Definition:Contradiction|contradicts]] the statement that $m \in J$ is the [[Definition:Smallest Element|smallest element]] of $J$ that is [[Definition:Strictly Positive Integer|(strictly) positive]].
+So $k > m$.
+Consider the set $T = \set {x \in m \Z: x > k}$.
+$T$ is not [[Definition:Empty Set|empty]], as $m \paren {k + 1}$, for example, is bound to be in $T$ (even if $m = 1$, the smallest it can be).
+By [[Set of Integers Bounded Below has Smallest Element]], $T$ has a [[Definition:Smallest Element|smallest element]], $m s$, say.
+Then:
+:$m \paren {s - 1} < k < m s$
+As $k > m$ it follows also that $m \paren {s - 1} > 0$.
+Equality cannot happen because $k \notin m \Z$.
+But we have that $k \in J$ and so:
+:$m s - k \in J$
+and:
+:$k - m \paren {s - 1} \in J$
+But $m s - k < m$, otherwise $m \paren {s - 1} > k$ which [[Definition:Contradiction|contradicts]] the statement that $m s \in J$ is the [[Definition:Smallest Element|smallest element]] of $T$
+Thus $m s - k \in J$ and $0 < m s - k < m$.
+This [[Definition:Contradiction|contradicts]] the statement that $m$ is the [[Definition:Smallest Element|smallest element]] of $J$ that is [[Definition:Strictly Positive Integer|(strictly) positive]].
+Hence by [[Proof by Contradiction]] there is no such $k$.
+Hence $J \subseteq m \Z$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference between Odd Squares is Divisible by 8}
+Tags: Square Numbers
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Odd Integer|odd integers]].
+Then $a^2 - b^2$ is [[Definition:Divisor of Integer|divisible]] by $8$.
+\end{theorem}
+
+\begin{proof}
+Let $a = 2 m + 1$, $b = 2 n + 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a^2 - b^2 
+      | r = \paren {2 m + 1}^2 - \paren {2 n + 1}^2
+      | c = 
+}}
+{{eqn | r = \paren {4 m^2 + 4 m + 1} - \paren {4 n^2 + 4 n - 1}
+      | c = 
+}}
+{{eqn | r = 4 \paren {m^2 - n^2} + 4 \paren {m - n}
+      | c = 
+}}
+{{eqn | r = 4 \paren {m + n} \paren {m - n} + 4 \paren {m - n}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = 4 \paren {m + n + 1} \paren {m - n}
+      | c = 
+}}
+{{end-eqn}}
+Suppose $m - n$ is [[Definition:Even Integer|even]] such that $m - n = 2 k$.
+Then:
+:$a^2 - b^2 = 4 \paren {2 k} \paren {m + n + 1} = 8 k \paren {m + n + 1}$
+and so is [[Definition:Divisor of Integer|divisible]] by $8$.
+Suppose $m - n$ is [[Definition:Odd Integer|odd]] such that $m - n = 2 k + 1$.
+Then:
+:$m + n + 1 = m + \paren {2 k + 1 + m} + 1 = 2 m + 2 k = 2 \paren {m + k}$
+and so:
+:$a^2 - b^2 = 4 \paren {2 k + 1} 2 \paren {m + k} = 8 \paren {2 k + 1} \paren {m + k}$
+and so is again [[Definition:Divisor of Integer|divisible]] by $8$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Norms Equivalent to Absolute Value on Rational Numbers/Necessary Condition}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\alpha \in \R_{\gt 0}$.
+Let $\norm{\,\cdot\,}:\Q \to \R$ be the [[Definition:Mapping|mapping]] defined by:
+:$\forall x \in \Q: \norm{x} = \size {x}^\alpha$
+where $\size {x}$ is the [[Definition:Absolute Value|absolute value]] of $x$ in $\Q$.
+Then:
+:$\norm{\,\cdot\,}$ is a [[Definition:Norm/Division Ring|norm]] on $\Q \implies \,\,\alpha \le 1$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Contrapositive|contrapositive]] is proved.
+Let $\alpha \gt 1$.
+The [[Definition:Norm/Division Ring|norm axiom (N3) (Triangle Inequality)]] is not satisfied:
+{{begin-eqn}}
+{{eqn|l= \norm {1 + 1}
+     |r= \size {1 + 1}^\alpha
+}}
+{{eqn|r= 2^\alpha
+}}
+{{eqn|o= \gt
+     |r= 2
+     |c= [[Power Function on Base Greater than One is Strictly Increasing/Real Number|Real Power Function on Base Greater than One is Strictly Increasing]]
+}}
+{{eqn|r= \size{1}^\alpha + \size{1}^\alpha
+}}
+{{eqn|r= \norm{1} + \norm{1}
+}}
+{{end-eqn}}
+By [[Rule of Transposition]] the result follows.
+\end{proof}<|endoftext|>
+\section{Norms Equivalent to Absolute Value on Rational Numbers/Sufficient Condition}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\alpha \in \R_{\gt 0}$.
+Let $\norm{\,\cdot\,}:\Q \to \R$ be the [[Definition:Mapping|mapping]] defined by:
+:$\forall x \in \Q: \norm{x} = \size {x}^\alpha$
+where $\size {x}$ is the [[Definition:Absolute Value|absolute value]] of $x$ in $\Q$.
+Then:
+:$\alpha \le 1 \implies \norm{\,\cdot\,}$ is a [[Definition:Norm/Division Ring|norm]] on $\Q$
+\end{theorem}
+
+\begin{proof}
+Suppose $\alpha \le 1$.
+It is shown that $\norm{\,\cdot\,}$ satisfies the [[Definition:Norm/Division Ring|norm axioms (N1)-(N3)]].
+==== (N1) Positive Definiteness ====
+Let $x \in \Q$.
+{{begin-eqn}}
+{{eqn|l= \norm {x} = 0
+     |o= \iff
+     |r= \size {x}^\alpha = 0
+     |c = Definition of $\norm{\,\cdot\,}$
+}}
+{{eqn|o= \iff
+     |r= \size {x} = 0
+     |c= Definition of $a$ to the [[Definition:Power (Algebra)|power]] of $r$ for $a \in \R_{\ge 0}$ and $r \in \R_{\gt 0}$
+}}
+{{eqn|o= \iff
+     |r= x = 0
+     |c= [[Absolute Value is Norm]] and [[Definition:Norm/Division Ring|norm axiom (N1) (Positive Definiteness)]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== (N2) Multiplicativity ====
+Let $x, y \in \Q$.
+Then:
+{{begin-eqn}}
+{{eqn|l= \norm {x y}
+     |r= \size {x y}^\alpha
+     |c = Definition of $\norm{\,\cdot\,}$
+}}
+{{eqn|r= \paren {\size {x} \size {y} }^\alpha
+     |c= [[Absolute Value is Norm]] and [[Definition:Norm/Division Ring|norm axiom (N2) (Multiplicativity)]]
+}}
+{{eqn|r= \size {x}^\alpha \size {y}^\alpha
+     |c= [[Exponent Combination Laws/Power of Product|Power of product]]
+}}
+{{eqn|r= \norm {x} \norm {y}
+     |c = Definition of $\norm{\,\cdot\,}$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+==== (N3) Triangle Inequality ====
+Let $x, y \in \Q$.
+{{WLOG}} let $\norm y \lt \norm x$.
+If $\norm x = 0$ then $\norm y = 0$.
+By (N1) above, $x = y = 0$.
+Hence:
+{{begin-eqn}}
+{{eqn|l= \norm{x + y}
+     |r= \norm 0
+}}
+{{eqn|r= 0
+}}
+{{eqn|r= 0 + 0
+}}
+{{eqn|r= \norm x + \norm y
+}}
+{{end-eqn}}
+If $\norm x \gt 0$ then:
+:$\norm x \gt 0 \iff \size {x}^\alpha \gt 0 \iff \size x \gt 0$
+Hence:
+{{begin-eqn}}
+{{eqn|l= \norm{x + y}
+     |r= \size{x + y}^\alpha
+}}
+{{eqn|o= \le
+     |r= \paren{\size x + \size y}^\alpha
+     |c= [[Definition:Norm/Division Ring|Norm axiom (N3) (Triangle Inequality)]]
+}}
+{{eqn|r= \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }^\alpha
+}}
+{{eqn|o= \le
+     |r= \size {x}^\alpha \paren{1 + \dfrac {\size y} {\size x} }
+     |c= [[Power Function on Base Greater than One is Strictly Increasing/Real Number|Real Power Function on base Greater than One is Strictly Increasing]]
+}}
+{{eqn|o= \le
+     |r= \size {x}^\alpha \paren{1 + \dfrac {\size {y}^\alpha} {\size {x}^\alpha} }
+     |c= [[Power Function on Base between Zero and One is Strictly Decreasing/Real Number|Real Power Function on base between Zero and One is Strictly Decreasing]]
+}}
+{{eqn|o= \le
+     |r= \size {x}^\alpha + \size {y}^\alpha
+}}
+{{eqn|r= \norm {x} + \norm {y}
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Lowest Common Multiple of Integers with Common Divisor}
+Tags: Lowest Common Multiple
+
+\begin{theorem}
+Let $b, d \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]]
+Then:
+:$\lcm \set {a b, a d} = a \lcm \set {b, d}$
+where:
+:$a \in \Z_{>0}$
+:$\lcm \set {b, d}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+{{begin-eqn}}
+{{eqn | l = b
+      | o = \divides
+      | r = \lcm \set {b, d}
+      | c = {{Defof|Lowest Common Multiple of Integers}}
+}}
+{{eqn | lo= \land
+      | l = d
+      | o = \divides
+      | r = \lcm \set {b, d}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r b
+      | r = \lcm \set {b, d}
+      | c = for some $r \in \Z$
+}}
+{{eqn | lo= \land
+      | l = s d
+      | r = \lcm \set {b, d}
+      | c = for some $s \in \Z$
+}}
+{{eqn | ll= \leadsto
+      | l = r \paren {a b}
+      | r = a \lcm \set {b, d}
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = s \paren {a d}
+      | r = a \lcm \set {b, d}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b
+      | o = \divides
+      | r = a \lcm \set {b, d}
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | lo= \land
+      | l = a d
+      | o = \divides
+      | r = a \lcm \set {b, d}
+      | c = 
+}}
+{{end-eqn}}
+Suppose $n \in \Z$ such that $a b \divides n$ and $a d \divides n$.
+It will be shown that $a \lcm \set {b, d} \divides n$.
+So:
+{{begin-eqn}}
+{{eqn | l = a b
+      | o = \divides
+      | r = n
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | lo= \land
+      | l = a d
+      | o = \divides
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a r b
+      | r = n
+      | c = for some $r \in \Z$
+}}
+{{eqn | lo= \land
+      | l = a s d
+      | r = n
+      | c = for some $s \in \Z$
+}}
+{{eqn | ll= \leadsto
+      | l = r b
+      | r = \dfrac n a
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = s d
+      | r = \dfrac n a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b
+      | o = \divides
+      | r = \dfrac n a
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | lo= \land
+      | l = d
+      | o = \divides
+      | r = \dfrac n a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lcm \set {b, d}
+      | o = \divides
+      | r = \dfrac n a
+      | c = [[LCM Divides Common Multiple]]
+}}
+{{eqn | lo= \land
+      | l = a \lcm \set {b, d}
+      | o = \divides
+      | r = n
+      | c = 
+}}
+{{end-eqn}}
+Thus we have:
+:$a b \divides a \lcm \set {b, d} \land a d \divides a \lcm \set {b, d}$
+and:
+:$a b \divides n \land a d \divides n \implies a \lcm \set {b, d} \divides n$
+It follows from [[LCM iff Divides All Common Multiples]] that:
+:$\lcm \set {a b, a d} = a \lcm \set {b, d}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Addition of Fractions}
+Tags: Lowest Common Multiple, Greatest Common Divisor, Fractions
+
+\begin{theorem}
+Let $a, b, c, d \in \Z$ such that $b d \ne 0$.
+Then:
+:$\dfrac a b + \dfrac c d = \dfrac {a D + B c} {\lcm \set {b, d} }$
+where:
+:$B = \dfrac b {\gcd \set {b, d} }$
+:$D = \dfrac d {\gcd \set {b, d} }$
+:$\lcm$ denotes [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]
+:$\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac a b + \dfrac c d
+      | r = \dfrac {a d} {b d} + \dfrac {b c} {b d}
+      | c = 
+}}
+{{eqn | r = \dfrac {a d + b c} {b d}
+      | c = 
+}}
+{{eqn | r = \dfrac {a d + b c} {\gcd \set {b, d} \lcm \set {b, d} }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \dfrac {a D \gcd \set {b, d} + B \gcd \set {b, d} c} {\gcd \set {b, d} \lcm \set {b, d} }
+      | c = substituting for $b$ and $d$ as defined above
+}}
+{{eqn | r = \dfrac {a D + B c} {\lcm \set {b, d} }
+      | c = dividing [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $\gcd \set {b, d}$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD of Sum and Difference of Integers}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+:$\gcd \set {a + b, a - b} \ge \gcd \set {a, b}$
+\end{theorem}
+
+\begin{proof}
+Let $d = \gcd \set {a, b}$.
+Then by definition of [[Definition:Greatest Common Divisor of Integers|greatest common divisor]]:
+:$d \divides a \land d \divides b$
+From [[Common Divisor Divides Integer Combination]]:
+:$d \divides \paren {a + b} \land d \divides \paren {a - b}$
+By definition of [[Definition:Common Divisor of Integers|common divisor]]:
+:$d \divides \gcd \set {a + b, a - b}$
+Hence from [[Absolute Value of Integer is not less than Divisors]]:
+:$d \le \gcd \set{a + b, a - b}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Common Divisor divides Lowest Common Multiple}
+Tags: Greatest Common Divisor, Lowest Common Multiple
+
+\begin{theorem}
+Let $a, b \in \Z$ such that $a b \ne 0$.
+Then:
+:$\gcd \set {a, b} \divides \lcm \set {a, b}$
+where:
+:$\lcm$ denotes [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]
+:$\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\gcd \set {a, b} \divides a$
+and:
+:$a \divides \lcm \set {a, b}$
+The result follows from [[Divisor Relation is Transitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Sets of Integer Multiples}
+Tags: Sets of Integer Multiples, Lowest Common Multiple, Intersection of Sets of Integer Multiples
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $m n \ne 0$.
+Let $m \Z$ denote the [[Definition:Set of Integer Multiples|set of integer multiples of $m$]].
+Then:
+:$m \Z \cap n \Z = \lcm \set {m, n} \Z$
+where $\lcm$ denotes [[Definition:Lowest Common Multiple of Integers|lowest common multiple]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in m \Z \cap n \Z$.
+Then by definition of [[Definition:Set Intersection|set intersection]]:
+:$m \divides x$ and $n \divides x$
+So from [[LCM Divides Common Multiple]]:
+:$\lcm \set {m, n} \divides x$
+and so $x \in \lcm \set {m, n} \Z$
+That is:
+:$m \Z \cap n \Z \subseteq \lcm \set {m, n} \Z$
+{{qed|lemma}}
+Now suppose $x \in \lcm \set {m, n} \Z$.
+Then $\lcm \set {m, n} \divides x$.
+Thus by definition of [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]:
+:$m \divides x$ 
+and:
+:$n \divides x$
+and so:
+:$x \in m \Z \land x \in n \Z$
+That is:
+:$x \in \Z \cap n \Z$
+and so:
+:$\lcm \set {m, n} \Z \subseteq m \Z \cap n \Z$
+{{qed|lemma}}
+The result follows by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Integer Multiples of GCD}
+Tags: Sets of Integer Multiples, Greatest Common Divisor
+
+\begin{theorem}
+Let $m, n \in \Z$.
+Let $m \Z$ denote the [[Definition:Set of Integer Multiples|set of integer multiples of $m$]]
+Then:
+:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
+where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in m \Z \cup n \Z$.
+Then either:
+:$m \divides x$
+or:
+:$n \divides x$
+In both cases:
+:$\gcd \set {m, n} \divides x$
+and so:
+:$x \in \gcd \set {m, n} \Z$
+Hence by definition of [[Definition:Subset|subset]]:
+:$m \Z \cup n \Z \subseteq \gcd \set {m, n} \Z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Join of Sets of Integer Multiples is Set of Integer Multiples of GCD}
+Tags: Sets of Integer Multiples, Greatest Common Divisor
+
+\begin{theorem}
+Let $m, n \in \Z$.
+Let $m \Z$ denote the [[Definition:Set of Integer Multiples|set of integer multiples of $m$]]
+Let $r \in \Z$ such that:
+:$m \Z \subseteq r \Z$
+and:
+:$n \Z \subseteq r \Z$
+Then:
+:$\gcd \set {m, n} \Z \subseteq r \Z$
+where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+\end{theorem}
+
+\begin{proof}
+From [[Set of Integer Multiples is Integral Ideal]], each of $m \Z$, $n \Z$, $r \Z$ and $\gcd \set {m, n} \Z$ are [[Definition:Integral Ideal|integral ideals]].
+Let $c \in \gcd \set {m, n} \Z$.
+By definition of [[Definition:Integral Ideal|integral ideal]]:
+:$\gcd \set {m, n} \divides c$
+By [[Set of Integer Combinations equals Set of Multiples of GCD]]:
+:$\exists x, y \in \Z: c = x m + y n$
+But as $m \Z \subseteq r \Z$ and $n \Z \subseteq r \Z$:
+:$m \in r \Z$ and $n \in \r Z$
+Thus by definition of [[Definition:Integral Ideal|integral ideal]]:
+:$x m + y n \in r \Z$
+So:
+:$c \in \gcd \set {m, n} \Z \implies c \in r \Z$
+and the result follows by definition of [[Definition:Subset|subset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD of Generators of General Fibonacci Sequence is Divisor of All Terms}
+Tags: Greatest Common Divisor, Fibonacci Numbers
+
+\begin{theorem}
+Let $\mathcal F = \sequence {a_n}$ be a [[Definition:General Fibonacci Sequence|general Fibonacci sequence]] generated by the parameters $r, s, t, u$:
+:$a_n = \begin{cases}
+r & : n = 0 \\
+s & : n = 1 \\
+t a_{n - 2} + u a_{n - 1} & : n > 1
+\end{cases}$
+Let:
+:$d = \gcd \set {r, s}$
+where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+Then:
+:$\forall n \in \Z_{>0}: d \divides a_n$
+\end{theorem}
+
+\begin{proof}
+From the construction of a [[Definition:General Fibonacci Sequence|general Fibonacci sequence]], $a_n$ is an [[Definition:Integer Combination|integer combination]] of $r$ and $s$.
+From [[Set of Integer Combinations equals Set of Multiples of GCD]], $a_n$ is [[Definition:Divisor of Integer|divisible]] by $\gcd \set {r, s}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD of Consecutive Integers of General Fibonacci Sequence}
+Tags: Greatest Common Divisor, Fibonacci Numbers
+
+\begin{theorem}
+Let $\mathcal F = \sequence {a_n}$ be a [[Definition:General Fibonacci Sequence|general Fibonacci sequence]] generated by the parameters $r, s, t, u$:
+:$a_n = \begin{cases}
+r & : n = 0 \\
+s & : n = 1 \\
+t a_{n - 2} + u a_{n - 1} & : n > 1
+\end{cases}$
+Let:
+:$d = \gcd \set {r, s}$
+where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+Let $f = \gcd \set {a_m, a_{m - 1} }$ for some $m \in \Z$.
+Let $\gcd \set {f, t} = 1$.
+Then:
+:$f \divides d$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Someone else feel free to take this on, I'm not up to the challenge tonight.}}
+\end{proof}<|endoftext|>
+\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 3}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm{\,\cdot\,}$,
+Let $x, y \in R$ and $\norm x \lt \norm y$.
+Then:
+:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \norm y$
+\end{theorem}
+
+\begin{proof}
+By [[Three Points in Ultrametric Space have Two Equal Distances/Corollary 2|Corollary 2]] then:
+:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} = \norm y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 4}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Normed Division Ring|normed division ring]] with [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm{\,\cdot\,}$,
+Let $x, y \in R$.
+Then:
+:* $\norm {x + y} \lt \norm y \implies \norm x = \norm y$
+:* $\norm {x - y} \lt \norm y \implies \norm x = \norm y$
+:* $\norm {y - x} \lt \norm y \implies \norm x = \norm y$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Contrapositive Statement|contrapositive statements]] are proved.
+Let $\norm x \ne \norm y$
+By [[Three Points in Ultrametric Space have Two Equal Distances/Corollary 2|Corollary 2]] then:
+:$\norm {x + y} = \norm {x - y} = \norm {y - x} = \max \set {\norm x, \norm y} \ge \norm y$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalent Norms are both Non-Archimedean or both Archimedean}
+Tags: Normed Division Rings, Non-Archimedean Norms
+
+\begin{theorem}
+Let $R$ be a [[Definition:Division Ring|division ring]] with [[Definition:Unity of Ring|unity]] $1_R$.
+Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent]] [[Definition:Norm/Division Ring|norms]] on $R$.
+Then $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ are either both [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] or both [[Definition:Archimedean Division Ring Norm|Archimedean]].
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\norm {\,\cdot\,}_1$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] $\iff \forall n \in \N_{>0}: \norm{n \cdot 1_R}_1 \le 1$.
+By the definition of [[Definition:Equivalent Division Ring Norms by Cauchy Sequence|norm equivalence]] then:
+:$\forall n \in \N: \norm {n \cdot 1_R}_1 \le 1 \iff \norm {n \cdot 1_R}_2 \le 1$
+Similarly, by [[Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\forall n \in \N_{>0}: \norm {n \cdot 1_R}_2 \le 1 \iff \norm {\,\cdot\,}_2$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Minimum Area of Triangle whose Vertices are Lattice Points}
+Tags: Areas of Triangles
+
+\begin{theorem}
+Let $T$ be a [[Definition:Triangle (Geometry)|triangle]] embedded in a [[Definition:Cartesian Plane|cartesian plane]].
+Let the [[Definition:Vertex of Polygon|vertices]] of $T$ be [[Definition:Lattice Point|lattice points]] which are not all on the same [[Definition:Straight Line|straight line]].
+Then the [[Definition:Area|area]] of $T$ is such that:
+:$\map \Area T \ge \dfrac 1 2$
+\end{theorem}
+
+\begin{proof}
+{{WLOG}} let one of the [[Definition:Vertex of Polygon|vertices]] of $T$ be at $\tuple {0, 0}$.
+Let the other $2$ [[Definition:Vertex of Polygon|vertices]] be at $\tuple {a, b}$ and $\tuple {x, y}$.
+By [[Area of Triangle in Determinant Form with Vertex at Origin]]:
+:$\map \Area T = \dfrac {\size {b y - a x} } 2$
+As the [[Definition:Vertex of Polygon|vertices]] of $T$ are non-[[Definition:Collinear Points|collinear]], $\map \Area T \ge 0$.
+Thus $\size {b y - a x} > 0$.
+As $\tuple {a, b}$ and $\tuple {x, y}$ are [[Definition:Lattice Point|lattice points]], all of $a, b, x, y \in \Z$.
+Thus $\size {b y - a x} \ge 1$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Perpendicular Distance from Straight Line in Plane to Origin}
+Tags: Perpendicular Distance from Straight Line in Plane to Point
+
+\begin{theorem}
+Let $L$ be the [[Definition:Straight Line|straight line]] embedded in the [[Definition:Cartesian Plane|cartesian plane]] whose [[Equation of Straight Line in Plane|equation]] is given as:
+:$a x + b y = c$
+Then the [[Definition:Perpendicular Distance|perpendicular distance]] $d$ between $L$ and $\tuple {0, 0}$ is given by:
+:$d = \size {\dfrac c {\sqrt {a^2 + b^2} } }$
+\end{theorem}
+
+\begin{proof}
+From [[Perpendicular Distance from Straight Line in Plane to Point]], the [[Definition:Perpendicular Distance|perpendicular distance]] $d$ between $L$ and the [[Definition:Point|point]] $\tuple {x_0, y_0}$ is gven by:
+:$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$
+The result follows by setting $x_0 = 0$ and $y_0 = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Necessary Condition for Integral Functional to have Extremum/Two Variables}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $D \subset \R^2$.
+Let $\Gamma$ be the [[Definition:Boundary (Geometry)|boundary]] of $D$.
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Real Function|real mappings]] such that:
+:$S = \set {\map z {x, y}: \paren {z: S_1 \subseteq \R^2 \to S_2 \subseteq \R}, \paren {\map z {x, y} \in \map {C^2}D}, \paren {\map z \Gamma = 0} }$
+Let $J \sqbrk z: S \to S_3 \subseteq \R$ be a [[Definition:Real Functional|functional]] of the form:
+:$\ds \iint_D \map F {x, y, z, z_x, z_y} \rd x \rd y$ 
+Then a [[Definition:Necessary Condition|necessary condition]] for $J \sqbrk y$ to have an [[Definition:Extremum of Functional|extremum]] (strong or weak) for a given function $\map z {x, y}$ is that $\map z {x, y}$ satisfy [[Definition:Euler's Equation for Vanishing Variation|Euler's equation]]:
+:$F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$
+\end{theorem}
+
+\begin{proof}
+From [[Condition for Differentiable Functional to have Extremum]] we have
+:$\bigvalueat {\delta J \sqbrk {z; h} } {z \mathop = \hat z} = 0$
+The variation exists if $J$ is a [[Definition:Differentiable Functional|differentiable functional]].
+Since $\map z \Gamma = 0$, $\map h {x, y}$ vanishes on the [[Definition:Boundary (Geometry)|boundary]] $\Gamma$:
+:$\bigvalueat h \Gamma = 0$
+From the definition of [[Definition:Differentiable Functional|increment of a functional]]:
+:{{begin-eqn}}
+{{eqn | l = \Delta J \sqbrk {z; h}
+      | r = J \sqbrk {z + h} - J \sqbrk z
+      | c = definition
+}}
+{{eqn | r = \iint_D \map F {x, y, z + h, z_x + h_x, z_y + h_y} \rd x \rd y - \iint_D \map F {x, y, z, z_x, z_y} \rd x \rd y
+      | c = form of considered [[Definition:Real Functional|functional]]
+}}
+{{eqn | r = \iint_D \paren {\map F {x, y, z + h, z_x + h_x, z_y + h_y} - \map F {x, y, z, z_x, z_y} } \rd x \rd y
+      | c = bringing under the same [[Definition:Definite Integral|integral]]
+}}
+{{end-eqn}}
+Using multivariate [[Taylor's Theorem|Taylor's theorem]], expand $\map F {x, y, z + h, z_x + h_x, z_y + h_y}$ {{WRT}} $h$, $h_x$ and $h_y$.
+Denote the [[Definition:Ordered Tuple|ordered tuples]] $\tuple {h, h_x, h_y}$ as $\mathbf h$ and $\tuple {z, z_x, z_y}$ as $\mathbf z$ respectively.
+Then:
+:$\ds \map F {x, y, \mathbf z + \mathbf h} = \valueat {\map F {x, y, \mathbf z + \mathbf h} } {\mathbf h \mathop = 0} + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z} } {\mathbf h \mathop = 0} h + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_x} } {\mathbf h \mathop = 0} z_x + \valueat {\frac {\partial {\map F {x, y, \mathbf z + \mathbf h} } } {\partial z_y} } {\mathbf h \mathop = 0} z_y + \map \OO {\mathbf h^2}$
+where the last term includes all terms of the order not lesser than 2 {{WRT}} the elements of $\mathbf h$.
+Substitute this back into the [[Definition:Definite Integral|integral]]: 
+:$\ds \Delta J \sqbrk {y; h} = \iint_D \paren {\map F {x, y, \mathbf h}_z h + \map F {x, y, \mathbf z}_{z_x} h_x + \map F {x, y, \mathbf z}_{z_y} h_y + \map \OO {\mathbf h^2} } \rd x \rd y$
+Terms in $\map \OO {\mathbf h^2}$ represent terms of order higher than 1 with respect to elements of $\mathbf h$.
+{{Explain|Precise way to drop higher order terms}}
+By definition, the [[Definition:Definite Integral|integral]] not counting in $\map \OO {\mathbf h^2}$ is a [[Definition:Differentiable Functional|variation of functional]]:
+{{begin-eqn}}
+{{eqn | l = \delta J \sqbrk {z; h}
+      | r = \iint_D \paren {F_z h + \dfrac \partial {\partial x} F_{z_x} h + \dfrac \partial {\partial y} F_{z_y} h } \rd x \rd y
+}}
+{{eqn | r = \iint_D \paren {F_z h + \dfrac \partial {\partial x} \paren {F_{z_x} h} + \dfrac \partial {\partial y} \paren {F_{z_y} h} } \rd x \rd y - \iint_D \paren {\dfrac \partial {\partial x} F_{z_x} + \dfrac \partial {\partial y} F_{z_y} } h \rd x \rd y
+}}
+{{eqn | r = \int_\Gamma \paren {F_{z_x} h \rd y - F_{z_y} h \rd x} - \iint_D \paren {\dfrac \partial {\partial x} F_{z_x} + \dfrac \partial {\partial y} F_{z_y} -F_z} h \rd x \rd y
+      | c = [[Green's Theorem]]
+}}
+{{eqn | r = \iint_D \paren {F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} } h \rd x \rd y
+      | c = $\map h \Gamma = 0$
+}}
+{{end-eqn}}
+Then, by the use of the [[Necessary Condition for Integral Functional to have Extremum/Two Variables/Lemma|lemma]]:
+:$F_z - \dfrac \partial {\partial x} F_{z_x} - \dfrac \partial {\partial y} F_{z_y} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Line in Plane is Straight iff Gradient is Constant}
+Tags: Straight Lines
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Curve|curve]] which can be embedded in [[Definition:The Plane|the plane]].
+Then $\mathcal L$ is a [[Definition:Straight Line|straight line]] {{iff}} it is of [[Definition:Constant|constant]] [[Definition:Gradient|gradient]].
+\end{theorem}
+
+\begin{proof}
+Let $L$ be embedded in the [[Definition:Cartesian Plane|cartesian plane]].
+The [[Definition:Slope|slope]] of $\mathcal L$ at a point $p = \tuple {x, y}$ is defined as being its [[Definition:Derivative|derivative]] at $p$ {{WRT|Differentiation}} $x$.
+$\grad p = \dfrac {\d y} {\d x}$
+:[[File:Gradient-of-Straight-Line.png|500px]]
+Let $\mathcal L$ be a [[Definition:Straight Line|straight line]].
+Let $\triangle ABC$ and $\triangle DEF$ be [[Definition:Right Triangle|right triangles]] constructed so that:
+:$A, B, D, E$ are on $\mathcal L$
+:$AC$ and $DF$ are [[Definition:Parallel Lines|parallel]] to the [[Definition:X-Axis|$x$-axis]]
+:$BC$ and $EF$ are [[Definition:Parallel Lines|parallel]] to the [[Definition:Y-Axis|$y$-axis]].
+From [[Parallelism implies Equal Corresponding Angles]]:
+:$\angle ABC = \angle DEF$
+and:
+:$\angle BAC = \angle EDF$
+Also we have that $\angle ACB = \angle DFE$ and are [[Definition:Right Angle|right angles]].
+Thus $\triangle ABC$ and $\triangle DEF$ are [[Definition:Similar Triangles|similar]].
+Thus:
+:$\dfrac {BC} {AC} = \dfrac {EF} {DF}$
+That is, the [[Definition:Slope|slope]] of $\mathcal L$ between $A$ and $B$ is the same as the [[Definition:Slope|slope]] of $\mathcal L$ between $D$ and $E$.
+The argument reverses.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/General Equation}
+Tags: Equations of Straight Lines in Plane
+
+\begin{theorem}
+A [[Definition:Straight Line|straight line]] $\mathcal L$ is the [[Definition:Set|set]] of all $\tuple {x, y} \in \R^2$, where:
+:$\alpha_1 x + \alpha_2 y = \beta$
+where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+Let $y = \map f x$ be the [[Definition:Equation of Geometric Figure|equation]] of a [[Definition:Straight Line|straight line]] $\mathcal L$.
+From [[Line in Plane is Straight iff Gradient is Constant]], $\mathcal L$ has [[Definition:Constant|constant]] [[Definition:Slope of Straight Line|slope]].
+Thus the [[Definition:Derivative|derivative]] of $y$ {{WRT|Differentiation}} $x$ will be of the form:
+:$y' = c$
+Thus:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \int c \rd x
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | r = c x + K
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+where $K$ is [[Definition:Arbitrary Constant (Calculus)|arbitrary]].
+Taking the equation:
+:$\alpha_1 x + \alpha_2 y = \beta$
+it can be seen that this can be expressed as:
+:$y = - \dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}$
+thus demonstrating that $\alpha_1 x + \alpha_2 y = \beta$ is of the form $y = c x + K$ for some $c, K \in \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Slope of Straight Line joining Points in Cartesian Plane}
+Tags: Straight Lines
+
+\begin{theorem}
+Let $p_1 := \tuple {x_1, y_1}$ and $p_2 := \tuple {x_2, y_2}$ be [[Definition:Point|points]] in a [[Definition:Cartesian Plane|cartesian plane]].
+Let $\mathcal L$ be the [[Definition:Straight Line|straight line]] passing through $p_1$ and $p_2$.
+Then the [[Definition:Slope of Straight Line|slope]] of $\mathcal L$ is given by:
+:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
+where $\theta$ is the [[Definition:Angle|angle]] made by $\mathcal L$ with the [[Definition:X-Axis|$x$-axis]].
+\end{theorem}
+
+\begin{proof}
+:[[File:Slope-of-Line-between-Points.png|500px]]
+The [[Definition:Slope of Straight Line|slope of a straight line]] is defined as the change in $y$ divided by the change in $x$.
+The change in $y$ from $p_1$ to $p_2$ is $y_2 - y_1$.
+The change in $x$ from $p_1$ to $p_2$ is $x_2 - x_1$.
+By definition of [[Definition:Tangent of Angle|tangent]] of $\theta$:
+:$\tan \theta = \dfrac {y_2 - y_1} {x_2 - x_1}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/Two-Point Form}
+Tags: Equations of Straight Lines in Plane
+
+\begin{theorem}
+Let $p_1 := \tuple {x_1, y_1}$ and $p_2 := \tuple {x_2, y_2}$ be [[Definition:Point|points]] in a [[Definition:Cartesian Plane|cartesian plane]].
+Let $\LL$ be the [[Definition:Straight Line|straight line]] passing through $p_1$ and $p_2$.
+Then $\LL$ can be described by the equation:
+:$\dfrac {y - y_1} {x - x_1} = \dfrac {y_2 - y_1} {x_2 - x_1}$
+or:
+:$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$
+\end{theorem}
+
+\begin{proof}
+From the [[Equation of Straight Line in Plane/Slope-Intercept Form|slope-intercept form]] of the equation of the [[Definition:Straight Line|straight line]]:
+:$(1): \quad y = m x + c$
+which is to be satisfied by both $\tuple {x_1, y_1}$ and $\tuple {x_2, y_2}$.
+We express $m$ and $c$ in terms of $\paren {x_1, y_1}$ and $\paren {x_2, y_2}$:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = m x_1 + c
+      | c = 
+}}
+{{eqn | l = y_2
+      | r = m x_2 + c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | r = y_1 - m x_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_2
+      | r = m x_2 + y_1 - m x_1
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = m
+      | r = \dfrac {y_2 - y_1} {x_2 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = m x_1 + c
+      | c = 
+}}
+{{eqn | l = y_2
+      | r = m x_2 + c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m
+      | r = \dfrac {y_2 - c} {x_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_1
+      | r = \dfrac {y_2 - c} {x_2} x_1 + c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 x_2
+      | r = x_1 y_2  + c \paren {x_2 - x_1}
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = c
+      | r = \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}
+      | c = 
+}}
+{{end-eqn}}
+Substituting for $m$ and $c$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = m x + c
+      | c = which is $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \dfrac {y_2 - y_1} {x_2 - x_1} x + \dfrac {y_1 x_2 - x_1 y_2} {x_2 - x_1}
+      | c = from $(2)$ and $(3)$
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {x_2 - x_1} + x_1 y_2
+      | r = x \paren {y_2 - y_1} + y_1 x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {x_2 - x_1} + x_1 y_2 - y_1 x_1
+      | r = x \paren {y_2 - y_1} + y_1 x_2 - x_1 y_1
+      | c = adding $y_1 x_1 = x_1 y_1$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {x_2 - x_1} - y_1 \paren {x_2 - x_1}
+      | r = x \paren {y_2 - y_1} - x_1 \paren {y_2 - y_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {y - y_1} \paren {x_2 - x_1} 
+      | r = \paren {x - x_1} \paren {y_2 - y_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {y - y_1} {x - x_1}
+      | r = \dfrac {y_2 - y_1} {x_2 - x_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {x - x_1} {x_2 - x_1}
+      | r = \dfrac {y - y_1} {y_2 - y_1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/Slope-Intercept Form}
+Tags: Equations of Straight Lines in Plane
+
+\begin{theorem}
+Let $\mathcal L$ be the [[Definition:Straight Line|straight line]] defined by the [[Equation of Straight Line in Plane/General Equation|general equation]]:
+:$\alpha_1 x + \alpha_2 y = \beta$
+Then $\mathcal L$ can be described by the equation:
+:$y = m x + c$
+where:
+{{begin-eqn}}
+{{eqn | l = m
+      | r = -\dfrac {\alpha_1} {\alpha_2}
+      | c = 
+}}
+{{eqn | l = c
+      | r = \dfrac {\beta} {\alpha_2}
+      | c = 
+}}
+{{end-eqn}}
+such that $m$ is the [[Definition:Slope of Straight Line|slope]] of $\mathcal L$ and $c$ is the [[Definition:Y-Intercept|$y$-intercept]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \alpha_1 x + \alpha_2 y
+      | r = \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha_2 y
+      | r = y_1 - \alpha_1 x + \beta
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = y
+      | r = -\dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}
+      | c = 
+}}
+{{end-eqn}}
+Setting $x = 0$ we obtain:
+:$y = \dfrac {\beta} {\alpha_2}$
+which is the [[Definition:Y-Intercept|$y$-intercept]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT|Differentiation}} $x$ gives:
+:$y' = -\dfrac {\alpha_1} {\alpha_2}$
+By definition, this is the [[Definition:Slope of Straight Line|slope]] of $\mathcal L$ and is seen to be [[Definition:Constant|constant]].
+The result follows by setting:
+{{begin-eqn}}
+{{eqn | l = m
+      | r = -\dfrac {\alpha_1} {\alpha_2}
+      | c = 
+}}
+{{eqn | l = c
+      | r = \dfrac {\beta} {\alpha_2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/Two-Intercept Form}
+Tags: Equations of Straight Lines in Plane
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Straight Line|straight line]] which [[Definition:Intercept|intercepts]] the [[Definition:X-Axis|$x$-axis]] and [[Definition:Y-Axis|$y$-axis]] respectively at $\tuple {a, 0}$ and $\tuple {0, b}$, where $a b \ne 0$.
+Then $\mathcal L$ can be described by the equation:
+:$\dfrac x a + \dfrac y a = 1$
+\end{theorem}
+
+\begin{proof}
+:[[File:Straight-line-double-intercept-form.png|400px]]
+From the [[Equation of Straight Line in Plane/General Equation|General Equation of Straight Line in Plane]], $\mathcal L$ can be expressed in the form:
+:$(1): \quad \alpha_1 x + \alpha_2 y = \beta$
+where $\alpha_1, \alpha_2, \beta \in \R$ are given, and not both $\alpha_1, \alpha_2$ are zero.
+Substituting for the two [[Definition:Point|points]] whose [[Definition:Cartesian Coordinate System|coordinates]] we know about:
+{{begin-eqn}}
+{{eqn | lo= x = a, y = 0:
+      | l = \alpha_1 \times a + \alpha_2 \times 0
+      | r = \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha_1
+      | r = \dfrac \beta a
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | lo= x = 0, y = b:
+      | l = \alpha_1 \times 0 + \alpha_2 \times b
+      | r = \beta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha_2
+      | r = \dfrac \beta b
+      | c = 
+}}
+{{end-eqn}}
+We know that $\beta \ne 0$ because none of $a, b, \alpha_1, \alpha_2$ are equal to $0$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \dfrac \beta a x + \dfrac \beta b y
+      | r = \beta
+      | c = substituting for $\alpha_1$ and $\alpha_2$ in $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac x a + \dfrac y a
+      | r = 1
+      | c = dividing both sides by $\beta$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/Normal Form}
+Tags: Equations of Straight Lines in Plane
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Straight Line|straight line]] such that:
+:the [[Definition:Perpendicular Distance|perpendicular distance]] from $\mathcal L$ to the [[Definition:Origin|origin]] is $p$
+:the [[Definition:Angle|angle]] made between that [[Definition:Perpendicular|perpendicular]] and the [[Definition:X-Axis|$x$-axis]] is $\alpha$.
+Then $\mathcal L$ can be defined by the [[Definition:Equation of Geometric Figure|equation]]:
+:$x \cos \alpha + y \sin \alpha = p$
+\end{theorem}
+
+\begin{proof}
+:[[File:Straight-line-normal-form.png|400px]]
+Let $A$ be the [[Definition:X-Intercept|$x$-intercept]] of $\mathcal L$.
+Let $B$ be the [[Definition:Y-Intercept|$y$-intercept]] of $\mathcal L$.
+Let $A = \tuple {a, 0}$ and $B = \tuple {0, b}$.
+From the [[Equation of Straight Line in Plane/Two-Intercept Form|Equation of Straight Line in Plane: Two-Intercept Form]], $\mathcal L$ can be expressed in the form:
+:$(1): \quad \dfrac x a + \dfrac y a = 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = a \cos \alpha
+      | c = {{Defof|Cosine of Angle}}
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = \dfrac p {\cos \alpha}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = p
+      | r = b \sin \alpha
+      | c = {{Defof|Sine of Angle}}
+}}
+{{eqn | ll= \leadsto
+      | l = b
+      | r = \dfrac p {\sin \alpha}
+      | c = 
+}}
+{{end-eqn}}
+Substituting for $a$ and $b$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = \dfrac x a + \dfrac y a
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {x \cos \alpha} p + \dfrac {y \sin \alpha} p
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \cos \alpha + y \sin \alpha
+      | r = p
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Perpendicular Distance from Straight Line in Plane to Point}
+Tags: Straight Lines, Perpendicular Distance from Straight Line in Plane to Point
+
+\begin{theorem}
+Let $\LL$ be a [[Definition:Straight Line|straight line]] embedded in a [[Definition:Cartesian Plane|cartesian plane]], given by the [[Equation of Straight Line in Plane/General Equation|equation]]:
+:$a x + b y = c$
+Let $P$ be a [[Definition:Point|point]] in the [[Definition:Cartesian Plane|cartesian plane]] whose [[Definition:Cartesian Coordinate System|coordinates]] are given by:
+:$P = \tuple {x_0, y_0}$
+Then the [[Definition:Perpendicular Distance|perpendicular distance]] $d$ from $P$ to $\LL$ is given by:
+:$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$
+\end{theorem}
+
+\begin{proof}
+We have that $\LL$ has the [[Equation of Straight Line in Plane/General Equation|equation]]:
+:$(1): \quad a x + b y + c$
+[[File:Distance-from-Straight-Line-to-Point.png|500px]]
+Let a [[Definition:Perpendicular|perpendicular]] be dropped from $P$ to $\LL$ at $Q$.
+The [[Definition:Perpendicular Distance|perpendicular distance]] $d$ that we are to find is then $PQ$.
+In order to simplify the algebra that will inevitably follow, we are to make a transformation as follows.
+Let $\MM$ be constructed [[Definition:Parallel Lines|parallel]] to $\LL$.
+Construct a [[Definition:Perpendicular|perpendicular]] from $\MM$ to pass through the [[Definition:Origin|origin]].
+Let this [[Definition:Perpendicular|perpendicular]] [[Definition:Intersection (Geometry)|intersect]] $\MM$ at $R$ and $\LL$ at $S$.
+We have that $PQSR$ is a [[Definition:Rectangle|rectangle]], and so $RS = PQ$.
+It remains to establish the [[Definition:Length of Line|length]] of $RS$.
+We can manipulate $(1)$ into [[Equation of Straight Line in Plane/Slope-Intercept Form|slope-intercept form]] as:
+:$y = -\dfrac a b x + \dfrac c b$
+Thus the [[Definition:Slope of Straight Line|slope]] of $\LL$ is $-\dfrac a b$.
+From [[Slope of Orthogonal Curves]], the  [[Definition:Slope of Straight Line|slope]] of $RS$ is then $\dfrac b a$.
+The next step is to find the [[Definition:Cartesian Coordinate System|coordinates]] of $R$ and $S$.
+From [[Equation of Straight Line in Plane/Point-Slope Form|Equation of Straight Line in Plane: Point-Slope Form]], the equation of $\MM$ can be given as:
+:$y - y_0 = -\dfrac a b \paren {x - x_0}$
+or:
+:$(2): \quad y = \dfrac {-a x + a x_0 + b y_0} b$
+From [[Equation of Straight Line in Plane/Slope-Intercept Form|Equation of Straight Line in Plane: Slope-Intercept Form]], the equation of $RS$ can be given as:
+:$(3): \quad y = \dfrac b a x$
+$\MM$ and $RS$ [[Definition:Intersection (Geometry)|intersect]] where these are equal:
+:$\dfrac b a x = \dfrac {-a x + a x_0 + b y_0} b$
+which gives us:
+:$x = \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}$
+Substituting back for $y$ in $3$, we find that:
+:$R = \tuple {\dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2}, \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }$
+Now to find the [[Definition:Cartesian Coordinate System|coordinates]] of $S$, which is the [[Definition:Intersection (Geometry)|intersection]] of $\LL$ and $RS$.
+We can express $\LL$ as:
+:$y = -\dfrac {a x + c} b$
+and so:
+:$\dfrac b a x = -\dfrac {a x + c} b$
+which leads to:
+:$x = -\dfrac {a c} {a^2 + b^2}$
+Substituting back for $y$ in $3$, we get (after algebra):
+:$S = \tuple {\dfrac {-a c} {a^2 + b^2}, \dfrac {-b c} {a^2 + b^2} }$
+It remains to find the [[Definition:Length of Line|length]] $d$ of $RS$.
+From the [[Distance Formula]]:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = \sqrt {\paren {\dfrac {-a c} {a^2 + b^2} - \dfrac {a \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 + \paren {\dfrac {-b c} {a^2 + b^2} - \dfrac {b \paren {a x_0 + b y_0} } {a^2 + b^2} }^2 }
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {\paren {-a \paren {a x_0 + b y_0 + c} }^2 + \paren {-b \paren {a x_0 + b y_0 + c} }^2} {\paren {a^2 + b^2}^2 } }
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {\paren {a^2 + b^2} \paren {a x_0 + b y_0 + c}^2} {\paren {a^2 + b^2}^2 } }
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {\paren {a x_0 + b y_0 + c}^2} {a^2 + b^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }
+      | c = as [[Definition:Length of Line|length]] is [[Definition:Positive Real Number|positive]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Straight Line in Plane/Point-Slope Form}
+Tags: Straight Lines
+
+\begin{theorem}
+Let $\mathcal L$ be a [[Definition:Straight Line|straight line]] embedded in a [[Definition:Cartesian Plane|cartesian plane]], given in [[Equation of Straight Line in Plane/Slope-Intercept Form|slope-intercept form]] as:
+:$y = m x + c$
+Let $\mathcal L$ pass through the [[Definition:Point|point]] $\tuple {x_0, y_0}$.
+Then $\mathcal L$ can be expressed by the equation:
+:$y - y_0 = m \paren {x - x_0}$
+\end{theorem}
+
+\begin{proof}
+As $\tuple {x_0, y_0}$ is on $\mathcal L$, it follows that:
+{{begin-eqn}}
+{{eqn | l = y_0
+      | r = m x_0 + c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | r = m x_0 - y_0
+      | c = 
+}}
+{{end-eqn}}
+Substituting back into the equation for $\mathcal L$:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = m x + \paren {m x_0 - y_0}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y - y_0
+      | r = m \paren {x - x_0}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Shortest Possible Distance between Lattice Points on Straight Line in Cartesian Plane}
+Tags: Straight Lines, Linear Diophantine Equations
+
+\begin{theorem}
+Let $\mathcal L$ be the [[Definition:Straight Line|straight line]] defined by the [[Equation of Straight Line in Plane/General Equation|equation]]:
+:$a x - b y = c$
+Let $p_1$ and $p_2$ be [[Definition:Lattice Point|lattice points]] on $\mathcal L$.
+Then the shortest possible [[Definition:Distance (Linear Measure)|distance]] $d$ between $p_1$ and $p_2$ is:
+:$d = \dfrac {\sqrt {a^2 + b^2} } {\gcd \set {a, b} }$
+where $\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor|greatest common divisor]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+Let $p_1 = \tuple {x_1, y_1}$ and $p_2 = \tuple {x_2, y_2}$ be on $\mathcal L$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a x_1 - b y_1
+      | r = c
+}}
+{{eqn | l = a x_2 - b y_2
+      | r = c
+      | c = 
+}}
+{{end-eqn}}
+From [[Solution of Linear Diophantine Equation]], it is [[Definition:Necessary and Sufficient|necessary and sufficient]] that:
+:$\gcd \set {a, b} \divides c$
+for such [[Definition:Lattice Point|lattice points]] to exist.
+Also from [[Solution of Linear Diophantine Equation]], all [[Definition:Lattice Point|lattice points]] on $\mathcal L$ are solutions to the equation:
+:$\forall k \in \Z: x = x_1 + \dfrac b m k, y = y_1 - \dfrac a m k$
+where $m = \gcd \set {a, b}$.
+So we have:
+{{begin-eqn}}
+{{eqn | l = x_2
+      | r = x_1 + \dfrac b m k
+}}
+{{eqn | l = y_2
+      | r = y_1 - \dfrac a m k
+      | c = 
+}}
+{{end-eqn}}
+for some $k \in \Z$ such that $k \ne 0$.
+The [[Definition:Distance (Linear Measure)|distance]] $d$ between $p_1$ and $p_2$ is given by the [[Distance Formula]]:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = \sqrt {\paren {x_1 - \paren {x_1 + \dfrac b m k} }^2 + \paren {y_1 - \paren {y_1 - \dfrac a m k} }^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {\paren {\dfrac {b k} m}^2 + \paren {\dfrac {a k} m}^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac {k^2 \paren {a^2 + b^2} } {m^2} }
+      | c = 
+}}
+{{eqn | r = k \dfrac {\sqrt {a^2 + b^2} } m
+      | c = 
+}}
+{{end-eqn}}
+This is a minimum when $k = 1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Decomposition into Even-Odd Integers is not always Unique}
+Tags: Even Integers
+
+\begin{theorem}
+For every [[Definition:Even Integer|even integer]] $n$ such that $n > 1$, if $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Even-Times Odd Integer|even-times odd integers]], it is not necessarily the case that this [[Definition:Integer Multiplication|product]] is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z$ be of the form $2^2 p q$ where $p$ and $q$ are [[Definition:Odd Prime|odd primes]].
+Then:
+:$n = \paren {2 p} \times \paren {2 q} = 2 \times \paren {2 p q}$
+A specific example that can be cited is $n = 60$:
+:$60 = 6 \times 10$
+and:
+:$60 = 2 \times 30$.
+Each of $2, 6, 10, 30$ are [[Definition:Even-Times Odd Integer|even-times odd integers]]:
+{{begin-eqn}}
+{{eqn | l = 2
+      | r = 2 \times 1
+}}
+{{eqn | l = 6
+      | r = 2 \times 3
+}}
+{{eqn | l = 10
+      | r = 2 \times 5
+}}
+{{eqn | l = 30
+      | r = 2 \times 15
+}}
+{{end-eqn}}
+Every $n \in \Z$ which has a [[Definition:Divisor of Integer|divisor]] in that same form $2^2 p q$ can similarly be decomposed non-uniquely into [[Definition:Even-Times Odd Integer|even-times odd integers]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Decomposition into Product of Power of 2 and Odd Integer is Unique}
+Tags: Integers
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Then $n$ can be decomposed into the [[Definition:Integer Multiplication|product]] of a [[Definition:Integer Power|power of $2$]] and an [[Definition:Odd Integer|odd integer]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exists $n \in \Z$ which can be decomposed into a [[Definition:Integer Power|power of $2$]] and an [[Definition:Odd Integer|odd integer]] in more than one way.
+That is:
+:$n = 2^a r$
+and:
+:$n = 2^b s$
+where:
+:$a, b \in \Z_{\ge 0}$
+:$r$ and $s$ are [[Definition:Odd Integer|odd integers]].
+:either $a \ne b$ or $r \ne s$.
+Suppose $r = s$.
+Then:
+:$\dfrac n r = \dfrac n s = 2^a = 2^b$
+which leads to:
+:$a = b$
+which [[Definition:Contradiction|contradicts]] our supposition that either $r \ne s$ or $a \ne b$.
+Similarly, suppose $a = b$.
+Then:
+:$\dfrac n {2^a} = \dfrac n {2^b} = r = s$
+which also [[Definition:Contradiction|contradicts]] our supposition that either $r \ne s$ or $a \ne b$.
+ 
+Thus both $r \ne s$ and $a \ne b$.
+{{WLOG}}, let $a > b$.
+Then $2^{a - b} = 2^c$ where $c \ge 1$.
+Thus:
+:$\dfrac n {2^b} = 2^c r = s$
+which means $s = 2 u$ for some $u \in \Z$.
+Thus $s$ is an [[Definition:Even Integer|even integer]].
+This [[Definition:Contradiction|contradicts]] our supposition that $r$ and $s$ are both [[Definition:Odd Integer|odd]].
+All possibilities have been investigated:
+:$a = b, r \ne s$
+:$a \ne b, r = s$
+:$a \ne b, r \ne s$
+All have been shown to lead to a [[Definition:Contradiction|contradiction]].
+Thus by [[Proof by Contradiction]] it can be seen that an decomposition of $n$ into the [[Definition:Integer Multiplication|product]] of a [[Definition:Integer Power|power of $2$]] and an [[Definition:Odd Integer|odd integer]] is [[Definition:Unique|unique]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Decomposition of Integer is Unique}
+Tags: Prime Decompositions
+
+\begin{theorem}
+Let $n$ be an [[Definition:Integer|integer]] such that $n > 1$.
+Then the [[Definition:Prime Decomposition|prime decomposition]] of $n$ is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+From [[Integer is Expressible as Product of Primes]], $n$ can be expressed as the [[Definition:Integer Multiplication|product]] of one or more [[Definition:Prime Number|primes]].
+Let $n = q_1 q_2 \dotsm q_s$ where $q_1, q_2, \ldots, q_s$ are all [[Definition:Prime Number|primes]] such that:
+:$(1): \quad n = q_1 \le q_2 \le \dotsb \le q_s$
+From [[Expression for Integer as Product of Primes is Unique]], the expression for $(1)$ is [[Definition:Unique|unique]].
+By the [[Fundamental Theorem of Equivalence Relations]], we can [[Definition:Set Partition|partition]] $\set {q_1, q_2, \dotsc, q_s}$ in $(1)$ according to equality.
+Thus the [[Definition:Equivalence Class|equivalence classes]] $\eqclass {q_j} =$ contain all repetitions of $q_j$.
+Hence the contribution of $q_j$ to $n$ is:
+:${q_j}^{k_j}$
+where $k_j = \card {\eqclass {q_j} =}$, the [[Definition:Cardinality|cardinality]] of $\eqclass {q_j} =$.
+Renaming the representative elements of the various $\eqclass {q_r} =$ as $p_1, p_2, \ldots, p_r$, where $r$ is the number of [[Definition:Equivalence Class|equivalence classes]].
+Hence:
+:$n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k^r}$
+As $n = q_1 \le q_2 \le \dotsb \le q_s$ is a unique representation, so is $n = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k^r}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Expression for Integers as Powers of Same Primes}
+Tags: Prime Decompositions
+
+\begin{theorem}
+Let $a, b \in \Z$ be [[Definition:Integer|integers]].
+Let their [[Definition:Prime Decomposition|prime decompositions]] be given by:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = {q_1}^{e_1} {q_2}^{e_2} \cdots {q_r}^{e_r}
+}}
+{{eqn | r = \prod_{\substack {q_i \mathop \divides a \\ \text {$q_i$ is prime} } } {q_i}^{e_i}
+}}
+{{eqn | l = b
+      | r = {s_1}^{f_1} {s_2}^{f_2} \cdots {s_u}^{f_u}
+}}
+{{eqn | r = \prod_{\substack {s_i \mathop \divides b \\ \text {$s_i$ is prime} } } {s_i}^{f_i}
+}}
+{{end-eqn}}
+Then there exist [[Definition:Prime Number|prime numbers]]:
+:$t_1 < t_2 < \dotsb < t_v$
+such that:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = a
+      | r = {t_1}^{g_1} {t_2}^{g_2} \cdots {t_v}^{g_v}
+}}
+{{eqn | n = 2
+      | l = b
+      | r = {t_1}^{h_1} {t_2}^{h_2} \cdots {t_v}^{h_v}
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+In the [[Definition:Prime Decomposition|prime decompositions]] $(1)$ and $(2)$, we have that:
+:$q_1 < q_2 < \dotsb < q_r$
+and:
+:$s_1 < s_2 < \dotsb < s_u$
+Hence we can define:
+{{begin-eqn}}
+{{eqn | l = E
+      | r = \set {q_1, q_2, \ldots, q_r}
+}}
+{{eqn | l = F
+      | r = \set {s_1, s_2, \ldots, s_u}
+}}
+{{end-eqn}}
+as all the $q_1, q_2, \dotsc, q_r$ are [[Definition:Distinct Elements|distinct]], and all the $s_1, s_2, \dotsc, s_u$ are [[Definition:Distinct Elements|distinct]].
+Then let:
+:$T = E \cup F$
+and let the [[Definition:Element|elements]] of $T$ be renamed as:
+:$T = \set {t_1, t_2, \ldots, t_v}$
+where all the $t_1, t_2, \dotsc, t_v$ are [[Definition:Distinct Elements|distinct]], and:
+:$t_1 < t_2 < \dotsb < t_v$
+Let $\iota: E \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]]:
+:$\forall q_i \in E: \map \iota {q_i} = q_i$
+Let $\iota: F \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]]:
+:$\forall s_i \in F: \map \iota {s_i} = s_i$
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = \prod_{i \mathop = 1}^r {q_i}^{e_i}
+      | c = 
+}}
+{{eqn | r = \prod_{q_i \mathop \in E} {q_i}^{e_i} \times \prod_{t_i \mathop \in T \mathop \setminus E} {t_i}^0
+      | c = 
+}}
+{{eqn | r = \prod_{t_j \mathop \in T} {t_j}^{g_j}
+      | c = where $g_j = \begin {cases} e_i & : t_j = q_i \\ 0 & : t_j \notin E \end{cases}$
+}}
+{{eqn | r = {t_1}^{g_1} {t_2}^{g_2} \dotsm {t_v}^{g_v} 
+      | c = for some $g_1, g_2, \dotsc, g_v \in \Z_{\ge 0}$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = \prod_{i \mathop = 1}^r {s_i}^{f_i}
+      | c = 
+}}
+{{eqn | r = \prod_{s_i \mathop \in F} {s_i}^{f_i} \times \prod_{t_i \mathop \in T \mathop \setminus F} {t_i}^0
+      | c = 
+}}
+{{eqn | r = \prod_{t_j \mathop \in T} {t_j}^{h_j}
+      | c = where $h_j = \begin {cases} f_i & : t_j = s_i \\ 0 & : t_j \notin F \end{cases}$
+}}
+{{eqn | r = {t_1}^{h_1} {t_2}^{h_2} \dotsm {t_v}^{h_v} 
+      | c = for some $h_1, h_2, \dotsc, h_v \in \Z_{\ge 0}$
+}}
+{{end-eqn}}
+Thus $a$ and $b$ can be expressed as the [[Definition:Integer Multiplication|product]] of [[Definition:Integer Power|powers]] of the same [[Definition:Prime Number|primes]], on the understanding that one or more of the [[Definition:Integer Power|powers]] in either [[Definition:Integer Multiplication|product]] may be [[Definition:Zero (Number)|zero]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Expression for Integers as Powers of Same Primes/General Result}
+Tags: Prime Decompositions
+
+\begin{theorem}
+Let $a_1, a_2, \dotsc, a_n \in \Z$ be [[Definition:Integer|integers]].
+Let their [[Definition:Prime Decomposition|prime decompositions]] be given by:
+:$\displaystyle a_i = \prod_{\substack {p_{i j} \mathop \divides a_i \\ \text {$p_{i j}$ is prime} } } {p_{i j} }^{e_{i j} }$
+Then there exists a [[Definition:Set|set]] $T$ of [[Definition:Prime Number|prime numbers]]:
+:$T = \set {t_1, t_2, \dotsc, t_v}$
+such that:
+:$t_1 < t_2 < \dotsb < t_v$
+:$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a [[Definition:Set|set]] $T = \set {t_1, t_2, \dotsc, t_v}$ of [[Definition:Prime Number|prime numbers]] such that $t_1 < t_2 < \dotsb < t_v$ such that:
+::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+there exist [[Definition:Prime Number|prime numbers]] $t_1 < t_2 < \dotsb < t_v$ such that:
+{{begin-eqn}}
+{{eqn | l = a_1
+      | r = \prod_{j \mathop = 1}^v {t_j}^{g_{1 j} }
+}}
+{{eqn | l = a_2
+      | r = \prod_{j \mathop = 1}^v {t_j}^{g_{2 j} }
+}}
+{{end-eqn}}
+This has been proved in [[Expression for Integers as Powers of Same Primes]].
+Thus $\map P 2$ is seen to hold.
+This is the [[Principle of Mathematical Induction#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 [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:for all $a_i \in \set {a_1, a_2, \ldots, a_k}$: there exists a [[Definition:Set|set]] $T = \set {t_1, t_2, \dotsc, t_v}$ of [[Definition:Prime Number|prime numbers]] such that $t_1 < t_2 < \dotsb < t_v$ such that:
+::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
+from which it is to be shown that:
+:for all $a_i \in \set {a_1, a_2, \ldots, a_{k + 1} }$: there exists a [[Definition:Set|set]] $T' = \set {t_1, t_2, \dotsc, t_w}$ of [[Definition:Prime Number|prime numbers]] such that $t_1 < t_2 < \dotsb < t_w$ such that:
+::$\displaystyle a_i = \prod_{j \mathop = 1}^w {t_j}^{g_{i j} }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = a_{k + 1}
+      | r = \prod_{\substack {q_i \mathop \divides a_{k + 1} \\ \text {$q_i$ is prime} } } {q_i}^{e_i}
+      | c = {{Defof|Prime Decomposition}}
+}}
+{{eqn | l = a_k
+      | r = \prod_{j \mathop = 1}^v {t_j}^{g_{k j} }
+      | c = by the [[Expression for Integers as Powers of Same Primes/General Result#Induction Hypothesis|induction hypothesis]]
+}}
+{{end-eqn}}
+Let $E = \set {q_i: q_i \divides a_{k + 1} , \text {$q_i$ is prime} }$.
+Then let:
+:$T' = E \cup T$
+and let the [[Definition:Element|elements]] of $T$ be renamed as:
+:$T' = \set {t_1, t_2, \ldots, t_w}$
+where all the $t_1, t_2, \dotsc, t_w$ are [[Definition:Distinct Elements|distinct]], and:
+:$t_1 < t_2 < \dotsb < t_w$
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = a_{k + 1}
+      | r = \prod_{q_i \mathop \in E} {q_i}^{e_i} \times \prod_{t_i \mathop \in T' \mathop \setminus E} {t_i}^0
+      | c = 
+}}
+{{eqn | r = \prod_{t_j \mathop \in T'} {t_j}^{g_j}
+      | c = where $g_j = \begin {cases} e_i & : t_j = q_i \\ 0 & : t_j \notin E \end{cases}$
+}}
+{{eqn | r = {t_1}^{g_1} {t_2}^{g_2} \dotsm {t_w}^{g_w} 
+      | c = for some $g_1, g_2, \dotsc, g_w \in \Z_{\ge 0}$
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore, for all $n \in \Z_{\ge 2}$:
+:for all $a_i \in \set {a_1, a_2, \ldots, a_n}$: there exists a [[Definition:Set|set]] $T = \set {t_1, t_2, \dotsc, t_v}$ of [[Definition:Prime Number|prime numbers]] such that $t_1 < t_2 < \dotsb < t_v$ such that:
+::$\displaystyle a_i = \prod_{j \mathop = 1}^v {t_j}^{g_{i j} }$
+\end{proof}<|endoftext|>
+\section{GCD from Prime Decomposition/General Result}
+Tags: Greatest Common Divisor, Prime Decompositions
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 2$.
+Let $\N_n$ be defined as:
+:$\N_n := \set {1, 2, \dotsc, n}$
+Let $A_n = \set {a_1, a_2, \dotsc, a_n} \subseteq \Z$ be a [[Definition:Set|set]] of $n$ [[Definition:Integer|integers]].
+From [[Expression for Integers as Powers of Same Primes]], let:
+:$\displaystyle \forall i \in \N_n: a_i = \prod_{p_j \mathop \in T} {p_j}^{e_{i j} }$
+where:
+:$T = \set {p_j: j \in \N_r}$
+such that:
+:$\forall j \in \N_{r - 1}: p_j < p_{j - 1}$
+:$\forall j \in \N_r: \exists i \in \N_n: p_j \divides a_i$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\displaystyle \map \gcd {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_n} }$
+where $\map \gcd {A_n}$ denotes the [[Definition:Greatest Common Divisor/Integers/General Definition|greatest common divisor]] of $a_1, a_2, \dotsc, a_n$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map \gcd {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_n} }$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+:$\displaystyle \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{1 j}, e_{2 j} } }$
+This is [[GCD from Prime Decomposition]].
+Thus $\map P 2$ is seen to hold.
+This is the [[Principle of Mathematical Induction#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 [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \gcd {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_k} }$
+from which it is to be shown that:
+:$\displaystyle \map \gcd {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_{k + 1} } }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \map \gcd {A_{k + 1} }
+      | r = \map \gcd {A_k \cup a_{k + 1} }
+      | c = 
+}}
+{{eqn | r = 
+      | c = 
+}}
+{{end-eqn}}
+{{finish|I think we need to prove $\gcd \set {a, b, c} {{=}} \gcd \set {\gcd \set {a, b}, c}$ }}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 2}: \displaystyle \map \gcd {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\min \set {e_{i j}: \, i \in \N_n} }$
+\end{proof}<|endoftext|>
+\section{LCM from Prime Decomposition/General Result}
+Tags: Lowest Common Multiple, Prime Decompositions
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]] such that $n \ge 2$.
+Let $\N_n$ be defined as:
+:$\N_n := \set {1, 2, \dotsc, n}$
+Let $A_n = \set {a_1, a_2, \dotsc, a_n} \subseteq \Z$ be a [[Definition:Set|set]] of $n$ [[Definition:Integer|integers]].
+From [[Expression for Integers as Powers of Same Primes]], let:
+:$\displaystyle \forall i \in \N_n: a_i = \prod_{p_j \mathop \in T} {p_j}^{e_{i j} }$
+where:
+:$T = \set {p_j: j \in \N_r}$
+such that:
+:$\forall j \in \N_{r - 1}: p_j < p_{j - 1}$
+:$\forall j \in \N_r: \exists i \in \N_n: p_j \divides a_i$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$
+where $\map \lcm {A_n}$ denotes the [[Definition:Greatest Common Divisor/Integers/General Definition|greatest common divisor]] of $a_1, a_2, \dotsc, a_n$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+:$\displaystyle \gcd \set {a_1, a_2} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{1 j}, e_{2 j} } }$
+This is [[GCD from Prime Decomposition]].
+Thus $\map P 2$ is seen to hold.
+This is the [[Principle of Mathematical Induction#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 [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map \lcm {A_k} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_k} }$
+from which it is to be shown that:
+:$\displaystyle \map \lcm {A_{k + 1} } = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_{k + 1} } }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \map \lcm {A_{k + 1} }
+      | r = \map \lcm {A_k \cup a_{k + 1} }
+      | c = 
+}}
+{{eqn | r = 
+      | c = 
+}}
+{{end-eqn}}
+{{finish|I think we need to prove $\lcm \set {a, b, c} {{=}} \lcm \set {\lcm \set {a, b}, c}$ }}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 2}: \displaystyle \map \lcm {A_n} = \prod_{j \mathop \in \N_r} {p_j}^{\max \set {e_{i j}: \, i \in \N_n} }$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{LCM of 3 Integers in terms of GCDs of Pairs of those Integers}
+Tags: Lowest Common Multiple, 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:
+:$\lcm \set {a, b, c} = \dfrac {a b c \gcd \set {a, b, c} } {d_1 d_2 d_3}$
+where:
+:$\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor]]
+:$\lcm$ denotes [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]
+:$d_1 = \gcd \set {a, b}$
+:$d_2 = \gcd \set {b, c}$
+:$d_3 = \gcd \set {a, c}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lcm \set {a, b, c}
+      | r = \lcm \set {a, \lcm \set {b, c} }
+}}
+{{eqn | r = \frac {a \lcm \set {b, c} } {\gcd \set {a, \lcm \set {b, c} } }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac 1 {\gcd \set {a, \lcm \set {b, c} } } }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac 1 {\lcm \set {\gcd \set {a, b}, \gcd \set {a, c} } } }
+      | c = [[GCD and LCM Distribute Over Each Other]]
+}}
+{{eqn | r = \frac {a b c} {\gcd \set {b, c} } \paren {\frac {\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } } {\gcd \set {a, b} \gcd \set {a, c} } }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {a b c \gcd \set {a, b, c} } {d_1 d_2 d_3}
+      | c = by [[LCM of 3 Integers in terms of GCDs of Pairs of those Integers/Lemma|Lemma]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Rule for Counting/General Theorem}
+Tags: Product Rule for Counting
+
+\begin{theorem}
+Suppose a process can be broken into $m$ successive, ordered, stages, with the $i$th stage having $r_i$ possible outcomes (for $i = 1, \ldots, m$).
+Let the number of outcomes at each stage be independent of the choices in previous stages
+Let the composite outcomes be all distinct.
+Then the total procedure has $\displaystyle \prod_{i \mathop = 1}^m r_i$ different composite outcomes.
+\end{theorem}<|endoftext|>
+\section{Alternating Summation of Binomial Coefficient of Summation of Binomial Coefficient of Sequence}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $\sequence a, \sequence b$ be [[Definition:Real Sequence|real sequences]] which satisfy the condition:
+:$a_n = \displaystyle \sum_{r \mathop = 0}^n \binom n r b_r$
+Then:
+:$\displaystyle \paren {-1}^n b_n = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s a_s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s a_s
+      | r = \sum_{s \mathop = 0}^n \binom n s \paren {-1}^s \paren {\sum_{r \mathop = 0}^s \binom s r b_r}
+      | c = 
+}}
+{{eqn | r = \sum_{s \mathop = 0}^n \sum_{\substack {r \mathop \le 0 \mathop \le n \\ s \mathop \ge r} } \binom n s \binom s r \paren {-1}^s b_r
+      | c = 
+}}
+{{eqn | r = \sum_{r \mathop = 0}^n b_r \sum_{s \mathop = r}^n \dfrac {n!} {s! \, \paren{n - s}!} \dfrac {s!} {r! \, \paren{s - r}!} \paren {-1}^s
+      | c = 
+}}
+{{eqn | r = \sum_{r \mathop = 0}^n \dfrac {n!} {r! \, \paren{n - r}!} b_r \sum_{s \mathop = r}^n \dfrac {\paren{n - r}! \paren {-1}^s} {\paren{n - s}! \paren{s - r}!}
+      | c = 
+}}
+{{eqn | r = \sum_{r \mathop = 0}^n \binom n r b_r \sum_{s \mathop = r}^n \binom {n - r} {s - r} \paren {-1}^s
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \sum_{r \mathop = 0}^n \binom n r b_r \sum_{s \mathop = 0}^{n - r} \binom {n - r} s \paren {-1}^{s + r}
+      | c = 
+}}
+{{end-eqn}}
+For $n - r > 1$ we can use the [[Binomial Theorem/Integral Index|Binomial Theorem]] with $x = 1$ and $y = -1$:
+:$0 = \paren {1 - 1}^{n - r} = \displaystyle \sum_{s \mathop = 0}^{n \mathop - r} \binom {n - r} s \paren {-1}^s$
+Hence all terms of $(1)$ vanish except for where $n - r$.
+That term is:
+:$\paren {-1}^n b_n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient}
+Tags: Fibonacci Numbers, Binomial Coefficients, Proofs by Induction
+
+\begin{theorem}
+Let $F_n$ denote the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]].
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall n \in \Z_{>0}:
+      | l = F_{2 n}
+      | r = \sum_{k \mathop = 1}^n \dbinom n k F_k
+      | c = 
+}}
+{{eqn | r = \dbinom n 1 F_1 + \dbinom n 2 F_2 + \dbinom n 3 F_3 + \dotsb + \dbinom n {n - 1} F_{n - 1} + \dbinom n n F_n
+      | c = 
+}}
+{{end-eqn}}
+where $\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = F_0
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^0 \dbinom 0 k F_k
+      | c = [[Definition:Vacuous Summation|vacuously]]
+}}
+{{end-eqn}}
+Thus $\map P 0$ is seen to hold.
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = F_2
+      | r = 1
+      | c = 
+}}
+{{eqn | r = \dbinom 1 1 F_1
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^1 \dbinom 1 k F_k
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$F_{2 m} = \displaystyle \sum_{k \mathop = 1}^m \dbinom m k F_k$
+from which it is to be shown that:
+:$F_{2 \paren {m + 1} } = \displaystyle \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^{m + 1} \dbinom {m + 1} k F_k
+      | r = \sum_{k \mathop = 1}^m \dbinom {m + 1} k F_k + \dbinom {m + 1} {m + 1} F_{m + 1}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^m \paren {\dbinom m {k - 1} + \dbinom m k} F_k + F_{m + 1}
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^m \dbinom m {k - 1} F_k  + F_{2 m} + F_{m + 1}
+      | c = [[Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^m \dbinom m {k - 1} \paren {F_{k - 1} + F_{k - 2} } + F_{2 m} + F_{m + 1}
+      | c = {{Defof|Fibonacci Numbers}}
+}}
+{{end-eqn}}
+{{finish|Can't see where this is going -- anyone care to have a go at this?}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 0}: F_{2 n} = \displaystyle \sum_{k \mathop = 1}^n \dbinom n k F_k$
+\end{proof}<|endoftext|>
+\section{Integer and Fifth Power have same Last Digit}
+Tags: Fifth Powers
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Then $n^5$ has the same last [[Definition:Digit|digit]] as $n$ when both are expressed in conventional [[Definition:Decimal Notation|decimal notation]].
+\end{theorem}
+
+\begin{proof}
+From [[Fermat's Little Theorem/Corollary 1|Fermat's Little Theorem: Corollary 1]]:
+:$n^5 \equiv n \pmod 5$
+Suppose $n \equiv 1 \pmod 2$.
+Then from [[Congruence of Powers]]:
+:$n^5 \equiv 1^5 \pmod 2$
+and so:
+:$n^5 \equiv 1 \pmod 2$
+Similarly, suppose $n \equiv 0 \pmod 2$.
+Then from [[Congruence of Powers]]:
+:$n^5 \equiv 0^5 \pmod 2$
+and so:
+:$n^5 \equiv 0 \pmod 2$
+Hence:
+:$n^5 \equiv n \pmod 2$
+So we have:
+:$n^5 \equiv n \pmod {2 \times 5}$
+and the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Sufficient Condition for 5 to divide n^2+1}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let:
+{{begin-eqn}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n - 1
+}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n
+}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n + 1
+}}
+{{end-eqn}}
+where $\nmid$ denotes non-[[Definition:Divisor of Integer|divisibility]].
+Then:
+:$5 \divides n^2 + 1$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+{{begin-eqn}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n - 1
+}}
+{{eqn | ll= \leadsto
+      | l = n - 1
+      | o = \not \equiv
+      | r = 0
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = \not \equiv
+      | r = 1
+      | rr= \pmod 5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = \not \equiv
+      | r = 0
+      | rr= \pmod 5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 5
+      | o = \nmid
+      | r = n + 1
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = \not \equiv
+      | r = 0
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = \not \equiv
+      | r = 4
+      | rr= \pmod 5
+}}
+{{end-eqn}}
+So either:
+:$n \equiv 2 \pmod 5$
+or:
+:$n \equiv 3 \pmod 5$
+and so:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \equiv
+      | r = 2
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n^2
+      | o = \equiv
+      | r = 4
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n^2 + 1
+      | o = \equiv
+      | r = 0
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = 5
+      | o = \divides
+      | r = n^2 + 1
+}}
+{{end-eqn}}
+or:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \equiv
+      | r = 3
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n^2
+      | o = \equiv
+      | r = 3^2
+      | rr= \pmod 5
+}}
+{{eqn | o = \equiv
+      | r = 4
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = n^2 + 1
+      | o = \equiv
+      | r = 0
+      | rr= \pmod 5
+}}
+{{eqn | ll= \leadsto
+      | l = 5
+      | o = \divides
+      | r = n^2 + 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Wilson's Theorem/Necessary Condition}
+Tags: Wilson's Theorem
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Then:
+:$\paren {p - 1}! \equiv -1 \pmod p$
+\end{theorem}<|endoftext|>
+\section{Wilson's Theorem/Sufficient Condition}
+Tags: Wilson's Theorem
+
+\begin{theorem}
+Let $p$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that:
+:$\paren {p - 1}! \equiv -1 \pmod p$
+Then $p$ is a [[Definition:Prime Number|prime number]].
+\end{theorem}
+
+\begin{proof}
+Assume $p$ is [[Definition:Composite Number|composite]], and $q$ is a [[Definition:Prime Number|prime]] such that $q \divides p$.
+Then both $p$ and $\paren {p - 1}!$ are [[Definition:Divisor of Integer|divisible]] by $q$.
+If the congruence $\paren {p - 1}! \equiv -1 \pmod p$ were satisfied, we would have $\paren {p - 1}! \equiv -1 \pmod q$.
+However, this amounts to $0 \equiv -1 \pmod q$, a contradiction.
+Hence for $p$ [[Definition:Composite Number|composite]], the [[Definition:Congruence Modulo Integer|congruence]] $\paren {p - 1}! \equiv -1 \pmod p$ cannot hold.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Necessary Condition for Integral Functional to have Extremum/Two Variables/Lemma}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $D \subset \R^2$.
+Let $\Gamma$ be the [[Definition:Boundary (Geometry)|boundary]] of $D$.
+Let $\alpha : D \to \R$ be a [[Definition:Continuous Real Function/Subset|continuous mapping]].
+Let $h : D \to \R$ be a twice [[Definition:Differentiability Class|differentiable mapping]] such that $\map h \Gamma = 0$.
+Suppose for every $h$ we have that:
+:$\displaystyle \iint_D \map \alpha {x, y} \map h {x,y} \rd x \rd y = 0$.
+Then:
+:$\displaystyle \forall x, y \in D : \map \alpha {x, y} = 0$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that:
+:$\displaystyle \exists x_0,y_0 \in D : \map \alpha {x_0,y_0} > 0$
+$\alpha$ is [[Definition:Continuous Real Function/Interval|continuous]] in $D$. 
+Therefore, there exists a [[Definition:Closed Ball|closed ball]] $B^-_{\epsilon}$ defined by:
+:$\map {B^-_{\epsilon}} {x_0, y_0} := \set {\tuple{x,y} \in D : \paren {x - x_0}^2 + \paren {y - y_0}^2 \le \epsilon^2}$
+such that:
+:$\forall x,y \in B^-_{\epsilon} : \map \alpha {x,y} > 0$
+Choose $\map h {x, y}$ in the following way:
+:$\map h {x,y} = \begin {cases} 0, \forall x, y \notin B^-_{\epsilon} \\ \sqbrk {\epsilon^2 - \paren {x - x_0}^2 - \paren {y - y_0}^2}^3, \forall x,y \in B^-_{\epsilon} \end{cases}$
+Such a choice for $\map h {x,y}$ satisfies the conditions of the [[Definition:Lemma|lemma]].
+But then both $\alpha$ and $h$ are [[Definition:Positive|positive]] inside this [[Definition:Closed Ball|ball]].
+Hence, the [[Definition:Definite Integral|integral]] is [[Definition:Positive|positive]].
+This [[Definition:Contradiction|contradicts]] assumptions of the [[Definition:Lemma|lemma]].
+Hence:
+:$\displaystyle \forall x, y \in D : \map \alpha {x,y} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Different n-gons that can be Inscribed in Circle}
+Tags: Stellated Polygons, Polygons, Combinatorics, Number of Different n-gons that can be Inscribed in Circle
+
+\begin{theorem}
+Let $C$ be a [[Definition:Circle|circle]] on whose [[Definition:Circumference of Circle|circumference]] $n$ [[Definition:Point|points]] are placed which divide $C$ into $n$ equal [[Definition:Arc of Circle|arcs]].
+The number of different [[Definition:Polygon|$n$-gons]] (either [[Definition:Stellated Polygon|stellated]] or otherwise) that can be described on $C$ whose [[Definition:Vertex of Polygon|vertices]] are those $n$ points is:
+:$S_n = \dfrac {\paren {n - 1}!} 2$
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Polygon|$n$-gon]] (either [[Definition:Stellated Polygon|stellated]] or otherwise) is constructed by connecting all $n$ [[Definition:Point|points]] in some order.
+From [[Number of Permutations]], there are ${}^n P_n = n!$ ways of ordering $n$ [[Definition:Point|points]].
+However, for each [[Definition:Polygon|$n$-gon]] formed in this way, we have:
+:$n$ ways of choosing the first [[Definition:Vertex of Polygon|vertex]]
+:$2$ different [[Definition:Side of Polygon|sides]] from that [[Definition:Vertex of Polygon|vertex]] to use as the first one (starting off either [[Definition:Clockwise|clockwise]] or [[Definition:Anticlockwise|anticlockwise]]).
+and all of these end up describing the same [[Definition:Polygon|$n$-gon]].
+Thus there are $\dfrac n 2 n! = \dfrac {\paren {n - 1}!} 2$ different [[Definition:Polygon|$n$-gons]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Regular Stellated Odd n-gons}
+Tags: Regular Stellated Polygons
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive]] [[Definition:Odd Integer|odd integer]].
+Then there are $\dfrac {n - 1} 2$ [[Definition:Distinct Elements|distinct]] [[Definition:Regular Stellated Polygon|regular stellated $n$-gons]].
+\end{theorem}
+
+\begin{proof}
+Let $P$ be a [[Definition:Regular Stellated Polygon|regular stellated $n$-gons]].
+Let the $n$ [[Definition:Vertex of Polygon|vertices]] of $P$ be $p_1, p_2, \dotsc, p_n$.
+These will be arranged on the [[Definition:Circumference of Circle|circumference]] of a [[Definition:Circle|circle]] $C$, dividing $C$ into $n$ [[Definition:Arc of Circle|arcs]] of equal [[Definition:Arc Length|length]].
+Once we have chosen the first [[Definition:Side of Polygon|side]] of $P$, the others are all the same [[Definition:Length of Line|length]] and are completely determined by that first [[Definition:Side of Polygon|side]].
+{{WLOG}}, the first [[Definition:Vertex of Polygon|vertex]] of $P$ is chosen to be $p_1$
+We can choose that first [[Definition:Side of Polygon|side]] as follows:
+:$p_1 p_2, p_1 n_3, \ldots, p_1 p_{n - 1}, p_1 p_n$
+But we have that:
+:$\size {p_1 p_2} = \size {p_1 p_n}$
+:$\size {p_1 p_3} = \size {p_1 p_{n - 1} }$
+and so on, down to:
+:$\size {p_1 p_{\paren {n - 1} / 2} }= \size {p_1 p_{\paren {n + 1} / 2} }$
+where $\size {p_a p_b}$ denotes the [[Definition:Length of Line|length]] of the [[Definition:Line Segment|line]] $p_1 p_b$.
+So for the $n - 1$ [[Definition:Line Segment|lines]] that are chosen for the first [[Definition:Side of Polygon|side]] of $P$, each is paired with another of the same [[Definition:Length of Line|length]].
+Hence there are $\dfrac {n - 1} 2$ ways of choosing the first [[Definition:Side of Polygon|side]] of $P$.
+Thus there are $\dfrac {n - 1} 2$ [[Definition:Distinct Elements|distinct]] [[Definition:Regular Stellated Polygon|regular stellated $n$-gons]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Wilson's Theorem/Corollary 1}
+Tags: Wilson's Theorem
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Then $p$ is the smallest [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] $\paren {p - 1}! + 1$.
+\end{theorem}
+
+\begin{proof}
+From [[Wilson's Theorem]], $p$ [[Definition:Divisor of Integer|divides]] $\paren {p - 1}! + 1$.
+Let $q$ be a [[Definition:Prime Number|prime number]] less than $p$.
+Then $q$ is a [[Definition:Divisor of Integer|divisor]] of $\paren {p - 1}!$ and so does not [[Definition:Divisor of Integer|divide]] $\paren {p - 1}! + 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes}
+Tags: Stellated Polygons, Set Partitions, Equivalence Classes, Partition of Non-Regular Prime Stellated Cyclic Polygons into Rotation Classes
+
+\begin{theorem}
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+Let $C$ be a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is $O$.
+Consider the [[Definition:Set|set]] $P$ of $p$ [[Definition:Point|points]] on the [[Definition:Circumference of Circle|circumference]] of $C$ dividing it into $p$ equal [[Definition:Arc of Circle|arcs]].
+Let $S$ be the [[Definition:Set|set]] of all non-[[Definition:Regular Stellated Polygon|regular]] [[Definition:Stellated Polygon|stellated $p$-gons]] whose [[Definition:Vertex of Polygon|vertices]] are the [[Definition:Element|elements]] of $P$.
+Let $\sim$ denote the [[Definition:Equivalence Relation|equivalence relation on $S$]] defined as:
+:$\forall \tuple {a, b} \in S \times S: a \sim b \iff$ there exists a [[Definition:Plane Rotation|plane rotation]] about $O$ transforming $a$ to $b$.
+Then the [[Definition:Equivalence Class|$\sim$-equivalence classes]] of $S$ into which $S$ can thereby be [[Definition:Set Partition|partitioned]] all have [[Definition:Cardinality|cardinality]] $p$.
+\end{theorem}<|endoftext|>
+\section{Square Modulo n Congruent to Square of Inverse Modulo n}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+:$a^2 \equiv \paren {n - a}^2 \pmod n$
+where the notation denotes [[Definition:Congruence Modulo Integer|congruence modulo $n$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {n - a}^2
+      | r = n^2 - 2 n - a^2
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = a^2
+      | rr= \pmod n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Modulo Arithmetic]]
+oaexkl44pkr2mnoa7jpv1g88idzc81s
+\end{proof}
\ No newline at end of file