diff --git "a/wiki/proofwiki/shard_34.txt" "b/wiki/proofwiki/shard_34.txt"
new file mode 100644--- /dev/null
+++ "b/wiki/proofwiki/shard_34.txt"
@@ -0,0 +1,14563 @@
+\section{Partition of Integer into Powers of 2 for Consecutive Integers}
+Tags: Partition Theory
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map b n$ denote the number of ways $n$ can be [[Definition:Integer Partition|partitioned]] into [[Definition:Integer Power|(integer) powers of $2$]].
+Then:
+:$\map b {2 n} = \map b {2 n + 1}$
+\end{theorem}
+
+\begin{proof}
+We prove the theorem by establishing a [[Definition:Bijection|bijection]] between the set of [[Definition:Integer Partition|partitions]] of $2 n$ with that of $2 n + 1$, under the constraint where each [[Definition:Integer Partition|partition]] is an [[Definition:Integer Power|integer power of $2$]].
+We have that $2^k$ is [[Definition:Even Integer|even]] for all $k > 0$.
+Also, we have that $2 n + 1$ is [[Definition:Odd Integer|odd]] for all $n$.
+So, for each [[Definition:Integer Partition|partition]] of $2 n + 1$ into [[Definition:Integer Power|integer powers of $2$]], there is at least one [[Definition:Part of Integer Partition|part]] of size $1$.
+Removing this [[Definition:Part of Integer Partition|part]] gives a [[Definition:Integer Partition|partition]] of $2 n$ into [[Definition:Integer Power|integer powers of $2$]].
+The [[Definition:Inverse Mapping|inverse]] of this is adding a [[Definition:Part of Integer Partition|part]] of size $1$ to a [[Definition:Integer Partition|partition]] of $2 n$.
+This will give a [[Definition:Integer Partition|partition]] of $2 n + 1$ into [[Definition:Integer Power|integer powers of $2$]].
+Hence both removing and adding are [[Definition:Well-Defined Mapping|well-defined mappings]].
+Thus they are [[Definition:Bijection|bijections]].
+Therefore:
+:$\map b {2 n} = \map b {2 n + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Partitions with no Multiple of 3 equals Number of Partitions where Parts appear No More than Twice}
+Tags: Examples of Integer Partitions
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\map t n$ denote the number of ways $n$ can be [[Definition:Integer Partition|partitioned]] into [[Definition:Part of Integer Partition|parts]] which are specifically not [[Definition:Integer Multiplication|multiples]] of $3$.
+Let $\map v n$ denote the number of ways $n$ can be [[Definition:Integer Partition|partitioned]] such that no [[Definition:Part of Integer Partition|part]] appears twice.
+Then:
+:$\forall n \in \Z_{>0}: \map t n = \map v n$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Chapter $12$ of {{BookReference|Number Theory|1971|George E. Andrews}} }}
+\end{proof}<|endoftext|>
+\section{Congruent Numbers are not necessarily Equal}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y, z \in \R$ be [[Definition:Real Number|real numbers]] such that:
+:$x \equiv y \pmod z$
+where $x \equiv y \pmod z$ denotes [[Definition:Congruence (Number Theory)|congruence modulo $z$]].
+Then it is not necessarily the case that $x = y$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+We have that:
+:$11 - 5 = 6 = 3 \times 2$
+and so by definition of [[Definition:Congruence (Number Theory)|congruence modulo $2$]]:
+:$10 \equiv 4 \pmod 2$
+But $11 \ne 5$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence Modulo Negative Number}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, b, c \in \R$ be [[Definition:Real Number|real numbers]].
+Then:
+:$a \equiv b \pmod c \iff a \equiv b \pmod {-c}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod c
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {a - b} 
+      | r = k c
+      | c = {{Defof|Congruence (Number Theory)|Congruence Modulo $c$}}: for some $k \in \Z$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {a - b} 
+      | r = -k \paren {-c}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod {-c}
+      | c = {{Defof|Congruence (Number Theory)|Congruence Modulo $-c$}}
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm and Absolute Value are Not Equivalent}
+Tags: Norm Theory, P-adic Number Theory, P-adic Norm and Absolute Value are Not Equivalent
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime number]] $p$.
+Let $\size{\,\cdot\,}$ be the [[Definition:Absolute Value|absolute value]] on the [[Definition:Rational Numbers|rationals $\Q$]].
+Then $\norm {\,\cdot\,}_p$ and $\size{\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+That is, the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_p$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\size {\,\cdot\,}$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
+:$\norm p_p = \frac 1 p < 1$
+By definition of the [[Definition:Absolute Value|absolute value]]:
+:$\size p = p > 1$
+By definition of [[Definition:Equivalent Division Ring Norms/Open Unit Ball Equivalent|open unit ball equivalence]], $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+By [[Equivalence of Definitions of Equivalent Division Ring Norms]] and the definition of [[Definition:Equivalent Division Ring Norms/Topologically Equivalent|topologically equivalent norms]] then the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_p$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\size{\,\cdot\,}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+It is noted that:
+:$\sup \set {\size n: n \in \Z} = +\infty$
+By a [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 3|corollary of Characterisation of Non-Archimedean Division Ring Norms]] then $\size {\,\cdot\,}$ is [[Definition:Archimedean Division Ring Norm|Archimedean]].
+By [[P-adic Norm is Non-Archimedean Norm]] then $\norm {\,\cdot\,}_p$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+By [[Equivalent Norms are both Non-Archimedean or both Archimedean]], $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm and Absolute Value are Not Equivalent/Proof 1}
+Tags: P-adic Norm and Absolute Value are Not Equivalent
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime number]] $p$.
+Let $\size {\,\cdot\,}$ be the [[Definition:Absolute Value|absolute value]] on the [[Definition:Rational Numbers|rationals $\Q$]].
+Then $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+That is, the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_p$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\size {\,\cdot\,}$.
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
+:$\norm p_p = \frac 1 p < 1$
+By definition of the [[Definition:Absolute Value|absolute value]]:
+:$\size p = p > 1$
+By definition of [[Definition:Equivalent Division Ring Norms/Open Unit Ball Equivalent|open unit ball equivalence]], $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+By [[Equivalence of Definitions of Equivalent Division Ring Norms]] and the definition of [[Definition:Equivalent Division Ring Norms/Topologically Equivalent|topologically equivalent norms]] then the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_p$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\size{\,\cdot\,}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm and Absolute Value are Not Equivalent/Proof 2}
+Tags: P-adic Norm and Absolute Value are Not Equivalent
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime number]] $p$.
+Let $\size {\,\cdot\,}$ be the [[Definition:Absolute Value|absolute value]] on the [[Definition:Rational Numbers|rationals $\Q$]].
+Then $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+That is, the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_p$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\size {\,\cdot\,}$.
+\end{theorem}
+
+\begin{proof}
+It is noted that:
+:$\sup \set {\size n: n \in \Z} = +\infty$
+By a [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 3|corollary of Characterisation of Non-Archimedean Division Ring Norms]] then $\size {\,\cdot\,}$ is [[Definition:Archimedean Division Ring Norm|Archimedean]].
+By [[P-adic Norm is Non-Archimedean Norm]] then $\norm {\,\cdot\,}_p$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+By [[Equivalent Norms are both Non-Archimedean or both Archimedean]], $\norm {\,\cdot\,}_p$ and $\size {\,\cdot\,}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norms are Not Equivalent}
+Tags: Norm Theory, P-adic Number Theory
+
+\begin{theorem}
+Let $p_1$ and $p_2$ be [[Definition:Prime Number|prime numbers]] such that $p_1 \neq p_2$.
+Let $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ be the [[Definition:P-adic Norm|$p$-adic norms]] on the [[Definition:Rational Numbers|rationals $\Q$]].
+Then $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+That is, the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_{p_1}$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_{p_2}$.
+\end{theorem}
+
+\begin{proof}
+Consider $p_1/p_2 \in \Q$. 
+With $\norm {\,\cdot\,}_{p_1}$:
+{{begin-eqn}}
+{{eqn| l = \norm{p_1/p_2}_{p_1}
+     | r = \norm{p_1}_{p_1} \norm{1/p_2}_{p_1}
+     | c = [[Definition:Norm/Division Ring|Norm axiom (M2) (Multiplicativity)]]
+}}
+{{eqn| r = \norm{p_1}_{p_1} \times 1
+     | c = $p_1$ does not divide $p_2$
+}}
+{{eqn| r = 1/{p_1}
+}}
+{{eqn| r = 1
+     | o = \lt
+}}
+{{end-eqn}}
+On the other hand, with $\norm {\,\cdot\,}_{p_2}$:
+{{begin-eqn}}
+{{eqn| l = \norm{p_1/p_2}_{p_2}
+     | r = \norm{p_1}_{p_2} \norm{1/p_2}_{p_2}
+     | c = [[Definition:Norm/Division Ring|Norm axiom (M2) (Multiplicativity)]]
+}}
+{{eqn| r = 1 \times \norm{1/p_2}_{p_2}
+     | c = $p_2$ does not divide $p_1$
+}}
+{{eqn| r = p_2
+}}
+{{eqn| r = 1
+     | o = \gt
+}}
+{{end-eqn}}
+By [[Definition:Equivalent Division Ring Norms/Open Unit Ball Equivalent|open unit ball equivalence]], $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+By [[Equivalence of Definitions of Equivalent Division Ring Norms]] and [[Definition:Equivalent Division Ring Norms/Topologically Equivalent|topological equivalence]] then the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_{p_1}$ does not equal the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by $\norm {\,\cdot\,}_{p_2}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomials of Congruent Integers are Congruent}
+Tags: Polynomial Theory, Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y, m \in \Z$ be [[Definition:Integer|integers]] where $m \ne 0$.
+Let:
+:$x \equiv y \pmod m$
+where the notation indicates [[Definition:Congruence Modulo Integer|congruence modlo $m$]].
+Let $a_0, a_1, \ldots, a_r$ be [[Definition:Integer|integers]].
+Then:
+:$\displaystyle \sum_{k \mathop = 0}^r a_k x^k \equiv \sum_{k \mathop = 0}^r a_k y^k \pmod m$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$x \equiv y \pmod m$
+From [[Congruence of Powers]]:
+:$x^k \equiv y^k \pmod m$
+From [[Modulo Multiplication is Well-Defined]]:
+:$\forall k \in \set {0, 2, \ldots, r}: a_k x^k \equiv a_k y^k \pmod m$
+The result follows from [[Modulo Addition is Well-Defined]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruent Integers less than Half Modulus are Equal}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $k \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $a, b \in \Z$ such that $\size a < \dfrac k 2$ and $\size b < \dfrac k 2$.
+Then:
+:$a \equiv b \pmod k \implies a = b$
+where $\equiv$ denotes [[Definition:Congruence Modulo Integer|congruence modulo $k$]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$-\dfrac k 2 < a < \dfrac k 2$
+and:
+:$-\dfrac k 2 < -b < \dfrac k 2$
+Thus:
+:$-k < a - b < k$
+Let $a \equiv b \pmod k$
+Then:
+:$a - b = n k$
+for some $n \in \Z$.
+But as $-k < n k < k$ it must be the case that $n = 0$.
+Thus $a - b = 0$ and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complete Residue System Modulo m has m Elements}
+Tags: Residue Classes
+
+\begin{theorem}
+Let $m \in \Z_{\ne 0}$ be a non-[[Definition:Zero (Number)|zero]] [[Definition:Integer|integer]].
+Let $S := \set {r_1, r_2, \dotsb, r_s}$ be a [[Definition:Complete Residue System|complete residue system modulo $m$]].
+Then $s = m$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$t_0 = 0, t_1 = 1, \dots, t_{m - 1} = m - 1$
+Let $n \in \Z$.
+Then from the [[Division Theorem]] there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q$ and $u$ such that:
+:$n = m q + u$
+such that $0 \le u < m$.
+That is:
+:$n \equiv u \pmod m$
+and $u$ is one of $t_0, t_1, \ldots, t_{m - 1}$.
+Also, since $\size {t_i - t_j} < m$, no two [[Definition:Element|elements]] of $\set {t_0, t_1, \ldots, t_{m - 1} }$ are [[Definition:Congruence Modulo Integer|congruent]].
+Thus $\set {t_0, t_1, \ldots, t_{m - 1} }$ is a [[Definition:Complete Residue System|complete residue system modulo $m$]].
+So each $r_i$ is [[Definition:Congruence Modulo Integer|congruent]] to [[Definition:Unique|exactly one]] [[Definition:Element|element]] of $\set {t_0, t_1, \ldots, t_{m - 1} }$.
+So:
+:$s \le m$
+And since $\set {r_1, r_2, \dotsb, r_s}$ forms a [[Definition:Complete Residue System|complete residue system modulo $m$]], every [[Definition:Element|element]] of $\set {t_0, t_1, \ldots, t_{m - 1} }$ is [[Definition:Congruence Modulo Integer|congruent]] to [[Definition:Unique|exactly one]] [[Definition:Element|element]] of $S$.
+So:
+:$m \le s$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Initial Segment of Natural Numbers forms Complete Residue System}
+Tags: Residue Classes
+
+\begin{theorem}
+Let $m \in \Z_{\ne 0}$ be a non-[[Definition:Zero (Number)|zero]] [[Definition:Integer|integer]].
+Let $\N_m = \set {0, 1, 2, \ldots, m - 1}$ denote the [[Definition:Initial Segment of Zero-Based Natural Numbers|initial segment of $\N$]]
+Then $\N_m$ is a [[Definition:Complete Residue System|complete residue system modulo $m$]].
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z$.
+From the [[Division Theorem]] there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q$ and $u$ such that:
+:$n = m q + u$
+such that $0 \le u < m$.
+That is:
+:$n \equiv u \pmod m$
+and $u$ is one of $0, 1, \ldots, m - 1$.
+Also, since $\forall i, j \in \N_m: \size {i - j} < m$, no two [[Definition:Element|elements]] of $\N_m$ are [[Definition:Congruence Modulo Integer|congruent]].
+Thus $\N_m = \set {0, 1, 2, \ldots, m - 1}$ is a [[Definition:Complete Residue System|complete residue system modulo $m$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence Modulo Power of p as Linear Combination of Congruences Modulo p}
+Tags: Residue Classes
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $S = \set {a_1, a_2, \ldots, a_p}$ be a [[Definition:Complete Residue System|complete residue system modulo $p$]].
+Then for all [[Definition:Integer|integers]] $n \in \Z$ and [[Definition:Non-Negative Integer|non-negative integer]] $s \in \Z_{\ge 0}$, there exists a [[Definition:Congruence Modulo Integer|congruence]] of the form:
+:$n \equiv \displaystyle \sum_{j \mathop = 0}^s b_j p^j \pmod {p^{s + 1} }$
+where $b_j \in S$.
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] on $s$:
+=== Basis for the Induction ===
+For $s = 0$, we apply the definition of a [[Definition:Complete Residue System|complete residue system modulo $p$]]:
+:$\forall n \in \Z: \exists a_i \in S: n \equiv a_i \pmod p$
+This is our [[Definition:Basis for the Induction|base case]].
+=== Induction Hypothesis ===
+This is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:For some $k \in \Z_{\ge 0}$, for all [[Definition:Integer|integers]] $n \in \Z$, there exists a [[Definition:Congruence Modulo Integer|congruence]] of the form:
+::$n \equiv \displaystyle \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$
+It is to be demonstrated that:
+:For all [[Definition:Integer|integers]] $n \in \Z$, there exists a [[Definition:Congruence Modulo Integer|congruence]] of the form:
+::$n \equiv \displaystyle \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+From $n \equiv \displaystyle \sum_{j \mathop = 0}^k b_j p^j \pmod {p^{k + 1} }$ we have:
+:$\exists q \in \Z: n = \displaystyle \sum_{j \mathop = 0}^k b_j p^j + q p^{k + 1}$
+From the definition of a [[Definition:Complete Residue System|complete residue system modulo $p$]]:
+:$\exists a_i \in S: q \equiv a_i \pmod p$
+This gives:
+:$\exists r \in \Z: q = a_i + r p$
+Substituting this to the original equation we have:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \sum_{j \mathop = 0}^k b_j p^j + \paren {a_i + r p} p^{k + 1}
+}}
+{{eqn | r = a_i p^{k + 1} + \sum_{j \mathop = 0}^k b_j p^j + r p^{k + 2}
+}}
+{{eqn | o = \equiv 
+      | r = a_i p^{k + 1} + \sum_{j \mathop = 0}^k b_j p^j
+      | rr = \pmod {p^{k + 2} }
+}}
+{{end-eqn}}
+This shows that $n$ can be expressed as the form:
+:$n \equiv \displaystyle \sum_{j \mathop = 0}^{k + 1} b_j p^j \pmod {p^{k + 2} }$
+By the [[Principle of Mathematical Induction]], the theorem is true for any $s$.
+Note that in the proof above, we did not use the fact that $p$ is a [[Definition:Prime Number|prime number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Non-Dividing Primes Less than n is Less than Euler Phi Function of n}
+Tags: Euler Phi Function, Prime Numbers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\map w n$ denote the number of [[Definition:Prime Number|primes]] strictly less than $n$ which are not [[Definition:Divisor of Integer|divisors]] of $n$.
+Let $\map \phi n$ denote the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $n$.
+Then:
+:$\map w n < \map \phi n$
+\end{theorem}
+
+\begin{proof}
+Let $P = \set {p < n: p \text { prime}, p \nmid n}$.
+Let $Q = \set {0 < q < n: q \perp n}$, where $q \perp n$ denotes that $q$ and $n$ are [[Definition:Coprime Integers|coprime]].
+Let $p \in P$.
+From [[Prime not Divisor implies Coprime]], $p$ is [[Definition:Coprime Integers|coprime]] to $n$.
+That is:
+:$p \in Q$
+So, by definition of [[Definition:Subset|subset]]:
+:$P \subseteq Q$
+From [[Integer is Coprime to 1]]:
+:$1 \in Q$
+But as [[One is not Prime]]:
+:$1 \notin P$
+Thus $P \subsetneq Q$ and so:
+:$\card P < \card Q$
+By definition of $\map w n$:
+:$\card P = \map w n$
+and by definition of [[Definition:Euler Phi Function|Euler $\phi$ function]]:
+:$\card Q = \map \phi n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Schatunowsky's Theorem}
+Tags: Euler Phi Function, Prime Numbers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\map w n$ denote the number of [[Definition:Prime Number|primes]] strictly less than $n$ which are not [[Definition:Divisor of Integer|divisors]] of $n$.
+Let $\map \phi n$ denote the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $n$.
+Then $30$ is the largest [[Definition:Integer|integer]] $n$ such that:
+:$\map w n = \map \phi n - 1$
+\end{theorem}
+
+\begin{proof}
+The above equation is equivalent to the property that all numbers greater than $1$ that are [[Definition:Coprime Integers|coprime]] to it but less are [[Definition:Prime Number|prime]].
+For an [[Definition:Integer|integer]] to have this property:
+If it is greater than $p^2$ for some [[Definition:Prime Number|prime]] $p$, then it must be [[Definition:Divisor of Integer|divisible]] by $p$.
+If not, it will be [[Definition:Coprime Integers|coprime]] to $p^2$, a [[Definition:Composite Number|composite number]].
+Let $p_n$ denote the $n$th [[Definition:Prime Number|prime]].
+Suppose $N$ has this property.
+By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_1 p_2 \cdots p_n \divides N$.
+By [[Absolute Value of Integer is not less than Divisors]], we have $p_1 p_2 \cdots p_n \le N$.
+[[Bertrand-Chebyshev Theorem]] asserts that there is a [[Definition:Prime Number|prime]] between $p_n$ and $2 p_n$.
+Thus we have $2 p_n > p_{n + 1}$.
+Hence for $n \ge 5$:
+{{begin-eqn}}
+{{eqn | l = N
+      | o = \ge
+      | r = p_1 p_2 \cdots p_n
+}}
+{{eqn | r = 2 \times 3 \times 5 p_4 \cdots p_n
+}}
+{{eqn | o = >
+      | r = 8 p_{n - 1} p_n
+}}
+{{eqn | o = >
+      | r = 4 p_n^2
+      | c = [[Bertrand-Chebyshev Theorem]]
+}}
+{{eqn | o = >
+      | r = p_{n + 1}^2
+      | c = [[Bertrand-Chebyshev Theorem]]
+}}
+{{eqn | o = \ge
+      | r = N
+      | c = From assumption
+}}
+{{end-eqn}}
+This is a [[Definition:Contradiction|contradiction]]. 
+Hence we must have $N \le p_5^2 = 121$.
+From the argument above we also have:
+:$2 \divides N$ for $4 < N \le 9$
+:$2, 3 \divides N$ for $9 < N \le 25$
+:$2, 3, 5 \divides N$ for $25 < N \le 49$
+:$2, 3, 5, 7 \divides N$ for $49 < N \le 121$
+So we end up with the list $N = 1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.
+This list is verified in [[Integers such that all Coprime and Less are Prime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Position of Card after n Modified Perfect Faro Shuffles}
+Tags: Cards
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck of cards]] $D$ of size $2 r$.
+Let $C$ be a [[Definition:Card|card]] in position $x$ of $D$.
+Let $n$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]] be performed on $C$.
+Then $C$ will be in position $w$, where:
+:$w \equiv 2^n x \pmod {2 r + 1}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:After $n$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]], $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod  {2 r + 1}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:After $1$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffle]], $C$ will be in position $a_1$ where $a_1 \equiv 2 x \pmod  {2 r + 1}$
+which is the definition of a [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffle]].
+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]]:
+:After $k$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]], $C$ will be in position $a_k$ where $a_k \equiv 2^k x \pmod  {2 r + 1}$
+from which it is to be shown that:
+:After $k + 1$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]], $C$ will be in position $a_{k + 1}$ where $a_{k + 1} \equiv 2^{k + 1} x \pmod  {2 r + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = a_{k + 1}
+      | o = \equiv
+      | r = 2 a_k
+      | rr= \pmod {2 r + 1}
+      | c = [[Position of Card after n Modified Perfect Faro Shuffles#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | o = \equiv
+      | r = 2 \paren {2^k x}
+      | rr= \pmod {2 r + 1}
+      | c = [[Position of Card after n Modified Perfect Faro Shuffles#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | o = \equiv
+      | r = 2^{k + 1} x
+      | rr= \pmod {2 r + 1}
+      | c = [[Congruence of Product]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>0}$: after $n$ [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]], $C$ will be in position $a_n$ where $a_n \equiv 2^n x \pmod  {2 r + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles}
+Tags: Deck of 52 Cards returns to Original Order after 52 Modified Perfect Faro Shuffles, Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, 52
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck]] of $52$ [[Definition:Card|cards]].
+Let $D$ be given a [[Definition:Sequence|sequence]] of [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]].
+Then after $52$ such [[Definition:Shuffle|shuffles]], the [[Definition:Card|cards]] of $D$ will be in the same [[Definition:Order of Cards|order]] they started in.
+\end{theorem}<|endoftext|>
+\section{Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order}
+Tags: Cards
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck]] of $2 m$ [[Definition:Card|cards]].
+Let $D$ be given a [[Definition:Sequence|sequence]] of [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]].
+Then the [[Definition:Card|cards]] of $D$ will return to their original [[Definition:Order of Cards|order]] after $n$ such [[Definition:Shuffle|shuffles]], where:
+:$2^n \equiv 1 \pmod {2 m + 1}$
+\end{theorem}
+
+\begin{proof}
+From [[Position of Card after n Modified Perfect Faro Shuffles]], after $n$ [[Definition:Shuffle|shuffles]] a [[Definition:Card|card]] in position $x$ will be in position $2^n x \pmod {m + 1}$.
+So for all $2 m$ [[Definition:Card|cards]] in $D$, we need to find $n$ such that:
+:$2^n x \equiv x \pmod {2 m + 1}$
+Because $2 m + 1$ is [[Definition:Odd Integer|odd]], we have:
+:$\gcd \set {2, 2 m + 1}$
+and so from [[Cancellability of Congruences]]:
+:$2^n \equiv 1 \pmod {2 m + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 62 Cards}
+Tags: Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, 62
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck]] of $62$ [[Definition:Card|cards]].
+Let $D$ be given a [[Definition:Sequence|sequence]] of [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]].
+Then after $6$ such [[Definition:Shuffle|shuffles]], the [[Definition:Card|cards]] of $D$ will be in the same [[Definition:Order of Cards|order]] they started in.
+\end{theorem}
+
+\begin{proof}
+From [[Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order]], the [[Definition:Card|cards]] of $D$ will return to their original [[Definition:Order of Cards|order]] after $n$ such [[Definition:Shuffle|shuffles]], where:
+:$2^n \equiv 1 \pmod {63}$
+We have that:
+:$63 = 2^6 - 1$
+and so:
+:$2^6 \equiv 1 \pmod {63}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 8 Cards}
+Tags: Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, 8
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck]] of $8$ [[Definition:Card|cards]].
+Let $D$ be given a [[Definition:Sequence|sequence]] of [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]].
+Then after $6$ such [[Definition:Shuffle|shuffles]], the [[Definition:Card|cards]] of $D$ will be in the same [[Definition:Order of Cards|order]] they started in.
+\end{theorem}
+
+\begin{proof}
+From [[Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order]], the [[Definition:Card|cards]] of $D$ will return to their original [[Definition:Order of Cards|order]] after $n$ such [[Definition:Shuffle|shuffles]], where:
+:$2^n \equiv 1 \pmod 9$
+Inspecting $2^n$ for $n$ from $1$:
+{{begin-eqn}}
+{{eqn | l = 2^1
+      | o = \equiv
+      | r = 2
+      | rr= \pmod 9
+      | c = 
+}}
+{{eqn | l = 2^2
+      | o = \equiv
+      | r = 4
+      | rr= \pmod 9
+      | c = 
+}}
+{{eqn | l = 2^3
+      | o = \equiv
+      | r = 8
+      | rr= \pmod 9
+      | c = 
+}}
+{{eqn | l = 2^4
+      | o = \equiv
+      | r = 7
+      | rr= \pmod 9
+      | c = 
+}}
+{{eqn | l = 2^5
+      | o = \equiv
+      | r = 7
+      | rr= \pmod 9
+      | c = 
+}}
+{{eqn | l = 2^6
+      | o = \equiv
+      | r = 1
+      | rr= \pmod 9
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order/Examples/Deck of 12 Cards}
+Tags: Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order, 12
+
+\begin{theorem}
+Let $D$ be a [[Definition:Deck of Cards|deck]] of $12$ [[Definition:Card|cards]].
+Let $D$ be given a [[Definition:Sequence|sequence]] of [[Definition:Modified Perfect Faro Shuffle|modified perfect faro shuffles]].
+Then after $6$ such [[Definition:Shuffle|shuffles]], the [[Definition:Card|cards]] of $D$ will be in the same [[Definition:Order of Cards|order]] they started in.
+\end{theorem}
+
+\begin{proof}
+From [[Number of Modified Perfect Faro Shuffles to return Deck of Cards to Original Order]], the [[Definition:Card|cards]] of $D$ will return to their original [[Definition:Order of Cards|order]] after $n$ such [[Definition:Shuffle|shuffles]], where:
+:$2^n \equiv 1 \pmod {13}$
+From [[Fermat's Little Theorem]]:
+:$2^{12} \equiv 1 \pmod {13}$
+so we know that $n$ is at most $12$.
+But $n$ may be smaller, so it is worth checking the values:
+Inspecting $2^n$ for $n$ from $1$:
+{{begin-eqn}}
+{{eqn | l = 2^1
+      | o = \equiv
+      | r = 2
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^2
+      | o = \equiv
+      | r = 4
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^3
+      | o = \equiv
+      | r = 8
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^4
+      | o = \equiv
+      | r = 3
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^6
+      | o = \equiv
+      | r = 6
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^6
+      | o = \equiv
+      | r = 12
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^7
+      | o = \equiv
+      | r = 11
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^8
+      | o = \equiv
+      | r = 9
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^9
+      | o = \equiv
+      | r = 5
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^{10}
+      | o = \equiv
+      | r = 10
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^{11}
+      | o = \equiv
+      | r = 7
+      | rr= \pmod {13}
+      | c = 
+}}
+{{eqn | l = 2^{12}
+      | o = \equiv
+      | r = 1
+      | rr= \pmod {13}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gamma Function/Examples/One Third}
+Tags: Examples of Gamma Function Values
+
+\begin{theorem}
+:$\map \Gamma {\dfrac 1 3} = 2 \cdotp 67893 \, 85347 \, 07747 \, 63 \ldots$
+\end{theorem}<|endoftext|>
+\section{Gamma Function/Examples/One Quarter}
+Tags: Examples of Gamma Function Values
+
+\begin{theorem}
+:$\map \Gamma {\dfrac 1 4} = 3 \cdotp 62560 \, 99082 \, 21908 \ldots$
+\end{theorem}<|endoftext|>
+\section{Difference of Two Fourth Powers}
+Tags: Fourth Powers, Difference of Two Fourth Powers, Examples of Use of Difference of Two Powers
+
+\begin{theorem}
+:$x^4 - y^4 = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^4 - y^4
+      | r = \paren {x^2}^2 - \paren {y^2}^2
+      | c = 
+}}
+{{eqn | r = \paren {x^2 - y^2} \paren {x^2 + y^2}
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \paren {x - y} \paren {x + y} \paren {x^2 + y^2}
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Two Odd Powers}
+Tags: Algebra, Polynomial Theory, Sum of Two Odd Powers
+
+\begin{theorem}
+Let $\F$ be one of the [[Definition:Standard Number System|standard number systems]], that is $\Z, \Q, \R$ and so on.
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = a^{2 n + 1} + b^{2 n + 1}
+      | r = \paren {a + b} \sum_{j \mathop = 0}^{2 n} \paren {-1}^j a^{2 n - j} b^j
+      | c = 
+}}
+{{eqn | r = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb - a b^{2 n - 1} + b^{2 n} }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a^{2 n + 1} + b^{2 n + 1}
+      | r = a^{2 n + 1} - \paren {-\paren {b^{2 n + 1} } }
+      | c = 
+}}
+{{eqn | r = a^{2 n + 1} - \paren {-b}^{2 n + 1}
+      | c = as $n$ is [[Definition:Odd Integer|odd]]
+}}
+{{eqn | r = \paren {a - \paren {-b} } \sum_{j \mathop = 0}^{2 n} a^{2 n - j} \paren {-b}^j
+      | c = [[Difference of Two Powers]]
+}}
+{{eqn | r = \paren {a + b} \sum_{j \mathop = 0}^{2 n} \paren {-1}^j a^{2 n - j} b^j
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Fifth Powers}
+Tags: Fifth Powers, Examples of Use of Difference of Two Powers
+
+\begin{theorem}
+:$x^5 - y^5 = \paren {x - y} \paren {x^4 + x^3 y + x^2 y^2 + x y^3 + y^4}$
+\end{theorem}
+
+\begin{proof}
+From [[Difference of Two Powers]]:
+:$\displaystyle a^n - b^n = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+The result follows directly by setting $n = 5$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Two Fifth Powers}
+Tags: Fifth Powers, Examples of Use of Sum of Two Odd Powers
+
+\begin{theorem}
+:$x^5 + y^5 = \paren {x + y} \paren {x^4 - x^3 y + x^2 y^2 - x y^3 + y^4}$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Two Odd Powers]]:
+:$a^{2 n + 1} + b^{2 n + 1} = \paren {a + b} \paren {a^{2 n} - a^{2 n - 1} b + a^{2 n - 2} b^2 - \dotsb + a b^{2 n - 1} + b^{2 n} }$
+We have that $5 = 2 \times 2 + 1$.
+The result follows directly by setting $n = 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Sixth Powers}
+Tags: Sixth Powers, Examples of Use of Difference of Two Powers
+
+\begin{theorem}
+:$x^6 - y^6 = \paren {x - y} \paren {x + y} \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^6 - y^6
+      | r = \paren {x^3}^2 - \paren {y^3}^2
+      | c = 
+}}
+{{eqn | r = \paren {x^3 - y^3} \paren {x^3 + y^3}
+      | c = 
+}}
+{{eqn | r = \paren {x - y} \paren {x^2 + x y + y^2} \paren {x^3 + y^3}
+      | c = [[Difference of Two Cubes]]
+}}
+{{eqn | r = \paren {x - y} \paren {x^2 + x y + y^2} \paren {x + y} \paren {x^2 - x y + y^2}
+      | c = [[Sum of Two Cubes]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Fourth Powers with Product of Squares}
+Tags: Fourth Powers, Square Function
+
+\begin{theorem}
+:$x^4 + x^2 y^2 + y^4 = \paren {x^2 + x y + y^2} \paren {x^2 - x y + y^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^6 - y^6
+      | r = \paren {x - y} \paren {x + y} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
+      | c = [[Difference of Two Sixth Powers]]
+}}
+{{eqn | r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x^6 - y^6
+      | r = \paren {x^2}^3 - \paren {y^2}^3
+      | c = 
+}}
+{{eqn | r = \paren {x^2 - y^2} \paren {\paren {x^2}^2 + \paren {x^2} \paren {y^2} + \paren {y^2}^2}
+      | c = [[Difference of Two Cubes]]
+}}
+{{eqn | r = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}
+      | c = simplifying
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {x^2 - y^2} \paren {x^4 + x^2 y^2 + y^4}
+      | r = \paren {x^2 - y^2} \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
+      | c = as both equal $x^6 - y^6$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x^4 + x^2 y^2 + y^4}
+      | r = \paren {x^2 + 2 x y + 2 y^2} \paren {x^2 - 2 x y + 2 y^2}
+      | c = cancelling $\paren {x^2 - y^2}$ from both sides
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Odd Powers}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $\mathbb F$ denote one of the [[Definition:Standard Number System|standard number systems]], that is $\Z$, $\Q$, $\R$ and $\C$.
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then for all $a, b \in \mathbb F$:
+{{begin-eqn}}
+{{eqn | l = a^{2 n + 1} - b^{2 n + 1}
+      | r = \paren {a - b} \sum_{j \mathop = 0}^{2 n} a^{2 n - j} b^j
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {a^{2 n} + a^{2 n - 1} b + a^{2 n - 2} b^2 + \dotsb + a b^{2 n - 1} + b^{2 n} }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+A direct application of [[Difference of Two Powers]]:
+{{begin-eqn}}
+{{eqn | l = a^n - b^n
+      | r = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+and setting $n \to 2 n + 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Even-Times Odd Powers}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $\F$ be one of the [[Definition:Standard Number System|standard number systems]], that is $\Z, \Q, \R$ and so on.
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive]] [[Definition:Odd Integer|odd integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = a^{2 n} - b^{2 n}
+      | r = \paren {a - b} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {a + b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} } \paren {a^{n - 1} - a^{n - 2} b + a^{n - 3} b^2 - \dotsb - a b^{n - 2} + b^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a^{2 n} - b^{2 n}
+      | r = \paren {a^n}^2 - \paren {b^n}^2
+      | c = 
+}}
+{{eqn | r = \paren {a^n - b^n} \paren {a^n + b^n}
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a^n + b^n}
+      | c = [[Difference of Two Powers]]
+}}
+{{eqn | r = \paren {a - b} \paren {\sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j} \paren {a + b} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j}
+      | c = [[Sum of Two Odd Powers]]
+}}
+{{end-eqn}}
+whence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Even Powers}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $\mathbb F$ denote one of the [[Definition:Standard Number System|standard number systems]], that is $\Z$, $\Q$, $\R$ and $\C$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then for all $a, b \in \mathbb F$:
+{{begin-eqn}}
+{{eqn | l = a^{2 n} - b^{2 n}
+      | r = \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \paren {a + b} \paren {a^{2 n - 2} + a^{2 n - 4} b^2 + a^{2 n - 6} b^4 + \dotsb + a^2 b^{2 n - 4} + b^{2 n - 2} }
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a^{2 n} - b^{2 n}
+      | r = \paren {a^2}^n - \paren {b^2}^n
+      | c = 
+}}
+{{eqn | r = \paren {a^2 - b^2} \paren {\sum_{j \mathop = 0}^{n - 1} \paren {a^2} ^{n - j - 1} \paren {b^2}^j}
+      | c = [[Difference of Two Powers]]
+}}
+{{eqn | r = \paren {a - b} \paren {a + b} \sum_{j \mathop = 0}^{n - 1} a^{2 \paren {n - j - 1} } b^{2 j}
+      | c = [[Difference of Two Squares]]
+}}
+{{end-eqn}}
+whence the result.
+{{qed}}
+[[Category:Difference of Two Powers]]
+5mm6ext2i6hvk5jyif69hx9pvofs2x2
+\end{proof}<|endoftext|>
+\section{Factors of Difference of Two Odd Powers}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = x^{2 n + 1} - y^{2 n + 1}
+      | r = \paren {x - y} \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 - 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}
+      | c = 
+}}
+{{eqn | r = \paren {x - y} \paren {x^2 - 2 x y \cos \dfrac {2 \pi} {2 n + 1} + y^2} \paren {x^2 - 2 x y \cos \dfrac {4 \pi} {2 n + 1} + y^2} \dotsm \paren {x^2 - 2 x y \cos \dfrac {2 n \pi} {2 n + 1} + y^2}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Factorisation of z^n-a|Factorisation of $z^n - a$]]:
+:$x^{2 n + 1} - y^{2 n + 1} = \displaystyle \prod_{k \mathop = 0}^{2 n} \paren {x - \alpha^k y}$
+where $\alpha$ is a [[Definition:Primitive Complex Root of Unity|primitive complex $2 n + 1$th roots of unity]], for example:
+{{begin-eqn}}
+{{eqn | l = \alpha
+      | r = e^{2 i \pi / \paren {2 n + 1} }
+      | c = 
+}}
+{{eqn | r = \cos \dfrac {2 \pi} {2 n + 1} + i \sin \dfrac {2 \pi} {2 n + 1}
+      | c = 
+}}
+{{end-eqn}}
+From [[Complex Roots of Unity occur in Conjugate Pairs]]:
+:$U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$
+where $U_{2 n + 1}$ denotes the [[Definition:Complex Roots of Unity|complex $2 n + 1$th roots of unity]]:
+:$U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$
+The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.
+Taking the [[Definition:Complex Multiplication|product]] of each of the remaining factors of $x^{2 n + 1} - y^{2 n + 1}$ in pairs:
+{{begin-eqn}}
+{{eqn | l = \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k + 1} y}
+      | r = \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}
+      | c = [[Complex Roots of Unity occur in Conjugate Pairs]]
+}}
+{{eqn | r = x^2 - x \paren {\alpha^k + \overline {\alpha^k} } y + \alpha^k y \, \overline {\alpha^k} y
+      | c = 
+}}
+{{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2
+      | c = [[Modulus in Terms of Conjugate]]
+}}
+{{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2
+      | c = [[Modulus of Complex Root of Unity equals 1]]
+}}
+{{eqn | r = x^2 - x y \paren {\cos \dfrac {2 k \pi} {2 n + 1} + i \sin \dfrac {2 k \pi} {2 n + 1} + \cos \dfrac {2 k \pi} {2 n + 1} - i \sin \dfrac {2 k \pi} {2 n + 1} } + y^2
+      | c = Definition of $\alpha$
+}}
+{{eqn | r = x^2 - 2 x y \cos \dfrac {2 k \pi} {2 n + 1} + y^2
+      | c = simplification
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Factors of Sum of Two Odd Powers}
+Tags: Sum of Two Odd Powers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+{{begin-eqn}}
+{{eqn | l = x^{2 n + 1} + y^{2 n + 1}
+      | r = \paren {x + y} \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}
+      | c = 
+}}
+{{eqn | r = \paren {x + y} \paren {x^2 + 2 x y \cos \dfrac {2 \pi} {2 n + 1} + y^2} \paren {x^2 + 2 x y \cos \dfrac {4 \pi} {2 n + 1} + y^2} \dotsm \paren {x^2 + 2 x y \cos \dfrac {2 n \pi} {2 n + 1} + y^2}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{2 n + 1} + y^{2 n + 1}
+      | r = x^{2 n + 1} - \paren {-\paren {y^{2 n + 1} } }
+      | c = 
+}}
+{{eqn | r = x^{2 n + 1} - \paren {-y}^{2 n + 1}
+      | c = 
+}}
+{{eqn | r = \paren {x - \paren {-y} } \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 - 2 x \paren {-y} \cos \dfrac {2 \pi k} {2 n + 1} + \paren {-y}^2}
+      | c = [[Factors of Difference of Two Odd Powers]]
+}}
+{{eqn | r = \paren {x + y} \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Factors of Difference of Two Even Powers}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+:$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \displaystyle \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$
+\end{theorem}
+
+\begin{proof}
+From [[Factorisation of z^n-a|Factorisation of $z^n - a$]]:
+:$z^{2 n} - y^{2 n} = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha^k y}$
+where $\alpha$ is a [[Definition:Primitive Complex Root of Unity|primitive complex $2 n$th roots of unity]], for example:
+{{begin-eqn}}
+{{eqn | l = \alpha
+      | r = e^{2 i \pi / \paren {2 n} }
+      | c = 
+}}
+{{eqn | r = \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}
+      | c = 
+}}
+{{eqn | r = \cos \dfrac \pi n + i \sin \dfrac \pi n
+      | c = 
+}}
+{{end-eqn}}
+From [[Complex Roots of Unity occur in Conjugate Pairs]]:
+:$U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$
+where $U_{2 n}$ denotes the [[Definition:Complex Roots of Unity|complex $2 n$th roots of unity]]:
+:$U_{2 n} = \set {z \in \C: z^{2 n} = 1}$
+The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.
+The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $x + y$.
+Taking the [[Definition:Complex Multiplication|product]] of each of the remaining factors of $x^{2 n} - y^{2 n}$ in pairs:
+{{begin-eqn}}
+{{eqn | l = \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k} y}
+      | r = \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}
+      | c = [[Complex Roots of Unity occur in Conjugate Pairs]]
+}}
+{{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k y \overline {\alpha^k} y
+      | c = 
+}}
+{{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2
+      | c = [[Modulus in Terms of Conjugate]]
+}}
+{{eqn | r = x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2
+      | c = [[Modulus of Complex Root of Unity equals 1]]
+}}
+{{eqn | r = x^2 - x y \paren {\cos \dfrac {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + y^2
+      | c = Definition of $\alpha$
+}}
+{{eqn | r = x^2 - 2 x y \cos \dfrac {k \pi} n + x y
+      | c = simplification
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Factors of Sum of Two Even Powers}
+Tags: Sum of Two Powers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+:$x^{2 n} + y^{2 n} = \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}$
+\end{theorem}
+
+\begin{proof}
+From [[Factorisation of z^n+a|Factorisation of $z^n + a$]]:
+:$z^{2 n} + y^{2 n} = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha_k y}$
+where $\alpha_k$ are the complex $2n$th roots of negative unity:
+{{begin-eqn}}
+{{eqn | l = \alpha_k
+      | r = e^{i \paren {2 k + 1} \pi / {2 n} }
+      | c = from [[Roots of Complex Number]]
+}}
+{{eqn | r = \map \cos {\dfrac {\paren {2 k + 1} \pi} {2 n} } + i \, \map \sin {\dfrac {\paren {2 k + 1} \pi} {2 n} }
+      | c = $k \in \set {0, 1, 2, \ldots, 2 n - 1}$
+}}
+{{end-eqn}}
+Then we have that:
+:$U_{2 n} = \set {\tuple {\alpha_0, \alpha_{2 n - 1} }, \tuple {\alpha_1, \alpha_{2 n - 2} }, \ldots, \tuple {\alpha_k, \alpha_{2 n - k - 1} }, \ldots, \tuple {\alpha_{n - 1}, \alpha_n } }$
+where $U_{2 n}$ denotes the complex $2n$th roots of negative unity:
+:$U_{2 n} = \set {z \in \C: z^{2 n} = -1}$
+Taking the [[Definition:Complex Multiplication|product]], $p_k$, of the factors of $x^{2 n} + y^{2 n}$ in pairs:
+{{begin-eqn}}
+{{eqn | l = p_k
+      | r = \paren {x - \alpha_k y} \paren {x - \alpha_{2 n - k - 1} y}
+      | c = 
+}}
+{{eqn | r = \paren {x - \alpha_k y} \paren {x - \overline {\alpha_k} y}
+      | c = [[Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs]]
+}}
+{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \alpha_k y \overline {\alpha_k} y
+      | c = 
+}}
+{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \cmod {\alpha_k}^2 y^2
+      | c = [[Modulus in Terms of Conjugate]]
+}}
+{{eqn | r = x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + y^2
+      | c = [[Modulus of Complex Root of Unity equals 1|Modulus of Complex Root of Negative Unity equals 1]]
+}}
+{{eqn | r = x^2 - x y \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + y^2
+      | c = Definition of $\alpha_k$
+}}
+{{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + y^2
+      | c = simplification
+}}
+{{end-eqn}}
+However
+{{begin-eqn}}
+{{eqn | l = p_{n - k}
+      | r = x^2 - 2 x y \cos \dfrac {\paren {2 n - 2 k + 1} \pi} {2 n} + y^2
+      | c =
+}}
+{{eqn | r = x^2 - 2 x y \cos \dfrac {\paren {-2 k + 1} \pi + 2 n \pi} {2 n} + y^2
+      | c =
+}}
+{{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {-2 k + 1} \pi} {2 n} + y^2
+      | c = [[Cosine of Angle plus Straight Angle]]: $\map \cos {x + \pi} = -\cos x$
+}}
+{{eqn | r = x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2
+      | c = [[Cosine Function is Even]]: $\map \cos {- x} = \cos x$
+}}
+{{end-eqn}}
+Consider the permutation:
+:$\sigma = \begin{pmatrix}
+  1 & 2     & \cdots & k     & \cdots & n - 1 & n \\
+  n & n - 1 & \cdots & n - k & \cdots & 2     & 1
+\end{pmatrix}$
+From [[Permutation of Indices of Product]]:
+:$\displaystyle \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = x^{2 n} + y^{2 n}
+      | r = \displaystyle \prod_{k \mathop = 1}^n p_k
+      | c =
+}}
+{{eqn | r = \displaystyle \prod_{k \mathop = 1}^n p_{n - k}
+      | c =
+}}
+{{eqn | r = \displaystyle \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2}
+      | c =
+}}
+{{end-eqn}}
+{{Proofread|I believe this to be correct, but would welcome a sanity check.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Properties of Non-Archimedean Division Rings/Spheres are Clopen}
+Tags: Topological Properties of Non-Archimedean Division Rings
+
+\begin{theorem}
+:The [[Definition:Sphere in Normed Division Ring|$r$-sphere of $x$]], $\map {S_r} x$, 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}
+We have:
+{{begin-eqn}}
+{{eqn | l = \map {S_r} x
+      | r = \set {y \in R : \norm {y - x} = r}
+      | c = {{Defof|Sphere in Normed Division Ring}}
+}}
+{{eqn | r = \set {y \in R : \norm {y - x} \le r} \cap \set {y \in R : \norm{y - x} \ge r}
+}}
+{{eqn | r = \map { {B_r}^-} x \cap \paren {R \setminus \map {B_r} x}
+      | c = {{Defof|Open Ball of Normed Division Ring}} and {{Defof|Closed Ball of Normed Division Ring}}
+}}
+{{end-eqn}}
+Let $d$ be the [[Definition:Metric Induced by Norm on Division Ring|metric induced by the norm]] $\norm {\,\cdot\,}$.
+By [[Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen|Open Balls Are Clopen]] then $\map {B_r} x$ is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in $d$.
+By [[Metric Induces Topology]] then $R \setminus \map {B_r} x$ is is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in $d$.
+By [[Topological Properties of Non-Archimedean Division Rings/Closed Balls are Clopen|Closed Balls Are Clopen]] then $\map { {B_r}^-} x$ is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in $d$.
+By [[Metric Induces Topology]] then the [[Definition:Intersection|intersection]] of a [[Definition:Finite Set|finite number]] of [[Definition:Open Set of Metric Space|open sets]] is [[Definition:Open Set of Metric Space|open]].   
+Hence $\map {S_r} x$ is [[Definition:Open Set of Metric Space|open]] in the [[Definition:Metric Space|metric space]] $\struct {R, d}$.
+By [[Intersection of Closed Sets is Closed in Topological Space]] then $\map {S_r} x$ is [[Definition:Closed Set of Metric Space|closed]] in $\struct {R, d}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Real Polynomial with no Real Root}
+Tags: Polynomial Theory, Analysis
+
+\begin{theorem}
+There exist [[Definition:Polynomial over Real Numbers|polynomials in real numbers $\R$]] which have no [[Definition:Root of Polynomial|roots]] in $\R$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]
+Take the [[Definition:Quadratic Equation|quadratic equation]]:
+:$(1): \quad x^2 + 1 = 0$
+From the [[Quadratic Formula]], the solution to $(1)$ is:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \dfrac {-0 \pm \sqrt {0^2 - 4 \times 1 \times 1} } {2 \times 1}
+      | c = 
+}}
+{{eqn | r = \pm \sqrt {-1}
+      | c = 
+}}
+{{end-eqn}}
+But there is no [[Definition:Real Number|real number]] $x$ such that:
+:$x^2 = -1$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Magnitude of Projection of Complex Number on Another}
+Tags: Complex Dot Product, Complex Modulus
+
+\begin{theorem}
+Let $z_1$ and $z_2$ denote [[Definition:Complex Number as Vector|complex numbers in vector form]].
+Let $\map {\pr_1} {z_1, z_2}$ denote the [[Definition:Projection (Analytic Geometry)|projection]] of $z_1$ on $z_2$.
+{{explain|We really need another page to explain the concept of [[Definition:Projection (Analytic Geometry)]] in the context of the [[Definition:Complex Plane]]}}
+Then:
+:$\cmod {\map {\pr_1} {z_1, z_2} } = \dfrac {\cmod {z_1 \circ z_2} } {\cmod {z_2} }$
+where:
+:$z_1 \circ z_2$ denotes [[Definition:Complex Dot Product|complex dot product]]
+:$\cmod {z_2}$ denotes [[Definition:Complex Modulus|complex modulus]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Needs a proper explanation of the nature of [[Definition:Projection (Analytic Geometry)]]}}
+\end{proof}<|endoftext|>
+\section{Non-Archimedean Division Ring is Totally Disconnected}
+Tags: Normed Division Rings, Non-Archimedean Norms, Totally Disconnected Spaces
+
+\begin{theorem}
+Let $\struct {R, \norm{\,\cdot\,} }$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]].
+Let $\tau$ be the [[Definition:Topology Induced by Division Ring Norm|topology induced]] by the [[Definition:Norm on Division Ring|norm]] $\norm{\,\cdot\,}$.
+Then the [[Definition:Topological Space|topological space]] $\struct {R, \tau}$ is [[Definition:Totally Disconnected Space|totally disconnected]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Subset|subset]] of $R$.
+Let $x, y \in S: x \ne y$
+Let $r \in \R_{>0} : r = \norm {x - y}$
+Consider the [[Definition:Open Ball of Normed Division Ring|open ball]] $\map {B_r} x$ such that:
+:$x \in \map {B_r} x$
+:$y \notin \map {B_r} x$
+By [[Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen|Open Balls are Clopen]] then $\map {B_r} x$ is both [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]].
+By [[Complement of Clopen Set is Clopen]] then $R \setminus \map {B_r} x$ is [[Definition:Open Set (Topology)|open]].
+Hence $S$ is not [[Definition:Connected Topological Space|connected]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Distance between Points in Complex Plane}
+Tags: Complex Addition, Geometry of Complex Plane
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Point|points]] in the [[Definition:Complex Plane|complex plane]] such that:
+:$A = \tuple {x_1, y_1}$
+:$B = \tuple {x_2, y_2}$
+Then the [[Definition:Distance (Linear Measure)|distance]] between $A$ and $B$ is given by:
+{{begin-eqn}}
+{{eqn | l = \size {AB}
+      | r = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}
+      | c = 
+}}
+{{eqn | r = \cmod {z_1 - z_2}
+      | c = 
+}}
+{{end-eqn}}
+where $z_1$ and $z_2$ are represented by the [[Definition:Complex Number as Vector|complex numbers]] $z_1$ and $z_2$ respectively.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = AB
+      | r = z_2 - z_1
+      | c = [[Geometrical Interpretation of Complex Subtraction]]
+}}
+{{eqn | r = \paren {x_2 + i y_2} - \paren {x_1 + i y_1}
+      | c = 
+}}
+{{eqn | r = \paren {x_2 - x_1} + \paren {y_2 - y_1} i
+      | c = {{Defof|Complex Subtraction}}
+}}
+{{eqn | ll= \leadsto
+      | l = \size {AB}
+      | r = \cmod {\paren {x_2 - x_1} + \paren {y_2 - y_1} i}
+      | c = 
+}}
+{{eqn | r = \sqrt {\paren {x_2 - x_1}^2 + \paren {y_2 - y_1}^2}
+      | c = {{Defof|Complex Modulus}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Non-Parallel Complex Numbers is Zero if Factors are Both Zero}
+Tags: Geometry of Complex Plane
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number as Vector|complex numbers expressed as vectors]] such taht $z_1$ is not [[Definition:Parallel Lines|parallel]] to $z_2$.
+Let $a, b \in \R$ be [[Definition:Real Number|real numbers]] such that:
+:$a z_1 + b z_2 = 0$
+Then $a = 0$ and $b = 0$.
+\end{theorem}
+
+\begin{proof}
+Suppose it is not the case that $a = b = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a z_1 + b z_2
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a \paren {x_1 + i y_1} + b \paren {x_2 + i y_2}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a x_1 + b x_2
+      | r = 0
+      | c = 
+}}
+{{eqn | l = a y_1 + b y_2
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a x_1
+      | r = -b x_2
+      | c = 
+}}
+{{eqn | l = a y_1
+      | r = -b y_2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \dfrac {y_1} {x_1}
+      | r = \dfrac {y_2} {x_2}
+      | c = 
+}}
+{{end-eqn}}
+and $z_1$ and $z_2$ are [[Definition:Parallel Lines|parallel]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Line through Two Points in Complex Plane/Formulation 1}
+Tags: Equation for Line through Two Points in Complex Plane
+
+\begin{theorem}
+$L$ can be expressed by the equation:
+:$\map \arg {\dfrac {z - z_1} {z_2 - z_1} } = 0$
+\end{theorem}
+
+\begin{proof}
+Let $z$ be a point on the $L$.
+Then:
+:$z - z_1 = b \paren {z - z_2}$
+where $b$ is some [[Definition:Real Number|real number]].
+Then:
+{{begin-eqn}}
+{{eqn | l = b
+      | r = \frac {z - z_1} {z - z_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \arg {\frac {z - z_1} {z_2 - z_1} }
+      | r = \arg b
+      | c = 
+}}
+{{eqn | r = 0
+      | c = as $b$ is [[Definition:Real Number|real]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Line through Two Points in Complex Plane/Parametric Form 1}
+Tags: Equation for Line through Two Points in Complex Plane
+
+\begin{theorem}
+$L$ can be expressed by the equation:
+:$z = z_1 + t \paren {z_2 - z_1}$
+or:
+:$z = \paren {1 - t} z_1 + t z_2$
+This form of $L$ is known as the '''parametric form''', where $t$ is the '''parameter'''.
+\end{theorem}
+
+\begin{proof}
+Let $z_1$ and $z_2$ be represented by the [[Definition:Point|points]] $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the [[Definition:Complex Plane|complex plane]].
+Let $z$ be an arbitrary [[Definition:Point|point]] on $L$ represented by the [[Definition:Point|point]] $P$.
+:[[File:Line-in-Complex-Plane-through-Two-Points.png|500px]]
+From [[Geometrical Interpretation of Complex Addition]]:
+{{begin-eqn}}
+{{eqn | l = OA + AP
+      | r = OP
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z_1 + AP
+      | r = z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = AP
+      | r = z - z_1
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = OA + AB
+      | r = OB
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z_1 + AB
+      | r = z_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = AB
+      | r = z_2 - z_1
+      | c = 
+}}
+{{end-eqn}}
+As $AP$ and $AB$ are collinear:
+:$AP = t AB$
+and so:
+:$z - z_1 = t \paren {z_2 - z_1}$
+The given expressions follow after algebra.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Line through Two Points in Complex Plane/Symmetric Form}
+Tags: Equation for Line through Two Points in Complex Plane
+
+\begin{theorem}
+$L$ can be expressed by the equation:
+:$z = \dfrac {m z_1 + n z_2} {m + n}$
+This form of $L$ is known as the '''symmetric form'''.
+\end{theorem}
+
+\begin{proof}
+Let $z_1$ and $z_2$ be represented by the [[Definition:Point|points]] $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the [[Definition:Complex Plane|complex plane]].
+Let $z$ be an arbitrary [[Definition:Point|point]] on $L$ represented by the [[Definition:Point|point]] $P$.
+:[[File:Line-in-Complex-Plane-through-Two-Points.png|500px]]
+As $AP$ and $AB$ are collinear:
+:$m AP = n PB$
+and so:
+:$m \paren {z - z_1} = n \paren {z_2 - z_1}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring of Non-Archimedean Division Ring is Subring}
+Tags: Normed Division Rings, Non-Archimedean Norms
+
+\begin{theorem}
+Let $\struct {R, \norm{\,\cdot\,}}$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]] with [[Definition:Ring Zero|zero]] $0_R$ and [[Definition:Unity of Ring|unity]] $1_R$.
+Let $\OO$ be the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring induced]] by the [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm {\,\cdot\,}$, that is:
+:$\OO = \set {x \in R : \norm{x} \le 1}$
+Then $\OO$ is a [[Definition:Subring|subring]] of $R$:
+:with a [[Definition:Ring with Unity|unity]]: $1_R$
+:in which there are no [[Definition:Proper Zero Divisor|(proper) zero divisors]], that is:
+:::$\forall x, y \in \OO: x \circ y = 0_R \implies x = 0_R \text{ or } y = 0_R$
+\end{theorem}
+
+\begin{proof}
+To show that $\OO$ is a [[Definition:Subring|subring]] the [[Subring Test]] is used by showing:
+:$(1): \quad \OO \ne \O$
+:$(2): \quad \forall x, y \in \OO: x + \paren {-y} \in \OO$
+:$(3): \quad \forall x, y \in \OO: x y \in \OO$
+'''(1)'''
+By [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity]],
+:$\norm{1_R} = 1$
+Hence:
+:$1_R \in \OO \ne \O$
+{{qed|lemma}}
+'''(2)'''
+Let $x, y \in \OO$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x + \paren{-y} } 
+      | o = \le
+      | r = \max \set {\norm x, \norm{-y} }
+      | c = {{NormAxiom|4}}
+}}
+{{eqn | r = \max \set {\norm x, \norm y}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]
+}}
+{{eqn | o = \le
+      | r = 1
+      | c = Since $x, y \in \OO$
+}}
+{{end-eqn}}
+Hence:
+:$x + \paren {-y} \in \OO$
+{{qed|lemma}}
+'''(3)'''
+Let $x, y \in \OO$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm{x y} 
+      | o = \le
+      | r = \norm x \norm y
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = \le
+      | r = 1
+      | c = Since $x, y \in \OO$
+}}
+{{end-eqn}}
+Hence:
+:$x y \in \OO$
+{{qed|lemma}}
+By [[Subring Test]] it follows that $\OO$ is a [[Definition:Subring|subring]] of $R$.
+Since $1_R \in S$ and $1_R$ is the [[Definition:Unity of Ring|unity]] of $R$ then $1_R$ is the [[Definition:Unity of Ring|unity]] of $\OO$.
+By [[Division Ring has No Proper Zero Divisors]] then $R$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+Hence $\OO$ has no [[Definition:Proper Zero Divisor|proper zero divisors]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ideal is Maximal Ideal of Induced Valuation Ring}
+Tags: Normed Division Rings, Non-Archimedean Norms
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,}}$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]] with [[Definition:Ring Zero|zero]] $0_R$ and [[Definition:Unity of Ring|unity]] $1_R$.
+Let $\OO$ be the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring induced]] by the [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm {\,\cdot\,}$, that is:
+:$\OO = \set{x \in R : \norm x \le 1}$
+Let $\PP$ be the [[Definition:Valuation Ideal Induced by Non-Archimedean Norm|valuation ideal induced]] by the [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm {\,\cdot\,}$, that is:
+:$\PP = \set{x \in R : \norm x < 1}$
+Then $\PP$ is an [[Definition:Ideal of Ring|ideal]] of $\OO$:
+:$(a):\quad \PP$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]]
+:$(b):\quad \PP$ is a [[Definition:Maximal Right Ideal of Ring|maximal right ideal]]
+:$(c):\quad$ the [[Definition:Quotient Ring|quotient ring]] $\OO / \PP$ is a [[Definition:Division Ring|division ring]].
+\end{theorem}
+
+\begin{proof}
+First it is shown that $\PP$ is an [[Definition:Ideal of Ring|ideal]] of $\OO$ by applying [[Test for Ideal]].
+That is, it is shown that:
+:$(1): \quad \PP \ne \O$
+:$(2): \quad \forall x, y \in \PP: x + \paren {-y} \in \PP$
+:$(3): \quad \forall x \in \PP, y \in \OO: x y \in \PP$
+'''(1)'''
+By {{NormAxiom|1}}:
+:$\norm {0_R} = 0$
+Hence:
+:$0_R \in \PP \ne \O$
+{{qed|lemma}}
+'''(2)'''
+Let $x, y \in \PP$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x + \paren{-y} } 
+      | o = \le
+      | r = \max \set {\norm x, \norm{-y} }
+      | c = {{NormAxiom|4}}
+}}
+{{eqn | r = \max \set {\norm x, \norm y}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]
+}}
+{{eqn | o = <
+      | r = 1
+      | c = Since $x, y \in \PP$
+}}
+{{end-eqn}}
+Hence:
+:$x + \paren {-y} \in \PP$
+{{qed|lemma}}
+'''(3)'''
+Let $x \in \PP, y \in \OO$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm{x y} 
+      | o = \le
+      | r = \norm x \norm y
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = <
+      | r = 1
+      | c = Since $x \in \PP, y \in \OO$
+}}
+{{end-eqn}}
+Hence:
+:$x y \in \PP$
+{{qed|lemma}}
+By [[Test for Ideal]] it follows that $\PP$ is an [[Definition:Ideal of Ring|ideal]] of $\OO$.
+By [[Maximal Left and Right Ideal iff Quotient Ring is Division Ring]] the statements '''(a)''', '''(b)''' and '''(c)''' above are [[Definition:Equivalent|equivalent]].
+It is now shown that statement '''(a)''' holds.
+Let $J$ be a [[Definition:Left Ideal of Ring|left ideal]] of $\OO$:
+:$\PP \subsetneq J \subset \OO$
+Let $x \in J \setminus \PP$, then:
+:$\norm x = 1$
+By [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]] then:
+:$\norm {x^{-1} } = 1 / \norm x = 1 / 1 = 1$ 
+Hence:
+:$x^{-1} \in \OO$
+Since $J$ is a [[Definition:Left Ideal of Ring|left ideal]] then:
+:$x^{-1} x = 1_R \in J$
+Thus:
+:$\forall y \in \OO: y \cdot 1_R = y \in J$
+That is, $J = \OO$
+Hence $\PP$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals}
+Tags: Examples of Convergent Complex Sequences, Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+Consider the [[Definition:Subset|subset]] $S$ of the [[Definition:Complex Plane|complex plane]] defined as:
+:$S := \set {\dfrac i n : n \in \Z_{>0} }$
+That is:
+:$S := \set {i, \dfrac i 2, \dfrac i 3, \dfrac i 4, \ldots}$
+where $i$ is the [[Definition:Imaginary Unit|imaginary unit]].
+\end{theorem}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Boundedness}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The set $S$ is [[Definition:Bounded Subset of Complex Plane|bounded in $\C$]].
+\end{theorem}
+
+\begin{proof}
+Let $z \in S$.
+Then, for example:
+:$\cmod z \le 2$
+That is, $S$ is contained entirely within a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $2$ whose [[Definition:Center of Circle|center]] is at the [[Definition:Origin|origin]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Limit Points}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The set $S$ has exactly one [[Definition:Limit Point (Complex Analysis)|limit point]], and that is $z = 0$.
+\end{theorem}
+
+\begin{proof}
+We have that $S$ is [[Definition:Countably Infinite Set|countably infinite]] and also [[Sequence of Imaginary Reciprocals/Boundedness|bounded]].
+Hence by the [[Bolzano-Weierstrass Theorem]] $S$ has at least one [[Definition:Limit Point (Complex Analysis)|limit point]].
+Let $\epsilon \in \R_{>0}$.
+Let $\map {N_\epsilon} z$ be the [[Definition:Neighborhood (Complex Analysis)|$\epsilon$-neighborhood]] of $z$.
+Let $n \in \N$ such that $n > \dfrac 1 \epsilon$.
+Then $\cmod {\dfrac i n} < \epsilon$ and so:
+: $\dfrac i n \in \map {N_\epsilon} z$
+where $\map {N_\epsilon} {z_0}$ denotes the [[Definition:Neighborhood (Complex Analysis)|$\epsilon$-neighborhood]] of $z$.
+Hence, by definition, $z$ is a [[Definition:Limit Point (Complex Analysis)|limit point]] of $S$.
+Now let $z_1 \in \C$ such that $z_1 \ne 0$.
+Let $z \in S$ such that $z \ne z_1$.
+Then:
+:$\cmod {z_1 - z} > 0$
+Let $\epsilon \in \R_{>0}$ such that $\epsilon < \cmod {z_1 - z}$.
+Then:
+:$z \notin \map {N_\epsilon} {z_1}$
+By making $\epsilon$ smaller than the minimum $\cmod {z_1 - z}$ for $z \in S$, it is seen that:
+:$\forall z \in S: z \ne z_1 \implies z \notin \map {N_\epsilon} {z_1}$
+and so $z_1$ is not a [[Definition:Limit Point (Complex Analysis)|limit point]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Closedness}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The set $S$ is not [[Definition:Closed Set (Complex Analysis)|closed]].
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Imaginary Reciprocals/Limit Points|Sequence of Imaginary Reciprocals: Limit Points]], $S$ has one [[Definition:Limit Point (Complex Analysis)|limit point]] $z = 0$.
+But:
+:$\nexists n \in \N: \dfrac i n = 0$
+so $0 \notin S$.
+As $S$ does not contain (all) its [[Definition:Limit Point (Complex Analysis)|limit point(s)]], it follows by definition that $S$ is not [[Definition:Closed Set (Complex Analysis)|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring of P-adic Norm on Rationals}
+Tags: P-adic Number Theory, Valuation Ring of P-adic Norm on Rationals
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
+The [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the [[Definition:Set|set]]:
+:$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$
+\end{theorem}
+
+\begin{proof}
+Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the [[Definition:P-adic Valuation|$p$-adic valuation]] on $\Q$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \OO
+      | r = \set {\dfrac a b \in \Q : \norm {\dfrac a b}_p \le 1}
+      | c = {{Defof|Valuation Ring Induced by Non-Archimedean Norm}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set{\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} \ge 0}
+      | c = {{Defof|P-adic Norm}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b \ge 0}
+      | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a \ge \map {\nu_p} b}
+}}
+{{end-eqn}}
+Let $\dfrac a b \in \Q$ be in [[Definition:Canonical Form of Rational Number|canonical form]]. 
+Then $a \perp b$
+Suppose $p \divides b$. 
+Then $p \nmid a$.
+Hence:
+:$\map {\nu_p} b \gt 0 = \map {\nu_p} a$
+Suppose $p \nmid b$. 
+Then:
+:$\map {\nu_p} a \ge 0 = \map {\nu_p} b$
+So: 
+:$\map {\nu_p} a \ge \map {\nu_p} b$ {{iff}} $p \nmid b$  
+Hence:
+:$\OO = \set {\dfrac a b \in \Q : p \nmid b }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ideal of P-adic Norm on Rationals}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
+The [[Definition:Valuation Ideal Induced by Non-Archimedean Norm|induced valuation ideal]] on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the [[Definition:Set|set]]:
+:$\PP = p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
+where $\Z_{\ideal p}$ is the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] on $\struct {\Q,\norm {\,\cdot\,}_p}$
+\end{theorem}
+
+\begin{proof}
+Let $\nu_p: \Q \to \Z \cup \set {+\infty}$ be the [[Definition:P-adic Valuation|$p$-adic valuation]] on $\Q$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \PP
+      | r = \set {\dfrac a b \in \Q : \norm{\dfrac a b}_p < 1}
+      | c = {{Defof|Valuation Ideal Induced by Non-Archimedean Norm}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} {\dfrac a b} > 0}
+      | c = {{Defof|P-adic Norm|$p$-adic Norm}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a - \map {\nu_p} b > 0}
+      | c = {{Defof|P-adic Valuation|$p$-adic Valuation on Rationals}}
+}}
+{{eqn | o = }}
+{{eqn | r = \set {\dfrac a b \in \Q : \map {\nu_p} a > \map {\nu_p} b}
+}}
+{{end-eqn}}
+Let $\dfrac a b \in \Q$ be in [[Definition:Canonical Form of Rational Number|canonical form]]. 
+Then $a \perp b$
+Suppose $p \divides a$. 
+Then $p \nmid b$.
+Hence:
+:$\map {\nu_p} a > 0 = \map {\nu_p} b$
+Suppose $p \nmid a$. 
+Then:
+:$\map {\nu_p} b \ge 0 = \map {\nu_p} a$
+So: 
+:$\map {\nu_p} a > \map {\nu_p} b$ {{iff}} $p \nmid b$ and $p \divides a$   
+Hence:
+:$\PP = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
+So:
+{{begin-eqn}}
+{{eqn | l = \dfrac a b \in \PP
+      | o = \leadstoandfrom
+      | r = p \nmid b, p \divides a
+}}
+{{eqn | o = \leadstoandfrom
+      | r = p \nmid b, \exists a' \in \Z: a = p a' 
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \exists a' \in \Z: a = p a',  \dfrac {a'} b \in \Z_{\ideal p}
+      | c = [[Valuation Ring of P-adic Norm on Rationals]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \dfrac a b \in p \Z_{\ideal p}
+}}
+{{end-eqn}}
+Hence:
+:$\PP = p \Z_{\ideal p}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue Field of P-adic Norm on Rationals}
+Tags: P-adic Number Theory, Residue Field of P-adic Norm on Rationals
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
+The [[Definition:Residue Division Ring Induced by Non-Archimedean Norm|induced residue field]] on $\struct {\Q,\norm {\,\cdot\,}_p}$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to the [[Ring of Integers Modulo Prime is Field|field $\F_p$ of integers modulo $p$]].
+\end{theorem}
+
+\begin{proof}
+By [[Valuation Ring of P-adic Norm on Rationals|Valuation Ring of P-adic Norm on Rationals]]:
+:$\Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b}$
+is the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] on $\struct {\Q,\norm {\,\cdot\,}_p}$.
+By [[Valuation Ideal of P-adic Norm on Rationals|Valuation Ideal of P-adic Norm on Rationals]]:
+:$p \Z_{\ideal p} = \set {\dfrac a b \in \Q : p \nmid b, p \divides a}$
+is the [[Definition:Valuation Ideal Induced by Non-Archimedean Norm|induced valuation ideal]] on $\struct {\Q,\norm {\,\cdot\,}_p}$.
+By definition, the [[Definition:Residue Division Ring Induced by Non-Archimedean Norm|induced residue field]] on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the [[Definition:Quotient Ring|quotient ring]] $\Z_{\ideal p} / p \Z_{\ideal p}$.
+By [[Quotient Ring of Integers with Principal Ideal]], $\F_p$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $\Z / p \Z$, where $p \Z$ is the [[Definition:Principal Ideal of Ring|principal ideal of $\Z$ generated by $p$]].
+To complete the proof it is sufficient to show that $\Z / p \Z$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $\Z_{\ideal p} / p \Z_{\ideal p}$.
+By [[Integers form Subring of Valuation Ring of P-adic Norm on Rationals]] then $\Z$ is a [[Definition:Subring|subring]] of $\Z_{\ideal p}$.
+Let $\phi : \Z \to \Z_{\ideal p} / p \Z_{\ideal p}$ be the [[Definition:Mapping|mapping]] defined by:
+:$\forall a \in \Z: \map \phi a = a + p \Z_{\ideal p}$
+=== [[Residue Field of P-adic Norm on Rationals/Lemma 1|Lemma 1]] ===
+{{:Residue Field of P-adic Norm on Rationals/Lemma 1}}{{qed|lemma}}
+=== [[Residue Field of P-adic Norm on Rationals/Lemma 2|Lemma 2]] ===
+{{:Residue Field of P-adic Norm on Rationals/Lemma 2}}{{qed|lemma}}
+=== [[Residue Field of P-adic Norm on Rationals/Lemma 3|Lemma 3]] ===
+{{:Residue Field of P-adic Norm on Rationals/Lemma 3}}{{qed|lemma}}
+Hence $\phi$ is a [[Definition:Ring Epimorphism|ring epimorphism]] with:
+:$p \Z = \map \ker \phi$
+By [[Quotient Ring of Kernel of Ring Epimorphism]] then $\Z / p \Z$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $\Z_{\ideal p} / p \Z_{\ideal p}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Boundary Points}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+Every point of $S$, along with the point $z = 0$, is a [[Definition:Boundary Point (Complex Analysis)|boundary point]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Consider the point $z = \dfrac i n \in S$.
+Let $\delta \in \R_{>0}$.
+Let $\map {N_\delta} z$ be the [[Definition:Neighborhood (Complex Analysis)|$\delta$-neighborhood]] of $z$.
+Then $\map {N_\delta} z$ contains at least one point of $S$ ($i / n$ itself) as well as points which are not in $S$.
+Hence, by definition, $z$ is a [[Definition:Boundary Point (Complex Analysis)|boundary point]] of $S$.
+Let $z = 0$.
+Similarly, let $\map {N_\delta} z$, 
+Let $\delta \in \R_{>0}$.
+Let $\map {N_\delta} z$ be the [[Definition:Neighborhood (Complex Analysis)|$\delta$-neighborhood]] of $z$.
+Let $n \in \N$ such that $n > \dfrac 1 \delta$.
+Then $\cmod {\dfrac i n} < \delta$ and so:
+: $\dfrac i n \in \map {N_\delta} z$
+Thus every [[Definition:Neighborhood (Complex Analysis)|$\delta$-neighborhood]] of $z = 0$ also contains points of $S$ and points not in $S$.
+Hence, by definition, $z = 0$ is also a [[Definition:Boundary Point (Complex Analysis)|boundary point]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Interior}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+No point of $S$ is an [[Definition:Interior Point (Complex Analysis)|interior point]].
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Imaginary Reciprocals/Boundary Points|Sequence of Imaginary Reciprocals: Boundary Points]], every $z \in S$ is a [[Definition:Boundary Point (Complex Analysis)|boundary point]] of $S$.
+Thus no $z \in S$ is an [[Definition:Interior Point (Complex Analysis)|interior point]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Openness}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+$S$ is not an [[Definition:Open Set (Complex Analysis)|open set]].
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Imaginary Reciprocals/Interior|Sequence of Imaginary Reciprocals: Interior]], no $z \in S$ is an [[Definition:Interior Point (Complex Analysis)|interior point]].
+Hence $S$ cannot be [[Definition:Open Set (Complex Analysis)|open]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Connectedness}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+$S$ is not [[Definition:Connected Set (Complex Analysis)|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $z_1 \in S$ and $z_2 \in S$ be joined by a [[Definition:Polygonal Path|polygonal path]] $P$.
+Then there are points of $P$ which are not in $S$.
+Hence, by definition, $S$ is not [[Definition:Connected Set (Complex Analysis)|connected]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Not an Open Region}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+$S$ is not an [[Definition:Open Region (Complex Analysis)|open region]].
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Imaginary Reciprocals/Openness|Sequence of Imaginary Reciprocals: Openness]], $S$ is not an [[Definition:Open Set (Complex Analysis)|open set]].
+From [[Sequence of Imaginary Reciprocals/Connectedness|Sequence of Imaginary Reciprocals: Connectedness]], $S$ is not [[Definition:Connected Set (Complex Analysis)|connected]].
+Hence, by definition, $S$ is not an [[Definition:Open Region (Complex Analysis)|open region]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Countability}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The set $S$ is [[Definition:Countably Infinite|countably infinite]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi: \N \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \N: \map \phi n = \dfrac i n$
+$\phi$ is a [[Definition:Bijection|bijection]].
+Hence the result by definition of [[Definition:Countably Infinite|countably infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Not Compact}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The set $S$ is not [[Definition:Compact Subset of Complex Plane|compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Imaginary Reciprocals/Boundedness|Sequence of Imaginary Reciprocals: Boundedness]], $S$ is [[Definition:Bounded Subset of Complex Plane|bounded in $\C$]].
+But from [[Sequence of Imaginary Reciprocals/Closedness|Sequence of Imaginary Reciprocals: Closedness]], $S$ is not [[Definition:Closed Set (Complex Analysis)|closed]].
+Hence the result by definition of [[Definition:Compact Subset of Complex Plane|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Imaginary Reciprocals/Closure is Compact}
+Tags: Sequence of Imaginary Reciprocals
+
+\begin{theorem}
+The [[Definition:Closure (Topology)|closure]] $S^-$ of the set $S$ is [[Definition:Compact Subset of Complex Plane|compact]].
+\end{theorem}
+
+\begin{proof}
+From [[Topological Closure is Closed]], $S^-$ is [[Definition:Closed Set (Complex Analysis)|closed]].
+From [[Sequence of Imaginary Reciprocals/Boundedness|Sequence of Imaginary Reciprocals: Boundedness]], $S$ is [[Definition:Bounded Subset of Complex Plane|bounded in $\C$]].
+It follows trivially that $S^-$ is also [[Definition:Bounded Subset of Complex Plane|bounded in $\C$]].
+Hence the result by definition of [[Definition:Compact Subset of Complex Plane|compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Complex Conjugates/Examples/3 Arguments}
+Tags: Product of Complex Conjugates
+
+\begin{theorem}
+Let $z_1, z_2, z_3 \in \C$ be [[Definition:Complex Number|complex numbers]].
+Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of the [[Definition:Complex Number|complex number]] $z$.
+Then:
+:$\overline {z_1 z_2 z_3} = \overline {z_1} \cdot \overline {z_2} \cdot \overline {z_3}$
+\end{theorem}<|endoftext|>
+\section{Valuation Ideal is Maximal Ideal of Induced Valuation Ring/Corollary 1}
+Tags: Normed Division Rings, Non-Archimedean Norms
+
+\begin{theorem}
+:$\OO$ is a [[Definition:Local Ring/Noncommutative|local ring]].
+\end{theorem}
+
+\begin{proof}
+Let $\PP$ be the [[Definition:Valuation Ideal Induced by Non-Archimedean Norm|valuation ideal induced]] by the [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] $\norm {\,\cdot\,}$, that is:
+:$\PP = \set{x \in R : \norm{x} \lt 1}$
+By [[Valuation Ideal is Maximal Ideal of Induced Valuation Ring|Valuation Ideal is Maximal Ideal of Induced Valuation Ring]] then:
+:$\PP$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] of $\OO$.
+Let $J \subsetneq \OO$ be any [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] of $\OO$.
+Let $x \in \OO \setminus \PP$.
+{{AimForCont}} $x \in J$.
+By [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]] then:
+:$\norm {x^{-1}} = 1 / \norm x = 1 / 1 = 1$ 
+Hence:
+:$x^{-1} \in \OO$
+Since $J$ is a [[Definition:Left Ideal|left ideal]] then:
+:$x^{-1} x = 1_R \in J$
+Thus:
+:$\forall y \in \OO: y \cdot 1_R = y \in J$
+That is: 
+:$J = \OO$
+This [[Definition:Contradiction|contradicts]] the assumption that $J \ne \OO$.
+So:
+:$x \notin J$
+Hence: 
+:$\paren {\OO \setminus \PP} \cap J = \O$
+That is:
+:$J \subseteq \PP$
+Since $J$ and $\PP$ are both [[Definition:Maximal Left Ideal of Ring|maximal left ideals]] then:
+:$J = \PP$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue Field of P-adic Norm on Rationals/Lemma 1}
+Tags: Residue Field of P-adic Norm on Rationals
+
+\begin{theorem}
+:$\phi$ is a [[Definition:Ring Homomorphism|homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \Z_{\paren p}$.
+Hence $\Z \subset \Z_{\paren p}$ is a [[Definition:Subring|subring]] of $\Z_{\paren p}$.
+Let $i: \Z \to \Z_{\paren p}$ be the [[Definition:Inclusion Mapping|inclusion mapping]] defined by:
+:$\map i a = a$
+By [[Inclusion Mapping is Monomorphism]] then $i$ is a [[Definition:Ring Monomorphism|ring monomorphism]].
+Let $q: \Z_{\paren p} \to \Z_{\paren p} / p \, \Z_{\paren p}$ be the [[Definition:Quotient Ring Epimorphism|quotient ring epimorphism]] from $\Z_{\paren p}$ to $\Z_{\paren p} / p \, \Z_{\paren p}$.
+By [[Quotient Ring Epimorphism is Epimorphism]], $q$ is a [[Definition:Ring Epimorphism|epimorphism]].
+By definition, $\phi = q \circ i$ is the [[Definition:Composition of Mappings|composition]] of $i$ with $q$.
+By [[Composition of Ring Homomorphisms is Ring Homomorphism]], $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue Field of P-adic Norm on Rationals/Lemma 2}
+Tags: Residue Field of P-adic Norm on Rationals
+
+\begin{theorem}
+:$p \Z = \map \ker \phi$
+\end{theorem}
+
+\begin{proof}
+Let $\map \ker \phi$ denote the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \in
+      | r = \map \ker \phi
+      | c =
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi a
+      | r = p \Z_{\ideal p}
+      | c = {{Defof|Kernel of Ring Homomorphism}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a + p \Z_{\ideal p}
+      | r = p \Z_{\ideal p}
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a = a - 0 
+      | o = \in
+      | r = p \Z_{\ideal p}
+      | c = [[Element in Right Coset iff Product with Inverse in Subgroup]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \exists a' \in \Z:
+      | l = a = p a'
+      | r = p a'
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a
+      | o = \in
+      | r = p \Z
+}}
+{{end-eqn}}
+Hence:
+:$p \Z = \map \ker \phi$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue Field of P-adic Norm on Rationals/Lemma 3}
+Tags: Residue Field of P-adic Norm on Rationals
+
+\begin{theorem}
+:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a [[Definition:Surjective|surjection]].
+\end{theorem}
+
+\begin{proof}
+Let $a / b \in \Z_{\paren p}$, where $a / b$ are in [[Definition:Canonical Form of Rational Number|canonical form]].
+Then $p \nmid b$
+Let $\F_p$ be the [[Ring of Integers Modulo Prime is Field|field of integers modulo $p$]].
+By the definition of a [[Definition:Field (Abstract Algebra)|field]]:
+:$\exists b' \in \Z: b b' \equiv 1 \pmod p$
+By the definition of [[Definition:Congruence Modulo Integer|congruence modulo $p$]]:
+:$p \divides b b' - 1$
+By [[Divisor Divides Multiple]]:
+:$\forall a \in \Z: p \divides a b b' - a$
+By [[Valuation Ideal of P-adic Norm on Rationals]] then:
+:$a b' - \dfrac a b = \dfrac {a b b' - a} b \in p \Z_{\paren p}$
+By [[Element in Left Coset iff Product with Inverse in Subgroup]]:
+:$\map \phi {a b'} = a b' + p \Z_{\paren p} = a / b + p \Z_{\paren p}$
+It follows that:
+:$\phi : \Z \to \Z_{\paren p} / p \Z_{\paren p}$ is a [[Definition:Surjective|surjection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Line through Two Points in Complex Plane/Parametric Form 2}
+Tags: Equation for Line through Two Points in Complex Plane
+
+\begin{theorem}
+$L$ can be expressed by the equations:
+{{begin-eqn}}
+{{eqn | l = x - x_1
+      | r = t \paren {x_2 - x_1}
+}}
+{{eqn | l = y - y_1
+      | r = t \paren {y_2 - y_1}
+}}
+{{end-eqn}}
+These are the '''parametric equations of $L$''', where $t$ is the [[Definition:Parameter|parameter]].
+\end{theorem}
+
+\begin{proof}
+From [[Equation for Line through Two Points in Complex Plane/Parametric Form 1|Equation for Line through Two Points in Complex Plane: Parametric Form 1]]:
+:$z = z_1 + t \paren {z_2 - z_1}$
+Letting:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x + i y
+}}
+{{eqn | l = z_1
+      | r = x_1 + i y_1
+}}
+{{eqn | l = z_2
+      | r = x_2 + i y_2
+}}
+{{end-eqn}}
+the parametric equations follow by equating [[Definition:Real Part|real parts]] and [[Definition:Imaginary Part|imaginary parts]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Perpendicular Bisector of Two Points in Complex Plane/Parametric Form 1}
+Tags: Equation for Perpendicular Bisector of Two Points in Complex Plane
+
+\begin{theorem}
+$L$ can be expressed by the equation:
+:$z = ...$
+or:
+:$z = ...$
+This form of $L$ is known as the '''parametric form''', where $t$ is the '''parameter'''.
+\end{theorem}
+
+\begin{proof}
+Let $z_1$ and $z_2$ be represented by the [[Definition:Point|points]] $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the [[Definition:Complex Plane|complex plane]].
+Let $z$ be an arbitrary [[Definition:Point|point]] on $L$ represented by the [[Definition:Point|point]] $P$.
+:[[File:Perpendicular Bisector of Two Points in Complex Plane.png|400px]]
+We have that $L$ passes through the [[Definition:Point|point]]:
+:$\dfrac {z_1 + z_2} 2$
+and is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Straight Line|straight line]]:
+:$z = z_1 + t \paren {z_2 - z_1}$
+{{ProofWanted}}
+[[Category:Equation for Perpendicular Bisector of Two Points in Complex Plane]]
+azrcyl0u9al6hkpf7oqidu3zb161zkd
+\end{proof}<|endoftext|>
+\section{Geodesic Equation/2d Surface Embedded in 3d Euclidean Space}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $\sigma: U \subset \R^2 \to V \subset \R^3$ be a [[Definition:Smooth Real Function|smooth]] [[Definition:Surface|surface]] specified by a [[Definition:Vector-Valued Function|vector-valued function]]:
+:$\mathbf r = \map {\mathbf r} {u, v}$
+Then a [[Definition:Geodesic Curve|geodesic]] of $\sigma$ satisfies the following [[Definition:System of Differential Equations|system of differential equations]]:
+:$\dfrac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t} \dfrac {2 \paren {E u' + F v'} } {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$
+:$\dfrac {E_v u'^2 + 2 F_v u' v' + G_v v'^2} {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } - \dfrac \d {\d t} \dfrac {2 \paren {F u' + G v'} } {\sqrt{E u'^2 + 2 F u' v' + G v'^2} } = 0$
+where $E, F, G$ are the [[Definition:Real Function|functions]] of the [[Definition:First Fundamental Form|first fundamental form]]:
+:$\displaystyle E = {\mathbf r}_u \cdot {\mathbf r}_u, F = {\mathbf r}_u \cdot {\mathbf r}_v, G = {\mathbf r}_v \cdot {\mathbf r}_v$
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Curve|curve]] on the [[Definition:Surface|surface]] $\mathbf r$ can be specified as $u = \map u t$, $v = \map v t$
+The [[Definition:Arc Length|arc length]] between the [[Definition:Point|points]] corresponding to $t_0$ and $t_1$ equals:
+:$\displaystyle J \sqbrk {u, v} = \int_{t_0}^{t_1} \sqrt {E u'^2 + 2 F u'v' + G v'^2} \rd t$
+The following [[Definition:Derivative|derivatives]] will appear in [[Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions|Euler's Equations]]:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial} {\partial u'} \sqrt {E u'^2 + 2 F u'v' + G v'^2}
+      | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u'} } {E u'^2 + 2 F u'v' + G v'^2}
+}}
+{{eqn | r = \frac {2 E u' + 2 F_u u'} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} }
+}}
+{{eqn | r = \frac {E u' + F u'} {\sqrt {E u'^2 + 2 F u'v' + G v'^2} }
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial} {\partial u} \sqrt {E u'^2 + 2 F u'v' + G v'^2}
+      | r = \frac 1 {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} } \map {\dfrac {\partial} {\partial u} } {E u'^2 + 2 F u'v' + G v'^2}
+}}
+{{eqn | r = \frac {E_u u'^2 + 2 F_u u' v' + G_u v'^2} {2 \sqrt {E u'^2 + 2 F u'v' + G v'^2} }
+}}
+{{end-eqn}}
+Analogous relations hold for [[Definition:Derivative|derivatives]] {{WRT|Differentiation}} $v$ and $v'$.
+[[Necessary Condition for Integral Functional to have Extremum for given function/Dependent on N Functions|Euler's Equation]] together with the results above yield the [[Definition:Geodesic Equation|geodesic equations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Magnitude and Direction of Equilibrant}
+Tags: Force, Equilibrants
+
+\begin{theorem}
+Let $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ be a [[Definition:Set|set]] of $n$ [[Definition:Force|forces]] acting on a [[Definition:Particle|particle]] $B$ at a [[Definition:Point|point]] $P$ in [[Definition:Ordinary Space|space]].
+The '''equilibrant''' $\mathbf E$ of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$ is:
+:$\mathbf E = -\displaystyle \sum_{k \mathop = 1}^n \mathbf F_k$
+That is, the [[Definition:Magnitude|magnitude]] and [[Definition:Direction|direction]] of $\mathbf E$ is such as to balance out the effect of $\mathbf F_1, \mathbf F_2, \ldots, \mathbf F_n$.
+\end{theorem}
+
+\begin{proof}
+From [[Newton's First Law of Motion]], the total [[Definition:Force|force]] on $B$ must equal [[Definition:Zero Vector|zero]] in order for $B$ to remain [[Definition:Stationary|stationary]].
+That is, $\mathbf E$ must be such that:
+:$\mathbf E + \displaystyle \sum_{k \mathop = 1}^n \mathbf F_k = \bszero$
+That is:
+:$\mathbf E = -\displaystyle \sum_{k \mathop = 1}^n \mathbf F_k$
+{{qed}}
+[[Category:Force]]
+[[Category:Equilibrants]]
+k6ckdj50bq38jsmydrpwo6cbv5ij6sz
+\end{proof}<|endoftext|>
+\section{Modulus of Exponential of i z where z is on Circle}
+Tags: Complex Modulus, Exponential Function, Circles, Modulus of Exponential of i z where z is on Circle
+
+\begin{theorem}
+Let $C$ be the [[Definition:Circle|circle]] embedded in the [[Definition:Complex Plane|complex plane]] given by the equation:
+:$z = R e^{i \theta}$
+Then:
+:$\cmod {e^{i z} } = e^{-R \sin \theta}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cmod {e^{i z} }
+      | r = \cmod {\map \exp {i R \, \map \exp {i \theta} } }
+      | c = 
+}}
+{{eqn | r = \cmod {\map \exp {i R \paren {\cos \theta + i \sin \theta} } }
+      | c = 
+}}
+{{eqn | r = \cmod {\map \exp {R \paren {-\sin \theta + i \cos \theta} } }
+      | c = 
+}}
+{{eqn | r = \cmod {\map \exp {- R \sin \theta} \, \map \exp {i \cos \theta} }
+      | c = 
+}}
+{{eqn | r = \map \exp {- R \sin \theta}
+      | c = [[Modulus and Argument of Complex Exponential]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Complex Numbers in Exponential Form/General Result}
+Tags: Complex Addition
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Positive Integer|positive integer]].
+For all $k \in \set {1, 2, \dotsc, n}$, let:
+:$z_k = r_k e^{i \theta_k}$
+be [[Definition:Complex Zero|non-zero]] [[Definition:Complex Number|complex numbers]] in [[Definition:Exponential Form of Complex Number|exponential form]].
+Let:
+:$r e^{i \theta} = \displaystyle \sum_{k \mathop = 1}^n z_k = z_1 + z_2 + \dotsb + z_k$
+Then:
+{{begin-eqn}}
+{{eqn | l = r
+      | r = \sqrt {\displaystyle \sum_{k \mathop = 1}^n r_k + \displaystyle \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \, \map \cos {\theta_j - \theta_k} }
+}}
+{{eqn | l = \theta
+      | r = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = r e^{i \theta}
+      | r = \displaystyle \sum_{k \mathop = 1}^n z_k
+      | c =
+}}
+{{eqn | r = z_1 + z_2 + \dotsb + z_k
+      | c =
+}}
+{{eqn | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2} + \dotsb + r_n \paren {\cos \theta_n + i \sin \theta_n}
+      | c = {{Defof|Complex Number}}
+}}
+{{eqn | r = r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}
+      | c = rerranging
+}}
+{{end-eqn}}
+By the definition of the [[Definition:Complex Modulus|complex modulus]], with $z = x + i y$, $r$ is defined as:
+:$r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}$
+Hence
+{{begin-eqn}}
+{{eqn | l = r
+      | r = \sqrt {\map {\Re^2} z + \map {\Im^2} z}
+      | c =
+}}
+{{eqn | l = r
+      | r = \sqrt {\paren {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n }^2 + \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n}^2 }
+      | c =
+}}
+{{end-eqn}}
+In the above we have two types of pairs of terms:
+{{begin-eqn}}
+{{eqn | n = 1
+      | lo= 1 \le k \le n:
+      | l = {r_k}^2 \cos^2 {\theta_k}^2 + {r_k}^2 \sin^2 {\theta_k}^2
+      | r = {r_k}^2 \paren {\cos^2 {\theta_k}^2 + \sin^2 {\theta_k}^2}
+      | c =
+}}
+{{eqn | r = {r_k}^2
+      | c =[[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | n = 2
+      | lo= 1 \le j < k \le n:
+      | l = 2 r_j r_k \cos \theta_j \cos \theta_k + 2 {r_j} {r_k} \sin \theta_j \sin \theta_k
+      | r = 2 r_j r_k \paren {\cos \theta_j \cos \theta_k + \sin \theta_j \sin \theta_k}
+      | c =
+}}
+{{eqn | r = 2 r_j r_k \, \map \cos {\theta_j - \theta_k}
+      | c = [[Cosine of Difference]]
+}}
+{{end-eqn}}
+Hence:
+:$r = \sqrt {\displaystyle \sum_{k \mathop = 1}^n r_k + \displaystyle \sum_{1 \mathop \le j \mathop < k \mathop \le n} 2 {r_j} {r_k} \, \map \cos {\theta_j - \theta_k} }$
+Note that $r > 0$ since $r_k > 0$ for all $k$.
+Hence we may safely assume that $r > 0$ when determining the argument below.
+By definition of the [[Definition:Argument of Complex Number|argument of a complex number]], with $z = x + i y$, $\theta$ is defined as any solution to the pair of equations:
+:$(1): \quad \dfrac x {\cmod z} = \map \cos \theta$
+:$(2): \quad \dfrac y {\cmod z} = \map \sin \theta$
+where $\cmod z$ is the [[Definition:Complex Modulus|modulus]] of $z$.
+As $r > 0$ we have that $\cmod z \ne 0$ by definition of [[Definition:Complex Modulus|modulus]].
+Hence we can [[Definition:Complex Division|divide]] $(2)$ by $(1)$, to get:
+{{begin-eqn}}
+{{eqn | l = \map \tan \theta
+      | r = \frac y x
+      | c =
+}}
+{{eqn | r = \frac {\map \Im z} {\map \Re z}
+      | c =
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \theta
+      | r = \map \arctan {\frac {\map \Im {r e^{i \theta} } } {\map \Re {r e^{i \theta} } } }
+      | c =
+}}
+{{eqn | r = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2 + \dotsb + r_n \sin \theta_n} {r_1 \cos \theta_1 + r_2 \cos \theta_2 + \dotsb + r_n \cos \theta_n} }
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bias of Sample Variance}
+Tags: Inductive Statistics
+
+\begin{theorem}
+Let $X_1, X_2, \ldots, X_n$ form a [[Definition:Random Sample|random sample]] from a population with [[Definition:Expectation|mean]] $\mu$ and [[Definition:Variance|variance]] $\sigma^2$.
+Let: 
+:$\displaystyle \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
+Then:
+:$\displaystyle \hat {\sigma^2} = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
+is a [[Definition:Bias of Estimator|biased]] [[Definition:Estimator|estimator]] of $\sigma^2$, with: 
+:$\displaystyle \operatorname{bias} \paren {\hat {\sigma ^2}} = -\frac {\sigma^2} n$
+\end{theorem}
+
+\begin{proof}
+If $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$, then: 
+:$\displaystyle \expect {\hat {\sigma^2} } \ne \sigma^2$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \expect {\hat {\sigma^2} }
+	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
+}}
+{{eqn	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu} - \paren {\bar X - \mu} }^2}
+	| c = writing $X_i - \bar X = X_i - \bar X - \mu + \mu$
+}}
+{{eqn	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren{\paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu} \paren {X_i -\mu} + \paren {\bar X - \mu}^2} }
+	| c = [[Square of Difference]]
+}}
+{{eqn	| r = \expect {\frac 1 n  \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}
+	| c = [[Summation is Linear]]
+}}
+{{end-eqn}}
+We have that: 
+{{begin-eqn}}
+{{eqn	| l = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}
+	| r = \frac 1 n \sum_{i \mathop = 1}^n X_i - \frac n n \mu
+	| c = from $\displaystyle \sum_{i \mathop = 1}^n 1 = n$, noting that $\mu$ is independent of $i$. 
+}}
+{{eqn	| r = \bar X - \mu
+	| c = by our definition of $\bar X$
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn	| l = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - \frac 2 n \paren {\bar X - \mu} \sum_{i \mathop = 1}^n \paren {X_i -\mu} + \frac 1 n \paren {\bar X - \mu}^2 \sum_{i \mathop = 1}^n 1}
+	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \mu}^2 - 2 \paren {\bar X - \mu}^2 + \paren {\bar X - \mu}^2}
+}}
+{{eqn	| r = \frac 1 n \expect {\sum_{i \mathop = 1}^n \paren {X_i - \mu}^2} - \expect {\paren {\bar X - \mu}^2}
+	| c = [[Linearity of Expectation Function]]
+}}
+{{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \expect {\paren {X_i - \mu}^2} - \var {\bar X}
+	| c = {{Defof|Variance}}, [[Linearity of Expectation Function]]
+}}
+{{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \var {X_i} - \frac {\sigma^2} n
+	| c = {{Defof|Variance}}, [[Variance of Sample Mean]]
+}}
+{{eqn	| r = \frac n n \sigma^2 - \frac {\sigma^2} n
+	| c = $\var {X_i} = \sigma^2$, $\displaystyle \sum_{i \mathop = 1}^n 1 = n$
+}}
+{{eqn	| r = \sigma^2 - \frac {\sigma^2} n
+}}
+{{eqn	| o = \ne
+	| r = \sigma^2
+}}
+{{end-eqn}}
+So $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$. 
+Further, we have: 
+:$\displaystyle \operatorname{bias} \paren {\hat {\sigma ^2}} = \sigma^2 - \frac {\sigma^2} n - \sigma^2 = -\frac {\sigma^2} n$
+{{qed}}
+[[Category:Inductive Statistics]]
+has08yzmpl19phi5wwcvio4oj4zzyo8
+\end{proof}<|endoftext|>
+\section{Bessel's Correction}
+Tags: Inductive Statistics
+
+\begin{theorem}
+Let $X_1, X_2, \ldots, X_n$ form a [[Definition:Random Sample|random sample]] from a population with [[Definition:Expectation|mean]] $\mu$ and [[Definition:Variance|variance]] $\sigma^2$.
+Let: 
+:$\displaystyle \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
+Then:
+:$\displaystyle \hat {\sigma^2} = \frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$
+is an [[Definition:Bias of Estimator|unbiased]] [[Definition:Estimator|estimator]] of $\sigma^2$.
+\end{theorem}
+
+\begin{proof}
+If $\hat{\sigma^2}$ is an unbiased estimator of $\sigma^2$, then: 
+:$\displaystyle \expect {\hat {\sigma^2}} = \sigma^2$
+In [[Bias of Sample Variance]], it is shown that: 
+:$\displaystyle \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2} = \paren {1 - \frac 1 n} \sigma^2$
+By [[Linearity of Expectation Function]]: 
+{{begin-eqn}}
+{{eqn	| l = n \expect {\frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
+	| r = \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
+}}
+{{eqn	| r = \paren {n - 1} \sigma^2
+}}
+{{end-eqn}}
+So: 
+{{begin-eqn}}
+{{eqn	| l = \sigma^2
+	| r = \frac 1 {n - 1} \expect {\sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
+}}
+{{eqn	| r = \expect {\frac 1 {n - 1} \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2}
+	| c = [[Linearity of Expectation Function]]
+}}
+{{eqn	| r = \expect {\hat {\sigma^2} }
+}}
+{{end-eqn}}
+So $\hat {\sigma^2}$ is an unbiased estimator of $\sigma^2$. 
+{{qed}}
+{{Namedfor|Friedrich Wilhelm Bessel|cat = Bessel}}
+[[Category:Inductive Statistics]]
+hi7tbzjldl7mmosk8kkdcus2epjmpi8
+\end{proof}<|endoftext|>
+\section{Sample Mean is Unbiased Estimator of Population Mean}
+Tags: Inductive Statistics
+
+\begin{theorem}
+Let $X_1, X_2, \ldots, X_n$ form a [[Definition:Random Sample|random sample]] from a population with [[Definition:Expectation|mean]] $\mu$ and [[Definition:Variance|variance]] $\sigma^2$.
+Then: 
+:$\displaystyle \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$
+is an [[Definition:Unbiased Estimator|unbiased estimator]] of $\mu$.
+\end{theorem}
+
+\begin{proof}
+If $\bar X$ is an unbiased estimator of $\mu$, then: 
+:$\displaystyle \expect {\bar X} = \mu$
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \expect {\bar X}
+	| r = \expect {\frac 1 n \sum_{i \mathop = 1}^n X_i}
+}}
+{{eqn	| r = \frac 1 n \sum_{i \mathop = 1}^n \expect {X_i}
+	| c = [[Linearity of Expectation Function]]
+}}
+{{eqn	| r = \frac 1 n \sum_{i = 1}^n \mu
+	| c = as $\expect {X_i} = \mu$
+}}
+{{eqn	| r = \frac n n \mu
+	| c = as $\displaystyle \sum_{i \mathop = 1}^n 1 = n$
+}}
+{{eqn	| r = \mu
+}}
+{{end-eqn}}
+So $\bar X$ is an unbiased estimator of $\mu$. 
+{{qed}}
+[[Category:Inductive Statistics]]
+osbe59f8es82v6b730jl34k4l86lymx
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Archimedean Norm}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}$ be a [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Archimedean Division Ring Norm|Archimedean]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Then $\norm {\, \cdot \,}$ is [[Definition:Equivalent Division Ring Norms|equivalent]] to the [[Definition:Absolute Value|absolute value]] $\size {\, \cdot \,}$.
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\exists n \in \N$ such that $\norm n > 1$
+Let $n_0 = \min \set {n \in \N : \norm n > 1}$
+By [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity]] then:
+:$n_0 > 1$
+Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$
+Since $n_0, \norm n_0 > 1$ then:
+:$\alpha > 0$
+==== [[Ostrowski's Theorem/Archimedean Norm/Lemma 1.1|Lemma 1.1]] ====
+{{:Ostrowski's Theorem/Archimedean Norm/Lemma 1.1}}{{qed|lemma}}
+==== [[Ostrowski's Theorem/Archimedean Norm/Lemma 1.2|Lemma 1.2]] ====
+{{:Ostrowski's Theorem/Archimedean Norm/Lemma 1.2}}{{qed|lemma}}
+Hence:
+:$\forall n \in \N: \norm n = n^\alpha = \size n^\alpha$
+By [[Equivalent Norms on Rational Numbers|Equivalent Norms on Rational Numbers]] then $\norm {\, \cdot \,}$ is [[Definition:Equivalent Division Ring Norms|equivalent]] to the [[Definition:Absolute Value|absolute value]] $\size {\, \cdot \,}$.
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Non-Archimedean Norm}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}$ be a [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Then $\norm {\, \cdot \,}$ is [[Definition:Equivalent Division Ring Norms|equivalent]] to the [[Definition:P-adic Norm|$p$-adic norm $\norm {\, \cdot \,}_p$]] for some [[Definition:Prime Number|prime]] $p$.
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\forall n \in \N: \norm n \le 1$
+==== [[Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1|Lemma 2.1]] ====
+{{:Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1}}{{qed|lemma}}
+Let $n_0 = \min \set {n \in N : \norm n < 1}$.
+==== [[Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.2|Lemma 2.2]] ====
+{{:Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.2}}{{qed|lemma}}
+Let $p = n_0$.
+Let $\alpha = - \dfrac {\log \norm p } {\log p}$ then:
+:$\norm p = p^{-\alpha} = \paren {p^{-1}}^\alpha = \norm p_p^\alpha$
+Let $b \in N$
+Let $p \nmid b$.
+Then $p$ and $b$ are [[Definition:Coprime|coprime]], that is:
+:$p \perp b$.
+By [[Three Points in Ultrametric Space have Two Equal Distances/Corollary 5|Corollary 5 of Three Points in Ultrametric Space have Two Equal Distances]]:
+:$\norm b = 1$
+By the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
+:$\norm b_p = 1$ 
+Hence:
+:$\norm b = 1 = 1^\alpha = \norm b_p^\alpha$
+Let $p \divides b$.
+Let $\nu = \map {\nu_p} b$ where $\nu_p$ is the [[Definition:P-adic Valuation/Integers|$p$-adic valuation on $\Z$]].
+Then:
+:$b = p^\nu c$
+where $p \nmid c$
+From the above:
+:$\norm c = 1$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm b
+      | r = \norm p^\nu \norm {c}
+      | c = {{Norm Axiom|2}}
+}}
+{{eqn | r = \norm p^\nu
+      | c = 
+}}
+{{eqn | r = \norm p_p^{\alpha \nu}
+}}
+{{eqn | r = \paren {p^{-1} }^{\alpha \nu}
+      | c = {{Defof|P-adic Norm|$p$-adic norm}}
+}}
+{{eqn | r = p^{-\alpha \nu}
+      | c = 
+}}
+{{eqn | r = \paren {p^{-\nu} }^\alpha
+      | c = 
+}}
+{{eqn | r = \norm b_p^\alpha
+      | c = {{Defof|P-adic Norm|$p$-adic norm}}
+}}
+{{end-eqn}}
+It has been shown:
+:$\forall b \in \N: \norm b = \norm b_p^\alpha$
+By [[Equivalent Norms on Rational Numbers]]:
+:$\norm {\, \cdot \,}$ is [[Definition:Equivalent Division Ring Norms|equivalent]] to the [[Definition:Absolute Value|absolute value]] $\norm {\, \cdot \,}_p$.
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Archimedean Norm/Lemma 1.1}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+:$\forall n \in N: \norm n \le n^\alpha$
+\end{theorem}
+
+\begin{proof}
+By the definition of $\alpha$ then:
+:$\norm {n_0} = n_0^\alpha$
+By the definition of $n_0$ then:
+:$n_0^\alpha > 1$
+Let $n \in \N$.
+By [[Basis Representation Theorem]] then $n$ can be written:
+:$n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$
+where $0 \le a_i < n_0$ and $a_s \ne 0$
+Since all of the $a_i < n_0$, by choice of $n_0$ then:
+:$\forall a_i: \norm {a_i} \le 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm n
+      | o = \le
+      | r = \norm {a_0} + \norm {a_1 n_0} + \norm {a_2 n_0^2} + \cdots + \norm {a_s n_0^s}
+      | c = {{NormAxiom|3}}
+}}
+{{eqn | r = \norm {a_0} + \norm {a_1} \norm {n_0} + \norm {a_2} \norm {n_0}^2 + \cdots + \norm {a_s} \norm {n_0}^s
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | r = \norm {a_0} + \norm {a_1} n_0^\alpha + \norm {a_2} n_0^{2 \alpha} + \cdots + \norm {a_s} n_0^{s\alpha}
+      | c = as $\norm {n_0} = n_0^\alpha$
+}}
+{{eqn | o = \le
+      | r = 1 + n_0^\alpha + n_0^{2 \alpha} + \cdots + n_0^{s \alpha}
+      | c = as $\forall a_i: \norm {a_i} \le 1$
+}}
+{{eqn | r = n_0^{s \alpha} \paren {1 + n_0^{-\alpha} + n_0^{-2 \alpha} + \cdots + n_0^{-s \alpha} }
+      | c = 
+}}
+{{eqn | o = \le
+      | r = n_0^{s \alpha} \paren {\sum_{i \mathop = 0}^\infty \paren{\frac 1 {n_0^\alpha} }^i} 
+      | c = 
+}}
+{{eqn | r = n_0^{s \alpha} \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }
+      | c = [[Sum of Infinite Geometric Progression]] (and as $n_0^\alpha > 1$)
+}}
+{{eqn | o = \le
+      | r = n^\alpha \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }
+      | c = as $n \ge n_0^s$
+}}
+{{end-eqn}}
+Let $C = \paren {\dfrac {n_0^\alpha} {n_0^\alpha - 1} }$
+Hence:
+:$\norm n \le C n^\alpha$
+As $n \in \N$ was arbitrary:
+:$\forall n \in N: \norm n \le C n^\alpha$
+Let $n, N \in N$
+Then:
+:$\norm {n^N} \le C \paren {n^N}^\alpha$
+Now:
+{{begin-eqn}}
+{{eqn | l = \norm {n^N} \le C \paren {n^N}^\alpha
+      | o = \leadsto
+      | r = \norm n^N \le C \paren {n^N}^\alpha
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = \leadsto
+      | r = \norm n^N \le C \paren {n^\alpha}^N
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \norm n \le \sqrt [N] C n^\alpha
+      | c = taking $N$th roots
+}}
+{{end-eqn}}
+By [[Limit of Root of Positive Real Number]]:
+:$\sqrt [N] C \to 1$ as $N \to \infty$
+By the [[Multiple Rule for Real Sequences]]:
+:$\sqrt [N] C n^\alpha \to n^\alpha$ as $N \to \infty$
+By [[Inequality Rule for Real Sequences]], letting $N \to \infty$ for fixed $n$:
+:$\norm n \le n^\alpha$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Archimedean Norm/Lemma 1.2}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+:$\forall n \in N: \norm n \ge n^\alpha$
+\end{theorem}
+
+\begin{proof}
+By the definition of $\alpha$:
+:$\norm {n_0} = n_0^\alpha$
+By the definition of $n_0$:
+:$n_0^\alpha > 1$
+Let $n \in \N$.
+By [[Basis Representation Theorem]] then $n$ can be written:
+:$n = a_0 + a_1 n_0 + a_2 n_0^2 + \cdots + a_s n_0^s$
+where $0 \le a_i < n_0$ and $a_s \ne 0$
+By [[Bounds for Integer Expressed in Base k]]:
+:$n_0^{s + 1} > n \ge n_0^s$
+By [[Ostrowski's Theorem/Archimedean Norm/Lemma 1.1|Lemma 1.1]]:
+:$\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm n
+      | o = \ge
+      | r = \norm {n_0^{s + 1} } - \norm {n_0^{s + 1} - n}
+      | c = [[Reverse Triangle Inequality/Normed Division Ring|Reverse Triangle Inequality]]
+}}
+{{eqn | r = \norm {n_0}^{s + 1} - \norm {n_0^{s + 1} - n}
+      | c = {{Norm Axiom|2}}
+}}
+{{eqn | r = n_0^{\alpha \paren {s + 1} } - \norm {n_0^{s + 1} - n}
+      | c = as $\norm {n_0} = n_0^\alpha$
+}}
+{{eqn | o = \ge
+      | r = n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n}^\alpha
+      | c = as $\norm {n_0^{s + 1} - n} \le \paren {n_0^{s + 1} - n}^\alpha$
+}}
+{{eqn | o = \ge
+      | r = n_0^{\alpha \paren {s + 1} } - \paren {n_0^{s + 1} - n_0^s}^\alpha
+      | c = as $n \ge n_0^s$
+}}
+{{eqn | r = n_0^{\alpha \paren {s + 1} } \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = n^\alpha \paren {1 - \paren {1 - \frac 1 {n_0} }^\alpha}
+      | c = as $n_0^{s+1}>n$
+}}
+{{end-eqn}}
+Let $C' = \paren{1 - \paren{1 - \frac 1 {n_0} }^\alpha}$.
+Then:
+:$\norm n \ge C' n^\alpha$
+As $n \in \N$ was arbitrary:
+:$\forall n \in N: \norm n \ge C' n^\alpha$
+Let $n, N \in N$.
+Then:
+:$\norm {n^N} \ge C' \paren {n^N}^\alpha$
+Now:
+{{begin-eqn}}
+{{eqn | l = \norm {n^N} \ge C' \paren {n^N}^\alpha
+      | o = \leadsto
+      | r = \norm n^N \ge C' \paren {n^N}^\alpha
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | o = \leadsto
+      | r = \norm n^N \ge C' \paren {n^\alpha}^N
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \norm n \ge \sqrt [N] {C'} n^\alpha
+      | c = taking $N$th roots
+}}
+{{end-eqn}}
+By [[Limit of Root of Positive Real Number]]:
+:$\sqrt [N] {C'} \to 1$ as $N \to \infty$
+By the [[Multiple Rule for Real Sequences]]:
+:$\sqrt [N] {C'} n^\alpha \to n^\alpha$ as $N \to \infty$
+By [[Inequality Rule for Real Sequences]], letting $N \to \infty$ for fixed $n$:
+:$\norm n \ge n^\alpha$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalent Norms on Rational Numbers}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Norm on Division Ring|norms]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are [[Definition:Equivalent Division Ring Norms|equivalent]] {{iff}}:
+:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalent Norms on Rational Numbers/Necessary Condition|Necessary Condition]] ===
+{{:Equivalent Norms on Rational Numbers/Necessary Condition}}{{qed|lemma}}
+=== [[Equivalent Norms on Rational Numbers/Sufficient Condition|Sufficient Condition]] ===
+{{:Equivalent Norms on Rational Numbers/Sufficient Condition}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+:$\exists n \in \N: 0 < \norm n < 1$.
+\end{theorem}
+
+\begin{proof}
+Because $\norm {\, \cdot \,}$ is [[Definition:Nontrivial Division Ring Norm|non-trivial]]:
+:$\exists \dfrac a b \in \Q : 0 < \norm {\dfrac a b} \mbox { and } \norm {\dfrac a b} \ne 1$
+By [[Properties of Norm on Division Ring/Norm of Inverse|Norm of Inverse]] then:
+:$\norm {\dfrac a b} > 1 \implies \norm {\dfrac b a} < 1$
+Hence either $\norm {\dfrac a b} < 1$ or $\norm {\dfrac b a} < 1$.
+{{WLOG}} assume $\norm {\dfrac a b} < 1$.
+By [[Properties of Norm on Division Ring/Norm of Quotient|Norm of Quotient]] then:
+:$\dfrac {\norm a} {\norm b} < 1$
+Hence:
+:${\norm a} < \norm b$
+Let $n = \size a$ and $m = \size b$ where $\size {\,\cdot\,}$ is the [[Definition:Absolute Value|absolute value]] on $\Q$.
+Then $n, m \in \N$
+By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]] then:
+:$\norm n = \norm a$
+:$\norm m = \norm b$
+Hence:
+:$\norm n < \norm m$
+Since $\norm {\, \cdot \,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] then:
+:$\norm m \le 1$
+Hence:
+:$\norm n < \norm m \le 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.2}
+Tags: Ostrowski's Theorem
+
+\begin{theorem}
+:$n_0$ is a [[Definition:Prime Number|prime number]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $n_0$ is a [[Definition:Composite Number|composite number]]. 
+Let $n_1, n_2 \in \N$ such that $n_1, n_2 < n_0$ and $n_0 = n_1 n_2$. 
+By the definition of $n_0$ then:
+:$\norm {n_1} = 1$ 
+:$\norm {n_2} = 1$ 
+By {{NormAxiom|2}}:
+:$\norm {n_0} = \norm {n_1 n_2} = \norm {n_1} \norm {n_2} = 1$
+This [[Definition:Contradiction|contradicts]] the assumption that $\norm {n_0} < 1$. 
+So $n_0$ must be a [[Definition:Prime Number|prime number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Cosines of k pi over 5}
+Tags: Complex 5th Roots of Unity
+
+\begin{theorem}
+:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = 144 \degrees
+      | r = 180 \degrees - 36 \degrees
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cos 36 \degrees
+      | r = -\cos 144 \degrees
+      | c = [[Cosine of Supplementary Angle]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = 108 \degrees
+      | r = 180 \degrees - 72 \degrees
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cos 72 \degrees
+      | r = -\cos 108 \degrees
+      | c = [[Cosine of Supplementary Angle]]
+}}
+{{end-eqn}}
+Thus:
+:$\cos 36 \degrees + \cos 72 \degrees + \cos 108 \degrees + \cos 144 \degrees = 0$
+:[[File:Sum of Cosines of k pi over 5.png|500px]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Products of nth Roots of Unity taken up to n-1 at a Time is Zero}
+Tags: Viète's Formulas, Complex Roots of Unity
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $U_n = \set {z \in \C: z^n = 1}$ be the [[Definition:Complex Roots of Unity|complex $n$th roots of unity]].
+Then the [[Definition:Sum (Addition)|sum]] of the [[Definition:Product (Algebra)|products]] of the [[Definition:Element|elements]] of $U_n$ taken $2, 3, 4, \dotsc n - 1$ at a time is [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Element|elements]] of $U_n = \set {z \in \C: z^n = 1}$ are the solutions to the equation:
+:$z^n - 1 = 0$
+Thus by definition the [[Definition:Coefficient|coefficients]] of the [[Definition:Power (Algebra)|powers]] of $z$:
+:$z^2, z^3, \ldots, z^{n - 1}$
+are all [[Definition:Zero (Number)|zero]].
+The result follows directly from [[Viète's Formulas]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value of Complex Dot Product is Commutative}
+Tags: Absolute Value Function, Complex Dot Product
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number|complex numbers]].
+Let $z_1 \circ z_2$ denote the [[Definition:Complex Dot Product|(complex) dot product]] of $z_1$ and $z_2$.
+Then:
+:$\size {z_1 \circ z_2} = \size {z_2 \circ z_1}$
+where $\size {\, \cdot \,}$ denotes the [[Definition:Absolute Value|absolute value function]].
+\end{theorem}
+
+\begin{proof}
+From [[Dot Product Operator is Commutative]]:
+:$z_1 \circ z_2 = z_2 \circ z_1$
+The result follows trivially.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value of Complex Cross Product is Commutative}
+Tags: Absolute Value Function, Complex Cross Product
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number|complex numbers]].
+Let $z_1 \times z_2$ denote the [[Definition:Complex Cross Product|(complex) cross product]] of $z_1$ and $z_2$.
+Then:
+:$\size {z_1 \times z_2} = \size {z_2 \times z_1}$
+where $\size {\, \cdot \,}$ denotes the [[Definition:Absolute Value|absolute value function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \size {z_2 \times z_1}
+      | r = \size {-z_1 \times z_2}
+      | c = [[Complex Cross Product is Anticommutative]]
+}}
+{{eqn | r = \size {z_1 \times z_2}
+      | c = {{Defof|Absolute Value}}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area of Quadrilateral in Determinant Form}
+Tags: Areas of Quadrilaterals, Area of Quadrilateral in Determinant Form
+
+\begin{theorem}
+Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$, $C = \tuple {x_3, y_3}$ and $D = \tuple {x_4, y_4}$ be [[Definition:Point|points]] in the [[Definition:Cartesian Plane|Cartesian plane]].
+Let $A$, $B$, $C$ and $D$ form the [[Definition:Vertex of Polygon|vertices]] of a [[Definition:Quadrilateral|quadrilateral]].
+The [[Definition:Area|area]] $\mathcal A$ of $\Box ABCD$ is given by:
+:$\mathcal A = \dfrac 1 2 \paren {\size {\paren {\begin{vmatrix}
+x_1 & y_1 & 1 \\
+x_2 & y_2 & 1 \\
+x_3 & y_3 & 1 \\
+\end{vmatrix} } } + \size {\paren {\begin{vmatrix}
+x_1 & y_1 & 1 \\
+x_4 & y_4 & 1 \\
+x_3 & y_3 & 1 \\
+\end{vmatrix} } } }$
+\end{theorem}
+
+\begin{proof}
+$\Box ABCD$ can be divided into $2$ [[Definition:Triangle (Geometry)|triangles]]: $\triangle ABC$ and $\triangle ADC$.
+Hence $\mathcal A$ is the [[Definition:Sum (Addition)|sum]] of the [[Definition:Area|areas]] of $\triangle ABC$ and $\triangle ADC$.
+From [[Area of Triangle in Determinant Form]]:
+{{begin-eqn}}
+{{eqn | l = \map \Area {\triangle ABC}
+      | r = \dfrac 1 2 \size {\paren {\begin{vmatrix}
+x_1 & y_1 & 1 \\
+x_2 & y_2 & 1 \\
+x_3 & y_3 & 1 \\
+\end{vmatrix} } }
+}}
+{{eqn | l = \map \Area {\triangle ADC}
+      | r = \dfrac 1 2 \size {\paren {\begin{vmatrix}
+x_1 & y_1 & 1 \\
+x_4 & y_4 & 1 \\
+x_3 & y_3 & 1 \\
+\end{vmatrix} } }
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Three Points in Ultrametric Space have Two Equal Distances/Corollary 5}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}$ be a [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Let $a, b \in \Z_{\ne 0}$ be [[Definition:Coprime|coprime]], $a \perp b$
+Then:
+:$\norm a = 1$ or $\norm b = 1$
+\end{theorem}
+
+\begin{proof}
+By [[Bézout's Identity]] then:
+:$\exists n, m \in \Z : m a + n b = 1$  
+By [[Properties of Norm on Division Ring/Norm of Unity|Norm of Unity]] then:
+:$\norm {m a + n b} = 1$
+By [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 5|Corollary 5 of Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\norm a, \norm b, \norm n, \norm m \le 1$
+Let $\norm a \lt 1$.
+By [[Definition:Norm on Division Ring|Norm axiom $(\text N 2)$: Multiplicativity]]:
+:$\norm {m a} = \norm m \norm a \lt 1$
+Hence:
+:$\norm {m a} < \norm {m a + n b}$
+By [[Three Points in Ultrametric Space have Two Equal Distances/Corollary 4|Corollary 4]]:
+:$\norm {n b} = \norm {m a + n b} = 1$
+By [[Definition:Norm on Division Ring|Norm axiom $(\text N 2)$: Multiplicativity]]:
+:$\norm n \norm b = 1$.
+Hence $\norm b = 1$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of 2 m i Arccotangent of p}
+Tags: Arccotangent Function
+
+\begin{theorem}
+:$\map \exp {2 m i \arccot p} \paren {\dfrac {p i + 1} {p i - 1} }^m = 1$
+\end{theorem}
+
+\begin{proof}
+Let $z = \arccot p$.
+Then:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = \frac {\cos z} {\sin z}
+      | c = {{Defof|Real Arccotangent}}
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | r = i \dfrac {\map \exp {i z} + \map \exp {-i z} } {\map \exp {i z} - \map \exp {-i z} }
+      | c = [[Cotangent Exponential Formulation]]
+}}
+{{eqn | r = \frac {i \paren {\map \exp {2 i z} + 1} } {\map \exp {2 i z} - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p \paren {\map \exp {2 i z} - 1}
+      | r = i \paren {\map \exp {2 i z} + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {p - i} \map \exp {2 i z}
+      | r = p + i
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \exp {2 i z}
+      | r = \frac {p + i} {p - i}
+      | c = 
+}}
+{{end-eqn}}
+So we have:
+{{begin-eqn}}
+{{eqn | l = \map \exp {2 m i \arccot p}
+      | r = \paren {\map \exp {2 i \arccot p} }^m
+      | c = 
+}}
+{{eqn | r = \paren {\frac {p + i} {p - i} }^m
+      | c = 
+}}
+{{end-eqn}}
+Now:
+{{begin-eqn}}
+{{eqn | l = \paren {\frac {p + i} {p - i} } \paren {\frac {p i + 1} {p i - 1} }
+      | r = \frac {p^2 i + p i^2 + p + i} {p^2 i - p i^2 - p + i}
+      | c = 
+}}
+{{eqn | r = \frac {p^2 i + i} {p^2 i + i}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus z - 1 Less than Modulus z + 1 iff Real z Greater than Zero}
+Tags: Complex Modulus
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Then:
+:$\cmod {z - 1} < \cmod {z + 1} \iff \map \Re z > 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \cmod {z - 1}
+      | o = <
+      | r = \cmod {z + 1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \cmod {z + 1}
+      | o = >
+      | r = \cmod {z - 1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x + 1}^2 + y^2
+      | o = >
+      | r = \paren {x - 1}^2 + y^2
+      | c = {{Defof|Complex Modulus}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x + 1}^2
+      | o = >
+      | r = \paren {x - 1}^2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x^2 + 2 x + 1
+      | o = >
+      | r = x^2 - 2 x + 1
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 4 x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \Re z
+      | o = >
+      | r = 0
+      | c = {{Defof|Real Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vertices of Equilateral Triangle in Complex Plane/Sufficient Condition}
+Tags: Vertices of Equilateral Triangle in Complex Plane
+
+\begin{theorem}
+Let $z_1$, $z_2$ and $z_3$ be [[Definition:Complex Number|complex numbers]].
+Let $z_1$, $z_2$ and $z_3$ represent on the [[Definition:Complex Plane|complex plane]] the [[Definition:Vertex of Polygon|vertices]] of an [[Definition:Equilateral Triangle|equilateral triangle]].
+Then:
+:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
+\end{theorem}
+
+\begin{proof}
+Let $T$ be the [[Definition:Equilateral Triangle|equilateral triangle]] whose [[Definition:Vertex of Polygon|vertices]] are $z_1$, $z_2$ and $z_3$.
+We have that $z_2 - z_1$ and $z_3 - z_1$ are two sides of $T$ which meet at $z_1$.
+From the geometry of $T$ it follows that $z_2 - z_1$ is at an angle of $\pi/3$ to $z_3 - z_1$.
+Similarly, $z_1 - z_3$ and $z_2 - z_3$ are two sides of $T$ which meet at $z_3$.
+From the geometry of $T$ it follows that $z_1 - z_3$ is at an angle of $\pi / 3$ to $z_2 - z_3$.
+From [[Complex Multiplication as Geometrical Transformation/Corollary]]:
+:$(1): \quad z_2 - z_1 = e^{i \pi / 3} \left({z_3 - z_1}\right)$
+:$(2): \quad z_1 - z_3 = e^{i \pi / 3} \left({z_2 - z_3}\right)$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {z_2 - z_1} {z_1 - z_3}
+      | r = \dfrac {z_3 - z_1} {z_2 - z_3}
+      | c = $(1)$ divided by $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {z_2 - z_1} \paren {z_2 - z_3}
+      | r = \paren {z_3 - z_1} \paren {z_1 - z_3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1
+      | r = - {z_1}^2 - {z_3}^2 + 2 z_3 z_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {z_1}^2 + {z_2}^2 + {z_3}^2
+      | r = z_1 z_2 + z_2 z_3 + z_3 z_1
+      | c = 
+}}
+{{end-eqn}}
+\end{proof}<|endoftext|>
+\section{Vertices of Equilateral Triangle in Complex Plane/Necessary Condition}
+Tags: Vertices of Equilateral Triangle in Complex Plane
+
+\begin{theorem}
+Let $z_1$, $z_2$ and $z_3$ be [[Definition:Complex Number|complex numbers]].
+Let $z_1$, $z_2$ and $z_3$ fulfil the condition:
+:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
+Then $z_1$, $z_2$ and $z_3$ represent on the [[Definition:Complex Plane|complex plane]] the [[Definition:Vertex of Polygon|vertices]] of an [[Definition:Equilateral Triangle|equilateral triangle]].
+\end{theorem}
+
+\begin{proof}
+Let:
+:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
+Then:
+{{begin-eqn}}
+{{eqn | l = {z_1}^2 + {z_2}^2 + {z_3}^2
+      | r = z_1 z_2 + z_2 z_3 + z_3 z_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {z_2}^2 - z_1 z_2 - z_2 z_3 + z_3 z_1
+      | r = - {z_1}^2 - {z_3}^2 + 2 z_3 z_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {z_2 - z_1} \paren {z_2 - z_3}
+      | r = \paren {z_3 - z_1} \paren {z_1 - z_3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {z_2 - z_1} {z_1 - z_3}
+      | r = \dfrac {z_3 - z_1} {z_2 - z_3}
+      | c = 
+}}
+{{end-eqn}}
+Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same [[Definition:Angle|angle]] to each other as $z_1 - z_3$ and $z_2 - z_1$.
+Similarly:
+{{begin-eqn}}
+{{eqn | l = {z_1}^2 + {z_2}^2 + {z_3}^2
+      | r = z_1 z_2 + z_2 z_3 + z_3 z_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = - {z_2}^2 - {z_1}^2 + 2 z_1 z_2
+      | r = {z_3}^2 - z_1 z_3 - z_2 z_3 + z_1 z_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {z_2 - z_1} \paren {z_1 - z_2}
+      | r = \paren {z_3 - z_1} \paren {z_3 - z_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {z_2 - z_1} {z_3 - z_2}
+      | r = \dfrac {z_3 - z_1} {z_1 - z_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same [[Definition:Angle|angle]] to each other as $z_1 - z_2$ and $z_3 - z_2$.
+Thus all three [[Definition:Angle|angles]]:
+:$\angle z_2 z_1 z_3$
+:$\angle z_1 z_3 z_2$
+:$\angle z_3 z_2 z_1$
+are equal.
+By definition, therefore, $\triangle z_1 z_2 z_3$ is [[Definition:Equilateral Triangle|equilateral]].
+\end{proof}<|endoftext|>
+\section{Squares of Diagonals of Parallelogram}
+Tags: Parallelograms
+
+\begin{theorem}
+Let $ABCD$ be a [[Definition:Parallelogram|parallelogram]].
+:[[File:DiameterOfParallelogram.png|400px]]
+Then:
+:$AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = AC^2
+      | r = AB^2 + BC^2 - 2 (AB) (BC) \cos \angle B
+      | c = [[Cosine Rule]]
+}}
+{{eqn | l = BD^2
+      | r = BC^2 + CD^2 - 2 (CD) (BC) \cos \angle C
+      | c = [[Cosine Rule]]
+}}
+{{eqn | r = DA^2 + CD^2 - 2 (AB) (CD) \cos \angle C
+      | c = as $AB = CD$ and $BC = DA$
+}}
+{{eqn | ll= \leadsto
+      | l = AC^2 + BD^2
+      | r = AB^2 + BC^2 + DA^2 + CD^2 - 2 (AB) (BC) \paren {\cos \angle B + \cos \angle C}
+      | c = 
+}}
+{{end-eqn}}
+But we have that $\angle C$ and $\angle B$ are [[Definition:Supplementary Angles|supplementary]]:
+:$\angle C = \angle B = 180 \degrees$
+Thus from [[Cosine of Supplementary Angle]]:
+:$\cos \angle B + \cos \angle C = 0$
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Geodesic Equation/2d Surface Embedded in 3d Euclidean Space/Cylinder}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Let $\sigma$ be the [[Definition:Surface|surface]] of a [[Definition:Cylinder|cylinder]].
+Let $\sigma$ be embedded in [[Definition:Real Euclidean Space|3-dimensional Euclidean space]].
+Let $\sigma$ be [[Definition:Parametric Equation|parameterised]] by $\tuple {\phi, z}$ as
+:$\mathbf r = \tuple {a \cos \phi, a \sin \phi, z}$
+where
+:$a > 0$
+and
+:$z, \phi \in \R$
+Then [[Definition:Geodesic Curve|geodesics]] on $\sigma$ are of the following form:
+:$z = C_1 \phi + C_2$
+where $C_1, C_2$ are [[Definition:Real Number|real]] [[Definition:Arbitrary Constant (Calculus)|arbitrary constants]].
+\end{theorem}
+
+\begin{proof}
+From the given [[Definition:Parametric Equation|parametrization]] it follows that:
+{{begin-eqn}}
+{{eqn | l = E
+      | r = \mathbf r_\phi \cdot \mathbf r_\phi
+}}
+{{eqn | r = \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {-a \sin \phi, a \cos \phi, 0}
+}}
+{{eqn | r = a^2
+}}
+{{eqn | l = F
+      | r = \mathbf r_\phi \cdot \mathbf r_z
+}}
+{{eqn | r = \tuple {-a \sin \phi, a \cos \phi, 0} \cdot \tuple {0, 0, 1}
+}}
+{{eqn | r = 0
+}}
+{{eqn | l = G
+      | r = \mathbf r_z \cdot \mathbf r_z
+}}
+{{eqn | r = \tuple {0, 0, 1} \cdot \tuple {0, 0, 1}
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+where $E, F, G$ are the [[Definition:Real Function|functions]] of the [[Definition:First Fundamental Form|first fundamental form]].
+Furthermore, all [[Definition:Differentiable Mapping|derivatives]] of $E, F, G$ {{WRT|Differentiation}} $\phi$ and $z$ vanish.
+Then [[Geodesic Equation/2d Surface Embedded in 3d Euclidean Space|geodesic equations]] read:
+:$\dfrac \d {\d t} \dfrac {a^2 \phi'} {\sqrt {a^2 \phi'^2 + z'^2} } = 0$
+:$\dfrac \d {\d t} \dfrac {z'} {\sqrt {a^2 \phi'^2 + z'^2} } = 0$
+Integrate these [[Definition:Differential Equation|differential equations]] once:
+:$\dfrac {a^2 \phi'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_1$
+:$\dfrac {z'} {\sqrt {a^2 \phi'^2 + z'^2} } = b_2$
+where $b_1, b_2$ are [[Definition:Real Number|real]] [[Definition:Arbitrary Constant (Calculus)|arbitrary constants]].
+Divide the first [[Definition:Differential Equation|equation]] by the second one:
+:$\dfrac {a^2 \phi'} {z'} = \dfrac {b_1} {b_2}$
+To solve this in terms of $z$ as a [[Definition:Mapping|function]] of $\phi$, define:
+:$C_1 = \dfrac {a^2 b_2} {b_1}$ 
+and use the [[Derivative of Composite Function/Corollary|chain rule]]:
+:$\dfrac {\d z} {\d \phi} = C_1$
+[[Definition:Integration|Integration]] {{WRT|Integration}} $\phi$ yields the desired result.
+In other words, [[Definition:Geodesic Curve|geodesics]] are helical lines.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary}
+Tags: Complex Modulus
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number|complex numbers]] such that:
+:$\cmod {z_1 + z_2} = \cmod {z_1 - z_2}$
+Then $\dfrac {z_2} {z_1}$ is [[Definition:Wholly Imaginary|wholly imaginary]].
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod {z_1 + z_2}
+      | r = \cmod {z_1 - z_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x_1 + x_2}^2 + \paren {y_1 + y_2}^2
+      | r = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2
+      | c = {{Defof|Complex Modulus}}
+}}
+{{eqn | ll= \leadsto
+      | l = {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2
+      | r = {x_1}^2 - 2 x_1 x_2 + {x_1}^2 + {y_1}^2 - 2 y_1 y_2 + {y_1}^2
+      | c = [[Square of Sum]], [[Square of Difference]]
+}}
+{{eqn | ll= \leadsto
+      | l = 4 x_1 x_2 + 4 y_1 y_2
+      | r = 0
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 x_2 + y_1 y_2
+      | r = 0
+      | c = simplifying
+}}
+{{end-eqn}}
+Now we have:
+{{begin-eqn}}
+{{eqn | l = \dfrac {z_1} {z_2}
+      | r = \frac {x_1 + i y_1} {x_2 + i y_2}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {x_1 + i y_1} \paren {x_2 - i y_2} } { {x_2}^2 + {y_2}^2}
+      | c = {{Defof|Complex Division}}
+}}
+{{eqn | r = \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + \frac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}
+      | c = {{Defof|Complex Multiplication}}
+}}
+{{end-eqn}}
+But we have:
+:$x_1 x_2 + y_1 y_2 = 0$
+Thus:
+:$\dfrac {z_1} {z_2} = \dfrac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}$
+which is [[Definition:Wholly Imaginary|wholly imaginary]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Even Powers of z + a and z - a}
+Tags: Difference of Two Powers
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]] such that $m > 1$.
+Then for all [[Definition:Complex Number|complex number]] $z$:
+:$\paren {z + a}^{2 m} - \paren {z - a}^{2 m} = 4 m a z \displaystyle \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \dfrac {k \pi} {2 m} }$
+\end{theorem}
+
+\begin{proof}
+From [[Factors of Difference of Two Even Powers]]:
+:$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \displaystyle \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$
+Substituting $z + a$ for $x$, $z - a$ for $y$, and $m$ for $n$ we get:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {z + a}^{2 m} - \paren {z - a}^{2 m}
+      | c = 
+}}
+{{eqn | r = \paren {\paren {z + a} - \paren {z - a} } \paren {\paren {z + a} + \paren {z - a} } \prod_{k \mathop = 1}^{m - 1} \paren {\paren {z + a}^2 - 2 \paren {z + a} \paren {z - a} \cos \frac {k \pi} n + \paren {z - a}^2}
+      | c = 
+}}
+{{eqn | r = \paren {2 a} \paren {2 z} \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + 2 a z + a^2 - 2 \paren {z^2 - a^2} \cos \frac {k \pi} m + \paren {z^2 - 2 a z + a^2} }
+      | c = 
+}}
+{{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 + 2 a^2 - 2 \paren {z^2 - a^2} \cos \frac {k \pi} m}
+      | c = 
+}}
+{{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {1 - \cos \frac {k \pi} m} + 2 a^2 \paren {1 + \cos \frac {k \pi} m} }
+      | c = 
+}}
+{{eqn | r = 4 a z \prod_{k \mathop = 1}^{m - 1} \paren {2 z^2 \paren {2 \sin^2 \frac {k \pi} {2 m} } + 2 a^2 \paren {2 \cos^2 \frac {k \pi} {2 m} } }
+      | c = [[Double Angle Formula for Cosine]]: [[Double Angle Formulas/Cosine/Corollary 1|Corollary 1]] and [[Double Angle Formulas/Cosine/Corollary 2|Corollary 2]]
+}}
+{{eqn | r = 4 a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} 4 \sin^2 \frac {k \pi} {2 m} }
+      | c = {{Defof|Cotangent|subdef = Real Function}}
+}}
+{{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = 1}^{m - 1} \map \sin {\pi - \frac {k \pi} {2 m} } }
+      | c = [[Sine of Supplementary Angle]]
+}}
+{{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = 1}^{m - 1} \sin \frac {\paren {2 m - k} \pi} {2 m} } \sin \frac {m \pi} {2 m}
+      | c = [[Sine of Right Angle]]
+}}
+{{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^m \sin \frac {k \pi} {2 m} } \paren {\prod_{k \mathop = m + 1}^{2 m - 1} \sin \frac {k \pi} {2 m} }
+      | c = [[Translation of Index Variable of Product]]
+}}
+{{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\prod_{k \mathop = 1}^{2 m - 1} \sin \frac {k \pi} {2 m} }
+}}
+{{eqn | r = 4^m a z \paren {\prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} } } \paren {\frac {2 m} {2^{2 m - 1} } }
+      | c = [[Product of Sines of Fractions of Pi]]
+}}
+{{eqn | r = 4 m a z \prod_{k \mathop = 1}^{m - 1} \paren {z^2 + a^2 \cot^2 \frac {k \pi} {2 m} }
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Part of Complex Product}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number|complex numbers]].
+Then:
+:$\map \Re {z_1 z_2} = \map \Re {z_1} \, \map \Re {z_2} - \map \Im {z_1} \, \map \Im {z_2}$
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
+By definition of [[Definition:Complex Multiplication|complex multiplication]]:
+:$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \Re {z_1 z_2}
+      | r = x_1 x_2 - y_1 y_2
+      | c = {{Defof|Real Part}}
+}}
+{{eqn | r = \map \Re {z_1} \, \map \Re {z_2} - \map \Im {z_1} \, \map \Im {z_2}
+      | c = {{Defof|Real Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Imaginary Part of Complex Product}
+Tags: Complex Multiplication
+
+\begin{theorem}
+Let $z_1$ and $z_2$ be [[Definition:Complex Number|complex numbers]].
+Then:
+:$\map \Im {z_1 z_2} = \map \Re {z_1} \, \map \Im {z_2} + \map \Im {z_1} \, \map \Re {z_2}$
+\end{theorem}
+
+\begin{proof}
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
+By definition of [[Definition:Complex Multiplication|complex multiplication]]:
+:$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$
+Then
+{{begin-eqn}}
+{{eqn | l = \map \Im {z_1 z_2}
+      | r = x_1 y_2 + x_2 y_1
+      | c = {{Defof|Imaginary Part}}
+}}
+{{eqn | r = \map \Re {z_1} \, \map \Im {z_2} + \map \Im {z_1} \, \map \Re {z_2}
+      | c = {{Defof|Imaginary Part}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation for Perpendicular Bisector of Two Points}
+Tags: Straight Lines
+
+\begin{theorem}
+Let $\tuple {x_1, y_1}$ and $\tuple {y_1, y_2}$ be two [[Definition:Point|points]] in the [[Definition:Cartesian Plane|cartesian plane]].
+Let $L$ be the [[Definition:Perpendicular Bisector|perpendicular bisector]] of the [[Definition:Straight Line|straight line]] through $z_1$ and $z_2$ in the [[Definition:Complex Plane|complex plane]].
+$L$ can be expressed by the equation:
+:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$
+\end{theorem}
+
+\begin{proof}
+Let $M$ be the [[Definition:Straight Line|straight line]] passing through $z_1$ and $z_2$.
+Let $Q$ be the [[Definition:Midpoint|midpoint]] of $M$.
+We have that:
+:$Q = \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}$
+The [[Definition:Slope|slope]] of $M$ is $\dfrac {y_2 - y_1} {x_2 - x_1}$.
+As $L$ is [[Definition:Perpendicular|perpendicular]] to the $M$, its [[Definition:Slope|slope]] is $\dfrac {x_1 - x_2} {y_2 - y_1}$.
+Thus by [[Equation of Straight Line in Plane/Point-Slope Form|Equation of Straight Line in Plane: Point-Slope Form]], the equation for $L$ is:
+:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$
+{{qed}}
+{{improve|I can find no actual page on the web anywhere which gives this result explicitly, so I don't know what the "standard form" may be for this line. Hence I have not tried to simplify it, as any such "simplification" only seems to make it more complicated and less intuitive.}}
+[[Category:Straight Lines]]
+2ecrrfe5hfj681ty310ptb9l4v0yfac
+\end{proof}<|endoftext|>
+\section{Element of Center in Group whose Order is Power of 2}
+Tags: Centers of Groups
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $2^n$.
+Let $x \in G$ be of [[Definition:Order of Group Element|order]] $2^{n - 1}$ in $G$.
+Then $x^{2^{n - 2} }$ is an [[Definition:Element|element]] of the [[Definition:Center of Group|center]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $H = \gen x$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $x$.
+We have that the [[Definition:Index of Subgroup|index]] of $H$ in $G$ is $2$.
+Then by [[Subgroup of Index 2 is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Let $y = x^{2^{n - 2} }$.
+We have that $y \in H$, by definition of the construction of $H$.
+Then the [[Definition:Order of Group Element|order]] of $y$ in $G$ is $2$.
+As $H$ is [[Definition:Normal Subgroup|normal]], we have:
+:$\forall g \in G: g y g^{-1} \in H$
+By [[Order of Conjugate Element equals Order of Element]], $g y g^{-1}$ is also of [[Definition:Order of Group Element|order]] $2$.
+We can write $g y g^{-1} = x^m$ for some $m \in \Z$.
+By [[Order of Power of Group Element]], we have:
+:$2 = \dfrac {2^{n - 1} } {\gcd \set {m, 2^{n - 1} } }$
+This leads to:
+:$\gcd \set {m, 2^{n - 1} } = 2^{n - 2}$
+We have $0 < m < 2^{n - 1}$ and $2^{n - 2} \divides m$.
+This forces $m = 2^{n - 2}$ and $g y g^{-1} = y$.
+It follows that $g y = y g$.
+Hence $y = x^{2^{n - 2} }$ is an [[Definition:Element|element]] of the [[Definition:Center of Group|center]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Abelian Subgroups is Normal Subgroup of Subgroup Generated by those Subgroups}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $L$ and $M$ be [[Definition:Abelian Group|abelian]] [[Definition:Subgroup|subgroups]] of $G$.
+Let $H = \gen {L, M}$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $L$ and $M$.
+Then $L \cap M$ is a [[Definition:Normal Subgroup|normal subgroup]] of $H$.
+\end{theorem}
+
+\begin{proof}
+From 
+We have that $L$ and $M$ are [[Definition:Abelian Group|abelian]].
+From [[Intersection of Subgroups is Subgroup]], we have that $L \cap M$ is a [[Definition:Subgroup|subgroup]] of both $L$ and $M$.
+From [[Subgroup of Abelian Group is Normal]], we also have that $L \cap M$ is a [[Definition:Normal Subgroup|normal subgroup]] of both $L$ and $M$.
+Thus from [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
+:$L \subseteq \map N {L \cap M}$
+and:
+:$M \subseteq \map N {L \cap M}$
+Hence:
+:$L \cup M \subseteq \map N {L \cap M}$
+But from [[Normalizer is Subgroup]], $\map N {L \cap M}$ is a [[Definition:Subgroup|subgroup]] of $G$.
+So it follows that:
+:$H \subseteq \map N {L \cap M}$
+Hence the result by [[Subgroup is Normal Subgroup of Normalizer]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterisation of Non-Archimedean Division Ring Norms/Corollary 5}
+Tags: Characterisation of Non-Archimedean Division Ring Norms
+
+\begin{theorem}
+If $\norm {\, \cdot \,}$ is [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]] then:
+:$\sup \set {\norm {n \cdot 1_R}: n \in \Z} = 1$
+where $n \cdot 1_R = 
+\begin{cases}
+\underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} } & : n > 0 \\
+0 & : n = 0 \\
+\\
+-\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$ times} } & : n < 0 \\
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+By [[Characterisation of Non-Archimedean Division Ring Norms/Corollary 1|Corollary 1 of Characterisation of Non-Archimedean Division Ring Norms]] then:
+:$\sup \set {\norm{n \cdot 1_R}: n \in \N_{> 0}} = 1$
+By [[Definition:Norm on Division Ring|Norm Axiom $(\text N 1)$ (Positive Definiteness)]] then:
+:$\norm {0 \cdot 1_R} = 0 \le 1$
+Let $n < 0$ then:
+{{begin-eqn}}
+{{eqn | l = \norm {n \cdot 1_R}
+      | r = \norm {-\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$ times} } }
+}}
+{{eqn | r = \norm {\underbrace {1_R + 1_R + \dots + 1_R}_{\text {$-n$ times} } }
+      | c = [[Norm of Ring Negative]]
+}}
+{{eqn | r = \norm {\paren {-n} \cdot 1_R}
+      | c = 
+}}
+{{eqn | o = \le
+      | r = 1
+      | c = [[Characterisation of Non-Archimedean Division Ring Norms]]
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup Containing all Squares of Group Elements is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ with the property that:
+:$\forall x \in G: x^2 \in H$
+Then $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {x h}^2 h^{-1} \paren {x^{-1} }^2
+      | r = x h x h h^{-1} x^{-1} x^{-1}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = x h x x^{-1} x^{-1}
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | r = x h x^{-1}
+      | c = {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both [[Definition:Element|elements]] of $H$.
+Thus:
+:$x h x^{-1} \in H$
+and so $H$ is [[Definition:Normal Subgroup|normal]] in $G$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Order Elements of Infinite Abelian Group form Normal Subgroup}
+Tags: Normal Subgroups, Infinite Groups, Abelian Groups
+
+\begin{theorem}
+Let $G$ be an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group| abelian group]].
+Let $H \subseteq G$ be the [[Definition:Subset|subset]] of $G$ defined as:
+:$H := \set {x \in G: x \text { is of finite order in } G}$
+Then $H$ forms a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $e$ be the [[Definition:Identity Element|identity element]] of $G$.
+From [[Identity is Only Group Element of Order 1]], $\order e = 1$ and so $H \ne \O$.
+Let $a \in H$.
+Then by [[Order of Group Element equals Order of Inverse]]:
+:$\order a = \order {a^{-1} }$
+and so $a \in H$.
+Let $a, b \in H$.
+From [[Product of Orders of Abelian Group Elements Divides LCM of Order of Product]]:
+:$\order {a b} \divides \lcm \set {\order a, \order b}$
+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]].
+Thus $a b$ is also of [[Definition:Finite Order Element|finite order]].
+Thus by definition:
+:$a b \in H$.
+By the [[Two-Step Subgroup Test]] it follows that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+By [[Subgroup of Abelian Group is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Commutator is Identity iff Elements Commute}
+Tags: Group Commutators
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x, y \in G$.
+Let $\sqbrk {x, y}$ denote the [[Definition:Commutator of Group Elements|commutator]] of $x$ and $y$.
+Then $\sqbrk {x, y} = e$ {{iff}} $x$ and $y$ [[Definition:Commuting Elements|commute]].
+\end{theorem}
+
+\begin{proof}
+As $G$ is a [[Definition:Group|group]], it is by definition a [[Definition:Monoid|monoid]].
+Hence [[Product of Commuting Elements with Inverses]] applies:
+:$x y x^{-1} y^{-1} = e = x^{-1} y^{-1} x y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Commutator of Quotient Group Elements}
+Tags: Group Commutators, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $\sqbrk {x, y}$ denote the [[Definition:Commutator of Group Elements|commutator]] of $x, y \in G$:
+:$\sqbrk {x, y} = x^{-1} y^{-1} x y$
+Then:
+:$\forall x, y \in G: \sqbrk {x N, y N} = \sqbrk {x, y} N$
+where $x N$ and $y N$ are [[Definition:Left Coset|left cosets]] of $N$, and so [[Definition:Element|elements]] of the [[Definition:Quotient Group|quotient group]] $G / N$ of $G$ by $N$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sqbrk {x N, y N}
+      | r = \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}
+      | c = {{Defof|Commutator of Group Elements}}
+}}
+{{eqn | r = \paren {x^{-1} N} \paren {y^{-1} N} \paren {x N} \paren {y N}
+      | c = [[Quotient Group is Group]]: [[Definition:Inverse Element|inverse]] of $x N$ is $x^{-1} N$
+}}
+{{eqn | r = \paren {x^{-1} y^{-1} N} \paren {x y N}
+      | c = {{Defof|Coset Product}}
+}}
+{{eqn | r = x^{-1} y^{-1} x y N
+      | c = {{Defof|Coset Product}}
+}}
+{{eqn | r = \sqbrk {x, y} N
+      | c = {{Defof|Commutator of Group Elements}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group is Abelian iff All Commutators in Divisor}
+Tags: Group Commutators, Quotient Groups, Abelian Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|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 the [[Definition:Quotient Group|quotient group]] $G / N$ is [[Definition:Abelian Group|abelian]] {{iff}}:
+:$\forall x, y \in G: \sqbrk {x, y} \in N$
+where $\sqbrk {x, y}$ denotes the [[Definition:Commutator of Group Elements|commutator]] of $x$ and $y$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \paren {x N} \paren {y N}
+      | r = \paren {y N} \paren {x N}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \sqbrk {x N, y N}
+      | r = N
+      | c = {{Defof|Commutator of Group Elements}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \sqbrk {x, y} N
+      | r = N
+      | c = [[Commutator of Quotient Group Elements]]
+}}
+{{end-eqn}}
+Let $G / N$ be [[Definition:Abelian Group|abelian]].
+Then by definition:
+:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
+and it follows from $(1)$ that:
+:$\forall x, y \in G: \sqbrk {x, y} \in N$
+Conversely, let:
+:$\forall x, y \in G: \sqbrk {x, y} \in N$
+Again it follows from $(1)$ that:
+:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
+That is, that $G / N$ is [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sufficient Condition for Quotient Group by Intersection to be Abelian}
+Tags: Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ and $K$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Let the [[Definition:Quotient Group|quotient groups]] $G / N$ and $G / K$ be [[Definition:Abelian Group|abelian]].
+Then the [[Definition:Quotient Group|quotient group]] $G / \paren {N \cap K}$ is also [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Normal Subgroups is Normal]], we have that $N \cap K$ is [[Definition:Normal Subgroup|normal]] in $G$.
+We are given that $G / N$ and $G / K$ are [[Definition:Abelian Group|abelian]].
+Hence:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y \in G:
+      | l = \sqbrk {x, y}
+      | o = \in
+      | r = N
+      | c = [[Quotient Group is Abelian iff All Commutators in Divisor]]
+}}
+{{eqn | lo= \land
+      | l = \sqbrk {x, y}
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | ll= \forall x, y \in G:
+      | l = \sqbrk {x, y}
+      | o = \in
+      | r = N \cap K
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | lll=\leadsto
+      | l = G / \paren {N \cap K}
+      | o = 
+      | r = \text {is abelian}
+      | c = [[Quotient Group is Abelian iff All Commutators in Divisor]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group by Intersection of Normal Subgroups not necessarily Cyclic if Quotient Groups are}
+Tags: Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N$ and $K$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Let the [[Definition:Quotient Group|quotient groups]] $G / N$ and $G / K$ be [[Definition:Cyclic Group|cyclic]].
+Then the [[Definition:Quotient Group|quotient group]] $G / \paren {N \cap K}$ is not necessarily [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $G = \set {e, a, b, c}$ be the [[Definition:Klein Four-Group|Klein $4$-group]] whose [[Definition:Identity Element|identity element]] is $e$.
+Let $N = \set {e, a}$ and $K = \set {e, b}$.
+By [[Subgroups of Klein Four-Group]], both $N$ and $K$ are [[Definition:Subgroup|subgroups]] of $G$.
+By [[Prime Group is Cyclic]], both $N$ and $K$ are [[Definition:Cyclic Group|cyclic]].
+By [[Subgroup of Abelian Group is Normal]], both $N$ and $K$ are [[Definition:Normal Subgroup|normal]] in $G$.
+Then we have that:
+:$N \cap K = e$
+and so by [[Trivial Quotient Group is Quotient Group]]:
+:$G / \paren {N \cap K} \cong G$
+But $G$ is the [[Definition:Klein Four-Group|Klein $4$-group]], which is not [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Boolean Group is Power of 2}
+Tags: Boolean Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Boolean Group|Boolean group]].
+Let $\order G$ denote the [[Definition:Order of Group|order]] of $G$.
+Then:
+:$\order G = 2^n$
+where $n \in \Z_{\ge 0}$ is a [[Definition:Positive Integer|positive integer]].
+\end{theorem}
+
+\begin{proof}
+The case where $n = 0$ is clear:
+:$\order {\set e} = 1$
+and $e^2 = e$.
+{{AimForCont}} $\order G = m \times 2^k$ for some [[Definition:Odd Integer|odd integer]] $m$.
+Then $m$ itself has an [[Definition:Odd Prime|odd prime]] $p$ as a [[Definition:Divisor of Integer|integer]] (which may of course equal $m$ if $m$ is itself [[Definition:Odd Prime|prime]]).
+Then by [[Cauchy's Lemma (Group Theory)]] there exists $g \in G$ such that $\order g = p$.
+Hence it is not the case that $g^2 = e$.
+Hence $\order G$ has no [[Definition:Prime Factor|prime factor]] which is [[Definition:Odd Integer|odd]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Index 2 contains all Squares of Group Elements}
+Tags: Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] is $2$.
+Then:
+:$\forall x \in G: x^2 \in H$
+\end{theorem}
+
+\begin{proof}
+By [[Subgroup of Index 2 is Normal]], $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Hence the [[Definition:Quotient Group|quotient group]] $G / H$ exists.
+Then we have:
+{{begin-eqn}}
+{{eqn | lo= \forall x \in G:
+      | l = \paren {x^2} H
+      | r = \paren {x H}^2
+      | c = 
+}}
+{{eqn | r = H
+      | c = as $G / H$ is of [[Definition:Order of Group|order]] $2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements}
+Tags: Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] is $3$.
+Then it is not necessarily the case that:
+:$\forall x \in G: x^3 \in H$
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Consider $S_3$, the [[Definition:Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
+From [[Subgroups of Symmetric Group on 3 Letters]], the [[Definition:Subgroup|subgroups]] of $S_3$ are:
+[[Definition:Subset|subsets]] of $S_3$ which form [[Definition:Subgroup|subgroups]] of $S_3$ are:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = S_3
+}}
+{{eqn | o = 
+      | r = \set e
+}}
+{{eqn | o = 
+      | r = \set {e, \tuple {123}, \tuple {132} }
+}}
+{{eqn | o = 
+      | r = \set {e, \tuple {12} }
+}}
+{{eqn | o = 
+      | r = \set {e, \tuple {13} }
+}}
+{{eqn | o = 
+      | r = \set {e, \tuple {23} }
+}}
+{{end-eqn}}
+One such [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Index of Subgroup|index]] is $3$ is $\set {e, \tuple {12} }$
+But $\set {e, \tuple {12} }$ does not contain $\tuple {123}$ or $\tuple {132}$, both of which are of [[Definition:Order of Group Element|order $3$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Center of Non-Abelian Group of Order pq is Trivial}
+Tags: Centers of Groups, Groups of Order p q
+
+\begin{theorem}
+Let $p$ and $q$ be [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|prime numbers]].
+Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p q$ whose [[Definition:Identity Element|identity]] is $e$.
+Then the [[Definition:Center of Group|center]] of $G$ is [[Definition:Trivial Subgroup|trivial]]:
+:$\map Z G = \set e$
+\end{theorem}
+
+\begin{proof}
+From [[Center of Group is Normal Subgroup]], $\map Z G$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+By [[Definition:Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Order of Group|order]] of $\map Z G$ is either $1$, $p$, $q$ or $p q$.
+Because $G$ is not [[Definition:Abelian Group|abelian]], $G \ne \map Z G$.
+Hence $\order {\map Z G} \ne p q$.
+From [[Quotient of Group by Center Cyclic implies Abelian]]:
+:$G / \map Z G$ cannot be a [[Definition:Cyclic Group|cyclic group]] which is non-[[Definition:Trivial Subgroup|trivial]].
+Then we have:
+:$\map C x \subset G$
+where $\map C x$ is the [[Definition:Centralizer of Group Element|centralizer]] of $x$.
+Hence by [[Prime Group is Cyclic]], $G / \map Z G$ cannot be [[Definition:Cyclic Group|cyclic]] of [[Definition:Order of Group|order]] $p$ or $q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Elements of Order p in Group of Order pq is Multiple of q}
+Tags: Groups of Order p q
+
+\begin{theorem}
+Let $p$ and $q$ be [[Definition:Distinct Elements|distinct]] [[Definition:Prime Number|prime numbers]].
+Let $G$ be a non-[[Definition:Abelian Group|abelian]] [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p q$.
+Then the number of [[Definition:Element|elements]] of $G$ of order $p$ is a [[Definition:Integer Multiple|multiple]] of $q$.
+\end{theorem}
+
+\begin{proof}
+Let $x$ be an [[Definition:Element|element]] of $G$ of order $p$.
+From [[Center of Non-Abelian Group of Order pq is Trivial]]:
+:$p \notin \map Z G$
+where $\map Z G$ denotes the [[Definition:Center of Group|center]] of $G$.
+As $x \notin \map Z G$:
+:$\map C x \subsetneq G$
+where $\map C x$ is the [[Definition:Centralizer of Group Element|centralizer]] of $x$.
+From [[Order of Element divides Order of Centralizer]]:
+:$p \divides \order {\map C x}$
+Then from [[Definition:Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], it follows directly that:
+:$\order {\map C x} = p$
+By [[Number of Conjugates is Number of Cosets of Centralizer]]:
+:$\card {\conjclass x} = \index G {\map {C_G} x}$
+where $\conjclass x$ denotes the [[Definition:Conjugacy Class|conjugacy class]] of $x$.
+It follows that:
+:$\card {\conjclass x} = q$
+and so the [[Definition:Element|element]] of $G$ of order $p$ come in [[Definition:Set|sets]] whose [[Definition:Cardinality|cardinality]] is a [[Definition:Divisor of Integer|divisor]] of $q$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Normal Subgroup with Center in p-Group}
+Tags: Normal Subgroups, P-Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]]
+Let $G$ be a [[Definition:P-Group|$p$-group]].
+Let $N$ be a [[Definition:Non-Trivial Subgroup|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
+Then:
+:$N \cap \map Z G$ is a [[Definition:Non-Trivial Subgroup|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+First we note that:
+:[[Center of Group is Normal Subgroup]]
+and from [[Intersection of Normal Subgroups is Normal]]:
+:$N \cap \map Z G$ is [[Definition:Normal Subgroup|normal]] in $G$.
+Suppose $G$ is [[Definition:Abelian Group|abelian]].
+From [[Group equals Center iff Abelian]]:
+:$\map Z G = G$
+Then:
+:$N \cap \map Z G = N$
+which is [[Definition:Non-Trivial Subgroup|non-trivial]].
+From [[Prime Group is Cyclic]] and [[Cyclic Group is Abelian]], this will always be the case for $r = 1$.
+So, suppose $G$ is non-[[Definition:Abelian Group|abelian]].
+From [[Center of Group of Prime Power Order is Non-Trivial]], $\map Z G$ is [[Definition:Non-Trivial Subgroup|non-trivial]].
+From [[Union of Conjugacy Classes is Normal]]:
+:$N = \displaystyle \bigcup_{x \mathop \in N} \conjclass x$
+Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the [[Definition:Conjugacy Class|conjugacy classes]] into which $N$ is [[Definition:Partition (Set Theory)|partitioned]].
+From [[Conjugacy Class of Element of Center is Singleton]], all of these will have more than one [[Definition:Element|element]].
+From [[Number of Conjugates is Number of Cosets of Centralizer]]:
+:$\order {\conjclass {x_j} } \divides \order G$
+for all $j \in \set {1, 2, \ldots, m}$.
+{{finish|Needs more thought.}}
+\end{proof}<|endoftext|>
+\section{Normal Subgroup of p-Group of Order p is Subset of Center}
+Tags: Normal Subgroups, Centers of Groups, P-Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:P-Group|$p$-group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$ of [[Definition:Order of Group|order]] $p$.
+Then:
+:$N \subseteq \map Z G$
+where $\map Z G$ denotes the [[Definition:Center of Group|center]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Normal Subgroup with Center in p-Group]]:
+:$\order {N \cap \map Z G} > 1$
+From [[Intersection of Subgroups is Subgroup]], $N \cap \map Z G$ is a [[Definition:Subgroup|subgroup]] of $N$.
+It follows from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]] that:
+:$\order {N \cap \map Z G} = p$
+and so:
+:$N \cap \map Z G = N$
+But from [[Intersection of Subgroups is Subgroup]], $N \cap \map Z G$ is a [[Definition:Subgroup|subgroup]] of $\map Z G$
+That is:
+:$N$ is a [[Definition:Subgroup|subgroup]] of $\map Z G$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p}
+Tags: P-Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a non-[[Definition:Abelian Group|abelian group]] of [[Definition:Order of Group|order]] $p^3$.
+Then $G$ contains [[Definition:Unique|exactly one]] [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $p$.
+\end{theorem}
+
+\begin{proof}
+From [[Center of Group of Prime Power Order is Non-Trivial]], $\map Z G$ is not the [[Definition:Trivial Subgroup|trivial subgroup]].
+From [[Quotient of Group by Center Cyclic implies Abelian]], $G / \map G Z$ cannot be [[Definition:Cyclic Group|cyclic]] and [[Definition:Non-Trivial Subgroup|non-trivial]].
+Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.
+Thus $\order {\map G Z} = p$.
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$ of [[Definition:Order of Group|order]] $p$.
+Then from [[Normal Subgroup of p-Group of Order p is Subset of Center]]:
+:$N \subseteq \map G Z$
+Thus there is no [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $p$ different from $\map G Z$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternion Group has Normal Subgroup without Complement}
+Tags: Subgroup Complements, Quaternion Group
+
+\begin{theorem}
+Let $Q$ denote the [[Definition:Quaternion Group|quaternion group]].
+There exists a [[Definition:Normal Subgroup|normal subgroup]] of $Q$ which has no [[Definition:Complement of Subgroup|complement]].
+\end{theorem}
+
+\begin{proof}
+From [[Subgroups of Quaternion Group]]:
+{{:Subgroups of Quaternion Group}}
+From [[Quaternion Group is Hamiltonian]], all these [[Definition:Subgroup|subgroups]] are [[Definition:Normal Subgroup|normal]].
+For two [[Definition:Subgroup|subgroups]] to be [[Definition:Complement of Subgroup|complementary]], they need to have an [[Definition:Set Intersection|intersection]] which is [[Definition:Trivial Subgroup|trivial]].
+However, apart from $\set e$ itself, all these [[Definition:Subgroup|subgroups]] contain $a^2$.
+Hence none of these [[Definition:Subgroup|subgroups]] has a [[Definition:Complement of Subgroup|complement]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Abelian Order 2p Group has Order p Element}
+Tags: Order of Group Elements, Groups of Order 2 p
+
+\begin{theorem}
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+Let $G$ be a non-[[Definition:Abelian Group|abelian group]] of [[Definition:Order of Structure|order $2 p$]].
+Then $G$ has at least one [[Definition:Element|element]] of [[Definition:Order of Group Element|order $p$]].
+\end{theorem}
+
+\begin{proof}
+By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], all the [[Definition:Element|elements]] of $G$ have [[Definition:Order of Group Element|orders]] $1$, $2$, $p$ or $2p$.
+From [[Identity is Only Group Element of Order 1]], $2 p - 1$ [[Definition:Element|elements]] of $G$ have [[Definition:Order of Group Element|orders]] greater than $1$.
+From [[Cyclic Group is Abelian]], $G$ is not the [[Definition:Cyclic Group|cyclic group]] $2 p$.
+If $g \in G$ was of [[Definition:Order of Group Element|order $2 p$]] then $g$ would [[Definition:Generator of Cyclic Group|generate]] the [[Definition:Cyclic Group|cyclic group]] $C_{2 p}$.
+Thus $G$ has no [[Definition:Element|element]] of [[Definition:Order of Group Element|order $2 p$]].
+So all elements of $G$ except the [[Definition:Identity Element|identity]] have [[Definition:Order of Group Element|orders]] $2$ or $p$.
+{{AimForCont}} $G$ has no [[Definition:Element|element]] of [[Definition:Order of Group Element|order $p$]].
+Then all [[Definition:Element|elements]] are of [[Definition:Order of Group Element|order $2$]].
+Then $G$ is a [[Definition:Boolean Group|Boolean group]].
+Then by [[Boolean Group is Abelian]], $G$ is [[Definition:Abelian Group|abelian]].
+This [[Definition:Contradiction|contradicts]] the assertion that $G$ is non-[[Definition:Abelian Group|abelian]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fourth Power Modulo 5}
+Tags: Fourth Powers
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Then:
+:$n^4 \equiv m \pmod 5$
+where $m \in \set {0, 1}$.
+\end{theorem}
+
+\begin{proof}
+By [[Congruence of Powers]]:
+:$a \equiv b \pmod 5 \iff a^4 \equiv b^4 \pmod 5$
+so it is sufficient to demonstrate the result for $n \in \set {0, 1, 2, 3, 4}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = 0^4
+      | m = 0
+      | mo= \equiv
+      | r = 0
+      | rr= \pmod 5
+      | c = 
+}}
+{{eqn | l = 1^4
+      | m = 1
+      | mo= \equiv
+      | r = 1
+      | rr= \pmod 5
+      | c = 
+}}
+{{eqn | l = 2^4
+      | m = 16
+      | mo= \equiv
+      | r = 1
+      | rr= \pmod 5
+      | c = 
+}}
+{{eqn | l = 3^4
+      | m = 81
+      | mo= \equiv
+      | r = 1
+      | rr= \pmod 5
+      | c = 
+}}
+{{eqn | l = 4^4
+      | m = 256
+      | mo= \equiv
+      | r = 1
+      | rr= \pmod 5
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+[[Category:Fourth Powers]]
+2a82dfb5jyv3nyxyy58zi9gjf88h0dg
+\end{proof}<|endoftext|>
+\section{Subgroup of Order p in Group of Order 2p is Normal/Corollary}
+Tags: Groups of Order 2 p
+
+\begin{theorem}
+Let $G$ be non-[[Definition:Abelian Group|abelian]].
+Every [[Definition:Element|element]] of $G \setminus K$ is of [[Definition:Order of Group Element|order]] $2$, and:
+:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$
+\end{theorem}
+
+\begin{proof}
+By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Element|elements]] of $G \setminus K$ can be of [[Definition:Order of Group Element|order]] $1$, $2$, $p$ or $2 p$.
+$1$ is not possible because [[Identity is Only Group Element of Order 1]].
+Then we have that $G$ is non-[[Definition:Abelian Group|abelian]].
+Hence from [[Cyclic Group is Abelian]], $G$ is not [[Definition:Cyclic Group|cyclic]].
+Thus $\order b \ne 2 p$.
+It remains to investigate $2$ and $p$.
+Let $b \in G \setminus K$.
+By [[Subgroup of Index 2 contains all Squares of Group Elements]]:
+:$b^2 \in K$
+{{AimForCont}} $b$ has [[Definition:Order of Group Element|order]] $p$.
+Then by [[Intersection of Subgroups of Prime Order]]:
+:$K \cap \gen b = e$
+which [[Definition:Contradiction|contradicts]] $b^2 \in K$.
+Thus by [[Proof by Contradiction]] $\order b \ne p$
+Hence it must follow that $\order b = 2$.
+{{qed|lemma}}
+We have that:
+:$b a \in G \setminus K$
+Thus:
+:$\paren {b a}^2 = e$
+{{begin-eqn}}
+{{eqn | l = b a
+      | o = \in
+      | r = G \setminus K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {b a}^2
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b a b
+      | r = \paren {b a}^2 a^{-1}
+      | c = 
+}}
+{{eqn | r = a^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b a b^{-1}
+      | r = a^{-1}
+      | c = as $b = b^{-1}$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Order p in Group of Order 2p is Normal}
+Tags: Groups of Order 2 p
+
+\begin{theorem}
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order $2 p$]].
+Let $a \in G$ be of [[Definition:Order of Group Element|order]] $p$.
+Let $K = \gen a$ be the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $a$.
+Then $K$ is [[Definition:Normal Subgroup|normal]] in $G$.
+\end{theorem}
+
+\begin{proof}
+The result [[Non-Abelian Order 2p Group has Order p Element]] demonstrates that such an [[Definition:Element|element]] $a$ exists in $G$.
+By definition of [[Definition:Generator of Cyclic Group|generator of cyclic group]], $K$ is the [[Definition:Cyclic Group|cyclic group]] $C_p$ of [[Definition:Order of Group Element|order]] $p$.
+By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Index of Subgroup|index]] of $K$ is:
+:$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$
+The result follows from [[Subgroup of Index 2 is Normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Automorphisms form Subgroup of Automorphism Group}
+Tags: Automorphism Groups, Inner Automorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then the [[Definition:Set|set]] $\Inn G$ of all [[Definition:Inner Automorphism|inner automorphisms]] of $G$ forms a [[Definition:Subgroup|normal subgroup]] of the [[Definition:Automorphism Group of Group|automorphism group]] $\Aut G$ of $G$:
+:$\Inn G \le \Aut G$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\kappa_x: G \to G$ be the [[Definition:Inner Automorphism|inner automorphism]] defined as:
+:$\forall g \in G: \kappa_x \paren g = x g x^{-1}$
+We see that:
+:$\Inn G \ne \O$
+as $\kappa_x$ is defined for all $x \in G$.
+We show that:
+:$\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$
+So:
+{{begin-eqn}}
+{{eqn | ll= \forall g \in G:
+      | l = \map {\paren {\kappa_x \circ \paren {\kappa_y}^{-1} } } g
+      | r = \map {\kappa_x} {\map {\paren {\kappa_y}^{-1} } g}
+      | c = 
+}}
+{{eqn | r = \map {\kappa_x} {\map {\kappa_{y^{-1} } } g}
+      | c = [[Inverse of Inner Automorphism]]
+}}
+{{eqn | r = \map {\kappa_x} {y^{-1} g y}
+      | c = 
+}}
+{{eqn | r = x y^{-1} g y x^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {x y^{-1} } g \paren {x y^{-1} }^{-1}
+      | c = [[Inverse of Product]]
+}}
+{{eqn | r = \map {\kappa_{x y^{-1} } } g
+      | c = 
+}}
+{{end-eqn}}
+As $x y^{-1} \in G$, it follows that:
+:$\kappa_{x y^{-1} } \in \Inn G$
+By the [[One-Step Subgroup Test]]:
+:$\Inn G \le \Aut G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Property of Group Automorphism which Fixes Identity Only}
+Tags: Group Automorphisms, Property of Group Automorphism which Fixes Identity Only
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\phi: G \to G$ be a [[Definition:Group Automorphism|group automorphism]].
+Let $\phi$ have the property that:
+:$\forall g \in G \setminus \set e: \map \phi t \ne t$
+That is, the only [[Definition:Fixed Element|fixed element]] of $\phi$ is $e$.
+Then:
+:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{-1} \, \map \phi x
+      | r = y^{-1} \, \map \phi y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi x
+      | r = x y^{-1} \, \map \phi y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi x \paren {\map \phi y}^{-1}
+      | r = x y^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi x \, \map \phi {y^{-1} }
+      | r = x y^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {x y^{-1} }
+      | r = x y^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y^{-1}
+      | r = e
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Property of Group Automorphism which Fixes Identity Only/Corollary 1}
+Tags: Property of Group Automorphism which Fixes Identity Only
+
+\begin{theorem}
+:$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$
+\end{theorem}
+
+\begin{proof}
+Let $\psi: G \to G$ be the [[Definition:Mapping|mapping]]:
+:$\forall x \in G: \map \psi x = x^{-1} \, \map \phi x$
+From [[Property of Group Automorphism which Fixes Identity Only]]:
+:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$
+That is, $\psi$ is an [[Definition:Injection|injection]].
+From [[Injection from Finite Set to Itself is Surjection]], it follows that $\psi$ is a [[Definition:Surjection|surjection]].
+That is:
+:$\forall x \in G: \exists g \in G: x = \map \psi g$
+which is a restatement of the hypothesis.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Automorphism Group of Cyclic Group is Abelian}
+Tags: Cyclic Groups, Automorphism Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
+Let $\Aut G$ denote the [[Definition:Automorphism Group|automorphism group]] of $G$.
+Then $\Aut G$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $G = \gen g$
+Let $\phi, \psi \in \Aut G$.
+As $G$ is [[Definition:Cyclic Group|cyclic]]:
+{{begin-eqn}}
+{{eqn | lo= \exists a \in \Z:
+      | l = \map \phi g 
+      | r = g^a
+}}
+{{eqn | lo= \exists b \in \Z:
+      | l = \map \psi g 
+      | r = g^b
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {\phi \circ \psi} g
+      | r = \paren {g^a}^b
+      | c = 
+}}
+{{eqn | r = g^{a b}
+      | c = 
+}}
+{{eqn | r = g^{b a}
+      | c = 
+}}
+{{eqn | r = \paren {g^b}^a
+      | c = 
+}}
+{{eqn | r = \map {\psi \circ \phi} g
+      | c = 
+}}
+{{end-eqn}}
+Thus in particular, $\phi \circ \psi$ and $\psi \circ \phi$ are equal on the [[Definition:Generator of Cyclic Group|generator]] $g$.
+Since $g$ generates $G$, they must be equal as [[Definition:Group Automorphism|automorphisms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Automorphism Group of Prime Group}
+Tags: Prime Groups, Automorphism Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p$.
+Let $\Aut G$ denote the [[Definition:Automorphism Group|automorphism group]] of $G$.
+Then:
+:$\order {\Aut G} = p - 1$
+where $\order {\, \cdot \,}$ denotes the [[Definition:Order of Group|order]] of a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+From [[Prime Group is Cyclic]] we have that $G$ is a [[Definition:Cyclic Group|cyclic group]].
+From [[Order of Automorphism Group of Cyclic Group]]:
+:$\order {\Aut G} = \map \phi p$
+where $\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+The result follows from [[Euler Phi Function of Prime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Order 2 Matrices over Real Numbers form Ring with Unity}
+Tags: Rings with Unity
+
+\begin{theorem}
+Let $S$ denote the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] whose [[Definition:Matrix Entry|entries]] are the [[Definition:Real Number|set of real numbers]].
+Then $S$ forms a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity|unity]] is the [[Definition:Square Matrix|matrix]] $\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$.
+\end{theorem}
+
+\begin{proof}
+This is an instance of [[Ring of Square Matrices over Field is Ring with Unity]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Division Ring}
+Tags: Division Rings
+
+\begin{theorem}
+{{TFAE|def = Division Ring}}
+A '''division ring''' is a [[Definition:Ring with Unity|ring with unity]] $\struct {R, +, \circ}$ with the following properties:
+\end{theorem}
+
+\begin{proof}
+In the following, let:
+:$0_R$ denote the [[Definition:Ring Zero|zero]] of $R$
+:$1_R$ denote the [[Definition:Unity of Ring|unity]] of $R$
+:$R^*$ denote the [[Definition:Set|set]] of [[Definition:Element|elements]] of $R$ without $0_R$.
+=== $(1)$ implies $(2)$ ===
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring/Definition 1|Division Ring by definition 1]].
+Then by definition:
+:$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$
+That is, $x^{-1}$ is the (unique) [[Definition:Product Inverse|product inverse]] of $x$.
+Thus, by definition, $x$ is a [[Definition:Unit of Ring|unit]] of $R$.
+Thus $\struct {R, +, \circ}$ is a [[Definition:Division Ring/Definition 2|Division Ring by definition 2]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring/Definition 2|Division Ring by definition 2]].
+Then by definition:
+:$\forall x \in \R^*: x$ is a [[Definition:Unit of Ring|unit]].
+Thus, by definition, $x$ has a [[Definition:Product Inverse|product inverse]] $x^{-1}$.
+From [[Product Inverse in Ring is Unique]] it follows that:
+:$\forall x \in R^*: \exists! x^{-1} \in R^*: x^{-1} \circ x = x \circ x^{-1} = 1_R$
+Thus $\struct {R, +, \circ}$ is a [[Definition:Division Ring/Definition 1|Division Ring by definition 1]].
+{{qed|lemma}}
+=== $(2)$ is equivalent to $(3)$ ===
+By definition, a [[Definition:Unit of Ring|unit]] of $R$ is an [[Definition:Element|element]] of $R$ which has a [[Definition:Product Inverse|product inverse]].
+Also by definition, a [[Definition:Proper Element of Ring|proper element]] of $R$ is a non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] of $R$ which does not have a [[Definition:Product Inverse|product inverse]].
+Hence:
+:a [[Definition:Ring with Unity|ring with unity]] whose non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] are all [[Definition:Unit of Ring|units]]
+and:
+:a [[Definition:Ring with Unity|ring with unity]] whose non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] are all not [[Definition:Proper Element of Ring|proper elements]]
+are the same thing.
+{{qed}}
+[[Category:Division Rings]]
+f8ydh0swavqutk5yqin3q01r478aqoo
+\end{proof}<|endoftext|>
+\section{Product Formula for Norms on Non-zero Rationals}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $\Q_{\ne 0}$ be the [[Definition:Set|set]] of non-zero [[Definition:Rational Numbers|rational numbers]].
+Let  $\Bbb P$ denote the [[Definition:Set|set]] of [[Definition:Prime Number|prime numbers]].
+Let $a \in \Q_{\ne 0}$.
+Then the following [[Definition:Infinite Product|infinite product]] [[Definition:Convergent Real Sequence|converges]]:
+:$\size a \times \displaystyle \prod_{p \mathop \in \Bbb P}^{} \norm a_p = 1$
+where:
+:$\size {\,\cdot\,}$ is the [[Definition:Absolute Value|absolute value]] on $\Q$
+:$\norm {\,\cdot\,}_p$ is the [[Definition:P-adic Norm|$p$-adic norm]] on $\Q$ for [[Definition:Prime Number|prime number $p$]]
+\end{theorem}
+
+\begin{proof}
+=== [[Product Formula for Norms on Non-zero Rationals/Lemma|Lemma]] ===
+{{:Product Formula for Norms on Non-zero Rationals/Lemma}}{{qed|lemma}}
+Let $a = \dfrac b c$, where $b, c \in \Z_{\ne 0}$.
+From the [[Product Formula for Norms on Non-zero Rationals/Lemma|Lemma]], the following [[Definition:Infinite Product|infinite products]] [[Definition:Convergent Real Sequence|converge]]:
+:$\size b \displaystyle \times \prod_{p \mathop \in \Bbb P} \norm b_p = 1$
+:$\size c \displaystyle \times \prod_{p \mathop \in \Bbb P} \norm c_p = 1$
+From [[Quotient Rule for Real Sequences]], the following [[Definition:Infinite Product|infinite product]] [[Definition:Convergent Real Sequence|converges]]:
+:$\displaystyle \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P } \dfrac {\norm b_p} {\norm c_p} = \dfrac 1 1 = 1 $
+We have:
+{{begin-eqn}}
+{{eqn | l = \size a \times \mathop \prod_{p \mathop \in \Bbb P} \norm a_p
+      | r = \size {\dfrac b c} \times \prod_{p \mathop \in \Bbb P} \norm {\dfrac b c}_p
+}}
+{{eqn | r = \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P} \dfrac {\norm b_p} {\norm c_p}
+      | c = [[Properties of Norm on Division Ring/Norm of Quotient|Norm of Quotient]]
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring is Subring of Itself}
+Tags: Subrings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Then $R$ is a [[Definition:Subring|subring]] of itself.
+\end{theorem}
+
+\begin{proof}
+$R$ is a [[Definition:Ring (Abstract Algebra)|ring]] and $R \subseteq R$.
+It follows by definition that $R$ is a [[Definition:Subring|subring]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Ring is Subring of Ring}
+Tags: Subrings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Then the [[Definition:Null Ring|null ring]] is a [[Definition:Subring|subring]] of $R$.
+\end{theorem}
+
+\begin{proof}
+From [[Null Ring is Ring]], the [[Definition:Null Ring|null ring]] $\struct {\set {0_R}, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+We also have that $\set {0_R}$ is a [[Definition:Subset|subset]] of $R$
+Hence the result by definition of [[Definition:Subring|subring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers form Subring of Reals}
+Tags: Integers, Real Numbers, Subrings
+
+\begin{theorem}
+The [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$ forms a [[Definition:Subring|subring]] of the [[Definition:Field of Real Numbers|field of real numbers]].
+\end{theorem}
+
+\begin{proof}
+We have that the [[Definition:Integer|set of integers]] $\Z$ are a [[Definition:Subset|subset]] of the [[Definition:Real Numbers|real numbers]] $\R$.
+The [[Definition:Field of Real Numbers|field of real numbers]] is, [[Definition:A Fortiori|a fortiori]], also a [[Definition:Ring (Abstract Algebra)|ring]].
+Hence the result, by definition of [[Definition:Subring|subring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unity of Subring is not necessarily Unity of Ring}
+Tags: Subrings
+
+\begin{theorem}
+Let $\struct {S, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_S$.
+Let $\struct {T, + \circ}$ be a [[Definition:Subring|subring]] of $\struct {S, + \circ}$ whose [[Definition:Unity of Ring|unity]] is $1_T$.
+Then it is not necessarily the case that $1_T = 1_S$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {S, + \times}$ be the [[Definition:Ring (Abstract Algebra)|ring]] formed by the [[Definition:Set|set]] of [[Definition:Order of Square Matrix|order $2$]] [[Definition:Square Matrix|square matrices]] over the [[Definition:Real Number|real numbers]] $R$ under [[Definition:Matrix Entrywise Addition|(conventional) matrix addition]] and [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]].
+Let $T$ be the [[Definition:Subset|subset]] of $S$ consisting of the [[Definition:Square Matrix|matrices]] of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
+We have that:
+:$\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$
+and so $T$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Matrix Subtraction|subtraction]].
+From [[Subsemigroup/Examples/2x2 Matrices with One Non-Zero Entry|Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$]] we have that $\struct {T, \times}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \times}$.
+It follows from the [[Subring Test]] that $\struct {T, + \circ}$ be a [[Definition:Subring|subbring]] of $\struct {S, + \circ}$.
+From (some result somewhere) $\struct {S, \times}$ has an [[Definition:Identity Element|identity]] $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
+However, note that:
+{{begin-eqn}}
+{{eqn | l = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
+      | r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
+      | c = 
+}}
+{{end-eqn}}
+demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the [[Definition:Identity Element|identity]] of $\struct {T, \times}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero of Subfield is Zero of Field}
+Tags: Subfields, Zero of Subfield is Zero of Field
+
+\begin{theorem}
+The [[Definition:Field Zero|zero]] of $\struct {K, +, \times}$ is also $0$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both [[Definition:Ring (Abstract Algebra)|rings]].
+Thus $\struct {K, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {F, +, \times}$
+The result follows from [[Zero of Subring is Zero of Ring]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
+By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K, +}$ and $\struct {F, +}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
+So, by definition, $\struct {K, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {F, +}$.
+By [[Identity of Subgroup]], the [[Definition:Identity Element|identity]] of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unity of Subfield is Unity of Field}
+Tags: Subfields
+
+\begin{theorem}
+The [[Definition:Unity of Ring|unity]] of $\struct {K, +, \times}$ is also $1$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
+By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
+So $\struct {K^*, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {F^*, \times}$.
+By [[Identity of Subgroup]], the [[Definition:Identity Element|identity]] of $\struct {F^*, \times}$, which is $1$, is also the identity of $\struct {K^*, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero of Subring is Zero of Ring}
+Tags: Subrings
+
+\begin{theorem}
+The [[Definition:Ring Zero|zero]] of $\struct {S, +, \times}$ is also $0$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {S, +, \times}$ is a [[Definition:Subset|subset]] of $R$ which is a [[Definition:Ring (Abstract Algebra)|ring]].
+By definition of [[Definition:Ring (Abstract Algebra)|ring]], $\struct {S, +}$ and $\struct {R, +}$ are [[Definition:Group|groups]] such that $S \subseteq R$.
+So, by definition, $\struct {S, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {R, +}$.
+By [[Identity of Subgroup]], the [[Definition:Identity Element|identity]] of $\struct {S, +}$, which is $0$, is also the identity of $\struct {R, +}$.
+{{qed}}
+[[Category:Subrings]]
+eu7v5l65ix0xuk1vhigy84hewdr458h
+\end{proof}<|endoftext|>
+\section{Zero of Subfield is Zero of Field/Proof 1}
+Tags: Zero of Subfield is Zero of Field
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0$.
+Let $\struct {K, +, \times}$ be a [[Definition:Subfield|subfield]] of $\struct {F, +, \times}$.
+{{:Zero of Subfield is Zero of Field}}
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both [[Definition:Ring (Abstract Algebra)|rings]].
+Thus $\struct {K, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {F, +, \times}$
+The result follows from [[Zero of Subring is Zero of Ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero of Subfield is Zero of Field/Proof 2}
+Tags: Zero of Subfield is Zero of Field
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0$.
+Let $\struct {K, +, \times}$ be a [[Definition:Subfield|subfield]] of $\struct {F, +, \times}$.
+{{:Zero of Subfield is Zero of Field}}
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {K, +, \times}$ is a [[Definition:Subset|subset]] of $F$ which is a [[Definition:Field (Abstract Algebra)|field]].
+By definition of [[Definition:Field (Abstract Algebra)|field]], $\struct {K, +}$ and $\struct {F, +}$ are [[Definition:Group|groups]] such that $K \subseteq F$.
+So, by definition, $\struct {K, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {F, +}$.
+By [[Identity of Subgroup]], the [[Definition:Identity Element|identity]] of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p > 3$.
+ 
+Then:
+:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a [[Definition:Complete Normed Division Ring|complete normed division ring]].
+That is, there exists a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not [[Definition:Convergent Sequence (Normed Division Ring)|converge]] to a [[Definition:Limit of Sequence (Normed Division Ring)|limit]] in $\Q$.
+\end{theorem}
+
+\begin{proof}
+Let $p > 3$.
+Then there exists $a \in \Z: 1 < a < p-1$.
+Consider the [[Definition:Sequence|sequence]] $\sequence {x_n} \subseteq \Q$ where $x_n = a^{p^n}$ for some $a \in \Z: 1 < a < p-1$.
+Let $n \in \N$.
+Then:
+:$\norm {a^{p^{n + 1} } - a^{p^n} }_p = \norm {a^{p^n} (a^{p^n \left({p - 1}\right)} - 1) }_p$
+From the [[Euler's Theorem/Corollary 1|corollary to Euler's Theorem]]:
+:$a^{p^n \left({p - 1}\right)} - 1 \equiv 0 \pmod {p^n}$
+so:
+:$\norm {a^{p^n} \paren {a^{p^n \paren {p - 1} } - 1} }_p \le p^{-n} \xrightarrow {n \to \infty} 0$
+That is:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_{n + 1} - x_n}_p = 0$
+By [[Characterisation of Cauchy Sequence in Non-Archimedean Norm]]
+:$\sequence {x_n }$ is a [[Definition:Cauchy Sequence|cauchy sequence]] in $\struct {\Q, \norm {\,\cdot\,}_p }$.
+{{AimForCont}} $\sequence {x_n}$ [[Definition:Convergent Sequence|converges]] to some $x \in \Q$.
+That is:
+:$x = \displaystyle \lim_{n \mathop \to \infty} x_n$
+By [[Modulus of Limit/Normed Division Ring|Modulus of Limit on a Normed Division Ring]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n }_p = \norm {x }_p$
+Since $\forall n, p \nmid a^{p^n} = x_n$, then:
+:$ \norm {x_n }_p = 1$
+So:
+:$\norm x_p =  \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}_p = 1$ 
+By Axiom (N1) of a [[Definition:Norm on Division Ring|norm on a division ring]] then:
+:$x \ne 0$.
+Since:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \lim_{n \mathop \to \infty} x_n
+      | c = Definition of $x$
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} x_{n + 1}
+      | c = [[Limit of Subsequence equals Limit of Sequence]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \paren {x_n}^p
+      | c = Definition of $x_n$
+}}
+{{eqn | r = \paren {\lim_{n \mathop \to \infty} x_n}^p
+      | c = [[Combination Theorem for Sequences/Normed Division Ring/Product Rule|Product rule for Normed Division Rings]]
+}}
+{{eqn | r = x^p
+      | c = Definition of $x$
+}}
+{{end-eqn}}
+and $x \ne 0$ then:
+:$x^{p-1} = 1$
+So:
+:$x = 1$ or $x = -1$
+and so $a-x$ is an integer:
+:$0 < a - x < p$
+It follows that:
+:$p \nmid \paren {a - x}$
+and so:
+:$\norm {x - a}_p = 1$
+Since $x_n \to x$ as $n \to \infty$ then:
+:$\exists N: \forall n > N: \norm {x_n - x}_p < \norm {x - a}_p$
+That is:
+:$\exists N: \forall n > N: \norm {a^{p^n} - x}_p < \norm {x - a}_p$
+Let $n > N$:
+{{begin-eqn}}
+{{eqn | l = \norm {x - a}_p
+      | r = \norm {x - a^{p^n} + a^{p^n} - a}_p
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {x - a^{p^n} }_p, \norm {a^{p^n} - a}_p}
+      | c = [[P-adic Norm is Non-Archimedean Norm]]
+}}
+{{end-eqn}}
+As $\norm {x - a^{p^n} }_p < \norm {x - a}_p$:
+{{begin-eqn}}
+{{eqn | l = \norm {x - a}_p
+      | r = \norm {a^{p^n} - a}_p
+      | c = [[Three Points in Ultrametric Space have Two Equal Distances]] 
+}}
+{{eqn | r = \norm a_p \norm {a^{p^n - 1} - 1}_p
+      | c = Axiom (N2) of a [[Definition:Norm on Division Ring|norm on a division ring]]
+}}
+{{eqn | r = \norm {a^{p^n - 1} - 1}_p
+      | c = as $\norm a_p = 1$
+}}
+{{eqn | o = <
+      | r = 1
+      | c = [[Fermat's Little Theorem/Corollary 4|corollary 4 to Fermat's Little Theorem]]
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the earlier assertion that $\norm {x - a}_p = 1$.
+In conclusion:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]] that does not [[Definition:Convergent Sequence|converge]] in $\struct {\Q, \norm {\,\cdot\,}_p }$.
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for $p = 2$ or $3$.
+ 
+Then:
+:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a [[Definition:Complete Normed Division Ring|complete normed division ring]].
+That is, there exists a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not [[Definition:Convergent Sequence in Normed Division Ring|converge]] to a [[Definition:Limit of Sequence (Normed Division Ring)|limit]] in $\Q$.
+\end{theorem}
+
+\begin{proof}
+{{WIP}}
+\end{proof}<|endoftext|>
+\section{Element in Integral Domain is Divisor iff Principal Ideal is Superset}
+Tags: Integral Domains, Principal Ideals, Factorization
+
+\begin{theorem}
+:$x \divides y \iff \ideal y \subseteq \ideal x$
+\end{theorem}
+
+\begin{proof}
+Let that $x \divides y$.
+Then by definition of [[Definition:Divisor of Ring Element|divisor]]: 
+{{begin-eqn}}
+{{eqn | l = x \divides y
+      | o = \leadsto
+      | r = \exists t \in D: y = t x
+      | c = {{Defof|Divisor of Ring Element}}
+}}
+{{eqn | o = \leadsto
+      | r = y \in \ideal x
+      | c = {{Defof|Principal Ideal of Ring}}
+}}
+{{eqn | o = \leadsto
+      | r = \ideal y \subseteq \ideal x
+      | c = {{Defof|Principal Ideal of Ring}}: $\ideal y$ is the smallest ideal containing $y$
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = \ideal y \subseteq \ideal x
+      | o = \leadsto
+      | r = y \in \ideal x
+      | c = as $y \in \ideal y$
+}}
+{{eqn | o = \leadsto
+      | r = \exists t \in D: y = t x
+      | c = {{Defof|Principal Ideal of Ring}}
+}}
+{{eqn | o = \leadsto
+      | r = x \divides y
+      | c = {{Defof|Divisor of Ring Element}}
+}}
+{{end-eqn}}
+So:
+:$x \divides y \iff \ideal y \subseteq \ideal x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element in Integral Domain is Unit iff Principal Ideal is Whole Domain}
+Tags: Integral Domains, Principal Ideals
+
+\begin{theorem}
+:$x \in U_D \iff \ideal x = D$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x \in U_D
+      | o = \leadsto
+      | r = \exists u \in U_D: u \in \ideal x
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \ideal x = D
+      | c = [[Ideal of Unit is Whole Ring]]
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = \ideal x = D
+      | o = \leadsto
+      | r = 1_D \in \ideal x
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \exists t \in D: 1 = t x
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = u \in U_D
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Residue of Gamma Function}
+Tags: Gamma Function
+
+\begin{theorem}
+Let $\Gamma$ be the [[Definition:Gamma Function]]. 
+Let $n$ be a non-[[Definition:Negative Integer|negative integer]].
+Then: 
+:$\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$
+\end{theorem}
+
+\begin{proof}
+By [[Poles of Gamma Function]], $\Gamma$ has [[Definition:Simple Pole|simple poles]] at the non-positive integers, so $-n$ is a simple pole of $\Gamma$. 
+Then:
+{{begin-eqn}}
+{{eqn	| l = \Res \Gamma {-n}
+	| r = \lim_{z \mathop \to -n} \paren {z + n} \map \Gamma z
+	| c = [[Residue at Simple Pole]]
+}}
+{{eqn	| r = \lim_{z \mathop \to -n} \paren {z + n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n} }
+}}
+{{eqn	| r = \lim_{z \mathop \to -n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n - 1} }
+}}
+{{eqn	| r = \frac {\map \Gamma 1} {-n \paren {-n + 1} \ldots \paren {-1} }
+}}
+{{eqn	| r = \frac 1 {\paren {-1}^n \paren {n \paren {n - 1} \ldots \paren 1} }
+	| c = $\map \Gamma 1 = 1$
+}}
+{{eqn	| r = \frac 1 {\paren {-1}^{n} n!}
+	| c = {{Defof|Factorial}}
+}}
+{{eqn	| r = \frac {\paren {-1}^n} {n!}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Gamma Function]]
+33l04x5xfrse8mv2w8ojn26725r27l2
+\end{proof}<|endoftext|>
+\section{Periodic Element is Multiple of Period}
+Tags: Periodic Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Periodic Function|real periodic function]] with [[Definition:Period of Function|period]] $P$.
+Let $L$ be a [[Definition:Periodic Element|periodic element]] of $f$.
+Then $P \divides L$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $P \nmid L$.
+Then by the [[Division Theorem/Real Number Index|Division Theorem]] we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$.
+And so:
+{{begin-eqn}}
+{{eqn | l = \map f {x + L}
+      | r = \map f {x + \paren {q P + r} }
+}}
+{{eqn | r = \map f {\paren {x + r} + q P}
+}}
+{{eqn | r = \map f {x + r}
+      | c = [[General Periodicity Property]]
+}}
+{{eqn | r = \map f x
+      | c = {{Defof|Periodic Element}}
+}}
+{{end-eqn}}
+But then $r$ is a [[Definition:Periodic Element|periodic element]] of $f$ that is less than $P$.
+Therefore $P$ cannot be the [[Definition:Period of Function|period]] of $f$.
+The result follows from [[Proof by Contradiction]].
+{{qed}}
+[[Category:Periodic Functions]]
+8ozj1819qzmmkun1rf2fdo4ry2m0wqc
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Associate in Integral Domain}
+Tags: Integral Domains, Associates, Equivalence of Definitions of Associate in Integral Domain
+
+\begin{theorem}
+{{TFAE|def = Associate in Integral Domain|context = Integral Domain|view = Associate}}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
+Let $x, y \in D$.
+\end{theorem}
+
+\begin{proof}
+=== [[Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2|$(1)$ is Equivalent to $(2)$]] ===
+{{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2}}
+=== [[Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3|$(1)$ is Equivalent to $(3)$]] ===
+{{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 3}}
+[[Category:Integral Domains]]
+[[Category:Associates]]
+[[Category:Equivalence of Definitions of Associate in Integral Domain]]
+lbb4s0ifmkn6vtxm5u7br1j16jo7mv3
+\end{proof}<|endoftext|>
+\section{Finite Set of Elements in Principal Ideal Domain has GCD}
+Tags: Principal Ideal Domains, Greatest Common Divisor
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $a_1, a_2, \dotsc, a_n$ be non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
+Then $a_1, a_2, \dotsc, a_n$ all have a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]].
+\end{theorem}
+
+\begin{proof}
+Let $0_D$ and $1_D$ be the [[Definition:Ring Zero|zero]] and [[Definition:Unity of Ring|unity]] respectively of $D$.
+Let $J$ be the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
+From [[Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal]]:
+:$J = \ideal x$
+for some $x \in D$, where $\ideal x$ denotes the [[Definition:Principal Ideal of Ring|principal ideal]] [[Definition:Generator of Ideal|generated]] by $x$.
+We have that each $a_i$ can be expressed as a [[Definition:Linear Combination|linear combination]] of $\set {a_1, a_2, \dotsc, a_n}$:
+:$a_i = 0_D a_1 + 0_d a_2 + \dotsb + 1_D a_i + \dotsb + 0_D a_n$
+Thus:
+:$\forall i \in \set {0, 1, \dotsc, n}: a_i \in J$
+and so by definition of $J$:
+:$\forall i \in \set {0, 1, \dotsc, n}: a_i = t_i x$
+for some $t_i \in D$.
+Thus $x$ is a [[Definition:Common Divisor of Ring Elements|common divisor]] of $a_1, a_2, \dotsc, a_n$.
+As $x \in \ideal x = J$, we have:
+:$x = c_1 a_1 + c_2 a_2 + \dotsb + c_n a_n$
+for some $c_1, c_2, \dotsc, c_n \in D$.
+Thus every [[Definition:Common Divisor of Ring Elements|common divisor]] of $a_1, a_2, \dotsc, a_n$ also is a [[Definition:Divisor of Ring Element|divisor]] of $x$.
+Thus $x$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $a_1, a_2, \dotsc, a_n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal}
+Tags: Principal Ideal Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $a_1, a_2, \dotsc, a_n$ be non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
+Let $J$ be the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] in $D$ of $\set {a_1, a_2, \dotsc, a_n}$
+Then for some $x \in D$:
+:$J = \ideal x$
+where $\ideal x$ denotes the [[Definition:Principal Ideal of Ring|principal ideal]] [[Definition:Generator of Ideal|generated]] by $x$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Unity of Ring|unity]] of $D$ be $1_D$.
+By definition of [[Definition:Principal Ideal of Ring|principal ideal]]:
+:$\ideal a = \displaystyle \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
+Let $x, y \in J$.
+By definition of [[Definition:Linear Combination|linear combination]]:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \sum_{i \mathop = 1}^n r_i \circ a_i
+      | c = for some $n \in \N$ and for some $r_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
+}}
+{{eqn | r = r_1 \circ a_1 + r_2 \circ a_2 + \dotsb + r_n \circ a_n
+      | c = for some $r_1, r_2, \dotsc, r_n \in D$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \sum_{i \mathop = 1}^n s_i \circ a_i
+      | c = for some $n \in \N$ and for some $s_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
+}}
+{{eqn | r = s_1 \circ a_1 + s_2 \circ a_2 + \dotsb + s_n \circ a_n
+      | c = for some $s_1, s_2, \dotsc, s_n \in R$
+}}
+{{eqn | ll= \leadsto
+      | l = -y
+      | r = -\sum_{i \mathop = 1}^n s_i \circ a_i
+      | c = 
+}}
+{{eqn | r = -1_D \times \sum_{i \mathop = 1}^n s_i \circ a_i
+      | c = [[Product with Ring Negative/Corollary|Product with Ring Negative: Corollary]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {-1_D} \times \paren {s_i \circ a_i}
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {-\paren {s_i \circ a_i} }
+      | c = [[Product with Ring Negative/Corollary|Product with Ring Negative: Corollary]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
+      | c = [[Product with Ring Negative]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = x + \paren {-y}
+      | r = \sum_{i \mathop = 1}^n r_i \circ a_i + \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + s_i \circ \paren {-a_i} }
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-\paren {s_i \circ a_i} } }
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-s_i} \circ a_i}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i + \paren {-s_i} } \circ a_i
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n t_i \circ a_i
+      | c = where $t_i - r_1 + \paren {-s_i}$
+}}
+{{eqn | o = \in
+      | r = J
+      | c = as $t_i \in D$
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = x \circ y
+      | r = \paren {\sum_{i \mathop = 1}^n r_i \circ a_i} \circ \paren {\sum_{i \mathop = 1}^n s_i \circ \a_i }
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {t_i \circ a_i}
+      | c = where $t_i \in D$ for $i \in \set {1, 2, \dotsc, n}$
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \paren {a_i \circ t_i}
+      | c = as $\circ$ is [[Definition:Commutative Operation|commutative]] in an [[Definition:Integral Domain|integral domain]]
+}}
+{{end-eqn}}
+{{explain|There exists (or ought to) some convolution result which proves the above -- I just haven't found it yet.}}
+Thus by the [[Test for Ideal]], $J$ is an [[Definition:Ideal of Ring|ideal]] of $D$.
+As $D$ is a [[Definition:Principal Ideal Domain|principal ideal domain]], it follows that $J$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
+Thus by definition of [[Definition:Principal Ideal of Ring|principal ideal]]:
+:$J = \ideal x$
+for some $x \in D$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Common Divisors in Principal Ideal Domain are Associates}
+Tags: Principal Ideal Domains, Greatest Common Divisor, Associates
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a [[Definition:Set|set]] of non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
+Let $y_1$ and $y_2$ be [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of $S$.
+Then $y_1$ and $y_2$ are [[Definition:Associate in Integral Domain|associates]].
+\end{theorem}
+
+\begin{proof}
+From [[Finite Set of Elements in Principal Ideal Domain has GCD]] we have that at least one such [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] exists.
+So, let $y_1$ and $y_2$ be [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisors]] of $S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | o = \divides
+      | r = y_2
+      | c = as $y_2$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]]
+}}
+{{eqn | l = y_2
+      | o = \divides
+      | r = y_1
+      | c = as $y_1$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]]
+}}
+{{end-eqn}}
+Thus we have:
+:$y_1 \divides y_2$ and $y_2 \divides y_1$
+where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
+Hence the result, by definition of [[Definition:Associate in Integral Domain|associates]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Common Divisor in Principal Ideal Domain is Expressible as Linear Combination}
+Tags: Principal Ideal Domains, Greatest Common Divisor
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a [[Definition:Set|set]] of non-[[Definition:Ring Zero|zero]] [[Definition:Element|elements]] of $D$.
+Let $y$ be a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $S$.
+Then $y$ is expressible in the form:
+:$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$
+where $d_1, d_2, \dotsc, d_n \in D$.
+\end{theorem}
+
+\begin{proof}
+From [[Finite Set of Elements in Principal Ideal Domain has GCD]] we have that at least one such [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] exists.
+So, let $y$ be a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $S$.
+Let $J$ be the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
+From [[Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal]]:
+:$J = \ideal x$
+for some $x \in D$, where $\ideal x$ denotes the [[Definition:Principal Ideal of Ring|principal ideal]] [[Definition:Generator of Ideal|generated]] by $x$.
+From [[Finite Set of Elements in Principal Ideal Domain has GCD]], $x$ is a [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $S$.
+From [[Greatest Common Divisors in Principal Ideal Domain are Associates]], $y$ is an [[Definition:Associate in Integral Domain|associate]] of $x$.
+By definition of [[Definition:Associate in Integral Domain|associate]]:
+:$\ideal y = \ideal x$
+Therefore:
+:$y \in J$
+and so by definition, $y$ is expressible in the form:
+:$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$
+where $d_1, d_2, \dotsc, d_n \in D$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complete Factorizations of Proper Element in Principal Ideal Domain are Equivalent}
+Tags: Factorization, Principal Ideal Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Let $x \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
+Let there be two [[Definition:Complete Factorization|complete factorizations]] of $x$:
+:$x = u_y \circ y_1 \circ y_2 \circ \cdots \circ y_m = F_1$
+:$x = u_z \circ z_1 \circ z_2 \circ \cdots \circ z_n = F_2$
+Then $F_1$ and $F_2$ are [[Definition:Equivalent Factorizations|equivalent]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Whitelaw leaves this unresolved at the end of $\S 62$ as an exercise for the student. I haven't read ahead that far, but it may be proved in the exercises. Will return to this later.}}#
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Poisson Distribution}
+Tags: Moment Generating Functions, Poisson Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]] for some $\lambda \in \R_{> 0}$.
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Poisson Distribution|Poisson distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
+:$\map \Pr {X = n} = \dfrac {\lambda^n e^{-\lambda} } {n!}$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^\infty \map \Pr {X = n} e^{t n}$
+So: 
+{{begin-eqn}}
+{{eqn	| l = \map {M_X} t
+	| r = \sum_{n \mathop = 0}^\infty \frac {\lambda^n e^{-\lambda} } {n!} e^{t n}
+}}
+{{eqn	| r = e^{-\lambda} \sum_{n \mathop = 0}^\infty \frac {\paren {\lambda e^t}^n} {n!}
+}}
+{{eqn	| r = e^{-\lambda} e^{\lambda e^t}
+	| c = [[Power Series Expansion for Exponential Function]]
+}}
+{{eqn	| r = e^{\lambda \paren {e^t - 1} }
+	| c = [[Exponential of Sum]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Poisson Distribution]]
+cqmgdzhn5lv8husq2qyzlg4f4qvlsvw
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Binomial Distribution}
+Tags: Moment Generating Functions, Binomial Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Binomial Distribution|binomial distribution with parameters $n$ and $p$]] for some $n \in \N$ and $0 \le p \le 1$:
+:$X \sim \Binomial n p$
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\map {M_X} t = \paren {1 - p + p e^t}^n$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition: Binomial Distribution|Binomial distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
+:$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^n \map \Pr {X = k} e^{t k}$
+So: 
+{{begin-eqn}}
+{{eqn	| l = \map {M_X} t 
+	| r = \sum_{k \mathop = 0}^n \binom n k p^k \paren {1 - p}^{n - k} e^{t k}
+}}
+{{eqn	| r = \sum_{k \mathop = 0}^n \binom n k \paren {p e^t}^k \paren {1 - p}^{n - k}
+}}
+{{eqn	| r = \paren {1 - p + p e^t}^n
+	| c = [[Binomial Theorem]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Binomial Distribution]]
+09me4821a56apv712qyp0ij2zps2dww
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Exponential Distribution}
+Tags: Moment Generating Functions, Exponential Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with an [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$ for some $\beta \in \R_{> 0}$.
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\displaystyle \map {M_X} t = \frac 1 {1 - \beta t}$
+for $t < \dfrac 1 \beta$, and is undefined otherwise.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Exponential Distribution|Exponential distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
+:$\displaystyle \map {f_X} x = \frac 1 \beta e^{-\frac x \beta}$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \int_0^\infty e^{t x} \map {f_X} x \rd x$
+Then: 
+{{begin-eqn}}
+{{eqn	| l = \map {M_X} t
+	| r = \frac 1 \beta \int_0^\infty e^{x \paren {-\frac 1 \beta + t} } \rd x
+	| c = [[Exponential of Sum]]
+}}
+{{eqn	| r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty
+	| c = [[Primitive of Exponential Function]]
+}}
+{{end-eqn}}
+Note that if $t > \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to \infty$ as $x \to \infty$ by [[Exponential Tends to Zero and Infinity]], so the integral diverges in this case. 
+If $t = \dfrac 1 \beta$ then the integrand is identically $1$, so the integral similarly diverges in this case. 
+If $t < \dfrac 1 \beta$, then $\displaystyle e^{x \paren {-\frac 1 \beta + t} } \to 0$ as $x \to \infty$ from [[Exponential Tends to Zero and Infinity]], so the integral converges in this case. 
+Therefore, the function is only well defined for $t < \dfrac 1 \beta$.
+Proceeding:
+{{begin-eqn}}
+{{eqn	| l = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \sqbrk {e^{x \paren {-\frac 1 \beta + t} } }_0^\infty
+	| r = \frac 1 {\beta \paren {-\frac 1 \beta + t} } \paren {0 - 1}
+	| c = [[Exponential Tends to Zero and Infinity]], [[Exponential of Zero]]
+}}
+{{eqn	| r = \frac 1 {\beta \paren {\frac 1 \beta - t} }
+}}
+{{eqn	| r = \frac 1 {1 - \beta t}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Exponential Distribution]]
+9omebrbdg9jyyp53lp3kisnerl3psqr
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Geometric Distribution}
+Tags: Moment Generating Functions, Geometric Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Geometric Distribution|geometric distribution with parameter $p$]] for some $0 < p \le 1$. 
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\displaystyle \map {M_X} t = \frac p {1 - \paren {1 - p} e^t}$
+for $t < -\map \ln {1 - p}$, and is undefined otherwise.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Geometric Distribution|geometric distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
+:$\displaystyle \map \Pr {X = k} = \paren {1 - p}^k p$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}$
+So:
+{{begin-eqn}}
+{{eqn	| l = \sum_{k \mathop = 0}^\infty \map \Pr {X = k} e^{k t}
+	| r = \sum_{k \mathop = 0}^\infty \paren {1 - p}^k p e^{k t}
+}}
+{{eqn	| r = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k
+}}
+{{end-eqn}}
+By [[Sum of Infinite Geometric Sequence]], for this sum to be convergent we must have: 
+:$\displaystyle \size {\paren {1 - p} e^t} < 1$
+In the case $p = 1$, this demand is satisfied immediately regardless of $t$.
+Otherwise, as both $e^t$ and $1 - p$ are positive: 
+:$\displaystyle e^t < \frac 1 {1 - p}$
+So, by [[Logarithm of Power]]: 
+:$\displaystyle t < -\map \ln {1 - p}$
+is the range of $t$ for which the expectation is well-defined.
+Now applying [[Sum of Infinite Geometric Sequence]], we have for this range of $t$: 
+:$\displaystyle \map {M_X} t = p \sum_{k \mathop = 0}^\infty \paren {\paren {1 - p} e^t}^k = \frac p {1 - \paren {1 - p} e^t}$
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Geometric Distribution]]
+j3aisrqpplynhjy8tsn3vgg9upy2uij
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Discrete Uniform Distribution}
+Tags: Moment Generating Functions, Discrete Uniform Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Discrete Uniform Distribution|discrete uniform distribution with parameter $n$]] for some $n \in \N$.
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\map {M_X} t = \dfrac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Discrete Uniform Distribution|Discrete Uniform distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
+:$\map \Pr {X = N} = \dfrac 1 n$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{N \mathop = 1}^n \map \Pr {X = N} e^{N t}$
+So: 
+{{begin-eqn}}
+{{eqn	| l = \map {M_X} t
+	| r = \frac 1 n \sum_{N \mathop = 1}^n \paren {e^t}^N
+}}
+{{eqn	| r = \frac {e^t} n \sum_{N \mathop = 0}^{n - 1} \paren {e^t}^N
+}}
+{{eqn	| r = \frac {e^t \paren {1 - e^{n t} } } {n \paren {1 - e^t} }
+	| c = [[Sum of Geometric Sequence]] with $r = e^t$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Discrete Uniform Distribution]]
+dnfdu3348s4qk7brcq03pglci688tqw
+\end{proof}<|endoftext|>
+\section{Maximal Ideal iff Quotient Ring is Field/Proof 1/Maximal Ideal implies Quotient Ring is Field}
+Tags: Maximal Ideal iff Quotient Ring is Field
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+Then the [[Definition:Quotient Ring|quotient ring]] $R / J$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+Let $J$ be a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+Because $J \subset R$, it follows from [[Quotient Ring of Commutative Ring is Commutative]] and [[Quotient Ring of Ring with Unity is Ring with Unity]] that $R / J$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+We now need to prove that every non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] of $\struct {R / J, +,  \circ}$ has a [[Definition:Product Inverse|product inverse]] in $R / J$.
+Let $x \in R$ such that $x + J \ne J$, that is: $x \notin J$.
+Thus $x + J \in R / J$ is not the [[Definition:Ring Zero|zero element]] of $R / J$.
+Take $K \subseteq R$ such that:
+:$K = \set {j + r \circ x: j \in J, r \in R}$
+that is, the [[Definition:Subset|subset]] of $R$ which can be expressed as a sum of an [[Definition:Element|element]] of $J$ and a [[Definition:Ring Product|product]] in $R$ of $x$.
+Now $0_R \in K$ as $0_R \in J$ and $0_R \in R$, giving $0_R + 0_R \circ x = 0_R$.
+So:
+:$(1): \quad K \ne \O$
+Now let $g, h \in K$.
+That is:
+:$g = j_1 + r_1 \circ x, h = j_2 + r_2 \circ x$
+Then:
+:$-h = -j_2 + \paren {-r_2} \circ x$
+But $j_1 - j_2 \in J$ from [[Test for Ideal]].
+Similarly $-r_2 \in R$.
+So $-h \in K$ and we have:
+:$(2) \quad g + \paren {-h} = \paren {j_1 - j_2} + \paren {r_1 - r_2} \circ x$
+Now consider $g \in J, y \in R$.
+Then:
+:$g \circ y = \paren {j_1 + r_1 \circ x} \circ y = \paren {j_1 \circ y} + \paren {r_1 \circ y} \circ x$
+which is valid by the fact that $R$ is a [[Definition:Commutative Ring|commutative ring]].
+But as $J$ is an [[Definition:Ideal of Ring|ideal]]:
+:$\paren {j_1 \circ y} \in J$
+while $r_1 \circ y \in R$.
+Thus:
+:$(3) \quad g \circ y \in K$
+and similarly:
+:$(3) \quad y \circ g \in K$
+So [[Test for Ideal]] can be applied to statements $(1)$ to $(3)$, and it is seen that $K$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+Now:
+{{begin-eqn}}
+{{eqn | l = j
+      | o = \in
+      | r = J
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = j + 0_R \circ x
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = j
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = J
+      | o = \subseteq
+      | r = K
+      | c = 
+}}
+{{end-eqn}}
+and because $x = 0_R + 1_R \circ x$ (remember $0_R \in J$), then $x \in K$ too.
+So, because $x \notin J$, $K$ is an [[Definition:Ideal of Ring|ideal]] such that $J \subset K \subseteq R$.
+Because $J$ is a [[Definition:Maximal Ideal|maximal ideal]], then $K = R$.
+Thus $1_R \in K$ and thus:
+:$\exists j_0 \in J, s \in R: 1_R = j_0 + s \circ x$
+So:
+:$1_R + \paren {-s \circ x} = j_0 \in J$
+Hence:
+:$1_R + J = s \circ x + J = \paren {s + J} \circ \paren {x + J}$
+So in the [[Definition:Commutative Ring|commutative ring]] $\struct {R / J, +,  \circ}$, the [[Definition:Product Inverse|product inverse]] of $x + J$ is $s + J$.
+The result follows.
+\end{proof}<|endoftext|>
+\section{Maximal Ideal iff Quotient Ring is Field/Proof 1/Quotient Ring is Field implies Ideal is Maximal}
+Tags: Maximal Ideal iff Quotient Ring is Field
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let the [[Definition:Quotient Ring|quotient ring]] $R / J$ be a [[Definition:Field (Abstract Algebra)|field]].
+Then $J$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+\end{theorem}
+
+\begin{proof}
+Let $R / J$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $K$ be a [[Definition:Left Ideal|left ideal]] of $R$ such that $J \subsetneq K \subseteq R$.
+We have that $J$ is the [[Definition:Ring Zero|zero]] of $R / J$.
+Let $x \in K \setminus J$.
+Because $x \notin J$ then $x + J \ne J$.
+Because $R / J$ is a [[Definition:Field (Abstract Algebra)|field]] then $x + J \in R / J$ has a [[Definition:Product Inverse|product inverse]], say $s + J$.
+Hence:
+:$1_R + J = \paren {s + J} \circ \paren {x + J} = \paren {s \circ x } + J$
+By [[Left Cosets are Equal iff Product with Inverse in Subgroup]]:
+:$1_R - s \circ x \in J \subsetneq K$
+By the definition of an [[Definition:Ideal of Ring|ideal]]:
+:$x \in K, s \in R \implies s \circ x \in K$
+:$1_R - s \circ x \in K, s \circ x \in K \implies \paren {1_R - s \circ x} + \paren {s \circ x} = 1_R \in K$
+:$1_R \in K \implies \forall y \in R:  y \circ 1_R = y \in K$
+Hence $K = R$.
+The result follows.
+\end{proof}<|endoftext|>
+\section{Ring of Polynomial Forms is not necessarily Isomorphic to Ring of Polynomial Functions}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $D$ be an [[Definition:Integral Domain|integral domain]].
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] in $X$ over $D$.
+Let $\map P D$ be the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] over $D$.
+Then it is not necessarily the case that $D \sqbrk X$ is [[Definition:Ring Isomorphism|isomorphic]] with $\map P D$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Consider the [[Definition:Integral Domain|integral domain]] $\struct {\Z_2, +, \times}$.
+From [[Ring of Integers Modulo Prime is Integral Domain]], it is seen that $\struct {\Z_2, +, \times}$ is indeed an [[Definition:Integral Domain|integral domain]].
+Consider the [[Definition:Ring of Polynomials in Ring Element|ring of polynomial forms]] $\Z_2 \sqbrk X$.
+This is an [[Definition:Infinite Set|infinite]] [[Definition:Ring (Abstract Algebra)|ring]], as it can be seen that $S \subseteq \Z_2 \sqbrk X$ where:
+:$S = \set {1, X, X^2, X^3, \dotsc}$
+But the [[Definition:Ring of Polynomial Functions|ring of polynomial functions]] $\map P D$ is [[Definition:Finite Set|finite]], as:
+:$\map P D \subseteq \Z_2^{Z_2}$
+where $\Z_2^{Z_2}$ is the [[Definition:Set|set]] of all [[Definition:Mapping|mappings]] from $\Z_2$ to $\Z_2$, and has $4$ [[Definition:Element|elements]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Moment Generating Function of Bernoulli Distribution}
+Tags: Moment Generating Functions, Bernoulli Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]] for some $0 \le p \le 1$. 
+Then the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\map {M_X} t = q + p e^t$
+where $q = 1 - p$.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Bernoulli Distribution|Bernoulli distribution]], $X$ has [[Definition:Probability Mass Function|probability mass function]]:
+:$\map \Pr {X = n} = \begin{cases}
+q & : n = 0 \\
+p & : n = 1 \\
+0 & : n \notin \set {0, 1} \\
+\end{cases}$
+From the definition of a [[Definition:Moment Generating Function|moment generating function]]:
+:$\displaystyle \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^1 \map \Pr {X = n} e^{t n}$
+So:
+{{begin-eqn}}
+{{eqn	| l = \map {M_X} t
+	| r = \map \Pr {X = 0} e^0 + \map \Pr {X = 1} e^t
+}}
+{{eqn	| r = q + p e^t
+	| c = [[Exponential of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Moment Generating Functions]]
+[[Category:Bernoulli Distribution]]
+5ivqydxhdfwngfzylmj8i7nvjxh7eef
+\end{proof}<|endoftext|>
+\section{Field of Quotients of Ring of Polynomial Forms on Reals that yields Complex Numbers}
+Tags: Polynomial Theory, Fields of Quotients, Real Numbers, Complex Numbers
+
+\begin{theorem}
+Let $\struct {\R, +, \times}$ denote the [[Definition:Field of Real Numbers|field of real numbers]].
+Let $X$ be [[Definition:Transcendental over Field|transcendental over $\R$]].
+Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials|ring of polynomials]] in $X$ over $F$.
+Consider the [[Definition:Field of Quotients|field of quotients]]:
+:$\R \sqbrk X / \ideal p$
+where:
+:$p = X^2 + 1$
+:$\ideal p$ denotes the [[Definition:Ideal of Ring|ideal]] [[Definition:Generator of Ideal|generated]] by $p$.
+Then $\R \sqbrk X / \ideal p$ is the [[Definition:Field of Complex Numbers|field of complex numbers]].
+\end{theorem}
+
+\begin{proof}
+It is taken as read that $X^2 + 1$ is [[Definition:Irreducible Element of Ring|irreducible]] in $\R \sqbrk X$.
+Hence by [[Polynomial Forms over Field form Principal Ideal Domain/Corollary 1|Polynomial Forms over Field form Principal Ideal Domain: Corollary 1]], $\R \sqbrk X / \ideal p$ is indeed a [[Definition:Field (Abstract Algebra)|field]].
+Let $\nu$ be the [[Definition:Quotient Ring Epimorphism|quotient epimorphism]] from $\R \sqbrk X$ onto $\R \sqbrk X / \ideal p$.
+From [[Quotient Ring Epimorphism is Epimorphism]]:
+:$\map \ker {\map \nu \R} = \R \cap \ideal p = \set 0$
+So $\map \nu \R$ is a [[Definition:Ring Monomorphism|monomorphism]] from $\R \sqbrk X$ to $\R \sqbrk X / \ideal p$.
+Thus $\map \nu \R$ is an [[Definition:Isomorphic Copy|isomorphic copy]] of $\R$ inside $\R \sqbrk X / \ideal p$.
+We identify this [[Definition:Isomorphic Copy|isomorphic copy]] of $\R$ with $\R$ itself, ignoring the difference between $\map \nu x$ and $x$ when $x = \R$.
+Hence:
+:$(1): \quad \R \subseteq \R \sqbrk X / \ideal p$
+Let $f \in \R \sqbrk X$ be arbitrary.
+By [[Division Theorem for Polynomial Forms over Field]]:
+:$\exists q, r \in \R \sqbrk X: f = q p + r$
+where $r =a + b X$ for some $a, b \in \R$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \nu f
+      | r = \map \nu q \, \map \nu p + \map \nu r
+      | c = as $\nu$ is a [[Definition:Ring Homomorphism|ring honomorphism]]
+}}
+{{eqn | r = \map \nu r
+      | c = as $p \in \map \ker \nu$
+}}
+{{eqn | r = \map \nu a + \map \nu b \, \map \nu X
+      | c = 
+}}
+{{eqn | r = a + b \, \map \nu X
+      | c = as we have identified $\map \nu \R$ with $\R$
+}}
+{{eqn | r = a + b i
+      | c = where $i := \map \nu X$
+}}
+{{end-eqn}}
+As $\nu$ is an [[Definition:Ring Epimorphism|epimorphism]], it is [[Definition:A Fortiori|a fortiori]] [[Definition:Surjection|surjection]].
+Hence:
+:$(2): \quad$ Every [[Definition:Element|element]] $w$ of $\R \sqbrk X / \ideal p$ can be expressed in the form:
+::$w = a + b i$
+:for some $a, b \in \R$.
+Because $X^2 + 1 \in \ker \nu$, we have:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \map \nu {X^2 + 1}
+      | c = 
+}}
+{{eqn | r = \paren {\map \nu X}^2 + \map \nu 1
+      | c = 
+}}
+{{eqn | r = i^2 + 1
+      | c = 
+}}
+{{end-eqn}}
+Hence we have that:
+:$(3): \quad$ In $\R \sqbrk X / \ideal p$, $i^2 = -1$
+Thus $(2)$ can be improved to:
+:$(4): \quad$ Every [[Definition:Element|element]] $w$ of $\R \sqbrk X / \ideal p$ can be expressed '''[[Definition:Unique|uniquely]]''' in the form:
+::$w = a + b i$
+:for some $a, b \in \R$.
+From $(1)$, $(3)$ and $(4)$, the [[Definition:Field (Abstract Algebra)|field]] $\R \sqbrk X / \ideal p$ is recognised as the [[Definition:Field of Complex Numbers|field of complex numbers]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Double of Antiperiodic Element is Periodic}
+Tags: Antiperiodic Functions
+
+\begin{theorem}
+Let $f: X \to X$ be a [[Definition:Function|function]], where $X$ is either $\R$ or $\C$.
+Let $L \in X_{\ne 0}$ be an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
+Then $2L$ is a [[Definition:Periodic Element|periodic element]] of $f$.
+In other words, every [[Definition:Antiperiodic Function|anti-periodic function]] is also [[Definition:Periodic Function|periodic]].
+\end{theorem}
+
+\begin{proof}
+Let $X = \C$.
+By [[Non-Zero Complex Numbers Closed under Multiplication]] we have that $2 L \in X_{\ne 0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {x + 2L}
+      | r = \map f {x + \paren {L + L} }
+}}
+{{eqn | r = \map f {\paren {x + L} + L}
+      | c = [[Complex Addition is Associative]]
+}}
+{{eqn | r = -\map f {x + L}
+}}
+{{eqn | r = \map f x
+      | c = [[Negative of Negative Complex Number]]
+}}
+{{end-eqn}}
+The proof for when $X = \R$ is nearly identical.
+{{qed}}
+[[Category:Antiperiodic Functions]]
+s2rr55kih3dzer6rke4k22dwmvhp3sz
+\end{proof}<|endoftext|>
+\section{Idempotent Ring has Characteristic Two/Corollary}
+Tags: Idempotent Rings
+
+\begin{theorem}
+:$\forall x \in R: -x = x$
+\end{theorem}
+
+\begin{proof}
+Let $0_R$ denote the [[Definition:Ring Zero|zero]] of $R$.
+Let $x \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x + x
+      | r = 0_R
+      | c = [[Idempotent Ring has Characteristic Two]]
+}}
+{{eqn | ll= \leadsto
+      | l = -x + x + x
+      | r = -x + 0_R
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = -x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Skewness of Gaussian Distribution}
+Tags: Skewness, Gaussian Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with a [[Definition:Gaussian Distribution|Gaussian distribution with parameters $\mu$ and $\sigma^2$]] for some $\mu \in \R$ and $\sigma \in \R_{> 0}$. 
+Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is equal to $0$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Skewness|skewness]]: 
+:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
+From the definition of the [[Definition:Gaussian Distribution|Gaussian distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
+:$\map {f_X} x = \dfrac 1 {\sigma \sqrt{2 \pi} } \, \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} }$
+So, from [[Expectation of Function of Continuous Random Variable]]:
+:$\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac {x - \mu} \sigma}^3 \map \exp {-\dfrac { \paren {x - \mu}^2} {2 \sigma^2} } \rd x$
+Making a [[Integration by Substitution|substitution]] of $u = x - \mu$:
+:$\displaystyle \gamma_1 = \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u$
+We have that: 
+:$\paren {-u}^3 \map \exp {-\dfrac {\paren {-u}^2} {2 \sigma^2} } = -u^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} }$
+So we can see that the integrand is [[Definition:Odd Function|odd]].
+So, by [[Definite Integral of Odd Function]]:
+:$\displaystyle \dfrac 1 {\sigma \sqrt{2 \pi} } \int_{-\infty}^\infty \paren {\dfrac u \sigma}^3 \map \exp {-\dfrac {u^2} {2 \sigma^2} } \rd u = 0$
+giving: 
+:$\gamma_1 = 0$ 
+{{qed}}
+[[Category:Skewness]]
+[[Category:Gaussian Distribution]]
+i7arvorvfenqsg3oqny9d8bnl36n5ql
+\end{proof}<|endoftext|>
+\section{Inverse of Unit in Centralizer of Ring is in Centralizer}
+Tags: Ring Theory, Subrings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $S$ be a [[Definition:Subset|subset]] of $R$.
+Let $\map {C_R} S$ denote the [[Definition:Centralizer of Ring Subset|centralizer]] of $S$ in $R$
+Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
+Then:
+:$u \in \map {C_R} S \implies u^{-1} \in \map {C_R} S$
+\end{theorem}
+
+\begin{proof}
+Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
+Let $u \in \map {C_R} S$.
+Then from [[Commutation with Inverse in Monoid]]:
+:$u^{-1} \in \map {C_R} S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Central Unit of Ring is in Center}
+Tags: Ring Theory, Subrings
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\map Z R$ denote the [[Definition:Center of Ring|center]] of $R$.
+Let $u \in R$ be a [[Definition:Unit of Ring|unit]] of $R$.
+Then:
+:$u \in \map Z R \implies u^{-1} \in \map Z R$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Center of Ring|center]] and [[Inverse of Unit in Centralizer of Ring is in Centralizer]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Skewness in terms of Non-Central Moments}
+Tags: Skewness
+
+\begin{theorem}
+Let $X$ be a [[Definition:Random Variable|random variable]] with [[Definition:Expectation|mean]] $\mu$ and [[Definition:Standard Deviation|standard deviation]] $\sigma$.
+Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by: 
+:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn	| l = \gamma_1
+	| r = \expect {\paren {\dfrac {X - \mu} \sigma}^3}
+	| c = {{Defof|Skewness}}
+}}
+{{eqn	| r = \frac {\expect {X^3 - 3 \mu X^2 + 3 \mu^2 X - \mu^3} } {\sigma^3} 
+	| c = [[Linearity of Expectation Function]], [[Cube of Difference]]
+}}
+{{eqn	| r = \frac {\expect {X^3} - 3 \mu \expect {X^2} + 3 \mu^2 \expect X - \mu^3} {\sigma^3}
+	| c = [[Linearity of Expectation Function]]
+}}
+{{eqn	| r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \mu \expect X} - \mu^3} {\sigma^3}
+}}
+{{eqn	| r = \frac {\expect {X^3} - 3 \mu \paren {\expect {X^2} - \paren {\expect X}^2} - \mu^3} {\sigma^3}
+	| c = $\mu = \expect X$
+}}
+{{eqn	| r = \frac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}
+	| c = [[Variance as Expectation of Square minus Square of Expectation]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Skewness]]
+pm850rumjl7z7uv4xmbn26t3vt0viye
+\end{proof}<|endoftext|>
+\section{Intersection of Ring Ideals is Ideal}
+Tags: Set Intersection, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
+Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Ideal of Ring|ideals]] of $R$.
+Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb L$ of the members of $\mathbb L$ is itself an [[Definition:Ideal  of Ring|ideal]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb L$.
+From [[Intersection of Subrings is Subring]], we have that $L$ is a [[Definition:Subring|subring]] of $R$.
+Let $x \in L$ and $y \in R$.
+Then:
+:$\forall T \in \bigcap \mathbb L: x \circ y \in T, y \circ x \in T$
+as every [[Definition:Element|element]] of $\bigcap \mathbb L$, including $T$, is an [[Definition:Ideal of Ring|ideal]] of $R$.
+If $y \in R$, then $x \circ y$ and $y \circ x$ are in every [[Definition:Element|element]] of $\mathbb L$, and hence in $L$.
+So $L$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Ring Elements forming Zero Product with given Element is Ideal}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $a \in R$ be an arbitrary [[Definition:Element|element]] of $R$.
+Let $A$ be the [[Definition:Subset|subset]] of $R$ defined as:
+:$A = \set {x \in R: x \circ a = 0_R}$
+Then $A$ is an [[Definition:Ideal of Ring|ideal]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Ring Zero|ring zero]]:
+:$\forall x \in R: x \circ 0_R = 0_R$
+Hence $0_R \in A$ and so $A \ne \O$.
+Let $a, b \in A$.
+{{begin-eqn}}
+{{eqn | ll= \forall x \in R:
+      | l = x \circ b
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = -\paren {x \circ b}
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = x \circ \paren {-b}
+      | r = 0_R
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | ll= \forall x \in R:
+      | l = x \circ a
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = x \circ \paren {-b}
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = \paren {x \circ a} + \paren {x \circ \paren {-b} }
+      | r = 0_R
+      | c = as $0_R$ is [[Definition:Identity Element|identity]] of $\struct {R, +}$
+}}
+{{eqn | lll=\leadsto
+      | l = x \circ \paren {a + \paren {-b} }
+      | r = 0_R
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | lll=\leadsto
+      | l = a + \paren {-b}
+      | o = \in
+      | r = A
+      | c = Definition of $A$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall x \in R:
+      | l = x \circ a
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = x \circ b
+      | r = 0_R
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = \paren {x \circ a} \circ b
+      | r = 0_R
+      | c = {{Defof|Ring Zero}}
+}}
+{{eqn | lll=\leadsto
+      | l = x \circ \paren {a \circ b}
+      | r = 0_R
+      | c = {{Ring-axiom|M1}}
+}}
+{{eqn | lll=\leadsto
+      | l = a \circ b
+      | o = \in
+      | r = A
+      | c = Definition of $A$
+}}
+{{end-eqn}}
+Hence the result, from [[Test for Ideal]]:
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Skewness of Bernoulli Distribution}
+Tags: Bernoulli Distribution, Skewness
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Bernoulli Distribution|Bernoulli distribution with parameter $p$]].
+Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by: 
+:$\gamma_1 = \dfrac {1 - 2 p} {\sqrt {p q} }$
+where $q = 1 - p$.
+\end{theorem}
+
+\begin{proof}
+From [[Skewness in terms of Non-Central Moments]]: 
+:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ 
+where $\mu$ is the [[Definition:Expectation|mean]] of $X$, and $\sigma$ the [[Definition:Standard Deviation|standard deviation]].
+We have, by [[Expectation of Bernoulli Distribution]]:
+:$\mu = p$
+By [[Variance of Bernoulli Distribution]], we also have: 
+:$\var X = \sigma^2 = p \paren {1 - p}$
+so: 
+:$\sigma = \sqrt {p \paren {1 - p} }$
+By [[Raw Moment of Bernoulli Distribution]], we have: 
+:$\expect {X^3} = p$
+So:
+{{begin-eqn}}
+{{eqn	| l = \gamma_1
+	| r = \frac {p - 3 p^2 \paren {1 - p} - p^3} {p^{3/2} \paren {1 - p}^{3/2} }
+}}
+{{eqn	| r = \frac {p \paren {1 - p^2} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
+}}
+{{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p} - 3 p^2 \paren {1 - p} } {p^{3/2} \paren {1 - p}^{3/2} }
+	| c = [[Difference of Two Squares]]
+}}
+{{eqn	| r = \frac {p \paren {1 - p} \paren {1 + p - 3p} } {p^{3/2} \paren {1 - p}^{3/2} }
+}}
+{{eqn	| r = \frac {1 - 2 p} {\sqrt {p \paren {1 - p} } }
+}}
+{{eqn	| r = \frac {1 - 2 p} {\sqrt {p q} }
+	| c = $q = 1 - p$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Bernoulli Distribution]]
+[[Category:Skewness]]
+sdipu0lqoiyvtqullidfadba1bjeyul
+\end{proof}<|endoftext|>
+\section{Intersection of Subrings is Subring}
+Tags: Set Intersection, Subrings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subring|subrings]] of $R$.
+Then the [[Definition:Set Intersection|intersection]] $\displaystyle \bigcap \mathbb L$ of the members of $\mathbb L$ is itself a [[Definition:Subring|subring]] of $R$.
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle L = \bigcap \mathbb L$.
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {R, +}$.
+By the [[One-Step Subgroup Test]]:
+:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
+By [[Intersection of Subsemigroups]], $\struct {L, \circ}$ a [[Definition:Subsemigroup|subsemigroup]] of $\struct {R, \circ}$.
+So by definition of [[Definition:Subsemigroup|subsemigroup]]:
+:$\forall x, y \in \struct {L, \circ}: x \circ y \in L$
+By the [[Subring Test]] it follows that $\struct {L, +, \circ}$ is a [[Definition:Subring|subring]] of $\struct {R, +, \circ}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Median of Exponential Distribution}
+Tags: Exponential Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] of the [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$ for some $\beta \in \R_{> 0}$. 
+Then the [[Definition:Median of Continuous Random Variable|median]] of $X$ is equal to $\beta \ln 2$.
+\end{theorem}
+
+\begin{proof}
+Let $M$ be the median of $X$.
+From the definition of the [[Definition:Exponential Distribution|exponential distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]: 
+:$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$
+Note that $f_X$ is non-zero, so the [[Definition:Median of Continuous Random Variable|median]] is unique. 
+We have by the definition of a [[Definition:Median of Continuous Random Variable|median]]: 
+:$\displaystyle \map \Pr {X < M} = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x = \frac 1 2$
+Evaluating this integral:
+{{begin-eqn}}
+{{eqn	| l = \frac 1 \beta \int_0^M e^{-\frac x \beta} \rd x
+	| r = \frac 1 \beta \sqbrk {-\beta e^{-\frac x \beta} }_0^M
+	| c = [[Primitive of Exponential Function]]
+}}
+{{eqn	| r = 1 - e^{-\frac M \beta}
+}}
+{{end-eqn}}
+So: 
+:$1 - e^{-\frac M \beta} = \dfrac 1 2$
+Then: 
+:$e^{-\frac M \beta} = \dfrac 1 2$
+giving: 
+:$-\dfrac M \beta = \map \ln {\dfrac 1 2}$
+So, by [[Logarithm of Reciprocal]]: 
+:$M = \beta \ln 2$
+{{qed}}
+[[Category:Exponential Distribution]]
+j5t0e96gx4xvfgvj4qdzsdgyhjfg3cf
+\end{proof}<|endoftext|>
+\section{Intersection of Ring Ideals is Largest Ideal Contained in all Ideals}
+Tags: Set Intersection, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
+Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Ideal of Ring|ideals]] of $R$.
+Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb L$ of the members of $\mathbb L$ is the largest [[Definition:Ideal of Ring|ideal]] of $R$ contained in each member of $\mathbb L$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb L$.
+From [[Intersection of Ring Ideals is Ideal]]  $L$ is indeed an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $L = \bigcap \mathbb L$.
+From [[Intersection of Subrings is Largest Subring Contained in all Subrings]], we have that $L$ is the largest [[Definition:Subring|subring]] of $R$ contained in each member of $\mathbb L$.
+As $L$ is the largest [[Definition:Subring|subring]] of $R$ contained in each member of $\mathbb L$, and it is an [[Definition:Ideal of Ring|ideal]] of $R$, there can be no larger [[Definition:Ideal of Ring|ideal]] as it would then not be a [[Definition:Subring|subring]].
+So $L$ is the largest [[Definition:Ideal of Ring|ideal]] of $R$ contained in each member of $\mathbb L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subrings is Largest Subring Contained in all Subrings}
+Tags: Set Intersection, Subrings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\mathbb L$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subring|subrings]] of $R$.
+Then the [[Definition:Set Intersection|intersection]] $\displaystyle \bigcap \mathbb L$ of the members of $\mathbb L$ is the largest [[Definition:Subring|subring]] of $R$ contained in each member of $\mathbb L$.
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle L = \bigcap \mathbb L$.
+From [[Intersection of Subrings is Subring]], $L$ is indeed a [[Definition:Subring|subring]] of $R$.
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is the largest [[Definition:Subgroup|subgroup]] of $\struct {R, +}$ contained in each member of $\mathbb L$.
+By [[Intersection of Subsemigroups]], $\struct {L, \circ}$ is the largest [[Definition:Subsemigroup|subsemigroup]] of $\struct {R, \circ}$ contained in each member of $\mathbb L$.
+Let $S$ be a [[Definition:Subring|subring]] of $R$ such that:
+:$\forall K \in \mathbb L: S \subseteq K$
+By [[Intersection is Largest Subset]], $S \subseteq L$.
+Thus $L$ is the largest [[Definition:Subring|subring]] of $R$ contained in each member of $\mathbb L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Median of Continuous Uniform Distribution}
+Tags: Continuous Uniform Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] which is [[Definition:Continuous Uniform Distribution|uniformly distributed]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$. 
+Then the [[Definition:Median of Continuous Random Variable|median]] $M$ of $X$ is given by: 
+:$M = \dfrac {a + b} 2$
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Continuous Uniform Distribution|continuous uniform distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
+:$\map {f_X} x = \dfrac 1 {b - a}$
+Note that $f_X$ is non-zero, so the [[Definition:Median of Continuous Random Variable|median]] is [[Definition:Unique|unique]]. 
+We have by the definition of a [[Definition:Median of Continuous Random Variable|median]]: 
+:$\displaystyle \map \Pr {X < M} = \frac 1 {b - a} \int_a^M \rd x = \frac 1 2$
+We have, by [[Primitive of Constant]]: 
+:$\dfrac {M - a} {b - a} = \dfrac 1 2$
+So: 
+{{begin-eqn}}
+{{eqn	| l = M
+	| r = a + \frac {b - a} 2
+}}
+{{eqn	| r = \frac {b - a + 2 a} 2
+}}
+{{eqn	| r = \frac {b + a} 2
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Continuous Uniform Distribution]]
+joatsn7dvulzmoat6ha7uljktgmfme4
+\end{proof}<|endoftext|>
+\section{Intersection of Ring Ideals Containing Subset is Smallest}
+Tags: Set Intersection, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]]
+Let $S \subseteq R$ be a [[Definition:Subset|subset]] of $R$.
+Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R$ containing $S$.
+Then $L$ is the smallest [[Definition:Ideal of Ring|ideal]] of $R$ containing $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Subrings Containing Subset is Smallest]], $L$ is the smallest [[Definition:Subring|subring]] of $R$ containing $S$.
+From [[Intersection of Ring Ideals is Ideal]], $L$ is an [[Definition:Ideal of Ring|ideal]] of $R$.
+As $L$ is the smallest [[Definition:Subring|subring]] of $R$ containing $S$, and it is an [[Definition:Ideal of Ring|ideal]] of $R$, there can be no smaller [[Definition:Ideal of Ring|ideal]] of $R$ containing $S$ as it would then not be a [[Definition:Subring|subring]].
+So the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Ideal of Ring|ideals]] of $R$ containing $S$ is the smallest [[Definition:Ideal of Ring|ideals]] of $R$ containing $S$. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subrings Containing Subset is Smallest}
+Tags: Set Intersection, Subrings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $S \subseteq R$ be a [[Definition:Subset|subset]] of $R$.
+Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Subring|subrings]] of $R$ containing $S$.
+Then $L$ is the smallest [[Definition:Subring|subring]] of $R$ containing $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Subrings is Subring]], $L$ is indeed a [[Definition:Subring|subring]] of $R$.
+Let $T$ be a [[Definition:Subring|subring]] of $R$ containing $S$.
+Let $x, y \in L$.
+By the [[Subring Test]], we have that:
+{{begin-eqn}}
+{{eqn | l = x - y
+      | o = \in
+      | r = L
+}}
+{{eqn | l = x \circ y
+      | o = \in
+      | r = L
+}}
+{{end-eqn}}
+By [[Intersection is Largest Subset]], it follows that $x, y \in T$.
+But $T$ is also a [[Definition:Subring|subring]] of $R$.
+So, by the [[Subring Test]] again, we have that:
+{{begin-eqn}}
+{{eqn | l = x - y
+      | o = \in
+      | r = T
+}}
+{{eqn | l = x \circ y
+      | o = \in
+      | r = T
+}}
+{{end-eqn}}
+So by definition of [[Definition:Subset|subset]], $L \subseteq T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Division Subrings is Division Subring}
+Tags: Division Subrings
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Division Subring|division subrings]] of $D$.
+Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb K$ of the members of $\mathbb K$ is itself a [[Definition:Division Subring|division subring]] of $D$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb K$.
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {D, +}$.
+By the [[One-Step Subgroup Test]]:
+:$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, \circ}$ a [[Definition:Subgroup|subgroup]] of $\struct {D, \circ}$.
+By the [[Two-Step Subgroup Test]]:
+:$\forall x, y \in \struct {L, \circ}: x \circ y \in L$
+:$\forall x \in \struct {L, \circ}: x^{-1} \in L$
+By the [[Division Subring Test]] it follows that $\struct {L, +, \circ}$ is a [[Definition:Division Subring|division subring]] of $\struct {D, +, \circ}$.
+{{qed}}
+[[Category:Division Subrings]]
+69ali8l00sj7xt1s2imaspbw72jzljt
+\end{proof}<|endoftext|>
+\section{Intersection of Subfields is Subfield}
+Tags: Set Intersection, Subfields
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subfield|subfields]] of $F$.
+Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb K$ of the members of $\mathbb K$ is itself a [[Definition:Subfield|subfield]] of $F$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb K$.
+A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Subring|division subring]].
+From [[Intersection of Division Subrings is Division Subring]], $L$ is itself a [[Definition:Division Subring|division subring]] of $F$.
+As $\struct {F, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]], $\circ$ is [[Definition:Commutative Operation|commutative]] on $F$.
+By [[Restriction of Commutative Operation is Commutative]], it follows that $\circ$ is also [[Definition:Commutative Operation|commutative]] on $L$.
+Thus $\struct {L, +, \circ}$ is a [[Definition:Division Ring|division ring]] where $\circ$ is [[Definition:Commutative Operation|commutative]].
+Thus, by definition, of $\struct {L, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Division Subrings is Largest Division Subring Contained in all Division Subrings}
+Tags: Division Subrings
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Division Subring|division subrings]] of $D$.
+Let $\bigcap \mathbb K$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Element|elements]] of $\mathbb K$.
+Then $\bigcap \mathbb K$ is the largest [[Definition:Division Subring|division subring]] of $D$ contained in each [[Definition:Element|element]] of $\mathbb K$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb K$.
+From [[Intersection of Division Subrings is Division Subring]], $\struct {L, +, \circ}$ is a [[Definition:Division Subring|division subring]] of $\struct {D, +, \circ}$.
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, +}$ is the largest [[Definition:Subgroup|subgroup]] of $\struct {D, +}$ contained in each [[Definition:Element|element]] of $\mathbb K$.
+By [[Intersection of Subgroups is Subgroup]], $\struct {L, \circ}$ is the largest [[Definition:Subgroup|subgroup]] of $\struct {D, \circ}$ contained in each [[Definition:Element|element]] of $\mathbb K$.
+Let $S$ be a [[Definition:Subring|subring]] of $D$ such that:
+:$\forall K \in \mathbb K: S \subseteq K$
+By [[Intersection is Largest Subset]], $S \subseteq L$.
+Thus $L$ is the largest [[Definition:Division Subring|division subring]] of $D$ contained in each [[Definition:Element|element]] of $\mathbb L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Division Subrings Containing Subset is Smallest}
+Tags: Division Subrings
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $S \subseteq D$ be a [[Definition:Subset|subset]] of $D$.
+Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $D$ containing $S$.
+Then $L$ is the smallest [[Definition:Division Subring|division subring]] of $D$ containing $S$.
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Division Subrings is Division Subring]], $L$ is indeed a [[Definition:Division Subring|division subring]] of $D$.
+Let $T$ be a [[Definition:Division Subring|division subring]] of $D$ containing $S$.
+Let $x, y \in L$.
+By the [[Division Subring Test]], we have that:
+{{begin-eqn}}
+{{eqn | l = x - y
+      | o = \in
+      | r = L
+}}
+{{eqn | l = x \circ y
+      | o = \in
+      | r = L
+}}
+{{eqn | l = x^{-1} \circ y
+      | o = \in
+      | r = L
+}}
+{{end-eqn}}
+By [[Intersection is Largest Subset]], it follows that $x, y \in T$.
+But $T$ is also a [[Definition:Division Subring|division subring]] of $D$.
+So, by the [[Division Subring Test]] again, we have that:
+{{begin-eqn}}
+{{eqn | l = x - y
+      | o = \in
+      | r = T
+}}
+{{eqn | l = x \circ y
+      | o = \in
+      | r = T
+}}
+{{eqn | l = x^{-1} \circ y
+      | o = \in
+      | r = T
+}}
+{{end-eqn}}
+So by definition of [[Definition:Subset|subset]], $L \subseteq T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subfields Containing Subset is Smallest}
+Tags: Set Intersection, Subfields
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $S \subseteq F$ be a [[Definition:Subset|subset]] of $F$.
+Let $L$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Subfield|subfields]] of $F$ containing $S$.
+Then $L$ is the smallest [[Definition:Subfield|subfield]] of $F$ containing $S$.
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Subring|division subring]].
+Thus $L$ is the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Division Subring|division subrings]] of $F$ containing $S$.
+From [[Intersection of Division Subrings Containing Subset is Smallest]], $L$ is the smallest [[Definition:Division Subring|division subring]] of $F$ containing $S$.
+From [[Intersection of Subfields is Subfield]], $L$ is a [[Definition:Subfield|subfield]] of $F$.
+As $L$ is the smallest [[Definition:Division Subring|division subring]] of $F$ containing $S$, and it is a [[Definition:Subfield|subfield]] of $F$, there can be no smaller [[Definition:Subfield|subfield]] of $F$ containing $S$ as it would then not be a [[Definition:Division Subring|division subring]].
+So the [[Definition:Set Intersection|intersection]] of the [[Definition:Set|set]] of all [[Definition:Subfield|subfields]] of $F$ containing $S$ is the smallest [[Definition:Subfield|subfield]] of $F$ containing $S$. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subfields is Largest Subfield Contained in all Subfields}
+Tags: Set Intersection, Subfields
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\mathbb K$ be a [[Definition:Non-Empty Set|non-empty set]] of [[Definition:Subfield|subfields]] of $F$.
+Let $\bigcap \mathbb K$ be the [[Definition:Set Intersection|intersection]] of the [[Definition:Element|elements]] of $\mathbb K$.
+Then $\bigcap \mathbb K$ is the largest [[Definition:Subfield|subfield]] of $F$ contained in each [[Definition:Element|element]] of $\mathbb K$.
+\end{theorem}
+
+\begin{proof}
+Let $L = \bigcap \mathbb K$.
+From [[Intersection of Subfields is Subfield]], $\struct {L, +, \circ}$ is itself a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$.
+A [[Definition:Field (Abstract Algebra)|field]] is by definition also a [[Definition:Division Subring|division subring]].
+Thus $\struct {L, +, \circ}$ is the largest [[Definition:Division Subring|division subring]] of $F$ contained in each [[Definition:Element|element]] of $\mathbb K$.
+But as $\struct {F, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]], $\circ$ is [[Definition:Commutative Operation|commutative]].
+So by [[Restriction of Commutative Operation is Commutative]], $\struct {L, +, \circ}$ is also a [[Definition:Field (Abstract Algebra)|field]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalent Norms on Rational Numbers/Necessary Condition}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Norm on Division Ring|norms]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent norms]].
+Then:
+:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
+\end{theorem}
+
+\begin{proof}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent]].
+By [[Definition:Equivalent Division Ring Norms/Norm is Power of Other Norm|Norm is Power of Other Norm]] then:
+:$\exists \alpha \in \R_{\gt 0}: \forall q \in \Q: \norm q_1 = \norm q_2^\alpha$
+In particular:
+:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
+\end{proof}<|endoftext|>
+\section{Equivalent Norms on Rational Numbers/Sufficient Condition}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ be [[Definition:Norm on Division Ring|norms]] on the [[Definition:Rational Numbers|rational numbers]] $\Q$.
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
+Then $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are [[Definition:Equivalent Division Ring Norms|equivalent]]
+\end{theorem}
+
+\begin{proof}
+Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
+:$\exists \alpha \in \R_{\gt 0}: \forall n \in \N: \norm n_1 = \norm n_2^\alpha$
+By [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]] then:
+:$\forall n \in \N: \norm {-n}_1 = \norm n_1 = \norm n_2^\alpha = \norm {-n}_2^\alpha$
+Hence:
+:$\forall k \in \Z: \norm k_1 = \norm k_2^\alpha$
+By [[Properties of Norm on Division Ring/Norm of Quotient|Norm of Quotient]] then:
+:$\forall \dfrac a b \in \Q: \norm {\dfrac a b}_1 = \dfrac {\norm a_1} {\norm b_1} = \dfrac {\norm a_2^\alpha} {\norm b_2^\alpha} = \norm {\dfrac a b}_2^\alpha$
+By [[Definition:Equivalent Division Ring Norms/Norm is Power of Other Norm|Norm is Power of Other Norm]] then  $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are [[Definition:Equivalent Division Ring Norms|equivalent]].
+\end{proof}<|endoftext|>
+\section{Commutative and Unitary Ring with 2 Ideals is Field}
+Tags: Commutative Algebra, Field Theory, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\struct {R, +, \circ}$ be such that the only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are:
+:$\set {0_R}$
+and:
+$\struct {R, +, \circ}$ itself.
+That is, such that $\struct {R, +, \circ}$ has no non-[[Definition:Null Ideal|null]] [[Definition:Proper Ideal|proper ideals]].
+Then $\struct {R, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+From [[Null Ring is Ideal]] and [[Ring is Ideal of Itself]], it is always the case that $\set {0_R}$ and $\struct {R, +, \circ}$ are [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$.
+Let $a \in R^*$, where $R^* := R \setminus \set {0_R}$.
+Let $\ideal a$ be the [[Definition:Principal Ideal of Ring|principal ideal of $R$ generated by $a$]].
+We have that $\ideal a$ is a [[Definition:Non-Null Ideal|non-null ideal]] and hence $\ideal a = R$.
+Thus $1_R \in \ideal a$.
+Thus $\exists x \in R: x \circ a = 1_R$ by the definition of [[Definition:Principal Ideal of Ring|principal ideal]].
+Therefore $a$ is [[Definition:Invertible Element|invertible]].
+As $a$ is arbitrary, it follows that all such $a$ are [[Definition:Invertible Element|invertible]].
+Thus by definition $\struct {R, +, \circ}$ is a [[Definition:Division Ring|division ring]] such that $\circ$ is [[Definition:Commutative Operation|commutative]].
+The result follows by definition of [[Definition:Field (Abstract Algebra)|field]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Field has 2 Ideals}
+Tags: Commutative Algebra, Field Theory, Ideal Theory
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Then the only [[Definition:Ideal of Ring|ideals]] of $\struct {F, +, \circ}$ are:$\struct {F, +, \circ}$ and $\set {0_F}$.
+That is, $\struct {F, +, \circ}$ has no non-[[Definition:Null Ideal|null]] [[Definition:Proper Ideal|proper ideals]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Field (Abstract Algebra)|field]] is a [[Definition:Division Ring|division ring]].
+From [[Null Ring is Ideal]] and [[Ring is Ideal of Itself]], it is always the case that $\set {0_F}$ and $\struct {F, +, \circ}$ are [[Definition:Ideal of Ring|ideals]] of $\struct {F, +, \circ}$.
+From [[Ideals of Division Ring]], it follows that the only [[Definition:Ideal of Ring|ideals]] of $\struct {F, +, \circ}$ are $\struct {F, +, \circ}$ and $\set {0_F}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Commutative Ring with Unity and 2 Ideals not necessarily Division Ring}
+Tags: Division Rings, Rings with Unity, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_F$ and whose [[Definition:Unity of Ring|unity]] is $1_F$.
+Let $\struct {R, +, \circ}$ specifically not be [[Definition:Commutative Ring|commutative]].
+Let $\struct {R, +, \circ}$ be such that the only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
+Then it is not necessarily the case that $\struct {R, +, \circ}$ is a [[Definition:Division Ring|division ring]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Square Matrix|square matrices]] of [[Definition:Order of Square Matrix|order $2$]] over the [[Definition:Real Number|real numbers]] $\R$.
+$S$ is not a [[Definition:Division Ring|division ring]], as for example:
+:$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$
+and so both $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ are [[Definition:Proper Zero Divisor|proper zero divisors]] of $S$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $S$ containing a non-[[Definition:Zero Matrix|zero]] [[Definition:Square Matrix|matrix]] $A$ whose $\tuple {r, s}$th [[Definition:Entry of Matrix|entry]] is $\lambda \ne 0$
+Let $E_{i j}$ be the [[Definition:Square Matrix|matrix]] of [[Definition:Order of Square Matrix|order $2$]] defined as:
+:$e_{a b} = \begin{cases} 1 & : a = i, b = j \\ 0 & : \text {otherwise} \end{cases}$
+Thus for example:
+:$E_{1 2} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$
+Then for all $i, j \in \set {1, 2}$:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 \lambda E_{i r} A E_{s j}
+      | r = \dfrac 1 \lambda E_{i r} A E_{s j}
+      | c = 
+}}
+{{eqn | r = E_{i j}
+      | c = 
+}}
+{{end-eqn}}
+{{explain|Prove the above}}
+It follows that:
+:$\forall i, j \in \set {1, 2}: E_{i j} \in J$
+It remains to be shown that $J$ is the whole of $S$. 
+{{finish}}.
+\end{proof}<|endoftext|>
+\section{General Antiperiodicity Property}
+Tags: Antiperiodic Functions
+
+\begin{theorem}
+Let $f: X \to X$ be an [[Definition:Antiperiodic Function|antiperiodic function]], where $X$ is either $\R$ or $\C$.
+Let $L$ be an [[Definition:Antiperiodic Element|antiperiodic element]] of $f$.
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+:If $n$ is [[Definition:Even Integer|even]], then $n L$ is a [[Definition:Periodic Element|periodic element]] of $f$.
+:If $n$ is [[Definition:Odd Integer|odd]], then $n L$ is an [[Definition:Antiperiodic Element|antiperiodic element]] of $f$.
+\end{theorem}
+
+\begin{proof}
+Suppose that $X = \C$.
+=== Case 1 ===
+If $n$ is [[Definition:Even Integer|even]], then:
+{{begin-eqn}}
+{{eqn | l = \map f {x + n L}
+      | r = \map f {x + \paren {2 k} L}
+      | c = for some $k \in \Z$
+}}
+{{eqn | r = \map f {x + \paren {k 2} L}
+      | c = [[Complex Multiplication is Commutative]]
+}}
+{{eqn | r = \map f {x + k \paren {2 L} }
+      | c = [[Complex Multiplication is Associative]]
+}}
+{{eqn | r = \map f x
+      | c = by [[Double of Antiperiodic Element is Periodic]] and the [[General Periodicity Property]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Case 2 ===
+If $n$ is [[Definition:Odd Integer|odd]], then:
+{{begin-eqn}}
+{{eqn | l = \map f {x + n L}
+      | r = \map f {x + \paren {2 k + 1} L}
+      | c = for some $k \in \Z$
+}}
+{{eqn | r = \map f {x + \paren {\paren {2 k} L + L} }
+      | c = [[Complex Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \map f {\paren {x + \paren {2 k} L} + L}
+      | c = [[Complex Addition is Associative]]
+}}
+{{eqn | r = -\map f {x + \paren {2 k} L}
+}}
+{{eqn | r = -\map f x
+      | c = by '''[[General Antiperiodicity Property#Case 1|Case 1]]'''
+}}
+{{end-eqn}}
+The proof for when $X = \R$ is nearly identical.
+{{qed}}
+[[Category:Antiperiodic Functions]]
+d6sgward9zyusbcx8pw9ly4hy5c3szv
+\end{proof}<|endoftext|>
+\section{Antiperiodic Element is Multiple of Antiperiod}
+Tags: Antiperiodic Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real anti-periodic function]] with [[Definition:Antiperiod|anti-period]] $A$.
+Let $L$ be an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
+Then $A \divides L$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $A \nmid L$.
+By the [[Division Theorem/Real Number Index|Division Theorem]] we have:
+:$\exists! q \in \Z, r \in \R: L = q A + r, 0 < r < A$
+By [[Even and Odd Integers form Partition of Integers]], it follows that $q$ must be either [[Definition:Even Integer|even]] or [[Definition:Odd Integer|odd]].
+=== Case 1 ===
+Suppose $q$ is [[Definition:Even Integer|even]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {x + L}
+      | r = \map f {x + \paren {q A + r} }
+}}
+{{eqn | r = \map f {\paren {x + r} + q A}
+}}
+{{eqn | r = \map f {x + r}
+      | c = [[General Antiperiodicity Property]]
+}}
+{{eqn | r = -\map f x
+}}
+{{end-eqn}}
+And so $r$ is an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$ that is less than $A$.
+But then $A$ cannot be the [[Definition:Antiperiod|anti-period]] of $f$.
+Therefore by [[Definition:Contradiction|contradiction]] $q$ cannot be even.
+=== Case 2 ===
+Suppose $q$ is [[Definition:Odd Integer|odd]].
+Then:
+{{begin-eqn}}
+{{eqn | l = - \map f x
+      | r = \map f {x + L}
+}}
+{{eqn | r = \map f {x + \paren {q A + r} }
+}}
+{{eqn | r = \map f {\paren {x + r} + q A}
+}}
+{{eqn | r = - \map f {x + r}
+      | c = [[General Antiperiodicity Property]]
+}}
+{{end-eqn}}
+So:
+:$-\map f x = - \map f {x + r} \implies \map f x = \map f {x + r}$
+It is seen that $r$ is a [[Definition:Periodic Element|periodic element]] of $f$ such that $0 < r < A$.
+But consider $0 < A - r < A$:
+{{begin-eqn}}
+{{eqn | l = \map f {x + \paren {A - r} }
+      | r = \map f {\paren {x + A} - r}
+}}
+{{eqn | r = \map f {x + A}
+}}
+{{eqn | r = -\map f x
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the fact that $A$ is the [[Definition:Antiperiod|anti-period]] of $f$.
+Hence the result.
+{{qed}}
+[[Category:Antiperiodic Functions]]
+6lspg960b2ut2ajzetaiu2r0qfdbb9m
+\end{proof}<|endoftext|>
+\section{Skewness of Poisson Distribution}
+Tags: Skewness, Poisson Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Discrete Random Variable|discrete random variable]] with a [[Definition:Poisson Distribution|Poisson distribution with parameter $\lambda$]].
+Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is given by: 
+:$\gamma_1 = \dfrac 1 {\sqrt \lambda}$
+\end{theorem}
+
+\begin{proof}
+From [[Skewness in terms of Non-Central Moments]]: 
+:$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$ 
+where $\mu$ is the [[Definition:Expectation|mean]] of $X$, and $\sigma$ the [[Definition:Standard Deviation|standard deviation]].
+We have, by [[Expectation of Poisson Distribution]]:
+:$\expect X = \lambda$
+By [[Variance of Poisson Distribution]]:
+:$\var X = \sigma^2 = \lambda$
+so:
+:$\sigma = \sqrt \lambda$
+To now calculate $\gamma_1$, we must calculate $\expect {X^3}$.
+We find this using the [[Definition:Moment Generating Function|moment generating function]] of $X$, $M_X$. 
+By [[Moment Generating Function of Poisson Distribution]], this is given by: 
+:$\displaystyle \map {M_X} t = e^{\lambda \paren {e^t - 1} }$
+From [[Moment in terms of Moment Generating Function]]:
+:$\expect {X^3} = \map {M_X'''} 0$
+In [[Variance of Poisson Distribution/Proof 3|Variance of Poisson Distribution: Proof 3]], it is shown that:
+:$\displaystyle \map {M_X''} t = \lambda \paren {\lambda e^t + 1} e^{\lambda \paren {e^t - 1} + t}$
+So:
+{{begin-eqn}}
+{{eqn	| l = \map {M_X'''} t
+	| r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1} \frac \d {\d t} \paren {\lambda \paren {e^t - 1} + t} \frac \d {\d \paren {\lambda \paren {e^t - 1} + t} } \paren {e^{\lambda \paren {e^t - 1} + t} }
+	| c = [[Product Rule]], [[Exponential of Sum]], [[Chain Rule for Derivatives]]
+}}
+{{eqn	| r = \lambda^2 e^{\lambda \paren {e^t - 1} + 2t} + \lambda \paren {\lambda e^t + 1}^2 e^{\lambda \paren {e^t - 1} + t}
+	| c = [[Derivative of Exponential Function]], [[Derivative of Power]]
+}}
+{{end-eqn}}
+Setting $t = 0$:
+{{begin-eqn}}
+{{eqn	| l = \expect {X^3}
+	| r = \lambda^2 e^{\lambda \paren {e^0 - 1} + 0} + \lambda \paren {\lambda e^0 + 1}^2 e^{\lambda \paren {e^0 - 1} + 0}
+}}
+{{eqn	| r = \lambda^2 + \lambda \paren {\lambda + 1}^2
+	| c = [[Exponential of Zero]]
+}}
+{{eqn	| r = \lambda^3 + 3 \lambda^2 + \lambda
+	| c = [[Square of Sum]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn	| l = \gamma_1
+	| r = \frac {\lambda^3 + 3 \lambda^2 + \lambda - 3 \lambda^2 - \lambda^3} {\lambda^{3/2} }
+}}
+{{eqn	| r = \frac \lambda {\lambda^{3/2} }
+}}
+{{eqn	| r = \frac 1 {\sqrt \lambda}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Skewness]]
+[[Category:Poisson Distribution]]
+oml15vhprg2hwwjnhtgtfbkhko1uqex
+\end{proof}<|endoftext|>
+\section{Periodic Element is Multiple of Antiperiod}
+Tags: Antiperiodic Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real anti-periodic function]] with [[Definition:Antiperiod|anti-period]] $A$.
+Let $L$ be a [[Definition:Periodic Element|periodic element]] of $f$.
+Then $A \divides L$.
+\end{theorem}
+
+\begin{proof}
+Consider $A + L$:
+{{begin-eqn}}
+{{eqn | l = \map f {x + \paren {A + L} }
+      | r = \map f {\paren {x + A} + L}
+}}
+{{eqn | r = \map f {x + A}
+}}
+{{eqn | r = -\map f x
+}}
+{{end-eqn}}
+Hence $A + L$ is an [[Definition:Antiperiodic Element|anti-periodic element]] of $f$.
+Combining [[Antiperiodic Element is Multiple of Antiperiod]], [[Divides is Reflexive]], and [[Common Divisor Divides Difference]] yields: 
+:$A \divides \paren {A + L} \land A \divides A \implies A \divides L$
+Hence the result.
+{{qed}}
+[[Category:Antiperiodic Functions]]
+hywff4b2rkdxz178xdjt5efld55qywm
+\end{proof}<|endoftext|>
+\section{Skewness of Continuous Uniform Distribution}
+Tags: Skewness, Continuous Uniform Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] which is [[Definition:Continuous Uniform Distribution|uniformly distributed]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$. 
+Then the [[Definition:Skewness|skewness]] $\gamma_1$ of $X$ is equal to $0$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Skewness|skewness]]: 
+:$\gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3}$
+where: 
+:$\mu$ is the [[Definition:Expectation|mean]] of $X$
+:$\sigma$ is the [[Definition:Standard Deviation|standard deviation]] of $X$. 
+From the definition of the [[Definition:Continuous Uniform Distribution|continuous uniform distribution]], $X$ has [[Definition:Probability Density Function|probability density function]]:
+:$\map {f_X} x = \dfrac 1 {b - a}$
+So, from [[Expectation of Function of Continuous Random Variable]]:
+:$\displaystyle \gamma_1 = \expect {\paren {\dfrac {X - \mu} \sigma}^3} = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x$
+Then: 
+{{begin-eqn}}
+{{eqn	| l = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \mu}^3 \rd x
+	| r = \frac 1 {\paren {b - a} \sigma^3} \int_a^b \paren {x - \frac {a + b} 2}^3 \rd x
+	| c = [[Expectation of Continuous Uniform Distribution]]
+}}
+{{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{a - \frac {a + b} 2}^{b - \frac {a + b} 2} u^3 \rd u
+	| c = [[Integration by Substitution|substituting]] $u = x - \dfrac {a + b} 2$
+}}
+{{eqn	| r = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u
+}}
+{{end-eqn}}
+We have: 
+:$\paren {-u}^3 = -u^3$
+so we can see the integrand is [[Definition:Odd Function|odd]]. 
+So, by [[Definite Integral of Odd Function]]: 
+:$\displaystyle \gamma_1 = \frac 1 {\paren {b - a} \sigma^3} \int_{-\frac {b - a} 2}^{\frac {b - a} 2} u^3 \rd u = 0$
+{{qed}}
+[[Category:Skewness]]
+[[Category:Continuous Uniform Distribution]]
+2a2owjs7yj6v2pzrxxl73wkltthj1ya
+\end{proof}<|endoftext|>
+\section{Double of Antiperiod is Period}
+Tags: Antiperiodic Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Antiperiodic Function|real antiperiodic function]] with an [[Definition:Antiperiod|anti-period]] of $A$.
+Then $f$ is also [[Definition:Real Periodic Function|periodic]] with a [[Definition:Period of Function|period]] of $2A$.
+\end{theorem}
+
+\begin{proof}
+Let $L_f$ be the [[Definition:Set|set]] of all [[Definition:Periodic Element|periodic elements]] of $f$.
+By [[Periodic Element is Multiple of Antiperiod]] and [[Absolute Value of Real Number is not less than Divisors]]:
+:$\forall p \in L_f: A \divides p \land A \le p$
+Suppose there is a $p \in L_f$ such that $p = A$. 
+Then: 
+{{begin-eqn}}
+{{eqn | l = \map f x
+      | r = \map f {x + A}
+      | c = 
+}}
+{{eqn | r = -\map f x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+which contradicts [[Constant Function has no Period]].
+Therefore $\forall p \in L_f: A \divides p \land A < p$.
+The [[Definition:Smallest Element of Subset|smallest number]] $r$ such that $A < r \land A \divides r$ is $2A$.
+But by [[Double of Antiperiodic Element is Periodic]], this is a [[Definition:Periodic Element|periodic element]] of $f$.
+Hence the result.
+{{qed}}
+[[Category:Antiperiodic Functions]]
+g6ijhsitxemjy4cby6j0aqeniqikh1o
+\end{proof}<|endoftext|>
+\section{Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power}
+Tags: Integral Domains, Principal Ideals
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $a \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
+Then:
+:$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
+where $\ideal x$ denotes the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $x$]].
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \ideal {a^{n + 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists r \in D:
+      | l = x
+      | r = r \circ a^{n + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \paren {r \circ a} \circ a^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \ideal {a^n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ideal {a^{n + 1} }
+      | o = \subseteq
+      | r = \ideal {a^n}
+      | c = 
+}}
+{{end-eqn}}
+It remains to be shown that $\ideal {a^{n + 1} } \ne \ideal {a^n}$.
+{{AimForCont}} $\ideal {a^{n + 1} } = \ideal {a^n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \ideal {a^{n + 1} }
+      | r = \ideal {a^n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^n
+      | o = \in
+      | r = \ideal {a^{n + 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x \in D:
+      | l = a^n
+      | r = x \circ a^{n + 1} 
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 1_D
+      | r = x \circ a 
+      | c = [[Cancellation Law of Ring Product of Integral Domain]]
+}}
+{{end-eqn}}
+That is, $a$ is a [[Definition:Unit of Ring|unit]] of $D$.
+This [[Definition:Contradiction|contradicts]] the assertion that $a$ is a [[Definition:Proper Element of Ring|proper element]] of $D$.
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Field Integral Domain has Infinite Number of Ideals}
+Tags: Integral Domains, Principal Ideals
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] which is not a [[Definition:Field (Abstract Algebra)|field]].
+Then $\struct {D, +, \circ}$ has an [[Definition:Infinite Set|infinite number]] of [[Definition:Distinct Elements|distinct]] [[Definition:Ideal of Ring|ideals]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in D$ be a [[Definition:Proper Element of Ring|proper element]] of $D$.
+Because $\struct {D, +, \circ}$ is not a [[Definition:Field (Abstract Algebra)|field]], such an [[Definition:Element|element]] is known to exist.
+From [[Principal Ideal in Integral Domain generated by Power Plus One is Subset of Principal Ideal generated by Power]]:
+:$\forall n \in \Z_{\ge 0}: \ideal {a^{n + 1} } \subsetneq \ideal {a_n}$
+where $\ideal x$ denotes the [[Definition:Principal Ideal of Ring|principal ideal of $D$ generated by $x$]].
+Hence the [[Definition:Set|set]]:
+:$S = \set {\ideal {1_D}, \ideal a, \ideal {a^2}, \ideal {a^3}, \dotsc}$
+is [[Definition:Infinite Set|infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Homomorphism from Ring with Unity to Integral Domain Preserves Unity}
+Tags: Ring Homomorphisms, Rings with Unity, Integral Domains
+
+\begin{theorem}
+Let $\struct {R, +_R, \circ_R}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\struct {D, +_D, \circ_D}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $\phi: R \to D$ be a [[Definition:Ring Homomorphism|ring homomorphism]] such that:
+:$\map \ker \phi \ne R$
+where $\map \ker \phi$ denotes the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$.
+Then $\map \phi {1_R} = 1_D$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\map \phi {1_R} = 0_D$.
+Let $x \in R$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi x
+      | r = \map \phi {x \circ_R 1_R}
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = \map \phi x \circ_D \map \phi {1_R}
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r = \map \phi x \circ_D 0_D
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = 0_D
+      | c = {{Defof|Ring Zero}}
+}}
+{{end-eqn}}
+But this [[Definition:Contradiction|contradicts]] the assertion that $\map \ker \phi \ne R$.
+It follows that $\map \phi {1_R} \ne 0_D$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = 1_D \circ_D \map \phi {1_R}
+      | r = \map \phi {1_R}
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = \map \phi {1_R \circ_R 1_R}
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = \map \phi {1_R} \circ_D \map \phi {1_R}
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | ll= \leadsto
+      | l = 1_D
+      | r = \map \phi {1_R}
+      | c = [[Cancellation Law of Ring Product of Integral Domain]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unity plus Negative of Nilpotent Ring Element is Unit}
+Tags: Nilpotent Ring Elements
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $x \in R$ be [[Definition:Nilpotent Ring Element|nilpotent]].
+Then $1_R - x$ is a [[Definition:Unit of Ring|unit]] of $R$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Nilpotent Ring Element|nilpotent element]]:
+:$x^n = 0_R$
+for some $n \in \Z_{>0}$.
+From [[Difference of Two Powers]]:
+{{begin-eqn}}
+{{eqn | l = a^n - b^n
+      | r = \paren {a - b} \circ \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} \circ b^j
+      | c = 
+}}
+{{eqn | r = \paren {a - b} \circ \paren {a^{n - 1} + a^{n - 2} \circ b + a^{n - 3} \circ b^2 + \dotsb + a \circ b^{n - 2} + b^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+for $a, b \in R$.
+Putting $a = 1_R$ and $b = x$, we have:
+{{begin-eqn}}
+{{eqn | l = \paren {1_R}^n - x^n
+      | r = \paren {1_R - x} \circ \paren {1_R^{n - 1} + 1_R^{1_R - 2} \circ x + 1_R^{n - 3} \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 1_R - 0_R
+      | r = \paren {1_R - x} \circ \paren {1_R + 1_R \circ x + 1_R \circ x^2 + \dotsb + 1_R \circ x^{n - 2} + x^{n - 1} }
+      | c = as $x^n = 0_R$, $1_R^k = 1_R$
+}}
+{{eqn | ll= \leadsto
+      | l = 1_R
+      | r = \paren {1_R - x} \circ \paren {1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1} }
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition $1_R - x$ has a [[Definition:Product Inverse|product inverse]] $1_R + x + x^2 + \dotsb + x^{n - 2} + x^{n - 1}$.
+Hence by definition $1_R - x$ is a [[Definition:Unit of Ring|unit]] of $R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Commutative Ring by Nilradical is Reduced}
+Tags: Nilpotent Ring Elements, Commutative Rings, Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\struct {N, +, \circ}$ denote the [[Definition:Nilradical of Ring|nilradical]] of $R$.
+The [[Definition:Quotient Ring|quotient ring]] $R / N$ is a [[Definition:Reduced Ring|reduced ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Nilpotent Elements of Commutative Ring form Ideal]], $\struct {N, +, \circ}$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$.
+Hence the [[Definition:Quotient Ring|quotient ring]] $R / N$ is defined.
+By definition of the [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$, $N$ is the [[Definition:Ring Zero|zero]] of $R / N$.
+Let $\paren {x + N}^n \in N$.
+Then:
+:$x^n \in N$
+and so:
+:$\paren {x^n}^m = 0_R$
+for some $m \in \Z_{>0}$.
+Hence:
+:$x \in N$
+and so:
+:$x + N = N$
+That is, if a [[Definition:Left Coset|(left) coset]] of $N$ is [[Definition:Nilpotent Ring Element|nilpotent]] in $R / N$ that [[Definition:Left Coset|(left) coset]] is $N$ itself.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Self-Inverse Element of Integral Domain is Unity or its Negative}
+Tags: Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $x \in D$ such that $x^2 = 1_D$.
+Then either $x = 1_D$ or $x = -1_D$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^2
+      | r = 1_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x + 1_D} \paren {x + \paren {-1_D} }
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + 1_D
+      | r = 0_D
+      | c = 
+}}
+{{eqn | lo= \lor
+      | l = x + \paren {-1_D}
+      | r = 0_D
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = -1_D
+      | c = 
+}}
+{{eqn | lo= \lor
+      | l = x
+      | r = 1_D
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Product of Units of Integral Domain with Finite Number of Units}
+Tags: Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let $D$ have a [[Definition:Finite Set|finite number]] of [[Definition:Unit of Ring|units]].
+Let $U_D$ be the [[Definition:Set|set]] of [[Definition:Unit of Ring|units]] of $\struct {D, +, \circ}$.
+Then:
+:$\displaystyle \prod_{x \mathop \in U_D} x = -1_D$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Set|set]] $S$ defined as:
+:$S = U_R \setminus \set {1_D, -1_D}$
+If $S$ has [[Definition:Even Integer|even]] [[Definition:Cardinality|cardinality]], it can be [[Definition:Set Partition|partitioned]] into [[Definition:Doubleton|doubletons]] of the form $\set {u, u^{-1} }$.
+Each of these [[Definition:Doubleton|doubletons]] has a [[Definition:Ring Product|product]] of $1_D$.
+The [[Definition:Ring Product|product]] of all these with $1_D$ and $-1_D$ is $-1_D$.
+It remains to be shown that $S$ cannot be of [[Definition:Odd Integer|odd]] [[Definition:Cardinality|cardinality]].
+{{AimForCont}} $S$ has an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]].
+Then after pairing off each $u \in D$ with its [[Definition:Product Inverse|product inverse]] $u^{-1}$, we are left with one [[Definition:Element|element]] of $S$ which is [[Definition:Self-Inverse Element|self-inverse]].
+But from [[Self-Inverse Element of Integral Domain is Unity or its Negative]], this [[Definition:Self-Inverse Element|self-inverse element]] is either $1_D$ or $-1_D$.
+These have already been counted.
+Hence there cannot be an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Ideals of Ring of Integers}
+Tags: Integers, Prime Ideals of Rings
+
+\begin{theorem}
+Let $\struct {\Z, +, \times}$ denote the [[Definition:Ring of Integers|ring of integers]].
+Let $J$ be a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $\Z$.
+Then either:
+:$J = \set 0$
+or:
+:$J = \ideal p$
+where:
+:$p$ is a [[Definition:Prime Number|prime number]]
+:$\ideal p$ denotes the [[Definition:Principal Ideal of Ring|principal ideal of $\Z$ generated by $p$]].
+\end{theorem}
+
+\begin{proof}
+From [[Prime Ideal iff Quotient Ring is Integral Domain]]:
+:$J$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $\Z$ {{iff}} $\Z / J$ is an [[Definition:Integral Domain|integral domain]].
+From [[Quotient Ring of Integers and Zero]]:
+:$\Z / \set 0 \cong \Z$
+As $\Z$ is an [[Definition:Integral Domain|integral domain]], it follows that $\set 0$ is a [[Definition:Prime Ideal of Commutative and Unitary Ring|prime ideal]] of $\Z$.
+From [[Quotient Ring of Integers with Principal Ideal]]:
+:$\struct {\Z, +, \times} / \ideal p$ is [[Definition:Ring Isomorphism|isomorphic]] to $\struct {\Z_p, +_p, \times_p}$, the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]].
+From [[Ring of Integers Modulo Prime is Integral Domain]]:
+:$\struct {\Z_n, +_n, \times_p}$ is an [[Definition:Integral Domain|integral domain]] {{iff}} $p$ is [[Definition:Prime Number|prime]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Power Mapping on Galois Field is Automorphism}
+Tags: Galois Fields
+
+\begin{theorem}
+Let $\GF$ be a [[Definition:Galois Field|Galois field]] whose [[Definition:Field Zero|zero]] is $0_\GF$ and whose [[Definition:Characteristic of Ring|characteristic]] is $p$.
+Let $\sigma: \GF \to \GF$ be defined as:
+:$\forall x \in \GF: \map \sigma x = x^p$
+Then $\sigma$ is an [[Definition:Field Automorphism|automorphism]] of $\GF$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \GF$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \sigma {x y}
+      | r = \paren {x y}^p
+      | c = Definition of $\sigma$
+}}
+{{eqn | r = x^p y^p
+      | c = [[Power of Product of Commutative Elements in Group]]
+}}
+{{eqn | r = \map \sigma x \, \map \sigma y
+      | c = Definition of $\sigma$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map \sigma {x + y}
+      | r = \paren {x + y}^p
+      | c = Definition of $\sigma$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^p \dbinom p k x^k y^{p - k}
+      | c = [[Binomial Theorem]]
+}}
+{{eqn | o = \equiv
+      | r = x^p + y^p
+      | rr= \pmod p
+      | c = [[Power of Sum Modulo Prime]]
+}}
+{{eqn | r = \map \sigma x + \map \sigma y
+      | c = Definition of $\sigma$
+}}
+{{end-eqn}}
+Thus it has been demonstrated that $\sigma$ is a [[Definition:Field Homomorphism|homomorphism]].
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \map \ker \sigma
+      | r = \set {x \in \GF: \map \sigma x = 0_\GF}
+      | c = {{Defof|Kernel of Ring Homomorphism}}
+}}
+{{eqn | r = \set {x \in \GF: x^p = 0_\GF}
+      | c = Definition of $\sigma$
+}}
+{{eqn | r = \set {x \in \GF: x = 0_\GF}
+      | c = [[Congruence of Powers]]
+}}
+{{eqn | r = \set {0_\GF}
+      | c = [[Congruence of Powers]]
+}}
+{{end-eqn}}
+From [[Kernel is Trivial iff Monomorphism]], $\sigma$ is a [[Definition:Ring Monomorphism|monomorphism]].
+That is, $\sigma$ is an [[Definition:Injection|injection]].
+Then from [[Injection from Finite Set to Itself is Surjection]], $\sigma$ is a [[Definition:Surjection|surjection]].
+Thus $\sigma$ is a [[Definition:Bijection|bijective]] [[Definition:Field Homomorphism|homomorphism]] to itself.
+The result follows by definition of [[Definition:Field Automorphism|automorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group and Multiplicative Group of Field are not Isomorphic}
+Tags: Field Theory, Group Isomorphisms
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $\struct {F, +}$ denote the [[Definition:Additive Group of Ring|additive group]] of $F$.
+Let $\struct {F_{\ne 0_F}, \times}$ denote the [[Definition:Multiplicative Group|multiplicative group]] of $F$.
+Then $\struct {F, +}$ and $\struct {F_{\ne 0_F}, \times}$ are not [[Definition:Group Isomorphism|isomorphic]] to each other.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an [[Definition:Group Isomorphism|isomorphism]].
+By definition:
+:$0_F$ is the [[Definition:Identity Element|identity]] of $\struct {F, +}$
+and
+:$1_F$ is the [[Definition:Identity Element|identity]] of $\struct {F_{\ne 0_F}, \times}$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = 0_F
+      | r = \map \phi {1_F}
+      | c = [[Epimorphism Preserves Identity]]
+}}
+{{eqn | r = \map \phi {\paren {-1_F} \times \paren {-1_F} }
+      | c = 
+}}
+{{eqn | r = \map \phi {-1_F} + \map \phi {-1_F}
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{end-eqn}}
+and so by definition $F$ has [[Definition:Characteristic of Field|characteristic]] $2$.
+Let $x \in \struct {F_{\ne 0_F}, \times}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x^2}
+      | r = \map \phi x + \map \phi x
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{eqn | r = 0_F
+      | c = as $F$ has [[Definition:Characteristic of Field|characteristic]] $2$
+}}
+{{end-eqn}}
+As $\phi$ is an [[Definition:Group Isomorphism|isomorphism]], it is also a [[Definition:Group Monomorphism|monoomorphism]].
+From [[Kernel is Trivial iff Monomorphism]]
+:$\map \ker \phi = \set {1_F}$
+and so $x^2 = 1$.
+Thus $x = 1$ and so $\order F = 2$.
+Thus $\order {F_{\ne 0_F} } = 1$.
+So $\phi$ is a [[Definition:Bijection|bijection]] from a [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $1$ to a [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $2$.
+So $\phi$ cannot be a [[Definition:Bijection|bijection]] and so cannot be an [[Definition:Group Isomorphism|isomorphism]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principal Ideal Domain cannot have Infinite Strictly Increasing Sequence of Ideals}
+Tags: Principal Ideal Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Then $D$ cannot have an [[Definition:Infinite Sequence|infinite sequence]] of [[Definition:Ideal of Ring|ideals]] $\sequence {j_n}_{n \mathop \in \N}$ such that:
+:$\forall n \in \N: J_n \subsetneq j_{n + 1}$
+\end{theorem}
+
+\begin{proof}
+Let $K = \displaystyle \bigcup_{n \mathop \in \N} J_n$.
+Then from [[Increasing Union of Sequence of Ideals is Ideal]], $K$ is an [[Definition:Ideal of Ring|ideal]] of $D$.
+We have that $D$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Hence there exists $a \in D$ such that:
+:$K = \ideal a$
+where $\ideal a$ is the [[Definition:Principal Ideal of Ring|principal ideal]] of $D$ generated by $a$.
+But $a \in J_m$ for some $m \in \N$.
+Thus $K \subseteq J_m$
+Thus it follows that $J_{m + 1} \subseteq J_m$ which [[Definition:Contradiction|contradicts]] our initial assertion that:
+:$\forall n \in \N: J_n \subsetneq j_{n + 1}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Norm of Complex Number is Positive Definite}
+Tags: Field Norm of Complex Number
+
+\begin{theorem}
+Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
+Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
+:$\forall z \in \C: \map N z = \cmod z^2$
+where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
+Then $N$ is [[Definition:Positive Definite (Ring)|positive definite]] on $\C$.
+\end{theorem}
+
+\begin{proof}
+First it is shown that $\map N z = 0 \iff z = 0$.
+{{begin-eqn}}
+{{eqn | l = z
+      | r = 0
+      | c = 
+}}
+{{eqn | r = 0 + 0 i
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map N z
+      | r = 0^2 + 0^2
+      | c = Definition of $N$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Let $z = x + i y$.
+{{begin-eqn}}
+{{eqn | l = \map N z
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map N {x + i y}
+      | r = 0
+      | c = Definition of $z$
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + y^2
+      | r = 0
+      | c = Definition of $N$
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = 0
+      | c = [[Square of Real Number is Non-Negative]]
+}}
+{{eqn | l = b
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = 0
+      | c = Definition of $z$
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \map N z
+      | r = \map N {x + i y}
+      | c = Definition of $z$
+}}
+{{eqn | r = x^2 + y^2
+      | c = Definition of $N$
+}}
+{{eqn | 0 = \ge
+      | r = 0
+      | c = [[Square of Real Number is Non-Negative]]
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Positive Definite (Ring)|positive definite]].
+{{qed}}
+[[Category:Field Norm of Complex Number]]
+fgrcmu847s49n7be08imferkgspbgaq
+\end{proof}<|endoftext|>
+\section{Field Norm of Complex Number is Multiplicative Function}
+Tags: Field Norm of Complex Number
+
+\begin{theorem}
+Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
+Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
+:$\forall z \in \C: \map N z = \cmod z^2$
+where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
+Then $N$ is a [[Definition:Multiplicative Function on Ring|multiplicative function]] on $\C$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N {z_1 z_2}
+      | r = \cmod {z_1 z_2}^2
+      | c = Definition of $N$
+}}
+{{eqn | r = \paren {\cmod {z_1} \cmod {z_2} }^2
+      | c = [[Complex Modulus of Product of Complex Numbers]]
+}}
+{{eqn | r = \cmod {z_1}^2 \cmod {z_2}^2
+      | c = 
+}}
+{{eqn | r = \map N {z_1} \map N {z_2}
+      | c = Definition of $N$
+}}
+{{end-eqn}}
+So $N$ is a [[Definition:Multiplicative Function on Ring|multiplicative function]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Norm of Complex Number is not Norm}
+Tags: Field Norm of Complex Number
+
+\begin{theorem}
+Let $\C$ denote the [[Definition:Complex Number|set of complex numbers]].
+Let $N: \C \to \R_{\ge 0}$ denote the [[Definition:Field Norm of Complex Number|field norm on complex numbers]]:
+:$\forall z \in \C: \map N z = \cmod z^2$
+where $\cmod z$ denotes the [[Definition:Complex Modulus|complex modulus]] of $z$.
+Then $N$ is not a [[Definition:Norm on Ring|norm]] on $\C$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $z_1 = z_2 = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map N {z_1 + z_2}
+      | r = \cmod {z_1 + z_2}^2
+      | c = Definition of $N$
+}}
+{{eqn | r = 2^2
+      | c = 
+}}
+{{eqn | r = 4
+      | c = 
+}}
+{{end-eqn}}
+But:
+{{begin-eqn}}
+{{eqn | l = \map N {z_1} + \map N {z_2}
+      | r = \cmod {z_1}^2 + \cmod {z_2}^2
+      | c = Definition of $N$
+}}
+{{eqn | r = 1^2 + 1^2
+      | c = 
+}}
+{{eqn | r = 2
+      | c = 
+}}
+{{end-eqn}}
+So we have that for these instances of $z_1$ and $z_2$:
+:$\map N {z_1 + z_2} > \map N {z_1} + \map N {z_2}$
+and so the [[Definition:Triangle Inequality|triangle inequality]] is not satisfied.
+Hence by definition $N$ is not a [[Definition:Norm on Ring|norm]] on $\C$.
+{{qed}}
+[[Category:Field Norm of Complex Number]]
+dofpceh1eri25d4elz5bhvmm0obtoo4
+\end{proof}<|endoftext|>
+\section{Units of 5th Cyclotomic Ring}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+The [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.
+\end{theorem}
+
+\begin{proof}
+Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
+Let $z_1 \in \Z \sqbrk {i \sqrt 5}$ be a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
+Thus by definition:
+:$\exists z_2 \in \Z \sqbrk {i \sqrt 5}: z_1 \times z_2 = 1$
+Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \map N 1
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | r = \map N {z_1 \times z_2}
+      | c = 
+}}
+{{eqn | r = \map N {z_1} \times \map N {z_2}
+      | c = 
+}}
+{{eqn | r = \paren { {x_1}^2 + 5 {y_1}^2} \paren { {x_2}^2 + 5 {y_2}^2}
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | ll= \leadsto
+      | l = {x_1}^2
+      | r = 1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = {x_2}^2
+      | r = 1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = y_1
+      | r = y_2 = 0
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Norm on 5th Cyclotomic Ring}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+Let $\alpha = a + i b \sqrt 5$ be an arbitrary [[Definition:Element|element]] of $\Z \sqbrk {i \sqrt 5}$.
+The [[Definition:Field Norm of Complex Number|field norm]] of $\alpha$ is given by:
+:$\map N \alpha = a^2 + 5 b^2$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N \alpha
+      | r = \cmod \alpha^2
+      | c = {{Defof|Field Norm of Complex Number}}
+}}
+{{eqn | r = \paren {\sqrt {a^2 + \paren {b \sqrt 5}^2} }^2
+      | c = {{Defof|Complex Modulus}}
+}}
+{{eqn | r = a^2 + 5 b^2
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+There are no [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose [[Definition:Field Norm of Complex Number|field norm]] is either $2$ or $3$.
+\end{theorem}
+
+\begin{proof}
+Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
+Let $z = x + i y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map N z
+      | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + 5 y^2
+      | r = 2
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^2
+      | r = 2
+      | c = as $y$ is an [[Definition:Integer|integer]] and has to equal $0$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+But [[Square Root of Prime is Irrational]] so $x \notin \Z$.
+Similarly for $\map N z = 3$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Irreducible Elements of 5th Cyclotomic Ring}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+The following [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are [[Definition:Irreducible Element of Ring|irreducible]]:
+:$2$
+:$3$
+:$1 + i \sqrt 5$
+:$1 - i \sqrt 5$
+\end{theorem}
+
+\begin{proof}
+{{TheoremWanted|For the concept of [[Definition:Irreducible Element of Ring|irreducibility]] to be defined, it needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an [[Definition:Integral Domain|integral domain]].}}
+Let $z = x + i y$ be an [[Definition:Element|element]] of $\Z \sqbrk {i \sqrt 5}$ in the [[Definition:Set|set]] $S$, where:
+:$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$
+Let $z$ have a [[Definition:Non-Trivial Factorization|non-trivial factorization]]:
+:$z = z_1 z_2$
+where neither $z_1$ nor $z_2$ are [[Definition:Unit of Ring|units]] of $\Z \sqbrk {i \sqrt 5}$.
+Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map N z
+      | r = \map N {z_1 z_2}
+      | c = 
+}}
+{{eqn | r = \map N {z_1} \map N {z_2}
+      | c = {{Defof|Field Norm of Complex Number}}
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \map N 2
+      | r = 2^2 + 5 \times 0^2
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | r = 4
+      | c = 
+}}
+{{eqn | l = \map N 3
+      | r = 3^2 + 5 \times 0^2
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | r = 9
+      | c = 
+}}
+{{eqn | l = \map N {1 + i \sqrt 5}
+      | r = 1^2 + 5 \times 1^2
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | r = 6
+      | c = 
+}}
+{{eqn | l = \map N {1 - i \sqrt 5}
+      | r = 1^2 + 5 \times 1^2
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | r = 6
+      | c = 
+}}
+{{end-eqn}}
+From [[Elements of 5th Cyclotomic Ring with Field Norm 1]], the only [[Definition:Element|elements]] of $\Z \sqbrk {i \sqrt 5}$ whose [[Definition:Field Norm of Complex Number|field norm]] is $1$ are the [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.
+From [[5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3]], none of $4$, $6$ and $9$ have [[Definition:Proper Divisor of Integer|proper divisors]] which are [[Definition:Field Norm of Complex Number|field norms]] of [[Definition:Element|elements]] of $\Z \sqbrk {i \sqrt 5}$.
+Thus either $z_1$ or $z_2$ is a [[Definition:Unit of Ring|unit]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
+So none of the [[Definition:Element|elements]] of $S$ has a [[Definition:Non-Trivial Factorization|non-trivial factorization]] in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
+Hence the result, by definition of [[Definition:Irreducible Element of Ring|irreducible]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Value of Field Norm on 5th Cyclotomic Ring is Integer}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+Let $\alpha = a + i b \sqrt 5$ be an arbitrary [[Definition:Element|element]] of $\Z \sqbrk {i \sqrt 5}$.
+Let $\map N \alpha$ denoted the [[Definition:Field Norm of Complex Number|field norm]] of $\alpha$.
+Then $\map N \alpha$ is an [[Definition:Integer|integer]].
+\end{theorem}
+
+\begin{proof}
+From [[Field Norm on 5th Cyclotomic Ring]]:
+:$\map N \alpha = a^2 + 5 b^2$
+From the definition of the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]]:
+:$\Z \sqbrk {i \sqrt 5} = \set {a + i \sqrt 5 b: a, b \in \Z}$
+That is, both $a$ and $b$ are [[Definition:Integer|integers]].
+Hence $a^2 + 5 b^2$ is also an [[Definition:Integer|integer]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Elements of 5th Cyclotomic Ring with Field Norm 1}
+Tags: Cyclotomic Rings
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+The only [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ whose [[Definition:Field Norm of Complex Number|field norm]] equals $1$ are the [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.
+\end{theorem}
+
+\begin{proof}
+From [[Units of 5th Cyclotomic Ring]], $1$ and $-1$ are the only [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
+Let $\map N z$ denote the [[Definition:Field Norm of Complex Number|field norm]] of $z \in \Z \sqbrk {i \sqrt 5}$.
+Let $z \in \Z \sqbrk {i \sqrt 5}$ such that $\map N z = 1$.
+Let $z = x + i y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map N z
+      | r = 1
+      | c = [[Field Norm on 5th Cyclotomic Ring]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + 5 y^2
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2
+      | r = 1
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = y
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{5th Cyclotomic Ring is not a Unique Factorization Domain}
+Tags: Cyclotomic Rings, Unique Factorization Domains
+
+\begin{theorem}
+Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the [[Cyclotomic Ring/Examples/5th|$5$th cyclotomic ring]].
+Then $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is not a [[Definition:Unique Factorization Domain|unique factorization domain]].
+The following [[Definition:Element|elements]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are [[Definition:Irreducible Element of Ring|irreducible]]:
+:$2$
+:$3$
+:$1 + i \sqrt 5$
+:$1 - i \sqrt 5$
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Unique Factorization Domain|unique factorization domain]] $D$ is an [[Definition:Integral Domain|integral domain]] with the properties that:
+:For all $x \in D$ such that $x$ is non-[[Definition:Ring Zero|zero]] and not a [[Definition:Unit of Ring|unit]] of $D$:
+::$(1): \quad x$ possesses a [[Definition:Complete Factorization|complete factorization]] in $D$
+::$(2): \quad$ Any two [[Definition:Complete Factorization|complete factorizations]] of $x$ are [[Definition:Equivalent Factorizations|equivalent]].
+A [[Definition:Complete Factorization|complete factorization]] is a [[Definition:Tidy Factorization|tidy factorization]] 
+:$x = u \circ y_1 \circ y_2 \circ \cdots \circ y_n$
+such that:
+:$u$ is a [[Definition:Unit of Ring|unit]] of $D$
+:all of $y_1, y_2, \ldots, y_n$ are [[Definition:Irreducible Element of Ring|irreducible]].
+{{TheoremWanted|It needs to be demonstrated that $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ is actually an [[Definition:Integral Domain|integral domain]].}}
+From [[Units of 5th Cyclotomic Ring]], the only [[Definition:Unit of Ring|units]] of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are $1$ and $-1$.
+From [[Irreducible Elements of 5th Cyclotomic Ring]], all of the [[Definition:Element|elements]] of the [[Definition:Set|set]] $S$, where:
+:$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$
+are [[Definition:Irreducible Element of Ring|irreducible]] in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = 6
+      | r = 1 \times 2 \times 3
+      | c = 
+}}
+{{eqn | r = 1 \times \paren {1 + i \sqrt 5} \paren {1 - i \sqrt 5} 
+      | c = 
+}}
+{{end-eqn}}
+So there are two [[Definition:Tidy Factorization|tidy factorizations]] of $6$ in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ which are not [[Definition:Equivalent Factorizations|equivalent]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Ring of Polynomials in Ring Element on Integral Domain by that Polynomial is that Domain}
+Tags: Polynomial Rings, Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
+Let $X \in R$ be [[Definition:Transcendental over Integral Domain|transcencental over $D$]].
+Let $D \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $D$.
+Let $D \sqbrk X / \ideal X$ denote the [[Definition:Quotient Ring|quotient ring]] of $D \sqbrk X$ by the [[Definition:Ideal of Ring|ideal]] of $D$ generated by $X$.
+Then:
+:$D \sqbrk X / \ideal X \cong D$
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z_{> 0}$ be arbitrary.
+Let $P = a_n X^n + a_{n - 1} X^{n - 1} + \dotsb + a_1 X + a_0$ be a [[Definition:Polynomial over Ring in One Variable|polynomial over $D$ in $X$]].
+Consider the [[Definition:Mapping|mapping]] $\phi: D \sqbrk X \to D$ defined as:
+:$\forall P \in D \sqbrk X: \map \phi P = a_0$
+Let:
+:$P_1 = a_m X^m + a_{m - 1} X^{m - 1} + \dotsb + a_1 X + a_0$
+:$P_2 = b_m X^m + b_{m - 1} X^{m - 1} + \dotsb + b_1 X + b_0$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map \phi {P_1 \times P_2}
+      | r = a_0 \times b_0
+      | c = 
+}}
+{{eqn | r = \map \phi {P_1} \times \map \phi {P_2}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map \phi {P_1 + P_2}
+      | r = a_0 + b_0
+      | c = 
+}}
+{{eqn | r = \map \phi {P_1} + \map \phi {P_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus it is seen that $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]].
+The [[First Isomorphism Theorem for Rings]] can then be used.
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Polynomials in Integers with Even Constant Term forms Ideal}
+Tags: Polynomial Rings, Integers
+
+\begin{theorem}
+Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
+Let $S \subseteq \Z \sqbrk X$ be the [[Definition:Set|set]] of [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] which have a [[Definition:Constant Term of Polynomial|constant term]] which is [[Definition:Even Integer|even]].
+Then $S$ is an [[Definition:Ideal of Ring|ideal]] of $\Z \sqbrk X$.
+\end{theorem}
+
+\begin{proof}
+For example, $X + 2$ is a [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] with an [[Definition:Even Integer|even]] [[Definition:Constant Term of Polynomial|constant term]].
+So $S$ is not [[Definition:Empty Set|empty]].
+Let $P_1 = \displaystyle \sum_{k \mathop = 0}^n a_k X^k$ and $P_2 = \displaystyle \sum_{k \mathop = 0}^n b_k X^k$ be [[Definition:Element|elements]] of $S$.
+We have:
+{{begin-eqn}}
+{{eqn | l = P_1 - P_2
+      | r = \sum_{k \mathop = 0}^n a_k X^k + \sum_{k \mathop = 0}^n b_k X^k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {a_k - b_k} X^k
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Constant Term of Polynomial|constant term]] of $P_1 - P_2$ is $a_0 - b_0$ which is [[Definition:Even Integer|even]].
+Thus $P_1 - P_2 \in S$.
+Let $P_3 = \sum_{k \mathop = 0}^s c_k X^k \in \Z \sqbrk X$.
+Then the [[Definition:Constant Term of Polynomial|constant term]] of $P_3 \times P_1$ is $c_0 \times a_0$.
+As $a_0$ is [[Definition:Even Integer|even]], so is $c_0 \times a_0$.
+The result follows by [[Test for Ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomials in Integers is not Principal Ideal Domain}
+Tags: Polynomial Rings, Integers, Principal Ideal Domains
+
+\begin{theorem}
+Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
+Then $\Z \sqbrk X$ is not a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+Let $J$ be the [[Definition:Ideal of Ring|ideal]] formed from the [[Definition:Set|set]] of [[Definition:Polynomial over Ring in One Variable|polynomials over $\Z$ in $X$]] which have a [[Definition:Constant Term of Polynomial|constant term]] which is [[Definition:Even Integer|even]].
+From [[Polynomials in Integers with Even Constant Term forms Ideal]], $J$ is indeed an [[Definition:Ideal of Ring|ideal]].
+{{AimForCont}} $J$ is a [[Definition:Principal Ideal of Ring|principal ideal of]] $\Z \sqbrk X$ such that $J = \ideal f$.
+But $2 \in J$, and so $2$ is a [[Definition:Multiple of Integer|multiple]] of $f$ in $\Z \sqbrk X$.
+So $f = \pm 2$ or $f = \pm 1$.
+But this [[Definition:Contradiction|contradicts]] the fact that $J = \ideal f$.
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomials in Integers is Unique Factorization Domain}
+Tags: Polynomial Rings, Integers, Unique Factorization Domains
+
+\begin{theorem}
+Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $\Z$.
+Then $\Z \sqbrk X$ is a [[Definition:Unique Factorization Domain|unique factorization domain]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Integers form Unique Factorization Domain]].
+The result follows from [[Gauss's Lemma on Unique Factorization Domains]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ideal of Ring of Polynomials over Field has Unique Monic Polynomial forming Principal Ideal}
+Tags: Principal Ideals, Polynomial Theory
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over $F$.
+Let $J$ be a non-[[Definition:Null Ideal|null]] [[Definition:Ideal of Ring|ideal]] of $F \sqbrk X$.
+Then there exists [[Definition:Unique|exactly one]] [[Definition:Monic Polynomial|monic polynomial]] $f \in F \sqbrk X$ such that:
+:$J = \ideal f$
+where $\ideal f$ is the [[Definition:Principal Ideal of Ring|principal ideal]] generated by $f$ in $F \sqbrk X$.
+\end{theorem}
+
+\begin{proof}
+Let $f_1$ and $f_2$ be [[Definition:Generator of Ideal|generators]] of $J$.
+Then $f_1$ and $f_2$ are unit multiples of each other.
+The [[Definition:Unit of Ring|units]] of $F \sqbrk X$ are the non-[[Definition:Field Zero|zero]] [[Definition:Element|elements]] of $F$.
+{{explain|The above needs to be properly derived.}}
+\end{proof}<|endoftext|>
+\section{Ring of Gaussian Integers is Principal Ideal Domain}
+Tags: Gaussian Integers, Principal Ideal Domains
+
+\begin{theorem}
+The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
+:$\struct {\Z \sqbrk i, +, \times}$
+forms a [[Definition:Principal Ideal Domain|principal ideal domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Gaussian Integers form Integral Domain]], we have that $\struct {\Z \sqbrk i, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+Let $a, d \in \Z \sqbrk i$ such that $d \ne 0$.
+Suppose $\cmod a \ge \cmod d$.
+Reference to an [[Definition:Argand Diagram|Argand diagram]] shows that one of:
+:$a + d, a - d, a + i d, a - i d$
+is closer to the [[Definition:Origin|origin]] than $a$ is.
+So it is possible to subtract [[Definition:Gaussian Integer|Gaussian integer]] [[Definition:Multiple of Ring Element|multiples]] of $d$ from $a$ until the [[Definition:Square Function|square]] of the [[Definition:Complex Modulus|modulus]] of the remainder drops below $\cmod d^2$.
+That remainder can only take [[Definition:Integer|integer]] values.
+Thus a [[Division Theorem]] result follows:
+:$\exists q, r \in \Z \sqbrk i: a = q d + r$
+where $\cmod r < \cmod d$.
+Let $J$ be an arbitrary non-[[Definition:Null Ideal|null]] [[Definition:Ideal of Ring|ideal]] of $\Z \sqbrk i$.
+Let $d$ be an [[Definition:Element|element]] of minimum [[Definition:Complex Modulus|modulus]] in $J$.
+Then the [[Division Theorem]] can be used to prove that $J = \ideal d$.
+{{finish|The above is the outline only.}}
+\end{proof}<|endoftext|>
+\section{Polynomial X^2 + 1 is Irreducible in Ring of Real Polynomials}
+Tags: Principal Ideals, Polynomial Theory
+
+\begin{theorem}
+Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in $X$ over the [[Definition:Real Number|real numbers]] $\R$.
+Then the [[Definition:Polynomial in Ring Element|polynomial]] $X^2 + 1$ is an [[Definition:Irreducible Element of Ring|irreducible element]] of $\R \sqbrk X$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $x^2 + 1$ has a [[Definition:Non-Trivial Factorization|non-trivial factorization]] in $\R \sqbrk X$.
+Then:
+:$\exists \alpha, \beta \in \R: \paren {X - \alpha} \paren {X - \beta}$
+and from the [[Polynomial Factor Theorem]]:
+:$\alpha^2 + 1 = 0$
+But that means:
+:$\alpha^2 = -1$
+and such an $\alpha$ does not exist in $\R$.
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Multiplication is not Commutative}
+Tags: Conventional Matrix Multiplication, Commutativity, Proofs by Induction, Matrix Multiplication is not Commutative
+
+\begin{theorem}
+Let $R$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $n \ne 1$.
+Let $\map {\MM_R} n$ denote the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
+Then [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] over $\map {\MM_R} n$ is not [[Definition:Commutative Operation|commutative]]:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
+If $R$ is specifically not [[Definition:Commutative Ring|commutative]], then the result holds when $n = 1$ as well.
+\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]]:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
+=== Edge Cases ===
+==== $n = 1$ ====
+Consider the case where $n = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf {A B}
+      | r = a_{11} b_{11}
+}}
+{{eqn | l = \mathbf {B A}
+      | r = b_{11} a_{11}
+      | c = 
+}}
+{{end-eqn}}
+and it follows that [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] over $\map {\MM_R} 1$ is [[Definition:Commutative Operation|commutative]] {{iff}} $R$ is a [[Definition:Commutative Ring|commutative ring]].
+==== $R$ not a [[Definition:Ring with Unity|Ring with Unity]] ====
+Consider the case where $R$ is not a [[Definition:Ring with Unity|ring with unity]], and is a general [[Definition:Ring (Abstract Algebra)|ring]].
+Let $R$ be the [[Definition:Trivial Ring|trivial ring]].
+From [[Matrix Multiplication on Square Matrices over Trivial Ring is Commutative]]:
+:$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$
+Hence the result does not follow for all [[Definition:Ring (Abstract Algebra)|rings]].
+It is not established at this point on exactly which [[Definition:Ring (Abstract Algebra)|rings]] [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] $\map {\MM_R} n$ [[Definition:Commutative Operation|commutes]].
+However, the existence of just one such [[Definition:Ring (Abstract Algebra)|ring]] (the [[Definition:Trivial Ring|trivial ring]]) warns us that we cannot apply the main result to ''all'' [[Definition:Ring (Abstract Algebra)|rings]].
+==== Matrices are not Square ====
+We note that $\mathbf A \mathbf B$ is defined when:
+:$\mathbf A = \sqbrk a_{m n}$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B = \sqbrk b_{n p}$ is an [[Definition:Matrix|$n \times p$ matrix]].
+Hence for both $\mathbf A \mathbf B$ and $\mathbf B \mathbf A$ to be defined, it is necessary that:
+:$\mathbf A = \sqbrk a_{m n}$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B = \sqbrk b_{n p}$ is an [[Definition:Matrix|$n \times m$ matrix]]
+for some $m, n \in \Z_{>0}$.
+But in this situation:
+:$\mathbf A \mathbf B$ is an [[Definition:Matrix|$m \times m$ matrix]]
+while:
+:$\mathbf B \mathbf A$ is an [[Definition:Matrix|$n \times n$ matrix]]
+and so if $\mathbf A$ and $\mathbf B$ are not [[Definition:Square Matrix|square matrices]], they cannot [[Definition:Commute|commute]].
+=== Basis for the Induction ===
+From [[Matrix Multiplication is not Commutative/Order 2 Square Matrices|Matrix Multiplication is not Commutative/Order 2 Square Matrices]], it is seen that there exist $2 \times 2$ matrices that don't commute under [[Definition:Matrix Product (Conventional)|matrix multiplication]], thus proving the result for $n = 2$.
+$\map P 2$ is the case:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} 2: \mathbf {A B} \ne \mathbf {B A}$
+This is demonstrated in [[Matrix Multiplication is not Commutative/Order 2 Square Matrices|Matrix Multiplication is not Commutative: Order $2$ Square Matrices]].
+Thus $\map P 2$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} k: \mathbf {A B} \ne \mathbf {B A}$
+from which it is to be shown that:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} {k + 1}: \mathbf {A B} \ne \mathbf {B A}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+From the [[Matrix Multiplication is not Commutative#Induction Hypothesis|induction hypothesis]], it is assumed that there exist $2$ [[Definition:Order of Square Matrix|order $k$]] [[Definition:Square Matrix|square matrices]] $\mathbf A$ and $\mathbf B$ such that $\mathbf {A B} \ne \mathbf {B A}$.
+For an [[Definition:Order of Square Matrix|order]] $n$ [[Definition:Square Matrix|square matrix]]  $\mathbf D$, let $\mathbf {D'}$ be the [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $n + 1$ defined as:
+:$d'_{i j} = \begin {cases} d_{i j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$
+Thus $\mathbf D'$ is just $\mathbf D$ with a [[Definition:Zero Row or Column|zero row]] and [[Definition:Zero Row or Column|zero column]] added at the ends.
+We have that $\mathbf D$ is a [[Definition:Submatrix|submatrix]] of $\mathbf D'$.
+Now:
+:$\paren {a' b'}_{i j} = \begin{cases} \displaystyle \sum_{r \mathop = 1}^{n + 1} \mathbf a'_{i r} b'_{r j} & : i < n + 1 \land j < n + 1 \\ 0 & : i = n + 1 \lor j = n + 1 \end{cases}$
+But:
+{{begin-eqn}}
+{{eqn | l = \sum_{r \mathop = 1}^{n + 1} a'_{i r} b'_{r j}
+      | r = a'_{i \paren {n + 1} } b'_{\paren {n + 1} i} + \sum_{r \mathop = 1}^n a'_{i r} b'_{r j}
+      | c = 
+}}
+{{eqn | r = \sum_{r \mathop = 1}^n a_{i r} b_{r j}
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \mathbf A' \mathbf B' \paren {n + 1, n + 1}
+      | r = \paren {\mathbf {A B} }' \paren {n + 1, n + 1}
+      | c = 
+}}
+{{eqn | r = \mathbf {A B}
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = \mathbf {B A}
+      | c = 
+}}
+{{eqn | r = \paren {\mathbf {B A} }' \paren {n + 1, n + 1}
+      | c = 
+}}
+{{eqn | r = \mathbf B' \mathbf A' \paren {n + 1; n + 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus it is seen that:
+:$\exists \mathbf A', \mathbf B' \in \MM_{n + 1 \times n + 1}: \mathbf A' \mathbf B' \ne \mathbf B' \mathbf A'$
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\exists \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} \ne \mathbf {B A}$
+and by definition [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] over $\map {\MM_R} n$ is not [[Definition:Commutative Operation|commutative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matrix Multiplication on Square Matrices over Trivial Ring is Commutative}
+Tags: Conventional Matrix Multiplication, Trivial Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be the [[Definition:Trivial Ring|trivial ring]] over an [[Definition:Underlying Set of Structure|underlying set]].
+Let $\map {\MM_R} n$ denote the [[Definition:Matrix Space|$n \times n$ matrix space]] over $R$.
+Then [[Definition:Matrix Product (Conventional)|(conventional) matrix multiplication]] is [[Definition:Commutative Operation|commutative]] over $\map {\MM_R} n$:
+:$\forall \mathbf A, \mathbf B \in \map {\MM_R} n: \mathbf {A B} = \mathbf {B A}$
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_n$ and $\mathbf B = \sqbrk b_n$ be [[Definition:Square Matrix|order $n$ square matrices]] over $R$.
+By definition of [[Definition:Matrix Product (Conventional)|matrix multiplication]], $\mathbf A \mathbf B = \mathbf C = \sqbrk c_n$ where:
+:$\displaystyle \forall i \in \closedint 1 n, j \in \closedint 1 n: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
+But by definition of the [[Definition:Trivial Ring|trivial ring]]:
+:$\forall a, b \in R: a \circ b = 0_R$
+where $0_R$ is the [[Definition:Ring Zero|zero]] of $R$.
+Thus $\mathbf A \mathbf B$ is the [[Definition:Zero Matrix|zero $n \times n$ matrix]].
+The same applies to $\mathbf B \mathbf A$, which is also the [[Definition:Zero Matrix|zero $n \times n$ matrix]].
+That is:
+:$\mathbf A \mathbf B = \mathbf B \mathbf A = \bszero_n$
+and the result follows by definition of [[Definition:Commutative Operation|commutative operation]].
+{{qed}}
+[[Category:Conventional Matrix Multiplication]]
+[[Category:Trivial Rings]]
+1t7trbnhmhdi2g676rctto7cmkgtb9k
+\end{proof}<|endoftext|>
+\section{Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle}
+Tags: Calculus of Variations, Isoperimetrical Problems
+
+\begin{theorem}
+Let $y$ be a [[Definition:Smooth Curve|smooth curve]], embedded in [[Definition:Dimension of Vector Space|$2$-dimensional]] [[Definition:Real Euclidean Space|Euclidean space]].
+Let $y$ have a total [[Definition:Length of Curve|length]] of $l$.
+Let it be contained in the upper halfplane with an exception of [[Definition:Endpoints of Directed Smooth Curve|endpoints]], which are on the x-[[Definition:Coordinate Axis|axis]] and are given.
+Suppose, $y$, together with a [[Definition:Line Segment|line segment]] connecting $y$'s [[Definition:Endpoints of Directed Smooth Curve|endpoints]], maximizes the enclosed [[Definition:Area|area]].
+Then $y$ is an [[Definition:Arc of Circle|arc]] of a [[Definition:Circle|circle]].
+\end{theorem}
+
+\begin{proof}
+{{WLOG}}, we choose our [[Definition:Point of Reference|point of reference]] such that $y$ intersect x-[[Definition:Coordinate Axis|axis]] at [[Definition:Point|points]] $\tuple {-a, 0}$ and $\tuple {a, 0}$ for some $a > 0$.
+The [[Definition:Area|area]] below the [[Definition:Curve|curve]] $y$ is a [[Definition:Real Functional|functional]] of the following form:
+:$\ds A \sqbrk y = \int_{-a}^a y \rd x$ 
+Furthermore, $y$ has to satisfy the following conditions:
+:$\map y {-a} = \map y a = 0$
+:$\ds L \sqbrk y = \int_{-a}^a \sqrt {1 + y'^2} \rd x = l$
+By [[Simplest Variational Problem with Subsidiary Conditions]], there exists a [[Definition:Constant Mapping|constant]] $\lambda$ such that the [[Definition:Real Functional|functional]]:
+:$\ds A \sqbrk y + \lambda L \sqbrk y = \int_{-a}^a \paren {y + \lambda \sqrt {1 + y'^2} } \rd x$
+is [[Definition:Extremum of Functional|extremized]] by the [[Definition:Mapping|mapping]] $y$.
+The corresponding [[Definition:Euler's Equation for Vanishing Variation|Euler's Equation]] reads:
+:$1 + \lambda \dfrac \d {\d x} \dfrac {y'} {\sqrt {1 + y'^2} } = 0$
+[[Definition:Primitive (Calculus)|Integrating]] {{WRT|Integration}} $x$ once yields:
+:$x + \lambda \dfrac {y'} {\sqrt {1 + y'^2} } = C_1$
+Solve this for $y'$:
+:$\ds y' = \pm \frac {c_1 - x} {\sqrt {\lambda^2 - c_1^2 + 2 c_1 x - x^2} }$
+[[Definition:Primitive (Calculus)|Integration]] yields:
+:$\paren {x - C_1}^2 + \paren {y - C_2}^2 = \lambda^2$
+This is an equation for a [[Definition:Circle|circle]] with [[Definition:Radius of Circle|radius]] $\lambda$ and [[Definition:Center of Circle|center]] $\tuple {C_1, C_2}$.
+To find $C_1, C_2, \lambda$, apply boundary conditions and the [[Definition:Length of Curve|length]] constraint.
+From the boundary conditions we have that:
+{{begin-eqn}}
+{{eqn | l = \paren {-a - C_1}^2 
+      | r = \lambda^2
+}}
+{{eqn | l = \paren {a - C_1}^2 
+      | r = \lambda^2
+}}
+{{end-eqn}}
+Take the [[Definition:Subtraction|difference]] of these two equations:
+:$4 a C_1 = 0 \implies C_1 = 0$
+because $a > 0$.
+Apply one of the boundary conditions again, that is, at $\tuple {a, 0}$:
+:$a^2 + C_2^2 = \lambda^2$
+Then:
+:$C_2 = \pm \sqrt {\lambda^2 - a^2}$.
+which can be used to get rid of $C_2$.
+The last [[Definition:Parameter of Differential Equation|parameter]] to find is $\lambda$.
+We have two cases: 
+:the [[Definition:Curve|curve]] is an [[Definition:Arc of Circle|arc]] of the upper [[Definition:Semicircle|semicirle]];
+:the [[Definition:Curve|curve]] is a [[Definition:Set Union|union]] of upper [[Definition:Semicircle|semicirle]] with two [[Definition:Arc of Circle|arcs]] of lower [[Definition:Semicircle|semicirle]].
+In the first case the [[Definition:Length of Curve|length]] constraint is:
+:$l = 2 \lambda \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$
+For real $\lambda$, if $\lambda \ge a$, then $l \in \R$.
+To find [[Definition:Extremum of Functional|extremal]] values, consider the [[Definition:Derivative|derivate]] $\dfrac {\d l} {\d \lambda}$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d l} {\d \lambda} 
+      | r = 2 \paren {\map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } - \frac 1 {\sqrt {\lambda^2 - 1} } } 
+}}
+{{eqn | o = < 
+      | r = 0
+      | c = [[Tangent Inequality]]
+}}
+{{end-eqn}}
+Hence the [[Definition:Domain of Variable|domain]] of $l$ is determined by boundary values.
+At the boundary of $\lambda = a$ we have:
+{{begin-eqn}}
+{{eqn | l = \lim_{\lambda \mathop \to a^+} l 
+      | r = \lim_{\lambda \mathop \to a^+} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
+}}
+{{eqn | r = 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
+      | c = [[Combination Theorem for Limits of Functions/Product Rule|Product Rule for Limits]]
+}}
+{{end-eqn}}
+To calculate the limit of such [[Definition:Composition of Mappings|composite function]] denote:
+{{begin-eqn}}
+{{eqn | l = \map f y
+      | r = \map \arctan y
+}}
+{{eqn | l = \map g {\lambda}
+      | r = \frac a {\sqrt {\lambda^2 - a^2} }
+}}
+{{end-eqn}}
+It follows that:
+:$\ds \lim_{\lambda \mathop \to a^+} \map g \lambda = + \infty$
+:$\ds \lim_{y \mathop \to \infty} \map f y = \frac \pi 2$
+[[Definition:Real Arctangent|Arctangent]] is [[Definition:Continuous on Interval|continuous]] for all $x \in \R$.
+Then, by [[Limit of Composite Function]]:
+:$\displaystyle 2 a \lim_{\lambda \mathop \to a^+} \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } = \pi a$
+At the boundary of $\lambda = + \infty$ we have:
+{{begin-eqn}}
+{{eqn | l = \lim_{\lambda \mathop \to \infty} l
+      | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } 
+}}
+{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \frac {\map \arctan {\frac a {\sqrt {\lambda^2 - a^2} } } } {\frac 1 \lambda} 
+      | c = Indeterminate limit $\infty \cdot 0$
+}}
+{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \frac {-\frac a {\lambda \sqrt {\lambda^2 - a^2} } } {-\frac 1 {\lambda^2} }
+      | c = [[L'Hôpital's Rule]]
+}}
+{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 a \frac \lambda {\sqrt {\lambda^2 - a^2} } 
+}}
+{{eqn | r = 2 a
+}}
+{{end-eqn}}
+In the second case the [[Definition:Length of Curve|length]] constraint is:
+:$l = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }$
+Similarly to the previous case:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d l} {\d \lambda} 
+      | r = 2 \pi - 2 \map \arctan {\frac 1 {\sqrt {\lambda^2 - 1} } } + \frac 1 {\sqrt {\lambda^2 - 1} } 
+}}
+{{eqn | o = >
+      | r = 0
+      | c = [[Tangent Inequality]]
+}}
+{{end-eqn}}
+Hence the [[Definition:Domain of Variable|domain]] of $l$ is determined by boundary values.
+At the boundary of $\lambda = a$ we have:
+{{begin-eqn}}
+{{eqn | l = \lim_{\lambda \mathop \to a^+} l
+      | r = 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }
+}}
+{{eqn | r = 2 \pi a - \pi a
+}}
+{{eqn | r = \pi a
+}}
+{{end-eqn}}
+As $\lambda$ approaches the infinity we have:
+{{begin-eqn}}
+{{eqn | l = \lim_{\lambda \mathop \to \infty} l
+      | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \paren {\pi - \arctan \dfrac a {\sqrt {\lambda^2 - a^2} } }
+}}
+{{eqn | r = \lim_{\lambda \mathop \to \infty} 2 \lambda \pi - 2 a
+}}
+{{eqn | r = \infty
+}}
+{{end-eqn}}
+Therefore:
+:$\forall l \ge 2 a: \exists \lambda \ge a$
+Hence, within these constraints the real solution maximizing the [[Definition:Area|area]] with fixed [[Definition:Endpoints of Directed Smooth Curve|endpoints]] is an [[Definition:Arc of Circle|arc]] of a [[Definition:Circle|circle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring Subtraction equals Zero iff Elements are Equal}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$
+Then:
+:$\forall a, b \in R: a - b = 0_R \iff a = b$
+where $a - b$ denotes [[Definition:Ring Subtraction|ring subtraction]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a - b
+      | r = 0_R
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a + \paren {-b}
+      | r = 0_R
+      | c = {{Defof|Ring Subtraction}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {a + \paren {-b} } + b
+      | r = 0_R + b
+      | c = [[Cancellation Laws]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a + \paren {b^{-1} + b}
+      | r = 0_R \circ b
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a
+      | r = b
+      | c = {{GroupAxiom|2}} and {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preordering of Products under Operation Compatible with Preordering}
+Tags: Preorder Theory, Compatible Relations
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $\precsim$ be a [[Definition:Preordering|preordering]] on $S$.
+Then $\precsim$ is [[Definition:Relation Compatible with Operation|compatible with $\circ$]] {{iff}}:
+:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$
+\end{theorem}
+
+\begin{proof}
+By definition, $\precsim$ is [[Definition:Relation Compatible with Operation|compatible with $\circ$]] {{iff}}:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y, z \in S:
+      | l = x \precsim y
+      | o = \implies
+      | r = \paren {x \circ z} \precsim \paren {y \circ z}
+}}
+{{eqn | l = x \precsim y
+      | o = \implies
+      | r = \paren {z \circ x} \precsim \paren {z \circ y}
+}}
+{{end-eqn}}
+=== Sufficient Condition ===
+Let $\precsim$ be [[Definition:Relation Compatible with Operation|compatible with $\circ$]].
+Then for all $x_1, x_2, y_1, y_2 \in S$:
+{{begin-eqn}}
+{{eqn | l = x_1 \precsim x_2
+      | o = \implies
+      | r = \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_1}
+      | c = {{Defof|Relation Compatible with Operation}}
+}}
+{{eqn | l = y_1 \precsim y_2
+      | o = \implies
+      | r = \paren {x_2 \circ y_1} \precsim \paren {x_2 \circ y_2}
+      | c = {{Defof|Relation Compatible with Operation}}
+}}
+{{end-eqn}}
+As $\precsim$ is a [[Definition:Preordering|preordering]] it is by definition [[Definition:Transitive Relation|transitive]].
+Thus it follows that:
+:$x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ y_1} \precsim \paren {x_2 \circ y_2}$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\precsim$ fulfil the property that:
+:$\forall x_1, x_2, y_1, y_2 \in S: x_1 \precsim x_2 \land y_1 \precsim y_2 \implies \paren {x_1 \circ x_2} \precsim \paren {y_1 \circ y_2}$
+As $\precsim$ is a [[Definition:Preordering|preordering]] it is by definition [[Definition:Reflexive Relation|reflexive]].
+That is:
+:$\forall z \in S: z \precsim z$
+Make the substitutions:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | o = \to
+      | r = x
+}}
+{{eqn | l = x_2
+      | o = \to
+      | r = y
+}}
+{{eqn | l = y_1
+      | o = \to
+      | r = z
+}}
+{{eqn | l = y_2
+      | o = \to
+      | r = z
+}}
+{{end-eqn}}
+It follows that:
+:$\forall x, y, z \in S: x \precsim y \implies \paren {x \circ z} \precsim \paren {y \circ z}$
+Similarly, make the substitutions:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | o = \to
+      | r = z
+}}
+{{eqn | l = x_2
+      | o = \to
+      | r = z
+}}
+{{eqn | l = y_1
+      | o = \to
+      | r = x
+}}
+{{eqn | l = y_2
+      | o = \to
+      | r = y
+}}
+{{end-eqn}}
+It follows that:
+:$\forall x, y, z \in S: x \precsim y \implies \paren {z \circ x} \precsim \paren {z \circ y}$
+So for all $x, y, z \in S$, both conditions are fulfilled:
+{{begin-eqn}}
+{{eqn | l = x \precsim y
+      | o = \implies
+      | r = \paren {x \circ z} \precsim \paren {y \circ z}
+}}
+{{eqn | l = x \precsim y
+      | o = \implies
+      | r = \paren {z \circ x} \precsim \paren {z \circ y}
+}}
+{{end-eqn}}
+for $\precsim$ to be [[Definition:Relation Compatible with Operation|compatible with $\circ$]].
+{{qed|lemma}}
+The result follows.
+{{qed}}
+[[Category:Preorder Theory]]
+[[Category:Compatible Relations]]
+9pogqcvs165m94664h0ve6uy63qtwsu
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+:$\exists x \in \Z_{>0}:  p \nmid x, x \ge \dfrac {p + 1} 2$
+\end{theorem}
+
+\begin{proof}
+Let $x = p + 1$.
+Then $p \nmid x$ and:
+:$x = p + 1 > p > 0$
+{{qed}}
+[[Category:P-adic Norm not Complete on Rational Numbers]]
+klt8vnjgaediwd7qt6bk8eururfgx8p
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 5}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+:$\displaystyle \lim_{n \mathop \to \infty} x_n^k = a$ in $\struct {\Q, \norm {\,\cdot\,}_p}$
+\end{theorem}
+
+\begin{proof}
+By assumption:
+:$\forall n \in \N: p^n \divides \paren {x_n^k - a}$
+By the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
+:$\forall n \in \N: \norm {x_n^k - a}_p \le \dfrac 1 {p^n}$
+By [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
+By [[Squeeze Theorem for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm {x_n^k - a}_p = 0$.
+By the definition of [[Definition:Convergent Sequence in Normed Division Ring|convergence]] in $\struct {\Q, \norm {\,\cdot\,}_p}$ then:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n^k = a$
+{{qed}}
+[[Category:P-adic Norm not Complete on Rational Numbers]]
+orpqibd6mumv5ml5n87nontx9ivpf76
+\end{proof}<|endoftext|>
+\section{Characterisation of Cauchy Sequence in Non-Archimedean Norm/Corollary 1}
+Tags: Metric Spaces, Normed Spaces, P-adic Number Theory
+
+\begin{theorem}
+Let $\norm {\,\cdot\,}_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]] for some [[Definition:Prime Number|prime]] $p$.
+Let $\sequence {x_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Integer|integers]] such that:
+:$\forall n: x_{n + 1} \equiv x_n \pmod {p^n}$
+Then:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence (Normed Division Ring)|Cauchy sequence]] in $\struct {\Q, \norm {\,\cdot\,}_p}$.
+\end{theorem}
+
+\begin{proof}
+By assumption:
+:$\forall n \in \N: p^n \divides \paren {x_{n + 1} - x_n}$
+By the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
+:$\forall n \in \N: \norm {x_{n + 1} - x_n}_p \le \dfrac 1 {p^n}$
+By [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \dfrac 1 {p^n} = 0$
+By the [[Squeeze Theorem for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \norm{x_{n+1} - x_n}_p = 0$.
+From [[P-adic Norm is Non-Archimedean Norm]], the [[Definition:P-adic Norm|$p$-adic norm]] is  [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean]].
+By [[Characterisation of Cauchy Sequence in Non-Archimedean Norm]]:
+:$\sequence {x_n}$ is a [[Definition:Cauchy Sequence (Normed Division Ring)|Cauchy sequence]] in $\struct {\Q, \norm {\,\cdot\,}_p}$.
+{{qed}}
+[[Category:Metric Spaces]]
+[[Category:Normed Spaces]]
+[[Category:P-adic Number Theory]]
+hec0ra2qr4h55bn1py1cwb2l55mknq0
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+:$a \in \Z_{> 0}: \nexists \,c \in \Z : c^k = a$
+\end{theorem}
+
+\begin{proof}
+Since $x, p > 0$ then $a > 0$.
+{{AimForCont}} for some $c \in \Z:c^k = a$.
+Since $c^k \in \Z$, by [[Nth Root of Integer is Integer or Irrational]] then:
+:$c \in \Z$
+Suppose $k$ is [[Definition:Odd Integer|odd]].
+Since $a > 0$, by [[Odd Power Function is Strictly Increasing]] then $c > 0$
+Hence $a = \size c^k$
+On the other hand, suppose $k$ is [[Definition:Even Integer|even]], that is $k = 2l$ for some $l \in Z_{> 0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = c^{2l}
+}}
+{{eqn | r = \paren {c^2}^l
+}}
+{{eqn | r = \paren {\size c^2}^l
+      | c = [[Equivalence of Definitions of Absolute Value Function]]
+}}
+{{eqn | r = \size c^{2l}
+}}
+{{eqn | r = \size c^k
+}}
+{{end-eqn}}
+In either case $\size c \in \Z_{> 0}$ and $\size c^k = a$
+Let $d = \size c$
+ 
+By the definition of $a$ it follows that $d^k = x^k + p$
+Hence:
+{{begin-eqn}}
+{{eqn | l = p 
+      | r = d^k - x^k
+}}
+{{eqn | r = \paren {d - x} \paren {d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + c x^{k - 2} + x^{k - 1} }
+      | c = [[Difference of Two Powers]]
+}}
+{{end-eqn}}
+Let $y = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}$
+Since $d, x \in \Z_{> 0}$ then $d - x \in \Z$ and $y \in \Z$
+So $d - x$ and $y$ are [[Definition:Factor|factors]] of $p$
+The [[Definition:Factor|factors]] of $p$ by definition are:
+:$\pm 1$ and $\pm p$
+Since $d, x \in \Z_{> 0}$ then:
+{{begin-eqn}}
+{{eqn | l = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}
+      | o = \ge 
+      | r = d^{k - 1} + x^{k - 1}
+}}
+{{eqn | o = \ge 
+      | r = d + x
+}}
+{{eqn | o = \ge 
+      | r = 1 + 1
+}}
+{{eqn | o = \ge 
+      | r = 2
+}}
+{{end-eqn}}
+Hence $y = p$
+Then:
+{{begin-eqn}}
+{{eqn | l = p 
+      | r = d^{k - 1} + d^{k - 2} x + d^{k - 3} x^2 + \dotsb + d x^{k - 2} + x^{k - 1}
+}}
+{{eqn | o = \ge 
+      | r = d^{k - 1} + x^{k - 1}
+}}
+{{eqn | o = \ge 
+      | r = d + x
+}}
+{{end-eqn}}
+It also follows that $d - x = 1$, that is, $d = x + 1$
+Then
+{{begin-eqn}}
+{{eqn | l = d + x
+      | r = \paren {x + 1} + x
+}}
+{{eqn | r = 2x + 1
+}}
+{{eqn | o = >
+      | r = 2x
+}}
+{{eqn | o = >
+      | r = p + 1
+}}
+{{eqn | o = >
+      | r = p
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the previous conclusion that $p \ge d + x$
+So:
+:$\nexists \,c \in \Z : c^k = a$
+{{qed}}
+[[Category:P-adic Norm not Complete on Rational Numbers]]
+phaadqtr1ve7gbf201hybvqikj8ieqs
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 3}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+:$\map f {x_1} \equiv 0 \pmod p$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map f {x_1}
+      | r = x_1^k - \paren {x_1^k + p}
+}}
+{{eqn | r = \paren {x_1^k - x_1^k} - p
+}}
+{{eqn | r = -p
+}}
+{{eqn | o = \equiv
+      | r = 0 \pmod p
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:P-adic Norm not Complete on Rational Numbers]]
+4y313yewjfwjz6avent2oo6zfqueega
+\end{proof}<|endoftext|>
+\section{P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 4}
+Tags: P-adic Norm not Complete on Rational Numbers
+
+\begin{theorem}
+:$\map {f'} {x_1} \not \equiv 0 \pmod p$
+\end{theorem}
+
+\begin{proof}
+By [[Euclid's Lemma for Prime Divisors]]:
+:$p \nmid k x_1^{k - 1}$
+Hence:
+:$k x_1^{k - 1} \not \equiv 0 \mod p$
+The [[Definition:Formal Derivative of Polynomial|formal derivative]] $\map {f'} X \in \Z \sqbrk X$ of $\map f X$ is by definition:
+:$k X^{k - 1}$
+Then:
+:$\map {f'} {x_1} = k x_1^{k - 1} \not \equiv 0 \pmod p$
+{{qed}}
+[[Category:P-adic Norm not Complete on Rational Numbers]]
+nvzcpl2u0gv7vtca4wdgk9kegsflq64
+\end{proof}<|endoftext|>
+\section{Surjection from Natural Numbers iff Countable/Corollary 1}
+Tags: Countable Sets, Surjections
+
+\begin{theorem}
+Let $T$ be a [[Definition:Countably Infinite Set|countably infinite set]].
+Let $S$ be a [[Definition:Non-Empty Set|non-empty set]].
+Then $S$ is [[Definition:Countable Set|countable]] {{iff}} [[Definition:Existential Quantifier|there exists]] a [[Definition:Surjection|surjection]] $f: T \to S$.
+\end{theorem}
+
+\begin{proof}
+Let $g: T \to \N$ be a [[Definition:Bijection|bijection]] from $T$ to $\N$.
+By [[Inverse of Bijection is Bijection]], $g^{-1}: \N \to T$ is a [[Definition:Bijection|bijection]] from $\N$ to $T$.
+=== Necessary Condition ===
+Let $f: T \to S$ be a [[Definition:Surjection|surjection]].
+By [[Composite of Surjections is Surjection]], $f \circ g^{-1}:\N \to S$ is a [[Definition:Surjection|surjection]].
+By [[Surjection from Natural Numbers iff Countable]], $S$ is a [[Definition:Countable Set|countable set]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $S$ is [[Definition:Countable Set|countable]] and [[Definition:Non-Empty Set|non-empty]].
+By [[Surjection from Natural Numbers iff Countable]], there exists a [[Definition:Surjection|surjection]] $f: \N \to S$.
+By [[Composite of Surjections is Surjection]], $f \circ g:T \to S$ is a [[Definition:Surjection|surjection]].
+{{qed}}
+[[Category:Countable Sets]]
+[[Category:Surjections]]
+c07x8v8ecbjipe10fv8564tehnr3ojm
+\end{proof}<|endoftext|>
+\section{Surjection from Natural Numbers iff Countable/Corollary 2}
+Tags: Countable Sets, Surjections
+
+\begin{theorem}
+Let $T$ be a [[Definition:Countably Infinite Set|countably infinite set]].
+Let $S$ be an [[Definition:Uncountable Set|uncountable set]].
+Let $f:T \to S$ be a [[Definition:Mapping|mapping]].
+Then $f$ is not a [[Definition:Surjection|surjection]].
+\end{theorem}
+
+\begin{proof}
+By [[Surjection from Natural Numbers iff Countable/Corollary 1|Corollary 1]] no [[Definition:Mapping|mapping]] from $T$ to $S$ is a [[Definition:Surjection|surjection]].
+{{qed}}
+[[Category:Countable Sets]]
+[[Category:Surjections]]
+220auysxelec96dq78xe8ymn2ex26jk
+\end{proof}<|endoftext|>
+\section{P-adic Numbers are Uncountable}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be any [[Definition:Prime Number|prime number]].
+The [[Definition:Set|set]] of [[Definition:P-adic Number|$p$-adic numbers]] $\Q_p$ is an [[Definition:Uncountable Set|uncountable set]].
+\end{theorem}
+
+\begin{proof}
+{{WIP}}
+\end{proof}<|endoftext|>
+\section{Ordered Integral Domain is Totally Ordered Ring}
+Tags: Ordered Integral Domains, Totally Ordered Rings
+
+\begin{theorem}
+Let $\struct {D, +, \times, \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]].
+Then $\struct {D, +, \times, \le}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]].
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {D, +, \times, \le}$ is an [[Definition:Integral Domain|integral domain]] endowed with a [[Definition:Strict Positivity Property|strict positivity property]].
+From [[Strict Positivity Property induces Total Ordering]], the [[Definition:Ordering|ordering]] $\le$ on $\struct {D, +, \times, \le}$ is a [[Definition:Total Ordering|total ordering]].
+Hence the result by definition of [[Definition:Totally Ordered Ring|totally ordered ring]].
+{{qed}}
+[[Category:Ordered Integral Domains]]
+[[Category:Totally Ordered Rings]]
+aunzqytajll4lim3grktfz3nxvzq9dr
+\end{proof}<|endoftext|>
+\section{Strict Negativity is equivalent to Strictly Preceding Zero}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+:$\map N a \iff a < 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N a
+      | o = \leadstoandfrom
+      | r = \map P {-a}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \map P {-a + 0}
+      | c = 
+}}
+{{eqn | o = \leadstoandfrom
+      | r = a < 0
+      | c = [[Strict Positivity Property induces Total Ordering]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Negativity is equivalent to Strict Positivity of Negative}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+:$\map P a \iff \map N {-a}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map P a
+      | o = \leadstoandfrom
+      | r = \map P {-\paren {-a} }
+      | c = 
+}}
+{{eqn | o = \leadstoandfrom
+      | r = \map N {-a}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Strictly Negative Elements is Strictly Negative}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+:$\map N a, \map N b \implies \map N {a + b}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N a, \map N b
+      | o = \leadsto
+      | r = \map P {-a}, \map P {-b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \map P {\paren {-a} + \paren {-b} }
+      | c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 1)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-\paren {a + b} }
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \map N {a + b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Two Strictly Negative Elements is Strictly Positive}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+:$\map N a, \map N b \implies \map P {a \times b}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N a, \map N b
+      | o = \leadsto
+      | r = \map P {-a}, \map P {-b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \map P {\paren {-a} \times \paren {-b} }
+      | c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 2)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \map P {a \times b}
+      | c = [[Product of Ring Negatives]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Strictly Negative Element with Strictly Positive Element is Strictly Negative}
+Tags: Ordered Integral Domains
+
+\begin{theorem}
+:$\map N a, \map P b \implies \map N {a \times b}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map N a, \map P b
+      | o = \leadsto
+      | r = \map P {-a}, \map P b
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{eqn | o = \leadsto
+      | r = \map P {\paren {-a} \times b}
+      | c = [[Definition:Strict Positivity Property|Strict Positivity Property: $(P \, 2)$]]
+}}
+{{eqn | o = \leadsto
+      | r = \map P {-\paren {a \times b} }
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | o = \leadsto
+      | r = \map N {a \times b}
+      | c = {{Defof|Strict Negativity Property}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Well-Ordered Integral Domain}
+Tags: Well-Ordered Integral Domains
+
+\begin{theorem}
+{{TFAE|def = Well-Ordered Integral Domain}}
+Let $\struct {D, +, \times \le}$ be an [[Definition:Ordered Integral Domain|ordered integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$.
+\end{theorem}
+
+\begin{proof}
+=== $(1)$ implies $(2)$ ===
+Let $\struct {D, +, \times \le}$ be a [[Definition:Well-Ordered Integral Domain/Definition 1|well-ordered integral domain by definition 1]].
+Then by definition the [[Definition:Total Ordering Induced by Strict Positivity Property|ordering $\le$]] is a [[Definition:Well-Ordering|well-ordering]] on the [[Definition:Set|set]] $P$ of [[Definition:Strictly Positive|(strictly) positive elements]] of $D$.
+Let $S \subseteq P$.
+Thus by definition of [[Definition:Well-Ordering|well-ordering]], $S$ has a [[Definition:Minimal Element|minimal element]].
+Thus $\struct {D, +, \times \le}$ is a [[Definition:Well-Ordered Integral Domain/Definition 2|well-ordered integral domain by definition 2]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $\struct {D, +, \times \le}$ be a [[Definition:Well-Ordered Integral Domain/Definition 2|well-ordered integral domain by definition 2]].
+Then by definition every [[Definition:Subset|subset]] $S$ of the [[Definition:Set|set]] $P$ of [[Definition:Strictly Positive|(strictly) positive elements]] of $D$ has a [[Definition:Minimal Element|minimal element]]:
+:$\forall S \subseteq D_{\ge 0_D}: \exists x \in S: \forall a \in S: x \le a$
+where $D_{\ge 0_D}$ denotes all the [[Definition:Element|elements]] $d \in D$ such that $\map P d$.
+Thus by definition $\le$ is a [[Definition:Well-Ordering|well-ordering]] of $P$.
+Thus $\struct {D, +, \times \le}$ is a [[Definition:Well-Ordered Integral Domain/Definition 1|well-ordered integral domain by definition 1]].
+{{qed}}
+[[Category:Well-Ordered Integral Domains]]
+3li6rwnx3jgfsghvnaajbh37busnvzg
+\end{proof}<|endoftext|>
+\section{Principle of Mathematical Induction/Zero-Based}
+Tags: Principle of Mathematical Induction
+
+\begin{theorem}
+Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N$.
+Suppose that:
+:$(1): \quad \map P 0$ is [[Definition:True|true]]
+:$(2): \quad \forall k \in \N: k \ge 0 : \map P k \implies \map P {k + 1}$
+Then:
+:$\map P n$ is [[Definition:True|true]] for all $n \in \N$.
+\end{theorem}
+
+\begin{proof}
+Consider $\N$ defined as a [[Definition:Peano Structure|Peano structure]].
+The result follows from [[Principle of Mathematical Induction for Peano Structure]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Mathematical Induction/One-Based}
+Tags: Principle of Mathematical Induction
+
+\begin{theorem}
+Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
+Suppose that:
+:$(1): \quad \map P 1$ is [[Definition:True|true]]
+:$(2): \quad \forall k \in \N_{>0}: k \ge 1 : \map P k \implies \map P {k + 1}$
+Then:
+:$\map P n$ is [[Definition:True|true]] for all $n \in \N_{>0}$.
+\end{theorem}<|endoftext|>
+\section{Principle of Finite Induction/Zero-Based}
+Tags: Principle of Finite Induction
+
+\begin{theorem}
+Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]].
+Suppose that:
+:$(1): \quad 0 \in S$
+:$(2): \quad \forall n \in \N : n \in S \implies n + 1 \in S$
+Then:
+:$S = \N$
+\end{theorem}
+
+\begin{proof}
+Consider $\N$ defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+The result follows directly from [[Principle of Mathematical Induction for Naturally Ordered Semigroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Finite Induction/One-Based}
+Tags: Principle of Finite Induction
+
+\begin{theorem}
+Let $S \subseteq \N_{>0}$ be a [[Definition:Subset|subset]] of the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
+Suppose that:
+:$(1): \quad 1 \in S$
+:$(2): \quad \forall n \in \N_{>0} : n \in S \implies n + 1 \in S$
+Then:
+:$S = \N_{>0}$
+\end{theorem}<|endoftext|>
+\section{Second Principle of Finite Induction/Zero-Based}
+Tags: Second Principle of Finite Induction
+
+\begin{theorem}
+Let $S \subseteq \N$ be a [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|natural numbers]].
+Suppose that:
+:$(1): \quad 0 \in S$
+:$(2): \quad \forall n \in \N: \paren {\forall k: 0 \le k \le n \implies k \in S} \implies n + 1 \in S$
+Then:
+:$S = \N$
+\end{theorem}
+
+\begin{proof}
+Define $T$ as:
+:$T = \set {n \in \N : \forall k: 0 \le k \le n: k \in S}$
+Since $n \le n$, it follows that $T \subseteq S$.
+Therefore, it will suffice to show that:
+:$\forall n \ge 0: n \in T$
+Firstly, we have that $0 \in T$ {{iff}} the following condition holds:
+:$\forall k: 0 \le k \le 0 \implies k \in S$
+Since $0 \in S$, it thus follows that $0 \in T$.
+Now suppose that $n \in T$; that is:
+:$\forall k: 0 \le k \le n \implies k \in S$
+By $(2)$, this implies:
+:$n + 1 \in S$
+Thus, we have:
+:$\forall k: 0 \le k \le n + 1 \implies k \in S$
+{{MissingLinks|[[Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor]] for $\N$}}
+Therefore, $n + 1 \in T$.
+Hence, by the [[Principle of Finite Induction]]:
+:$\forall n \ge 0: n \in T$
+That is:
+:$T = \N$
+and as $S \subseteq \N$ it follows that:
+:$S = N$
+{{Qed}}
+[[Category:Second Principle of Finite Induction]]
+4mwpf2fajzm72tpav4jdn67u9l0guw1
+\end{proof}<|endoftext|>
+\section{Second Principle of Finite Induction/One-Based}
+Tags: Second Principle of Finite Induction
+
+\begin{theorem}
+Let $S \subseteq \N_{>0}$ be a [[Definition:Subset|subset]] of the [[Definition:1-Based Natural Numbers|$1$-based natural numbers]].
+Suppose that:
+:$(1): \quad 1 \in S$
+:$(2): \quad \forall n \in \N_{>0}: \paren {\forall k: 1 \le k \le n \implies k \in S} \implies n + 1 \in S$
+Then:
+:$S = \N_{>0}$
+\end{theorem}
+
+\begin{proof}
+Define $T$ as:
+:$T = \set {n \in \N_{>0}: \forall k: 1 \le k \le n: k \in S}$
+Since $n \le n$, it follows that $T \subseteq S$.
+Therefore, it will suffice to show that:
+:$\forall n \ge 1: n \in T$
+Firstly, we have that $1 \in T$ {{iff}} the following condition holds:
+:$\forall k: 1 \le k \le 1 \implies k \in S$
+Since $1 \in S$, it thus follows that $1 \in T$.
+Now suppose that $n \in T$; that is:
+:$\forall k: 1 \le k \le n \implies k \in S$
+By $(2)$, this implies:
+:$n + 1 \in S$
+Thus, we have:
+:$\forall k: 1 \le k \le n + 1 \implies k \in S$
+{{MissingLinks|[[Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor]] for $\N$}}
+Therefore, $n + 1 \in T$.
+Hence, by the [[Principle of Finite Induction]]:
+:$\forall n \ge 1: n \in T$
+That is:
+:$T = \N_{>0}$
+and as $S \subseteq \N_{>0}$ it follows that:
+:$S = N_{>0}$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Second Principle of Mathematical Induction/Zero-Based}
+Tags: Second Principle of Mathematical Induction
+
+\begin{theorem}
+Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N$.
+Suppose that:
+:$(1): \quad \map P 0$ is [[Definition:True|true]]
+:$(2): \quad \forall k \in \N: \map P 0 \land \map P 1 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
+Then:
+:$\map P n$ is [[Definition:True|true]] for all $n \in \N$.
+\end{theorem}
+
+\begin{proof}
+For each $n \in \N$, let $\map {P'} n$ be defined as:
+:$\map {P'} n := \map P 0 \land \dots \land \map P n$
+It suffices to show that $\map {P'} n$ is true for all $n \in \N$.
+It is immediate from the assumption $\map P 0$ that $\map {P'} 0$ is [[Definition:True|true]].
+Now suppose that $\map {P'} n$ holds.
+By $(2)$, this implies that $\map P {n + 1}$ holds as well.
+Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds.
+Thus by the [[Principle of Mathematical Induction]]:
+:$\map {P'} n$ holds for all $n \in \N$
+as desired.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Second Principle of Mathematical Induction/One-Based}
+Tags: Second Principle of Mathematical Induction
+
+\begin{theorem}
+Let $\map P n$ be a [[Definition:Propositional Function|propositional function]] depending on $n \in \N_{>0}$.
+Suppose that:
+:$(1): \quad \map P 1$ is [[Definition:True|true]]
+:$(2): \quad \forall k \in \N_{>0}: \map P 1 \land \map P 2 \land \ldots \land \map P {k - 1} \land \map P k \implies \map P {k + 1}$
+Then:
+:$\map P n$ is [[Definition:True|true]] for all $n \in \N_{>0}$.
+\end{theorem}<|endoftext|>
+\section{Null Sequences form Maximal Left and Right Ideal/Corollary 1}
+Tags: Null Sequences form Maximal Left and Right Ideal
+
+\begin{theorem}
+Then $\NN$ is a [[Definition:Maximal Ideal of Ring|maximal ring ideal]] of $\CC$.
+\end{theorem}
+
+\begin{proof}
+By [[Null Sequences form Maximal Left and Right Ideal]] then $\NN$ is a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] of $\CC$. 
+A [[Definition:Field (Abstract Algebra)|field]] is by definition a [[Definition:Commutative Ring|commutative ring]].
+In a [[Definition:Commutative Ring|commutative ring]], a [[Definition:Maximal Left Ideal of Ring|maximal left ideal]] is by definition a [[Definition:Maximal Ideal of Ring|maximal ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Integer has Finite Number of Divisors}
+Tags: Number Theory, Divisors
+
+\begin{theorem}
+Let $n \in \Z_{\ne 0}$ be a non-[[Definition:Zero (Number)|zero]] [[Definition:Integer|integer]].
+Then $n$ has a [[Definition:Finite Set|finite number]] of [[Definition:Divisor of Integer|divisors]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be the [[Definition:Set|set]] of all [[Definition:Divisor of Integer|divisors]] of $n$.
+Then from [[Absolute Value of Integer is not less than Divisors]]:
+:$\forall m \in S: -n \le m \le n$
+Thus $S$ is [[Definition:Finite Set|finite]].
+{{qed}}
+[[Category:Number Theory]]
+[[Category:Divisors]]
+dhwii31ikzevtdctqmctmo6ixevhs9s
+\end{proof}<|endoftext|>
+\section{Coprimality Relation is Non-Reflexive}
+Tags: Coprime Integers
+
+\begin{theorem}
+:$\perp$ is [[Definition:Non-Reflexive Relation|non-reflexive]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+We have from [[GCD of Integer and Divisor]]:
+:$\gcd \set {n, n} = n$
+and so, for example:
+:$\gcd \set {2, 2} = 2$
+and so:
+:$2 \not \perp 2$
+Hence $\perp$ is not [[Definition:Reflexive Relation|reflexive]].
+But we also note that:
+:$\gcd \set {1, 1} = 1$
+and so:
+:$1 \perp 1$
+demonstrating that $\perp$ is not [[Definition:Antireflexive Relation|antireflexive]] either.
+The result follows by definition of [[Definition:Non-Reflexive Relation|non-reflexive relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Coprimality Relation is Symmetric}
+Tags: Coprime Integers
+
+\begin{theorem}
+:$\perp$ is [[Definition:Symmetric Relation|symmetric]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \perp
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {x, y}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {y, x}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | o = \perp
+      | r = x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Symmetric Relation|symmetric relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Coprimality Relation is not Antisymmetric}
+Tags: Coprime Integers
+
+\begin{theorem}
+:$\perp$ is not [[Definition:Antisymmetric Relation|antisymmetric]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+We have:
+:$\gcd \set {3, 5} = 1 = \gcd \set {5, 3}$
+and so:
+:$3 \perp 5$ and $5 \perp 3$
+However, it is not the case that $3 = 5$.
+The result follows by definition of [[Definition:Antisymmetric Relation|antisymmetric relation]].
+{{qed}}
+[[Category:Coprime Integers]]
+44ni75ofq0kkckmeabv42zb9lt25kkl
+\end{proof}<|endoftext|>
+\section{Coprimality Relation is Non-Transitive}
+Tags: Coprime Integers
+
+\begin{theorem}
+:$\perp$ is [[Definition:Non-Transitive Relation|non-transitive]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+We have:
+{{begin-eqn}}
+{{eqn | l = \gcd \set {2, 3}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \gcd \set {3, 4}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \gcd \set {2, 4}
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+Hence we have:
+:$2 \perp 3$ and $3 \perp 4$
+However, it is not the case that $2 \perp 4$.
+Thus $\perp$ is not [[Definition:Transitive Relation|transitive]].
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \gcd \set {2, 3}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \gcd \set {3, 5}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \gcd \set {2, 5}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+:$2 \perp 3$ and $3 \perp 5$
+and also:
+:$2 \perp 5$
+Thus $\perp$ is not [[Definition:Antitransitive Relation|antitransitive]] either.
+The result follows by definition of [[Definition:Non-Transitive Relation|non-transitive relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Polynomials over Reals is not Field}
+Tags: Polynomial Rings, Field Theory
+
+\begin{theorem}
+Let $\R \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] in an [[Definition:Indeterminate|indeterminate]] $X$ over $\R$.
+Then $\R \sqbrk X$ is not a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Polynomial over Real Numbers|polynomial]] $x + 1$ in $\R \sqbrk X$.
+There exists no [[Definition:Polynomial over Real Numbers|polynomial]] $\map f x$ such that:
+:$\paren {x + 1} \map f x = 1$
+This is because the {{LHS}} has [[Definition:Degree of Polynomial|degree]] $1$, and the {{RHS}} has [[Definition:Degree of Polynomial|degree]] $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Integers form Commutative Ring}
+Tags: Integers, Commutative Rings
+
+\begin{theorem}
+Let $2 \Z$ be the [[Definition:Even Integer|set of even integers]].
+Then $\struct {2 \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+However, $\struct {2 \Z, +, \times}$ is not an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Integer Multiples form Commutative Ring]], $\struct {2 \Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+As $2 \ne 1$, we also have from [[Integer Multiples form Commutative Ring]] that $\struct {2 \Z, +, \times}$ has no [[Definition:Unity of Ring|unity]].
+Hence by definition it is not an [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subrings of Integers are Sets of Integer Multiples/Examples/Even Integers}
+Tags: Subrings of Integers are Sets of Integer Multiples
+
+\begin{theorem}
+Let $2 \Z$ be the [[Definition:Even Integer|set of even integers]].
+Then $\struct {2 \Z, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+From [[Subrings of Integers are Sets of Integer Multiples]], a [[Definition:Ring (Abstract Algebra)|ring]] of the form $\struct {n \Z, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Z, +, \times}$ when $n \ge 1$.
+$\struct {2 \Z, +, \times}$ is such an example.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gaussian Integers does not form Subfield of Complex Numbers}
+Tags: Subfields, Gaussian Integers
+
+\begin{theorem}
+The [[Definition:Ring of Gaussian Integers|ring of Gaussian integers]]:
+:$\struct {\Z \sqbrk i, +, \times}$
+is not a [[Definition:Subfield|subfield]] of $\C$.
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+We have that:
+:$2 + 0 i \in \Z \sqbrk i$
+However there is no $z \in \Z \sqbrk i$ such that:
+:$x \paren {2 + 0 i} = 1 + 0 i$
+So, by definition, $\Z \sqbrk i$ is not a [[Definition:Field (Abstract Algebra)|field]].
+Thus $\Z \sqbrk i$ is not a [[Definition:Subfield|subfield]] of $\C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ideals of Ring of Integers Modulo m}
+Tags: Ring of Integers Modulo m, Ideal Theory, Ideals of Ring of Integers Modulo m
+
+\begin{theorem}
+Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\struct {\Z_m, +, \times}$ denote the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+The [[Definition:Ideal of Ring|ideals]] of $\struct {\Z_m, +, \times}$ are of the form:
+:$d \Z / m \Z$
+where $d$ is a [[Definition:Divisor of Integer|divisor]] of $m$.
+\end{theorem}
+
+\begin{proof}
+{{MissingLinks}}
+{{proofread}}
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $\struct {\Z_m, +, \times}$.
+$\struct {J, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +}$.
+Let $\struct {G, +}$ be a [[Definition:Subgroup|subgroup]] of $\struct {\Z_m, +}$.
+Then $\struct {G, +}$ is a [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] generated by $\gen d$, where $d \divides m$.
+We know that for a [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $k$]], the [[Definition:Order of Group|order]] of every [[Definition:Subgroup|subgroup]] is a [[Definition:Divisor of Integer|divisor]] of $k$.
+Also there is exactly one [Definition:Subgroup|subgroup]] for each [[Definition:Divisor of Integer|divisor]].
+It follows that all [[Definition:Ideal of Ring|ideals]] of $\struct {\Z_m, +, \times}$ are of form $\gen d$, where $d$ is a [[Definition:Strictly Positive Integer|positive]] [[Definition:Divisor of Integer|divisor]] of $m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+Then the [[Definition:Quotient Ring|quotient ring]] $\CC \,\big / \NN$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+By [[Quotient Ring of Cauchy Sequences is Division Ring]] then $\CC \,\big / \NN$ is a [[Definition:Division Ring|division ring]].
+By [[Cauchy Sequences form Ring with Unity/Corollary|Corollary to Cauchy Sequences form Ring with Unity]] then $\CC$ is a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
+By [[Quotient Ring of Commutative Ring is Commutative]] then $\CC \,\big / \NN$ is a [[Definition:Commutative Ring|commutative]] [[Definition:Division Ring|division ring]], that is, a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Cauchy Sequences is Normed Division Ring/Corollary 1}
+Tags: Completion of Normed Division Ring
+
+\begin{theorem}
+Then $\struct {\CC \,\big / \NN, \norm {\, \cdot \,}_1 }$ is a [[Definition:Valued Field|valued field]].
+\end{theorem}
+
+\begin{proof}
+By [[Quotient Ring of Cauchy Sequences is Normed Division Ring]] then $\CC \,\big / \NN$ is a [[Definition:Normed Division Ring|normed division ring]].
+By [[Quotient Ring of Cauchy Sequences is Division Ring/Corollary 1|Corollary to Quotient Ring of Cauchy Sequences is Normed Division Ring]] then $\CC \,\big / \NN$ is a [[Definition:Field (Abstract Algebra)|field]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy Sequence Converges Iff Equivalent to Constant Sequence}
+Tags: Normed Division Rings
+
+\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]].
+Let $\\CC \,\big / \NN$ be the [[Quotient Ring of Cauchy Sequences is Division Ring|quotient ring of Cauchy sequences]] of $\CC$ by the [[Null Sequences form Maximal Left and Right Ideal|maximal ideal]] $\NN$.
+Let $\sequence {x_n} \in \CC$.
+Then $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] in $\struct {R, \norm {\,\cdot\,} }$ {{iff}}
+:$\exists a \in R: \sequence {x_n} \in \sequence {a, a, a, \dotsc} + \NN$
+where $\sequence {a, a, a, \dotsc} + \NN$ is the [[Definition:Left Coset|left coset]] in $\CC \, \big / \NN$ that contains the constant [[Definition:Sequence|sequence]] $\sequence {a, a, a, \dotsc}$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $a \in R$ {{iff}} $\displaystyle \lim_{n \to \infty} \norm {x_n - a} = 0$
+Then:
+{{begin-eqn}}
+{{eqn | l = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n - a} = 0
+      | o = \leadstoandfrom
+      | r = \sequence {x_n - a} \in \NN
+      | c = {{Defof|Null Sequence in Normed Division Ring}}
+}}
+{{eqn | r = \sequence {x_n} - \sequence {a, a, a, \dotsc} \in \NN
+      | o = \leadstoandfrom
+      | c = {{Defof|Ring of Cauchy Sequences|Addition in Ring of Cauchy Sequences}}
+}}
+{{eqn | r = \sequence {x_n} \in \sequence {a, a, a, \dotsc} + \NN
+      | o = \leadstoandfrom
+      | c = [[Element in Left Coset iff Product with Inverse in Subgroup]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Normed Division Rings]]
+kyl41kgiab0qw0p446n6xt5ds8zq1ne
+\end{proof}<|endoftext|>
+\section{Homomorphism of Ring Subtraction}
+Tags: Ring Homomorphisms
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Then:
+:$\forall a, b \in R_1: \map \phi {a -_1 b} = \map \phi a -_2 \map \phi b$
+where $a -_1 b$ denotes [[Definition:Ring Subtraction|subtraction]] of $b$ from $a$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \phi {a -_1 b}
+      | r = \map \phi {a +_1 \paren {-b} }
+      | c = {{Defof|Ring Subtraction}}
+}}
+{{eqn | r = \map \phi a +_2 \map \phi {-b}
+      | c = {{Defof|Ring Homomorphism}}
+}}
+{{eqn | r = \map \phi a +_2 \paren {-\map \phi b}
+      | c = [[Ring Homomorphism Preserves Negatives]]
+}}
+{{eqn | r = \map \phi a -_2 \map \phi b
+      | c = {{Defof|Ring Subtraction}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Modulo Operation/Limit 1}
+Tags: Limit of Modulo Operation
+
+\begin{theorem}
+Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
+Then $\displaystyle \lim_{y \mathop \to 0} x \bmod y = 0$.
+\end{theorem}
+
+\begin{proof}
+By [[Range of Modulo Operation for Positive Modulus]] and [[Range of Modulo Operation for Negative Modulus]] we have:
+:$- \size y \lt x \bmod y \lt \size y$
+The result follows from the [[Squeeze Theorem/Functions|Squeeze Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Modulo Operation/Limit 2}
+Tags: Limit of Modulo Operation
+
+\begin{theorem}
+Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
+Then $\displaystyle \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$.
+\end{theorem}
+
+\begin{proof}
+As $y \to \infty$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | m = x
+      | mo= <
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | m = \frac x y
+      | mo= <
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | o =
+      | m = \floor {\frac x y}
+      | mo= =
+      | r = 0
+}}
+{{end-eqn}}
+Therefore by the definition of [[Definition:Modulo Operation|modulo operation]]:
+{{begin-eqn}}
+{{eqn | l = \lim_{y \mathop \to \infty} x \bmod y
+      | r = \lim_{y \mathop \to \infty} x - y \floor {\dfrac x y}
+}}
+{{eqn | r = \lim_{y \mathop \to \infty} x - y \cdot 0
+}}
+{{eqn | r = x
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Modulo Operation}
+Tags: Limits of Functions, Limit of Modulo Operation
+
+\begin{theorem}
+Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]].
+Then holding $x$ fixed gives:
+:$\displaystyle \lim_{y \mathop \to 0} x \bmod y = 0$
+:$\displaystyle \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$
+{{expand|What about the limits with respect to $x$?}}
+\end{theorem}
+
+\begin{proof}
+=== [[Limit of Modulo Operation/Limit 1|Limit 1]] ===
+{{:Limit of Modulo Operation/Limit 1}}
+=== [[Limit of Modulo Operation/Limit 2|Limit 2]] ===
+{{:Limit of Modulo Operation/Limit 2}}
+[[Category:Limits of Functions]]
+[[Category:Limit of Modulo Operation]]
+44gsv3kmkf9ww8992bf3ug38ugwi974
+\end{proof}<|endoftext|>
+\section{Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary}
+Tags: Normed Division Rings
+
+\begin{theorem}
+Let $\struct {R, \norm{\,\cdot\,} }$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]] with [[Definition:Ring Zero|zero]] $0_R$
+Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] such that $\sequence {x_n}$ does not [[Definition:Convergent Sequence in Normed Division Ring|converge]] to $0_R$.
+Then:
+:$\exists N \in \N: \forall n, m \ge N: \norm {x_n} = \norm {x_m}$
+\end{theorem}
+
+\begin{proof}
+By [[Cauchy Sequence Is Eventually Bounded Away From Non-Limit]] then:
+:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
+Since $\sequence {x_n}$ is a [[Definition:Cauchy Sequence inNormed Division Ring|Cauchy sequence]] then:
+:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$
+Let $N = \max \set {N_1, N_2}$.
+Let $n, m \ge N$.
+Then:
+:$\norm {x_n - x_m} < C < \norm {x_n}$
+By [[Three Points in Ultrametric Space have Two Equal Distances/Corollary 4|Corollary to Three Points in Ultrametric Space have Two Equal Distances]] then:
+:$\norm {x_n} = \norm {x_m}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Unit of Ring}
+Tags: Units of Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
+{{TFAE|def = Unit of Ring}}
+\end{theorem}
+
+\begin{proof}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
+=== $(1)$ implies $(2)$ ===
+Let $x \in R$ be a [[Definition:Unit of Ring/Definition 1|unit of $\struct {R, +, \circ}$ by definition 1]].
+Then by definition:
+:$\exists y \in R: x \circ y = 1_R = y \circ x$
+That is, by definition of [[Definition:Divisor of Ring Element|divisor]]:
+:$x \divides 1_R$
+Thus $x$ is a [[Definition:Unit of Ring/Definition 2|unit of $\struct {R, +, \circ}$ by definition 2]].
+{{qed|lemma}}
+=== $(2)$ implies $(1)$ ===
+Let $x \in R$ be a [[Definition:Unit of Ring/Definition 2|unit of $\struct {R, +, \circ}$ by definition 2]].
+Then by definition:
+:$x \divides 1_R$
+By definition of [[Definition:Divisor of Ring Element|divisor]]:
+:$\exists t \in R: 1_R = t \circ x$
+Thus $x$ is a [[Definition:Unit of Ring/Definition 1|unit of $\struct {R, +, \circ}$ by definition 1]].
+{{qed}}
+[[Category:Units of Rings]]
+l4gkmgz1c66ajs0aoygzxdbic0xbwud
+\end{proof}<|endoftext|>
+\section{Bézout's Lemma/Euclidean Domain}
+Tags: Euclidean Domains, Greatest Common Divisor, Bézout's Lemma
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$ and whose [[Definition:Unity of Ring|unity]] is $1$.
+Let $\nu: D \setminus \set 0 \to \N$ be the [[Definition:Euclidean Valuation|Euclidean valuation]] on $D$.
+Let $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.
+Let $\gcd \set {a, b}$ be the [[Definition:Greatest Common Divisor of Ring Elements|greatest common divisor]] of $a$ and $b$.
+Then:
+:$\exists x, y \in D: a \times x + b \times y = \gcd \set {a, b}$
+such that $\gcd \set {a, b}$ is the [[Definition:Element|element]] of $D$ such that:
+:$\forall c = a \times x + b \times y \in D: \map \nu {\gcd \set {a, b} } \le \map \nu c$
+\end{theorem}
+
+\begin{proof}
+{{Proofread}}
+We are given that $a, b \in D$ such that $a$ and $b$ are not both equal to $0$.
+{{WLOG}}, suppose specifically that $b \ne 0$.
+Let $S \subseteq D$ be the [[Definition:Set|set]] defined as:
+:$S = \set {x \in D_{\ne 0}: x = m \times a + n \times b: m, n \in D}$
+where $D_{\ne 0}$ denotes $D \setminus 0$.
+Setting $m = 0$ and $n = 1$, for example, it is noted that $b \in S$.
+Therefore $S \ne \O$.
+By definition, $\nu$ has the properties:
+:$(1): \quad \forall a, b \in D, b \ne 0: \exists q, r \in D$ such that $\map \nu r < \map \nu b$, or $r = 0$, such that:
+::::$a = q \times b + r$
+:$(2): \quad \forall a, b \in D, b \ne 0$:
+::::$\map \nu a \le \map \nu {a \times b}$
+Let $\nu \sqbrk S$ denote the [[Definition:Image of Subset under Mapping|image]] of $S$ under $\nu$.
+We have that:
+:$\nu \sqbrk S \subseteq \N$
+Hence by the [[Well-Ordering Principle]] $\nu \sqbrk S$ has a [[Definition:Minimal Element|smallest element]].
+Let $d \in S$ be such that $\map \nu d$ is that [[Definition:Minimal Element|smallest element]] of $\nu \sqbrk S$.
+By definition of $S$, we have that:
+:$d = u \times a + v \times b$
+for some $u, v \in D$.
+Let $x \in S$.
+By $(2)$ above:
+:$x = q \times d + r$
+such that either:
+:$\map \nu r < \map \nu d$
+or:
+:$r = 0$
+{{AimForCont}} $r \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | lo= \exists m, n \in D:
+      | l = x
+      | r = m \times a + n \times b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r
+      | r = x - q \times d
+      | c = 
+}}
+{{eqn | r = \paren {m \times a + n \times b} - q \paren {u \times a + v \times b}
+      | c = 
+}}
+{{eqn | r = \paren {m - q \times u} a + \paren {n - q \times v} b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | o =
+      | r = \paren {r \in S} \land \paren {\map \nu r < \map \nu d}
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Contradiction|contradicts]] the choice of $d$ as the [[Definition:Element|element]] of $S$ such that $\map \nu d$ is the [[Definition:Minimal Element|smallest element]] of $\nu \sqbrk S$.
+Therefore:
+:$\forall x \in S: x = q \times d$
+for some $q \in D$.
+That is:
+:$\forall x \in S: d \divides x$
+where $\divides$ denotes [[Definition:Divisor of Ring Element|divisibility]].
+In particular:
+:$d \divides a = 1 \times a + 0 \times b$
+:$d \divides b = 0 \times a + 1 \times b$
+Thus:
+:$d \divides a \land d \divides b \implies \map \nu 1 \le \map \nu d \le \map \nu {\gcd \set {a, b} }$
+However, note that as $\gcd \set {a, b}$ also divides $a$ and $b$ (by definition), we have:
+{{begin-eqn}}
+{{eqn | l = \gcd \set {a, b}
+      | o = \divides
+      | r = \paren {u \times a + v \times b} = d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {a, b}
+      | o = \divides
+      | r = d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \nu {\gcd \set {a, b} }
+      | o = \le
+      | r = \map \nu d
+      | c = 
+}}
+{{end-eqn}}
+Since $d$ is the [[Definition:Element|element]] of $S$ such that $\map \nu d$ is the [[Definition:Minimal Element|smallest element]] of $\nu \sqbrk S$:
+:$\gcd \set {a, b} = d = u \times a + v \times b$
+{{qed}}
+{{Namedfor|Étienne Bézout|cat = Bézout}}
+\end{proof}<|endoftext|>
+\section{Euclidean Valuation of Non-Unit is less than that of Product}
+Tags: Euclidean Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$, and [[Definition:Unity of Ring|unity]] is $1$.
+Let the [[Definition:Euclidean Valuation|valuation function]] of $D$ be $\nu$.
+Let $b, c \in D_{\ne 0}$.
+Then:
+:If $c$ is not a [[Definition:Unit of Ring|unit]] of $D$ then $\map \nu b < \map \nu {b c}$
+\end{theorem}
+
+\begin{proof}
+By [[Euclidean Domain is Principal Ideal Domain|principal ideal domain]], $D$ is a [[Definition:Principal Ideal Domain|principal ideal domain]]
+Consider the [[Definition:Principal Ideal of Ring|principal ideal]] $U = \ideal b$ of $D$ generated by $b$.
+As $\nu$ is a [[Definition:Euclidean Valuation|valuation function]], we have:
+:$\forall x \in D, x \ne 0: \map \nu b \le \map \nu {b \times x}$
+As $U$ is a [[Definition:Principal Ideal of Ring|principal ideal]]:
+:$\forall a \in \ideal b: \exists x \in D: a = x \times b$
+and so:
+:$\forall a \in U: \map \nu b \le \map \nu a$
+Let $\map \nu b = \map \nu {b c}$.
+Then from above:
+:$\forall a \in U: \map \nu {b c} \le \map \nu a$
+From the argument in [[Euclidean Domain is Principal Ideal Domain]], we have that $U$ is also the [[Definition:Generator of Ideal|ideal generated by $b c$]].
+{{explain|The above statement is not very clear and needs to be expanded.}}
+Thus $a b$ is a [[Definition:Divisor of Ring Element|divisor]] of $b$ itself, that is:
+:$\exists y \in D: b = b c y$
+Thus:
+:$c y = 1$
+and so $c$ is a [[Definition:Unit of Ring|unit]] of $D$.
+So if $c$ is not a [[Definition:Unit of Ring|unit]], then it must be the case that $\map \nu b < \map \nu {b c}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element is Unit iff its Euclidean Valuation equals that of 1}
+Tags: Euclidean Domains
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be a [[Definition:Euclidean Domain|Euclidean domain]] whose [[Definition:Ring Zero|zero]] is $0$, and [[Definition:Unity of Ring|unity]] is $1$.
+Let the [[Definition:Euclidean Valuation|valuation function]] of $D$ be $\nu$.
+Let $a \in D$.
+Then:
+:$a$ is a [[Definition:Unit of Ring|unit]] of $D$ {{iff}} $\map \nu a = \map \nu 1$
+\end{theorem}
+
+\begin{proof}
+For $a \in D$ we have that:
+:$\map \nu 1 \le \map \nu {1 a} = \map \nu a$
+by definition of [[Definition:Euclidean Valuation|Euclidean valuation]].
+Let $a$ be a [[Definition:Unit of Ring|unit]] of $D$.
+Then:
+:$\exists b \in D: a b = 1$
+Then:
+:$\map \nu a \le \map \nu {a b} = \map \nu 1$
+and so:
+:$\map \nu a = \map \nu 1$
+{{qed|lemma}}
+Let $\map \nu a = \map \nu 1$.
+We can write this as:
+:$\map \nu {1 a} = \map \nu 1$
+and it follows from [[Euclidean Valuation of Non-Unit is less than that of Product]] that $a$ is a [[Definition:Unit of Ring|unit]] of $D$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gauss's Lemma on Primitive Polynomials over Ring}
+Tags: Polynomial Theory, Gauss's Lemma (Polynomial Theory)
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring with Unity|commutative ring with unity]].
+Let $f, g \in R \sqbrk X$ be [[Definition:Primitive Polynomial over Ring|primitive polynomials]].
+{{explain|[[Definition:Primitive Polynomial over Ring]]}}
+Then $f g$ is [[Definition:Primitive Polynomial over Ring|primitive]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
+[[Category:Polynomial Theory]]
+[[Category:Gauss's Lemma (Polynomial Theory)]]
+flfxswga8xeuuwnzrf368tz9ymcj507
+\end{proof}<|endoftext|>
+\section{Rational Polynomial is Content Times Primitive Polynomial/Existence}
+Tags: Rational Polynomial is Content Times Primitive Polynomial
+
+\begin{theorem}
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map f X \in \Q \sqbrk X$.
+Then:
+:$\map f X = \cont f \, \map {f^*} X$
+where:
+:$\cont f$ is the [[Definition:Content of Rational Polynomial|content]] of $\map f X$
+:$\map {f^*} X$ is a [[Definition:Primitive Polynomial (Ring Theory)|primitive polynomial]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Polynomial Coefficient|coefficients]] of $f$ expressed as [[Definition:Rational Number|fractions]].
+Let $k$ be any [[Definition:Positive Integer|positive integer]] that is [[Definition:Divisor of Integer|divisible]] by the [[Definition:Denominator|denominators]] of all the [[Definition:Polynomial Coefficient|coefficients]] of $f$.
+Such a number is bound to exist: just multiply all those denominators together, for example.
+Then $\map f X$ is a polynomial equal to $\dfrac 1 k$ multiplied by a polynomial with integral coefficients.
+Let $d$ be the [[Definition:Greatest Common Divisor of Integers|GCD]] of all these integral coefficients.
+Then $\map f X$ is equal to $\dfrac h k$ multiplied by a [[Definition:Primitive Polynomial (Ring Theory)|primitive polynomial]].
+\end{proof}<|endoftext|>
+\section{Rational Polynomial is Content Times Primitive Polynomial/Uniqueness}
+Tags: Rational Polynomial is Content Times Primitive Polynomial
+
+\begin{theorem}
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map f X \in \Q \sqbrk X$ be given.
+Then there exist [[Definition:Unique|unique]] [[Definition:Content of Rational Polynomial|content]] $\cont f$ of $\map f X$ and [[Definition:Primitive Polynomial (Ring Theory)|primitive polynomial]] $\map {f^*} X$ such that:
+:$\map f X = \cont f \, \map {f^*} X$
+\end{theorem}
+
+\begin{proof}
+Suppose that $a \cdot \map f X = b \cdot \map g X$ where $a, b \in \Q$ and $f, g$ are [[Definition:Primitive Polynomial (Ring Theory)|primitive]].
+Then:
+:$\map g X = \dfrac a b \map f X$
+where $\dfrac a b$ is some [[Definition:Rational Number|rational number]] which can be expressed as $\dfrac m n$ where $m$ and $n$ are [[Definition:Coprime Integers|coprime]].
+Then:
+:$\map g X = \dfrac m n \map f X$
+that is:
+:$m \cdot \map f X = n \cdot \map g X$
+Suppose $m > 1$.
+Then from [[Euclid's Lemma]] $m$ has a [[Definition:Divisor of Integer|divisor]] $p$ which does not divide $n$ (as $m \perp n$).
+So $m$ must divide every coefficient of $g$.
+But this can not be so, as $g$ is [[Definition:Primitive Polynomial (Ring Theory)|primitive]], so $m = 1$.
+In a similar way, $n = 1$.
+So $f = g$ and $a = b$, so demonstrating [[Definition:Unique|uniqueness]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Content of Polynomial in Dedekind Domain is Multiplicative}
+Tags: Content of Polynomial, Gauss's Lemma (Polynomial Theory)
+
+\begin{theorem}
+Let $R$ be a [[Definition:Dedekind Domain|Dedekind domain]].
+Let $f, g \in R \sqbrk X$ be [[Definition:Polynomial over Ring in One Variable|polynomials]].
+Let $\cont f$ denote the [[Definition:Content of Polynomial|content]] of $f$.
+Then $\cont {f g} = \cont f \cont g$ is the [[Definition:Product of Ideals of Ring|product]] of $\cont f$ and $\cont g$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Content of Polynomial]]
+[[Category:Gauss's Lemma (Polynomial Theory)]]
+6r0v780lamw3ass7o2x8pu1nvetx5f8
+\end{proof}<|endoftext|>
+\section{Factors of Polynomial with Integer Coefficients have Integer Coefficients}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map h X \in \Q \sqbrk X$ have [[Definition:Polynomial Coefficient|coefficients]] all of which are [[Definition:Integer|integers]].
+Let it be possible to express $\map h X$ as:
+:$\map h X = \map f X \, \map g X$
+where $\map f X, \map g X \in \Q \sqbrk X$.
+Then it is also possible to express $\map h X$ as:
+:$\map h X = \map {f'} X \, \map {g'} X$
+where:
+:$\map {f'} X, \map {g'} X \in \Q \sqbrk X$
+:the [[Definition:Polynomial Coefficient|coefficients]] of $\map {f'} X$ and $\map {g'} X$ are all [[Definition:Integer|integers]]
+:$\map {f'} X = a \map f X$ and $\map {g'} X = b \map f X$, for $a, b \in \Q$.
+\end{theorem}
+
+\begin{proof}
+Let $\cont h$ denote the [[Definition:Content of Rational Polynomial|content]] of $\map h X$.
+From [[Polynomial has Integer Coefficients iff Content is Integer]]:
+:$\cont h \in \Z$
+Let $\map h X = \map f X \, \map g X$ as suggested.
+Then from [[Rational Polynomial is Content Times Primitive Polynomial]]:
+{{begin-eqn}}
+{{eqn | l = \map h X
+      | r = \cont f \cont g \cdot \map {f'} X \, \map {g'} X
+      | c = [[Rational Polynomial is Content Times Primitive Polynomial]]
+}}
+{{eqn | r = \cont h \cdot \map {f'} X \, \map {g'} X
+      | c = [[Content of Rational Polynomial is Multiplicative]]
+}}
+{{end-eqn}}
+From the above, $\map {f'} X$ and $\map {g'} X$ are [[Definition:Primitive Polynomial (Ring Theory)|primitive]].
+Hence by definition:
+:$\cont {f'} = \cont {g'} = 1$
+From [[Polynomial has Integer Coefficients iff Content is Integer]], both $\map {f'} X$ and $\map {g'} X$ have [[Definition:Polynomial Coefficient|coefficients]] which are all [[Definition:Integer|integers]].
+We also have by definition of [[Definition:Content of Rational Polynomial|content]] that $\cont f$ and $\cont g$ are [[Definition:Rational Number|rational numbers]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial which is Irreducible over Integers is Irreducible over Rationals}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\Z \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Integer|integers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\Q \sqbrk X$ be the [[Definition:Ring of Polynomial Forms|ring of polynomial forms]] over the [[Definition:Field of Rational Numbers|field of rational numbers]] in the [[Definition:Indeterminate (Polynomial Theory)|indeterminate]] $X$.
+Let $\map f X \in \Z \sqbrk X$ be [[Definition:Irreducible Polynomial|irreducible]] in $\Z \sqbrk X$.
+Then $\map f X$ is also [[Definition:Irreducible Polynomial|irreducible]] in $\Q \sqbrk X$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.
+[[Definition:By Hypothesis|By hypothesis]]:
+:$\map f X \in \Z \sqbrk X$
+and so by definition has [[Definition:Polynomial Coefficient|coefficients]] all of which are [[Definition:Integer|integers]].
+But from [[Factors of Polynomial with Integer Coefficients have Integer Coefficients]] it follows that $\map f X$ can be expressed as:
+:$\map f X = \map {g'} X \, \map {h'} X$
+where both $\map {g'} X$ and $\map {h'} X$ are [[Definition:Element|elements]] of $\Q \sqbrk X$ which have [[Definition:Polynomial Coefficient|coefficients]] all of which are [[Definition:Integer|integers]].
+That is:
+:$\map {g'} X \in \Z \sqbrk X$
+and:
+:$\map {h'} X \in \Z \sqbrk X$
+This [[Definition:Contradiction|contradicts]] the statement that $\map f X$ is [[Definition:Irreducible Polynomial|irreducible]] in $\Z \sqbrk X$.
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dirichlet Function is Periodic}
+Tags: Periodic Functions
+
+\begin{theorem}
+Let $D: \R \to \R$ be a [[Definition:Dirichlet Function|Dirichlet function]]:
+:$\forall x \in \R: \map D x = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$
+Then $D$ is [[Definition:Real Periodic Function|periodic]].
+Namely, every non-[[Definition:Zero (Number)|zero]] [[Definition:Rational Number|rational number]] is a [[Definition:Periodic Element|periodic element]] of $D$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R$.
+Let $L \in \Q$.
+If $x \in \Q$, then:
+{{begin-eqn}}
+{{eqn | l = \map D {x + L}
+      | r = c
+      | c = [[Rational Addition is Closed]]
+}}
+{{eqn | r = \map D x
+}}
+{{end-eqn}}
+If $x \notin \Q$, then:
+{{begin-eqn}}
+{{eqn | l = \map D {x + L}
+      | r = d
+      | c = [[Rational Number plus Irrational Number is Irrational]]
+}}
+{{eqn | r = \map D x
+}}
+{{end-eqn}}
+Combining the above two shows that:
+:$\forall x \in \R: \map D x = \map D {x + L}$
+Hence the result.
+{{qed}}
+[[Category:Periodic Functions]]
+tl67hs3t48g1041uc6cl3wfo3tmenr4
+\end{proof}<|endoftext|>
+\section{Dirichlet Function has no Period}
+Tags: Periodic Functions
+
+\begin{theorem}
+The [[Definition:Dirichlet Function|Dirichlet functions]] are [[Definition:Real Periodic Function|periodic]] by [[Dirichlet Function is Periodic]].
+However, they do not admit a [[Definition:Period of Function|period]].
+That is, there does not exist a [[Definition:Smallest Element|smallest]] value $L \in \R_{> 0}$ such that:
+:$\forall x \in \R: \map D x = \map D {x + L}$
+\end{theorem}
+
+\begin{proof}
+Let $D: \R \to \R$ be a [[Definition:Dirichlet Function|Dirichlet function]].
+In proving that the [[Dirichlet Function is Periodic]], it was shown that every [[Definition:Nonzero|non-zero]] [[Definition:Rational Number|rational number]] is a [[Definition:Periodic Element|periodic element]] of $D$.
+Therefore, the [[Definition:Period of Function|period]] of $D$ must be the [[Definition:Smallest Element|smallest element]] of $\Q_{> 0}$.
+But from [[Rational Numbers are not Well-Ordered under Conventional Ordering]] there cannot exist such an element.
+Hence the result.
+{{qed}}
+[[Category:Periodic Functions]]
+gbw385yz4t49ue86fmt3m04luu36wc7
+\end{proof}<|endoftext|>
+\section{Existence of Nonconstant Periodic Function with no Period}
+Tags: Periodic Functions
+
+\begin{theorem}
+There exists a [[Definition:Real Function|real]], [[Definition:Nonconstant Function|non-constant function]] $f$ such that:
+:$(1): \quad f$ is [[Definition:Real Periodic Function|periodic]].
+:$(2): \quad f$ does '''not''' have a [[Definition:Period of Function|period]].
+\end{theorem}
+
+\begin{proof}
+By [[Dirichlet Function is Periodic]] and [[Dirichlet Function has no Period]], it is seen that the [[Definition:Dirichlet Function|Dirichlet functions]] are such an example.
+{{qed}}
+[[Category:Periodic Functions]]
+h8bvusygz1495o4a7vflouyy4m8j6jr
+\end{proof}<|endoftext|>
+\section{Vector Augend plus Addend equals Augend implies Addend is Zero}
+Tags: Vector Spaces
+
+\begin{theorem}
+Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] $F$.
+Let $\mathbf a, \mathbf b \in V$.
+Let $\mathbf a + \mathbf b = \mathbf a$.
+Then:
+:$\mathbf b = \bszero$
+where $\bszero$ is the [[Definition:Zero Vector|zero vector]] of $V$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf a + \mathbf b
+      | r = \mathbf a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-\mathbf a} + \paren {\mathbf a + \mathbf b}
+      | r = \paren {-\mathbf a} + \mathbf a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {-\mathbf a} + \mathbf a} + \mathbf b
+      | r = \paren {-\mathbf a} + \mathbf a
+      | c = {{Vector-space-axiom|2}}
+}}
+{{eqn | ll= \leadsto
+      | l = \bszero + \mathbf b
+      | r = \bszero
+      | c = {{Vector-space-axiom|4}}
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf b
+      | r = \bszero
+      | c = {{Vector-space-axiom|3}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Space of All Mappings is Vector Space}
+Tags: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $S$ be a set.
+Let $\struct {G^S, +_G', \circ}_R$ be the [[Definition:Vector Space of All Mappings|vector space of all mappings]] from $S$ to $G$.
+Then $\struct {G^S, +_G', \circ}_K$ is a [[Definition:Vector Space|$K$-vector space]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Module of All Mappings is Module]] and the definition of [[Definition:Vector Space|vector space]].
+\end{proof}<|endoftext|>
+\section{Unitary Module of All Mappings is Unitary Module}
+Tags: Examples of Unitary Modules
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {G^S, +_G', \circ}_R$ be the [[Definition:Module of All Mappings|module of all mappings]] from $S$ to $G$.
+Then $\struct {G^S, +_G', \circ}_R$ is a [[Definition:Unitary Module|unitary module]].
+\end{theorem}
+
+\begin{proof}
+From [[Module of All Mappings is Module]], we have that $\struct {G^S, +_G', \circ}_R$ is an [[Definition:Module|$R$-module]].
+To show that $\struct {G^S, +_G', \circ}_R$ is a [[Definition:Unitary Module|unitary $R$-module]], we verify the following:
+:$\forall f \in G^S: 1_R \circ f = f$
+Let $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Then:
+:$\forall x \in G: 1_R \circ x = x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {1_R \circ f} } x
+      | r = 1_R \circ \paren {\map f x}
+      | c = 
+}}
+{{eqn | r = \map f x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Direct Product of Unitary Modules is Unitary Module}
+Tags: Unitary Modules, Direct Products, Finite Direct Product of Unitary Modules is Unitary Module
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Unitary Module|unitary $R$-modules]].
+Let:
+:$\displaystyle G = \prod_{k \mathop = 1}^n G_k$
+be their [[Definition:Module Direct Product|direct product]].
+Then $G$ is a [[Definition:Unitary Module|unitary module]].
+\end{theorem}
+
+\begin{proof}
+This is a special case of [[Direct Product of Unitary Modules is Unitary Module]].
+\end{proof}
+
+\begin{proof}
+From [[Finite Direct Product of Modules is Module]] we have that $G$ is a [[Definition:Module|module]].
+It remains to be shown that:
+:$\forall x \in G: 1_R \circ x = x$
+Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 1_R \circ x
+      | r = 1_R \circ \tuple {x_1, x_2, \ldots, x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {1_R \circ x_1, 1_R \circ x_2, \ldots, 1_R \circ x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {x_1, x_2, \ldots, x_n}
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nonconstant Periodic Function with no Period is Discontinuous Everywhere}
+Tags: Periodic Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Periodic Function|real periodic function]] that does not have a [[Definition:Period of Function|period]].
+Then $f$ is either [[Definition:Constant Function|constant]] or [[Definition:Discontinuous Real Function/Everywhere|discontinuous everywhere]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ be a [[Definition:Real Periodic Function|real periodic function]] that does not have a [[Definition:Period of Function|period]].
+Let $x \in \R$.
+If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by [[Constant Function has no Period]].
+If $f$ is [[Definition:Nonconstant Function|non-constant]], then let $y$ be such a value.
+Let $\mathcal L_{f > 0}$ be the [[Definition:Set|set]] of all [[Definition:Positive Real Number|positive]] [[Definition:Periodic Element|periodic elements]] of $f$.
+This set is [[Definition:Non-Empty Set|non-empty]] by [[Absolute Value of Periodic Element is Peroidic]].
+It is seen that $<$ forms a [[Definition:Right-Total Relation|right-total relation]] on $\mathcal L_{f > 0}$, for if not then $f$ would have a [[Definition:Period of Function|period]].
+By the [[Axiom:Axiom of Dependent Choice|Axiom of Dependent Choice]] there must exist a [[Definition:Strictly Decreasing Real Sequence|strictly-decreasing sequence]] $\sequence {L_n}$ in $\mathcal L_{f > 0}$.
+Since this [[Definition:Sequence|sequence]] is [[Definition:Bounded Below Real Sequence|bounded below]] by zero, it follows via [[Monotone Convergence Theorem (Real Analysis)|Monotone Convergence Theorem]] that this sequence [[Definition:Convergent Real Sequence|converges]].
+Also, by [[Convergent Sequence is Cauchy Sequence]] the sequence is [[Definition:Real Cauchy Sequence|Cauchy]].
+Since $\sequence {L_n}$ is Cauchy, the sequence $\sequence {d_n}_{n \mathop \ge 1}$ formed by taking $d_n = L_n - L_{n - 1}$ is [[Definition:Real Null Sequence|null]].
+From [[Combination of Periodic Elements]] it follows that $\sequence {d_n}_{n \mathop \ge 1}$ is contained in $\mathcal L_{f > 0}$, for every $d_n$ is a [[Definition:Periodic Element|periodic element]] of $f$.
+Consider the [[Definition:Real Sequence|sequence]] $\sequence {\paren {x - y} \bmod d_n}_{n \mathop \ge 1}$.
+It is seen by [[Limit of Modulo Operation]] that this sequence is also [[Definition:Real Null Sequence|null]].
+Let $\epsilon \in \R_{> 0}$ such that $\epsilon < \size {\map f x - \map f y}$.
+For any $\delta \in \R_{> 0}$, there is a $n \in \N_{> 0}$ such that:
+{{begin-eqn}}
+{{eqn | l = \paren {x - y} \bmod d_n
+      | r = \paren {x - y} - d_n \floor {\frac {x - y} {d_n} }
+      | c = {{Defof|Modulo Operation}}
+}}
+{{eqn | r = x - \paren {y + d_n \floor {\frac {x - y} {d_n} } }
+}}
+{{eqn | o = <
+      | r = \delta
+}}
+{{end-eqn}}
+But:
+{{begin-eqn}}
+{{eqn | l = \size {\map f x - \map f {y + d_n \floor {\frac {x - y} {d_n} } } }
+      | r = \size {\map f x - \map f y}
+      | c = [[General Periodicity Property]]
+}}
+{{eqn | o = >
+      | r = \epsilon
+}}
+{{end-eqn}}
+And so $f$ is [[Definition:Discontinuous Real Function at Point|not continuous]] at $x$.
+But our choice of $x$ was completely arbitrary.
+Hence the result.
+{{qed}}
+{{ADC}}
+[[Category:Periodic Functions]]
+t87r6owlmp5wcgtu8s0a2c4rqgdqan3
+\end{proof}<|endoftext|>
+\section{Module on Cartesian Product is Module}
+Tags: Examples of Modules, Direct Products, Module on Cartesian Product
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $n \in \N_{>0}$.
+Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
+Then $\struct {R^n, +, \times}_R$ is an [[Definition:Module|$R$-module]].
+\end{theorem}
+
+\begin{proof}
+This is a special case of [[Direct Product of Modules is Module]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Module on Cartesian Product of Ring with Unity is Unitary Module}
+Tags: Examples of Unitary Modules, Direct Products, Module on Cartesian Product
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]].
+Let $n \in \N_{>0}$.
+Let $\struct {R^n, +, \times}_R$ be the '''[[Definition:Module on Cartesian Product|$R$-module $R^n$]]'''.
+Then $\struct {R^n, +, \times}_R$ is a [[Definition:Unitary Module|unitary $R$-module]].
+\end{theorem}<|endoftext|>
+\section{Vector Space on Cartesian Product is Vector Space}
+Tags: Examples of Vector Spaces, Vector Space on Cartesian Product
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $n \in \N_{>0}$.
+Let $\struct {K^n, +, \times}_K$ be the '''[[Definition:Vector Space on Cartesian Product|$K$-vector space $K^n$]]'''.
+Then $\struct {K^n, +, \times}_K$ is a [[Definition:Vector Space|$K$-vector space]].
+\end{theorem}
+
+\begin{proof}
+This is a special case of the [[Definition:Vector Space of All Mappings|Vector Space of All Mappings]], where $S$ is the set $\closedint 1 n \subset \N^*$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This is a special case of a [[Definition:Direct Product of Vector Spaces|direct product of vector spaces]] where each of the $G_k$ is the [[Definition:Vector Space|$K$-vector space]] $K$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Polynomial Forms over Field is Vector Space}
+Tags: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $F \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] over $F$.
+Then $F \sqbrk X$ is an [[Definition:Vector Space|vector space over $F$]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Binary Operation|operation]] $\times': F \to F \sqbrk X$ be defined as follows.
+Let $x \in F$.
+Let $\mathbf y \in F \sqbrk X$ be defined as:
+:$\mathbf y = \displaystyle \sum_{k \mathop = 0}^n y_k X^k$
+where $n = \map \deg {\mathbf y}$ denotes the [[Definition:Degree of Polynomial|degree]] of $\mathbf y$
+Thus:
+:$x \times' \mathbf y := \displaystyle x \sum_{k \mathop = 0}^n y_k X^k = \displaystyle \sum_{k \mathop = 0}^n \paren {x \times y_k} X^k$
+We have that $\times': F \to F \sqbrk X$ is an instance of [[Definition:Polynomial Multiplication|polynomial multiplication]] where the [[Definition:Multiplier|multiplier]] $x$ is a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $0$.
+Hence, let the supposed [[Definition:Vector Space|vector space over $F$]] in question be denoted in full as:
+:$\mathbf V = \struct {F \sqbrk X, +', \times'}_F$
+where:
+:$+': F \sqbrk X \to F \sqbrk X$ denotes [[Definition:Polynomial Addition|polynomial addition]]
+:$\times': F \to F \sqbrk X$ denotes the operation as defined above. 
+We already have that $F \sqbrk X$ is an [[Definition:Integral Domain|integral domain]].
+Thus [[Definition:Vector Space Axioms|vector space axioms]] $\text V 0$ to $\text V 4$ are fulfilled.
+By definition of $\times'$, it is seen that the remaining [[Definition:Vector Space Axioms|vector space axioms]] are fulfilled as follows:
+Let $\lambda, \mu \in F$.
+Let $\mathbf x, \mathbf y \in F \sqbrk X$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
+=== {{Vector-space-axiom|5}} ===
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda + \mu} \times' \mathbf x
+      | r = \paren {\lambda + \mu} \times' \sum_{k \mathop = 0}^m x_k X^k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^m \paren {\lambda + \mu} \times x_k X^k
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^m \paren {\lambda \times x_k X^k + \mu \times x_k X^k}
+      | c = {{Field-axiom|D}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^m \lambda \times x_k X^k + \sum_{k \mathop = 0}^m \mu \times x_k X^k
+      | c = 
+}}
+{{eqn | r = \lambda \times' \mathbf x + \mu \times' \mathbf x
+      | c = Definition of $\times'$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== {{Vector-space-axiom|6}} ===
+{{begin-eqn}}
+{{eqn | l = \lambda \times' \paren {\mathbf x + \mathbf y}
+      | r = \lambda \times' \paren {\sum_{j \mathop = 0}^m x_j X^j + \sum_{k \mathop = 0}^n y_k X^k}
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^m \lambda \times x_j X^j + \sum_{k \mathop = 0}^n \lambda \times y_k X^k
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {\lambda \times x_k + \lambda \times y_k} X^k
+      | c = {{Defof|Polynomial Addition}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \lambda \times \paren {x_k + y_k} X^k
+      | c = {{Field-axiom|D}}
+}}
+{{eqn | r = \lambda \times' \mathbf x + \lambda \times' \mathbf y
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== {{Vector-space-axiom|7}} ===
+{{begin-eqn}}
+{{eqn | l = \lambda \times' \paren {\mu \times' \mathbf x}
+      | r = \lambda \times' \paren {\mu \times' \sum_{k \mathop = 0}^n x_k X^k}
+      | c = 
+}}
+{{eqn | r = \lambda \times' \paren {\sum_{k \mathop = 0}^n \mu \times x_k X^k}
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \lambda \times \paren {\mu \times x_k} X^k
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \paren {\lambda \times \mu} \times x_k X^k
+      | c = {{Field-axiom|M1}}
+}}
+{{eqn | r = \paren {\lambda \times \mu} \times' \sum_{k \mathop = 0}^n x_k X^k
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \paren {\lambda \times \mu} \times' \mathbf x
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== {{Vector-space-axiom|8}} ===
+{{begin-eqn}}
+{{eqn | l = 1_F \times' \mathbf x
+      | r = 1_F \times' \sum_{k \mathop = 0}^n x_k X^k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n 1_F \times x_k X^k
+      | c = Definition of $\times'$
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n x_k X^k
+      | c = {{Field-axiom|M3}}
+}}
+{{eqn | r = \mathbf x
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+All [[Definition:Vector Space Axioms|vector space axioms]] are hence seen to be fulfilled.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{No Non-Trivial Norm on Rational Numbers is Complete}
+Tags: Normed Division Rings, Complete Metric Spaces
+
+\begin{theorem}
+No [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Norm on Division Ring|norm]] on the [[Definition:Set|set]] of the [[Definition:Rational Number|rational numbers]] is [[Definition:Complete Normed Division Ring|complete]].
+\end{theorem}
+
+\begin{proof}
+By [[P-adic Norm not Complete on Rational Numbers]], no [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on the [[Definition:Set|set]] of the [[Definition:Rational Number|rational numbers]], for any [[Definition:Prime Number|prime]] $p$, is [[Definition:Complete Normed Division Ring|complete]].
+By [[Rational Number Space is not Complete Metric Space]], the [[Definition:Absolute Value|absolute value]] $\size{\,\cdot\,}$ on the [[Definition:Set|set]] of the [[Definition:Rational Number|rational numbers]] is not [[Definition:Complete Normed Division Ring|complete]].
+By [[Norm is Complete Iff Equivalent Norm is Complete]], no [[Definition:Norm on Division Ring|norm]] is [[Definition:Complete Normed Division Ring|complete]] if it is [[Definition:Equivalent Division Ring Norms|equivalent]] to either the [[Definition:Absolute Value|absolute value]] $\size{\,\cdot\,}$ or the [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ for some [[Definition:Prime Number|prime]] $p$.
+By [[Ostrowski's Theorem]], every [[Definition:Nontrivial Division Ring Norm|non-trivial]] [[Definition:Norm on Division Ring|norm]] is [[Definition:Equivalent Division Ring Norms|equivalent]] to either the [[Definition:Absolute Value|absolute value]] $\size{\,\cdot\,}$ or the [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ for some [[Definition:Prime Number|prime]] $p$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Norm is Complete Iff Equivalent Norm is Complete}
+Tags: Normed Division Rings, Complete Metric Spaces
+
+\begin{theorem}
+Let $R$ be a [[Definition:Division Ring|division ring]].
+Let $\norm {\,\cdot\,}_1$ and $\norm {\,\cdot\,}_2$ be [[Definition:Equivalent Division Ring Norms|equivalent norms]] on $R$.
+Then:
+:$\struct {R,\norm {\,\cdot\,}_1}$ is [[Definition:Complete Normed Division Ring|complete]] {{iff}} $\struct {R,\norm {\,\cdot\,}_2}$ is [[Definition:Complete Normed Division Ring|complete]].
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Equivalent Division Ring Norms by Cauchy Sequence|Cauchy Sequence Equivalence]], 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 $\norm{\,\cdot\,}_2$.
+By [[Definition:Equivalent Division Ring Norms by Convergence|Convergent Equivalence]], for all sequences $\sequence {x_n}$ in $R$:
+:$\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] in $\norm{\,\cdot\,}_1$ {{iff}} $\sequence {x_n}$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] in $\norm{\,\cdot\,}_2$.
+Hence:
+:every [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm{\,\cdot\,}_1$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] in $\norm{\,\cdot\,}_1$ {{iff}} every [[Definition:Cauchy Sequence in Normed Division Ring|Cauchy sequence]] in $\norm{\,\cdot\,}_2$ [[Definition:Convergent Sequence in Normed Division Ring|converges]] in $\norm{\,\cdot\,}_2$.
+The result follows.
+{{qed}}
+[[Category:Normed Division Rings]]
+[[Category:Complete Metric Spaces]]
+ekkypuxvfwisc4mndifp6qmzk0hcwwq
+\end{proof}<|endoftext|>
+\section{Ring of Polynomial Forms over Field is Vector Space/Corollary}
+Tags: Examples of Vector Spaces
+
+\begin{theorem}
+Let $S \subseteq F \sqbrk X$ denote the [[Definition:Subset|subset]] of $F \sqbrk X$ defined as:
+:$S = \set {\mathbf x \in F \sqbrk X: \map \deg {\mathbf x} < d}$
+for some $d \in \Z_{>0}$.
+Then $S$ is an [[Definition:Vector Space|vector space over $F$]].
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Polynomial Forms over Field is Vector Space]] we note that $\struct {F, +, \times}$ is a [[Definition:Vector Space|vector space over $F$]].
+The remaining question is that $S$ remains [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Polynomial Addition|polynomial addition]] and [[Definition:Scalar Multiplication on Vector Space|scalar multiplication]].
+Let $\mathbf x, \mathbf y \in S$ such that $\map \deg {\mathbf x} = m$ and $\map \deg {\mathbf y} = n$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \mathbf x + \mathbf y
+      | r = \sum_{j \mathop = 0}^m x_j X^j + \sum_{k \mathop = 0}^n y_k X^k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{\max \set {m, n} } \paren {x_k + y_k} X^k
+      | c = {{Defof|Polynomial Addition}}
+}}
+{{end-eqn}}
+As both $m < d$ and $n < d$ by dint of them belonging to $S$, it follows that $\max \set {m, n} < d$.
+Hence $\mathbf x + \mathbf y \in S$ and so $S$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Polynomial Addition|polynomial addition]].
+Let the [[Definition:Binary Operation|operation]] $\times': F \to F \sqbrk X$ be defined as follows.
+Let $\lambda \in F$.
+Let $\mathbf x \in F \sqbrk X$ be defined as:
+:$\mathbf x = \displaystyle \sum_{k \mathop = 0}^n x_k X^k$
+where $n = \map \deg {\mathbf x}$ denotes the [[Definition:Degree of Polynomial|degree]] of $\mathbf x$.
+Thus:
+:$\lambda \times' \mathbf x := \displaystyle \lambda \times' \sum_{k \mathop = 0}^n x_k X^k = \displaystyle \sum_{k \mathop = 0}^n \paren {\lambda \times x_k} X^k$
+We have that $\times': F \to F \sqbrk X$ is an instance of [[Definition:Polynomial Multiplication|polynomial multiplication]] where the [[Definition:Multiplier|multiplier]] $\lambda$ is a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $0$.
+By definition of [[Definition:Polynomial Multiplication|polynomial multiplication]]:
+{{begin-eqn}}
+{{eqn | l = \map \deg {\lambda \times' \mathbf x}
+      | r = \map \deg \lambda + \map \deg {\mathbf x}
+      | c = 
+}}
+{{eqn | r = 0 + \map \deg {\mathbf x}
+      | c = 
+}}
+{{eqn | o = <
+      | r = d
+      | c = as $\mathbf x \in S$
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Norm of p-adic Number is Power of p}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $x \in \Q_p: x \ne 0$.
+Then:
+:$\exists v \in \Z: \norm x_p = p^{-v}$
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:P-adic Number|$p$-adic numbers]] $\Q_p$, the [[Definition:Rational Number|rational numbers]] $\Q$ are [[Definition:Everywhere Dense|dense]] in $\Q_p$.
+By the definition of a [[Definition:Everywhere Dense|dense subset]] then $\map \cl \Q = \Q_p$.
+By [[Closure of Subset of Metric Space by Convergent Sequence]] then:
+:there exists a [[Definition:Rational Sequence|sequence]] $\sequence {x_n} \subseteq \Q$ that [[Definition:Convergent Sequence in Normed Division Ring|converges]] to $x$.
+That is:
+:$\displaystyle \lim_{n \mathop \to \infty} x_n = x$
+By [[Convergent Sequence in Normed Division Ring is Cauchy Sequence]], $\sequence {x_n}$ is a [[Definition:Cauchy Sequence (Normed Division Ring)|Cauchy sequence]].
+Since $\sequence {x_n}$ does not [[Definition:Convergent Sequence in Normed Division Ring|converge]] to $0$, by [[Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary]] then:
+:$\exists N \in \N: \forall n, m > N: \norm {x_n}_p = \norm {x_m}_p$
+By definition of the [[Definition:P-adic Norm|$p$-adic norm]] on $\Q$:
+:$\exists v \in \Z: \norm {x_{N + 1} }_p = p^{-v}$
+Hence:
+:$\forall n > N: \norm {x_n}_p = p^{-v}$
+Let $\sequence {y_n}$ be the [[Definition:Subsequence|subsequence]] of $\sequence {\norm {x_n}}$ defined by:
+:$\forall n: y_n = \norm {x_{N + n} }$
+Then $\sequence {y_n}$ is the constant [[Definition:Sequence|sequence]] $\tuple {p^{-v}, p^{-v}, p^{-v}, \dotsc}$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \norm x_p
+      | r = \displaystyle \lim_{n \mathop \to \infty} \norm {x_n}
+      | c = [[Modulus of Limit/Normed Division Ring|Modulus of Limit]]
+}}
+{{eqn | r = \displaystyle \lim_{n \mathop \to \infty} y_n
+      | c = [[Limit of Subsequence equals Limit of Real Sequence]]
+}}
+{{eqn | r = p^{-v}
+      | c = [[Constant Sequence Converges to Constant in Normed Division Ring]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fisher-Tippet-Gnedenko Theorem}
+Tags: 
+
+\begin{theorem}
+In general, for iid $\set {X_i} _{i\mathop = 1}^n$ with cdf $\map F x$ we have $\mathbb P \sqbrk {\max_{i \mathop = 1}^n \set {X_i} > x} = \map {F^n} x$.
+Let $x_* := \sup \set {x: \map F x < 1}$.
+Then as $n \to \infty$:
+:$\displaystyle \max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} \stackrel {prob} {\to} x_*$
+We normalize it by constants $a_n > 0, b_n$ such that:
+:$\dfrac {\displaystyle\max_{i \mathop = 1, \mathop \ldots, n} \set {X_i} - b_n} {a_n} \stackrel {dis} {\to} X_*$
+for some non-degenerate $X_*$ with distribution $\map G x$.
+We have the following three possible limits
+* Frechet: If $x > 0$, $\map G x = \map \exp {-x^{-a} }$ for some $a > 0$. If $x \le 0$, $\map G x = 0$.
+* Gumbel: If $x \in \R$, $\map G x = \map \exp {-e^{-\frac {x - \mu} {\sigma} } }$ for $\mu\in \R, \sigma>0$. 
+* Weibull: If $x \ge 0$, $\map G x = 1 - \map \exp {-\paren {\dfrac x \lambda}^k}$ for $\lambda, k > 0$. If $x < 0$ , $\map G x = 0$.
+'''Fisher-Tippet-Gnedenko theorem'''
+:$(1): \quad \map G x$ will be Frechet {{iff}} $\map F x < 1$ and $\displaystyle \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {t x} } {1 - \map F t} = x^{-\theta}$ for $x > 0$ and $\theta > 0$. Here we can set $b_n := 0$ and $a_n := F^{-1} \paren {1 - \dfrac 1 n}$.
+:$(2): \quad \map G x$ will be Weibull {{iff}} $x_* : = \sup \set {x: \map F x < 1} < \infty$ and $\displaystyle \lim_{t \mathop \to +\infty} \dfrac {1 - \map F {x_* + t x} } {1 - \map F {x_* - t} } = \paren {-x}^{-\theta}$ for $x < 0$ and $\theta > 0$. Here we can set $b_n := x_*$ and $a_n := x_* - \map {F^{-1} } {1 - \dfrac 1 n}$.
+:$(3): \quad G$ will be Gumbel if the density $\map f x := \dfrac \d {\d x} \map F x > 0$ and it is differentiable in $\openint {x_1} {x_*}$ for some $x_1$, and $\displaystyle \lim_{x \mathop \to x_*} \map {\dfrac \d {\d x} } {\dfrac {1 - \map F x} {\map f x} } = 0$. Here we can set $b_n := \map {F^{-1} } {1 - \dfrac 1 n}$ and $a_n := \dfrac 1 {n \map f {b_n} }$.
+\end{theorem}
+
+\begin{proof}
+Proof under construction
+n2eiud6qu49ofk9cwyc34h4b2znlr35
+\end{proof}<|endoftext|>
+\section{Borel-TIS inequality}
+Tags: 
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $\sequence {f_t}_{t \mathop \in T}$ be a centred (i.e. mean zero) [[Definition:Gaussian Process|Gaussian process]] on $T$, such that:
+:$\norm f_T := \sup_{t \mathop \in T} \size {f_t}$
+is [[Definition:Almost Surely|almost surely]] finite.
+Let:
+:$\sigma_T^2 := \sup_{t \mathop \in T} \operatorname E \size {f_t}^2$
+Then $\map {\operatorname E} {\norm f_T}$ and $\sigma_T$ are both finite, and, for each $u > 0$:
+:$\map {\operatorname P} {\norm f_T > \map {\operatorname E} {\norm f_T} + u} \le \map \exp {\dfrac {-u^2} {2 \sigma_T^2} }$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+7fa10a8tcu293r3c6w0vu2rraqqpq7j
+\end{proof}<|endoftext|>
+\section{Gaussian Isoperimetric Inequality}
+Tags: Measure Theory
+
+\begin{theorem}
+Let $A$ be a [[measurable]] subset of $\R^n$ endowed with the standard Gaussian measure $\gamma^n$ with the density $\dfrac {\map \exp {-\norm x^2 / 2} } {\paren {2 \pi}^{n/2} }$
+Denote by:
+:$A_\epsilon = \set {x \in \R^n: \map d {x, A} \le \epsilon}$
+the $\epsilon$-extension of $A$.
+The '''Gaussian isoperimetric inequality''' states that:
+:$\displaystyle \liminf_{\epsilon \mathop \to +0} \epsilon^{-1} \set {\map {\gamma^n} {A_\epsilon} - \map {\gamma^n} A} \ge \map \varphi {\map {\Phi^{-1} } {\map {\gamma^n} A } }$
+where:
+:$\map \varphi t = \dfrac {\map \exp {-t^2 / 2} } {\sqrt {2 \pi} }$
+and:
+:$\map \Phi t = \displaystyle \int_{-\infty}^t \map \varphi s \rd s$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Carl Friedrich Gauss|cat = Gauss}}
+[[Category:Measure Theory]]
+2ghil591y8y2ir7m0jdvvr7w2csie69
+\end{proof}<|endoftext|>
+\section{Negation of Propositional Function in Two Variables}
+Tags: Universal Quantifier, Existential Quantifier
+
+\begin{theorem}
+Let $\map P {x, y}$ be a [[Definition:Propositional Function|propositional function]] of two [[Definition:Variable|Variables]].
+Then:
+:$\neg \forall x: \exists y: \map P {x, y} \iff \exists x: \forall y: \neg \map P {x, y}$
+That is:
+:''It is not the case that for all $x$ a [[Definition:Value of Variable|value]] of $y$ can be found to [[Definition:Satisfaction|satisfy]] $\map P {x, y}$''
+means the same thing as:
+:''There exists at least one [[Definition:Value of Variable|value]] of $x$ such that for all $y$ it is not possible to [[Definition:Satisfaction|satisfy]] $\map P {x, y}$''
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | lo= \neg \forall x:
+      | l = \exists y
+      | o = :
+      | r = \map P {x, y}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \exists x:
+      | l = \neg \exists y
+      | o = :
+      | r = \map P {x, y}
+      | c = [[Denial of Universality]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \exists x:
+      | l = \forall y
+      | o = :
+      | r = \neg \map P {x, y}
+      | c = [[Denial of Existence]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ideal of P-adic Numbers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+The [[Definition:Valuation Ideal Induced by Non-Archimedean Norm|valuation ideal induced by norm]] $\norm {\,\cdot\,}_p$ is the [[Definition:Principal Ideal of Ring|principal ideal]]:
+:$p \Z_p = \set {x \in \Q_p: \norm x_p < 1}$
+\end{theorem}
+
+\begin{proof}
+By [[Principal Ideal from Element in Center of Ring]] then $p \Z_p$ is a [[Definition:Principal Ideal of Ring|principal ideal]].
+Now:
+{{begin-eqn}}
+{{eqn | l = \norm x_p
+      | o = <
+      | r = 1
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = \norm x_p
+      | o = \le
+      | r = \dfrac 1 p
+      | c = [[P-adic Norm of p-adic Number is Power of p]]
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = p \norm x_p
+      | o = \le
+      | r = 1
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = \dfrac {\norm x_p} {\norm p_p}
+      | o = \le
+      | r = 1
+      | c = as ${\norm p_p} = \dfrac 1 p$
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = \norm {\dfrac x p}_p
+      | o = \le
+      | r = 1
+      | c = [[Properties of Norm on Division Ring/Norm of Quotient|Norm of quotient]]
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = \dfrac x p
+      | o = \in
+      | r = \Z_p
+      | c = {{Defof|P-adic Integers|$p$-adic integers}}
+}}
+{{eqn | ll= \leadstoandfrom 
+      | l = x
+      | o = \in
+      | r = p \Z_p
+}}
+{{end-eqn}}
+Hence:
+:$p \Z_p = \set {x \in \Q_p: \norm x_p < 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers are Arbitrarily Close to P-adic Integers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $\Z_p$ be the [[Definition:P-adic Integers|$p$-adic integers]].
+Let $x \in \Z_p$.
+Then for $n \in \N$ there exists [[Definition:Unique|unique]] $\alpha \in \Z$:
+:$(1): \quad 0 \le \alpha \le p^n - 1$
+:$(2): \quad \norm { x -\alpha}_p \le p^{-n}$
+\end{theorem}
+
+\begin{proof}
+Let $n \in \N$.
+By definition of the [[Definition:P-adic Numbers|$p$-adic numbers]], the [[Definition:Rational Number|rational numbers]] are [[Definition:Everywhere Dense|dense]] in $\Q_p$.
+So there exists:
+:$\dfrac a b \in \Q: \norm {x - \dfrac a b}_p \le p^{-n}$
+From [[Unique Integer Close to Rational in Valuation Ring of P-adic Norm]],  there exists  [[Definition:Unique|unique]] $\alpha \in \Z$ such that:
+:$\norm {\dfrac a b - \alpha}_p \le p^{-n}$
+:$0 \le \alpha \le p^n - 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x - \alpha}_p
+      | r = \norm {\paren {x - \dfrac a b} + \paren {\dfrac a b - \alpha} }_p
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {x - \dfrac a b}_p, \: \norm {\dfrac a b - \alpha}_p }
+      | c = [[Definition:Non-Archimedean Division Ring Norm|Norm axiom (N4) (Ultrametric Inequality)]]
+}}
+{{eqn | o = \le
+      | r = p^{-n}
+}}
+{{end-eqn}}
+Now suppose $\beta \in \Z$ also satisfies conditions $(1)$ and $(2)$, that is:
+:$0 \le \beta \le p^n - 1$
+:$\norm { x - \beta}_p \le p^{-n}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm{\dfrac a b - \beta}_p
+      | r = \norm{\paren{\dfrac a b - x} + \paren{x - \beta} }_p
+}}
+{{eqn | o = \le
+      | r = \max \set{\norm{\dfrac a b - x}_p,\:\norm{x - \beta}_p}
+      | c = [[Definition:Non-Archimedean Division Ring Norm|Norm axiom (N4) (Ultrametric Inequality)]]
+}}
+{{eqn | o = \le
+      | r = \max \set{\norm{x - \dfrac a b}_p,\:\norm{x - \beta}_p}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of Negative]]
+}}
+{{eqn | o = \le
+      | r = p^{-n}
+}}
+{{end-eqn}}
+From [[Unique Integer Close to Rational in Valuation Ring of P-adic Norm]],  $\alpha \in \Z$ was [[Definition:Unique|unique]], so:
+:$\beta = \alpha$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers are Dense in P-adic Integers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ are [[Definition:Everywhere Dense|dense]] in the [[Definition:Metric Space|metric space]] $\struct{\Z_p, d_p}$.
+\end{theorem}
+
+\begin{proof}
+By [[Open Ball Characterization of Denseness]] it is sufficient to show that every [[Definition:Open Ball|open ball]] of $\struct {\Z_p, d_p}$ contains an [[Definition:Element|element]] of $\Z$.
+Let $x \in \Z_p$ and $\epsilon \in \R_{>0}$.
+By definition the [[Definition:Open Ball|open ball]] $\map {B_\epsilon} x$ is:
+:$\map {B_\epsilon} x = \set {y \in \Z_p: \norm y_p < \epsilon}$
+By [[Sequence of Powers of Number less than One]] then:
+:$\displaystyle \lim_{n \mathop \to \infty} p^{-n} = 0$
+Hence there exists $N \in \N$:
+:$\forall n \ge N: p^{-n} < \epsilon$
+Consider the [[Definition:Open Ball|open ball]] $\map {B_{p^{-N} } } x$.
+Since $0 < p^{-n} < \epsilon$ then:
+:$\map {B_{p^{-N} } } x \subseteq \map {B_\epsilon} x$.
+By [[Integers are Arbitrarily Close to P-adic Integers|Integers are Arbitrarily Close to P-adic Integers]] then:
+:$\exists \alpha \in \Z: \alpha \in \map {B_{p^{-N} } } x$
+Hence $\alpha \in \map {B_\epsilon} x$.
+Since $x$ and $\epsilon$ were arbitrary then every [[Definition:Open Ball|open ball]] of $\struct {\Z_p, d_p}$ contains an [[Definition:Element|element]] of $\Z$.
+By [[Open Ball Characterization of Denseness]] then $\Z$ is [[Definition:Everywhere Dense|dense]] in the [[Definition:Metric Space|metric space]] $\struct {\Z_p, d_p}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Integer is Limit of Unique Coherent Sequence of Integers}
+Tags: P-adic Number Theory, P-adic Integer is Limit of Unique Coherent Sequence of Integers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $\Z_p$ be the [[Definition:P-adic Integers|$p$-adic integers]].
+Let $x \in \Z_p$.
+Then there exists a [[Definition:Unique|unique]] [[Definition:Coherent Sequence|coherent sequence]] $\sequence {\alpha_n}$:
+:$\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$
+\end{theorem}
+
+\begin{proof}
+By the definition of a [[Definition:Coherent Sequence|coherent sequence]] it needs to be proved that there exists a [[Definition:Unique|unique]] [[Definition:Integer Sequence|integer sequence]] $\sequence {\alpha_n}$:
+:$(1): \quad \forall n \in \N: \alpha_n \in \Z$ and $0 \le \alpha_n \le p^{n + 1} - 1$
+:$(2): \quad \forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
+:$(3): \quad \displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$
+By [[Integers are Arbitrarily Close to P-adic Integers|Integers are Arbitrarily Close to P-adic Integers]] then for all $n \in \N$ there exists $\alpha_n \in \Z$:
+::$\text {(a)}: \quad 0 \le \alpha_n \le p^{n + 1} - 1$
+::$\text {(b)}: \quad \norm {x - \alpha_n}_p \le p^{-\paren{n + 1} }$
+Hence the [[Definition:Sequence|sequence]] $\sequence {\alpha_n}$ satisfies $(1)$ above.
+=== [[P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1|Lemma 1]] ===
+{{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1}}
+Hence the [[Definition:Sequence|sequence]] $\sequence {\alpha_n}$ satisfies $(2)$ above.
+=== [[P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 2|Lemma 2]] ===
+{{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 2}}
+Hence the [[Definition:Sequence|sequence]] $\sequence {\alpha_n}$ satisfies $(3)$ above.
+=== [[P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 3|Lemma 3]] ===
+{{:P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 3}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Dimension of Vector Space on Cartesian Product}
+Tags: Vector Space on Cartesian Product, Dimension of Vector Space
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $n \in \N_{>0}$.
+Let $\mathbf V := \struct {K^n, +, \times}_K$ be the '''[[Definition:Vector Space on Cartesian Product|$K$-vector space $K^n$]]'''.
+Then the [[Definition:Dimension of Vector Space|dimension]] of $\mathbf V$ is $n$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Unity of Ring|unity]] of $K$ be $1$, and the [[Definition:Ring Zero|zero]] of $K$ be $0$.
+Consider the [[Definition:Vector (Linear Algebra)|vectors]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf e_1
+      | o = :=
+      | r = \underbrace {\tuple {1, 0, \ldots, 0} }_{n \text { coordinates} }
+      | c = 
+}}
+{{eqn | l = \mathbf e_2
+      | o = :=
+      | r = \underbrace {\tuple {0, 1, \ldots, 0} }_{n \text { coordinates} }
+      | c = 
+}}
+{{eqn | o = \vdots
+      | c = 
+}}
+{{eqn | l = \mathbf e_n
+      | o = :=
+      | r = \underbrace {\tuple {0, 0, \ldots, 1} }_{n \text { coordinates} }
+      | c = 
+}}
+{{end-eqn}}
+Thus $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the [[Definition:Standard Ordered Basis on Vector Space|standard ordered basis]] of $\mathbf V$.
+From [[Standard Ordered Basis is Basis]], $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is a [[Definition:Basis of Vector Space|basis]] of $\mathbf V$.
+The result follows from [[Linearly Independent Set is Basis iff of Same Cardinality as Dimension]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dimension of Vector Space of Polynomial Functions}
+Tags: Dimension of Vector Space, Polynomial Theory
+
+\begin{theorem}
+Let $\struct {F, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $F_n \sqbrk X$ be the [[Definition:Ring of Polynomials in Ring Element|ring of polynomials]] over $F$ whose [[Definition:Degree of Polynomial|degree]] is less than $n$.
+Then the [[Definition:Dimension of Vector Space|dimension]] of the [[Definition:Vector Space|vector space]] $F_n \sqbrk X$ is $n$.
+\end{theorem}
+
+\begin{proof}
+Let $B$ be the [[Definition:Set|set]] of all the [[Definition:Identity Mapping|identity functions]] $I^k$ on $F_n \sqbrk X$ where $n \in \Z_{\ge 0}$.
+By definition, every element of $F_n \sqbrk X$ is a [[Definition:Linear Combination|linear combination]] of $B$.
+Suppose:
+:$\displaystyle \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$
+Then by [[Definition:Differentiation|differentiating]] $m$ times, we obtain from [[Nth Derivative of Nth Power]]:
+:$m! \alpha_m = 0$
+whence $\alpha_m = 0$ which is a contradiction.
+Hence $B$ is [[Definition:Linearly Independent Set|linearly independent]] and therefore is a [[Definition:Basis of Vector Space|basis]] for $F_n \sqbrk X$.
+The result follows from [[Linearly Independent Set is Basis iff of Same Cardinality as Dimension]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Independent Set is Contained in some Basis/Proof 1}
+Tags: Linearly Independent Set is Contained in some Basis
+
+\begin{theorem}
+Let $E$ be a [[Definition:Vector Space|vector space]] of $n$ [[Definition:Dimension of Vector Space|dimensions]].
+Let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $E$.
+{{:Linearly Independent Set is Contained in some Basis}}
+\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}<|endoftext|>
+\section{Linearly Independent Set is Contained in some Basis/Proof 2}
+Tags: Linearly Independent Set is Contained in some Basis
+
+\begin{theorem}
+Let $E$ be a [[Definition:Vector Space|vector space]] of $n$ [[Definition:Dimension of Vector Space|dimensions]].
+Let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $E$.
+{{:Linearly Independent Set is Contained in some Basis}}
+\end{theorem}
+
+\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{Vector Space over Division Subring is Vector Space}
+Tags: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $S$ be a [[Definition:Division Subring|division subring]] of $R$, such that $1_R \in S$.
+The [[Definition:Vector Space over Division Subring|vector space $\struct {R, +, \circ_S}_S$ over $\circ_S$]] is a [[Definition:Vector Space|$S$-vector space]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Vector Space|vector space]] is by definition a [[Definition:Unitary Module|unitary module]] over a [[Definition:Division Ring|division ring]]. 
+$S$ is a [[Definition:Division Ring|division ring]] by assumption.
+$\struct {R, +, \circ_S}_S$ is a [[Definition:Unitary Module|unitary module]] by [[Subring Module/Special Case]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Space on Field Extension is Vector Space}
+Tags: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {K, +, \times}$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $L / K$ be a [[Definition:Field Extension|field extension]] over $K$.
+Let $\struct {L, +, \times}_K$ be the a [[Definition:Vector Space on Field Extension|vector space of $L$]] over $K$.
+Then $\struct {L, +, \times}_K$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+We have that $L$ is a 
+By definition, $L / K$ is a [[Definition:Field Extension|field extension]] over $K$.
+Thus, by definition, $K$ is a [[Definition:Subfield|subfield]] of $L$.
+Thus, also by definition, $K$ is a [[Definition:Division Subring|division subring]] of $L$.
+The result follows by [[Vector Space over Division Subring is Vector Space]].
+{{qed}}
+[[Category:Examples of Vector Spaces]]
+1xsw69v077em48g99y2v1buy7i14d4k
+\end{proof}<|endoftext|>
+\section{Sheldon Conjecture}
+Tags: 73, Prime Numbers, Named Theorems
+
+\begin{theorem}
+There is only $1$ [[Definition:Sheldon Prime|Sheldon prime]], and that is $73$.
+\end{theorem}<|endoftext|>
+\section{Condition for Linear Divisor of Polynomial}
+Tags: Polynomial Theory
+
+\begin{theorem}
+Let $\map P x$ be a [[Definition:Polynomial|polynomial]] in $x$.
+Let $a$ be a [[Definition:Constant|constant]].
+Then $x - a$ is a [[Definition:Divisor of Polynomial|divisor]] of $\map P x$ {{iff}} $a$ is a [[Definition:Root of Polynomial|root]] of $P$.
+\end{theorem}
+
+\begin{proof}
+From the [[Little Bézout Theorem]], the [[Definition:Remainder (Polynomial Long Division)|remainder]] of $\map P x$ when divided by $x - a$ is equal to $\map P a$.
+=== Sufficient Condition ===
+Let $x - a$ be a [[Definition:Divisor of Polynomial|divisor]] of $\map P x$.
+From the [[Little Bézout Theorem]], the [[Definition:Remainder (Polynomial Long Division)|remainder]] of $\map P x$ when divided by $x - a$ is equal to $\map P a$.
+By definition of [[Definition:Divisor of Polynomial|divisor]], the [[Definition:Remainder (Polynomial Long Division)|remainder]] of $\map P x$ when divided by $x - a$ equals $0$.
+That is:
+:$\map P a = 0$
+It follows by definition that $a$ is a [[Definition:Root of Polynomial|root]] of $P$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $a$ be a [[Definition:Root of Polynomial|root]] of $P$.
+From the [[Little Bézout Theorem]], and by [[Polynomial Long Division]], we have:
+:$\map P x = \paren {x - a} \map Q x + \map P a$
+where $\map Q x$ is a [[Definition:Polynomial|polynomial]] in $x$ of [[Definition:Degree of Polynomial|degree]] one less than $\map P x$.
+By definition of [[Definition:Root of Polynomial|root of polynomial]]:
+:$\map P a = 0$
+So we have:
+:$\map P x = \paren {x - a} \map Q x$
+and the result follows by definition of [[Definition:Divisor of Polynomial|divisor]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Number is Algebraic over Real Numbers}
+Tags: Algebraic Numbers
+
+\begin{theorem}
+Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
+Then $z$ is [[Definition:Algebraic Number over Field|algebraic over $\R$]].
+\end{theorem}
+
+\begin{proof}
+Let $z = a + i b$.
+Let $z = a - i b$ be the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+From [[Product of Complex Number with Conjugate]]:
+:$z \overline z = a^2 + b^2$
+From [[Sum of Complex Number with Conjugate]]:
+:$z + \overline z = 2 a$
+Thus from [[Viète's Formulas]], both $z$ and $\overline z$ are [[Definition:Root of Polynomial|roots of the polynomial]]:
+:$z^2 - 2 a + \paren {a^2 + b^2}$
+Hence the result by definition of [[Definition:Algebraic Number over Field|algebraic over $\R$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial}
+Tags: Minimal Polynomials
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\map P x$ be a [[Definition:Polynomial|polynomial]] in $F$.
+Let $z$ be a [[Definition:Root of Polynomial|root]] of $\map P x$.
+Then $\map P x$ is a [[Definition:Multiple of Ring Element|multiple]] of the [[Definition:Minimal Polynomial|minimal polynomial]] $\map m x$ in $z$ over $F$.
+\end{theorem}
+
+\begin{proof}
+For $z$ to be a [[Definition:Root of Polynomial|root]] of $F$, $z$ must be [[Definition:Algebraic Number over Field|algebraic over $F$]].
+Let us write:
+:$\map P x = \map m x \, \map q x + \map r x$
+where $\map q x$ and $\map r x$ are [[Definition:Polynomial|polynomials]] in $F$.
+Then either $\map r x = 0$ or $\map \deg {\map r x} < \map \deg {\map m x}$.
+Then:
+:$\map P z = \map m z \, \map q z + \map r z$
+But as $z$ is a [[Definition:Root of Polynomial|root]] of both $\map P x$ and $\map m x$, we have that:
+:$\map P z = \map m z = 0$
+and so:
+:$\map r z = 0$
+So if $\map r x \ne 0$ we have that $\map r x$ is a [[Definition:Polynomial|polynomial]] of smaller [[Definition:Degree of Polynomial|degree]] than $\map m x$.
+This [[Definition:Contradiction|contradicts]] the [[Definition:Minimal Polynomial|minimality]] of $\map m x$.
+Thus $\map r x = 0$ and so $\map P x$ is a [[Definition:Multiple of Ring Element|multiple]] of $\map m x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simple Algebraic Field Extension consists of Polynomials in Algebraic Number}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\theta \in \C$ be [[Definition:Algebraic Number over Field|algebraic over $F$]].
+Let $\map F \theta$ be the [[Definition:Simple Algebraic Field Extension|simple field extension]] of $F$ by $\theta$.
+Then $\map F \theta$ consists of [[Definition:Polynomial in One Variable|polynomials]] that can be written in the form $\map f \theta$, where $\map f x$ is a [[Definition:Polynomial in One Variable|polynomial]] over $F$.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be the [[Definition:Set|set]] of all [[Definition:Number|numbers]] which can be written in the form $\map f \theta$.
+We have that:
+:$H$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Polynomial Addition|addition]] and [[Definition:Polynomial Multiplication|multiplication]].
+:$H$ contains $0$ and $1$
+:For every [[Definition:Element|element]] of $H$, $H$ also contains its [[Definition:Field Negative|negative]].
+Let $\map f \theta \ne 0$.
+Then $\theta$ is not a [[Definition:Root of Polynomial|root]] of $\map f x$.
+Hence from [[Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial]]:
+:the [[Definition:Minimal Polynomial|minimal polynomial]] $\map m x$ in $\theta$ does not [[Definition:Divisor|divide]] $\map f x$.
+From [[Minimal Polynomial is Irreducible]], the [[Definition:Greatest Common Divisor|GCD]] of $\map m x$ and $\map f x$ is $1$.
+Therefore:
+:$\exists \map s x, \map t x: \map s x \map m x + \map t x \map f x = 1$
+Substituting for $\theta$:
+:$\map s \theta \, \map m \theta + \map t \theta \, \map f \theta = 1$
+Because $\map m \theta = 0$ it follows that:
+:$\map t \theta \, \map f \theta = 1$
+We have that $\map t \theta \in H$.
+Thus $\map t \theta$ is the [[Definition:Product Inverse|product inverse]] of $\map f x$ in $H$.
+Thus $H$ is a [[Definition:Field (Abstract Algebra)|field]].
+A [[Definition:Field (Abstract Algebra)|field]] containing $F$ and $\theta$ must contain $1$ and all the [[Definition:Power (Algebra)|powers]] of $\theta$ for [[Definition:Positive Integer|positive integer]] index.
+Hence such a [[Definition:Field (Abstract Algebra)|field]] also contains all [[Definition:Linear Combination|linear combinations]] of these, with [[Definition:Polynomial Coefficient|coefficients]] in $F$.
+So a [[Definition:Field (Abstract Algebra)|field]] containing $F$ and $\theta$ contains all the [[Definition:Element|elements]] of $H$:
+:$H \subseteq \map F \theta$
+But by definition, $\map F \theta$ is the [[Definition:Smallest Field containing Subfield and Complex Number|smallest field containing $F$ and $\theta$]].
+That is:
+:$\map F \theta \subseteq H$
+Thus:
+:$\map F \theta = H$
+and the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Element of Simple Algebraic Field Extension of Degree n is Polynomial in Algebraic Number of Degree Less than n}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\theta \in \C$ be [[Definition:Algebraic Number over Field|algebraic over $F$]] of [[Definition:Degree of Algebraic Number over Field|degree $n$]].
+Let $\map F \theta$ be the [[Definition:Simple Algebraic Field Extension|simple field extension]] of $F$ by $\theta$.
+Then any [[Definition:Element|element]] of $\map F \theta$ can be written as $\map f \theta$, where $\map f x$ is a [[Definition:Polynomial in One Variable|polynomial]] over $F$ of [[Definition:Degree of Polynomial|degree]] at most $n - 1$.
+\end{theorem}
+
+\begin{proof}
+From [[Simple Algebraic Field Extension consists of Polynomials in Algebraic Number]], an arbitrary [[Definition:Element|element]] of $\map F \theta$ can be written as $\map f \theta$.
+But:
+:$\map f x = \map m x \, \map q x + \map r x$
+where:
+:$\map m x$ is [[Definition:Minimal Polynomial|minimal polynomial]] in $\theta$
+:$\map q x$ is a [[Definition:Polynomial|polynomial]] in $\map F \theta$
+:$\map r x$ is a [[Definition:Polynomial|polynomial]] in $\map F \theta$ such that either:
+::$\map \deg {\map r x} < \map \deg {\map m x}$
+:or:
+::$\map r x = 0$
+Thus:
+:$\map f \theta = \map m \theta \, \map q \theta + \map r \theta$
+and as $\map m \theta = 0$ we have:
+:$\map f \theta = \map r \theta$
+So $\map f \theta$ can be expressed as $\map r \theta$ instead, which is of [[Definition:Degree of Polynomial|degree]] strictly less than that of $\map m \theta$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Simple Algebraic Field Extension equals Degree of Algebraic Number}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\theta \in \C$ be [[Definition:Algebraic Number over Field|algebraic over $F$]] of [[Definition:Degree of Algebraic Number over Field|degree $n$]].
+Let $\map F \theta$ be the [[Definition:Simple Algebraic Field Extension|simple field extension]] of $F$ by $\theta$.
+Then $\map F \theta$ is a [[Definition:Finite Field Extension|finite extension]] of $F$ whose [[Definition:Degree of Field Extension|degree]] is:
+:$\index {\map F \theta} F = n$
+\end{theorem}
+
+\begin{proof}
+Considered as a [[Definition:Vector Space on Field Extension|vector space]] over $F$, $\map F \theta$ is [[Definition:Generated Field Extension|generated]] by the [[Definition:Set|set]] $B$, where:
+:$B := \set {1, \theta, \theta^2, \ldots, \theta^{n - 1} }$
+But $S$ is [[Definition:Linearly Independent Set|linearly independent]] over $F$, because otherwise:
+:$c_0 1 + c_1 \theta + c_2 \theta^2 + \dotsb + c_{n - 1} \theta^{n - 1} = 0$
+with all the $c$s non-zero.
+That would mean $\theta$ was the [[Definition:Root of Polynomial|root]] of a [[Definition:Polynomial|polynomial]] whose [[Definition:Degree of Polynomial|degree]] was less than that of the [[Definition:Minimal Polynomial|minimal polynomial]] $\map m x$ of $\theta$, whose [[Definition:Degree of Polynomial|degree]] equals $n$.
+Therefore $B$ is a [[Definition:Basis of Vector Space|basis]] of $\map F \theta$ over $F$.
+Thus $\map F \theta$ is of [[Definition:Dimension of Vector Space|dimension]] $\size B = n$.
+Hence the $\index {\map F \theta} F = n$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Degree of Element of Finite Field Extension divides Degree of Extension}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0$ and whose [[Definition:Unity of Field|unity]] is $1$.
+Let $K / F$ be a [[Definition:Finite Field Extension|finite field extension]] of [[Definition:Degree of Field Extension|degree]] $n$.
+Let $\alpha \in K$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $F$.
+Then the [[Definition:Degree of Algebraic Element|degree]] of $\alpha$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
+\end{theorem}
+
+\begin{proof}
+Let $\alpha \in K$.
+The [[Definition:Dimension of Vector Space|dimension]] of $K / F$ considered as a [[Definition:Vector Space on Field Extension|vector space]] is $n$.
+Let $S$ be the [[Definition:Set|set]] defined as:
+:$S := \set {1, \alpha, \alpha^2, \ldots, \alpha_n}$
+We have that:
+:$\card S = n + 1$
+From [[Cardinality of Linearly Independent Set is No Greater than Dimension]]:
+:$S$ is [[Definition:Linearly Dependent Set|linearly dependent]] on $F$.
+Hence there are [[Definition:Scalar (Vector Space)|scalars]] $c_0, c_1, \ldots, c_n \in F$, not all [[Definition:Field Zero|zero]], such that:
+:$c_0 + c_1 \alpha + c_2 \alpha^2 + \dotsb + c_n \alpha_n = 0$
+which is a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$.
+So $\alpha$ is [[Definition:Algebraic Element of Field Extension|algebraic]] over $F$ with [[Definition:Degree of Algebraic Element|degree]] no greater than $n$.
+Consider $\map F \alpha$, the [[Definition:Simple Algebraic Field Extension|simple algebraic extension of $F$ on $\alpha$]].
+We have that:
+:$F \subseteq \map F \alpha \subseteq K$
+Let $\alpha$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $F$ with [[Definition:Degree of Algebraic Element|degree]] $m$.
+Then $\map F \alpha$ is a [[Definition:Finite Field Extension|finite extension]] of $F$ with [[Definition:Degree of Field Extension|degree]] $\index {\map F \alpha} F = m$.
+We also have that $K$ is a [[Definition:Finite Field Extension|finite extension]] of $F$ with [[Definition:Degree of Field Extension|degree]] $n$.
+So by [[Degree of Field Extensions is Multiplicative]]:
+:$\index K F = \index K {\map F \alpha} \index {\map F \alpha} F$
+But $\index K F = n$ and $\index {\map F \alpha} F = m$
+so it follows that:
+:$m \divides n$
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2}
+Tags: Field Extensions
+
+\begin{theorem}
+Let $K / F$ be a [[Definition:Finite Field Extension|finite field extension]] of [[Definition:Degree of Field Extension|degree]] $2^m$.
+Let $\alpha \in K$ be [[Definition:Algebraic Element of Field Extension|algebraic]] over $F$ with [[Definition:Degree of Algebraic Element|degree]] $3$.
+Then $\alpha \notin K$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\alpha \in K$.
+From [[Degree of Element of Finite Field Extension divides Degree of Extension]]:
+:$\map \deg \alpha \divides \map \deg {K / F}$
+But:
+:$3 \nmid 2^m$
+From this [[Definition:Contradiction|contradiction]], it follows that $\alpha \notin K$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Point in Cartesian Plane with Rational Coordinates}
+Tags: Analytic Geometry
+
+\begin{theorem}
+Let $\CC$ be a [[Definition:Cartesian Plane|Cartesian plane]].
+Let $P = \tuple {x, y}$ be a [[Definition:Rational Point in Plane|rational point]] in $\CC$.
+Then $P$ is [[Definition:Constructible Point in Plane|constructible]] using a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]].
+\end{theorem}
+
+\begin{proof}
+Let $x = \dfrac m n$ where $m, n \in \Z_{\ne 0}$ are non-zero [[Definition:Integer|integers]].
+Let $O$ denote the [[Definition:Point|point]] $\tuple {0, 0}$.
+Let $A$ denote the [[Definition:Point|point]] $\tuple {1, 0}$.
+Let $M$ denote the [[Definition:Point|point]] $\tuple {0, m}$.
+Let $N$ denote the [[Definition:Point|point]] $\tuple {0, n}$.
+The [[Definition:X-Axis|$x$-axis]] is identified with the [[Definition:Straight Line|straight line]] through $O$ and $A$.
+The [[Definition:Y-Axis|$y$-axis]] is constructed as the [[Definition:Perpendicular|line perpendicular]] to $OA$ through $O$.
+Using [[Construction of Lattice Point in Cartesian Plane]], the [[Definition:Point|points]] $M$ and $\N$ are constructed.
+The [[Definition:Straight Line|line]] $NA$ is drawn.
+Using [[Construction of Parallel Line]], $MQ$ is drawn, where $Q$ is the [[Definition:Point|point]] of [[Definition:Intersection (Geometry)|intersection]] of $MQ$ with the [[Definition:X-Axis|$x$-axis]].
+We have that $\dfrac {OM} {ON} = \dfrac m n$
+As $\triangle ONA$ is [[Definition:Similar Triangles|similar]] to $\triangle OMQ$, it follows that $\dfrac {OM} {ON} = \dfrac {OQ} {OA}$
+Therefore $Q$ is the [[Definition:Point|point]] $\tuple {\dfrac m n, 0}$.
+:[[File:Construction of (m over n, 0).png|400px]]
+Let $y = \dfrac r s$ where $r, s \in \Z_{\ne 0}$ are non-zero [[Definition:Integer|integers]].
+Using [[Construction of Lattice Point in Cartesian Plane]], the [[Definition:Point|points]] $\tuple {r, 0}$ and $\tuple {s, 0}$ are constructed.
+Using the same technique as above, [[Definition:Mutatis Mutandis|mutatis mutandis]], the point $\tuple {0, \dfrac r s}$ is constructed.
+Using [[Construction of Parallel Line]], a [[Definition:Straight Line|straight line]] is drawn through $\tuple {\dfrac m n, 0}$ [[Definition:Parallel Lines|parallel]] to the [[Definition:Y-Axis|$y$-axis]].
+Using [[Construction of Parallel Line]], a [[Definition:Straight Line|straight line]] is drawn through $\tuple {0, \dfrac r s}$ [[Definition:Parallel Lines|parallel]] to the [[Definition:X-Axis|$x$-axis]].
+By definition of [[Definition:Cartesian Plane|Cartesian plane]], their [[Definition:Intersection (Geometry)|intersection]] is at $\tuple {\dfrac m n, \dfrac r s}$, which is the required [[Definition:Point|point]] $P = \tuple {x, y}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Integer Multiple of Line Segment}
+Tags: Euclidean Geometry
+
+\begin{theorem}
+Let $AB$ be a [[Definition:Line Segment|line segment]] in the [[Definition:Plane|plane]].
+Let $AC$ be a [[Definition:Line Segment|line segment]] in the [[Definition:Plane|plane]] through a [[Definition:Point|point]] $C$
+Let $D$ be a [[Definition:Point|point]] on $AC$ such that $AD = n AB$ for some $n \in \Z$.
+Then $AD$ is [[Definition:Constructible Point in Plane|constructible]] using a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]].
+\end{theorem}
+
+\begin{proof}
+Let $AB$ be given.
+Let the [[Definition:Straight Line|straight line]] through $AC$ be constructed.
+Let $D_1$ be constructed on $AC$ such that $A D_1 = AB$ by constructing a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $A$ and whose [[Definition:Radius of Circle|radius]] is $B$.
+Let $D_0$ be identified as the [[Definition:Point|point]] $A$.
+For each $k \in \set {1, 2, \ldots, n - 1}$, construct a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $D_k$ and whose [[Definition:Radius of Circle|radius]] is $D_{k - 1} D_k$.
+We have that $D_k D_{k + 1} = D_{k - 1} D_k = AB$.
+Thus $D_0 D_{k + 1}$ is ${k + 1}$ times $D_0 D_1 = AB$.
+When $k = n - 1$ we have that $D_{k + 1} = D_n = D$ is $n$ times $AB$.
+The diagram below illustrates the construction for $n = 4$.
+:[[File:Construction of Integer Multiple of Line Segment.png|800px]]
+{{qed}}
+[[Category:Euclidean Geometry]]
+tle54cgwus4m4tar0cwy8pqxhhhf5ah
+\end{proof}<|endoftext|>
+\section{Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2}
+Tags: Analytic Geometry
+
+\begin{theorem}
+Let $\CC$ be a [[Definition:Cartesian Plane|Cartesian plane]].
+Let $S$ be a [[Definition:Set|set]] of [[Definition:Point|points]] in $\CC$.
+Let $F$ be the smallest [[Definition:Field (Abstract Algebra)|field]] containing all the [[Definition:Cartesian Coordinate System|coordinates]] of the [[Definition:Point|points]] in $S$.
+Let $P = \tuple {a, b}$ be a [[Definition:Point|point]] in $\CC$.
+Then:
+:$P$ is [[Definition:Constructible Point in Plane|constructible]] from $S$ using a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]]
+{{iff}}:
+:the [[Definition:Cartesian Coordinate System|coordinates]] of $P$ are contained in a [[Definition:Finite Field Extension|finite extension]] $K$ of $F$ whose [[Definition:Degree of Field Extension|degree]] is given by:
+::$\index K F = 2^m$
+:for some $m \in \Z_{\ge 0}$.
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Point|point]] $P$ is constructed in a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]] from one of $3$ basic operations:
+:$(1): \quad$ the [[Definition:Intersection (Geometry)|intersection]] of $2$ [[Definition:Straight Line|straight lines]]
+:$(2): \quad$ the [[Definition:Intersection (Geometry)|intersection]] of a [[Definition:Straight Line|straight line]] and the [[Definition:Circumference of Circle|circumference]] of a [[Definition:Circle|circle]]
+:$(3): \quad$ the [[Definition:Intersection (Geometry)|intersection]] of the [[Definition:Circumference of Circle|circumferences]] of $2$ [[Definition:Circle|circle]].
+Let $A$, $B$, $C$ and $D$ be [[Definition:Point|points]] in $S$.
+In case $(1)$, the equations defining the [[Definition:Straight Line|straight lines]] $AB$ and $CD$ are [[Definition:Polynomial|polynomials]] of [[Definition:Degree of Polynomial|degree]] $1$.
+Hence the [[Definition:Cartesian Coordinate System|coordinates]] of $P$ can be found by solving the [[Definition:Linear Simultaneous Equations|linear simultaneous equations]] defining $AB$ and $CD$.
+It follows that the [[Definition:Solution to System of Simultaneous Equations|solution]] is in $F$.
+{{qed|lemma}}
+In case $(2)$:
+:the equation defining the [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $A$ and whose [[Definition:Radius of Circle|radius]] is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$.
+while:
+:the equation defining the [[Definition:Straight Line|straight line]] $CD$ is a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $1$.
+The [[Definition:X Coordinate|$x$-coordinate]] of $P$ is then the [[Definition:Root of Polynomial|root]] of a [[Definition:Quadratic Equation|quadratic equation]] with [[Definition:Polynomial Coefficient|coefficients]] in $F$.
+The [[Definition:Y Coordinate|$y$-coordinate]] of $P$ is then found by solving a [[Definition:Linear Equation|linear equation]].
+Thus the [[Definition:Cartesian Coordinate System|coordinates]] of $P$ are [[Definition:Element|elements]] of either a [[Definition:Field Extension|field extension]] of $F$ of [[Definition:Degree of Field Extension|degree $2$]] or of $F$ itself.
+In case $(3)$:
+:the equation defining the [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $A$ and whose [[Definition:Radius of Circle|radius]] is $AB$ is of the form $x^2 + y^2 + 2 f x + 2 g y + c = 0$, where $c, f, g \in F$
+:the equation defining the [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $C$ and whose [[Definition:Radius of Circle|radius]] is $CD$ is of the form $x^2 + y^2 + 2 f' x + 2 g' y + c' = 0$, where $c', f', g' \in F$.
+The [[Definition:Solution to System of Simultaneous Equations|solution]] of these equations is the same as the [[Definition:Solution to System of Simultaneous Equations|solution]] of either [[Definition:Circle|circle]] and the [[Definition:Straight Line|straight line]] $2 \paren {f - f'} x + 2 \paren {g - g'} y + \paren {c - c'} = 0$, which is the same as case $(2)$.
+We have that the [[Definition:Cartesian Coordinate System|coordinates]] of the various [[Definition:Point|points]] belong to the [[Definition:Field (Abstract Algebra)|field]] $F = F_0$, say.
+Suppose $P$ is constructed using $k$ of the basic constructions $(1)$, $(2)$ and $(3)$.
+Let $F_i$ be the [[Definition:Smallest Field containing Subfield and Complex Number/General Definition|smallest field]] containing all the [[Definition:Point|points]] so far obtained by $i$ such constructions, for $i = 1, 2, \ldots, k$.
+Either $F_i = F_{i - 1}$ or $F_i$ is an [[Definition:Finite Field Extension|finite extension]] of $F_i$ of [[Definition:Degree of Field Extension|degree]] $2$.
+Thus for each $i$, either:
+:$\index {F_i} {F_{i - 1} } = 1$
+or:
+:$\index {F_i} {F_{i - 1} } = 2$
+Therefore:
+:$\index {F_k} {F_0} = \index {F_k} {F_{k - 1} } \index {F_{k - 1} } {F_{k - 2} } \dotsm  {F_1} {F_0} = 2^m$
+where $0 \le m \le k$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Lattice Point in Cartesian Plane}
+Tags: Analytic Geometry
+
+\begin{theorem}
+Let $\CC$ be a [[Definition:Cartesian Plane|Cartesian plane]].
+Let $P = \tuple {a, b}$ be a [[Definition:Lattice Point|lattice point]] in $\CC$.
+Then $P$ is [[Definition:Constructible Point in Plane|constructible]] using a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]].
+\end{theorem}
+
+\begin{proof}
+Let $O$ denote the [[Definition:Point|point]] $\tuple {0, 0}$.
+Let $A$ denote the [[Definition:Point|point]] $\tuple {1, 0}$.
+The [[Definition:X-Axis|$x$-axis]] is identified with the [[Definition:Straight Line|straight line]] through $O$ and $A$.
+The [[Definition:Y-Axis|$y$-axis]] is constructed as the [[Definition:Perpendicular|line perpendicular]] to $OA$ through $O$.
+From [[Construction of Integer Multiple of Line Segment]], the [[Definition:Point|point]] $\tuple {a, 0}$ is constructed.
+Drawing a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $O$ and whose [[Definition:Radius of Circle|radius]] is $OA$ the point $A'$ is constructed on the [[Definition:Y-Axis|$y$-axis]] where $OA' = OA$.
+Thus $A'$ is the [[Definition:Point|point]] $\tuple {0, 1}$.
+From [[Construction of Integer Multiple of Line Segment]], the [[Definition:Point|point]] $\tuple {0, b}$ is constructed.
+Using [[Construction of Parallel Line]], a [[Definition:Straight Line|straight line]] is drawn through $\tuple {a, 0}$ [[Definition:Parallel Lines|parallel]] to the [[Definition:Y-Axis|$y$-axis]].
+Using [[Construction of Parallel Line]], a [[Definition:Straight Line|straight line]] is drawn through $\tuple {0, b}$ [[Definition:Parallel Lines|parallel]] to the [[Definition:X-Axis|$x$-axis]].
+By definition of [[Definition:Cartesian Plane|Cartesian plane]], their [[Definition:Intersection (Geometry)|intersection]] is at $\tuple {a, b}$, which is the required [[Definition:Point|point]] $P$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Regular Heptagon by Compass and Straightedge Construction is Impossible}
+Tags: Trisecting the Angle
+
+\begin{theorem}
+There is no [[Definition:Compass and Straightedge Construction|compass and straightedge construction]] for a [[Definition:Regular Heptagon|regular heptagon]].
+\end{theorem}
+
+\begin{proof}
+Construction of a [[Definition:Regular Heptagon|regular heptagon]] is the equivalent of constructing the [[Definition:Point|point]] $\tuple {\cos \dfrac {2 \pi} 7, \sin \dfrac {2 \pi} 7}$ from the [[Definition:Point|points]] $\tuple {0, 0}$ and $\tuple {1, 0}$
+Let $\epsilon = \map \exp {\dfrac {2 \pi} 7}$.
+Then $\epsilon$ is a [[Definition:Root of Polynomial|root]] of $x^7 - 1$.
+We have:
+{{begin-eqn}}
+{{eqn | l = x^7 - 1
+      | r = \paren {x - 1} \paren {x^6 + x^5 + x^4 + x^3 + x^2 + x + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^6 + \epsilon^5 + \epsilon^4 + \epsilon^3 + \epsilon^2 + \epsilon + 1
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^3 + \epsilon^2 + \epsilon + 1 + \epsilon^{-1} + \epsilon^{-2} + \epsilon^{-3}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+But we have:
+{{begin-eqn}}
+{{eqn | l = \epsilon
+      | r = \cos \dfrac {2 \pi} 7 + i \sin \dfrac {2 \pi} 7
+}}
+{{eqn | l = \epsilon^{-1}
+      | r = \cos \dfrac {2 \pi} 7 - i \sin \dfrac {2 \pi} 7
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon + \epsilon^{-1}
+      | r = 2 c
+      | c = where $c = \cos \dfrac {2 \pi} 7$
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^2 + \epsilon^{-2} + 2
+      | r = 4 c^2
+      | c = squaring
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^3 + \epsilon^{-3} + 3 \paren {\epsilon + \epsilon^{-1} }
+      | r = 8 c^3
+      | c = cubing
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^2 + \epsilon^{-2}
+      | r = 4 c^2 - 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon^3 + \epsilon^{-3}
+      | r = 8 c^3 - 6 c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {8 c^3 - 6 c} + \paren {4 c^2 - 2} + 2 c - 1
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 8 c^3 + 4 c^2 - 4 c - 1
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Thus $2 c$ is a [[Definition:Root of Polynomial|root]] of the [[Definition:Polynomial|polynomial]] $x^3 + x^2 - 2 x - 1$
+But from [[Irreducible Polynomial/Examples/x^3 + x^2 - 2 x - 1 in Rationals|Irreducible Polynomial: $x^3 + x^2 - 2 x - 1$ in Rationals]]:
+:$x^3 + x^2 - 2 x - 1$ is [[Definition:Irreducible Polynomial|irreducible]] over $\Q$.
+Thus by [[Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2]], $\cos \dfrac {2 \pi} 7$ is not an [[Definition:Element|element]] of any [[Definition:Field Extension|extension]] of $\Q$ of [[Definition:Degree of Field Extension|degree]] $2^m$.
+The result follows from [[Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers with Denominator Power of Two form Integral Domain}
+Tags: Rational Numbers, Integral Domains
+
+\begin{theorem}
+Let $\Q$ denote the [[Definition:Rational Number|set of rational numbers]].
+Let $S \subseteq \Q$ denote the [[Definition:Set|set]] of [[Definition:Rational Number|set of rational numbers]] of the form $\dfrac p q$ where $q$ is a [[Definition:Integer Power|power of $2$]]:
+:$S = \set {\dfrac p q: p \in \Z, q \in \set {2^m: m \in \Z_{\ge 0} } }$
+Then $\struct {S, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Rational Numbers form Integral Domain]] we have that $\struct {\Q, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+Hence to demonstrate that $\struct {S, +, \times}$ is an [[Definition:Integral Domain|integral domain]], we can use the [[Subdomain Test]].
+We have that the [[Definition:Unity of Ring|unity]] of $\struct {\Q, +, \times}$ is $1$.
+Then we note:
+:$1 = \dfrac 1 1$
+and:
+:$1 = 2^0$
+and so $1 \in S$.
+Thus property $(2)$ of the [[Subdomain Test]] is fulfilled.
+It remains to demonstrate that $\struct {S, +, \times}$ is a [[Definition:Subring|subring]] of $\struct {\Q, +, \times}$, so fulfilling property $(2)$ of the [[Subdomain Test]].
+Hence we use the [[Subring Test]].
+We note that $S \ne \O$ as $1 \in S$.
+This fulfils property $(1)$ of the [[Subring Test]].
+Let $x, y \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x + \paren {-y}
+      | r = \dfrac a {2^p} + \dfrac b {2^q}
+      | c = for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$
+}}
+{{eqn | r = \dfrac {a 2^q - b 2^p} {2^p 2^q}
+      | c = {{Defof|Rational Addition}}
+}}
+{{eqn | r = \dfrac {a 2^q - b 2^p} {2^{p + q} }
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+This fulfils property $(2)$ of the [[Subring Test]].
+Then: 
+{{begin-eqn}}
+{{eqn | l = x \times y
+      | r = \dfrac a {2^p} \times \dfrac b {2^q}
+      | c = for some $a, b \in \Z$ and $p, q \in \Z_{\ge 0}$
+}}
+{{eqn | r = \dfrac {a b} {2^p 2^q}
+      | c = {{Defof|Rational Multiplication}}
+}}
+{{eqn | r = \dfrac {a b} {2^{p + q} }
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+This fulfils property $(3)$ of the [[Subring Test]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal Subgroup of Order 25 in Group of Order 100}
+Tags: Groups of Order 100
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $100$.
+Then $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $25$.
+\end{theorem}
+
+\begin{proof}
+Let $r$ be the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of [[Definition:Order of Group|order]] $5^2 = 25$
+The [[First Sylow Theorem]] guarantees existence, so $r \ge 1$.
+From the [[Fourth Sylow Theorem]]:
+:$r \equiv 1 \pmod 5$
+That is:
+:$r \in \set {1, 6, 11, 16, \ldots}$
+From the [[Fifth Sylow Theorem]]:
+:$r \divides \dfrac {100} {25} = 4$
+from which it follows that $r = 1$.
+From [[Sylow p-Subgroup is Unique iff Normal]], this [[Definition:Unique|unique]] [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] must be [[Definition:Normal Subgroup|normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal p-Subgroup contained in All Sylow p-Subgroups}
+Tags: P-Groups, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$ which is a [[Definition:P-Group|$p$-group]].
+Then $H$ is a [[Definition:Subset|subset]] of every [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+By [[Second Sylow Theorem]], $H$ is a [[Definition:Subset|subset]] of a [[Definition:Conjugate of Group Subset|conjugate]] of $P$.
+Then:
+:$\exists g \in G: H \subseteq g P g^{-1}$.
+This implies:
+{{begin-eqn}}
+{{eqn | l = H
+      | r = g^{-1} H g
+      | c = [[Subgroup equals Conjugate iff Normal]]
+}}
+{{eqn | o = \subseteq
+      | r = g^{-1} \paren {g P g^{-1} } g
+      | c = [[Subset Relation is Compatible with Subset Product/Corollary 2]]
+}}
+{{eqn | r = P
+}}
+{{end-eqn}}
+Since $P$ is arbitrary, $H$ is a [[Definition:Subset|subset]] of every [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Subgroup whose Index is Prime Power}
+Tags: P-Groups, Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$ which has a [[Definition:Finite Index|finite index]] in $G$.
+Let:
+:$p^k \divides \index G H$
+where:
+:$p$ is a [[Definition:Prime Number|prime number]]
+:$k \in \Z_{>0}$ is a [[Definition:Strictly Positive Integer|(strictly) positive integer]]
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then $G$ contains a [[Definition:Subgroup|subgroup]] $K$ such that:
+:$\index K H = p^k$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Order of Group|order]] $\order {G / H}$ of the [[Definition:Quotient Group|quotient group]] $G / H$ is $\index G H$.
+Hence $p^k$ [[Definition:Divisor of Integer|divides]] $\order {G / H}$.
+By [[Group has Subgroups of All Prime Power Factors]], $G / H$ has a [[Definition:Subgroup|subgroup]] of order $p^k$.
+By [[Correspondence Theorem (Group Theory)|Correspondence Theorem]], this [[Definition:Subgroup|subgroup]] is in the form $K / H$ where $H \le K \le G$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order p^2 q is not Simple}
+Tags: Groups of Order p^2 q, Simple Groups
+
+\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 $G$ is not [[Definition:Simple Group|simple]].
+\end{theorem}
+
+\begin{proof}
+From [[Group of Order p^2 q has Normal Sylow p-Subgroup|Group of Order $p^2 q$ has Normal Sylow $p$-Subgroup]], $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order $p^2$]].
+Hence the result, by definition of [[Definition:Simple Group|simple group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simple Group of Order Less than 60 is Prime}
+Tags: Simple Groups, Groups of Order 60
+
+\begin{theorem}
+Let $G$ be a [[Definition:Simple Group|simple group]].
+Let $\order G < 60$, where $\order G$ denotes the [[Definition:Order of Group|order]] of $G$.
+Then $G$ is a [[Definition:Prime Group|prime group]].
+\end{theorem}
+
+\begin{proof}
+First it is noted that [[Prime Group is Simple]].
+We also note from [[Alternating Group is Simple]] that the [[Definition:Alternating Group|alternating group $A_5$]], which is of [[Definition:Order of Group|order]] $60$, is [[Definition:Simple Group|simple]].
+Hence the motivation for the result.
+It remains to be shown that all [[Definition:Group|groups]] of [[Definition:Composite Number|composite]] [[Definition:Order of Group|order]] such that $\order G < 60$ are not [[Definition:Simple Group|simple]].
+Let $S$ be the [[Definition:Set|set]]:
+:$S = \set {n \in \Z: 0 < n < 60: \text { there exists a simple group of order $n$ such that $n$ is composite} }$
+The aim is to show that $S$ is [[Definition:Empty Set|empty]].
+From [[Abelian Group is Simple iff Prime]], all [[Definition:Abelian Group|abelian groups]] of [[Definition:Composite Number|composite]] [[Definition:Order of Group|order]] are not [[Definition:Simple Group|simple]].
+Thus any [[Definition:Simple Group|simple group]] must be non-[[Definition:Abelian Group|abelian]].
+Let $p$ and $q$ be [[Definition:Prime Number|prime]].
+From [[Prime Power Group has Non-Trivial Proper Normal Subgroup]], no [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p^n$ is [[Definition:Simple Group|simple]].
+Thus:
+:$\forall n \in \Z_{>0}: p^n \notin S$
+Thus, with the [[Definition:Prime Number|primes]] and [[Definition:Prime Power|prime powers]] eliminated, we have:
+:$S \subseteq {6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58}$
+From [[Group of Order p q has Normal Sylow p-Subgroup|Group of Order $p q$ has Normal Sylow $p$-Subgroup]]:
+:$p q \notin S$
+The [[Definition:Set|set]] of non-[[Definition:Square Number|square]] [[Definition:Semiprime|semiprimes]] less than $60$ is:
+:$\set {6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58}$
+Thus we have so far:
+:$S \subseteq {12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56}$
+From [[Group of Order p^2 q is not Simple|Group of Order $p^2 q$ is not Simple]]:
+:$p^2 q \notin S$
+This eliminates:
+{{begin-eqn}}
+{{eqn | l = 12
+      | r = 2^2 \times 3
+}}
+{{eqn | l = 18
+      | r = 3^2 \times 2
+}}
+{{eqn | l = 20
+      | r = 2^2 \times 5
+}}
+{{eqn | l = 28
+      | r = 2^2 \times 7
+}}
+{{eqn | l = 44
+      | r = 2^2 \times 11
+}}
+{{eqn | l = 45
+      | r = 3^2 \times 5
+}}
+{{eqn | l = 50
+      | r = 5^2 \times 2
+}}
+{{eqn | l = 52
+      | r = 2^2 \times 13
+}}
+{{end-eqn}}
+Thus we are left with:
+:$S \subseteq {24, 30, 36, 40, 42, 48, 54, 56}$
+From [[Normal Subgroup of Group of Order 24]], a [[Definition:Group|group]] of [[Definition:Order of Group|order $24$]] has a [[Definition:Normal Subgroup|normal subgroup]] either of [[Definition:Order of Group|order $4$]] or [[Definition:Order of Group|order $8$]].
+Hence $24 \notin S$.
+From [[Group of Order 56 has Unique Sylow 2-Subgroup or Unique Sylow 7-Subgroup]], a [[Definition:Group|group]] of [[Definition:Order of Group|order $56$]] has at least one [[Definition:Normal Subgroup|normal subgroup]].
+Hence $56 \notin S$.
+From [[Group of Order 30 is not Simple]]:
+:$30 \notin S$
+Thus we are left with:
+:$S \subseteq {36, 40, 42, 48, 54}$
+$40$ is eliminated by [[Group of Order 40 has Normal Subgroup of Order 5]].
+$42$ is eliminated by [[Group of Order 42 has Normal Subgroup of Order 7]].
+$54$ is eliminated by [[Group of Order 54 has Normal Subgroup of Order 27]].
+Thus remains:
+:$S \subseteq {36, 48}$
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Group of Order 42 has Normal Subgroup of Order 7}
+Tags: Groups of Order 42
+
+\begin{theorem}
+Let $G$ be of [[Definition:Order of Group|order]] $42$.
+Then $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $7$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$42 = 2 \times 3 \times 7$
+From the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]], which is of [[Definition:Order of Group|order]] $7$.
+Let $n_7$ denote the number of [[Definition:Sylow p-Subgroup|Sylow $7$-subgroups]] of $G$.
+From the [[Fourth Sylow Theorem]]:
+:$n_7 \equiv 1 \pmod 7$
+and from the [[Fifth Sylow Theorem]]:
+:$n_7 \divides 6$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_7 = 1$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], this [[Definition:Sylow p-Subgroup|Sylow $7$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+[[Category:Groups of Order 42]]
+qzh3u1kuc9cl8kk2l8oa9yookmzd6a5
+\end{proof}<|endoftext|>
+\section{Group of Order 54 has Normal Subgroup of Order 27}
+Tags: Groups of Order 54
+
+\begin{theorem}
+Let $G$ be of [[Definition:Order of Group|order]] $54$.
+Then $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $27$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$54 = 2 \times 3^3$
+From the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]], which is of [[Definition:Order of Group|order]] $3^3 = 27$.
+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]]:
+:$n_3 \divides 2$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_3 = 1$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], this [[Definition:Sylow p-Subgroup|Sylow $3$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+[[Category:Groups of Order 54]]
+49bi4fhqoq3ujnnbiungpqlspagce6y
+\end{proof}<|endoftext|>
+\section{Group of Order 40 has Normal Subgroup of Order 5}
+Tags: Groups of Order 40
+
+\begin{theorem}
+Let $G$ be of [[Definition:Order of Group|order]] $40$.
+Then $G$ has a [[Definition:Normal Subgroup|normal subgroup]] of [[Definition:Order of Group|order]] $5$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$40 = 2^3 \times 5$
+From the [[First Sylow Theorem]], $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]], which is of [[Definition:Order of Group|order]] $5$.
+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]]:
+:$n_5 \divides 8$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that $n_5 = 1$.
+From [[Sylow p-Subgroup is Unique iff Normal|Sylow $p$-Subgroup is Unique iff Normal]], this [[Definition:Sylow p-Subgroup|Sylow $5$-subgroup]] is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+[[Category:Groups of Order 40]]
+djhna8lndx7fauabsnev82th3in9f1m
+\end{proof}<|endoftext|>
+\section{Finite Group with One Sylow p-Subgroup per Prime Divisor is Isomorphic to Direct Product}
+Tags: Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Group|order]] is $n$ and whose [[Definition:Identity Element|identity element]] is $e$.
+Let $G$ be such that it has exactly $1$ [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] for each [[Definition:Prime Divisor|prime divisor]] of $n$.
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] to the [[Definition:Internal Group Direct Product|internal direct product]] of all its [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]].
+\end{theorem}
+
+\begin{proof}
+If each of the [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] are [[Definition:Unique|unique]], they are all [[Definition:Normal Subgroup|normal]].
+As the [[Definition:Order of Group|order]] of each one is [[Definition:Coprime Integers|coprime]] to each of the others, their [[Definition:Set Intersection|intersection]] is $\set e$.
+{{finish|It remains to be shown that the direct product is what is is}}
+\end{proof}<|endoftext|>
+\section{Number of Subgroups of Prime Power Order is Congruent to 1 modulo Prime}
+Tags: Sylow p-Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Order of Group|order]] is $n$.
+Let $p$ be a [[Definition:Prime Number|prime number]] such that $p^k$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
+Then the number of [[Definition:Subgroup|subgroups]] of [[Definition:Order of Group|order]] $p^k$ is [[Definition:Congruence Modulo Integer|congruent]] to $1$ modulo $p$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Similar to the [[First Sylow Theorem]] at first glance. The book suggests this proof is due to Frobenius, and is long.}}
+\end{proof}<|endoftext|>
+\section{Count of Distinct Homomorphisms between Additive Groups of Integers Modulo m}
+Tags: Additive Groups of Integers Modulo m
+
+\begin{theorem}
+Let $m, n \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
+Let $\struct {\Z_m, +}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+The number of distinct [[Definition:Group Homomorphism|homomorphisms]] $\phi: \struct {\Z_m, +} \to \struct {\Z_n, +}$ is $\gcd \set {m, n}$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|This was just a guess. It seems plausible from experimentation.}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Additive Group Modulo 16 and Multiplicative Group Modulo 17}
+Tags: Additive Groups of Integers Modulo m, Multiplicative Groups of Reduced Residues, Groups of Order 16, Groups of Order 17
+
+\begin{theorem}
+Let $\struct {\Z_{16}, +}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $16$]].
+Let $\struct {\Z'_{17}, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $17$]].
+Let $\phi: \struct {\Z_{16}, +} \to \struct {\Z'_{17}, \times}$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \eqclass k {16} \in \struct {\Z_{16}, +}: \map \phi {\eqclass k {16} } = \eqclass {3^k} {17}$
+Then $\phi$ is a [[Definition:Group Isomorphism|group isomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $\eqclass x {16}, \eqclass y {16} \in \struct {\Z_{16}, +}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\eqclass x {16} } \times \map \phi {\eqclass y {16} }
+      | r = \map \phi {x + 16 m_1} \times \map \phi {y + 16 m_2}
+      | c = {{Defof|Residue Class}}: for some representative $m_1, m_2 \in \Z$
+}}
+{{eqn | r = 3 \uparrow \paren {x + 16 m_1} \times 3 \uparrow \paren {y + 16 m_2}
+      | c = using [[Definition:Knuth Uparrow Notation|Knuth uparrow notation]] $3 \uparrow k := 3^k$
+}}
+{{eqn | r = 3 \uparrow \paren {x + 16 m_1 + y + 16 m_2}
+      | c = [[Product of Powers]]
+}}
+{{eqn | r = 3 \uparrow \paren {\paren {x + y} + 16 \paren {m_1 + m_2} }
+      | c = 
+}}
+{{eqn | r = 3 \uparrow \paren {\eqclass {x + y} {16} }
+      | c = {{Defof|Residue Class}} and {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \map \phi {\eqclass x {16} + \eqclass y {16} }
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Thus it is seen that $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
+{{qed|lemma}}
+It remains to be seen that $\phi$ is a [[Definition:Bijection|bijection]].
+Because $17$ is [[Definition:Prime Number|prime]]:
+$\forall x \in \Z, 1 \le x < 17: x \perp 17$
+where $\perp$ denotes [[Definition:Coprime Integers|coprimality]].
+Thus by definition of [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $17$]]:
+:$\order {\struct {\Z'_{17}, \times} } = 16$
+where $\order {\, \cdot \,}$ denotes the [[Definition:Order of Group|order]] of a [[Definition:Group|group]].
+Similarly, by definition of [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $16$]]:
+:$\order {\struct {\Z_{16}, +} } = 16$
+So:
+:$\order {\struct {\Z'_{17}, \times} } = \order {\struct {\Z_{16}, +} }$
+which is a [[Definition:Necessary Condition|necessary condition]] for [[Definition:Group Isomorphism|group isomorphism]].
+{{qed|lemma}}
+Now we have:
+{{begin-eqn}}
+{{eqn | l = 16
+      | o = \equiv
+      | r = 0
+      | rr= \pmod {16}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {\eqclass {16} {16} }
+      | r = \map \phi {\eqclass 0 {16} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \phi {\eqclass {16} {16} }
+      | r = \eqclass 1 {17}
+      | c = [[Group Homomorphism Preserves Identity]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = 3^{16}
+      | o = \equiv
+      | r = 1
+      | rr= \pmod {17}
+      | c = Definition of $\phi$
+}}
+{{end-eqn}}
+Now let $\eqclass x {16}, \eqclass y {16} \in \Z_{16}$ such that $\map \phi {\eqclass x {16} } = \map \phi {\eqclass y {16} }$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \map \phi {\eqclass x {16} }
+      | r = \map \phi {\eqclass y {16} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall m_1, m_2 \in \Z
+      | l = \map \phi {x + 16 m_1}
+      | r = \map \phi {y + 16 m_2}
+      | c = {{Defof|Residue Class}}
+}}
+{{eqn | ll= \leadsto
+      | l = 3 \uparrow \paren {x + 16 m_1}
+      | r = 3 \uparrow \paren {y + 16 m_2}
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = 3^x \paren {3^{16} }^{m_1}
+      | r = 3^y \paren {3^{16} }^{m_2}
+      | c = [[Product of Powers]], [[Power of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = 3^x \times 1^{m_1}
+      | r = 3^y \times 1^{m_2}
+      | c = as $3^{16} = 1 \pmod {17}$ from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = 3^x
+      | r = 3^y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\phi$ is an [[Definition:Injection|injection]].
+From [[Equivalence of Mappings between Sets of Same Cardinality]] it follows that $\phi$ is a [[Definition:Bijection|bijection]].
+{{qed|lemma}}
+Thus $\phi$ is a [[Definition:Bijection|bijective]] [[Definition:Group Homomorphism|group homomorphism]].
+Hence the result by definition of [[Definition:Group Isomorphism|group isomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 1}
+Tags: P-adic Integer is Limit of Unique Coherent Sequence of Integers
+
+\begin{theorem}
+:$\forall n \in \N: \alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1}}$
+\end{theorem}
+
+\begin{proof}
+For any $n \in \N$ then:
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha_{n + 1} - \alpha_n }_p
+      | r = \norm {\paren {\alpha_{n + 1} - x} + \paren {x - \alpha_n } }_p
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {\alpha_{n + 1} - x}_p, \: \norm {x - \alpha_n }_p }
+      | c = {{NormAxiom|4}}
+}}
+{{eqn | o = \le
+      | r = \max \set {\norm {x - \alpha_{n + 1} }_p, \: \norm {x - \alpha_n }_p}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of negative]]
+}}
+{{eqn | o = \le
+      | r = \max \set {p^{-\paren{n + 2} } , p^{-\paren{n + 1} } }
+}}
+{{eqn | r = p^{-\paren{n + 1} }
+      | c = Since $p^{-n - 1} < p^{-n}$
+}}
+{{end-eqn}}
+Hence:
+:$p^{n + 1} \divides \paren {\alpha_{n + 1} - \alpha_n} $
+or equivalently:
+:$\alpha_{n + 1} \equiv \alpha_n \pmod {p^{n + 1} }$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 3}
+Tags: P-adic Integer is Limit of Unique Coherent Sequence of Integers
+
+\begin{theorem}
+$\sequence {\alpha_n}$ is a [[Definition:Unique|unique]] [[Definition:Sequence|sequence]] satisfying properties $(1)$, $(2)$ and $(3)$ above.
+\end{theorem}
+
+\begin{proof}
+Suppose that there exists a [[Definition:Sequence|sequence]] $\sequence {\alpha'_n}$ with:
+:$(1'): \quad \forall n \in \N: \alpha'_n \in \Z$ and $0 \le \alpha'_n \le p^{n + 1} - 1$
+:$(2'): \quad \forall n \in \N: \alpha'_{n + 1} \equiv \alpha'_n \pmod {p^{n + 1} }$
+:$(3'): \quad \displaystyle \lim_{n \mathop \to \infty} \alpha'_n = x$
+{{AimForCont}}:
+:$\alpha'_N \ne \alpha_N$ for some $N \in \N$
+By [[Initial Segment of Natural Numbers forms Complete Residue System]]:
+:$\alpha'_N \not \equiv \alpha_N \pmod {p^{N + 1}}$
+Then for all $n > N$:
+{{begin-eqn}}
+{{eqn | l = \alpha'_n
+      | o = \equiv
+      | r = \alpha'_N \pmod {p^{N + 1} }
+      | c = by $(2)$ above
+}}
+{{eqn | o = \not \equiv
+      | r = \alpha_N \pmod {p^{N + 1} }
+}}
+{{eqn | o = \equiv
+      | r = \alpha_n \pmod {p^{N + 1} }
+      | c = by $(2)$ above
+}}
+{{end-eqn}}
+That is, for all $n > N$:
+:$\alpha'_n \not \equiv \alpha_n \pmod {p^{N + 1}}$
+Hence for all $n > N$:
+:$\norm {\alpha'_n - \alpha_n}_p > p^{-\paren{N + 1}}$
+By $(3)$ the [[Definition:Limit of Sequence (Normed Division Ring)|limit]] of $\sequence{\alpha_n}$ is $x$:
+:$\exists N_1 \in \N: \forall n \ge N_1: \norm {x - \alpha_n}_p \le p^{-\paren{N + 1}}$
+Similarly for $\sequence{\alpha'_n}$:
+:$\exists N_2 \in \N: \forall n \ge N_2: \norm {x - \alpha'_n}_p \le p^{-\paren{N + 1}}$
+Let $M = \max \set {N+1, N_1, N_2}$.
+Then:
+:$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren{N + 1}}$
+:$\norm {x - \alpha_M} _p\le p^{-\paren{N + 1}}$
+:$\norm {x - \alpha'_M}_p \le p^{-\paren{N + 1}}$
+But:
+{{begin-eqn}}
+{{eqn | l = \norm {\alpha'_M - \alpha_M}_p
+      | r = \norm {\paren {\alpha'_M - x} + \paren {x - \alpha_M} }_p
+}}
+{{eqn | o = \le
+      | r = \max \set{\norm{\alpha'_M - x}_p, \: \norm {x - \alpha_M}_p}
+      | c = [[Definition:Non-Archimedean Division Ring Norm|Norm axiom (N4) (Ultrametric Inequality)]]
+}}
+{{eqn | r = \max \set{\norm{x - \alpha'_M}_p, \: \norm {x - \alpha_M}_p}
+      | c = [[Properties of Norm on Division Ring/Norm of Negative|Norm of negative]]
+}}
+{{eqn | o = \le
+      | r = \max \set{p^{-\paren{N + 1} } , p^{-\paren{N + 1} } }
+}}
+{{eqn | r = p^{-\paren{N + 1} }
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the previous assertion that:
+:$\norm {\alpha'_M - \alpha_M}_p > p^{-\paren{N + 1}}$
+Hence:
+:$\sequence{\alpha'_n} = \sequence{\alpha_n}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Integer is Limit of Unique Coherent Sequence of Integers/Lemma 2}
+Tags: P-adic Integer is Limit of Unique Coherent Sequence of Integers
+
+\begin{theorem}
+:$\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$
+\end{theorem}
+
+\begin{proof}
+From [[Sequence of Powers of Number less than One]]:
+:$\displaystyle \lim_{n \mathop \to \infty} p^{-n} = 0$
+From [[Multiple Rule for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} p^{-\paren{n + 1}} = 0$
+By the [[Squeeze Theorem for Real Sequences]] :
+:$\displaystyle \lim_{n \mathop \to \infty} \norm{x - \alpha_n}_p = 0$.
+Hence the [[Definition:Limit of Sequence (Normed Division Ring)|limit]] of $\sequence {\alpha_n}$ is by definition:
+:$\displaystyle \lim_{n \mathop \to \infty} \alpha_n = x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Types of Order Prime Squared}
+Tags: Groups of Order p^2
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $p^2$.
+Then $G$ is [[Definition:Group Isomorphism|isomorphic]] either to $\Z_{p^2}$ or to $\Z_p \times \Z_p$, where $\Z_p$ denotes the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $p$]].
+\end{theorem}
+
+\begin{proof}
+From [[Group of Order Prime Squared is Abelian]], $G$ is an [[Definition:Abelian Group|abelian group]].
+From [[Abelian Group of Prime-power Order is Product of Cyclic Groups]], $G$ is either:
+:the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $p^2$
+:the [[Definition:Group Direct Product|direct product]] of the [[Definition:Cyclic Group|cyclic group]]  of [[Definition:Order of Group|order]] $p$ with itself.
+The result follows from [[Finite Cyclic Group is Isomorphic to Integers under Modulo Addition]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers}
+Tags: Examples of Groups: x+y over 1+xy
+
+\begin{theorem}
+$\struct {G, \circ}$ is [[Definition:Isomorphism|isomorphic]] to the [[Definition:Additive Group of Real Numbers|additive group of real numbers]] $\struct {\R, +}$.
+\end{theorem}<|endoftext|>
+\section{Open Set Characterization of Denseness/Analytic Basis}
+Tags: Denseness
+
+\begin{theorem}
+Let $\mathcal B \subseteq \tau$ be an [[Definition:Analytic Basis|analytic basis]] for $\tau$.
+Then $S$ is [[Definition:Everywhere Dense|(everywhere) dense]] in $X$ {{iff}} every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $\mathcal B$ contains an [[Definition:Element|element]] of $S$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $S$ be [[Definition:Everywhere Dense|everywhere dense]] in $X$.
+By [[Open Set Characterization of Denseness]] then every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set contains an [[Definition:Element|element]] of $S$.
+Every [[Definition:Non-Empty Set|non-empty]] set of an [[Definition:Analytic Basis|analytic basis]] is an [[Definition:Open Set (Topology)|open set]] by definition.
+Hence every [[Definition:Non-Empty Set|non-empty]] set of [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $\mathcal B$ contains an [[Definition:Element|element]] of $S$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set of $\mathcal B$ contains an [[Definition:Element|element]] of $S$.
+Let $U$ be any [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set.
+Let $x \in U$.
+By the definition of an [[Definition:Analytic Basis|analytic basis]] then:
+:$\exists B \in \mathcal B: x \in B \subseteq U$
+By [[Definition:Assumption|assumption]] $B$ contains an [[Definition:Element|element]] $s$ of $S$.
+Hence $s \in U$.
+Since $U$ was an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set then every [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set (Topology)|open]] set contains an [[Definition:Element|element]] of $S$
+By [[Open Set Characterization of Denseness]], $S$ is [[Definition:Everywhere Dense|everywhere dense]] in $X$.
+{{qed}}
+[[Category:Denseness]]
+9ml9xc4pcclv46blz9n2x6euv1yrene
+\end{proof}<|endoftext|>
+\section{Open Set Characterization of Denseness/Open Ball}
+Tags: Denseness
+
+\begin{theorem}
+Let $\struct{X, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\tau_d$ be the [[Definition:Topology Induced by Metric|topology induced]] by the [[Definition:Metric Space|metric]] $d$.
+Let $S \subseteq X$.
+Then $S$ is [[Definition:Everywhere Dense|(everywhere) dense]] in $\struct{X, \tau_d}$ {{iff}} every [[Definition:Open Ball|open ball]] contains an [[Definition:Element|element]] of $S$.
+\end{theorem}
+
+\begin{proof}
+By [[Open Balls form Basis for Open Sets of Metric Space]], the [[Definition:Set|set]] of [[Definition:Open Ball|open balls]] are an [[Definition:Analytic Basis|analytic basis]] for the [[Definition:Topology|topology]] $\tau_d$.
+By [[Analytic Basis Characterization of Denseness|Analytic Basis Characterization of Denseness]] then:
+:$S$ is [[Definition:Everywhere Dense|(everywhere) dense]] in $\struct{X, \tau_d}$ {{iff}} every [[Definition:Open Ball|open ball]] contains an [[Definition:Element|element]] of $S$.
+{{qed}}
+[[Category:Denseness]]
+5c156wij1viyrf4bkv3dd0c590r9jv0
+\end{proof}<|endoftext|>
+\section{P-adic Valuation Extends to P-adic Numbers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\nu_p^\Q: \Q \to \Z \cup \set {+\infty}$ be the [[Definition:P-adic Valuation/Rational Numbers|$p$-adic valuation]] on the [[Definition:Set|set]] of  [[Definition:Rational Number|rational numbers]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ be defined by:
+:$\forall x \in \Q_p : \map {\nu_p} x = \begin {cases}
+-\log_p \norm {x}_p : x \ne 0 \\
++\infty : x = 0
+\end{cases}$
+Then $\nu_p: \Q_p \to \Z \cup \set {+\infty}$ is a [[Definition:Valuation|valuation]] that [[Definition:Extension of Mapping|extends]] $\nu_p^\Q$ from $\Q$ to $\Q_p$.
+\end{theorem}
+
+\begin{proof}
+It needs to be shown that $\nu_p$:
+:$(1): \quad \nu_p$ is a [[Definition:Mapping|mapping]] into $\Z \cup \set {+\infty}$
+:$(2): \quad \nu_p$ satisfies the [[Definition:Valuation Axioms|valuation axioms $\text V 1$, $\text V 2$ and $\text V 3$]]
+:$(3): \quad \nu_p$ [[Definition:Extension of Mapping|extends]] $\nu_p^\Q$.
+Let $x, y \in \Q_p$.
+=== $\nu_p$ is a [[Definition:Mapping|mapping]] into $\Z \cup \set {+\infty}$ ===
+If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.
+Let $x \ne 0$.
+By [[P-adic Norm of p-adic Number is Power of p]] then:
+:$\exists v \in \Z: \norm x_p = p^{-v}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} x
+      | r = -\log_p \norm {x}_p
+      | c = Since $x \ne 0$
+}}
+{{eqn | r = -\log_p p^{-v}
+      | c = Definition of $v$
+}}
+{{eqn | r = -\paren {-v}
+      | c = {{Defof|Real General Logarithm}}
+}}
+{{eqn | r = v
+}}
+{{eqn | o = \in
+      | r = \Z
+      | c = Definition of $v$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $\nu_p$ satisfies $(\text V 1)$ ===
+If $x = 0$ then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {0 \cdot y}
+      | r = \map {\nu_p} 0
+}}
+{{eqn | r = +\infty
+      | c = Definition of $\nu_p$
+}}
+{{eqn | r = +\infty \cdot \map {\nu_p} y
+      | c = {{Defof|Extended Real Multiplication}}
+}}
+{{eqn | r = \map {\nu_p} 0 \cdot \map {\nu_p} y
+      | c = Definition of $\nu_p$
+}}
+{{end-eqn}}
+Similarly, if $y = 0$ then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {x \cdot 0}
+      | r = \map {\nu_p} x \cdot \map {\nu_p} 0
+}}
+{{end-eqn}}
+If $x \ne 0, y \ne 0$ then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {x y}
+      | r = -\log \norm {x y}_p
+      | c = $x y \ne 0$
+}}
+{{eqn | r = -\log \norm x_p \norm y_p
+      | c = {{NormAxiom|2}}
+}}
+{{eqn | r = -\paren {\log \norm x_p + \log \norm y_p}
+      | c = [[Sum of Logarithms/General Logarithm|Sum of General Logarithms]]
+}}
+{{eqn | r = \paren {-\log \norm x_p} + \paren {-\log \norm y_p}
+      | c = 
+}}
+{{eqn | r = \map {\nu_p} x + \map {\nu_p} y
+      | c = Definition of $\nu_p$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $\nu_p$ satisfies $(\text V 2)$ ===
+If $x = 0$ then $\map {\nu_p} x = +\infty$ by definition.
+If $x \ne 0$ then $\map {\nu_p} x \in \Z$ by the above.
+Hence:
+:$\map {\nu_p} x = +\infty \iff x = 0$ 
+{{qed|lemma}}
+=== $\nu_p$ satisfies $(\text V 3)$ ===
+Suppose $x = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {0 + y}
+      | r = \map {\nu_p} y
+}}
+{{eqn | o = \ge
+      | r = \min \set {\map {\nu_p} 0, \map {\nu_p} y}
+      | c = {{Defof|Min Operation}}
+}}
+{{end-eqn}}
+Similarly, if $y = 0$ then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {x + 0}
+      | o = \ge
+      | r = \min \set {\map {\nu_p} x, \map {\nu_p} 0}
+}}
+{{end-eqn}}
+Suppose $x + y = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} {x + y}
+      | r = \map {\nu_p} 0
+}}
+{{eqn | r = +\infty
+      | c = Definition of $\nu_p$
+}}
+{{eqn | o = \ge
+      | r = \map {\nu_p} x
+      | c = {{Defof|Extended Real Number Line}}
+}}
+{{eqn | o = \ge
+      | r = \min \set {\map {\nu_p} x, \map {\nu_p} y}
+      | c = {{Defof|Min Operation}}
+}}
+{{end-eqn}}
+Suppose $x \ne 0, y \ne 0, x + y \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \norm {x + y}
+      | o = \le
+      | r = \max \set {\norm x_p, \norm y_p}
+      | c = {{NormAxiom|4}}
+}}
+{{eqn | o = }}
+{{eqn | ll = \leadsto
+      | l = \log \norm {x + y}
+      | o = \le
+      | r = \log \max \set {\norm x_p, \norm y_p}
+      | c = [[Logarithm is Strictly Increasing]]
+}}
+{{eqn | r = \max \set {\log \norm x_p, \log \norm y_p}
+      | c = [[Logarithm is Strictly Increasing]]
+}}
+{{eqn | o = }}
+{{eqn | ll = \leadsto
+      | l = -\log \norm {x + y}
+      | o = \ge
+      | r = -\max \set {\log \norm x_p, \log \norm y_p}
+      | c = [[Inversion Mapping Reverses Ordering in Ordered Group]]
+}}
+{{eqn | r = \min \set {-\log \norm x_p, -\log \norm y_p}
+      | c = 
+}}
+{{eqn | o = }}
+{{eqn | ll = \leadsto
+      | l = \map {\nu_p} {x + y}
+      | o = \ge
+      | r = \min \set {\map {\nu_p} x, \map {\nu_p} y}
+      | c = Definition of $\nu_p$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== $\nu_p$ extends $\nu_p^\Q$ ===
+Let $x \in \Q$.
+If $x = 0$ then $\map {\nu_p} 0 = +\infty = \map {\nu_p^\Q} 0$.
+Let $x \ne 0$.
+The [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on [[Definition:P-adic Numbers|$p$-adic numbers]] $\Q_p$ is an [[Definition:Extension of  Mapping|extension]] of the [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on [[Definition:Rational Number|rational numbers]] $\Q$ by definition.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map {\nu_p} x
+      | r = -\log \norm x_p
+      | c = Definition of $\nu_p$
+}}
+{{eqn | r = \map {\nu_p^\Q} x
+      | c = {{Defof|P-adic Norm|$p$-adic norm}} $\norm{\,\cdot\,}_p$ on [[Definition:Rational Number|rational numbers]] $\Q$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Integers is Metric Completion of Integers}
+Tags: P-adic Number Theory
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $d$ be the [[Definition:Metric Subspace|subspace metric]] of the [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]] on the [[Definition:P-adic Integers|$p$-adic integers]] $\Z_p$.
+Then $\struct {\Z_p, d}$ is the [[Definition:Completion (Metric Space)|metric completion]] of the [[Definition:Integer|integers]] $\Z$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Integer|integers]] $\Z$ are a [[Definition:Subring|subring]] of the [[Definition:P-adic Integers|$p$-adic integers]] $Z_p$ by [[Integers form Subring of P-adic Integers]].
+Hence $\Z \subseteq \Z_p$.
+The [[Definition:Set|set]] of [[Definition:P-adic Integers|$p$-adic integers]] $\Z_p$ is [[Definition:Closed Set of Metric Space|closed]] in the [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]] by [[Set of P-adic Integers is Clopen in P-adic Numbers]].
+By [[Closure of Subset of Closed Set of Metric Space is Subset]] then the [[Definition:Closure (Metric Space)|closure]] of $\Z$ is [[Definition:Contain|contained]] in $\Z_p$:
+:$\map \cl \Z \subseteq \Z_p$
+By [[P-adic Integer is Limit of Unique Coherent Sequence of Integers]] then:
+:for all $x \in \Z_p$ there exists a [[Definition:Sequence|sequence]] $\sequence {x_n}$ in $\Z$ such that $\displaystyle \lim_{n \to \infty} x_n = x$
+By [[Closure of Subset of Metric Space by Convergent Sequence]] then:
+:$\Z_p \subseteq \map \cl \Z$ 
+Hence $\map \cl \Z = \Z_p$.
+By [[Metric Subspace Induces Subspace Topology]] then the [[Definition:Topology Induced by Metric|topology induced]] by $d$ on $\Z_p$ is the [[Definition:Subspace Topology|subspace topology]].
+By [[Closure in Subspace]] then $\map \cl \Z = \Z_p$ in the [[Definition:Metric Subspace|subspace metric]] $d$ on $\Z_p$.
+It follows that $\struct {\Z_p, d}$ is the [[Definition:Completion (Metric Space)|metric completion]] of $\Z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{P-adic Metric on P-adic Numbers is Non-Archimedean Metric}
+Tags: P-adic Metrics
+
+\begin{theorem}
+Let $p \in \N$ be a [[Definition:Prime Number|prime]].
+Let $\norm{\,\cdot\,}_p: \Q_p \to \R_{\ge 0}$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:P-adic Number|$p$-adic numbers]] $\Q_p$.
+Let $d_p$ be the [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]] on $\Q_p$:
+:$\forall x, y \in \Q_p: \map {d_p} {x, y} = \norm{x - y}_p$
+Then $d_p$ is a [[Definition:Non-Archimedean Metric|non-Archimedean metric]] that [[Definition:Extension of Mapping|extends]] the [[Definition:P-adic Metric|$p$-adic metric]] on the [[Definition:Rational Number|rationals]] $\Q$ to $\Q_p$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]] on $\Q_p$ is defined as the [[Definition:Metric Induced by Norm|metric induced]] by the [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]] on $\Q_p$.
+It follows from [[Metric Induced by Norm is Metric]] that $d_p$ is a [[Definition:Metric|metric]].
+By definition of the [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]] on $\Q_p$, $\norm{\,\cdot\,}_p$ is a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]].
+From [[Non-Archimedean Norm iff Non-Archimedean Metric]], then $d_p$ is a [[Definition:Non-Archimedean Metric|non-Archimedean metric]].
+The [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on the [[Definition:P-adic Number|$p$-adic numbers]] $\Q_p$ is defined to be an [[Definition:Extension of Mapping|extension]] of the [[Definition:P-adic Norm|$p$-adic norm]] $\norm{\,\cdot\,}_p$ on $\Q$.
+By the definition of the [[Definition:Metric Induced by Norm|metric induced]] by the [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]], it follows that $d_p$ is an [[Definition:Extension of Mapping|extension]] of the [[Definition:P-adic Metric|$p$-adic metric]] on the [[Definition:Rational Number|rationals]] $\Q$.
+{{qed}}
+[[Category:P-adic Metrics]]
+la84i9e4l0koi4phd8626sogono0596
+\end{proof}<|endoftext|>
+\section{Kochen-Stone Borel-Cantelli}
+Tags: 
+
+\begin{theorem}
+Let $A_n$ be a sequence of events with $\displaystyle \sum \map \Pr {A_n} = \infty$ and:
+:$\displaystyle \liminf_{k \mathop \to \infty} \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} < \infty$
+Then there is a positive probability that $A_n$ occur infinitely often.
+\end{theorem}
+
+\begin{proof}
+Fix $\ell < k$.
+Let $\displaystyle X = \sum_{n \mathop = \ell}^k 1_{A_n}$. It follows that:
+:$\displaystyle \expect X = \sum_{n \mathop = \ell}^k \Pr(A_n)$
+and:
+:$\displaystyle \expect {X^2} = \sum_{\ell \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m}$
+Using Paley-Zygmund inequality for $\theta = 0$, we obtain:
+{{begin-eqn}}
+{{eqn | l = \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}
+      | r = \map \Pr {X > 0}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \frac {\displaystyle \paren {\sum_{n \mathop = \ell}^k \map \Pr {A_n} }^2} {\displaystyle \sum_{\ell \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} }
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} - \sum_{1 \mathop \le m, n \mathop < \ell} \map \Pr {A_n \cap A_m} }
+      | c = 
+}}
+{{end-eqn}}
+Now, it holds that:
+:$\displaystyle \map \Pr {A_n \text { occurs i.o.} } = \lim_{\ell \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$
+We have:
+:$\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} - \sum_{1 \mathop \le m, n \mathop < \ell} \map \Pr {A_n \cap A_m} } \ge \frac {\displaystyle \left(\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \Pr (A_n)\right)^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} }$
+Fix $l \in \N_1$.
+Since $\displaystyle \lim_{k \mathop \to \infty}\sum_{n \mathop = 1}^k \map \Pr {A_n} = \infty$, by assumption, if $k$ is sufficiently large:
+:$\displaystyle \paren {\sum_{n \mathop = 1}^ k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2 \approx \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2$
+So if $k$ is sufficiently large:
+:$\displaystyle \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} - \sum_{n \mathop = 1}^{\ell - 1} \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } \approx \frac {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } = \frac 1 {\frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} }$
+Suppose:
+:$\displaystyle \liminf_{k \mathop \to \infty} \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} = c < \infty$
+This means that no matter how large $k$ is, there is always some $k' \ge k$, such that:
+:$\displaystyle \frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k'} \map \Pr {A_n \cap A_m} } {\displaystyle \paren {\sum_{n \mathop = 1}^{k'} \map \Pr {A_n} }^2} \approx c$
+So for infinitely many $k$s:
+:$\displaystyle \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 {\frac {\displaystyle \sum_{1 \mathop \le m, n \mathop \le k} \map \Pr {A_n \cap A_m} } {\paren {\displaystyle \sum_{n \mathop = 1}^k \map \Pr {A_n} }^2} } \approx \frac 1 c$
+Since $\displaystyle \map \Pr {\bigcup_{n \mathop = \ell}^k A_n}$ is decreasing in $k$, we have, approximately:
+:$\displaystyle \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c$
+Since $l$ was arbitrary:
+:$\displaystyle \lim_{l \mathop \to \infty} \lim_{k \mathop \to \infty} \map \Pr {\bigcup_{n \mathop = \ell}^k A_n} \ge \frac 1 c > 0$
+{{qed}}
+arrinj0mrefo48o28yku8w5tdyftnuu
+\end{proof}<|endoftext|>
+\section{Order of Automorphism Group of Cyclic Group}
+Tags: Automorphism Groups, Cyclic Groups
+
+\begin{theorem}
+Let $C_n$ denote the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order]] $n$.
+Let $\Aut {C_n}$ denote the [[Definition:Automorphism Group|automorphism group]] of $C_n$.
+Then:
+:$\order {\Aut {C_n} } = \map \phi n$
+where:
+:$\order {\, \cdot \,}$ denotes the [[Definition:Order of Group|order]] of a [[Definition:Group|group]]
+:$\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+Let $g$ be a [[Definition:Generator of Cyclic Group|generator]] of $C_n$.
+Let $\varphi$ be an [[Definition:Group Automorphism|automorphism]] on $C_n$.
+By [[Homomorphic Image of Cyclic Group is Cyclic Group]], $\map \varphi g$ is a [[Definition:Generator of Cyclic Group|generator]] of $C_n$.
+By [[Homomorphism of Generated Group]], $\varphi$ is uniquely determined by $\map \varphi g$.
+By [[Finite Cyclic Group has Euler Phi Generators]], there are $\map \phi n$ possible values for $\map \varphi g$.
+Therefore there are $\map \phi n$ [[Definition:Group Automorphism|automorphisms]] on $C_n$:
+:$\order {\Aut {C_n} } = \map \phi n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Automorphism Group of Dihedral Group}
+Tags: Automorphism Groups, Dihedral Groups
+
+\begin{theorem}
+Let $D_n$ denote the [[Definition:Dihedral Group|dihedral group]] of [[Definition:Order of Group|order]] $n$.
+Let $\Aut {D_n}$ denote the [[Definition:Automorphism Group|automorphism group]] of $D_n$.
+Then:
+:$\order {\Aut {D_n} } = 2 \map \phi n$
+where:
+:$\order {\, \cdot \,}$ denotes the [[Definition:Order of Group|order]] of a [[Definition:Group|group]]
+:$\map \phi n$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Not even sure the statement is correct at this stage, but it seems plausible}}
+\end{proof}<|endoftext|>
+\section{Kernel of Group Homomorphism is not Empty}
+Tags: Kernels of Group Homomorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identity elements]] are $e_G$ and $e_H$ respectively.
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|homomorphism]] from $G$ to $H$.
+Let $\map \ker \phi$ denote the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+Then:
+:$\map \ker \phi \ne \O$
+where $\O$ denotes the [[Definition:Empty Set|empty set]].
+\end{theorem}
+
+\begin{proof}
+From [[Identity is in Kernel of Group Homomorphism]] we have that:
+:$e_G \in \map \ker \phi$
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring of P-adic Norm on Rationals/Corollary 1}
+Tags: Valuation Ring of P-adic Norm on Rationals
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is a [[Definition:Subring|subring]] of $\OO$.
+\end{theorem}
+
+\begin{proof}
+By [[Valuation Ring of P-adic Norm on Rationals]], the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] $\OO$ is the [[Definition:Set|set]]:
+:$\OO = \Z_{\paren p} = \set {\dfrac a b \in \Q : p \nmid b}$
+Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \OO$.
+Hence $\Z \subseteq \OO$.
+By [[Valuation Ring of Non-Archimedean Division Ring is Subring]] then $\OO$ is a [[Definition:Subring|subring]] of $\Q$.
+By [[Integers form Subdomain of Rationals]] then $\Z$ is a [[Definition:Subring|subring]] of $\Q$.
+By [[Intersection of Subrings is Largest Subring Contained in all Subrings]] then $\Z \cap \OO = \Z$ is a [[Definition:Subring|subring]] of $\OO$. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring of P-adic Norm is Subring of P-adic Integers}
+Tags: P-adic Number Theory, Valuation Ring of P-adic Norm is Subring of P-adic Integers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Let $\Z_p$ be the [[Definition:P-adic Integers|$p$-adic integers]].
+Let $\Z_{\ideal p}$ be the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] on $\struct {\Q,\norm {\,\cdot\,}_p}$.
+Then:
+:$(1): \quad \Z_{\ideal p} = \Q \cap \Z_p$.
+:$(2): \quad \Z_{\ideal p}$ is a [[Definition:Subring|subring]] of $\Z_p$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:P-adic Integers|$p$-adic integers]] is defined as:
+:$\Z_p = \set {x \in \Q_p: \norm x_p \le 1}$
+The [[Definition:Valuation Ring Induced by Non-Archimedean Norm|induced valuation ring]] on $\struct {\Q,\norm {\,\cdot\,}_p}$ is defined as:
+:$\Z_{\ideal p} = \set {x \in \Q: \norm x_p \le 1}$
+By definition, the [[Definition:P-adic Norm on P-adic Numbers|$p$-adic norm]] $\norm {\,\cdot\,}_p$ on $\Q_p$ is an [[Definition:Extension of Mapping|extension]] of the [[Definition:P-adic Norm|$p$-adic norm]] $\norm {\,\cdot\,}_p$ on $\Q$.
+It follows that $\Z_{\ideal p} = \Q \cap \Z_p$.
+This proves $(1)$ above.
+By [[Valuation Ring of Non-Archimedean Division Ring is Subring]] then $\Z_p$ is a [[Definition:Subring|subring]] of $\Q_p$.
+By definition of [[Definition:P-adic Integers|$p$-adic integers]] then $\Q$ is a [[Definition:Subring|subring]] of $\Q_p$.
+By [[Intersection of Subrings is Largest Subring Contained in all Subrings]] then $\Z_{\paren p}$ is a [[Definition:Subring|subring]] of $\Z_p$.
+This proves $(2)$ above.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Valuation Ring of P-adic Norm is Subring of P-adic Integers/Corollary 1}
+Tags: Valuation Ring of P-adic Norm is Subring of P-adic Integers
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is a [[Definition:Subring|subring]] of $\Z_p$.
+\end{theorem}
+
+\begin{proof}
+Let $\Z_{\paren p}$ be the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring induced]] by $\norm {\,\cdot\,}_p$ on $\Q$.
+By [[Integers form Subring of Valuation Ring of P-adic Norm on Rationals]] then:
+:$\Z$ is a [[Definition:Subring|subring]] of $\Z_{\paren p}$
+By [[Valuation Ring of P-adic Norm is Subring of P-adic Integers]] then:
+:$\Z_{\paren p}$ is a [[Definition:Subring|subring]] of $\Z_p$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of 1}
+Tags: Laplace Transforms, Laplace Transform of 1
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|function]] defined as:
+:$\forall t \in \R: \map f t = 1$
+Then the [[Definition:Laplace Transform|Laplace transform]] of $\map f t$ is given by:
+:$\laptrans {\map f t} = \dfrac 1 s$
+for $\map \Re s > 0$.
+\end{theorem}<|endoftext|>
+\section{Primitive of Exponential of a x by Hyperbolic Sine of b x}
+Tags: Primitives involving Exponential Function, Primitives involving Hyperbolic Sine Function
+
+\begin{theorem}
+:$\displaystyle \int e^{a x} \sinh b x \rd x = \frac {e^{a x} \paren {a \sinh b x - b \cosh b x} } {a^2 - b^2} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int e^{a x} \sinh b x \rd x
+      | r = \int e^{a x} \paren {\frac {e^{b x} - e^{-b x} } 2} \rd x
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} - e^{-b x} } \rd x
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} - e^{a x} e^{-b x} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} - e^{a x - b x} } \rd x
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x - \frac 1 2 \int e^{a x - b x} \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x - \frac 1 2 \int e^{\paren {a - b} x} \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} - \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} - \frac 1 2 \frac {e^{a x - b x} } {a - b} + C
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} - \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} \paren {a - b} } {\paren {a + b} \paren {a - b} } - \frac 1 2 \frac {e^{a x} e^{-b x} \paren {a + b} } {\paren {a - b} \paren {a + b} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} - e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a + b} \paren {a - b} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} - e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a^2 - b^2} } + C
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \frac {a e^{a x} e^{b x} - b e^{a x} e^{b x} - a e^{a x} e^{-b x} - b e^{a x} e^{-b x} } {2 \paren {a^2 - b^2} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {\frac {a e^{b x} - b e^{b x} - a e^{-b x} - b e^{-b x} } 2} + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^{b x} - e^{-b x} } 2 - b \frac {e^{b x} + e^{-b x} } 2} + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^b x - e^{-b} x} 2 - b \cosh b x} + C
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \sinh b x - b \cosh b x } + C
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac {e^{a x} \paren {a \sinh b x - b \cosh b x} } {a^2 - b^2} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Primitive of Exponential of a x by Hyperbolic Cosine of b x}
+Tags: Primitives involving Exponential Function, Primitives involving Hyperbolic Cosine Function
+
+\begin{theorem}
+:$\displaystyle \int e^{a x} \cosh b x \rd x = \frac {e^{a x} \paren {a \cosh b x + b \sinh b x} } {a^2 - b^2} + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int e^{a x} \cosh b x \rd x
+      | r = \int e^{a x} \paren {\frac {e^{b x} + e^{-b x} } 2} \rd x
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \frac 1 2 \int e^{a x} \paren {e^{b x} + e^{-b x} } \rd x
+      | c = [[Primitive of Constant Multiple of Function]]
+}}
+{{eqn | r = \frac 1 2 \int \paren {e^{a x} e^{b x} + e^{a x} e^{-b x} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int \paren {e^{a x + b x} + e^{a x - b x} } \rd x
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \frac 1 2 \int e^{a x + b x} \rd x + \frac 1 2 \int e^{a x - b x} \rd x
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 1 2 \int e^{\paren {a + b} x} \rd x + \frac 1 2 \int e^{\paren {a - b} x} \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \frac {e^{\paren {a + b} x} } {a + b} + \frac 1 2 \frac {e^{\paren {a - b} x} } {a - b} + C
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x + b x} } {a + b} + \frac 1 2 \frac {e^{a x - b x} } {a - b} + C
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} } {a + b} + \frac 1 2 \frac {e^{a x} e^{-b x} } {a - b} + C
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \frac 1 2 \frac {e^{a x} e^{b x} \paren {a - b} } {\paren {a + b} \paren {a - b} } + \frac 1 2 \frac {e^{a x} e^{-b x} \paren {a + b} } {\paren {a - b} \paren {a + b} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} + e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a + b} \paren {a - b} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} e^{b x} \paren {a - b} + e^{a x} e^{-b x} \paren {a + b} } {2 \paren {a^2 - b^2} } + C
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = \frac {a e^{a x} e^{b x} - b e^{a x} e^{b x} + a e^{a x} e^{-b x} + b e^{a x} e^{-b x} } {2 \paren {a^2 - b^2} } + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {\frac {a e^{b x} - b e^{b x} + a e^{-b x} + b e^{-b x} } 2} + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^{b x} + e^{-b x} } 2 + b \frac {e^{b x} - e^{-b x} } 2} + C
+      | c = 
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \frac {e^b x + e^{-b} x} 2 + b \sinh b x} + C
+      | c = {{Defof|Hyperbolic Sine}}
+}}
+{{eqn | r = \frac {e^{a x} } {\paren {a^2 - b^2} } \paren {a \cosh b x + b \sinh b x } + C
+      | c = {{Defof|Hyperbolic Cosine}}
+}}
+{{eqn | r = \frac {e^{a x} \paren {a \cosh b x + b \sinh b x} } {a^2 - b^2} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform Exists if Function Piecewise Continuous and of Exponential Order}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is:
+:[[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous]] in every [[Definition:Closed Real Interval|closed interval]] $\closedint 0 N$
+:of [[Definition:Exponential Order to Real Index|exponential order $\gamma$]] for $t > N$
+Then the [[Definition:Laplace Transform|Laplace transform]] $\map F s$ of $\map f t$ exists for all $s > \gamma$.
+\end{theorem}
+
+\begin{proof}
+For all $N \in \Z_{>0}$:
+:$\displaystyle \int_0^\infty e^{-s t} \map f t \rd t = \int_0^N e^{-s t} \map f t \rd t + \int_N^\infty e^{-s t} \map f t \rd t$
+We have that $f$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous]] in every [[Definition:Closed Real Interval|closed interval]] $\closedint 0 N$.
+Hence the first of the integrals on the {{RHS}} exists.
+Also, as $\map f t$ is of [[Definition:Exponential Order to Real Index|exponential order $\gamma$]] for $t > N$, so does the second integral on the {{RHS}}.
+Indeed:
+{{begin-eqn}}
+{{eqn | l = \size {\int_N^\infty e^{-s t} \map f t \rd t}
+      | o = \le
+      | r = \int_N^\infty \size {e^{-s t} \map f t} \rd t
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \int_0^\infty e^{-s t} \size {\map f t} \rd t
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \int_0^\infty e^{-s t} M e^{\gamma t} \rd t
+      | c = 
+}}
+{{eqn | r = \dfrac M {s - \gamma}
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{end-eqn}}
+Thus the [[Definition:Laplace Transform|Laplace transform]] exists for $s > \gamma$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Derivative/Discontinuity at t = 0}
+Tags: Laplace Transforms of Derivatives
+
+\begin{theorem}
+Let $f$ fail to be [[Definition:Continuous Mapping|continuous]] at $t = 0$, but let:
+:$\displaystyle \lim_{t \mathop \to 0} \map f t = \map f {0^+}$
+exist.
+Then $\laptrans f$ exists for $\map \Re s > a$, and:
+:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f {0^+}$
+\end{theorem}<|endoftext|>
+\section{Laplace Transform of Derivative/Discontinuity at t = a}
+Tags: Laplace Transforms of Derivatives
+
+\begin{theorem}
+Let $f$ have a [[Definition:Jump Discontinuity|jump discontinuity]] at $t = a$.
+Then:
+:$\laptrans {\map {f'} t} = s \laptrans {\map f t} - \map f 0 - e^{a s} \paren {\map f {a^+} - \map f {a^-} }$
+\end{theorem}<|endoftext|>
+\section{Laplace Transform of Integral}
+Tags: Laplace Transforms
+
+\begin{theorem}
+:$\displaystyle \laptrans {\int_0^t \map f u \rd u} = \dfrac {\map F s} s$
+wherever $\laptrans f$ exists.
+\end{theorem}
+
+\begin{proof}
+Let $\map g t = \displaystyle \int_0^t \map f u \rd u$.
+Then:
+:$\map {g'} t = \map f t$
+and:
+:$\map g 0 = 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {g'} t}
+      | r = s \laptrans {\map g t} - \map g 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = s \laptrans {\map g t}
+      | c = 
+}}
+{{eqn | r = \map F s
+      | c = as $\map F s = \laptrans {\map f t}$
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map g t}
+      | r = \dfrac {\map F s} s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\int_0^t \map f u \rd u}
+      | r = \dfrac {\map F s} s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+:$\displaystyle \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$
+wherever $\displaystyle \lim_{t \mathop \to 0} \dfrac {\map f t} t$ and $\laptrans f$ exist.
+\end{theorem}
+
+\begin{proof}
+Let $\map g t := \dfrac {\map f t} t$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f t
+      | r = t \, \map g t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \laptrans {t \, \map g t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = -\dfrac \d {\d s} \laptrans {\map g t}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map F s
+      | r = -\dfrac {\d G} {\d s}
+      | c = $\map F s := \laptrans {\map f t}$, and so on
+}}
+{{eqn | ll= \leadsto
+      | l = \map G s
+      | r = -\int_{-\infty}^s \map f u \rd u
+      | c =
+}}
+{{eqn | r = \int_s^\infty \map f u \rd u
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping on Subring is Homomorphism}
+Tags: Ring Homomorphisms, Inclusion Mappings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {S, +{\restriction_S}, \circ {\restriction_S}}$ be a [[Definition:Subring|subring]] of $R$.
+Let $i_S: S \to R$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $S$ to $R$.
+Then ${i_S}$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {i_S} x + \map {i_S} y
+      | r = x + y
+      | c = {{Defof|Inclusion Mapping}}
+}}
+{{eqn | r = x \mathbin{ + {\restriction_S} } y
+      | c = as $x, y \in S$
+}}
+{{eqn | r = \map {i_S} {x \mathbin{ + {\restriction_S} } y}
+      | c = as $x \mathbin{ + {\restriction_S} } y \in S$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map {i_S} x \circ \map {i_S} y
+      | r = x \circ y
+      | c = {{Defof|Inclusion Mapping}}
+}}
+{{eqn | r = x \mathbin{\circ {\restriction_S} } y
+      | c = as $x, y \in S$
+}}
+{{eqn | r = \map {i_S} {x \mathbin{ \circ {\restriction_S} } y}
+      | c = as $x \mathbin{ \circ {\restriction_S} } y \in S$
+}}
+{{end-eqn}}
+Hence ${i_S}$ is a [[Definition:Ring Homomorphism|ring homomorphism]] by definition.
+{{qed}}
+[[Category:Ring Homomorphisms]]
+[[Category:Inclusion Mappings]]
+n1i0aoahsj9suxon5donbspmxfj84sr
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping on Subring is Monomorphism}
+Tags: Group Monomorphisms, Inclusion Mappings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {S, +{\restriction_S}, \circ {\restriction_S} }$ be a [[Definition:Subring|subring]] of $R$.
+Let $i_S: S \to R$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $S$ to $R$.
+Then $i_S$ is a [[Definition:Ring Monomorphism|ring monomorphism]].
+\end{theorem}
+
+\begin{proof}
+By [[Inclusion Mapping on Subring is Homomorphism]], $i_S$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+By [[Inclusion Mapping is Injection]], $i_S$ is an [[Definition:Injection|injection]].
+The result follows by definition of [[Definition:Ring Monomorphism|(ring) monomorphism]].
+{{qed}}
+[[Category:Group Monomorphisms]]
+[[Category:Inclusion Mappings]]
+h5ohqc6l5mbb3zoxs5vxcw9butsp61s
+\end{proof}<|endoftext|>
+\section{Negative of Subring is Negative of Ring}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+For each $x \in R$ let $-x$ denote the [[Definition:Ring Negative|ring negative]] of $x$ in $R$.
+Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a [[Definition:Subring|subring]] of $R$.
+For each $x \in S$ let $\mathbin \sim x$ denote the [[Definition:Ring Negative|ring negative]] of $x$ in $S$.
+Then:
+:$\forall x \in S: \mathbin \sim x = -x$
+\end{theorem}
+
+\begin{proof}
+Let $i_S: S \to R$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $S$ to $R$.
+By [[Inclusion Mapping on Subring is Homomorphism]], $i_S$ is a [[Definition:Ring Homomorphism|ring homomorphism]].
+By [[Ring Homomorphism of Addition is Group Homomorphism]], $i_S$ is a [[Definition:Group Homomorphism|group homomorphism]] on [[Definition:Ring Addition|ring addition]] $+$.
+Let $x \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbin \sim x
+      | r = \map {i_S} {\mathbin \sim x}
+      | c = as $\mathbin \sim x \in S$
+}}
+{{eqn | r = -x
+      | c = [[Group Homomorphism Preserves Inverses]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ring Theory]]
+m4kp7cy9zr5f3hq4oksoxoc5bccqsta
+\end{proof}<|endoftext|>
+\section{Subtraction of Subring is Subtraction of Ring}
+Tags: Ring Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be an [[Definition:Ring (Abstract Algebra)|ring]].
+For each $x, y \in R$ let $x - y$ denote the [[Definition:Ring Subtraction|subtraction]] of $x$ and $y$ in $R$.
+Let $\struct {S, + {\restriction_S}, \circ {\restriction_S}}$ be a [[Definition:Subring|subring]] of $R$.
+For each $x, y \in S$ let $x \sim y$ denote the [[Definition:Ring Subtraction|subtraction]] of $x$ and $y$ in $S$.
+Then:
+:$\forall x, y \in S: x \sim y = x - y$
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+Let $-x$ denote the [[Definition:Ring Negative|ring negative]] of $x$ in $R$.
+Let $\mathbin \sim x$ denote the [[Definition:Ring Negative|ring negative]] of $x$ in $S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \sim y
+      | r = x \mathbin {+ {\restriction_S} } \paren {\mathbin \sim y}
+      | c = {{Defof|Ring Subtraction}}
+}}
+{{eqn | r = x + \paren {\mathbin \sim y}
+      | c = {{Defof|Subring|Addition on Subring}}
+}}
+{{eqn | r = x + \paren {-y}
+      | c = [[Negative of Subring is Negative of Ring]]
+}}
+{{eqn | r = x - y
+      | c = {{Defof|Ring Subtraction}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ring Theory]]
+qt1dwlw2zy1nwl6tzgkgzzos92fs2na
+\end{proof}<|endoftext|>
+\section{Limit to Infinity of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\displaystyle \lim_{s \mathop \to \infty} \map F s = 0$
+\end{theorem}<|endoftext|>
+\section{Initial Value Theorem of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\displaystyle \lim_{t \mathop \to 0} \map f t = \lim_{s \mathop \to \infty} s \, \map F s$
+if those limits exist.
+\end{theorem}
+
+\begin{proof}
+We have that $\map {f'} t$ is [[Definition:Piecewise Continuous Function with One-Sided Limits|piecewise continuous with one-sided limits]] and of [[Definition:Exponential Order|exponential order]].
+Hence:
+:$\displaystyle \lim_{s \mathop \to \infty} \int_0^\infty e^{-s t} \map {f'} t \rd t = 0$
+Suppose that $f$ is [[Definition:Continuous Real Function at Point|continuous]] at $t = 0$.
+From [[Laplace Transform of Derivative]]:
+:$(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$
+Taking the [[Definition:Limit of Real Function|limit]] as $s \to \infty$ in $(1)$, where it is assumed that $\map f t$ is [[Definition:Continuous Real Function at Point|continuous]] at $t = 0$:
+:$0 = \displaystyle \lim_{s \mathop \to \infty} s \, \map F s - \map f 0$
+or:
+:$\displaystyle \lim_{s \mathop \to \infty} s \, \map F s = \map f 0 = \lim_{t \mathop \to 0} \map f t$
+{{qed|lemma}}
+Suppose that $f$ is not [[Definition:Continuous Real Function at Point|continuous]] at $t = 0$.
+From [[Laplace Transform of Derivative with Discontinuity at Zero]]:
+:$\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$
+which means:
+:$(2): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$
+Similarly taking the [[Definition:Limit of Real Function|limit]] as $s \to \infty$ in $(2)$, where it is assumed that $\map f t$ is [[Definition:Continuous Real Function at Point|continuous]] at $t = 0$:
+:$0 = \displaystyle \lim_{s \mathop \to \infty} s \, \map F s - \lim_{u \mathop \to 0} \map f u$
+and so:
+:$\displaystyle \lim_{s \mathop \to \infty} s \, \map F s = \lim_{u \mathop \to 0} \map f u = \lim_{t \mathop \to 0} \map f t$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Final Value Theorem of Laplace Transform}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\displaystyle \lim_{t \mathop \to \infty} \map f t = \lim_{s \mathop \to 0} s \, \map F s$
+if those limits exist.
+\end{theorem}
+
+\begin{proof}
+From [[Laplace Transform of Derivative]]:
+:$(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}
+      | r = \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \map {f'} t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^\infty \map {f'} t \rd t
+      | c = 
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \int_0^L \map {f'} t \rd t
+      | c = 
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\map f L - \map f 0}
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | n = 2
+      | r = \lim_{t \mathop \to \infty} \map f t - \map f 0
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}
+      | r = \lim_{s \mathop \to 0} s \, \map F s - \map f 0
+      | c = from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{t \mathop \to \infty} \map f t - \map f 0
+      | r = \lim_{s \mathop \to 0} s \, \map F s - \map f 0
+      | c = from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{t \mathop \to \infty} \map f t
+      | r = \lim_{s \mathop \to 0} s \, \map F s
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Suppose that $f$ is not [[Definition:Continuous Real Function at Point|continuous]] at $t = 0$.
+From [[Laplace Transform of Derivative with Discontinuity at Zero]]:
+:$\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$
+which means:
+:$(3): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}
+      | r = \lim_{u \mathop \to 0} \paren {\lim_{s \mathop \to 0} \int_u^\infty e^{-s t} \map {f'} t \rd t}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \lim_{u \mathop \to 0} \paren {\int_u^\infty \map {f'} t \rd t}
+      | c = 
+}}
+{{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \int_u^L \map {f'} t \rd t}
+      | c = 
+}}
+{{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \paren {\map f L - \map f u} }
+      | c = [[Fundamental Theorem of Calculus]]
+}}
+{{eqn | r = \lim_{u \mathop \to 0} \paren {\lim_{t \mathop \to \infty} \map f t - \map f u}
+      | c = 
+}}
+{{eqn | n = 4
+      | r = \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to 0} \laptrans {\map {f'} t}
+      | r = \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u
+      | c = from $(3)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u
+      | r = \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u
+      | c = from $(4)$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{t \mathop \to \infty} \map f t
+      | r = \lim_{s \mathop \to 0} s \, \map F s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Initial Value Theorem of Laplace Transform/General Result}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $\displaystyle \lim_{t \mathop \to 0} \dfrac {\map f t} {\map g t} = 1$.
+Then:
+:$\displaystyle \lim_{s \mathop \to \infty} \dfrac {\map F s} {\map G s} = 1$
+if those limits exist.
+\end{theorem}<|endoftext|>
+\section{Final Value Theorem of Laplace Transform/General Result}
+Tags: Laplace Transforms
+
+\begin{theorem}
+Let $\displaystyle \lim_{t \mathop \to \infty} \dfrac {\map f t} {\map g t} = 1$.
+Then:
+:$\displaystyle \lim_{s \mathop \to 0} \dfrac {\map F s} {\map G s} = 1$
+if those limits exist.
+\end{theorem}<|endoftext|>
+\section{Evaluation of Integral using Laplace Transform}
+Tags: Laplace Transforms, Integral Calculus
+
+\begin{theorem}
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|real function]] $f$.
+Then:
+:$\displaystyle \int_0^{\to \infty} \map f t \rd t = \map F 0$
+assuming the integral is [[Definition:Convergent Integral|convergent]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Laplace Transform|Laplace transform]]:
+:$\displaystyle \int_0^{\to \infty} e^{-s t} \map f t \rd t = \map F s$
+The result follows by taking the [[Definition:Limit of Real Function|limit]] as $s \to 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area between Smooth Curve and Line is Maximized by Semicircle}
+Tags: Isoperimetrical Problems
+
+\begin{theorem}
+Let $y$ be a [[Definition:Smooth Curve|smooth curve]], embedded in $2$-[[Definition:Dimension of Vector Space|dimensional]] [[Definition:Real Euclidean Space|Euclidean space]].
+Let $y$ have a total [[Definition:Length of Curve|length]] of $l$.
+Let it be contained in the upper [[Definition:Half-Plane|half-plane]] with an exception of [[Definition:Endpoints of Directed Smooth Curve|endpoints]], which are on the [[Definition:X-Axis|$x$-axis]].
+Suppose, $y$, together with a [[Definition:Line Segment|line segment]] connecting $y$'s [[Definition:Endpoints of Directed Smooth Curve|endpoints]], maximizes the enclosed [[Definition:Area|area]].
+Then $y$ is a [[Definition:Semicircle|semicircle]].
+\end{theorem}
+
+\begin{proof}
+By [[Area between Smooth Curve and Line with Fixed Endpoints is Maximized by Arc of Circle]] the maximizing [[Definition:Curve|curve]] is an [[Definition:Arc of Circle|arc of a circle]].
+It is described as follows:
+:If $\dfrac l \pi \le \lambda < \infty$ then:
+::$y = \sqrt {\lambda^2 - x^2} - \sqrt {\lambda^2 - a^2}$
+:where:
+::$l = 2 \lambda \, \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } }$
+:If $\dfrac l {2 \pi} \le \lambda < \dfrac l \pi$ then:
+::$y = \sqrt{\lambda^2 - a^2} - \sqrt{\lambda^2 - x^2}$
+:where:
+::$l = 2 \lambda \paren {\pi - \map \arctan {\dfrac a {\sqrt {\lambda^2 - a^2} } } }$
+In the first case the [[Definition:Area|area]] is a [[Definition:Definite Integral|definite integral]] between $x = -a$ and $x = a$:
+{{begin-eqn}}
+{{eqn | l = A
+      | r = \int_{-a}^a \paren {\sqrt {\lambda^2 - x^2} - \sqrt {\lambda^2 - a^2} } \rd x
+}}
+{{eqn | r = \intlimits {\frac 1 2 \paren {x \sqrt {\lambda^2 - a^2} + \lambda^2 \arctan {\dfrac x {\sqrt {\lambda^2 - x^2} } } } - x \sqrt {\lambda^2 - a^2} } {-a} a
+      | c = [[Primitive of Root of a squared minus x squared]]
+}}
+{{eqn | r = \frac 1 2 \paren {\lambda^2 \arctan \frac a {\sqrt {\lambda^2 - a^2} } - a \sqrt {\lambda^2 - a^2} }
+}}
+{{end-eqn}}
+Solve the [[Definition:Length of Curve|length]] constraint for $a$.
+The solution, satisfying [[Definition:Strictly Positive|strict positivity]] of $l$ and the [[Definition:Domain of Variable|allowed values]] of $\lambda$, is:
+:$a = \dfrac {\lambda \map \tan {\frac l {2 \lambda} } } {\sqrt {1 + \map {\tan^2} {\frac l {2 \lambda} } } }$
+Substitution of this into the expression for the [[Definition:Area|area]] leads to:
+:$A = \dfrac \lambda 2 \paren {l - \lambda \sin \dfrac l \lambda}$
+$l$ is a [[Definition:Constant Mapping|constant]], while $\lambda$ is a [[Definition:Free Variable|free variable]].
+For the allowed values of $\lambda$, the [[Definition:Area|area]] [[Definition:Positive Real Function|function]] is a [[Definition:Continuous Real Function|continuous real function]].
+By [[Definition:Local Maximum|Definition of Local Maximum]], the [[Definition:Maximum Value|maximum]] is either in a [[Definition:Subinterval|subinterval]] of [[Definition:Domain of Variable|domain]] of $A$ {{WRT}} $\lambda$ or at one of the [[Definition:Boundary (Topology)|boundary]] values.
+By [[Derivative at Maximum or Minimum]], we have to find the extremum of $A$:
+:$\dfrac {\d A} {\d \lambda} = \dfrac \lambda 2 \paren {\dfrac l \lambda + \dfrac l \lambda \cos \dfrac l \lambda - 2 \sin \dfrac l \lambda}$
+This vanishes at $\dfrac l \lambda = \paren {2 k + 1} \pi$ and $\dfrac l \lambda = 0$, where $k \in \Z$.
+Then the [[Definition:Area|area]] at these extremums acquires the following values:
+:$\map A {\dfrac l \lambda = 0} = 0$;
+:$\map A {\dfrac l \lambda = \paren {2 k + 1} \pi} = \dfrac {l^2} {\paren {1 + 2 k} 2 \pi}$
+This is maximized for $k = 0$, or $\dfrac l \lambda = \pi$.
+Incidentally, these are also [[Definition:Boundary (Topology)|boundary]] values of the [[Definition:Area|area]] [[Definition:Positive Real Function|function]].
+This concludes the maximization of [[Definition:Area|area]] when the [[Definition:Arc of Curve|arc]] is being varied from a [[Definition:Straight Line|straight line]] to a [[Definition:Semicircle|semicirle]].
+The second part considers the variation of the [[Definition:Curve|curve]] from a [[Definition:Semicircle|semicirle]] to a [[Definition:Circle|circle]].
+In this case the [[Definition:Area|area]] is that of a [[Definition:Circle/Semicircle|semicirle]] plus a part of the lower [[Definition:Semicircle|semicirle]]:
+{{begin-eqn}}
+{{eqn | l = A 
+      | r = \pi \frac {\lambda^2} 2 + \int_{-\lambda}^{-a} \sqbrk {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x + \int_{-a}^a \sqbrk {\sqrt {\lambda^2 - a^2} } \rd x + \int_a^\lambda \sqbrk {\sqrt {\lambda^2 - a^2} - \paren {\sqrt {\lambda^2 - a^2} - \sqrt {\lambda^2 - x^2} } } \rd x
+}} 
+{{eqn | r = \pi \lambda^2 + a \sqrt {\lambda^2 - a^2} - \lambda^2 \arctan \paren {\frac a {\sqrt {\lambda^2 - a^2} } }
+      | c = [[Primitive of Root of a squared minus x squared]]
+}}
+{{end-eqn}}
+Like in the previous case, solve the [[Definition:Length of Curve|length]] constraint for $a$, while satisfying positivity and [[Definition:Domain of Variable|range]] conditions:
+:$a = \dfrac {\tan {\frac {2 \pi \lambda - l} {2 \lambda} } } {\sqrt {1 + \tan^2 {\frac {2 \pi \lambda - l} {2 \lambda} } } }$
+Substitution into the [[Definition:Area|area]] expression leads to:
+:$A = \dfrac {\lambda} 2 \paren {l - \lambda \sin \dfrac l \lambda}$
+To find the extremum, compute its [[Definition:Derivative of Real Function|derivative]] {{WRT}} $\lambda$:
+:$\dfrac {\d A} {\d \lambda} = \dfrac 1 2 \paren {l + l \cos \dfrac l \lambda - 2 \lambda \sin \dfrac l \lambda}$
+It vanishes if $\dfrac l \lambda = 0$ or $\dfrac l \lambda = \pi \paren {1 + 2 k}$, with $k \in \Z$.
+From these solutions the one satisfying the [[Definition:Domain of Variable|range]] of $\lambda$ is $\dfrac l \lambda = \pi$.
+The [[Definition:Area|area]] for this value is $\dfrac {\pi \lambda^2} 2$.
+For completeness we have to check the other [[Definition:Boundary (Topology)|boundary]] value in this [[Definition:Domain of Variable|range]], namely, $\lambda = \dfrac l {2 \pi}$.
+:$\map A {\lambda = \dfrac l {2\pi} } = \pi \lambda^2$.
+Since we have [[Definition:Length of Curve|length]] as an input for this problem, express both [[Definition:Area|areas]] in terms of [[Definition:Length of Curve|length]].
+:$\map A {l = \pi \lambda} = \dfrac {l^2} {2 \pi}$
+:$\map A {l = 2 \pi \lambda} = \dfrac {l^2} {4 \pi}$
+Hence, the [[Definition:Area|area]] is maximized when the [[Definition:Curve|curve]] $y$ is a [[Definition:Circle/Semicircle|semicirle]].
+{{qed}}
+[[Category:Isoperimetrical Problems]]
+nxo4guskqb1siiysuw57hgyo75ow037
+\end{proof}<|endoftext|>
+\section{Valuation Ring of Non-Archimedean Division Ring is Clopen}
+Tags: Normed Division Rings, Valuation Ring of Non-Archimedean Division Ring is Clopen
+
+\begin{theorem}
+Let $\struct {R, \norm {\,\cdot\,} }$ be a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean normed division ring]] with [[Definition:Ring Zero|zero]] $0_R$.
+Let $\mathcal O$ be [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring induced]] by $\norm{\,\cdot\,}$.
+Then $\mathcal O$ is a 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}
+The [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring]] $\mathcal O$ Is the [[Definition:Open Ball|open ball]] ${B_1}^- \paren {0_R}$ by definition.
+By [[Topological Properties of Non-Archimedean Division Rings/Open Balls are Clopen|Open Balls of Non-Archimedean Division Rings are Clopen]] then $\mathcal O$ 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 $\norm {\,\cdot\,}$.
+{{qed}}
+[[Category:Normed Division Rings]]
+[[Category:Valuation Ring of Non-Archimedean Division Ring is Clopen]]
+3mq7moxsruz5gxr2j57b2c8kntv45zv
+\end{proof}<|endoftext|>
+\section{Valuation Ring of Non-Archimedean Division Ring is Clopen/Corollary 1}
+Tags: Valuation Ring of Non-Archimedean Division Ring is Clopen
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:P-adic Number|$p$-adic numbers]].
+Then the [[Definition:P-adic Integers|$p$-adic integers]] $\Z_p$ is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in the [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:P-adic Integers|$p$-adic integers]] $\Z_p$ is the [[Definition:Valuation Ring Induced by Non-Archimedean Norm|valuation ring induced]] by $\norm {\,\cdot\,}_p$ by definition.
+By [[Valuation Ring of Non-Archimedean Division Ring is Clopen|Valuation Ring of Non-Archimedean Division Ring is Clopen]] then the [[Definition:P-adic Integers|$p$-adic integers]] $\Z_p$ is both [[Definition:Open Set of Metric Space|open]] and [[Definition:Closed Set of Metric Space|closed]] in the [[Definition:P-adic Metric on P-adic Numbers|$p$-adic metric]].
+{{qed}}
+[[Category:Valuation Ring of Non-Archimedean Division Ring is Clopen]]
+qhjehmabhkaqu8whhsdtkl33zp63q8w
+\end{proof}<|endoftext|>
+\section{Bessel Function of the First Kind of Negative Integer Order}
+Tags: Bessel Functions
+
+\begin{theorem}
+:$\map {J_{-n} } x = \paren {-1}^n \map {J_n} x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {J_{-n} } x 
+      | r = \dfrac 1 \pi \int_0^\pi \map \cos {-n \theta - x \sin \theta} \rd \theta
+      | c = [[Integral Representation of Bessel Function of the First Kind/Integer Order]]
+}}
+{{eqn | r = \dfrac 1 \pi \int_0^\pi \map \cos {-n \paren {\pi - \theta} - x \sin \paren {\pi - \theta} } \rd \paren {\pi - \theta}
+      | c = substitution of $\pi - \theta$
+}}
+{{eqn | r = -\dfrac 1 \pi \int_\pi^0 \map \cos {n \theta - x \sin \theta - n \pi} \rd \theta
+      | c = [[Sine of Supplementary Angle]]
+}}
+{{eqn | r = \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta - n \pi} \rd \theta
+      | c = [[Reversal of Limits of Definite Integral]]
+}}
+{{eqn | r = \paren {-1}^{-n} \dfrac 1 \pi \int_0^\pi \map \cos {n \theta - x \sin \theta} \rd \theta
+      | c = [[Cosine of Angle plus Integer Multiple of Pi]]
+}}
+{{eqn | r = \paren {-1}^n \map {J_n} x
+      | c = [[Integral Representation of Bessel Function of the First Kind/Integer Order]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Series Expansion of Bessel Function of the First Kind}
+Tags: Bessel Functions
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \map {J_n} x
+      | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+We employ Frobenius's Method to find the solutions to the [[Definition:Bessel's Equation|Bessel's Equation]]:
+:$x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y = 0$
+for $n \ge 0$, in the form:
+:$\displaystyle \map y x = \sum_{k \mathop = 0}^\infty A_k x^{k + r}$
+defined on $x > 0$, for some constants $r, A_i$, with $A_0 \neq 0$, which are to be determined.
+Differentiating the expression [[Definition:Derivative/Real Function/With Respect To|with respect to]] $x$:
+{{begin-eqn}}
+{{eqn | l = \map {y'} x
+      | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1}
+}}
+{{eqn | l = \map {y''} x
+      | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2}
+}}
+{{end-eqn}}
+Substituting $y, y', y''$ into [[Definition:Bessel's Equation|Bessel's Equation]]:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y
+}}
+{{eqn | r = x^2 \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2} + x \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1} + \paren {x^2 - n^2} \sum_{k \mathop = 0}^\infty A_k x^{k + r}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2} - n^2 \sum_{k \mathop = 0}^\infty A_k x^{k + r}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r} \paren {k + r - 1} + \paren {k + r} - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 2}^\infty A_{k - 2} x^{k + r}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^k + \sum_{k \mathop = 2}^\infty A_{k - 2} x^k
+      | c = $x^r \neq 0$
+}}
+{{end-eqn}}
+Comparing the constant term on both sides:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = A_0 \paren {r^2 - n^2}
+}}
+{{eqn | ll = \leadsto
+      | l = r
+      | r = \pm n
+}}
+{{end-eqn}}
+Take $r = n$. Comparing the rest of the [[Definition:Coefficient of Polynomial|coefficients]]:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = A_1 \paren {\paren{n + 1}^2 - n^2}
+}}
+{{eqn | l = 0
+      | r = A_1 \paren {2 n + 1}
+}}
+{{eqn | ll = \leadsto
+      | l = A_1
+      | r = 0
+}}
+{{eqn | l = 0
+      | r = A_k \paren {\paren{n + k}^2 - n^2} + A_{k - 2}
+      | c = for $k \ge 2$
+}}
+{{eqn | l = 0
+      | r = A_k k \paren {2 n + k} + A_{k - 2}
+}}
+{{eqn | ll = \leadsto
+      | l = A_k
+      | r = - \dfrac 1 {k \paren {2 n + k} } A_{k - 2}
+}}
+{{end-eqn}}
+From the recurrence relation above, we see that $A_k = 0$ for odd $k$, and:
+{{begin-eqn}}
+{{eqn | l = A_{2 k}
+      | r = - \dfrac 1 {2 k \paren {2 n + 2 k} } A_{2 k - 2}
+}}
+{{eqn | r = \dfrac {- 1} {2^2 k \paren {n + k} } A_{2 k - 2}
+}}
+{{eqn | r = \dfrac {\paren {- 1}^2} {2^4 k \paren {k - 1} \paren {n + k}  \paren {n + k - 1} } A_{2 k - 4}
+}}
+{{eqn | r = \cdots
+}}
+{{eqn | r = \dfrac {\paren {- 1}^k} {2^{2 k} k \paren {k - 1} \cdots \paren 1 \paren {n + k} \paren {n + k - 1} \cdots \paren {n + 1} } A_0
+}}
+{{eqn | r = \dfrac {\paren {- 1}^k} {2^{2 k} k! \dfrac {\map \Gamma {n + k + 1} } {\map \Gamma {n + 1} } } A_0
+}}
+{{eqn | r = \dfrac {\paren {- 1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} }
+      | c = By choice of $A_0 = \dfrac {2^{- n} } {\map \Gamma {n + 1} }$
+}}
+{{end-eqn}}
+Substituting this result to our original equation: 
+{{begin-eqn}}
+{{eqn | l = \map y x
+      | r = \sum_{k \mathop = 0}^\infty A_k x^{k + n}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty A_{2 k} x^{2 k + n}
+      | c = since all even terms vanish
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {- 1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} } x^{2 k + n}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {- 1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{2 k + n}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Recurrence Formula for Bessel Function of the First Kind}
+Tags: Bessel Functions
+
+\begin{theorem}
+Let $\map {J_n} x$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order $n$]].
+Then:
+:$\map {J_{n + 1} } x = \dfrac {2 n} x \map {J_n} x - \map {J_{n - 1} } x$
+And: 
+:$\map {J_{n + 1} } x = -2 \map {J_n'} x + \map {J_{n - 1} } x$
+{{refactor|page per result}}
+\end{theorem}
+
+\begin{proof}
+From [[Generating Function for Bessel Function of the First Kind of Order n of x]] we have:
+:$\displaystyle \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} } = \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n$
+Differentiating both sides of the equation with respect to $t$:
+{{begin-eqn}}
+{{eqn | l = \dfrac x 2 \paren {1 + \dfrac 1 {t^2} } \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} }
+      | r = \sum_{m \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac x 2 \paren {1 + \dfrac 1 {t^2} } \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n
+      | r = \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}
+      | c = [[Generating Function for Bessel Function of the First Kind of Order n of x]]
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac x 2 \paren {\sum_{n \mathop = - \infty}^\infty \map {J_n} x t^n + \sum_{n \mathop = - \infty}^\infty \map {J_n} x t^{n - 2} }
+      | r = \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}
+      | c = 
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac x 2 \paren {\sum_{n \mathop = - \infty}^\infty \map {J_{n - 1} } x t^{n - 1} + \sum_{n \mathop = - \infty}^\infty \map {J_{n + 1} } x t^{n - 1} }
+      | r = \sum_{n \mathop = - \infty}^\infty n \map {J_n} x t^{n - 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac x 2 \paren {\map {J_{n - 1} } x + \map {J_{n + 1} } x}
+      | r = n \map {J_n} x
+      | c = by comparing coefficients
+}}
+{{eqn | ll = \leadsto
+      | l = \map {J_{n - 1} } x + \map {J_{n + 1} } x
+      | r = \dfrac {2n} x \map {J_n} x
+}}
+{{eqn | ll = \leadsto
+      | l = \map {J_{n + 1} } x
+      | r = \dfrac {2n} x \map {J_n} x - \map {J_{n - 1} } x
+}}
+{{end-eqn}}
+This is the first recurrence formula.
+{{qed|lemma}}
+We prove the second recurrence formula by differentiating both sides of the original equation with respect to $x$:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 \paren {t - \dfrac 1 t} \map \exp {\dfrac x 2 \paren {t - \dfrac 1 t} }
+      | r = \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac 1 2 \paren {t - \dfrac 1 t} \sum_{m \mathop = - \infty}^\infty \map {J_n} x t^n
+      | r = \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac 1 2 \paren {\sum_{m \mathop = - \infty}^\infty \map {J_n} x t^{n + 1} - \sum_{m \mathop = - \infty}^\infty \map {J_n} x t^{n - 1} } 
+      | r = \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac 1 2 \paren {\sum_{m \mathop = - \infty}^\infty \map {J_{n - 1} } x t^n - \sum_{m \mathop = - \infty}^\infty \map {J_{n + 1} } x t^n} 
+      | r = \sum_{m \mathop = - \infty}^\infty \map {J_n'} x t^n
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac 1 2 \paren {\map {J_{n - 1} } x - \map {J_{n + 1} } x}
+      | r = \map {J_n'} x
+      | c = by comparing coefficients
+}}
+{{eqn | ll = \leadsto
+      | l = \map {J_{n - 1} } x - \map {J_{n + 1} } x
+      | r = 2 \map {J_n'} x
+}}
+{{eqn | ll = \leadsto
+      | l = \map {J_{n + 1} } x
+      | r = - 2 \map {J_n'} x + \map {J_{n - 1} } x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of x^n by Bessel Function of the First Kind of Order n of x}
+Tags: Bessel Functions
+
+\begin{theorem}
+Let $\map {J_n} x$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order $n$]].
+Then:
+:$\map {\dfrac \d {\d x} } {x^n \map {J_n} x} = x^n \map {J_{n - 1} } x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\frac \d {\d x} } {x^n \map {J_n} x}
+      | r = n x^{n - 1} \map {J_n} x + x^n \map {J_n'} x
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\map {J_{n + 1} } x - \map {J_{n - 1} } x} 2}
+      | c = [[Recurrence Formula for Bessel Function of the First Kind]]
+}}
+{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\frac {2 n} x \map {J_n} x - 2 \map {J_{n - 1} } x} 2}
+      | c = [[Recurrence Formula for Bessel Function of the First Kind]]
+}}
+{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac n x \map {J_n} x - \map {J_{n - 1} } x}
+}}
+{{eqn | r = n x^{n - 1} \map {J_n} x - n x^{n - 1} \map {J_n} x + x^n \map {J_{n - 1} } x
+}}
+{{eqn | r = x^n \map {J_{n - 1} } x
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generating Function for Bessel Function of the First Kind of Order n of x}
+Tags: Bessel Functions
+
+\begin{theorem}
+Let $\map {J_n} x$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order $n$]].
+Then:
+:$\map \exp {\dfrac {x \paren {t - \frac 1 t} } 2} = \displaystyle \sum_{n \mathop = -\infty}^\infty \map {J_n} x t^n$
+\end{theorem}<|endoftext|>
+\section{Series Expansion of Bessel Function of the First Kind/Negative Index}
+Tags: Bessel Functions
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \map {J_{-n} } x
+      | r = \dfrac {x^{-n} } {2^{-n} \, \map \Gamma {1 - n} } \paren {1 - \dfrac {x^2} {2 \paren {2 - 2 n} } + \dfrac {x^4} {2 \times 4 \paren {2 - 2 n} \paren {4 - 2 n} } - \cdots}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1 - n} } \paren {\dfrac x 2}^{2 k - n}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Series Expansion of Bessel Function of the First Kind]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map {J_n} x
+      | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}
+      | c = 
+}}
+{{end-eqn}}
+The result follows by substituting $-n$ for $n$ in $1$ and simplifying.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bessel Function of the First Kind for Imaginary Argument}
+Tags: Bessel Functions
+
+\begin{theorem}
+Let $\map {J_n} x$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order $n$]].
+Then:
+:$\map {J_n } {i x} = i^{-n} \, \map {I_n} x$
+where:
+:$i$ denotes the [[Definition:Imaginary Unit|imaginary unit]]
+:$\map {I_n} x$ denotes the [[Definition:Modified Bessel Function of the First Kind|modified Bessel function of the first kind]] of [[Definition:Order of Modified Bessel Function|order $n$]].
+\end{theorem}<|endoftext|>
+\section{Integral to Infinity of Dirac Delta Function}
+Tags: Dirac Delta Function
+
+\begin{theorem}
+Let $\map \delta x$ denote the [[Definition:Dirac Delta Function|Dirac delta function]].
+Then:
+:$\displaystyle \int_0^{+ \infty} \map \delta x \rd x = 1$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\map \delta x = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
+where:
+:$\map {F_\epsilon} x = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end{cases}$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \rd x
+      | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \int_\epsilon^\infty 0 \rd x
+      | c = Definition of $\map {F_\epsilon} x$
+}}
+{{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \dfrac 1 \epsilon \paren {\epsilon - 0} + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
+      | c = [[Definite Integral of Constant]]
+}}
+{{eqn | r = 1 + \lim_{L \mathop \to \infty} 0
+      | c = simplification
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \int_0^{+ \infty} \map \delta x \rd x
+      | r = \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \rd x
+      | c = {{Defof|Dirac Delta Function}}
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} 1
+      | c = from above
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral to Infinity of Dirac Delta Function by Continuous Function}
+Tags: Dirac Delta Function
+
+\begin{theorem}
+Let $\map \delta x$ denote the [[Definition:Dirac Delta Function|Dirac delta function]].
+Let $g$ be a [[Definition:Continuous Real Function|continuous real function]].
+Then:
+:$\displaystyle \int_0^{+ \infty} \map \delta x \, \map g x \rd x = \map g 0$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\map \delta x = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
+where:
+:$\map {F_\epsilon} x = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end{cases}$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x
+      | r = \int_0^\epsilon \dfrac 1 \epsilon \, \map g x \rd x + \int_\epsilon^\infty 0 \times \map g x \rd x
+      | c = Definition of $\map {F_\epsilon} x$
+}}
+{{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \, \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \times \map g x \rd x
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
+      | c = 
+}}
+{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
+      | c = [[Definite Integral of Constant]]
+}}
+{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} 0
+      | c = simplification
+}}
+{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x
+      | c = 
+}}
+{{end-eqn}}
+From [[Upper and Lower Bounds of Integral]]:
+:$\displaystyle m \paren {\epsilon - 0} \le \int_0^\epsilon \map g x \rd x \le M \paren {\epsilon - 0}$
+where:
+:$M$ is the [[Definition:Maximal Element|maximum]] of $\map g x$
+:$m$ is the [[Definition:Minimal Element|minimum]] of $\map g x$
+on $\closedint 0 \epsilon$.
+Hence:
+:$\displaystyle m \epsilon \le \int_0^\epsilon \map g x \rd x \le M \epsilon$
+and so dividing by $\epsilon$:
+:$\displaystyle m \le \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x \le M$
+Then:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} M = m = \map g 0$
+and so by the [[Squeeze Theorem]]:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \map g 0$
+But by definition of the [[Definition:Dirac Delta Function|Dirac delta function]]:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \displaystyle \int_0^{+ \infty} \map \delta x \, \map g x \rd x$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral to Infinity of Shifted Dirac Delta Function by Continuous Function}
+Tags: Dirac Delta Function
+
+\begin{theorem}
+Let $\map \delta x$ denote the [[Definition:Dirac Delta Function|Dirac delta function]].
+Let $g$ be a [[Definition:Continuous Real Function|continuous real function]].
+Let $a \in \R_{\ge 0}$ be a [[Definition:Positive Real Number|positive real number]].
+Then:
+:$\displaystyle \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x = \map g a$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\map \delta {x - a} = \displaystyle \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
+where:
+:$\map {F_\epsilon} x = \begin{cases} 0 & : x < a \\ \dfrac 1 \epsilon & : a \le x \le a + \epsilon \\ 0 & : x > a + \epsilon \end{cases}$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x
+      | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \, \map g x \rd x + \int_{a + \epsilon}^\infty 0 \times \map g x \rd x
+      | c = Definition of $\map {F_\epsilon} x$
+}}
+{{eqn | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \, \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \times \map g x \rd x
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \int_0^a 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \rd x
+      | c = 
+}}
+{{eqn | r = \paren {0 \times \paren {a - 0} } + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \paren {a + \epsilon} } }
+      | c = [[Definite Integral of Constant]]
+}}
+{{eqn | r = 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} 0
+      | c = simplification
+}}
+{{eqn | r = \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x
+      | c = 
+}}
+{{end-eqn}}
+From [[Upper and Lower Bounds of Integral]]:
+:$\displaystyle m \paren {\paren {a + \epsilon} - a} \le \int_a^{a + \epsilon} \map g x \rd x \le M \paren {\paren {a + \epsilon} - 0}$
+where:
+:$M$ is the [[Definition:Maximal Element|maximum]] of $\map g x$
+:$m$ is the [[Definition:Minimal Element|minimum]] of $\map g x$
+on $\closedint a {a + \epsilon}$.
+Hence:
+:$\displaystyle m \epsilon \le \int_a^{a + \epsilon} \map g x \rd x \le M \epsilon$
+and so dividing by $\epsilon$:
+:$\displaystyle m \le \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x \le M$
+Then:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} M = m = \map g a$
+and so by the [[Squeeze Theorem]]:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \map g a$
+But by definition of the [[Definition:Dirac Delta Function|Dirac delta function]]:
+:$\displaystyle \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \, \map g x \rd x = \displaystyle \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x$
+Hence the result.
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Function which is Zero except on Countable Set of Points is Null}
+Tags: Null Functions
+
+\begin{theorem}
+Let $S \subseteq \R$ be a [[Definition:Subset|subset]] of $\R$ such that $S$ is [[Definition:Countable Set|countable]], either [[Definition:Finite Set|finite]] or [[Definition:Countably Infinite|countably infinite]].
+Let $f: \R \to \R$ be a [[Definition:Real Function|real function]] such that:
+:$\forall x \in \R \setminus S: \map f x = 0$
+That is, except perhaps for the [[Definition:Element|elements]] of $S$, the [[Definition:Image of Element under Mapping|value]] of $f$ is [[Definition:Zero (Number)|zero]].
+Then $f$ is a [[Definition:Null Function|null function]].
+\end{theorem}
+
+\begin{proof}
+This is an instance of [[Integrable Function Zero A.E. iff Absolute Value has Zero Integral]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Bessel Function of the First Kind of Order Zero}
+Tags: Laplace Transforms of Bessel Functions, Laplace Transform of Bessel Function of the First Kind of Order Zero
+
+\begin{theorem}
+Let $J_0$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order]] $0$.
+Then the [[Definition:Laplace Transform|Laplace transform]] of $J_0$ is given as:
+:$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
+\end{theorem}
+
+\begin{proof}
+From [[Bessel Function of the First Kind of Order Zero]]:
+{{begin-eqn}}
+{{eqn | l = \map {J_0} t
+      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
+      | c = 
+}}
+{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {J_0} t}
+      | r = \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
+      | c = {{Defof|Binomial Coefficient}}: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {\sqrt {s^2 + 1} }
+      | r = \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}
+      | c = [[Binomial Coefficient of Minus Half]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }
+      | c = rearranging, and bringing $\dfrac 1 s$ inside the [[Definition:Summation|summation]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
+      | c = further rearrangement
+}}
+{{end-eqn}}
+The two expressions match, and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Bessel Function of the First Kind of Order Zero]]:
+{{begin-eqn}}
+{{eqn | l = \map {J_0} t
+      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
+      | c = 
+}}
+{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {J_0} t}
+      | r = \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {s^2 + 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]], $\map {J_0} t$ satisfies [[Definition:Bessel's Equation|Bessel's equation]]:
+{{begin-eqn}}
+{{eqn | l = t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+From [[Laplace Transform of Derivative]]:
+:$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$
+and from [[Laplace Transform of Second Derivative]]:
+:$\laptrans {\map {J_0''} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$
+From [[Bessel Function of the First Kind of Order Zero|Bessel Function of the First Kind of Order $0$]]:
+:$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
+from which it follows immediately that:
+:$\map {J_0} 0 = 1$
+From [[Derivative of Bessel Function of the First Kind of Order 0|Derivative of Bessel Function of the First Kind of Order $0$]] we have:
+{{ProofWanted|Derivative of Bessel Function of the First Kind of Order 0}}
+from which it follows that:
+:$\map {J'_0} 0 = 0$
+Then from [[Derivative of Laplace Transform]]:
+{{begin-eqn}}
+{{eqn | l = \laptrans {t \, \map {J_0''} t}
+      | r = -\map {\dfrac \d {\d s} } {\map {J_0''} t}
+      | c = 
+}}
+{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}
+      | c = 
+}}
+{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}
+      | c = $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$
+}}
+{{end-eqn}}
+and:
+:$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
+Thus by taking the [[Definition:Laplace Transform|Laplace transform]] of $(1)$ we have:
+Let $y = \laptrans {\map {J_0} t}$.
+{{begin-eqn}}
+{{eqn | l = \laptrans {t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t}
+      | r = \laptrans 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}
+      | r = 0
+      | c = setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y} {\d s}
+      | r = -\dfrac {s y} {s^2 + 1}
+      | c = rearranging and simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \int \dfrac {\d y} y
+      | r = -\int \dfrac {s \rd s} {s^2 + 1}
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \dfrac c {\sqrt {s^2 + 1} }
+      | c = for some [[Definition:Constant|constant]] $c$
+}}
+{{end-eqn}}
+Now we have that:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to \infty} s \, \map y s
+      | r = \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }
+      | c = 
+}}
+{{eqn | r = c
+      | c = 
+}}
+{{end-eqn}}
+and:
+:$\displaystyle \lim_{t \mathop \to 0} \map {J_0} t = 1$
+Thus by the [[Initial Value Theorem of Laplace Transform]]:
+:$c = 1$
+and so:
+:$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Bessel Function of the First Kind}
+Tags: Laplace Transforms of Bessel Functions, Laplace Transform of Bessel Function of the First Kind
+
+\begin{theorem}
+Let $J_n$ denote the [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]] of [[Definition:Order of Bessel Function|order]] $n$.
+Then the [[Definition:Laplace Transform|Laplace transform]] of $J_n$ is given as:
+:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$
+\end{theorem}
+
+\begin{proof}
+From [[Bessel Function of the First Kind of Order Zero]]:
+{{begin-eqn}}
+{{eqn | l = \map {J_0} t
+      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
+      | c = 
+}}
+{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {J_0} t}
+      | r = \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
+      | c = {{Defof|Binomial Coefficient}}: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {\sqrt {s^2 + 1} }
+      | r = \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}
+      | c = [[Binomial Coefficient of Minus Half]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }
+      | c = rearranging, and bringing $\dfrac 1 s$ inside the [[Definition:Summation|summation]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }
+      | c = further rearrangement
+}}
+{{end-eqn}}
+The two expressions match, and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Bessel Function of the First Kind of Order Zero]]:
+{{begin-eqn}}
+{{eqn | l = \map {J_0} t
+      | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}
+      | c = 
+}}
+{{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {J_0} t}
+      | r = \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 {\sqrt {s^2 + 1} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Bessel Function of the First Kind|Bessel function of the first kind]], $\map {J_0} t$ satisfies [[Definition:Bessel's Equation|Bessel's equation]]:
+{{begin-eqn}}
+{{eqn | l = t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+From [[Laplace Transform of Derivative]]:
+:$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$
+and from [[Laplace Transform of Second Derivative]]:
+:$\laptrans {\map {J_0''} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$
+From [[Bessel Function of the First Kind of Order Zero|Bessel Function of the First Kind of Order $0$]]:
+:$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$
+from which it follows immediately that:
+:$\map {J_0} 0 = 1$
+From [[Derivative of Bessel Function of the First Kind of Order 0|Derivative of Bessel Function of the First Kind of Order $0$]] we have:
+{{ProofWanted|Derivative of Bessel Function of the First Kind of Order 0}}
+from which it follows that:
+:$\map {J'_0} 0 = 0$
+Then from [[Derivative of Laplace Transform]]:
+{{begin-eqn}}
+{{eqn | l = \laptrans {t \, \map {J_0''} t}
+      | r = -\map {\dfrac \d {\d s} } {\map {J_0''} t}
+      | c = 
+}}
+{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}
+      | c = 
+}}
+{{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}
+      | c = $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$
+}}
+{{end-eqn}}
+and:
+:$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$
+Thus by taking the [[Definition:Laplace Transform|Laplace transform]] of $(1)$ we have:
+Let $y = \laptrans {\map {J_0} t}$.
+{{begin-eqn}}
+{{eqn | l = \laptrans {t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t}
+      | r = \laptrans 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}
+      | r = 0
+      | c = setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y} {\d s}
+      | r = -\dfrac {s y} {s^2 + 1}
+      | c = rearranging and simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \int \dfrac {\d y} y
+      | r = -\int \dfrac {s \rd s} {s^2 + 1}
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \dfrac c {\sqrt {s^2 + 1} }
+      | c = for some [[Definition:Constant|constant]] $c$
+}}
+{{end-eqn}}
+Now we have that:
+{{begin-eqn}}
+{{eqn | l = \lim_{s \mathop \to \infty} s \, \map y s
+      | r = \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }
+      | c = 
+}}
+{{eqn | r = c
+      | c = 
+}}
+{{end-eqn}}
+and:
+:$\displaystyle \lim_{t \mathop \to 0} \map {J_0} t = 1$
+Thus by the [[Initial Value Theorem of Laplace Transform]]:
+:$c = 1$
+and so:
+:$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + (t^2-\alpha^2)x} s 
+      | r = 0 
+      | c = [[Definition:Laplace Transform|Laplace Transform]] of [[Definition:Bessel's Equation|Bessel's Equation]]
+}}
+{{eqn | l = \frac {\d^2} {\d s^2}\mathcal{L}[x''] - \frac \d {\d s} \mathcal{L}[x'] + \frac {\d^2} {\d s^2} \mathcal{L}[x] - \alpha^2 \mathcal{L} [x]
+      | r = 0
+      | c = setting the initial conditions as $x(0)=1,\ x'(0)=0$
+}}
+{{eqn | l = (s^2+1) \mathcal{L}''[x] + 3 s \mathcal {L}'[x] + (1 - \alpha^2) \mathcal{L}[x]
+      | r = 0
+      | c = Make the following change of variable $u = \sqrt {s^2 + 1} \mathcal {L} [x]$
+}}
+{{eqn | l = u'' \sqrt {s^2 + 1} + \dfrac s {\sqrt {s^2 + 1} } u'
+      | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }
+      | c = 
+}}
+{{eqn | l = (u' \sqrt {s^2 + 1} )'
+      | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }
+      | c = multiplying both sides by $u' \sqrt {s^2 + 1}$
+}}
+{{eqn | l = \frac 1 2 ( (u' \sqrt {s^2 + 1} )^2)'
+      | r = \frac 1 2 \alpha^2 (u^2)'
+      | c = 
+}}
+{{eqn | l = u' \sqrt {s^2 + 1}
+      | r = -\alpha u
+      | c = [[Definition:Arbitrary Constant|Constant of Integration]] removed by the [[Final Value Theorem of Laplace Transform]]
+}}
+{{eqn | l = \int \frac 1 u \rd u
+      | r = \int -\frac \alpha {\sqrt {s^2 + 1} } \rd s
+      | c = 
+}}
+{{eqn | l = \map \ln u
+      | r = \alpha \ln (\sqrt {s^2 + 1} - s)
+      | c = 
+}}
+{{eqn | l = u
+      | r = (\sqrt {s^2 + 1} - s)^\alpha
+      | c = reverting the substitution
+}}
+{{eqn | l = \mathcal{L} [x]
+      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \mathcal{L}[J_\alpha (t)](s)
+      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Sine of Root}
+Tags: Laplace Transforms, Sine Function, Laplace Transform of Sine of Root
+
+\begin{theorem}
+:$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
+where $\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sin \sqrt t
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1}!}
+      | c = {{Defof|Real Sine Function}}
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}!} t^{n + \frac 1 2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\sin \sqrt t}
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1}! s^{n + \frac 3 2} }
+      | c = [[Laplace Transform of Power]], [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1}! s^n}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n + 1} \paren {2 n}!} {2^{2 n} n! \paren {2 n + 1}! s^n}
+      | c = [[Gamma Function of Positive Half-Integer]]
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2^{2 n} n! s^n}
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {-\frac 1 {2^2 s} }^n
+      | c = 
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } e^{-1/\paren {2^2 s} }
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}}
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Cosine of Root over Root}
+Tags: Laplace Transforms, Cosine Function, Laplace Transform of Cosine of Root over Root
+
+\begin{theorem}
+:$\laptrans {\dfrac {\cos \sqrt t} {\sqrt t} } = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }$
+where $\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$.
+\end{theorem}
+
+\begin{proof}
+Let $\map f t = \sin \sqrt t$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = \dfrac {\cos \sqrt t} {2 \sqrt t}
+      | c = 
+}}
+{{eqn | l = \map f 0
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {f'} t}
+      | r = \dfrac 1 2 \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
+      | c = 
+}}
+{{eqn | r = s \, \map F s - \map f 0
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | r = \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }
+      | c = [[Laplace Transform of Sine of Root]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
+      | r = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Error Function}
+Tags: Laplace Transforms, Error Function
+
+\begin{theorem}
+:$\laptrans {\map \erf t} = \dfrac 1 s \, \map \exp {\dfrac {s^2} 4} \, \map \erfc {\dfrac s 2}$
+where:
+:$\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$
+:$\erf$ denotes the [[Definition:Error Function|error function]]
+:$\erfc$ denotes the [[Definition:Complementary Error Function|complementary error function]]
+:$\exp$ denotes the [[Definition:Exponential Function|exponential function]].
+\end{theorem}
+
+\begin{proof}
+By [[Derivative of Error Function]], we have: 
+:$\displaystyle \frac \d {\d t} \paren {\map \erf t} = \frac 2 {\sqrt \pi} e^{-t^2}$
+By [[Primitive of Exponential Function]], we have: 
+:$\displaystyle \int e^{-s t} \rd t = -\frac {e^{-s t} } s$
+So: 
+{{begin-eqn}}
+{{eqn	| l = \laptrans {\map \erf t}
+	| r = \int_0^\infty e^{-s t} \map \erf t \rd t
+	| c = {{Defof|Laplace Transform}}
+}}
+{{eqn	| r = \intlimits {-\frac 1 s e^{-s t} \map \erf t} 0 \infty - \int_0^\infty \paren {-\frac 2 {\sqrt \pi} \frac {e^{-s t} } s e^{-t^2} } \rd t
+	| c = [[Integration by Parts]]
+}}
+{{eqn	| r = -\frac 1 s \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t} + \frac 1 s e^0 \erf 0 + \frac 2 {s \sqrt \pi} \int_0^\infty \exp \paren {-s t - t^2} \rd t
+}}
+{{end-eqn}}
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t}
+	| r = \paren {\lim_{t \mathop \to \infty} e^{-s t} } \paren {\lim_{t \mathop \to \infty} \map \erf t}
+	| c = [[Combination Theorem for Limits of Functions/Product Rule|Combination Theorem for Limits of Functions: Product Rule]]
+}}
+{{eqn	| r = 0 \times 1
+	| c = [[Exponential Tends to Zero and Infinity]], [[Limit to Infinity of Error Function]]
+}}
+{{eqn	| r = 0
+}}
+{{end-eqn}}
+We also have: 
+{{begin-eqn}}
+{{eqn	| l = \frac 1 s e^0 \erf 0
+	| r = \frac 1 s \int_0^0 e^{-t^2} \rd t
+	| c = [[Exponential of Zero]], {{Defof|Error Function}}
+}}
+{{eqn	| r = 0
+	| c = [[Definite Integral on Zero Interval]]
+}}
+{{end-eqn}}
+Therefore: 
+{{begin-eqn}}
+{{eqn	| l = \laptrans {\map \erf t} 
+	| r = \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {t^2 + s t} } \rd t
+}}
+{{eqn	| r = \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {\paren {t + \frac s 2}^2 - \frac {s^2} 4} } \rd t
+	| c = completing the square
+}}
+{{eqn	| r = \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_0^\infty \map \exp {-\paren {t + \frac s 2}^2} \rd t
+	| c = [[Exponential of Sum]]
+}}
+{{eqn	| r = \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u
+	| c = [[Integration by Substitution|substituting]] $u = t + \dfrac s 2$
+}}
+{{eqn	| r = \frac 1 s \map \exp {\frac {s^2} 4} \paren {\frac 2 {\sqrt \pi} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u}
+}}
+{{eqn	| r = \frac 1 s \map \exp {\frac {s^2} 4} \map \erfc {\frac s 2}
+	| c = {{Defof|Complementary Error Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map \erf {\sqrt t} }
+      | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \map \exp {-u^2} \rd u}
+      | c = {{Defof|Error Function}}
+}}
+{{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^{2 n} } {n!} \rd u} }
+      | c = {{Defof|Exponential Function/Real|Real Exponential Function|subdef = Sum of Series}}
+}}
+{{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1} n!} } }
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1} n!} \laptrans {t^{n + \frac 1 2} }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1} n! s^{n + \frac 3 2} } 
+      | c = [[Laplace Transform of Real Power]]
+}}
+{{eqn | r = \frac 2 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1} n! s^n}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n} 
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{end-eqn}}
+We have: 
+{{begin-eqn}}
+{{eqn | l = \map \Gamma {n + \frac 1 2}
+      | r = \frac \pi {\map \sin {\pi \paren {\frac 1 2 - n} } \map \Gamma {\frac 1 2 - n} }
+      | c = [[Euler's Reflection Formula]]
+}}
+{{eqn | r = \frac \pi {\map \cos {-n \pi} \map \Gamma {\frac 1 2 - n} }
+      | c = [[Sine of Complement equals Cosine]]
+}}
+{{eqn | r = \frac {\paren {-1}^n \pi} {\map \Gamma {\frac 1 2 - n} }
+      | c = [[Cosine of Integer Multiple of Pi]]
+}}
+{{end-eqn}}
+So: 
+{{begin-eqn}}
+{{eqn | l = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n} 
+      | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {-1}^n \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\sqrt \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {-\frac 1 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {-\frac 1 2 - n + 1} s^n}
+      | c = [[Gamma Function of One Half]]
+}}
+{{eqn | r = \dfrac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \binom {-\frac 1 2} n \frac 1 {s^n}
+      | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}}
+}}
+{{eqn | r = \dfrac 1 {s^{3/2} } \paren {1 + \dfrac 1 s}^{-1/2}
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 {s \sqrt {s + 1} }
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Error Function of Root}
+Tags: Laplace Transforms, Error Function, Laplace Transform of Error Function of Root
+
+\begin{theorem}
+:$\laptrans {\map \erf {\sqrt t} } = \dfrac 1 {s \sqrt {s + 1} }$
+where:
+:$\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$
+:$\erf$ denotes the [[Definition:Error Function|error function]]
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map \erf {\sqrt t} }
+      | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \map \exp {-u^2} \rd u}
+      | c = {{Defof|Error Function}}
+}}
+{{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^{2 n} } {n!} \rd u} }
+      | c = {{Defof|Exponential Function/Real|Real Exponential Function|subdef = Sum of Series}}
+}}
+{{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1} n!} } }
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1} n!} \laptrans {t^{n + \frac 1 2} }
+      | c = [[Linear Combination of Laplace Transforms]]
+}}
+{{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1} n! s^{n + \frac 3 2} } 
+      | c = [[Laplace Transform of Real Power]]
+}}
+{{eqn | r = \frac 2 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1} n! s^n}
+      | c = [[Gamma Difference Equation]]
+}}
+{{eqn | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n} 
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{end-eqn}}
+We have: 
+{{begin-eqn}}
+{{eqn | l = \map \Gamma {n + \frac 1 2}
+      | r = \frac \pi {\map \sin {\pi \paren {\frac 1 2 - n} } \map \Gamma {\frac 1 2 - n} }
+      | c = [[Euler's Reflection Formula]]
+}}
+{{eqn | r = \frac \pi {\map \cos {-n \pi} \map \Gamma {\frac 1 2 - n} }
+      | c = [[Sine of Complement equals Cosine]]
+}}
+{{eqn | r = \frac {\paren {-1}^n \pi} {\map \Gamma {\frac 1 2 - n} }
+      | c = [[Cosine of Integer Multiple of Pi]]
+}}
+{{end-eqn}}
+So: 
+{{begin-eqn}}
+{{eqn | l = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n} 
+      | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {-1}^n \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\sqrt \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}
+      | c = 
+}}
+{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {-\frac 1 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {-\frac 1 2 - n + 1} s^n}
+      | c = [[Gamma Function of One Half]]
+}}
+{{eqn | r = \dfrac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \binom {-\frac 1 2} n \frac 1 {s^n}
+      | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}}
+}}
+{{eqn | r = \dfrac 1 {s^{3/2} } \paren {1 + \dfrac 1 s}^{-1/2}
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \dfrac 1 {s \sqrt {s + 1} }
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Sine Integral Function}
+Tags: Laplace Transforms, Sine Integral Function, Laplace Transform of Sine Integral Function
+
+\begin{theorem}
+:$\laptrans {\map \Si t} = \dfrac 1 s \arctan \dfrac 1 s$
+where:
+:$\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$
+:$\Si$ denotes the [[Definition:Sine Integral Function|sine integral function]]
+\end{theorem}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = \dfrac {\sin t} t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = t \map {f'} t
+      | r = \sin t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {t \map {f'} t}
+      | r = \laptrans {\sin t}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {s^2 + 1}
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}
+      | r = \dfrac 1 {s^2 + 1}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}
+      | r = -\dfrac 1 {s^2 + 1}
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = -\int \dfrac 1 {s^2 + 1} \rd s
+      | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = -\arctan s + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+By the [[Initial Value Theorem of Laplace Transform]]:
+:$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
+which leads to:
+:$c = \dfrac \pi 2$
+Thus:
+{{begin-eqn}}
+{{eqn | l = s \laptrans {\map f t}
+      | r = \dfrac \pi 2 - \arctan s
+      | c = 
+}}
+{{eqn | r = \arccot s
+      | c = [[Sum of Arctangent and Arccotangent]]
+}}
+{{eqn | r = \arctan \dfrac 1 s
+      | c = [[Arctangent of Reciprocal equals Arccotangent]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map \Si t
+      | r = \int_0^t \dfrac {\sin u} u \rd u
+      | c = {{Defof|Sine Integral Function}}
+}}
+{{eqn | r = \int_0^t \dfrac 1 u \paren {u - \dfrac {u^3} {3!} + \dfrac {u^5} {5!} - \dfrac {u^7} {7!} + \dotsb} \rd u
+      | c = {{Defof|Real Sine Function}}
+}}
+{{eqn | r = t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb
+      | c = [[Primitive of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map \Si t}
+      | r = \laptrans {t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 \times 3!} \dfrac {3!} {s^4} + \dfrac 1 {5 \times 5!} \dfrac {5!} {s^6} - \dfrac 1 {7 \times 7!} \dfrac {7!} {s^8} + \dotsb
+      | c = [[Laplace Transform of Positive Integer Power]]
+}}
+{{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 s^4} + \dfrac 1 {5 s^6} - \dfrac 1 {7 s^8} + \dotsb
+      | c = simplifying
+}}
+{{eqn | r = \dfrac 1 s \paren {\dfrac {\paren {1 / s} } 1 - \dfrac {\paren {1 / s}^3} 3 + \dfrac {\paren {1 / s}^5} 5 - \dfrac {\paren {1 / s}^7} 7 + \dotsb}
+      | c = rearranging
+}}
+{{eqn | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = [[Power Series Expansion for Real Arctangent Function]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$.
+Then:
+:$\map f 0 = 0$
+and:
+{{begin-eqn}}
+{{eqn | l = \map \Si t
+      | r = \int_0^t \dfrac {\sin u} u \rd u
+      | c = {{Defof|Sine Integral Function}}
+}}
+{{eqn | r = \int_0^1 \dfrac {\sin t v} v \rd v
+      | c = [[Integration by Substitution]] $u = t v$
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map \Si t}
+      | r = \laptrans {\int_0^1 \dfrac {\sin t v} v \rd v}
+      | c = 
+}}
+{{eqn | r = \int_0^\infty e^{-s t} \paren {\int_0^1 \dfrac {\sin t v} v \rd v} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^1 \dfrac 1 v \paren {\int_0^\infty e^{-s t} \sin t v \rd t} \rd v
+      | c = exchanging order of integration
+}}
+{{eqn | r = \int_0^1 \dfrac {\laptrans {\sin t v} } v \rd v
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^1 \dfrac {\d v} {s^2 + v^2}
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | r = \intlimits {\dfrac 1 s \arctan \dfrac v s} 0 1
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \dfrac 1 s \arctan \dfrac 1 s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Cosine Integral Function}
+Tags: Laplace Transforms, Cosine Integral Function, Laplace Transform of Cosine Integral Function
+
+\begin{theorem}
+:$\laptrans {\map \Ci t} = \dfrac {\map \ln {s^2 + 1} } {2 s}$
+where:
+:$\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$
+:$\Ci$ denotes the [[Definition:Cosine Integral Function|cosine integral function]].
+\end{theorem}
+
+\begin{proof}
+Let $\map f t := \map \Ci t = \displaystyle \int_t^\infty \dfrac {\cos u} u \rd u$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = -\dfrac {\cos t} t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = t \map {f'} t
+      | r = -\cos t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {t \map {f'} t}
+      | r = -\laptrans {\cos t}
+      | c = 
+}}
+{{eqn | r = -\dfrac s {s^2 + 1}
+      | c = [[Laplace Transform of Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}
+      | r = -\dfrac s {s^2 + 1}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}
+      | r = \dfrac s {s^2 + 1}
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \int \dfrac s {s^2 + 1} \rd s
+      | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \dfrac 1 2 \map \ln {s^2 + 1} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+By the [[Initial Value Theorem of Laplace Transform]]:
+:$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
+which leads to:
+:$c = 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = s \laptrans {\map f t}
+      | r = \dfrac 1 2 \map \ln {s^2 + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \dfrac {\map \ln {s^2 + 1} } {2 s}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Exponential Integral Function}
+Tags: Laplace Transforms, Exponential Integral Function, Laplace Transform of Exponential Integral Function
+
+\begin{theorem}
+:$\laptrans {\map \Ei t} = \dfrac {\map \ln {s + 1} } s$
+where:
+:$\laptrans f$ denotes the [[Definition:Laplace Transform|Laplace transform]] of the [[Definition:Real Function|function]] $f$
+:$\Ei$ denotes the [[Definition:Exponential Integral Function|exponential integral function]].
+\end{theorem}
+
+\begin{proof}
+Let $\map f t := \map \Ei t = \displaystyle \int_t^\infty \dfrac {e^{-u} } u \rd u$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f'} t
+      | r = -\dfrac {e^{-t} } t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = t \map {f'} t
+      | r = -e^{-t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {t \map {f'} t}
+      | r = -\laptrans {e^{-t} }
+      | c = 
+}}
+{{eqn | r = -\dfrac 1 {s + 1}
+      | c = [[Laplace Transform of Exponential]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}
+      | r = -\dfrac 1 {s + 1}
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}
+      | r = \dfrac 1 {s + 1}
+      | c = [[Laplace Transform of Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \int \dfrac 1 {s + 1} \rd s
+      | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$
+}}
+{{eqn | ll= \leadsto
+      | l = s \laptrans {\map f t}
+      | r = \map \ln {s + 1} + C
+      | c = [[Primitive of Reciprocal of a x + b|Primitive of $\dfrac 1 {a x + b}$]]
+}}
+{{end-eqn}}
+By the [[Initial Value Theorem of Laplace Transform]]:
+:$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$
+which leads to:
+:$c = 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = s \laptrans {\map f t}
+      | r = \map \ln {s + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\map f t}
+      | r = \dfrac {\map \ln {s + 1} } s
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Heaviside Step Function}
+Tags: Laplace Transforms, Heaviside Step Function, Laplace Transform of Heaviside Step Function
+
+\begin{theorem}
+Let $\map {u_c} t$ denote the [[Definition:Heaviside Step Function|Heaviside step function]]:
+:$\map {u_c} t = \begin{cases}
+1 & : t > c \\
+0 & : t < c
+\end{cases}$
+The [[Definition:Laplace Transform|Laplace transform]] of $\map {u_c} t$ is given by:
+:$\laptrans {\map {u_c} t} = \dfrac {e^{-s c} } s$
+for $\map \Re s > c$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {u_c} t}
+      | r = \int_0^{\to +\infty} \map {u_c} t e^{-s t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^c \map {u_c} t e^{-s t} \rd t + \int_c^{\to +\infty} \map {u_c} t e^{-s t} \rd t
+      | c = [[Sum of Integrals on Adjacent Intervals for Integrable Functions]]
+}}
+{{eqn | r = \int_0^c 0 \times e^{-s t} \rd t + \int_c^{\to +\infty} 1 \times e^{-s t} \rd t
+      | c = {{Defof|Heaviside Step Function}}
+}}
+{{eqn | r = \int_c^{\to +\infty} e^{-s t} \rd t
+      | c = 
+}}
+{{eqn | r = \lim_{L \mathop \to +\infty} \int_c^L e^{-s t} \rd t
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lim_{L \mathop \to +\infty} \intlimits {\dfrac {e^{-s t} } {-s} } c L
+      | c = [[Primitive of Exponential of a x|Primitive of $e^{a x}$]]
+}}
+{{eqn | r = \lim_{L \mathop \to +\infty} \dfrac {e^{-s L} } {-s} - \dfrac {e^{-s c} } {-s}
+      | c = 
+}}
+{{eqn | r = 0 + \dfrac {e^{-s c} } s
+      | c = simplification
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Dirac Delta Function}
+Tags: Laplace Transforms, Dirac Delta Function, Laplace Transform of Dirac Delta Function
+
+\begin{theorem}
+Let $\map \delta t$ denote the [[Definition:Dirac Delta Function|Dirac delta function]].
+The [[Definition:Laplace Transform|Laplace transform]] of $\map \delta t$ is given by:
+:$\laptrans {\map \delta t} = 1$
+\end{theorem}
+
+\begin{proof}
+=== [[Laplace Transform of Dirac Delta Function/Lemma|Lemma]] ===
+{{:Laplace Transform of Dirac Delta Function/Lemma}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map \delta t}
+      | r = \lim_{\epsilon \mathop \to 0} \laptrans {\map {F_\epsilon} t}
+      | c = {{Defof|Dirac Delta Function}}
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {1 - e^{-s \epsilon} } {\epsilon s}
+      | c = [[Laplace Transform of Dirac Delta Function/Lemma|Lemma]]
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac 1 {\epsilon s} \paren {1 - \paren {1 - s \epsilon + \dfrac {s^2 \epsilon^2} {2!} - \dotsb} }
+      | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}}
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \paren {1 - \dfrac {s \epsilon} {2!} + \dotsb}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== [[Laplace Transform of Dirac Delta Function/Lemma|Lemma]] ===
+{{:Laplace Transform of Dirac Delta Function/Lemma}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map \delta t}
+      | r = \lim_{\epsilon \mathop \to 0} \laptrans {\map {F_\epsilon} t}
+      | c = {{Defof|Dirac Delta Function}}
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {1 - e^{-s \epsilon} } {\epsilon s}
+      | c = [[Laplace Transform of Dirac Delta Function/Lemma|Lemma]]
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {\map {\dfrac \d {\d s} } {1 - e^{-s \epsilon} } } {\map {\dfrac \d {\d s} } {\epsilon s} }
+      | c = [[L'Hôpital's Rule]]
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {\paren {-\epsilon} \times \paren {-e^{-s \epsilon} } } \epsilon
+      | c = 
+}}
+{{eqn | r = \lim_{\epsilon \mathop \to 0} e^{-s \epsilon}
+      | c = simplification
+}}
+{{eqn | r = 1
+      | c = [[Exponential of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Shifted Dirac Delta Function}
+Tags: Laplace Transforms, Dirac Delta Function, Laplace Transform of Shifted Dirac Delta Function
+
+\begin{theorem}
+Let $\map \delta t$ denote the [[Definition:Dirac Delta Function|Dirac delta function]].
+The [[Definition:Laplace Transform|Laplace transform]] of $\map \delta {t - a}$ is given by:
+:$\laptrans {\map \delta {t - a} } = e^{-a s}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map \delta {t - a} }
+      | r = \int_0^{\to +\infty} e^{-s t} \map \delta {t - a} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = e^{-s \times a}
+      | c = [[Integral to Infinity of Shifted Dirac Delta Function by Continuous Function]]
+}}
+{{eqn | r = e^{-a s}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Null Function}
+Tags: Laplace Transforms, Null Functions
+
+\begin{theorem}
+Let $\mathcal N: \R \to \R$ be a [[Definition:Null Function|null function]].
+The [[Definition:Laplace Transform|Laplace transform]] of $\map {\mathcal N} t$ is given by:
+:$\laptrans {\map {\mathcal N} t} = 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map {\mathcal N} t}
+      | r = \int_0^{\to +\infty} e^{-s t} \map {\mathcal N} t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{finish}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Function of t minus a/Proof 1}
+Tags: Laplace Transform of Function of t minus a
+
+\begin{theorem}
+Let $f$ be a [[Definition:Function|function]] such that $\laptrans f$ exists.
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of $f$.
+Let $a \in \C$ or $\R$ be [[Definition:Constant|constant]].
+{{:Laplace Transform of Function of t minus a}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map f {t - a} }
+      | r = \int_0^{\to + \infty} e^{-s t} \map f {t - a} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^{\to + \infty} e^{-s \paren {t - a} } e^{-a s} \map f {t - a} \rd \paren {t - a}
+      | c = 
+}}
+{{eqn | r = e^{-a s} \int_0^{\to + \infty} e^{-s \paren {t - a} } \map f {t - a} \rd \paren {t - a}
+      | c = 
+}}
+{{eqn | r = e^{-a s}\map F s
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Function of t minus a/Proof 2}
+Tags: Laplace Transform of Function of t minus a
+
+\begin{theorem}
+Let $f$ be a [[Definition:Function|function]] such that $\laptrans f$ exists.
+Let $\laptrans {\map f t} = \map F s$ denote the [[Definition:Laplace Transform|Laplace transform]] of $f$.
+Let $a \in \C$ or $\R$ be [[Definition:Constant|constant]].
+{{:Laplace Transform of Function of t minus a}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\map g t}
+      | r = \int_0^\infty e^{-s t} \map g t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \int_0^a e^{-s t} \map g t \rd t + \int_a^\infty e^{-s t} \map g t \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^a 0 \times e^{-s t} \rd t + \int_a^\infty e^{-s t} \map f {t - a} \rd t
+      | c = Definition of $\map g t$
+}}
+{{eqn | r = \int_a^\infty e^{-s t} \map f {t - a} \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^\infty e^{-s \paren {u + a} } \map f u \rd u
+      | c = [[Integration by Substitution]]: $t = u + a$
+}}
+{{eqn | r = e^{-a s} \int_0^\infty e^{-s u} \map f u \rd u
+      | c = 
+}}
+{{eqn | r = e^{-a s} \map F s
+      | c = {{Defof|Laplace Transform}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Sine of t over t}
+Tags: Laplace Transforms, Sine Function, Laplace Transform of Sine
+
+\begin{theorem}
+Let $\sin$ denote the [[Definition:Real Sine Function|real sine function]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {\dfrac {\sin t} t} = \arctan \dfrac 1 s$
+\end{theorem}
+
+\begin{proof}
+From [[Limit of Sine of X over X]]:
+:$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
+From [[Laplace Transform of Sine]]:
+:$(1): \quad \displaystyle \laptrans {\sin t} = \dfrac 1 {s^2 + 1}$
+From [[Laplace Transform of Integral]]:
+:$(2): \quad \displaystyle \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\dfrac {\sin t} t}
+      | r = \int_s^{\to \infty} \dfrac 1 {u^2 + 1} \rd u
+      | c = $(1)$ and $(2)$
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \int_s^L \dfrac 1 {u^2 + 1} \rd u
+      | c = {{Defof|Improper Integral}}
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \bigintlimits {\arctan u} s L
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \lim_{L \mathop \to \infty} \paren {\arctan L - \arctan s}
+      | c = 
+}}
+{{eqn | r = \dfrac \pi 2 - \arctan s
+      | c = 
+}}
+{{eqn | r = \arccot s
+      | c = [[Sum of Arctangent and Arccotangent]]
+}}
+{{eqn | r = \arctan \dfrac 1 s
+      | c = [[Arctangent of Reciprocal equals Arccotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of Sine of t over t/Corollary}
+Tags: Laplace Transforms, Sine Function, Laplace Transform of Sine
+
+\begin{theorem}
+:$\laptrans {\dfrac {\sin a t} t} = \arctan \dfrac a s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\dfrac {\sin t} t}
+      | r = \arctan \dfrac 1 s
+      | c = [[Laplace Transform of Sine of t over t]]
+}}
+{{eqn | ll= \leadsto
+      | l = \laptrans {\dfrac {\sin a t} {a t} }
+      | r = \dfrac 1 a \arctan \dfrac 1 {s / a}
+      | c = [[Laplace Transform of Constant Multiple]]
+}}
+{{eqn | r = \dfrac 1 a \arctan \dfrac a s
+      | c = 
+}}
+{{end-eqn}}
+But:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\dfrac {\sin a t} {a t} }
+      | r = \int_0^\infty e^{-s t} \dfrac {\sin a t} {a t} \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \dfrac 1 a \int_0^\infty e^{-s t} \dfrac {\sin a t} t \rd t
+      | c = 
+}}
+{{eqn | r = \dfrac 1 a \laptrans {\dfrac {\sin a t} t}
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of t by Sine a t}
+Tags: Laplace Transforms, Sine Function, Laplace Transform of t by Sine a t
+
+\begin{theorem}
+Let $\sin$ denote the [[Definition:Real Sine Function|real sine function]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {t \sin a t} = \dfrac {2 a s} {\paren {s^2 + a^2}^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {t \sin a t}
+      | r = -\map {\dfrac \d {\d s} } {\laptrans {\sin a t} }
+      | c = [[Derivative of Laplace Transform]]
+}}
+{{eqn | r = -\map {\dfrac \d {\d s} } {\dfrac a {s^2 + a^2} }
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | r = \dfrac {2 a s} {\paren {s^2 + a^2}^2}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = \laptrans {\cos a t}
+      | r = \int_0^\infty e^{-s t} \cos a t \rd t
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | n = 1
+      | r = \dfrac s {s^2 + a^2}
+      | c = [[Laplace Transform of Cosine]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \dfrac \d {\d a} \int_0^\infty e^{-s t} \cos a t \rd t
+      | r = \int_0^\infty e^{-s t} \paren {\map {\dfrac \partial {\partial a} } {\cos a t} } \rd t
+      | c = [[Derivative of Integral]]
+}}
+{{eqn | r = \int_0^\infty e^{-s t} \paren {-t \sin a t} \rd t
+      | c = [[Derivative of Cosine Function]]
+}}
+{{eqn | r = -\laptrans {t \sin a t}
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | r = \map {\dfrac \d {\d a} } {\dfrac s {s^2 + a^2} }
+      | c = from $(1)$
+}}
+{{eqn | r = -\dfrac {2 a s} {\paren {s^2 + a^2}^2}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{end-eqn}}
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Laplace Transform of t^2 by Cosine a t}
+Tags: Laplace Transforms, Cosine Function
+
+\begin{theorem}
+Let $\sin$ denote the [[Definition:Real Sine Function|real sine function]].
+Let $\laptrans f$ denote the [[Definition:Laplace Transform|Laplace transform]] of a [[Definition:Real Function|real function]] $f$.
+Then:
+:$\laptrans {t^2 \cos a t} = \dfrac {2 s^3 - 6 a^2 s} {\paren {s^2 + a^2}^3}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {t^2 \cos a t}
+      | r = -\map {\dfrac {\d^2} {\d s^2} } {\laptrans {\cos a t} }
+      | c = [[Higher Order Derivatives of Laplace Transform]]
+}}
+{{eqn | r = -\map {\dfrac {\d^2} {\d s^2} } {\dfrac a {s^2 + a^2} }
+      | c = [[Laplace Transform of Sine]]
+}}
+{{eqn | r = \dfrac {2 s^3 - 6 a^2 s} {\paren {s^2 + a^2}^3}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral to Infinity of Function over Argument}
+Tags: Laplace Transforms
+
+\begin{theorem}
+:$\displaystyle \int_0^\infty {\dfrac {\map f t} t} = \int_0^{\to \infty} \map F u \rd u$
+provided the integrals converge.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \laptrans {\dfrac {\map f t} t}
+      | r = \int_s^{\to \infty} \map F u \rd u
+      | c = [[Integral of Laplace Transform]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int_0^\infty e^{-s t} {\dfrac {\map f t} t} \rd t
+      | r = \int_s^{\to \infty} \map F u \rd u
+      | c = {{Defof|Laplace Transform}}
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \dfrac {\map f t} t \rd t
+      | r = \lim_{s \mathop \to 0} \int_s^{\to \infty} \map F u \rd u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int_0^\infty \dfrac {\map f t} t \rd t
+      | r = \int_0^{\to \infty} \map F u \rd u
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
\ No newline at end of file