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+\section{Range of Modulo Operation for Positive Modulus}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := \begin{cases}
+x - y \floor {\dfrac x y} & : y \ne 0 \\
+x & : y = 0
+\end{cases}$
+where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfrac x y$.
+Let $y > 0$.
+Then:
+:$0 \le x \bmod y < y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | m = \frac {x \bmod y} y
+      | mo= <
+      | r = 1
+      | c = [[Quotient of Modulo Operation with Modulus]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | m = \frac {x \bmod y} y \times y
+      | mo= <
+      | r = 1 \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | m = x \bmod y
+      | mo= <
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Range of Modulo Operation for Negative Modulus}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := \begin{cases}
+x - y \left \lfloor {\dfrac x y}\right \rfloor & : y \ne 0 \\
+x & : y = 0
+\end{cases}$
+where $\left \lfloor {\dfrac x y}\right \rfloor$ denotes the [[Definition:Floor Function|floor]] of $\dfrac x y$.
+Let $y < 0$.
+Then:
+:$0 \ge x \bmod y > y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \le
+      | m = \frac {x \bmod y} y
+      | mo= <
+      | r = 1
+      | c = [[Quotient of Modulo Operation with Modulus]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \ge
+      | m = \frac {x \bmod y} y \times y
+      | mo= >
+      | r = 1 \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \ge
+      | m = x \bmod y
+      | mo= >
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero is Integer Multiple of Zero}
+Tags: Number Theory
+
+\begin{theorem}
+[[Definition:Zero (Number)|Zero]] is an [[Definition:Integer Multiple|integer multiple]] of [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$0 \times 0 = 0$
+The result follows by definition of [[Definition:Integer Multiple|integer multiple]].
+{{qed}}
+[[Category:Number Theory]]
+cr4qeo16keg2mqnhx82y0t2xypme5gz
+\end{proof}<|endoftext|>
+\section{Number minus Modulo is Integer Multiple}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := \begin{cases}
+x - y \left \lfloor {\dfrac x y}\right \rfloor & : y \ne 0 \\
+x & : y = 0
+\end{cases}$
+where $\left \lfloor {\dfrac x y}\right \rfloor$ denotes the [[Definition:Floor Function|floor]] of $\dfrac x y$.
+Let $y < 0$.
+Then:
+:$x - \left({x \bmod y}\right)$
+is an [[Definition:Integer Multiple|integer multiple]] of $y$.
+\end{theorem}
+
+\begin{proof}
+When $y = 0$ we have:
+:$x \bmod y := x$
+Thus:
+:$x - \left({x \bmod y}\right) = 0$
+From [[Zero is Integer Multiple of Zero]] it follows that:
+:$x - \left({x \bmod y}\right)$
+is an [[Definition:Integer Multiple|integer multiple]] of $y$.
+Let $y \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \bmod y
+      | r = x - y \left \lfloor {\dfrac x y}\right \rfloor
+      | c = Definition of [[Definition:Modulo Operation|Modulo Operation]]
+}}
+{{eqn | ll= \leadsto
+      | l = x - \left({x \bmod y}\right)
+      | r = y \left \lfloor {\dfrac x y}\right \rfloor
+      | c = Definition of [[Definition:Modulo Operation|Modulo Operation]]
+}}
+{{end-eqn}}
+From [[Floor Function is Integer]], $\left \lfloor {\dfrac x y}\right \rfloor$ is an [[Definition:Integer|integer]].
+Thus:
+:$\exists n \in \Z: x - \left({x \bmod y}\right) = n y$
+where in the case $n = \left \lfloor {\dfrac x y}\right \rfloor$
+Hence the result by definition of [[Definition:Integer Multiple|integer multiple]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Operation/Examples/5 mod 3}
+Tags: Examples of Modulo Operation
+
+\begin{theorem}
+:$5 \bmod 3 = 2$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$
+for $y \ne 0$.
+We have:
+:$\dfrac 5 3 = 1 + \dfrac 2 3$
+and so:
+:$\left\lfloor{\dfrac 5 3}\right\rfloor = 1$
+Thus:
+:$5 \bmod 3 = 5 - 3 \times \left\lfloor{\dfrac 5 3}\right\rfloor = 5 - 3 \times 1 = 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Operation/Examples/18 mod 3}
+Tags: Examples of Modulo Operation
+
+\begin{theorem}
+:$18 \bmod 3 = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$
+for $y \ne 0$.
+We have:
+:$\dfrac {18} 3 = 6 + \dfrac 0 3$
+and so:
+:$\left\lfloor{\dfrac {18} 3}\right\rfloor = 6$
+Thus:
+:$18 \bmod 3 = 18 - 3 \times \left\lfloor{\dfrac {18} 3}\right\rfloor = 18 - 3 \times 6 = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Operation/Examples/-2 mod 3}
+Tags: Examples of Modulo Operation
+
+\begin{theorem}
+:$-2 \bmod 3 = 1$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := x - y \left \lfloor {\dfrac x y}\right \rfloor$
+for $y \ne 0$.
+We have:
+:$\dfrac {-2} 3 = -1 + \dfrac 1 3$
+and so:
+:$\left\lfloor{\dfrac {-2} 3}\right\rfloor = -1$
+Thus:
+:$-2 \bmod 3 = -2 - 3 \times \left\lfloor{\dfrac {-2} 3}\right\rfloor = -2 - 3 \times \left({-1}\right) = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number is Divisor iff Modulo is Zero}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $x \bmod y$ denote the [[Definition:Modulo Operation|modulo operation]]:
+:$x \bmod y := \begin {cases}
+x - y \floor {\dfrac x y} & : y \ne 0 \\
+x & : y = 0
+\end {cases}$
+where $\floor {\dfrac x y}$ denotes the [[Definition:Floor Function|floor]] of $\dfrac x y$.
+Then $x \bmod y = 0$ {{iff}} $x$ is an [[Definition:Integer Multiple|integer multiple]] of $y$.
+\end{theorem}<|endoftext|>
+\section{Correctness of Definition of Increasing Mappings Satisfying Inclusion in Lower Closure}
+Tags: Order Theory
+
+\begin{theorem}
+Let $R = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let ${\it Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
+Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$
+Let 
+:$M = \left({F, \preccurlyeq}\right)$
+where
+:$F = \left\{ {f: S \to {\it Ids}\left({R}\right): f}\right.$ is [[Definition:Increasing Mapping|increasing mapping]] $\left.{\land \forall x \in S: f\left({x}\right) \subseteq x^\preceq}\right\}$
+and
+:$\preccurlyeq$ is [[Definition:Ordering on Mappings|ordering on mappings]] generated by $\precsim$
+where $x^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $x$.
+Then
+:$M$ is an [[Definition:Ordered Set|ordered set]].
+\end{theorem}
+
+\begin{proof}
+=== Reflexivity ===
+Let $f \in F$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\forall x \in S: f\left({x}\right) \precsim f\left({x}\right)$
+Thus by definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$f \preccurlyeq f$
+{{qed|lemma}}
+=== Transitivity ===
+Let $f, g, h \in F$ such that
+:$f \preccurlyeq g \preccurlyeq h$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \precsim h\left({x}\right)$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$
+Thus by definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$f \preccurlyeq h$
+{{qed|lemma}}
+=== Antisymmetry ===
+Let $f, g \in F$ such that
+:$f \preccurlyeq g$ and $g \preccurlyeq f$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$\forall x \in S: f\left({x}\right) \precsim g\left({x}\right) \land g\left({x}\right) \precsim f\left({x}\right)$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\forall x \in S: f\left({x}\right) = g\left({x}\right)$
+Thus by [[Equality of Mappings|equality of mappings]]:
+:$f = g$
+{{qed|lemma}}
+Thus by definition:
+:$M$ is an [[Definition:Ordered Set|ordered set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number less than Integer iff Floor less than Integer}
+Tags: Floor Function
+
+\begin{theorem}
+:$\left \lfloor{x}\right \rfloor < n \iff x < n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x < n$.
+By definition of the [[Definition:Floor Function|floor]] of $x$:
+:$\left \lfloor {x} \right \rfloor \le x$
+Hence:
+:$\left \lfloor {x} \right \rfloor < n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\left \lfloor{x}\right \rfloor < n$.
+We have that:
+:$\forall m, n \in \Z: m < n \iff m + 1 \le n$
+and so:
+:$(1): \quad \left \lfloor{x}\right \rfloor + 1 \le n$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = \left \lfloor{x}\right \rfloor + 1
+      | c = Definition of [[Definition:Floor Function|Floor Function]]
+}}
+{{eqn | o = \le
+      | r = n
+      | c = from $(1)$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Hence the result:
+:$\left \lfloor{x}\right \rfloor < n \iff x < n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number not less than Integer iff Floor not less than Integer}
+Tags: Floor Function
+
+\begin{theorem}
+:$x \ge n \iff \left \lfloor{x}\right \rfloor \ge n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\left \lfloor{x}\right \rfloor \ge n$.
+By definition of the [[Definition:Floor Function|floor]] of $x$:
+:$x \ge \left \lfloor{x}\right \rfloor$
+Hence:
+:$x \ge n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x \ge n$.
+{{AimForCont}} $\left \lfloor{x}\right \rfloor < n$.
+We have that:
+:$\forall m, n \in \Z: m < n \iff m + 1 \le n$
+Hence:
+:$\left \lfloor{x}\right \rfloor + 1 \le n$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$\left \lfloor{x}\right \rfloor + 1 \le x$
+This [[Definition:Contradiction|contradicts]] the definition of the [[Definition:Floor Function|floor]] of $x$:
+:$\left \lfloor{x}\right \rfloor + 1 > x$
+Thus by [[Proof by Contradiction]]:
+:$\left \lfloor{x}\right \rfloor \ge n$
+{{qed|lemma}}
+Hence the result:
+:$\left \lfloor{x}\right \rfloor \ge n \iff x \ge n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number not greater than Integer iff Ceiling not greater than Integer}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\left \lceil{x}\right \rceil \le n \iff x \le n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\left \lceil{x}\right \rceil \le n$.
+By [[Number is between Ceiling and One Less]]:
+:$x \le \left \lceil{x}\right \rceil$
+Hence:
+:$x \le n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x \le n$.
+{{AimForCont}} $\left \lceil{x}\right \rceil > n$.
+We have that:
+:$\forall m, n \in \Z: m < n \iff m \le n - 1$
+Hence:
+:$\left \lceil{x}\right \rceil - 1 \ge n$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$\left \lceil{x}\right \rceil - 1 \ge x$
+This [[Definition:Contradiction|contradicts]] the result [[Number is between Ceiling and One Less]]:
+:$\left \lceil{x}\right \rceil - 1 < x$
+Thus by [[Proof by Contradiction]]:
+:$\left \lceil{x}\right \rceil \le n$
+{{qed|lemma}}
+Hence the result:
+:$\left \lceil{x}\right \rceil \le n \iff x \le n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number greater than Integer iff Ceiling greater than Integer}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\left \lceil{x}\right \rceil > n \iff x > n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x > n$.
+By [[Number is between Ceiling and One Less]]:
+:$\left \lceil{x}\right \rceil \ge x$
+Hence:
+:$\left \lceil{x}\right \rceil > n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\left \lceil{x}\right \rceil > n$.
+We have that:
+:$\forall m, n \in \Z: m < n \iff m \le n - 1$
+and so:
+:$(1): \quad \left \lceil{x}\right \rceil - 1 \ge n$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = >
+      | r = \left \lceil{x}\right \rceil - 1
+      | c = [[Number is between Ceiling and One Less]]
+}}
+{{eqn | o = \ge
+      | r = n
+      | c = from $(1)$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Hence the result:
+:$\left \lceil{x}\right \rceil > n \iff x > n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer equals Floor iff between Number and One Less}
+Tags: Floor Function
+
+\begin{theorem}
+:$\floor x = n \iff x - 1 < n \le x$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x - 1 < n \le x$.
+From $n \le x$, we have by [[Number not less than Integer iff Floor not less than Integer]]:
+:$n \le \floor x$
+From $x - 1 < n$:
+:$x < n + 1$
+Hence by [[Number less than Integer iff Floor less than Integer]]:
+:$\floor x < n + 1$
+We have that:
+:$\forall m, n \in \Z: m \le n \iff m < n + 1$
+and so:
+:$\floor x \le n$
+Thus as:
+:$n \le \floor x$
+and:
+:$\floor x \le n$
+it follows that:
+:$\floor x = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\floor x = n$.
+Then:
+:$\floor x \ge n$
+By [[Number not less than Integer iff Floor not less than Integer]]:
+:$n \le x$
+By definition of the [[Definition:Floor Function|floor]] of $x$:
+:$x < \floor x + 1$
+and so subtracting $1$ from both sides:
+:$x - 1 < \floor x$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$x - 1 < n$
+So:
+:$\floor x = n \implies x - 1 < n \le x$
+{{qed|lemma}}
+Hence the result:
+:$\floor x = n \iff x - 1 < n \le x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer equals Floor iff Number between Integer and One More}
+Tags: Floor Function
+
+\begin{theorem}
+:$\floor x = n \iff n \le x < n + 1$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $n \le x < n + 1$.
+From [[Number not less than Integer iff Floor not less than Integer]]:
+:$n \le x \implies n \le \floor x$
+By definition of the [[Definition:Floor Function|floor]] of $x$:
+:$\floor x \le x$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$\floor x < n + 1$
+We have that:
+:$\forall m, n \in \Z: m \le n \iff m < n + 1$
+and so:
+:$\floor x \le n$
+So we have:
+:$n \le \floor x$
+and:
+:$\floor x \le n$
+Thus:
+:$n \le x < n + 1 \implies \floor x = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\floor x = n$.
+By definition of the [[Definition:Floor Function|floor]] of $x$:
+:$\floor x \le x$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$n \le x$
+Also by definition of the [[Definition:Floor Function|floor]] of $x$:
+:$x < \floor x + 1$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$x < n + 1$
+Thus:
+:$\floor x = n \implies n \le x < n + 1$.
+{{qed|lemma}}
+Hence the result:
+:$\floor x = n \iff n \le x < n + 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer equals Ceiling iff between Number and One More}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x \le n < x + 1$.
+From $x \le n$, we have by [[Number not greater than Integer iff Ceiling not greater than Integer]]:
+:$\left \lceil{x}\right \rceil \le n$
+From $n < x + 1$:
+:$n - 1 < x$
+Hence by [[Number greater than Integer iff Ceiling greater than Integer]]:
+:$n - 1 < \left \lceil{x}\right \rceil$
+We have that:
+:$\forall m, n \in \Z: m < n \iff m \le n - 1$
+and so:
+:$n \le \left \lceil{x}\right \rceil$
+Thus as:
+:$\left \lceil{x}\right \rceil \le n$
+and:
+:$\left \lceil{x}\right \rceil \ge n$
+it follows that:
+:$\left \lceil{x}\right \rceil = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\left \lceil{x}\right \rceil = n$.
+Then:
+:$\left \lceil{x}\right \rceil \le n$
+By [[Number not greater than Integer iff Ceiling not greater than Integer]]:
+:$x \le n$
+From [[Number is between Ceiling and One Less]]:
+:$\left \lceil{x}\right \rceil - 1 < x$
+and so adding $1$ to both sides:
+:$\left \lceil{x}\right \rceil < x + 1$
+and so [[Definition:By Hypothesis|by hypothesis]]:
+:$n < x + 1$
+So:
+:$\left \lceil{x}\right \rceil = n \implies x \le n < x + 1$
+{{qed|lemma}}
+Hence the result:
+:$\left \lceil{x}\right \rceil = n \iff x \le n < x + 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer equals Ceiling iff Number between Integer and One Less}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\ceiling x = n \iff n - 1 < x \le n$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $n - 1 < x \le n$.
+From [[Number not greater than Integer iff Ceiling not greater than Integer]]:
+:$x \le n \implies \ceiling x \le n$
+From [[Number is between Ceiling and One Less]]:
+:$x \le \ceiling x$
+and so:
+:$n - 1 < \ceiling x$
+We have that:
+:$\forall m, n \in \Z: m - 1 < n \iff m \le n$
+and so:
+:$n \le \ceiling x$
+So we have:
+:$\ceiling x \le n$
+and:
+:$n \le \ceiling x$
+Thus:
+:$n - 1 < x \le n \implies \ceiling x = n$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\ceiling x = n$.
+From [[Number is between Ceiling and One Less]]:
+:$x \le \ceiling x$
+and so:
+:$x \le n$
+Also from [[Number is between Ceiling and One Less]]:
+:$\ceiling x - 1 < x$
+and so:
+:$n - 1 < x$
+Thus:
+:$\ceiling x = n \implies n - 1 < x \le n$.
+{{qed|lemma}}
+Hence the result:
+:$\ceiling x = n \iff n - 1 < x \le n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of Root of Floor equals Floor of Root}
+Tags: Floor Function
+
+\begin{theorem}
+:$\displaystyle \floor {\sqrt {\floor x} } = \floor {\sqrt x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = n
+      | m = \floor {\sqrt x}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | o = \le
+      | m = \sqrt x
+      | mo= <
+      | r = n + 1
+      | c = [[Integer equals Floor iff Number between Integer and One More]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n^2
+      | o = \le
+      | m = x
+      | mo= <
+      | r = \paren {n + 1}^2
+      | c = [[Order is Preserved on Positive Reals by Squaring]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n^2
+      | o = \le
+      | m = \floor x
+      | mo= <
+      | r = \paren {n + 1}^2
+      | c = [[Number not less than Integer iff Floor not less than Integer]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | o = \le
+      | m = \sqrt {\floor x}
+      | mo= <
+      | r = n + 1
+      | c = [[Order is Preserved on Positive Reals by Squaring]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | m = \floor {\sqrt {\floor x} }
+      | c = [[Integer equals Floor iff Number between Integer and One More]]
+}}
+{{end-eqn}}
+{{qed}}
+{{expand|Add a proof based on [[McEliece's Theorem (Integer Functions)]]}}
+\end{proof}<|endoftext|>
+\section{Ceiling of Root of Ceiling equals Ceiling of Root}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\displaystyle \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = n
+      | m = \ceiling {\sqrt x}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n - 1
+      | o = <
+      | m = \sqrt x
+      | mo= \le
+      | r = n
+      | c = [[Integer equals Ceiling iff Number between Integer and One Less]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {n - 1}^2
+      | o = <
+      | m = x
+      | mo= \le
+      | r = n^2
+      | c = [[Order is Preserved on Positive Reals by Squaring]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {n - 1}^2
+      | o = <
+      | m = \ceiling x
+      | mo= \le
+      | r = n^2
+      | c = [[Number not greater than Integer iff Ceiling not greater than Integer]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n - 1
+      | o = <
+      | m = \sqrt {\ceiling x}
+      | mo= \le
+      | r = n
+      | c = [[Order is Preserved on Positive Reals by Squaring]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | m = \ceiling {\sqrt {\ceiling x} }
+      | c = [[Integer equals Ceiling iff Number between Integer and One Less]]
+}}
+{{end-eqn}}
+{{qed}}
+{{expand|Add a proof based on [[McEliece's Theorem (Integer Functions)]]}}
+\end{proof}<|endoftext|>
+\section{Congruence by Divisor of Modulus/Integer Modulus}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $r, s \in \Z$ be [[Definition:Integer|integers]].
+Let $a, b \in \Z$ such that $a$ is [[Definition:Congruence Modulo Integer|congruent modulo $r s$]] to $b$, that is:
+:$a \equiv b \pmod {r s}$
+Then:
+:$a \equiv b \pmod r$
+and:
+:$a \equiv b \pmod s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod {r s}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = a - b
+      | r = q r s
+      | c = {{Defof|Congruence Modulo Integer}}
+}}
+{{eqn | ll= \implies
+      | l = a - b
+      | r = \paren {q r} s
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = a - b
+      | r = \paren {q s} r
+      | c = 
+}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod r
+      | c = {{Defof|Congruence Modulo Integer}}: $q s$ is an [[Definition:Integer|integer]]
+}}
+{{eqn | lo= \land
+      | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod s
+      | c = {{Defof|Congruence Modulo Integer}}: $q r$ is an [[Definition:Integer|integer]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition is Well-Defined/Real Modulus}
+Tags: Modulo Addition, Modulo Addition is Well-Defined
+
+\begin{theorem}
+Let $z \in \R$ be a [[Definition:Real Number|real number]].
+Let:
+:$a \equiv b \pmod z$
+and:
+:$x \equiv y \pmod z$
+where $a, b, x, y \in \R$.
+Then:
+: $a + x \equiv b + y \pmod z$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod z
+      | c =
+}}
+{{eqn | l = c
+      | o = \equiv
+      | r = d
+      | rr= \pmod z
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = a \bmod z
+      | r = b \bmod z
+      | c = {{Defof|Congruence (Number Theory)|Congruence}}
+}}
+{{eqn | l = x \bmod z
+      | r = y \bmod z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists k_1 \in \Z:
+      | l = a - b
+      | r = k_1 z
+      | c = 
+}}
+{{eqn | lo= \exists k_2 \in \Z:
+      | l = x - y
+      | r = k_2 z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a + x} - \paren {b + y}
+      | r = \paren {k_1 + k_2} z
+      | c = {{Defof|Integer Addition}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + x
+      | o = \equiv
+      | r = b + y
+      | rr= \pmod z
+      | c = {{Defof|Congruence (Number Theory)|Congruence}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication is Well-Defined/Warning}
+Tags: Modulo Multiplication
+
+\begin{theorem}
+Let $z \in \R$ be a [[Definition:Real Number|real number]].
+Let:
+:$a \equiv b \pmod z$
+and:
+:$x \equiv y \pmod z$
+where $a, b, x, y \in \R$.
+Then it does '''not''' necessarily hold that:
+: $a x \equiv b y \pmod z$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | l = x
+      | o = \equiv
+      | r = y
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \bmod m
+      | r = b \bmod m
+      | c = Definition of [[Definition:Congruence (Number Theory)|Congruence]]
+}}
+{{eqn | l = x \bmod m
+      | r = y \bmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists k_1 \in \Z:
+      | l = a
+      | r = b + k_1 z
+      | c = 
+}}
+{{eqn | lo= \exists k_2 \in \Z:
+      | l = x
+      | r = y + k_2 z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a x
+      | r = \left({b + k_1 z}\right) \left({y + k_2 z}\right)
+      | c = Definition of [[Definition:Multiplication|Multiplication]]
+}}
+{{eqn | r = b y + b k_2 z + y k_1 z + k_1 k_2 z^2
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{eqn | r = b y + \left({b k_2 + y k_1 + k_1 k_2 z}\right) z
+      | c = 
+}}
+{{end-eqn}}
+But it is not necessarily the case that:
+:$b k_2 + y k_1 + k_1 k_2 z$
+is an [[Definition:Integer|integer]].
+In fact, $b k_2 + y k_1 + k_1 k_2 z$ can only be ''guaranteed'' to be an [[Definition:Integer|integer]] if each of $b, y, z \in \Z$.
+Hence $a b$ is not necessarily congruent to $x y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence by Product of Moduli/Real Modulus}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, b, z \in \R$.
+Let $a \equiv b \pmod z$ denote that [[Definition:Congruence (Number Theory)|$a$ is congruent to $b$ modulo $z$]].
+Then $\forall y \in \R, y \ne 0$:
+:$a \equiv b \pmod z \iff y a \equiv y b \pmod {y z}$
+\end{theorem}
+
+\begin{proof}
+Let $y \in \R: y \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod z
+      | c =
+}}
+{{eqn | ll= \iff
+      | l = a \bmod z
+      | r = b \bmod z
+      | c = {{Defof|Congruence (Number Theory)|Congruence}}
+}}
+{{eqn | ll= \iff
+      | l = y \left({a \bmod z}\right)
+      | r = y \left({b \bmod z}\right)
+      | c = Left hand implication valid only when $y \ne 0$
+}}
+{{eqn | ll= \iff
+      | l = \left({y a}\right) \bmod \left({y z}\right)
+      | r = \left({y b}\right) \bmod \left({y z}\right)
+      | c = [[Product Distributes over Modulo Operation]]
+}}
+{{eqn | l = y a
+      | o = \equiv
+      | r = y b
+      | rr= \pmod {y z}
+      | c = {{Defof|Congruence (Number Theory)|Congruence}}
+}}
+{{end-eqn}}
+Hence the result.
+Note the invalidity of the third step when $y = 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Segment of Auxiliary Relation is Subset of Lower Closure}
+Tags: Auxiliary Relations, Lower Closures
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $R$ be [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
+Let $x \in S$.
+Then
+:$x^R \subseteq x^\preceq$
+where
+:$x^R$ denotes the [[Definition:Relation Segment|$R$-segment]] of $x$,
+:$x^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in x^R$.
+By definition of [[Definition:Relation Segment|$R$-segment]] of $x$:
+:$\left({a, x}\right) \in R$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$a \preceq x$
+Thus by definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$a \in x^\preceq$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function is Completely Multiplicative/Integers}
+Tags: Completely Multiplicative Functions, Number Theory
+
+\begin{theorem}
+Let $c \in \Z$ be an [[Definition:Integer|integer]].
+Let $f_c: \Z \to \Z$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \Z: \map {f_c} n = n^c$
+Then $f_c$ is [[Definition:Completely Multiplicative Function|completely multiplicative]].
+\end{theorem}
+
+\begin{proof}
+Let $r, s \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f_c} {r s}
+      | r = \paren {r s}^c
+      | c = 
+}}
+{{eqn | r = r^c s^c
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = \map {f_c} r \map {f_c} s
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Preceding implies Inclusion of Segments of Auxiliary Relation}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $R$ be an [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
+Let $x, y \in S$ such that
+:$x \preceq y$
+Then
+:$x^R \subseteq y^R$
+where $x^R$ denotes the [[Definition:Relation Segment|$R$-segment]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $a \in x^R$.
+By definition of [[Definition:Relation Segment|$R$-segment]] of $x$:
+:$\left({a, x}\right) \in R$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$a \preceq a$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({a, y}\right) \in R$
+Thus by definition of [[Definition:Relation Segment|$R$-segment]] of $y$:
+:$a \in y^R$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristic Function of Square-Free Integers is Multiplicative}
+Tags: Multiplicative Functions, Number Theory, Square-Free Integers
+
+\begin{theorem}
+Let $S \subseteq \Z$ be the [[Definition:Set|set]] of [[Definition:Positive Integer|positive integers]] defined as:
+:$S := \set {n \in \Z: \forall k \in \Z_{>1}: k^2 \nmid n}$
+That is, let $S$ be the [[Definition:Set|set]] of all [[Definition:Square-Free Integer|square-free]] [[Definition:Positive Integer|positive integers]].
+Let $\chi_S: \N \to \Z$ denote the [[Definition:Characteristic Function of Set|characteristic function]] of $S$:
+:$\forall n \in \Z: \map {\chi_S} n = \sqbrk {n \in S}$
+where $\sqbrk {n \in S}$ is [[Definition:Iverson's Convention|Iverson's convention]].
+Then $\chi_S$ is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+\end{theorem}
+
+\begin{proof}
+Let $r, s \in \Z$ such that $r \perp s$.
+=== Case 1: Either factor is not [[Definition:Square-Free Integer|square-free]] ===
+Let either $r \notin S$ or $s \notin S$ or both.
+Then either:
+:$\exists k \in \Z_{>1}: k^2 \divides r$
+or:
+:$\exists k \in \Z_{>1}: k^2 \divides s$
+Thus either:
+:$\map {\chi_S} r = 0$
+or:
+:$\map {\chi_S} s = 0$
+and so:
+:$\map {\chi_S} r \, \map {\chi_S} s = 0$
+But then:
+:$\exists k \in \Z_{>1}: k^2 \divides r s$
+and so:
+:$\map {\chi_S} {r s} = 0$
+demonstrating that:
+:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
+{{qed|lemma}}
+=== Case 1: Both factors are [[Definition:Square-Free Integer|square-free]] ===
+Let $r \in S$ and $s \in S$.
+Thus:
+:$\nexists k \in \Z_{>1}: k^2 \divides r$
+and:
+:$\nexists k \in \Z_{>1}: k^2 \divides s$
+Hence:
+:$\map {\chi_S} r = 1$
+and:
+:$\map {\chi_S} s = 1$
+and so:
+:$\map {\chi_S} r \, \map {\chi_S} s = 1$
+Because $r \perp s$:
+:$\nexists k \in \Z_{>1}: k \divides r, k \divides s$
+Hence there can be no $k \in \Z_{>1}$ whose multiplicity in $r s$ is greater than $1$.
+Thus:
+:$\nexists k \in \Z_{>1}: k^2 \divides {r s}$
+and so:
+:$\map {\chi_S} {r s} = 1$
+once more demonstrating that:
+:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
+{{qed|lemma}}
+In both cases:
+:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
+Hence the result, by definition of [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Segment of Auxiliary Relation Mapping is Increasing}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $R = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let ${\it Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
+Let $L = \left({ {\it Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\precsim \mathop = \subseteq\restriction_{ {\it Ids}\left({R}\right) \times {\it Ids}\left({R}\right)}$
+Let $r$ be an [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
+Let $f: S \to {\it Ids}\left({R}\right)$ be a [[Definition:Mapping|mapping]] such that
+:$\forall x \in S: f\left({x}\right) = x^r$
+where $x^r$ denotes the [[Definition:Relation Segment|$r$-segment]] of $x$.
+Then
+:$f$ is [[Definition:Increasing Mapping|increasing mapping]].
+\end{theorem}
+
+\begin{proof}
+$f$ is well-defined because by [[Relation Segment of Auxiliary Relation is Ideal]]:
+:$\forall x \in S: x^r$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$
+Let $x, y \in S$ such that
+:$x \preceq y$
+By [[Preceding implies Inclusion of Segments of Auxiliary Relation]]:
+:$x^r \subseteq y^r$
+Thus by definitions of $\precsim$ and $f$:
+:$f\left({x}\right) \precsim f\left({y}\right)$
+Thus by definition:
+:$f$ is [[Definition:Increasing Mapping|increasing mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constant to Power of Number of Distinct Prime Divisors is Multiplicative Function}
+Tags: Multiplicative Functions, Number Theory
+
+\begin{theorem}
+Let $c \in \R$ be a [[Definition:Constant|constant]].
+Let $f: \N \to \R$ denotes the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \N: f \left({n}\right) = c^k$
+where $k$ is [[Definition:Number|number]] of [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]] that [[Definition:Divisor of Integer|divide]] $n$.
+Then $f$ is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+\end{theorem}
+
+\begin{proof}
+Let $r, s \in \Z$ such that $r \perp s$.
+Let $r$ be composed of $p$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]]:
+:$r_1, r_2, \ldots r_p$
+Thus:
+:$f \left({r}\right) = c^p$
+Let $s$ be composed of $q$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]]:
+:$s_1, s_2, \ldots s_q$
+Thus:
+:$f \left({s}\right) = c^q$
+As $r \perp s$, all the $r_k$ and $s_k$ are [[Definition:Distinct|distinct]].
+Thus $r s$ is composed of:
+:the $p$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]] $r_1, r_2, \ldots r_p$
+and:
+:the $q$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]] $s_1, s_2, \ldots s_q$
+which is a total of $p + q$ [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = f \left({r s}\right)
+      | r = c^{p + q}
+      | c = Definition of $f$
+}}
+{{eqn | r = c^p c^q
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{eqn | r = f \left({r}\right) f \left({s}\right)
+      | c = from above
+}}
+{{end-eqn}}
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Product of Multiplicative Functions is Multiplicative}
+Tags: Multiplicative Functions
+
+\begin{theorem}
+Let $f: \N \to \C$ and $g: \N \to \C$ be [[Definition:Multiplicative Arithmetic Function|multiplicative functions]].
+Then their [[Definition:Pointwise Multiplication of Integer-Valued Functions|pointwise product]]:
+:$f \times g: \Z \to \Z: \forall s \in S: \map {\paren {f \times g} } s := \map f s \times \map g s$
+is also [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+\end{theorem}
+
+\begin{proof}
+Let $f$ and $g$ be [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+Let $m \perp n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {f \times g} {m \times n}
+      | r = \map f {m \times n} \times \map g {m \times n}
+      | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}}
+}}
+{{eqn | r = \map f m \times \map f n \times \map g m \times \map g n
+      | c = {{Defof|Multiplicative Arithmetic Function}}
+}}
+{{eqn | r = \map f m \times \map g m \times \map f n \times \map g n
+      | c = [[Integer Multiplication is Commutative]]
+}}
+{{eqn | r = \paren {\map {f \times g} n} \times \paren {\map {f \times g} n}
+      | c = {{Defof|Pointwise Multiplication of Integer-Valued Functions}}
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $R = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\mathit{Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
+Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$.
+Let $r$ be an [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
+Let $M = \left({F, \preccurlyeq}\right)$ be the [[Definition:Increasing Mappings Satisfying Inclusion in Lower Closure|ordered set of increasing mappings $g$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$]].
+Let $f: S \to \mathit{Ids}\left({R}\right)$ be a [[Definition:Mapping|mapping]] such that:
+:$\forall x \in S: f\left({x}\right) = x^r$
+where $x^r$ denotes the [[Definition:Relation Segment|$r$-segment]] of $x$.
+Then:
+:$(1): \quad f \in F$
+Let $h: S \to \mathit{Ids}\left({R}\right): x \mapsto x^\preceq$
+Then:
+:$(2): \quad h \in F \land f \preccurlyeq h$
+Let $k: S \to \mathit{Ids}\left({R}\right): x \mapsto \left\{ {\bot}\right\}$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
+Then:
+:$(3): \quad k \in F$ and $k \preccurlyeq f$
+\end{theorem}
+
+\begin{proof}
+=== Condition $(1)$ ===
+By [[Segment of Auxiliary Relation Mapping is Increasing]]:
+:$f$ is an [[Definition:Increasing Mapping|increasing mapping]].
+By [[Segment of Auxiliary Relation is Subset of Lower Closure]]:
+:$\forall x \in S: x^r \subseteq x^\preceq$
+By definition of $f$:
+:$\forall x \in S: f\left({x}\right) \subseteq x^\preceq$
+Thus
+:$f \in F$
+{{qed|lemma}}
+=== Condition $(2)$ ===
+By [[Lower Closure is Increasing]]:
+:$\forall x, y \in S: x \preceq y \implies x^\preceq \subseteq y^\preceq$
+By definitions of $h$ and $\precsim$:
+:$\forall x, y \in S: x \preceq y \implies h\left({x}\right) \precsim h\left({y}\right)$
+By definition:
+:$h$ is an [[Definition:Increasing Mapping|increasing mapping]].
+By definition of $h$:
+:$\forall x \in S: h\left({x}\right) \subseteq x^\preceq$
+Thus
+:$h \in F$
+By [[Segment of Auxiliary Relation is Subset of Lower Closure]]:
+:$\forall x \in S: x^r \subseteq x^\preceq$
+By definitions of $f$ and $h$:
+:$\forall x \in S: f\left({x}\right) \subseteq h\left({x}\right)$
+By definition of $\precsim$:
+:$\forall x \in S: f\left({x}\right) \precsim h\left({x}\right)$
+Thus by definition of [[Definition:Increasing Mappings Satisfying Inclusion in Lower Closure|ordering of $M$]]:
+:$f \preccurlyeq h$
+{{qed|lemma}}
+=== Condition $(3)$ ===
+$k$ is well-defined because by [[Singleton of Bottom is Ideal]]:
+:$\left\{ {\bot}\right\}$ is an [[Definition:Ideal in Ordered Set|ideal]] in $R$.
+By [[Mapping is Constant iff Increasing and Decreasing]]:
+:$k$ is an [[Definition:Increasing Mapping|increasing mapping]].
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\forall x \in S: \bot \preceq x$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\forall x \in S: \bot \in x^\preceq$
+By definition of [[Definition:Subset|subset]]:
+:$\forall x \in S: \left\{ {\bot}\right\} \subseteq x^\preceq$
+By definition of $k$:
+:$\forall x \in S: k\left({x}\right) \subseteq x^\preceq$
+Thus
+:$k \in F$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\forall x \in S: \left({\bot, x}\right) \in r$
+By definition of [[Definition:Relation Segment|$r$-segment]]:
+:$\forall x \in S: \bot \in x^r$
+By definition of [[Definition:Subset|subset]]:
+:$\forall x \in S: \left\{ {\bot}\right\} \subseteq x^r$
+By definitions of $k$ and $f$:
+:$\forall x \in S: k\left({x}\right) \subseteq f\left({x}\right)$
+By definition of $\precsim$:
+:$\forall x \in S: k\left({x}\right) \precsim f\left({x}\right)$
+Thus by definition of [[Definition:Increasing Mappings Satisfying Inclusion in Lower Closure|ordering of $M$]]:
+:$k \preccurlyeq f$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton of Bottom is Ideal}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Ordered Set|ordered set]].
+Then
+:$\set \bot$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $S$.
+\end{theorem}
+
+\begin{proof}
+=== Non-empty ===
+By definition of [[Definition:Singleton|singleton]]:
+:$\bot \in \set \bot$
+By definition:
+:$\set \bot$ is a [[Definition:Non-Empty Set|non-empty set]].
+{{qed|lemma}}
+=== Directed ===
+Thus by [[Singleton is Directed and Filtered Subset]]:
+:$\set \bot$ is [[Definition:Directed Subset|directed]].
+{{qed|lemma}}
+=== Lower ===
+Let $x \in \set \bot, y \in S$ such that
+:$y \preceq x$
+By definition of [[Definition:Singleton|singleton]]:
+:$x = \bot$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\bot \preceq y$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$y = \bot$
+Thus by definition of [[Definition:Singleton|singleton]]:
+:$y \in \set \bot$
+Thus by definition:
+:$\set \bot$ is a [[Definition:Lower Set|lower set]].
+{{qed|lemma}}
+Thus by definition:
+:$\set \bot$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Digits in Factorial}
+Tags: Factorials
+
+\begin{theorem}
+Let $n!$ denote the [[Definition:Factorial|factorial]] of $n$.
+The number of [[Definition:Digit|digits]] in $n!$ is approximately:
+:$1 + \left\lfloor{\dfrac 1 2 \left({\log_{10} 2 + \log_{10} \pi}\right) + \dfrac 1 2 \log_{10} n + n \left({\log_{10} n - \log_{10} e}\right)}\right\rfloor$
+when $n!$ is shown in [[Definition:Decimal Notation|decimal notation]].
+This evaluates to:
+:$1 + \left\lfloor{\left({n + \dfrac 1 2}\right) \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934}\right\rfloor$
+\end{theorem}
+
+\begin{proof}
+From [[Stirling's Formula]]:
+:$n! \sim \sqrt {2 \pi n} \left({\dfrac n e}\right)^n$
+from which the result can be calculated.
+To count the number of [[Definition:Digit|digits]]:
+{{begin-eqn}}
+{{eqn | l = \log_{10} n!
+      | o = \sim
+      | r = \log_{10} \left({\sqrt {2 \pi n} \left({\dfrac n e}\right)^n}\right)
+      | c = [[Stirling's Formula]]
+}}
+{{eqn | r = \log_{10} \left({\sqrt {2 \pi n} }\right) + \log_{10} \left({\left({\dfrac n e}\right)^n}\right)
+      | c = [[Sum of Logarithms]]
+}}
+{{eqn | r = \frac 1 2 \log_{10} 2 \pi n + n \log_{10} \left({\dfrac n e}\right)
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | r = \frac 1 2 \left({\log_{10} 2 + \log_{10} \pi}\right) + \frac 1 2 \log_{10} n + n \left({\log_{10} n - \log_{10} e}\right)
+      | c = [[Sum of Logarithms]], [[Difference of Logarithms]]
+}}
+{{end-eqn}}
+We have:
+: [[Common Logarithm of 2|Common Logarithm of $2$]]: $\ln 2 \approx 0.30102 \ 9996$
+: [[Common Logarithm of Pi|Common Logarithm of $\pi$]]: $\ln \pi \approx 0.49714 \ 9873$
+: [[Common Logarithm of e|Common Logarithm of $e$]]: $\ln e \approx 0.43429 \ 4481$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \log_{10} n!
+      | o = \sim
+      | r = \frac 1 2 \left({0.30102 \ 9996 + 0.49714 \ 9873}\right) + \frac 1 2 \log_{10} n + n \left({\log_{10} n - 0.43429 \ 4481}\right)
+      | c = 
+}}
+{{eqn | r = \left({n + \frac 1 2}\right) \log_{10} n - 0.43429 \ 4481 \, n + 0.39908 \ 9934
+      | c = 
+}}
+{{end-eqn}}
+Hence the result from [[Number of Digits in Number]].
+{{qed}}
+[[Category:Factorials]]
+boax76h7gvqf13y3w9tvz95nqxlydm3
+\end{proof}<|endoftext|>
+\section{Rising Factorial in terms of Falling Factorial}
+Tags: Falling Factorials, Rising Factorials
+
+\begin{theorem}
+:$x^{\overline n} = \left({x + n - 1}\right)^{\underline n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\overline n}
+      | r = \prod_{j \mathop = 0}^{n - 1} \left({x + j}\right)
+      | c = Definition of [[Definition:Rising Factorial|Rising Factorial]]
+}}
+{{eqn | r = x \left({x + 1}\right) \cdots \left({x + n - 1}\right)
+      | c = Definition of [[Definition:Product Notation (Algebra)|Product Notation]]
+}}
+{{eqn | r = \left({x + n - 1}\right) \left({x + n - 2}\right) \cdots \left({x + 1}\right) x
+      | c = reversing the order
+}}
+{{eqn | r = \prod_{j \mathop = 0}^{n - 1} \left({\left({x + n - 1}\right) - j}\right)
+      | c = Definition of [[Definition:Falling Factorial|Falling Factorial]]
+}}
+{{eqn | r = \left({x + n - 1}\right)^{\underline n}
+      | c = Definition of [[Definition:Product Notation (Algebra)|Product Notation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rising Factorial in terms of Falling Factorial of Negative}
+Tags: Falling Factorials, Rising Factorials
+
+\begin{theorem}
+:$x^{\overline k} = \paren {-1}^k \paren {-x}^{\underline k}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {-x}^{\underline k}
+      | r = \prod_{j \mathop = 0}^{k - 1} \paren {-x - j}
+      | c = {{Defof|Falling Factorial}}
+}}
+{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \paren {x + j}
+      | c = 
+}}
+{{eqn | r = \prod_{j \mathop = 0}^{k - 1} \paren {-1} \prod_{j \mathop = 0}^{k - 1} \paren {x + j}
+      | c = [[Product of Products]]
+}}
+{{eqn | r = \paren {-1}^k \prod_{j \mathop = 0}^{k - 1} \paren {x + j}
+      | c = 
+}}
+{{eqn | r = \paren {-1}^k x^{\overline k}
+      | c = {{Defof|Rising Factorial}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-1}^k \paren {-x}^{\underline k}
+      | r = \paren {-1}^{2 k} x^{\overline k}
+      | c = multiplying both sides by $\paren {-1}^k$
+}}
+{{eqn | r = x^{\overline k}
+      | c = $-1$ to an even power is $1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Falling Factorial as Quotient of Factorials}
+Tags: Falling Factorials, Factorials
+
+\begin{theorem}
+:$x^{\underline n} = \dfrac {x!} {\paren {x - n}!} = \dfrac {\map \Gamma {x + 1} } {\map \Gamma {x - n + 1} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\underline n}
+      | r = \prod_{j \mathop = 0}^{n - 1} \paren {x - j}
+      | c = {{Defof|Falling Factorial}}
+}}
+{{eqn | r = x \paren {x - 1} \paren {x - 2} \dotsm \paren {x - n + 1}
+      | c = 
+}}
+{{eqn | r = \dfrac {x!} {\paren {x - n}!}
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = \dfrac {\map \Gamma {x + 1} } {\map \Gamma {x - n + 1} }
+      | c = {{Defof|Gamma Function}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One to Integer Rising is Integer Factorial}
+Tags: Rising Factorials, Factorials
+
+\begin{theorem}
+:$1^{\overline n} = n!$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1^{\overline n}
+      | r = \dfrac {\paren {1 + n - 1}!} {\paren {1 - 1}!}
+      | c = [[Rising Factorial as Quotient of Factorials]]
+}}
+{{eqn | r = \dfrac {n!} {0!}
+      | c = 
+}}
+{{eqn | r = n!
+      | c = [[Factorial of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Rising Factorials]]
+[[Category:Factorials]]
+4dcqx0irfqi9tnnvf7ysu9tdczqi1vu
+\end{proof}<|endoftext|>
+\section{Number to Power of One Rising is Itself}
+Tags: Rising Factorials
+
+\begin{theorem}
+:$x^{\overline 1} = x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\overline 1}
+      | r = \dfrac {\Gamma \left({x + 1}\right)}{\Gamma \left({x}\right)}
+      | c = [[Rising Factorial as Quotient of Factorials]]
+}}
+{{eqn | r = x
+      | c = [[Gamma Difference Equation]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Rising Factorials]]
+aeu50xkg5nt93tq2vghn7q97cbnuqvj
+\end{proof}<|endoftext|>
+\section{Integer to Power of Itself Falling is Factorial}
+Tags: Falling Factorials
+
+\begin{theorem}
+:$n^{\underline n} = n!$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = n^{\underline n}
+      | r = \dfrac {n!} {\paren {n - n}!}
+      | c = [[Falling Factorial as Quotient of Factorials]]
+}}
+{{eqn | r = \dfrac {n!} {0!}
+      | c = 
+}}
+{{eqn | r = n!
+      | c = [[Factorial of Zero]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Falling Factorials]]
+opk3w0uaoprxmpgvd9rzvqako48cfx9
+\end{proof}<|endoftext|>
+\section{Number to Power of One Falling is Itself}
+Tags: Falling Factorials
+
+\begin{theorem}
+:$x^{\underline 1} = x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\underline 1}
+      | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x - 1 + 1}\right)}
+      | c = [[Falling Factorial as Quotient of Factorials]]
+}}
+{{eqn | r = \dfrac {\Gamma \left({x + 1}\right)} {\Gamma \left({x}\right)}
+      | c = 
+}}
+{{eqn | r = x
+      | c = [[Gamma Difference Equation]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Falling Factorials]]
+jjmyfhhwutxv5dyo2xhk3s6plb6w4nx
+\end{proof}<|endoftext|>
+\section{Number of Permutations of One Less}
+Tags: Permutation Theory, Number of Permutations of One Less
+
+\begin{theorem}
+:${}^{n - 1} P_n = {}^n P_n$
+where ${}^k P_n$ denotes the number of [[Definition:Permutation (Ordered Selection)|ordered selections of $k$ objects from $n$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = {}^{n - 1} P_n
+      | r = \dfrac {n!} {\paren {n - \paren {n - 1} }!}
+      | c = [[Number of Permutations]]
+}}
+{{eqn | r = \dfrac {n!} {1!}
+      | c = 
+}}
+{{eqn | r = n!
+      | c = 
+}}
+{{eqn | r = {}^n P_n
+      | c = [[Number of Permutations]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = {}^{n - 1} P_n
+      | r = n^{\underline {n - 1} }
+      | c = [[Number of Permutations]]: $n^{\underline {n - 1} }$ denotes [[Definition:Falling Factorial|Falling Factorial]]
+}}
+{{eqn | r = n!
+      | c = [[Integer to Power of Itself Less One Falling is Factorial]]
+}}
+{{eqn | r = {}^n P_n
+      | c = [[Number of Permutations]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $R = \left({S, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathit{Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
+Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$.
+Let $M = \left({F, \preccurlyeq}\right)$ be the [[Definition:Increasing Mappings Satisfying Inclusion in Lower Closure|ordered set of increasing mappings $g:S \to \mathit{Ids}\left({R}\right)$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$]].
+Let $f \in F$.
+Then
+:there exists an [[Definition:Auxiliary Relation|auxiliary relation]] $\mathcal R$ on $S$ such that
+::$\forall x \in S:f\left({x}\right) = x^{\mathcal R}$
+where $x^{\mathcal R}$ denotes the [[Definition:Relation Segment|$\mathcal R$-segment]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Define [[Definition:Relation|relation]] $\mathcal R$ on $S$:
+:$\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$
+We will prove that
+:$(i): \quad \forall x, y \in S: \left({x, y}\right) \in \mathcal R \implies x \preceq y$
+Let $x, y \in S$ such that
+: $\left({x, y}\right) \in \mathcal R$
+By definitions of $\mathcal R$ and $F$:
+:$x \in f\left({y}\right) \subseteq y^\preceq$
+By definition of [[Definition:Subset|subset]];
+:$x \in y^\preceq$
+Thus be definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$x \preceq y$
+{{qed|lemma}}
+We will prove that
+:$(ii): \quad \forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u \implies \left({x, u}\right) \in \mathcal R$
+Let $x, y, z, u \in S$ such that
+: $x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u$
+By definition of $\mathcal R$:
+:$y \in f\left({z}\right)$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$f\left({z}\right) \precsim f\left({u}\right)$
+By definition of $\precsim$:
+:$f\left({z}\right) \subseteq f\left({u}\right)$
+By definition of [[Definition:Subset|subset]]:
+:$y \in f\left({u}\right)$
+By definition of $F$:
+:$f\left({u}\right)$ is an [[Definition:Ideal in Ordered Set|ideal]] in $R$.
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$f\left({u}\right)$ is a [[Definition:Lower Set|lower set]].
+By definition of [[Definition:Lower Set|lower set]]:
+:$x \in f\left({u}\right)$
+Thus by definition of $\mathcal R$:
+:$\left({x, u}\right) \in \mathcal R$
+{{qed|lemma}}
+We will prove that
+:$(iii): \quad \forall x, y, z \in S: \left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \implies \left({x \vee y, z}\right) \in \mathcal R$
+Let $x, y, z \in S$ such that
+: $\left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R$
+By definition of $\mathcal R$:
+:$x \in f\left({z}\right)$ and $y \in f\left({z}\right)$
+By definition of $F$:
+:$f\left({z}\right)$ is an [[Definition:Ideal in Ordered Set|ideal]] in $R$.
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$f\left({z}\right)$ is a [[Definition:Directed Subset|directed]] [[Definition:Lower Set|lower set]].
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists d \in f\left({z}\right): x \preceq d \land y \preceq d$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$x \vee y \preceq d$
+By definition of [[Definition:Lower Set|lower set]]:
+:$x \vee y \in f\left({z}\right)$
+Thus by definition of $\mathcal R$:
+:$\left({x \vee y, z}\right) \in \mathcal R$
+{{qed|lemma}}
+We will prove that
+:$(iv): \quad \forall x \in S: \left({\bot, x}\right) \in \mathcal R$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $R$.
+Let $x \in S$.
+By definition of $F$:
+:$f\left({x}\right)$ is an [[Definition:Ideal in Ordered Set|ideal]] in $R$.
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$f\left({z}\right)$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Lower Set|lower set]].
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists z: z \in f\left({x}\right)$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\bot \preceq z$
+By definition of [[Definition:Lower Set|lower set]]:
+:$\bot \in f\left({x}\right)$
+Thus by definition of $\mathcal R$:
+:$\left({\bot, x}\right) \in \mathcal R$
+{{qed|lemma}}
+By definition:
+:$\mathcal R$ is an [[Definition:Auxiliary Relation|auxiliary relation]] on $S$.
+Thus by definitions of $\mathcal R$ and [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$\forall x \in S: f\left({x}\right) = x^{\mathcal R}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient with Two/Corollary}
+Tags: Examples of Binomial Coefficients, Triangular Numbers
+
+\begin{theorem}
+:$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$
+where $T_n$ is the [[Definition:Triangular Number|$n$th triangular number]].
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+:$\dbinom n 2 = \dfrac {n!} {2! \paren {n - 2}!}$
+The result follows directly from the definition of the [[Definition:Factorial|factorial]]:
+:$2! = 1 \times 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Increasing Mappings Satisfying Inclusion in Lower Closure is Isomorphic to Auxiliary Relations}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $R = \left({S, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathit{Ids}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $R$.
+Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$.
+Let $M = \left({F, \preccurlyeq}\right)$ be the [[Definition:Increasing Mappings Satisfying Inclusion in Lower Closure|ordered set of increasing mappings $g:S \to \mathit{Ids}\left({R}\right)$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$]].
+Let $\mathit{Aux}\left({R}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
+Let $P = \left({ \mathit{Aux}\left({R}\right), \precsim'}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\precsim' \mathop = \subseteq\restriction_{\mathit{Aux}\left({R}\right) \times \mathit{Aux}\left({R}\right)}$.
+Then
+:there exists an [[Definition:Order Isomorphism|order isomorphism]] between $P$ and $M$
+\end{theorem}
+
+\begin{proof}
+By [[Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure]] define $G:\mathit{Aux}\left({R}\right) \to F$
+:$\forall \mathcal R \in \mathit{Aux}\left({R}\right): G\left({\mathcal R}\right) = \left({S \ni x \mapsto x^{\mathcal R} }\right)$
+We will prove by [[Order Isomorphism is Surjective Order Embedding]] that
+:$G$ is an [[Definition:Order Isomorphism|order isomorphism]].
+=== Surjection ===
+By [[Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation]]:
+:$\forall f \in F: \exists \mathcal R \in \mathit{Aux}\left({R}\right): f = \left({S \ni x \mapsto x^{\mathcal R} }\right)$
+By definition of [[Definition:Image of Mapping|image of mapping]]:
+:$\forall f \in F: f \in \operatorname{Im}\left({G}\right)$
+Thus by definition:
+:$G$ is a [[Definition:Surjection|surjection]].
+{{qed|lemma}}
+=== Order Embedding ===
+Let $\mathcal R, \mathcal Q \in \mathit{Aux}\left({R}\right)$
+We will prove that
+:$\mathcal R \precsim' \mathcal Q \implies G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$
+Let
+:$\mathcal R \precsim' \mathcal Q$
+By definition of $\precsim'$:
+:$\mathcal R \subseteq \mathcal Q$
+Let $x \in S$.
+By [[Relation Segment is Increasing]]
+:$x^{\mathcal R} \subseteq x^{\mathcal Q}$
+By [[Relation Segment of Auxiliary Relation is Ideal]]:
+:$x^{\mathcal R} \in \mathit{Ids}\left({R}\right)$ and $x^{\mathcal Q} \in \mathit{Ids}\left({R}\right)$
+By definition of $\precsim$:
+:$x^{\mathcal R} \precsim x^{\mathcal Q}$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$\left({S \ni x \mapsto x^{\mathcal R} }\right) \precsim \left({S \ni x \mapsto x^{\mathcal Q} }\right)$
+Thus by definitions of $G$ and $\preccurlyeq$:
+:$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$
+{{qed|lemma}}
+We will prove that
+:$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right) \implies \mathcal R \precsim' \mathcal Q$
+Let
+:$G\left({\mathcal R}\right) \preccurlyeq G\left({\mathcal Q}\right)$
+By definition of $G$ and $\preccurlyeq$:
+:$\left({S \ni x \mapsto x^{\mathcal R} }\right) \precsim \left({S \ni x \mapsto x^{\mathcal Q} }\right)$
+Let $x, y \in S$ such that
+:$\left({x, y}\right) \in \mathcal R$
+By definition of [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$x \in y^{\mathcal R}$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$y^{\mathcal R} \precsim y^{\mathcal Q}$
+By definition of $\precsim$:
+:$y^{\mathcal R} \subseteq y^{\mathcal Q}$
+By definition of [[Definition:Subset|subset]]:
+:$x \in y^{\mathcal Q}$
+Thus by definition of [[Definition:Relation Segment|$\mathcal Q$-segment]]:
+:$\left({x, y}\right) \in \mathcal Q$
+By definition of [[Definition:Subset|subset]]:
+:$\mathcal R \subseteq \mathcal Q$
+Thus by definition of $\precsim'$:
+:$\mathcal R \precsim' \mathcal Q$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Theorem/Abel's Generalisation}
+Tags: Binomial Coefficients, Binomial Theorem
+
+\begin{theorem}
+:$\displaystyle \left({x + y}\right)^n = \sum_k \binom n k x \left({x - k z}\right)^{k - 1} \left({y + k z}\right)^{n - k}$
+\end{theorem}<|endoftext|>
+\section{Sum over k of r Choose m+k by s Choose n+k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $s \in \R, r \in \Z_{\ge 0}, m, n \in \Z$.
+Then:
+:$\displaystyle \sum_k \binom r {m + k} \binom s {n + k} = \binom {r + s} {r - m + n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_k \binom r {m + k} \binom s {n + k}
+      | r = \sum_k \binom r {r - m - k} \binom s {s - n - k}
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \sum_k \binom r k \binom s {s - n - \left({r - m - k}\right)}
+      | c = [[Change of Index Variable of Summation]]
+}}
+{{eqn | r = \sum_k \binom r k \binom s {r - m - k + n}
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \sum_k \binom r k \binom s {\left({r - m + n}\right) - k}
+      | c = 
+}}
+{{eqn | r = \binom {r + s} {r - m + n}
+      | c = [[Chu-Vandermonde Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r Choose k by s+k Choose n by -1^r-k}
+Tags: Binomial Coefficients, Sum over k of r Choose k by s+k Choose n by -1^r-k
+
+\begin{theorem}
+Let $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
+Then:
+:$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $r$.
+For all $r \in \Z_{>0}$, let $P \left({r}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \binom 0 k \binom {s + k} n \left({-1}\right)^{0 - k}
+      | r = \delta_{0 k} \binom {s + k} n \left({-1}\right)^{0 - k}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \binom s n
+      | c = All terms vanish but for $k = 0$
+}}
+{{eqn | r = \binom s {n - 0}
+      | c = 
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 0$, then it logically follows that $P \left({m + 1}\right)$ is true.
+This is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} = \binom s {n - m}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \binom {m + 1} k \binom {s + k} n \left({-1}\right)^{m + 1 - k} = \binom s {n - \left({m + 1}\right)}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {m + 1} k \binom {s + k} n \left({-1}\right)^{m + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left({\binom m k + \binom m {k - 1} }\right) \binom {s + k} n \left({-1}\right)^{m + 1 - k}
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m + 1 - k} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k}
+      | c = 
+}}
+{{eqn | r = -\sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k}
+      | c = 
+}}
+{{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k} n \left({-1}\right)^{m + 1 - k} 
+      | c = [[Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \left({\binom {s + k - 1} n + \binom {s + k - 1} {n - 1} }\right) \left({-1}\right)^{m + 1 - k}
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = -\binom s {n - m} + \sum_k \binom m {k - 1} \binom {s + k - 1} n \left({-1}\right)^{m - \left({k - 1}\right)}
+      | c = 
+}}
+{{eqn | o = 
+      | r = + \sum_k \binom m {k - 1} \binom {s + k - 1} {n - 1} \left({-1}\right)^{m - \left({k - 1}\right)}
+      | c = 
+}}
+{{eqn | r = -\binom s {n - m} + \sum_k \binom m k \binom {s + k} n \left({-1}\right)^{m - k} + \sum_k \binom m k \binom {s + k} {n - 1} \left({-1}\right)^{m - k}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = -\binom s {n - m} + \binom s {n - m} + \binom s {n - 1 - m}
+      | c = [[Sum over k of r Choose k by s+k Choose n by -1^r-k/Proof 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \binom s {n - \left({m + 1}\right)}
+      | c = simplifying
+}}
+{{end-eqn}}
+So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \sum_k \binom r k \binom {s + k} n \left({-1}\right)^{r - k} = \binom s {n - r}$
+for all $s \in \R, r \in \Z_{\ge 0}, n \in \Z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r-k Choose m by s Choose k-t by -1^k-t}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $s \in \R, r, t, m \in \Z_{\ge 0}$.
+Then:
+:$\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom s {k - t} \paren {-1}^{k - t} = \binom {r - t - s} {r - t - m}$
+\end{theorem}<|endoftext|>
+\section{Sum over k of r-k Choose m by s+k Choose n}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $m, n, r, s \in \Z_{\ge 0}$ such that $n \ge s$.
+Then:
+:$\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n = \binom {r + s + 1} {m + n + 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom {-\left({n + 1}\right)} {s + k - m} \left({-1}\right)^{r - m + s - m}
+      | c = [[Moving Top Index to Bottom in Binomial Coefficient]]
+}}
+{{eqn | r = \binom {-\left({m + 1}\right) - \left({n + 1}\right)} {r - m + s - m} \left({-1}\right)^{r - m + s - m}
+      | c = [[Chu-Vandermonde Identity]]
+}}
+{{eqn | r = \binom {r + s + 1} {m + n + 1}
+      | c = [[Moving Top Index to Bottom in Binomial Coefficient]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk}
+Tags: Binomial Coefficients, Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
+
+\begin{theorem}
+Let $r, s, t \in \R, n \in \Z$.
+Then:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+\end{theorem}
+
+\begin{proof}
+For all $n \in \Z$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+Let the {{LHS}} of this equation be denoted $\tuple {r, s, t, n}$.
+Let $n = 0$.
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \tuple {r, s, t, 0}
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + t k} {- k} \frac r {r - t k} \delta_{k 0}
+      | c = [[N Choose Negative Number is Zero]]: $\dbinom {s + t k} {- k} = 0$ for all $k > 0$
+}}
+{{eqn | r = \binom r 0 \binom s 0
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = 1
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = \binom {r + s} 0
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = \binom {r + s - t n} n
+      | c = putting $n = 0$
+}}
+{{end-eqn}}
+Thus $\map P 0$ holds.
+Let $n < 0$.
+Let $n = -m$ where $m > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \tuple {r, s, t, n}
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {-m - k} } {-m - k} \frac r {r - t k}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = [[N Choose Negative Number is Zero]]: $\dbinom {s - t \paren {-m - k} } {-m - k} = 0$
+}}
+{{eqn | r = \binom {r + s + t m} {-m}
+      | c = [[N Choose Negative Number is Zero]]: $\dbinom {r + s + t m} {-m} = 0$ for all $m$
+}}
+{{end-eqn}}
+Thus $\map P n$ holds for all $n < 0$.
+It remains to demonstrate that $\map P n$ holds for all $n > 0$.
+The proof continues by [[Second Principle of Mathematical Induction|strong induction]] on $n$.
+The procedure is to substitute $n - r + n t + m$ for the variable $s$ and establish that the identity holds for all $m \ge 0$.
+For all $m \in \Z_{\ge 0}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {n - r + n t + m} - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + \paren {n - r + n t + m} - t n} n$
+That is:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$
+=== [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction|Basis for the Induction]] ===
+Consider the special case where $s = n - 1 - r + n t$.
+{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction}}{{qed|lemma}}
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $0 \le j \le m$, then it logically follows that $\map P {m + 1}$ is true.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + t k} {n - k} \frac r {r - t k} = \binom {n + m} n$
+from which it is to be shown that:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - r + m + 1 + t k} {n - k} \frac r {r - t k} = \binom {n + m + 1} n$
+=== [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma|Lemma]] ===
+First a [[Definition:Lemma|lemma]]:
+{{:Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma}}
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+We have that $\tuple {r, n - 1 - r + n t, t, n}$ holds.
+We have also determined that if:
+:$\tuple {r, s, t, n}$ holds
+and:
+:$\tuple {r, s - t, t, n - 1}$ holds
+then:
+:$\tuple {r, s + 1, t, n}$ holds.
+So $\map P m \implies \map P {m + 1}$.
+Thus $\tuple {r, s, t, n}$ is shown to hold for [[Definition:Infinite Set|infinitely many]] $s$.
+As both the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $s$ it follows that $\tuple {r, s, t, n}$ holds for all $s$.
+Therefore:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+for all $r, s, t \in \R, n \in \Z$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
+:$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$
+where:
+: $A_n \left({x, t}\right)$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree $n$]] defined as:
+::$A_n \left({x, t}\right) := \dbinom {x - n t} n \dfrac x {x - n t}$
+:for $x \ne n t$
+:$z = x^{t + 1} - x^t$.
+From [[Sum over k of r-kt choose k by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $z^k$]]:
+:$\displaystyle \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\left({t + 1}\right)x - t}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k \dbinom {s - t k} k z^k
+      | r = \frac {x^r x^{s + 1} } {\left({t + 1}\right)x - t}
+      | c = 
+}}
+{{eqn | r = \frac {x^{\left({r + s}\right) + 1} } {\left({t + 1}\right)x - t}
+      | c = 
+}}
+{{eqn | r = \sum_k \dbinom {r + s - t k} k z^k
+      | c = 
+}}
+{{end-eqn}}
+Equating coefficients of $z$:
+{{begin-eqn}}
+{{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k \dbinom {s - \left({n - k}\right) t} k z^{n - k}
+      | r = \binom {r + s - n t} n z^n
+      | c = 
+}}
+{{eqn | l = \sum_k \dbinom {r - k t} k \dfrac r {r - k t} z^k \dbinom {s - \left({n - k}\right) t} k z^{n - k}
+      | r = \binom {r + s - n t} n z^n
+      | c = 
+}}
+{{eqn | l = \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} k \dfrac r {r - k t}
+      | r = \binom {r + s - t n} n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r Choose k by s Choose k by k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+:$\displaystyle \sum_k \binom r k \binom s k k = \binom {r + s - 1} {r - 1} s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_k \binom r k \binom s k k
+      | r = \sum_k \binom r k \binom {s - 1} {k - 1} s
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = s \sum_k \binom r k \binom {s - 1} {k - 1}
+      | c = as $s$ is [[Definition:Constant|constant]]
+}}
+{{eqn | r = s \binom {r + s - 1} {r - 1}
+      | c = [[Sum over k of r Choose m+k by s Choose n+k|Sum over $k$ of $\dbinom r {m + k} \dbinom s {n + k}$]]: $s \gets s - 1$, $m \gets 0$, $n \gets -1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Segment is Increasing}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\RR, \QQ$ be [[Definition:Relation|relations]] on $S$ such that
+:$\RR \subseteq \QQ$
+Let $x \in S$.
+Then
+:$x^\RR \subseteq x^\QQ$
+where $x^\RR$ denotes the [[Definition:Relation Segment|$\RR$-segment]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in x^\RR$.
+By definition of [[Definition:Relation Segment|$\RR$-segment]]:
+:$\tuple {y, x} \in \RR$
+By definition of [[Definition:Subset|subset]]:
+:$\tuple {y, x} \in \QQ$
+Thus by definition of [[Definition:Relation Segment|$\QQ$-segment]]:
+:$y \in x^\QQ$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1}
+Tags: Binomial Coefficients, Sum over k of n+k Choose 2 k by 2 k Choose k by -1^k over k+1
+
+\begin{theorem}
+:$\displaystyle \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\paren {-1}^k} {k + 1} = \sqbrk {n = 0}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | r = \sum_k \binom {n + k} k \binom {n + 1} {k + 1} \frac {\left({-1}\right)^k} {n + 1}
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = \frac 1 {n + 1} \sum_{k \mathop \ge 0} \binom {n + k} n \binom {n + 1} {k + 1} \left({-1}\right)^k
+      | c = [[Symmetry Rule for Binomial Coefficients]] and rearranging
+}}
+{{eqn | r = -\frac 1 {n + 1} \sum_{k \mathop \ge 1} \binom {n - 1 + k} n \binom {n + 1} k \left({-1}\right)^k
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = -\frac 1 {n + 1} \sum_{k \mathop \ge 0} \binom {n - 1 + k} n \binom {n + 1} k \left({-1}\right)^k + \frac 1 {n + 1} \binom {n - 1} n
+      | c = separating out the case where $k = 0$
+}}
+{{eqn | r = -\frac 1 {n + 1} \left({-1}\right)^{n + 1} \binom {n - 1} {-1} + \frac 1 {n + 1} \binom {n - 1} n
+      | c = [[Sum over k of r Choose k by s+k Choose n by -1^r-k|Sum over $k$ of $\dbinom r k \dbinom {s + k} n (-1)^{r - k}$]]
+}}
+{{eqn | r = \frac 1 {n + 1} \binom {n - 1} n
+      | c = as $\dbinom {n - 1} {-1} = 0$
+}}
+{{eqn | r = \left[{n = 0}\right]
+      | c = as $\dbinom {n - 1} n = 1 \iff n = 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | r = \sum_k \binom {n + k} {2 k} \binom {2 k} k \frac {\left({-1}\right)^k} {k + 1}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \binom {n + k} k \binom n k \frac {\left({-1}\right)} {k + 1}
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | r = \sum_k \binom {n + k} k \binom {n + 1} {k + 1} \frac {\left({-1}\right)^k} {n + 1}
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = \frac 1 {n + 1} \sum_k \binom {- \left({n + 1}\right)} k \binom {n + 1} {k + 1}
+      | c = [[Negated Upper Index of Binomial Coefficient]]
+}}
+{{eqn | r = \frac 1 {n + 1} \binom {n + 1 - \left({n + 1}\right)} {n + 1 - 1 + 0}
+      | c = [[Sum over k of r Choose m+k by s Choose n+k|Sum over $k$ of $\dbinom r {m + k} \dbinom s {n + k}$]]
+}}
+{{eqn | r = \frac 1 {n + 1} \binom 0 n
+      | c = 
+}}
+{{eqn | r = \left[{n = 0}\right]
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x \in S$.
+Then
+:$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$
+where $\mathit{Ids}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+\end{theorem}
+
+\begin{proof}
+By [[Supremum of Lower Closure of Element]]:
+:$\sup \left({x^\preceq}\right) = x$
+By [[Lower Closure of Element is Ideal]]:
+:$x^\preceq \in \mathit{Ids}$
+Then by definition of [[Definition:Reflexivity|reflexivity]]:
+:$x^\preceq \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+We will prove that
+:$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} \subseteq x^\ll$
+Let $z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+By definition of [[Definition:Intersection of Set of Sets|intersection]]:
+:$\forall I \in \mathit{Ids}: x \preceq \sup I \implies z \in I$
+By [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$z \ll x$
+Thus by definition of [[Definition:Way Below Closure|way below closure]]:
+:$z \in x^\ll$
+{{qed|lemma}}
+It by definition of [[Definition:Set Equality|set equality]] remains to prove opposite inclusion.
+Let $z \in x^\ll$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$z \ll x$
+We will prove that
+:$\forall Y \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}: z \in Y$
+Let $Y \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+Then
+:$Y \in \mathit{Ids}$ and $x \preceq \sup Y$
+Thus [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$z \in Y$
+{{qed|lemma}}
+Thus by definition of [[Definition:Intersection of Set of Sets|intersection]]:
+:$z \in \displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Closure is Ideal in Bounded Below Join Semilattice}
+Tags: Join and Meet Semilattices, Way Below Relation
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $x \in S$.
+Then
+:$x^\ll$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+\end{theorem}
+
+\begin{proof}
+By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
+:$x^\ll$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Directed Subset|directed set]].
+Let $y \in x^\ll, z \in S$ such that
+:$z \preceq y$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$
+By [[Preceding and Way Below implies Way Below]]:
+:$z \ll x$
+Thus by definition of [[Definition:Way Below Closure|way below closure]]:
+:$z \in x^\ll$
+Thus by definition
+:$x^\ll$ is a [[Definition:Lower Set|lower set]].
+Thus by definition
+:$x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t}
+Tags: Binomial Coefficients, Sum over k of r-kt Choose k by r over r-kt by s-(n-k)t Choose n-k by s over s-(n-k)t
+
+\begin{theorem}
+For $n \in \Z_{\ge 0}$:
+:$\displaystyle \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right) = A_n \left({r + s, t}\right)$
+where $A_n \left({x, t}\right)$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n$ defined as:
+:$A_n \left({x, t}\right) = \dbinom {x - n t} n \dfrac x {x - n t}$
+where $x \ne n t$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\displaystyle S = \sum_k A_k \left({r, t}\right) A_{n - k} \left({s, t}\right)$
+Both sides of the statement of the theorem are [[Definition:Polynomial over Real Numbers|polynomials]] in $r$, $s$ and $t$.
+Therefore it can be assumed that $r \ne k t \ne s$ for $0 \le k \le n$ or something will become undefined.
+By replacing the [[Definition:Polynomial over Real Numbers|polynomials]] $A_n$ with their [[Definition:Binomial Coefficient|binomial coefficient]] definitions, the theorem can be expressed as:
+:$\displaystyle S = \sum_k \dbinom {r - k t} k  \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} \dfrac s {s - \left({n - k}\right) t}$
+Using the technique of [[Definition:Partial Fractions|partial fractions]]:
+:$\dfrac 1 {r - k t} \dfrac 1 {s - \left({n - k}\right) t} = \dfrac 1 {r + s - n t} \left({\dfrac 1 {r - k t} + \dfrac 1 {s - \left({n - k}\right) t} }\right)$
+Thus:
+:$\displaystyle S = \frac s {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac r {r - k t} + \frac r {r + s - n t} \sum_k \dbinom {r - k t} k \dbinom {s - \left({n - k}\right) t} {n - k} \dfrac s {s - \left({n - k}\right) t}$
+From [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk|Sum over $k$ of $\dbinom {r - t k} k \dbinom {s - t \left({n - k}\right)} {n - k} \dfrac r {r - t k}$]]:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+for $r, s, t \in \R, n \in \Z$.
+Thus we have:
+:$S = \dfrac s {r + s - n t} \dbinom {r + s - t n} n + \dfrac r {r + s - n t} \dbinom {s + r - t n} n$
+after changing $k$ to $n - k$ in the second term.
+That is:
+:$S = \dbinom {r + s - n t} n \dfrac {r + s} {r + s - n t}$
+which is $A_n \left({r + s, t}\right)$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
+:$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$
+and:
+:$\displaystyle \sum_k A_k \left({s, t}\right) z^k = x^s$
+Hence:
+{{begin-eqn}}
+{{eqn | l = x^{r + s}
+      | r = \sum_k A_k \left({r, t}\right) z^k \sum_k A_k \left({s, t}\right) z^k
+      | c = 
+}}
+{{eqn | r = \sum_k A_k \left({r + s, t}\right) z^k
+      | c = 
+}}
+{{end-eqn}}
+Taking the $z^n$ coefficient:
+{{begin-eqn}}
+{{eqn | l = \sum_k A_k \left({r, t}\right) z^k \sum_k A_{n - k} \left({s, t}\right) z^{n - k}
+      | r = A_n \left({r + s, t}\right) z^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k A_k \left({r, t}\right) \sum_k A_{n - k} \left({s, t}\right)
+      | r = A_n \left({r + s, t}\right)
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r Choose k by -1^r-k by Polynomial}
+Tags: Binomial Coefficients, Sum over k of r Choose k by -1^r-k by Polynomial
+
+\begin{theorem}
+Let $r \in \Z_{\ge 0}$.
+Then:
+:$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
+where:
+:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ of [[Definition:Degree of Polynomial|degree]] $r$.
+\end{theorem}
+
+\begin{proof}
+From the [[Sum over k of r Choose k by s+k Choose n by -1^r-k/Corollary|corollary to Sum over $k$ of $\dbinom r k \dbinom {s + k} n \paren {-1}^{r - k}$]]:
+:$\displaystyle \sum_k \binom r k \binom k n \paren {-1}^{r - k} = \delta_{n r}$
+where $\delta_{n r}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Thus when $n \ne r$:
+:$\displaystyle \sum_k \binom r k \binom k n \paren {-1}^{r - k} = 0$
+and so:
+:$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} \paren {c_0 \binom k 0 + c_1 \binom k 1 + \cdots + c_m \binom k m} = c_r$
+as the only term left standing is the $r$th one.
+Choosing the [[Definition:Polynomial Coefficient|coefficients]] $c_i$ as appropriate, a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ can be expressed as a [[Definition:Summation|summation]] of [[Definition:Binomial Coefficient|binomial coefficients]] in the form:
+:$c_0 \dbinom k 0 + c_1 \dbinom k 1 + \cdots + c_m \dbinom k m$
+Thus we can rewrite such a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ as:
+:$b_0 + b_1 k + \cdots + b_r k^r$
+Since each $c_m \dbinom k m$ is a [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $m$, it follows that the only one with a non-zero [[Definition:Degree of Polynomial|degree]] $r$ term is $c_r \dbinom k r$.
+The coefficient of $k^r$ in $c_r \dbinom k r$ must be equal to $b_r$, that is:
+:$b_r = \dfrac {c_r}{r!}$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Summation of Powers over Product of Differences]]:
+:$\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
+Now we have:
+{{begin-eqn}}
+{{eqn | l = \frac {\left({-1}\right)^k \dbinom n k} {n!}
+      | r = \frac {\left({-1}\right)^k } {k! \left({n - k}\right)!}
+      | c = 
+}}
+{{eqn | r = \frac {\left({-1}\right)^k } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({k - j}\right)}
+      | c = 
+}}
+{{end-eqn}}
+{{proof wanted|in progress}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r Choose k by s-kt Choose r by -1^k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $r \in \Z_{\ge 0}$.
+Then:
+:$\displaystyle \sum_k \binom r k \binom {s - k t} r \left({-1}\right)^k = t^r$
+where $\dbinom r k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]].
+\end{theorem}
+
+\begin{proof}
+From [[Sum over k of r Choose k by -1^r-k by Polynomial|Sum over $k$ of $\dbinom r k \left({-1}\right)^k$ by Polynomial]]:
+:$\displaystyle \sum_k \binom r k \left({-1}\right)^{r - k} P_r \left({k}\right) = r! \, b_r$
+where:
+:$P_r \left({k}\right) = b_0 + b_1 k + \cdots + b_r k^r$ is a [[Definition:Polynomial over Real Numbers|polynomial]] in $k$ of [[Definition:Degree of Polynomial|degree]] $r$.
+{{proof wanted}}
+\end{proof}<|endoftext|>
+\section{Sum over k of -2 Choose k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 0}^n \binom {-2} k = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$
+where:
+:$\dbinom {-2} k$ is a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$\left\lceil {x}\right\rceil$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^n \binom {-2} k
+      | r = \sum_{k \mathop = 0}^n \left({-1}\right) \binom {k - \left({-2}\right) - 1} k
+      | c = [[Negated Upper Index of Binomial Coefficient]]
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \left({-1}\right) \binom {k + 1} k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n \left({-1}\right) \left({k + 1}\right)
+      | c = [[Binomial Coefficient with Self minus One]]
+}}
+{{eqn | r = 1 - 2 + 3 - 4 + \cdots \pm \left({n + 1}\right)
+      | c = 
+}}
+{{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + m}\right)
+      | c = where $m = n$ or $m = n + 1$ according to whether $n$ is [[Definition:Odd Integer|odd]] or [[Definition:Even Integer|even]]
+}}
+{{end-eqn}}
+When $n$ is [[Definition:Even Integer|even]], we have:
+{{begin-eqn}}
+{{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + n}\right)
+      | o = 
+      | c = 
+}}
+{{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 4 \left({1 + 2 + 3 + 4 + \cdots + \frac n 2}\right)
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac n 2} k
+      | c = 
+}}
+{{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - 4 \frac {\frac n 2 \left({\frac n 2 + 1}\right)} 2
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - \frac {n \left({n + 2}\right)} 2
+      | c = simplifying
+}}
+{{eqn | r = \frac {\left({n + 2}\right) \left({n + 1 - n}\right)} 2
+      | c = simplifying
+}}
+{{eqn | r = \frac {n + 2} 2
+      | c = simplifying
+}}
+{{end-eqn}}
+As $n$ is [[Definition:Even Integer|even]], $n + 1$ is [[Definition:Odd Integer|odd]], and so:
+:$\dfrac {n + 2} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$
+{{qed|lemma}}
+When $n$ is [[Definition:Odd Integer|odd]], we have:
+{{begin-eqn}}
+{{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 2 \times \left({2 + 4 + 6 + 8 + \cdots + n + 1}\right)
+      | o = 
+      | c = 
+}}
+{{eqn | r = \left({1 + 2 + 3 + 4 + \cdots + \left({n + 1}\right)}\right) - 4 \left({1 + 2 + 3 + 4 + \cdots + \frac {n + 1} 2}\right)
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 2}^{n + 1} k - 4 \sum_{k \mathop = 1}^{\frac {n + 1} 2} k
+      | c = 
+}}
+{{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - 4 \frac {\frac {n + 1} 2 \left({\frac {n + 1} 2 + 1}\right)} 2
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = \frac {\left({n + 1}\right) \left({n + 2}\right)} 2 - \frac {\left({n + 1}\right) \left({n + 3}\right)} 2
+      | c = simplifying
+}}
+{{eqn | r = \frac {\left({n + 1}\right) \left({\left({n + 2}\right) - \left({n + 3}\right)}\right)} 2
+      | c = simplifying
+}}
+{{eqn | r = \left({-1}\right) \frac {n + 1} 2
+      | c = simplifying
+}}
+{{end-eqn}}
+As $n$ is [[Definition:Odd Integer|odd]], $n + 1$ is [[Definition:Even Integer|even]], and so $\dfrac {n + 1} 2$ is an [[Definition:Integer|integer]].
+Thus from [[Real Number is Integer iff equals Ceiling]]:
+:$\left({-1}\right) \dfrac {n + 1} 2 = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$
+{{qed|lemma}}
+Thus:
+:$\displaystyle \sum_{k \mathop = 0}^n \binom {-2} k = \left({-1}\right)^n \left\lceil {\dfrac {n + 1} 2}\right\rceil$
+whether $n$ is [[Definition:Odd Integer|odd]] or [[Definition:Even Integer|even]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of m Choose k by k minus m over 2}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 0}^n \binom m k \left({k - \dfrac m 2}\right) = -\dfrac m 2 \binom {m - 1} n$
+where $\dbinom m k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^n \binom m k \left({k - \dfrac m 2}\right)
+      | r = \sum_{k \mathop = 0}^n k \binom m k - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 0}^n m \binom {m - 1} {k - 1} - \dfrac m 2 \sum_{k \mathop = 0}^n \binom m k
+      | c = [[Factors of Binomial Coefficient]]
+}}
+{{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({2 \binom {m - 1} {k - 1} - \binom m k}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({2 \binom {m - 1} {k - 1} - \left({\binom {m - 1} {k - 1} + \binom {m - 1} k}\right)}\right)
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \dfrac m 2 \sum_{k \mathop = 0}^n \left({\binom {m - 1} {k - 1} - \binom {m - 1} k}\right)
+      | c = simplifying
+}}
+{{eqn | r = \dfrac m 2 \left({\binom {m - 1} {-1} - \binom {m - 1} n}\right)
+      | c = as the series [[Definition:Telescoping Series|telescopes]]
+}}
+{{eqn | r = -\dfrac m 2 \binom {m - 1} n
+      | c = {{Defof|Binomial Coefficient}} for negative lower index
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient is instance of Gaussian Binomial Coefficient}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $\dbinom r m_q$ denote the [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
+Then:
+:$\displaystyle \lim_{q \mathop \to 1^-} \dbinom r m_q = \dbinom r m$
+where $\dbinom r m$ denotes the conventional [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+We have by definition of [[Definition:Gaussian Binomial Coefficient|Gaussian binomial coefficient]]:
+:$\displaystyle \dbinom r m_q = \prod_{k \mathop = 0}^{m - 1} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }$
+Consider a typical [[Definition:Factor (Algebra)|factor]] of this [[Definition:Product Notation (Algebra)|product]]:
+{{begin-eqn}}
+{{eqn | l = \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }
+      | r = \dfrac {\paren {1 - q^{r - k} } / \paren {1 - q} } {\paren {1 - q^{k + 1} } / \paren {1 - q} }
+      | c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $1 - q$
+}}
+{{eqn | r = \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{q \mathop \to 1^-} \dfrac {1 - q^{r - k} } {1 - q^{k + 1} }
+      | r = \lim_{q \mathop \to 1^-} \dfrac {\sum_{j \mathop = 0}^{r - k - 1} q^j} {\sum_{j \mathop = 0}^k q^j}
+      | c = 
+}}
+{{eqn | r = \dfrac {\sum_{j \mathop = 0}^{r - k - 1} 1} {\sum_{j \mathop = 0}^k 1}
+      | c = 
+}}
+{{eqn | r = \dfrac {r - k} {k + 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \lim_{q \mathop \to 1^-} \dbinom r k_q
+      | r = \prod_{k \mathop = 0}^{m - 1} \dfrac {r - k} {k + 1}
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = 1}^m \dfrac {r - k - 1} k
+      | c = [[Translation of Index Variable of Product]]
+}}
+{{eqn | r = \dbinom r m
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multinomial Coefficient expressed as Product of Binomial Coefficients}
+Tags: Binomial Coefficients, Multinomial Coefficients
+
+\begin{theorem}
+:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
+where:
+:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m}$ denotes a [[Definition:Multinomial Coefficient|multinomial coefficient]]
+:$\dbinom {k_1 + k_2} {k_1}$ etc. denotes [[Definition:Binomial Coefficient|binomial coefficients]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $m \in \Z_{> 1}$, let $P \left({m}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \dbinom {k_1 + k_2} {k_1, k_2}
+      | r = \frac {\left({k_1 + k_2}\right)!} {k_1! \, k_2!}
+      | c = Definition of [[Definition:Multinomial Coefficient|Multinomial Coefficient]]
+}}
+{{eqn | r = \frac {\left({k_1 + k_2}\right)!} {k_1! \, \left({\left({k_1 + k_2}\right) - k_1}\right)!}
+      | c = 
+}}
+{{eqn | r = \dbinom {k_1 + k_2} {k_1}
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{end-eqn}}
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r} {k_1 + k_2 + \cdots + k_{r - 1} }$
+from which it is to be shown that:
+:$\dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1, k_2, \ldots, k_r, k_{r + 1} } = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1, k_2, \ldots, k_r, k_{r + 1} }
+      | o = 
+}}
+{{eqn | r = \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {k_1! \, k_2! \, \cdots k_r! \, k_{r + 1}!}
+      | c =Definition of [[Definition:Multinomial Coefficient|Multinomial Coefficient]]
+}}
+{{eqn | r = \frac {\left({k_1 + k_2 + \cdots + k_r}\right)!} {k_1! \, k_2! \, \cdots k_r!} \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {\left({k_1 + k_2 + \cdots + k_r}\right)! k_{r + 1} }
+      | c =  multiplying [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!$
+}}
+{{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} \frac {\left({k_1 + k_2 + \cdots + k_r + k_{r + 1} }\right)!} {\left({k_1 + k_2 + \cdots + k_r}\right)! k_{r + 1} }
+      | c = Definition of [[Definition:Multinomial Coefficient|Multinomial Coefficient]]
+}}
+{{eqn | r = \dbinom {k_1 + k_2 + \cdots + k_r} {k_1, k_2, \ldots, k_r} \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{eqn | r = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r} {k_1 + k_2 + \cdots + k_{r - 1} } \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}
+      | c = [[Multinomial Coefficient expressed as Product of Binomial Coefficients#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_r + k_{r + 1} } {k_1 + k_2 + \cdots + k_r}
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m \in \Z_{>1}: \dbinom {k_1 + k_2 + \cdots + k_m} {k_1, k_2, \ldots, k_m} = \dbinom {k_1 + k_2} {k_1} \dbinom {k_1 + k_2 + k_3} {k_1 + k_2} \cdots \dbinom {k_1 + k_2 + \cdots + k_m} {k_1 + k_2 + \cdots + k_{m - 1} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Semantic Equivalence for Predicate Logic}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\mathbf A, \mathbf B$ be [[Definition:WFF of Predicate Logic|WFFs of predicate logic]].
+{{TFAE|def = Semantic Equivalence (Predicate Logic)|semantic equivalence}}
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\sigma$ be an [[Definition:Assignment for Formula|assignment]] for $\mathbf A \iff \mathbf B$ in $\mathcal A$.
+Then the [[Definition:Value of Formula under Assignment|value of $\mathbf A \iff \mathbf B$ under $\sigma$]] is given by:
+:$f^\leftrightarrow \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] }\right)$
+and from the [[Definition:Biconditional/Truth Function|definition of $f^\leftrightarrow$]] we see that $\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf A \iff \mathbf B$ [[Definition:Iff|iff]]:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right]$
+Because the possible values are just $T$ and $F$, this is equivalent to:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = T$ [[Definition:Iff|iff]] $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] = T$
+which by definition of $\mathrm{PL_A}$-[[Definition:Model of Formula (Predicate Logic)|model]] amounts to:
+:$\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf A$ [[Definition:Iff|iff]] $\mathcal A, \sigma \models_{\mathrm{PL_A}} \mathbf B$
+Because $\mathcal A$ and $\sigma$ were arbitrary, the above equivalence holds for all such $\mathcal A$ and $\sigma$.
+The result follows by definition of [[Definition:Tautology (Predicate Logic)|tautology]].
+{{qed}}
+[[Category:Predicate Logic]]
+c1t9ipnxtcw46n3dqnc3gnf56vucb8l
+\end{proof}<|endoftext|>
+\section{Universal Closures are Semantically Equivalent}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+Let $\mathbf B, \mathbf B'$ be [[Definition:Universal Closure of Well-Formed Formula|universal closures]] of $\mathbf A$.
+Then $\mathbf B$ and $\mathbf B'$ are [[Definition:Semantic Equivalence (Predicate Logic)|semantically equivalent]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\mathbf B$ be any [[Definition:Universal Closure of Well-Formed Formula|universal closure]] of $\mathbf A$.
+Then $\mathbf B$ is a [[Definition:Sentence|sentence]] of the form:
+:$\forall x_1: \cdots \forall x_n: \mathbf A$
+By definition of the [[Definition:Model of Sentence (Predicate Logic)|models relation]]:
+:$\mathcal A \models_{\mathrm{PL}} \mathbf B$ [[Definition:Iff|iff]] $\operatorname{val}_{\mathcal A} \left({\mathbf B}\right) = T$ 
+Hence, recursively applying the [[Definition:Value of Formula under Assignment|definition of $\mathop{ \operatorname{val}_{\mathcal A} \left({\cdot}\right) } \left[{\sigma}\right]$]], we see:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\varnothing}\right] = T$ [[Definition:Iff|iff]] $\forall a_1, \ldots, a_n \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right] = T$
+where $\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n}$ denotes the iterated [[Definition:Extension of Assignment|extension of an assignment]].
+By [[Value of Formula under Assignment Determined by Free Variables]] $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right]$ only depends on the $a_i$ for the [[Definition:Free Variable|free variables]] $x_i$ in $\mathbf A$.
+Because we check all possible $a_i \in A$ and all [[Definition:Free Variable|free variables]] $x_i$ in $\mathbf A$ are quantified over in $\mathbf B$, it follows that:
+:$\forall a_1, \ldots, a_n \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_n} {a_n} }\right] = T$
+[[Definition:Iff|iff]]:
+:$\forall a_1, \ldots, a_k \in A: \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\frac{x_1} {a_1} + \ldots + \frac{x_k} {a_k} }\right] = T$
+where $x_1, \ldots, x_k$ are the [[Definition:Free Variable|free variables]] of $\mathbf A$.
+But this last condition does not depend on $\mathbf B$ beyond that it be a [[Definition:Universal Closure of Well-Formed Formula|universal closure]].
+Hence, for any two [[Definition:Universal Closure of Well-Formed Formula|universal closures]] $\mathbf B, \mathbf B'$ of $\mathbf A$:
+:$\mathcal A \models_{\mathrm{PL}} \mathbf B$ [[Definition:Iff|iff]] $\mathcal A \models_{\mathrm{PL}} \mathbf B$
+The result follows by definition of [[Definition:Semantic Equivalence (Predicate Logic)|semantic equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Then
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]
+{{iff}}
+:$\forall x \in S$, [[Definition:Directed Subset|directed subset]] $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $L$ be [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Let $x$ be an [[Definition:Element|element]] of $S$, $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$x \preceq \sup D$
+Thus
+{{begin-eqn}}
+{{eqn | l = x
+      | r = x \wedge \sup D
+      | c = [[Preceding iff Meet equals Less Operand]]
+}}
+{{eqn | r = \sup \set {x \wedge d: d \in D}
+      | c = {{Defof|Meet-Continuous Lattice}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Let:
+:$\forall x \in S$, [[Definition:Directed Subset|directed subset]] $D$ of $S: x \preceq \sup D \implies x = \sup \set {x \wedge d: d \in D}$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\forall x \in S$, [[Definition:Directed Subset|directed subset]] $D$ of $S: x \preceq \sup D \implies x \preceq \sup \set {x \wedge d: d \in D}$
+By [[Meet is Directed Suprema Preserving]]:
+:$\wedge$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]] as a [[Definition:Mapping|mapping]] from [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] $\struct {S \times S, \precsim}$ of $L$ and $L$ into $L$.
+Thus by [[Meet-Continuous iff Meet Preserves Directed Suprema]]
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Lattice is Meet-Continuous}
+Tags: Meet-Continuous Lattices, Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Then $L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$, $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$x \preceq \sup D$
+By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ is [[Definition:Up-Complete|up-complete]] and satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$x^\ll$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Lower Closure of Element is Ideal]] and definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$ admits a [[Definition:Supremum of Set|supremum]].
+We will prove that
+:$x^\ll \subseteq \left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$
+Let $a \in x^\ll$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$a \ll x$
+By [[Way Below implies Preceding]]:
+:$a \preceq x$
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d \in D: a \preceq d$
+By [[Meet is Idempotent]] and [[Meet Semilattice is Ordered Structure]]:
+:$a = a \wedge a \preceq x \wedge d \in \left\{ {x \wedge y: y \in D}\right\}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \left\{ {x \wedge y: y \in D}\right\}$ is [[Definition:Upper Bound of Set|upper bound]] for $\left\{ {x \wedge y: y \in D}\right\}$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$x \wedge d \preceq \sup \left\{ {x \wedge y: y \in D}\right\}$
+By definition of [[Definition:Transitivity|transitivity]]
+:$a \preceq \sup \left\{ {x \wedge y: y \in D}\right\}$
+Thus by definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$a \in \left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq$
+{{qed|lemma}}
+By [[Supremum of Subset]]:
+:$\sup \left({x^\ll}\right) \preceq \sup \left({\left({\sup \left\{ {x \wedge d: d \in D}\right\} }\right)^\preceq}\right)$
+By [[Definition:Axiom of Approximation|axiom of approximation]] and [[Supremum of Lower Closure of Element]]:
+:$x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$
+By definition:
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $\left\{ {x \wedge d: d \in D}\right\}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \left\{ {x \wedge d: d \in D}\right\} \preceq x$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$x = \sup \left\{ {x \wedge d: d \in D}\right\}$
+Thus by [[Meet-Continuous iff if Element Precedes Supremum of Directed Subset then Element equals Supremum of Meet of Element by Directed Subset]]:
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number to Power of Zero Rising is One}
+Tags: Rising Factorials
+
+\begin{theorem}
+:$x^{\overline 0} = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\overline 0}
+      | r = \prod_{j \mathop = 0}^{-1} \paren {x + j}
+      | c = {{Defof|Rising Factorial}}
+}}
+{{eqn | r = 1
+      | c = Product is [[Definition:Vacuous Product|Vacuous]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Rising Factorials]]
+1tfwsij49sdapn8zmyovlleahasydr4
+\end{proof}<|endoftext|>
+\section{Number to Power of Zero Falling is One}
+Tags: Falling Factorials
+
+\begin{theorem}
+:$x^{\underline 0} = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\underline 0}
+      | r = \prod_{j \mathop = 0}^{-1} \paren {x - j}
+      | c = {{Defof|Falling Factorial}}
+}}
+{{eqn | r = 1
+      | c = Product is [[Definition:Vacuous Product|Vacuous]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Falling Factorials]]
+0thb0bl3pugwn0hxx3fa89xkjtu0dbp
+\end{proof}<|endoftext|>
+\section{Product of Number by its Falling Factorial}
+Tags: Falling Factorials
+
+\begin{theorem}
+Let $x^{\underline n}$ denote the [[Definition:Falling Factorial|$n$th falling factorial power]] of $x$.
+Then:
+:$x x^{\underline n} = x^{\underline {n + 1} } + n x^{\underline n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x x^{\underline n}
+      | r = \left({x - n + n}\right) x^{\underline n}
+      | c = 
+}}
+{{eqn | r = \left({x - n}\right) x^{\underline n} + n x^{\underline n}
+      | c = 
+}}
+{{eqn | r = \left({x - n}\right) \prod_{j \mathop = 0}^{n - 1} \left({x - j}\right) + n x^{\underline n}
+      | c = Definition of [[Definition:Falling Factorial|Falling Factorial]]
+}}
+{{eqn | r = \prod_{j \mathop = 0}^n \left({x - j}\right) + n x^{\underline n}
+      | c = Definition of [[Definition:Product Notation (Algebra)|Product]]
+}}
+{{eqn | r = x^{\underline {n + 1} } + n x^{\underline n}
+      | c = Definition of [[Definition:Falling Factorial|Falling Factorial]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Falling Factorials]]
+9vaivzg0rwkuq3jc54wr842zac0fg21
+\end{proof}<|endoftext|>
+\section{Product of Number by its Rising Factorial}
+Tags: Rising Factorials
+
+\begin{theorem}
+Let $x^{\overline n}$ denote the [[Definition:Rising Factorial|$n$th rising factorial power]] of $x$.
+Then:
+:$x x^{\overline n} = x^{\overline {n + 1} } - n x^{\overline n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x x^{\overline n}
+      | r = \paren {x + n - n} x^{\overline n}
+      | c = 
+}}
+{{eqn | r = \paren {x + n} x^{\overline n} - n x^{\overline n}
+      | c = 
+}}
+{{eqn | r = \paren {x + n} \prod_{j \mathop = 0}^{n - 1} \paren {x + j} - n x^{\overline n}
+      | c = {{Defof|Rising Factorial}}
+}}
+{{eqn | r = \prod_{j \mathop = 0}^n \paren {x + j} - n x^{\overline n}
+      | c = {{Defof|Product Notation (Algebra)|Product}}
+}}
+{{eqn | r = x^{\overline {n + 1} } - n x^{\overline n}
+      | c = {{Defof|Rising Factorial}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Rising Factorials]]
+hstzvqnlhk5v5epvgxwerkmj8pk0fv8
+\end{proof}<|endoftext|>
+\section{Value of Formula under Assignment Determined by Free Variables}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\sigma, \sigma'$ be [[Definition:Assignment for Formula|assignments for $\mathbf A$ in $\mathcal A$]] such that:
+:For each [[Definition:Free Variable|free variable]] $x$ of $\mathbf A$, $\sigma \left({x}\right) = \sigma' \left({x}\right)$
+Then:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]$
+where $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right]$ is the [[Definition:Value of Formula under Assignment|value of $\mathbf A$ under $\sigma$]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] applied to the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]].
+If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = p_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma}\right] }\right)$
+Because $\mathbf A$ contains no [[Definition:Quantifier|quantifiers]], all its variables are [[Definition:Free Variable|free]], and hence are in the [[Definition:Domain of Mapping|domain]] of $\sigma, \sigma'$ as [[Definition:Assignment for Formula|assignments]].
+Thus $\sigma, \sigma'$ are [[Definition:Assignment for Term|assignments]] for each $\tau_i$, and by [[Value of Term under Assignment Determined by Variables]]:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau_i}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_i}\right) } \left[{\sigma'}\right]$
+for each $\tau_i$.
+It is immediate that:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]$
+If $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$, then:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right]
+     |r = f^\neg \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f^\neg \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma'}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma'$]]
+}}
+{{end-eqn}}
+If $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right]
+     |r = f^\circ \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B'}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f^\circ \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma'}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B'}\right) } \left[{\sigma'}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma'$]]
+}}
+{{end-eqn}}
+If $\mathbf A = \exists x: \mathbf B$ or $\mathbf A = \forall x : \mathbf B$, and the induction hypothesis applies to $\mathbf B$, then from the definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right]$
+is determined by the [[Definition:Value of Formula under Assignment|values]]:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right)}\right]$
+where $a$ ranges over $\mathcal A$, and $\sigma + \left({x / a}\right)$ is the [[Definition:Extension of Assignment|extension]] of $\sigma$ mapping $x$ to $a$.
+Now, for a [[Definition:Free Variable|free variable]] $y$ of $\mathbf B$:
+{{begin-eqn}}
+{{eqn|l = \left({\sigma + \left({x / a}\right)}\right) \left({y}\right)
+     |r = \begin{cases} a &: \text{if $y = x$} \\ \sigma \left({y}\right) &: \text{otherwise} \end{cases}
+     |c = Definition of [[Definition:Extension of Assignment|Extension]]
+}}
+{{eqn|r = \begin{cases} a &: \text{if $y = x$} \\ \sigma' \left({y}\right) &: \text{otherwise} \end{cases}
+     |c = Assumption on $\sigma, \sigma'$
+}}
+{{eqn|r = \left({\sigma' + \left({x / a}\right)}\right) \left({y}\right)
+     |c = Definition of [[Definition:Extension of Assignment|Extension]]
+}}
+{{end-eqn}}
+Hence, by the induction hypothesis:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right)}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma' + \left({x / a}\right)}\right]$
+It follows that:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma'}\right]$
+The result follows from the [[Principle of Structural Induction]].
+{{qed}}
+[[Category:Predicate Logic]]
+72pyybgluhk8a7ghrccyz86nqc2qh1q
+\end{proof}<|endoftext|>
+\section{First Inversion Formula for Stirling Numbers}
+Tags: Stirling Numbers
+
+\begin{theorem}
+For all $m, n \in \Z_{\ge 0}$:
+:$\displaystyle \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$
+where:
+:$\displaystyle \left[{n \atop k}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\displaystyle \left\{ {k \atop m}\right\}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$\delta_{m n}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{0 \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{0 - k}
+      | r = \sum_k \delta_{k 0} \left\{ {k \atop m}\right\} \left({-1}\right)^{-k}
+      | c = Definition of [[Definition:Unsigned Stirling Numbers of the First Kind|Unsigned Stirling Numbers of the First Kind]]
+}}
+{{eqn | r = \left\{ {0 \atop m}\right\}
+      | c = as all other terms vanish by $\delta_{k 0}$
+}}
+{{eqn | r = \delta_{m 0}
+      | c = Definition of [[Definition:Stirling Numbers of the Second Kind|Stirling Numbers of the Second Kind]]
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ has been shown to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({j}\right)$ is true for all $j$ such $0 \le j \le r$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \forall j: 0 \le j \le r: \forall m \in \Z_{\ge 0}: \sum_k \left[{j \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{j - k} = \delta_{m j}$
+from which it is to be shown that:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{ {r + 1} \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} = \delta_{m \left({r + 1}\right)}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | r = \sum_k \left[{ {r + 1} \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \left({\left[{r \atop k - 1}\right] + r \left[{r \atop k}\right]}\right) \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k}
+      | c = Definition of [[Definition:Unsigned Stirling Numbers of the First Kind|Unsigned Stirling Numbers of the First Kind]]
+}}
+{{eqn | r = \sum_k \left[{r \atop k - 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} - r \sum_k \left[{r \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left[{r \atop k - 1}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{r + 1 - k} - r \delta_{m r}
+      | c = [[First Inversion Formula for Stirling Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_k \left[{r \atop k - 1}\right] \left({\left\{ {k - 1 \atop m - 1}\right\} + m \left\{ {k - 1 \atop m}\right\} }\right) \left({-1}\right)^{r + 1 - k} - r \delta_{r m}
+      | c = Definition of [[Definition:Stirling Numbers of the Second Kind|Stirling Numbers of the Second Kind]]
+}}
+{{eqn | r = \sum_k \left[{r \atop k - 1}\right] \left\{ {k - 1 \atop m - 1}\right\} \left({-1}\right)^{r - \left({k - 1}\right)} + m \sum_k \left[{r \atop k - 1}\right] \left\{ {k - 1 \atop m}\right\} \left({-1}\right)^{r - \left({k - 1}\right)} - r \delta_{r m}
+      | c = 
+}}
+{{eqn | r = \delta_{r \left({m - 1}\right)} + m \delta_{r m} - r \delta_{r m}
+      | c = [[First Inversion Formula for Stirling Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \delta_{m \left({r + 1}\right)} + \left({m - r}\right) \delta_{r m}
+      | c = 
+}}
+{{eqn | r = \delta_{m \left({r + 1}\right)}
+      | c = as $\left({m - r}\right) \delta_{r m}$ vanishes for all $r$ and $m$
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \left\{ {k \atop m}\right\} \left({-1}\right)^{n - k} = \delta_{m n}$
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of 1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\displaystyle {1 \brack n} = \delta_{1 n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = {1 \brack n}
+      | r = 0 \times {0 \brack n} + {0 \brack n - 1}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = {0 \brack n - 1}
+      | c = 
+}}
+{{eqn | r = \delta_{0 \paren {n - 1} }
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = \delta_{1 n}
+      | c = $0 = n - 1 \iff 1 = n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of 1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$s \left({1, n}\right) = \delta_{1 n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = s \left({1, n}\right)
+      | r = s \left({0, {n - 1} }\right) - 0 \times s \left({0, n}\right)
+      | c = Definition of [[Definition:Signed Stirling Numbers of the First Kind|Signed Stirling Number of the First Kind]]
+}}
+{{eqn | r = s \left({0, {n - 1} }\right)
+      | c = 
+}}
+{{eqn | r = \delta_{0 \left({n - 1}\right)}
+      | c = Definition of [[Definition:Signed Stirling Numbers of the First Kind|Signed Stirling Number of the First Kind]]
+}}
+{{eqn | r = \delta_{1 n}
+      | c = $0 = n - 1 \iff 1 = n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of 1}
+Tags: Stirling Numbers
+
+\begin{theorem}
+:$\displaystyle {1 \brace n} = \delta_{1 n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = {1 \brace n}
+      | r = n \times {0 \brace n} + {0 \brace n - 1}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = n \times \delta_{0 n} + \delta_{0 \paren {n - 1} }
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = \delta_{0 \paren {n - 1} }
+      | c = as $n \times \delta_{0 n}$ is either $n \times 0$ or $0 \times 1$ and so always $0$
+}}
+{{eqn | r = \delta_{1 n}
+      | c = $0 = n - 1 \iff 1 = n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Inversion Formula for Stirling Numbers}
+Tags: Stirling Numbers
+
+\begin{theorem}
+For all $m, n \in \Z_{\ge 0}$:
+:$\displaystyle \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$
+where:
+:$\displaystyle \left\{ {n \atop k}\right\}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$\displaystyle \left[{k \atop m}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\delta_{m n}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Second Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left\{ {0 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{0 - k}
+      | r = \sum_k \delta_{k 0} \left[{k \atop m}\right] \left({-1}\right)^{-k}
+      | c = Definition of [[Definition:Stirling Numbers of the Second Kind|Stirling Numbers of the Second Kind]]
+}}
+{{eqn | r = \left[{0 \atop m}\right]
+      | c = as all other terms vanish by $\delta_{k 0}$
+}}
+{{eqn | r = \delta_{m 0}
+      | c = Definition of [[Definition:Unsigned Stirling Numbers of the First Kind|Unsigned Stirling Numbers of the First Kind]]
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ has been shown to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true for all $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {r \atop k}\right\}\left[{k \atop m}\right]  \left({-1}\right)^{r - k} = \delta_{m r}$
+from which it is to be shown that:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ { {r + 1} \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} = \delta_{m \left({r + 1}\right)}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | r = \sum_k \left\{ { {r + 1} \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \left({k \left\{ {r \atop k}\right\} + \left\{ {r \atop k - 1}\right\} }\right) \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = Definition of [[Definition:Stirling Numbers of the Second Kind|Stirling Numbers of the Second Kind]]
+}}
+{{eqn | r = \sum_k k \left\{ {r \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k k \left\{ {r \atop k}\right\} \frac 1 k \left({\left[{k + 1 \atop m}\right] - \left[{k \atop m - 1}\right]}\right) \left({-1}\right)^{r + 1 - k}
+      | c = Definition of [[Definition:Unsigned Stirling Numbers of the First Kind|Unsigned Stirling Numbers of the First Kind]]
+}}
+{{eqn | o = 
+      | r = + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | o = 
+      | r = + \sum_k \left\{ {r \atop k - 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | o = 
+      | r = + \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k + 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k} - \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | o = 
+      | r = - \sum_k \left\{ {r \atop k}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r \atop k}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{r - k}
+      | c = Simplification
+}}
+{{eqn | r = \delta_{\left({m - 1}\right) r}
+      | c = [[Second Inversion Formula for Stirling Numbers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \delta_{m \left({r + 1}\right)}
+      | c = Definition of [[Definition:Kronecker Delta|Kronecker Delta]]
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {n \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{n - k} = \delta_{m n}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Choose n}
+Tags: Examples of Binomial Coefficients
+
+\begin{theorem}
+:$\dbinom 0 n = \delta_{0 n}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+:$\dbinom m n = \begin{cases}
+\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\
+& \\
+0 & : \text { otherwise } \end{cases}$
+Thus when $n > 0$:
+:$\dbinom 0 n = 0$
+and when $n = 0$:
+:$\dbinom 0 0 = \dfrac {0!} {0! \paren {0 - 0}!} = 1$
+by definition of [[Definition:Factorial|factorial]].
+Hence the result by definition of the [[Definition:Kronecker Delta|Kronecker delta]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of 0}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\displaystyle \left[{0 \atop n}\right] = \delta_{0 n}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
+$\displaystyle x^{\underline 0} = \sum_k \left({-1}\right)^{0 - k} \left[{0 \atop k}\right] x^k$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = x^{\underline 0}
+      | r = 1
+      | c = [[Number to Power of Zero Falling is One]]
+}}
+{{eqn | r = x^0
+      | c = Definition of [[Definition:Integer Power|Integer Power]]
+}}
+{{end-eqn}}
+Thus, in the expression:
+:$\displaystyle x^{\underline 0} = \sum_k \left({-1}\right)^{-k} \left[{0 \atop k}\right] x^k$
+we have:
+:$\displaystyle \left[{0 \atop 0}\right] = 1$
+and for all $k \in \Z_{>0}$:
+:$\displaystyle \left[{0 \atop k}\right] = 0$
+That is:
+:$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Values of Signed Stirling Numbers of the First Kind}
+Tags: Stirling Numbers, Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+This page gathers together some particular values of [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling numbers of the first kind]].
+\end{theorem}<|endoftext|>
+\section{Particular Values of Unsigned Stirling Numbers of the First Kind}
+Tags: Stirling Numbers, Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+This page gathers together some particular values of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling numbers of the first kind]].
+\end{theorem}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of 0}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$s \left({0, n}\right) = \delta_{0 n}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]]:
+$\displaystyle x^{\underline 0} = \sum_k s \left({0, k}\right) x^k$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = x^{\underline 0}
+      | r = 1
+      | c = [[Number to Power of Zero Falling is One]]
+}}
+{{eqn | r = x^0
+      | c = Definition of [[Definition:Integer Power|Integer Power]]
+}}
+{{end-eqn}}
+Thus, in the expression:
+:$\displaystyle x^{\underline 0} = \sum_k s \left({0, k}\right) x^k$
+we have:
+:$s \left({0, 0}\right) = 1$
+and for all $k \in \Z_{>0}$:
+:$s \left({0, k}\right) = 0$
+That is:
+:$s \left({0, k}\right) = \delta_{0 k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Particular Values of Stirling Numbers of the Second Kind}
+Tags: Stirling Numbers, Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+This page gathers together some particular values of [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]].
+\end{theorem}<|endoftext|>
+\section{Stirling Number of the Second Kind of 0}
+Tags: Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+:$\displaystyle {0 \brace n} = \delta_{0 n}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]]:
+$\displaystyle x^{\underline 0} = \sum_k {0 \brace k} x^k$
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = x^0
+      | r = 1
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | r = x^{\underline 0}
+      | c = [[Number to Power of Zero Falling is One]]
+}}
+{{end-eqn}}
+Thus, in the expression:
+:$\displaystyle x^0 = \sum_k {0 \brace k} x^{\underline k}$
+we have:
+:$\displaystyle {0 \brace 0} = 1$
+and for all $k \in \Z_{>0}$:
+:$\displaystyle {0 \brace k} = 0$
+That is:
+:$\displaystyle {0 \brace k} = \delta_{0 k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of Number with Self}
+Tags: Stirling Numbers
+
+\begin{theorem}
+:$\displaystyle \left[{n \atop n}\right] = 1$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \left[{n \atop n}\right] = 1$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \left[{0 \atop 0}\right]
+      | r = \delta_{0 0}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = 1
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \left[{k \atop k}\right] = 1$
+from which it is to be shown that:
+:$\displaystyle \left[{ {k + 1} \atop {k + 1} }\right] = 1$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left[{ {k + 1} \atop {k + 1} }\right]
+      | r = k \left[{k \atop {k + 1} }\right] + \left[{k \atop k}\right]
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = k \times 0 + \left[{k \atop k}\right]
+      | c = [[Stirling Number of Number with Greater]]
+}}
+{{eqn | r = \left[{k \atop k}\right]
+      | c = 
+}}
+{{eqn | r = 1
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Self#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \left[{n \atop n}\right] = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of Number with Greater}
+Tags: Stirling Numbers, Unsigned Stirling Number of the First Kind of Number with Greater
+
+\begin{theorem}
+Let $\displaystyle {n \brack k}$ denote an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
+Then:
+:$\displaystyle {n \brack k} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition, [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\displaystyle \left[{n \atop k}\right]$ which satisfy the equation:
+:$\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$
+where $x^{\underline n}$ denotes the [[Definition:Falling Factorial|$n$th falling factorial of $x$]].
+Both of the [[Definition:Expression|expressions]] on the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $x$ of [[Definition:Degree of Polynomial|degree $n$]].
+Hence the [[Definition:Polynomial Coefficient|coefficient]] $\displaystyle \left[{n \atop k}\right]$ of $x^k$ where $k > n$ is $0$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle k > n \implies \left[{n \atop k}\right] = 0$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+:$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$
+from [[Unsigned Stirling Number of the First Kind of 0]].
+So by definition of [[Definition:Kronecker Delta|Kronecker delta]]:
+:$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle \left[{0 \atop k}\right] = 0$
+and so $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $0 \le r$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle k > r \implies \left[{r \atop k}\right] = 0$
+from which it is to be shown that:
+:$\displaystyle k > r + 1 \implies \left[{r + 1 \atop k}\right] = 0$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left[{r + 1 \atop k}\right]
+      | r = r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]
+      | c = 
+}}
+{{eqn | r = r \times 0 + \left[{r \atop k - 1}\right]
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \times 0 + 0
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]: $k > r + 1 \implies k - 1 > r$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies \left[{n \atop k}\right] = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of Number with Greater}
+Tags: Stirling Numbers
+
+\begin{theorem}
+Let $\map s {n, k}$ denote a [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]].
+Then:
+:$\map s {n, k} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling numbers of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\map s {n, k}$ which satisfy the equation:
+:$\displaystyle x^{\underline n} = \sum_k \map s {n, k} x^k$
+where $x^{\underline n}$ denotes the [[Definition:Falling Factorial|$n$th falling factorial of $x$]].
+Both of the [[Definition:Expression|expressions]] on the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $x$ of [[Definition:Degree of Polynomial|degree $n$]].
+Hence the [[Definition:Polynomial Coefficient|coefficient]] $s \left({n, k}\right)$ of $x^k$ where $k > n$ is $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of Number with Greater}
+Tags: Stirling Numbers, Stirling Number of the Second Kind of Number with Greater
+
+\begin{theorem}
+Let $\displaystyle {n \brace k}$ denote a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
+Then:
+:$\displaystyle {n \brace k} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]] are defined as the [[Definition:Coefficient|coefficients]] $\displaystyle {n \brace k}$ which satisfy the equation:
+:$\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$
+where $x^{\underline k}$ denotes the [[Definition:Falling Factorial|$k$th falling factorial of $x$]].
+Both of the [[Definition:Expression|expressions]] on the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $x$ of [[Definition:Degree of Polynomial|degree $n$]].
+Hence the [[Definition:Polynomial Coefficient|coefficient]] $\displaystyle {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle k > n \implies {n \brace k} = 0$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+:$\displaystyle {0 \brace k} = \delta_{0 k}$
+from [[Stirling Number of the Second Kind of 0]].
+So by definition of [[Definition:Kronecker Delta|Kronecker delta]]:
+:$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle {0 \brace k} = 0$
+and so $\map P 0$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle k > r \implies {r \brace k} = 0$
+from which it is to be shown that:
+:$\displaystyle k > r + 1 \implies {r + 1 \brace k} = 0$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {r + 1 \brace k}
+      | r = k {r \brace k} + {r \brace k - 1}
+      | c = 
+}}
+{{eqn | r = r \times 0 + {r \brace k - 1}
+      | c = [[Stirling Number of the Second Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \times 0 + 0
+      | c = [[Stirling Number of the Second Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]: $k > r + 1 \implies k - 1 > r$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of Number with Greater}
+Tags: Stirling Numbers
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$.
+Let $k > n$.
+\end{theorem}<|endoftext|>
+\section{Stirling Number of the Second Kind of Number with Self}
+Tags: Stirling Numbers
+
+\begin{theorem}
+:$\displaystyle {n \brace n} = 1$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle {n \brace n} = 1$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = {0 \brace 0}
+      | r = \delta_{0 0}
+      | c = [[Stirling Number of the Second Kind of 0]]
+}}
+{{eqn | r = 1
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+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 $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle {k \brace k} = 1$
+from which it is to be shown that:
+:$\displaystyle {k + 1 \brace k + 1} = 1$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {k + 1 \brace k + 1}
+      | r = \paren {k + 1} {k \brace k + 1} + {k \brace k}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = \paren {k + 1} \times 0 + {k \brace k}
+      | c = [[Stirling Number of Number with Greater]]
+}}
+{{eqn | r = {k \brace k}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = [[Stirling Number of the Second Kind of Number with Self#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: {n \brace n} = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Value of Term under Assignment Determined by Variables}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]].
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\sigma, \sigma'$ be [[Definition:Assignment for Formula|assignments for $\tau$ in $\mathcal A$]] such that:
+:For each [[Definition:Variable (Logic)|variable]] $x$ [[Definition:Occurrence (Predicate Logic)|occurring]] in $\tau$, $\sigma \left({x}\right) = \sigma' \left({x}\right)$
+Then:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]$
+where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the [[Definition:Value of Term under Assignment|value of $\tau$ under $\sigma$]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] applied to the [[Definition:Term (Predicate Logic)|definition of a term]].
+If $\tau = x$, then:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]
+     |r = \sigma \left({x}\right)
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = \sigma' \left({x}\right)
+     |c = Assumption on $\sigma, \sigma'$
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma$]]
+}}
+{{end-eqn}}
+as desired.
+If $\tau = f \left({\tau_1, \ldots, \tau_n}\right)$ and the induction hypothesis applies to each $\tau_i$, then:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]
+     |r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma'}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma'}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma'}\right]
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma'$]]
+}}
+{{end-eqn}}
+The result follows from the [[Principle of Structural Induction]].
+{{qed}}
+[[Category:Predicate Logic]]
+10ma1jy4vmkw964isbnsakdzd6mfpgc
+\end{proof}<|endoftext|>
+\section{Relation between Signed and Unsigned Stirling Numbers of the First Kind}
+Tags: Stirling Numbers
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
+Then:
+:$\displaystyle {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
+where:
+:$\displaystyle {n \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\map s {n, m}$ denotes a [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_k \map s {n, k} x^k
+      | r = x^{\underline n}
+      | c = {{Defof|Signed Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = \sum_k \paren {-1}^{n - k} {n \brack k} x^k
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | ll = \leadsto
+      | l = \map s {n, m}
+      | r = \paren {-1}^{n - m} {n \brack m}
+      | c = Comparing coefficients of $x^m$
+}}
+{{eqn | ll = \leadsto
+      | l = {n \brack m}
+      | r = \paren {-1}^{m - n} \map s {n, m}
+}}
+{{eqn | r = \paren {-1}^{n + m} \map s {n, m}
+      | c = as $\paren {-1}^{2 n} = 1$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Stirling Numbers]]
+hnondew9tstatc6ggzwls4frlmt9tv5
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of Number with Greater/Proof 1}
+Tags: Unsigned Stirling Number of the First Kind of Number with Greater
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
+{{:Unsigned Stirling Number of the First Kind of Number with Greater}}
+\end{theorem}
+
+\begin{proof}
+By definition, [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]] are defined as the [[Definition:Polynomial Coefficient|polynomial coefficients]] $\displaystyle \left[{n \atop k}\right]$ which satisfy the equation:
+:$\displaystyle x^{\underline n} = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k$
+where $x^{\underline n}$ denotes the [[Definition:Falling Factorial|$n$th falling factorial of $x$]].
+Both of the [[Definition:Expression|expressions]] on the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $x$ of [[Definition:Degree of Polynomial|degree $n$]].
+Hence the [[Definition:Polynomial Coefficient|coefficient]] $\displaystyle \left[{n \atop k}\right]$ of $x^k$ where $k > n$ is $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2}
+Tags: Unsigned Stirling Number of the First Kind of Number with Greater
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
+{{:Unsigned Stirling Number of the First Kind of Number with Greater}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle k > n \implies \left[{n \atop k}\right] = 0$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+:$\displaystyle \left[{0 \atop k}\right] = \delta_{0 k}$
+from [[Unsigned Stirling Number of the First Kind of 0]].
+So by definition of [[Definition:Kronecker Delta|Kronecker delta]]:
+:$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle \left[{0 \atop k}\right] = 0$
+and so $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $0 \le r$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle k > r \implies \left[{r \atop k}\right] = 0$
+from which it is to be shown that:
+:$\displaystyle k > r + 1 \implies \left[{r + 1 \atop k}\right] = 0$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left[{r + 1 \atop k}\right]
+      | r = r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]
+      | c = 
+}}
+{{eqn | r = r \times 0 + \left[{r \atop k - 1}\right]
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \times 0 + 0
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]: $k > r + 1 \implies k - 1 > r$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies \left[{n \atop k}\right] = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of Number with Greater/Proof 1}
+Tags: Stirling Number of the Second Kind of Number with Greater
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
+{{:Stirling Number of the Second Kind of Number with Greater}}
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]] are defined as the [[Definition:Coefficient|coefficients]] $\displaystyle {n \brace k}$ which satisfy the equation:
+:$\displaystyle x^n = \sum_k {n \brace k} x^{\underline k}$
+where $x^{\underline k}$ denotes the [[Definition:Falling Factorial|$k$th falling factorial of $x$]].
+Both of the [[Definition:Expression|expressions]] on the {{LHS}} and {{RHS}} are [[Definition:Polynomial over Real Numbers|polynomials]] in $x$ of [[Definition:Degree of Polynomial|degree $n$]].
+Hence the [[Definition:Polynomial Coefficient|coefficient]] $\displaystyle {n \brace k}$ of $x^{\underline k}$ where $k > n$ is $0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of Number with Greater/Proof 2}
+Tags: Stirling Number of the Second Kind of Number with Greater
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$ such that $k > n$.
+{{:Stirling Number of the Second Kind of Number with Greater}}
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle k > n \implies {n \brace k} = 0$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+:$\displaystyle {0 \brace k} = \delta_{0 k}$
+from [[Stirling Number of the Second Kind of 0]].
+So by definition of [[Definition:Kronecker Delta|Kronecker delta]]:
+:$\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle {0 \brace k} = 0$
+and so $\map P 0$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle k > r \implies {r \brace k} = 0$
+from which it is to be shown that:
+:$\displaystyle k > r + 1 \implies {r + 1 \brace k} = 0$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {r + 1 \brace k}
+      | r = k {r \brace k} + {r \brace k - 1}
+      | c = 
+}}
+{{eqn | r = r \times 0 + {r \brace k - 1}
+      | c = [[Stirling Number of the Second Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \times 0 + 0
+      | c = [[Stirling Number of the Second Kind of Number with Greater/Proof 2#Induction Hypothesis|Induction Hypothesis]]: $k > r + 1 \implies k - 1 > r$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of Number with Self}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$s \left({n, n}\right) = 1$
+\end{theorem}
+
+\begin{proof}
+From [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
+:$\displaystyle \left[{n \atop n}\right] = \left({-1}\right)^{n + n} s \left({n, n}\right)$
+We have that:
+:$\left({-1}\right)^{n + n} = \left({-1}\right)^{2 n} = 1$
+and so:
+:$\displaystyle \left[{n \atop n}\right] = s \left({n, n}\right)$
+The result follows from [[Unsigned Stirling Number of the First Kind of Number with Self]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of n with n-1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\displaystyle \left[{n \atop n - 1}\right] = \binom n 2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+=== Basis for the Induction ===
+For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \left[{n \atop n - 1}\right] = \binom n 2$
+$P \left({1}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \left[{1 \atop 0}\right]
+      | r = \delta_{1 0}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \binom 1 2
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \left[{k \atop k - 1}\right] = \binom k 2$
+from which it is to be shown that:
+:$\displaystyle \left[{ {k + 1} \atop k}\right] = \binom {k + 1} 2$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left[{ {k + 1} \atop k}\right]
+      | r = k \left[{k \atop k}\right] + \left[{k \atop k - 1}\right]
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = k \times 1 + \left[{k \atop k - 1}\right]
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Self]]
+}}
+{{eqn | r = k + \binom k 2
+      | c = [[Unsigned Stirling Number of the First Kind of n with n-1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = k + \frac {k \left({k - 1}\right)} 2
+      | c = [[Binomial Coefficient with Two]]
+}}
+{{eqn | r = \frac {k \left({k - 1}\right) + 2 k} 2
+      | c = 
+}}
+{{eqn | r = \frac {k^2 - k + 2 k} 2
+      | c = 
+}}
+{{eqn | r = \frac {\left({k + 1}\right) k} 2
+      | c = 
+}}
+{{eqn | r = \binom {k + 1} 2
+      | c = [[Binomial Coefficient with Two]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 1}: \left[{n \atop n - 1}\right] = \binom n 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of n with n-1}
+Tags: Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+:$\displaystyle {n \brace n - 1} = \binom n 2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle {n \brace n - 1} = \binom n 2$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = {1 \brace 0}
+      | r = \delta_{1 0}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \binom 1 2
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle {k \brace k - 1} = \binom k 2$
+from which it is to be shown that:
+:$\displaystyle {k + 1 \brace k} = \binom {k + 1} 2$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {k + 1 \brace k}
+      | r = k {k \brace k} + {k \brace k - 1}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = k \times 1 + {k \brace k - 1}
+      | c = [[Stirling Number of the Second Kind of Number with Self]]
+}}
+{{eqn | r = k + \binom k 2
+      | c = [[Stirling Number of the Second Kind of n with n-1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = k + \frac {k \paren {k - 1} } 2
+      | c = [[Binomial Coefficient with Two]]
+}}
+{{eqn | r = \frac {k \paren {k - 1} + 2 k} 2
+      | c = 
+}}
+{{eqn | r = \frac {k^2 - k + 2 k} 2
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1} k} 2
+      | c = 
+}}
+{{eqn | r = \binom {k + 1} 2
+      | c = [[Binomial Coefficient with Two]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 1}: {n \brace n - 1} = \binom n 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of n with n-1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\map s {n, n - 1} = -\dbinom n 2$
+\end{theorem}
+
+\begin{proof}
+From [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
+:$\displaystyle {n \brack n - 1} = \paren {-1}^{n + n - 1} \map s {n, n - 1}$
+where $\displaystyle {n \brack n - 1}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
+We have that:
+:$\paren {-1}^{n + n - 1} = \paren {-1}^{2 n - 1} = -1$
+and so:
+:$\displaystyle {n \brack n} = -\map s {n, n}$
+The result follows from [[Unsigned Stirling Number of the First Kind of Number with Self]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of n+1 with 0}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\displaystyle {n + 1 \brack 0} = 0$
+\end{theorem}
+
+\begin{proof}
+We are given that $k = 0$.
+So by definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
+:$\displaystyle {n \brack k} = \delta_{n k}$
+where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \ge
+      | r = 0
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \delta_{\paren {n + 1} 0}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of n+1 with 0}
+Tags: Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+:$\displaystyle {n + 1 \brace 0} = 0$
+\end{theorem}
+
+\begin{proof}
+We are given that $k = 0$.
+So by definition of [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]:
+:$\displaystyle {n \brace k} = \delta_{n k}$
+where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Thus
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \ge
+      | r = 0
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \delta_{\paren {n + 1} 0}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of n+1 with 0}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\map s {n + 1, 0} = 0$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Signed Stirling Numbers of the First Kind|signed Stirling number of the first kind]]:
+:$\map s {n, k} = \delta_{n k}$
+where $\delta_{n k}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Thus
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \ge
+      | r = 0
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n + 1
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \delta_{\paren {n + 1} 0}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unsigned Stirling Number of the First Kind of n+1 with 1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$\displaystyle \left[{n + 1 \atop 1}\right] = n!$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \left[{n + 1 \atop 1}\right] = n!$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \left[{1 \atop 1}\right]
+      | r = \delta_{1 1}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = 1
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = 0!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \left[{k + 1 \atop 1}\right] = k!$
+from which it is to be shown that:
+:$\displaystyle \left[{k + 2 \atop 1}\right] = \left({k + 1}\right)!$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left[{k + 2 \atop 1}\right]
+      | r = \left({k + 1}\right) \left[{k + 1 \atop 1}\right] + \left[{k + 1 \atop 0}\right]
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = \left({k + 1}\right) \left[{k + 1 \atop 1}\right] + 0
+      | c = [[Unsigned Stirling Number of the First Kind of n+1 with 0]]
+}}
+{{eqn | r = \left({k + 1}\right) \times k!
+      | c = [[Unsigned Stirling Number of the First Kind of n+1 with 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({k + 1}\right)!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \left[{n + 1 \atop 1}\right] = n!$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of n+1 with 1}
+Tags: Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+:$\displaystyle {n + 1 \brace 1} = 1$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle {n + 1 \brace 1} = 1$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = {1 \brace 1}
+      | r = \delta_{1 1}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = 1
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+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 $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle {k + 1 \brace 1} = 1$
+from which it is to be shown that:
+:$\displaystyle {k + 2 \brace 1} = 1$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {k + 2 \brace 1}
+      | r = 1 \times {k + 1 \brace 1} + {k + 2 \brace 0}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = {k + 1 \brace 1} + 0
+      | c = [[Stirling Number of the Second Kind of n+1 with 0]]
+}}
+{{eqn | r = 1
+      | c = [[Stirling Number of the Second Kind of n+1 with 1#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 1} = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Signed Stirling Number of the First Kind of n+1 with 1}
+Tags: Examples of Stirling Numbers of the First Kind
+
+\begin{theorem}
+:$s \left({n + 1, 1}\right) = \left({-1}\right)^n n!$
+\end{theorem}
+
+\begin{proof}
+By [[Relation between Signed and Unsigned Stirling Numbers of the First Kind]]:
+:$\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^{n + 1 + 1} s \left({n + 1, 1}\right)$
+where $\displaystyle \left[{n + 1 \atop 1}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
+We have that:
+:$\left({-1}\right)^{n + 1 + 1} = \left({-1}\right)^n$
+and so:
+:$\displaystyle \left[{n + 1 \atop 1}\right] = \left({-1}\right)^n s \left({n + 1, 1}\right)$
+The result follows from [[Unsigned Stirling Number of the First Kind of Number with Self]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Stirling Number of the Second Kind of n+1 with 2}
+Tags: Examples of Stirling Numbers of the Second Kind
+
+\begin{theorem}
+:$\displaystyle {n + 1 \brace 2} = 2^n - 1$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle {n + 1 \brace 2} = 2^n - 1$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = {1 \brace 2}
+      | r = \delta_{1 2}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = 0
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = 2^0 - 1
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+So the result holds for $\map P 0$.
+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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle {k + 1 \brace 2} = 2^k - 1$
+from which it is to be shown that:
+:$\displaystyle {k + 2 \brace 2} = 2^{k + 1} - 1$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = {k + 2 \brace 2}
+      | r = 2 \times {k + 1 \brace 2} + {k + 1 \brace 1}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = 2 \times {k + 1 \brace 2} + 1
+      | c = [[Stirling Number of the Second Kind of n+1 with 1]]
+}}
+{{eqn | r = 2 \times {2^k - 1} + 1
+      | c = [[Stirling Number of the Second Kind of n+1 with 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 2^{k + 1} - 2 + 1
+      | c = 
+}}
+{{eqn | r = 2^{k + 1} - 1
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: {n + 1 \brace 2} = 2^n - 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$
+where:
+:$\displaystyle \left[{n \atop k}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\dbinom k m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{0 \atop k}\right] \binom k m
+      | r = \sum_k \delta_{0 k} \binom k m
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = \binom 0 m
+      | c = all terms vanish except for $k = 0$
+}}
+{{eqn | r = \delta_{0 m}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \delta_{1 \left({m + 1}\right)}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \left[{1 \atop m + 1}\right]
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = \left[{0 + 1 \atop m + 1}\right]
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left[{r \atop k}\right] \binom k m = \left[{r + 1 \atop m + 1}\right]$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left[{r + 1 \atop k}\right] \binom k m = \left[{r + 2 \atop m + 1}\right]$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left[{r + 1 \atop k}\right] \binom k m
+      | c = 
+}}
+{{eqn | r = \sum_k \left({r \left[{r \atop k}\right] + \left[{r \atop k - 1}\right]}\right) \binom k m
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = r \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k - 1}\right] \binom k m
+      | c = 
+}}
+{{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom k m
+      | c = [[Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \left({\binom {k - 1} m + \binom {k - 1} {m - 1} }\right)
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} m + \sum_k \left[{r \atop k - 1}\right] \binom {k - 1} {m - 1}
+      | c = 
+}}
+{{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \sum_k \left[{r \atop k}\right] \binom k m + \sum_k \left[{r \atop k}\right] \binom k {m - 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = r \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]
+      | c = [[Sum over k of Unsigned Stirling Numbers of the First Kind of n with k by k choose m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r + 1 \atop m + 1}\right] + \left[{r + 1 \atop m}\right]
+      | c = 
+}}
+{{eqn | r = \left[{r + 2 \atop m + 1}\right]
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] \binom k m = \left[{n + 1 \atop m + 1}\right]$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Semantically Equivalent Terms are Equal}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\tau_1, \tau_2$ be [[Definition:Term (Predicate Logic)|terms]].
+Suppose that they are [[Definition:Relative Semantic Equivalence of Terms|semantically equivalent]] with respect to the [[Definition:Empty Set|empty set]].
+Then $\tau_1 = \tau_2$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|While an exercise, this basically requires construction of the syntactic model in order to achieve full rigour}}
+\end{proof}<|endoftext|>
+\section{Down Mapping is Generated by Approximating Relation}
+Tags: Meet-Continuous Lattices, Approximating Relations
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous]] [[Definition:Meet Semilattice|meet semilattice]].
+Let $\mathit{Ids}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+Let $I$ be an [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+Let $f: S \to \mathit{Ids}\left({L}\right)$ be a [[Definition:Mapping|mapping]] such that
+:$\forall x \in S: x \preceq \sup I \implies f\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$
+and
+:$\forall x \in S: x \npreceq \sup I \implies f\left({x}\right) = x^\preceq$
+where $x^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $x$.
+Then
+:there exists an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal R$ on $S$ such that
+::$\forall s \in S: f\left({s}\right) = s^{\mathcal R}$
+where $s^{\mathcal R}$ denotes the [[Definition:Relation Segment|$\mathcal R$-segment]] of $s$.
+\end{theorem}
+
+\begin{proof}
+Define [[Definition:Relation|relation]] $\mathcal R$ on $S$:
+:$\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$
+By definition of [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$\forall x \in S: f\left({x}\right) = x^{\mathcal R}$
+We will prove that
+:$\mathcal R$ is an [[Definition:Approximating Relation|approximating relation]].
+Let $x \in S$.
+Suppose the case holds that:
+:$x \preceq \sup I$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sup \left({x^{\mathcal R} }\right)
+      | r = \sup f \left({x}\right)
+}}
+{{eqn | r = \sup \left\{ {x \wedge d: d \in I}\right\}
+      | c = definition of $f$
+}}
+{{eqn | r = x \wedge \sup I
+      | c = {{Defof|Meet-Continuous Lattice}}
+}}
+{{eqn | r = x
+      | c = [[Preceding iff Meet equals Less Operand]]
+}}
+{{end-eqn}}
+Suppose the case holds that
+:$x \npreceq \sup I$
+Thus
+{{begin-eqn}}
+{{eqn | l = \sup \left({x^{\mathcal R} }\right)
+      | r = \sup f \left({x}\right)
+}}
+{{eqn | r = \sup \left({x^\preceq}\right)
+      | c = definition of $f$
+}}
+{{eqn | r = x
+      | c = [[Supremum of Lower Closure of Element]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding is Approximating Relation}
+Tags: Approximating Relations
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Then $\preceq$ is an [[Definition:Approximating Relation|approximating relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Define $\mathcal R := \mathord\preceq$.
+By definitions of [[Definition:Lower Closure of Element|lower closure of element]] and [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$x^\preceq = x^{\mathcal R}$
+where:
+:$x^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $x$
+:$x^{\mathcal R}$ denotes the [[Definition:Relation Segment|$\mathcal R$-segment]] of $x$
+Thus by [[Supremum of Lower Closure of Element]]:
+:$x = \sup \left({x^\preceq}\right) = \sup \left({ x^{\mathcal R} }\right)$
+Hence $\preceq$ is an [[Definition:Approximating Relation|approximating relation]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Substitution Instance of Term is Term}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\beta, \tau$ be [[Definition:Term (Predicate Logic)|terms of predicate logic]].
+Let $x \in \operatorname {VAR}$ be a [[Definition:Variable (Logic)|variable]].
+Let $\map \beta {x \gets \tau}$ be the [[Definition:Substitution Instance of Term|substitution instance of $\beta$ substituting $\tau$ for $x$]].
+Then $\map \beta {x \gets \tau}$ is a [[Definition:Term (Predicate Logic)|term]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] on the [[Definition:Term (Predicate Logic)|definition of term]], applied to $\beta$.
+If $\beta = y$ for some [[Definition:Variable (Logic)|variable]] $y$, then:
+:$\map \beta {x \gets \tau} = \begin{cases} \tau & : \text {if $y = x$} \\ y &: \text {otherwise} \end{cases}$
+In either case, $\map \beta {x \gets \tau}$ is a [[Definition:Term (Predicate Logic)|term]].
+If $\beta = \map f {\tau_1, \ldots, \tau_n}$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then:
+:$\map \beta {x \gets \tau} = \map f {\map {\tau_1} {x \gets \tau}, \ldots, \map {\tau_n} {x \gets \tau} }$
+By the induction hypothesis, each $\map {\tau_i} {x \gets \tau}$ is a [[Definition:Term (Predicate Logic)|term]].
+Hence so is $\map \beta {x \gets \tau}$.
+The result follows by the [[Principle of Structural Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Substitution Theorem for Terms}
+Tags: Named Theorems, Predicate Logic
+
+\begin{theorem}
+Let $\beta, \tau$ be [[Definition:Term (Predicate Logic)|terms]].
+Let $x \in \mathrm{VAR}$ be a [[Definition:Variable (Logic)|variable]].
+Let $\beta \left({x \gets \tau}\right)$ be the [[Definition:Substitution Instance of Term|substitution instance of $\beta$ substituting $\tau$ for $x$]].
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\sigma$ be an [[Definition:Assignment for Term|assignment]] for $\beta$ and $\tau$.
+Suppose that:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = a$
+where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the [[Definition:Value of Term under Assignment|value of $\tau$ under $\sigma$]].
+Then:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta}\right) } \left[{\sigma + \left({x / a}\right)}\right]$
+where $\sigma + \left({x / a}\right)$ is the [[Definition:Extension of Assignment|extension of $\sigma$ by mapping $x$ to $a$]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] on the [[Definition:Term (Predicate Logic)|definition of term]], applied to $\beta$.
+If $\beta = y$ for some [[Definition:Variable (Logic)|variable]] $y$, then:
+:$\beta \left({x \gets \tau}\right) = \begin{cases} \tau &: \text{if $y = x$} \\ y &: \text{otherwise} \end{cases}$
+In the first case:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau }\right) } \left[{\sigma}\right]
+}}
+{{eqn|r = a
+     |c = Definition of $a$
+}}
+{{eqn|r = \left({\sigma + \left({x / a}\right)}\right) \left({x}\right)
+     |c = Definition of [[Definition:Extension of Assignment|Extension of Assignment]]
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta }\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+In the second case:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = \mathop{ \operatorname{val}_{\mathcal A} \left({y}\right) } \left[{\sigma}\right]
+}}
+{{eqn|r = \sigma \left({y}\right)
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = \left({\sigma + \left({x / a}\right)}\right) \left({y}\right)
+     |c = Definition of [[Definition:Extension of Assignment|Extension of Assignment]]
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta }\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+as desired.
+If $\beta = f \left({\tau_1, \ldots, \tau_n}\right)$ and the induction hypothesis holds for $\tau_1, \ldots, \tau_n$, then:
+:$\beta \left({x \gets \tau}\right) = f \left({ \tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right) }\right)$
+Now:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = \mathop{ \operatorname{val}_{\mathcal A} \left({f \left({ \tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right) }\right) }\right) } \left[{\sigma}\right]
+}}
+{{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1 \left({x \gets \tau}\right)}\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n \left({x \gets \tau}\right)}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma + \left({x / a}\right)}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma + \left({x / a}\right)}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\beta}\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Term under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+as desired.
+The result follows by the [[Principle of Structural Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Substitution Instance of WFF is WFF}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]].
+Let $x \in \mathrm{VAR}$ be a [[Definition:Variable (Logic)|variable]].
+Let $\mathbf A \left({x \gets \tau}\right)$ be the [[Definition:Substitution Instance of Well-Formed Formula|substitution instance of $\mathbf A$ substituting $\tau$ for $x$]].
+Then $\mathbf A \left({x \gets \tau}\right)$ is a [[Definition:WFF of Predicate Logic|WFF]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] on the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]], applied to $\mathbf A$.
+If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
+:$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\right)$
+where $\tau_i \left({x \gets \tau}\right)$ is the [[Definition:Substitution Instance of Term|substitution instance]] of $\tau_i$.
+By [[Substitution Instance of Term is Term]], each such $\tau_i \left({x \gets \tau}\right)$ is again a [[Definition:Term (Predicate Logic)|term]].
+It follows that $\mathbf A \left({x \gets \tau}\right)$ is again a [[Definition:WFF of Predicate Logic|WFF]].
+If $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$, then:
+:$\mathbf A \left({x \gets \tau}\right) = \neg \mathbf B \left({x \gets \tau}\right)$
+and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+If $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$:
+:$\mathbf A \left({x \gets \tau}\right) = \mathbf B \left({x \gets \tau}\right) \circ \mathbf B' \left({x \gets \tau}\right)$
+and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+If $\mathbf A = \exists x: \mathbf B$, and the induction hypothesis applies to $\mathbf B$:
+:$\mathbf A \left({x \gets \tau}\right) = \exists x: \mathbf B \left({x \gets \tau}\right)$
+and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+If $\mathbf A = \forall x : \mathbf B$, and the induction hypothesis applies to $\mathbf B$:
+:$\mathbf A \left({x \gets \tau}\right) = \forall x: \mathbf B \left({x \gets \tau}\right)$
+and it follows from the induction hypothesis that $\mathbf A \left({x \gets \tau}\right)$ is a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+The result follows by the [[Principle of Structural Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \left[{n + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{n \atop m}\right]$
+where:
+:$\displaystyle \left[{n + 1 \atop k + 1}\right]$ etc. denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\dbinom k m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left[{n + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{n \atop m}\right]$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{1 \atop k + 1}\right] \binom k m \left({-1}\right)^{- m}
+      | r = \sum_k \delta_{1 \left({k + 1}\right)} \binom k m \left({-1}\right)^{- m}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = \sum_k \delta_{0 k} \binom k m \left({-1}\right)^{- m}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \binom 0 m \left({-1}\right)^{- m}
+      | c = all terms vanish except for $k = 0$
+}}
+{{eqn | r = \delta_{0 m} \left({-1}\right)^{- m}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \left[{0 \atop m}\right] \left({-1}\right)^{- m}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{end-eqn}}
+But when $m \ne 0$ we have that:
+:$\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- m} = 0 = \left[{0 \atop m}\right]$
+and when $m = 0$ we have that:
+:$\displaystyle \left[{0 \atop m}\right] \left({-1}\right)^{- 0} = 1 = \left[{0 \atop m}\right]$
+So $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{r \atop m}\right]$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left[{r + 2 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{r + 1 \atop m}\right]$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left[{r + 2 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | r = \sum_k \left({\left({r + 1}\right) \left[{r + 1 \atop k + 1}\right] + \left[{r + 1 \atop k}\right]}\right) \binom k m \left({-1}\right)^{k - m}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = \left({r + 1}\right) \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k}\right] \binom k m \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \binom k m \left({-1}\right)^{k - m}
+      | c = [[Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \left({\binom {k - 1} m + \binom {k - 1} {m - 1} }\right) \left({-1}\right)^{k - m}
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k}\right] \binom {k - 1} m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k}\right] \binom {k - 1} {m - 1} \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k + 1 - m} + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k {m - 1} \left({-1}\right)^{k + 1 - m}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] - \sum_k \left[{r + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} + \sum_k \left[{r + 1 \atop k + 1}\right] \binom k {m - 1} \left({-1}\right)^{k - \left({m - 1}\right)}
+      | c = 
+}}
+{{eqn | r = \left({r + 1}\right) \left[{r \atop m}\right] - \left[{r \atop m}\right] + \left[{r \atop m - 1}\right]
+      | c = [[Sum over k of Unsigned Stirling Numbers of the First Kind of n+1 with k+1 by k choose m by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r \left[{r \atop m}\right] + \left[{r \atop m - 1}\right]
+      | c = 
+}}
+{{eqn | r = \left[{r + 1 \atop m}\right]
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left[{n + 1 \atop k + 1}\right] \binom k m \left({-1}\right)^{k - m} = \left[{n \atop m}\right]$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Substitution Theorem for Well-Formed Formulas}
+Tags: Named Theorems, Predicate Logic
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+Let $x \in \mathrm{VAR}$ be a [[Definition:Variable (Logic)|variable]].
+Let $\tau$ be a [[Definition:Term (Predicate Logic)|term of predicate logic]] which is [[Definition:Freely Substitutable|freely substitutable]] for $x$ in $\mathbf A$.
+Let $\mathbf A \left({x \gets \tau}\right)$ be the [[Definition:Substitution Instance of Well-Formed Formula|substitution instance of $\mathbf A$ substituting $\tau$ for $x$]].
+Let $\mathcal A$ be a [[Definition:Structure for Predicate Logic|structure for predicate logic]].
+Let $\sigma$ be an [[Definition:Assignment for Structure|assignment]] for $\mathbf A$ and $\tau$.
+Suppose that:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = a$
+where $\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right]$ is the [[Definition:Value of Term under Assignment|value of $\tau$ under $\sigma$]].
+Then:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]$
+where $\sigma + \left({x / a}\right)$ is the [[Definition:Extension of Assignment|extension of $\sigma$ by mapping $x$ to $a$]].
+\end{theorem}
+
+\begin{proof}
+Proceed by the [[Principle of Structural Induction]] on the [[Definition:Bottom-Up Specification of Predicate Logic|bottom-up specification of predicate logic]], applied to $\mathbf A$.
+If $\mathbf A = p \left({\tau_1, \ldots, \tau_n}\right)$, then:
+:$\mathbf A \left({x \gets \tau}\right) = p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\right)$
+where $\tau_i \left({x \gets \tau}\right)$ is the [[Definition:Substitution Instance of Term|substitution instance]] of $\tau_i$.
+Now:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = \mathop{ \operatorname{val}_{\mathcal A} \left({p \left({\tau_1 \left({x \gets \tau}\right), \ldots, \tau_n \left({x \gets \tau}\right)}\right) }\right) } \left[{\sigma}\right]
+}}
+{{eqn|r = p_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1 \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = p_{\mathcal A} \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_1}\right) } \left[{\sigma + \left({x / a}\right)}\right], \ldots, \mathop{ \operatorname{val}_{\mathcal A} \left({\tau_n}\right) } \left[{\sigma + \left({x / a}\right)}\right] }\right)
+     |c = [[Substitution Theorem for Terms]]
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+Suppose $\mathbf A = \neg \mathbf B$ and the induction hypothesis applies to $\mathbf B$.
+Then since $\tau$ is also [[Definition:Freely Substitutable|free]] for $x$ in $\mathbf B$:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = f^\neg \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f^\neg \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right)}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+Suppose $\mathbf A = \mathbf B \circ \mathbf B'$ for $\circ$ one of $\land, \lor, \implies, \iff$ and the induction hypothesis applies to $\mathbf B, \mathbf B'$.
+Then since $\tau$ is also [[Definition:Freely Substitutable|free]] for $x$ in $\mathbf B$ and $\mathbf B'$:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]
+     |r = f^\circ \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma}\right] }\right)
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]]
+}}
+{{eqn|r = f^\circ \left({ \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right)}\right], \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right)}\right] }\right)
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]
+     |c = Definition of [[Definition:Value of Formula under Assignment|value under $\sigma + \left({x / a}\right)$]]
+}}
+{{end-eqn}}
+Suppose $\mathbf A = \exists y: \mathbf B$ or $\mathbf A = \forall y: \mathbf B$, and the induction hypothesis applies to $\mathbf B$.
+Because $\tau$ is [[Definition:Freely Substitutable|free]] for $x$ in $\mathbf A$, it must be that either $x$ does not [[Definition:Free Occurrence|occur freely]] in $\mathbf A$, or $y$ does not occur in $\tau$.
+In the first case:
+:$\mathbf A \left({x \gets \tau}\right) = \mathbf A$
+and by [[Value of Formula under Assignment Determined by Free Variables]]:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A }\right) } \left[{\sigma + \left({x / a}\right)}\right]$
+Now consider the case where $y$ does not occur in $\tau$.
+From the definition of [[Definition:Value of Formula under Assignment|value under $\sigma$]], $\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right]$ is determined by:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B \left({x \gets \tau}\right) }\right) } \left[{\sigma + \left({y / a'}\right) }\right]$
+where $a'$ ranges over $\mathcal A$.
+Now from [[Value of Term under Assignment Determined by Variables]], since $y$ does not occur in $\tau$:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma + \left({y / a'}\right)}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\tau}\right) } \left[{\sigma}\right] = a$
+for all $a'$.
+Hence the induction hypothesis also applies to the [[Definition:Assignment for Structure|assignment]] $\sigma + \left({y / a'}\right)$.
+Thus, for all $a'$:
+{{begin-eqn}}
+{{eqn|l = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B \left({x \gets \tau}\right) }\right) } \left[{\sigma + \left({y / a'}\right) }\right]
+     |r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({y / a'}\right) + \left({x / a}\right) }\right]
+     |c = Induction Hypothesis
+}}
+{{eqn|r = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf B}\right) } \left[{\sigma + \left({x / a}\right) + \left({y / a'}\right) }\right]
+     |c = Definition of [[Definition:Extension of Assignment|Extension of Assignment]]
+}}
+{{end-eqn}}
+from which we infer:
+:$\mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A \left({x \gets \tau}\right) }\right) } \left[{\sigma}\right] = \mathop{ \operatorname{val}_{\mathcal A} \left({\mathbf A}\right) } \left[{\sigma + \left({x / a}\right)}\right]$
+as desired.
+The result follows by the [[Principle of Structural Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$
+where:
+:$\displaystyle \left\{ {k \atop m}\right\}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left\{ {k \atop m}\right\} \binom 0 k
+      | r = \sum_k \left\{ {k \atop m}\right\} \delta_{0 k}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \left\{ {0 \atop m}\right\}
+      | c = all terms vanish except for $k = 0$
+}}
+{{eqn | r = \delta_{0 m}
+      | c = [[Stirling Number of the Second Kind of 0]]
+}}
+{{eqn | r = \delta_{1 \left({m + 1}\right)}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \left\{ {1 \atop m + 1}\right\}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = \left\{ {0 + 1 \atop m + 1}\right\}
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom r k = \left\{ {r + 1 \atop m + 1}\right\}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left\{ {k \atop m}\right\} \binom {r + 1} k = \left\{ {r + 2 \atop m + 1}\right\}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left\{ {k \atop m}\right\} \binom {r + 1} k
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {k \atop m}\right\} \left({\binom r k + \binom r {k - 1} }\right)
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \sum_k \left\{ {k \atop m}\right\} \binom r k + \sum_k \left\{ {k \atop m}\right\} \binom r {k - 1}
+      | c = 
+}}
+{{eqn | r = \left\{ {r + 1 \atop m + 1}\right\} + \sum_k \left\{ {k \atop m}\right\} \binom r {k - 1}
+      | c = [[Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left\{ {r + 1 \atop m + 1}\right\} + \sum_k \left({m \left\{ {k - 1 \atop m}\right\} + \left\{ {k - 1 \atop m - 1}\right\} }\right) \binom r {k - 1}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = \left\{ {r + 1 \atop m + 1}\right\} + m \sum_k \left\{ {k - 1 \atop m}\right\} \binom r {k - 1} + \sum_k \left\{ {k - 1 \atop m - 1}\right\} \binom r {k - 1}
+      | c = 
+}}
+{{eqn | r = \left\{ {r + 1 \atop m + 1}\right\} + m \sum_k \left\{ {k \atop m}\right\} \binom r k + \sum_k \left\{ {k \atop m - 1}\right\} \binom r k
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \left\{ {r + 1 \atop m + 1}\right\} + m \left\{ {r + 1 \atop m + 1}\right\} + \left\{ {r + 1 \atop m}\right\}
+      | c = [[Sum over k of Stirling Numbers of the Second Kind of k with m by n choose k#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({m + 1}\right) \left\{ {r + 1 \atop m + 1}\right\} + \left\{ {r + 1 \atop m}\right\}
+      | c = 
+}}
+{{eqn | r = \left\{ {r + 2 \atop m + 1}\right\}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {k \atop m}\right\} \binom n k = \left\{ {n + 1 \atop m + 1}\right\}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom n k \left({-1}\right)^{n - k} = \left\{ {n \atop m}\right\}$
+where:
+:$\displaystyle \left\{ {k + 1 \atop m + 1}\right\}$ etc. denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom n k \left({-1}\right)^{n - k} = \left\{ {n \atop m}\right\}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | o =
+      | r = \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom 0 k \left({-1}\right)^{- k}
+}}
+{{eqn | r = \sum_k \left\{ {k + 1 \atop m + 1}\right\} \delta_{0 k} \left({-1}\right)^{- k}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \left\{ {1 \atop m + 1}\right\}
+      | c = all terms vanish except for $k = 0$
+}}
+{{eqn | r = \delta_{1 \left({m + 1}\right)}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = \delta_{0 m}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \left\{ {0 \atop m}\right\}
+      | c = [[Stirling Number of the Second Kind of 0]]
+}}
+{{end-eqn}}
+So $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r k \left({-1}\right)^{r - k} = \left\{ {r \atop m}\right\}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom {r + 1} k \left({-1}\right)^{r + 1 - k} = \left\{ {r + 1 \atop m}\right\}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom {r + 1} k \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {k + 1 \atop m + 1}\right\} \left({\binom r k + \binom r {k - 1} }\right) \left({-1}\right)^{r + 1 - k}
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = -\sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r k \left({-1}\right)^{r - k} + \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)}
+      | c = 
+}}
+{{eqn | r = -\left\{ {r \atop m}\right\} + \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)}
+      | c = [[Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = -\left\{ {r \atop m}\right\} + \sum_k \left({\left({m + 1}\right) \left\{ {k \atop m + 1}\right\} + \left\{ {k \atop m }\right\} }\right) \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = -\left\{ {r \atop m}\right\} + \left({m + 1}\right) \sum_k \left\{ {k \atop m + 1}\right\} \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)}
+      | c = 
+}}
+{{eqn | o = 
+      | r = + \sum_k \left\{ {k \atop m }\right\} \binom r {k - 1} \left({-1}\right)^{r - \left({k - 1}\right)}
+      | c = 
+}}
+{{eqn | r = -\left\{ {r \atop m}\right\} + \left({m + 1}\right) \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom r k \left({-1}\right)^{r - k} + \sum_k \left\{ {k + 1 \atop m }\right\} \binom r k \left({-1}\right)^{r - k}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = -\left\{ {r \atop m}\right\} + \left({m + 1}\right) \left\{ {r \atop m}\right\} + \left\{ {r \atop m - 1}\right\}
+      | c = [[Sum over k of Stirling Numbers of the Second Kind of k+1 with m+1 by n choose k by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = m \left\{ {r \atop m}\right\} + \left\{ {r \atop m - 1}\right\}
+      | c = 
+}}
+{{eqn | r = \left\{ {r + 1 \atop m}\right\}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {k + 1 \atop m + 1}\right\} \binom n k \left({-1}\right)^{n - k} = \left\{ {n \atop m}\right\}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Instantiation/Informal Statement}
+Tags: Predicate Logic
+
+\begin{theorem}
+Suppose we have a [[Definition:Universal Statement|universal statement]]:
+:$\forall x: \map P x$
+where $\forall$ is the [[Definition:Universal Quantifier|universal quantifier]] and $\map P x$ is a [[Definition:Propositional Function|propositional function]].
+Then we can deduce:
+:$\map P {\mathbf a}$
+where $\mathbf a$ is any arbitrary [[Definition:Object|object]] we care to choose in the [[Definition:Universe of Discourse|universe of discourse]].
+In [[Definition:Natural Language|natural language]]:
+:''Suppose $P$ is true of everything in the [[Definition:Universe of Discourse|universe of discourse]].''
+:''Let $\mathbf a$ be an element of the [[Definition:Universe of Discourse|universe of discourse]]."
+:''Then $P$ is true of $\mathbf a$.''
+\end{theorem}
+
+\begin{proof}
+In the language of [[Definition:Symbolic Logic|symbolic logic]]:
+{{begin-eqn}}
+{{eqn | lo= \forall x:
+      | l = \map P x
+      | o =
+      | c = 
+}}
+{{eqn | ll= \therefore
+      | l = \map P {\mathbf a}
+      | o =
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Universal Instantiation/Model}
+Tags: Predicate Logic
+
+\begin{theorem}
+Let $\map {\mathbf A} x$ be a [[Definition:WFF of Predicate Logic|WFF of predicate logic]].
+Let $\tau$ be a [[Definition:Term (Predicate Logic)|term]] which is [[Definition:Freely Substitutable|freely substitutable]] for $x$ in $\mathbf A$.
+Then $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a [[Definition:Tautology (Predicate Logic)|tautology]].
+\end{theorem}
+
+\begin{proof}
+Let $\AA$ be a [[Definition:Structure for Predicate Logic|structure]] on a [[Definition:Set|set]] $A$, and let $\sigma$ be an [[Definition:Assignment for Formula|assignment]] for $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$.
+Define:
+:$a_\tau := \mathop{ \operatorname{val}_\AA \left({\tau}\right) } \sqbrk \sigma$
+the [[Definition:Value of Term under Assignment|value]] of $\tau$ under $\sigma$.
+From the definition of [[Definition:Value of Formula under Assignment|value]] under $\sigma$:
+:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau} \sqbrk \sigma = \map {f^\to} {\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma, \map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma}$
+where $f^\to$ is the [[Definition:Conditional/Truth Function|truth function]] of $\implies$.
+We thus need to ascertain that if:
+:$\map {\mathrm {val}_\AA} {\forall x: \map {\mathbf A} x} \sqbrk \sigma = T$
+then also:
+:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = T$
+By definition of [[Definition:Value of Formula under Assignment|value]] under $\sigma$, the former amounts to:
+:$\map {\mathrm {val}_\AA} {\map {\mathbf A} x} \sqbrk {\sigma + \paren {x / a} } = T$
+for all $a \in A$.
+By the [[Substitution Theorem for Well-Formed Formulas]]:
+:$\map {\mathrm {val}_\AA} {\map {\mathbf A} \tau} \sqbrk \sigma = \map {\mathrm {val}_\AA} {\mathbf A} \sqbrk {\sigma + \paren {x / a_\tau} }$
+Since $a_\tau \in A$, the conclusion follows, and $\forall x: \map {\mathbf A} x \implies \map {\mathbf A} \tau$ is a [[Definition:Tautology (Predicate Logic)|tautology]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existential Generalisation/Informal Statement}
+Tags: Predicate Logic
+
+\begin{theorem}
+:$P \left({\mathbf a}\right) \vdash \exists x: P \left({x}\right)$
+Suppose we have the following:
+: We can find an arbitrary [[Definition:Object|object]] $\mathbf a$ in our [[Definition:Universe of Discourse|universe of discourse]] which has the [[Definition:Property|property]] $P$.
+Then we may infer that:
+: there exists in that universe ''at least one'' object $x$ which has that property $P$.
+This is called the '''Rule of Existential Generalisation''' and often appears in a proof with its abbreviation '''EG'''.
+\end{theorem}<|endoftext|>
+\section{Way Below is Approximating Relation}
+Tags: Approximating Relations, Continuous Lattices, Way Below Relation
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Then $\ll$ is an [[Definition:Approximating Relation|approximating relation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Define $\mathcal R := \mathord\ll$.
+By definitions of [[Definition:Way Below Closure|way below closure]] and [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$x^\ll = x^{\mathcal R}$
+where:
+:$x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$
+:$x^{\mathcal R}$ denotes the [[Definition:Relation Segment|$\mathcal R$-segment]] of $x$
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+Thus by [[Definition:Axiom of Approximation|axiom of approximation]]:
+:$x = \sup \left({x^\ll}\right) = \sup \left({ x^{\mathcal R} }\right)$
+Hence $\ll$ is an [[Definition:Approximating Relation|approximating relation]] on $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Relation Segments of Approximating Relations equals Way Below Closure}
+Tags: Approximating Relations, Meet-Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
+Let $\mathit{App}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relations]] on $S$.
+Let $x \in S$.
+Then
+: $\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = x^\ll$
+\end{theorem}
+
+\begin{proof}
+By [[Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element]]:
+:$\displaystyle \bigcap \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\} = x^\ll$
+where $\mathit{Ids}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+For all $I \in \mathit{Ids}$ define a [[Definition:Mapping|mapping]] $m_I: S \to \mathit{Ids}$:
+:$\forall x \in S: x \preceq \sup I \implies m_I\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$
+and
+:$\forall x \in S: x \npreceq \sup I \implies m_I\left({x}\right) = x^\preceq$
+By [[Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element]]:
+:$\forall x \in S: \displaystyle \bigcap \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} = x^\ll$
+We will prove that
+:$\left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\} \subseteq \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$
+Let $a \in \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}$
+Then
+:$\exists I \in \mathit{Ids}: a = m_I\left({x}\right)$
+By [[Down Mapping is Generated by Approximating Relation]]:
+:$\exists \mathcal R \in \mathit{App}\left({L}\right): \forall s \in S: m_I(s) = s^{\mathcal R}$
+Then
+:$a = x^{\mathcal R}$
+Thus
+:$a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$
+{{qed|lemma}}
+By [[Intersection of Family is Subset of Intersection of Subset of Family]]:
+:$\displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq x^\ll$
+We will prove that
+:$\left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} \subseteq \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+Let $a \in \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$
+Then
+:$\exists \mathcal R \in \mathit{App}\left({L}\right): a = x^{\mathcal R}$
+By definition of [[Definition:Approximating Relation|approximating relation]]:
+:$x = \sup a$
+By [[Relation Segment of Auxiliary Relation is Ideal]]:
+:$a \in \mathit{Ids}$
+Thus by definition of [[Definition:Reflexivity|reflexivity]]:
+:$a \in \left\{ {I \in \mathit{Ids}: x \preceq \sup I}\right\}$
+{{qed|lemma}}
+By [[Intersection of Family is Subset of Intersection of Subset of Family]]:
+:$x^\ll \subseteq \displaystyle \bigcap \left\{ {x^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\}$
+Hence the result by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Monotone Real Function/Increasing}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Increasing Real Function|increasing]] and [[Definition:Bounded Above Mapping|bounded above]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let the [[Definition:Supremum of Real-Valued Function|supremum]] of $f$ on $\openint a b$ be $L$.
+Then:
+:$\displaystyle \lim_{x \mathop \to b^-} \map f x = L$
+where $\displaystyle \lim_{x \mathop \to b^-} \map f x$ is the [[Definition:Limit from Left|limit of $f$ from the left]] at $b$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$.
+That is, that $L - \epsilon < \map f x < L + \epsilon$.
+As $L$ is an [[Definition:Upper Bound of Set|upper bound]] for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically happens.
+Since $L - \epsilon$ is ''not'' an upper bound for $f$ on $\openint a b$, it follows that:
+:$\exists y \in \openint a b: \map f y > L - \epsilon$
+But $f$ [[Definition:Increasing Real Function|increases]] on $\openint a b$.
+So:
+:$\forall x: y < x < b: L - \epsilon < \map f y \le \map f x$
+We choose $\delta = b - y$ and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Monotone Real Function/Decreasing}
+Tags: Limits of Functions
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Decreasing Real Function|decreasing]] and [[Definition:Bounded Below Mapping|bounded below]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let the [[Definition:Infimum of Real-Valued Function|infimum]] of $f$ on $\openint a b$ be $l$.
+Then:
+:$\displaystyle \lim_{x \mathop \to b^-} \map f x = l$
+where $\displaystyle \lim_{x \mathop \to b^-} \map f x$ is the [[Definition:Limit from Left|limit of $f$ from the left]] at $b$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon > 0$.
+We have to find a value of $\delta > 0$ such that:
+:$\forall x: b - \delta < x < b: \size {\map f x - l} < \epsilon$
+That is:
+:$l - \epsilon < \map f x < l + \epsilon$
+As $l$ is a [[Definition:Lower Bound of Set|lower bound]] for $f$ on $\openint a b$:
+:$l - \epsilon < \map f x$
+automatically happens
+Because $l + \epsilon$ is ''not'' a [[Definition:Lower Bound of Set|lower bound]] for $f$ on $\openint a b$:
+:$\exists y \in \openint a b: \map f y < l + \epsilon$
+But $f$ [[Definition:Decreasing Real Function|decreases]] on $\openint a b$.
+So:
+:$\forall x: y < x < b: \map f x \le \map f y < l + \epsilon$
+We choose $\delta = y - a$ and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of m choose k by -1^m-k by k to the n}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
+where:
+:$\dbinom m k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$\displaystyle {n \brace m}$ etc. denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$m!$ denotes a [[Definition:Factorial|factorial]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
+For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | r = \sum_k \binom 0 k \paren {-1}^{0 - k} k^n
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \delta_{0 k} \paren {-1}^{- k} k^n
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \paren {-1}^{- 0} 0^n
+      | c = all terms vanish except for $k = 0$
+}}
+{{eqn | r = \delta_{0 n}
+      | c = $0^n = 0$ except when $n = 0$, as $0^0 = 1$
+}}
+{{eqn | r = 0! {n \brace 0}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{end-eqn}}
+So $\map P 0$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \binom r k \paren {-1}^{r - k} k^n = r! {n \brace r}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n = \paren {r + 1}! {n \brace r + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {r + 1} k \paren {-1}^{r + 1 - k} k^n
+      | c = 
+}}
+{{eqn | r = \sum_k \paren {\binom r k + \binom r {k - 1} } \paren {-1}^{r + 1 - k} k^n
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = \sum_k \binom r k \paren {-1}^{r + 1 - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r + 1 - k} k^n
+      | c = 
+}}
+{{eqn | r = -\sum_k \binom r k \paren {-1}^{r - k} k^n + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n
+      | c = 
+}}
+{{eqn | r = -r! {n \brace r} + \sum_k \binom r {k - 1} \paren {-1}^{r - \paren {k - 1} } k^n
+      | c = [[Sum over k of m choose k by -1^m-k by k to the n#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = -r! {n \brace r} + \sum_k \binom r k \paren {-1}^{r - k} \paren {k + 1}^n
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{end-eqn}}
+{{finish|The next step to take is not clear.}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom m k \paren {-1}^{m - k} k^n = m! {n \brace m}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of m-n choose m+k by m+n choose n+k by Stirling Number of the Second Kind of m+k with k}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
+where:
+:$\dbinom {m - n} {m + k}$ etc. denote [[Definition:Binomial Coefficient|binomial coefficients]]
+:$\displaystyle {m + k \brace k}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]]
+:$\displaystyle {n \brack n - m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
+For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brace k}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \binom {- n} k \binom n {n + k}
+      | c = [[Stirling Number of the Second Kind of Number with Self]]
+}}
+{{eqn | r = \sum_k \binom {- n} k \delta_{0 k}
+      | c = as $\dbinom n {n + k} = 0$ for $k = 0$, and [[Binomial Coefficient with Self]]
+}}
+{{eqn | r = \binom {- n} 0
+      | c = All terms but where $k = 0$ vanish
+}}
+{{eqn | r = 1
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = {n \brack n - 0}
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Self]]
+}}
+{{end-eqn}}
+So $\map P 0$ is seen to hold.
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{finish|Needs to be established whether the case $\map P 1$ actually needs to be proved separately, or whether we can get away with using $\map P 0$ as our base case.}}
+Thus $\map P 1$ is seen to hold for all $m$.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brace k} = {n \brack n - r}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}
+      | c = 
+}}
+{{eqn | r = \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brace k} = {n \brack n - r + 1}
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+{{finish|Lots to do, not sure if this is a workable approach.}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brace k} = {n \brack n - m}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of m-n choose m+k by m+n choose n+k by Unsigned Stirling Number of the First Kind of m+k with k}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
+where:
+:$\dbinom {m - n} {m + k}$ etc. denote [[Definition:Binomial Coefficient|binomial coefficients]]
+:$\displaystyle {m + k \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\displaystyle {n \brace n - m}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $m$.
+For all $m \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\forall n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {0 - n} {0 + k} \binom {0 + n} {n + k} {0 + k \brack k}
+      | c = 
+}}
+{{eqn | r = \sum_k \binom {- n} k \binom n {n + k}
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Self]]
+}}
+{{eqn | r = \sum_k \binom {- n} k \delta_{0 k}
+      | c = as $\dbinom n {n + k} = 0$ for $k = 0$, and [[Binomial Coefficient with Self]]
+}}
+{{eqn | r = \binom {- n} 0
+      | c = All terms but where $k = 0$ vanish
+}}
+{{eqn | r = 1
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = {n \brace n - 0}
+      | c = [[Stirling Number of the Second Kind of Number with Self]]
+}}
+{{end-eqn}}
+So $\map P 0$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \binom {r - n} {r + k} \binom {r + n} {n + k} {r + k \brack k} = {n \brace n - r}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {r + 1 - n} {r + 1 + k} \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}
+      | c = 
+}}
+{{eqn | r = \sum_k \paren {\binom {r - n} {r + 1 + k} + \binom {r - n} {r + k} } \binom {r + 1 + n} {n + k} {r + 1 + k \brack k} = {n \brace n - r + 1}
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+{{finish|Lots to do, not sure if this is a workable approach.}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall m, n \in \Z_{\ge 0}: \displaystyle \sum_k \binom {m - n} {m + k} \binom {m + n} {n + k} {m + k \brack k} = {n \brace n - m}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m}
+Tags: Stirling Numbers, Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$
+where:
+:$\dbinom n m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$\displaystyle \left[{k \atop m}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\displaystyle \left\{ {n + 1 \atop k + 1}\right\}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $P \left({m}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{\ge 0}: \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | r = \sum_k \left\{ {0 + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | o = 
+      | c = 
+}}
+{{eqn | r = \sum_k \delta_{1 \left({k + 1}\right)} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{eqn | r = \sum_k \delta_{0 k} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{eqn | r = \left[{0 \atop m}\right] \left({-1}\right)^{- m}
+      | c = All terms but where $k = 0$ vanish
+}}
+{{eqn | r = \delta_{0 m} \left({-1}\right)^{- m}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = \delta_{0 m}
+      | c = multiplier irrelevant when $m \ne 0$
+}}
+{{eqn | r = \binom 0 m
+      | c = [[Zero Choose n]]
+}}
+{{end-eqn}}
+So $P \left({0}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom r m$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left\{ {r + 2 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom {r + 1} m$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left\{ {r + 2 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | r = \sum_k \left({\left({k + 1}\right) \left\{ {r + 1 \atop k + 1}\right\} + \left\{ {r + 1 \atop k}\right\} }\right) \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = [[Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \left({-1}\right)^{r + 1 - m} \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k}
+      | c = 
+}}
+{{eqn | r = \binom r m + \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} +  \left({-1}\right)^{r + 1 - m} \delta_{m \left({r + 1}\right)}
+      | c = [[Second Inversion Formula for Stirling Numbers]]
+}}
+{{eqn | r = \sum_k k \left\{ {r + 1 \atop k + 1}\right\} \left({\frac 1 k \left[{k + 1 \atop m}\right] - \frac 1 k \left[{k \atop m - 1}\right]}\right) \left({-1}\right)^{k - m}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{k - m} + \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k \atop m - 1}\right] \left({-1}\right)^{k - m}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r + 1 \atop k + 1}\right\} \left[{k + 1 \atop m}\right] \left({-1}\right)^{k - m} + \binom r {m - 1}
+      | c = [[Sum over k of Stirling Number of the Second Kind of n+1 with k+1 by Unsigned Stirling Number of the First Kind of k with m by -1^k-m#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = 
+}}
+{{eqn | r = \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - 1 - m} + \binom r {m - 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = 
+}}
+{{eqn | r = \left({-1}\right)^{r - m} \sum_k \left\{ {r + 1 \atop k}\right\} \left[{k \atop m}\right] \left({-1}\right)^{r + 1 - k} + \binom r {m - 1}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = 
+}}
+{{eqn | r = \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)} + \binom r {m - 1} + \binom r m - \left({-1}\right)^{r - m} \delta_{m \left({r + 1}\right)}
+      | c = [[Second Inversion Formula for Stirling Numbers]]
+}}
+{{eqn | r = \binom r {m - 1} + \binom r m
+      | c = simplifying
+}}
+{{eqn | r = \binom {r + 1} m
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_k \left\{ {n + 1 \atop k + 1}\right\} \left[{k \atop m}\right] \left({-1}\right)^{k - m} = \binom n m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k to n of Unsigned Stirling Number of the First Kind of k with m by n factorial over k factorial}
+Tags: Stirling Numbers, Factorials
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
+where:
+:$\displaystyle {k \brack m}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$ n!$ denotes a [[Definition:Factorial|factorial]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | r = \sum_{k \mathop \le 0} {k \brack m} \frac {0!} {k!}
+      | o = 
+      | c = 
+}}
+{{eqn | r = {0 \brack m} \frac 1 {0!}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = \delta_{0 m}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = \delta_{1 \paren {m + 1} }
+      | c = 
+}}
+{{eqn | r = {1 \brack m + 1}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{end-eqn}}
+So $\map P 0$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} = {r + 1 \brack m + 1}$
+from which it is to be shown that:
+:$\displaystyle \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!} = {r + 2 \brack m + 1}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{k \mathop \le r + 1} {k \brack m} \frac {\paren {r + 1}!} {k!}
+      | c = 
+}}
+{{eqn | r = {r + 1 \brack m} \frac {\paren {r + 1}!} {\paren {r + 1}!} + \paren {r + 1} \paren {\sum_{k \mathop \le r} {k \brack m} \frac {r!} {k!} }
+}}
+{{eqn | r = {r + 1 \brack m} + \paren {r + 1} {r + 1 \brack m + 1}
+}}
+{{eqn | r = {r + 2 \brack m + 1}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{\ge 0}: \sum_{k \mathop \le n} {k \brack m} \frac {n!} {k!} = {n + 1 \brack m + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Applications of Down Mappings at Element equals Way Below Closure of Element}
+Tags: Way Below Relation, Meet-Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
+Let $\mathit{Ids}$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+Let for all $I \in \mathit{Ids}$: $m_I: S \to \mathit{Ids}$ be a [[Definition:Mapping|mapping]]:
+:$\forall x \in S: x \preceq \sup I \implies m_I\left({x}\right) = \left\{ {x \wedge i: i \in I}\right\}$
+and
+:$\forall x \in S: x \npreceq \sup I \implies m_I\left({x}\right) = x^\preceq$
+where $x^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $x$.
+Let $x \in S$.
+Then
+:$\displaystyle \bigcap \left\{ {m_I \left({x}\right): I \in \mathit{Ids} }\right\} = x^\ll$
+where $x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}
+      | r = \left\{ {m_I\left({x}\right): I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {m_I\left({x}\right): I \in \mathit{Ids} \land x \npreceq \sup I}\right\}
+}}
+{{eqn | r = \left\{ {\left\{ {x \wedge i: i \in I}\right\}: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}
+      | c = definition of $m_I$
+}}
+{{eqn | r = \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cup \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}
+      | c = [[Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal]]
+}}
+{{end-eqn}}
+Thus
+{{begin-eqn}}
+{{eqn | l = \bigcap \left\{ {m_I\left({x}\right): I \in \mathit{Ids} }\right\}
+      | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap \bigcap \left\{ {x^\preceq: I \in \mathit{Ids} \land x \npreceq \sup I}\right\}
+      | c = 
+}}
+{{eqn | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap \bigcap \left\{ {x^\preceq}\right\}
+}}
+{{eqn | r = \bigcap \left\{ {\left({x^\preceq}\right) \cap I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap x^\preceq
+      | c = [[Intersection of Singleton]]
+}}
+{{eqn | r = \left({x^\preceq}\right) \cap \bigcap \left\{ {I: I \in \mathit{Ids} \land x \preceq \sup I}\right\} \cap x^\preceq
+      | c = [[Intersection Distributes over Intersection]]
+}}
+{{eqn | r = \left({x^\preceq}\right) \cap \left({x^\ll}\right) \cap x^\preceq
+      | c = [[Intersection of Ideals with Suprema Succeed Element equals Way Below Closure of Element]]
+}}
+{{eqn | r = x^\ll
+      | c = [[Intersection with Subset is Subset]] as $x^\ll \subseteq x^\preceq$ by [[Way Below implies Preceding]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k to n of Stirling Number of the Second Kind of k with m by m+1^n-k}
+Tags: Stirling Numbers
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+:$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$
+where $\displaystyle {k \brace m}$ etc. denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$
+=== Basis for the Induction ===
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | r = \sum_{k \mathop \le 0} {k \brace m} \paren {m + 1}^{0 - k}
+      | o = 
+      | c = 
+}}
+{{eqn | r = {0 \brace m} \paren {m + 1}^{0 - 0}
+}}
+{{eqn | r = \delta_{0 m}
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{eqn | r = \delta_{1 \paren {m + 1} }
+}}
+{{eqn | r = {1 \brace {m + 1} }
+      | c = [[Stirling Number of the Second Kind of 1]]
+}}
+{{end-eqn}}
+So $\map P 0$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} = {{r + 1} \brace {m + 1}}$
+from which it is to be shown that:
+:$\displaystyle \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k} = { {r + 2} \brace {m + 1} }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o =
+      | r = \sum_{k \mathop \le {r + 1} } {k \brace m} \paren {m + 1}^{r + 1 - k}
+}}
+{{eqn | r = { {r + 1} \brace m} \paren {m + 1}^{r + 1 - r - 1} + \paren {m + 1} \paren {\sum_{k \mathop \le r} {k \brace m} \paren {m + 1}^{r - k} }
+      | c = 
+}}
+{{eqn | r = { {r + 1} \brace m} + \paren {m + 1} { {r + 1} \brace {m + 1} }
+}}
+{{eqn | r = { {r + 2} \brace {m + 1} }
+      | c = {{Defof|Stirling Numbers of the Second Kind}}
+}}
+{{end-eqn}}
+So $\map P r \implies \map P {r + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \sum_{k \mathop \le n} {k \brace m} \paren {m + 1}^{n - k} = {{n + 1} \brace {m + 1}}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Choose Zero}
+Tags: Examples of Binomial Coefficients, Zero Choose Zero
+
+\begin{theorem}
+:$\dbinom 0 0 = 1$
+\end{theorem}
+
+\begin{proof}
+By [[Binomial Coefficient with Zero]]:
+:$\forall r \in \R: \dbinom r 0 = 1$
+Hence directly:
+:$\dbinom 0 0 = 1$
+\end{proof}
+
+\begin{proof}
+By [[Zero Choose n]]:
+:$\dbinom 0 n = \delta_{0 n}$
+where:
+:$\dbinom 0 n$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$\delta_{0 n}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+Hence directly:
+:$\dbinom 0 0 = 1$
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient/Examples/Number of Bridge Hands}
+Tags: Examples of Binomial Coefficients
+
+\begin{theorem}
+The total number $N$ of possible different hands for a game of [https://en.wikipedia.org/wiki/Contract_bridge bridge] is:
+:$N = \dfrac {52!} {13! \, 39!} = 635 \ 013 \ 559 \ 600$
+\end{theorem}
+
+\begin{proof}
+The total number of cards in a standard deck is $52$.
+The number of cards in a single bridge hand is $13$.
+Thus $N$ is equal to the number of ways $13$ things can be chosen from $52$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = N
+      | r = \dbinom {52} {23}
+      | c = [[Cardinality of Set of Subsets]]
+}}
+{{eqn | r = \frac {52!} {13! \left({52 - 13}\right)!}
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = \frac {52!} {13! \, 39!}
+      | c = 
+}}
+{{eqn | r = 635 \ 013 \ 559 \ 600
+      | c = after calculation
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Lower Closure of Element with Ideal equals Meet of Element and Ideal}
+Tags: Lower Closures
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $I$ be an [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$.
+Let $x \in S$.
+Then:
+:$\paren {x^\preceq} \cap I = \set {x \wedge i: i \in I}$
+where $x^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+=== First Inclusion ===
+Let $z \in \paren {x^\preceq} \cap I$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$z \in x^\preceq$ and $z \in I$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$z \preceq x$
+By [[Preceding iff Meet equals Less Operand]]:
+:$x \wedge z = z$
+Thus:
+:$z \in \set {x \wedge i: i \in I}$
+{{qed|lemma}}
+=== Second Inclusion ===
+Let $z \in \set {x \wedge i: i \in I}$
+Then
+:$\exists i \in I: z = x \wedge i$
+By [[Meet Precedes Operands]]:
+:$z \preceq x$ and $z \preceq i$
+By definition of [[Definition:Ideal in Ordered Set|ideal in ordered set]]:
+:$I$ is a [[Definition:Lower Set|lower set]].
+By definitions of [[Definition:Lower Set|lower set]] and [[Definition:Lower Closure of Element|lower closure of element]]:
+:$z \in I$ and $z \in x^\preceq$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$z \in \struct {x^\preceq} \cap I$
+{{qed|lemma}}
+Thus the result by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Extended Pascal's Triangle}
+Tags: Definitions: Binomial Coefficients
+
+\begin{theorem}
+$\begin{array}{r|rrrrrrrrrr}
+n & \binom n 0 & \binom n 1 & \binom n 2 & \binom n 3 & \binom n 4 & \binom n 5 & \binom n 6 & \binom n 7 & \binom n 8 & \binom n 9 & \binom n {10} & \binom n {11} & \binom n {12} \\
+\hline
+-3 & 1 & -3 &  6 & -10 &  15 & -21 &  28 & -36 &  45 & -55  &  66 & -78 &  91 \\
+-2 & 1 & -2 &  3 &  -4 &   5 &  -6 &   7 &  -8 &   9 & -10  &  11 & -12 &  13 \\
+-1 & 1 & -1 &  1 &  -1 &   1 &  -1 &   1 &  -1 &   1 &  -1  &   1 &  -1 &   1 \\
+ 0 & 1 &  0 &  0 &   0 &   0 &   0 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 1 & 1 &  1 &  0 &   0 &   0 &   0 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 2 & 1 &  2 &  1 &   0 &   0 &   0 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 3 & 1 &  3 &  3 &   1 &   0 &   0 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 4 & 1 &  4 &  6 &   4 &   1 &   0 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 5 & 1 &  5 & 10 &  10 &   5 &   1 &   0 &   0 &   0 &   0  &   0 &   0 &   0 \\
+ 6 & 1 &  6 & 15 &  20 &  15 &   6 &   1 &   0 &   0 &   0  &   0 &   0 &   0 \\
+\end{array}$
+\end{theorem}<|endoftext|>
+\section{Condition on Lower Coefficient for Binomial Coefficient to be Maximum}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $n, k \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]].
+Let $\dbinom n k$ denote the [[Definition:Binomial Coefficient|binomial coefficient]] of $n$ choose $k$.
+Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a [[Definition:Maximum Element|maximum]] is:
+:$k_\max = \floor {\dfrac n 2}$
+and also:
+:$k_\max = \ceiling {\dfrac n 2}$
+\end{theorem}
+
+\begin{proof}
+From [[Condition for Increasing Binomial Coefficients]]:
+:$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$
+Let $n$ be an [[Definition:Even Integer|even integer]], such that:
+:$n = 2 r$
+Then:
+:$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
+and:
+:$\forall k < r: \dbinom n r > \dbinom n k$
+From [[Symmetry Rule for Binomial Coefficients]]:
+:$\forall k < r: \dbinom n r > \dbinom n {n - k}$
+and so:
+:$\forall k > r: \dbinom n r > \dbinom n k$
+We also have:
+:$\dbinom n r = \dbinom n {n - r} = \dbinom n {n / 2}$
+Thus for all $k \ne r$:
+:$\dbinom n k < \dbinom n {n / 2}$
+and so:
+:$k_\max = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
+{{qed|lemma}}
+Let $n$ be an [[Definition:Odd Integer|odd integer]], such that:
+:$n = 2 r + 1$
+Then:
+:$r = \floor {\dfrac n 2}$
+As $r < n$:
+:$\forall k < r: \dbinom n r > \dbinom n k$
+When $k = r$ we also have:
+:$\dbinom n r = \dbinom n {n - r} = \dbinom n {r + 1}$
+Then from [[Symmetry Rule for Binomial Coefficients]]:
+:$\forall k < r: \dbinom n r > \dbinom n {n - k}$
+and so:
+:$\forall k > r + 1: \dbinom n r > \dbinom n k$
+We have that:
+:$r = \floor {\dfrac n 2}$
+and:
+:$r + 1 = \ceiling {\dfrac n 2}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Row in Pascal's Triangle forms Palindromic Sequence}
+Tags: Pascal's Triangle
+
+\begin{theorem}
+Each of the [[Definition:Row of Pascal's Triangle|rows]] of [[Definition:Pascal's Triangle|Pascal's triangle]] forms a [[Definition:Palindrome|palindromic]] [[Definition:Sequence|sequence]].
+\end{theorem}
+
+\begin{proof}
+The $n$th [[Definition:Row of Pascal's Triangle|row]] of [[Definition:Pascal's Triangle|Pascal's triangle]] consists of the [[Definition:Finite Sequence|finite sequence]]:
+:$\dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n$
+By the [[Symmetry Rule for Binomial Coefficients]]:
+:$\dbinom n m = \dbinom n {n - m}$
+Hence we can write the $n$th [[Definition:Row of Pascal's Triangle|row]] in reverse order:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \dbinom n n, \dbinom n {n - 1}, \dbinom n {n - 2}, \ldots, \dbinom n 2, \dbinom n 1, \dbinom n 0
+      | c = 
+}}
+{{eqn | r = \dbinom n {n - n}, \dbinom n {n - \left({n - 1}\right)}, \dbinom n {n - \left({n - 2}\right)}, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n {n - 0}
+      | c = 
+}}
+{{eqn | r = \dbinom n 0, \dbinom n 1, \dbinom n 2, \ldots, \dbinom n {n - 2}, \dbinom n {n - 1}, \dbinom n n
+      | c = 
+}}
+{{end-eqn}}
+and the [[Definition:Finite Sequence|sequences]] are seen to be the same.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{N Choose Negative Number is Zero}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Let $k \in \Z_{<0}$ be a [[Definition:Strictly Negative Integer|(strictly) negative integer]].
+Then:
+:$\dbinom n k = 0$
+\end{theorem}
+
+\begin{proof}
+From [[Pascal's Rule]] we have:
+:$\forall n, k \in \Z: \dbinom n {k - 1} = \dbinom {n + 1} k - \dbinom n k$
+Thus it is sufficient to prove that:
+:$\forall n \in \Z: \dbinom n {-1} = 0$
+So:
+{{begin-eqn}}
+{{eqn | l = \dbinom n {-1}
+      | r = \dbinom {n + 1} 0 - \dbinom n 0
+      | c = [[Pascal's Rule]]
+}}
+{{eqn | r = 1 - 1
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Binomial Coefficients]]
+jilju9t3idp64zejhlfrap6ma23dwna
+\end{proof}<|endoftext|>
+\section{Kummer's Theorem}
+Tags: Binomial Coefficients, Number Bases, Prime Numbers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $a, b \in \Z_{\ge 0}$.
+Let:
+:$p^n \divides \dbinom {a + b} b$
+but
+:$p^{n + 1} \nmid \dbinom {a + b} b$
+where:
+:$\divides$ denotes [[Definition:Divisor of Integer|divisibility]]
+:$\nmid$ denotes non-[[Definition:Divisor of Integer|divisibility]]
+:$\dbinom {a + b} b$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+Then $n$ equals the number of [[Definition:Carry Digit|carries]] that occur when $a$ is added to $b$ using the [[Definition:Classical Addition Algorithm|classical addition algorithm]] in [[Definition:Number Base|base $p$]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|We need to establish the [[Definition:Classical Algorithm|classical algorithms]] and build up the knowledge of exactly what a [[Definition:Carry Digit|carry]] is before tackling this delightful little number.}}
+{{Namedfor|Ernst Eduard Kummer|cat = Kummer}}
+\end{proof}<|endoftext|>
+\section{Numbers in Row of Pascal's Triangle all Odd iff Row number 2^n - 1}
+Tags: Pascal's Triangle
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then the non-zero elements of the $n$th [[Definition:Row of Pascal's Triangle|row]] of [[Definition:Pascal's Triangle|Pascal's triangle]] are all [[Definition:Odd Integer|odd]] {{iff}}:
+:$n = 2^m - 1$
+for some $m \in \Z_{\ge 0}$.
+{{:Definition:Pascal's Triangle}}
+As can be seen, the entries in rows $0, 1, 3, 7$ are all [[Definition:Odd Integer|odd]].
+\end{theorem}
+
+\begin{proof}
+The statement:
+:$\dbinom n k$ is [[Definition:Odd Integer|odd]]
+is equivalent to:
+:$\dbinom n k \equiv 1 \pmod p$
+The [[Lucas' Theorem/Corollary|corollary to Lucas' Theorem]] gives:
+:$\displaystyle \dbinom n k \equiv \prod_{j \mathop = 0}^r \dbinom {a_j} {b_j} \pmod p$
+where:
+: $n, k \in \Z_{\ge 0}$ and $p$ is [[Definition:Prime Number|prime]]
+: the [[Definition:Number Base|representations of $n$ and $k$ to the base $p$]] are given by:
+::$n = a_r p^r + \cdots + a_1 p + a_0$
+::$k = b_r p^r + \cdots + b_1 p + b_0$
+When $p = 2$, the [[Definition:Digit|digits]] $a_j, b_j$ are either $0$ or $1$.
+In order for $\dbinom n k \equiv 1 \pmod p$, it is [[Definition:Necessary and Sufficient|necessary and sufficient]] that $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$.
+In order for this to happen, $\dbinom {a_j} {b_j} = \dbinom 0 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 0$ or $\dbinom {a_j} {b_j} = \dbinom 1 1$.
+Suppose $a_i = 0$ for some $i \in \left\{ {0, 1, 2, \ldots, r}\right\}$.
+Then if $b_i = 1$:
+:$\dbinom {a_j} {b_j} = \dbinom 0 1 = 0$
+and so:
+: $\dbinom n k \equiv 0 \pmod p$
+for whichever $k$ (and there will be at least one) has [[Definition:Digit|digit]] $b_i = 1$.
+So the only way it can be assured that all $\dbinom {a_j} {b_j} \equiv 1 \pmod p$ for all $k \in \left\{ {0, 1, 2, \ldots, n}\right\}$ is for $a_j = 0$ for all $j \in \left\{ {0, 1, 2, \ldots, r}\right\}$.
+That is, for $n = 2^m - 1$ for some $m \in \Z_{\ge 0}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Fourth Powers}
+Tags: Sums of Sequences, Fourth Powers, Sum of Sequence of Fourth Powers
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 0}^n k^4 = \dfrac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Powers of Positive Integers]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n i^p
+      | r = 1^p + 2^p + \cdots + n^p
+      | c = 
+}}
+{{eqn | r = \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}
+      | c = 
+}}
+{{end-eqn}}
+where $B_k$ are the [[Definition:Bernoulli Numbers|Bernoulli numbers]].
+Setting $p = 4$:
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n i^4
+      | r = \frac {n^{4 + 1} } {4 + 1} + \sum_{k \mathop = 1}^4 \frac {B_k \, 4^{\underline {k - 1} } \, n^{4 - k + 1} } {k!}
+      | c = 
+}}
+{{eqn | r = \frac {n^5} 5 + \frac {B_1 \, 4^{\underline 0} \, n^4} {1!} + \frac {B_2 \, 4^{\underline 1} \, n^3} {2!} + \frac {B_3 \, 4^{\underline 2} \, n^2} {3!} + \frac {B_4 \, 4^{\underline 3} \, n} {4!}
+      | c = 
+}}
+{{eqn | r = \frac {n^5} 5 + \frac 1 2 \frac {n^4} {1!} + \frac 1 6 \frac {4 n^3} {2!} + 0 \frac {4 \times 3 n^2} {3!} - \frac 1 {30} \frac {4 \times 3 \times 2 n} {4!}
+      | c = {{Defof|Bernoulli Numbers}} and {{Defof|Falling Factorial}}
+}}
+{{eqn | r = \frac {n^5} 5 + \frac {n^4} 2 + \frac {n^3} 3 - \frac n {30}
+      | c = simplifying
+}}
+{{eqn | r = \frac {6 n^5 + 15 n^4 + 10 n^3 - n} {30}
+      | c = 
+}}
+{{eqn | r = \frac {n \paren {n + 1} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1} } {30}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+{{begin-eqn}}
+{{eqn | l = \binom k 4
+      | r = \frac {k \paren {k - 1} \paren {k - 2} \paren {k - 3} } {4 !}
+      | c = 
+}}
+{{eqn | l = \binom k 3
+      | r = \frac {k \paren {k - 1} \paren {k - 2} } {3 !}
+      | c = 
+}}
+{{eqn | l = \binom k 2
+      | r = \frac {k \paren {k - 1} } {2 !}
+      | c = 
+}}
+{{eqn | l = \binom k 1
+      | r = k 
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 24 \binom k 4
+      | r = k^4 - 6 k^3 + 11 k^2 - 6 k
+      | c = 
+}}
+{{eqn | l = 36 \binom k 3
+      | r = 6 k^3 - 18 k^2 + 12 k
+      | c = 
+}}
+{{eqn | l = 14 \binom k 2
+      | r = 7 k^2 - 7 k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k^4
+      | r = 24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^n k^4
+      | r = \sum_{k \mathop = 0}^n \paren {24 \binom k 4 + 36 \binom k 3 + 14 \binom k 2 + \binom k 1}
+      | c = 
+}}
+{{eqn | r = 24 \binom {n + 1} 5 + 36 \binom {n + 1} 4 + 14 \binom {n + 1} 3 + \binom {n + 1} 2
+      | c = [[Sum of Binomial Coefficients over Upper Index]]
+}}
+{{eqn | r = 24 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} \paren {n - 3} } {5 !}
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = 36 \frac {\paren {n + 1} n \paren {n - 1} \paren {n - 2} } {4 !}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = 14 \frac {\paren {n + 1} n \paren {n - 1} } {3 !}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac {\paren {n + 1} n } {2 !}
+      | c = 
+}}
+{{eqn | r = \paren {n + 1} n \paren {\frac {\paren {n - 1} \paren {n - 2} \paren {n - 3} } 5 + \frac {3 \paren {n - 1} \paren {n - 2} } 2 + \frac {7 \paren {n - 1} } 3 + \frac 1 2}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 \paren {n^3 - 6 n^2 + 11 n - 6 } + 45 \paren {n^2 - 3 n + 2} + 70 \paren {n - 1} + 15}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 n^3 - 36 n^2 + 66 n - 36 + 45 n^2 - 135 n + 90 + 70 n - 70 + 15}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {n + 1} n} {30} \paren {6 n^3 + 9 n^2 + n - 1}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {n + 1} n} {30} \paren {2 n + 1} \paren {3 n^2 + 3 n - 1}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {n + 1} n \paren {n + \frac 1 2} \paren {3 n^2 + 3 n - 1} } {15}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complete Lattice is Bounded}
+Tags: Bounded Lattices, Complete Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Then
+:$L$ is [[Definition:Bounded Ordered Set|bounded]]
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$\varnothing$ admits a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
+By [[Infimum of Empty Set is Greatest Element]]:
+:$\forall x \in S: x \preceq \inf \varnothing$
+Thus by definition:
+:$L$ is [[Definition:Bounded Above Set|bounded above]].
+By [[Supremum of Empty Set is Smallest Element]]:
+:$\forall x \in S: \sup \varnothing \preceq x$
+Thus by definition:
+:$L$ is [[Definition:Bounded Below Set|bounded below]].
+Hence $L$ is [[Definition:Bounded Ordered Set|bounded]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constant Function is Primitive Recursive/General Case}
+Tags: Primitive Recursive Functions, Constant Mappings
+
+\begin{theorem}
+The [[Definition:Constant Mapping|constant function]] of $k$ variables: $f_c^k: \N^k \to \N$, defined as:
+:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = c$ where $c \in \N$
+is [[Definition:Primitive Recursive Function|primitive recursive]].
+\end{theorem}
+
+\begin{proof}
+For $k \ge 1$, let $f_c^k$ be the [[Definition:Constant Mapping|constant function]] of $k$ variables with value $c$.
+We know from [[Constant Function is Primitive Recursive]] that $f_c^1$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+Now:
+:$\map {f_c^k} {n_1, n_2, \ldots, n_k} = \map {f_c^1} {n_1} = \map {f_c^1} {\map {\pr_1^k} {n_1, n_2, \ldots, n_k} }$
+where $\pr_1^k$ is a [[Definition:Projection Function|projection function]] which is [[Definition:Basic Primitive Recursive Function|basic primitive recursive]].
+So $f_c^k$ is obtained from the [[Definition:Primitive Recursive Function|primitive recursive function]] $f_c^1$ and the [[Definition:Basic Primitive Recursive Function|basic primitive recursive function]] $\pr_1^k$ by [[Definition:Substitution (Mathematical Logic)|substitution]].
+Hence by definition, $f_c^k$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+{{qed}}
+[[Category:Primitive Recursive Functions]]
+[[Category:Constant Mappings]]
+p7103pmh89wjmd8e0mfrg551rg2r1jz
+\end{proof}<|endoftext|>
+\section{Value of Multiplicative Function at One}
+Tags: Multiplicative Functions
+
+\begin{theorem}
+Let $f: \N \to \C$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
+If $f$ is not identically zero, then $\map f 1 = 1$.
+\end{theorem}
+
+\begin{proof}
+If $f$ is not identically zero, then:
+:$\exists m \in \Z: \map f m \ne 0$
+Then:
+:$\map f m = \map f {1 \times m} = \map f 1 \, \map f m$
+Hence $\map f 1 = 1$.
+{{qed}}
+[[Category:Multiplicative Functions]]
+3yg93q06vyrng6lsj376wzn4zqpmn45
+\end{proof}<|endoftext|>
+\section{Value of Multiplicative Function is Product of Values of Prime Power Factors}
+Tags: Multiplicative Functions
+
+\begin{theorem}
+Let $f: \N \to \C$ be a [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
+Let $n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$ be the [[Definition:Prime Decomposition|prime decomposition]] of $n$.
+Then:
+:$\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
+We also have:
+:$\forall i, j \in \closedint 1 n: i \ne j \implies p_i^{k_i} \perp p_j^{k_j}$
+So:
+:$\map f {p_i^{k_i} p_j^{k_j} } = \map f {p_i^{k_i} } \, \map f {p_j^{k_j} }$
+It is a simple [[Principle of Mathematical Induction|inductive process]] to show that $\map f n = \map f {p_1^{k_1} } \, \map f {p_2^{k_2} } \dotsm \map f {p_r^{k_r} }$.
+{{handwaving}}
+{{qed}}
+[[Category:Multiplicative Functions]]
+2wrphk1b2zl7ohc8fzbupknu3hs00xb
+\end{proof}<|endoftext|>
+\section{-1^n by -n choose k-1 equals -1^k by -k choose n-1}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+:$\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {-1}^n \dbinom {-n} {k - 1}
+      | r = \paren {-1}^n \paren {\paren {-1}^{-n - \paren {k - 1} } \binom {- \paren {k - 1 + 1} } {- n - \paren {k - 1} } }
+      | c = [[Moving Top Index to Bottom in Binomial Coefficient]]
+}}
+{{eqn | r = \paren {-1}^{- \paren {k - 1} } \binom {- k} {- \paren {n - 1} - k}
+      | c = 
+}}
+{{eqn | r = \paren {-1}^k \binom {- k} {- k - \paren {n - 1} }
+      | c = 
+}}
+{{eqn | r = \paren {-1}^k \binom {- k} {n - 1}
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{end-eqn}}
+{{questionable|Can someone explain the disappearance of a factor of $-1$? Seems to have to do with a negative bottom index}}
+\end{proof}<|endoftext|>
+\section{One Choose n}
+Tags: Examples of Binomial Coefficients
+
+\begin{theorem}
+:$\dbinom 1 n = \begin{cases} 1 & : n \in \left\{ {0, 1}\right\} \\ 0 & : \text {otherwise} \end{cases}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+:$\dbinom m n = \begin{cases}
+\dfrac {m!} {n! \left({m - n}\right)!} & : 0 \le n \le m \\
+& \\
+0 & : \text { otherwise } \end{cases}$
+Thus when $n > 1$:
+:$\dbinom 1 n = 0$
+and when $n < 0$:
+:$\dbinom 1 n = 0$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dbinom 1 0
+      | r = \dfrac {1!} {0! \left({0 - 1}\right)!}
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{eqn | r = 1
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \dbinom 1 1
+      | r = \dbinom 1 {1 - 1}
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \dbinom 1 0
+      | c = 
+}}
+{{eqn | r = 1
+      | c = from above
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of Binomial Coefficients]]
+rq46fe27b1uakci157m84qzw75tysil
+\end{proof}<|endoftext|>
+\section{Summations of Products of Binomial Coefficients}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+This page gathers together some identities concerning [[Definition:Summation|summations]] of products of [[Definition:Binomial Coefficient|binomial coefficients]].
+In the following, unless otherwise specified:
+:$k, m, n \in \Z$
+:$r, s, t \in \R$.
+\end{theorem}<|endoftext|>
+\section{Separable Discrete Space is Countable}
+Tags: Separable Spaces, Discrete Topology, Separable Discrete Space is Countable
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Discrete Space|discrete]] [[Definition:Topological Space|topological space]].
+Let $T$ be [[Definition:Separable Space|separable]].
+Then $S$ is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be [[Definition:Separable Space|separable]].
+By [[Space is Separable iff Density not greater than Aleph Zero]]:
+:$\map d T \le \aleph_0$
+where:
+:$\map d T$ denotes the [[Definition:Density of Topological Space|density]] of $T$
+:$\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]].
+By definition of [[Definition:Density of Topological Space|density]]:
+:$\exists A \subseteq S: A$ is [[Definition:Everywhere Dense|dense]] $\land \map d T = \card A$
+where $\card A$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+By definition of [[Definition:Everywhere Dense|dense set]]:
+:$A^- = S$
+where $A^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A$.
+By [[Set in Discrete Topology is Clopen]]:
+:$A$ is [[Definition:Closed Set (Topology)|closed]]
+Then by [[Set is Closed iff Equals Topological Closure]]:
+:$A^- = A$
+Thus by [[Countable iff Cardinality not greater than Aleph Zero]]:
+:$S$ is [[Definition:Countable Set|countable]]
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $T$ be [[Definition:Separable Space|separable]].
+{{AimForCont}} $S$ is [[Definition:Uncountable Set|uncountable]].
+Then by [[Uncountable Discrete Space is not Separable]], $T$ is not [[Definition:Separable Space|separable]].
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Basis for the Induction}
+Tags: Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
+
+\begin{theorem}
+Let $r, s, t \in \R, n \in \Z$.
+Consider the equation:
+:$\displaystyle (1): \quad \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+where $\dbinom {r - t k} k$ etc. are [[Definition:Binomial Coefficient|binomial coefficients]].
+Then equation $(1)$ holds for the special case where $s = n - 1 - r + n t$.
+\end{theorem}
+
+\begin{proof}
+Substituting $n - 1 - r + n t$ for $s$ in the {{RHS}}:
+{{begin-eqn}}
+{{eqn | l = \binom {r + s - t n} n
+      | r = \binom {r + \paren {n - 1 - r + n t} - t n} n
+      | c = 
+}}
+{{eqn | r = \binom {r + n - 1 - r + n t - t n} n
+      | c = 
+}}
+{{eqn | r = \binom {n - 1} n
+      | c = 
+}}
+{{end-eqn}}
+Substituting $n - 1 - r + n t$ for $s$ in the {{LHS}}:
+{{begin-eqn}}
+{{eqn | o =
+      | r = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \dfrac {\paren {r - t k}! \paren {n - 1 - r + t k}! \, r} {k! \paren {r - t k - k}! \paren {n - k}! \paren {k - 1 - r + t k}! \paren {r - t k} }
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \dfrac {\paren {r - t k - 1}! \paren {n - 1 - r + t k}!} {\paren {r - t k - k}! \paren {k - 1 - r + t k}!}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \prod_{0 \mathop < j \mathop < k} \paren {r - t k - j} \prod_{0 \mathop < j \mathop < n \mathop - k} \paren {n - 1 - r + t k - j}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop \ge 0} \frac r {n!} \binom n k \paren {-1}^{k - 1} \prod_{0 \mathop < j \mathop < k} \paren {-r + t k + j} \prod_{k \mathop \le j \mathop < n} \paren {- r + t k + j}
+      | c = 
+}}
+{{end-eqn}}
+The two products give a [[Definition:Polynomial|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n - 1$ in $k$.
+Hence the sum for all $k$ is $0$.
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {n - 1 - r + t k} {n - k} \frac r {r - t k}
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \binom {n - 1} n
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+Thus the equation indeed holds for the special case where $s = n - 1 - r + n t$.
+\end{proof}<|endoftext|>
+\section{Continuous iff Meet-Continuous and There Exists Smallest Auxiliary Approximating Relation}
+Tags: Approximating Relations, Continuous Lattices, Meet-Continuous Lattices
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
+Then:
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{iff}}
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] and there exists the [[Definition:Smallest Element|smallest]] [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$
+That is:
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{iff}}
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] and there exists an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal R$ on $S$
+::for every [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal Q$ on $S$: $\mathcal R \subseteq \mathcal Q$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
+Thus by [[Continuous Lattice is Meet-Continuous]]:
+:$L$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Thus by [[Way Below is Approximating Relation]] and [[Way Below Relation is Auxiliary Relation]]:
+:$\ll$ is [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
+Thus by [[Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation]]:
+:for every [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal Q$ on $S$: $\ll \subseteq \mathcal Q$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $L$ be [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Assume that
+:there exists an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal R$ on $S$
+::for every [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] $\mathcal Q$ on $S$: $\mathcal R \subseteq \mathcal Q$
+Let $x \in S$.
+By [[Intersection of Relation Segments of Approximating Relations equals Way Below Closure]]:
+: $\displaystyle \bigcap \set {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)} = x^\ll$
+where
+:$x^{\mathcal Q}$ denotes the [[Definition:Relation Segment|$\mathcal Q$-segment]] of $x$, 
+:$\mathit{App}\left({L}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relations]] on $S$.
+By [[Intersection is Subset/General Result]]:
+:$x^\ll \subseteq x^{\mathcal R}$
+By definition of [[Definition:Approximating Relation|approximating relation]]:
+:$x = \sup \left({x^{\mathcal R} }\right)$
+We will prove that
+:$\forall a \in \set {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}: x^{\mathcal R} \subseteq a$
+Let $a \in \left\{ {x^{\mathcal Q}: \mathcal Q \in \mathit{App}\left({L}\right)}\right\}$
+Then
+:$\exists \mathcal Q \in \mathit{App}\left({L}\right): a = x^{\mathcal Q}$
+By assumption:
+:$\mathcal R \subseteq \mathcal Q$
+Thus by [[Relation Segment is Increasing]]:
+:$x^{\mathcal R} \subseteq a$
+{{qed|lemma}}
+By [[Intersection is Largest Subset/General Result]]:
+:$x^{\mathcal R} \subseteq x^\ll$
+By definition of [[Definition:Set Equality|set equality]]:
+:$x^{\mathcal R} = x^\ll$
+Thus
+:$x = \sup \left({x^\ll}\right)$
+By definition:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]]
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$L$ is [[Definition:Up-Complete|up-complete]].
+Hence $L$ is [[Definition:Continuous Ordered Set|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Lattice iff Auxiliary Approximating Relation is Superset of Way Below Relation}
+Tags: Approximating Relations, Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $\mathit{App}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
+Then
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{iff}}
+:$\forall \mathcal R \in \mathit{App}\left({L}\right): \ll \subseteq \mathcal R$ and $\ll$ is an [[Definition:Approximating Relation|approximating relation]]
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
+Let $\mathcal R \in \mathit{App}\left({L}\right)$
+Let $\left({a, b}\right) \in S \times S$ such that
+:$a \ll b$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$a \in b^\ll$
+where $b^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $b$.
+By [[Complete Lattice is Bounded]] and [[Continuous Lattice is Meet-Continuous]]:
+$L$ is a [[Definition:Bounded Below Set|bounded below]] [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
+By [[Intersection of Relation Segments of Approximating Relations equals Way Below Closure]]:
+: $\displaystyle \bigcap \left\{ {b^{\mathcal R}: \mathcal R \in \mathit{App}\left({L}\right)}\right\} = b^\ll$
+where $b^{\mathcal R}$ denotes the [[Definition:Relation Segment|$\mathcal R$-segment]] of $b$.
+By [[Intersection is Subset/General Result]]:
+:$b^\ll \subseteq b^{\mathcal R}$
+By definition of [[Definition:Subset|subset]]:
+:$a \in b^{\mathcal R}$
+By definition of [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$\left({a, b}\right) \in \mathcal R$
+Thus by definition of [[Definition:Subset|subset]]:
+:$\ll \subseteq \mathcal R$
+Thus by [[Way Below is Approximating Relation]]:
+:$\ll$ is an [[Definition:Approximating Relation|approximating relation]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$\forall \mathcal R \in \mathit{App}\left({L}\right): \ll \subseteq \mathcal R$ and $\ll$ is an [[Definition:Approximating Relation|approximating relation]]
+Define $\mathcal R := \mathord \ll$.
+By definition of [[Definition:Approximating Relation|approximating relation]]:
+:$\forall x \in S: x = \sup \left({x^{\mathcal R} }\right)$
+By definitions of [[Definition:Way Below Closure|way below closure]] and [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$\forall x \in S: x = \sup \left({x^\ll}\right)$
+By definition:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+By [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]]
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$L$ is [[Definition:Up-Complete|up-complete]].
+Hence $L$ is [[Definition:Continuous Ordered Set|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Completely Additive Function is Additive}
+Tags: Number Theory, Ring Theory
+
+\begin{theorem}
+Let $f: \N \to \C$ be a [[Definition:Completely Additive Function|completely additive function]].
+Then $f$ is also [[Definition:Additive Arithmetic Function|additive]].
+\end{theorem}
+
+\begin{proof}
+Let $m, n$ be [[Definition:Coprime Integers|coprime integers]].
+Then in particular, $m, n \in \Z$.
+Hence, since $f$ is [[Definition:Completely Additive Function|completely additive]]:
+:$f \left({m \times n}\right) = f \left({m}\right) + f \left({n}\right)$
+and $f$ is [[Definition:Additive Arithmetic Function|additive]], as desired.
+{{qed}}
+[[Category:Number Theory]]
+[[Category:Ring Theory]]
+okn9cbt94w16hdr1f8wcq1ox7gqmhnb
+\end{proof}<|endoftext|>
+\section{Real Logarithm is Completely Additive}
+Tags: Logarithms, Number Theory
+
+\begin{theorem}
+Let $\log_b: \R_{>0} \to \R$ be the [[Definition:Real General Logarithm|real logarithm]] to [[Definition:Base of Logarithm|base]] $b$.
+Then $\log_b$ is [[Definition:Completely Additive Function|completely additive]].
+\end{theorem}
+
+\begin{proof}
+From [[Sum of General Logarithms]]:
+:$\log_b x + \log_b y = \log_b \left({x y}\right)$
+The result follows from the definition of [[Definition:Completely Additive Function|complete additivity]].
+{{qed}}
+[[Category:Logarithms]]
+[[Category:Number Theory]]
+8wryrdf91qouep0q87julzna5febp3i
+\end{proof}<|endoftext|>
+\section{Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk/Proof 1/Lemma}
+Tags: Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk
+
+\begin{theorem}
+Let this hold for $\tuple {r, s, t, n}$:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+and also for $\tuple {r, s - t, t, n - 1}$.
+Then it also holds for $\tuple {r, s + 1, t, n}$.
+\end{theorem}
+
+\begin{proof}
+Evaluating the equation for $\tuple {r, s - t, t, n - 1}$:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {\paren {s - t} - t \paren {\paren {n - 1} - k} } {\paren {n - 1} - k} \frac r {r - t k}
+      | r = \binom {r + \paren {s - t} - t \paren {n - 1} } {n - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t - t n + t + t k} {n - k - 1} \frac r {r - t k}
+      | r = \binom {r + s - t - t n + 1} {n - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \paren {n - k} } {n - k - 1} \frac r {r - t k}
+      | r = \binom {r + s - t n} {n - 1}
+      | c = 
+}}
+{{end-eqn}}
+Adding the equation in $\tuple {r, s, t, n}$:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \paren {\binom {s - t \paren {n - k} } {n - k - 1} + \binom {s - t \paren {n - k} } {n - k} } \frac r {r - t k}
+      | r = \binom {r + s - t n} {n - 1} + \binom {r + s - t n} n
+      | c = 
+}}
+{{eqn | l = \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s + 1 - t \paren {n - k} } {n - k } \frac r {r - t k}
+      | r = \binom {r + s + 1 - t n} n
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+Hence the equation holds for $\tuple {r, s + 1, t, n}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r-kt choose k by r over r-kt by z^k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Let $A_n \left({x, t}\right)$ be the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree $n$]] defined as:
+:$A_n \left({x, t}\right) := \dbinom {x - n t} n \dfrac x {x - n t}$
+for $x \ne n t$.
+Let $z = x^{t + 1} - x^t$.
+Then:
+:$\displaystyle \sum_k A_k \left({r, t}\right) z^k = x^r$
+for [[Definition:Sufficiently Small|sufficiently small]] $z$.
+\end{theorem}
+
+\begin{proof}
+From [[Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt|Sum over $k$ of $\left({-1}\right)^k$ by $\dbinom n k$ by $\dbinom {r - k t} n$ by $\dfrac r {r - k t}$]] and renaming variables:
+:$\displaystyle \sum_j \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} = \delta_{k 0}$
+where $\delta_{k 0}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Thus:
+:$\displaystyle \sum_{j, k} \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k = 1$
+We have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{j, k} \left({-1}\right)^j \dbinom k j \dbinom {r - j t} k \dfrac r {r - j t} w^k
+      | c = 
+}}
+{{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \sum_k \dbinom k j \dbinom {r - j t} k w^k
+      | c = 
+}}
+{{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \sum_k \dbinom {r - j t} j \dbinom {r - j t - j} {k - j} w^k 
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | r = \sum_j \left({-1}\right)^j \dfrac r {r - j t} \dbinom {r - j t} j \sum_k \dbinom {r - j t - j} {k - j} w^k  
+      | c = 
+}}
+{{eqn | r = \sum_j \left({-1}\right)^j A_j \left({r, t}\right) \sum_k \dbinom {r - j t - j} {k - j} w^k  
+      | c = Definition of $A_j \left({r, t}\right)$
+}}
+{{eqn | r = \sum_j \left({-1}\right)^j A_j \left({r, t}\right) \left({1 + w}\right)^{r - j t - j} w^j
+      | c = [[Binomial Theorem]]
+}}
+{{end-eqn}}
+Now let:
+:$x = \dfrac 1 {1 + w}$
+{{begin-eqn}}
+{{eqn | l = x
+      | o = :=
+      | r = \dfrac 1 {1 + w}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = -\frac w {\left({1 + w}\right)^{1 + t} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left({1 + w}\right)^{r - j k - j} w^j \left({-1}\right)^j
+      | r = z^r z^j
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_j A_j \left({r, t}\right) z^j \left({1 + w}\right)^r
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_j A_j \left({r, t}\right) z^j
+      | r = \left({1 + w}\right)^{- r}
+      | c = 
+}}
+{{eqn | r = x^r
+      | c = 
+}}
+{{end-eqn}}
+{{finish|Establish the fact that we have convergence by ratio test and estimates for large $k$. Or use complex variable theory to establish that the function is analytic around $x {{=}} 1$.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Not Preceding implies Approximating Relation and not Preceding}
+Tags: Approximating Relations, Complete Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x, y \in S$ such that
+:$x \npreceq y$
+Let $\mathcal R$ be an [[Definition:Approximating Relation|approximating relation]] on $S$.
+Then
+:$\exists u \in S: \left({u, x}\right) \in \mathcal R \land u \npreceq y$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Approximating Relation|approximating relation]]:
+:$x = \sup \left({x^{\mathcal R} }\right)$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $x^{\mathcal R} \implies x \preceq y$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\exists u \in S: u \in x^{\mathcal R} \land u \npreceq y$
+Thus by definition of [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$\exists u \in S: \left({u, x}\right) \in \mathcal R \land u \npreceq y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Auxiliary Approximating Relation has Quasi Interpolation Property}
+Tags: Approximating Relations, Complete Lattices, Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x, z \in S$.
+Let $\mathcal R$ be an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$ such that
+:$\left({x, z}\right) \in \mathcal R \land x \ne z$
+Then
+:$\exists y \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land x \ne y$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$x \preceq z$
+By definition of $\prec$:
+:$x \prec z$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$z \nprec x$
+Then
+:$z \npreceq x$
+By [[Not Preceding implies Approximating Relation and not Preceding]]:
+:$\exists u \in S: \left({u, z}\right) \land u \npreceq x$
+Define $y = x \vee u$.
+Thus by [[Join Succeeds Operands]]:
+:$x \preceq y$
+Thus by definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({y, z}\right) \in \mathcal R$
+Thus by [[Preceding iff Join equals Larger Operand]]:
+:$x \ne y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding iff Join equals Larger Operand}
+Tags: Join and Meet, Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $x, y \in S$.
+Then
+:$x \preceq y$ {{iff}} $x \vee y = y$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$x \preceq y$
+By definition of [[Definition:Join (Order Theory)|join]]:
+:$x \vee y = \sup \left\{ {x, y}\right\}$
+By definitions of [[Definition:Upper Bound of Set|upper bound]] and [[Definition:Reflexivity|reflexivity]]:
+:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $\left\{ {x, y}\right\}$
+and
+:$\forall z \in S: z$ is [[Definition:Upper Bound of Set|upper bound]] for $\left\{ {x, y}\right\} \implies y \preceq z$
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$y = \sup \left\{ {x, y}\right\} = x \vee y$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$x \vee y = y$
+Thus by [[Join Succeeds Operands]]:
+:$x \preceq y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of -1^k by n choose k by r-kt choose n by r over r-kt}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+:$\displaystyle \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
+where $\delta_{n 0}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left({-1}\right)^k \dbinom 0 k \dbinom {r - k t} 0 \dfrac r {r - k t}
+      | r = \sum_k \left({-1}\right)^k \delta_{0 k} \dbinom {r - k t} 0 \dfrac r {r - k t}
+      | c = [[Zero Choose n]]
+}}
+{{eqn | r = \dbinom r 0 \dfrac r r
+      | c = All terms but for $k = 0$ vanish
+}}
+{{eqn | r = 1
+      | c = [[Binomial Coefficient with Zero]]
+}}
+{{eqn | r = \delta_{0 0}
+      | c = 
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ is seen to hold.
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({j}\right)$ is true, where $j \ge 0$, then it logically follows that $P \left({j + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left({-1}\right)^k \dbinom j k \dbinom {r - k t} j \dfrac r {r - k t} = \delta_{j 0}$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left({-1}\right)^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t} = \delta_{\left({j + 1}\right) 0}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \left({-1}\right)^k \dbinom {j + 1} k \dbinom {r - k t} {j + 1} \dfrac r {r - k t}
+      | c = 
+}}
+{{eqn | r = \sum_k \left({-1}\right)^k \left({\dbinom j k + \dbinom j {k - 1} }\right) \dbinom {r - k t} {j + 1} \dfrac r {r - k t}
+      | c = {{Defof|Binomial Coefficient}}
+}}
+{{end-eqn}}
+{{finish|brain not working tonight. Ought to be able to split binomial coeffs and reduce but it didn't do what it should. Perhaps try to solve it by reducing to factorials.}}
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0} \sum_k \left({-1}\right)^k \dbinom n k \dbinom {r - k t} n \dfrac r {r - k t} = \delta_{n 0}$
+\end{proof}<|endoftext|>
+\section{Sum over k of r-kt choose k by z^k}
+Tags: Binomial Coefficients, Sum over k of r-kt choose k by z^k
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
+Then:
+:$\displaystyle \sum_k \dbinom {r - t k} k z^k = \frac {x^{r + 1} } {\left({t + 1}\right)x - t}$
+where $\dbinom {r - t k} k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+From  [[Sum over k of r Choose k by s-kt Choose r by -1^k|Sum over $k$ of $\dbinom r k$ by $\dbinom {s - k t} r$ by $\paren {-1}^k$]] and renaming variables:
+:$\displaystyle \sum_j \paren {-1}^j \binom k j \binom {r - j t} k = t^k$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_{j, k} \binom k j \binom {r - j t} k \paren {-1}^j
+      | r = \sum_{k \mathop \ge 0} t^k
+      | c = when $k < 0$ we have $\dbinom {r - j t} k = 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_j \paren {-1}^j \sum_k \binom k j \binom {r - j t} k
+      | r = \frac 1 {1 - t}
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_j \paren {-1}^j \sum_k \binom {r - j t} j \binom {r - j t - k} {j - k}
+      | r = \frac 1 {1 - t}
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_j \paren {-1}^j \binom {r - j t} j \sum_k \binom {r - j t - k} {j - k}
+      | r = \frac 1 {1 - t}
+      | c = 
+}}
+{{end-eqn}}
+{{finish|... and so on}}
+\end{proof}
+
+\begin{proof}
+From [[Sum over k of r-kt choose k by r over r-kt by z^k|Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$]]:
+:$\displaystyle (1): \quad \sum_k \map {A_k} {r, t} z^k = x^r$
+where:
+:$\map {A_n} {x, t}$ is the [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree $n$]] defined as:
+::$\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
+:for $x \ne n t$
+:$z = x^{t + 1} - x^t$.
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT|Differentiation}} $z$:
+{{begin-eqn}}
+{{eqn | l = \sum_k \map {A_k} {r, t} k z^{k - 1}
+      | r = \frac \d {\d z} x^r
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \map {A_k} {r, t} k z^k
+      | r = z \frac \d {\d z} x^r
+      | c = 
+}}
+{{eqn | r = \paren {x^{t + 1} - x^t} r x^{r - 1} \frac {\d x} {\d z}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = x^{t + 1} - x^t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = \paren {t + 1} x^t - t x^{t - 1}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d z}
+      | r = \frac 1 {\paren {t + 1} x^t - t x^{t - 1} }
+      | c = [[Derivative of Inverse Function]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sum_k k \map {A_k} {r, t} z^k
+      | r = \frac {\paren {x^{t + 1} - x^t} r x^{r - 1} } {\paren {t + 1} x^t - t x^{t - 1} }
+      | c = 
+}}
+{{eqn | r = \frac {\paren {x - 1} r x^r} {\paren {t + 1} x - t}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \map {A_k} {r, t} z^k - \frac t r \sum_k k \map {A_k} {r, t} z^k
+      | r = x^r - \frac t r \frac {\paren {x - 1} r x^r} {\paren {t + 1} x - t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \paren {1 - \frac t r k} \map {A_k} {r, t} z^k 
+      | r = \frac {\paren {\paren {t + 1} x - t} x^r - t \paren {x - 1} x^r} {\paren {t + 1} x - t}
+      | c = 
+}}
+{{eqn | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \frac {r - t k} r \dbinom {r - k t} k \dfrac r {r - k t} z^k 
+      | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}
+      | c = substituting for $\map {A_k} {r, t}$
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \dbinom {r - k t} k z^k 
+      | r = \frac {x^{r + 1} } {\paren {t + 1} x - t}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of r+tk choose k by s-tk choose n-k}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
+Then:
+:$\displaystyle \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} = \sum_{k \mathop \ge 0} \dbinom {r + s - k} {n - k} t^k$
+where $\dbinom {r + t k} k$ etc. denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+Let $f \left({r, s, t, n}\right)$ be the [[Definition:Real-Valued Function|function]] defined as:
+:$\displaystyle f \left({r, s, t, n}\right) := \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}$
+We have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k}
+      | c = 
+}}
+{{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {r + t k} {r + t k}
+      | c = 
+}}
+{{eqn | r = \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac r {r + t k} + \sum_k \dbinom {r + t k} k \dbinom {s - t k} {n - k} \frac {t k} {r + t k}
+      | c = 
+}}
+{{end-eqn}}
+{{proof wanted|in progress}}
+\end{proof}<|endoftext|>
+\section{Auxiliary Approximating Relation has Interpolation Property}
+Tags: Approximating Relations, Complete Lattices, Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x, z \in S$ such that
+:$x \ll z \land x \ne z$
+Let $\mathcal R$ be an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
+Then
+:$\exists y \in S: \left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \land x \ne y$
+\end{theorem}
+
+\begin{proof}
+Define $I := \left\{ {u \in S: \exists y \in S: \left({u, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R}\right\}$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({\bot, \bot}\right) \in \mathcal R$ and $\left({\bot, z}\right) \in \mathcal R$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
+Then
+:$\bot \in I$
+By definition:
+:$I$ is a [[Definition:Non-Empty Set|non-empty set]].
+We will prove that
+:$I$ is a [[Definition:Lower Set|lower set]].
+Let $a \in I, b \in S$ such that
+:$b \preceq a$
+By definition of $I$:
+:$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$
+By definitions of [[Definition:Auxiliary Relation|auxiliary relation]] and [[Definition:Reflexivity|reflexivity]]:
+:$\left({b, s}\right) \in \mathcal R$
+Thus
+:$b \in I$
+{{qed|lemma}}
+We will prove that
+:$I$ is [[Definition:Directed Subset|directed]].
+Let $a, b \in I$.
+By definition of $I$:
+:$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$
+and
+:$\exists t \in S: \left({b, t}\right) \in \mathcal R \land \left({t, z}\right) \in \mathcal R$
+By [[Auxiliary Relation is Congruent]]:
+:$\left({a \vee b, s \vee t}\right) \in \mathcal R$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({s \vee t, z}\right) \in \mathcal R$
+Thus by definition of $I$:
+:$a \vee b \in I$
+Thus by [[Join Succeeds Operands]]:
+:$a \preceq a \vee b$ and $b \preceq a \vee b$
+{{qed|lemma}}
+By definition:
+:$I$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+We will prove that
+:$I \subseteq z^{\mathcal R}$
+Let $a \in I$.
+By definition of $I$:
+:$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$a \preceq s$
+Again by definition of [[Definition:Auxiliary Relation|auxiliary relation]] and [[Definition:Reflexivity|reflexivity]]:
+:$\left({a, z}\right) \in \mathcal R$
+Thus by definition of [[Definition:Relation Segment|$\mathcal R$-segment]]:
+:$a \in z^{\mathcal R}$
+{{qed|lemma}}
+By [[Supremum of Subset]] and definition of [[Definition:Approximating Relation|approximating relation]]:
+:$\sup I \preceq \sup \left({z^{\mathcal R} }\right) = z$
+We will prove that
+:$\sup I = z$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$\sup I \ne z$
+By definition of $\prec$:
+:$\sup I \prec z$
+Then by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$z \npreceq \sup I$
+By [[Not Preceding implies Approximating Relation and not Preceding]]:
+:$\exists y \in S: \left({y, z}\right) \in \mathcal R \land y \npreceq \sup I$
+Again by [[Not Preceding implies Approximating Relation and not Preceding]]:
+:$\exists u \in S: \left({u, y}\right) \in \mathcal R \land u \npreceq \sup I$
+By definition of $I$:
+:$u \in I$
+This contradicts $u \preceq \sup I$ by definition of [[Definition:Supremum of Set|supremum]].
+{{qed|lemma}}
+By [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$x \in I$
+By definition of $I$:
+:$\exists s \in S: \left({x, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$x \preceq s$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]] and [[Definition:Reflexivity|reflexivity]]:
+:$\left({x, z}\right) \in \mathcal R$
+By [[Auxiliary Approximating Relation has Quasi Interpolation Property]]:
+:$\exists y \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land x \ne y$
+By definition of $\prec$:
+:$ x \prec y$
+Define $Y := s \vee y$
+By [[Join Succeeds Operands]]:
+:$s \preceq Y$ and $y \preceq Y$
+Then 
+:$x \ne Y$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({x, Y}\right) \in \mathcal R$
+Again by definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({Y, z}\right) \in \mathcal R$
+Hence
+:$\exists y \in S: \left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \land x \ne y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below has Strong Interpolation Property}
+Tags: Approximating Relations, Continuous Lattices, Way Below Relation
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Let $x, z \in S$ such that
+:$x \ll z \land x \ne z$
+Then
+:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
+\end{theorem}
+
+\begin{proof}
+By [[Way Below is Approximating Relation]] and [[Way Below Relation is Auxiliary Relation]]:
+:$\ll$ is an [[Definition:Auxiliary Relation|auxiliary]] [[Definition:Approximating Relation|approximating relation]] on $S$.
+Thus by [[Auxiliary Approximating Relation has Interpolation Property]]:
+:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below has Interpolation Property}
+Tags: Continuous Lattices, Way Below Relation
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Let $x, z \in S$ such that
+:$x \ll z$
+Then
+:$\exists y \in S: x \ll y \land y \ll z$
+\end{theorem}
+
+\begin{proof}
+Case $x \ne z$:
+By [[Way Below has Strong Interpolation Property]]:
+:$\exists y \in S: x \ll y \land y \ll z \land x \ne y$
+Thus
+:$\exists y \in S: x \ll y \land y \ll z$
+Case $x = z$:
+Define $y = x$
+Thus
+:$x \ll y \land y \ll z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Left Coset Equals Subgroup iff Element in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+:$x H = H \iff x \in H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x H
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x H
+      | r = e H
+      | c = [[Left Coset by Identity]]: $e H = H$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x e^{-1}
+      | o = \in
+      | r = H
+      | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = H
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Right Coset Equals Subgroup iff Element in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+:$H x = H \iff x \in H$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = H x
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = H x
+      | r = H e
+      | c = [[Right Coset by Identity]]: $H = H e$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x e^{-1}
+      | o = \in
+      | r = H
+      | c = [[Right Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = H
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Elements in Same Left Coset iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+: $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$ {{iff}} $x^{-1} y \in H$.
+\end{theorem}
+
+\begin{proof}
+Let: $x H$ denote the [[Definition:Left Coset|left coset]] of $H$ by $x$
+First we note that, from [[Congruence Class Modulo Subgroup is Coset]], we have that the [[Definition:Left Coset|left cosets]] of $G$ form a [[Definition:Partition (Set Theory)|partition]] of $G$.
+=== Sufficient Condition ===
+Suppose $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$.
+It follows from [[Congruence Class Modulo Subgroup is Coset]] that:
+: $x \in y H$ and $y \in x H$ {{iff}} $x H = y H$
+From [[Cosets are Equal iff Product with Inverse in Subgroup]], we have that:
+: $x H = y H \iff x^{-1} y \in H$
+So if $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$ then $x^{-1} y \in H$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that $x^{-1} y \in H$.
+From [[Left Cosets are Equal iff Product with Inverse in Subgroup]], we have that:
+: $x H = y H \iff x^{-1} y \in H$
+Again, it follows from [[Congruence Class Modulo Subgroup is Coset]] that:
+: $x \in y H$ and $y \in x H$ iff $x H = y H$
+and so $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elements in Same Right Coset iff Product with Inverse in Subgroup}
+Tags: Cosets
+
+\begin{theorem}
+: $x, y$ are in the same [[Definition:Right Coset|right coset]] of $H$ {{iff}} $x y^{-1} \in H$
+\end{theorem}
+
+\begin{proof}
+Let $H x$ denote the [[Definition:Right Coset|right coset]] of $H$ by $x$.
+First we note that, from [[Congruence Class Modulo Subgroup is Coset]], we have that the [[Definition:Right Coset|right cosets]] of $G$ form a [[Definition:Partition (Set Theory)|partition]] of $G$.
+=== Sufficient Condition ===
+Suppose $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$.
+It follows from [[Congruence Class Modulo Subgroup is Coset]] that:
+: if $x \in H y$ and $y \in H x$ iff $H x = H y$
+From [[Cosets are Equal iff Product with Inverse in Subgroup]], we have that:
+: $H x = H y \iff x y^{-1} \in H$
+So if $x, y$ are in the same [[Definition:Right Coset|right coset]] of $H$ then $x y^{-1} \in H$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Now suppose that $x y^{-1} \in H$.
+From [[Right Cosets are Equal iff Product with Inverse in Subgroup]], we have that:
+: $H x = H y \iff x y^{-1} \in H$
+It follows from [[Congruence Class Modulo Subgroup is Coset]] that:
+: $x \in H y$ and $y \in H x$ iff $H x = H y$
+and so:
+: $x, y$ are in the same [[Definition:Left Coset|left coset]] of $H$
+Hence $x, y$ are in the same [[Definition:Right Coset|right coset]] of $H$ if $x y^{-1} \in H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of m+r+s Choose k by n+r-s Choose n-k by r+k Choose m+n}
+Tags: Binomial Coefficients
+
+\begin{theorem}
+Let $m, n \in \Z_{\ge 0}$.
+Then:
+:$\displaystyle \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m - n} = \binom r m \binom s n$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \binom {r + k} {m - n}
+      | c = 
+}}
+{{eqn | r = \sum_k \binom {m - r + s} k \binom {n + r - s} {n - k} \left({\sum_j \binom r {m + n - j} \binom k j}\right)
+      | c = [[Chu-Vandermonde Identity]]
+}}
+{{eqn | r = \sum_k \sum_j \binom {m - r + s} j \binom {m - r + s - j} {k - j} \binom {n + r - s} {n - k} \binom r {m + n - j}
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \left({\sum_k \binom {m - r + s - j} {k - j} \binom {n + r - s} {n - k} }\right)
+      | c = 
+}}
+{{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \binom {m + n - j} {n - j}
+      | c = [[Chu-Vandermonde Identity]]
+}}
+{{eqn | r = \sum_j \binom {m - r + s} j \binom r {m + n - j} \binom {m + n - j} m
+      | c = [[Symmetry Rule for Binomial Coefficients]]
+}}
+{{eqn | r = \sum_j \binom {m - r + s} j \binom r m \binom {r - m} {n - j}
+      | c = [[Product of r Choose m with m Choose k|Product of $\dbinom r m$ with $\dbinom m k$]]
+}}
+{{eqn | r = \binom r m \left({\sum_j \binom {m - r + s} j \binom {r - m} {n - j} }\right)
+      | c = 
+}}
+{{eqn | r = \binom r m \binom s n
+      | c = [[Chu-Vandermonde Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Unsigned Stirling Numbers of First Kind by x^k}
+Tags: Stirling Numbers, Rising Factorials, Examples of Power Series
+
+\begin{theorem}
+:$\displaystyle \sum_k \left[{n \atop k}\right] x^k = x^{\overline n}$
+where:
+:$\displaystyle \left[{n \atop k}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$x^{\overline n}$ denotes [[Definition:Rising Factorial|$x$ to the $n$ rising]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] x^k
+      | r = x^{\underline n}
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \left({-1}\right)^{n - k} \left[{n \atop k}\right] \left({-x}\right)^k
+      | r = \left({-x}\right)^{\underline n}
+      | c = putting $-x$ for $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \left({-1}\right)^n \sum_k \left[{n \atop k}\right] x^k
+      | r = \left({-x}\right)^{\underline n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_k \left[{n \atop k}\right] x^k
+      | r = \left({-1}\right)^n \left({-x}\right)^{\underline n}
+      | c = 
+}}
+{{eqn | r = x^{\overline n}
+      | c = [[Rising Factorial in terms of Falling Factorial of Negative]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Capelli's Sum}
+Tags: Rising Factorials
+
+\begin{theorem}
+:$\displaystyle \left({x + y}\right)^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
+where:
+:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$x^{\overline k}$ denotes [[Definition:Rising Factorial|$x$ to the $k$ rising]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \left({x + y}\right)^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
+=== Basis for the Induction ===
+$P \left({1}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \left({x + y}\right)^{\overline 1}
+      | r = x^{\overline 1} + y^{\overline 1}
+      | c = [[Number to Power of One Rising is Itself]]
+}}
+{{eqn | r = x^{\overline 0} y^{\overline 1} + x^{\overline 1} y^{\overline 0}
+      | c = [[Number to Power of Zero Rising is One]]
+}}
+{{eqn | r = \binom 1 0 x^{\overline 0} y^{\overline 1} + \binom 1 1 x^{\overline 1} y^{\overline 0}
+      | c = [[One Choose n]]
+}}
+{{eqn | r = \sum_k \binom 1 k x^{\overline k} y^{\overline {1 - k} }
+      | c = 
+}}
+{{end-eqn}}
+Thus $P \left({1}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 1$, then it logically follows that $P \left({m + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \left({x + y}\right)^{\overline m} = \sum_k \binom m k x^{\overline k} y^{\overline {m - k} }$
+from which it is to be shown that:
+:$\displaystyle \left({x + y}\right)^{\overline {m + 1} } = \sum_k \binom {m + 1} k x^{\overline k} y^{\overline {m + 1 - k} }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \left({x + y}\right)^{\overline {m + 1} }
+      | r = \left({x + y}\right)^{\overline m} \left({x + y + m}\right)
+      | c = Definition of [[Definition:Rising Factorial|Rising Factorial]]
+}}
+{{eqn | r = \left({\sum_k \binom m k x^{\overline k} y^{\overline {m - k} } }\right) \left({x + y + m}\right)
+      | c = [[Capelli's Sum#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({\sum_k \binom m k x^{\overline k} y^{\overline {m - k} } }\right) \left({x + k + y + m - k}\right)
+      | c = 
+}}
+{{eqn | r = \sum_k \binom m k x^{\overline k} \left({x + k}\right) y^{\overline {m - k} } + \sum_k \binom m k x^{\overline k} y^{\overline {m - k} } \left({y + m - k}\right)
+      | c = 
+}}
+{{eqn | r = \sum_k \binom m k x^{\overline {k + 1} } y^{\overline {m - k} } + \sum_k \binom m k x^{\overline k} y^{\overline {m - k + 1} }
+      | c = Definition of [[Definition:Rising Factorial|Rising Factorial]]
+}}
+{{eqn | r = \sum_k \binom m k x^{\overline {k + 1} } y^{\overline {m - k} } + \sum_k \binom m {k + 1} x^{\overline {k + 1} } y^{\overline {m - k} }
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \sum_k \binom {m + 1} k x^{\overline {k + 1} } y^{\overline {m - k} }
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{> 0}: \left({x + y}\right)^{\overline n} = \sum_k \binom n k x^{\overline k} y^{\overline {n - k} }$
+\end{proof}<|endoftext|>
+\section{Torelli's Sum}
+Tags: Rising Factorials
+
+\begin{theorem}
+:$\displaystyle \left({x + y}\right)^{\overline n} = \sum_k \binom n k x \left({x - k z + 1}\right)^{\overline {k - 1} } \left({y + k z}\right)^{\overline {n - k} }$
+where:
+:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
+:$x^{\overline k}$ denotes [[Definition:Rising Factorial|$x$ to the $k$ rising]].
+\end{theorem}
+
+\begin{proof}
+From [[Rising Factorial as Factorial by Binomial Coefficient]]:
+:$\left({x + y}\right)^{\overline n} = n! \dbinom {x + y + n - 1} n$
+Recall [[Sum over k of r-tk Choose k by s-t(n-k) Choose n-k by r over r-tk|Sum over $k$ of $\dbinom {r - t k} k$ by $\dbinom {s - t \left({n - k}\right)} {n - k}$ by $\dfrac r {r - t k}$]]:
+:$\displaystyle \sum_{k \mathop \ge 0} \binom {r - t k} k \binom {s - t \left({n - k}\right)} {n - k} \frac r {r - t k} = \binom {r + s - t n} n$
+Let the following substitutions be made:
+:$r \gets x$
+:$t \gets -\left({1 - z}\right)$
+:$s \gets y - 1 + n z$
+and so to obtain:
+:$\displaystyle \dbinom {x + y + n - 1} n = \sum_k \dbinom {x + \left({1 - z}\right) k} k \dbinom {y - 1 + n z + \left({n - k}\right) \left({1 - z}\right)} {n - k} \dfrac x {x + \left({1 - z}\right) k}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dbinom {x + \left({1 - z}\right) k} k
+      | r = \dfrac {\left({x - k z + 1}\right)^{\overline k} } {k!}
+      | c = [[Rising Factorial as Factorial by Binomial Coefficient]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac x {x + \left({1 - z}\right) k} \dbinom {x + \left({1 - z}\right) k} k
+      | r = \dfrac {x \left({x - k z + 1}\right)^{\overline {k - 1} } } {k!}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \dbinom {y - 1 + n z + \left({n - k}\right) \left({1 - z}\right)} {n - k}
+      | r = \dbinom {y - 1 + n z + n - k - n z + k z} {n - k}
+      | c = 
+}}
+{{eqn | r = \dbinom {y + k z \left({n - k}\right) - 1} {n - k}
+      | c = 
+}}
+{{eqn | r = \frac 1 {\left({n - 1}\right)!} \left({y + k z}\right)^{\overline {n - k} }
+      | c = [[Rising Factorial as Factorial by Binomial Coefficient]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = n! \dbinom {x + y + n - 1} n
+      | r = \sum_k \frac {n!} {k! \left({n - k}\right)!} x \left({x - k z + 1}\right)^{\overline {k - 1} } \left({y + k z}\right)^{\overline {n - k} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left({x + y}\right)^{\overline n}
+      | r = \sum_k \binom n k x \left({x - k z + 1}\right)^{\overline {k - 1} } \left({y + k z}\right)^{\overline {n - k} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Ruggiero Torelli|cat = Torelli}}
+\end{proof}<|endoftext|>
+\section{Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element}
+Tags: Way Below Relation, Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Let $x, y$ be [[Definition:Element|elements]] of $S$.
+Then
+:$x \ll y$
+{{iff}}
+:for every [[Definition:Directed Subset|directed subset]] $D$ of $S$ such that $y \preceq \sup D$
+::there exists an [[Definition:Element|element]] $d$ of $D$: $x \ll d$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \ll y$.
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$y \preceq \sup D$
+By [[Way Below has Interpolation Property]]:
+:$\exists x' \in S: x \ll x' \land x' \ll y$
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d \in D: x' \preceq d$
+Thus by [[Preceding and Way Below implies Way Below]] and definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \ll d$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every [[Definition:Directed Subset|directed subset]] $D$ of $S$ such that $y \preceq \sup D$
+::there exists an [[Definition:Element|element]] $d$ of $D$: $x \ll d$
+By [[Way Below implies Preceding]]:
+:for every [[Definition:Directed Subset|directed subset]] $D$ of $S$ such that $y \preceq \sup D$
+::there exists an [[Definition:Element|element]] $d$ of $D$: $x \preceq d$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous iff Way Below Closure is Ideal and Element Precedes Supremum}
+Tags: Continuous Lattices, Way Below Relation, Up-Complete Semilattices
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Then
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{iff}}
+:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
+::for every [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
+where $x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
+Let $x \in S$.
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+By [[Way Below Closure is Lower Set]]:
+:$x^\ll$ is a [[Definition:Lower Set|lower set]].
+Thus by definition:
+:$x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+By definition of [[Definition:Axiom of Approximation|axiom of approximation]]:
+:$x = \sup \left({x^\ll}\right)$
+Thus by definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq \sup \left({x^\ll}\right)$
+Let $I$ be an [[Definition:Ideal in Ordered Set|ideal]] in $L$ such that
+:$x \preceq \sup I$
+We will prove that
+:$x^\ll \subseteq I$
+Let $y \in x^\ll$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$
+Thus by [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$y \in I$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
+::for every [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
+Let $x \in S$.
+By assumption:
+:$x \preceq \sup \left({x^\ll}\right)$
+By [[Way Below implies Preceding]]:
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \left({x^\ll}\right) \preceq x$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$x = \sup \left({x^\ll}\right)$
+Thus by definition:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+Thus by assumption and definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]].
+Hence $L$ is [[Definition:Continuous Ordered Set|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed}
+Tags: Characterization of Euclidean Borel Sigma-Algebra
+
+\begin{theorem}
+Let $\mathcal O^n$, $\mathcal C^n$ and $\mathcal K^n$ be the collections of [[Definition:Open Set (Topology)|open]] and [[Definition:Closed Set (Topology)|closed]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\left({\R^n, \tau}\right)$, respectively.
+Then:
+:$\sigma \left({\mathcal O^n}\right) = \sigma \left({\mathcal C^n}\right)$
+where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+Recall that a [[Definition:Closed Set (Topology)|closed set]] is by definition the [[Definition:Relative Complement|relative complement]] of an [[Definition:Open Set (Topology)|open set]].
+Hence [[Sigma-Algebra Generated by Complements of Generators]] applies to yield:
+:$\sigma \left({\mathcal O^n}\right) = \sigma \left({\mathcal C^n}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact}
+Tags: Characterization of Euclidean Borel Sigma-Algebra
+
+\begin{theorem}
+Let $\mathcal C^n$ and $\mathcal K^n$ be the collections of [[Definition:Closed Set (Topology)|closed]] and [[Definition:Compact (Real Analysis)|compact]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\left({\R^n, \tau}\right)$, respectively.
+Then:
+:$\sigma \left({\mathcal C^n}\right) = \sigma \left({\mathcal K^n}\right)$
+where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+By the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]], $\mathcal K^n \subseteq \mathcal C^n$.
+Thus from [[Generated Sigma-Algebra Preserves Subset]], $\sigma \left({\mathcal K^n}\right) \subseteq \sigma \left({\mathcal C^n}\right)$.
+Next, let, for all $n \in \N$, $B^- \left({\mathbf 0; n}\right)$ be the [[Definition:Closed Ball|closed ball]] of radius $n$ around $\mathbf 0$ in $\R^n$.
+Observe that $\R^n = \displaystyle \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right)$.
+Now let $U \in \mathcal C^n$ be a [[Definition:Closed Set (Topology)|closed]] subset of $\R^n$.
+Then from [[Intersection with Subset is Subset]] and [[Intersection Distributes over Union]]:
+:$\displaystyle U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right) = \bigcup_{n \mathop \in \N} \left({U \cap B^- \left({\mathbf 0; n}\right)}\right)$
+From [[Intersection of Closed Sets is Closed in Topological Space]], $U \cap B^- \left({\mathbf 0; n}\right)$ is [[Definition:Closed Set (Topology)|closed]] for all $n \in \N$.
+By definition, $B^- \left({\mathbf 0; n}\right)$ is [[Definition:Bounded Metric Space|bounded]].
+Thus, by the [[Heine-Borel Theorem/Euclidean Space|Heine-Borel Theorem]], $U \cap B^- \left({\mathbf 0; n}\right)$ is [[Definition:Compact|compact]].
+Thus, any [[Definition:Closed Set (Topology)|closed set]] is the [[Definition:Countable Union|countable union]] of [[Definition:Compact|compact]] sets.
+By the third axiom for a [[Definition:Sigma-Algebra|$\sigma$-algebra]], this means that $\mathcal C^n \subseteq \sigma \left({\mathcal K^n}\right)$.
+Now the definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]] ensures that $\sigma \left({\mathcal C^n}\right) \subseteq \sigma \left({\mathcal K^n}\right)$.
+Hence the result, by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Euclidean Borel Sigma-Algebra/Open equals Rectangle}
+Tags: Characterization of Euclidean Borel Sigma-Algebra
+
+\begin{theorem}
+Let $\mathcal O^n$ be the collection of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of the [[Definition:Euclidean Space|Euclidean space]] $\left({\R^n, \tau}\right)$.
+Let $\mathcal J_{ho}^n$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$.
+Then:
+:$\sigma \left({\mathcal O^n}\right) = \sigma \left({\mathcal J_{ho}^n}\right)$
+where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+Let $\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal J^n_{ho}$.
+Then:
+:$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) = \left(({-\infty \,.\,.\, \mathbf b}\right)) \cap \left[[{\mathbf a \,.\,.\, +\infty}\right))$
+provides a way of writing this [[Definition:Half-Open Rectangle|half-open $n$-rectangle]] as an [[Definition:Set Intersection|intersection]] of an [[Definition:Open Set (Topology)|open]] and a [[Definition:Closed Set (Topology)|closed set]].
+By [[Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed]], these are both in $\mathcal B \left({\R^n}\right)$, and so [[Sigma-Algebra Closed under Intersection]] yields:
+:$\left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \sigma \left({\mathcal O^n}\right)$
+Hence, by definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]]:
+:$\sigma \left({\mathcal J^n_{ho}}\right) \subseteq \sigma \left({\mathcal O^n}\right)$
+Denote $\mathbf 1 = \left({1, \ldots, 1}\right) \in \R^n$.
+Define, for all $k \in \N$, $\mathcal S \left({k}\right)$ by:
+:$\mathcal S \left({k}\right) := \left\{{ \left[\left[{2^{-k}\mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) : \mathbf j \in \Z^n}\right\}$
+It is immediate that $\displaystyle \bigcup \mathcal S \left({k}\right) = \R^n$ and $\mathcal S \left({k}\right) \subseteq \mathcal J^n_{ho}$.
+Also, $\mathcal S \left({k}\right)$ is [[Definition:Countable|countable]] from [[Cartesian Product of Countable Sets is Countable]].
+Now define, again for all $k \in \N$, $U_k$ by:
+:$\displaystyle U_k := \bigcup \, \left\{{S \in \mathcal S \left({k}\right): S \subseteq U}\right\}$
+From [[Set Union Preserves Subsets]], $U_k \subseteq U$.
+Also, $U_k \in \sigma \left({\mathcal J^n_{ho}}\right)$ since the [[Definition:Set Union|union]] is [[Definition:Countable|countable]].
+It follows that also $\displaystyle \bigcup_{k \mathop \in \N} U_k \in \sigma \left({\mathcal J^n_{ho}}\right)$.
+Next, it is to be shown that $\displaystyle \bigcup_{k \mathop \in \N} U_k = U$.
+Note that [[Set Union Preserves Subsets]] ensures $\displaystyle \bigcup_{k \mathop \in \N} U_k \subseteq U$.
+For the converse, let $\mathbf x \in U$.
+As $U$ is [[Definition:Open Set (Topology)|open]], there exists an $\epsilon > 0$ such that the [[Definition:Open Ball|open ball]] $B \left({\mathbf x; \epsilon}\right)$ is contained in $U$.
+Fix $k \in \N$ such that $\sqrt n \, 2^{-k} < \epsilon$, and find $\mathbf j \in \Z^n$ such that:
+:$\mathbf x \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$
+Now it is to be shown that:
+:$\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$
+To this end, observe that for any $\mathbf y \in \left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)$, it holds that:
+:$d \left({\mathbf x, \mathbf y}\right) \le \operatorname{diam} \left({\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right)}\right)$
+by definition of [[Definition:Diameter of Bounded Metric Subspace|diameter]].
+Now from [[Diameter of Rectangle]], the right-hand side equals:
+:$\left\Vert{2^{-k} \left({\mathbf j + \mathbf 1}\right) - 2^{-k} \mathbf j}\right\Vert = \left\Vert{2^{-k} \mathbf 1}\right\Vert = \sqrt{n} \, 2^{-k}$
+which is smaller than $\epsilon$ by the way $k$ was chosen.
+Hence:
+:$\left[\left[{2^{-k} \mathbf j \,.\,.\, 2^{-k} \left({\mathbf j + \mathbf 1}\right)}\right)\right) \subseteq B \left({\mathbf x; \epsilon}\right)$
+and so every $\mathbf x \in U$ is contained in some $U_k$.
+Thus it follows that $U \subseteq \displaystyle \bigcup_{k \mathop \in \N} U_k$.
+Thereby we have shown that:
+:$\sigma \left({\mathcal J^n_{ho}}\right) = \sigma \left({\mathcal O^n}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle}
+Tags: Characterization of Euclidean Borel Sigma-Algebra
+
+\begin{theorem}
+Let $\mathcal J_{ho}^n$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$.
+Let $\mathcal J^n_{ho, \text{rat}}$ be the collection of [[Definition:Half-Open Rectangle|half-open rectangles]] in $\R^n$ with [[Definition:Rational Number|rational]] endpoints.
+Then:
+:$\sigma \left({\mathcal J_{ho}^n}\right) = \sigma \left({\mathcal J^n_{ho, \text{rat}}}\right)$
+where $\sigma$ denotes [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+\end{theorem}
+
+\begin{proof}
+From [[Generated Sigma-Algebra Preserves Subset]]:
+:$\sigma \left({\mathcal J_{ho, \text{rat}}^n}\right) \subseteq \sigma \left({\mathcal J_{ho}^n}\right)$
+For the converse, it will suffice to show:
+:$\mathcal J_{ho}^n \subseteq \sigma \left({\mathcal J_{ho, \text{rat}}^n}\right)$
+by definition of [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]].
+So, let $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$ be a [[Definition:Half-Open Rectangle|half-open $n$-rectangle]].
+Let $\left({\mathbf a_m}\right)_{m \in \N}$ be a [[Definition:Sequence|sequence]] in $\Q^n$ with [[Definition:Limit of Sequence|limit]] $\mathbf a$.
+Also, let this [[Definition:Sequence|sequence]] be such that $m_1 > m_2 \implies \mathbf a_{m_1} > \mathbf a_{m_2}$, in the [[Definition:Componentwise Ordering|component-wise ordering]].
+Also, choose $\mathbf b' \in \Q^n$ such that $\mathbf b' > \mathbf b$, again in the [[Definition:Componentwise Ordering|component-wise ordering]].
+Then, for any $m \in \N$:
+:$\left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right) \in \mathcal J_{ho, \text{rat}}^n$
+By [[Sigma-Algebra Closed under Countable Intersection]], it follows that:
+:$\displaystyle \bigcap_{m \mathop \in \N} \left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right) \in \sigma \left({\mathcal J_{ho, \text{rat}}^n}\right)$
+Now observe, for $\mathbf x \in \R^n$:
+{{begin-eqn}}
+{{eqn | l = \mathbf x
+      | o = \in 
+      | r = \bigcap_{m \mathop \in \N} \left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right)
+}}
+{{eqn | ll= \iff
+      | l = \forall m \in \N: \mathbf x
+      | o = \in
+      | r = \left[\left[{\mathbf a_m \,.\,.\, \mathbf b'}\right)\right)
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | ll= \iff
+      | l = \forall m \in \N: \mathbf x
+      | o = \ge
+      | r = \mathbf a_m \land \mathbf x < \mathbf b'
+      | c = {{Defof|Half-Open Rectangle}}
+}}
+{{eqn | ll= \iff
+      | l = \mathbf x
+      | o = \ge
+      | r = \mathbf a \land \mathbf x < \mathbf b'
+      | c = $\left({\mathbf a_m}\right)_{m \in \N}$ is [[Definition:Increasing Sequence|increasing]] with [[Definition:Limit of Sequence|limit]] $\mathbf a$
+}}
+{{eqn | ll= \iff
+      | l = \mathbf x
+      | o = \in
+      | r = \left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right)
+      | c = {{Defof|Half-Open Rectangle}}
+}}
+{{end-eqn}}
+Next, let $\left({\mathbf b_m}\right)_{m \in \N}$ be an [[Definition:Increasing Sequence|increasing sequence]] in $\Q^n$ with [[Definition:Limit of Sequence|limit]] $\mathbf b$.
+Also, let $\mathbf a' \in \Q^n$ be such that $\mathbf a' < \mathbf a$.
+Again, it follows that $\left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right) \in \mathcal J_{ho, \text{rat}}^n$.
+Thus, by the third axiom for a [[Definition:Sigma-Algebra|$\sigma$-algebra]]:
+:$\displaystyle \bigcup_{m \mathop \in \N} \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right) \in \sigma \left({\mathcal J^n_{ho}}\right)$
+Similar to the above approach, for any $\mathbf x \in \R^n$:
+{{begin-eqn}}
+{{eqn | l = \mathbf x
+      | o = \in
+      | r = \bigcup_{m \mathop \in \N} \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right)
+}}
+{{eqn | ll= \iff
+      | l = \exists m \in \N: \mathbf x
+      | o = \in 
+      | r = \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right)
+      | c = {{Defof|Set Union}}
+}}
+{{eqn | ll= \iff
+      | l = \exists m \in \N: \mathbf x
+      | o = <
+      | r = \mathbf b_m \land \mathbf x \ge \mathbf a'
+      | c = {{Defof|Half-Open Rectangle}}
+}}
+{{eqn | ll= \iff
+      | l = \mathbf x
+      | o = <
+      | r = \mathbf b \land \mathbf x \ge \mathbf a'
+      | c = $\left({\mathbf b_m}\right)_{m \in \N}$ is [[Definition:Increasing Sequence|increasing]] with [[Definition:Limit of Sequence|limit]] $\mathbf b$
+}}
+{{eqn | ll= \iff
+      | l = \mathbf x
+      | o = \in
+      | r = \left[\left[{\mathbf a' \,.\,.\, \mathbf b}\right)\right)
+      | c = {{Defof|Half-Open Rectangle}}
+}}
+{{end-eqn}}
+Hence, it follows that:
+:$\displaystyle \bigcup_{m \mathop \in \N} \left[\left[{\mathbf a' \,.\,.\, \mathbf b_m}\right)\right) = \left[\left[{\mathbf a' \,.\,.\, \mathbf b}\right)\right)$
+whence the latter is in $\sigma \left({\mathcal J^n_{ho, \text{rat}}}\right)$.
+Hence by [[Sigma-Algebra Closed under Intersection]]:
+:$\left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right) \cap \left[\left[{\mathbf a' \,.\,.\, \mathbf b}\right)\right) \in \sigma \left({\mathcal J^n_{ho, \text{rat}}}\right)$
+and finally (the proof of) [[Half-Open Rectangles Closed under Intersection]] yields:
+:$\left[\left[{\mathbf a \,.\,.\, \mathbf b'}\right)\right) \cap \left[\left[{\mathbf a' \,.\,.\, \mathbf b}\right)\right) = \left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)$
+since $\mathbf a' < \mathbf a$ and $\mathbf b < \mathbf b'$, thus finishing the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rising Factorial as Factorial by Binomial Coefficient}
+Tags: Binomial Coefficients, Factorials, Rising Factorials
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+:$x^{\overline n} = n! \dbinom {x + n - 1} n$
+where:
+:$x^{\overline n}$ denotes [[Definition:Rising Factorial|$x$ to the $n$ rising]]
+:$n!$ denotes the [[Definition:Factorial|factorial of $n$]]
+:$\dbinom {x + n - 1} n$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Rising Factorial|$x$ to the $n$ rising]]:
+:$x^{\overline n} = x \left({x + 1}\right) \cdots \left({x + n - 1}\right)$
+By definition of the [[Definition:Binomial Coefficient/Real Numbers|binomial coefficient of a real number]]:
+:$\dbinom {x + n - 1} n = \dfrac {\left({x + n - 1}\right) \left({x + n - 2}\right) \cdots x} {n!}$
+Hence the result.
+{{qed}}
+[[Category:Binomial Coefficients]]
+[[Category:Factorials]]
+[[Category:Rising Factorials]]
+281ldsrexhi9ewe5sjbi9fuerv0gh89
+\end{proof}<|endoftext|>
+\section{Ramus's Identity}
+Tags: Binomial Coefficients, Ramus's Identity
+
+\begin{theorem}
+Let $k, m, n \in \Z_{\ge 0}$ be [[Definition:Positive Integer|positive integers]] such that $0 \le k < m$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop \ge 0} \binom n {j m + k}
+      | r = \dbinom n k + \dbinom n {m + k} +  \dbinom n {2 m + k} + \cdots 
+      | c = 
+}}
+{{eqn | r = \dfrac 1 m \sum_{0 \mathop \le j \mathop < m} \paren {2 \cos \dfrac {j \pi} m}^n \cos \dfrac {j \paren {n - 2 k} \pi} m
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Let $\omega := e^{2 \pi i / m}$.
+Then by the [[Binomial Theorem]]:
+:$(1): \quad \displaystyle \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k} = \sum_t \sum_{0 \mathop \le j \mathop < m} \binom n t \omega^{j \paren {t - k} }$
+By [[Sum of Geometric Sequence]]:
+:$\displaystyle \sum_{0 \mathop \ge j \mathop < m} \omega^{j \paren {t - k} } = \begin{cases} \dfrac {1 - \omega^{m \paren {t - k} } } {1 - \omega^{t - k} } & : t - k \not \equiv 0 \pmod m \\ m & : t \equiv k \pmod m \end{cases}$
+We have that:
+:$\omega = \exp \dfrac {2 \pi i} m \implies \omega^m = 1$
+and so:
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le j \mathop < m} \omega^{r j}
+      | r = \frac {1 - \omega^{r m} } {1 - \omega^r}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = m \sqbrk {r \equiv 0 \pmod m}
+      | c = where $\sqbrk {\, \cdots \,}$ is [[Definition:Iverson's Convention|Iverson's convention]]
+}}
+{{end-eqn}}
+Thus the [[Definition:Summation|summation]] on the {{RHS}} of $(1)$ is:
+:$m \displaystyle \sum_{t \bmod m \mathop = k} \binom n t$
+The [[Definition:Summation|summation]] on the {{LHS}} of $(1)$ is:
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le j \mathop < m} \paren {1 + \omega^j}^n \omega^{-j k}
+      | r = \sum_{0 \mathop \le j \mathop < m} \paren {\omega^{-j / 2} + \omega^{j / 2} }^n \omega^{j \paren {n / 2} - k}
+      | c = 
+}}
+{{eqn | r = \sum_{0 \mathop \le j \mathop < m} \paren {2 \cos \dfrac {j \pi} m}^n \omega^{j \paren {n / 2} - k}
+      | c = 
+}}
+{{end-eqn}}
+Because the {{RHS}} is [[Definition:Wholly Real|wholly real]], so must the {{LHS}} be.
+So, taking the [[Definition:Real Part|real parts]] of the {{LHS}} and equating it to the {{RHS}} reveals the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over Lower Index of Unsigned Stirling Numbers of the First Kind}
+Tags: Stirling Numbers, Factorials
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+:$\displaystyle \sum_k \left[{n \atop k}\right] = n!$
+where:
+:$\displaystyle \left[{n \atop k}\right]$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$n!$ denotes the [[Definition:Factorial|factorial of $n$]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_k \left[{n \atop k}\right] = n!$
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{0 \atop k}\right]
+      | r = \sum_k \delta_{0 k}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = 1
+      | c = all terms vanish but for $k = 0$
+}}
+{{eqn | r = 0!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+Thus $P \left({0}\right)$ is seen to hold.
+=== Basis for the Induction ===
+$P \left({1}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{1 \atop k}\right]
+      | r = \sum_k \delta_{1 k}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = 1
+      | c = all terms vanish but for $k = 1$
+}}
+{{eqn | r = 1!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+Thus $P \left({1}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({m}\right)$ is true, where $m \ge 2$, then it logically follows that $P \left({m + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \left[{m \atop k}\right] = m!$
+from which it is to be shown that:
+:$\displaystyle \sum_k \left[{m + 1 \atop k}\right] = \left({m + 1}\right)!$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_k \left[{m + 1 \atop k}\right]
+      | r = \sum_k \left({m \left[{m \atop k}\right] + \left[{m \atop k - 1}\right]}\right)
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k - 1}\right]
+      | c = 
+}}
+{{eqn | r = m \sum_k \left[{m \atop k}\right] + \sum_k \left[{m \atop k}\right]
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \left({m + 1}\right) \sum_k \left[{m \atop k}\right]
+      | c = 
+}}
+{{eqn | r = \left({m + 1}\right) m!
+      | c = [[Summation over Lower Index of Unsigned Stirling Numbers of the First Kind#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \left({m + 1}\right)!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+So $P \left({m}\right) \implies P \left({m + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \left[{n \atop k}\right] = n!$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs}
+Tags: Stirling Numbers, Factorials
+
+\begin{theorem}
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+:$\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
+where:
+:$\displaystyle {n \brack k}$ denotes an [[Definition:Unsigned Stirling Numbers of the First Kind|unsigned Stirling number of the first kind]]
+:$\delta_{n 0}$ denotes the [[Definition:Kronecker Delta|Kronecker delta]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \paren {-1}^k {0 \brack k}
+      | r = \sum_k \delta_{0 k}
+      | c = [[Unsigned Stirling Number of the First Kind of 0]]
+}}
+{{eqn | r = 1
+      | c = all terms vanish but for $k = 0$
+}}
+{{eqn | r = \delta_{0 0} - \delta_{0 1}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+Thus $\map P 0$ is seen to hold.
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \paren {-1}^k {1 \brack k}
+      | r = \sum_k \paren {-1}^k \delta_{1 k}
+      | c = [[Unsigned Stirling Number of the First Kind of 1]]
+}}
+{{eqn | r = -1
+      | c = all terms vanish but for $k = 1$
+}}
+{{eqn | r = \delta_{1 0} - \delta_{1 1}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_k \paren {-1}^k {2 \brack k}
+      | r = - {2 \brack 1} + {2 \brack 2}
+      | c = {{Defof|Summation}}
+}}
+{{eqn | r = - {2 \brack 1} + 1
+      | c = [[Unsigned Stirling Number of the First Kind of Number with Self]]
+}}
+{{eqn | r = -\binom 2 2 + 1
+      | c = [[Unsigned Stirling Number of the First Kind of n with n-1]]
+}}
+{{eqn | r = -1 + 1
+      | c = [[Binomial Coefficient with Self]]
+}}
+{{eqn | r = 0
+      | c = [[Binomial Coefficient with Self]]
+}}
+{{eqn | r = \delta_{2 0} - \delta_{2 1}
+      | c = {{Defof|Kronecker Delta}}
+}}
+{{end-eqn}}
+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 $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_k \paren {-1}^k {m \brack k} = \delta_{m 0} - \delta_{m 1} = 0$
+from which it is to be shown that:
+:$\displaystyle \sum_k \paren {-1}^k {m + 1 \brack k} = \delta_{\paren {m + 1} 0} - \delta_{\paren {m + 1} 1} = 0$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_k \paren {-1}^k {m + 1 \brack k}
+      | r = \sum_k \paren {-1}^k \paren {m {m \brack k} + {m \brack k - 1} }
+      | c = {{Defof|Unsigned Stirling Numbers of the First Kind}}
+}}
+{{eqn | r = m \sum_k \paren {-1}^k {m \brack k} + \sum_k \paren {-1}^k {m \brack k - 1}
+      | c = 
+}}
+{{eqn | r = m \sum_k \paren {-1}^k {m \brack k} - \sum_k \paren {-1}^k {m \brack k}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \paren {m - 1} \sum_k {m \brack k}
+      | c = 
+}}
+{{eqn | r = \paren {m + 1} \times 0
+      | c = [[Summation over Lower Index of Unsigned Stirling Numbers of the First Kind with Alternating Signs#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $\map P m \implies \map P {m + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_k \paren {-1}^k {n \brack k} = \delta_{n 0} - \delta_{n 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Beta Function of Real Number with 1}
+Tags: Beta Function
+
+\begin{theorem}
+:$\Beta \left({x, 1}\right) = \Beta \left({1, x}\right) = \dfrac 1 x$
+\end{theorem}
+
+\begin{proof}
+By definition:
+{{begin-eqn}}
+{{eqn | l = \Beta \left({x, 1}\right)
+      | r = \int_0^1 t^{x - 1} \left({1 - t}\right)^{1 - 1} \rd t
+      | c = {{Defof|Beta Function}}
+}}
+{{eqn | r = \int_0^1 t^{x - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^1 t^{x - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \left[{\frac {t^x} x}\right]_0^1
+      | c = 
+}}
+{{eqn | r = \frac 1 x - \frac 0 x
+      | c = 
+}}
+{{eqn | r = \frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+{{begin-eqn}}
+{{eqn | l = \Beta \left({1, x}\right)
+      | r = \int_0^1 t^{1 - 1} \left({1 - t}\right)^{x - 1} \rd t
+      | c = {{Defof|Beta Function}}
+}}
+{{eqn | r = \int_0^1 \left({1 - t}\right)^{x - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \int_1^0 -\left({z}\right)^{x - 1} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = -\left[{\frac {z^x} x}\right]_1^0
+      | c = 
+}}
+{{eqn | r = -\frac 0 x + \frac 1 x
+      | c = 
+}}
+{{eqn | r = \frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Closure is Lower Set}
+Tags: Join and Meet Semilattices, Way Below Relation
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then
+:$x^\ll$ is a [[Definition:Lower Set|lower set]].
+\end{theorem}
+
+\begin{proof}
+Let $y \in x^\ll, z \in S$ such that:
+:$z \preceq y$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$
+By [[Preceding and Way Below implies Way Below]]:
+:$z \ll x$
+Thus by definition of [[Definition:Way Below Closure|way below closure]]:
+:$z \in x^\ll$
+Thus by definition:
+:$x^\ll$ is a [[Definition:Lower Set|lower set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous iff For Every Element There Exists Ideal Element Precedes Supremum}
+Tags: Continuous Lattices, Up-Complete Semilattices
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Then
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{iff}}
+:for every [[Definition:Element|element]] $x$ of $S$ there exists [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$:
+::$x \preceq \sup I$ and for every [[Definition:Ideal in Ordered Set|ideal]] $J$ in $L: x \preceq \sup J \implies I \subseteq J$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
+By [[Continuous iff Way Below Closure is Ideal and Element Precedes Supremum]]:
+:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
+::for every [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
+where $x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$.
+Thus
+:for every [[Definition:Element|element]] $x$ of $S$ there exists [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$:
+::$x \preceq \sup I$ and for every [[Definition:Ideal in Ordered Set|ideal]] $J$ in $L: x \preceq \sup J \implies I \subseteq J$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every [[Definition:Element|element]] $x$ of $S$ there exists [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$:
+::$x \preceq \sup I$ and for every [[Definition:Ideal in Ordered Set|ideal]] $J$ in $L: x \preceq \sup J \implies I \subseteq J$
+Let $x \in S$.
+There exists [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$:
+:$x \preceq \sup I$ and for every [[Definition:Ideal in Ordered Set|ideal]] $J$ in $L: x \preceq \sup J \implies I \subseteq J$
+We will prove that
+:$I \subseteq x^\ll$
+where $x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$.
+Let $y \in I$.
+Let $J$ be an [[Definition:Ideal in Ordered Set|ideal]] in $L$ such that
+:$x \preceq \sup J$
+Then $I \subseteq J$
+Thus by definition of [[Definition:Subset|subset]]:
+:$y \in J$
+By [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$y \ll x$
+Thus by definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \in x^\ll$
+{{qed|lemma}}
+By [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]
+:$\forall y \in x^\ll: y \in I$
+By definition of [[Definition:Subset|subset]]:
+:$x^\ll \subseteq I$
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\forall x \in S: x^\ll$ is an [[Definition:Ideal in Ordered Set|ideal]] in $L$ and $x \preceq \sup \left({x^\ll}\right)$ and
+::for every [[Definition:Ideal in Ordered Set|ideal]] $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$
+Hence by [[Continuous iff Way Below Closure is Ideal and Element Precedes Supremum]]:
+:$L$ is [[Definition:Continuous Ordered Set|continuous]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Beta Function of x+1 with y plus Beta Function of x with y+1}
+Tags: Beta Function
+
+\begin{theorem}
+:$\Beta \left({x + 1, y}\right) + \Beta \left({x, y + 1}\right) = \Beta \left({x, y}\right)$
+\end{theorem}
+
+\begin{proof}
+By definition:
+{{begin-eqn}}
+{{eqn | l = \Beta \left({x + 1, y}\right) + \Beta \left({x, y + 1}\right)
+      | r = \int_0^1 t^x \left({1 - t}\right)^{y - 1} \rd t + \int_0^1 t^{x - 1} \left({1 - t}\right)^y \rd t
+      | c = {{Defof|Beta Function}}
+}}
+{{eqn | r = \int_0^1 \left({t^x \left({1 - t}\right)^{y - 1} + t^{x - 1} \left({1 - t}\right)^y}\right) \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^1 t^{x - 1} \left({1 - t}\right)^{y - 1} \left({t + 1 - t}\right) \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^1 t^{x - 1} \left({1 - t}\right)^{y - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \Beta \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Beta Function of x with y+1 by x+y over y}
+Tags: Beta Function
+
+\begin{theorem}
+:$\map \Beta {x, y} = \dfrac {x + y} y \map \Beta {x, y + 1}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Beta Function|Beta function]]:
+:$\displaystyle \Beta \left({x + 1, y}\right) = \int_0^1 t^x \left({1 - t}\right)^{y - 1} \rd t$
+With a view to expressing the [[Definition:Primitive (Calculus)|primitive]] in the form:
+:$\displaystyle \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = t x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d x}
+      | r = x t^{x - 1}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\d v} {\d x}
+      | r = \paren {1 - t}^{y - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = -\frac {\paren {1 - t}^y} y
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int_0^1 t^{x - 1} \paren {1 - t}^{y - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \sqbrk {t^x \frac {-\paren {1 - t}^y} y}_0^1 - \int_0^1 x t^{x - 1} \frac {-\paren {1 - t}^y} y \rd t
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac x y \int_0^1 t^{x - 1} \paren {1 - t}^y \rd t
+      | c = as the left hand term vanishes
+}}
+{{eqn | r = \frac x y \map \Beta {x, y + 1}
+      | c = {{Defof|Beta Function}}
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \Beta {x, y}
+      | r = \map \Beta {x + 1, y} + \map \Beta {x, y + 1}
+      | c = [[Beta Function of x+1 with y plus Beta Function of x with y+1]]
+}}
+{{eqn | r = \frac x y \map \Beta {x, y + 1} + \map \Beta {x, y + 1}
+      | c = from above
+}}
+{{eqn | r = \frac {x + y} y \map \Beta {x, y + 1}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Gamma Function expressed as Integral}
+Tags: Gamma Function
+
+\begin{theorem}
+Let $m \in \Z_{\ge 1}$.
+Let $\map {\Gamma_m} x$ denote the [[Definition:Partial Gamma Function|partial Gamma function]], defined as:
+:$\map {\Gamma_m} x := \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + m} }$
+Then:
+:$\displaystyle \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
+for $x > 0$.
+\end{theorem}
+
+\begin{proof}
+=== [[Partial Gamma Function expressed as Integral/Lemma|Lemma]] ===
+First we establish:
+{{:Partial Gamma Function expressed as Integral/Lemma}}
+The proof continues by [[Principle of Mathematical Induction|induction]] on $m$.
+For all $m \in \Z_{\ge 1}$, let $\map P m$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map {\Gamma_m} x = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t
+      | r = \int_0^1 \paren {1 - t} t^{x - 1} \rd t
+      | c = 
+}}
+{{eqn | r = \int_0^1 t^{x - 1} \rd t - \int_0^1 t^x \rd t
+      | c = 
+}}
+{{eqn | r = \sqbrk {\frac {t^x} x}_0^1 - \sqbrk {\frac {t^{x + 1} } {x + 1} }_0^1
+      | c = 
+}}
+{{eqn | r = \frac 1 x - \frac 1 {x + 1}
+      | c = 
+}}
+{{eqn | r = \frac {x + 1 - x} {x \paren {x + 1} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {x \paren {x + 1} }
+      | c = 
+}}
+{{eqn | r = \frac {1^x 1!} {x \paren {x + 1} }
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map {\Gamma_k} x = k^x \int_0^1 \paren {1 - t}^k t^{x - 1} \rd t$
+from which it is to be shown that:
+:$\displaystyle \map {\Gamma_{k + 1} } x = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+With a view to expressing the [[Definition:Primitive (Calculus)|primitive]] in the form:
+:$\displaystyle \int u \frac {\rd v} {\rd x} \rd x = u v - \int v \frac {\rd u} {\rd x} \rd x$
+let:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \paren {1 - t}^{k + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\rd u} {\rd t}
+      | r = -\paren {k + 1} \paren {1 - t}^k
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+and let:
+{{begin-eqn}}
+{{eqn | l = \frac {\rd v} {\rd t}
+      | r = t^{x - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \frac {t^x} x
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | r = \paren {k + 1}^x \int_0^1 \paren {1 - t}^{k + 1} t^{x - 1} \rd t
+      | o = 
+      | c = 
+}}
+{{eqn | r = \paren {k + 1}^x \sqbrk {-\paren {k + 1} \paren {1 - t}^k \frac {t^x} x}_0^1 - \paren {k + 1}^x \int_0^1 \paren {-\paren {k + 1} \frac {t^x} x} \paren {\paren {1 - t}^k} \rd t
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \int_0^1 \paren {1 - t}^k t^x \rd t
+      | c = $u v$ term vanishes
+}}
+{{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \frac {\map {\Gamma_k} {x + 1} } {k^{x + 1} }
+      | c = [[Partial Gamma Function expressed as Integral#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \frac {\paren {k + 1}^{x + 1} } x \frac {k^{x + 1} k!} {k^{x + 1} \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + 1 + k} }
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1}^{x + 1} k!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }
+      | c = 
+}}
+{{eqn | r = \frac {\paren {k + 1}^x \paren {k + 1}!} {x \paren {x + 1} \paren {x + 2} \cdots \paren {x + k} \paren {x + k + 1} }
+      | c = 
+}}
+{{eqn | r = \map {\Gamma_{k + 1} } x
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m \in \Z_{\ge 1}: \map {\Gamma_m} x = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t$
+for $x > 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Gamma Function expressed as Integral/Lemma}
+Tags: Gamma Function
+
+\begin{theorem}
+:$(1): \quad \displaystyle \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t$
+for $x > 0$.
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \frac t m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d t} {\d z}
+      | r = m
+      | c = 
+}}
+{{end-eqn}}
+Recalculating the limits:
+{{begin-eqn}}
+{{eqn | l = t
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = 0
+      | c = 
+}}
+{{eqn | l = t
+      | r = m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - \frac t m}^m
+      | r = \paren {1 - z}^m
+      | c = 
+}}
+{{eqn | l = t^{x - 1}
+      | r = \paren {m z}^{x - 1}
+      | c = 
+}}
+{{eqn | r = m^{x - 1} z^{x - 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus $(1)$ can be written:
+{{begin-eqn}}
+{{eqn | l = \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t
+      | r = m^x \int_0^1 \paren {1 - z}^m z^{x - 1} \rd z
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = m^x \int_0^1 \paren {1 - t}^m t^{x - 1} \rd t
+      | c = changing the name of the dummy variable
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral Form of Gamma Function equivalent to Euler Form}
+Tags: Gamma Function, Integral Form of Gamma Function equivalent to Euler Form, Equivalence of Definitions of Gamma Function
+
+\begin{theorem}
+{{TFAENocat|def = Gamma Function}}
+=== [[Definition:Integral Form of Gamma Function|Integral Form]] ===
+{{:Definition:Integral Form of Gamma Function}}
+=== [[Definition:Euler Form of Gamma Function|Euler Form]] ===
+{{:Definition:Euler Form of Gamma Function}}
+\end{theorem}
+
+\begin{proof}
+It is taken for granted that the [[Definition:Gamma Function|Gamma function]] [[Definition:Strictly Increasing Mapping|increases monotonically]] on $\R_{\ge 1}$.
+{{handwaving|Replace the above with a link to a proof}}
+We begin with an inequality that can easily be verified using cross multiplication.
+{{handwaving|... so verify it.}}
+Let $x$ be a [[Definition:Real Number|real number]] between $0$ and $1$.
+Let $n$ is a [[Definition:Positive Integer|positive integer]].
+Then:
+:$\displaystyle \frac {\log \Gamma \left({n - 1}\right) - \log \Gamma \left({n}\right)} {\left({n - 1}\right) - n} \le \frac {\log \Gamma \left({x + n}\right) - \log \Gamma \left({n}\right)} {\left({x + n}\right) - n} \le \frac {\log \Gamma \left({n + 1}\right) - \log \Gamma \left({n}\right)}{\left({n + 1}\right) - n}$
+Since n is a [[Definition:Positive Integer|positive integer]], we can make use of the identity:
+:$\Gamma \left({n}\right) = \left({n - 1}\right)!$
+Simplifying, we get:
+:$\log \left({n - 1}\right) \le \dfrac {\log \Gamma \left({x + n}\right) - \log \left({\left({n - 1}\right)!}\right)} x \le \log \left({n}\right)$
+We now make use of the identity:
+:$\displaystyle \Gamma \left({x + n}\right) = \prod_{k \mathop = 1}^n \left({x + n - k}\right) \Gamma \left({x}\right)$
+along with the fact that the [[Gamma Function is Log-Convex]], to simplify the inequality:
+:$\displaystyle \left({n - 1}\right)^x \left({n - 1}\right)! \prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1} \le \Gamma \left({x}\right) \le n^x \left({n - 1}\right)!\prod_{k \mathop = 1}^n \left({x + n - k}\right)^{-1}$
+Taking the limit as $n$ goes to infinity and using the [[Squeeze Theorem]]:
+:$\displaystyle \Gamma \left({x}\right) = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \left({x + n - k}\right)^{-1}$
+which is another representation of [[Definition:Gamma Function/Euler Form|Euler's form]].
+This proves equivalence for $x$ between $0$ and $1$.
+The result follows from the [[Gamma Difference Equation]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+First we present a [[Integral Form of Gamma Function equivalent to Euler Form/Lemma|lemma]]:
+{{:Integral Form of Gamma Function equivalent to Euler Form/Lemma}}
+Recall the definition of the [[Definition:Partial Gamma Function|partial Gamma function]]:
+:$\displaystyle \Gamma_m \left({x}\right) := \frac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$
+We have that:
+{{begin-eqn}}
+{{eqn | r = \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)
+      | o = 
+      | c = 
+}}
+{{eqn | r = \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right)
+      | c = 
+}}
+{{eqn | r = \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \int_0^m \left({1 - \frac t m}\right)^m t^{x - 1} \rd t
+      | c = [[Partial Gamma Function expressed as Integral]]
+}}
+{{eqn | r = \int_m^\infty e^{-t} t^{x - 1} \rd t + \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t
+      | c = 
+}}
+{{end-eqn}}
+We have that for large $t$:
+:$t^{x - 1} < t^{t / 2}$
+and so as $m \to \infty$:
+:$\displaystyle \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \int_0^m \left({e^t - \left({1 - \frac t m}\right)^m t^{x - 1} }\right) \rd t
+      | c = 
+}}
+{{eqn | o = \le
+      | r = \int_0^m \frac {t^2 e^{-t} } m e^{-t} t^{x - 1} \rd t
+      | c = [[Integral Form of Gamma Function equivalent to Euler Form/Lemma|Lemma]]
+}}
+{{eqn | r = \frac 1 m \int_0^m t^{x + 1} e^{-t} \rd t
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t
+      | c = 
+}}
+{{end-eqn}}
+Now we have that as $m \to \infty$:
+:$\displaystyle \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$
+so:
+:$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t - \Gamma_m \left({x}\right) = 0$
+leading to:
+:$\displaystyle \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \left({x + 1}\right) \left({x + 2}\right) \cdots \left({x + m}\right)}$
+as was to be demonstrated.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Composite Function/Second Derivative}
+Tags: Differential Calculus
+
+\begin{theorem}
+:$D_x^2 w = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u$
+\end{theorem}
+
+\begin{proof}
+For ease of understanding, let [[Definition:Leibniz's Notation for Derivatives|Leibniz's notation]] be used:
+:$\dfrac {\d^k u} {\d x^k} := D_x^k u$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = D_x^2 w
+      | r = \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d x} }
+      | c = 
+}}
+{{eqn | r = \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d u} \dfrac {\d u} {\d x} }
+      | c = [[Derivative of Composite Function]]
+}}
+{{eqn | r = \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d x} } {\dfrac {\d w} {\d u} } + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \dfrac {\d u} {\d x} \map {\dfrac {\d} {\d u} } {\dfrac {\d w} {\d u} } \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}
+      | c = [[Derivative of Composite Function]]
+}}
+{{eqn | r = \map {\dfrac {\d^2 w} {\d u^2} } {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2}
+      | c = 
+}}
+{{eqn | r = D_u^2 w \paren {D_x^1 u}^2 + D_u^1 w D_x^2 u
+      | c = {{Defof|Leibniz's Notation for Derivatives}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula}
+Tags: Differential Calculus, Faà di Bruno's Formula
+
+\begin{theorem}
+Let $D_x^k u$ denote the [[Definition:Nth Derivative|$k$th derivative]] of a [[Definition:Real Function|function]] $u$ {{WRT|Differentiation}} $x$.
+Then:
+:$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+The [[Faà di Bruno's Formula/Example/0|case for $n = 0$]] is:
+{{:Faà di Bruno's Formula/Example/0}}
+which is consistent with the definition of the [[Definition:Zeroth Derivative|zeroth derivative]].
+The [[Faà di Bruno's Formula/Example/1|case for $n = 1$]] is:
+{{:Faà di Bruno's Formula/Example/1}}
+which is consistent with [[Derivative of Composite Function]].
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the [[Faà di Bruno's Formula/Example/2|case]]:
+{{:Faà di Bruno's Formula/Example/2}}
+This is consistent with [[Derivative of Composite Function/Second Derivative|Derivative of Composite Function: Second Derivative]].
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle D_x^r w = \sum_{j \mathop = 0}^r D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } r! \prod_{m \mathop = 1}^r \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+from which it is to be shown that:
+:$\displaystyle D_x^{r + 1} w = \sum_{j \mathop = 0}^{r + 1} D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = r \mathop + 1 \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } \left({r + 1}\right)! \prod_{m \mathop = 1}^{r + 1} \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Note that when $k_m = 0$:
+:$\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$
+which shows that any contribution to the [[Definition:Summation|summation]] where $k_m = 0$ can be disregarded.
+Let $j = 0$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 0$
+:$1 \times k_1 + 2 k_2 + \cdots = r$
+:$k_1, k_2, \ldots \ge 0$
+It is apparent by inspection that, for all $r > 0$, no set of $k_p$ can fulfil these conditions.
+Therefore when $j = 0$ the [[Definition:Summation|summation]] is [[Definition:Vacuous Summation|vacuous]].
+Also note that from:
+:$1 \times k_1 + 2 k_2 + \cdots = r$
+it follows that:
+:$k_{r + 1} = k_{r + 2} = \cdots = 0$
+Thus, while there are only [[Definition:Finite Set|finitely many]] $k$'s, their upper limit need not be explicitly considered.
+Let $c \left({r, j, k_1, k_2, \ldots}\right)$ be the [[Definition:Coefficient|coefficient]] of $D_u^j w$ in $D_x^r w$.
+We establish some [[Definition:Lemma|lemmata]]:
+[[Faà di Bruno's Formula/Lemma 1|Lemma 1]]:
+{{:Faà di Bruno's Formula/Lemma 1}}
+[[Faà di Bruno's Formula/Lemma 2|Lemma 2]]:
+{{:Faà di Bruno's Formula/Lemma 2}}
+{{finish}}
+By [[Definition:Differentiation|differentiating]] {{WRT|Differentiation}} $x$:
+{{begin-eqn}}
+{{eqn | l = c \left({r + 1, j, k_1, k_2, \ldots}\right)
+      | r = c \left({r, j - 1, k_1 - 1, k_2, \ldots}\right)
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \left({k_1 + 1}\right) c \left({r, j, k_1 + 1, k_2 - 1, k_3, \ldots}\right)
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \left({k_2 + 1}\right) c \left({r, j, k_1, k_2 + 1, k_3 - 1, k_4, \ldots}\right)
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \cdots
+      | c = 
+}}
+{{end-eqn}}
+The equations:
+:$k_1 + k_2 + \cdots = j$
+and:
+:$k_1 + 2 k_2 + \cdots = r$
+are preserved by this [[Proof by Mathematical Induction#Induction Step|induction step]].
+Thus it is possible to factor out:
+:$\dfrac {r!} {k_1! \left({1!}\right)^{k_1} \cdots k_r! \left({r!}\right)^{k_r} }$
+from each term on the {{RHS}} of the equation for $c \left({r + 1, j, k_1, k_2, \ldots}\right)$.
+Thus we are left with:
+:$k_1 + 2 k_2 + 3 k_3 + \cdots = r + 1$
+So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: D_x^n w = \sum_{j \mathop = 0}^n D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } n! \prod_{m \mathop = 1}^n \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+\end{proof}
+
+\begin{proof}
+For convenience, let:
+:$w_j := D_u^j w$
+:$u_k := D_x^k u$
+Then:
+{{begin-eqn}}
+{{eqn | l = D_x \left({w_j}\right)
+      | r = w_{j + 1} u_1
+      | c = [[Derivative of Composite Function]]
+}}
+{{eqn | l = D_x \left({u_k}\right)
+      | r = u_{k + 1}
+      | c = {{Defof|Higher Derivative}}
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = D_x^1 w
+      | r = w_1 u_1
+      | c = [[Derivative of Composite Function]]
+}}
+{{eqn | l = D_x^2 w
+      | r = \left({w_2 u_1 u_1 + w_1 u_2}\right)
+      | c = [[Derivative of Composite Function/Second Derivative|Derivative of Composite Function: Second Derivative]]
+}}
+{{eqn | l = D_x^3 w
+      | r = \left({\left({w_3 u_1 u_1 u_1 + w_2 u_2 u_1 + w_2 u_1 u_2}\right) + \left({w_2 u_1 u_2 + w_1 u_3}\right)}\right)
+      | c = [[Derivative of Composite Function/Third Derivative|Derivative of Composite Function: Third Derivative]] or directly
+}}
+{{end-eqn}}
+Analogously, let a corresponding tableau be set up of [[Definition:Set Partition|set partitions]] as follows:
+{{begin-eqn}}
+{{eqn | l = \mathcal D^1
+      | r = \set 1
+      | c = 
+}}
+{{eqn | l = \mathcal D^2
+      | r = \set {\set 2 \mid \set 1} + \set {2, 1}
+      | c = 
+}}
+{{eqn | l = \mathcal D^3
+      | r = \set {\set 3 \mid \set 2 \mid \set 1} + \set {\set {3, 2} \mid \set 1} + \set {\set 2 \mid \set {3, 1} } + \set {\set 3 \mid \set {2, 1} } + \set {3, 2, 1}
+      | c = 
+}}
+{{end-eqn}}
+Thus inspired, let $a_1 a_2 \cdots a_j$ denote a [[Definition:Set Partition|partition]] of the [[Definition:Set|set]] $\set {1, 2, \ldots, n}$.
+Let:
+{{begin-eqn}}
+{{eqn | l = \mathcal D a_1 a_2 \cdots a_j
+      | o = :=
+      | r = \set n a_1 a_2 \cdots a_j
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = \left({a_1 \cup \set n}\right) a_2 \cdots a_j 
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = a_1 \left({a_2 \cup \set n}\right) \cdots a_j 
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = \cdots
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = a_1 a_2 \cdots \left({a_j \cup \set n}\right)
+      | c = 
+}}
+{{end-eqn}}
+This rule is isomorphic to:
+{{begin-eqn}}
+{{eqn | l = D_x \left({w_j u_{r_1} u_{r_2} \ldots u_{r_j} }\right)
+      | r = w_{j + 1} u_1 u_{r_1} u_{r_2} \ldots u_{r_j}
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = w_j u_1 u_{r_1 + 1} u_{r_2} \ldots u_{r_j}
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = w_j u_1 u_{r_1} u_{r_2 + 1} \ldots u_{r_j}
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = \cdots
+      | c = 
+}}
+{{eqn | ro= +
+      | o = 
+      | r = w_j u_1 u_{r_1} u_{r_2} \ldots u_{r_j + 1}
+      | c = 
+}}
+{{end-eqn}}
+where the term $w_j u_{r_1} u_{r_2} \ldots u_{r_j}$ corresponds to a [[Definition:Set Partition|partition]] $a_1 a_2 \cdots a_j$.
+Thus there exists a [[Definition:Bijection|bijection]] from $\mathcal D^n$ to $D_x^n w$.
+{{explain|Link this to [[Number of Set Partitions by Number of Components]]}}
+Thus, collecting like terms in $D_x^n w$, we obtain a sum of terms:
+:$c \left({k_1, k_2, \ldots}\right) w_j u_i^{k_1} u_2^{k_2} \ldots$
+where:
+:$j = k_1 + k_2 + \cdots$
+and:
+:$n = k_1 + 2 k_2 + \cdots$
+and where:
+:$c \left({k_1, k_2, \ldots}\right)$ is the number of [[Definition:Set Partition|partitions]] of $\set {1, 2, \ldots, n}$ into $j$ [[Definition:Subset|subsets]] where there are $k_t$ [[Definition:Subset|subsets]] with $t$ [[Definition:Element|elements]].
+Consider an [[Definition:Array|array]] of $k_t$ boxes of capacity $t$.
+The number of ways to put $n$ different [[Definition:Element|elements]] into these boxes is the [[Definition:Multinomial Coefficient|multinomial coefficient]]:
+:$\dbinom n {1, 1, \ldots, 1, 2, 2, \ldots, 2, 3, 3, \ldots, 3, 4, \ldots} = \dfrac {n!} {\left({1!}\right)^{k_1} \left({2!}\right)^{k_2} \left({3!}\right)^{k_3} \cdots}$
+It remains to divide by $k_1! \, k_2! \, k_3! \ldots$ corresponding to the number of ways each group of $k_t$ can be [[Definition:Permutation (Ordered Selection)|permuted]]
+Hence:
+:$c \left({k_1, k_2, \ldots}\right) = \dfrac {n!} {k_1! \left({1!}\right)^{k_1} \, k_2! \left({2!}\right)^{k_2} k_3! \left({3!}\right)^{k_3} \cdots}$
+and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that:
+: $\dfrac {D_x^k u} {k!}$ is the coefficient of $z^k$ in $u \left({x + z}\right)$
+: $\dfrac {D_u^j w} {j!}$ is the coefficient of $y^j$ in $w \left({u + y}\right)$.
+Hence the coefficient of $z^n$ in $w \left({u \left({x + z}\right)}\right)$ is:
+:$\dfrac {D_x^n w} {n!} = \displaystyle \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = n \\ \forall p \mathop \ge 1: k_p \mathop \ge 0} } \dfrac {j!} {k_1! \, k_2! \cdots k_n!} \left({\dfrac {D_x^1 u} {1!} }\right)^{k_1} \left({\dfrac {D_x^2 u} {2!} }\right)^{k_2} \cdots \left({\dfrac {D_x^n u} {n!} }\right)^{k_n}$
+{{finish|plenty of details to be filled in}}
+\end{proof}
+
+\begin{proof}
+$D_x^n$ can be expressed as a [[Definition:Determinant of Matrix|determinant]]:
+:$D_x^n = \begin{vmatrix}
+\dbinom {n - 1} 0 u_1 & \dbinom {n - 1} 1 u_2 & \dbinom {n - 1} 2 u_3 & \cdots & \dbinom {n - 1} {n - 2} u_{n - 1} & \dbinom {n - 1} {n - 1} u_n       \\
+                   -1 & \dbinom {n - 2} 0 u_1 & \dbinom {n - 2} 1 u_2 & \cdots & \dbinom {n - 2} {n - 3} u_{n - 2} & \dbinom {n - 2} {n - 2} u_{n - 1} \\
+                    0 &                    -1 & \dbinom {n - 3} 0 u_1 & \cdots & \dbinom {n - 3} {n - 4} u_{n - 3} & \dbinom {n - 3} {n - 3} u_{n - 2} \\
+               \vdots &                \vdots &                \vdots & \ddots &                            \vdots &                            \vdots \\
+                    0 &                     0 &                     0 & \cdots &                                -1 &                   \dbinom 0 0 u_1
+\end{vmatrix}$
+where $u_j := \paren {D_x^j u} D_u$.
+Both sides of this equation are differential operators which are to be applied to $w$.
+{{finish|Fill in the details}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Reduced Residue System}
+Tags: Reduced Residue Systems, Euler Phi Function
+
+\begin{theorem}
+Let $n \ge 2$.
+Let $\Z'_n$ be the [[Definition:Reduced Residue System|reduced residue system modulo $n$]].
+Then:
+:$\card {\Z'_n} = \map \phi n$
+where $\map \phi n$ is the [[Definition:Euler Phi Function|Euler phi function]].
+\end{theorem}
+
+\begin{proof}
+Recall the [[Definition:Reduced Residue System|definition]] of $\Z'_n$:
+:$\Z'_n = \set {\eqclass k n \in \Z_n: k \perp n}$
+and the [[Definition:Euler Phi Function|definition]] of $\map \phi n$:
+:$\map \phi n = \card {\set {k: 1 \le k \le n, k \perp n} }$
+The result follows from [[Integer is Congruent to Integer less than Modulus]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Norm of Linear Functional/Corollary}
+Tags: Linear Transformations on Hilbert Spaces, Definition Equivalences
+
+\begin{theorem}
+For all $h \in H$, the following inequality holds:
+:$\left|{Lh}\right| \le \left\|{L}\right\| \left\|{h}\right\|$
+\end{theorem}
+
+\begin{proof}
+If $h = \mathbf 0$ there is nothing to prove.
+Let $h \ne \mathbf 0$.
+By the definition of the [[Definition:Supremum of Set|supremum]]:
+: $\dfrac{\left|{Lh}\right|} {\left\Vert{h}\right\Vert} \le \left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert$
+whence:
+: $\left|{Lh}\right| \le \left\|{L}\right\| \left\|{h}\right\|$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula/Example/0/Proof}
+Tags: Faà di Bruno's Formula
+
+\begin{theorem}
+Consider [[Faà di Bruno's Formula]]:
+{{:Faà di Bruno's Formula}}
+When $n = 0$ we have:
+{{:Faà di Bruno's Formula/Example/0}}
+\end{theorem}
+
+\begin{proof}
+In the [[Definition:Summation|summation]]:
+:$\displaystyle \sum_{j \mathop = 0}^0 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 0} k_p \mathop = j \\ \sum_{p \mathop \ge 0} p k_p \mathop = 0 \\ \forall p \ge 0: k_p \mathop \ge 0} } 0! \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
+the only element appearing is for $j = 0$, and the [[Definition:Product Notation (Algebra)|product]] is [[Definition:Vacuous Product|vacuous]].
+Thus:
+:$\displaystyle \prod_{m \mathop = 0}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$
+and we are left with:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^0 D_u^j w
+      | r = D_u^0 w
+      | c = 
+}}
+{{eqn | r = w
+      | c = {{Defof|Zeroth Derivative}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula/Example/1/Proof}
+Tags: Faà di Bruno's Formula
+
+\begin{theorem}
+Consider [[Faà di Bruno's Formula]]:
+{{:Faà di Bruno's Formula}}
+When $n = 1$ we have:
+{{:Faà di Bruno's Formula/Example/1}}
+\end{theorem}
+
+\begin{proof}
+In the [[Definition:Summation|summation]]:
+:$\displaystyle \sum_{j \mathop = 0}^1 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 1 \\ \forall p \ge 1: k_p \mathop \ge 0} } 1! \prod_{m \mathop = 1}^1 \dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} }$
+we need to consider $j = 0$ and $j = 1$.
+Let $j = 0$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 0$
+:$1 \times k_1 + 2 k_2 + \cdots = 1$
+:$k_1, k_2, \ldots \ge 0$
+It is apparent by inspection that no set of $k_p$ can fulfil these conditions.
+Therefore when $j = 0$ the [[Definition:Summation|summation]] is [[Definition:Vacuous Summation|vacuous]]
+Let $j = 1$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 1$
+:$1 \times k_1 + 2 k_2 + \cdots = 1$
+:$k_1, k_2, \ldots \ge 0$
+By inspection, it is seen that these can be satisfied only by:
+:$k_1 = 1$
+and all other $k_p = 0$.
+When $k_m = 0$:
+:$\dfrac {\paren {D_x^m u}^{k_m} } {k_m! \paren {m!}^{k_m} } = 1$
+by definition of [[Definition:Zeroth Derivative|zeroth derivative]] and [[Definition:Factorial|factorial of $0$]].
+Thus any contribution to the [[Definition:Summation|summation]] where $k_m = 0$ can be disregarded.
+Thus we have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{j \mathop = 0}^1 \sum_{\substack {k_1 \mathop = j \\ k_1 \mathop = 1 \\ k_1 \mathop \ge 0} } D_u^j w \dfrac {1!} {k_1! \paren {1!}^{k_1} } \paren {D_x^1 u}^{k_1}
+      | c = 
+}}
+{{eqn | r = \sum_{k_1 \mathop = 1} D_u^1 w \dfrac {1!} {k_1! \paren {1!}^{k_1} } \paren {D_x^1 u}^{k_1}
+      | c = simplifying: $j = 0$ is [[Definition:Vacuous Summation|vacuous]]
+}}
+{{eqn | r = D_u w D_x u
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula/Example/2/Proof}
+Tags: Faà di Bruno's Formula
+
+\begin{theorem}
+Consider [[Faà di Bruno's Formula]]:
+{{:Faà di Bruno's Formula}}
+When $n = 2$ we have:
+{{:Faà di Bruno's Formula/Example/2}}
+\end{theorem}
+
+\begin{proof}
+In the [[Definition:Summation|summation]]:
+:$\displaystyle \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {\sum_{p \mathop \ge 1} k_p \mathop = j \\ \sum_{p \mathop \ge 1} p k_p \mathop = 2 \\ \forall p \ge 1: k_p \mathop \ge 0} } 2! \prod_{m \mathop = 1}^2 \dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} }$
+we need to consider $j = 0, j = 1$ and $j = 2$.
+Note that when $k_m = 0$:
+:$\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } = 1$
+by definition of [[Definition:Zeroth Derivative|zeroth derivative]] and [[Definition:Factorial|factorial of $0$]].
+Thus any contribution to the [[Definition:Summation|summation]] where $k_m = 0$ can be disregarded.
+Let $j = 0$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 0$
+:$1 \times k_1 + 2 k_2 + \cdots = 2$
+:$k_1, k_2, \ldots \ge 0$
+It is apparent by inspection that no set of $k_p$ can fulfil these conditions.
+Therefore when $j = 0$ the [[Definition:Summation|summation]] is [[Definition:Vacuous Summation|vacuous]]
+Let $j = 1$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 1$
+:$1 \times k_1 + 2 k_2 + \cdots = 2$
+:$k_1, k_2, \ldots \ge 0$
+By inspection, it is seen that these can be satisfied only by:
+:$k_2 = 1$
+and all other $k_p = 0$.
+Let $j = 2$.
+Consider the set of $k_p$ such that:
+:$k_1 + k_2 + \cdots = 2$
+:$1 \times k_1 + 2 k_2 + \cdots = 2$
+:$k_1, k_2, \ldots \ge 0$
+By inspection, it is seen that these can be satisfied only by:
+:$k_1 = 2$
+and all other $k_p = 0$.
+Thus we have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{j \mathop = 0}^2 D_u^j w \sum_{\substack {k_1 \mathop + k_2 \mathop = j \\ k_1 \mathop + 2 k_2 \mathop = 2 \\ k_1, k_2 \mathop \ge 0} } 2! \dfrac {\left({D_x^1 u}\right)^{k_1} \left({D_x^2 u}\right)^{k_2} } {k_1! \left({1!}\right)^{k_1} k_2! \left({2!}\right)^{k_2} } 
+      | c = 
+}}
+{{eqn | r = D_u^1 w \dfrac {2! \left({D_x^1 u}\right)^0 \left({D_x^2 u}\right)^1} {0! \left({1!}\right)^0 1! 2^1} + D_u^2 w \dfrac {2! \left({D_x^1 u}\right)^2 \left({D_x^2 u}\right)^0} {2! \left({1!}\right)^2 0! \left({2!}\right)^0} 
+      | c = substituting for $k_1$ and $k_2$ for each case
+}}
+{{eqn | r = D_u^1 w \left({D_x^2 u}\right) + D_u^2 w \left({D_x^1 u}\right)^2
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula/Lemma 1}
+Tags: Faà di Bruno's Formula
+
+\begin{theorem}
+:$D_x \left({\left({D_x^m u}\right)^{k_m} }\right) = k_m \left({D_x^m u}\right)^{k_m - 1} D_x^{m + 1} u$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = D_x \left({\left({D_x^m u}\right)^{k_m} }\right)
+      | r = \left({k_m \left({D_x^m u}\right)^{k_m - 1} }\right) D_x \left({D_x^m u}\right)
+      | c = [[Derivative of Power of Function]]
+}}
+{{eqn | r = k_m \left({\left({D_x^m u}\right)^{k_m - 1} }\right) D_x^{m + 1} u
+      | c = {{Defof|Higher Derivative}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Faà di Bruno's Formula]]
+lxvuaw0ammntxkyzm0qexwv6cw3pn3e
+\end{proof}<|endoftext|>
+\section{Faà di Bruno's Formula/Lemma 2}
+Tags: Faà di Bruno's Formula
+
+\begin{theorem}
+:$\displaystyle D_x \left({\prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) }\right) = \prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | r = D_x \left({\prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) }\right)
+      | o = 
+      | c = 
+}}
+{{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) D_x \left({\prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right) }\right)
+      | c = 
+}}
+{{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({D_x \left({\left({D_x^m u}\right)^{k_m} }\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right) }\right)
+      | c = [[Product Rule for Derivatives/General Result|Product Rule for Derivatives: General Result]]
+}}
+{{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r \left({\left({k_m \left({D_x^m u}\right)^{k_m - 1} D_x^{m + 1} u}\right) \prod_{s \mathop \ne m} \left({\left({D_x^s u}\right)^{k_s} }\right)}\right)
+      | c = [[Faà di Bruno's Formula/Lemma 1|Faà di Bruno's Formula: Lemma 1]]
+}}
+{{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac 1 {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u} \prod_{m \mathop = 1}^r \left({\left({D_x^m u}\right)^{k_m} }\right)
+      | c = 
+}}
+{{eqn | r = \prod_{m \mathop = 1}^r \left({\dfrac {\left({D_x^m u}\right)^{k_m} } {k_m! \left({m!}\right)^{k_m} } }\right) \sum_{m \mathop = 1}^r k_m \dfrac {D_x^{m + 1} u} {D_x^m u}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Faà di Bruno's Formula]]
+56df0zm1pg9z1et79yyuorg5w41mfb0
+\end{proof}<|endoftext|>
+\section{Supremum of Ideals is Upper Adjoint}
+Tags: Continuous Lattices, Galois Connections
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathit{Ids}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
+Let $f: \mathit{Ids}\left({L}\right) \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
+Then $f$ is an [[Definition:Galois Connection|upper adjoint of Galois connection]].
+\end{theorem}
+
+\begin{proof}
+Define $d: S \to \mathit{Ids}\left({L}\right)$
+:$\forall t \in S: d\left({t}\right) =  \inf \left({f^{-1}\left[{t^\succeq}\right]}\right)$
+where
+:$t^\succeq$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $t$,
+:$f^{-1}\left[{t^\succeq}\right]$ denotes the [[Definition:Image of Subset under Relation|image]] of $t^\succeq$ over $f^{-1}$.
+We will prove that
+:$\forall t \in S: d\left({t}\right) =  \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$
+Let $t \in S$.
+By [[Continuous iff For Every Element There Exists Ideal Element Precedes Supremum]]:
+:there exists an [[Definition:Ideal in Ordered Set|ideal]] $J$ in $L$ such that
+::$t \preceq \sup J$ and for every [[Definition:Ideal in Ordered Set|ideal]] $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$
+We will prove that
+:$\forall K \in \mathit{Ids}\left({L}\right): K$ is [[Definition:Lower Bound of Set|lower bound]] for $f^{-1}\left[{t^\succeq}\right] \implies K \precsim J$
+Let $K \in \mathit{Ids}\left({L}\right)$ such that
+:$K$ is [[Definition:Lower Bound of Set|lower bound]] for $f^{-1}\left[{t^\succeq}\right]$
+By definition of $f$:
+:$t \preceq f\left({J}\right)$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$f\left({J}\right) \in t^\succeq$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$J \in f^{-1}\left[{t^\succeq}\right]$
+Thus by definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$K \precsim J$
+{{qed|lemma}}
+We will prove that
+:$J$ is [[Definition:Lower Bound of Set|lower bound]] for $f^{-1}\left[{t^\succeq}\right]$
+Let $K \in \mathit{Ids}\left({L}\right)$ such that
+:$K \in f^{-1}\left[{t^\succeq}\right]$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$f\left({K}\right) \in t^\succeq$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$t \preceq f\left({K}\right)$
+By definition of $f$:
+:$t \preceq \sup K$
+Then
+:$J \subseteq K$
+Thus by definition of $\precsim$:
+:$J \precsim K$
+{{qed|lemma}}
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$t \preceq \sup \left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$
+By definition of $f$:
+:$t \preceq f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right)$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$f\left({\inf \left({f^{-1}\left[{t^\succeq}\right]}\right)}\right) \in t^\succeq$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\inf \left({f^{-1}\left[{t^\succeq}\right]}\right) \in f^{-1}\left[{t^\succeq}\right]$
+Thus by definition of [[Definition:Smallest Element|smallest element]]:
+:$d\left({t}\right) = \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$
+{{qed|lemma}}
+By [[Supremum of Ideals is Increasing]]:
+:$f$ is an [[Definition:Increasing Mapping|increasing mapping]].
+By [[Galois Connection is Expressed by Minimum]]:
+:$\left({f, d}\right)$ is a [[Definition:Galois Connection|Galois connection]].
+Hence $f$ is an [[Definition:Galois Connection|upper adjoint of Galois connection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Ideals is Increasing}
+Tags: Increasing Mappings
+
+\begin{theorem}
+Let $L = \left({S, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Ordered Set|ordered set]].
+Let $\mathit{Ids}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
+Let $f: \mathit{Ids}\left({L}\right) \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
+Then $f$ is an [[Definition:Increasing Mapping|increasing mapping]].
+\end{theorem}
+
+\begin{proof}
+Let $I, J \in \mathit{Ids}\left({L}\right)$ such that
+:$I \precsim J$
+By definition of $\precsim$:
+:$I \subseteq J$
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$I$ and $J$ admit [[Definition:Supremum of Set|suprema]] in $L$.
+By [[Supremum of Subset]]:
+:$\sup I \preceq \sup J$
+Thus by definition of $f$:
+:$f\left({I}\right) \preceq f\left({J}\right)$
+Hence $f$ is an [[Definition:Increasing Mapping|increasing mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Composite Function/Third Derivative}
+Tags: Differential Calculus
+
+\begin{theorem}
+:$D_x^3 w = D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u$
+\end{theorem}
+
+\begin{proof}
+For ease of understanding, let [[Definition:Leibniz's Notation for Derivatives|Leibniz's notation]] be used:
+:$\dfrac {\d^k u} {\d x^k} := D_x^k u$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = D_x^3 w
+      | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d x^2} }
+      | c = 
+}}
+{{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d w} {\d u} \dfrac {\d^2 u} {\d x^2} }
+      | c = [[Derivative of Composite Function/Second Derivative|Derivative of Composite Function: Second Derivative]]
+}}
+{{eqn | r = \map {\dfrac \d {\d x} } {\dfrac {\d^2 w} {\d u^2} } \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} \map {\dfrac \d {\d x} } {\paren {\dfrac {\d u} {\d x} }^2}
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \map {\dfrac \d {\d x} } {\dfrac {\d w} {\d u} } \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \map {\dfrac \d {\d x} } {\dfrac {\d^2 u} {\d x^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {\d^3 w} {\d u^3} \dfrac {\d u} {\d x} \paren {\dfrac {\d u} {\d x} }^2 + \dfrac {\d^2 w} {\d u^2} 2 \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2}
+      | c = [[Derivative of Composite Function]]
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \dfrac {\d^2 w} {\d u^2} \dfrac {\d u} {\d x} \dfrac {\d^2 u} {\d x^2} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }
+      | c = 
+}}
+{{eqn | r = \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 2 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x}
+      | c = simplifying
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }
+      | c = 
+}}
+{{eqn | r = \dfrac {\d^3 w} {\d u^3} \paren {\dfrac {\d u} {\d x} }^3 + 3 \dfrac {\d^2 w} {\d u^2} \dfrac {\d^2 u} {\d x^2} \dfrac {\d u} {\d x} + \dfrac {\d w} {\d u} \paren {\dfrac {\d^3 u} {\d x^3} }
+      | c = simplifying
+}}
+{{eqn | r = D_u^3 w \paren {D_x^1 u}^3 + 3 D_u^2 w D_x^2 u D_x^1 u + D_u^1 w D_x^3 u
+      | c = [[Definition:Leibniz's Notation for Derivatives|Leibniz's Notation]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Differential Calculus]]
+kiwj42mfwm8i3qy33qkqwbvx9nxljff
+\end{proof}<|endoftext|>
+\section{Number of Set Partitions by Number of Components}
+Tags: Combinatorics, Stirling Numbers, Number of Set Partitions by Number of Components, Set Partitions
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|(finite) set]] whose [[Definition:Cardinality|cardinality]] is $n$.
+Let $\map f {n, k}$ denote the number of different ways $S$ can be [[Definition:Set Partition|partitioned]] into $k$ [[Definition:Component of Partition|components]].
+Then:
+:$\displaystyle \map f {n, k} = {n \brace k}$
+where $\displaystyle {n \brace k}$ denotes a [[Definition:Stirling Numbers of the Second Kind|Stirling number of the second kind]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \map f {n, k} = {n \brace k}$
+$\map P 0$ is the [[Definition:Degenerate Case|degenerate case]]:
+:$\displaystyle \map f {0, k} = \delta_{0 k} = {0 \brace k}$
+That is: the [[Definition:Empty Set|empty set]] can be [[Definition:Set Partition|partitioned]] one and only one way: into $0$ [[Definition:Subset|subsets]].
+Thus $\map P 0$ is seen to hold.
+The remainder of the proof considers $n \in \Z_{> 0}$.
+First we note that when $k < 1$ or $k > n$:
+:$\displaystyle \map f {n, k} = 0 = {n \brace k}$
+Hence, throughout, we consider only such $k$ as $1 \le k \le n$.
+We define the representative [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $n$ to be:
+:$S_n := \set {1, 2, \ldots, n}$
+=== Basis for the Induction ===
+$\map P 1$ is the case $\map f {1, 1}$.
+There is exactly one way to [[Definition:Set Partition|partition]] $\set 1$, and that is:
+:$\set {\set 1}$
+From [[Stirling Number of the Second Kind of Number with Self]]:
+:$\displaystyle {1 \brace 1} = 1$
+Thus $\map P 1$ is seen to hold.
+This is the [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
+So this is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map f {m, k} = {m \brace k}$
+from which it is to be shown that:
+:$\displaystyle \map f {m + 1, k} = {m + 1 \brace k}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+By definition, the number of [[Definition:Set Partition|partitions]] of $S_m$ into $k$ [[Definition:Subset|subsets]] is $\map f {m, k}$.
+A [[Definition:Set Partition|partition]] of $S_{m + 1}$ can be generated by adding [[Definition:Element|element]] $m + 1$ into one of the existing [[Definition:Set Partition|partitions]] of $S_m$.
+There are two ways this can be done:
+:$(1): \quad$ The [[Definition:Subset|subset]] $\set {m + 1}$ may be added, in one way, to one of the [[Definition:Set Partition|partitions]] of $S_m$ into $k - 1$ [[Definition:Subset|subsets]].
+:$(2): \quad$ The [[Definition:Element|element]] $m + 1$ may be added to any one of the $k$ [[Definition:Subset|subsets]] in one of the [[Definition:Set Partition|partitions]] of $S_m$ into $k$ [[Definition:Subset|subsets]].
+Option $(1)$ gives $1$ [[Definition:Set Partition|partition]] of $S_{m + 1}$ for each [[Definition:Set Partition|partition]] of $S_m$ into $k - 1$ [[Definition:Subset|subsets]], that is: $\map f {m, k - 1}$.
+Option $(2)$ gives $k$ [[Definition:Set Partition|partitions]] of $S_{m + 1}$ for each [[Definition:Set Partition|partition]] of $S_m$ into $k$ [[Definition:Subset|subsets]], that is: $k \map f {m, k}$.
+Thus:
+:$\map f {m + 1, k} = k \map f {m, k} + \map f {m, k - 1}$
+By the [[Number of Set Partitions by Number of Components#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \map f {m + 1, k} = k {m \brace k} + {m  \brace k - 1}$
+So by definition of [[Definition:Stirling Numbers of the Second Kind|Stirling numbers of the second kind]]:
+:$\displaystyle \map f {m + 1, k} = {m + 1 \brace k}$
+So $\map P m \implies \map P {m + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n, k \in \Z_{\ge 0}: \map f {n, k} = {n \brace k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of Summation over Divisors of Function of Two Variables}
+Tags: Summations, Divisors
+
+\begin{theorem}
+Let $c, d, n \in \Z$.
+Then:
+:$\displaystyle \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
+where $c \divides d$ denotes that $c$ is a [[Definition:Divisor of Integer|divisor]] of $d$.
+\end{theorem}
+
+\begin{proof}
+From [[Exchange of Order of Summation with Dependency on Both Indices]]:
+{{:Exchange of Order of Summation with Dependency on Both Indices}}
+We have that:
+:$\map R d$ is the [[Definition:Propositional Function|propositional function]]:
+::$d \divides n$
+:$\map S {d, c}$ is the [[Definition:Propositional Function|propositional function]]:
+::$c \divides d$
+Thus $\map {R'} {d, c}$ is the [[Definition:Propositional Function|propositional function]]:
+::Both $d \divides n$ and $c \divides d$
+This is the same as:
+::$c \divides n$ and $\dfrac d c \divides \dfrac n c$
+Similarly, $\map {S'} c$ is the [[Definition:Propositional Function|propositional function]]:
+::$\exists d$ such that both $d \divides n$ and $c \divides d$
+This is the same as:
+::$c \divides n$
+This gives:
+:$\displaystyle \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{\paren {d / c} \mathrel \divides \paren {n / c} } \map f {c, d}$
+Replacing $d / c$ with $d$:
+:$\displaystyle \sum_{d \mathop \divides n} \sum_{c \mathop \divides d} \map f {c, d} = \sum_{c \mathop \divides n} \sum_{d \mathop \divides \paren {n / c} } \map f {c, c d}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of Half of n+m plus Floor of Half of n-m+1}
+Tags: Floor Function
+
+\begin{theorem}
+Let $n, m \in \Z$ be [[Definition:Integer|integers]].
+:$\floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2} = n$
+where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Either $n + m$ or $n - m + 1$ is [[Definition:Even Integer|even]].
+Thus:
+:$\dfrac {n + m} 2 \bmod 1 + \dfrac {n - m + 1} 2 \bmod 1 = \dfrac 1 2 < 1$
+and so:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac {n + m} 2} + \floor {\dfrac {n - m + 1} 2}
+      | r = \floor {\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}
+      | c = [[Sum of Floors not greater than Floor of Sum]]
+}}
+{{eqn | r = \floor {\dfrac {n + m + n - m + 1} 2}
+      | c = 
+}}
+{{eqn | r = \floor {n + \dfrac 1 2}
+      | c = 
+}}
+{{eqn | r = n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling of Half of n+m plus Ceiling of Half of n-m+1}
+Tags: Ceiling Function
+
+\begin{theorem}
+Let $n, m \in \Z$ be [[Definition:Integer|integers]].
+:$\left\lceil{\dfrac {n + m} 2}\right\rceil + \left\lceil{\dfrac {n - m + 1} 2}\right\rceil = n + 1$
+where $\left\lceil{x}\right\rceil$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Either $n + m$ or $n - m + 1$ is [[Definition:Even Integer|even]].
+Thus either $\dfrac {n + m} 2$ or $\dfrac {n - m + 1} 2$ is an [[Definition:Integer|integer]].
+So:
+{{begin-eqn}}
+{{eqn | l = \left\lceil{\dfrac {n + m} 2}\right\rceil + \left\lceil{\dfrac {n - m + 1} 2}\right\rceil
+      | r = \left\lceil{\dfrac {n + m} 2 + \dfrac {n - m + 1} 2}\right\rceil
+      | c = [[Sum of Ceilings not less than Ceiling of Sum]]
+}}
+{{eqn | r = \left\lceil{\dfrac {n + m + n - m + 1} 2}\right\rceil
+      | c = 
+}}
+{{eqn | r = \left\lceil{n + \dfrac 1 2}\right\rceil
+      | c = 
+}}
+{{eqn | r = n + 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of Non-Integer}
+Tags: Floor Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $x \notin \Z$.
+Then:
+:$\left\lfloor{x}\right\rfloor < x$
+where $\left\lfloor{x}\right\rfloor$ denotes the [[Definition:Floor Function|floor]] of $x$.
+\end{theorem}
+
+\begin{proof}
+From [[Floor is between Number and One Less]]:
+:$\left\lfloor{x}\right\rfloor \le x$
+From [[Real Number is Integer iff equals Floor]]:
+:$x = \left \lfloor {x} \right \rfloor \iff x \in \Z$
+But we have $x \notin \Z$.
+So:
+:$\left \lfloor {x} \right \rfloor \ne x$
+and so:
+:$\left \lfloor {x} \right \rfloor < x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Logarithm of Base}
+Tags: Logarithms
+
+\begin{theorem}
+Let $b \in \R_{>0}$ such that $b \ne 1$.
+Then:
+:$\log_b b = 1$
+where $\log_b$ denotes the [[Definition:General Logarithm|logarithm]] to [[Definition:Base of Logarithm|base $b$]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in \R_{>0}$ such that $a \ne 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \log_b b
+      | r = \dfrac {\log_a b} {\log_a b}
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Logarithms]]
+8p8v7a9h3w1b8jvsqgyq3scehh3d4y0
+\end{proof}<|endoftext|>
+\section{Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x}
+Tags: Logarithms, Floor Function, Conditions for Floor of Log base b of x to equal Floor of Log base b of Floor of x
+
+\begin{theorem}
+Let $b \in \R$ be a [[Definition:Real Number|real number]].
+:$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$
+where $\left\lfloor{x}\right\rfloor$ denotes the [[Definition:Floor Function|floor of $x$]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let:
+:$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor$
+Let $x = b$.
+Then:
+:$\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$
+{{begin-eqn}}
+{{eqn | l = \left\lfloor{\log_b b}\right\rfloor
+      | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left\lfloor{1}\right\rfloor
+      | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor
+      | c = [[Logarithm of Base]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | r = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor
+      | c = [[Real Number is Integer iff equals Floor]]
+}}
+{{end-eqn}}
+{{AimForCont}} $b \notin \Z$.
+{{begin-eqn}}
+{{eqn | l = \left\lfloor{b}\right\rfloor
+      | o = <
+      | r = b
+      | c = [[Floor of Non-Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \log_b \left\lfloor{b}\right\rfloor
+      | o = <
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor
+      | o = <
+      | r = \left\lfloor{\log_b b}\right\rfloor
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor
+      | o = \ne
+      | r = \left\lfloor{\log_b b}\right\rfloor
+      | c = which [[Definition:Contradiction|contradicts]] $\left\lfloor{\log_b b}\right\rfloor = \left\lfloor{\log_b \left\lfloor{b}\right\rfloor}\right\rfloor$
+}}
+{{end-eqn}}
+Thus by [[Proof by Contradiction]]:
+:$b \in \Z$
+But for $\log_b$ to be defined, $b > 0$ and $b \ne 1$.
+Hence:
+:$b \in \Z_{> 1}$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $b \in \Z_{> 1}$.
+Let $\left\lfloor{\log_b x}\right\rfloor = n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n
+      | m = \left\lfloor{\log_b x}\right\rfloor
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = n
+      | o = \le
+      | m = \log_b x
+      | mo= <
+      | r = n + 1
+      | c = [[Integer equals Floor iff Number between Integer and One More]]
+}}
+{{eqn | ll= \iff
+      | l = b^n
+      | o = \le
+      | m = x
+      | mo= <
+      | r = b^{n + 1}
+      | c = [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals]]
+}}
+{{eqn | ll= \iff
+      | l = b^n
+      | o = \le
+      | m = \left \lfloor {x}\right \rfloor
+      | mo= <
+      | r = b^{n + 1}
+      | c = [[Number not less than Integer iff Floor not less than Integer]]
+}}
+{{eqn | ll= \iff
+      | l = n
+      | o = \le
+      | m = \log_b \left\lfloor{x}\right\rfloor
+      | mo= <
+      | r = n + 1
+      | c = [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals]]
+}}
+{{eqn | ll= \iff
+      | l = n
+      | m = \left \lfloor {\log_b \left\lfloor{x}\right\rfloor}\right \rfloor
+      | c = [[Integer equals Floor iff Number between Integer and One More]]
+}}
+{{eqn | ll= \iff
+      | l = \left\lfloor{\log_b x}\right\rfloor
+      | m = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor 
+      | c = Definition of $n$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have that:
+:[[Logarithm is Strictly Increasing]]
+and:
+:[[Natural Logarithm Function is Continuous]]
+Suppose that $\log_b x \in \Z$: let $\log_b x = n$, say.
+Then:
+:$x = b^n$
+It follows that $x \in \Z \iff b \in \Z$.
+Thus by [[McEliece's Theorem (Integer Functions)]]:
+:$\forall x \in \R_{\ge 1}: \left\lfloor{\log_b x}\right\rfloor = \left\lfloor{\log_b \left\lfloor{x}\right\rfloor}\right\rfloor \iff b \in \Z_{> 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of x+m over n}
+Tags: Floor Function, Floor of x+m over n
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $n > 0$.
+Let $x \in \R$.
+Then:
+:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
+where $\floor x$ denotes the [[Definition:Floor Function|floor of $x$]].
+\end{theorem}
+
+\begin{proof}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: f \left({x}\right) = \dfrac {x + m} n$
+It is clear that $f$ is both [[Definition:Strictly Increasing Real Function|strictly increasing]] and [[Definition:Continuous Real Function|continuous]] on the whole of $\R$.
+Let $\dfrac {x + m} n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \exists s \in \Z: \dfrac {x + m} n
+      | r = s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + m
+      | r = n s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = n s - m
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Z
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\forall x \in \R: f \left({x}\right) \in \Z \implies x \in \Z$
+So the conditions are fulfilled for [[McEliece's Theorem (Integer Functions)]] to be applied:
+:$\left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor \iff \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+:$y = x - \floor x$
+:$M = m + \floor x$
+We now have:
+:$(1): \quad 0 \le y < 1$
+and
+:$\floor y = 0$
+Write:
+:$(2): \quad M = k n + r$
+with $k \in \Z$ and $0 \le r \le n - 1$.
+By $(1)$ and $(2)$:
+:$(3): \quad 0 \le r + y < n - 1 + 1 = n$
+We have:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac {y + M} n}
+      | r = \floor {\dfrac {y + r} n + k}
+      | c = 
+}}
+{{eqn | r = k + \floor {\dfrac {r + y} n}
+      | c = 
+}}
+{{eqn | r = k
+      | c = from $(3)$
+}}
+{{eqn | r = \floor {\dfrac M n}
+      | c = 
+}}
+{{end-eqn}}
+Substituting $y$ and $M$, we obtain:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac {x - \floor x + m + \floor x} n }
+      | r = \floor {\dfrac {x + m} n }
+      | c = 
+}}
+{{eqn | r = \floor {\dfrac {m + \floor x} n}
+      | c = 
+}}
+{{end-eqn}}
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{McEliece's Theorem (Integer Functions)}
+Tags: Floor Function, Ceiling Function
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Continuous Real Function|continuous]], [[Definition:Strictly Increasing Real Function|strictly increasing real function]] defined on an [[Definition:Real Interval|interval]] $A$.
+Let:
+:$\forall x \in A: \left \lfloor{x}\right \rfloor \in A, \left \lceil{x}\right \rceil \in A$
+where:
+:$\left \lfloor{x}\right \rfloor$ denotes the [[Definition:Floor Function|floor]] of $x$
+:$\left \lceil{x}\right \rceil$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$
+Then:
+:$\forall x \in A: \left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor \iff \left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil \iff \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$
+\end{theorem}
+
+\begin{proof}
+Let $x \in A$.
+Hence {{hypothesis}} we have that both $\left \lfloor{x}\right \rfloor \in A$ and $\left \lceil{x}\right \rceil \in A$.
+=== Necessary Condition ===
+Let $f \left({x}\right) \in \Z$.
+Let:
+:$\left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$
+Then:
+{{begin-eqn}}
+{{eqn | l = f \left({x}\right)
+      | r = \left \lfloor{f \left({x}\right)}\right \rfloor
+      | c = [[Real Number is Integer iff equals Floor]]
+}}
+{{eqn | r = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor
+      | c = {{hypothesis}}
+}}
+{{end-eqn}}
+{{AimForCont}} $x \notin \Z$.
+{{begin-eqn}}
+{{eqn | l = \left \lfloor{x}\right \rfloor
+      | o = <
+      | r = x
+      | c = [[Floor of Non-Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = f \left({\left \lfloor{x}\right \rfloor}\right)
+      | o = <
+      | r = f \left({x}\right)
+      | c = as $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]
+}}
+{{eqn | ll= \leadsto
+      | l = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor
+      | o = <
+      | r = \left \lfloor{f \left({x}\right)}\right \rfloor
+      | c = as $\left \lfloor{f \left({x}\right)}\right \rfloor = f \left({x}\right)$ by [[Real Number is Integer iff equals Floor]]
+}}
+{{eqn | ll= \leadsto
+      | l = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor
+      | o = \ne
+      | r = \left \lfloor{f \left({x}\right)}\right \rfloor
+      | c = which [[Definition:Contradiction|contradicts]] $\left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$
+}}
+{{end-eqn}}
+Thus by [[Proof by Contradiction]]:
+:$x \in \Z$
+Similarly, let:
+:$\left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$
+Then:
+{{begin-eqn}}
+{{eqn | l = f \left({x}\right)
+      | r = \left \lceil{f \left({x}\right)}\right \rceil
+      | c = [[Real Number is Integer iff equals Ceiling]]
+}}
+{{eqn | r = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil
+      | c = {{hypothesis}}
+}}
+{{end-eqn}}
+{{AimForCont}} $x \notin \Z$.
+{{begin-eqn}}
+{{eqn | l = \left \lceil{x}\right \rceil
+      | o = >
+      | r = x
+      | c = [[Ceiling of Non-Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = f \left({\left \lceil{x}\right \rceil}\right)
+      | o = >
+      | r = f \left({x}\right)
+      | c = as $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]]
+}}
+{{eqn | ll= \leadsto
+      | l = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil
+      | o = >
+      | r = \left \lceil{f \left({x}\right)}\right \rceil
+      | c = as $\left \lceil{f \left({x}\right)}\right \rceil = f \left({x}\right)$ by [[Real Number is Integer iff equals Ceiling]]
+}}
+{{eqn | ll= \leadsto
+      | l = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil
+      | o = \ne
+      | r = \left \lceil{f \left({x}\right)}\right \rceil
+      | c = which [[Definition:Contradiction|contradicts]] $\left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$
+}}
+{{end-eqn}}
+Thus by [[Proof by Contradiction]]:
+:$x \in \Z$
+Thus:
+:$\forall x \in A: \left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor \implies \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$
+and:
+:$\forall x \in A: \left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil \implies \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $f$ be such that:
+:$f \left({x}\right) \in \Z \implies x \in \Z$
+{{AimForCont}} there exists $x \in A$ such that:
+:$\left \lfloor{f \left({x}\right)}\right \rfloor \ne \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$
+We have by definition of the [[Definition:Floor Function|floor function]] that $x \ge \left \lfloor{x}\right \rfloor$.
+As $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]], it cannot be the case that $\left \lfloor{f \left({x}\right)}\right \rfloor < \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$.
+So it must be that:
+:$\left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor < \left \lfloor{f \left({x}\right)}\right \rfloor$
+Because $f$ is [[Definition:Continuous Real Function|continuous]]:
+:$\exists y \in A: \left \lfloor{x}\right \rfloor < y \le x$
+such that $f \left({y}\right) \in \Z$
+But by definition of the [[Definition:Floor Function|floor function]], $y$ cannot be an [[Definition:Integer|integer]].
+Thus by [[Proof by Contradiction]]:
+:$\left({f \left({x}\right) \in \Z \implies x \in \Z}\right) \implies \left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$
+{{qed|lemma}}
+{{AimForCont}}, similarly, that there exists $x \in A$ such that:
+:$\left \lceil{f \left({x}\right)}\right \rceil \ne \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$
+We have by definition of the [[Definition:Ceiling Function|ceiling function]] that $x \le \left \lceil{x}\right \rceil$.
+As $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]], it cannot be the case that $\left \lceil{f \left({x}\right)}\right \rceil > \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$.
+So it must be that:
+:$\left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil > \left \lceil{f \left({x}\right)}\right \rceil$
+Because $f$ is [[Definition:Continuous Real Function|continuous]]:
+:$\exists y \in A: \left \lceil{x}\right \rceil > y \ge x$
+such that $f \left({y}\right) \in \Z$
+But by definition of the [[Definition:Ceiling Function|ceiling function]], $y$ cannot be an [[Definition:Integer|integer]].
+Thus by [[Proof by Contradiction]]:
+:$\left({f \left({x}\right) \in \Z \implies x \in \Z}\right) \implies \left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$
+{{qed|lemma}}
+Thus we have:
+:$\left({f \left({x}\right) \in \Z \implies x \in \Z}\right) \iff \left \lfloor{f \left({x}\right)}\right \rfloor = \left \lfloor{f \left({\left \lfloor{x}\right \rfloor}\right)}\right \rfloor$
+and
+:$\left({f \left({x}\right) \in \Z \implies x \in \Z}\right) \iff \left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil$
+Hence the result.
+{{qed}}
+{{Namedfor|Robert James McEliece|cat = McEliece}}
+\end{proof}<|endoftext|>
+\section{Ceiling of Non-Integer}
+Tags: Ceiling Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $x \notin \Z$.
+Then:
+:$\left \lceil{x}\right \rceil > x$
+where $\left \lceil{x}\right \rceil$ denotes the [[Definition:Ceiling Function|ceiling ]] of $x$.
+\end{theorem}
+
+\begin{proof}
+From [[Ceiling is between Number and One More]]:
+:$\left \lceil{x}\right \rceil \ge x$
+From [[Real Number is Integer iff equals Ceiling]]:
+:$x = \left \lceil {x} \right \rceil \iff x \in \Z$
+But we have $x \notin \Z$.
+So:
+:$\left \lceil {x} \right \rceil \ne x$
+and so:
+:$\left \lceil {x} \right \rceil > x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling of x+m over n}
+Tags: Ceiling Function, Ceiling of x+m over n
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $n > 0$.
+Let $x \in \R$.
+Then:
+:$\ceiling {\dfrac {x + m} n} = \ceiling {\dfrac {\ceiling x + m} n}$
+where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling of $x$]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left \lceil{\dfrac {x + m} n}\right \rceil
+      | r = -\left \lfloor{-\dfrac {x + m} n}\right \rfloor
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{eqn | r = -\left \lfloor{\dfrac {- x - m} n}\right \rfloor
+      | c = 
+}}
+{{eqn | r = -\left \lfloor{\dfrac {\left \lfloor{- x}\right \rfloor - m} n}\right \rfloor
+      | c = [[Floor of x+m over n|Floor of $\dfrac{x + m} n$]]
+}}
+{{eqn | r = -\left \lfloor{\dfrac {-\left \lceil{x}\right \rceil - m} n}\right \rfloor
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{eqn | r = -\left \lfloor{-\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \left \lceil{\dfrac {\left \lceil{x}\right \rceil + m} n}\right \rceil 
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: f \left({x}\right) = \dfrac {x + m} n$
+It is clear that $f$ is both [[Definition:Strictly Increasing Real Function|strictly increasing]] and [[Definition:Continuous Real Function|continuous]] on the whole of $\R$.
+Let $\dfrac {x + m} n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \exists s \in \Z: \dfrac {x + m} n
+      | r = s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + m
+      | r = n s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = n s - m
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \Z
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\forall x \in \R: f \left({x}\right) \in \Z \implies x \in \Z$
+So the conditions are fulfilled for [[McEliece's Theorem (Integer Functions)]] to be applied:
+:$\left \lceil{f \left({x}\right)}\right \rceil = \left \lceil{f \left({\left \lceil{x}\right \rceil}\right)}\right \rceil \iff \left(f \left({x}\right) \in \Z \implies x \in \Z)\right)$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over k of Floor of k over 2}
+Tags: Floor Function, Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \floor {\dfrac {n^2} 4}$
+\end{theorem}
+
+\begin{proof}
+By [[Permutation of Indices of Summation]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \sum_{k \mathop = 1}^n \floor {\dfrac {n + 1 - k} 2}$
+and so:
+:$\displaystyle \sum_{k \mathop = 1}^n \floor {\dfrac k 2} = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }$
+First take the case where $n$ is [[Definition:Even Integer|even]].
+For $k$ [[Definition:Odd Integer|odd]]:
+:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
+and:
+:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
+      | r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k - 1 + n + 1 - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac n 2
+      | c = 
+}}
+{{end-eqn}}
+For $k$ [[Definition:Even Integer|even]]:
+:$\floor {\dfrac k 2} = \dfrac k 2$
+and:
+:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2 = \dfrac {n - k} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
+      | r = \dfrac k 2 + \dfrac {n - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + n - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac n 2
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
+      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\dfrac n 2}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 n \dfrac n 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n^2} 4
+      | c = 
+}}
+{{eqn | r = \floor {\dfrac {n^2} 4}
+      | c = as $\dfrac {n^2} 4$ is an [[Definition:Integer|integer]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Next take the case where $n$ is [[Definition:Odd Integer|odd]].
+For $k$ [[Definition:Odd Integer|odd]]:
+:$\floor {\dfrac k 2} = \dfrac k 2 - \dfrac 1 2$
+and:
+:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2 - \dfrac 1 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
+      | r = \dfrac k 2 - \dfrac 1 2 + \dfrac {n + 1 - k} 2 - \dfrac 1 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k - 1 + n + 1 - k - 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n - 1} 2
+      | c = 
+}}
+{{end-eqn}}
+For $k$ [[Definition:Even Integer|even]]:
+:$\floor {\dfrac k 2} = \dfrac k 2$
+and:
+:$\floor {\dfrac {n + 1 - k} 2} = \dfrac {n + 1 - k} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2}
+      | r = \dfrac k 2 + \dfrac {n - k + 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + n - k + 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n + 1} 2
+      | c = 
+}}
+{{end-eqn}}
+Let $n = 2 t + 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \floor {\dfrac k 2}
+      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \paren {\floor {\dfrac k 2} + \floor {\dfrac {n + 1 - k} 2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \paren {\floor {\dfrac k 2} + \floor {\dfrac {2 t + 2 - k} 2} }
+      | c = 
+}}
+{{eqn | r = \dfrac t 2 \dfrac {\paren {2 t + 1} + 1} 2 + \dfrac {t + 1} 2 \dfrac {\paren {2 t + 1} - 1} 2
+      | c = there are $t$ [[Definition:Even Integer|even]] terms and $t + 1$ [[Definition:Odd Integer|odd]] terms 
+}}
+{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 2 t} 4
+      | c = multiplying out 
+}}
+{{eqn | r = \dfrac {4 t^2 + 4 t} 4 + \dfrac 1 4 -  \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {2 t + 1}^2} 4 - \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \dfrac {n^2} 4 - \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \floor {\dfrac {n^2} 4}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over k of Ceiling of k over 2}
+Tags: Ceiling Function, Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil$
+\end{theorem}
+
+\begin{proof}
+By [[Permutation of Indices of Summation]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \sum_{k \mathop = 1}^n \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil$
+and so:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)$
+First take the case where $n$ is [[Definition:Even Integer|even]].
+For $k$ [[Definition:Odd Integer|odd]]:
+:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$
+and:
+:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
+      | r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + 1 + n + 1 - k} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n + 2} 2
+      | c = 
+}}
+{{end-eqn}}
+For $k$ [[Definition:Even Integer|even]]:
+:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$
+and:
+:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2 = \dfrac {n - k + 2} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
+      | r = \dfrac k 2 + \dfrac {n - k + 2} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + n - k + 2} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n + 2} 2
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil
+      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\dfrac {n + 2} 2}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 n \left({\dfrac {n + 2} 2}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {n \left({n + 2}\right)} 4
+      | c = 
+}}
+{{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil
+      | c = as $\dfrac {n \left({n + 2}\right)} 4$ is an [[Definition:Integer|integer]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Next take the case where $n$ is [[Definition:Odd Integer|odd]].
+For $k$ [[Definition:Odd Integer|odd]]:
+:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2 + \dfrac 1 2$
+and:
+:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2 + \dfrac 1 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
+      | r = \dfrac k 2 + \dfrac 1 2 + \dfrac {n + 1 - k} 2 + \dfrac 1 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + 1 + n + 1 - k + 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n + 3} 2
+      | c = 
+}}
+{{end-eqn}}
+For $k$ [[Definition:Even Integer|even]]:
+:$\left \lceil{\dfrac k 2}\right \rceil = \dfrac k 2$
+and:
+:$\left \lceil{\dfrac {n + 1 - k} 2}\right \rceil = \dfrac {n + 1 - k} 2$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil
+      | r = \dfrac k 2 + \dfrac {n - k + 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {k + n - k + 1} 2
+      | c = 
+}}
+{{eqn | r = \dfrac {n + 1} 2
+      | c = 
+}}
+{{end-eqn}}
+Let $n = 2 t + 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \left \lceil{\dfrac k 2}\right \rceil
+      | r = \dfrac 1 2 \sum_{k \mathop = 1}^n \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {n + 1 - k} 2}\right \rceil}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac 1 2 \sum_{k \mathop = 1}^{2 t + 1} \left({\left \lceil{\dfrac k 2}\right \rceil + \left \lceil{\dfrac {2 t + 2 - k} 2}\right \rceil}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac t 2 \dfrac {\left({2 t + 1}\right) + 1} 2 + \dfrac {t + 1} 2 \dfrac {\left({2 t + 1}\right) + 3} 2
+      | c = there are $t$ [[Definition:Even Integer|even]] terms and $t + 1$ [[Definition:Odd Integer|odd]] terms 
+}}
+{{eqn | r = \dfrac {2 t^2 + 2 t} 4 + \dfrac {2 t^2 + 6 t + 4} 4
+      | c = multiplying out 
+}}
+{{eqn | r = \dfrac {4 t^2 + 8 t + 3} 4 + \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \dfrac {\left({2 t + 1}\right) \left({2 t + 3}\right)} + \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \dfrac {n \left({n + 2}\right)} 4 + \dfrac 1 4
+      | c = 
+}}
+{{eqn | r = \left \lceil{\dfrac {n \left({n + 2}\right)} 4}\right \rceil
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over k of Floor of mk+x over n}
+Tags: Floor Function, Summations
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $n > 0$.
+Let $x \in \R$.
+Then:
+:$\displaystyle \sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {m k + x} n}\right \rfloor = \dfrac {\left({m - 1}\right) \left({n - 1}\right)} 2 + \dfrac {d - 1} 2 + d \left \lfloor{\dfrac x d}\right \rfloor$
+where:
+:$\left \lfloor{x}\right \rfloor$ denotes the [[Definition:Floor Function|floor]] of $x$
+:$d$ is the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Modulo 1|modulo 1]]:
+:$\displaystyle \sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {m k + x} n}\right \rfloor = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n - \sum_{0 \mathop \le k \mathop < n} \left\{ {\dfrac {m k + x} n}\right\}$
+where $\left\{ {y}\right\}$ in this context denotes the [[Definition:Fractional Part|fractional part]] of $y$.
+First we have:
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \dfrac {m k + x} n
+      | r = \frac m n \sum_{0 \mathop \le k \mathop < n} k + \sum_{0 \mathop \le k \mathop < n} \dfrac x n
+      | c = 
+}}
+{{eqn | r = \frac m n \frac {n \left({n - 1}\right)} 2 + n \dfrac x n
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = \frac {m \left({n - 1}\right)} 2 + x
+      | c = 
+}}
+{{end-eqn}}
+Let $S$ be defined as:
+:$\displaystyle S := \sum_{0 \mathop \le k \mathop < n} \left\{ {\dfrac {m k + x} n}\right\}$
+Thus:
+:$\displaystyle (1): \quad \sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {m k + x} n}\right \rfloor = \dfrac {m \left({n - 1}\right)} 2 + x - S$
+Let $d = \gcd \left\{ {m, n}\right\}$.
+Let:
+{{begin-eqn}}
+{{eqn | l = t
+      | r = \frac n d
+}}
+{{eqn | l = u
+      | r = \frac m d
+}}
+{{eqn | ll= \leadsto
+      | l = \frac m n
+      | r = \frac u t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m t
+      | r = u n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u
+      | r = \frac {m t} n
+      | c = 
+}}
+{{end-eqn}}
+We have that:
+{{begin-eqn}}
+{{eqn | l = \left\{ {\dfrac {m k + x} n}\right\}
+      | r = \left\{ {\dfrac {m k + x} n + u}\right\}
+      | c = {{Defof|Fractional Part}}: $u$ is an [[Definition:Integer|integer]]
+}}
+{{eqn | r = \left\{ {\dfrac {m k + x} n + \frac {m t} n}\right\}
+      | c = 
+}}
+{{eqn | r = \left\{ {\dfrac {m \left({k + t}\right) + x} n}\right\}
+      | c = 
+}}
+{{end-eqn}}
+Thus $S$ consists of $d$ copies of the same [[Definition:Summation|summation]]:
+{{begin-eqn}}
+{{eqn | l = S
+      | r = \sum_{0 \mathop \le k \mathop < n} \left\{ {\dfrac {m k + x} n}\right\}
+      | c = 
+}}
+{{eqn | r = d \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac {m k + x} n}\right\}
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac {m k + x} n}\right\}
+      | r = \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac x n + \dfrac {u k} t}\right\}
+      | c = substituting $\dfrac u t$ for $\dfrac m n$
+}}
+{{eqn | r = \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac {x \bmod d} n + \dfrac k t}\right\}
+      | c = as $t \perp u$
+}}
+{{eqn | r = \sum_{0 \mathop \le k \mathop < t} \dfrac {x \bmod d} n + \dfrac k t
+      | c = as $\dfrac {x \bmod d} n < \dfrac 1 t$
+}}
+{{eqn | r = t \dfrac {x \bmod d} n + \frac 1 t \sum_{0 \mathop \le k \mathop < t} k
+      | c = 
+}}
+{{eqn | r = \dfrac {t \left({x \bmod d}\right)} n + \frac 1 t \frac {t \left({t - 1}\right)} 2
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = \dfrac {t \left({x \bmod d}\right)} n + \frac {t - 1} 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = S
+      | r = d \left({\dfrac {t \left({x \bmod d}\right)} n + \frac {t - 1} 2}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {n \left({x \bmod d}\right)} n + \frac {n - d} 2
+      | c = as $n = d t$
+}}
+{{eqn | r = x \bmod d + \frac {n - d} 2
+      | c = 
+}}
+{{end-eqn}}
+{{explain|Greater detail needed as to why $\displaystyle \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac x n + \dfrac {u k} t}\right\} {{=}} \sum_{0 \mathop \le k \mathop < t} \left\{ {\dfrac {x \bmod d} n + \dfrac k t}\right\}$}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {m k + x} n}\right \rfloor
+      | r = \frac {m \left({n - 1}\right)} 2 + x - S
+      | c = from $(1)$
+}}
+{{eqn | r = \frac {m \left({n - 1}\right)} 2 + x - d \left({\dfrac {t \left({x \bmod d}\right)} n + \frac {t - 1} 2}\right)
+      | c = 
+}}
+{{eqn | r = \frac {m \left({n - 1}\right)} 2 + x - x \bmod d - \frac {n - d} 2
+      | c = 
+}}
+{{eqn | r = \frac {m \left({n - 1}\right)} 2 + x - x + d \left \lfloor{\frac x d}\right \rfloor - \frac {n - 1} 2 + \frac {d - 1} 2
+      | c = {{Defof|Modulo Operation}} and algebra
+}}
+{{eqn | r = \frac {\left({m - 1}\right) \left({n - 1}\right)} 2 + \frac {d - 1} 2 + d \left \lfloor{\frac x d}\right \rfloor 
+      | c = simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over k of Floor of x plus k over y}
+Tags: Floor Function, Summations
+
+\begin{theorem}
+Let $x, y \in \R$ such that $y > 0$.
+Then:
+:$\displaystyle \sum_{0 \mathop \le k \mathop < y} \left \lfloor{x + \dfrac k y}\right \rfloor = \left \lfloor{x y + \left \lfloor{x + 1}\right \rfloor \left({\left \lceil{y}\right \rceil - y}\right)}\right \rfloor$
+\end{theorem}
+
+\begin{proof}
+When $x$ increases by $1$, both sides increase by $\left \lceil{y}\right \rceil$.
+So we can assume $0 \le x < 1$.
+When $x = 0$, both sides are equal to $0$.
+When $x$ increases past $1 - \dfrac k y$ for $0 \le k < y$, both sides increase by $1$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation over k of Ceiling of mk+x over n}
+Tags: Ceiling Function, Summations
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $n > 0$.
+Let $x \in \R$.
+Then:
+:$\displaystyle \sum_{0 \mathop \le k \mathop < n} \left \lceil{\dfrac {m k + x} n}\right \rceil = \dfrac {\left({m + 1}\right) \left({n - 1}\right)} 2 - \dfrac {d - 1} 2 + d \left \lceil{\dfrac x d}\right \rceil$
+where:
+:$\left \lceil{x}\right \rceil$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$
+:$d$ is the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{0 \mathop \le k \mathop < n} \left \lceil{\dfrac {m k + x} n}\right \rceil
+      | r = \sum_{0 \mathop \le k \mathop < n} \left({-\left \lfloor{-\dfrac {m k + x} n}\right \rfloor}\right)
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{eqn | r = -\sum_{0 \mathop \le k \mathop < n} \left \lfloor{\dfrac {-m k - x} n}\right \rfloor
+      | c = 
+}}
+{{eqn | r = -\left({\dfrac {\left({\left({-m}\right) - 1}\right) \left({n - 1}\right)} 2 + \dfrac {d - 1} 2 + d \left \lfloor{\dfrac {\left({-x}\right)} d}\right \rfloor}\right)
+      | c = [[Summation over k of Floor of mk+x over n|Summation over $k$ of $\left \lfloor{\dfrac {m k + x} n}\right \rfloor$]]
+}}
+{{eqn | r = \dfrac {\left({m + 1}\right) \left({n - 1}\right)} 2 - \dfrac {d - 1} 2 - d \left \lfloor{- \dfrac x d}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \dfrac {\left({m + 1}\right) \left({n - 1}\right)} 2 - \dfrac {d - 1} 2 + d \left \lceil{\dfrac x d}\right \rceil
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Semantic Consequence preserved in Supersignature}
+Tags: Predicate Logic, Model Theory
+
+\begin{theorem}
+Let $\mathcal L, \mathcal L'$ be [[Definition:Signature for Predicate Logic|signatures]] for the [[Definition:Language of Predicate Logic|language of predicate logic]].
+Let $\mathcal L'$ be a [[Definition:Supersignature|supersignature]] of $\mathcal L$.
+Let $\mathbf A$ be an [[Definition:Sentence|$\mathcal L$-sentence]].
+Let $\Sigma$ be a [[Definition:Set|set]] of [[Definition:Sentence|$\mathcal L$-sentences]].
+Then the following are [[Definition:Logically Equivalent|equivalent]]:
+:$\mathcal A \models_{\mathrm{PL}} \mathbf A$ for all [[Definition:Structure for Predicate Logic|$\mathcal L$-structure]] $\mathcal A$ for which $\mathcal A \models_{\mathrm{PL}} \Sigma$
+:$\mathcal A' \models_{\mathrm{PL}} \mathbf A$ for all [[Definition:Structure for Predicate Logic|$\mathcal L'$-structure]] $\mathcal A'$ for which $\mathcal A' \models_{\mathrm{PL}} \Sigma$
+where $\models_{\mathrm{PL}}$ denotes the [[Definition:Model (Predicate Logic)|models]] relation.
+That is to say, the notion of [[Definition:Semantic Consequence (Predicate Logic)|semantic consequence]] is preserved in passing to a [[Definition:Supersignature|supersignature]].
+\end{theorem}
+
+\begin{proof}
+{{finish|
+* every $\mathcal A$ arises as the reduct of some $\mathcal A'$;
+* every $\mathcal A'$ has the same valuations as its reduct.
+}}
+\end{proof}<|endoftext|>
+\section{Satisfiability preserved in Supersignature}
+Tags: Predicate Logic, Model Theory
+
+\begin{theorem}
+Let $\mathcal L, \mathcal L'$ be [[Definition:Signature for Predicate Logic|signatures]] for the [[Definition:Language of Predicate Logic|language of predicate logic]].
+Let $\mathcal L'$ be a [[Definition:Supersignature|supersignature]] of $\mathcal L$.
+Let $\Sigma$ be a [[Definition:Set|set]] of [[Definition:Sentence|$\mathcal L$-sentences]].
+Then the following are [[Definition:Logically Equivalent|equivalent]]:
+:$\mathcal A \models_{\mathrm{PL}} \Sigma$ for some [[Definition:Structure for Predicate Logic|$\mathcal L$-structure]] $\mathcal A$
+:$\mathcal A' \models_{\mathrm{PL}} \Sigma$ for some [[Definition:Structure for Predicate Logic|$\mathcal L'$-structure]] $\mathcal A'$
+where $\models_{\mathrm{PL}}$ is the [[Definition:Model (Predicate Logic)|models]] relation.
+That is to say, the notion of [[Definition:Satisfiable Set of Formulas|satisfiability]] is preserved in passing to a [[Definition:Supersignature|supersignature]].
+\end{theorem}
+
+\begin{proof}
+{{finish|* every $\mathcal A$ arises as the reduct of some $\mathcal A'$;
+* every $\mathcal A'$ has the same valuations as its reduct.
+}}
+\end{proof}<|endoftext|>
+\section{Floor Function is Replicative}
+Tags: Floor Function, Replicative Functions
+
+\begin{theorem}
+The [[Definition:Floor Function|floor function]] is a [[Definition:Replicative Function|replicative function]] in the sense that:
+:$\displaystyle \forall n \in \Z_{> 0}: \sum_{k \mathop = 0}^{n - 1} \left \lfloor{x + \frac k n}\right \rfloor = \left \lfloor{n x}\right \rfloor$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \left \lfloor{x + \dfrac k n}\right \rfloor
+      | r = \left \lfloor{x n + \left \lfloor{x + 1}\right \rfloor \left({\left \lceil{n}\right \rceil - n}\right)}\right \rfloor
+      | c =  [[Summation over k of Floor of x plus k over y|Summation over $k$ of $\left \lfloor{x + \dfrac k y}\right \rfloor$]], setting $y = n$
+}}
+{{eqn | r = \left \lfloor{x n + \left \lfloor{x + 1}\right \rfloor \left({n - n}\right)}\right \rfloor
+      | c = [[Real Number is Integer iff equals Ceiling]]
+}}
+{{eqn | r = \left \lfloor{n x + \left \lfloor{x + 1}\right \rfloor \times 0}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \left \lfloor{n x}\right \rfloor
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subtract Half is Replicative Function}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = x - \dfrac 1 2$
+Then $f$ is a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
+      | r = \sum_{k \mathop = 0}^{n - 1} \paren {x - \frac 1 2 + \frac k n}
+      | c = 
+}}
+{{eqn | r = n x - \frac n 2 + \frac 1 n \sum_{k \mathop = 0}^{n - 1} k
+      | c = 
+}}
+{{eqn | r = n x - \frac n 2 + \frac 1 n \frac {n \paren {n - 1} } 2
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = n x - \frac n 2 + \frac n 2 - \frac 1 2
+      | c = 
+}}
+{{eqn | r = n x - \frac 1 2
+      | c = 
+}}
+{{eqn | r = \map f {n x}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Replicative Function|replicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Membership of Set of Integers is Replicative Function}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = \sqbrk {x \in \Z}$
+where $\sqbrk {\cdots}$ is [[Definition:Iverson's Convention|Iverson's convention]].
+Then $f$ is a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+First note that the interval between $x$ and $x + \dfrac {n - 1} n$ is less than $1$.
+Thus there can be no more than one $k$ such that $0 \le k < n$ such that:
+:$x + \dfrac k n \in \Z$
+Hence:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} \le 1$
+First let:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = x + \frac k n
+      | o = \in 
+      | r = \Z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n \paren {x + \frac k n}
+      | o = \in 
+      | r = \Z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n x + k
+      | o = \in 
+      | r = \Z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n x
+      | o = \in 
+      | r = \Z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqbrk {n x \in \Z}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 0$
+{{AimForCont}} $\sqbrk {n x \in \Z} = 1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n x
+      | o = \in
+      | r = \Z
+      | c = 
+}}
+{{eqn | lll= \leadsto
+      | ll= \forall k \in \Z:
+      | l = n x + k
+      | o = \in
+      | r = \Z
+      | c = 
+}}
+{{end-eqn}}
+But at least one $n x + k$ such that $0 \le k < n$ is a [[Definition:Multiple of Integer|multiple]] of $n$.
+Hence:
+:$\exists k \in \Z: 0 \le k < n: \sqbrk {x + \dfrac k n \in \Z} = 1$
+So:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z} = 1$
+which [[Definition:Contradiction|contradicts]] the supposition that $\sqbrk {n x \in \Z} = 1$.
+Hence the result by definition of [[Definition:Replicative Function|replicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Membership of Set of Strictly Positive Integers is Replicative Function}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = \sqbrk {x \in \Z_{> 0} }$
+where $\sqbrk \cdots$ is [[Definition:Iverson's Convention|Iverson's convention]].
+Then $f$ is a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R$ such that $x > 0$.
+Then for all $k \in \Z$ such that $0 \le k < n$:
+:$x + \dfrac k n \in \Z_{> 0}$
+and so from [[Membership of Set of Integers is Replicative Function]]:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = \sqbrk {n x \in \Z_{> 0} }$
+Let $x \le 0$.
+Then for all $k \in \Z$ such that $0 \le k < n$:
+:$x + \dfrac k n < 1$
+and so:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \sqbrk {x + \frac k n \in \Z_{> 0} } = 0 = \sqbrk {n x \in \Z_{> 0} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Membership of Equivalence Class of m mod pi is Replicative Function}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = \sqbrk {\exists r \in \Q, \exists m \in \Z: x = r \pi + m}$
+where $\sqbrk {\cdots}$ is [[Definition:Iverson's Convention|Iverson's convention]].
+Then $f$ is a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+Let $\map f x = 1$.
+Then:
+:$\exists r \in \Q, \exists m \in \Z: x = r \pi + m$
+and:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = r \pi + m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n x
+      | r = \paren {r n} \pi + m n
+      | c = 
+}}
+{{end-eqn}}
+But $r n \in Q$ and $m n \in \Z$.
+So:
+:$\map f x = 1 \implies \map f {n x} = 1$
+Let $n \in \Z_{>0}$.
+Let $k \in \Z: 0 \le k < n$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x + \frac k n
+      | r = r \pi + m + \frac k n 
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + \frac k n
+      | r = \paren {r + \frac {k \pi} n} \pi + m
+      | c = 
+}}
+{{end-eqn}}
+Thus, when $0 \le k < n$ it follows that the [[Definition:Coefficient|coefficient]] of $\pi$ cannot be [[Definition:Rational Number|rational]].
+Thus:
+:$\map f x = 1 \implies \map f {x + \dfrac k n} = 0$
+and so:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = \map f {n x}$
+{{qed|lemma}}
+Now let $\map f {n x} = 1$.
+Then:
+:$r_n \pi + m_n = n x$
+for some [[Definition:Rational Number|rational]] $r_n$ and [[Definition:Integer|integer]] $m_n$.
+Thus exactly one of $m_n + k$ where $0 \le n < k$ is a [[Definition:Multiple of Integer|multiple]] of $n$.
+Therefore there exists exactly one $k \in \Z$ such that $0 \le n < k$ such that:
+:$\map f {x + \dfrac k n} = 1$
+Therefore:
+:$\displaystyle \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n} = 1 = \map f {n x}$
+Hence the result by definition of [[Definition:Replicative Function|replicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Logarithm of Absolute Value of 2 times Sine of pi x is Replicative Function}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\forall x \in \R: \map f x = \log \, \size {2 \sin \pi x}$
+Then $f$ is a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \map f {x + \frac k n}
+      | r = \sum_{k \mathop = 0}^{n - 1} \log \, \size {2 \sin \pi \paren {x + \frac k n} }
+      | c = 
+}}
+{{eqn | r = \log \prod_{k \mathop = 0}^{n - 1} \size {2 \sin \pi \paren {x + \frac k n} }
+      | c = [[Sum of Logarithms]]
+}}
+{{end-eqn}}
+Thus to demonstrate that $f$ is [[Definition:Replicative Function|replicative]], it is sufficient to demonstrate that:
+:$\displaystyle \prod_{k \mathop = 0}^{n - 1} \left({2 \sin \pi \left({x + \frac k n}\right)}\right) = 2 \sin \pi n x$
+By [[Product Formula for Sine]]:
+{{begin-eqn}}
+{{eqn | l = 2 \sin \pi n x
+      | r = 2^n \prod_{k \mathop = 0}^{n - 1} \paren {\map \sin {\pi x + \frac {k \pi} n} }
+}}
+{{eqn | r = \prod_{k \mathop = 0}^{n - 1} \paren {2 \sin \pi \paren {x + \frac k n} }
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Ideals is Upper Adjoint implies Lattice is Continuous}
+Tags: Continuous Lattices, Galois Connections
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Let $\mathit{Ids}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $L$.
+Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$
+Let $f: \mathit{Ids}\left({L}\right) \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall I \in \mathit{Ids}\left({L}\right): f\left({I}\right) = \sup I$
+Let $f$ be an [[Definition:Galois Connection|upper adjoint of a Galois connection]].
+Then $L$ is [[Definition:Continuous Ordered Set|continuous]].
+\end{theorem}
+
+\begin{proof}
+We will prove that
+:$\forall x \in S: \exists I \in \mathit{Ids}\left({L}\right): x \preceq \sup I \land \forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$ 
+Let $x \in S$.
+Define $I := \inf \left({f^{-1}\left[{x^\succeq}\right]}\right)$.
+By definition of $P$:
+:$I \in \mathit{Ids}\left({L}\right)$
+We will prove that
+:$\forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$
+Let $J \in \mathit{Ids}\left({L}\right)$ such that
+:$x \preceq \sup J$
+By definition of $f$:
+:$x \preceq f\left({J}\right)$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$f\left({J}\right) \in x^\succeq$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$J \in f^{-1}\left[{x^\succeq}\right]$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$I \precsim J$
+Hence by definition of $\precsim$:
+:$I \subseteq J$
+{{qed|lemma}}
+By definition of [[Definition:Galois Connection|upper adjoint of a Galois connection]]:
+:there exists a [[Definition:Mapping|mapping]] $d: S \to \mathit{Ids}\left({L}\right)$: $\left({f, d}\right)$ is a [[Definition:Galois Connection|Galois connection]].
+By [[Galois Connection is Expressed by Minimum]]
+:$d\left({x}\right) = \min \left({f^{-1} \left[{x^\succeq}\right]}\right)$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$I \in f^{-1} \left[{x^\succeq}\right]$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$f\left({I}\right) \in x^\succeq$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$x \preceq f\left({I}\right)$
+Thus by definition of $f$:
+:$x \preceq \sup I$
+Hence
+:$\exists I \in \mathit{Ids}\left({L}\right): x \preceq \sup I \land \forall J \in \mathit{Ids}\left({L}\right): x \preceq \sup J \implies I \subseteq J$ 
+{{qed|lemma}}
+Hence by [[Continuous iff For Every Element There Exists Ideal Element Precedes Supremum]]:
+:$L$ is [[Definition:Continuous Ordered Set|continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Replicative Functions is Replicative}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ and $g: \R \to \R$ be [[Definition:Real Function|real functions]].
+Let $f$ and $g$ both be [[Definition:Replicative Function|replicative functions]].
+Then the [[Definition:Pointwise Sum|pointwise sum]] of $f$ and $g$ is also a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} \left({f + g}\right) \left({x + \frac k n}\right)
+      | r = \sum_{k \mathop = 0}^{n - 1} \left({f \left({x + \frac k n}\right) + g \left({x + \frac k n}\right)}\right)
+      | c = {{Defof|Pointwise Sum}}
+}}
+{{eqn | r = \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right) + \sum_{k \mathop = 0}^{n - 1} g \left({x + \frac k n}\right)
+      | c = 
+}}
+{{eqn | r = f \left({n x}\right) + g \left({n x}\right)
+      | c = {{Defof|Replicative Function}}
+}}
+{{eqn | r = \left({f + g}\right) \left({n x}\right)
+      | c = {{Defof|Pointwise Sum}}
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Replicative Function|replicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constant Multiple of Replicative Function is Replicative}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
+Let $f$ be a [[Definition:Replicative Function|replicative function]].
+Let $c \in \R$ be a [[Definition:Constant|constant]].
+Let $g: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$g \left({x}\right) = c f \left({x}\right)$
+Then $g$ is also a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} g \left({x + \frac k n}\right)
+      | r = \sum_{k \mathop = 0}^{n - 1} c f \left({x + \frac k n}\right)
+      | c = 
+}}
+{{eqn | r = c \sum_{k \mathop = 0}^{n - 1} f \left({x + \frac k n}\right)
+      | c = 
+}}
+{{eqn | r = c f \left({n x}\right)
+      | c = {{Defof|Replicative Function}}
+}}
+{{eqn | r = g \left({n x}\right)
+      | c = 
+}}
+{{end-eqn}}
+Hence the result by definition of [[Definition:Replicative Function|replicative function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Replicative Function of x minus Floor of x is Replicative}
+Tags: Replicative Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
+Let $f$ be a [[Definition:Replicative Function|replicative function]].
+Let $g: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$g \left({x}\right) = f \left({x - \left \lfloor{x}\right \rfloor}\right)$
+Then $g$ is also a [[Definition:Replicative Function|replicative function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 0}^{n - 1} g \left({x + \frac k n}\right)
+      | r = \sum_{k \mathop = 0}^{n - 1} \left({f \left({x + \frac k n - \left \lfloor{x + \frac k n}\right \rfloor}\right)}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Closed Form for Sequence 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...}
+Tags: Sequences, Ceiling Function, Closed Forms
+
+\begin{theorem}
+Let $a_1, a_2, a_3, \ldots$ be the sequence:
+:$\left \langle{a_n}\right \rangle = 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots$
+Then:
+:$a_n = \left \lceil{\dfrac {\sqrt {1 + 8 n} - 1} 2}\right \rceil$
+\end{theorem}
+
+\begin{proof}
+From [[Closed Form for Triangular Numbers]], for $n = 1, 3, 6, 10, \ldots$:
+:$n = \dfrac {a_n \left({a_n + 1}\right)} 2$
+Thus by the [[Quadratic Formula]]:
+$a_n = \dfrac {-1 \pm \sqrt {1 + 8 n} } 2$
+In this context it is the [[Definition:Positive Real Number|positive]] root that is required.
+The result follows by definition of [[Definition:Ceiling Function|ceiling function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence as Summation of Difference of Adjacent Terms}
+Tags: Summations, Sum of Sequence as Summation of Difference of Adjacent Terms
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n a_k
+      | r = \sum_{k \mathop = 1}^n \left({k - \left({k - 1}\right)}\right) a_k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 1}^n \left({k - 1}\right) a_k
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n k a_k - \sum_{k \mathop = 0}^{n - 1} k a_{k + 1}
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = n a_n + \sum_{k \mathop = 1}^{n - 1} k a_k - \sum_{k \mathop = 1}^{n - 1} k a_{k + 1} + 0 \times a_0
+      | c = extracting the end terms
+}}
+{{eqn | r = n a_n + \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right)
+      | c = 
+}}
+{{eqn | r = n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Abel's Lemma: Formulation 2]], after renaming and reassigning variables:
+:$\displaystyle \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} A_k \left({a_k - a_{k + 1} }\right) + A_n a_n$
+where:
+:$\left \langle {a} \right \rangle$ and $\left \langle {b} \right \rangle$ be [[Definition:Sequence|sequences]] in $\R$
+:$\displaystyle A_n = \sum_{k \mathop = 1}^n {b_k}$ be the [[Definition:Partial Sum|partial sum]] of $\left \langle {b} \right \rangle$ from $1$ to $n$.
+Let $\left \langle{b}\right \rangle$ be defined as:
+:$\forall k: b_k = 1$
+Thus:
+:$\displaystyle A_n = \sum_{k \mathop = 1}^n 1 = n$
+and so:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n a_k
+      | r = \sum_{k \mathop = 1}^{n - 1} k \left({a_k - a_{k + 1} }\right) + n a_n
+      | c = 
+}}
+{{eqn | r = n a_n - \sum_{k \mathop = 1}^{n - 1} k \left({a_{k + 1} - a_k}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Lattice and Way Below implies Preceding implies Preceding}
+Tags: Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Continuous Ordered Set|continuous]] [[Definition:Complete Lattice|complete lattice]].
+Let $a, b \in S$.
+Let
+:$\forall c \in S: c \ll a \implies c \preceq b$
+Then $a \preceq b$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$\forall c \in a^\ll: c \preceq b$
+By definition:
+:$b$ is an [[Definition:Upper Bound of Set|upper bound]] for $a^\ll$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \left({a^\ll}\right) \preceq b$
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+Thus by [[Definition:Axiom of Approximation|axiom of approximation]]:
+:$a = \sup \left({a^\ll}\right) \preceq b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Floor of Log base b of k}
+Tags: Summations, Floor Function, Logarithms
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $b \in \Z$ such that $b \ge 2$.
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\log_b k}\right \rfloor = \left({n + 1}\right) \left \lfloor{\log_b n}\right \rfloor - \dfrac {b^{\left \lfloor{\log_b n}\right \rfloor + 1} - b} {b - 1}$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Sequence as Summation of Difference of Adjacent Terms]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\log_b k}\right \rfloor = n \left \lfloor{\log_b n}\right \rfloor - \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor}\right)$
+Let $S$ be defined as:
+:$\displaystyle S := \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor}\right)$
+As $b \ge 2$, we have that:
+:$\log_b {k + 1} - \log_b k < 1$
+As $b$ is an [[Definition:Integer|integer]]:
+:$\left \lfloor{\log_b {k + 1} }\right \rfloor - \left \lfloor{\log_b k}\right \rfloor = 1$
+{{iff}} $k + 1$ is a [[Definition:Integer Power|power]] of $b$.
+So:
+{{begin-eqn}}
+{{eqn | l = S
+      | r = \sum_{k \mathop = 1}^{n - 1} k \left[{k + 1 \text{ is a power of } b}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le t \mathop < \log_b n} \left({b^t - 1}\right)
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le t \mathop < \log_b n} b^t - \left \lfloor{\log_b n}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \frac {b^{\left \lfloor{\log_b n}\right \rfloor + 1} - 1} {b - 1} - \left \lfloor{\log_b n}\right \rfloor
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Floor of Root k}
+Tags: Summations, Floor Function
+
+\begin{theorem}
+Let $n \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $b \in \Z$ such that $b \ge 2$.
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\sqrt k}\right \rfloor = \left \lfloor{\sqrt n}\right \rfloor \left({n - \dfrac {\left({2 \left \lfloor{\sqrt n}\right \rfloor + 5}\right) \left({\left \lfloor{\sqrt n}\right \rfloor - 1}\right)} 6 }\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Sequence as Summation of Difference of Adjacent Terms]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \left \lfloor{\sqrt k}\right \rfloor = n \left \lfloor{\sqrt n}\right \rfloor - \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor}\right)$
+Let $S$ be defined as:
+:$\displaystyle S := \sum_{k \mathop = 1}^{n - 1} k \left({\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor}\right)$
+We have that:
+:$\sqrt {k + 1} - \sqrt k < 1$
+and so:
+:$\left \lfloor{\sqrt {k + 1} }\right \rfloor - \left \lfloor{\sqrt k}\right \rfloor = 1$
+{{iff}} $k + 1$ is a [[Definition:Square Number|square number]].
+So:
+{{begin-eqn}}
+{{eqn | l = S
+      | r = \sum_{k \mathop = 1}^{n - 1} k \left[{k + 1 \text{ is a square number} }\right]
+      | c = 
+}}
+{{eqn | l = S
+      | r = \sum_{k \mathop = 2}^{n} (k - 1) \left[{k \text{ is a square number} }\right]
+      | c = 
+}}
+{{eqn | r = \sum_{2 \mathop \le t \mathop \le \sqrt n} \left({t^2 - 1}\right)
+      | c = $1$ is not included in this [[Definition:Summation|summation]]
+}}
+{{eqn | r = \sum_{1 \mathop \le t \mathop \le \sqrt n} t^2 - 1 - \sum_{2 \mathop \le t \mathop \le \sqrt n} 1
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le t \mathop \le \sqrt n} t^2 - \sum_{1 \mathop \le t \mathop \le \sqrt n} 1
+      | c = 
+}}
+{{eqn | r = \frac {\left \lfloor{\sqrt n}\right \rfloor \left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right)} 6 - \left \lfloor{\sqrt n}\right \rfloor
+      | c = [[Sum of Sequence of Squares]] 
+}}
+{{eqn | r = \frac {\left \lfloor{\sqrt n}\right \rfloor \left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right) - 6 \left \lfloor{\sqrt n}\right \rfloor} 6
+      | c = 
+}}
+{{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \frac {\left({\left \lfloor{\sqrt n}\right \rfloor + 1}\right) \left({2 \left \lfloor{\sqrt n}\right \rfloor + 1}\right) - 6} 6
+      | c = 
+}}
+{{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \frac {2 \left \lfloor{\sqrt n}\right \rfloor^2 + 3 \left \lfloor{\sqrt n}\right \rfloor - 5} 6
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{k \mathop = 1}^n \left \lfloor{\sqrt k}\right \rfloor
+      | r = n \left \lfloor{\sqrt n}\right \rfloor - \left \lfloor{\sqrt n}\right \rfloor \frac {\left({2 \left \lfloor{\sqrt n}\right \rfloor + 5}\right) \left({\left \lfloor{\sqrt n}\right \rfloor - 1}\right)} 6
+      | c = 
+}}
+{{eqn | r = \left \lfloor{\sqrt n}\right \rfloor \left({n - \dfrac {\left({2 \left \lfloor{\sqrt n}\right \rfloor + 5}\right) \left({\left \lfloor{\sqrt n}\right \rfloor - 1}\right)} 6 }\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Top is Meet Irreducible}
+Tags: Order Theory, Meet Irreducible
+
+\begin{theorem}
+Let $\left({S, \wedge, \preceq}\right)$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Meet Semilattice|meet semilattice]].
+Then $\top$ is [[Definition:Meet Irreducible|meet irreducible]]
+where $\top$ denotes the [[Definition:Greatest Element|greatest element]] in $S$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ such that
+:$\top = x \wedge y$
+By [[Meet Precedes Operands]]
+:$\top \preceq x$ and $\top \preceq y$
+By definition of [[Definition:Greatest Element|greatest element]]:
+:$x \preceq \top$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\top = x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Irreducible iff Finite Infimum equals Element}
+Tags: Order Theory, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $x \in S$.
+Then
+:$x$ is [[Definition:Meet Irreducible|meet irreducible]]
+{{iff}}
+:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S$: $x = \inf A \implies x \in A$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x$ be [[Definition:Meet Irreducible|meet irreducible]].
+We will prove the result by [[Principle of Mathematical Induction|induction]] on [[Definition:Cardinality|cardinality]] of $A$.
+==== Base case ====
+:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $A$ of $S$: $\left\vert{A}\right\vert = 1 \land x = \inf A \implies x \in A$
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ such that
+:$\left\vert{A}\right\vert = 1 \land x = \inf A$
+Then
+:$A = \left\{ {a}\right\}$
+By [[Infimum of Singleton]]:
+:$x = a$
+Thus by definition of [[Definition:Singleton|singleton]]:
+:$x \in A$
+{{qed|lemma}}
+==== Induction Hypothesis ====
+:$n \ge 1$ and for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $A$ of $S$: $\left\vert{A}\right\vert = n \land x = \inf A \implies x \in A$
+==== Induction Step ====
+:$n \ge 1$ and for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] $A$ of $S$: $\left\vert{A}\right\vert = n+1 \land x = \inf A \implies x \in A$
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$ such that
+:$\left\vert{A}\right\vert = n+1 \land x = \inf A$
+Then
+:$A = \left\{ {a_1, \dots, a_n, a_{n+1} }\right\}$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\inf A = \left({a_1 \wedge \dots \wedge a_n}\right) \wedge a_{n+1}$
+By definition of [[Definition:Meet Irreducible|meet irreducible]]:
+:$x = a_1 \wedge \dots \wedge a_n$ or $x = a_{n+1}$
+By Induction Hypothesis:
+:$x \in \left\{ {a_1, \dots, a_n}\right\}$ or $x = a_{n+1}$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \in A$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S$: $x = \inf A \implies x \in A$
+Let $y, z \in S$ such that
+:$x = y \wedge z$
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$x = \inf \left\{ {y, z}\right\}$
+By assumption:
+:$x \in \left\{ {y, z}\right\}$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x = y$ or $x = z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over k of Sum over j of Floor of n + jb^k over b^k+1}
+Tags: Summations, Floor Function
+
+\begin{theorem}
+Let $n, b \in \Z$ such that $n \ge 0$ and $b \ge 2$.
+Then:
+:$\displaystyle \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor = n$
+\end{theorem}
+
+\begin{proof}
+We have that $\left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor$ is in the form $\left \lfloor{\dfrac {m k + x} n}\right \rfloor$ so that:
+{{begin-eqn}}
+{{eqn | l = \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor
+      | r = \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {j + \frac n {b^k} } b}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \sum_{0 \mathop \le j \mathop < b} \left \lfloor{\dfrac {j + \frac n {b^k} } b}\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor
+      | c = 
+}}
+{{eqn | r = \dfrac {\left({1 - 1}\right) \left({b - 1}\right)} 2 + \dfrac {\left({1 - 1}\right)} 2 + 1 \left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor
+      | c = (find link)
+}}
+{{eqn | r = \left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor
+      | c = 
+}}
+{{end-eqn}}
+{{LinkWanted}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop \ge 0} \sum_{1 \mathop \le j \mathop < b} \left \lfloor{\dfrac {n + j b^k} {b^{k + 1} } }\right \rfloor
+      | r = \sum_{k \mathop \ge 0} \left({\left \lfloor{\dfrac n {b^k} }\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor}\right)
+      | c = 
+}}
+{{eqn | r = \lim_{k \mathop \to \infty} \left \lfloor{\dfrac n 1}\right \rfloor - \left \lfloor{\dfrac n {b^{k + 1} } }\right \rfloor
+      | c = [[Definition:Telescoping Series|Telescoping Series]]
+}}
+{{eqn | r = n
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum over j of Function of Floor of mj over n}
+Tags: Summations, Floor Function
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]].
+Then:
+:$\displaystyle \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} } = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} - \map f r} + n \map f {m - 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = r
+      | m = \floor{\dfrac {m j} n}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r
+      | o = \le
+      | m = \dfrac {m j} n
+      | mo= <
+      | r = r + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {r n} m
+      | o = \le
+      | m = j
+      | mo= <
+      | r = \dfrac {\paren {r + 1} } m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ceiling {\dfrac {r n} m}
+      | o = \le
+      | m = j
+      | mo= <
+      | r = \ceiling {\dfrac {\paren {r + 1} } m}
+      | c = as $j$ is an [[Definition:Integer|integer]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \sum_{0 \mathop \le j \mathop < n} \map f {\floor {\dfrac {m j} n} }
+      | c = 
+}}
+{{eqn | r = \sum_{0 \mathop \le \ceiling {\frac {r n} m} \mathop < n} \map f r
+      | c = 
+}}
+{{eqn | r = \sum_{0 \mathop \le r \mathop < m} \map f r \paren {\ceiling {\dfrac {\paren {r + 1} } m} - \ceiling {\dfrac {r n} m} }
+      | c = 
+}}
+{{eqn | r = \map f 0 \ceiling {\dfrac n m} + \map f 1 \paren {\ceiling {\dfrac {2 n} m} - \ceiling {\dfrac n m} } + \cdots + \map f {m - 1} \paren {\ceiling {\dfrac {m n} m} - \ceiling {\dfrac {\paren {m - 1} n} m} }
+      | c = 
+}}
+{{eqn | r = \ceiling {\dfrac n m} \paren {\map f 0 + \map f 1} + \ceiling {\dfrac {2 n} m} \paren {\map f 0 + \map f 1} + \cdots + \ceiling {\dfrac {\paren {m - 1} n} m} \paren {\map f {m - 2} + \map f {m - 1} } + n \map f {m - 1}
+      | c = 
+}}
+{{eqn | r = \sum_{0 \mathop \le r \mathop < m} \ceiling {\dfrac {r n} m} \paren {\map f {r - 1} + \map f r} + n \map f {m - 1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{General Reciprocity Law}
+Tags: Summations, Floor Function
+
+\begin{theorem}
+:$\displaystyle \sum_{0 \mathop \le j \mathop < \alpha n} f \left({\left \lfloor{\dfrac {m j} n}\right \rfloor}\right) = \sum_{0 \mathop \le r \mathop < \alpha m} \left \lceil{\dfrac {r n} m}\right \rceil \left({f \left({r - 1}\right) - f \left({r}\right)}\right) + \left \lceil{\alpha n}\right \rceil f \left({\left \lceil{\alpha m}\right \rceil - 1}\right)$
+for $\alpha \in \R$.
+\end{theorem}<|endoftext|>
+\section{Equivalence of Definitions of Legendre Symbol}
+Tags: Legendre Symbol
+
+\begin{theorem}
+Let $p$ be an [[Definition:Odd Prime|odd prime]].
+Let $a \in \Z$.
+{{TFAE|def = Legendre Symbol}}
+\end{theorem}
+
+\begin{proof}
+From [[Integer to Power of p-1 over 2 Modulo p|Integer to Power of $\dfrac {p - 1} 2$ Modulo $p$]], one of the following cases holds:
+:$(1): \quad a^{\frac{(p - 1)} 2} \bmod p = 0$
+:$(2): \quad a^{\frac{(p - 1)} 2} \bmod p = 1$
+:$(3): \quad a^{\frac{(p - 1)} 2} \bmod p = p - 1$
+Let $\paren {\dfrac a p}$ be defined as the [[Definition:Legendre Symbol/Definition 1|Legendre symbol by definition 1]]:
+{{:Definition:Legendre Symbol/Definition 1}}
+By [[Negative Number is Congruent to Modulus minus Number]] we have:
+:$p - 1 \pmod p = -1$
+Thus $\paren {\dfrac a p}$ is the [[Definition:Legendre Symbol/Definition 2|Legendre symbol by definition 2]]:
+{{:Definition:Legendre Symbol/Definition 2}}
+{{qed}}
+[[Category:Legendre Symbol]]
+6mmfykrvmn3tw8fytzql9zcxr16oxlz
+\end{proof}<|endoftext|>
+\section{Upper Closure of Element without Element is Filter implies Element is Meet Irreducible}
+Tags: Order Theory, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Let $x \in S$.
+Let
+: $x^\succeq \setminus \left\{ {x}\right\}$ be a [[Definition:Filter in Ordered Set|filter]] in $L$.
+Then $x$ is [[Definition:Meet Irreducible|meet irreducible]].
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in S$.
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x = a \wedge b$ and $x \ne a$ and $x \ne b$
+By [[Meet Precedes Operands]]:
+:$x \preceq b$ and $x \preceq a$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$b, a \in x^\succeq$
+By definitions of [[Definition:Singleton|singleton]] and [[Definition:Set Difference|difference]]:
+:$b, a \in x^\succeq \setminus \left\{ {x}\right\}$
+By definition of [[Definition:Filtered Subset|filtered]]:
+:$\exists z \in x^\succeq \setminus \left\{ {x}\right\}: z \preceq a \land z \preceq b$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$z \preceq x$
+By definition of [[Definition:Upper Set|upper set]]:
+:$x \in x^\succeq \setminus \left\{ {x}\right\}$
+Thus this contradicts $x \in \left\{ {x}\right\}$ by definition of [[Definition:Singleton|singleton]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximal Element of Complement of Filter is Meet Irreducible}
+Tags: Order Theory, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Let $F$ be a [[Definition:Filter in Ordered Set|filter]] in $L$.
+Let $p \in S$.
+Let $p = \max \complement_S\left({F}\right)$.
+Then $p$ is [[Definition:Meet Irreducible|meet irreducible]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$p = x \wedge y$ and $p \ne x$ and $p \ne y$
+By [[Meet Precedes Operands]]:
+:$p \preceq x$ and $p \preceq y$
+By definition of $\prec$:
+:$p \prec x$ and $p \prec y$
+We will prove that
+:$x \notin F$ or $y \notin F$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x \in F \land y \in F$
+By definition of [[Definition:Filtered Subset|filtered]]:
+:$\exists z \in F: z \preceq x \land z \preceq y$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$z \preceq p$
+By definition of [[Definition:Upper Set|upper set]]:
+:$p \in F$
+Thus this contradicts $p \in \complement_S\left({F}\right)$ by definition of [[Definition:Greatest Element|greatest element]].
+{{qed|lemma}}
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in \complement_S\left({F}\right)$ or $y \in \complement_S\left({F}\right)$
+Thus by definition of [[Definition:Greatest Element|greatest element]]:
+this contradicts $p \prec x$ and $p \prec y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{De Polignac's Formula/Technique}
+Tags: De Polignac's Formula
+
+\begin{theorem}
+When calculating $\mu$, the easiest way to calculate the next term is simply to divide the previous term by $p$ and discard the remainder:
+:$\floor {\dfrac n {p^{k + 1} } } = \floor {\floor {\dfrac n {p^k} } / p}$
+\end{theorem}
+
+\begin{proof}
+From [[Floor of x+m over n/Corollary|Floor of $\dfrac {x + m} n$: Corollary]]:
+:$\floor {\dfrac {x + m} n} = \floor {\dfrac {\floor x + m} n}$
+which is valid for all [[Definition:Integer|integers]] $m, n$ such that $n > 0$.
+In this instance, $m = 0$ and $n = p$, while $x = \dfrac n {p^k}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Not Preceding implies There Exists Meet Irreducible Element Not Preceding}
+Tags: Order Theory, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Let $x, y \in S$ such that
+:$y \npreceq x$
+Then
+:$\exists p \in S: p$ is [[Definition:Meet Irreducible|meet irreducible]] and $x \preceq p$ and $y \npreceq p$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]]
+and
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]].
+By [[Axiom of Approximation in Up-Complete Semilattice]]
+:$\exists u \in S: u \ll y \land u \npreceq x$
+By [[Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure]]:
+:there exists [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] $F$ in $L$: $y \in F \subseteq u^\gg$
+By [[Way Above Closure is Subset of Upper Closure of Element]]:
+:$u^\gg \subseteq u^\succeq$
+By definition of [[Definition:Upper Set|upper set]]:
+:$x \notin F$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in \complement_S\left({F}\right)$
+By [[Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement]]:
+:$\exists m \in \complement_S\left({F}\right): x \preceq m \land m = \max \complement_S\left({F}\right)$
+Thus by [[Maximal Element of Complement of Filter is Meet Irreducible]]:
+:$m$ is [[Definition:Meet Irreducible|meet irreducible]].
+Thus
+:$x \preceq m$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$y \preceq m$
+By definition of [[Definition:Upper Set|upper set]]:
+:$m \in F$
+This contradicts $m \in \complement_S\left({F}\right)$ by definition of [[Definition:Greatest Element|greatest element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Termial on Real Numbers is Extension of Integers}
+Tags: Termial Function
+
+\begin{theorem}
+The [[Definition:Termial|termial function]] as [[Definition:Termial/Real Numbers|defined]] on the [[Definition:Real Number|real numbers]] is an [[Definition:Extension of Mapping|extension]] of its definition on the [[Definition:Integer|integers]] $\Z$.
+\end{theorem}
+
+\begin{proof}
+From the definition of the [[Definition:Termial|termial function]] on the [[Definition:Integer|integers]]:
+:$\displaystyle n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$
+From [[Closed Form for Triangular Numbers]], we have that:
+:$\displaystyle \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$
+This [[Definition:Agreement of Mappings|agrees]] with the definition of the [[Definition:Termial/Real Numbers|termial function]] on the [[Definition:Real Number|real numbers]].
+Hence the result, by definition of [[Definition:Extension of Mapping|extension]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of m+n-1 over n}
+Tags: Floor Function
+
+\begin{theorem}
+Let $m, n \in \Z$ such that $n > 0$.
+Then:
+:$\floor {\dfrac {m + n - 1} n} = \ceiling {\dfrac m n}$
+The identity does not necessarily apply for $n < 0$.
+\end{theorem}
+
+\begin{proof}
+First let $n > 0$ as stated.
+Suppose $n \divides m$. 
+Then $m = k n$ for some $k \in \Z$.
+It follows that:
+:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 - \dfrac 1 n} = k$
+and:
+:$\ceiling {\dfrac m n} = k$
+Now suppose $n \nmid m$.
+Since $n > 0$, we have $m = k n + r$ for some $k \in\Z$ and $r \in \N$, $0 < r < n$.
+Therefore:
+:$\floor {\dfrac {m + n - 1} n} = \floor {k + 1 + \dfrac {r - 1} n} = k + 1$
+and:
+:$\ceiling {\dfrac m n} = k + 1$
+{{qed|lemma}}
+Setting $m = 1, n = -2$ we have:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac {m + n - 1} n}
+      | r = \floor {\dfrac {1 + \paren {-2} - 1} {\left({-2}\right)} }
+      | c = 
+}}
+{{eqn | r = \ceiling 1
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \ceiling {\dfrac 1 {\paren {-2} } }
+      | c = 
+}}
+{{eqn | r = \ceiling {\dfrac m n}
+      | c = 
+}}
+{{end-eqn}}
+Thus, as stated, it is confirmed that the identity does not hold for $n < 0$.
+It is noted that when $n = 0$ the expressions on either side are not defined.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristics of Floor and Ceiling Function}
+Tags: Floor Function, Ceiling Function
+
+\begin{theorem}
+Let $f: \R \to \Z$ be an [[Definition:Integer-Valued Function|integer-valued function]] which satisfies both of the following:
+:$(1): \quad \map f {x + 1} = \map f x + 1$
+:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
+Then either:
+:$\forall x \in \Q: \map f x = \floor x$
+or:
+:$\forall x \in \Q: \map f x = \ceiling x$
+\end{theorem}
+
+\begin{proof}
+From $(1)$, by [[Principle of Mathematical Induction|induction]] we have:
+:$\forall n \in \N: \map f {x + n} = \map f x + n$
+and
+:$\forall n \in\N: \map f {x - n} = \map f x - n$
+and therefore, in particular:
+:$(3): \quad \forall n \in \Z: \map f n = \map f 0 + n$
+From $(2)$, we get
+{{begin-eqn}}
+{{eqn | l = \map f 0
+      | r = \map f {\map f 0}
+}}
+{{eqn | r = \map f 0 + \map f 0
+      | c = using $(3)$ and $f$ being [[Definition:Integer-Valued Function|integer valued]]
+}}
+{{end-eqn}}
+Hence
+:$\map f 0 = 0$
+Thus from $(3)$ it follows that:
+:$\forall n \in \Z: \map f n = n$
+Suppose that $\map f {\dfrac 1 2} = k \le 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = k
+      | r = \map f {\dfrac 1 {1 - 2 k} \map f {\dfrac 1 2 - k} }
+      | c = 
+}}
+{{eqn | r = \map f {\dfrac 1 {1 - 2 k} \paren {\map f {\dfrac 1 2} - k} }
+      | c = 
+}}
+{{eqn | r = \map f 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+We show that in this case, by [[Principle of Mathematical Induction|induction]]:
+:$\map f {\dfrac 1 n} = 0$ for all $n \in \N$
+Induction hypothesis:
+:$\map f {\dfrac 1 {n - 1} } = 0$
+Then from $(1)$:
+:$\map f {\dfrac n {n - 1} } = \map f {\dfrac 1 {n - 1} + 1} = 0 + 1 = 1$
+so:
+{{begin-eqn}}
+{{eqn | l = \map f {\dfrac 1 n}
+      | r = \map f {\dfrac 1 n \map f {\dfrac n {n - 1} } }
+      | c = 
+}}
+{{eqn | r = \map f {\dfrac 1 {n - 1} }
+      | c = using $(2)$
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Finally, we show by induction on $m$ that even:
+:$\map f {\dfrac m n} = 0$
+for $m \in \set {1, \ldots, n - 1}$.
+Above we have shown this for $m = 1$.
+Let $1 \le m < n$. 
+If $m \divides n$, then
+:$\map f {\dfrac m n} = \map f {\dfrac 1 {n / m} } = 0$
+Otherwise, write:
+:$n = \paren {k - 1} m + r$
+Then:
+:$k > 1$
+and:
+:$1 \le r \le m - 1$
+We therefore have:
+:$k m = n + m - r$
+so by the induction hypothesis:
+:$\map f {\dfrac {k m} n} = \map f {1 + \dfrac {m - r} n} = 1 + \map f {\dfrac {m - r} n} = 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map f {\dfrac m n}
+      | r = \map f {\dfrac 1 m \map f {\dfrac {m k} n} }
+      | c = 
+}}
+{{eqn | r = \map f {\dfrac 1 m}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+By $(1)$, this shows that
+:$\map f {\dfrac 1 2} \le 0 \implies \map f x = \floor x$
+for all [[Definition:Rational Number|rational]] $x$.
+Suppose that $\map f {\dfrac 1 2} = k > 0$.
+Then the [[Definition:Integer-Valued Function|integer-valued function]] $g: \R \to \Z$ satisfies:
+:$\map g x = -\map f {-x}$
+satisfies $(1)$ and $(2)$, and also:
+{{begin-eqn}}
+{{eqn | l = \map g {\dfrac 1 2}
+      | r = 1 - \map f {\dfrac 1 2}
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map f x
+      | r = -\map g {-x}
+      | c = 
+}}
+{{eqn | r = -\floor {-x}
+      | c = 
+}}
+{{eqn | r = \ceiling x
+      | c = [[Floor of Negative equals Negative of Ceiling]]
+}}
+{{end-eqn}}
+Thus:
+:$\map f {\dfrac 1 2} > 0 \implies \map f x = \ceiling x$
+for all [[Definition:Rational Number|rational]] $x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characteristics of Floor and Ceiling Function/Real Domain}
+Tags: Floor Function, Ceiling Function
+
+\begin{theorem}
+Let $f: \R \to \Z$ be an [[Definition:Integer-Valued Function|integer-valued function]] which satisfies both of the following:
+:$(1): \quad \map f {x + 1} = \map f x + 1$
+:$(2): \quad \forall n \in \Z_{> 0}: \map f x = \map f {\dfrac {\map f {n x} } n}$
+Then it is not necessarily the case that either:
+:$\forall x \in \R: \map f x = \floor x$
+or:
+:$\forall x \in \R: \map f x = \ceiling x$
+\end{theorem}
+
+\begin{proof}
+Let $h: \R \to \R$ be a [[Definition:Real Function|real function]] such that for all $x, y \in \R$:
+{{begin-eqn}}
+{{eqn | n = 3
+      | l = \map h 1
+      | r = 1
+      | c = 
+}}
+{{eqn | n = 4
+      | l = \map h x + \map h y
+      | r = \map h {x + y}
+      | c = 
+}}
+{{end-eqn}}
+Consider the [[Definition:Integer-Valued Function|integer-valued function]] $f: \R \to \Z$ defined as:
+:$\map f x = \floor {\map h x}$
+We claim that $f$ satisfies $(1)$ and $(2)$.
+Proof for $(1)$:
+We have that:
+:$\map h {x + 1} = \map h x + \map h 1 = \map h x + 1$ 
+by $(4)$ and $(3)$.
+It follows that:
+{{begin-eqn}}
+{{eqn | l = \map f {x + 1}
+      | r = \floor {\map h {x + 1} }
+      | c = 
+}}
+{{eqn | r = \floor {\map h x + 1}
+      | c = 
+}}
+{{eqn | r = \floor {\map h x} + 1
+      | c = 
+}}
+{{eqn | r = \map f x + 1
+      | c = 
+}}
+{{end-eqn}}
+Thus $h$ satisfies $(1)$.
+{{qed|lemma}}
+Proof for $(2)$:
+Since $h$ satisfies $(4)$, it is an [[Definition:Additive Function|additive function]].
+By $(3)$ and since [[Additive Function is Linear for Rational Factors]], this implies
+:$(5): \quad \map h x = x$
+for all $x \in \Q$.
+Let $x \in \R$. 
+Define $\alpha$ and $\beta$ by
+:$(6): \quad \alpha := \floor {\map h x} = \map f x$
+:$(7): \quad \beta := \map h x - \alpha$
+Then:
+{{begin-eqn}}
+{{eqn | n = 8
+      | l = 0
+      | o = \le
+      | r = \beta < 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | r = n \beta < n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = \le
+      | r = \floor {n \beta} \le n - 1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map f {\dfrac {\map f n x} n}
+      | r = \floor {\map h {\dfrac {\floor {\map h {n x} } } n} }
+}}
+{{eqn | r = \floor {\dfrac {\floor {n \map h x} } n}
+      | c = $(5)$, [[Additive Function is Linear for Rational Factors]]
+}}
+{{eqn | r = \floor {\dfrac {\floor {n \paren {\alpha + \beta} } } n}
+      | c = $(6)$, $(7)$
+}}
+{{eqn | r = \floor {\dfrac {n \alpha + \floor {n \beta} } n}
+      | c = as $\alpha \in \Z$
+}}
+{{eqn | r = \alpha + \floor {\dfrac 1 n \floor {n \beta} }
+      | c = 
+}}
+{{eqn | r = \floor {\map h x}
+      | c = $(8)$
+}}
+{{eqn | r = \map f x
+      | c = Definition of $\map f x$
+}}
+{{end-eqn}}
+Thus $h$ satisfies $(2)$.
+{{qed|lemma}}
+We have that:
+:[[Rational Numbers form Subfield of Real Numbers]]
+:[[Vector Space on Field Extension is Vector Space]]
+Thus we can consider $\R$ as a [[Definition:Vector Space|vector space]] over $\Q$.
+We also have that [[Square Root of 2 is Irrational]]
+Hence the set $\set {1, \sqrt 2}$ is a [[Definition:Linearly Independent|linearly independent set]] in the [[Definition:Vector Space|vector space]] $\R$.
+From [[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]]:
+:there exists a [[Definition:Basis (Linear Algebra)|basis]] $B$ of $\R$ which includes $1$ and $\sqrt 2$.
+Then each $x \in \R$ can be written as a finite [[Definition:Summation|sum]]:
+:$x := \displaystyle \sum_{i \mathop = 1}^n b_i x_i$
+where $b_i \in B$, $x_i \in \Q$ and $n$ depends on $x$.
+Let $f$ be defined as:
+:$\map f x = \displaystyle \sum_{i \mathop = 1}^n \map f {b_i} x_i$
+From [[Expression of Vector as Linear Combination from Basis is Unique]], we have:
+:$\map f x + \map f y = \map f {x + y}$
+no matter how $\map f b$ is defined for $b \in B$.
+Let $f$ be further defined as:
+:$\map f 1 = 1$
+and, for example:
+:$\map f {\sqrt 2} = 4$
+Then $f$ satisfies $(1)$ and $(2)$.
+But:
+:$\map f {\sqrt 2} \notin \set {1, 2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Replicative Function}
+Tags: Replicative Functions, Continuous Functions
+
+\begin{theorem}
+Let $f: \R \to \R$ be a [[Definition:Real Function|real function]].
+Let $f$ be [[Definition:Continuous Real Function|continuous]] on $\R$.
+Let $f$ also be a [[Definition:Replicative Function|replicative function]].
+Then $f$ is of the form:
+:$f \left({x}\right) = \left({x - \dfrac 1 2}\right) a$
+where $a \in \R$.
+\end{theorem}
+
+\begin{proof}
+Let $f$ be a [[Definition:Replicative Function|replicative function]].
+Then:
+:$\forall n > 0: f \left({n x + 1}\right) - f \left({n x}\right) = f \left({x + 1}\right) - f \left({x}\right)$
+If $f$ is then also [[Definition:Continuous Real Function|continuous]]:
+:$\forall x \in \R: f \left({x + 1}\right) - f \left({x}\right)$
+and so:
+:$g \left({x}\right) = f \left({x}\right) - c \left \lfloor{x}\right \rfloor$
+is both [[Definition:Replicative Function|replicative]] and [[Definition:Periodic Function|periodic]].
+We have:
+:$\displaystyle \int_0^1 e^{2 \pi i n x} g \left({x}\right) \mathrm d x = \dfrac 1 n \int_0^1 e^{2 \pi y} g \left({y}\right) \mathrm d y$
+Expanding in a [[Definition:Fourier Series|Fourier series]] shows:
+:$g \left({x}\right) = \left({x - \dfrac 1 2}\right) a$
+for $0 < x < 1$.
+Thus it follows that:
+:$f \left({x}\right) = \left({x - \dfrac 1 2}\right) a$
+{{qed}}
+{{MissingLinks|The above needs to be linked to the results used.}}
+\end{proof}<|endoftext|>
+\section{Sum over k to p over 2 of Floor of 2kq over p}
+Tags: Prime Numbers
+
+\begin{theorem}
+Let $p \in \Z$ be an [[Definition:Odd Prime|odd prime]].
+Let $q \in \Z$ be an [[Definition:Odd Integer|odd integer]].
+Then:
+:$\displaystyle \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {2 k q} p}r \equiv  \sum_{0 \mathop \le k \mathop < p / 2} \floor {\dfrac {k q} p} \pmod 2$
+\end{theorem}
+
+\begin{proof}
+When $k < \dfrac p 4$ we have:
+{{begin-eqn}}
+{{eqn | l = \floor {\dfrac {\paren {p - 1 - 2 k} q} p}
+      | r = q - \ceiling {\dfrac {\paren {2 k + 1} q} p}
+      | c = 
+}}
+{{eqn | r = q - 1 - \floor {\dfrac {\paren {2 k + 1} q} p}
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = \floor {\dfrac {\paren {2 k + 1} q} p}
+      | rr= \pmod 2
+      | c = 
+}}
+{{end-eqn}}
+Thus it is possible to replace the last terms:
+:$\floor {\dfrac {\paren {p - 1} q} p}, \floor {\dfrac {\paren {p - 3} q} p}, \ldots$
+by:
+:$\floor {\dfrac q p}, \floor {\dfrac {3 q} p}, \ldots$
+The result follows.
+{{qed}}
+{{MissingLinks|Links needed to results that back up the reasoning.}}
+\end{proof}<|endoftext|>
+\section{Way Below implies There Exists Way Below Open Filter Subset of Way Above Closure}
+Tags: Continuous Lattices
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]] [[Definition:Lattice|lattice]].
+Let $x, y \in S$ such that
+:$x \ll y$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+Then there exists a [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] in $L$: $y \in F \land F \subseteq x^\gg$
+where $x^\gg$ denotes the [[Definition:Way Above Closure|way above closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+We will prove that
+:$x^\gg$ is [[Definition:Way Below Open|way below open]].
+Let $z \in x^\gg$.
+By definition of [[Definition:Way Above Closure|way above closure]]:
+:$x \ll z$
+By [[Way Below has Interpolation Property]]:
+:$\exists x' \in S: x \ll x' \land x' \ll z$
+Thus by definition of [[Definition:Way Above Closure|way above closure]]:
+:$x' \in x^\gg$
+Thus
+:$x' \ll z$
+{{qed|lemma}}
+Then
+:$\forall z \in x^\gg: \exists y \in x^\gg: y \ll z$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $f: x^\gg \to x^\gg$:
+:$\forall z \in x^\gg: f\left({z}\right) \ll z$
+By definition of [[Definition:Way Above Closure|way above closure]]:
+:$y \in x^\gg$
+Define $V := \left\{ {z^\succeq: \exists n \in \N: z = f^n\left({y}\right)}\right\}$
+We will prove that
+:$\forall X, Y \in V: \exists Z \in V: X \cup Y \subseteq Z$
+Let $X, Y \in V$.
+By definition of $V$:
+:$\exists z_1 \in S: X = z_1^\succeq \land \exists n_1 \in \N: z_1 = f^{n_1}\left({y}\right)$
+and
+:$\exists z_2 \in S: Y = z_2^\succeq \land \exists n_2 \in \N: z_2 = f^{n_2}\left({y}\right)$
+We will prove that
+:$\forall n, k \in \N: f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$
+Let $n \in \N$.
+'''Base Case''':
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$f^{n+0}\left({y}\right) \preceq f^n\left({y}\right)$
+'''Induction Hypothesis''':
+:$f^{n+k}\left({y}\right) \preceq f^n\left({y}\right)$
+'''Induction Step''':
+By definition of $f$:
+:$f^{n+k+1}\left({y}\right) \ll f^{n+k}\left({y}\right)$
+By [[Way Below implies Preceding]]:
+:$f^{n+k+1}\left({y}\right) \preceq f^{n+k}\left({y}\right)$
+By Induction Hypothesis and definition of [[Definition:Transitivity|transitivity]]:
+:$f^{n+k+1}\left({y}\right) \preceq f^{n}\left({y}\right)$
+{{qed|lemma}}
+[[Definition:WLOG|WLOG]]: suppose $n_1 \le n_2$
+Then
+:$\exists k \in \N: n_2 = n_1+k$
+Then
+:$z_2 \preceq z_1$
+By [[Preceding iff Meet equals Less Operand]]:
+:$z_1 \wedge z_2 = z_2$
+By definition of $V$:
+:$Z := \left({z_1 \wedge z_2}\right)^\succeq \in V$
+By [[Meet Precedes Operands]]:
+:$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$
+By [[Upper Closure is Decreasing]]:
+:$z_1^\succeq \subseteq Z$ and $z_2^\succeq \subseteq Z$
+Thus by [[Union of Subsets is Subset]]:
+:$X \cup Y \subseteq Z$
+{{qed|lemma}}
+Define $F := \bigcup V$.
+We will prove that
+:$F$ is [[Definition:Way Below Open|way below open]].
+Let $u \in F$.
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$\exists Y \in V: u \in Y$
+By definition of $V$:
+:$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$
+By definition of $f$:
+:$z \in x^\gg$
+By definition of $f$:
+:$f\left({z}\right) \ll z$
+Then
+:$z' := f\left({z}\right) = f^{n+1}\left({y}\right)$
+By definition of $V$:
+:${z'}^\succeq \in V$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$z' \preceq z'$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$z' \in {z'}^\succeq$
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$z' \in F$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$z \preceq u$
+By [[Preceding and Way Below implies Way Below]]:
+:$z' \ll u$
+Hence
+:$\exists g \in F: g \ll u$
+{{qed|lemma}}
+By [[Upper Closure of Element is Filter]]:
+:$\forall X \in V: X$ is a [[Definition:Filtered Subset|filtered]] [[Definition:Upper Set|upper set]].
+By [[Union of Upper Sets is Upper]]:
+:$F$ is an [[Definition:Upper Set|upper set]].
+By [[Union of Filtered Sets is Filtered]]:
+:$F$ is [[Definition:Filtered Subset|filtered]].
+By definition of [[Definition:Filter in Ordered Set|filter]]:
+:$F$ is a [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] in $L$.
+Then
+:$f^0\left({y}\right) = y$
+By definition of $V$:
+:$y^\succeq \in V$
+By definitions of [[Definition:Upper Closure/Element|upper closure of element]] and [[Definition:Reflexivity|reflexivity]]:
+:$y \in y^\succeq$
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$y \in F$
+It remains to prove that
+:$F \subseteq x^\gg$
+Let $u \in F$.
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$\exists Y \in V: u \in Y$
+By definition of $V$:
+:$\exists z \in S: Y = z^\succeq \land \exists n \in \N: z = f^n\left({y}\right)$
+By definition of $f$:
+:$z \in x^\gg$
+By definition of [[Definition:Way Above Closure|way above closure]]:
+:$x \ll z$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$z \preceq u$ and $x \preceq x$
+By [[Preceding and Way Below implies Way Below]]:
+:$x \ll u$
+Thus by definition of [[Definition:Way Above Closure|way above closure]]:
+:$u \in x^\gg$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Group Elements/Sum of Indices/Additive Notation}
+Tags: Group Theory, Index Laws
+
+\begin{theorem}
+:$\forall m, n \in \Z: m g + n g = \paren {m + n} g$
+\end{theorem}
+
+\begin{proof}
+All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Sum of Indices|Index Laws for Monoids: Sum of Indices]]:
+:$\forall m, n \in \Z: g^m \circ g^n = g^{m + n}$
+where in this context the [[Definition:Group Operation|group operation]] is $+$ and [[Definition:Power of Group Element|$n$th power of $g$]] is denoted $n g$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Group Elements/Negative Index/Additive Notation}
+Tags: Group Theory, Index Laws
+
+\begin{theorem}
+:$\forall n \in \Z: -\paren {n g} = \paren {-n} g = n \paren {-g}$
+\end{theorem}
+
+\begin{proof}
+All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Sum of Indices|Index Laws for Monoids: Sum of Indices]]:
+:$\forall n \in \Z: \paren {g^n}^{-1} = g^{-n} = \paren {g^{-1} }^n$
+where in this context:
+:the [[Definition:Group Operation|group operation]] is $+$
+:the [[Definition:Power of Group Element|$n$th power of $g$]] is denoted $n g$
+:the [[Definition:Inverse Element|inverse]] of $g$ is $-g$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Group Elements/Product of Indices/Additive Notation}
+Tags: Group Theory, Index Laws
+
+\begin{theorem}
+:$\forall m, n \in \Z: n \paren {m g} = \paren {m n} g = m \paren {n g}$
+\end{theorem}
+
+\begin{proof}
+All elements of a [[Definition:Group|group]] are [[Definition:Invertible Element|invertible]], so we can directly use the result from [[Index Laws for Monoids/Product of Indices|Index Laws for Monoids: Product of Indices]]:
+:$\forall m, n \in \Z: g^{m n} = \paren {g^m}^n = \paren {g^n}^m$
+where in this context:
+:the [[Definition:Group Operation|group operation]] is $+$
+:the [[Definition:Power of Group Element|$n$th power of $g$]] is denoted $n g$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Product in Abelian Group/Additive Notation}
+Tags: Abelian Groups
+
+\begin{theorem}
+: $k \left({x + y}\right) = k x + k y$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Abelian Group|abelian group]], $x$ and $y$ [[Definition:Commute|commute]].
+That is:
+:$x + y = y + x$
+The result follows from [[Power of Product of Commutative Elements in Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Idempotent Element}
+Tags: Idempotence
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $s \in S$ be an [[Definition:Idempotent Element|idempotent element]] with respect to $\circ$.
+Then:
+:$\forall n \in \Z_{> 0}: s^n = s$
+where $s^n$ is defined as:
+:$s^n = \begin{cases} s & : n = 1 \\
+s^{n - 1} \circ s & : n > 1 \end{cases}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$s^n = s$
+$P \left({1}\right)$ is the case:
+:$s^1 = s$
+which holds by definition.
+Thus $P \left({1}\right)$ is seen to hold.
+=== Basis for the Induction ===
+$P \left({2}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = s^2
+      | r = s^1 \circ s
+      | c = Definition of $s^n$
+}}
+{{eqn | r = s \circ s
+      | c = Definition of $s^1$
+}}
+{{eqn | r = s
+      | c = {{Defof|Idempotent Element}}
+}}
+{{end-eqn}}
+Thus $P \left({2}\right)$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$s^k = s$
+from which it is to be shown that:
+:$s^{k + 1} = s$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = s^{k + 1}
+      | r = s^k \circ s
+      | c = 
+}}
+{{eqn | r = s \circ s
+      | c = [[Power of Idempotent Element#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = s
+      | c = [[Power of Idempotent Element#Basis for the Induction|Basis for the Induction]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{> 0}: s^n = s$
+</onlyinclude>
+[[Category:Idempotence]]
+d7rj1kcs86m65c7frk9092f545vlpib
+\end{proof}<|endoftext|>
+\section{Powers of Group Element Commute}
+Tags: Group Theory, Powers (Abstract Algebra), Commutativity
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $g \in G$.
+Let $m, n \in \N_{>0}$.
+Then:
+:$\forall m, n \in \N_{>0}: g^n \circ g^m = g^m \circ g^n$
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Group|group]] is also a [[Definition:Semigroup|semigroup]].
+The result follows as a special case of [[Powers of Semigroup Element Commute]]
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Positive Real Numbers under Multiplication form Subgroup of Non-Zero Real Numbers}
+Tags: Real Multiplication, Subgroups
+
+\begin{theorem}
+Let $\R_{>0}$ be the set of [[Definition:Strictly Positive Real Number|strictly positive real numbers]], that is:
+:$\R_{>0} = \set {x \in \R: x > 0}$
+The [[Definition:Algebraic Structure|structure]] $\struct {\R_{>0}, \times}$ forms a [[Definition:Subgroup|subgroup]] of $\struct {\R_{\ne 0}, \times}$, where $\R_{\ne 0}$ is the [[Definition:Set|set]] of [[Definition:Real Number|real numbers]] without [[Definition:Zero (Number)|zero]], that is:
+:$\R_{\ne 0} = \R \setminus \set 0$
+\end{theorem}
+
+\begin{proof}
+From [[Non-Zero Real Numbers under Multiplication form Abelian Group]] we have that $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Group|group]].
+We know that $\R_{>0} \ne \O$, as (for example) $1 \in \R_{>0}$.
+Now, verify that the conditions for [[Two-Step Subgroup Test]] are satisfied:
+=== Closure under $\times$ ===
+Let $a, b \in \R_{>0}$.
+We take on board the fact that the [[Real Numbers form Ordered Integral Domain]].
+Then:
+:$a b \in \R_{\ne 0}$
+From [[Positive Elements of Ordered Ring]] :
+:$a \times b > 0$
+so $a b \in \R_{>0}$.
+{{qed|lemma}}
+=== Closure under Inverse ===
+Let $a \in \R_{>0}$. 
+Then $a^{-1} = \dfrac 1 a \in \R_{>0}$.
+{{qed|lemma}}
+Hence, by the [[Two-Step Subgroup Test]], $\struct {\R_{>0}, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R_{\ne 0}, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Positive Rational Numbers under Multiplication form Subgroup of Non-Zero Rational Numbers}
+Tags: Rational Multiplication, Subgroups
+
+\begin{theorem}
+Let $\Q_{> 0}$ be the set of [[Definition:Strictly Positive Rational Number|strictly positive rational numbers]], that is $\Q_{> 0} = \set { x \in \Q: x > 0}$.
+The [[Definition:Algebraic Structure|structure]] $\struct {\Q_{> 0}, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{\ne 0}, \times}$, where $\Q_{\ne 0}$ is the set of [[Definition:Rational Number|rational numbers]] without [[Definition:Zero (Number)|zero]]: $\Q_{\ne 0} = \Q \setminus \set 0$.
+\end{theorem}
+
+\begin{proof}
+From [[Non-Zero Rational Numbers under Multiplication form Abelian Group]] we have that $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Group|group]].
+We know that $\Q_{> 0} \ne \O$, as (for example) $1 \in \Q_{> 0}$.
+Let $a, b \in \Q_{> 0}$.
+Then:
+:$a b \in \Q_{\ne 0}$ and $ab > 0$
+Hence:
+:$a b \in \Q_{> 0}$
+Let $a \in \Q_{> 0}$.
+Then:
+:$a^{-1} = \dfrac 1 a \in \Q_{> 0}$
+So, by the [[Two-Step Subgroup Test]], $\struct {\Q_{> 0}, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{\ne 0}, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Upper Sets is Upper}
+Tags: Upper Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $A$ be a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$ such that
+:$\forall X \in A: X$ is an [[Definition:Upper Set|upper set]].
+Then:
+:$\bigcup A$ is also an [[Definition:Upper Set|upper set]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in \bigcup A, y \in S$ such that
+:$x \preceq y$
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$\exists Y \in A: x \in Y$
+By assumption:
+:$Y$ is an [[Definition:Upper Set|upper set]].
+By definition of [[Definition:Upper Set|upper set]]:
+:$y \in Y$
+Thus by definition of [[Definition:Set Union/General Definition|union]]:
+:$y \in \bigcup A$
+Hence
+:$\bigcup A$ is an [[Definition:Upper Set|upper set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Rotations in Space about Fixed Point forms Infinite Group}
+Tags: Examples of Groups
+
+\begin{theorem}
+Let $\SS$ be a [[Definition:Rigid|rigid]] [[Definition:Body|body]] in [[Definition:Ordinary Space|space]].
+Let $O$ be a fixed [[Definition:Point|point]] in [[Definition:Ordinary Space|space]].
+The [[Definition:Set|set]] of all [[Definition:Space Rotation|rotations]] of $\SS$ through some [[Definition:Straight Line|line]] through $O$ forms an [[Definition:Infinite Group|infinite group]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Needs the mathematical definition of a rotation in space to be defined before we start.}}
+\end{proof}<|endoftext|>
+\section{Group of Rotations about Fixed Point is not Abelian}
+Tags: Examples of Groups
+
+\begin{theorem}
+Let $\SS$ be a [[Definition:Rigid|rigid]] [[Definition:Body|body]] in [[Definition:Ordinary Space|space]].
+Let $O$ be a fixed [[Definition:Point|point]] in [[Definition:Ordinary Space|space]].
+Let $\GG$ be the [[Definition:Group|group]] of all [[Definition:Space Rotation|rotations]] of $\SS$ around $O$.
+Then $\GG$ is not an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Let $\SS$ be a [[Definition:Square (Geometry)|square]] [[Definition:Lamina|lamina]].
+Let $O$ be the center of $\SS$.
+Recall the definition of the [[Definition:Symmetry Group of Square|symmetry group of the square]] $D_4$:
+{{:Definition:Symmetry Group of Square}}
+We have that:
+:Reflection $t_x$ can be achieved by a [[Definition:Space Rotation|rotation]] of $\SS$ of $\pi$ [[Definition:Radian|radians]] about $x$.
+:Reflection $t_y$ can be achieved by a [[Definition:Space Rotation|rotation]] of $\SS$ of $\pi$ [[Definition:Radian|radians]] about $y$.
+Thus $D_4$ forms a [[Definition:Subgroup|subgroup]] of $\GG$.
+From [[Symmetry Group of Square is Group]] we have that $D_4$ is not [[Definition:Abelian Group|abelian]].
+From [[Subgroup of Abelian Group is Abelian]] it follows that $\GG$ is also not [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Filtered Sets is Filtered}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $A$ be a [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$ such that
+:$\forall X \in A: X$ is [[Definition:Filtered Subset|filtered]]
+and
+:$\forall X, Y \in A: \exists Z \in A: X \cup Y \subseteq Z$
+Then:
+:$\bigcup A$ is also [[Definition:Filtered Subset|filtered]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \bigcup A$.
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$\exists X \in A: x \in X$
+and
+:$\exists Y \in A: y \in Y$
+By assumption:
+:$\exists Z \in A: X \cup Y \subseteq Z$
+By definition of [[Definition:Set Union|union]]:
+:$x, y \in X \cup Y$
+By definition of [[Definition:Subset|subset]]:
+:$x, y \in Z$
+By assumption:
+:$Z$ is [[Definition:Filtered Subset|filtered]].
+By definition of [[Definition:Filtered Subset|filtered]]:
+:$\exists z \in Z: z \preceq x \land z \preceq y$
+Thus by definition of [[Definition:Set Union/General Definition|union]]:
+:$z \in \bigcup A$
+Hence
+:$\bigcup A$ is [[Definition:Filtered Subset|filtered]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Special Orthogonal Group is Group}
+Tags: Orthogonal Groups
+
+\begin{theorem}
+Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
+The $n$th [[Definition:Special Orthogonal Group|orthogonal group]] on $k$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+A direct corollary of [[Special Orthogonal Group is Subgroup of Orthogonal Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Matrix is its own Inverse}
+Tags: Unit Matrices, Inverse Matrices
+
+\begin{theorem}
+The [[Definition:Inverse Matrix|inverse]] of the [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is $\mathbf I_n$.
+That is, a [[Definition:Unit Matrix|unit matrix]] it its own [[Definition:Inverse Matrix|inverse]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Unit Matrix|unit matrix]] is a [[Definition:Diagonal Matrix|diagonal matrix]].
+From [[Inverse of Diagonal Matrix]], the [[Definition:Inverse Matrix|inverse]] of a [[Definition:Diagonal Matrix|diagonal matrix]]:
+:$\mathbf D = \begin{bmatrix}
+a_{11} & 0 & \cdots & 0 \\
+0 & a_{22} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & a_{nn} \\
+\end{bmatrix}$
+is the [[Definition:Diagonal Matrix|diagonal matrix]]:
+: $\mathbf D^{-1} = \begin{bmatrix}
+\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
+0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
+\end{bmatrix}$
+When $\mathbf D$ is the [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$, all elements $a_{kk}$ are equal to $1$:
+:$\mathbf I_n = \begin{bmatrix}
+1 & 0 & \cdots & 0 \\
+0 & 1 & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & 1 \\
+\end{bmatrix}$
+Hence:
+:$\mathbf I_n^{-1} = \begin{bmatrix}
+\dfrac 1 1 & 0 & \cdots & 0 \\
+0 & \dfrac 1 1 & \cdots & 0 \\
+\vdots & \vdots & \ddots & \vdots \\
+0 & 0 & \cdots & \dfrac 1 1 \\
+\end{bmatrix}$
+Hence the result.
+{{qed}}
+[[Category:Unit Matrices]]
+[[Category:Inverse Matrices]]
+rgio764lguz6z2g6wb4yn16z1brin39
+\end{proof}<|endoftext|>
+\section{Way Above Closure is Subset of Upper Closure of Element}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then $x^\gg \subseteq x^\succeq$
+where
+:$x^\gg$ denotes the [[Definition:Way Above Closure|way above closure]] of $x$,
+:$x^\succeq$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in x^\gg$.
+By definition of [[Definition:Way Above Closure|way above closure]]:
+:$x \ll y$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+By [[Way Below implies Preceding]]:
+:$x \preceq y$
+Thus by definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$y \in x^\succeq$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Matrix is Symmetric}
+Tags: Symmetric Matrices, Diagonal Matrices
+
+\begin{theorem}
+Let $D$ be a [[Definition:Diagonal Matrix|diagonal matrix]].
+Then $D$ is [[Definition:Symmetric Matrix|symmetric]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Diagonal Matrix|diagonal matrix]]:
+:$\forall j, k: j \ne k \implies a_{jk} = 0 = a_{kj}$
+So by definition of [[Definition:Transpose of Matrix|transpose]] of $D$:
+:$D = D^\intercal$
+where $D^\intercal$ denotes the [[Definition:Transpose of Matrix|transpose]].
+Hence the result, by definition of [[Definition:Symmetric Matrix|symmetric matrix]].
+{{qed}}
+[[Category:Symmetric Matrices]]
+[[Category:Diagonal Matrices]]
+dv7n9kzvzryg6tjumxlrly9jp6semgo
+\end{proof}<|endoftext|>
+\section{Upper Way Below Open Subset Complement is Non Empty implies There Exists Maximal Element of Complement}
+Tags: Complete Lattices
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $X$ be [[Definition:Upper Set|upper]] [[Definition:Way Below Open|way below open]] [[Definition:Subset|subset]] of $S$.
+Let $x \in S$ such that
+:$x \in \relcomp S X$
+Then
+:$\exists m \in S: x \preceq m \land m = \max \relcomp S X$
+\end{theorem}
+
+\begin{proof}
+Define $A := \set {C \in \map {\mathit {Chains} } L: C \subseteq \relcomp S X \land x \in C}$
+where $\map {\mathit {Chains} } L$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Chain (Set Theory)|chains]] of $L$.
+We will prove that
+:$\forall Z: Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X} \implies \bigcup Z \in A$
+Let $Z$ such that
+:$Z \ne \O \land Z \subseteq A \land \paren {\forall X, Y \in Z: X \subseteq Y \lor Y \subseteq X}$
+We will prove that
+:$\bigcup Z$ is a [[Definition:Chain (Set Theory)|chain]] of $L$
+Let $a, b \in \bigcup Z$
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$\exists Y_1 \in Z: a \in Y_1$
+and
+:$\exists Y_2 \in Z: b \in Y_2$
+By assumption:
+:$Y_1 \subseteq Y_2$ or $Y_2 \subseteq Y_1$
+By definition of [[Definition:Subset|subset]]:
+:$a, b \in Y_1$ or $a, b \in Y_2$
+By definition of $A$:
+:$Y_1, Y_2 \in \map {\mathit {Chains} } L$
+Thus by definition of [[Definition:Connected Relation|connected relation]]
+:$a \preceq b$ or $b \preceq a$
+{{qed|lemma}}
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists Y: Y \in Z$
+By definition of $A$:
+:$x \in Y$
+By definition of [[Definition:Set Union/General Definition|union]]:
+:$x \in \bigcup Z$
+By definition of $A$:
+:$\forall Y \in Z: Y \subseteq \relcomp S X$
+By [[Union of Subsets is Subset/Set of Sets]]:
+:$\bigcup Z \subseteq \relcomp S X$
+Thus by definition of $A$
+:$\bigcup Z \in A$
+{{qed|lemma}}
+By [[Singleton is Chain]]:
+:$\set x$ is a [[Definition:Chain (Set Theory)|chain]] of $L$.
+By definition of $A$:
+:$\set x \in A$
+By [[Zorn's Lemma]]:
+:$\exists Y \in A: Y$ is a [[Definition:Maximal Element|maximal element]] of $A$.
+By definition of [[Definition:Maximal Element|maximal element]]:
+:$\forall Z \in A: Y \subseteq Z \implies Y = Z$
+By definition of $A$:
+:$Y \in \map {\mathit {Chains} } L \land Y \subseteq \relcomp S X \land x \in Y$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup Y$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$.
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$x \preceq \sup Y$
+We will prove that
+:$\lnot \exists y \in S: y \in \relcomp S X \land y \succ \sup Y$
+{{AimForCont}} that
+:$\exists y \in S: y \in \relcomp S X \land y \succ \sup Y$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$y \notin Y$
+By definition of $\succ$
+:$\sup Y \preceq y$
+We wiil prove that
+:$Y \cup \set y$ is a [[Definition:Chain (Set Theory)|chain]] of $L$.
+Let $a, b \in Y \cup \set y$
+Case $a, b \in Y$.
+Thus by definition of [[Definition:Connected Relation|connected relation]]:
+:$a \preceq b$ or $b \preceq a$
+Case $a \in Y \land b \in \set y$
+By definition of [[Definition:Singleton|singleton]]:
+:$b = y$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$a \preceq \sup Y$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$a \preceq b$
+Thus
+:$a \preceq b$ or $b \preceq a$
+Case $a \in \set y \land b \in Y$
+Analogical case as previous.
+Case $a, b \in \set y$
+By definition of [[Definition:Singleton|singleton]]:
+:$ a = y$ and $b = y$
+Be definition of [[Definition:Reflexivity|reflexivity]]:
+:$a \preceq b$
+Thus
+:$a \preceq b$ or $b \preceq a$
+{{qed|lemma}}
+By definitions of [[Definition:Singleton|singleton]] and [[Definition:Subset|subset]]:
+:$\set y \subseteq \relcomp S X$
+By [[Union of Subsets is Subset]]:
+:$Y \cup \set y \subseteq \relcomp S X$
+By definition of [[Definition:Set Union|union]]:
+:$x \in Y \cup \set y$
+By definition of $A$:
+:$Y \cup \set y \in A$
+By [[Set is Subset of Union]]:
+:$Y \subseteq Y \cup \set y$
+Then
+:$Y = Y \cup \set y$
+By definitions of [[Definition:Set Union|union]] and [[Definition:Singleton|singleton]]:
+:$y \in Y$
+This contradicts $y \notin Y$.
+{{qed|lemma}}
+We will prove that
+:$\sup Y \in \relcomp S X$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$\sup Y \in X$
+By definition of [[Definition:Way Below Open|way below open]]:
+:$\exists y \in X: y \ll \sup Y$
+By [[Chain is Directed]]:
+:$Y$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d \in Y: y \preceq d$
+By definition of [[Definition:Upper Set|upper set]]:
+:$d \in X$
+Thus it contradicts $d \in \relcomp S X$ by definition of [[Definition:Subset|subset]].
+{{qed|lemma}}
+By definition of [[Definition:Maximal Element|maximal element]]
+:$\sup Y = \max \relcomp S X$
+Hence
+:$\exists m \in S: x \preceq m \land m = \max \relcomp S X$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Orthogonal}
+Tags: Unit Matrices, Orthogonal Matrices
+
+\begin{theorem}
+The [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is [[Definition:Orthogonal Matrix|orthogonal]].
+\end{theorem}
+
+\begin{proof}
+By [[Unit Matrix is its own Inverse]] the [[Definition:Inverse Matrix|inverse]] $I_n^{-1}$ of $I_n$ is $I_n$.
+By definition a [[Definition:Unit Matrix|unit matrix]] is a [[Definition:Diagonal Matrix|diagonal matrix]].
+Hence by [[Diagonal Matrix is Symmetric]]:
+:$I_n = I_n^\intercal$
+where $I_n^\intercal$ is the [[Definition:Transpose of Matrix|transpose]] of $I_n$.
+Thus:
+:$I_n^{-1} = I_n^\intercal$
+and the result follows by definition of [[Definition:Orthogonal Matrix|orthogonal]].
+{{qed}}
+[[Category:Unit Matrices]]
+[[Category:Orthogonal Matrices]]
+thfuv0wq4j5hsyxq6htlcafz1z4aotx
+\end{proof}<|endoftext|>
+\section{Unit Matrix is Proper Orthogonal}
+Tags: Unit Matrices, Orthogonal Matrices
+
+\begin{theorem}
+The [[Definition:Unit Matrix|unit matrix]] $\mathbf I_n$ of [[Definition:Order of Square Matrix|order]] $n$ is [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
+\end{theorem}
+
+\begin{proof}
+By [[Unit Matrix is Orthogonal]], $\mathbf I_n$ is an [[Definition:Orthogonal Matrix|orthogonal matrix]].
+By [[Determinant of Unit Matrix]], the [[Definition:Determinant of Matrix|determinant]] of $\mathbf I_n$ is equal to $1$.
+The result follows by definition of [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
+{{qed}}
+[[Category:Unit Matrices]]
+[[Category:Orthogonal Matrices]]
+jukceu1rs4hk2cse2pnlfcn8ztarw3v
+\end{proof}<|endoftext|>
+\section{Inverse of Orthogonal Matrix is Orthogonal}
+Tags: Orthogonal Matrices, Inverse Matrices
+
+\begin{theorem}
+Let $\mathbf A$ be an [[Definition:Orthogonal Matrix|orthogonal matrix]].
+Then its [[Definition:Inverse Matrix|inverse]] $\mathbf A^{-1}$ is also [[Definition:Orthogonal Matrix|orthogonal]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Orthogonal Matrix|orthogonal matrix]]:
+:$\mathbf A^\intercal = \mathbf A^{-1}$
+where $\mathbf A^\intercal$ is the [[Definition:Transpose of Matrix|transpose]] of $\mathbf A$.
+By [[Inverse of Inverse of Matrix]]:
+:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
+By [[Transpose of Transpose of Matrix]]:
+:$\paren {\mathbf A^\intercal}^\intercal = \mathbf A$
+Thus we have:
+:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A = \paren {\mathbf A^\intercal}^\intercal = \paren {\mathbf A^{-1} }^\intercal$
+and so by definition $\mathbf A^{-1}$ is [[Definition:Orthogonal Matrix|orthogonal]].
+{{qed}}
+[[Category:Orthogonal Matrices]]
+[[Category:Inverse Matrices]]
+8qgfo01t3z6dht0dvrmkf741k1527ht
+\end{proof}<|endoftext|>
+\section{Inverse of Inverse of Matrix}
+Tags: Inverse Matrices
+
+\begin{theorem}
+Let $\mathbf A$ be an [[Definition:Invertible Matrix|invertible matrix]].
+Then:
+:$\paren {\mathbf A^{-1} }^{-1} = \mathbf A$
+That is, an [[Definition:Invertible Matrix|invertible matrix]] equals the [[Definition:Inverse Matrix|inverse]] of its [[Definition:Inverse Matrix|inverse]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Inverse Matrix|inverse matrix]]:
+:$\mathbf A^{-1} \mathbf A = \mathbf I$
+where $\mathbf I$ is the [[Definition:Unit Matrix|unit matrix]].
+Thus the [[Definition:Inverse Matrix|inverse]] of $\mathbf A^{-1}$ is $\mathbf A$.
+Hence the result.
+{{qed}}
+[[Category:Inverse Matrices]]
+c92664gz3mew2c1geggrxuvcjmj5c4x
+\end{proof}<|endoftext|>
+\section{Inverse of Proper Orthogonal Matrix is Proper Orthogonal}
+Tags: Orthogonal Matrices, Inverse Matrices
+
+\begin{theorem}
+Let $\mathbf A$ be a [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]].
+Then its [[Definition:Inverse Matrix|inverse]] $\mathbf A^{-1}$ is also [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]]:
+:$\mathbf A$ is [[Definition:Orthogonal Matrix|orthogonal]]
+:the [[Definition:Determinant of Matrix|determinant]] of $\mathbf A$ is equal to $1$.
+By [[Inverse of Orthogonal Matrix is Orthogonal]], $\mathbf A^{-1}$ is [[Definition:Orthogonal Matrix|orthogonal]].
+By [[Determinant of Inverse Matrix]]:
+:$\det \mathbf A^{-1} = \dfrac 1 {\det \mathbf A} = 1$
+Hence, by definition, $\mathbf A^{-1}$ is [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
+{{qed}}
+[[Category:Orthogonal Matrices]]
+[[Category:Inverse Matrices]]
+sxpg2zjfvvqirrj1e87ohv80ib8a4k4
+\end{proof}<|endoftext|>
+\section{Special Orthogonal Group is Subgroup of Orthogonal Group}
+Tags: Orthogonal Groups
+
+\begin{theorem}
+Let $k$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\map {\operatorname O} {n, k}$ be the $n$th [[Definition:Orthogonal Group|orthogonal group]] on $k$.
+Let $\map {\operatorname {SO} } {n, k}$ be the $n$th [[Definition:Special Orthogonal Group|special orthogonal group]] on $k$.
+Then $\map {\operatorname {SO} } {n, k}$ is a [[Definition:Subgroup|subgroup]] of $\map {\operatorname O} {n, k}$.
+\end{theorem}
+
+\begin{proof}
+We have that [[Unit Matrix is Proper Orthogonal]], so $\map {\operatorname {SO} } {n, k}$ is not [[Definition:Empty Set|empty]].
+Let $\mathbf A, \mathbf B \in \map {\operatorname {SO} } {n, k}$.
+Then, by definition, $\mathbf A$ and $\mathbf B$ are [[Definition:Proper Orthogonal Matrix|proper orthogonal]].
+Then by [[Inverse of Proper Orthogonal Matrix is Proper Orthogonal]]:
+:$\mathbf B^{-1}$ is a [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]].
+By [[Product of Proper Orthogonal Matrices is Proper Orthogonal Matrix]]:
+:$\mathbf A \mathbf B^{-1}$ is a [[Definition:Proper Orthogonal Matrix|proper orthogonal matrix]].
+Thus by definition of [[Definition:Special Orthogonal Group|special orthogonal group]]:
+:$\mathbf A \mathbf B^{-1} \in \map {\operatorname {SO} } {n, k}$
+Hence the result by [[One-Step Subgroup Test]].
+{{qed}}
+[[Category:Orthogonal Groups]]
+c84yl527lbmgzjz3uiuiri8mcwzpl8t
+\end{proof}<|endoftext|>
+\section{Negative Matrix is Inverse for Hadamard Product}
+Tags: Negative Matrices, Hadamard Product
+
+\begin{theorem}
+Let $\struct {G, \cdot}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\map {\MM_G} {m, n}$ be a [[Definition:Matrix Space|$m \times n$ matrix space]] over $\struct {G, \cdot}$.
+Let $\mathbf A$ be an [[Definition:Element|element]] of $\map {\MM_G} {m, n}$.
+Let $-\mathbf A$ be the [[Definition:Negative Matrix/General Group|negative]] of $\mathbf A$.
+Then $-\mathbf A$ is the [[Definition:Inverse Element|inverse]] for the operation $\circ$, where $\circ$ is the [[Definition:Hadamard Product|Hadamard product]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A = \sqbrk a_{m n} \in \map {\MM_G} {m, n}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \mathbf A \circ \paren {-\mathbf A}
+      | r = \sqbrk a_{m n} \circ \paren {-\sqbrk a_{m n} }
+      | c = Definition of $\mathbf A$
+}}
+{{eqn | r = \sqbrk a_{m n} \circ \sqbrk {a^{-1} }_{m n}
+      | c = {{Defof|Negative Matrix|subdef = General Group}}
+}}
+{{eqn | r = \sqbrk {a \cdot \paren {a^{-1} } }_{m n}
+      | c = {{Defof|Hadamard Product}}
+}}
+{{eqn | r = \sqbrk e_{m n}
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf A \circ \paren {-\mathbf A}
+      | r = \mathbf e
+      | c = {{Defof|Zero Matrix over General Monoid}}
+}}
+{{end-eqn}}
+The result follows from [[Zero Matrix is Identity for Hadamard Product]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellable Infinite Semigroup is not necessarily Group}
+Tags: Semigroups, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] whose [[Definition:Underlying Set|underlying set]] is [[Definition:Infinite Set|infinite]].
+Let $\struct {S, \circ}$ be such that all [[Definition:Element|elements]] of $S$ are [[Definition:Cancellable Element|cancellable]].
+Then it is not necessarily the case that $\struct {S, \circ}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Semigroup|semigroup]] $\struct {\N, +}$.
+From [[Natural Numbers under Addition form Commutative Semigroup]], $\struct {\N, +}$ forms a [[Definition:Semigroup|semigroup]].
+From [[Natural Numbers are Infinite]], the [[Definition:Underlying Set|underlying set]] of $\struct {\N, +}$ is [[Definition:Infinite Set|infinite]].
+From [[Natural Number Addition is Cancellable]], all elements of $\struct {\N, +}$ are [[Definition:Cancellable Element|cancellable]].
+But from [[Natural Numbers under Addition do not form Group]], $\struct {\N, +}$ is not a [[Definition:Group|group]].
+\end{proof}<|endoftext|>
+\section{Left Regular Representation wrt Left Cancellable Element on Finite Semigroup is Bijection}
+Tags: Semigroups, Regular Representations, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
+Let $a \in S$ be [[Definition:Left Cancellable Element|left cancellable]].
+Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ of $\struct {S, \circ}$ with respect to $a$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+By [[Left Cancellable iff Left Regular Representation Injective]], $\lambda_a$ is an [[Definition:Injection|injection]].
+[[Definition:By Hypothesis|By hypothesis]], $S$ is [[Definition:Finite Set|finite]].
+From [[Injection from Finite Set to Itself is Surjection]], $\lambda_a$ is a [[Definition:Surjection|surjection]].
+Thus $\lambda_a$ is [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], and therefore a [[Definition:Bijection|bijection]].
+{{qed}}
+[[Category:Semigroups]]
+[[Category:Regular Representations]]
+[[Category:Cancellability]]
+lkm98i64qmrup50xs7e5y34akkjnvud
+\end{proof}<|endoftext|>
+\section{Singleton is Chain}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then $\left\{ {x}\right\}$ is a [[Definition:Chain (Set Theory)|chain]] of $\left({S, \preceq}\right)$.
+\end{theorem}
+
+\begin{proof}
+It suffices to prove that
+:$\left\{ {x}\right\}$ is [[Definition:Connected Relation|connected]]
+Let $y, z \in \left\{ {x}\right\}$
+By definition of [[Definition:Singleton|singleton]]:
+:$y = x$ and $z = x$
+By definition of [[Definition:Reflexivity|reflexivity]];
+:$y \preceq z$
+Thus
+:$y \preceq z$ or $z \preceq y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity Elements occupy Diagonal of Cayley Table in Inverse Row Form}
+Tags: Cayley Tables
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]].
+Let $\CC$ be a [[Definition:Cayley Table|Cayley table]] for $\struct {G, \circ}$ presented in [[Definition:Inverse Row form of Cayley Table for Group|inverse row form]].
+Then all the [[Definition:Entry of Cayley Table|entries]] in the [[Definition:Main Diagonal|main diagonal]] of $\CC$
+are instances of the [[Definition:Identity Element|identity element]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Inverse Row form of Cayley Table for Group|inverse row form]], the rows of $\CC$ are headed by the [[Definition:Inverse Element|inverse elements]] of the [[Definition:Element|elements]] which head the corresponding columns.
+The [[Definition:Entry of Cayley Table|entries]] in the [[Definition:Main Diagonal|main diagonal]] of $\CC$ have the same column number as row number.
+Let $\sqbrk c_{k k}$ denote the [[Definition:Entry of Cayley Table|entry]] of $\CC$ corresponding to the [[Definition:Element|element]] where the $k$th row intersects the $k$th column.
+Let $a$ be the [[Definition:Element|element]] which heads column $k$.
+Then, by definition, $a^{-1}$ is the [[Definition:Element|element]] which heads row $k$.
+Thus the [[Definition:Element|element]] which occupies [[Definition:Entry of Cayley Table|entry]] $\struct c_{k k}$ is $a^{-1} \circ a = e$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Latin Square is not necessarily Cayley Table of Group}
+Tags: Cayley Tables
+
+\begin{theorem}
+While it is true that the [[Definition:Cayley Table|Cayley table]] of a [[Definition:Finite Group|(finite) group]] is in the form of a [[Definition:Latin Square|Latin square]] it is not necessarily the case that a [[Definition:Latin Square|Latin square]] is the [[Definition:Cayley Table|Cayley table]] of a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $\struct {S, \circ}$ be the [[Definition:Algebraic Structure|algebraic structure]] defined by the following [[Definition:Cayley Table|Cayley table]]:
+:$\begin{array}{c|ccccc}
+\circ & e & a & b & c & d \\
+\hline
+e & e & a & b & c & d  \\
+a & a & e & d & b & c  \\
+b & b & c & e & d & a  \\
+c & c & d & a & e & b  \\
+d & d & b & c & a & e  \\
+\end{array}$
+By inspection it can be seen that the [[Definition:Cayley Table|Cayley table]] for $\struct {S, \circ}$ is a [[Definition:Latin Square|Latin square]].
+However, we have that:
+:$\paren {a \circ b} \circ c = d \circ c = a$
+:$a \circ \paren {b \circ c} = a \circ d = c$
+Thus $\circ$ is not an [[Definition:Associative Operation|associative operation]].
+So, by definition, $\struct {S, \circ}$ is not a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Scaling preserves Modulo Addition}
+Tags: Modulo Addition
+
+\begin{theorem}
+Let $m \in \Z_{> 0}$.
+Let $x, y, c \in \Z$.
+Let $x \equiv y \pmod m$.
+Then:
+:$c x \equiv c y \pmod m$
+\end{theorem}
+
+\begin{proof}
+Let $x \equiv y \pmod m$.
+Then by definition of [[Definition:Congruence Modulo Integer|congruence]]:
+:$\exists k \in Z: x - y = k m$
+Hence:
+:$c x - c y = c k m$
+and so by definition of [[Definition:Congruence Modulo Integer|congruence]]:
+:$c x \equiv c y \pmod m$
+{{qed}}
+[[Category:Modulo Addition]]
+5ro93s15864yankv4wl8fdte2utm0ix
+\end{proof}<|endoftext|>
+\section{Modulo Addition is Linear}
+Tags: Modulo Addition
+
+\begin{theorem}
+Let $m \in \Z_{> 0}$.
+Let $x_1, x_2, y_1, y_2, c_1, c_2 \in \Z$.
+Let:
+:$x_1 \equiv y_1 \pmod m$
+:$x_2 \equiv y_2 \pmod m$
+Then:
+:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$
+\end{theorem}
+
+\begin{proof}
+By [[Scaling preserves Modulo Addition]]:
+:$c_1 x_1 \equiv c_1 y_1 \pmod m$
+:$c_2 x_2 \equiv c_2 y_2 \pmod m$
+and so by [[Modulo Addition is Well-Defined]]:
+:$c_1 x_1 + c_2 x_2 \equiv c_1 y_1 + c_2 y_2 \pmod m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler Phi Function of Prime}
+Tags: Euler Phi Function, Prime Numbers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]] $p > 1$.
+Then:
+:$\map \phi p = p - 1$
+where $\phi: \Z_{>0} \to \Z_{>0}$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+From the definition of a [[Definition:Prime Number|prime number]], the only [[Definition:Strictly Positive Integer|(strictly) positive integer]] less than or equal to a [[Definition:Prime Number|prime]] $p$ which is ''not'' [[Definition:Coprime Integers|prime to]] $p$ is $p$ itself.
+Thus it follows directly that:
+:$\map \phi p = p - 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Chain is Directed}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $C$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Chain (Set Theory)|chain]] of $S$.
+Then $C$ is [[Definition:Directed Subset|directed]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in C$.
+By definition of [[Definition:Connected Relation|connected relation]]:
+:$x \preceq y$ or $y \preceq x$
+{{WLOG}}, suppose that
+:$x \preceq y$
+Define $z = y$.
+Thus by definition of [[Definition:Reflexivity|reflexivity]]
+:$x \preceq z$ and $y \preceq z$
+Hence $C$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduced Residue System Modulo Prime}
+Tags: Reduced Residue Systems, Prime Numbers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+The [[Definition:Reduced Residue System|reduced residue system modulo $p$]] contains $p - 1$ [[Definition:Element|elements]]:
+:$Z'_p = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {p - 1} m}$
+and so can be defined as:
+:$Z'_p = Z_p \setminus \set {\eqclass 0 m}$
+\end{theorem}
+
+\begin{proof}
+From [[Prime not Divisor implies Coprime]], each of $1, 2, \ldots, p - 1$ is [[Definition:Coprime Integers|coprime]] to $p$.
+The result follows by definition of [[Definition:Reduced Residue System|reduced residue system modulo $p$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication on Reduced Residue System is Closed}
+Tags: Modulo Multiplication, Reduced Residue Systems
+
+\begin{theorem}
+Let $m \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
+:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
+Let $S = \struct {\Z'_m, \times_m}$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of $\Z'_m$ under the [[Definition:Modulo Multiplication|modulo multiplication]].
+Then $S$ is [[Definition:Closed Algebraic Structure|closed]], in the sense that:
+:$\forall a, b \in \Z'_m: a \times_m b \in \Z'_m$
+\end{theorem}
+
+\begin{proof}
+Let $\eqclass r m, \eqclass s m \in \Z'_m$.
+Then by definition of [[Definition:Reduced Residue System|reduced residue system]]:
+:$r, s \perp m$
+By [[Bézout's Lemma]]:
+:$\exists u_1, v_1 \in \Z: u_1 r + v_1 m = 1$
+:$\exists u_2, v_2 \in \Z: u_2 s + v_2 m = 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {u_1 r + v_1 m} \paren {u_2 s + v_2 m}
+      | r = u_1 u_2 r s + v_1 u_2 s m + u_1 v_2 r m + v_1 v_2 m^2
+      | c = 
+}}
+{{eqn | r = \paren {u_1 u_2} r s + \paren {v_1 u_2 s + u_1 v_2 r + v_1 v_2 m} m
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+So, again by [[Bézout's Lemma]], $r s$ is [[Definition:Coprime Integers|coprime to $m$]].
+So the [[Definition:Product Element|product]] of two [[Definition:Element|elements]] of $\struct {\Z'_m, \times_m}$ is again in $\struct {\Z'_m, \times_m}$.
+That is, $\struct {\Z'_m, \times_m}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication on Reduced Residue System is Cancellable}
+Tags: Modulo Multiplication, Reduced Residue Systems
+
+\begin{theorem}
+Let $m \in \Z_{> 0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
+:$\Z'_m = \set {\eqclass k m \in \Z_m: k \perp m}$
+Let $S = \struct {\Z'_m, \times_m}$ be the [[Definition:Algebraic Structure|algebraic structure]] consisting of $\Z'_m$ under [[Definition:Modulo Multiplication|modulo multiplication]].
+Then $\times_m$ is [[Definition:Cancellable Operation|cancellable]], in the sense that:
+:$\forall a, b, c \in \Z'_m: a \times_m c = b \times_m c \implies a = b$
+and:
+:$\forall a, b, c \in \Z'_m: c \times_m a = c \times_m b \implies a = b$
+\end{theorem}
+
+\begin{proof}
+Let $a, b, c \in \Z'_m$ such that $a \times_m c = b \times_m c$
+Let $p, q, r$ be [[Definition:Integer|integers]] such that:
+:$p \in a$
+:$q \in b$
+:$r \in c$
+By definition of [[Definition:Residue Class|residue class]], this means:
+{{begin-eqn}}
+{{eqn | l = p r
+      | o = \equiv
+      | r = q r
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | o = \equiv
+      | r = q
+      | rr= \pmod m
+      | c = [[Cancellability of Congruences/Corollary 1|Cancellability of Congruences: Corollary 1]]
+}}
+{{end-eqn}}
+Thus as $p \in a$ and $q \in b$ it follows that $a = b$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order Generating iff Every Element is Infimum}
+Tags: Complete Lattices, Order Generating
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then
+:$X$ is [[Definition:Order Generating|order generating]]
+{{iff}}
+:$\forall x \in S: \exists Y \subseteq X: x = \inf Y$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $X$ be [[Definition:Order Generating|order generating]].
+Let $x \in S$.
+By definition of [[Definition:Order Generating|order generating]]:
+:$x = \map \inf {x^\succeq \cap X}$
+Thus
+:$\exists Y \subseteq X: x = \inf Y$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$\forall x \in S: \exists Y \subseteq X: x = \inf Y$
+Let $x \in S$.
+By assumption:
+:$\exists Y \subseteq X: x = \inf Y$
+Define $Z := x^\succeq \cap X$.
+We will prove that
+:$\forall b \in S: b$ is [[Definition:Lower Bound of Set|lower bound]] for $Z \implies b \preceq x$
+Let $b \in S$ such that
+:$b$ is [[Definition:Lower Bound of Set|lower bound]] for $Z$.
+Let $c \in Y$.
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$x$ is [[Definition:Lower Bound of Set|lower bound]] for $Y$.
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$x \preceq c$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$c \in x^\succeq$
+By definition of [[Definition:Subset|subset]]:
+:$c \in X$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$c \in Z$
+Thus by definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$b \preceq c$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$b$ is [[Definition:Lower Bound of Set|lower bound]] for $Y$
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$b \preceq x$
+{{qed|lemma}}
+We will prove that
+:$x$ is [[Definition:Lower Bound of Set|lower bound]] for $Z$
+Let $a \in Z$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$a \in x^\succeq$
+Thus by definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$x \preceq a$
+{{qed|lemma}}
+Thus by definitions of [[Definition:Complete Lattice|complete lattice]] and [[Definition:Infimum of Set|infimum]]:
+:$x^\succeq \cap X$ admits an [[Definition:Infimum of Set|infimum]] and $x = \map \inf {x^\succeq \cap X}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weierstrass Factorization Theorem}
+Tags: Complex Analysis, Infinite Products, Entire Functions
+
+\begin{theorem}
+Let $f$ be an [[Definition:Entire Function|entire function]].
+Let $0$ be a zero of $f$ of [[Definition:Multiplicity (Analytic Function)|multiplicity]] $m \ge 0$.
+Let the [[Definition:Sequence|sequence]] $\sequence {a_n}$ consist of the nonzero [[Definition:Zero of Function|zeroes]] of $f$, repeated according to [[Definition:Multiplicity (Complex Analysis)|multiplicity]].
+\end{theorem}
+
+\begin{proof}
+From [[Weierstrass Product Theorem]], the [[Definition:Complex Function|function]]:
+:$\displaystyle \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
+defines an [[Definition:Entire Function|entire function]] that has the same [[Definition:Zero of Function|zeroes]] as $f$ counting [[Definition:Multiplicity (Complex Analysis)|multiplicity]]. 
+Thus $f / h$ is both an [[Definition:Entire Function|entire function]] and non-vanishing.
+As $f / h$ is both [[Definition:Holomorphic Function|holomorphic]] and nowhere zero there exists a [[Definition:Holomorphic Function|holomorphic function]] $g$ such that:
+:$e^g = f / h$
+Therefore:
+:$f = e^g h$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Axiom Schemata for Finite Group}
+Tags: Finite Groups
+
+\begin{theorem}
+The following [[Definition:Axiom Schema|axiom schemata]] for the definition of a [[Definition:Finite Group|finite group]] are [[Definition:Logical Equivalence|logically equivalent]]:
+\end{theorem}
+
+\begin{proof}
+Let $S$ be an [[Definition:Algebraic Structure|algebraic structure]] of [[Definition:Finite Order (Structure)|finite order]] which fulfils the [[Definition:Group Axioms|group axioms]].
+It is to be shown that $S$ also fulfils all the [[Definition:Finite Group Axioms|finite group axioms]].
+We have that $\text {FG} 0$ and $\text {FG} 1$ are the same as $\text G 0$ and $\text G 1$ and so [[Definition:A Priori|a priori]] $\text {FG} 0$ and $\text {FG} 1$ are fulfilled.
+[[Definition:By Hypothesis|By hypothesis]] $S$ is of [[Definition:Finite Order (Structure)|finite order]].
+That is, there exists some [[Definition:Strictly Positive Integer|(strictly) positive integer]] $n$ such that $s$ has exactgly $n$ [[Definition:Element|elements]].
+That is, $\text {FG} 2$ is fulfilled.
+As $S$ fulfils the [[Definition:Group Axioms|group axioms]], it is by definition a [[Definition:Group|group]].
+Therefore the [[Cancellation Laws]] hold:
+:$b a = c a \implies b = c$
+:$a b = a c \implies b = c$
+That is, $\text {FG} 3$ is fulfilled.
+Thus, $S$ fulfils all the [[Definition:Finite Group Axioms|finite group axioms]].
+{{qed|lemma}}
+Let $S$ be an [[Definition:Algebraic Structure|algebraic structure]] which fulfils the [[Definition:Finite Group Axioms|finite group axioms]].
+It is to be shown that $S$ also fulfils all the [[Definition:Group Axioms|group axioms]].
+We have that $\text G 0$ and $\text G 1$ are the same as $\text {FG} 0$ and $\text {FG} 1$ and so [[Definition:A Priori|a priori]] $\text G 0$ and $\text G 1$ are fulfilled.
+By definition, an [[Definition:Algebraic Structure|algebraic structure]] which fulfils [[Definition:Group Axioms|group axioms]] $\text G 0$ and $\text G 1$ is a [[Definition:Semigroup|semigroup]].
+From [[Cancellable Finite Semigroup is Group]], it then follows that $S$ is a [[Definition:Group|group]].
+That is, $S$ fulfils all the [[Definition:Group Axioms|group axioms]] $\text G 0$, $\text G 1$, $\text G 2$ and $\text G 3$.
+Hence the result.
+{{qed}}
+[[Category:Finite Groups]]
+j7vmvepggdd73ql70gasxnpgizbyf6k
+\end{proof}<|endoftext|>
+\section{Order Generating iff Every Superset Closed on Infima is Whole Space}
+Tags: Complete Lattices, Order Generating
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then
+:$X$ is [[Definition:Order Generating|order generating]]
+{{iff}}
+:$\forall Y \subseteq S: Y \supseteq X \land \left({\forall Z \subseteq Y: \inf Z \in Y}\right) \implies Y = S$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $X$ be [[Definition:Order Generating|order generating]].
+Let $Y \subseteq S$ such that
+:$Y \supseteq X \land \left({\forall Z \subseteq Y: \inf Z \in Y}\right)$
+We will prove that
+:$S \subseteq Y$
+Let $s \in S$.
+By [[Order Generating iff Every Element is Infimum]]:
+:$\exists Z \subseteq X: s = \inf Z$
+By [[Subset Relation is Transitive]]:
+:$Z \subseteq Y$
+Thus by assumption:
+:$s \in Y$
+{{qed|lemma}}
+By definition of [[Definition:Set Equality|set equality]]:
+:$Y = S$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall Y \subseteq S: Y \supseteq X \land \left({\forall Z \subseteq Y: \inf Z \in Y}\right) \implies Y = S$
+Define $Y := \left\{ {\inf Z: Z \subseteq X}\right\}$
+By definition of [[Definition:Subset|subset]]:
+:$Y \subseteq S$
+We will prove that
+:$X \subseteq Y$
+Let $x \in X$.
+By definitions of [[Definition:Singleton|singleton]] and [[Definition:Subset|subset]]:
+:$\left\{ {x}\right\} \subseteq X$
+By [[Infimum of Singleton]]:
+:$\inf \left\{ {x}\right\} = x$
+Thus by definition of $Y$:
+:$x \in Y$
+{{qed|lemma}}
+We will prove that
+:$\forall x \in S: \exists Z \subseteq X: x = \inf Z$
+Let $x \in S$.
+We will prove that
+:$\forall Z \subseteq Y: \inf Z \in Y$
+Let $Z \subseteq Y$.
+Define $T := \bigcup \left\{ {A \subseteq X: \inf A \in Z}\right\}$
+By [[Union of Subsets is Subset]]:
+:$T \subseteq X \subseteq S$
+Define $N := \left\{ {\inf A: A \subseteq X \land \inf A \in Z}\right\}$
+We will prove that
+:$Z \subseteq N$
+Let $y \in Z$.
+By definition of [[Definition:Subset|subset]]:
+:$y \in Y$
+By definition of $Y$:
+:$\exists Z \subseteq X: y = \inf Z$
+Thus by definition of $N$:
+:$y \in N$
+{{qed|lemma}}
+By definition of [[Definition:Subset|subset]]:
+:$N \subseteq Z$
+By definition of [[Definition:Set Equality|set equality]]:
+:$Z = N$
+By [[Infimum of Infima]]:
+:$\inf Z = \inf T$
+Thus
+:$\inf Z \in Y$
+{{qed|lemma}}
+By assumption:
+:$Y = S$
+Then
+:$x \in Y$
+Thus by definition of $Y$:
+:$\exists Z \subseteq X: x = \inf Z$
+{{qed|lemma}}
+Hence $X$ is [[Definition:Order Generating|order generating]] by [[Order Generating iff Every Element is Infimum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Isomorphism Preserves Inverses}
+Tags: Group Isomorphisms, Inverse Elements, Group Isomorphism Preserves Inverses
+
+\begin{theorem}
+Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Isomorphism|group isomorphism]].
+Let:
+:$e_G$ be the [[Definition:Identity Element|identity]] of $\struct {G, \circ}$
+:$e_H$ be the [[Definition:Identity Element|identity]] of $\struct {H, *}$.
+Then:
+: $\forall g \in G: \map \phi {g^{-1} } = \paren {\map \phi g}^{-1}$
+\end{theorem}
+
+\begin{proof}
+An [[Definition:Group Isomorphism|group isomorphism]] is by definition a [[Definition:Group Epimorphism|group epimorphism]].
+The result follows from [[Epimorphism Preserves Inverses]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $g \in G$.
+{{begin-eqn}}
+{{eqn | l = \phi \left({g}\right) * \phi \left({g^{-1} }\right)
+      | r = \phi \left({g \circ g^{-1} }\right)
+      | c = {{Defof|Group Isomorphism}}
+}}
+{{eqn | r = \phi \left({e_G}\right)
+      | c = {{Defof|Inverse Element}}
+}}
+{{eqn | r = e_H
+      | c = [[Group Isomorphism Preserves Identity]]
+}}
+{{end-eqn}}
+It follows from [[Inverse in Group is Unique]] that $\phi \left({g^{-1} }\right)$ is the [[Definition:Unique|unique]] [[Definition:Inverse Element|inverse element]] of $\phi \left({g}\right)$ in $\left({H, *}\right)$.
+That is:
+: $\forall g \in G: \phi \left({g^{-1}}\right) = \left({\phi \left({g}\right)}\right)^{-1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Composite Relation}
+Tags: Composite Relations
+
+\begin{theorem}
+Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations]].
+Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]].
+Then the [[Definition:Preimage of Relation|preimage]] of $\RR_2 \circ \RR_1$ is given by:
+:$\Preimg {\RR_2 \circ \RR_1} = \Preimg {\Img {\RR_1} \cap \Preimg {\RR_2} }$
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Composite Relations]]
+6lh1ebouq13pwncjkma96kpqrizpu1t
+\end{proof}<|endoftext|>
+\section{Image of Composite Relation}
+Tags: Composite Relations
+
+\begin{theorem}
+Let $\RR_1 \subseteq S_1 \times T_1$ and $\RR_2 \subseteq S_2 \times T_2$ be [[Definition:Relation|relations]].
+Let $\RR_2 \circ \RR_1 \subseteq S_1 \times T_2$ be the [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]].
+Then the [[Definition:Image of Relation|image]] of $\RR_2 \circ \RR_1$ is given by:
+:$\Img {\RR_2 \circ \RR_1} = \Img {\Img {\RR_1} \cap \Preimg {\RR_2} }$
+\end{theorem}
+
+\begin{proof}
+We have by definition of [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]]:
+:$\RR_2 \circ \RR_1 := \set {\tuple {x, z} \in S_1 \times T_2: \exists y \in S_2 \cap T_1: \tuple {x, y} \in \RR_1 \land \tuple {y, z} \in \RR_2}$
+Let $\left({x, z}\right) \in \RR_2 \circ \RR_1$.
+By definition of [[Definition:Image of Relation|image]]:
+:$z \in \Img {\RR_2 \circ \RR_1}$
+By definition of [[Definition:Composition of Relations|composition of $\RR_1$ and $\RR_2$]]:
+:$z \in \Img {\RR_2}$
+But also:
+:$z \in \Img {\Img {\RR_1} }$
+We also have that $\Img {\RR_2} = \Img {\Preimg {\RR_2} }$
+That is:
+$z \in \Img {\Img {\RR_1} } \cap \Img {\Preimg {\RR_2} }$
+{{finish|Not convinced this is going to work.}}
+[[Category:Composite Relations]]
+b5ge9pk84j3e46bqba3xyq39i1lm3y0
+\end{proof}<|endoftext|>
+\section{Weierstrass Product Theorem}
+Tags: Infinite Products
+
+\begin{theorem}
+Let $\sequence {a_k}$ be a [[Definition:Sequence|sequence]] of non-zero [[Definition:Complex Number|complex numbers]] such that:
+:$\cmod {a_n} \to \infty$ as $n \to \infty$
+Let $\sequence {p_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Non-Negative Integer|non-negative integers]] for which the [[Definition:Series|series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \size {\dfrac r {a_n} }^{1 + p_n}$
+[[Definition:Convergent Series|converges]] for every $r \in \R_{> 0}$.
+Let:
+:$\displaystyle \map f z = \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$
+where $E_{p_n}$ are [[Definition:Weierstrass Elementary Factor|Weierstrass elementary factors]].
+Then $f$ is [[Definition:Entire Function|entire]] and its [[Definition:Zero of Function|zeroes]] are the points $a_n$, counted with [[Definition:Multiplicity (Complex Analysis)|multiplicity]].
+\end{theorem}
+
+\begin{proof}
+By:
+:[[Locally Uniformly Absolutely Convergent Product is Locally Uniformly Convergent]]
+:[[Infinite Product of Analytic Functions is Analytic]]
+:[[Zeroes of Infinite Product of Analytic Functions]]
+it suffices to show that the product $\displaystyle \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$ [[Definition:Local Uniform Absolute Convergence of Product|converges locally uniformly absolutely]].
+By [[Bounds for Weierstrass Elementary Factors]] and [[Weierstrass M-Test]], this is the case.
+{{stub}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Gaussian Integer Units and Rotation Matrices Order 4}
+Tags: Examples of Group Isomorphisms: Order 4, Group of Rotation Matrices Order 4, Group of Gaussian Integer Units
+
+\begin{theorem}
+Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
+Let $\struct {R_4, \times}$ be the [[Definition:Group of Rotation Matrices Order 4|group of rotation matrices of order $4$]] under [[Definition:Modulo Addition|modulo addition]].
+Then $\struct {U_\C, \times}$ and $\struct {R_4, \times}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $f: U_C \to R_4$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map f 1
+      | r = r_0
+}}
+{{eqn | l = \map f i
+      | r = r_1
+}}
+{{eqn | l = \map f {-1}
+      | r = r_2
+}}
+{{eqn | l = \map f {-i}
+      | r = r_3
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Group of Gaussian Integer Units/Cayley Table|Cayley Table of Gaussian Integer Units]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {U_\C, \times}$ is as follows:
+{{:Group of Gaussian Integer Units/Cayley Table}}</onlyinclude>
+=== [[Group of Rotation Matrices Order 4/Cayley Table|Group of Rotation Matrices Order $4$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {R_4, \times}$ is as follows:
+{{:Group of Rotation Matrices Order 4/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Gaussian Integer Units and Reduced Residue System Modulo 5 under Multiplication}
+Tags: Examples of Group Isomorphisms: Order 4, Multiplicative Group of Reduced Residues Modulo 5, Group of Gaussian Integer Units
+
+\begin{theorem}
+Let $\struct {U_\C, \times}$ be the [[Definition:Group of Gaussian Integer Units|group of Gaussian integer units]] under [[Definition:Complex Multiplication|complex multiplication]].
+Let $\struct {\Z'_5, \times_5}$ be the [[Multiplicative Group of Reduced Residues Modulo 5|multiplicative group of reduced residues modulo $5$]].
+Then $\struct {U_\C, \times}$ and $\struct {\Z'_5, \times_5}$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $f: U_C \to \Z'_5$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map f 1
+      | r = \eqclass 1 5
+}}
+{{eqn | l = \map f i
+      | r = \eqclass 2 5
+}}
+{{eqn | l = \map f {-1}
+      | r = \eqclass 4 5
+}}
+{{eqn | l = \map f {-i}
+      | r = \eqclass 3 5
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $f$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Group of Gaussian Integer Units/Cayley Table|Cayley Table of Gaussian Integer Units]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {U_\C, \times}$ is as follows:
+{{:Group of Gaussian Integer Units/Cayley Table}}</onlyinclude>
+=== [[Multiplicative Group of Reduced Residues Modulo 5/Cayley Table|Multiplicative Group of Reduced Residues Modulo $5$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $\struct {\Z'_5, \times_5}$ is as follows:
+{{:Multiplicative Group of Reduced Residues Modulo 5/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Group Generated by Reciprocal of z and 1 minus z}
+Tags: Groups of Order 6, Group Generated by Reciprocal of z and 1 minus z
+
+\begin{theorem}
+Let $\struct {S, \circ}$ denote the '''[[Definition:Group Generated by Reciprocal of z and 1 minus z|group generated by $\dfrac 1 z$ and $1 - z$]]'''.
+Then $\struct {S, \circ}$ is a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order $6$]].
+\end{theorem}
+
+\begin{proof}
+By definition:
+:$S = \set {f_1, f_2, f_3, f_4, f_5, f_6}$
+where $f_1, f_2, \ldots, f_6$ are [[Definition:Complex Function|complex functions]] defined for all $z \in \C \setminus \set {0, 1}$ as:
+{{begin-eqn}}
+{{eqn | l = \map {f_1} z
+      | r = z
+}}
+{{eqn | l = \map {f_2} z
+      | r = \dfrac 1 {1 - z}
+}}
+{{eqn | l = \map {f_3} z
+      | r = \dfrac {z - 1} z
+}}
+{{eqn | l = \map {f_4} z
+      | r = \dfrac 1 z
+}}
+{{eqn | l = \map {f_5} z
+      | r = 1 - z
+}}
+{{eqn | l = \map {f_6} z
+      | r = \dfrac z {z - 1}
+}}
+{{end-eqn}}
+Also by definition, $\circ$ denotes [[Definition:Composition of Mappings|composition of functions]].
+First it is necessary to establish the [[Definition:Cayley Table|Cayley table]] for $\struct {S, \circ}$.
+First note that it is apparent by definition of [[Definition:Composition of Mappings|composition of functions]] that:
+:$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$
+It remains to establish the rest of the [[Definition:Composition of Mappings|compositions]]
+{{begin-eqn}}
+{{eqn | l = f_2 \circ f_2
+      | r = \dfrac 1 {1 - \dfrac 1 {1 - z} }
+}}
+{{eqn | r = \dfrac {1 - z} {1 - z - 1}
+}}
+{{eqn | r = \dfrac {z - 1} z
+}}
+{{eqn | r = f_3
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_2 \circ f_3
+      | r = \dfrac 1 {1 - \dfrac {z - 1} z}
+}}
+{{eqn | r = \dfrac z {z - \paren {z - 1} }
+}}
+{{eqn | r = z
+}}
+{{eqn | r = f_1
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_2 \circ f_4
+      | r = \dfrac 1 {1 - \dfrac 1 z}
+}}
+{{eqn | r = \dfrac z {z - 1}
+}}
+{{eqn | r = f_6
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_2 \circ f_5
+      | r = \dfrac 1 {1 - \paren {1 - z} }
+}}
+{{eqn | r = \dfrac 1 z
+}}
+{{eqn | r = f_4
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_2 \circ f_6
+      | r = \dfrac 1 {1 - \dfrac z {z - 1} }
+}}
+{{eqn | r = \dfrac {z - 1} {z - 1 - z}
+}}
+{{eqn | r = 1 - z
+}}
+{{eqn | r = f_5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_3 \circ f_2
+      | r = \dfrac {\dfrac 1 {1 - z} - 1} {\dfrac 1 {1 - z} }
+}}
+{{eqn | r = \paren {1 - z} \dfrac {1 - \paren {1 - z} } {1 - z}
+}}
+{{eqn | r = z
+}}
+{{eqn | r = f_1
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_3 \circ f_3
+      | r = \dfrac {\dfrac {z - 1} z - 1} {\dfrac {z - 1} z}
+}}
+{{eqn | r = \dfrac {z - 1 - z} z \dfrac z {z - 1}
+}}
+{{eqn | r = \dfrac 1 {1 - z}
+}}
+{{eqn | r = f_2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_3 \circ f_4
+      | r = \dfrac {\dfrac 1 z - 1} {\dfrac 1 z}
+}}
+{{eqn | r = z \dfrac {1 - z} z
+}}
+{{eqn | r = 1 - z
+}}
+{{eqn | r = f_5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_3 \circ f_5
+      | r = \dfrac {\paren {1 - z} - 1} {1 - z}
+}}
+{{eqn | r = \dfrac z {z - 1}
+}}
+{{eqn | r = f_6
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_3 \circ f_6
+      | r = \dfrac {\dfrac z {z - 1} - 1} {\dfrac z {z - 1} }
+}}
+{{eqn | r = \dfrac {z - \paren {z - 1} } {z - 1} \dfrac {z - 1} z
+}}
+{{eqn | r = \dfrac 1 z
+}}
+{{eqn | r = f_4
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_4 \circ f_2
+      | r = \dfrac 1 {\dfrac 1 {1 - z} }
+}}
+{{eqn | r = 1 - z
+}}
+{{eqn | r = f_5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_4 \circ f_3
+      | r = \dfrac 1 {\dfrac {z - 1} z}
+}}
+{{eqn | r = \dfrac z {z - 1}
+}}
+{{eqn | r = f_6
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_4 \circ f_4
+      | r = \dfrac 1 {1 / z}
+}}
+{{eqn | r = z
+}}
+{{eqn | r = f_1
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_4 \circ f_5
+      | r = \dfrac 1 {1 - z}
+}}
+{{eqn | r = f_2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_4 \circ f_6
+      | r = \dfrac 1 {\dfrac z {z - 1} }
+}}
+{{eqn | r = \dfrac {z - 1} z
+}}
+{{eqn | r = f_3
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_5 \circ f_2
+      | r = 1 - \dfrac 1 {1 - z}
+}}
+{{eqn | r = \dfrac {1 - z - 1} {1 - z}
+}}
+{{eqn | r = \dfrac z {z - 1}
+}}
+{{eqn | r = f_6
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_5 \circ f_3
+      | r = 1 - \dfrac {z - 1} z
+}}
+{{eqn | r = \dfrac {z - \paren {z - 1} } z
+}}
+{{eqn | r = \dfrac 1 z
+}}
+{{eqn | r = f_4
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_5 \circ f_4
+      | r = 1 - \dfrac 1 z
+}}
+{{eqn | r = \dfrac {z - 1} z
+}}
+{{eqn | r = f_3
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_5 \circ f_5
+      | r = 1 - \paren {1 - z}
+}}
+{{eqn | r = z
+}}
+{{eqn | r = f_1
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_5 \circ f_6
+      | r = 1 - \dfrac z {z - 1}
+}}
+{{eqn | r = \dfrac {z - 1 - z} {z - 1}
+}}
+{{eqn | r = \dfrac 1 {1 - z}
+}}
+{{eqn | r = f_2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_6 \circ f_2
+      | r = \dfrac {\dfrac 1 {1 - z} } {\dfrac 1 {1 - z} - 1}
+}}
+{{eqn | r = \dfrac 1 {1 - z} \dfrac {1 - z} {1 - \paren {1 - z} }
+}}
+{{eqn | r = \dfrac 1 z
+}}
+{{eqn | r = f_4
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_6 \circ f_3
+      | r = \dfrac {\dfrac {z - 1} z} {\dfrac {z - 1} z - 1}
+}}
+{{eqn | r = \dfrac {z - 1} z \dfrac z {z - 1 - z}
+}}
+{{eqn | r = 1 - z
+}}
+{{eqn | r = f_5
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_6 \circ f_4
+      | r = \dfrac {\dfrac 1 z} {\dfrac 1 z - 1}
+}}
+{{eqn | r = \dfrac 1 z \dfrac z {1 - z}
+}}
+{{eqn | r = \dfrac 1 {1 - z}
+}}
+{{eqn | r = f_2
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_6 \circ f_5
+      | r = \dfrac {1 - z} {1 - z - 1}
+}}
+{{eqn | r = \dfrac {z - 1} z
+}}
+{{eqn | r = f_3
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = f_6 \circ f_6
+      | r = \dfrac {\dfrac z {z - 1} } {\dfrac z {z - 1} - 1}
+}}
+{{eqn | r = \dfrac z {z - 1} \dfrac {z - 1} {z - 1 - z}
+}}
+{{eqn | r = z
+}}
+{{eqn | r = f_1
+}}
+{{end-eqn}}
+Hence the [[Group Generated by Reciprocal of z and 1 minus z/Cayley Table|Cayley table]] is established:
+{{:Group Generated by Reciprocal of z and 1 minus z/Cayley Table}}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+By inspection it can be seen that $\struct {S, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+From [[Composition of Mappings is Associative]], it follows that $\struct {S, \circ}$ is [[Definition:Associative Operation|associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+We have that:
+:$\forall n: f_1 \circ f_n = f_n = f_n \circ f_1$
+Thus $\struct {S, \circ}$ has an [[Definition:Identity Element|identity element]].
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+From the above analysis:
+{{begin-eqn}}
+{{eqn | l = f_1 \circ f_1
+      | r = f_1
+}}
+{{eqn | l = f_2 \circ f_3
+      | r = f_1
+}}
+{{eqn | l = f_3 \circ f_2
+      | r = f_1
+}}
+{{eqn | l = f_4 \circ f_4
+      | r = f_1
+}}
+{{eqn | l = f_5 \circ f_5
+      | r = f_1
+}}
+{{eqn | l = f_6 \circ f_6
+      | r = f_1
+}}
+{{end-eqn}}
+Thus every element of $\struct {S, \circ}$ has an [[Definition:Inverse Element|inverse]].
+{{qed|lemma}}
+All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a [[Definition:Group|group]].
+$\struct {S, \circ}$ has $6$ [[Definition:Element|elements]] and so is a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order $6$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Generated by Reciprocal of z and Minus z is Klein Four-Group}
+Tags: Examples of Group Isomorphisms: Order 4, Klein Four-Group, Group Generated by Reciprocal of z and Minus z
+
+\begin{theorem}
+Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
+Let $S$ be the [[Group Generated by Reciprocal of z and Minus z|group generated by $1 / z$ and $-z$]].
+Then $K_4$ and $S$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to S$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map \phi e
+      | r = z
+}}
+{{eqn | l = \map \phi a
+      | r = -z
+}}
+{{eqn | l = \map \phi b
+      | r = \dfrac 1 z
+}}
+{{eqn | l = \map \phi c
+      | r = -\dfrac 1 z
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Klein Four-Group/Cayley Table|Cayley Table of Klein $4$-Group]] ===
+The [[Definition:Cayley Table|Cayley table]] for $K_4$ is as follows:
+{{:Klein Four-Group/Cayley Table}}</onlyinclude>
+=== [[Group Generated by Reciprocal of z and Minus z/Cayley Table|Group Generated by $1 / z$ and $-z$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $S$ is as follows:
+{{:Group Generated by Reciprocal of z and Minus z/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Group of Reflection Matrices Order 4 is Klein Four-Group}
+Tags: Examples of Group Isomorphisms: Order 4, Klein Four-Group, Group of Reflection Matrices Order 4
+
+\begin{theorem}
+Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
+Let $R_4$ be the [[Group of Reflection Matrices Order 4|Group of Reflection Matrices Order $4$]].
+Then $K_4$ and $R_4$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to R_4$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map \phi e
+      | r = r_0
+}}
+{{eqn | l = \map \phi a
+      | r = r_1
+}}
+{{eqn | l = \map \phi b
+      | r = r_2
+}}
+{{eqn | l = \map \phi c
+      | r = r_3
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Klein Four-Group/Cayley Table|Cayley Table of Klein $4$-Group]] ===
+The [[Definition:Cayley Table|Cayley table]] for $K_4$ is as follows:
+{{:Klein Four-Group/Cayley Table}}</onlyinclude>
+=== [[Group of Reflection Matrices Order 4/Cayley Table|Group of Reflection Matrices Order $4$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $S$ is as follows:
+{{:Group of Reflection Matrices Order 4/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Reduced Residues Modulo 8 is Klein Four-Group}
+Tags: Examples of Group Isomorphisms: Order 4, Klein Four-Group, Multiplicative Group of Reduced Residues Modulo 8
+
+\begin{theorem}
+Let $K_4$ denote the [[Definition:Klein Four-Group|Klein $4$-group]].
+Let $R_4$ be the [[Multiplicative Group of Reduced Residues Modulo 8|multiplicative group of reduced residues Modulo $8$]].
+Then $K_4$ and $R_4$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $\phi: K_4 \to R_4$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map \phi e
+      | r = \eqclass 1 8
+}}
+{{eqn | l = \map \phi a
+      | r = \eqclass 3 8
+}}
+{{eqn | l = \map \phi b
+      | r = \eqclass 5 8
+}}
+{{eqn | l = \map \phi c
+      | r = \eqclass 7 8
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Klein Four-Group/Cayley Table|Cayley Table of Klein $4$-Group]] ===
+The [[Definition:Cayley Table|Cayley table]] for $K_4$ is as follows:
+{{:Klein Four-Group/Cayley Table}}</onlyinclude>
+=== [[Multiplicative Group of Reduced Residues Modulo 8/Cayley Table|Multiplicative Group of Reduced Residues Modulo $8$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $S$ is as follows:
+{{:Multiplicative Group of Reduced Residues Modulo 8/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic}
+Tags: Examples of Group Isomorphisms: Order 4, Klein Four-Group, Cyclic Group of Order 4, Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic
+
+\begin{theorem}
+The [[Definition:Klein Four-Group|Klein $4$-group]] $K_4$ and the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $4$]] $C_4$ are not [[Definition:Group Isomorphism|isomorphic]].
+\end{theorem}
+
+\begin{proof}
+Recall the [[Klein Four-Group/Cayley Table|Cayley table]] for $K_4$:
+{{:Klein Four-Group/Cayley Table}}
+From [[Finite Cyclic Group is Isomorphic to Integers under Modulo Addition]], $C_4$ can be exemplified using the [[Definition:Additive Group of Integers Modulo 4|additive group of integers modulo $4$]].
+Recall the [[Modulo Addition/Cayley Table/Modulo 4|Cayley table]] for $\struct {\Z_4, +_4}$:
+{{:Modulo Addition/Cayley Table/Modulo 4}}
+Note that all [[Definition:Element|elements]] of $K_4$ are [[Definition:Self-Inverse Element|self-inverse]].
+However, for example:
+:$\eqclass 1 4 +_4 \eqclass 1 4 = \eqclass 2 4$
+and so $\eqclass 1 4$ is not [[Definition:Self-Inverse Element|self-inverse]].
+Thus there can be no [[Definition:Bijection|bijection]] between $K_4$ and $\struct {\Z_4, +_4}$ such that:
+:$\forall a, b \in K_4: \map \phi a +_4 \map \phi b = \map \phi {a b}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding}
+Tags: Complete Lattices, Order Generating
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then
+:$X$ is [[Definition:Order Generating|order generating]]
+{{iff}}
+:$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}\right)$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $X$ be [[Definition:Order Generating|order generating]].
+Let $x, y \in S$ such that
+:$y \npreceq x$
+By [[Order Generating iff Every Element is Infimum]]:
+:$\exists P \subseteq X: x = \inf P$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$y$ is [[Definition:Lower Bound of Set|lower bound]] for $P \implies y \preceq x$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$\exists p \in P: y \npreceq p$
+Thus by definition of [[Definition:Subset|subset]]:
+:$p \in X$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$x$ is [[Definition:Lower Bound of Set|lower bound]] for $P$.
+Thus by definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$x \preceq p$
+Thus
+:$y \npreceq p$
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that
+:$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p}\right)$
+Let $x \in S$.
+Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
+:$x^\succeq \cap X$ admits an [[Definition:Infimum of Set|infimum]].
+Define $y := \inf \left({x^\succeq \cap X}\right)$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$y$ is [[Definition:Lower Bound of Set|lower bound]] for $x^\succeq \cap X$
+We will prove that
+:$x$ is [[Definition:Lower Bound of Set|lower bound]] for $x^\succeq \cap X$
+Let $a \in x^\succeq \cap X$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$a \in x^\succeq$
+Thus by definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$x \preceq a$
+{{qed|lemma}}
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$x \preceq y$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$y \nprec x$
+It remains to prove that
+:$y = x$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$y \ne x$
+By definition of $\prec$
+:$y \npreceq x$ or $y = x$
+Case $y \npreceq x$
+By assumption:
+:$\exists p \in X: x \preceq p \land y \npreceq p$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$p \in x^\succeq$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$p \in x^\succeq \cap X$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$y \preceq p$
+Thus this contradicts $y \npreceq p$
+Case $y = x$
+Thus this contradicts $y \ne x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Permutation Induces Equivalence Relation/Corollary}
+Tags: Symmetric Groups, Equivalence Relations
+
+\begin{theorem}
+:$i \mathrel {\mathcal R_\pi} j$ {{iff}} $i$ and $j$ are in the same [[Definition:Cyclic Permutation|cycle]] of $\pi$.
+\end{theorem}
+
+\begin{proof}
+We have that [[Permutation Induces Equivalence Relation]].
+The [[Definition:Equivalence Class|equivalence classes]] of that [[Definition:Equivalence Relation|equivalence relation]] are the [[Definition:Cyclic Permutation|cycles]] of $\pi$.
+Hence the result by definition of [[Definition:Equivalence Class|equivalence class]].
+{{qed}}
+[[Category:Symmetric Groups]]
+[[Category:Equivalence Relations]]
+r1yzoalv7uf4e0a7t9ld6j0e4d4r65j
+\end{proof}<|endoftext|>
+\section{Order of Group Element not less than Order of Power}
+Tags: Order of Group Elements, Order of Group Element not less than Order of Power
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$ be an [[Definition:Element|element]] of $g$.
+Let $\left\lvert{g}\right\rvert$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
+Then:
+:$\forall m \in \Z: \left\lvert{g^m}\right\rvert \le \left\lvert{g}\right\rvert$
+where $g^m$ denotes the [[Definition:Power of Group Element|$m$th power of $g$ in $G$]].
+\end{theorem}
+
+\begin{proof}
+Let $\left\lvert{g}\right\rvert = n$.
+Then from [[Order of Power of Group Element]]:
+:$\forall m \in \Z: \left\lvert{g^m}\right\rvert = \dfrac n {\gcd \left\{ {m, n}\right\} }$
+where $\gcd \left\{ {m, n}\right\}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
+The result follows from [[Greatest Common Divisor is at least 1]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $g^n = e$.
+Let $h = g^m$.
+Then:
+:$h^n = g^{m n} = \paren {g^n}^m = e^m = e$
+Hence by definition of [[Definition:Order of Group Element|order of group element]]:
+:$\order h \le n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Group Element equals Order of Coprime Power}
+Tags: Order of Group Elements, Order of Group Element equals Order of Coprime Power
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$ be an [[Definition:Element|element]] of $g$.
+Let $\order g$ denote the [[Definition:Order of Group Element|order]] of $g$ in $G$.
+Then:
+:$\forall m \in \Z: \order {g^m} = \order g \iff m \perp \order g$
+where:
+:$g^m$ denotes the [[Definition:Power of Group Element|$m$th power of $g$ in $G$]]
+:$\perp$ denotes [[Definition:Coprime Integers|coprimality]].
+\end{theorem}
+
+\begin{proof}
+Let $\order g = n$.
+Then from [[Order of Power of Group Element]]:
+:$\forall m \in \Z: \order {g^m} = \dfrac n {\gcd \set {m, n}}$
+where $\gcd \set {m, n}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $m$ and $n$.
+Thus:
+:$\order {g^m} = \order g \iff \gcd \set {m, n} = 1$
+The result follows by definition of [[Definition:Coprime Integers|coprime integers]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\order g = n$.
+=== Necessary Condition ===
+Let $m \perp n$.
+Then by [[Bézout's Lemma]]:
+:$\exists x, y \in \Z: x m + y n = 1$
+Let $h = g^m$.
+Then:
+:$h^x = g^{m x} = g^{1 - y n} = g g^{- y n} = g e = g$
+and so $g$ is also a [[Definition:Power of Group Element|power]] of $h$.
+Hence from [[Order of Group Element not less than Order of Power]]:
+:$\order g \le \order h \le \order g$
+and it follows that:
+:$\order g = \order h$
+{{qed|lemma}}
+=== Sufficient Condition ===
+We prove the [[Definition:Contrapositive Statement|contrapositive]].
+Suppose $m \not \perp n$.
+Then:
+:$\exists d \in \Z_{>1}: \exists a, b \in \Z: m = a d \land n = b d$
+Let $h = g^m$.
+Then:
+:$h^b = \paren {g^m}^{n/d} = g^{a n} = \paren {g^n}^a = e^a = e$
+Hence:
+:$\order h \le b < n$
+and it follows that:
+:$\order g \ne \order h$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unique Composition of Group Element whose Order is Product of Coprime Integers}
+Tags: Order of Group Elements
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$ be an [[Definition:Element|element]] of $g$.
+Let:
+:$\order g = m n$
+where:
+:$\order g$ denotes the [[Definition:Order of Group Element|order]] of $g$ in $G$
+:$m$ and $n$ are [[Definition:Coprime Integers|coprime integers]].
+Then $g$ can be expressed [[Definition:Unique|uniquely]] as the [[Definition:Product Element|product]] of two [[Definition:Commuting Elements|commuting elements]] $a$ and $b$ of [[Definition:Order of Group Element|order]] $m$ and $n$ respectively.
+\end{theorem}
+
+\begin{proof}
+Let $g_1 = g^n$ and $g_2 = g^m$.
+By [[Powers of Group Element Commute]], $g_1$ and $g_2$ [[Definition:Commuting Elements|commute]].
+We have:
+:${g_1}^m = g^{n m} = e$
+:${g_2}^n = g^{m n} = e$
+It follows that:
+:the [[Definition:Order of Group Element|order]] of $g_1$ is $m$
+:the [[Definition:Order of Group Element|order]] of $g_2$ is $n$
+otherwise if either $g_1$ or $g_2$ were of a smaller [[Definition:Order of Group Element|order]] then $g$ would also be of a smaller [[Definition:Order of Group Element|order]].
+By [[Bézout's Lemma]]:
+:$\exists u, v \in \Z: u n + v m = 1$
+as $m \perp n$.
+Thus:
+:$g = g^{u n + v m} = \paren {g^n}^u \paren {g^m}^v = {g_1}^u {g_2}^v$
+Also by [[Bézout's Lemma]]:
+:$u \perp m$
+and:
+:$v \perp n$
+Thus by [[Order of Group Element equals Order of Coprime Power]]:
+:$\order { {g_1}^u} = m$
+and:
+:$\order { {g_2}^v} = n$
+We have that $g_1$ and $g_2$ [[Definition:Commute|commute]].
+So by [[Commutativity of Powers in Group]], ${g_1}^u$ and ${g_2}^v$ also [[Definition:Commuting Elements|commute]].
+Putting $a = g_1^u$ and $b = g_2^v$, we have $a$ and $b$ which satisfy the required conditions.
+It remains to prove [[Definition:Unique|uniqueness]].
+Suppose that:
+:$(1): \quad g = r_1 r_2 = s_1 s_2$
+where:
+:$r_1$ and $r_2$ [[Definition:Commuting Elements|commute]]
+:$s_1$ and $s_2$ [[Definition:Commuting Elements|commute]]
+:$\order {r_1} = \order {s_1} = m$
+:$\order {r_2} = \order {s_2} = n$
+Raising $(1)$ to the [[Definition:Power of Group Element|$n u$th power]]:
+:$g^{n u} = {r_1}^{n u} {r_2}^{n u} = {s_1}^{n u} {s_2}^{n u}$
+and so:
+{{begin-eqn}}
+{{eqn | l = {r_1}^{n u} {r_2}^{n u}
+      | r = {s_1}^{n u} {s_2}^{n u}
+      | c = 
+}}
+{{eqn | l = {r_1}^{n u} \paren { {r_2}^n}^u
+      | r = {s_1}^{n u} \paren { {s_2}^n}^u
+      | c = 
+}}
+{{eqn | l = {r_1}^{n u} e^u
+      | r = {s_1}^{n u} e^u
+      | c = as $\order {r_2} = \order {s_2} = n$
+}}
+{{eqn | ll= \leadsto
+      | l = {r_1}^{n u}
+      | r = {s_1}^{n u}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {r_1}^{1 - m v}
+      | r = {s_1}^{1 - m v}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_1 \paren { {r_1}^m}^{-v}
+      | r = s_1 \paren { {s_1}^m}^{-v}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_1 e^{-v}
+      | r = s_1 e^{-v}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_1
+      | r = s_1
+      | c = 
+}}
+{{end-eqn}}
+It follows directly from $(1)$ that $r_2 = s_2$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Cyclic Group equals Order of Generator}
+Tags: Order of Group Elements, Order of Groups, Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated]] by $a \in G$.
+Then:
+:$\order a = \order G$
+where:
+:$\order a$ denotes the [[Definition:Order of Group Element|order]] of $a$ in $G$
+:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $\order a = n$.
+From [[List of Elements in Finite Cyclic Group]]:
+:$G = \set {a_0, a_1, \ldots, a_{n - 1} }$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group whose Order equals Order of Element is Cyclic}
+Tags: Order of Group Elements, Order of Groups, Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group Element|order]] $n$.
+Let $g \in G$ have [[Definition:Order of Group Element|order]] $n$.
+Then $G$ is a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated]] by $g$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Subgroup|subgroup]] of $G$ [[Definition:Generator of Subgroup|generated]] by $g$ is:
+:$\gen g = \set {g^0, g^1, g^2, \ldots, g^{n - 1} }$
+This contains $n$ [[Definition:Element|elements]].
+Thus:
+:$\gen g = G$
+and the result follows by definition of [[Definition:Cyclic Group|cyclic group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Powers of 2 under Multiplication form Infinite Abelian Group}
+Tags: Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]] defined as:
+:$S = \set {2^k: k \in \Z}$
+Then $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+It can be seen by inspection that $S \subseteq \Q_{>0}$.
+That is, all the [[Definition:Element|elements]] of $S$ are [[Definition:Strictly Positive Rational Number|strictly positive rational numbers]].
+From [[Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group]]:
+:$\struct {\Q_{>0}, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+It is noted that $S$ is an [[Definition:Infinite Set|infinite set]] and so trivially $S \ne \varnothing$.
+Let $a, b \in S$.
+Then:
+:$\exists k_1 \in \Z: a = 2^{k_1}$
+:$\exists k_2 \in \Z: b = 2^{k_2}$
+We have that:
+:$\dfrac 1 {2^{k_2} } 2^{k_2} = 1$
+and so $\dfrac 1 {2^{k_2} } = 2^{-k_2} = b^{-1}$ is the [[Definition:Inverse Element|inverse]] of $b \in \struct {\Q_{>0}, \times}$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = a \times b^{-1}
+      | r = 2^{k_1} \times \dfrac 1 {2^{k_2} }
+      | c = 
+}}
+{{eqn | r = 2^{k_1 - k_2}
+      | c = 
+}}
+{{end-eqn}}
+We have that:
+:$k_1 + k_2 \in \Z$
+and so $a \times b^{-1} \in \S$.
+Hence by the [[One-Step Subgroup Test]], $\struct {S, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{>0}, \times}$.
+It has been established that $S$ is an [[Definition:Infinite Set|infinite set]].
+Hence by definition $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite group]].
+Finally, from [[Subgroup of Abelian Group is Abelian]], $\struct {S, \times}$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Numbers of form 1 + 2m over 1 + 2n form Infinite Abelian Group under Multiplication}
+Tags: Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Integer|integers]] defined as:
+:$S = \set {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}$
+Then $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Let $k \in \Z$.
+Then $1 + 2 k \ne 0$.
+Thus:
+:$\forall x \in S: x \in \Q_{\ne 0}$
+Thus by definition of [[Definition:Subset|subset]]:
+:$S \subseteq \Q_{\ne 0}$
+From [[Non-Zero Rational Numbers under Multiplication form Abelian Group]]:
+:$\struct {\Q_{\ne 0}, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+It is noted that $S$ is an [[Definition:Infinite Set|infinite set]] and so trivially $S \ne \varnothing$.
+Consider $a = \dfrac {1 + 2 m} {1 + 2 n} \in S$.
+We have that:
+:$\dfrac {1 + 2 m} {1 + 2 n} \times \dfrac {1 + 2 n} {1 + 2 m} = 1$
+and so $\dfrac {1 + 2 n} {1 + 2 m}$ is the [[Definition:Inverse Element|inverse]] of $b \in \struct {\Q_{>0}, \times}$.
+By inspection it can be seen that $b^{-1} \in S$.
+Let $a, b \in S$.
+Then:
+:$\exists m_1, n_1 \in \Z: a = \dfrac {1 + 2 m_1} {1 + 2 n_1}$
+:$\exists m_2, n_2 \in \Z: b = \dfrac {1 + 2 m_2} {1 + 2 n_2}$
+Then we have:
+{{begin-eqn}}
+{{eqn | l = a \times b
+      | r = \dfrac {1 + 2 m_1} {1 + 2 n_1} \dfrac {1 + 2 m_2} {1 + 2 n_2}
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {1 + 2 m_1} \paren {1 + 2 m_2} } {\paren {1 + 2 n_1} \paren {1 + 2 n_2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {1 + 2 m_1 + 2 m_2 + 4 m_1 m_2} {1 + 2 n_1 + 2 n_2 + 4 n_1 n_2}
+      | c = 
+}}
+{{eqn | r = \dfrac {1 + 2 \paren {m_1 + m_2 + 2 m_1 m_2} } {1 + 2 \paren {n_1 + n_2 + 2 n_1 n_2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {1 + 2 m} {1 + 2 n}
+      | c = where $m = m_1 + m_2 + 2 m_1 m_2, n = n_1 + n_2 + 2 n_1 n_2$
+}}
+{{end-eqn}}
+Hence by the [[Two-Step Subgroup Test]], $\struct {S, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q_{>0}, \times}$.
+It has been established that $S$ is an [[Definition:Infinite Set|infinite set]].
+Hence by definition $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite group]].
+Finally, from [[Subgroup of Abelian Group is Abelian]], $\struct {S, \times}$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Meet Irreducible Elements Excluded Top is Order Generating}
+Tags: Continuous Lattices, Order Generating, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Continuous Ordered Set|continuous]] [[Definition:Complete Lattice|complete lattice]].
+Let $X = \mathit{IRR}\left({L}\right) \setminus \left\{ {\top}\right\}$
+where $\mathit{IRR}\left({L}\right)$ denotes the [[Definition:Set|set]] of all [[Definition:Meet Irreducible|meet irreducible]] [[Definition:Element|element]] of $S$,
+:$\top$ denotes the [[Definition:Top (Lattice Theory)|top]] of $L$.
+Then $X$ is [[Definition:Order Generating|order generating]].
+\end{theorem}
+
+\begin{proof}
+We will prove that
+:$\forall x, y \in S: \left({ y \npreceq x \implies \exists p \in X: x \preceq p \land y \npreceq p }\right)$
+Let $x, y \in S$ such that
+:$y \npreceq x$
+By [[Not Preceding implies There Exists Meet Irreducible Element Not Preceding]]
+:$\exists p \in S: p$ is [[Definition:Meet Irreducible|meet irreducible]] and $x \preceq p$ and $y \npreceq p$
+By definition of [[Definition:Greatest Element|greatest element]]:
+:$p \ne \top$ and $p \in \mathit{IRR}\left({L}\right)$
+By definitions of [[Definition:Set Difference|difference]] and [[Definition:Singleton|singleton]]:
+:$p \in X$
+Thus
+:$\exists p \in X: x \preceq p \land y \npreceq p$
+{{qed|lemma}}
+Hence by [[Order Generating iff Not Preceding implies There Exists Element Preceding and Not Preceding]]:
+:$X$ is [[Definition:Order Generating|order generating]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Order Generating is Order Generating}
+Tags: Complete Lattices, Order Generating
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $X, Y$ be [[Definition:Subset|subsets]] of $S$ such that
+:$X$ is [[Definition:Order Generating|order generating]]
+and
+:$X \subseteq Y$
+Then $Y$ is [[Definition:Order Generating|order generating]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in S$.
+Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
+:$x^\succeq \cap Y$ admits an [[Definition:Infimum of Set|infimum]].
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$x^\succeq$ admits an [[Definition:Infimum of Set|infimum]]
+and
+:$x^\succeq \cap X$ admits an [[Definition:Infimum of Set|infimum]].
+By [[Intersection is Subset]]:
+:$x^\succeq \cap Y \subseteq x^\succeq$
+By [[Infimum of Subset]]:
+:$\inf \left({x^\succeq}\right) \preceq \inf \left({x^\succeq \cap Y}\right)$
+By [[Infimum of Upper Closure of Element]]:
+:$x \preceq \inf \left({x^\succeq \cap Y}\right)$
+By [[Set Intersection Preserves Subsets/Corollary]]:
+:$x^\succeq \cap X \subseteq x^\succeq \cap Y$
+By [[Infimum of Subset]]:
+:$\inf \left({x^\succeq \cap Y}\right) \preceq \inf \left({x^\succeq \cap X}\right)$
+By definition of [[Definition:Order Generating|order generating]]:
+:$\inf \left({x^\succeq \cap Y}\right) \preceq  x$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$x = \inf \left({x^\succeq \cap Y}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus 1 Rational Argument Complex Numbers under Multiplication form Infinite Abelian Group}
+Tags: Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] defined as:
+:$S = \set {\cos \theta + i \sin \theta: \theta \in \Q}$
+Then the [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Polar Form of Complex Number|polar form of complex numbers]], the [[Definition:Element|elements]] of $S$ are also [[Definition:Element|elements]] of the [[Definition:Circle Group|circle group]] $\struct {K, \times}$:
+:$K = \set {z \in \C: \cmod z = 1}$
+$S$ is [[Definition:Infinite Set|infinite]] by construction.
+Thus $S \subseteq C$ and trivially $S \ne \O$.
+Let $a, b \in S$.
+Then:
+:$a = \cos \theta_1 + i \sin \theta_1$
+and:
+:$b = \cos \theta_2 + i \sin \theta_2$
+for some $\theta_1, \theta_2 \in \Q$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = a \times b
+      | r = \paren {\cos \theta_1 + i \sin \theta_1} \paren {\cos \theta_2 + i \sin \theta_2}
+      | c = 
+}}
+{{eqn | r = \map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2}
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = b^{-1}
+      | r = \cos \theta_2 - i \sin \theta_2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+Hence by the [[Two-Step Subgroup Test]], $\struct {S, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {K, \times}$.
+It has been established that $S$ is an [[Definition:Infinite Set|infinite set]].
+Hence by definition $\struct {S, \times}$ is an [[Definition:Infinite Group|infinite group]].
+Finally, from [[Subgroup of Abelian Group is Abelian]], $\struct {S, \times}$ is an [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inversion Mapping is Automorphism iff Group is Abelian}
+Tags: Inversion Mappings, Abelian Groups, Group Automorphisms
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\iota: G \to G$ be the [[Definition:Inversion Mapping|inversion mapping]] on $G$, defined as:
+:$\forall g \in G: \map \iota g = g^{-1}$
+Then $\iota$ is an [[Definition:Group Automorphism|automorphism]] {{iff}} $G$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+From [[Inversion Mapping is Permutation]], $\iota$ is a [[Definition:Permutation|permutation]].
+It remains to be shown that $\iota$ has the [[Definition:Morphism Property|morphism property]] {{iff}} $G$ is [[Definition:Abelian Group|abelian]].
+=== Sufficient Condition ===
+Suppose $\iota$ is an [[Definition:Group Automorphism|automorphism]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \forall x, y \in G: \map \iota {x \circ y}
+      | r = \map \iota x \circ \map \iota y
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x \circ y}^{-1}
+      | r = x^{-1} \circ y^{-1}
+      | c = Definition of $\iota$
+}}
+{{end-eqn}}
+Thus from [[Inverse of Commuting Pair]], $x$ [[Definition:Commute|commutes]] with $y$.
+This holds for all $x, y \in G$.
+So $\struct {G, \circ}$ is [[Definition:Abelian Group|abelian]] by definition.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\struct {G, \circ}$ be [[Definition:Abelian Group|abelian]].
+{{begin-eqn}}
+{{eqn | l = \forall x, y \in G: \paren {x \circ y}^{-1}
+      | r = x^{-1} \circ y^{-1}
+      | c = [[Inverse of Commuting Pair]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \iota {x \circ y}
+      | r = \map \iota x \circ \map \iota y
+      | c = Definition of $\iota$
+}}
+{{end-eqn}}
+Thus $\iota$ has the [[Definition:Morphism Property|morphism property]] and is therefore an [[Definition:Group Automorphism|automorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Even Order Group has Odd Number of Order 2 Elements}
+Tags: Finite Groups, Order of Group Elements
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $G$ be of even [[Definition:Order of Structure|order]].
+Then $G$ has an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]] of [[Definition:Order of Group Element|order $2$]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {x \in G: \order x > 2}$ be the [[Definition:Set|set]] of all [[Definition:Element|elements]] of $G$ whose [[Definition:Order of Group Element|order]] is strictly greater than $2$.
+Let $h \in S$.
+Then from [[Order of Group Element equals Order of Inverse]]:
+:$h^{-1} \in S$.
+Because $\order h > 2$ it follows that $h^{-1} \ne h$ by definition of [[Definition:Self-Inverse Element|self-inverse]].
+Thus every [[Definition:Element|element]] in $S$ can be paired off with its [[Definition:Inverse Element|inverse]] which is also in $S$.
+It follows that $S$ has an [[Definition:Even Integer|even number]] of [[Definition:Element|elements]].
+Thus $G \setminus S$ also contains an [[Definition:Even Integer|even number]] of [[Definition:Element|elements]].
+From [[Identity is Only Group Element of Order 1]], there is precisely $1$ [[Definition:Element|element]] of $G$ whose [[Definition:Order of Group Element|order]] is less than $2$.
+This leaves $G$ with an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]] of order $2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Roots of Resolvent of Cubic}
+Tags: Cubic Equations
+
+\begin{theorem}
+Let $P$ be the [[Definition:Cubic Equation|cubic equation]]:
+: $a x^3 + b x^2 + c x + d = 0$ with $a \ne 0$
+Let $\map r P$ be the [[Definition:Cubic Equation/Resolvent|resolvent equation]] of $P$, given by:
+: $u^6 - 2 R u^3 - Q^3$
+Let the [[Definition:Root of Polynomial|roots]] of $P$ be $\alpha_1, \alpha_2, \alpha_3$.
+Then the [[Definition:Root of Polynomial|roots]] of $\map r P$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \frac 1 3 \paren {\alpha_1 + \omega \alpha_2 + \omega^2 \alpha_3}
+}}
+{{eqn | l = \omega v
+      | r = \frac 1 3 \paren {\alpha_3 + \omega \alpha_1 + \omega^2 \alpha_2}
+      | c = 
+}}
+{{eqn | l = \omega^2 v
+      | r = \frac 1 3 \paren {\alpha_2 + \omega \alpha_3 + \omega^2 \alpha_1}
+      | c = 
+}}
+{{eqn | l = u
+      | r = \frac 1 3 \paren {\alpha_1 + \omega \alpha_3 + \omega^2 \alpha_2}
+}}
+{{eqn | l = \omega u
+      | r = \frac 1 3 \paren {\alpha_2 + \omega \alpha_1 + \omega^2 \alpha_3}
+      | c = 
+}}
+{{eqn | l = \omega^2 u
+      | r = \frac 1 3 \paren {\alpha_3 + \omega \alpha_2 + \omega^2 \alpha_1}
+      | c = 
+}}
+{{end-eqn}}
+where $\omega = -\dfrac {-1 + \sqrt {-3} } 2$ is one of the [[Primitive Complex Cube Roots of Unity|primitive (complex) cube roots of $1$]].
+\end{theorem}<|endoftext|>
+\section{Roots of Complex Number/Examples/Cube Roots}
+Tags: Examples of Complex Roots
+
+\begin{theorem}
+Let $z := \polar {r, \theta}$ be a [[Definition:Polar Form of Complex Number|complex number expressed in polar form]], such that $z \ne 0$.
+Then the [[Definition:Complex Root|complex cube roots]] of $z$ are given by:
+:$z^{1 / 3} = \set {r^{1 / 3} \paren {\map \cos {\dfrac {\theta + 2 \pi k} 3} + i \, \map \sin {\dfrac {\theta + 2 \pi k} 3} }: k \in \set {0, 1, 2} }$
+There are $3$ [[Definition:Distinct Elements|distinct]] such [[Definition:Complex Root|complex cube roots]].
+These can also be expressed as:
+:$z^{1 / 3} = \set {r^{1 / 3} e^{i \paren {\theta + 2 \pi k} / 3}: k \in \set {0, 1, 2} }$
+or:
+:$z^{1 / 3} = \set {r^{1 / 3} e^{i \theta / 3} \omega^k: k \in \set {0, 1, 2} }$
+where $\omega = e^{2 i \pi / 3} = -\dfrac 1 2 + \dfrac {i \sqrt 3} 2$ is the [[Definition:First Complex Root of Unity|first cube root of unity]].
+\end{theorem}
+
+\begin{proof}
+An example of [[Roots of Complex Number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Roots of Complex Number/Corollary}
+Tags: Complex Roots
+
+\begin{theorem}
+Let $z := \polar {r, \theta}$ be a [[Definition:Polar Form of Complex Number|complex number expressed in polar form]], such that $z \ne 0$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $w$ be one of the [[Definition:Complex Root|complex $n$th roots]] of $z$.
+Then the [[Definition:Complex Root|$n$th roots]] of $z$ are given by:
+:$z^{1 / n} = \set {w \epsilon^k: k \in \set {1, 2, \ldots, n - 1} }$
+where $\epsilon$ is a [[Definition:Primitive Root of Unity|primitive $n$th root of unity]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Primitive Complex Root of Unity|primitive complex $n$th root of unity]]:
+:$\omega = e^{2 m i \pi k}$
+for some $m \in \Z: 1 \le m < n$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {w \omega^k}^n
+      | r = w^n \paren {e^{2 m i \pi k / n} }^n
+      | c = 
+}}
+{{eqn | r = z e^{2 m i \pi k}
+      | c = 
+}}
+{{eqn | r = z \paren {e^{2 i \pi} }^{m k}
+      | c = 
+}}
+{{eqn | r = z \times 1^{m k}
+      | c = 
+}}
+{{eqn | r = z
+      | c = 
+}}
+{{end-eqn}}
+This demonstrates that $w \omega^k$ is one of the [[Definition:Complex Root|complex $n$th roots]] of $z$.
+All of the [[Definition:Complex Roots of Unity|complex $n$th roots of unity]] are represented by powers of $\omega$.
+Thus it follows from [[Roots of Complex Number]] that:
+:$z^{1 / n} = \set {w \omega^k: k \in \set {1, 2, \ldots, n - 1} }$
+are the $n$ [[Definition:Complex Root|complex $n$th roots]] of $z$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Top is Prime Element}
+Tags: Prime Elements
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Meet Semilattice|meet semilattice]].
+Then $\top$ is a [[Definition:Prime Element (Order Theory)|prime element]]
+where $\top$ denotes the [[Definition:Greatest Element|greatest element]] of $L$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ such that
+:$x \wedge y \preceq \top$
+Thus by definition of [[Definition:Greatest Element|greatest element]]:
+:$x \preceq \top$ or $y \preceq \top$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Prime Element in Meet Semilattice}
+Tags: Prime Elements
+
+\begin{theorem}
+Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $p \in S$,
+Then:
+:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
+{{iff}}:
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] $A$ of $S$:
+::if $\inf A \preceq p$, then there exists [[Definition:Element|element]] $x$ of $A$ such that $x \preceq p$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $p$ be [[Definition:Prime Element (Order Theory)|prime element]].
+Let $A$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $S$.
+Define
+:$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$
+where $X \subseteq S$.
+We will prove that
+:$\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$
+Let $x \in A, B \subseteq A$ such that
+:$\map P B$
+and
+:$B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \preceq p$
+Case $B = \O$.
+By [[Union with Empty Set]]:
+:$B \cup \set x = \set x$
+By [[Infimum of Singleton]]:
+:$\inf \set x = x$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \set x$
+Thus
+:$\exists z \in B \cup \set x: z \preceq p$
+{{qed|lemma}}
+Case $B \ne \O$.
+By [[Subset of Finite Set is Finite]]:
+:$B$ is [[Definition:Finite Set|finite]].
+By [[Existence of Non-Empty Finite Infima in Meet Semilattice]]:
+:$B, \set x$ admit [[Definition:Infimum of Set|infima]].
+By [[Infimum of Infima]]:
+:$\map \inf {B \cup \set x} = \inf B \wedge \inf \set x$
+By definition of [[Definition:Prime Element (Order Theory)|prime element]]:
+:$\inf B \preceq p$ or $x \preceq p$
+Case $\inf B \preceq p$.
+By assumption:
+:$\exists z \in B: z \preceq p$
+By definition of [[Definition:Set Union|union]]:
+:$z \in B \cup \set x$
+Thus
+:$\exists z \in B \cup \set x: z \preceq p$
+{{qed|lemma}}
+Case $x \preceq p$.
+By definitions of [[Definition:Set Union|union]] and [[Definition:Singleton|singleton]]:
+:$x \in B \cup \set x$
+Thus
+:$\exists z \in B \cup \set x: z \preceq p$
+{{qed|lemma}}
+By definition of [[Definition:Empty Set|empty set]]:
+:$\map P \O$
+By [[Induction of Finite Set]]:
+:$\map P A$
+Thus
+:if $\inf A \preceq p$, then there exists [[Definition:Element|element]] $x$ of $A$ such that $x \preceq p$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] $A$ of $S$:
+::if $\inf A \preceq p$, then there exists [[Definition:Element|element]] $x$ of $A$ such that $x \preceq p$.
+Let $x, y \in S$ such that
+:$x \wedge y \preceq p$
+Define $A := \set {x, y}$
+By definitions of [[Definition:Non-Empty Set|non-empty set]] and [[Definition:Finite Set|finite set]]:
+:$A$ is [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$\inf A = x \wedge y$
+By assumption:
+:$\exists z \in A: z \preceq p$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \preceq p$ or $y \preceq p$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism between Group Generated by Reciprocal of z and 1 minus z and Symmetric Group on 3 Letters}
+Tags: Examples of Group Isomorphisms: Order 6, Symmetric Group on 3 Letters, Group Generated by Reciprocal of z and 1 minus z
+
+\begin{theorem}
+Let $S_3$ denote the [[Symmetric Group on 3 Letters|symmetric group on $3$ letters]].
+Let $G$ be the [[Definition:Group Generated by Reciprocal of z and 1 minus z|group generated by $1 / z$ and $1 - z$]].
+Then $S_3$ and $G$ are [[Definition:Isomorphic Algebraic Structures|isomorphic algebraic structures]].
+\end{theorem}
+
+\begin{proof}
+Establish the [[Definition:Mapping|mapping]] $\phi: S_3 \to G$ as follows:
+{{begin-eqn}}
+{{eqn | l = \map \phi e
+      | r = z
+}}
+{{eqn | l = \map \phi {\paren {123} }
+      | r = \dfrac 1 {1 - z}
+}}
+{{eqn | l = \map \phi {\paren {132} }
+      | r = \dfrac {z - 1} z
+}}
+{{eqn | l = \map \phi {\paren {23} }
+      | r = \dfrac 1 z
+}}
+{{eqn | l = \map \phi {\paren {13} }
+      | r = 1 - z
+}}
+{{eqn | l = \map \phi {\paren {12} }
+      | r = \dfrac z {z - 1}
+}}
+{{end-eqn}}
+From [[Isomorphism by Cayley Table]], the two [[Definition:Cayley Table|Cayley tables]] can be compared by eye to ascertain that $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]:
+=== [[Symmetric Group on 3 Letters/Cayley Table|Cayley Table of Symmetric Group on $3$ Letters]] ===
+The [[Definition:Cayley Table|Cayley table]] for $S_3$ is as follows:
+{{:Symmetric Group on 3 Letters/Cayley Table}}</onlyinclude>
+=== [[Group Generated by Reciprocal of z and 1 minus z/Cayley Table|Group Generated by $1 / z$ and $1 - z$]] ===
+The [[Definition:Cayley Table|Cayley table]] for $S$ is as follows:
+{{:Group Generated by Reciprocal of z and 1 minus z/Cayley Table}}
+\end{proof}<|endoftext|>
+\section{Prime Element is Meet Irreducible}
+Tags: Prime Elements, Meet Irreducible
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Let $p \in S$.
+Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]] of $L$.
+Then $p$ is [[Definition:Meet Irreducible|meet irreducible]] in $L$.
+\end{theorem}
+
+\begin{proof}
+Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]].
+Let $x, y \in S$ such that
+:$p = x \wedge y$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \wedge y \preceq p$
+By definition of [[Definition:Prime Element (Order Theory)|prime element]]:
+:$x \preceq p$ or $y \preceq p$
+By [[Meet Precedes Operands]]:
+:$p \preceq x$ and $p \preceq y$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$p = x$ or $p = y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Element iff Meet Irreducible in Distributive Lattice}
+Tags: Prime Elements, Meet Irreducible
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Distributive Lattice|distributive lattice]].
+Let $p \in S$.
+Then $p$ is a [[Definition:Prime Element (Order Theory)|prime element]] {{iff}} $p$ is [[Definition:Meet Irreducible|meet irreducible]].
+\end{theorem}
+
+\begin{proof}
+By [[Prime Element is Meet Irreducible]]:
+:if $p$ is a [[Definition:Prime Element (Order Theory)|prime element]], then $p$ is [[Definition:Meet Irreducible|meet irreducible]].
+Assume that:
+:$p$ is [[Definition:Meet Irreducible|meet irreducible]].
+Let $x, y \in S$ such that:
+:$x \wedge y \preceq p$
+{{begin-eqn}}
+{{eqn | l = p
+      | r = p \vee \paren {x \wedge y}
+      | c = [[Preceding iff Join equals Larger Operand]]
+}}
+{{eqn | r = \paren {p \vee x} \wedge \paren {p \vee y}
+      | c = {{Defof|Distributive Lattice}}
+}}
+{{end-eqn}}
+By definition of [[Definition:Meet Irreducible|meet irreducible]]:
+:$p = p \vee x$ or $p = p \vee y$
+Thus by [[Preceding iff Join equals Larger Operand]]:
+:$x \preceq p$ or $y \preceq p$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiple Equilibrium Points all have Equal Payoffs}
+Tags: Game Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
+Let $G$ have more than one [[Definition:Equilibrium Point|equilibrium point]].
+Then every [[Definition:Equilibrium Point|equilibrium point]] of $G$ has the same [[Definition:Payoff|payoff]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Stated without proof, or even an explanatory comment on its scope of applicability, in Davis. This work is too vague to be useful. I am going to have to use a different work.}}
+\end{proof}<|endoftext|>
+\section{Definition:Information of Game}
+Tags: Definitions: Game Theory
+
+\begin{theorem}
+The '''information''' of a [[Definition:Game|game]] is the knowledge each [[Definition:Player|player]] has of the history of all the [[Definition:Move|moves]] of all the [[Definition:Player|players]] of the [[Definition:Game|game]].
+\end{theorem}<|endoftext|>
+\section{Non-Zero-Sum Game as Zero-Sum Game}
+Tags: Zero-Sum Games
+
+\begin{theorem}
+Let $G$ be a [[Definition:Non-Zero-Sum Game|non-zero-sum game]] for $n$ [[Definition:Player|players]].
+Then $G$ can be modelled as a [[Definition:Zero-Sum Game|zero-sum game]] for $n + 1$ [[Definition:Player|players]].
+\end{theorem}
+
+\begin{proof}
+At each [[Definition:Outcome of Game|outcome]], the total [[Definition:Payoff|payoff]] of $G$ will be an amount which will (for at least one [[Definition:Outcome of Game|outcome]]) not be [[Definition:Zero (Number)|zero]]
+Let an $n + 1$th [[Definition:Player|player]] be introduced to $G$ who has one [[Definition:Move|move]]:
+:$(1): \quad$ Select any [[Definition:Player|player]] $m$.
+:$(2): \quad$ If the total [[Definition:Payoff|payoff]] of $G$ is $+k$, receive in payment $k$ from [[Definition:Player|player]] $m$.
+:$(3): \quad$ If the total [[Definition:Payoff|payoff]] of $G$ is $-k$, [[Definition:Pay|pay]] $k$ to [[Definition:Player|player]] $m$.
+This new [[Definition:Game|game]] is the same as $G$ but with an extra (dummy) [[Definition:Player|player]], and is now [[Definition:Zero-Sum Game|zero-sum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two-Person Zero-Sum Game is Non-Cooperative}
+Tags: Two-Person Zero-Sum Games, Cooperative Games
+
+\begin{theorem}
+A [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] necessarily has to be [[Definition:Non-Cooperative Game|non-cooperative]].
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Cooperative Game|cooperative game]] is one where [[Definition:Player|players]] form [[Definition:Coalition|coalitions]] against the other [[Definition:Player|players]].
+If the [[Definition:Player|players]] in a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] were to form a [[Definition:Coalition|coalition]], there would be no other [[Definition:Player|players]] to form it against.
+Further, as the total [[Definition:Payoff|payoff]] is [[Definition:Zero (Number)|zero]], there would be no benefit in collaborating on using one [[Definition:Strategy|strategy]] over another, as when they pool their [[Definition:Payoff|payoffs]] they are back where they started.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simple Graph with Finite Vertex Set is Finite}
+Tags: Graph Theory, Simple Graphs
+
+\begin{theorem}
+Let $G$ be a [[Definition:Simple Graph|simple graph]].
+Suppose that the [[Definition:Vertex Set|vertex set]] of $G$ is [[Definition:Finite Set|finite]].
+Then $G$ is a [[Definition:Finite Graph|finite graph]].
+That is to say, its [[Definition:Edge Set|edge set]] is also [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+Since $G$ is [[Definition:Simple Graph|simple]], it can have at most one [[Definition:Edge of Graph|edge]] between each two [[Definition:Vertex of Graph|vertices]].
+Therefore, the [[Definition:Mapping|mapping]]:
+:$v: \map E G \to \powerset {\map V G}, \map v e = \set {a_e, b_e}$
+which assigns to each [[Definition:Edge of Graph|edge]] its [[Definition:Endvertex|endvertices]] $a_e, b_e$, is an [[Definition:Injection|injection]].
+From [[Power Set of Finite Set is Finite]] and [[Domain of Injection Not Larger than Codomain]], it follows that $\map E G$ is [[Definition:Finite Set|finite]].
+Hence $G$ is [[Definition:Finite Graph|finite]].
+{{qed}}
+[[Category:Graph Theory]]
+[[Category:Simple Graphs]]
+itanc072i6il71lvytg9hxwb3ysbn34
+\end{proof}<|endoftext|>
+\section{Multigraph with Finite Vertex Set may not be Finite}
+Tags: Graph Theory, Multigraphs
+
+\begin{theorem}
+Let $G$ be a [[Definition:Multigraph|multigraph]].
+Let the [[Definition:Vertex Set|vertex set]] of $G$ be [[Definition:Finite Set|finite]].
+Then it is not necessarily the case that $G$ is also [[Definition:Finite Graph|finite]].
+\end{theorem}
+
+\begin{proof}
+{{finish|two vertices, $\N$ as edges}}
+[[Category:Graph Theory]]
+[[Category:Multigraphs]]
+4694d4g1dnmmomeps1l1rkeh713gspp
+\end{proof}<|endoftext|>
+\section{Modulus Larger than Real Part}
+Tags: Complex Modulus
+
+\begin{theorem}
+:$\cmod z \ge \size {\map \Re z}$
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Complex Number|definition of a complex number]], we have:
+:$z = \map \Re z + i \map \Im z$
+Then:
+{{begin-eqn}}
+{{eqn | l = \cmod z
+      | r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
+      | c = {{Defof|Complex Modulus}}
+}}
+{{eqn | o = \ge
+      | r = \sqrt {\paren {\map \Re z}^2 }
+      | c = [[Square of Real Number is Non-Negative]], as $\map \Im z$ is [[Definition:Real Number|real]]
+}}
+{{eqn | r = \cmod {\map \Re z}
+      | c = [[Square of Real Number is Non-Negative]], as $\map \Re z$ is [[Definition:Real Number|real]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulus Larger than Imaginary Part}
+Tags: Complex Modulus
+
+\begin{theorem}
+:$\cmod z \ge \size {\map \Im z}$
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Complex Number|definition of a complex number]], we have:
+:$z = \map \Re z + i \map \Im z$
+{{begin-eqn}}
+{{eqn | l = \cmod z
+      | r = \sqrt {\paren {\map \Re z}^2 + \paren {\map \Im z}^2}
+      | c = {{Defof|Complex Modulus}}
+}}
+{{eqn | o = \ge
+      | r = \sqrt {\paren {\map \Im z}^2 }
+      | c = [[Square of Real Number is Non-Negative]], as $\map \Re z$ is [[Definition:Real Number|real]]
+}}
+{{eqn | r = \size {\map \Im z}
+      | c = [[Square of Real Number is Non-Negative]], as $\map \Im z$ is [[Definition:Real Number|real]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Space is T0 iff Factor Spaces are T0/General Result}
+Tags: T0 Spaces, Product Spaces
+
+\begin{theorem}
+Let $\SS = \family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Topological Space|topological spaces]] where $I$ is an arbitrary [[Definition:Indexing Set|index set]].
+Let $\displaystyle T = \struct{S, \tau} = \prod \struct {S_\alpha, \tau_\alpha}$ be the [[Definition:Product Space of Topological Spaces|product space]] of $\SS$.
+Then $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] {{iff}} each of $\struct{S_\alpha, \tau_\alpha}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Suppose $\exists \beta: \struct {S_\beta, \tau_\beta}$ is not a [[Definition:Kolmogorov Space|$T_0$ space]].
+Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, either $a, b \in U_\beta$ or $a, b \notin U_\beta$.
+Consider the elements $y, z \in S$ defined as:
+:$y = \family {x_\alpha} : x_\alpha = \begin {cases}
+s_\alpha & : \alpha \ne \beta \\
+a & : \alpha = \beta \end {cases}$
+:$z = \family {x_\alpha} : x_\alpha = \begin {cases}
+s_\alpha & : \alpha \ne \beta \\
+b & : \alpha = \beta \end {cases}$
+That is, $y$ and $z$ match (arbitrarily) on all ordinates except that for $\beta$.
+Let $H \subseteq S: y \in H$.
+Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H : b \in U_\beta$
+Similarly, let $K \subseteq S: z \in K$.
+Then $y \in K$ as $\forall U_\beta \in \map {\pr_\beta} H : a \in U_\beta$
+So $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose $T$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, either $a, b \in U$ or $a, b \notin U$.
+Then $a$ and $b$ are different in at least one ordinate.
+Suppose, $a_m = p, b_m = q$ for some ordinate $m$.
+Then either $a_m, b_m \in U_m$ or $a_m, b_m \notin U_m$.
+It follows that $\struct{S_m, \tau_m}$ is not a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Separation Properties Preserved under Topological Product/Corollary}
+Tags: Separation Axioms, Product Spaces
+
+\begin{theorem}
+$T = \struct {S, \tau}$ has one of the following properties {{iff}} each of $\struct {S_i, \tau_i}$ has the same property:
+:[[Definition:Regular Space|Regular Property]]
+:[[Definition:Tychonoff Space|Tychonoff (Completely Regular) Property]]
+If $T = \struct {S, \tau}$ has one of the following properties then each of $\struct {S_i, \tau_i}$ has the same property:
+:[[Definition:Normal Space|Normal Property]]
+:[[Definition:Completely Normal Space|Completely Normal Property]]
+but the [[Definition:Converse Statement|converse]] does not necessarily hold.
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Regular Space|regular space]] is a [[Definition:Topological Space|topological space]] which is both a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] and a [[Definition:T3 Space|$T_3$ space]].
+Hence from:
+:[[Product Space is T0 iff Factor Spaces are T0|Product Space is $T_0$ iff Factor Spaces are $T_0$]]
+and
+:[[Product Space is T3 iff Factor Spaces are T3|Product Space is $T_3$ iff Factor Spaces are $T_3$]]
+it follows that $T$ is a [[Definition:Regular Space|regular space]] {{iff}} each of $\struct {S_i, \tau_i}$ is a [[Definition:Regular Space|regular space]].
+{{qed|lemma}}
+A [[Definition:Tychonoff Space|Tychonoff (completely regular) space]] is a [[Definition:Topological Space|topological space]] which is both a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]] and a [[Definition:T3 1/2 Space|$T_{3 \frac 1 2}$ space]].
+Hence from:
+:[[Product Space is T0 iff Factor Spaces are T0|Product Space is $T_0$ iff Factor Spaces are $T_0$]]
+and
+:[[Product Space is T3 1/2 iff Factor Spaces are T3 1/2|Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$]]
+it follows that $T$ is a [[Definition:Tychonoff Space|Tychonoff space]] {{iff}} each of $\struct {S_i, \tau_i}$ is a [[Definition:Tychonoff Space|Tychonoff space]].
+{{qed|lemma}}
+A [[Definition:Normal Space|normal space]] is a [[Definition:Topological Space|topological space]] which is both a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] and a [[Definition:T4 Space|$T_4$ space]].
+Hence from:
+:[[Product Space is T1 iff Factor Spaces are T1|Product Space is $T_1$ iff Factor Spaces are $T_1$]]
+and
+:[[Factor Spaces are T4 if Product Space is T4|Factor Spaces are $T_4$ if Product Space is $T_4$]]
+it follows that if $T$ is a [[Definition:Normal Space|normal space]] then each of $\struct {S_i, \tau_i}$ is a [[Definition:Normal Space|normal space]].
+{{qed|lemma}}
+A [[Definition:Completely Normal Space|completely normal space]] is a [[Definition:Topological Space|topological space]] which is both a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]] and a [[Definition:T5 Space|$T_5$ space]].
+Hence from:
+:[[Product Space is T1 iff Factor Spaces are T1|Product Space is $T_1$ iff Factor Spaces are $T_1$]]
+and
+:[[Factor Spaces are T5 if Product Space is T5|Factor Spaces are $T_5$ if Product Space is $T_5$]]
+it follows that if $T$ is a [[Definition:Completely Normal Space|completely normal space]] then each of $\struct {S_i, \tau_i}$ is a [[Definition:Completely Normal Space|completely normal space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Distance to Nearest Integer Function}
+Tags: Real Analysis
+
+\begin{theorem}
+The following definitions of the [[Definition:Distance to Nearest Integer Function|distance to nearest integer function]] $\norm \cdot: \R \to \closedint 0 {\dfrac 1 2}$ are [[Definition:Logical Equivalence|equivalent]]:
+\end{theorem}
+
+\begin{proof}
+Let $\alpha \in \R$, $n \in \Z$.
+From [[Real Number is between Floor Functions]]:
+:$\floor \alpha \le \alpha < \floor \alpha + 1$
+For any $n < \floor \alpha \le \alpha$:
+{{begin-eqn}}
+{{eqn | l = \size {n - \alpha}
+      | r = \alpha - n
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | o = >
+      | r = \alpha - \floor \alpha
+}}
+{{eqn | r = \size {\alpha - \floor \alpha}
+      | c = {{Defof|Absolute Value}}
+}}
+{{end-eqn}}
+For any $n > \floor \alpha + 1 > \alpha$:
+{{begin-eqn}}
+{{eqn | l = \size {n - \alpha}
+      | r = n - \alpha
+      | c = {{Defof|Absolute Value}}
+}}
+{{eqn | o = >
+      | r = \floor \alpha + 1 - \alpha
+      | c = 
+}}
+{{eqn | r = \size {\floor \alpha + 1 - \alpha}
+      | c = {{Defof|Absolute Value}}
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \min \set {\size {n - \alpha}: n \in \Z}
+      | r = \min \set {\size {\floor \alpha - \alpha}, \size {\floor \alpha + 1 - \alpha} }
+      | c = Other $n$'s are disregarded by above
+}}
+{{eqn | r = \min \set {\alpha - \floor \alpha, \floor \alpha + 1 - \alpha}
+}}
+{{eqn | r = \min \set {\set \alpha, 1 - \set \alpha}
+      | c = {{Defof|Fractional Part}}
+}}
+{{end-eqn}}
+which shows that the definitions are indeed equivalent.
+{{qed}}
+[[Category:Real Analysis]]
+mc6dwrx70vuuipe3s97te11mdqlyvuo
+\end{proof}<|endoftext|>
+\section{Quotient and Remainder to Number Base/General Result}
+Tags: Number Theory
+
+\begin{theorem}
+:$\left \lfloor {\dfrac n {b^s}} \right \rfloor = \left[{r_m r_{m-1} \ldots r_{s+1} r_s}\right]_b$
+:$\displaystyle n \,\bmod\, {b^s} = \sum_{j \mathop = 0}^{s-1} {r_j b^j} = \left[{r_{s-1} r_{s-2} \ldots r_1 r_0}\right]_b$
+\end{theorem}
+
+\begin{proof}
+{{finish|Induction}}
+[[Category:Number Theory]]
+bc0ep9i9xz0twlu6e5kb3yv9p430m0e
+\end{proof}<|endoftext|>
+\section{Remainder is Primitive Recursive}
+Tags: Primitive Recursive Functions
+
+\begin{theorem}
+Let $m, n \in \N$ be [[Definition:Natural Numbers|natural numbers]].
+Let us define the [[Definition:Function|function]] $\operatorname{rem}: \N^2 \to \N$:
+:$\map \rem {n, m} = \begin{cases}
+\text{the remainder when } n \text{ is divided by } m & : m \ne 0 \\
+0 & : m = 0
+\end{cases}$
+where the $\text{remainder}$ is as defined in the [[Division Theorem]]:
+:If $n = m q + r$, where $0 \le r < m$, then $r$ is the [[Definition:Remainder|remainder]].
+Then $\rem$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+\end{theorem}
+
+\begin{proof}
+Let $\map \rem {n, m} = r$.
+We see that as $n$ increases by $1$, then so does $r$, except when $n = m-1$ in which case increasing $n$ by $1$ makes $r$ go to $0$.
+We also take into account the case where $m = 0$:
+So we can define $\rem$ by cases:
+:$\map \rem {n, m} = \begin{cases}
+0 & : \map \rem {n - 1, m} = m - 1 \text { or } m = 0 \text { or } n = 0\\
+\map \rem {n - 1, m} + 1 & :  \text {otherwise} \\
+\end{cases}$
+We remind ourselves of the following [[Definition:Primitive Recursive Function|primitive recursive functions]]:
+* [[Signum Function is Primitive Recursive|Signum function $\sgn$]]
+* [[Equality Relation is Primitive Recursive|Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$]]
+* [[Cut-Off Subtraction is Primitive Recursive|Cut-off subtraction $m \, \dot - \, n$]].
+We have:
+* $\map \sgn m = 1 \iff m \ne 0$
+* $\map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} } = 1 \iff \map \rem {n, m} \ne m \, \dot - \, 1$.
+So:
+:$\map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} }  = 1 \iff m > 0 \land \map \rem {n, m} \ne m \, \dot - \, 1$.
+So we see that:
+:$\map \rem {n + 1, m} = \paren {\map \rem {n, m} + 1} \, \map \sgn m \, \map {\overline \sgn} {\map {\chi_{\operatorname{eq} } } {\map \rem {n, m}, m \, \dot - \, 1} }$
+So $\rem$ is obtained by [[Definition:Primitive Recursion|primitive recursion]] (over the first variable, which is allowed by [[Permutation of Variables of Primitive Recursive Function]]) from the [[Definition:Primitive Recursive Function|primitive recursive functions]]:
+* [[Signum Function is Primitive Recursive|Signum function $\sgn$]]
+* [[Equality Relation is Primitive Recursive|Characteristic function of equality relation $\map {\chi_{\operatorname{eq} } } {n, m}$]]
+* [[Cut-Off Subtraction is Primitive Recursive|Cut-off subtraction $m \, \dot - \, n$]]
+* [[Addition is Primitive Recursive|Addition]]
+* [[Multiplication is Primitive Recursive|Multiplication]].
+Thus we can use [[Definition by Cases is Primitive Recursive]] and it follows that $\rem$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+{{qed}}
+[[Category:Primitive Recursive Functions]]
+9zoo0k2s3rxz1h0j34qmq5ur3u5iph3
+\end{proof}<|endoftext|>
+\section{Quotient is Primitive Recursive}
+Tags: Primitive Recursive Functions
+
+\begin{theorem}
+Let $m, n \in \N$ be [[Definition:Natural Numbers|natural numbers]].
+Let us define the [[Definition:Function|function]] $\operatorname {quot}: \N^2 \to \N$:
+:$\map {\operatorname {quot} } {n, m} = \begin{cases}
+\text{the quotient when } n \text{ is divided by } m & : m \ne 0 \\
+0 & : m = 0
+\end{cases}$
+where the $\text {quotient}$ and $\text {remainder}$ are as defined in the [[Division Theorem]]:
+:If $n = m q + r$, where $0 \le r < m$, then $q$ is the [[Definition:Quotient (Algebra)|quotient]].
+Then $\operatorname {quot}$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+\end{theorem}
+
+\begin{proof}
+We note that if $m \ne 0$ and $n = m q + r$, we have:
+:$\dfrac n m = q + \frac r m$
+Also note that $\dfrac n m$ and $\dfrac r m$ are [[Definition:Rational Number|rational numbers]] and not necessarily [[Definition:Natural Numbers|natural numbers]].
+Indeed, we have:
+:$0 \le \dfrac r m < 1$
+So if $m > 0$ then $\map {\operatorname {quot} } {n, m}$ is the [[Definition:Floor Function|floor]] $\floor {\dfrac n m}$ of $\dfrac n m$.
+So we have:
+:$\map {\operatorname {quot} } {n, m} = \begin{cases}
+0 & : m = 0 \\
+\floor {\dfrac n m} & : m \ne 0
+\end{cases}$
+Then we see that for $m \ne 0$:
+:$\floor {\dfrac {n + 1} m} = \begin {cases}
+\floor {\dfrac n m} + 1 & : m \divides \paren {n + 1} \\
+\floor {\dfrac n m} & : \text {otherwise}
+\end {cases}$
+So for $m \ne 0$:
+:$\map {\operatorname {quot} } {n + 1, m} = \begin{cases}
+\map {\operatorname {quot} } {n, m} + 1 & : \map \rem {n + 1, m} = 0 \\
+\map {\operatorname {quot} } {n, m} & : \map \rem {n + 1, m} \ne 0
+\end{cases}$
+Now note that:
+:$\map {\overline {\sgn} } {\map \rem {n + 1, m} } = \begin{cases}
+1 & : \map \rem {n + 1, m} = 0 \\
+0 & : \map \rem {n + 1, m} \ne 0
+\end{cases}$
+So the $\operatorname {quot}$ is defined as:
+:$\map {\operatorname {quot} } {0, m} = 0$
+:$\map {\operatorname {quot} } {n + 1, m} = \map \sgn m \map {\operatorname {quot} } {n, m} + \map {\overline {\sgn} } {\map \rem {n + 1, m} }$
+(note that the factor $\map \sgn m$ is needed to cover the case where $m = 0$).
+Thus $\operatorname {quot}$ is obtained  by [[Definition:Primitive Recursion|primitive recursion]] (over the first variable, which is allowed by [[Permutation of Variables of Primitive Recursive Function]]) from the [[Definition:Primitive Recursive Function|primitive recursive functions]]:
+:[[Signum Function is Primitive Recursive|Signum function $\sgn$]]
+:[[Remainder is Primitive Recursive|Remainder $\rem$]]
+:[[Addition is Primitive Recursive|Addition]]
+:[[Multiplication is Primitive Recursive|Multiplication]].
+So it follows that $\operatorname {quot}$ is [[Definition:Primitive Recursive Function|primitive recursive]].
+{{qed}}
+[[Category:Primitive Recursive Functions]]
+mtvzwief8480pk2jxmsosxrnawnk465
+\end{proof}<|endoftext|>
+\section{Tarski's Geometry is Complete/Corollary}
+Tags: Tarski's Geometry
+
+\begin{theorem}
+[[Definition:Tarski's Geometry|Tarski's geometry]] does not contain [[Definition:Minimal Arithmetic|minimal arithmetic]].
+\end{theorem}
+
+\begin{proof}
+Immediate from [[Tarski's Geometry is Complete]] and the [[Gödel's First Incompleteness Theorem/Corollary|corollary to Gödel's First Incompleteness Theorem]].
+{{qed}}
+[[Category:Tarski's Geometry]]
+ew232a0m24zm6l7im6b8stu8ykwizrx
+\end{proof}<|endoftext|>
+\section{Symmetric Closure of Symmetric Relation}
+Tags: Relation Theory
+
+\begin{theorem}
+Let $\mathcal R$ be a [[Definition:Relation|relation]] on a [[Definition:Set|set]] $S$.
+Let $\mathcal R^\leftrightarrow$ be the [[Definition:Symmetric Closure|symmetric closure]] of $\mathcal R$.
+Then $\mathcal R = \mathcal R^\leftrightarrow$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn|l = \mathcal R^\leftrightarrow 
+     |r = \mathcal R \cup \mathcal R^{-1}
+     |c = 
+Definition of [[Definition:Symmetric Closure|Symmetric Closure]]
+}}
+{{eqn|r = \mathcal R \cup \mathcal R
+     |c = [[Inverse of Symmetric Relation is Symmetric]]
+}}
+{{eqn|r = \mathcal R
+     |c = [[Union is Idempotent]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Relation Theory]]
+ntp7xd7skcscvnph1k9bn2m9iccb53e
+\end{proof}<|endoftext|>
+\section{Not Every Two-Person Zero-Sum Game has Saddle Point}
+Tags: Two-Person Zero-Sum Games
+
+\begin{theorem}
+Not every [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] has a [[Definition:Saddle Point (Game Theory)|saddle point]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Matching Pennies|game of Matching Pennies]].
+Recall its [[Definition:Matching Pennies/Payoff Table|payoff table]]:
+{{:Definition:Matching Pennies/Payoff Table}}
+Trivially, by inspection, this has no [[Definition:Entry in Payoff Table|entry]] which is the smallest [[Definition:Entry in Payoff Table|entry]] in its row and the largest [[Definition:Entry in Payoff Table|entry]] in its column.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eluding Game has no Saddle Point}
+Tags: Two-Person Zero-Sum Games
+
+\begin{theorem}
+The [[Definition:Eluding Game|eluding game]] has no [[Definition:Saddle Point (Game Theory)|saddle point]].
+\end{theorem}
+
+\begin{proof}
+Recall the [[Definition:Eluding Game/Payoff Table|payoff table]]:
+{{:Definition:Eluding Game/Payoff Table}}
+Trivially, by inspection, this has no [[Definition:Entry in Payoff Table|entry]] which is the smallest [[Definition:Entry in Payoff Table|entry]] in its row and the largest [[Definition:Entry in Payoff Table|entry]] in its column.
+\end{proof}<|endoftext|>
+\section{Two-Person Zero-Sum Game with Multiple Solutions}
+Tags: Examples of Two-Person Zero-Sum Games
+
+\begin{theorem}
+There exists a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]] with more than one [[Definition:Solution of Game|solution]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Game|game]] defined by the following [[Definition:Payoff Table|payoff table]]:
+{{PayoffTable|table = $\begin{array} {r {{|}} c {{|}} c {{|}} c {{|}} }
+& B_1 & B_2 & B_3 \\
+\hline A_1 & 1 & 2 & 4 \\
+\hline A_2 & 4 & 2 & 1 \\
+\hline \end{array}$}}
+This has two [[Definition:Solution of Game|solutions]]:
+:$(1): \quad A: \left({1/3, 2/3}\right), B: \left({0, 1, 0}\right)$
+:$(2): \quad A: \left({2/3, 1/3}\right), B: \left({0, 1, 0}\right)$
+{{explain|The reason why the above are solutions is not indicated.}}
+It follows that the [[Definition:Strategy|strategies]]:
+:$\left({\dfrac t 3, \dfrac {2 \left({1 - t}\right)} 3}\right), B: \left({0, 1, 0}\right)$
+for all $0 \le t \le 1$ are also [[Definition:Solution of Game|solutions]].
+{{explain|Again, the explanation of this needs to wait till more learning has been done.}}
+\end{proof}<|endoftext|>
+\section{Two-Person Zero-Sum Game with Finite Strategies has Solution}
+Tags: Two-Person Zero-Sum Games
+
+\begin{theorem}
+Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
+Let each [[Definition:Player|player]] of $G$ have a [[Definition:Finite Set|finite set]] of [[Definition:Strategy|strategies]] available.
+Then $G$ has at least one [[Definition:Solution of Game|solution]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Proof given later in book. Chapter III gets technical.}}
+\end{proof}<|endoftext|>
+\section{Card Game with Bluffing is Completely Mixed Game}
+Tags: Examples of Two-Person Zero-Sum Games
+
+\begin{theorem}
+The [[Definition:Game|game]] of [[Definition:Card Game with Bluffing|cards with bluffing]] is a [[Definition:Completely Mixed Game|completely mixed game]].
+\end{theorem}
+
+\begin{proof}
+From [[Solution to Card Game with Bluffing]], the optimum [[Definition:Strategy|strategy]] for the [[Definition:Card Game with Bluffing|card game with bluffing, is:
+:For $A$, to use both [[Definition:Pure Strategy|pure strategies]] but favour $A_1$ by $2/3$ to $1/3$.
+:For $B$, to use both [[Definition:Pure Strategy|pure strategies]] but favour $B_1$ by $2/3$ to $1/3$.
+Hence the result by definition of [[Definition:Completely Mixed Game|completely mixed game]].
+\end{proof}<|endoftext|>
+\section{Matching Pennies is Completely Mixed Game}
+Tags: Examples of Two-Person Zero-Sum Games
+
+\begin{theorem}
+The [[Definition:Game|game]] of [[Definition:Matching Pennies|matching pennies]] is a [[Definition:Completely Mixed Game|completely mixed game]].
+\end{theorem}
+
+\begin{proof}
+The optimum [[Definition:Strategy|strategy]] for [[Definition:Matching Pennies|matching pennies]], for both [[Definition:Player|players]], is to randomise both their [[Definition:Pure Strategy|pure strategies]].
+Hence the result by definition of [[Definition:Completely Mixed Game|completely mixed game]].
+\end{proof}<|endoftext|>
+\section{Dominated Strategy may be Optimal}
+Tags: Game Theory
+
+\begin{theorem}
+A [[Definition:Dominating Strategy|dominated strategy]] of a [[Definition:Game|game]] may be the optimal [[Definition:Strategy|strategy]] for a [[Definition:Player|player]] of that [[Definition:Game|game]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Game|game]] defined by the following [[Definition:Payoff Table|payoff table]]:
+{{PayoffTable|table = $\begin{array} {r {{|}} c {{|}} }
+& B_1 & B_2 \\
+\hline A_1 & 1 & 2 \\
+\hline A_2 & 1 & 3  \\
+\hline \end{array}$}}
+This has two [[Definition:Solution of Game|solutions]]:
+:$(1): \quad A: \left({1, 0}\right), B: \left({1, 0}\right)$
+:$(2): \quad A: \left({0, 1}\right), B: \left({1, 0}\right)$
+{{explain|The reason why the above are solutions is not indicated.}}
+Thus both [[Definition:Pure Strategy|pure strategies]] for $A$ are optimal, but $A_1$ is [[Definition:Dominating Strategy|dominated]] by $A_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Value of Skew-Symmetric Game is Zero}
+Tags: Two-Person Zero-Sum Games
+
+\begin{theorem}
+Let $G$ be a [[Definition:Two-Person Zero-Sum Game|two-person zero-sum game]].
+Let $G$ be represented by a [[Definition:Payoff Table|payoff table]] that is [[Definition:Skew-Symmetric|skew-symmetric]].
+Then the [[Definition:Value|value]] of $G$ is zero.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|Once we have a definition for skew-symmetric we can start thinking about how to present this.}}
+\end{proof}<|endoftext|>
+\section{Prime Element iff Complement of Lower Closure is Filter}
+Tags: Prime Elements
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Bounded Above Set|bounded above]] [[Definition:Lattice|lattice]].
+Let $p \in S$ such that
+:$p \ne \top$
+where $\top$ denotes the [[Definition:Top (Lattice Theory)|top]] of $L$.
+Then:
+:$p$ is a [[Definition:Prime Element (Order Theory)|prime element]]
+{{iff}}
+:$\relcomp S {p^\preceq}$ is [[Definition:Filter in Ordered Set|filter]] in $L$
+where
+:$p^\preceq$ denotes the [[Definition:Lower Closure of Element|lower closure]] of $p$
+:$\relcomp S {p^\preceq}$ denotes the [[Definition:Relative Complement|relative complement]] of $p^\preceq$ relative to $S$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that $p$ is a [[Definition:Prime Element (Order Theory)|prime element]].
+By definition of the [[Definition:Greatest Element|greatest element]]:
+:$p \preceq \top$
+By definition of $\prec$:
+:$p \prec \top$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\top \npreceq p$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$\top \notin p^\preceq$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$\top \in \relcomp S {p^\preceq}$
+By definition
+:$X := \relcomp S {p^\preceq}$ is a [[Definition:Non-Empty Set|non-empty set]]
+We will prove that
+:$X$ is [[Definition:Filtered Subset|filtered]].
+Let $x, y \in X$.
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin p^\preceq$ and $y \notin p^\preceq$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$x \npreceq p$ and $y \npreceq p$
+Then by definition of [[Definition:Prime Element (Order Theory)|prime element]]:
+:$x \wedge y \npreceq p$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$x \wedge y \notin p^\preceq$
+Thus by definition of [[Definition:Relative Complement|relative complement]]:
+:$x \wedge y \in X$
+Thus by [[Meet Precedes Operands]]:
+:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
+Thus
+:$\exists z \in X: z \preceq x \land z \preceq y$
+{{qed|lemma}}
+We will prove that $X$ is an [[Definition:Upper Set|upper set]].
+Let $x \in X, y \in S$ such that:
+:$x \preceq y$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin p^\preceq$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$x \npreceq p$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$y \npreceq p$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$y \notin p^\preceq$
+Thud by definition of [[Definition:Relative Complement|relative complement]]:
+:$y \in X$
+{{qed|lemma}}
+Hence, by definition, $X$ is [[Definition:Filter in Ordered Set|filter]] in $L$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that $\relcomp S {p^\preceq}$ is [[Definition:Filter in Ordered Set|filter]] in $L$.
+Let $x, y \in S$ such that:
+:$x \wedge y \preceq p$
+{{AimForCont}} that:
+:$x \npreceq p$ and $y \npreceq p$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+$x \notin p^\preceq$ and $y \notin p^\preceq$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in \relcomp S {p^\preceq}$ and $y \in \relcomp S {p^\preceq}$
+By definition of [[Definition:Filtered Subset|filtered]]:
+:$\exists z \in \relcomp S {p^\preceq}: z \preceq x \text { and } z \preceq y$
+Then:
+$z \preceq x \wedge y$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$z \preceq p$
+By definition of [[Definition:Upper Set|upper set]]:
+:$p \in \relcomp S {p^\preceq}$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$p \preceq p$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$p \in p^\preceq$
+This [[Definition:Contradiction|contradicts]] $p \in \relcomp S {p^\preceq}$.
+Hence it is not the case that $x \npreceq p$ and $y \npreceq p$.
+That is:
+:$x \preceq p$ or $y \preceq p$
+and so, by definition, $p$ is a [[Definition:Prime Element (Order Theory)|prime element]] of $\struct {S, \vee, \wedge, \preceq}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood)}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
+Let $x$ be a [[Definition:Element|point]] in $D$.
+Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhoods]] of $x$.
+Let $E_x$ be the [[Definition:Set|set]] of [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhoods]] of $x$.
+Let $\map {\omega_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$ based on $N_x$:
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+where $\map {\omega_f} I$ is the [[Definition:Oscillation/Real Space/Oscillation on Set|oscillation]] of $f$ on $I$:
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\map {\omega^E_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$ based on $E_x$:
+:$\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$
+Then:
+:$\map {\omega_f} x \in \R$ {{iff}} $\map {\omega^E_f} x \in \R$
+and, if $\map {\omega_f} x$ and $\map {\omega^E_f} x$ exist as [[Definition:Real Number|real numbers]]:
+:$\map {\omega_f} x = \map {\omega^E_f} x$
+\end{theorem}
+
+\begin{proof}
+=== Lemma 1 ===
+Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
+Let $x$ be a [[Definition:Element|point]] in $D$.
+Let $N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhoods]] of $x$.
+Let $E_x$ be the [[Definition:Set|set]] of [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhoods]] of $x$.
+Let $\map {\omega_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$ based on $N_x$:
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+where $\map {\omega_f} I$ is the [[Definition:Oscillation/Real Space/Oscillation on Set|oscillation]] of $f$ on $I$:
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\map {\omega^E_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$ based on $E_x$:
+:$\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$
+Let $\map {\omega_f} x \in \R$.
+Let $\map {\omega^E_f} x \in \R$.
+Then $\map {\omega_f} x = \map {\omega^E_f} x$.
+=== Proof ===
+We have:
+:$\map {\omega_f} x \in \R$
+:$\map {\omega^E_f} x \in \R$
+We need to prove that:
+:$\map {\omega_f} x = \map {\omega^E_f} x$
+Let:
+:$N = \set {\map {\omega_f} I: I \in N_x}$
+:$E = \set {\map {\omega_f} I: I \in E_x}$
+:$\map {\omega_f} x = \inf N$
+:$\map {\omega^E_f} x = \inf E$
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+We have:
+:$\inf N \in \R$ as $\map {\omega_f} x \in \R$
+:$\inf E \in \R$ as $\map {\omega^E_f} x \in \R$
+Let:
+:$NR = N \cap \R$
+Let $I \in N_x$.
+Therefore, $x \in I$.
+[[Oscillation on Set is an Extended Real Number]] gives that $\map {\omega_f} I$ is an [[Definition:Extended Real Number Line|extended real number]].
+Therefore:
+:$N$ is a [[Definition:Set|set]] of [[Definition:Extended Real Number Line|extended real numbers]] as $N = \set {\map {\omega_f} I: I \in N_x}$.
+Also, we have that $N$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]] (in $\R$) as $\inf N \in \R$.
+This gives by [[Infimum of Real Subset]]:
+:$\inf NR \in \R$ as $\inf N \in \R$
+:$\inf NR = \inf N$ as $\inf NR \in \R$ and $\inf N \in \R$
+Let:
+:$ER = E \cap \R$
+We observe by the definitions of $E_x$ and $N_x$ that every $I$ in $E_x$ is also an [[Definition:Element|element]] of $N_x$.
+Therefore, $E_x$ is a [[Definition:Subset|subset]] of $N_x$.
+Accordingly:
+:$E$ is a [[Definition:Subset|subset]] of $N$ by the definitions of $E$ and $N$
+Therefore:
+:$ER$ is a [[Definition:Subset|subset]] of $NR$ by [[Set Intersection Preserves Subsets/Corollary]]
+We have that $N$ is a [[Definition:Set|set]] of [[Definition:Extended Real Number Line|extended real numbers]].
+Also, we have that $E$ is a [[Definition:Subset|subset]] of $N$.
+Therefore, $E$ is a [[Definition:Set|set]] of [[Definition:Extended Real Number Line|extended real numbers]].
+Also, we have that $E$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]] (in $\R$) as $\inf E \in \R$.
+This gives by [[Infimum of Real Subset]]:
+:$\inf ER \in \R$ as $\inf E \in \R$
+:$\inf ER = \inf E$ as $\inf ER \in \R$ and $\inf E \in \R$
+Assume that:
+:$s$ is a [[Definition:Real Number|real number]] in $N$
+Then an $I \in N_x$ exists such that:
+:$\map {\omega_f} I = s$ as $N = \set {\map {\omega_f} {I'}: I' \in N_x}$
+We have that $\map {\omega_f} I \in \R$ as $s \in \R$.
+The real set $I$ is an [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhood]] of $x$ as $I \in N_x$.
+This means that $I$ contains an [[Definition:Open Set/Real Analysis|open]] [[Definition:Subset|subset]] that contains (as an [[Definition:Element|element]]) $x$.
+Therefore, an $h \in \R_{>0}$ exists such that $\openint {x - h} {x + h}$ is a [[Definition:Subset|subset]] of $I$.
+Observe that $\openint {x - h} {x + h} \in N_x$.
+We have:
+:$I \in N_x$
+:$\map {\omega_f} I \in \R$
+:$\openint {x - h} {x + h} \in N_x$
+:$\openint {x - h} {x + h} \subset I$
+This gives by [[Oscillation on Subset]]:
+:$\map {\omega_f} {\openint {x - h} {x + h} } \in \R$
+:$\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$
+We have that $\openint {x - h} {x + h} \in E_x$ as $\openint {x - h} {x + h}$ is an [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhood]] of $x$.
+Therefore:
+:$\map {\omega_f} {\openint {x - h} {x + h} } \in E$ as $E = \set {\map {\omega_f} {I'}: I' \in E_x}$
+Let:
+:$t = \map {\omega_f} {\openint {x - h} {x + h} }$
+We have $t \in E$ as $\map {\omega_f} {\openint {x - h} {x + h} } \in E$.
+Also, $t \in \R$ as $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$.
+Accordingly:
+:$t \in ER$ as $ER = E \cap \R$
+We have $t = \map {\omega_f} {\openint {x - h} {x + h} }$ and $s = \map {\omega_f} I$.
+Therefore:
+:$t \le s$ as $\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$
+We have assumed that $s$ is a [[Definition:Real Number|real number]] in $N$.
+Therefore:
+:$s \in NR$ as $NR = N \cap \R$
+We have:
+{{begin-eqn}}
+{{eqn | l = \forall s \in NR: \exists t \in ER: s
+      | o = \ge
+      | r = t
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s
+      | o = \ge
+      | r = t \land \epsilon > 0
+      | c = as $\epsilon > 0$ is true since $\epsilon \in \R_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s
+      | o = \ge
+      | r = t \land s + \epsilon > s
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon
+      | o = >
+      | r = s \ge t
+}}
+{{eqn | ll= \leadsto
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon
+      | o = >
+      | r = t
+}}
+{{eqn | ll= \leadsto
+      | l = \inf NR
+      | r = \inf ER
+      | c = by [[Condition for Infimum of Subset to equal Infimum of Set]] as $ER$ is a [[Definition:Subset|subset]] of $NR$, $\inf NR \in \R$, and $\inf ER \in \R$
+}}
+{{eqn | ll= \leadsto
+      | l = \inf N
+      | r = \inf ER
+      | c = as $\inf N = \inf NR$
+}}
+{{eqn | ll= \leadsto
+      | l = \inf N
+      | r = \inf E
+      | c = as $\inf E = \inf ER$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\omega_f} x
+      | r = \map {\omega^E_f} x
+      | c = definitions
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Let:
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+:$\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\map {\omega_f} x \in \R$.
+We need to prove:
+:$\map {\omega^E_f} x \in \R$
+:$\map {\omega^E_f} x = \map {\omega_f} x$
+First, we intend to prove that $\map {\omega^E_f} x \in \R$.
+We have $\map {\omega_f} x \in \R$.
+Therefore, $\set {\map {\omega_f} I: I \in N_x}$ contains a [[Definition:Real Number|real number]] by [[Infimum of Set of Oscillations on Set]].
+Accordingly, an $I \in N_x$ exists such that $\map {\omega_f} I$ is a [[Definition:Real Number|real number]].
+$I$ is an [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhood]] of $x$ as $I \in N_x$.
+This means that $I$ contains an [[Definition:Open Set/Real Analysis|open]] [[Definition:Subset|subset]] that contains (as an [[Definition:Element|element]]) $x$.
+Therefore, an $h \in \R_{>0}$ exists such that $\openint {x - h} {x + h}$ is a [[Definition:Subset|subset]] of $I$.
+Observe that $\openint {x - h} {x + h} \in N_x$.
+We have:
+:$I \in N_x$
+:$\map {\omega_f} I \in \R$
+:$\openint {x - h} {x + h} \in N_x$
+:$\openint {x - h} {x + h} \subset I$
+This gives by [[Oscillation on Subset]]:
+:$\map {\omega_f} {\openint {x - h} {x + h} } \in \R$
+We observe that $\openint {x - h} {x + h}$ is an [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhood]] of $x$.
+Therefore, $\openint {x - h} {x + h} \in E_x$.
+Accordingly, $\map {\omega_f} {\openint {x - h} {x + h} } \in \set {\map {\omega_f} I: I \in E_x}$.
+Therefore, $\set {\map {\omega_f} I: I \in E_x}$ contains a [[Definition:Real Number|real number]] as $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$.
+From this follows that $\map {\omega^E_f} x \in \R$ by [[Infimum of Set of Oscillations on Set]].
+This is the first statement that we intended to prove.
+Next, we need to prove that $\map {\omega^E_f} x = \map {\omega_f} x$.
+This result follows by [[#Lemma 1|Lemma 1]] as $\map {\omega_f} x$ and $\map {\omega^E_f} x$ exist as [[Definition:Real Number|real numbers]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let:
+:$N = \set {\map {\omega_f} I: I \in N_x}$
+:$E = \set {\map {\omega_f} I: I \in E_x}$
+:$\map {\omega_f} x = \inf N$
+:$\map {\omega^E_f} x = \inf E$
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\map {\omega^E_f} x \in \R$.
+We need to prove:
+:$\map {\omega_f} x \in \R$
+:$\map {\omega_f} x = \map {\omega^E_f} x$
+First, we intend to prove that $\map {\omega_f} x \in \R$.
+We have $\map {\omega^E_f} x \in \R$.
+Therefore,  $E$ contains a [[Definition:Real Number|real number]] by [[Infimum of Set of Oscillations on Set]].
+We observe by the definitions of $E_x$ and $N_x$ that every $I$ in $E_x$ is also an [[Definition:Element|element]] of $N_x$.
+Therefore, $E_x$ is a [[Definition:Subset|subset]] of $N_x$.
+Accordingly:
+:$E$ is a [[Definition:Subset|subset]] of $N$ by the definitions of $E$ and $N$
+We have:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = E \text { contains a real number}
+}}
+{{eqn | o = \leadsto
+      | r = N \text { contains a real number}
+      | c = as $E$ is a [[Definition:Subset|subset]] of $N$
+}}
+{{eqn | o = \leadsto
+      | r = \map {\omega_f} x \in \R
+      | c = [[Infimum of Set of Oscillations on Set]]
+}}
+{{end-eqn}}
+This is the first statement that we intended to prove.
+Next, we need to prove that $\map {\omega_f} x = \map {\omega^E_f} x$.
+This result follows by [[#Lemma 1|Lemma 1]] as $\map {\omega^E_f} x$ and $\map {\omega_f} x$ exist as [[Definition:Real Number|real numbers]].
+{{qed}}
+[[Category:Real Analysis]]
+n8wxkt68gjl4mpocw12jqmyoddgztw7
+\end{proof}<|endoftext|>
+\section{Prime Element iff Element Greater is Top}
+Tags: Prime Elements
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Boolean Lattice|Boolean lattice]].
+Let $p \in S$ such that
+:$p \ne \top$
+Then
+:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
+{{iff}}
+:$\forall x \in S: \left({ p \prec x \implies x = \top }\right)$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Suppose that
+:$p$ is [[Definition:Prime Element (Order Theory)|prime element]]
+Let $x \in S$ such that
+:$p \prec x$
+By definition of [[Definition:Boolean Lattice|Boolean lattice]]:
+:$L$ is [[Definition:Complemented Lattice|complemented]] [[Definition:Distributive Lattice|distributive lattice]].
+By definition of [[Definition:Complemented Lattice|complemented]]:
+:$\exists y \in S: y$ is [[Definition:Complement (Lattice Theory)|complement]] of $x$.
+By definition of [[Definition:Complement (Lattice Theory)|complement]]
+:$x \wedge y = \bot$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$x \wedge y \preceq p$
+By definition of [[Definition:Prime Element (Order Theory)|prime element]]:
+:$x \preceq p$ or $y \preceq p$
+Then by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$y \prec x$
+By definition of $\prec$:
+:$y \preceq x$
+By definition of [[Definition:Complement (Lattice Theory)|complement]]:
+:$x \vee y = \top$
+Thus by [[Preceding iff Join equals Larger Operand]]:
+:$x = \top$
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that
+:$\forall x \in S: \left({ p \prec x \implies x = \top }\right)$
+Let $x, y \in S$ such that
+:$x \wedge y \preceq p$
+{{begin-eqn}}
+{{eqn | l = p
+| r = p \vee \left({x \wedge y}\right)
+| c = [[Preceding iff Join equals Larger Operand]]
+}}
+{{eqn | r = \left({p \vee x}\right) \wedge \left({p \vee y}\right)
+| c = definition of [[Definition:Distributive Lattice|distributive lattice]]
+}}
+{{end-eqn}}
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x \npreceq p$ and $y \npreceq p$
+By [[Preceding iff Join equals Larger Operand]]:
+:$p \ne p \vee y$ and $p \ne p \vee x$
+By [[Join Succeeds Operands]]:
+:$p \preceq p \vee y$ and $p \preceq p \vee x$
+By definition of $\prec$:
+:$p \prec p \vee y$ and $p \prec p \vee x$
+By assumption:
+:$p \vee y = \top$ and $p \vee x = \top$
+By [[Meet is Idempotent]]:
+:$p = \top$
+This contradicts $p \ne \top$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Element iff There Exists Way Below Open Filter which Complement has Maximum}
+Tags: Prime Elements
+
+\begin{theorem}
+Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Bounded Below Set|bounded below]] [[Definition:Continuous Ordered Set|continuous]]
+[[Definition:Distributive Lattice|distributive lattice]].
+Let $p \in S$ such that
+:$p \ne \top$
+where $\top$ denotes the [[Definition:Top (Lattice Theory)|top]] of $L$.
+Then
+:$p$ is a [[Definition:Prime Element (Order Theory)|prime element]]
+{{iff}}
+:there exists a [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] $F$ in $L$: $p = \max \left({ \complement_S\left({F}\right) }\right)$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $p$ be a [[Definition:Prime Element (Order Theory)|prime element]].
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]]
+We will prove that
+:$\forall x \in S: \left({ x \in \complement_S\left({p^\preceq}\right) \implies \exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x }\right)$
+Let $x \in S$ such that
+:$x \in \complement_S\left({p^\preceq}\right)$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin p^\preceq$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$x \npreceq p$
+By [[Axiom of Approximation in Up-Complete Semilattice]]:
+:$\exists y \in S: y \ll x \land y \npreceq p$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$y \notin p^\preceq$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$y \in \complement_S\left({p^\preceq}\right)$
+Thus
+:$\exists y \in S: y \in \complement_S\left({p^\preceq}\right) \land y \ll x$
+{{qed|lemma}}
+By [[Prime Element iff Complement of Lower Closure is Filter]]:
+:$F := \complement_S\left({p^\preceq}\right)$ is [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] in $L$.
+By definitions of [[Definition:Antisymmetric Relation|antisymmetry]] and [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\lnot \exists y \in S: y \in F \land y \prec p$
+By [[Relative Complement of Relative Complement]]::
+:$\complement_S\left({F}\right) = p^\preceq$
+By definitions of [[Definition:Reflexivity|reflexivity]] and [[Definition:Lower Closure/Element|lower closure of element]]:
+:$p \in \complement_S\left({F}\right)$
+Thus by definition of [[Definition:Greatest Element|greatest element]]:
+:$p = \max \left({\complement_S\left({F}\right)}\right)$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:there exists a [[Definition:Way Below Open|way below open]] [[Definition:Filter in Ordered Set|filter]] $F$ in $L$: $p = \max \left({ \complement_S\left({F}\right) }\right)$
+By [[Maximal Element of Complement of Filter is Meet Irreducible]]:
+:$p$ is [[Definition:Meet Irreducible|meet irreducible]].
+Thus by [[Prime Element iff Meet Irreducible in Distributive Lattice]]:
+:$p$ is a [[Definition:Prime Element (Order Theory)|prime element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Space has Unique Additive Identity}
+Tags: Linear Algebra, Vector Spaces, Vector Space has Unique Additive Identity
+
+\begin{theorem}
+A vector space has unique additive identity.
+\end{theorem}<|endoftext|>
+\section{Characterization of Prime Ideal}
+Tags: Prime Ideals (Order Theory)
+
+\begin{theorem}
+Let $L = \struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $I$ be an [[Definition:Proper Subset|proper]] [[Definition:Ideal (Order Theory)|ideal]] in $L$.
+Then
+:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]]
+{{iff}}
+:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that
+:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+Let $x, y \in S$ such that
+:$x \wedge y \in I$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \wedge y \notin \relcomp S I$
+By definition of [[Definition:Prime Ideal (Order Theory)|prime ideal]]:
+:$\relcomp S I$ is [[Definition:Filter|filter]] in $L$.
+By [[Filtered in Meet Semilattice]]:
+:$x \notin \relcomp S I$ or $y \notin \relcomp S I$
+Thus by definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in I$ or $y \in I$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall x, y \in S: \paren {x \wedge y \in I \implies x \in I \lor y \in I}$
+Define $F := \relcomp S I$.
+By definition of [[Definition:Proper Subset|proper subset]]:
+:$\exists x_0 \in S: x_0 \notin I$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x_0 \in F$
+Thus by definition
+:$F$ is a [[Definition:Non-Empty Set|non-empty set]].
+We will prove that
+:$F$ is [[Definition:Filtered Subset|filtered]].
+Let $x, y \in F$.
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin I$ and $y \notin I$
+By assumption:
+:$x \wedge y \notin I$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \wedge y \in F$
+By [[Meet Precedes Operands]]:
+:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
+Thus
+:$\exists z \in F: z \preceq x \land z \preceq y$
+{{qed|lemma}}
+We will prove that:
+:$F$ is an [[Definition:Upper Set|upper set]].
+Let $x \in F, y \in S$ such that
+:$x \preceq y$
+By definition of [[Definition:Lower Set|lower set]]:
+:$y \in I \implies x \in I$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin I$
+Then
+:$y \notin I$
+Thus by definition of [[Definition:Relative Complement|relative complement]]:
+:$y \in F$
+{{qed|lemma}}
+Thus by definition:
+:$F$ is a [[Definition:Filter in Ordered Set|filter]] in $L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Matching Pennies is Strictly Competitive}
+Tags: Examples of Two-Person Games
+
+\begin{theorem}
+The [[Definition:Game|game]] of [[Definition:Matching Pennies|Matching Pennies]] is [[Definition:Strictly Competitive Game|strictly competitive]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Strictly Competitive Game|strictly competitive game]] is a [[Definition:Game|game]] in which the interests of each [[Definition:Player|player]] are diametrically opposed.
+Recall the [[Definition:Matching Pennies/Payoff Table|payoff table]] of [[Definition:Matching Pennies|Matching Pennies]]:
+{{:Definition:Matching Pennies/Payoff Table}}
+It can be seen by inspection that exchanging $\text A$ and $\text B$ results in exactly the same [[Definition:Entry in Payoff Table|entries]] in the [[Definition:Matching Pennies/Payoff Table|payoff table]].
+From the nature of a [[Definition:Matching Pennies/Payoff Table|payoff table]] it follows that the fortunes of each [[Definition:Player|player]] are opposite and equal.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection Condition for Direct Sum of Subspaces}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $U$ and $W$ be subspaces of a vector space $V$.
+Then $U+W$ is a direct sum if and only if $U\cap W = 0$.
+\end{theorem}
+
+\begin{proof}
+We must first prove that if $U+W$ is a direct sum, then $U\cap W = 0$
+Suppose $U+W$ is a direct sum. If a vector $v \in U\cap W$, then $0 = v + (-v)$, where $v \in U$ and $-v \in W$. 
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Second Price Auction has Inefficient Equilibria}
+Tags: Auctions
+
+\begin{theorem}
+A [[Definition:Second Price Auction|second price auction]] has [[Definition:Nash Equilibrium|Nash equilibria]] which are [[Definition:Inefficient Nash Equilibrium|inefficient]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Awaiting a reliable definition of inefficient Nash equilibrium}}
+\end{proof}<|endoftext|>
+\section{Expected-Utility Maximization Theorem}
+Tags: Game Theory, Named Theorems
+
+\begin{theorem}
+Let $P$ be a [[Definition:Rational Decision-Maker|rational decision-maker]].
+Let $A$ be a [[Definition:Set|set]] of possible [[Definition:Move|moves]] available to $P$.
+Then there exists a way of assigning values to a [[Definition:Payoff Function|payoff function]] $u: A \to \R$
+so that $P$ will always choose the [[Definition:Move|move]] such that $u$ is a maximum.
+\end{theorem}<|endoftext|>
+\section{Epic Equalizer is Isomorphism}
+Tags: Category Theory
+
+\begin{theorem}
+Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
+Let $e: E \to C$ be the [[Definition:Equalizer|equalizer]] of two [[Definition:Morphism|morphisms]] $f, g: C \to D$.
+Let $e$ be an [[Definition:Epimorphism (Category Theory)|epimorphism]].
+Then $e$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+As $e$ [[Definition:Equalizer|equalises]] $f$ and $g$, $f \circ e = g \circ e$.
+Since $e$ is [[Definition:Epimorphism (Category Theory)|epic]], it follows that $f = g$.
+Then in the equaliser diagram: 
+:::$\begin{xy}\xymatrix{
+ E 
+  \ar[r]^*{e}
+&
+ A
+  \ar[r]<2pt>^*{f}
+  \ar[r]<-2pt>_*{g} 
+&
+ B
+\\
+ A
+  \ar@{.>}[u]^*{k}
+  \ar[ur]_*{\operatorname{id}_A}
+}\end{xy}$
+We have that:
+:$f \circ 1_A = g \circ 1_A$
+so there is a unique $k$ with $e \circ k = \operatorname{id}_A$.
+Then:
+{{begin-eqn}}
+{{eqn | l = e \circ k \circ e
+      | r = \operatorname {id}_A \circ e
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{eqn | r = e
+      | c = e \circ \operatorname{id}_E
+}}
+{{end-eqn}}
+By [[Equalizer is Monomorphism]], it follows that $k \circ e = \operatorname{id}_E$.
+This gives $k$ as an [[Definition:Inverse Morphism|inverse]] to $e$.
+Thus $e$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Subset of Extended Real Numbers is Arbitrarily Close}
+Tags: Extended Real Numbers
+
+\begin{theorem}
+Let $A \subseteq \overline \R$ be a [[Definition:Subset|subset]] of the [[Definition:Extended Real Number Line|extended real numbers]].
+Let $b$ be an [[Definition:Infimum of Subset of Real Numbers|infimum]] (in $\R$) of $A$.
+Let $\epsilon \in \R_{>0}$.
+Then:
+:$\exists x \in \paren {A \cap \R}: x - b < \epsilon$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$A$ is a a [[Definition:Set|set]] of [[Definition:Extended Real Number Line|extended real numbers]]
+:$A$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]] (in $\R$) as the [[Definition:Real Number|real number]] $b$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] for $A$.
+From this follows by [[Infimum of Real Subset]]:
+:$\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$
+Let $\epsilon \in \R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \inf {A \cap \R}
+      | o = \in
+      | r = \R
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x
+      | o = \in
+      | r = \paren {A \cap \R}: x - b < \epsilon
+      | c = [[Infimum of Subset of Real Numbers is Arbitrarily Close]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Extended Real Numbers]]
+35yrzle13fpml1vc5r1lhrs1egxujts
+\end{proof}<|endoftext|>
+\section{Oscillation on Set is an Extended Real Number}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
+Let $x$ be a [[Definition:Element|point]] in $D$.
+Let $I$ be a [[Definition:Real Number|real set]] that contains (as an [[Definition:Element|element]]) $x$.
+Let $\omega_f \left({I}\right)$ be the [[Definition:Oscillation/Real Space/Oscillation on Set|oscillation]] of $f$ on $I$:
+:$\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$
+Then:
+:$\omega_f \left({I}\right) \in \overline \R_{\ge 0}$
+and:
+:$\omega_f \left({I}\right) = \begin{cases}
+\text{a positive real number} & \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \text{is bounded above} \\
+\infty & \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \text{is not bounded above}
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert = 0$ for $y = z = x$.
+Therefore:
+:$0 \in \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ as $x \in I \cap D$
+Accordingly:
+:$\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Non-Empty Set|non-empty]]
+We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a [[Definition:Real Number|real set]] as $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.
+Every [[Definition:Real Number|real number]] is less than $\infty$.
+Therefore:
+:$\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] in $\overline \R$
+There are two cases: either $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] (in $\R$), or it is not.
+First, assume that:
+:$\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] (in $\R$)
+We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] and [[Definition:Non-Empty Set|non-empty]].
+Therefore:
+:$\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ exists as a [[Definition:Real Number|real number]] by the [[Continuum Property]]
+We know that $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is greater than or equal to every [[Definition:Element|element]] of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.
+Also $0 \in \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.
+Therefore, $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \ge 0$.
+We also have that $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ exists as a [[Definition:Real Number|real number]].
+Therefore, $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a [[Definition:Positive/Real Number|positive real number]].
+In other words:
+:$\omega_f \left({I}\right)$ is a [[Definition:Positive/Real Number|positive real number]] as $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$
+Next, assume that:
+:$\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is not [[Definition:Bounded Above Subset of Real Numbers|bounded above]] (in $\R$)
+We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] in $\overline \R$.
+Therefore, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ has $\infty$ as an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]].
+A number that is less than $\infty$ is a [[Definition:Real Number|real number]] or equal to $-\infty$.
+No [[Definition:Real Number|real number]] is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ as $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is not [[Definition:Bounded Above Subset of Real Numbers|bounded above]] in $\R$.
+$-\infty$ is not an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ because then every [[Definition:Real Number|real number]] would be an upper bound for $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.
+Therefore, $\infty$ is the only [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.
+Accordingly, $\infty$ is the [[Definition:Infimum of Subset of Real Numbers|least upper bound]] of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.
+In other words:
+:$\omega_f \left({I}\right) = \infty$ as $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$
+In either case, whether $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] or not:
+:$\omega_f \left({I}\right) \in \overline \R_{\ge 0}$.
+{{qed}}
+[[Category:Real Analysis]]
+6239qfs8yo8h8ss0iwvicjq6s56z8l2
+\end{proof}<|endoftext|>
+\section{Oscillation on Subset}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f: D \to \R$ be a [[Definition:Real Function|real function]] where $D \subseteq \R$.
+Let $x$ be a [[Definition:Element|point]] in $D$.
+Let $S_x$ be a [[Definition:Set|set]] of [[Definition:Real Number|real sets]] that contain (as an [[Definition:Element|element]]) $x$.
+Let $\map {\omega_f} I$ be the [[Definition:Oscillation on Real Subset|oscillation]] of $f$ on a [[Definition:Set|set]] $I$ in $S_x$:
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $I \in S_x$.
+Let $\map {\omega_f} I \in \R$.
+Let $J \in S_x$ be a [[Definition:Subset|subset]] of $I$.
+Then:
+:$\map {\omega_f} J \in \R$
+:$\map {\omega_f} J \le \map {\omega_f} I$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$I, J \in S_x$
+:$J \subset I$
+:$\map {\omega_f} I \in \R$
+where:
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+We need to prove:
+:$\map {\omega_f} J \in \R$
+:$\map {\omega_f} J \le \map {\omega_f} I$
+We intend to prove that $\map {\omega_f} J \in \R$.
+We start by proving that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] and [[Definition:Non-Empty Set|non-empty]].
+We have that $J$ is a [[Definition:Subset|subset]] of $I$.
+Therefore:
+:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a [[Definition:Subset|subset]] of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
+The statement $\map {\omega_f} I \in \R$ means that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]].
+Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]].
+Accordingly:
+:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]] as $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a [[Definition:Subset|subset]] of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.
+Therefore, $0 \in \set {\size {\map f y - \map f z}: y, z \in J \cap D}$ as $x \in J \cap D$.
+Accordingly:
+:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is [[Definition:Non-Empty Set|non-empty]]
+We have that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a [[Definition:Real Number|real set]] as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$.
+We have shown that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is [[Definition:Non-Empty Set|non-empty]] and [[Definition:Bounded Above Subset of Real Numbers|bounded above]].
+Therefore, $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]] by the [[Continuum Property]].
+In other words:
+{{begin-eqn}}
+{{eqn | l = \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}
+      | o = \in
+      | r = \R
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map {\omega_f} J
+      | o = \in
+      | r = \R
+      | c = definition
+}}
+{{end-eqn}}
+We finished proving that $\map {\omega_f} J \in \R$.
+It remains to prove that $\map {\omega_f} J \le \map {\omega_f} I$.
+We have:
+:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a [[Definition:Subset|subset]] of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]]
+:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a [[Definition:Supremum of Subset of Real Numbers|supremum]]
+Then:
+{{begin-eqn}}
+{{eqn | l = \sup \set {\size {\map f y - \map f z}: y, z \in J \cap D}
+      | o = \le
+      | r = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}
+      | c = [[Supremum of Subset]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map {\omega_f} J
+      | o = \le
+      | r = \map {\omega_f} I
+      | c = definitions
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Real Analysis]]
+gnz0ox5scu4o4yjgcwbre4uui50klar
+\end{proof}<|endoftext|>
+\section{Generating Fraction for Lucas Numbers}
+Tags: Lucas Numbers
+
+\begin{theorem}
+The [[Definition:Fraction|fraction]]:
+:$\dfrac {199} {9899}$
+has a [[Definition:Decimal Expansion|decimal expansion]] which contains within it the start of the [[Definition:Lucas Number|Lucas sequence]]:
+:$0 \cdotp 02010 \, 30407 \, 11 \ldots$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Long Division|long division]]:
+<pre>
+         0.0201030407111829
+     ----------------------
+9899 ) 199.0000000000000000
+       197 98
+       --- --
+         1 0200
+           9899
+         - ----
+            30100
+            29697
+            -----
+              40300
+              39596
+              -----
+                70400
+                69293
+                -----
+                 11070
+                  9899
+                 -----
+                  11710
+                   9899
+                  -----
+                   18110
+                    9899
+                   -----
+                    82110
+                    79192
+                    -----
+                     29180
+                     19798
+                     -----
+                      93820
+                      89091
+                      -----
+                       4729
+</pre>
+By [[Generating Function for Lucas Numbers]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop \ge 0} L_k 10^{-2 k - 2}
+      | r = \frac {2 - 10^{-2} } {1 - 10^{-2} - 10^{-4} } \times 10^{-2}
+}}
+{{eqn | r = \frac {2 \times 10^2 - 1} {10^4 - 10^2 - 1}
+}}
+{{eqn | r = \frac {199} {9899}
+}}
+{{end-eqn}}
+The first few terms are contained in the decimal expansion, as long as $L_{k + 1} < 100$, where there is no carry.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transcendental Numbers are Uncountable}
+Tags: Transcendental Numbers
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Transcendental Number|transcendental]] [[Definition:Real Number|real numbers]] is [[Definition:Uncountable Set|uncountable]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Transcendental Number|transcendental number]] (in this context) is a [[Definition:Real Number|real number]] which is not an [[Definition:Algebraic Number|algebraic number]].
+Recall that the [[Real Numbers are Uncountable]].
+Also recall that the [[Algebraic Numbers are Countable]].
+The result follows from [[Uncountable Set less Countable Set is Uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Almost All Real Numbers are Transcendental}
+Tags: Transcendental Numbers
+
+\begin{theorem}
+[[Definition:Almost All/Set Theory/Uncountable|Almost all]] [[Definition:Real Number|real numbers]] are [[Definition:Transcendental Number|transcendental]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Transcendental Number|transcendental number]] (in this context) is a [[Definition:Real Number|real number]] which is not an [[Definition:Algebraic Number|algebraic number]].
+Recall that the [[Real Numbers are Uncountable]].
+Also recall that the [[Algebraic Numbers are Countable]].
+Thus the [[Definition:Subset|subset]] of all [[Definition:Real Number|real numbers]] which are not [[Definition:Transcendental Number|transcendental]] is [[Definition:Countable Set|countable]].
+The result follows by the definition of '''[[Definition:Almost All/Set Theory/Uncountable|almost all]]'''.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Champernowne Constant is Transcendental}
+Tags: Transcendental Numbers
+
+\begin{theorem}
+The [[Definition:Champernowne Constant|Champernowne constant]]:
+:$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
+is [[Definition:Transcendental|transcendental]].
+\end{theorem}<|endoftext|>
+\section{Champernowne Constant is Normal}
+Tags: Normal Numbers
+
+\begin{theorem}
+The [[Definition:Champernowne Constant|Champernowne constant]]:
+:$0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
+is [[Definition:Normal Real Number|normal]] with respect to [[Definition:Number Base|base $10$]].
+\end{theorem}<|endoftext|>
+\section{Anomalous Cancellation on 2-Digit Numbers}
+Tags: Anomalous Cancellation
+
+\begin{theorem}
+There are exactly four [[Definition:Anomalous Cancellation|anomalously cancelling]] [[Definition:Vulgar Fraction|vulgar fractions]] having two-digit [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] when expressed in [[Definition:Decimal Notation|base $10$ notation]]:
+{{:Anomalous Cancellation on 2-Digit Numbers/Examples}}
+\end{theorem}
+
+\begin{proof}
+Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an [[Definition:Anomalous Cancellation|anomalously cancelling]] [[Definition:Vulgar Fraction|vulgar fraction]].
+Then we have:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = a, b, x \le 9
+      | o =
+      | c = as they are digits in [[Definition:Decimal Notation|base $10$]]
+}}
+{{eqn | n = 2
+      | l = a < b
+      | o =
+      | c = this fraction is a [[Definition:Vulgar Fraction|vulgar fraction]]
+}}
+{{eqn | n = 3
+      | l = \dfrac a b = \dfrac {\sqbrk {a x} } {\sqbrk {x b} }
+      | o =
+      | c = this fraction is [[Definition:Anomalous Cancellation|anomalously cancelling]]
+}}
+{{end-eqn}}
+From $(3)$:
+{{begin-eqn}}
+{{eqn | l = \dfrac a b
+      | r = \dfrac {10 a + x} {10 x + b}
+      | c = 
+}}
+{{eqn | ll = \leadsto
+      | l = a \paren {10 x + b}
+      | r = b \paren {10 a + x}
+      | c = 
+}}
+{{eqn | ll = \leadsto
+      | l = 10 a x + a b
+      | r = 10 a b + b x
+      | c = 
+}}
+{{eqn | ll = \leadsto
+      | l = 10 a x - b x
+      | r = 9 a b
+      | c = 
+}}
+{{eqn | ll = \leadsto
+      | l = x
+      | r = \dfrac {9 a b} {10 a - b}
+      | c = 
+}}
+{{end-eqn}}
+Therefore $\dfrac {9 a b} {10 a - b}$ must be an [[Definition:Integer|integer]].
+{{AimForCont}} $10 a - b$ is [[Definition:Divisor of Integer|divisible]] by a [[Definition:Prime Number|prime]] $p$ such that $p > 9$.
+We have that:
+:$\paren {10 a - b} \divides 9 a b$
+and so:
+:$p \divides 9 a b$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+By [[Euclid's Lemma for Prime Divisors]], that means one of the [[Definition:Element|elements]] of $\set {9, a, b}$ must be [[Definition:Divisor of Integer|divisible]] by $p$.
+Hence from [[Absolute Value of Integer is not less than Divisors]], either $a$ or $b$ must be greater than $9$.
+This [[Definition:Contradiction|contradicts]] $(1)$ above.
+Hence by [[Proof by Contradiction]] it cannot be the case that $10 a - b$ is [[Definition:Divisor of Integer|divisible]] by a [[Definition:Prime Number|prime]] greater than $9$.
+We continue:
+{{begin-eqn}}
+{{eqn | l = \dfrac {9 a b} {10 a - b}
+      | o = \le
+      | r = 9
+      | c = from $(1)$: $x \le 9$
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac {a b} {10 a - b}
+      | o = \le
+      | r = 1
+      | c =
+}}
+{{eqn | ll = \leadsto
+      | l = a b
+      | o = \le
+      | r = 10 a - b
+      | c = as $10 a - b \ge 10 - b > 0$
+}}
+{{eqn | ll = \leadsto
+      | l = b
+      | o = \le
+      | r = 10 a - a b
+      | c = adding $b - a b$ to both sides
+}}
+{{eqn | ll = \leadsto
+      | l = a
+      | o = \ge
+      | r = \dfrac b {10 - b} 
+      | c = as $10 - b > 0$
+}}
+{{end-eqn}}
+It remains for us to check whether $x = \dfrac {9 a b} {10 a - b}$ is an [[Definition:Integer|integer]] for $\dfrac b {10 - b} \le a < b$, when $b$ ranges from $2$ to $9$.
+We note first that the following [[Definition:Ordered Pair|pairs]] $\tuple {a, b}$ are such that $\dfrac b {10 - b} > a$:
+{{begin-eqn}}
+{{eqn | lo= b = 6:
+      | l = \dfrac b {10 - b}
+      | r = \dfrac 3 2 > 1
+      | c = which eliminates $\tuple {1, 6}$
+}}
+{{eqn | lo= b = 7:
+      | l = \dfrac b {10 - b}
+      | r = \dfrac 7 3 > 2
+      | c = which eliminates $\tuple {1, 7}$ and $\tuple {2, 7}$
+}}
+{{eqn | lo= b = 8:
+      | l = \dfrac b {10 - b}
+      | r = 4 > 3
+      | c = which eliminates $\tuple {1, 8}$, $\tuple {2, 8}$ and $\tuple {3, 8}$
+}}
+{{eqn | lo= b = 9:
+      | l = \dfrac b {10 - b}
+      | r = 9
+      | c = which eliminates all possible such solutions such that $b = 9$
+}}
+{{end-eqn}}
+Then we eliminate the [[Definition:Ordered Pair|pairs]] $\tuple {a, b}$ where $10 a - b$ has a [[Definition:Prime Factor|prime factor]] greater than $9$:
+{{begin-eqn}}
+{{eqn | lo= \tuple {3, 4}:
+      | l = 10 a - b
+      | r = 26 = 13 \times 2
+      | c = 
+}}
+{{eqn | lo= \tuple {4, 6}:
+      | l = 10 a - b
+      | r = 34 = 17 \times 2
+      | c = 
+}}
+{{eqn | lo= \tuple {5, 6}:
+      | l = 10 a - b
+      | r = 44 = 11 \times 4
+      | c = 
+}}
+{{eqn | lo= \tuple {3, 7}:
+      | l = 10 a - b
+      | r = 23
+      | c = 
+}}
+{{eqn | lo= \tuple {4, 7}:
+      | l = 10 a - b
+      | r = 33 = 11 \times 3
+      | c = 
+}}
+{{eqn | lo= \tuple {5, 7}:
+      | l = 10 a - b
+      | r = 43
+      | c = 
+}}
+{{eqn | lo= \tuple {6, 7}:
+      | l = 10 a - b
+      | r = 53
+      | c = 
+}}
+{{eqn | lo= \tuple {6, 8}:
+      | l = 10 a - b
+      | r = 52 = 13 \times 4
+      | c = 
+}}
+{{eqn | lo= \tuple {7, 8}:
+      | l = 10 a - b
+      | r = 62 = 31 \times 2
+      | c = 
+}}
+{{end-eqn}}
+Thus it remains to check whether $\dfrac {9 a b} {10 a - b}$ is an [[Definition:Integer|integer]] for the following [[Definition:Ordered Pair|pairs]] $\tuple {a, b}$:
+:$\tuple {1, 2}$, $\tuple {1, 3}$, $\tuple {2, 3}$, $\tuple {1, 4}$, $\tuple {2, 4}$, $\tuple {1, 5}$, $\tuple {2, 5}$, $\tuple {3, 5}$, $\tuple {4, 5}$, $\tuple {2, 6}$, $\tuple {3, 6}$, $\tuple {4, 8}$, $\tuple {5, 8}$
+{{begin-eqn}}
+{{eqn | lo= \tuple {1, 2}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {18} 8
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {1, 3}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {27} 7
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {2, 3}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {54} {16}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {1, 4}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {36} 6 = 6
+      | c = which leads to the solution $\dfrac {16} {64}$
+}}
+{{eqn | lo= \tuple {2, 4}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {72} {16}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {1, 5}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {45} 5 = 9
+      | c = which leads to the solution $\dfrac {19} {95}$
+}}
+{{eqn | lo= \tuple {2, 5}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {90} {15} = 6
+      | c = which leads to the solution $\dfrac {26} {65}$
+}}
+{{eqn | lo= \tuple {3, 5}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {135} {25}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {4, 5}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {180} {35}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {2, 6}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {108} {14}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {3, 6}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {108} {14}
+      | c = not an integer
+}}
+{{eqn | lo= \tuple {4, 8}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {288} {32} = 9
+      | c = which leads to the solution $\dfrac {49} {98}$
+}}
+{{eqn | lo= \tuple {5, 8}:
+      | l = \dfrac {9 a b} {10 a - b}
+      | r = \dfrac {360} {42}
+      | c = not an integer
+}}
+{{end-eqn}}
+So the only solutions for $\tuple {a, x, b}$ are:
+:$\tuple {1, 6, 4}$, $\tuple {1, 9, 5}$, $\tuple {2, 6, 5}$ and $\tuple {4, 9, 8}$
+These correspond to the [[Definition:Fraction|fractions]]:
+:$\dfrac {16} {64}$, $\dfrac {19} {95}$, $\dfrac {26} {65}$ and $\dfrac {49} {98}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Anomalous Cancellation/Variants/37 + 13 over 37 + 24}
+Tags: Anomalous Cancellation
+
+\begin{theorem}
+:$\dfrac {37 + 13} {37 + 24} = \dfrac {37^3 + 13^3} {37^3 + 24^3}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 37^3 + 13^3
+      | r = \paren {37 + 13}^3 - \paren {3 \times 37^2 \times 13 + 3 \times 37 \times 13^2}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 13}^3 - 3 \times 13 \times 37 \paren {37 + 13}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 13} \paren {\paren {37 + 13}^2 - 3 \times 13 \times 37}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 13} \paren {2500 - 1443}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 13} \times 1057
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 37^3 + 24^3
+      | r = \paren {37 + 24}^3 - \paren {3 \times 37^2 \times 24 + 3 \times 37 \times 24^2}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 24}^3 - 3 \times 24 \times 37 \paren {37 + 24}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 24} \paren {\paren {37 + 24}^2 - 3 \times 24 \times 37}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 24} \paren {3721 - 2664}
+      | c = 
+}}
+{{eqn | r = \paren {37 + 24} \times 1057
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\dfrac {37^3 + 13^3} {37^3 + 24^3} = \dfrac {1057 \paren {37 + 13} } {1057 \paren {37 + 24} } = \dfrac {37 + 13} {37 + 24}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39}
+Tags: Anomalous Cancellation
+
+\begin{theorem}
+:$\dfrac {3 + 25 + 38} {7 + 20 + 39} = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}
+      | r = \dfrac {81 + 390625 + 2085136} {2401 + 160000 + 2313441}
+      | c = 
+}}
+{{eqn | r = \dfrac {2475842} {2475842}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \dfrac {3 + 25 + 38} {7 + 20 + 39}
+      | r = \dfrac {66} {66}
+      | c = 
+}}
+{{eqn | r = \dfrac {2475842} {2475842}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Anomalous Cancellation/Examples/3544 over 7531}
+Tags: Anomalous Cancellation
+
+\begin{theorem}
+:$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \frac {355 \cdots 544} {755 \cdots 531}
+      | r = \paren {3 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {\sum_{i \mathop = 2}^{n - 1} 5 \times 10^i} + 31}
+      | c = 
+}}
+{{eqn | r = \paren {3 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \sum_{i \mathop = 0}^{n - 3} \times 10^i} + 31}
+      | c = 
+}}
+{{eqn | r = \paren {3 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 44} \Big / \paren {7 \times 10^n + \paren {500 \times \frac {10^{n - 2} - 1} {10 - 1} } + 31}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = \dfrac {3 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 44 \times \paren {10 - 1} } {7 \times 10^n \times \paren {10 - 1} + 500 \times \paren {10^{n - 2} - 1} + 31 \times \paren {10 - 1} }
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {27 \times 10^n + 500 \times 10^{n - 2} - 104} {63 \times 10^n + 500 \times 10^{n - 2} - 221}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {2700 \times 10^{n - 2} + 500 \times 10^{n - 2} - 104} {6300 \times 10^{n - 2} + 500 \times 10^{n - 2} - 221}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {3200 \times 10^{n - 2} - 104} {6800 \times 10^{n - 2} - 221}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {8 \times \paren {400 \times 10^{n - 2} - 13} } {17 \times \paren {400 \times 10^{n - 2} - 13} }
+      | c = factoring
+}}
+{{eqn | r = \dfrac 8 {17}
+      | c = factoring
+}}
+{{eqn | r = \dfrac {8 \times 43} {17 \times 43}
+      | c = 
+}}
+{{eqn | r = \dfrac {344} {731}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Anomalous Cancellation/Examples/143 185 over 17 018 560}
+Tags: Anomalous Cancellation
+
+\begin{theorem}
+:$\dfrac {1435} {170 \, 560} = \dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {14 \, 318 \, 185} {1 \, 701 \, 818 \, 560} = \cdots$
+\end{theorem}
+
+\begin{proof}
+Let $q = \dfrac r s = \dfrac {1431818 \cdots 185} {1701818 \cdots 18560}$.
+Let the number of [[Definition:Digit|digits]] in $r$ be $n + 2$.
+Then the number of [[Definition:Digit|digits]] in $s$ is $n + 4$.
+By inspection, it is seen that $n$ is [[Definition:Even Integer|even]].
+Then it is seen that:
+{{begin-eqn}}
+{{eqn | l = r
+      | r = 5 + 18 \times 10 + 18 \times 1000 + \cdots + 18 \times 10^{n - 5} + 18 \times 10^{n - 3} + 143 \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 + 10 \times \paren {18 + 18 \times 10^2 + \cdots + 18 \times 10^{n - 6} + 18 \times 10^{n - 4} } + 143 \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 + 10 \times \paren {18 \times \paren{10^2}^0 + 18 \times \paren{10^2}^1 + \cdots + 18 \times \paren{10^2}^{n / 2 - 3} + 18 \times \paren{10^2}^{n / 2 - 2} } + 143 \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 + 10 \paren {\sum_{i \mathop = 0}^{n / 2 - 2} 18 \times 10^{2 i} } + 143 \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 + 10 \times 18 \paren {\frac {10^{n - 2} - 1} {10^2 - 1} } + 143 \times 10^{n - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = 5 \times \paren {10^2 - 1} + 10 \times 18 \paren {10^{n - 2} - 1} + 143 \times \paren {10^2 - 1} \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 \times 99 + 180 \times 10^{n - 2} - 180 + 143 \times 99 \times 10^{n - 1}
+      | c = 
+}}
+{{eqn | r = 5 \times 99 + 180 \times 10^{n - 2} - 180 + 143 \times 99 \times 10 \times 10^{n - 2}
+      | c = 
+}}
+{{eqn | r = 315 + 141750 \times 10^{n - 2}
+      | c = 
+}}
+{{eqn | r = 35 \times \paren {9 + 4050 \times 10^{n - 2} }
+      | c = 
+}}
+{{end-eqn}}
+and similarly:
+{{begin-eqn}}
+{{eqn | l = s
+      | r = 560 + 18 \times 1000 + 18 \times 100 \, 000 + \cdots + 18 \times 10^{3 + \paren {n - 3} } + 18 \times 10^{3 + \paren {n - 1} } + 170 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 + 1000 \times \paren {18 + 18 \times 10^2 + \cdots + 18 \times 10^{n - 6} + 18 \times 10^{n - 4} } + 170 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 + 1000 \times \paren {18 \times \paren {10^2}^0 + 18 \times \paren {10^2}^1 + \cdots + 18 \times \paren {10^2}^{n / 2 - 3} + 18 \times \paren {10^2}^{n / 2 - 2} } + 170 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 + 1000 \paren {\sum_{i \mathop = 0}^{n / 2 - 2} 18 \times 10^{2 i} } + 170 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 + 1000 \times 18 \paren {\frac {10^{n - 2} - 1} {10^2 - 1} } + 170 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 \times \paren {10^2 - 1} + 1000 \times 18 \paren {10^{n - 2} - 1} + 170 \times \paren {10^2 - 1} \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 560 \times 99 + 18000 \times 10^{n - 2} - 18000 + 170 \times 99 \times 10^{n + 1}
+      | c = 
+}}
+{{eqn | r = 55440 + 18000 \times 10^{n - 2} - 18000 + 16830 \times 1000 \times 10^{n - 2}
+      | c = 
+}}
+{{eqn | r = 37440 + 16848000 \times 10^{n - 2}
+      | c = 
+}}
+{{eqn | r = 4160 \times \paren {9 + 4050 \times 10^{n - 2} }
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = q = \dfrac r s
+      | r = \dfrac {35 \times \paren {9 + 4050 \times 10^{n - 2} } } {4160 \times \paren {9 + 4050 \times 10^{n - 2} } }
+      | c = 
+}}
+{{eqn | r = \dfrac {35 \times 41} {4160 \times 41}
+      | c = 
+}}
+{{eqn | r = \dfrac {1435} {170 \, 560}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Digits in Power of 2}
+Tags: Logarithms, Powers of 2, Number of Digits in Power of 2
+
+\begin{theorem}
+Let $n$ be a [[Definition:Positive Integer|positive integer]].
+Expressed in conventional [[Definition:Decimal Notation|decimal notation]], the number of [[Definition:Digit|digits]] in the [[Definition:Integer Power|$n$th power]] of $2$:
+:$2^n$
+is equal to:
+:$\ceiling {n \log_{10} 2}$
+where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $2^n$ have $m$ [[Definition:Digit|digits]] when expressed in [[Definition:Decimal Notation|decimal notation]].
+By the [[Basis Representation Theorem]] and its implications, a [[Definition:Positive Integer|positive integer]] $x$ has $m$ [[Definition:Digit|digits]] {{iff}}:
+:$10^{m - 1} \le x < 10^m$
+Thus:
+{{begin-eqn}}
+{{eqn | l = 10^{m - 1}
+      | o = \le
+      | m = 2^n
+      | mo= <
+      | r = 10^m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m - 1
+      | o = \le
+      | m = \log_{10} \paren {2^n}
+      | mo= <
+      | r = m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m - 1
+      | o = \le
+      | m = n \log_{10} 2
+      | mo= <
+      | r = m
+      | c = 
+}}
+{{end-eqn}}
+Because a [[Definition:Integer Power|power]] of $2$ cannot equal a [[Definition:Integer Power|power]] of $10$, it will always be the case that:
+:$m - 1 < n \log_{10} 2 < m$
+and so:
+:$m - 1 < n \log_{10} 2 \le m$
+Hence from [[Integer equals Ceiling iff Number between Integer and One Less]]:
+:$m = \ceiling {n \log_{10} 2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{One Third as Quotient of Sequences of Odd Numbers}
+Tags: One Third
+
+\begin{theorem}
+:$\dfrac 1 3 = \dfrac {1 + 3} {5 + 7} = \dfrac {1 + 3 + 5} {7 + 9 + 11} = \dfrac {1 + 3 + 5 + 7} {9 + 11 + 13 + 15} = \cdots$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = n + 1}^{2 n} \paren {2 k - 1}
+      | r = \sum_{k \mathop = 1}^{2 n} \paren {2 k - 1} - \sum_{k \mathop = 1}^n \paren {2 k - 1}
+      | c = 
+}}
+{{eqn | r = \paren {2 n}^2 - n^2
+      | c = [[Odd Number Theorem]]
+}}
+{{eqn | r = 3 n^2
+      | c = 
+}}
+{{eqn | r = 3 \sum_{k \mathop = 1}^n \paren {2 k - 1}
+      | c = [[Odd Number Theorem]]
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Number as Sum of Egyptian Fractions}
+Tags: Euler's Number
+
+\begin{theorem}
+The [[Reciprocal/Examples/Euler's Number|reciprocal of Euler's number $e$]] can be approximated by the following sequence of [[Definition:Egyptian Fraction|Egyptian fractions]]:
+:$\dfrac 1 e = \dfrac 1 3 + \dfrac 1 {29} + \dfrac 1 {15 \, 786} + \dfrac 1 {513 \, 429 \, 610} + \cdots$
+{{OEIS|A006526}}
+\end{theorem}
+
+\begin{proof}
+We have by definition of the [[Reciprocal/Examples/Euler's Number|reciprocal of $e$]] that:
+:$\dfrac 1 e \approx 0 \cdotp 36787 \, 94411 \, 71442 \, 32159 \, 55237 \, 70161 \, 46086 \, 74458 \, 11131 \, 031 \ldots$
+By inspection:
+:$\dfrac 1 3 < \dfrac 1 e < \dfrac 1 2$
+Thus:
+:$\dfrac 1 e - \dfrac 1 3 \approx 0 \cdotp 03454 \, 61078 \, 38109 \, 08826 \, 21904 \, 36828 \, 12753 \, 41124 \, 77797 \, 70 \ldots$
+Then:
+:$\dfrac 1 {29} = 0 \cdotp 03448 \, 27586 \, 20689 \, 65517 \, 24137 \, 931 \ldots$ repeating
+and:
+:$\dfrac 1 {28} = 0 \cdotp 03571 \, 42857 \, 14285 \ldots$ repeating
+and so:
+:$\dfrac 1 e - \dfrac 1 3 - \dfrac 1 {29} \approx 0 \cdotp 00006 \, 33482 \, 17499 \, 43308 \, 97766 \, 33724 \, 67925 \, 82504 \, 08832 \, 19 \cdots$
+ 
+Thus one can generate this [[Definition:Integer Sequence|sequence]] of [[Definition:Denominator|denominators]] $\sequence {D_n}$ by:
+:$D_n = \ceiling {\paren {\dfrac 1 e - \displaystyle \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$
+\end{proof}<|endoftext|>
+\section{One Half as Pandigital Fraction}
+Tags: One Half, Pandigital Fractions
+
+\begin{theorem}
+There are $12$ ways $\dfrac 1 2$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 2 = \dfrac {6729} {13 \, 458}$
+:$\dfrac 1 2 = \dfrac {6792} {13 \, 584}$
+:$\dfrac 1 2 = \dfrac {6927} {13 \, 854}$
+:$\dfrac 1 2 = \dfrac {7269} {14 \, 538}$
+:$\dfrac 1 2 = \dfrac {7293} {14 \, 586}$
+:$\dfrac 1 2 = \dfrac {7329} {14 \, 658}$
+:$\dfrac 1 2 = \dfrac {7692} {15 \, 384}$
+:$\dfrac 1 2 = \dfrac {7923} {15 \, 846}$
+:$\dfrac 1 2 = \dfrac {7932} {15 \, 864}$
+:$\dfrac 1 2 = \dfrac {9267} {18 \, 534}$
+:$\dfrac 1 2 = \dfrac {9273} {18 \, 546}$
+:$\dfrac 1 2 = \dfrac {9327} {18 \, 654}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Seventh as Pandigital Fraction}
+Tags: One Seventh, Pandigital Fractions
+
+\begin{theorem}
+There are $7$ ways $\dfrac 1 7$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 2 = \dfrac {2394} {16 \, 758}$
+:$\dfrac 1 2 = \dfrac {2637} {18 \, 459}$
+:$\dfrac 1 2 = \dfrac {4527} {31 \, 689}$
+:$\dfrac 1 2 = \dfrac {5274} {36 \, 918}$
+:$\dfrac 1 2 = \dfrac {5418} {37 \, 926}$
+:$\dfrac 1 2 = \dfrac {5976} {41 \, 832}$
+:$\dfrac 1 2 = \dfrac {7614} {53 \, 298}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Third as Pandigital Fraction}
+Tags: One Third, Pandigital Fractions
+
+\begin{theorem}
+There are $2$ ways $\dfrac 1 3$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 3 = \dfrac {5823} {17469}$
+:$\dfrac 1 3 = \dfrac {5832} {17496}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Quarter as Pandigital Fraction}
+Tags: One Quarter, Pandigital Fractions
+
+\begin{theorem}
+There are $4$ ways $\dfrac 1 4$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 4 = \dfrac {3942} {15768}$
+:$\dfrac 1 4 = \dfrac {4392} {17568}$
+:$\dfrac 1 4 = \dfrac {5796} {23184}$
+:$\dfrac 1 4 = \dfrac {7956} {31824}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Fifth as Pandigital Fraction}
+Tags: One Fifth, Pandigital Fractions
+
+\begin{theorem}
+There are $12$ ways $\dfrac 1 5$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 5 = \dfrac {2697} {13485}$
+:$\dfrac 1 5 = \dfrac {2769} {13845}$
+:$\dfrac 1 5 = \dfrac {2937} {14685}$
+:$\dfrac 1 5 = \dfrac {2967} {14835}$
+:$\dfrac 1 5 = \dfrac {2973} {14865}$
+:$\dfrac 1 5 = \dfrac {3297} {16485}$
+:$\dfrac 1 5 = \dfrac {3729} {18645}$
+:$\dfrac 1 5 = \dfrac {6297} {31485}$
+:$\dfrac 1 5 = \dfrac {7629} {38145}$
+:$\dfrac 1 5 = \dfrac {9237} {46185}$
+:$\dfrac 1 5 = \dfrac {9627} {48135}$
+:$\dfrac 1 5 = \dfrac {9723} {48615}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Sixth as Pandigital Fraction}
+Tags: One Sixth, Pandigital Fractions
+
+\begin{theorem}
+There are $3$ ways $\dfrac 1 6$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 6 = \dfrac {2943} {17658}$
+:$\dfrac 1 6 = \dfrac {4653} {27918}$
+:$\dfrac 1 6 = \dfrac {5697} {34182}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Eighth as Pandigital Fraction}
+Tags: One Eighth, Pandigital Fractions
+
+\begin{theorem}
+There are $46$ ways $\dfrac 1 8$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 8 = \dfrac {3187} {25496}$
+:$\dfrac 1 8 = \dfrac {4589} {36712}$
+:$\dfrac 1 8 = \dfrac {4591} {36728}$
+:$\dfrac 1 8 = \dfrac {4689} {37512}$
+:$\dfrac 1 8 = \dfrac {4691} {37528}$
+:$\dfrac 1 8 = \dfrac {4769} {38152}$
+:$\dfrac 1 8 = \dfrac {5237} {41896}$
+:$\dfrac 1 8 = \dfrac {5371} {42968}$
+:$\dfrac 1 8 = \dfrac {5789} {46312}$
+:$\dfrac 1 8 = \dfrac {5791} {46328}$
+:$\dfrac 1 8 = \dfrac {5839} {46712}$
+:$\dfrac 1 8 = \dfrac {5892} {47136}$
+:$\dfrac 1 8 = \dfrac {5916} {47328}$
+:$\dfrac 1 8 = \dfrac {5921} {47368}$
+:$\dfrac 1 8 = \dfrac {6479} {51832}$
+:$\dfrac 1 8 = \dfrac {6741} {53928}$
+:$\dfrac 1 8 = \dfrac {6789} {54312}$
+:$\dfrac 1 8 = \dfrac {6791} {54328}$
+:$\dfrac 1 8 = \dfrac {6839} {54712}$
+:$\dfrac 1 8 = \dfrac {7123} {56984}$
+:$\dfrac 1 8 = \dfrac {7312} {58496}$
+:$\dfrac 1 8 = \dfrac {7364} {58912}$
+:$\dfrac 1 8 = \dfrac {7416} {59328}$
+:$\dfrac 1 8 = \dfrac {7421} {59368}$
+:$\dfrac 1 8 = \dfrac {7894} {63152}$
+:$\dfrac 1 8 = \dfrac {7941} {63528}$
+:$\dfrac 1 8 = \dfrac {8174} {65392}$
+:$\dfrac 1 8 = \dfrac {8179} {65432}$
+:$\dfrac 1 8 = \dfrac {8394} {67152}$
+:$\dfrac 1 8 = \dfrac {8419} {67352}$
+:$\dfrac 1 8 = \dfrac {8439} {67512}$
+:$\dfrac 1 8 = \dfrac {8932} {71456}$
+:$\dfrac 1 8 = \dfrac {8942} {71536}$
+:$\dfrac 1 8 = \dfrac {8953} {71624}$
+:$\dfrac 1 8 = \dfrac {8954} {71632}$
+:$\dfrac 1 8 = \dfrac {9156} {73248}$
+:$\dfrac 1 8 = \dfrac {9158} {73264}$
+:$\dfrac 1 8 = \dfrac {9182} {73456}$
+:$\dfrac 1 8 = \dfrac {9316} {74528}$
+:$\dfrac 1 8 = \dfrac {9321} {74568}$
+:$\dfrac 1 8 = \dfrac {9352} {74816}$
+:$\dfrac 1 8 = \dfrac {9416} {75328}$
+:$\dfrac 1 8 = \dfrac {9421} {75368}$
+:$\dfrac 1 8 = \dfrac {9523} {76184}$
+:$\dfrac 1 8 = \dfrac {9531} {76248}$
+:$\dfrac 1 8 = \dfrac {9541} {76328}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{One Ninth as Pandigital Fraction}
+Tags: One Ninth, Pandigital Fractions
+
+\begin{theorem}
+There are $3$ ways $\dfrac 1 9$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]]:
+:$\dfrac 1 9 = \dfrac {6381} {57429}$
+:$\dfrac 1 9 = \dfrac {6471} {58239}$
+:$\dfrac 1 9 = \dfrac {8361} {75249}$
+\end{theorem}
+
+\begin{proof}
+Can be verified by brute force.
+\end{proof}<|endoftext|>
+\section{Four Fifths as Pandigital Fraction}
+Tags: One Fifth, Pandigital Fractions
+
+\begin{theorem}
+The [[Definition:Fraction|fraction]] $\dfrac 4 5$ can be expressed as a [[Definition:Pandigital Fraction|pandigital fraction]] in the following interesting way:
+:$\dfrac 4 5 = \dfrac {9876} {12 \, 345}$
+\end{theorem}
+
+\begin{proof}
+Can be found by brute force.
+\end{proof}<|endoftext|>
+\section{Probability of Two Random Integers having no Common Divisor}
+Tags: Riemann Zeta Function, Coprime Integers
+
+\begin{theorem}
+Let $a$ and $b$ be [[Definition:Integer|integers]] chosen at random.
+The [[Definition:Probability|probability]] that $a$ and $b$ are [[Definition:Coprime Integers|coprime]] is given by:
+:$\map \Pr {a \perp b} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$
+where $\zeta$ denotes the [[Definition:Riemann Zeta Function|zeta function]].
+The [[Definition:Decimal Expansion|decimal expansion]] of $\dfrac 1 {\map \zeta 2}$ starts:
+:$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$
+{{OEIS|A059956}}
+\end{theorem}
+
+\begin{proof}
+Let $a$ and $b$ be two [[Definition:Integer|integers]] chosen at random.
+For $a$ and $b$ to be [[Definition:Coprime Integers|coprime]], it is [[Definition:Necessary and Sufficient|necessary and sufficient]] that no [[Definition:Prime Number|prime number]] divides both of them.
+The probability that any particular [[Definition:Integer|integer]] is [[Definition:Divisor of Integer|divisible]] by a [[Definition:Prime Number|prime number]] $p$ is $\dfrac 1 p$.
+Hence the probability that both $a$ and $b$ are [[Definition:Divisor of Integer|divisible]] by $p$ is $\dfrac 1 {p^2}$.
+The probability that either $a$ or $b$ or both is not [[Definition:Divisor of Integer|divisible]] by $p$ is therefore $1 - \dfrac 1 {p^2}$.
+Whether or not $a$ is [[Definition:Divisor of Integer|divisible]] by $p$ or [[Definition:Divisor of Integer|divisible]] by another [[Definition:Prime Number|prime number]] $q$ is independent of both $p$ and $q$.
+Thus by the [[Product Rule for Probabilities]], the probability that $a$ and $b$ are not both [[Definition:Divisor of Integer|divisible]] by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$.
+This independence extends to all [[Definition:Prime Number|prime numbers]].
+That is, the probability that $a$ and $b$ are not both [[Definition:Divisor of Integer|divisible]] by any [[Definition:Prime Number|prime number]] is equal to the [[Definition:Multiplication|product]] of $1 - \dfrac 1 {p^2}$ over all [[Definition:Prime Number|prime numbers]]:
+:$\map \Pr {a \perp b} = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
+From [[Sum of Reciprocals of Powers as Euler Product]]:
+:$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
+from which:
+:$\displaystyle \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
+where $\map \zeta 2$ is the [[Definition:Riemann Zeta Function|Riemann $\zeta$ (zeta) function]] evaluated at $2$.
+The result follows from [[Riemann Zeta Function of 2]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Probability of Random Integer being Square-Free}
+Tags: Riemann Zeta Function, Square-Free Integers
+
+\begin{theorem}
+Let $a$ be an [[Definition:Integer|integer]] chosen at random.
+The [[Definition:Probability|probability]] that $a$ is [[Definition:Square-Free Integer|square-free]] is given by:
+:$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \dfrac 1 {\map \zeta 2} = \dfrac 6 {\pi^2}$
+where $\zeta$ denotes the [[Definition:Riemann Zeta Function|zeta function]].
+The [[Definition:Decimal Expansion|decimal expansion]] of $\dfrac 1 {\map \zeta 2}$ starts:
+:$\dfrac 1 {\map \zeta 2} = 0 \cdotp 60792 \, 71018 \, 54026 \, 6 \ldots$
+{{OEIS|A059956}}
+\end{theorem}
+
+\begin{proof}
+Let $a$ be an [[Definition:Integer|integer]] chosen at random.
+For $a$ to be [[Definition:Square-Free Integer|square-free]], it is [[Definition:Necessary and Sufficient|necessary and sufficient]] that for all [[Definition:Prime Number|prime numbers]] $p$, it is not the case that $p^2$ is a  [[Definition:Divisor of Integer|divisor]] of $a$.
+The probability that any particular [[Definition:Integer|integer]] is [[Definition:Divisor of Integer|divisible]] by $p^2$ is $\dfrac 1 {p^2}$.
+The probability that $a$ is not [[Definition:Divisor of Integer|divisible]] by $p^2$ is therefore $1 - \dfrac 1 {p^2}$.
+Whether or not $a$ is [[Definition:Divisor of Integer|divisible]] by $p^2$ or [[Definition:Divisor of Integer|divisible]] by $q^2$ for another [[Definition:Prime Number|prime number]] $q$ is independent of both $p$ and $q$.
+Thus by the [[Product Rule for Probabilities]], the probability that $a$ is not [[Definition:Divisor of Integer|divisible]] by either $p^2$ or $q^2$ is $\paren {1 - \dfrac 1 {p^2} } \paren {1 - \dfrac 1 {q^2} }$.
+This independence extends to all [[Definition:Prime Number|prime numbers]].
+That is, the probability that $a$ is not [[Definition:Divisor of Integer|divisible]] by the [[Definition:Square (Algebra)|square]] of any [[Definition:Prime Number|prime number]] is equal to the [[Definition:Multiplication|product]] of $1 - \dfrac 1 {p^2}$ over all [[Definition:Prime Number|prime numbers]]:
+:$\map \Pr {\neg \exists b \in \Z: b^2 \divides a} = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
+From [[Sum of Reciprocals of Powers as Euler Product]]:
+:$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
+from which:
+:$\displaystyle \dfrac 1 {\map \zeta 2} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^2} }$
+where $\map \zeta 2$ is the [[Definition:Riemann Zeta Function|Riemann $\zeta$ (zeta) function]] evaluated at $2$.
+The result follows from [[Riemann Zeta Function of 2]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kepler's Conjecture}
+Tags: Hilbert 23, Spheres
+
+\begin{theorem}
+The densest packing of identical [[Definition:Sphere (Geometry)|spheres]] in [[Definition:Ordinary Space|space]] is obtained when the [[Definition:Sphere (Geometry)|spheres]] are arranged with their [[Definition:Center of Sphere|centers]] at the points of a [[Definition:Face-Centered Cubic Lattice|face-centered cubic lattice]].
+This obtains a density of $\dfrac \pi {3 \sqrt 2} = \dfrac \pi {\sqrt {18} }$:
+:$\dfrac \pi {\sqrt {18} } = 0 \cdotp 74048 \ldots$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Johannes Kepler|cat = Kepler}}
+\end{proof}<|endoftext|>
+\section{Prime Ideal is Prime Element}
+Tags: Prime Ideals (Order Theory), Prime Elements
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice|lattice]].
+Let $I$ be an [[Definition:Ideal (Order Theory)|ideal]] in $L$.
+Then:
+:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]]
+{{iff}}:
+:$I$ is a [[Definition:Prime Element (Order Theory)|prime element]] in $\struct {\map {\mathit {Ids} } L, \precsim}$
+where:
+:$\map {\mathit {Ids} } L$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal (Order Theory)|ideals]] in $L$
+:$\mathord \precsim := \mathord \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $I$ be a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+Let $x, y \in \map {\mathit {Ids} } L$ such that
+:$x \wedge y \precsim I$
+By definition of $\precsim$:
+:$x \wedge y \subseteq I$
+By [[Meet is Intersection in Set of Ideals]]:
+:$x \cap y \subseteq I$
+{{AimForCont}}:
+:$x \not \precsim I$ and $y \not \precsim I$
+By definition of $\precsim$:
+$x \nsubseteq I$ and $y \nsubseteq I$
+By definition of [[Definition:Subset|subset]]:
+:$\exists a \in x: a \notin I$
+and
+:$\exists b \in y: b \notin I$
+By [[Meet Precedes Operands]]:
+:$a \wedge b \preceq a$ and $a \wedge b \preceq b$
+By definition [[Definition:Lower Set|lower set]]:
+:$a \wedge b \in x$ and $a \wedge b \in y$
+By definitions [[Definition:Subset|subset]] and [[Definition:Set Intersection|intersection]]:
+:$a \wedge b \in I$
+By [[Characterization of Prime Ideal]]:
+:$a \in I$ or $b \in I$
+This [[Definition:Contradiction|contradicts]] $a \notin I$ and $b \notin I$
+Hence the assumption:
+:$x \not \precsim I$ and $y \not \precsim I$
+was false.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $I$ be [[Definition:Prime Element (Order Theory)|prime element]] in $\struct {\map {\mathit {Ids} } L, \precsim}$
+Let $x, y \in S$ such that:
+:$x \wedge y \in I$
+By [[Lower Closure of Element is Ideal]]:
+:$X := x^\preceq$ and $Y := y^\preceq$ are [[Definition:Ideal (Order Theory)|ideals]] in $L$
+By [[Meet is Intersection in Set of Ideals]]:
+:$X \cap Y = X \wedge Y$
+We will prove that:
+:$X \wedge Y \subseteq I$
+Let $a \in X \wedge Y$.
+By definition of [[Definition:Set Intersection|intersection]]:
+:$a \in X$ and $a \in Y$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$a \preceq x$ and $a \preceq y$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$a \preceq x \wedge y$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$a \in I$
+{{qed|lemma}}
+By definition of $\precsim$:
+:$X \wedge Y \precsim I$
+By definition of [[Definition:Prime Element (Order Theory)|prime element]]:
+:$X \precsim I$ or $Y \precsim I$
+By definition of $\precsim$:
+:$X \subseteq I$ or $Y \subseteq I$
+By definition of [[Definition:Reflexive Relation|reflexivity]]:
+:$x \preceq x$ and $y \preceq y$
+By definition of [[Definition:Lower Closure of Element|lower closure of element]]:
+:$x \in X$ and $y \in Y$
+Thus by definition of [[Definition:Subset|subset]]:
+:$x \in I$ or $y \in I$
+Hence by [[Characterization of Prime Ideal]]:
+:$I$ is [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Probability of Three Random Integers having no Common Divisor}
+Tags: Riemann Zeta Function, Coprime Integers
+
+\begin{theorem}
+Let $a, b$ and $c$ be [[Definition:Integer|integers]] chosen at random.
+The [[Definition:Probability|probability]] that $a, b$ and $c$ have no [[Definition:Common Divisor of Integers|common divisor]]:
+:$\map \Pr {\map \perp {a, b, c} } = \dfrac 1 {\map \zeta 3}$
+where $\zeta$ denotes the [[Definition:Riemann Zeta Function|zeta function]]:
+:$\map \zeta 3 = \dfrac 1 {1^3} + \dfrac 1 {2^3} + \dfrac 1 {3^3} + \dfrac 1 {4^3} + \cdots$
+The [[Definition:Decimal Expansion|decimal expansion]] of $\dfrac 1 {\map \zeta 3}$ starts:
+:$\dfrac 1 {\map \zeta 3} = 0 \cdotp 83190 \, 73725 \, 80707 \ldots$
+{{OEIS|A088453}}
+\end{theorem}
+
+\begin{proof}
+Let $a$, $b$ and $c$ be three [[Definition:Integer|integers]] chosen at random.
+For $a$, $b$ and $c$ to be [[Definition:Coprime Integers|coprime]], it is [[Definition:Necessary and Sufficient|necessary and sufficient]] that no [[Definition:Prime Number|prime number]] divides all of them.
+The probability that any particular [[Definition:Integer|integer]] is [[Definition:Divisor of Integer|divisible]] by a [[Definition:Prime Number|prime number]] $p$ is $\dfrac 1 p$.
+Hence the probability that $a$, $b$ and $c$ are [[Definition:Divisor of Integer|divisible]] by $p$ is $\dfrac 1 {p^3}$.
+The probability that at least one of $a$, $b$ and $c$ is not [[Definition:Divisor of Integer|divisible]] by $p$ is therefore $1 - \dfrac 1 {p^3}$.
+Whether or not $a$ is [[Definition:Divisor of Integer|divisible]] by $p$ or [[Definition:Divisor of Integer|divisible]] by another [[Definition:Prime Number|prime number]] $q$ is independent of both $p$ and $q$.
+Thus by the [[Product Rule for Probabilities]], the probability that $a$, $b$ and $c$ are not all [[Definition:Divisor of Integer|divisible]] by either $p$ or $q$ is $\paren {1 - \dfrac 1 {p^3} } \paren {1 - \dfrac 1 {q^3} }$.
+This independence extends to all [[Definition:Prime Number|prime numbers]].
+That is, the probability that $a$, $b$ and $c$ are not all [[Definition:Divisor of Integer|divisible]] by any [[Definition:Prime Number|prime number]] is equal to the [[Definition:Multiplication|product]] of $1 - \dfrac 1 {p^3}$ over all [[Definition:Prime Number|prime numbers]]:
+:$\map \Pr {\map \perp {a, b, c} } = \displaystyle \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$
+From [[Sum of Reciprocals of Powers as Euler Product]]:
+:$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$
+from which:
+:$\displaystyle \dfrac 1 {\map \zeta 3} = \prod_{\text {$p$ prime} } \paren {1 - \dfrac 1 {p^3} }$
+where $\map \zeta 3$ is the [[Definition:Riemann Zeta Function|Riemann $\zeta$ (zeta) function]] evaluated at $3$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Densest Packing of Identical Circles}
+Tags: Geometry
+
+\begin{theorem}
+The densest packing of identical [[Definition:Circle|circles]] in [[Definition:The Plane|the plane]] obtains a density of $\dfrac \pi {2 \sqrt 3} = \dfrac \pi {\sqrt {12} }$:
+:$\dfrac \pi {2 \sqrt 3} = 0 \cdotp 90689 \, 96821 \ldots$
+{{OEIS|A093766}}
+This happens when they are packed together in a [[Definition:Regular Hexagon|hexagonal]] array, with each [[Definition:Circle|circle]] touching $6$ others.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|It remains to be proved that the hexagonal packing is in fact the densest that can be achieved.}}
+Consider the [[Definition:Rectangle|rectangular area]] $ABCD$ of the densest packing of [[Definition:Circle|circles]].
+:[[File:Close-Packed-Circles.png|300px]]
+Let the [[Definition:Radius|radius]] of one [[Definition:Circle|circle]] be $1$.
+The [[Definition:Length of Line|length]] $AB$ is $2$.
+The [[Definition:Length of Line|length]] $AC$ is $2 \sqrt 3$.
+Thus, from [[Area of Rectangle]], the [[Definition:Area|area]] of $\Box ABCD$ is $4 \sqrt 3$.
+Within $ABCD$ there is one complete [[Definition:Circle|circle]] and one [[Definition:Quarter|quarter]] of each of $4$ other [[Definition:Circle|circles]].
+That makes a total of $2$ [[Definition:Circle|circles]].
+Thus, from [[Area of Circle]], the area of $ABCD$ which is inside a [[Definition:Circle|circle]] is $2 \pi$.
+So the density is:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\text {Area of Circles} } {\text {Area of Rectangle} }
+      | r = \dfrac {2 \pi} {4 \sqrt 3}
+      | c = 
+}}
+{{eqn | r = \dfrac {\pi} {2 \sqrt 3}
+      | c = 
+}}
+{{eqn | r = \dfrac {\pi} {12}
+      | c = 
+}}
+{{end-eqn}}
+as required.
+\end{proof}<|endoftext|>
+\section{Anning's Theorem}
+Tags: Number Theory
+
+\begin{theorem}
+In any [[Definition:Number Base|base]] greater than $1$, the [[Definition:Fraction|fraction]]:
+:$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$
+has the property that if the two $1$'s in the center of the [[Definition:Numerator|numerator]] and the [[Definition:Denominator|denominator]] are replaced by the same [[Definition:Odd Integer|odd number]] of $1$'s, the value of the [[Definition:Fraction|fraction]] remains the same.
+For example:
+:$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011} = \dfrac {1 \, 010 \, 111 \, 110 \, 101} {1 \, 100 \, 111 \, 110 \, 011} = \dfrac {9091} {9901}$ (in [[Definition:Decimal Notation|base $10$)]].
+\end{theorem}
+
+\begin{proof}
+Let $b$ be the [[Definition:Number Base|base]] in question.
+Let $F = \dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$.
+Then:
+:$F = \dfrac {1 + b^2 + b^4 + b^6 + b^8} {1 + b + b^4 + b^7 + b^8}$
+It is necessary to prove that for all $k \in \Z_{>0}$:
+:$F = \dfrac {1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6} } {1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6} }$
+This is done by:
+: multiplying the [[Definition:Numerator|numerator]] of one by the [[Definition:Denominator|denominator]] of the other
+and then:
+: multiplying the [[Definition:Denominator|denominator]] of one by the [[Definition:Numerator|numerator]] of the other
+and checking that they are equal.
+Thus we proceed:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 1 + b^2 + b^4 + b^6 + b^8
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}
+      | c = 
+}}
+{{eqn | o = =
+      | r = 1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^2 + b^3 + b^6 + b^7 + \cdots + b^{2 k + 4} + b^{2 k + 7} + b^{2 k + 8}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^4 + b^5 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 9} + b^{2 k + 10}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^6 + b^7 + b^{10} + b^{11} + \cdots + b^{2 k + 8} + b^{2 k + 11} + b^{2 k + 12}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^8 + b^9 + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 13} + b^{2 k + 14}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 1 + b + b^4 + b^7 + b^8
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}
+      | c = 
+}}
+{{eqn | o = =
+      | r = 1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b + b^3 + b^5 + b^6 + \cdots + b^{2 k + 3} + b^{2 k + 5} + b^{2 k + 7}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^4 + b^6 + b^8 + b^9 + \cdots + b^{2 k + 6} + b^{2 k + 8} + b^{2 k + 10}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^7 + b^9 + b^{11} + b^{12} + \cdots + b^{2 k + 9} + b^{2 k + 11} + b^{2 k + 13}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = b^8 + b^{10} + b^{12} + b^{13} + \cdots + b^{2 k + 10} + b^{2 k + 12} + b^{2 k + 14}
+      | c = 
+}}
+{{end-eqn}}
+Equality can be demonstrated.
+{{qed}}
+{{Namedfor|Norman Herbert Anning|cat = Anning}}
+\end{proof}<|endoftext|>
+\section{Meet is Intersection in Set of Ideals}
+Tags: Join and Meet
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \wedge, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $\map {\mathit {Ids} } {\mathscr S}$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+Let $P = \struct {\map {\mathit {Ids} } {\mathscr S}, \precsim}$ be an [[Definition:Ordered Set|ordered set]] where $\mathord \precsim = \mathord \subseteq \restriction_{\map {\mathit {Ids} } {\mathscr S} \times \map {\mathit {Ids} } {\mathscr S} }$
+Let $I_1, I_2$ be [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+Then
+:$I_1 \wedge_P I_2 = I_1 \cap I_2$
+\end{theorem}
+
+\begin{proof}
+By [[Intersection of Semilattice Ideals is Ideal]]:
+:$I_1 \cap I_2 \in \map {\mathit {Ids} } {\mathscr S}$
+We will prove that
+:$I_1 \cap I_2$ is [[Definition:Lower Bound of Set|lower bound]] for $\set {I_1, I_2}$
+Let $x \in \set {I_1, I_2}$.
+Then by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x = I_1$ or $x = I_2$
+By [[Intersection is Subset]]
+:$I_1 \cap I_2 \subseteq x$
+Thus by definition of $\precsim$:
+:$I_1 \cap I_2 \precsim x$
+{{qed|lemma}}
+We will prove that
+:$\forall I \in \map {\mathit {Ids} } {\mathscr S}: I$ is [[Definition:Lower Bound of Set|lower bound]] for $\set {I_1, I_2} \implies I \precsim I_1 \cap I_2$
+Let $I \in \map {\mathit {Ids} } {\mathscr S}$ such that
+:$I$ is [[Definition:Lower Bound of Set|lower bound]] for $\set {I_1, I_2}$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$I \precsim I_1$ and $I \precsim I_2$
+By definition of $\precsim$:
+:$I \subseteq I_1$ and $I \subseteq I_2$
+By [[Intersection is Largest Subset]]:
+:$I \subseteq I_1 \cap I_2$
+Thus by definition of $\precsim$
+:$I \precsim I_1 \cap I_2$
+{{qed|lemma}}
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\inf \set {I_1, I_2} = I_1 \cap I_2$
+Hence by definition of [[Definition:Meet (Order Theory)|meet]]:
+:$I_1 \wedge I_2 = I_1 \cap I_2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of Prime Ideal by Finite Infima}
+Tags: Prime Ideals (Order Theory)
+
+\begin{theorem}
+Let $L = \left({S, \wedge, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $I$ be an [[Definition:Ideal (Order Theory)|ideal]] in $L$.
+Then
+:$I$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]]
+{{iff}}
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S: \left({ \inf A \in I \implies \exists a \in A: a \in I}\right)$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $I$ be a [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+Define $\mathcal P\left({X}\right) :\equiv X \ne \varnothing \land \inf X \in I \implies \exists x \in X: x \in I$
+where $X$ is [[Definition:Subset|subset]] of $S$.
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+By definition of [[Definition:Empty Set|empty set]]:
+:$\mathcal P\left({\varnothing}\right)$
+We will prove that
+:$\forall x \in A, B \subseteq A: \mathcal P\left({B}\right) \implies \mathcal P\left({B \cup \left\{ {x}\right\} }\right)$
+Let $x \in A, B \subseteq A$ such that
+:$\mathcal P\left({B}\right)$ (Induction Hypothesis)
+Assume that
+:$B \cup \left\{ {x}\right\} \ne \varnothing$ and $\inf \left({B \cup \left\{ {x}\right\} }\right) \in I$
+Case $B = \varnothing$:
+By [[Union with Empty Set]]:
+:$B \cup \left\{ {x}\right\} = \left\{ {x}\right\}$
+By [[Infimum of Singleton]]:
+:$\inf \left\{ {x}\right\} = x$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \left\{ {x}\right\}$
+Thus
+:$\exists a \in B \cup \left\{ {x}\right\}: a \in I$
+{{qed|lemma}}
+Case $B \ne \varnothing$:
+By [[Subset of Finite Set is Finite]]:
+:$B$ is [[Definition:Finite Set|finite]].
+By [[Existence of Non-Empty Finite Infima in Meet Semilattice]]:
+:$B$ admits an [[Definition:Infimum of Set|infimum]].
+By [[Infimum of Singleton]]:
+:$\left\{ {x}\right\}$ admits an [[Definition:Infimum of Set|infimum]].
+{{begin-eqn}}
+{{eqn | l = \inf\left({B \cup \left\{ {x}\right\} }\right)
+| r = \inf\left({\bigcup\left\{ {B, \left\{ {x}\right\} }\right\} }\right)
+| c = definition of [[Definition:Set Union|union]]
+}}
+{{eqn | r = \inf \left\{ {\inf B, x}\right\}
+| c = [[Infimum of Infima]]:
+}}
+{{eqn | r = \left({\inf B}\right) \wedge x
+| c = definition of [[Definition:Meet (Order Theory)|meet]]
+}}
+{{end-eqn}}
+By [[Characterization of Prime Ideal]]:
+:$\inf B \in I$ or $x \in I$
+Case $\inf B \in I$:
+By Induction Hypothesis:
+:$\exists a \in B: a \in I$
+By definition of [[Definition:Set Union|union]]:
+:$a \in B \cup \left\{ {x}\right\}$
+Thus
+:$\exists a \in B \cup \left\{ {x}\right\}: a \in I$
+{{qed|lemma}}
+Case $x \in I$:
+By definition of [[Definition:Set Union|union]]:
+:$x \in B \cup \left\{ {x}\right\}$
+Thus
+:$\exists a \in B \cup \left\{ {x}\right\}: a \in I$
+{{qed|lemma}}
+By [[Induction of Finite Set]]:
+:$\mathcal P\left({A}\right)$
+Thus the result.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S: \left({ \inf A \in I \implies \exists a \in A: a \in I}\right)$
+Let $x, y \in S$ such that
+:$x \wedge y \in I$
+By [[Unordered Pair is Finite]]:
+:$\left\{ {x, y}\right\}$ is a [[Definition:Finite Set|finite set]].
+By definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \in \left\{ {x, y}\right\}$
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\left\{ {x, y}\right\}$ is a [[Definition:Non-Empty Set|non-empty set]].
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$\inf \left\{ {x, y}\right\} = x \wedge y$
+By assumption:
+:$\exists a \in \left\{ {x, y}\right\}: a \in I$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \in I$ or $y \in I$
+Hence by [[Characterization of Prime Ideal]]:
+:$I$ is [[Definition:Prime Ideal (Order Theory)|prime ideal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Doubling the Cube/Archytas Curve}
+Tags: Doubling the Cube
+
+\begin{theorem}
+The [[Definition:Archytas Curve|Archytas curve]] can be used for [[Doubling the Cube]].
+\end{theorem}<|endoftext|>
+\section{Doubling the Cube/Intersection of Parabolas}
+Tags: Doubling the Cube
+
+\begin{theorem}
+The problem of [[Doubling the Cube]] can be solved by finding the [[Definition:Intersection (Geometry)|intersection]] of two [[Definition:Parabola|parabolas]].
+\end{theorem}<|endoftext|>
+\section{Doubling the Cube/Intersection of Parabola and Hyperbola}
+Tags: Doubling the Cube
+
+\begin{theorem}
+The problem of [[Doubling the Cube]] can be solved by finding the [[Definition:Intersection (Geometry)|intersection]] of a [[Definition:Parabola|parabola]] and a [[Definition:Hyperbola|hyperbola]].
+\end{theorem}<|endoftext|>
+\section{Unordered Pair is Finite}
+Tags: Set Theory
+
+\begin{theorem}
+Let $x, y$ be arbitrary.
+Then $\left\{ {x, y}\right\}$ is [[Definition:Finite Set|finite]].
+\end{theorem}
+
+\begin{proof}
+By [[Pair is Union of Singletons]]:
+:$\left\{ {x, y}\right\} = \left\{ {x}\right\} \cup \left\{ {y}\right\}$
+By [[Singleton is Finite]]:
+:$\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ are [[Definition:Finite Set|finite]].
+Thus by [[Union of Finite Sets is Finite]]:
+:$\left\{ {x, y}\right\}$ is [[Definition:Finite Set|finite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pair is Union of Singletons}
+Tags: Set Theory, Singletons
+
+\begin{theorem}
+Let $x, y$ be arbitrary.
+{{questionable|Need to state that $x \ne y$.}}
+Then $\left\{ {x, y}\right\} = \left\{ {x}\right\} \cup \left\{ {y}\right\}$
+\end{theorem}
+
+\begin{proof}
+Straightforward from [[Union of Unordered Tuples]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Doubling the Cube/Conchoid of Nicomedes}
+Tags: Doubling the Cube
+
+\begin{theorem}
+The problem of [[Doubling the Cube]] can be solved by using a [[Definition:Conchoid of Nicomedes|conchoid of Nicomedes]].
+\end{theorem}<|endoftext|>
+\section{Doubling the Cube/Cissoid of Diocles}
+Tags: Doubling the Cube
+
+\begin{theorem}
+The problem of [[Doubling the Cube]] can be solved by using a [[Definition:Cissoid of Diocles|cissoid of Diocles]].
+\end{theorem}<|endoftext|>
+\section{Length of Diagonal of Unit Square}
+Tags: Squares
+
+\begin{theorem}
+The [[Definition:Length (Linear Measure)|length]] of a [[Definition:Diagonal of Quadrilateral|diagonal]] of a [[Definition:Square (Geometry)|square]] of [[Definition:Side of Polygon|side]] [[Definition:Length (Linear Measure)|length]] $1$ is $\sqrt 2$) (the [[Definition:Square Root|square root]] of $2$).
+\end{theorem}
+
+\begin{proof}
+:[[File:DiagonalOfSquare.png|300px]]
+Two adjacent [[Definition:Side of Polygon|sides]] $AB$, $BC$ and the [[Definition:Diagonal of Quadrilateral|diagonal]] $AC$ of [[Definition:Square (Geometry)|square]] $ABCD$ form a [[Definition:Right Triangle|right triangle]].
+The [[Definition:Hypotenuse|hypotenuse]] of [[Definition:Triangle (Geometry)|triangle]] $\triangle ABC$ can be found by using [[Pythagoras's Theorem]]:
+:$AC^2 = AB^2 + BC^2$
+from which:
+:$AC^2 = 2$
+and so:
+:$AC = \sqrt 2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequence of Best Rational Approximations to Square Root of 2}
+Tags: Square Root of 2
+
+\begin{theorem}
+A [[Definition:Sequence|sequence]] of [[Definition:Best Rational Approximation|best rational approximations]] to the [[Definition:Square Root|square root]] of $2$ starts:
+:$\dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
+where:
+:the [[Definition:Numerator|numerators]] are half of the [[Definition:Pell-Lucas Numbers|Pell-Lucas numbers]], $\dfrac 1 2 Q_n$
+:the [[Definition:Denominator|denominators]] are the [[Definition:Pell Numbers|Pell numbers]] $P_n$
+starting from $\dfrac {\tfrac12 Q_1} {P_1}$.
+{{OEIS-Numerators|A001333}}
+{{OEIS-Denominators|A000129}}
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {a_0, a_1, \ldots}$ be the [[Definition:Continued Fraction Expansion of Real Number|continued fraction expansion]] of $\sqrt 2$.
+By [[Continued Fraction Expansion of Root 2]]:
+:$\sqrt 2 = \sqbrk {1, \sequence 2} = \sqbrk {1, 2, 2, 2, \ldots}$
+From [[Convergents are Best Approximations]], the [[Definition:Convergent of Continued Fraction|convergents]] of $\sqbrk {1, \sequence 2}$ are the [[Definition:Best Rational Approximation|best rational approximations]] of $\sqrt 2$.
+Let $\sequence {p_n}_{n \mathop \ge 0}$ and $\sequence {q_n}_{n \mathop \ge 0}$ be the [[Definition:Numerators and Denominators of Continued Fraction|numerators and denominators]] of the continued fraction expansion of $\sqrt 2$.
+Then $\dfrac {p_n} {q_n}$ is the $n$th [[Definition:Convergent of Continued Fraction|convergent]] of $\sqbrk {1, \sequence 2}$.
+By [[Convergents of Simple Continued Fraction are Rationals in Canonical Form]], $p_n$ and $q_n$ are [[Definition:Coprime Integers|coprime]] for all $n \ge 0$.
+It remains to show that for all $n \ge 1$:
+:$Q_n = 2 p_{n - 1}$
+:$P_n = q_{n - 1}$
+It is sufficient to prove that they satisfy the same [[Definition:Recurrence Relation|recurrence relation]].
+By definition:
+{{begin-eqn}}
+{{eqn | l = p_0
+      | r = a_0
+}}
+{{eqn | r = 1
+}}
+{{eqn | l = q_0
+      | r = 1
+}}
+{{end-eqn}}
+so that:
+:$\tuple {Q_1, P_1} = \tuple {2, 1} = \tuple {2 p_0, q_0}$
+{{begin-eqn}}
+{{eqn | l = p_1
+      | r = a_0 a_1 + 1
+}}
+{{eqn | r = 1 \times 2 + 1
+}}
+{{eqn | r = 3
+}}
+{{eqn | l = q_1
+      | r = a_1
+}}
+{{eqn | r = 2
+}}
+{{end-eqn}}
+so that:
+:$\tuple {Q_2, P_2} = \tuple {6, 2} = \tuple {2 p_1, q_1}$
+{{begin-eqn}}
+{{eqn | l = p_k
+      | r = a_k p_{k - 1} + p_{k - 2}
+}}
+{{eqn | r = 2 p_{k - 1} + p_{k - 2}
+}}
+{{eqn | l = q_k
+      | r = a_k q_{k - 1} + q_{k - 2}
+}}
+{{eqn | r = 2 q_{k - 1} + q_{k - 2}
+}}
+{{end-eqn}}
+The result follows by definition of [[Definition:Pell Numbers|Pell numbers]] and [[Definition:Pell-Lucas Numbers|Pell-Lucas numbers]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference between Adjacent Convergents of Simple Continued Fraction}
+Tags: Continued Fractions
+
+\begin{theorem}
+Then for $k \ge 1$:
+:$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$
+That is:
+:$C_k - C_{k - 1} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 1} } {q_{k - 1} } = \dfrac {\paren {-1}^{k + 1} } {q_k q_{k - 1} }$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z: n \ge 2$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$p_n q_{n - 1} - p_{n - 1} q_n = \paren {-1}^{n + 1}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$p_0 = a_0, p_1 = a_0 a_1 + 1, q_0 = 1, q_1 = a_1$
+So:
+:$p_1 q_0 - p_0 q_1 = \paren {a_0 a_1 + 1} \times 1 - a_0 a_1 = 1 = \paren {-1}^2$
+So $P(2)$ is seen to hold.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$p_k q_{k - 1} - p_{k - 1} q_k = \paren {-1}^{k + 1}$
+Then we need to show:
+:$p_{k + 1} q_k - p_k q_{k + 1} = \paren {-1}^{k + 2}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Consider $\sqbrk {a_0, a_1, a_2, \ldots, a_k, a_{k + 1} }$.
+Its final [[Definition:Numerators and Denominators of Continued Fraction|numerator and denominator]] are by definition:
+:$p_{k + 1} = a_{k + 1} p_k + p_{k - 1}, q_{k + 1} = a_{k + 1} q_k + q_{k - 1}$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = p_{k + 1} q_k - p_k q_{k + 1}
+      | r = \paren {a_{k + 1} p_k + p_{k - 1} } q_k - p_k \paren {a_{k + 1} q_k + q_{k - 1} }
+      | c = 
+}}
+{{eqn | r = p_{k - 1} q_k - p_k q_{k - 1}
+      | c = 
+}}
+{{eqn | r = \paren {-1} \paren {p_k q_{k - 1} - p_{k - 1} q_k}
+      | c = 
+}}
+{{eqn | r = \paren {-1} \paren {-1}^{k + 1}
+      | c = [[Difference between Adjacent Convergents of Simple Continued Fraction#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {-1}^{k + 2}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \ge 1: p_n q_{n-1} - p_{n-1} q_n = \paren {-1}^{n + 1}$
+{{qed}}
+[[Category:Continued Fractions]]
+adgeudgpif0gfqi87f5hkm6uavpfct8
+\end{proof}<|endoftext|>
+\section{Difference between Adjacent Convergents But One of Simple Continued Fraction}
+Tags: Continued Fractions
+
+\begin{theorem}
+For $k \ge 2$:
+:$p_k q_{k - 2} - p_{k - 2} q_k = \paren {-1}^k a_k$
+That is:
+:$C_k - C_{k-2} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 2} } {q_{k - 2} } = \dfrac {\paren {-1}^k a_k} {q_k q_{k - 2} }$
+\end{theorem}
+
+\begin{proof}
+Let $k \ge 2$.
+{{begin-eqn}}
+{{eqn | l = p_k q_{k - 2} - p_{k - 2} q_k
+      | r = \paren {a_k p_{k - 1} + p_{k - 2} } q_{k - 2} - p_{k - 2} \paren {a_k q_{k - 1} + q_{k - 2} }
+      | c = 
+}}
+{{eqn | r = a_k \paren {p_{k - 1} q_{k - 2} - p_{k - 2} q_{k - 1} }
+      | c = 
+}}
+{{eqn | r = a_k \paren {-1}^k
+      | c = [[Difference between Adjacent Convergents of Simple Continued Fraction]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Continued Fractions]]
+cr2z76m7r823waql0ap13l9jameiac2
+\end{proof}<|endoftext|>
+\section{Convergents of Simple Continued Fraction are Rationals in Canonical Form}
+Tags: Continued Fractions
+
+\begin{theorem}
+For all $k \ge 1$, $\dfrac {p_k} {q_k}$ is in [[Definition:Canonical Form of Rational Number|canonical form]]:
+:$p_k$ and $q_k$ are [[Definition:Coprime Integers|coprime]]
+:$q_k > 0$.
+\end{theorem}
+
+\begin{proof}
+Let $k \ge 1$.
+Let $d = \gcd \set {p_k, q_k}$.
+From [[Common Divisor Divides Integer Combination]]:
+:$p_k q_{k - 1} - p_{k - 1} q_k$ is a [[Definition:Divisor of Integer|multiple]] of $d$.
+From [[Difference between Adjacent Convergents of Simple Continued Fraction]]:
+:$d \divides \paren {-1}^{k + 1}$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+It follows that:
+:$d = 1$
+By:
+:$q_0 = 1$
+:[[Denominators of Simple Continued Fraction are Strictly Increasing]]
+It follows that $q_k > 0$ for all $k \ge 0$.
+{{qed}}
+[[Category:Continued Fractions]]
+kpjquz66bpv06cdtbedgi40mne9f7t2
+\end{proof}<|endoftext|>
+\section{Characterization of Prime Filter by Finite Suprema}
+Tags: Prime Ideals (Order Theory)
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \preceq}$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $F$ be a [[Definition:Filter|filter]] in $L$.
+Then
+:$F$ is a [[Definition:Prime Filter (Order Theory)|prime filter]]
+{{iff}}:
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $F$ be a [[Definition:Prime Filter (Order Theory)|prime ideal]].
+Define $\map \PP X: \equiv X \ne \O \land \sup X \in F \implies \exists x \in X: x \in F$
+where $X$ is [[Definition:Subset|subset]] of $S$.
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+By definition of [[Definition:Empty Set|empty set]]:
+:$\map \PP \O$
+We will prove that:
+:$\forall x \in A, B \subseteq A: \map \PP B \implies \map \PP {B \cup \set x}$
+Let $x \in A, B \subseteq A$ such that:
+:$\map \PP B$ (Induction Hypothesis)
+Assume that:
+:$B \cup \set x \ne \O$ and $\map \sup {B \cup \set x} \in F$
+Case $B = \O$:
+By [[Union with Empty Set]]:
+:$B \cup \set x = \set x$
+By [[Supremum of Singleton]]:
+:$\sup \set x = x$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \set x$
+Thus
+:$\exists a \in B \cup \set x: a \in F$
+{{qed|lemma}}
+Case $B \ne \O$:
+By [[Subset of Finite Set is Finite]]:
+:$B$ is [[Definition:Finite Set|finite]].
+By [[Existence of Non-Empty Finite Suprema in Join Semilattice]]:
+:$B$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Supremum of Singleton]]:
+:$\set x$ admits a [[Definition:Supremum of Set|supremum]].
+{{begin-eqn}}
+{{eqn | l = \map \sup {B \cup \set x}
+      | r = \map \sup {\bigcup \set {B, \set x} }
+      | c = {{Defof|Set Union}}
+}}
+{{eqn | r = \sup \set {\sup B, x}
+      | c = [[Supremum of Suprema]]:
+}}
+{{eqn | r = \paren {\sup B} \vee x
+      | c = {{Defof|Join (Order Theory)}}
+}}
+{{end-eqn}}
+By definition of [[Definition:Prime Filter (Order Theory)|prime filter]]:
+:$\sup B \in F$ or $x \in F$
+Case $\sup B \in F$:
+By Induction Hypothesis:
+:$\exists a \in B: a \in F$
+By definition of [[Definition:Set Union|union]]:
+:$a \in B \cup \set x$
+Thus:
+:$\exists a \in B \cup \set x: a \in F$
+{{qed|lemma}}
+Case $x \in F$:
+By definition of [[Definition:Set Union|union]]:
+:$x \in B \cup \set x$
+Thus:
+:$\exists a \in B \cup \set x: a \in F$
+{{qed|lemma}}
+By [[Induction of Finite Set]]:
+:$\map \PP A$
+Thus the result.
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose that
+:for all [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$
+Let $x, y \in S$ such that
+:$x \vee y \in F$
+By [[Unordered Pair is Finite]]:
+:$\set {x, y}$ is a [[Definition:Finite Set|finite set]].
+By definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \in \set {x, y}$
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\set {x, y}$ is a [[Definition:Non-Empty Set|non-empty set]].
+By definition of [[Definition:Join (Order Theory)|join]]:
+:$\sup \set {x, y} = x \vee y$
+By assumption:
+:$\exists a \in \set {x, y}: a \in F$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$x \in F$ or $y \in F$
+Hence $I$ is [[Definition:Prime Filter (Order Theory)|prime filter]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dual of Preordered Set is Preordered Set}
+Tags: Dual Ordered Sets
+
+\begin{theorem}
+Let $P = \left({S, \preceq}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Then [[Definition:Dual Ordered Set|dual]] of $P$, $P^{-1} = \left({S, \succeq}\right)$ is also a [[Definition:Preordered Set|preordered set]].
+{{finish|Dual Ordered Set $\ne$ Dual Preordered Set}}
+\end{theorem}
+
+\begin{proof}
+By [[Inverse of Reflexive Relation is Reflexive]]:
+:$\succeq$ is [[Definition:Reflexive Relation|reflexive]].
+By [[Inverse of Transitive Relation is Transitive]]:
+:$\succeq$ is [[Definition:Transitive Relation|transitive]].
+Hence $\succeq$ is a [[Definition:Preordering|preordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Dual Ordered Set is Ordered Set}
+Tags: Dual Ordered Sets
+
+\begin{theorem}
+Let $P = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Then its [[Definition:Dual Ordered Set|dual]], $P^{-1} = \left({S, \succeq}\right)$, is also an [[Definition:Ordered Set|ordered set]].
+\end{theorem}
+
+\begin{proof}
+Immediate from [[Dual Ordering is Ordering]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Value of Finite Continued Fraction equals Numerator Divided by Denominator}
+Tags: Continued Fractions, Proofs by Induction
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $\tuple {a_0, a_1, \ldots, a_n}$ be a [[Definition:Finite Continued Fraction|finite continued fraction]] of [[Definition:Length of Continued Fraction|length]] $n \ge 0$.
+Let $p_n$ and $q_n$ be its $n$th [[Definition:Numerator and Denominator of Continued Fraction|numerator and denominator]].
+Then the [[Definition:Value of Continued Fraction|value]] $\sqbrk {a_0, a_1, \ldots, a_n}$ equals $\dfrac {p_n} {q_n}$.
+\end{theorem}
+
+\begin{proof}
+We will use a proof by [[Principle of Mathematical Induction|induction]] on the [[Definition:Length of Continued Fraction|length]] $n$.
+For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\sqbrk {a_0, a_1, \ldots, a_n} = \dfrac {p_n} {q_n}$
+=== Basis for the Induction ===
+$\map P 0$ is the case: 
+:$\sqbrk {a_0} = \dfrac {a_0} 1 = \dfrac {p_0} {q_0}$
+This holds for any [[Definition:Continued Fraction|continued fraction]].
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {a_0, a_1}
+      | r = a_0 + \frac 1 {a_1}
+}}
+{{eqn | r = \frac {a_0 a_1 + 1} {a_1}
+}}
+{{eqn | r = \frac {p_1} {q_1}
+}}
+{{end-eqn}}
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\sqbrk {a_0, a_1, \ldots, a_k} = \dfrac {p_k} {q_k}$
+Then we need to show:
+:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Consider the [[Definition:Continued Fraction|continued fraction]]:
+:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$
+The [[Definition:Numerator|numerators]] are:
+:$p_0, p_1, \ldots, p_k, p_{k + 1}$
+and the [[Definition:Denominator|denominators]] are:
+:$q_0, q_1, \ldots, q_k, q_{k + 1}$
+By definition of [[Definition:Value of Finite Continued Fraction|value of a finite continued fraction]]:
+:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$
+where $a_k' = a_k + \dfrac 1 {a_{k + 1} }$
+Consider the {{RHS}}.
+Take the [[Definition:Continued Fraction|continued fraction]]:
+:$\sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$
+Its [[Definition:Numerator|numerators]] are:
+:$p_0, p_1, \ldots, p_{k-1}$ and $p_k'$
+where $p_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2}$ by definition.
+Its [[Definition:Denominator|denominators]] are:
+:$q_0, q_1, \ldots, p_{k - 1}$ and $q_k'$
+where $q_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2}$ by definition.
+As it has just $k$ [[Definition:Partial Quotient|partial quotients]], the [[Value of Finite Continued Fraction equals Numerator Divided by Denominator#Induction Hypothesis|induction hypothesis]] tells us that its value is:
+:$\dfrac {p_k'} {q_k'}$
+So:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }
+      | r = \frac {p_k'} {q_k'}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {a_k + \frac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2} } {\paren {a_k + \frac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2} }
+      | c = 
+}}
+{{eqn | r = \frac {a_{k + 1} \paren {a_k p_{k - 1} + p_{k - 2} } + p_{k - 1} } {a_{k + 1} \paren {a_k q_{k - 1} + q_{k - 2} } + q_{k - 1} }
+      | c = 
+}}
+{{eqn | r = \frac {a_{k + 1} p_k + p_{k - 1} } {a_{k + 1} q_k + q_{k - 1} }
+      | c = 
+}}
+{{eqn | r = \frac {p_{k + 1} } {q_{k + 1} }
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$
+{{qed}}
+[[Category:Continued Fractions]]
+[[Category:Proofs by Induction]]
+koba2avyl8b5h1kyqcgccqhkffnryi2
+\end{proof}<|endoftext|>
+\section{Simple Infinite Continued Fraction Converges}
+Tags: Continued Fractions
+
+\begin{theorem}
+Let $C = (a_0, a_1, \ldots)$ be a [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]] in $\R$.
+Then $C$ [[Definition:Convergent Continued Fraction|converges]].
+\end{theorem}
+
+\begin{proof}
+We need to show that for any [[Definition:Simple Infinite Continued Fraction|SICF]] its [[Definition:Sequence|sequence]] of [[Definition:Convergent of Continued Fraction|convergents]] $\sequence {C_n}$ always tends to a [[Definition:Limit of Sequence (Number Field)|limit]].
+Several techniques can be used here, but a quick and easy one is to show that $\sequence {C_n}$ is a [[Definition:Cauchy Sequence|Cauchy sequence]].
+Let $\epsilon > 0$.
+For $m > n \ge \max \set {5, \dfrac 1 \epsilon}$:
+{{begin-eqn}}
+{{eqn | l = \size {C_m - C_n}
+      | o = \le
+      | r = \size {C_m - C_{m - 1} } + \cdots + \size {C_{n + 1} - C_n}
+      | c = [[Triangle Inequality]]
+}}
+{{eqn | r = \frac 1 {q_m q_{m - 1} } + \cdots + \frac 1 {q_{n + 1} q_n}
+      | c = [[Difference between Adjacent Convergents of Simple Continued Fraction]]
+}}
+{{eqn | o = <
+      | r = \frac 1 {m \paren {m - 1} } + \cdots + \frac 1 {\paren {n + 1} n}
+      | c = [[Lower Bounds for Denominators of Simple Continued Fraction]]
+}}
+{{eqn | r = \sum_{k \mathop = n}^{m - 1} \frac 1 {k \paren {k + 1} }
+}}
+{{eqn | r = \sum_{k \mathop = n}^{m - 1} \paren {\frac 1 k - \frac 1 {k + 1} }
+}}
+{{eqn | r = \frac 1 n - \frac 1 m
+      | c = [[Telescoping Series/Example 1]]
+}}
+{{eqn | o = <
+      | r = \frac 1 n
+}}
+{{eqn | o = \le
+      | r = \epsilon
+}}
+{{end-eqn}}
+So $\sequence {C_n}$ is indeed a [[Definition:Cauchy Sequence|Cauchy sequence]].
+Hence the result.
+{{qed}}
+[[Category:Continued Fractions]]
+rv4r4kw0uqzbis22s4urrf4nk1jzvbp
+\end{proof}<|endoftext|>
+\section{Bound for Difference of Irrational Number with Convergent}
+Tags: Continued Fractions
+
+\begin{theorem}
+Let $x$ be an [[Definition:Irrational Number|irrational number]].
+Let $\sequence {C_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Convergent of Continued Fraction|convergents]] of the [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]] of $x$.
+Then $\forall n \ge 1$:
+:$C_n < x < C_{n + 1}$ or $C_{n + 1} < x < C_n$
+:$\size {x - C_n} < \dfrac 1 {q_n q_{n + 1} }$
+\end{theorem}
+
+\begin{proof}
+{{questionable|this needs [[Correspondence between Irrational Numbers and Simple Infinite Continued Fractions]]}}
+Immediate.
+Note that:
+:$\size {x - C_n} < \size {C_{n + 1} - C_n} = \dfrac 1 {q_n q_{n + 1} }$
+{{qed}}
+[[Category:Continued Fractions]]
+3uolekaum9vp63qh9xj7w2yks1jnw2a
+\end{proof}<|endoftext|>
+\section{Continued Fraction Identities/First/Infinite}
+Tags: Continued Fractions
+
+\begin{theorem}
+Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a [[Definition:Simple Infinite Continued Fraction|simple infinite continued fraction]].
+Then:
+:$\left[{a_1, a_2, a_3, \ldots}\right] = a_1 + \dfrac 1 {\left[{a_2, a_3, \ldots}\right]}$
+\end{theorem}
+
+\begin{proof}
+{{proof wanted|use continuity}}
+{{Qed}}
+[[Category:Continued Fractions]]
+dp9ldks9dr7mcpnf5dtn1nk3q23f457
+\end{proof}<|endoftext|>
+\section{Relation between Adjacent Best Rational Approximations to Root 2}
+Tags: Square Root of 2
+
+\begin{theorem}
+Consider the [[Sequence of Best Rational Approximations to Square Root of 2]]:
+:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
+Let $\dfrac {p_n} {q_n}$ and $\dfrac {p_{n + 1} } {q_{n + 1} }$ be adjacent [[Definition:Term of Sequence|terms]] of $\sequence S$.
+Then:
+:$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \dfrac {p_2} {q_2}
+      | r = \dfrac 3 2
+      | c = 
+}}
+{{eqn | r = \dfrac {1 + 2 \times 1} {1 + 1}
+      | c = 
+}}
+{{eqn | r = \dfrac {p_1 + 2 \times q_1} {p_1 + q_1}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$
+from which it is to be shown that:
+:$\dfrac {p_{k + 2} } {q_{k + 2} } = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \dfrac {p_{k + 2} } {q_{k + 2} }
+      | r = \dfrac {2 p_{k + 1} + p_k} {2 q_{k + 1} + q_k}
+      | c = {{Defof|Numerators and Denominators of Continued Fraction}}
+}}
+{{eqn | r = \dfrac {2 \paren {p_k + 2 q_k} + p_k} {2 \paren {p_k + q_k} + q_k}
+      | c = [[Relation between Adjacent Best Rational Approximations to Root 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \dfrac {3 p_k + 4 q_k} {2 p_k + 3 q_k}
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {p_k + 2 q_k} + 2 \paren {p_k + q_k} } {\paren {p_k + q_k} + \paren {p_k + 2 q_k} }
+      | c = 
+}}
+{{eqn | r = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }
+      | c = [[Relation between Adjacent Best Rational Approximations to Root 2#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>0}: \dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity of Pell Numbers}
+Tags: Pell Numbers
+
+\begin{theorem}
+Consider the [[Definition:Pell Numbers|Pell numbers]] $P_0, P_1, P_2, \ldots$
+:$0, 1, 2, 5, 12, 29, \ldots$
+$P_n$ has the same [[Definition:Parity of Integer|parity]] as $n$.
+That is:
+:if $n$ is [[Definition:Odd Integer|odd]] then $P_n$ is [[Definition:Odd Integer|odd]]
+:if $n$ is [[Definition:Even Integer|even]] then $P_n$ is [[Definition:Even Integer|even]].
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$P_n$ has the same [[Definition:Parity of Integer|parity]] as $n$.
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+:$P_0 = 0$
+which is [[Definition:Even Integer|even]].
+$P \left({1}\right)$ is the case:
+:$P_1 = 1$
+which is [[Definition:Odd Integer|odd]].
+Thus $P \left({n}\right)$ holds for $n = 0$ and $n = 1$.
+These comprise the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $P \left({j}\right)$ is true for all $j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\forall j \le k: P_j$ has the same [[Definition:Parity of Integer|parity]] as $j$
+from which it is to be shown that:
+:$P_{k + 1}$ has the same [[Definition:Parity of Integer|parity]] as $k + 1$.
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+By definition of [[Definition:Pell Numbers|Pell numbers]]:
+:$P_{k + 1} = 2 P_k + P_{k - 1}$
+$2 P_k$ is [[Definition:Even Integer|even]] by definition.
+By [[Sum of Even Integers is Even]], if $P_{k - 1}$ is [[Definition:Even Integer|even]] then so is $P_{k + 1}$.
+Similarly, if $P_{k - 1}$ is [[Definition:Odd Integer|odd]] then so is $P_{k + 1}$.
+Thus the [[Definition:Parity of Integer|parity]] of $P_{k + 1}$ matches the [[Definition:Parity of Integer|parity]] of $P_{k - 1}$.
+But by the [[Parity of Pell Numbers#Induction Hypothesis|induction hypothesis]], the [[Definition:Parity of Integer|parity]] of $P_{k - 1}$ matches $k - 1$.
+We have that:
+:$k + 1 = 2 + k - 1$
+and so the [[Definition:Parity of Integer|parity]] of $k + 1$ matches the [[Definition:Parity of Integer|parity]] of $k - 1$.
+Thus the [[Definition:Parity of Integer|parity]] of $P_{k + 1}$ matches $k + 1$.
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 0}: P_n$ has the same [[Definition:Parity of Integer|parity]] as $n$.
+{{qed}}
+[[Category:Pell Numbers]]
+t7sq0mgq4zfifnhkla1415qrce7jefg
+\end{proof}<|endoftext|>
+\section{Parity of Best Rational Approximations to Root 2}
+Tags: Square Root of 2
+
+\begin{theorem}
+Consider the [[Sequence of Best Rational Approximations to Square Root of 2]]:
+:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
+where $S_1 := \dfrac 1 1$.
+The [[Definition:Numerator|numerators]] of the [[Definition:Term of Sequence|terms]] of $\sequence S$ are all [[Definition:Odd Integer|odd]].
+For all $n$, the [[Definition:Parity of Integer|parity]] of the [[Definition:Denominator|denominator]] of [[Definition:Term of Sequence|term]] $S_n$ is the same as the [[Definition:Parity of Integer|parity]] of $n$.
+\end{theorem}
+
+\begin{proof}
+First the [[Definition:Parity of Integer|parity]] of the [[Definition:Numerator|numerators]] of the [[Definition:Term of Sequence|terms]] of $\sequence S$ is established.
+Let $\dfrac {p_n} {q_n}$ be a general [[Definition:Term of Sequence|term]] of $\sequence S$.
+By [[Relation between Adjacent Best Rational Approximations to Root 2]]:
+:$p_{n + 1} = p_n + 2 q_n$
+Thus if $p_n$ is [[Definition:Odd Integer|odd]] then so is $p_{n + 1}$.
+But $p_1 = 1$ is [[Definition:Odd Integer|odd]].
+So $p_n$ is [[Definition:Odd Integer|odd]] for all $n$, by [[Principle of Mathematical Induction]].
+The [[Definition:Denominator|denominators]] are the [[Definition:Pell Numbers|Pell numbers]].
+The result follows from [[Parity of Pell Numbers]].
+{{qed}}
+[[Category:Square Root of 2]]
+i9dhdw5j9ztimw3f2vzovtljpv3jeui
+\end{proof}<|endoftext|>
+\section{Best Rational Approximations to Root 2 generate Pythagorean Triples}
+Tags: Square Root of 2, Pythagorean Triples
+
+\begin{theorem}
+Consider the [[Sequence of Best Rational Approximations to Square Root of 2]]:
+:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
+Every other [[Definition:Term of Sequence|term]] of $\sequence S$ can be expressed as:
+:$\dfrac {2 a + 1} b$
+such that:
+:$a^2 + \left({a + 1}\right)^2 = b^2$
+:$b$ is [[Definition:Odd Integer|odd]].
+\end{theorem}
+
+\begin{proof}
+From [[Parity of Best Rational Approximations to Root 2]]:
+:The [[Definition:Numerator|numerators]] of the [[Definition:Term of Sequence|terms]] of $\sequence S$ are all [[Definition:Odd Integer|odd]].
+:For all $n$, the [[Definition:Parity of Integer|parity]] of the [[Definition:Denominator|denominator]] of [[Definition:Term of Sequence|term]] $S_n$ is the same as the [[Definition:Parity of Integer|parity]] of $n$.
+Thus it follows that every other [[Definition:Term of Sequence|term]] of $\sequence S$ has a [[Definition:Numerator|numerator]] and a [[Definition:Denominator|denominator]] which are both [[Definition:Odd Integer|odd]].
+{{ProofWanted|To be completed once several other results about Pell numbers established. Proof by {{AuthorRef|Artemas Martin}} establishes the triple $2 P_n P_{n + 1}, P_{n + 1}^2 - P_n^2, P_{n + 1}^2 + P_n^2 + P_{2 n + 1}$.}}
+\end{proof}<|endoftext|>
+\section{Prime Ideal is Prime Filter in Dual Lattice}
+Tags: Prime Ideals (Order Theory), Dual Ordered Sets
+
+\begin{theorem}
+Let $L = \struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Lattice|lattice]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then
+:$X$ is a [[Definition:Prime Ideal (Order Theory)|prime ideal]] in $L$
+{{iff}}:
+:$X$ is a [[Definition:Prime Filter (Order Theory)|prime filter]] in $L^{-1}$
+where $L^{-1} = \struct {S, \succeq}$ denotes the [[Definition:Dual Ordered Set|dual]] of $L$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $X$ be a [[Definition:Prime Ideal (Order Theory)|prime ideal]] in $L$.
+Then
+:$X$ is an [[Definition:Ideal (Order Theory)|ideal]] in $L$.
+By [[Ideal is Filter in Dual Ordered Set]]:
+:$X$ is a [[Definition:Filter|filter]] in $L^{-1}$.
+Let $x, y \in S$ such that
+:$x \vee' y \in X$
+where $\vee'$ denotes the [[Definition:Join (Order Theory)|join]] in $L^{-1}$.
+By [[Join is Dual to Meet]]:
+:$x \wedge y \in X$
+Thus by [[Characterization of Prime Ideal]]:
+:$x \in X$ or $y \in X$
+Hence $X$ is a [[Definition:Prime Filter (Order Theory)|prime filter]]
+{{qed|lemma}}
+=== Necessary Condition ===
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ideal is Filter in Dual Ordered Set}
+Tags: Order Theory, Dual Ordered Sets
+
+\begin{theorem}
+Let $P = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Then
+:$X$ is [[Definition:Ideal (Order Theory)|ideal]] in $P$
+{{iff}}
+:$X$ is [[Definition:Filter|filter]] in $P^{-1}$
+where $P^{-1} = \left({S, \succeq}\right)$ denotes the [[Definition:Dual Ordered Set|dual]] of $P$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $X$ be [[Definition:Ideal (Order Theory)|ideal]] in $P$.
+By definition of [[Definition:Ideal in Ordered Set|ideal in ordered set]]:
+:$X$ is [[Definition:Non-Empty Set|non-empty]] [[Definition:Directed Subset|directed]] [[Definition:Lower Set|lower]].
+By definition of [[Definition:Directed Subset|directed]]:
+:$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
+Then
+:$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
+By definition
+:$X$ is [[Definition:Filtered Subset|filtered]] in $P^{-1}$
+By definition of [[Definition:Lower Set|lower set]]:
+:$\forall x \in X, y \in S: y \preceq x \implies y \in X$
+Then
+:$\forall x \in X, y \in S: x \succeq y \implies y \in X$
+By definition
+:$X$ is an [[Definition:Upper Set|upper set]] in $P^{-1}$.
+Thus by definition of [[Definition:Filter in Ordered Set|filter in ordered set]]
+:$X$ is a [[Definition:Filter|filter]] in $P^{-1}$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $X$ be a [[Definition:Filter|filter]] in $P^{-1}$.
+By definition of [[Definition:Filter in Ordered Set|filter in ordered set]]
+:$X$ is [[Definition:Non-Empty Set|non-empty]], [[Definition:Filtered Subset|filtered]] in $P^{-1}$, and [[Definition:Upper Set|upper]] in $P^{-1}$.
+By definition of [[Definition:Filtered Subset|filtered]] in $P^{-1}$: 
+:$\forall x, y \in X: \exists z \in X: z \succeq x \land z \succeq y$
+Then
+:$\forall x, y \in X: \exists z \in X: x \preceq z \land y \preceq z$
+By definition
+:$X$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Upper Set|upper set]] in $P^{-1}$:
+:$\forall x \in X, y \in S: x \succeq y \implies y \in X$
+Then
+:$\forall x \in X, y \in S: y \preceq x \implies y \in X$
+By definition
+:$X$ is a [[Definition:Lower Set|lower set]].
+Thus by definition of [[Definition:Ideal in Ordered Set|ideal in ordered set]]:
+:$X$ is an [[Definition:Ideal (Order Theory)|ideal]] in $P$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of 2 as Sum of Egyptian Fractions}
+Tags: Square Root of 2
+
+\begin{theorem}
+The [[Definition:Square Root|square root]] of $2$ can be approximated by the following sequence of [[Definition:Egyptian Fraction|Egyptian fractions]]:
+:$\sqrt 2 = 1 + \dfrac 1 3 + \dfrac 1 {253} + \dfrac 1 {218 \, 201} + \dfrac 1 {61 \, 323 \, 543 \, 802} + \cdots$
+{{OEIS|A006487}}
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|If anyone wants to do the arithmetic they are welcome to complete this.}}
+\end{proof}<|endoftext|>
+\section{Sprague's Property of Root 2}
+Tags: Square Root of 2, Beatty Sequences
+
+\begin{theorem}
+Let $S = \sequence {s_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Fraction|fractions]] defined as follows:
+Let the [[Definition:Numerator|numerator]] of $s_n$ be:
+:$\floor {n \sqrt 2}$
+where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$.
+Let the [[Definition:Denominator|denominators]] of the [[Definition:Term of Sequence|terms]] of $S$ be the [[Definition:Strictly Positive Integer|(strictly) positive integers]] missing from the [[Definition:Numerator|numerators]] of $S$:
+:$S := \dfrac 1 3, \dfrac 2 6, \dfrac 4 {10}, \dfrac 5 {13}, \dfrac 7 {17}, \dfrac 8 {20}, \ldots$
+Then the difference between the [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] of $s_n$ is equal to $2 n$.
+\end{theorem}
+
+\begin{proof}
+Denote the [[Definition:Numerator|numerators]] of the [[Definition:Term of Sequence|terms]] of $S$ as $\sequence {N_n}$.
+Denote the [[Definition:Denominator|denominators]] of the [[Definition:Term of Sequence|terms]] of $S$ as $\sequence {D_n}$.
+From the definition:
+:$\sequence {N_n}$ is a [[Definition:Beatty Sequence|Beatty sequence]], where $\sequence {N_n} = \BB_{\sqrt 2} = \sequence{\floor{n \sqrt 2} }_{n \mathop \in \Z_{> 0} }$
+:$\sequence {N_n}$ and $\sequence {D_n}$ are [[Definition:Beatty Sequence/Complementary|complementary Beatty sequences]].
+Then by [[Beatty's Theorem]], $\sequence {D_n}$ is a [[Definition:Beatty Sequence|Beatty sequence]].
+Define $\sequence {D_n} = \BB_y = \sequence{\floor{n y} }_{n \mathop \in \Z_{> 0} }$.
+Then we have: 
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 {\sqrt 2} + \dfrac 1 y
+      | r = 1
+      | c = [[Beatty's Theorem]]
+}}
+{{eqn | ll = \leadsto
+      | l = \dfrac 1 y
+      | r = 1 - \dfrac 1 {\sqrt 2}
+}}
+{{eqn | r = \dfrac {\sqrt 2 - 1} {\sqrt 2}
+}}
+{{eqn | ll = \leadsto
+      | l = y
+      | r = \dfrac {\sqrt 2} {\sqrt 2 - 1} 
+}}
+{{eqn | r = \dfrac {\sqrt 2 \paren {\sqrt 2 + 1} } {\paren {\sqrt 2 - 1} \paren {\sqrt 2 + 1} }
+}}
+{{eqn | r = 2 + \sqrt 2
+}}
+{{end-eqn}}
+So we have $s_n = \dfrac {N_n} {D_n} = \dfrac {\floor {n \sqrt 2} } {\floor {n \paren {2 + \sqrt 2} } }$.
+The difference between the [[Definition:Numerator|numerator]] and [[Definition:Denominator|denominator]] of $s_n$ is $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$.
+From [[Integer equals Floor iff Number between Integer and One More]]:
+{{begin-eqn}}
+{{eqn | l = n \paren {2 + \sqrt 2} 
+      | o = <
+      | m = \floor {n \paren {2 + \sqrt 2} }
+      | mo = <
+      | r = n \paren {2 + \sqrt 2} + 1
+}}
+{{eqn | l = n \sqrt 2
+      | o = <
+      | m = \floor {n \sqrt 2}
+      | mo = <
+      | r = n \sqrt 2 + 1
+}}
+{{eqn | ll = \leadsto
+      | l = n \paren {2 + \sqrt 2} - \paren {n \sqrt 2 + 1}
+      | o = <
+      | m = \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}
+      | mo = <
+      | r = n \paren {2 + \sqrt 2} + 1 - n \sqrt 2
+}}
+{{eqn | ll = \leadsto
+      | l = 2 n - 1
+      | o = <
+      | m = \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}
+      | mo = <
+      | r = 2 n + 1
+}}
+{{end-eqn}}
+The only [[Definition:Integer|integer]] [[Definition:Strictly Between|strictly between]] $2 n - 1$ and $2 n + 1$ is $2 n$, and $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$ is an [[Definition:Integer|integer]].
+Therefore $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2} = 2 n$.
+{{qed}}
+{{Namedfor|Roland Percival Sprague|cat = Sprague, Roland}}
+\end{proof}
\ No newline at end of file