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+\section{Archimedes' Limits to Value of Pi}
+Tags: Analysis
+
+\begin{theorem}
+The value of $\pi$ lies between $3 \frac {10} {71}$ and $3 \frac 1 7$:
+:$3 \dfrac {10} {71} < \pi < 3 \dfrac 1 7$
+\end{theorem}<|endoftext|>
+\section{Surface Area of Sphere}
+Tags: Spheres, Area Formulas
+
+\begin{theorem}
+The [[Definition:Surface Area|surface area]] $A$ of a [[Definition:Sphere (Geometry)|sphere]] whose [[Definition:Radius of Sphere|radius]] $a$ is given by:
+:$A = 4 \pi a^2$
+\end{theorem}<|endoftext|>
+\section{Upper Bound for Number of Grains of Sand to fill Universe}
+Tags: Number Theory
+
+\begin{theorem}
+The number of grains of sand that would fill the [[Definition:Physical Universe|Universe]], to the extent it was believed to exist in the days of {{AuthorRef|Archimedes of Syracuse|Archimedes}}, is [[Definition:Bounded Above|bounded above]] by $10^{63}$.
+\end{theorem}<|endoftext|>
+\section{Upper Adjoint Preserves All Infima}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $g: S \to T$ be an [[Definition:Galois Connection|upper adjoint of Galois connection]].
+Then $g$ [[Definition:Mapping Preserves Infimum/All|preserves all infima]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Galois Connection|upper adjoint]]
+:$\exists d: T \to S: \left({g, d}\right)$ is a [[Definition:Galois Connection|Galois connection]]
+Let $X$ be a [[Definition:Subset|subset]] of $S$ such that
+:$X$ admits an [[Definition:Infimum of Set|infimum]].
+We will prove as lemma 1 that
+:$\forall t \in T: t$ is [[Definition:Lower Bound of Set|lower bound]] for $g^\to\left({X}\right) \implies t \precsim g\left({\inf X}\right)$
+Let $t \in T$ such that
+:$t$ is [[Definition:Lower Bound of Set|lower bound]] for $g^\to\left({X}\right)$
+We will prove as sublemma that
+:$d\left({t}\right)$ is [[Definition:Lower Bound of Set|lower bound]] for $X$
+Let $s \in X$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$g\left({s})\right) \in g^\to\left({X}\right)$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$t \precsim g\left({s}\right)$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$d\left({t}\right) \preceq s$
+This ends the proof of sublemma.
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$d\left({t}\right) \preceq \inf X$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$t \precsim g\left({\inf X}\right)$
+This ends the proof of lemma 1.
+We will prove as lemma 2 that
+:$g\left({\inf X}\right)$ is [[Definition:Lower Bound of Set|lower bound]] for $g^\to\left({X}\right)$
+Let $t \in g^\to\left({X}\right)$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists s \in S: s \in X \land g\left({s}\right) = t$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\inf X$ is [[Definition:Lower Bound of Set|lower bound]] for $X$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$\inf X \preceq s$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$g$ is [[Definition:Increasing Mapping|increasing mapping]].
+Thus by definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$g\left({\inf X}\right) \precsim t$
+This ends the proof of lemma 2.
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$g^\to\left({X}\right)$ admits an [[Definition:Infimum of Set|infimum]]
+and
+:$\inf\left({g^\to\left({X}\right)}\right) = g\left({\inf X}\right)$
+Thus by definition:
+:$g$ [[Definition:Mapping Preserves Infimum/Subset|preserves infimum]] on $X$
+Thus by definition:
+:$g$ [[Definition:Mapping Preserves Infimum/All|preserves all infima]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Adjoint Preserves All Suprema}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $d: T \to S$ be an [[Definition:Galois Connection|lower adjoint of Galois connection]].
+Then $d$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Galois Connection|lower adjoint]]
+:$\exists g: S \to T: \left({g, d}\right)$ is a [[Definition:Galois Connection|Galois connection]]
+Let $X$ be a [[Definition:Subset|subset]] of $T$ such that
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+We will prove as lemma 1 that
+:$\forall s \in S: s$ is [[Definition:Upper Bound of Set|upper bound]] for $d^\to\left({X}\right) \implies d\left({\sup X}\right) \preceq s$
+Let $s \in S$ such that
+:$s$ is [[Definition:Upper Bound of Set|upper bound]] for $d^\to\left({X}\right)$
+We will prove as sublemma that
+:$g\left({s}\right)$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+Let $t \in X$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$d \left({t})\right) \in d^\to \left({X}\right)$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$d\left({t}\right) \preceq s$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$t \precsim g\left({s}\right)$
+This ends the proof of sublemma.
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup X \precsim g\left({t}\right)$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$d\left({\sup X}\right) \preceq t$
+This ends the proof of lemma 1.
+We will prove as lemma 2 that
+:$d\left({\sup X}\right)$ is [[Definition:Upper Bound of Set|upper bound]] for $d^\to\left({X}\right)$
+Let $s \in d^\to\left({X}\right)$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists t \in T: t \in X \land d\left({t}\right) = s$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup X$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$t \precsim \sup X$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$d$ is [[Definition:Increasing Mapping|increasing mapping]].
+Thus by definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$s \preceq d\left({\sup X}\right)$
+This ends the proof of lemma 2.
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$d^\to\left({X}\right)$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$\sup\left({d^\to\left({X}\right)}\right) = d\left({\sup X}\right)$
+Thus by definition:
+:$g$ [[Definition:Mapping Preserves Supremum/Subset|preserves supremum]] on $X$
+Thus by definition:
+:$g$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Oscillation at Point (Infimum) equals Oscillation at Point (Limit)}
+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 $\map {\omega_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Infimum|oscillation]] of $f$ at $x$ as defined by:
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+where $\map {\omega_f} I$ is the [[Definition:Oscillation on Real Subset|oscillation]] of $f$ on a [[Definition:Real Number|real]] [[Definition:Set|set]] $I$:
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\map {\omega^L_f} x$ be the [[Definition:Oscillation/Real Space/Oscillation at Point/Limit|oscillation]] of $f$ at $x$ as defined by:
+:$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$
+Then:
+:$\map {\omega_f} x \in \R$ {{iff}} $\map {\omega^L_f} x \in \R$
+and, if $\map {\omega_f} x$ and $\map {\omega^L_f} x$ exist as [[Definition:Real Number|real numbers]]:
+:$\map {\omega_f} x = \map {\omega^L_f} x$
+\end{theorem}
+
+\begin{proof}
+=== [[Oscillation at Point (Infimum) equals Oscillation at Point (Limit)/Lemma|Lemma]] ===
+{{:Oscillation at Point (Infimum) equals Oscillation at Point (Limit)/Lemma}} {{qed|lemma}}
+=== Necessary Condition ===
+Let $\map {\omega_f} x \in \R$.
+We need to prove:
+:$\map {\omega^L_f} x \in \R$
+:$\map {\omega^L_f} x = \map {\omega_f} x$
+where:
+:$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$
+:$\map {\omega_f} {\openint {x - h} {x + h} } = \sup \set {\size {\map f y - \map f z}: y, z \in \openint {x - h} {x + h} \cap D}$
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+Let $\epsilon \in \R_{>0}$.
+Then an $I \in N_x$ exists such that:
+:$\map {\omega_f} I - \map {\omega_f} x < \epsilon$ by [[Infimum of Set of Oscillations on Set is Arbitrarily Close]]
+Let $I$ be such an [[Definition:Element|element]] of $N_x$.
+We observe in particular that $\map {\omega_f} I \in \R$.
+A [[Definition:Neighborhood (Real Analysis)/Open Subset|neighborhood]] in $N_x$ contains an [[Definition:Open Set/Real Analysis|open]] [[Definition:Subset|subset]] that contains the [[Definition:Element|point]] $x$.
+So, $I$ [[Definition:Subset|contains]] such an [[Definition:Open Set/Real Analysis|open]] [[Definition:Subset|subset]] as $I \in N_x$.
+Therefore, a $\delta \in \R_{>0}$ exists such that $\openint {x - \delta} {x + \delta}$ is a [[Definition:Subset|subset]] of $I$.
+Let $h$ be a [[Definition:Real Number|real number]] that satisfies: $0 < h < \delta$.
+We observe that $\openint {x - h} {x + h} \subset I$.
+Also, $\openint {x - h} {x + h} \in N_x$.
+We have:
+:$I \in N_x$
+:$\openint {x - h} {x + h} \in N_x$
+:$\openint {x - h} {x + h} \subset I$
+:$\map {\omega_f} I \in \R$
+from which follows 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:
+:$\map {\omega_f} {\openint {x - h} {x + h} } \in \set {\map {\omega_f} {I'}: I' \in N_x}$
+as $\openint {x - h} {x + h} \in N_x$.
+Also, $\map {\omega_f} x$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] for $\set {\map {\omega_f} {I'}: I' \in N_x}$ by the definition of $\map {\omega_f} x$.
+Therefore:
+:$\map {\omega_f} {\openint {x - h} {x + h} } \ge \map {\omega_f} x$
+We find:
+{{begin-eqn}}
+{{eqn | l = \map {\omega_f} x \le \map {\omega_f} {\openint {x - h} {x + h} }
+      | o = \le
+      | r = \map {\omega_f} I
+}}
+{{eqn | ll= \leadsto
+      | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x
+      | o = \le
+      | r = \map {\omega_f} I - \map {\omega_f} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x
+      | o = \le
+      | r = \map {\omega_f} I - \map {\omega_f} x < \epsilon
+      | c = as $\map {\omega_f} I - \map {\omega_f} x < \epsilon$ is true
+}}
+{{eqn | ll= \leadsto
+      | l = 0 \le \map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {\map {\omega_f} {\openint {x - h} {x + h} } - \map {\omega_f} x}
+      | o = <
+      | r = \epsilon
+      | c = 
+}}
+{{end-eqn}}
+which means that $\displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$ exists and equals $\map {\omega_f} x$ by the definition of [[Definition:Limit from Right|limit]].
+In other words, $\map {\omega^L_f} x \in \R$ and $\map {\omega^L_f} x = \map {\omega_f} x$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\map {\omega^L_f} x \in \R$.
+We need to prove:
+:$\map {\omega_f} x \in \R$
+:$\map {\omega_f} x = \map {\omega^L_f} x$
+where:
+:$\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
+:$\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
+:$\map {\omega^L_f} x = \displaystyle \lim_{h \mathop \to 0^+} \map {\omega_f} {\openint {x - h} {x + h} }$
+We have:
+:$\displaystyle \lim_{h \mathop \to 0^+} \map{\omega_f} {\openint {x - h} {x + h} } \in \R$ as $\map {\omega^L_f} x \in \R$
+Therefore, $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$ for a small enough $h$ in $\R_{>0}$ by the definition of [[Definition:Limit from Right|limit]].
+Let $h$ be such a [[Definition:Real Number|real number]].
+We observe that $\openint {x - h} {x + h}$ is a [[Definition:Neighborhood (Real Analysis)/Open Subset|neighborhood]] in $N_x$.
+We have:
+:$\openint {x - h} {x + h} \in N_x$
+:$\map {\omega_f} {\openint {x - h} {x + h} } \in \R$
+Accordingly:
+:$\map {\omega_f} x \in \R$ by [[Infimum of Set of Oscillations on Set]]
+$\map {\omega_f} x = \map {\omega^L_f} x$ follows by [[Oscillation at Point (Infimum) equals Oscillation at Point (Limit)/Lemma|Lemma]].
+This finishes the proof of the theorem.
+{{qed}}
+[[Category:Real Analysis]]
+9fanzs4g1wiajx7jhvn2b5rk2aq0wir
+\end{proof}<|endoftext|>
+\section{Trisecting the Angle/Parabola}
+Tags: Trisecting the Angle
+
+\begin{theorem}
+Let $\alpha$ be an [[Definition:Angle|angle]] which is to be [[Definition:Trisection|trisected]].
+This can be achieved by means of a [[Definition:Parabola|parabola]].
+\end{theorem}
+
+\begin{proof}
+{{tidy|eqn template}}
+{{MissingLinks}}
+First, notice that since $A$ lies on $\CC_1$, then $A = \tuple {\cos \angle POQ, \sin \angle POQ}$.
+This means that $B = \tuple {0, \sin \angle POQ}$.
+Since $C$ is the [[Definition:Line/Midpoint|midpoint]] of $AB$, we have that $C = \tuple {\dfrac {\cos \angle POQ} 2, \sin \angle POQ}$.
+Because $D$ lies on $\CC_1$, then $D = \tuple {0, 1}$ which makes $E = \tuple {\dfrac {\cos \angle POQ} 2, 1}$.
+From [[Equation of Circle]], $C_2$ has the equation:
+:$\paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {y -1}^2 = \dfrac {\cos^2 \angle POQ} 4 + 1$
+Because $F$ lies on both $\CC_2$ and $\PP$, we can use substitution to solve for the [[Definition:X Coordinate|$x$-coordinate]] of $F$:
+{{begin-eqn}}
+{{eqn | l = \paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {2 x^2 - 1}^2
+      | r = \dfrac {\cos^2 \angle POQ} 4 + 1
+}}
+{{eqn | ll = \leadsto
+      | l = x^2 - x \cos \angle POQ + \dfrac {\cos^2 \angle POQ} 4 + 4 x^4 - 4 x^2 + 1
+      | r = \dfrac {\cos^2 \angle POQ} 4 + 1
+}}
+{{eqn | ll = \leadsto
+      | l = x^2 - x \cos \angle POQ + 4 x^4 - 4 x^2
+      | r = 0
+}}
+{{eqn | ll = \leadsto
+      | l = 4 x^4 - 3 x^2 - x \cos \angle POQ
+      | r = 0
+}}
+{{eqn | ll = \leadsto
+      | l = x \paren {4 x^3 - 3 x - \cos \angle POQ}
+      | r = 0
+}}
+{{end-eqn}}
+Thus $x = 0$ or $4 x^3 - 3 x - \cos \angle POQ = 0$.
+This confirms what is obvious by the construction, namely that $\CC_2$ and $\PP$ intersect at the origin.
+We can also see that the solution we are after must lie in the first quadrant also, where $x \neq 0$.
+So there must be one positive $x$ such that $4 x^3 - 3 x = \cos \angle POQ$
+By the [[Triple Angle Formula for Cosine]] $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$, it is clear that one solution that works is:
+:$x = \map \cos {\dfrac {\angle POQ} 3}$
+Since, by construction, there is a vertical line through $F$ and $H$ and $H$ lies on the unit circle $\CC_1$, then $H = \tuple {\cos \dfrac {\angle POQ} 3, \sin \dfrac {\angle POQ} 3}$
+Therefore, $\angle POH = \dfrac {\angle POQ} 3$
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{All Infima Preserving Mapping is Upper Adjoint of Galois Connection}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $\left({T, \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $g: S \to T$ be [[Definition:Mapping Preserves Infimum/All|all infima preserving mapping]].
+Then there exists a [[Definition:Mapping|mapping]] $d: T \to S$ such that $\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]] and
+:$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+where
+:$\min$ denotes the [[Definition:Smallest/Ordered Set/Subset|minimum]]
+:$g^{-1}\left[{t^\succsim}\right]$ denotes the [[Definition:Image of Subset under Mapping|image]] of $t^\succsim$ under relation $g^{-1}$
+:$t^\succsim$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $t$
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Mapping|mapping]] $d: T \to S$:
+:$\forall t \in T:d\left({t}\right) := \inf\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+We will prove as lemma 1 that
+:$g$ is an [[Definition:Increasing Mapping|increasing mapping]].
+Let $x, y \in S$ such that
+:$x \preceq y$
+By [[Upper Closure is Decreasing]]:
+:$y^\succeq \subseteq x^\succeq$
+By [[Infimum of Upper Closure of Element]]:
+:$\inf\left({x^\succeq}\right) = x$ and $\inf\left({y^\succeq}\right) = y$
+By definition of [[Definition:Mapping Preserves Infimum/All|mapping preserves all infima]]:
+:$g$ [[Definition:Mapping Preserves Infimum/Subset|preserves the infimum]] on $x^\succeq$ and $g$ [[Definition:Mapping Preserves Infimum/Subset|preserves the infimum]] on $y^\succeq$
+By definition of [[Definition:Mapping Preserves Infimum/Subset|mapping preserves the infimum of set]]:
+:$\inf\left({g^\to\left({x^\succeq}\right)}\right) = g\left({x}\right)$ and $\inf\left({g^\to\left({y^\succeq}\right)}\right) = g\left({y}\right)$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$g^\to\left({y^\succeq}\right) \subseteq g^\to\left({x^\succeq}\right)$
+Thus by [[Infimum of Subset]]:
+:$g\left({x}\right) \precsim g\left({y}\right)$
+This ends the proof of lemma 1.
+We will prove as lemma 2 that
+:$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+Let $t \in T$.
+By definition of $d$:
+:$d\left({t}\right) = \inf\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+By [[Image of Inverse Image]]:
+:$g\left[{g^{-1}\left[{t^\succsim}\right]}\right] \subseteq t^\succsim$
+By [[Infimum of Subset]] and [[Infimum of Upper Closure of Element]]:
+:$t = \inf\left({t^\succsim}\right) \precsim \inf\left({g\left[{g^{-1}\left[{t^\succsim}\right]}\right]}\right)$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$\inf\left({g\left[{g^{-1}\left[{t^\succsim}\right]}\right]}\right) \in t^\succsim$
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$g^{-1}\left[{t^\succsim}\right]$ admits an [[Definition:Infimum of Set|infimum]]
+By definitions [[Definition:Mapping Preserves Infimum|mapping preserves the infimum]]:
+:$\inf\left({g\left[{g^{-1}\left[{t^\succsim}\right]}\right]}\right) = g\left({d\left({t}\right)}\right)$
+Thus by definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$d\left({t}\right) \in g^{-1}\left[{t^\succsim}\right]$
+Thus
+:$d\left({t}\right) = \min \left({ g^{-1}\left[{t^\succsim}\right] }\right)$
+This ends the proof of lemma 2.
+Thus by [[Galois Connection is Expressed by Minimum]]:
+:$\left({g, d}\right)$ is a [[Definition:Galois Connection|Galois connection]].
+Thus by lemma 2:
+:$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Theorem/General Binomial Theorem/Historical Note}
+Tags: Historical Notes
+
+\begin{theorem}
+The [[General Binomial Theorem]] was first conceived by {{AuthorRef|Isaac Newton}} during the years $1665$ to $1667$ when he was living in his home in Woolsthorpe.
+He announced the result formally, in letters to [https://en.wikipedia.org/wiki/Henry_Oldenburg Henry Oldenburg] on $13$th June $1676$ and $24$th October $1676$ but did not provide a proper proof (at that time the need for the appropriate level of rigor had not been recognised).
+{{AuthorRef|Leonhard Paul Euler}} made an incomplete attempt in $1774$, but the full proof had to wait for {{AuthorRef|Carl Friedrich Gauss}} to provide it in $1812$.
+This was, in fact, the first time anything about [[Definition:Infinite Summation|infinite summations]] was proved adequately.
+\end{theorem}<|endoftext|>
+\section{Image of Inverse Image}
+Tags: Mapping Theory
+
+\begin{theorem}
+Let $S, T$ be [[Definition:Set|sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let $X$ be a [[Definition:Subset|subset]] of $T$.
+Then:
+:$f\left[{f^{-1}\left[{X}\right]}\right] \subseteq X$
+where:
+:$f^{-1}\left[{X}\right]$ denotes the [[Definition:Image of Subset under Relation|image]] of $X$ under the [[Definition:Relation|relation]] $f^{-1}$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in f \left[{f^{-1} \left[{X}\right]}\right]$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists y \in S: y \in f^{-1} \left[{X}\right] \land x = f \left({y}\right)$
+By definition of [[Definition:Image of Subset under Relation|image of set under relation]]:
+:$f \left({y}\right) \in X$
+Thus $x \in X$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{All Suprema Preserving Mapping is Lower Adjoint of Galois Connection}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $\struct {T, \precsim}$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $d: T \to S$ be [[Definition:Mapping Preserves Supremum/All|all suprema preserving mapping]].
+Then there exists a [[Definition:Mapping|mapping]] $g: S \to T$ such that $\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]] and
+:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+where
+:$\min$ denotes the [[Definition:Greatest Element of Subset|maximum]]
+:$d^{-1} \sqbrk {s^\preceq}$ denotes the [[Definition:Image of Subset under Relation|image]] of $s^\preceq$ under relation $d^{-1}$
+:$s^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $t$
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Mapping|mapping]] $d: T \to S$:
+:$\forall s \in S: \map g s := \map \sup {d^{-1} \sqbrk {s^\preceq} }$
+We will prove as lemma 1 that:
+:$d$ is an [[Definition:Increasing Mapping|increasing mapping]].
+Let $x, y \in T$ such that
+:$x \precsim y$
+By [[Lower Closure is Increasing]]:
+:$x^\precsim \subseteq y^\precsim$
+By [[Supremum of Lower Closure of Element]]:
+:$\map \sup {x^\precsim} = x$ and $\map \sup {y^\precsim} = y$
+By definition of [[Definition:Mapping Preserves Supremum/All|mapping preserves all suprema]]:
+:$d$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] on $x^\precsim$ and $d$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] on $y^\precsim$
+By definition of [[Definition:Mapping Preserves Supremum/Subset|mapping preserves the supremum on set]]:
+:$\map \sup {\map {d^\to} {x^\precsim} } = \map d x$ and $\map \sup {\map {d^\to} {y^\precsim} } = \map d y$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$\map {d^\to} {x^\precsim} \subseteq \map {d^\to} {y^\precsim}$
+Thus by [[Supremum of Subset]]:
+:$\map d x \preceq \map d y$
+This ends the proof of lemma 1.
+We will prove as lemma 2 that
+:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+Let $s \in S$.
+By definition of $g$:
+:$\map g s = \map \sup {d^{-1} \sqbrk {s^\preceq} }$
+By [[Image of Inverse Image]]:
+:$d \sqbrk {d^{-1} \sqbrk {s^\preceq} } \subseteq s^\preceq$
+By [[Supremum of Subset]] and [[Supremum of Lower Closure of Element]]:
+:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \preceq \map \sup {s^\preceq} = s$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \in s^\preceq$
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$d^{-1} \sqbrk {s^\preceq}$ admits a [[Definition:Supremum of Set|supremum]]
+By definition of [[Definition:Mapping Preserves Supremum|mapping preserves the supremum]]:
+:$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } = \map d {\map g s}$
+Thus by definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\map g s \in d^{-1} \sqbrk {s^\preceq}$
+Thus
+:$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+This ends the proof of lemma 2.
+Thus by [[Galois Connection is Expressed by Maximum]]:
+:$\tuple {g, d}$ is a [[Definition:Galois Connection|Galois connection]].
+Thus by lemma 2:
+:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative at Point of Inflection}
+Tags: Differential Calculus
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice]] [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\left({a \,.\,.\, b}\right)$.
+Let $f$ have a [[Definition:Point of Inflection|point of inflection]] at $\xi \in \left({a \,.\,.\, b}\right)$.
+Then:
+:$f'' \left({\xi}\right) = 0$
+where $f'' \left({\xi}\right)$ denotes the [[Definition:Second Derivative|second derivative]] of $f$ at $\xi$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Point of Inflection|point of inflection]], $f'$ has either a [[Definition:Local Maximum|local maximum]] or a [[Definition:Local Minimum|local minimum]] at $\xi$.
+From [[Derivative at Maximum or Minimum]], it follows that the [[Definition:Derivative|derivative]] of $f'$ at $\xi$ is zero, that is:
+:$f'' \left({\xi}\right) = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Galois Connection Implies Order on Mappings}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be [[Definition:Ordered Set|ordered sets]].
+Let $g: S \to T$ and $d: T \to S$ be [[Definition:Mapping|mappings]] such that
+:$\tuple {g, d}$ is [[Definition:Galois Connection|Galois connection]].
+Then $d \circ g \preceq I_S$ and $I_T \precsim g \circ d$
+where
+:$\preceq, \precsim$ denote the [[Definition:Ordering on Mappings|orderings on mappings]],
+:$I_S$ denotes the [[Definition:Identity Mapping|identity mapping]] of $S$
+\end{theorem}
+
+\begin{proof}
+Let $s \in S$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\map g s \precsim \map g s$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$\map d {\map g s} \preceq s$
+By definition of [[Definition:Composition of Mappings|composition]]:
+:$\map {\paren {d \circ g} } s \preceq s$
+By definition of [[Definition:Identity Mapping|identity mapping]]:
+:$\map {\paren {d \circ g} } s \preceq \map {I_S} s$
+Thus by definition of [[Definition:Order on Mappings|order on mappings]]:
+:$d \circ g \preceq I_S$
+Let $t \in T$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\map d t \preceq \map d t$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$t \precsim \map g {\map d t}$
+By definition of [[Definition:Composition of Mappings|composition]]:
+:$t \precsim \map {\paren {g \circ d} } t$
+By definition of [[Definition:Identity Mapping|identity mapping]]:
+:$\map {I_T} t \precsim \map {\paren {g \circ d} } t$
+Thus by definition of [[Definition:Order on Mappings|order on mappings]]:
+:$I_T \precsim g \circ d$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering on Mappings Implies Galois Connection}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $g: S \to T$ and $d: T \to S$ be [[Definition:Mapping|mappings]] such that:
+:$g$ and $d$ are [[Definition:Increasing Mapping|increasing mappings]] 
+and
+:$d \circ g \preceq I_S$ and $I_T \precsim g \circ d$
+Then
+:$\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]].
+where
+:$\preceq, \precsim$ denote the [[Definition:Ordering on Mappings|orderings on mappings]]
+:$I_S$ denotes the [[Definition:Identity Mapping|identity mapping]] of $S$
+:$\circ$ denotes the [[Definition:Composition of Mappings|composition of mappings]].
+\end{theorem}
+
+\begin{proof}
+We will prove that
+:$\forall s \in S, t \in T: t \precsim g\left({s}\right) \iff d\left({t}\right) \preceq s$
+Let $s \in S, t \in T$.
+First implication:
+Let
+:$t \precsim g\left({s}\right)$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$d\left({t}\right) \preceq d\left({g\left({s}\right)}\right)$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$\left({d \circ g}\right)\left({s}\right) \preceq I_S\left({s}\right)$
+By definition of [[Definition:Composition of Mappings|composition]]:
+:$d\left({g\left({s}\right)}\right) \preceq I_S\left({s}\right)$
+By definition of [[Definition:Identity Mapping|identity mapping]]:
+:$d\left({g\left({s}\right)}\right) \preceq s$
+Thus by definition of [[Definition:Transitivity (Relation Theory)|transitivity]]:
+:$d\left({t}\right) \preceq s$
+Second implication:
+Let
+:$d\left({t}\right) \preceq s$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$g\left({d\left({t}\right)}\right) \precsim g\left({s}\right)$
+By definition of [[Definition:Ordering on Mappings|ordering on mappings]]:
+:$I_T\left({t}\right) \precsim \left({g \circ d}\right)\left({t}\right)$
+By definition of [[Definition:Composition of Mappings|composition]]:
+:$I_T\left({t}\right) \precsim g\left({d\left({t}\right)}\right)$
+By definition of [[Definition:Identity Mapping|identity mapping]]:
+:$t \precsim g \left({d\left({t}\right)}\right)$
+Thus by definition of [[Definition:Transitivity (Relation Theory)|transitivity]]:
+:$t \precsim g \left({s}\right)$
+Thus by definition:
+:$\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Liouville's Theorem (Hamiltonian Mechanics)}
+Tags: Hamiltonian Mechanics
+
+\begin{theorem}
+[[Definition:Volume Integral|Volume integrals]] are [[Definition:Time Invariant|time invariant]] in [[Definition:Phase Space|phase space]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Joseph Liouville|cat = Liouville}}
+\end{proof}<|endoftext|>
+\section{Primitive of exp (-x^2) has no Solution in Elementary Functions}
+Tags: Primitives involving Exponential Function
+
+\begin{theorem}
+The [[Definition:Primitive (Calculus)|primitive]]:
+:$\displaystyle \int \map \exp {-x^2} \rd x$
+cannot be expressed in terms of a [[Definition:Finite Set|finite number]] of [[Definition:Elementary Function|elementary functions]].
+\end{theorem}<|endoftext|>
+\section{Primitive of exp x over x has no Solution in Elementary Functions}
+Tags: Primitives involving Exponential Function
+
+\begin{theorem}
+The [[Definition:Primitive (Calculus)|primitive]]:
+:$\displaystyle \int \frac {e^{a x} \, \mathrm d x} x$
+cannot be expressed in terms of a [[Definition:Finite Set|finite number]] of [[Definition:Elementary Function|elementary functions]].
+\end{theorem}<|endoftext|>
+\section{Primitive of Sine of x over x has no Solution in Elementary Functions}
+Tags: Primitives involving Sine Function
+
+\begin{theorem}
+The [[Definition:Primitive (Calculus)|primitive]]:
+:$\displaystyle \int \frac {\sin x \, \mathrm d x} x$
+cannot be expressed in terms of a [[Definition:Finite Set|finite number]] of [[Definition:Elementary Function|elementary functions]].
+\end{theorem}<|endoftext|>
+\section{Primitive of Reciprocal of Logarithm of x has no Solution in Elementary Functions}
+Tags: Primitives involving Logarithm Function
+
+\begin{theorem}
+The [[Definition:Primitive (Calculus)|primitive]]:
+:$\displaystyle \int \frac {\mathrm d x} {\ln x}$
+cannot be expressed in terms of a [[Definition:Finite Set|finite number]] of [[Definition:Elementary Function|elementary functions]].
+\end{theorem}<|endoftext|>
+\section{Pi is Transcendental}
+Tags: Transcendental Number Theory, Pi
+
+\begin{theorem}
+[[Definition:Pi|$\pi$ (pi)]] is [[Definition:Transcendental Number|transcendental]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Contradiction]]:
+{{AimForCont}} $\pi$ is not [[Definition:Transcendental Number|transcendental]].
+Hence by definition, $\pi$ is [[Definition:Algebraic Number|algebraic]].
+Let $\pi$ be the root of a non-zero [[Definition:Polynomial (Analysis)|polynomial]] with [[Definition:Rational Number|rational]] coefficients, namely $\map f x$.
+Then, $\map g x := \map f {i x} \map f {-i x}$ is also a non-zero [[Definition:Polynomial (Analysis)|polynomial]] with [[Definition:Rational Number|rational]] coefficients such that:
+:$\map g {i \pi} = 0$
+Hence, $i \pi$ is also [[Definition:Algebraic Number|algebraic]].
+From the [[Hermite-Lindemann-Weierstrass Theorem/Weaker|Weaker Hermite-Lindemann-Weierstrass Theorem]], $e^{i \pi}$ is [[Definition:Transcendental Number|transcendental]].
+However, from [[Euler's Identity]]:
+:$e^{i \pi} = -1$
+which is the [[Definition:Root of Polynomial|root]] of $\map h z = z + 1$, and so is [[Definition:Algebraic Number|algebraic]].
+This [[Definition:Contradiction|contradicts]] the conclusion that $e^{i \pi}$ is [[Definition:Transcendental Number|transcendental]].
+Hence by [[Proof by Contradiction]] it must follow that $\pi$ is [[Definition:Transcendental Number|transcendental]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gelfond's Constant is Transcendental}
+Tags: Exponential Function, Gelfond's Constant
+
+\begin{theorem}
+'''[[Definition:Gelfond's Constant|Gelfond's constant]]''':
+:$e^\pi$
+is [[Definition:Transcendental|transcendental]].
+\end{theorem}
+
+\begin{proof}
+From the [[Gelfond-Schneider Theorem]]:
+If:
+:$\alpha$ and $\beta$ are [[Definition:Algebraic Number|algebraic numbers]] such that $\alpha \notin \set {0, 1}$ 
+:$\beta$ is either [[Definition:Irrational Number|irrational]] or not [[Definition:Wholly Real|wholly real]]
+then $\alpha^\beta$ is [[Definition:Transcendental Number|transcendental]].
+We have that:
+{{begin-eqn}}
+{{eqn | l = i^{-2 i}
+      | r = \paren {e^{i \pi / 2} }^{- 2 i}
+      | c = 
+}}
+{{eqn | r = e^{-\pi i^2}
+      | c = 
+}}
+{{eqn | r = e^\pi
+      | c = 
+}}
+{{end-eqn}}
+As:
+:$i$ is [[Definition:Algebraic Number|algebraic]]
+:$-2 i$ is [[Definition:Algebraic Number|algebraic]] and not [[Definition:Wholly Real|wholly real]]
+the conditions of the [[Gelfond-Schneider Theorem]] are fulfilled.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Qunitic Equation using Elliptic Functions}
+Tags: Polynomial Equations
+
+\begin{theorem}
+A [[Definition:Quintic Equation|quintic equation]] can be solved using [[Definition:Elliptic Function|elliptic functions]].
+\end{theorem}<|endoftext|>
+\section{Riemann Surface is Path-Connected}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+A [[Definition:Riemann Surface|Riemann surface]] is [[Definition:Path-Connected Space|path-connected]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Riemann Surface|Riemann surface]] is a [[Definition:Complex Manifold|complex manifold]].
+Hence it is [[Definition:Connected Topological Space|connected]] and [[Definition:Locally Path-Connected Space|locally path-connected]].
+{{explain|It needs to be established that a complex manifold indeed has both of these properties.}}
+{{ProofWanted|A link needed to a page which derives the fact of path-connectedness from connectedness and local path-connectedness.}}
+[[Category:Riemann Surfaces]]
+euw14tot4aq803m48uc9qnzb4b31o5i
+\end{proof}<|endoftext|>
+\section{Properties of Riemann Surface}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+This page gathers some results about [[Definition:Riemann Surface|Riemann surfaces]].
+\end{theorem}<|endoftext|>
+\section{Radó's Theorem (Riemann Surfaces)}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+A [[Definition:Riemann Surface|Riemann surface]] is [[Definition:Second-Countable Space|second countable]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Tibor Radó|cat = Radó T}}
+\end{proof}<|endoftext|>
+\section{Conformal Isomorphism of Universal Cover of Riemann Surface}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+The [[Definition:Universal Cover|universal cover]] of a [[Definition:Riemann Surface|Riemann surface]] is [[Definition:Conformal Isomorphism|conformally isomorphic]] to either:
+:the [[Definition:Riemann Sphere|Riemann sphere]]
+:the [[Definition:Complex Plane|complex plane]]
+or
+:the [[Definition:Unit Disk|unit disk]].
+\end{theorem}
+
+\begin{proof}
+By the [[Riemann Uniformization Theorem]].
+{{ProofWanted|Expand into a proof.}}
+[[Category:Riemann Surfaces]]
+h3pzxxsv4n5hnxjbbi8d4y8v7d91ypu
+\end{proof}<|endoftext|>
+\section{Riemann Surface is Metrizable}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+A [[Definition:Riemann Surface|Riemann surface]] is [[Definition:Metrizable Space|metrizable]].
+\end{theorem}
+
+\begin{proof}
+Follows [[Definition:A Priori|a priori]] from [[Riemann Surface admits Metric of Constant Curvature]].
+{{qed}}
+[[Category:Riemann Surfaces]]
+66rch1oa45nh22ktrt9tgypjhors75a
+\end{proof}<|endoftext|>
+\section{Riemann Surface admits Metric of Constant Curvature}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+A [[Definition:Riemann Surface|Riemann surface]] admits a [[Definition:Metric|metric]] of [[Definition:Constant Curvature Metric|constant curvature]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Riemann Surfaces]]
+m5kylsa3axmq372rmkcc91uk2txeovp
+\end{proof}<|endoftext|>
+\section{Riemann Sphere is only Elliptic Riemann Surface}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+The [[Definition:Riemann Sphere|Riemann sphere]] is the only [[Definition:Elliptic Riemann Surface|elliptic Riemann surface]] (up to [[Definition:Conformal Isomorphism|conformal isomorphism]]).
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Riemann Surfaces]]
+cdq7310y4pazt8ulnwuf5vy8g31pgq1
+\end{proof}<|endoftext|>
+\section{Parabolic Riemann Surface is Plane, Punctured Plane or Torus}
+Tags: Riemann Surfaces
+
+\begin{theorem}
+A [[Definition:Parabolic Riemann Surface|parabolic Riemann surface]] is [[Definition:Conformal Isomorphism|conformally isomorphic]] to either:
+:the [[Definition:Complex Plane|complex plane]]
+:the [[Definition:Punctured Complex Plane|punctured complex plane]] $\C \setminus \set 0$
+or:
+:a [[Definition:Torus (Topology)|torus]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Riemann Surfaces]]
+a1rl0eigrhpgy6ha6rrzg3u3wsrszus
+\end{proof}<|endoftext|>
+\section{Riemann's Rearrangement Theorem}
+Tags: Series
+
+\begin{theorem}
+Let $S$ be an [[Definition:Series of Numbers|infinite series]] which is [[Definition:Conditionally Convergent Series|conditionally convergent]].
+Then its [[Definition:Term of Sequence|terms]] can be arranged in a [[Definition:Permutation|permutation]] so that the new [[Definition:Series of Numbers|series]] [[Definition:Conditionally Convergent Series|converges]] to any given value, or [[Definition:Divergent Series|diverges]].
+\end{theorem}<|endoftext|>
+\section{Riemann-Christoffel Tensor in Two Dimensions is Gaussian Curvature}
+Tags: Riemannian Geometry
+
+\begin{theorem}
+Let $M$ be a [[Definition:Riemannian Manifold|Riemannian manifold]] of [[Definition:Dimension of Riemannian Manifold|dimension]] $2$.
+Then the [[Definition:Riemann-Christoffel Tensor|Riemann-Christoffel tensor]] on $M$ reduces to the [[Definition:Gaussian Curvature|Gaussian curvature]] on $M$.
+\end{theorem}<|endoftext|>
+\section{Riemannian Manifold has Zero Gaussian Curvature iff Euclidean}
+Tags: Riemannian Geometry
+
+\begin{theorem}
+Let $M$ be a [[Definition:Riemannian Manifold|Riemannian manifold]] of [[Definition:Dimension of Riemannian Manifold|dimension]] $2$.
+The [[Definition:Gaussian Curvature|Gaussian curvature]] on $M$ is zero {{iff}} the [[Definition:Riemannian Metric|Riemannian metric]] on $\mathcal M$ is the same as the [[Definition:Euclidean Metric|Euclidean metric]].
+\end{theorem}<|endoftext|>
+\section{Particle on Curved Surface under no Force moves along Geodesic}
+Tags: Calculus of Variations
+
+\begin{theorem}
+Consider a [[Definition:Particle|particle]] $P$ which is constrained to move on a [[Definition:Curved Surface|curved surface]] $C$.
+Let $P$ be such that no [[Definition:Force|force]] acts upon it.
+Then $P$ moves along a [[Definition:Geodesic Curve|geodesic]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|A consequence of [[Hamilton's Principle]] in the [[Definition:Calculus of Variations|calculus of variations]].}}
+\end{proof}<|endoftext|>
+\section{Sum of Reciprocals of Powers as Euler Product}
+Tags: Riemann Zeta Function, Sum of Reciprocals of Powers as Euler Product
+
+\begin{theorem}
+Let $\zeta$ be the [[Definition:Riemann Zeta Function|Riemann zeta function]].
+Let $s \in \C$ be a [[Definition:Complex Number|complex number]] with [[Definition:Real Part|real part]] $\sigma > 1$.
+Then:
+:$\displaystyle \map \zeta s = \prod_{\text {$p$ prime} } \frac 1 {1 - p^{-s} }$
+where the [[Definition:Infinite Product|infinite product]] runs over the [[Definition:Prime Number|prime numbers]].
+\end{theorem}
+
+\begin{proof}
+From [[Euler Product]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z} = \prod_p \frac 1 {1 - a_p p^{-z} }$
+{{iff}} $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]].
+For all $n \in \Z_{\ge 1}$, let $a_n = 1$.
+From [[Convergence of P-Series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]]
+{{iff}}:
+:$\cmod z \ge 1$
+It then follows that:
+:$\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {n^z} = \prod_p \frac 1 {1 - p^{-z} }$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sum of Geometric Sequence]]:
+:$\dfrac 1 {1 - p^{-z} } = 1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \cdots$
+From [[Convergence of P-Series]]:
+:$\displaystyle \sum_{n \mathop = 1}^\infty n^{-z}$ is [[Definition:Absolutely Convergent Series|absolutely convergent]]
+{{iff}}:
+:$\cmod z \ge 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sum_p \dfrac 1 {1 - p^{-z} }
+      | r = \sum_p \paren {1 + \dfrac 1 {p^z} + \dfrac 1 {p^{2 z} } + \dfrac 1 {p^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | r = \paren {1 + \dfrac 1 {2^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {2^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \paren {1 + \dfrac 1 {3^z} + \dfrac 1 {3^{2 z} } + \dfrac 1 {3^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \paren {1 + \dfrac 1 {5^z} + \dfrac 1 {5^{2 z} } + \dfrac 1 {5^{3 z} } + \cdots}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= \times
+      | r = \cdots
+      | c = 
+}}
+{{eqn | r = 1 + \dfrac 1 {2^z} + \dfrac 1 {3^z} + \dfrac 1 {2^{2 z} } + \dfrac 1 {5^z}
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \dfrac 1 {2^z 3^z} + \dfrac 1 {7^z} + \dfrac 1 {2^{3 z} } + \dfrac 1 {3^{2 z} }
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \cdots
+      | c = 
+}}
+{{end-eqn}}
+The result follows from the [[Fundamental Theorem of Arithmetic]].
+{{qed}}
+{{handwaving}}
+\end{proof}<|endoftext|>
+\section{Shift Mapping is Lower Adjoint iff Appropriate Maxima Exist}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Then the following two conditions are equivalent:
+$(1): \quad \forall x \in S, f:S \to S: \left({\forall s \in S: f\left({s}\right) = x \wedge s}\right) \implies f$ is [[Definition:Galois Connection|lower adjoint]]
+$(2): \quad \forall x, t \in S: \max \left\{ {s \in S: x \wedge s \preceq t}\right\}$ exists.
+\end{theorem}
+
+\begin{proof}
+=== $(1) \implies (2)$ ===
+Assume that
+:$\forall x \in S, f:S \to S: \left({\forall s \in S: f\left({s}\right) = x \wedge s}\right) \implies f$ is [[Definition:Galois Connection|lower adjoint]]
+Let $x, t \in S$.
+Define $f: S \to S$:
+:$\forall s \in S: f\left({s}\right) = x \wedge s$
+By assumption:
+:$f$ is [[Definition:Galois Connection|lower adjoint]]
+By definition of [[Definition:Galois Connection|lower adjoint]]:
+:$\exists g: S \to S: \left({g, f}\right)$ is [[Definition:Galois Connection|Galois connection]]
+By [[Galois Connection is Expressed by Maximum]]:
+:$\forall s \in S: g\left({s}\right) = \max\left({f^{-1}\left[{s^\preceq}\right]}\right)$
+Then
+:$\max\left({f^{-1}\left[{t^\preceq}\right]}\right)$ exists.
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\max\left\{ {s \in S: f\left({s}\right) \in t^\preceq}\right\}$ exists.
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\max \left\{ {s \in S: x \wedge s \preceq t}\right\}$ exists.
+{{qed|lemma}}
+=== $(2) \implies (1)$ ===
+Assume that
+:$\forall x, t \in S: \max \left\{ {s \in S: x \wedge s \preceq t}\right\}$ exists.
+Let $x \in S, f: S \to S$ such that
+:$\forall s \in S: f\left({s}\right) = x \wedge s$
+As maxima exist define a [[Definition:Mapping|mapping]] $g: S \to S$:
+:$\forall s \in S: g\left({s}\right) = \max\left({f^{-1}\left[{s^\preceq}\right]}\right)$
+We will prove that
+:$f$ is an [[Definition:Increasing Mapping|increasing mapping]]
+Let $y, z \in S$ such that
+:$y \preceq z$
+By definition of $f$:
+:$f\left({y}\right) = x \wedge y$ and $f\left({z}\right) = x \wedge z$
+Thus by [[Meet Semilattice is Ordered Structure]]:
+:$f\left({y}\right) \preceq f\left({z}\right)$
+By [[Galois Connection is Expressed by Maximum]]:
+:$\left({g, f}\right)$ is [[Definition:Galois Connection|Galois connection]]
+By definition
+:$f$ is a [[Definition:Galois Connection|lower adjoint]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Brouwerian Lattice iff Shift Mapping is Lower Adjoint}
+Tags: Galois Connections, Brouwerian Lattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Lattice|lattice]].
+Then $\left({S, \preceq}\right)$ is a [[Definition:Brouwerian Lattice|Brouwerian lattice]] {{iff}}
+:$\forall x \in S, f: S \to S: \left({\forall s \in S: f \left({s}\right) = x \wedge s}\right) \implies f$ is a [[Definition:Galois Connection|lower adjoint]]
+\end{theorem}
+
+\begin{proof}
+:$\left({S, \preceq}\right)$ is a [[Definition:Brouwerian Lattice|Brouwerian lattice]]
+{{iff}}
+:$\forall x, y \in S: x$ has [[Definition:Relative Pseudocomplement|relative pseudocomplement]] with respect to $y$ by definition of [[Definition:Brouwerian Lattice|Brouwerian lattice]]
+{{iff}}
+:$\forall x, y \in S: \max \left\{ {s \in S: x \wedge s \preceq y}\right\}$ exists by definition of [[Definition:Relative Pseudocomplement|relative pseudocomplement]]
+{{iff}}
+:$\forall x \in S, f: S \to S: \left({\forall s \in S: f \left({s}\right) = x \wedge s}\right) \implies f$ is a [[Definition:Galois Connection|lower adjoint]] by [[Shift Mapping is Lower Adjoint iff Appropriate Maxima Exist]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Brouwerian Lattice is Distributive}
+Tags: Distributive Lattices, Brouwerian Lattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+Then $\left({S, \preceq}\right)$ is a [[Definition:Distributive Lattice|distributive lattice]]
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in S$.
+By [[Brouwerian Lattice iff Shift Mapping is Lower Adjoint]]:
+:$\forall x \in S, f: S \to S: \left({\forall s \in S: f \left({s}\right) = x \wedge s}\right) \implies f$ is a [[Definition:Galois Connection|lower adjoint]]
+Define a [[Definition:Mapping|mapping]] $f: S \to S$:
+:$\forall s \in S: f \left({s}\right) = x \wedge s$
+Then:
+:$f$ is a [[Definition:Galois Connection|lower adjoint]]
+By [[Lower Adjoint Preserves All Suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]]
+By definition of [[Definition:Mapping Preserves Supremum/All|preserves all suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] of $\left\{ {y, z}\right\}$
+By definition of [[Definition:Lattice|lattice]]:
+:$\left\{ {y, z}\right\}$ admits a [[Definition:Supremum of Set|supremum]]
+By [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum of set]]:
+:$\sup \left({f^\to\left({\left\{ {y, z}\right\} }\right)}\right) = f \left({\sup \left\{ {y, z}\right\} }\right)$
+Thus
+{{begin-eqn}}
+{{eqn | l = x \wedge \left({y \vee z}\right)
+      | r = f \left({y \vee z}\right)
+      | c = definition of $f$
+}}
+{{eqn | r = f \left({\sup \left\{ {y, z}\right\} }\right)
+      | c = definition of [[Definition:Join (Order Theory)|join operation]]
+}}
+{{eqn | r = \sup \left({f^\to\left({\left\{ {y, z}\right\} }\right)}\right)
+}}
+{{eqn | r = \sup \left({\left\{ {f\left({y}\right), f\left({z}\right)}\right\} }\right)
+      | c = definition of [[Definition:Image of Subset under Mapping|image of set]]
+}}
+{{eqn | r = \sup \left\{ {x \wedge y, x \wedge z}\right\}
+      | c = definition of $f$
+}}
+{{eqn | r = \left({x \wedge y}\right) \vee \left({x \wedge z}\right)
+      | c = definition of [[Definition:Join (Order Theory)|join operation]]
+}}
+{{end-eqn}}
+Thus by definition:
+:$\left({S, \preceq}\right)$ is a [[Definition:Distributive Lattice|distributive lattice]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definite Integral of Reciprocal of Root of a Squared minus x Squared}
+Tags: Definite Integrals, Arcsine Function, Reciprocals
+
+\begin{theorem}
+:$\displaystyle \int_0^x \frac {\d t} {\sqrt{1 - t^2} } = \arcsin x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \int_0^x \frac {\d t} {\sqrt{1 - t^2} }
+      | r = \intlimits {\arcsin \frac t 1} 0 x
+      | c = [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$]], {{Defof|Definite Integral}}
+}}
+{{eqn | r = \arcsin x - \arcsin 0  
+      | c = 
+}}
+{{eqn | r = \arcsin x 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meromorphic Function is Elliptic iff Doubly Periodic}
+Tags: Elliptic Functions
+
+\begin{theorem}
+Let $f: \C \to \C$ be a [[Definition:Meromorphic Function|meromorphic function]]. 
+Then $\map f z$ is an [[Definition:Elliptic Function|elliptic function]] {{iff}} it is also [[Definition:Doubly Periodic Function|doubly periodic]].
+\end{theorem}<|endoftext|>
+\section{Entire Function is Transcendental iff Power Series Expansion is Infinite}
+Tags: Entire Functions
+
+\begin{theorem}
+Let $f$ be an [[Definition:Entire Function|entire function]].
+Then $f$ is '''[[Definition:Transcendental Entire Function|transcendental]]''' {{iff}} the [[Definition:Power Series|power series expansion]] of $f$ has an [[Definition:Infinite|infinite]] number of [[Definition:Polynomial Coefficient|coefficients]] which are non-zero.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Entire Functions]]
+miflckvdvsw7isipgspnhc4j3jjh42b
+\end{proof}<|endoftext|>
+\section{Entire Function is Transcendental iff not Complex Polynomial Function}
+Tags: Entire Functions
+
+\begin{theorem}
+Let $f$ be an [[Definition:Entire Function|entire function]].
+Then $f$ is '''[[Definition:Transcendental Entire Function|transcendental]]''' {{iff}} it is not a [[Definition:Complex Polynomial Function|complex polynomial function]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Entire Functions]]
+gyuedjao26bgt0dgqyejduxfigo72tg
+\end{proof}<|endoftext|>
+\section{Complex Function is Entire iff it has Everywhere Convergent Power Series}
+Tags: Entire Functions
+
+\begin{theorem}
+Let $f: \C \to \C$ be a [[Definition:Complex Function|complex function]].
+Then $f$ is an '''[[Definition:Entire Function|entire function]]''' {{iff}} $f$ can be given by an [[Definition:Everywhere Convergence|everywhere convergent]] [[Definition:Power Series|power series]]:
+:$\displaystyle \map f z = \sum_{n \mathop = 0}^\infty a_n z^n; \quad \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+From [[Holomorphic Function is Analytic]], if $f$ is [[Definition:Holomorphic Complex Function|holomorphic]] on some [[Definition:Open Ball|open ball]] $D$ in $\C$, then $f$ is [[Definition:Analytic Complex Function|complex analytic]] on $D$. 
+However, as, by the definition of an [[Definition:Entire Function|entire function]], $f$ is [[Definition:Holomorphic Complex Function|holomorphic]] everywhere, the radius of $D$ can be made arbitrarily large, meaning that $f$ is [[Definition:Analytic Complex Function|complex analytic]] everywhere.
+This implies that an [[Definition:Entire Function|entire function]] is [[Definition:Analytic Complex Function|complex analytic]] everywhere.
+{{qed|lemma}}
+=== Necessary Condition ===
+It remains to prove that if $f$ is [[Definition:Analytic Complex Function|complex analytic]] everywhere, $f$ is [[Definition:Entire Function|entire]]. 
+By [[Power Series is Termwise Differentiable within Radius of Convergence]], if $f$ is [[Definition:Analytic Complex Function|complex analytic]] everywhere, $f$ is [[Definition:Holomorphic Complex Function|holomorphic]] everywhere. 
+That is, a function [[Definition:Analytic Complex Function|complex analytic]]everywhere is [[Definition:Entire Function|entire]]. 
+Note that by the [[Cauchy-Hadamard Theorem/Complex Case|Cauchy-Hadamard theorem]]: 
+:$\displaystyle \lim_{n \mathop \to \infty} \sqrt [n] {\size {a_n} } = 0$
+{{iff}} this series has an infinite [[Definition:Radius of Convergence|radius of convergence]]. 
+That is, {{iff}} the series converges for all $z \in \C$, satisfying our second demand. 
+{{qed}}
+[[Category:Entire Functions]]
+dowwf37pjjl19876udtgduezr6puxvk
+\end{proof}<|endoftext|>
+\section{Brouwerian Lattice is Upper Bounded}
+Tags: Bounded Lattices, Brouwerian Lattices
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+Then $S$ is [[Definition:Bounded Above Set|upper bounded]].
+\end{theorem}
+
+\begin{proof}
+By assumption:
+:$S \ne \O$
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists s: s \in S$
+By definition of [[Definition:Brouwerian Lattice|Brouwerian lattice]]:
+:$s$ has [[Definition:Relative Pseudocomplement|relative pseudocomplement]] with respect to $s$
+By definition of [[Definition:Relative Pseudocomplement|relative pseudocomplement]]:
+:$\max \set {x \in S: s \wedge x \preceq s}$ exists and equals $s \to s$
+Let $x \in S$.
+By [[Meet Precedes Operands]]:
+:$s \wedge x \preceq s$
+Then:
+:$x \in \set {x \in S: s \wedge x \preceq s}$
+Thus by definition of [[Definition:Greatest Element of Subset|maximum]]:
+:$x \preceq s \to s$
+Thus by definition:
+:$s \to s$ is an [[Definition:Upper Bound of Set|upper bound]] for $S$
+Thus by definition:
+:$S$ is [[Definition:Bounded Above Set|upper bounded]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Extreme Value Theorem/Real Function}
+Tags: Real Analysis, Continuous Functions, Real Functions
+
+\begin{theorem}
+Let $f$ be [[Definition:Continuous on Interval|continuous]] in a [[Definition:Closed Real Interval|closed real interval]] $\left[{a \,.\,.\, b}\right]$.
+Then:
+:$\forall x \in \left[{a \,.\,.\, b}\right]: \exists x_s \in \left[{a \,.\,.\, b}\right]: f \left({x_s}\right) \le f \left({x}\right)$
+:$\forall x \in \left[{a \,.\,.\, b}\right]: \exists x_n \in \left[{a \,.\,.\, b}\right]: f \left({x_n}\right) \ge f \left({x}\right)$
+\end{theorem}
+
+\begin{proof}
+First it is shown that $f \left({x}\right)$ is bounded in the [[Definition:Closed Real Interval|closed real interval]] $\left[{a \,.\,.\, b}\right]$. 
+{{AimForCont}} $f \left({x}\right)$ has no [[Definition:Upper Bound of Real-Valued Function|upper bound]].
+Then for all $x \in \left[{a \,.\,.\, b}\right]$:
+:$\forall N \in \R: \exists x_n \in \left[{a \,.\,.\, b}\right]: f \left({\left \langle {x_n} \right \rangle}\right) > N$
+We have that:
+:$\N \subset \R$
+{{WLOG}}, we can consider $N \in \N$.
+Consider the [[Definition:Sequence|sequence]]:
+:$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N}$
+and the [[Definition:Sequence|sequence]]:
+:$\left \langle {N} \right \rangle_{N \mathop \in \N}$
+Because:
+:$f \left({\left \langle {x_n} \right \rangle}\right) > N$
+and:
+:$\left \langle {N} \right \rangle_{N \mathop \in \N} \to \infty$
+it follows that:
+:$\left\langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \mathop \in \N} > \infty \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle_{n \in \N} \to \infty$
+{{explain|$> \infty$?}}
+Consider the [[Definition:Sequence|sequence]]:
+:$\left \langle {x_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$
+It is [[Definition:Bounded Real Sequence|bounded]]: [[Definition:Bounded Above Real Sequence|bounded above]] by $b$ and [[Definition:Bounded Below Real Sequence|bounded below]] by $a$.
+So by the [[Bolzano-Weierstrass Theorem]] there exists a [[Definition:Subsequence|subsequence]] of $\left \langle {x_n} \right \rangle $:
+:$\left \langle {g_n} \right \rangle \to d$
+Since $\left[{a \,.\,.\, b}\right]$ is [[Definition:Closed Real Interval|closed]]:
+:$d \in \left[{a \,.\,.\, b}\right]$
+By [[Continuity of Mapping between Metric Spaces by Convergent Sequence]], then:
+:$f \left({x}\right)$ is [[Definition:Continuous Real Function|continuous]] in $d$
+and:
+:$f \left({\left \langle {g_n} \right \rangle}\right) = f \left({d}\right)$
+But our first conlusion indicates that:
+:$\left \langle {f \left({x_n}\right)} \right \rangle_{n \in \N} \to \infty$
+which [[Definition:Contradiction|contradicts]]:
+:$\left \langle {f \left({g_n}\right)} \right \rangle_{n \in \N} \to d$
+A similar argument can be used to prove the [[Definition:Lower Bound of Real-Valued Function|lower bound]].
+{{qed|lemma}}
+It remains to be proved that:
+:$\exists d \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) \le f \left({d}\right)$
+where $f \left({d}\right) = N$ (the maximum).
+It will be shown that:
+:$\forall n \in \R_{\ne 0}: N - 1/n < f \left({\left \langle {x_n} \right \rangle }\right) \le N$
+as follows;
+{{WLOG}}, consider $n \in \N$.
+Let $I$ denote the codomain of $f \left[{a \,.\,.\, b}\right]$.
+Because $N$ is its maximum and $N - 1/n < N$:
+:$\forall n \in \N: \exists y_n \in \left[{a \,.\,.\, b}\right]: N - 1/n < y_n < N$
+But $y_n \in I$, so:
+:$\forall n \in \N: \exists x_n \in \left[{a \,.\,.\, b}\right]$
+That means:
+:$f \left({\left \langle {x_n} \right \rangle}\right) = y_n \implies  N - 1/n < f \left({x_n}\right) \le \ N$
+{{qed|lemma}}
+It follows that:
+:$\left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to N$
+Considering:
+:$\left \langle {N - 1/n} \right \rangle \to \ N$ as $n \to \infty$
+and:
+:$\forall n \in \N: \left \langle {N} \right \rangle \to N$
+by:
+: $\left \langle {N - 1/n} \right \rangle < \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \le N \implies \left \langle {f \left({\left \langle {x_n} \right \rangle}\right)} \right \rangle \to \left \langle {N} \right \rangle$
+{{explain|The grammatical sense of the above is unclear.}}
+Consider $\left \langle {x_n} \right \rangle$.
+Because it is bounded, by [[Bolzano-Weierstrass Theorem]] there exists a subsequence:
+:$\left \langle {s_n} \right \rangle$
+that converges.
+Let it converge to $l$.
+Because $\left \langle {s_n} \right \rangle \in \left[{a \,.\,.\, b}\right]$ it follows that:
+:$l \in \left[{a \,.\,.\, b}\right]$.
+Finally, $f \left({x}\right) $ is continuous in $\left[{a \,.\,.\, b}\right]$.
+So, by [[Continuity of Mapping between Metric Spaces by Convergent Sequence]]:
+:$\left \langle {s_n} \right \rangle \to d \implies f \left({\left \langle {s_n} \right \rangle}\right) \to f \left({d}\right)$
+But:
+:$\left \langle {f \left({x_n}\right)} \right \rangle \to N \implies \left \langle {f \left({s_n}\right)} \right \rangle_{n \in \N} \to N \iff f \left({d}\right) = N$
+{{explain|Clarify, in that last compound proposition, what is being proved and how that conclusion is reached.}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Galileo's Paradox}
+Tags: Veridical Paradoxes
+
+\begin{theorem}
+The [[Definition:Natural Number|natural numbers]] $\N$ are in [[Definition:One-to-One Correspondence|one-to-one correspondence]] with their [[Definition:Square Number|squares]]:
+That is, the [[Definition:Mapping|mapping]]:
+:$\forall n \in \N: f: n \mapsto n^2$
+is a [[Definition:Bijection|bijection]].
+Hence the [[Definition:Set|set]] of [[Definition:Square Number|square numbers]] is [[Definition:Set Equivalence|equinumerous]] to the [[Definition:Set|set]] of [[Definition:Natural Number|natural numbers]].
+That is, a [[Definition:Set|set]] is [[Definition:Set Equivalence|equinumerous]] to one of its [[Definition:Proper Subset|proper subsets]].
+\end{theorem}<|endoftext|>
+\section{Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice}
+Tags: Galois Connections, Brouwerian Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+Let $a$ be an [[Definition:Element|element]] of $S$.
+Let $g, d: S \to S$ be [[Definition:Mapping|mappings]] such that
+:$\forall s \in S: g\left({s}\right) = a \to s$
+and
+:$\forall s \in S: d\left({s}\right) = a \wedge s$
+Then $\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]].
+\end{theorem}
+
+\begin{proof}
+By [[Brouwerian Lattice iff Shift Mapping is Lower Adjoint]]:
+:$d$ is [[Definition:Galois Connection|lower adjoint]]
+By definition of [[Definition:Galois Connection|lower adjoint]]:
+:$\exists g': S \to S: \left({g', d}\right)$ is [[Definition:Galois Connection|Galois connection]]
+By [[Galois Connection is Expressed by Maximum]]:
+:$\forall s \in S: g'\left({s}\right) = \max\left({d^{-1}\left[{s^\preceq}\right]}\right)$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: d\left({x}\right) \in s^\preceq}\right\} }\right)$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: d\left({x}\right) \preceq s}\right\} }\right)$
+By definition of $d$:
+:$\forall s \in S: g'\left({s}\right) = \max\left({\left\{ {x \in S: a \wedge x \preceq s}\right\} }\right)$
+By definition of [[Definition:Relative Pseudocomplement|relative pseudocomplement]]:
+:$\forall s \in S: g'\left({s}\right) = a \to s = g\left({s}\right)$
+By [[Equality of Mappings]]:
+:$g = g'$
+Hence
+: $\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice}
+Tags: Brouwerian Lattices
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \preceq}$ be s [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+Let $a, x, y \in S$.
+Then
+:$a \wedge x \preceq y$ {{iff}} $x \preceq a \to y$
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Mapping|mapping]] $d: S \to S$:
+:$\forall s \in S: \map d s = a \wedge s$
+Define a [[Definition:Mapping|mapping]] $g: S \to S$:
+:$\forall s \in S: \map g s = a \to s$
+By [[Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice]]:
+:$\tuple {g, d}$ is [[Definition:Galois Connection|Galois connection]].
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$x \preceq \map g y$ {{iff}} $\map d x \preceq y$
+Thus by definitions of $g$ and $d$:
+:$a \wedge x \preceq y$ {{iff}} $x \preceq a \to y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Weierstrass Approximation Theorem}
+Tags: Real Analysis, Weierstrass Approximation Theorem
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\Bbb I$.
+Then $f$ can be [[Definition:Uniform Approximation|uniformly approximated]] on $\Bbb I$ by a [[Definition:Real Polynomial Function|polynomial function]] to any given degree of accuracy.
+\end{theorem}
+
+\begin{proof}
+Let $\map f t: \Bbb I = \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
+Introduce $\map x t$ with a rescaled [[Definition:Domain (Mapping Theory)|domain]]:
+:$\map f t \mapsto \map x {a + t \paren {b - a} } : \closedint a b \to \closedint 0 1$.
+From now on we will work with $x: \closedint 0 1 \to \R$, which is also [[Definition:Continuous Real Function|continuous]].
+Let $n \in \N$.
+For $t \in \closedint 0 1$ consider the [[Definition:Bernstein Polynomial|Bernstein polynomial]]:
+:$\displaystyle \map {B_n x} t = \sum_{k \mathop = 0}^n \map x {\frac k n} \binom n k t^k \paren {1 - t}^{n - k}$
+For $t \in \closedint 0 1$, $0 \le k \le n$, let:
+:$\displaystyle \map {p_{n, k} } t := \binom n k t^k \paren {1 - t}^{n - k}$
+By the [[Binomial Theorem/Integral Index|binomial theorem]]:
+:$\displaystyle \sum_{k \mathop = 0}^n \map {p_{n, k} } t = 1$
+=== [[Weierstrass Approximation Theorem/Lemma 1|Lemma 1]] ===
+{{:Weierstrass Approximation Theorem/Lemma 1}}{{qed|lemma}}
+=== [[Weierstrass Approximation Theorem/Lemma 2|Lemma 2]] ===
+{{:Weierstrass Approximation Theorem/Lemma 2}}{{qed|lemma}}
+Now we construct the estimates.
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t
+      | o = \le
+      | r = \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \map {p_{n, k} } t \frac {\paren {k - n t}^2} {\delta^2 n^2}
+      | c = $\displaystyle \frac {\paren {k - n t}^2} {\delta^2 n^2} \ge 1$
+}}
+{{eqn | o = \le
+      | r = \frac 1 {n^2 \delta^2} \sum_{k \mathop = 0}^n \paren{k - n t}^2 \map {p_{n, k} } t
+}}
+{{eqn | r = \frac {t \paren {1 - t} } {n \delta^2}
+      | c = [[Weierstrass Approximation Theorem/Lemma 2|Lemma 2]]
+}}
+{{eqn | o = \le
+      | r = \frac 1 {4 n \delta^2}
+      | c = $\forall t \in \closedint 0 1: \frac 1 4 \ge t \paren {1 - t}$
+}}
+{{end-eqn}}
+Here $\displaystyle \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n$ denotes the [[Definition:Summation|summation]] over those values of $k \in \N$, $k \le n$, which satisfy the [[Definition:Inequality|inequality]] $\displaystyle \size {\frac k n - t} \ge \delta$.
+For some $\delta > 0$ denote:
+:$\displaystyle \map {\omega_\delta} x := \sup_{\size {t - s} < \delta} \size {\map x s - \map x t}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \size {\map {B_n x} t - \map x t}
+      | r = \size {\map {B_n x} t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t}
+}}
+{{eqn | r = \size {\sum_{k \mathop = 0}^n \map x {\frac k n} \map {p_{n, k} } t - \map x t \sum_{k \mathop = 0}^n \map {p_{n, k} } t}
+}}
+{{eqn | o = \le
+      | r = \sum_{k \mathop = 0}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t
+      | c = as $\map {p_{n, k} } t \ge 0$
+}}
+{{eqn | r = \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t + \sum_{k \mathop : \size {\frac k n \mathop - t} \ge \delta}^n \size {\map x {\frac k n}  - \map x t} \map {p_{n, k} } t
+}}
+{{eqn | o = \le
+      | r = \map {\omega_\delta} x \sum_{k \mathop : \size {\frac k n \mathop - t} < \delta}^n \map {p_{n, k} } t + 2 \norm x_\infty \frac 1 {4 n \delta^2}
+}}
+{{eqn | o = \le 
+      | r = \map {\omega_\delta} x \cdot 1 + \frac {\norm x_\infty} {2 n \delta^2}
+}}
+{{end-eqn}}
+where $\norm {\,\cdot \,}_\infty$ denotes the [[Definition:Supremum Norm/Continuous on Closed Interval Real-Valued Function|supremum norm]].
+Let $\epsilon > 0$.
+By [[Continuous Function on Closed Interval is Uniformly Continuous]], $\map x t$ is [[Definition:Uniformly Continuous Real Function|uniformly continuous]].
+We choose $\delta > 0$ such that $\displaystyle \map {\omega_\delta} x < \frac \epsilon 2$.
+Choose $\displaystyle n > \frac {\norm x_\infty} {\epsilon \delta^2}$
+Then:
+:$\displaystyle \norm {\map {B_n x} t - \map x t}_\infty < \epsilon$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Right Angle by Stretched Rope}
+Tags: Pythagoras's Theorem, Classic Problems
+
+\begin{theorem}
+Let a rope be knotted at regular intervals so that there are $12$ equal segments separated by knots.
+Tie the rope in a loop consisting of those $12$ segments.
+Fix one of the knots to the ground at the point you want the [[Definition:Right Angle|right angle]] to be placed.
+Stretch the rope tightly in the direction of one of the legs of the [[Definition:Right Angle|right angle]] you want to create and fix the fourth knot to the ground.
+Stretch the rope tightly in the direction of the other leg of the [[Definition:Right Angle|right angle]] you want to create and fix the third knot in the ground, at the same time making sure the remaining segment of $5$ knots is also tight.
+The point where the $3$-knot section and the $4$-knot section meet, the rope will be bent at a [[Definition:Right Angle|right angle]].
+:[[File:RightAngleRopeConstruction.png|400px]]
+\end{theorem}
+
+\begin{proof}
+The rope will be stretched in the form of a [[Definition:Triangle (Geometry)|triangle]] with sides $3$, $4$ and $5$.
+We have that:
+:$3^2 + 4^2 = 9 + 16 = 25 = 5^2$
+and so by [[Pythagoras's Theorem]] the [[Definition:Triangle (Geometry)|triangle]] is [[Definition:Right Triangle|right angled]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Top equals to Relative Pseudocomplement in Brouwerian Lattice}
+Tags: Brouwerian Lattices
+
+\begin{theorem}
+Let $\struct {S, \vee, \wedge, \preceq}$ be a [[Definition:Brouwerian Lattice|Brouwerian lattice]] with [[Definition:Greatest Element of Subset|greatest element]] $\top$.
+Let $a, b \in S$.
+Then
+:$\top = a \to b$ {{iff}} $a \preceq b$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let:
+:$\top = a \to b$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\top \preceq a \to b$
+By [[Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice]]:
+:$\top \wedge a \preceq b$
+Thus
+:$a \preceq b$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let:
+:$a \preceq b$
+Then:
+:$a \wedge \top \preceq b$
+By [[Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice]]:
+:$\top \preceq a \to b$
+By definition of [[Definition:Greatest Element of Subset|greatest element]]:
+:$a \to b \preceq \top$
+By definition of [[Definition:Ordered Set|ordered set]]:
+:$\top = a \to b$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Up-Complete Product}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq_1}$, $\struct {T, \preceq_2}$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Ordered Set|ordered sets]].
+Let $\struct {S \times T, \preceq}$ be [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$.
+\end{theorem}
+
+\begin{proof}
+=== [[Up-Complete Product/Lemma 1|Lemma 1]] ===
+{{:Up-Complete Product/Lemma 1}}{{qed|lemma}}
+=== [[Up-Complete Product/Lemma 2|Lemma 2]] ===
+{{:Up-Complete Product/Lemma 2}}{{qed|lemma}}
+=== Sufficient Condition ===
+Assume that:
+:$\struct {S \times T, \preceq}$ is [[Definition:Up-Complete|up-complete]].
+Let $X$ be a [[Definition:Directed Subset|directed]] [[Definition:Subset|subset]] of $S$.
+By assumption:
+:$T \ne \O$
+By definition by [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists t: t \in T$
+By [[Singleton is Directed and Filtered Subset]]:
+:$\set t$ is [[Definition:Directed Subset|directed]]
+By [[Up-Complete Product/Lemma 1|Lemma 1]]:
+:$X \times \set t$ is a [[Definition:Directed Subset|directed subset]] of $S \times T$
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$X \times \set t$ admits a [[Definition:Supremum of Set|supremum]]
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\exists \tuple {a, b} \in S \times T: \tuple {a, b}$ is [[Definition:Upper Bound of Set|upper bound]] for $X \times \set t$
+and:
+:$\forall \tuple {c, d} \in S \times T: \tuple {c, d}$ is [[Definition:Upper Bound of Set|upper bound]] for $X \times \set t \implies \tuple {a, b} \preceq \tuple {c, d}$
+We will prove that
+:$a$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+Let $s \in X$.
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+:$\tuple {s, t} \in X \times \set t$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\tuple {s, t} \preceq \tuple {a, b}$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$s \preceq_1 a$
+We will prove that
+:$\forall x \in S: x$ is [[Definition:Upper Bound of Set|upper bound]] for $X \implies a \preceq_1 x$
+Let $x \in S$ such that
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+We will prove as sublemma that:
+:$\tuple {x, b}$ is [[Definition:Upper Bound of Set|upper bound]] for $X \times \set t$
+Let $\tuple {y, z} \in X \times \set t$.
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+:$y \in X$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\tuple {y, z} \preceq \tuple {a, b}$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$z \preceq_2 b$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$y \preceq_1 x$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\tuple {y, z} \preceq \tuple {x, b}$
+This ends the proof of sublemma.
+{{qed|lemma}}
+Then:
+:$\tuple {a, b} \preceq \tuple {x, b}$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$x \preceq_1 a$
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$X$ admits a [[Definition:Supremum of Set|supremum]]:
+Thus by definition:
+:$\struct {S, \preceq_1}$ is [[Definition:Up-Complete|up-complete]].
+By [[Definition:Mutatis Mutandis|mutatis mutandis]]
+:$\struct {T, \preceq_2}$ is [[Definition:Up-Complete|up-complete]]
+=== Necessary Condition ===
+Assume that
+:$\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ are [[Definition:Up-Complete|up-complete]].
+Let $X$ be [[Definition:Directed Subset|directed subset]] of $S \times T$.
+By [[Up-Complete Product/Lemma 2|Lemma 2]]:
+:$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ are [[Definition:Directed Subset|directed]]
+where 
+:$\pr_1$ denotes the [[Definition:First Projection|first projection]] on $S \times T$
+:$\pr_2$ denotes the [[Definition:Second Projection|second projection]] on $S \times T$
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ admit [[Definition:Supremum of Set|suprema]]
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\exists a \in S: a$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_1^\to} X$
+and:
+:$\forall c \in S: c$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_1^\to} X \implies a \preceq_1 c$
+and
+:$\exists b \in T: b$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_2^\to} X$
+and:
+:$\forall c \in T: c$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_2^\to} X \implies b \preceq_2 c$
+We will prove that
+:$\tuple {a, b}$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+Let $\tuple {x, y} \in X$.
+By definitions of [[Definition:Image of Subset under Mapping|image of set]] and [[Definition:Projection (Mapping Theory)|projections]]:
+:$x \in \map {\pr_1^\to} X$ and $y \in \map {\pr_2^\to} X$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$x \preceq_1 a$ and $y \preceq_2 b$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\tuple {x, y} \preceq \tuple {a, b}$
+We will prove that
+:$\forall \tuple {x, y} \in S \times T: \tuple {x, y}$ is [[Definition:Upper Bound of Set|upper bound]] for $X \implies \tuple {a, b} \preceq \tuple {x, y}$
+Let $\tuple {x, y} \in S \times T$ such that:
+:$\tuple {x, y}$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+We will prove as sublemma that
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_1^\to} X$
+Let $c \in \map {\pr_1^\to} X$.
+By definitions of [[Definition:Image of Subset under Mapping|image of set]] and [[Definition:Projection (Mapping Theory)|projections]]:
+:$\exists d: \tuple {c, d} \in X$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\tuple {c, d} \preceq \tuple {x, y}$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$c \preceq_1 x$
+By [[Definition:Mutatis Mutandis|mutatis mutandis]]
+:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $\map {\pr_2^\to} X$
+Then:
+:$a \preceq_1 x$ and $b \preceq_2 y$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\tuple {a, b} \preceq \tuple {x, y}$
+By definition
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+Thus by definition
+:$\struct {S \times T, \preceq}$ is [[Definition:Up-Complete|up-complete]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cartesian Product of Ordered Sets is Ordered Set}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $\left({S_1 \times S_2, \preceq}\right)$ be the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$.
+Then $\left({S_1 \times S_2, \preceq}\right)$ is also an [[Definition:Ordered Set|ordered set]].
+\end{theorem}
+
+\begin{proof}
+=== Reflexivity ===
+Let $\left({s, t}\right) \in S_1 \times S_2$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$s \preceq_1 s$ and $t \preceq_2 t$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product]]:
+:$\left({s, t}\right) \preceq \left({s, t}\right)$
+=== Transitivity ===
+Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right), \left({s_3, t_3}\right) \in S_1 \times S_2$ such that
+:$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right) \preceq \left({s_3, t_3}\right)$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product]]:
+:$s_1 \preceq_1 s_2 \preceq_1 s_3$ and $t_1 \preceq_2 t_2 \preceq_2 t_3$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$s_1 \preceq_1 s_3$ and $t_1 \preceq_2 t_3$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product]]:
+:$\left({s_1, t_1}\right) \preceq \left({s_3, t_3}\right)$
+=== Antisymmetry ===
+Let $\left({s_1, t_1}\right), \left({s_2, t_2}\right) \in S_1 \times S_2$ such that
+:$\left({s_1, t_1}\right) \preceq \left({s_2, t_2}\right)$ and $\left({s_2, t_2}\right) \preceq \left({s_1, t_1}\right)$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product]]:
+:$s_1 \preceq_1 s_2$ and $s_2 \preceq_1 s_1$ and $t_1 \preceq_2 t_2$ and $t_2 \preceq_2 t_1$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$s_1 = s_2$ and $t_1 = t_2$
+Thus:
+:$\left({s_1, t_1}\right) = \left({s_2, t_2}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Motion of Rocket in Outer Space}
+Tags: Rocket Science
+
+\begin{theorem}
+Let $B$ be a [[Definition:Rocket|rocket]] travelling in outer space.
+Let the [[Definition:Velocity|velocity]] of $B$ at [[Definition:Time|time]] $t$ be $\mathbf v$.
+Let the [[Definition:Mass|mass]] of $B$ at [[Definition:Time|time]] $t$ be $m$.
+Let the [[Definition:Exhaust Velocity|exhaust velocity]] of $B$ be constant at $\mathbf b$.
+Then the equation of motion of $B$ is given by:
+:$m \dfrac {\mathrm d \mathbf v} {\mathrm d t} = - \mathbf b \dfrac {\mathrm d m} {\mathrm d t}$
+\end{theorem}
+
+\begin{proof}
+From [[Motion of Body with Variable Mass]]:
+:$(1): \quad \mathbf w \dfrac {\mathrm d m} {\mathrm d t} + \mathbf F = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$
+where:
+:$\mathbf F$ is the external [[Definition:Force|force]] being applied
+:$\mathbf w$ is the [[Definition:Velocity|velocity]] of the added [[Definition:Mass|mass]] relative to $B$.
+In this scenario:
+:there is no external [[Definition:Force|force]] and so $\mathbf F = \mathbf 0$
+:the [[Definition:Velocity|velocity]] of the added [[Definition:Mass|mass]] relative to $B$ is $-\mathbf b$.
+Thus $(1)$ becomes:
+:$-\mathbf b \dfrac {\mathrm d m} {\mathrm d t} = m \dfrac {\mathrm d \mathbf v} {\mathrm d t}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Newton's Second Law of Motion]]:
+:$\mathbf F = \dfrac \d {\d t} \paren {m \mathbf v}$
+At time $t + \Delta t$, let:
+: the [[Definition:Mass|mass]] of $B$ be $m + \Delta m$
+: the [[Definition:Velocity|velocity]] of $B$ be $\mathbf v + \Delta \mathbf v$.
+The fuel is being consumed, so the increase in [[Definition:Mass|mass]] of the fuel during time $\Delta t$ is $-\Delta m$.
+Thus the exhaust products, therefore of mass $-\Delta m$, are expelled at a [[Definition:Velocity|velocity]] $-\mathbf b$ relative to $B$.
+Thus this material is actually moving at a [[Definition:Velocity|velocity]] $\mathbf v - \mathbf b$.
+By [[Conservation of Momentum]], the total [[Definition:Linear Momentum|linear momentum]] is [[Definition:Constant|constant]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = m \mathbf v
+      | r = \paren {m + \Delta m} \paren {\mathbf v + \Delta \mathbf v} + \paren {-\Delta m} \paren {\mathbf v - \mathbf b}
+      | c = 
+}}
+{{eqn | r = m \mathbf v + m \Delta \mathbf v + \paren {\Delta m} \mathbf v + \Delta m \Delta \mathbf v - \paren {-\Delta m} \mathbf v + \Delta m \mathbf b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m \Delta \mathbf v
+      | r = - \Delta m \paren {\mathbf b + \Delta \mathbf v}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m \frac {\Delta \mathbf v} {\Delta t}
+      | r = - \frac {\Delta m} {\Delta t} \paren {\mathbf b + \Delta \mathbf v}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = m \frac {\d \mathbf v} {\d t}
+      | r = - \mathbf b \frac {\d m} {\d t}
+      | c = letting $\Delta t \to 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Velocity of Rocket in Outer Space}
+Tags: Rocket Science
+
+\begin{theorem}
+Let $B$ be a [[Definition:Rocket|rocket]] travelling in outer space.
+Let the [[Definition:Velocity|velocity]] of $B$ at [[Definition:Time|time]] $t$ be $\mathbf v$.
+Let the [[Definition:Mass|mass]] of $B$ at [[Definition:Time|time]] $t$ be $m$.
+Let the [[Definition:Exhaust Velocity|exhaust velocity]] of $B$ be constant at $\mathbf b$.
+Then the [[Definition:Velocity|velocity]] of $B$ at [[Definition:Time|time]] $t$ is given by:
+:$\map {\mathbf v} t = \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}$
+where $\map {\mathbf v} 0$ and $\map m 0$ are the [[Definition:Velocity|velocity]] and [[Definition:Mass|mass]] of $B$ at [[Definition:Time|time]] $t = 0$.
+\end{theorem}
+
+\begin{proof}
+From [[Motion of Rocket in Outer Space]], the equation of motion of $B$ is given by:
+:$m \dfrac {\d \mathbf v} {\d t} = -\mathbf b \dfrac {\d m} {\d t}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int_0^t \dfrac {\d \mathbf v} {\d t} \rd t
+      | r = -\int_0^t \mathbf b \frac 1 m \dfrac {\d m} {\d t} \rd t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int_{t \mathop = 0}^t \rd \mathbf v
+      | r = -\int_{t \mathop = 0}^t \mathbf b \dfrac {\d m} m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \bigintlimits {\mathbf v} 0 t
+      | r = -\mathbf b \bigintlimits {\ln m} 0 t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\mathbf v} t - \map {\mathbf v} 0
+      | r = -\mathbf b \paren {\map {\ln m} t - \map {\ln m} 0}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\mathbf v} t
+      | r = \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Acceleration of Rocket in Outer Space}
+Tags: Rocket Science
+
+\begin{theorem}
+Let $B$ be a [[Definition:Rocket|rocket]] travelling in outer space.
+Let the [[Definition:Velocity|velocity]] of $B$ at [[Definition:Time|time]] $t$ be $\mathbf v$.
+Let the [[Definition:Mass|mass]] of $B$ at [[Definition:Time|time]] $t$ be $m$.
+Let the [[Definition:Exhaust Velocity|exhaust velocity]] of $B$ be constant at $\mathbf b$.
+Then the [[Definition:Acceleration|acceleration]] of $B$ at [[Definition:Time|time]] $t$ is given by:
+:$\mathbf a = \dfrac 1 m \left({- \mathbf b \dfrac {\mathrm d m} {\mathrm d t} }\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Motion of Rocket in Outer Space]], the equation of motion of $B$ is given by:
+:$(1): \quad m \dfrac {\mathrm d \mathbf v} {\mathrm d t} = - \mathbf b \dfrac {\mathrm d m} {\mathrm d t}$
+By definition, the [[Definition:Acceleration|acceleration]] of $B$ is its [[Definition:Rate of Change|rate of change]] of [[Definition:Velocity|velocity]]:
+:$\mathbf a = \dfrac {\mathrm d \mathbf v} {\mathrm d t}$
+The result follows by substituting $\mathbf a$ for $\dfrac {\mathrm d \mathbf v} {\mathrm d t}$ and dividing by $m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Up-Complete Product/Lemma 2}
+Tags: Order Theory
+
+\begin{theorem}
+Let $X$ be a [[Definition:Directed Subset|directed subset]] of $S \times T$.
+Then
+:$\map {\pr_1^\to} X$ and $\map {\pr_2^\to} X$ are [[Definition:Directed Subset|directed]]
+where 
+:$\pr_1$ denotes the [[Definition:First Projection|first projection]] on $S \times T$
+:$\pr_2$ denotes the [[Definition:Second Projection|second projection]] on $S \times T$
+:$\map {\pr_1^\to} X$ denotes the [[Definition:Image of Subset under Mapping|image]] of $X$ under $\pr_1$
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \map {\pr_1^\to} X$.
+By definitions of [[Definition:Image of Subset under Mapping|image of set]] and [[Definition:Projection (Mapping Theory)|projections]]:
+:$\exists x' \in T: \tuple {x, x'} \in X$
+and
+:$\exists y' \in T: \tuple {y, y'} \in X$
+By definition of [[Definition:Directed Subset|directed]]:
+:$\exists \tuple {a, b} \in X: \tuple {x, x'} \preceq \tuple {a, b} \land \tuple {y, y'} \preceq \tuple {a, b}$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\exists a \in \map {\pr_1^\to} X: x \preceq_1 a \land y \preceq_1 a$
+Thus by definition
+:$\map {\pr_1^\to} X$ is [[Definition:Directed Subset|directed]].
+By [[Definition:Mutatis Mutandis|mutatis mutandis]]:
+:$\map {\pr_2^\to} X$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Cosines of Multiples of Angle}
+Tags: Cosine Function, Telescoping Series
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \frac 1 2 + \sum_{k \mathop = 1}^n \cos \left({k x}\right)
+      | r = \frac 1 2 + \cos x + \cos 2 x + \cos 3 x + \cdots + \cos n x
+      | c = 
+}}
+{{eqn| r = \frac {\sin \left({\left({2 n + 1}\right) x / 2}\right)} {2 \sin \left({x / 2}\right)}
+      | c = 
+}}
+{{end-eqn}}
+where $x$ is not an [[Definition:Integer Multiple|integer multiple]] of $2 \pi$.
+\end{theorem}
+
+\begin{proof}
+By the [[Simpson's Formula for Cosine by Sine]]:
+:$2 \cos \alpha \sin \beta = \sin \left({\alpha + \beta}\right) - \sin \left({\alpha - \beta}\right)$
+Thus we establish the following sequence of identities:
+{{begin-eqn}}
+{{eqn | l = 2 \cdot \frac 1 2 \sin \frac x 2
+      | r = \sin \frac x 2
+      | c = 
+}}
+{{eqn | l = 2 \cos x \sin \frac x 2
+      | r = \sin \frac {3 x} 2 - \sin \frac x 2
+      | c = 
+}}
+{{eqn | l = 2 \cos 2 x \sin \frac x 2
+      | r = \sin \frac {5 x} 2 - \sin \frac {3 x} 2
+      | c = 
+}}
+{{eqn | o = \cdots
+      | c = 
+}}
+{{eqn | l = 2 \cos n x \sin \frac x 2
+      | r = \sin \frac {\left({2 n + 1}\right) x} 2 - \sin \frac {\left({2 n - 1}\right) x} 2
+      | c = 
+}}
+{{end-eqn}}
+Summing the above:
+:$\displaystyle 2 \sin \frac x 2 \left({\frac 1 2 + \sum_{k \mathop = 1}^n \cos \left({k x}\right)}\right) = \sin \frac {\left({2 n + 1}\right) x} 2$
+as the sums on the {{RHS}} form a [[Definition:Telescoping Series|telescoping series]].
+The result follows by dividing both sides by $2 \sin \dfrac x 2$.
+It is noted that when $x$ is a multiple of $2 \pi$ then:
+:$\sin \dfrac x 2 = 0$
+leaving the {{RHS}} undefined.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eratosthenes' Measurement of Earth}
+Tags: Geodesy
+
+\begin{theorem}
+The [[Definition:Circumference of Circle|circumference]] of [[Definition:Earth|Earth]] was measured by {{AuthorRef|Eratosthenes of Cyrene}} to be of the order of $45 \, 000 \, \mathrm{km}$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CircumferenceOfEarth.png|500px]]
+{{finish}}
+{{Namedfor|Eratosthenes of Cyrene|cat = Eratosthenes}}
+\end{proof}<|endoftext|>
+\section{Existence of Prime-Free Sequence of Natural Numbers}
+Tags: Number Theory, Prime Numbers
+
+\begin{theorem}
+Let $n$ be a [[Definition:Natural Number|natural number]].
+Then there exists a [[Definition:Sequence|sequence]] of consecutive [[Definition:Natural Number|natural numbers]] of length $n$ which are all [[Definition:Composite Number|composite]].
+\end{theorem}
+
+\begin{proof}
+Consider the number:
+:$N := \paren {n + 1}!$
+where $!$ denotes the [[Definition:Factorial|factorial]].
+Let $i \in I$ where:
+:$I = \set {i \in \N: 2 \le i \le n + 1}$
+We have that $\size I = n$.
+Then for all $i \in I$:
+{{begin-eqn}}
+{{eqn | l = i
+      | o = \divides
+      | r = N
+      | c = where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]]
+}}
+{{eqn | ll= \leadsto
+      | l = i
+      | o = \divides
+      | r = N + i
+      | c = 
+}}
+{{end-eqn}}
+and so $N + i$ is [[Definition:Composite Number|composite]].
+That is:
+:$N + 2, N + 3, \ldots, N + n, N + n + 1$
+are all [[Definition:Composite Number|composite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Reciprocals of Twin Primes}
+Tags: Prime Numbers
+
+\begin{theorem}
+The sum of the [[Definition:Reciprocal|reciprocals]] of all the [[Definition:Twin Primes|twin primes]]:
+:$\dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 {11} + \dfrac 1 {13} + \dfrac 1 {17} + \dfrac 1 {19} + \dfrac 1 {29} + \dfrac 1 {31} + \cdots$
+is either [[Definition:Finite|finite]] or [[Definition:Convergent Series|convergent]].
+\end{theorem}<|endoftext|>
+\section{Approximate Value of Nth Prime Number}
+Tags: Prime Number Theorem, Prime Numbers, Analytic Number Theory
+
+\begin{theorem}
+The $n$th [[Definition:Prime Number|prime number]] is approximately $n \ln n$.
+\end{theorem}
+
+\begin{proof}
+This will be demonstrated by showing that:
+:$\displaystyle \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n} = 1$
+where $p_n$ denotes the $n$th [[Definition:Prime Number|prime number]].
+By definition of [[Definition:Prime-Counting Function|prime-counting function]]:
+:$\map \pi {p_n} = n$
+The [[Prime Number Theorem]] gives:
+:$\displaystyle \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x} = 1$
+Thus:
+:$\displaystyle \lim_{x \mathop \to \infty} \dfrac n {p_n / \ln p_n} = 1$
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to \infty} \dfrac {\map \pi x} {x / \ln x}
+      | r = 1
+      | c = [[Prime Number Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \dfrac n {p_n / \ln p_n}
+      | r = 1
+      | c = Definition of [[Definition:Prime-Counting Function|Prime-Counting Function]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \paren {\ln p_n - \ln n - \ln \ln p_n}
+      | r = 0
+      | c = [[Definition:Logarithm|Logarithm]] of both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} - \frac {\ln \ln p_n} {\ln p_n} }
+      | r = 0
+      | c = multiplying argument through by $\ln p_n$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \ln p_n \paren {1 - \frac {\ln n} {\ln p_n} }
+      | r = 0
+      | c = as $\dfrac {\ln n} n \to 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \paren {1 - \frac {\ln n} {\ln p_n} }
+      | r = 0
+      | c = as $\ln p_n \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \frac {\ln n} {\ln p_n}
+      | r = 1
+      | c = as $\ln p_n \ne 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln n} 
+      | r = \lim_{n \mathop \to \infty} \dfrac {p_n} {n \ln p_n} \frac {\ln p_n} {\ln n}
+      | c = from $(1)$
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Number Theorem in Eulerian Logarithmic Integral Form}
+Tags: Prime Number Theorem, Analytic Number Theory
+
+\begin{theorem}
+The [[Prime Number Theorem]] is equivalent to:
+:$\displaystyle \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x} = 1$
+where:
+:$\map \pi x$ is the [[Definition:Prime-Counting Function|prime-counting function]]
+:$\map \Li x$ is the [[Definition:Eulerian Logarithmic Integral|Eulerian logarithmic integral]]:
+::$\displaystyle \map \Li x := \int_2^x \dfrac {\d t} {\ln t}$
+\end{theorem}
+
+\begin{proof}
+Using [[Integration by Parts]]:
+{{begin-eqn}}
+{{eqn | l = \map \Li x
+      | r = \int_2^x \dfrac {\d t} {\ln t}
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \dfrac x {\ln x} - \dfrac 2 {\ln 2} + \int_2^x \dfrac {\d t} {\paren {\ln t}^2}
+      | c = 
+}}
+{{end-eqn}}
+We have that $\dfrac 1 {\paren {\ln t}^2}$ is [[Definition:Positive Real Function|positive]] and [[Definition:Decreasing Real Function|decreasing]] for $t > 1$.
+Let $x \ge 4$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | r = \int_2^x \dfrac {\d t} {\paren {\ln t}^2}
+      | c = 
+}}
+{{eqn | r = \int_2^{\sqrt x} \dfrac {\d t} {\paren {\ln t}^2} + \int_{\sqrt x}^x \dfrac {\d t} {\paren {\ln t}^2}
+      | c = 
+}}
+{{eqn | o = <
+      | r = \dfrac {\sqrt x - 2} {\paren {\ln 2}^2} + \dfrac {x - \sqrt x} {\paren {\ln \sqrt x}^2}
+      | c = 
+}}
+{{eqn | o = <
+      | r = \dfrac {\sqrt x} {\paren {\ln 2}^2} + \dfrac {4 x} {\paren {\ln \sqrt x}^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}
+      | c = 
+}}
+{{eqn | o = <
+      | r = \dfrac {\ln x} {\sqrt x \paren {\ln 2}^2} + \frac 4 {\ln x}
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \lim_{x \mathop \to \infty} \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\map \Li x} {x / \ln x}
+      | r = \dfrac {\dfrac x {\ln x} - \dfrac 2 {\ln 2} + \displaystyle \int_2^x \dfrac {\d t} {\paren {\ln t}^2} } {x / \ln x}
+      | c = from $(1)$
+}}
+{{eqn | r = 1 - \dfrac {2 \ln x} {x \ln 2} + \dfrac {\displaystyle \int_2^x \frac {\d t} {\paren {\ln t}^2} } {x / \ln x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}
+      | r = 1 - \lim_{x \mathop \to \infty} \dfrac 2 {\ln 2} \dfrac x {\ln x} - 0
+      | c = from $(2)$
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \lim_{x \mathop \to \infty} \frac {\map \Li x} {x / \ln x}
+      | r = 1
+      | c = as $\displaystyle \lim_{x \mathop \to \infty} \dfrac x {\ln x} = 0$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to \infty} \frac {\map \pi x} {x / \ln x}
+      | r = \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x} \frac {\map \Li x} {x / \ln x}
+      | c = 
+}}
+{{eqn | r = \lim_{x \mathop \to \infty} \frac {\map \pi x} {\map \Li x}
+      | c = from $(3)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of Rational Number is Irrational}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $r$ be a [[Definition:Rational Number|rational number]] such that $r \ne 0$.
+Then:
+:$e^r$ is [[Definition:Irrational Number|irrational]]
+where $e$ is [[Definition:Euler's Number|Euler's number]].
+\end{theorem}
+
+\begin{proof}
+Let $r = \dfrac p q$ be [[Definition:Rational Number|rational]] such that $r \ne 0$.
+{{AimForCont}} $e^r$ is [[Definition:Rational Number|rational]].
+Then $\paren {e^r}^q = e^p$ is also [[Definition:Rational Number|rational]].
+Then if $e^{-p}$ is [[Definition:Rational Number|rational]], it follows that $e^p$ is [[Definition:Rational Number|rational]].
+It is therefore sufficient to derive a [[Definition:Contradiction|contradiction]] from the supposition that $e^p$ is [[Definition:Rational Number|rational]] for every $p \in \Z_{>0}$.
+Let $e^p = \dfrac a b$ for $a, b \in \Z_{>0}$.
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $\map f x$ be the [[Definition:Real Function|function]] defined as:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map f x
+      | o = :=
+      | r = \frac {x^n \paren {1 - x}^n} {n!}
+      | c = 
+}}
+{{eqn | r = \frac 1 {n!} \sum_{k \mathop = n}^{2 n} c_k x^k
+      | c = where the $c_k$'s are some [[Definition:Integer|integers]]
+}}
+{{end-eqn}}
+Note that:
+:$(2): \quad 0 < x < 1 \implies 0 < \map f x < \dfrac 1 {n!}$
+We also have:
+:$\map f 0 = 0$
+:$\map {f^{\paren m} } 0 = 0$ if $m < n$ or $m > 2 n$
+and:
+:$n \le m \le 2 n \implies \map {f^{\paren m} } 0 = \dfrac {m!} {n!} c_m$
+and this number is an [[Definition:Integer|integer]].
+Thus at $x = 0$, $\map f x$ and all its [[Definition:Higher Derivative|derivatives]] are [[Definition:Integer|integers]].
+Since $\map f {1 - x} = \map f x$, the same is true for $x = 1$.
+Let $\map F x$ be the [[Definition:Real Function|function]] defined as:
+{{begin-eqn}}
+{{eqn | n = 3
+      | l = \map F x
+      | o = :=
+      | r = \sum_{k \mathop = 0}^{2 n} \paren {-1}^k p^{2 n - k} \map {f^{\paren k} } x
+      | c = 
+}}
+{{eqn | r = p^{2 n} \map f x - p^{2 n - 1} \map {f'} x + p^{2 n - 2} \map {f''} x - \cdots - p \map {f^{\paren {2 n - 1} } } x + \map {f^{\paren {2 n} } } x
+      | c = 
+}}
+{{end-eqn}}
+Because of the properties of $\map f x$ and its [[Definition:Higher Derivative|derivatives]] above, $\map F 0$ and $\map F 1$ are [[Definition:Integer|integers]].
+Next we have:
+{{begin-eqn}}
+{{eqn | n = 4
+      | l = \map {\frac {\d} {\d x} } {e^{p x} \map F x}
+      | r = e^{p x} \paren {\map {F'} x + p \, \map F x}
+      | c = 
+}}
+{{eqn | r = p^{2 n + 1} e^{p x} \map f x
+      | c = from $\map {F'} x + p \, \map F x$ based on $(3)$
+}}
+{{end-eqn}}
+$(4)$ leads to:
+{{begin-eqn}}
+{{eqn | l = b \int_0^1 p^{2 n + 1} e^{p x} \map f x \d x
+      | r = b \bigintlimits {e^{p x} \map F x} 0 1
+      | c = 
+}}
+{{eqn | r = a \, \map F 1 - b \, \map F 0
+      | c = 
+}}
+{{end-eqn}}
+which is an [[Definition:Integer|integer]].
+But from $(2)$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | r = b \int_0^1 p^{2 n + 1} e^{p x} \map f x \rd x
+      | c = 
+}}
+{{eqn | o = <
+      | r = \frac {b p^{2 n + 1} e^p} {n!}
+      | c = 
+}}
+{{eqn | r = b p e^p \frac {\paren {p^2}^n e^p} {n!}
+      | c = 
+}}
+{{end-eqn}}
+The expression on the {{RHS}} tends to $0$ as $n$ tends to $\infty$.
+Hence:
+:$0 < a \, \map F 1 - b \, \map F 0 < 1$
+for sufficiently large $n$.
+But there exists no [[Definition:Integer|integer]] strictly between $0$ and $1$.
+From this [[Definition:Contradiction|contradiction]] it follows that our original assumption, that is, that $e^r$ is [[Definition:Rational Number|rational]], must have been false.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Points on Graph of Exponential Function}
+Tags: Exponential Function
+
+\begin{theorem}
+Consider the [[Definition:Graph of Mapping|graph]] $f$ of the [[Definition:Real Exponential Function|exponential function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
+:$f := \set {\tuple {x, y} \in \R^2: y = e^x}$
+The only [[Definition:Rational Point in Plane|rational point]] of $f$ is $\tuple {0, 1}$.
+\end{theorem}
+
+\begin{proof}
+From [[Exponential of Rational Number is Irrational]]:
+:$r \in \Q_{\ne 0} \implies e^r \in \R - \Q$
+Thus, apart from the point $\tuple {0, 1}$, when $x$ is [[Definition:Rational Number|rational]], $e^x$ is not.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Points on Graph of Logarithm Function}
+Tags: Logarithms
+
+\begin{theorem}
+Consider the [[Definition:Graph of Mapping|graph]] of the [[Definition:Logarithm|logarithm function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
+:$f := \set {\tuple {x, y} \in \R^2: y = \ln x}$
+The only [[Definition:Rational Point in Plane|rational point]] of $f$ is $\tuple {1, 0}$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Graph of Mapping|graph]] of the [[Definition:Real Exponential Function|exponential function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
+:$g := \set {\tuple {x, y} \in \R^2: y = e^x}$
+From [[Rational Points on Graph of Exponential Function]], the only [[Definition:Rational Point in Plane|rational point]] of $g$ is $\tuple {0, 1}$.
+By definition of the [[Definition:Exponential Function/Real/Inverse of Natural Logarithm|exponential function]], $f$ and $g$ are [[Definition:Inverse Mapping|inverses]].
+Thus:
+:$\tuple {x, y} \in g \iff \tuple {y, x} \in f$
+Thus for $\tuple {x, y} \in g$ to be [[Definition:Rational Point in Plane|rational]], $\tuple {y, x} = \tuple {0, 1}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Points on Graph of Sine Function}
+Tags: Sine Function
+
+\begin{theorem}
+Consider the [[Definition:Graph of Mapping|graph]] of the [[Definition:Sine Function|sine function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
+:$f := \set {\tuple {x, y} \in \R^2: y = \sin x}$
+The only [[Definition:Rational Point in Plane|rational point]] of $f$ is $\tuple {0, 0}$.
+\end{theorem}<|endoftext|>
+\section{Rational Points on Graph of Cosine Function}
+Tags: Sine Function
+
+\begin{theorem}
+Consider the [[Definition:Graph of Mapping|graph]] of the [[Definition:Cosine Function|cosine function]] in the [[Definition:Cartesian Plane|real Cartesian plane]] $\R^2$:
+:$f := \left\{ {\left({x, y}\right) \in \R^2: y = \cos x}\right\}$
+The only [[Definition:Rational Point in Plane|rational point]] of $f$ is $\left({0, 1}\right)$.
+\end{theorem}<|endoftext|>
+\section{Meet-Continuous iff Ideal Supremum is Meet Preserving}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Let $f: \map {\it Ids} {\mathscr S} \to S$ be a [[Definition:Mapping|mapping]] such that:
+:$\forall I \in \map {\it Ids} {\mathscr S}: \map f I = \sup_{\mathscr S} I$
+where 
+:$\map {\it Ids} {\mathscr S}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$
+Then
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]
+{{iff}}
+:$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]] as a [[Definition:Mapping|mapping]] from $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$ into $\mathscr S$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+We will prove that
+:for every [[Definition:Directed Subset|directed subsets]] $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
+Let $D_1, D_2$ be [[Definition:Directed Subset|directed subsets]] of $S$.
+we will prove as sublemma that
+:for every an [[Definition:Element|element]] $x$ of $S$, a [[Definition:Directed Subset|directed subset]] $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \set {x \wedge d: d \in D}$
+Let $x \in S$, $D$ 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}}
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:for every [[Definition:Element|element]] $x$ of $S$, a [[Definition:Directed Subset|directed subset]] $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \set {x \wedge d: d \in D}$
+Define a [[Definition:Mapping|mapping]] $g: S \times S \to S$:
+:$\forall \tuple {s, t} \in S \times S: \map g {s, t} = s \wedge t$
+By [[Meet is Directed Suprema Preserving]]:
+:$g$ [[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 $\mathscr S$ and $\mathscr S$ into $\mathscr S$.
+Thus by [[Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets]]: 
+:$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
+By exemplification:
+:for every [[Definition:Ideal in Ordered Set|ideals]] $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$
+Thus by [[Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving]]:
+:$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]]
+Thus
+:$\mathscr S$ is [[Definition:Up-Complete|up-complete]]
+We will prove that
+:for every [[Definition:Ideal in Ordered Set|ideals]] $I_1, I_2$ in $\mathscr S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$
+Let $I_1, I_2$ be [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+Thus
+{{begin-eqn}}
+{{eqn | l = \paren {\sup I_1} \wedge \paren {\sup I_2}
+      | r = \map f {I_1} \wedge \map f {I_2}
+      | c = Definition of $f$
+}}
+{{eqn | r = \map f {I_1 \wedge I_2}
+      | c = {{Defof|Mapping Preserves Infimum|subdef = Meet}}
+}}
+{{eqn | r = \map \sup {I_1 \wedge I_2}
+      | c = Definition of $f$
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}
+      | c = [[Meet in Set of Ideals]]
+}}
+{{end-eqn}}
+We will prove that
+:for every [[Definition:Directed Subset|directed subsets]] $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
+Let $D_1, D_2$ be [[Definition:Directed Subset|directed subsets]] of $S$.
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D_1$ and $D_2$ admit [[Definition:Supremum of Set|suprema]]
+By [[Supremum of Lower Closure of Set]]:
+:$D_1^\preceq$ and $D_2^\preceq$ admit [[Definition:Supremum of Set|suprema]]
+Thus
+{{begin-eqn}}
+{{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}
+      | r = \paren {\sup D_1^\preceq} \wedge \paren {\sup D_2^\preceq}
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}
+      | c = $D_1^\preceq$ is [[Definition:Ideal in Ordered Set|ideal]] in $\mathscr S$
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq
+      | c = [[Lower Closure of Meet of Lower Closures]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{end-eqn}}
+It remains to prove (MC) of definition of [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Let $x$ be an [[Definition:Element|element]] of $S$.
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$.
+By [[Singleton is Directed and Filtered Subset]]:
+:$\set x$ is [[Definition:Directed Subset|directed]].
+Thus
+{{begin-eqn}}
+{{eqn | l = x \wedge \sup D
+      | r = \paren {\sup \set x} \wedge \sup D
+      | c = [[Supremum of Singleton]]
+}}
+{{eqn | r = \sup \set {d_1 \wedge d_2: d_1 \in \set x, d_2 \in D}
+}}
+{{eqn | r = \sup \set {x \wedge d: d \in D}
+      | c = {{Defof|Singleton}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Number is Algebraic}
+Tags: Algebraic Numbers, Rational Numbers
+
+\begin{theorem}
+Let $r \in \Q$ be a [[Definition:Rational Number|rational number]].
+Then $r$ is also an [[Definition:Algebraic Number|algebraic number]].
+\end{theorem}
+
+\begin{proof}
+Let $r$ be expressed in the form:
+:$r = \dfrac p q$
+Consider the [[Definition:Linear Polynomial|linear polynomial]] in $x$:
+:$q x - p = 0$
+which has the solution:
+:$x = \dfrac p q$
+Hence the result, by definition of [[Definition:Algebraic Number|algebraic number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Number is Algebraic of Degree 1}
+Tags: Algebraic Numbers, Rational Numbers
+
+\begin{theorem}
+Let $r \in \Q$ be a [[Definition:Rational Number|rational number]].
+Then $r$ is an [[Definition:Algebraic Number|algebraic number]] of [[Definition:Degree of Algebraic Number|degree]] $1$.
+\end{theorem}
+
+\begin{proof}
+Let $r$ be expressed in the form:
+:$r = \dfrac p q$
+By [[Rational Number is Algebraic]], $r$ can be expressed as the [[Definition:Root of Polynomial|root]] of the [[Definition:Linear Polynomial|linear polynomial]]:
+:$q x - p = 0$
+A [[Definition:Polynomial (Analysis)|polynomial]] of [[Definition:Polynomial of Degree Zero|degree zero]] is a [[Definition:Constant Polynomial|constant polynomial]], and has no [[Definition:Root of Polynomial|roots]].
+Hence the result by definition of [[Definition:Degree of Algebraic Number|degree of algebraic number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square Root of 2 is Algebraic of Degree 2}
+Tags: Algebraic Numbers, Square Root of 2
+
+\begin{theorem}
+The [[Definition:Square Root|square root]] of $2$ is an [[Definition:Algebraic Number|algebraic number]] of [[Definition:Degree of Algebraic Number|degree]] $2$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\sqrt 2$ could be expressed as the [[Definition:Root of Polynomial|root]] of the [[Definition:Linear Polynomial|linear polynomial]]:
+:$a_1 x + a_0 = 0$
+for some $a_0, a_1 \in \Q$.
+Then:
+:$\sqrt 2 = -\dfrac {a_0} {a_1}$
+and would be [[Definition:Rational Number|rational]].
+But as [[Square Root of 2 is Irrational]], this is not the case.
+However, $\sqrt 2$ is a [[Definition:Root of Polynomial|root]] of the [[Definition:Polynomial (Analysis)|polynomial]] of [[Definition:Degree of Polynomial|degree]] $2$:
+:$x^2 - 2 = 0$
+Hence the result by definition of [[Definition:Degree of Algebraic Number|degree of algebraic number]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Constructible Length with Compass and Straightedge}
+Tags: Euclidean Geometry
+
+\begin{theorem}
+Let $L$ be a [[Definition:Line Segment|line segment]] in a [[Definition:Euclidean Space|Eucldiean space]].
+Let the [[Definition:Length (Linear Measure)|length]] of $L$ be $d$.
+Let $L'$ be a [[Definition:Line Segment|line segment]] of [[Definition:Length (Linear Measure)|length]] $d'$ constructed from $L$ using a [[Definition:Compass and Straightedge Construction|compass and straightedge construction]].
+Then:
+:$d' = q d$
+where $q$ is an [[Definition:Algebraic Number|algebraic number]] whose [[Definition:Degree of Algebraic Number|degree]] is at most $2$.
+\end{theorem}<|endoftext|>
+\section{Gelfond-Schneider Constant is Transcendental}
+Tags: Real Analysis
+
+\begin{theorem}
+The [[Definition:Gelfond-Schneider Constant|Gelfond-Schneider constant]]:
+:$2^{\sqrt 2}$
+is [[Definition:Transcendental|transcendental]].
+\end{theorem}
+
+\begin{proof}
+From the [[Gelfond-Schneider Theorem]]:
+If:
+:$\alpha$ and $\beta$ are [[Definition:Algebraic Number|algebraic numbers]] such that $\alpha \notin \set {0, 1}$
+:$\beta$ is either [[Definition:Irrational Number|irrational]] or not [[Definition:Wholly Real|wholly real]]
+then $\alpha^\beta$ is [[Definition:Transcendental Number|transcendental]].
+From [[Rational Number is Algebraic]]:
+:$2$ is [[Definition:Algebraic Number|algebraic]].
+From [[Square Root of 2 is Algebraic of Degree 2]]:
+:$\sqrt 2$ is [[Definition:Algebraic Number|algebraic]].
+From [[Square Root of 2 is Irrational]]:
+:$\sqrt 2$ is [[Definition:Irrational Number|irrational]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Bernoulli Numbers by Binomial Coefficients Vanishes}
+Tags: Bernoulli Numbers, Binomial Coefficients, Sum of Bernoulli Numbers by Binomial Coefficients Vanishes
+
+\begin{theorem}
+:$\forall n \in \Z_{>1}: \displaystyle \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$
+where $B_k$ denotes the $k$th [[Definition:Bernoulli Numbers|Bernoulli number]].
+\end{theorem}
+
+\begin{proof}
+Take the definition of [[Definition:Bernoulli Numbers/Generating Function|Bernoulli numbers]]:
+:$\displaystyle \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
+From the definition of the [[Definition:Exponential Function/Real/Sum of Series|exponential function]]:
+{{begin-eqn}}
+{{eqn | l = e^x
+      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
+      | c = 
+}}
+{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {e^x - 1} x
+      | r = \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!}
+      | c = 
+}}
+{{eqn | r = 1 + \frac x {2!} + \frac {x^2} {3!} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }
+      | c = 
+}}
+{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{n} } {\paren{n + 1 }!} }
+      | c = Both [[Definition:Series|series]] start at [[Definition:Zero|zero]]
+}}
+{{end-eqn}}
+By [[Product of Absolutely Convergent Series]], we will let:
+{{begin-eqn}}
+{{eqn | l = a_n
+      | r = \frac {B_n x^n} {n!}
+      | c = 
+}}
+{{eqn | l = b_n
+      | r = \frac {x^{n} } {\paren {n + 1 }!} 
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
+      | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n }
+      | rrr = =1
+      | c = 
+}}
+{{eqn | l = c_n
+      | r = \sum_{k \mathop = 0}^n a_k b_{n - k}
+      | c = 
+}}
+{{eqn | l = c_0
+      | r = \frac {B_0 x^0} {0!} \frac {x^{0} } {\paren {0 + 1 }!} 
+      | rrr = = 1
+      | c = $c_0 = \paren {a_0 } \paren {b_{0 - 0 } } = \paren {a_0 } \paren {b_0 }$
+}}
+{{eqn | lll = \leadsto
+      | l = \sum_{n \mathop = 1}^\infty c_n
+      | r = \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n } - a_0 b_0
+      | rrr = =0
+      | c = [[Definition:Subtraction|Subtract]] [[1]] from both sides of the [[Definition:Equation|equation]].
+}}
+{{eqn | r = c_1x + c_2x^2 + c_3x^3 + \cdots
+      | rrr = =0
+}}
+{{end-eqn}}
+$\leadsto \forall n \in \Z_{\gt 0}$, $c_n = 0$.
+{{begin-eqn}}
+{{eqn | l = c_1
+      | r = \frac {B_0 x^0} {0!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{0} } {\paren {0 + 1 }!}
+      | rrr = = 0
+      | c = $= a_0 b_1 + a_1 b_0$
+}}
+{{eqn | l = c_2
+      | r = \frac {B_0 x^0} {0!} \frac {x^{2} } {\paren {2 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_2 x^2} {2!} \frac {x^{0} } {\paren {0 + 1 }!}
+      | rrr = = 0
+      | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$
+}}
+{{eqn | l = \cdots
+      | r = \cdots
+      | rrr = = 0
+}}
+{{eqn | l = c_n
+      | r = \frac {B_0 x^0} {0!} \frac {x^{n} } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {x^{0} } {\paren {0 + 1 }!}
+      | rrr = = 0
+      | c = $= a_0 b_n + a_1 b_{n - 1 } + a_2 b_{n - 2 } + \cdots + a_n b_0$
+}}
+{{end-eqn}}
+Multiplying $c_n$ through by $\paren {n + 1 }!$ gives:
+{{begin-eqn}}
+{{eqn | l = \paren {n + 1 }! c_n
+      | r = \frac {B_0 x^0} {0!} \frac {\paren {n + 1 }! x^n } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {\paren {n + 1 }! x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {\paren {n + 1 }! x^{0} } {\paren {0 + 1 }!}
+      | rrr = = 0
+      | c = 
+}}
+{{eqn | r = x^n \paren {\frac {\paren {n + 1 }! } {0! \paren {n + 1 }!} B_0 + \frac {\paren {n + 1 }! } {1! \paren {n - 1 + 1 }!} B_1 + \cdots + \frac {\paren {n + 1 }! } {n! \paren {0 + 1 }!} B_n }
+      | rrr = = 0
+      | c = factoring out $x^n$
+}}
+{{end-eqn}}
+But those coefficients are the [[Definition:Binomial Coefficient|binomial coefficients]]:
+{{begin-eqn}}
+{{eqn | l = \paren {n + 1 }! c_n
+      | r = \dbinom {n + 1 } 0 B_0 + \dbinom {n + 1 } 1 B_1 + \dbinom {n + 1 } 2 B_2 + \cdots + \dbinom {n + 1 } n B_n
+      | rrr = = 0
+      | c = 
+}}
+{{eqn | l = n! c_{n-1 }
+      | r = \dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1}
+      | rrr = = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli Numbers are Rational}
+Tags: Bernoulli Numbers
+
+\begin{theorem}
+The [[Definition:Bernoulli Numbers|Bernoulli numbers]] are [[Definition:Rational Number|rational]].
+\end{theorem}
+
+\begin{proof}
+By the [[Definition:Bernoulli Numbers/Recurrence Relation|recurrence relation for the Bernoulli numbers]]:
+:$B_n = \begin{cases} 1 & : n = 0 \\
+\displaystyle - \sum_{k \mathop = 0}^{n-1} \binom n k \frac {B_k} {n - k + 1} & : n > 0
+\end{cases}$
+:$B_0$ is [[Definition:Rational Number|rational]].
+:$B_n$ is a finite sum of the products of a [[Definition:Binomial Coefficient|binomial coefficient]] with a [[Definition:Bernoulli Numbers|Bernoulli number]] earlier in the [[Definition:Sequence|sequence]] divided by an [[Definition:Integer|integer]].
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$B_n$ is [[Definition:Rational Number|rational]].
+$P \left({0}\right)$ is true, as this just says:
+:$B_0$ is [[Definition:Rational Number|rational]].
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+:$B_0$ is [[Definition:Rational Number|rational]].
+This is true, as $B_0 = 1$ which is [[Definition:Rational Number|rational]].
+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, where $0 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\forall j \in \N: 0 \le j \le k: B_j$ is [[Definition:Rational Number|rational]]
+from which it is to be shown that:
+:$B_{k + 1}$ is [[Definition:Rational Number|rational]].
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+By the [[Definition:Bernoulli Numbers/Recurrence Relation|recurrence relation for the Bernoulli numbers]]:
+{{begin-eqn}}
+:$\displaystyle B_{k + 1} = - \sum_{j \mathop = 0}^k \binom {k + 1} j \frac {B_j} {k - j + 2}$
+We have that:
+:From [[Binomial Coefficient is Integer]], $\dbinom {k + 1} j$ is an [[Definition:Integer|integer]] throughout.
+:From the [[Bernoulli Numbers are Rational#Induction Hypothesis|Induction Hypothesis]], $\forall j \in \N: 0 \le j \le k: B_j$ is [[Definition:Rational Number|rational]].
+:$k - j + 2$ is an [[Definition:Integer|integer]].
+Thus as $B_{k + 1}$ is the sum of a [[Definition:Finite Set|finite number]] of [[Definition:Rational Number|rational numbers]], it follows that $B_{k + 1}$ it itself [[Definition:Rational Number|rational]].
+The result follows by the [[Second Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: B_n$ is [[Definition:Rational Number|rational]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Tangent Function/Proof of Convergence}
+Tags: Power Series Expansion for Tangent Function
+
+\begin{theorem}
+The [[Definition:Radius of Convergence of Real Power Series|radius of convergence]] of the [[Power Series Expansion for Tangent Function]]:
+:$\displaystyle \tan x = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1}$
+where $B_{2 n}$ denotes the [[Definition:Bernoulli Numbers|Bernoulli numbers]], is given as:
+:$\size x < \dfrac \pi 2$
+\end{theorem}
+
+\begin{proof}
+By [[Combination Theorem for Limits of Functions]] we can deduce the following.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \lim_{n \mathop \to \infty} \size {\frac {\frac {\paren {-1}^n 2^{2 n + 2} \paren {2^{2 n + 2} - 1} B_{2 n + 2} } {\paren {2 n + 2}!} x^{2 n + 1} } {\frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1} } }
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {2^{2 n + 2} - 1} } {\paren {2^{2 n} - 1} } \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } 4 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 2} - 1} {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n} } {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n} - 1} - \frac 1 {2^{2 n} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } { B_{2 n} } } 8 x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 8 x^2
+      | c = [[Asymptotic Formula for Bernoulli Numbers]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n } \paren {\frac {n + 1} n}^{2 n} } \frac 8 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n}^{2 n} } \frac 4 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 4 {\pi^2 e^2} x^2
+      | c =
+}}
+{{eqn | r = \frac {4 e^2} {\pi^2 e^2} x^2
+      | c = {{Defof|Euler's Number/Limit of Sequence|Euler's Number}}
+}}
+{{eqn | r = \frac 4 {\pi^2} x^2
+      | c =
+}}
+{{end-eqn}}
+This is less than $1$ {{iff}}:
+:$\size x < \dfrac \pi 2$
+Hence by the [[Ratio Test]], the series [[Definition:Convergent Series|converges]] for $\size x < \dfrac \pi 2$.
+\end{proof}<|endoftext|>
+\section{Power Series Expansion for Tangent Function/Sequence}
+Tags: Power Series Expansion for Tangent Function
+
+\begin{theorem}
+The [[Power Series Expansion for Tangent Function]] begins:
+:$\tan x = x + \dfrac 1 3 x^3 + \dfrac 2 {15} x^5 + \dfrac {17} {315} x^7 + \dfrac {62} {2835} x^9 + \cdots$
+\end{theorem}
+
+\begin{proof}
+From [[Power Series Expansion for Tangent Function]]:
+{{begin-eqn}}
+{{eqn | l = \tan x
+      | r = \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n - 1} 2^{2 n} \paren {2^{2 n} - 1} B_{2 n} } {\paren {2 n}!} x^{2 n - 1}
+      | c = 
+}}
+{{eqn | r = \frac {2^2 \paren {2^2 - 1} B_2} {2!} x - \frac {2^4 \paren {2^4 - 1} B_4} {4!} x^3 + \frac {2^6 \paren {2^6 - 1} B_6} {6!} x^5 - \frac {2^8 \paren {2^8 - 1} B_8} {8!} x^7 + \frac {2^{10} \paren {2^{10} - 1} B_{10} } {10!} x^9 - \cdots
+      | c = 
+}}
+{{end-eqn}}
+Enumerating the [[Definition:Sequence of Bernoulli Numbers|Bernoulli numbers]]:
+{{begin-eqn}}
+{{eqn | l = B_2
+      | r = \dfrac 1 6
+}}
+{{eqn | l = B_4
+      | r = -\dfrac 1 {30}
+}}
+{{eqn | l = B_6
+      | r = \dfrac 1 {42}
+}}
+{{eqn | l = B_8
+      | r = -\dfrac 1 {30}
+}}
+{{eqn | l = B_{10}
+      | r = \dfrac 5 {66}
+}}
+{{end-eqn}}
+Thus the appropriate arithmetic is performed on each coefficient:
+{{begin-eqn}}
+{{eqn | l = \frac {2^2 \paren {2^2 - 1} B_2} {2!}
+      | r = \frac {4 \times 3} {2} \times \dfrac 1 6
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = -\frac {2^4 \paren {2^4 - 1} B_4} {4!}
+      | r = -\dfrac {16 \times 15} {24} \times \dfrac {-1} {30}
+}}
+{{eqn | r = \dfrac {\paren {2^4} \times \paren {3 \times 5} } {\paren {2^3 \times 3} \times \paren {2 \times 3 \times 5} }
+}}
+{{eqn | r = \dfrac 1 3
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac {2^6 \paren {2^6 - 1} B_6} {6!}
+      | r = \dfrac {64 \times 63} {720} \times \dfrac 1 {42}
+}}
+{{eqn | r = \dfrac {\paren {2^6} \times \paren {3^2 \times 7} } {\paren {2^4 \times 3^2 \times 5} \times \paren {2 \times 3 \times 7} }
+}}
+{{eqn | r = \dfrac 2 {15}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = -\frac {2^8 \paren {2^8 - 1} B_8} {8!}
+      | r = -\dfrac {256 \times 255} {40 \, 320} \times \dfrac {-1} {30}
+}}
+{{eqn | r = \dfrac {\paren {2^8} \times \paren {3 \times 5 \times 17} } {\paren {2^7 \times 3^2 \times 5 \times 7} \times \paren {2 \times 3 \times 5} }
+}}
+{{eqn | r = \dfrac {17} {3^2 \times 5 \times 7}
+}}
+{{eqn | r = \dfrac {17} {315}
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \frac {2^{10} \paren {2^{10} - 1} B_{10} } {10!}
+      | r = \dfrac {1024 \times 1023} {3 \, 628 \, 800} \times \dfrac 5 {66}
+}}
+{{eqn | r = \dfrac {\paren {2^{10} } \times \paren {3 \times 11 \times 31} \times 5} {\paren {2^8 \times 3^4 \times 5^2 \times 7} \times \paren {2 \times 3 \times 11} }
+}}
+{{eqn | r = \dfrac {2 \times 31} {\paren {3^4 \times 5 \times 7} }
+}}
+{{eqn | r = \dfrac {62} {2835}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Partial Fractions Expansion of Cotangent}
+Tags: Cotangent Function, Partial Fractions Expansion of Cotangent
+
+\begin{theorem}
+Let $x \in \R - \Z$ , that is such that $x$ is a [[Definition:Real Number|real number]] that is not an [[Definition:Integer|integer]].
+Then:
+:$\pi \cot \pi x = \dfrac 1 2 + 2 x \displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2}$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\cot \pi x = \dfrac {\cos \pi x} {\sin \pi x}$
+has a [[Definition:Denominator|denominator]] which is $0$ at $x = 0, \pm 1, \pm 2, \ldots$.
+Hence the limitation on the [[Definition:Domain of Mapping|domain]] of $x \cot \pi x$ to exclude [[Definition:Integer|integer]] $x$.
+Having established that, we should be able to express $\cot \pi x$ in the form:
+:$\cot \pi x = \dfrac a x + \displaystyle \sum_{n \mathop = 1}^\infty \left({\frac {b_n} {x - n} + \frac {c_n} {x + n} }\right)$
+using a [[Definition:Partial Fractions Expansion|partial fractions expansion]].
+By evaluating the coefficients $b_n$ and $c_n$ in the usual manner, they are found to be:
+{{explain|Link to what this usual manner is}}
+{{finish|Do the work}}
+:$\forall n \in \N: b_n = c_n = \dfrac 1 \pi$
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the [[Euler Formula for Sine Function]]:
+:$\displaystyle \frac {\sin x} x = \prod_{n \mathop = 1}^\infty \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$
+Taking the [[Definition:Logarithm|logarithm]] of both sides:
+:$\displaystyle \ln \frac {\sin x} x = \sum_{n \mathop = 1}^\infty \ln \left({1 - \frac {x^2} {n^2 \pi^2}}\right)$
+and [[Definition:Differentiation|differentiating]] {{WRT|Differentiation}} $x$:
+{{begin-eqn}}
+{{eqn | l = \cot x - \frac 1 x
+      | r = \sum_{n \mathop = 1}^\infty \frac {- 2 x} {n^2 \pi^2 - x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cot x
+      | r = \frac 1 x + 2 x \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \pi x \cot \pi x
+      | r = 1 + 2 x^2 \sum_{n \mathop = 1}^\infty \frac 1 {x^2 - n^2 \pi^2}
+      | c = putting $\pi x$ for $x$ and multiplying by $\pi x$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent to Cycloid is Vertical at Cusps}
+Tags: Cycloids
+
+\begin{theorem}
+The [[Definition:Tangent to Curve|tangent]] to the [[Definition:Cycloid|cycloid]] whose [[Definition:Locus|locus]] is given by:
+: $x = a \paren {\theta - \sin \theta}$
+: $y = a \paren {1 - \cos \theta}$
+is [[Definition:Vertical|vertical]] at the [[Definition:Cusp of Cycloid|cusps]].
+\end{theorem}
+
+\begin{proof}
+From [[Slope of Tangent to Cycloid]], the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $C$ at the point $\tuple {x, y}$ is given by:
+:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
+At the [[Definition:Cusp of Cycloid|cusps]], $\theta = 2 n \pi$ for $n \in \Z$.
+Thus at the [[Definition:Cusp of Cycloid|cusps]], the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $C$ is $\cot n \pi$.
+From [[Shape of Cotangent Function]]:
+:$\displaystyle \lim_{\theta \mathop \to n \pi} \cot \theta \to \infty$
+Hence the result by definition of [[Definition:Vertical Tangent Line|vertical tangent line]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent to Cycloid passes through Top of Generating Circle}
+Tags: Cycloids
+
+\begin{theorem}
+Let $C$ be a [[Definition:Cycloid|cycloid]] generated by the equations:
+:$x = a \paren {\theta - \sin \theta}$
+:$y = a \paren {1 - \cos \theta}$
+Then the [[Definition:Tangent to Curve|tangent]] to $C$ at a point $P$ on $C$ passes through the top of the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$.
+\end{theorem}
+
+\begin{proof}
+From [[Tangent to Cycloid]], the equation for the [[Definition:Tangent to Curve|tangent]] to $C$ at a point $P = \tuple {x, y}$ is given by:
+:$(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$
+From [[Equation of Cycloid]], the point at the top of the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$ has coordinates $\tuple {2 a, a \theta}$.
+Substituting $x = 2 a$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = y - a \paren {1 - \cos \theta}
+      | r = \dfrac {\sin \theta} {1 - \cos \theta} \paren {a \theta - a \theta + a \sin \theta}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = a \paren {1 - \cos \theta} + \dfrac {\sin \theta} {1 - \cos \theta} a \sin \theta
+      | c = 
+}}
+{{eqn | r = \frac {a \paren {1 - \cos \theta}^2 + a \sin^2 \theta} {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a \paren {1 - 2 \cos \theta + \cos^2 \theta} + a \sin^2 \theta} {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a \paren {1 - 2 \cos \theta + 1} } {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a \paren {2 - 2 \cos \theta} } {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = 2 a
+      | c = 
+}}
+{{end-eqn}}
+That is, the [[Definition:Tangent to Curve|tangent]] to $C$ passes through $\tuple {a \theta, 2 a}$ as was required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent to Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Let $C$ be a [[Definition:Cycloid|cycloid]] generated by the equations:
+: $x = a \left({\theta - \sin \theta}\right)$
+: $y = a \left({1 - \cos \theta}\right)$
+Then the [[Definition:Tangent to Curve|tangent]] to $C$ at a point $\left({x, y}\right)$ on $C$ is given by the equation:
+:$y - a \left({1 - \cos \theta}\right) = \dfrac {\sin \theta} {1 - \cos \theta} \left({x - a \theta + a \sin \theta}\right)$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Derivative of Curve at Point]], the [[Definition:Tangent to Curve|tangent]] to $C$ at the point $\tuple {x, y}$ is the [[Definition:Derivative of Real Function at Point|derivative]] of its equation at that point.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = a \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = a \sin \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {\d y} {\d \theta} / \frac {\d x} {\d \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a \sin \theta} {a \paren {1 - \cos \theta} }
+      | c = 
+}}
+{{eqn | r = \frac {\sin \theta} {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac 1 {\map \tan {\theta / 2} }
+      | c = [[Half Angle Formulas/Tangent/Corollary 2|Half Angle Formula for Tangent: Corollary 2]]
+}}
+{{eqn | r = \cot \frac \theta 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider a [[Definition:Polygon|polygon]] $ABCD$ being rolled along a [[Definition:Straight Line|straight line]] in the same way as the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$.
+Let $A', B', C', D'$ be the [[Definition:Point|points]] around which the $ABCD$ rotates while rolling.
+:[[File:TangentToCycloid-construction.png|600px]]
+The point $A$ traces out in succession several [[Definition:Arc of Circle|arcs of circles]] with [[Definition:Center of Circle|centers]] $B', C', D'$.
+The [[Definition:Tangent to Curve|tangent]] to each of these [[Definition:Arc of Circle|arcs]] is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Straight Line|line]] joining the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]] to the corresponding [[Definition:Point|point]] of rotation.
+Consider a [[Definition:Circle|circle]] as being a [[Definition:Polygon|polygon]] with an [[Definition:Infinity|infinite]] number of [[Definition:Side of Polygon|sides]].
+It follows that in the limit the [[Definition:Tangent to Curve|tangent]] to $C$ at any given [[Definition:Point|point]] is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Straight Line|line]] joining the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]] to the corresponding [[Definition:Point|point]] where the [[Definition:Generating Circle of Cycloid|generating circle]] touches the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+:[[File:TangentToCycloid.png|600px]]
+Let $O$ be the [[Definition:Center of Circle|center]] of the [[Definition:Generating Circle of Cycloid|generating circle]].
+Let $P$ be the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]].
+Let $Q$ be the [[Definition:Point|point]] where the [[Definition:Generating Circle of Cycloid|generating circle]] touches the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+Let $S$ be the [[Definition:Point|point]] where the [[Definition:Tangent to Curve|tangent]] to $C$ meets the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+As $OP$ and $OQ$ are both [[Definition:Radius of Circle|radii]] of the [[Definition:Generating Circle of Cycloid|generating circle]]:
+:$OP = OQ = a$
+and so $\triangle POQ$ is [[Definition:Isosceles Triangle|isosceles]].
+Let $R$ be dropped [[Definition:Perpendicular|perpendicular]] to $PQ$ at $R$.
+As $\triangle POQ$ is [[Definition:Isosceles Triangle|isosceles]], $\triangle OPR = \triangle OQR$ and so $PR = RQ$.
+Thus $\angle POR = \angle QOR = \dfrac \theta 2$.
+As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is [[Definition:Similar Triangles|similar]] to $\triangle QPS$.
+Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$.
+By definition, $\tan \angle PSQ$ is the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] $\dfrac {\d y} {\d x}$ to $C$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = \tan \angle PSQ
+      | c = 
+}}
+{{eqn | r = \map \tan {\dfrac \pi 2 - \dfrac \theta 2}
+      | c = 
+}}
+{{eqn | r = \cot \frac \theta 2
+      | c = [[Tangent of Complement equals Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Slope of Tangent to Cycloid]], the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $C$ at the point $\left({x, y}\right)$ is given by:
+:$\dfrac {\mathrm d y} {\mathrm d x} = \cot \dfrac \theta 2$
+This [[Definition:Tangent to Curve|tangent]] to $C$ also passes through the point $\left({a \left({\theta - \sin \theta}\right),  a \left({1 - \cos \theta}\right)}\right)$.
+{{Finish|Find (or write) the result which gives the equation of a line from its slope and a point through which it passes.}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Hypocycloid}
+Tags: Hypocycloids
+
+\begin{theorem}
+Let a [[Definition:Circle|circle]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ roll without slipping around the inside of a [[Definition:Circle|circle]] $C_2$ of (larger) [[Definition:Radius of Circle|radius]] $a$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let $P$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of $C_1$.
+Let $C_1$ be initially positioned so that $P$ is its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located at [[Definition:Point|point]] $A = \tuple {a, 0}$ on the [[Definition:X-Axis|$x$-axis]].
+Let $H$ be the [[Definition:Hypocycloid|hypocycloid]] traced out by the [[Definition:Point|point]] $P$.
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the equations:
+:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
+:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
+\end{theorem}
+
+\begin{proof}
+:[[File:Hypocycloid.png|500px]]
+Let $C_1$ have rolled so that the [[Definition:Straight Line|line]] $OB$ through the [[Definition:Radius of Circle|radii]] of $C_1$ and $C_2$ is at [[Definition:Angle|angle]] $\theta$ to the [[Definition:X-Axis|$x$-axis]].
+Let $C_1$ have turned through an angle $\phi$ to reach that point.
+By definition of [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]], $P = \tuple {x, y}$ is defined by:
+:$x = \paren {a - b} \cos \theta + b \map \cos {\phi - \theta}$
+:$y = \paren {a - b} \sin \theta - b \map \sin {\phi - \theta}$
+The [[Definition:Arc of Circle|arc]] of $C_1$ between $P$ and $B$ is the same as the [[Definition:Arc of Circle|arc]] of $C_2$ between $A$ and $B$.
+Thus by [[Arc Length of Sector]]:
+:$ a \theta = b \phi$
+Thus:
+:$\phi - \theta = \paren {\dfrac {a - b} b} \theta$
+whence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet in Set of Ideals}
+Tags: Join and Meet Semilattices, Join and Meet
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let ${\it Ids}\left({\mathscr S}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+Let $I_1, I_2$ be [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+Then
+:$I_1 \wedge I_2 = \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$
+where
+:$I_1 \wedge I_2$ denotes the [[Definition:Meet (Order Theory)|meet]] in $\left({ {\it Ids}\left({\mathscr S}\right), \subseteq}\right)$
+\end{theorem}
+
+\begin{proof}
+By [[Meet is Intersection in Set of Ideals]]:
+:$I_1 \wedge I_2 = I_1 \cap I_2$
+=== ($\subseteq$) ===
+Let $x \in I_1 \cap I_2$.
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in I_1 \land x \in I_2$
+By [[Meet is Idempotent]]:
+:$x = x \wedge x$
+Thus
+:$x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$
+{{qed|lemma}}
+=== ($\supseteq$) ===
+Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}\right\}$.
+Then
+:$\exists i_1 \in I_1, i_2 \in I_2: x = i_1 \wedge i_2$
+By [[Meet Precedes Operands]]:
+:$x \preceq i_1 \land x \preceq i_2$
+By definition of [[Definition:Lower Set|lower set]]:
+:$x \in I_1 \land x \in I_2$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$x \in I_1 \cap I_2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Semilattice Ideals is Ideal}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $I_1, I_2$ be [[Definition:Ideal in Ordered Set|ideals]] in $\struct {S, \preceq}$.
+Then $I_1 \cap I_2$ is [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$
+\end{theorem}
+
+\begin{proof}
+=== Directed Subset ===
+Let $x, y \in I_1 \cap I_2$.
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x, y \in I_1$ and $x, y \in I_2$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists z_1 \in I_1: x \preceq z_1 \land y \preceq z_1$
+and
+:$\exists z_2 \in I_2: x \preceq z_2 \land y \preceq z_2$
+By [[Meet Precedes Operands]]:
+:$z_1 \wedge z_2 \preceq z_1$ and $z_1 \wedge z_2 \preceq z_2$
+By definition of [[Definition:Lower Set|lower set]]:
+:$z_1 \wedge z_2 \in I_1$ and $z_1 \wedge z_2 \in I_2$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$z_1 \wedge z_2 \in I_1 \cap I_2$
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$z_1 \wedge z_2 = \inf \set {z_1, z_2}$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$x \preceq z_1 \wedge z_2$ and $y \preceq z_1 \wedge z_2$
+Thus by definition:
+:$I_1 \cap I_2$ is [[Definition:Directed Subset|directed]].
+{{qed|lemma}}
+=== Lower Set ===
+Let $x, y \in S$ such that:
+:$x \in I_1 \cap I_2$ and $y \preceq x$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in I_1$ and $x \in I_2$
+By definition of [[Definition:Lower Set|lower set]]:
+:$y \in I_1$ and $y \in I_2$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$y \in I_1 \cap I_2$
+Thus by definition:
+:$I_1 \cap I_2$ is a [[Definition:Lower Set|lower set]].
+{{qed|lemma}}
+=== Non-Empty Set ===
+By assumption:
+:$I_1 \ne \O$ and $I_2 \ne \O$
+By definition of [[Definition:Non-Empty Set|non-empty]]:
+:$\exists x: x \in I_1$ and $\exists y: y \in I_2$
+By [[Meet Precedes Operands]]:
+:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
+By definition of [[Definition:Lower Set|lower set]]:
+:$x \wedge y \in I_1$ and $x \wedge y \in I_2$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \wedge y \in I_1 \cap I_2$
+Thus by definition:
+:$I_1 \cap I_2$ is [[Definition:Non-Empty Set|non-empty]].
+{{qed|lemma}}
+Thus by definition:
+:$I_1 \cap I_2$ is [[Definition:Ideal in Ordered Set|ideal]] in $\struct {S, \preceq}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii}
+Tags: Hypocycloids
+
+\begin{theorem}
+Let $H$ be a [[Definition:Hypocycloid|hypocycloid]] $H$ generated by a [[Definition:Circle|circle]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ rolling within a [[Definition:Circle|circle]] $C_2$ of (larger) [[Definition:Radius of Circle|radius]] $a$.
+Let $a = n b$ where $n$ is an [[Definition:Integer|integer]].
+Then $H$ has $n$ [[Definition:Cusp of Hypocycloid|cusps]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Arc Length|length]] of the [[Definition:Arc of Circle|arc]] of $C_2$ between two adjacent [[Definition:Cusp of Hypocycloid|cusps]] of $H$ is $2 \pi b$.
+The total length of the [[Definition:Circumference of Circle|circumference]] of $C_1$ is $2 \pi a$.
+Thus the total number of [[Definition:Cusp of Hypocycloid|cusps]] of $H$ is:
+:$\dfrac {2 \pi a} {2 \pi b} = \dfrac {2 \pi n b} {2 \pi b} = n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Cusps of Hypocycloid from Rational Ratio of Circle Radii}
+Tags: Hypocycloids
+
+\begin{theorem}
+Consider the [[Definition:Hypocycloid|hypocycloid]] $H$ generated by a [[Definition:Circle|circle]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ rolling within a [[Definition:Circle|circle]] $C_2$ of (larger) [[Definition:Radius of Circle|radius]] $a$.
+Let $k = \dfrac a b$ be a [[Definition:Rational Number|rational number]].
+Let $k$ be expressed in [[Definition:Canonical Form of Rational Number|canonical form]]:
+:$k = \dfrac p q$
+where $p$ and $q$ are [[Definition:Coprime Integers|coprime]].
+Then $H$ has $p$ [[Definition:Cusp of Hypocycloid|cusps]].
+\end{theorem}<|endoftext|>
+\section{Cusps of Hypocycloid from Irrational Ratio of Circle Radii}
+Tags: Hypocycloids
+
+\begin{theorem}
+Consider the [[Definition:Hypocycloid|hypocycloid]] $H$ generated by a [[Definition:Rotor of Hypocycloid|rotor]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ rolling within a [[Definition:Stator of Hypocycloid|stator]] $C_2$ of (larger) [[Definition:Radius of Circle|radius]] $a$.
+Let $k = \dfrac a b$ be an [[Definition:Irrational Number|irrational number]].
+Then $H$ has an [[Definition:Infinite Set|infinite number]] of [[Definition:Cusp of Hypocycloid|cusps]], which are evenly and densely distributed around the [[Definition:Circumference of Circle|circumference]] of $C_2$.
+\end{theorem}<|endoftext|>
+\section{Equation of Astroid}
+Tags: Astroids
+
+\begin{theorem}
+Let $H$ be the [[Definition:Astroid|astroid]] [[Definition:Generator of Hypocycloid|generated]] by the [[Definition:Rotor of Hypocycloid|rotor]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ rolling without slipping around the inside of a [[Definition:Stator of Hypocycloid|stator]] $C_2$ of [[Definition:Radius of Circle|radius]] $a = 4 b$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let $P$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of $C_1$.
+Let $C_1$ be initially positioned so that $P$ is its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located at [[Definition:Point|point]] $A = \tuple {a, 0}$ on the [[Definition:X-Axis|$x$-axis]].
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+=== [[Equation of Astroid/Parametric Form|Parametric Form]] ===
+{{:Equation of Astroid/Parametric Form}}
+=== [[Equation of Astroid/Cartesian Form|Cartesian Form]] ===
+{{:Equation of Astroid/Cartesian Form}}
+\end{theorem}<|endoftext|>
+\section{Tangent to Astroid between Coordinate Axes has Constant Length}
+Tags: Hypocycloids
+
+\begin{theorem}
+Let $C_1$ be a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $b$ roll without slipping around the inside of a [[Definition:Circle|circle]] $C_2$ of [[Definition:Radius of Circle|radius]] $a = 4 b$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let $P$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of $C_1$.
+Let $C_1$ be initially positioned so that $P$ is its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located at [[Definition:Point|point]] $A = \tuple {a, 0}$ on the [[Definition:X-Axis|$x$-axis]].
+Let $H$ be the [[Definition:Astroid|astroid]] formed by the [[Definition:Locus|locus]] of $P$.
+The [[Definition:Line Segment|segment]] of the [[Definition:Tangent to Curve|tangent]] to $H$ between the [[Definition:X-Axis|$x$-axis]] and the [[Definition:Y-Axis|$y$-axis]] is [[Definition:Constant|constant]], immaterial of the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]].
+\end{theorem}
+
+\begin{proof}
+:[[File:AstroidTangent.png|400px]]
+From [[Equation of Astroid]], $H$ can be expressed as:
+:$\begin{cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+Thus the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $H$ at $\tuple {x, y}$ is:
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = \frac {3 a \sin^2 \theta \cos \theta \rd \theta} {-3 a \cos^2 \theta \sin \theta \rd \theta}
+      | c = 
+}}
+{{eqn | r = -\tan \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus the equation of the [[Definition:Tangent to Curve|tangent]] to $H$ is given by:
+:$y - a \sin^3 \theta = -\tan \theta \paren {x - a \cos^3 \theta}$
+{{explain|Find, or post up, the equation of a line of given tangent passing through point $\tuple {x, y}$ as this is what is needed here}}
+The [[Definition:X-Intercept|$x$-intercept]] is found by setting $y = 0$ and solving for $x$:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \cos^3 \theta + a \sin^2 \theta \cos \theta
+      | c = 
+}}
+{{eqn | r = a \cos \theta \paren {\cos^2 \theta + \sin^2 \theta}
+      | c = 
+}}
+{{eqn | r = a \cos \theta
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Similarly, the [[Definition:Y-Intercept|$y$-intercept]] is found by setting $x = 0$ and solving for $y$, which gives:
+:$y = a \sin \theta$
+The [[Definition:Length (Linear Measure)|length]] of the part of the [[Definition:Tangent to Curve|tangent]] to $H$ between the [[Definition:X-Axis|$x$-axis]] and the [[Definition:Y-Axis|$y$-axis]] is given by:
+{{begin-eqn}}
+{{eqn | l = \sqrt {a^2 \cos^2 \theta + a^2 \sin^2 \theta}
+      | r = a \sqrt {\cos^2 \theta + \sin^2 \theta}
+      | c = 
+}}
+{{eqn | r = a
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+which is [[Definition:Constant|constant]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pendulum Contained by Cycloid moves along Cycloidal Path}
+Tags: Cycloids, Pendulums
+
+\begin{theorem}
+Let a [[Definition:Pendulum|pendulum]] with a [[Definition:Flexible Chain|flexible]] [[Definition:Rod|rod]] be suspended from a point $P$.
+Let the [[Definition:Rod|rod]] be contained by a pair of bodies shaped as the [[Definition:Arc of Cycloid|arcs]] of a [[Definition:Cycloid|cycloid]] such that $P$ is the [[Definition:Cusp of Cycloid|cusp]] between those two [[Definition:Arc of Cycloid|arcs]].
+:[[File:CycloidPendulum.png|400px]]
+Then the [[Definition:Pendulum Bob|bob]] is constrained to move such that its path traces the [[Definition:Arc of Cycloid|arc]] of a [[Definition:Cycloid|cycloid]].
+\end{theorem}
+
+\begin{proof}
+From [[Evolute of Cycloid is Cycloid]], the [[Definition:Evolute|evolute]] of a [[Definition:Cycloid|cycloid]] is another [[Definition:Cycloid|cycloid]].
+From [[Curve is Involute of Evolute]], the [[Definition:Involute|involute]] of a [[Definition:Cycloid|cycloid]] is another [[Definition:Cycloid|cycloid]] as well.
+But by the definition of [[Definition:Involute|involute]], the path defined by the [[Definition:Pendulum|pendulum]] as described is the [[Definition:Involute|involute]] of the [[Definition:Cycloid|cycloid]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure of Meet of Lower Closures}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $D_1, D_2$ be [[Definition:Subset|subsets]] of $S$.
+Then
+:$\left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq = \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$
+where
+:$D_1^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $D_1$.
+\end{theorem}
+
+\begin{proof}
+By [[Lower Closure is Closure Operator]]:
+:$D_1 \subseteq D_1^\preceq$ and $D_2 \subseteq D_2^\preceq$
+Then
+:$\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\} \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}$
+By [[Lower Closure is Closure Operator]]:
+:$\left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$
+By definition of [[Definition:Set Equality|set equality]] it remains to prove that
+:$\left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq \subseteq \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$
+Let $x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\}^\preceq$.
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists y \in \left\{ {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}\right\} \land x \preceq y$
+Then:
+:$\exists i_1 \in D_1^\preceq, i_2 \in D_2^\preceq: y = i_1 \wedge i_2$
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists d_1 \in D_1: i_1 \preceq d_1$
+and
+:$\exists d_2 \in D_2: i_2 \preceq d_2$
+By [[Meet Semilattice is Ordered Structure]]:
+:$y \preceq d_1 \wedge d_2$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq d_1 \wedge d_2$
+:$d_1 \wedge d_2 \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}$
+Thus by definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$x \in \left\{ {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}\right\}^\preceq$
+Thus the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Cycloid in Cartesian Coordinates}
+Tags: Cycloids
+
+\begin{theorem}
+Consider a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $a$ rolling without slipping along the [[Definition:X-Axis|$x$-axis]] of a [[Definition:Cartesian Plane|cartesian plane]].
+Consider the [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]] which is at the [[Definition:Origin|origin]] when its [[Definition:Center of Circle|center]] is on the [[Definition:Y-Axis|y-axis]].
+Consider the [[Definition:Cycloid|cycloid]] traced out by the [[Definition:Point|point]] $P$.
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the equation:
+:$a \sin^{-1} \paren {\dfrac {\sqrt {2 a y - y^2} } a} = \sqrt {2 a y - y^2} + x$
+\end{theorem}
+
+\begin{proof}
+From [[Equation of Cycloid]], the [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the equations:
+:$x = a \paren {\theta - \sin \theta}$
+:$y = a \paren {1 - \cos \theta}$
+Expressing $\theta$ and $\sin \theta$ in terms of $y$:
+{{begin-eqn}}
+{{eqn | l = \cos \theta
+      | r = 1 - \frac y a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sin \theta
+      | r = \sqrt {1 - \paren {1 - \frac y a}^2}
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac {2 y} a - \frac {y^2} {a^2} }
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt {2 a y - y^2} } a 
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \theta
+      | r = \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}
+      | c = 
+}}
+{{end-eqn}}
+Substituting for $\theta$ and $\sin \theta$ in the expression for $x$:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \paren {\sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a} - \frac 1 a \sqrt {2 a y - y^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \sin^{-1} \paren {\frac {\sqrt {2 a y - y^2} } a}
+      | r = \sqrt {2 a y - y^2} + x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Derivative of Locus of Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Consider a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $a$ rolling without slipping along the [[Definition:X-Axis|x-axis]] of a [[Definition:Cartesian Plane|cartesian plane]].
+Consider the [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]] which is at the [[Definition:Origin|origin]] when its [[Definition:Center of Circle|center]] is on the [[Definition:Y-Axis|y-axis]].
+Consider the [[Definition:Cycloid|cycloid]] traced out by the [[Definition:Point|point]] $P$.
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The [[Definition:Second Derivative|second derivative]] of the [[Definition:Locus|locus]] of $P$ is given by:
+:$y'' = -\dfrac a {y^2}$
+\end{theorem}
+
+\begin{proof}
+From [[Equation of Cycloid]]:
+:$x = a \paren {\theta - \sin \theta}$
+:$y = a \paren {1 - \cos \theta}$
+From [[Slope of Tangent to Cycloid]]:
+{{begin-eqn}}
+{{eqn | l = y'
+      | r = \cot \dfrac \theta 2
+      | c = [[Slope of Tangent to Cycloid]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y'} {\d x}
+      | r = \dfrac {\d} {\d \theta} \cot \dfrac \theta 2 \frac {\d \theta} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = -\dfrac 1 2 \csc^2 \dfrac \theta 2 / \dfrac {\d x} {\d \theta}
+      | c = [[Derivative of Cotangent Function]]
+}}
+{{eqn | r = -\dfrac 1 2 \csc^2 \dfrac \theta 2 \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | r = -\dfrac 1 {2 \sin^2 \dfrac \theta 2} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }
+      | c = {{Defof|Cosecant}}
+}}
+{{eqn | r = -\dfrac 1 {1 - \cos \theta} \paren {\dfrac 1 {a \paren {1 - \cos \theta} } }
+      | c = [[Double Angle Formulas/Cosine/Corollary 2|Double Angle Formulas for Cosine]]
+}}
+{{eqn | r = -\dfrac a {y^2}
+      | c = from $y = a \paren {1 - \cos \theta}$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Arc of Cycloid is Concave}
+Tags: Cycloids
+
+\begin{theorem}
+Consider a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $a$ rolling without slipping along the [[Definition:X-Axis|x-axis]] of a [[Definition:Cartesian Plane|cartesian plane]].
+Consider the [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]] which is at the [[Definition:Origin|origin]] when its [[Definition:Center of Circle|center]] is on the [[Definition:Y-Axis|y-axis]].
+Consider the [[Definition:Cycloid|cycloid]] traced out by the [[Definition:Point|point]] $P$.
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+Then the [[Definition:Locus|locus]] of $P$ is [[Definition:Concave Real Function|concave]].
+\end{theorem}
+
+\begin{proof}
+From [[Second Derivative of Locus of Cycloid]]:
+:$y'' = -\dfrac a {y^2}$
+As $y \ge 0$ throughout, then $y'' < 0$ wherever $y \ne 0$, which is at the [[Definition:Cusp of Cycloid|cusps]].
+The result follows from [[Second Derivative of Concave Real Function is Non-Positive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal to Cycloid passes through Bottom of Generating Circle}
+Tags: Cycloids, Normal to Cycloid passes through Bottom of Generating Circle
+
+\begin{theorem}
+Let $C$ be a [[Definition:Cycloid|cycloid]] generated by the equations:
+: $x = a \left({\theta - \sin \theta}\right)$
+: $y = a \left({1 - \cos \theta}\right)$
+Then the [[Definition:Normal to Curve|normal]] to $C$ at a point $P$ on $C$ passes through the bottom of the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$.
+\end{theorem}
+
+\begin{proof}
+From [[Normal to Cycloid]], the equation for the [[Definition:Normal to Curve|normal]] to $C$ at a point $P = \left({x, y}\right)$ is given by:
+{{ProofWanted|We don't even have the definition of a normal to a curve posted up yet, so there's a long way to go here.}}
+\end{proof}
+
+\begin{proof}
+From [[Tangent to Cycloid passes through Top of Generating Circle]], the [[Definition:Tangent to Curve|tangent]] to $C$ at a point $P = \left({x, y}\right)$ passes through the top of the [[Definition:Generating Circle of Cycloid|generating circle]].
+By definition, the [[Definition:Normal to Curve|normal]] to $C$ at $P$ is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$.
+From [[Thales' Theorem]], the [[Definition:Normal to Curve|normal]], the [[Definition:Tangent to Curve|tangent]] and the [[Definition:Diameter of Circle|diameter]] of the [[Definition:Generating Circle of Cycloid|generating circle]] form a [[Definition:Right Triangle|right triangle]].
+Thus the [[Definition:Normal to Curve|normal]] to $C$ at $P$ meets the [[Definition:Generating Circle of Cycloid|generating circle]] at the opposite end of the [[Definition:Diameter of Circle|diameter]] to the [[Definition:Tangent to Curve|tangent]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rate of Change of Cartesian Coordinates of Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Let a [[Definition:Circle|circle]] $C$ of [[Definition:Radius of Circle|radius]] $a$ roll without slipping along the [[Definition:X-Axis|x-axis]] of a [[Definition:Cartesian Plane|cartesian plane]] at a constant [[Definition:Speed|speed]] such that the [[Definition:Center of Circle|center]] moves with a [[Definition:Velocity|velocity]] $\mathbf v_0$ in the direction of increasing $x$.
+Consider a [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]].
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+Then the [[Definition:Rate of Change|rate of change]] of $x$ and $y$ can be expresssed as:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d t}
+      | r = \mathbf v_0 \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d t}
+      | r = \mathbf v_0 \sin \theta
+      | c = 
+}}
+{{end-eqn}}
+where $\theta$ is the [[Definition:Angle|angle]] turned by $C$ after time $t$.
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Center of Circle|center]] of $C$ be $O$.
+{{WLOG}}, let $P$ be at the [[Definition:Origin|origin]] at time $t = t_0$.
+By definition, $P$ traces out a [[Definition:Cycloid|cycloid]].
+From [[Equation of Cycloid]], $P = \tuple {x, y}$ is described by:
+:$(1): \quad \begin{cases} x & = a \paren {\theta - \sin \theta} \\
+y & = a \paren {1 - \cos \theta} \end{cases}$
+Let $\tuple {x_c, y_c}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $O$ at time $t$.
+We have that:
+:$y_c$ is [[Definition:Constant|constant]]: $y_c = a$
+From [[Body under Constant Acceleration/Velocity after Time|Body under Constant Acceleration: Velocity after Time]]
+:$x_c = \mathbf v_0 t$
+as the [[Definition:Acceleration|acceleration]] of $O$ is zero.
+But $x_c$ is equal to the [[Definition:Arc Length|length]] of the [[Definition:Arc of Circle|arc]] of $C$ that has rolled along the [[Definition:X-Axis|$x$-axis]].
+By [[Arc Length of Sector]]:
+:$x_c = a \theta$
+and so:
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = \theta
+      | r = \frac {\mathbf v_0 t} a
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = \frac {\d \theta} {\d t}
+      | r = \frac {\mathbf v_0} a
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \paren {\frac {\mathbf v_0 t} a - \sin \theta}
+      | c = substituting for $\theta$ from $(2)$
+}}
+{{eqn | l = x
+      | r = \mathbf v_0 t - a \sin \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d t}
+      | r = \mathbf v_0 - a \cos \theta \frac {\d \theta} {\d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \mathbf v_0 - a \cos \theta \frac {\mathbf v_0} a
+      | c = from $(3)$
+}}
+{{eqn | r = \mathbf v_0 \paren {1 - \cos \theta}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = a \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d t}
+      | r = a \sin \theta \frac {\d \theta} {\d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = a \sin \theta \frac {\mathbf v_0} a
+      | c = from $(3)$
+}}
+{{eqn | r = \mathbf v_0 \sin \theta
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximum Rate of Change of X Coordinate of Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Let a [[Definition:Circle|circle]] $C$ of [[Definition:Radius of Circle|radius]] $a$ roll without slipping along the [[Definition:X-Axis|x-axis]] of a [[Definition:Cartesian Plane|cartesian plane]] at a constant [[Definition:Speed|speed]] such that the [[Definition:Center of Circle|center]] moves with a [[Definition:Velocity|velocity]] $\mathbf v_0$ in the direction of increasing $x$.
+Consider a [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]].
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The maximum [[Definition:Rate of Change|rate of change]] of $x$ is $2 \mathbf v_0$, which happens when $P$ is at the top of the [[Definition:Circle|circle]] $C$.
+\end{theorem}
+
+\begin{proof}
+From [[Rate of Change of Cartesian Coordinates of Cycloid]], the [[Definition:Rate of Change|rate of change]] of $x$ is given by:
+:$\dfrac {\d x} {\d t} = \mathbf v_0 \paren {1 - \cos \theta}$
+This is a maximum when $1 - \cos \theta$ is a maximum.
+That happens when $\cos \theta$ is at a minimum.
+That happens when $\cos \theta = -1$.
+That happens when $\theta = \pi, 3 \pi, \ldots$
+That is, when $\theta = \paren {2 n + 1} \pi$ where $n \in \Z$.
+That is, when $P$ is at the top of the [[Definition:Circle|circle]] $C$.
+When $\cos \theta = -1$ we have:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d t}
+      | r = \mathbf v_0 \paren {1 - \paren {-1} }
+      | c = 
+}}
+{{eqn | r = 2 \mathbf v_0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Maximum Rate of Change of Y Coordinate of Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Let a [[Definition:Circle|circle]] $C$ of [[Definition:Radius of Circle|radius]] $a$ roll without slipping along the [[Definition:X-Axis|x-axis]] of a [[Definition:Cartesian Plane|cartesian plane]] at a constant [[Definition:Speed|speed]] such that the [[Definition:Center of Circle|center]] moves with a [[Definition:Velocity|velocity]] $\mathbf v_0$ in the direction of increasing $x$.
+Consider a [[Definition:Point|point]] $P$ on the [[Definition:Circumference of Circle|circumference]] of this [[Definition:Circle|circle]].
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The maximum [[Definition:Rate of Change|rate of change]] of $y$ is $\mathbf v_0$, which happens when $\theta = \dfrac \pi 2 + 2 n \pi$ where $n \in \Z$.
+\end{theorem}
+
+\begin{proof}
+From [[Rate of Change of Cartesian Coordinates of Cycloid]], the [[Definition:Rate of Change|rate of change]] of $y$ is given by:
+:$\dfrac {\d y} {\d t} = \mathbf v_0 \sin \theta$.
+This is a maximum when $\sin \theta$ is a maximum.
+That happens when $\sin \theta = 1$.
+That happens when $\theta = \dfrac \pi 2 + 2 n \pi$ where $n \in \Z$.
+When $\sin \theta = 1$ we have:
+:$\dfrac {\d y} {\d t} = \mathbf v_0$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure is Closure Operator}
+Tags: Lower Closures, Closure Operators
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Then
+:[[Definition:Lower Closure/Set|lower closure of set]] is a [[Definition:Closure Operator|closure operator]].
+\end{theorem}
+
+\begin{proof}
+=== [[Definition:Inflationary Mapping|Inflationary]] ===
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Let $x \in X$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$
+Thus by definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$x \in X^\preceq$
+Thus by definition of [[Definition:Subset|subset]]:
+:$X \subseteq X^\preceq$
+{{qed|lemma}}
+=== [[Definition:Increasing Mapping|Increasing]] ===
+Let $X, Y$ be [[Definition:Subset|subsets]] of $S$ such that
+:$X \subseteq Y$
+Let $x \in X^\preceq$.
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists y \in X: x \preceq y$
+By definition of [[Definition:Subset|subset]]:
+:$y \in Y$
+Thus by definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$x \in Y^\preceq$
+Thus by definition of [[Definition:Subset|subset]]:
+:$X^\preceq \subseteq Y^\preceq$
+{{qed|lemma}}
+=== [[Definition:Idempotent Mapping|Idempotent]] ===
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+Let $x \in \left({X^\preceq}\right)^\preceq$.
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists y \in X^\preceq: x \preceq y$
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists z \in X: y \preceq z$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq z$
+Thus by definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$x \in X^\preceq$
+Thus by definition of [[Definition:Subset|subset]]:
+:$\left({X^\preceq}\right)^\preceq \subseteq X^\preceq$
+By definition of [[Definition:Inflationary Mapping|inflationary]]:
+:$X \subseteq X^\preceq$
+By definition of [[Definition:Increasing Mapping|increasing]]:
+:$X^\preceq \subseteq \left({X^\preceq}\right)^\preceq$
+Thus definition of [[Definition:Set Equality|set equality]];
+:$\left({X^\preceq}\right)^\preceq = X^\preceq$
+{{qed|lemma}}
+Thus by definition:
+:[[Definition:Lower Closure/Set|lower closure of set]] is a [[Definition:Closure Operator|closure operator]].
+{{qed}}
+[[Category:Lower Closures]]
+[[Category:Closure Operators]]
+qovjrkspfmlam6i4h6nxxkjwqgeavjr
+\end{proof}<|endoftext|>
+\section{Area inside Astroid}
+Tags: Astroids
+
+\begin{theorem}
+The [[Definition:Area|area]] inside an [[Definition:Astroid|astroid]] $H$ constructed within a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $a$ is given by:
+:$\AA = \dfrac {3 \pi a^2} 8$
+\end{theorem}
+
+\begin{proof}
+Let $H$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its center at the [[Definition:Origin|origin]] and its [[Definition:Cusp of Hypocycloid|cusps]] positioned on the [[Definition:Coordinate Axis|axes]].
+:[[File:AstroidArea.png|400px]]
+By [[Definition:Symmetry|symmetry]], it is sufficient to evaluate the [[Definition:Area|area]] shaded yellow and to multiply it by $4$.
+By [[Equation of Astroid]]:
+:$\begin{cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \AA
+      | r = 4 \int_0^a y \rd x
+      | c = 
+}}
+{{eqn | r = 4 \int_{x \mathop = 0}^{x \mathop = a} y \frac {\d x} {\d \theta} \rd \theta
+      | c = 
+}}
+{{eqn | r = 4 \int_{x \mathop = 0}^{x \mathop = a} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta
+      | c = [[Definition:Differentiation|differentiating]] $a \cos^3 \theta$ {{WRT|Differentiation}} $\theta$
+}}
+{{eqn | r = 4 \int_{\theta \mathop = \pi / 2}^{\theta \mathop = 0} a \sin^3 \theta \, 3 a \cos^2 \theta \sin \theta \paren {-\sin \theta} \rd \theta
+      | c = $x = 0$ when $\theta = \pi / 2$, $x = a$ when $\theta = 0$
+}}
+{{eqn | r = 12 a^2 \int_0^{\pi / 2} \sin^4 \theta \cos^2 \theta \rd \theta
+      | c = simplifying
+}}
+{{end-eqn}}
+Simplifying the [[Definition:Integrand|integrand]]:
+{{begin-eqn}}
+{{eqn | l = \sin^4 \theta \cos^2 \theta
+      | r = \frac {\paren {2 \sin \theta \cos \theta}^2} 4 \frac {2 \sin^2 \theta} 2
+      | c = 
+}}
+{{eqn | r = \frac {\sin^2 2 \theta} 4 \frac {2 \sin^2 \theta} 2
+      | c = [[Double Angle Formula for Sine]]
+}}
+{{eqn | r = \frac {\sin^2 2 \theta} 4 \frac {1 - \cos 2 \theta} 2
+      | c = [[Square of Sine]]
+}}
+{{eqn | r = \frac {\sin^2 2 \theta - \sin^2 2 \theta \cos 2 \theta} 8
+      | c = 
+}}
+{{eqn | r = \frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8
+      | c = [[Square of Sine]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \AA
+      | r = 12 a^2 \int_0^{\pi / 2} \paren {\frac {1 - \cos 4 \theta} {16} - \frac {\sin^2 2 \theta \cos 2 \theta} 8} \rd \theta
+      | c = 
+}}
+{{eqn | r = \frac 3 4 a^2 \int_0^{\pi / 2} \paren {1 - \cos 4 \theta} \rd \theta - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta
+      | c = [[Linear Combination of Integrals]]
+}}
+{{eqn | r = \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \int_0^{\pi / 2} \sin^2 2 \theta \cos 2 \theta \rd \theta
+      | c = [[Primitive of Cosine of a x|Primitive of $\cos a x$]]
+}}
+{{eqn | r = \frac 3 4 a^2 \intlimits {\theta - \frac {\sin 4 \theta} 4} 0 {\pi / 2} - \frac 3 2 a^2 \intlimits {\frac {\sin^3 2 \theta} 6} 0 {\pi / 2}
+      | c = [[Primitive of Power of Sine of a x by Cosine of a x|Primitive of $\sin^n a x \cos a x$]]
+}}
+{{eqn | r = \frac {3 \pi a^2} 8 - \frac {3 a^2} {16} \sin 2 \pi  - \frac {3 a^2} {12} \sin^3 \pi
+      | c = evaluating limits of integration
+}}
+{{eqn | r = \frac {3 \pi a^2} 8
+      | c = $\sin 2 \pi = 0$ and $\sin^3 \pi = 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Arc of Astroid}
+Tags: Astroids
+
+\begin{theorem}
+The total [[Definition:Arc Length|length of the arcs]] of an [[Definition:Astroid|astroid]] constructed within a [[Definition:Stator of Hypocycloid|stator]] of [[Definition:Radius of Circle|radius]] $a$ is given by:
+:$\LL = 6 a$
+\end{theorem}
+
+\begin{proof}
+Let $H$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its center at the [[Definition:Origin|origin]] and its [[Definition:Cusp of Hypocycloid|cusps]] positioned on the [[Definition:Coordinate Axis|axes]].
+:[[File:Astroid.png|400px]]
+We have that $\LL$ is $4$ times the [[Definition:Arc Length|length]] of one [[Definition:Arc of Hypocycloid|arc]] of the [[Definition:Astroid|astroid]].
+From [[Arc Length for Parametric Equations]]:
+:$\ds \LL = 4 \int_{\theta \mathop = 0}^{\theta \mathop = \pi/2} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
+where, from [[Equation of Astroid]]:
+:$\begin{cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = -3 a \cos^2 \theta \sin \theta
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = 3 a \sin^2 \theta \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}
+      | r = \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }
+      | c = 
+}}
+{{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }
+      | c = 
+}}
+{{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 3 a \sin \theta \cos \theta
+      | c = 
+}}
+{{eqn | r = \frac {3 a \sin 2 \theta} 2
+      | c = [[Double Angle Formula for Sine]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \LL
+      | r = 4 \frac {3 a} 2 \int_0^{\pi / 2} \sin 2 \theta \rd \theta
+      | c = 
+}}
+{{eqn | r = 6 a \intlimits {\frac {-\cos 2 \theta} 2} 0 {\pi / 2}
+      | c = [[Primitive of Sine of a x|Primitive of $\sin a x$]]
+}}
+{{eqn | r = 6 a \paren {-\frac {\cos \pi} 2 + \frac {\cos 0} 2}
+      | c = evaluating limits of integration
+}}
+{{eqn | r = 6 a \frac {-\paren {-1} + 1} 2
+      | c = 
+}}
+{{eqn | r = 6 a
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Area of Surface of Revolution from Astroid}
+Tags: Astroids, Surfaces of Revolution
+
+\begin{theorem}
+Let $H$ be the [[Definition:Astroid|astroid]] constructed within a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $a$.
+The [[Definition:Surface of Revolution|surface of revolution]] formed by rotating $H$ around the [[Definition:X-Axis|$x$-axis]]:
+:[[File:AstroidSurfaceOfRevolution.png|400px]]
+evaluates to:
+:$\mathcal S = \dfrac {12 \pi a^2} 5$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Symmetry|symmetry]], it is sufficient to calculate the [[Definition:Surface of Revolution|surface of revolution]] of $H$ for $0 \le x \le a$.
+From [[Area of Surface of Revolution]], this [[Definition:Surface of Revolution|surface of revolution]] is given by:
+:$\displaystyle \mathcal S = 2 \int_0^{\pi / 2} 2 \pi y \, \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
+From [[Equation of Astroid]]:
+:$\begin{cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+so:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = -3 a \cos^2 \theta \sin \theta
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = 3 a \sin^2 \theta \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2}
+      | r = \sqrt {9 a^2 \paren {\sin^4 \theta \cos^2 \theta + \cos^4 \theta \sin^2 \theta} }
+      | c = 
+}}
+{{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta \paren {\sin^2 \theta + \cos^2 \theta} }
+      | c = 
+}}
+{{eqn | r = 3 a \sqrt {\sin^2 \theta \cos^2 \theta}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 3 a \sin \theta \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \mathcal S
+      | r = 2 \int_0^{\pi / 2} 2 \pi a \sin^3 \theta \, 3 a \sin \theta \cos \theta \rd \theta
+      | c = 
+}}
+{{eqn | r = 12 \pi a^2 \int_0^{\pi / 2} \sin^4 \theta \cos \theta \rd \theta
+      | c = 
+}}
+{{eqn | r = 12 \pi a^2 \intlimits {\frac {\sin^5 \theta} 5} 0 {\pi / 2}
+      | c = [[Primitive of Power of Sine of a x by Cosine of a x|Primitive of $\sin^n a x \cos a x$]]
+}}
+{{eqn | r = \frac {12 \pi a^2} 5 \paren {\sin^5 \theta \frac \pi 2 - \sin^5 \theta 0}
+      | c = evaluating limits of integration
+}}
+{{eqn | r = \frac {12 \pi a^2} 5 \paren {1 - 0}
+      | c = 
+}}
+{{eqn | r = \frac {12 \pi a^2} 5
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tusi Couple is Diameter of Stator}
+Tags: Tusi Couples
+
+\begin{theorem}
+A [[Definition:Tusi Couple|Tusi couple]] is a [[Definition:Degenerate Case|degenerate case]] of the [[Definition:Hypocycloid|hypocycloid]] whose form is a [[Definition:Straight Line|straight line]] that forms a [[Definition:Diameter of Circle|diameter]] of the [[Definition:Stator of Hypocycloid|stator]].
+\end{theorem}
+
+\begin{proof}
+Let $C_1$ be a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $b$ rolling without slipping around the inside of a [[Definition:Circle|circle]] $C_2$ of [[Definition:Radius of Circle|radius]] $a$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let $P$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of $C_1$.
+Let $C_1$ be initially positioned so that $P$ is its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located at [[Definition:Point|point]] $A = \tuple {a, 0}$ on the [[Definition:X-Axis|$x$-axis]].
+Let $H$ be the [[Definition:Hypocycloid|hypocycloid]] formed by the [[Definition:Locus|locus]] of $P$.
+From [[Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii]] we have that $H$ will have $2$ [[Definition:Cusp of Hypocycloid|cusps]] {{iff}}:
+:$a = 2 b$
+By [[Equation of Hypocycloid]] a [[Definition:Hypocycloid|hypocycloid]] can be expressed in parametric form as:
+:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
+:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \paren {2 b - b} \cos \theta + b \map \cos {\paren {\dfrac {2 b - b} b} \theta}
+      | c = 
+}}
+{{eqn | r = b \cos \theta + b \cos \theta
+      | c = 
+}}
+{{eqn | r = 2 b \cos \theta 
+      | c = 
+}}
+{{end-eqn}}
+Thus the [[Definition:X Coordinate|$x$ coordinate]] of the $2$ [[Definition:Cusp of Hypocycloid|cusp]] [[Definition:Hypocycloid|hypocycloid]] has a [[Definition:Image of Mapping|range]] $\closedint {-b} b$.
+Similarly:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \paren {2 b - b} \sin \theta - b \map \sin {\paren {\dfrac {2 b - b} b} \theta}
+      | c = 
+}}
+{{eqn | r = b \sin \theta - b \sin \theta
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Thus the [[Definition:Y Coordinate|$y$ coordinate]] of the $2$ [[Definition:Cusp of Hypocycloid|cusp]] [[Definition:Hypocycloid|hypocycloid]] is fixed at $y = 0$.
+Thus the $2$ [[Definition:Cusp of Hypocycloid|cusp]] [[Definition:Hypocycloid|hypocycloid]] consists of the [[Definition:Line Segment|line segment]]:
+:$x \in \closedint {-b} b, y = 0$.
+which is a [[Definition:Diameter of Circle|diameter]] of the containing [[Definition:Circle|circle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Preserves Directed Suprema}
+Tags: Meet-Continuous Lattices, Up-Complete Semilattices
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]] such that
+:$\forall x \in S$, a [[Definition:Directed Subset|directed subset]] $D$ of $S$: $x \preceq \sup D \implies x \preceq \sup \set {x \wedge y: y \in D}$
+\end{theorem}
+
+\begin{proof}
+=== [[Meet Preserves Directed Suprema/Lemma 2|Lemma 2]] ===
+{{:Meet Preserves Directed Suprema/Lemma 2}}{{qed|lemma}}
+Let $X$ be a [[Definition:Directed Subset|directed subset]] of $S \times S$ such that
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Up-Complete Product]]:
+:the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\mathscr S$ and $\mathscr S$ is [[Definition:Up-Complete|up-complete]].
+By [[Up-Complete Product/Lemma 2]]:
+:$X_1 := \map {\pr_1^\to} X$ is [[Definition:Directed Subset|directed]]
+and
+:$X_2 := \map {\pr_2^\to} X$ is [[Definition:Directed Subset|directed]]
+where
+:$\pr_1$ denotes the [[Definition:First Projection|first projection]] on $S \times S$
+:$\pr_2$ denotes the [[Definition:Second Projection|second projection]] on $S \times S$
+:$\map {\pr_1^\to} X$ denotes the [[Definition:Image of Subset under Mapping|image]] of $X$
+We will prove that
+:$(1): \quad \set {x \wedge \sup X_2: x \in X_1} = \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$
+'''First inclusion:'''
+Let $a \in \set {x \wedge \sup X_2: x \in X_1}$.
+Then
+:$\exists x \in X: a = x \wedge \sup X_2$
+By [[Meet Precedes Operands]]:
+:$x \wedge \sup X_2 \preceq \sup X_2$
+By assumption:
+:$x \wedge \sup X_2 \preceq \sup \set {x \wedge \sup X_2 \wedge y: y \in X_2}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\forall y \in X_2: y \preceq \sup X_2$
+By [[Preceding iff Meet equals Less Operand]]:
+:$\forall y \in X_2: \sup X_2 \wedge y = y$
+By [[Meet is Associative]]:
+:$x \wedge \sup X_2 \preceq \sup \set {x \wedge y: y \in X_2}$
+By [[Meet Semilattice is Ordered Structure]]:
+:$\forall y \in X_2: x \wedge y \preceq x \wedge \sup X_2$
+By definition:
+:$x \wedge \sup X_2$ is [[Definition:Upper Bound of Set|upper bound]] for $\set {x \wedge y: y \in X_2}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \set {x \wedge y: y \in X_2} \preceq x \wedge \sup X_2$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$a = \sup \set {x \wedge y: y \in X_2}$
+Thus
+:$a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$
+{{qed|lemma}}
+'''Second inclusion'''
+Let $a \in \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}$.
+Then
+:$\exists x \in X_1: a = \sup \set {x \wedge y: y \in X_2}$
+Analogically to first inclusion
+:$a = x \wedge \sup X_2$
+Thus
+:$a \in \set {x \wedge \sup X_2: x \in X_1}$
+{{qed|lemma}}
+We will prove as lemma that:
+:$\paren {\map {f^\to} X}$ is [[Definition:Directed Subset|directed]].
+Let $x, y \in \paren {\map {f^\to} X}$.
+By [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists \tuple {a, b} \in X: x = \map f {a, b}$
+and
+:$\exists \tuple {c, d} \in X: y = \map f {c, d}$
+By definition of $f$:
+:$x = a \wedge b$ and $y = c \wedge d$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists \tuple {g, h} \in X: \tuple {a, b} \precsim \tuple {g, h} \land \tuple {c, d} \precsim \tuple {g, h}$
+By [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$a \preceq g$, $b \preceq h$, $c \preceq g$, and $d \preceq h$
+By [[Meet Semilattice is Ordered Structure]]:
+:$x \preceq g \wedge h$ and $y \preceq g \wedge h$
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$g \wedge h = \map f {g, h} \in \map {f^\to} X$
+Thus:
+:$\exists z \in \map {f^\to} X: x \preceq z \land y \preceq z$
+This ends the proof of lemma.
+Thus by definition of [[Definition:Up-Complete|up-complete]]:
+:$\paren {\map {f^\to} X}$ admits a [[Definition:Supremum of Set|supremum]].
+Thus
+{{begin-eqn}}
+{{eqn | l = \map \sup {\map {f^\to} X}
+      | r = \sup \set {x \wedge y: x \in X_1, y \in X_2}
+      | c = [[Supremum of Meet Image of Directed Set]]
+}}
+{{eqn | r = \sup \bigcup \set {\set {x \wedge y: y \in X_2}: x \in X_1}
+}}
+{{eqn | r = \sup \set {\sup \set {x \wedge y: y \in X_2}: x \in X_1}
+      | c = [[Supremum of Suprema]]
+}}
+{{eqn | r = \sup \set {x \wedge \sup X_2: x \in X_1}
+      | c = the equality $(1)$
+}}
+{{eqn | r = \paren {\sup X_1} \wedge \paren {\sup X_2}
+      | c = [[Meet Preserves Directed Suprema/Lemma 2|Lemma 2]]
+}}
+{{eqn | r = \map f {\sup X_1, \sup X_2}
+      | c = Definition of $f$
+}}
+{{eqn | r = \map f {\sup X}
+      | c = [[Supremum by Suprema of Directed Set in Cartesian Product]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Astroid/Parametric Form}
+Tags: Astroids
+
+\begin{theorem}
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the [[Definition:Parametric Equation|parametric equation]]:
+:$\begin{cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+where $\theta$ is the [[Definition:Angle|angle]] between the [[Definition:X-Axis|$x$-axis]] and the [[Definition:Straight Line|line]] joining the [[Definition:Origin|origin]] to the [[Definition:Center of Circle|center]] of $C_1$.
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Astroid|astroid]] is a [[Definition:Hypocycloid|hypocycloid]] with $4$ [[Definition:Cusp of Hypocycloid|cusps]].
+:[[File:Astroid.png|400px]]
+By [[Equation of Hypocycloid]], the equation of $H$ is given by:
+:$\begin{cases}
+x & = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta} \\
+y & = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}
+\end{cases}$
+From [[Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii]], this can be generated by a [[Definition:Rotor of Hypocycloid|rotor]] $C_1$ of [[Definition:Radius of Circle|radius]] $\dfrac 1 4$ the [[Definition:Radius of Circle|radius]] of the [[Definition:Stator of Hypocycloid|stator]].
+Thus $a = 4 b$ and the equation of $H$ is now given by:
+:$\begin{cases}
+x & = 3 b \cos \theta + b \cos 3 \theta \\
+y & = 3 b \sin \theta - b \sin 3 \theta
+\end{cases}$
+From [[Triple Angle Formula for Cosine]]:
+:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
+and from [[Triple Angle Formula for Sine]]:
+:$\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$
+Thus $H$ can be expressed as:
+:$\begin{cases}
+x & = 4 b \cos^3 \theta = a \cos^3 \theta \\
+y & = 4 b \sin^3 \theta = a \sin^3 \theta
+\end{cases}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Astroid/Cartesian Form}
+Tags: Astroids
+
+\begin{theorem}
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the equation:
+:$x^{2/3} + y^{2/3} = a^{2/3}$
+\end{theorem}
+
+\begin{proof}
+By definition, an [[Definition:Astroid|astroid]] is a [[Definition:Hypocycloid|hypocycloid]] with $4$ [[Definition:Cusp of Hypocycloid|cusps]].
+:[[File:Astroid.png|400px]]
+From the [[Equation of Astroid/Parametric Form|parametric form of the equation of an astroid]], $H$ can be expressed as:
+:$\begin{cases}
+x & = 4 b \cos^3 \theta = a \cos^3 \theta \\
+y & = 4 b \sin^3 \theta = a \sin^3 \theta
+\end{cases}$
+[[Definition:Squaring|Squaring]], taking [[Definition:Cube Root|cube roots]] and adding:
+{{begin-eqn}}
+{{eqn | l = x^{2/3} + y^{2/3}
+      | r = a^{2/3} \paren {\cos^2 \theta + \sin^2 \theta}
+      | c = 
+}}
+{{eqn | r = a^{2/3}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding iff Meet equals Less Operand}
+Tags: Join and Meet, Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $x, y \in S$.
+Then
+:$x \preceq y$ {{iff}} $x \wedge y = x$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$x \preceq y$
+By definition of [[Definition:Meet (Order Theory)|meet]]:
+:$x \wedge y = \inf \left\{ {x, y}\right\}$
+By definitions of [[Definition:Lower Bound of Set|lower bound]] and [[Definition:Reflexivity|reflexivity]]:
+:$x$ is [[Definition:Lower Bound of Set|lower bound]] for $\left\{ {x, y}\right\}$
+and
+:$\forall z \in S: z$ is [[Definition:Lower Bound of Set|lower bound]] for $\left\{ {x, y}\right\} \implies z \preceq x$
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$x = \inf \left\{ {x, y}\right\} = x \wedge y$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$x \wedge y = x$
+Thus by [[Meet Precedes Operands]]:
+:$x \preceq y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Simple Harmonic Motion of Point on Tusi Couple}
+Tags: Tusi Couples
+
+\begin{theorem}
+Let $C_1$ and $C_2$ be the [[Definition:Rotor of Hypocycloid|rotor]] and [[Definition:Stator of Hypocycloid|stator]] respectively of a [[Definition:Tusi Couple|Tusi couple]] $H$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let the [[Definition:Center of Circle|center]] of $C_1$ move at a constant [[Definition:Angular Velocity|angular velocity]] $\omega$ around the [[Definition:Center of Circle|center]] of $C_2$.
+Let $P$ be the [[Definition:Point|point]] on the [[Definition:Circumference|circumference]] of $C_1$ whose [[Definition:Locus|locus]] is $H$.
+Let $C_1$ be initially positioned so that $P$ its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located on the [[Definition:X-Axis|$x$-axis]].
+Then $P$ moves back and forward on the [[Definition:X-Axis|$x$-axis]] with [[Definition:Simple Harmonic Motion|simple harmonic motion]] with [[Definition:Period of Simple Harmonic Motion|period]] $\dfrac {2 \pi} \omega$ and maximum [[Definition:Speed|speed]] $a \omega$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Straightforward but tedious. I'll get back to it in due course.}}
+\end{proof}<|endoftext|>
+\section{Maximum Rate of Change of Y Coordinate of Astroid}
+Tags: Hypocycloids
+
+\begin{theorem}
+Let $C_1$ and $C_2$ be the [[Definition:Rotor of Hypocycloid|rotor]] and [[Definition:Stator of Hypocycloid|stator]] respectively of an [[Definition:Astroid|astroid]] $H$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let the [[Definition:Center of Circle|center]] $C$ of $C_1$ move at a constant [[Definition:Angular Velocity|angular velocity]] $\omega$ around the [[Definition:Center of Circle|center]] of $C_2$.
+Let $P$ be the [[Definition:Point|point]] on the [[Definition:Circumference|circumference]] of $C_1$ whose [[Definition:Locus|locus]] is $H$.
+Let $C_1$ be positioned at [[Definition:Time|time]] $t = 0$ so that $P$ its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located on the [[Definition:X-Axis|$x$-axis]].
+Let $\theta$ be the [[Definition:Angle|angle]] made by $OC$ to the [[Definition:X-Axis|$x$-axis]] at [[Definition:Time|time]] $t$.
+Then the maximum [[Definition:Rate of Change|rate of change]] of the [[Definition:Y Coordinate|$y$ coordinate]] of $P$ in the [[Definition:First Quadrant|first quadrant]] occurs when $P$ is at the point where:
+:$x = a \paren {\dfrac 1 3}^{3/2}$
+:$y = a \paren {\dfrac 2 3}^{3/2}$
+\end{theorem}
+
+\begin{proof}
+:[[File:Astroid.png|400px]]
+The [[Definition:Rate of Change|rate of change]] of $\theta$ is given by:
+:$\omega = \dfrac {\d \theta} {\d t}$
+From [[Equation of Astroid/Parametric Form|Equation of Astroid: Parametric Form]], the [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the [[Definition:Parametric Equation|parametric equation]]:
+:$\begin {cases}
+x & = a \cos^3 \theta \\
+y & = a \sin^3 \theta
+\end{cases}$
+The [[Definition:Rate of Change|rate of change]] of $y$ is given by:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d t}
+      | r = \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = 3 a \omega \sin^2 \theta \cos \theta
+      | c = [[Power Rule for Derivatives]], [[Derivative of Sine Function]], [[Chain Rule for Derivatives]]
+}}
+{{end-eqn}}
+By [[Derivative at Maximum or Minimum]], when $\dfrac {\d y} {\d t}$ is at a [[Definition:Maximum Element|maximum]]:
+:$\dfrac {\d^2 y} {\d t^2} = 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d^2 y} {\d t^2}
+      | r = \map {\dfrac \d {\d \theta} } {3 a \omega \sin^2 \theta \cos \theta} \dfrac {\d \theta} {\d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = 3 a \omega^2 \paren {2 \sin \theta \cos^2 \theta - \sin^3 \theta}
+      | c = [[Product Rule for Derivatives]] and others
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d^2 y} {\d t^2}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 3 a \omega^2 \paren {2 \sin \theta \cos^2 \theta - \sin^3 \theta}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \sin \theta \cos^2 \theta
+      | r = \sin^3 \theta
+      | c = 
+}}
+{{end-eqn}}
+We can assume $\sin \theta \ne 0$ because in that case $\theta = 0$ and so $\dfrac {\d y} {\d t} = 0$.
+Thus when $\sin \theta = 0$, $y$ is not a [[Definition:Maximum Element|maximum]].
+So we can divide by $\sin \theta$ to give:
+{{begin-eqn}}
+{{eqn | l = 2 \cos^2 \theta
+      | r = \sin^2 \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tan^2 \theta
+      | r = 2
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | ll= \leadsto
+      | l = \tan \theta
+      | r = \sqrt 2
+      | c = 
+}}
+{{end-eqn}}
+We have:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \cos^3 \theta
+      | c = 
+}}
+{{eqn | l = y
+      | r = a \sin^3 \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac y x
+      | r = \tan^3 \theta
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = 2^{3/2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = 2^{3/2} x
+      | c = 
+}}
+{{end-eqn}}
+From [[Equation of Astroid/Cartesian Form|Equation of Astroid: Cartesian Form]]:
+:$x^{2/3} + y^{2/3} = a^{2/3}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = x^{2/3} + \paren {2^{3/2} x}^{2/3}
+      | r = a^{2/3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{2/3} \paren {1 + 2}
+      | r = a^{2/3}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \paren {\frac {a^{2/3} } 3}^{3/2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = a \paren {\frac 1 3}^{3/2}
+      | c = 
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \frac y x
+      | r = 2^{3/2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac 1 {2^{3/2} } y
+      | c = 
+}}
+{{eqn | l = \paren {\frac 1 {2^{3/2} } y}^{2/3} + y^{2/3}
+      | r = a^{2/3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^{2/3} \paren {1 + \frac 1 2}
+      | r = a^{2/3}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = y^{2/3}
+      | r = \frac 2 3 a^{2/3}
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \paren {2 \frac {a^{2/3} } 3}^{3/2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = a \paren {\frac 2 3}^{3/2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Deltoid}
+Tags: Deltoids
+
+\begin{theorem}
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the [[Definition:Parametric Equation|parametric equation]]:
+:$\begin{cases}
+x & = 2 b \cos \theta + b \cos 2 \theta \\
+y & = 2 b \sin \theta - b \sin 2 \theta
+\end{cases}$
+where $\theta$ is the [[Definition:Angle|angle]] between the [[Definition:X-Axis|$x$-axis]] and the [[Definition:Straight Line|line]] joining the [[Definition:Origin|origin]] to the [[Definition:Center of Circle|center]] of $C_1$.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Deltoid|deltoid]] is a [[Definition:Hypocycloid|hypocycloid]] with $3$ [[Definition:Cusp of Hypocycloid|cusps]].
+:[[File:Deltoid.png|400px]]
+By [[Equation of Hypocycloid]], the equation of $H$ is given by:
+:$\begin{cases}
+x & = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta} \\
+y & = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}
+\end{cases}$
+From [[Number of Cusps of Hypocycloid from Integral Ratio of Circle Radii]], this can be generated by a [[Definition:Rotor of Hypocycloid|rotor]] $C_1$ of [[Definition:Radius of Circle|radius]] $\dfrac 1 3$ the [[Definition:Radius of Circle|radius]] of the [[Definition:Stator of Hypocycloid|stator]].
+Thus $a = 3 b$ and the equation of $H$ is now given by:
+:$\begin{cases}
+x & = 2 b \cos \theta + b \cos 2 \theta \\
+y & = 2 b \sin \theta - b \sin 2 \theta
+\end{cases}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Arc of Deltoid}
+Tags: Deltoids
+
+\begin{theorem}
+The total [[Definition:Arc Length|length of the arcs]] of a [[Definition:Deltoid|deltoid]] constructed within a [[Definition:Stator of Hypocycloid|stator]] of [[Definition:Radius of Circle|radius]] $a$ is given by:
+:$\LL = \dfrac {16 a} 3$
+\end{theorem}
+
+\begin{proof}
+Let $H$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its center at the [[Definition:Origin|origin]] and one of its [[Definition:Cusp of Hypocycloid|cusps]] positioned at $\tuple {a, 0}$.
+:[[File:Deltoid.png|400px]]
+We have that $\LL$ is $3$ times the [[Definition:Arc Length|length]] of one [[Definition:Arc of Hypocycloid|arc]] of the [[Definition:Deltoid|deltoid]].
+From [[Arc Length for Parametric Equations]]:
+:$\ds \LL = 3 \int_{\theta \mathop = 0}^{\theta \mathop = 2 \pi/3} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
+where, from [[Equation of Deltoid]]:
+:$\begin{cases}
+x & = 2 b \cos \theta + b \cos 2 \theta \\
+y & = 2 b \sin \theta - b \sin 2 \theta
+\end{cases}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = -2 b \sin \theta - 2 b \sin 2 \theta
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = 2 b \cos \theta - 2 b \cos 2 \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
+      | c = 
+}}
+{{eqn | r = \paren {-2 b \sin \theta - 2 b \sin 2 \theta}^2 + \paren {2 b \cos \theta - 2 b \cos 2 \theta}^2
+      | c = 
+}}
+{{eqn | r = 4 b^2 \paren {\paren {-\sin \theta - \sin 2 \theta}^2 + \paren {\cos \theta - \cos 2 \theta}^2}
+      | c = 
+}}
+{{eqn | r = 4 b^2 \paren {\sin^2 \theta + 2 \sin \theta \sin 2 \theta + \sin^2 2 \theta + \cos^2 \theta - 2 \cos \theta \cos 2 \theta + \cos^2 2 \theta}
+      | c = 
+}}
+{{eqn | r = 4 b^2 \paren {2 + 2 \sin \theta \sin 2 \theta - 2 \cos \theta \cos 2 \theta}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 8 b^2 \paren {1 + \sin \theta \sin 2 \theta - \cos \theta \cos 2 \theta}
+      | c = 
+}}
+{{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \cos 2 \theta}
+      | c = [[Double Angle Formula for Sine]]
+}}
+{{eqn | r = 8 b^2 \paren {1 + 2 \sin^2 \theta \cos \theta - \cos \theta \paren {1 - 2 \sin^2 \theta} }
+      | c = [[Double Angle Formulas/Cosine/Corollary 2|Double Angle Formula for Cosine: Corollary 2]]
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \sin^2 \theta \cos \theta}
+      | c = simplifying
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 - \cos^2 \theta} }
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta + 4 \cos \theta \paren {1 + \cos \theta} \paren {1 - \cos \theta} }
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta \paren {1 + \cos \theta} }
+      | c = 
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 4 \cos \theta + 4 \cos^2 \theta}
+      | c = 
+}}
+{{eqn | r = 8 b^2 \paren {1 - \cos \theta} \paren {1 + 2 \cos \theta}^2
+      | c = 
+}}
+{{eqn | r = 8 b^2 \paren {2 \sin^2 \frac \theta 2} \paren {1 + 2 \cos \theta}^2
+      | c = [[Half Angle Formula for Sine]]
+}}
+{{eqn | r = 16 b^2 \sin^2 \frac \theta 2 \paren {1 + 2 \cos \theta}^2
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 4 b \sin \dfrac \theta 2 \size {1 + 2 \cos \theta}$
+In the range $0$ to $2 \pi / 3$, $1 + 2 \cos \theta$ is not less than $0$, and so:
+:$\ds \LL = 3 \int_0^{2 \pi / 3} 4 b \sin \dfrac \theta 2 \paren {1 + 2 \cos \theta} \rd \theta$
+Put:
+:$u = \cos \dfrac \theta 2$
+so:
+:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
+As $\theta$ increases from $0$ to $\dfrac {2 \pi} 3$, $u$ decreases from $1$ to $\dfrac 1 2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 1 + 2 \cos \theta
+      | r = 1 + 2 \paren {2 \paren {\cos \dfrac \theta 2}^2 - 1}
+      | c = [[Half Angle Formula for Cosine]]
+}}
+{{eqn | r = 4 u^2 - 1
+      | c = 
+}}
+{{end-eqn}}
+Substituting:
+:$2 \dfrac {\d u} {\d \theta} = -\sin \dfrac \theta 2$
+and the limits of integration:
+:$u = 1$ for $\theta = 0$
+:$u = \dfrac 1 2$ for $\theta = \dfrac {2 \pi} 3$
+we obtain, after simplifying the sign:
+{{begin-eqn}}
+{{eqn | l = \LL
+      | r = 12 b \int_1^{1/2} \paren {1 - 4 u^2} 2 \rd u
+      | c = [[Integration by Substitution]]
+}}
+{{eqn | r = 24 b \intlimits {u - \frac 4 3 u^3} 1 {1/2}
+      | c = 
+}}
+{{eqn | r = 24 b \paren {\paren {\frac 1 2 - \frac 4 3 \frac 1 {2^3} } - \paren {1 - \frac 4 3} }
+      | c = 
+}}
+{{eqn | r = 24 b \paren {\frac 1 2 - \frac 1 6 - 1 + \frac 4 3}
+      | c = 
+}}
+{{eqn | r = 24 b \paren {\frac 1 3 + \frac 1 3}
+      | c = 
+}}
+{{eqn | r = 16 b
+      | c = 
+}}
+{{eqn | r = \frac {16 a} 3
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Epicycloid}
+Tags: Epicycloids
+
+\begin{theorem}
+Let a [[Definition:Circle|circle]] $C_1$ of [[Definition:Radius of Circle|radius]] $b$ roll without slipping around the outside of a [[Definition:Circle|circle]] $C_2$ of [[Definition:Radius of Circle|radius]] $a$.
+Let $C_2$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its [[Definition:Center of Circle|center]] $O$ located at the [[Definition:Origin|origin]].
+Let $P$ be a [[Definition:Point|point]] on the [[Definition:Circumference of Circle|circumference]] of $C_1$.
+Let $C_1$ be initially positioned so that $P$ is its point of [[Definition:Tangent Circles|tangency]] to $C_2$, located at [[Definition:Point|point]] $A = \tuple {a, 0}$ on the [[Definition:X-Axis|$x$-axis]].
+Let $H$ be the [[Definition:Epicycloid|epicycloid]] traced out by the [[Definition:Point|point]] $P$.
+Let $\tuple {x, y}$ be the [[Definition:Cartesian Coordinate Pair|coordinates]] of $P$ as it travels over [[Definition:The Plane|the plane]].
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the equations:
+:$x = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}$
+:$y = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}$
+\end{theorem}
+
+\begin{proof}
+:[[File:Epicycloid.png|600px]]
+Let $C_1$ have rolled so that the [[Definition:Straight Line|line]] $OC$ through the [[Definition:Radius of Circle|radii]] of $C_1$ and $C_2$ is at [[Definition:Angle|angle]] $\theta$ to the [[Definition:X-Axis|$x$-axis]].
+Let $C_1$ have turned through an angle $\phi$ to reach that point.
+By definition of [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]], $P = \tuple {x, y}$ is defined by:
+:$x = \paren {a + b} \cos \theta - b \map \cos {\phi + \theta}$
+:$y = \paren {a + b} \sin \theta - b \map \sin {\phi + \theta}$
+The [[Definition:Arc of Circle|arc]] of $C_1$ between $P$ and $B$ is the same as the [[Definition:Arc of Circle|arc]] of $C_2$ between $A$ and $B$.
+Thus by [[Arc Length of Sector]]:
+:$a \theta = b \phi$
+Thus:
+:$\phi + \theta = \paren {\dfrac {a + b} b} \theta$
+whence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation between Equations for Hypocycloid and Epicycloid}
+Tags: Hypocycloids, Epicycloids
+
+\begin{theorem}
+Consider the [[Definition:Hypocycloid|hypocycloid]] defined by the [[Equation of Hypocycloid|equations]]:
+:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
+:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
+By replacing $b$ with $-b$, this converts to the [[Equation of Epicycloid|equations]] which define an [[Definition:Epicycloid|epicycloid]]:
+:$x = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}$
+:$y = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \paren {a - \paren {-b} } \cos \theta + \paren {-b} \map \cos {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}
+      | c = putting $-b$ for $b$
+}}
+{{eqn | r = \paren {a + b} \cos \theta - b \map \cos {-\paren {\dfrac {a + b} b} \theta}
+      | c = 
+}}
+{{eqn | r = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}
+      | c = [[Cosine Function is Even]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = y
+      | r = \paren {a - \paren {-b} } \sin \theta - \paren {-b} \map \sin {\paren {\dfrac {a - \paren {-b} } {\paren {-b} } } \theta}
+      | c = putting $-b$ for $b$
+}}
+{{eqn | r = \paren {a + b} \sin \theta + b \map \sin {-\paren {\dfrac {a + b} b} \theta}
+      | c = 
+}}
+{{eqn | r = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}
+      | c = [[Sine Function is Odd]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equation of Nephroid}
+Tags: Nephroids
+
+\begin{theorem}
+The [[Definition:Point|point]] $P = \tuple {x, y}$ is described by the [[Definition:Parametric Equation|parametric equation]]:
+:$\begin{cases}
+x & = 3 b \cos \theta - b \cos 3 \theta \\
+y & = 3 b \sin \theta - b \sin 3 \theta
+\end{cases}$
+where $\theta$ is the [[Definition:Angle|angle]] between the [[Definition:X-Axis|$x$-axis]] and the [[Definition:Straight Line|line]] joining the [[Definition:Origin|origin]] to the [[Definition:Center of Circle|center]] of $C_1$.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Nephroid|nephroid]] is an [[Definition:Epicycloid|epicycloid]] with $2$ [[Definition:Cusp of Hypocycloid|cusps]].
+:[[File:Nephroid.png|600px]]
+By [[Equation of Epicycloid]], the equation of $H$ is given by:
+:$\begin{cases}
+x & = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta} \\
+y & = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}
+\end{cases}$
+When $a = 2 b$ the equation of $H$ is now given by:
+:$\begin{cases}
+x & = 3 b \cos \theta - b \cos 3 \theta \\
+y & = 3 b \sin \theta - b \sin 3 \theta
+\end{cases}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Arc of Nephroid}
+Tags: Nephroids
+
+\begin{theorem}
+The total [[Definition:Arc Length|length of the arcs]] of a [[Definition:Nephroid|nephroid]] constructed around a [[Definition:Stator of Epicycloid|stator]] of [[Definition:Radius of Circle|radius]] $a$ is given by:
+:$\LL = 12 a$
+\end{theorem}
+
+\begin{proof}
+Let a [[Definition:Nephroid|nephroid]] $H$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]] with its center at the [[Definition:Origin|origin]] and its [[Definition:Cusp of Hypocycloid|cusps]] positioned at $\tuple {\pm a, 0}$.
+:[[File:Nephroid.png|600px]]
+We have that $\LL$ is $2$ times the [[Definition:Arc Length|length]] of one [[Definition:Arc of Epicycloid|arc]] of the [[Definition:Nephroid|nephroid]].
+From [[Arc Length for Parametric Equations]]:
+:$\ds \LL = 2 \int_{\theta \mathop = 0}^{\theta \mathop = \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$
+where, from [[Equation of Nephroid]]:
+:$\begin{cases}
+x & = 3 b \cos \theta - b \cos 3 \theta \\
+y & = 3 b \sin \theta - b \sin 3 \theta
+\end{cases}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = -3 b \sin \theta + 3 b \sin 3 \theta
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = 3 b \cos \theta - 3 b \cos 3 \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
+      | c = 
+}}
+{{eqn | r = \paren {-3 b \sin \theta + 3 b \sin 3 \theta}^2 + \paren {3 b \cos \theta - 3 b \cos 3 \theta}^2
+      | c = 
+}}
+{{eqn | r = 9 b^2 \paren {\paren {-\sin \theta + \sin 3 \theta}^2 + \paren {\cos \theta - \cos 3 \theta}^2}
+      | c = 
+}}
+{{eqn | r = 9 b^2 \paren {\sin^2 \theta - 2 \sin \theta \sin 3 \theta + \sin^2 3 \theta + \cos^2 \theta - 2 \cos \theta \cos 3 \theta + \cos^2 3 \theta}
+      | c = 
+}}
+{{eqn | r = 9 b^2 \paren {2 - 2 \sin \theta \sin 3 \theta - 2 \cos \theta \cos 3 \theta}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 18 b^2 \paren {1 - \sin \theta \sin 3 \theta - \cos \theta \cos 3 \theta}
+      | c = 
+}}
+{{eqn | r = 18 b^2 \paren {1 - \sin \theta \paren {3 \sin \theta - 4 \sin^3 \theta} - \cos \theta \cos 2 \theta}
+      | c = [[Triple Angle Formula for Sine]]
+}}
+{{eqn | r = 18 b^2 \paren {1 - \sin \theta \paren {3 \sin \theta - 4 \sin^3 \theta} - \cos \theta \paren {4 \cos^3 \theta - 3 \cos \theta} }
+      | c = [[Triple Angle Formula for Cosine]]
+}}
+{{eqn | r = 18 b^2 \paren {1 - 3 \sin^2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta + 3 \cos^2 \theta}
+      | c = simplifying
+}}
+{{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta}
+      | c = [[Double Angle Formula for Cosine]]
+}}
+{{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 x + \cos 4 x} 2 - 4 \cos^4 \theta}
+      | c = [[Power Reduction Formula for 4th Power of Sine|Power Reduction Formula for $\sin^4$]]
+}}
+{{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 \theta + \cos 4 \theta} 2 - \dfrac {3 + 4 \cos 2 \theta + \cos 4 \theta} 2}
+      | c = [[Power Reduction Formula for 4th Power of Cosine|Power Reduction Formula for $\cos^4$]]
+}}
+{{eqn | r = 18 b^2 \paren {1 - \cos 2 \theta}
+      | c = simplifying
+}}
+{{eqn | r = 18 b^2 \paren {2 \sin^2 \theta}
+      | c = [[Square of Sine]]
+}}
+{{eqn | r = 36 b^2 \sin^2 \theta
+      | c = simplifying
+}}
+{{end-eqn}}
+Thus:
+:$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 6 b \sin \theta$
+So:
+{{begin-eqn}}
+{{eqn | l = \LL
+      | r = 2 \int_0^\pi 6 b \sin \theta \rd \theta
+      | c = 
+}}
+{{eqn | r = 12 b \int_0^\pi \sin \theta \rd \theta
+      | c = 
+}}
+{{eqn | r = 12 b \bigintlimits {-\cos \theta} 0 \pi
+      | c = 
+}}
+{{eqn | r = 12 b \paren {-\cos \pi - \paren {-\cos 0} }
+      | c = 
+}}
+{{eqn | r = 12 b \paren {-\paren {-1} - \paren {-1} }
+      | c = 
+}}
+{{eqn | r = 24 b
+      | c = 
+}}
+{{eqn | r = 12 a
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Meet Image of Directed Set}
+Tags: Up-Complete Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Let $f: S \times S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall \tuple {x, y} \in S \times S: \map f {x, y} = x \wedge y$
+Let $D$ be [[Definition:Directed Subset|directed subset]] of $S \times S$ in [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] $\struct {S \times S, \precsim}$ of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
+Then:
+:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
+where
+:$\pr_1$ denotes the [[Definition:First Projection|first projection]] on $S \times S$
+:$\pr_2$ denotes the [[Definition:Second Projection|second projection]] on $S \times S$
+:$\map {\pr_1^\to} D$ denotes the [[Definition:Image of Subset under Mapping|image]] of $D$ under $\pr_1$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\map {f^\to} D = \set {x \wedge y: \tuple {x, y} \in D}$
+By definition of [[Definition:Subset|subset]]:
+:$\map {f^\to} D \subseteq \set {x \wedge y: x \in \map {\pr_1^\to} D, y \in \map {\pr_2^\to} D}$
+By [[Up-Complete Product/Lemma 2]]:
+:$D_1 := \map {\pr_1^\to} D$ is [[Definition:Directed Subset|directed]]
+and
+:$D_2 := \map {\pr_2^\to} D$ is [[Definition:Directed Subset|directed]].
+By [[Meet of Directed Subsets is Directed]]:
+:$\set {x \wedge y: x \in D_1, y \in D_2}$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\set {x \wedge y: x \in D_1, y \in D_2}$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Meet is Increasing]]:
+:$f$ is [[Definition:Increasing Mapping|increasing mapping]].
+By [[Image of Directed Subset under Increasing Mapping is Directed]]:
+:$\map {f^\to} D$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\map {f^\to} D$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Supremum of Subset]]:
+:$\map \sup {\map {f^\to} D} \preceq \sup \set {x \wedge y: x \in D_1, y \in D_2}$
+We will prove that
+:$\set {x \wedge y: x \in D_1, y \in D_2} \subseteq \paren {\map {f^\to} D}^\preceq$
+where
+:$\paren {\map {f^\to} D}^\preceq$ denotes the [[Definition:Lower Closure of Subset|lower closure]] of $\map {f^\to} D$.
+Let $z \in \set {x \wedge y: x \in D_1, y \in D_2}$.
+Then
+:$\exists x \in D_1, y \in D_2: z = x \wedge y$
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists \tuple {a, b} \in D: \map {\pr_1} {a, b} = x$
+and
+:$\exists \tuple {c, d} \in D: \map {\pr_2} {c, d} = y$
+By definition of [[Definition:First Projection|first projection]] and [[Definition:Second Projection|second projection]]:
+:$a = x$ and $d = y$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists \tuple {g, h} \in D: \tuple {x, b} \precsim \tuple {g, h} \land \tuple {c, y} \precsim \tuple {g, h}$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$x \preceq g$ and $y \preceq h$
+By [[Meet Semilattice is Ordered Structure]]:
+:$x \wedge y \preceq g \wedge h \in \set {x \wedge y: \tuple {x, y} \in D}$
+Thus by definition of [[Definition:Lower Closure of Subset|lower closure]]:
+:$z \in \paren {\map {f^\to} D}^\preceq$
+{{qed|lemma}}
+By [[Supremum of Lower Closure of Set]]:
+:$\paren {\map {f^\to} D}^\preceq$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$\map \sup {\map {f^\to} D}^\preceq = \map \sup {\map {f^\to} D}$
+By [[Supremum of Subset]]:
+:$\sup \set {x \wedge y: x \in D_1, y \in D_2} \preceq \map \sup {\map {f^\to} D}$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\map \sup {\map {f^\to} D} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet is Increasing}
+Tags: Join and Meet
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $f: S \times S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall s, t \in S: f\left({s, t}\right) = s \wedge t$
+Then:
+:$f$ is [[Definition:Increasing Mapping|increasing]] as a [[Definition:Mapping|mapping]] from [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] $\left({S\times S, \precsim}\right)$ of $\left({S, \preceq}\right)$ and $\left({S, \preceq}\right)$ into $\left({S, \preceq}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $\left({x, y}\right)$, $\left({z, t}\right) \in S \times S$ such that
+:$\left({x, y}\right) \precsim \left({z, t}\right)$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$x \preceq z$ and $y \preceq t$
+By [[Meet Semilattice is Ordered Structure]]:
+:$x \wedge y \preceq z \wedge t$
+By definition of $f$:
+:$f\left({x, y}\right) \preceq f\left({z, t}\right)$
+Thus by definition
+:$f$ is [[Definition:Increasing Mapping|increasing mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet Preserves Directed Suprema/Lemma 2}
+Tags: Up-Complete Semilattices
+
+\begin{theorem}
+Let $x$ be an [[Definition:Element|element]] of $S$, $D$ be a [[Definition:Directed Subset|directed subset]] of $S$.
+Then:
+:$\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$
+\end{theorem}
+
+\begin{proof}
+By [[Meet Precedes Operands]]:
+:$\paren {\sup D} \wedge x \preceq \sup D$
+By assumption:
+:$\paren {\sup D} \wedge x \preceq \sup \set {\paren {\sup D} \wedge x \wedge d: d \in D}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\forall d \in D: d \preceq \sup D$
+By [[Preceding iff Meet equals Less Operand]]:
+:$\forall d \in D: d \wedge \sup D = d$
+By [[Meet is Associative]] and [[Meet is Commutative]]:
+:$\paren {\sup D} \wedge x \preceq \sup \set {d \wedge x: d \in D}$
+By [[Meet Semilattice is Ordered Structure]]:
+:$\forall d \in D: d \wedge x \preceq \paren {\sup D} \wedge x$
+By definition:
+:$\paren {\sup D} \wedge x$ is [[Definition:Upper Bound of Set|upper bound]] for $\set {d \wedge x: d \in D}$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup \set {d \wedge x: d \in D} \preceq \paren {\sup D} \wedge x$
+if $\set {d \wedge x: d \in D}$ admits a [[Definition:Supremum of Set|supremum]]
+We will prove that:
+:$\set {d \wedge x: d \in D}$ is [[Definition:Directed Subset|directed]]
+Let $y, z \in \set {d \wedge x: d \in D}$.
+Then
+:$\exists d_1 \in D: y = d_1 \wedge x$
+and
+:$\exists d_2 \in D: z = d_2 \wedge x$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists d \in D: d_1 \preceq d \land d_2 \preceq d$
+By [[Meet Semilattice is Ordered Structure]]:
+:$y \preceq d \wedge x$ and $z \preceq d \wedge x$
+Thus
+:$d \wedge x \in \set {d \wedge x: d \in D}$
+Hence by definition:
+:$\set {d \wedge x: d \in D}$ is [[Definition:Directed Subset|directed]].
+{{qed|lemma}}
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\set {d \wedge x: d \in D}$ admits a [[Definition:Supremum of Set|supremum]]
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\paren {\sup D} \wedge x = \sup \set {d \wedge x: d \in D}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum by Suprema of Directed Set in Cartesian Product}
+Tags: Up-Complete Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Let $\struct {S \times S, \precsim}$ be the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S \times S$.
+Then:
+:$\sup D = \tuple {\map \sup {\map {\pr_1^\to} D}, \map \sup {\map {\pr_2^\to} D} }$
+\end{theorem}
+
+\begin{proof}
+By [[Up-Complete Product]]:
+:$\struct {S \times S, \precsim}$ is [[Definition:Up-Complete|up-complete]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+:$\exists d_1, d_2 \in S: \sup D = \tuple {d_1, d_2}$
+By [[Up-Complete Product/Lemma 2]]:
+:$D_1 := \map {\pr_1^\to} D$ is [[Definition:Directed Subset|directed]]
+and
+:$D_2 := \map {\pr_2^\to} D$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D_1$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$D_2$ admits a [[Definition:Supremum of Set|supremum]]
+We will prove that
+:$d_2$ is [[Definition:Upper Bound of Set|upper bound]] for $D_2$
+Let $x \in D_2$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists \tuple {a, b} \in D: \map {\pr_2} {a, b} = x$
+By definition of [[Definition:Second Projection|second projection]]:
+$b = x$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\tuple {d_1, d_2}$ is [[Definition:Upper Bound of Set|upper bound]] for $D$.
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\tuple {a, x} \precsim \tuple {d_1, d_2}$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$x \preceq d_2$
+{{qed|lemma}}
+Analogically we have that
+:$d_1$ is [[Definition:Upper Bound of Set|upper bound]] for $D_1$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup D_1 \preceq d_1$ and $\sup D_2 \preceq d_2$
+By definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\tuple {\sup D_1, \sup D_2} \precsim \sup D$
+By [[Up-Complete Product/Lemma 1]]:
+:$D_1 \times D_2$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D_1 \times D_2$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Subset|subset]]:
+:$D \subseteq D_1 \times D_2$
+By [[Supremum of Subset]]:
+:$\sup D \precsim \map \sup {D_1 \times D_2}$
+By [[Supremum of Cartesian Product]]:
+:$\sup D \precsim \tuple {\sup D_1, \sup D_2}$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$\sup D = \tuple {\sup D_1, \sup D_2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Evolute of Circle is its Center}
+Tags: Evolutes, Circles
+
+\begin{theorem}
+The [[Definition:Evolute|evolute]] of a [[Definition:Circle|circle]] is a single [[Definition:Point|point]]: its [[Definition:Center of Circle|center]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Evolute|evolute]] of $C$ is the [[Definition:Locus|locus]] of the [[Definition:Center of Curvature|centers of curvature]] of each [[Definition:Point|point]] on $C$
+{{WLOG}}, take the [[Definition:Circle|circle]] $C$ of [[Definition:Radius of Circle|radius]] $a$ whose [[Definition:Center of Circle|center]] is positioned at the [[Definition:Origin|origin]] of a [[Definition:Cartesian Plane|cartesian plane]].
+From [[Equation of Circle]], $C$ has the equation:
+:$x^2 + y^2 = a^2$
+From the definition of [[Definition:Curvature/Cartesian Form|curvature in cartesian form]]:
+:$k = \dfrac {y''} {\left({1 + y'^2}\right)^{3/2} }$
+Here we have:
+{{begin-eqn}}
+{{eqn | l = 2 x + 2 y \frac {\mathrm d y} {\mathrm d x}
+      | r = 0
+      | c = [[Definition:Differentiation|differentiating]] {{WRT|Differentiation}} $x$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\mathrm d y} {\mathrm d x}
+      | r = -\frac x y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\mathrm d^2 y} {\mathrm d x^2}
+      | r = \frac {x \frac {\mathrm d y} {\mathrm d x} - y} {y^2}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {y - x \left({-\frac x y}\right) } {y^2}
+      | c = substituting for $\dfrac {\mathrm d y} {\mathrm d x}$
+}}
+{{eqn | r = \frac {y^2 + x^2} {y^3}
+      | c = simplifying
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = k
+      | r = \dfrac {\frac {y^2 + x^2} {y^3} } {\left({1 + \left({-\frac x y}\right)^2}\right)^{3/2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {y^2 + x^2} {y^3 \left({1 + \frac {x^2} {y^2} }\right)^{3/2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {y^2 + x^2} {y^3 \left({\frac {y^2 + x^2} {y^2} }\right)^{3/2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {y^2 + x^2} {\left({y^2 + x^2}\right)^{3/2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {\left({y^2 + x^2}\right)^{1/2} }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 a
+      | c = as $x^2 + y^2 = a^2$
+}}
+{{end-eqn}}
+Thus the [[Definition:Curvature|curvature]] of $C$ is constant.
+The [[Definition:Radius of Curvature|radius of curvature]] of $C$ is likewise constant:
+:$\rho = a$
+From [[Radius at Right Angle to Tangent]], the [[Definition:Normal to Curve|normal]] to $C$ at all [[Definition:Point|points]] on $C$ passes through the [[Definition:Center of Circle|center]] of $C$.
+We have that $a$ is the [[Definition:Distance (Linear Measure)|distance]] from $C$ to the [[Definition:Center of Circle|center]] of $C$.
+Thus it follows that the [[Definition:Center of Curvature|center of curvature]] of $C$ is the [[Definition:Center of Circle|center]] of $C$ at all points.
+Hence the result by definition of [[Definition:Evolute|evolute]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Radius of Curvature in Cartesian Form}
+Tags: Analytic Geometry
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] defined by a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice]] [[Definition:Differentiable Real Function|differentiable]].
+Let $C$ be embedded in a [[Definition:Cartesian Plane|cartesian plane]].
+The '''[[Definition:Radius of Curvature|radius of curvature]]''' $\rho$ of $C$ at a [[Definition:Point|point]] $P = \tuple {x, y}$ is given by:
+:$\rho = \dfrac {\paren {1 + y'^2}^{3/2} } {\size {y''} }$
+where:
+:$y' = \dfrac {\d y} {\d x}$ is the [[Definition:Derivative|derivative]] of $y$ {{WRT|Differentiation}} $x$ at $P$
+:$y'' = \dfrac {\d^2 y} {\d x^2}$ is the [[Definition:Second Derivative|second derivative]] of $y$ {{WRT|Differentiation}} $x$ at $P$.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Radius of Curvature|radius of curvature]] $\rho$ is given by:
+:$\rho = \dfrac 1 {\size \kappa}$
+where $\kappa$ is the [[Definition:Curvature|curvature]], given in [[Definition:Curvature/Cartesian Form|Cartesian form]] as:
+:$\kappa = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$
+As $\paren {1 + y'^2}^{3/2}$ is [[Definition:Positive Real Number|positive]], it follows that:
+:$\size {\dfrac {y''} {\paren {1 + y'^2}^{3/2} } } = \dfrac {\size {y''} } {\paren {1 + y'^2}^{3/2} }$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parametric Equations for Evolute/Formulation 1}
+Tags: Evolutes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] expressed as the [[Definition:Locus|locus]] of an [[Definition:Equation|equation]] $\map f {x, y} = 0$.
+The [[Definition:Parametric Equation|parametric equations]] for the [[Definition:Evolute|evolute]] of $C$ can be expressed as:
+:$\begin{cases}
+X = x - \dfrac {y' \paren {1 + y'^2} } {y''} \\
+Y = y + \dfrac {1 + y'^2} {y''}
+\end{cases}$
+where:
+:$\left({x, y}\right)$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on $C$
+:$\left({X, Y}\right)$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on the [[Definition:Evolute|evolute]] of $C$
+:$y'$ and $y''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $y$ {{WRT|Differentiation}} $x$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CenterOfCurvature.png|400px]]
+Let $P = \tuple {x, y}$ be a general [[Definition:Point|point]] on $C$.
+Let $Q = \tuple {X, Y}$ be the [[Definition:Center of Curvature|center of curvature]] of $C$ at $P$.
+From the above diagram:
+:$x - X = \pm \rho \sin \psi$
+:$Y - y = \pm \rho \cos \psi$
+where:
+:$\rho$ is the [[Definition:Radius of Curvature|radius of curvature]] of $C$ at $P$
+:$\psi$ is the [[Definition:Angle|angle]] between the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$ and the [[Definition:X-Axis|$x$-axis]].
+Whether the sign is plus or minus depends on whether the curve is [[Definition:Convex Real Function|convex]] or [[Definition:Concave Real Function|concave]].
+By definition of [[Definition:Radius of Curvature|radius of curvature]]:
+:$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\
+Y - y = \dfrac 1 k \cos \psi \end {cases}$
+where $k$ is the [[Definition:Curvature|curvature]] of $C$ at $P$, given by:
+:$k = \dfrac {y''} {\paren {1 + y'^2}^{3/2} }$
+We have that:
+:$\sin \psi = \dfrac {\d y} {\d s} = \dfrac {y'} {\sqrt {1 + y'^2} }$
+:$\cos \psi = \dfrac {\d x} {\d s} = \dfrac 1 {\sqrt {1 + y'^2} }$
+Substituting for $k$ and $\psi$ in $(1)$ gives:
+{{begin-eqn}}
+{{eqn | l = x - X
+      | r = \dfrac {\paren {1 + y'^2}^{3/2} } {y''} \dfrac {y'} {\sqrt {1 + y'^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {y' \paren {1 + y'^2} } {y''}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = X
+      | r = x - \dfrac {y' \paren {1 + y'^2} } {y''}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = Y - y
+      | r = \dfrac {\paren {1 + y'^2}^{3/2} } {y''} \dfrac 1 {\sqrt {1 + y'^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {1 + y'^2} {y''}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = Y
+      | r = y + \dfrac {1 + y'^2} {y''}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parametric Equations for Evolute/Formulation 2}
+Tags: Evolutes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] expressed as the [[Definition:Locus|locus]] of an [[Definition:Equation|equation]] $\map f {x, y} = 0$.
+The [[Definition:Parametric Equation|parametric equations]] for the [[Definition:Evolute|evolute]] of $C$ can be expressed as:
+:$\begin{cases}
+X = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''} \\
+Y = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+\end{cases}$
+where:
+:$\tuple {x, y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on $C$
+:$\tuple {X, Y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on the [[Definition:Evolute|evolute]] of $C$
+:$x'$ and $x''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $x$ {{WRT|Differentiation}} $t$
+:$y'$ and $y''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $y$ {{WRT|Differentiation}} $t$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CenterOfCurvature.png|400px]]
+Let $P = \tuple {x, y}$ be a general [[Definition:Point|point]] on $C$.
+Let $Q = \tuple {X, Y}$ be the [[Definition:Center of Curvature|center of curvature]] of $C$ at $P$.
+From the above diagram:
+:$x - X = \pm \rho \sin \psi$
+:$Y - y = \pm \rho \cos \psi$
+where:
+:$\rho$ is the [[Definition:Radius of Curvature|radius of curvature]] of $C$ at $P$
+:$\psi$ is the [[Definition:Angle|angle]] between the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$ and the [[Definition:X-Axis|$x$-axis]].
+Whether the sign is plus or minus depends on whether the curve is [[Definition:Convex Real Function|convex]] or [[Definition:Concave Real Function|concave]].
+By definition of [[Definition:Radius of Curvature|radius of curvature]]:
+:$(1): \quad \begin {cases} x - X = \dfrac 1 k \sin \psi \\
+Y - y = \dfrac 1 k \cos \psi \end {cases}$
+where $k$ is the [[Definition:Curvature|curvature]] of $C$ at $P$, given by:
+:$k = \dfrac {x' y'' - y' x''} {\paren {x'^2 + y'^2}^{3/2} }$
+We have that:
+:$\sin \psi = \dfrac {\d y} {\d s} = \dfrac {y'} {\sqrt {x'^2 + y'^2} }$
+:$\cos \psi = \dfrac {\d x} {\d s} = \dfrac {x'} {\sqrt {x'^2 + y'^2} }$
+Substituting for $k$ and $\psi$ in $(1)$ gives:
+{{begin-eqn}}
+{{eqn | l = x - X
+      | r = \dfrac {\paren {x'^2 + y'^2}^{3/2} } {x' y'' - y' x''} \dfrac {y'} {\sqrt {x'^2 + y'^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = X
+      | r = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = Y - y
+      | r = \dfrac {\paren {x'^2 + y'^2}^{3/2} } {x' y'' - y' x''} \dfrac {x'} {\sqrt {x'^2 + y'^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = Y
+      | r = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Evolute of Parabola}
+Tags: Evolutes, Parabolas
+
+\begin{theorem}
+The [[Definition:Evolute|evolute]] of the [[Definition:Parabola|parabola]] $y = x^2$ is the [[Definition:Curve|curve]]:
+:$27 X^2 = 16 \paren {Y - \dfrac 1 2}^3$
+\end{theorem}
+
+\begin{proof}
+From [[Parametric Equations for Evolute/Formulation 1|Parametric Equations for Evolute: Formulation 1]]:
+:$\begin {cases}
+X = x - \dfrac {y' \paren {1 + y'^2} } {y''} \\
+Y = y + \dfrac {1 + y'^2} {y''}
+\end{cases}$
+where:
+:$\tuple {x, y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on $C$
+:$\tuple {X, Y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on the [[Definition:Evolute|evolute]] of $C$
+:$y'$ and $y''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $y$ {{WRT|Differentiation}} $x$.
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = y'
+      | r = 2 x
+      | c = 
+}}
+{{eqn | l = y''
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = x - \dfrac {2 x \paren {1 + \paren {2 x}^2} } 2
+      | c = 
+}}
+{{eqn | r = x - x - 4 x^3
+      | c = 
+}}
+{{eqn | r = -4 x^3
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = Y
+      | r = x^2 + \dfrac {1 + \paren {2 x}^2} 2
+      | c = 
+}}
+{{eqn | r = \frac {2 x^2 + 1 + 4 x^2} 2
+      | c = 
+}}
+{{eqn | r = \frac {6 x^2 + 1} 2
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = Y
+      | r = \frac {6 \paren {-\frac X 4}^{2/3} + 1} 2
+      | c = substituting for $X$
+}}
+{{eqn | l = Y
+      | r = 3 \paren {-\frac X 4}^{2/3} + \frac 1 2
+      | c = substituting for $X$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 3 \paren {Y - \frac 1 2}
+      | r = \paren {\frac {X^2} {16} }^{1/3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 27 X^2
+      | r = 16 \paren {Y - \dfrac 1 2}^3
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Parabola|parabola]] (blue) and its [[Definition:Evolute|evolute]] (red) are illustrated below:
+[[File:ParabolaAndEvolute.png|400px]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normal to Curve is Tangent to Evolute}
+Tags: Evolutes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] defined by a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice]] [[Definition:Differentiable Real Function|differentiable]].
+Let the [[Definition:Curvature|curvature]] of $C$ be non-[[Definition:Constant|constant]].
+Let $P$ be a [[Definition:Point|point]] on $C$.
+Let $Q$ be the [[Definition:Center of Curvature|center of curvature]] of $C$ at $P$.
+The [[Definition:Normal to Curve|normal]] to $C$ at $P$ is [[Definition:Tangent to Curve|tangent]] to the [[Definition:Evolute|evolute]] $E$ of $C$ at $Q$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CenterOfCurvature.png|400px]]
+Let $P = \left({x, y}\right)$ be a general [[Definition:Point|point]] on $C$.
+Let $Q = \left({X, Y}\right)$ be the [[Definition:Center of Curvature|center of curvature]] of $C$ at $P$.
+From the above diagram:
+:$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\
+Y - y = \pm \rho \cos \psi \end{cases}$
+where:
+:$\rho$ is the [[Definition:Radius of Curvature|radius of curvature]] of $C$ at $P$
+:$\psi$ is the [[Definition:Angle|angle]] between the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$ and the [[Definition:X-Axis|$x$-axis]].
+Whether the sign is plus or minus depends on whether the curve is [[Definition:Convex Real Function|convex]] or [[Definition:Concave Real Function|concave]].
+For simplicity, let it be assumed that the [[Definition:Curvature|curvature]] $k$ at each point under consideration on $C$ is [[Definition:Positive Real Number|positive]].
+The case for $k < 0$ can then be treated similarly.
+Thus we have $k > 0$ and so $(1)$ can be written:
+:$(2): \quad \begin{cases} X = x - \rho \sin \psi \\
+Y = y +\rho \cos \psi \end{cases}$
+By definition of [[Definition:Curvature|curvature]]:
+:$k = \dfrac {\mathrm d \psi} {\mathrm d s}$
+and:
+:$\rho = \dfrac 1 k = \dfrac {\mathrm d s} {\mathrm d \psi}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \rho \sin \psi
+      | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d y} {\mathrm d s}
+      | c = 
+}}
+{{eqn | n = 3
+      | r = \dfrac {\mathrm d y} {\mathrm d \psi}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \rho \cos \psi
+      | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d x} {\mathrm d s}
+      | c = 
+}}
+{{eqn | n = 4
+      | r = \dfrac {\mathrm d x} {\mathrm d \psi}
+      | c = 
+}}
+{{end-eqn}}
+[[Definition:Differentiation|Differentiating]] $(2)$ {{WRT|Differentiation}} $\psi$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d X} {\mathrm d \psi}
+      | r = \dfrac {\mathrm d x} {\mathrm d \psi} - \rho \cos \psi - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi
+      | c = 
+}}
+{{eqn | r = - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi
+      | c = from $(3)$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d Y} {\mathrm d \psi}
+      | r = \dfrac {\mathrm d y} {\mathrm d \psi} - \rho \sin \psi + \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi
+      | c = 
+}}
+{{eqn | r = \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi
+      | c = from $(4)$
+}}
+{{end-eqn}}
+By assumption, $\dfrac {\mathrm d \rho} {\mathrm d \psi} \ne 0$ on $C$.
+Hence we have:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d Y} {\mathrm d X}
+      | r = -\dfrac {\cos \psi} {\sin \psi}
+      | c = 
+}}
+{{eqn | r = -\dfrac 1 {\tan \psi}
+      | c = 
+}}
+{{end-eqn}}
+We have [[Slope of Normal is Minus Reciprocal of Tangent]].
+Thus the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $E$ equals the [[Definition:Slope|slope]] of the [[Definition:Normal to Curve|normal]].
+The result follows.
+Note the case when $\dfrac {\mathrm d \rho} {\mathrm d \psi} \ne 0$ on $C$.
+In this case $C$ is a [[Definition:Circle|circle]].
+By [[Evolute of Circle is its Center]] its [[Definition:Evolute|evolute]] is a single [[Definition:Point|point]] and so has no [[Definition:Tangent to Curve|tangent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Arc of Evolute equals Difference in Radii of Curvature}
+Tags: Evolutes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] defined by a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice]] [[Definition:Differentiable Real Function|differentiable]].
+Let the [[Definition:Curvature|curvature]] of $C$ be non-[[Definition:Constant|constant]].
+The [[Definition:Arc Length|length of arc]] of the [[Definition:Evolute|evolute]] $E$ of $C$ between any two [[Definition:Point|points]] $Q_1$ and $Q_2$ of $C$ is equal to the difference between the [[Definition:Radius of Curvature|radii of curvature]] at the corresponding [[Definition:Point|points]] $P_1$ and $P_2$ of $C$.
+\end{theorem}
+
+\begin{proof}
+:[[File:CenterOfCurvature.png|400px]]
+Let $P = \left({x, y}\right)$ be a general [[Definition:Point|point]] on $C$.
+Let $Q = \left({X, Y}\right)$ be the [[Definition:Center of Curvature|center of curvature]] of $C$ at $P$.
+From the above diagram:
+:$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\
+Y - y = \pm \rho \cos \psi \end{cases}$
+where:
+:$\rho$ is the [[Definition:Radius of Curvature|radius of curvature]] of $C$ at $P$
+:$\psi$ is the [[Definition:Angle|angle]] between the [[Definition:Tangent to Curve|tangent]] to $C$ at $P$ and the [[Definition:X-Axis|$x$-axis]].
+Whether the sign is plus or minus depends on whether the curve is [[Definition:Convex Real Function|convex]] or [[Definition:Concave Real Function|concave]].
+For simplicity, let it be assumed that the [[Definition:Curvature|curvature]] $k$ at each point under consideration on $C$ is [[Definition:Positive Real Number|positive]].
+The case for $k < 0$ can then be treated similarly.
+Thus we have $k > 0$ and so $(1)$ can be written:
+:$(2): \quad \begin{cases} X = x - \rho \sin \psi \\
+Y = y +\rho \cos \psi \end{cases}$
+By definition of [[Definition:Curvature|curvature]]:
+:$k = \dfrac {\mathrm d \psi} {\mathrm d s}$
+and:
+:$\rho = \dfrac 1 k = \dfrac {\mathrm d s} {\mathrm d \psi}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \rho \sin \psi
+      | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d y} {\mathrm d s}
+      | c = 
+}}
+{{eqn | n = 3
+      | r = \dfrac {\mathrm d y} {\mathrm d \psi}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \rho \cos \psi
+      | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d x} {\mathrm d s}
+      | c = 
+}}
+{{eqn | n = 4
+      | r = \dfrac {\mathrm d x} {\mathrm d \psi}
+      | c = 
+}}
+{{end-eqn}}
+[[Definition:Differentiation|Differentiating]] $(2)$ {{WRT|Differentiation}} $\psi$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d X} {\mathrm d \psi}
+      | r = \dfrac {\mathrm d x} {\mathrm d \psi} - \rho \cos \psi - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi
+      | c = 
+}}
+{{eqn | n = 5
+      | r = - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi
+      | c = from $(3)$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d Y} {\mathrm d \psi}
+      | r = \dfrac {\mathrm d y} {\mathrm d \psi} - \rho \sin \psi + \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi
+      | c = 
+}}
+{{eqn | n = 6
+      | r = \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi
+      | c = from $(4)$
+}}
+{{end-eqn}}
+Let $S$ be the [[Definition:Arc Length|length of arc]] of the [[Definition:Evolute|evolute]] $E$ from a fixed [[Definition:Point|point]] $Q_0$ on $E$ to the variable [[Definition:Point|point]] $Q$ corresponding to $P$.
+We have that:
+:$\mathrm d S^2 = \mathrm d X^2 + \mathrm d Y^2$
+and so:
+{{begin-eqn}}
+{{eqn | l = \left({\dfrac {\mathrm d S} {\mathrm d \psi} }\right)^2
+      | r = \left({\dfrac {\mathrm d X} {\mathrm d \psi} }\right)^2 + \left({\dfrac {\mathrm d Y} {\mathrm d \psi} }\right)^2
+      | c = 
+}}
+{{eqn | r = \left({- \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi}\right)^2 + \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi}\right)^2
+      | c = from $(5)$ and $(6)$
+}}
+{{eqn | r = \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} }\right)^2 \left({\sin^2 \psi + \cos^2 \psi}\right)
+      | c = 
+}}
+{{eqn | r = \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} }\right)^2 
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Choosing the direction of $S$ so that $S$ and $\rho$ increase together, this tells us:
+:$\dfrac {\mathrm d S} {\mathrm d \psi} = \dfrac {\mathrm d \rho} {\mathrm d \psi}$
+[[Definition:Integration|Integrating]] {{WRT|Integration}} $\psi$ gives:
+:$S = \rho + c$
+where $c$ is a [[Definition:Constant|constant]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Curve is Involute of Evolute}
+Tags: Evolutes, Involutes
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] defined by a [[Definition:Real Function|real function]] which is [[Definition:Second Derivative|twice]] [[Definition:Differentiable Real Function|differentiable]].
+Let the [[Definition:Curvature|curvature]] of $C$ be non-[[Definition:Constant|constant]].
+Let $E$ be the [[Definition:Evolute|evolute]] $C$.
+Then the [[Definition:Involute|involute]] of $E$ is $C$.
+\end{theorem}
+
+\begin{proof}
+From [[Length of Arc of Evolute equals Difference in Radii of Curvature]]:
+:the [[Definition:Arc Length|length of arc]] of the [[Definition:Evolute|evolute]] $E$ of $C$ between any two [[Definition:Point|points]] $Q_1$ and $Q_2$ of $C$ is equal to the difference between the [[Definition:Radius of Curvature|radii of curvature]] at the corresponding [[Definition:Point|points]] $P_1$ and $P_2$ of $C$.
+Thus $C$ exhibits precisely the property of the [[Definition:Involute|involute]] of $E$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Evolute of Cycloid is Cycloid}
+Tags: Evolutes, Cycloids
+
+\begin{theorem}
+The [[Definition:Evolute|evolute]] of a [[Definition:Cycloid|cycloid]] is another [[Definition:Cycloid|cycloid]].
+\end{theorem}
+
+\begin{proof}
+Let $C$ be the [[Definition:Cycloid|cycloid]] defined by the equations:
+:$\begin {cases}
+x = a \paren {\theta - \sin \theta} \\
+y = a \paren {1 - \cos \theta}
+\end {cases}$
+From [[Parametric Equations for Evolute/Formulation 2|Parametric Equations for Evolute: Formulation 2]]:
+:$(1): \quad \begin{cases}
+X = x - \dfrac {y' \paren {x'^2 + y'^2} } {x' y'' - y' x''} \\
+Y = y + \dfrac {x' \paren {x'^2 + y'^2} } {x' y'' - y' x''}
+\end{cases}$
+where:
+:$\tuple {x, y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on $C$
+:$\tuple {X, Y}$ denotes the [[Definition:Cartesian Coordinate System|Cartesian coordinates]] of a general [[Definition:Point|point]] on the [[Definition:Evolute|evolute]] of $C$
+:$x'$ and $x''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $x$ {{WRT|Differentiation}} $\theta$
+:$y'$ and $y''$ denote the [[Definition:Derivative|derivative]] and [[Definition:Second Derivative|second derivative]] respectively of $y$ {{WRT|Differentiation}} $\theta$.
+Thus we have:
+{{begin-eqn}}
+{{eqn | l = x'
+      | r = a \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | l = x''
+      | r = a \sin \theta
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = y'
+      | r = a \sin \theta
+      | c = 
+}}
+{{eqn | l = y''
+      | r = a \cos \theta
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \dfrac {x'^2 + y'^2} {x' y'' - y' x''}
+      | r = \dfrac {a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta} {a \paren {1 - \cos \theta} a \cos \theta - a \sin \theta \, a \sin \theta}
+      | c = substituting for $x'$, $x''$, $y'$, $y''$
+}}
+{{eqn | r = \dfrac {\paren {1 - 2 \cos \theta + \cos^2 \theta} + \sin^2 \theta} {\cos \theta - \cos^2 \theta - \sin^2 \theta}
+      | c = multiplying out, cancelling out $a^2$
+}}
+{{eqn | r = \dfrac {2 - 2 \cos \theta} {\cos \theta - 1}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | n = 2
+      | r = -2
+      | c = dividing [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $1 - \cos \theta$
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = X
+      | r = a \paren {\theta - \sin \theta} - a \sin \theta \paren {-2}
+      | c = substituting for $x$ and $y'$ and from $(2)$ in $(1)$
+}}
+{{eqn | r = a \paren {\theta + \sin \theta}
+      | c = simplifying
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = Y
+      | r = a \paren {1 - \cos \theta} + a \paren {1 - \cos \theta} \paren {-2}
+      | c = substituting for $y$ and $x'$ and from $(2)$ in $(1)$
+}}
+{{eqn | r = -a \paren {1 - \cos \theta}
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Cycloid|cycloid]] $C$ (blue) and its [[Definition:Evolute|evolute]] (red) are illustrated below:
+[[File:CycloidAndEvolute.png|600px]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Body behaves as Particle under Gravitation}
+Tags: Gravity
+
+\begin{theorem}
+A [[Definition:Body|body]] of [[Definition:Finite|finite]] [[Definition:Volume|volume]] under the [[Definition:Force of Gravity|force of gravity]] behaves identically to a [[Definition:Point Mass|point mass]] under the same [[Definition:Force of Gravity|force]].
+\end{theorem}<|endoftext|>
+\section{Meet of Directed Subsets is Directed}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $D_1, D_2$ be [[Definition:Directed Subset|directed subset]] of $S$.
+Then
+:$\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ is [[Definition:Directed Subset|directed subset]] of $S$
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$.
+Then
+:$\exists x \in D_1, y \in D_2: a = x \wedge y$
+and
+:$\exists z \in D_1, t \in D_2: b = z \wedge t$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists g \in D_1: x \preceq g \land z \preceq g$
+and
+:$\exists h \in D_2: y \preceq h \land t \preceq h$
+By [[Meet Semilattice is Ordered Structure]]:
+:$x \wedge y \preceq g \wedge h$ and $z \wedge t \preceq g \wedge h$
+:$g \wedge h \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$
+Thus
+:$\exists c \in \left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}: a \preceq c \land b \preceq c$
+Hence by definition
+:$\left\{ {x \wedge y: x \in D_1, y \in D_2}\right\}$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Directed Subset under Increasing Mapping is Directed}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $f: S \to T$ be an [[Definition:Increasing Mapping|increasing mapping]].
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$.
+Then
+:$f^\to \left({D}\right)$ is also a [[Definition:Directed Subset|directed subset]] of $T$
+where
+:$f^\to \left({D}\right)$ denotes the [[Definition:Image of Subset under Mapping|image]] of $D$ under $f$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in f^\to\left({D}\right)$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists a \in D: x = f \left({a}\right)$
+and
+:$\exists b \in D: y = f \left({b}\right)$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists c \in D: a \preceq c \land b \preceq c$
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$f\left({c}\right) \in f^\to\left({D}\right)$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$x \precsim f\left({c}\right)$ and $y \precsim f\left({c}\right)$
+Thus by definition
+:$f^\to\left({D}\right)$ is a [[Definition:Directed Subset|directed subset]] of $T$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mass of Sun from Gravitational Constant}
+Tags: Celestial Mechanics
+
+\begin{theorem}
+Let the [[Definition:Gravitational Constant|gravitational constant]] be known.
+Let the [[Definition:Mean|mean]] [[Definition:Distance (Linear Measure)|distance]] from the [[Definition:Earth|Earth]] to the [[Definition:Sun|sun]] be known.
+Then it is possible to calculate the [[Definition:Mass|mass]] of the [[Definition:Sun|sun]].
+\end{theorem}
+
+\begin{proof}
+From [[Kepler's Third Law of Planetary Motion]]:
+:$T^2 = \left({\dfrac {4 \pi^2} {G M} }\right) a^3$
+where:
+:$T$ is the [[Definition:Orbital Period|orbital period]] of the [[Definition:Planet|planet]] in question (in this case, the [[Definition:Earth|Earth]])
+:$a$ is the [[Definition:Distance (Linear Measure)|distance]] from the [[Definition:Planet|planet]] (in this case, the [[Definition:Earth|Earth]]) to the [[Definition:Sun|sun]]
+:$M$ is the [[Definition:Mass|mass]] of the [[Definition:Sun|sun]]
+:$G$ is the [[Definition:Gravitational Constant|gravitational constant]]
+In [[Definition:MKS|MKS units]]:
+:$T = 60 \times 60 \times 24 \times 365.24219 \, \mathrm s$ by definition of [[Definition:Year|year]]
+:$a = 149 \, 597 \, 870 \, 700 \, \mathrm m$ by definition of [[Definition:Astronomical Unit|astronomical unit]]
+:$G = 6.674 \times 10^{-11} \, \mathrm N \, \mathrm m^2 \, \mathrm{kg}^{-2}$ by measurement, as [[Definition:By Hypothesis|by hypothesis]].
+:$M$ will be the result in [[Definition:Kilogram|kilograms]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = M
+      | r = \dfrac {4 \pi^2 a^3} {T^2 G}
+      | c = 
+}}
+{{eqn | r = \frac {4 \times \left({3.14159}\right)^2 \times \left({149.60 \times 10^9}\right)^3} {\left({31.557 \times 10^6}\right) \times 6.674 \times 10^{-11} }
+      | c = 
+}}
+{{end-eqn}}
+The calculation has been left as an exercise for anyone who has the patience.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets}
+Tags: Up-Complete Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Let $\struct {S \times S, \precsim}$ be the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\struct {S, \preceq}$ and $\struct {S, \preceq}$.
+Let $f: S \times S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall s, t \in S: \map f {s, t} = s \wedge t$
+and
+:$f$ [[Definition:Mapping Preserves Directed Supremum|preserves directed suprema]].
+Let $D_1, D_2$ be [[Definition:Directed Subset|directed subsets]] of $S$.
+Then
+:$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {x \wedge y: x \in D_1, y \in D_2}$
+\end{theorem}
+
+\begin{proof}
+By [[Up-Complete Product]]:
+:$\struct {S \times S, \precsim}$ is [[Definition:Up-Complete|up-complete]].
+By [[Up-Complete Product/Lemma 1]]:
+:$D_1 \times D_2$ is [[Definition:Directed Subset|directed subsets]] of $S \times S$
+By definition of [[Definition:Mapping Preserves Directed Supremum|mapping preserves directed suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] of $D_1 \times D_2$
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D_1 \times D_2$ admits a [[Definition:Supremum of Set|supremum]]
+and
+$D_1$ and $D_2$ admit [[Definition:Supremum|suprema]].
+Thus
+{{begin-eqn}}
+{{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}
+      | r = \map f {\sup D_1, \sup D_2}
+      | c = Definition of $f$
+}}
+{{eqn | r = \map f {\map \sup {D_1 \times D_2} }
+      | c = [[Supremum of Cartesian Product]]
+}}
+{{eqn | r = \map \sup {\map {f^\to} {D_1 \times D_2} }
+      | c = [[Definition:Mapping Preserves Supremum/Subset|Mapping Preserves Supremum]]
+}}
+{{eqn | r = \sup \set {\map f {x, y}: \tuple {x, y} \in D_1 \times D_2}
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{eqn | r = \sup \set {x \wedge y: x \in D_1, y \in D_2}
+      | c = Definition of $f$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving}
+Tags: Up-Complete Semilattices
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Let $f: \map {\it Ids} {\mathscr S} \to S$ be a [[Definition:Mapping|mapping]] such that:
+:$\forall I \in \map {\it Ids} {\mathscr S}: \map f I = \sup_{\mathscr S} I$
+where 
+:$\map {\it Ids} {\mathscr S}$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$
+Let
+:$\forall I_1, I_2 \in \map {\it Ids} {\mathscr S}: \paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i \wedge j: i \in I_1, j \in I_2}$
+Then:
+$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]] as a [[Definition:Mapping|mapping]] from $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$ into $\mathscr S$
+\end{theorem}
+
+\begin{proof}
+Let $I, J \in \map {\it Ids} {\mathscr S}$ such that
+:$\set {I, J}$ admits an [[Definition:Infimum of Set|infimum]] in $\struct {\map {\it Ids} {\mathscr S}, \subseteq}$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\map {f^\to} {\set {I, J} } = \set {\map f I, \map f J}$
+Thus by definition of [[Definition:Meet Semilattice|meet semilattice]]:
+:$\map {f^\to} {\set {I, J} }$ admits an [[Definition:Infimum of Set|infimum]] in $\mathscr S$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map \inf {\map {f^\to} {\set {I, J} } }
+      | r = \map f I \wedge \map f J
+      | c = {{Defof|Meet (Order Theory)}}
+}}
+{{eqn | r = \paren {\sup I} \wedge \paren {\sup J}
+      | c = Definition of $f$
+}}
+{{eqn | r = \sup \set {x \wedge y: x \in I, y \in J}
+      | c = Assumption
+}}
+{{eqn | r = \map \sup {I \wedge J}
+      | c = [[Meet in Set of Ideals]]
+}}
+{{eqn | r = \map f {I \wedge J}
+      | c = definition of $f$
+}}
+{{eqn | r = \map f {\inf \set {I, J} }
+      | c = {{Defof|Meet (Order Theory)}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quaternion Multplication is not Commutative}
+Tags: Quaternions
+
+\begin{theorem}
+The operation of [[Definition:Quaternion Multiplication|multplication]] on the [[Definition:Quaternion|quaternions]] $H$ is not [[Definition:Commutative Operation|commutative]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Quaternion Multiplication|multplication]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf i \times \mathbf j
+      | r = \mathbf k
+      | c = 
+}}
+{{eqn | l = \mathbf j \times \mathbf i
+      | r = -\mathbf k
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Norm of Quaternion is not Norm}
+Tags: Quaternions, Norm Theory
+
+\begin{theorem}
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a [[Definition:Quaternion|quaternion]].
+Let $\overline {\mathbf x}$ be the [[Definition:Conjugate Quaternion|conjugate]] of $\mathbf x$.
+The [[Definition:Field Norm of Quaternion|field norm]] of $\mathbf x$:
+:$\map n {\mathbf x} := \size {\mathbf x \overline {\mathbf x} }$
+is not a [[Definition:Norm on Division Ring|norm]] in the [[Definition:Abstract Algebra|abstract algebraic]] context of a [[Definition:Division Ring|division ring]].
+\end{theorem}
+
+\begin{proof}
+Each of the [[Definition:Norm Axioms|norm axioms]] is examined in turn:
+=== $N1$: Positive Definiteness ===
+This is proved in [[Field Norm of Quaternion is Positive Definite]].
+{{qed|lemma}}
+=== $N2$: Multiplicativity ===
+This is proved in [[Field Norm of Quaternion is Multiplicative]].
+{{qed|lemma}}
+=== $N3$: Triangle Inequality ===
+For example:
+:$\map n {1 + 1} = 4 > 2 = \map n 1  + \map n 1$
+and so $N3$ is not [[Definition:Satisfaction|satisfied]].
+{{qed|lemma}}
+Not all the [[Definition:Norm Axioms|norm axioms]] are fulfilled.
+Hence the result.
+{{qed}}
+[[Category:Quaternions]]
+[[Category:Norm Theory]]
+q26q9bmf8pl0w3600m2bv534c6iiqun
+\end{proof}<|endoftext|>
+\section{Field Norm of Quaternion is Positive Definite}
+Tags: Quaternions
+
+\begin{theorem}
+Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a [[Definition:Quaternion|quaternion]].
+Let $\overline {\mathbf x}$ be the [[Definition:Conjugate Quaternion|conjugate]] of $\mathbf x$.
+The [[Definition:Field Norm of Quaternion|field norm]] of $\mathbf x$:
+:$\map n {\mathbf x} := \cmod {\mathbf x \overline {\mathbf x} }$
+is [[Definition:Positive Definite (Ring)|positive definite]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map n {\mathbf x}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \cmod {\mathbf x \overline {\mathbf x} }
+      | r = 0
+      | c ={{Defof|Field Norm of Quaternion}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a^2 + b^2 + c^2 + d^2
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a = 0, b = 0, c = 0, d = 0
+      | o = 
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k
+      | r = \mathbf 0
+      | c = 
+}}
+{{end-eqn}}
+Hence $n$ is [[Definition:Positive Definite (Ring)|positive definite]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field Norm of Quaternion is Multiplicative}
+Tags: Quaternions
+
+\begin{theorem}
+Let $\mathbf x$ be a [[Definition:Quaternion|quaternion]].
+Let $\overline {\mathbf x}$ be the [[Definition:Conjugate Quaternion|conjugate]] of $\mathbf x$.
+The [[Definition:Field Norm of Quaternion|field norm]] of $\mathbf x$:
+:$\map n {\mathbf x} := \size {\mathbf x \overline {\mathbf x} }$
+is a [[Definition:Multiplicative Arithmetic Function|multiplicative function]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map n {\mathbf x \mathbf y}
+      | r = \mathbf x \, \mathbf y \ \overline {\mathbf x \, \mathbf y}
+      | c = {{Defof|Field Norm of Quaternion}}
+}}
+{{eqn | r = \mathbf x \, \mathbf y \paren {\overline {\mathbf y} \, \overline {\mathbf x} }
+      | c = [[Product of Quaternion Conjugates]]
+}}
+{{eqn | r = \mathbf x \paren {\mathbf y \paren {\overline {\mathbf y} \, \overline {\mathbf x} } }
+      | c = 
+}}
+{{eqn | r = \mathbf x \paren {\paren {\mathbf y \, \overline {\mathbf y} } \overline {\mathbf x} }
+      | c = 
+}}
+{{eqn | r = \mathbf x \paren {\map n {\mathbf y} \overline {\mathbf x} }
+      | c = 
+}}
+{{eqn | r = \map n {\mathbf y} \paren {\mathbf x \, \overline {\mathbf x} }
+      | c = 
+}}
+{{eqn | r = \map n {\mathbf y} \, \map n {\mathbf x}
+      | c = 
+}}
+{{eqn | r = \map n {\mathbf x} \, \map n {\mathbf y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Oscillation at Point for Real Functions}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Real Number|real]] [[Definition:Set|sets]].
+Let $f: X \to Y$ be a [[Definition:Real Function|real function]].
+Let $x \in X$.
+{{TFAE|def = Oscillation/Real Space/Oscillation at Point|view = Oscillation at a Point}}
+\end{theorem}
+
+\begin{proof}
+=== Definitions 1 and 2 are equivalent ===
+We reformulate Definition 1 into Definition 1' by:
+:substituting the definition of $\omega_f \left({U \cap X}\right)$ into the definition of $\omega_f \left({x}\right)$
+Definition 1':
+:$\displaystyle \omega_f \left({x}\right) := \inf_{U \mathop \in \mathcal N_x} \left({\sup_{y, z \mathop \in U \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+We reformulate Definition 2 into Definition 2' by:
+:substituting the definition of $\omega_f \left({\left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X}\right)$ into the definition of $\omega_f \left({x}\right)$
+:changing set condition into sup tag
+Definition 2':
+:$\displaystyle \omega_f \left({x}\right) := \inf_{\epsilon \mathop \in \R_{>0}} \left({\sup_{y, z \mathop \in \left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+The theorem text of [[Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood)]] also contains two definitions of [[Definition:Oscillation/Real Space/Oscillation at Point|Oscillation at a Point]] for a [[Definition:Real Function|real function]] $f$.
+We call them definitions a and b.
+Definition a:
+Let $\mathcal N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhoods]] of $x$.
+:$\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in \mathcal N_x}\right\}$
+where:
+:$\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap X}\right\}$
+Definition b:
+:$\omega^E_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$
+where:
+:$E_x$ is the [[Definition:Set|set]] of [[Definition:Epsilon-Neighborhood (Real Number Line)|$\epsilon$-neighborhoods]] of $x$
+We reformulate Definition a into Definition a' by:
+:substituting the definition of $\omega_f \left({I}\right)$ into the definition of $\omega_f \left({x}\right)$
+:changing set conditions into sup tags
+:renaming parameter set $I$ to $U$
+Definition a':
+:$\displaystyle \omega_f \left({x}\right) = \inf_{U \mathop \in \mathcal N_x} \left({\sup_{y, z \mathop \in U \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+We reformulate Definition b into Definition b' by:
+:substituting the definition of $\omega_f \left({I}\right)$ into the definition of $\omega^E_f \left({x}\right)$
+:changing set conditions into sup tags
+:using $\epsilon$ as parameter instead of $I$
+Definition b':
+:$\displaystyle \omega^E_f \left({x}\right) = \inf_{\epsilon \mathop \in \R_{>0}} \left({\sup_{y, z \mathop \in \left({x - \epsilon \,.\,.\, x + \epsilon}\right) \cap X} \left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert}\right)$
+We observe that definitions 1' and a' are the same, so they are equivalent.
+Therefore:
+:definitions 1 and a are equivalent
+Also, definitions 2' and b' are the same, so they are equivalent.
+Therefore:
+:definitions 2 and b are equivalent
+We have:
+:definitions 1 and a are equivalent
+:definitions a and b are equivalent by [[Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood)]]
+:definitions b and 2 are equivalent
+Therefore:
+:definitions 1 and 2 are equivalent
+{{qed|lemma}}
+=== Definitions 1 and 3 are equivalent ===
+We reformulate Definition 1 into Definition 1' by:
+:substituting the definition of $\omega_f \left({U \cap X}\right)$ into the definition of $\omega_f \left({x}\right)$
+Definition 1':
+:$\displaystyle \omega_f \left({x}\right) := \inf_{U \mathop \in \mathcal N_x} \left({\sup_{y, z \mathop \in U \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+We reformulate Definition 3 into Definition 3' by:
+:substituting the definition of $\omega_f \left({\left({x - h \,.\,.\, x + h}\right) \cap X}\right)$ into the definition of $\omega_f \left({x}\right)$
+Definition 3':
+:$\displaystyle \omega_f \left({x}\right) := \lim_{h \mathop \to 0^+} \left({\sup_{y, z \mathop \in \left({x - h \,.\,.\, x + h}\right) \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+The theorem text of [[Oscillation at Point (Infimum) equals Oscillation at Point (Limit)]] also contains two definitions of [[Definition:Oscillation/Real Space/Oscillation at Point|Oscillation at a Point]] for a [[Definition:Real Function|real function]] $f$.
+We call them definitions a and b.
+Definition a:
+Let $\mathcal N_x$ be the [[Definition:Set|set]] of [[Definition:Neighborhood (Real Analysis)/Open Subset|open subset neighborhoods]] of $x$.
+:$\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in \mathcal N_x}\right\}$
+where:
+:$\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap X}\right\}$
+Definition b:
+:$\omega_f \left({x}\right) = \displaystyle \lim_{h \mathop \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$
+We reformulate Definition a into Definition a' by:
+:substituting the definition of $\omega_f \left({I}\right)$ into the definition of $\omega_f \left({x}\right)$
+:changing set conditions into sup tags
+:renaming parameter set $I$ to $U$
+Definition a':
+:$\displaystyle \omega_f \left({x}\right) = \inf_{U \mathop \in \mathcal N_x} \left({\sup_{y, z \mathop \in U \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+We reformulate Definition b into Definition b' by:
+:substituting the definition of $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ (by using the definition of $\omega_f \left({I}\right)$) into the definition of $\omega_f \left({x}\right)$
+:changing set condition into sup tag
+Definition b':
+:$\displaystyle \omega_f \left({x}\right) = \lim_{h \mathop \to 0^+} \left({\sup_{y, z \mathop \in \left({x - h \,.\,.\, x + h}\right) \cap X} \left\lvert{f \left({y}\right) - f \left({z}\right)}\right\rvert}\right)$
+We observe that definitions 1' and a' are the same, so they are equivalent.
+Therefore:
+:definitions 1 and a are equivalent
+Also, definitions 3' and b' are the same, so they are equivalent.
+Therefore:
+:definitions 3 and b are equivalent
+We have:
+:definitions 1 and a are equivalent
+:definitions a and b are equivalent by [[Oscillation at Point (Infimum) equals Oscillation at Point (Limit)]]
+:definitions b and 3 are equivalent
+Therefore:
+:definitions 1 and 3 are equivalent
+{{qed|lemma}}
+=== Definitions 2 and 3 are equivalent ===
+From:
+:Definitions 2 and 1 are equivalent
+:Definitions 1 and 3 are equivalent
+follows:
+:Definitions 2 and 3 are equivalent
+{{qed|lemma}}
+Thus all definitions listed in the theorem text are equivalent.
+{{qed}}
+[[Category:Real Analysis]]
+0lnoe98f2nec430ufqyzytt4ue8dcan
+\end{proof}<|endoftext|>
+\section{Complex Numbers cannot be Extended to Algebra in Three Dimensions with Real Scalars}
+Tags: Algebras
+
+\begin{theorem}
+It is not possible to extend the [[Definition:Complex Number|complex numbers]] to an [[Definition:Algebra over Field|algebra]] of $3$ [[Definition:Dimension (Linear Algebra)|dimensions]] with [[Definition:Real Number|real]] [[Definition:Scalar (Vector Space)|scalars]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $\set {1, i, j}$ forms a [[Definition:Basis (Linear Algebra)|basis]] for an [[Definition:Algebra over Field|algebra]] of $3$ [[Definition:Dimension (Linear Algebra)|dimensions]] with [[Definition:Real Number|real]] [[Definition:Scalar (Vector Space)|scalars]].
+Let $1$ and $i$ have their usual properties as they do as [[Definition:Complex Number|complex numbers]]:
+:$\forall a: 1 a = a 1 = a$
+:$i \cdot i = -1$
+Then:
+:$i j = a_1 + a_2 i + a_3 j$
+for some $a_1, a_2, a_3 \in \R$.
+Multiplying through by $i$:
+:$(1): \quad i \paren {i j} = \paren {i i} j = -j$
+and:
+{{begin-eqn}}
+{{eqn | l = i \paren {a_1 + a_2 i + a_3 j}
+      | r = a_1 i - a_2 + a_3 i j
+      | c = 
+}}
+{{eqn | r = a_1 i - a_2 + a_3 \paren {a_1 + a_2 i + a_3 j}
+      | c = 
+}}
+{{eqn | r = a_1 i - a_2 + a_1 a_3 + a_2 a_3 i + {a_3}^2 j
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = \paren {a_1 a_3 - a_2} + \paren {a_1 + a_2 a_3} i + \paren { {a_3}^2 + 1} j
+      | c = from $(1)$
+}}
+{{end-eqn}}
+But this implies that ${a_3}^2 = -1$, which contradicts our supposition that $a_3 \in \R$.
+Hence the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Cartesian Product}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $\left({S_1 \times S_2, \precsim}\right)$ be the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\left({S_1, \preceq_1}\right)$ and $\left({S_2, \preceq_2}\right)$.
+Let $X_1$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S_1$, $X_2$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S_2$ such that
+:$X_1$ and $X_2$ admit [[Definition:Supremum of Set|suprema]].
+Then
+:$X_1 \times X_2$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$\sup \left({X_1 \times X_2}\right) = \left({\sup X_1, \sup X_2}\right)$
+\end{theorem}
+
+\begin{proof}
+We will prove that
+:$\left({\sup X_1, \sup X_2}\right)$ is [[Definition:Upper Bound of Set|upper bound]] for $X_1 \times X_2$
+Let $\left({a, b}\right) \in X_1 \times X_2$.
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+:$a \in X_1$ and $b \in X_2$
+By definitions of [[Definition:Supremum of Set|supremum]] and [[Definition:Upper Bound of Set|upper bound]]:
+:$a \preceq_1 \sup X_1$ and $b \preceq_2 \sup X_2$
+Thus by definition [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\left({a, b}\right) \precsim \left({\sup X_1, \sup X_2}\right)$
+{{qed|lemma}}
+We will prove that
+:$\forall \left({a, b}\right) \in S1 \times S_2: \left({a, b}\right)$ is [[Definition:Upper Bound of Set|upper bound]] for $X_1 \times X_2 \implies \left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$
+Let $\left({a, b}\right) \in S1 \times S_2$ such that
+:$\left({a, b}\right)$ is [[Definition:Upper Bound of Set|upper bound]] for $X_1 \times X_2$
+We will prove as sublemma that
+:$a$ is [[Definition:Upper Bound of Set|upper bound]] for $X_1$
+Let $c \in X_1$.
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists d: d \in X_2$
+By definition of [[Definition:Cartesian Product|Cartesian product]]:
+:$\left({c, d}\right) in X_1 \times X_2$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\left({c, d}\right) \precsim \left({a, b}\right)$
+Thus be definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$c \preceq_1 a$
+This ends the proof of sublemma.
+Analogically we have that
+:$b$ is [[Definition:Upper Bound of Set|upper bound]] for $X_2$
+By definition of [[Definition:Supremum of Set|supremum]];
+:$\sup X_1 \preceq_1 a$ and $\sup X_2 \preceq_2 b$
+Thus by definition of [[Definition:Cartesian Product of Ordered Sets|Cartesian product of ordered sets]]:
+:$\left({\sup X_1, \sup X_2}\right) \precsim \left({a, b}\right)$
+{{qed|lemma}}
+Thus by definition
+:$X_1 \times X_2$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$\sup \left({X_1 \times X_2}\right) = \left({\sup X_1, \sup X_2}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vectors in Three Dimensional Space with Cross Product forms Lie Algebra}
+Tags: Lie Algebras, Vector Cross Product
+
+\begin{theorem}
+Let $S$ be the [[Definition:Set|set]] of [[Definition:Vector (Linear Algebra)|vectors]] in $3$ [[Definition:Dimension (Linear Algebra)|dimensional]] [[Definition:Euclidean Space|Euclidean space]].
+Let $\times$ denote the [[Definition:Vector Cross Product|vector cross product]] on $S$.
+Then $\struct {S, \times}$ is a [[Definition:Lie Algebra|Lie algebra]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Lie Algebra|Lie algebra]], it suffices to prove two properties:
+:$(1): \forall a \in S: a \times a = 0$
+:$(2): \forall a, b, c \in S: a \times \paren {b \times c} + b \times \paren {c \times a} + c \times \paren {a \times b} = 0$
+=== Proof of $(1)$ ===
+[[Cross Product of Vector with Itself is Zero]]
+{{Qed|lemma}}
+=== Proof of $(2)$ ===
+[[Vector Cross Product satisfies Jacobi Identity]]
+{{Qed|lemma}}
+Both properties hold, and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nonassociative Division Algebra with Real Scalars has Dimension of Power of 2}
+Tags: Algebras
+
+\begin{theorem}
+Let $A$ be a [[Definition:Nonassociative Algebra|nonassociative]] [[Definition:Division Algebra|division algebra]] with [[Definition:Real Number|real]] [[Definition:Scalar (Vector Space)|scalars]].
+Then the [[Definition:Dimension (Linear Algebra)|dimension]] of $A$ is a [[Definition:Power|power]] of $2$.
+\end{theorem}<|endoftext|>
+\section{Bott-Milnor-Kervaire 1,2,4,8 Theorem}
+Tags: Algebras
+
+\begin{theorem}
+Let $A$ be a [[Definition:Division Algebra|division algebra]] with [[Definition:Real Number|real]] [[Definition:Scalar (Vector Space)|scalars]].
+Then the [[Definition:Dimension (Linear Algebra)|dimension]] of $A$ is either:
+:$1$: the [[Definition:Real Number|real numbers]] $\R$
+:$2$: the [[Definition:Complex Number|complex numbers]] $\C$
+:$4$: the [[Definition:Quaternion|quaternions]] $\Bbb H$
+or:
+:$8$: the [[Definition:Octonion|octonions]] $\Bbb O$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Raoul Bott|name2 = John Willard Milnor|name3 = Michel André Kervaire|cat = Bott|cat2 = Milnor|cat3 = Kervaire}}
+\end{proof}<|endoftext|>
+\section{Montel's Theorem}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $U \subseteq \C$ be an [[Definition:Open Set (Complex Analysis)|open subset]] of the [[Definition:Complex Number|complex numbers]].
+Let $\mathcal H \left({U}\right)$ be the space of [[Definition:Holomorphic Function|holomorphic mappings]] on $U$.
+Then a [[Definition:Indexed Family|family]] of [[Definition:Mapping|mappings]] $\mathcal F \subseteq \mathcal H \left({U}\right)$ is [[Definition:Normal Family|normal]] {{iff}} $\mathcal F$ is [[Definition:Locally Bounded Family of Mappings|locally bounded]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}} 
+{{Namedfor|Paul Antoine Aristide Montel|cat = Montel}}
+\end{proof}<|endoftext|>
+\section{Vitali's Convergence Theorem}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $U$ be an [[Definition: Open Set of Metric Space | open]], [[Definition: Connected | connected]] subset of $\C$.
+Let $S \subseteq U$ contain a [[Definition:Limit Point (Complex Analysis)|limit point]] $\sigma$.
+Let $\sequence {f_n}_{n \mathop \in \N}$ be a [[Definition: Normal Family | normal family]] of [[Definition:Analytic Complex Function|holomorphic]] [[Definition:Mapping|mappings]] $f_n : U \to \C$.
+Let $\sequence {f_n}_{n \mathop \in \N}$ [[Definition:Pointwise Convergence | converge]] to some [[Definition:Holomorphic Complex Function|holomorphic mapping]] $f : U \to \C$ at $\sigma$.
+Then $f_n$ [[Definition:Uniform Convergence/Metric Space|converges uniformly]] to $f$ on all [[Definition:Compact Metric Space|compact]] [[Definition:Subset|subsets]] of $U$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} there exists some compact subset $K$ of $U$ such that $f_n$ does not converge uniformly to $f$ on $K$.
+Consider $K^* := K \cup \set \sigma$.
+From [[Subsets Inherit Uniform Convergence]], $f_n$ does not converge uniformly to $f$ on $K^*$.
+From [[Uniformly Convergent iff Difference Under Supremum Norm Vanishes]], the above is [[Definition:Logical Equivalence|equivalent]] to:
+:$\exists \epsilon > 0 : \forall N \in \N : \exists n \ge N : \norm {f_n - f}_{K^*} \ge \epsilon$
+where $\norm {\cdot}_{K^*}$ denotes the [[Definition:Supremum Norm|supremum norm]] over $K^*$.
+From [[Finite Union of Compact Sets is Compact]], $K^{*}$ is [[Definition:Compact Metric Space|compact]].
+Since $\sequence {f_n}$ is a [[Definition: Normal Family | normal family]], there is some [[Definition:Subsequence|subsequence]] $\sequence {f_{n_r} }$ of $\sequence {f_n}$ and some mapping $g \in \map {\mathcal H} U$ such that:
+:$\sequence {f_{n_r} }$ [[Definition:Uniform Convergence/Metric Space|converges uniformly]] to $g$ on $K^*$.
+Further:
+{{begin-eqn}}
+{{eqn | l = \map f \sigma
+      | r = \lim_{n \mathop \to \infty} \map {f_n} \sigma
+      | c = Definition of $f$ at $\sigma$
+}}
+{{eqn | r = \lim_{r \mathop \to \infty} \map {f_{n_r} } \sigma
+      | c = [[Limit of Subsequence equals Limit of Sequence]]
+}}
+{{eqn | r = \map g \sigma
+      | c = Definition of $g$ at $\sigma$
+}}
+{{end-eqn}}
+From the [[Identity Theorem]], $f$ and $g$ [[Definition:Agreement of Mappings|agree]] on $U$.
+From [[Uniformly Convergent iff Difference Under Supremum Norm Vanishes]]:
+:$\exists N \in \N: r \ge N \implies \norm {f_{n_r} - f}_{K^*} < \epsilon$
+This [[Definition:Contradiction|contradicts]] the result that:
+:$\forall N \in \N: \exists n \ge N: \norm {f_n - f}_{K^*} \ge \epsilon$
+Hence the result, by [[Proof by Contradiction]].
+{{qed}}
+{{Namedfor|Giuseppe Vitali|cat = Vitali}}
+\end{proof}<|endoftext|>
+\section{Identity Theorem}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $U$ be an [[Definition:Open Set (Complex Analysis)|open]] [[Definition:Connected Set (Topology)|connected]] [[Definition:Subset|subset]] of the [[Definition:Complex Plane|complex plane]] $\C$.
+Let $f$ and $g$ be [[Definition:Complex Function|complex functions]] whose [[Definition:Domain of Mapping|domain]] is $U$.
+Let $S = \left\{{z \in U: f \left({z}\right) = g \left({z}\right)}\right\}$.
+Let $f$ and $g$ be [[Definition:Analytic Function|analytic]] on $U$.
+Let $S$ have a [[Definition:Limit Point (Complex Analysis)|limit point]] in $U$.
+Then:
+:$\forall z \in U : f \left({z}\right) = g \left({z}\right)$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Complex Analysis]]
+oobe91rts02awiadl93obkey8ah8g7e
+\end{proof}<|endoftext|>
+\section{Exponential Sequence is Uniformly Convergent on Compact Sets}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $\mathcal E = \left \langle{E_n}\right \rangle$ denote the sequence of functions $E_n: \C \to \C$ defined as:
+:$E_n \left({z}\right) = \left({1 + \dfrac z n}\right)^n$
+Let $K$ be a [[Definition:Compact Subset of Complex Plane|compact subset]] of $\C$.
+Then $\mathcal E$ is [[Definition:Uniform Convergence/Metric Space|uniformly convergent]] on $K$.
+\end{theorem}
+
+\begin{proof}
+=== $\mathcal E$ is Uniformly Bounded on an open space containing $K$ ===
+First, from [[Equivalence of Definitions of Complex Exponential Function]]  we see that $\mathcal E$ is [[Definition:Pointwise Convergence|pointwise convergent]] to $\exp$.
+{{refactor|extract this result from that page}}
+From [[Combination Theorem for Continuous Functions]], $E_n$ is [[Definition:Continuous Complex Function|continuous]] for each $n \in \N$.
+From [[Compact Subspace of Metric Space is Bounded]], $K$ is [[Definition:Bounded Metric Space|bounded]] by some [[Definition:Real Number|real number]] $M \in \R$.
+Let:
+: $M' = M + 1$
+: $U = \left \{{z \in \C: \left \vert{z}\right \vert < M'}\right \}$
+: $z \in U$
+: $n \in \N$
+From the [[Triangle Inequality for Complex Numbers]]:
+:$\left \vert{1 + \dfrac z n}\right \vert \le 1 + \dfrac {\left \vert{z}\right \vert} n$
+Thus, from [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals: Natural Exponent]]:
+:$\left \vert{1 + \dfrac z n}\right \vert^n \le \left({1 + \dfrac {\left \vert{z}\right \vert} n}\right)^n$
+Call this result $(1)$.
+Also:
+{{begin-eqn}}
+{{eqn | l = \left \vert{z}\right \vert
+      | o = <
+      | r = M'
+}}
+{{eqn | ll= \implies
+      | l = 1 + \frac {\left \vert{z}\right \vert} n
+      | o = <
+      | r = 1 + \frac {M'} n
+      | c = Divide both sides by $n$, add $1$ to both sides
+}}
+{{eqn | n = 2
+      | ll= \implies 
+      | l = \left({1 + \frac {\left \vert{z}\right \vert} n}\right)^n
+      | o = <
+      | r = \left({1 + \frac {M'} n}\right)^n
+      | c = [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals: Natural Exponent]]
+}}
+{{end-eqn}}
+Thus it is sufficient to show that $\left({1 + \dfrac {M'} n}\right)^n$ is [[Definition:Bounded Real Sequence|bounded]].
+From [[Exponential Sequence is Eventually Increasing]]:
+:$\exists N \in \N: n \geq N \implies  \left({1 + \dfrac {M'} n}\right)^n \le \left({1 + \dfrac {M'} {n + 1}}\right)^{n + 1}$
+So, for $n \geq N$:
+{{begin-eqn}}
+{{eqn | l = \left({1 + \frac {M'} n}\right)^n
+      | o = \leq 
+      | r = \lim_{n \mathop \to \infty} \left({1 + \frac {M'} n}\right)^n
+      | c = [[Limit of Bounded Convergent Sequence is Bounded]]
+}}
+{{eqn | r = \exp M'
+}}
+{{end-eqn}}
+Now, for each $n \in \left \{{1, 2, \ldots, N-1}\right \}$:
+:$\exists M_n \in \R: \left({1 + \dfrac {M'} n}\right)^n \leq M_n$
+by [[Continuous Function on Compact Subspace of Euclidean Space is Bounded]].
+So:
+:$\forall n \in \N: \left({1 + \dfrac {M'} n}\right)^n \le \max \left \{{M_1, M_2, \ldots, M_{N-1}, \exp M'}\right \}$
+Call this result $(3)$.
+Hence, for $z \in U$; that is, for $\left \vert{z}\right \vert < {M'}$:
+{{begin-eqn}}
+{{eqn | l = \forall n \in \N: \left \vert{\left({1 + \frac z n}\right)^n}\right \vert
+      | r = \left \vert{1 + \dfrac z n}\right \vert^n
+      | c = [[Complex Modulus of Product of Complex Numbers]]
+}}
+{{eqn | o = \le
+      | r = \left({1 + \dfrac {\left \vert{z}\right \vert} n}\right)^n
+      | c = From $(1)$
+}}
+{{eqn | o = \le
+      | r = \left({1 + \frac {M'} n}\right)^n
+      | c = From $(2)$
+}}
+{{eqn | o = \le
+      | r = \max \left \{ {M_1, M_2, \ldots, M_{N-1}, \exp M'}\right \}
+      | c = From $(3)$
+}}
+{{end-eqn}}
+That is, $\mathcal E$ is [[Definition:Uniformly Bounded|uniformly bounded]] on $U$.
+=== $\mathcal E$ is Uniformly Convergent on $K$ ===
+Note that, from [[Combination Theorem for Complex Derivatives]]:
+:$\forall n \in \N: E_n \left({z}\right)$ is [[Definition:Holomorphic Function|holomorphic]] on $U$.
+From [[Uniformly Bounded Family is Locally Bounded]], $\mathcal E$ is [[Definition:Locally Bounded Family of Mappings|locally bounded]] on $U$.
+From [[Montel's Theorem]], $\mathcal E$ is a [[Definition:Normal Family|normal family]].
+From [[Vitali's Convergence Theorem]], $\mathcal E$ [[Definition:Uniform Convergence|converges uniformly]] on any [[Definition:Compact Subset of Complex Plane|compact subset]] of $U$.
+In particular, $\mathcal E$ [[Definition:Uniform Convergence|converges uniformly]] on $K$.
+Hence the result.
+{{qed}}
+[[Category:Exponential Function]]
+nt96mtpfcinreoq2gfzfitnm9x5wqg5
+\end{proof}<|endoftext|>
+\section{Uniformly Convergent Sequence Evaluated on Convergent Sequence}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be [[Definition:Complete Metric Space|metric spaces]].
+{{explain|[[Definition:Complete Metric Space]] is invoked, but it is called just "metric space". Analyse to determine whether it needs to be complete here or not.}}
+Let $K$ be a [[Definition:Metric Subspace|subspace]] of $X$.
+Let $\FF = \sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Continuous Mapping (Metric Space)|continuous mappings]] $f_n: X \to Y$ [[Definition: Uniform Convergence|uniformly convergent]] on $K$.
+Let $\sequence {a_n}$ be a [[Definition:Convergent Sequence in Metric Space|Convergent Sequence]] in $K$ with limit $a \in K$.
+Then $\sequence {\map {f_n} {a_n} }$ is [[Definition:Convergent Sequence in Metric Space|convergent]] such that:
+:$\displaystyle \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$
+\end{theorem}
+
+\begin{proof}
+We want to show that:
+:$\size {\map {f_n} {a_n} - \map f a} \to 0$ as $n \to \infty$
+From the [[Triangle Inequality for Complex Numbers|Triangle Inequality]]:
+:$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a}$
+Now fix $\epsilon \in \R^{>0}$.
+Since $f$ is [[Definition:Continuous Mapping (Metric Space)|continuous]] and $a_n \to a$, [[Sequential Continuity is Equivalent to Continuity in Metric Space]] tells us that:
+:$\exists M \in \N: n \ge M \implies \size {\map {f_n} {a_n} - \map f {a_n} } < \dfrac \epsilon 2$
+Since $\sequence {f_n}$ is [[Definition: Uniform Convergence|uniformly convergent]] on $K$:
+:$\exists N \in \N: n \ge N \implies \size {\map {f_n} {a_n} \to \map f {a_n} } < \dfrac \epsilon 2$
+So if $n \ge \max \set {M, N}$:
+:$\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a} < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$
+This completes the proof.
+{{qed}}
+[[Category:Metric Spaces]]
+gdg71i7b3w7cjce1vw37x9kwe8x2sr3
+\end{proof}<|endoftext|>
+\section{Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Then
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] {{iff}}
+:for every [[Definition:Ideal in Ordered Set|ideals]] $I, J$ in $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\mathscr S$ be [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Define $\II$, the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$
+Define a [[Definition:Mapping|mapping]] $f: \II \to S$ such that
+:$\forall I \in \II: \map f I = \sup I$
+By [[Meet-Continuous iff Ideal Supremum is Meet Preserving]]:
+:$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]].
+Let $I, J \in \II$.
+By definition of [[Definition:Mapping Preserves Infimum/Meet|mapping preserves meet]]:
+:$f$ [[Definition:Mapping Preserves Infimum/Subset|preserves the infimum]] of $\set {I, J}$
+Thus
+{{begin-eqn}}
+{{eqn | l = \paren {\sup I} \wedge \paren {\sup J}
+      | r = \map f I \wedge \map f J
+      | c = Definition of $f$
+}}
+{{eqn | r = \inf \set {\map f I, \map f J}
+      | c = {{Defof|Meet (Order Theory)}}
+}}
+{{eqn | r = \map \inf {\map {f^\to} {\set {I, J} } }
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{eqn | r = \map f {\inf \set {I, J} }
+      | c = {{Defof|Mapping Preserves Infimum|subdef = Subset}}
+}}
+{{eqn | r = \map f {I \wedge J}
+      | c = {{Defof|Meet (Order Theory)}}
+}}
+{{eqn | r = \map \sup {I \wedge J}
+      | c = Definition of $f$
+}}
+{{eqn | r = \sup \set {i \wedge j: i \in I, j \in J}
+      | c = [[Meet in Set of Ideals]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that for every [[Definition:Ideal in Ordered Set|ideals]] $I, J$ in $\mathscr S$:
+:$\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$
+By [[Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving]]:
+:$f$ [[Definition:Mapping Preserves Infimum/Meet|preserves meet]].
+Thus by [[Meet-Continuous iff Ideal Supremum is Meet Preserving]]:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ellipse is Mechanical Curve}
+Tags: Ellipses
+
+\begin{theorem}
+The [[Definition:Ellipse|ellipse]] is a [[Definition:Mechanical Curve|mechanical curve]].
+That is, it cannot be constructed using a [[Definition:Compass-and-Straightedge Construction|compass and straightedge]].
+However, as many points as is desired can be so constructed that lie on any given [[Definition:Ellipse|ellipse]].
+\end{theorem}<|endoftext|>
+\section{Parabola is Mechanical Curve}
+Tags: Parabolas
+
+\begin{theorem}
+The [[Definition:Parabola|parabola]] is a [[Definition:Mechanical Curve|mechanical curve]].
+That is, it cannot be constructed using a [[Definition:Compass-and-Straightedge Construction|compass and straightedge]].
+However, as many points as is desired can be so constructed that lie on any given [[Definition:Parabola|parabola]].
+\end{theorem}<|endoftext|>
+\section{Hyperbola is Mechanical Curve}
+Tags: Hyperbolas
+
+\begin{theorem}
+The [[Definition:Hyperbola|hyperbola]] is a [[Definition:Mechanical Curve|mechanical curve]].
+That is, it cannot be constructed using a [[Definition:Compass-and-Straightedge Construction|compass and straightedge]].
+However, as many points as is desired can be so constructed that lie on any given [[Definition:Hyperbola|hyperbola]].
+\end{theorem}<|endoftext|>
+\section{Area between Radii and Whorls of Archimedean Spiral}
+Tags: Archimedean Spirals
+
+\begin{theorem}
+Let $S$ be the [[Definition:Archimedean Spiral|Archimedean spiral]] defined by the equation:
+:$r = a \theta$
+Let $\theta = \theta_1$ and $\theta = \theta_2$ be the two [[Definition:Ray (Geometry)|rays]] from the [[Definition:Pole (Polar Coordinates)|pole]] at [[Definition:Angle|angles]] $\theta_1$ and $\theta_b$ to the [[Definition:Polar Axis|polar axis]] respectively.
+Let $R$ be the [[Definition:Geometric Figure|figure]] enclosed by:
+:$\theta_1$ and $\theta_2$
+:the $n$th turn of $S$ and the $n+1$th turn of $S$
+The [[Definition:Area|area]] $\mathcal A$ of $R$ is given by:
+:$\mathcal A = a^2 \pi \left({\theta_2 - \theta_1}\right) \left({\theta_2 + \theta_1 + 2 \pi \left({2 n + 1}\right)}\right)$
+{{Proofread|My algebra may have failed me}}
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Straight Line|straight line]] boundaries of $R$ are given as $\theta_1$ and $\theta_2$.
+The corners of $R$ are located where:
+:$\theta = \theta_1 + 2 n \pi$
+:$\theta = \theta_2 + 2 n \pi$
+:$\theta = \theta_1 + 2 \left({n + 1}\right) \pi$
+:$\theta = \theta_2 + 2 \left({n + 1}\right) \pi$
+[[File:ArchimedeanSpiralAreaBetweenRadii.png|500px]]
+{{begin-eqn}}
+{{eqn | l = \mathcal A
+      | r = \int_{\theta_1 + 2 \left({n + 1}\right) \pi}^{\theta_2 + 2 \left({n + 1}\right) \pi} \frac {\left({a \theta}\right)^2 \, \mathrm d \theta} 2 - \int_{\theta_1 + 2 n \pi}^{\theta_2 + 2 n \pi} \frac {\left({a \theta}\right)^2 \, \mathrm d \theta} 2
+      | c = [[Area between Radii and Curve in Polar Coordinates]]
+}}
+{{eqn | r = \left[{\frac {a^2 \theta^3} 6}\right]_{\theta_1 + 2 \left({n + 1}\right) \pi}^{\theta_2 + 2 \left({n + 1}\right) \pi} - \left[{\frac {a^2 \theta^3} 6}\right]_{\theta_1 + 2 n \pi}^{\theta_2 + 2 n \pi}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {a^2} 6 \left({\left({\theta_2 + 2 \left({n + 1}\right) \pi}\right)^3 - \left({\theta_1 + 2 \left({n + 1}\right) \pi}\right)^3 - \left({\theta_2 + 2 n \pi}\right)^3 + \left({\theta_1 + 2 n \pi}\right)^3}\right)
+      | c = 
+}}
+{{eqn | r = \frac {a^2} 6 \left({\theta_2^3 + 3 \left({2 \left({n + 1}\right) \pi}\right)\theta_2^2 + 3 \left({2 \left({n + 1}\right) \pi}\right)^2 \theta_2 + \left({2 \left({n + 1}\right) \pi}\right)^3}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \frac {a^2} 6 \left({\theta_1^3 + 3 \left({2 \left({n + 1}\right) \pi}\right)\theta_1^2 + 3 \left({2 \left({n + 1}\right) \pi}\right)^2 \theta_1 + \left({2 \left({n + 1}\right) \pi}\right)^3}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = \frac {a^2} 6 \left({\theta_2^3 + 3 \left({2 n \pi}\right)\theta_2^2 + 3 \left({2 n \pi}\right)^2 \theta_2 + \left({2 n \pi}\right)^3}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \frac {a^2} 6 \left({\theta_1^3 + 3 \left({2 n \pi}\right)\theta_1^2 + 3 \left({2 n \pi}\right)^2 \theta_1 + \left({2 n \pi}\right)^3}\right)
+      | c = 
+}}
+{{eqn | r = a^2 \pi \left({\left({n + 1}\right) \theta_2^2 + 2 \pi \left({n + 1}\right)^2 \theta_2 - \left({n \theta_2^2 + 2 \pi n^2 \theta_2}\right)}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = a^2 \pi \left({\left({n + 1}\right)\theta_1^2 + 2 \pi \left({n + 1}\right)^2 \theta_1 - \left({n \theta_1^2 + 2 \pi n^2 \theta_1}\right)}\right)
+      | c = 
+}}
+{{eqn | r = a^2 \pi \left({\theta_2^2 + 2 \pi \left({2 n + 1}\right) \theta_2}\right)
+      | c = 
+}}
+{{eqn | o = 
+      | ro= -
+      | r = a^2 \pi \left({\theta_1^2 + 2 \pi \left({2 n + 1}\right) \theta_1}\right)
+      | c = 
+}}
+{{eqn | r = a^2 \pi \left({\theta_2 - \theta_1}\right) \left({\theta_2 + \theta_1 + 2 \pi \left({2 n + 1}\right)}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{proofread}}
+\end{proof}<|endoftext|>
+\section{Derivative of Sequence of Holomorphic Functions}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $U$ be an [[Definition: Open Set of Metric Space | open]], [[Definition: Connected | connected]] subset of $\C$.
+Let $\left \langle{f_n}\right \rangle$ be a [[Definition:Sequence|sequence]] of [[Definition: Analytic Function on Complex Plane | holomorphic mappings]] $f_n: U \to \C$.
+Let $\left \langle{f_n}\right \rangle$ [[Definition:Pointwise Convergence | converge pointwise]] to some function $f: U \to \C$.
+Let $\left \langle{f_n}\right \rangle$ [[Definition:Uniform Convergence | converge uniformly]] on [[Definition:Compact Subset of Complex Plane | compact]] subsets of $U$.
+Then $f$ is holomorphic on $U$.
+Further, the sequence of [[Definition:Derivative of Complex Function | derivatives]] $\left \langle{f_n'}\right \rangle$ converges to $f'$ on $U$. 
+This convergence is uniform on compact subsets of $U$.
+\end{theorem}<|endoftext|>
+\section{Uniformly Convergent Sequence of Bounded Functions is Uniformly Bounded}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $X = \left({A, d}\right)$ and $Y = \left({B, \rho}\right)$ be [[Definition:Metric Space|metric spaces]].
+Let $\left \langle{f_i}\right \rangle_{i \in I}$ be a [[Definition:Uniform Convergence|uniformly convergent]] [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] $f_i: X \to Y$.
+$\forall i \in I$, let $f_i$ be [[Definition:Bounded Mapping to Metric Space|bounded]].
+Then $\left \langle{f_i}\right \rangle$ is [[Definition:Uniformly Bounded|uniformly bounded]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Metric Spaces]]
+7d5j0zwent5gjtl0jwe89zyrpjqlp4t
+\end{proof}<|endoftext|>
+\section{Product of Uniformly Convergent Sequences of Bounded Functions is Uniformly Convergent}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be [[Definition:Metric Space|metric spaces]].
+Let $\sequence {f_n}$ and $\sequence {g_n}$ be [[Definition:Sequence|sequences]] of [[Definition:Mapping|mappings]] from $X$ to $Y$.
+Let $\sequence {f_n}$ and $\sequence {g_n}$ be [[Definition:Uniform Convergence|uniformly convergent]] on some [[Definition:Metric Subspace|subspace]] $S$ of $X$. 
+$\forall n \in \N$, let $f_n$ and $g_n$ be [[Definition:Bounded Mapping to Metric Space|bounded]].
+Then the sequence $\sequence {f_n g_n}$ is uniformly convergent on $S$.
+{{explain|for the product to be defined, we probably want $Y$ to have some more structure}}
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Metric Spaces]]
+i5xrjd237orij2ge9yvscxqz4r2g3zo
+\end{proof}<|endoftext|>
+\section{Area between Radii and Curve in Polar Coordinates}
+Tags: Area Formulas
+
+\begin{theorem}
+Let $C$ be a [[Definition:Curve|curve]] expressed in [[Definition:Polar Coordinates|polar coordinates]] $\polar {r, \theta}$ as:
+:$r = \map g \theta$
+where $g$ is a [[Definition:Real Function|real function]].
+Let $\theta = \theta_a$ and $\theta = \theta_b$ be the two [[Definition:Ray (Geometry)|rays]] from the [[Definition:Pole (Polar Coordinates)|pole]] at [[Definition:Angle|angles]] $\theta_a$ and $\theta_b$ to the [[Definition:Polar Axis|polar axis]] respectively.
+Then the [[Definition:Area|area]] $\mathcal A$ between $\theta_a$, $\theta_b$ and $C$ is given by:
+:$\displaystyle \mathcal A = \int \limits_{\theta \mathop = \theta_a}^{\theta \mathop = \theta_b} \frac {\paren {\map g \theta}^2 \rd \theta} 2$
+as long as $\paren {\map g \theta}^2$ is [[Definition:Integrable Function|integrable]].
+\end{theorem}
+
+\begin{proof}
+:[[File:AreaPolarIntegral.png|600px]]
+{{ProofWanted}}
+[[Category:Area Formulas]]
+scbe5odf2woggmrc8v3geciapz1wfz9
+\end{proof}<|endoftext|>
+\section{Derivative of Exponential Function/Complex}
+Tags: Derivatives involving Exponential Function
+
+\begin{theorem}
+The [[Definition:Complex Exponential Function|complex exponential function]] is its own [[Definition:Derivative of Complex Function|derivative]].
+That is:
+:$\map {D_z} {\exp z} = \exp z$
+\end{theorem}<|endoftext|>
+\section{Exponential Sequence is Eventually Increasing}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $\sequence {E_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $E_n: \R \to \R$ defined as:
+:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
+Then, for sufficiently large $n \in \N$, $\sequence {\map {E_n} x}$ is [[Definition:Increasing Real Sequence|increasing]] with respect to $n$.
+That is:
+:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x \le \map {E_{n + 1} } x$
+where $\ceiling x$ denotes the [[Definition:Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Fix $x \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \ge
+      | r = \ceiling {\size x}
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = >
+      | r = -x
+      | c = [[Real Number is between Ceiling Functions]] and [[Negative of Absolute Value]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = >
+      | r = \frac {-x} n
+      | c = Divide both sides by $n$
+}}
+{{eqn | ll= \leadsto
+      | l = 1 + \frac x n
+      | o = >
+      | r = 0
+}}
+{{end-eqn}}
+So we may apply the [[Cauchy's Mean Theorem|AM-GM inequality]], with $x_1 := 1$ and $x_2 := \ldots := x_{n + 1} = 1 + \dfrac x n$, to obtain that:
+{{explain|The above sentence is unclear.}}
+:$\dfrac {1 + n \paren {1 + \dfrac x n} } {n + 1} > \paren {\paren {1 + \dfrac x n}^n}^{1 / \paren {n + 1} }$
+After simplification:
+:$1 + \dfrac x {n + 1} > \paren {1 + \dfrac x n}^{n / \paren {n + 1} }$
+From [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals: Natural Exponent]]:
+:$\paren {1 + \dfrac x {n + 1} }^{n + 1} > \paren {1 + \dfrac x n}^n$
+where we have raised both sides to the [[Definition:Integer Power|power of $n + 1$]].
+Hence the result.
+{{qed}}
+[[Category:Exponential Function]]
+c9b26890rtj1exfkejzxhbfk25q4mt5
+\end{proof}<|endoftext|>
+\section{Euler's Number: Limit of Sequence implies Base of Logarithm}
+Tags: Logarithms
+
+\begin{theorem}
+Let $e$ be [[Definition: Euler's Number | Euler's number]] defined by:
+:$\displaystyle e := \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right) ^n$
+Then $e$ is the [[Definition:Unique|unique]] solution to the equation $\ln \left({x}\right) = 1$.
+That is:
+:$\ln \left({x}\right) = 1 \iff x = e$
+\end{theorem}
+
+\begin{proof}
+First we prove that $e$ is a solution to $\ln \left({x}\right) = 1$:
+{{begin-eqn}}
+{{eqn | l = \ln \left({e}\right)
+      | r = \ln \left({\lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n}\right)
+      | c = [[Definition:Euler's Number/Limit of Sequence]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \left({\ln \left({1 + \frac 1 n}\right)^n}\right)
+      | c = [[Sequential Continuity is Equivalent to Continuity in Metric Space]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \left({n \ln \left({1 + \frac 1 n}\right)}\right)
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / n}
+      | c = [[Inverse of Group Inverse]]
+}}
+{{eqn | r = \lim_{x \mathop \to \infty} \frac {\ln \left({1 + \frac 1 n}\right)} {1 / x}
+      | c = [[Limit of Sequence is Limit of Real Function]]
+}}
+{{eqn | r = \lim_{x \mathop \to \infty} \frac{\left({-\frac 1 {x^2} }\right) \left({1 + \frac 1 x}\right)} {-\frac 1 {x^2} }
+      | c = [[L'Hôpital's Rule]]
+}}
+{{eqn | r = \lim_{x \mathop \to \infty} \frac x {x + 1}
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+{{improve|Find a better link than [[Inverse of Group Inverse]] -- [[Reciprocal of Reciprocal]] is worth creating. Then it will remove the need to invoke group-theoretical results in what is a purely real-analytical theorem.}}
+From [[Logarithm is Strictly Increasing]], $\ln$ is [[Definition:Strictly Monotone Mapping|strictly monotone]].
+By [[Strictly Monotone Mapping with Totally Ordered Domain is Injective]] it follows that $\ln$ is an [[Definition:Injection|injection]].
+So the solution to $\ln \left({x}\right) = 1$ is [[Definition:Unique|unique]].
+Hence the result.
+{{qed}}
+[[Category:Logarithms]]
+4pwnm4m0tgj1lbyhjni276cjdx9ospj
+\end{proof}<|endoftext|>
+\section{Diophantine Equation y cubed equals x squared plus 2}
+Tags: Diophantine Equations
+
+\begin{theorem}
+The [[Definition:Indeterminate Equation|indeterminate]] [[Definition:Diophantine Equation|Diophantine equation]]:
+:$y^3 = x^2 + 2$
+has only one [[Definition:Integer|integer]] solution:
+:$x = 5, y = 3$
+\end{theorem}<|endoftext|>
+\section{Continuous Extension from Dense Subset}
+Tags: Real Analysis, Metric Spaces
+
+\begin{theorem}
+Let $X$ be a [[Definition:Metric Space|metric space]].
+Let $D$ be a [[Definition:Everywhere Dense|dense]] subset of $X$.
+Let $f: D \to \R$ be a [[Definition:Uniformly Continuous Mapping (Metric Spaces)|uniformly continuous mapping]].
+Then there exists a unique [[Definition:Continuous Extension|continuous extension]] of $f$ to $X$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Real Analysis]]
+[[Category:Metric Spaces]]
+jde7psdzy42exqmb2yjik6axp2bc7ye
+\end{proof}<|endoftext|>
+\section{Sum of two Fourth Powers cannot be Fourth Power}
+Tags: Number Theory
+
+\begin{theorem}
+$\forall a, b, c \in \Z_{>0}$, the equation $a^4 + b^4 = c^4$ has no solutions.
+\end{theorem}
+
+\begin{proof}
+This is a direct consequence of [[Fermat's Right Triangle Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Moore-Osgood Theorem}
+Tags: Real Analysis, Metric Spaces
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition: Metric Space|metric spaces]].
+Let $S$ be a [[Definition:Metric Subspace|subspace]] of $X$.
+Let $c$ be a [[Definition:Limit Point (Metric Space)|limit point]] of $S$.
+Let $\left \langle{f_n}\right \rangle$ be a [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] $f_n : X \to Y$.
+Suppose that:
+:$(1): \quad \left \langle{f_n}\right \rangle$ is [[Definition:Uniform Convergence/Metric Space|uniformly convergent]] on $S$
+:$(2): \quad \displaystyle \forall n \in \N : \lim_{x \to c} f_n \left({ x }\right)$ exists
+Then:
+:$\displaystyle \lim_{x \to c} \lim_{n \to \infty} f_n \left({ x }\right) = \lim_{n \to \infty} \lim_{x \to c} f_n \left({ x }\right)$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Eliakim Hastings Moore|name2 = William Fogg Osgood|cat = Moore, Eliakim|cat2 = Osgood}}
+[[Category:Real Analysis]]
+[[Category:Metric Spaces]]
+n656qnqkt2s8y8w98xftde3efbhvri2
+\end{proof}<|endoftext|>
+\section{Cycloid is Best Shape for Arch}
+Tags: Architecture
+
+\begin{theorem}
+The optimal shape for an [[Definition:Arch|arch]] is a [[Definition:Cycloid|cycloid]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|... and of course an elaboration of what "optimal" means would be useful.}}
+\end{proof}<|endoftext|>
+\section{Center of Gravity of Cycloid}
+Tags: Cycloids
+
+\begin{theorem}
+Let $C$ be a [[Definition:Cycloid|cycloid]].
+The [[Definition:Center of Gravity|center of gravity]] of an [[Definition:Arc of Cycloid|arc]] of $C$ is at
+{{stub|research needed}}
+\end{theorem}<|endoftext|>
+\section{Ore Graph is Connected}
+Tags: Graph Theory
+
+\begin{theorem}
+Let $G = \left({V, E}\right)$ be an [[Definition:Ore Graph|Ore graph]].
+Then $G$ is [[Definition:Connected Graph|connected]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be an [[Definition:Ore Graph|Ore graph]].
+{{AimForCont}} $G$ is not [[Definition:Connected Graph|connected]].
+Then it has at least two [[Definition:Component of Graph|components]].
+Call these [[Definition:Component of Graph|components]] $C_1$ and $C_2$.
+Thus, there exist non-[[Definition:Adjacent Vertices (Undirected Graph)|adjacent vertices]] $u$ and $v$ such that $u$ is in $C_1$ and $v$ is in $C_2$.
+Let $m_1$ and $m_2$ be the number of vertices in $C_1$ and $C_2$ respectively.
+It is clear that:
+:$m_1 + m_2 \le n$
+By definition of [[Definition:Ore Graph|Ore graph]], $G$ is [[Definition:Simple Graph|simple]].
+Thus it follows that:
+:$\deg_G \left({u}\right) \le m_1 - 1$ and $\deg_G \left({v}\right) \le m_2 - 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \deg_G \left({u}\right) + \deg_G \left({v}\right)
+      | o = \le
+      | r = m_1 - 1 + m_2 - 1
+      | c = 
+}}
+{{eqn | r = m_1 + m_2 - 2
+      | c = 
+}}
+{{eqn | o = <
+      | r = m_1 + m_2
+      | c = 
+}}
+{{eqn | o = \le
+      | r = n
+      | c = 
+}}
+{{end-eqn}}
+That is:
+:$\deg_G \left({u}\right) + \deg_G \left({v}\right) < n$
+But by definition of [[Definition:Ore Graph|Ore graph]]:
+:$\deg_G \left({u}\right) + \deg_G \left({v}\right) \ge n$
+By [[Proof by Contradiction]], it follows that our assumption that $G$ is not [[Definition:Connected Graph|connected]] was [[Definition:False|false]].
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Power Function to Rational Power permits Unique Continuous Extension}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{> 0}$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f q = a^q$
+where $a^q$ denotes [[Definition:Rational Power|$a$ to the power of $q$]].
+Then there exists a unique [[Definition:Continuous Extension|continuous extension]] of $f$ to $\R$.
+\end{theorem}
+
+\begin{proof}
+Consider $I_k := \left({-k \,.\,.\, k }\right)$ for arbitrary $k \in \N$.
+Let $I_k' = I_k \cap \Q$.
+Note that, for all $x, y \in I_k'$:
+{{begin-eqn}}
+{{eqn | l = \size {a^x - a^y}
+      | r = \size {a^{x - y + y} - a^y}
+}}
+{{eqn | r = \size {a^{x - y} a^y - a^y}
+      | c = [[Powers of Group Elements]]
+}}
+{{eqn | r = \size {a^y} \size {a^{x - y} - 1}
+      | c = [[Absolute Value Function is Completely Multiplicative]]
+}}
+{{eqn | r = a^y \size {a^{x - y} - 1}
+      | c = [[Power of Positive Real Number is Positive over Rationals]]
+}}
+{{eqn | o = \le
+      | r = M \size {a^{x - y} - 1}
+      | c = [[Power Function is Monotone over Rationals]]
+}}
+{{end-eqn}}
+where $M = \map \max {a^{-k}, a^k}$.
+Fix $\epsilon \in \R_{> 0}$.
+From [[Power Function tends to One as Power tends to Zero/Rational Number|Power Function tends to One as Power tends to Zero: Rational Number]]:
+{{begin-eqn}}
+{{eqn | ll= \exists \delta \in \R_{> 0}:
+      | l = 0
+      | o = <
+      | m = \left\vert{x - y}\right\vert
+      | mo= <
+      | r = \delta
+}}
+{{eqn | lll= \leadsto
+      | o = 
+      | m = \size {a^{x - y} - a^0}
+      | mo= <
+      | r = \dfrac \epsilon M
+}}
+{{eqn | lll= \leadsto
+      | o = 
+      | m = \size {a^{x - y} - 1}
+      | mo= <
+      | r = \dfrac \epsilon M
+      | c = {{Defof|Integer Power|Power of Zero}}
+}}
+{{end-eqn}}
+where $M$ is defined as above.
+Therefore, $\exists \delta \in \R_{> 0}$ such that:
+{{begin-eqn}}
+{{eqn | l = \size {x - y}
+      | o = <
+      | r = \delta
+}}
+{{eqn | ll= \leadsto
+      | l = \size {a^x - a^y}
+      | o = \le
+      | r = M \size {a^{x - y} - 1}
+}}
+{{eqn | o = <
+      | r = M \dfrac \epsilon M
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+That is, $f$ is [[Definition:Uniform Continuity|uniformly continuous]] on $I_k'$.
+From [[Rationals are Everywhere Dense in Topological Space of Reals]], $I_k'$ is [[Definition:Everywhere Dense|everywhere dense]] in $I_k$.
+From [[Continuous Extension from Dense Subset]], there exists a unique [[Definition:Continuous Extension|continuous extension]] of $f$ to $I_k$.
+Call this function $f_k$.
+Define:
+: $\displaystyle F = \bigcup \set {f_k : k \in \N}$
+Note that $\sequence {I_k}$ is an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]] in $\R$.
+By the [[Union of Functions Theorem]], $F$ defines a function $\R \to \R$
+From the [[Pasting Lemma]], $F$ is continuous on $\R$.
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+4z686v4j71hiajz8ium9ffl6m5v7wuj
+\end{proof}<|endoftext|>
+\section{Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \struct {S, \vee, \wedge, \preceq}$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Then:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]
+{{iff}}:
+:for every [[Definition:Directed Subset|directed subsets]] $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\mathscr S$ be [[Definition:Meet-Continuous Lattice|meet-continuous]].
+By [[Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals]]:
+:for every [[Definition:Ideal in Ordered Set|ideals]] $I, J$ in $\mathscr S$: $\paren {\sup I} \wedge \paren {\sup J} = \sup \set {i \wedge j: i \in I, j \in J}$
+Let $D_1, D_2$ [[Definition:Directed Subset|directed subsets]] of $S$.
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$D_1$ and $D_2$ admit [[Definition:Supremum of Set|suprema]]
+By [[Supremum of Lower Closure of Set]]:
+:$D_1^\preceq$ and $D_2^\preceq$ admit [[Definition:Supremum of Set|suprema]]
+where
+:$D_1^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $D_1$.
+Thus
+{{begin-eqn}}
+{{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}
+      | r = \paren {\sup D_1^\preceq} \wedge \paren {\sup D_2^\preceq}
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}
+      | c = $D_1^\preceq$ is [[Definition:Ideal in Ordered Set|ideal]] in $\mathscr S$
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq
+      | c = [[Lower Closure of Meet of Lower Closures]]
+}}
+{{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every [[Definition:Directed Subset|directed subsets]] $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$
+By exemplification:
+:for every [[Definition:Ideal in Ordered Set|ideals]] $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$
+Thus by [[Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Ideals]]:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Brouwerian Lattice iff Meet-Continuous and Distributive}
+Tags: Brouwerian Lattices, Meet-Continuous Lattices, Distributive Lattices, Complete Lattices
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Then
+:$\mathscr S$ is a [[Definition:Brouwerian Lattice|Brouwerian lattice]]
+{{iff}}
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] and [[Definition:Distributive Lattice|distributive]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\mathscr S$ be a [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+By [[Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice]]:
+:for every $x \in S$, a [[Definition:Mapping|mapping]] $f: S \to S$ if $\forall y \in S: f\left({y}\right) = x \wedge y$, then $f$ is [[Definition:Galois Connection|lower adjoint of Galois connection]]
+By [[Shift Mapping is Lower Adjoint iff Meet is Distributive with Supremum]]:
+:$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \left\{ {x \wedge y: y \in X}\right\}$
+Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
+:$\mathscr S$ is [[Definition:Up-Complete|up-complete]].
+Thus
+:for every an [[Definition:Element|element]] $x \in S$ and a [[Definition:Directed Subset|directed subset]] $D$ of $S$: $x \wedge \sup D = \sup \left\{ {x \wedge d: d \in D}\right\}$
+Thus by definition
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+Thus by [[Brouwerian Lattice is Distributive]]:
+:$\mathscr S$ is [[Definition:Distributive Lattice|distributive]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]] and [[Definition:Distributive Lattice|distributive]].
+Let $x \in S$.
+Define a [[Definition:Mapping|mapping]] $f: S \to S$:
+:$\forall y \in S: f\left({y}\right) = x \wedge y$
+By [[Meet-Continuous and Distributive implies Shift Mapping Preserves Finite Suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/Finite|preserves finite suprema]].
+By [[Meet-Continuous implies Shift Mapping Preserves Directed Suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+By [[Mapping Preserves Finite and Directed Suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]].
+By [[All Suprema Preserving Mapping is Lower Adjoint of Galois Connection]]:
+:$f$ is [[Definition:Galois Connection|lower adjoint of Galois connection]].
+Thus by [[Brouwerian Lattice iff Shift Mapping is Lower Adjoint]]:
+:$\mathscr S$ is [[Definition:Brouwerian Lattice|Brouwerian lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power is Well-Defined/Integer}
+Tags: Powers
+
+\begin{theorem}
+Let $x$ be a [[Definition:Zero (Number)|non-zero]] [[Definition:Real Number|real number]].
+Let $k$ be an [[Definition:Integer|integer]].
+Then $x^k$ is [[Definition:Well-Defined Operation|well-defined]].
+\end{theorem}
+
+\begin{proof}
+Fix $x \in \R \setminus \left\{{ 0 }\right\}$.
+=== Positive Integers ===
+We first prove the theorem for positive integers by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+: $x^n$ is [[Definition:Well-Defined Operation|well-defined]]
+==== Basis for the Induction ====
+$P(1)$ is true, since $x^1 = 1$ by definition.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+==== Induction Hypothesis ====
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$x^k$ is [[Definition:Well-Defined Operation|well-defined]]
+Then we need to show:
+:$x^{k + 1}$ is [[Definition:Well-Defined Operation|well-defined]]
+==== Induction Step ====
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+By definition of [[Definition:Integer Power|integer power]]:
+:$x^{k + 1} = x \cdot x^k$
+From [[Real Multiplication is Well-Defined]] and the [[Power is Well-Defined/Integer#Induction Hypothesis|Induction Hypothesis]]:
+:$x \cdot x^k$ is [[Definition:Well-Defined Operation|well-defined]]
+So $x^{k + 1}$ is [[Definition:Well-Defined Operation|well-defined]].
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+: $\displaystyle \forall n \in \N: x^n$ is [[Definition:Well-Defined Operation|well-defined]]
+{{qed|lemma}}
+=== Non-Positive Integers ===
+Fix $k \in \Z_{< 0}$
+First:
+:$k \in \Z_{< 0} \implies -k \in \N$
+From the development above, $x^{-k}$ is [[Definition:Well-Defined Operation|well-defined]].
+From [[Powers of Group Elements/Negative Index]], $\dfrac 1 {x^k}$ is [[Definition:Well-Defined Operation|well-defined]].
+From [[Reciprocal of Real Number is Unique]], $x^k$ is [[Definition:Well-Defined Operation|well-defined]].
+Finally:
+:$\forall x \in \R \setminus \left\{{0}\right\} : x^0 = 1$
+So $x^k$ is [[Definition:Well-Defined Operation|well-defined]] for $k = 0$.
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+2aqjx1fi6383i67ea8ix5h7kbhd52jk
+\end{proof}<|endoftext|>
+\section{Power is Well-Defined/Rational}
+Tags: Powers
+
+\begin{theorem}
+Let $x \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $q$ be a [[Definition:Rational Number|rational number]].
+Then $x^q$ is [[Definition:Well-Defined Operation | well-defined]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R_{>0}$ be fixed.
+Let $q \in \Q \setminus \set 0$.
+Let $\dfrac r s$ and $\dfrac t u$ be two representations of $q$.
+That is, $r, s, t, u$ are [[Definition:Zero (Number)|non-zero]] [[Definition:Integer|integers]].
+We now show that:
+: $\dfrac r s = \dfrac t u \implies x^{r / s} = x^{t / u}$
+So:
+{{begin-eqn}}
+{{eqn | l = \dfrac r s = \dfrac t u
+      | o = \implies
+      | r = r u = t s
+}}
+{{eqn | o = \implies
+      | r = x^{r u} = x^{s t}
+      | c = [[Power is Well-Defined/Integer|Power is Well-Defined: Integer]]
+}}
+{{eqn | o = \implies
+      | r = \paren {x^r}^u = \paren {x^t}^s
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | o = \implies
+      | r = \sqrt [u] {\paren {x^r}^u} = \sqrt [u] {\paren {x^t}^s}
+      | c = [[Existence and Uniqueness of Positive Root of Positive Real Number]]
+}}
+{{eqn | o = \implies
+      | r = x^r = \sqrt [u] {\paren {x^t}^s}
+      | c = {{Defof|Root (Analysis)|Root}}
+}}
+{{eqn | o = \implies
+      | r = \sqrt [s] {\paren {x^r} } = \sqrt [s] {\sqrt [u] {\paren {x^t}^s} }
+      | c = [[Existence and Uniqueness of Positive Root of Positive Real Number]]
+}}
+{{eqn | o = \implies
+      | r = \sqrt [s] {\paren {x^r} } = \sqrt [u] {\sqrt [s] {\paren {x^t}^s} }
+      | c = [[Root is Commutative]]
+}}
+{{eqn | o = \implies
+      | r = \sqrt [s] {\paren {x^r} } = \sqrt [u] {\paren {x^t} }
+      | c = {{Defof|Root (Analysis)|Root}}
+}}
+{{eqn | o = \implies
+      | r = x^{r / s} = x^{t / u}
+      | c = {{Defof|Rational Power}}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+7xrtveddkhi280uir0dbcnwyfxcq8l1
+\end{proof}<|endoftext|>
+\section{Plane Figure with Maximum Area for given Perimeter is Circle}
+Tags: Isoperimetrical Problems
+
+\begin{theorem}
+Let $F$ be a [[Definition:Plane Figure|plane figure]].
+Let $P$ be the [[Definition:Length (Linear Measure)|length]] of the [[Definition:Perimeter|perimeter]] of $F$.
+Let the [[Definition:Area|area]] of $F$ be the largest of all the [[Definition:Plane Figure|plane figures]] whose [[Definition:Perimeter|perimeters]] are of [[Definition:Length (Linear Measure)|length]] $P$.
+Then $F$ is a [[Definition:Circle|circle]].
+\end{theorem}<|endoftext|>
+\section{Real Number between Zero and One is Greater than Power/Natural Number}
+Tags: Powers, Real Numbers, Inequalities, Real Number between Zero and One is Greater than Power
+
+\begin{theorem}
+Let $x \in \R$.
+Let $0 < x < 1$.
+Let $n$ be a [[Definition:Natural Number|natural number]].
+Then:
+: $0 < x^n \le x$
+\end{theorem}<|endoftext|>
+\section{Power Function is Strictly Increasing over Positive Reals/Natural Exponent}
+Tags: Powers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Let $f: \R_{>0} \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$f \left({x}\right) = x^n$
+where $x^n$ denotes [[Definition:Integer Power|$x$ to the power of $n$]].
+Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] on $n$:
+Let $x, y \in \R_{>0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$x < y \implies x^n < y^n$
+=== Basis for the Induction ===
+$P \left({1}\right)$ is true, as this just says:
+{{begin-eqn}}
+{{eqn | l = x^1
+      | r = x
+      | c = Definition of [[Definition:Integer Power|integer power]]
+}}
+{{eqn | o = <
+      | r = y
+      | c = By [[Definition:Assumption|assumption]]
+}}
+{{eqn | r = y^1
+      | c = Definition of [[Definition:Integer Power|integer power]]
+}}
+{{end-eqn}}
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$x < y \implies x^k < y^k$
+Then we need to show:
+:$x < y \implies x^{k+1} < y^{k+1}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+First:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = y
+}}
+{{eqn | ll = \implies
+      | l = x^k
+      | o = <
+      | r = y^k
+      | c = [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll = \implies
+      | l = x^{k+1}
+      | o = <
+      | r = x \times y^k
+      | c = Multiply both sides by $x > 0$
+}}
+{{eqn | o = <
+      | r = y \times y^k
+      | c = Multiply both sides of $x < y$ by $y^k > 0$
+}}
+{{eqn | r = y^{k+1}
+      | c = Definition of [[Definition:Integer Power|integer power]]
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>0}: x < y \implies x^n < y^n$
+{{qed}}
+[[Category:Powers]]
+dggzpwgfijhmelesjckhfrz3l7zc2u3
+\end{proof}<|endoftext|>
+\section{Root is Strictly Increasing}
+Tags: Roots of Numbers
+
+\begin{theorem}
+Let $x \in \R_{> 0}$.
+Let $n \in \N$.
+Let $f: \R_{> 0} \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map f x = \sqrt [n] x$
+where $\sqrt [n] x$ denotes the [[Definition:Root (Analysis)|$n$th root]] of $x$.
+Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$ such that $0 < x < y$.
+{{AimForCont}} that:
+:$\sqrt [n] x \ge \sqrt [n] y$
+We have:
+{{begin-eqn}}
+{{eqn | l = \sqrt [n] x
+      | o = \ge
+      | r = \sqrt [n] y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\sqrt [n] x}^n
+      | o = \ge
+      | r = \paren {\sqrt [n] y}^n
+      | c = [[Power Function is Strictly Increasing over Positive Reals/Natural Exponent|Power Function is Strictly Increasing over Positive Reals: Natural Exponent]]
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \ge
+      | r = y
+      | c ={{Defof|Root (Analysis)}}
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the hypothesis that $x < y$.
+Therefore, by [[Proof by Contradiction]]:
+:$0 < x < y \implies \sqrt [n] x \le \sqrt [n] y$
+Hence the result by definition of [[Definition:Strictly Increasing Real Function|strictly increasing real function]].
+{{qed}}
+[[Category:Roots of Numbers]]
+dayc12o1tyy9uobmrjhoftknykfuqxl
+\end{proof}<|endoftext|>
+\section{Shift Mapping is Lower Adjoint iff Meet is Distributive with Supremum}
+Tags: Galois Connections, Complete Lattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Then
+:$\forall x \in S, f: S \to S:\left({\forall y \in S: f\left({y}\right) = x \wedge y}\right) \implies f$ is [[Definition:Galois Connection|lower adjoint of Galois connection]]
+{{iff}}
+:$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \left\{ {x \wedge y: y \in X}\right\}$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that
+:$\forall x \in S, f: S \to S:\left({\forall y \in S: f\left({y}\right) = x \wedge y}\right) \implies f$ is [[Definition:Galois Connection|lower adjoint of Galois connection]]
+Let $x \in S, X \subseteq S$.
+Define a [[Definition:Mapping|mapping]] $f: S \to S$:
+:$\forall y \in S: f\left({y}\right) := x \wedge y$
+By assumption:
+:$f$ is [[Definition:Galois Connection|lower adjoint of a Galois connection]].
+By [[Lower Adjoint Preserves All Suprema]]:
+:$f$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]].
+By definition of [[Definition:Mapping Preserves Supremum/All|mapping preserves all suprema]]
+:$f$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] of $X$.
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+Thus
+{{begin-eqn}}
+{{eqn | l = x \wedge \sup X
+      | r = f\left({\sup X}\right)
+      | c = definition of $f$
+}}
+{{eqn | r = \sup \left({f^\to\left({X}\right)}\right)
+      | c = definition of [[Definition:Mapping Preserves Supremum/Subset|mapping preserves the supremum of set]]
+}}
+{{eqn | r = \sup \left\{ {f\left({y}\right): y \in X}\right\}
+      | c = definition of [[Definition:Image of Subset under Mapping|image of set]]
+}}
+{{eqn | r = \sup \left\{ {x \wedge y: y \in X}\right\}
+      | c = definition of $f$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition === 
+Assume that
+:$\forall x \in S, X \subseteq S: x \wedge \sup X = \sup \left\{ {x \wedge y: y \in X}\right\}$
+Let $x \in S$, $f: S \to S$ such that
+:$\forall y \in S: f\left({y}\right) = x \wedge y$
+We will prove that
+:$f$ [[Definition:Mapping Preserves Supremum/All|preserves all suprema]].
+Let $X \subseteq S$ such that
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
+:$f^\to\left({X}\right)$ admits a [[Definition:Supremum of Set|supremum]].
+Thus
+{{begin-eqn}}
+{{eqn | l = \sup \left({f^\to\left({X}\right)}\right)
+      | r = \sup \left\{ {f\left({y}\right): y \in X}\right\}
+      | c = definition of [[Definition:Image of Subset under Mapping|image of set]]
+}}
+{{eqn | r = \sup \left\{ {x \wedge y: y \in X}\right\}
+      | c = definition of $f$
+}}
+{{eqn | r = x \wedge \sup X
+      | c = assumption
+}}
+{{eqn | r = f\left({\sup X}\right)
+      | c = definition of $f$
+}}
+{{end-eqn}}
+Thus by definition
+:$f$ [[Definition:Mapping Preserves Supremum/Subset|preserves the supremum]] of $X$.
+{{qed|lemma}}
+Thus by [[All Suprema Preserving Mapping is Lower Adjoint of Galois Connection]]:
+:$f$ is [[Definition:Galois Connection|lower adjoint of a Galois connection]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Common Divisor is Associative}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b, c \in \Z$.
+Then:
+:$\gcd \left\{ {a, \gcd \left\{{b, c}\right\} }\right\} = \gcd \left\{ {\gcd \left\{{a, b}\right\}, c}\right\}$
+where $\gcd$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[GCD from Prime Decomposition]] and [[Max and Min are Associative|Min is Associative]].
+\end{proof}<|endoftext|>
+\section{Lowest Common Multiple is Associative}
+Tags: Lowest Common Multiple
+
+\begin{theorem}
+Let $a, b, c \in \Z$.
+Then:
+:$\lcm \left\{{a, \lcm \left\{{ b , c }\right\} }\right\} = \lcm \left\{{\lcm \left\{{a, b}\right\}, c}\right\}$
+where $\lcm$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[LCM from Prime Decomposition]] and [[Max and Min are Associative|Max is Associative]]
+\end{proof}<|endoftext|>
+\section{LCM of Coprime Integers}
+Tags: Lowest Common Multiple, Coprime Integers
+
+\begin{theorem}
+Let $a, b \in \Z_{>0}$ be [[Definition:Coprime Integers|coprime integers]].
+Then:
+:$\lcm \set {a, b} = a b$
+where $\lcm$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lcm \set {a, b}
+      | r = \frac {a b} {\gcd \set {a, b} }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {a b} 1
+      | c = {{Defof|Coprime Integers}}
+}}
+{{eqn | r = a b
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Lowest Common Multiple]]
+[[Category:Coprime Integers]]
+fcl75w8sx5a69le8gxkj1593ls9hup4
+\end{proof}<|endoftext|>
+\section{GCD with One Fixed Argument is Multiplicative Function}
+Tags: Greatest Common Divisor, Examples of Multiplicative Functions
+
+\begin{theorem}
+Let $a, b, c \in \Z: b \perp c$
+where $b \perp c$ denotes that $b$ is [[Definition:Coprime Integers|coprime]] to $c$.
+Then:
+:$\gcd \set {a, b} \gcd \set {a, c} = \gcd \set {a, b c}$
+That is, [[Definition:Greatest Common Divisor of Integers|GCD]] is [[Definition:Multiplicative Arithmetic Function|multiplicative]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \gcd \set {a, b c}
+      | r = \gcd \set {a, \lcm \set {b, c} }
+      | c = [[LCM of Coprime Integers]]
+}}
+{{eqn | r = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }
+      | c = [[GCD and LCM Distribute Over Each Other]]
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {\gcd \set {a, b}, \gcd \set {a, c} } }
+      | c = [[Product of GCD and LCM]]
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {b, \gcd \set {a, c} } } }
+      | c = [[Greatest Common Divisor is Associative]]
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {\gcd \set {b, c}, a} } }
+      | c = [[Greatest Common Divisor is Associative]]
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, \gcd \set {1, a} } }
+      | c = Definition of [[Definition:Coprime Integers|coprimality]]
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } {\gcd \set {a, 1} }
+}}
+{{eqn | r = \frac {\gcd \set {a, b} \gcd \set {a, c} } 1
+}}
+{{eqn | r = \gcd \set {a, b} \gcd \set {a, c}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Greatest Common Divisor]]
+[[Category:Examples of Multiplicative Functions]]
+t5bzldcx2jtjo5r2g2bsb49f38qlacr
+\end{proof}<|endoftext|>
+\section{Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer}
+Tags: Power Function on Base between Zero and One is Strictly Decreasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $0 < a < 1$.
+Let $f: \Z_{\ge 0} \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f n = a^n$
+where $a^n$ denotes [[Definition:Integer Power|$a$ to the power of $n$]].
+Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]] on $n$:
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$a^{n + 1} < a^n$
+$\map P 0$ is the case:
+{{begin-eqn}}
+{{eqn | l = a^1
+      | r = a
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | o = <
+      | r = 1 
+      | c = 
+}}
+{{eqn | r = a^0
+      | c = {{Defof|Integer Power}}
+}}
+{{end-eqn}}
+=== Basis for the Induction ===
+$\map P 1$ is true, since:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = <
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | l = a \times a
+      | o = <
+      | r = 1 \times a
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^2
+      | o = <
+      | r = a^1
+      | c = {{Defof|Integer Power}}
+}}
+{{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]]:
+:$a^{k + 1} < a^k$
+Then we need to show:
+:$a^{k + 2} < a^{k + 1}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = a^{k + 1}
+      | o = < 
+      | r = a^k
+      | c = [[Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = a \times a^{k + 1} 
+      | o = <
+      | r = a \times a^k
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^{k + 2}
+      | o = <
+      | r = a^{k + 1}
+      | c = {{Defof|Integer Power}}
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 0}: a^{n + 1} < a^n$
+Hence the result, by definition of [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+{{qed}}
+[[Category:Power Function on Base between Zero and One is Strictly Decreasing]]
+0t61o4y9snf2r829wdmop57akew0xjl
+\end{proof}<|endoftext|>
+\section{Power Function on Base between Zero and One is Strictly Decreasing/Rational Number}
+Tags: Power Function on Base between Zero and One is Strictly Decreasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $0 < a < 1$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function |real-valued function]] defined as:
+:$\map f q = a^q$
+where $a^q$ denotes [[Definition:Rational Power|$a$ to the power of $q$]].
+Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z, s, u \in \Z_{>0}$.
+Let $\dfrac r s < \dfrac t u$.
+Then:
+{{begin-eqn}}
+{{eqn | l = r u
+      | o = <
+      | r = t s
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^{r u}
+      | o = >
+      | r = a^{t s}
+      | c = [[Power Function on Base between Zero and One is Strictly Decreasing/Integer|Power Function on Base between Zero and One is Strictly Decreasing: Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^r}^u 
+      | o = > 
+      | r = \paren {a^t}^s
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [u] {\paren {a^r}^u} 
+      | o = >
+      | r = \sqrt [u] {\paren {a^t}^s }
+      | c = [[Root is Strictly Increasing]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^r 
+      | o = > 
+      | r = \sqrt [u] {\paren {a^t}^s }
+      | c = {{Defof|Root (Analysis)|$u$th Root}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [s] {\paren {a^r} } 
+      | o = > 
+      | r = \sqrt [s] {\sqrt [u] {\paren {a^t}^s } }
+      | c = [[Root is Strictly Increasing]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [s] {\paren {a^r} } 
+      | o = > 
+      | r = \sqrt [u] {\sqrt [s] {\paren {a^t}^s } }
+      | c = [[Root is Commutative]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [s] {\paren {a^r} } 
+      | o = > 
+      | r = \sqrt [u] {\paren {a^t} }
+      | c = {{Defof|Root (Analysis)|$s$th Root}}
+}}
+{{eqn | ll= \leadsto
+      | l = a^{r / s}
+      | o = > 
+      | r = a^{t / u}
+      | c = {{Defof|Rational Power}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Power Function on Base between Zero and One is Strictly Decreasing]]
+byhapr90wzqu0vwr4t9p7jhfjdogro1
+\end{proof}<|endoftext|>
+\section{Power Function on Base Greater than One is Strictly Increasing/Positive Integer}
+Tags: Power Function on Base Greater than One is Strictly Increasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $f: \Z_{\ge 0} \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f n = a^n$
+where $a^n$ denotes [[Definition:Integer Power|$a$ to the power of $n$]].
+Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+Fix $n \in \Z_{\ge 0}$.
+From [[Ordering of Reciprocals]]:
+:$0 < \dfrac 1 a < 1$
+From [[Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer|Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer]]:
+:$\paren {\dfrac 1 a}^{n + 1} < \paren {\dfrac 1 a}^n$
+From [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]:
+:$\dfrac 1 {a^{n + 1} } < \dfrac 1 {a^n}$
+From [[Ordering of Reciprocals]]:
+:$a^n < a^{n + 1}$
+Hence the result.
+{{qed}}
+[[Category:Power Function on Base Greater than One is Strictly Increasing]]
+9ty6jg9ov8atkj9p3yw8oxqprdu02lk
+\end{proof}<|endoftext|>
+\section{Root of Reciprocal is Reciprocal of Root}
+Tags: Roots of Numbers
+
+\begin{theorem}
+Let $x \in \R_{\ge 0}$.
+Let $n \in \N$.
+Let $\sqrt [n] x$ denote the [[Definition:Root (Analysis)|$n$th root]] of $x$.
+Then:
+:$\sqrt [n] {\dfrac 1 x} = \dfrac 1 {\sqrt [n] x}$
+\end{theorem}
+
+\begin{proof}
+Let $y = \sqrt [n] {\dfrac 1 x}$.
+{{begin-eqn}}
+{{eqn | l = \sqrt [n] {\dfrac 1 x} = y
+      | o = \iff
+      | r = \dfrac 1 x = y^n
+      | c = {{Defof|Root (Analysis)|$n$th root}}
+}}
+{{eqn | o = \iff
+      | r = x = \dfrac 1 {y^n}
+      | c = [[Reciprocal of Real Number is Unique]]
+}}
+{{eqn | o = \iff
+      | r = x = \left({\dfrac 1 y}\right)^n
+      | c = [[Powers of Group Elements/Negative Index]]
+}}
+{{eqn | o = \iff
+      | r = \sqrt [n] x  = \dfrac 1 y
+      | c = {{Defof|Root (Analysis)|$n$th root}}
+}}
+{{eqn | o = \iff
+      | r = \dfrac 1 {\sqrt [n] x} = y
+      | c = [[Reciprocal of Real Number is Unique]]
+}}
+{{eqn | o = \iff
+      | r = \dfrac 1 {\sqrt [n] x} = \sqrt [n] {\dfrac 1 x}
+      | c = Definition of $y$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Roots of Numbers]]
+qfegueuaev3965956uzlxnpaxmnozjt
+\end{proof}<|endoftext|>
+\section{Power Function on Base Greater than One is Strictly Increasing/Rational Number}
+Tags: Power Function on Base Greater than One is Strictly Increasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f q = a^q$
+where $a^q$ denotes [[Definition:Rational Power|$a$ to the power of $q$]].
+Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+Let $\dfrac r s, \dfrac t u \in \Q$, where $r, t \in \Z$ are [[Definition:Integer|integers]] and $s, u \in \Z_{>0}$ are [[Definition:Strictly Positive Integer|strictly positive integers]].
+Let $\dfrac r s < \dfrac t u$.
+From [[Ordering of Reciprocals]]:
+:$0 < \dfrac 1 a < 1$
+So:
+{{begin-eqn}}
+{{eqn | l = \paren {\frac 1 a}^{t / u} 
+      | o = < 
+      | r = \paren {\frac 1 a}^{r / s}
+      | c = [[Power Function on Base between Zero and One is Strictly Decreasing/Rational Number|Power Function on Base between Zero and One is Strictly Decreasing: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [u] {\paren {\frac 1 a}^t}
+      | o = < 
+      | r = \sqrt [s] {\paren {\frac 1 a}^r}
+      | c =  {{Defof|Rational Power}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [u] {\paren {\frac 1 {a^t} } } 
+      | o = < 
+      | r = \sqrt [s] {\paren {\frac 1 {a^r} } }
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {\sqrt [u] {a^t} }
+      | o = <
+      | r = \frac 1 {\sqrt [s] {a^r} }
+      | c = [[Root of Reciprocal is Reciprocal of Root]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {a^{t / u} }
+      | o = <
+      | r = \frac 1 {a^{r / s} }
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | ll= \leadsto
+      | l = a^{r / s}
+      | o = <
+      | r = a^{t / u}
+      | c = [[Ordering of Reciprocals]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Power Function on Base Greater than One is Strictly Increasing]]
+lm6ydcb44fyxnfu932l30zdwqxas4s4
+\end{proof}<|endoftext|>
+\section{Existence of Square Roots of Positive Real Number}
+Tags: Roots of Numbers, Square Roots, Existence of Square Roots of Positive Real Number
+
+\begin{theorem}
+Let $r \in \R_{\ge 0}$ be a [[Definition:Positive Real Number|positive real number]].
+Then:
+:$\exists y_1 \in \R_{\ge 0}: {y_1}^2 = r$
+:$\exists y_2 \in \R_{\le 0}: {y_2}^2 = r$
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {x \in \Q: x^2 < r}$.
+As $0 \in S$, it follows that $S$ is [[Definition:Non-Empty Set|non-empty]].
+To show that $S$ is [[Definition:Bounded Above Subset of Real Numbers|bounded above]], we note that $n + 1$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]]:
+:$y > n + 1 \implies y^2 > n^2 + 2 n + 1 > n$
+and so $y \notin S$.
+Thus $x \in S \implies x < n + 1$.
+By the [[Completeness Axiom]], $S$ has a [[Definition:Supremum of Subset of Real Numbers|supremum]], say:
+:$u = \sup S$
+We already have that $u \ge 1$, as $0 \in S$ as seen.
+It remains to demonstrate that $u^2 = r$.
+{{AimForCont}} $u^2 \ne r$.
+Then either $u^2 > r$ or $u^2 < r$.
+Suppose that $u^2 > r$.
+Then:
+:$\dfrac {u^2 - r} {2 u} > 0$
+So there exists $n \in \N$ such that:
+:$0 < \dfrac 1 n < \dfrac {u^2 - r} {2 u}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {u - \dfrac 1 n}^2
+      | r = u^2 - \dfrac {2 u} n + \dfrac 1 {n^2}
+      | c = 
+}}
+{{eqn | o = >
+      | r = u^2 - \dfrac {2 u} n
+      | c = 
+}}
+{{eqn | o = >
+      | r = u^2 - \paren {u^2 - r}
+      | c = 
+}}
+{{eqn | r = r
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2
+      | o = <
+      | r = r
+      | c = 
+}}
+{{eqn | o = <
+      | r = \paren {u - \dfrac 1 n}^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = <
+      | r = u - \dfrac 1 n
+      | c = 
+}}
+{{end-eqn}}
+which [[Definition:Contradiction|contradicts]] the leastness of $u$.
+Suppose instead that $u^2 < r$.
+Then $\exists n \in \N$ such that:
+:$0 < \dfrac 1 n \le \dfrac {r - u^2} {4 u}$
+and:
+:$\dfrac 1 n < 2 u$
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {u + \dfrac 1 n}^2
+      | r = u^2 + \dfrac {2 u} n + \dfrac 1 {n^2}
+      | c = 
+}}
+{{eqn | o = <
+      | r = u^2 + \dfrac {2 u} n + \dfrac {2 u} n
+      | c = as $\dfrac 1 n < 2 u$
+}}
+{{eqn | o = \le
+      | r = u^2 + r - u^2
+      | c = as $r - u^2 \ge \dfrac {4 u} n$
+}}
+{{eqn | r = r
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$u + \dfrac 1 n \in S$
+which [[Definition:Contradiction|contradicts]] the fact that $u$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] of $S$.
+Hence by [[Proof by Contradiction]] it follows that $u^2 = r$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inequality iff Difference is Positive}
+Tags: Inequalities, Real Numbers
+
+\begin{theorem}
+Let $x, y \in \R$.
+Then the following are equivalent:
+:$(1): \quad x < y$
+:$(2): \quad y - x > 0$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x < y
+      | o = \leadstoandfrom
+      | r = y > x
+      | c = {{Defof|Dual Ordering}}
+}}
+{{eqn | o = \leadstoandfrom
+      | r = y + \paren {-x} > x + \paren {-x}
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = y + \paren {-x} > 0
+      | c = [[Definition:Real Number Axioms|Real Number Axioms: $\R \text A 4$: Inverse Elements]]
+}}
+{{eqn | o = \leadstoandfrom
+      | r = y - x > 0
+      | c = {{Defof|Field Subtraction}}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function on Base between Zero and One is Strictly Decreasing/Integer}
+Tags: Power Function on Base between Zero and One is Strictly Decreasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $0 < a < 1$.
+Let $f: \Z \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f k = a^k$
+where $a^k$ denotes [[Definition:Integer Power|$a$ to the power of $k$]].
+Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Let $0 < a < 1$.
+By [[Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer|Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer]], the theorem is already proven for [[Definition:Positive Integer|positive integers]].
+It remains to be proven over the [[Definition:Negative Integer|negative integers]].
+Let $i, j$ be [[Definition:Integer|integers]] such that $i < j < 0$.
+From [[Order of Real Numbers is Dual of Order of their Negatives]]:
+:$0 < -j < -i$
+So:
+{{begin-eqn}}
+{{eqn | l = a^{-j}
+      | o = > 
+      | r = a^{-i}
+      | c = [[Power Function on Base between Zero and One is Strictly Decreasing/Positive Integer|Power Function on Base between Zero and One is Strictly Decreasing: Positive Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {a^j}
+      | o = >
+      | r = \frac 1 {a^i}
+      | c = [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^i
+      | o = >
+      | r = a^j
+      | c = [[Ordering of Reciprocals]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Power Function on Base between Zero and One is Strictly Decreasing]]
+rtbdssg7n11m3w8hg6poyxc849q5hfi
+\end{proof}<|endoftext|>
+\section{Meet-Continuous iff Meet Preserves Directed Suprema}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Lattice|lattice]].
+Let $\left({S \times S, \precsim}\right)$ be the [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] of $\mathscr S$ and $\mathscr S$.
+Let $f: S \times S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall x, y \in S: f\left({x, y}\right) = x \wedge y$
+Then
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]]
+{{iff}}
+:$f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]]
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+We will prove that
+:for every an [[Definition:Element|element]] $x$ of $S$, a [[Definition:Directed Subset|directed subset]] $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \left\{ {x \wedge d: d \in D}\right\}$
+Let $x \in S$, $D$ 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 \left\{ {x \wedge d: d \in D}\right\}
+      | c = {{Defof|Meet-Continuous Lattice}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:for every an [[Definition:Element|element]] $x$ of $S$, a [[Definition:Directed Subset|directed subset]] $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \left\{ {x \wedge d: d \in D}\right\}$
+By [[Meet is Directed Suprema Preserving]]:
+:$f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that:
+:$f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+By [[Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets]]: 
+:$\left({\sup D_1}\right) \wedge \left({\sup D_2}\right) = \sup \left\{ {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}\right\}$
+Thus by [[Meet-Continuous iff Meet of Suprema equals Supremum of Meet of Directed Subsets]]:
+:$\mathscr S$ is [[Definition:Meet-Continuous Lattice|meet-continuous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function on Base Greater than One is Strictly Increasing/Integer}
+Tags: Power Function on Base Greater than One is Strictly Increasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $f: \Z \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f k = a^k$
+where $a^k$ denotes [[Definition:Integer Power|$a$ to the power of $k$]].
+Then $f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Let $a > 1$.
+By [[Power Function on Base Greater than One is Strictly Increasing/Positive Integer|Power Function on Base Greater than One is Strictly Increasing: Positive Integer]], the theorem is already proven for [[Definition:Positive Integer|positive integers]].
+It remains to be proven over the [[Definition:Strictly Negative Integer|strictly negative integers]].
+Let $i, j$ be [[Definition:Integer|integers]] such that $i < j < 0$.
+From [[Order of Real Numbers is Dual of Order of their Negatives]]:
+:$0 < -j < -i$
+So:
+{{begin-eqn}}
+{{eqn | l = a^{-j}
+      | o = < 
+      | r = a^{-i}
+      | c = [[Power Function on Base Greater than One is Strictly Increasing/Positive Integer|Power Function on Base Greater than One is Strictly Increasing: Positive Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {a^j}
+      | o = < 
+      | r = \frac 1 {a^i}
+      | c = [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^i
+      | o = < 
+      | r = a^j
+      | c = [[Ordering of Reciprocals]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Power Function on Base Greater than One is Strictly Increasing]]
+sjlmfdufptgv70wy09w64dxget1hs7b
+\end{proof}<|endoftext|>
+\section{Real Star-Algebra is Commutative}
+Tags: Algebra
+
+\begin{theorem}
+Let $A = \left( A_F , \oplus \right)$ be a [[Definition:Real Star-Algebra|real *-algebra]] whose [[Definition:Conjugation (Abstract Algebra)|conjugation]] is denoted as $*$.
+Then:
+:$\forall a,b \in A, a \oplus b = b \oplus a$
+That is, [[Definition:Real Star-Algebra|real *-algebra]] is [[Definition:Commutative|commutative]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a \oplus b
+      | r = \left( a \oplus b \right) ^{*}
+      | c = Definition of [[Definition:Real Star-Algebra|real *-algebra]]
+}}
+{{eqn | r = b^{*} \oplus a^{*}
+      | c = Definition of [[Definition:Conjugation (Abstract Algebra)|conjugation]]
+}}
+{{eqn | r = b \oplus a
+      | c = Definition of [[Definition:Real Star-Algebra|real *-algebra]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Algebra]]
+i3wknk52uzvwo406a5i4lui4ojcz541
+\end{proof}<|endoftext|>
+\section{Meet-Continuous implies Shift Mapping Preserves Directed Suprema}
+Tags: Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Meet-Continuous Lattice|meet-continuous lattice]].
+Let $x \in S$.
+Let $f: S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall y \in S: f\left({y}\right) = x \wedge y$
+Then
+:$f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+\end{theorem}
+
+\begin{proof}
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$D$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Singleton is Directed and Filtered Subset]]:
+:$\left\{ {x}\right\}$ is [[Definition:Directed Subset|directed]].
+By [[Up-Complete Product/Lemma 1]]:
+:$\left\{ {x}\right\} \times D$ is [[Definition:Directed Subset|directed]] in [[Definition:Cartesian Product of Ordered Sets|Cartesian product]] $\left({S \times S, \precsim}\right)$ of $\mathscr S$ and $\mathscr S$.
+Define a [[Definition:Mapping|mapping]] $g: S \times S \to S$:
+:$\forall s, t \in S: g\left({s, t}\right) = s \wedge t$
+By [[Meet-Continuous iff Meet Preserves Directed Suprema]]:
+:$g$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+By definition of [[Definition:Meet-Continuous Lattice|meet-continuous]]:
+:$\mathscr S$ is [[Definition:Up-Complete|up-complete]].
+By [[Up-Complete Product]]:
+:$\left({S \times S, \precsim}\right)$ is [[Definition:Up-Complete|up-complete]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\left\{ {x}\right\} \times D$ admits a [[Definition:Supremum of Set|supremum]].
+Thus
+{{begin-eqn}}
+{{eqn | l = \sup \left({f^\to \left({D}\right)}\right)
+      | r = \sup \left\{ {f \left({d}\right): d \in D}\right\}
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{eqn | r = \sup \left\{ {x \wedge d: d \in D}\right\}
+      | c = definition of $f$
+}}
+{{eqn | r = \sup \left\{ {g \left({y, d}\right): y \in \left\{ {x}\right\}, d \in D}\right\}
+      | c = definition of $g$
+}}
+{{eqn | r = \sup \left({g^\to \left({\left\{ {x}\right\} \times D}\right)}\right)
+      | c = {{Defof|Image of Subset under Mapping}}
+}}
+{{eqn | r = g \left({\sup \left({\left\{ {x}\right\} \times D}\right)}\right)
+      | c = definition of [[Definition:Mapping Preserves Supremum/Subset|mapping preserves the supremum of set]]
+}}
+{{eqn | r = g \left({\sup \left\{ {x}\right\}, \sup D}\right)
+      | c = [[Supremum of Cartesian Product]]
+}}
+{{eqn | r = g \left({x, \sup D}\right)
+      | c = [[Supremum of Singleton]]
+}}
+{{eqn | r = x \wedge \sup D
+      | c = Definition of $g$
+}}
+{{eqn | r = f \left({\sup D}\right)
+      | c = Definition of $f$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Strictly Positive Real Number is Strictly Positive/Positive Integer}
+Tags: Power of Positive Real Number is Positive
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+:$x^n > 0$
+where $x^n$ denotes the [[Definition:Integer Power|$n$th power of $x$]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Proof by Mathematical Induction|Mathematical Induction]]:
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\forall x \in \R_{>0}: x^n > 0$
+$\map P 0$ is true, as this just says:
+{{begin-eqn}}
+{{eqn | l = x^0
+      | r = 1
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | o = >
+      | r = 0
+}}
+{{end-eqn}}
+=== Basis for the Induction ===
+$\map P 1$ true, as this just says:
+{{begin-eqn}}
+{{eqn | l = x^1
+      | r = x
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | o = >
+      | r = 0
+}}
+{{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 0 $, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\forall x \in \R_{>0}: x^k > 0$
+Then we need to show:
+:$\forall x \in \R_{>0}: x^{k + 1} > 0$
+=== Inductive Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = x^k
+      | o = >
+      | r = 0
+      | c = [[Power of Strictly Positive Real Number is Strictly Positive/Positive Integer#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^{k + 1}
+      | o = >
+      | r = 0
+      | c = Multiply both sides by $x > 0$
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$
+{{qed}}
+[[Category:Power of Positive Real Number is Positive]]
+tmg67c2ywdvxvu01dumrsl4rv3rxrj3
+\end{proof}<|endoftext|>
+\section{Power of Positive Real Number is Positive/Integer}
+Tags: Power of Positive Real Number is Positive
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Then:
+:$x^n > 0$
+where $x^n$ denotes the [[Definition:Integer Power|$n$th power of $x$]].
+\end{theorem}
+
+\begin{proof}
+By [[Power of Positive Real Number is Positive/Natural Number|Power of Positive Real Number is Positive: Natural Number]], the theorem is already proven for [[Definition:Non-Negative Integer|non-negative integers]].
+Suppose $n \in \Z_{< 0}$.
+When $n < 0$, by [[Real Number Ordering is Compatible with Multiplication/Negative Factor|Real Number Ordering is Compatible with Multiplication: Negative Factor]]:
+:$-n > 0$
+Then, by [[Power of Positive Real Number is Positive/Natural Number|Power of Positive Real Number is Positive: Natural Number]]:
+:$x^{-n} > 0$
+Therefore, by [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]:
+:$x^n = \dfrac 1 {x^{-n} } > 0$
+{{qed}}
+[[Category:Power of Positive Real Number is Positive]]
+q6y71g607yse61jxsqnx1yreip7soz1
+\end{proof}<|endoftext|>
+\section{Power of Positive Real Number is Positive/Rational Number}
+Tags: Power of Positive Real Number is Positive
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $q \in \Q$ be a [[Definition:Rational Number|rational number]].
+Then:
+:$x^q > 0$
+where $x^q$ denotes the [[Definition:Rational Power|$x$ to the power of $q$]].
+\end{theorem}
+
+\begin{proof}
+Let $q = \dfrac r s$, where $r \in \Z$, $s \in \Z \setminus \set 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x > 0
+      | o = \leadsto
+      | r = x^r > 0
+      | c = [[Power of Positive Real Number is Positive/Integer|Power of Positive Real Number is Positive: Integer]]
+}}
+{{eqn | o = \leadsto
+      | r = \sqrt [s] {\paren {x^r} } > 0
+      | c = [[Existence of Positive Root of Positive Real Number]]
+}}
+{{eqn | o = \leadsto
+      | r = x^{r / s}
+      | c = {{Defof|Rational Power|Rational Power}}
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Power of Positive Real Number is Positive]]
+82zj1qzhj73vryu2ve2f4ricm3ohqr4
+\end{proof}<|endoftext|>
+\section{Power Function is Monotone/Rational Number}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{>0}$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f r = a^r$
+where $a^r$ denotes [[Definition:Rational Power|$a$ to the power of $r$]].
+Then $f$ is [[Definition:Monotone Real Function|monotone]].
+Further, $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]] unless $a = 1$.
+\end{theorem}
+
+\begin{proof}
+=== Case 1: $a > 1$ ===
+Let $a > 1$.
+Then by [[Power Function on Base Greater than One is Strictly Increasing/Rational Number|Power Function on Base Greater than One is Strictly Increasing]]:
+:$f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+By [[Strictly Increasing Mapping is Increasing]]:
+:$f$ is [[Definition:Increasing Real Function|increasing]].
+=== Case 2 : $a = 1$ ===
+Let $a = 1$.
+Then:
+:$\forall r \in \Q: a^r = 1$
+and so $f$ is [[Definition:Constant Mapping|constant]].
+From [[Mapping is Constant iff Increasing and Decreasing]]:
+:$f$ is both [[Definition:Increasing Real Function|increasing]] and [[Definition:Decreasing Real Function|decreasing]].
+=== Case 3: $0 < a < 1$ ===
+Let $0 < a < 1$.
+Then by [[Power Function on Base between Zero and One is Strictly Decreasing/Rational Number|Power Function on Base between Zero and One is Strictly Decreasing]]:
+:$f$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+By [[Strictly Decreasing Mapping is Decreasing]]:
+:$f$ is [[Definition:Decreasing Real Function|decreasing]].
+=== Conclusion ===
+It has been shown that in all cases $f$ is either [[Definition:Increasing Real Function|increasing]] or [[Definition:Decreasing Real Function|decreasing]] or both.
+Thus by definition of [[Definition:Monotone Real Function|monotone function]], $f$ is [[Definition:Monotone Real Function|monotone]].
+In either of the cases where $a \ne 1$, It has been shown that in all cases $f$ is either [[Definition:Strictly Increasing Real Function|strictly increasing]] or [[Definition:Strictly Decreasing Real Function|strictly decreasing]].
+Thus by definition of [[Definition:Strictly Monotone Real Function|strictly monotone function]], $f$ is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+gr1gfcbgcpbpct12u4hd79apbasu0nh
+\end{proof}<|endoftext|>
+\section{Power Function tends to One as Power tends to Zero/Rational Number}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{> 0}$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f q = a^q$
+where $a^q$ denotes [[Definition:Rational Power|$a$ to the power of $q$]].
+Then:
+:$\displaystyle \lim_{x \mathop \to 0} \map f x = 1$
+\end{theorem}
+
+\begin{proof}
+=== Case 1: $a > 1$ ===
+If $a > 1$, then:
+:$\displaystyle \lim_{x \mathop \to 0} \map f x = 1$
+from [[Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number]].
+=== Case 2: $a = 1$ ===
+If $a = 1$, then:
+{{begin-eqn}}
+{{eqn | l = \lim_{x \mathop \to 0} \map f x
+      | r = \lim_{x \mathop \to 0} 1^x
+      | c = Definition of $f$
+}}
+{{eqn | r = \lim_{x \mathop \to 0} 1
+      | c = [[Exponential with Base One is Constant/Rational Number]]
+}}
+{{eqn | r = 1
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{end-eqn}}
+=== Case 3: $0 < a < 1$ ===
+If $0 < a < 1$, then:
+:$\displaystyle \lim_{x \mathop \to 0} \map f x = 1$
+from [[Power Function on Base between Zero and One Tends to One as Power Tends to Zero/Rational Number]].
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+4cfp0fmmaqxg4pnvnse02adeir3ajq5
+\end{proof}<|endoftext|>
+\section{Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $r \in \Q_{> 0}$ be a [[Definition:Strictly Positive Rational Number|strictly positive rational number]] such that $r < 1$.
+Then:
+:$1 < a^r < 1 + a r$
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Real Function|real function]] $g_r: \R_{> 0} \to \R$ as:
+:$\map {g_r} a = 1 + a r - a^r$
+Then [[Definition:Differentiation|differentiating]] {{WRT|Differentiation}} $a$ gives:
+:$D_a \map {g_r} a = r \paren {1 - a^{r - 1} }$
+We show now that the [[Definition:Derivative|derivative]] of $g_r$ is [[Definition:Positive Real Function|positive]] for all $a > 1$:
+{{begin-eqn}}
+{{eqn | l = r
+      | o = <
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | l = r - 1
+      | o = <
+      | r = 0
+      | c = Subtract $1$ from both sides
+}}
+{{eqn | ll= \leadsto
+      | l = a^{r - 1}
+      | o = <
+      | r = a^0
+      | c = [[Power Function on Base Greater than One is Strictly Increasing/Rational Number|Power Function on Base Greater than One is Strictly Increasing: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = a^{r - 1}
+      | o = <
+      | r = 1
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | ll= \leadsto
+      | l = -1 
+      | o = <
+      | r = -a^{r - 1}
+      | c = [[Order of Real Numbers is Dual of Order of their Negatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = 0 
+      | o = <
+      | r = 1 - a^{r - 1}
+      | c = adding $1$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = 0 
+      | o = <
+      | r = r \left({1 - a^{r - 1} }\right)
+      | c = multiplying both sides by $r > 0$
+}}
+{{eqn | ll= \leadsto
+      | l = 0 
+      | o = <
+      | r = D_a \map {g_r} a
+      | c = from the formula found for $D_a \map {g_r} a$ above
+}}
+{{end-eqn}}
+So $D_a \map {g_r} a$ is [[Definition:Positive Real Number|positive]] for all $a > 1$.
+Whence, by [[Derivative of Monotone Function]], $g_r$ is [[Definition:Increasing Real Function|increasing]] for all $a > 1$.
+Now:
+:$\map {g_r} 1 = r > 0$
+So $\map {g_r} a$ is [[Definition:Positive Real Number|positive]] for all $a > 1$.
+That is:
+:$0 < 1 + a r - a^r$
+Adding $a^r$ to both sides of the above yields:
+:$a^r < 1 + a r$
+Finally:
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = <
+      | r = r
+}}
+{{eqn | ll= \leadsto
+      | l = a^0
+      | o = <
+      | r = a^r
+      | c = [[Power Function on Base Greater than One is Strictly Increasing/Rational Number|Power Function on Base Greater than One is Strictly Increasing: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = <
+      | r = a^r
+      | c = {{Defof|Integer Power}}
+}}
+{{end-eqn}}
+So, for $0 < r < 1$:
+:$1 < a^r < 1 + a r$
+{{qed}}
+[[Category:Powers]]
+ee733l081xaewexuydgpzitbjpw99in
+\end{proof}<|endoftext|>
+\section{Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]] such that $a > 1$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f r = a^r$
+where $a^r$ denotes [[Definition:Rational Power|$a$ to the power of $r$]].
+Then:
+:$\displaystyle \lim_{r \mathop \to 0} \map f r = 1$
+\end{theorem}
+
+\begin{proof}
+We start by treating the [[Definition:Limit from Right|right-sided limit]].
+Let $0 < r < 1$.
+=== [[Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma|Lemma]] ===
+{{:Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma}}{{qed|lemma}}
+From the [[Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma|lemma]]:
+:$1 < a^r < 1 + a r$
+Also:
+{{begin-eqn}}
+{{eqn | l = \lim_{r \mathop \to 0} \paren {1 + a r}
+      | r = 1 + a \times 0
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{eqn | r = 1
+      | c = [[Real Zero is Zero Element]]
+}}
+{{eqn | r = \lim_{r \mathop \to 0} 1
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{end-eqn}}
+So from the [[Squeeze Theorem/Functions|Squeeze Theorem]]:
+:$\displaystyle \lim_{r \mathop \to 0^+} a^r = 1$
+We now treat the [[Definition:Limit from Left|left-sided limit]].
+Let $-1 < r < 0$.
+{{begin-eqn}}
+{{eqn | l = -1 
+      | o = < 
+      | m = r
+      | mo= < 
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = < 
+      | m = -r
+      | mo= < 
+      | r = 1
+      | c = [[Order of Real Numbers is Dual of Order of their Negatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = < 
+      | m = a^{-r}
+      | mo= < 
+      | r = 1 - a r
+      | c = [[Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma|Lemma]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = < 
+      | m = \frac 1 {a^r}
+      | mo= < 
+      | r = 1 - a r
+      | c = [[Real Number to Negative Power/Rational Number|Real Number to Negative Power: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {1 - a r}
+      | o = <
+      | m = a^r
+      | mo= <
+      | r = 1
+      | c = [[Ordering of Reciprocals]]
+}}
+{{end-eqn}}
+Also:
+{{begin-eqn}}
+{{eqn | l = \lim_{r \mathop \to 0} \dfrac 1 {1 - a r}
+      | r = \frac 1 {\displaystyle \lim_{r \mathop \to 0} \paren {1 - a r} }
+      | c = [[Quotient Rule for Limits of Functions]]
+}}
+{{eqn | r = \frac 1 {1 - a \times 0}
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{eqn | r = 1
+      | c = [[Real Zero is Zero Element]]
+}}
+{{eqn | r = \lim_{r \mathop \to 0} 1
+      | c = [[Real Polynomial Function is Continuous]]
+}}
+{{end-eqn}}
+So from the [[Squeeze Theorem/Functions|Squeeze Theorem]]:
+:$\displaystyle \lim_{r \mathop \to 0^-} a^r = 1$
+From [[Limit iff Limits from Left and Right]]:
+:$\displaystyle \lim_{r \mathop \to 0} a^r = 1$
+Hence the result.
+{{qed}}
+[[Category:Powers]]
+7akolrmpim9lbuhon3cb87n7o01jav3
+\end{proof}<|endoftext|>
+\section{Power Function on Base between Zero and One Tends to One as Power Tends to Zero/Rational Number}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]] such that $0 < a < 1$.
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f r = a^r$
+where $a^r$ denotes [[Definition:Rational Power|$a$ to the power of $r$]].
+Then:
+:$\displaystyle \lim_{r \mathop \to 0} \map f r = 1$
+\end{theorem}
+
+\begin{proof}
+From [[Ordering of Reciprocals]]:
+:$0 < a < 1 \implies 1 < \dfrac 1 a$
+So:
+{{begin-eqn}}
+{{eqn | l = \lim_{r \mathop \to 0} \paren {\frac 1 a}^r
+      | r = 1
+      | c = [[Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number|Power Function on Base greater than One tends to One as Power tends to Zero: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{r \mathop \to 0} \frac 1 {a^r}
+      | r = 1
+      | c = [[Exponent Combination Laws/Power of Quotient/Rational Numbers|Power of Quotient: Rational Numbers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {\displaystyle \lim_{r \mathop \to 0} a^r}
+      | r = 1
+      | c = [[Quotient Rule for Limits of Functions]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{r \mathop \to 0} a^r 
+      | r = 1
+      | c = taking reciprocal of each side
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Powers]]
+cy9oml498lz0mynci3npt3k51qjck6y
+\end{proof}<|endoftext|>
+\section{Euler's Number to Rational Power permits Unique Continuous Extension}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $e$ be [[Definition:Euler's Number|Euler's number]].
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$f \left({q}\right) = e^q$
+where $e^q$ denotes [[Definition:Rational Power| $e$ to the power of $q$]].
+Then there exists a [[Definition:Unique|unique]] [[Definition:Continuous Extension|continuous extension]] of $f$ to $\R$.
+\end{theorem}
+
+\begin{proof}
+Since $e > 0$, we may apply [[Power Function to Rational Power permits Unique Continuous Extension]].
+Hence the result.
+{{qed}}
+[[Category:Exponential Function]]
+j3aj4ucitdfj9juy3c0f2ti85dhfkmm
+\end{proof}<|endoftext|>
+\section{Open Unit Interval is Proper Subset of Closed Unit Interval}
+Tags: Real Intervals
+
+\begin{theorem}
+The [[Definition:Open Unit Interval|open unit interval]]:
+:$I_o = \left({0 \,.\,.\, 1}\right)$
+is a [[Definition:Proper Subset|proper subset]] of the [[Definition:Closed Unit Interval|closed unit interval]]:
+:$I_c = \left[{0 \,.\,.\, 1}\right]$
+\end{theorem}
+
+\begin{proof}
+Let $x \in I_o$.
+Then by definition:
+:$0 < x < 1$
+and so:
+:$0 \le x \le 1$
+and so:
+:$x \in I_c$.
+Thus:
+:$I_o \subseteq I_c$
+Consider:
+:$0 \in I_c$
+by definition of [[Definition:Closed Real Interval|closed interval]].
+But it is not the case that $0 < 0$.
+So $0 \notin I_o$ and so $I_c \not \subseteq I_o$.
+Hence the result by definition of [[Definition:Proper Subset|proper subset]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Cross Product satisfies Jacobi Identity}
+Tags: Algebra, Vector Cross Product
+
+\begin{theorem}
+Let $\mathbf a, \mathbf b, \mathbf c$ be [[Definition:Vector (Linear Algebra)|vectors]] in $3$ [[Definition:Dimension (Linear Algebra)|dimensional]] [[Definition:Euclidean Space|Euclidean space]].
+Let $\times$ denotes the [[Definition:Cross Product|cross product]].
+Then:
+:$\mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b} = \mathbf 0$
+That is, the [[Definition:Cross Product|cross product operation]] satisfies the [[Definition:Jacobi Identity|Jacobi identity]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b}
+      | r = \paren {\mathbf {a \cdot c} } \mathbf b - \paren {\mathbf {a \cdot b} } \mathbf c
+      | c = 
+}}
+{{eqn | o = 
+      | ro= +
+      | r = \paren {\mathbf {b \cdot a} } \mathbf c - \paren {\mathbf {b \cdot c} } \mathbf a
+      | c = 
+}}
+{{eqn | o =
+      | ro= +
+      | r = \paren {\mathbf {c \cdot b} } \mathbf a - \paren {\mathbf {c \cdot a} } \mathbf b
+      | c = [[Lagrange's Formula]]
+}}
+{{eqn | r = \mathbf 0
+      | c = [[Dot Product Operator is Commutative]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Algebra]]
+[[Category:Vector Cross Product]]
+6ec8ad0ti7cpssidwo3r3rb9zcdunlc
+\end{proof}<|endoftext|>
+\section{Meet-Continuous and Distributive implies Shift Mapping Preserves Finite Suprema}
+Tags: Distributive Lattices, Meet-Continuous Lattices, Complete Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Meet-Continuous Lattice|meet-continuous]] [[Definition:Distributive Lattice|distributive]] [[Definition:Complete Lattice|complete lattice]].
+Let $x \in S$.
+Let $f: S \to S$ be a [[Definition:Mapping|mapping]] such that
+:$\forall y \in S: f \left({y}\right) = x \wedge y$
+Then
+:$f$ [[Definition:Mapping Preserves Supremum/Finite|preserves finite suprema]]
+\end{theorem}
+
+\begin{proof}
+Let $X$ be [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$ such that:
+:$X$ admits a [[Definition:Supremum of Set|supremum]]
+By definition of [[Definition:Complete Lattice|complete lattice]]:
+:$f \left[{X}\right]$ admits a [[Definition:Supremum of Set|supremum]]
+We will prove the result by [[Principle of Mathematical Induction|induction]] on the [[Definition:Cardinality|cardinality]] of $X$.
+=== Basis Case ===
+:$\forall X \subseteq S: \left\vert{X}\right\vert = 0 \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$
+where:
+:$\left\vert{X}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $X$
+Let $X \subseteq S$ such that:
+:$\left\vert{X}\right\vert = 0$
+By [[Cardinality of Empty Set]]:
+:$X = \varnothing$
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$f \left[{X}\right] = \varnothing$
+By definitions of [[Definition:Bottom (Lattice Theory)/Definition 1|bottom]] and [[Definition:Smallest Element|smallest element]]:
+:$\bot \preceq x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sup \left({f \left[{X}\right]}\right)
+      | r = \bot
+      | c = [[Supremum of Empty Set is Smallest Element]]
+}}
+{{eqn | r = x \wedge \bot
+      | c = [[Preceding iff Meet equals Less Operand]]
+}}
+{{eqn | r = x \wedge \sup X
+      | c = [[Supremum of Empty Set is Smallest Element]]
+}}
+{{eqn | r = f \left({\sup X}\right)
+      | c = definition of $f$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Induction Hypothesis ===
+:$\forall X \subseteq S: \left\vert{X}\right\vert = n \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$
+=== Induction Step ===
+:$\forall X \subseteq S: \left\vert{X}\right\vert = n + 1 \implies \sup \left({f \left[{X}\right]}\right) = f \left({\sup X}\right)$
+Let $X \subseteq S$ such that:
+:$\left\vert{X}\right\vert = n + 1$
+Then $X = \left\{ {x_1, \dots, x_n, x_{n+1} }\right\}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sup \left({ f \left[{X}\right] }\right)
+      | r = \sup \left({ f \left[{\left\{ {x_1, \dots, x_n}\right\} \cup \left\{ {x_{n+1} }\right\} }\right] }\right)
+      | c = [[Union of Unordered Tuples]]
+}}
+{{eqn | r = \sup \left({ f \left[{ \left\{ {x_1, \dots, x_n}\right\} }\right] \cup f \left[{ \left\{ {x_{n+1} }\right\} }\right] }\right)
+      | c = [[Image of Union under Mapping]]
+}}
+{{eqn | r = \sup \left({f \left[{ \left\{ {x_1, \dots, x_n}\right\} }\right] }\right) \vee \sup \left({ f \left[{ \left\{ {x_{n+1} }\right\} }\right] }\right)
+      | c = [[Supremum of Suprema]]
+}}
+{{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \sup \left({ f \left[{\left\{ {x_{n+1} }\right\} }\right] }\right)
+      | c = Induction Hypothesis
+}}
+{{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \sup \left\{ { f \left({ x_{n+1} }\right) }\right\}
+      | c = [[Image of Singleton under Mapping]]
+}}
+{{eqn | r = f \left({ \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee f \left({ x_{n+1} }\right)
+      | c = [[Supremum of Singleton]]
+}}
+{{eqn | r = \left({ x \wedge \sup \left\{ {x_1, \dots, x_n}\right\} }\right) \vee \left({ x \wedge x_{n+1} }\right)
+      | c = Definition of $f$
+}}
+{{eqn | r = x \wedge \left({ \sup \left\{ {x_1, \dots, x_n}\right\} \vee x_{n+1} }\right)
+      | c = {{Defof|Distributive Lattice}}
+}}
+{{eqn | r = x \wedge \left({ \sup \left\{ {x_1, \dots, x_n}\right\} \vee \sup \left\{ {x_{n+1} }\right\} }\right)
+      | c = [[Supremum of Singleton]]
+}}
+{{eqn | r = x \wedge \sup X
+      | c = [[Supremum of Suprema]]
+}}
+{{eqn | r = f \left({\sup X}\right)
+      | c = definition of $f$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Union is not Cancellable}
+Tags: Set Union
+
+\begin{theorem}
+[[Definition:Set Union|Set union]] is not a [[Definition:Cancellable Operation|cancellable operation]].
+That is, for a given $A, B, C \subseteq S$ for some $S$, it is not always the case that:
+:$A \cup B = A \cup C \implies B = C$
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S = \set {a, b}$.
+Let:
+:$A = \set {a, b}$
+:$B = \set a$
+:$C = \set b$
+Then:
+{{begin-eqn}}
+{{eqn | l = A \cup B
+      | r = \set {a, b}
+      | c = 
+}}
+{{eqn | r = A \cup C
+      | c = 
+}}
+{{end-eqn}}
+but:
+{{begin-eqn}}
+{{eqn | l = B
+      | r = \set a
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = \set b
+      | c = 
+}}
+{{eqn | r = C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set Intersection is not Cancellable}
+Tags: Set Intersection
+
+\begin{theorem}
+[[Definition:Set Intersection|Set intersection]] is not a [[Definition:Cancellable Operation|cancellable operation]].
+That is, for a given $A, B, C \subseteq S$ for some $S$, it is not always the case that:
+:$A \cap B = A \cap C \implies B = C$
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Let $S = \left\{ {a, b, c}\right\}$.
+Let:
+:$A = \left\{ {a}\right\}$
+:$B = \left\{ {a, b}\right\}$
+:$C = \left\{ {a, c}\right\}$
+Then:
+{{begin-eqn}}
+{{eqn | l = A \cap B
+      | r = \left\{ {a}\right\}
+      | c = 
+}}
+{{eqn | r = A \cap C
+      | c = 
+}}
+{{end-eqn}}
+but:
+{{begin-eqn}}
+{{eqn | l = B
+      | r = \left\{ {a, b}\right\}
+      | c = 
+}}
+{{eqn | o = \ne
+      | r = \left\{ {a, c}\right\}
+      | c = 
+}}
+{{eqn | r = C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Real Intervals is Real Interval}
+Tags: Real Intervals
+
+\begin{theorem}
+Let $I_1$ and $I_2$ be [[Definition:Real Interval|real intervals]].
+Then $I_1 \cap I_2$ is also a [[Definition:Real Interval|real interval]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in I_1 \cap I_2$. 
+From the definition of a [[Definition:Real Interval/Definition 1|real interval]], it suffices to show that:
+:for each $z \in \R$ with $x \le z \le y$ we have $z \in I_1 \cap I_2$.
+Let $z$ be a [[Definition:Real Number|real number]] with:
+:$x \le z \le y$
+Since $x, y \in I_1$, we have: 
+:$z \in I_1$
+from the definition of a [[Definition:Real Interval/Definition 1|real interval]].
+Similarly, since $x, y \in I_2$, we have:
+:$z \in I_2$
+from the definition of a [[Definition:Real Interval|real interval]].
+So:
+:$z \in I_1 \cap I_2$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Real Intervals is not necessarily Real Interval}
+Tags: Real Intervals
+
+\begin{theorem}
+Let $I_1$ and $I_2$ be [[Definition:Real Interval|real intervals]].
+Then $I_1 \cup I_2$ is not necessarily a [[Definition:Real Interval|real interval]].
+\end{theorem}
+
+\begin{proof}
+[[Proof by Counterexample]]:
+Consider the [[Definition:Real Interval|real intervals]]:
+:$I_1 = \left({0 \,.\,.\, 2}\right)$
+:$I_2 = \left({4 \,.\,.\, 6}\right)$
+Then we have that:
+:$1 < 3 < 5$
+where:
+:$1 \in I_1 \cup I_2$
+:$5 \in I_1 \cup I_2$
+but:
+:$3 \notin I_1 \cup I_2$
+Thus $I_1 \cup I_2$ is not a [[Definition:Real Interval|real interval]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pasting Lemma/Corollary 1}
+Tags: Continuous Mappings
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Closed Set (Topology)|closed]] in $X$.
+Let $f : A \to Y$ and $g : B \to Y$ be [[Definition:Continuous Mapping on Set|continuous mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
+Then the mapping $f \cup g : A \cup B \to Y$ is [[Definition:Continuous Mapping on Set|continuous]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Continuous Mappings]]
+3r4ikb27sdc32d03jsirntf697vfqiq
+\end{proof}<|endoftext|>
+\section{Pasting Lemma/Corollary 2}
+Tags: Continuous Mappings
+
+\begin{theorem}
+Let $\mathcal A = \left\{ {A_i: i \in I} \right\}$ be a [[Definition:Set of Sets|set of sets]] that are [[Definition:Open Set (Topology)|open]] in $X$.
+Let $f: \bigcup \mathcal A \to Y$ be a [[Definition:Mapping|mapping]] such that:
+:$\forall i \in I : f \restriction A_i$ is [[Definition:Continuous Mapping on Set|continuous]]
+Then $f$ is [[Definition:Continuous Mapping on Set|continuous]] on $\bigcup \mathcal A$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Continuous Mappings]]
+fbvoxqmu3eco70lwyandhew9rgrxv66
+\end{proof}<|endoftext|>
+\section{Pasting Lemma}
+Tags: Continuous Mappings, Named Theorems
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
+Let $A$ and $B$ be [[Definition:Open Set (Topology)|open]] in $X$.
+Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Continuous Mapping on Set|continuous mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
+Let $f \cup g$ be the [[Definition:Set Union|union]] of the [[Definition:Mapping|mappings]] $f$ and $g$:
+:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$
+Then the mapping $f \cup g : A \cup B \to Y$ is [[Definition:Continuous Mapping on Set|continuous]].
+\end{theorem}
+
+\begin{proof}
+First it is shown that $f \cup g$ is [[Definition:Well-Defined Mapping|well-defined]].
+It is sufficient to check that $f \cup g$ maps any $x \in A \cap B$ to a single value in $Y$.
+Let $x \in A \cap B$.
+Because $x \in A$:
+:$\map {f \cup g} x = \map f x \in Y$
+Because $x \in B$:
+:$\map {f \cup g} x = \map g x \in Y$
+But by definition of [[Definition:Agreement of Mappings|agreement of mappings]]:
+:$\map f x = \map g x$
+so $f \cup g$ maps $x$ to a unique value in $Y$, as was to be shown.
+{{qed|lemma}}
+Next it is shown that $f \cup g$ is [[Definition:Continuous Mapping on Set|continuous]]:
+We show that $\map {\paren {f \cup g}^{-1} } U$ is [[Definition:Open Set (Topology)|open]] in $A \cup B$ for every [[Definition:Open Set (Topology)|open]] $U$ in $Y$.
+{{tidy|The following needs to be progressive rather than regressive.}}
+Observe first that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$.
+For if $x \in (f \cup g)^{-1}(U) \subseteq A \cup B$, then $(f \cup g)(x) \in U$; but if $x \in A$, then $f(x) = (f \cup g)(x) \in U$ i.e. $x \in f^{-1}(U)$ or if $x \in B$, then $g(x) = (f \cup g)(x) \in U$ i.e. $x \in g^{-1}(U)$. In both cases, $x \in f^{-1}(U) \cup g^{-1}(U)$. Conversely if $x \in f^{-1}(U) \cup g^{-1}(U)$, then without loss of generality consider the case of $x \in f^{-1}(U) \subseteq A$; in this case $(f \cup g)(x) = f(x) \in U$ so that $x \in (f \cup g)^{-1}(U)$.
+Having established that $(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U)$, note that by continuity of $f$ and $g$, we have that $f^{-1}(U)$ is open in $A$ and $g^{-1}(U)$ is open in $B$. So there are an open sets $P$ and $Q$ of $X$ such that $f^{-1}(U) = P \cap A$ and $g^{-1}(U) = Q \cap B$. Since $A$ and $B$ are open in $X$ and intersections of open sets are open, this implies that $f^{-1}(U)$ and $g^{-1}(U)$ are open in $X$. Furthermore, note that $P \cap A$ and $Q \cap B$ are both subsets of $A \cup B$. Hence, $f^{-1}(U) = f^{-1}(U) \cap (A \cup B)$ and $g^{-1}(U) = g^{-1}(U) \cap (A \cup B)$.
+Thus:
+:$(f \cup g)^{-1}(U) = f^{-1}(U) \cup g^{-1}(U) = (f^{-1}(U) \cup g^{-1}(U)) \cap (A \cup B)$
+demonstrating that $(f \cup g)^{-1}(U)$ is open in the subspace topology of $A \cup B$.
+{{qed}}
+[[Category:Continuous Mappings]]
+[[Category:Named Theorems]]
+2oyqi031kxgrr5k967kl1gpj0jfzjph
+\end{proof}<|endoftext|>
+\section{Domain of Real Square Function}
+Tags: Square Function
+
+\begin{theorem}
+The [[Definition:Domain of Real Function|domain]] of the [[Definition:Real Square Function|real square function]] is the entire [[Definition:Real Number|set of real numbers]] $\R$.
+\end{theorem}
+
+\begin{proof}
+The operation of [[Definition:Real Multiplication|real multiplication]] is defined on all [[Definition:Real Number|real numbers]].
+Thus:
+:$\forall x \in \R: \exists y \in \R: x^2 = y$
+Hence the result by definition of [[Definition:Domain of Real Function|domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Real Square Function}
+Tags: Square Function
+
+\begin{theorem}
+The [[Definition:Image of Mapping|image]] of the [[Definition:Real Square Function|real square function]] is the entire [[Definition:Positive Real Number|set of positive real numbers]] $\R_{\ge 0}$.
+\end{theorem}
+
+\begin{proof}
+From [[Square of Real Number is Non-Negative]], the [[Definition:Image of Mapping|image]] of $f$ is $\R_{\ge 0}$.
+From [[Positive Real has Real Square Root]]:
+:$\forall x \in \R: \exists y \in \R: x^2 = y$
+Hence the result by definition of [[Definition:Image of Mapping|image]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Domain of Real Square Root Function}
+Tags: Real Functions
+
+\begin{theorem}
+The [[Definition:Domain of Real Function|domain]] of the [[Definition:Real Square Root Function|real square root function]] is the [[Definition:Positive Real Number|set of positive real numbers]] $\R_{\ge 0}$:
+:$\left\{ {x \in \R: x \ge 0}\right\}$
+\end{theorem}
+
+\begin{proof}
+From [[Square of Real Number is Non-Negative]]:
+:$\forall x \in \R: x^2 \ge 0$
+Hence the result by definition of [[Definition:Domain of Real Function|domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Real Square Root Function}
+Tags: Real Functions
+
+\begin{theorem}
+The [[Definition:Image of Mapping|image]] of the [[Definition:Real Square Root Function|real square root function]] is the entire [[Definition:Positive Real Number|set of positive real numbers]] $\R_{\ge 0}$.
+\end{theorem}<|endoftext|>
+\section{Exponent Combination Laws/Product of Powers/Proof 1}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $a \in \R_{> 0}$ be a [[Definition:Positive Real Number|positive real number]].
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $a^x$ be defined as [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then:
+:$a^x a^y = a^{x + y}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a^{x + y}
+      | r = \map \exp {\paren {x + y} \ln a}
+      | c = {{Defof|Power to Real Number}}
+}}
+{{eqn | r = \map \exp {x \ln a + y \ln a}
+      | c = 
+}}
+{{eqn | r = \map \exp {x \ln a} \, \map \exp {y \ln a}
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = a^x a^y
+      | c = Definition of [[Definition:Power to Real Number|Power to Real Number]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{2197 is Cube of 13}
+Tags: Numbers
+
+\begin{theorem}
+:$13^3 = 2197$
+\end{theorem}
+
+\begin{proof}
+By [[Definition:Long Multiplication|long multiplication]]:
+<pre>
+   13
+ x 13
+-----
+   39
+  130
+-----
+  169
+-----
+</pre>
+then:
+<pre>
+ x 169
+    13
+------
+   507
+  1690
+-----
+  2197
+------
+</pre>
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inequality of Product of Unequal Numbers}
+Tags: Inequalities, Real Multiplication
+
+\begin{theorem}
+Let $a, b, c, d \in \R$.
+Then:
+:$0 < a < b \land 0 < c < d \implies 0 < a c < b d$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 1
+      | o = 
+      | r = 0 < a < b
+      | c = 
+}}
+{{eqn | n = 2
+      | o = \leadsto
+      | r = 0 < b
+      | c = [[Definition:Ordering/Definition 1|Ordering]] is [[Definition:Transitive|Transitive]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | n = 3
+      | o = 
+      | r = 0 < c < d
+      | c = 
+}}
+{{eqn | n = 4
+      | o = \leadsto
+      | r = 0 < c
+      | c = [[Definition:Ordering/Definition 1|Ordering]] is [[Definition:Transitive|Transitive]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | o = \leadsto
+      | r = 0 < a c < b c
+      | c = $(1)$ and $(4)$: [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication]]
+}}
+{{eqn | o = 
+      | r = 0 < b c < b d
+      | c = $(2)$ and $(3)$: [[Definition:Real Number Axioms|Real Number Axioms: $\R \text O 2$: Usual ordering is compatible with multiplication]]
+}}
+{{eqn | o = \leadsto
+      | r = 0 < a c < b d
+      | c = [[Definition:Ordering/Definition 1|Ordering]] is [[Definition:Transitive|Transitive]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Inequalities]]
+[[Category:Real Multiplication]]
+8tqah01rau1ebolhjvbwd6gc0gn332y
+\end{proof}<|endoftext|>
+\section{Negative of Absolute Value/Corollary 3}
+Tags: Negative of Absolute Value
+
+\begin{theorem}
+Let $y \in \R_{\ge 0}$.
+Let $z \in \R$.
+Then:
+:$\left\vert{x - z}\right\vert < y \iff z - y < x < z + y$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left\vert{x - z}\right\vert
+      | o = <
+      | m = y
+      | c = 
+}}
+{{eqn | ll= \iff
+      | l = -y
+      | o = <
+      | m = x - z
+      | mo= <
+      | r = y
+      | c = [[Negative of Absolute Value/Corollary 1|Negative of Absolute Value: Corollary 1]]
+}}
+{{eqn | ll= \iff
+      | l = z - y
+      | o = <
+      | m = x
+      | mo= <
+      | r = z + y
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Negative of Absolute Value]]
+bectnqs04v7m7wndbdbq7o8oomwoumw
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Product of Powers/Proof 2/Lemma}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $x_1, x_2, y_1, y_2 \in \R_{>0}$ be [[Definition:Strictly Positive Real Number|strictly positive real numbers]].
+Let $\epsilon \in \openint 0 {\min \set {y_1, y_2, 1} }$.
+Then:
+:$\size {x_1 - y_1} < \epsilon \land \size {x_2 - y_2} < \epsilon \implies \size {x_1 x_2 - y_1 y_2} < \epsilon \paren {y_1 + y_2 + 1}$
+\end{theorem}
+
+\begin{proof}
+First:
+{{begin-eqn}}
+{{eqn | l = \epsilon
+      | o = <
+      | r = \min \set {y_1, y_2, 1}
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon
+      | o = <
+      | r = y_1
+      | c = {{Defof|Min Operation}}
+}}
+{{eqn | ll= \leadsto
+      | l = \epsilon - \epsilon
+      | o = <
+      | r = y_1 - \epsilon
+      | c = subtracting $\epsilon$ from both sides
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = y_1 - \epsilon
+}}
+{{end-eqn}}
+The same logic, [[Definition:Mutatis Mutandis|mutatis mutandis]], shows that $0 < y_2 - \epsilon$.
+From [[Negative of Absolute Value/Corollary 3|Negative of Absolute Value: Corollary 3]]:
+{{begin-eqn}}
+{{eqn | l = \size {x_1 - y_1} < \epsilon
+      | o = \implies
+      | r = y_1 - \epsilon < x_1 < y_1 - \epsilon
+}}
+{{eqn | l = \size {x_2 - y_2} < \epsilon
+      | o = \implies
+      | r = y_2 - \epsilon < x_2 < y_2 - \epsilon
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {y_1 - \epsilon} \paren {y_2 - \epsilon}
+      | o = <
+      | m = x_1 x_2
+      | mo= <
+      | r = \paren {y_1 + \epsilon} \paren {y_1 + \epsilon}
+      | c = [[Inequality of Product of Unequal Numbers]]: from $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 y_2 - \epsilon \paren {y_1 + y_2} + \epsilon^2
+      | o = <
+      | m = x_1 x_2
+      | mo= <
+      | r = y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2
+      | c = [[Multiplication of Numbers Distributes over Addition]]
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 y_2 - \epsilon \paren {y_1 + y_2} - \epsilon^2
+      | o = <
+      | m = x_1 x_2
+      | mo= <
+      | r = y_1 y_2 + \epsilon \paren {y_1 + y_2} + \epsilon^2
+      | c = [[Square of Non-Zero Real Number is Strictly Positive]]
+}}
+{{end-eqn}}
+Subtracting $y_1 y_2$ from all sections of the inequality:
+:$-\epsilon \paren {y_1 + y_2} - \epsilon^2 < x_1 x_2 - y_1 y_2 < \epsilon \paren {y_1 + y_2} + \epsilon^2$
+If follows that:
+{{begin-eqn}}
+{{eqn | l = \size {x_1 x_2 - y_1 y_2}
+      | o = <
+      | r = \epsilon \paren {y_1 + y_2} + \epsilon^2
+      | c = [[Negative of Absolute Value/Corollary 1|Negative of Absolute Value: Corollary 1]]
+}}
+{{eqn | o = <
+      | r = \epsilon \paren {y_1 + y_2} + \epsilon
+      | c = Since $\epsilon < \min \size {y_1, y_2, 1} < 1$, we may apply [[Real Number between Zero and One is Greater than Square]]
+}}
+{{eqn | o = <
+      | r = \epsilon \paren {y_1 + y_2 + 1}
+      | c = [[Multiplication of Numbers Distributes over Addition]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Exponent Combination Laws]]
+j6l3ndjrgtgkvp631xr58058fdbl8x6
+\end{proof}<|endoftext|>
+\section{Union of Unordered Tuples}
+Tags: Set Theory
+
+\begin{theorem}
+Let $x_1, \dots, x_n, x_{n+1}, \dots, x_m$ be arbitrary.
+Then
+:$\set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m} = \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$
+\end{theorem}
+
+\begin{proof}
+Let $a$ be arbitrary.
+:$a \in \set {x_1, \dots, x_n} \cup \set {x_{n + 1}, \dots, x_m}$
+{{iff}}
+:$a \in \set {x_1, \dots, x_n}$ or $a \in \set {x_{n + 1}, \dots, x_m}$ by definition of [[Definition:Set Union|union]]
+{{iff}}
+:$a = x_1 \lor \dots \lor a = x_n$ or $a = x_{n + 1} \lor \dots \lor a = x_m$ by definition of [[Definition:Unordered Tuple|unordered tuple]]
+{{iff}}
+:$a = x_1 \lor \dots \lor a = x_n \lor a = x_{n + 1} \lor \dots \lor a = x_m$
+{{iff}}
+:$a \in \set {x_1, \dots, x_n, x_{n + 1}, \dots, x_m}$ by definition of [[Definition:Unordered Tuple|unordered tuple]]
+Thus result follows by definition of [[Definition:Set Equality|set equality]].
+{{qed}}
+[[Category:Set Theory]]
+e5gxlqozltwspxusogyf4rkscz2r779
+\end{proof}<|endoftext|>
+\section{Gauss-Lucas Theorem}
+Tags: Complex Analysis
+
+\begin{theorem}
+Let $P$ be a (nonconstant) polynomial with complex coefficients.
+Then all [[Definition:Root of Function|zeros]] of its [[Definition:Derivative|derivative]] $P'$  belong to the convex hull of the set of zeros of $P$.
+\end{theorem}
+
+\begin{proof}
+Over the complex numbers, $P$ is a product of prime factors:
+:$\displaystyle \map P z = \alpha \prod_{i \mathop = 1}^n (z-a_i)$
+where:
+:the complex numbers $a_1, a_2, \ldots, a_n$ are the (not necessary distinct) zeros of the polynomial $P$
+:the complex number $\alpha$ is the leading coefficient of $P$
+:$n$ is the degree of $P$.
+Let $z$ be any complex number for which $\map P z \ne 0$.
+Then we have for the [[Definition:Logarithmic Derivative|logarithmic derivative]]:
+:$\displaystyle \frac {\map {P'} z} {\map P z} = \sum_{i \mathop = 1}^n \frac 1 {z - a_i}$
+In particular, if $z$ is a zero of $P'$ and $\map P z \ne 0$, then:
+:$\displaystyle \sum_{i \mathop = 1}^n \frac 1 {z - a_i} = 0$
+or:
+:$\displaystyle \sum_{i \mathop = 1}^n \frac {\overline z - \overline {a_i} } {\size {z - a_i}^2} = 0$
+This may also be written as:
+ 
+:$\displaystyle \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} } \overline z = \paren {\sum_{i \mathop = 1}^n \frac 1 {\size {z - a_i}^2} \overline {a_i} }$
+Taking their conjugates, we see that $z$ is a weighted sum with positive coefficients that sum to one, or the [[Definition:Barycentric Coordinates (Astronomy)|barycenter on affine coordinates]], of the complex numbers $a_i$ (with different mass assigned on each root whose weights collectively sum to $1$).
+If $\map P z = \map {P'} z = 0$, then:
+:$z = 1 \cdot z + 0 \cdot a_i$
+and is still a [[Definition:Convex Combination|convex combination]] of the roots of $P$.
+{{qed}}
+{{Namedfor|Carl Friedrich Gauss|name2 = François Édouard Anatole Lucas|cat = Gauss|cat2 = Lucas}}
+[[Category:Complex Analysis]]
+pbpg6q87ucyirl7spobfu56x1sv47pf
+\end{proof}<|endoftext|>
+\section{Way Below implies Preceding}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x, y \in S$ such that
+:$x \ll y$
+where $\ll$ denotes [[Definition:Element is Way Below|element is way below second element]].
+Then
+:$x \preceq y$
+\end{theorem}
+
+\begin{proof}
+By [[Singleton is Directed and Filtered Subset]]:
+:$\left\{ {y}\right\}$ is [[Definition:Directed Subset|directed]].
+By [[Supremum of Singleton]]:
+:$\left\{ {y}\right\}$ admits a [[Definition:Supremum of Set|supremum]] and $\sup \left\{ {y}\right\} = y$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$y \preceq \sup \left\{ {y}\right\}$
+By definition of [[Definition:Element is Way Below|way below]]:
+:$\exists d \in \left\{ {y}\right\}: x \preceq d$
+Thus by definition of [[Definition:Singleton|singleton]]:
+:$x \preceq y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding and Way Below implies Way Below}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $u, x, y, z \in S$ such that
+:$u \preceq x \ll y \preceq z$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+Then
+:$u \ll z$
+\end{theorem}
+
+\begin{proof}
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$D$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$z \preceq \sup D$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$y \preceq \sup D$
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d \in D: x \preceq d$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$\exists d \in D: u \preceq d$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+:$u \ll z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Join is Way Below if Operands are Way Below}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, vee, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $x, y, z \in S$ such that
+:$x \ll z$ and $y \ll z$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+Then
+:$x \vee y \ll z$
+\end{theorem}
+
+\begin{proof}
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$D$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$z \preceq \sup D$
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d_1 \in D: x \preceq d_1$
+and
+:$\exists d_2 \in D: y \preceq d_2$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists d \in D: d_1 \preceq d$ and $d_2 \preceq d$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq d$ and $y \preceq d$
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$x \vee y \preceq d$
+Thus by definition [[Definition:Element is Way Below|way below relation]]:
+:$x \vee y \ll z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Relation is Transitive}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x, y, z \in S$ such that
+:$x \ll y \ll z$
+Then
+:$x \ll z$
+\end{theorem}
+
+\begin{proof}
+By [[Way Below implies Preceding]]:
+:$x \preceq y$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$z \preceq z$
+Thus by [[Preceding and Way Below implies Way Below]]:
+:$x \ll z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Relation is Antisymmetric}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $x, y \in S$ such that
+:$x \ll y$ and $y \ll x$
+Then
+:$x = y$
+\end{theorem}
+
+\begin{proof}
+By [[Way Below implies Preceding]]:
+:$x \preceq y$ and $y \preceq x$
+Thus by definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$x = y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continued Fraction Expansion of Golden Mean}
+Tags: Golden Mean, Continued Fractions
+
+\begin{theorem}
+The [[Definition:Golden Mean|golden mean]] has the simplest possible [[Definition:Continued Fraction Expansion of Irrational Number|continued fraction expansion]], namely $\sqbrk {1, 1, 1, 1, \ldots}$:
+:$\phi = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$x = 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots} } }$
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 1 + \frac 1 x
+      | c = substituting for $x$
+}}
+{{eqn | ll= \leadsto
+      | l = x^2
+      | r = x + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 - x - 1
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+The result follows from [[Golden Mean as Root of Quadratic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fibonacci Number less than Golden Section to Power less One}
+Tags: Fibonacci Numbers, Golden Mean
+
+\begin{theorem}
+For all $n \in \N_{> 0}$:
+:$F_n \le \phi^{n - 1}$
+where:
+:$F_n$ is the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]]
+:$\phi$ is the [[Definition:Golden Section|golden section]]: $\phi = \dfrac {1 + \sqrt 5} 2$
+\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]]:
+:$F_n \le \phi^{n - 1}$
+=== Basis for the Induction ===
+$P \left({1}\right)$ is true, as this just says:
+:$F_1 = 1 = \phi^0 = \phi^{1 - 1}$
+It is also necessary to demonstrate $P \left({2}\right)$ is true:
+:$F_2 = 1 \le \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{2 - 1}$
+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, for all $1 \le k \le n$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$F_k \le \phi^{k - 1}$
+from which it is to be shown that:
+:$F_{k + 1} \le \phi^k$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = F_{k + 1}
+      | r = F_{k - 1} + F_k
+      | c = Definition of [[Definition:Fibonacci Numbers|Fibonacci Numbers]]
+}}
+{{eqn | o = \le
+      | r = \phi^{k - 2} + \phi^{k - 1}
+      | c = [[Fibonacci Number less than Golden Section to Power less One#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \phi^{k - 2} \left({1 + \phi}\right)
+      | c = 
+}}
+{{eqn | r = \phi^{k - 2} \left({\phi^2}\right)
+      | c = [[Square of Golden Mean equals One plus Golden Mean]]
+}}
+{{eqn | r = \phi^k
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Second Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N_{> 0}: F_n \le \phi^{n - 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Square of Golden Mean equals One plus Golden Mean}
+Tags: Golden Mean
+
+\begin{theorem}
+:$\phi^2 = \phi + 1$
+where $\phi$ denotes the [[Definition:Golden Mean|golden mean]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \phi
+      | r = \frac 1 {\phi - 1}
+      | c = Definition of [[Definition:Golden Mean/Definition 3|Golden Mean]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \phi \paren {\phi - 1}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \phi^2
+      | r = \phi + 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fibonacci Number greater than Golden Section to Power less Two}
+Tags: Fibonacci Numbers, Golden Mean
+
+\begin{theorem}
+For all $n \in \N_{\ge 2}$:
+:$F_n \ge \phi^{n - 2}$
+where:
+:$F_n$ is the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]]
+:$\phi$ is the [[Definition:Golden Section|golden section]]: $\phi = \dfrac {1 + \sqrt 5} 2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{\ge 2}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$F_n \ge \phi^{n - 2}$
+=== Basis for the Induction ===
+$P \left({2}\right)$ is true, as this just says:
+:$F_2 = 1 = \phi^0 = \phi^{2 - 2}$
+It is also necessary to demonstrate $P \left({3}\right)$ is true:
+:$F_3 = 2 \ge \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{3 - 1}$
+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, for all $1 \le k \le n$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$F_k \ge \phi^{k - 2}$
+from which it is to be shown that:
+:$F_{k + 1} \ge \phi^{k - 1}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = F_{k + 1}
+      | r = F_{k - 1} + F_k
+      | c = Definition of [[Definition:Fibonacci Numbers|Fibonacci Numbers]]
+}}
+{{eqn | o = \ge
+      | r = \phi^{k - 3} + \phi^{k - 2}
+      | c = [[Fibonacci Number greater than Golden Section to Power less Two#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \phi^{k - 3} \left({1 + \phi}\right)
+      | c = 
+}}
+{{eqn | r = \phi^{k - 3} \left({\phi^2}\right)
+      | c = [[Square of Golden Mean equals One plus Golden Mean]]
+}}
+{{eqn | r = \phi^{k - 1}
+      | c = 
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Second Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N_{\ge 2}: F_n \ge \phi^{n - 2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Triangular Number as Alternating Sum and Difference of Squares}
+Tags: Triangular Numbers, Square Numbers
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | ll= \forall n \in \N: 
+      | l = \frac {n \paren {n + 1} } 2
+      | r = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2
+      | c = 
+}}
+{{eqn | r = n^2 - \paren {n - 1}^2 + \paren {n - 2}^2 - \cdots + \paren {-1}^{n - 1} 
+      | c = 
+}}
+{{end-eqn}}
+Thus the $n$th [[Definition:Triangular Number|triangular number]] can be expressed as the alternating sum and difference of [[Definition:Square Number|squares]]:
+So:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = 1^2
+      | c = 
+}}
+{{eqn | l = 3
+      | r = 2^2 - 1^2
+      | c = 
+}}
+{{eqn | l = 6
+      | r = 3^2 - 2^2 + 1^2
+      | c = 
+}}
+{{eqn | l = 10
+      | r = 4^2 - 3^2 + 2^2 - 1^2
+      | c = 
+}}
+{{end-eqn}}
+and so on.
+\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]]:
+:$\displaystyle \frac {n \paren {n + 1} } 2 = \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j \paren {n - j}^2$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \frac {n \paren {n + 1} } 2
+      | r = 1
+      | c = 
+}}
+{{eqn | r = 1^2
+      | c = 
+}}
+{{end-eqn}}
+which 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]]:
+:$\displaystyle \frac {k \paren {k + 1} } 2 = \sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2$
+from which it is to be shown that:
+:$\displaystyle \frac {\paren {k + 1} \paren {k + 2} } 2 = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^k \paren {-1}^j \paren {k - j + 1}^2
+      | r = \paren {k + 1}^2 - \paren {\sum_{j \mathop = 0}^{k - 1} \paren {-1}^j \paren {n - j}^2}
+      | c = 
+}}
+{{eqn | r = \paren {k + 1}^2 - \frac {k \paren {k + 1} } 2
+      | c = [[Triangular Number as Alternating Sum and Difference of Squares#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \frac {2 k^2 + 4 k + 2 - k^2 - k} 2
+      | c = multiplying everything out
+}}
+{{eqn | r = \frac {k^2 + 3 k + 2} 2
+      | c = simplifying
+}}
+{{eqn | r = \frac {\paren {k + 1} \paren {k + 2} } 2
+      | c = factoring
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Inequality/Corollary}
+Tags: Real Analysis, Inequalities, Bernoulli's Inequality
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $0 < x < 1$.
+Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
+Then:
+:$\left({1 - x}\right)^n \ge 1 - n x$
+\end{theorem}<|endoftext|>
+\section{2 to the n is Greater than n Cubed when n is 10 and above}
+Tags: Number Theory, 2, 10
+
+\begin{theorem}
+:$\forall n \in \Z, n \ge 10: 2^n > n^3$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z$ such that $n \ge 10$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$2^n > n^3$
+We note that:
+:$2^9 = 512 < 729 = 9^3$
+so when $n < 10$ the [[Definition:Proposition|proposition]] does not hold.
+=== Basis for the Induction ===
+$P \left({10}\right)$ is the case:
+:$2^{10} = 1024 > 1000 = 10^3$
+so $P \left({10}\right)$ is seen to hold.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 10$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$2^k > k^3$
+We need to show that:
+:$2^{k + 1} > \left({k + 1}\right)^3$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+We note that when $k \ge 10$:
+{{begin-eqn}}
+{{eqn | l = \left({1 + \dfrac 1 k}\right)^3
+      | r = 1 + \frac 3 k + \frac 3 {k^2} + \frac 1 {k^3}
+      | c = [[Binomial Theorem/Integral Index|Binomial Theorem]]
+}}
+{{eqn | o = \le
+      | r = 1 + \frac 3 {10} + \frac 3 {100} + \frac 1 {1000}
+      | c = 
+}}
+{{eqn | r = 1 + \frac {300 + 30 + 1} {1000}
+      | c = 
+}}
+{{eqn | n = 1
+      | o = <
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = 2^{k + 1}
+      | r = 2 \times 2^k
+}}
+{{eqn | o = >
+      | r = \left({1 + \dfrac 1 k}\right)^3 2^k
+      | c = from $(1)$
+}}
+{{eqn | o = >
+      | r = \left({1 + \dfrac 1 k}\right)^3 k^3
+      | c = [[2 to the n is Greater than n Cubed when n is 10 and above#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = k^3 + \frac {3 k^3} k + \frac {3 k^3} {k^2} +  \frac {k^3} {k^3}
+}}
+{{eqn | r = k^3 + 3 k^2 + 3 k + 1
+}}
+{{eqn | r = \left({k + 1}\right)^3
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+{{qed}}
+</onlyinclude>
+\end{proof}<|endoftext|>
+\section{Set of Integers is not Well-Ordered by Usual Ordering}
+Tags: Integers, Well-Orderings
+
+\begin{theorem}
+The [[Definition:Integer|set of integers]] $\Z$ is not [[Definition:Well-Ordered Set|well-ordered]] under the [[Definition:Usual Ordering|usual ordering]] $\le$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\Z$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+Then by definition, all [[Definition:Subset|subsets]] of $\Z$ has a [[Definition:Smallest Element|smallest element]].
+But take $\Z$ itself.
+Suppose $x \in \Z$ is a [[Definition:Smallest Element|smallest element]].
+Then $x - 1 \in \Z$.
+But $x - 1 < x$, which [[Definition:Contradiction|contradicts]] the supposition that $x \in \Z$ is a [[Definition:Smallest Element|smallest element]].
+Hence there can be no such [[Definition:Smallest Element|smallest element]].
+So by [[Proof by Contradiction]], $\Z$ is not [[Definition:Well-Ordered Set|well-ordered]] by $\le$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Integers can be Well-Ordered}
+Tags: Integers, Well-Orderings
+
+\begin{theorem}
+The [[Definition:Integer|set of integers]] $\Z$ can be [[Definition:Well-Ordered Set|well-ordered]] with an appropriately chosen [[Definition:Ordering|ordering]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Ordering|ordering]] $\preccurlyeq \subseteq \Z \times \Z$ defined as:
+:$x \preccurlyeq y \iff \left({\left\vert{x}\right\vert < \left\vert{y}\right\vert}\right) \lor \left({\left\vert{x}\right\vert = \left\vert{y}\right\vert \land x \le y}\right)$
+{{finish|It remains to be shown that $\preccurlyeq$ is an ordering, and also that it is a well-ordering.}}
+\end{proof}<|endoftext|>
+\section{Set of Non-Negative Real Numbers is not Well-Ordered by Usual Ordering}
+Tags: Real Numbers, Well-Orderings
+
+\begin{theorem}
+The [[Definition:Positive Real Number|set of non-negative real numbers]] $\R_{\ge 0}$ is not [[Definition:Well-Ordered Set|well-ordered]] under the [[Definition:Usual Ordering|usual ordering]] $\le$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\R_{\ge 0}$ is a [[Definition:Well-Ordered Set|well-ordered set]].
+Then by definition, all [[Definition:Subset|subsets]] of $\R_{\ge 0}$ has a [[Definition:Smallest Element|smallest element]].
+Take the subset $\R_{> 0}$:
+:$\R_{> 0} = \left\{ {x \in \R_{\ge 0}: x > 0}\right\} = \R_{\ge 0} \setminus \left\{ {0}\right\}$
+Suppose $x \in \R_{> 0}$ is a [[Definition:Smallest Element|smallest element]].
+Then $\dfrac x 2 \in \R_{> 0}$.
+But $\dfrac x 2 < x$, which [[Definition:Contradiction|contradicts]] the supposition that $x \in \R_{> 0}$ is a [[Definition:Smallest Element|smallest element]].
+Hence there can be no such [[Definition:Smallest Element|smallest element]].
+So by [[Proof by Contradiction]], $\R_{\ge 0}$ is not [[Definition:Well-Ordered Set|well-ordered]] by $\le$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Digital Root}
+Tags: Number Bases
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
+Let $b \in \N$ such that $b \ge 2$ also be a [[Definition:Natural Number|natural number]].
+Let $n$ be expressed in [[Definition:Number Base|base $b$]].
+Then the [[Definition:Digital Root|digital root]] [[Definition:Number Base|base $b$]] exists for $n$.
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Digital Root|digital root]] [[Definition:Number Base|base $b$]] for $n$ is the single [[Definition:Digit|digit]] resulting from:
+:adding up the [[Definition:Digit|digits]] in $n$, and expressing the result in [[Definition:Number Base|base $b$]]
+:adding up the [[Definition:Digit|digits]] in that result, and again expressing the result in [[Definition:Number Base|base $b$]]
+:repeating until down to one [[Definition:Digit|digit]].
+Let $n = d_1 + b d_2 + \dotsb + b^{m-1} d_m$ where, for all $i$, $0 \le d_i < b$.
+Let $S \left({n}\right)$ be the [[Definition:Digit Sum|digit sum]] of $n$.
+Then:
+:$S \left({n}\right) = d_1 + d_2 + \dotsb + d_m$
+Thus:
+:$S \left({n}\right) < n$
+unless $d_2, d_3, \dotsb, d_m = 0$
+in which case $n$ is a one [[Definition:Digit|digit]] number.
+Similarly: 
+:$S \left({S \left({n}\right)}\right) < S \left({n}\right)$
+Every time the [[Definition:Digit Sum|digit sum]] is taken, the result is at least one less than the previous [[Definition:Digit Sum|digit sum]].
+As $n$ is [[Definition:Finite|finite]], it will take a finite number of steps to reduce the result to a one [[Definition:Digit|digit]] number.
+Hence the result.
+{{qed}}
+[[Category:Number Bases]]
+gcq9t3x04n76dwoaqu1kohobad3hhmd
+\end{proof}<|endoftext|>
+\section{Golden Mean is Irrational}
+Tags: Golden Mean
+
+\begin{theorem}
+The [[Definition:Golden Mean|golden mean]] $\phi$ is [[Definition:Irrational Number|irrational]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Golden Mean|golden mean]]:
+:$\phi = \dfrac {1 + \sqrt 5} 2$
+By [[Square Root of Prime is Irrational]]:
+:$\sqrt 5$ is [[Definition:Irrational Number|irrational]].
+By [[Rational Number plus Irrational Number is Irrational]]:
+:$1 + \sqrt 5$ is [[Definition:Irrational Number|irrational]].
+By [[Irrational Number divided by Rational Number is Irrational]]:
+:$\dfrac {1 + \sqrt 5} 2$ is [[Definition:Irrational Number|irrational]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below iff Preceding Finite Supremum}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x, y \in S$.
+Then $x \ll y$ {{iff}}
+:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+where
+:$\ll$ denotes the [[Definition:Element is Way Below|way below relation]],
+:${\it Fin}\left({X}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \ll y$
+Let $X \subseteq S$ such that
+:$y \preceq \sup X$
+Define $F := \left\{ {\sup A: A \in {\it Fin}\left({X}\right)}\right\}$
+By definition of [[Definition:Set Union|union]]:
+:$X = \bigcup {\it Fin}\left({X}\right)$
+By [[Supremum of Suprema]]:
+:$\sup X = \sup F$
+We will prove that
+:$F$ is [[Definition:Directed Subset|directed]]
+Let $a, b \in F$
+By definition of $F$:
+:$\exists A \in {\it Fin}\left({X}\right): a = \sup A$
+and
+:$\exists B \in {\it Fin}\left({X}\right): b = \sup B$
+By [[Union of Subsets is Subset]]:
+:$A \cup B \subseteq X$
+By [[Finite Union of Finite Sets is Finite]]:
+:$A \cup B$ is [[Definition:Finite Set|finite]]
+By definition of {\it Fin}:
+:$A \cup B \in {\it Fin}\left({X}\right)$
+By definition of $F$:
+:$\sup\left({A \cup B}\right) \in F$
+By [[Set is Subset of Union]]:
+:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
+By [[Supremum of Subset]]:
+:$a \preceq \sup\left({A \cup B}\right)$ and $b \preceq \sup\left({A \cup B}\right)$
+Thus by definition
+:$F$ is [[Definition:Directed Subset|directed]]
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists d \in F: x \preceq d$
+Thus by definition of $F$:
+:$\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$y \preceq \sup D$
+By assumption:
+:$\exists A \in {\it Fin}\left({D}\right): x \preceq \sup A$
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists h \in D: \forall a \in A: a \preceq h$
+By definition:
+:$h$ is [[Definition:Upper Bound of Set|upper bound]] for $A$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup A \preceq h$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq h$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Product of Powers/Proof 2}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $a \in \R_{> 0}$ be a [[Definition:Positive Real Number|positive real number]].
+Let $x, y \in \R$ be [[Definition:Real Number|real numbers]].
+Let $a^x$ be defined as [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then:
+: $a^x a^y = a^{x + y}$
+{{proofread}}
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$.
+From [[Rational Sequence Decreasing to Real Number]], there exist [[Definition:Rational Sequence|rational sequences]] $\sequence {x_n}$ and $\sequence {y_n}$ [[Definition:Convergent Real Sequence|converging]] to $x$ and $y$, respectively.
+Then, since [[Power Function on Strictly Positive Base is Continuous/Real Power|Power Function on Strictly Positive Base is Continuous: Real Power]]:
+{{begin-eqn}}
+{{eqn | l = a^{x + y}
+      | r = a^{\displaystyle \paren {\lim_{n \mathop \to \infty} x_n + \lim_{n \mathop \to \infty} y_n} }
+}}
+{{eqn | r = a^{\displaystyle \paren {\lim_{n \mathop \to \infty} \paren {x_n + y_n} } }
+      | c = [[Sum Rule for Real Sequences]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} a^{x_n + y_n}
+      | c = [[Sequential Continuity is Equivalent to Continuity in the Reals]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} \paren {a^{x_n} a^{y_n} }
+      | c = [[Sum of Indices of Real Number/Rational Numbers|Sum of Indices of Real Number: Rational Numbers]]
+}}
+{{eqn | r = \paren {\lim_{n \mathop \to \infty} a^{x_n} } \paren {\lim_{n \mathop \to \infty} a^{y_n} }
+      | c = [[Product Rule for Real Sequences]]
+}}
+{{eqn | r = a^x a^y
+      | c = [[Sequential Continuity is Equivalent to Continuity in the Reals]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Power of Power/Proof 1}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+{{:Exponent Combination Laws/Power of Power}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a^{x y}
+      | r = \map \exp {x y \ln a}
+      | c = {{Defof|Power to Real Number}}
+}}
+{{eqn | r = \map \exp {y \, \map \ln {a^x} }
+      | c = [[Logarithms of Powers]]
+}}
+{{eqn | r = \paren {a^x}^y
+      | c = {{Defof|Power to Real Number}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Power of Power/Proof 2}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+{{:Exponent Combination Laws/Power of Power}}
+\end{theorem}
+
+\begin{proof}
+We will show that:
+:$\forall \epsilon \in \R_{>0}: \size {a^{x y} - \paren {a^x}^y} < \epsilon$
+{{WLOG}}, suppose that $x < y$.
+Consider $I := \closedint x y$.
+Let $I_\Q = I \cap \Q$.
+Let $M = \max \set {\size x, \size y}$
+Fix $\epsilon \in \R_{>0}$.
+From [[Real Polynomial Function is Continuous]]:
+:$\exists \delta' \in \R_{>0}: \size {a^x - a^{x'} } < \delta' \leadsto \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } < \dfrac \epsilon 4$
+From [[Power Function on Strictly Positive Base is Continuous]]:
+{{begin-eqn}}
+{{eqn | lll=\exists \delta_1 \in \R_{>0} :
+      | ll= \size {x x' - y y'}
+      | lo= <
+      | l = \delta_1
+      | o = \leadsto
+      | m = \size {a^{x x'} - a^{x y'} }
+      | mo= <
+      | r = \dfrac \epsilon 4
+}}
+{{eqn | lll=\exists \delta_2 \in \R_{>0} :
+      | ll= \size {x y' - x' y'}
+      | lo= <
+      | l = \delta_2
+      | o = \leadsto
+      | m = \size {a^{x y'} - a^{x' y'} }
+      | mo= <
+      | r = \dfrac \epsilon 4
+}}
+{{eqn | lll=\exists \delta_3 \in \R_{>0} :
+      | ll= \size {x' - x}
+      | lo= <
+      | l = \delta_3
+      | o = \leadsto
+      | m = \size {a^{x'} - a^x}
+      | mo= <
+      | r = \delta'
+}}
+{{eqn | o = \leadsto
+      | m = \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} }
+      | mo = <
+      | r = \dfrac \epsilon 4
+}}
+{{eqn | lll=\exists \delta_4 \in \R_{>0} :
+      | ll= \size {y' - y}
+      | lo= <
+      | l = \delta_4
+      | o = \leadsto
+      | m = \size {\paren {a^x}^{y'} - \paren {a^x}^{y} }
+      | mo= <
+      | r = \dfrac \epsilon 4
+}}
+{{end-eqn}}
+Further:
+{{begin-eqn}}
+{{eqn | l = \size {y - y'}
+      | o = <
+      | r = \frac {\delta_1} {\size x}
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x y - x y'}
+      | r = \size x \size {y - y'}
+      | c = [[Absolute Value Function is Completely Multiplicative]]
+}}
+{{eqn | o = <
+      | r = \size x \frac {\delta_1} {\size x}
+      | c = multiplying both sides by $\size x \ge 0$
+}}
+{{eqn | r = \delta_1
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \size {x - x'}
+      | o = <
+      | r = \frac {\delta_2} M
+}}
+{{eqn | ll= \leadsto
+      | l = \size {x y' - x'y'}
+      | r = \size {y'} \size {x - x'}
+      | c = [[Absolute Value Function is Completely Multiplicative]]
+}}
+{{eqn | o = \le
+      | r = M \size {x - x'}
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | o = <
+      | r = M \frac {\delta_1} M
+      | c = multiplying both sides by $M \ge 0$
+}}
+{{eqn | r = \delta_2
+}}
+{{end-eqn}}
+Let $\delta = \max \set {\dfrac {\delta_1} {\size x}, \dfrac {\delta_2} M, \delta_3, \delta_4}$.
+From [[Closure of Rational Interval is Closed Real Interval]]:
+:$\exists r, s \in I_\Q: \size {x - r} < \delta \land \size {y - s} < \delta$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \size {a^{x y} - \paren {a^x}^y}
+      | o = \le
+      | r = \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {a^{r s} - \paren {a^r}^s} + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}
+      | c = [[Triangle Inequality for Real Numbers]]
+}}
+{{eqn | r = \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}
+      | c = [[Product of Indices of Real Number/Rational Numbers|Product of Indices of Real Number: Rational Numbers]]
+}}
+{{eqn | o = < 
+      | r = \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4
+      | c = Definition of $r$ and $s$
+}}
+{{eqn | r = \epsilon
+}}
+{{end-eqn}}
+Hence the result, by [[Real Plus Epsilon]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Sequence Increasing to Real Number}
+Tags: Limits of Sequences, Rational Sequences
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then there exists some [[Definition:Increasing Sequence|increasing]] [[Definition:Rational Sequence|rational sequence]] that [[Definition:Convergent Real Sequence|converges]] to $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ denote the [[Definition:Sequence|sequence]] defined as:
+:$\forall n \in \N: x_n = \dfrac {\floor {n x} } n$
+where $\floor {n x}$ denotes the [[Definition:Floor Function|floor]] of $n x$.
+From [[Floor Function is Integer]], $\floor {n x}$ is an [[Definition:Integer|integer]].
+Hence by definition of [[Definition:Rational Number|rational number]], $\sequence {x_n}$ is a [[Definition:Rational Sequence|rational sequence]].
+From [[Real Number is between Floor Functions]]:
+:$n x - 1 < \floor {n x} \le nx $
+Thus:
+:$\dfrac {n x - 1} n < \dfrac {\floor {n x} } n \le x$
+Further:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \frac {n x - 1} n
+      | r = \lim_{n \mathop \to \infty} \paren {x - \frac 1 n}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} x - \lim_{n \mathop \to \infty} \frac 1 n
+      | c = [[Combined Sum Rule for Real Sequences]]
+}}
+{{eqn | r = x - 0
+      | c = [[Quotient Rule for Limits of Functions]]
+}}
+{{eqn | r = x
+}}
+{{end-eqn}}
+From the [[Squeeze Theorem for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {\floor {n x} } n = x$
+From [[Peak Point Lemma]], there is a [[Definition:Monotone Sequence|monotone]] [[Definition:Subsequence|subsequence]] $\sequence {x_{n_k} }$ of $\sequence {x_n}$.
+We have that $\sequence {x_n}$ is [[Definition:Bounded Above Real Sequence|bounded above]] by $x$.
+Hence $\sequence {x_{n_k} }$ is [[Definition:Increasing Sequence|increasing]].
+From [[Limit of Subsequence equals Limit of Real Sequence]], $\sequence {x_{n_k} }$ [[Definition:Convergent Real Sequence|converges]] to $x$.
+Hence the result.
+{{qed}}
+[[Category:Limits of Sequences]]
+[[Category:Rational Sequences]]
+j50mi1ifpkb72l8s4brjklid4wkap2x
+\end{proof}<|endoftext|>
+\section{Rational Sequence Decreasing to Real Number}
+Tags: Limits of Sequences, Rational Sequences
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then there exists some [[Definition:Decreasing Sequence|decreasing]] [[Definition:Rational Sequence|rational sequence]] that [[Definition:Convergent Real Sequence|converges]] to $x$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {x_n}$ denote the [[Definition:Real Sequence|sequence]] defined as:
+:$\forall n \in \N : x_n = \dfrac {\ceiling {n x} } n$
+where $\ceiling {n x}$ denotes the [[Definition:Ceiling Function|ceiling]] of $n x$.
+From [[Ceiling Function is Integer]], $\ceiling {n x}$ is an [[Definition:Integer|integer]].
+Hence by definition of [[Definition:Rational Number|rational number]], $\sequence {x_n}$ is a [[Definition:Rational Sequence|rational sequence]].
+From [[Real Number is between Ceiling Functions]]:
+:$n x < \ceiling {n x} \le n x + 1$
+Thus:
+:$x < \dfrac {\ceiling {n x} } n \le \dfrac {n x + 1} n$
+Further:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} \frac {nx + 1} n
+      | r = \lim_{n \mathop \to \infty} \paren {x + \frac 1 n}
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} x + \lim_{n \mathop \to \infty} \frac 1 n
+      | c = [[Combined Sum Rule for Real Sequences]]
+}}
+{{eqn | r = x + 0
+      | c = [[Quotient Rule for Limits of Functions]]
+}}
+{{eqn | r = x
+}}
+{{end-eqn}}
+Thus, from the [[Squeeze Theorem for Real Sequences]]:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac {\ceiling {n x} } n = x$
+From [[Peak Point Lemma]], there is a [[Definition:Monotone Sequence|monotone]] [[Definition:Subsequence|subsequence]] $\sequence {x_{n_k} }$ of $\sequence {x_n}$.
+We have that $\sequence {x_n}$ is [[Definition:Bounded Below Real Sequence|bounded below]] by $x$.
+Hence $\sequence {x_{n_k} }$ is [[Definition:Decreasing Sequence|decreasing]].
+From [[Limit of Subsequence equals Limit of Real Sequence]], $\sequence {x_{n_k} }$ [[Definition:Convergent Real Sequence|converges]] to $x$.
+Hence the result.
+{{qed}}
+[[Category:Limits of Sequences]]
+[[Category:Rational Sequences]]
+b7mtd4vcalkk2zn65niysnv74saecmo
+\end{proof}<|endoftext|>
+\section{Power Function on Base Greater than One is Strictly Increasing/Real Number}
+Tags: Power Function on Base Greater than One is Strictly Increasing
+
+\begin{theorem}
+Let $a \in \R$ be a [[Definition:Real Number|real number]] such that $a > 1$.
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map f x = a^x$
+where $a^x$ denotes [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then $f$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$ be such that $x < y$.
+Let $\delta = \dfrac {y - x} 2$.
+From [[Rational Sequence Decreasing to Real Number]], there is some [[Definition:Rational Sequence|rational sequence]] $\sequence {x_n}$ that [[Definition:Decreasing Sequence|decreases]] to $x$.
+From [[Rational Sequence Increasing to Real Number]], there is some [[Definition:Rational Sequence|rational sequence]] $\sequence {y_n}$ that [[Definition:Increasing Sequence|increases]] to $y$.
+From [[Convergent Real Sequence is Bounded]]:
+:$\exists N_1 \in \N: n \ge N_1 \implies x - \delta < x_n < x + \delta$
+Since $\sequence {x_n}$ is [[Definition:Decreasing Sequence|decreasing]]:
+:$n \ge N_1 \implies x \le x_n < x + \delta$
+From [[Convergent Real Sequence is Bounded]]:
+:$\exists N_2 \in \N : n \ge N_2 \implies y - \delta < y_n < y + \delta$
+Since $\sequence {y_n}$ is [[Definition:Increasing Sequence|increasing]]:
+:$n \ge N_2 \implies y - \delta < y_n \le y$
+Let $N = \max \set {N_1, N_2}$.
+Then, for $n \ge N$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \le
+      | r = x_n
+}}
+{{eqn | o = <
+      | r = x + \delta
+}}
+{{eqn | r = y - \delta
+}}
+{{eqn | o = <
+      | r = y_n
+}}
+{{eqn | o = \le
+      | r = y
+}}
+{{end-eqn}}      
+From [[Power Function on Strictly Positive Base is Continuous]] and [[Sequential Continuity is Equivalent to Continuity in the Reals]]:
+:$x_n \to x \implies a^{x_n} \to a^x$
+:$y_n \to y \implies a^{y_n} \to a^y$
+Also, from [[Power Function on Base Greater than One is Strictly Increasing/Rational Number|Power Function on Base Greater than One is Strictly Increasing: Rational Number]]:
+:$\sequence {a^{x_n} }$ [[Definition:Decreasing Sequence|decreases]] to $a^x$
+and:
+:$\sequence {a^{y_n} }$ [[Definition:Increasing Sequence|increases]] to $a^y$.
+So, for $n \ge N$:
+{{begin-eqn}}
+{{eqn | l = a^x
+      | o = \le
+      | r = a^{x_n}
+      | c = as $\sequence {a^{x_n} }$ [[Definition:Decreasing Sequence|decreases]] to $a^x$
+}}
+{{eqn | o = <
+      | r = a^{y_n}
+      | c = [[Power Function on Base Greater than One is Strictly Increasing/Rational Number|Power Function on Base Greater than One is Strictly Increasing: Rational Number]]
+}}
+{{eqn | o = \le
+      | r = a^y
+      | c = as $\sequence {a^{y_n} }$ [[Definition:Increasing Sequence|increases]] to $a^y$
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Power Function on Base Greater than One is Strictly Increasing]]
+dtnnkqxpiiafd566txd5qdaq52eo9sp
+\end{proof}<|endoftext|>
+\section{Power Function on Strictly Positive Base is Continuous/Real Power}
+Tags: Powers
+
+\begin{theorem}
+Let $f : \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map f x = a^x$
+where $a^x$ denotes [[Definition:Power (Algebra)/Real Number/Definition 2|$a$ to the power of $x$]].
+Then $f$ is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+By definition, $a^x$ is the [[Definition:Unique|unique]] [[Definition:Real Continuous Extension|continuous extension]] of $a^r$, for [[Definition:Rational Number|rational]] $r$.
+By definition, [[Definition:Real Continuous Extension|continuous extensions]] are [[Definition:Continuous Real Function|continuous]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function on Strictly Positive Base is Continuous/Rational Power}
+Tags: Powers
+
+\begin{theorem}
+Let $f: \Q \to \R$ be the [[Definition:Real-Valued Function|real-valued function]] defined as:
+:$\map f x = a^x$
+where $a^x$ denotes [[Definition:Rational Power|$a$ to the power of $x$]].
+Then $f$ is [[Definition:Continuous Real Function|continuous]].
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Powers]]
+5jr6gc6hq8d6y5t1kgyelpak9nq74ce
+\end{proof}<|endoftext|>
+\section{Defining Sequence of Natural Logarithm is Convergent}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $\left\langle{ f_n }\right\rangle$ be the [[Definition:Sequence|sequence]] of mappings $f_n : \R_{>0} \to \R$ defined as:
+:$f_n \left({ x }\right) = n \left({ \sqrt[n]{ x } - 1 }\right)$
+Then $\left\langle{ f_n }\right\rangle$ is [[Definition:Pointwise Convergence|pointwise convergent]].
+\end{theorem}
+
+\begin{proof}
+Fix $x \in \R_{>0}$.
+From [[Defining Sequence of Natural Logarithm is Strictly Decreasing]], $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is [[Definition:Strictly Decreasing/Sequence | strictly decreasing]].
+From [[Lower Bound of Natural Logarithm/Proof 3]], $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is [[Definition:Bounded Below Real Sequence|bounded below]].
+From [[Monotone Convergence Theorem (Real Analysis)|Monotone Convergence Theorem]], $\left\langle{ f_n \left({ x }\right) }\right\rangle$ is [[Definition:Convergent Real Sequence|convergent]]
+Hence the result, by definition of [[Definition:Pointwise Convergence|pointwise convergence]].
+{{qed}}
+[[Category:Natural Logarithms]]
+g613kd5i1wa3ssljn1w1bfk4gqetbha
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Positive Integers}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $r \in \R_{>0}$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
+\end{theorem}<|endoftext|>
+\section{Sum of Indices of Real Number/Rational Numbers}
+Tags: Sum of Indices of Real Number
+
+\begin{theorem}
+Let $x, y \in \Q$ be [[Definition:Rational Number|rational numbers]].
+Let $r^x$ be defined as [[Definition:Rational Power|$r$ to the power of $n$]].
+Then:
+: $r^{x + y} = r^x \times r^y$
+\end{theorem}
+
+\begin{proof}
+Let $x = \dfrac p q, y = \dfrac u v$.
+Then:
+{{begin-eqn}}
+{{eqn | l = r^\paren {x + y}
+      | r = r^\paren {\paren {p / q} + \paren {u / v} }
+      | c = 
+}}
+{{eqn | r = r^\paren {\paren {p v + u q} / q v}
+      | c = 
+}}
+{{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v + u q}
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = \paren {r^\paren {1 / q v} }^\paren {p v} \times \paren {r^\paren {1 / q v} }^\paren {u q}
+      | c = [[Sum of Indices of Real Number/Integers|Sum of Indices of Real Number: Integers]]
+}}
+{{eqn | r = r^\paren {p v / q v} \times r^\paren {u q / q v}
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = r^\paren {p / q} \times r^\paren {u / v}
+      | c = 
+}}
+{{eqn | r = r^x \times r^y
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponent Combination Laws/Rational Numbers}
+Tags: Exponent Combination Laws
+
+\begin{theorem}
+Let $r \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+\end{theorem}<|endoftext|>
+\section{Product of Indices of Real Number/Rational Numbers}
+Tags: Product of Indices of Real Number
+
+\begin{theorem}
+Let $x, y \in \Q$ be [[Definition:Rational Number|rational numbers]].
+Let $r^x$ be defined as [[Definition:Rational Power|$r$ to the power of $x$]].
+Then:
+:$\paren {r^x}^y = r^{x y}$
+\end{theorem}
+
+\begin{proof}
+Let $x = \dfrac p q, y = \dfrac u v$.
+Consider $\paren {\paren {r^x}^y}^{q v}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {\paren {r^x}^y}^{q v}
+      | r = \paren {\paren {r^\paren {p / q} }^\paren {u / v} }^{q v}
+      | c = 
+}}
+{{eqn | r = \paren {\paren {\paren {r^\paren {p / q} }^\paren {u / v} }^v}^q
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | r = \paren {\paren {r^\paren {p / q} }^u}^q
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = \paren {r^\paren {p / q} }^{q u}
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | r = \paren {\paren {r^\paren {p / q} }^q}^u
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | r = \paren {r^p}^u
+      | c = {{Defof|Rational Power}}
+}}
+{{eqn | r = r^{p u}
+      | c = [[Product of Indices of Real Number/Integers|Product of Indices of Real Number: Integers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {r^\paren {p / q} }^\paren {u / v} }
+      | r = r^{\paren {p u} / \paren {q v} }
+      | c = taking $q v$th [[Definition:Root (Analysis)|root]] of both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {r^x}^y
+      | r = r^{x y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Indices of Real Number}
+Tags: Powers, Sum of Indices of Real Number
+
+\begin{theorem}
+Let $r \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+\end{theorem}<|endoftext|>
+\section{Product of Indices of Real Number}
+Tags: Powers, Product of Indices of Real Number
+
+\begin{theorem}
+Let $r \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+\end{theorem}<|endoftext|>
+\section{Sum of Indices of Real Number/Integers}
+Tags: Sum of Indices of Real Number
+
+\begin{theorem}
+Let $n, m \in \Z$ be [[Definition:Integer|integers]].
+Let $r^n$ be defined as [[Definition:Integer Power|$r$ to the power of $n$]].
+Then:
+:$r^{n + m} = r^n \times r^m$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Indices of Real Number/Positive Integers|Sum of Indices of Real Number: Positive Integers]], we have that:
+:$m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
+It remains to be shown that:
+:$\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
+The proof will proceed by [[Principle of Mathematical Induction|induction]] on $m$.
+As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.
+For all $p \in \Z_{>0}$, let $\map P p$ be the [[Definition:Proposition|proposition]]:
+:$\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$
+that is:
+:$\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$
+=== Basis for the Induction ===
+$\map P 1$ is true, as follows:
+When $n > 0$:
+{{begin-eqn}}
+{{eqn | l = r^{n - 1} \times r
+      | r = r^n
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | ll= \leadsto
+      | l = r^{n - 1}
+      | r = r^n \dfrac 1 r
+      | c = multiplying both sides by $\dfrac 1 r$
+}}
+{{eqn | r = r^n \times r^{-1}
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{end-eqn}}
+When $n \le 0$:
+{{begin-eqn}}
+{{eqn | l = r^{n - 1}
+      | r = \dfrac {r^n} r
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | r = r^n \times r^{-1}
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{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]]:
+:$\forall n \in \Z: r^{n - k} = r^n \times r^{- k}$
+Then we need to show:
+:$\forall n \in \Z: r^{n - \paren {k + 1} } = r^n \times r^{-\paren {k + 1} }$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = r^n \times r^{-\paren {k + 1} }
+      | r = r^n \times \dfrac {r^{-k} } r
+      | c = {{Defof|Integer Power}}
+}}
+{{eqn | r = \paren {r^n \times r^{-k} } \times \dfrac 1 r
+      | c = [[Real Multiplication is Associative]]
+}}
+{{eqn | r = r^{n - k} \times \dfrac 1 r
+      | c = [[Sum of Indices of Real Number/Integers#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = r^{n - k} \times r^{-1}
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{eqn | r = r^{n - k - 1}
+      | c = [[Sum of Indices of Real Number/Integers#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = r^{n - \paren {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:
+:$\forall n, m \in \Z: r^{n + m} = r^n \times r^m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Indices of Real Number/Integers}
+Tags: Product of Indices of Real Number
+
+\begin{theorem}
+Let $n, m \in \Z$ be [[Definition:Positive Integer|positive integers]].
+Let $r^n$ be defined as [[Definition:Integer Power|$r$ to the power of $n$]].
+Then:
+:$\paren {r^n}^m = r^{n m}$
+\end{theorem}
+
+\begin{proof}
+From [[Product of Indices of Real Number/Positive Integers|Product of Indices of Real Number: Positive Integers]], we have that:
+:$m \in \Z_{\ge 0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$
+It remains to be shown that:
+:$\forall m \in \Z_{<0}: \forall n \in \Z: \paren {r^n}^m = r^{n m}$
+As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {r^n}^{- p}
+      | r = \dfrac 1 {\paren {r^n}^p}
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{eqn | r = \dfrac 1 {r^{n p} }
+      | c = from $(1)$
+}}
+{{eqn | r = r^{-n p}
+      | c = [[Real Number to Negative Power/Integer|Real Number to Negative Power: Integer]]
+}}
+{{end-eqn}}
+Hence the result, by replacing $-p$ with $m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Uniformly Convergent iff Difference Under Supremum Metric Vanishes}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $X$ and $Y$ be [[Definition:Metric Space|metric spaces]]
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] defined on $X$.
+Let $f: X \to Y$ be a [[Definition:Mapping|mapping]].
+Let $d_S: S \times S \to Y$ denote the [[Definition:Supremum Metric|supremum metric]] on $S \subseteq X$.
+Then $\sequence {f_n}$ [[Definition:Uniform Convergence|converges uniformly]] to $f$ on $S$ {{iff}} $\map {d_S} {f_n, f} \to 0$ as $n \to \infty$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Metric Spaces]]
+r6cgon333ra1ffa0lfmk1box36jrp0s
+\end{proof}<|endoftext|>
+\section{Uniform Convergence is Hereditary}
+Tags: Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Mapping|mappings]] defined on $A$.
+Let $\sequence {f_n}$ be [[Definition:Uniform Convergence|uniformly convergent]] on $S \subseteq A$.
+Then $\sequence {f_n}$ is [[Definition:Uniform Convergence|uniformly convergent]] on every [[Definition:Metric Subspace|metric subspace]] of $S$.
+That is, [[Definition:Uniform Convergence|uniform convergence]] is a [[Definition:Hereditary Property (Topology)|hereditary property]] of a [[Definition:Metric Space|metric space]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Metric Spaces]]
+3bkrpgwg5yp9kn5et5d2op3zfqsnlf6
+\end{proof}<|endoftext|>
+\section{Real Number to Negative Power}
+Tags: Powers
+
+\begin{theorem}
+Let $r \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+\end{theorem}<|endoftext|>
+\section{Real Number to Negative Power/Integer}
+Tags: Powers
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Let $r^n$ be defined as [[Definition:Integer Power|$r$ to the power of $n$]].
+Then:
+:$r^{-n} = \dfrac 1 {r^n}$
+\end{theorem}
+
+\begin{proof}
+Let $n \in \Z_{\ge 0}$.
+Then from [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]:
+:$r^{-n} = \dfrac 1 {r^n}$
+It remains to show that this holds when $n < 0$.
+Let $n \in \Z_{<0}$.
+Then $n = - m$ for some $m \in \Z_{> 0}$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = r^{-m}
+      | r = \dfrac 1 {r^m}
+      | c = [[Real Number to Negative Power/Positive Integer|Real Number to Negative Power: Positive Integer]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 1 {r^{-m} }
+      | r = \dfrac 1 {1 / r^m}
+      | c = taking [[Definition:Reciprocal|reciprocal]] of both sides
+}}
+{{eqn | r = r^m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 1 {r^n}
+      | r = r^{-n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Powers]]
+rcdclehr2jtyzdkxmwo7chsg18957rf
+\end{proof}<|endoftext|>
+\section{Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Ordered Set|ordered set]].
+Let $x, y \in S$.
+Then $x \ll y$ {{iff}}
+:$\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$
+where
+:$\ll$ denotes the [[Definition:Element is Way Below|way below relation]],
+:${\it Ids}\left({\mathscr S}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \ll y$
+Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that
+:$y \preceq \sup I$
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$I$ is [[Definition:Directed Subset|directed]] and [[Definition:Lower Set|lower]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$I$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists i \in I: x \preceq i$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$x \in I$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$D$ admits a [[Definition:Supremum of Set|supremum]] and $y \preceq \sup D$
+By [[Supremum of Lower Closure of Set]]:
+:$D^\preceq$ admits a [[Definition:Supremum of Set|supremum]] and $\sup D^\preceq = \sup D$
+where $D^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $D$.
+By [[Lower Closure of Directed Subset is Ideal]]:
+:$D^\preceq$ is an [[Definition:Ideal in Ordered Set|ideal]].
+By assumption:
+:$x \in D^\preceq$
+Thus by definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists d \in D: x \preceq d$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure of Directed Subset is Ideal}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$.
+Then
+:$D^\preceq$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\mathscr S$
+where $D^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $D$.
+\end{theorem}
+
+\begin{proof}
+By [[Directed iff Lower Closure Directed]]:
+:$D^\preceq$ is [[Definition:Directed Subset|directed]].
+By [[Lower Closure is Lower Set]]:
+:$D^\preceq$ is a [[Definition:Lower Set|lower set]].
+Thus by definition
+:$D^\preceq$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\mathscr S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Geometric Sequence/Corollary 2}
+Tags: Geometric Sequences, Sum of Geometric Sequence
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 0}^{n - 1} j x^j = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}
+      | r = \sum_{j \mathop = 0}^{n - 1} D_x x^j
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | r = D_x \sum_{j \mathop = 0}^{n - 1} x^j
+      | c = [[Sum Rule for Derivatives]]
+}}
+{{eqn | r = D_x \frac {x^n - 1} {x - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = \frac {\paren {n x^{n - 1} } \paren {x - 1} - \paren 1 \paren {x^n - 1} } {\paren {x - 1}^2}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 0}^{n - 1} j x^j
+      | r = x \sum_{j \mathop = 0}^{n - 1} j x^{j - 1}
+      | c = [[Multiplication of Numbers Distributes over Addition]]
+}}
+{{eqn | r = x \frac {\paren {n - 1} x^n - n x^{n - 1} + 1} {\paren {x - 1}^2}
+      | c = by the above result
+}}    
+{{eqn | r = \frac {\paren {n - 1} x^{n + 1} - n x^n + x} {\paren {x - 1}^2}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Geometric Sequences]]
+[[Category:Sum of Geometric Sequence]]
+eprp4tj6qovtipfbbjp6fic4ipk970e
+\end{proof}<|endoftext|>
+\section{Defining Sequence of Natural Logarithm is Strictly Decreasing}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $\sequence {f_n}$ be the [[Definition:Sequence|sequence]] of mappings $f_n : \R_{>0} \to \R$ defined as:
+:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
+Then $\forall x \in \R_{>0}: \sequence {\map {f_n} x}$ is [[Definition:Strictly Decreasing Sequence|strictly decreasing]].
+\end{theorem}
+
+\begin{proof}
+Fix $t \in R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n \paren {t^{n + 1} - 1} - \paren {n + 1} \paren {t^n - 1}
+      | r = \paren {t - 1}^2 \paren {1 + 2 t + 3 t^2 + \ldots + n t^{n - 1} }
+      | c = [[Sum of Geometric Sequence/Corollary 2|Sum of Geometric Sequence: Corollary 2]]
+}}
+{{eqn | o = >
+      | r = 0
+      | c = [[Product of Real Numbers is Positive iff Numbers have Same Sign]]
+}}
+{{end-eqn}}
+Thus:
+:$n \paren {t^{n + 1} - 1} > \paren {n + 1} \paren {t^n - 1}$
+Fix $x \in \R_{>0}$.
+From [[Power of Positive Real Number is Positive/Rational Number|Power of Positive Real Number is Positive: Rational Number]]:
+:$\forall n \in \N : x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \forall t \in \R_{>0} : n \paren {t^{n + 1} - 1}
+      | o = >
+      | r = \paren {n + 1} \paren {t^n - 1}
+      | c = by the result obtained above
+}}
+{{eqn | ll= \leadsto
+      | l = n \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^{n + 1} - 1}
+      | o = > 
+      | r = \paren {n + 1} \paren {\paren {x^{1 / \paren {n \paren {n + 1} } } }^n - 1 }
+      | c = as $x^{1 / \paren {n \paren {n + 1} } } \in \R_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = n \paren {x^{1 / n} - 1}
+      | o = >
+      | r = \paren {n + 1} \paren {x^{1 / \paren {n + 1} } - 1}
+      | c = [[Product of Indices of Real Number/Rational Numbers|Product of Indices of Real Number: Rational Numbers]]
+}}
+{{eqn | ll= \leadsto
+      | l = n \paren {\sqrt [n] x - 1}
+      | o = >
+      | r = \paren {n + 1} \paren {\map {f_n} x - 1 }
+      | c = {{Defof|Definition:Root (Analysis)|$n$th root}}
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Strictly Decreasing Sequence|strictly decreasing sequence]].
+{{qed}}
+[[Category:Natural Logarithms]]
+iyqyhorbip7hkcij2oytplaw2931uc1
+\end{proof}<|endoftext|>
+\section{Logarithm Base 10 of 2 is Irrational}
+Tags: Common Logarithms, 2, 10, Logarithm Base 10 of 2 is Irrational
+
+\begin{theorem}
+The [[Common Logarithm of 2|common logarithm of $2$]]:
+:$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
+is [[Definition:Irrational Number|irrational]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\log_{10} 2$ is [[Definition:Rational Number|rational]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \log_{10} 2
+      | r = \frac p q
+      | c = for some $p, q \in \Z_{\ne 0}$
+}}
+{{eqn | ll= \leadsto
+      | l = 2
+      | r = 10^{p / q}
+      | c = {{Defof|General Logarithm}}
+}}
+{{eqn | ll= \leadsto
+      | l = 2^q
+      | r = 10^p
+      | c = raising both sides to the [[Definition:Integer Power|power of $q$]]
+}}
+{{end-eqn}}
+Both $10^p$ and $2^q$ are [[Definition:Integer|integers]], by construction.
+But $10^p$ is [[Definition:Divisor of Integer|divisible]] by $5$, while $2^p$, which has only $2$ as a [[Definition:Prime Factor|prime factor]], is not.
+So $10^p \ne 2^q$.
+So, by [[Proof by Contradiction]], it follows that $\log_{10} 2$ is [[Definition:Irrational Number|irrational]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Because $5$ is a [[Definition:Divisor of Integer|divisor]] of $10$ but not $2$, it cannot be the case that $2^a = 10^b$ for $a, b \in \Z_{>0}$.
+Hence this is a special case of [[Irrationality of Logarithm]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Nth Root of 1 plus x not greater than 1 plus x over n}
+Tags: Inequalities
+
+\begin{theorem}
+Let $x \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then:
+:$\sqrt [n] {1 + x} \le 1 + \dfrac x n$
+\end{theorem}
+
+\begin{proof}
+From [[Bernoulli's Inequality]]:
+:$\left({1 + y}\right)^n \ge 1 + n y$
+which holds for:
+:$y \in \R$ where $y > -1$
+:$n \in \Z_{\ge 0}$
+Thus it holds for $y \in \R_{> 0}$ and  $n \in \Z_{> 0}$.
+So:
+{{begin-eqn}}
+{{eqn | l = 1 + n y
+      | o = \le
+      | r = \left({1 + y}\right)^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 1 + n \frac x n
+      | o = \le
+      | r = \left({1 + \frac x n}\right)^n
+      | c = substituting $y = \dfrac x n$
+}}
+{{eqn | ll= \leadsto
+      | l = 1 + x
+      | o = \le
+      | r = \left({1 + \frac x n}\right)^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [n] {1 + x}
+      | o = \le
+      | r = 1 + \dfrac x n
+      | c = [[Root is Strictly Increasing]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reciprocal of Logarithm}
+Tags: Logarithms, Reciprocals, Reciprocal of Logarithm
+
+\begin{theorem}
+Let $x, y \in \R_{> 0}$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
+Then:
+:$\dfrac 1 {\log_x y} = \log_y x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \log_x y \log_y x
+      | r = \log_y y
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \log_y x
+      | r = \dfrac 1 {\log_x y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discontinuity of Monotonic Function is Jump Discontinuity}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $X$ be an [[Definition:Open Set (Real Analysis)|open subset]] of $\R$.
+Let $f: X \to Y$ be a [[Definition:Monotone Real Function|monotone real function]].
+Then $f$ is [[Definition:Discontinuous|discontinuous]] at a point $c \in X$ {{iff}} $c$ is a [[Definition:Jump Discontinuity|jump discontinuity]] of $f$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Real Analysis]]
+bnv80om909h6k6w797snv2wwpj77mwr
+\end{proof}<|endoftext|>
+\section{Surjective Monotone Function is Continuous}
+Tags:  Real Analysis,  Continuity
+
+\begin{theorem}
+Let $X$ be an [[Definition:Open Set (Real Analysis)|open subset]] of $\R$.
+Let $Y$ be a [[Definition:Real Interval|real interval]].
+Let $f: X \to Y$ be a [[Definition:Surjection|surjective]] [[Definition:Monotone Real Function|monotone real function]].
+Then $f$ is [[Definition:Continuous on Interval|continuous]] on $X$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category: Real Analysis]]
+[[Category: Continuity]]
+p2sptxjqmmw0vpoq2o1507prvllb297
+\end{proof}<|endoftext|>
+\section{Upper Bound of Natural Logarithm/Corollary}
+Tags: Upper Bound of Natural Logarithm
+
+\begin{theorem}
+:$\forall s \in \R_{>0}: \ln x \le \dfrac {x^s} s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = s \ln x
+      | r = \ln {x^s}
+      | c = [[Logarithm of Power]]
+}}
+{{eqn | o = \le
+      | r = x^s - 1
+      | c = [[Upper Bound of Natural Logarithm]]
+}}
+{{eqn | o = \le
+      | r = x^s
+      | c =
+}}
+{{end-eqn}}
+The result follows by dividing both sides by $s$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers Drown Logarithms/Corollary}
+Tags: Logarithms, Powers
+
+\begin{theorem}
+:$\displaystyle \lim_{y \mathop \to 0_+} y^r \ln y = 0$
+\end{theorem}
+
+\begin{proof}
+Put $y = \dfrac 1 x$ in the [[Powers Drown Logarithms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Logarithm of Logarithm in terms of Natural Logarithms}
+Tags: Logarithms, Natural Logarithms
+
+\begin{theorem}
+Let $b, x \in \R_{>0}$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
+Then:
+:$\map {\log_b} {\log_b x} = \dfrac {\map \ln {\ln x} - \map \ln {\ln b} } {\ln b}$
+where $\ln x$ denotes the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map {\log_b} {\log_b x}
+      | r = \map {\log_b} {\frac {\ln x} {\ln b} }
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = \map {\log_b} {\ln x} - \map {\log_b} {\ln b}
+      | c = [[Difference of Logarithms]]
+}}
+{{eqn | r = \frac {\map \ln {\ln x} } {\ln b} - \frac {\map \ln {\ln b} } {\ln b}
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = \frac {\map \ln {\ln x} - \map \ln {\ln b} } {\ln b}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Limit of Bounded Convergent Sequence is Bounded}
+Tags: Limits of Sequences, Real Analysis
+
+\begin{theorem}
+Let $\left \langle {x_n} \right \rangle$, $\left \langle {a_n} \right \rangle$, and $\left \langle {b_n} \right \rangle$ be [[Definition:Convergent Real Sequence|convergent sequences in $\R$]].
+Let $\left \langle {x_n} \right \rangle$, $\left \langle {a_n} \right \rangle$, and $\left \langle {b_n} \right \rangle$ [[Definition:Convergent Real Sequence|converge]] to $x, a, b \in \R$, respectively.
+Suppose that:
+:$\exists N \in \N : n \geq N \implies a_n \leq x_n \leq b_n$
+Then:
+:$ a \le x \le b$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} that $x < a$.
+Let $\epsilon = \dfrac{a - x} 2 > 0$
+From the convergence of $\left \langle {x_n} \right \rangle$:
+:$\exists M_1 \in \N : n \ge M \implies x - \epsilon < x_n < x + \epsilon$
+Or, [[Definition:Logical Equivalence|equivalently]]:
+:$\exists M_1 \in \N : n \ge M \implies \dfrac {3x - a} 2 < x_n < \dfrac{x + a} 2$
+From the convergence of $\left \langle {a_n} \right \rangle$:
+:$\exists M_2 \in \N : n \ge M \implies a - \epsilon < a_n < a + \epsilon$
+Or, [[Definition:Logical Equivalence|equivalently]]:
+:$\exists M_2 \in \N : n \ge M \implies \dfrac {x + a} 2 < a_n < \dfrac{3a - x} 2$
+Let $M = \max\left\{ {N, M_1, M_2} \right\}$
+Then, for any $n \ge M$:
+{{begin-eqn}}
+{{eqn | l = x_n
+      | o = <
+      | r = \dfrac {x + a} 2
+}}
+{{eqn | o = <
+      | r = a_n
+}}
+{{end-eqn}}
+This [[Definition:Contradiction|contradicts]] the hypothesis that:
+:$\forall n \ge N : a_n \le x_n$
+The same argument, [[Definition:Mutatis Mutandis|mutatis mutandis]], brings us to a [[Definition:Contradiction|contradiction]] if we suppose $x > b$.
+Hence the result, by [[Proof by Contradiction]].
+{{qed}}
+[[Category:Limits of Sequences]]
+[[Category:Real Analysis]]
+20djjxluat320apfe4oqvxm3pt6pi9b
+\end{proof}<|endoftext|>
+\section{Logarithm of Power/Natural Logarithm/Natural Power}
+Tags: Logarithm of Power
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $n \in \R$ be any [[Definition:Natural Numbers|natural number]].
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+Then:
+:$\ln \left({x^n}\right) = n \ln x$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Proof by Mathematical Induction|Mathematical Induction]]:
+For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$
+=== Basis for the Induction ===
+$P \left({0}\right)$ is the case:
+:$\forall x \in \R_{>0}: \ln \left({ 1 }\right) = 0 $
+from [[Logarithm of 1 is 0]]
+This is our [[Proof by Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0 $, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\forall x \in \R_{>0}: \ln \left({ x^k }\right) = k\ln \left({ x }\right)$
+Then we need to show:
+:$\forall x \in \R_{>0}: \ln \left({ x^{k+1} }\right) = \left({ k + 1 }\right) \ln \left({ x }\right)$
+=== Inductive Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Fix $x \in \R_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \ln \left({ x^{k+1} }\right)
+      | r = \ln \left({ x^{k}x }\right)
+}}
+{{eqn | r = \ln \left({ x^{k} }\right) + \ln \left({ x }\right)
+      | c = [[Sum of Logarithms/Natural Logarithm]]
+}}
+{{eqn | r = k\ln \left({ x }\right) + \ln \left({ x }\right)
+      | c = [[Logarithm of Power/Natural Logarithm/Natural Power#Induction Step|induction step]]
+}}
+{{eqn | r = \left({ k + 1 }\right) \ln \left({ x }\right)
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: \ln \left({ x^n }\right) = n\ln \left({ x }\right)$
+{{qed}}
+[[Category:Logarithm of Power]]
+aymbmtbcoj8pq4ppxvizt7oih7bi9el
+\end{proof}<|endoftext|>
+\section{Logarithm of Power/Natural Logarithm/Integer Power}
+Tags: Logarithm of Power
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $n \in \R$ be any [[Definition:Integer|integer]].
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+Then:
+:$\ln \left({ x^n }\right) = n \ln x$
+\end{theorem}
+
+\begin{proof}
+From [[Logarithm of Power/Natural Logarithm/Natural Power]], the theorem is already proven for positive integers.
+Let $j \in \Z_{<0}$.
+Let $-j = k \in Z_{>0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \ln \left({ 1 }\right)
+      | c = [[Logarithm of 1 is 0]]
+}}
+{{eqn | r = \ln \left({ x^{k}x^{-k} }\right)
+}}
+{{eqn | r = \ln \left({ x^{k} }\right) + \ln \left({ x^{-k} }\right)
+}}
+{{eqn | r = k\ln \left({ x }\right) + \ln \left({ x^{-k} }\right)
+      | c = [[Logarithm of Power/Natural Logarithm/Natural Power]]
+}}
+{{eqn | ll = \implies
+      | l = \ln \left({ x^{-k} }\right)
+      | r = -k\ln \left({ x }\right)
+}}
+{{eqn | ll = \implies
+      | l = \ln \left({ x^{j} }\right)
+      | r = j\ln \left({ x }\right)
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Logarithm of Power]]
+qs8ggz27tpoxo5iyij4hi21m0oymec7
+\end{proof}<|endoftext|>
+\section{Logarithm of Power/Natural Logarithm/Rational Power}
+Tags: Logarithm of Power
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $r \in \R$ be any [[Definition:Rational Number|rational number]].
+Let $\ln x$ be the [[Definition:Natural Logarithm|natural logarithm]] of $x$.
+Then:
+:$\ln \left({ x^r }\right) = r \ln x$
+\end{theorem}
+
+\begin{proof}
+Let $r = \dfrac s t$, where $s \in \Z$ and $t \in \Z_{>0}$
+First:
+{{begin-eqn}}
+{{eqn | l = \ln \left({ x }\right)
+      | r = \ln \left({ x^{t/t} }\right)
+}}
+{{eqn | r = \ln \left({ \left({ x^{1/t} }\right)^{t} }\right)
+      | c = [[Product of Indices of Real Number/Rational Numbers]]
+}}
+{{eqn | r = t\ln \left({ x^{1/t} }\right)
+      | c = [[Logarithm of Power/Natural Logarithm/Integer Power]]
+}}
+{{eqn | ll= \implies
+      | l = \ln \left({ x^{1/t} }\right)
+      | r = \frac 1 t \ln \left({ x }\right)
+      | c = Dividing both sides by $t$
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln \left({x^{s/t} }\right)
+      | r = \ln \left({\left({x^{1/t} }\right)^s}\right)
+      | c = [[Product of Indices of Real Number/Rational Numbers]]
+}}
+{{eqn | r = s \ln \left({ x^{1/t} }\right)
+      | c = [[Logarithm of Power/Natural Logarithm/Integer Power]]
+}}
+{{eqn | r = \frac s t \ln \left({ x }\right)
+      | c = From the result above
+}}
+{{eqn | r = r \ln \left({ x }\right)
+      | c = From the definition of $s$ and $t$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Logarithm of Power]]
+j9zx0kfhcn7erra58301rrstw4qzbh0
+\end{proof}<|endoftext|>
+\section{Dini's Theorem}
+Tags:  Real Analysis
+
+\begin{theorem}
+Let $K \subseteq \R$ be [[Definition:Compact Metric Space|compact]].
+Let $\sequence {f_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Continuous Real Function|continuous real functions]] defined on $K$.
+Let $\sequence {f_n}$ [[Definition:Pointwise Convergence|converge pointwise]] to a [[Definition:Continuous Real Function|continuous function]] $f$.
+Suppose that:
+:$\forall x \in K : \sequence {\map {f_n} x}$ is [[Definition:Monotone Sequence|monotone]].
+Then the convergence of $\sequence {f_n}$ to $f$ is [[Definition:Uniform Convergence|uniform]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Ulisse Dini|cat = Dini}}
+[[Category: Real Analysis]]
+ckn079isrwizqchag6lje5vugut4uy1
+\end{proof}<|endoftext|>
+\section{Way Below in Meet-Continuous Lattice}
+Tags: Way Below Relation, Meet-Continuous Lattices
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Meet-Continuous Lattice|meet-continuous]] [[Definition:Bounded Below|bounded below]] [[Definition:Lattice|lattice]].
+Let $x, y \in S$.
+Then $x \ll y$ {{iff}}
+:$\forall I \in {\it Ids}\left({\mathscr S}\right): y = \sup I \implies x \in I$
+where
+:$\ll$ denotes the [[Definition:Element is Way Below|way below relation]],
+:${\it Ids}\left({\mathscr S}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Ideal in Ordered Set|ideals]] in $\mathscr S$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \ll y$
+Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that
+:$y = \sup I$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$y \preceq \sup I$
+By definition of [[Definition:Meet-Continuous Lattice|meet-continuous]]:
+:$\mathscr S$ is [[Definition:Up-Complete|up-complete]].
+Thus by [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$x \in I$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall I \in {\it Ids}\left({\mathscr S}\right): y = \sup I \implies x \in I$
+We will prove that
+:$\forall I \in {\it Ids}\left({\mathscr S}\right): y \preceq \sup I \implies x \in I$
+Let $I \in {\it Ids}\left({\mathscr S}\right)$ such that
+:$y \preceq \sup I$
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$I$ is [[Definition:Directed Subset|directed]] and [[Definition:Lower Set|lower]].
+Define $Y := \left\{ {y \wedge d: d \in I}\right\}$
+By [[Set of Finite Suprema is Directed]]:
+:$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}$ is [[Definition:Directed Subset|directed]].
+where ${\it Fin}\left({Y}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $Y$.
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Set Union|union]]:
+:$Y = \bigcup {\it Fin}\left({Y}\right)$
+{{begin-eqn}}
+{{eqn | l = y
+      | r = y \wedge \sup I
+      | c = [[Preceding iff Meet equals Less Operand]]
+}}
+{{eqn | r = \sup Y
+      | c = definition of [[Definition:Meet-Continuous Lattice|meet-continuous]]
+}}
+{{eqn | r = \sup \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}
+      | c = [[Supremum of Suprema]]
+}}
+{{eqn | r = \sup \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{end-eqn}}
+where $^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure of set]].
+By [[Lower Closure of Directed Subset is Ideal]]:
+:$\left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq$ is an [[Definition:Ideal in Ordered Set|ideal]] in $\mathscr S$
+By assumption:
+:$x \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}^\preceq$
+By definition of [[Definition:Lower Closure/Set|lower closure of set]]:
+:$\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({Y}\right) \land A \ne \varnothing}\right\}: x \preceq z$
+Then
+:$\exists A \in {\it Fin}\left({Y}\right): z = \sup A \land A \ne \varnothing$
+We will prove as a sublemma that:
+:$A \subseteq I$
+Let $a \in A$.
+By definition of [[Definition:Subset|subset]]:
+:$a \in Y$
+By definition of $Y$:
+:$\exists i \in I: a = y \wedge i$
+By [[Meet Precedes Operands]]:
+:$a \preceq i$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$a \in I$
+This ends the proof of sublemma.
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists h \in I: \forall a \in A: a \preceq h$
+By definition:
+:$h$ is [[Definition:Upper Bound of Set|upper bound]] for $A$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$z = \sup A \preceq h$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq h$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$x \in I$
+{{qed|lemma}}
+Thus by [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Approximation to Binary Logarithm from Natural and Common Logarithm}
+Tags: Logarithms
+
+\begin{theorem}
+The [[Definition:Binary Logarithm|binary logarithm]] $\lg x$ can be approximated, to within $1 \%$, by the expression:
+:$\lg x \approx \ln x + \log_{10} x$
+That is, by the sum of the [[Definition:Natural Logarithm|natural logarithm]] and [[Definition:Common Logarithm|common logarithm]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \lg x 
+      | r = \frac {\ln x} {\ln 2}
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = \frac {\log_{10} x} {\log_{10} 2}
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lg x
+      | r = \frac {\ln x + \log_{10} x} {\ln 2 + \log_{10} 2}
+      | c = 
+}}
+{{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 + \frac {\ln 2} {\ln 10} }
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \frac 1 {\ln 10} }\right)}
+      | c = 
+}}
+{{eqn | r = \frac {\ln x + \log_{10} x} {\ln 2 \left({1 + \log_{10} e}\right)}
+      | c = [[Reciprocal of Logarithm]]
+}}
+{{end-eqn}}
+We have that:
+:$\ln 2 = 0 \cdotp 69347 \ 1805 \ldots$
+:$\log_{10} e = 0 \cdotp 43429 \ 44819 \ldots$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln 2 \left({1 + \log_{10} e}\right)
+      | r = 0 \cdotp 69347 \ 1805 \ldots \left({1 + 0 \cdotp 43429 \ 44819 \ldots}\right)
+      | c = 
+}}
+{{eqn | r = 0 \cdotp 99462 \ldots
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$\lg x + \log_{10} x \approx 0 \cdotp 994 \ln x$
+That is, the approximation is $99.4 \%$ accurate.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Value of b for b by Logarithm Base b of x to be Minimum}
+Tags: Logarithms
+
+\begin{theorem}
+Let $x \in \R_{> 0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
+Consider the [[Definition:Real Function|real function]] $f: \R_{> 0} \to \R$ defined as:
+:$f \left({b}\right) := b \log_b x$
+$f$ attains a [[Definition:Local Minimum|minimum]] when
+:$b = e$
+where $e$ is [[Definition:Euler's Number|Euler's number]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative at Maximum or Minimum]], when $f$ is at a [[Definition:Local Minimum|minimum]], its [[Definition:Derivative|derivative]] $\dfrac \d {\d b} f$ will be zero.
+Let $y = f \left({b}\right)$.
+We have:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = b \log_b x
+      | c = 
+}}
+{{eqn | r = \frac {b \ln x} {\ln b} 
+      | c = [[Change of Base of Logarithm]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d b}
+      | r = \frac {\ln b \frac \d {\d b} \left({b \ln x}\right) - b \ln x \frac \d {\d b} \ln b} {\left({\ln b}\right)^2}
+      | c = [[Quotient Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\ln b \ln x - b \ln x \frac 1 b} {\left({\ln b}\right)^2}
+      | c = [[Derivative of Natural Logarithm]], [[Derivative of Identity Function]]
+}}
+{{eqn | r = \frac {\ln x} {\ln b} \left({1 - \frac 1 {\ln b} }\right)
+      | c = simplifying
+}}
+{{eqn | r = \frac {\ln x} {\ln b^2} \left({\ln b - 1}\right)
+      | c = simplifying
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d b} 
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\ln x} {\ln b}
+      | r = \frac {\ln x} {\left({\ln b}\right)^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ln b
+      | r = 1
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = b
+      | r = e
+      | c = Definition of [[Definition:Natural Logarithm|Natural Logarithm]]
+}}
+{{end-eqn}}
+To determine that $f$ is a [[Definition:Local Minimum|minimum]] at this point, we [[Definition:Differentiation|differentiate]] again {{WRT|Differentiation}} $b$:
+{{begin-eqn}}
+{{eqn | l = \frac {\d^2 y} {\d b^2}
+      | r = \frac \d {\d b} \left({\frac {\ln x} {\ln b^2} \left({\ln b - 1}\right) }\right)
+      | c = 
+}}
+{{eqn | r = \frac {\ln x} b \left({\frac {\ln b - 2 \left({\ln b - 1}\right)} {\left({\ln b}\right)^3} }\right)
+      | c = 
+}}
+{{end-eqn}}
+Setting $b = e$ gives:
+:$\dfrac {\d^2 y} {\d b^2} \Bigg \vert_{b \mathop = e} = \dfrac {\ln x} e \dfrac {\left({1 - 2 \left({0}\right)}\right)} 1$
+which works out to be [[Definition:Strictly Positive Real Number|(strictly) positive]].
+From [[Second Derivative of Strictly Convex Real Function is Strictly Positive]], $f$ is [[Definition:Strictly Convex Real Function|strictly convex]] at this point.
+Thus $f$ is a [[Definition:Local Minimum|minimum]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monotone Real Function with Everywhere Dense Image is Continuous}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $I$ and $J$ be [[Definition:Real Interval|real intervals]].
+Let $f: I \to J$ be a [[Definition:Monotone Real Function|monotone real function]].
+Let $f \sqbrk I$ be [[Definition:Everywhere Dense|everywhere dense]] in $J$, where $f \sqbrk I$ denotes the [[Definition:Image of Mapping|image]] of $I$ under $f$.
+Then $f$ is [[Definition:Continuous Real Function|continuous]] on $I$.
+\end{theorem}
+
+\begin{proof}
+Let $f$ be [[Definition:Increasing Real Function|increasing]].
+{{AimForCont}} $f$ is [[Definition:Discontinuous|discontinuous]] at $c \in I$.
+Let:
+:$L^-$ denote $\displaystyle \lim_{x \mathop \to c^-} \map f x$
+:$L$ denote $\map f c$
+:$L^+$ denote $\displaystyle \lim_{x \mathop \to c^+} \map f x$
+From [[Discontinuity of Monotonic Function is Jump Discontinuity]], $L^-$ and $L^+$ are [[Definition:Real Number|(finite) real numbers]].
+Since $f$ is increasing:
+:$L^- \le L \le L^+$
+Let $S = \openint {L^-} {L^+}$.
+If $L^- = L = L^+$, then $c$ is not a [[Definition:Discontinuous|discontinuity]], a [[Definition:Contradiction|contradiction]].
+Thus $S$ is [[Definition:Non-Degenerate Connected Set|non-degenerate]].
+From [[Non-Degenerate Real Interval is Infinite]], $S$ is [[Definition:Infinite Set|infinite]].
+From [[Relative Difference between Infinite Set and Finite Set is Infinite]], $S \setminus \set L$ is [[Definition:Non-Empty Set | non-empty]].
+From [[Finite Subset of Metric Space is Closed]], $\set L$ is [[Definition:Closed Set (Real Analysis)|closed]].
+From [[Open Set minus Closed Set is Open]], $S \setminus \set L$ is [[Definition:Open Set (Real Analysis)|open]].
+=== [[Monotone Real Function with Everywhere Dense Image is Continuous/Lemma|Lemma]] ===
+{{:Monotone Real Function with Everywhere Dense Image is Continuous/Lemma}}{{qed|lemma}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = S \cap f \sqbrk I
+      | o = \subseteq
+      | r = \set L
+      | c =  [[Monotone Real Function with Everywhere Dense Image is Continuous/Lemma|Lemma]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {S \cap f \sqbrk I} \setminus \set L
+      | o = \subseteq
+      | r = \set L \setminus \set L
+      | c = [[Set Difference over Subset]]
+}}
+{{eqn | r = \O
+      | c = [[Set Difference with Self is Empty Set]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {S \cap f \sqbrk I} \setminus \set L
+      | r = \O
+      | c = [[Subset of Empty Set iff Empty]]
+}}
+{{eqn | ll= \leadsto
+      | l = \O
+      | r = \paren {S \cap f \sqbrk I \setminus \set L}
+      | c = [[Equality is Symmetric]]
+}}
+{{eqn | r = \paren { S \setminus \set L} \cap f \sqbrk I
+      | c = [[Intersection with Set Difference is Set Difference with Intersection]]
+}}
+{{eqn | r = \paren {\paren {\openint {L^-} L \cup \set L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I
+      | c = Break $S$ into $3$ [[Definition:Disjoint Sets | disjoint]] [[Definition:Open Real Interval|open intervals]]
+}}
+{{eqn | r = \paren {\paren {\openint {L^-} L \cup \openint L {L^+} } \setminus \set L} \cap f \sqbrk I
+      | c = [[Set Difference with Union is Set Difference]]
+}}
+{{eqn | r = \paren {\openint {L^-} L \cup \openint L {L^+} } \cap f \sqbrk I
+      | c = [[Set Difference with Disjoint Set]]
+}}
+{{end-eqn}}
+Thus there exists a [[Definition:Non-Empty Set|non-empty]] [[Definition:Open Set of Real Numbers|open]] [[Definition:Subset|subset]] of $f \sqbrk I$ that is [[Definition:Disjoint Sets|disjoint]] from $Y$.
+From [[Open Set Characterization of Denseness]], $f \sqbrk I$ is not [[Definition:Everywhere Dense|everywhere dense]] in $J$.
+This [[Definition:Contradiction|contradicts]] the [[Definition:Hypothesis|hypothesis]] that $f \sqbrk I$ is everywhere dense in $J$.
+Therefore $I$ contains no [[Definition:Discontinuous|discontinuities]], by [[Proof by Contradiction]].
+The same argument, [[Definition:Mutatis Mutandis|mutatis mutandis]], proves the case where $f$ is [[Definition:Decreasing Real Function|decreasing]]
+Hence the result, by [[Proof by Cases]].
+{{qed}}
+[[Category:Real Analysis]]
+s0183mib65aiwfu78bwvbiwqtjxc830
+\end{proof}<|endoftext|>
+\section{Relative Difference between Infinite Set and Finite Set is Infinite}
+Tags: Set Difference, Infinite Sets
+
+\begin{theorem}
+Let $S$ be an [[Definition:Infinite Set|infinite set]].
+Let $T$ be a [[Definition:Finite Set|finite set]].
+Then $S \setminus T$ is an [[Definition:Infinite Set|infinite set]].
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $S \setminus T$ is a [[Definition:Finite Set|finite set]].
+Then:
+{{begin-eqn}}
+{{eqn | l = S
+      | o = \subseteq
+      | r = S \cup T
+      | c = [[Set is Subset of Union]]
+}}
+{{eqn | r = \paren {S \setminus T} \cup T
+      | c = [[Set Difference Union Second Set is Union]]
+}}
+{{end-eqn}}
+By [[Union of Finite Sets is Finite]], $S \cup T$ is a [[Definition:Finite Set|finite set]].
+By [[Subset of Finite Set is Finite]], $S$ must also be a [[Definition:Finite Set|finite set]].
+But $S$ is an [[Definition:Infinite Set|infinite set]].
+This is a [[Definition:Contradiction|contradiction]].
+Therefore $S \setminus T$ is an [[Definition:Infinite Set|infinite set]].
+{{qed}}
+[[Category:Set Difference]]
+[[Category:Infinite Sets]]
+ktntts6ukzwzcluwrgimxns9p2ils5a
+\end{proof}<|endoftext|>
+\section{Set Difference over Subset}
+Tags: Set Difference
+
+\begin{theorem}
+Let $A$, $B$, and $S$ be [[Definition:Set|sets]].
+Let $A \subseteq B$.
+Then:
+:$A \setminus S \subseteq B \setminus S$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = A \setminus S
+      | r = A \cap \map \complement S
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{eqn | o = \subseteq
+      | r = B \cap \map \complement S
+      | c = [[Set Intersection Preserves Subsets/Corollary|Corollary to Set Intersection Preserves Subsets]]
+}}
+{{eqn | r = B \setminus S
+      | c = [[Set Difference as Intersection with Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Set Difference]]
+jn9zvcha8xojzr8k7s5voxtb7nljqts
+\end{proof}<|endoftext|>
+\section{Subset of Empty Set iff Empty}
+Tags: Empty Set
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\O$ denote the [[Definition:Empty Set|empty set]].
+Then $S \subseteq \O$ {{iff}} $S = \O$.
+\end{theorem}
+
+\begin{proof}
+Suppose $x \in S$.
+Then since $S \subseteq \O$, it follows that $x \in \O$.
+Hence $x \notin S$.
+That is, $S = \O$.
+{{qed}}
+[[Category:Empty Set]]
+h7lpbqjhov78wp3cpmrvdho7zc9riec
+\end{proof}<|endoftext|>
+\section{Finite Subset of Metric Space is Closed}
+Tags: Closed Sets, Metric Spaces
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $S \subseteq A$ be [[Definition:Finite Set|finite]].
+Then $S$ is [[Definition:Closed Set (Metric Space)|closed]] in $M$.
+\end{theorem}
+
+\begin{proof}
+From [[Metric Space is Hausdorff]], $M$ is [[Definition:Hausdorff Space|Hausdorff]].
+From [[Finite Subspace of Hausdorff Space is Closed]], $S$ is [[Definition:Closed Set (Metric Space)|closed]].
+{{qed}}
+[[Category:Closed Sets]]
+[[Category:Metric Spaces]]
+o2ifexi1cunsg6pq68v9xoyi184umf8
+\end{proof}<|endoftext|>
+\section{Monotone Real Function with Everywhere Dense Image is Continuous/Lemma}
+Tags: Real Analysis
+
+\begin{theorem}
+:$\displaystyle \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c+ } \map f x} \cap f \sqbrk I \subseteq \set {\map f c}$
+\end{theorem}
+
+\begin{proof}
+From [[Discontinuity of Monotonic Function is Jump Discontinuity]], $\displaystyle \lim_{x \mathop \to c^-} \map f x$ and $\displaystyle \lim_{x \mathop \to c^+} \map f x$ are [[Definition:Real Number|finite]].
+Since $f$ is [[Definition:Increasing Real Function|increasing]]:
+:$\displaystyle \lim_{x \mathop \to c^-} \map f x < \lim_{x \mathop \to c^+} \map f x$
+Suppose $z \in \displaystyle \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I$.
+Then:
+:$\displaystyle \exists t \in I : \map f t \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}$
+=== Case 1 : $t < c$ ===
+Suppose that $t < c$.
+{{begin-eqn}}
+{{eqn | l = t
+      | o = <
+      | r = c
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | o = \le
+      | r = \lim_{x \mathop \to c^-} \map f x
+      | c = {{Defof|Increasing Real Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | o = \notin
+      | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+So we may discard this case.
+{{qed|lemma}}
+=== Case 2 : $t = c$ ===
+Suppose that $t = c$.
+{{begin-eqn}}
+{{eqn | l = t
+      | r = c
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | r = \map f c
+      | c = {{Defof|Mapping}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | o = \in
+      | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+So the proposition holds in this case.
+{{qed|lemma}}
+=== Case 3 : $c < t$ ===
+Suppose that $t > c$.
+{{begin-eqn}}
+{{eqn | l = t
+      | o = >
+      | r = c
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | o = \ge
+      | r = \lim_{x \mathop \to c^+} \map f x
+      | c = {{Defof|Increasing Real Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \map f t
+      | o = \notin
+      | r = \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+So we may discard this case.
+{{qed|lemma}}
+So $\map f t = c$, by [[Proof by Cases]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \map f t
+      | c = Definition of $t$
+}}
+{{eqn | r = c
+      | c = from above
+}}
+{{eqn | ll = \leadsto
+      | l = z
+      | o = \in
+      | r = \set {\map f c}
+}}
+{{end-eqn}}
+Thus:
+:$\displaystyle z \in \openint {\lim_{x \mathop \to c^-} \map f x} {\lim_{x \mathop \to c^+} \map f x} \cap f \sqbrk I \implies z \in \set {\map f c}$
+Hence the result, by definition of [[Definition:Subset|subset]].
+{{qed}}
+[[Category:Real Analysis]]
+4ghkaxwvk86zqyav0esgqfzxchrawyp
+\end{proof}<|endoftext|>
+\section{Defining Sequence of Natural Logarithm is Uniformly Convergent on Compact Sets}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]] such that $x > 0$.
+Let $\sequence {f_n}$ be the [[Definition:Sequence|sequence]] of mappings $f_n : \R_{>0} \to \R$ defined as:
+:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
+Let $K \subseteq \R_{>0}$ be [[Definition:Compact Subset of Real Numbers|compact]].
+Then $\sequence {f_n}$ is [[Definition:Uniform Convergence|uniformly convergent]] on $K$.
+\end{theorem}
+
+\begin{proof}
+From [[Continuity of Root Function]] and [[Combination Theorem for Continuous Functions]]:
+:$\forall n \in \N : \map {f_n} x$ is [[Definition:Continuous Real Function|continuous]] on $K$
+From [[Defining Sequence of Natural Logarithm is Convergent]], $\sequence {f_n}$ is [[Definition:Pointwise Convergence|pointwise convergent]] to $\ln$.
+From [[Natural Logarithm Function is Continuous]], $\ln$ is continuous on $K$.
+From [[Defining Sequence of Natural Logarithm is Strictly Decreasing]]:
+:$\forall x \in K: \sequence {\map {f_n} x}$ is [[Definition:Monotone Sequence|monotone]].
+From [[Dini's Theorem]], $\sequence {f_n}$ is [[Definition:Uniform Convergence|uniformly convergent]] to $\ln$.
+Hence the result.
+{{qed}}
+[[Category:Natural Logarithms]]
+ezp28yikfargl93un1yx8ymacc4q5zj
+\end{proof}<|endoftext|>
+\section{Multiplication of Positive Number by Real Number Greater than One}
+Tags: Real Multiplication, Inequalities
+
+\begin{theorem}
+Let $x$ and $y$ be [[Definition:Real Number|real numbers]].
+Let $x > 1$.
+Let $y > 0$.
+Then $\dfrac y x < y$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x
+      | o = <
+      | r = 1
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 x
+      | o = <
+      | r = 1
+      | c = [[Ordering of Reciprocals]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac y x
+      | o = <
+      | r = y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Real Multiplication]]
+[[Category:Inequalities]]
+aylhlcergqss4bgexgl50e85n07nkrf
+\end{proof}<|endoftext|>
+\section{Set of Finite Suprema is Directed}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then
+:$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is [[Definition:Directed Subset|directed]].
+where ${\it Fin}\left({X}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$.
+\end{theorem}
+
+\begin{proof}
+By [[Existence of Non-Empty Finite Suprema in Join Semilattice]]:
+:for every $A \in {\it Fin}\left({S}\right)$ if $A \ne \varnothing$, then $A$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists a: a \in X$
+By definitions of [[Definition:Subset|subset]] and [[Definition:Singleton|singleton]]:
+:$\left\{ {x}\right\} \subseteq X$
+By [[Singleton is Finite]]:
+:$\left\{ {x}\right\}$ is [[Definition:Finite Set|finite]]
+By definitions of [[Definition:Non-Empty Set|non-empty set]] and [[Definition:Singleton|singleton]]:
+:$\left\{ {x}\right\} \ne \varnothing$
+By definition of ${\it Fin}$:
+:$\sup \left\{ {x}\right\} \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+Thus by definition:
+:$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is a [[Definition:Non-Empty Set|non-empty set]].
+Let $x, y \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+Then
+:$\exists A \in {\it Fin}\left({X}\right): x = \sup A$ and $A \ne \varnothing$
+and
+:$\exists B \in {\it Fin}\left({X}\right): y = \sup B$ and $B \ne \varnothing$
+By [[Finite Union of Finite Sets is Finite]]:
+:$A \cup B$ is [[Definition:Finite Set|finite]]
+By [[Union of Subsets is Subset]]:
+:$A \cup B \subseteq X$
+By definitions of [[Definition:Non-Empty Set|non-empty set]] and [[Definition:Set Union|union]]:
+:$A \cup B \ne \varnothing$
+By definition of $\it Fin$:
+:$\sup \left({A \cup B}\right) \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+By [[Set is Subset of Union]]:
+:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
+Thus by [[Supremum of Subset]]:
+:$x \preceq \sup \left({A \cup B}\right)$ and $y \preceq \sup \left({A \cup B}\right)$
+Thus by definition:
+:$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Change of Index Variable of Summation}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{R \left({i}\right)} a_i = \sum_{R \left({j}\right)} a_j$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Too obvious?}}
+\end{proof}<|endoftext|>
+\section{Singleton is Finite}
+Tags: Singletons
+
+\begin{theorem}
+Let $x$ be arbitrary.
+Then $\set x$ is a [[Definition:Finite Set|finite set]].
+\end{theorem}
+
+\begin{proof}
+Define a [[Definition:Mapping|mapping]] $f: \set x \to \N_{< 1}$:
+:$\map f x = 0$
+By definition of [[Definition:Singleton|singleton]]:
+:$\forall y, z \in \set x: \map f y = \map f z \implies y = z$
+By definition:
+:$f$ is an [[Definition:Injection|injection]].
+By definition of [[Definition:Initial Segment of Natural Numbers|initial segment of natural numbers]]:
+:$\N_{< 1} = \set 0$
+By definition of $f$:
+:$\forall n \in N_{< 1}: \exists z \in \set x: \map f z = n$
+By definition:
+:$f$ is a [[Definition:Surjection|surjection]].
+By definition:
+:$f$ is a [[Definition:Bijection|bijection]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:$\set x \sim \N_{< 1}$
+Thus by definition:
+:$\set x$ is a [[Definition:Finite Set|finite set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T4 Property Preserved in Closed Subspace/Corollary}
+Tags: Closed Sets, Normal Spaces, Topological Subspaces
+
+\begin{theorem}
+If $T$ is a [[Definition:Normal Space|normal space]] then $T_K$ is also a [[Definition:Normal Space|normal space]].
+That is, the property of being a [[Definition:Normal Space|normal space]] is [[Definition:Weakly Hereditary Property|weakly hereditary]].
+\end{theorem}
+
+\begin{proof}
+From the definition, $T = \struct {S, \tau}$ is a [[Definition:Normal Space|normal space]] {{iff}}:
+:$\struct {S, \tau}$ is a [[Definition:T4 Space|$T_4$ space]]
+:$\struct {S, \tau}$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
+From [[Separation Properties Preserved in Subspace]], any [[Definition:Topological Subspace|subspace]] of a [[Definition:Fréchet Space (Topology)|$T_1$ space]] is also a [[Definition:Fréchet Space (Topology)|$T_1$ space]].
+From [[T4 Property Preserved in Closed Subspace]], any [[Definition:Closed Set (Topology)|closed]] [[Definition:Topological Subspace|subspace]] of a [[Definition:T4 Space|$T_4$ space]] is also a [[Definition:T4 Space|$T_4$ space]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Uniformly Convergent Sequence of Differentiable Functions}
+Tags: Differential Calculus, Convergence, Continuity
+
+\begin{theorem}
+Let $J$ be a [[Definition:Bounded Real Interval|bounded interval]].
+Let $\left\langle{f_n}\right\rangle$ be a [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $f_n: J \to \R$.
+Let each of $\left\langle{f_n}\right\rangle$ be [[Definition:Differentiable Real-Valued Function|differentiable]] on $J$.
+Let $\left\langle{ f_n \left({ x_0 }\right) }\right\rangle$ be [[Definition:Convergent Real Sequence|convergent]] for some $x_0 \in J$.
+Let the sequence of derivatives $\left\langle{f_n'}\right\rangle$ [[Definition:Uniform Convergence|converge uniformly]] on $J$ to a function $g : J \to \R$.
+Then $\left\langle{f_n}\right\rangle$ [[Definition:Uniform Convergence|converge uniformly]] on $J$ to a [[Definition:Differentiable Real-Valued Function|differentiable function]] $f: J \to \R$ and $D_x f = g$.
+\end{theorem}<|endoftext|>
+\section{Derivative of Natural Logarithm Function/Proof 4/Lemma}
+Tags: Derivative of Natural Logarithm Function
+
+\begin{theorem}
+Let $\sequence {f_n}_n$ be the [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $f_n: \R_{>0} \to \R$ defined as:
+:$\map {f_n} x = n \paren {\sqrt [n] x - 1}$
+Let $k \in \N$.
+Let $J = \closedint {\dfrac 1 k} k$.
+Then the [[Definition:Sequence|sequence]] of [[Definition:Derivative of Real Function|derivatives]] $\sequence { {f_n}'}_n$ [[Definition:Uniform Convergence|converges uniformly]] to some [[Definition:Real Function|real function]] $g: J \to \R$.
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Nth Root|Derivative of $n$th Root]] and [[Combination Theorem for Sequences]]:
+:$\forall n \in \N: \forall x \in J : D_x \map {f_n} x = \dfrac {\sqrt [n] x} x$
+From [[Closed Bounded Subset of Real Numbers is Compact]], $J$ is [[Definition:Compact (Real Analysis)|compact]].
+Thus:
+{{begin-eqn}}
+{{eqn | l = \lim_{n \mathop \to \infty} D_x \map {f_n} x
+      | r = \lim_{n \mathop \to \infty} \dfrac {\sqrt [n] x} x
+      | c = from above
+}}
+{{eqn | r = \frac 1 x \lim_{n \mathop \to \infty} \sqrt [n] x
+      | c = [[Multiple Rule for Real Sequences]]
+}}
+{{eqn | r = \frac 1 x \lim_{n \mathop \to \infty} x^{1/n}
+      | c = {{Defof|Root (Analysis)|$n$th Root}}
+}}
+{{eqn | r = \frac 1 x
+      | c = [[Power Function tends to One as Power tends to Zero/Rational Number|Power Function tends to One as Power tends to Zero: Rational Number]]
+}}
+{{end-eqn}}
+In particular, $\sequence { {f_n}'}_n$ is [[Definition:Pointwise Convergence|pointwise convergent]] to a [[Definition:Continuous on Interval|continuous function]] on $J$.
+For $x \in \closedint {\frac 1 k} 1$:
+{{begin-eqn}}
+{{eqn | l = x^{1/n}
+      | o = \le
+      | r = x^{1 / \paren {n + 1} }
+      | c = [[Exponential is Monotonic/Rational Number|Exponential is Monotonic: Rational Number]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt [n] x
+      | o = \le
+      | r = \sqrt [n + 1] x
+      | c = {{Defof|Root (Analysis)|$n$th Root}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\sqrt [n] x} x
+      | o = \le
+      | r = \frac {\sqrt [n + 1] x} x
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{end-eqn}}
+So $\sequence {\map { {f_n}'} x}_n$ is [[Definition:Increasing Real Sequence|increasing]] when $x \in \closedint {\dfrac 1 k} 1$.
+For $x \in \closedint 1 k$:
+{{begin-eqn}}
+{{eqn | l = x^{1/n}
+      | o = \ge
+      | r = x^{1 / \paren {n + 1} }
+      | c = [[Exponential is Monotonic/Rational Number|Exponential is Monotonic: Rational Number]]
+}}
+{{eqn | ll= \implies
+      | l = \sqrt [n] x
+      | o = \ge
+      | r = \sqrt [n + 1] x
+      | c = {{Defof|Root (Analysis)|$n$th Root}}
+}}
+{{eqn | ll= \implies
+      | l = \frac {\sqrt [n] x} x
+      | o = \ge
+      | r = \frac {\sqrt [n + 1] x} x
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{end-eqn}}
+So $\sequence {\map { {f_n}'} x}_n$ is [[Definition:Decreasing Real Sequence|decreasing]] when $x \in \closedint {\dfrac 1 k} 1$.
+Thus $\sequence {\map { {f_n}'} x}_n$ is [[Definition:Monotone Real Sequence|monotone]] for all $x \in \closedint {\frac 1 k} 1 \cup \closedint 1 k = J$.
+From [[Dini's Theorem]], $\sequence { {f_n}'}_n$ [[Definition:Uniform Convergence|converges uniformly]] to $\dfrac 1 x$ on $J$.
+Hence the result.
+{{qed}}
+[[Category:Derivative of Natural Logarithm Function]]
+4zsg9gzbbr954dzh89w5jzi2ibj1hy3
+\end{proof}<|endoftext|>
+\section{Exponential Function is Well-Defined/Real}
+Tags: Exponential Function, Exponential Function is Well-Defined
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\exp x$ be the [[Definition:Real Exponential Function|exponential of $x$]].
+Then $\exp x$ is [[Definition:Well-Defined Mapping|well-defined]].
+\end{theorem}<|endoftext|>
+\section{Tail of Convergent Sequence}
+Tags: Convergence, Sequences
+
+\begin{theorem}
+Let $\left\langle{a_n}\right\rangle$ be a [[Definition:Real Sequence|real sequence]].
+Let $N \in \N$ be a [[Definition:Natural Numbers|natural number]].
+Let $a \in R$ be a [[Definition:Real Number|real number]].
+Then:
+:$a_n \to a$
+{{iff}}:
+:$a_{n + N} \to a$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Convergence]]
+[[Category:Sequences]]
+m0bqneer6zq8hok8abolxrgexgzfs2v
+\end{proof}<|endoftext|>
+\section{Union of Inverses of Mappings is Inverse of Union of Mappings}
+Tags: Inverse Relations, Set Union, Indexed Families
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\family {f_i: i \in I}$ be an [[Definition:Indexed Family|indexed family]] of [[Definition:Mapping|mappings]].
+For each $i \in I$, let $f^{-1}$ denote the [[Definition:Inverse of Mapping|inverse of $f$]].
+Then the [[Definition:Inverse Relation|inverse]] of the [[Definition:Union of Family|union]] of $\family {f_i: i \in I}$ is the [[Definition:Union of Family|union]] of the [[Definition:Inverse Relation|inverses]] of $f_i, i \in I$.
+That is:
+:$\ds \paren {\bigcup \family {f_i: i \in I} }^{-1} = \bigcup \family {f_i^{-1}: i \in I}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \tuple {y, x}
+      | o = \in
+      | r = \paren {\bigcup \family {f_i: i \in I} }^{-1}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \tuple {x, y}
+      | o = \in
+      | r = \family {f_i: i \in I}
+      | c = {{Defof|Inverse Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists i \in I: \tuple {x, y}
+      | o = \in
+      | r = f_i
+      | c = {{Defof|Union of Family}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists i \in I: \tuple {y, x}
+      | o = \in
+      | r = f_i^{-1}
+      | c = {{Defof|Inverse Relation}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \tuple {y, x}
+      | o = \in
+      | r = \bigcup \family {f_i^{-1}: i \in I}
+      | c = {{Defof|Union of Family}}
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Inverse Relations]]
+[[Category:Set Union]]
+[[Category:Indexed Families]]
+21stoju5jrleqtp3rfx69i97ysmc2xi
+\end{proof}<|endoftext|>
+\section{Union of Functions Theorem/Corollary}
+Tags: Inverse Mappings
+
+\begin{theorem}
+For each $i \in \N$, let $g_i : X_i \to Y$ be [[Definition:Inverse Mapping|invertible]].
+Then $\displaystyle \bigcup \left\{{g_i: i \in \N}\right\}$ is [[Definition:Inverse Mapping|invertible]] and:
+:$\displaystyle \left({\bigcup \left\{{g_i: i \in \N}\right\} }\right)^{-1} = \bigcup \left\{{g_i^{-1}: i \in \N}\right\}$
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Inverse Mappings]]
+4nly5dl7wx1v3mmqfls3pyimb5rvy22
+\end{proof}<|endoftext|>
+\section{Permutation of Indices of Summation}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{\map R {\map \pi j} } a_{\map \pi j}
+      | r = \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
+      | c = {{Defof|Summation by Iverson's Convention}}
+}}
+{{eqn | r = \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sum_{j \mathop \in \Z} \sqbrk {i = \map \pi j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i}
+      | c = 
+}}
+{{eqn | r = \sum_{\map R i} a_i
+      | c = 
+}}
+{{eqn | r = \sum_{\map R j} a_j
+      | c = [[Change of Index Variable of Summation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Translation of Index Variable of Summation}
+Tags: Summations
+
+\begin{theorem}
+Let $R: \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
+Let $\displaystyle \sum_{R \left({j}\right)} a_j$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
+Then:
+:$\displaystyle \sum_{R \left({j}\right)} a_j = \sum_{R \left({c \mathop + j}\right)} a_{c \mathop + j} = \sum_{R \left({c \mathop - j}\right)} a_{c \mathop - j}$
+where $c$ is an [[Definition:Integer|integer]] [[Definition:Constant|constant]] which is not dependent upon $j$.
+\end{theorem}
+
+\begin{proof}
+First we consider the case where the [[Definition:Fiber of Truth|fiber of truth]] of $R$ is [[Definition:Finite Set|finite]].
+{{ProofWanted}}
+=== [[Translation of Index Variable of Summation/Infinite Series|Infinite case]] ===
+Now consider the case where the [[Definition:Fiber of Truth|fiber of truth]] of $R$ is [[Definition:Infinite Set|infinite]].
+{{:Translation of Index Variable of Summation/Infinite Series}}
+\end{proof}<|endoftext|>
+\section{Axiom of Approximation in Up-Complete Semilattice}
+Tags: Way Below Relation, Up-Complete Semilattices
+
+\begin{theorem}
+Let $\mathscr S = \left({S, \wedge, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Meet Semilattice|meet semilattice]].
+Let
+:$\forall x \in S: x^\ll$ is [[Definition:Directed Subset|directed]]
+Then
+:$\mathscr S$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]]
+{{iff}}
+:$\forall x, y \in S: x \npreceq y \implies \exists u \in S: u \ll x \land u \npreceq y$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$\mathscr S$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+Let $x, y \in S$ such that
+:$x \npreceq y$
+By assumption:
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$x^\ll$ admits a [[Definition:Supremum of Set|supremum]].
+By definition of [[Definition:Axiom of Approximation|axiom of approximation]]:
+:$x = \sup \left({x^\ll}\right)$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:if $y$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$, then $x \preceq y$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\exists u \in x^\ll: u \npreceq y$
+Thus by definition of [[Definition:Way Below Closure|way below closure]]:
+:$\exists u \in S: u \ll x \land u \npreceq y$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$\forall x, y \in S: x \npreceq y \implies \exists u \in S: u \ll x \land u \npreceq y$
+Let $x \in S$.
+By assumption:
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Up-Complete|up-complete]]:
+:$x^\ll$ admits a [[Definition:Supremum of Set|supremum]].
+By [[Operand is Upper Bound of Way Below Closure]]:
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$
+We will prove that
+:$\forall y \in S: y$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll \implies x \preceq y$
+Let $y \in S$ such that
+:$y$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$
+Aiming at a [[Definition:Contradiction|contradiction]] suppose that
+:$x \npreceq y$
+By assumption:
+:$\exists u \in S: u \ll x \land u \npreceq y$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$u \in x^\ll$
+By definition of [[Definition:Upper Bound of Set|upper bound]]
+:$u \preceq y$
+This contradicts $u \npreceq y$
+{{qed|lemma}}
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$x = \sup \left({x^\ll}\right)$
+Thus by definition:
+:$\mathscr S$ satisfies [[Definition:Axiom of Approximation|axiom of approximation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Operand is Upper Bound of Way Below Closure}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$
+where $x^\ll$ denotes the [[Definition:Way Below Closure|way below closure]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $y \in x^\ll$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+Thus by [[Way Below implies Preceding]]:
+:$y \preceq x$
+Thus by definition:
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $x^\ll$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exchange of Order of Summation}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({j}\right)} a_{i j} = \sum_{S \left({j}\right)} \sum_{R \left({i}\right)} a_{i j}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}
+      | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{R \left({i}\right)}\right] \left[{S \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{R \left({i}\right) \land S \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{S' \left({j}\right)}\right] \left[{R' \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{S' \left({j}\right)} \sum_{R' \left({i, j}\right)} a_{i j}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Summations equals Summation of Sum/Infinite Sequence}
+Tags: Summations
+
+\begin{theorem}
+Let $R: \Z \to \set {\mathrm T, \mathrm F}$ be a [[Definition:Propositional Function|propositional function]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$.
+Let $\displaystyle \sum_{\map R i} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Infinite Set|infinite]].
+Let $\displaystyle \sum_{\map R i} b_i$ and $\displaystyle \sum_{\map R i} c_i$ be [[Definition:Convergent Real Sequence|convergent]].
+Then:
+:$\displaystyle \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
+\end{theorem}<|endoftext|>
+\section{Exchange of Order of Summation with Dependency on Both Indices}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j} = \sum_{S' \left({j}\right)} \sum_{R' \left({i, j}\right)} a_{i j}$
+where:
+:$S' \left({j}\right)$ denotes the [[Definition:Propositional Function|propositional function]]:
+::there exists an $i$ such that both $R \left({i}\right)$ and $S \left({i, j}\right)$ hold
+:$R' \left({i, j}\right)$ denotes the [[Definition:Propositional Function|propositional function]]:
+::both $R \left({i}\right)$ and $S \left({i, j}\right)$ hold.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}
+      | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{R \left({i}\right)}\right] \left[{S \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{R \left({i}\right) \land S \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{S' \left({j}\right)}\right] \left[{R' \left({i, j}\right)}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{S' \left({j}\right)} \sum_{R' \left({i, j}\right)} a_{i j}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of i from 1 to n of Summation of j from 1 to i}
+Tags: Summations, Summation of i from 1 to n of Summation of j from 1 to i
+
+\begin{theorem}
+:$\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$
+\end{theorem}
+
+\begin{proof}
+:$\displaystyle \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$
+can be expressed as:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j}$
+where:
+:$R \left({i}\right)$ is the [[Definition:Propositional Function|propositional function]] $1 \le i \le n$
+:$S \left({i, j}\right)$ is the [[Definition:Propositional Function|propositional function]] $1 \le j \le i$
+We wish to find a [[Definition:Propositional Function|propositional function]] $S' \left({j}\right)$ which is to be:
+:there exists an $i$ such that both $1 \le i \le n$ and $1 \le j \le i$
+This is satisfied by the [[Definition:Propositional Function|propositional function]]:
+:$S' \left({j}\right) := 1 \le j \le n$
+Next we wish to find a [[Definition:Propositional Function|propositional function]] $R' \left({i, j}\right)$ which is to be:
+:both $1 \le i \le n$ and $1 \le j \le i$
+This is satisfied by the [[Definition:Propositional Function|propositional function]]:
+:$R' \left({i, j}\right) := j \le i \le n$
+Hence the result, from [[Exchange of Order of Summation with Dependency on Both Indices]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j}
+      | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le i \le n}\right] \left[{1 \le j \le i}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le i \le n}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le n}\right] \left[{j \le i \le n}\right]
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exchange of Order of Summation/Infinite Series}
+Tags: Summations
+
+\begin{theorem}
+Let the [[Definition:Fiber of Truth|fiber of truth]] of both $R$ and $S$ be [[Definition:Infinite Set|infinite]].
+Let:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({j}\right)} \left\vert{a_{i j} }\right\vert$
+exist.
+Then:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({j}\right)} a_{i j} = \sum_{S \left({j}\right)} \sum_{R \left({i}\right)} a_{i j}$
+\end{theorem}<|endoftext|>
+\section{Exchange of Order of Summation with Dependency on Both Indices/Infinite Series}
+Tags: Summations
+
+\begin{theorem}
+Let the [[Definition:Fiber of Truth|fiber of truth]] of both $R$ and $S$ be [[Definition:Infinite Set|infinite]].
+Let:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} \left\vert{a_{i j} }\right\vert$
+exist.
+Then:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({i, j}\right)} a_{i j} = \sum_{S' \left({j}\right)} \sum_{R' \left({i, j}\right)} a_{i j}$
+where:
+:$S' \left({j}\right)$ denotes the [[Definition:Propositional Function|propositional function]]:
+::there exists an $i$ such that both $R \left({i}\right)$ and $S \left({i, j}\right)$ hold
+:$R' \left({i, j}\right)$ denotes the [[Definition:Propositional Function|propositional function]]:
+::both $R \left({i}\right)$ and $S \left({i, j}\right)$ hold.
+\end{theorem}<|endoftext|>
+\section{Exchange of Order of Summation/Finite and Infinite Series}
+Tags: Summations
+
+\begin{theorem}
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Infinite Set|infinite]].
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $S$ be [[Definition:Finite Set|finite]].
+For all $j$ in the [[Definition:Fiber of Truth|fiber of truth]] of $S$, let $\displaystyle \sum_{R \left({i}\right)} a_{i j}$ be [[Definition:Convergent Real Sequence|convergent]].
+Then:
+:$\displaystyle \sum_{R \left({i}\right)} \sum_{S \left({j}\right)} a_{i j} = \sum_{S \left({j}\right)} \sum_{R \left({i}\right)} a_{i j}$
+\end{theorem}<|endoftext|>
+\section{Permutation of Indices of Summation/Infinite Series}
+Tags: Summations
+
+\begin{theorem}
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Infinite Set|infinite]].
+Let $\displaystyle \sum_{\map R i} a_i$ be [[Definition:Absolutely Convergent Series|absolutely convergent]].
+Then:
+:$\displaystyle \sum_{\map R j} a_j = \sum_{\map R {\map \pi j} } a_{\map \pi j}$
+\end{theorem}
+
+\begin{proof}
+This is a restatemtent of [[Manipulation of Absolutely Convergent Series/Permutation|Manipulation of Absolutely Convergent Series: Permutation]] in the context of [[Definition:Summation by Propositional Function|summations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Summations equals Summation of Sum}
+Tags: Summations
+
+\begin{theorem}
+Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
+Let $\displaystyle \sum_{\map R i} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Finite Set|finite]].
+Then:
+:$\displaystyle \sum_{\map R i} \paren {b_i + c_i} = \sum_{\map R i} b_i + \sum_{\map R i} c_i$
+\end{theorem}
+
+\begin{proof}
+Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{\map R i} \paren {b_i + c_i}
+      | r = \sum_{\map R i} \paren {a_{i 1} + a_{i 2} }
+      | c = Definition of $b_i$ and $c_i$
+}}
+{{eqn | r = \sum_{\map R i} \paren {\sum_{1 \mathop \le j \mathop \le 2} a_{i j} }
+      | c = {{Defof|Summation}}
+}}
+{{eqn | r = \sum_{1 \mathop \le j \mathop \le 2} \paren {\sum_{\map R i} a_{i j} }
+      | c = [[Exchange of Order of Summation]]
+}}
+{{eqn | r = \sum_{\map R i} a_{i 1} + \sum_{\map R i} a_{i 2}
+      | c = {{Defof|Summation}}
+}}
+{{eqn | r = \sum_{\map R i} b_i + \sum_{\map R i} c_i
+      | c = Definition of $b_i$ and $c_i$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Summations over Overlapping Domains}
+Tags: Summations, Sum of Summations over Overlapping Domains
+
+\begin{theorem}
+:$\displaystyle \sum_{R \left({j}\right)} a_j + \sum_{S \left({j}\right)} a_j = \sum_{R \left({j}\right) \mathop \lor S \left({j}\right)} a_j + \sum_{R \left({j}\right) \mathop \land S \left({j}\right)} a_j$
+where $\lor$ and $\land$ signify [[Definition:Disjunction|logical disjunction]] and [[Definition:Conjunction|logical conjunction]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{\map R j} a_j + \sum_{\map S j} a_j
+      | r = \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j} + \sum_{j \mathop \in \Z} a_j \sqbrk {\map R j}
+      | c = {{Defof|Summation by Iverson's Convention}}
+}}
+{{eqn | r = \sum_{j \mathop \in \Z} a_j \paren {\sqbrk {\map R j} + \sqbrk {\map R j} }
+      | c = 
+}}
+{{end-eqn}}
+Let:
+:$A := \set {j \in \Z: \map R j}$
+:$B := \set {j \in \Z: \map R j}$
+The result then follows from [[Cardinality of Set Union]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2}
+Tags: Summation of Products of n Numbers taken m at a time with Repetitions
+
+\begin{theorem}
+:$\displaystyle \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 + \paren {\sum_{i \mathop = a}^b {x_i}^2} }$
+\end{theorem}<|endoftext|>
+\section{Rules for Manipulating Summations}
+Tags: Summations
+
+\begin{theorem}
+Let $R: \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ and $S: \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ be [[Definition:Propositional Function|propositional functions]] on the set of [[Definition:Integer|integers]].
+Let $S: \Z \times \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ be a [[Definition:Propositional Function|propositional functions]] on the [[Definition:Cartesian Product|Cartesian product]] of the [[Definition:Integer|set of integers]] with itself.
+Let $\displaystyle \sum_{R \left({i}\right)} x_i$ denote a [[Definition:Summation by Propositional Function|summation]] over $R$.
+Let $\pi$ be a [[Definition:Permutation|permutation]] on the [[Definition:Fiber of Truth|fiber of truth]] of $R$.
+\end{theorem}<|endoftext|>
+\section{Uniqueness of Continuously Differentiable Solution to Initial Value Problem}
+Tags: First Order ODEs
+
+\begin{theorem}
+Let $D \subseteq \R^2$ be a [[Definition:Region of Plane|region]] containing $\tuple {a, b}$.
+Let $f: D \to \R$ be [[Definition:Real Function of Two Variables|real-valued mapping]] such that $f$ and $\dfrac {\partial f} {\partial x}$ are [[Definition:Continuous Real-Valued Vector Function|continuous]] on $D$.
+Consider the [[Definition:Initial Value Problem|initial value problem]]:
+:$\dfrac {\d x} {\d t} = \map f {t, x}$
+:$\map x a = b$
+Suppose the [[Definition:Initial Value Problem|initial value problem]] above has a [[Definition:Solution of Differential Equation|solution]] $\phi$ for all $x$ in some [[Definition:Real Interval|interval]] $J$ containing $a$.
+Then the solution $\phi$ is [[Definition:Unique | unique]] on $J$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|Is this the same as Picard's Little Theorem?}}
+\end{proof}<|endoftext|>
+\section{Exhausting Sequence of Sets on the Strictly Positive Real Numbers}
+Tags: Real Numbers
+
+\begin{theorem}
+For each $k \in \N$, let $S_k = \openint {\dfrac 1 k} k$.
+Then $\sequence {S_k}_k$ is an [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]] on $\R_{>0}$.
+\end{theorem}
+
+\begin{proof}
+To prove that $\sequence {S_k}_k$ is [[Definition:Exhausting Sequence of Sets|exhausting]] $\R_{>0}$, it is sufficient to show:
+:$(1): \quad \forall k \in \N: S_k \subseteq S_{k + 1}$
+:$(2): \quad \ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$
+=== $\sequence {S_k}_k$ is [[Definition:Increasing Sequence of Sets|increasing]] ===
+Let $k \in \N$.
+Let $k = 1$.
+Then:
+:$S_k = \openint {\dfrac 1 k} k = \O$
+Thus, by [[Empty Set is Subset of All Sets]]:
+:$\openint {\dfrac 1 k} k \subseteq \openint {\dfrac 1 {k + 1} } {k + 1}$
+If $k \ge 2$:
+{{begin-eqn}}
+{{eqn | o = 
+      | m = x
+      | mo= \in
+      | r = S_k
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 k
+      | o = <
+      | m = x
+      | mo= <
+      | r = k
+      | c = {{Defof|Open Real Interval}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {k + 1}
+      | o = <
+      | m = x
+      | mo= <
+      | r = k
+      | c = [[Ordering of Reciprocals]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {k + 1}
+      | o = <
+      | m = x
+      | mo= <
+      | r = k + 1
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | m = x
+      | mo= \in
+      | r = S_{k + 1}
+      | c = {{Defof|Open Real Interval}}
+}}
+{{end-eqn}}
+It follows that $\sequence {S_k}_k$ is [[Definition:Increasing Sequence of Sets|increasing]].
+{{qed|lemma}}
+=== $\ds \bigcup_{k \mathop \in \N} S_k = \R_{>0}$ ===
+Let $x \in \R_{>0}$.
+==== Case 1: $1 < x$ ====
+Let $1 < x$.
+Let $k = \ceiling x$.
+From [[Real Number is between Ceiling Functions]]:
+:$x < k$
+From [[Ordering of Reciprocals]]:
+:$1 < k \implies \dfrac 1 k < 1$
+So:
+:$\dfrac 1 k < x < k$
+and so $x \in S_k$.
+==== Case 2: $x = 1$ ====
+Let $x = 1$.
+Let $k = 2$.
+Then:
+:$\dfrac 1 2 < 1 < 2$
+and hence $x \in S_2$.
+==== Case 3: $0 < x < 1$ ====
+Let $0 < x < 1$.
+Let $k = \ceiling {\dfrac 1 x}$.
+From [[Ordering of Reciprocals]]:
+:$0 < x < 1 \implies 1 < \dfrac 1 x$
+From [[Real Number is between Ceiling Functions]]:
+:$1 < \dfrac 1 x < k$
+From [[Ordering of Reciprocals]]:
+:$\dfrac 1 k < x < 1$
+So:
+:$\dfrac 1 k < x < k$
+and so $x \in S_k$.
+Hence:
+:$\forall x \in \R_{>0} : \exists k \in \N : x \in S_k$
+{{qed|lemma}}
+The result follows from the definition of [[Definition:Exhausting Sequence of Sets|exhausting sequence of sets]].
+{{qed}}
+[[Category:Real Numbers]]
+i0x820ehi7pux4ldqc3afafq0cwwykx
+\end{proof}<|endoftext|>
+\section{Way Below in Ordered Set of Topology}
+Tags: Way Below Relation, Topology
+
+\begin{theorem}
+Let $\left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $\left({\tau, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
+Let $x, y \in \tau$.
+Then
+:$x \ll y$ in $\left({\tau, \preceq}\right)$
+{{iff}}
+:for every [[Definition:Set of Sets|set]] $F$ of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$: if $y \subseteq \bigcup F$, then
+::there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $G$ of $F$: $x \subseteq \bigcup G$
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$x \ll y$
+Let $F$ be a [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$ such that
+:$y \subseteq \bigcup F$
+By definition of [[Definition:Subset|subset]]:
+:$F \subseteq \tau$
+By proof of [[Topology forms Complete Lattice]]:
+:$\sup F = \bigcup F$
+By assumption:
+:$y \preceq \sup F$
+By [[Topology forms Complete Lattice]]:
+:$\left({\tau, \preceq}\right)$ is [[Definition:Complete Lattice|complete lattice]].
+By [[Way Below in Complete Lattice]]:
+:$\exists G \in {\it Fin}\left({F}\right): x \preceq \sup G$
+where ${\it Fin}\left({F}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $F$.
+Thus by proof of [[Topology forms Complete Lattice]]:
+:$x \subseteq \bigcup G$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every [[Definition:Set of Sets|set]] $F$ of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$: if $y \subseteq \bigcup F$, then
+::there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $G$ of $F$: $x \subseteq \bigcup G$
+Let $I$ be an [[Definition:Ideal in Ordered Set|ideal]] in $\left({\tau, \preceq}\right)$ such that
+:$y \preceq \sup I$
+By proof of [[Topology forms Complete Lattice]]:
+:$y \subseteq \bigcup I$
+By assumption:
+:there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $G$ of $I$: $x \subseteq \bigcup G$
+By proof of [[Topology forms Complete Lattice]]:
+:$x \preceq \sup I$
+By definition of [[Definition:Ideal in Ordered Set|ideal]]:
+:$I$ is [[Definition:Directed Subset|directed]] and [[Definition:Lower Set|lower]].
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists u \in I: u$ is [[Definition:Upper Bound of Set|upper bound]] for $G$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup G \preceq u$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq u$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$x \in I$
+Thus by [[Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Inverse Theorem}
+Tags: Real Analysis, Continuity, Named Theorems
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $f$ be [[Definition:Strictly Monotone Real Function|strictly monotone]] and [[Definition:Continuous on Interval|continuous]] on $I$.
+Let $g$ be the [[Definition:Inverse of Mapping|inverse mapping]] to $f$.
+Let $J := f \left[{I}\right]$ be the [[Definition:Image of Mapping|image]] of $I$ under $f$.
+Then $g$ is [[Definition:Strictly Monotone Real Function|strictly monotone]] and [[Definition:Continuous on Interval|continuous]] on $J$.
+\end{theorem}
+
+\begin{proof}
+From [[Strictly Monotone Real Function is Bijective]], $f$ is a [[Definition:Bijection|bijection]].
+From [[Inverse of Strictly Monotone Function]], $f^{-1} : J \to I$ exists and is [[Definition:Strictly Monotone Real Function|strictly monotone]].
+From [[Surjective Monotone Function is Continuous]], $f^{-1}$ is [[Definition:Continuous on Interval|continuous]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Logarithm is Strictly Concave}
+Tags: Logarithms
+
+\begin{theorem}
+:$\ln x: x > 0$ [[Definition:Strictly Concave Real Function|strictly concave]].
+\end{theorem}
+
+\begin{proof}
+From [[Logarithm is Strictly Increasing]], $\ln x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]] on $x > 0$.
+From [[Second Derivative of Natural Logarithm Function]]:
+: $D^2 \ln x = -\dfrac 1 {x^2}$
+Thus $D^2 \ln x$ is [[Definition:Strictly Negative|strictly negative]] on $x > 0$ (in fact is [[Definition:Strictly Negative|strictly negative]] for all $x \ne 0$).
+Thus from [[Derivative of Monotone Function]], $D \dfrac 1 x$ is [[Definition:Strictly Decreasing Real Function|strictly decreasing]] on $x > 0$.
+So from [[Real Function is Strictly Concave iff Derivative is Strictly Decreasing]], $\ln x$ is [[Definition:Strictly Concave Real Function|strictly concave]] on $x > 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential is Strictly Convex}
+Tags: Exponential Function
+
+\begin{theorem}
+:The [[Definition:Real Function|function]] $f \left({x}\right) = \exp x$ is [[Definition:Strictly Convex Real Function|strictly convex]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Real Exponential Function|exponential function]] is the [[Definition:Inverse Mapping|inverse]] of the [[Definition:Natural Logarithm|natural logarithm function]].
+From [[Logarithm is Strictly Increasing]], $\ln x$ is [[Definition:Strictly Increasing Real Function|strictly increasing]].
+From [[Logarithm is Strictly Concave]], $\ln x$ is [[Definition:Strictly Concave Real Function|strictly concave]].
+The result follows from [[Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex]].
+{{qed}}
+[[Category:Exponential Function]]
+pa119jqwoaw6rc4k41qso9nu8eayuv8
+\end{proof}<|endoftext|>
+\section{Way Below in Complete Lattice}
+Tags: Way Below Relation, Complete Lattices
+
+\begin{theorem}
+Let $\left({S, \vee, \wedge, \preceq}\right)$ be a [[Definition:Complete Lattice|complete lattice]].
+Let $x, y \in S$.
+Then
+:$x \ll y$
+{{iff}}
+:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+where
+:$\ll$ denotes the [[Definition:Element is Way Below|way below relation]],
+:${\it Fin}\left({X}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$x \ll y$
+Let $X \subseteq S$ such that
+:$y \preceq \sup X$
+By [[Set of Finite Suprema is Directed]]:
+:$\left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ is [[Definition:Directed Subset|directed]].
+By definition of [[Definition:Set Union|union]]:
+:$\bigcup {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\} = X$
+By [[Supremum of Suprema]]:
+:$\sup X = \sup \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+By definition of [[Definition:Element is Way Below|way below relation]]:
+:$\exists z \in \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}: x \preceq z$
+Thus
+:$\exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$\forall X \subseteq S: y \preceq \sup X \implies \exists A \in {\it Fin}\left({X}\right): x \preceq \sup A$
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$y \preceq \sup D$
+By assumption:
+:$\exists A \in {\it Fin}\left({D}\right): x \preceq \sup A$
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists u \in D: u$ is [[Definition:Upper Bound of Set|upper bound]] for $A$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup A \preceq u$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$x \preceq u$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Continuous Midpoint-Convex Function is Convex}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f$ be a [[Definition:Real Function|real function]] which is defined on a [[Definition:Real Interval|real interval]] $I$.
+Let $f$ be [[Definition:Midpoint-Convex|midpoint-convex]] and [[Definition:Continuous on Interval|continuous]] on $I$.
+Then $f$ is [[Definition:Convex Real Function|convex]].
+\end{theorem}<|endoftext|>
+\section{Way Below Compact is Topological Compact}
+Tags: Way Below Relation, Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $L = \left({\tau, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]] where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
+Let $x \in \tau$.
+Then
+:$x$ is [[Definition:Compact Element|compact]] in $L$ (it means: $x \ll x$)
+{{iff}}
+:$T_x$ is [[Definition:Compact Topological Subspace|compact (topologically)]]
+where $T_x = \left({x, \tau_x}\right)$ denotes the [[Definition:Topological Subspace|topological subspace]] of $x$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let
+:$x \ll x$
+Let $F \subseteq  \tau_x$ be a [[Definition:Open Cover|open cover]] of $x$.
+By definition of [[Definition:Cover of Set|cover]]:
+:$x \subseteq \bigcup F$
+By definition of [[Definition:Topological Space|topological space]]:
+:$\forall y \in \tau: x \cap y \in \tau$
+By definition of [[Definition:Subset|subset]]:
+:$\tau_x \subseteq \tau$
+By [[Subset Relation is Transitive]]:
+:$F \subseteq \tau$
+By [[Way Below in Ordered Set of Topology]]:
+:there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $G$ of $F$: $x \subseteq \bigcup G$
+By definition:
+:$G$ is [[Definition:Cover of Set|cover]] of $x$
+Thus by definition:
+:$x$ has [[Definition:Finite Subcover|finite subcover]].
+Thus by definition:
+:$T_x$ is [[Definition:Compact Topological Subspace|compact]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$T_x$ is [[Definition:Compact Topological Subspace|compact]].
+Let $F$ be a [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$ such that
+:$x \subseteq \bigcup F$
+Define $Y := \left\{ {x \cap y: y \in F}\right\}$
+By definition of [[Definition:Subset|subset]]:
+:$Y \subseteq \tau_x$
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$x \subseteq \bigcup Y$
+By definition
+:$Y$ is [[Definition:Open Cover|open cover]] of $x$.
+By definition of [[Definition:Compact Topological Subspace|compact]]:
+:$x$ has [[Definition:Finite Subcover|finite subcover]] $G$.
+By [[Axiom:Axiom of Choice|Axiom of Choice]]:
+:$\exists g:G \to F: \forall y \in G: g\left({y}\right) \cap x = y$
+By [[Cardinality of Image of Set not greater than Cardinality of Set]]:
+:$\left\vert{g^\to\left({G}\right)}\right\vert \le \left\vert{G}\right\vert$
+where $\left\vert{G}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $G$.
+Thus
+:$g^\to\left({G}\right)$ is [[Definition:Finite Set|finite]].
+Thus by definitions of [[Definition:Subset|subset]] and [[Definition:Image of Subset under Mapping|image of set]]:
+:$g^\to\left({G}\right) \subseteq F$
+By [[Intersection is Subset]]:
+:$\forall y \in G: y \subseteq g\left({y}\right)$
+By [[Set Union Preserves Subsets/Families of Sets]]:
+:$\displaystyle \bigcup G = \bigcup_{y \mathop \in G}y \subseteq \bigcup_{y \mathop \in G} g \left({y}\right) = \bigcup g^\to \left({G}\right)$
+By definition of [[Definition:Cover of Set|cover]]:
+:$x \subseteq \bigcup G$
+Thus by [[Subset Relation is Transitive]]:
+:$x \subseteq \bigcup g^\to\left({G}\right)$
+Thus by [[Way Below in Ordered Set of Topology]];
+:$x \ll x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power Function on Strictly Positive Base is Continuous}
+Tags: Powers
+
+\begin{theorem}
+Let $a \in \R_{>0}$.
+\end{theorem}<|endoftext|>
+\section{Continuous Strictly Midpoint-Concave Function is Strictly Concave}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f$ be [[Definition:Strictly Midpoint-Concave|strictly midpoint-concave]] and [[Definition:Continuous on Interval|continuous]] on $I$.
+Then $f$ is [[Definition:Strictly Concave Real Function|strictly concave]].
+\end{theorem}<|endoftext|>
+\section{Continuous Midpoint-Concave Function is Concave}
+Tags:  Real Analysis
+
+\begin{theorem}
+Let $f$ be [[Definition:Midpoint-Concave|midpoint-concave]] and [[Definition:Continuous on Interval|continuous]] on $I$.
+Then $f$ is [[Definition: Concave Real Function|concave]].
+\end{theorem}<|endoftext|>
+\section{Continuous Strictly Midpoint-Convex Function is Strictly Convex}
+Tags:  Real Analysis
+
+\begin{theorem}
+Let $f$ be [[Definition:Strictly Midpoint-Convex|strictly midpoint-convex]] and [[Definition:Continuous on Interval|continuous]] on $I$.
+Then $f$ is [[Definition:Strictly Convex Real Function|strictly convex]].
+\end{theorem}<|endoftext|>
+\section{Power Function on Strictly Positive Base is Convex}
+Tags: Real Analysis, Powers
+
+\begin{theorem}
+Let $a \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|strictly positive real number]].
+Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
+:$\map f x = a^x$
+where $a^x$ denotes [[Definition:Power to Real Number|$a$ to the power of $x$]].
+Then $f$ is [[Definition:Convex Real Function|convex]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$.
+Note that, from [[Power of Positive Real Number is Positive/Real Number|Power of Positive Real Number is Positive: Real Number]]:
+:$\forall t \in \R: a^t > 0$.
+So:
+{{begin-eqn}}
+{{eqn | l = a^{\paren {x + y} / 2}
+      | r = \sqrt {a^{x + y} }
+      | c = [[Exponent Combination Laws/Power of Power/Proof 2|Exponent Combination Laws: Power of Power: Proof 2]]
+}}
+{{eqn | r = \sqrt {a^x a^y}
+      | c = [[Exponent Combination Laws/Product of Powers/Proof 2|Exponent Combination Laws: Product of Powers: Proof 2]]
+}}
+{{eqn | o = \le
+      | r = \frac {a^x + a^y} 2
+      | c = [[Cauchy's Mean Theorem]]
+}}
+{{end-eqn}}
+Hence $a^x$ is [[Definition:Midpoint-Convex|midpoint-convex]].
+Further, from [[Power Function on Strictly Positive Base is Continuous/Real Power|Power Function on Strictly Positive Base is Continuous: Real Power]], $a^x$ is [[Definition:Continuous on Interval|continuous]].
+Thus, from [[Continuous Midpoint-Convex Function is Convex]], $a^x$ is [[Definition:Convex Real Function|convex]].
+{{qed}}
+[[Category:Real Analysis]]
+[[Category:Powers]]
+ozrx4xqacs1l03nudbcq362bc9qbixd
+\end{proof}<|endoftext|>
+\section{Natural Logarithm as Derivative of Exponential at Zero}
+Tags: Natural Logarithms
+
+\begin{theorem}
+Let $\ln: \R_{>0}$ denote the [[Definition:Real Natural Logarithm|real natural logarithm]].
+Then:
+:$\displaystyle \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$
+\end{theorem}
+
+\begin{proof}
+Fix $x \in \R_{>0}$.
+Let $x > 1$.
+From [[Power Function on Strictly Positive Base is Convex]], $x^h$ is [[Definition:Convex Real Function|convex]].
+Thus for $0 < h < s$:
+{{begin-eqn}}
+{{eqn | l = \frac {x^h - a^0} {h - 0}
+      | o = \le
+      | m = \frac {x^s - a^0} {s - 0}
+      | c = {{Defof|Convex Real Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^h - 1} h
+      | o = \le
+      | m = \frac {x^s - 1} s
+}}
+{{end-eqn}}
+Further, $0 < \dfrac 1 x < 1$.
+So, for $h < s < 0 \iff 0 < -s < -h$:
+{{begin-eqn}}
+{{eqn | l = \frac {\paren {\frac 1 x}^{-s} - \paren {\frac 1 x}^0} {-s - 0}
+      | o = \le
+      | m = \frac {\paren {\frac 1 x}^{-h} - \paren {\frac 1 x}^0} {-h - 0}
+      | c = [[Power Function on Strictly Positive Base is Convex]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^s - 1} {-s}
+      | o = \le
+      | m = \frac {x^h - 1} {-h}
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^h - 1} h
+      | o = \le
+      | m = \frac {x^s - 1} s
+      | c = [[Order of Real Numbers is Dual of Order of their Negatives]]
+}}
+{{end-eqn}}
+Hence $\dfrac {x^h - 1} h$ is [[Definition:Increasing Real Function|increasing]] on $\R \setminus \set 0$.
+Next:
+{{begin-eqn}}
+{{eqn | l = h
+      | o = >
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = x^h
+      | o = >
+      | r = 1
+      | c = [[Power Function on Base Greater than One is Strictly Increasing]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^h - 1
+      | o = >
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^h - 1} h
+      | o = >
+      | r = 0
+      | c = [[Order of Real Numbers is Dual of Order of their Negatives]]
+}}
+{{end-eqn}}
+So $\dfrac {x^h - 1} h$ is [[Definition:Strictly Positive Real Function|strictly positive]] on $\R_{>0}$.
+In particular:
+:$\dfrac {x^h - 1} h$ is [[Definition:Bounded Below Real-Valued Function|bounded below]] (by $0$) and [[Definition:Increasing Real Function|increasing]] on $\openint 0 \to$
+:$\dfrac {x^h - 1} h$ is [[Definition:Bounded Above Real-Valued Function|bounded above]] (by $\displaystyle \inf_{h \mathop > 0} \frac {x^h - 1} h$) and [[Definition:Increasing Real Function|increasing]] on $\openint \gets 0$
+So from [[Limit of Increasing Function]], $\displaystyle \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ and $\displaystyle \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exist.
+Further:
+{{begin-eqn}}
+{{eqn | l = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h
+      | r = \lim_{h \mathop \to \infty} h \paren {x^{1 / h} - 1}
+      | c = [[Limit of Composite Function]]
+}}
+{{eqn | r = \lim_{n \mathop \to \infty} n \paren {x^{1 / n} - 1}
+      | c = [[Limit of Sequence is Limit of Real Function]]
+}}
+{{eqn | r = \ln x
+      | c = [[Definition:Natural Logarithm/Positive Real/Definition 3|Sequential definition of natural logarithm]]
+}}
+{{end-eqn}}
+where $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ is now a [[Definition:Real Sequence|real sequence]].
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h
+      | r = \lim_{h \mathop \to 0^+} -\frac {x^{-h} - 1} h
+      | c = [[Limit of Composite Function]]
+}}
+{{eqn | r = -\lim_{h \mathop \to \infty} h \paren {x^{-1 / h} - 1}
+      | c = [[Limit of Composite Function]]
+}}
+{{eqn | r = -\lim_{h \mathop \to \infty} h \paren {\paren {\frac 1 x}^{1 / h} - 1}
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{eqn | r = -\lim_{n \mathop \to \infty} n \paren {\paren {\frac 1 x}^{1 / n} - 1}
+      | c = [[Limit of Sequence is Limit of Real Function]]
+}}
+{{eqn | r = -\ln \frac 1 x
+      | c = {{Defof|Natural Logarithm|subdef = Positive Real|index = 3}}
+}}
+{{eqn | r = \ln x
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{end-eqn}}
+Thus, for $x > 1$:
+{{begin-eqn}}
+{{eqn | l = \lim_{h \mathop \to 0^-} \frac {x^h - 1} h
+      | r = \ln x
+      | c = 
+}}
+{{eqn | r = \lim_{h \mathop \to 0^+} \frac {x^h - 1} h
+      | c = 
+}}
+{{end-eqn}}
+So from [[Limit iff Limits from Left and Right]], for $x > 1$:
+:$\displaystyle \lim_{h \mathop \to 0} \frac {x^h - 1} h = \ln x$
+Suppose instead that $0 < x < 1$.
+From [[Ordering of Reciprocals]]:
+:$\dfrac 1 x > 1$
+Thus, from above:
+{{begin-eqn}}
+{{eqn | l = \ln \frac 1 x
+      | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h
+}}
+{{eqn | ll= \leadsto
+      | l = \ln x
+      | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^h - 1} h
+      | c = [[Logarithm of Reciprocal]]
+}}
+{{eqn | r = -\lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} {-h}
+      | c = [[Limit of Composite Function]]
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {\paren {\frac 1 x}^{-h} - 1} h
+}}
+{{eqn | r = \lim_{h \mathop \to 0} \frac {x^h - 1} h
+      | c = [[Exponent Combination Laws/Negative Power|Exponent Combination Laws: Negative Power]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Natural Logarithms]]
+mwvx3z0ccwp1mp2c0tgi4kvsnp6qn4t
+\end{proof}<|endoftext|>
+\section{Product of Increasing Positive Functions is Increasing}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] defined on an [[Definition:Real Interval|interval]] $I$.
+Let $f$ and $g$ be [[Definition:Increasing Mapping|increasing]] and [[Definition:Positive Real Function|positive]] on $I$.
+Then the [[Definition:Pointwise Multiplication of Real-Valued Functions|product]] $f g$ is [[Definition:Increasing Mapping|increasing]] on $I$.
+\end{theorem}<|endoftext|>
+\section{Derivative of Exponential Function/Proof 5/Lemma}
+Tags: Derivative of Exponential Function
+
+\begin{theorem}
+:$\forall x \in \R : n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is [[Definition:Increasing Real Sequence|increasing]].
+\end{theorem}
+
+\begin{proof}
+First:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = >
+      | r = \left\lceil{\left\vert{x}\right\vert}\right\rceil
+}}
+{{eqn | ll= \implies
+      | l = n 
+      | o = >
+      | r = -x
+      | c = [[Negative of Absolute Value]] and [[Real Number is between Ceiling Functions]]
+}}
+{{eqn | ll= \implies
+      | l = \frac n {n + x}
+      | o = >
+      | r = 0
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = >
+      | m = \left\lceil{\left\vert{x}\right\vert}\right\rceil
+}}
+{{eqn | ll= \implies
+      | l = n
+      | o = >
+      | m = \left\vert{x}\right\vert
+      | c = [[Real Number is between Ceiling Functions]]
+}}
+{{eqn | ll= \implies
+      | l = \left\vert{\frac x n}\right\vert
+      | o = <
+      | m = 1
+      | c = dividing both sides by $n$
+}}
+{{eqn | ll= \implies
+      | l = -1
+      | o = <
+      | m = \frac x n
+      | mo= <
+      | r = 1
+      | c = [[Negative of Absolute Value/Corollary 1|Negative of Absolute Value: Corollary 1]]
+}}
+{{eqn | ll= \implies
+      | l = 0
+      | o = <
+      | m = 1 + \frac x n
+}}
+{{eqn | ll= \implies
+      | l = 0
+      | o = <
+      | m = \left({1 + \frac x n}\right)^n
+      | c = [[Power of Positive Real Number is Positive/Natural Number|Power of Positive Real Number is Positive: Natural Number]]
+}}
+{{end-eqn}}
+So, for $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$:
+:$\left\langle{\dfrac n {n + x} }\right\rangle$
+and:
+:$\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$
+are [[Definition:Positive Real Number| positive]].
+Now let $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$.
+Suppose first that $x \in \R_{\ge 0}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac n {n + x}
+      | o = \le
+      | r = \frac {n + 1} {n + x + 1}
+}}
+{{eqn | ll= \iff
+      | l = n \left({n + x + 1}\right)
+      | o = \le
+      | r = \left({n + 1}\right) \left({n + x}\right)
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \iff
+      | l = n^2 + n x + n
+      | o = \le
+      | r = n^2 + n x + n + x
+}}
+{{eqn | ll= \iff
+      | l = 0
+      | o = \le
+      | r = x
+}}
+{{end-eqn}}
+So $\left\langle{\dfrac n {n + x} }\right\rangle$ is [[Definition:Increasing Real Sequence|increasing]].
+Further, from [[Exponential Sequence is Eventually Increasing]]:
+:$\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is [[Definition:Increasing Real Sequence|increasing]].
+Hence, from [[Product of Positive Increasing Functions is Increasing]]:
+:$n \ge \left\lceil{\left\vert{x}\right\vert }\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is [[Definition:Increasing Real Sequence|increasing]].
+Suppose instead that $x \in \R_{<0}$.
+{{AimForCont}} that:
+:$\left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is [[Definition:Decreasing Real Sequence|decreasing]].
+From above: $\left\langle{1 + \dfrac x n}\right\rangle = \left\langle{\dfrac {n + x} n}\right\rangle$ is [[Definition:Decreasing Real Sequence|decreasing]].
+Thus, from [[Product of Positive Increasing and Decreasing Functions is Decreasing]]:
+:$\left\langle{\dfrac {n + x} n \dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle = \left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is [[Definition:Decreasing Real Sequence|decreasing]].
+This [[Definition:Contradiction|contradicts]] [[Exponential Sequence is Eventually Increasing]].
+Hence the result, by [[Proof by Contradiction]].
+{{qed}}
+[[Category:Derivative of Exponential Function]]
+ckqbhfr7zj6nuyrsadq9uq2v596uvtk
+\end{proof}<|endoftext|>
+\section{Permutation of Indices}
+Tags: Summations, Products
+
+\begin{theorem}
+Let $R: \Z \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Finite Set|finite]].
+Let  $\pi$ is a [[Definition:Permutation|permutation]] on the [[Definition:Fiber of Truth|fiber of truth]] of $R$.
+Then:
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{\map R {\map \pi j} } a_{\map \pi j}
+      | r = \sum_{j \mathop \in \Z} a_{\map \pi j} \sqbrk {\map R {\map \pi j} }
+      | c = {{Defof|Summation by Iverson's Convention}}
+}}
+{{eqn | r = \sum_{j \mathop \in \Z} \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sqbrk {i = \map \pi j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i} \sum_{j \mathop \in \Z} \sqbrk {i = \map \pi j}
+      | c = 
+}}
+{{eqn | r = \sum_{i \mathop \in \Z} a_i \sqbrk {\map R i}
+      | c = 
+}}
+{{eqn | r = \sum_{\map R i} a_i
+      | c = 
+}}
+{{eqn | r = \sum_{\map R j} a_j
+      | c = [[Change of Index Variable of Summation]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Change of Index Variable of Product}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{R \left({i}\right)} a_i = \prod_{R \left({j}\right)} a_j$
+\end{theorem}
+
+\begin{proof}
+Let $S = \left\{ {i \in \Z: R \left({i}\right)}\right\}$.
+Let $T = \left\{ {j \in \Z: R \left({j}\right)}\right\}$.
+Let $i \in S$.
+Then $R \left({i}\right)$.
+Let $j = i$.
+By [[Definition:Leibniz's Law|Leibniz's Law]], $R \left({j}\right)$.
+Thus $i \in T$.
+By definition of [[Definition:Subset|subset]], $S \subseteq T$.
+Similarly, let $j \in T$.
+Then $R \left({j}\right)$.
+Let $i = j$.
+By [[Definition:Leibniz's Law|Leibniz's Law]], $R \left({i}\right)$.
+Thus $j \in S$.
+By definition of [[Definition:Subset|subset]], $T \subseteq S$.
+Thus by definition of [[Definition:Set Equality|set equality]]:
+$S = T$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \prod_{R \left({i}\right)} a_i
+      | r = \prod a_i \left[{R \left({i}\right)}\right]
+      | c = {{Defof|Iverson's Convention}}
+}}
+{{eqn | r = \prod a_i \times \chi_S
+      | c = {{Defof|Characteristic Function of Set}}
+}}
+{{eqn | r = \prod a_j \times \chi_T
+      | c = as $S = T$
+}}
+{{eqn | r = \prod a_j \left[{R \left({j}\right)}\right]
+      | c = {{Defof|Iverson's Convention}}
+}}
+{{eqn | r = \prod_{R \left({j}\right)} a_j
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Translation of Index Variable of Summation/Corollary}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = m}^n a_j = \sum_{j \mathop = m + c}^{n + c} a_{j - c}$
+where $c$ is an [[Definition:Integer|integer]] [[Definition:Constant|constant]] which is not dependent upon $j$.
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+[[Category:Summations]]
+lbyqh95b2mddftv6wlct6ggm79eelp0
+\end{proof}<|endoftext|>
+\section{Permutation of Indices of Product}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{\map R j} a_j = \prod_{\map R {\map \pi j} } a_{\map \pi j}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \prod_{\map R {\map \pi j} } a_{\map \pi j}
+      | r = \prod_{j \mathop \in \Z} {a_{\map \pi j} }^{\sqbrk {\map R {\map \pi j} } }
+      | c = {{Defof|Product by Iverson's Convention}}
+}}
+{{eqn | r = \prod_{j \mathop \in \Z} \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} \sqbrk {i = \map \pi j} }
+      | c = 
+}}
+{{eqn | r = \paren {\prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} } } \uparrow \paren {\prod_{j \mathop \in \Z} \sqbrk {i = \map \pi j} }
+      | c = using [[Definition:Knuth Uparrow Notation|Knuth uparrow notation]]
+}}
+{{eqn | r = \prod_{i \mathop \in \Z} {a_i}^{\sqbrk {\map R i} }
+      | c = 
+}}
+{{eqn | r = \prod_{\map R i} a_i
+      | c = 
+}}
+{{eqn | r = \prod_{\map R j} a_j
+      | c = [[Change of Index Variable of Product]]
+}}
+{{end-eqn}}
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Translation of Index Variable of Product}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{R \left({j}\right)} a_j = \prod_{R \left({c \mathop + j}\right)} a_{c \mathop + j} = \prod_{R \left({c \mathop - j}\right)} a_{c \mathop - j}$
+where:
+:$\displaystyle \prod_{R \left({j}\right)} a_j$ denotes the [[Definition:Product by Propositional Function|product]] over $a_j$ for all $j$ that satisfy the [[Definition:Propositional Function|propositional function]] $R \left({j}\right)$
+:$c$ is an [[Definition:Integer|integer]] [[Definition:Constant|constant]] which is not dependent upon $j$.
+\end{theorem}<|endoftext|>
+\section{Exchange of Order of Product}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{R \left({i}\right)} \prod_{S \left({j}\right)} a_{i j} = \prod_{S \left({j}\right)} \prod_{R \left({i}\right)} a_{i j}$
+\end{theorem}<|endoftext|>
+\section{Product of Products over Overlapping Domains}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{R \left({j}\right)} a_j \prod_{S \left({j}\right)} a_j = \left({\prod_{R \left({j}\right) \mathop \lor S \left({j}\right)} a_j}\right) \left({\prod_{R \left({j}\right) \mathop \land S \left({j}\right)} a_j}\right)$
+where $\lor$ and $\land$ signify [[Definition:Disjunction|logical disjunction]] and [[Definition:Conjunction|logical conjunction]] respectively.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \prod_{R \left({j}\right)} a_j \times \prod_{S \left({j}\right)} a_j
+      | r = \prod_{j \mathop \in \Z} a_j^{\left[{R \left({j}\right)}\right]} \times \prod_{j \mathop \in \Z} a_j^{\left[{S \left({j}\right)}\right]}
+      | c = {{Defof|Product by Iverson's Convention}}
+}}
+{{eqn | r = \prod_{j \mathop \in \Z} a_j^{\left({\left[{R \left({j}\right)}\right] + \left[{S \left({j}\right)}\right]}\right)}
+      | c = [[Exponent Combination Laws/Product of Powers|Exponent Combination Laws: Product of Powers]]
+}}
+{{end-eqn}}
+Let:
+:$A := \left\{ {j \in \Z: R \left({j}\right)}\right\}$
+:$B := \left\{ {j \in \Z: S \left({j}\right)}\right\}$
+The result then follows from [[Cardinality of Set Union]].
+\end{proof}<|endoftext|>
+\section{Topology is Locally Compact iff Ordered Set of Topology is Continuous}
+Tags: Continuous Lattices, Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $L = \left({\tau, \preceq}\right)$ be the [[Definition:Ordered Set|ordered set]] where $\preceq$ is the [[Definition:Inclusion Relation|inclusion relation]].
+Then
+:$(1): \quad T$ is [[Definition:Locally Compact Space|locally compact]] implies $L$ is [[Definition:Continuous Ordered Set|continuous]]
+:$(2): \quad T$ is [[Definition:T3 Space|$T_3$ space]] and $L$ is [[Definition:Continuous Ordered Set|continuous]] implies $T$ is [[Definition:Locally Compact Space|locally compact]]
+\end{theorem}
+
+\begin{proof}
+=== Condition $(1)$ ===
+Let $T$ be [[Definition:Locally Compact Space|locally compact]].
+By [[Topology forms Complete Lattice]]:
+:$L$ is [[Definition:Complete Lattice|complete lattice]].
+Thus by [[Way Below Closure is Directed in Bounded Below Join Semilattice]]:
+:$\forall x \in \tau: x^\ll$ is [[Definition:Directed Subset|directed]]
+Thus by definition:
+:$L$ is [[Definition:Up-Complete|up-complete]].
+Let $x \in \tau$.
+We will prove that
+:$x \subseteq \sup \left({x^\ll}\right)$
+Let $a \in x$.
+By definition:
+:$x$ is [[Definition:Open Set (Topology)|open set]] in $T$
+By definition of [[Definition:Locally Compact Space|locally compact]]:
+:there exists [[Definition:Local Basis|local basis]] $\mathcal B$ of $a$ such that all [[Definition:Element|elements]] are [[Definition:Compact|compact]].
+By definition of [[Definition:Local Basis|local basis]]:
+:$\exists y \in \mathcal B: y \subseteq x$
+Then
+:$y$ is [[Definition:Compact|compact]].
+By [[Way Below if Between is Compact Set in Ordered Set of Topology]]:
+:$y \ll x$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \in x^\ll$
+By [[Set is Subset of Union/Set of Sets]]:
+:$\displaystyle y \subseteq \bigcup \left({x^\ll}\right)$
+By definition of [[Definition:Subset|subset]]:
+:$\displaystyle a \in \bigcup \left({x^\ll}\right)$
+Thus by proof of [[Topology forms Complete Lattice]]:
+:$a \in \sup \left({x^\ll}\right)$
+{{qed|lemma}}
+According to definition of [[Definition:Set Equality|set equality]] it remains to prove that:
+:$\sup \left({x^\ll}\right) \subseteq x$
+Let $a \in \sup \left({x^\ll}\right)$.
+By proof of [[Topology forms Complete Lattice]]:
+:$\displaystyle a \in \bigcup \left({x^\ll}\right)$
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$\exists y \in x^\ll: a \in y$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$
+By [[Way Below implies Preceding]]:
+:$y \preceq x$
+Then
+:$y \subseteq x$
+Thus by definition of [[Definition:Subset|subset]]:
+:$a \in x$
+By definition:
+:$L$ satisfies the [[Definition:Axiom of Approximation|axiom of approximation]].
+Thus by definition
+:$L$ is [[Definition:Continuous Ordered Set|continuous]].
+{{qed|lemma}}
+=== Condition $(2)$ ===
+Let $T$ be a [[Definition:T3 Space|$T_3$ space]].
+Let $L$ be [[Definition:Continuous Ordered Set|continuous]].
+Let $x \in S, A \in \tau$ such that:
+:$x \in A$
+By definition of [[Definition:Continuous Ordered Set|continuous]]:
+:$L$ satisfies the [[Definition:Axiom of Approximation|axiom of approximation]].
+{{begin-eqn}}
+{{eqn | l = A
+      | r = \sup \left({A^\ll}\right)
+      | c = {{Defof|Axiom of Approximation}}
+}}
+{{eqn | r = \bigcup \left({A^\ll}\right)
+      | c = [[Topology forms Complete Lattice]]
+}}
+{{end-eqn}}
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$\exists y \in A^\ll: x \in y$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll A$
+By definition of [[Definition:T3 Space/Definition 2]]:
+:$\exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists V \in \tau: x \in V \subseteq N_x \subseteq y$
+By definition of [[Definition:Interior (Topology)|interior]]:
+:$x \in N_x^\circ$
+where $N_x^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $N_x$.
+We will prove that $N_x$ is [[Definition:Compact|topologically compact]].
+Let $\mathcal F$ be a [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$ such that:
+:$\mathcal F$ is a [[Definition:Cover of Set|cover]] of $N_x$
+Define a [[Definition:Set of Sets|set]] of [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $S$:
+:$\mathcal F' := \mathcal F \cup \left\{ {\complement_S\left({N_x}\right)}\right\}$
+{{begin-eqn}}
+{{eqn | l = \bigcup \mathcal F'
+      | r = \left({\bigcup \mathcal F}\right) \cup \bigcup \left\{ {\complement_S\left({N_x}\right)}\right\}
+      | c = [[Union Distributes over Union/Sets of Sets]]
+}}
+{{eqn | r = \left({\bigcup \mathcal F}\right) \cup \complement_S\left({N_x}\right)
+      | c = [[Union of Singleton]]
+}}
+{{eqn | o = \supseteq
+      | r = N_x \cup \complement_S\left({N_x}\right)
+      | c = {{{Defof|Cover of Set}}
+}}
+{{eqn | r = S
+      | c = {{Defof|Relative Complement}}
+}}
+{{eqn | o = \supseteq
+      | r = A
+}}
+{{end-eqn}}
+Then
+:$A \subseteq \bigcup \mathcal F'$
+By [[Way Below in Ordered Set of Topology]]:
+:there exists a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $\mathcal G'$ of $\mathcal F'$: $y \subseteq \bigcup \mathcal G'$
+Define a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\mathcal F$:
+:$\mathcal G := \mathcal G' \setminus \left\{ {\complement_S\left({N_x}\right)}\right\}$
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$N_x \subseteq \bigcup \mathcal G$
+Thus by definition:
+:$N_x$ has [[Definition:Finite Subcover|finite subcover]] of $\mathcal F$
+Thus by definition:
+:$N_x$ is [[Definition:Compact|compact]].
+{{qed|lemma}}
+Thus by [[Locally Compact iff Open Neighborhood contains Compact Set]];
+:$T$ is [[Definition:Locally Compact Space|locally compact]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Sequence of Power by Index}
+Tags: Arithmetic-Geometric Sequences, Sum of Sequence of Power by Index
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 0}^n j x^j = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}$
+for $x \ne 1$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^n j x^j
+      | r = x \sum_{1 \mathop \le j \mathop \le n}^n j x^{j - 1}
+      | c = 
+}}
+{{eqn | r = x \sum_{0 \mathop \le j \mathop \le n - 1} \paren {j + 1} x^j
+      | c = 
+}}
+{{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} j x^j} + \paren {x \sum_{0 \mathop \le j \mathop \le n - 1} x^j}
+      | c = 
+}}
+{{eqn | r = \paren {x \sum_{0 \mathop \le j \mathop \le n} j x^j} - n x^{n + 1} + \paren {\sum_{0 \mathop \le j \mathop \le n - 1} x^{j + 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x - 1} \sum_{0 \mathop \le j \mathop \le n} j x^j
+      | r = n x^{n + 1} - \paren {\sum_{1 \mathop \le j \mathop \le n} x^j}
+      | c = 
+}}
+{{eqn | r = n x^{n + 1} - \paren {\sum_{0 \mathop \le j \mathop \le n} x^j} + 1
+      | c = 
+}}
+{{eqn | r = n x^{n + 1} - \paren {\frac {x^{n - 1} - 1} {x - 1} } + 1
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 0}^n j x^j
+      | r = \frac {n x^{n + 2} - \paren {n + 1} x^{n + 1} + x} {\paren {x - 1}^2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Sum of Arithmetic-Geometric Sequence]]:
+:$\ds \sum_{j \mathop = 0}^n \paren {a + j d} x^j = \frac {a \paren {1 - x^{n + 1} } } {1 - x} + \frac {x d \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^n j x^j
+      | r = \frac {0 \paren {1 - x^{n + 1} } } {1 - x} + \frac {x \times 1 \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}
+      | c = putting $a = 0, d = 1$
+}}
+{{eqn | r = \frac {x \paren {1 - \paren {n + 1} x^n + n x^{n + 1} } } {\paren {1 - x}^2}
+      | c = initial simplification
+}}
+{{eqn | r = \frac {x - \paren {n + 1} x^{n + 1} + n x^{n + 2} } {\paren {x - 1}^2}
+      | c = further simplification
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering of Series of Ordered Sequences}
+Tags:  Series
+
+\begin{theorem}
+Let $\sequence {a_n}$ and $\sequence {b_n}$ be two [[Definition:Real Sequence|sequences]].
+Let $\displaystyle \sum_{n \mathop = 1}^{\infty} a_n$ and $\displaystyle \sum_{n \mathop = 1}^\infty b_n$ be [[Definition:Convergent Series|convergent series]].
+For each $n \in \N$, let $a_n < b_n$.
+Then:
+:$\displaystyle \sum_{n \mathop = 0}^\infty a_n < \displaystyle \sum_{n \mathop = 0}^\infty b_n$
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {\epsilon_n}$ be the [[Definition:Real Sequence|sequence]] defined by:
+:$\forall n \in \N : b_n - a_n$
+From [[Linear Combination of Convergent Series]], $\displaystyle \sum_{n \mathop = 0}^\infty \epsilon_n$ is [[Definition:Convergent Series|convergent]] with sum $\epsilon > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{n \mathop = 0}^\infty b_n - \sum_{n \mathop = 0}^\infty a_n
+      | r = \sum_{n \mathop = 0}^\infty \paren {a_n + \epsilon_n} - \sum_{n \mathop = 0}^\infty a_n
+}}
+{{eqn | r = \sum_{n \mathop = 0}^\infty \epsilon_n
+      | c = [[Linear Combination of Convergent Series]]
+}}
+{{eqn | r = \epsilon
+}}
+{{eqn | o = > 
+      | r = 0
+}}
+{{end-eqn}}
+Hence the result, by definition of [[Definition:Less Than (Real Numbers)|inequality]].
+{{qed}}
+[[Category: Series]]
+l15hent94uzzn1e41bci3z6k85rp6oh
+\end{proof}<|endoftext|>
+\section{Reciprocal of Real Exponential}
+Tags: Exponential Function, Reciprocals
+
+\begin{theorem}
+:$\dfrac 1 {\map \exp x} = \map \exp {-x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \exp 0
+      | r = 1
+      | c = [[Exponential of Zero]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \exp {x - x}
+      | r = 1
+      | c = as $\forall x \in \R : x - x = 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \map \exp {-x} \, \map \exp x
+      | r = 1
+      | c = [[Exponential of Sum/Real Numbers|Exponential of Sum: Real Numbers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \exp {-x}
+      | r = \dfrac 1 {\exp x}
+      | c = dividing both sides by $\exp x$, which is non-zero by [[Exponential of Real Number is Strictly Positive]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Exponential Function]]
+[[Category:Reciprocals]]
+9qdlcogsx46aqmeilw6mncowfgb3vsk
+\end{proof}<|endoftext|>
+\section{Telescoping Series/Example 2}
+Tags: Telescoping Series
+
+\begin{theorem}
+Let $\left \langle {b_n} \right \rangle$ be a [[Definition:Real Sequence|sequence in $\R$]].
+Let $\left \langle {a_n} \right \rangle$ be a [[Definition:Real Sequence|sequence]] whose [[Definition:Term of Sequence|terms]] are defined as:
+:$a_k = b_k - b_{k - 1}$
+Then:
+:$\displaystyle \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \displaystyle \sum_{k \mathop = m}^n a_k
+      | r = \sum_{k \mathop = m}^n \left({b_k - b_{k - 1} }\right)
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m}^n b_{k - 1}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = m}^n b_k - \sum_{k \mathop = m - 1}^{n - 1} b_k
+      | c = [[Translation of Index Variable of Summation]]
+}}
+{{eqn | r = \left({\sum_{k \mathop = m}^{n - 1} b_k + b_n}\right) - \left({b_{m - 1} + \sum_{k \mathop = m}^{n - 1} b_k}\right)
+      | c = 
+}}
+{{eqn | r = b_n - b_{m - 1}
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Repunit Integer as Product of Base - 1 by Increasing Digit Integer}
+Tags: Repunits
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = 9 \times 1 + 2
+      | r = 11
+}}
+{{eqn | l = 9 \times 12 + 3
+      | r = 111
+}}
+{{eqn | l = 9 \times 123 + 4
+      | r = 1111
+}}
+{{eqn | l = 9 \times 1234 + 5
+      | r = 11111
+}}
+{{eqn | l = 9 \times 12345 + 6
+      | r = 11111
+}}
+{{eqn | o = \ldots
+}}
+{{end-eqn}}
+That is:
+:$\displaystyle 9 \sum_{j \mathop = 0}^n \left({n - j}\right)10^j + n + 1 = \sum_{j \mathop = 0}^n 10^j$
+\end{theorem}
+
+\begin{proof}
+A specific instance of the [[Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result|general result]]:
+{{:Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result}}
+where $b = 10$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result}
+Tags: Recreational Mathematics
+
+\begin{theorem}
+:$\displaystyle \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j}b^j + n + 1 = \sum_{j \mathop = 0}^n b^j$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j}b^j + n + 1
+      | c = 
+}}
+{{eqn | r = n \paren {b - 1} \sum_{j \mathop = 0}^n b^j - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1
+      | c = 
+}}
+{{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | r = n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \frac {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} {\paren {b - 1}^2} + n + 1
+      | c = [[Sum of Sequence of Power by Index]]
+}}
+{{eqn | r = \frac {n \paren {b - 1} \paren {b^{n + 1} - 1} - \paren {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} + \paren {b - 1} n + \paren {b - 1} } {b - 1}
+      | c = elementary simplification
+}}
+{{eqn | r = \frac {n b^{n + 2} - n b^{n - 1} - n b + n - n b^{n + 2} + n b^{n - 1} + b^{n + 1} - b + n b - n + b - 1} {b - 1}
+      | c = multiplying out
+}}
+{{eqn | r = \frac {b^{n + 1} - 1} {b - 1}
+      | c = simplification
+}}
+{{eqn | r = \sum_{j \mathop = 0}^n b^j
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential of Real Number is Strictly Positive/Proof 5/Lemma}
+Tags: Exponential of Real Number is Strictly Positive
+
+\begin{theorem}
+:$\forall x \in \R: \exp x \ne 0$
+\end{theorem}
+
+\begin{proof}
+This proof assumes the [[Definition:Exponential Function/Real/Differential Equation|definition of $\exp$ as the solution to an initial value problem]].
+That is, suppose $\exp$ satisfies:
+:$(1): \quad D_x \exp x = \exp x$
+:$(2): \quad \map \exp 0 = 1$
+on $\R$.
+{{AimForCont}} that $\exists \alpha \in \R: \exp \alpha = 0$.
+Suppose that $\alpha > 0$.
+Let $J = \closedint 0 \alpha$.
+From [[Exponential Function is Continuous/Real Numbers/Proof 3|Exponential Function is Continuous]], $\exp$ is [[Definition:Continuous Real Function on Subset|continuous]] on $J$.
+From [[Max and Min of Function on Closed Real Interval]]:
+:$\exists K \in \R: \forall x \in J: \size {\exp x} < K$
+Then, $\forall n \in \N : \exists c_n \in J$ such that:
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \exp 0
+      | c = by $(2)$
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac {\map {\exp^j} \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n
+      | c = [[Taylor's Theorem/One Variable|Taylor's Theorem for Univariate Functions]]
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac {\map \exp \alpha} {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n
+      | c = by $(1)$
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{n - 1} \frac 0 {j!} \paren {-\alpha}^j + \frac {\exp c_n} {n!} \paren {-\alpha}^n
+      | c = from our assumption aiming at a contradiction
+}}
+{{eqn | r = \frac {\exp c_n} {n!} \paren {-\alpha}^n
+}}
+{{end-eqn}}
+So:
+:$\forall n \in \N: 1 \le K \dfrac {\alpha^n} {n!}$
+That is, dividing both sides by $K$:
+:$\forall n \in \N: \dfrac 1 K \le \dfrac {\alpha^n} {n!}$
+But from [[Power over Factorial]], $\dfrac {\alpha^n} {n!} \to 0$.
+This [[Definition:Contradiction|contradicts]] our assumption.
+The same argument, [[Definition:Mutatis Mutandis|mutatis mutandis]] proves the result for $\alpha < 0$.
+By hypothesis $(2)$:
+{{begin-eqn}}
+{{eqn | lll=\alpha = 0
+      | ll= \implies
+      | l = \exp \alpha 
+      | r = 1
+}}
+{{eqn | o = \ne
+      | r = 0
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+[[Category:Exponential of Real Number is Strictly Positive]]
+bwx8xkcv36pi3vwjjp1lnqhh642foz1
+\end{proof}<|endoftext|>
+\section{Product of Products}
+Tags: Products
+
+\begin{theorem}
+Let $R: \Z \to \set {\mathrm T, \mathrm F}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
+Let $\displaystyle \prod_{R \paren i} x_i$ denote a [[Definition:Product by Propositional Function|product]] over $R$.
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Finite Set|finite]].
+Then:
+:$\displaystyle \prod_{R \paren i} \paren {b_i c_i} = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}$
+\end{theorem}
+
+\begin{proof}
+Let $b_i =: a_{i 1}$ and $c_i =: a_{i 2}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \prod_{R \paren i} \paren {b_i c_i}
+      | r = \prod_{R \paren i} \paren {a_{i 1} a_{i 2} }
+      | c = by definition
+}}
+{{eqn | r = \prod_{R \paren i} \paren {\prod_{1 \mathop \le j \mathop \le 2} a_{i j} }
+      | c = {{Defof|Product by Propositional Function}}
+}}
+{{eqn | r = \prod_{1 \mathop \le j \mathop \le 2} \paren {\prod_{R \paren i} a_{i j} }
+      | c = [[Exchange of Order of Product]]
+}}
+{{eqn | r = \paren {\prod_{R \paren i} a_{i 1} } \paren {\prod_{R \paren i} a_{i 2} }
+      | c = {{Defof|Product by Propositional Function}}
+}}
+{{eqn | r = \paren {\prod_{R \paren i} b_i} \paren {\prod_{R \paren i} c_i}
+      | c = by definition
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of General Logarithms}
+Tags: Summations, Products, Logarithms
+
+\begin{theorem}
+Let $R: \Z \to \left\{ {\mathrm T, \mathrm F}\right\}$ be a [[Definition:Propositional Function|propositional function]] on the set of [[Definition:Integer|integers]].
+Let $\displaystyle \prod_{R \left({i}\right)} a_i$ denote a [[Definition:Product by Propositional Function|product]] over $R$.
+Let the [[Definition:Fiber of Truth|fiber of truth]] of $R$ be [[Definition:Finite Set|finite]].
+Then:
+:$\displaystyle \log_b \left({\prod_{R \left({i}\right)} a_i}\right) = \sum_{R \left({i}\right)} \log_b a_i$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+First let:
+:$S := \left\{ {a_i: R \left({i}\right)}\right\}$
+We have that $S$ is [[Definition:Finite Set|finite]].
+Hence the contents of $S$ can be [[Definition:Well-Ordered Set|well-ordered]], by [[Finite Totally Ordered Set is Well-Ordered]].
+Let $S$ have $m$ [[Definition:Element|elements]], identified as:
+:$S = \left\{ {s_1, s_2, \ldots, s_m}\right\}$
+For all $n \in \Z_{\ge 0}$ such that $n \le m$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \log_b \left({\prod_{i \mathop = 1}^0 s_i}\right)
+      | r = \log_b 1
+      | c = {{Defof|Vacuous Product}}
+}}
+{{eqn | r = 0
+      | c = [[Logarithm of 1 is 0]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^n \log_b s_i
+      | c = {{Defof|Vacuous Summation}}
+}}
+{{end-eqn}}
+=== Basis for the Induction ===
+$P \left({1}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \log_b \left({\prod_{i \mathop = 1}^1 s_i}\right)
+      | r = \log_b s_1
+      | c = {{Defof|Product Notation (Algebra)|Product}}
+}}
+{{eqn | r = \sum_{i \mathop = 1}^1 \log_b s_i
+      | c = {{Defof|Summation}}
+}}
+{{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 $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \log_b \left({\prod_{i \mathop = 1}^k s_i}\right) = \sum_{i \mathop = 1}^k \log_b s_i$
+from which it is to be shown that:
+:$\displaystyle \log_b \left({\prod_{i \mathop = 1}^{k + 1} s_i}\right) = \sum_{i \mathop = 1}^{k + 1} \log_b s_i$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \log_b \left({\prod_{i \mathop = 1}^{k + 1} s_i}\right)
+      | r = \log_b \left({\left({\prod_{i \mathop = 1}^k s_i}\right) \times s_{k + 1} }\right)
+      | c = {{Defof|Product Notation (Algebra)|Product}}
+}}
+{{eqn | r = \log_b \left({\prod_{i \mathop = 1}^k s_i}\right) + \log_b s_{k + 1}
+      | c = [[Sum of General Logarithms]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^k \log_b s_i + \log_b s_{k + 1}
+      | c = [[Summation of General Logarithms#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^{k + 1} \log_b s_i
+      | c = {{Defof|Summation}}
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \log_b \left({\prod_{i \mathop = 1}^n s_i}\right) = \sum_{i \mathop = 1}^n \log_b s_i$
+Hence, by definition of $S$:
+:$\displaystyle \log_b \left({\prod_{R \left({i}\right)} a_i}\right) = \sum_{R \left({i}\right)} \log_b a_i$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Closure is Directed in Bounded Below Join Semilattice}
+Tags: Join and Meet Semilattices, Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $x \in S$.
+Then
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+\end{theorem}
+
+\begin{proof}
+By [[Bottom is Way Below Any Element]]:
+:$\bot \ll x$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$\bot \in x^\ll$
+Thus by definition:
+:$x^\ll$ is a [[Definition:Non-Empty Set|non-empty set]].
+Let $y, z \in x^\ll$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \ll x$ and $z \ll x$
+By [[Join is Way Below if Operands are Way Below]]
+:$y \vee z \ll x$
+By definition of [[Definition:Way Below Closure|way below closure]]:
+:$y \vee z \in x^\ll$
+By [[Join Succeeds Operands]]:
+:$y \preceq y \vee z$ and $z \preceq y \vee z$
+Thus by definition
+:$x^\ll$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bottom is Way Below Any Element}
+Tags: Way Below Relation
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Ordered Set|ordered set]].
+Let $x \in S$.
+Then
+:$\bot \ll x$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $S$.
+\end{theorem}
+
+\begin{proof}
+Let $D$ be [[Definition:Directed Subset|directed subset]] of $S$ such that
+:$D$ admits a [[Definition:Supremum of Set|supremum]] and $x \preceq \sup D$
+Because $D$ is [[Definition:Non-Empty Set|non-empty]], therefore
+:$\exists a: a \in D$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\bot \preceq a$
+Thus by definition of [[Definition:Element is Way Below|way below relation]];
+:$\bot \ll x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Real Exponential Function/Proof 2}
+Tags: Equivalence of Definitions of Exponential Function
+
+\begin{theorem}
+The following definitions of the [[Definition:Real Exponential Function|exponential function]] are [[Definition:Logical Equivalence|equivalent]].
+\end{theorem}
+
+\begin{proof}
+From [[Derivative of Exponential Function]] and [[Exponential of Zero]], each definition of $\exp$ satisfies the following:
+:$ (1): \quad D_x \exp = \exp$
+:$ (2): \quad \exp \left({ 0 }\right) = 1$
+on $\R$.
+From [[Exponential Function is Well-Defined/Real/Proof 5 | Exponential Function is Well-Defined]], such a solution is [[Definition:Unique | unique]].
+Thus they all are all [[Definition:Logical Equivalence | equivalent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product to n of Product to Index}
+Tags: Products
+
+\begin{theorem}
+:$\displaystyle \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j = \prod_{i \mathop = 0}^n {a_i}^{n + 2}$
+\end{theorem}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = P_1
+      | o = :=
+      | r = \prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j
+}}
+{{eqn | l = P_2
+      | o = :=
+      | r = \prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = P_1 P_2
+      | r = \left({\prod_{i \mathop = 0}^n \prod_{j \mathop = 0}^i a_i a_j}\right) \left({\prod_{i \mathop = 0}^n \prod_{j \mathop = i}^n a_i a_j}\right)
+      | c = 
+}}
+{{eqn | r = \prod_{i \mathop = 0}^n \left({\left({\prod_{j \mathop = 0}^i a_i a_j}\right) \left({\prod_{j \mathop = i}^n a_i a_j}\right)}\right)
+      | c = 
+}}
+{{eqn | r = \prod_{i \mathop = 0}^n \left({\left({\prod_{j \mathop = 0}^n a_j a_j}\right) {a_i}^2}\right)
+      | c = [[Product of Products over Overlapping Domains]]
+}}
+{{eqn | r = \prod_{i \mathop = 0}^n \left({ {a_i}^{n + 1} \left({\prod_{j \mathop = 0}^n a_j}\right) {a_i}^2}\right)
+      | c = 
+}}
+{{eqn | r = \left({\prod_{i \mathop = 0}^n {a_i}^{n + 1} }\right) \left({\prod_{j \mathop = 0}^n \left({\prod_{j \mathop = 0}^n a_j}\right) {a_i}^2}\right)
+      | c = 
+}}
+{{eqn | r = \left({\prod_{i \mathop = 0}^n {a_i}^{n + 1} }\right) \left({\prod_{j \mathop = 0}^n {a_i}^{n + 1} {a_i}^2}\right)
+      | c = 
+}}
+{{eqn | r = \prod_{i \mathop = 0}^n {a_i}^{2 n + 4}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = P_1
+      | r = \prod_{i \mathop = 0}^n {a_i}^{n + 2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Inequality/Corollary/General Result}
+Tags: Summations, Products, Bernoulli's Inequality
+
+\begin{theorem}
+For all $n \in \Z_{\ge 0}$:
+:$\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$
+where $0 < a_j < 1$ for all $j$.
+\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]]:
+:$\displaystyle \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$
+$P \left({0}\right)$ is the case:
+{{begin-eqn}}
+{{eqn | l = \prod_{j \mathop = 1}^0 \left({1 - a_j}\right)
+      | r = 1
+      | c = Definition of [[Definition:Vacuous Product|Vacuous Product]]
+}}
+{{eqn | o = \ge
+      | r = 1 - 0
+      | c = 
+}}
+{{eqn | r = 1 - \sum_{j \mathop = 1}^0 a^j
+      | c = Definition of [[Definition:Vacuous Summation|Vacuous Summation]]
+}}
+{{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 = \prod_{j \mathop = 1}^1 \left({1 - a_j}\right)
+      | r = 1 - a_j
+      | c = 
+}}
+{{eqn | r = 1 - \sum_{j \mathop = 1}^1 a^j
+      | 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \prod_{j \mathop = 1}^k \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^k a^j$
+from which it is to be shown that:
+:$\displaystyle \prod_{j \mathop = 1}^{k + 1} \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^{k + 1} a^j$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \prod_{j \mathop = 1}^{k + 1} \left({1 - a_j}\right)
+      | r = \left({\prod_{j \mathop = 1}^k \left({1 - a_j}\right)}\right) \left({1 - a_{k + 1} }\right)
+      | c = Definition of [[Definition:Product Notation (Algebra)|Product]]
+}}
+{{eqn | o = \ge
+      | r = \left({1 - \sum_{j \mathop = 1}^k a^j}\right) \left({1 - a_{k + 1} }\right)
+      | c = [[Bernoulli's Inequality/Corollary/General Result#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = 1 - \sum_{j \mathop = 1}^k a^j - a_{k + 1} + a_{k + 1} \sum_{j \mathop = 1}^k a^j
+      | c = 
+}}
+{{eqn | r = 1 - \sum_{j \mathop = 1}^{k + 1} a^j + a_{k + 1} \sum_{j \mathop = 1}^k a^j
+      | c = 
+}}
+{{eqn | o = \ge
+      | r = 1 - \sum_{j \mathop = 1}^{k + 1} a^j
+      | 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:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \prod_{j \mathop = 1}^n \left({1 - a_j}\right) \ge 1 - \sum_{j \mathop = 1}^n a^j$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Sequence of 1 minus Reciprocal of Squares}
+Tags: Products, Product of Sequence of 1 minus Reciprocal of Squares
+
+\begin{theorem}
+For all $n \in \Z_{\ge 1}$:
+:$\displaystyle \prod_{j \mathop = 2}^n \paren {1 - \dfrac 1 {j^2} } = \dfrac {n + 1} {2 n}$
+\end{theorem}
+
+\begin{proof}
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \prod_{j \mathop = 2}^n \paren {1 - \frac 1 {j^2} }
+	| r = \prod_{j \mathop = 2}^n \paren {\frac {\paren {j - 1} \paren {j + 1} } {j^2} }
+	| c = [[Difference of Two Squares]]
+}}
+{{eqn	| r = \paren {\prod_{j \mathop = 2}^n \paren {j - 1} } \paren {\prod_{j \mathop = 2}^n \paren {j + 1} } \paren {\prod_{j \mathop = 2}^n \frac 1 j}^2
+	| c = [[Product of Products]]
+}}
+{{eqn	| r = \paren {\prod_{j \mathop = 1}^{n - 1} j} \paren {\prod_{j \mathop = 3}^{n + 1} j} \frac 1 {\paren {\prod_{j \mathop = 2}^n j}^2}
+}}
+{{eqn	| r = \paren {n - 1}! \paren {\frac 1 2 \prod_{j \mathop = 2}^{n + 1} j} \frac 1 {\paren {n!}^2}
+	| c = {{Defof|Factorial}}
+}}
+{{eqn	| r = \frac {\paren {n - 1}! \paren {n + 1}!} {2 \times n! \times n!}
+	| c = {{Defof|Factorial}}
+}}
+{{eqn	| r = \frac {\paren {n - 1}! n! \paren {n + 1} } {2 \paren {n - 1}! \times n \times n!}
+	| c = [[Gamma Difference Equation]]
+}}
+{{eqn	| r = \frac {n + 1} {2 n}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3}
+Tags: Summation of Products of n Numbers taken m at a time with Repetitions
+
+\begin{theorem}
+:$\displaystyle \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \dfrac { {S_1}^3} 6 + \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
+where:
+:$\displaystyle S_r := \sum_{k \mathop = a}^b {x_k}^r$
+\end{theorem}<|endoftext|>
+\section{Summation of Product of Differences}
+Tags: Summations
+
+\begin{theorem}
+:$\displaystyle \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({u_j - u_k}\right) \left({v_j - v_k}\right) = n \sum_{j \mathop = 1}^n u_j v_j - \sum_{j \mathop = 1}^n u_j \sum_{j \mathop = 1}^n v_j$
+\end{theorem}
+
+\begin{proof}
+Take the [[Binet-Cauchy Identity]]:
+:$\displaystyle \left({\sum_{i \mathop = 1}^n a_i c_i}\right) \left({\sum_{j \mathop = 1}^n b_j d_j}\right) = \left({\sum_{i \mathop = 1}^n a_i d_i}\right) \left({\sum_{j \mathop = 1}^n b_j c_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({a_i b_j - a_j b_i}\right) \left({c_i d_j - c_j d_i}\right)$
+Make the following assignments:
+{{begin-eqn}}
+{{eqn | ll= 1 \le i \le n:
+      | l = a_i
+      | o = :=
+      | r = u_i
+}}
+{{eqn | ll= 1 \le i \le n:
+      | l = c_i
+      | o = :=
+      | r = v_i
+}}
+{{eqn | ll= 1 \le j \le n:
+      | l = b_j
+      | o = :=
+      | r = 1
+}}
+{{eqn | ll= 1 \le j \le n:
+      | l = d_j
+      | o = :=
+      | r = 1
+}}
+{{end-eqn}}
+Then we have:
+:$\displaystyle \left({\sum_{i \mathop = 1}^n u_i v_i}\right) \left({\sum_{j \mathop = 1}^n 1 \times 1}\right) = \left({\sum_{i \mathop = 1}^n u_i \times 1}\right) \left({\sum_{j \mathop = 1}^n 1 \times v_j}\right) + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \left({u_i \times 1 - u_j \times 1}\right) \left({v_i \times 1 - v_j \times 1}\right)$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Summations}
+Tags: Summations, Products
+
+\begin{theorem}
+:$\displaystyle \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j} = \sum_{1 \mathop \le i_1, \mathop \ldots \mathop , i_n \le m} a_{i_1 1} \cdots a_{i_n n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j}
+      | r = \sum_{i \mathop = 1}^m a_{i 1} \sum_{i \mathop = 1}^m a_{i 2} \cdots  \sum_{i \mathop = 1}^m a_{i n}
+      | c = Definition of [[Definition:Product Notation (Algebra)|Product]]
+}}
+{{eqn | r = \sum_{i_1 \mathop = 1}^m a_{i_1 1} \sum_{i_2 \mathop = 1}^m a_{i_2 2} \cdots  \sum_{i_n \mathop = 1}^m a_{i_n n}
+      | c = 
+}}
+{{eqn | r = \sum_{i_1 \mathop = 1}^m \sum_{i_2 \mathop = 1}^m \cdots  \sum_{i_n \mathop = 1}^m a_{i_1 1} a_{i_2 2} \cdots a_{i_n n}
+      | c = [[General Distributivity Theorem]]
+}}
+{{eqn | r = \sum_{1 \mathop \le i_1, \mathop \ldots \mathop , i_n \le m} a_{i_1 1} \cdots a_{i_n n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of Powers over Product of Differences}
+Tags: Summations, Products, Summation of Powers over Product of Differences
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
+\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]]:
+:$\displaystyle S_n := \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
+=== Edge Cases ===
+$\map P 0$ is a [[Definition:Vacuous Summation|vacuous summation]]:
+:$\displaystyle S_0 := \sum_{j \mathop = 1}^0 \paren {\dfrac { {x_j}^0} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 0 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = 0 = \sum_{j \mathop = 1}^0 x_0$
+which is seen to hold.
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^1 \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | r = \dfrac { {x_1}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne 1} } \paren {x_1 - x_k} }
+      | c = 
+}}
+{{eqn | r = \dfrac { {x_1}^r} 1
+      | c = as the [[Definition:Product Notation (Algebra)|product]] is [[Definition:Vacuous Product|vacuous]]
+}}
+{{eqn | r = {x_1}^r
+      | c = simplification
+}}
+{{eqn | r = \begin{cases} 1 & : r = 0 \\ x_1 & : r = 1 \end{cases}
+      | c = 
+}}
+{{end-eqn}}
+This is also seen to hold.
+=== Basis for the Induction ===
+$\map P 2$ is the case where $n = 2$:
+{{begin-eqn}}
+{{eqn | l = S_2
+      | r = \sum_{j \mathop = 1}^2 \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le 2 \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | c = 
+}}
+{{eqn | r = \frac { {x_1}^r} {x_1 - x_2} + \frac { {x_2}^r} {x_2 - x_1}
+      | c = 
+}}
+{{eqn | r = \frac { {x_1}^r - {x_2}^r} {x_1 - x_2}
+      | c = 
+}}
+{{end-eqn}}
+When $0 \le r < n - 1$, it must be that $r = 0$:
+{{begin-eqn}}
+{{eqn | l = S_2
+      | r = \frac {1 - 1} {x_1 - x_2}
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+When $r = n - 1 = 1$:
+{{begin-eqn}}
+{{eqn | l = S_2
+      | r = \frac {x_1 - x_2} {x_1 - x_2}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{end-eqn}}
+When $r = n = 2$:
+{{begin-eqn}}
+{{eqn | l = S_2
+      | r = \frac { {x_1}^2 - {x_2}^2} {x_1 - x_2}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {x_1 + x_2} \paren {x_1 - x_2} } {x_1 - x_2}
+      | c = 
+}}
+{{eqn | r = x_1 + x_2
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^2 x_j
+      | c = 
+}}
+{{end-eqn}}
+Thus in all cases, $\map P 2$ 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]]:
+:$\displaystyle \sum_{j \mathop = 1}^m \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m - 1 \\ 1 & : r = m - 1 \\ \displaystyle \sum_{j \mathop = 1}^m x_j & : r = m \end{cases}$
+from which it is to be shown that:
+:$\displaystyle \sum_{j \mathop = 1}^{m + 1} \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le m + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < m \\ 1 & : r = m \\ \displaystyle \sum_{j \mathop = 1}^{m + 1} x_j & : r = {m + 1} \end{cases}$
+=== Induction Step ===
+This is the [[Definition:Induction Step|induction step]]:
+For $n > 2$, let the formula be rewritten:
+{{begin-eqn}}
+{{eqn | l = S_n
+      | r = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 {x_n - x_{n - 1} } \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_n - x_{n - 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 {x_n - x_{n - 1} } \paren {\sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_{n - 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = S_{n + 1}
+      | r = \frac 1 {x_{n + 1} - x_n} \paren {\sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_{n + 1} } } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } }
+      | c = 
+}}
+{{eqn | r = \frac 1 {x_{n + 1} - x_n} \paren {\sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }
+      | c = as both parts are the original sum
+}}
+{{end-eqn}}
+When $r < n$, both parts are equal to $0$ or $1$ by the [[Summation of Powers over Product of Differences/Proof 1#Induction Hypothesis|induction hypothesis]].
+Thus either:
+:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {1 - 1} = 0$
+or:
+:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {0 - 0} = 0$
+and so $\map P {m + 1}$ holds for $r < n$.
+When $r = n$:
+{{begin-eqn}}
+{{eqn | l = \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } }
+      | r = \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } x_j
+      | c = 
+}}
+{{eqn | l = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | r = \sum_{j \mathop = 1}^n x_j
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {x_{n + 1} - x_n} = 1$
+and so $\map P {m + 1}$ holds for $r = n$.
+Now:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^n \paren {\dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_j - x_k} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | r = 0
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n - \paren {x_1 + \cdots + x_n} {x_j}^{n + 1} + \map P {x_j} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | c = where $\map P {x_j}$ is a [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n - 2$
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }
+      | r = \sum_{l \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_l \paren {\dfrac { {x_j}^{n - 1} } {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }
+      | c = 
+}}
+{{eqn | r = \sum_{l \mathop = 1}^n x_l
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P {m + 1}$ holds for $r = n + 1$.
+So $\map P m \implies \map P {m + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the [[Residue Theorem]]:
+:$\displaystyle \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \left({x_j - x_k}\right)} } \end{pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\left\lvert{z}\right\rvert \mathop = R} \dfrac {z^r \rd z} {\left({z - z_1}\right) \cdots \left({z - z_n}\right)}$
+where $R > \left\lvert{z_1}\right\rvert, \ldots, \left\lvert{z_n}\right\rvert$.
+The [[Definition:Laurent Series|Laurent series]] of the [[Definition:Integrand|integrand]] [[Definition:Uniform Convergence|converges uniformly]] on $\left\lvert{z}\right\rvert = R$:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = z^{r - n} \left({\dfrac 1 {1 - x_1 / z} }\right) \cdots \left({\dfrac 1 {1 - x_n / z} }\right)
+      | c = 
+}}
+{{eqn | r = z^{r - n} + \left({x_1 + \cdots x_n}\right) z^{r - n - 1} + \left({x_1^2 + x_1 x_2 + \cdots}\right) z^{r - n - 2} + \cdots
+      | c = 
+}}
+{{end-eqn}}
+On [[Definition:Integration|integrating]] term my term, everything vanishes except the coefficient of $z^{-1}$.
+Thus:
+:$\displaystyle \sum_{\substack {j_1 \mathop + \mathop \cdots \mathop + j_n \mathop = r \mathop - n \mathop + 1 \\ j_1, \mathop \ldots j_n \mathop \ge 0} } {x_1}^{j_1} {x_2}^{j_2} \cdots {x_n}^{j_n}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Definition ===
+{{begin-eqn}}
+{{eqn | l = \map p x 
+      | r = \prod_{k \mathop = 1}^n \paren {x - x_k}
+}}
+{{eqn | l = \map {p_m} x
+      | r = \prod_{k \mathop = 1, \, k \mathop \ne m}^n \paren {x - x_k},\quad 1 \le m \le n
+}}
+{{eqn | l = A 
+      | r = \begin{pmatrix}
+  1      & x_1    & \cdots & x_1^{n - 2} & x_1^{n - 1} \\
+  \vdots & \vdots & \cdots &   \vdots    & \vdots \\
+  1      & x_n    & \cdots & x_n^{n - 2} & x_n^{n - 1} \\
+  \end{pmatrix}
+      | c = Set $\set {x_1, \ldots, x_n}$ assumed to have distinct elements.
+}}
+{{eqn | l = A_{\vec Y}
+      | r = \begin{pmatrix}
+  1      & x_1    & \cdots & x_1^{n - 2} & y_1 \\
+  \vdots & \vdots & \cdots &   \vdots    & \vdots \\
+  1      & x_n    & \cdots & x_n^{n  -2} & y_n \\
+\end{pmatrix}
+      | c = Vector $\vec Y$ has entries $y_1,\ldots,y_n$.
+}}
+{{end-eqn}}
+Symbol $\map {\mathbf {Cof } } {M, i, j}$ denotes [[Definition:Cofactor of Element|cofactor]] $M_{i j}$ of matrix $M$
+=== Lemma 1 ===
+{{begin-eqn}}
+{{eqn | l = \map \det A
+      | r = \map {p_m} {x_m} \map {\mathbf {Cof} } {A, m, n}
+}}
+{{end-eqn}}
+'''Proof of Lemma 1'''
+By [[Effect of Elementary Row Operations on Determinant]], the determinant of $A$ is unchanged by adding a linear combination of the first $n - 1$ columns to the last column.
+Let $\map f x$ be the monic polynomial $\map {p_m} x = x^{n - 1} + $ lower power terms.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \det {\begin{pmatrix}
+  1      & x_1    & \cdots & x_1^{n - 2} & x_1^{n - 1} \\
+  \vdots & \vdots & \cdots &   \vdots    & \vdots \\
+  1      & x_n    & \cdots & x_n^{n - 2} & x_n^{n - 1} \\
+  \end{pmatrix} }
+      | r = \map \det {\begin{pmatrix}
+  1      & x_1    & \cdots & x_1^{n - 2} & \map f {x_1} \\
+  \vdots & \vdots & \cdots &   \vdots    & \vdots \\
+  1      & x_n    & \cdots & x_n^{n - 2} & \map f {x_n} \\
+ \end{pmatrix} }
+}}
+{{end-eqn}}
+The last column on the right has all components zero except $m$th entry $\map {p_m} {x_m}$.
+Apply [[Expansion Theorem for Determinants|cofactor expansion]] along column $n$.
+{{qed|lemma}}
+=== Lemma 2 ===
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \map \det A 
+      | r = \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}
+}}   
+{{eqn | n = 2
+      | l = \map {\mathbf {Cof} } {A, m, n}
+      | r = \paren {-1}^{m + n} \prod_{1 \mathop \le i \mathop < j \mathop \le n, \, i \mathop \ne m, \, j \mathop \ne m} \paren {x_j - x_i}  
+}}   
+{{end-eqn}}
+'''Proof of Lemma 2''':
+[[Vandermonde Determinant]] establishes $(1)$.
+To prove (2), let:
+:$\set {y_1, \ldots, y_{n - 1} } = \set {x_1, \ldots, x_n} \setminus \set {x_m}$
+then apply $(1)$ with $\set {x_1, \ldots, x_n}$ replaced by $\set {y_1, \ldots, y_{n - 1} }$.
+{{qed|lemma}}
+'''Theorem Details''':
+The Lemmas change the Theorem's equation to:
+{{begin-eqn}}
+{{eqn | n = 3
+      | l = \sum_{m \mathop = 1}^n {x_m^r} \, \dfrac {\map {\mathbf {Cof} } {A, m, n} } {\map \det A}
+      | r = \begin{cases}
+0 & : 0 \mathop \le r \mathop < n - 1 \\
+1 & : r = n - 1 \\
+\sum_{j \mathop = 1}^n x_j & : r = n \\
+    \end{cases}
+}}
+{{end-eqn}}
+[[Expansion Theorem for Determinants|Cofactor expansion]] changes identity $(3)$ to:
+{{begin-eqn}}
+{{eqn | n = 4
+      | l = \det \begin{pmatrix}
+1      & x_1    & \cdots & x_1^{n - 2} & x_1^r  \\
+\vdots & \vdots & \cdots &   \vdots    & \vdots \\
+1      & x_n    & \cdots & x_n^{n - 2} & x_n^r  \\
+    \end{pmatrix}
+      | r = \map \det A \begin{cases}
+0 & : 0 \mathop \le r \mathop < n - 1 \\
+1 & : r = n - 1 \\
+\sum_{j \mathop = 1}^n x_j & : r = n \\
+    \end{cases}
+}}
+{{end-eqn}}
+The proof is divided into three cases.
+'''Case 1''': $0 \mathop \le r \mathop \le n - 2$:
+The determinant on the left in $(4)$ is zero by [[Square Matrix with Duplicate Rows has Zero Determinant]].
+'''Case 2''': $r = n - 1$:
+The determinant on the left in $(4)$ is $\map \det A$.
+'''Case 3''': $r = n$:
+Expand $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$ by [[Taylor's Theorem]] with remainder $R$ of degree $n - 2$:
+:$\map p x = x^n - \paren {x_1 + \cdots + x_n} x^{n - 1} + \map R x$
+Let $x = x_k$ for $1 \mathop \le k \mathop \le n$.
+Then $x$ is a root of $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \map p {x_k}
+}}
+{{eqn | r = x_k^n - \paren {x_1 + \cdots + x_n} x_k^{n - 1} + \map R {x_k}
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = c
+      | r = x_1 + \cdots + x_n
+}}
+{{eqn | l = \vec u
+      | r = \begin {pmatrix} x_1^n \\ \vdots \\ x_n^n \end{pmatrix}
+      | c = Column $n$ of the left side of $(4)$
+}}
+{{eqn | l = \vec z
+      | r = c \begin {pmatrix} x_1^{n - 1} \\ \vdots \\ x_n^{n - 1} \end{pmatrix}
+      | c = Constant $c$ times column $n$ of $A$
+}}
+{{eqn | l = \vec w
+      | r = \begin{pmatrix} -\map R {x_1} \\ \vdots \\ -\map R {x_n} \end{pmatrix}
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | n = 5
+      | l = \vec u
+      | r = \vec z + \vec w
+      | c = by equation $x_k^n = c x_k^{n - 1} - \map R {x_k}$
+}}
+{{eqn | n = 6
+      | l = \map \det {A_{\vec z} }
+      | r = c \map \det A
+      | c = [[Effect of Elementary Row Operations on Determinant]]: factoring $c$ from last column of $A_{\vec z}$
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | n = 7
+      | l = \det \paren { A_{\vec w} } 
+      | r = 0
+      | c = [[Effect of Elementary Row Operations on Determinant]]: $\vec w$ is a linear combination of columns $1$ to $n - 1$
+}}
+{{end-eqn}}
+The left side of $(4)$:
+{{begin-eqn}}
+{{eqn | l = \map \det {A_{\vec u} } 
+      | r = \map \det {A_{\vec z} } + \map \det {A_{\vec w} }
+      | c = by $(5)$ and [[Determinant as Sum of Determinants]]
+}}
+{{eqn | r = c \map \det A + 0
+      | c = by $(6)$ and $(7)$
+}}
+{{eqn | r = \map \det A \displaystyle \sum_{i \mathop = 1}^n x_i
+      | c = definition of $c$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation of Powers over Product of Differences/Example}
+Tags: Summations, Products
+
+\begin{theorem}
+{{begin-eqn}}
+{{eqn | l = \frac 1 {\paren {a - b} \paren {a - c} } + \frac 1 {\paren {b - a} \paren {b - c} } + \frac 1 {\paren {c - a} \paren {c - b} }
+      | r = 0
+}}
+{{eqn | l = \frac a {\paren {a - b} \paren {a - c} } + \frac b {\paren {b - a} \paren {b - c} } + \frac c {\paren {c - a} \paren {c - b} }
+      | r = 0
+}}
+{{eqn | l = \frac {a^2} {\paren {a - b} \paren {a - c} } + \frac {b^2} {\paren {b - a} \paren {b - c} } + \frac {c^2} {\paren {c - a} \paren {c - b} }
+      | r = 1
+}}
+{{eqn | l = \frac {a^3} {\paren {a - b} \paren {a - c} } + \frac {b^3} {\paren {b - a} \paren {b - c} } + \frac {c^3} {\paren {c - a} \paren {c - b} }
+      | r = a + b + c
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Instance of [[Summation of Powers over Product of Differences]]:
+:$\displaystyle \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases}0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$
+where:
+:$n = 3$
+:$x_1 = a$
+:$x_2 = b$
+:$x_3 = c$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r}
+Tags: Summations, Products
+
+\begin{theorem}
+:$\displaystyle \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) = 1$
+where $1 \le m \le n$ and $x$ is arbitrary.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)
+      | r = \sum_{k \mathop = 1}^n \left({\dfrac {k^{n - 1} + P \left({k}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)
+      | c = where $P \left({k}\right)$ is a [[Definition:Polynomial over Real Numbers|polynomial]] of [[Definition:Degree of Polynomial|degree]] $n - 2$
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \left({\dfrac {k^{n - 1} } {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) + \sum_{k \mathop = 1}^n \left({\dfrac {P \left({k}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right)
+      | c = 
+}}
+{{eqn | r = 1 + 0
+      | c = [[Summation of Powers over Product of Differences]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r/Example}
+Tags: Summations, Products
+
+\begin{theorem}
+:$\dfrac {x \left({x - 2}\right) \left({x - 3}\right)} {\left({-1}\right) \left({-2}\right) \left({-3}\right)} + \dfrac {\left({x + 1}\right) \left({x - 1}\right) \left({x - 2}\right)} {\left({1}\right) \left({-1}\right) \left({-2}\right)} + \dfrac {\left({x + 2}\right) x \left({x - 1}\right)} {\left({2}\right) \left({1}\right) \left({-1}\right)} + \dfrac {\left({x + 3}\right) \left({x + 1}\right) x} {\left({3}\right) \left({2}\right) \left({1}\right)} = 1$
+\end{theorem}
+
+\begin{proof}
+This is an example of [[Summation by k of Product by r of x plus k minus r over Product by r less k of k minus r]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \left({\dfrac {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne m} } \left({x + k - r}\right)} {\displaystyle \prod_{\substack {1 \mathop \le r \mathop \le n \\ r \mathop \ne k} } \left({k - r}\right)} }\right) = 1$
+where $n = 4$ and $m = 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Cauchy Matrix}
+Tags: Cauchy Matrix
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Square Matrix|square]] [[Definition:Cauchy Matrix|Cauchy matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
+:$C_n = \begin{bmatrix}
+\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2} & \cdots & \dfrac 1 {x_1 + y_n} \\
+\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2} & \cdots & \dfrac 1 {x_2 + y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2} & \cdots & \dfrac 1 {x_n + y_n} \\
+\end{bmatrix}$
+Then its [[Definition:Inverse Matrix|inverse]] $C_n^{-1} = \sqbrk b_n$ can be specified as:
+:$\begin{bmatrix} b_{ij} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\displaystyle \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - y_k} } } \end{bmatrix}$
+\end{theorem}
+
+\begin{proof}
+'''Preliminaries''':
+[[Vandermonde Matrix Identity for Cauchy Matrix]] supplies matrix equation
+:$\displaystyle (1)\quad - C = PV_x^{-1} V_y Q^{-1}$
+:Definitions of symbols:
+::$\displaystyle V_x=\paren {\begin{smallmatrix}
+1         & 1         & \cdots & 1 \\
+x_1       & x_2       & \cdots & x_n \\
+\vdots    & \vdots    & \ddots & \vdots \\
+x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\
+\end{smallmatrix} },\quad 
+V_y=\paren {\begin{smallmatrix}
+1         & 1         & \cdots & 1 \\
+y_1       & y_2       & \cdots & y_n \\
+\vdots    & \vdots    & \ddots & \vdots \\
+y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\
+\end{smallmatrix} }$ [[Definition:Vandermonde Matrix|Vandermonde matrices]]
+::$\displaystyle P= \paren {\begin{smallmatrix}
+p_1(x_1) &  \cdots & 0 \\
+\vdots   & \ddots  & \vdots \\
+0        & \cdots  & p_n(x_n) \\
+\end{smallmatrix} }, \quad
+Q= \paren {\begin{smallmatrix}
+p(y_1)  & \cdots  & 0 \\
+\vdots  & \ddots  & \vdots \\
+0       & \cdots  & p(y_n) \\
+\end{smallmatrix} }$  [[Definition:Diagonal Matrix|Diagonal matrices]]
+::$\displaystyle 
+p(x) = \prod_{i \mathop = 1}^n \paren {x - x_i},
+\quad
+\displaystyle 
+p_k(x) = \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i},
+\quad
+1 \mathop \le k \mathop \le n$ [[Definition:Polynomial/Complex Numbers|Polynomials]]
+'''Compute the inverse''' $C^{-1}$ for:
+{{begin-eqn}}
+{{eqn | l = C
+      | r = \begin{bmatrix}
+            \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2} & \cdots & \dfrac 1 {x_1 - y_n} \\
+            \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2} & \cdots & \dfrac 1 {x_2 - y_n} \\
+            \vdots               & \vdots               & \ddots & \vdots \\
+            \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2} & \cdots & \dfrac 1 {x_n - y_n} \\
+            \end{bmatrix}
+      | c = Assume $\set {x_1,\ldots,x_n,y_1,\ldots,y_n}$ has distinct elements.
+}}
+{{end-eqn}}
+Replacement $y_k \to -y_k$ then gives the inverse $C_n^{-1}$ in the theorem.
+[[Inverse of Matrix Product]] applied to equation (1) gives: 
+:$ C^{-1} = -Q V_y^{-1} V_x P^{-1}$ 
+Let ${\vec K}_1,\ldots,{\vec K}_n$ denote the columns of the $n\times n$ identity matrix.
+Then $n\times n$ matrix $B = C^{-1}$ has entries $b_{ij} = {\vec K}_i^T C^{-1} {\vec K}_j$.
+Hold fixed until the end of the proof the row and column index symbols $i$ and $j$.
+Define column vectors $\vec A$, $\vec B$ so that $b_{ij} = {\vec A}^T \vec B$:
+:$\displaystyle 
+   \vec A = \paren { {\vec K}_i^T \, Q \, V_y^{-1} }^T,
+   \quad  
+   \vec B = -V_x \, P^{-1} \, {\vec K}_j
+$
+ 
+Define $u = x_j$ and simplify:
+{{begin-eqn}}
+{{eqn | l = \vec A 
+      | r = {\map p {y_i} } \, \paren { V_y^{-1} }^T \, {\vec K}_i 
+      | c = [[Transpose of Matrix Product]]
+}}
+{{eqn | r = \dfrac{ \map p {y_i} }{\det \paren {V_y} } \paren { \adj {V_y} }^T {\vec K}_i
+      | c = [[Matrix Product with Adjugate Matrix]]
+}}
+{{eqn | r = \dfrac{ \map p {y_i} }{\det \paren {V_y} }
+        \begin{bmatrix} 
+          {\mathbf {Cofactor} } \paren {V_y,1,i } \\ 
+          \vdots \\ 
+          {\mathbf {Cofactor} } \paren {V_y,n,i } \\ 
+        \end{bmatrix}
+      | c = ${\mathbf {Cofactor} } \paren {M,r,s}$ denotes [[Definition:Cofactor|cofactor]] $M_{rs}$
+}}
+{{eqn | l = \vec B
+      | r = -\dfrac{1}{\map {p_j} {x_j} } \, V_x \, {\vec K}_j
+}}
+{{eqn | r = -\dfrac{1}{\map {p_j} {x_j} } \, 
+        \begin{bmatrix} 
+          1 \\ 
+          u \\ 
+          \vdots \\ 
+          u^{n-1} \\ 
+        \end{bmatrix}
+      | c = Symbol $u = x_j$.
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = A^T\, B
+      | r = \dfrac
+            { -\map p {y_i}  }
+            {\map {p_j} {x_j} \det \paren {V_y} } 
+             \sum_{k \mathop = 1}^n {\mathbf {Cofactor} } \paren {V_y,k,i } \, u^{k-1}
+      | c = A [[Expansion Theorem for Determinants|Cofactor expansion]] 
+along column $i$.
+}}
+{{eqn | r = \dfrac
+            { -\map p {y_i}  }
+            {\map {p_j} {x_j} \det \paren {V_y} } 
+        \det \paren { \begin{matrix} 
+        1         & \cdots & 1             & 1         & 1             & \cdots & 1 \\
+        y_1       & \cdots & y_{i-1}       & u         & y_{i+1}       & \cdots & y_n \\
+        \vdots    & \cdots & \vdots        &\vdots     & \vdots        & \cdots & \vdots \\
+        y_1^{n-1} & \cdots & y_{i-1}^{n-1} & u^{n-1}   & y_{i+1}^{n-1} & \cdots & y_n^{n-1} \\
+        \end{matrix}  } 
+}}
+{{eqn | r = \dfrac
+            { -\map p {y_i}  }
+            {\map {p_j} {x_j} }
+            \,\, 
+            \dfrac
+            {
+              \left.
+                \paren { \det \paren {V_y} }
+              \right \vert_{y_i \mathop = u}
+            }
+            { \det \paren {V_y} }
+}}
+{{end-eqn}}
+Simplify the fraction on the right:
+{{begin-eqn}}
+{{eqn | l = (2)\quad \displaystyle
+            \dfrac
+            { \displaystyle
+              \left. 
+                \paren { \det \paren {V_y} } 
+              \right\vert_{y_i \mathop = u} 
+            }
+            { \displaystyle \det \paren {V_y} }
+      | r = \displaystyle
+            \dfrac
+            {\displaystyle
+              \left. 
+                \paren { \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren { y_k - y_m  } } 
+              \right\vert_{ y_i \mathop = u }
+            }
+            {\displaystyle
+              \prod_{1 \mathop \le m \mathop \lt k \mathop \le n} \paren { y_k - y_m  }
+            }
+      | c = [[Vandermonde Determinant]]
+}}
+{{end-eqn}}
+Define sets $D,A,B,C$ with disjoint decomposition $D = A \cup B \cup C$:
+{{begin-eqn}}
+{{eqn | l = D 
+      | r = \set { \paren { m,k } : 1 \le m \lt k \le n }
+}}
+{{eqn | l = A 
+      | r = \set { \paren { m,k } \in D : m \neq i \mbox{ and } k \neq i }
+}}
+{{eqn | l = B
+      | r = \set { \paren { m,k } \in D : m = i }
+}}
+{{eqn | l = C
+      | r = \set { \paren { m,k } \in D : k = i }
+}}
+{{end-eqn}}
+Use $\prod_{D} = \prod_{A} \prod_{B} \prod_{C}$ to convert the numerator and denominator in
+the right side of (2).
+Then:
+{{begin-eqn}}
+{{eqn | l = \displaystyle
+            \dfrac
+            { \displaystyle
+              \left. 
+                \paren { \det \paren {V_y} } 
+              \right\vert_{y_i \mathop = u} 
+            }
+            { \displaystyle \det \paren {V_y} }
+      | r = \displaystyle
+            \dfrac
+            {\displaystyle 
+              \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { u - y_k } 
+            }
+            {\displaystyle 
+              \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { y_i - y_k }  
+            }
+      | c = Common factors canceled in (2). 
+}}
+{{end-eqn}}
+Replacement of ${ \map p {y_i}  }$ and ${ \map {p_j} {x_j} }$ by products gives:
+{{begin-eqn}}
+{{eqn | l = (3)\quad \displaystyle
+            b_{ij} 
+      | r = 
+            \paren {-1}
+            \,
+            \dfrac
+            { \displaystyle \prod_{k \mathop = 1}^n \paren {y_i - x_k } }
+            { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq j}^n \paren { x_j - x_k } } 
+            \,\,
+            \prod_{k \mathop = 1,\, k \mathop \neq i}^n 
+            \paren { 
+              \dfrac { x_j - y_k }{ y_i - y_k }
+            }
+      | c = Replaced $u = x_j$.
+}}
+{{end-eqn}}
+ 
+After replacement $y_k \to -y_k$ and
+canceling a common factor, ''Knuth's original identity'' (1997) becomes:
+{{begin-eqn}}
+{{eqn | l = \displaystyle (4)\quad b_{ij}
+      | r = \displaystyle
+  \dfrac
+  { \displaystyle \prod_{k \mathop = 1}^n \paren { x_k - y_i } }
+  { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq j}^n \paren { x_j - x_k } }
+  \dfrac
+  { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { x_j - y_k } }
+  { \displaystyle \prod_{k \mathop = 1,\, k \mathop \neq i}^n \paren { - y_i + y_k } }
+}}
+{{end-eqn}}
+In (3), factor $(-1)^{n+1}$ from the numerator and $(-1)^{n-1}$ from the denominator.
+Then $(-1)^{n+1} = (-1)^{n-1}$ verifies that (3) matches (4).{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Elements in Inverse of Combinatorial Matrix}
+Tags: Combinatorial Matrix
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Combinatorial Matrix|combinatorial matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
+:$C_n = \begin{bmatrix}
+x + y & y & \cdots & y \\
+y & x + y & \cdots & y \\
+\vdots & \vdots & \ddots & \vdots \\
+y & y & \cdots & x + y
+\end{bmatrix}$
+Let $C_n^{-1}$ be its [[Definition:Inverse Matrix|inverse]], from [[Inverse of Combinatorial Matrix]]:
+:$b_{i j} = \dfrac {-y + \delta_{i j} \left({x + n y}\right)} {x \left({x + n y}\right)}$
+where $\delta_{i j}$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+The sum of all the [[Definition:Element of Matrix|elements]] of $C_n^{-1}$ is:
+:$\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \dfrac n {x + n y}$
+\end{theorem}
+
+\begin{proof}
+All $n^2$ [[Definition:Element of Matrix|elements]] of $C_n^{-1}$ have a term $\dfrac {-y} {x \left({x + n y}\right)}$.
+Further to this, the $n$ elements on the [[Definition:Main Diagonal|main diagonal]] contribute an extra $\dfrac {x + n y} {x \left({x + n y}\right)}$ to the total.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
+      | r = n^2 \dfrac {-y} {x \left({x + n y}\right)} + n \dfrac {x + n y} {x \left({x + n y}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac {- n^2 y + n \left({x + n y}\right)} {x \left({x + n y}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac {- n^2 y + n x + n^2 y} {x \left({x + n y}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac {n x} {x \left({x + n y}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac n {x + n y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Elements in Inverse of Vandermonde Matrix}
+Tags: Vandermonde Matrices
+
+\begin{theorem}
+Let $V_n$ be the [[Definition:Vandermonde Matrix|Vandermonde matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
+:$V_n = \begin{bmatrix}
+  x_1 & x_2 & \cdots & x_n \\
+  x_1^2 & x_2^2 & \cdots & x_n^2 \\
+\vdots & \vdots & \ddots & \vdots \\
+  x_1^n & x_2^n & \cdots & x_n^n
+\end{bmatrix}$
+Let $V_n^{-1}$ be its [[Definition:Inverse Matrix|inverse]], from [[Inverse of Vandermonde Matrix]]:
+:$b_{i j} = \begin{cases}
+\paren {-1}^{j - 1} \paren {\dfrac{\displaystyle \sum_{\substack {1 \mathop \le m_1 \mathop < \ldots \mathop < m_{n - j} \mathop \le n \\ m_1, \ldots, m_{n - j} \mathop \ne i} } x_{m_1} \cdots x_{m_{n - j} } } {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_m - x_i} } } & : 1 \le  j < n \\
+\qquad \qquad \qquad \dfrac 1 {x_i \displaystyle \prod_{\substack {1 \mathop \le m \mathop \le n \\ m \mathop \ne i} } \paren {x_i - x_m} } & : j = n
+\end{cases}$
+The sum of all the [[Definition:Element of Matrix|elements]] of $V_n^{-1}$ is:
+:$\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = 1 - \prod_{k \mathop = 1}^n \paren {1 - \dfrac 1 {x_k} }$
+\end{theorem}
+
+\begin{proof}
+[[Sum of Elements of Invertible Matrix]] implies the sum of elements in $V_n^{-1}$ is:
+:$ 1 - \det \paren { V_n^{-1} } \det \paren { V_n - J_n } $.
+The plan is to expand $\det \paren { V_n - J_n } $ and simplify.
+The method is efficiently communicated in the $3\times 3$ case:
+{{begin-eqn}}
+{{eqn
+  | l = \det \paren { V_3 - J_3 } =
+  | r = \det \paren { \begin{smallmatrix}
+                    x_1 - 1 & x_2 - 1 & x_3 - 1 \\
+                    x_1^2 - 1 & x_2^2 - 1 & x_3^2 - 1 \\
+                    x_1^3 - 1 & x_2^3 - 1 & x_3^3 - 1 \\
+                    \end{smallmatrix} }
+  | c = $J_3$ is the $3\times 3$ ones matrix.
+}}
+{{eqn
+  | l =
+  | r =  \paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1}
+         \det \paren { \begin{smallmatrix}
+                     1 & 1 & 1 \\
+                     x_1+1 & x_2+1 & x_3+1 \\
+                     x_1^2+x+1+1 & x_2^2+x_2+1 & x_3^2+x_3+1 \\
+                     \end{smallmatrix} } 
+  | c =
+Use $a^n - b^n$ factorization.
+[[Effect of Elementary Row Operations on Determinant]] collects common column factors outside the determinant.
+}}
+{{eqn
+  | l =
+  | r =\paren {x_1 - 1} \paren {x_2 - 1} \paren {x_3 - 1}
+       \det \paren { \begin{smallmatrix}
+       1 & 1 & 1 \\
+       x_1 & x_2 & x_3 \\
+       x_1^2 & x_2^2 & x_3^2 \\
+       \end{smallmatrix} } 
+  | c = [[Effect of Elementary Row Operations on Determinant]] simplifies the determinant.
+}}
+{{end-eqn}}
+The proof is completed by '''induction''' using the lemmas below.
+=== Lemma 1 ===
+Let:
+:$\displaystyle A_n = \det \paren { \begin{smallmatrix}
+1 & \cdots & 1 \\
+x_1 & \cdots & x_n \\
+\vdots & \cdots & \vdots \\
+x_1^{n-1} & \cdots & x_n^{n-1} \\
+\end{smallmatrix} }$ 
+Then:
+:$\displaystyle \det \paren { V_n } =  \paren { \prod_{k \mathop = 1}^n x_k } \det \paren { A_n }$  [[Effect of Elementary Row Operations on Determinant]]
+=== Lemma 2 ===
+:$\displaystyle \det(V_n - J_n) = \prod_{k \mathop = 1}^n \paren { x_k - 1 } 
+\det \paren { A_n }$
+=== Lemma 3 ===
+:$\displaystyle \det \paren { V_n^{-1} } \det(V_n - J_n) = 
+\dfrac
+{ \prod_{k \mathop = 1}^n \paren { x_k - 1 }   }
+{ \prod_{k \mathop = 1}^n { x_k }  }$  by Lemmas 1 and 2.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Elements in Inverse of Cauchy Matrix}
+Tags: Summations, Products, Summation of Powers over Product of Differences, Cauchy Matrix
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Cauchy Matrix|Cauchy matrix]] of [[Definition:Order of Square Matrix|order $n$]] given by:
+:$C_n = \begin{bmatrix}
+\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\
+\dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\
+\vdots & \vdots & \ddots & \vdots \\
+\dfrac 1 {x_m + y_1} & \dfrac 1 {x_m + y_2 } & \cdots & \dfrac 1 {x_m + y_n} \\
+\end{bmatrix}$
+Let $C_n^{-1}$ be its [[Definition:Inverse Matrix|inverse]], from [[Inverse of Cauchy Matrix]]:
+:$b_{i j} = \dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\displaystyle \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } }$
+The sum of all the [[Definition:Element of Matrix|elements]] of $C_n^{-1}$ is:
+:$\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$
+\end{theorem}
+
+\begin{proof}
+It suffices to prove the Theorem for Cauchy matrix:
+{{begin-eqn}}                        
+{{eqn | l = C
+      | r = \begin{pmatrix}
+            \dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\
+            \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n} \\
+            \vdots               & \vdots                & \ddots & \vdots \\
+            \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2 } & \cdots & \dfrac 1 {x_n - y_n} \\
+            \end{pmatrix}
+      | c = Distinct values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ required.
+}}
+{{end-eqn}}
+The sum $S$ of elements in $C^{-1}$ will be shown to be a scalar product of two vectors $\vec A$ and $\vec B$.
+The statement of the Theorem is obtained by replacing $y_k \to -y_k$ in $C$ and in the sum $S$.
+Let:
+{{begin-eqn}}
+{{eqn | l = \vec \bsone
+      | r = \paren { \begin{smallmatrix} 1 \\ \vdots \\ 1 \end{smallmatrix}  }
+      | c = Vector of all ones.
+}}
+{{end-eqn}}
+The sum $S$ of elements in $C^{-1}$ is the matrix product:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l =  S
+      | r =  { \vec \bsone }^T C^{-1} {\vec \bsone }
+}}
+{{end-eqn}}
+To identify vectors $\vec A$ and $\vec B$, the tool is:
+{{begin-eqn}}
+{{eqn | l =  C 
+      | r = -PV_x^{-1} V_y Q^{-1}
+      | c = [[Vandermonde Matrix Identity for Cauchy Matrix]]
+}}
+{{end-eqn}}
+Definitions of symbols:
+{{begin-eqn}}
+{{eqn | l = V_x
+      | r = 
+            \paren {\begin{smallmatrix}
+            1         & 1         & \cdots & 1 \\
+            x_1       & x_2       & \cdots & x_n \\
+            \vdots    & \vdots    & \ddots & \vdots \\
+            x_1^{n-1} & x_2^{n-1} & \cdots & x_n^{n-1} \\
+            \end{smallmatrix} },
+            \quad 
+            V_y = 
+            \paren {\begin{smallmatrix}
+            1         & 1         & \cdots & 1 \\
+            y_1       & y_2       & \cdots & y_n \\
+            \vdots    & \vdots    & \ddots & \vdots \\
+            y_1^{n-1} & y_2^{n-1} & \cdots & y_n^{n-1} \\
+            \end{smallmatrix} }
+      | c = [[Definition:Vandermonde Matrix|Vandermonde matrices]]
+}}
+{{eqn | l = \map p {x}
+      | r = 
+            \prod_{i \mathop = 1}^n \paren {x - x_i},
+            \quad
+            \map {p_k} {x} = 
+            \prod_{i \mathop = 1,i \mathop \ne k}^n \, \paren {x - x_i}
+            ,\quad 
+            1 \mathop \le k \mathop \le n 
+      | c = [[Definition:Polynomial/Complex Numbers|Polynomials]]  
+}}
+{{eqn | l = P
+      | r = \paren {\begin{smallmatrix}
+                    \map {p_1} {x_1} &  \cdots & 0 \\
+                    \vdots           & \ddots  & \vdots \\
+                    0                & \cdots  & \map {p_n} {x_n} \\
+                    \end{smallmatrix} }, 
+                    \quad
+            Q = 
+            \paren {\begin{smallmatrix}
+                    \map p {y_1}  & \cdots  & 0 \\
+                    \vdots        & \ddots  & \vdots \\
+                    0             & \cdots  & \map p {y_n} \\
+                    \end{smallmatrix} }
+      | c = [[Definition:Diagonal Matrix|Diagonal matrices]]
+}}
+{{end-eqn}}
+'''Compute the sum''' $S$:
+{{begin-eqn}}
+{{eqn | l = C^{-1}
+      | r = - Q V_y^{-1} V_x P^{-1}
+      | c = [[Inverse of Matrix Product]]
+}}
+{{eqn | l = S 
+      | r = { \vec {\bsone} }^T 
+            C^{-1} 
+            \vec {\bsone}
+      | c = Equation (1). 
+}}
+{{eqn | r = \paren { 
+              - { \vec {\bsone} }^T 
+              Q 
+              V_y^{-1}
+            } 
+            \,\, 
+            \paren { 
+              V_x 
+              P^{-1} 
+              \vec {\bsone} 
+            }
+}}
+{{end-eqn}}
+Define vectors $\vec A$ and $\vec B$:
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = {\vec A }^T
+      | r = - { \vec {\bsone} }^T Q V_y^{-1}
+}}
+{{eqn | n = 3
+      | l =  \vec B
+      | r  = V_x P^{-1} {\vec {\bsone} }
+      | c = Sum $S$ equals scalar product ${ \vec A }^T \vec B$.
+}}
+{{end-eqn}}
+'''Compute''' $\vec A$:
+{{begin-eqn}}
+{{eqn | l = {\vec A }
+      | r = -\paren { { \vec \bsone }^T Q V_y^{-1}  }^T
+      | c = Equation (2).
+}}
+{{eqn | r = -\paren { V_y^T }^{-1} Q^T { \vec \bsone }
+      | c = [[Definition:Inverse Matrix]] and  [[Transpose of Matrix Product ]]
+}}
+{{end-eqn}}
+Then $\vec A$ '''uniquely''' solves the problem 
+{{begin-eqn}}
+{{eqn | l = \begin{pmatrix}  
+            1 & \cdots & y_1^{n-1} \\
+             \vdots & \cdots & \vdots \\
+            1 & \cdots & y_n^{n-1} \\ 
+            \end{pmatrix} 
+            \vec A 
+      |r = -\begin{pmatrix}
+            \map p {y_1} \\
+            \vdots \\
+            \map p {y_n} \\
+            \end{pmatrix}
+      | c = Coefficient matrix $V_y^T$.
+}}
+{{end-eqn}}
+The two polynomials
+{{begin-eqn}}
+{{eqn | l = \map p {x} 
+      | r = \prod_{j \mathop = 1}^n \paren { x - x_j } 
+}}
+{{eqn | r = x^n + \sum_{i \mathop = 1}^{n} a_i x^{i-1} 
+}}
+{{eqn | l = \map q {y} 
+      | r = \prod_{j \mathop = 1}^n \paren { y - y_j }
+}}
+{{eqn | r = y^n + \sum_{i \mathop = 1}^{n} b_i y^{i-1}
+}}
+{{end-eqn}}
+have coefficients expressed by [[Definition:Elementary Symmetric Function|symmetric functions]]:
+{{begin-eqn}}
+{{eqn | l = a_{i}
+      | r = \paren {-1}^{n+1-i} e_{n+1-i} \paren { \set {x_1,\ldots,x_n} }
+      | c = [[Viète's Formulas]]
+}}
+{{eqn | l = b_{i}
+      | r = \paren {-1}^{n+1-i} e_{n+1-i} \paren {\set {y_1,\ldots y_n} }
+      | c = [[Viète's Formulas]]
+}}
+{{end-eqn}}
+Solve in equation $\map q {y_j} = 0$ for highest power $y_j^n$ and replace in the equation for $\map p {y_j}$:
+{{begin-eqn}}
+{{eqn | l = \map p {y_j}
+      | r =\sum_{i \mathop = 1}^{n} \paren {a_i - b_i } y_j^{i-1}
+      | c = for  $j = 1,\ldots,n$
+}}
+{{end-eqn}}
+Vector $\paren { \begin{smallmatrix}
+-\map p {y_1} \\
+\vdots \\
+-\map p {y_n} \\
+\end{smallmatrix} }$ is a linear combination of the column vectors of matrix  $V_y^T$ with weights (coefficients) $b_i - a_i$. Then:
+{{begin-eqn}}
+{{eqn | l = \vec A
+      | r = \begin{pmatrix}
+            b_1 - a_1  \\
+            \vdots \\
+            b_{n} - a_{n} \\
+            \end{pmatrix}
+}}
+{{eqn | r = \begin{pmatrix}
+            \paren {-1}^n \paren { e_n \paren {\set {y_1,\dots,y_n} } - e_n \paren {\set {x_1,\dots,x_n} } }  \\
+            \vdots \\
+            \paren {-1}^1 \paren {e_1 \paren {\set {y_1,\dots,y_n} } - e_1 \paren {\set {x_1,\dots,x_n} } }  \\
+            \end{pmatrix}
+      | c = The last entry contains the sum of the elements in $C^{-1}$:
+$S = x_1 + \cdots + x_n - \paren {y_1 + \cdots + y_n}$
+}}
+{{end-eqn}}
+'''Compute''' $\vec B$:
+{{begin-eqn}}
+{{eqn | l = \vec B
+      | r = V_x P^{-1} \vec { \bsone }
+      | c = Equation (3). 
+}}
+{{eqn | r = V_x 
+            \begin{pmatrix} 
+            \dfrac {1}{ \map {p_1} {x_1} } \\
+            \vdots \\
+            \dfrac {1}{ \map {p_n} {x_n} } \\
+            \end{pmatrix} 
+}}
+{{eqn | r = \begin{pmatrix} 
+            \displaystyle\sum_{k \mathop = 1}^n \frac {1}{ \map {p_k} {x_k} } \\
+            \vdots \\
+            \displaystyle\sum_{k \mathop = 1}^n \frac { x_n^{n-1} }{ \map {p_n} {x_n} } \\
+            \end{pmatrix}
+}}
+{{eqn | r = \paren { 
+              \begin{smallmatrix} 
+              0 \\
+              0 \\
+              \vdots \\
+              1 \\
+              \end{smallmatrix}
+            } 
+      | c = [[Summation of Powers over Product of Differences]] 
+}}
+{{end-eqn}}
+'''Evaluate''' $S$:
+{{begin-eqn}}
+{{eqn | l = S
+      | r = { \vec A }^T {\vec B}
+}}
+{{eqn | r = \paren { 
+              \begin{smallmatrix} 
+              * \\
+              * \\
+              \vdots \\
+              x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n }  \\
+              \end{smallmatrix} 
+            }^T 
+            \paren { 
+              \begin{smallmatrix} 
+              0 \\
+              0 \\
+              \vdots \\
+              1 \\
+              \end{smallmatrix} 
+            }
+      | c = Entries $*$ don't contribute because of matching zero in $\vec B$.
+}}                                                                                                     
+{{eqn | r = x_1 + \cdots + x_n - \paren { y_1 + \cdots + y_n}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Hilbert Matrix is Cauchy Matrix}
+Tags: Hilbert Matrix, Cauchy Matrix
+
+\begin{theorem}
+A [[Definition:Hilbert Matrix|Hilbert matrix]] is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Hilbert Matrix|Hilbert matrix]], the [[Definition:Element of Matrix|element]] $a_{i j}$ is:
+:$a_{i j} = \dfrac 1 {i + j - 1}$
+For all $i, j \in \Z$ such that $1 \le i \le n$ and $1 \le j \le n$, let:
+:$x_i = i$
+:$y_j = j - 1$
+Then:
+:$a_{i j} = \dfrac 1 {x_i + y_j}$
+The result follows by definition of a [[Definition:Cauchy Matrix|Cauchy matrix]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Hilbert Matrix}
+Tags: Hilbert Matrix
+
+\begin{theorem}
+Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
+:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
+Then its [[Definition:Inverse Matrix|inverse]] $H_n^{-1} = \sqbrk b_n$ can be specified as:
+:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - j}! \paren {n - i}! \paren {i + j - 1} } \end{bmatrix}$
+\end{theorem}
+
+\begin{proof}
+From [[Hilbert Matrix is Cauchy Matrix]], $H_n$ is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]]:
+:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
+where:
+:$x_i = i$
+:$y_j = j - 1$
+From [[Inverse of Cauchy Matrix]], the [[Definition:Inverse Matrix|inverse]] of the [[Definition:Square Matrix|square]] [[Definition:Cauchy Matrix|Cauchy matrix]] of [[Definition:Order of Square Matrix|order $n$]] is:
+:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\displaystyle \paren {x_j + y_i} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {y_i - x_k} } } \end{bmatrix}$
+Thus $H_n^{-1}$ can be specified as:
+:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\displaystyle \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} } {\displaystyle \paren {i + j - 1} \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k} } \paren {\prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} } } \end{bmatrix}$
+First, from [[Product of Products]]:
+:$\displaystyle \prod_{k \mathop = 1}^n \paren {i + k - 1} \paren {j + k - 1} = \prod_{k \mathop = 1}^n \paren {i + k - 1} \prod_{k \mathop = 1}^n \paren {j + k - 1}$
+We address in turn the various [[Definition:Factor|factors]] of this [[Definition:Expression|expression]] for $b_{i j}$.
+{{begin-eqn}}
+{{eqn | l = \prod_{k \mathop = 1}^n \paren {i + k - 1}
+      | r = \prod_{k \mathop = 0}^{n - 1} \paren {i + k}
+      | c = [[Translation of Index Variable of Product]]
+}}
+{{eqn | r = i^{\overline n}
+      | c = Definition of [[Definition:Rising Factorial|Rising Factorial]]
+}}
+{{eqn | r = \frac {\paren {i + n - 1}!} {\paren {i - 1}!}
+      | c = [[Rising Factorial as Quotient of Factorials]]
+}}
+{{end-eqn}}
+and similarly:
+:$\displaystyle \prod_{k \mathop = 1}^n \paren {j + k - 1} = \frac {\paren {j + n - 1}!} {\paren {j - 1}!}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne i} } \paren {i - k}
+      | r = \paren {\prod_{1 \mathop \le k \mathop < i} \paren {i - k} } \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }
+      | c = 
+}}
+{{eqn | r = \paren {i - 1}! \paren {\prod_{i \mathop < k \mathop \le n} \paren {i - k} }
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = \paren {i - 1}! \paren {\prod_{0 \mathop < k \mathop \le n - i} \paren {-k} }
+      | c = [[Translation of Index Variable of Product]]
+}}
+{{eqn | r = \paren {i - 1}! \paren {-1}^{n - i} \paren {\prod_{0 \mathop < k \mathop \le n - i} k}
+      | c = 
+}}
+{{eqn | r = \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}!
+      | c = {{Defof|Factorial}}
+}}
+{{end-eqn}}
+and similarly:
+:$\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {j - k} = \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!$
+Thus we can write:
+{{begin-eqn}}
+{{eqn | l = \begin{bmatrix} b_{i j} \end{bmatrix}
+      | r = \frac {\paren {\dfrac {\paren {i + n - 1}!} {\paren {i - 1}!} } \paren {\dfrac {\paren {j + n - 1}!} {\paren {j - 1}!} } } {\paren {i + j - 1} \paren {i - 1}! \paren {-1}^{n - i} \paren {n - i}! \paren {j - 1}! \paren {-1}^{n - j} \paren {n - j}!}
+      | c = 
+}}
+{{eqn | r = \frac {\paren {-1}^{i + j} \paren {i + n - 1}! \paren {j + n - 1}!} {\paren {\paren {i - 1}!}^2 \paren {\paren {j - 1}!}^2 \paren {n - i}! \paren {n - j}! \paren {i + j - 1} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Elements of Inverse of Hilbert Matrix are Integers}
+Tags: Hilbert Matrix
+
+\begin{theorem}
+Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
+:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
+Consider its [[Definition:Inverse Matrix|inverse]] $H_n^{-1}$.
+All the [[Definition:Element of Matrix|elements]] of $H_n^{-1}$ are [[Definition:Integer|integers]].
+\end{theorem}
+
+\begin{proof}
+From [[Inverse of Hilbert Matrix]], $H_n^{-1} = \left[{b}\right]_n$ can be specified as:
+:$\begin{bmatrix} b_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac {\left({-1}\right)^{i + j} \left({i + n - 1}\right)! \left({j + n - 1}\right)!} {\left({\left({i - 1}\right)!}\right)^2 \left({\left({j - 1}\right)!}\right)^2 \left({n - i}\right)! \left({n - j}\right)! \left({i + j - 1}\right)} \end{bmatrix}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = b_{i j}
+      | r = \frac {\left({-1}\right)^{i + j} \left({i + n - 1}\right)! \left({j + n - 1}\right)!} {\left({\left({i - 1}\right)!}\right)^2 \left({\left({j - 1}\right)!}\right)^2 \left({n - i}\right)! \left({n - j}\right)! \left({i + j - 1}\right)}
+      | c = 
+}}
+{{eqn | r = \left({\frac {\left({-1}\right)^{i + j} } {i + j - 1} }\right) \left({\frac {\left({i + n - 1}\right)!} {\left({i - 1}\right)! \, n!} }\right) \left({\frac {\left({j + n - 1}\right)!} {\left({j - 1}\right)! \, n!} }\right) \left({\frac {n! \, i} {i! \, \left({n - i}\right)!} }\right) \left({\frac {n! \, j} {j! \, \left({n - j}\right)! } }\right)
+      | c = 
+}}
+{{eqn | r = \frac {\left({-1}\right)^{i + j} i j} {i + j - 1} \binom {i + n - 1} n \binom {j + n - 1} n \binom n i \binom n j
+      | c = Definition of [[Definition:Binomial Coefficient|Binomial Coefficient]]
+}}
+{{eqn | r = \left({-1}\right)^{i + j} j \binom {i + n - 1} {i - i} \binom {j + n - 1} {n - 1} \binom {i + j - 2} {n - i} \binom n j
+      | c = 
+}}
+{{end-eqn}}
+All of the factors of the above expression are [[Definition:Integer|integers]], from [[Binomial Coefficient is Integer]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Elements in Inverse of Hilbert Matrix}
+Tags: Cauchy Matrix
+
+\begin{theorem}
+Let $H_n$ be the [[Definition:Hilbert Matrix|Hilbert matrix]] of [[Definition:Order of Square Matrix|order $n$]]:
+:$\begin{bmatrix} a_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {i + j - 1} \end{bmatrix}$
+Consider its [[Definition:Inverse Matrix|inverse]] $H_n^{-1}$.
+All the [[Definition:Element of Matrix|elements]] of $H_n^{-1}$ are [[Definition:Integer|integers]].
+The sum of all the [[Definition:Element of Matrix|elements]] $b_{i j}$ of $H_n^{-1}$ is:
+:$\displaystyle \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = n^2$
+\end{theorem}
+
+\begin{proof}
+From [[Hilbert Matrix is Cauchy Matrix]], $H_n$ is a special case of a [[Definition:Cauchy Matrix|Cauchy matrix]]:
+:$\begin{bmatrix} c_{i j} \end{bmatrix} = \begin{bmatrix} \dfrac 1 {x_i + y_j} \end{bmatrix}$
+where:
+:$x_i = i$
+:$y_j = j - 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j}
+      | r = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k
+      | c = [[Sum of Elements in Inverse of Cauchy Matrix]]
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n k + \sum_{k \mathop = 1}^n \paren {k - 1}
+      | c = 
+}}
+{{eqn | r = 2 \sum_{k \mathop = 1}^n k - n
+      | c = 
+}}
+{{eqn | r = n \paren {n + 1} - n
+      | c = [[Closed Form for Triangular Numbers]]
+}}
+{{eqn | r = n^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Suprema of two Real Sets}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Real Number|real sets]].
+Let $S$ and $T$ admit [[Definition:Supremum of Subset of Real Numbers|suprema]].
+Then:
+: $\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\sup S \le \sup T$.
+The aim is to establish that:
+:$\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
+Let $\epsilon \in \R_{>0}$.
+Let:
+:$S' = \set {s \in S : \exists t \in T : s \le t}$
+There are two cases for an $s \in S$:
+:$s \in S'$, in other words: $\exists t \in T : s \le t$
+and:
+:$s \in S \setminus S'$, in other words: $\nexists t \in T : s \le t$
+First, consider the case that $s \in S'$.
+Let $s \in S'$.
+Then $t \in T$ exists such that:
+{{begin-eqn}}
+{{eqn | l = s
+      | o = \le
+      | r = t
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | o = \le
+      | r = t < t + \epsilon
+      | c = as $t < t + \epsilon$ is true
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | o = <
+      | r = t + \epsilon
+      | c = 
+}}
+{{end-eqn}}
+We have found:
+: $\forall \epsilon \in \R_{>0}: \forall s \in S': \exists t \in T: s < t + \epsilon$
+Next, consider the second case, that $s \in S \setminus S'$.
+Let $s \in S \setminus S'$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \nexists t \in T : s
+      | o = \le
+      | r = t
+}}
+{{eqn | ll= \leadsto
+      | l = \forall t \in T : s
+      | o = >
+      | r = t
+      | c = so, $s$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $T$
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | o = \ge
+      | r = \sup T
+      | c = as (a) $s$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $T$ and (b) $\sup T$ is the [[Definition:Supremum of Subset of Real Numbers|least upper bound]] of $T$
+}}
+{{eqn | ll= \leadsto
+      | l = \sup S
+      | o = \ge
+      | r = s \ge \sup T
+      | c = as $\sup S \ge s$ is true since (a) $\sup S$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $S$ and (b) $s \in S$
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = \ge
+      | r = \sup S \ge s \ge \sup T
+      | c = as $\sup S \le \sup T$ is true
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = \ge
+      | r = s \ge \sup T
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | r = \sup T
+}}
+{{end-eqn}}
+So, $s = \sup T$ is the only [[Definition:Element|element]]  of $S \setminus S'$.
+A $t \in T$ exists such that:
+{{begin-eqn}}
+{{eqn | l = \sup T - t
+      | o = <
+      | r = \epsilon
+      | c = by [[Supremum of Subset of Real Numbers is Arbitrarily Close]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = <
+      | r = t + \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = s
+      | o = <
+      | r = t + \epsilon
+      | c = as $s = \sup T$
+}}
+{{end-eqn}}
+We have found:
+: $\forall \epsilon \in \R_{>0}: \forall s \in S \setminus S': \exists t \in T: s < t + \epsilon$
+We conclude by combining the results for the two cases:
+: $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.
+The aim is to establish that $\sup S \le \sup T$.
+Let $\epsilon \in \R_{>0}$.
+An $s \in S$ exists such that:
+{{begin-eqn}}
+{{eqn | l = \sup S - s
+      | o = <
+      | r = \epsilon
+      | c = by [[Supremum of Subset of Real Numbers is Arbitrarily Close]]
+}}
+{{eqn | ll = \iff
+      | l = s
+      | o = >
+      | r = \sup S - \epsilon
+}}
+{{end-eqn}}
+Let $s$ be such an [[Definition:Element|element]] of $S$.
+A $t \in T$ exists such that:
+{{begin-eqn}}
+{{eqn | l = s
+      | o = <
+      | r = t + \epsilon
+      | c = as $\forall \epsilon' \in \R_{>0}: \forall s' \in S: \exists t \in T: s' < t + \epsilon'$
+}}
+{{eqn | ll = \iff
+      | l = t
+      | o = >
+      | r = s - \epsilon
+}}
+{{end-eqn}}
+Let $t$ be such an [[Definition:Element|element]] of $T$.
+Then:
+{{begin-eqn}}
+{{eqn | l = t
+      | o = >
+      | r = s - \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = t
+      | o = >
+      | r = s - \epsilon \text { and } s > \sup S - \epsilon 
+      | c = as $s > \sup S - \epsilon$ is true
+}}
+{{eqn | ll= \leadsto
+      | l = t
+      | o = >
+      | r = s - \epsilon \text{ and } s - \epsilon > \sup S - 2 \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = t
+      | o = >
+      | r = s - \epsilon > \sup S - 2 \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = t
+      | o = >
+      | r = \sup S - 2 \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = \ge
+      | r = t > \sup S - 2 \epsilon
+      | c = as $\sup T \ge t$ is true since (a) $\sup T$ is an [[Definition:Upper Bound of Subset of Real Numbers|upper bound]] for $T$ and (b) $t \in T$
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = >
+      | r = \sup S - 2 \epsilon
+}}
+{{eqn | ll= \leadsto
+      | l = \sup T
+      | o = \ge
+      | r = \sup S
+      | c = as $\epsilon$ is an arbitrary [[Definition:Element|element]] of $\R_{>0}$
+}}
+{{eqn | ll= \leadsto
+      | l = \sup S
+      | o = \le
+      | r = \sup T
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sequential Continuity is Equivalent to Continuity in the Reals}
+Tags: Real Analysis, Continuity, Sequential Continuity is Equivalent to Continuity in the Reals
+
+\begin{theorem}
+Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
+Let $c \in A$.
+Let $f : A \to \R$ be a [[Definition:Real Function|real function]].
+Then $f$ is [[Definition:Continuous Real Function|continuous]] at $c$ {{iff}}:
+:for each [[Definition:Real Sequence|sequence]] $\sequence {x_n}$ in $A$ that [[Definition:Convergent Real Sequence|converges]] to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$.
+\end{theorem}
+
+\begin{proof}
+=== [[Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition|Sufficient Condition]] ===
+{{:Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition}}
+=== [[Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition|Necessary Condition]] ===
+{{:Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition}}
+\end{proof}<|endoftext|>
+\section{Rising Factorial as Quotient of Factorials}
+Tags: Rising Factorials, Factorials
+
+\begin{theorem}
+:$x^{\overline n} = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!} = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x^{\overline n}
+      | r = \prod_{j \mathop = 0}^{n - 1} \paren {x + j}
+      | c = {{Defof|Rising Factorial}}
+}}
+{{eqn | r = x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + n - 1}
+      | c = 
+}}
+{{eqn | r = \dfrac {\paren {x + n - 1}!} {\paren {x - 1}!}
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = \dfrac {\map \Gamma {x + n} } {\map \Gamma x}
+      | c = [[Gamma Function Extends Factorial]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalent Definition for Alternating Bilinear Mapping}
+Tags: Algebra
+
+\begin{theorem}
+Let $\left({A_R, \oplus}\right)$ be an algebra over a ring $R$ with the property that $\operatorname{Char}\left({R}\right) \neq 2$.
+Then the following definitions for [[Definition:Alternating Bilinear Mapping|alternating bilinear maps]] are equivalent:
+* $\oplus$ is an alternating bilinear map {{iff}} for all $a \in A_R$, $a \oplus a = 0$.
+* $\oplus$ is an alternating bilinear map {{iff}} for all $a, b \in A_R$, $a \oplus b + b \oplus a = 0$.
+\end{theorem}
+
+\begin{proof}
+Let $\oplus$ be a bilinear map with the property that for all $a \in A_R$, $a \oplus a = 0$.
+Then for all $u, v \in A_R$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = \left({u + v}\right) \oplus \left({u + v}\right)
+}}
+{{eqn | r = u \oplus \left({u + v}\right) + v \oplus \left({u + v}\right)
+}}
+{{eqn | r = u \oplus v + v \oplus u = 0 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Let $\oplus$ be a bilinear map with the property that for all $a, b \in A_R$, $a \oplus b + b \oplus a = 0$.
+Let $u \in A_R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = u \oplus u + u \oplus u
+}} 
+{{eqn | r = 2 \left({u \oplus u}\right)
+}}
+{{end-eqn}}
+Since the [[Definition:Characteristic of Ring|characteristic]] of $R$ is not $2$, thus:
+:$u \oplus u = 0$
+{{qed}}
+[[Category:Algebra]]
+kvu4p5d5p00w9uwep4sjnbm6jkg7k3j
+\end{proof}<|endoftext|>
+\section{Way Below if Between is Compact Set in Ordered Set of Topology}
+Tags: Way Below Relation, Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $L = \left({\tau, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\preceq \mathop = \subseteq\restriction_{\tau \times \tau}$
+Let $x, y \in \tau$ such that
+:$\exists H \subseteq S: x \subseteq H \subseteq y \land H$ is [[Definition:Compact|compact]]
+Then
+:$x \ll y$
+\end{theorem}
+
+\begin{proof}
+Let $D$ be a [[Definition:Directed Subset|directed subset]] of $\tau$ such that
+:$y \preceq \sup D$
+By proof of [[Topology forms Complete Lattice]]:
+:$y \subseteq \bigcup D$
+By [[Subset Relation is Transitive]]:
+:$H \subseteq \bigcup D$
+By definition:
+:$D$ is [[Definition:Open Cover|open cover]] of $H$
+By definition of [[Definition:Compact|compact]]:
+:$H$ has [[Definition:Finite Subcover|finite subcover]] $\mathcal G$ of $D$
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists d \in D: d$ is [[Definition:Upper Bound of Set|upper bound]] for $\mathcal G$
+By definitions of [[Definition:Upper Bound of Set|upper bound]] and of $\preceq$:
+:$\forall z \in \mathcal G: z \subseteq d$
+By [[Union is Smallest Superset/Set of Sets]]:
+:$\bigcup \mathcal G \subseteq d$
+By definition of [[Definition:Cover of Set|cover]]:
+:$H \subseteq \bigcup \mathcal G$
+By [[Subset Relation is Transitive]]:
+:$x \subseteq d$
+Thus by definition of $\preceq$:
+:$x \preceq d$
+Thus by definition of [[Definition:Element is Way Below|way below relation]]:
+:$x \ll y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Locally Compact iff Open Neighborhood contains Compact Set}
+Tags: Locally Compact Spaces
+
+\begin{theorem}
+Let $T = \struct{S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Then
+:$T$ is [[Definition:Locally Compact Space|locally compact]]
+{{iff}}
+:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
+where $K^\circ$ denotes the [[Definition:Interior (Topology)|interior]] of $K$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $T$ be [[Definition:Locally Compact Space|locally compact]].
+Let $U \in \tau, x \in U$.
+Since $T$ is [[Definition:Locally Compact Space|locally compact]] there exists a [[Definition:Neighborhood Basis|neighborhood basis]] $\mathcal B$ of $x$ consisting of [[Definition:Compact Topological Subspace|compact sets]]. 
+From [[Set is Open iff Neighborhood of all its Points]], $U$ is a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
+By the definition of a [[Definition:Neighborhood Basis|neighborhood basis]]:
+:$\exists K \in \mathcal B : K \subseteq U$.
+By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$:
+:$\exists V \in \tau : x \in V \subseteq K$
+By the definition of the [[Definition:Interior (Topology)|interior]] of a [[Definition:Subset|subset]] then:
+:$x \in V \subseteq K^\circ$
+Thus
+:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
+{{qed|lemma}}
+=== Necessary Condition ===
+Let
+:$\forall U \in \tau, x \in U: \exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
+Let $\mathcal B$ be the [[Definition:Set|set]] of all [[Definition:Compact Topological Subspace|compact]] [[Definition:Neighborhood (Topology)|neighborhoods]] of $x$.
+It is shown that $\mathcal B$ is a [[Definition:Neighborhood Basis|neighborhood basis]] of $x$.
+Let $N$ be a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
+By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$:
+:$\exists U \in \tau : x \in U \subseteq N$
+By assumption:
+:$\exists K \subseteq S: x \in K^\circ \land K \subseteq U \land K$ is [[Definition:Compact Topological Subspace|compact]]
+By the definition of the [[Definition:Interior (Topology)|interior]] of a [[Definition:Subset|subset]] $K^\circ$ is [[Definition:Open Set/Topology|open]].
+By the definition of a [[Definition:Neighborhood (Topology)|neighborhood]] then $K$ is a [[Definition:Neighborhood (Topology)|neighborhood]] of $x$.
+Hence $K \in \mathcal B$ and $K \subseteq U \subseteq N$.
+As $N$ was arbitrary, then $B$ is a [[Definition:Neighborhood Basis|neighborhood basis]] of $x$ consisting of [[Definition:Compact Topological Subspace|compact sets]].
+\end{proof}<|endoftext|>
+\section{Cauchy-Binet Formula}
+Tags: Matrix Theory, Determinants, Cauchy-Binet Formula
+
+\begin{theorem}
+Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
+Let $\mathbf B$ be an [[Definition:Matrix|$n \times m$ matrix]].
+Let $1 \le j_1, j_2, \ldots, j_m \le n$.
+Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+Let $\mathbf B_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
+Then:
+:$\displaystyle \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
+where $\det$ denotes the [[Definition:Determinant of Matrix|determinant]].
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {k_1, k_2, \ldots, k_m}$ be an [[Definition:Ordered Tuple|ordered $m$-tuple]] of [[Definition:Integer|integers]].
+Let $\map \epsilon {k_1, k_2, \ldots, k_m}$ denote the [[Definition:Sign of Ordered Tuple|sign]] of $\tuple {k_1, k_2, \ldots, k_m}$.
+Let $\tuple {l_1, l_2, \ldots, l_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ except for $k_i$ and $k_j$ having been transposed.
+Then from [[Transposition is of Odd Parity]]:
+:$\map \epsilon {l_1, l_2, \ldots, l_m} = -\map \epsilon {k_1, k_2, \ldots, k_m}$
+Let $\tuple {j_1, j_2, \ldots, j_m}$ be the same as $\tuple {k_1, k_2, \ldots, k_m}$ by arranged into non-decreasing order.
+That is:
+:$j_1 \le j_2 \le \cdots \le j_m$
+Then it follows that:
+:$\map \det {\mathbf B_{k_1 \cdots k_m} } = \map \epsilon {k_1, k_2, \ldots, k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \det {\mathbf A \mathbf B}
+      | r = \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \map \epsilon {l_1, \ldots, l_m} \paren {\sum_{k \mathop = 1}^n a_{1 k} b_{k l_1} } \cdots \paren {\sum_{k \mathop = 1}^n a_{m k} b_{k l_m} }
+      | c = {{Defof|Determinant of Matrix}}
+}}
+{{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \sum_{1 \mathop \le l_1, \mathop \ldots \mathop , l_m \mathop \le m} \map \epsilon {l_1, \ldots, l_m} b_{k_1 l_1} \cdots b_{k_m l_m} 
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \cdots a_{m k_m} \map \det {\mathbf B_{k_1 \cdots k_m} }
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le k_1, \mathop \ldots \mathop , k_m \mathop \le n} a_{1 k_1} \map \epsilon {k_1, \ldots, k_m} \cdots a_{m k_m} \map \det {\mathbf B_{j_1 \cdots j_m} }
+      | c = 
+}}
+{{eqn | r = \sum_{1 \mathop \le j_1 \mathop \le j_2 \mathop \le \cdots \mathop \le j_m \le n} \map \det {\mathbf A_{j_1 \cdots j_m} } \map \det {\mathbf B_{j_1 \cdots j_m} }
+      | c = 
+}}
+{{end-eqn}}
+If two $j$s are equal:
+:$\map \det {\mathbf A_{j_1 \cdots j_m} } = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Auxiliary Relation is Congruent}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathcal R$ be [[Definition:Relation|relation]] on $S$ satisfying conditions $(ii)$ and $(iii)$ of [[Definition:Auxiliary Relation|auxiliary relation]].
+Then
+:$\forall x, y, z, u \in S: \left({x, z}\right) \in \mathcal R \land \left({y, u}\right) \in \mathcal R \implies \left({x \vee y, z \vee u}\right) \in \mathcal R$
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z, u \in S$ such that
+:$\left({x, z}\right) \in \mathcal R \land \left({y, u}\right) \in \mathcal R$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$ and $y \preceq y$
+By [[Join Succeeds Operands]]:
+:$z \preceq z \vee u$ and $u \preceq z \vee u$
+By condition $(ii)$ of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({x, z \vee u}\right) \in \mathcal R$ and $\left({y, z \vee u}\right) \in \mathcal R$
+Thus by condition $(iii)$ of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({x \vee y, z \vee u}\right) \in \mathcal R$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Auxiliary Relation is Transitive}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $\left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathcal R$ be [[Definition:Relation|relation]] on $S$ satisfying conditions $(i)$ and $(ii)$ of [[Definition:Auxiliary Relation|auxiliary relation]].
+Then
+:$\mathcal R$ is a [[Definition:Transitive Relation|transitive relation]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y, z \in S$ such that
+:$\left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$z \preceq z$ 
+By condition $(i)$ of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$x \preceq y$
+Thus by condition $(ii)$ of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({x, z}\right) \in \mathcal R$
+Thus by definition
+:$\mathcal R$ is a [[Definition:Transitive Relation|transitive relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Binet Formula/Example/m equals 1}
+Tags: Determinants, Cauchy-Binet Formula
+
+\begin{theorem}
+Let $\mathbf A = \left[{a}\right]_{1 n}$ be a [[Definition:Row Matrix|row matrix]] with $n$ [[Definition:Column of Matrix|columns]].
+and $\mathbf B = \left[{b}\right]_{n 1}$ be a [[Definition:Column Matrix|column matrix]] with $n$ [[Definition:Row of Matrix|rows]].
+Let $\mathbf A \mathbf B$ be the [[Definition:Matrix Product (Conventional)|(conventional) matrix product]] of $\mathbf A$ and $\mathbf B$.
+Then:
+:$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{j \mathop = 1}^n a_j b_j$
+where:
+:$a_j$ is [[Definition:Element of Matrix|element]] $a_{1 j}$ of $\mathbf A$
+:$b_j$ is [[Definition:Element of Matrix|element]] $b_{j 1}$ of $\mathbf B$.
+\end{theorem}
+
+\begin{proof}
+The [[Cauchy-Binet Formula]] gives:
+:$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$
+where:
+:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B$ is an [[Definition:Matrix|$n \times m$ matrix]].
+:For $1 \le j_1, j_2, \ldots, j_m \le n$:
+::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
+When $m = 1$, the relation:
+:$1 \le j_1 < j_2 < \cdots < j_m \le n$
+degenerates to:
+:$1 \le j \le n$
+By definition of [[Definition:Determinant of Order 1|order $1$ determinant]]:
+:$\det \left({\left[{a_j}\right]}\right) = a_j$
+:$\det \left({\left[{b_j}\right]}\right) = b_j$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding is 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]].
+Then
+:$\preceq$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+\end{theorem}
+
+\begin{proof}
+:$\forall x, y \in S: x \preceq y \implies x \preceq y$
+Then condition $(i)$ of [[Definition:Auxiliary Relation|auxiliary relation]] is satisfied.
+By definition of [[Definition:Transitivity|transitivity]]:
+:$\forall x, y, z, u \in S: x \preceq y \preceq z \preceq u \implies x \preceq u$
+Then the condition $(ii)$ of [[Definition:Auxiliary Relation|auxiliary relation]] is satisfied.
+By  definition of [[Definition:Supremum of Set|supremum]]:
+:$\forall x, y, z \in S: x \preceq z \land y \preceq z \implies x \vee y \preceq z$
+Then the condition $(iii)$ of [[Definition:Auxiliary Relation|auxiliary relation]] is satisfied.
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\forall x \in S: \bot \preceq x$
+Then the condition $(iv)$ of [[Definition:Auxiliary Relation|auxiliary relation]] is satisfied.
+Thus the result by definition [[Definition:Auxiliary Relation|auxiliary relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Binet Formula/Example/Matrix by Transpose}
+Tags: Cauchy-Binet Formula
+
+\begin{theorem}
+Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
+Let $\mathbf A^\intercal$ be the [[Definition:Transpose of Matrix|transpose]] $\mathbf A$.
+Let $1 \le j_1, j_2, \ldots, j_m \le n$.
+Let $\mathbf A_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+Let $\mathbf A^\intercal_{j_1 j_2 \ldots j_m}$ denote the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf A^\intercal$.
+Then:
+:$\displaystyle \det \left({\mathbf A \mathbf A^\intercal}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \left({\det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right)}\right)^2$
+where $\det$ denotes the [[Definition:Determinant of Matrix|determinant]].
+\end{theorem}
+
+\begin{proof}
+The [[Cauchy-Binet Formula]] gives:
+:$\displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$
+where:
+:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B$ is an [[Definition:Matrix|$n \times m$ matrix]].
+:For $1 \le j_1, j_2, \ldots, j_m \le n$:
+::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
+From the definition of [[Definition:Transpose of Matrix|transpose]] $\mathbf A^\intercal$ is an [[Definition:Matrix|$n \times m$ matrix]].
+Hence the [[Cauchy-Binet Formula]] can be applied directly:
+:$\displaystyle \det \left({\mathbf A \mathbf A^\intercal}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right) \det \left({\mathbf A^\intercal_{j_1 j_2 \ldots j_m} }\right)$
+Note that by construction:
+:$\mathbf A_{j_1 j_2 \ldots j_m}$ is a [[Definition:Square Matrix|square matrix]]
+Also, by definition of [[Definition:Transpose of Matrix|transpose]]:
+:$\mathbf A^\intercal_{j_1 j_2 \ldots j_m} = \left({\mathbf A_{j_1 j_2 \ldots j_m} }\right)^\intercal$
+The result follows from [[Determinant of Transpose]]:
+:$\det \left({\mathbf A}\right) = \det \left({\mathbf A^\intercal}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preceding is Top in Ordered Set of Auxiliary Relations}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let ${\it Aux}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
+Let $P = \left({ {\it Aux}\left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\precsim \mathop = \subseteq\restriction_{{\it Aux}\left({L}\right) \times {\it Aux}\left({L}\right)}$
+Then
+:$\preceq \mathop = \top_P$
+where $\top_P$ denotes the [[Definition:Greatest Element|greatest element]] in $P$.
+\end{theorem}
+
+\begin{proof}
+By [[Preceding is Auxiliary Relation]]:
+:$\preceq \mathop \in {\it Aux}\left({L}\right)$
+By definition
+:$\preceq$ is [[Definition:Lower Bound of Set|lower bound]] for $\varnothing$ in $P$
+We will prove that
+:$\forall R \in {\it Aux}\left({L}\right): R$ is [[Definition:Lower Bound of Set|lower bound]] for $\varnothing \implies R \mathop \precsim \preceq$
+Let $R \in {\it Aux}\left({L}\right)$
+By condition $(i)$ of definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$R \mathop \subseteq \preceq$
+Thus by definition of $\precsim$:
+:$R \mathop \precsim \preceq$
+{{qed|lemma}}
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\preceq \mathop = \inf_P \varnothing$
+Thus by [[Infimum of Empty Set is Greatest Element]]:
+:$\preceq \mathop = \top_P$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bottom Relation is Auxiliary Relation}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $B = \left\{ {\left({\bot, x}\right): x \in S}\right\}$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
+Then
+:$B$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\forall x \in S: \bot \preceq x$
+Thus by definition of $B$:
+:$\forall x, y \in S: \left({x, y}\right) \in B \implies x \preceq y$
+We will prove that
+:$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in B \land z \preceq u \implies \left({x, u}\right) \in B$
+Let $x, y, z, u \in S$ such that
+:$x \preceq y \land \left({y, z}\right) \in B \land z \preceq u$
+By definition of $B$:
+:$y = \bot$
+By definition of [[Definition:Smallest Element|smallest element]]:
+:$\bot \preceq x$
+By definition of [[Definition:Antisymmetric Relation|antisymmetry]]:
+:$x = \bot$
+Thus by definition of $B$:
+:$\left({x, u}\right) \in B$
+{{qed|lemma}}
+We will prove that
+:$\forall x, y, z \in S: \left({x, z}\right) \in B \land \left({y, z}\right) \in B \implies \left({x \vee y, z}\right) \in B$
+Let $x, y, z \in S$ such that
+: $\left({x, z}\right) \in B \land \left({y, z}\right) \in B$
+By definition of $B$:
+:$x = y = \bot$
+By [[Join is Idempotent]]:
+:$x \vee y = \bot$
+Thus by definition of $B$:
+:$\left({x \vee y, z}\right) \in B$
+{{qed|lemma}}
+Thus by definition of $B$:
+:$\forall x \in S: \left({\bot, x}\right) \in B$
+Thus by definition:
+:$B$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bottom Relation is Bottom in Ordered Set of Auxiliary Relations}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let ${\it Aux}\left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
+Let $P = \left({ {\it Aux}\left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\precsim \mathop = \subseteq\restriction_{{\it Aux}\left({L}\right) \times {\it Aux}\left({L}\right)}$
+Let $B = \left\{ {\left({\bot_L, x}\right): x \in S}\right\}$
+where $\bot_L$ denotes [[Definition:Smallest Element|smallest element]] in $L$.
+Then
+:$B \mathop = \bot_P$
+\end{theorem}
+
+\begin{proof}
+By [[Bottom Relation is Auxiliary Relation]]:
+:$B \mathop \in {\it Aux}\left({L}\right)$
+By definition
+:$B$ is [[Definition:Upper Bound of Set|upper bound]] for $\varnothing$ in $P$
+We will prove that
+:$\forall R \in {\it Aux}\left({L}\right): R$ is [[Definition:Upper Bound of Set|upper bound]] for $\varnothing \implies B \mathop \precsim R$
+Let $R \in {\it Aux}\left({L}\right)$
+By condition $(iv)$ of definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$B \mathop \subseteq R$
+Thus by definition of $\precsim$:
+:$B \mathop \precsim R$
+{{qed|lemma}}
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$B \mathop = \sup_P \varnothing$
+Thus by [[Supremum of Empty Set is Smallest Element]]:
+:$B \mathop = \bot_P$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Conditional is Converse of Contrapositive}
+Tags: Implication
+
+\begin{theorem}
+Let $p \implies q$ be a [[Definition:Conditional|conditional]].
+Then the [[Definition:Inverse Statement|inverse]] of $p \implies q$ is the [[Definition:Converse Statement|converse]] of its [[Definition:Contrapositive Statement|contrapositive]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
+:$\neg p \implies \neg q$
+The [[Definition:Contrapositive Statement|contrapositive]] of $p \implies q$ is:
+:$\neg q \implies \neg p$
+The [[Definition:Converse Statement|converse]] of $\neg q \implies \neg p$ is:
+:$\neg p \implies \neg q$
+The two are seen to be equal.
+{{qed}}
+[[Category:Implication]]
+3evau7fkxx96gwhtt8hqi0rpefw4w1z
+\end{proof}<|endoftext|>
+\section{Inverse of Conditional is Contrapositive of Converse}
+Tags: Implication
+
+\begin{theorem}
+Let $p \implies q$ be a [[Definition:Conditional|conditional]].
+Then the [[Definition:Inverse Statement|inverse]] of $p \implies q$ is the [[Definition:Contrapositive Statement|contrapositive]] of its [[Definition:Converse Statement|converse]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
+:$\neg p \implies \neg q$
+The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
+:$q \implies p$
+The [[Definition:Contrapositive Statement|contrapositive]] of $q \implies p$ is:
+:$\neg p \implies \neg q$
+The two are seen to be equal.
+{{qed}}
+[[Category:Implication]]
+9b8kfzy0f04jach47afrtwkv29o2wg3
+\end{proof}<|endoftext|>
+\section{Converse of Conditional is Contrapositive of Inverse}
+Tags: Implication
+
+\begin{theorem}
+Let $p \implies q$ be a [[Definition:Conditional|conditional]].
+Then the [[Definition:Converse Statement|converse]] of $p \implies q$ is the [[Definition:Contrapositive Statement|contrapositive]] of its [[Definition:Inverse Statement|inverse]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
+:$q \implies p$
+The [[Definition:Inverse Statement|inverse]] of $p \implies q$ is:
+:$\neg p \implies \neg q$
+The [[Definition:Contrapositive Statement|contrapositive]] of $\neg p \implies \neg q$ is:
+:$\neg\neg q \implies \neg\neg p$
+By [[Double Negation/Formulation 1|Double Negation]], the two are seen to be equal.
+{{qed}}
+[[Category:Implication]]
+plj00q6y1603076aolfow75v0tbe9y3
+\end{proof}<|endoftext|>
+\section{Converse of Conditional is Inverse of Contrapositive}
+Tags: Implication
+
+\begin{theorem}
+Let $p \implies q$ be a [[Definition:Conditional|conditional]].
+Then the [[Definition:Converse Statement|converse]] of $p \implies q$ is the [[Definition:Inverse Statement|inverse]] of its [[Definition:Contrapositive Statement|contrapositive]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Converse Statement|converse]] of $p \implies q$ is:
+:$q \implies p$
+The [[Definition:Contrapositive Statement|contrapositive]] of $p \implies q$ is:
+:$\neg q \implies \neg p$
+The [[Definition:Inverse Statement|inverse]] of $\neg q \implies \neg p$ is:
+:$\neg\neg q \implies \neg\neg p$
+By [[Double Negation/Formulation 1|Double Negation]], the two are seen to be equal.
+{{qed}}
+[[Category:Implication]]
+3mmpmjh5n6hqtx9aqgmij19sqj146rm
+\end{proof}<|endoftext|>
+\section{Intersection of Auxiliary Relations is Auxiliary Relation}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\mathcal F$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set]] of [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
+Then
+:$\bigcap \mathcal F$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$\exists \mathcal R: \mathcal R \in \mathcal F$
+We will prove that
+:$\forall x, y \in S: \left({x, y}\right) \in \bigcap \mathcal F \implies x \preceq y$
+Let $x, y \in S$ such that
+:$\left({x, y}\right) \in \bigcap \mathcal F$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$\left({x, y}\right) \in \mathcal R$
+Thus by definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$x \preceq y$
+{{qed|lemma}}
+We will prove that
+:$\forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in \bigcap \mathcal F \land z \preceq u \implies \left({x, u}\right) \in \bigcap \mathcal F$
+Let $x, y, z, u \in S$ such that
+:$x \preceq y \land \left({y, z}\right) \in \bigcap \mathcal F \land z \preceq u$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$\forall r \in \mathcal F: \left({y, z}\right) \in r$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\forall r \in \mathcal F: \left({x, u}\right) \in r$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$\left({x, u}\right) \in \bigcap \mathcal F$
+{{qed|lemma}}
+We will prove that
+:$\forall x, y, z \in S: \left({x, z}\right) \in \bigcap \mathcal F \land \left({y, z}\right) \in \bigcap \mathcal F \implies \left({x \vee y, z}\right) \in \bigcap \mathcal F$
+Let $x, y, z \in S$ such that
+:$\left({x, z}\right) \in \bigcap \mathcal F \land \left({y, z}\right) \in \bigcap \mathcal F$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$\forall r \in \mathcal F: \left({x, z}\right) \in r \land \left({y, z}\right) \in r$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\forall r \in \mathcal F: \left({x \vee y, z}\right) \in r$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$\left({x \vee y, z}\right) \in \bigcap \mathcal F$
+{{qed|lemma}}
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\forall x \in S: \forall r \in \mathcal F: \left({\bot, x}\right) \in r$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$\forall x \in S: \left({\bot, x}\right) \in \bigcap \mathcal F$
+Thus by definition:
+:$\bigcap \mathcal F$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordered Set of Auxiliary Relations is Complete Lattice}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Let $\operatorname {Aux} \left({L}\right)$ be the [[Definition:Set of Sets|set]] of all [[Definition:Auxiliary Relation|auxiliary relations]] on $S$.
+Let $P = \left({\operatorname {Aux} \left({L}\right), \precsim}\right)$ be an [[Definition:Ordered Set|ordered set]]
+where $\precsim \mathop = \subseteq \restriction_{\operatorname {Aux} \left({L}\right) \times \operatorname {Aux} \left({L}\right)}$
+Then
+:$P$ is a [[Definition:Complete Lattice|complete lattice]].
+\end{theorem}
+
+\begin{proof}
+Let $X \subseteq \operatorname {Aux} \left({L}\right)$
+In the case when $X \ne \O$:
+By [[Intersection of Auxiliary Relations is Auxiliary Relation]]:
+:$\bigcap X \in \operatorname {Aux} \left({L}\right)$
+By [[Intersection is Largest Subset]], $\bigcap X$ is the [[Definition:Infimum of Set|infimum]] of $X$.
+In case when $X = \O$:
+By proof of [[Preceding is Top in Ordered Set of Auxiliary Relations]]:
+:$\O$ admits an [[Definition:Infimum of Set|infimum]] in $P$
+Then:
+:$X$ ([[Definition:Empty Set|empty]] or [[Definition:Non-Empty Set|non-empty]]) admits an [[Definition:Infimum of Set|infimum]] in $P$
+By duality of [[Lattice is Complete iff it Admits All Suprema]]:
+:$P$ is a [[Definition:Complete Lattice|complete lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy-Binet Formula/Example/m greater than n}
+Tags: Cauchy-Binet Formula
+
+\begin{theorem}
+Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]].
+Let $\mathbf B$ be an [[Definition:Matrix|$n \times m$ matrix]].
+Let $m > n$.
+Then:
+:$\displaystyle \det \left({\mathbf A \mathbf B}\right) = 0$
+\end{theorem}
+
+\begin{proof}
+The [[Cauchy-Binet Formula]] gives:
+:$(1): \quad \displaystyle \det \left({\mathbf A \mathbf B}\right) = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \det \left({\mathbf A_{j_1 j_2 \ldots j_m}}\right) \det \left({\mathbf B_{j_1 j_2 \ldots j_m}}\right)$
+where:
+:$\mathbf A$ is an [[Definition:Matrix|$m \times n$ matrix]]
+:$\mathbf B$ is an [[Definition:Matrix|$n \times m$ matrix]].
+:For $1 \le j_1, j_2, \ldots, j_m \le n$:
+::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Column of Matrix|columns]] $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
+::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the [[Definition:Matrix|$m \times m$ matrix]] consisting of [[Definition:Row of Matrix|rows]] $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
+But here $m > n$.
+Therefore the [[Definition:Set|set]] $\left\{ {j_1, j_2, \ldots, j_m}\right\}$ such that:
+:$1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n$
+is the [[Definition:Empty Set|empty set]].
+Thus the {{RHS}} of $(1)$ is a [[Definition:Vacuous Summation|vacuous summation]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Krattenthaler's Identity}
+Tags: Determinants
+
+\begin{theorem}
+:$\begin{vmatrix}
+\left({x + q_2}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + p_2}\right) \\
+\left({y + q_2}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + p_2}\right) \\
+\left({z + q_2}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + p_2}\right)
+\end{vmatrix} = \left({x - y}\right) \left({x - z}\right) \left({y - z}\right) \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right)$
+where $\left\vert{\, \cdot \,}\right\vert$ denotes [[Definition:Determinant of Matrix|determinant]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \begin{vmatrix}
+\left({x + q_2}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + q_3}\right) & \left({x + p_1}\right) \left({x + p_2}\right) \\
+\left({y + q_2}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + q_3}\right) & \left({y + p_1}\right) \left({y + p_2}\right) \\
+\left({z + q_2}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + q_3}\right) & \left({z + p_1}\right) \left({z + p_2}\right)
+\end{vmatrix}
+      | c = 
+}}
+{{eqn | r = \begin{vmatrix}
+\left({x + q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_2}\right) \left({x + q_3}\right) & \left({p_1 - q_3}\right) \left({x + p_2}\right) \\
+\left({y + q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_2}\right) \left({y + q_3}\right) & \left({p_1 - q_3}\right) \left({y + p_2}\right) \\
+\left({z + q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_2}\right) \left({z + q_3}\right) & \left({p_1 - q_3}\right) \left({z + p_2}\right)
+\end{vmatrix}
+      | c = [[Multiple of Row Added to Row of Determinant]]
+}}
+{{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix}
+\left({x + q_2}\right) \left({x + q_3}\right) & x + q_3 & x + p_2 \\
+\left({y + q_2}\right) \left({y + q_3}\right) & y + q_3 & y + p_2 \\
+\left({z + q_2}\right) \left({z + q_3}\right) & z + q_3 & z + p_2
+\end{vmatrix}
+      | c = [[Determinant with Row Multiplied by Constant]]
+}}
+{{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \begin{vmatrix}
+x \left({x + q_3}\right) & x + q_3 & p_2 - q_3 \\
+y \left({y + q_3}\right) & y + q_3 & p_2 - q_3 \\
+z \left({z + q_3}\right) & z + q_3 & p_2 - q_3
+\end{vmatrix}
+      | c = [[Multiple of Row Added to Row of Determinant]]
+}}
+{{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix}
+x \left({x + q_3}\right) & x + q_3 & 1\\
+y \left({y + q_3}\right) & y + q_3 & 1\\
+z \left({z + q_3}\right) & z + q_3 & 1
+\end{vmatrix}
+      | c = [[Determinant with Row Multiplied by Constant]]
+}}
+{{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \begin{vmatrix}
+x^2 & x & 1\\
+y^2 & y & 1\\
+z^2 & z & 1
+\end{vmatrix}
+      | c = [[Multiple of Row Added to Row of Determinant]]
+}}
+{{eqn | r = \left({p_1 - q_2}\right) \left({p_1 - q_3}\right) \left({p_2 - q_3}\right) \left({x - y}\right) \left({y - z}\right) \left({x - z}\right)
+      | c = [[Vandermonde Determinant]]
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Christian Friedrich Krattenthaler|cat = Krattenthaler}}
+\end{proof}<|endoftext|>
+\section{Floor Function/Examples/Floor of Root 2}
+Tags: Examples of Floor Function
+
+\begin{theorem}
+:$\floor {\sqrt 2} = 1$
+\end{theorem}
+
+\begin{proof}
+The [[Square Root of 2|decimal expansion of $\sqrt 2$]] is:
+:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
+Thus:
+:$1 \le \sqrt 2 < 2$
+Hence $1$ is the [[Definition:Floor Function|floor]] of $\sqrt 2$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling Function/Examples/Ceiling of Root 2}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\ceiling {\sqrt 2} = 2$
+\end{theorem}
+
+\begin{proof}
+The [[Square Root of 2|decimal expansion of $\sqrt 2$]] is:
+:$\sqrt 2 \approx 1.41421 \ 35623 \ 73095 \ 0488 \ldots$
+Thus:
+:$1 < \sqrt 2 \le 2$
+Hence $2$ is the [[Definition:Ceiling Function|ceiling]] of $\sqrt 2$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor Function/Examples/Floor of One Half}
+Tags: Examples of Floor Function
+
+\begin{theorem}
+:$\floor {\dfrac 1 2} = 0$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$0 \le \dfrac 1 2$
+and:
+:$1 = \dfrac 2 2 > \dfrac 1 2$
+Hence $0$ is the [[Definition:Floor Function|floor]] of $\dfrac 1 2$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling Function/Examples/Ceiling of Minus One Half}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\ceiling {- \dfrac 1 2} = 0$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$0 \ge - \dfrac 1 2$
+and:
+:$-1 = - \dfrac 2 2 < - \dfrac 1 2$
+Hence $0$ is the [[Definition:Ceiling Function|ceiling]] of $- \dfrac 1 2$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor Function/Examples/Floor of Minus One Half}
+Tags: Examples of Floor Function
+
+\begin{theorem}
+:$\floor {-\dfrac 1 2} = -1$
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$-1 = -\dfrac 2 2 \le -\dfrac 1 2$
+and:
+:$0 > -\dfrac 1 2$
+Hence $-1$ is the [[Definition:Floor Function|floor]] of $-\dfrac 1 2$ by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infima of two Real Sets}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]] of [[Definition:Real Number|real numbers]].
+Let $S$ and $T$ admit [[Definition:Infimum of Subset of Real Numbers|infima]].
+Then:
+: $\inf S \ge \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$
+\end{theorem}
+
+\begin{proof}
+Let:
+: $-S = \left\{{-s: s \in S}\right\}$
+: $-T = \left\{{-t: t \in T}\right\}$
+Observe that:
+: $s \in S \iff -s \in -S$
+: $t \in T \iff -t \in -T$
+We know that $\inf S$ and $\inf T$ exist.
+The [[Definition:Expression|expression]] $\inf S \ge \inf T$ exists as $\inf S$ and $\inf T$ exist.
+In other words, for fixed [[Definition:Set|sets]] $S$ and $T$, $\inf S \ge \inf T$ is either [[Definition:True|true]] or [[Definition:False|false]].
+We find:
+{{begin-eqn}}
+{{eqn | l = \inf S
+      | o = \ge
+      | r = \inf T
+}}
+{{eqn | ll= \iff
+      | l = -\sup -S
+      | o = \ge
+      | r = -\sup -T
+      | c = by the lemma
+}}
+{{eqn | ll= \iff
+      | l = \sup -S
+      | o = \le
+      | r = \sup -T
+}}
+{{eqn | ll= \iff
+      | l = \forall \epsilon \in \R_{>0}: \forall x \in -S: \exists y \in -T: x
+      | o = <
+      | r = y + \epsilon
+      | c = [[Suprema of two Real Sets]]
+}}
+{{eqn | ll= \iff
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \left({x: = -s}\right): \exists t \in T: \left({y: = -t}\right): x
+      | o = <
+      | r = y + \epsilon
+      | c = as $s \in S \iff -s \in −S$ and $t \in T \iff -t \in −T$
+}}
+{{eqn | ll= \iff
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: -s
+      | o = <
+      | r = -t + \epsilon
+}}
+{{eqn | ll= \iff
+      | l = \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon
+      | o = >
+      | r = t
+}}
+{{end-eqn}}
+=== Lemma ===
+Let $X$ be a [[Definition:Set|set]] of [[Definition:Real Number|real numbers]].
+Let $X$ admit an [[Definition:Infimum of Subset of Real Numbers|infimum]].
+Let $-X = \left\{{-x: x \in X}\right\}$.
+Then:
+: $ \sup -X = -\inf X$
+=== Proof ===
+Because $X$ admits an [[Definition:Infimum of Subset of Real Numbers|infimum]], it follows that it is [[Definition:Non-Empty Set|not empty]].
+The result follows by [[Negative of Infimum is Supremum of Negatives]].
+{{qed|lemma}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor Function is Integer}
+Tags: Floor Function
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+Then the [[Definition:Floor Function|floor function]] of $x$ is an [[Definition:Integer|integer]]:
+:$\floor x \in \Z$
+\end{theorem}
+
+\begin{proof}
+This is by definition of the [[Definition:Floor Function|floor function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor of Number plus Integer}
+Tags: Floor Function
+
+\begin{theorem}
+:$\forall n \in \Z: \floor x + n = \floor {x + n}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \floor {x + n}
+      | o = \le
+      | m = x + n 
+      | mo= <
+      | r = \floor {x + n} + 1
+      | c = {{Defof|Floor Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \floor {x + n} - n
+      | o = \le
+      | m = x
+      | mo= <
+      | r = \floor {x + n} - n + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \floor x
+      | r = \floor {x + n} - n
+      | c = {{Defof|Floor Function}}
+}}
+{{eqn | ll= \leadsto
+      | l = \floor {x + n}
+      | r = \floor x + n
+      | c = adding $n$ to both sides
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor is between Number and One Less}
+Tags: Floor Function
+
+\begin{theorem}
+:$x - 1 < \left\lfloor{x}\right\rfloor \le x$
+where $\left\lfloor{x}\right\rfloor$ is the [[Definition:Floor Function|floor]] of $x$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Floor Function|floor function]]:
+:$\left\lfloor{x}\right\rfloor \le x < \left\lfloor{x}\right\rfloor + 1$
+Thus by subtracting $1$:
+:$x - 1 < \left({\left\lfloor{x}\right\rfloor + 1}\right) - 1 = \left\lfloor{x}\right\rfloor$
+So:
+:$\left\lfloor{x}\right\rfloor \le x$
+and:
+:$x - 1 < \left\lfloor{x}\right\rfloor$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Floor Function}
+Tags: Floor Function
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+{{TFAE|def = Floor Function}}
+\end{theorem}
+
+\begin{proof}
+=== Definition 1 equals Definition 2 ===
+Follows from [[Supremum of Set of Integers equals Greatest Element]].
+{{qed|lemma}}
+=== Definition 1 equals Definition 3 ===
+Let $S$ be the [[Definition:Set|set]]:
+:$S = \set {m \in \Z: m \le x}$
+Let $n = \sup S$.
+By [[Supremum of Set of Integers is Integer]], $n \in \Z$.
+By [[Supremum of Set of Integers equals Greatest Element]], $n\in S$.
+Because $n \in S$, we have $n \le x$.
+Because $n + 1 > n$, we have by definition of [[Definition:Supremum of Subset of Real Numbers|supremum]]:
+:$n + 1 \notin S$
+Thus $n + 1 > x$.
+Thus $n$ is an [[Definition:Integer|integer]] such that:
+:$n \le x < n + 1$
+So $n$ is the [[Definition:Floor Function/Definition 3|floor function by definition 3]].
+{{qed|lemma}}
+=== Definition 3 equals Definition 2 ===
+Let $n$ be an [[Definition:Integer|integer]] such that:
+:$n \le x < n + 1$
+We show that $n$ is the [[Definition:Greatest Element|greatest element]] of the [[Definition:Set|set]]:
+:$S = \set {m \in \Z: m \le x}$
+Let $m \in \Z$ such that $m \le x$.
+We show that $n \ge m$.
+{{AimForCont}} $m > n$.
+By [[Weak Inequality of Integers iff Strict Inequality with Integer plus One]]:
+:$m \ge n + 1$
+and so from the definition of $g$ it follows that $m > x$.
+By [[Proof by Contradiction]] it follows that $m \le n$.
+Because $m \in S$ was arbitrary, $n$ is the [[Definition:Greatest Element|greatest element]] of $S$.
+Thus $n$ is the [[Definition:Floor Function/Definition 2|floor function by definition 2]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Floor Function}
+Tags: Floor Function
+
+\begin{theorem}
+This page gathers together some basic propeties of the [[Definition:Floor Function|floor function]].
+\end{theorem}<|endoftext|>
+\section{Properties of Ceiling Function}
+Tags: Ceiling Function
+
+\begin{theorem}
+This page gathers together some basic propeties of the [[Definition:Ceiling Function|ceiling function]].
+\end{theorem}<|endoftext|>
+\section{Ceiling Function is Integer}
+Tags: Ceiling Function
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+Then the [[Definition:Ceiling Function|ceiling function]] of $x$ is an [[Definition:Integer|integer]]:
+:$\ceiling x \in \Z$
+\end{theorem}
+
+\begin{proof}
+This is by definition of the [[Definition:Ceiling Function|ceiling function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number is Integer iff equals Ceiling}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$x = \ceiling x \iff x \in \Z$
+\end{theorem}
+
+\begin{proof}
+Let $x = \ceiling x$.
+As $\ceiling x \in \Z$, then so must $x$ be.
+Now let $x \in \Z$.
+We have:
+:$\ceiling x = \inf \set {m \in \Z: m \ge x}$
+As $x \in \inf \set {m \in \Z: m \ge x}$, and there can be no lesser $n \in \Z$ such that $n \in \inf \set {m \in \Z: m \ge x}$, it follows that:
+:$x = \ceiling x$
+{{qed}}
+[[Category:Ceiling Function]]
+3987x4r1nfwbzj69zb29b5k9c9szjdv
+\end{proof}<|endoftext|>
+\section{Ceiling Function is Idempotent}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\ceiling {\ceiling x} = \ceiling x$
+\end{theorem}
+
+\begin{proof}
+Let $y = \ceiling x$.
+By [[Ceiling Function is Integer]], $y$ is an [[Definition:Integer|integer]].
+Then from [[Real Number is Integer iff equals Ceiling]]:
+:$\ceiling y = y$
+So:
+:$\ceiling {\ceiling x} = \ceiling x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Way Below Relation is Auxiliary Relation}
+Tags: Auxiliary Relations, Way Below Relation
+
+\begin{theorem}
+Lrt $L = \left({S, \vee, \preceq}\right)$ be a [[Definition:Bounded Below|bounded below]] [[Definition:Join Semilattice|join semilattice]].
+Then
+:$\ll$ is [[Definition:Auxiliary Relation|auxiliary relation]]
+where $\ll$ denotes the [[Definition:Element is Way Below|way below relation]].
+\end{theorem}
+
+\begin{proof}
+By [[Way Below implies Preceding]]:
+:$\forall x, y \in S: x \ll y \implies x \preceq y$
+By [[Preceding and Way Below implies Way Below]]:
+:$\forall x, y, z, u \in S: x \preceq y \ll z \preceq u \implies x \ll u$
+By [[Join is Way Below if Operands are Way Below]]:
+:$\forall x, y, z \in S: x \ll z \land y \ll z \implies x \vee y \ll z$
+By [[Bottom is Way Below Any Element]]:
+:$\forall x: \bot \ll x$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
+Thus by definition:
+:$\ll$ is [[Definition:Auxiliary Relation|auxiliary relation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential Sequence is Eventually Strictly Positive}
+Tags: Exponential Function
+
+\begin{theorem}
+Let $\sequence {E_n}$ be the [[Definition:Sequence|sequence]] of [[Definition:Real Function|real functions]] $E_n: \R \to \R$ defined as:
+:$\map {E_n} x = \paren {1 + \dfrac x n}^n$
+Then, for each $x \in \R$ and for sufficiently large $n \in \N$, $\map {E_n} x$ is [[Definition:Positive Real Number|positive]].
+That is:
+:$\forall x \in \R: \forall n \in \N: n \ge \ceiling {\size x} \implies \map {E_n} x > 0$
+where $\ceiling x$ denotes the [[Definition: Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Fix $x \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n
+      | o = \ge
+      | r = \ceiling {\size x}
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = >
+      | r = -x
+      | c = [[Real Number is between Ceiling Functions]] and [[Negative of Absolute Value]]
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | o = >
+      | r = \frac {-x} n
+      | c = dividing both sides by $n$
+}}
+{{eqn | ll= \leadsto
+      | l = 1 + \frac x n
+      | o = >
+      | r = 0
+      | c = adding $\dfrac {-x} n$ to both sides
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {1 + \frac x n}^n
+      | o = >
+      | r = 0
+      | c = [[Power of Strictly Positive Real Number is Strictly Positive/Positive Integer|Power of Strictly Positive Real Number is Strictly Positive: Positive Integer]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Exponential Function]]
+2f98hvb2ixed1gcy50znmnpjs04sny9
+\end{proof}<|endoftext|>
+\section{Number is between Ceiling and One Less}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\ceiling x - 1 < x \le \ceiling x$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Ceiling Function|ceiling]] of $x$:
+:$\forall x \in \R: \ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
+By definition of [[Definition:Infimum of Subset of Real Numbers|infimum]]:
+:$\ceiling x \ge x$
+Also by definition of [[Definition:Infimum of Subset of Real Numbers|infimum]]:
+:$\ceiling x - 1 \not \ge x$
+as $\ceiling x$ is the smallest [[Definition:Integer|integer]] with that property.
+That is:
+:$x > \ceiling x - 1$
+Hence the result.
+{{qed}}
+[[Category:Ceiling Function]]
+l75iq9rcega0z865kapwqgmw7httq2r
+\end{proof}<|endoftext|>
+\section{Ceiling is between Number and One More}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$x \le \ceiling x < x + 1$
+where $\ceiling x$ is the [[Definition:Ceiling Function|ceiling]] of $x$.
+\end{theorem}
+
+\begin{proof}
+From [[Number is between Ceiling and One Less]]:
+:$\ceiling x - 1 < x \le \ceiling x$
+Thus by adding $1$:
+:$x + 1 > \paren {\ceiling x - 1} + 1 = \ceiling x$
+So:
+:$x \le \ceiling x$
+and:
+:$\ceiling x < x + 1$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling of Number plus Integer}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\forall n \in \Z: \left \lceil {x} \right \rceil + n = \left \lceil {x + n} \right \rceil$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left \lceil {x + n} \right \rceil - 1
+      | o = <
+      | m = x + n 
+      | mo= \le
+      | r = \left \lceil {x + n} \right \rceil
+      | c = [[Number is between Ceiling and One Less]]
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x + n} \right \rceil - n - 1
+      | o = <
+      | m = x
+      | mo= \le
+      | r = \left \lceil {x + n} \right \rceil - n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x} \right \rceil
+      | r = \left \lceil {x + n} \right \rceil - n
+      | c = [[Number is between Ceiling and One Less]]
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x + n} \right \rceil
+      | r = \left \lceil {x} \right \rceil + n
+      | c = adding $n$ to both sides
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Relation Segment of Auxiliary Relation is Ideal}
+Tags: Auxiliary Relations
+
+\begin{theorem}
+Let $L = \left({S, \vee, \preceq}\right)$ 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
+:$x^R$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$
+where $x^R$ denotes the $R$[[Definition:Relation Segment|-segment]] of $x$.
+\end{theorem}
+
+\begin{proof}
+=== Non-empty ===
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({\bot, x}\right) \in R$
+where $\bot$ denotes the [[Definition:Smallest Element|smallest element]] in $L$.
+By definition of [[Definition:Relation Segment|relation segment]]:
+:$\bot \in x^R$
+Thus by definition:
+:$x^R$ is [[Definition:Non-Empty Set|non-empty]].
+{{qed|lemma}}
+=== Directed ===
+Let $y, z \in x^R$.
+By definition of [[Definition:Relation Segment|relation segment]]:
+:$\left({y, x}\right) \in R$ and $\left({z, x}\right) \in R$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({y \vee z, x}\right) \in R$
+By definition of [[Definition:Relation Segment|relation segment]]:
+:$y \vee z \in x^R$
+Thus by [[Join Succeeds Operands]]:
+:$y \preceq y \vee z$ and $z \preceq y \vee z$
+Thus by definition:
+:$x^R$ is [[Definition:Directed Subset|directed]].
+{{qed|lemma}}
+=== Lower ===
+Let $y \in x^R, z \in S$ such that
+:$z \preceq y$
+By definition of [[Definition:Relation Segment|relation segment]]:
+:$\left({y, x}\right) \in R$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$x \preceq x$
+By definition of [[Definition:Auxiliary Relation|auxiliary relation]]:
+:$\left({z, x}\right) \in R$
+Thus by definition of [[Definition:Relation Segment|relation segment]]:
+:$z \in x^R$
+Thus by definition:
+:$x^R$ is [[Definition:Lower Set|lower set]].
+{{qed|lemma}}
+Thus by definition:
+:$x^R$ is [[Definition:Ideal in Ordered Set|ideal]] in $L$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Modulo Operation with 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 $y \ne 0$.
+Then:
+:$0 \le \dfrac x y - \left \lfloor {\dfrac x y}\right \rfloor = \dfrac {x \bmod y} y < 1$
+\end{theorem}
+
+\begin{proof}
+From [[Real Number minus Floor]]:
+:$\dfrac x y - \left \lfloor {\dfrac x y}\right \rfloor \in \left[{0 \,.\,.\, 1}\right)$
+Thus by definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
+:$0 \le \dfrac x y - \left \lfloor {\dfrac x y}\right \rfloor < 1$
+Then:
+{{begin-eqn}}
+{{eqn | l = x \bmod y
+      | r = x - y \left \lfloor {\frac x y}\right \rfloor
+      | c = Definition of [[Definition:Modulo Operation|Modulo Operation]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x \bmod y} y
+      | r = \frac {x - y \left \lfloor {\dfrac x y}\right \rfloor} y
+      | c = 
+}}
+{{eqn | r = \frac x y - \frac y y \left \lfloor {\dfrac x y}\right \rfloor
+      | c = 
+}}
+{{eqn | r = \frac x y - \left \lfloor {\dfrac x y}\right \rfloor
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}
\ No newline at end of file