hcju/proofwikiqa huggingface dataset:
=== Definition 1 implies Definition 2 === Let $H_n$ be a hexagonal number by definition 1. Let $n = 0$. By definition: :$H_0 = 0$ By vacuous summation: :$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \left({4 \left({i - 1}\right) + 1}\right) = 0$ By definition of summation: {{begin-eqn}} {{eqn | l = H_{n-1}       | r = \sum_{i \mathop = 1}^{n - 1} \left({4 \left({i - 1}\right) + 1}\right)       | c =  }} {{eqn | r = 1 + 5 + \cdots + 4 \left({n - 2}\right) + 1       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = H_n       | r = H_{n-1} + 4 \left({n - 1}\right) + 1       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \left({4 \left({i - 1}\right) + 1}\right)       | c =  }} {{eqn | r = 1 + 5 + \cdots + 4 \left({n - 1}\right) + 1       | c =  }} {{end-eqn}} Thus $H_n$ is a hexagonal number by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $H_n$ be a hexagonal number by definition 2. Then: {{begin-eqn}} {{eqn | l = H_n       | r = \sum_{i \mathop = 1}^n \left({4 \left({i - 1}\right) + 1}\right)       | c =  }} {{eqn | r = 1 + 5 + \cdots + \left({4 \left({n - 2}\right) + 1}\right) + \left({4 \left({n - 1}\right) + 1}\right)       | c =  }} {{eqn | r = H_{n-1} + 4 \left({n - 1}\right) + 1       | c =  }} {{end-eqn}} Then: :$\displaystyle H_0 = \sum_{i \mathop = 1}^0 \left({4 \left({i - 1}\right) + 1}\right)$ is a vacuous summation and so: :$H_0 = 0$ Thus $H_n$ is a hexagonal number by definition 1. {{qed|lemma}} === Definition 1 equivalent to Definition 3 === We have by definition that $H_0 = 0 = P \left({6, 0}\right)$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = P \left({6, n}\right)       | r = P \left({6, n - 1}\right) + \left({6 - 2}\right) \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({6, n - 1}\right) + 4 \left({n - 1}\right) + 1       | c =  }} {{end-eqn}} Thus $P \left({6, n}\right)$ and $H_n$ are generated by the same recurrence relation. {{qed}} Category:Hexagonal Numbers 4z6nw8koag388772jwqimoofsui33jd
=== Definition 1 equivalent to Definition 3 === By the Odd Number Theorem: :$\displaystyle \sum_{j \mathop = 1}^n \left({2 j - 1}\right) = n^2$ {{qed|lemma}} === Definition 2 equivalent to Definition 3 === By the Corollary to the Odd Number Theorem: :$S_n = \displaystyle \sum_{j \mathop = 1}^{n - 1} + 2 n - 1$ and so by Definition 2: :$\displaystyle \sum_{j \mathop = 1}^n \left({2 j - 1}\right) = S_{n-1} + 2 n - 1$ {{qed|lemma}} === Definition 2 equivalent to Definition 4 === We have by definition that $S_n = 0 = P \left({4, n}\right)$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = P \left({4, n}\right)       | r = P \left({4, n - 1}\right) + \left({4 - 2}\right) \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({4, n - 1}\right) + 2 \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({4, n - 1}\right) + 2 n - 1       | c =  }} {{end-eqn}} Thus $P \left({4, n}\right)$ and $S_n$ are generated by the same recurrence relation. {{qed}} Category:Square Numbers cpzbpkty93fld7jqtuvk4ki71xftns8
=== $(1)$ {{iff}} $(2)$ === From the Division Theorem, setting $b = 2$: :$\forall a \in \Z: \exists! q, r \in \Z: a = 2 q + r, 0 \le r < 2$ Thus, either: :$\exists q \in \Z: n = 2 q$ or: :$\exists q \in \Z: n = 2 q + 1$ When $n = 2 q$, $n$ is even by definition. When $n$ is not even, $n = 2 q + 1$ Likewise, when $n = 2 q + 1$ it follows that $n$ is not even. Hence both definitions of odd integer are equivalent. {{qed|lemma}} === $(2)$ {{iff}} $(3)$ === By definition of congruence modulo $2$: :$x \equiv y \pmod 2 \iff \exists r \in \Z: x - y = 2 r$ Setting $y = 1$: :$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x - y = 2 r$ from which: :$x \equiv 1 \pmod 2 \iff \exists r \in \Z: x = 2 r + 1$ Thus definition 2 is logically equivalent to definition 3. {{qed}}
Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ even numbers. By definition of even number, this can be expressed as: :$S = \set {2 s_1, 2 s_2, \ldots, 2 s_n}$ where: :$\forall k \in \closedint 1 n: r_k = 2 s_k$ Then: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n r_k       | r = \sum_{k \mathop = 1}^n 2 s_k       | c =  }} {{eqn | r = 2 \sum_{k \mathop = 1}^n s_k       | c =  }} {{end-eqn}} Thus, by definition, $\displaystyle \sum_{k \mathop = 1}^n r_k$ is even. {{qed}}
Proof by induction: For all $n \in \N$, let $P \left({n}\right)$ be the proposition: :The sum of $n$ even integers is an even integer. $P(1)$ is trivially true, as this just says: :The sum of $1$ even integers is an even integer. The sum of $0$ even integers is understood, from the definition of a vacuous summation, to be $0$, which is even. So $P(0)$ is also true. === Basis for the Induction === $P(2)$ is the case: :The sum of any two even integers is itself even. Consider two even integers $x$ and $y$. Since they are even, they can be written as $x = 2a$ and $y = 2b$ respectively for integers $a$ and $b$. Therefore, the sum is: :$x + y = 2 a + 2 b = 2 \left({a + b}\right)$ Thus $x + y$ has $2$ as a divisor and is even by definition. This is our basis for the induction. {{qed|lemma}} === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :The sum of any $k$ even integers is itself even. Then we need to show: :The sum of any $k+1$ even integers is itself even. === Induction Step === This is our induction step: Consider the sum of any $k + 1$ even integers. This is the sum of: : $k$ even integers (which is even by the induction hypothesis) and: : another even integer. That is, it is the sum of two even integers. By the basis for the induction, this is also even. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. That is: :The sum of any finite number of even integers is itself even. {{qed}}
=== Sufficient Condition === Let $\tuple {g, d}$ be a Galois connection. Thus by definition of Galois connection: :$d$ is an increasing mapping. Let $s \in S$. By definition of reflexivity: :$\map g s \preceq \map g s$ By definition of Galois connection: :$\map d {\map g s} \preceq s$ By definition of lower closure: :$\map d {\map g s} \in s^\preceq$ By definition of image of set under relation: :$\map g s \in d^{-1} \sqbrk {s^\preceq}$ By definition of upper bound: :$\forall t \in T: t$ is an upper bound for $d^{-1} \sqbrk {s^\preceq} \implies \map g s \precsim t$ We will prove that: :$\map g s$ is an upper bound for $d^{-1} \sqbrk {s^\preceq}$ Let $t \in d^{-1} \sqbrk {s^\preceq}$. By definition of image of set: :$\map d t \in s^\preceq$ By definition of lower closure of element: :$\map d t \preceq s$ Thus by definition of Galois connection: :$t \precsim \map g s$ By definition of supremum: :$d^{-1} \sqbrk {s^\preceq}$ admits a supremum and :$\map \sup {d^{-1} \sqbrk {s^\preceq} } = \map g s$ Thus: :$\map g s = \map \min {d^{-1} \sqbrk {s^\preceq} }$ {{qed|lemma}} === Necessary Condition === Let $d: T \to S$ be an increasing mapping. Let: :$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$ We will prove that: :$g$ is an increasing mapping. Let $x, y \in S$ such that: :$x \preceq y$ By Lower Closure is Increasing: :$x^\preceq \subseteq y^\preceq$ By Image of Subset under Relation is Subset of Image/Corollary 3: :$d^{-1} \sqbrk {x^\preceq} \subseteq d^{-1} \sqbrk {y^\preceq}$ By assumption: :$\map g s = \map \max {d^{-1} \sqbrk {x^\preceq} } = \map \sup {d^{-1} \sqbrk {x^\preceq} }$ and: :$\map g y = \map \max {d^{-1} \sqbrk {y^\preceq} } = \map \sup {d^{-1} \sqbrk {y^\preceq} }$ By Supremum of Subset: :$\map g s \preceq \map g y$ Thus by definition: :$g$ is an increasing mapping. Thus: :$d$ is an increasing mapping. We will prove that: :$\forall s \in S, t \in T:  t \precsim \map g s \iff \map d t \preceq s$ Let $s \in S, t \in T$. Second implication: Let $\map d t \preceq s$. By definition of lower closure of element: :$\map d t \in s^\preceq$ By definition of image of set: :$t \in d^{-1} \sqbrk {s^\preceq}$ By assumption: :$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} } = \map \sup {d^{-1} \sqbrk {s^\preceq} }$ By definition of supremum: :$\map g s$ is an upper bound for $d^{-1} \sqbrk {s^\preceq}$ Thus by definition of upper bound: :$t \precsim \map g s$ First implication: Let $t \precsim \map g s$ By assumption: :$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$ By greatest element of set: :$\map g s \in d^{-1} \sqbrk {s^\preceq}$ By definition of image of set: :$\map d {\map g s} \in s^\preceq$ By definition of lower closure of element: :$\map d {\map g s} \preceq s$ By definition of increasing mapping: :$\map d t \preceq \map d {\map g s}$ Thus by definition of transitivity: :$\map d t \preceq s$ Thus by definition: :$\tuple {g, d}$ is a Galois connection. {{qed}}
By definition, the regular heptagon has $7$ sides. $7$ is a prime number which is not a fermat prime. The result follows Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime. {{qed}}
=== Necessary Condition === Let $n^{\size m} \notin \Z$. We have that: :$n^m = n^{u/v}$ for some $u \in\Z$ and $v \in \Z_{\ne 0}$ Then it follows from the definition of a rational power and the existence of a real $v$th root of $n^u$ that: :$n^m = n^{u/v} = \paren {n^u}^{1/v} \in \R$ {{AimForCont}} $n^m \in \Q$. We have that: :$\size m = 0 \implies n^{\size m} \in \Z$ It follows that: :$\size m > 0$ Then: :$\size m > 0$ :$m \in \Q$ :$n^{\size m} > 0$ :$n^m \in \Q$ imply that :$n^{\size m} = n^{p/q} = \dfrac r s$ for some $\tuple {p, q, r, s} \in \Z_{>0}^4$ where $r$ and $s$ have no common prime factors. Raising both sides of $n^{p/q} = \dfrac r s$ to the power of $q$ yields: :$n^p = \dfrac {r^q} {s^q}$ We have that: :$\dfrac {r^q} {s^q} = n^p \in \Z$ Hence $r^q$ is divisible by $s^q$. By the Fundamental Theorem of Arithmetic, $r^q$ and $s^q$ have the same prime factors as $r$ and $s$ respectively. But we have that: :$s^q \divides r^q$ and: :$s \ne 1$ imply that $r$ and $s$ have a common prime factor. But this would contradict the fact that $r$ and $s$ have no common prime factors. Therefore $s = 1$. Then: :$n^{\size m} = \dfrac r s$ and: :$s = 1$ imply that: :$r = n^{\size m} \notin \Z$ From this contradiction it follows that: :$n^m \notin \Q$ and since $n^m \in \R$: :$n^m \in \R \setminus \Q$ That is, $n^m$ is an irrational number. {{qed|lemma}} === Sufficient Condition === Now let $n^m$ be an irrational number. That is: :$n^m \in \R \setminus \Q$ Then because $n^{-m}$ is the reciprocal of $n^m$: :$n^{-m} \in \R \setminus \Q$ So: :$n^{\size m} \in \R \setminus \Q$ and so since $\Z \subseteq \Q$: :$n^{\size m} \notin \Z$ {{qed}} Category:Number Theory Category:Irrational Numbers na3dwtd25a4dmxxztwcd54gbp631ts3
{{begin-eqn}} {{eqn | l = \mathcal A       | r = \int_0^{2 \pi} \frac {\left({a \theta}\right)^2} 2 \, \mathrm d \theta       | c = Area between Radii and Curve in Polar Coordinates }} {{eqn | r = \left[{\frac {a^2 \theta^3} 6}\right]_0^{2 \pi}       | c = Primitive of Power }} {{eqn | r = \frac {\left({2 \pi}\right)^3 a^2 \theta^3} 6       | c =  }} {{eqn | r = \frac {4 \pi^3 a^2} 3       | c = after simplification }} {{end-eqn}} {{qed}}
Let the equation of the Archimedean spiral be $r = a \theta$. Then: {{begin-eqn}} {{eqn | l = \angle DOB       | r = \frac {OD} a }} {{eqn | r = \frac {OA/3} a }} {{eqn | r = \frac 1 3 \angle AOB }} {{end-eqn}} {{qed}}
By definition of upper adjoint :$\exists d: T \to S: \left({g, d}\right)$ is a Galois connection Let $X$ be a subset of $S$ such that :$X$ admits an infimum. We will prove as lemma 1 that :$\forall t \in T: t$ is lower bound for $g^\to\left({X}\right) \implies t \precsim g\left({\inf X}\right)$ Let $t \in T$ such that :$t$ is lower bound for $g^\to\left({X}\right)$ We will prove as sublemma that :$d\left({t}\right)$ is lower bound for $X$ Let $s \in X$. By definition of image of set: :$g\left({s})\right) \in g^\to\left({X}\right)$ By definition of lower bound: :$t \precsim g\left({s}\right)$ Thus by definition of Galois connection: :$d\left({t}\right) \preceq s$ This ends the proof of sublemma. By definition of infimum: :$d\left({t}\right) \preceq \inf X$ Thus by definition of 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 lower bound for $g^\to\left({X}\right)$ Let $t \in g^\to\left({X}\right)$. By definition of image of set: :$\exists s \in S: s \in X \land g\left({s}\right) = t$ By definition of infimum: :$\inf X$ is lower bound for $X$ By definition of lower bound: :$\inf X \preceq s$ By definition of Galois connection: :$g$ is increasing mapping. Thus by definition of increasing mapping: :$g\left({\inf X}\right) \precsim t$ This ends the proof of lemma 2. Thus by definition of infimum: :$g^\to\left({X}\right)$ admits an infimum and :$\inf\left({g^\to\left({X}\right)}\right) = g\left({\inf X}\right)$ Thus by definition: :$g$ preserves infimum on $X$ Thus by definition: :$g$ preserves all infima. {{qed}}
By definition of lower adjoint :$\exists g: S \to T: \left({g, d}\right)$ is a Galois connection Let $X$ be a subset of $T$ such that :$X$ admits a supremum. We will prove as lemma 1 that :$\forall s \in S: s$ is upper bound for $d^\to\left({X}\right) \implies d\left({\sup X}\right) \preceq s$ Let $s \in S$ such that :$s$ is upper bound for $d^\to\left({X}\right)$ We will prove as sublemma that :$g\left({s}\right)$ is upper bound for $X$ Let $t \in X$. By definition of image of set: :$d \left({t})\right) \in d^\to \left({X}\right)$ By definition of upper bound: :$d\left({t}\right) \preceq s$ Thus by definition of Galois connection: :$t \precsim g\left({s}\right)$ This ends the proof of sublemma. By definition of supremum: :$\sup X \precsim g\left({t}\right)$ Thus by definition of 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 upper bound for $d^\to\left({X}\right)$ Let $s \in d^\to\left({X}\right)$. By definition of image of set: :$\exists t \in T: t \in X \land d\left({t}\right) = s$ By definition of supremum: :$\sup X$ is upper bound for $X$ By definition of upper bound: :$t \precsim \sup X$ By definition of Galois connection: :$d$ is increasing mapping. Thus by definition of increasing mapping: :$s \preceq d\left({\sup X}\right)$ This ends the proof of lemma 2. Thus by definition of supremum: :$d^\to\left({X}\right)$ admits a supremum and :$\sup\left({d^\to\left({X}\right)}\right) = d\left({\sup X}\right)$ Thus by definition: :$g$ preserves supremum on $X$ Thus by definition: :$g$ preserves all suprema. {{qed}}
{{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 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 $x$-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}}
Define a 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 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 mapping preserves all infima: :$g$ preserves the infimum on $x^\succeq$ and $g$ preserves the infimum on $y^\succeq$ By definition of 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 upper closure of element: :$\inf\left({g\left[{g^{-1}\left[{t^\succsim}\right]}\right]}\right) \in t^\succsim$ By definition of complete lattice: :$g^{-1}\left[{t^\succsim}\right]$ admits an infimum By definitions 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 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 Galois connection. Thus by lemma 2: :$\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$ {{qed}}
Let $x \in f \left[{f^{-1} \left[{X}\right]}\right]$. By definition of image of set: :$\exists y \in S: y \in f^{-1} \left[{X}\right] \land x = f \left({y}\right)$ By definition of image of set under relation: :$f \left({y}\right) \in X$ Thus $x \in X$ {{qed}}
Define a 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 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 mapping preserves all suprema: :$d$ preserves the supremum on $x^\precsim$ and $d$ preserves the supremum on $y^\precsim$ By definition of 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 lower closure of element: :$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } \in s^\preceq$ By definition of complete lattice: :$d^{-1} \sqbrk {s^\preceq}$ admits a supremum By definition of mapping preserves the supremum: :$\map \sup {d \sqbrk {d^{-1} \sqbrk {s^\preceq} } } = \map d {\map g s}$ Thus by definition of 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 Galois connection. Thus by lemma 2: :$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$ {{qed}}
By definition of point of inflection, $f'$ has either a local maximum or a local minimum at $\xi$. From Derivative at Maximum or Minimum, it follows that the derivative of $f'$ at $\xi$ is zero, that is: :$f'' \left({\xi}\right) = 0$ {{qed}}
Let $s \in S$. By definition of reflexivity: :$\map g s \precsim \map g s$ By definition of Galois connection: :$\map d {\map g s} \preceq s$ By definition of composition: :$\map {\paren {d \circ g} } s \preceq s$ By definition of identity mapping: :$\map {\paren {d \circ g} } s \preceq \map {I_S} s$ Thus by definition of order on mappings: :$d \circ g \preceq I_S$ Let $t \in T$. By definition of reflexivity: :$\map d t \preceq \map d t$ By definition of Galois connection: :$t \precsim \map g {\map d t}$ By definition of composition: :$t \precsim \map {\paren {g \circ d} } t$ By definition of identity mapping: :$\map {I_T} t \precsim \map {\paren {g \circ d} } t$ Thus by definition of order on mappings: :$I_T \precsim g \circ d$ {{qed}}
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 increasing mapping: :$d\left({t}\right) \preceq d\left({g\left({s}\right)}\right)$ By definition of ordering on mappings: :$\left({d \circ g}\right)\left({s}\right) \preceq I_S\left({s}\right)$ By definition of composition: :$d\left({g\left({s}\right)}\right) \preceq I_S\left({s}\right)$ By definition of identity mapping: :$d\left({g\left({s}\right)}\right) \preceq s$ Thus by definition of transitivity: :$d\left({t}\right) \preceq s$ Second implication: Let :$d\left({t}\right) \preceq s$ By definition of increasing mapping: :$g\left({d\left({t}\right)}\right) \precsim g\left({s}\right)$ By definition of ordering on mappings: :$I_T\left({t}\right) \precsim \left({g \circ d}\right)\left({t}\right)$ By definition of composition: :$I_T\left({t}\right) \precsim g\left({d\left({t}\right)}\right)$ By definition of identity mapping: :$t \precsim g \left({d\left({t}\right)}\right)$ Thus by definition of transitivity: :$t \precsim g \left({s}\right)$ Thus by definition: :$\left({g, d}\right)$ is Galois connection. {{qed}}
Proof by Contradiction: {{AimForCont}} $\pi$ is not transcendental. Hence by definition, $\pi$ is algebraic. Let $\pi$ be the root of a non-zero polynomial with rational coefficients, namely $\map f x$. Then, $\map g x := \map f {i x} \map f {-i x}$ is also a non-zero polynomial with rational coefficients such that: :$\map g {i \pi} = 0$ Hence, $i \pi$ is also algebraic. From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{i \pi}$ is transcendental. However, from Euler's Identity: :$e^{i \pi} = -1$ which is the root of $\map h z = z + 1$, and so is algebraic. This contradicts the conclusion that $e^{i \pi}$ is transcendental. Hence by Proof by Contradiction it must follow that $\pi$ is transcendental. {{qed}}