hcju/proofwikiqa huggingface dataset:
Let $z \in y^\succeq$. By definition of upper closure of element: :$y \preceq z$ By definition of ordering, $\preceq$ is transitive. From $x \preceq y$ and $y \preceq z$: :$x \preceq z$ Thus again by definition of upper closure of element: :$z \in x^\succeq$ {{qed}}
Let $z \in x^\preceq$. By definition of lower closure of element: :$z \preceq x$ By definition of transitivity: :$z \preceq y$ Thus again by definition of lower closure of element: :$z \in y^\preceq$ {{qed}}
Let $x, y \in S$ such that: :$x \preceq y$ By Supremum of Singleton: :$\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ admit suprema in $\left({S, \preceq}\right)$ By Supremum of Lower Closure of Set: :$\left\{ {x}\right\}^\preceq$ and $\left\{ {y}\right\}^\preceq$ admit suprema in $\left({S, \preceq}\right)$ where $\left\{ {x}\right\}^\preceq$ denotes the lower closure of $\left\{ {x}\right\}$ By Lower Closure of Singleton: :$x^\preceq$ and $y^\preceq$ admit suprema in $\left({S, \preceq}\right)$ By Lower Closure of Element is Ideal: :$x^\preceq$ and $y^\preceq$ are ideals in $\left({S, \preceq}\right)$ By assumption and definition of mapping preserves the supremum on subset: :$f^\to \left({x^\preceq}\right)$ and $f^\to \left({y^\preceq}\right)$ admit suprema in $\left({T, \precsim}\right)$ and :$\sup \left({f^\to \left({x^\preceq}\right)}\right) = f \left({\sup \left({x^\preceq}\right)}\right)$ and $\sup \left({f^\to \left({y^\preceq}\right)}\right) = f \left({\sup \left({y^\preceq}\right)}\right)$ By Supremum of Lower Closure of Element: :$\sup \left({x^\preceq}\right) = x$ and $\sup \left({y^\preceq}\right) = y$ By Lower Closure is Increasing: :$x^\preceq \subseteq y^\preceq$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$f^\to \left({x^\preceq}\right) \subseteq f^\to \left({y^\preceq}\right)$ Thus by Supremum of Subset: :$f\left({x}\right) \precsim f\left({y}\right)$ Thus by definition: :$f$ is increasing. {{qed}}
{{begin-eqn}} {{eqn | l = i^i       | r = \left({e^{i \pi / 2} }\right)^i       | c = Euler's Formula }} {{eqn | r = e^{\pi i^2 / 2}       | c =  }} {{eqn | r = e^{-\pi / 2}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = i^i       | r = \exp \paren {i \ln \paren i}       | c = {{Defof|Complex Power}} }} {{eqn | r = \exp \paren {i \paren {\ln 1 + i \paren {\dfrac \pi 2 + 2 k \pi} } }       | c = {{Defof|Complex Natural Logarithm}}: for all $k \in \Z$ }} {{eqn | r = \exp \paren {i \paren {i \paren {\dfrac \pi 2 + 2 k \pi} } }        | c = Logarithm of 1 is 0 }} {{eqn | r = \exp \paren {i^2 \dfrac \pi 2 + 2 k \pi i^2}       | c =  }} {{eqn | r = \exp \paren {-\dfrac \pi 2 - 2 k \pi}       | c =  }} {{eqn | r = \exp \paren {\dfrac {3 \pi} 2 + 2 k \pi}       | c = as $k$ ranges over all integers }} {{eqn | r = \exp \paren {\dfrac {4 k + 3} 2 \pi}       | c =  }} {{end-eqn}} {{qed}}
By Fourier Series of Fourth Power of x, for $x \in \closedint {-\pi} \pi$: :$\displaystyle x^4 = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map \cos {n \pi} \, \map \cos {n x}$ Setting $x = \pi$:  {{begin-eqn}} {{eqn | l = \pi^4        | r = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map {\cos^2} {n \pi} }} {{eqn | ll= \leadsto       | l = \frac {4 \pi^4} 5        | r = \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2} {n^4} - \sum_{n \mathop = 1}^\infty \frac {48} {n^4}       | c = Cosine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi^4} 5       | r = 2 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4} }} {{eqn | r = \frac {\pi^4} 3 - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}       | c = Basel Problem }} {{eqn | ll= \leadsto       | l = 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}        | r = \frac {\pi^4} 3 - \frac {\pi^4} 5       | c = rearranging }} {{eqn | r = \frac {2 \pi^4} {15} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^4}       | r = \frac {\pi^4} {90} }} {{end-eqn}} {{qed}}
By Fourier Series of x squared, for $x \in \left[{- \pi \,.\,.\, \pi}\right]$: :$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos n x}\right)$ Hence: {{begin-eqn}} {{eqn | l = \frac 1 \pi \int_{-\pi}^\pi x^4 \, \mathrm d x       | r = \frac 1 2 \left({\frac {2 \pi^2} 3}\right)^2 + \sum_{n \mathop = 1}^\infty \left({\frac {4 \left({-1}\right)^n} {n^2} }\right)^2       | c = Parseval's Theorem }} {{eqn | ll= \leadsto       | l = \frac 2 \pi \int_0^\pi x^4 \, \mathrm d x        | r = \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}        | c = Definite Integral of Even Function  }} {{eqn | ll= \leadsto       | l = \frac {2 \pi^4} 5        | r = \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4}  }} {{eqn | ll= \leadsto       | l = \frac {8 \pi^4} {45}        | r = \sum_{n \mathop = 1}^\infty \frac {16} {n^4} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^4}        | r = \frac {\pi^4} {90} }} {{end-eqn}} {{qed}}
{{ProofWanted|$\displaystyle \int_0^1\int_0^1\int_0^1\int_0^1 \frac 1 {1 - x_1x_2x_3x_4} \mathrm dx_1 \mathrm dx_2 \mathrm dx_3 \mathrm dx_4$. Use Beukers-Calabi-Kolk sub.}}
{{begin-eqn}} {{eqn | l = \map \zeta 4       | r = \paren {-1}^3 \dfrac {B_4 2^3 \pi^4} {4!}       | c = Riemann Zeta Function at Even Integers }} {{eqn | r = \paren {-1}^3 \paren {-\dfrac 1 {30} } \dfrac {2^3 \pi^4} {4!}       | c = {{Defof|Sequence of Bernoulli Numbers}} }} {{eqn | r = \paren {\dfrac 1 {30} } \paren {\dfrac 8 {24} } \pi^4       | c = {{Defof|Factorial}} }} {{eqn | r = \dfrac {\pi^4} {90}       | c = simplifying }} {{end-eqn}} {{qed}}
From the Basel Problem we have: :$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$ Therefore: {{begin-eqn}} {{eqn | l = \map \zeta 2       | r = \paren {\frac 1 {1^2 } + \frac 1 {2^2 } + \frac 1 {3^2 } + \frac 1 {4^2 } + \cdots }       | c =  }} {{eqn | r = \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }       | c =  }} {{end-eqn}} Taking the square of $\map \zeta 2$, we have: {{begin-eqn}} {{eqn | l = \displaystyle \paren {\map \zeta 2}^2       | r = \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots } \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }       | c = }} {{eqn | r = \paren {\sum_{i \mathop = 1}^{\infty} {\frac 1 {i^2 } } } \paren {\sum_{j \mathop = 1}^{\infty} {\frac 1 {j^2 } } }       | c = }} {{eqn | r = \sum_{i \mathop = 1}^{\infty} \sum_{j \mathop = 1}^{\infty} {\frac 1 {i^2 } } {\frac 1 {j^2 } }       | c = Product of Absolutely Convergent Series }} {{eqn | r = \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4 } } + \sum_{i \mathop = 2}^{\infty} \sum_{j \mathop = 1}^{i-1} {\frac 1 {i^2 } } {\frac 1 {j^2 } } + \sum_{j \mathop = 2}^{\infty} \sum_{i \mathop = 1}^{j - 1} {\frac 1 {i^2 } } {\frac 1 {j^2 } }       | c = $\paren {i = j } + \paren {j \lt i } + \paren {j \gt i }$ }} {{eqn | r = \map \zeta 4  + 2  \dfrac {\pi^4} {5!}       | c = The sums $\paren {j < i }$ and $\paren {j > i }$ are symmetric and each equal to the coefficient of the 5th power term in the sin(x) expansion See Basel_Problem/Proof_2 }} {{end-eqn}} :$\begin{array}{r|cccccccccc} \displaystyle \paren {\map \zeta 2}^2 & \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}  & \cdots \\ \hline \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {1}}^2 & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {4}}  & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}}^2 & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {9}}  & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}}^2 & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {16}} & \cdots \\ \paren {\dfrac {1} {16}}  & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}^2 & \cdots \\ \cdots  & \cdots & \cdots & \cdots & \cdots & \cdots \\ \end{array}$ {{begin-eqn}} {{eqn | l = \map \zeta 4       | r = \paren {\map \zeta 2}^2  - 2  \dfrac {\pi^4} {5!}       | c = Rearranging }} {{eqn | r = \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}       | c = simplifying }} {{eqn | r = \dfrac {\pi^4} {90}       | c = simplifying }} {{end-eqn}} {{qed}}
By Fourier Series: $x^6$ over $-\pi$ to $\pi$, for $x \in \closedint {-\pi} \pi$: :$\displaystyle x^6 = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map \cos {n \pi} \, \map \cos {n x}$ Setting $x = \pi$:  {{begin-eqn}} {{eqn | l = \pi^6        | r = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map {\cos^2} {n \pi} }} {{eqn | ll= \leadsto       | l = \frac { 6 \pi^6} 7        | r = \sum_{n \mathop = 1}^\infty \frac {12 \pi^4} {n^2} - \sum_{n \mathop = 1}^\infty \frac {240 \pi^2} {n^4} + \sum_{n \mathop = 1}^\infty \frac {1440} {n^6}       | c = Cosine of Multiple of Pi }} {{eqn | ll= \leadsto       | l = \frac {\pi^6} 7       | r = 2 \pi^4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 40 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^4} + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6} }} {{eqn | r = - \frac {\pi^6} 9 + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}       | c = Basel Problem and Riemann Zeta Function of 4 }} {{eqn | ll= \leadsto       | l = 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}        | r = \frac {\pi^6} 9 + \frac {\pi^6} 7       | c = rearranging }} {{eqn | r = \frac {16 \pi^4} {63} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^6}       | r = \frac {\pi^6} {945} }} {{end-eqn}} {{qed}}
Let $F$ be a filtered subset of $S$ such that: :$F$ admits an infimum in $\struct {S, \preceq}$. By Filtered iff Upper Closure Filtered: :$F^\succeq$ is filtered where $F^\succeq$ denotes the upper closure of $F$. By Upper Closure is Upper Set: :$F^\succeq$ is upper. Because filtered is non-empty, by definition: :$F^\succeq$ is filter in $\struct {S, \preceq}$. By Infimum of Upper Closure of Set: :$F^\succeq$ admits an infimum in $\struct {S, \preceq}$ and :$\map \inf {F^\succeq} = \inf F$ By assumption and mapping preserves the infimum on subset: :$\map {f^\to} {F^\succeq}$ admits an infimum in $\struct {T, \precsim}$ and :$\map \inf {\map {f^\to} {F^\succeq} } = \map f {\map \inf {F^\succeq} }$ By Upper Closure is Closure Operator: :$F \subseteq F^\succeq$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$\map {f^\to} F \subseteq \map {f^\to} {F^\succeq}$ By definition of infimum: :$\map f {\inf F}$ is lower bound for $\map {f^\to} {F^\succeq}$ By Lower Bound is Lower Bound for Subset: :$\map f {\inf F}$ is a lower bound for $\map {f^\to} F$ We will prove that: :for every element $x$ of $T$: ::if $x$ is lower bound for $\map {f^\to} F$, then $x \precsim \map f {\inf F}$ Let $x \in T$ such that: :$x$ is a lower bound for $\map {f^\to} \F$ We will prove as a sublemma that: :$x$ is a lower bound for $\map {f^\to} {F^\succeq}$ Let $y \in \map {f^\to} {F^\succeq}$. By definition of image of set under mapping: :$\exists a \in S: a \in F^\succeq \land y = \map f a$ By definition of upper closure of set: :$\exists b \in F: b \preceq a$ By Infima Preserving Mapping on Filters is Increasing: :$f$ is increasing. By definition of increasing: :$\map f b \precsim \map f a$ By definition of image of set under mapping: :$\map f b \in \map {f^\to} F$ By definition of lower bound of set: :$x \precsim \map f b$ Thus by definition of transitivity: :$x \precsim y$ Thus by definition: :$x$ is a lower bound for $\map {f^\to} {F^\succeq}$ This ends the proof of the sublemma. Thus by definition of infimum: :$x \precsim \map f {\inf F}$ Thus again by definition of infimum: :$\map {f^\to} F$ admits an infimum in $\struct {T, \precsim}$ and :$\map \inf {\map {f^\to} F} = \map f {\inf F}$ Thus result follows by definition of mapping preserves filtered infima. {{qed}}
This follows by mutatis mutandis of the proof of Infima Preserving Mapping on Filters Preserves Filtered Infima. {{qed}}
Let $s = a + b i$, where $a, b \in \R$, $a<0$. {{begin-eqn}} {{eqn | l = \cmod {\lim_{t \mathop \to + \infty} t^s}       | r = \lim_{t \mathop \to +\infty} \cmod {t^s}       | c =  }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{s \ln t} }       | c = {{Defof|Power to Complex Number}} }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{\paren {a + b i} \ln t} }       | c =  }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} e^{i b \ln t} }       | c =  }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} } \cmod {e^{i b \ln t} }       | c = Modulus of Product }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} }       | c = Modulus of Exponential of Imaginary Number is One }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {t^a}       | c = {{Defof|Power to Complex Number}} }} {{eqn | r = \lim_{t \mathop \to +\infty} \cmod {\dfrac 1 {t^{-a} } }       | c =  }} {{eqn | r = 0       | c = Sequence of Powers of Reciprocals is Null Sequence }} {{eqn | ll= \leadsto       | l = \lim_{t \mathop \to +\infty} t^s       | r = 0       | c =  }} {{end-eqn}} {{qed}} Category:Heaviside Step Function fftjekx69m6362u7vuu4botdlc11iua
Let $X$ be a subset of $S$. In the case when $X = \varnothing$: by definition of lower bounded: :$\exists L \in S: L$ is lower bound for $S$. By definition of empty set: :$L$ is upper bound for $X$. By definition of lower bound: :$\forall x \in S: x$ is upper bound for $X \implies L \preceq x$ Thus by definition :$L$ is a supremum of $X$. Thus: :$X$ admint a supremum. {{qed|lemma}} In the case when $X \ne \varnothing$: Define :$Y := \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ where ${\it Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$. By Existence of Non-Empty Finite Suprema in Join Semilattice :all suprema in $Y$ exist, By definition of non-empty set: :$Y$ is a non-empty set. We will prove that :$Y$ is directed Let $x, y \in Y$. By definition of $Y$: :$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: x = \sup A$ and :$\exists B \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \sup B$ By Finite Union of Finite Sets is Finite: :$A \cup B$ is finite :$A \cup B \ne \varnothing$ By Union is Smallest Superset: :$A \cup B \subseteq X$ By definition of $Y$: :$\sup\left({A \cup B}\right) \in Y$ 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: :$Y$ is directed. By definition up-complete: :$Y$ admits a supremum By definition of supremum :$\sup Y$ is upper bound for $Y$ We will prove that :$X \subseteq Y$ Let $x \in X$. By definitions of subset and singleton: :$\left\{ {x}\right\} \subseteq X$ :$\left\{ {x}\right\}$ is finite :$\left\{ {x}\right\} \ne \varnothing$ By definition of $Y$: :$\sup\left\{ {x}\right\} \in Y$ Thus by Supremum of Singleton: :$x \in Y$ By Upper Bound is Upper Bound for Subset: :$\sup Y$ is upper bound for $X$ We will prove that :$\forall x \in S: x$ is upper bound for $X \implies \sup Y \preceq x$ Let $x \in S$ such that :$x$ is upper bound for $X$ We will prove as sublemma that :$x$ is upper bound for $Y$ Let $y \in Y$. By definition of $Y$: :$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \sup A$ By definition of ${\it Fin}$: :$A \subseteq X$ By Upper Bound is Upper Bound for Subset :$x$ is upper bound for $A$ Thus by definition of supremum: :$y \preceq x$ Thus by definition :$x$ is upper bound for $Y$ This ends the proof of sublemma. Thus by definition of supremum: :$\sup Y \preceq x$ This ends the proof of lemma. By definition :$\sup Y$ is a supremum of $X$ and thus: :$X$ admits a supremum. {{qed|lemma}} Thus result follows by Lattice is Complete iff it Admits All Suprema. {{qed}}
{{begin-eqn}} {{eqn | l = 1       | r = \frac {\sec^2 \theta} {\sec^2 \theta}       | c =  }} {{eqn | l =        | r = \sec^2 \theta \times \frac 1 {1 - \paren {-\tan^2 \theta} }       | c =  }} {{eqn | l =        | r = \sec^2 \theta \times \sum_{n \mathop = 0}^\infty \paren {-\tan^2 \theta}^n       | c = Sum of Infinite Geometric Sequence }} {{eqn | l =        | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \sec^2 \theta \tan^{2 n} \theta       | c = }} {{end-eqn}} By the root test the radius of convergence is $-\dfrac \pi 4 \le \theta \le \dfrac \pi 4$. {{begin-eqn}} {{eqn | l = \int \paren {1} \rd \theta       | r = \int \paren {\sum_{n \mathop = 0} ^ \infty \paren {-1} ^ n \sec ^ 2 \theta \tan ^ {2 n} \theta} \rd \theta       | c = Integrating both sides }} {{eqn | l = \int \rd \theta       | r = \sum_{n \mathop = 0}^\infty \paren {-1} ^ n \int  \paren { \sec ^ 2 \theta \tan ^ {2 n} \theta} \rd \theta       | c =  }} {{eqn | l = \theta       | r = \sum_{n \mathop = 0} ^ \infty \frac {\paren {-1} ^ n}{2 n + 1} \tan ^ {2 n + 1} \theta       | c = Primitive of Power of Tangent of a x by Square of Secant of a x }} {{end-eqn}} {{qed}}
=== Sufficient Condition === Assume that $\left({S, \preceq}\right)$ is a complete lattice. Thus by definition of complete lattice: :$\forall X \subseteq S: X$ admits a supremum. {{qed|lemma}} === Necessary Condition === Assume that :$\forall X \subseteq S: X$ admits a supremum. We will prove that :$\forall X \subseteq S: X$ admits both a supremum and an infimum. Let $X \subseteq S$. Thus by assumption :$X$ admits a supremum. Define :$Y := \left\{ {s \in S: s}\right.$ id lower bound for $\left.{X}\right\}$ By assumption :$Y$ admits a supremum We will prove that :$\sup Y$ is lower bound for $X$ Let $x \in X$. By definition of lower bound: :$x$ is upper bound for $Y$ By definition of supremum: :$\sup Y \preceq x$ thus by definition :$\sup Y$ is lower bound for $X$ We will prove that :$\forall s \in S: s$ is lower bound for $X \implies s \preceq \sup Y$ Let $s \in S$ such that :$s$ is lower bound for $X$. By definition of $Y$: :$s \in Y$ By definition of supremum: :$\sup Y$ is upper bound for $Y$. Thus by definition of upper bound: :$s \preceq \sup Y$ Thus by definition :$X$ admits an infimum By definition :$\left({S, \preceq}\right)$ is a lattice. Thus the result follows by definition of complete lattice. {{qed}}
=== Definition 1 implies Definition 2 === Let $T_n$ be a triangular number by definition 1. Let $n = 0$. By definition: :$T_0 = 0$ By vacuous summation: :$\displaystyle T_0 = \sum_{i \mathop = 1}^0 i = 0$ By definition of summation: :$\displaystyle T_{n-1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \left({n - 1}\right)$ and so: {{begin-eqn}} {{eqn | l = T_n       | r = T_{n-1} + n       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \left({n - 1}\right) + n       | c =  }} {{end-eqn}} Thus $T_n$ is a triangular number by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $T_n$ be a triangular number by definition 2. :$\displaystyle T_n = \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + \left({n - 1}\right) + n$ Thus: :$\displaystyle T_{n-1} = \sum_{i \mathop = 1}^{n - 1} i = 1 + 2 + \cdots + \left({n - 1}\right)$ and so: :$T_n = T_{n - 1} + n$ Then: :$\displaystyle T_0 = \sum_{i \mathop = 1}^0 i$ is a vacuous summation and so: :$T_0 = 0$ Thus $T_n$ is a triangular number by definition 1. {{qed|lemma}} === Definition 1 equivalent to Definition 3 === We have by definition that $T_n = 0 = P \left({3, n}\right)$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = P \left({3, n}\right)       | r = P \left({3, n - 1}\right) + \left({n - 1}\right) + 1       | c =  }} {{eqn | r = P \left({3, n - 1}\right) + n       | c =  }} {{end-eqn}} Thus $P \left({3, n}\right)$ and $T_n$ are generated by the same recurrence relation. {{qed}} Category:Triangular Numbers nrr86ljdnp2cwhwl0c34h408tavtsid
Pentagonal numbers are $k$-gonal numbers where $k = 5$. From Closed Form for Polygonal Numbers we have that: :$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$ Hence: {{begin-eqn}} {{eqn | l = P_n       | r = \frac n 2 \paren {\paren {5 - 2} n - 5 + 4}       | c = Closed Form for Polygonal Numbers }} {{eqn | r = \dfrac {n \paren {3 n - 1} } 2       | c =  }} {{end-eqn}} Hence the result. {{qed}}
=== Definition 1 implies Definition 2 === Let $P_n$ be a pentagonal number by definition 1. Let $n = 0$. By definition: :$P_0 = 0$ By vacuous summation: :$P_0 = \displaystyle \sum_{i \mathop = 1}^0 \paren {3 i - 2} = 0$ By definition of summation: {{begin-eqn}} {{eqn | l = P_{n - 1}       | r = \sum_{i \mathop = 1}^{n - 1} \paren {3 i - 2}       | c =  }} {{eqn | r = 1 + 4 + \cdots + 3 \paren {n - 1} - 2       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = P_n       | r = P_{n - 1} + 3 n - 2       | c =  }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {3 i - 2}       | c =  }} {{eqn | r = 1 + 4 + \cdots + 3 n - 2       | c =  }} {{end-eqn}} Thus $P_n$ is a pentagonal number by definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $P_n$ be a pentagonal number by definition 2. Then: {{begin-eqn}} {{eqn | l = P_n       | r = \sum_{i \mathop = 1}^n \paren {3 i - 2}       | c =  }} {{eqn | r = 1 + 4 + \cdots + \paren {3 \paren {n - 1} - 2} + \paren {3 n - 2}       | c =  }} {{eqn | r = P_{n - 1} + 3 n - 2       | c =  }} {{end-eqn}} Then: :$P_0 = \displaystyle \sum_{i \mathop = 1}^0 \paren {3 n - 2}$ is a vacuous summation and so: :$P_0 = 0$ Thus $P_n$ is a pentagonal number by definition 1. {{qed|lemma}} === Definition 1 equivalent to Definition 3 === We have by definition that $P_n = 0 = \map P {5, n}$. Then: {{begin-eqn}} {{eqn | lo= \forall n \in \N_{>0}:       | l = \map P {5, n}       | r = \map P {5, n - 1} + \paren {5 - 2} \paren {n - 1} + 1       | c =  }} {{eqn | r = \map P {5, n - 1} + 3 \paren {n - 1} + 1       | c =  }} {{eqn | r = \map P {5, n - 1} + 3 n - 2       | c =  }} {{end-eqn}} Thus $\map P {5, n}$ and $P_n$ are generated by the same recurrence relation. {{qed}} Category:Pentagonal Numbers o67u2fnxyv5xfp0f312rk3sihwqp0sh
Hexagonal numbers are $k$-gonal numbers where $k = 6$. From Closed Form for Polygonal Numbers we have that: :$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$ Hence: {{begin-eqn}} {{eqn | l = H_n       | r = \frac n 2 \paren {\paren {6 - 2} n - 6 + 4}       | c = Closed Form for Polygonal Numbers }} {{eqn | r = n \paren {2 n - 1}       | c =  }} {{end-eqn}} Hence the result. {{qed}}