hcju/proofwikiqa huggingface dataset:
=== Forward implication === Let $z_1, z_2, \ldots, z_n$ be linearly independent over the rational numbers $\Q$. That is, if $q_1, q_2, \ldots, q_n$ are rational numbers such that: :$q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$ then: :$q_1 = q_2 = \cdots = q_n = 0$ Let $a_1, a_2, \ldots, a_n$ be integers such that: :$a_1 z_1 + a_2 z_2 + \cdots + a_n z_n = 0$ Then, by Integers form Subdomain of Rationals and the assumption, it follows that: :$a_1 = a_2 = \cdots = a_n = 0$ Hence it follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the integers $\Z$. {{qed|lemma}} === Backward implication === Let $z_1, z_2, \ldots, z_n$ be linearly independent over the integers $\Z$. That is, if $a_1, a_2, \ldots, a_n$ are integers such that: :$a_1 z_1 + \cdots + a_n z_n = 0$ then: :$a_1 = \cdots = a_n = 0$ Let $q_1, q_2, \ldots, q_n$ be rational numbers such that: :$q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$ Let the lowest common multiple of the denominators of $q_1, q_2, \ldots, q_n$ be $m$. Then, by the definition of lowest common multiple: :$m q_1, m q_2, \ldots, m q_n \in \Z$ Thus: :$m q_1 z_1 + m q_2 z_2 + \cdots + m q_n z_n = 0$ where the coefficients of $z_1, z_2, \ldots, z_n$ are all integers. Hence, by the assumption: :$m q_1 = m q_2 = \cdots = m q_n = 0$ Since $m$ is non-zero, it follows that: :$q_1 = q_2 = \cdots = q_n = 0$ It follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the rational numbers $\Q$. {{qed}} Category:Complex Analysis 1x4vc8eendh9yty7n1l49dv6nkli3k6
From the definition of Lambert W function: :$a = W \left({a}\right) e^{W \left({a}\right)}$ {{AimForCont}} $W \left({a}\right)$ is not transcendental. Hence, $W \left({a}\right)$ is algebraic. From the weaker Hermite-Lindemann-Weierstrass theorem, $e^{W \left({a}\right)}$ is transcendental. However, from the equation above, it is also equal to: :$\dfrac a {W \left({a}\right)}$ which must be algebraic, considering the fact that $a$ and $W \left({a}\right)$ are both algebraic. From this contradiction, it follows that $W \left({a}\right)$ must be transcendental whenever $a$ is a non-zero algebraic number. {{qed}} Category:Lambert W Function Category:Transcendental Number Theory 802x6rh25dp5ck883xvcfzybyt700pr
A slightly modified proof of Pi is Irrational/Proof 2 also proves it for $\pi^2$: {{AimForCont}} $\pi^2$ is rational. Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$. Note that $\paren {q \pi}^2 = q^2 \paren {\dfrac p q} = p q$ is an integer. Now let: :$\displaystyle A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x $ Integration by Parts twice gives: {{begin-eqn}} {{eqn | l = A_n       | r = \frac {q^n} {n!} \int_0^\pi n \paren {x \paren {\pi - x} }^{n - 1} \paren {\pi - 2 x} \cos x \rd x       | c =  }} {{eqn | r = \frac {q^n} {n!} \int_0^\pi 2 n \paren {x \paren {\pi - x} }^{n - 1} \sin x - n \paren {n - 1} \paren {x \paren {\pi - x} }^{n - 2} \paren {\pi - 2 x}^2 \sin x \rd x       | c =  }} {{end-eqn}} Writing: :$\paren {\pi - 2 x}^2 = \pi^2 - 4 x \paren {\pi - x}$ gives the result: :$(1): \quad A_n = \paren {4 n - 2} q A_{n - 1} - \paren {q \pi}^2 A_{n - 2}$  We will deduce that $A_n$ is an integer for all $n$. First confirm by direct integration that $A_0$ and $A_1$ are integers: :$\displaystyle A_0 = \int_0^\pi \sin x \rd x = 2$ :$\displaystyle A_1 = q \int_0^\pi x \paren {\pi - x} \sin x \rd x = 4 q$ Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers.  Then by $(1)$ and the assumption that $q$ and $\paren {q \pi}^2$ are integers, $A_k$ is also an integer.  So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction. For $x \in \closedint 0 \pi$, we have: :$0 \le \sin x \le 1$ and: :$0 \le x \paren {\pi - x} \le \pi^2 / 4$ hence: :$0 < A_n < \pi \dfrac {\paren {q \pi^2 / 4}^n} {n!}$ From Power over Factorial: :$\displaystyle \lim_{n \mathop \to \infty} \frac {\paren {q \pi^2 / 4}^n} {n!} = 0$ It follows from the Squeeze Theorem that: :$\displaystyle \lim_{n \mathop \to \infty} A_n = 0$ Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$. This contradicts the supposition that $A_n$ is an integer. It follows that $\pi^2$ is irrational. {{qed}}
=== Sufficient Condition === Let $x \ll y$ Let $Y$ be a set of subsets of $X$ such that :$y \subseteq \bigcup Y$ By definitions of power set and subset: :$Y \subseteq \mathcal P\left({X}\right)$ By the proof of Power Set is Complete Lattice: :$\bigcup Y = \sup Y$ By definition of $\preceq$: :$y \preceq \sup Y$ By Way Below in Complete Lattice: :there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$ By the proof of Power Set is Complete Lattice: :$\bigcup Z = \sup Z$ Thus by definition of $\preceq$: :there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$ {{qed|lemma}} === Necessary Condition === Suppose :for every a set $Y$ of subsets of $X$ such that $y \subseteq \bigcup Y$ ::then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$ We will prove that :for every a subset $Y$ of $\mathcal P\left({X}\right)$ such that $y \preceq \sup Y$ ::then there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$ Let $Y$ be a subset of $\mathcal P\left({X}\right)$ such that :$y \preceq \sup Y$ By definition of power set: :$Y$ is a set of subsets of $X$. By the proof of Power Set is Complete Lattice: :$\bigcup Y = \sup Y$ By definition of $\preceq$: :$y \subseteq \bigcup Y$ By assumption: :there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$ By the proof of Power Set is Complete Lattice: :$\bigcup Z = \sup Z$ Thus by definition of $\preceq$: :there exists a finite subset $Z$ of $Y$: $x \preceq \sup Z$ {{qed|lemma}} Thus by Way Below in Complete Lattice: :$x \ll y$ {{qed}}
Let $N$ be divisible by $10$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod {10} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod {10} }} {{eqn | ll= \leadstoandfrom       | l = a_0 + 10 \sum_{k \mathop = 1}^n a_k 10^{k - 1}       | o = \equiv       | r = 0 \pmod {10} }} {{eqn | ll= \leadstoandfrom       | l = a_0       | o = \equiv       | r = 0 \pmod {10}       | c = as $10 \equiv 0 \pmod {10}$ }} {{end-eqn}} {{qed}}
Let $N$ be divisible by $10^r$. Then: {{begin-eqn}} {{eqn | l = N       | o = \equiv       | r = 0 \pmod {10^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^n a_k 10^k       | o = \equiv       | r = 0 \pmod {10^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r + \sum_{k \mathop = r + 1}^n a_k 10^k       | o = \equiv       | r = 0 \pmod {10^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r + 10^r \sum_{k \mathop = r + 1}^n a_k 10^{k - r}       | o = \equiv       | r = 0 \pmod {10^r} }} {{eqn | ll= \leadstoandfrom       | l = \sum_{k \mathop = 0}^r a_k 10^r       | o = \equiv       | r = 0 \pmod {10^r}       | c = as $10^r \equiv 0 \pmod {10^r}$ }} {{end-eqn}} Hence the result. {{qed}}
Let $n = a^2 - b^2$ for some $a, b \in \Z$. By Square Modulo 4, both $a$ and $b$ are of the form $4 k$ or $4 k + 1$ for some integer $k$. There are $4$ cases: ;$a \equiv b \equiv 0 \pmod 4$: Then: :$a^2 - b^2 \equiv 0 \pmod 4$ and so $n$ is in the form $4 k$. ;$a \equiv 0 \pmod 4$, $b \equiv 1 \pmod 4$: Then: :$a^2 - b^2 \equiv -1 \pmod 4 \equiv 3 \pmod 4$ and so $n$ is in the form $4 k + 3$. ;$a \equiv 1 \pmod 4$, $b \equiv 0 \pmod 4$: Then: :$a^2 - b^2 \equiv 1 \pmod 4$ and so $n$ is in the form $4 k + 1$. ;$a \equiv b \equiv 1 \pmod 4$: Then: :$a^2 - b^2 \equiv 0 \pmod 4$ and so $n$ is in the form $4 k$. Thus it is never the case that $a^2 - b^2 = 4 k + 2$. {{qed}}
=== $\subseteq$ === Let $y \in x^{\mathrm{compact} }$. By definition of compact closure: :$y \preceq x$ and $y$ is compact. By definition of $\preceq$: :$y \subseteq x$ By Element is Finite iff Element is Compact in Lattice of Power Set: "$y$ is a finite set. Thus by definition of $\mathit{Fin}$: :$y \in \mathit{Fin}\left({x}\right)$ {{qed|lemma}} === $\supseteq$ === Let $y \in \mathit{Fin}\left({x}\right)$. By definition of $\mathit{Fin}$: :$y \subseteq x$ and $y$ is finite. By definition of $\preceq$: :$y \preceq x$ By Element is Finite iff Element is Compact in Lattice of Power Set: :$y$ is compact. Thus by definition of compact closure: :$y \in x^{\mathrm{compact} }$ {{qed}}
{{:Closed Form for Triangular Numbers}} for $n \in \Z_{\ge 0}$. The expression for the $n$th odd square number is: :$4 n^2 + 4 n + 1$ again, for $n \in \Z_{\ge 0}$. Therefore the closed-form expression for the $n$th sum of two consecutive odd squares is: :$4 n^2 + 4 n + 1 + 4 \paren {n + 1}^2 + 4 \paren {n + 1} + 1$ This simplifies to: :$8 n^2 + 16 n + 10$ Equate the two with a variable replacing $n$: :$8 x^2 + 16 x + 10 = \dfrac {y \paren {y + 1} } 2$ This simplifies to: :$16 x^2 + 32 x + 20 = y^2 + y$ We then apply Solutions to Diophantine Equation $16 x^2 + 32 x + 20 = y^2 + y$: {{:Solutions to Diophantine Equation 16x^2+32x+20 = y^2+y}} Due to the restrictions on the variables, solutions with negative inputs are invalid. This leaves one solution: :$\tuple {0, 4}$ as follows: {{begin-eqn}} {{eqn | l = 8 \paren 0^2 + 16 \paren 0 + 10       | r = \frac {\paren 4 \paren {\paren 4 + 1} } 2       | c =  }} {{eqn | ll= \leadsto       | l = 1^2 + 3^2       | r = 10       | c =  }} {{end-eqn}} {{qed}}
Consider $2$ elements $a, b$ of $S$ which share a common divisor $d$. Then $d \divides \size {a - b}$ and so $d < 10$. Now from the Fundamental Theorem of Arithmetic, $d$ must have a prime factor which is strictly less than $10$. So for $a$ and $b$ to have a common divisor, at least one such common divisor is in $\set {2, 3, 5, 7}$. There are exactly $5$ elements of $S$ which have a common divisor of $2$. There are either $3$ or $4$ elements of $S$ common divisor of $3$. The case where there are $4$ happens when $n = 3 k$ and $n + 9 = 3 \paren {k + 3}$. Of these $3$ or $4$, no more than $2$ are odd and so have not been accounted for. There are exactly $2$ elements of $S$ which have $5$ as a common divisor. One of those is even and so has been counted already. There are at most $2$ elements of $S$ which have $7$ as a common divisor. One of those is even and so has been counted already. Thus we have a count of how many elements of $S$ which can possibly share a common divisor with another element of $S$: :$5$ are divisible by $2$ :$2$ or fewer are divisible by $3$ and not $2$ :$1$ at the most is divisible by $5$ and not by $2$ or $3$ :$1$ at the most is divisible by $7$ and not by $2$, $3$ or $5$. That makes a total of $9$ elements of $S$ which are divisible by $2$, $3$, $5$ or $7$. Thus there exists (at least) $1$ element of $S$ which is not divisible by $2$, $3$, $5$ or $7$ and so cannot share a common divisor with another element of $S$. Hence the result. {{qed}}
{{begin-eqn}} {{eqn | l = H_n       | r = \sum_{k \mathop = 1}^n T_k       | c = {{Defof|Tetrahedral Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \frac {n \paren {n + 1} } 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\frac 1 2 n^2 + \frac 1 2 n}       | c =  }} {{eqn | r = \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 2 \sum_{k \mathop = 1}^n n       | c = Summation is Linear }} {{eqn | r = \frac 1 2 \sum_{k \mathop = 1}^n n^2 + \frac 1 2 T_n       | c = {{Defof|Triangular Number|index = 2}} }} {{eqn | r = \frac 1 2 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 + \frac 1 2 \frac {n \paren {n + 1} } 2       | c = Sum of Sequence of Squares, Closed Form for Triangular Numbers }} {{eqn | r = \frac {n \paren {n + 1} } {12} \paren {\paren {2 n + 1} + 3}       | c =  }} {{eqn | r = \frac {n \paren {n + 1} \paren {n + 2} } 6       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({\left({n!}\right)!}\right)!       | r = \left({n!}\right)! \times \left({\left({n!}\right)! - 1}\right)!       | c = Factorial as Product of Two Factorials }} {{eqn | r = n! \times \left({n! - 1}\right)! \times \left({\left({n!}\right)! - 1}\right)!       | c = Factorial as Product of Two Factorials }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac z {\paren {1 - z}^3}       | r = \paren {1 - z}^{-3}       | c = Exponent Combination Laws for Negative Power }} {{eqn | r = z \sum_{n \mathop = 0}^\infty \dbinom {-3} n \paren {-z}^n       | c = General Binomial Theorem }} {{eqn | r = z \sum_{n \mathop = 0}^\infty \paren {-1}^n \dbinom {n + 2} n \paren {-z}^n       | c = Negated Upper Index of Binomial Coefficient }} {{eqn | r = z \sum_{n \mathop = 0}^\infty \dbinom {n + 2} n z^n       | c =  }} {{eqn | r = z \sum_{n \mathop = 0}^\infty \dbinom {n + 2} 2 z^n       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \sum_{n \mathop = 0}^\infty \dbinom {n + 2} 2 z^{n + 1}       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty \dbinom {n + 1} 2 z^n       | c =  }} {{eqn | r = \sum_{n \mathop = 1}^\infty T_n z^n       | c = Corollary to Binomial Coefficient with Two }} {{eqn | r = 0 z^0 + \sum_{n \mathop = 1}^\infty T_n z^n       | c =  }} {{eqn | r = \sum_{n \mathop = 0}^\infty T_n z^n       | c = $T_0 = 0$ by {{Defof|Triangular Number|index = 1}} }} {{eqn | r = \map G z       | c = {{Defof|Generating Function}} }} {{end-eqn}} {{qed}} Category:Triangular Numbers Category:Generating Functions jhmucsupi9z1iyohf3u4nlwz97m09c1
We will prove that :$\forall x \in \mathcal P\left({X}\right): x^{\mathrm{compact} }$ is directed. Let $x \in \mathcal P\left({X}\right)$. By Empty is Bottom of Lattice of Power Set: :$\varnothing = \bot$ where $\bot$ denotes the bottom of $L$. By Bottom is Way Below Any Element: :$\bot \ll \bot$ where $\ll$ is the way below relation. By definition: :$\bot$ is compact. By definition of the smallest element: :$\bot \preceq x$ By definition of compact closure: :$\bot \in x^{\mathrm{compact} }$ By definition: :$x^{\mathrm{compact} }$ is non-empty. Thus by Non-Empty Compact Closure is Directed: :$x^{\mathrm{compact} }$ is directed. {{qed|lemma}} By Power Set is Complete Lattice: :$L$ is complete lattice. Thus by definition of complete lattice: :$L$ is up-complete. It remains to prove that :$L$ satisfies axiom of K-approximation. Let $x \in \mathcal P\left({X}\right)$. We will prove that :$\forall a \in \mathcal P\left({X}\right): a$ is upper bound for $x^{\mathrm{compact} } \implies x \preceq a$ Let $a \in \mathcal P\left({X}\right)$ such that :$a$ is upper bound for $x^{\mathrm{compact} }$ We will prove that :$x \subseteq a$ Let $t \in x$. By definition of power set: :$x \subseteq X$ By definitions of subset and singleton: :$\left\{ {t}\right\} \subseteq X$ and $\left\{ {t}\right\} \subseteq x$ By definition of power set: :$\left\{ {t}\right\} \in \mathcal P\left({X}\right)$ By Singleton is Finite: :$\left\{ {t}\right\}$ is finite. By definition of $\mathit{Fin}$: :$\left\{ {t}\right\} \in \mathit{Fin}\left({x}\right)$ where $\mathit{Fin}\left({x}\right)$ denotes the set of all finite subset of $x$. By Compact Closure is Set of Finite Subsets in Lattice of Power Set: :$\left\{ {t}\right\} \in x^{\mathrm{compact} }$ By definition of upper bound: :$\left\{ {t}\right\} \preceq a$ By definition of $\preceq$: :$\left\{ {t}\right\} \subseteq a$ Thus by definitions of subset and singleton: :$t \in a$ {{qed|lemma}} Thus by definition of $\preceq$: :$x \preceq a$ {{qed|lemma}} By definition of compact closure: :$\forall y \in x^{\mathrm{compact} }: y \preceq x$ By definition of upper bound: :$x$ is upper bound for $x^{\mathrm{compact} }$ Thus by definition of supremum: :$x = \sup \left({x^{\mathrm{compact} } }\right)$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \left({a + b}\right)^2 - \left({a - b}\right)^2       | c =  }} {{eqn | r = \left({a^2 + 2 a b + b^2}\right) - \left({a^2 - 2 a b + b^2}\right)       | c = Square of Sum and Square of Difference }} {{eqn | r = a^2 + 2 a b + b^2 - a^2 + 2 a b - b^2       | c =  }} {{eqn | r = 2 a b + 2 a b }} {{eqn | r = 4 a b       | c =  }} {{end-eqn}} {{qed}}
Thus by assumption: :$x^{\mathrm{compact} }$ is a non-empty set. Let $y, z \in x^{\mathrm{compact} }$. By definition of compact closure: :$y \preceq x$, $z \preceq x$, and $y$ and $z$ are compact. By definition of compact element: :$y \ll y$ and $z \ll z$ where $\ll$ denotes the way below relation. By Way Below is Congruent for Join: :$y \vee z \ll y \vee z$ By definition: :$y \vee z$ is a compact element By definition of supremum: :$y \vee z \preceq x$ Thus by definition of compact closure: :$y \vee z \in x^{\mathrm{compact} }$ Thus by Join Succeeds Operands: :$y \preceq y \vee z$ and $z \preceq y \vee z$ {{qed}}
By Finite iff Cardinality Less than Aleph Zero: :$\left\vert{Y}\right\vert < \aleph_0$ Then :$\left\vert{X}\right\vert < \aleph_0$ Thus by Finite iff Cardinality Less than Aleph Zero: :$X$ is a finite set. {{qed}}
=== Sufficient Condition === Let $X$ be finite. By definition of finite set: :$\exists n \in \N: X \sim \N_n$ where: :$\sim$ denotes the set equivalence :$\N_n$ denotes the initial segment of natural numbers less than $n$. By the von Neumann construction of natural numbers: :$\N_n = n$ By definition of cardinality: :$\card X = n$ By the von Neumann construction of natural numbers: :$\forall i \in \N: i \subseteq \N$ Then: :$n + 1 \subseteq \N$ By Subset implies Cardinal Inequality: :$n + 1 = \card {n + 1} \le \card \N = \aleph_0$ Also: :$n < n + 1$ Thus: :$\card X < \aleph_0$ {{qed|lemma}} === Necessary Condition === Let $\card X < \aleph_0$. By definition of aleph mapping: :$\aleph_0 = \omega$ By the von Neumann construction of natural numbers: :$\N = \omega$ By definition of ordinal: :$\card X \in \N$ By definition of cardinal: :$\exists n \in \N: X \sim n$ By the von Neumann construction of natural numbers: :$\exists n \in \N: X \sim \N_n$ Thus by definition: :$X$ is finite. {{qed}}
We have: :$x^n - y^n = \paren {x - y} \paren {x^{n - 1} + \cdots + y^{n - 1} }$ We have to prove that: :$p \nmid x^{n - 1} + \cdots + y^{n - 1}$ Let $\map P u = u^n - y^n$. If $p \divides x^{n - 1} + \cdots + y^{n - 1}$, then $x$ would be a double root of $P$ modulo $p$. By Double Root of Polynomial is Root of Derivative (or a version of this in modular arithmetic): :$p \divides \map {P'} x = n x^{n - 1}$ which is impossible. Therefore: :$p \nmid x^{n - 1} + \cdots + y^{n - 1}$ {{finish}}
{{begin-eqn}} {{eqn | l = \int_\mathcal C \nabla f \cdot \rd \mathbf r       | r = \int_a^b \nabla f \cdot \mathbf r' \left({t}\right) \rd t }} {{eqn | r = \int_a^b \frac {\partial f} {\partial x} \frac {\rd x} {\rd t} + \frac {\partial f} {\partial y} \frac {\rd y} {\rd t} + \frac {\partial f} {\partial z} \frac {\rd z} {\rd t} \rd t}} {{eqn | r = \int_a^b \frac \rd {\rd t} f \left({\mathbf r \left({t}\right)}\right) \rd t       | c = Chain Rule for Real-Valued Functions }} {{eqn | r = f \left({\mathbf r \left({b}\right)}\right) - f \left({\mathbf r \left({a}\right)}\right)       | c = Fundamental Theorem of Calculus }} {{end-eqn}} {{qed}} Category:Vector Calculus iefbddltrzh0ku75we12ousbhq7ujp1
Let $P$ be a palindromic number with $2 n$ digits: :$P = \sqbrk {a_{2 n - 1} a_{2 n - 2} \ldots a_2 a_1 a_0}_{10}$ Thus: :$P = \displaystyle \sum_{j \mathop = 0}^{n - 1} a_j + 10^{2 n - 1 - j}$ Consider the summation: :$S = \displaystyle \sum_{k \mathop = 0}^{2 n - 1} \paren {-1}^k a_k$ As $a_k = a_{2 n - 1 - k}$ we have: {{begin-eqn}} {{eqn | l = S       | r = \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k a_k + \paren {-1}^{2 n - 1 - k} a_k       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^{n - 1} a_k \paren {\paren {-1}^k + \paren {-1}^{2 n - 1 - k} }       | c =  }} {{end-eqn}} As $2 n - 1$ is odd, it follows that $k$ and $2 n - 1 - k$ are of opposite parity. Thus: :$\paren {-1}^k = -\paren {-1}^{2 n - 1 - k}$ and it follows that: :$S = 0$ It follows by Divisibility by 11 that $P$ is  divisible by $11$. Thus, except for $11$ itself, a palindromic number with an even number of digits cannot be prime. The result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \lim_{n \mathop \to \infty} \frac {L_{n + 1} } {L_n}       | r = \lim_{n \mathop \to \infty} \frac {\phi^{n + 1} + \paren {-\phi^{-1} }^{n + 1} } {\phi^n + \paren {-\phi^{-1} }^n}       | c = Closed Form for Lucas Numbers }} {{eqn | r = \lim_{n \mathop \to \infty} \frac {\phi - \phi^{-1} \paren {-\phi^{-2} }^n } {1 + \paren {-\phi^{-2} }^n}       | c = Dividing both numerator and denominator by $\phi^n$ }} {{eqn | r = \frac \phi 1       | c = Quotient Rule for Limits of Functions }} {{eqn | r = \phi }} {{end-eqn}} {{qed}}
By definition of ordered set of all mappings: :$L^X = \struct {S^X, \precsim}$ where :$\forall f, g \in S^X: f \precsim g \iff f \preceq g$ :$\preceq$ denotes the ordering on mappings, :$S^X$ denotes the set of all mappings from $X$ into $S$. === Reflexivity === Let $f \in S^X$. By definition of reflexivity: :$\forall x \in X: \map f x \preceq \map f x$ By definition of ordering on mappings: :$f \preceq f$ Thus by definition of $\precsim$: :$f \precsim f$ {{qed|lemma}} === Transitivity === Let $f, g, h \in S^X$ such that :$f \precsim g$ and $g \precsim h$ By definition of $\precsim$: :$f \preceq g$ and $g \preceq h$ By definition of ordering on mappings: :$\forall x \in X: \map f x \preceq \map g x$ and :$\forall x \in X: \map g x \preceq \map h x$ By definition of transitivity: :$\forall x \in X: \map f x \preceq \map h x$ By definition of ordering on mappings: :$f \preceq h$ Thus by definition of $\precsim$: :$f \precsim h$ {{qed|lemma}} === Antisymmetry === Let $f, g \in S^X$ such that :$f \precsim g$ and $g \precsim f$ By definition of $\precsim$: :$f \preceq g$ and $g \preceq f$ By definition of ordering on mappings: :$\forall x \in X: \map f x \preceq \map g x$ and :$\forall x \in X: \map g x \preceq \map f x$ By definition of antisymmetry: :$\forall x \in X: \map f x = \map g x$ Thus by Equality of Mappings: :$f = g$ {{qed}}
Proof by induction: For all $n \in \N$, let $\map P n$ be the proposition: :$L_n = \phi^n + \paren {1 - \phi}^n = \paren {\dfrac {1 + \sqrt 5} 2})^n + \paren {\dfrac {1 - \sqrt 5} 2}^n = \phi^n + \hat \phi^n$ === Basis for the Induction === $\map P 0$ is true, as this just says: :$\phi^0 + \hat \phi^0 = 1 + 1 = 2 = L_0$ $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \phi^1 + \paren {1 - \phi}^1       | r = \phi + 1 - \phi       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = L_1       | c =  }} {{end-eqn}} This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true. So this is our induction hypothesis: :$\forall 0 \le j \le k + 1: L_j = \phi^j + \hat \phi^j$ Then we need to show: :$L_{k + 2} = \phi^{k + 2} + \hat \phi^{k + 2}$ === Induction Step === This is our induction step: We observe that we have the following two identities: :$\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$ :$\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$ This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$. Thus: {{begin-eqn}} {{eqn | l = \phi^{k + 2} + \hat \phi^{k + 2}       | r = \paren {1 + \phi} \phi^k + \paren {1 + \hat \phi} \hat \phi^k }} {{eqn | r = \paren {\phi^k + \hat \phi^k} + \paren {\phi^{k + 1} + \hat \phi^{k + 1} } }} {{eqn | r = L_k + L_{k + 1}       | c = Induction Hypothesis }} {{eqn | r = L_{k + 2}       | c = {{Defof|Lucas Number}} }} {{end-eqn}} The result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \N: L_n = \phi^n + \hat \phi^n$ {{qed}}
=== Definition 1 implies Definition 2 === Let $\sequence {L_n}$ be the sequence defined as in definition 1. It follows from Lucas Number as Sum of Fibonacci Numbers that $\sequence {L_n}$ is the sequence defined as in definition 2. {{qed|lemma}} === Definition 2 implies Definition 1 === Let $\sequence {L_n}$ be the sequence defined as in definition 2. It follows from Lucas Number as Element of Recursive Sequence that $\sequence {L_n}$ is the sequence defined as in definition 1. {{qed}} Category:Lucas Numbers 3l0wne1w9qvrad9wbjgkhegv7qynhia