hcju/proofwikiqa huggingface dataset:
=== 1 iff 2 === Follows directly from Equivalence of Definitions of Absolute Convergence of Product. {{qed}} === 2 implies 3 === By Terms in Convergent Series Converge to Zero, there exists $n_0\in\N$ such that $|a_n|\leq \frac12$ for $n>n_0$. Thus $a_n\neq-1$ for $n>n_0$. By Bounds for Complex Logarithm, $|\log (1+a_n)| \leq \frac32|a_n|$ for $n>n_0$. By the Comparison Test, $\displaystyle \sum_{n \mathop = n_0+1}^\infty \log(1+a_n)$ is absolutely convergent. {{qed}} === 3 implies 2 === By Terms in Convergent Series Converge to Zero, $\log(1+a_n)\to 0$. By Complex Exponential is Continuous, $1+a_n\to1$. That is, $a_n\to0$. Let $n_1\in\N$ be such that $|a_n|\leq\frac12$ for $n>n_1$. By Bounds for Complex Logarithm, $\frac12|a_n| \leq |\log (1+a_n)|$ for $n>\max(n_0,n_1)$. By the Comparison Test, $\displaystyle \sum_{n \mathop = 1}^\infty a_n$ is absolutely convergent. {{qed}}
Let $P_n$ and $Q_n$ denote the $n$th partial products of $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ and $\ds \prod_{n \mathop = 1}^\infty \paren {1 + \norm {a_n} }$ respectively. We show that $\sequence {P_n}$ is Cauchy. We have, for $m > n$: {{begin-eqn}} {{eqn | l = \norm {P_m - P_n}       | r = \prod_{k \mathop = 1}^n \norm {1 + a_k} \cdot \norm {\prod_{k \mathop = n + 1}^m \paren {1 + a_k} - 1}       | c =  }} {{eqn | o = }} {{eqn | o = \le       | r = \prod_{k \mathop = 1}^n \paren {1 + \norm {a_k} } \paren {\prod_{k \mathop = n + 1}^m \paren {1 + \norm {a_k} } - 1}       | c =  }} {{eqn | o = }} {{eqn | r = Q_m - Q_n       | c =  }} {{end-eqn}} Because $\sequence {Q_n}$ converges, $\sequence {Q_n}$ is Cauchy. By the above inequality, $\sequence {P_n}$ is Cauchy. Because $\mathbb K$ is complete, $\sequence {P_n}$ converges to some $a \in \mathbb K$. By Absolutely Convergent Product Does not Diverge to Zero, the product converges. {{qed}} Category:Infinite Products m4zibqgdhcfj6y9sr2er8czd27twrc6
Let $D$ be a directed subset of $S$ such that :$\sup D \in \complement_S\left({X}\right)$ By definition of relative complement: :$\sup D \notin X$ By definition of closed under directed suprema: :$D \nsubseteq X$ By Complement of Complement: :$D \nsubseteq \complement_S\left({\complement_S\left({X}\right)}\right)$ Thus by Empty Intersection iff Subset of Relative Complement: :$D \cap \complement_S\left({X}\right) \ne \varnothing$ {{qed}}
This follows mutatis mutandis from the proof of Complement of Lower Set is Upper Set. {{qed}}
=== Sufficient Condition === Assume that :$A$ is closed. By definition of closed set: :$\complement_S\left({A}\right) \in \tau$ By definition of Scott topology: :$\complement_S\left({A}\right)$ is upper and inaccessible by directed suprema. By Complement of Complement: :$\complement_S\left({\complement_S\left({A}\right)}\right) = A$ Thus by Complement of Upper Set is Lower Set: :$A$ is a lower set. Thus by Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema: :$A$ is closed under directed suprema. {{qed|lemma}} === Necessary Condition === Assume that $A$ is lower and closed under directed suprema. By Complement of Lower Set is Upper Set and Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema: :$\complement_S\left({A}\right)$ is upper and inaccessible by directed suprema. By definition of Scott topology: :$\complement_S\left({A}\right) \in \tau$ Thus by definition: :$A$ is closed set. {{qed}}
Let $D$ be a directed subset of $S$ such that :$D \subseteq \complement_S\left({X}\right)$ By Empty Intersection iff Subset of Relative Complement: :$D \cap X = \varnothing$ By definition of inaccessible by directed suprema: :$\sup D \notin X$ Thus by definition of relative complement: :$\sup D \in \complement_S\left({X}\right)$ {{qed}}
Suppose $r_n \to r$ and $\theta_n + 2 k_n \pi \to \theta$. We have, by Complex Modulus of Difference of Complex Numbers: {{begin-eqn}} {{eqn  | l = \cmod {z_n - z}^2        | r = r_n^2 + r^2 - 2 r r_n \, \map \cos {\theta_n + 2 k_n \pi - \theta}        | c = }} {{end-eqn}} Because Cosine Function is Continuous: :$\map \cos {\theta_n + 2 k_n \pi - \theta} \to 1$ It follows that: :$\cmod {z_n - z}^2 \to 0$ Conversely, suppose $z_n \to z$. By Modulus of Limit, $r_n \to r$. We have, by Complex Modulus of Difference of Complex Numbers: {{begin-eqn}} {{eqn  | l = \map \cos {\theta_n - \theta}        | r = \frac {r_n^2 + r^2 - \cmod {z_n - z}^2} {2 r r_n} \to 1        | c = }} {{end-eqn}} By Convergence of Cosine of Sequence, there exists a sequence $\sequence {k_n}$ of integers such that $\theta_n + 2 k_n \pi$ converges to $\theta$. {{qed}} Category:Complex Analysis rr4ie42o8rulamqw9yn0tbeajutdxgj
We have: {{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2}^2       | r = \paren {z_1 + z_2} \paren {\overline {z_1} + \overline {z_2} }       | c = Modulus in Terms of Conjugate and Sum of Complex Conjugates }} {{eqn | r = z_1 \overline {z_1} + z_2 \overline {z_2} + z_1\overline {z_2} + \overline {z_1} z_2       | c =  }} {{eqn | r = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \, \map \Re {z_1 \overline {z_2} }       | c = Modulus in Terms of Conjugate and Sum of Complex Number with Conjugate }} {{eqn | r = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \, \cmod{ z_1 } \cmod{ z_2 } \map \cos {\theta_1 - \theta_2}       | c = Product of Complex Numbers in Polar Form and Argument of Conjugate of Complex Number }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cmod {z_1 + z_2}^2       | r = \paren {\cmod {z_1} \cos \theta_1 + \cmod {z_2} \cos \theta_2}^2 + \paren {\cmod {z_1} \sin \theta_1 + \cmod {z_2} \sin \theta_2}^2       | c = {{Defof|Complex Modulus}} }} {{eqn | r = 2 \cmod {z_1} \cmod {z_2} \cos \theta_1 \cos \theta_2 + \cmod {z_1}^2 \cos^2 \theta_1 + \cmod {z_2}^2 \cos^2 \theta_2       | c =  }} {{eqn | o =        | ro= +       | r = 2 \cmod {z_1} \cmod {z_2} \sin \theta_1 \sin \theta_2 + \cmod {z_1}^2 \sin^2 \theta_1 + \cmod {z_2}^2 \sin^2 \theta_2       | c =  }} {{eqn | r = 2 \cmod {z_1} \cmod {z_2} \paren {\cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2} + \cmod {z_1}^2 + \cmod {z_2}^2       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \cmod {z_1}^2 + \cmod {z_2}^2 + 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}       | c = Cosine of Difference }} {{end-eqn}} {{qed}}
By Complex Argument of Additive Inverse, $\theta_2 + \pi$ is an argument of $-z_2$. We have: {{begin-eqn}} {{eqn | l = \cmod {z_1 - z_2}^2       | r = \cmod {z_1}^2 + \cmod {-z_2}^2 + 2 \cmod {z_1} \cmod {-z_2} \, \map \cos {\theta_1 - \theta_2 - \pi}       | c = Complex Modulus of Sum of Complex Numbers }} {{eqn | r = \cmod {z_1}^2 + \cmod {z_2}^2 - 2 \cmod {z_1} \cmod {z_2} \, \map \cos {\theta_1 - \theta_2}       | c = Complex Modulus of Additive Inverse }} {{end-eqn}} {{qed}}
=== Lemma === {{:Uniform Absolute Convergence of Infinite Product of Complex Functions/Lemma}} ---- Because $\displaystyle \sum_{n \mathop = 1}^\infty f_n$ converges uniformly, there exists $n_0>0$ such that $|f_n(x)|<\frac12$ for $n\geq n_0$ and $x\in X$. Then $\Re(1+f_n(x))>0$. By Bounds for Complex Logarithm: :$|\log(1 + f_n(x))|\leq \frac32 |f_n(x)|$ :$|\log(1 + |f_n(x)|)|\leq \frac32 |f_n(x)|$ for $n\geq n_0$. By Comparison Test for Uniformly Convergent Series, :$g = \displaystyle \sum_{n \mathop = n_0}^\infty \log(1+f_n)$ and :$h = \displaystyle \sum_{n \mathop = n_0}^\infty \log(1 + |f_n|)$ converge uniformly. By Uniform Limit Theorem, $g$ and $h$ are continuous. By Continuous Function on Compact Subspace of Euclidean Space is Bounded, $g$ and $h$ are bounded By Exponential of Series Equals Infinite Product: :$\exp(g) = \displaystyle \prod_{n \mathop = n_0}^\infty (1+f_n)$ :$\exp(h) = \displaystyle \prod_{n \mathop = n_0}^\infty (1+|f_n|)$ By the lemma, the products converge uniformly. By Uniform Limit Theorem, $\exp(g)$ is continuous. Because the second product converges uniformly, the first converges uniformly absolutely. By Image of Complex Exponential Function, $\exp(g)$ has no zeroes. {{qed}} Category:Infinite Products ppkl55j54z3deis1y8471y1iew0ro4t
:$A \cap B = \varnothing$ {{iff}} :$\forall x \in S: x \notin A \cap B$ by Empty Set as Subset {{iff}} :$\forall x \in S:\lnot\left({x \in A \land x \in B}\right)$ by definition of intersection {{iff}} :$\forall x \in S: x \notin A \lor x \notin B$ by De Morgan's Laws (Logic)/Disjunction of Negations {{iff}} :$\forall x \in S: x \in A \implies x \notin B$ by Rule of Material Implication {{iff}} :$\forall x \in S: x \in A \implies x \in \complement_S\left({B}\right)$ by definition of relative complement {{iff}} :$A \subseteq \complement_S\left({B}\right)$ by Subset in Subsets. {{qed}}
=== Sufficient Condition === Follows by definition of subset. === Necessary Condition === Let: :$\forall x \in S: x \in A \implies x \in B$ Let $x \in A$. By definition of subset: :$x \in S$ Thus by assumption: :$x \in B$ The result follows by definition of subset. {{qed}} Category:Subsets iba73mvv9c2l5u0iu3nfd7s3nmb17dl
Sufficient condition follows by definition of empty set. For necessary condition assume that: :$\forall x \in S: x \notin A$ Let $x$ be arbitrary. {{AimForCont}} that: :$x \in A$ By definition of subset: :$x \in S$ By assumption: :$x \notin A$ Thus this contradicts: :$x \in A$ {{qed}} Category:Empty Set Category:Subsets g8jdogbcwe1pt3da9kpvz5dgt1njjio
From Pythagorean Triangle whose Area is Half Perimeter, the area of the $3-4-5$ triangle is $6$, which is trivially repdigit. The next Pythagorean triangles in area are: :the $6-8-10$ triangle, which has area $\dfrac {6 \times 8} 2 = 24$ :the $5-2-13$ triangle, which has area $\dfrac {5 \times 12} 2 = 30$ So there are no more Pythagorean triangles whose areas consist of a single digit. We have that the $693-1924-2045$ triangle is Pythagorean. Then its area $A$ is given by: {{begin-eqn}} {{eqn | l = A       | r = \dfrac {693 \times 1924} 2       | c =  }} {{eqn | r = \dfrac {\paren {3^2 \times 7 \times 11} \times \paren {2^2 \times 13 \times 37} } 2       | c = 693 and 1924 }} {{eqn | r = \paren {2 \times 3} \times \paren {3 \times 37} \times \paren {7 \times 11 \times 13}       | c =  }} {{eqn | r = 6 \times 111 \times 1001       | c = 6, 111 and 1001 }} {{eqn | r = 666 \, 666       | c =  }} {{end-eqn}} {{qed}}
Let $S_n$ be the $n$th partial sum of $\displaystyle \sum_{n \mathop = 1}^\infty z_n$. Let $P_n$ be the $n$th partial product of $\displaystyle \prod_{n \mathop = 1}^\infty \exp \left({z_n}\right)$. By Exponential of Sum, $\exp \left({S_n}\right) = P_n$ for all $n \in \N$. By Exponential Function is Continuous, $\displaystyle \lim_{n \mathop \to \infty} \exp \left({S_n}\right) = \exp z$. By Exponential of Complex Number is Nonzero, $\exp z \ne 0$. Thus $\displaystyle \lim_{n \mathop \to \infty} P_n = \exp z \ne 0$. {{qed}}
Let: : the $n$th partial sum of $\sequence {z_n}$ be denoted $Z_n$ : the $n$th partial sum of $\sequence {\Re \paren {z_n} }$ be denoted $U_n$ : the $n$th partial sum of $\sequence {\Im \paren {z_n} }$ be denoted $V_n$ Then: : $Z_n = U_n + i V_n$ Let: : $\lim_{n \mathop \to \infty} U_n = U$ : $\lim_{n \mathop \to \infty} V_n = V$ By definition of convergent complex sequence: :$\lim_{n \mathop \to \infty} Z_n = \lim_{n \mathop \to \infty} U_n + i \lim_{n \mathop \to \infty} V_n$ and so: :$\lim_{n \mathop \to \infty} Z_n = U + i V$ and the result follows. {{Qed}}
Let $D$ be a directed subset of $S$ such that :$D \subseteq x^\preceq$ By Lower Closure of Element is Ideal: :$x^\preceq$ is directed. By definition of up-complete: :$D$ and $x^\preceq$ admit suprema. By Supremum of Subset: :$\sup D \preceq \map \sup {x^\preceq}$ By Supremum of Lower Closure of Element: :$\sup D \preceq x$ Thus by definition of lower closure of element: :$\sup D \in x^\preceq$ {{qed}}
We have from Solutions of Pythagorean Equation that the set of all primitive Pythagorean triples is generated by: :$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m, n \in \Z$ such that: :$m, n \in \Z_{>0}$ are (strictly) positive integers :$m \perp n$, that is, $m$ and $n$ are coprime :$m$ and $n$ are of opposite parity :$m > n$. === Necessary Condition === Let the generator $G$ for $T$ be of the form: :$G = \set {n, n + 1}$ Recall that from Solutions of Pythagorean Equation, $T$ is of the form: :$\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m, n \in \Z$ such that: :$m, n \in \Z_{>0}$ are (strictly) positive integers :$m \perp n$, that is, $m$ and $n$ are coprime :$m$ and $n$ are of opposite parity :$m > n$. We are given that $n \in \Z_{>0}$ such that $m = n + 1$. We have that Consecutive Integers are Coprime. They are also of opposite parity. Therefore the Pythagorean triple generated by $n$ and $n + 1$ is primitive. So: {{begin-eqn}} {{eqn | l = 2 m n       | r = 2 n \paren {n + 1}       | c =  }} {{eqn | l = m^2 - n^2       | r = \paren {n + 1}^2 - n^2       | c =  }} {{eqn | r = n^2 + 2 n + 1 - n^2       | c =  }} {{eqn | r = 2 n + 1       | c =  }} {{eqn | l = m^2 + n^2       | r = n^2 + 2 n + 1 + n^2       | c =  }} {{eqn | r = 2 n^2 + 2 n + 1       | c =  }} {{eqn | r = 2 n \paren {n + 1} + 1       | c =  }} {{end-eqn}} Thus it is seen that $P$ is a  Pythagorean triangle where: :one legs is of length $2 n \paren {n + 1}$ :the hypotenuse is of  length $2 n \paren {n + 1} + 1$ Thus $P$ is a Pythagorean triangle whose  hypotenuse of $P$ is $1$ greater than one of its legs. {{qed|lemma}} === Sufficient Condition === Let the hypotenuse of $P$ be $1$ greater than one of its legs. Let the legs of $P$ be $a$ and $b$. Let the hypotenuse of $P$ be $h$ such that $h = b + 1$. By Consecutive Integers are Coprime, $h$ is coprime to $b$. Then by Elements of Primitive Pythagorean Triple are Pairwise Coprime, $P$ is primitive. Because $h$ and $b$ are of opposite parity, it follows that $b$ is even. Thus by Solutions of Pythagorean Equation: :$a = m^2 - n^2$ :$b = 2 m n$ :$h = m^2 + n^2$ for some $m, n \in \Z_{>0}$ where $m > n$. Thus: {{begin-eqn}} {{eqn | l = h       | r = b + 1       | c =  }} {{eqn | ll= \leadsto       | l = m^2 + n^2       | r = 2 m n + 1       | c =  }} {{eqn | ll= \leadsto       | l = m^2 - 2 m n + n^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \paren {m - n}^2       | r = 1       | c = Square of Difference }} {{eqn | ll= \leadsto       | l = m - n       | r = 1       | c = {{Defof|Square Root}}, and we have that $m > n$ }} {{eqn | ll= \leadsto       | l = m       | r = n + 1       | c =  }} {{end-eqn}} Hence the result. {{qed}}
By definition of almost isosceles: :$\size {a - b} = 1$ First note that, from Consecutive Integers are Coprime, an almost isosceles Pythagorean triangle is a primitive Pythagorean triangle. Hence $T$ and $T'$ are primitive Pythagorean triples. Thus it is established that by Solutions of Pythagorean Equation, both $P$ and $P'$ are of the form: :$\tuple {2 m n, m^2 + n^2, m^2 - n^2}$ for some $m, n \in \Z_{>0}$ where: :$m > n$ :$m \perp n$ :$m$ and $n$ are of opposite parity. === Necessary Condition === Let $\tuple {2 m + n, m}$ be the generator for the Pythagorean triple $T'$ of the almost isosceles Pythagorean triangle $P'$. Let $p$ and $q$ be the legs of $P'$. By Solutions of Pythagorean Equation: :$p = 2 \paren {2 m + n} m$ :$q = \paren {2 m + n}^2 - m^2$ Thus: {{begin-eqn}} {{eqn | l = \size {2 \paren {2 m + n} m - \paren {\paren {2 m + n}^2 - m^2} }       | r = 1       | c = {{Defof|Almost Isosceles Pythagorean Triangle}} }} {{eqn | ll= \leadsto       | l = \size {4 m^2 + 2 m n - \paren {4 m^2 + 4 m n + n^2 - m^2} }       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \size {\paren {m^2 - n^2} - 2 m n}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \size {a - b}       | r = 1       | c = Solutions of Pythagorean Equation }} {{end-eqn}} That is, $P$ is almost isosceles. {{qed|lemma}} === Sufficient Condition === Let $P$ be almost isosceles. {{WLOG}}, by Solutions of Pythagorean Equation: :$a = 2 m n$ :$b = m^2 - n^2$ Thus: {{begin-eqn}} {{eqn | l = \size {a - b}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \size {\paren {m^2 - n^2} - 2 m n}       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \size {4 m^2 + 2 m n - \paren {4 m^2 + 4 m n + n^2 - m^2} }       | r = 1       | c = working the above sequence of equations in reverse }} {{eqn | ll= \leadsto       | l = \size {2 \paren {2 m + n} m - \paren {\paren {2 m + n}^2 - m^2} }       | r = 1       | c = Solutions of Pythagorean Equation }} {{end-eqn}} By Solutions of Pythagorean Equation, $2 \paren {2 m + n} m$ and $\paren {2 m + n}^2 - m^2$ are the legs of a Pythagorean triangle $P'$ whose generator is $\paren {2 m + n, m}$. But from the above, these legs differ by $1$. Hence, by definition, $P'$ is an almost isosceles Pythagorean triangle $P'$. {{qed}}
By Lower Closure of Element is Closed under Directed Suprema: :$x^\preceq$ is closed under directed suprema. By Lower Closure of Singleton: :$\set x^\preceq = x^\preceq$ By Lower Closure is Lower Set: :$x^\preceq$ is a lower set. By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set: :$x^\preceq$ is closed. By definitions of lower closure of element and reflexivity: :$x \in x^\preceq$ By Singleton of Element is Subset: :$\set x \subseteq x^\preceq$ We will prove that: :for every a closed subset $C$ of $S$: $\set x \subseteq C \implies x^\preceq \subseteq C$ Let $C$ be a closed subset of $S$ such that: :$\set x \subseteq C$ By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set: :$C$ is a lower set. Let $y \in x^\preceq$. By definition of lower closure of element: :$y \preceq x$ By definitions of subset and singleton: :$x \in C$ Thus by definition of lower set: :$y \in C$ {{qed|lemma}} Thus by definition of topological closure: :$\set x^- = x^\preceq$ {{qed}}
{{begin-eqn}} {{eqn | l = \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)       | r = \left({a c + b d}\right)^2 + \left({a d - b c}\right)^2       | c = Brahmagupta-Fibonacci Identity }} {{eqn | ll= \leadsto       | l = \left({a^2 + \left({-b}\right)^2}\right) \left({c^2 + d^2}\right)       | r = \left({a c + \left({-b}\right) d}\right)^2 + \left({a d - \left({-b}\right) c}\right)^2       | c = substituting $-b$ for $b$ }} {{eqn | ll= \leadsto       | l = \left({a^2 + b^2}\right) \left({c^2 + d^2}\right)       | r = \left({a c - b d}\right)^2 + \left({a d + b c}\right)^2       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Brahmagupta|name2 = Leonardo Fibonacci}}
Let $x, y \in S$ such that :$x \ne y$ By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice: :$\left\{ {x}\right\}^- = x^\preceq$ and $\left\{ {y}\right\}^- = y^\preceq$ Thus by Lower Closures are Equal implies Elements are Equal: :$\left\{ {x}\right\}^- \ne \left\{ {y}\right\}^-$ Hence by Characterization of T0 Space by Distinct Closures of Singletons: :$T$ is $T_0$ space. {{qed}}
By definitions of lower closure of element and reflexivity: :$x \in x^\preceq$ and $y \in y^\preceq$ By definition of lower closure of element: :$x \preceq y$ and $y \preceq x$ Thus by definition of antisymmetry: :$x = y$ {{qed}}
By Fermat's Two Squares Theorem, a Pythagorean prime, $p$ can be expressed in the form: :$p = m^2 + n^2$ where $m$ and $n$ are (strictly) positive integers. :$(1): \quad m \ne n$, otherwise $p$ would be of the form $2 m^2$ and so even and therefore not a prime. :$(2): \quad m \perp n$, otherwise $\exists c \in \Z: p = c^2 \paren {p^2 + q^2}$ for some $p, q \in \Z$ and therefore not a prime. :$(3): \quad m$ and $n$ are of opposite parity, otherwise $p = m^2 + n^2$ is  even and therefore not a prime. {{WLOG}}, let $m > n$. From Solutions of Pythagorean Equation, the triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ forms a primitive Pythagorean triple such that $m^2 + n^2$ forms the hypotenuse of a primitive Pythagorean triangle. Thus we have that $p^2 = r^2 + s^2$ for some (strictly) positive integers $r$ and $s$. Similarly, we have that $r \ne s$, otherwise $p^2$ would be of the form $2 r^2$ and so not a square number. {{WLOG}}, let $r > s$. From Solutions of Pythagorean Equation, the triple $\tuple {2 r s, r^2 - s^2, r^2 + s^2}$ forms a Pythagorean triple such that $p^2 = r^2 + s^2$ forms the hypotenuse of a Pythagorean triangle. {{qed}}
Let $a$ and $b$ both be even. Then: {{begin-eqn}} {{eqn | l = a       | r = 2 n       | c =  }} {{eqn | l = b       | r = 2 \paren {n + 1}       | c = for some $n \in \Z_{>0}$ }} {{eqn | ll= \leadsto       | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 {2 n} + \dfrac 1 {2 \paren {n + 1} }       | c =  }} {{eqn | r = \dfrac {n + \paren {n + 1} } {2 n \paren {n + 1} }       | c =  }} {{eqn | r = \dfrac {2 n + 1} {2 n \paren {n + 1} }       | c =  }} {{eqn | r = \dfrac {n^2 + 2 n + 1 - n^2} {2 n \paren {n + 1} }       | c =  }} {{eqn | r = \dfrac {\paren {n + 1}^2 - n^2} {2 n \paren {n + 1} }       | c =  }} {{end-eqn}} From Solutions of Pythagorean Equation, $\tuple {n, n + 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = n + 1$. The legs of the resulting primitive Pythagorean triangle are $2 n \paren {n + 1}$ and $\paren {n + 1}^2 - n^2$. Let $a$ and $b$ both be odd. Then: {{begin-eqn}} {{eqn | l = a       | r = 2 n + 1       | c =  }} {{eqn | l = b       | r = 2 \paren {n + 1} + 1       | c = for some $n \in \Z_{>0}$ }} {{eqn | ll= \leadsto       | l = \dfrac 1 a + \dfrac 1 b       | r = \dfrac 1 {2 n + 1} + \dfrac 1 {2 \paren {n + 1} + 1}       | c =  }} {{eqn | r = \dfrac {4 n + 4} {4 n^2 + 8 n + 3}       | c =  }} {{eqn | r = \dfrac {2 \paren {2 n + 2} \times 1} {\paren {2 n + 2}^2 - 1}       | c =  }} {{end-eqn}} From Solutions of Pythagorean Equation, $\tuple {2 n + 2, 1}$ form the generator for the primitive Pythagorean triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ where $m = 2 n + 2$ and $n = 1$. The legs of the resulting primitive Pythagorean triangle are $2 \paren {2 n + 2}$ and $\paren {2 n + 2}^2 - 1$. {{qed}}
By Pythagoras's Theorem, the square of the hypotenuse of $P_1$ is $a^2 + b^2$. By Solutions of Pythagorean Equation, the sides of $P_2$ can be expressed as $\tuple {2 a b, a^2 - b^2, a^2 + b^2}$, where the hypotenuse is $a^2 + b^2$. {{qed}}
{{AimForCont}} $z \in \Q$. By definition of rational numbers: :$\exists a, b \in \Z, b \ne 0: x = \dfrac a b$ :$\exists c, d \in \Z, d \ne 0: z = \dfrac c d$ Then: {{begin-eqn}} {{eqn | l = x + y       | r = z       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac a b + y       | r = \dfrac c d       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac {b c - a d} {b d}       | c = by rearrangement }} {{end-eqn}} This shows that $y$ is  rational, which is a contradiction. By Proof by Contradiction, $z$ is irrational. {{qed}} Category:Rational Numbers Category:Irrational Numbers oewkqy1szldvm50y77fqhvs2pevj8ei
By general variation of integral functional with $n = 1$: :$\ds \delta J \sqbrk{y; h} = \int_{x_0}^{x_1} \intlimits {\paren {F_y - \dfrac \d {\d x} F_{y'} } \map h x + F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$ Since the curve $y = \map y x$ is an extremum of $\map J y$, the integral term vanishes: :$\ds \delta J \sqbrk {y; h} = \bigintlimits {F_{y'} \delta y} {x \mathop = x_0} {x \mathop = x_1} + \bigintlimits {\paren {F - y'F_{y'} } \delta x} {x \mathop = x_0} {x \mathop = x_1}$ According to Taylor's Theorem: :$\delta y_0 = \paren {\map {\phi'} {x_0} + \epsilon_0} \delta x_0$ :$\delta y_1 = \paren {\map {\phi'} {x_1} + \epsilon_1} \delta x_1$ {{explain|what do square brackets mean?}} where $\epsilon_0 \to 0$ as $\delta x_0 \to 0$ and $\epsilon_1 \to 0$ as $\delta x_1 \to 0$. {{Stub| add a figure; explain better}} Substitution of $\delta y_0$ and $\delta y_1$ into $\delta J$ leads to: :$\delta J \sqbrk {y; h} = \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_1} \delta x_1 - \bigvalueat {\paren {F_{y'} \psi' + F - y' F_{y'} } } {x \mathop = x_0} \delta x_0$ $y = \map y x$ is an extremal of $J \sqbrk y$, thus $\delta J = 0$. $\delta x_0$ and $\delta x_1$ are independent increments.  Hence both remaining terms in $\delta J$ vanish independently. {{qed}}