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= Welcome to MWiki = | = Welcome to MWiki = | ||
− | == | + | == Theorems of the month == |
+ | === Leibniz' differentiation rule === | ||
− | === | + | For <math>f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math>, and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright D a(x) = a(\curvearrowright B x)</math> and <math>\curvearrowright D b(x) = b(\curvearrowright B x)</math> it holds that<div style="text-align:center;"><math>\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).</math></div> |
− | + | ==== Proof: ==== | |
− | + | <div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right) &={\left( {\uparrow}_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t)){\downarrow}Dt}+{\uparrow}_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt}-{\uparrow}_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t){\downarrow}Dt} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}Dt}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned}</math></div> | |
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− | + | === Beal's theorem === | |
+ | Equation <math>a^m + b^n = c^k</math> where <math>a, b, c \in \mathbb{N}^{*}</math> and <math>k, m, n \in \mathbb{N}_{\ge 3}</math> implies gcd<math>(a, b, c) > 1.</math> | ||
==== Proof: ==== | ==== Proof: ==== | ||
− | + | Proof: Real points between <math>r, s \in {}^{\omega}\mathbb{Q}</math> do not avoid that every nontrivial equation for <math>c^k > 1</math> is given by <math>(a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s)</math> where all relations <math>a^{m-\hat{r}} = b^{n-\hat{s}}</math> then imply <math>p \mid</math> gcd<math>(a, b, c)</math> as well as the claim for <math>p \in {}^{\omega} \mathbb{P}.\square</math> | |
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− | + | === Conclusion: === | |
+ | The preceding theorem enables an infinite descent because of gcd<math>(a, b, c) > 1</math> such that no <math>n \in {}^{\omega}\mathbb{N}_{\ge 3}</math> satisfies <math>a^n + b^n = c^n</math> for arbitrary <math>a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square</math> | ||
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== Recommended reading == | == Recommended reading == | ||
Revision as of 16:58, 1 March 2023
Welcome to MWiki
Theorems of the month
Leibniz' differentiation rule
For [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math], and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright D a(x) = a(\curvearrowright B x) }[/math] and [math]\displaystyle{ \curvearrowright D b(x) = b(\curvearrowright B x) }[/math] it holds that
Proof:
Beal's theorem
Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] where [math]\displaystyle{ a, b, c \in \mathbb{N}^{*} }[/math] and [math]\displaystyle{ k, m, n \in \mathbb{N}_{\ge 3} }[/math] implies gcd[math]\displaystyle{ (a, b, c) \gt 1. }[/math]
Proof:
Proof: Real points between [math]\displaystyle{ r, s \in {}^{\omega}\mathbb{Q} }[/math] do not avoid that every nontrivial equation for [math]\displaystyle{ c^k \gt 1 }[/math] is given by [math]\displaystyle{ (a^{m-r} + ib^{n-s})(a^r - ib^s) =c^k +i(a^rb^{n-s} - a^{m-r}b^s) }[/math] where all relations [math]\displaystyle{ a^{m-\hat{r}} = b^{n-\hat{s}} }[/math] then imply [math]\displaystyle{ p \mid }[/math] gcd[math]\displaystyle{ (a, b, c) }[/math] as well as the claim for [math]\displaystyle{ p \in {}^{\omega} \mathbb{P}.\square }[/math]
Conclusion:
The preceding theorem enables an infinite descent because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] such that no [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 3} }[/math] satisfies [math]\displaystyle{ a^n + b^n = c^n }[/math] for arbitrary [math]\displaystyle{ a, b, c \in {}^{\omega}\mathbb{N}^{*}.\square }[/math]