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__NOTOC__
 
__NOTOC__
 
= Welcome to MWiki =
 
= Welcome to MWiki =
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== Theorem of the month ==
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== Theorems of the month ==
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=== Prime number theorem ===
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=== Leibniz' differentiation rule ===
  
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For <math>\pi(x) := |\{p \in {}^{\omega}{\mathbb{P}} : p \le x \in {}^{\omega}{\mathbb{R}}\}|</math> holds <math>\pi(\omega) = \widehat{{_e}\omega}\omega + \mathcal{O}({_e}\omega\sqrt{\omega})</math>.
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For <math>f: {}^{(\omega)}\mathbb{K}^{\grave{n}} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright x := {(s, {x}_{2}, ..., {x}_{n})}^{T}</math> and <math>s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\}</math>, choosing <math>\curvearrowright a(x) = a(\curvearrowright x)</math> and <math>\curvearrowright b(x) = b(\curvearrowright 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}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).</math></div>
  
 
==== Proof: ====
 
==== Proof: ====
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In the sieve of Eratosthenes, the number of prime numbers decreases almost regularly. From intervals of fix length <math>y \in {}^{\omega}{\mathbb{R}_{&gt;0}}, \hat{2}y</math> set-2-tuples of prime numbers are formed such that the first interval has the unchanged representative prime number density and the second interval is empty, then the interval with the second most prime number density is followed by the second least one etc.
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<div style="text-align:center;"><math>\begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned}</math></div>
  
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For induction basis <math>n = 2</math> resp. 3, the induction hypothesis is that the first interval contains <math>x_n/{_e}x_n</math> prime numbers for <math>n \in {}^{\omega}{\mathbb{N}_{\ge2}}</math> and arbitrary <math>x_4 \in [2, 4[</math>. Then the induction step from <math>x_n</math> to <math>x_n^2</math> by considering the prime gaps of prime <math>p\# /q + 1</math> for <math>p, q \in {}^{\omega}\mathbb{P}</math> proves that there are <math>\pi(x_n^2) = \pi(x_n) x_n/2</math> prime numbers only from <math>\pi(x_n) = x_n/{_e}x_n</math>. The average distance between the prime numbers is <math>{_e}x_n</math> and the maximal <math>x_n^2</math> to <math>x_n</math> behaves like <math>\omega</math> to <math>\sqrt{\omega}.\square</math>
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=== Beal's theorem ===
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Equation <math>a^m + b^n = c^k</math> for <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>
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==== Proof: ====
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For <math>b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr}</math>, the function <math>f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0</math> is continuous in <math>q, r \in {}^{\omega} \mathbb{R}_{>0}</math> and <math>(q_0, r_0) = \left(\check{1}, \check{1}\right)</math> solves the equation. Every further solution in fractions yields after exponentiation gcd<math>(a, c) > 1</math> and thus proves the claim.<math>\square</math>
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=== Conclusion: ===
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The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd<math>(a, b, c) > 1</math> 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>
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 08:15, 4 March 2024

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 x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] and [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math], choosing [math]\displaystyle{ \curvearrowright a(x) = a(\curvearrowright x) }[/math] and [math]\displaystyle{ \curvearrowright b(x) = b(\curvearrowright x) }[/math], it holds that

[math]\displaystyle{ \tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)). }[/math]

Proof:

[math]\displaystyle{ \begin{aligned}\tfrac{{\downarrow} }{{\downarrow} {{x}_{1}}}\left( {\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right) &={\left( {\uparrow}_{a(\curvearrowright x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{b(x)}{f(x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\left( {\uparrow}_{a(x)}^{b(x)}{(f(\curvearrowright x,t)-f(x,t)){\downarrow}t}+{\uparrow}_{b(x)}^{b(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t}-{\uparrow}_{a(x)}^{a(\curvearrowright x)}{f(\curvearrowright x,t){\downarrow}t} \right)}/{{\downarrow} {{x}_{1}}}\; \\ &={\uparrow}_{a(x)}^{b(x)}{\tfrac{{\downarrow} f(x,t)}{{\downarrow} {{x}_{1}}}{\downarrow}t}+\tfrac{{\downarrow} b(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,b(x))-\tfrac{{\downarrow} a(x)}{{\downarrow} {{x}_{1}}}f(\curvearrowright x,a(x)).\square\end{aligned} }[/math]

Beal's theorem

Equation [math]\displaystyle{ a^m + b^n = c^k }[/math] for [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:

For [math]\displaystyle{ b^n = (c^{kq}-a^{mr})\left(\tilde{c}^{k\acute{q}} + \tilde{a}^{m\acute{r}}\right) = c^k - a^m + c^{kq} \tilde{a}^{m\acute{r}} - \tilde{c}^{k\acute{q}} a^{mr} }[/math], the function [math]\displaystyle{ f(q,r) := c^{k(\hat{q}-1)} - a^{m(\hat{r}-1)} = 0 }[/math] is continuous in [math]\displaystyle{ q, r \in {}^{\omega} \mathbb{R}_{\gt 0} }[/math] and [math]\displaystyle{ (q_0, r_0) = \left(\check{1}, \check{1}\right) }[/math] solves the equation. Every further solution in fractions yields after exponentiation gcd[math]\displaystyle{ (a, c) \gt 1 }[/math] and thus proves the claim.[math]\displaystyle{ \square }[/math]

Conclusion:

The Fermat-Catalan conjecture can be proven analogously and an infinite descent implies because of gcd[math]\displaystyle{ (a, b, c) \gt 1 }[/math] 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]

Recommended reading

Nonstandard Mathematics

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