Difference between revisions of "Main Page"

From MWiki
Jump to: navigation, search
(Greatest-prime Criterion and Transcendence of Euler's Constant)
(Proof:)
(43 intermediate revisions by the same user not shown)
Line 2: Line 2:
 
= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
=== Greatest-prime Criterion ===
+
=== Leibniz' differentiation rule ===
  
If a real number may be represented as an irreducible fraction <math>\widehat{ap}b \pm \hat{s}t</math>, where <math>a, b, s</math>, and <math>t</math> are natural numbers, <math>abst \ne 0</math>, <math>a + s &gt; 2</math>, and the (second-)greatest prime number <math>p \in {}^{\omega }\mathbb{P}, p \nmid b</math> and <math>p \nmid s</math>, then <math>r</math> is <math>\omega</math>-transcendental.
+
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: ====
 
==== Proof: ====
The denominator <math>\widehat{ap s} (bs \pm apt)</math> is <math>\ge 2p \ge 2\omega - \mathcal{O}({_e}\omega\sqrt{\omega}) &gt; \omega</math> by the prime number theorem.<math>\square</math>
+
<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>
  
=== Transcendence of Euler's Constant ===
+
=== 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>
For <math>x \in {}^{\omega }{\mathbb{R}}</math>, let be <math>s(x) := \sum\limits_{n=1}^{\omega}{\hat{n}{{x}^{n}}}</math> and <math>\gamma := s(1) - {_e}\omega = \int\limits_{1}^{\omega}{\left( \widehat{\left\lfloor x \right\rfloor} - \hat{x} \right)dx}</math> Euler's constant, where rearranging shows <math>\gamma \in \; ]0, 1[</math>.
 
 
 
If <math>{_e}\omega = s(\hat{2})\;{_2}\omega</math> is accepted, <math>\gamma \in {}^{\omega }\mathbb{T}_{\mathbb{R}}</math> is true with a precision of <math>\mathcal{O}({2}^{-\omega}\hat{\omega}\;{_e}\omega)</math>.
 
  
 
==== Proof: ====
 
==== Proof: ====
The (exact) integration of the geometric series yields <math>-{_e}(-\acute{x}) = s(x) + \mathcal{O}(\hat{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx</math> for <math>x \in [-1, 1 - \hat{\nu}]</math> and <math>t(x) \in {}^{\omega }{\mathbb{R}}</math> such that <math>|t(x)| &lt; {\omega}</math>.
+
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>
  
After applying Fermat's little theorem to the numerator of <math>\hat{p}(1 - 2^{-p}\,{_2}\omega)</math> for <math>p = \max\, {}^{\omega}\mathbb{P}</math>, the greatest-prime criterion yields the claim.<math>\square</math>
+
=== 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>
  
 
== Recommended reading ==
 
== Recommended reading ==

Revision as of 15:58, 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

[math]\displaystyle{ \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]

Proof:

[math]\displaystyle{ \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]

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]

Recommended reading

Nonstandard Mathematics