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(Greatest-prime Criterion and Transcendence of Euler's Constant)
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= Welcome to MWiki =
 
= Welcome to MWiki =
== Theorems of the month ==
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== Theorem of the month ==
=== Greatest-prime Criterion ===
 
  
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.
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=== Universal multistep theorem ===
  
==== Proof: ====
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For <math>n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, d_{\curvearrowright B} x \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, f : [a, b] \times {}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\curvearrowright B x) := g_{\acute{k}}(x)</math> and <math>g_0(a) = f((\curvearrowleft B)a, y_0, ... , y_{\acute{n}})</math>, the Taylor series of the initial value problem <math>y^\prime(x) = f(x, y((\curvearrowright B)^0 x), ... , y((\curvearrowright B)^{\acute{n}} x))</math> of order <math>n</math> implies <div style="text-align:center;"><math>y(\curvearrowright B x) = y(x) - d_{\curvearrowright B}x\sum\limits_{k=1}^{p}{i^{2k} g_{p-k}((\curvearrowright B) x)\sum\limits_{m=k}^{p}{\widehat{m!}\binom{\acute{m}}{\acute{k}}}} + \mathcal{O}((d_{\curvearrowright B} x)^{\grave{p}}).\square</math></div>
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>
 
  
=== Transcendence of Euler's Constant ===
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=== Goldbach’s theorem ===
  
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>.
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Every even whole number greater than 2 is the sum of two primes.
  
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>.
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==== Proof: ====
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Induction over all prime gaps until the maximally possible one each time.<math>\square</math>
  
==== Proof: ====
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=== Foundation theorem ===
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>.
 
  
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>
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Only the postulation of the axiom of foundation that every nonempty subset <math>X \subseteq Y</math> contains an element <math>x_0</math> such that <math>X</math> und <math>x_0</math> are disjoint guarantees cycle freedom.
  
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==== Proof: ====
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Set <math>X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\}</math> and <math>x_{\acute{n}} := \{x_n\}</math> for <math>m \in {}^{\omega}\mathbb{N}</math> and <math>n \in {}^{\omega}\mathbb{N}_{\ge 2}\}</math> .<math>\square</math>
 
== Recommended reading ==
 
== Recommended reading ==
  

Revision as of 01:05, 1 June 2021

Welcome to MWiki

Theorem of the month

Universal multistep theorem

For [math]\displaystyle{ n \in {}^{\nu}\mathbb{N}_{\le p}, k, m, p \in {}^{\nu}\mathbb{N}^{*}, d_{\curvearrowright B} x \in\, ]0, 1[, x \in [a, b] \subseteq {}^{\omega}\mathbb{R}, y : [a, b] \rightarrow {}^{\omega}\mathbb{R}^q, f : [a, b] \times {}^{\omega}\mathbb{R}^{q \times n} \rightarrow {}^{\omega}\mathbb{R}^q, g_k(\curvearrowright B x) := g_{\acute{k}}(x) }[/math] and [math]\displaystyle{ g_0(a) = f((\curvearrowleft B)a, y_0, ... , y_{\acute{n}}) }[/math], the Taylor series of the initial value problem [math]\displaystyle{ y^\prime(x) = f(x, y((\curvearrowright B)^0 x), ... , y((\curvearrowright B)^{\acute{n}} x)) }[/math] of order [math]\displaystyle{ n }[/math] implies

[math]\displaystyle{ y(\curvearrowright B x) = y(x) - d_{\curvearrowright B}x\sum\limits_{k=1}^{p}{i^{2k} g_{p-k}((\curvearrowright B) x)\sum\limits_{m=k}^{p}{\widehat{m!}\binom{\acute{m}}{\acute{k}}}} + \mathcal{O}((d_{\curvearrowright B} x)^{\grave{p}}).\square }[/math]

Goldbach’s theorem

Every even whole number greater than 2 is the sum of two primes.

Proof:

Induction over all prime gaps until the maximally possible one each time.[math]\displaystyle{ \square }[/math]

Foundation theorem

Only the postulation of the axiom of foundation that every nonempty subset [math]\displaystyle{ X \subseteq Y }[/math] contains an element [math]\displaystyle{ x_0 }[/math] such that [math]\displaystyle{ X }[/math] und [math]\displaystyle{ x_0 }[/math] are disjoint guarantees cycle freedom.

Proof:

Set [math]\displaystyle{ X := \{x_m : x_0 := \{\emptyset\}, x_{\omega} := \{x_1\} }[/math] and [math]\displaystyle{ x_{\acute{n}} := \{x_n\} }[/math] for [math]\displaystyle{ m \in {}^{\omega}\mathbb{N} }[/math] and [math]\displaystyle{ n \in {}^{\omega}\mathbb{N}_{\ge 2}\} }[/math] .[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics