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(Counter-directional and Archimedes' theorem)
(Greatest-prime Criterion and Transcendence of Euler's Constant)
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= Welcome to MWiki =
 
= Welcome to MWiki =
 
== Theorems of the month ==
 
== Theorems of the month ==
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=== Counter-directional theorem ===
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=== Greatest-prime Criterion ===
  
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If the path <math>\gamma: [a, b[ \, \cap \, C \rightarrow V</math> with <math>C \subseteq \mathbb{R}</math> passes the edges of every <math>n</math>-cube of side length <math>\iota</math> in the <math>n</math>-volume <math>V \subseteq {}^{(\omega)}\mathbb{R}^{n}</math> with <math>n \in \mathbb{N}_{\ge 2}</math> exactly once, where the opposite edges in all two-dimensional faces of every <math>n</math>-cube are traversed in reverse direction, but uniformly, then, for <math>D \subseteq \mathbb{R}^{2}, B \subseteq {V}^{2}, f = ({f}_{1}, ..., {f}_{n}): V \rightarrow {}^{(\omega)}\mathbb{R}^{n}, \gamma(t) = x, \gamma(\curvearrowright t) = \curvearrowright x</math> and <math>{V}_{\curvearrowright } := \{\curvearrowright x \in V: x \in V, \curvearrowright x \ne \curvearrowleft x\}</math>, it holds that
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If a real number may be represented as an irreducible fraction <math>\widetilde{ap}b \pm \tilde{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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==== Proof: ====
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The denominator <math>\widetilde{ap s} (bs \pm apt)</math> is <math>\ge \hat{p} \ge \hat{\omega} - \mathcal{O}({_e}\omega{\omega}^{\tilde{2}}) &gt; \omega</math> by the prime number theorem.<math>\square</math>
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=== Transcendence of Euler's Constant ===
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For <math>x \in {}^{\omega }{\mathbb{R}}</math>, let be <math>s(x) := {+}_{n=1}^{\omega}{\tilde{n}{{x}^{n}}}</math> and <math>\gamma := s(1) - {_e}\omega = {\uparrow}_{1}^{\omega}{\left( \widetilde{\left\lfloor x \right\rfloor} - \tilde{x} \right)\downarrow x}</math> Euler's constant, where rearranging shows <math>\gamma \in \; ]0, 1[</math>.
  
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<div style="text-align:center;"><math>\uparrow_{t \in [a,b[ \; \cap \; C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)\downarrow t}=\uparrow_{\begin{smallmatrix} (x,\curvearrowright x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)\downarrow x}=\uparrow_{\begin{smallmatrix} t \in [a,b[ \; \cap \; C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)\downarrow t}.</math></div>
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If <math>{_e}\omega = s(\tilde{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}\tilde{\omega}\;{_e}\omega)</math>.
  
 
==== Proof: ====
 
==== Proof: ====
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If two arbitrary squares are considered with common edge of length <math>\iota</math> included in one plane, then only the edges of <math>V\times{V}_{\curvearrowright}</math> are not passed in both directions for the same function value. They all, and thus the path to be passed, are exactly contained in <math>{\partial}^{\acute{n}}V.\square</math>
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The (exact) integration of the geometric series yields <math>-{_e}(-\acute{x}) = s(x) + \mathcal{O}(\tilde{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx</math> for <math>x \in [-1, 1 - \tilde{\nu}]</math> and <math>t(x) \in {}^{\omega }{\mathbb{R}}</math> such that <math>|t(x)| &lt; {\omega}</math>.
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After applying Fermat's little theorem to the numerator of <math>\tilde{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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=== Archimedes' theorem ===
 
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There exists <math>m \in {}^{\nu}\mathbb{N}</math> such that <math>a < bm</math> if and only if <math>a < b\nu</math> whenever <math>a > b</math> for <math>a, b \in {\mathbb{R}}_{>0}</math>, since <math>\nu = \max {}^{\nu}\mathbb{N}</math> holds.<math>\square</math>
 
 
== Recommended reading ==
 
== Recommended reading ==
  

Revision as of 00:39, 1 October 2023

Welcome to MWiki

Theorems of the month

Greatest-prime Criterion

If a real number may be represented as an irreducible fraction [math]\displaystyle{ \widetilde{ap}b \pm \tilde{s}t }[/math], where [math]\displaystyle{ a, b, s }[/math], and [math]\displaystyle{ t }[/math] are natural numbers, [math]\displaystyle{ abst \ne 0 }[/math], [math]\displaystyle{ a + s > 2 }[/math], and the (second-)greatest prime number [math]\displaystyle{ p \in {}^{\omega }\mathbb{P}, p \nmid b }[/math] and [math]\displaystyle{ p \nmid s }[/math], then [math]\displaystyle{ r }[/math] is [math]\displaystyle{ \omega }[/math]-transcendental.

Proof:

The denominator [math]\displaystyle{ \widetilde{ap s} (bs \pm apt) }[/math] is [math]\displaystyle{ \ge \hat{p} \ge \hat{\omega} - \mathcal{O}({_e}\omega{\omega}^{\tilde{2}}) > \omega }[/math] by the prime number theorem.[math]\displaystyle{ \square }[/math]

Transcendence of Euler's Constant

For [math]\displaystyle{ x \in {}^{\omega }{\mathbb{R}} }[/math], let be [math]\displaystyle{ s(x) := {+}_{n=1}^{\omega}{\tilde{n}{{x}^{n}}} }[/math] and [math]\displaystyle{ \gamma := s(1) - {_e}\omega = {\uparrow}_{1}^{\omega}{\left( \widetilde{\left\lfloor x \right\rfloor} - \tilde{x} \right)\downarrow x} }[/math] Euler's constant, where rearranging shows [math]\displaystyle{ \gamma \in \; ]0, 1[ }[/math].

If [math]\displaystyle{ {_e}\omega = s(\tilde{2})\;{_2}\omega }[/math] is accepted, [math]\displaystyle{ \gamma \in {}^{\omega }\mathbb{T}_{\mathbb{R}} }[/math] is true with a precision of [math]\displaystyle{ \mathcal{O}({2}^{-\omega}\tilde{\omega}\;{_e}\omega) }[/math].

Proof:

The (exact) integration of the geometric series yields [math]\displaystyle{ -{_e}(-\acute{x}) = s(x) + \mathcal{O}(\tilde{\omega}{x}^{\grave{\omega}}/\acute{x}) + t(x)dx }[/math] for [math]\displaystyle{ x \in [-1, 1 - \tilde{\nu}] }[/math] and [math]\displaystyle{ t(x) \in {}^{\omega }{\mathbb{R}} }[/math] such that [math]\displaystyle{ |t(x)| < {\omega} }[/math].

After applying Fermat's little theorem to the numerator of [math]\displaystyle{ \tilde{p}(1 - 2^{-p}\,{_2}\omega) }[/math] for [math]\displaystyle{ p = \max\, {}^{\omega}\mathbb{P} }[/math], the greatest-prime criterion yields the claim.[math]\displaystyle{ \square }[/math]

Recommended reading

Nonstandard Mathematics

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