Difference between revisions of "Main Page"

From MWiki
Jump to: navigation, search
m (→‎Proof:)
(69 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
__NOTOC__
 
__NOTOC__
 
= Welcome to MWiki =
 
= Welcome to MWiki =
−
== Theorem of the month ==
+
== Theorems of the month ==
−
=== Fermat's Last Theorem ===
+
=== Leibniz' differentiation rule ===
  
−
For all <math>p \in {}^{\omega }{\mathbb{P}_{\ge 3}}</math> and <math>x, y, z \in {}^{\omega }{\mathbb{N}^{*}}</math>, always <math>x^p + y^p \ne z^p</math> holds and thus for all <math>m \in {}^{\omega }{\mathbb{N}_{\ge 3}}</math> instead of <math>p</math>.
+
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: ====
−
Because of [[w:Fermat's little theorem|<span class="wikipedia">Fermat's little theorem</span>]], rewritten, <math>f_{akp}(n) := (2n + a - kp)^p - n^p - (n + a)^p \ne 0</math> is to show for <math>a, k, n \in {}^{\omega }{\mathbb{N}^{*}}</math> where <math>kp &lt; n</math>.
+
<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>
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
 
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content">From <math>x := n, y:= n + a</math> and <math>z := 2n + a + d</math> where <math>d \in {}^{\omega }{\mathbb{N}^{*}}</math>, it follows due to <math>z^p \equiv y, y^p \equiv y</math> and <math>z^p \equiv z</math> first <math>d \equiv 0 \mod p</math>, then <math>d = \pm kp</math>. Since <math>x + y = 2n + a &gt; z</math> is required, <math>f_{akp}(n)</math> is chosen properly.</div></div>
 
  
−
[[w:Mathematical induction|<span class="wikipedia">Induction</span>]] for <math>n</math> implies the claim due to the case <math>m = 4</math><ref name="Ribenboim">[[w:Paulo Ribenboim|<span class="wikipedia">Ribenboim, Paulo</span>]]: ''Thirteen Lectures on Fermat's Last Theorem'' : 1979; Springer; New York; ISBN 9780387904320, p. 35 - 38.</ref> and <math>y &gt; x &gt; p</math><ref name="loccit">loc. cit., p. 226.</ref>:
+
=== Beal's theorem ===
 +
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>
  
−
'''Induction basis''' <math>(n \le p): f_{akp}(n) \ne 0</math> for all <math>a, k</math> and <math>p</math>. Let <math>r \in {}^{\omega }{\mathbb{N}_{&lt; p}}</math>.  
+
==== Proof: ====
 +
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>
  
−
'''Induction step''' <math>\,(n = q + r \; \rightarrow \; n^{*} = n + p):</math> Let <math>f_{akp}(n^{*}) \ge 0</math>, but <math>f_{akp}(n) &lt; 0</math>, since <math>f_{akp}(n)</math> is [[w:Monotonic function|<span class="wikipedia">strictly monotonically increasing</span>]] and otherwise nothing to prove.
+
=== Conclusion: ===
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
+
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>
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content">The strict monotonicity follows from (continuously) differentiating by <math>n</math> such that <math>f_{akp}(n)' = p(2(2n + a - kp)^{p - 1} - n^{p - 1} - (n + a)^{p - 1}) &gt; 0</math>.</div></div>
 
−
 
 
−
It holds <math>f_{akp}(n^{*}) = (\int_0^{n^{*}}{f_{akp}(v)}dv)' \ne 0</math>, since <math>(n^{*})^{p + 1} + (n^{*} + a)^{p + 1}</math> does not divide <math>((n^{*})^p + (n^{*} + a)^p)^2</math> after separating the positive factor as [[w:Polynomial long division|<span class="wikipedia">polynomial division</span>]] shows.<math>\square</math>
 
−
<div class="toccolours mw-collapsible mw-collapsed" style="width:100%; overflow:auto;">
 
−
<div style="font-weight:bold;line-height:1.6;">Proof details</div>
 
−
<div class="mw-collapsible-content"><math>\int_0^{n^{*}}{f_{akp}(v)}dv = ((2n^{*} + a - kp)^{p + 1} / 2 - (n^{*})^{p + 1} - (n^{*} + a)^{p + 1})/(p + 1) + t = ((2n^{*} + a - kp)^{(p + 1)/2} \pm \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})^2/(2p + 2) + t</math> for <math>t \in {}^{\omega}{\mathbb{Q}}</math> where the third binomial formula <math>r^2 - s^2 = (r \pm s)^2 := (r + s)(r - s)</math> was used. After separating the negligible factor <math>\hat{2}((2n^{*} + a - kp)^{(p + 1)/2} + \sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})/(p + 1)</math>, the derivative is just <math>(\hat{2}(2n^{*} + a - kp)^{(p - 1)/2} - \hat{2}((n^{*})^p + (n^{*} + a)^p)/\sqrt{2(n^{*})^{p + 1} + 2(n^{*} + a)^{p + 1}})</math>. After squaring the terms, the polynomial division gives <math>(n^{*})^{p - 1} + (n^{*} + a)^{p - 1} + a^2(n^{*})^{p - 1}(n^{*} + a)^{p - 1}/((n^{*})^{p + 1} + (n^{*} + a)^{p + 1})</math> as recalculating by multiplication confirms.</div></div>
 
  
 
== Recommended reading ==
 
== Recommended reading ==
  
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
 
[https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
−
 
−
== References ==
 
−
<references />
 
  
 
[[de:Hauptseite]]
 
[[de:Hauptseite]]

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

Retrieved from "https://en.mwiki.de/index.php?title=Main_Page&oldid=297"