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Theorem of the month

Counter-directional theorem

If the path [math]\displaystyle{ \gamma: [a, b[ \, \cap \, C \rightarrow V }[/math] with [math]\displaystyle{ C \subseteq \mathbb{R} }[/math] passes the edges of every [math]\displaystyle{ n }[/math]-cube of side length d0 in the [math]\displaystyle{ n }[/math]-volume [math]\displaystyle{ V \subseteq {}^{(\omega)}\mathbb{R}^{n} }[/math] with [math]\displaystyle{ n \in \mathbb{N}_{\ge 2} }[/math] exactly once, where the opposite edges in all two-dimensional faces of every [math]\displaystyle{ n }[/math]-cube are traversed in reverse direction, but uniformly, then, for [math]\displaystyle{ 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 D t) = \curvearrowright B x }[/math] and [math]\displaystyle{ {V}_{\curvearrowright } := \{\curvearrowright B x \in V: x \in V, \curvearrowright B x \ne \curvearrowleft B x\} }[/math], it holds that

[math]\displaystyle{ \int\limits_{t \in [a,b[ \, \cap \, C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}=\int\limits_{\begin{smallmatrix} (x,\curvearrowright B\,x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)dBx}=\int\limits_{\begin{smallmatrix} t \in [a,b[ \, \cap \, C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}. }[/math]

Proof:

If two arbitrary squares are considered with common edge of length d0 included in one plane, then only the edges of [math]\displaystyle{ 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]\displaystyle{ {\partial}^{\acute{n}}V.\square }[/math]

@@ Line 2: / Line 2: @@
 = Welcome to MWiki =
 == Theorem of the month ==
-=== Fermat's Last Theorem ===
+=== Counter-directional theorem ===
-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>.
+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 d0 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 D t) = \curvearrowright B x</math> and <math>{V}_{\curvearrowright } := \{\curvearrowright B x \in V: x \in V, \curvearrowright B x \ne \curvearrowleft B x\}</math>, it holds that
-==== 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 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>:
+<div style="text-align:center;"><math>\int\limits_{t \in [a,b[ \, \cap \, C}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}=\int\limits_{\begin{smallmatrix} (x,\curvearrowright B\,x) \\ \in V\times {{V}_{\curvearrowright}} \end{smallmatrix}}{f(x)dBx}=\int\limits_{\begin{smallmatrix} t \in [a,b[ \, \cap \, C, \\ \gamma | {\partial{}^{\acute{n}}} V \end{smallmatrix}}{f(\gamma (t)){{{{\gamma }'}}_{\curvearrowright }}(t)dDt}.</math></div>
-'''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: ====
+If two arbitrary squares are considered with common edge of length d0 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>
-'''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.
-<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">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 ==
 [https://en.calameo.com/books/003777977258f7b4aa332 Nonstandard Mathematics]
-== References ==
-<references />
 [[de:Hauptseite]]

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