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= Welcome to MWiki = | = Welcome to MWiki = | ||
== Theorems of the month == | == Theorems of the month == | ||
− | === | + | === Counter-directional theorem === |
− | + | 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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− | + | <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> | |
− | === | + | ==== Proof: ==== |
− | + | 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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− | === | + | === Archimedes' theorem === |
− | + | 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 19:45, 31 August 2023
Welcome to MWiki
Theorems 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 [math]\displaystyle{ \iota }[/math] 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 t) = \curvearrowright x }[/math] and [math]\displaystyle{ {V}_{\curvearrowright } := \{\curvearrowright x \in V: x \in V, \curvearrowright x \ne \curvearrowleft x\} }[/math], it holds that
Proof:
If two arbitrary squares are considered with common edge of length [math]\displaystyle{ \iota }[/math] 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]
Archimedes' theorem
There exists [math]\displaystyle{ m \in {}^{\nu}\mathbb{N} }[/math] such that [math]\displaystyle{ a \lt bm }[/math] if and only if [math]\displaystyle{ a \lt b\nu }[/math] whenever [math]\displaystyle{ a \gt b }[/math] for [math]\displaystyle{ a, b \in {\mathbb{R}}_{\gt 0} }[/math], since [math]\displaystyle{ \nu = \max {}^{\nu}\mathbb{N} }[/math] holds.[math]\displaystyle{ \square }[/math]