Difference between revisions of "User:Foxjwill/Proofs"
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==A theorem== | ==A theorem== | ||
− | '''THEOREM.''' Let <math>C</math> be a circle of radius <math>r</math>, let <math>A</math> be the set of chords of <math>C</math>, for all <math>p\in \mathbb{R}^+</math>, let <math>S_p\equiv \{B\subset A|B\mbox{ is finite}, \sum_{b\in B}|b|=p\}</math>. <!--Additionally, let <math>\Theta\equiv \{\mbox{angle }\theta| \mbox{for all }a\in</math>--> Then for all <math>p\in \mathbb{R}^+</math>, there exists an angle <math>\phi\in \mathbb{R}^+</math> such that for all <math>B\in S_p</math>, there exists a positive integer <math>k</math> such that for all sets <math>\Theta\equiv \{\mbox{angle }\theta| \theta \in \Theta\mbox{ iff there exists a }b\in B \mbox{ that cuts },\theta\mbox{ and for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\},</math> | + | '''THEOREM.''' Let <math>C</math> be a circle of radius <math>r</math>, let <math>A</math> be the set of chords of <math>C</math>, for all <math>p\in \mathbb{R}^+</math>, let <math>S_p\equiv \{B\subset A|B\mbox{ is finite}, \sum_{b\in B}|b|=p\}</math>. <!--Additionally, let <math>\Theta\equiv \{\mbox{angle }\theta| \mbox{for all }a\in</math>--> Then for all <math>p\in \mathbb{R}^+</math>, there exists an angle <math>\phi\in \mathbb{R}^+</math> such that for all <math>B\in S_p</math>, there exists a positive integer <math>k</math> such that for all sets <math>\Theta\equiv \{\mbox{angle }\theta| \theta \in \Theta\mbox{ iff there exists a } b\in B \mbox{ that cuts },\theta\mbox{ and for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\},</math> |
<cmath> | <cmath> | ||
\sum_{\theta\in \Theta}\theta = \phi | \sum_{\theta\in \Theta}\theta = \phi | ||
</cmath> | </cmath> |
Revision as of 13:16, 13 January 2009
Proof that , where is prime, is irrational
- Assume that is rational. Then such that is coprime to and .
- It follows that , and that .
- So, by the properties of exponents along with the unique factorization theorem, divides both and .
- Factoring out from (2), we have for some .
- Therefore divides .
- But this contradicts the assumption that and are coprime.
- Therefore .
- Q.E.D.
A theorem
THEOREM. Let be a circle of radius , let be the set of chords of , for all , let . Then for all , there exists an angle such that for all , there exists a positive integer such that for all sets