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Difference between revisions of "User:Foxjwill/Proofs"

(→‎Proof that \sqrt{2} is irrational)
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# Therefore <math>p^{1/n}\not\in \mathbb{Q}</math>.
 
# Therefore <math>p^{1/n}\not\in \mathbb{Q}</math>.
 
::'''''Q.E.D.'''''
 
::'''''Q.E.D.'''''
 +
 +
==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| \mbox{for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\}</math>,
 +
\[
 +
\sum_{\theta\in \Theta}\theta = \phi
 +
\]

Revision as of 13:12, 13 January 2009

Proof that $p^{1/n}$, where $p$ is prime, is irrational

  1. Assume that $p^{1/n}$ is rational. Then $\exists a,b \in \mathbb{Z}$ such that $a$ is coprime to $b$ and $p^{1/n}={a \over b}$.
  2. It follows that $p = {a^n \over b^n}$, and that $a^n=pb^n$.
  3. So, by the properties of exponents along with the unique factorization theorem, $p$ divides both $a^n$ and $a$.
  4. Factoring out $p$ from (2), we have $a^n=p^{n-1}b'$ for some $b'\in \mathbb{Z}$.
  5. Therefore $p$ divides $a$.
  6. But this contradicts the assumption that $a$ and $b$ are coprime.
  7. Therefore $p^{1/n}\not\in \mathbb{Q}$.
Q.E.D.

A theorem

THEOREM. Let $C$ be a circle of radius $r$, let $A$ be the set of chords of $C$, for all $p\in \mathbb{R}^+$, let $S_p\equiv \{B\subset A|B\mbox{ is finite}, \sum_{b\in B}|b|=p\}$. Then for all $p\in \mathbb{R}^+$, there exists an angle $\phi\in \mathbb{R}^+$ such that for all $B\in S_p$, there exists a positive integer $k$ such that for all sets $\Theta\equiv \{\mbox{angle }\theta| \mbox{for all }b\in B,b\mbox{ cuts exactly one element in }\Theta\}$, \[ \sum_{\theta\in \Theta}\theta = \phi \]

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