User:Foxjwill/Proofs

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Revision as of 13:18, 13 January 2009 by Foxjwill (talk | contribs) (A theorem)

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 $\Theta$ consisting only of elements $\theta$ such that for each $b\in B$, there is exactly one $\theta \in \Theta$ cut by $b$, \[\sum_{\theta\in \Theta}\theta = \phi\]