User:Foxjwill/Proofs

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

DEFINITION. Let $a$ be a chord of some circle $C$. Then the small angle of $a$, denoted $S(a)$, is the smaller of the two angles cut by $a$.

THEOREM. Let $p\in \mathbb{R}^+$, and let $C$ be a circle. Then there exists a $\theta\in \mathbf{R}^+$ such that for every set A of chords of $C$ with lengths adding to $p$, \[\sum_{a\in A}S(a) = \theta.\]