User:Foxjwill/Proofs

< User:Foxjwill
Revision as of 14:49, 8 July 2008 by Foxjwill (talk | contribs) (Proof that \sqrt{2} is irrational)

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.