User:Foxjwill/Proofs

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Revision as of 14:48, 8 July 2008 by Foxjwill (talk | contribs) (New page: ==Proof that <math>\sqrt{2}</math> is irrational== # Assume that <math>p^{1/n}</math> is rational. Then <math>\exists a,b \in \mathbb{Z}</math> such that <math>a</math> is coprime to <math...)
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Proof that $\sqrt{2}$ 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.