User:Foxjwill/Proofs

Proof that $\sqrt{2}$ is irrational

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

Q.E.D.