Prove that $2^{1/n}$ is irrational Proof by contradiction, Assume $2^{1/n}$ is rational so:
$$2^{1/n} = \frac ab $$
where a,b have no common factors.
$$2 = \frac{a^n}{b^n}$$
$2$ divides LHS, therefore $2$ divides RHS
so $2$ divides $a^n$ or $2$ divides $b^n$ which implies $2$ divides $a$ or $2$ divides $b$. 
Stuck on what to do next.

Let $n \geq 2$; let there be some rational $a,b > 0$ such that $2^{1/n} = a/b$ with $\gcd (a,b) = 1$. Note that $2^{1/n} = a/b$ iff $2b^{n} = a^{n}$; so $2$ divides $a^{n}$, and hence $a$ must be even. Then $n \geq 2$ implies that $4$ divides $a^{n}$; so $2$ divides $b^{n}$ and hence $b$ is even too. But this is a contradiction. 
