TITLE: Irrationality of $\sqrt[n]2$
QUESTION [5 upvotes]: I know how to prove the result for $n=2$ by contradiction, but does anyone know a proof for general integers $n$ ?
Thank you for your answers.
Marcus

REPLY [5 votes]: FLT is quite a cannon to settle the proof. One can also follow the "classical" approach: assuming WLOG $\gcd(p,q)=1$, if $p^n=2q^n$, then $2 \mid p^n$, implying $2 \mid p$, say $p=2r$. But then $2^{n-1}r^n=q^n$, and since $n \gt 2$, we see that $2 \mid q^n$. It follows that $2 \mid q$, contracting the fact that $p$ and $q$ are relatively prime.