TITLE: Different ways to prove $\sqrt p$ irrational for $p$ prime.
QUESTION [8 upvotes]: I know this fact can be proved by contradiction(reductio ad absurdum) but please give proofs by different methods.

REPLY [2 votes]: By Dirichlet's theorem on primes in arithmetic progressions and by quadratic reciprocity (!!), there exists a prime $\ell$ such that $$\left(\frac{p}{\ell}\right) = -1.$$
By Gauss's lemma, any rational root of $x^2−p$ must be an integer. But if $p$ were the square of an integer, it would certainly be a square mod $\ell$ also. Thus, $x^2-p$ has no rational root.