Does anyone know of a simple direct proof that if p is prime, then $\sqrt{p}$ is irrational?
I have always seen this proved by contradiction and have been trying unsuccessfully to prove it constructively. I searched this site and could not find the question answered without using contradiction.
Maybe an elementary proof that I gave in a more general context, but I can't find it on the site, so I'll adapt it to this case.
Set $n=\lfloor \sqrt p\rfloor$. Suppose $\sqrt p$ is rational and let $m$ be the smallest positive integer such that $m\sqrt p$ is an integer. Consider $m'=m(\sqrt p-n)$; it is an integer, and $$ m'\sqrt p=m(\sqrt p-n)\sqrt p=mp-nm\sqrt p $$ is an integer too.
However, since $0\le \sqrt p-n <1$, we have $0\le m' <m $. Since $m$ is the smallest positive integer such that $ m\sqrt p$ is an integer, it implies $m'=0$, which means $\sqrt p=n$, hence $p=n^2$, which contradicts $p$ being prime.
You could first prove that all square numbers have a prime decomposition with all primes of even powers.
For example if $k$ is square, and,
$k=\prod_{k=1}^{n}p_{k}^{e_{k}}$,
then each of $e_{k}$ must necessarily be even.
From there it is enough to show that primes have an odd power (namely 1) in their decomposition.
For specific primes you can prove $x^2-p$ is irreducible over the rationals by proving it is irreducible modulo some other prime $q$. This means that $\sqrt{p}$ cannot be rational. This can probably be done in general using quadratic reciprocity.
If you have the Rational Root Theorem available to you (https://en.wikipedia.org/wiki/Rational_root_theorem) you can apply it to $x^2-p$.
