real analysis - Direct proof that if p is prime, then $sqrt{p}$ is irrational.

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.

Answer

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' 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.