I need to use the fundamental theorem of arithmetic to show that
if $p$ is prime then $\sqrt p$ is irrational.
So far I've stated that $\sqrt p=m/n$ where $m,n$ are positive integers, then $pn^2=m^2$. Now I can factor $m$ and $n$ into primes but I don't know where to go on from there.
Given a prime $p$ and some $n\in\mathbb{N}^*$, let we define: $$\nu_p(n)=\max\{k\in\mathbb{N}: p^k\mid n\}.\tag{1}$$ Since $\mathbb{Z}$ is a UFD we have $\nu_p(ab)=\nu_p(a)+\nu_p(b)$. In particular, $\nu_p$ of a square is always an even number. If we assume that $\sqrt{p}=\frac{m}{n}$ with $m,n\in\mathbb{N}^*$, we also have $$ p n^2 = m^2. \tag{2} $$ However, such identity cannot hold, since $\nu_p(\text{RHS})$ is an even number and $\nu_p(\text{LHS})$ is an odd number. It follows that $\sqrt{p}\not\in\mathbb{Q}$ as wanted.
Hint. What can you say about the number of times $p$ appears in the prime factorization of $m^2$? Of $pn^2$?
