If $P$ and $Q$ are distinct primes, how to prove that $\sqrt{PQ}$ is irrational?

If $$ \sqrt{pq}=\frac{m}{n}, \quad (m,n)=1, $$ then $$ n^2pq=m^2, \tag{$\star$} $$ which means that $p\mid m^2$ and hence $p\mid m$. Thus $m=pm_1$, and $(\star)$ becomes $$ n^2q=pm_1^2. $$ But this means that $p\mid qn^2$, and as $p\ne q$ and hence $p\not\mid q$, then $p\mid n^2$, and thus $p\mid n$. Therefore, $n=pn_1$.

This is a contradiction, since $p\mid m$ and $p\mid n$, and we had assumed that $(m,n)=1$.