If $P$ and $Q$ are distinct primes, how to prove that $\sqrt{PQ}$ is irrational? $P$ and $Q$ are two distinct prime numbers. How can I prove that $\sqrt{PQ}$ is an irrational number?
 A: 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 $(m,n)=1$.
A: Proof: Assume, to the contrary, that $\sqrt{pq}$ is rational. Then $\sqrt{pq}=\frac{x}{y}$ for two integers $x$ and $y$ and we further assume that $gcd(x,y)=1$. Observe that $pqy^2=(qy^2)p=x^2$. Since $qy^2$ is an integer, $p\mid x^2$ and by Euclid's Lemma, $p\mid x$. Thus $x=pd$ for some integer $d$ and so $pqy^2=x^2=(pd)^2$ and so $qy^2=p(d^2)$. Hence, $p\mid qy^2$. Since $p,q$ are distinct primes, $p \neq q$ and so $p \nmid q$, which means $p\mid y^2$. Thus, $p\mid y$. This contradicts our assumption that $gcd(x,y)=1$.
A: If you follow through the usual proof that $\sqrt 2 $ is irrational, it goes through in this case as well.  One of them (for $\sqrt 3$) is here, but a search for irrational+sqrt will find many choices
