Let $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $\sqrt{pq}$ is also irrational. Progress:

Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cdot\sqrt{k}=\sqrt{pq}\cdot\sqrt{pq}=k$$
  Which implies that $\sqrt{pq}$ is irrational, since, $k$ is irrational and that $p^2\cdot{q^2}$ cannot be a proper square being $p^2$ and $q^2$ relatively prime and that concludes the proof.

The above lines are my attempt to prove the assertion. Is the proof correct? If not then how can I improve or disprove it. 
Regrads  

Suppose $\sqrt{pq}$ is rational, say $\sqrt{pq} = \dfrac mn$ where $m$ and $m$ are positive, relatively prime integers. Then $p^2 q^2 = \dfrac{m^4}{n^4}$ where $p^2q^2$ is an integer. Since $m$ and $n$ are relatively prime integers, then $m^4$ and $n^4$ are also relatively prime integers. It follows that $n^4=1$ and hence $n=1$. So we find that $p^2q^2 = m^4$ for some positive integer $m$. 
Since $p^2$ and $q^2$ are relatively prime integers,then any prime divisor of $m$ cannot be a common divisor of both $p^2$ and $q^2$. It follow that there must be relatively prime positive integers $a$ and $b$ such that $m = ab$, $\;p^2 = a^4$, and $q^2 = b^4$. Hence $p=a^2$ and $q=b^2$, which implies that $p$ and $q$ are rational (integers). By contradiction, $\sqrt{pq}$ must be irrational.
