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  

The first part is correct, but you could clarify it a little bit:

Since $p$ and $q$ are irrational, they are not integers
Since $p$ and $q$ are not integers, $p^2$ and $q^2$ are not perfect squares
Since in addition to that $p^2$ and $q^2$ are relatively prime, $p^2q^2$ is not a perfect square
Since $p^2q^2$ is not a perfect square, $\sqrt{p^2q^2}=pq$ is irrational


The second part is somewhat obscure, but you could simply use the following argument instead:

Since $pq$ is irrational, it is not a multiple of two integers
Since $pq$ is not a multiple of two integers, it is not a perfect square
Since $pq$ is not a perfect square, $\sqrt{pq}$ is irrational

