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  

Except that you haven't proved $p^2 * q^2 $ cannot be perfect square, rest of the proof is correct.
Since $p^2 $ and $q^2$ are relatively prime, they do not have common prime factors. And they are not perfect squares. Else p,q would be rational numbers. This is necessary since 25, 4 are relatively prime but their product is perfect square.
Hence atleast one of the prime factors of p^2 and q^2 will be raised to an odd power in prime factorisation and those primes will be distinct. So   $p^2 * q^2 $ cannot be perfect square
