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

Answer

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

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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