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

Question

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 1

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

Answer 2

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.

Answer 3

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