Prove that the square root of pq when p and q are distinct primes is irrational

Question

I am trying to prove the following:

If p and q are distinct primes, then $\sqrt{pq}$ $\notin$ $\Bbb{Q}$.

Here is my proof thus far:

1. Suppose towards a contradiction that if p and q are distinct primes, that $\sqrt{pq}$ $\in$ $\Bbb{Q}$. If so, there exists some m , n $\in$ $\Bbb{Z}$ such that $\sqrt{pq}$ = $\frac mn$.

2. Squaring both sides, we see that pq = $\left(\frac{m^2}{n^2}\right)$ .

3. Multiplying by $\left(\frac nm\right)$, we see that $\left(\frac{m}{n}\right)$ = $\left(\frac{npq}{m}\right)$.

4. This implies that n|m and m|npq. But, since m, n share no common factors, n $\not\mid$ m.

5. Hence we have reached a contradiction.

**My concern is step 4 of my proof. Does it follow that since n $\not\mid$ m , $\sqrt{pq}$ $\notin$ $\Bbb{Q}$ ?

Is this "enough" to render a contradiction? It feels fishy to me but I'm not sure. If anybody could help that would be greatly appreciated! I am trying to develop my intuition in regards to this.

Answer 1

The method you use in step 4 is correct but, we need to assume at the start that wlog $m,n$ are coprime, i.e. $m/n$ is in lowest terms. See here or here for further details

However, this does not yet yield a contradiction. Rather, it yields that $n = 1,$ so if the radical is rational then it is an integer. It remains to finish the proof.

Note $ $ The links are to answers with complete elementary proofs, including John Conway's favorite simple proof. Though there I also append remarks on more advanced ways to conceptualize these elementary proofs, I emphasize that those proofs do not require any knowledge of these more advanced ideas.

Answer 2

You have $n^2pq=m^2$, but the exponents of $p$ and $q$ are odd.