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

elementary-number-theoryrationality-testing

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.

Best Answer

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

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