Prove $\sqrt{\frac32}$ is irrational

Question

I first suppose for contradiction that if it's rational, then it can be written as a form of $\frac{p}{q}$, where it's in the lowest term. Then it becomes $\sqrt{\dfrac{3}{2}}=\dfrac{p}{q}$. Then I square both side and get $\frac{3}{2}=\frac{p^2}{q^2}$. I'm confused about what I should do next to get the contradiction. Can anyone please give me a hint?

Answer 1

Since $2p^2=3q^2$ we have $3q^2$ is even, which means that $q$ is even, $q=2r$. Then $2p^2 =3 (2r)^2$, then $p^2=2(3r^2)$, so $p$ is even. This contradicts the fact that the fraction is in its lower terms.

Answer 2

Hint

Write $2p^2=3q^2$. Now take a prime factor, $p_1$, of $p$. What can you say about $p_1$? You already know that it can not divide $q$.

After you find out the condition for $p_1$, do the same for a prime factor, $q_1$, of $q$.