rational numbers - Square root of 6 proof rationality

I was proving $\sqrt 6 \notin \Bbb Q$, by assuming its negation and stating that: $\exists (p,q) \in \Bbb Z \times \Bbb Z^*/ \gcd(p,q) = 1$, and $\sqrt 6 = (p/q)$.

$\implies p^2 = 2 \times 3q^2 \implies \exists k \in \Bbb Z; p = 2k \implies 2k^2 = 3q^2$ and found two possible cases, either $q$ is even or odd, if even we get contradiction that $\gcd(p, q) \neq 1$, if odd we get contradiction that $2k^2 = 3q^2$.

Is it a right path for reasoning it?