Could someone verify the correctness of this proof for the irrationality of $\sqrt{15}$?
Assume $\sqrt{15}\in\mathbf{Q}$, then $\sqrt{15}=\frac{p}{q}$ with $p,q\in\mathbf{Z}$ ($q\ne0$ and $\gcd(p,q)=1$). $\implies 15q^2=p^2 \implies 15\mid p^2 \implies 3\mid p^2 \implies 3\mid p$ (Euclid's Lemma)
Now we write $p=3k$ for $k\in\mathbf{Z}$, then we have $15q^2=9k^2 \implies 5q^2=3k^2 \implies 3\mid 5q^2$. Since $\gcd(3,5)=1$ (this lemma: if $a\mid bc$ and $\gcd(a,b)=1$ then $a\mid c$) this gives $3\mid q^2 \implies 3\mid q$ (Euclid's Lemma).
Therefore $3\mid p$ and $3\mid q$. Contradiction.