How to prove that $\forall x\in \Bbb{Q}:\ x\ne 0\implies [\exists a,\ b\in \Bbb{I}: x=a\cdot b]$ if $\Bbb{I}$ is set of irrational numbers?

Question

I initially thought contrapositive would be easier, so I wrote $\forall x\in \Bbb{Q}:\ [\forall a,\ b\in \Bbb{I}: x\ne a\cdot b]$ $\implies x=0$. But I still had no idea how to start. Could someone help?

Answer 1

Let $a=x\sqrt 2$ and $b=1/\sqrt 2$.