Does an irrational number $C$ exist such that $C \cdot \sqrt 2 \in \Bbb{Q}$?

Question

Does an irrational number $C$ exist such that $C \cdot \sqrt 2 \in \Bbb{Q}$, where $\sqrt2 \not\mid C$?

I just thought of this, I'm trying to find answers that aren't of the form $C=a\sqrt2, a\in\Bbb{Q}$.

Answer 1

If $C\cdot\sqrt 2\in\mathbb Q$, then $C\cdot\sqrt 2=q$ for some $q\in\mathbb Q$, hence $C=\frac q{\sqrt 2}=\frac q2\cdot\sqrt 2$.

Answer 2

Obviously not: if $C \cdot \sqrt 2 \in \mathbb Q$, then there exist integers $p,q$ such that $C \cdot \sqrt 2 = \frac p q$, i.e. $C=\frac p {2q} \sqrt 2$, which is exactly what you said you do not want.