$\exists r\in \mathbb{Q}$ such that $\forall s\in\overline{\mathbb{Q}}$, $rs\in\mathbb{Q}$

Question

False,

If $r = 2$ and $s = \pi$ then $rs = 2\pi$ which is irrational.

Counter-example, since a rational multiplied by an irrational is always irrational. Or am I mistaken?

Answer 1

Just let $r=0$, which indicates that for each $s \in \mathbb{Q}^c$, $rs=0 \in \mathbb{Q}$. But it is false when $r \neq 0$, since $s = \frac{rs}{r}$ is a rational number.