Product of irrational and nonzero rational

Suppose $r\in\mathbb{Q}\setminus\{0\}$, $x\notin\mathbb{Q}$. By contradiction, if $rx=q\in\mathbb{Q}$, then $x=q/r\in\mathbb{Q}$ (since $r\neq0$), contradicting $x\notin\mathbb{Q}$.

Conclusion. $rx$ is irrational.

Application. $2\sqrt2$ is irrational.