# Proof the following using denseness in reals

If $\frac{x}{y}$ is rational, then $y=x\left(\frac{x}{y}\right)^{-1}$ is rational too. ( $\left(\frac{x}{y}\right)^{-1}$ exists as $x\ne0$.) Contrapositively, if $y$ is irrational, $\frac{x}{y}$ must be irrational.