Contradiction in proof by contraposition and counterexample $\forall x\in Q (\forall y \in R(y \notin Q \to \frac{x}{y} \notin Q))$
counterexample: let $x=0, y=\sqrt2$, then $\frac{0}{\sqrt2} = 0 = \frac{0}{1} \in Q$,
but by contrapositive: 
$\forall x\in Q (\forall y \in R(\frac{x}{y} \in Q \to y \in Q ))$
Suppose $\frac{x}{y}$ is rational, then $x \in Z$ and $y \in Z$ by definition of rational and any integer is rational, then y = $\frac{a}{b}$ for some integers $a$ and $b$.
So the original statement is false because there is a counterexample, but its contrapositive is true and hence the statement is true too.
What am I missing here?

Let do this:
$$\frac{x}{y} \in \mathbb{Q} \Longrightarrow \exists q \in \mathbb{Q} : \frac{x}{y}=q \Longrightarrow x=qy \Longrightarrow y=\frac{x}{q} \in \mathbb{Q}$$
That the last conclusion (dividing by $q$) is allowed iff $q \neq 0$, that need $x \neq 0$!
