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?


Suppose $\frac{x}{y}$ is rational, then $x \in Z$ and $y \in Z$ by definition of rational

False. If $x=y=\frac12$, then $\frac{x}{y}$ is rational, but $x\notin \mathbb Z$ and $y\notin\mathbb Z$.
