Proving that rational numbers cannot have denominator zero (without referring to division by zero)

Question

Given the set $S$ of all pairs $(a, b),$ with $a, b \in \mathbb{Z},$ the relation $Q$ on $S$ is defined by $(a,b)Q(c,d) \iff ad=bc$. How can I prove that $b$ cannot be equal to zero, without using the fact that you cannot divide by zero?

I'm guessing it's something to do with $Q$ being an equivalence relation, and perhaps using the property of transitivity?

Thanks!

Answer 1

You know that $Q \neq \mathbb Z^2$, because, for example, $1 \neq 2$ and so $(1,1)$ and $(1,2)$ are not related by $Q$.

Now, if we allow $b=0$ (and likewise $d$), then from $0y = 0x$ we get $(0,0) Q (x,y)$, for all $x,y \in \mathbb Z$.
Therefore $Q$ would not be an equivalence relation.