(Help with) A simple yet specific argument to prove Q is countable I was asked to prove that $\mathbb{Q}$ is countable. Though there are several proofs to this I want to prove it through a specific argument.
Let $\mathbb{Q} = \{x|n.x+m=0; n,m\in\mathbb{Z}\}$
I would like to go with the following argument: given that we know $\mathbb{Z}$ is countable, there are only countable many n and countable many m , therefore there can only be countable many equations AND COUNTABLE MANY SOLUTIONS.
The instructor told me that though he liked the argument, it doesn't follow directly that there can only be countable many solutions to those equations. Is there any way of proving that without it being a traditional proof of "$\mathbb{Q}$ is countable"?

This is simply isomorphic to the Cartesian product of two copies of $\mathbb{Z}$ .
$\mathbb{Z}\times\mathbb{Z}$ is countable and can be ordered by the lexicographic order so it is countable.
