# Showing countability without explicit bijection [duplicate]

(The questions are below the question heading)

I've seen this image posted:

I actually agree with it - that is not to say that $\mathbb{Q}$ isn't countable.

Reasoning
The definition of bijection is a useful starting point, and it is easy to see that $\mathbb{Z}$ is countable. I shall denote this as $\mathbb{Z}\sim\mathbb{N}$ for simplicity.

I then take: $$f:\mathbb{Z}\times\mathbb{Z}\rightarrow\mathbb{Q}\text{ by }\ f:(a,b)\mapsto\left\{\begin{array}{lr}\frac{a}{b} & \text{if }b\ne 0\\ 1 & \text{otherwise}\end{array}\right.$$