Show that the rationals $\mathbb{Q}$ are a countable set

Question

Show that rationals $\mathbb{Q}$ are a countable set.

This is how I showed it, but I am not confident about my proof. I think I got the right idea at the beginning but I messed up at the end. Hence, I need someone to correct my proof. Can someone help me on this? Appreciate your help and support.

See also: https://math.stackexchange.com/questions/7643/produce-an-explicit-bijection-between-rationals-and-naturals — Thomas Andrews, Mar 08 '20 at 05:07

score 1 · Accepted Answer · answered Mar 08 '20 at 07:53

Your map $\Bbb Q\to \Bbb Z \times (\Bbb Z\setminus\{0\})$ is not well-defined: $\frac{1}{2}$ maps to $(1,2)$ but $\frac{1}{2}=\frac{2}{4}$ so it should be mapped to $(2,4)$ too, etc. A rational number is an equivalence class of pairs of integers, and to define a function on it, it should not depend on the representative of the class.

The other way around is better: send $(n,m) \in \Bbb Z \times (\Bbb Z\setminus\{0\}) \to \frac{n}{m} \in \Bbb Q$. This at least is well-defined and surjective (all rationals have such a representation, in fact many of them).

A set $A$ can be shown to be countable by either

giving a bijection with a set $N$ you already know to be countable, or
a surjection from such an $N$ onto $A$, or
an injection from $A$ into such an $N$.

We've now done the second and so we're done, assuming you know that $\Bbb Z \times (\Bbb Z\setminus\{0\})$ is countable (product of two countable sets).

Or, you know, map a rational to its reduced form with smallest positive denominator. — Asaf Karagila, Mar 08 '20 at 07:58
@AsafKaragila That is an injection, which is not surjective so achieves 3. Also possible. — Henno Brandsma, Mar 08 '20 at 07:59
Yes, it is an injection into an infinite subset of a countable set. So that's more or less it. — Asaf Karagila, Mar 08 '20 at 08:00

Show that the rationals $\mathbb{Q}$ are a countable set

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