Proving all rational numbers including negatives are countable

Solution 1:

You said "irrationals" in your last sentence by mistake - but "rationals" in the body of the problem. Yes, the argument is correct. Here's an even more direct method (without splitting into two cases)

Order everything in the usual positive matrix, and put the negative number immediately after the corresponding positive number:

$ \\ \frac{1}{1}, -\frac{1}{1}, \frac{1}{2}, -\frac{1}{2}, \ldots \\ \frac{2}{1}, -\frac{2}{1}, \frac{2}{2}, -\frac{2}{2}, \ldots \\ \vdots \\ \frac{n}{1}, -\frac{n}{1}, \frac{n}{2}, -\frac{n}{2}, \ldots \\ \vdots $

or use your favourite matrix (this one is missing $0$, by the way).