The set $Q$ of rational numbers is countable: that is, there is a bijection between $Q$ and the set $N$ of natural numbers.

Proof

By the Schröder-Bernstein theorem, it is enough to find an injection $N\to Q$ and an injection $Q\to N$.

The former is easy, because $N$ is a subset of $Q$ so the identity injection $n\mapsto \frac{n}{1}$ works.

For the latter, we may define a function $Q\to N$ as follows. Take any rational in its lowest terms, as $\frac{p}{q}$, say. ^[1] At most one of $p$ and $q$ is negative (if both are negative, we may just cancel $-1$ from the top and bottom of the fraction); by multiplying by $\frac{-1}{-1}$ if necessary, assume without loss of generality that $q$ is positive. If $p=0$ then take $q=1$.

Define $s$ to be $1$ if $p$ is positive, and $2$ if $p$ is negative.

Then produce the natural number $2^p{3}^q{5}^s$.

The function $f:\frac{p}{q}\mapsto 2^p{3}^q{5}^s$ is injective, because prime factorisations are unique so if $f\left(\frac{p}{q}\right)=f\left(\frac{a}{b}\right)$ (with both fractions in their lowest terms, and $q$ positive) then $|p|=|a|,q=b$ and the sign of $p$ is equal to the sign of $a$. Hence the two fractions were the same after all.

1. ^{^︎}

   That is, the GCD of the numerator $p$ and denominator $q$ is $1$.