The set 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 $f (\frac{p}{q}) = f (\frac{a}{b})$ (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.

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