Prove that the Rationals are Countably Infinite

Question

I was assigned to Prove that $\mathbb{Q}$ is countably infinite

I did the following:

We define $\mathbb{Q}= \lbrace \frac{a}{b} \mid a, b \in \mathbb{Z}_{>0} \rbrace$. Also define $\mathbb{Q} \xrightarrow{a} \mathbb{Z}_{>0} \times \mathbb{Z}_{>0}$ as $\forall a, b \in \mathbb{Z}_{>0}$ where $\frac{a}{b} \in \mathbb{Q}$, $q(\frac{a}{b})=(a,b)$. Then since $\mathbb{Z}_{>0}$ is countably infinite, and by the fact that the Cartesian product of two countably infinite sets is countably infinite, $\mathbb{Z}_{>0} \times \mathbb{Z}_{>0}$ is countably infinite with some mapping $\mathbb{Z}_{>0} \times \mathbb{Z}_{>0} \xrightarrow{b} \mathbb{Z}_{>0}$. Then the composite function $a \circ b$ is a bijection $\mathbb{Q} \xrightarrow{a \circ b} \mathbb{Z}_{>0}$ proving $\mathbb{Q}$ is countably infinite.

My question: is there any alternative way to prove this or is there any way to strengthen my proof or change it?

Answer 1

In your proof you defnie $q(\frac{a}{b})=(a,b)$, but then what should the value of $q(0.5)$ be? is it $(1,2)$ or perhaps $(50,100)$? You need to adopt some convention as to how to uniquely represent fractions. Next, you claim that the composite will yield a bijection between $\mathbb Q$ and $\mathbb Z$, but the function $q$ you constructed (after the required modification) is not surjective.

Your proof idea is basically correct, but you need to be more careful. Remember that to prove that $\mathbb Q$ is countably infinite it suffices to construct an injection into a countable set (make sure you understand why this is true), and the $q$ you defined can quite easily be turned into a precise function that maps into $\mathbb Z \times \mathbb Z$.

There are other ways to establish the result. For instance, you can show in various ways that the rationals are a countable union of finite sets (or a countable union of countable sets) and that suffices as well.