The instruction for proving that the set of rational numbers, $\mathbb{Q}^{+}$, is countable, is as follows:
- First assume each $x\in \mathbb{Q}^{+}$ is in reduced form.
- Construct a one-to-one function $p: \omega \times \omega \rightarrow \omega$.
- Using $p$ to show that $\mathbb{Q}^{+}$ is countable.
Following this path, I constructed the following proof (step 2. is done earlier in another exercise, so I skip this step):
Assume 1. holds, we have that $(\forall x\in \mathbb{Q}^{+})(x\text{ in reduced form })\rightarrow (\exists!n\in \mathbb{N})(\exists!m\in \mathbb{N})(x=n/m)$ so the function $g: \mathbb{Q}^{+}\rightarrow \omega \times \omega$ defined by $g=\{\langle x, \langle n, m\rangle \rangle: (\exists x\in \mathbb{Q}^{+}) \wedge(x\text{ in reduced form}) \wedge (x=n/m)\}$ is one-to-one. So $p\circ g:\mathbb{Q}^{+}\rightarrow \omega$ is also one-to-one. So $\mathbb{Q}^{+}$ consists of all $x$ in reduced form is countable.
This proof explicitly assumes that $1/2$ and $2/4$ represent the same number $.5$ in $\mathbb{Q}^{+}$ and only one of them is counted not both.
However, I found another proof based on Cantor's construction of rational numbers which does count both $1/2$ and $2/4$ separately. So, I'm quite confused over whether $1/2$ and $2/4$ should be counted as two different elements of $\mathbb{Q}^{+}$ or not?