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I came across this proof of the countability of $\mathbb{Q}$ in a textbook:

Each rational can be expressed as a fraction $a/b$, where $a$ and $b$ are integers. We know that the set $\{(a,b) : a,b\in\mathbb{Z}\}$ is countable, thus $\mathbb{Q}$ is countable.

My question is this: Clearly each rational can be uniquely expressed in lowest terms as a fraction $a/b$. However, each fraction can be expressed in a multitude of different ways that are in the set $S=\{(a,b) : a,b\in\mathbb{Z}\}$, specifically $(na)/(nb)$, where $n\in\mathbb{Z}$. Thus we only know from this that $|\mathbb{Q}|\leq|S|$, and not that $|\mathbb{Q}|=|S|$. Because of this, does this mean that we are assuming CH?

*As a side note: I'm not questioning whether or not the rationals are countable, I'm completely sure that they are. I am just asking if this proof is assuming CH or not.

Thank you all in advance for the help.

高田航
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    No one is assuming CH. – Asaf Karagila Jan 08 '20 at 18:12
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    Your argument gives an injection from $\mathbb Q$ to $\mathbb Z\times \mathbb Z$. Easy to get an injection the other way. Now use Cantor Schroeder Bernstein. – lulu Jan 08 '20 at 18:14
  • @AsafKaragila whoops, did not mean to put CH... I was just curious if this proof was acceptable given that a clear bijection between the rationals and the set $S$ had not been shown. – 高田航 Jan 08 '20 at 18:14
  • You don't have to write down bijection, many things happens behind the scene. For example @lulu proof depends on some other proofs, and in these other proofs, there is clear bijection (for example if there is injection from $A$ to $B$ and there is surjection from $A$ to $B$ then you can find a bijection (maybe not construct, but find), or if there is injection $A\to B$ and injection $B\to A$ then there exists a bijection). – pem Jan 08 '20 at 18:19

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