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This question already has an answer here:

$\Bbb{R} / \Bbb{Q}$ is a quotient set of $\Bbb{R}$ with the following equivalence relation $\sim$ :

$$r \sim s \Longleftrightarrow r-s \in \Bbb{Q}$$

Then is there a bijection between $\Bbb{R}$ and $\Bbb{R} / \Bbb{Q}$?

I know that, with Axiom of Choice, there exists an injection from $\Bbb{R} / \Bbb{Q}$ to $\Bbb{R}$.

Thus $|\Bbb{R} / \Bbb{Q}| \leq |\Bbb{R}|$.

But I'm not certain that there exists an injection from $\Bbb{R}$ to $\Bbb{R} / \Bbb{Q}$.

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marked as duplicate by Asaf Karagila set-theory Oct 28 '18 at 8:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/questions/512397/… $\endgroup$ – Aaron Zolotor Oct 28 '18 at 3:32
  • $\begingroup$ Yes, I know, this [new] duplicate is not what this question is asking. But practically all the answers there prove that there is a subset of $\Bbb R$ which has size continuum and is $\Bbb Q$-independent, which is a far stronger result than just having distinct $\Bbb{R/Q}$ equivalence classes. $\endgroup$ – Asaf Karagila Oct 28 '18 at 8:30
  • $\begingroup$ (See also math.stackexchange.com/questions/1719458/… and the discussion in the comments there.) $\endgroup$ – Asaf Karagila Oct 28 '18 at 8:31
  • $\begingroup$ @RhythmInk: $\Bbb{R/Q\neq R\setminus Q}$. In fact, it is consistent with the failure of the axiom of choice that one of these sets is strictly larger than the other. $\endgroup$ – Asaf Karagila Oct 28 '18 at 8:32

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