# Is there a bijection between $\Bbb{R}$ and $\Bbb{R} / \Bbb{Q}$? [duplicate]

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 $$\Bbb{R} / \Bbb{Q}$$.

## marked as duplicate by Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); 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.

• math.stackexchange.com/questions/512397/… – Aaron Zolotor Oct 28 '18 at 3:32
• 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. – Asaf Karagila Oct 28 '18 at 8:30
• (See also math.stackexchange.com/questions/1719458/… and the discussion in the comments there.) – Asaf Karagila Oct 28 '18 at 8:31
• @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. – Asaf Karagila Oct 28 '18 at 8:32