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I am trying to better understand the group $\mathbb{R/Q}$. It's unclear to me when two irrational numbers will give the same coset of $\mathbb{Q}$, but I know that this must happen since, for example $\pi+1 = (\pi-1)+2$, meaning the cosets $\pi\mathbb{Q}$ and $(\pi−1)\mathbb{Q}$ share an element and thus are equivalent. Can we describe a set of irrational numbers that give each coset of $\mathbb{Q}$ exactly once in a way that, given an irrational number, we would be able to say whether or not it's in the set?

Edit: I am also interested in understanding this group in other ways. What is it's order? Are there any groups it's isomorphic to?

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    $\begingroup$ By the axiom of choice, it is mathematically possible to do so. Since I have to say something like "by the axiom of choice", there is no nice (i.e. countably constructive) way to do so. $\endgroup$ – Omnomnomnom May 31 '16 at 12:56
  • $\begingroup$ @Omnomnomnom you should make that comment an answer, because that is, indeed, the answer to the question. $\endgroup$ – Emre May 31 '16 at 13:00
  • $\begingroup$ But it isn't a helpful answer (in its own right) to anyone who is encountering the axiom of choice for the first time. I'm hoping someone follows my comment up with a more robust answer, which I find that I am too lazy to put together. $\endgroup$ – Omnomnomnom May 31 '16 at 13:02
  • $\begingroup$ Should I take this to mean that, assuming AC, we can show that such a set exists, but can't actually write it out or say which irrational numbers are in it? $\endgroup$ – user52969 May 31 '16 at 13:11
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$\mathbb{R/Z}$ is isomorphic to the unit circle $S^1$, the set of all complex numbers of absolute value $1$, seen as a multiplicative group.

$\mathbb{Q/Z}$ is isomorphic to the torsion subgroup of $S^1$, formed by the elements of finite order, that is, all roots of unity.

$\mathbb{R/Q}$ is thus isomorphic to $S^1/tor(S^1)$, a very large and complicated group. For instance, the powers of every element form a dense subset.


Another approach is to consider $\mathbb{R}$ as a vector space over $\mathbb{Q}$. The dimension here is the cardinality of $\mathbb{R}$ and so $\mathbb{R/Q}$ is essentially the same as $\mathbb{R}$.

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  • $\begingroup$ Do you know the size of this group? Also, is there a nice, constructible subset of $S^1$ that gives each coset of $tor(S^1)$ only once? $\endgroup$ – user52969 May 31 '16 at 14:33
  • $\begingroup$ The cardinality is the same as that of ${\mathbb R}$, which is $2^{\aleph_0}$. $\endgroup$ – Derek Holt May 31 '16 at 14:40

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