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What is the quotient group of the real numbers over the rational numbers seen as a quotient of groups? Is there a common way to represent the quotient $\frac{\mathbb{R}}{\mathbb{Q}}$?

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    $\begingroup$ I don't know if this is in common use anywhere. The only place I've seen a construction like this is in a proof using the axiom of choice to prove that there are non-measurable sets (Let $X$ consist of one representative between $0$ and $1$ for each coset in $\Bbb R/\Bbb Q$. Observe that you can cover the interval from countably infinitely many translated copies of $X$ (wrapping around when you pass $1$). Arrive at the conclusion that $X$ can have neither $0$ nor positive measure. $\endgroup$
    – Arthur
    Mar 22, 2014 at 22:59

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There's no easy way to describe $\Bbb{R/Q}$, except the trivial way.

This group is not an ordered group, and it's quite easy to prove that: for every $x<y$ there is some rational number $q$ such that $y<x+q$.

But we can say even more, without assuming some fragment of the axiom of choice, we cannot even prove that the set $\Bbb{R/Q}$ can be linearly ordered (let alone well-ordered). The axiom of choice comes in from time to time in this group's context because if we consider the canonical homomorphism $x\mapsto x+\Bbb Q$, and then consider an injective inverse, the range of that inverse is a non-measurable subset of $\Bbb R$.

If we assume the axiom of choice, and we think about $\Bbb R$ as a $\Bbb Q$-vector space, then this quotient group is isomorphic to $\Bbb R$ again. To see this, note that $\Bbb R$ is has dimension $2^{\aleph_0}$, therefore $\Bbb{R/Q}$ has dimension $2^{\aleph_0}-1=2^{\aleph_0}$. Since two vector spaces with the same dimension are isomorphic, their additive groups are also isomorphic.

But of course the existence of that isomorphism requires using the axiom of choice.

Another thread on this topic: Visualizing quotient groups: $\mathbb{R/Q}$

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  • $\begingroup$ Thank you very much. I was worried it was some essential group I was forgetting about. $\endgroup$
    – yess
    Mar 22, 2014 at 23:46

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