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$\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}$ Looking at the group of real numbers under addition $(\R, +)$ it contains the (normal) subgroup of rational numbers $(\Q, +)$. I am wondering how to describe the cosets of $\R / \Q$.

I know from looking at the cardinality of the sets that because $\R$ is uncountable and $\Q$ is countable that $\R / \Q$ is uncountable. I am also thinking of $\R / \Q$ containing a "representative" of each irrational number. I am also aware that $\Q$ is dense in $\R$, so that each member of $\R$ is the limit of a sequence of numbers in $\Q$.

Both $\R$ and $\Q$ are ordered. But is there a natural order on $\R / \Q$? What else can we determine about $\R / \Q$?

Background: I am investigating functions satisfying $f(a + b) = f(a)f(b)$ for all $a,b\in\R$.

If $f$ is required to be continuous then $f(x) = \exp(A x)$

but if $f$ is not required to be continuous then I think I can define $f(x) = \exp(A_t x)$ where $x$ in $\Q$ where $t$ in some coset $\R / \Q$ and $t \Q = {t + q \text{ where } q\in \Q}$ and $A_t$ is different for each coset. This makes for quite an interesting function!!

Michaela Light
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  • No, there is no natural order on $\mathbb{R}/\mathbb{Q}$. The cosets consist precisely of the elements ${a+q\mid q\in\mathbb{Q}}$, with $a$ a fixed real number. One cannot exhibit a complete set of coset representatives, though. – Arturo Magidin May 31 '12 at 23:52
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    The existence of a basis for $\mathbb{R}$ as a $\mathbb{Q}$ vector space requires a certain amount of the Axiom of Choice; moreover, the existence of such a basis is equivalent to the existence of a complete set of representatives $A_t$ of the cosets. Nonzero functions satisfying $f(a+b)=f(a)f(b)$ correspond to $\mathbb{Q}$-linear functions on $\mathbb{R}$, which can be described in terms of a given basis. – Arturo Magidin May 31 '12 at 23:55
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    And that there is no natural order (as @ArturoMagidin points out) is a fact: In models of the axiom of determinacy, for example, it is impossible to linearly order this set. In natural models of determinacy, a set $X$ is not linearly orderable iff there is an injection of this quotient into $X$. – Andrés E. Caicedo May 31 '12 at 23:57
  • I have edited your question trying to improve how the formulas are displayed. Please review if everything is in the correct place. One thing we can say is that one the cosets is $\Bbb Q$ itself. The others only have irrational numbers. This is a classic application of such cosets. – leo Jun 01 '12 at 00:10
  • @Arturo - thanks for your insights. I will think about the Q-linear function approach as I have been reading about functional analysis and distributions recently. – Michaela Light Jun 01 '12 at 00:27
  • @Leo - thanks for your edits - I wasn't quiet sure how to add in Tex. Your edits are correct and I added a few more clean ups too – Michaela Light Jun 01 '12 at 00:27
  • @Andres: I have to ask, what are unnatural models of AD? Things which are not $L(\mathbb R)$ (under sufficiently strong assumptions of course)? – Asaf Karagila Jun 01 '12 at 18:19
  • @AsafKaragila A natural model is one of that satisfies $V=L({\mathcal P}({\mathbb R}),A)$ for some class $A\subseteq Ord$. – Andrés E. Caicedo Jun 01 '12 at 21:32

4 Answers4

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Suppose $V$ is a set of coset representatives, i.e. a set containing one member of each coset. We may assume $V \subset [0,1]$. Then $V$ is what is called a Vitali set. One interesting thing about it is that $V$ is non-measurable. See http://en.wikipedia.org/wiki/Vitali_set

Robert Israel
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By definition, $r\equiv s$ in the quotient iff $r-s\in \mathbb{Q}$. Said another way, given any real number $r$, the coset $\{r+q\mid q\in \mathbb{Q}\}$ is an element in the quotient.

This seems rather difficult to describe in terms of coset representatives! No obvious candidate for sensible coset representatives jumps to mind for me, because no matter what you pick, you can always strip one (or finitely many) decimal places away to get another rep. For example $\pi\equiv 0.14159...\equiv0.04159\equiv 0.00159... $ etc.

rschwieb
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    that idea of stripping out all leading digits in the decimal representation of a coset representative reminds me of infinitesimals in non-standard analysis. Just a thought here - perhaps the costs of R / Q are homeomorphic to the set of infintestimal numbers that have zero real part. – Michaela Light Jun 01 '12 at 00:11
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    @MichaelSmith Yes, they are infinitesimal in some sort of way :) The infinite "tail" is all that matters, but none of the initial finite segments matter. – rschwieb Jun 01 '12 at 00:15
  • This has little to do with infinitessimals... For example, we also have $\pi\equiv\pi+10^{10^{10}}$. – Mariano Suárez-Álvarez Jun 01 '12 at 16:14
  • @MarianoSuárez-Alvarez The largeness or the smallness of the representative is not the idea I had in mind. The idea is that the identity of the class is somehow contained in the infinite string of decimals to the right. Of course you can change finitely many digits to make the representative smaller (or larger, as you did) but the essense of the class is still buried in the infinte string to the right. This may have nothing to do with infinitesimals of course, but subjectively it kind of has that flavor :) – rschwieb Jun 01 '12 at 16:28
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    To help get a better intuitive handle on R / Q I thought to look at R / Z. In this case we get representatives as the interval [0,1) or equivalently as the circle Lie group. This is where we only look at the digits of the real number after the decimal point. Now look at the subgroup Yn = 1/n Z. R / Yn = [0, 1/n). We are looking at digits to the right of a certain point. We can consider R / Q as a limit of R / Yn - the circles get infinitesimally small. This reminds me of some modern physical models of space-time with 11 dimensions where the non-space time dimensions curl in on themselves – Michaela Light Jun 02 '12 at 15:33
  • Since $\mathbb{R/Q}$ and $\mathbb R$ are isomorphic additive groups ($\Bbb Q$-vector spaces of same dimension), can it be useful to understand better the cosets in $\mathbb{R/Q}$ ? – Alphonse Aug 20 '16 at 14:16
  • @Alphonse: the previous comment is related to http://math.stackexchange.com/questions/238854/showing-that-frac-mathbb-r-mathbb-q-cong-mathbb-r-both-over-mathbb-q/238857#238857 :-) – Alphonse Sep 01 '16 at 12:46
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There is no natural ordering on $\mathbb{R/Q}$. Of course if you choose a set of representatives then you have a natural ordering inherited from $\mathbb R$, however it is easy to see that this order would be very dependent on the choice of the representatives.

In fact this answer on MathOverflow explains why in some models without the axiom of choice $\mathbb{R/Q}$ cannot be linearly ordered at all.

Community
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Asaf Karagila
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Think it this way: two irrational number $\,r,s\,$ are in the same coset iff $$r+\mathbb{Q}=s+\mathbb{Q}\Longleftrightarrow r-s\in\mathbb{Q}$$

DonAntonio
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