What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?

$\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}$ 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!!

Solution 1:

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

Solution 2:

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.

Solution 3:

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.