Is $\mathbb{R}$ the disjoint union of finitely many congruent dense sets? Both the set of rational numbers $\mathbb{Q}$ and its complement are dense in $\mathbb{R}$, but the relationship between them is very asymmetric. For instance, the rationals are countable and have Lebesgue measure 0, whereas the irrationals are uncountable and have infinite Lebesgue measure. Is it possibly to decompose the real numbers into dense subsets in a more symmetric way, so that $\mathbb{R}$ can be written as a union of finitely many disjoint sets which can be mapped into each other by translation or reflection (i.e. are congruent)?
 A: $$A=\bigcup_{n\in\mathbb Z}[2n,2n+1)$$
$$B=\bigcup_{n\in\mathbb Z}[2n+1,2n+2)$$
$$\mathbb R= [(A\cap\mathbb Q)\cup(B\setminus\mathbb Q)] \cup [(B\cap\mathbb Q)\cup(A\setminus\mathbb Q)] $$
The translation $x\mapsto x+1$ maps $[(A\cap\mathbb Q)\cup(B\setminus\mathbb Q)]$ onto $[(B\cap\mathbb Q)\cup(A\setminus\mathbb Q)].$
A: Yes, this is doable, via a construction like that of the Vitali set but for integers. Indeed, with Choice we can get an uncountably dense example! (Note that bof's answer solves the problem as stated, without using Choice at all.)
For $x, y\in\mathbb{R}$, let $x\sim y$ if $x-y\in\mathbb{Z}$. Now via Choice we can get an uncountably dense transversal $T$ for $\sim$ - that is, $T$ is uncountably dense and contains exactly one real from each $\sim$-class. (Note that $[0, 1)$ is a non-dense transversal - the existence of a transversal, full stop, does not require choice.)
Now let $$A=\{t+2k: t\in T, k\in\mathbb{Z}\},\quad B=\{t+2k+1: t\in T, k\in\mathbb{Z}\}.$$ It's not hard to see that $B$ is gotten by shifting $A$ one unit (in either direction!), and that $A$ and $B$ are disjoint and cover $\mathbb{R}$.
