Ernst Zermelo's counterexample According to the book Real Analysis by Royden page 6(=1+2+3):

Given an equivalence relation on a set X, it is often necessary to choose a subset C 
  of X which consists of exactly one member from each equivalence class. Is it obvious that 
  there is such a set? Ernst Zermelo called attention to this question regarding the choice of 
  elements from collections of sets. Suppose, for instance, we define two real numbers to be 
  rationally equivalent provided their difference is a rational number. It is easy to check that 
  this is an equivalence relation on the set of real numbers. But it is not easy to identify a set 
  of real numbers that consists of exactly one member from each rational equivalence class. 

I think this question has an answer. A rational equivalence class can be one of the following types: 
1- As a set of "sum of a given irrational and all possible rationals", say ${\{\sqrt{2}+r \ | \ r\in \mathbb {Q}}\}$. Suppose a representative of this set is the given irrational, $\sqrt{2}$ in this example. 
2- As a set of "sum of a given rational and all possible rationals", say ${\{1+r \ | \ r\in \mathbb {Q}}\}$. Suppose a representative of this set is the given rational, $1$ in this example.
So the subset $C$ of $\mathbb R$ which consists of exactly one member from each equivalence class is the collection of irrational and rational numbers and thus $C = \mathbb Q \cup \mathbb I = \mathbb R $. 
Is this argument correct?
 A: This does not work. 
Zermelo's problem is to find a set of reals $C$ such that for each real $r$, there is exactly one real $s\in C$ such that $r\sim s$ (that is, $r-s\in\mathbb{Q}$).
If we replace "exactly" with "at least," then this is easy: e.g. just take $C=\mathbb{R}$. Similarly, if we replace "exactly" with "at most," we can just take $C=\emptyset$ (:P).
The difficulty comes when we try to hit each rational equivalence class exactly once. Given a rational equivalence class $W$, we might hope that one of the following will work:


*

*Pick the smallest positive element of $W$.

*Pick the unique $r$ such that $W$ has the form $\{r+q: q\in\mathbb{Q}\}$.
The former doesn't work because each rational equivalence class is dense; the latter doesn't work since any $r\in W$ will have this property. 
Finally, we might hope something like the following will work:


*

*Pick the simplest element of $W$.


For instance, if $W$ is the rational equivalence class of $\sqrt{2}$, then clearly $\sqrt{2}$ is the simplest element in $W$. $17+\sqrt{2}$ is also in $W$, but is more complicated.
The problem is: what does "simplest" mean? Any attempt to pin this down will quickly run into trouble. If you play around with this for a while, you'll see what I mean . . .
