Defining an equivalence relation For the set $ \Bbb R$, define two elements in $ \Bbb R$ to be equivalent if their difference belongs to $ \Bbb Q$.
I can prove that this defines an equivalence relation. (see Verify an equivalence relation.)
However, I would like to be able to define the class and the partition.  Is it possible to place an element into a class without comparing it to other members of the class?
A similar problem, where the difference belongs to $ \Bbb Z$, any element r $\in \Bbb R$ can be classified by setting up class Am such that m = r - g(r) where g(r) returns the nearest integer less than r.  As a result, any r $\in \Bbb R$ can be put into some class Am, $0<=m<1$.  There may be other indexing schemes.
But I can't come up with anything for the difference belonging to $ \Bbb Q$.
This problem is from Pinter's A Book of Abstract Algebra.
thanks.
 A: We have:
$$
x \sim y \iff x - y=q \in \mathbb{Q}
$$
Consider the equivalence class of $0$, this is the set of real numbers $x$ such that $ x-0=q \in\mathbb{Q}$ so the class $[0]$ is the set of rational numbers.
Now consider an irrational number $z$, since $z-0$ is irrational we have $z\notin [0]$ (it is not rational), So it represents another equivalence class $[z]$ that contains all the irrational numbers $y$ such that $y=z+q$ with $q \in \mathbb{Q}$.
For another irrational number $y \notin [z]$  ( i.e. $y-z \notin \mathbb{Q}$), we have another equivalence class $[y]$ .. and so on.
So, The equivalence classes are represented by $[0]$ and $[z_1],[z_2],\ldots$ where $z_i$ are all irrational numbers that differ by an irrational number, and there are uncountable infinitely many such classes.
A: 1st example
The equivalence relation is:
$$
x \sim y \iff x - y \in \mathbb{Q}
$$
For every $x \in \mathbb{R}$, let $[x]_\sim$ be its equivalence class:
$$
[x]_{\sim} = \{y \in \mathbb{R} \mid x \sim y\}
$$
Then the partition, or quotient, is just:
$$
\mathbb{R}/\!\!\sim \; = \{[x]_{\sim} \mid x \in \mathbb{R} \}
$$
Two reals $x, y$ are $\sim$-equivalent when their difference is a rational. Notice, then, that the equivalence class of $0$ is $\mathbb{Q}$: all rationals are $\sim$-equivalent and none are equivalent to any irrational. The equivalence class of $\pi$ is just $\{\pi + q \mid q \in \mathbb{Q}\}$, and does not contain, for example, $e$, or $e + 17$. 
In general, the equivalence classes are of the form $r + \mathbb{Q} = \{r + q \mid q \in \mathbb{Q}\}$.
Each equivalence class is countable, so there are uncountably many equivalence classes.
2nd example
Here, differences are in $\mathbb{Z}$, and every equivalence class has a unique representative in $[0,1)$. The equivalence class of $\frac 1 2$ is $\{\dots - \frac 3 2, -\frac 1 2, \frac 1 2, \frac 3 2, \dots \}$; the equivalence class of $\pi$ is $\{\dots \pi - 4, \pi - 3, \pi - 2, \dots \}$. 
Again, each equivalence class is countable, so there are uncountably many equivalence classes.
