How do I describe the equivalence classes in $\mathbf{R}$ of the relation $x\mathsf{R}y$ if $x - y \in \mathbf{Z}$? I have the relation $x\mathsf{R}y$ if $x - y \in \mathbf{Z}$ where $x, y$ are real numbers. I want to describe the equivalence classes of the relation.
I have proven to myself that $x-y$ is an equivalence relation if $x-y \in \mathbf{Z}$.  But, how does this set partition out?  Since an equivalence class of a set S containing a is defined as:  $[a] = \{x \in S : x\mathsf{R}a\}$ isn't this just the set of all real numbers?  i.e.  doesn't this simply partition out into $\mathbf{R}$?
I guess I'm confused on how the definition for an equivalence class works.  I read the questions and answers related to this question before I submitted it.  I think, unfortunately, I need a bit more of an explanation for a complete understanding as I am a complete beginner in abstract algebra.  Aren't there an infinite number of equivalence classes here?
Thank you in advance, very much appreciated.
 A: $y$ is equivalent to $x$ if $x-y =k$ for some integer $k$, right?  That means $x = y + k$ for some integer $k$. 
So, the equivalence class of $y$ is the set of elements of the form $y+k$, where $k$ ranges over the integers, since it is exactly this set that satisfies the above property of $x-y=k$. 
So, start with an element $y$. To get its equivalence class, simply shift $y$ by $k$ for each integer $k$, and you have the equivalence class of $y$. 
In math, you would write $[y] = \{y +k | k \in \Bbb Z \}$. 
A: Pick any real number $x$. Its equivalence class will consist of elements $y$ such that $y-x=k$ for some $k\in\mathbb{Z}$, so $y=x+k$. This tells you that $[x]=x+\mathbb{Z}:=\{x+k:k\in\mathbb{Z}\}$. The quotient space is the set of all equivalence classes $\mathbb{R}\big/\mathcal{R}=\{[x]:x\in\mathbb{R}\}$, but for every equivalence class there must be one representative belonging to $[0,1)$ --why? think of the fractional part function-- so you can write $\mathbb{R}\big/\mathcal{R}$ without repetitions as
$$\mathbb{R}\big/\mathcal{R}=\{[x]:x\in[0,1)\}=\{x+\mathbb{Z}:x\in[0,1)\},$$
which is indeed an infinite set.
A: Here $x$R$y$ if $x-y \in Z$. So if we consider equivalence classes for $a$, then it contain $a-1,a,a+1$, but it does not contain any number in between $(a,a+1)$. Also any two numbers in $[a,a+1)$ are not in equivalence relation to each other as their difference is not an integer. The equivalence classes are $\{[x]| x \in [a,a+1)\}$ where $a \in R$ is any real number. For example if we take $a=0$ we get  the equivalence classes as $\{[x]| x \in [0,1)\}$ and this number of equivalence classes is uncountable.
A: Note that $x R y$ iff $ \{x\} R \{y\}$ where $\{x\}=x-\lfloor x\rfloor$
The equivalence classes that partitions $\mathbb R$ are $\{class(x): x\in [0,1)\}$
