elementary algebra (Equivalence) I can't totally understand this statement:
"congruence modulo $n$ is a Equivalence relation for the set $\mathbb{Z}$ which exactly has $n$ Equivalence class."
I tried to give myself an example like this:
assume there exist congruence modulo $6$ and I want to put all the elements with the definition $\bar x=\left\{\forall a\in \mathbb{Z}: a\sim x\right\}$ and $x=6k$ into a class, so we will have $\bar x=\left\{\ldots,-6,0,+6,\ldots\right\}$, now it's clear that we just have 1 class not 6 class.
 A: Well, each integer $a$ lies in the same class as its remainder modulo $n$, since by division $a=qn+r$ with $0\leq r<n$ and $n$ divides $a-r$, so $a$ and $r$ are equivalent.
Moreover, different remainders $r,r'$ modulo $n$ lie in different equivalence classes, since if $r'\equiv r\mod n$ with $0\leq r,r'<n$, then $n$ divides $r'-r$. But $-n<r'-r<n$ and so this can only happen if $r'=r$. Done.
Thus there are exactly $n$ remainders modulo $n$ given by the remainder modulo $n$.
In your case, the remainders are $0,\ldots,5$.
A: It is clear that you have $6$ classes then. You have the class $[0] = \lbrace \dots,-6,0,6,\dots \rbrace$, which contains all multiples of $6$. Not every integer is a multiple of $6$ though. That is why we have more classes. For example the class $[1] = \lbrace \dots,-5,1,7,\dots \rbrace = \lbrace 1+6k \mid k \in \mathbb{Z} \rbrace$. Analogously for $2$, $3$ etc. until you find the class of $6$ again. In total you will get $\mathbb{Z}/6\mathbb{Z} = \lbrace [0],[1],\dots, [5] \rbrace$.
A: A given integer is of the form $6k + r$, where $0\leq r<6$. The number of choices we have for $r$ coincides with the number of equivalence classes modulo $6$.
A: That's $1$ class and it gives you all the integers where $x = 6k$.
But what about the other integers?  What about $x = 6k +1$ and $x = 6k +2$ and $x = 6k +3$ and $x = 6k + 4$ and $x = 6k + 5$.
EVERY integer has to be in some class and no integer can be in more than one class (that's the definition of equivalence classes and  of partioning $\mathbb Z$).
You have six classes.
$[0] = \{.....,-6,0, 6,12,.....\}$ 
$[1] = \{.....,-5,1, 7,13,.....\}$ 
$[2] = \{.....,-4,2, 8,14,.....\}$ 
$[3] = \{.....,-3,3, 9,15,.....\}$
$[4] = \{.....,-2,4, 10,16,.....\}$ 
$[5] = \{.....,-7,-1, 5,11,.....\}$ 
That way $\mathbb Z = \cup_{i=0...5}[i]$ and $[i]\cap [j] = \emptyset$ if $i \ne j$. And $\mathbb Z$ is partitioned into $6$ different disjoint sets (or classes).
A: You may visualise the classes by listing the integers in the following way:
$$\begin{matrix}
&&\vdots&&&\\
-12,&-11,&-10,&-9,&-8,&-7,\\
-6, & -5, &-4,&-3,&-2,&-1,\\
0,&1,&2,&3,&4,&5,\\
6,&7,&8,&9,&10,&11,\\
12,&13,&14,&15,&16,&17,\\
18,&19,&20,&21,&22,&23,\\
&&\vdots&&&
\end{matrix}$$
Now each column is a class.
