What are the equivalence classes for the relation "congruence modulo 5?"

Question

I'm still a little mixed up on equivalence classes, so I'm trying to make some connections. I need to be specific of how many there are and what is in each.

Here's what I have:

Let $\mathscr R$ be the relation "congruence modulo $5$" on a set $A$, and let $a ∈ A$.
Then $[a] = \{ x\in \mathbb Z : 5\mid(x-a)\}$.

From my understanding, there are infinitely many equivalence classes:
$[0] = \{\dotsc, -5, 0, 5, 10, 15,\dotsc\}$
$[1] = \{\dotsc, -4, 1, 6, \dotsc\}$
...
$[5] = \{\dotsc, -5, 0, 5, 10, 15\dotsc\}$
...

Am I correct? How can I describe exactly what these classes contain using relations?

Answer 1

There would be 5 distinct equivalence classes for congruence modulo 5.These would be [0],[1],[2],[3],[4]. Notice how these classes together will cover all the integers. This is because the congruence class for

...[0]=[5]=[10]=...

...[1]=[6]=[11]=...

...[2]=[7]=[12]=...

...[3]=[8]=[13]=...

...[4]=[9]=[14]=...

Answer 2

There are $5$ equivalence classes. You described all of them, but a lot of times.

The classes are $[0],[1],[2],[3],[4]$. Note that $[0]=[5]=[10]=\dots=[5k]$ for every integer $k$, and the same goes for the others.

You can see that these are indeed all the equivalence classes, because from the division algorithm, you know that for every integer $n$ there's a unique $0\leq r<|b|$ such that $n=qb+r$ (in this case $b=5$, and we've covered the five possible $r$'s).

For your second question: $$[0]=\{n\in \Bbb Z: n\equiv 0 \pmod 5\}$$

And more generally: $$[k]=\{n\in \Bbb Z: n\equiv k \pmod 5\}$$

It's obvious that these five classes are disjoint because (Why?).

Answer 3

There are five classes. The most common choice to represent the classes is $[0],[1],[2],[3],[4]$, but note that you are not restricted to this choice. Another convenient choice is $[-2],[-1],[0],[1],[2]$. If you really wanted to, you could use $[190],[501],[2352],[1463],[14369]$. (Check for yourself that these classes are distinct and that their union is $\mathbb Z$.)