can someone explain me the following please?:
$\mathbb{Z} /_5 = \{ [0]_5, [1]_5, [2]_5, [3]_5, [4]_5\}$
Is that correct?
$[0]_5 = \{ \dotsc, -15, -10, -5, 0, 5, 10, 15, \dotsc\}$
$[1]_5 = \{ \dotsc, -14, -9, -4, 1, 6, 11, 16, \dotsc\}$
$[2]_5 = \{ \dotsc, -13, -8, -3, 2, 7, 12, 17, \dotsc\}$
and does $/_5$ mean that the set has 5 equivalence classes?
Here, $\mathbb Z/_5 = \mathbb Z_5$ is used to denote the group of integers under addition, modulo $5$. It consists of the equivalence classes (also known as residue classes $[r_i]$), which represent the integers $z_i$ which are all congruent to $r_i$ modulo 5, where $r_i$ denotes the remainder in the Euclidean Algorithm $$z_i = 5q + r_i,\;\;0\leq r_i \lt 5$$
You are correct that $\mathbb Z/_5 = \{[0]_5, [1]_5, [2]_5, [3]_5, [4]_5 \}$, which essentially means that the set of integers are partitioned into $5$ equivalence classes, every integer belonging to one, and only one, equivalence class.
We can also represent your equivalence classes as follows:
$[0]_5 = \{5k\mid k \in \mathbb Z\}$
$[1]_5 = \{5k + 1 \mid k \in \mathbb Z\}$
$[2]_5= \{5k + 2 \mid k\in \mathbb Z\}$
$[3]_5 = \{5k + 3 \mid k \in \mathbb Z\}$
$[4]_5 = \{5k + 4 \mid k\in \mathbb Z\}$
Indeed, for $\mathbb Z_n$, there will always be exactly $n$ such equivalence classes, one for each $r$ such that $0 \leq r \lt n$. So in your case, that means five equivalence classes.