Which of the following equivalence classes are equal? Let R be the relation of congruence modulo 3. Which of the following equivalence classes are equal. 
[7], [-4], [-6], [17], [4], [27], [19]

The answer is: [7]=[4]=[19],[−4]=[17],[−6]=[27]. I just don't understand why the answer is this. How and why do we create relationships based on these numbers? 
 A: By definition, $a \equiv b \mod{3} \iff a - b = 3n, n \in \mathbb{Z}$.  What this means is that two integers are equivalent modulo $3$ if they have the same remainder when divided by $3$.  
By definition, $[7] = \{k \in \mathbb{Z} \mid 7 - k = 3n, n \in \mathbb{Z}\}$.  Thus, the numbers in this equivalence class differ from $7$ by a multiple of $3$.  Hence, 
$$[7] = \{\ldots, -11, -8, -5, -2, 1, 4, 7, 10, 13, 16, 19, 22, 25, \ldots\}$$
Notice that $[7]$ is the set of all integers that have remainder $1$ when divided by $3$.  The same is true for $[4]$ and $[19]$.  Therefore, $[4] = [7] = [19] = [1]$.  
The relationship $a \equiv b \pmod{3}$ partitions all integers into three equivalence classes, those with remainder $0$ when divided by $3$, those with remainder $1$ when divided by $3$, and those with remainder $2$ when divided by $3$.
Notice that those numbers in the equivalence classes $[-6]$ and $[27]$ have remainder $0$ when divided by $3$.  Hence, $[-6] = [27] = [0]$.  
The numbers in the equivalence classes $[-4]$ and $[17]$ have remainder $2$ when divided by $3$.  Hence, $[-4] = [17] = [2]$.
