Take any $n \in \mathbb{Z}.$
Then for each such $n$ there is an integer $k$ such that one of the following equations are satisfied. $$n=6k + 0$$ $$n=6k+1$$ $$n=6k+2$$ $$n=6k+3$$ $$n=6k+4$$
$$n=6k+5$$
Can any integer satisfy more than one of the above equalities?
Can any integer NOT satisfy any of of the above equalities?
Then we can describe the equivalence classes of the equivalence relation $\equiv_{6}$ (congruence, mod $6$) as they relate to the division algorithm:
Given any $d \in \mathbb{Z}, d>0$, we know that $\forall n \in \mathbb{Z}$, there exist unique integers $q, r$ such that $n = dq + r$ where $0\leq r< d$.
In this problem, we have the divisor $d = 6$ of a given $n$:
- $k = q\in \mathbb{Z}$ is the unique corresponding quotient which results when dividing $n$ by $d = 6$, and
- $r$ is the unique corresponding remainder, $0\le r < d = 6$, left after dividing that $n$ by $6$.
Then the corresponding equivalence classes can be defined in terms of the the remainder $r$:
$$[r]_6 \in \{[0]_6,[1]_6, [2]_6, [3]_6, [4]_6, [5]_6\} \;\text{ where}$$
$$ [0]_6 = \{...,-12,-6,0,6,12,...\}$$
$$[1]_6 = \{...-11,-5,1,7,13,...\}$$
$$[2]_6 = \{...-10,-4,2, 8, 14...\}$$
$$ \vdots$$
$$[5]_6 = \{...-7,-1,5, 11, 17,...\}$$
Your intuition is correct. To describe the equivalence classes explicitly, it often helps to find the equivalence classes of numbers that are easy to work with.
Let's find, for example, the equivalence classes $[0]$ and $[1]$.
$0+a$ is even for which integers $a$? All these will form $[0]$.
$1+b$ is even for which integers $b$? All these will form $[1]$.
Are there any other equivalence classes? (You can be sure you've found them all when your equivalence classes form a partition of $\Bbb Z$.)
Best Answer
Note that the equivalence classes must be subsets of $\Bbb Z$. Using set-builder notation, the equivalence classes are $$S_a=\{\,a+5k\mid k\in{\Bbb Z}\,\}$$ for $a=0,1,2,3,4$.