Different ways of writing congruence classes of Zn

Question

Congruence classes of, say, $\Bbb Z_5$ can be written as $\Bbb Z_5 = \{[0],[1],[2],[3],[4]\}$, but can also be written as $\Bbb Z_5 = \{[-2],[-1],[0],[1],[2]\}$. Why? How does it work?

Answer 1

Remember that equivalence classes are sets of values whose union is the set you are partitioning into classes. Particularly, you are partitioning the set of integers into $5$ sets using the equivalence relation $x \equiv y \; (\bmod \; 5)$, and any two elements in the same set are equivalent to each other under this relation.

The reason the two representations are equivalent is that under the equivalence relation, we have: $$[n + 5k] = [n]$$ which comes directly from the definition of equivalence modulo $5$. Therefore, in your second representation $[-2] = [3]$, and $[-1] = [4]$, which makes your two representations equivalent.

In this case, we actually have the following general representation: $$\mathbb{Z_5} = \{[5 n_0],[5 n_1+1],[5 n_2+2],[5 n_3+3],[5 n_4+4]\}$$ for some $n_0,n_1,n_2,n_3,n_4 \in \mathbb{Z}$ since elements from distinct classes are equivalent to a different residue ($0 \leq r < 5$) modulo $5$. The elements of the class $[5 n_0]$, for example, are multiples of $5$ (equivalent to $0$ modulo $5$), and the elements of $[5 n_2 + 2]$ are equivalent to $2$ modulo $5$.

It is also worth noting that any two equivalence classes have an empty intersection; as a matter of fact, this is true for equivalence classes in general, regardless of the equivalence relation used.

Answer 2

Take an integer $n$, divide it by $5$ to get a remainder $r$. Now the number $n+1$ will give a remainder $r+1$, $n+2$ will give $r+2$ and so on. One minor issue is $r+k$ becoming larger than 4. THen take further remainer.

The upshot of this is that 5 consecutive integers will give 5 distinct remainders.