Produce the set of all Equivalent Classes

Question

Consider the equivalence relation $C_m = \{(x,y)\in\mathbf{Z}\times\mathbf{Z}|x\equiv y$ (mod $m$)$\}$ where $m\in\mathbf{Z}^{+}$

We are required to determine the set of all equivalence classes for $C_m$

This is what i think $\mathcal{F}_m = \{i\in\mathbf{Z}|\{i\equiv j$ (mod $m$)$|j\in\{0,\pm1,\pm2,...,\pm m-1\}\}\}$.

Would it be correct to state that $|\mathcal{F}_m| = 2m-1$ ?

Answer 1

A couple of ideas for you to fill in the details:

$$\text{Prove that}\;\;[0],\,[1],\ldots,[m-1]\;\;\text{are all different classes in}\;\;\Bbb Z$$

$$\text{Prove that for any}\;k\in\Bbb Z\;\text{ there exists}\;\; \;0\le i\le m-1\;\;s.t.\;\;k\in[i]\;$$

Answer 2

When $\equiv$ is an equivalence relation on a set $S$ and $x\in S,$ the equivalence class of $x,$ usually denoted by $[x]_{\equiv}$, is $\{y\in S:y\equiv x\}.$

In this case we have $y\equiv x \iff \exists n\in \mathbb Z \;(y=mn+x)\iff y\in \{mn+x:n\in \mathbb Z\}.$ Therefore $[x]_{\equiv}=\{mn+x:n\in \mathbb Z\}.$

Example: When $m=2$ we have $[0]_{\equiv}\;=\{2n:n\in \mathbb Z\}$ and $[1]_{\equiv}\;=\{2n+1:n\in \mathbb Z\}$.

$F_m$ has exactly $m$ members because every $y\in \mathbb Z$ is equivalent modulo $m$ to exactly one member of the set $\{0,...,m-1\}$. For example with $m=8$ we have $-5\equiv 3\pmod 8$.