Find all equivalence classes of $5\mid(x-y)$

Question

Find all equivalence classes of $$x\mathcal{R}y\iff5\mid(x-y),$$ where $x,y\in\Bbb{Z}$.

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I know that $5\mid(x-y)$ means that there exists an integer $k$ such that $x-y=5k$. Hence,

$\mathrm{cl}(1)=[1]=\{y\in\Bbb{Z}\colon1\mathcal{R}y\}=\{y\in\Bbb{Z}\colon1-y=5k,\;k\in\Bbb{Z}\}=\{y\in\Bbb{Z}\colon y=1-5k,\;k\in\Bbb{Z}\}=\{1-5k,\;k\in\Bbb{Z}\}.$

$[0]=\{y\in\Bbb{Z}\colon0\mathcal{R}y\}=\{-5k,\;k\in\Bbb{Z}\}.$

$[2]=\{y\in\Bbb{Z}\colon2\mathcal{R}y\}=\{2-5k,\;k\in\Bbb{Z}\}.$

$[3]=\{y\in\Bbb{Z}\colon3\mathcal{R}y\}=\{3-5k,\;k\in\Bbb{Z}\}.$

$[4]=\{y\in\Bbb{Z}\colon4\mathcal{R}y\}=\{4-5k,\;k\in\Bbb{Z}\}.$

$[5]=\{y\in\Bbb{Z}\colon5\mathcal{R}y\}=\{5-5k,\;k\in\Bbb{Z}\}=[0].$

$[6]=\{y\in\Bbb{Z}\colon6\mathcal{R}y\}=\{6-5k,\;k\in\Bbb{Z}\}=[1].$

$\vdots$

So can I say that the equivalence classes are $[0]$, $[1]$, $[2]$, $[3]$ and $[4]$?

And the quotient set is $\Bbb{Z}/\mathcal{R}=\{[0],[1],[2],[3],[4]\}=\Bbb{Z}_{5}$?

Thanks!

Answer 1

Yup, that looks good! To prove that these were the only classes, you'd appeal to the principle of mathematical induction, by taking $n=0,1,2,3,4$ as your base cases.