Equivalence relation - Help to formulate my proof

Question

so I got this question that I am stuck on: So: Consider the set $$S = \{0,\pm1,\pm2,\pm3,\pm4,\pm5,\pm6,\pm7,\pm8\}$$ Consider the relation $R$ on $S$ defined by $(a,b)$ part of $R$ if and only if $[a]4 = [b]4$ in $\Bbb Z_4$

Now the way that I understand this question is that $[a]4$ is a remainder in the congruence class $4$. So basically $a$ and $b$ needs to have the same remainders.

I find this by doing $n | a - b$ , and let $n = 4$

I find the subsets of

$[0] = {0,\pm4,\pm8}\\ [1] = {-7,-3,1,5}\\ [2] = {\pm2,\pm6}\\ [3] = {-5,-1,3,7}$

So, now I am going to prove that this is a equivalence class

First I have to prove that $x \sim x$, which it is because $[x]4 = [x]4,$ $R$ is reflexive

Symmetric property $x \sim y \Rightarrow y \sim x$ I am not sure how to show this, I know if i take for example $[2]4$ I get the $[2] = {\pm2,\pm6}$ and no matter how I put them I will always be able to get the same remainder.

Transitivity $x \sim y, y \sim z,$ then $x \sim z$ Not sure how to show this either, as I cannot use the numbers in the subsets I've created to show.

Could anyone help me out here? Not sure if this is against the rules to ask for this type of guidance. Thank you!

/ Novice in math

Answer 1

In this case, for symmetry and transitivity, we can actually say that symmetry and transitivity are inherited from equality. Because, if we assume $[a]_4 = [b]_4$, then clearly $[b]_4 = [a]_4$ (this is symmetry of equality) so symmetry holds. And if we assume $[a]_4 = [b]_4$ and $[b]_4 = [c]_4$, then $[a]_4 = [c]_4$ (this is transitivity of equality) so transitivity holds.

Answer 2

Regarding symmetry, observe that if $x-y$ is a multiple of $4$, then so is $y-x$.

For transitivity, let $x-y = 4k$ for some $k \in \mathbb{Z}$ and let $y-z = 4\ell$ for some $\ell \in \mathbb{Z}$. Then $x-z = (x-y)+(y-z) = 4(k+\ell)$, which is also a multiple of $4$.