Prove, that $M=\{(a,b)\mid a,b \in \mathbb{N_0} \land (a-b) \text{ mod } 4 = 0\} $ is an equivalence-relation.
Refl.: $a-a=0 \text{ mod } 4 =0$
Sym.: $\forall x,y \in M: (x,y) \implies (y,x)$ (Not sure about this.)
Trans.: $\left((a-b)+(b-c)\right)\text{ mod } 4 = 0 \iff (a-c)\text{ mod } 4 =0$
Can you help me with proving, that the relation is symmetric and is everything else correct?
I'll assume the notation "$(a - b) \mod 4 = 0$" is equivalent to the more commonplace $a \equiv b \pmod 4$. Both essentially, to my understanding anyhow, mean that $4$ divides $(a-b)$.
To show symmetry, suppose $(x,y) \in M$. Thus, $x \equiv y \pmod 4$. Thus, there exists an integer $k$ such that $x-y = 4k$. Now, notice, if we multiply through by $-1$ that $y-x = -4k=4(-k)$. Since $k$ is an integer, $-k$ also is. Thus, there exists an integer - namely $m=-k$ - such that $y-x = 4m$. But this means $y \equiv x \pmod 4$, and thus $(y,x) \in M$.
Using this framing, the other two requisites are also easy to prove.
Reflexive: Does $a \equiv a \pmod 4$ hold? Yes, because $4 \mid (a-a)$, i.e. $4 \mid 0$. Thus, $(a,a) \in M$.
Transitivity: Let $(a,b),(b,c) \in M$. Thus, $a \equiv b \pmod 4$ and $b \equiv c \pmod 4$. Then there exist integers $p,q$ such that $a-b = 4p$ and $b-c=4q$. Adding these equations together we see that $a-b+b-c=4p+4q$, i.e. $a-c = 4(p+q)$. Since $p,q$ are integers, so is $p+q$, and thus there exists an integer $(p+q)$ such that $(a-c)/4$ is that integer. Therefore, $a \equiv c \pmod 4$ and $(a,c) \in M$, giving transitivity.