For reflexive: Let $a\in\mathbb{Z}$. This means $a \equiv a(mod n)$, which can be written as $n\mid a-a$. There is an integer $k$ such that $a-a=nk$.
I know we need to find if there is a $k \in \mathbb{Z}$ for integer $a$, when $a-a=nk$. I know $k$ must be zero for this to be true, but I'm not sure how to write the proof.
For symmetric: Let $a,b \in \mathbb{Z}$. This means when $a\equiv b(mod n)$, then $a-b=nk$ for some integer $k$, so $b-a=nr$ for some integer $r$ and hence $b\equiv a(mod n)$.
Again I'm not sure how to write the proof.
For transitive: Let $a,b,c \in\mathbb{Z}$. This means when $a\equiv b(mod n)$, then $a-b=nk$ for some integer $k$. We also know $b\equiv c(mod n)$, so $b-c=nm$ for some integer $m$. There must be some integer $r$ such that $a-c=nr$.
For symmetry
Let $ (a,b)\in \Bbb Z^2.$
$$a\equiv b \mod n \implies$$
$$\exists k\in \Bbb Z \;\; : \;\; a-b=kn \implies$$
$$\exists k\in \Bbb Z \;: b-a=(-k)n \implies $$ $$\exists r(=-k) \in \Bbb Z \;\;:\; b-a=rn \implies$$
$$b\equiv a \mod n$$
For the transitivity, observe that $$(a-b) + (b-c) = (a-c)$$