I want to prove $a \equiv b\;(\text{mod} \;n)$ is an equivalence relation then would it be ok to write,
Reflexive as, for all $a$, $a \equiv a\;(\text{mod} \;n)$
Symmetric as, $a \equiv b\;(\text{mod} \;n)$ which implies $b \equiv a\;(\text{mod} \;n)$
Transitive as if $a \equiv b\;(\text{mod} \;n)$ and $b \equiv c\;(\text{mod} \;n)$ this implies $a \equiv c\;(\text{mod} \;n)$
I dont know if this has actually proved equivalence. Also the set on which this relatiton acts on was not specified so for reflexivity is it ok to say for all $a$. Thanks.
It is not ok, because what you did is just wrote what you have to prove, not really proved it. First of all, the relation is on $\mathbb{Z}$. Next, what is the relation? $a\equiv b$(mod $n$) by definition means that $n|(a-b)$. So that is the relation: $a,b\in\mathbb{Z}$ are related if $n$ divides $a-b$. Now you have to prove it is an equivalence relation. For example, it is reflexive because for each $a\in\mathbb{Z}$ we have $a-a=0$ and obviously $n|0$ because $0\times n=0$. Now try to show the other properties.