Z = { ..., -2, -1, 0, 1, 2, ... } Relation ~ is defined such that a~b <=> a evenly divides b.
I said that this function IS reflexive because a evenly divides a. I said that this function is NOT symmetric because if a divides b, b does not necessarily divide a. This function would be transitive because if a~b and b~c then a~c.
Is that how I do this?
Yes, you are basically correct (except that when you write 'function' you mean 'relation'). Also, you need to give some argument to support your arguments, if the course you are taking is expecting proofs, rather than just claims. For instance, you can prove that the relation is transitive as follows. Suppose $a~b$ and b~c. That means that $b=ka$ and $c=mb$ for some $k,m\in \mathbb Z$. Thus $c=mb=mka$ and thus $a$ evenly divides $c$, so that $a~c$.