The question is:
Define a relation $\rho$ on $\mathbb{Z}$ such that $a\ \rho \ b\ $ iff $a=b$ or $a,b$ are positive. Show that the relation is an equivalence and find the equivalence classes.
My solution:
Let us write $\mathbb{Z}$ as $\mathbb{Z^+ \cup {Z^-}^*}$ . If $a \in \mathbb{Z^+}$ then $a \ \rho \ b$ implies $b \in \mathbb{Z^+}$. Similarly for other set. We then proceed by showing the reflexivity [ $a \ \rho\ a \ \forall a \in \mathbb{Z}$ since if $a \in \mathbb{Z}$ then a is positive and relates to itself and if $a \in \mathbb{{Z^-}^*}$, $a=a$ ] , symmetry [ leaving out the case when $a \in \mathbb{Z^+}$ and $b \in \mathbb{{Z^-}^*}$ as in this case a does not relate to b ], and transitivity [ following similar methods].
The disjoint classes I end up with are $\mathbb{Z^+}$, $\{i:i \in \mathbb{{Z^-}^*}$} i.e $\{0\}, \{-1\}...,\{-n\},...$
First of all, a relation can be thought as a "boolean", which can be either TRUE or FALSE. For example, equality (=) is also a relation and we have $2 =2$ TRUE but $2 = 3$ FALSE.
According to that, when we say $a\ \rho\ b$ iff $a =b$ or $a,b \in \mathbb{Z^+}$, we actually say that in order for $a\ \rho\ b$ to be TRUE, it is enough to have only $a = b$, only $a,b \in \mathbb{Z}^+$ or both (truth table of an "or" statement can give more concrete explanation).
Now, in order to prove $\rho$ is an equivalence relation, we need to check whether it is reflexive, symmetric and transitive:
Reflexivity: We have $a\ \rho\ a$ for all $a \in \mathbb{Z}$ since we have $a = a$ for all $a \in \mathbb{Z}$. And by above information, is is enough to check equality since it already holds for everything in $\mathbb{Z}$.
Symmetry: For both equality and positivity part, it is obvious. For negative numbers, since they only satisfy the equality condition, it can be said that symmetry holds for negative numbers as well.
Transitivity: For equality and positivity, it is obvious. For negative numbers, it is similar to symmetry condition.
After that, equivalence classes that you mentioned are correct.