How many equivalence classes in the equivalence relation 
Consider the equivalence relation defined on the set A = Z \ {0}, where a~b if an only if ab > 0.

I assume this means that A is the set of all integers except 0.

How many equivalence classes are there in the above equivalence relation? Describe each of the equivalence classes.

This is my first equivalence relation assignment, and I am not sure I understand equivalence relations and classes yet. But I assume that to "form" a class, I must select all pairs (a,b) that make ab = 1 (for equivalence class 1), all those that make ab = 2 (for equivalence class 2), etc.
However, would that not mean there are infinite equivalence classes since each integer number must have a class of its own?
 A: You are reading the relation incorrectly. Write $a \sim b$, if $ab > 0$. E.g. $1 \sim 2$ since $1\cdot 2 > 0$, in fact $1 \sim a$, with any $a > 0$, since $1a > 0$. What about $a < 0$?
A: An equivalence relation on a set $S$ "divides" that set into disjoint subsets, called equivalence classes. Now, suppose $\sim$ is an equivalence relation and $a \in S$. The equivalence class of $a$ under $\sim$, denoted $[a]$ is defined by
$$[a] := \{ b \in S \,|\, a \sim b\} $$
So, in your example, given a non-zero integer $n$, its equivalence class is the set
$$[n]:= \{ m \in \mathbb{Z}\setminus\{0\} \, \mid \, nm > 0\}$$
By definition of an equivalence relation, $\sim$ is reflexive, which means that  $a \sim a$ for every $a \in S$, so $a \in [a].$
Using this property for $2 \in \mathbb{Z}\setminus \{0\},$ we see that $ 2 \in [2]$, which is true since $2 \cdot 2 >0$.
$3$ is also a member of $[2]$, since $2 \cdot 3 > 0$, but $-2 \not\in [2]$, since $2 \cdot (-2) < 0,$ so we have at least two distinct equivalence classes. 
Hope this helps.
