Solution
= Solution
Every nonzero integer $a$ has $a^2>0$, proving <reflexivity>. Because $ab=ba$, the relation is <symmetric>. If $ab>0$ and $bc>0$, then $ac=(ab)(bc)/b^2>0$, proving <transitivity>. Thus this is an <equivalence relation>, whose \b[two <equivalence classes> are the positive integers and the negative integers]:
$$
\{1,2,3,\ldots\},\qquad\{-1,-2,-3,\ldots\}.
$$