With $A = \mathbb{Z}$ and $B = \mathbb{Z}-\{0\}$, I'm trying to prove that the relation $R$ defined on $A\times B\,$ by $(a,b)R(c,d)\,$ iff $\,ad=bc$, is an equivalent relations.
While I do see why reflexivity and symmetry hold, I have troubles with transitivity. Could you please give me a clue?
Let $(a,b),(c,d), (e,f)\in A\times B$ and assume that $(a,b)R (c,d)$ and $(c,d)R(e,f)$. The aim is to show that $(a,b)R(e,f)$, i.e. that $af = be$. For, note that $$ad = bc$$ and
$$cf = de$$ and $b$ and $c$ are non-zero. Now observe that
$$ adf = (ad)f = (bc)f = b(cf) = b(de) = bde, $$
and so $adf = bde$, which can be rearranged to $$ d(af-be)=0. $$ Since $d\in B$, it follows that $d \neq 0$. Therefore, the zero product property for integers implies that $af-be=0$, which implies $(a,b)R(e,f)$.