∀(a,b)∈N×N∗,ab=ab
So (a,b)R(a,b) and R is reflexive.
Assume now that (a,b)R(c,d) i.e ad=bc.
Multiplication is commutative so
cb=da and
(c,d)R(a,b).
The relation is symmetric
Now take (a,b)R(c,d) and (c,d)R(e,f) ;
this means ad=bc and cf=de.
Multiply the first equality by f≠0 to get afd=bcf
and the second by b≠0 to get bcf=bed.
So we have afd=bed and
d≠0 we have
af=be
i.e (a,b)R(e,f) and the relation is transitive
It is therefore an equivalence relation
Now suppose (a,b)≃(c,d) and (c,d)≃(e,f)
Then ad=bc and cf=de
Thus
(ad)(cf)=(bc)(de)
and by cancelling from both sides
af=be
Accordingly (a,b)≃(e,f) and R is transitive.
R is an equivalence relation