Question

R is a relation on N*N defined by (a, b) R(c, d) for all a, b, c, d belongs to N. Show that R is an equivalence relation. ​

Answer 1

∀(a,b)∈N×N∗,ab=ab∀(a,b)∈N×N∗,ab=ab

So (a,b)R(a,b)(a,b)R(a,b) and RR is reflexive.

Assume now that (a,b)R(c,d)(a,b)R(c,d) i.e ad=bcad=bc. Multiplication is commutative so we can write cb=dacb=da and this gives (c,d)R(a,b)(c,d)R(a,b). The relation is symmetric

Now take (a,b)R(c,d)(a,b)R(c,d) and (c,d)R(e,f)(c,d)R(e,f) ; this means ad=bcad=bc and cf=decf=de. Multiply the first equality by f≠0f≠0 to get afd=bcfafd=bcf and the second by b≠0b≠0 to get bcf=bedbcf=bed. So we have afd=bedafd=bed and keeping in mind d≠0d≠0we have af=beaf=be i.e (a,b)R(e,f)(a,b)R(e,f) and the relation is transitive

It is therefore an equivalence relation