Let N denotes the set of all natural numbers. On NxN define R as follows: (a,b),(c,d)∈N×N (a, b) R (c, d) if ad(b + c) = bc(a + d), then

ad(b+c)=bc(a+d) ⇔ b+cbc=a+dad ⇔ 1b+1c=1a+1d ⇔ 1a−1b=1c−1dR is reflexiveLet (a,b)∈N×N 1a−1b=1a−1b⇒(a,b)R(a,b)R is reflexive. R is symmetric. Suppose (a,b),(c,d)∈N×N and (a,b)R(c,d)⇔1a−1b=1c−1d⇔1c−1d=1a−1b⇔(c,d)R(a,b)R is symmetric. R is transitive.Suppose (a,b),(c,d),(e,f)∈N×N and (a, b) R (c, d), (c, d) R (e,f) ⇒ 1a−1b=1c−1dand1c−1d=1e−1f ⇒ 1a−1b=1c−1f⇒(a,b)R(e,f)R is transitive. R is an equivalence relation.