Solution:A relation R on a set N is said to an equivalence relation it is Reflexive, symmetric and transitive.
Reflexive: For any (a,b) in N × N, we nee to show that (a, b) R (a, b).
ad (b + a) = ab (a + d), which is true since both sides are equal.
Symmetric: For any (a, b) and (c, d) in N × N,
If (a, b) R (c, d) then (c, d) R (a, b).
if ad (b + c) = bc (a + d), then bc (a + d) = ad (b + c), which is the same relation, so it's symmetric.
Transitive: For any (a, b) (c, d) and (e, f) in N × N, if (a, b)R (c, d) and (c, d) R (e, f) then (a, b) R (e, f). If ad (b + c) = bc (a + d) and ce (d + f) = de (c + e), then ac (b + f ) = be (a + f), which means (a, b) R (e, f).
Since R satisfies all three properties, it is indeed an equivalence relation.