Let N denote the set of all natural numbers and R be the relation on N × N defined by (a, b) R (c, d) if ad(b + c) = bc(a + d). Show that R is an equivalence relation.

A relation is said to be equivalence if it is reflexive, symmetric and transitive.

To prove: R an equivalence relation we need to prove that it is reflexive, symmetric and transitive.

Proof for Reflexive:

As, (a, b) R (c, d) if ad(b + c) = bc(a + d)

Let c = a and d = b

∴ ad(b + c) = bc(a + d) becomes

ab(b+a) = ba(a + b)

as the condition is satisfied.

∴ we can say that –

(a, b) R (a, b)

∴ R is reflexive

Proof for symmetricity:

Let, (a,b),(c,d) ∈ N × N

Let (a, b) R (c, d) ad(b + c) = bc(a + d)

⇒ da(c+b) = cb(d+a)

⇒ (c, d) R (a, b)

∴ R is symmetric

Proof for transitivity:

Let (a,b), (c,d), (e,f) ∈ N × N

As (a,b) R (c,d)

∴ ad(b + c) = bc(a + d)

Dividing by abcd both sides.

We have-

∴ we can say that if (a,b) R (c,d) ⟺ …(1)

Similarly (c,d)R(e,f) ⟺ …(2)

Adding equation 1 and 2 we get-

⇒

⇒ (a, b) R (e, f)

∴ R is transitive.

Hence we showed that R is reflexive, symmetric and transitive.

∴ We can say that R is an equivalence relation. …proved