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.

Given that, R be the relation in N ×N defined by (a, b) R (c, d) if ad(b + c) = bc(a + d) for (a, b), (c, d) in N ×N.

Reflexivity:

Let (a,b) R (a,b)

⇒ ab(a+b) =ba( a+b)

which is true since multiplication is commutative on N.

⇒ R is reflexive.

Symmetricity:

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

⇒ ad(b + c) = bc(a + d)

⇒ da(c + b) = cb(d + a)

[since addition and multiplication is commutative on N]

⇒ cb(d + a) = da(c + b)

⇒ (c,d) R (a,b)

∴ (a,b) R (c,d) ⇒ (c,d) R (a,b) ∀ (a,b), (c,d), (e,f) ∈ N ×N.

⇒ R is symmetric.

Transitivity:

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

⇒ ad(b + c) = bc(a + d)

also (c,d) R (e,f)

and cf(d + e) = de(c + f)

Adding (i) and (ii) , we get

⇒ af(b + e) = be(a + f)

⇒ (a,b) R (e,f)

Thus , (a,b) R (c,d) and (c,d) R (e,f) ⇒ (a,b) R (e,f) ∀ (a,b), (c,d), (e,f) ∈ N ×N.

⇒ R is transitive.

Hence, R is an equivalence relation.