Given: 

Set N = {1, 2, 3, 4, …} 

= Set of natural numbers 

A relation R in N × N is defined as 

(a, b) R(c, d) ⇔ ad = bc 

where a, b,c, d ∈ N ∀ (a, b), (c, d) ∈ N × N 

Here, to prove R is a equivalence relation, we have to show that R is reflexive, symmetric and transitive. 

(i) Reflexivity: 

Let (a, b) ∈ N × N 

{a, b) ∈ N × N 

⇒ a.b = ba 

(Commutative law of multiplication) 

⇒ (a, b) R(a, b) ∀ (a, b) ∈ N × N 

R is reflexive relation. 

(ii) Symmetricity: 

Let (a, b) (c, d) ∈ N × N is in this way 

(a, b) R(c, d) (a, b) R(c, d) 

⇒ ad = bc 

⇒ bc = ad 

⇒ c.b = d.a 

⇒ (c, d) R(a, b) 

So, (a, b) R(c, d) 

⇒ (c, d) R(a, b) ∀ (a, b)(c, d) ∈ N × N 

R is a symmetric relation. 

(iii) Transitivity: 

Let (a, b), (c, d), (e, f) ∈ N × N is in this way 

(a, b) R(c, d) and (c, d) R(e, f) 

(a, b) R(c, d) ⇒ ad = bc 

(c, d) R(e, f) ⇒ cf = de 

(a, b)R (c, d) and (c, d)R (e, f) 

⇒ (ad) (cf) = (bc)(de) (on multiplication) 

⇒ af = be 

⇒ (a.b) R(e, f) 

So, (a, b) and R(c, d) and (c, f) R(e, f) 

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

R is a transitive relation. 

Hence, according to (i), (ii) and (iii), the given relation is an equivalence relation. 

Hence Proved.