Let N denote the set of all natural numbers and R be the relation on N x N defined by (a, b) R (c, d ) ⇔ a d (b + c) = b c (a + d). Check whether R is an equivalence relation on N x N.
(i) Let (a, b) be any element of N x N
Now (a, b) ∈ N x N ⇒ a, b ∈ N ∴ a b (b + a) = b a (a + b)
⇒ (a, b) R (a, b)
But (a, b) is any element of N x N ∴ (a, b) R (a, b) ∀ (a, b) ∈ N x N ∴ R is reflexive on N x N.
(ii) Let (a, b), (c, d ) ∈ N x N such that (a, b) R (c, d)
Now (a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
⇒ c b (d + a) = d a (c + b)
⇒ (c, d) R (a, b)
∴ (a, b) R (c, d ) ⇒ (c, d) R (a, b) ∀ (a, b), (c, d) ∈ N x N ∴ R is symmetric on N x N.
(iii) Let (a, b), (c, d ), (e, f) ∈ N x N such that
(a, b) R (c, d ) and (c, d ) R (e, f)
(a, b) R (c, d) ⇒ a d (b + c) = b c (a + d)
Also (c, d) R (e, f) cf(d + c) = d e (c + f)
Adding (1) and (22), we get
⇒ a f (b + e) = b e (a + f) ⇒ (a, b) R (e,f)
∴ (a, b) R (c, d) and (c, d) R (e.f) ⇒ (a, b) R (e, f) ∀ (a, b), (b, c), (c, d) ∈ N x N ∴ R is transitive on N x N ∴ R is reflexive, symmetric and transitive ∴ R is an equivalence relation on N x N