Question

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

Answer 1

1.  (a, b) R (a, b). For a + b = b + a Therefore, R is reflexive.

2.  (a, b) R (c, d) ⇒ a + d = b + c

⇒ c + b = d + a ⇒ (c, d) R (a, b)

Therefore, R is symmetric.

3.  (a, b) R (c, d) and (c, d) R (e, f)

 ⇒ a + d = b + c and c + f = d + e

⇒ a + d + c + f = b + c + d + e 

 ⇒ a + f = b + e ⇒ (a, b) R (e, f)

Therefore, R is transitive.

Thus, R is an equivalence relation on N x N.