Consider the relations R 1 and R 2 defined as a R 1 b ⇔ a 2 + b 2 = 1 for all a , b , ∈ R and a , b R 2 c , d ⇔ a + d = b + c for all a , b , c , d ∈ N × N . Then

Step-by-step solution

Given, a R 1 b ⇔ a 2 + b 2 = 1 ; a , b ∈ R R 1 is not reflexive as a , a ∉ a R 1 b So, it is not equivalence. Now, solving a , b R 2 c , d ⇔ a + d = b + c ; a , b , c , d ∈ N Reflexive: a + b = b + a → True Symmetric: a , b R 2 c , d ⇒ a + d = b + c ⇒ d + a = c + b ⇒ c + b = d + a ⇒ c , d R 2 a , b Transitive: a , b R 2 c , d ⇒ a + d = b + c . . . i c , d R 2 e , f ⇒ c + f = d + e . . . i i Now, adding above equation we get, ⇒ a + f = b + e ⇒ a , b R 2 e , f So, R 2 is reflexive, symmetric and transitive Hence only