Correct Answer is (A) symmetric but neither reflexive nor transitive

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and a² + b² = 1 }

Formula

For a relation R in set A

Reflexive

The relation is reflexive if (a , a) ∈ R for every a ∈ A

Symmetric

The relation is Symmetric if (a , b) ∈ R , then (b , a) ∈ R

Transitive

Relation is Transitive if (a , b) ∈ R & (b , c) ∈ R , then (a , c) ∈ R

Equivalence

If the relation is reflexive , symmetric and transitive , it is an equivalence relation.

Check for reflexive

Consider , (a,a)

∴ a² + a² = 1 which is not always true

Ex_if a=2

∴ 2² + 2² = 1 ⇒ 4 + 4 = 1 which is false.

Therefore , R is not reflexive ……. (1)

Check for symmetric

a R b ⇒ a² + b² = 1

b R a ⇒ b² + a² = 1

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

a R b ⇒ a² + b² = 1

b R c ⇒ b² + c² = 1

∴ a² + c² = 1 will not always be true

Ex _a=-1 , b= 0 and c= 1

∴ (-1)² + 0² = 1 , 0² + 1² = 1 are true

But (-1)² + 1² = 1 is false.

Therefore , R is not transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (A)