Correct Answer is (A) reflexive and symmetric but not transitive

Given set S = {…….,-2,-1,0,1,2 …..}

And R = {(a, b) : a,b ∈ S and (1 + ab) > 0 }

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)

∴ (1 + a×a) > 0 which is always true because a×a will always be positive.

Ex_if a=2

∴ (1 + 4) > 0 ⇒ (5) > 0 which is true.

Therefore , R is reflexive ……. (1)

Check for symmetric

a R b ⇒ (1 + ab) > 0

b R a ⇒ (1 + ba) > 0

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

Ex _ If a=2 and b=1

∴ (1 + 2×1) > 0 is true and (1+1×2) > which is also true.

Therefore , R is symmetric ……. (2)

Check for transitive

a R b ⇒ (1 + ab) > 0

b R c ⇒ (1 + bc) > 0

∴(1 + ac) > 0 will not always be true

Ex _a=-1 , b= 0 and c= 2

∴ (1 + -1×0) > 0 , (1 + 0×2) > 0 are true

But (1 + -1×2) > 0 is false.

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

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

Correct option will be (A)