Q. 115.0( 1 Vote )

# Mark the tick against the correct answer in the following:Let S be the set of all real numbers and let R be a relation on S, defined by a R b ⇔ (1 + ab) > 0. Then, R isA. reflexive and symmetric but not transitiveB. reflexive and transitive but not symmetricC. symmetric and transitive but not reflexiveD. none of these

According to the question ,

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)

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