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 is
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)