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)

1