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 |a – b| ≤ 1. Then, R is


According to the question ,


Given set S = {…….,-2,-1,0,1,2 …..}


And R = {(a, b) : a,b S and |a – b| ≤ 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)


|a – a| ≤ 1 and which is always true.


Ex_if a=2


|2-2| ≤ 1 0 ≤ 1 which is true.


Therefore , R is reflexive ……. (1)


Check for symmetric


a R b |a – b| ≤ 1


b R a |b – a| ≤ 1


Both can be true.


Ex _ If a=2 and b=1


|2 – 1| ≤ 1 is true and |1–2| ≤ 1 which is also true.


Therefore , R is symmetric ……. (2)


Check for transitive


a R b |a – b| ≤ 1


b R c |b – c| ≤ 1


|a – c| ≤ 1 will not always be true


Ex _a=-5 , b= -6 and c= -7


|6-5| ≤ 1 , |7 – 6| ≤ 1 are true But |7 – 5| ≤ 1 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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