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)