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. 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 }
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 and which is not always true.
Ex_if a=-2
∴ |-2| ≤ -2 ⇒ 2 ≤ -2 which is false.
Therefore , R is not reflexive ……. (1)
Check for symmetric
a R b ⇒ |a| ≤ b
b R a ⇒ |b| ≤ a
Both cannot be true.
Ex _ If a=-2 and b=-1
∴ 2 ≤ -1 is false and 1 ≤ -2 which is also false.
Therefore , R is not symmetric ……. (2)
Check for transitive
a R b ⇒ |a| ≤ b
b R c ⇒ |b| ≤ c
∴ |a| ≤ c
Ex _a=-5 , b= 7 and c=9
∴ 5 ≤ 7 , 7 ≤ 9 and hence 5 ≤ 9
Therefore , R is transitive ……. (3)
Now , according to the equations (1) , (2) , (3)
Correct option will be (C)