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 ⇔ a2 + b2 = 1. Then, R is
According to the question ,
Given set S = {…….,-2,-1,0,1,2 …..}
And R = {(a, b) : a,b ∈ S and a2 + b2 = 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)
∴ a2 + a2 = 1 which is not always true
Ex_if a=2
∴ 22 + 22 = 1 ⇒ 4 + 4 = 1 which is false.
Therefore , R is not reflexive ……. (1)
Check for symmetric
a R b ⇒ a2 + b2 = 1
b R a ⇒ b2 + a2 = 1
Both the equation are the same and therefore will always be true.
Therefore , R is symmetric ……. (2)
Check for transitive
a R b ⇒ a2 + b2 = 1
b R c ⇒ b2 + c2 = 1
∴ a2 + c2 = 1 will not always be true
Ex _a=-1 , b= 0 and c= 1
∴ (-1)2 + 02 = 1 , 02 + 12 = 1 are true
But (-1)2 + 12 = 1 is false.
Therefore , R is not transitive ……. (3)
Now , according to the equations (1) , (2) , (3)
Correct option will be (A)