Q. 135.0( 1 Vote )

# 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 isA. symmetric but neither reflexive nor transitiveB. reflexive but neither symmetric nor transitiveC. transitive but neither reflexive nor symmetricD. none of these

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)

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