Question

A relation S in the set of real numbers is defined as xSy ⇒ x – y+ √3 is an irrational number, then relation S is

Answer 1

SOLUTION

TO DETERMINE

A relation S in the set of real numbers is defined as xSy ⇒ x – y + √3 is an irrational number, then relation S

EVALUATION

Here the given relation is

xSy ⇒ x – y + √3 is an irrational number

CHECKING FOR REFLEXIVE

Let x ∈ R

Since x – x + √3 = √3 is an irrational number

So (x, x) ∈ S

So S is Reflexive

CHECKING FOR SYMMETRIC

√3 , 0 ∈ R

Now (√3 , 0) ∈ S as √3 - 0 + √3 = 2√3 is an irrational number

But (0,√3 ) ∉ S as 0 - √3 + √3 = 0 is a rational number

Thus (√3 , 0) ∈ S but (0,√3 ) ∉ S

So S is not symmetric

CHECKING FOR TRANSITIVE

Here √3 , 0 ∈ R

Now

(√3 , 0) ∈ S as √3 - 0 + √3 = 2√3 is an irrational number

Also ( 0, 2√3 ) ∈ S as 0 - 2√3 + √3 = - √3 is an irrational number

But (√3, 2√3 ) ∉ S as √3 - 2√3 + √3 = 0 is a rational number

Thus (√3 , 0) ∈ S and ( 0, 2√3 ) ∈ S but (√3, 2√3 ) ∉ S

S is not transitive

FINAL ANSWER

S is reflexive but neither symmetric nor transitive

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