For real number x and y we have xRy ⇔ \(\rm y-x+\sqrt{3}\) is an i

1.	For real number x and y we have xRy ⇔ \(\rm y-x+\sqrt{3}\) is an irrational number then, the relation R is1. Reflexive2. Symmetric3. Transitive4. None of these
Answer» Correct Answer - Option 1 : Reflexive Concept: A relation R in a set A is called Reflexive, if (a, a) ∈ R, for every a ∈ A. Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈A. Transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b, c ∈A. Calculation: Given: R = {(x, y) : y - x + √3 is an irrational number, where x, y ∈ R} Reflexive: If xRx ⇔ x - x + √3 = √3 and we know that √3 is irrational number. Hence, relation R is reflexive. Symmetric: If x = 1, y = √3 and we can see that y - x + √3 = 2√3 - 1 which is an irrational number. So, (x, y) ∈ R. But we can see that, x - y + √3 = 1 and 1 is not an irrational number. So, (y, x) does not belongs to R. Hence, relation R is not symmetric. Transitive: As we can see that, (√3, √2) ∈ R, (√2, 1) ∈ R but (√3, 1) does not belongs to R Hence, relation R is not transitive.

For real number x and y we have xRy ⇔ \(\rm y-x+\sqrt{3}\) is an irrational number then, the relation R is1. Reflexive2. Symmetric3. Transitive4. None of these

Answer» Correct Answer - Option 1 : Reflexive

Concept:

A relation R in a set A is called

Reflexive, if (a, a) ∈ R, for every a ∈ A.
Symmetric, if (a, b) ∈ R implies that (b, a) ∈ R, for all a, b ∈A.
Transitive, if (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R, for all a, b, c ∈A.

Calculation:

Given: R = {(x, y) : y - x + √3 is an irrational number, where x, y ∈ R}

Reflexive:

If xRx ⇔ x - x + √3 = √3 and we know that √3 is irrational number.

Hence, relation R is reflexive.

Symmetric:

If x = 1, y = √3 and we can see that y - x + √3 = 2√3 - 1 which is an irrational number. So, (x, y) ∈ R.

But we can see that, x - y + √3 = 1 and 1 is not an irrational number. So, (y, x) does not belongs to R.

Hence, relation R is not symmetric.

Transitive:

As we can see that, (√3, √2) ∈ R, (√2, 1) ∈ R but (√3, 1) does not belongs to R

Hence, relation R is not transitive.