For real numbers x and y, define xRy iff is an irrational number. Then the relation R is

reflexive Explanation: 1. Reflexivity A relation is reflexive if for all . Test: Since is an irrational number, holds true for all real . Result: is reflexive . 2. Symmetry A relation is symmetric if . Let and . (Irrational). So, is true. Now check : . Since is a rational number, is false. Result: is not symmetric . 3. Transitivity A relation is transitive if and . Let , , and . (Irrational). True. (Irrational). True. Now check : . Since is rational, is false. Result: is not transitive . Final Conclusion: The relation is reflexive but neither symmetric nor transitive .

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JAMIA 2022 — Mathematics PYQ

JAMIA | Mathematics | 2022

For real numbers x and y, define xRy iff $x - y + 2$ is an irrational number. Then the relation R is

Choose the correct answer:

B.

symmetric

C.
transitive

D.
None

Explanation

1. Reflexivity

A relation is reflexive if $xRx$ for all $x \in \mathbb{R}$ .

Test: $x - x + \sqrt{2} = \sqrt{2}$

Since $\sqrt{2}$ is an irrational number, $xRx$ holds true for all real $x$ .

Result: $R$ is reflexive.

2. Symmetry

A relation is symmetric if $xRy \implies yRx$ .

Let $x = \sqrt{2}$ and $y = 0$ .

$xRy \implies \sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}$ (Irrational). So, $\sqrt{2}R0$ is true.

Now check $yRx$ : $0 - \sqrt{2} + \sqrt{2} = 0$ .

Since $0$ is a rational number, $0R\sqrt{2}$ is false.

Result: $R$ is not symmetric.

3. Transitivity

A relation is transitive if $xRy$ and $yRz \implies xRz$ .

Let $x = \sqrt{2}$ , $y = 1$ , and $z = 2\sqrt{2}$ .

$xRy \implies \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$ (Irrational). True.

$yRz \implies 1 - 2\sqrt{2} + \sqrt{2} = 1 - \sqrt{2}$ (Irrational). True.

Now check $xRz$ : $\sqrt{2} - 2\sqrt{2} + \sqrt{2} = 0$ .

Since $0$ is rational, $xRz$ is false.

Result: $R$ is not transitive.

Final Conclusion:

The relation $R$ is reflexive but neither symmetric nor transitive.