COMEDK 2025 Afternoon Solved Paper

Concept:A relation is reflexive if every element is related to itself; symmetric if $xRy$ implies $yRx$; transitive if $xRy$ and $yRz$ imply $xRz$.Explanation:For reflexivity, let $x$ be any real number. Then $x - x + \sqrt{2} = \sqrt{2}$, which is irrational. Hence $xRx$ holds for all $x$, so the relation is reflexive.For symmetry, choose $x$ and $y$ such that $x - y = \sqrt{2} - 1$. Then $x - y + \sqrt{2} = 2\sqrt{2} - 1$ (irrational), so $xRy$. But $y - x + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$ (rational). Thus $yRx$ is false, so the relation is not symmetric.For transitivity, let $x - y + \sqrt{2} = \sqrt{3}$ (irrational) and $y - z + \sqrt{2} = \sqrt{2} - \sqrt{3}$ (irrational). Then $x - z + \sqrt{2} = \sqrt{3} + (\sqrt{2} - \sqrt{3}) - \sqrt{2} = 0$, a rational number. Hence $xRy$ and $yRz$ hold but $xRz$ does not, so the relation is not transitive.The relation satisfies reflexivity only.Answer:Option A: Reflexive