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(A) $1$. Reflexivity: For any $x \in \mathbb{R}$,we have $x - x + \sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number,$xRx$ holds for all $

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Reflexive

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(A) $1$. Reflexivity: For any $x \in \mathbb{R}$,we have $x - x + \sqrt{2} = \sqrt{2}$. Since $\sqrt{2}$ is an irrational number,$xRx$ holds for all $x \in \mathbb{R}$. Thus,$R$ is reflexive.$2$. Symmetry: For $R$ to be symmetric,$xRy$ must imply $yRx$. Let $x = \sqrt{2}$ and $y = 1$. Then $x - y + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$,which is irrational. So,$\sqrt{2}R1$ is true. However,for $yRx$,we check $y - x + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$,which is a rational number. Thus,$1
ot R\sqrt{2}$. Therefore,$R$ is not symmetric.$3$. Transitivity: For $R$ to be transitive,$xRy$ and $yRz$ must imply $xRz$. Let $x = \sqrt{2}$,$y = 1$,and $z = 0$. Then $xRy = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$ (irrational) and $yRz = 1 - 0 + \sqrt{2} = 1 + \sqrt{2}$ (irrational). However,$xRz = \sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}$ (irrational). Let us try another set: $x = \sqrt{2}$,$y = 1$,$z = 1 + \sqrt{2}$. Here $xRy$ is irrational and $yRz = 1 - (1 + \sqrt{2}) + \sqrt{2} = 0$,which is rational. Since $yRz$ is not irrational,transitivity fails. Thus,$R$ is not transitive.

For real numbers $x$ and $y$,we define the relation $R$ as $xRy$ if $x - y + \sqrt{2}$ is an irrational number. Then the relation $R$ is:

Similar Questions

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