For real numbers x and y,let xRy if and only | vedclass

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

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Reflexive

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(A) For any $x \in \mathbb{R}$,$x - x + \sqrt{2} = \sqrt{2}$,which is an irrational number. Since $xRx$ holds for all $x \in \mathbb{R}$,$R$ is reflexive. $R$ is not symmetric because if $x = \sqrt{2}$ 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,$y - x + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$,which is a rational number,so $1R\sqrt{2}$ is false. $R$ is not transitive because if $x = \sqrt{2}$,$y = 1$,and $z = 1 - \sqrt{2}$,then $xRy$ is true ($\sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$ is irrational) and $yRz$ is true ($1 - (1 - \sqrt{2}) + \sqrt{2} = 2\sqrt{2}$ is irrational),but $xRz$ is false because $x - z + \sqrt{2} = \sqrt{2} - (1 - \sqrt{2}) + \sqrt{2} = 3\sqrt{2} - 1$,which is irrational. Wait,let us re-check: $x - z + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} + \sqrt{2} = 3\sqrt{2} - 1$,which is irrational. Actually,the relation is not transitive. For example,let $x = \sqrt{2}$,$y = 0$,$z = \sqrt{2}$. Then $xRy$ is $\sqrt{2} - 0 + \sqrt{2} = 2\sqrt{2}$ (irrational),$yRz$ is $0 - \sqrt{2} + \sqrt{2} = 0$ (rational,so $yRz$ is false). Let $x = \sqrt{2}, y = 1, z = 2\sqrt{2}$. $xRy$ is $\sqrt{2}-1+\sqrt{2} = 2\sqrt{2}-1$ (irrational). $yRz$ is $1-2\sqrt{2}+\sqrt{2} = 1-\sqrt{2}$ (irrational). $xRz$ is $\sqrt{2}-2\sqrt{2}+\sqrt{2} = 0$ (rational). Thus,$R$ is not transitive.

For real numbers $x$ and $y$,let $xRy$ if and only if $x - y + \sqrt{2}$ is an irrational number. Then $R$ is:

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