On the set of real numbers R,a relation rho i | vedclass

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(C) $1$. Reflexivity: For any $x \in R$,$x - x = 0$. Since $0$ is allowed,$x \rho x$ holds. Thus,$\rho$ is reflexive. $2$. Symmetry: If $x \rho y$,the

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$\rho$ is reflexive and symmetric but not transitive

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(C) $1$. Reflexivity: For any $x \in R$,$x - x = 0$. Since $0$ is allowed,$x \rho x$ holds. Thus,$\rho$ is reflexive. $2$. Symmetry: If $x \rho y$,then $x - y$ is $0$ or irrational. Since $y - x = -(x - y)$,if $x - y$ is $0$,$y - x$ is $0$. If $x - y$ is irrational,$y - x$ is also irrational. Thus,$y \rho x$ holds. $\rho$ is symmetric. $3$. Transitivity: Let $x = \sqrt{2} + 1$,$y = \sqrt{2}$,and $z = 0$. Here,$x - y = 1$ (rational,not zero or irrational),so $x \rho y$ is false. Let us test $x = \sqrt{2}$,$y = 0$,$z = -\sqrt{2}$. Then $x - y = \sqrt{2}$ (irrational) and $y - z = \sqrt{2}$ (irrational). However,$x - z = 2\sqrt{2}$ (irrational). Consider $x = \sqrt{2} + 1$,$y = \sqrt{2}$,$z = 1$. $x - y = 1$ (not allowed). Consider $x = 1 + \sqrt{2}$,$y = 1$,$z = \sqrt{2}$. $x - y = \sqrt{2}$ (irrational) and $y - z = 1 - \sqrt{2}$ (irrational). But $x - z = 1$ (rational,not zero). Thus,$x \rho y$ and $y \rho z$ hold,but $x \rho z$ does not hold. Therefore,$\rho$ is not transitive.

On the set of real numbers $R$,a relation $\rho$ is defined by $x \rho y$ if and only if $x-y$ is zero or an irrational number. Then:

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