Let the relation rho be defined on mathbb{R} | vedclass

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(B) $1$. Reflexivity: For any $a \in \mathbb{R}$, $a-a = 0$. Since $0$ is allowed, $a \rho a$ holds for all $a$. Thus, $\rho$ is reflexive.$2$. Symmet

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$\rho$ is reflexive and symmetric but is not transitive

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(B) $1$. Reflexivity: For any $a \in \mathbb{R}$, $a-a = 0$. Since $0$ is allowed, $a \rho a$ holds for all $a$. Thus, $\rho$ is reflexive.$2$. Symmetry: If $a \rho b$, then $a-b = 0$ or $a-b$ is irrational. If $a-b=0$, then $b-a=0$. If $a-b$ is irrational, then $b-a = -(a-b)$ is also irrational. Thus, $b \rho a$ holds. So, $\rho$ is symmetric.$3$. Transitivity: Let $a = 2 + \sqrt{2}$, $b = 2$, and $c = \sqrt{2}$.Then $a-b = (2+\sqrt{2}) - 2 = \sqrt{2}$ (irrational), so $a \rho b$.And $b-c = 2 - \sqrt{2}$ (irrational), so $b \rho c$.However, $a-c = (2+\sqrt{2}) - \sqrt{2} = 2$ (rational and non-zero), so $a \rho c$ is false.Therefore, $\rho$ is not transitive.

Let the relation $\rho$ be defined on $\mathbb{R}$ by $a \rho b$ if and only if $a-b$ is zero or irrational. Then:

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