On R, the relation rho is defined by 'x rho y | vedclass

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(B) We have $x \rho y \iff x-y \in \{0\} \cup \mathbb{I}$, where $\mathbb{I}$ is the set of irrational numbers.$1$. Reflexivity: For any $x \in R$, $x

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$\rho$ is reflexive and symmetric but not transitive

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(B) We have $x ho y \iff x-y \in \{0\} \cup \mathbb{I}$, where $\mathbb{I}$ is the set of irrational numbers.$1$. Reflexivity: For any $x \in R$, $x-x = 0$. Since $0$ is zero, $(x, x) \in ho$. Thus, $ho$ is reflexive.$2$. Symmetry: If $(x, y) \in ho$, then $x-y$ is $0$ or irrational. Then $y-x = -(x-y)$ is also $0$ or irrational. Thus, $(y, x) \in ho$. Therefore, $ho$ is symmetric.$3$. Transitivity: Consider $x = 2, y = \sqrt{3}, z = 4$. $(x, y) = (2, \sqrt{3}) \in ho$ because $2-\sqrt{3}$ is irrational.$(y, z) = (\sqrt{3}, 4) \in ho$ because $\sqrt{3}-4$ is irrational.However, $(x, z) = (2, 4) 
otin ho$ because $2-4 = -2$, which is a rational number (not zero or irrational).Therefore, $ho$ is not transitive.

On $R$, the relation $\rho$ is defined by '$x \rho y$ holds if and only if $x-y$ is zero or irrational'. Then,

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