Let $\displaystyle \mathrm{R}$ be a relation on set of real numbers R defined as $\displaystyle \{(\mathrm{x}, y): \mathrm{x}-y+\sqrt{3}$ is an irrational number, $\displaystyle \mathrm{x}, y \in \mathbb{\mathrm{R}}\}$ Verify $\displaystyle \mathrm{R}$ for reflexivity, symmetry and transitivity.

Let $\displaystyle \mathrm{x} \in \mathbb{\mathrm{R}}$. Then we know that $\displaystyle \mathrm{x}-\mathrm{x}+\sqrt{3}=\sqrt{3}$, which is an irrational number. \[\Rightarrow(\mathrm{x}, \mathrm{x}) \in \mathrm{R} \] Hence, R is reflexive. We have $\displaystyle \sqrt{3}, 2 \in \mathbb{\mathrm{R}}$ such that $\displaystyle \sqrt{3}-2+\sqrt{3}=2(\sqrt{3}-1)$, which is an irrational number $\displaystyle \Rightarrow(\sqrt{3}, 2) \in \mathrm{R}$. But, $\displaystyle 2-\sqrt{3}+\sqrt{3}=2$, which is a rational number. Hence, $\displaystyle \Rightarrow(2, \sqrt{3}) \notin \mathrm{R}$. - Therefore, R is not symmetric. Let $\displaystyle -\sqrt{3}, \sqrt{3}, 2 \in \mathbb{\mathrm{R}}$ such that $\displaystyle (-\sqrt{3}, \sqrt{3}),(\sqrt{3}, 2) \in \mathbb{\mathrm{R}}$. But, $\displaystyle (-\sqrt{3}, 2) \notin \mathrm{R}$ Therefore, R is not transitive.