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(B) The relation $r$ on $R$ is defined as $r = \{(a,b) \in R 	imes R \mid a - b + \sqrt{3} \in R \setminus Q \}$.$1$. Reflexivity: For any $a \in R$,

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reflexive only

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(B) The relation $r$ on $R$ is defined as $r = \{(a,b) \in R 	imes R \mid a - b + \sqrt{3} \in R \setminus Q \}$.$1$. Reflexivity: For any $a \in R$,$aRa \iff a - a + \sqrt{3} = \sqrt{3}$. Since $\sqrt{3}$ is an irrational number,$aRa$ holds for all $a \in R$. Thus,$r$ is reflexive.$2$. Symmetry: For $r$ to be symmetric,$aRb \implies bRa$. Let $a = \sqrt{3}$ and $b = 0$. Then $a - b + \sqrt{3} = \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$,which is irrational. So,$(\sqrt{3}, 0) \in r$. However,$b - a + \sqrt{3} = 0 - \sqrt{3} + \sqrt{3} = 0$,which is a rational number. So,$(0, \sqrt{3}) 
otin r$. Thus,$r$ is not symmetric.$3$. Transitivity: For $r$ to be transitive,$aRb$ and $bRc \implies aRc$. Consider $a = \sqrt{3}$,$b = 0$,and $c = 2\sqrt{3}$. $aRb \implies \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$ (irrational).$bRc \implies 0 - 2\sqrt{3} + \sqrt{3} = -\sqrt{3}$ (irrational).$aRc \implies \sqrt{3} - 2\sqrt{3} + \sqrt{3} = 0$ (rational).Since $aRc$ is not irrational,$r$ is not transitive.Therefore,$r$ is reflexive only.

Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = \{(a,b) \mid a,b \in R \text{ and } a - b + \sqrt{3} \text{ is an irrational number} \}$. The relation $r$ is

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