Let R be a relation on mathbb{R},given by R = | vedclass

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(A) Check for reflexivity:For any $a \in \mathbb{R}$,$3a - 3a + \sqrt{7} = \sqrt{7}$. Since $\sqrt{7}$ is an irrational number,$(a, a) \in R$. Thus,$R

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Reflexive but neither symmetric nor transitive

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(A) Check for reflexivity:For any $a \in \mathbb{R}$,$3a - 3a + \sqrt{7} = \sqrt{7}$. Since $\sqrt{7}$ is an irrational number,$(a, a) \in R$. Thus,$R$ is reflexive.Check for symmetry:Let $(a, b) \in R$. Then $3a - 3b + \sqrt{7} = I_1$,where $I_1$ is an irrational number.For symmetry,we need $(b, a) \in R$,which means $3b - 3a + \sqrt{7}$ must be irrational.Note that $3b - 3a + \sqrt{7} = -(3a - 3b - \sqrt{7}) = -(I_1 - 2\sqrt{7}) = 2\sqrt{7} - I_1$.If we choose $a = \frac{\sqrt{7}}{3}$ and $b = 0$,then $3(\frac{\sqrt{7}}{3}) - 3(0) + \sqrt{7} = 2\sqrt{7}$ (irrational),so $(a, b) \in R$.However,for $(b, a)$,we have $3(0) - 3(\frac{\sqrt{7}}{3}) + \sqrt{7} = 0$,which is rational. Thus,$(b, a) 
otin R$. $R$ is not symmetric.Check for transitivity:Let $(a, b) \in R$ and $(b, c) \in R$. Then $3a - 3b + \sqrt{7} = I_1$ and $3b - 3c + \sqrt{7} = I_2$,where $I_1, I_2$ are irrational.For transitivity,$(a, c) \in R$ implies $3a - 3c + \sqrt{7}$ must be irrational.Adding the two relations: $(3a - 3b + \sqrt{7}) + (3b - 3c + \sqrt{7}) = 3a - 3c + 2\sqrt{7} = I_1 + I_2$.So,$3a - 3c + \sqrt{7} = I_1 + I_2 - \sqrt{7}$.If we take $a = \frac{\sqrt{7}}{3}, b = 1, c = \frac{2\sqrt{7}}{3}$,then $(a, b) \in R$ and $(b, c) \in R$,but $3a - 3c + \sqrt{7} = 3(\frac{\sqrt{7}}{3}) - 3(\frac{2\sqrt{7}}{3}) + \sqrt{7} = \sqrt{7} - 2\sqrt{7} + \sqrt{7} = 0$,which is rational. Thus,$(a, c) 
otin R$. $R$ is not transitive.

Let $R$ be a relation on $\mathbb{R}$,given by $R = \{(a, b) : 3a - 3b + \sqrt{7} \text{ is an irrational number} \}$. Then $R$ is

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