Let $R$ be a relation on $\mathbb{R}$, given by $R = \{(a, b) : 3a - 3b + \sqrt{7}\}$ is an irrational…
Let $R$ be a relation on $\mathbb{R}$, given by $R = \{(a, b) : 3a - 3b + \sqrt{7}\}$ is an irrational number. Then $R$ is
Reflexive but neither symmetric nor transitive
Reflexive and transitive but not symmetric
Reflexive and symmetric but not transitive
An equivalence relation
Solution
Given: $R = \{(a, b) : 3a - 3b + \sqrt{7} \text{ is irrational}\}$
Reflexive: $3a - 3a + \sqrt{7} = \sqrt{7}$ is irrational. So, $(a, a) \in R$, hence $R$ is reflexive.
Given: $R = \{(a, b) : 3a - 3b + \sqrt{7} \text{ is irrational}\}$
Symmetric: Let $(a, b) \in R$, then $3a - 3b + \sqrt{7}$ is irrational. So, $3b - 3a + \sqrt{7}$ is not necessarily irrational, since if $3a = \sqrt{7}$ and $3b \in \mathbb{Z}$, then $3a - 3b + \sqrt{7}$ is irrational, but $3b - 3a + \sqrt{7}$ is not irrational. So, $(a, b) \in R \nRightarrow (b, a) \in R$, hence $R$ is not symmetric.
Transitive: Take $(a, b)$ as $\left(\frac{\sqrt{7}}{3}, 1\right)$ and $(b, c)$ as $(1, \frac{2\sqrt{7}}{3})$. So, $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$ which means relation is not transitive.