Explanation

(b)
Given, $r = \left\{ (a,\ b) \right|a,\ b \in R$ and $a - b + \sqrt{3}$ is an irrational number}
(i) Reflexive
$ara = a - a + \sqrt{3} = \sqrt{3}$ which is irrational number.
(ii) Symmetric
Now, $2r\sqrt{3} = 2 - \sqrt{3} + \sqrt{3} = 2$
Which is not an irrational.
Also, $\sqrt{3}r2 = \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3} - 2$ which is an irrational.
$2r\sqrt{3} \neq \sqrt{3}r2$
Which is not symmetric.
(iii) Transitive
Now, $\sqrt{3}r2$ and $2r4\sqrt{5},\ ie,$
$\sqrt{3} - 2 + \sqrt{3} + 2 - 4\sqrt{5} + \sqrt{3}$
$\ \ \ \ \ \ \ \ \ \ = 2\sqrt{3} - 4\sqrt{5} + \sqrt{3} \neq \sqrt{3}r4\sqrt{5}$
$\therefore$ It is not transitive.