Let R _ 1 and R _ 2 be two relations defined as follows : R _ 1 = \...

$a{R}_{1}b ⇒ a^{2} + b^{2} = \left(1 + \sqrt{2}\right)2 + \left(1 - \sqrt{2}\right)2 = 6 \in Q$

$b{R}_{1}c ⇒ b^{2} + c^{2} = \left(1 - \sqrt{2}\right)2 + \left(8\dfrac{1}{4}\right)2 = 3 \in Q$

$a{R}_{1}c ⇒ a^{2} + c^{2} = \left(1 + \sqrt{2}\right)2 + \left(8\dfrac{1}{4}\right)2 = 3 + 4\sqrt{2} \notin Q$

$∴{R}_{1}$~is not transitive.

For& ${R}_{2}$~let& $a = 1 + \sqrt{2}, b = \sqrt{2}, c = 1 - \sqrt{2}$

$a{R}_{2}b ⇒ a^{2} + b^{2} = \left(1 + \sqrt{2}\right)2 + \left(\sqrt{2}\right)2 = 5 + 2\sqrt{2} \notin Q$

$b{R}_{2}c ⇒ b^{2} + c^{2} = \left(\sqrt{2}\right)2 + \left(1 - \sqrt{2}\right)2 = 5 - 2\sqrt{2} \notin Q$

$a{R}_{2}c ⇒ a^{2} + c^{2} = \left(1 + \sqrt{2}\right)2 + \left(1 - \sqrt{2}\right)2 = 6 \in Q$

$∴{R}_{2}$& is not transitive.