JEE Main 2020 — Mathematics Algebra

For ${R}_{1}$R1​ let $a=1+\sqrt{2},b=1-\sqrt{2},c={8}^{\frac{1}{4}}$a=1+2​,b=1−2​,c=841​$a{R}_{1}b\Rightarrow {a}^{2}+{b}^{2}={(1+\sqrt{2})}^{2}+{(1-\sqrt{2})}^{2}=6\in Q$aR1​b⇒a2+b2=(1+2​)2+(1−2​)2=6∈Q

$$b{R}_{1}c\Rightarrow {b}^{2}+{c}^{2}={(1-\sqrt{2})}^{2}+{({8}^{\frac{1}{4}})}^{2}=3\in Q$$bR1​c⇒b2+c2=(1−2​)2+(841​)2=3∈Q

$$a{R}_{1}c\Rightarrow {a}^{2}+{c}^{2}={(1+\sqrt{2})}^{2}+{({8}^{\frac{1}{4}})}^{2}=3+4\sqrt{2}\notin Q$$aR1​c⇒a2+c2=(1+2​)2+(841​)2=3+42​∈/Q

$\therefore$∴${R}_{1}$R1​ is not transitive.

For ${R}_{2}$R2​ let $a=1+\sqrt{2},b=\sqrt{2},c=1-\sqrt{2}$a=1+2​,b=2​,c=1−2​

$$a{R}_{2}b\Rightarrow {a}^{2}+{b}^{2}={(1+\sqrt{2})}^{2}+{(\sqrt{2})}^{2}=5+2\sqrt{2}\notin Q$$aR2​b⇒a2+b2=(1+2​)2+(2​)2=5+22​∈/Q

$$b{R}_{2}c\Rightarrow {b}^{2}+{c}^{2}={(\sqrt{2})}^{2}+{(1-\sqrt{2})}^{2}=5-2\sqrt{2}\notin Q$$bR2​c⇒b2+c2=(2​)2+(1−2​)2=5−22​∈/Q

$$a{R}_{2}c\Rightarrow {a}^{2}+{c}^{2}={(1+\sqrt{2})}^{2}+{(1-\sqrt{2})}^{2}=6\in Q$$aR2​c⇒a2+c2=(1+2​)2+(1−2​)2=6∈Q

$\therefore$∴${R}_{2}$R2​ is not transitive.