Verified Solution • 2025–26 Session moderate

QJEEM20Ma8287: JEE Main 2020 Mathematics - Class 11 JEEM20Ma8287 Solution: Let R 1 and R 2 be two relation defined as follows : R 1 = {(a, …

Let R1 and R2 be two relation defined as follows :
R1 = {(a, b) ∈\in R2 : a2 + b2 ∈\in Q} and
R2 = {(a, b) ∈\in R2 : a2 + b2 ∉\notin Q},
where Q is the set of all rational numbers. Then :

Multiple Choice Options

  • A Neither R1 nor R2 is transitive.
  • B R2 is transitive but R1 is not transitive.
  • C R1 and R2 are both transitive.
  • D R1 is transitive but R2 is not transitive.

Step-by-Step Verified Solution

For R1 :

Let a = 1 + 2\sqrt 2 , b = 1 −- 2\sqrt 2 , c = 814{8^{{1 \over 4}}}

aR1b : a2 + b2 = 6 ∈\in Q

bR1c : b2 + c2 = 3 −- 2 2\sqrt 2 + 2 2\sqrt 2 = 3 ∈\in Q

aR1c : a2 + c2 = 3 + 2 2\sqrt 2 + 2 2\sqrt 2 ∉\notin Q

∴\therefore R1 is not transitive.

For R2 :

Let a = 1 + 2\sqrt 2 , b = 2\sqrt 2 , c = 1 −- 2\sqrt 2

aR2b : a2 + b2 = 5 + 2 2\sqrt 2 ∉\notin Q

bR2c : b2 + c2 = 5 −- 2 2\sqrt 2 ∉\notin Q

aR2c : a2 + c2 = 3 + 2 2\sqrt 2 + 3 −- 2 2\sqrt 2 = 6 ∈\in Q

∴\therefore R2 is not transitive.

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