JEE Main 2020 — Sets And Relations Question with Solution
From: JEE Main 2020 (Online) 3rd September Evening Slot
Question
Let R1
and R2
be two relation defined as
follows :
R1 = {(a, b) R2 : a2 + b2 Q} and
R2 = {(a, b) R2 : a2 + b2 Q},
where Q is the set of all rational numbers. Then :
R1 = {(a, b) R2 : a2 + b2 Q} and
R2 = {(a, b) R2 : a2 + b2 Q},
where Q is the set of all rational numbers. Then :
Choose an option
Show full solutionCorrect option: A
Correct answer
ANeither R1
nor R2
is transitive.
Step-by-step explanation
For R1 :
Let a = 1 + , b = 1 , c =
aR1b : a2 + b2 = 6 Q
bR1c : b2 + c2 = 3 2 + 2 = 3 Q
aR1c : a2 + c2 = 3 + 2 + 2 Q
R1 is not transitive.
For R2 :
Let a = 1 + , b = , c = 1
aR2b : a2 + b2 = 5 + 2 Q
bR2c : b2 + c2 = 5 2 Q
aR2c : a2 + c2 = 3 + 2 + 3 2 = 6 Q
R2 is not transitive.
Let a = 1 + , b = 1 , c =
aR1b : a2 + b2 = 6 Q
bR1c : b2 + c2 = 3 2 + 2 = 3 Q
aR1c : a2 + c2 = 3 + 2 + 2 Q
R1 is not transitive.
For R2 :
Let a = 1 + , b = , c = 1
aR2b : a2 + b2 = 5 + 2 Q
bR2c : b2 + c2 = 5 2 Q
aR2c : a2 + c2 = 3 + 2 + 3 2 = 6 Q
R2 is not transitive.
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