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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 :

JEE · Math · previous-year question

  1. A.Neither R1 nor R2 is transitive.correct
  2. B.R2 is transitive but R1 is not transitive.
  3. C.R1 and R2 are both transitive.
  4. D.R1 is transitive but R2 is not transitive.

Answer

A. Neither R1 nor R2 is transitive.

Explanation

For R1 : Let a = 1 + $$\sqrt 2 $$, b = 1 $$-$$ $$\sqrt 2 $$, c = $${8^{{1 \over 4}}}$$ aR1b : a2 + b2 = 6 $$ \in $$ Q bR1c : b2 + c2 = 3 $$-$$ 2$$\sqrt 2 $$ + 2$$\sqrt 2 $$ = 3 $$ \in $$ Q aR1c : a2 + c2 = 3 + 2$$\sqrt 2 $$ + 2$$\sqrt 2 $$ $$ \notin $$ Q $$ \therefore $$ R1 is not transitive. For R2 : Let a = 1 + $$\sqrt 2 $$, b = $$\sqrt 2 $$, c = 1 $$-$$ $$\sqrt 2 $$ aR2b : a2 + b2 = 5 + 2$$\sqrt 2 $$ $$ \notin $$ Q bR2c : b2 + c2 = 5 $$-$$ 2$$\sqrt 2 $$ $$ \notin $$ Q aR2c : a2 + c2 = 3 + 2$$\sqrt 2 $$ + 3 $$-$$ 2$$\sqrt 2 $$ = 6 $$ \in $$ Q $$ \therefore $$ R2 is not transitive.

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