JEE MAIN - Mathematics (2020 - 3rd September Evening Slot - No. 4)
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 :
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 :
Neither R1
nor R2
is transitive.
R2
is transitive but R1
is not transitive.
R1
and R2
are both transitive.
R1
is transitive but R2
is not 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.
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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