Let R1 and R2 be two relation defined as follows : R1 = (a, b) [ ] R2 : a2 + b2 [ ] Q and …

Question

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 :

Accepted Answer

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.

Solutions · 1