Let R _{1} and R _{2} be two relations define | vedclass

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Question

Let $a^{2}+b^{2} \in Q \& b^{2}+c^{2} \in Q$

eg. $\quad a =2+\sqrt{3} \& b =2-\sqrt{3}$

$a^{2}+b^{2}=14 \in Q$

Let $\quad c =(1+2 \sq

Vedclass · Accepted Answer

Neither $R _{1}$ nor $R _{2}$ is transitive

Vedclass · Answer

Let $a^{2}+b^{2} \in Q \& b^{2}+c^{2} \in Q$

eg. $\quad a =2+\sqrt{3} \& b =2-\sqrt{3}$

$a^{2}+b^{2}=14 \in Q$

Let $\quad c =(1+2 \sqrt{3})$

$b ^{2}+ c ^{2}=20 \in Q$

But $\quad a^{2}+c^{2}=(2+\sqrt{3})^{2}+(1+2 \sqrt{3})^{2} 
otin Q$

for $R _{2}$ Let $a ^{2}=1, b ^{2}=\sqrt{3} \& c ^{2}=2$

$a^{2}+b^{2} 
otin Q \& b^{2}+c^{2} 
otin Q$

But $a^{2}+c^{2} \in Q$

Let $R _{1}$ and $R _{2}$ be two relations defined as follows :

$R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$

where $Q$ is the set of all rational numbers. Then

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Let $R _{1}$ and $R _{2}$ be two relations defined as follows : $R _{1}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \in Q \right\}$ and $R _{2}=\left\{( a , b ) \in R ^{2}: a ^{2}+ b ^{2} \notin Q \right\}$ where $Q$ is the set of all rational numbers. Then