Question

For real numbers $x$ and $y,$ a relation $R$ is defined as $xRy\iff x-y+\sqrt{2}$ is an irrational number. Then the relation $R$ is

• A. reflexive and symmetric only
• B. not reflexive but symmetric
• C. reflexive but not symmetric and transitive
• D. reflexive and transitive but not symmetric

Answer 1

The correct option is C reflexive but not symmetric and transitive

$xRy\iff x-y+\sqrt{2}\in Q^C$

Reflexive :
Let $x\in R$
$xRx$ as $x-x+\sqrt{2}=\sqrt{2}\in Q^C$
$\therefore R$ is reflexive.

Symmetric :
By taking example,
$\sqrt{2}R1$ as $\sqrt{2}-1+\sqrt{2}=2\sqrt{2}-1\in Q^C$
But $1\mathrm{R/}\sqrt{2}$ as $1-\sqrt{2}+\sqrt{2}=1\notin Q^C$
$\therefore R$ is not symmetric.

Transitive :
Again, by taking example,
$\sqrt{2}R1$ as $\sqrt{2}-1+\sqrt{2}=2\sqrt{2}-1\in Q^C$
and $1R2\sqrt{2}$ as $1-2\sqrt{2}+\sqrt{2}=1-\sqrt{2}\in Q^C$
But $\sqrt{2}\mathrm{R/}2\sqrt{2}$ as $\sqrt{2}-2\sqrt{2}+\sqrt{2}=0\notin Q^C$
$\therefore R$ is not transitive.