For real numbers x and y, we write xRy $\Rightarrow$ $-y+\sqrt{2}$ is an irrational number. Then the relation R is :

Question

For real numbers $x$ and $y$, we write $x \mathrm{R} y \Rightarrow-y+\sqrt{2}$ is an irrational number. Then the relation $R$ is :

(A) Reflexive

(B) Symmetric

(C) Transitive

(D) None of these

Answer 1

SOLUTION : For any $x \in R$, we have $x-x+\sqrt{2}=\sqrt{2}$ an irrational number $\Rightarrow \quad x R x$ for all $x$. So, $R$ is reflexive.

$R$ is not symmetric, because $\sqrt{2} R 1$ but $1 R \sqrt{2}, R$ is not transitive also because $\sqrt{2} R 1$ and $1 R 2 \sqrt{2}$ but $\sqrt{2} \times 2 \sqrt{2}$.