Question

For real numbers $x$ and $y$ we write ${}_x{R}_y$ iff $x-y+\sqrt{2}$ is an irrational numbers. Then relation $R$ is reflexive and symmetric also.

Answer 1

For every value of x belongs to R, $x-x+\sqrt{2}$

i.e., $\sqrt{2}$ is an irrational number.

Therefore, it is reflexive.

Now Let's say $x=\sqrt{2}$ and $y=2$ then $x-y+\sqrt{2}=2\sqrt{2}-2$ which is irrational

but when $y=\sqrt{2}$ and $x=2$,

$x-y+\sqrt{2}$ is not irrational;

$\therefore$ it is not symmetric.