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  4. For real numbers x and y, we write xRy Leftrightarrow x-y+√2 is an irrational number. Then, the relation R is

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

Relations and Functions - Part 2

Solution:

Reflexive : For any $x \in R$, we have $x-x+\sqrt{2}=\sqrt{2}$
is an irrational number $\Rightarrow xRx$ for all $x$. So, $R$ is reflexive.
Symmetric : $R$ is not symmetric, because $\sqrt{2}\,R1$ but $1 \not R \sqrt{2}$.
Transitive : $R$ is not transitive also, because $\sqrt{2}\,R1$ and $1R2\sqrt{2}$ but $\sqrt{2}\, \not R\, 2 \sqrt{2}$.