For x, y $\in$R, define a relation R by xRy if and only if $\mathrm{x}-\mathrm{y}+\sqrt{2}$ is an irrational number. Then R is

Since $\mathrm{x}-\mathrm{x}+\sqrt{2}=\sqrt{2}$ which is an irrational number so xRx for all $\mathrm{x}\in \mathrm{R}$. Hence R is reflexive .

$$\begin{gathered} \text{ Now Let's say }x=\sqrt{2}\text{ and }y=2\text{ then } \\ x-y+\sqrt{2}=2\sqrt{2}-2\text{ which is irrational } \\ \text{ but when }y=\sqrt{2}\text{ and }x=2 \\ \mathrm{x}-\mathrm{y}+\sqrt{2}\text{ is not irrational; } \\ \therefore \text{ it is not symmetric. } \\ \mathrm{let} x=1;y=2\sqrt{2}\mathrm{\&}z=\sqrt{2} \\ then x-y+\sqrt{2}\text{ is irrational \&} \\ y-z+\sqrt{2}\text{ is irrational } \\ \text{ but }x-z+\sqrt{2}\text{ is not irrational, } \\ \mathrm{so},\mathrm{the} \mathrm{relation} \mathrm{is} \mathrm{not} \mathrm{transitive}. \end{gathered}$$