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

$\mathrm{R}=\left\{\right(\mathrm{x},\mathrm{y}):\mathrm{x}-\mathrm{y}+\sqrt{2}$ Is an irrational number ,$\mathrm{x},\mathrm{y}\in \mathrm{R}\}$ $\text{As, }\mathrm{x}-\mathrm{x}+\sqrt{2}=\sqrt{2}$ 
Which an irrational number $\Rightarrow (x,x)\in R$ So, R is reflexive relation 
Since, $(\sqrt{2},2)\in \text{ R i.e. }\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2$

Which is an irrational number 
But $2-\sqrt{2}+\sqrt{2}=2$ Which is a rational number $\Rightarrow (2,\sqrt{2})\notin \mathrm{R}$ 

So, R is not symmetric realtion
Also, $(\sqrt{2},2)\in \mathrm{R}\text{ and }(2,2\sqrt{2})\in \mathrm{R}\text{ i.e. }\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-2$

Which is also an irrational number
But $\sqrt{2}-2\sqrt{2}+\sqrt{2}=0$ Which is a rational number $\Rightarrow (\sqrt{2},2\sqrt{2})\notin \mathrm{R}$ So, R is not transitive relation
Hence R is reflexive.