On set of all real numbers a relation $R$ is defined by $xRy$ if $x-y+\sqrt{5}$ is an irrational number. The relation $R$ is

Let $x\in R\Rightarrow x-x+\sqrt{5}=\sqrt{5}$ 

$\therefore (x,x)\in R\therefore \mathrm{R}\text{ is reflexive }(\sqrt{5},1)\in R\text{ because }$

$\sqrt{5}-1+\sqrt{5}=2\sqrt{5}-1,$

Which is an irrational number.

But $(1,\sqrt{5})\notin R \therefore \mathrm{R}$ is not symmetric

We have, $(\sqrt{5},1),(1,2\sqrt{5})\in R$ because

$\sqrt{5}-1+\sqrt{5}=2\sqrt{5}-1\text{ if }$

$1-2\sqrt{5}+\sqrt{5}=1-\sqrt{5}$

is an irrational number

Also, $(\sqrt{5},2\sqrt{5})\in R,\text{ and }\sqrt{5}-2\sqrt{5}+\sqrt{5}=0$

Which is not an irrational numbers.

$\therefore (\sqrt{5},2\sqrt{5})\notin R$

$\therefore$ R is not transitive