For real numbers $$x$$ and $$y,$$ we define $$xRy$$  iff $$x - y + \sqrt 5 $$   is an irrational number. The relation $$R$$ is :

Solution :
$$\eqalign{ & x\, \in \,R \Rightarrow x - x + \sqrt 5 = \sqrt 5 {\text{ is an irrational number}}{\text{.}} \cr & \therefore \,\left( {x,\,x} \right) \in \,R \cr & \therefore \,R{\text{ is reflexive}}{\text{.}} \cr & \left( {\sqrt 5 ,\,1} \right) \in \,R{\text{ because}} \cr & \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1{\text{ which is an irrational number}}{\text{.}} \cr & \therefore \,\left( {1,\,\sqrt 5 } \right) \notin \,R \cr & \therefore \,R{\text{ is not symmetric}}{\text{.}} \cr & {\text{We have, }}\left( {\sqrt 5 ,\,1} \right),\,\left( {1,\,2\sqrt 5 } \right) \in \,R{\text{ because}} \cr & \sqrt 5 - 1 + \sqrt 5 = 2\sqrt 5 - 1{\text{ if }}1 - 2\sqrt 5 + \sqrt 5 = 1 - \sqrt 5 \cr & {\text{are irrational number}}{\text{.}} \cr & {\text{Also, }}\left( {\sqrt 5 ,\,2\sqrt 5 } \right) \in \,R{\text{ and}} \cr & \sqrt 5 - 2\sqrt 5 + \sqrt 5 = 0\,{\text{which is not an irrational number}}{\text{.}} \cr & \therefore \,\left( {\sqrt 5 ,\,2\sqrt 5 } \right) \notin \,R \cr & \therefore \,R{\text{ is not transitive}}{\text{.}} \cr} $$