x ∊ R ⇒ x - x + √5 = √5 is an irrational number ∴ (x, x) ∊ R , so R is reflexive (√5 , 1) ∊ R because , √5 - 1 + √5 = 2√5 - 1 which is an irrational number Also, (1, √5) ∊ R if 1 - √5 + √5 = 1 , which is not an irrational number ∴ (1, √5) ∉ R ∴ R is not symmetric We have (√5 , 1) , (1 , 2√5) ∊ R because √5 - 1 + √5 = 2 √5 and 1 - 2√5 + √5 = 1 - √5 are irrational numbers Also (√5 , 2√5) ∉ R . So, R is not transitive