Given that aRb⇔|a−b|≥1 Now, aRa⇔|a−a|=0≤1 Therefore, R is reflexive Again,aRb⇔|a−b|≤1 Then,bRa⇔|b−a|≤1 ⇒|a−b|≤1, which is true. Therefore, R is symmetric Take a b = = 1 2 , . Then,|a−b|=|1−2|=1=1 Take b c = = 2 3 and . Then,|b−c|=|2−3|=1 But aRc |a−c|=|1−3|=2>1 Therefore, R not transitive Hence, R is reflexive and symmetric but not transitive.