Let ( S ) be the set of all real numbers. A relation ( R ) has been defined on ( S ) by ( a R b |a-b| 1 ), then ( R ) is

Step-by-step solution

Given that a R b |a-b| 1 Now, a R a |a-a|=0 1 Therefore, R is reflexive Again, a R b |a-b| 1 Then, b R a |b-a| 1 |a-b| 1 , which is true. Therefore, R is symmetric Take a b ==12 , Then, |a-b|=|1-2|=1=1 Take b c ==23 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.