Let S be the set of all real numbers. A relation R has been defined on S by aRb $\Leftrightarrow$ |a-b| $\le$ 1, then R is

The correct answer is Reflexive and Symmetric but not Transitive

Given that aRb $\Leftrightarrow$ |a-b| $\le$ 1

∴ aRa $\Leftrightarrow$ |a-a| = 0 $\le$ 1

Therefore, R is reflexive.


Again, aRb $\Leftrightarrow$ |a-b| $\le$ 1

∴ bRa $\Leftrightarrow$ |b-a| $\le$ 1

⇒ |a-b| $\le$ 1, which is true

Therefore, R is symmetric.


Take a b = = 1 2, then |a-b| = |1-2| = 1

Take b c = = 2 3, then |b-c| = |2-3| = 1

But aRc = |a-c| = |1-3| = 2 > 1

Therefore, R is not transitive.


Hence R is reflexive and symmetric but not transitive.