BITSAT 2013 Paper

Since, | a - a |= 0 ≤ 1 ,so aRa, for all a ∊ R
∴ R is reflexive.
Now, aRb ⇒ |a - b| ≤ 1 ⇒ |b - a| ≤ 1 ⇒ bRa
∴ R is symmetric.
But R is not transitive as
1R2, 2R3 but 1 ≠ R ≠ 3
[Since |1 – 3| = 2 > 1]