Let S be the set of all real numbers and let R be a relation on S, defined by \( aRb\Leftrightarrow|a-b|\le 1 \) . Then, R is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → reflexive and symmetric but not transitive.
Reflexive: for every real a, \( |a-a|=0\le 1 \), so aRa. R is reflexive.
Symmetric: if aRb, then \( |b-a|=|a-b|\le 1 \), so bRa. R is symmetric.
Transitive: a single counter-example is enough to show that it fails. Take a = 0, b = 1, c = 2:
\( \displaystyle |0-1|=1\le 1\ \text{so 0R1} \)
\( \displaystyle |1-2|=1\le 1\ \text{so 1R2} \)
\( \displaystyle |0-2|=2\gt 1\ \text{so 0 is not related to 2} \)
R is not transitive, so R is reflexive and symmetric but not transitive.
Tip: "Distance at most 1" relations are the standard example of a relation that is not transitive: two short steps can add up to a long one.