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Let S be the set of all real numbers and let R be a relation on S, defined by aRb⇔∣a−b∣≤1aRb \Leftrightarrow |a-b| \leq 1. Then, R is

Solution

✅ Correct Option: 1

∣a−a∣=0≤1|a-a|=0\le1, so R is reflexive; ∣a−b∣=∣b−a∣|a-b|=|b-a|, so it is symmetric. It fails transitivity: ∣1−2∣≤1|1-2|\le1 and ∣2−3∣≤1|2-3|\le1 but ∣1−3∣=2>1|1-3|=2>1. Hence reflexive and symmetric but not transitive.

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