<p>If \(S\) is the set of all real numbers. A relation \(R\) has been defined on \(S\) by \(aRb \Longleftrightarrow |a - b| \leq 1\), then \(R\) is:</p>
<p>(a) symmetric and transitive but not reflexive</p>
<p>(b) reflexive and transitive but not symmetric</p>
<p>(c) reflexive and symmetric but not transitive</p>
<p>(d) an equivalence relation</p>
Step-by-Step Solution
Key Concept: A relation is reflexive if every element relates to itself, symmetric if aRb implies bRa, and transitive if aRb and bRc implies aRc. Here we must check all three properties: reflexivity is immediate (|a-a|=0≤1), symmetry follows from |a-b|=|b-a|, but transitivity fails because |a-b|≤1 and |b-c|≤1 doesn't guarantee |a-c|≤1 (counterexample: a=0, b=1, c=2 gives 1≤1 and 1≤1 but 2≰1).
<p><strong>Step 1 (Reflexivity):</strong> For any a∈S, |a-a|=0≤1 ✓. So R is reflexive.</p><p><strong>Step 2 (Symmetry):</strong> If aRb, then |a-b|≤1. Since |a-b|=|b-a|, we have |b-a|≤1, so bRa ✓. So R is symmetric.</p><p><strong>Step 3 (Transitivity):</strong> Suppose aRb and bRc, so |a-b|≤1 and |b-c|≤1. Does |a-c|≤1 follow? <br/>Counterexample: Let a=0, b=1, c=2. Then |0-1|=1≤1 ✓ and |1-2|=1≤1 ✓, but |0-2|=2≰1 ✗. So R is NOT transitive.</p><p><strong>Step 4 (Conclusion):</strong> R is reflexive and symmetric but NOT transitive. Therefore R is not an equivalence relation.</p><p>∴ Answer: <strong>C (Reflexive and Symmetric but not Transitive)</strong></p>
Correct Answer: C