Sets, Relations & Functions
Relations
Grade 11
Question:
<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: Check reflexivity (a relates to itself), symmetry (if aRb then bRa), and transitivity (if aRb and bRc then aRc) systematically. Transitivity fails because |a-b| ≤ 1 and |b-c| ≤ 1 don't guarantee |a-c| ≤ 1.
<p><strong>Step 1: Check Reflexivity</strong></p><p>For any a ∈ S: |a - a| = 0 ≤ 1 ✓</p><p>So aRa holds for all a ∈ S. <strong>R is reflexive.</strong></p><p><strong>Step 2: Check Symmetry</strong></p><p>If aRb, then |a - b| ≤ 1</p><p>Since |a - b| = |b - a|, we have |b - a| ≤ 1, so bRa ✓</p><p><strong>R is symmetric.</strong></p><p><strong>Step 3: Check Transitivity</strong></p><p>Consider a = 0, b = 1, c = 2:</p><p>• |0 - 1| = 1 ≤ 1 ✓ (so 0R1)</p><p>• |1 - 2| = 1 ≤ 1 ✓ (so 1R2)</p><p>• |0 - 2| = 2 > 1 ✗ (so 0 ⊭ 2)</p><p><strong>R is NOT transitive.</strong></p><p>∴ Answer: <strong>R is Reflexive and Symmetric but NOT Transitive</strong> (Option C)</p>
Correct Answer: C