<p>Let \(n\) be a positive integer and \(R = \{(a, b) \in \mathbb{Z} \times \mathbb{Z} \mid a - b = nm\text{, where } m \in \mathbb{Z} \text{ and } m \neq 0\}\). Then \(R\) is</p>
Step-by-Step Solution
Key Concept: Recognize that R defines pairs where a - b is a non-zero multiple of n. To verify if R is an equivalence relation, check reflexivity (fails: a - a = 0, but m ≠ 0), symmetry, and transitivity. The condition m ≠ 0 breaks reflexivity, making R NOT an equivalence relation.
<p><strong>Step 1: Check Reflexivity</strong></p><p>For R to be reflexive, (a,a) ∈ R for all a ∈ ℤ.</p><p>This requires a - a = 0 = nm where m ≠ 0.</p><p>But 0 = nm with m ≠ 0 implies n = 0, which contradicts n being positive.</p><p>∴ R is <strong>NOT reflexive</strong>.</p><p><strong>Step 2: Check Symmetry</strong></p><p>If (a,b) ∈ R, then a - b = nm₁ for some m₁ ≠ 0.</p><p>Then b - a = -nm₁ = n(-m₁) where -m₁ ≠ 0.</p><p>∴ (b,a) ∈ R. R is <strong>symmetric</strong>.</p><p><strong>Step 3: Check Transitivity</strong></p><p>If (a,b) ∈ R and (b,c) ∈ R, then a - b = nm₁ and b - c = nm₂ (m₁, m₂ ≠ 0).</p><p>Adding: a - c = n(m₁ + m₂) where m₁ + m₂ ≠ 0 (since both are non-zero integers).</p><p>∴ (a,c) ∈ R. R is <strong>transitive</strong>.</p><p><strong>Conclusion:</strong> R is symmetric and transitive but NOT reflexive, so it is <strong>NOT an equivalence relation</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D