<p>Let \( R \) be a relation defined on \( \mathbb{Z} \) by \( (a,b) \in R \Rightarrow a - b = nm \) where \( m \in \mathbb{Z} \) and \( m \neq 0 \). Which of the following is true?</p>
<p>\( R \) is reflexive and transitive but not symmetric</p>
<p>\( R \) is symmetric but not reflexive and not transitive</p>
<p>\( R \) is reflexive and symmetric but not transitive</p>
<p>\( R \) is an equivalence relation</p>
Step-by-Step Solution
Key Concept: For a fixed m ≠ 0, the relation partitions ℤ into equivalence classes based on remainders modulo |m|. You must recognize that (a,b) ∈ R means a and b differ by a multiple of m, making R an equivalence relation with respect to modular arithmetic on a fixed m.
<p><strong>Step 1:</strong> Interpret the relation: (a,b) ∈ R ⟺ a - b = nm for some m ∈ ℤ, m ≠ 0.</p><p><strong>Step 2:</strong> For a FIXED non-zero m, check if R is an equivalence relation:</p><p><strong>Reflexivity:</strong> a - a = 0 = 0·m ✓ (So (a,a) ∈ R)</p><p><strong>Symmetry:</strong> If a - b = nm, then b - a = -nm = n'·m where n' = -n ∈ ℤ ✓</p><p><strong>Transitivity:</strong> If a - b = n₁m and b - c = n₂m, then a - c = (n₁ + n₂)m ✓</p><p><strong>Step 3:</strong> R partitions ℤ into |m| equivalence classes: {... , -2|m|, -|m|, 0, |m|, 2|m|, ...}, {... , 1-2|m|, 1-|m|, 1, 1+|m|, ...}, etc.</p><p><strong>Step 4:</strong> R is reflexive, symmetric, and transitive, hence an EQUIVALENCE RELATION (on a fixed m).</p><p>∴ Answer: B</p>
Correct Answer: B