<p><strong>51.</strong> If \(A\) and \(B\) are two equivalence relations defined on set \(C\), then which of the following is always true?</p>
<p>(a) \(A \cap B\) is an equivalence relation</p>
<p>(b) \(A \cap B\) is not an equivalence relation</p>
<p>(c) \(A \cup B\) is an equivalence relation</p>
<p>(d) \(A \cup B\) is not an equivalence relation</p>
Step-by-Step Solution
Key Concept: The intersection of two equivalence relations is always an equivalence relation because reflexivity, symmetry, and transitivity are preserved under set intersection. However, the union of two equivalence relations is NOT necessarily an equivalence relation because transitivity may fail.
<p><strong>Step 1:</strong> Recall that an equivalence relation must satisfy reflexivity, symmetry, and transitivity.</p><p><strong>Step 2:</strong> Check if A ∩ B is always an equivalence relation:</p><ul><li><strong>Reflexivity:</strong> If (x,x) ∈ A and (x,x) ∈ B for all x ∈ C, then (x,x) ∈ A ∩ B ✓</li><li><strong>Symmetry:</strong> If (x,y) ∈ A ∩ B, then (x,y) ∈ A and (x,y) ∈ B, so (y,x) ∈ A and (y,x) ∈ B, thus (y,x) ∈ A ∩ B ✓</li><li><strong>Transitivity:</strong> If (x,y) ∈ A ∩ B and (y,z) ∈ A ∩ B, then both pairs are in A and in B, so (x,z) ∈ A and (x,z) ∈ B, thus (x,z) ∈ A ∩ B ✓</li></ul><p><strong>Step 3:</strong> Check if A ∪ B is always an equivalence relation: Counterexample exists where transitivity fails. For instance, if (a,b) ∈ A only and (b,c) ∈ B only, then (a,b) ∈ A ∪ B and (b,c) ∈ A ∪ B, but (a,c) may not be in A ∪ B.</p><p><strong>Conclusion:</strong> A ∩ B is always an equivalence relation.</p><p>∴ Answer: A</p>
Correct Answer: A