Case Study: Decoding Relations
Topic: Relations and Functions (Class 12 Mathematics)
Detailed Solution & Analysis
(i) Answer: R₄
Analysis: For A = {1, 2, 3}, R₄ is:
- Reflexive: Includes (1,1), (2,2), and (3,3).
- Not Symmetric: Includes (1,2) but NOT (2,1).
- Transitive: No chains (a,b) and (b,c) exist that break the (a,c) rule; it is “vacuously” transitive.
(ii) Answer: R₅
Analysis: R₅ contains (1,1), (2,2), (3,3) (Reflexive) and all pairs have opposites (Symmetric). However, it contains (1,2) and (2,3) but is missing (1,3), making it NOT transitive.
(iii) (a) Answer: R₁
Analysis: R₁ is symmetric because for (2,3) it has (3,2). It fails reflexivity (no 1,1) and transitivity (has 2,3 and 3,2 but no 2,2).
(iii) (b) Answer: Add {(2,2), (3,3), (3,1), (3,2), (2,3)}
Analysis: To make R₃ an equivalence relation on {1,2,3}, we must ensure it is reflexive (add 2,2 and 3,3) and that all connections between elements are balanced and linked.
⚠️ Audit Your Thinking: Silly Mistake Alert!
The “Proof by Example” Trap: Never assume a relation is transitive just because you can’t see a “break.” In Part (i), R₄ is transitive because the condition “If (a,b) and (b,c) exist…” is never triggered. This is a vacuous truth.