This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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a(i): Prove n(A B) = n(A) + n(B) - n(A B). Step 1: Elements in A B are in A only, B only, or both. n(A B) = n(A B) + n(B A) + n(A B) Step 2: Substitute n(A B) = n(A) - n(A B). n(A B) = [n(A) - n(A B)] + n(B A) + n(A B) Step 3: Substitute n(B A) = n(B) - n(A B). n(A B) = [n(A) - n(A B)] + [n(B) - n(A B)] + n(A B) Step 4: Simplify. n(A B) = n(A) - n(A B) + n(B) - n(A B) + n(A B) = n(A) + n(B) - n(A B) n(A B) = n(A) + n(B) - n(A B) a(ii): Same as a(i), standard inclusion-exclusion principle for two sets. n(A B) = n(A) + n(B) - n(A B) a(iii): Prove using rearrangement from a(i). From n(A B) = n(A) + n(B) - n(A B), rearrange. n(A B) = n(A) + n(B) - n(A B) But listed as for union, perhaps alternative form. n(A B) = n(A) + n(B) - n(A B) b(i): Prove n(A B) = n(A) + n(B) - n(A B). Step 1: n(A) = n(A B) + n(A B). Step 2: n(B) = n(B A) + n(A B). Step 3: Add the formulas. n(A) + n(B) = n(A B) + n(A B) + n(B A) + n(A B) n(A) + n(B) = n(A B) + n(B A) + 2n(A B) Step 4: Note n(A B) = n(A B) + n(B A) + n(A B). Step 5: Substitute into sum. n(A) + n(B) = [n(A B) - n(A B)] + 2n(A B) = n(A B) + n(A B) Step 6: Solve for intersection. n(A B) = n(A) + n(B) - n(A B) n(A B) = n(A) + n(B) - n(A B) c(i): Likely n(A B) = n(A B) + n(A B) + n(B A), which is equivalent to a(i). Step 1: Partition A B. n(A B) = n(A B) + n(A B) + n(B A) n(A B) = n(A B) + n(A B) + n(B A) d(i): Prove n(A - B) = n(A) - n(A B). Step 1: A - B = A B = A B^c, elements in A not in B. Step 2: n(A) = n(A B) + n(A B). Step 3: Solve for difference. n(A B) = n(A) - n(A B) n(A - B) = n(A) - n(A B) e: Both sides equal: n(A - B) = n(A) - n(A B). Verified as in d(i). n(A - B) = n(A) - n(A B) f: Both: Alternative proof n(A - B) = n(A B^c). Assuming finite sets, count elements in A excluding A B. Same as d(i). n(A - B) = n(A) - n(A B) g: Of the two: n(A - B) = n(A B'). Yes, by definition A - B = A B^c, where B' is complement relative to universe, but cardinality same. n(A - B) = n(A B') h: Prove n(A B C) = n(A)+n(B)+n(C)-n(A B)-n(A C)-n(B C)+n(A B C). Step 1: Use inclusion-exclusion principle for three sets. Divide into 7 disjoint regions: Only A: n(A (B C)) Only B: n(B (A C)) Only C: n(C (A B)) A and B only: n((A B) C) A and C only: n((A C) B) B and C only: n((B C) A) All three: n(A B C) Step 2: n(A B C) = sum of all 7. Step 3: Express each: n(A (B C)) = n(A) - n(A B) - n(A C) + n(A B C) (Subtract pairwise, add back triple.) Similarly for others. Step 4: Add all: Full expansion: n(A) + n(B) + n(C) - 2n(A B) - 2n(A C) - 2n(B C) + 3n(A B C) No, standard derivation: From two-set: n((A B) C) = n(A B) + n(C) - n((A B) C) Step 5: n(A B) = n(A)+n(B)-n(A B) Step 6: n((A B) C) = n( (A C) (B C) ) = n(A C) + n(B C) - n(A B C) Step 7: Substitute. n(A B C) = [n(A)+n(B)-n(A B)] + n(C) - [n(A C) + n(B C) - n(A B C)] Step 8: Simplify. = n(A) + n(B) + n(C) - n(A B) - n(A C) - n(B C) + n(A B C) n(A B C) = n(A) + n(B) + n(C) - n(A B) - n(A C) - n(B C) + n(A B C)