A∩B — A Intersection B — Set Theory Explained

Definition: A∩B = {x : x ∈ A and x ∈ B} (The set of all elements x such that x belongs to A AND x belongs to B)

Formula (number of elements): n(A∩B) = number of elements common to both A and B

Addition Rule (very important formula): n(A∪B) = n(A) + n(B) − n(A∩B) Or: n(A∩B) = n(A) + n(B) − n(A∪B)

Examples:

Example 1: A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6, 7} A∩B = {3, 4, 5} (elements in both A and B)

Example 2: A = {vowels} = {a, e, i, o, u} B = {first 5 letters} = {a, b, c, d, e} A∩B = {a, e}

Example 3 (disjoint sets): A = {1, 2, 3}, B = {4, 5, 6} A∩B = {} = ∅ (empty set) (No common elements → disjoint sets)

Properties of set intersection:

1. Commutative Law: A∩B = B∩A (Order doesn't matter)

2. Associative Law: A∩(B∩C) = (A∩B)∩C

3. Identity Law: A∩U = A (intersection with Universal set = A) A∩∅ = ∅ (intersection with empty set = empty set)

4. Idempotent Law: A∩A = A

5. Distributive Law: A∩(B∪C) = (A∩B)∪(A∩C) A∪(B∩C) = (A∪B)∩(A∪C)

6. De Morgan's Laws: (A∩B)' = A'∪B' (A∪B)' = A'∩B'

Intersection vs Union: • A∩B (intersection): elements in BOTH A and B (AND) • A∪B (union): elements in A OR B or both (OR) • If A = {1,2,3}, B = {2,3,4}: A∩B = {2,3} (common elements) A∪B = {1,2,3,4} (all elements combined)

Problem 1: In a class of 40 students, 25 play cricket and 15 play football. 10 play both. Find n(A∩B) and the number who play neither.

Solution: n(A) = 25, n(B) = 15, n(A∩B) = 10 n(A∪B) = n(A) + n(B) − n(A∩B) = 25 + 15 − 10 = 30 Neither = Total − n(A∪B) = 40 − 30 = 10 students

Problem 2: If n(A) = 20, n(B) = 30, n(A∪B) = 40, find n(A∩B).

Solution: n(A∩B) = n(A) + n(B) − n(A∪B) = 20 + 30 − 40 = 10

Problem 3: A = {x : x is a factor of 12}, B = {x : x is a factor of 18} Find A∩B.

Solution: Factors of 12: {1, 2, 3, 4, 6, 12} Factors of 18: {1, 2, 3, 6, 9, 18} A∩B = {1, 2, 3, 6} (common factors = factors of HCF)

Note: A∩B = factors of HCF(12, 18) = factors of 6 = {1, 2, 3, 6} ✓

Venn Diagram description: • Draw two overlapping circles — left circle = A, right circle = B • Overlapping (shaded) region = A∩B • Total area of both circles (including overlap) = A∪B