Complements and De Morgan's laws

The complement of A is everything in U that is not in A: A′ = {x : x ∈ U and x ∉ A} = U – A. In a Venn diagram it is the part of the rectangle outside circle A.

Example: U = {1, 2, ..., 12} and A = {3, 6, 9, 12}, the multiples of 3. Then A′ = {1, 2, 4, 5, 7, 8, 10, 11}.

Complement laws:

• A ∪ A′ = U and A ∩ A′ = φ.
• (A′)′ = A: taking the complement twice brings A back.
• φ′ = U and U′ = φ.

De Morgan's laws: the complement of a union is the intersection of the complements, and the other way round. ∪ and ∩ swap, and each set gets its own dash:

• (A ∪ B)′ = A′ ∩ B′
• (A ∩ B)′ = A′ ∪ B′

Check with the same U and A, and B = {2, 4, 6, 8, 10, 12}. A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}, so (A ∪ B)′ = {1, 5, 7, 11}. And A′ ∩ B′ = {1, 2, 4, 5, 7, 8, 10, 11} ∩ {1, 3, 5, 7, 9, 11} = {1, 5, 7, 11}. The same.

A very useful link: A – B = A ∩ B′. 'Minus B' is the same as 'and not in B'. Shade both in a Venn diagram and you get the same crescent of A.

With three sets, fill the Venn diagram from the centre outwards. Of 60 students, 30 play football, 25 cricket, 20 hockey; 10 play football and cricket, 8 cricket and hockey, 7 football and hockey, and 4 all three.

• Centre: 4. Then the pairs only: football and cricket 10 – 4 = 6, cricket and hockey 4, football and hockey 3.
• Then each game only: football 30 – 6 – 3 – 4 = 17, cricket 25 – 6 – 4 – 4 = 11, hockey 20 – 4 – 3 – 4 = 9.
• At least one game: 17 + 11 + 9 + 6 + 4 + 3 + 4 = 54, so 6 play none, the complement of the union. Exactly one game: 17 + 11 + 9 = 37.

The same total comes from n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C) = 30 + 25 + 20 – 10 – 8 – 7 + 4 = 54.