Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From Year 10, Standard: for A = {1, 2, 3} and B = {3, 4}, list A ∪ B.
Answer
{1, 2, 3, 4}. 3 is in both sets and is listed once. - From Year 10, Standard: with the same sets, list A ∩ B.
Answer
{3}. Only 3 is in both. - From Year 8: write “not greater than 7” as an inequality in x.
Answer
x ≤ 7. Not greater than 7 means 7 or less, so 7 itself is allowed.
Words used in this lesson
- universal set
- the set of everything being considered, written U. For example: all the students in a year group.
- complement
- all the elements of U that are not in the set, written with a prime. For example: the students who do not play chess, C′.
- De Morgan's laws
- the rules for a complement of a union or intersection. For example: rewriting (P ∪ Q)′ without a bracket.
- distributive law
- one operation spread over another inside a bracket. For example: X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z).
- proof by an element
- following one general element x through the definitions to show two sets are equal. For example: x ∈ P ∩ Q means x ∈ P and x ∈ Q.
- disjoint
- having no element in common. For example: P ∩ Q = ∅.
Not this, or not that?
A school library in Nairobi sorts its members by what they borrowed last term: A for fiction, B for non-fiction and C for poetry. The librarian wants every member who borrowed neither fiction nor non-fiction. One assistant writes (A ∪ B)′. Another writes A′ ∪ B′.
Predict first: do the two assistants name the same set of members? Which of the two expressions is equal to A′ ∩ B′? Then test your prediction on the figure, which names every expression that shades the same regions.
A ∩ B shades 2 of the 8 regions, holding 4 of the 36 elements. Same set as: none of the other eight.
Show me the steps
- A ∩ B: the two regions inside both A and B are shaded, A and B only (3 elements) and all three (1 element). That is 2 regions and 4 elements. No other expression in the list shades just these.
- Slide to (A ∪ B)′: the regions outside both circles are shaded, C only (6) and none (8). That is 14 elements, and A′ ∩ B′ shades exactly the same two regions.
- Slide to A′ ∪ B′: six regions are shaded, everything except the two in A ∩ B, so 32 elements. It matches (A ∩ B)′ and does not match (A ∪ B)′.
- Slide to A ∩ (B ∪ C): three regions, holding 3 + 2 + 1 = 6 elements, and it matches (A ∩ B) ∪ (A ∩ C). A ∪ (B ∩ C) shades five regions (17 elements) and matches (A ∪ B) ∩ (A ∪ C).
De Morgan's laws. (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′. To push a complement through a bracket, break the line and flip the sign: ∪ becomes ∩, and ∩ becomes ∪.
Distributive laws. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C), and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Proof by an element. To show two sets are equal, follow one element x. Say what “x is in” means for each side, turn “not”, “and” and “or” into logic step by step, and reach the other side. A shaded diagram suggests a law; the argument about x proves it.
The prime does not go through a bracket unchanged. Take U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3} and B = {3, 4}. Then (A ∪ B)′ = {5, 6, 7, 8, 9}, but A′ ∪ B′ = {1, 2, 4, 5, 6, 7, 8, 9}. The element 1 is in A, so it is in A ∪ B and out of (A ∪ B)′; yet it is not in B, so it is in B′ and therefore in A′ ∪ B′.
Worked examples
1. Write “students who play football or chess but not both” as a set expression.
- Let F be football and C be chess. “Football or chess” is F ∪ C.
- “Both” is F ∩ C, and “not both” is (F ∩ C)′.
- Both conditions together: (F ∪ C) ∩ (F ∩ C)′. It shades F only and C only, which is also (F ∩ C′) ∪ (F′ ∩ C).
2. Simplify (A′ ∪ B)′.
- Break the line and flip the sign: (A′)′ ∩ B′.
- A complement of a complement cancels: (A′)′ = A.
- So (A′ ∪ B)′ = A ∩ B′: in A and not in B.
3. Prove that (A ∩ B)′ = A′ ∪ B′.
- x ∈ (A ∩ B)′ means x is not in A ∩ B, that is, it is not true that (x ∈ A and x ∈ B).
- If “A and B” fails, then A fails or B fails: x ∉ A or x ∉ B.
- x ∉ A means x ∈ A′, and x ∉ B means x ∈ B′, so x ∈ A′ or x ∈ B′, which is x ∈ A′ ∪ B′. Every step can be reversed, so the sets are equal.
Your go. The first step is given. Prove that A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
- x ∈ A ∩ (B ∪ C) means x ∈ A and (x ∈ B or x ∈ C).
- Now split the “or” and carry the “x ∈ A” with each part.
Check the answer
Either (x ∈ A and x ∈ B) or (x ∈ A and x ∈ C). The first is x ∈ A ∩ B and the second is x ∈ A ∩ C, so x ∈ (A ∩ B) ∪ (A ∩ C). The steps reverse, so the two sets are equal.
Check before you go on
1. Which expression is equal to (P ∪ Q ∪ R)′?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. A universal set has 90 elements. n(A′ ∪ B′) = 75. Find n(A ∩ B).
3. In a three-set Venn diagram the regions hold: A only 9, B only 6, C only 8, A and B only 4, A and C only 5, B and C only 3, all three 2, none 7. Find n(A ∩ (B ∪ C)).
Level 2 · Building
4. Use the same diagram: A only 9, B only 6, C only 8, A and B only 4, A and C only 5, B and C only 3, all three 2, none 7. Find n(A′ ∩ (B ∪ C)).
5. The set (A ∩ B)′ ∩ C is shaded on a three-set Venn diagram with eight regions. How many of the eight regions are shaded?
Level 3 · Stretch
6. n(U) = 50, n(A′ ∪ B′) = 34 and n(A ∪ B) = 41. Find n(A) + n(B).
7. A survey of 70 people has n(A ∪ B ∪ C) = 3x + 6 and n(A′ ∩ B′ ∩ C′) = x. Find x.
Something to think about
Two sets A and B are disjoint, so A ∩ B = ∅. What do De Morgan's laws say about (A ∪ B)′ and about (A ∩ B)′ in that case? Then find a pair of sets for which A ∪ (B ∩ C) and (A ∪ B) ∩ C are not equal, to show that the distributive law cannot be moved around freely.
One way to think about it
With A ∩ B = ∅ the second law gives (A ∩ B)′ = U, so A′ ∪ B′ is everything: any element is outside A or outside B. The first law still gives (A ∪ B)′ = A′ ∩ B′, which is U with both sets removed. For the second part take A = {1}, B = {2} and C = {3}: A ∪ (B ∩ C) = {1}, but (A ∪ B) ∩ C = {1, 2} ∩ {3} = ∅. Moving the bracket changes which operation is done first: the distributive law carries A into both parts, (A ∪ B) ∩ (A ∪ C), while (A ∪ B) ∩ C leaves A out of the second part.
Exit ticket
8. Which expression is equal to (A ∩ (B ∪ C))′?