At a glance. One lesson of about 60 minutes. It opens the unit and assumes union, intersection and complement on two sets. Students shade and count three-set expressions, apply De Morgan's laws and the distributive laws, and prove two expressions equal by following an element.
Calculator: no calculator. No calculator is needed. Everything is a count or a short argument, and a calculator adds nothing to a proof.
What students will learn
Know
- (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′, for any number of sets.
- A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
- A complement of a complement cancels: (A′)′ = A.
Understand
- Symbols turn a sentence into an object that can be operated on and checked.
- A diagram suggests that two expressions are equal; an argument about one element proves it.
- “Not” changes “and” into “or”, which is why the sign flips.
Do
- Translate a sentence into a set expression and shade it.
- Simplify a nested complement with De Morgan's laws.
- Prove two set expressions equal by following one element.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: applying De Morgan's laws, reading three-set regions and counting (questions 1 to 7).
- Criterion B. Investigating patterns: the figure. Students find which expressions always shade the same regions and state the rule.
- Criterion C. Communicating: correct set notation, and a proof that states what “x is in” means at each step (worked example 3 and the faded example).
- Criterion D. Applying mathematics in real-life contexts: the library hook, where two assistants write different expressions for one list of members.
The one thing to do with the figure
Ask for a prediction before every slide. With A′ ∪ B′ chosen, ask which other expression will shade the same regions before reading the readout. Then ask which is the odd one out between (A ∪ B)′ and A′ ∪ B′.
Finally slide through the last four expressions and ask students to state the distributive law in words, then to say why the diagram alone does not prove it.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the library. Collect predictions on the two assistants. |
| 12 min | The figure with predictions; the key idea; settle the hook. |
| 10 min | Warning card and worked examples; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose A, return to break the line and flip the sign. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | {1, 2, 3, 4}; {3}; x ≤ 7. |
|---|---|
| Hook | No. (A ∪ B)′ is outside both circles, which equals A′ ∩ B′, while A′ ∪ B′ is everything except A ∩ B. |
| Worked examples | (F ∪ C) ∩ (F ∩ C)′; A ∩ B′; the proof that (A ∩ B)′ = A′ ∪ B′. |
| Faded example | x ∈ (A ∩ B) ∪ (A ∩ C), with every step reversible. |
| 1. De Morgan with three sets | B. A keeps the union sign; C leaves R alone; D flips only some of the signs. |
| 2. De Morgan to count an intersection | 15. 90 − 75. |
| 3. Reading a three-set diagram | 11. 4 + 5 + 2. |
| 4. Complement intersected with a union | 17. 6 + 8 + 3. |
| 5. Counting shaded regions | 3. C only, A and C only, B and C only. |
| 6. De Morgan with the counting rule | 57. 41 + 16. |
| 7. Setting up an equation from De Morgan | 16. 4x + 6 = 70. |
| 8. Nested brackets with De Morgan | A. B keeps both signs; C flips only the outer sign; D flips only the inner sign (it is (A ∪ B ∪ C)′). |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, C or D in question 1 | Kept the union sign, left one set uncomplemented, or flipped only some of the signs. | Test an element that is in P only and ask which expressions contain it. |
| 75 or 165 in question 2 | Gave n(A′ ∪ B′) itself, or added it to the universal set. | Say that A′ ∪ B′ is the same set as (A ∩ B)′, then take it away from 90. |
| 9, 20 or 14 in question 3 | Gave A only, the whole of A, or included B and C only. | Colour the three regions inside A that touch B or C. |
| 28, 14 or 24 in question 4 | Did not remove the part in A, missed B and C only, or kept the region outside all three. | Cross out every region that touches A, then every region outside B and C. |
| 4, 1 or 6 in question 5 | Counted all of C, only C only, or forgot the ∩ C. | Write (A ∩ B)′ as A′ ∪ B′ and list the regions. |
| 41, 16 or 25 in question 6 | Gave the union, the intersection, or subtracted instead of adding. | Write n(A ∪ B) = n(A) + n(B) − n(A ∩ B) and rearrange. |
| 54 or 19 in question 7 | Gave the union, or moved the 6 across with the wrong sign. | Write (3x + 6) + x = 70 first and solve it. |
| B, C or D in question 8 | Kept both signs, flipped only the outer sign, or flipped only the inner one. | Break the outer line first and write A′ ∪ (B ∪ C)′, then break the inner line. |
Questions worth asking
- Why does a shaded diagram not prove a set law? It shows one picture with particular sets; an argument about a general element x covers every possible set.
- Why does “not (and)” become “or”? For “A and B” to fail, only one of them needs to fail.
- Does De Morgan work for four sets? Yes: (A ∪ B ∪ C ∪ D)′ = A′ ∩ B′ ∩ C′ ∩ D′, by applying the two-set law repeatedly.
- Is A ∪ (B ∩ C) the same as (A ∪ B) ∩ C? Not in general: take A = {1}, B = {2}, C = {3}. The left side is {1} and the right is empty.
Differentiation
A way in for anyone stuck
- Give a three-set Venn template with the eight regions pre-labelled, and ask students to colour one expression at a time.
- Keep a card with the words: or is ∪, and is ∩, not is ′, and the sentence “break the line, flip the sign”.
- For the proof, give the first and last lines and let students write the middle step.
Stretch
- Prove (A ∪ B)′ = A′ ∩ B′ by following one element, and say which step is the one that changes “or” into “and”.
- Show that A \ B (in A but not in B) equals A ∩ B′, and write the symmetric difference of two sets in two ways.
Students learning in English as an additional language
“Complement” means “everything in the universal set that is not in the set”. Say “A prime” for A′, “A union B” for A ∪ B and “A intersection B” for A ∩ B. “Equal sets” have exactly the same elements.
| Word | Plain meaning | Example |
|---|---|---|
| universal set | the set of everything being considered, written U | all the students in a year group |
| complement | all the elements of U that are not in the set, written with a prime | the students who do not play chess, C′ |
| De Morgan's laws | the rules for a complement of a union or intersection | rewriting (P ∪ Q)′ without a bracket |
| distributive law | one operation spread over another inside a bracket | 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 | x ∈ P ∩ Q means x ∈ P and x ∈ Q |
| disjoint | having no element in common | P ∩ Q = ∅ |
Sentence starters for explaining aloud or in writing:
- x is in this set means …
- These two expressions shade the same regions because …
- To complement a bracket, break the line and …
Two things not to say
“The prime goes straight through the brackets.” It also flips the sign inside.
“The diagram shows it, so it is proved.” A diagram is one example; the argument about x covers every set.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- ExpressWrite “in X or Y but not in Z” as a set expression.
(X ∪ Y) ∩ Z′. - Earlier work (Standard Unit 1)List {x | x ∈ ℤ, −2 ≤ x < 2}.
{−2, −1, 0, 1}.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- SimplifySimplify (X′ ∩ Y′)′ and say what it means in words.
X ∪ Y: in X or in Y or in both. - Countn(U) = 60, n(P) = 25, n(Q) = 30 and n(P′ ∩ Q′) = 14. Find n(P ∩ Q).
n(P ∪ Q) = 60 − 14 = 46, so n(P ∩ Q) = 25 + 30 − 46 = 9.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningProve that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) by following one element.
x ∈ A, or (x ∈ B and x ∈ C). If x ∈ A it is in both brackets. Otherwise it is in B and in C, so in both brackets. Conversely, if it is in both brackets and not in A, it must be in B and in C. - CounterexampleGive sets for which (X ∪ Y) ∩ Z ≠ X ∪ (Y ∩ Z).
X = {1}, Y = {2}, Z = {3}: the left side is empty, the right side is {1}.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
The figure works with the keyboard (tab to the slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No calculator is needed. A printed three-set Venn template helps in the proof section.