Problems on Sets (Senior 3 Mathematics): free lessons | Ganzaa

Problems on Sets

Lessons

Senior 3 · Mathematics

Problems on Sets

4 lessons

Lesson 1 of 4

Reading and shading Venn diagrams

24 min

One region, several names

For one week the librarian of a school in Rubavu keeps two lists: $N$ , the learners who borrowed a novel, and $S$ , the learners who borrowed a science book. The head teacher asks how many learners borrowed neither. The librarian can describe that group as $(N \cup S)'$ , "not in the union of the two lists", or as $N' \cap S'$ , "not on the novel list and not on the science list". The two descriptions look different, yet they name the same learners.

In S1 you drew Venn diagrams and shaded a region described in symbols. In S3 questions often run the other way: a region is shaded and you must write it in set notation, or two expressions are given and you must decide whether they are equal. This lesson gives you one reliable method for all of these: work with the regions of the diagram.

Regions: the building blocks of every set expression

Two overlapping circles inside a rectangle cut the universal set $U$ into four regions. Number them once and use the same numbers all through this unit.

The four regions of a two-set Venn diagram.

Region	In $A$ ?	In $B$ ?	The region as a set
1	yes	no	$A \cap B'$
2	yes	yes	$A \cap B$
3	no	yes	$A' \cap B$
4	no	no	$A' \cap B'$

Each letter is a list of regions: $A = \{1, 2\}$ , $B = \{2, 3\}$ , and $U = \{1, 2, 3, 4\}$ . Any set you can build from $A$ and $B$ with $\cup$ , $\cap$ and $^{'}$ is just a collection of these regions.

Definition

Region (of a Venn diagram): one of the separate parts into which the circles divide the rectangle. A region is fixed by saying, for every set, whether its elements are inside or outside that set.

Notation: number the regions; write a set expression as the list of its regions.

Example: in a two-set diagram, region 3 is "in $B$ but not in $A$ ", which is the set $A' \cap B$ , also written $B - A$ .

Definition

Set identity: a statement that two set expressions are equal for every choice of the sets.

Notation: an ordinary equals sign between two expressions, for example $A - B = A \cap B'$ .

Example: $A \cap B = B \cap A$ is an identity. To prove an identity with a Venn diagram, show that both sides cover exactly the same regions. To disprove one, find one region that is in one side but not the other.

Turning symbols into regions

Work from the inside of the brackets outwards, using three rules on the region lists.

Operation	Rule on the lists	Example with $A = \{1, 2\}$ , $B = \{2, 3\}$
Union $\cup$	keep every region that is in either list	$A \cup B = \{1, 2, 3\}$
Intersection $\cap$	keep only the regions that are in both lists	$A \cap B = \{2\}$
Complement $^{'}$	keep the regions of $U$ that are not in the list	$B' = \{1, 4\}$

For example, $A \cup B' = \{1, 2\} \cup \{1, 4\} = \{1, 2, 4\}$ : everything except region 3. The six diagrams below show some sets that appear again and again in examinations.

Six shaded sets. $A \,\Delta\, B$ (the symmetric difference) holds the elements in exactly one of the two sets.

Worked example 1: naming a shaded region in two ways

In the diagram, regions 1, 3 and 4 are shaded. Write the shaded set in two different ways.

Everything is shaded except the overlap.

Solution

Way 1: look at what is NOT shaded. Only region 2, which is $A \cap B$ , is left white. So the shaded set is the complement of that region: $(A \cap B)'$ .

Way 2: build it from pieces. $A' = \{3, 4\}$ and $B' = \{1, 4\}$ . Their union is $\{1, 3, 4\}$ , exactly the shaded regions. So the shaded set is also $A' \cup B'$ .

Both answers are correct, and together they show that $(A \cap B)' = A' \cup B'$ .

Important

De Morgan's laws. For any two sets $A$ and $B$ in a universal set $U$ :

(A \cup B)' = A' \cap B' \qquad\qquad (A \cap B)' = A' \cup B'

In words: the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. The laws are named after the British mathematician Augustus De Morgan (1806–1871).

Why: use the region lists. $A \cup B = \{1, 2, 3\}$ , so $(A \cup B)' = \{4\}$ ; and $A' \cap B' = \{3, 4\} \cap \{1, 4\} = \{4\}$ . Equal. Next, $A \cap B = \{2\}$ , so $(A \cap B)' = \{1, 3, 4\}$ ; and $A' \cup B' = \{3, 4\} \cup \{1, 4\} = \{1, 3, 4\}$ . Equal. The four regions cover every possible element, so the laws hold for all sets.

De Morgan's first law: the overlap of $A^{'}$ and $B^{'}$ is region 4 only, the same region as $(A \cup B)'$ .

Worked example 2: checking De Morgan's laws with lists

$U = \{1, 2, 3, \ldots, 12\}$ , $A$ is the set of even numbers in $U$ and $B$ is the set of multiples of 3 in $U$ . Verify both of De Morgan's laws by listing.

Solution $A = \{2, 4, 6, 8, 10, 12\}$ and $B = \{3, 6, 9, 12\}$ , so $A' = \{1, 3, 5, 7, 9, 11\}$ and $B' = \{1, 2, 4, 5, 7, 8, 10, 11\}$ .

First law. $A \cup B = \{2, 3, 4, 6, 8, 9, 10, 12\}$ , so $(A \cup B)' = \{1, 5, 7, 11\}$ . The elements common to $A^{'}$ and $B^{'}$ are also $\{1, 5, 7, 11\}$ . ✓

Second law. $A \cap B = \{6, 12\}$ , so $(A \cap B)' = \{1, 2, 3, 4, 5, 7, 8, 9, 10, 11\}$ . Joining $A^{'}$ and $B^{'}$ gives $\{1, 2, 3, 4, 5, 7, 8, 9, 10, 11\}$ . ✓

Notice that $(A \cup B)'$ is the set of numbers from 1 to 12 that are divisible by neither 2 nor 3.

Three sets: the same method with eight regions

Keep the S1 numbering of the eight regions. The key facts are the region lists of the three letters:

A = \{1, 4, 5, 7\}, \qquad B = \{2, 4, 6, 7\}, \qquad C = \{3, 5, 6, 7\}, \qquad U = \{1, 2, \ldots, 8\}.

The eight regions: 1, 2, 3 are in one set only; 4, 5, 6 are in exactly two; 7 is in all three; 8 is in none.

De Morgan's laws extend to three sets: $(A \cup B \cup C)' = A' \cap B' \cap C'$ (region 8 only) and $(A \cap B \cap C)' = A' \cup B' \cup C'$ (every region except 7).

Worked example 3: shading from symbols

Shade (a) $(A \cap B)' \cap C$ and (b) $A' \cap (B \cup C)$ .

Solution

(a) $A \cap B = \{4, 7\}$ , so $(A \cap B)' = \{1, 2, 3, 5, 6, 8\}$ . Intersect with $C = \{3, 5, 6, 7\}$ : the common regions are $\{3, 5, 6\}$ . This is "in $C$ , but not in both $A$ and $B$ ".

(b) $A' = \{2, 3, 6, 8\}$ and $B \cup C = \{2, 3, 4, 5, 6, 7\}$ . The common regions are $\{2, 3, 6\}$ : "in $B$ or $C$ , but not in $A$ ".

(a) regions 3, 5, 6 (b) regions 2, 3, 6

Worked example 4: writing a shaded region in symbols

Write each shaded set in set notation.

(a) regions 4, 5, 6, 7 (b) regions 3 and 8

Solution

(a) These are the elements in at least two of the sets. Each pairwise intersection gives two of the regions: $A \cap B = \{4, 7\}$ , $A \cap C = \{5, 7\}$ , $B \cap C = \{6, 7\}$ . Their union is $\{4, 5, 6, 7\}$ , so the shaded set is

(A \cap B) \cup (A \cap C) \cup (B \cap C).

(b) Regions 3 and 8 are the parts outside both $A$ and $B$ : $A \cup B = \{1, 2, 4, 5, 6, 7\}$ and its complement is ${3, 8}$ . So the shaded set is $(A \cup B)'$ , which by De Morgan's law is also $A' \cap B'$ . Circle $C$ plays no part.

When the circles do not overlap in the usual way

A problem may tell you that one set lies inside another ("every learner in the choir is also in the music club") or that two sets have nothing in common ("no learner is in both the morning and the afternoon shift"). Draw the circles to match: a circle inside a circle for a subset, and separate circles for disjoint sets.

Left: the usual overlapping position. Centre: $B \subseteq A$ . Right: $A$ and $B$ are disjoint.

Position	What follows
$B \subseteq A$	$A \cap B = B$ , $A \cup B = A$ , so $n(A \cup B) = n(A)$ and $n(A \cap B) = n(B)$
$A \cap B = \varnothing$ (disjoint)	$n(A \cup B) = n(A) + n(B)$ , because nothing is counted twice

Warning

The dash cannot just be copied onto each letter. $(A \cap B)'$ is not $A' \cap B'$ . Check with regions: $(A \cap B)' = \{1, 3, 4\}$ , but $A' \cap B' = \{4\}$ . When the complement moves inside the brackets, the operation between the sets must change as well.

Tip

"Break the line, change the sign." Think of the dash over a whole bracket as a line. When you break it into a dash on each letter, change $\cup$ into $\cap$ , or $\cap$ into $\cup$ .

Mathispower4u: De Morgan's Laws with Venn Diagrams (5 min). Watch how each side of a law is shaded in its own diagram and the two results are compared. Books write the complement of A in several ways, such as A′, Aᶜ or with a bar over the A; they all mean the same set.

✦

Practice

Use region numbers for every question. The last three are harder.

For two sets, list the regions of (a) $A^{'}$ (b) $A \cup B'$ (c) $A' \cap B$ .
Write two different names for region 1 of a two-set diagram.
$U = \{1, 2, \ldots, 10\}$ , $A = \{1, 2, 3, 4, 5\}$ and $B = \{4, 5, 6, 7\}$ . List $(A \cup B)'$ and $A' \cap B'$ , and say what you notice.
For the sets in question 3, find $n\big((A \cap B)'\big)$ in two different ways.
For three sets, list the regions of (a) $A \cap B'$ (b) $(B \cup C)'$ (c) $A \cup (B \cap C)$ .
Write in set notation the set of elements that are in exactly one of the three sets $A$ , $B$ and $C$ .
Use regions to decide whether $A - (B \cup C) = (A - B) \cap (A - C)$ is an identity.
$B \subseteq A$ , $n (U) = 30$ , $n (A) = 20$ and $n (B) = 8$ . Find $n(A \cap B)$ , $n(A \cup B)$ and $n(A' \cap B')$ .

Show answersHide answers

(a) 3, 4 (b) 1, 2, 4 (c) 3
For example $A \cap B'$ and $A - B$ (also $(A' \cup B)'$ ).
$A \cup B = \{1, \ldots, 7\}$ , so $(A \cup B)' = \{8, 9, 10\}$ . $A' = \{6, 7, 8, 9, 10\}$ , $B' = \{1, 2, 3, 8, 9, 10\}$ , so $A' \cap B' = \{8, 9, 10\}$ . They are equal: De Morgan's first law.
$A \cap B = \{4, 5\}$ , so $n\big((A \cap B)'\big) = 10 - 2 = 8$ . Or $A' \cup B' = \{1, 2, 3, 6, 7, 8, 9, 10\}$ , which also has 8 elements.
(a) $\{1, 4, 5, 7\} \cap \{1, 3, 5, 8\} = \{1, 5\}$ (b) $B \cup C = \{2, 3, 4, 5, 6, 7\}$ , so the complement is ${1, 8}$ (c) $\{1, 4, 5, 7\} \cup \{6, 7\} = \{1, 4, 5, 6, 7\}$
$(A \cap B' \cap C') \cup (A' \cap B \cap C') \cup (A' \cap B' \cap C)$ : regions 1, 2 and 3.
Left: $A \cap (B \cup C)' = \{1, 4, 5, 7\} \cap \{1, 8\} = \{1\}$ . Right: $A - B = \{1, 5\}$ , $A - C = \{1, 4\}$ , and their intersection is ${1}$ . Same regions, so it is an identity.
$n(A \cap B) = n(B) = 8$ ; $n(A \cup B) = n(A) = 20$ ; $n(A' \cap B') = n\big((A \cup B)'\big) = 30 - 20 = 10$ .

Sources and credits

1.3: Understanding Venn Diagrams, from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
1.5: Set Operations with Three Sets (including De Morgan's laws), from Contemporary Mathematics by Donna Kirk and contributors, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The region-list method, the three-set extension of De Morgan's laws and the table of subset and disjoint positions. Written by Ganzaa.
Video: De Morgan's Laws with Venn Diagrams, Mathispower4u, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1One region, several names

2Regions: the building blocks of every set expression

3Turning symbols into regions

4Three sets: the same method with eight regions

5When the circles do not overlap in the usual way

One region, several names

Regions: the building blocks of every set expression

Turning symbols into regions

Three sets: the same method with eight regions

When the circles do not overlap in the usual way