Maths · 1 Number

1.2 Sets

Before you start

The types of number from 1.1: natural numbers (starting at $0$ ), integers, primes, factors, multiples and square numbers.
The inequality signs: $x < 3$ means $x$ is less than 3, and $x ⩽ 3$ means $x$ is less than or equal to 3.
Solving a linear equation with the unknown on both sides, such as $3 x - 2 = 6 - x$ .
Collecting like terms, e.g. $(10 - x) + (7 - x) + x = 17 - x$ .
That the straight line $y = m x + c$ has gradient $m$ and crosses the $y$ -axis at $c$ , and that two lines with the same gradient are parallel.

Syllabus objectives covered (1)

Understand and use set language, notation and Venn diagrams to describe sets and represent relationships between sets.

Part 1 of 7

Sets, elements and counting them

A set is a collection of things, and each thing in it is an element (or member) of the set. Name a set with a capital letter and list its elements inside curly brackets, separated by commas.
$A = {2, 4, 6, 8}$
Order does not matter, and an element is either in the set or not in it. ${2, 4, 6, 8}$ and ${8, 6, 4, 2}$ are the same set, and writing an element twice adds nothing. List each element once.
The symbol $\in$ means 'is an element of', so $4 \in A$ . The symbol $\notin$ means 'is not an element of', so $5 \notin A$ . On the left of either symbol goes one single element, never a set in curly brackets.
$n (A)$ means the number of elements in $A$ . Here $n (A) = 4$ . Notice that $n (A)$ is a number, not a set. Write $n (A) = 4$ , never $n (A) = {4}$ and never $n (A) = {2, 4, 6, 8}$ .
The empty set, written $⌀$ , is the set with no elements, so $n (⌀) = 0$ . For example, the set of even prime numbers greater than 2 is $⌀$ , because 2 is the only even prime.
The trap: ${0}$ is not the empty set. It has one element, the number $0$ , so $n ({0}) = 1$ . Empty means nothing at all inside the brackets.

Part 2 of 7

Three ways to describe a set

You can list the elements, as in $D = {a, b, c, d}$ . Three dots mean the pattern carries on, so ${1, 3, 5, \dots}$ is every odd natural number. A set that goes on for ever like this is called infinite.
You can describe the elements in words inside the brackets. $P = {prime numbers less than 10}$ means $P = {2, 3, 5, 7}$ .
Or you can use set-builder notation, which gives a rule instead of a list.
$A = {x | x is a natural number}$
Read the vertical bar as 'such that'. This says ' $A$ is the set of all $x$ such that $x$ is a natural number'. To the left of the bar is what an element looks like. To the right is the rule it must obey.
Natural numbers start at $0$ in this syllabus (see 1.1), so this $A$ is ${0, 1, 2, 3, \dots}$ . That matters when you list a finite piece of it: ${x | x is a natural number, x < 5} = {0, 1, 2, 3, 4}$ , which has 5 elements, not 4.
A rule can be an inequality. $C = {x | a ⩽ x ⩽ b}$ is every number from $a$ to $b$ , both ends included. If the rule does not say that $x$ is an integer, $x$ can be any number in that range. So ${x | 2 ⩽ x ⩽ 5}$ contains $2.5$ , $π$ and $\sqrt{10}$ as well as $2, 3, 4, 5$ . It is infinite, and you cannot list it.
Once the rule says 'integer', you can list it. Check each end on its own: $<$ leaves the end out, and $⩽$ keeps it in.
${x | x is an integer, - 2 < x ⩽ 3} = {- 1, 0, 1, 2, 3}$
Here $- 2$ is left out because $- 2 < - 2$ is false, and $3$ is kept because $3 ⩽ 3$ is true. Count the result slowly: zero is an element too, so there are 5 elements.
The elements of a set can be coordinate points. $B = {(x, y) | y = m x + c}$ is the set of every point $(x, y)$ that fits the equation. That is every point on the straight line $y = m x + c$ .
To test a point, substitute it. For $L = {(x, y) | y = 2 x + 3}$ : $2 (1) + 3 = 5$ , so $(1, 5) \in L$ . But $2 (2) + 3 = 7$ , not $6$ , so $(2, 6) \notin L$ .

Part 3 of 7

The universal set, complements and subsets

The universal set $U$ is everything being considered in the question. Every other set in the question is made from elements of $U$ . For example, $U = {x | x is an integer, 1 ⩽ x ⩽ 10}$ means the numbers 1 to 10 and nothing else.
The complement of $A$ , written $A^{'}$ , is every element of $U$ that is not in $A$ . With that $U$ and $A = {2, 4, 6, 8, 10}$ , you get $A^{'} = {1, 3, 5, 7, 9}$ .
The complement depends on $U$ . If $U$ were the integers 1 to 12, the same $A$ would have $A^{'} = {1, 3, 5, 7, 9, 11, 12}$ . So read $U$ before you write any complement.
Every element of $U$ is in $A$ or in $A^{'}$ , and never in both. So the two counts add up to the whole. Here $5 + 5 = 10$ .
$n (A) + n (A^{'}) = n (U)$
$A \subseteq B$ means $A$ is a subset of $B$ : every element of $A$ is also an element of $B$ . ${2, 4} \subseteq {1, 2, 3, 4}$ , because 2 and 4 are both in the bigger set.
$A ⊈ B$ means $A$ is not a subset of $B$ : at least one element of $A$ is missing from $B$ . ${2, 5} ⊈ {1, 2, 3, 4}$ , because 5 is not in it. One element outside is enough.
Two edge cases follow straight from the definition. Every set is a subset of itself, $B \subseteq B$ , because each element of $B$ is in $B$ . And $⌀ \subseteq B$ for every set $B$ , because $⌀$ has no element that could be missing from $B$ .
Keep $\in$ and $\subseteq$ apart. $\in$ joins an element to a set. $\subseteq$ joins a set to a set. So $3 \in {1, 2, 3}$ and ${3} \subseteq {1, 2, 3}$ are both true. But ${3} \in {1, 2, 3}$ is false, because the elements of ${1, 2, 3}$ are numbers and ${3}$ is a set.

Part 4 of 7

Union and intersection

The intersection $A \cap B$ is the set of elements that are in $A$ and in $B$ , in both at once.
Take $A = {1, 2, 3, 4, 6, 12}$ , the factors of 12, and $B = {1, 2, 3, 6, 9, 18}$ , the factors of 18. Go through $A$ and keep only what is also in $B$ . These are the common factors from 1.1, written in a new way.
$A \cap B = {1, 2, 3, 6}$
The union $A \cup B$ is the set of elements in $A$ or $B$ or both. Write each element once, even if it is in both sets. (The symbol $\cup$ looks like the U of Union.)
$A \cup B = {1, 2, 3, 4, 6, 9, 12, 18}$
Here 'or' includes 'both'. That is why $1, 2, 3$ and $6$ are in $A \cup B$ : being in both sets still counts. A union is never smaller than either set, and an intersection is never bigger than either set.
If $A \cap B = ⌀$ , the two sets have no element in common. The odd numbers and the even numbers are an example.
With brackets, work out the bracket first and then take the complement. ${(A \cup B)}^{'}$ means: build $A \cup B$ , then take everything in $U$ outside it. That is the elements in neither $A$ nor $B$ .
For two sets of points, $A \cap B$ is every point on both lines. Two lines with different gradients cross at one point, so the intersection has one element. You find it by solving the two equations together.
Two lines with the same gradient and different values of $c$ are parallel. They never meet, so their intersection is $⌀$ . (If $m$ and $c$ are both the same, the two equations describe one line, and $A \cap B = A$ .)

Part 5 of 7

Two-set Venn diagrams: four regions

A Venn diagram draws the sets. A rectangle stands for $U$ , and each set is a circle inside it. Two overlapping circles cut the rectangle into exactly four regions.
graph diagram — not drawn yet
A rectangle labelled U containing two overlapping circles labelled A (left) and B (right). Each of the four regions carries a label: 'A ∩ B′' in the part of A outside B, 'A ∩ B' in the overlap, 'A′ ∩ B' in the part of B outside A, and '(A ∪ B)′' in the space inside the rectangle but outside both circles.
Four regions, and every element of U sits in exactly one of them, including the space outside both circles.
Name each region with two yes-or-no questions: is it in $A$ ? is it in $B$ ? Write $A$ for yes and $A^{'}$ for no, and join the answers with $\cap$ . In $A$ but not $B$ is $A \cap B^{'}$ . In both is $A \cap B$ . In $B$ but not $A$ is $A^{'} \cap B$ . In neither is $A^{'} \cap B^{'}$ , which is the same region as ${(A \cup B)}^{'}$ .
To place an element, ask the same two questions about it. Every element of $U$ goes in exactly one region. The elements in neither set go in the rectangle, outside both circles. Leaving them off the diagram is the same as forgetting part of $U$ .
To shade an expression, decide which regions it contains, then shade exactly those. $A \cup B$ is three regions: $A$ only, both, and $B$ only. $A^{'}$ is two regions: $B$ only, and outside both circles.
graph diagram — not drawn yet
Three small two-set Venn diagrams side by side, each a rectangle U with overlapping circles A (left) and B (right). Left, titled 'A ∪ B': both circles fully shaded, the outside unshaded. Centre, titled 'A′': the part of B outside A and the space outside both circles shaded, all of circle A unshaded. Right, titled 'A′ ∪ B': everything shaded except the part of A outside B.
Shade region by region. A′ includes the outside, and A′ ∪ B leaves out only 'A only'.
The awkward ones join a complement to a set. $A^{'} \cup B$ means 'not in $A$ , or in $B$ '. Test each region. 'Both' is in $B$ , so it passes. ' $B$ only' and 'outside' are not in $A$ , so they pass. Only ' $A$ only' fails both tests. So you shade everything except the part of $A$ outside $B$ .
Two relationships have their own picture. If $A \subseteq B$ , draw the circle for $A$ completely inside the circle for $B$ . If $A \cap B = ⌀$ , draw the two circles apart, with no overlap.
graph diagram — not drawn yet
Two rectangles, each labelled U, side by side. Left: a small circle A drawn completely inside a larger circle B, with the label 'A ⊆ B' under the rectangle. Right: two circles A and B that do not touch or overlap, with the label 'A ∩ B = ∅' under the rectangle.
A subset sits inside the other circle. Sets with nothing in common do not overlap.
To describe a shaded region in notation, go the other way. List which regions are shaded, then build an expression that gives exactly those. Test it against all four regions, including the outside, before you write it down.

Part 6 of 7

Venn diagrams with numbers

A region can hold a count, how many elements are in it, instead of the elements themselves. All of the arithmetic depends on one fact: $n (A)$ is the total for the whole circle $A$ , overlap included.
So the region ' $A$ only' is not $n (A)$ . It is $n (A)$ with the overlap taken away.
$n (A \cap B^{'}) = n (A) - n (A \cap B)$
Work from the inside out. Write the overlap first. Then find each 'only' region by subtraction. Find the outside region last, as $n (U)$ minus everything inside the circles.
Adding $n (A)$ and $n (B)$ counts the overlap twice, once in each circle. Take it away once and you get the number in the union.
$n (A \cup B) = n (A) + n (B) - n (A \cap B)$
A quick test: $n (A) = 12$ , $n (B) = 9$ and $n (A \cap B) = 4$ . The formula gives $n (A \cup B) = 12 + 9 - 4 = 17$ . The regions are $12 - 4 = 8$ , then $4$ , then $9 - 4 = 5$ , and $8 + 4 + 5 = 17$ agrees.
When the overlap is not given, call it $x$ . Write every region in terms of $x$ , then use the fact that all the regions together make $n (U)$ . That gives an equation in $x$ to solve.
Before you give a final answer, add up every region. The total must be $n (U)$ , and every region must be a whole number that is not negative. A negative region means a subtraction went the wrong way.

Part 7 of 7

Three-set Venn diagrams

Three overlapping circles $A$ , $B$ and $C$ cut the rectangle into eight regions: one in all three sets, three in exactly two, three in exactly one, and one outside all three.
graph diagram — not drawn yet
A rectangle labelled U with three overlapping circles A (top left), B (top right) and C (bottom), making eight regions. Label each region: 'A ∩ B ∩ C' in the centre; 'A ∩ B ∩ C′', 'A ∩ B′ ∩ C' and 'A′ ∩ B ∩ C' in the three regions where exactly two circles overlap; 'A ∩ B′ ∩ C′', 'A′ ∩ B ∩ C′' and 'A′ ∩ B′ ∩ C' in the three regions inside one circle only; '(A ∪ B ∪ C)′' outside all three circles.
Eight regions: 1 in all three, 3 in exactly two, 3 in exactly one, 1 outside.
Name a region with three yes-or-no questions. In $A$ , in $B$ , not in $C$ is $A \cap B \cap C^{'}$ . That is the part of the overlap of $A$ and $B$ that lies outside $C$ .
Counts for three sets overlap each other. $n (A \cap B)$ counts everything in both $A$ and $B$ , including what is also in $C$ . So fill the centre, $n (A \cap B \cap C)$ , before anything else.
Next fill the three 'exactly two' regions. Each one is a two-set total minus the centre.
$n (A \cap B \cap C^{'}) = n (A \cap B) - n (A \cap B \cap C)$
Then fill the three 'one set only' regions. Each one is the circle's total minus the three regions already inside that circle: two 'exactly two' regions and the centre.
Last, the outside: $n (U)$ minus the sum of the seven regions inside the circles. Then check that all eight regions add up to $n (U)$ .
If the centre is unknown, call it $x$ , exactly as with two sets. Each 'exactly two' region becomes a total minus $x$ , each 'one set only' region becomes an expression in $x$ , and the eight regions summed equal $n (U)$ .
Read the wording exactly. 'Exactly one' means the three one-set-only regions added. 'At least two' means the three 'exactly two' regions plus the centre. 'Not in $A$ ' is $A^{'}$ , and that includes the region outside all three circles.

Worked examples

Example 1. $U = {x | x is an integer, 1 ⩽ x ⩽ 12}$ , $A = {x | x is a multiple of 3}$ and $B = {x | x is a factor of 12}$ . (a) Write down the elements of $A$ and of $B$ . (b) Write down $A \cap B$ and $A \cup B$ . (c) Write down $A^{'}$ . (d) Find $n ({(A \cup B)}^{'})$ . (e) Use set notation to describe the set ${9}$ . (f) Is $B \subseteq A$ ? Give a reason.

Show the working

Start with $U$ . It is ${1, 2, 3, \dots, 12}$ . Every set in the question is made from these 12 numbers, so $A$ stops at 12.
(a) The multiples of 3 up to 12, and the factors of 12.
$A = {3, 6, 9, 12} B = {1, 2, 3, 4, 6, 12}$
(b) For the intersection, go through $A$ and keep what is also in $B$ . 9 is not a factor of 12, so it goes.
$A \cap B = {3, 6, 12}$
For the union, take everything in either set, each element once.
$A \cup B = {1, 2, 3, 4, 6, 9, 12}$
(c) $A^{'}$ is every element of $U$ that is not in $A$ .
$A^{'} = {1, 2, 4, 5, 7, 8, 10, 11}$
(d) ${(A \cup B)}^{'}$ is everything in $U$ outside $A \cup B$ , which is ${5, 7, 8, 10, 11}$ . The question asks for $n$ , so the answer is the number 5. Check: $n (A \cup B) = 7$ and $7 + 5 = 12 = n (U)$ .
(e) Ask the two questions about 9. Is it in $A$ ? Yes. Is it in $B$ ? No. So 9 is in $A \cap B^{'}$ , and it is the only element there.
graph diagram — not drawn yet
A rectangle labelled U with overlapping circles A (left) and B (right). The overlap contains 3, 6, 12. The part of A outside B contains 9. The part of B outside A contains 1, 2, 4. Outside both circles, inside the rectangle: 5, 7, 8, 10, 11.
All twelve numbers, each written once. 9 is alone in A ∩ B′.
(f) To show a set is not a subset, find one element of $B$ that is not in $A$ . $1 \in B$ but $1 \notin A$ , so $B ⊈ A$ . One counter-example is the whole reason.

Answer: (a) $A = {3, 6, 9, 12}$ , $B = {1, 2, 3, 4, 6, 12}$ (b) $A \cap B = {3, 6, 12}$ , $A \cup B = {1, 2, 3, 4, 6, 9, 12}$ (c) $A^{'} = {1, 2, 4, 5, 7, 8, 10, 11}$ (d) $5$ (e) $A \cap B^{'}$ (f) No: $B ⊈ A$ , because $1 \in B$ but $1 \notin A$ .

Example 2. In a class of $40$ students, $25$ study French, $21$ study Spanish and $4$ study neither language. Find the number of students who study both languages, and the number who study exactly one.

Show the working

Let $x$ be the number who study both. Put $x$ in the overlap first, because the other regions are built from it.
French only is the French total minus the overlap, $25 - x$ . Spanish only is $21 - x$ . The 4 who study neither go outside both circles.
The four regions together make the whole class.
$(25 - x) + x + (21 - x) + 4 = 40$
Collect terms: $50 - x = 40$ , so $x = 10$ .
Now the 'only' regions: French only $= 25 - 10 = 15$ and Spanish only $= 21 - 10 = 11$ . Exactly one language means French only plus Spanish only: $15 + 11 = 26$ .
Check that the regions make the class: $15 + 10 + 11 + 4 = 40$ . Correct.
The union formula gives the same result. $n (F \cup S) = 40 - 4 = 36$ , and $36 = 25 + 21 - x$ , so $x = 10$ .

Answer: $10$ students study both languages, and $26$ study exactly one.

Example 3. $50$ members of a sports club are asked whether they play football ( $F$ ), hockey ( $H$ ) or tennis ( $T$ ). $3$ play all three. $8$ play football and hockey, $6$ play football and tennis, and $5$ play hockey and tennis. $20$ play football, $18$ play hockey and $15$ play tennis. (a) Draw a Venn diagram to show this information. (b) Find the number of members who play none of the three. (c) Find the number who play exactly one of the sports. (d) Find $n (F \cap H^{'})$ .

Show the working

The 8 who play football and hockey include the 3 who play all three. So start in the centre with 3, before any other region.
Exactly-two regions: each pair total minus the centre.
$F \cap H only = 8 - 3 = 5 F \cap T only = 6 - 3 = 3 H \cap T only = 5 - 3 = 2$
One-sport regions: each circle's total minus the three regions already inside it.
$F only = 20 - 5 - 3 - 3 = 9 H only = 18 - 5 - 2 - 3 = 8 T only = 15 - 3 - 2 - 3 = 7$
(b) Inside the circles: $3 + 5 + 3 + 2 + 9 + 8 + 7 = 37$ . Outside all three: $50 - 37 = 13$ .
graph diagram — not drawn yet
A rectangle labelled U with three overlapping circles F (top left), H (top right) and T (bottom). Centre, in all three circles: 3. In F and H only: 5. In F and T only: 3. In H and T only: 2. In F only: 9. In H only: 8. In T only: 7. Inside the rectangle, outside all three circles: 13.
Filled from the centre out. The eight numbers add up to 50.
(c) Exactly one sport is the three one-sport regions added: $9 + 8 + 7 = 24$ .
(d) $F \cap H^{'}$ means plays football and does not play hockey. Inside the $F$ circle, two regions are outside $H$ : football only ( $9$ ) and football-and-tennis only ( $3$ ). So $n (F \cap H^{'}) = 9 + 3 = 12$ .
Check (d): the members of $F$ either play hockey or do not, so $n (F \cap H^{'}) + n (F \cap H) = 12 + 8 = 20 = n (F)$ . Correct.

Answer: (a) Centre $3$ ; $F$ and $H$ only $5$ , $F$ and $T$ only $3$ , $H$ and $T$ only $2$ ; $F$ only $9$ , $H$ only $8$ , $T$ only $7$ ; outside $13$ (b) $13$ (c) $24$ (d) $12$

Example 4. (a) $P = {x | x is an integer, - 3 < x ⩽ 2}$ . Write down the elements of $P$ , and $n (P)$ . (b) $L = {(x, y) | y = 2 x + 1}$ and $M = {(x, y) | y = 7 - x}$ . Find $L \cap M$ . (c) $N = {(x, y) | y = 2 x - 4}$ . Write down $L \cap N$ and give a reason.

Show the working

(a) $P$ is the integers greater than $- 3$ and at most $2$ . $- 3$ is out, because the sign is $<$ . $2$ is in, because the sign is $⩽$ .
$P = {- 2, - 1, 0, 1, 2}, n (P) = 5$
Zero is one of the elements. Listing $- 2, - 1, 1, 2$ and stopping would give $n (P) = 4$ , which is wrong.
(b) An element of $L \cap M$ is a point on both lines, so its $y$ fits both equations. Set the two expressions for $y$ equal to each other.
$2 x + 1 = 7 - x$
Add $x$ to both sides and subtract 1: $3 x = 6$ , so $x = 2$ . Then $y = 2 (2) + 1 = 5$ . Check in the other equation: $7 - 2 = 5$ . Correct.
The intersection is a set, and its only element is the point $(2, 5)$ .
$L \cap M = {(2, 5)}$
(c) $L$ and $N$ both have gradient 2, but they cross the $y$ -axis at different places, $1$ and $- 4$ . So the lines are parallel, no point is on both, and $L \cap N = ⌀$ .
If you try to solve anyway, $2 x + 1 = 2 x - 4$ gives $1 = - 4$ , which is never true. An equation with no solution is the algebra telling you the intersection is empty.

Answer: (a) $P = {- 2, - 1, 0, 1, 2}$ , $n (P) = 5$ (b) $L \cap M = {(2, 5)}$ (c) $L \cap N = ⌀$ , because the lines have the same gradient, 2, and different intercepts, so they are parallel and never meet.

Where people go wrong

Mistake: Writing a set's total in its 'only' region, e.g. putting $n (A) = 25$ in the part of circle $A$ outside the overlap.

Instead: Fill the overlap first. The ' $A$ only' region is $n (A) - n (A \cap B)$ . With three sets, subtract all three other regions inside that circle.

$n (A)$ counts the whole circle, overlap included. Writing it in one region counts the overlap twice, and the regions no longer add up to $n (U)$ .

Mistake: Swapping $\cup$ and $\cap$ , or leaving the overlap out of $A \cup B$ .

Instead: $A \cap B$ ('and') is only what is in both. $A \cup B$ ('or') is everything in either set, overlap included, each element written once.

In sets, 'or' includes 'both'. A union is never smaller than either set, and an intersection is never bigger than either set, so a quick size check catches a swap.

Mistake: Answering an $n (\dots)$ question with a set, or a set question with a number, e.g. $n (A^{'}) = {1, 3, 5}$ , or $A \cap B = 3$ .

Instead: $n (\dots)$ asks how many, so the answer is one number. An expression such as $A \cap B$ is a set, so list its elements in curly brackets.

A set and a number are different kinds of object. ${3}$ is a set with one element, and $n ({3}) = 1$ , not 3.

Mistake: Leaving the elements that are in neither set off the Venn diagram, or out of a complement.

Instead: Every element of $U$ belongs somewhere. Those in no circle go in the rectangle outside the circles, and they are part of $A^{'}$ , of $B^{'}$ and of ${(A \cup B)}^{'}$ .

$A^{'}$ means 'in $U$ but not in $A$ ', not 'in $B$ but not in $A$ '. Without the outside region, $n (A) + n (A^{'})$ would not equal $n (U)$ .

Mistake: Listing ${x | 1 ⩽ x ⩽ 4}$ as ${1, 2, 3, 4}$ .

Instead: List only when the rule says $x$ is an integer (or a natural number). Without that, the set is every number from 1 to 4, including $1.5$ and $\sqrt{2}$ , and it cannot be listed.

A letter such as $x$ stands for any number unless the rule restricts it. The word 'integer' is the restriction.

Check yourself

$A = {2, 4, 6, 8}$ and $B = {1, 2, 3, 4}$ . Write down $A \cap B$ and $A \cup B$ .
Answer
$A \cap B = {2, 4}$ and $A \cup B = {1, 2, 3, 4, 6, 8}$ .
$U = {x | x is an integer, 1 ⩽ x ⩽ 10}$ and $A = {1, 3, 5, 7, 9}$ . Write down $A^{'}$ and $n (A^{'})$ .
Answer
$A^{'} = {2, 4, 6, 8, 10}$ and $n (A^{'}) = 5$ .
Write down the elements of ${x | x is an integer, - 2 < x ⩽ 3}$ . Can you do the same for ${x | - 2 < x ⩽ 3}$ ?
Answer
${- 1, 0, 1, 2, 3}$ . No: without 'integer', $x$ can be any number in the interval, such as $0.5$ , so the set is infinite and cannot be listed.
Fill each gap with $\in$ , $\notin$ , $\subseteq$ or $⊈$ . (a) $7$ ... ${odd numbers}$ (b) ${2, 9}$ ... ${prime numbers}$ (c) $⌀$ ... ${1, 2}$ (d) $0$ ... $⌀$
Answer
(a) $\in$ (b) $⊈$ , because 9 is not prime (c) $\subseteq$ , because the empty set is a subset of every set (d) $\notin$ , because the empty set has no elements.
$n (A) = 18$ , $n (B) = 12$ and $n (A \cup B) = 25$ . Find $n (A \cap B)$ .
Answer
$25 = 18 + 12 - n (A \cap B)$ , so $n (A \cap B) = 30 - 25 = 5$ .
In a two-set Venn diagram, which of the four regions make up $A^{'} \cup B$ ?
Answer
Three regions: both, $B$ only, and outside both circles. Only ' $A$ only' is left out, because it is the one region that is in $A$ and not in $B$ .
In a three-set Venn diagram for $A$ , $B$ and $C$ , write in set notation the region that is inside $C$ only.
Answer
$A^{'} \cap B^{'} \cap C$ : not in $A$ , not in $B$ , in $C$ .
$A = {(x, y) | y = x + 1}$ and $B = {(x, y) | y = 5 - x}$ . Find $A \cap B$ .
Answer
$x + 1 = 5 - x$ , so $2 x = 4$ , $x = 2$ and $y = 3$ . $A \cap B = {(2, 3)}$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. $U = {x | x is an integer, 1 ⩽ x ⩽ 15}$ , $P = {x | x is a prime number}$ and $M = {x | x is a multiple of 3}$ .

(a) Write down the elements of $P$ . 1 mark No calculator

Mark scheme

B1 {2, 3, 5, 7, 11, 13} cao (1 is not prime)

(b) Write down $n (M)$ . 1 mark No calculator

Mark scheme

B1 5 cao (M = {3, 6, 9, 12, 15}); a list of the elements instead of a number is not the answer

(c) Write down the elements of $P \cap M^{'}$ . 1 mark No calculator

Mark scheme

B1 {2, 5, 7, 11, 13} cao

(d) Write down $n ({(P \cup M)}^{'})$ . 1 mark No calculator

Mark scheme

B1 5 cao ((P ∪ M)′ = {1, 4, 8, 10, 14})

(e) Write the set ${1, 4, 8, 10, 14}$ in set notation, using $P$ and $M$ . 1 mark No calculator

Mark scheme

B1 (P ∪ M)′ oe, e.g. P′ ∩ M′

(f) Complete each statement using one of the symbols $\in$ , $\notin$ , $\subseteq$ or $⊈$ . (i) $9$ ... $P$ (ii) ${6, 12}$ ... $M$ 2 marks No calculator

Mark scheme

B1 (i) ∉
B1 (ii) ⊆

Question 2. In a group of $50$ students, each student was asked whether they do athletics ( $A$ ), swimming ( $S$ ) or cycling ( $C$ ). $x$ students do all three. $10$ do athletics and swimming, $7$ do athletics and cycling, and $9$ do swimming and cycling. $22$ do athletics, $25$ do swimming and $20$ do cycling. $5$ students do none of the three.

(a) Draw a Venn diagram to show this information. Write an expression in terms of $x$ in each region. 3 marks No calculator

Mark scheme

B1 x in the centre, with 10 − x, 7 − x and 9 − x in the correct exactly-two regions
B1 5 + x, 6 + x and 4 + x in the correct one-set-only regions (A only, S only, C only) oe, e.g. 22 − (10 − x) − (7 − x) − x
B1 5 inside the rectangle, outside all three circles

(b) Show that $x = 4$ . 2 marks No calculator

Mark scheme

M1 the sum of their eight regions equated to 50, e.g. x + (10 − x) + (7 − x) + (9 − x) + (5 + x) + (6 + x) + (4 + x) + 5 = 50; FT their (a); or a correct alternative method, e.g. 22 + 25 + 20 − 10 − 7 − 9 + x = 50 − 5
A1 x + 46 = 50 oe, leading to x = 4 with no errors seen

(c) Find the number of students who do exactly one of the three activities. 2 marks No calculator

Mark scheme

M1 (5 + 4) + (6 + 4) + (4 + 4) oe, i.e. 9 + 10 + 8
A1 27 cao

(d) Find $n (A^{'} \cap (S \cup C))$ . 1 mark No calculator

Mark scheme

B1 23 cao (S only 10, C only 8, S and C only 5)

Question 3. $A = {(x, y) | y = 3 x - 2}$ , $B = {(x, y) | x + y = 6}$ and $C = {(x, y) | y = 3 x + 5}$ .

(a) Find $A \cap B$ . 3 marks No calculator

Mark scheme

M1 3x − 2 = 6 − x oe, or y = 3x − 2 substituted into x + y = 6
A1 x = 2
A1 {(2, 4)} oe; accept (2, 4) or x = 2, y = 4

(b) Explain why $A \cap C = ⌀$ . 2 marks No calculator

Mark scheme

B1 both lines have gradient 3 (the same gradient) and different y-intercepts, −2 and 5
B1 so the lines are parallel and no point lies on both; or 3x − 2 = 3x + 5 leads to −2 = 5, which is impossible, so no point is in both sets

How this is examined

Filling a three-set Venn diagram from totals that overlap, then answering 'how many' questions from it: A context with three groups (sports, subjects, clubs) giving each set's total, each pair's total and the number in all three or none. The centre may be an unknown $x$ . You complete the diagram, may be asked to Show that $x$ has a given value, then find counts such as 'exactly one'. Centre first; each exactly-two region as the pair total minus the centre; each one-set region as the circle total minus the three regions already in it. For a Show that: an equation with all eight regions summing to $n (U)$ , simplified line by line to the given value. Each count answered as a single number, checked by adding all eight regions.
Finding an unknown overlap in a two-set problem: A context giving $n (U)$ , $n (A)$ , $n (B)$ and the number in neither set, asking how many are in both, then how many are in one set only An equation such as $(25 - x) + x + (21 - x) + 4 = 40$ , or $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ with the numbers substituted, then the value of $x$ . Any 'only' region is then the set total minus $x$ .
Turning set-builder notation into a list, and finding where two lines written as sets meet: A set such as ${x | x is an integer, - 3 < x ⩽ 2}$ to Write down in full, or two sets of points ${(x, y) | y = \dots}$ whose intersection you Find For a list: every integer in the interval, each end checked against $<$ or $⩽$ , and $0$ not missed. For lines: the two expressions for $y$ set equal, solved, and the point given as the single element, e.g. ${(2, 5)}$ . Parallel lines (same gradient, different intercept) give $⌀$ , with that reason stated.
Reading set notation and answering in it: A run of 1-mark Write down parts on one universal set: the elements of a set such as $A \cap B^{'}$ , a value of $n (\dots)$ , a gap to fill with $\in$ , $\notin$ , $\subseteq$ or $⊈$ , or a given list of elements to describe in notation Exactly the elements asked for, in curly brackets, with nothing extra. A single number for $n (\dots)$ . The outside region included in any complement. $\in$ only between an element and a set, and $\subseteq$ only between two sets.
Shading the region an expression describes: A blank two-set or three-set Venn diagram and an expression such as $A^{'} \cup B$ or ${(A \cup B)}^{'}$ to shade, 1 mark per diagram Exactly the right regions shaded and no others. Decide region by region with yes-or-no questions (in $A$ ? in $B$ ?) before you shade anything, and do not forget the region outside the circles.

Words to know

Set: A collection of objects, called elements, written in curly brackets, e.g. ${2, 4, 6}$ .
Element: One member of a set. $x \in A$ means $x$ is an element of $A$ , and $x \notin A$ means it is not.
n(A): The number of elements in set $A$ . It is always a number, never a set.
Empty set: The set with no elements, written $⌀$ , so $n (⌀) = 0$ . The set ${0}$ is not empty: it has one element.
Universal set: The set $U$ of everything being considered in a question. It is the rectangle in a Venn diagram.
Complement: $A^{'}$ is the set of elements of $U$ that are not in $A$ .
Subset: $A \subseteq B$ means every element of $A$ is also in $B$ . $A ⊈ B$ means at least one element of $A$ is not in $B$ .
Intersection: $A \cap B$ is the set of elements in both $A$ and $B$ .
Union: $A \cup B$ is the set of elements in $A$ or $B$ or both.
Set-builder notation: A set described by a rule, e.g. ${x | x is an integer, 1 ⩽ x ⩽ 5}$ . The bar $|$ is read 'such that'.
Infinite set: A set with infinitely many elements, such as ${0, 1, 2, 3, \dots}$ or ${x | 2 ⩽ x ⩽ 5}$ . It cannot be listed in full.
Venn diagram: A diagram with a rectangle for $U$ and a circle for each set. Overlapping circles show elements in more than one set.
Region: One of the separate areas a Venn diagram is cut into: four for two sets, eight for three sets.

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