Ideas beyond Number
Teacher’s guide to
Activity worksheets
Learning objectives
To explore reasoning, logic and proof through practical, experimental, structured and
formalised methods of communication for others to follow.
To have FUN exploring PROOF
To present exciting examples for others to follow.
Activity sheet 1
These questions were taken from the Level 6-8 SAT paper 2003, and we will use them as a
warm-up to revisit True/False statements and Counter-Examples as methods of PROOF.
Task 1
Some students started to solve this equation in different ways:
For each statement tick True or False:
+ =
=
+
+ =
+
=
+
+
+ =
=
+
+
+ =
+
+ =
+
+ =
+ =
=
+
− =−
Task 2: Counter-examples – The exception disproves the rule
Find a COUNTER-EXAMPLE to show that each of these CONJECTURES is FALSE:
1. Square numbers only end in , ,
or .
2. Cube numbers can end in any digit except .
3. The product of two numbers is greater than either of the two numbers.
4. The square of the number is greater than the number.
5. Adding two numbers and then squaring them gives the same result as squaring them and then adding
them.
6. Division always results in a smaller number.
7. Every whole number is either a cube number or is the sum of , ,
8. The sum of two numbers is greater than their difference.
9. The sum of two numbers is always greater than zero.
10. The product of two numbers is greater than their sum.
Teacher’s guide to
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Activity worksheets
or
cube numbers.
Lots of counterexamples, but here are some.
1.
2
3.
4.
5.
, whereas
6.
7. Huh? This seems to be true!
8
, whereas
9.
10.
, whereas
Task 3: Developing algebraic reasoning
Decide whether each of the following are:
♦
always true
♦
sometimes true
♦
never true.
Convince a friend
1. Three consecutive numbers sum to a multiple of three.
2. Three consecutive numbers sum to a multiple of 6.
3. For three consecutive numbers, the product of the first and the last is 1 less than the middle squared.
4. The sum of two consecutive triangle numbers is a square number.
5. An odd number added to an odd number always gives an even number.
6. When you add together consecutive odd numbers you get a square number.
7. Add 1 to the square of an odd number. You always get an odd number.
Hint: the technique is:
A.
Try some examples and see if you can find cases that comply – then the conjecture is true sometimes.
B.
Then to look for a counter example and if you find one then the conjecture is not always true.
C.
If you cannot find a counter example fairly quickly, then in these cases algebra should provide a more
formal PROOF.
1. This one’s always true. Proof:
2. This one’s sometimes true. For example,
, which is a multiple of
. but
, which isn’t.
3. This one’s always true. Proof:
, whereas
.
4. This one’s always true. Proof: how much do they know about triangular numbers?
5. This one’s always true. Proof:
6. This one’s sometimes true. For example,
.
, but
.
7. This one’s never true. Actually, an odd number squared is always odd, so when you add 1
to it, you always get an even number.
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Activity worksheets
Activity sheet 2
Further development of algebraic reasoning
Task 1
represents a square number;
is an integer.
1. Think about the expression
. Explain how you know there are values of
for which this expression
does not represent a square number.
2. Explain why the expression
must represent a square number.
Answer
1. Counterexample
, which is not square.
2. 16 is a square number, and so is
.
is equal to
.
Task 2
Kate is solving the inequality
. She says: “
whenever
is less than .”
Kate is not correct. Explain why.
Answer
Counterexample:
equal to
Teacher’s guide to
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.
will do just as well.
Activity worksheets
, which is not less than . Any number less than or
Task 3: Teacher-led activity
Your teacher will give you a statement – investigate it! Is it true? If not, can you find a counterexample? If so,
can you prove it?
Present the statement to pupils in either of the following (equivalent) ways:
“All prime numbers greater than
can be expressed in the form
, where
is a
positive whole number.”
or
“Every prime number greater than
is either one more than or one less than a
multiple of .”
Give students some time to investigate the above statement (preferably using Excel) and
encourage them to come up with a proof for this proposition.
Investigation in Excel:
n
6n - 2
0
1
2
3
4
5
6
7
8
9
10
11
12
6n
6n - 1
4
10
16
22
28
34
40
46
52
58
64
70
5
11
17
23
29
35
41
47
53
59
65
71
6n + 1
6
12
18
24
30
36
42
48
54
60
66
72
6n + 2
6n + 3
1
7
13
19
25
31
37
43
49
55
61
67
73
2
8
14
20
26
32
38
44
50
56
62
68
74
3
9
15
21
27
33
39
45
51
57
63
69
75
The results show that:
(i) All prime numbers so far tested (apart from
and ) lie either in the
column or the
column. (There are no primes in the other columns.)
(ii) There are, however, non-primes in the
column and the
column.
So far all we can see is that some examples fit the conjecture. We have not yet proved that
all primes fit the conjecture. This would take forever!
Suggested proof:
Start by noticing that every whole number can be expressed in the form
,
Teacher’s guide to
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or
(as in the Excel spreadsheet above).
Activity worksheets
,
,
,
Then, notice the following facts:
(i)
is always divisible by , for all values of
(so none of the numbers in this column can
be prime).
(ii)
is always divisible by , for all values of
can be prime either, except
itself).
(iii) The same is true of
.
(iv)
is always divisible by , for all values of
can be prime, except
(so none of the numbers in this column
(so none of the numbers in this column
itself).
So all the primes greater than
must lie in the
and
columns.
Talking point:
This is quite a good opportunity to introduce the idea of a proposition and its converse. Here,
the proposition is that all primes greater than
are of the form
proposition in reverse: that all numbers of the form
. The converse is the
are prime. The converse, here, is
false: it’s easy to find numbers of this form that aren’t prime.
Suggested extension:
What’s so special about
? Will any other formula do this job for us? Investigate using
Excel.
The answer seems to be “Yes, but mostly less well”. For example, the statement “All primes
greater than
are of the form
or
” is true, and worth the students’ investigating,
but when you really think about it, it amounts to nothing more than saying “All primes greater
than
are odd.”
This includes all the primes all right, but does a worse job of excluding the non-primes. The
formula
fits two out of every six numbers, about 33%. The formula “
or
”
fits four out of every eight, fully 50%. It’s doing less well at pinning the primes down.
We did find the following improvement, though: “All primes greater than
,
,
or
are of the form
” This fits eight out of every 30 numbers, about 27%. A
very good pinning-down…
Further questions that occurred to us:
Is this really an improvement? What are its disadvantages?
What’s so special about
and
? How could we push this even further?
How about more complicated formulae? (This is a really rich area to mess around in.)
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Activity sheet 3
Task1: The Wason Test
Your teacher will introduce you to the Wason Test.
How did you do on the Wason Test? How did your classmates do?
What do you think this shows about human beings and logic?
The Wason Test
The Wason Test is a warm-up activity that can be used to get students thinking about logical
reasoning. It comes in two versions. It might be worth giving half the class version 1 to do
followed by version 2, and the other half the other way round. It requires the Wason Cards:
you can find instructions for making the Wason Cards on the Harrow Maths website.
Version 1 of the Wason Test
Four cards are put in front of the student with the following sides facing up.
COKE
19 17
WINE
Each of these cards represents a customer in a bar. As the customer enters the bar they
write down their age and what they are going to drink. The owner needs to check that:
only persons over 18 are drinking alcohol.
Those over 18 can drink soft drinks if they want, but under-18s can drink only non-alcoholic
drinks. By turning over only two cards, how can the owner decide if the law is being
upheld?
Correct answer
The owner should turn over the cards with the 17 facing up and the WINE facing up (asking:
what’s the underage person drinking, and who’s got the wine?)
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Version 2 of the Wason Test
Four cards are put in front of the student. Each card, the students are told, has a letter on
one side and a number on the other. The cards appear as follows:
A
3
4
D
Tell the students they need to test whether:
every card with a D on one side has a 3 on the other.
By turning over only two cards, how can we decide whether this conjecture is true?
Correct answer
In order to test the conjecture students need to turn over the card with the D facing up and
the card with the 4 facing up.
Talking points
The Psychology literature suggests that most people find Version 2 of the Wason test very
difficult, and Version 1 much easier (even though logically speaking they’re equivalent
problems). It will be interesting to see whether your students are typical! Perhaps the order
the test is given in matters (this isn’t strictly maths, but it’s fascinating).
The usual error people make with Version 2 is to turn over the 3 instead of the 4. The reason
this is an error is that we don’t care, for the purposes of this problem, what letter the “3” card
has on its other side. We’re checking whether:
every card with a D on one side has a 3 on the other,
and for that we need to do just two things: check all the Ds to make sure they’re 3s, and
check every number that isn’t a 3 to make sure it isn’t a D.
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Activity worksheets
If we turn over the 3 card, we’re not checking this rule at all. We’re checking its converse:
every card with a 3 on one side has a D on the other.
And of course, we haven’t been asked to check the converse!
This is a well-known error in logical argument, known as Affirming the Consequent. The
Wason Test shows how easy a mistake this is to make (at least in certain circumstances – in
others, as Version 1 is meant to show, we find it rather easier).
Task 2: Three digits
In groups of three?
1. Take any 3 digits.
2. From these digits make six different number pairs.
3. Add them up.
4. Add the three original digits.
5. Divide the larger of the last two answers by the smaller.
6. Write down the result.
7. Choose another set of three digits and repeat.
8. Write down briefly what you find.
9. Try to prove it will always be true.
Examples of answers for the Three Digits task
Step 1.
2, 6, 5
1, 7, 8
Step 2.
25, 26, 52, 56, 62, 65
17, 18, 71, 78, 81, 87
Step 3.
286
352
Step 4.
13
16
Step 5.
22
22
Proof that the answer is always 22
Let the three single-digit numbers be ,
and .
Then our six two-digit numbers are
,
The total is
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,
,
,
and
.
Task 3: Consecutive sums
In groups of three?
Investigate.
Are there some whole numbers that cannot be
expressed as the sum of two or more
consecutive numbers?
Which are they? Why?
etc
This is a rather testing activity. Finding the pattern is relatively straightforward – but this is a
module on proving, and students will need some support with that process.
We can spot a pattern readily in the first few cases:
can’t be expressed as a consecutive sum
can’t be expressed as a consecutive sum
can’t be expressed as a consecutive sum
etc
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Conjecture
The numbers that can’t be expressed as consecutive sums are the powers of two.
Proof strategies
This conjecture is indeed true, but proving it is difficult. Here are some suggestions for
offering students support.
(i) Encourage students to think about the set of numbers that can be expressed as
consecutive sums. Of course, one way of describing this set is “everything that isn’t a power
of 2”; but encourage them to think of a more positive way of describing it.
Now, the thing about the powers of two is that all their factors, other than 1, are even. So the
non-powers of 2 are numbers that have odd factors other than 1. In other words, all the
multiples of all the odd numbers other than 1.
(ii) Once you’ve reached agreement on that, suggest that the students draw up a table of
multiples of odd numbers, like this one.
×
1
2
3
4
5
6
7
8
9
3
3
6
9
12
15
18
21
24
27
5
5
10
15
20
25
30
35
40
45
7
7
14
21
28
35
42
49
56
63
9
9
18
27
36
45
54
63
72
81
11
11
22
33
44
55
66
77
88
99
13
13
26
39
52
65
78
91
104
117
15
15
30
45
60
75
90
105
120
135
17
17
34
51
68
85
102
119
136
153
19
19
38
57
76
95
114
133
152
171
(This one was done in Excel: once you’ve got the row and column that appear in bold—row 1
and column A, that is—type the formula =B$1*$A2 into cell B2, and copy into all the other
cells of the table.)
(iii) The 2-sums are:
etc. This gives us all odd numbers apart from . Get the students to indicate these numbers
in the table, say by shading:
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Activity worksheets
×
1
2
3
4
5
6
7
8
9
3
3
6
9
12
15
18
21
24
27
5
5
10
15
20
25
30
35
40
45
7
7
14
21
28
35
42
49
56
63
9
9
18
27
36
45
54
63
72
81
11
11
22
33
44
55
66
77
88
99
13
13
26
39
52
65
78
91
104
117
15
15
30
45
60
75
90
105
120
135
17
17
34
51
68
85
102
119
136
153
19
19
38
57
76
95
114
133
152
171
(iv) The 3-sums are:
etc. This gives us all multiples of
apart from
itself—but of course
is already covered as
a 2-sum. Again, we can show this in the table by shading:
×
1
2
3
4
5
6
7
8
9
3
3
6
9
12
15
18
21
24
27
5
5
10
15
20
25
30
35
40
45
7
7
14
21
28
35
42
49
56
63
9
9
18
27
36
45
54
63
72
81
11
11
22
33
44
55
66
77
88
99
13
13
26
39
52
65
78
91
104
117
15
15
30
45
60
75
90
105
120
135
17
17
34
51
68
85
102
119
136
153
19
19
38
57
76
95
114
133
152
171
(v) The 4-sums are:
etc. This gives us all odd multiples of
apart from
and
; but
is already a 3-sum.
Shading:
×
1
2
3
4
5
6
7
8
9
Teacher’s guide to
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3
3
6
9
12
15
18
21
24
27
5
5
10
15
20
25
30
35
40
45
Activity worksheets
7
7
14
21
28
35
42
49
56
63
9
9
18
27
36
45
54
63
72
81
11
11
22
33
44
55
66
77
88
99
13
13
26
39
52
65
78
91
104
117
15
15
30
45
60
75
90
105
120
135
17
17
34
51
68
85
102
119
136
153
19
19
38
57
76
95
114
133
152
171
(iv) The 5-sums are:
etc. This gives us all multiples of
and
apart from
and
; but of course
is already a 2-sum,
is already a 4-sum. Shading:
×
1
2
3
4
5
6
7
8
9
3
3
6
9
12
15
18
21
24
27
5
5
10
15
20
25
30
35
40
45
7
7
14
21
28
35
42
49
56
63
9
9
18
27
36
45
54
63
72
81
11
11
22
33
44
55
66
77
88
99
13
13
26
39
52
65
78
91
104
117
15
15
30
45
60
75
90
105
120
135
17
17
34
51
68
85
102
119
136
153
19
19
38
57
76
95
114
133
152
171
And so on.
In general, what we can say is:
1. If
is odd, then an -sum is always a multiple of . If
an odd multiple of
2. If
is even, then an -sum is always
. Either way, the sum always has an odd factor (other than ).
is odd (and not equal to ), then the -sums will give us all multiples of , except
.
But n is already a 2-sum, 2n is already a 4-sum,
which is already an
sum. So all multiples of
is already a 6-sum, etc—up to
are covered.
In other words, every consecutive sum has an odd factor (other than ), and all numbers with
odd factors (other than ) are consecutive sums. This shows that the numbers that can’t be
consecutive sums are, precisely, the numbers without odd factors (other than )—that is, the
powers of .
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Activity worksheets
Activity sheet 4
Consecutive sums revisited
What is the next line?
What is the
th
line?
Can you prove it?
The students might find it helpful if you start leading them through the next few. The next
couple are:
and:
Things students might notice about this pattern:
1. The first number in each line is a square number.
rd
2. The number at the start of the 3 line is
3. There is always one more term on the left than there is on the right.
4. The numbers increase consecutively.
5. The first number in the next line is always one more than the last number in the previous
line.
6. On the left side, the last term is always the first term
.
7. The first term on the right side is always one more than the last term on the left.
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Activity worksheets
The
th
line is
.
To prove this: the left hand side has got
added to a number between
terms in it, each of which consists of
and . It is therefore equal to
lots of
, plus
Its value is therefore:
.
The right hand side has got
terms in it, each of which consists of
added to a number
between 1 and . Its value is therefore:
.
Now, even if your students don’t know what
adds up to (which they may), this
sum appears on both sides, and therefore cancels. So all we really have to prove is that:
,
which is actually pretty easy…
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Activity worksheets
Activity sheet 5
However you slice it… (challenging proof activity)
Draw a circle, and place two points on its circumference. Join these points with a line. The interior of the circle is
then divided into two regions.
Now draw another, and place three points on its circumference. Join each of these points to the other two. The
interior of the circle is then divided into four regions.
3
1
1
2
2
4
How many regions are formed if you place four points on the circumference? Five? Six? ? Make a conjecture.
Test your conjecture.
The false pattern
If you draw up a table of results, it looks like this:
Number of points on circumference
Number of regions
1
2
3
4
5
It looks very much, then, as though the number of internal regions doubles each time, and
that if the number of points is , the number of regions is
.
The first counterexample
However, if the number of points is , the number of regions is not
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but
.
1
2
3
5
4
7
9
6
11
8
15
13
12
10
16
14
17
21
20
18
19
22
23
24
27
25
28
29
26
30
31
So, the truth is more complicated than it appears! This underlines, beautifully, one of the
reasons why we need proofs at all in mathematics – how many cases do you need to check
before you regard a conjecture as confirmed? The answer, as this example demonstrates, is
“all of them” – but when, as here, the number of cases is infinite, that’s not an option.
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Refine your conjecture
First, some definitions:
An external node is a point on the circumference of the circle.
A cut is a line joining two external nodes.
external node
external node
cut
An internal node is the intersection of two cuts.
An external arc is part of the circumference of the circle joining two external nodes.
An internal arc is any other line joining two nodes.
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Activity worksheets
external arc
internal arc
internal node
internal arc
A region is an enclosed area.
Draw up a table relating the number of external nodes to the number of external arcs, cuts, internal nodes,
internal arcs and regions. Look for patterns in this table. Frame some conjectures. Can you prove them?
Things you might find useful along the way
1. Getting choosy
The number of ways of choosing two objects from three is written
, and said “three choose ”. Its value is .
Similarly, the number of ways of choosing two objects from four is written
, and said “four choose ”. Its value
is .
.
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Activity worksheets
Questions
1. What are the values of
choose
and
2. What are the values of 6 choose ,
choose ?
choose
and
choose ?
2. Pascal’s triangle
This is Pascal’s triangle:
To make each number in Pascal’s triangle, add together the two directly above it: for example,
.
3. What’s the connection between Pascal’s triangle and the “choose” numbers?
4. What’s the connection between Pascal’s triangle and the “nodes, arcs and regions” table?
3. Euler’s formula for plane graphs
A plane graph is a collection of points (nodes) connected by lines (arcs) and dividing the page into regions.
2
1
3
In this example, there are 4 nodes, 6 arcs and 3 regions. Euler’s formula says that if there are
and R regions, then in every case,
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Activity worksheets
.
nodes,
arcs
5. Does Euler’s formula apply to this example?
6. Check Euler’s formula on some plane graphs of your own. Does it work?
7. Our “nodes and regions” diagrams are plane graphs. Does Euler’s formula work for them?
The true formula here is a little less simple – and proving it is rather involved. A good first
step is for students to draw up a more detailed table, like the one below.
ext. nodes ext. arcs
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
cuts
0
1
3
6
10
15
21
28
int. nodes int. arcs
regions
0
0
1
0
1
2
0
3
4
1
8
8
5
20
16
15
45
31
35
91
57
70
168
99
Conjectures that are available from the table include, in roughly ascending order of
difficulty:
(i) ‘The number of external arcs is always equal to the number of external nodes.’
(ii) ‘The number of cuts is given by the triangular numbers, , , ,
, etc, which can be
generated by taking the third number in the corresponding row of Pascal’s triangle.’
(iii) ‘The number of internal nodes is given by the fifth number in the corresponding row of
Pascal’s triangle.’
(iv) ‘The number of internal arcs is always the number of cuts, plus twice the number of
internal nodes.’
And finally, our theorem:
(v) ‘The number of regions is , plus the number of cuts, plus the number of internal nodes:
that is, the sum of the first, third and fifth numbers in the corresponding row of Pascal’s
triangle: that is,
.’
Students will need help making some of these conjectures – and quite a lot of help proving
them.
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Activity worksheets
Here’s what we need to know in order to prove this formula.
1. The number of ways of choosing r items from
is
2. Euler’s formula for plane graphs:
graph,
.
, where
is the number of regions and
is the number of nodes in a
is the number of arcs.
3. Two lines that intersect in a node become four arcs (so every intersection of two lines
adds two to the number of arcs):
2
3
Now, with
1
4
points on the outside of the circle:
(i) The number of external nodes is .
(ii) The number of external arcs is also .
(iii) Each line crossing the circle corresponds to two external nodes: the number of such lines
is therefore
2.
(iv) Each internal node corresponds to four external nodes: the number of internal nodes is
therefore
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(v) The trickiest bit. The number of internal arcs is equal to the number of lines crossing the
circle plus
× the number of times they cross. (Because every time they cross, two arcs are
added: see above!) So the number of internal arcs is
.
OK, time to tot up.
total number of nodes
total number of arcs
external nodes + internal nodes
external arcs + internal arcs
By the Euler formula, the number of regions is
Teacher’s guide to
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.
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