Number

0 Intro

1 Place value1.1 Ancient number systems1.2 Finger maths1.3 Zeroes and ones

1.3.1 Napier’s location arithmetic1.3.2 Binary1.3.3 Binary codes1.3.4 Binary fractions1.3.5 Negabinary

2 Calculating 2.1 Multiplying methods

2.1.1 Grid method2.1.2 Per crocetta2.1.3 Napier’s bones2.1.4 Egyptian method2.1.5 Russian method2.1.6 Line method2.1.7 Lucas rulers2.1.8 Abacus method

2.2 Division2.2.1 Divisibility tests2.2.2 Remainders2.2.3 Power remainders*

2.3 Negatives and zero2.3.1 Some history2.3.2 Multiplying with negatives2.3.3 Adding and subtracting negatives2.3.4 Dividing by zero

3 Fractions3.1 Babylonian fractions3.2 Egyptian fractions3.3 Fibonacci’s fractions3.4 Galileo’s odd fractions3.5 Fraction trees3.6 Leibniz’s fractions3.7 Euler’s totient numbers3.8 Recurring decimals*3.9 Greek ladders3.10 Problem of the points3.11 Pigeonhole fractons

4 Primes

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4.1 Numbers as shapes4.2 Finding primes4.3 Prime games

4.3.1 Last prime4.3.2 Prime Nim

4.4 Prime factorization4.4.1 Factor trees4.4.2 Uniqueness4.4.3 Defining primes

4.5 Greatest Common Divisors4.5.1 Relative primes and the GCD4.5.2 Prime stars4.5.3 Diophantine equations

4.6 Prime sequences4.6.1 Linear prime sequences4.6.2 Quadratic prime sequences

5 Number sequences5.1 Triangle numbers

5.1.1 Counting crossings5.1.2 Triangle patterns5.1.3 Triangle cakes5.1.4 Multiplication tables5.1.5 Triangles from triangles

5.2 Squares and beyond5.2.1 Square sort5.2.2 Sums of squares5.2.3 Summing sequences

5.3 The great Gauss5.3.1 Quadratic residues*5.3.2 Wilson’s theorem5.3.3 Eight queens problem

5.4 Towers of Hanoi5.5 Some unusual sequences

5.5.1 Stirling numbers of the 2nd kind5.5.2 Stirling numbers of the 1st kind5.5.3 Moessner sequences5.5.4 Frequency sequences5.5.5 Thue-Morse sequence

5.6 Pascal’s triangle5.6.1 Blob pyramid5.6.2 Rook moves5.6.3 Lines in the plane5.6.4 Catalan numbers

5.7 Fibonacci’s sequence 5.7.1 Steps5.7.2 Partitions5.7.3 Fibonacci reflections

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5.7.4 Dominoes5.7.5 Cumulative Fibonacci sequence5.7.6 Fibonacci identities5.7.7 Fibonacci remainders5.7.8 Number bracelets5.7.9 Fibonacci factors5.7.10 Fibonacci Nims5.7.11 Lucas numbers

6 Parity6.1 Cogs6.2 Invariants

6.2.1 Odd blobs6.2.2 Odd cups6.2.3 Binary strings

6.3 Coins6.3.1 Coin tricks6.3.2 Coins in a row

6.4 Fair game6.4.1 Choco choice6.4.2 Poisoned square6.4.3 Clobber6.4.4 Stomp!6.4.5 Seat swap

6.5 Colouring in6.5.1 Two-colourable maps6.5.2 Three-colourable maps6.5.3 Four-colourable maps

6.6 In and out6.6.1 Bridges6.6.2 Inside/outside6.6.3 Odd triangle6.6.4 Weaving

6.7 Graphs6.7.1 Ghost house6.7.2 Domino loop6.7.3 Vertex game

6.8 Number puzzles6.8.1 Number circles6.8.2 Oddly even6.8.3 Magic squares

7 The Maths of voting7.1 What makes a good voting system?7.2 Some voting systems

7.2.1 Plurality7.2.2 Borda Count7.2.3 Pairwise voting

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7.2.4 Alternative Vote7.2.5 Approval voting

7.3 Arrow’s Theorem7.3.1 Independence of Irrelevant Alternatives7.3.2 Arrow’s Theorem

7.4 UK 20107.4.1 How do UK elections work?7.4.2 Who won?7.4.3 Seats v Votes7.4.4 Referendum 20117.4.5 Single Transferable Vote7.4.6 Equality in the House of Commons7.4.7 Voter apathy

7.5 Weighted systems7.6 Banzhaf power7.7 Electoral College

7.7.1 How do US elections work?7.7.2 US 20007.7.3 Who would have won under AV?7.7.4 How are electoral votes allocated?

8 Maths and music8.1 The standard scale8.2 Frequency8.3 Pythagorean tuning8.4 Just intonation8.5 Equal temperament8.6 Chords8.7 Duration of notes8.8 Time signatures

Glossary

Bibliography

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0 Introduction

Why did I write this book?

The main reason I have written this book is to collect in one place some of the mathematics that I, and my pupils, have found intriguing; I hope you find them as intriguing as we do.

This book is designed for the interested mathematician, whether they are 14 years old or 114. I would like to think that there is equally as much of interest for the secondary school student of Maths as the secondary school teacher.

What Maths is in this book?

I have tried to take some of the more interesting areas of number and make them accessible for the younger student. I want younger students to gain exposure to some of the beautiful mathematics usually reserved for older students.

You will find little algebra, and no geometry here, not because I don’t love these areas of Mathematics (I do!) but because these investigations form the basis of the Number aspect of the Key Stage 3 scheme of work I have developed at Greenwich Free School.

Curriculum

With regards to the curriculum, most of the areas under the number strand of the UK National Curriculum at Key Stage 3 (age 11-14) are covered here: place value, multiplication, division, fractions, decimals, percentages, ratio, negative numbers, factors, multiples, primes, sequences, and patterns as well as using and applying Maths. In that sense it can be used as part of a scheme of work that follows the National Curriculum.

That said, the focus is not on curriculum content, but rather investigating and exploring Maths; the activities are intended to allow the reader to explore how numbers work. The aim is for pupils to practice and apply basic skills in a non-basic context.

Questions are posed in red boxes, with (mostly) full worked solutions beneath. To get the most out of this book, the reader should attempt the questions before reading on. Only by fully investigating the problem can the beauty of the solution be fully appreciated. I hope the solutions give enlightening ways on how to go about solving mathematical problems.

How much Maths do I need to know?

The level of maths involved is generally no higher than that required for Key Stage 3, although some of the investigations are a bit harder and may be more suitable for older pupils (these are marked with a *). It is my hope that most readers can access

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most of this book. There is a Glossary at the back of the book that contains some basic skills and definitions you might need if you didn’t already know them; terms in the Glossary are in bold.

Well, if you are a student of Maths, I hope you enjoy trying some of the investigations, and if you are a teacher I hope you enjoy sharing some of the investigations with your kids. Most of all, I hope this book will give anyone reading it the same pleasure that I get from doing Maths.

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1 Place value

Understanding and reflecting on how our number system works is the fundamental building block to being able to work with numbers with fluency.

This section takes us on a whistle-stop tour of some ancient number systems, gives some methods of working with our fingers, and explores some other number systems such as binary.

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1.1 Ancient number systems

Before we invented the number system we use today, which is now used almost everywhere in the world, humans came up with lots of different ideas for writing numbers.

The first known methods of counting, possibly around 30,000BC, involved making tally marks, one mark for every number, often in bone or stone. This is OK for small numbers but it would take a while to write bigger numbers!

Around 20,000 years later there is evidence the Sumerian civilization (modern Iraq) were using different shaped clay tokens to represent different numbers and objects (some of which can be found in the Louvre):

The Ancient Babylonians developed these ideas into a more sophisticated number system, written on clay tablets. They originally started with lots of symbols for different numbers, based on the shapes of the clay tokens seen above. For example, they used symbols like these for 60, 600, 3600 and 36000:

They also used tally marks to represent 1 and 10 like this:

However, the true genius of the Babylonian number system was their invention of place value. Putting the symbols in different places would result in them being of different value, which meant they didn’t need lots of symbols for different sized numbers. Gradually decided they could get by with only the two symbols for 1 and 10, and scrapped the other symbols.

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Now, the Babylonians used a base-60 system, so each column would be worth sixty times more than the one before. So, using blobs, they might have written 736 = 12.60 + 16 like this:

3600 60 units

You can see why they invented an extra symbol for 10 units - you could have up to 59 blobs in one place! They would write 736 like this:

Although this is lovely and simple, only using two symbols, it is still quite confusing as it is not totally clear which symbol has which value. In around 500BC they came up with a much nicer solution, using a couple of triangle blobs to separate the two places like this:

Using our numerals, we often write Babylonian numbers using a comma to separate the different columns, so we might write 736 like this: 12, 16.

Here are some Babylonian numbers found on a clay tablet.

1 11,2,1 1,11,2,3,2,1 1,1,1

In our numbers, the table reads:

1 13721 6113402921 3661

The numbers on the left are the squares of the numbers on the right.

Why do you think Babylonians used base 60?

Translate these numbers into our numbers. What do you think this represents?

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Around the same time as the Babylonians were developing their number system, the Ancient Egyptians were inventing one of their own. It was not as sophisticated as the Babylonian one, as it did not use place value, but it was based on 10 instead of 60.

This meant they had to invent lots of symbols for different sizes of numbers. Here are a few early ones they used:

One advantage of this is that it didn’t really matter where they put the symbols in the number as they had a fixed value.

Here’s how they might have written 136:

The Egyptians came up with an ingenious method for multiplying (and dividing).

Here is their method for calculating 19 x 23:

1 x 23 = 232 x 23 = 464 x 23 = 928 x 23 = 18416 x 23 = 368

So we have 23 + 46 + 368 = 437.

Compare the Egyptian and Babylonian systems. Which do you prefer and why? How could you make a system with the best of both?

Babylonian mathematics was much more sophisticated than just multiplying and dividing. They could solve quadratic and higher order equations. There is evidence to suggest they had discovered Pythagoras’ Theorem long before Pythagoras, with tablets containing Pythagorean triples and an approximation to the square root of 2.

Before reading on, can you see how this works? Will it always work?

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In this method we start at 1 x 23 and just double the left hand number until we get to 19. Then we only include the numbers on the left that add to make 19. Here, 19 = 1 + 2 + 16 so we just use those calculations. This method will always work because all numbers can be expressed using powers of 2.

They also used this for division. To divide 135 by 15 (say) they would just write out the multiplication for 15, stopping when they got to the target number 135:

1 x 15 = 152 x 15 = 304 x 15 = 608 x 15 = 120

Then they just found the numbers that add to 135 and took the multipliers, in this 1 + 8 = 9.

Around 500BC the Ancient Chinese came up with a base-10 system using bamboo rods, with numbers 1 to 9 represented by:

The rods were placed on a counting board (like a chess board) where each box represents a different value. Also, in each adjacent box the rods were alternately placed horizontally and vertically like this:

They did not have a numeral for zero because they did not really need one! However, in later years they used a stone from the game ‘Go’ to represent a zero. This number system was replaced by the abacus, which first became used in China around 200BC, although it was still used in parts of Japan for working with algebra.

The Ancient Egyptians also discovered more than just multiplying and dividing. They could solve complex practical problems, including geometrical ones involving area and volume.

Ancient Chinese mathematics was highly advanced, although it had little influence on mathematics as we know it today

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The base-10 system we use today came to Europe via Indian and Islamic mathematics. Indian mathematicians came up with the idea of expressing each number using one symbol each, as we do today. The development of these symbols happened over more than 1,000 years into something resembling our current symbols 0-9. Here is an image of Devanagari numerals from around 1000AD:

Note that zero appears in the above numerals. The birth of zero as a number, and its rules for using it in calculations, can be traced to one of the most famous Indian mathematicians, Brahmagupta (c. 700AD) – more about him later.

Find out more about other ancient number systems.

I have not included anything about Islamic mathematics in this book – why not? Find out what branches of mathematics Islamic mathematicians developed.

Indian mathematics had a huge influence on the development of Islamic mathematics. Unlike the Greeks, they did not study arithmetic in order to solve geometric problems; they studied it as a subject in its own right. The result of this is that they developed solely numerical procedures for solving problems, many of which we use today, as well as paving the way for the use of zero, negative and irrational numbers.

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1.2 Finger maths

People have probably always counted on their fingers. You can definitely count up to 10 on two hands, but is this as far as you can go…?

Put your hands in front of you like you are about to play the piano. Each finger has the value as shown in the picture below:

By ‘playing’ the index finger of your right hand you count 1. Leaving your fingers in the table as you go, you can start counting to 4. When you get to 4, all of the fingers on your right hand should now be on the table.

Now, to count 5, put your right thumb on the table, and at the same time, lift all your fingers from the table. To count 6, now put your (right) index finger down again (and keep your thumb down), giving 5+1 = 6. Now in the same way as before you can count up to 9 on your right hand – when you get to 9, all your fingers and thumb on your right hand are on the table.

To count to 10, you put your left index finger on the table, and at the same time, lift all your right hand from the table. Now you can count to 99! You should practice counting up and down before you try the next bit.

Try doing some addition or subtraction using your hands. Can you find any methods for making these calculations quicker?

Try some times tables. What does the 2x table look like on your fingers? Can you see any patterns? What about other times tables?

Which times tables are the easiest? Which are the hardest? Can you work out which times table this is a picture of?

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This is based on the Japanese abacus called the Soroban. You can buy one of these quite cheaply online. Here is a picture of one:

You can see that your fingers are like the earth beads (1 to 4) and your thumbs are like the heaven beads (fives).

To count a number on the Soroban, move the beads so that they touch the beam (like putting your fingers on the table). For example, here is the number 736.

To add 654 + 213 we start with 634 on the abacus and then add the 2 to the 6, 1 to the 5, and 3 to the 4, giving 867. Notice that we work from left to right.

This was a simple example as there were enough beads to do each column separately without carrying, but what happens when we need to do carrying? For this we need to use the idea of complementary numbers. Complementary numbers are pairs that add up to 10, so the complementary number for 8 (say) is 2.

Let’s see an example. To work out 13 + 58, set 13 on the Soroban and try adding left to right. The tens column is OK as there are enough beads, but in the second (units) column there are not enough beads (3+8), so we subtract the complement of 8 which is 2 (giving 1 in the units column) and add 1 to the tens column (giving 71).

To subtract on a Soroban, we do the opposite. To calculate 53 – 26, going left to right, we try the tens column first which is OK (5-2=3), but there are not enough beads in the second column. So, doing the opposite to addition, we take one from the tens column (leaving 2) and then add the complement of 6 (which is 4) to the 3 already in the units column, which gives 27.

You can actually count up to much more than 99 on your hands. To see how, let’s give each finger a value double to the previous one, like this:

Practice some calculations like these yourself! Why not explore abacus methods for multiplying.

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Now you can count much further! For example, to show the number 21 you just hold up the fingers 16, 1 and 4.

Practice counting using this method; what is the biggest number you can count up to? What do times tables look like now? Why not try some calculations?

Can you see why this works?

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Here is another way of showing numbers using fingers: British Sign Language (BSL).

This is one system that only uses one hand:

Find out more about other mathematical signs in BSL. Can you think of some signs for mathematical terms and concepts that do not exist in BSL?

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1.3 Zeroes and Ones

1.3.1 Napier’s Location Arithmetic

Let’s look a bit closer at the double-hand number system above.

In the 16th century a Scottish mathematician called Napier used letters instead of fingers, with a = 1, b = 2, c = 4, and so on. So, for example, he would write 21 as ace.

Napier allowed repeats to be used. For example, the word add represents 17. However, you might have noticed that the word add can be shortened to ae; Napier called this abbreviation.

Napier used counters on grids to perform calculations. Consider the sum 29 + 8. Place counters on 29 (= 16 + 8 + 4 + 1 = acde) and 9 (= 8 + 1 = ad) like this:

You can think of abbreviating as replacing two counters in any column with one counter in the next column up (to the left). So we abbreviate as follows:

And one final abbreviation leads to:

Try writing some other numbers using Napier’s method. What are the advantages and disadvantages of this method?

What is the abbreviated answer to the sum add + ace?

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Multiplication can be performed on a square grid like this:

If each grey square contains the result of multiplying the two letters together, can you fill in the values of all the grey squares?

What do you notice?

You may have noticed that the diagonals of the grid contain the same values;

Napier exploited this fact to perform multiplications on this grid using counters.

Here is an example showing the calculation of 5 x 6. The word form of this sum is ac x bc, so place counters on cells where these letters ‘intersect’ like this:

Then, as all values on the diagonal are the same, slide the counters along the diagonals to the bottom row as shown below and read off the result (abbreviating if necessary) to give the answer bcde which is 2 + 4 + 8 + 16 = 30.

Try some more addition using these grids. How would you adapt this method to perform subtraction?

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Napier said of this

method: “…it might be well described as more of a lark than a labor, for it carries out addition, subtraction, multiplication and division purely by moving counters from place to place.”

Try some of these calculations yourself. Can you see how to use it for division? Make a larger grid and have a play with these methods for larger numbers!

John Napier was a Scottish mathematician born around 1550AD who had many interesting ideas - he predicted the machine gun and submarine around 300 years before they were invented. However, he is most famous for his work in mathematics. He once famously said:

There is nothing more troublesome in mathematics than multiplications, divisions of great numbers which involve a tedious expenditure of time, as well as being subject to ‘slippery errors’

I agree! Apart from the discoveries here, he also invented the logarithm (log), considered by some to be the most important discovery of the Renaissance as it helped in calculations with large numbers. We will find out a bit more about logs in the section on Maths and music.

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1.3.2 Binary

Instead of using letters for each finger, let’s have a column for each finger, like our usual columns for units, tens, hundreds and so on:

There are lots of patterns you might have spotted. One that interests me is that the last columns alternates blob, then blank, then blob… which is equivalent to saying all odd numbers end in a blob.

In the 17th century, the German mathematician Leibniz starting using this system to do calculations. He represented the blobs by 1s and the blanks with 0s. For example, the number 5 would be 101. This is now called binary, or base -2.

Here is the addition 3 + 8 performed using binary. Change 3 and 8 into binary (11 and 1000) and then do the addition in the usual way:

I definitely think that some calculations are easier using binary, although there is a often lot of carrying involved!

Carry on with this table – can you see any patterns?

Try doing some more calculations in binary – don’t just stick to addition… Do you think it is easier or harder than our normal number system?

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If we write them in binary we get 11, 110, 101, 1001, 1100, 10001, 10010, and so on. Now you might have noticed that these are all the numbers with exactly 2 binary digits.

You can make all the numbers from 1 to 25 with these five digits; in fact, 25 is the highest total you can make.

However, you can make a ‘better’ set of 5 numbers that will make all the numbers up to 25 and a few more. Notice that 3 is kind of ‘wasted’ as you can make this out of 1+2, so why not have the next number as 4 instead of 3? Now we can make all the numbers up to 7, which suggests 8 as the next best choice… But now you are probably starting to notice that there are all the doubles (powers of 2), which is a nice demonstration that you can make all the numbers by summing powers of 2.

Investigation 1: Here is a sequence of numbers: 3, 5, 6, 9, 10, 12, 17, 18, … what is special about this sequence of numbers? Can you find a similar interesting sequence of numbers?

Investigation 2: Try making all the numbers 1 to 25 by adding some or all of the five numbers 1, 2, 3, 7 and 12 (without repeats). Is it possible? What is the highest total you can make? Is there a ‘better’ set of 5 numbers you can use?

Gottfried Wilhelm Leibniz (born 1646) is most famous for discovering calculus at the same time as Newton (and having a dispute with him about who found it first). It is Leibniz’s notation that we use today, not Newton’s.

He was the first European mathematician to use matrices to solve systems of equations and asked the question whether all polynomials can be factorized into linear and quadratic factors.

He is considered the most important logician since Aristotle, and believed that human thought could be reduced to a series of calculations. He also developed one of the first mechanical calculators that could perform multiplication and division, which used binary.

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1.3.3 Binary codes

Here is the number 5 (101) represented as a binary ‘picture’:

We could use this idea to create more interesting binary pictures; here is a picture for the ‘code’ {5,2,5}, where each number represents a row in the picture:

What is binary used for? In the 19th century English mathematicians George Boole built on Leibniz’s and Babbage’s (see More Prime Sequences) ideas that Maths was just a branch of general logical thought.

He developed a system of algebra for working with mathematical logic called Boolean Algebra, which was considered relatively useless in his lifetime; its importance was not fully realized until US mathematician Claude Shannon linked it to electrical circuits and used it in the design of circuit boards for computers.

Binary is now used in electrical circuits everywhere, in our computers and phones, to store and transmit information. Information can be thought of as a long binary code where each digit represents a bit of data.

When transmitting data across long distances, there is often ‘noise’, which creates errors in the signal. In the 1970s, methods were developed by NASA to transmit a code that could detect and correct its own errors. We are going to have a look at how this might be possible:

Find out more about electrical circuits, logic gates, Boolean algebra and the use of binary in logic and computers.

What picture does the code {17,27,21,17} represent? Explore other binary pictures.

George Boole’s father was a carpenter, at that time considered a job for the lower classes. In these days (and now?) the upper classes looked down on the lower classes. Mathematics was forbidden in working class schools, being considered a ‘gentlemanly pursuit’.

George Boole broke this mold, initially being taught by his dad, then getting into a charity school until the age of 16 (usually the poor worked from the age of 8 or 9). He taught himself a number of European languages so he could read classic maths textbooks.

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Before reading on, consider the following problem:

One idea is to use 00, 01, 10 and 11 for up, down, left and right. Or perhaps you decided to use 0 for forward and 1 for a turn of 90 degrees (clockwise).

So a transmission using this code would be (looking down from above, starting facing to the right): 01001110100111001110001110.

Then comes the problem: what if there is an error in the transmission? If (say) any 0 turns to 1, the robot will not reach its goal. How can we devise a code that recognizes that an error has been made?

One solution is to use something called a Hamming Code, invented by US mathematician Richard Hamming in 1950.

Suppose we wish to transmit (say) a 4-digit symbol, such as 1101. Then we can add extra digits to the end (called a parity check) to check whether there is an error in the transmission. Here is an example:

Each digit is written in a section of the Venn diagram. So the digits in sections 1 to 4 here are our codeword 1101.

How are the other digits in sections 5, 6 and 7 chosen? They are chosen to make each circle contain an even number of ones (hence the name parity check).

So here we transmit the word 1101010, which can then be checked for errors by the receiver using the Venn diagram.

But this code not only detects errors, it can also correct them! Once we have found where the error is, we can also change the digit to the correct (opposite) value, and work out what the codeword should have been!

Can you create a code to transmit to this robot so that it can navigate through the maze?

You must only use zeroes and ones; no spaces, commas, ...

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For example, if this word was transmitted but we received 1001010, then we could put these in the Venn diagram and realize that the error must be in section 2.

Can you find the error in the codeword 1001110 using this method? Find out more about error-correcting codes.

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1.3.4 Binary fractions

Here are the binary place values, including fractions:

Some are easy like 1/2 = 0.1, 1/4 = 0.01 and so on. How about 1/3?

Let’s just recap how we would work out 1/3 in base-10. 1/3 is the same as 1 divided by 3, or ‘how many threes in one?’, and we use long division:

There are no threes in 1 (0 remainder 1), and we carry the remainder into the next column (1/10ths), where it counts as 10 (ten tenths is 1 unit). We usually write this like this:

Now we do threes into 10, which is 3 remainder 1, and we write the answer (3) above the line, and carry the remainder 1 into the next column, where it counts as 10 again:

This will carry on forever, giving the recurring decimal 0.333… We can write this process like this:

1 = 0 x 3 + 1 10 = 3 x 3 + 1 10 = 3 x 3 + 1 …

Now let’s think about what is happening here: the remainders become tens in the next column s because we are working in base-10. What will happen in binary? The remainders will become two in the next column!

Try writing some binary fractions.

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Let’s try working out 1/3 in binary:

1 = 0 x 3 + 1 2 = 0 x 3 + 2 4 = 1 x 3 + 1 2 = 0 x 3 + 2 4 = 1 x 3 + 1

You can see the remainders doubling as we move into the next column… So our binary-mal here is recurring 0.0101…

Here are the first few binary-mals, with recurring parts underlined:

1/2 = 0.11/3 = 0.011/4 = 0.011/5 = 0.00111/6 = 0.0011/7 = 0.0011/8 = 0.0011/9 = 0.0001111/10 = 0.000111/11 = 0.00010111011/12 = 0.0001

There are a few patterns I notice here. The most obvious one is that 1/2, 1/4, 1/8, … are the only terminating (not recurring) decimals. Can you explain why?

You might also have noticed that the fractions with denominators one less than a ‘double’ (1/3, 1/7, …) follow the pattern 0.01, 0.001, 0.0001, … When you work them out you can see why.

Also, you might have noticed that one you have worked (say) 1/3, you can easily write down all the fractions that are half as big (1/6, 1/12, …) because they just have one extra zero after the decimal point. So 1/3 = 0.01 implies that 1/6 = 0.001, 1/12 = 0.0001 and so on. This is because they are made of fractions that are half as big, so all 1s are shifted one place to the right.

More than this, we can see all fractions such as 1/3 can be written as an (infinite) sum of fractions. For example, 1/3 = 0.010101… can be written as

1/3=1 /4+1 /16+1 /64+…

You can type these into a calculator and see that including more terms gets us closer and closer to 1/3. You could also use these to find infinite series for other fractions. For example, doubling gives 2/3=1/2+1 /8+1/32+… For more on infinite series, see the section Leibniz’s fractions.

Try using this method to work out other binary-mals. Any patterns?

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I personally like base-12 as a number system as it has lots of factors, which means it would be good for writing fractions. Before we start using base 12 we need to invent two more number symbols (called numerals) to represent our 10 and 11, so I’m going to use A and B.

So the numerals for base-12 are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, then 12 would be 10 and so on.

Here are some common fractions written using base-12:

It takes a bit of practice working them out but after a while it’s OK.

How can you check these are correct? For example, 1/8 = 1/12 + 6/144, which is the same as 12/144 + 6/144, which is 18/144, which simplifies to 1/8.

Unfortunately we still have recurring ‘decimals’, we can’t get away from them!

To work them out we need long division, remembering to carry 12 over for each remainder into the next column. Here are some I worked out:

1/5 = 0.249724972… 1/10 = 0.124962497… 1/11 = 0.111……and our old friend 7, gives us 1/7 = 0.186A35186…

See the sections Babylonian fractions and Recurring decimals for more about this.

Do you think base-12 is a good idea?

Try writing fractions in other number bases. Which do you like best? Why?

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1.3.5 Negabinary

Before finding out about negabinary, let’s look a bit closer at normal binary. Before starting this section, you might want to look at the section on multiplying negatives.

We have seen that the place values (columns) in binary are 1, 2, 4, 8, … which are the powers of 2: 20 = 1, 21 = 2, 22 = 4, 23 = 8, and so on.

Carrying on the pattern above, we have the 1/2 must be 2-1, 1/4 is 2-2 and so on. When talking about powers, we say the big number at the bottom is the base, and the little number at the top is the power or the index. All the columns in binary all have a base of 2, hence the name base 2 (and in the same way, all the columns in our normal number system have base 10).

US mathematician Donald Knuth investigated the properties of an alternative/extension to binary called negabinary, which can be thought of as base -2.

Well, let’s concentrate on the non-fractional place values. So we have:

(-2)0 = 1 (anything to the power 0 is 1)(-2)1 = -2 (-2)2 = (-2) x (-2) = +4(-2)3 = (-2) x (-2) x (-2) = -8

The place values are like binary, but with alternating negative signs:

The first few numbers in negabinary are:

What are the powers for the fractional columns we saw in binary fractions?

What are the place values in negabinary?

What are the first few numbers in negabinary? What numbers can be written in negabinary?

Can you see any patterns?

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1 = 110 = 1 x (-2) = -211 = 1 x (-2) + 1 x 1 = -1100 = 1 x 4 = 4101 = 1 x 4 + 1 x 1 = 5110 = 1 x 4 + 1 x (-2) = 2111 = 1 x 4 + 1 x (-2) + 1 x 1 = 31000 = 1 x (-8) = -81001 = -71010 = -101011 = -91100 = -41101 = -31110 = -61111 = -510000 = 16

It turns out they can. Here are the positive and negative numbers represented in negabinary:

Positives Negatives

1 11110 10111 1101100 1100101 111111010 111011011 1001…

If you hadn’t noticed it already, you can see that the positives have an odd number of digits and the negative numbers have an even number. You might not have noticed some other patterns, such as multiples of 3 have 3 ones.

You might have also found a quick way of working out the negabinary representation of a number. For example, to work out what 7 looks like, note that 7 can be made from 16 + (-9) = 10000 + 1011 = 11011.

Do you think all positive and negative numbers can be represented in negabinary?

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If we want to count in negabinary, we note that two blobs in the (units) column turn into two blobs in the next two columns, like this:

I’ll leave you to explore adding, subtracting and multiplying!

What happens if we use counters to count with negabinary? How does adding, subtracting and multiplying work?

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2 Calculating

This section is about boring old times tables and long division. Not really, but we will have a look at some more interesting aspects of multiplication and division. Although I have given some mention as to the origins of these methods, there is considerable debate over many of them.

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2.1 Multiplying methods

2.1.1 Grid method

The most common that is taught at school is the grid or area method; here is 23 x 34. We split the two numbers up and calculate (20 + 3) x (30 + 4). Then we just add the results together to get 782.

This method is good because it is clear and has a meaningful interpretation (area). Also, it is useful for understanding BODMAS and for multiplying algebraic expressions. Although this method is the one I would recommend to use, there are more interesting methods.

2.1.2 Per crocetta

A different method, possibly invented by Indian mathematicians, and which appears in Italian maths books around 1500AD, is called per crocetta (by the cross), and is shown by this diagram.

I have started at the 20, and gone round the arrows multiplying each pair as I go along and writing it in red. You can see it is really just the grid method in a different form.

2.1.3 Napier’s Bones

John Napier (mentioned in the previous chapter, which incidentally included another method for multiplying) invented another multiplication method called Napier’s Bones. Napier’s bones are basically just times tables written on strips, like these ones for the 2 and 7 times tables below.

Note how each box splits the product into tens and units. Now, we can use these strips to calculate the 27 times tables easily. For example, to work out 6 x 27, just look at the 6th row (circled) you can see 6 x 2 = 12 and 6 x 7 = 42. Putting these together on the right hand side and adding along the diagonals gives the answer 162.

Can you make a similar cross for multiplying two 3-digit numbers together?

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Of course, you don’t need the strips to work out this calculation; you could have just written down the boxes on the right straight away. This method is sometimes called Gelosia.

2.1.4 Egyptian method

This was described in the previous section but is included again here for completeness. Here is the method for calculating 19 x 23:

1 x 23 = 232 x 23 = 464 x 23 = 928 x 23 = 18416 x 23 = 368

So we have 23 + 46 + 368 = 437.

2.1.5 Russian method

This is a similar method as above. Here is the calculation for 19 x 23:

19 x 23 9 x 46 4 x 92 2 x 1841 x 368

Again adding the red numbers we can see that the answer is 437. This time we have halved the left number and doubled the right number. We then cross out the ones in which the left number is even.

How does this method work?

Why not make some Napier’s rods of your own!

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2.1.6 Line method

I’m not sure where this method comes from, but here is the calculation for 23 x 34:

Now we have 12 units, 17 tens and 6 hundreds, which add together to make 782.

Compare this with the grid method – can you see how it works?

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2.1.7 Lucas rulers

Here is a set of strips, a bit like Napier’s Bones, that are used for multiplication. They were devised by French mathematician Eduoard Lucas in the 19th century. He was also famous for the Lucas numbers (see the section on Fibonacci) and inventing the puzzle Towers of Hanoi.

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Cutting out the strips, we can put them together to work out calculations. For example, to work out 6 x 27 we put the strips for 2 and 7 together like this, along with the ‘index’ strip:

Now we calculate the product 6 x 27 by looking at the 6th row.

First, take the top number in the furthest right column (which is a 2).

We then follow the triangle arrow to the left, which is pointing to a 6 in the next column.

Following the next triangle, it is pointing to a 1.

And we have the answer = 162!

2.1.8 Abacus method

This method is basically the long multiplication method that used to be taught in schools, but is adapted to work on an abacus.

First, choose a column of the abacus to be the units column. Then enter the number you want to multiply (367, shown in blue here) two columns to the left of this column. Now multiply each digit of this number and put the answer in the two spaces to the right of the number you just multiplied.

Here is the calculation for 6 x 367. First we calculated 6 x 7 and put it in the two spaces to the right of the 7, then 6 x 6 in the two spaces to the right of the 6, then finally 6 x 3 in the two spaces to the right of the three. Adding then gives the answer, 2202.

Test this out for some other numbers...How does this work?

Test this out on an abacus.

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2.2 Division

2.2 1 Divisibility tests

There are lots of divisibility tests, which people get more or less excited about. Some of the divisibility tests require the idea of a reduced number (or digital root), which is where you add the digits of a number together. For example, 38 reduces to 3 + 8 = 11, which reduces to 1 + 1 = 2. So the reduced number of 38 is 2.

For example, here is the 4x table recorded on a 9-dot circle:

You may have noticed that the 9x table always reduces to 9 (and in fact the final reduced number is the remainder on division by 9), and the 3x table always reduces to 3, 6 or 9. This gives a quick way of checking whether a number is divisible by 3 or 9.

Here’s a proof for divisibility by 9. Consider a 3-digit number abc, where a is the hundreds digit, and so on. So the value of this number can be thought of as 100a + 10b + c.

Now the reduced number is a + b + c. The difference between 100a + 10b + c and a + b + c is 99a + 9b. Now this is definitely a multiple of 9 (why?) so if a + b + c is also a multiple of 9, them so is 100a + 10b + c. Think about this for a while and try it out with some numbers if you are not sure why. You can probably adapt this proof to see why the test for divisibility by 3 works too.

Here is a fun puzzle that uses the divisibility test for 3:

Before reading on, investigate the reduced number sequences of some basic sequences, such as times tables, square/cube numbers, triangle numbers, ...

Try recording your results on 9-dot circles. What patterns can you find?

Can you prove why this works?

Suppose there are three piles of sweets, one with 1 sweet, another with 2 sweets and a third with 3 sweets.

You can take sweets according to these two rules: You can either take the same number of sweets from each, or you can divide any even pile into two and move half into another heap (leaving the other half where it was).

The aim is to take all the sweets from the table: can you work out how to take all the sweets?

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This is not too tricky; take 1 from each, leaving 0-1-2. The split the third pile into half, putting one half into pile 1, giving 1-1-1 and we are done.

Did you work systematically to find which ones were possible and which were not?

It turns out that any piles of sweets that are a multiple of 3 will be possible. To see why this might be true, consider the final position you want to get to, where all piles have the same number of sweets; the total must be a multiple of 3. If you think about this, the piles must always contain a total multiple of 3 otherwise we can never take them all. So just use your divisibility test for 3 to see if the puzzle will be possible!

Here is a trick that involves divisibility by 9, taken from Martin Gardner’s 2nd Scientific American Book of Mathematical Puzzles and Diversions:

First of all, draw a spiral on a piece of paper and put it in an envelope. Then ask a friend to take their phone number and then scramble the numbers into a different order. Then ask them to subtract the smallest number from the biggest one and work out the reduced number of the answer.

The reduced number of the answer is always 9, which is why we end at the spiral (if we start at the star – you could vary this to make the puzzle harder to figure out).

Let’s see how it works through an example. My phone number is 237115, and I could scramble this into (say) 751213. Then subtracting we have 751213 – 237115 = 514098, which has reduced number 9. In fact, the answer will always be 9 because the phone number and its scrambled version will have the same reduced number. So subtracting them will cancel out the reduced numbers, leaving the same reduced number. Why is it 9? Consider the digit 7 in the number above: it contributes 100000 x 7 – 1000 x 7 = 99000 x 7 to the final reduced number, which is a multiple of 9. A similar argument proves that all the digits will contribute a multiple of 9 to the answer.

Now, let’s turn our attention to divisibility tests for other numbers. You already know some other divisibility tests, such as those for the 10x, 5x and 2x tables (by just

Try for different starting piles of sweets. Can you always take all the sweets? If not, when is it not possible? Can you explain why?

Starting at the star = 1 on the picture below and counting round this reduced number, where do you end? Can you explain how this trick works?

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looking at the last digit). There are similar tests for the 4x and 8x tables (by looking at the last 2 or 3 digits).

We can combine the tests for the 2x and 3x tables to make a test for the 6x table.

How about a test for divisibility by 7? This is much less obvious. One test goes as follows: Take the unit digit of the number and double it, then subtract this from the rest of the number. Carry on until you get down to a small number. If it is a multiple of 7 then the whole number is.

For example: 2849 --> 284 - 2x9 = 266 --> 26 - 2x6 = 14 which is a multiple of 7, so 2849 is.

There is also a nice test for divisibility by 11: add alternate digits to make two totals, for example 154,737 gives 1+4+3=8 and 5+7+7=19. Then if the difference between these totals is a multiple of 11 then the number is divisible by 11.

Let’s take the number abc = 100a + 10b + c again. Now we have a + c in one total and b in the other. The difference between these totals is a – b + c. Now comparing 100a + 10b + c with a – b + c we have a a difference of 99a + 11b which is a multiple of 11. So if a – b + c is also a multiple of 11, so is abc.

Finally, here’s an interesting test for divisibility by 7, 11 and 13. Take a large number such as 174572528. Split the number into 3-digit segments 174, 572 and 528. Add together alternate ones, as above, so here we have 174 + 528 = 702 and 572. Now take the difference of the two totals, which is 702 – 572 = 130. As this is a multiple of 13, this number is divisible by 13.

To see how this works, consider a number abcdefghi. Now comparing 1000000xabc + 1000def + ghi with abc – def + ghi we have a difference of 999999abc + 1001def, which is a multiple of 1001, and further to this, 1001 is a multiple of 7, 11 and 13. So if our final number is a multiple of any of these, so is the original.

Can you prove why this works? Use a similar argument to the one for divisibility by 9.

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2.2.2 Remainders

For me, the most interesting thing about division is the patterns in remainders. They help you find patterns in sequences that you might not have noticed at first sight.

For example, if you look at the remainders of the square numbers when divided by 2, you get 1, 0, 1, 0, 1, 0, … which if you think about it, is just telling us that the square numbers alternate odd, even, odd, even, … (why?)

Here are a few more investigations to get you started. If you can, try and explain why you find the patterns you do. The first investigation is from the excellent book Points of Departure, available through the ATM.

Start with any number and continue doubling it… what are the remainders on division by different numbers?

Here is what happens for the numbers 0 to 6 on division by 7…

Here are some more open-ended investigations of remainders for different sequences:

We are going to look at these questions in more detail in the next section. But notice that even a sequence with seemingly little pattern, such as the prime numbers, may have some pattern when studying its remainders. For example, here are the remainders of the first few prime numbers when divided by 6:

2, 3, 1, 5, 1, 5, 1, 5, …

Now, you might be forgiven for thinking that this sequence carries on alternating between 1 and 5 like this forever. If only! In fact, the next few remainders are 5, 5, 1, 1, 5, 1, 5, … and there is no discernable pattern to find the next one, apart from the fact that it will either be 1 or 5.

We saw above how the square numbers have a pattern on division by 2. Investigate the remainders of the square, triangle and cube numbers with different divisors. How about the prime numbers?

Investigate for other divisors.

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2.2.3 Power remainders*

We are going to look at patterns in the remainders of squares, cubes and so on. The French mathematician Pierre de Fermat was the first to study this question in great depth.

In this section, we are going to use modular arithmetic, a description of which can be found in the Glossary.

Using the language of modular arithmetic, the square number sequence 1, 4, 9, 16, 25, … is congruent to 1, 0, 1, 0, 1, … mod 2.

Did you look at the cubes mod 2 in the previous investigations? Here is a table of the whole numbers (integers), squares and cubes mod 2:

Integers 1 0 1 0 1 0 …Squares 1 0 1 0 1 0 …Cubes 1 0 1 0 1 0 …

It would seem that all sequences of powers become 1, 0, 1, 0, ... mod 2 (check this if you want, or better still can you explain why?).

Will all sequences of powers be the same for other divisors? No: here are the integers, squares and cubes mod 3 – they are different:

Integers 1 2 0 1 2 0 …Squares 1 1 0 1 1 0 …Cubes 1 2 0 1 2 0 …

Why do prime numbers bigger than 3 only have remainders 1 or 5 when divided by 6? [Hint: consider all the other possibilities and see why they can’t be true]

Pierre de Fermat, born in 1601, was perhaps the greatest French mathematician of the Renaissance period, which is no mean feat considering the competition.

He made many discoveries in number theory, which he shared with others through letters. He also laid the foundations of probability along with Pascal (again through a series of letters between them), is credited with inventing co-ordinates (and analytical geometry in general) before Descartes, and he invented methods that were later developed by Newton and Leibniz in the calculus.

All this by an ‘amateur’ mathematician!

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You might have noticed that the integers and squares are the same mod 2, and the integers and cubes are the same mod 3… so will the integers and powers of 4 be the same mod 4, and so on? Let’s see:

If our conjecture is true, we would have expected that a4 (mod 4 ) would be the same as the integers mod 4, that is: 1, 2, 3, 0, 1, 2, 3 … but this is not the case. However, it is true for a5(mod 5).

Investigate further we find that it’s also true for modulo 7 and 11:

It appears to be true if n is a prime number! It turns out that this is indeed the case: this result is known as Fermat’s Little Theorem and is often written

a≡ap(mod p)

for any prime p. Fermat’s Little Theorem is exciting; it gives us a rule for prime numbers, which are few and far between. But is the converse true? That is, is it true that if a≡an(mod n) then we know that n is prime? If this is true, it will give us a way of finding primes – which would be great! Let’s investigate with a = 2 in the table on the right. We want to find out whether 2≡2n(mod n) means that n is prime.

It would appear that it is true… the only 2n that are congruent to 2 are when n is prime. So it seems we have

What do you notice about the last two tables?

What conjectures would you like to make before investigating further?

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found a way of finding primes… and this is what mathematicians thought in the 17th century.

Unfortunately, in the 18th century it was found that if we carry on going we see that 2341 is congruent to 2 mod 341, but 341 is not a prime (341 = 11 x 31). So we have found an exception to the rule, which means it can’t be true.

So Fermat’s Little Theorem is not true both ways; all we can say is that if p is prime, then a≡ap(mod p), but we can’t say that if a≡ap(mod p) then p is prime!

One final note: can we ‘divide’ this equation by a like this:

1≡ap−1(mod p)

Is this allowed? Let’s look at a p−1(mod p) to see if it is true:

It appears to be true, apart from when a is a multiple of p. This is indeed the case, so Fermat’s Little Theorem becomes

1≡ap−1(mod p)

if p is prime and a and p are relatively prime.

What is the relevance of Fermat’s Little Theorem? It is one of the most important theorems in a branch of maths called Number Theory, which studies the properties of the integers.

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2.3 Negatives and zero

Negative numbers can be hard to get your head around. This is not surprising, as most humans didn’t really accept them as proper numbers until around 500 years ago. Here is a brief account of how they came about and how to calculate with them.

2.3.1 Some history

The first people to work with negative numbers may have been the ancient Chinese, who used different coloured rods when dealing with negatives. Recall from chapter one that they used red rods on a counting board to show numbers like this:

There is evidence they used black rods to represent negative numbers, and did calculations with them, perhaps as early as 200BC. They did not need zero, as it was a blank space on the counting board, so they did not invent it.

The birth of zero probably happened between 400 and 600AD in India. The Indian mathematician Brahmagupta was one of the first to try and write down the rules for working with negative numbers and zero. Here is an extract from his book Brahmasphutasiddhanta, written in 628AD:

He describes positive numbers as fortunes and negatives numbers as debts.

2.3.2 Multiplying with negatives

Check through his rules and see if they make sense (use a calculator if it helps).

In Brahmasphutasiddhantajust like those we use today. In addition, he gives algebraic methods for finding square roots and solving quadratic equations, as well as method for summing series.

The book also contains astronomical calculations of the planets and eclipses that involve sophisticated use of trigonometry.

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He did really well to get all of these rules correct. Let’s take a bit of time to look at the rules involving products (and quotients). Most people will agree that the product of two fortunes is a fortune. A bit of thought will convince you that the product of a fortune with a debt is a debt (for example, 3 debts is just more debt).

How about the rule ‘the product of two debts is a fortune’ (i.e. two negative numbers multiply to make a positive number)?

If you are not convinced why this is true, try this ‘proof’ using the grid method for multiplication. Consider the product 27 x 38 (for example); we know the answer is 1026, so let’s use the grid method to work it out, using 27 x 38 = (30 – 3) x (40 – 2):

So far we have got 1020, using the rules we are intuitively happy with. Now we are left with -2 x -3, which must be +6 in order to make 1026.

Here’s a fun investigation that involves multiplying negatives. Look at the following grid; each square is going to change according to the following rule:

Multiply the signs of all the squares that share an edge with that square, and change that square to that sign.

For example, looking at the middle top square (in red on the left) it shares an edge with three other squares (in green). Multiplying these green signs together, we get positive, so the red square changes to a positive.

Here is what happens after working out the new sign for each square.

I am not going to give this one away, except to give you the hint that squares can’t remain negative forever… 2.3.3 Adding and subtracting negatives

Now, time for a moment of caution! In my years of teaching secondary school pupils, they always want to jump to nice easy rules. The classic with negative numbers is that two negatives make a positive. We have just seen that it is true that two

Do this again on this new grid and see what happens. What do you notice?

Try for other starting grids… can you explain what is happening?

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negatives multiply (and divide) to make a positive. It is also true that subtracting a negative number is equivalent to adding a positive (taking away a debt is like giving someone a fortune).

However, it is definitely not true that adding two negatives make a positive! To see why, consider the hours of rain and sunshine added over two days. If we have 10 hours of rain one day and 10 hours of rain the next, this is 20 hours of rain (and a pretty bad couple of days) – it does not miraculously turn into twenty hours of sunshine!

Another way of thinking about this is using holes and piles, as described by James Tanton in his excellent book Mathematics Galore.

There aren’t any rules for adding and subtracting positives and negatives, it just depends on how many you have of each.

This is shown expertly in this diagram by one of my pupils:

And here is another way how my pupils think of adding positives and negatives.

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Here’s a puzzle that requires you to be able to add postives and negatives:

Write down any 4 positive or negative numbers in a circle that have a sum of +1. Now find the sums of all the chains of adjacent numbers.

So, for example, with the (blue) numbers in the circle shown on the right the adjacent pair sums are shown in red and the adjacent triple sums are shown in green.

To see how these have been calculated, the red +7 is the sum of the adjacent blue pair +3 and +4. The green +6 is the sum of the adjacent blue triple -1, +3 and +4.

In the diagram shown, there are 6 positive numbers and 6 negative numbers. The blue numbers sum to +1, the reds to +2 and the greens to +3.

In order to explain why, consider the fact that if certain coloured numbers are positive, the others of that colour must be negative.

Investigate for other blue numbers of your own. What do you notice? Can you explain why?

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2.3.4 Dividing by zero

The Indian Mathematician Brahmagupta also came up with rules for dividing with zero. His rules are show below:

• Numbers when divided by zero are fractions with zero as the denominator• Zero divided by a number is either zero or a fraction with zero as numerator• Zero divided by zero is zero

You may have noticed that the first rule doesn’t really mean anything! The second rule is correct, but the third rule is definitely questionable!

It took another 500 years for another Indian mathematician Bhaskara (1100AD) to get closer to an answer to the problem of dividing by zero. He considered dividing 1 by smaller and smaller fractions.

Starting with 1 divided by a half = 2, we then go for 1 divided by a third = 3, and so on... with the answer getting larger and larger. He concluded that 1 / 0 = infinity and so n / 0 = infinity.

One way to check whether this makes sense is to rearrange the calculation into a multiplication. For example, 6 divided by 3 = 2 makes sense because 2 x 3 = 6.

So Bhaskara’s rule 1 divided by 0 = infinity implies that infinity x 0 = 1, which doesn’t really make sense – so he probably should have concluded that you shouldn’t divide by zero.

And what about 0 divided by 0? Let’s try the multiplication test on this: Suppose the answer is some number, N. Then we are saying 0 divided by 0 = N, so that 0 x N = 0. This seems fine, except that the number N could be anything! This time we have too many possible answers, and we say that 0 divided by 0 is not defined. (Of course, we could define it to be whatever we want as mathematicians, but it should have to make sense!)

Check these rules: Do they make sense? Are they correct?

What is something divided by zero? How about zero divided by zero?

Do you agree with his conclusion?

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While we are talking about Bhaskara, here is a problem from his famous book Lilavata (which means ‘arithmetic’), which is about fractions not zero. I found this in the excellent book From Five Fingers to Infinity edited by Frank Swetz:

I wish all questions were written like this! So what’s the answer?

Well, we can call the total number of bees B. So we have B/5 going to the lotus flower, plus B/3 going to the banana tree and 3 x (B/3 – B/5) going to the Codaga tree. Adding the undecided one, we then have a total of:

B=B5

+ B3

+3 (B3 – B5 )+1Adding the two fractions B/5+B/3=8 B/15 (why?) and subtracting the other two B/3– B/5=2 B/15 so we have

B=8B15

+3× 2B15

+1

So B = 14B/15 + 1, so that there must be 15 bees in the swarm.

One fifth of a swarm of bees flew towards a lotus flower, one third towards a banana tree. A number equal to three times the difference between the two preceding amounts, O my beauty with the eyes of a gazelle, flew towards the Codaga tree (whose bitter bark provides a substitute for quinine). Finally, one other bee, undecided, flew hither and thither equally attracted by the delicious perfume of the jasmine and the pandanus. Tell me, O charming one, how many bees were there?

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3 Fractions

Everybody loves fractions… not! I’m not sure why, but most pupils I teach start off hating fractions, probably due to previous failures when trying to work with them. However, I think if they realize how long it took the human race to understand fractions, they won’t feel as bad.

Before starting this section, you should read the section on fractions in the Glossary to remind yourself (or learn for the first time) how to calculate with fractions.

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3.1 Babylonian fractions

Before starting this section, you should make sure you are happy with the link between fractions and decimals, and have tried the section binary fractions.

The Babylonians used fractions in their calculations. They were using base-60, so the columns in their number system can be extended to include fractions like this:

3600 60 units 1/60 1/3600

Notice how their first ‘fractions’ column is 1/60ths compared with our 1/10ths.

Let’s try writing some numbers using this sexigesimal system. We will represent the gaps between columns with a comma, and the decimal point will be represented by a semi-colon. So the number above would be:

12, 16; 4 = 12 x 60 + 16 x 1 + 4 x 1/60 = 736 and 4/60

Now, note that 4/60 is the same as 1/15. In the Babylonian system it is easy to write this fraction, it’s just 0; 4, but in our number system its comes out to be a long recurring decimal 0.0666…

You can see that the many common fractions can be written simply in sexigesimal but are recurring decimals in our number system.

For example, 1/3 is the same as 20/60 which is just 0; 20. Compare this with the decimal for 1/3, which is 0.333….. This is because 60 has many factors, and so many fractions can be represented nicely out of 60.

The Babylonians were definitely onto something using base-60.

Explore some other common fractions in sexigesimal and decimal.

Do you think the Babylonian number system is good for working with fractions? Can you explain why? [Hint: consider the number 60]

What would be other good numbers to use as a base? Try using different bases to write fractions – which one is the most convenient? Convince a friend!

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Now, let’s explore Babylonian fractions a bit further. To do this we need to know about the reciprocal of a fraction: we find the reciprocal of a fraction by flipping it upside down, so 2/3 and 3/2 are reciprocals.

Using 2/3 and 3/2, we get 2/3 x 3/2 = 6 / 6 = 1. In fact, any pair of reciprocals will multiply to give 1; this is an alternative way of defining them.

What we need to multiply by 2 to make 1 is 1/2. This also makes sense if we think of 2 as 2/1. The reciprocal of any whole number n is 1/n.

The Babylonians kept tables of reciprocals like this:

2 0; 303 0; 204 0; 15

They used these for dividing. For example, instead of dividing by 5 they would multiply by 1/5. So the calculation 75 divided by 5 became 75 x 1/5, or in Babylonian notation we have 1, 15 x 0; 12. Ignoring the sexigesimal point (the semi-colon) for a moment, 1, 15 x 12 = 12, 180 which is the same as 15, 0 (why?). Now we reintroduce the sexigesimal point to give 15; 0, or just 15.

I mentioned earlier that the Babylonians found an approximation to the square root of 2. How did they do this?

They did it by successive approximations. An initial estimate of 1; 30 is too large (what is it in our numbers?) so they divided 2 by 1; 30.

The reciprocal of 1; 30 is 0; 40 and 2 multiplied by 0; 40 is 1; 20 which is too small. The average of the two is 1; 25. Repeating this with 1; 25 as the initial approximation gives their approximation of 1; 24, 51, 10.

While we are talking about reciprocals, can you solve this problem:

Write down some other reciprocals. What do you get if you multiply reciprocal pairs together?

What is the reciprocal the whole number 2? Or of any whole number n?

Find the Babylonian reciprocals for 5, 6 and 10. How about 8 and 9?

What is the reciprocal of 1; 30? Can you use this to work out 2 divided by 1; 30?

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Let’s try a few, such as 1/2, which gives 1/2 + 2/1 = 3/2 (no) and 1/3 + 3/1 = 10/3 (no). We can see this approach is not going to be very effective.

Let’s go for a bit of algebra: Let the fraction be a /b, so we are adding a /b+b /a which

gives a2+b2

ab.

Now, for this to be a whole number, we need ab to divide into a2 + b2 so that a must divide into b2 (why?) and also b must divide into a2. The only way this is possible is if a and b are 1, so the only fraction we can have is 1/1!

Add any fraction to its reciprocal. For which fractions is the answer a whole number?

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3.2 Egyptian fractions

The Ancient Egyptians were one of the first civilizations to really start thinking about fractions. However, they only used unit fractions (of the form 1/n) and would write them like this:

The symbol is called rho, and denotes a (unit) fraction.

They would write them as sums of unit fractions. However, they did not take the easy way out! You might have thought they would just write 2/5 as 1/5 + 1/5 but for some reason they didn’t – they had strict rules about how fractions like this were written.

They created a table of 2/n fractions on something called the Rhind Papyrus, which can be seen at the British Museum.

It turns out they wrote 2/5 as 1/3 + 1/15:

You could write 2/5 in lots of different ways, but the first unit fraction (1/3) is the largest one that is smaller than 2/5, so 2/5 = 1/3 + 1/15 seems quite sensible. We could draw it like this:

This picture suggests a method for finding other 2/n fractions:

How do you think Egyptians wrote non-unit fractions, like 2/5?

Is this the only way of writing 2/5 using unit fractions? How do you think they decided which unit fractions to use? How might the Egyptians have written 2/7?

Can you find a pattern that will help you write other fractions of the form 2/n? Or 3/n?

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So we can write 2/7 = 1/4 + 1/28.

Following on this pattern we have 2/9 = 1/5 + 1/45, 2/11 = 1/6 + 1/66 and so on…

Strangely, the Egyptians did not use this pattern to write 2/n fractions. They had a range of criteria for writing them. For example, the Rhind Papyrus gives 2/9 = 1/6 + 1/18 and 2/15 = 1/10 + 1/30 which suggests the use of the formula 2/3n = 1/2n + 1/6n.

Can you a quick way of finding these? Can you continue it? Can you adapt your method for finding fractions of the form 3/n?

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3.3 Fibonacci’s fractions

In the Book of the Abacus, Fibonacci explains how to write any fraction as the sum of unit fraction (like the Egyptians) using something called the Greedy Algorithm. The algorithm says that you repeatedly take the largest unit fraction away from any fraction, you will always eventually get an expression of the fraction as a sum of unit fractions.

Let’s try it for the fraction 3/7. First we need to find the largest unit fraction less than 3/7; it is 1/3 (why?).

Then we subtract it: 3/7 –1 /3=9 /21 –7 /21=2/21. Now we apply the same process to 2/21.

Using equivalent fractions, 1/10 is the same as 2/20 (which is larger than 2/21), whereas 1/11 is the same as 2/22 which is slightly smaller than 2/21, so the fraction we’re looking for is 1/11.

Now 2/21– 1/11=1/231 (using the pattern from Egyptian fractions above) and we have the final answer 3/7=1/3+1/11+1/231.

Consider the fraction a /b, and subtract the largest unit fraction 1/N that is less than

a /b. If we do this we get Na−bbN .

Now a /b>1/N implies that Na > b and so the numerator of this fraction is positive.

Fibonacci, or Leonardo of Pisa, lived during the 13th century. Today he is mostly famous for his sequence, of which we will learn more in chapter 6, but he did much more important than that.

He travelled with his merchant father through many Islamic countries and came back to Europe and wrote Book of the Abacus, which explains many methods for calculation he found there, and advocates use of the Hindu-Arabic numerals we use today. He is possibly the main link between Islamic mathematics and modern Europe.

The book also contains examples of algebraic manipulation and solutions to Diophantine equations.

What is the largest unit fraction less than 2/21?

Try some more applications of the Greedy algorithm. Can you spot any more patterns? Can you explain why it always works (this is difficult)?

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Also a /b<1/(N−1) by our choice of N, so that a (N−1)<b which means that Na– b<a. What this means is that if we repeat this process then we will get a series of fractions that are positive with ever smaller numerators, so eventually we will get to a numerator of 1, and we will always get a sum of distinct unit fractions.

I’ll leave this one to you! Next we are going to learn about a method for writing and working with fractions invented by Fibonacci. This is taken from the excellent book, A History of Mathematics by Jeff Suzuki.

First of all, he wrote mixed numbers back to front, so instead of 5 ½ he would write ½ 5.

He also wrote fractions as sums of products. For example, he wrote 82123 to

represent 812.3

+ 23 . The reason he did this was to make it easier to work with the

imperial measurements in use at the time.

Consider the measurement 5 yards, 2 feet and 8 inches. As there are 3 feet in a yard,

and 12 inches in a foot, he would write this as 82123

5 yards.

His method also simplified division. Suppose we wanted to work out 326 divided by 24. He would split 24 into a product of factors, let’s say 3 x 8. We could write 1/24 as

1/3.8 which in his notation would be 1038 . Now he would proceed as follows:

Divide the first factor, 3, into 326 giving 108 remainder 2. He would write the remainder 2 above the 3 in his answer. Then he would divide the second factor, 8, into 108 giving 13 remainder 4. The 4 is then written above the 8, and his answer

would be 243813.

We can check this is correct: 243813 is the same as 13 + 2/3.8 + 4/8 which is 13 and

7/12, which is the correct answer.

Try Fibonacci’s method to work out some other division sums.

Can you explain how it works?

What happens if we use this process to write 1 as the sum of (distinct) unit fractions?

1 = 1/2 + 1/3 + 1/7 + …

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3.4 Galileo’s odd fractions

You probably know that 9/15 and 6/10 are equivalent to 3/5. In other words, these are all different ‘names’ for the same (rational) number; we say that 3/5 is irreducible, or in simplest terms. To find the simplified fraction we divide top and bottom by any factors common to both until they are relatively prime (have no common factors).

Here are some surprising equivalent fractions discovered by the famous physicist Galileo:

13=1+35+7

= 1+3+57+9+11

=…

Here is a picture that might help explain what is happening:

Talking of ‘odd’ fractions, here are some equivalent fractions:

I hope you found they are all equivalent to ½.

Now, here’s a very unusual method for ‘proving’ this to be true – note: this is NOT how to simplify fractions!

Check that these are equivalent fractions. Can you explain why this works?

Check that these are all equivalent.

Why does this ‘work’?

Can you find other sequences of fractions that behave like this?

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3.5 Fraction trees

One German mathematician, Moritz Stern, and one French clockmaker Achille Brocot, independently discovered our first fraction tree, the Stern-Brocot tree, in the 19th century. Brocot used the tree to design gear systems for clock designs.

Starting with 0/1 and 1/0 (ignoring the fact that 1/0 is not allowed), we construct a new fraction called the mediant from this by just adding the tops and bottoms (note that this is NOT how to add fractions!).

So here we construct 1/1, and write the mediant between the previous two in sequence: 0/1, 1/1, 1/0. Now we insert mediants in the two gaps between these fractions to give 0/1, 1/2, 1/1, 2/1, 1/0.

Here is the sequence of fractions shown in a tree:

You can see that no fraction is repeated, in that there are no equivalent fractions, and that all fractions are simplified.

What is more, it can be shown that this procedure will create all the positive fractions!

To see why there are no equivalent fractions is easy; the mediants are always between two existing fractions by the way they are constructed.

To see why they are all simplified fractions is slightly tricky but if you’d like to then here it is. Consider the first two fractions 0/1 and 1/0. Calling these a/b and c/d we have:

bc – ad = 1

Now if a and b are relatively prime (have no common factors) then this equation means that so are c and d. This is an example of something called a Diophantine equation: more of these later. For now, you can just check this with some numbers of your own.

Now, the next fraction we create will be (a+c)/(b+d) and putting this in place of a/b in this equation we get (b+d)c – (a+c)d = bc – ad = 1. So the top and bottom of any

Find the next set of mediants between these fractions, then the next. What do you notice about the fractions this procedure is creating?

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new fraction we create are also relatively prime – so all the fractions we create are simplified!

It’s much more tricky to show that we also don’t miss any out; can you convince yourself this is true?

A different way of thinking about the generating the next rows in the Stern-Brocot tree is to consider how the ‘parent’ and ‘siblings’ are related.

Calling the parent a/b, you might have noticed that the left sibling is a/(a+b) and the right sibling is (a+b)/b. This gives us an easy way of generating the next rows of the tree.

Interestingly, the Stern-Brocot tree gives us a different way of representing all the fractions. If we use 0 or 1 to represent travelling left or right down the tree, then each fraction can be represented by a binary string of zeroes and ones, for example 010 = 3/5.

Further than this, we can also construct the binary string from the fraction:

Starting with 3/5, if top is smaller than the bottom this is equivalent to a move Left and we subtract the top from the bottom to give 3/2. Now the top is bigger than the bottom, which is a move Right. This time we subtract bottom from top to give 1/2, which is a Left, giving the root of the tree, 1/1.

Here’s another interesting property of this tree: for each fraction a/b in the tree, calculate 1/ab. Then add together all these values for each row and see what happens. For example, in the third row we have 1/3, 2/3, 3/2 and 3/1. Calculating 1/ab for each of these gives 1/3, 1/6, 1/6 and 1/3. Now try this for some more rows… anything interesting?

Find the strings for other fractions, and construct the fractions for other strings. Can you find any patterns?

Can you find any other patterns in this tree?

Consider the parent 3/2 and its siblings 3/5 and 5/2. Can you think of a rule that tells you how to create the siblings from the parent?

Use this method to generate the next row of the tree and compare it to the mediant method – are they the same?

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For more on the Stern-Brocot tree, see the book Concrete Mathematics by Graham, Knuth and Patashnik.

The famous Farey series is another set of fractions constructed in the same way but with starting numbers 0/1 and 1/1. This creates all the fractions from 0 to 1. It has some interesting properties but I am not going to go into them here – they are freely available on the internet.

Another interesting fraction tree is the Calkin-Wilf tree that has a similar construction:

Considering parents and siblings, if the parent is a/b then each left sibling is made from a/(a+b) as for the Stern-Brocot tree, and each right sibling is b/(a+b), which is the reciprocal (the upside down version) of the Stern-Brocot tree.

This tree also produces the fractions between 0 and 1.

I’ll leave this one open to you!

Find out more about the Farey series.

Can you see how this has been constructed? Can you see any interesting patterns in this sequence?

Do you think it produces all the fractions between 0 and 1? Can you convince yourself this is true?

One last thing; notice how we started with different top numbers in the two fraction trees. What happens with other starting numbers, or perhaps the same starting number in each tree; can you find any links between them?

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3.6 Leibniz’s fractions

We are going to look at another one of Leibniz’s discoveries called the harmonic triangle but before we do, a few investigations to get you started.

First let’s try adding the whole number sequence 1, 2, 3, 4, … together one by one like this: starting with 1, then 1 + 2 = 3, then 1 + 2 + 3 = 6, then 1 + 2 + 3 + 4 = 10, …

These totals are called the partial sums of the sequence. The partial sums of the whole numbers are the triangle numbers, which makes sense if you think about it (draw a picture if you’re not sure why)!

Here are some more to try:

You may have noticed that the partial sums of the odd numbers are in fact the square numbers. Here is a ‘proof without words’ that shows why this is true:

The partial sums of the even numbers are double the triangle numbers (which also makes sense if you think about it).

The partial sums of the doubling sequence are one less than the powers of 2 (think back to the section on finger maths).If we carry on calculating these partial sums forever (called the sum to infinity), we will get bigger and bigger numbers that get closer and closer to infinity. Must this always happen with the partial sums of any sequence?

What do you notice about the totals? Have you seen this sequence of numbers before?

Find the partial sums of the odd number sequence 1, 3, 5, 7, …

Find the partial sums of the even numbers 2, 4, 6, 8, …

Find the partial sums of the doubling sequence 1, 2, 4, 8, 16, …

Try some more of your own. What do you notice about any of them? Can you explain why?

Find the sum to infinity of the sequence 1, -1, 1, -1, 1, …

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What do you think the answer to this one is? In Leibniz’s time this was a cause for great debate. Some people though 1, others thought -1, some went for 0, while others went for 1/2! In fact, it can be anything you want it to be. For example, if I put brackets like this:

( 1 + -1 ) + ( 1 + -1 ) + ( 1 + -1 ) + …

then we could justify the answer being 0, but if we put brackets like this:

1 + ( -1 + 1 ) + ( -1 + 1 ) + …

then we can justify the answer being 1. It’s probably safest to say this one doesn’t really have a sum to infinity.

Things get more exciting when we look at series involving fractions!

We calculate the partial sums to be 1/2, 3/4, 7/8, 15/16, 31/32 …

If we keep doing this forever, we see that the top number is always one less than the bottom number, so the partial sums are getting closer and closer to 1.

In other words, the sum to infinity of this series is 0.999…. or is it 1? Well, a mathematician would say they are the same thing!

Here’s one way I think about it. What is 1/3 as a decimal? I guess you would probably say 0.333… so let’s multiply this by 3: 3 x 1/3 is definitely 1, and 3 x 0.333… is 0.999… so they are the same thing…?

These questions about infinity have troubled mathematicians for thousands of years so don’t worry if you are slightly troubled too. A classic example of someone who was troubled by infinity is Greek mathematician/philosopher Zeno.

Find the partial sums for the halving sequence: 1/2, 1/4, 1/8, 1/16, … What do you think the sum to infinity is?

Do you agree with the statement that 0.999…. is the same as 1? Explain why you believe what you do!

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He came up with lots of paradoxes about infinity; here is one of them, called the Dichotomy:

Suppose you are in your house somewhere and want to get to the kitchen. To get there, you must go halfway. Then once you have got halfway, you must go half of what remains (by now you have gone 3/4 of the way), and then you must go half of what remains again (starting to look familiar?). If we keep doing this forever, you will never reach the kitchen therefore motion is impossible.

Do you think motion is impossible? Or alternatively do you believe that infinite series can have finite sums, that 0.999… is the same as 1?

So where were we? OK… we have seen that the sum to infinity of the previous sequence 1/2 + 1/4 + 1/8 + … = 1. Does this mean that all fraction sequences that get smaller like this have a sum to infinity?

You might have realized that A has a sum to infinity = 1/2. Here is a proof without words to show this is true:

You might recognize B from the section binary fractions; it does have a sum to infinity of 1/3!

How about series C and D? Well, this is where Leibniz comes in… Series C is called the harmonic sequence, and he discovered it in his harmonic triangle:

Before reading on, can you work out how it is constructed?

Can you see the harmonic sequence?

Can you find any other interesting patterns in the triangle?

Can you see how it is related to Pascal’s triangle?

Find out about some more paradoxes about infinity, such as Hilbert’s Hotel or Thomson’s Lamp.

Find the partial sums for these sequences and try to work out their sums to infinity:

A. 1/3 + 1/9 + 1/27 + ...B. 1/4 + 1/16 + 1/64 + ...C. 1/1 + 1/2 + 1/3 + 1/4 + ...D. 1/1 + 1/3 + 1/6 + 1/10 + …

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Every fraction in the triangle is calculated by adding the two below it.

For example, 1/12 + 1/4 = 1/3.

Or you could think of this as 1/3 – 1/12 = 1/4.

Now let’s concentrate on the harmonic sequence – it is there in the first diagonal. Let’s try finding its sum to infinity.

Starting with the first two we have 1/1 + 1/2 = 3/2, then we have 1/1 + 1/2 + 1/3 = 11/6 (which is just a bit less than 2), then we have 1/1 + 1/2 + 1/3 + 1/4 = 50/24 (which is a bit more than 2)…

In fact, the sum to infinity of the harmonic series is… infinity! Here is a proof:

1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 +...

> 1/1 + 1/2 + (1/4 + 1/4) + (1/4 + 1/4 + 1/4 + 1/4) + ... (why?)

= 1/1 + 1/2 + 1/2 + 1/2 + ...

We say that the harmonic series is divergent, compared with the previous ones, which had a (finite) sum to infinity and are called convergent.

Finding partial sums we get 1/2, then 2/3 (=1/2 + 1/6) then 3/4 (=1/2 + 1/6 + 1/12) then 4/5 and so on… and it looks like the top number is always one less than the bottom one (can you prove this?). This is just like the halving sequence, and so we should be safe to say its sum to infinity is 1.

Finally, How about the last one, series D: 1/1 + 1/3 + 1/6 + 1/10 + …? For this one, look closely again at the harmonic triangle:

What is the sum to infinity for the second diagonal 1/2, 1/6, 1/12, 1/20, ...?

What do you think is the sum to infinity of the harmonic series? Can you prove it?

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To find this sum, Leibniz noticed that the sum of the differences of any increasing sequence a, b, c, d, e, ... , z is just the last term subtract the first term, that is z - a.

By the way it is constructed, the numbers in the second diagonal of the harmonic triangle are the differences of the first diagonal. So the sum of the differences 1/2 + 1/6 + 1/12 + ... is the same as the last term subtract the first term of the first diagonal, namely 1/1 - 0 = 1.

Now we can see that series D: 1/1 + 1/3 + 1/6 + ... is double this series, so must have sum to infinity = 2.

Leibniz's exploration with infinite series and the harmonic triangle led to his discovery of the calculus, one of the most important mathematical developments in history.

Try this out to see if it is true.

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3.7 Euler’s Totient Numbers

Here is one of the questions he pondered:

For how many of these fractions is n the least possible denominator?

1n , 2n , … , n−1n

For example, with n = 4 we have ¼ and ¾ where 4 is the least possible denominator, and 2/4 where it is not. So there are 2 fractions with 4 the least possible denominator.

Here is a table of results for denominators bigger than 2:

Find the number of fractions for n up to 12. Any patterns?

Leonard Euler was possibly one of the most famous mathematicians of all time. He lived during the 18th century and made many discoveries about mathematics, in areas such as calculus, analysis, trigonometry and number theory, proving many results that had not previously been proved. He also invented whole new branches of mathematics such as graph theory.

He also invented much of the notation we used today, including function notation and complex numbers, and has a number named after him: e = 2.71828….

I am a particular fan of his formula V – E + F = 2 which gives a relation between the vertices, edges and faces of any convex polyhedron, but most people prefer his other equatione iπ=1.

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You might have spotted a few interesting things, such as the number of fractions is always even. Will this always be true, and if so, why?

Also, for prime numbers, the number of fractions is one less than the denominator. This makes sense because all fractions do not simplify if the denominator is prime. So this clearly has something to do with prime-ness.

We can think of it in a different way as being the number of numbers that are relatively prime to the denominator. This number is called Euler’s totient function, usually written using the Greek letter phi ϕ (n). So we would say that ϕ (12 )=4.

Here are some more exciting things about it.

You might have spotted that ϕ (12 )=4=2×2=ϕ (4 )×ϕ (3 ), or that ϕ (15 )=8=4×2=ϕ (5 )×ϕ (3 ). In fact, it is always true that ϕ (mn )=ϕ (m )×ϕ (n ) if m and n are relatively prime (this is called a multiplicative function).

If we try really hard, we might be able to find a formula for working out ϕ (n ). Let’s consider ϕ (100 ).

Well, numbers that have common factors are those with a factor of 2 or a factor of 5 (why?). So of the 100 numbers we can get rid of 1/2 of them (for the factors of 2), which is 50, and then another 1/5 of those left (for the factors of 5) which is another 10, leaving 40. So ϕ (100 )=40.

We could have calculated this as 100 x 1/2 x 4/5 = 40. This is leading us to conjecture a formula.

Don’t worry if you couldn’t follow in the footsteps of the great Euler. It turns out that the formula is:

ϕ (n )=n×(1−1p )×(1−1q )×(1−1r )×…Where p, q, r, … are the different prime factors of n. We can test this formula out for ones we know, like this:

Find ϕ (n ) for some more non-prime numbers n. Can you see any connection between ϕ (n ) and the totient numbers for the factors of n?

What is ϕ (100 )? Can you think of any systematic ways of working it out without having to list everything?

Try some other calculations like this. Can you come up with a formula yourself?

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ϕ (15 )=15×(1−13 )×(1−15 )=15× 23× 45=8

One last amazing thing about Euler’s totient function is that if we add up all the totient numbers for the factors of some number n, it will equal the number itself. For example:

ϕ (10 )=ϕ (1 )+ϕ (2 )+ϕ (5 )+ϕ (10 )=1+1+4+4=10

Test the formula out for some other numbers to check it works.

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3.8 Recurring decimals*

You should try the section on binary fractions and Babylonian fractions before trying this section.

Have you ever wondered why some fractions have nice well-behaved decimals like 1/4 = 0.25, which stop after a couple of decimal places (terminating), while others are more badly behaved like 1/7 = 0.14285714… which go on forever (recurring)?

Well, after looking at binary and Babylonian fractions you may have already realized it has something to do with our number system. For example, 1/3 = 0.333… is badly behaved for us, but for the Babylonians it equals 0;20 which is nice!). So what exactly makes a fraction recur?

After a bit of investigating you may have noticed that terminating fractions have a denominator that is a multiple of 2 or 5. Why exactly does this mean that they terminate?

Let’s look at 1/4, which means 1 divided by 4, which means ‘how many 4s are there in 1?’ Using long division we get the terminating decimal 0.25. Looking at the way long division works, for a decimal to terminate we need a zero remainder at some point in the calculation:

Investigate different decimal fractions on a calculator. Put them into two lists of terminating and recurring ones. Can you describe exactly what makes a fraction terminate or recur?

For the recurring ones, can you see any further patterns?

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Another way of thinking about this is to say that 4 doesn’t go into 1 or 10, but does go exactly into 100, which is exactly why this terminates after 2 decimal places. To use the terminology of modular arithmetic, we can say that 100≡0 (mod 4 ). So really this is about powers of ten. If any power of ten is congruent to 0 for the given denominator then it will terminate.

Let’s look at the congruences of powers of 10 under different mod n, where n can be thought of as the denominator of some fraction.

mod 2 3 4 5 6 7 8 9

100 1 1 1 1 1 1 1 1

101 0 1 2 0 4 3 2 1

102 0 1 0 0 4 2 4 1

103 0 1 0 0 4 6 0 1

104 0 1 0 0 4 4 0 1

105 0 1 0 0 4 5 0 1

106 0 1 0 0 4 1 0 1

107 0 1 0 0 4 3 0 1

108 0 1 0 0 4 2 0 1

These are basically the remainders you get when you do the long divisions for each one (try them).

You might have noticed a few things from this table:

10 is congruent to 0 mod 2 and 5, which means 1/2 and 1/5 have only one decimal place in their decimals.

100 is congruent to 0 mod 4 as we saw earlier, which means 1/4 has two decimal places in its decimal.

1000 is congruent to 0 mod 8, so 1/8 terminates after 3 decimal places (=0.125)

The fractions 1/3, 1/6, 1/7 and 1/9 never terminate as they never have 0 remainders.

The red squares are examples of Fermat’s Little Theorem in action: 1≡10p−1(mod p) for p = 3 and 7.

Let’s look at the recurring decimals in more detail now. We can see from the remainders how the decimals recur. For example, all the remainders for 3 and 9 are the same, which is why they have the same number in their recurring decimal. Similarly, 6 = 0.16666 has the same remainder (4) apart from the first one.

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The most interesting number here modulo is 7. It cycles through all the remainders 1 to 6 before going back to the start (remainder 1). We say it has cycle length 6.

Here are the steps in the long division of 1/7 set out one after another:

1 = 0 x 7 + 1

10 = 1 x 7 + 3

30 = 4 x 7 + 2

20 = 2 x 7 + 6

60 = 8 x 7 + 440 = 5 x 7 + 550 = 7 x 7 + 1 ≡ 0 x 7 + 1

We can see by step 6 we are back to where we started, with a remainder of 1, which is why it has a cycle of 6. If we think about this for a bit longer, it’s fairly obvious that the longest cycle we can ever have is one less than the denominator, and this only happens if we cycle through all the possible remainders.

It turns out the next fraction with cycle length 1 less than the denominator is 1/17. You might have conjectured that 1/13 = 0.07692307 was going to be one, but it only has a cycle length of 6.

You might have also have noticed that all the cycle lengths are factors of one less than the denominator (6 is a factor of one less than 13).

At this point you have probably realized that it has something to do with primality. Well, yes it has; all numbers that are relatively prime to 2 or 5 will not terminate as they will never have 0 remainders.

Putting this all together, we can finally say that 1/n will be a recurring decimal with a cycle length a factor of n-1 if n is relatively prime to 10.

Investigate the cycle lengths of other decimal fractions. Can you find any fractions with cycle lengths one less than the denominator like 7? What can you say about the cycle lengths of different fractions in general?

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3.9 Greek Ladders

Here is something called a Greek ladder from around 100AD, although there is evidence it may have been known to Indian mathematicians 500 years earlier as it appears in their Sulba-Sultras, religious texts that contained instructions on how to construct altars to high levels of precision. The Sulba-sultras gave a way of constructing squares equal to the sum of two smaller squares, also showing that they were aware of Pythagoras' theorem before the Greeks.

A B1 12 35 712 1729 4170 99

One pattern you might have noticed is that each row is created from the one above it as (A+B, 2A + B).

You can get more of an idea what the numbers might represent by squaring each number like this:

1 14 925 49144 289841 1681… …

Now you might see that B2 /A2 is approximately 2. So B /A is an approximation for the square root of 2. The approximation gets better as the ladder goes down. We can never use this method to get the exact value of √2 as it can’t be written down as a fraction – it is called an irrational number (fractions are called rational numbers).

Write down as many patterns as you can see in these numbers.

What do you think these numbers represent? [Hint: try writing A and B as a fraction]

Try typing 90/77 into your calculator. How close is it to √2? Find the next few numbers on the ladder and see how close they are to √2.

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A different way of thinking about the numbers for A and B are as solutions to the equation B2−2 A2=±1, which is an example of a Pell’s equation, named after English mathematician John Pell.

A different way of making the ladder is to start with A = 1, 3, … and B = 1, 2, … then use the rule that the next number is 2x the one before it plus the one before that.

So the A numbers are 1, 3, 7, 17, 41, … and the B numbers are 1, 2, 5, 12, 29, …

This method is linked to something called continued fractions.

The continued fraction of root 2 is given by the (infinitely) repeated fraction on the right.

Understanding continued fractions is a bit tricky; if you want to skip this bit that’s OK. If not, then here goes:

Starting with 1, we then have 1+12=32 , followed by

1+ 1

2+ 12

=1+ 152

=1+ 25=75 and so

on.

Another way of thinking about this is to choose different rules for creating fractions. For the first Greek Ladder, we could have created these numbers using the rule (a+2b)/(a+b), so starting with 1/1, we then get 3/2, then 7/5 and so on.

Here is a different Greek ladder:

1 11 23 54 711 1915 26

Can you work out what it is for?

What fractions does the rule (a+3b)/(a+b) create? Investigate other rules like this and see what sequences of fractions you can create. Why not try different starting numbers too?

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If you start with 1/1 and do (a+3b)/(a+b) you get the sequence of fractions 1/1, 4/2, 10/6, 28/16, 76/44 which at first sight doesn’t look familiar, but if you simplify each fraction you get the Greek ladder above, which is an approximation for √3.

Why not find out more about irrational numbers, Pell’s equations and continued fractions?

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3.10 Problem of the points

Fractions are also used to work out probabilities. Here is a little taste:

In 1654, two French mathematicians Pascal and Fermat sent each other several letters about an old problem studied by Luca Pacioli (a mathematician friend of Leonardo da Vinci) as early as 1494. The problem was this:

Two players are playing a game of chance (let’s say throwing a coin) and the winner is the first to get to an agreed number of points (heads). How should the winnings be divided fairly if the game is interrupted before the end?

Pacioli’s solution was to divide the winnings according to the points the players had scored before the game was interrupted. However, Pascal and Fermat noticed that this lead to unfair results; suppose the score was 1-0 and it was first to (say) 10. Then player 1 would get all the money but it is by no means certain that they are going to win.

So they devised a way of sharing the winnings that was based on the probabilities of what might happen in the future.

Suppose it is first to 10 and the score is 9-8. Can we work out the probability of each player winning? Yes we can, using a probability tree like this:

It is clear from the tree that player 1 has a ¾ chance of winning, compared with player 2’s chances of ¼. So the winnings should be shared in the ratio 3:1 in this case.

Suppose it is first to 10 and the score is 9-8. How would you share out the winnings of (say) £100 if the game is interrupted at this stage?

Calculate the ratios for sharing the winnings for some other scores. Can you find an easy way to work out the ratios? [Hint: it is connected to Pascal’s triangle]

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In the example given, the score is 9-8 and the winner is the first to get 10 points. Player 1 requires only 1 point (head) whereas player 2 requires 2 points (tails). We showed (using a tree diagram) that the probabilities of winning were ¾ and ¼ respectively, so the winnings should be shared in the ratio 3:1.

Instead of drawing a tree diagram, we could have just looked at the number of ways of getting 1 or more heads (player 1 win) in two tosses of a coin compared with the number of ways of getting 2 tails (player 2 win). Note that although the game ends if the first throw is a head, we can still analyse it in this way.

So player 1 wins if the ‘two’ tosses are HH, HT or TH compared with TT only for player 2, and we have the ratio 3:1.

We could work out this ratio even quicker using Pascal’s triangle as this is exactly what the numbers represent. Looking at the third row we can work out the fair way to share the money by doing 1+2:1

To see the ease of this method, suppose that the score is 8-7 so that player 1 requires 2 points and player 2 requires 3 points. If you draw the (rather large) tree diagram or list all the possible outcomes you will see that the winnings should be split in the ratio 11:5.

However, a glance at the fifth row of Pascal’s triangle reveals this answer immediately:

We will find out more about Pascal’s life and discoveries later.

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3.11 Pigeonhole fractions

Pick any 5 numbers from 1 to 30 and make all the possible fractions out of them.

For example, suppose we choose 2, 15, 24, 6 and 12. Then we have failed because 15/12 and 24/15 are both between 1 and 2, and 24/12 is equal to 2.

One way to think about this is to say if we choose 1, then we must choose 3, 7, 15 and our fifth number would have to be 31, which we haven’t got. These are the Mersenne numbers.

A different way of thinking about this is to use something called the Pigeonhole principle. Suppose we split the number 1 to 30 into groups that contain up to double the first number, then we have:

{1, 2}, {3, 4, 5, 6}, {7, 8, 9, 10, 11, 12, 13, 14} and {15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30}.

Now we can’t choose two from one of these groups, but as we need 5 numbers and there are only 4 groups, there is no possible solution.

Can you choose 5 numbers so that none of the possible fractions are between 1 and 2 inclusive?

OK, so it seems like it might not be possible – can you prove why?

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4 Primes

Why are many mathematicians obsessed with prime numbers?

Prime numbers are the building blocks of all the numbers – in the 19th century Gauss proved that all (whole) numbers can be written as a unique product of primes. But there is no simple formula that can tell us the nth prime, or whether a given number is prime. Arguably the most important unsolved problem in Pure Mathematics, the Riemann Hypothesis, concerns the distribution of the primes.

That said, there are some patterns in the primes; this section is designed to be a gentle introduction to prime numbers and seeking patterns in them.

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4.0 Prime Clothing

There are numerous items of clothing that have designs based on the primes. I particularly like this prime factorization jumper created by Sondra Eklund:

Can you see how it works? If you can’t quite work it out, here’s the pattern with the first few numbers on it:

You can see that each prime factor has a colour code (2 = blue, 3 = red, etc.) so each square shows which prime factors, and how many of them, occur in each prime factorization.

Can you spot any patterns in the columns or diagonals of this pattern?

Can you create your own prime factorization pattern?

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If we put them in 6 columns we get:

You can clearly see the primes in columns 1 and 5.

We can also see the multiples of 2 and 3 in columns and multiples of 5 and 7 in the diagonals.

Here is another t-shirt that is interesting!

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4.1 Numbers as shapes

What shape is a number? For example, what shape is the number 4? Here are a few different representations:

If we stick with squares and rectangles, then we get the following

We could think of the prime numbers as 1xp rectangles.

Using blobs, we could have something like this:

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Notice how the primes appear as circles as they can’t be grouped like the other numbers.

The online applet ‘Primitives’ groups the first few numbers like this:

These pictures give a beautiful way of showing prime factors; notice how the primes are colour coded.

Explore some other ways of representing numbers as shapes.

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4.2 Finding primes

It is not easy to find out if a number is prime; you have to divide it by all the numbers up to its square root and see if any of them divide into it.

The Sieve of Eratosthenes gives a systematic way of doing this; make a grid of numbers, say 1 to 100, and cross out all the numbers that are in all n times tables (apart from n itself).

Here is an example:

I quite like drawing this like this:

One important aspect of prime numbers is that all the ones bigger than 3 are in columns 1 and 5 (we say they are of the form 6k+1 or 6k-1, or that they are congruent to 1 or 5 mod 6).

What can you say about prime numbers by looking at this grid?

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We can eliminate all the other possibilities one by one. Numbers in the 6th column are in the 6x table, numbers of the form 6k+2 and 6k+4 (columns 2 and 4) are divisible by 2 and numbers of the form 6k+3 (column 3) are divisible by 3.

As primes either 6k+1 or 6k-1, the often appear in twins one more and one less than a multiple of 6. The twin prime conjecture appears true but has remained a conjecture for thousands of years, but recently (2013) there have been considerable advances towards proving it. It is particularly remarkable that these advances were made using reasonably simple mathematics by a relatively unknown mathematician.

The first prime triple is 3, 5, 7. There are no more because that would mean we would have to have another prime in the third column (of the form 6k+3), but 3 can be the only one. Here is another way to find primes. Look at this image of the 2x table drawn as a curve through the number line:

When we add the 3x table, we get:

Notice how some numbers (such as 6) have been crossed twice. Now let’s add the 4x table:

Find out more about the work towards a proof of the twin prime conjecture.

A prime triple is three primes that are 2 apart, such as 3, 5 and 7. What does this picture tell us about other triple primes?

Can you explain why this is true?

The twin prime conjecture says that there are an infinite number of primes that are 2 apart (such as 5 and 7). Find some twin primes. Why does the above discussion suggest this might be the case?

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When you have drawn your curves, pick a number and highlight all the curves that go through it, like this one for 6:

The number of crossings will tell us the number of factors (excluding 1 if we don’t complete the 1x table).

Note also that all the numbers will be crossed an odd number of times (if we exclude 1) apart from the square numbers (why?).

Incidentally, here is another interesting ‘sieve’ discovered by relatively unknown Indian mathematician S. P Sundaram in the early 20th century.

Using the numbers 4, 7, 10, 13, … in the first row, the numbers in each column go up by 3, 5, 7, … and so on like this:

Make one of these of your own and draw in all the times tables up to say 15.

What does the number of curves through a given number tell us about that number? What more can we say about how many times each number is crossed?

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Doubling the first few numbers and adding one gives 9, 15, 21, 25, 27, 33, 35, and so on. These numbers are obviously odd (why?) but what else can you say about them?

You have probably noticed by now that if a number k is in this sieve, then 2k + 1 will not be a prime number.

We can prove this with a little algebra; ignore this if it’s too difficult:

Let m represent the row number, and let n represent the column number. Then any given number k in the sieve is given by the expression (1+3m) + (n-1)(2m+1). Now we can work out 2k + 1 and after a bit of algebra we get 2k + 1 = (2n+1)(2m+1) which is not prime; it is the product of two odd numbers.

Before reading on, try doubling all the numbers and then adding one. What do you notice?

Have a look at all the odd numbers that are not in this list; what can you say?

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4.3 Prime games

4.3.1 Last prime

Here is a Nim-like game to help you get familiar with the prime numbers called Last prime. Before starting this game you should read about Nim in the Glossary.

Two players alternatively hold up any number of fingers from one to five. The cumulative total is noted down. The object is to keep the total prime. The first player unable to raise the total to a higher prime is the loser.

1 hand: You should hold up 5 fingers on your first go. Then the game is fixed to follow this pattern: 5, 7, 11, 13, 17 and then there is a gap of 6 to the next prime so player 1 is the winner.

2 hands: This follows a similar principle but is just a bit longer. The first primes that are more than 10 apart are 113 and 127. So you are aiming to be the first to 113. Working backwards, you can create safe positions: 113, 101, 89, 73, 61, 47, 31, 19 and so you should start with 7 fingers.

It turns out that primes go on forever; there is no last, or largest, prime. Greek mathematician Euclid (more about him later) proved this with the following argument:

Suppose there is a largest prime, P. Then make the number N = 2.3.5.7…P + 1 which is made by multiplying all the primes together and adding 1. This number must have a prime divisor Q (as all numbers are divisible by primes). Now, Q isn’t one of the 2, 3, 5, 7, … , P primes otherwise it would divide 2.3.5.7…P and therefore it would divide 1 = N – 2.3.5.7…P (why?). This doesn’t make sense; Q can’t divide 1 because it’s bigger than 1, so we have a contradiction – our assumption that there is a largest prime P must be false. So we conclude there are infinitely many primes.

This is called proof by contradiction, or Reductio ad Absurdam, and is a common method of proof.

Why all this fuss, you might say? Can’t we just say that the number N = 2.3.5.7…P + 1 is just a prime bigger than P (as it clearly doesn’t divide by any of 2, 3, 5, 7, …, P) and we have proved there are infinitely many primes straight away?

If you play first, how many fingers should you hold up?

Play this again but with both hands (10 fingers). What should you start with this time?

Oh, by the way, is there a last prime, or do they go on forever?

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The first few are:

2 + 1 = 32.3 + 1 = 72.3.5 + 1 = 312.3.5.7 + 1 = 2112.3.5.7.11 + 1 = 2311

So far so good, but unfortunately the next one isn’t prime: 2.3.5.7.11.13 + 1 = 30011 = 59 x 509. This is an instance of what mathematician Richard Guy calls The Strong Law of Small Numbers, which basically says you should be really careful about making conclusion by just looking at the first few (small) numbers in a pattern.

In that article, he mentions a related pattern that may always produce primes: Subtract each product from the next prime after the numbers in the sequence above. For example, the first few are:

5 – 2 = 311 – (2 x 3) = 537 – (2 x 3 x 5) = 7 [e.g. 37 is the next prime after 31 above]223 – (2 x 3 x 5 x 7) = 13

Unfortunately, a proof of whether this is true does not exist – can you be the first to prove it??

There are many other proofs that there are infinitely many primes. Here is one that uses Fermat numbers; it’s quite tricky but let’s have a go at it anyway. Fermat numbers are of the form 2n+1 where n is a number from the ‘doubling’ sequence 1, 2, 4, 8, 16, …

The first few Fermat numbers are:

What are the first few Fermat numbers? What do you notice about them?

Work out 2.3.5…P + 1 for different values of P. Is it always prime?

Find the next few. Do you believe this sequence will keep producing primes?

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F0 = 21 + 1 = 3F1 = 22 + 1 = 5F2 = 24 + 1 = 17F3 = 28 + 1 = 257F4 = 216 + 1 = 65,537F5 = 232 + 1 = 4,294,967,297

You might have noticed that they are all odd, which is definitely true (why?), and you may have conjectured that they are all prime. If you did, you are in good company; Fermat himself believed them all to be prime, but could not prove it. In fact, it took none other that Euler to prove that they are not all prime, by showing that 4,294,967,297 is divisible by 641! For a proof of how he did this, see the book Journey Through Genius by William Dunham.

Now, back to our proof there are infinite primes. First of all, the fact they are all odd is going to be important later on. However, if you are really switched on, you might have also noticed that each one is made by multiplying all the ones before, and adding 2. For example, 5 = 3 + 2, 17 = 3.5 + 2 and 257 = 3.5.17 + 2. It turns out this is always true, which we can prove using a technique called mathematical induction (see the Glossary for this bit).

Now this means that if some number divides any two Fermat numbers, then it must also divide 2 (why?) so any common divisor of two Fermat numbers must be 1 or 2. But 2 doesn’t divide any Fermat number as they are all odd! So any common divisor of two Fermat numbers must be 1. This means all Fermat numbers are relatively prime.

Why does this mean there are infinity many primes? Well, the Fermat numbers definitely go on forever – we can just keep making them. This, coupled with the fact that are relatively prime means that each new one will always have new prime factors, so there are infinitely many primes!

For n = 1, 2, 3, etc we have 3, 5, 9, 17, 33, 65, 129, 257, and so on. You might have noticed that all the odd ones are divisible by 3.

To see why we first note that 22 is congruent to 1 (mod 3), so that then any power of this will also be congruent to 1 (mod 3). So any even power of 2 is congruent to 1 (mod 3). Now multiplying this by 2 gives an odd power of 2, which must be congruent to 2 (mod 3). Now if we add one to this, we are back to 0 (mod 3) – i.e. 2n + 1 is divisible by 3 for odd n.

Investigate other numbers of the form 2n + 1. What can you say about them?

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4.3.2 Prime Nim

Here is a more complicated variation on the game Nim. Start with one pile of counters (of any size). Each player can only take 1 or a prime number of counters.

Trivially, if we start with a prime number of counters, player 1 wins straight away.

If we don’t start with a prime number of counters, a few games reveal that we are trying to get to the safe position of 4 counters, from which our opponent must take 1, 2 or 3 leaving us with a win.

We can do this as player 1 by getting the safe position of a multiple of 4 on each go (by taking 1, 2 or 3) unless we start on a multiple of 4 in which case player 2 would win by the same strategy.

Let’s look at the game with three piles (1 still included). In the first version of the game about, we simplified the game by only considering the remainder the pile on division by 4, as we can always reduce piles by 4 by taking 1, 2 or 3 counters.

We can also do this with three piles. In the case of 1, 5 and 10 counters, we can simplify these to remainders 1, 1 and 2. Now it is clear that we should take 2 from the large pile to get to the standard safe position of an even number of binary digits.

When we play with 1 not included, this game is much more difficult to analyze because we can no longer take 1 and use the multiple of 4 strategy.

Playing the games a few times we realize that sometimes we will get to positions where 1 counter is left. I suppose we should then change the rules to say the last person who can take any counters is the winner (so leaving your opponent with 1 counter constitutes a win). So if we are on 4 (say), a winning move is to take 3.

The safe positions I have found so far are 9, 10, 18, 24, 30, … For example: 9 is safe because your opponent can only take 2 (leaving 7), 3 (leaving 6 which is a winning position), 5 (leaving 4 which is a winning position) or 7 (leaving 2).

It might seem safe positions have something to do with multiples of 6, but not all of them are; for example 42 isn’t safe because your opponent could take 23 leaving you with 9 which is a safe (losing) position. It seems that there is no simple formula for safe positions in this game, which is probably to be expected if we are dealing with primes!

What is a good strategy for this game?

Try changing the rules so maybe there is more than one pile.

What happens if we do not include 1 as a prime?

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4.4 Prime factorization

4.4.1 Factor trees

The fact that any whole number has a unique prime factorization is very important.

Usually prime factors are found using a factor tree. Here are lots of different ways of prime-factorizing 300:

However you do it, you probably notice that they all come up with the same answer, 2 x 2 x 3 x 5 x 5.

But did you also notice that:

- They all have 4 pairs of factors too?

- There is one even pair of factors in each?

- There is one pair of factors that are both divisible by 5. (In the third one above, the even pair is also the pair that is divisible by 5)?

- The rest of the pairs are relatively prime?

- Why do all the factorizations of 300 have 4 pairs of factors?

One way of looking at this is to think of the prime factors as squares of chocolate that we are going to snap off into single blocks. Moving along a branch of the factor tree is like snapping the bar of chocolate along a line. So the factorization of 300 on the right below could be represented by breaking a chocolate bar like this:

Can you explain any of these facts? Investigate similar facts for other prime factorizations.

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We will always need 4 breaks to break this chocolate bar into single squares. In general we will need n-1 breaks to break up a chocolate bar with n squares (see the game Choco Choice in the section on Parity).

Another way of thinking about this is join pairs of prime factors together (reversing the factorizing process) to make composites like this:

Then we can see that there must be four pairings to get 5 numbers to 1.

- Why is there one even pair of factors in all factorizations of 300?

If we look at the even and odd factors in the above diagram, we can see that pairing two odds created another odd and we pair an odd and even, we create another even; in each case the number of evens does not change. If we pair two evens then we reduce the number of evens by one. So as there is only two even prime factors in 300, we can only have one even pairing; as soon as we pair them (whenever we do it) we only have one even left.

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Generally, pairing two even numbers reduces the evens by one, so if we have n even factors we will have n-1 even pairs.

This is also the case with the multiples of 5. So for 300 we have one even pair, one multiple of 5 pair, and the rest are relatively prime as they are contain combinations of different factors.

Fermat came up with an algorithm for factorizing large numbers. Starting with the number N, let m be the smallest number such that m2 is bigger than N. For example, if we have N = 187, then m = 14 as 142 = 196.

Now work out m2 – N, (m+1)2 – N and so on until a perfect square is reached. In this case we have 196 – 187 = 9, and we have found a perfect square straight away.

Now as m2 – N = x2 (in this case with x = 3) we can rearrange to give N = m2 – x2 and you might recognize this as the difference between two squares. It is a fact that we can write the difference between two squares as (m+x)(m-x), in this case (14+3)(14-3) and we have our factorization: 187 = 17 x 11.

First of all, note that m = 97, then successively calculate 972 – 9271, 982 – 9271 until you find a square. You should have got the sequence 138, 333, 530, 729, … (this sequence is increasing by 2 more each time) and then realizing 729 is a square we have 1002 – 9271 = 272 and so 9271 = 1002 – 272 = (100+27)(100-27) = 127 x 73.

Does this work for all numbers? Let’s try 300. We have m = 18, and so 182 – 300 = 24, then 192 – 300 = 61, 202 – 300 = 100, which is a perfect square, so 300 = 202 – 102 = (20+10)(20-10) = 30 x 10. Then we perform the same calculation on 30 and 10 until we get to the prime factorization!

Check the formula given for the difference between two squares m2 – x2 = (m+x)(m-x) works for some other numbers m and x.

Now try Fermat’s factorization method for N = 9271. [Hint: 729 = 272)

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4.4.2 Uniqueness

Have you ever thought about the uniqueness of prime factorization?

To prove that the prime factors of a number are unique, we need something called Euclid’s Lemma, which says that if a prime p divides a composite number r x s, then either p divides r or p divides s.

Suppose there were two different prime factorizations of 300. We know one factorization is 2.2.3.5.5; the other one could either have 5 different factors, or a different number of factors altogether.

Suppose there are two different prime factorizations of 300. One is 2.2.3.5.5, the other is p.q.r.s.t.u…. for some other primes p, q, r, …

By repeated applications of Euclid’s Lemma, the prime p must divide one of the factors 2, 3 or 5, in which case it must be one of 2, 3 or 5. If we carry on this logic, the other factors q, r, etc. must also be one of each of the other factors of 2.2.3.5.5 and so all prime factorizations are unique.

Here is a different proof. Suppose we know that all the numbers up to n have a unique prime factorization. If n is composite then let’s suppose it can be written as the product of primes in two different ways as above. These two factorizations can’t share a prime otherwise we would cancel by this prime and we have a number smaller than n that can be written as the sum of two primes, which is not true by our assumption.

Let p be the smallest prime in the first factorization, and P be the smallest prime in the second one. As n is composite (so has at least one other factor) then n > p2 or n > P2 and we definitely have n > pP. Now consider n-pP, which has a unique factorization (by assumption, as it is smaller than n). As p divides n, it must divide n – pP, as must P. So pP divides n – pP, which means that pP divides n.

But this can’t happen. If pP divides n, then P must be one of the prime factors in the first factorization of n, which we have said at the start can’t happen. So our assumption that n has two different factorizations is false. Now because it’s true for all numbers less than, and true for n, we can say by mathematical induction that it is true for all n.

Let the other factorization be p.q.r.s.t.u…. for some other primes p, q, r, … How can you use Euclid’s Lemma to prove that the factorization 2.2.3.5.5 is unique?

Try this out for yourself.

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4.4.3 Defining primes

One common definition is to say that a prime is any number with exactly two factors, 1 and itself. A different way of saying this is that if a prime number (greater than 1) is written as the product of two numbers, exactly one of them must be 1.

We haven’t even got the number 1 anymore, so we definitely have to change these definitions! To start with, we must define the basic unit of our number system – for normal numbers it is 1 (or -1), but for the even numbers it might make sense to say the unit is 2.

If we stick with the usual rule that the unit can’t be a prime, we then have the first few primes for the even numbers: 4, 6, 8, 10, 12 and 14. The first non-prime is 16, which can be made from 4 x 4. Notice that some of the numbers don’t even have a product any more, like 6, as there is no number 3.

Strictly speaking, our normal definition of primes is actually called irreducibility.

Primes are more exactly defined in an alternative way: a prime is any number p such that if p divides any number made by multiplying two numbers together (a x b), then p must divide one of these numbers.

In Hilbert numbers, 693 can be written as 9 x 77 and 21 x 33. Each one of these factors is irreducible; factorization into irreducibles is not necessarily unique.

How would you define a prime number?

Consider a world in which only even numbers existed. Would these definitions of primes be any good now? How would you define primes now? Which numbers would be prime under your definitions?

What are the primes in the world of odds, assuming 1 to be the unit?

Investigate other worlds of numbers, such as a world where the only numbers are 1, 5, 9, 13, 17,…? (Let’s call these Hilbert numbers, after the mathematician David Hilbert who invented this example, and make the unit 1).

Are these two definitions the same thing? If not, how are they different?

Can you find an example where a number has more than one factorization into irreducibles? [Hint: use the Hilbert numbers]Consider a different world where the only numbers are 1 and the evens. If 1 is the unit, what are the primes now? 96

4.5 Greatest Common Divisors

4.5.1 Relative primes and the GCD

Two numbers are relatively prime if they do not share any common (prime) factors. This is strongly tied up to the idea of a greatest common divisor (GCD). If two numbers are relatively prime then their GCD is 1.

Two numbers being relatively prime is important in maths. We have already seen that if two numbers a and b are relatively prime, then we will not be able to simplify the fraction a/b.

A good way of presenting numbers and their prime factors is through a Venn diagram, named after English mathematician John Venn (although they were possibly invented by Leonard Euler).

Here are the prime factors of 300 = 2.2.3.5.5 and 36 = 2.2.3.3 placed in a Venn diagrams.

We can see that they are not relatively prime as they share common factors.

In fact, we can say a bit more than this. The GCD of these two numbers is the product of the factors in the middle, which here is 2.2.3 = 12.

A further by-product of this method is that you get the Lowest Common Multiple (LCM) for free! If you take the product of all the numbers in the Venn diagram you get the LCM, so here it is 2.2.3.3.5.5.

The Venn diagram method is good for small numbers; we would not want to use it for large numbers as the prime factorization takes a long time.

Find some pairs of numbers that are relatively prime.

What are the drawbacks of this method? Try it for two larger numbers like 3108 and 5291.

This method suggests that the GCD and LCM are closely connected – can you see how?

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Using the example shown, 300 = 2.2.3.5.5 and 36 = 2.2.3.3 and we have that GCD = 2.2.3 and LCM = 2.2.3.3.5.5.

How are these connected? Well, if we do GCD x LCM we get 2.2.2.2.3.3.3.5.5 which is the same as 300 x 36. This is indeed always the case, so we have the rule:

lcm (a , b )×gcd (a ,b)=a×b

This gives us a handy way of finding the LCM if we know the GCD (and vice versa).

One thing we can conclude from this is that if two numbers are relatively prime, then gcd(a,b) = 1 and then we have lcm (a ,b )=a×b.

There are other ways of finding the GCD. The Greek mathematician Euclid came up with an algorithm for finding the GCD to two numbers.

Euclid noticed that any number that divides two numbers (let’s call them a and b) also divides their difference (can you see why?). So a, b and a - b are all divisible by GCD(a,b). We can use this to find the GCD by successively taking the smallest from the largest, until we get two numbers the same like this:

GCD(20,12) = GCD(8,12) = GCD(8,4) = GCD(4,4) = 4.

This is a shortened version of the full algorithm that is usually used; here is the usual form of Euclid’s algorithm to find GCD(5291, 3108):

Euclid took all the Greek work on maths, tidied it all up and put it in his book The Elements in around 300BC. Little is known about Euclid, and there are some people who do think Euclid may not have been one person, but rather a group of scholars.

The Elements is possibly the most important work in the history of maths. For the first time we are given a rigorous basis of axioms, or definitions, from which many theorems are proved using sequences of logical steps. This process forms the basis of mathematics today. The Elements contained many proofs about geometry and number theory, and culminated in a full description of the Platonic solids.

The famous first words of The Elements are:

• A point is that which has no part.

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5291 = 1 x 3108 + 21833108 = 1 x 2183 + 9252183 = 2 x 925 + 333925 = 2 x 333 + 259333 = 1 x 259 + 74259 = 3 x 74 + 3774 = 2 x 37 + 0

The last non-zero remainder is 37, and this is the greatest common divisor of 5291 and 3108.

We can get an idea of how it works by ‘undoing’ the calculation from the bottom up. From the bottom line, we can see that GCD(74,37) = 37. Now, moving one line up, as 37 is the GCD of 37 and 74, it must also be the GCD of 259. We can carry on moving up the calculation like this until we can conclude that 37 is the GCD of 5291 and 3108.

Here is another interesting method for finding GCDs:

If

you pick a co-ordinate on one of the lines, say (12,8), and go along its line starting at the origin, we pass through 4 lattice points (integer co-ordinates). This corresponds to the fact that the greatest common divisor of 12 and 8 is 4. This works for any lattice point.

Euclid took all the Greek work on maths, tidied it all up and put it in his book The Elements in around 300BC. Little is known about Euclid, and there are some people who do think Euclid may not have been one person, but rather a group of scholars.

The Elements is possibly the most important work in the history of maths. For the first time we are given a rigorous basis of axioms, or definitions, from which many theorems are proved using sequences of logical steps. This process forms the basis of mathematics today. The Elements contained many proofs about geometry and number theory, and culminated in a full description of the Platonic solids.

The famous first words of The Elements are:

• A point is that which has no part.

Look at these lines. How is the number of lattice points (integer co-ordinates) connected to the GCD?

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Now, let’s consider standing at the point (0,0) and looking out at the ‘forest’ of lattice points. Some of them will be ‘visible’ from where we are standing and some will not. For example, (6,4) is obscured by (3,2), so (6,4) is not visible, while (3,2) is.

From the previous discussion you will probably have realized that the visible points are all those that are relatively prime (with GCD = 1).

Here are the visible points on axes 0 to 10:

There are some simple patterns here, such as symmetry in the diagonal (45 degree) line from (0,0) (why?). You may have also noticed that the horizontal and vertical

What can you say about the co-ordinates of the visible points?

Find all the visible lattice points for axes numbered 0 to 10 (say). Any patterns?

Are there any lines that contain all visible points?

Are there any lines that contain no lattice points at all?

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‘prime’ have lots of visible points, the only non-visible ones being multiples of the number (why).

Perhaps more exciting than this is the fact that there are lines that contain all visible lattice points, shown in the diagram on the right.

You might have noticed that these lines all go through prime numbers on the axes.

Why do they contain only visible points? If we add the co-ordinates (x, y) of each point on these lines, this equals the prime number they go through, so we have x + y = prime. Now, there can be no number (apart from 1) that divides into x and y, otherwise it would also divide into the prime which is not possible (why?). So x and y are relatively prime, and so they are visible.

There are lines that never touch any lattice points, such as the one shown:

This is the line y / x=√2. This number is called irrational; y and x can never be whole numbers (lattice points) on this line.

There is a proof that √2 is irrational in the introduction to the section on Parity.

We can also link this to our work on Greek Ladders in chapter 3.

Look at the visible points that come close to the line. They are (2,3) and (5,7), which are numbers in the Greek Ladder for √2. If we extended the line further up, we would find that it gets nearer and nearer to lattice points (12,17), (29, 41) and so on, but never actually goes over one.

Euclid’s algorithm can also be used to write fractions as continued fractions:

Can you explain what is special about these lines, and why they contain only visible points?

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For example, here is the algorithm for 67 and 24, to help us find the continued fraction for 67/24:

67 = 2 x 24 + 1924 = 1 x 19 + 519 = 3 x 5 + 45 = 1 x 4 + 14 = 4 x 1 + 0

So we can see that these numbers are relatively prime (the last non-zero remainder is 1). Now if we divide through by the middle number on each line we get:

67/24 = 2 + 19/2424/19 = 1 + 5/1919/5 = 3 + 4/55/4 = 1 + 1/4

Starting at the bottom line, we have 5/4 = 1 + 1/4, the reciprocal of which (4/5)

appears in the next line up, so we have 195

=3+ 1

1+ 14

We can carry on to the next line up to get

2419

=1+ 1

3+ 1

1+ 14

And finally we have

6724

=2+ 1

1+ 1

3+ 1

1+ 14

which is nice.

Why are continued fractions useful? Well, we have already seen that they can be used to approximate irrational numbers (see Greek Ladders), but they can also be used to solve Diophantine equations – see the next section!

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4.5.2 Prime stars

You have probably made some star doodles at some point. If you haven’t, draw some dots (roughly) evenly round a circle and join them missing a few each time. Here are a few:

The first number is the number of dots and the second number is how many missed each time.

Which ones make good stars? Well, they’re all good, but my favourite are the ones that go back to the start without taking pen from paper (in one go).

I was thinking it was something to do with the first number being prime (5 and 7 here), so I drew a few more:

As you can see, it is not as simple as the first number being a prime number, as (8,3) and (10,3) both work.

Find some other stars that can go back to the start without taking pen from paper. What can you say about the ones that go back to the start?

So what is it about?

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If the number of dots and the size of jump are relatively prime, we get a star that can be drawn without taking pen from paper.

Generally, the number of stars we get is the greatest common divisor of the two numbers.

So if they are relatively prime, the GCD is 1, and we get a star that can be drawn without taking the pen from paper.

We can see that for GCD(9,3) = 3, and we get 3 separate stars.

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4.5.3 Diophantine equations

GCDs are important in the theory of Diophantine equations, named after the first person to study them, the Greek mathematician Diophantus. They are concerned with finding integer (whole numbers including the negatives) solutions to equations and are still being studied by mathematicians to this day.

To start with, we are going to look at Diophantine equations of the form ax+by=c. Here is a classic puzzle:

We can weigh the 1kg elephant like this:

This means that we can any integer-weight elephant by putting on multiples of 2kg and 3kg weights. For example, we could weigh a 100kg elephant with 200 x 2kg weights on the left pan, and 100 x 3kg weights on the right pan.

So to work out the weight of any elephant (x kg), we just place weights in the ratio 2:1 as shown until we get a balance.

How can you weigh any object (that weighs a whole number of kilograms) on a pair of balancing scales with a (infinite) set of 2kg and 3kg weights?

How can we weight this elephant if she weighs 1kg? How about 100kg?

This problem is the same as trying to find two integers x and y that solve 2 x+3 y=1. How many different solutions can you find?

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As we saw above, one solution to the equation 2 x+3 y=1 is x = 2, y = -1, where the negative value represents putting a weight on the opposite scale. There are infinitely many others (such as x=5, y=-3).

More generally, we can solve any equation 2 x+3 y=c with x = 2c, y = -c.

After a bit of trial and error you will have realized that you can’t weigh any odd weighted elephants.

There are no integer solutions to the equation 2 x+4 y=1 for the same reason as above. Substituting different values of x and y into the expression 2 x+4 y, we soon realize that we can only generate even numbers.

Why? Substituting values for x and y into expression mx+ny, we can only generate multiples of the GCD of m and n. So if m and n are relatively prime (as in our first example with m = 2 and n = 3) then we have a GCD of 1 and we can generate any multiple of this i.e. any integer. But with m = 2 and n = 4 we can only generate multiples of GCD(2,4) = 2

Based on this we can work out if there are solutions to the following Diophantine equations:

(a) 2 x+3 y=2 has (infinite) solutions as GCD(2,3) = 1

(b) 3 x+6 y=99 has (infinite) solutions as 99 is a multiple of GCD(3,6) = 3

(c) 4 x+6 y=99 has no solutions as 99 is not a multiple of GCD(4,6) = 2

Now, as promised in the previous section, here is how continued fractions can be used to solve Diophantine equations.

Let’s use the example 67 /24 to solve a Diophantine equation involving 67 and 24, let’s say 67 x+24 y=1. We know this is solvable from the previous section (why?).

Now suppose we had 2kg and 4kg weights instead; can we weigh any elephant now? Which elephants can we not weigh? Can you weigh a 1kg elephant?

Now try solving 2 x+4 y=1. Is it possible?

Now, can you tell by looking at these Diophantine equations if they will have solutions?

(a) 2 x+3 y=2 (b) 3 x+6 y=99 (c) 4 x+6 y=99

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Starting with the finished continued fraction, we have

6724

=2+ 1

1+ 1

3+ 1

1+ 14

Now we can snip off the very last fraction to get another fraction that is very close to

67/24, so we have: 2+ 1

1+ 1

3+ 11

=2+ 1

1+ 14

=2+ 154

=2+ 45=145 .

Subtracting gives 6724

– 145

=67×5−14×24120

=−1120

This is exciting, but did you notice that we have just solved our Diophantine equation? Look again at the top line in our calculations: it shows that 67 x 5 – 14 x 24 = -1, or 14 x 24 – 67 x 5 = 1. If we compare this with the equation we are trying to solve, we have one solution for x and y, namely x = -5 and y = 14.

How can we use this to find other solutions? Well, now we know one solution, we can use this trick:

Let a general solution be x = -5 + 24t and y = 14 – 67t. Putting this into our equation, we know that 67(−5+24 t)+24 (14−67 t)=1 because 67 x -5 + 24 x 14 = 1 and 67 x 24t – 24 x 67t = 0. So choosing any integer for t gives us other solutions!

Now let’s turn our attention to other Diophantine equations of other types. Pythagoras’ Theorem a2+b2=c2 is a Diophantine equation with infinite solutions, whereas Fermat’s Last Theorem says that an+bn=cn has no solutions for n > 2.

The first equation has 4 pairs of solutions: x = 1, y = 0, x = -1 y = 0 and x = 0, y = 1 and x = 0, y = -1. How about the second one, x2− y2=2? Trying numbers for x and y, it seems unlikely that there are any solutions.

We could prove this by noticing that the expression x2− y2 is the expression for the difference of two squares. So we can rewrite the second equation as (x+y)(x-y) = 2 which has one solution x + y = 2 and x – y = 1 (or vice versa). Neither of these has

How close is 14 /5to 67 /24?

What (integer) solutions can you find to the equation x2− y2=1? How about x2− y2=2?

For which values of c does the equation x2− y2=c have solutions?

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integer solutions; if we add these two equations together, we get 2x = 3 which gives x = 3/2 (not an integer).

A bit of experimentation leads us to realize that all odd values for c have solutions, which we can see by noticing that the odd numbers are the differences between the squares numbers 1, 4, 9, 16, ....

How about the even numbers? We have no solutions for 2, solutions for 4 (what are they?), no solutions for 6, solutions for 8, … which leads us to think that there are solutions if c is a multiple of 4.

To see why, consider the factors of multiples of 4; we can always find two even factors (why?) so that (x+y) and (x-y) are both even. When we combine the two equations together we get 2x = even, so x (and therefore y) have integer solutions.

Why not explore other Diophantine equations, or find out more about Fermat’s Last Theorem?

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4.6 Prime sequences

4.6.1 Linear prime sequences

Linear (arithmetic) sequences are sequences that go up by the same amount each time, such as 5, 9, 13, 17, 21, … (this one is has nth term 4n+1).

You can see the primes in this sequence by looking at this grid; the primes are coloured in red:

1 5 9 13 17 21 25 292 6 10 14 18 22 26 30 3 7 11 15 19 23 27 314 8 12 16 20 24 28 32

The primes bigger than 2 only appear in rows 1 and 3 (why?).

There are an infinite number of primes, and it was shown by French mathematician Dirichlet that there are an infinite number of primes in the sequence 4n + 1 and 4n +

3 (and indeed any other linear sequence an + b where a and b are relatively prime).

Famous English mathematician John Littlewood proved that leader of the race between 4n+1 and 4n+3 primes changes infinitely often. It has also been shown that the sequence 4n+1 contains around half the primes, the other half being in 4n+3 (apart from 2 of course).

The sequence 4n+1 can only ever contain at most two primes in a row. To see why, consider two consecutive primes in this sequence; they must be of the form 6n+1 then 6n+5 (like 13 and 17) as they are 4 apart. But then the next number in this sequence will be 6n+9, which is not prime. The sequence 4n+3 starts with 3 primes in a row (3,7,11) but, for similar reasons, this never happens again.

You may have noticed that this sequence contains quite a few primes. Roughly what percentage of the primes do you think are in this sequence?

The sequence at the start of this section has 2 primes in a row (13 and 17). Does this sequence ever have more than 2 primes in a row?

Can you find a sequence with 3 primes in a row? Or more? Find some other linear sequences that contain a lot of primes [hint: look at the previous section!].

There are 6 primes in the third row (4n+3) and only 4 in the first row (4n+1) in the table above. The 4n+3 primes are leading the race so far; do you think they always will?

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From the previous section we know that two other linear sequences containing all the primes between them (apart from 2 and 3) are 6n+1 and 6n+5. The sequence 6n+5 starts with a run of 5 primes (5, 11, 17, 23, 29) but this is as good as it will ever get.

I’ll leave this one to you! Here’s one last question about 4n+1 and 4n-1 numbers:

The famous French mathematician Adrien-Marie Legendre proved that there will always be more 4n+1 factors! In fact, he showed more than this; he showed that the number of ways of writing a number as the sum of two squares is 4(F+1 – F+3) where F+1 is the number of 4n+1 factors and F+3 is the number of 4n+3 ones.

So for example, 21 can not be written as the sum of two squares as F+1 = F+3. However, 45 can be written as the sum of two squares in 8 ways (noting that order is important!

45 = (6)2 + (3)2 (3)2 + (6)2

(6)2 + (-3)2 (-3)2 + (6)2

(-6)2 + (3)2 (3)2 + (-6)2

(-6)2 + (-3)2 (-3)2 + (-6)2

Can you prove why? [Hint: use the sieve of Eratosthenes above, or use an algebraic argument as above for 4n+1].

Choose any odd number. Work out whether it has more 4n+1 (including 1) factors or more 4n+3 factors. What do you think is the probability it has more 4n+1 ones?

For example, 21 has an equal number (1 and 21 compared with 3 and 7), whereas 45 has more 4n+1 ones (1, 5, 9 and 45 compared with 3 and 15).

How many ways are there of writing 65 as the sum of two squares?

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4.6.2 Quadratic prime sequences

The spiral on the right is called the Ulam spiral, named after Polish-US mathematician Stanislaw Ulam.

You will probably have noticed that most of the primes between 41 and 100 lay on the diagonal.

We can work out the function that generates this sequence by working out the differences like this:

Now let’s halve the bottom number and call that a (so a = 1). Subtract this from the middle number and call this b (so b = 2 – 1 = 1) and call the top number c (= 41).

Now we put these into the expression a .n2+b .n+c to give 1.n2+1.n+41; this is the function we need.

Putting n = 0, n = 1, n = 2, … into the function gives 41, 43, 47, … so it works.

Here is a larger image of the Ulam spiral showing where the primes occur:

Continue the spiral and then circle all the prime numbers. What do you notice? Why do you think this is?

Why have I started at 41 instead of 1? Explore spirals starting with different numbers.

What is the function that generates this diagonal sequence 41, 43, 47, 53, 61, ...?

Check that this function produces the sequence given.

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Notice how many of the primes seem to lie on diagonals.

The triangle below is called the Klauber triangle, named after Laurence Klauber, an American amateur mathematician and expert in rattlesnakes (!):

Here is the Klauber triangle with the primes in red:

12 3 4

5 6 7 8 910 11 12 13 14 15 16

17 18 19 20 21 22 23 24 2526 27 28 29 30 31 32 33 34 35 36

Continue the triangle and circle the primes. Can you find any interesting patterns? What happens if you start the Klauber triangle with a different number at the top, like maybe 41?

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Notice how primes seem to appear in columns; the column starting with 5 looks promising as a prime sequence.

What function is it? Let’s extend the columns upwards to create a rectangle like this

-4 -3 -2 -1 0 1 2 3 4 5 6-2 -1 0 1 2 3 4 5 6 7 82 3 4 5 6 7 8 9 10 11 128 9 10 11 12 13 14 15 16 17 1816 17 18 19 20 21 22 23 24 25 2626 27 28 29 30 31 32 33 34 35 36

Using the method above, the three numbers we need are -1 (top), 2 (first difference, middle) and 2 (difference of the differences, bottom). Our method gives the function n2+n−1. [You can test this by putting n=0, n=1, n=2, etc into it and check you get -1, 1, 5, 11, …]

Unfortunately this function does not produce primes for much longer; the next one is 41 (prime) but then the next one is 55 (not prime).

As we move along the columns to the right the sequences are all one higher than before. So the middle column which is two to the right of our original one is the function n2+n+1, which does quite well at generating primes too.

If we keep going along to the right, we’ll eventually get to the function n2+n+41, which we now generates lots of primes. Here is a larger image of Klauber’s triangle showing where primes occur, with this function shown as a vertical red line:

The function n2+n+41 was first discovered by Euler.

Can you work out the function for the column we are interested in?

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This function generates primes until n = 40. Putting n = 40 into the expression gives 402+40+41=40 (40+1 )+41=40×41+41=412 = 1681.

So this polynomial generates 40 primes. Other quadratics that generate lots of primes are n2+n+17 and 2n2+29. Christian Goldbach, a German mathematician, proved that there is no function like this (polynomials with integer coefficients) that will generate primes forever.

There are two ways to work out functions. You can substitute n=0, n=1, n=2 into the function (as above). Or, supposing I gave you the first few terms of the sequence 41, 43, 47, 53, … then you might notice that the differences go 2, 4, 6, … Carrying this on you can guess the differences will carry on 8, 10, … and can work out the next numbers in the sequence.

This idea was used by Charles Babbage when building his Difference Engine. To see how it worked, look at following sequence and its differences:

Given the function n2+n+41 he would work out the first few terms by hand, which he put into the machine. The machine would then calculate the differences, and the differences of the differences (called the 2nd differences) and so on. When it got to a set of differences that stayed the same (here, the 2nd differences are all 2), the computer then used these to calculate all the rest of the numbers in the sequence (or values of the function) like this:

The difference engine could find the value of any polynomial function just by using differences and repeated addition (which was easier to do than multiplication). Polynomials were important as they were used to approximate logarithms and trigonometric functions, which were used in engineering.

Does this function generate primes forever? If not, when does it stop?

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Whilst giving a speech about his new machine, mathematician Ada Lovelace translated his words and added twice as many notes of her own, and in doing so she created the first ever computer program.

A man called Leslie Comrie who was head of the Nautical Almanac Office in Greenwich used a Babbage machine to compute motions of the moon for working out tidal patterns. In 1928 an American man who was doing the same job by hand came over to Greenwich and found what he had been doing. This man went back to the USA and told some mathemticians about it, including someone called John von Neumann, who then built what is regarded as the first ever computer as we know it today.

First let’s look at n2 + 1. The first few terms, with the primes in red, are 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, … This sequence seems to generate quite a few primes. It is not yet known whether it will continue to produce primes forever.

Now let’s look at n2 – 1. The first few terms are 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, … and we can see that none of these are prime! To see why, consider the difference of two squares, n2 – 1 = (n+1)(n-1). This suggests why these numbers are generally not prime; they can be split into factors.

We see that when n is odd we have 0, 8, 24, 80, 120, 168, …You may have noticed

Charles Babbage was an English mathematician and inventor/engineer who lived in the 19th century. Among other things, he analyzed the post and recommended that all mail go at the same rate (called the ‘penny post’). He also invented railway signals and the speedometer, and devised a system of lighthouse signalling. He was also good at picking locks and deciphering codes.

Babbage was very passionate about mathematics. He was one of the first people to realise that mathematics was really a branch of logic. This lead to his invention of the Difference Engine, and later the Analytical Engine.

He came up with the idea when correcting log tables - incorrect log tables were hugely costly to engineers and navigators – and realized that a machine could do the calculations much more efficiently. He spent most of his life dedicated to building this machine although he never saw it built.

Investigate some other quadratic sequences such as n2 + 1 or n2 – 1. Do they contain primes?

Look at the odd terms in this sequence. What can you say about them? Can you explain why?

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that all of them are divisible by 8, and all alternate ones are divisible by 3 as well, so are divisible by 24.

To see why, use the difference of two squares again. We are looking at odd terms so that n is odd, which means n – 1 and n + 1 are consecutive even numbers, which also means that one of them is a multiple of 4 (why?). So n2 – 1 = (n+1)(n-1) is made by multiplying a multiple of 2 and a multiple of 4, which is a multiple of 8.

For the alternate ones, n is not divisible by 3 (n = 1, 5, 9, etc) so either n – 1 or n + 1 must be (why?). This means that n2 – 1 = (n+1)(n-1) is also a multiple of 3.

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5 Number sequences

Following on from our investigation into primes, and thinking of numbers as shapes, here are some further investigations into shapes and pattern in numbers.

There will be some number patterns that you have seen before (Pascal’s triangle and Fibonacci’s sequence) but hopefully with some new insights, and some patterns you may not have seen before; enjoy!

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5.1 Triangle numbers

Triangle numbers appear in the strangest of places…

5.1.1 Counting crossings

Here is a picture of one of my pupils trying to solve this problem on a mini-whiteboard:

You can see the pattern 1, 3, 6, 10, … the triangle numbers.

So we can see the next picture will have 15 dots (crossings). So how does this answer our question for how many crossings will be possible with any number (n) of lines?

One way to see this is if we double each triangle number, giving 2, 6, 12, 20, 30, … and then noticing that these numbers are 1x2, 2x3, 3x4, 4x5, … so we can say that triangle numbers are half of these numbers, giving the rule ½ x n x (n + 1).

Another way to see this is to consider two triangles put together like this:

We can see that each triangle is half a rectangle.

So the 1st triangle number is ½ x 1 x 2, the second is ½ x 2 x 3 and so on until we have the nth triangle number is ½ x n x (n + 1) for any number n. We can check this formula for the 5th triangle number = ½ x 5 x (5 + 1) = ½ x 5 x 6 = ½ x 30 = 15.

Triangle numbers pop up all over the place: Here they are in Pascal’s triangle.

You should definitely read about Pascal’s triangle in the Glossary, and maybe go online to find out even more; it is going to pop up in lots of the investigations

How many crossing are there when a number of straight lines cross each other? What is the maximum number of crossings with 3 straight lines? How about 4? How about n?

Can you convince yourself with a good reason why this is happening?

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we are going to do here! Make sure you understand the comments about combinations in there before moving on.

Ready for a second puzzle about counting crossings? OK, here goes:

There are 0, 1 and 5 crossings in these diagrams. What is this pattern? What comes next? The 6-dot diagram on the right reveals that there are 15 crossings.

You might have spotted these numbers in the 5th

diagonal of Pascal’s triangle above! You might want to check the 7-dot diagram has 35 crossings, but then again you might not!

Why are these numbers in this diagonal of Pascal’s triangle? Well, let’s think carefully about what the numbers in Pascal’s triangle represent; they are the numbers of ways of choosing things. The fifth diagonal represents the number of ways of choosing 4 things…

So each crossing matches to one way of choosing 4 things. What are these 4 things in this puzzle? They are the four dots from which we draw lines to make each crossing!

Here is a picture showing one of the ways of choosing 4 dots, and the crossing that comes from it.

Let’s dig a bit deeper:

How many crossings are there in these diagrams (not counting the dots themselves)? Can you see a pattern? Can you predict how many are in the next pattern with 6 dots?

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No prizes for guessing that the number of lines follows the triangle numbers!

As for regions, the first diagram has 4 regions, the second has 8 and the third one has 16. From this you might conjecture that the next one has 32, but it doesn’t.

Counting them in the 6-dot picture, there are 31 – no, you haven’t miscounted and no, there isn’t a way of drawing it so there are 32. Assuming you draw them so there are no double crossings (thus giving the maximum number of triangles possible), you will always get 31.

I bet you didn’t guess there would be 57? If you draw it correctly, this is how many there should be. So where are these numbers coming from? To see how, notice something extraordinary - that the number of regions is always one more than the number of lines plus the number of crossings.

So we have a pattern in the lines (3, 6, 10, 15, …) and a pattern in the number of crossings (0, 1, 5, 15, …), giving the sequence in the number of regions:

4 = 3 + 0 + 18 = 6 + 1 + 116 = 10 + 5 + 131 = 15 + 15 + 157 = 21 + 35 + 1

This is just another example of Richard Guy’s Strong Law of Small Numbers – don’t jump to conclusions! You can’t tell just by looking!

We have to dig deeper to understand why before we can start making predictions!

How many lines are there in these diagrams (ignoring the lines that make the circle?

How many regions (spaces) are there inside the circle for each of the diagrams?

Can you see any patterns? Can you see any connections between the crossings, lines and regions?

So the sequence goes 4, 8, 16, 31, … Can you predict how many regions will be in the 7-dot circle?

Now draw it nice and big, without any double crossings, and see if your prediction was correct.

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Here’s one last puzzle that involves counting crossings, taken from James Tanton’s fantastic book Math Without Words.

Draw two lines and make a few points on each (here I’ve done 2 at the top and 3 at the bottom). Now connect each top dot to each of the bottom ones and count the crossings (making sure no crossings overlap): here the number of crossing is 3.

Using * to represent the number of crossings, we have 2 * 3 = 6 for the diagram. Working systematically, here are a few more:

2 * 2 = 1 3 * 3 = 9 4 * 4 = 362 * 3 = 3 3 * 4 = 18 4 * 5 = 602 * 4 = 6 3 * 5 = 30

It seems that there are triangles of triangles going on here. For example, for 2 * n we have just the triangle numbers, for 3* n we have 3 x triangle numbers, and for 4 * n we have 6 x triangle numbers.

So generalizing to m * n, I think we can go for a formula something like:

m * n = ∆m−1×∆n−1

where ∆m is the mth triangle number from 1, 3, 6, 10, ....

It seems to; for example 3 * 4 = ∆2×∆3 = 3 x 6 = 18.

Can you come up with a general rule for this puzzle with any number of dots on the top and bottom? Can you explain it?

Test the formula to see if it works!

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5.1.2 Triangle patterns

So, what other rules and patterns can we find about triangle numbers 1, 3, 6, 10, …?

You will have noticed that any two consecutive triangle numbers add to make a square number (eg 6 + 10 = 16).

More precisely, the 3rd and 4th triangle numbers sum to make the 4th

square number, which you could write like this:

Δ3+Δ4=⊡4

Or more generally, Δn−1+Δn=⊡n.

If we add triangle numbers that are two terms apart, we get a picture like this, which suggests the identity:

Δn−2+Δn=2. Δn−1+1

This basically says that if you add triangle numbers two terms apart, you get double the one between them plus 1 - check this to see if it works.

There are lots of other identities you can find by dividing the triangles up in different ways, like this:

This one suggests the identity: Δ2n=3. Δn−Δn−1

Or perhaps: Δ2n=2. Δn+⊡n

You can put triangles together in other ways to give some more interesting identities… this one is a classic:

This suggests that the (odd) squares can be made by adding 8 triangles + 1, or more precisely:

⊡2n+1=8. Δn+1

What happens if you add two consecutive triangle numbers together?

How about adding triangle numbers 2 terms apart? Or 3 terms apart?

Try squaring triangle numbers… explore what happens when you add/subtract consecutive triangles squared?

What do you get if you do 9. Δn+1? Can you explain what is happening using a diagram?

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Squaring triangle numbers leads to some interesting identities too. Here is the squared triangle sequence: 1, 9, 36, 100, …

The difference between the terms gives to the sequence 8, 27, 64, which is the cube numbers, which we could write as ∆n

2−∆n−12 =n3

Here is a diagram showing how this might work for n = 3. Try making the same diagram for n = 4 and convince yourself it is works.

Can you see any interesting patterns in this sequence?

Can you use a diagram (or better still, multi-link cubes) to explain why this is happening?

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Here’s another way of thinking about it: Make a square from two triangles as above, using Δn+Δn−1=⊡n. We need n of these squares to make a cube; here’s a picture of this for n = 3:

Now, we know that n=Δn−Δn−1 so that we have:

n (Δn+Δn−1)=(Δn−Δn−1 ) (Δn+ Δn−1 )=n3

Multiplying out the brackets (see difference of two squares) gives ∆n2−∆n−1

2 =n3 as required.

If all this wasn’t enough, here’s one last pattern involving triangle numbers:

If you just count the black and white ones you have sums of consecutive triangle numbers, which we know gives the square numbers.

Hold on, but what about all the other sized triangles? For example, in the second one you have the 4 triangles plus the 1 large (2x2) one, which makes 5 in total.

So then looking at the third triangle again we have 9 small ones, 3 2x2 ones, and one 3x3 one, which makes 13.

How many triangles are in each one?

Now can you come up with a rule the total number of triangles in each one?

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If all this counting triangles is not enough for you, try this one:

The number of triangles in these 3 diagrams is 1, 8 and 35. To see how there are 8 in the second one, there are two ‘types’ of triangles (red and green in the diagram below) and in the third one there are three types of triangles (red, green, blue).

To see how we get 35 in the third one, there are 5 reds (points of the pentagram star), 20 blues (4 triangles are made by the crossings of 4 points, and there are 5 ways of choosing these 4 dots), and 10 greens (there are 10 ways of choosing 3 dots to make a triangle).

We can use this idea to visualize the 111 triangles in the 6-dot example:

Where do the

numbers 20, 15, 6 and 1 come from? Well, they are the numbers of ways of choosing

How many triangles are there in each of these diagrams (you might recognize them from counting crossings)? Can you find a pattern? Can you explain it?

How many triangles will be in the next one with 6 dots?

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3, 4, 5 and 6 dots from the total of 6 - see the notes on Pascal’s triangle in the Glossary for more on this.

Just another example of Richard Guy’s Strong Law of Small Numbers!

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5.1.3 Triangle cakes

Here is a puzzle about triangle numbers:

This is the order in which he eats the cakes: 0, 1, 3, 6, 10, 15, 21, 3(already eaten), 11, 20, 5, 16, and back to 3(still already eaten).

You may have noticed something strange: at all the places he arrives from now on, he has already eaten the cake. In fact, he goes back to all the cakes in reverse, so he only gets to eat 11 of the cakes!

This time Danny gets to eat cakes 0, 1, 4, 9, 16, 11, 24, 14, 6, 21 and 19 – still only 11 cakes!

This happens because there is a 1-to-1 match between triangle numbers and squares numbers, given by the identity ⊡2n+1=8. Δn+1.

This is linked to quadratic residues – more about these later.

Danny likes eating cake, and he likes triangle numbers. He comes to a round table with 25 types of cake around it numbered 0 to 24. He decides to go round the table trying every triangle numbered cake. So he starts by eating cake number 0, then cake number 1, then cake number 3, then numbers 6, 10, 15 and 21.

However, he doesn’t stop when he gets to 24, he keeps going as though cake 0 was cake number 25, cake 1 is now cake 26 and so on, so next he would have eaten cake 3 (or cake 28 depending on how you look at it) if only he hadn’t eaten it already!

What is the next cake he will eat?

If he carries on in this way, will he eat all the cakes? Will he eat half the cakes?Suppose Danny had decided to eat the square number cakes instead… would he have had more success? Can you explain why?

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5.1.4 Multiplication tables

This investigation is derived from the excellent book What to Solve? by Judita Cofman.

Everyone loves multiplication tables.

What did you find out? You might have spotted that the sum of the numbers in each increasingly large square are 1, 9, 36, 100, ... This is the squared triangle numbers we found earlier on.

The sums of the numbers in each L-shape are the cube numbers, which, as we found before, are the differences between the squared triangle numbers.

You could prove this algebraically if you wanted to (do you?).

Now let’s investigate some other squares on this multiplication grid.

In the square shown, the oranges sum to 12, and the purples sum to 13. If we take the difference of the purples we get 5.

It turns out that this creates some of the Pythagorean triples, numbers that satisfy Pythagoras’ Theorem a2+b2=c2.

Here is the classic 3-4-5 Pythagorean triple:

We can show why this works with a bit of algebra. Let’s call the rows and columns we are interested in m and n, like this:

The two purple numbers are square numbers, m2 and n2, and the orange numbers are both mn.

Investigate numbers given by the squares and L-shapes shown.

What is interesting about these numbers?

Explore some other square sums in this grid – can you explain what is happening?

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So we have:

Sum of oranges = 2mn.Sum of the purples = m2+n2

Difference of purples = m2−n2

These are formulae for generating Pythagorean triples, where m and n are chosen to be one odd and one even and where m and n are relatively prime.

There are lots of patterns in primitive Pythagorean triples, such as that one of the numbers is odd and the others are always even. If we take one generated by a 2x2 square in the original multiplication grid (i.e. with m and n consecutive) we get two consecutive numbers in the triple, but we don’t necessarily need to.

One last interesting pattern: In the 3-4-5 triple (n = 1, m = 2), we have the two shortest sides consecutive numbers. The next time this happens is the 20-21-29 triple, with n = 2 and m = 5:

The next time this happens is 119-120-169, when n = 5 and m = 12.

If we put these in a table we start to see some familiar numbers:

1 2 difference = 12 5 difference = 35 12 difference = 7

These numbers are related to those in the Greek ladders in the section on fractions! This gives us a way of finding Pythagorean triples with consecutive small numbers if we needed to!

Create some more Pythagorean triples and check they satisfy Pythagoras’ Theorem. What interesting patterns do you notice in Pythagorean triples?

Apart from being an important mathematician and philosopher, Pythagoras (around 500BC) was one of the first people to recognize that maths wasn't just something you would study to help you do the shopping or measure a field. He saw it had beauty and value in its own right, so he set up one of the first schools for studying mathematics.

He is most famous for his theorem, and although we have seen that he didn't

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5.1.5 Triangles from triangles

Mike Ollerton first introduced me to this curiosity.

Consider making triangles from matchsticks. There is only one triangle you can make from 3 matchsticks, the 1-1-1 triangle.

There are 3 triangles you can make from 4 matchsticks: 1-1-2, 1-2-1 and 2-1-1, which are all considered different for this investigation.

From 5 matchsticks you can also make 3 triangles, and from 6 matchsticks you can make only 1. Note that you can’t make 1-2-3 triangles (why not?).

I have tabulated the next few solutions:

matchsticks number types5 3 1-2-26 1 2-2-27 6 1-3-3, 2-2-38 3 2-3-39 10 1-4-4, 2-3-4 (6 of these), 3-3-3

10 6 3-3-4, 2-4-411 …12 …

You can start to see that the numbers of triangles for odd matchsticks are following the triangle numbers, as are the even matchsticks (but a bit behind).

Why might this be the case? Let’s look a bit more systematically at the triangles made from (say) 10 matchsticks, starting with the first matchstick. We can’t have 1-a-b as the only logical choices for a and b are 4 and 5 which are not valid. So we have:

2-4-4 3-3-5 4-2-43-5-3 4-3-3

4-4-2

We can start to see the triangle structure. It turns out that the number of ways of writing the blue numbers for each red number is always one less than that number.

This is still far from satisfactory; I hope you explore this further for yourself. If you have any further insights, please let me know!

Apart from being an important mathematician and philosopher, Pythagoras (around 500BC) was one of the first people to recognize that maths wasn't just something you would study to help you do the shopping or measure a field. He saw it had beauty and value in its own right, so he set up one of the first schools for studying mathematics.

He is most famous for his theorem, and although we have seen that he didn't

How many triangles can you make from 5 matchsticks? 6 matchsticks?

Can you find a pattern? Can you explain why?

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5.2 Squares and beyond

Here are some investigations about square numbers and beyond.

5.2.1 Square sort

Suppose there are (say) 25 people sitting in a row. Number each person 1 to 25. Now, you are going to call out each times table, one after another. So, starting with the 1 x table, call out: “1, 2, 3, … , 25”. Then call out the 2 x table: “2, 4, 6, …” and so on.

The rules are this: If a person hears their number called, they must stand if they were sitting, or sit if they were standing.

So everyone will stand when the 1x table is called out. Then when the 2x table is called out, all the even numbers will sit down (and the odd numbers remain standing).

Continue in the way until all the times tables up to 25 have been called out.

[Hint: consider how many times each person changes position]

You will find that the square numbers should be left standing… but why?

Factors usually come in pairs (such as 2x3 = 6) so numbers usually have an even number of factors. But a square number has one factor that does not have a pair – its square root – so will have an odd number of factors.

So the square numbered people will change position an odd number of times. If they began in a seated position they will be standing at the end.

Non-square numbers have an even number of factors because they come in pairs, so will return to a seated position.

Who will be left standing at the end? Can you explain why?

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Number Factors Number of Factors

1 1 1

2 1,2 2

3 1,3 2

4 1,2,4 3

5 1,5 2

6 1,2,3,6 4

7 1,7 2

8 1,2,4,8 4

9 1,3,9 3

10 1,2,5,10 4

Well clearly, all prime numbers have 2 factors. And we have just found out that square numbers have an odd number of factors.

What about all the other numbers? Is there a rule to find out how many factors any integer will have, without just finding them all?

Well, yes there is – let’s consider the number 12. It has 6 factors (1, 2, 3, 4, 6, 12).

If we look at each of these factors more closely they share something with 12 – prime factors! 12 can be written as a product of prime factors like this:

12=22×31

We can use this to generate all the factors by changing the value of the indices like this:

1=20×30 2=21×30 4=22×30

3=20×31 6=21×31 12=22×31

We have 3 choices (0, 1 or 2) for the power of 2, and 2 choices (0 or 1) for the power of 3, so there are 3 x 2 = 6 choices (factors) in all.

We just look at the indices of the prime factorization and this tells us the number of factors, generally for any number n:

n=p1k1× p2

k2×…× prk r

Then the number of factors = (k 1+1 )× (k2+1 )×…× (kr+1 )

If we now consider the values of k for a square number, they will always be even (why?), so the number of factors will be the product of odd numbers, so will always be odd.

Can you find a pattern in the how many factors a number has? Or better still, a formula? [Hint: Explore the prime factors of each number]

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5.2.2 Sums of squares

You should find that the most square numbers we need for any number between 1 and 10 is 4, which is 7 = 4 + 1 + 1 + 1. We can some of them in more than one way, such as 9 = 4 + 4 + 1 or just 9 = 9.

Here are 1 to 20 written using the minimum number of squares possible:

1 = 12 = 1 + 13 = 1 + 1 + 14 = 45 = 4 + 16 = 4 + 1 + 17 = 4 + 1 + 1 + 18 = 4 + 49 = 910 = 9 + 111 = 9 + 1 + 112 = 9 + 1 + 1 + 113 = 9 + 414 = 9 + 4 + 115 = 9 + 4 + 1 + 116 = 1617 = 16 + 118 = 9 + 919 = 9 + 9 + 120 = 16 + 4

You can see that we start of with a kind of pattern, where numbers need increasingly more squares until the ‘pressure’ is relieved by a new square number, or one that only needs two squares. However, the pattern seems to break down a bit at the end.

Can you write all the numbers from 1 to 10 as the sum of square numbers?

Is there more than one way to do them all?

What is the minimum number of squares needed for each one?

What about numbers bigger than 10?

If we limit ourselves to a maximum of 4 squares, can we write all the numbers from 11 to 20? How about higher than this?

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Fermat was one of the first mathematicians to work out which numbers needed which numbers of squares, although he left it to others to prove his results. Interestingly, the results have a lot to do with prime numbers, so let’s look at those first:

If you investigate this question, you will probably notice that some of the primes behave ‘nicely’ and some don’t. By nicely, I mean that they seem to only need a couple of primes: these are 2, 5, 13, and 17. The badly behaved ones need 3 or 4 squares.

Apart from 2 (which is nearly always a special case!) these nice primes are all one more than the four times table (i.e. of the form 4k+1). Let’s explore this further by looking at other 4k+1 primes:

5 = 4 + 113 = 9 + 417 = 16 + 129 = 25 + 437 = 36 + 141 = 25 + 16

It seems that all 4k+1 primes can be written using only two primes, and what’s more than this, there is only one way of doing it. This is in fact true, and was proved by Leonard Euler in the 18th century. Of the other primes, one less than the 4 times table (4k-3), all need at least three primes.

To see why we can’t write any of the 4k+3 primes as the sum of two squares, consider all the possibilities, using modular arithmetic:

n(mod 4) n2(mod 4)0 01 12 03 1

This table tells us that square numbers are all of the form 0 or 1 (mod 4). Therefore adding any two square numbers can only be of the form 0, 1 or 2 (mod 4). So no sum of two square numbers can be of the form 3 (mod 4), or 4k+3.

Do you think that 4 squares will always be enough?

What other patterns do you notice?

Look at the prime numbers from 1 to 20. How many squares are needed for each? How many different ways are there of writing them?

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Although the proof that all 4k+3 primes can’t be written as the sum of two squares is easy, the proof that 4k+1 primes can be written as the (unique) sum of two squares is quite complicated and is not included here.

Ignoring the squares themselves, we have:

6 = 4 + 1 + 18 = 4 + 410 = 9 + 112 = 9 + 1 + 1 + 114 = 9 + 4 + 115 = 9 + 4 + 1 + 118 = 9 + 920 = 16 + 4

Again, some of them are nicely behaved and some are not. Let’s list the nicely behaved ones that only need two squares and see if we can spot anything about them: they are 8, 10, 18 and 20. Perhaps it has something to do with 2? But then 6 and 14 are not included, so let’s look a bit more deeply.

Let’s suppose it has something to do with primes; we know that the ‘good’ primes are 2, and the 4k+1 ones. What are the prime factors of 8, 10, 18 and 20?

8 = 2 x 2 x 210 = 2 x 518 = 2 x 3 x 320 = 2 x 2 x 5

We can see that most of these are made from 2 or the 5 apart from 18, so this doesn’t quite back up our conjecture that the good ones are made from (2 or) 4k+1 primes. How about the ones that can’t be written as the sum of two squares:

6 = 2 x 312 = 2 x 2 x 314 = 2 x 7

There are definitely plenty of 4k+3 primes here, so we might be on to something after all; let’s keep looking for a pattern.

22 = 2 x 11 = 9 + 9 + 4 24 = 2 x 2 x 2 x 3 = 16 + 4 + 4 26 = 2 x 13 = 25 + 1 (OK)28 = 2 x 2 x 7 = 16 + 9 + 1 30 = 2 x 3 x 5 = 25 + 4 + 1

What about non-primes? How many squares are needed for these?

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Again, the only one that can be written using two squares is one with a 4k+1 prime, and all the bad ones contain a 4k+3 prime. So generally the numbers with 4k+3 prime factors can’t be written as sums of two squares… let’s revisit our special case, 18, which contains two bad primes… but wait, of course this can be written as two squares as 18 = 2 x 3 x 3 = 3 x 3 + 3 x 3 = 9 + 9…

If we have an even number of bad primes, then maybe we can split them into two like for 18. This is definitely true if we have a prime factor of 2 in our number (why?) but what if we don’t? Let’s try the number 5 x 3 x 3 = 45… can this be written as the sum of two squares? It can! 45 = 36 + 9. So it doesn’t matter whether 2 is a factor; we conjecture that any number with an even number (including zero) of 4k+3 primes can be written as the sum of two squares, which is indeed the case, although we won’t prove it here.

Let’s test it out. First choose a prime factorization that has an even number of 4k+3 primes, such as 23.32.5 = 360. Can this be written as the sum of two squares?

360 = 9 x 40 = 9 x (36 + 4) = 324 + 36 = 182 + 62 so it works for this one.

Finally, we have seen that some numbers need 3 or 4 squares, but is 4 always enough? Surprisingly, it is! The French mathematician Joseph-Louis Lagrange proved this in the 18th century.

Finally, there is another way of finding the number of ways of writing a number as the sum of two squares; I included it in Prime Sequence, but here it is again for completeness

French mathematician Adrien-Marie Legendre showed that the number of ways of writing a number as the sum of two squares is 4(F+1 – F+3) where F+1 is the number of 4n+1 factors and F+3 is the number of 4n+3 ones.

So 45 can be written as the sum of two squares in 8 ways (noting that order is important!

45 = (6)2 + (3)2 (3)2 + (6)2

(6)2 + (-3)2 (-3)2 + (6)2

Thinking about this, can you come up with a rule for which non-primes can be written as the sum of two squares?

Explore some more numbers with and without an even number of 4k+3 primes and test the conjecture.

Test out Lagrange’s Four Square Theorem for yourself.

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(-6)2 + (3)2 (3)2 + (-6)2

(-6)2 + (-3)2 (-3)2 + (-6)2

5.2.3 Summing sequences

We have seen that the triangle numbers go 1, 3, 6, 10, …

We can think of these as being the sums of the whole numbers:

1, 1+2, 1+2+3, 1+2+3+4, …

As we have found the nth term of the sequence to be ½.n.(n+1), we can easily use this to find the sum of (say) the first 100 whole numbers by just putting in n = 100 into the formula, which gives ½.100.101 = 50.101 = 5,050.

To find the sum of the first 100 even numbers is easy if you think about it. You just need to double the answer for the first 100 whole numbers, as each number is doubled! So we have 5,050 x 2 = 10,100, and the formula for this is just n(n+1).

Even more ingeniously, we can also use this to find the formula for the first 100 odd numbers. If we take one off for every even number we just take 100 off to get 10,000.

Using algebra, we could just take n from n(n+1), which is:n(n+1) – n = n2 + n – n = n2. So the sum of the first n odd numbers is the nth square number! If this is surprising, have a look at this diagram, which shows why it is true!

We might have guessed this if we remembered that the differences of the square numbers 1, 4, 9, 16 are the odd numbers 3, 5, 7, … So the square numbers are also sums 1, 1+3, 1+3+5, 1+3+5+7, …

We can extend this idea further like this:

What is the sum of the first 100 odd numbers? How about the first 100 even numbers? Can we find the sum of the first 100 numbers of different sequences?

Do any other sums of sequences like this have an interpretation as a shape?

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The pentagonal numbers go 1, 5, 12, 22, … and are also sums of sequences: 1, 1+4, 1+4+7, … just like the triangle and square numbers.

We are now going to find a formula for the nth pentagonal number; it involves a bit of algebra so if you want to skip this bit don’t worry.

The pentagonal numbers are sums of arithmetic sequences, which means they go up by the same amount each time, so we can use a trick to find them. First, add the first and last term together and then divide by 2. This is the same as finding the middle (average) number of the sequence. Then all we have to do is multiply by how many numbers we’ve got.

For example, to work out 1+4+7+10, we just do 1+10 = 11, divide this by 2 to get 5.5, which is the ‘middle’ number, then multiply by 4 (because there are 4 of them) to give the total of 5.5 x 4 = 22, which is the 4th pentagonal number.

We can use this to work out the nth term for the pentagonal numbers. The sequence 1,4,7,10,.. has itself got nth term 3n-2 (as it’s 2 less than the 3x table), so we havefirst term plus last term = 1 + (3n – 2) = 3n – 1, then divide this by 2 and multiply by how many we’ve got (n) to give the nth pentagonal number = ½.n.(3n-1).

A different way to find this rule is to consider the identity for triangle numbers, which is Δn=n+Δn−1. We can then think of square numbers as being made from the previous two triangle numbers like this:

An identity for this would be ⊡n=n+2. Δn−1. You can check that substituting the n-1th

term for triangle numbers, which is ½.(n-1).n, gives ⊡n=n2.

Can you write down an identity for this pattern?

Before reading on, can you find the rules for generating these shape/number sequences?

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This suggests that the next type of number in this sequence might be n+3. Δn−1, which is like saying pentagons are made from 3 triangles, which makes sense!

You can check that putting the formula for the n-1th triangle number into this identity gives the correct formula for the nth pentagonal number if you want!There are other types of shape-numbers you can investigate. The last one I am going to include here is the hex-numbers (not to be confused with the hexagonal numbers above).

The hex numbers go like this:

Each hex number is made from 6 triangle numbers + 1, so for example 7 = 6 x 1 + 1, then 19 = 6 x 3 + 1, so the next one is 6 x 6 + 1 = 37. You can check this by drawing it as shown on the left

An identity for this might look like this:

¿n=6. Δn−1+1 ¿

Finally, notice that summing the hex numbers leads to the cube numbers. To see why this is true, imagine them as three sides of a cube, and then try to imagine putting the smaller ones inside the next largest one, like this:

What interesting identities can you find for the hex numbers?

How are they connected to the triangle numbers? Or the cube numbers?

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If you can’t see how, try making them with multi-link cubes!

Here are some interesting sums of squares:

12 – 22 + 32 – 42 = -(1 + 2 + 3 + 4)

12 – 22 + 32 – 42 + 52 = 1 + 2 + 3 + 4 + 5

They are indeed true. To see why, you could use the difference of two squares.

Writing the first one as (1+2)(1-2) + (3+4)(3-4) = (1+2)(-1) + (3+4)(-1) and the second one as 1 + (3+2)(3-2) + (5+4)(5-4) = 1 + (3+2) + (5+4) we can see these patterns will continue.

Summing the odd squares gives:

12 = 1 = 1 x 2 x 3/612 + 32 = 10 = 3 x 4 x 5 / 612 + 32 + 52 = 35 = 5 x 6 x 7 / 6

Confirm these are both true. Does this pattern always work? Can you explain why?

What patterns do you get if you sum the odd squares only? How about the even squares?

Can you explain this one using a diagram!?

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5.3 The great Gauss

Here are a few investigations that involve some of the mathematics that German mathematician Carl Friedrich Gauss worked on.

5.3.1 Quadratic Residues*

Gauss was the first to study Quadratic Residues in depth. Quadratic Residues are just the square numbers mod n for different n. We found the QRs mod 25 in the second part of the investigation triangle cakes.

For example, if you take the square numbers mod 5 you will find that you only get 1 and 4 (as well as 0 which we are not interested in here). We say that 1 and 4 are quadratic residues (QR) mod 5, and 2 and 3 are non-residues (NR).

Here are the quadratic residues for different modulo n with prime n:

1 4 9 16 25 36 49 6

4 81 100 121 144

mod 3 1 1 0 1 1 0 1 1 0 1 1 1mod 5 1 4 4 1 0 1 4 4 1 0 0 0mod 7 1 4 2 2 4 1 0 1 4 2 2 2mod 11 1 4 9 5 3 3 5 9 4 1 0 1

mod 13 1 4 9 3 12 10 10 1

2 3 9 4 1

If we put these into a table (and add in 17 and 19 as well) we can see that half of the numbers are QR and half are NR:

n QR NR3 1 25 1, 4 2, 37 1, 2, 4 3, 5, 6

Investigate patterns in the square numbers for other mod n before reading on!

Investigate what happens for prime n in particular.

No history of numbers can be complete without mention of Gauss. Many people consider him to be the greatest mathematician that ever lived.

He was a German mathematician who lived during the 19th century. He made contributions to many areas of maths, but for a long period of his life his main passion was Number Theory. He developed the mathematics of modular arithmetic and proved the law of quadratic reciprocity, which he called the ‘golden theorem’. We are going to look at quadratic residues, which form the basis of the law of quadratic reciprocity, described at the end of the section.

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11 1, 3, 4, 5, 9 2, 6, 7, 8, 1013 1, 3, 4, 9, 10, 12 2, 5, 6, 7, 8, 1117 1, 2, 4, 8, 9, 13, 15, 16 3, 5, 6, 7, 10, 11, 12, 1419 1, 4, 5, 6, 7, 9, 11, 16, 17 2, 3, 8, 10, 12, 13, 14, 15, 18

These numbers look fairly random; are there any more patterns here? Well, if we look at some of the values of n (5, 13, 17 – those of the form 4k+1), we can see that n-1 is a QR and for the others (of the form 4k+3) it is a NR. This suggests that there is something different about primes of the form 4k+1 and those of the form 4k+3.

Looking only at the 4k+1 primes we have:

n QR NR5 1, 4 2, 3

13 1, 3, 4, 9, 10, 12 2, 5, 6, 7, 8, 1117 1, 2, 4, 8, 9, 13, 15, 16 3, 5, 6, 7, 10, 11, 12, 14

Closer inspection reveals that the QRs come in pairs. For example, for 13 we can add pairs together to make 13 (1+12, 3+10, 4+9). We can also see that NRs also come in pairs.

Is this true for the 4k+3 primes too?

n QR NR3 1 27 1, 2, 4 3, 5, 6

11 1, 3, 4, 5, 9 2, 6, 7, 8, 1019 1, 4, 5, 6, 7, 9, 11, 16, 17 2, 3, 8, 10, 12, 13, 14, 15, 18

It is not; another difference between 4k+1 and 4k+3 primes! So what patterns are there in the 4k+3 primes?

In fact, they come in QR/NR pairs. Looking at 11 as an example, we have 1+10, 3+8, 4+7, 5+6 and 9+2. This is true for all of these 4k+3 primes.

When you have found your QRs for different modulo n, try multiplying them together. For example, here is what happens for mod 5:

QR = {1,4} NR = {2,3}

If we multiply any two QRs together we get 1 or 4 (mod 5) so we might conjecture that QR x QR = QR for any mod n.

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Let’s continue to use mod 5 as an example. What about QR x NR? Well, we have (working in mod 5):

1 x 2 = 2 1 x 3 = 34 x 2 = 3 4 x 3 = 2

So we can conjecture that QR x NR = NR.

What about NR x NR? Trying it out reveals that NR x NR = QR.

We have a full set of rules for multiplying QR and NR:

QR x QR = QRQR x NR = NRNR x NR = QR

Of course, these are the same rules as multiplying positive and negatives if we think of QR as positive and NR as negatives.

This helps us solve quadratic congruence equations like this:

x2≡3(mod 5)

In fact, you didn’t even need to try; knowing that 3 is a NR mod 5 means that there is no solution to this equation.

Now, the law of quadratic reciprocity (LQR) states that if either of the numbers 3 or 5 is a 4k+1 prime, then if x2≡3(mod 5) has (no) solutions, then so does x2≡5(mod 3). In the other case where neither of the numbers is a 4k+1 prime (both 4k+3 primes) then if one equation has a solution, the other one doesn’t.

This can be particularly useful when the numbers are quite large. For example, does x2≡37(mod 53) have solutions? Well, as 37 is a 4k+1 prime, LQR says that if x2≡37(mod 53) has a solution then so does x2≡53(mod 37).

Why is this useful to know? Well, because we can note that 53≡16 (mod 37). So solving x2≡53 (mod 37 ) is the same as solving x2≡16(mod 37) which we can see has a solution, x≡4 (mod 37). So then LQR tells us the original equation also has a solution.

Try this for other QRs mod n. Does it seem true?

What other rules can you come up with for multiplying QRs and NRs together? For example, what is QR x NR or NR x NR?

Can you find a value for x2 that is congruent to 3 mod 5?

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5.3.2 Wilson’s Theorem

Gauss was the first person to explore something now called Wilson’s theorem. He decided to use modular arithmetic to see if there was a pattern in the factorials.

What is the remainder of 4! (= 24) for different modulo n? Here is a table:

mod n 1 2 3 4 5 6 7 8 9 10 114! 0 0 0 0 4 0 3 0 6 4 2

Nothing interesting here… or is there? First of all, we note that the first non-zero remainder happens at mod 5 (because 4! = 4 x 3 x 2 x 1). But is it a coincidence that the first non-zero remainder of 4! is also 4?

If you look at n! (mod n+1) you get the following results:

2 !(mod 3)≡23 !(mod 4)≡24 !(mod 5)≡45 !(mod 6)≡06 ! (mod 7)≡6

It would appear that n !(modn+1)≡n in the cases where n + 1 is prime.

This is Wilson’s Theorem. I am not going to prove it here but will give an example that will give a suggestion of the proof. Let’s consider 6! = 1 x 2 x 3 x 4 x 5 x 6.

We can take these factors in pairs that are congruent to 1 (mod 7) like this:

2×4=8≡1(mod 7)

3×5=15≡1(mod 7)

We have 1 and 6 left over. So 6 ! (mod 7 )=1×2×3×4×5×6≡1×6≡6(mod 7).

This pairing is always possible for any n factorial under prime modulo n + 1 – try it!

Investigate the first non-zero remainder for other factorials. Can you make a conjecture?

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5.3.3 Eight Queens Problem

While we are still talking about Gauss, let’s look at a different problem he investigated:

Here is a solution for the 4 queens problem, and below is a way of representing this arrangement:

1 2 3 42 4 1 3

If you look at the solution to the 4 queens problem given you can see that each column is like a co-ordinate for each queen.

1 2 3 42 4 1 3

If you add the columns you will notice that each number is different. This means that each queen is on a different diagonal. To see why, consider the co-ordinate sums – they are the same on the downward diagonals.

So we could think of the n queens problem as finding 4 pairs of co-ordinates with different sums.

But unfortunately there isn’t all! We also need to check the other (upward) diagonals. Gauss came up with an ingenious way of doing this.

If we reverse the order of the top row and plot the queens we get the following solution:

4 3 2 12 4 1 3

Can you arrange n queens on a n x n chessboard so that no queen is attacking any other?

Start by trying to to arrange 3 queens on a 3 x 3 board, then 4 queens on a 4 x 4 board and so on.

Can you see what the numbers represent? How might this help you solve more difficult queens problems? [Hint: try adding the columns]

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Q

Q

Q

Q

Q

Q

Q

Q

5 6 7 8

4 5 6 7

3 4 5 6

2 3 4 5

Q

Q

Q

Q

This has effectively reflected the board in a vertical line of symmetry. All the upward diagonals are now downward diagonals, and all we have to do now is check the column sums for this new arrangement, thus checking upward and downward diagonals!

Here’s one that I found:

1 2 3 4 5 6 7 85 3 1 7 2 8 6 46 5 4 11 7 14 13 12

8 7 6 5 4 3 2 15 3 1 7 2 8 6 413 10 7 12 6 11 8 5

Now can you find a solution to the 8 queens problem?

Apparently there are 11 different ones (ignoring rotations and reflections)… Can you find them?

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Q

Q

Q

Q

Q

Q

Q

Q

5.4 Towers of Hanoi

This seems as good a place as any to slip in another famous puzzle with lots of interesting maths attached.

The Towers of Hanoi puzzle was invented by French Mathematician Eduoard Lucas in the 19th century. The puzzle is this:

Arrange differently sized discs on three pegs as shown:

Your task is to get the discs from this peg to the end peg, one at a time, but with the condition that you can never place a larger disc on top of a smaller one, in the minimum number of moves possible.

Numbering the discs 1 (smallest) to 4 (largest) the sequence of moves that takes the discs to the end peg is 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 which is 15 moves.

Experimenting with different numbers leads to the following sequences of moves:

2 discs 1 2 1 3 moves3 discs 1 2 1 3 1 2 1 7 moves

Adding this to the result for 4 discs gives a pattern in the minimum number of moves of 1, 3, 7, 15 (4 discs), 31, and so on. These are the powers of 2 minus 1 (called the Mersenne numbers), or alternatively each number is double the last. We can see this structure in the sequence of moves. The previous sequence of moves is repeated either side of moving the largest disc.

You might have spotted this sequence when looking at the binary numbers:

You can see from the diagram that the position of the right-most 1 in each number follows the pattern 1 2 1 3 1 2 1 4 1 …

Investigate with different numbers of discs. Can you see a pattern? Can you describe a sequence of moves that will give the minimum number of moves?

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Here are a couple of fun ways of thinking about the solution to the puzzle.

Consider the odd numbered discs as moving clockwise around 3 pegs arranged in a circle, and even numbered discs as moving anti-clockwise, with the smallest available peg moving at all times but never twice in a row, and you have an algorithm for solving the problem. I have shown this in the picture below; note that the first move determines which peg you finish on with an odd number of discs, and vice versa for an even number of discs.

You can also think a way of visting all the vertices on a cube:

Starting at the bottom left corner, we can think of each move as a move along an edge in each dimension, giving 1213121.

This is known as a Hamiltonian circuit.

We can extend this idea further by thinking about the 4-disc example as a Hamiltonian circuit around a hypercube!

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5.5 Some unusual sequences

Here are some investigations into some number sequences you might not have seen before.

5.5.1 Stirling numbers… of the second kind

James Stirling was a Scottish mathematician who lived in the 18th century. He is possibly most famous for his discovery of a formula that approximates factorials. However, he also discovered the lesser know Stirling numbers, which are becoming increasingly important in computer science.

There are two types of Stirling numbers, the first and second kind. The second kind is probably the easiest to understand, so we’ll start there.

Using letters, we have 3 ways: {A,B} and {C}{A,C} and {B}{B,C} and {A}

The number of ways of putting 3 objects in 2 piles is the Stirling number {32} = 3.

The curly brackets are the standard way of showing sets in maths, and this is

probably why they are used here; {nk } means the number of ways of putting n objects

into k sets.

One way of finding {42 } is to put the fourth letter (D) into either the first or second

set in the list above, giving:

{A,B,D} and {C} {A,B} and {C,D}{A,C,D} and {B} {A,C} and {B,D}{B,C,D} and {A} {B,C} and {A,D}

And finally we have a 7th arrangement, which is {A,B,C} and {D}, so {42 } = 7.

To find {43 } we have

Consider putting three objects into two different piles (neither of these piles being empty). How many ways are there of doing this? (order doesn’t matter)

Can you find a systematic way of working out the Stirling numbers {42 } and {43 }?

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{A,B},{C},{D} {B,C},{A},{D} {C,D},{A},{B}{A,C},{B},{D} {B,D},{A},{C}{A,D},{B},{C}

We can put all the Stirling numbers we have found into a triangle, a bit like Pascal’s triangle:

It’s a bit harder to see, but you can find the numbers in the next row of the table.

We found {42 } using {32} systematically, which was:

{42 }=2 {32}+{31} = 2 x 3 + 1 = 7.

You can see that here in the table:

The blue number is 2 x green + orange.

Looking closely we can see that the 2 represents the column the green number is in, which we can write as a general relationship like this:

{nk }=k {n−1k }+{n−1k−1}.

Use this approach to find the Stirling numbers {5k } for different values of k.

Can you find any connections between successive rows of the table, like in Pascal’s triangle?

Can you adapt this idea to see how the numbers in the bottom row of the table could be calculated from the numbers in the row above?

What other patterns can you see in the table?

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You might have noticed the good old triangle numbers make their appearance in the

table. To see why, have a look at the way I arranged {43 } when calculating it above.

We could draw this like this:

You can see that this is equivalent to the triangle numbers by noticing that A is connected to the other three letters, then B is connected to the other two and C is connected to D = 3 + 2 + 1. (You might recognize the picture on the right as the complete graph K4).

Also, in the second column (n objects into 2 groups), you might have noticed that the

numbers are one less than a power of 2. To see why, consider {42 } again.

Fixing one of the objects, say A, in one group (it can’t be empty), we have choices of whether to put each of the other 3 letters in or out of that group. This gives 23 ways of making this group. However, we have to subtract 1 as we are not allowed {A,B,C,D} as that would leave the other group empty.

This sequence of 1 less than a power of 2 appears often in counting problems like this, such as the Towers of Hanoi puzzle from the previous section.

Finally, if you sum the rows of this table you get what are called the Bell numbers, named after mathematician and author of detective novels Eric Bell.

So, for example, b4 = 15 represents the number of ways of grouping 4 objects into different groups:

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5.5.2 Stirling numbers… of the first kind

Now let’s turn our attention to Stirling numbers of the first kind. They are similar to the numbers above, but slightly different.

The number [nk ] now represents the number of ways of putting n objects into k

cycles. A cycle is like a loop or a bracelet (see number bracelets) like this:

These are different cycles as they are arranged in a different order. Written as a list, we could write the one on the left in four different ways: [ABCD], [BCDA], [CDAB] and [DABC] but these are all really the same cycle, or the same order when put in a loop. None of these are the same as the second cycle, which could be written [ACBD].

Using this notation from now on, let’s find out if there’s anything interesting about these Stirling numbers. First, let’s work out some values; starting like before, what is

[32]? We have 3 different ways of putting 3 objects into 2 cycles:

[AB] and [C][AC] and [B][BC] and [A]

So far, they are the same as the other numbers. Can wo

Of course it turns out they are different. For example, [31] = 2, which is different to

{31}=1 because the order matters.

Can you work out [31]? Is this the same as {31}? Explain why.

Can you find a systematic way of working out [42 ] and [43 ]?Are these the same as the other Stirling numbers?

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You should have also found that [42 ] = 11 and [43 ] = 6, so they are generally different.

Here are the first few Stirling numbers of the first kind in a table:

You might have noticed that triangle numbers are still there, and perhaps you noticed that the factorial numbers are in the first column.

To see why, consider [41 ] = 6.

Fixing one of the letters, say A, we have 3! ways of positioning the other letters like this:

[ABCD] [ACBD] [ADBC][ABDC] [ACDB] [ADCB]

In fact there is a very similar relationship as above. Looking at the table we see that the blue number is 3 x green + orange.

Closer inspection reveals that it is the row number that we multiply green by this time, instead of the column number.

So this gives us the general relationship:

[nk ]=(n−1 )[n−1k ]+[n−1k−1]

You get the factorials! To see why, consider the cycle arrangements as different permutations of letters.

To see what I mean, consider the two different arrangements of letters ABC and ACB. We can interpret this as saying B has swapped places with C, and A has stayed in the

Work out some more of these Stirling numbers and put them in a table. Can you find any patterns?

Is there a way of working out successive rows from previous rows?

What numbers do you get if you sum the rows of this table?

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same place. In cycle notation, we can write this as [A][BC] which is one of our

arrangements in [32]. Here is a list of all the possible ways of writing three letters

from the ‘starting position’ of ABC, and their corresponding cycle arrangements:

ABC [A][B][C] nothing has changedACB [A][BC] BC swappedBAC [AB][C] AB swappedBCA [ACB] cycled anticlockwise CAB [ABC] cycled clockwiseCBA [AC][B] AC swapped

Now you can see all the different cycle structures are here, and we know that the number of ways of arranging 3 objects is 3!, which shows why the sums of the rows of the table are the factorials.

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5.5.3 Moessner sequences

Mathematician Alfred Moessner found the following pattern in the 1950s. Starting with the numbers in a row 1 2 3 4 … take out all the even numbers (leaving the odds) 1 3 5 7 … Now if we work out the cumulative sum (adding up the total as we go along) it should be no surprise to you that we get 1, 4, 9, 16, … the square numbers.

But what happens if we take out every third number? We get the sequence 1 2 4 5 7 8 10 11 13 14… and adding these up cumulatively we get 1 3 7 12 19 27 37 48… which are known as the ‘three quarter squares’ (can you see why?)… but more interestingly, if we now cross out every second number we get 1 7 19 37 and adding cumulatively again we get 1 8 27 64 … the cube numbers!

We can show this with this diagram:

It’s kind of tricky to get right, but if you did it correctly you should have got the fourth powers 1, 16, 81, ...

Here’s what happens when you cross out the triangle numbers:

Try deleting every fourth number, then add, then delete every third number, then add, and so on like above. What numbers do you get at the end?

Try crossing out some other sequences of numbers, like the triangle number or the square numbers, and following the process. What do you get?

What is special about the bottom number in each section 2, 6, 24, 120, …?

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5.5.4 Frequency sequences

Here is the square number sequence: 1, 4, 9, 16, …

Here is my table:

We can see that there is a pattern to this sequence 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, … There is one 0, three 1s, five 3s and so on (why?). Now let’s ask the same question about this sequence:

It may or may not have surprised you to find out that the less than sequence for the less than sequence takes us back to the square numbers!

I’m going to leave this one open for you to investigate. The square number and ‘less than’ sequences are called inverse sequences, meaning that if you follow this procedure you get back to the original sequence.

However, there is something more interesting about these sequences.

Take the square number sequence and to each number, add its position in the sequence. So for square numbers we have:

I’ve called this new sequence F.

Here are the calculations for the less than sequence:

How many numbers in the sequence are less than 1? Less than 2? Less than 3? Less than 4? Make a table and complete these values for up to, say, 10.

Can you see any patterns?

How many numbers in the ‘less than’ sequence are less than 1? Less than 2?

Try this for other sequences. Is it always true?

Now do the same for the less than sequence. What do you notice?

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I’ve called this new sequence G. You can see that F and G do not share the same numbers. In fact, G contains all the numbers not contained in F!

Explore this for other sequences such as the triangle numbers.

Does it always work? Can you explain why?

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5.5.5 Thue-Morse sequence

This is a strange sequence. It starts with the numbers 0, 01, …

The next number is made by appending the reverse of the previous number to itself. For example, the next number is 0110.

The first few numbers of the sequence are 0, 01, 0110, 01101001, 0110100110010110, 01101001100101101001011001101001, …

Here is one way of representing the next number in the sequence if we read line by line with white = 0 and red = 1:

You might have noticed things like there are never more three of the same digits in a row, or that each odd term in the sequence is a palindrome (reads the same backwads).

We can think of the number 01101001100101101001011001101001… as a sequence in itself; it is known as the Thue-Morse sequence, discovered around the start of the 20th century named by Norwegian mathematician Axel Thue and US mathematician Marston Morse.

There is a way of working out the nth term in this sequence. If you want to find the nth term, just count how many 1s are in the binary number for n. If there is an even number of 1s, the nth term is a 0, else it is a 1.

For example, if we wanted the 5th term, we know 5 is 101 in binary and there are an even number of 1s, so we conclude the 5th term is a 0 (which it is, remembering the first term is the 0th term).

Mathematician John Conway calls the numbers corresponding to 1s odious (odd) numbers and those corresponding to 0s evil numbers. So the evil numbers are 0, 3, 5, 8, 9, …

You may also have spotted some other patterns, such as the 1st term is the same as the 2nd term. In fact, it is always true that the nth term is the same as the 2nth term. Or you may have spotted that every odd term is opposite the one before it.

Find the next few numbers in this sequence. Can you see any patterns in the sequence?

If we call the first 0 the 0th term, 1 the 1st term, 1 the 2nd term and so on, what observations can you make?

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One interesting application of the sequence is that it gives a way of picking two teams more fairly (or of sharing something between two people). It is like taking turns taking turns…

If 0 represents a pick for Team Zero, and 1 represents a pick for Team One, then we can use the sequence to produce a fairer way of choosing players than just 0101010101…

One thing you should definitely do after this is to find out more about mathematician John Conway. He has written some amazing books, such as The Book of Numbers, Winning Ways For Your Mathematical Plays, and The Symmetries of Things.

Do you think this is a fairer way of picking teams? Can you think of a better way?

Find out more about ways of sharing things fairly.

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5.6 Pascal’s triangle

These investigations explore some of the patterns behind Pascal’s triangle.

5.6.1 Blob pyramid

Here are the rules for creating a blob pyramid…

and here are the first few rows:

You should get something like this:Continue the pattern using the rules. What do you notice? Can you explain why?

Blaise Pascal was another great French mathematician who lived during the 17th century. He was most famous for his work in the applied sciences, and built one of the first mechanical calculators. He also made some valuable contributions to maths; he laid the foundations for probability theory (with Fermat), made contributions to projective geometry and the study of cycloids, and is famous for his book ‘Treatise on the Arithmetical Triangle’.

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If you carry this on for a while you will get something that resembles Sierpinski’s triangle.

This is equivalent to colouring the odd numbers in Pascal’s triangle white and the even numbers black.

This is the same as colouring the remainders on division by 2 (mod 2).

This ‘Pascal’s cube’ was made by one of my pupils; each face is Pascal’s triangle mod 5, with each remainder given a chosen colour, then put together to give the 3D illusion.

Explore the patterns in remainders of Pascal’s triangle with other divisors.

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5.6.2 Rook moves

Consider a rook placed at square A1 on a chessboard.

For example, there are 3 shortest ways of getting to square C2 as shown below:

We can see that Pascal’s triangle has popped up again.

So the number of ways of getting to

H8 is… 3432, or (147 ).

Here’s a related puzzle:

You can make the word SUM in 4 different ways in this pattern as shown.

How many ways can you make the word MATHS in the one below?

M A T H S

A T H S

T H S

H S

S

How many ways can it get to various squares on the board using the shortest possible route only? How many ways are there of getting to square H8?

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You will notice that there are 4 ways of making the word SUM in the example puzzle. If you break this down into how many ways of getting to each M, we have 1 + 2 + 1.

It is no coincidence that these are the numbers in row 3 of Pascal’s triangle. You might have conjectured that there are 1 + 4 + 6 + 4 + 1 = 16 ways of making the word MATHS in the puzzle given and this is indeed the case.

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5.6.3 Lines in the plane

Draw lines across a rectangle like this so that every new line divides the rectangle into as many regions as possible:

The number of regions follows this sequence: 2, 4, 7, 11, 16, …

This is just the triangle numbers plus 1.

So these numbers appear in Pascal’s triangle as shown.

This is because each line creates one extra region that the last line.

Here’s a little puzzle involving this line puzzle:

This is possible if you think the sum of the numbers is 28, so we need 14 either side of each line, with one 14 made out of 3 of the numbers, the other out of 4.

The possible sets of three numbers that might work are {1, 6, 7}, {2, 5, 7}, {3, 4, 7} and {3, 5, 6}. Trying these out gives us three solutions, two of which work and one of them doesn’t work (which one?).

How does this pattern continue? Can you find a rule? Can you explain it?

How is this connected to the activities in this section?

Can you place the numbers 1, 2, 3, 4, 5, 6, 7 into the third picture on the right above so that there is an equal sum on either side of each of the 3 lines?

Can you find more than one different way of doing it?

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5.6.4 Catalan numbers

Here are some interesting patterns, many of which are taken from the excellent Book of Numbers by Conway and Guy.

You can see that the number of triangles, squares and pentagons is 1, 2 and 5, with each shape partitioned into triangles in a different way. If you continued drawing the hexagons you should have found 14 of them.

This sequence of trees also has 1, 2 and 5 different drawings, and the next set with 4 buds does indeed have 14 different trees.

If you try superimposing the tree on the shapes, you can see how they are connected like this:

Can you work out how to continue this sequence without any instructions?

Can you see how and why these two seemingly different sequences are connected? [Hint: think of each bud as being a triangle and each branch as an edge]

Now consider this sequence of trees; what comes next? How many trees with 4 ‘buds’ are there?

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These are different representations of the sequence of numbers 1, 2, 5, 14, 42, … discovered by Belgian mathematician Eugene Catalan in the 19th century. Catalan numbers appear in lots of combinatorial problems, and also have applications in computer science.

To see how, let’s look at the tree again. This is called a binary tree in computing (as each bud grows two branches) and is often used as a structure for storing and searching data.

Now, suppose each bud (node) on the binary tree corresponds to a 0 and each tip of a branch to a 1. Then imagine a snail crawling up the tree from the root, calling out the numbers of the nodes and tips as it goes along, but not calling out the same node/tip twice. If the snail went round this tree clockwise (left before right) we would have the sequence 0100111.

All the possibilities for 3 nodes are 0001111, 0010111, 0011011, 0101011 and 0100111 (above).

Now have a look at these three different sequences (in columns). Can you make the connections between them (and the representations above)?

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These are called Dyck words; typically we ignore the final 1 as they all end in a 1. So the 6-letter Dyck words are 000111, 001011, 001101, 010101 and 010011.

The Dyck words are all even length. The Dyck word of length 2 is 01, and of length 4 are 0011 and 0101. You will probably have noticed that they all contain an equal number of ones and zeroes.

You might not have noticed that if you take the first n letters of any Dyck word, there will always be more 0’s than 1’s.

We could think of Dyck words as representing votes in an election between two candidates (0 and 1) where both candidates get an equal number of votes, but where 0 always leads 1.

I have also seen this considered as the number of ways of drunkard can exit a pub door, stagger around outside backwards and forwards (without re-entering the pub), and return to exactly where he started! For example, there is only one way with 2 steps (forward, back), two ways with 4 steps (FFBB and FBFB), 5 ways with six steps (FFFBBB, FFBFBB, FFBBFB, FBFFBB and FBFBFB) and so on.

Here’s another representation of Catalan numbers: Consider the number of ways of getting from the bottom left of this square to the top right without crossing the green diagonal:

Find all the 8 letter Dyck words and confirm that they follow the Catalan sequence.

Check the number of paths for a 4x4 square is 14. Can you make a link between this representation and the others in this section?

Find the Dyck words corresponding to other binary trees. What can you say about Dyck words in general?

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Finally, here’s an interesting one.

Look at this diagram, which shows the numbers of ways of arranging pennies so that every penny in the row above must touch two adjacent pennies in the row below:

Now, if you rearrange this pennies so that we get all the ones with 1 on the bottom row, then 2, and so on, we get the following picture:

I can definitely see a link between this representation and the ‘mountain’ arrangement in the earlier part of this section.

So, why are Catalan numbers in this section on Pascal’s triangle? Well, Catalan numbers appear in Pascal’s triangle! If you look at the numbers in the (vertical) spine of the triangle, you can see 1, 2, 6, 20, 70, 252, … and if we divide these by their position in this sequence we get the Catalan numbers (1), 1, 2, 5, 14, 42, …

The amazing Euler found a nice pattern in the Catalan numbers:

22!

=1 2×63 !

=2 2×6×104 !=5 2×6×10×145 !

=14 …

How many ways are there of arranging 7 pennies? Can you find a formula for the terms in this sequence?

Do you think this will follow the Catalan sequence? If so, can you explain why?

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5.7 Fibonacci’s sequence

We are going to investigate the Fibonacci sequence through a few investigations.

5.7.1 Steps

Suppose you walk up a flight of steps either one or two steps at a time.

5.7.2 Partitions

Here are all the ways of partitioning the first few counting numbers using only 1 and 2:

1 = 1 2 = 1 + 1 3 = 1 + 1 + 1 4 = 1 + 1 + 1 + 12 = 2 3 = 1 + 2 4 = 1 + 1 + 2

3 = 2 + 1 4 = 1 + 2 + 14 = 2 + 1 + 14 = 2 + 2

You will probably have realized these are just two versions of the same problem. You may also have realized that they give the Fibonacci sequence... but why?

Consider the partitions; if you want to find the (8) ways of partitioning 5 using 1s and 2s just + 2 to all the ways of partitioning 3, and + 1 to all the ways of partitioning 4 like this:

1 + 1 + 1 + 2 1 + 1 + 1 + 1 + 11 + 2 + 2 1 + 1 + 2 + 12 + 1 + 2 1 + 2 + 1 + 1

2 + 1 + 1 + 12 + 2 + 1

Equivalently, to get up a flight of 5 stairs we can take a step of 2 from the third step, or take a step of 1 from the fourth step.

How many ways are there of walking up a flight of 4 steps? Or 5 steps? Or n steps? Can you find a pattern?

Find the number of partitions for 5, 6 and so on. What do you notice?

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There are a lot more interesting results about partitions, not related to Fibonacci’s sequence, but I will mention them here as it’s as good a place as any:

First, let’s consider how many ways there are if order is important (called ordered partitions, or compositions). For the first few numbers we have:

2 3 41 + 1 2 + 1 3 + 1

1 + 2 1 + 31 + 1 + 1 2 + 2

2 + 1 + 11 + 2 + 11 + 1 + 21 + 1 + 1 + 1

It appears that the number of partitions is doubling each time.

In fact, there is a very elegant explanation. Consider the number 3 made up of three blobs . We can partition these blobs in 4 different ways as follows:

| | ||

is the same as choosing whether or not to put a line in the two ‘gaps’ between the three blobs, and there are 4 ways of doing this; 2 choices (line or no line) for the first gap and 2 choices for the second gap.

With 4 blobs, we have 2 x 2 x 2 choices of where to put the lines, and so on, confirming our belief that the number of ordered partitions follows the doubling pattern.

For unordered partitions, where order is not important, it is a much more complicated story. The first few numbers of partitions are 1, 2, 3, 5, 7, 11, … and there is no known formula for these numbers, although Euler did find that the partitions into different (distinct) numbers is the same as the partitions into odd numbers, using some beautiful and not too complicated mathematics (see the excellent book Euler: The Master of Us All by William Dunham).

How many ways are there of partitioning numbers generally if order is important?

What if order isn’t important?

Being sure not to fall into the trap of making assumptions, can you explain why this might be true?

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Let’s look a bit closer at these partitions; is there anything else we can say? Let’s use the number of unordered partitions of 6 as an example:

6 5 + 1 4 + 2 3 + 3 2 + 2 + 2 1 + 1 + 1 + 1 + 1 + 14 + 1 + 1 3 + 2 + 1 2 + 2 + 1 + 1

3 + 1 + 1 + 1 2 + 1 + 1 + 1 + 1

Let’s rearrange these in order of how many numbers are in each partition:

6 5 + 1 4 + 1 + 1 3 + 1 + 1 + 1 2 + 1 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 + 14 + 2 3 + 2 + 1 2 + 2 + 1 + 13 + 3 2 + 2 + 2

Let’s take one partition and look more closely at it, say 4 + 2. This could be drawn using squares like this, called a Ferrer’s (or Young’s) diagram.

By looking at it in rows and then columns, you can see that the partitions 4 + 2 and 2 + 2 + 1 + 1 are related (called conjugate pairs). Now other connections become clear, for example: The number of partitions with maximum number 4 (n) is the same as the number of partitions with 4 (n) numbers.

Another similar one is that the number of partitions into less than n parts is the same as the number of partitions into parts less than n. So for example, the number of partitions less than 3 parts is the same as the number of partitions with parts less than 3. In our example, there is a match between 6, 5 + 1, 4 + 2 and 3 + 3 with their respective conjugates 1 + 1 + 1 + 1 + 1 + 1, 2 + 1 + 1 + 1 + 1, 2 + 2 + 1 + 1 and 2 + 2 + 2.

Here’s a puzzle that involves partitions:

First of all, note that the sums around A, D and F make up 9 of the circles, with the centre circle making the 10th. So if the sum around each triangle is S, and the middle number is M, then 3S + M = 45 (why?) which gives us limited choices for S and M.

Can you see a connection between these two ways of arranging these partitions?

Can you find any other rules like this?

Can you place the numbers 0, 1, 2, … , 9 (no repeats) in the ten circles shown so that the sums around the triangles A to F are all equal?

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As 3S is a multiple of 3, we must choose M to be a multiple of 3, so M must be 0, 3, 6 or 9. This leaves S to be 15, 14, 13 or 12 respectively.

If we put M = 0, it will be a part of three triangles: B, C and E. But there are only two possible sums including 0, namely 9+6+0 and 8+7+0. So this won’t work; so if we consider M = 3, this gives S = 14.

We have:

9 + 5 + 0 8 + 6 + 0 7 + 6 + 1 6 + 5 + 39 + 4 + 1 8 + 5 + 1 7 + 5 + 29 + 3 + 2 8 + 4 + 2 7 + 4 + 2

Notice that there are 3 partitions containing 3, so these could form the triangles B, C and E.

Trial and error would probably be OK from here, but notice also that the three corner triangles A, D and F will not contain 3, and also will not share any numbers. There are only two sets of partitions that fit this bill: 7 + 5 + 2, 8 + 6 + 0, 9 + 4 + 1 and 7 + 6 + 1, 8 + 4 + 2, 9 + 5 + 0.

Only the second set of these works, and we have the solution shown:

What are the possible choices for S and M?

Try putting M = 0. Why can this not be a solution?

What are the partitions of 14 into 3 distinct digits from 0 to 9?

Are you going to go for trial and error from here, or is there some additional logic you can use to make the task easier?

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Can you find another completely different solution?

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5.7.3 Other representations

Here is an interesting diagram showing the number of different ways that rays of light can reflect between two plates of glass (each horizontal strip is a plate of glass seen side on).

The way I think of this is to consider the action of the ray of light at the middle line only. If it bounces off the middle line, count this as a 2, and if it passes through the middle line, count this as a 1. Bounces off the edges don’t matter.

So, using the 3 bounces as an example, and only looking at the middle line, we have:

2 + 2 2 bounces off the middle1 + 2 + 1 pass, bounce, pass1 + 1 + 1 + 1 pass, pass, pass, pass2 + 1 + 1 bounce, pass, pass1 + 1 + 2 pass, pass, bounce

Here we have the 5 partitions of 4 into 1s and 2s as we saw in the previous section!

Here is a ‘tree’ that is growing according to the Fibonacci sequence. Can you see how the branches are generated? Continue growing the tree.

Why does this sequence generate the Fibonacci numbers?

Can you think of any other interesting ways of representing the Fibonacci sequence?

What do you notice about the number of new (white) and old (blue) buds at each step? Experiment with different ‘growth rules’. For example, what sequence do you get if

you allow new branches to form every third day instead of every second day? Or perhaps two branches could grow each time… Or?

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Each ‘bud’ grows a branch every day apart from the first day it is alive. This is basically the growth pattern for Fibonacci’s rabbits.

Experimenting with different growth patterns leads to some other Fibonacci-type sequences:

If a bud doesn’t branch until the third day we get the sequence: 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, … which has the recursion relationship Fn=Fn−1+Fn−3

Alternatively, if we keep the growth period to 2 days but allow 2 branches to form each time, we have the sequence: 1, 1, 3, 5, 11, 21, 43, … and the recursion rule becomes Fn=Fn−1+2 Fn−2

Finally, here is a bee that wants to work out the number of ways of getting to each cell in this honeycomb:

It only travels to honeycombs of a higher number than the one it is currently on, so for example the different ways of getting to cell 3 are 0123, 013 and 023.

Can you generalize these rules? Do you have any other interesting ideas for growing trees?

Check this leads to the Fibonacci sequence. Can you explain why?

Can you think of any other interesting ways of representing the Fibonacci sequence?

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5.7.4 Dominoes

Here are the first few values of 1 x n strips:

Surprise, surprise: the number of ways of placing dominoes on the strips follow the Fibonacci sequence! To see why, we can analyze how to get the next number in the pattern in a similar way to steps and partitions. Look at the picture below:

We can get the number of ways for the 1x4 strip by placing a single (empty) square at the end of the three 1x3 strips, and a domino at the end of the two 1x2 strips.

5.7.5 Cumulative Fibonacci sequence

Continue the sequence for 1x 5 and 1x6 grids. What do you notice? Can you explain why?

How many ways are there for putting 1 x 2 dominoes on various 1 x n strips?

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Make a sequence by adding up the numbers in the Fibonacci sequence cumulatively. Here are the first few:

1 = 1, 1 + 1 = 2, 1 + 1 + 2 = 4, 1 + 1 + 2 + 3 = 7, …

[Hint: for finding rules: compare terms to the Fibonacci numbers, or squares of Fibonacci numbers, or products of consecutive Fibonacci numbers, or near-squares, or near-products, or half-products, or…]

The cumulative Fibonacci sequence goes 1, 2, 4, 7, 12, 20, 33, … A bit of careful inspection leads us to the identity Fn+2=F0+F1+…+Fn+1 where Fn is the nth Fibonacci number with F0=1 and F1=1.

But why? Using dominoes on strips again, consider different positions for the last domino on the strip. Here is an example for F4:

The cumulative sequence of the even Fibonacci numbers leads to the identity F2n+1=F0+F2+…+F2n.

5.7.6 Fibonacci identities

Find the next few… what do you notice? Can you create a rule that is always true? Can you explain why it is true?

Now just add up the even terms cumulatively, or perhaps just the odd terms. Can you come up with some more rules?

How about exploring sums of every third Fibonacci number? Or every fourth…?

Can you prove why this one is true?

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We saw a nice identity for the Fibonacci sequence in the previous section. Squaring Fibonacci numbers leads to some more interesting identities:

Have a look at the three numbers circled. Can you find a relationship (identity) between numbers in these relative positions in this table?

If you multiply the two top numbers together (3x8) you get one less than the bottom number (25). But for the next position along, the two top numbers (5 x 13) are one more than the bottom number (64).

This leads to Cassini’s identity: Fn2=Fn−1 . Fn+1+ (−1 )n

To prove Cassini’s identity we will go back to dominoes. Consider two 1 x n strips offset by one place like this:

We are going to explore faults in the strips – these are the positions not covered by dominoes as shown. A moments’ thought reveals that the fault pattern actually defines an arrangement of dominoes.

Now, define a tail of a strip as the bit after the last fault. So here the top strip has a tail of 3 squares and the bottom strip has tail of 4 squares.

Now we are going to swap the tails over like this:

Note that the top strip has become one longer and the bottom strip has become one shorter.

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Now for the proof:

There are Fn×Fn ways of arranging dominoes on the two 1 x n strips (top picture). This is one more than the number of fault patterns as we do not have a fault pattern for the case where dominoes cover all squares (when n is even) as shown here on the right:

There are Fn+1× Fn−1 ways of arranging dominoes in the second picture (after swapping tails) – and this is exactly equal to the number of fault patterns (as they are now odd there is no all-domino pattern possible).

Putting these two ways of counting the fault patterns together, we can say Fn×Fn=Fn+ 1×Fn−1+1 (if n is even). A few moments more reflection reveals that the extra arrangement with no faults occurs in the second picture if n is odd, and we have a proof of Cassini’s identity!

Try adding the Fibonacci squares cumulatively. Here are the first few:

1, 1 + 1 = 2, 1 + 1 + 4 = 6, 1 + 1 + 4 + 9 = 15, …

Here is a nice identity for the cumulative Fibonacci squares sequence:

1 = 11 + 1 = 2 = 1 x 21 + 1 + 4 = 6 = 2 x 31 + 1 + 4 + 9 = 15 = 3 x 51 + 1 + 4 + 9 + 25 = 40 = 5 x 8

This suggests the identity: F12+…+Fn

2=Fn .Fn+1. Here’s a proof:

Continue this pattern. What do you notice? Can you come up with another identity?

How about just adding or subtracting pairs of Fibonacci squares?

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You can see this pattern by looking at this picture of the Fibonacci spiral:

If you enjoyed all this, here are a couple more identities involving Fibonacci squares to try and prove:

Fn2+Fn+1

2 =F2n+1

Fn+12 −Fn−1

2 =F2n

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5.7.7 Fibonacci remainders

Let’s look at the patterns in the remainders of the Fibonacci sequence. The Fibonacci sequence starts 1, 1, and then every number after this is worked out by adding the previous two. So the next number is 2, and then 3, then 5, then 8, … giving:

1, 1, 2, 3, 5, 8, 13, …

If you tried dividing this by different numbers you would have found some interesting patterns. Here are the first few remainders of the Fibonacci sequence on dividing by 3 (in modular arithmetic, we would say mod 3):

Sequence 1 1 2 3 5 8 13 21 …Remainder 1 1 2 0 2 2 1 0 …

To spot patterns, it might help to draw the remainders like this:

Here’s a bit of maths that explains why – if you don’t follow the bit with Fnright away, don’t worry! Let’s start by systematically looking at the sequences of remainders for the first few divisors:

Div 2: 1, 1, 0, 1, 1, …

Div 3: 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, …

Div 4: 1, 1, 2, 3, 1, 0, 1, 1,

Div 5: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, 4, 3, 2, 0, 2, 2, 4, 1, 0, 1, 1, …

So there are loops of various lengths, with zeroes every n places. Is there any way of predicting the loop lengths or occurrences of zeroes? Do you think that the loop for dividing by 4 seems different to the others? Perhaps this has something to do with prime numbers (2, 3 and 5), or maybe something else? Let’s keep looking…

Div 6: 1, 1, 2, 3, 5, 2, 1, 3, 4, 1, 5, 0, 5, 5, 4, 3, 1, 4, 5, 3, 2, 5, 1, 0, 1, 1, …

Div 7: 1, 1, 2, 3, 5, 1, 6, 0, 6, 6, 5, 4, 2, 6, 1, 0, 1, 1, …

What patterns can you see? What can you say about where the zeros occur? Can you find any other patterns? Can you explain why?

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Div 8: 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, …

Some of the most behaved ones seem to be the ones that are divided by the Fibonacci numbers. If you think about it for a while, this could be expected – when we get to the Fibonacci number we are dividing by, it will of course turn to 0. So we can say:

Div Fn: F0 , F1 , … , Fn−1 , 0 , Fn−1 , Fn−1 , …

where Fn is the nth Fibonacci number.

But why does it loop back to zero and repeat every n terms? To find out, we are going to do something sneaky to write this last Fn−1in a different way.

By the definition of Fibonacci numbers, we have that Fn=Fn−1+Fn−2, which rearranges to give Fn−1=Fn−Fn−2. Now, on dividing by Fn this changes to Fn−1=0−Fn−2 so we can rewrite the last Fn−1 as −Fn−2 giving:

Div Fn: F0 , F1 , … , Fn−1 , 0 , Fn−1 , −Fn−2 , …

Now we can use the definition of Fibonacci numbers again to find the next number. We have Fn−1=Fn−2+Fn−3 so that we can rearrange to give Fn−1−Fn−2=Fn−3 . So now we have:

Div Fn: F0 , F1 , … , Fn−1 , 0 , Fn−1 , −Fn−2 , Fn−3 , …

Maybe now you can see what is happening… the Fibonacci sequence is going back ‘down’ again (with alternating signs) to zero. This suggests why there is a zero every n digits and a repeating pattern for Div Fn.

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5.7.8 Number bracelets

Consider the units digits only of the Fibonacci sequence. So in place of 1, 1, 2, 3, 5, 8, 13, 21, 34, … we have 1, 1, 2, 3, 5, 8, 3, 1, 4, … (This is the same as considering the Fibonacci sequence mod 10).

This activity is called number bracelets because you always get loops and so can arrange them like bracelets.

Here is the bracelet when starting with 1 and 3.

There are 100 possible (ordered) pairs of numbers for the initial investigation (why?). 12 of the possible pairs are part of this bracelet. What about the other 88?

It turns out there are only a few possible bracelets (noting that ‘starting with’ doesn’t really make sense!):

Starting with Length of loop

(0,0) 1

(1,1) 60

(0,2) 20

(1,3) 12

(2,6) 4

(0,5) 3

Continue this sequence. Does it carry on forever or will it loop back to the beginning? Note: we require two ones (not just one) in order for it to loop.

Why not investigate other Fibonacci-type rules (such as adding the 3 prior numbers instead of 2) or other mod arithmetics?

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5.7.9 Fibonacci factors

We call the position of a number in a sequence its index. For example, the index of 13 in the Fibonacci sequence (assuming we start with 1, 1, …) is 7.

You should have noticed that every third number has 2 as a factor; that is, all numbers with an index 3n have a factor of 2.

The table of Fibonacci factors when completed looks like this:

Factor 2 3 5 8 13 …Index 3n 4n 5n 6n 7n …

We explored this idea already in Fibonacci remainders; consider the Fibonacci numbers modulo 5: 1, 1, 2, 3, 0, 3, 3, 1, 4, 0, 4, … every fifth number is a zero.

The fact that there are no consecutive zeroes for any mod n leads us to conclude that consecutive Fibonacci numbers do not have common factors (are relatively prime).

Looking at the table above, it looks like prime Fibonacci numbers might have prime index (apart from 3).

Prime Fibonacci number

2 3 5 13 89 233 …

Index 3 4 5 7 11 13 …

What can you say about the index of numbers in the Fibonacci sequence that have 2 as a factor? Investigate before reading on.

Explore for other factors.

Complete this table for the Fibonacci factors:

Factor 2 3 5 8 13 …Index 3n ? ? ? ? …

Now, consider consecutive pairs of Fibonacci numbers. Do they have any factors in common? How does the preceding investigation help us explain why?

Furthermore, it would appear that prime Fibonacci numbers have prime index – for example, 13 has index 7. Is this always true? Nearly always true?

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This turns out to be true! If you’re up for a tough challenge, try proving it!

Here is a table of the prime factors of Fibonacci numbers:

Index Fn Prime factorisation1 1 12 1 13 2 24 3 35 5 56 8 2x2x27 13 138 21 3x79 34 2x17

10 55 5x1111 89 8912 144 2x2x2x2x3x313 233 233… … …

Each prime factorization of successive Fibonacci numbers contains a new prime apart from F 1 , F2 (boring) and F6 and F12 . This is known as Carmichael’s Theorem.

And finally… investigate the prime factors of Fibonacci numbers - any patterns?

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5.7.10 Fibonacci Nims

Here are two variations of Nim that involve the Fibonacci sequence.

Variation 1: Players take it in turns to take a Fibonacci number of counters from a large pile. The player who makes the last move (takes the last counters) wins the game.

In any game of Nim we aim to move to safe positions; what are the safe positions for Fibonacci Nim? Well, looking at each number of counters in turn, we can see that 4 is a safe position – if we can reduce the pile to 4 counters then we will definitely win - any legal move from the other player (1, 2 or 3 counters) results in a win for us.

We can reduce the pile to the safe position of 4 counters from 5, 6, 7, and 9 counters, and of course 8 is a straight win. What about 10 counters? Well, we can’t take 6… so this is a safe position too. Further analysis like this suggests that the safe positions are of the form 10n and 10n + 4… they are like stepping stones to get to 4. But is this always true, even for very large piles of counters?

Variation 2: Players take it in turns to take counters from one pile according to the following rule: they can take up to double the number of counters taken by the previous person. The only other restriction is that the first player can’t take all the counters on the first go! The winner is the one to take the last counter(s) from the pile.

This one is a bit more interesting!

Before we analyze this game, first note the perhaps surprising fact that all numbers can be written as the sum of 2 or more Fibonacci numbers, for example 12 = 8 + 3 + 1. Before going any further, convince yourself this is true! Note that this expression may or may not be unique. For example, 15 can be written as 13 + 2 or 8 + 5 + 2. For this activity we will assume that the expression containing the largest Fibonacci number is the one we want (13 + 2 here).

We are now in a position to invent a number system based on Fibonacci numbers called Fibonacci binary! Just like any place value number system (like the base 10 one we use) we are going to give each place a value. Let’s give each place a Fibonacci number value (in the top row of the table below).

Here are 12 and 15 in Fibonacci binary:

Who wins this game? How?

Should you go first or second? What is a good strategy? Why is this in this section on Fibonacci numbers?

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… 13 8 5 3 2 112 1 0 1 0 115 1 0 0 0 1 0

Now, let’s analyze the game. What are the safe positions for Double Nim?

Well, they are very closely related to the Fibonacci numbers – hence the name! In fact, it turns out that we can move from one safe position to another by removing the smallest number in the Fibonacci binary expression of the current number of counters.

Here’s an example game:

Starting with 12 counters (say), we will be player 1 and follow this strategy:

Notice how this strategy moves towards simpler Fibonacci sums. But why does this work? Here is a list of the first few numbers expressed in Fibonacci binary:

… 13 8 5 3 2 11 12 1 03 1 0 04 1 0 15 1 0 0 0

Start: 12 = 8 + 3 + 1 = 10101 Take one counter, the smallest number in the Fibonacci sum

11 = 8 + 3 Player 2 takes 2, say

9 = 8 + 1 = 10001 Take one counter

8 = 8 Player 2 takes 1, say

7 = 5 + 2 = 1010 Take two counters, the smallest number in the Fibonacci sum.

5 = 5 Player 2 must take 1 (so as not to lose on the next turn).

4 = 3 + 1 = 101 Take one counter.

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6 1 0 0 17 1 0 1 08 1 0 0 0 09 1 0 0 0 1

10 1 0 0 1 0

You may notice that all Fibonacci binary numbers do not have any adjacent ones (why?). Also note that Fn+2>2Fn. So, roughly speaking, removing the smallest 1 from a Fibonacci binary number means your opponent can’t remove the next smallest 1, and you then remove it on the next go. In this way you have control over removing the 1s from the Fibonacci binary numbers and move between subsequent Fibonacci numbers until you get to the winning position of taking 1 to get to 3.

Start: 12 = 8 + 3 + 1 = 10101 Take one counter, the smallest number in the Fibonacci sum

11 = 8 + 3 Player 2 takes 2, say

9 = 8 + 1 = 10001 Take one counter

8 = 8 Player 2 takes 1, say

7 = 5 + 2 = 1010 Take two counters, the smallest number in the Fibonacci sum.

5 = 5 Player 2 must take 1 (so as not to lose on the next turn).

4 = 3 + 1 = 101 Take one counter.

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5.7.11 Lucas numbers

Here is a sequence that is closely related to the Fibonacci sequence, first discovered by the French mathematician Edouard Lucas:

1, 3, 4, 7, 11, 18, …

It is generated in the same way as the Fibonacci sequence, but with different starting numbers.

[Hints: - Try adding pairs of Fibonacci numbers or pairs of Lucas numbers and

comparing them. - Try multiplying or dividing Lucas and Fibonacci numbers.]

A bit of exploration leads to this identity linking Fibonacci and Lucas numbers Fn+1+Fn−1=Ln.

Did you spot this one:Ln+1+Ln−1=5 Fn? Or this one: Fn . Ln=F2n ?

Explore identities for the Lucas numbers in the same way as we have for Fibonacci numbers.

Can you find any identities linking Lucas and Fibonacci numbers?

What other identities did you find?

Finally, what do you get if you divide LnFn

for large n? [Hint: try squaring the

amounts you get]

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6 Parity

Parity is an important idea that helps us solve many problems in Maths. Often we can unlock a difficult problem with a very simple argument involving oddness or evenness, or an ingenious way of counting something, which is why these activities are in this booklet.

In order to solve these problems, try and keep a track of how the parity of the situation changes as you go along. Sometimes you might notice that the parity of some part of the problem stays the same (called an invariant). Often this will give you insight into the solution to the problem.

You have probably used parity in the past without realizing it. Here is a classic proof that uses parity in a very powerful way.

I promise that this is it pretty much it for algebra for the rest of this section; these activities are designed to be intriguing and fun, without the need for much algebra!

Suppose you can write √2= nm , where n and m are relatively prime (so that n and m is

simplified as far as possible.

Squaring both sides gives 2= n2

m2 which we can rearrange to give 2m2=n2 [*]

So n2 must be even, and so n must be even (as even x even = even).

So we can write n = 2k for some integer k.

Now we can substitute this into [*] to give 2m2=(2 k )2=4 k2

Dividing both sides of this equation by 2 gives m2=2k2 which means that m is also even.

But this contradicts the original assumption that √2= nm where n and m are relatively

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6.1 Cogs and chains

This section is about cogs…

Look at the picture on the right. Which way will the smaller cog turn?

Consider a row of cogs like the one below…

Now consider cogs joined by chains like this:

Clearly the two cogs rotate in the same direction…

Here are some more cog/chain loops:

Any cog will rotate in the opposite direction to its neighbor. So

in a closed loop of cogs, the ‘odd-numbered’ cogs rotate one direction (say anti-clockwise) while the even numbered cogs rotate in the other direction. So if we put a 5th cog in a loop, we have a problem.

Why? Because the 5th cog and the 1st cog are neighbours in the loop so should turn in opposite directions… but they are also odd-numbered cogs so want to turn in the

Suppose you can write √2= nm , where n and m are relatively prime (so that n and m is

simplified as far as possible.

Squaring both sides gives 2= n2

m2 which we can rearrange to give 2m2=n2 [*]

So n2 must be even, and so n must be even (as even x even = even).

So we can write n = 2k for some integer k.

Now we can substitute this into [*] to give 2m2=(2 k )2=4 k2

Dividing both sides of this equation by 2 gives m2=2k2 which means that m is also even.

But this contradicts the original assumption that √2= nm where n and m are relatively

How is this related to the (closed) loops of cogs above?

Explore other systems of cogs and chains…

What if we cross the chain over in the middle something like this?

Which way will each cog turn in these pictures (if at all)?

Which way will each cog turn? Can you come up with a rule?

What will happen if we put another cog between the first and last one to make a closed loop?

Investigate for other cog loops…

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same direction, resulting in a jam! We can conclude that we need an even number of cogs in a loop for the system to rotate freely.

In fact, the ‘crossed chains’ problems are equivalent to the cogs as neighbouring cogs rotate in opposite directions, so we need an even number of crossed chains in a loop for it to work! We can combine systems of crossed and non-crossed chains…

Note that it is not the number of cogs that matters here, but rather the number of crosses…

Will the one on the right rotate freely?

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6.2 Invariants

Here are some puzzles that can be solved using the idea of an invariant – a quantity that doesn’t change through the course of an investigation. Invariance is a useful tool for solving parity problems.

6.2.1 Odd blobs

Here are the rules for this investigation:

Here’s an example of how it works; starting with some set of black and white blobs (here I have chosen 2 white, 3 black):

[Hint: keep a record of the number of black blobs at each step of your investigations… what do you notice? Why do you think this is? How does this help you come up with a rule?]

One way of thinking about this is to replace the black blobs with ones and the white blobs with zeroes. Then my starting position has a sum of 3. If you keep a track of the sum it will remain odd throughout the cancelling. The sum is an example of an invariant.

Remember Blob Pyramid? Complete the pyramid using the same rules as above…

Try cancelling different pairs for this example. Do you still get the same outcome?

Try some other combinations of black and white blobs. Can you find any patterns? Can you work out what the final outcome will be at the start?

What patterns do you see in the pyramid you have created?

What if you change the rules for combining blobs?

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6.2.2 Odd cups

Start with 3 cups in this position:

Now try the same thing for different numbers of cups arranged in the same up down up down pattern…

…..

[Hint: like in odd blobs, keep a record of the number of cups facing up at each step of your investigations… are there any invariants?]

Here is a table of results:

This is all a bit confusing… you might have thought it was going to be possible for all even ones after trying 3 and 4 cups…

However, a bit of further investigation of different starting positions reveals the true structure of the problem (the up down up down positions we have been looking at are in red):

If you follow the hint and keep a record of the number of up-facing cups at each stage you will see that each flip of 2 cups maintains the parity of up-facing cups…

Turn over 2 cups at a time… can you get them all the right way up?

Can you come up with any rules for which arrangements of cups can be turned face up?

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For example:

You can only have an even number of cups facing up, so we will never have 3 facing up. A similar analysis will show which ones will be possible.

Note that flipping 3 cups at a time reverses the parity, which suggests that it will be possible to flip any number of cups in any starting position…

6.2.3 Binary strings

This investigation is adapted from the excellent book The Inquisitive Problem Solver by Vaderlind, Guy and Larson.

Suppose we start with chains of binary strings and wish to change them to something else. There are three rules for changing strings:

01 ↔ 10010 ↔ 11111 ↔ 000

So for example we could change 010 to 111 using the following sequence of changes:

010 → 1000 → 111

I am not going to give the game away yet, just to hint that you might want to keep a track of invariants while carrying out your investigations.

You can also change 01 into 10; here is the first couple of steps you get you started:

Try different combinations of cups, or try turning different numbers of cups each time… can you come up with a general rule?

Which other three digit strings can you change 010 into? How about 110?

Investigate other 3-digit strings, then 4-digit strings, then…

Can you make any conclusions about which ones are possible and which are not?

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01 → 100 → 1110 → …

I don’t think you can change either 01 or 10 to 11 though (can you?).

OK, now here’s the spoiler! You may have noticed that any change maintains the parity of 1s in the string, so we can only change between strings with the same number of 1s.

It turns out it is possible to change 10 into any string with an odd number of 1s and 110 into any string with an even number of 1s.

Using this fact, you can just replace 01 in any string with 10! Does this idea make your investigations easier?

Can you change 01 to 1000? Or 10000? How about 11111?

Make up some problems of your own.

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6.3 Coins

6.3.1 Coin tricks

Here’s are some variations on a coin trick.

Ask a volunteer to put some coins on the table like this:

I am going to explain how these tricks work, but see if you can work it out for yourself first.

Before performing these tricks, quickly count the number of heads (here it’s 3).

Version 1: ask the volunteer to turn over 2 at a time silently while you turn your back. Then before you turn back round, ask them to cover one of the coins with their hand.

Version 2: Now ask them to cover 2 coins…

Version 3: How about if they turn over one at a time, calling out the word FLIP each time they do it…

Version 4: How about if they turn over 1 then 2 then 3 coins, then 1-2-3 again, the 1-2-3 again…

These tricks are based on the same ideas as the odd cups puzzle. If you practice this trick on your own and keep track of the number of heads after each flip you will see that if the number of heads starts odd (even), it will remain odd (even).

Here’s how these puzzles work:

Version 1: supposing there is an odd number of heads at the start, when you turn back round there will still be an odd number of heads facing up.

A quick count of the heads showing will tell you what is hidden under their hand (here it would be a head).

T

TH

H

H

T

TH

H

H

You should be able to tell whether the coin they have covered is heads of tails… but how?

How can you decide whether the 2 coins are the same or different?

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Version 2: If they cover two coins you can determine whether they are the same or different in the same way (although if they are the same you will not be able to tell whether they are heads or tails!).

So here, the covered ones could be two heads or two tails, but they are definitely the same!

Version 3: If they call out FLIP each time then you need to count the number of times they call FLIP. If it is even, then the parity of heads will be the same. If it is odd, the parity of heads will have changed and you can perform the trick in the same way.

I’ll leave you to puzzle out how version 4 works for yourself!

T

TH

H

H

Can you invent any other versions of this trick?

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6.3.2 Coins in a row

Now let’s try a different kind of coin trick: Put a row of any number of coins of different denominations on the table, for example:

Now, two players take it in turns to take a coin from either end of the row. The aim of the game is to get the most money.

The original game (as shown) should be won by player 2. The best move for player 1 is to take the 5p. Then player 2 should take the 10p. Player 1 will then take the £1, player 2 the 10p, player 1 the other £1, giving player 2 the £2 and finally player 1 takes the 1p. So player 2 wins by £2.20 to £2.06.

Now try changing one of the coins, for example changing the 1p on the right end to £1. It might seem like player 1 would now win this game… but again it turns out to be a player 2 win… try it!

If we choose another 7 random coins and play again, player 2 usually wins, although it depends on the denominations (consider the case where all the coins are the same value). Play it a few times and see! Generally, all we can say is that the winner of the game with an odd number of coins depends on the arrangement of coins.

However, if we add a coin so that there are an even number of coins, player 1 can at least draw in all cases! If they are all the same, it will be a draw. If at least one of the coins is different to the others, player 1 will definitely win… to see why consider a simple example such as 2 or 4 coins and generalize.

5p 10p £1 £1 £2 1p10p

What is the best move to make? Who will win this game?

Change one of the coins… does it change who wins this game? Try adding an extra coin on one of the ends…

Investigate for different numbers of coins and denominations. Can you make a general rule?

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6.4 Fair game?

Here are some games of strategy, which do no involve any luck. What are the best strategies for each game? Are the games fair?

6.4.1 Choco choice

Consider a game where players take it in turns to break a (large) bar of chocolate into squares as shown, making one break along the lines:

The winner is the last person to make a break… and they get to eat all the chocolate.

Let’s consider the 3x4 example. This game might seem quite complicated to analyse at first sight… But playing it a few times you start to realize it doesn’t matter what you do because player 1 always wins.

To see why, consider each break – it creates one extra piece. We must eventually end up with 3x4 = 12 pieces. So there must be 11 breaks, that is 11 moves... so player 1 must always win!

We can generalize this to say that if the number of squares is even (for any shape), player 1 will win and if not then player 2 will win… not much of a game at all!

6.4.2 Poisoned square

Now suppose one of the squares is poisoned (shown in red).

The loser of this game is the one left with the poisoned square at the end.

Make a move by choosing a (non-poisoned) square and breaking off all the squares above and to the right:

What is the best strategy to win this game with the 3 x 4 bar shown? How about any m x n bar of chocolate? Or any shaped bar?

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Now, player 2 might move by choosing the green square:

This game will always be won by player 1, assuming they play optimally (i.e. always do the right thing). Why? Because any move player 2 can make, this could have been played earlier by player 1. So suppose a certain move is the winning move, then player 1 always has the opportunity to play it first! Unfortunatley this does not tell us what this winning move is…!

6.4.3 Clobber

The normal starting position for the game of Clobber is:

Blue starts… by moving horizontally or vertically to take a red counter. If you can’t take an opposition counter you lose!

So an example start of a game might be:

Who should win this game and why?

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The first move made by blue in the game shown turns out to be a bad one… red can win by playing a ‘mirroring’ strategy. So if blue plays the move shown, red can mirror it (consider the board rotated 180 degrees)…

Try playing this strategy… you will see that red will win this game as it can always ‘copy’ any move blue makes.

Now consider this different starting clobber position:

The mirroring strategy will also work for the 2x4 board, for example:

Play the game a few times… what is a good strategy?

Would you have played the same moves as in this game above?

Who should win this game, player 1 or player 2? How?

Investigate other similar 2xn boards?

How is this similar to 2-pile Nim?

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…and red wins. A bit of investigation will reveal that red will win on a 2xn board set up like this for n even if you use this strategy. However, this mirroring strategy is not possible if n is odd (why not?) and it would appear that red has the advantage now.

Answers on a postcard please! Note that this game is similar to 2-pile Nim; if the 2 piles have the same number of counters then player 2 wins by playing a mirroring strategy.

6.4.4 Stomp!

This puzzle is loosely based on the game Whack-a-Mole. Consider a square ‘garden’ made from 4 x 4 squares, any of which may contain a mole. The aim is to eradicate the moles from the garden by squashing them with your foot. But if you step on a blank square then a mole will pop up when you lift your foot up.

Let’s start by considering what happens with a ‘foot’ shaped like a domino 1 by 2 squares like this:

So, let’s suppose you come out one morning and there is one mole in the garden in some square and you step on it. This is what happens:

Who has the advantage on an m x n board for different values of m and n?

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After a while you will notice that the single mole puzzle is not possible with the 1x2 foot. To see why, keep a record of the number of moles after any step – it is always odd, so we can never get to zero moles (as zero is even). This is the case with any foot with an even number of squares.

Now try different shaped feet like these:

You might think this would mean it would be possible with the 1x3 foot, but it isn’t! To see why, consider a colouring of the garden like this:

Now putting a 1x3 foot in any position on this board will cover exactly one of each colour, therefore changing the parity of the number of moles on each colour with every stomp.

If we start with (say) 1 mole on a white square (and 0 on blue or red squares) then if we stomp on the ‘white’ mole with a 1x3 foot we will now have white = 0 , blue = 1 , red = 1 … and the parity of each colour has changed.

So if we start with differing parity on the different coloured squares then the puzzle will be impossible. Note, however, that this does not mean that if the parity is the same on all the coloured squares then the puzzle is definitely possible!

In fact, the single mole puzzle is only possible with the L-shaped foot. To see how, consider the following sequence of moves:

So if we can cycle through different numbers of moles like this we conclude that all puzzles are solvable with an L-shaped foot.

Can you completely eradicate the mole from this garden? If not, can you explain why not?

Explore different positions and numbers of moles in this garden with this 1x2 foot. Which ones are possible and which are not?

Can you remove a single mole from the garden with any of these shaped feet?

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Using the T-shape, you can move moles one square horizontally (or vertically) using this sequence of moves:

Moving a mole on top of another one will ‘cancel it out’… so this same sequence of moves will solve the 2-mole puzzle given.

6.4.5 Seat swap

This is not really a game, more of a puzzle. Suppose you have a 25 people sat in a 5x5 square.

Using the T-shape, can you work out a sequence of moves that will ‘move’ a mole in any direction, as shown in the diagram below?

Make up puzzles of your own and try and solve them.

Can you make up any rules about which numbers of moles can be eradicated with different shaped feet?

Why not experiment with different garden shapes...

Can you use technique this to eradicate the moles from this garden?

Can all the people move from their seat to an adjacent seat, so that they all finish one seat away (horizontally or vertically, not diagonally) from where they started?

If not, can you give a convincing argument why?

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It is not possible to move the people so that they are all one seat away. To see why this is the case, we are going to use a colouring argument like that above.

Suppose the chairs are coloured like a chessboard. Then each person moves from a black to a white square or vice versa. But there are 13 black squares and 12 white squares, so it will not be possible for everyone to move.

The knight’s puzzle on this board is not possible for the same reasons as above – a knight’s move changes parity.

The 2-move puzzle is not possible on this board either. Consider the people sat at the black squares: there is a 3x3 ‘sub-board’ inside the larger board, so will not be possible.

This puzzle will only be possible on m x n boards if either m or n is even, so that the number of black and white squares are even (see equal rectangles). This is the same puzzle as placing 1x2 dominos on a chessboard.

Can they move so that they are 2 squares away from where they started? How about if they make moves like a knight?

For which m x n grids is the 1-move puzzle possible and which grids is it not?

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6.5 Colouring in

Here are some puzzles about map colourings. Map colouring problems are famous in mathematics; the general aim is to colour adjacent regions (regions sharing an edge, but not a corner) using the minimum possible different number of colours.

6.5.1 Two-colourable maps

Before we get to colouring maps, consider the Venn diagram shown.

You will probably notice that adjacent regions have opposite parity.

Now make a weird Venn diagram however you want like this one – lets call it a map - and count overlapping shapes again…

In the map here, the same is still true of adjacent regions. Try some more and see if the same is true for yours.

To see why, consider walking across a border. As you leave one region, you go to a region with either one less overlap or one more overlap, so the overlap count changes parity.

Now try colouring this map (so that no adjacent regions have the same colour).

You should have been able to colour any of these maps using only 2 colours, like the one shown.

How many circles overlap in each region?

What do you notice about any two adjacent sections (with their edges touching)?

What is the minimum number of colours needed? Can you prove why?

Can you explain why?

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To explain why, consider colouring odd sections black and even sections white…

Now pick two points in separate regions.

The shortest line between these two points passes through an even number of region ‘borders’.

Any line between these two points passes through an even number of borders…

Look at the example with 2 points shown; the paths go from an odd-numbered region to another odd region.

By the above argument every time we cross a border we move from an even to an odd region. Equivalently, we move from a region of one colour to a different colour.

So to travel between two regions of the same parity, we must cross an even number of borders, and return to a region of the same colour. Any line between these points will cross through an even number of regions for the same reason.

Any line between regions of different parity will cross through an odd number of regions. So every region will either be one colour or another.

6.5.2 Three-colourable maps

Make another map like this one, where every vertex (points where borders meet) is of degree 3 (ie has 3 edges coming out of it). Let’s call this a degree-3 map.

Investigate for other pairs of points.

Why does this also show that these maps are 2-colourable?

What is the minimum number of colours you need to colour this map so that no adjacent regions are the same colour?

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Here is another degree-3 map.

The first map needs a minimum of 4 colours. You can see here that the middle region will need a 4th colour…

This is because we need 3 colours for the outer loop. We can’t colour it in only 2 colours because there are an odd number of regions in the loop.

This suggests that if there is an even number of regions in the outer loop then we would only need 3 colours, which is indeed the case.

The outer loop only requires two colours, and the inside can be coloured with a third.

If we analyse this a bit further, we can think of an ‘odd loop’ as being equivalent to the central region having an odd number of edges (5 in the first map above). So having regions with an odd number of edges will lead to 4 colours being needed.

However, if all the regions in the degree-3 map have an even number of edges (and so are surrounded by even loops), we could conjecture that only 3 colours are needed.

Can you make a degree-3 map that can be coloured in fewer colours than this one? Can you make one that needs more colours?

What is the minimum number of colours needed for this one?

Can you come up with any rules about degree-3 maps?

Is this logic sound? Test it out!

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6.5.3 Four-colourable maps

Here’s a bit of history about the famous Four-Colour Theorem.

In 1852, Francis Guthrie, then a student at University College London, was colouring a map of England and found that he required only four colours to colour this map so that no two counties with a common border have the same colour.

By common border we mean that the common border is a line, not a point. Here are some examples of acceptable map colourings:

He conjectured that for any map, only four colours would be sufficient. Was this true? Finding a proof for this simple conjecture has tantalized mathematicians ever since; surely such a simple conjecture must have a simple, elegant proof?

Many great mathematicians of this era such as Augustus De Morgan and Arthur Cayley could not find a proof. However, 27 years later, Sir Alfred Kempe found a more convincing proof, which he published in the journal Nature. An outline of the proof is as follows:

Suppose there exists at least one map that requires five colours; let’s call the one with the fewest regions M.

Now, it can be shown (using Euler’s Theorem) that any map must have at least one region with 5 edges or less. As every map must contain at least member from this set of regions, it is called an unavoidable set.

Consider the case where M contains a region with 3 or fewer edges. We can reduce this region to a point, creating a map with fewer countries. By our assumption, this map must be four-colourable as it is smaller than M.

Now we reinstate the triangle (by expanding this point). As it has only three neighbours, we can colour it using a fourth colour – meaning that M is four-colourable. We say that the triangle is a reducible configuration and conclude that M cannot contain a triangle (otherwise M would be four-colourable).

Kempe then went on to use similar arguments (using an idea called Kempe chains) to show that the square and pentagon are also reducible.

So as M must contain (at least) one region from this unavoidable set, and each member of this set is reducible, then our assumption is false; there is no such map M, and so four colours are sufficient for all maps.

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This proof soon became part of undergraduate courses in mathematics and remained so for 11 years until Percy Heawood found a flaw in the proof – the pentagon is not reducible in the way shown by Kempe. However, Kempe’s idea was sound: if we could find any unavoidable set of reducible configurations, then the conjecture would be proven true.

In the early 20th century the search began for other unavoidable sets made from more complex reducible configurations of regions. However, the size of unavoidable sets appeared to be much larger than the number of reducible configurations and it was hard to see if a match would ever be found.

Heinrich Heesch, who began work on the conjecture in 1936, was perhaps the first mathematician to publicly state that it would be possible to find an unavoidable set of reducible configurations. He developed a method called discharging to eliminate irreducible configurations, and was the first mathematician to use a computer algorithm to check reducibility in 1965. He conjectured that there might be an unavoidable set of around 10,000 reducible configurations. Unfortunately, computer power at the time was not sufficient to check this number of configurations.

Two US mathematicians Kenneth Appel and Wolfgang Haken finally proved the conjecture to be true in 1976. They developed the methods of Kempe and Heesch, and invented other probabilistic methods to predict which configurations might be reducible. Their final unavoidable set still contained 1,936 configurations (which was later reduced to 1,482), which they checked for reducibility using a computer algorithm.

The checking procedure was only feasible due to their ingenuity and incredibly hard work, and to the increasing power of computers. To check all configurations at the time took around 1,200 computer-hours; they estimated that a single configuration would take around 10,000 man-hours to check by hand. The proof contains 100 pages of summary, and hundreds more pages of detail. As Kenneth Appel stated on declaring the proof:

“It was terribly tedious with no intellectual stimulation. There is no simple elegant answer, and we had to make an absolutely horrendous case analysis of every possibility. I hope it will encourage mathematicians to realize that there are some problems still to be solved, where there is no God-given answer, and which can only be solved by this kind of detailed work.”

Their proof was initially treated with skepticism as it relied on the use of computers and could not be checked by hand. The proof has since been independently verified it is now widely accepted that the proof is valid, although it raises questions about what constitutes a mathematical proof.

It has since been shown that there is an unavoidable set containing 663 reducible configurations, and there have been improvements to the algorithms developed by Appel and Haken. However, the search for an elegant proof continues.

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6.6 In and out

Here are three investigations about things that intertwine.

6.6.1 Bridges

Make any closed loop with a pencil (without taking your pen off the paper) then add in ‘over’ and ‘under’ bridges like this:

You will notice that the bridges alternate between over and under bridges.

Consider the simple loop on the right with the bridges numbered. Now tracing a route along the path, we pass the bridges in the sequence 1 2 3 1 2 3 1 …

Now look a bit closer at this pattern. If a bridge is an over bridge the first time we pass it, it will be an under bridge the next time… This is a consequence of the fact that we pass an even number of bridges before we come back to between occurrences of any bridge in this sequence.

To see why, note that it is always the case that there is an even number of bridges between repeats.

Let’s number the bridges in the main example: Tracing a route round the path we get

1 2 3 4 5 3 6 1 7 7 2 6 4 5 …

Check this for your bridge pattern – is it the same? Can you work out why this happens?

Is it always possible to make an over-under pattern like this for any closed loop? Why?

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You can see that there is always an even number of bridges between each pair.

To see why, consider the path as made of two separate loops made between occurrences of bridge 1.

The first (blue) loop is 1 2 3 4 5 3 6 1 and the second (red) loop is 1 7 7 2 6 4 5 1.

Consider (say) the red loop. It crosses through itself at bridge 7 (so passing bridge 7 twice) and then crosses the blue loop four times.

So it crosses 6 bridges before it returns to 1. This is an even number. Will this always be even?

Well, there are two types of crossings. When a loop crosses itself, this will always contribute an even number (2) to the total.

The other type of crossing is when we cross the other half of the loop. To see why this is also even, consider the diagram on the right.

As we are passing in and out of the blue section, there will be an even number of crossings here too. So there will be an even number of crossings in total, and we will always get the over-under pattern.

Here’s a related game… Ask someone to make a closed loop like these ones and label the crossings (bridges) in the same way. Then ask them to trace a path once around the loop and read out the bridge numbers in the order they pass them… but… ask them to try and trick you and tell a lie somewhere by swapping the order of just two of the bridges.

Without looking at their picture, how can you work out where they lied?

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6.6.2 Inside/outside

Draw any closed loop, any shape, and put a few vertices on it like this:

You might have found that it’s not possible with this one as you can see.

You will find that it is only possible to draw the second (red) closed loop if you have an even number of vertices (and hence segments).

If you consider the original closed loop as having an inside and outside then if you start from the outside, you must cross the line an even number of times to return to the outside (and so create a closed loop).

This is related to the Jordan Curve Theorem, which basically states that a closed loop in the plane has an inside and an outside.

Now draw any line and put vertices on it in a similar way. This time, label each vertex with a zero or one however you choose. :

Now calculate the value of each segment as the difference between its end points, so in this example we will have:

Is it possible to draw another closed loop that passes through each segment (between vertices) once only?

Why not?

Under what conditions is it possible?

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Adding the green values in this example we get 3.

You might have found that the sum of the green values will be odd if the end vertices have different values (such as the line shown) and even otherwise.

If you think of this as being 0 = outside and 1 = inside then getting from a 0 at one end to a 1 at the other end requires an odd number of changes from 0 to 1, and so an odd sum.

If you ‘open out’ the loop into a line, then it will have the same value at each end, so the green sum will always be even.

Do this for lots of different lines and vertices… what do you notice? Can you explain why? [Hint: consider the values of the end vertices]

Now do this numbering for your closed loops like this one below… what do you notice?

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6.6.3 Odd triangle

We can extend this idea. Draw any triangle, label each corner A, B and C. Then make as many points on the edges and inside the triangle. Triangulate between these points. Label the points A, B or C however you want (in red).

The green numbers inside the triangles have been worked out according to the following rules:

You will have noticed that the total is always odd. To see why, notice that lines connecting different letters have a value of 1, and lines connecting the same letters have value 0.

Now, lines inside the triangle contribute 2 to the total, because they form part of 2 triangles, so will contribute an even amount in total.

From the previous investigation, the lines on the edge of the main triangle will contribute an odd number to the total (why?) and they are only counted once. So the overall total will be even + odd = odd.

Add up all the green numbers in my triangle. Then make some of your own.

What do you notice? Can you explain why?

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6.6.4 Weaving

Consider weaving two threads so they intertwine with each other like this, and then return to their starting positions:

Notice that it takes an even number of crossings (swaps) for the threads to return to their starting positions. Convince yourself that you will always need an even number of crossings to return to the starting positions for 2 threads…

Here is a set of three threads returning to their starting positions:

We are going work out what is happening in the 3-thread example.

Will always be true for 3 threads, or 4 threads, or more?

How many crossings are there here? Is it odd or even? Will this always be the case for 3 threads? Can you explain why?

Can you find a rule for any number of threads?

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The first change switches the position of all three threads. You could think of it as doing 2 swaps in one go (R with B then R with G), and it will always give 2 crossings.

A single swap can either result in one crossing:

…or sometimes three crossings:

So if the number of swaps is even/odd, then so are the crossings. This table records what happened in the example given:

Order Swaps Number of swaps

Number of crossings

Letters in the original cycle

order?

Start: (G,B,R) - 0 0 Y

(R,G,B) (R,B) (R,G) 2 2 Y

(B,G,R) (R,B) 3 5 N

(B,R,G) (G,R) 4 6 Y

End: (G,B,R) (B,G) (R,G) 6 8 Y

The final column in the table records whether the letters are in the original ‘cyclic order’ (G,B,R) as shown on the right:

A single swap changes the cyclic order (basically it reverses the direction of the arrows). Every time an even number of swaps/crossings has passed, we are back to the original cyclic order. So to return to the original starting position we must eventually go through an even number of swaps or crossings.

6.7 Graphs

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These puzzles involve a branch of mathematics known as graph theory. They are not graphs like you might be used to, but have edges and vertices (nodes) and are used to represent things.

6.7.1 Ghost house

You are allowed to start and finish in different rooms.

We can represent the house as a network with rooms as vertices and doors as edges like this:

It is a well-known fact of graph theory that we can traverse a graph (travel along each edge exactly once) and return back to the start if there is an even number of edges coming out of each vertex. This is because if we enter a vertex by one edge, we need another edge out of the vertex to carry on our path. For our puzzle, this is equivalent to requiring an even number of doors in a room.

If there are an odd number of doors in a room then eventually we will enter a room by one door and there will be no door to come out of…. Does this happen in our

Is it possible to walk through this house using each door exactly once?

A B

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house? Well, yes – at rooms A and B. But if we start at (say) room A then as long as we finish at room B it will be OK.

We could represent the ghost-house using a graph too, this time with wall segments as edges. If you do this you will notice that are 4 rooms with an odd number of wall segments. This time we can’t do the same ‘trick’ of starting and finishing at the ‘odd’ rooms because there are too many of them, so we can’t perform a ghost-tour of this house.

6.7.2 Domino loop

We can represent a set of 0-6 dominoes by this diagram, known as the complete graph K7.

To see how, let the vertices represent one ‘half’ of the domino, so for example the edge between 0 and 1 represents the 0-1 domino.

Then finding a loop containing all the dominoes is equivalent to finding a path that traverses all

If you were a ghost (and so pass through walls instead of doors) could you pass through every wall segment (between two vertices) exactly once?

It is possible to place a set of ‘normal’ 0-6 dominoes in a complete loop?

Now take out all the dominoes with a blank (zero) on. Is it possible now?

Now take out all the dominoes with a 0 and/or 1 on… is it possible now? Any conjectures? Can you explain why?

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edges in this network. So this will be possible for any set of dominoes with even vertices (the 0-6 set but not the 1-6 set).

6.7.3 Vertex game

This game is for 2 players.

Draw any graph, like the one shown here.

Each player takes it in turns to delete an even vertex (one with an even number of edges attached) and also any of the edges that are attached to it.

There are three even vertices on this graph. It is not a very interesting game because player 1 can win straight away with this move:

[Hint: you might want to consider how many (and what type of) vertices there are after each move]

Let’s look at a more interesting game:

Play the game with some more interesting graphs. Can you make any conjectures? Can you prove them?

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Player 1 wins this game. It seems like player 1 wins all the time… is this true?

In fact, it depends on how many vertices we start with. But first, some graph theory. If we count the number of odd and even-degree vertices in this game we have:

Vertices 7 6 5 4 3 2

Evens 3 4 3 2 1 2

Odds 4 2 2 2 2 0

Notice how the number of odd-degree vertices is always even. This is a famous result of graph theory known as the ‘handshaking lemma’. It is a consequence of the fact that the sum of all vertex degrees is always even (as each edge has 2 vertices at the end). So if we have odd-degree vertices there must be an even number of them to contribute an even amount to this total.

So if we have a graph with an odd number of vertices then as the number of odd-degree vertices is always even, the number of even-degree vertices must be odd.

Vertices 7 6 5 4 3 2

Evens 3 4 3 2 1 2

Odds 4 2 2 2 2 0

Now, suppose we start with an odd number of vertices. If we follow the number of even-degree vertices throughout the game, there must always be an odd number on player ones turn. So player 1 always has a move, and will win in this case.

If there is an even number of vertices at the start then the same is true for player 2 and they will always win!

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6.8 Number puzzles

Here are a couple of number investigations.

6.8.1 Number circles

In the sequence below, each outer circle contains the differences of the numbers in the inner circle:

It would appear that we always end up with four zeroes:

Investigate what happens if you continue this pattern.

Explore different starting numbers. Can you find any rules

Agood way to start investigating might be to just use zeroes and ones.

Or perhaps investigate odds and evens?

What happens if you start with 3, or 5, central numbers?

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Why? Well, we could consider all the different combinations of zeroes and ones. In fact, they are all encapsulated in this example:

Each circle contains one of the possible combinations of zeroes and ones, which means that any combination of zeroes and ones will end with 4 zeroes.

(Note that I am considering three zeroes and a one to be equivalent to three ones and a zero)

Does this constitute a proof that all numbers will end with 4 zeroes? Not yet; I think we need to analyze things a bit further.

How about if we analyze odds and evens? The pattern is the same as that above (consider changing all the zeroes to evens and all the ones to odds), which does suggest that all circles eventually end up as 4 evens (and so will stay as 4 evens) in at most 4 goes; let’s call these 2a, 2b, 2c and 2d. Now consider the halves of these evens (a, b, c, d); whatever they are, they too will end up 4 evens in at most 4 goes.

Now as the halves (a, b, c, d) also end up as evens, then our original numbers (2a, 2b, 2c, 2d) will end up as multiples of 4! If we carry on this logic, we will get numbers that are divisible by larger and larger powers of 2. But this can’t carry on forever, so we must conclude that eventually all the numbers go to zero.

If you can, send them to me!

Now, if we consider three or five numbers, it appears that there are no combinations that go to all zeroes (unless we start with all the same number of course) as they get stuck in infinite loops.

Again, answers on a postcard please!

Are you convinced by this proof?

Can you think of a better way of proving this is true?

Can you prove why this is true?

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6.8.2 Oddly even

This is a daunting task! I wouldn’t like to try trial and error for this one, so how could we go about it? Let’s simplify it!

We can’t simplify it to only 3 numbers as it won’t be possible, which makes me think 25 is an important factor in it being possible. So let’s some other numbers instead.

With 4 numbers we have 1, 0, 1, 1 as one possible solution. With 5 we have 0, 1, 1, 0, 1. There are others ways too, such as 1, 0, 1, 1, 0. It does not seem possible with 6 numbers. Why do you think this is?

Looking at the above solutions to the 4 and 5-number problem, they have a similar structure, something like E, O, O, E, O, O, … which suggests that any problem that does not require a multiple of 3 numbers will be possible. Using this structure for the 25 number problem reveals that it is indeed possible.

Can you arrange any 25 (whole) numbers, not necessarily different, so that the sum of any 3 successive numbers is even but the sum of all 25 is odd?

Can you do it with only 4, 5 or 6 numbers? In how many ways? What do you notice about the sequence of odds and evens?

For which numbers will this problem be possible? Will it be possible for the original question of 25 numbers?

Can you solve the 25-number problem if we require all sets of 5 successive numbers to have an even sum?

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6.8.3 Magic squares

Good old magic squares. Is there anything interesting left to do with them?

Explore the parity of the numbers in these magic squares:

Here is the parity pattern for the first 3x3 square:

I have coloured the even numbers in red.

The row/column total of a 3x3 magic square is 15, which is odd, so each row and column must contain an even amount of even numbers (0 or 2).

This suggests that there are no other possible arrangements for a 3x3 magic square.

For the 4x4 magic square, the parity pattern is on the right:

This time the row/column total is 34, which is even, so there needs to be an even amount of evens in each line.

Here is a 3x3 magic square. What patterns can you see in the numbers? What can you say about parity in this magic square?

Can you make a different 3x3 magic square to this one? Is it really different?

Can you make 4x4 and 5x5 magic squares with different ‘parity patterns’? What types of symmetry do they have?

Can you find another possible parity pattern for a 4x4 square?

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4 3 8

9 5 1

2 7 6

16 3 2 13

5 10 11 8

9 6 7 12

4 15 14 1

23 6 19 2 15

4 12 25 8 16

10 18 1 14 22

11 24 7 20 3

17 5 13 21 94 3 8

9 5 1

2 7 6

Here is the parity pattern for the 5x5 square:

Notice that in fact all magic squares have an even amount of even numbers in each row/column/diagonal.

What else do you notice?

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6.8.4 Golomb’s rulers

Golomb’s rulers are named after mathematician Solomon Golomb (who also wrote the book on Polyominoes). This investigation is based on the information given in Martin Gardner’s book Wheels, Life and Other Mathematical Amusements.

Before we find out what they are, here’s a puzzle:

This is not too tricky; here’s the solution:

After a bit of trial and error, you might have found this is impossible. The ‘best’ we can do is put ten numbers from 1 to 11 in here, so there will be one number ‘missing’. By best, I mean the lowest possible numbers we can use.

Here is the solution for numbers 1 to 11 in the 10-triangle:

Can you put the numbers 1 to 6 in the shaded squares given so that:

- Each number in row 2 is the sum of the numbers above it

- The number in row 3 is the sum of all the numbersin row 1.

Can you put the numbers 1 to 10 in this triangle, where the numbers in third row are the sum of the three numbers in the first row as shown, and the number in the fourth row is the sum of all the numbers in the first row.

Can you work out how to do this? If not, what is the best we can do for a triangle of 15?

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I am going to leave the 15-triangle open for you to try.

Golomb thought of this puzzle as being the way in which you can make a ruler with a certain number of marks so that it can measure the most distances possible. To see how this works, consider the ruler with 3 marks, at 1, 3 and 2 as shown:

We could use this ruler to measure distances of 1 to 6. Can you see how this is effectively the same as the triangle puzzle? The 10-triangle problem leads to this ruler with 4 marks:

Golomb discovered this problem when investigating the ‘distances’ between points on complete graphs.

Here is the 3-mark ruler represented in yet another different way, on the graph K4. The edges are labelled with the difference (distances) between the two end vertices.

You can see how the all the distances 1 to 6 are present between the numbered vertices.

Here is a famous graph called the Petersen graph:

Check this ruler can measure all distances from 1 to 11. Compare this with the 10-triangle problem above; which distance can it not measure?

What is the ‘best’ ruler with 5 marks?

Can you draw the corresponding graph for K5?

Golomb found that it was possible to create a graph (ruler) of distances 1 to 9 by removing one edges of K5. Can you draw this?

Can you label the graph so that all the distances 1 to 15 are represented?

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7 The Maths of voting

This section is based on a series of lessons I taught on voting systems. I devised the lessons with two objectives in mind: to improve students understanding of percentages and working with data, and to show how maths can be meaningfully applied to real life situations.

Some of the ideas here come from the excellent book The Mathematics of Voting and Elections by Jonathan Hodge and Richard Klima, part of the amazing AMS Mathematical World series.

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7.1 What makes a good voting system?

There are many features of a good voting system, many of which we will discuss as we go through this section. However, to start with, we are going to analyze some basic mathematical requirements.

To see what these requirements are, consider these voting systems:

Dictatorship: A dictatorship can be thought of as an election in which only 1 person's vote counts - whoever that person votes for wins. Clearly all voters are not equal - a property we probably do not want our voting system to have!

A system in which all voters are equal is called anonymous.

Imposed Rule: With imposed rule, it doesn't matter how anyone votes, so all candidates are not equal (as opposed to voters). We could think of this as saying it doesn't matter how voters swap their votes, the winner will not change - again, this is a not property we want our voting system to have.

A system in which the order of two candidates changes if all voters swap their votes for these candidates is called neutral.

Minority Rule: Suppose the winner of an election was the one that got the least votes (which would be a bit weird). Then as a candidate gets more votes they would eventually lose the election. Clearly this is not desirable.

A system in which increased votes = an increased chance of winning is called monotone.

These three properties are a good start for any voting system. May's Theorem (named after US mathematician Kenneth May) says that the only two-candidate system that has all these three properties is Majority Rule: the winner is the one with most votes (greater than 50%).

So far, so simple! But what if there are more than two candidates? As we introduce more candidates it is unlikely that any one candidate will gain a majority. For example, the last time a UK party had a majority vote was in 1935!

We are going to look at some multi-candidate systems that only have one winner, and then we are going to look at some systems that have many winners in smaller regions, leading to one overall winner, such as the UK and US elections.

What do you think is important in a good voting system?

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7.2 Some voting systems

7.2.1 Plurality

Plurality is the system in which the winning candidate receives the most (first place) votes, regardless of whether they gain a majority. You might think this is the perfect way of choosing a winner.

However, simple plurality is not used in the UK or US national elections. For example, in the 2000 US election, Al Gore received more votes across the whole country than George W Bush, but Bush won the election.

The US system uses a system called Electoral College, which we will explore in more depth later. In simple terms, the winning candidate is chosen in each state using plurality, which then counts as a certain number of votes in the House of Representatives and Senate, the number of votes depending on the size of the state.

The UK uses a mixed system called First Past The Post (FPTP): each winning MP is chosen using plurality, which then counts as 1 seat towards the party total. We will also explore the UK system in more depth.

This example will demonstrate some of the problems with using simple plurality to elect a candidate. Consider the following table in which voters chose their first second and third preferences from three candidates Norm, Skip and Jesse. The numbers at the top are the number of voters who chose that particular order of preference, so for example 35 people chose Norm first, Skip second and Jesse third.

Under plurality, Jesse wins this election with more first place votes.

This is a representation of the situation that happened in the US Minnesota Governor elections in 1998, when Jesse ‘The Body’ Ventura was elected into government.

What do you think might be the problems with plurality? How about the mixed version of plurality used by the US and UK?

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You have probably spotted that although Jesse got more first place votes (37) than the others, he was also the last place vote of the rest (63).

Perhaps you think it would be better to award points for first, second and third? Or perhaps there should be some kind of knockout competition? These are all ideas we are going to explore, but for the time being let’s just say it would appear there are problems with plurality.

Do you think is the best and fairest representation of the public preferences? If not, why not?

Can you think of a better method for electing a winner?

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7.2.2 Borda Count

The Borda Count is a system invented by the French mathematician, physicist and naval officer Jean Charles de Borda in the 18th century. He was most famous for his work on fluids, disproving some of Isaac Newton’s theories, as well as inventing navigational instruments. It is argued by some that his is the most logical choice for a simple voting system.

It works by allocating points to candidates according to the voter’s order of preference, so taking account of other places apart from first. Consider this example of an election between Fred, George, Harriet and India:

According to this system, 0 points are awarded for last place, increasing up to 3 points awarded for a first place. So Fred gains 41 points ( = 12 x 3 + 7 x 0 + 5 x 1 + 3 x 0 ).

We can calculate that George would win with 48 points, even though he would lose under plurality; he only got 7 first place votes compared with Fred’s 12.

You could argue that it is, as it takes more information into account than plurality, and we have seen how flawed plurality can be. In fact, the Borda Count satisfies all the conditions we have stated so far. Further to this, it does not allow for irrational conclusions; the candidates are ranked in order of top to bottom based on the points. We will see an irrational conclusion might look like in the next section.

An undesirable property of the Borda Count is that a candidate with a majority can actually lose under the Borda Count.

Who wins under this system? Do you agree with the winner under this system? Who would win under plurality?

Can you create an example where this happens?

Do you think this is fair?

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7.2.3 Pairwise voting

This system is a knockout system. To see how it works, consider the election between Fred, George, Harriet and India again:

Let's set up a knockout competition, where George 'plays' Harriet, the winner plays Fred and the winner of that plays India. This ordering is called an agenda and is written GHFI.

Under this agenda, G beats H by 19 votes to 8, then F beats G by 17 votes to 10, then I beats F by 15 votes to 12. So India is the winner. But this doesn't seem fair – she would lose under plurality! And we know that George won under the Borda Count.

Fred would win under the agenda IGHF! It is possible that all the candidates could with under certain agendas. Surely we can't have a system whose outcomes are so dependent on the ordering.

Under neutrality, by swapping Harriet’s votes for India’s, we would expect Harriet to win in place of India, but this doesn't happen; this time George wins. So this method fails the criterion of neutrality.

Who wins under this system? Do you think it is fair? What are the problems with it? What criterion does it not satisfy? Try different agendas; is the outcome still the same?

Who would win if the agenda was changed to IGHF? Can you create agendas so that each candidate can win?

Now switch all the votes for Harriet and India and use the agenda GHFI again. What happens? Why does this highlight another problem with this system?

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Who would win the Minnesota governor elections under this system? Does it seem a fair method for this election?

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In the 18th century another French mathematician Marquis de Condorcet came up with another criterion that a good voting system should have, called the Condorcet Winner Criterion (CWC). In a system that satisfies CWC, the Condorcet Winner must beat all other candidates when compared head-to-head, so pairwise voting would definitely meet this criterion.

Well, as it is tied in with pairwise voting, which seems flawed, we might gather it is not a sensible criterion. Consider the example with Fred et al.; George and Harriet win two of the head-to-head comparisons, whilst Fred and India wins one; there is no Condorcet Winner here.

We can see this more clearly with this simple example:

There isn’t one! A beats B, B beats C and C beats A, and we have no winner.

This example shows that pairwise voting, and the Condercet Winner Criterion, allows for something called non-transitivity. Transitivity is an important mathematical concept. For example, if we say a = b and b = c, then it seems logical to say a = c; we say that equality is a transitive relation.

Here is an example to suggest why non-transitivity is irrational. Suppose you were asked to state your preferences for crisp flavours; would it seem rational to say you prefer salt and vinegar to ready salted, and ready salted to cheese and onion, but that you preferred cheese and onion to salt and vinegar?

The fact that pairwise voting allows for non-transitive conclusions is the biggest argument against its use.

Do you think this is a reasonable criterion or not? Explain why.

Who is the Condorcet Winner for this election?

Can you think of any other transitive relations?

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I have a set of dice that display non-transitivity. You can make a set from these nets:

Suppose we play a game where you choose a dice, then I choose a dice, roll them ten times and the most highest scores wins.

A logical thing to do would be to add up the dots on each one and choose the one with the most; unfortunately they all have 21 dots. At this point you might go for the one with the highest numbers, but it’s quite hard to say… Green looks like a good bet, so let’s suppose you went for green.

Ah, but look at blue; all the 5s beats all of your numbers, and the 2s beat the 1. Maybe I have a better chance of winning…

Here is a table showing all the possible outcomes. My odds of winning are 21:15, or more simply 7:5. Alternatively you can say the probability of me winning is 21 out of 36, or 21/36 = 7/12, and for you it is only 5/12.

So you decide that you are going to choose blue instead.

Another moments thought and you realize that I am going to choose red, which will again give me a 7:5 chance of winning (why?). So of course, you should choose red, but this is the worst possible thing you could do, because then I will choose green!

Which of these dice would you choose to play with? Why?

Which one should I choose? Why?

Can you work out my chances of winning if you choose green and I choose blue?

What dice should I choose now?

What are my chances of winning now?

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Explore these other sets of non-transitive dice:

Can you make a set of your own?

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7.2.4 Alternative Vote

This method is called Alternative Vote (AV) in the UK and Instant Run-off in the US. AV is used in the London Mayoral elections.

This is how it works: Voters submit their preference orders (first to last place). The candidate with the least 1st place votes is eliminated, and the ordering is adjusted accordingly. Then the candidate with the next least 1st place votes in eliminated and so on until we have a winner.

Consider this example:

Under AV, D would be eliminated first, leading to this adjusted table:

B would be eliminated next, leaving A with 11 first place votes and victory. Note how B would have won under Borda Count.

This system fails the criterion of monotonicity. If the 2 voters in the last column change swap their preferences for A and C, this actually leaves A in a worse position and B now wins!

Also, have a look at the following example:

Can you now work out who wins this election?

Try swapping the preference between A and C for the 2 voters in the last column; what happens now? Based on this, on which of our criteria does this system fail?

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A would lose under AV, even though she is the Condorcet Winner, so AV also fails CWC.

We will find out more about AV later when we look at the UK and US election systems.

Who wins under this election under AV? Who is the Condorcet winner?

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7.2.5 Approval voting

Approval voting is a relatively recent method used to elect the Secretary General of the UN, among other things. In this system, voters can choose one or more candidates they approve of using a tick. They leave candidates they do not approve of blank.

Here is a simple example in which David would win:

It would make sure that unpopular candidates did not get voted in. It also satisfies all the criteria we have stated so far apart from CWC

Approval voting has been criticized by Fairvote, a US organisation that advocates voting reform. They have said it is susceptible to certain practical flaws such as tactical voting, as well as the theoretical flaw that it can allow candidates to win who may not be the majority choice.

What do you think are the pros and cons of this method? What criteria does this method satisfy? Which ones does it fail?

Do you think this method could be successfully used in UK elections?

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7.3 Arrow’s Theorem

This section looks at a famous result in the analysis of voting systems called Arros’s Theorem

7.3.1 Independence of Irrelevant Alternatives (IIA)

As well as the criterion for voting systems we have already looked at, there are many other features we could consider. One of these is something called Independence of Irrelevant Alternatives (IIA). This means that if you remove a third (or lower) placed candidate, the result should stay the same.

Consider the following election results:

Under (say) the Borda Count, we can see that B wins and A comes second. But if we remove the third place candidate C, then A beats B. So the Borda system fails the IIA criterion.

In fact, it turns out that the only one we have studied so far that meets the IIA criterion is Approval Voting.

Often third candidates can have a huge effect on the outcomes of an election. As we will see later (Electoral College) in the 2000 US elections, Bush won Florida State by just 537 votes. The third place independent candidate Ralph Nader got nearly 100,000 votes, along with 40,000 votes for other candidates; if only 538 of these voters had voted for Gore, he would have been president.

How about the other systems we have been looking at; do they meet the IIA criterion?

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At this point, you might find it interesting to investigate other voting systems, such as:

• Voting for who will hold the Olympic games• The French presidential elections• The Oscars• Game shows like Weakest Link, Survivor or Pop Idol or...?

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7.3.2 Arrow’s Theorem

It appears so far that it is hard to create a system that meets all our requirements, namely anonymity, neutrality, monotonicity, CWC and IIA. Is there a system that meets all these requirements or not?

In 1972, an American economist called Kenneth Arrow won the Nobel Prize for his work on voting systems.

He was playing around with the different voting systems we have played with here, and was coming to the same conclusions: no voting system is perfect. He decided to try and prove that there was no voting system that would satisfy certain key criteria. His first 4 criteria (in our words) for a good voting system are that it should be:

• Anonymous (every voter is equal)• Neutral (if voters swapped votes, the results would also swap accordingly)• Monotone (increase in votes should not result in losing from a winning position)• Independence of Irrelevant Alternatives

We have seen that these are reasonable criteria for a good system. His fifth criteria was called Universality, which more or less says that voters should be able to vote for who they wish in whatever order, without restriction. He did not include CWC in his criteria (we have seen this is a rather flawed criterion).

However, unfortunately, his ‘Impossibility Theorem’ states that it is impossible for any voting system (with more than two candidates) to satisfy all five of Arrow’s criteria!

This theorem has interesting consequences. For example, we have already seen that the Borda Count fails IIA through an example.

We can state that the Borda Count does not satisfy IIA because it satisfies the other 4 criteria (it clearly satisfies universality – why?), and so by Arrow’s Theorem it must fail IIA.

We can say that plurality also fails IIA for the same reasons. We can’t conclude anything about whether pairwise voting and AV satisfy IIA from Arrow’s Theorem, as they fail other criteria.

How could you use Arrow’s Theorem to prove that the Borda Count does not satisfy IIA?

How about our other methods; can we say anything about them as a result of Arrow's Theorem?

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Approval voting does not really fit with the criteria of Arrow’s Theorem, as voters are not allowed to rate candidates specifically; it can be argued that for this reason it fails Arrow’s criterion of universality.

There is no correct answer to this question; it is dependent on the purposes of the elections.

Of course, these are not the only five criteria; there are many other (mathematical) criteria that can be used for deciding between voting systems.

Bearing in mind it is impossible to satisfy all five criteria, which ones do you think a good voting system should aim for?

Find out more about criteria for judging voting systems.

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7.4 UK 2010

7.4.1 How do UK elections work?

The UK elections use a voting system called First-Past-The-Post (FPTP). There are 650 constituencies (regions) in the UK. Everybody who is over 18 years old (and has lived in the UK for a number of years) votes for an MP (Member of Parliament) in the constituency where they live. The person with the most votes in the constituency wins a seat in Parliament.

Each MP is a member of a political party; all the MPs across the country are added up and the party with the most seats wins. If they get more than half of the seats they have a majority in the House of Commons and get to run the country.

We are going to have a look at this voting system and see how it works, and if it is a good system.

You could argue that Party A should win because they won 2 seats. But you could also argue that Party C should win because they got the most votes.

According to the UK system, Party A would win 2 seats (and gain a majority in the House of Commons), Party B would win 1 seat, and Party C would get no seats, and therefore no power. This is an example of one of the key flaws of the UK voting system.

Here’s a simplified UK election with only 3 seats; who do you think should win this election?

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7.4.2 Who won UK 2010?

Here are the results of the 2010 election:

You can see that the Conservative party gained 47% of the seats, and therefore 47% of the power. The Labour party gained 40% of the seats, and the Liberal Democrats gained 9% of the seats. As no party gained a majority of seats (for the first time since 1974), the Conservatives and the Liberal Democrats decided to form a coalition (joint) government in order to gain a majority in the House of Commons.

Here is a map showing where the different seats were won.

It seems that conservatives gained more seats in the South of England (except for London), perhaps as these areas are more affluent?

What percentage of seats did each party win?

Did any party gain a majority (more than half the votes)?

What do you notice about the map?

Why do you think this is?

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As I live and teach in London, let’s have a look at the spread of seats across London:

It would appear that Inner London is mostly Labour, apart from West London which is Conservative, with some Liberal Democrat regions.

Here is the full London scorecard:

What can you say about the spread of Conservative and Labour seats across London?

What else can you say about votes in London from this data?

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7.4.3 Seats v votes

Now have a look at the table on the right, which shows how many votes each party won in the election.

The number of people who voted (called the turnout) was 29,691,380.

The percentage of the turnout who voted for the Conservative party is 36%, compared with 29% for Labour, and 23% for the Liberal Democrats. You can see there is a large difference between the percentage of seats and the percentage of votes.

This is possibly because the Liberal Democrats have to beat both the Conservative and the Labour parties to win any seat. Can you think of any other reasons?

The Conservatives gained 34,940 votes per MP (=10726614 / 307), Labour 33,370 but the Liberal Democrats required 119,944 votes for each of their MPs.

What percentage of the turnout voted for each party? Compare these percentages of seats above. Do you think the number of seats is given out fairly?

Why do you think this might be?

You can see the difference between votes and seats even more clearly if you work out the votes per MP for each party.

How many votes did each party win for each MP?

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7.4.4 Referendum 2011

Not surprisingly, following this election the Liberal Democrat leader Nick Clegg called for a referendum of the UK voting system. He proposed that we change the UK system to the Alternative Vote system described earlier, in order to share the seats more fairly in line with the percentage of votes. The 2011, the UK public voted to keep FPTP by 13,013,123 to 6,152,607.

The idea of sharing seats in line with votes is called Proportional Representation (PR). AV is not a straight PR system, but it does attempt to match seats to votes more accurately. It has been projected that under AV, the seat share in 2010 would have been Conservative 282, Labour 264 and Liberals Democrats 74.

Working out the percentages, you can see that it might have been possible that the Labour and Liberal Democrat form a coalition instead, as together they gained 338 seats (52%).

Although it has its flaws, FPTP has the benefit that it creates more outright winners of elections (although not in 2010). Under PR, there would be less outright winners of elections and more coalitions, which may be a negative thing.

There are other issues with a PR system. Let’s look at one of the more extreme parties in the UK. The BNP party is a far right party that, to put it politely, has questionable policies, especially regarding immigration.

This is around just under 2% of the vote, so they would have gained around 12 seats in the House of Commons; thankfully, they did not gain any under PR. This also raises a second flaw of PR; if no constituency voted for the BNP (or any other party), where are these 12 seats to be allocated? Would you be happy with an MP you didn’t vote for?

How might history have been different under AV?

Do you changing the UK voting system to AV, or straight PR would be a good idea?

In the 2010 election they gained 564,331 votes; what percentage of the vote was this? How many seats would they gain under straight PR?

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7.4.5 Single Transferable Vote (STV)

What other options are there apart from FPTP and AV? Single Transferable Vote (STV) is used in Australian and Irish parliamentary elections, and is preferred by the Electoral Reform Society in the UK and Fairvote in the US, two leading analysts of electoral systems.

How does it work? There are many variations, but generally voters rank the candidates, in a similar way to AV (when there is only one winner it is the same as AV). Any candidate who attains above the quota of 1st place votes is elected.

The quota is calculated using the according to the Droop formula, which uses the ceiling function, usually written as ⌈ ⌉. The ceiling function takes any decimal and rounds it up to the nearest whole number.

The Droop formula is: votesneeded ¿win=⌈ total votesnumber of seats+1

+1 ⌉. This ensures that

there are not more winning candidates than there are seats.

Consider the example where there are 10 seats to fill and 1000 votes are cast. Then

the formula gives votesneeded ¿win=⌈ 100010+1

+1⌉=⌈ 91.909…⌉=92. So each

candidate needs at least 92 votes to win. So we can only have at most 10 winners; we can’t have 11 winners as 11 x 92 = 1012.

After counting first preferences, there will probably still be seats left to fill. For example, if (say) 200 votes were cast for 5 candidates in the example above we would only have 5 seats filled. In this case, the votes for the winning candidates are removed and voter’s second preferences become their first preferences. Any candidates now attaining the quota are elected and so on until the seats are filled.

The main argument for STV is that it will give a fairer (more proportional) representation of overall votes. The arguments against are that it is complicated, but it is not complicated for the voter – they just have to write their preferences on the ballot!

It is estimated by the Electoral Reform Society that in the under STV, the UK 2010 election results (seats) may have looked like this:

What do you think are the benefits and problems with this method?

How does this formula ensure there are not more winning candidates than there are seats?

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7.4.6 Equality in the House of Commons

Last but not least, here is a graphic showing the split between male and female MPs in the main UK parties:

First of all, note that this graphic is actually rather misleading; it exaggerates the differences between men and women.

You have probably spotted that the heights of the figures represent the numbers given, but that the figures are also wider in the same proportions. For example, the red male is not only around twice as high as the red female (to represent twice as many MPs), he is also twice as wide; so he is 4 times bigger by area!

I will leave you to calculate some of these figures. Here is some information about black and minority ethnic (BME) MPs in recent elections:

Based on the figures, work out the percentage of male and female MP in each party and in total. Which party has the most equal gender balance? Is the gender balance of MPs representative of the UK as a whole?

Can you see why? How could you make this graphic more accurate?

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The percentage of BME MPs in 2010 was 26/650 = 4%. This is an increase of around (26-15)/15 = 73% on the previous election, so it is increasing.

Is the percentage of BME MPs representative of the UK population? We need more data on the percentage of BME people in the UK. Here is a map of England (sorry I didn’t have data for the whole of the UK) with percentages of White-British residents, available from The Guardian data site:

What can we say about ethnicities across the UK by looking at this picture? Well, it appears there are low percentages of BME people in the most Northern parts of England, and there are patches where 0-50% of people are White-British, most notably London. Here is the numerical data for the main areas of England, in thousands:

How would you summarize these figures? What percentage of MPs is currently BME? How has the number of BME MPs increased over the years?

Is the percentage of BME MPs representative of the UK population?

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You can calculate that the percentage of White-British (and Irish) people living in England is around 43,451,000/51,810,000 = 84%, so there are 16% BME people living in England. Looking at London in more detail, we can calculate that the percentage of BME people living in Inner London is 43%. Looking even more closely at where I live and teach, here is the split for Greenwich & Woolwich:

However you look at it, the number of BME MPs in the UK does not represent the population as a whole.

Can you work out the split by ethnicity for England? How about the area in which you live (if you live in England!)?

Why do you think this might be?

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7.4.7 Voter apathy

Many people in the UK feel as though their vote does not count, and many do not vote in elections (called voter apathy). Here is a graph of voter turnout since 1945:

The turnout was 29,692,380 / 45,684,501 = 65%. This is one of the lowest turnouts in recent history, although not as low as 2001. If you are interesting, why not find out more about the UK political climate in 2001. Why do you think was turnout so low?

The number of people who voted Conservative in 2010 was 10,726,614 out of 45,684,501, which gives 23% percentage. So less than a quarter of the UK population voted for the current Prime Minister.

This may explain some voter apathy, but of course, if you don’t vote, you can’t moan

Confirm the calculation for 2010. The turnout was 29,692,380 from the total electorate of 45,684,501.

What percentage of the total electorate voted for a Conservative Prime Minister in 2010?

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about who is in charge!

Consider the votes in in Greenwich & Woolwich (shown here). If you wanted to vote for Conservative, your vote is unlikely to have much impact on the result in Greenwich & Woolwich (sometimes called a wasted vote).

Furthermore, these votes will not count towards the general election result.

There are other issues that lead to voter apathy. A general discontentment with politics in the UK (lack of real choice in the main parties, perceived corruption) might lead people to not vote. There may also be feelings of lack of representation (see the section above on BME MPs for example).

I’ll leave this to you to think about…

How does the FPTP system contribute to voter apathy? What other reasons might there be?

What are the consequences if more and more people did not vote due to these reasons?

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7.5 Weighted systems

In this section we are going to turn our attention to weighted voting systems, in preparation for a look at the US voting system. Here we relax the condition of anonymity, so that all voters are not equal. As an example of where this might be the case, consider the following shareholders of a company:

Suppose the shareholders wish to vote on some motion, and that they decide it is fair that the weight of their vote matches their stake in the company. For example, a vote by David counts for 101 of the 200 total ‘votes’.

If we set the quota (the winning amount) to 100, then we have a dictatorship - only David's vote counts. So where should we set the quota?

Perhaps we could set it at 102. Then David would need another voter to agree with him in order to form a winning coalition. So with this quota, the sets of winning coalitions are {David, Ed}, {David, Nick} and {David, Ed, Nick}.

In fact, it doesn't matter whether the quota is 102 or 103, the sets of winning coalitions stay the same, and we say that these systems are isomorphic. We will use the notation [102: 101,97,2] to describe this system, in which the quota is 102 and the weights are 101, 97 and 2.

We can see that [4: 2, 2, 1] and [5: 3, 2, 1] are isomorphic, as the winning coalitions for both are either {A,B} or {A,B,C}, as are [4: 3, 2, 1] and [5: 3, 2, 2]. Notice how the

How should we decide whether the motion should pass? Should it be based on a majority? If not, how else should we decide?

Which of these five systems are isomorphic to each other? You will need to work out the winning coalitions for each system.

[4: 2, 2, 1] [4: 3, 2, 1][5: 3, 2, 1][5: 3, 2, 2][5: 3, 3, 2]

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fifth one is different; it is the only one in which B and C can form a winning coalition.Now, if a voter is in every winning coalition, we say that voter has veto power.

The sets of winning coalitions are {David, Ed}, {David, Nick} and {David, Ed, Nick}, so David has veto power – if he doesn’t vote for something, it doesn’t happen.

However, this doesn’t mean that David has all the power. We will answer the question of how power is shared between the members in the next section

UN Security Council

The UN Security Council has 5 permanent members and 10 temporary members who change every two years. In order for a motion to pass, all five of the permanent members must agree on the motion, as well as at least four of the temporary members.

To calculate this, let's suppose that the weight of vote for each temporary member is 1 and for each permanent member is X. For a motion to pass, we need the total weight of votes to exceed the quota, which we will call Q.

Well, the minimal way in which a motion can pass is if 5X + 4 >= Q. Can you think of a 'maximal' way in which a motion does not pass and express this as an inequality?

A motion does not pass is 4X + 10 < Q. Putting these two inequalities together gives Q = 39 and X = 7, so we can express this system as [39: 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1].

Which of these five systems are isomorphic to each other? You will need to work out the winning coalitions for each system.

[4: 2, 2, 1] [4: 3, 2, 1][5: 3, 2, 1][5: 3, 2, 2][5: 3, 3, 2]

Who has veto power in the original example with David, Ed and Nick?

More generally, how do you think the power is shared between the three members?

Who has veto power in this system? What is the quota for this system? What are the weights of the members?

Can you express this information in an inequality?

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7.6 Banzhaf Power

Let's consider the weighted system with weights below and a quota of 102.

We saw in the previous section that the winning coalitions for this system are {D,E}, {D,N} and {D,E,N}, and we saw that David has veto power - no coalition can win without him... but can we go a bit further an work out exactly how much power David has?

There is no correct answer to this question, but one way of measuring the power of each voter is to use Banzhof Power. It is calculated as follows:

• D is critical in all three of the winning coalitions - without him they would not win.• E is critical in only one of the winning coalitions {D,E}.• N is critical in only one of the winning coalitions {D,N}.

Now, the ratio of critical appearances for D:E:N is 3:1:1. Each of these numbers is called the Banzhaf Power of each voter. We can also calculate the Banzhaf Power Index (BPI), which is just each voter’s fraction of the total Banzhaf Power of the system. Here David's BPI is 3/5, and Ed and Nick each have a BPI of 1/5.

Here's a more complex example. Suppose there are 6 board members with the following shares in a company:

There are 115 votes in total, so 58 is a majority. There are 32

What do you think? Do you think he has all the power? Do you think he has (say) 50 times more power than Nick? Do you think Ed has 48 times more power than Nick?

Do you think this is a good measure of power in this voting system? Why/why not?

Suppose the quota is 58 (why?). What is the BPI of each shareholder?

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winning coalitions (did you find them all), each of which contain either A and B, or A and C, or B and C (why?). To find these easily, take these pairs and find all the combinations of others that go with them:

So we have AB, ABC, ABD, ABE, ABF, ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ABCDE, ABCDF, ABCEF, ABDEF, ABCDEF, then AC, ACD, ACE, ACF, ACDE, ACDF, ACEF, ACDEF and finally BC, BCD, BCE, BCF, BCDE, BCDF, BCEF, BCDEF.

The critical voters are in bold. The ratio of critical votes for A to F is 16:16:16:0:0:0 so that A, B and C share all the power (1/3 each).

This example is actually taken from a real example in a county election in the US where the 'shares' (weights) were allocated according to the population of various districts within a county New York. They realized that the three smaller districts had no power.

Following an analysis of the BPI of each district, the system was changed to [65: 30, 28, 22, 15, 7, 6].

Did you work them all out? Good for you! If you did, you will have found that even the smallest district has some power (around 2%); although this does not represent the populations exactly, it does lead to a fairer voting system.

If you really want a challenge you might want to work out the BPIs for the members of the UN Security Council using the weighting system in the previous section!

How did this alleviate the earlier problems? Do you think it is fair to allocate larger weights even though the smaller districts have smaller populations?

If this has interested you, why not find out about some other measures of power, such as the Shapley-Shubik Index?

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7.7 Electoral College

7.7.1 How do US elections work?

The Electoral College is the name of the voting system in the US. This section gives a slightly simplified analysis of this system and explores some of the historical consequences.

Electoral College is similar to the FPTP system used in the UK, in that the US is divided into states (constituencies) and the winner of each state is decided by plurality. However, instead of each constituency contributing one seat to the parliament as in the UK, each state contributes a number of members to congress according to its population.

There are 538 electoral votes (seats) in the current US system, made up of the House of Representatives (435), the Senate (100) and Washington DC (3). The electoral votes in the House of Representatives are allocated according to population; as populations change, so do the allocations. Here are the allocations in the House of Representatives as at 2013:

You can see that California has the largest allocation of votes (53), and some have only 1. We will look at how these seats are allocated below.

The 100 seats in the Senate are fixed at 2 for each state. This gives a bit of extra weight in the system to the smallest states - you can't have a total of less than 3 representatives in congress (1 in the House of Representatives + 2 in the Senate).

Washington DC is a slightly odd one out; it has 3 members (total) in congress.

Now, when the election happens, people in (say) California decide who they want to be president, from the list of candidates. Now, as there are 55 electoral votes (seats) in California, they could divide these according to the votes cast. However, there is an unwritten rule in US politics that is obeyed by (nearly) all states called winner-

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takes-it-all (just like FPTP) that says that the candidate with the plurality of votes in that state gains all the seats for that state.

As we have seen from our analysis of the UK FPTP system, this could result in the candidate with the most overall votes losing the election.

7.7.2 US 2000

Now let's look at the US 2000 election. It was won by George W Bush with 271 electoral votes to Al Gore's 266 (with 1 abstention).

The key state turned out to be Florida; it had a population in 2000 of 15,982,378 (what is this as a percentage of the population of the US?) and was allocated 25 electoral votes. The votes in Florida were as follows:

You can see that Bush gained 537 more votes than Al Gore in this state, but gained 25 electoral votes to Gore's 0 in this state.

Now, as you might have guessed, here are the total votes:

You can see the flaws in the system here. If the US election was based on the number of votes, history might have been very different. A tiny difference in the preference of voters (a swap of 269 from Bush to Gore here) would have resulted in a different US President!

Do you think this is fair?

What might be the consequences of the winner-takes-it-all system?

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It is actually (mathematically) possible for a candidate to win the US election with as few as 11 votes - if only one person turned up to vote for them in each of the 11 largest states (which would give a majority of electoral votes)! Of course this wouldn't happen, but it is possible to win the US election with a relatively small number of votes.

There is a further flaw in the system. Candidates who have little chance of winning in certain states do not campaign there. For example, in 2000 Bush did not campaign in California as he had little chance of winning there (and similarly for Gore in Texas).

7.7.3 Who would have won US 2000 under AV (Instant Run-off)?

Let’s suppose that Gore was the 2nd choice of all the Browne (Libertarian) voters, and Bush was the second choice of the Buchanan (Reform) voters, and let's say the other 2nd votes were split 50-50.

In the first two rounds of the AV process, all of Browne’s votes would go to Gore and Buchanan’s votes would go to Bush, with others split between them 50-50, leaving the total votes as:

Now, the 2nd choices of Nader voters decide the election.

Have another look at the map of allocations between states in the House of Representatives. Mathematically speaking, what is the minimum number of votes with which someone could win the US presidential election?

With these assumptions, who would win this election under AV?

What percentage of these voters would have to prefer Bush for him to win the election?

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The difference between Bush and Gore at this stage is 479,431. So Bush would win this election if this number of Nader voters preferred him to Gore.

So the percentage, p, that Bush would need is given by 2,882,955 x p – 2,882,955 x (1 – p) > 479,431 which gives p around 58%.

Again this stresses the importance of Spoiler candidates, but perhaps more importantly it stresses again the importance of casting your vote!

7.7.4 How are electoral votes allocated?

The allocation of electoral votes (seats) in the US is reasonably complicated. Consider the following example from the first US population census in 1790:

In 1790, the House of Representatives wanted to allocate 105 seats between the 15 states fairly.

Assuming that you calculated each allocation based on the proportion of the overall population in that state and then rounded in the normal way, you will find that this method requires 106 seats.

In 1790, a mathematician called Hamilton suggested that the fractional allocation of seats should not be rounded in the usual way, but instead should be rounded down (so each allocation is the integer part of the fractional allocation); the seats are then

Find out more about the (2000) US elections. What type of people voted for Bush and Gore? Was Bush a popular president?

Who was Ralph Nader, and what type of voters who voted for him?

Do you think it is more likely that Bush or Gore would have won this election under AV?

How would you do it? Work out the figures using your method; does it work?

So how can you fairly allocate the seats? Who would you take that last seat away from?

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allocated accordingly.

Using Hamilton’s method you will find that 96 seats are allocated, leaving 9 spares. Hamilton then suggested that the remaining 9 surplus seats are allocated to the states with the largest decimal parts left over.

Here are my calculations:

Is this method fair? Consider Delaware, which has a fractional allocation of 1.594, but misses out under Hamilton's method. Compare this with Maryland, which gets one of the surplus seats with a fractional allocation of 8.622. Do you think it is fairer to allocate the seat to Maryland or Delaware?

Try out Hamilton’s method. How many seats are left over? How do you think you could allocate these spare seats fairly?

Who would get the 9 remaining seats under this method? Do you think Hamilton's method is fair?

Can you think of a fairer apportionment method? Find out more about other apportionment methods that are/have been used.

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8 Maths and music

This section was designed to give an introduction to the ideas of ratio, along with practice with fractional indices and working with fractions.

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8.1 The standard scale

The diatonic scale (sometimes called the standard scale) is given by the series of intervals tone-tone-semitone-tone-tone-tone-semitone. If we start with the note C, the sequence is then C-D-E-F-G-A-B-C. Each (whole) tone is worth 2 semitones, so the intervals between E-F and B-C are semitones.

We can write this on a normal musical staff like this:

The symbol on the left is called a treble clef (or G-clef) as the note G is on the line through the ‘centre’ of the symbol. Each line on the staff represents two whole notes, so these notes represent the standard scale C-D-E-F-G-A-B-C.

If we start with a different starting note (called the tonic) and keep the same scale, we will get a different sequence of notes. For example, with tonic D we have D-E-F♯-G-A-B-C♯-D. The symbol ♯ is called a sharp, and denotes that the note is a semitone higher (so F♯ is a semitone higher than an F). The opposite of a sharp is a flat, and is written with the symbol ♭. The sharps and flats (called accidentals) are the black keys on the keyboard.

This describes the key signature for the key of D, which has two sharps on F and C like this:

The bottom staff has a bass clef, or F-clef, which is centred on F instead of G.

For the key of E we have E-F♯-G♯-A-B-C♯-D♯-E, which has 4 sharps. Note that each letter appears once, so we need sharps not flats to make that happen. The key of F is F-G-A-B♭-C-D-E-F, which has one flat. Here are the key signatures for these:

What sequence of notes do we get with tonic E? Or F? Can you write the key signatures for these keys?

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The notes are often given names, according to their position on the standard scale. For example, the fifth note is called the (perfect) fifth. This is G in the key of C, and is 7 semitones above C. Other names for the notes we will be talking about here are: Minor third (3 semitones), major third (4 semitones), perfect fourth (5 semitones), major sixth (9 semitones), minor seventh (10 semitones), major seventh (11 semitones) and octave (12 semitones). We will see where these names come from later.

If we start at C and go up a fifth (7 semitones), we get to G, which gives the same note (albeit an octave apart) as going down a fourth (5 semitones). We could think of this in the language of modular arithmetic by noting that 7≡−5(mod 12).

There are other scales that are used, some of which are cycles of the standard scale above. For example, the Aeolian scale is tone-semitone-tone-tone-tone-semitone-tone, which in the key of C is C-D-E♭-F-G-A-B♭-C.

As the Aeolian scale is shifted one place to the left, we also shift the notes one place to the left to give D-E-F-G-A-B-C-D, so it’s the key of D.

We will find out in the next few sections.

Which key in the Aeolian scale is the same (all white keys) as the key of C in the standard scale?

Where did this musical scale come from? Why are there 12 notes, and why are they arranged in this way?

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8.2 Frequency

The history of musical instruments date back to the beginning of human culture, and it is impossible to say where the first instruments were created. It has been known for thousands of years that shortening the length of a string and plucking it (by holding it down on a fret-board) changes the frequency of vibrations and creates different notes.

For example, shortening the string by half its length creates a note that sounds twice as high pitched; we say the note doubles in frequency. Shortening a string to 1/3 its length creates a note with three times the frequency.

This suggests that length and frequency are inversely proportional, or we could say that the change in frequency is the reciprocal of the change in length; as length gets shorter, frequency gets higher. We can write this using the sign for proportion:

frequency∝ 1length

Noting the relationship frequency∝ 1length from the previous section, if we wanted

to multiply the frequency by 3/2, we can find the change in length by working out the reciprocal to give 2/3; the length of the string must be reduced to 2/3 its original length.

It has also been known for many years that some notes sound harmonious when played together (consonant) and some less so (dissonant). What is it about some combinations of notes that make them sound more harmonious than others? It must be connected to the frequency, or more accurately the relative frequency, of the notes.

What exactly is the frequency of a note? It is the number of times it pulsates in a second, and is measured in Hertz (Hz, cycles per second). The note A4 has a frequency of 440Hz, where the subscript 4 denotes where it is in the grand scale of notes; A5 is the same note but is an octave higher (double the frequency). We can

By what fraction would we need to shorten a string to play a note 3/2 times higher in frequency than the open string?

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say the ratio of the frequencies A5 : A4 is 2 :1, or A5/ A4 = 2 / 1 = 2, and we can calculate that A5 has frequency 880Hz.

It has long been known that notes with relative frequencies written using ‘small’ numbers are the most harmonious. To see why, we need to think about what is physically happening when we hear a sound. A simplified account of what happens (given by Galileo) is to consider sound as waves with a certain frequency; here is a picture showing two waves an octave apart:

We can see the peaks and troughs of the waves are aligned, so making the sound more consonant. It turns out this is an oversimplification of what happens, but we will stick with this explanation for now.

To simplify this even further, we can think of as ‘pockets’ of air hitting our ear every fraction of a second. For example, pockets of sound travelling with a frequency of 440Hz hit our ear every 1/440 = 0.00227 seconds.

A5 has twice the frequency, so would hit our ear twice as often, around every 0.00114 seconds. We can plot the arrival of these air pockets on a graph:

You can see that the two sounds hit our ear in the ratio 2:1 as described by the frequency.

Now let’s explore frequencies for notes between the octave, i.e. between 1/1 and 2/1.

It seems obvious that it is 3/2 (one and a half). A note with frequency in the ratio 3:2 to A4 would have frequency 3/2 x 440, which is 660Hz. We can check this by noting that the ratio of the frequencies is 660:440 which is equivalent to 3:2.

How often would the note A5 hit our ear?

What is the simplest ratio (with the lowest whole numbers) between 1/1 and 2/1?

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The pockets of air hit our ear every 1/660 = 0.00152 seconds. The graph for the new sound (the top line, in black) compared to A4 (below, red) looks like this:

You can see that the notes coincide in the ratio 3:2 as expected. This is why they sound pleasing to the ear.

We are looking for ratios with as low numbers as possible. Here are the ratios with

denominators 5 or less: 32 ,43, 53, 54, 74, 65, 75, 85, 95 .

The easiest way to do this (without a calculator!) is to change them into 60ths (60 = 2 x 2 x 3 x 5 is the lowest common multiple of 2, 3, 4 and 5). Here they are in 60ths: 90/60, 80/60, 100/60, 75/60, 105/60, 72/60, 84/60, 96/60, 108/60, and so here they

are in order: 65, 54, 43, 75, 32, 85, 53, 74, 95 .

How does this relate to our musical scale? You may have noticed that there are 9 ‘notes’ here that sound harmonious to the tonic (10 if you include the octave), but

How often would the sound pockets hit out ear for this new sound? Can you draw a similar graph to that above to show how the pockets of sound hit our ear?

seconds

What other ratios between 1 and 2 might sound good to the ear?

Can you put these fractions in order of size, smallest to largest?

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our musical scale contains 12 notes. How do they match with to these frequencies? This is the tricky part – it all depends on how you tune your instrument!

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8.3 Pythagorean tuning

The Ancient Greeks, and in particular Pythagoras, believed that the ratio 3:2 represented the perfection of the universe and based his musical scale upon it.

How did Pythagoras use the ratio 3:2 to make a whole scale of notes? If we use the example of the key of C, his scale starts with two notes in the ratio 3:2, meaning that the new note has frequency 3/2 times the tonic. We could write this ratio 3/2 : 1, and adding in the octave gives three notes in the frequency ratio 1 :3/2:2 (lowest to highest).

The ratios can be written equivalently as 2 :3: 4, so that the ratio of the third and second notes is 4 :3. This ratio was then used to create a new note, at a ratio 4:3 above the tonic, giving a four note scale: 1 :4 /3:3 /2:2.

The lowest common multiple of the denominators is 6, so multiplying through to give denominators of 6 gives the ratio 6: 8: 9: 12.

Pythagoras did not want a scale with just 4 notes (that would make for some boring music!), so he used the ratio between the middle two notes to create another interval.

We can see from the above calculations that the ratio between the middle notes is 9 :8. Pythagoras used this fact to find the ratios for all the other notes in the standard scale.

Starting with the first note as 1, we have the next new note as 9/8. Using this ratio again, we can create another new note by just multiplying this one by 9/8 again, giving 9/8 x 9/8 = 81/64.

Now, if we multiply by 9/8 again, we get another new note (729/512) but Pythagoras decided against this, because the ratio between the numbers is high, producing a discordant note. Also, 729/512 is higher than 4/3, which we already have, so he left

Can you write these ratios as whole numbers?

What is the ratio between the second and third notes?

Can you write these ratios in the simplest form (with the lowest possible integers)?

What is the ratio between of the middle two notes?

Can you work this out how to do this?

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4/3 as the next (fourth) note, and 3/2 as the fifth. Then starting again with the interval 9/8, he multiplied to give the sixth and seventh notes 27/16 and 243/128.

So here are the frequency ratios given by Pythagorean tuning:

1 98816443322716243128

2

This scale forms the basis of the musical scale we use today; the seven notes are the major notes in our standard scale. In the key of C we would call these C-D-E-F-G-A-B-C, with the ratio 9/8 representing a whole tone.

First of all, we can see that only two of the notes have low-integer ratios, so some of the notes might sound discordant. Let’s compare this to the low-integer ratios we

found earlier: 65, 54, 43, 75, 32, 85, 53, 74, 95 .

The Pythagorean third 81/64 is close to 5/4, the 27/16 is close to 4/3 - Pythagorean tuning comes close to some of the lowest integer ratios.

However, there is another important problem with Pythagorean tuning. The ratio for a whole tone is 9/8, but what is the ratio for the semitone?

Looking at the scale above, we can find the semitone ratio by working out 81/64 x something = 4/3. So we need to work out 4/3 divided by 81/64, which by the rules of dividing fractions, is the same as 4/3 x 64/81 and we get 256/243.

Two semitones should make a whole tone, so using the ratios above, it should be true that 256/243 x 256/243 = 9/8, but it isn’t!

What do you think are the main problems with the Pythagorean tuning?

Can you see a problem with this?

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A different way of thinking about Pythagorean tuning is to use the circle of fifths:

If we start at the top (C) and go round (clockwise) in intervals of fifths, we get the notes shown.

[Notice also how the key signatures of each key increase with one extra sharp (one less flat) each time.]

How is this related to Pythagorean tuning? We can tune according to Pythagoras’ ‘perfect’ ratio 3:2. Starting at C, we move clockwise to G and multiply frequency by 3/2.

Moving one more step round the circle to get D we multiply 3/2 by 3/2 again, which gives 9/4. This is bigger than 2 so is more than an octave above the starting C. We can adjust this to be in the same octave by dividing the frequency by 2 (why?) to give the tuning of D as 9/8, which matches our Pythagorean tuning.

The next note in the circle, A, is then found by calculating 9/8 x 3/2 = 27/16 which again matches the tuning given above.

The frequencies do indeed match the tunings given above, but the problem comes when we try and tunes notes above the octave. For example, what is the note that is a seventh above A?

Starting at 27/16, we multiply by the ratio for a seventh to give 27/16 x 243/128 = 6561/2048, which is a new note not on our scale (it doesn’t simplify to anything on our scale).

Here is a diagram showing the problems of Pythagorean tuning:

Keep going round the circle of fifths and check the tunings match those given above. Can you suggest why this might be problematic?

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You can see the discrepancy between A♭and G♯. Thus the circle of fifths becomes a spiral of fifths, not quite meeting to make a circle!

We can think of this in a different way. There are 12 fifths (12 x 7 semitones) in 7 octaves (7 x 12 semitones), so these tunings should match.

Using the Pythagorean tuning of 3/2 for a fifth we have 12 fifths = (3/2 )12= 129.75 whereas 7 octaves is 27 = 128. So they do not match.

If we think about this a bit more, we could have known this by noticing (3/2 )n≠2m for all n and m as 3n≠2m for any values of n and m.

During the thousand years after the Greeks, Pythagorean tuning was the most common in use. Due to the dissonance between the tonic and thirds and sixths, these notes were not generally used in music during this time, and notes were often not played simultaneously. Also, music was not played across many octaves due to the problems with transposition.

Between around 900AD – 1300AD musicians started to play 2 notes simultaneously from the limited selection of octave, fifth, fourth, octave + fifth, octave + fourth and double octave.

Work out the Pythagorean ratio for 12 fifths compared with that for 7 octaves. What do you notice?

Can you work out the ratios for octave + fifth and octave + fourth?

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Octave + fifth is the ratio 3/1 = 2/1 x 3/2 and octave + fourth is 8/3 = 2/1 x 4/3.

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8.4 Just intonation

People tried to find a solution to these problems with the Pythagorean tuning for hundreds of years. In 1558, a musician called Guiseppe Zarlino proposed a different tuning. He kept the Pythagorean ratios for the tonic, fifth and octave 1: 3/2 : 2 (2:3:4), which also gives 4/3 for the fourth. He then fixed the ratios between the tonic, third and fifth to be 1: 5/4 : 3/2 (or 4:5:6).

Using the ratio between the fourth and fifth for the whole tone, we get 9/8 again. Also, using the thirds ratio, we can use the fourth note to derive the ratio of the sixth to be 4/3 x 5/4 = 5/3, and the fifth note to derive the ratio of the seventh to be 3/2 x 5/4 = 15/8, giving the whole scale C-D-E-F-G-A-B-C as:

1 :9/8 :5/4 :4 /3:3 /2:5 /3 :15/8 :2

You can see that this has much lower ratios than the Pythagorean tuning.

The first thing to note is that the interval for a whole tone depends on which whole tone we are talking about! For example, the first whole tone is 9/8 but the second one is given by 9/8 x something = 5/4. So, 5/4 divided by 9/8 = 5/4 x 8/9 = 10/9. Working out intervals for all successive notes gives whole tones the alternate between 9/8 and 10/9, with a semitone of 16/15. Zarlino called the larger whole tone (9/8) the major tone, and the smaller one (10/9) the minor tone.

As you have probably guessed, this creates a problem. Consider transposing from D to A, up a fifth. D is 9/8 and a fifth is 3/2, so we get 9/8 x 3/2 = 27/16, which is not the correct ratio for A (it should be 5/3).

This happens because the interval for a fifth in the scale above comprises of major tone, minor tone, semitone, major tone. However, the interval from D to A is minor tone, semitone, major tone, minor tone; this has a different number of major and minor tones! The difference between these two intervals is 27/16 divided by 5/3, which is 27/16 x 3/5 = 81/80, which is called the ‘syntonic comma’.

That said, just intonation brought about more freedom in music, most notably increasing the use of thirds and sixths, and allowing combinations of 3 or more notes to be played together. 8.5 Equal temperament

Can you derive the whole (7 note) musical scale that comes from these ratios?

Can you see any problems with this tuning? Consider the ratios between successive notes, and think about the other problems with the Pythagorean tuning – do they apply here?

Can you explain exactly why this happens?

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All of these problems with tuning come down to the fact that it is not possible to have a tuning system based on rational numbers (fractions) such as the ratio 3/2 for a fifth. This is due to the fact that 3n≠2m for any whole numbers n and m.

In the 16th century, a Flemish Mathematician called Simon Stevin suggested making the interval between a semitone 21 /12. This means the twelfth root of 2; if we raise this number to the power 12, we will get 2. This system of tuning is known as equal temperament. It was not well regarded for some time but gained favour in the 19th

century as it allows freedom of transposition across many octaves, and is now used for tuning all keyboard instruments.

The second root is known as the square root. So the second root of 49 is 491/2=√49=7. The third root is the cube root, so 1251 /3= 3√125=5.

The main advantage of equal temperament is its simplicity. We can transpose notes in any way we like without the problems encountered by Pythagorean tuning or just intonation. If we want to transpose a note by a fifth (7 semitones), we just multiply the frequency by 27 /12.

You are probably thinking, “What about the harmonious ratios? Surely we can’t have harmonious low-integer ratios now?” and you are right. 21 /12 is not a nice ratio; in fact, it can not be written using a ratio at all – it is an irrational number. To the first few decimal places, 21 /12 is about 1.059463… so why might this system be used?

The answer lies in the fact that it is a good approximation to low-integer ratios, most notably the fifth.

It turns out that 27 /12 is about 1.498…, which is really close to 3/2, and this is one of the main reasons why equal temperament ‘works’. In fact, our ears can’t differentiate between the two notes – the difference has to be around 0.1 for our ears to notice the difference.

What is the second root more commonly known as? What is the second root of 49? How about the third root of 125?

What do you think might be the advantages and disadvantages of equal temperament?

What is the difference between the ratio 3/2 and 27 /12?

Explore the difference between other intervals in equal temperament (2n /12¿ and low-integer ratios. Which ones are close?

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Looking at the other notes in the standard scale under equal temperament, they are all reasonably close to harmonious ratios:

2nd D 22 /12≈1.12≈9/83rd E 24 /12≈1.26≈5/44th F 25 /12≈1.33≈4 /35th G 27 /12≈1.50≈3/26th A 29 /12≈1.68≈5/37th B 211/12≈1.89≈…

In addition, we have the minor third (3 semitones), which has the ratio 23 /12≈1.22≈6/5, the augmented 4th (diminished 5th) with approximate ratio 7/5, the minor 6th, which is around 8/5 and the minor 7th which is approximately 9/5.

Remember that the frequency of notes is inversely proportional to the length of the string played, hence using the reciprocals.

So you can see that equal temperament with 12 semitones seems like a reasonably good idea; it is reasonably simple and allows transposition, at the expense of some consonance.

The division of the octave into 12 semitones is probably due to a mixture of historical reasons and practical simplicity. Other divisions of the octave into 19ths, 27ths or 31sts that would match these key ratios more accurately.

Although the division into 19ths provides a better match to the harmonious ratios, that this scale has not been adopted, presumably because it would be too complicated. It turns out that a division into 53rds gives an almost exact approximation to the perfect fifth, a fact known by Ancient Chinese musicians around the time of the Ancient Greeks.

Why are there 12 semitones? Why not some other division of the octave?

Investigate other divisions of the octave: can you find one that is closer to the

harmonious ratios: 65, 54, 43, 75, 32, 85, 53, 74, 95 ?

Find the ratios for 19ths; how many of the 9 harmonious ratios are met by this division of the octave?

If you can get your hands on a guitar, compare the distances between each of the frets from the bridge with the reciprocal of these ratios. Do they match?

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You can solve this by trial and error to give the value n = 31. However, we could use the ‘inverse’ of powers, which are called logarithms (logs for short), invented by Scottish Mathematician John Napier in the 16th century.

The inverse of powers of 2 is called log2, by which I mean if we apply log2 to 2n /53 we just get the power, n /53. So applying logs to both sides of this equation we get n /53≈ log23/2, and multiplying by 53 gives n≈53× log23/2 = 31.003, which is very close to 31.

Logarithms are useful in talking about frequencies as they allow us to add instead of multiply thanks to the log rule log a×b=log a+log b. They also allow us to create a one-to-one correspondence between numbers and intervals.

Suppose we have a tonic; let’s associate this with the number 1. Then we have already seen that we can think of the octave as being 2 (twice the frequency) and that a perfect fifth matches the value 3/2.

Using log2 we can directly match the value to the number of semitones above the tonic using 12×log2 (3/2 ) = 7.02, showing that the ratio 3/2 is just over 7 semitones (a fifth) in our 12-semitone equal temperament scale.

For example, 12×log2 (4 /3 )=4.98 which tells us this ratio is close to 5 semitones, or a fourth, in our scale.

You may have noticed that the integer 4 represents 2 octaves above the tonic. You can think of this as doubling twice, or alternatively 12×log2 (4 )=24 semitones. Perhaps you also realized that 8 represents 3 octaves, 16 = 4 octaves and so on.

What happens with integers that are not powers of 2? Let’s start with 3; we can use

the log rule log a×b=log a+log b by noting that log23=log22×32=log22+log2

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and we can see that this is one octave and a fifth.

Taking this further we can use prime factorization to find some intervals matching other integers, such as log26=log22×3=log22+log23 which we know is octave + octave + fifth.

How many ‘semitones’ (n) do we need so that 2n /53≈3 /2?

Work out 12×log2 (a /b ) for other low-integer ratios and check they match the number of notes given above.

How about intervals larger than an octave? What intervals match the integers 3, 4, and so on?

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Some intervals are not particularly near to a whole number of semi-tones. Consider 12×log25=27.86. This is discernibly flat; for this particular reason, equal temperament was disregarded for many years!

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8.6 Chords

Chords are when more than one note is played at the same time. Some combinations of notes give harmonious chords and others don’t; we have seen that this is due to the relative frequencies of the notes.

The most common chord is the major chord, which is given by the intervals (4,3,5). What this means is that the interval between the root and the 2nd note of the chord is 4 semitones, the interval from the 2nd to 3rd note in the chord is 3 semitones. For completeness, we give the interval from the 3rd back to the root note as 5 semitones.

So the C major chord (written as C) is given by C-E-G, and is written on the musical staff as shown.

The major chord comprises of the major third and the fifth.

There are many other types of chords, each with different intervals invoking different moods; here are a few of them:

Minor chord (3,4,5), usually written with a small m like this: Cm Diminished chord (3,3,6), written Cdim or C0. Augmented chord (4,4,4), written Caug or C+. Seventh chord (4,3,3,2), written C7. Minor 7th (3,4,3,2), written Cm7. Major 7th (4,3,4,1), written CM7. Diminished 7th (3,3,3,3), written Cdim7. Augmented 7th (3,3,4,2), written Caug7.

For example, minor chords have a slightly dissonant, sad quality. My favourite chord is the minor 7th (3,4,3,2) which comprises of the minor third, perfect fifth and minor seventh with approximate ratios 1 :6/5 :3/2: 9/5 = 10: 12: 15: 18. Perhaps this is something to do with low ratios between pairs of notes?

The structure and feel of a song is determined by the combination of chords used, called a chord progression. Just as different notes sound harmonious when played together, the same can be said for chords; the circle of fifths gives harmonious chord progressions. For example, a simple chord progression that is often used in popular music is A-D-G-A.

If you can get your hands on a keyboard, translate these into the correct notes and try playing them. How does each chord make you feel? Can you explain this with regards to their frequencies?

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8.7 Duration of notes

So far we have looked at the melodic aspect of music, but what about the rhythmic aspect? The timing of the notes in a melody is as equally important as the pitch of the notes.

The duration of notes are based on powers of 2. Here are some notes and their durations:

Going from left to right, each note is half the duration of the one before. Adding more flags to the stem of the note will make shorter notes. Sometimes beams are used to join notes with flags together like this:

There is an equivalent set of rests:

We can make notes of different durations to these powers of two by using a dot, like this:

The dot increases the duration of the note by half its original value. So the half note with a dot shown here now has duration of 1/2 + 1/4 = 3/8.

Adding a dot to an eighth note produces 1/8 + 1/16 = 3/16. Two dots adds a half and a quarter of a note, so adding another dot to an eighth note would produce a note of duration 1/8 + 1/16 + 1/32 = 7/32.

There are many note lengths that we can’t make using powers of 2. For example, how do we make a note of length 1/3? To do this we need tuplets; there are numerous ways of writing these:

What does adding a dot to the eighth note produce? What do you think happens if you add two dots to a note? Can we produce notes of all durations in this way?

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The 3-tuplet (or triplet) on the left comprises of three quarter notes. Usually three quarter notes would have a total duration 3/4 but generally in a triplet they have duration 1/2, so that each note is 1/3 of 1/2, which is 1/3 x ½ = 1/6.

Each note in the central triplet has duration 1/3 of 1/4, which is 1/12. The quintuplet has total duration 4 x 1/16 = 1/4, so each note has duration 1/5 of 1/4 = 1/20.

What do you think is the duration of each note in the central triplet? How about the quintuplet on the right?

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8.8 Time signatures

A time signature of a piece of music determines its rhythm.

Time signatures are written with two numbers. In simple time signatures, the top number represents the number of beats in each bar and the bottom number represents the note that is worth one beat (called the beat unit).

So in the time signature 34we have 3 beats to each bar, and one quarter note is worth

one beat; we can think of this as a measure having three quarter beats.

There are more complex time signatures called compound signatures, such as 98. You

can tell these from the fact that the top number is a multiple of 3; in compound time signatures the number of beat notes is the top number divided by 3.

In 98 there are 9 eighth notes in each bar (in groups of 3) with three beat notes of

value 3/8; they will be dotted quarter notes.

Here is a fun investigation that requires clapping different rhythms; you need at least two people to do it:

While one person repeatedly claps the rhythm shown on the right, the second person explores other rhythms, such as 1 2 . 4 . 6 . 8

What is the duration of the beat note in 98 time?

Why not investigate other clapping rhythms with different time signatures?

Try writing them using the correct musical notation.

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Glossary

Arithmetic sequences

Any sequence that goes up by the same amount each time is called an arithmetic sequence. So 2, 4, 6, 8, … is one, and so is 1, 4, 7, 10, …

The nth term for any arithmetic sequence is an + b where a and b are numbers we have to find. So for example, the nth term of 1, 4, 7, 10 is 3n – 2. This is because it is 2 less than the 3x table.

To find the sum of an arithmetic series, just add the first and last term and divide by 2 to get an ‘average’ or ‘middle’ term, then multiply by how many terms there are.

So for example, to sum 1 + 4 + 7 + 10 + 13 easily we can do 1 + 13 = 14, divide by 2 is 7, then multiply by 5 (number of terms) is 35.

We can use this to find the sum for the first n terms of a series, for example: 1 + 4 + … + (3n-2). First term plus last term is 1 + (3n-2) = 3n-1, then divide by 2 and multiply by the number of terms (n) to give ½.n.(3n-1).

Base-n number system

Our number system is based around the number 10. It is called base-10 because the value of each place in our number system is a power of 10. Units can also be written 100, tens can be written 101, hundreds can be written 102 and so on, each one 10 times bigger than the last. Moving from right to left for any number, each digit has 10 times the value of the previous one.

This compares with other number systems, which have different bases, such as the Babylonian system (base-60) and binary.

Binary

Binary is a base-2 number system. Each column in binary is two times larger than the one to the right, like this:

So for example the number 1011 in binary is 8 + 2 + 1 = 11 in base-10. Sometimes people write binary numbers like this 10112 so it is clear we are working in binary.

You can think of this as writing numbers as sums of powers of two, so for example 1011 can be written as 8 + 2 + 1 = 23 + 21 + 20. All numbers can be written in this way.

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Complete graphs

A complete graph is one in which all vertices are joined with one edge to every other vertex. They are called Kn where n represents the number of vertices; here is K7 that we encountered in domino loops:

Continued fractions

Fractions can be written as continued fractions, like this:

6724

=2+ 1

1+ 1

3+ 1

1+ 14

We can show these are the same as follows. Starting at the bottom corner, we have

1+ 14=54 . Substituting this into the fraction above we then have

2+ 1

1+ 1

3+ 154

. Now, 154

is the reciprocal of 5/4 which is 4/5. So we have 2+ 1

1+ 1

3+ 45

.

Now, again in the bottom corner, we have 3+45=195 so we can rewrite the continued

fraction as 2+ 1

1+ 1195

and flipping the last fraction again we have 2+ 1

1+ 519

.

One more time: 1+519

=2419 so we have

2+ 12419

=2+ 1924

=6724

To find the continued fraction for a given fraction, follow Euclid’s algorithm. Using the example in the text, here is the algorithm for 67 and 24:

67 = 2 x 24 + 1924 = 1 x 19 + 519 = 3 x 5 + 45 = 1 x 4 + 1

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4 = 4 x 1 + 0

Now if we divide through by the middle number on each line we get:

67/24 = 2 + 19/2424/19 = 1 + 5/1919/5 = 3 + 4/55/4 = 1 + 1/4

This forms the continued fraction.

Difference of two squares

The difference of two squares is exactly what it says it is: it is the difference between two square numbers. For example, 5 is the difference of the two square number 4 and 9, and can be calculated 32 – 22 = 9 – 4 = 5.

Algebraically, the difference of two squares is M2 – N2, which can be factorized as (M+N)(M-N). To see why, let’s multiply these two brackets together using the grid method:

You can see that the two terms +MN and –MN cancel out, leaving M2 – N2.

Diophantine equations

For Diophatine equations we are only interested in integer solutions. The most famous Diophantine equation is Fermat’s Last Theorem, which states that there are no integer solutions to the Diophantine equation xn + yn = zn for n > 2. If n = 2, then we have Pythagoras’ equation; there are infinite solutions to this.

Factorials

These numbers are made by calculating 1, 1 x 2, 1 x 2 x 3, 1 x 2 x 3 x 4, and so on, and are usually written as 1!, 2!, 3!, 4!, …

A different way of thinking about factorials is that they represent the number of ways of arranging n objects. So for example 3! Is the number of ways of arranging 3 letters:

abc, acb, bac, bca, cab, cba

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Fibonacci numbers

Each number in the sequence is the sum of the previous two, so starting with (0th term = 0) 1st term = 1 and 2nd term = 1, we get the famous sequence:

1, 1, 2, 3, 5, 8, 13, 21, …

This is sometimes written using a recurrence relation un+1 = un + un-1 which basically says the next term is the sum of the previous two.

Fractions – calculation methods

Equivalent fractions: When first getting used to equivalent fractions, you might find it easier to use a fraction wall to work these out like this.

However, you soon realize that multiplying or dividing the top and bottom of a fraction by the same number maintains the value of the fraction.

Dividing a fraction is called simplifying. If we keep dividing until we can’t go any further (without making decimals) then we say the fraction is in its simplest form. So 6/8 is not in its simplest form, but dividing top and bottom by 2 gives 3/4, which is in its simplest form.

Adding/subtracting fractions: Once you have mastered equivalent fractions, you will be able to add and subtract them. Notice that it is easy to add/subtract fractions when the bottom numbers are the same, such as 2/4 + 1/4 = 3/4 and 2/4 – 1/4 = 1/4.

So in order to add/subtract a pair of fractions, we just need to make the bottom numbers the same, which we can do using equivalent fractions. So in order to work out 1/6 + 3/4 we can just change it to 2/12 + 9/12, which is equal to 11/12.

Multiplying fractions: This is easier than adding/subtracting; all you have to do is multiply the tops together to get the top number and the bottoms together to get the bottom. For example, 2/3 x 3/4 = 6/12 (which simplifies to 1/2).

Dividing fractions: What does a divide b mean? It means ‘how many b’s are in a?’. So for example, 3 divided by ½ means ‘how many halves are in 3’, to which the answer is 6. How could we have calculated this? Noting that dividing by ½ is equivalent to multiplying by 2, we seek a connection between dividing and multiplying.

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Continuing this further, we note that dividing by 1/3 is the same as multiplying by 3 and so on, so to divide by 1/n, we multiply by n. Taking this one step further, we recognize this as the reciprocal, suggesting that dividing by a/b is the same as multiplying by b/a, which is indeed the case. So m/n divided by a/b is equivalent to m/n multiplied by b/a. For example, 4/3 divided by 81/64 = 4/3 x 64/81 = 256/243.

Fractions, decimals and percentages

Remember that percent (%) means out of a hundred… So 20% means 20 out of a hundred, or 20/100.

We can draw this like this:

From the picture you might also be to see that we could write 20/100 as 2/10 (or even 1/5).

So we can write any percentage as a fraction really easily.

We can change percentages and fractions to decimals using place value: 1% or 1/100 is a 1 in the hundredths column:

We usually write this as 0.01. Now we can keep going until we get to 9% = 9/100 = 0.09:

But when we get to 10% = 10/100, we can’t put all 10 in the hundredths column so we carry the one up into the next columns (tenths), so it looks like this:

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So we can write 10% = 10/100 = 0.10. After a while you notice that the two decimal numbers (in the tenths and hundredths columns) are just the percentage! So for example, 71% = 0.71.

Functions

Functions have an input and an output. You put numbers into a function, and other numbers come out. For example, if we put x = 3 into the function x2 we get the output 9. We write this as function as f(x) = x2, and if we put 3 into it, we write f(3) = 32 = 9.

Functions can be thought of as another way of writing the nth term for a sequence. For example, the sequence 1, 4, 7, 10, has nth term 3n – 2, but we could equally describe it as the outputs of the function f(n) = 3n – 2 after inputting numbers n=1, n=2, etc…

Putting n=5 in the function gives the 5th number in the sequence 3x5 – 2 = 13, so performing the same job as the nth term.

GCD

The greatest common divisor GCD, sometimes called the highest common factor (HCF) of two numbers is the largest number that divides into both of them. So for example, the GCD of 24 and 30 is 6.

Integer

Integers and the positive and negative whole numbers, including zero. The integer part of a decimal is the whole number part.

Logarithm (log)

Logs are functions that act as the inverse to powers. Consider the statement 32=9. We can write this using logs as log39=2. The subscript of the log is the base, and matches the base in the first statement. The log gives you the power you have to raise the base by, in order to get the value 9. So if we wanted to know the power that satisfies the equation 32=10, we would calculate log310.

Logs were invented by Napier to help with large calculations as they make multiplications into additions due to the log rule: log a×b=log a+log b. In the days before calculators they would have large books of log calculations called log tables.

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To see how this could help with large multiplication sums, consider the calculation 256 x 128. You could write log2256×128=log2256+log2128 which they could read from a log table as 8 + 7 = 15. Then they could look in the tables again to find which log matches the answer 15, which is 215 = 32768.

Lowest common multiple

The lowest common multiple (LCM) of two numbers is the lowest multiple shared by both numbers. For example, the LCM of 6 and 8 is 24. If the two numbers are relatively prime, the LCM is the product of the two numbers.

Mathematical induction

Mathematical induction is used to prove statements are true. I will show it using the example of Fermat numbers, to prove the ‘statement’ F0 x F1 x … x Fn-1 x Fn + 2 = Fn+1.

Remembering that the nth Fermat number is Fn = 2m+1 for some number m from the doubling sequence 1, 2, 4, 8, 16, … so the first few Fermat numbers are 3, 5, 17, 257, …

Step 1: Prove it is true for the first case. Here, we want F0 + 2 = F1 which is true as 3 + 2 = 5.

Step 2: Assume the statement to be true for the nth number. So we assume F0 x F1 x … x Fn-1 + 2 = Fn.

Step 3: Now, using our assumption, prove it to be true for the (n+1)th number. So we are trying to use Fn = F0 x F1 x … x Fn-1 + 2 to show F0 x F1 x … x Fn-1 x Fn + 2 = Fn+1.

Well, we have

F0 x F1 x … x Fn-1 x Fn + 2 = ( Fn – 2 ) x Fn + 2 (using our assumption) = (2m + 1 – 2) x (2m + 1) + 2 (for some number m from the doubles)= (2m - 1) x (2m + 1) + 2 = 22m + 1 (multiplying out the brackets)= Fn+1. (2m is the next number in the doubles)

So because the statement is true for the first case (step 1), and is true for every subsequent case (steps 2 and 3) then it is always true.

Mersenne numbers

Mersenne numbers, named after the French monk Marin Mersenne, are numbers of the form 2n – 1, so we have 1, 3, 7, 15, 31, 63, …

There are lots of prime Mersenne numbers (3, 7, 31, …) which occur at prime terms (i.e. n = prime) in the sequence (2nd, 3rd, 5th, …) and it can be shown that any Mersenne

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primes have prime n; this is used to find large primes such as the current largest prime 257,885,161 − 1.

Note that the Mersenne numbers are 1, 11, 111, … in binary.

Modular arithmetic (clock arithmetic)

Modular arithmetic is the study of remainders of numbers. They often reveal patterns in numbers we might not originally recognize.

We use modular arithmetic every day, when telling the time. As the hours get to 11, then 12, we do not then go to 13 o’clock, we instead reset to 1 o’clock. Hours of the day work in modulo 12, or mod 12 for short. This is the same as saying the number 13 has remainder 1 on division by 12.

We say a number is congruent to x (mod n) if that number has remainder x when divided by n. So the number 13 is congruent to 1 (mod 12). This is sometimes written as: 13≡1(mod 12).

nth term

Functions and sequences are closely related. Consider the sequence 2, 4, 6, 8, … this is also known as the even numbers, or the two times table. Each number in the sequence is called a term.

We could give a rule for working out numbers in this sequence: it is 2 x something. So the first term is 2 x 1, the second term is 2 x 2, the 5th term is 2 x 5 = 10.

Using any letter (say n) instead of the word something we could write this as 2 x n, or even more simply as 2n (ignoring the x sign). This is sometimes called the nth term of a sequence.

If we choose numbers for n we get that term in the sequence. For example, if we wanted the 5th term of the sequence, choose n to be 5 and then we have 2 x 5 = 10.

Nim

In standard two-player Nim, there are three piles of counters. Players take it in turns to take any counters from one of the piles. The winner is the one who takes the last counter.

The best way to discover how to win at Nim is to play it. As you play, you realize there are positions you want to get to on your turn, so that you leave your opponent in a losing situation; these are called safe positions.

In normal Nim we can get to safe positions by looking at the binary digits of each pile and getting an even number of each binary digit. For example, with 1, 5 and 10 counters (say) we have binary digits:

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8 4 2 1 1= 0 0 0 15= 0 1 0 110 = 1 0 1 0

Now there are an odd number of binary digits in the columns 2, 4 and 8. So as player 1 we can move to a safe position by taking 6 counters from the pile with 10 in, to give:

8 4 2 1 1= 0 0 0 15= 0 1 0 14 = 0 1 0 0

Now there is an even number of binary digits in each column and we will win from here if we play perfectly by continuing to adopt this strategy. Of course, if you’re opponent has found a safe position first then there is not much you can do about it!

Pascal’s triangle

Although Pascal didn’t invent the triangle (it was known to the Ancient Chinese), he wrote a book about it and it was given his name.

The numbers in each row are made by adding the two numbers in the row above.

There are many other patterns in there, such as the triangle numbers (1, 3, 6, 10, …) and tetrahedral numbers (1, 4, 10, 20, …) and the Fibonacci sequence can be found by summing along skew diagonals.

Pascal’s triangle occurs in lots of problems as it represents the numbers of ways of choosing things. For example, in the section on triangle numbers, I stated that the number of ways of choosing 3 (dots) from 6 is 20; you can test this by drawing or listing them all.

These numbers are often called combinations, and are usually written like this:

(63)=20, spoken as ‘6 choose 3’.

Place value

We use a place value number system; it is one of the first things you learn at school. Place value means that the place of each digit in a number represents its value.

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Using a blob to represent 1, here is the number 736 in our number system:

Hundreds Tens Units

This is a more efficient way of writing 736 than writing seven hundred and thirty-six blobs. Every column has a different value. The 6 in the units column represents 6 single objects. Moving to the left, the 3 represents three tens, or thirty blobs, and then the 7 represents seven hundred blobs. So this represents 700 + 30 + 6 = 736.

Note that each column can contain up to 9 blobs. When we get to 10 blobs in a column, this becomes one blob in the next column to the left. Of course, we don’t use blobs, we use 10 numerals 0 to 9 to represent 0 to 9 blobs in any column.

This compares with the Egyptian number system, which does not use place value.

Powers

Powers, or indices, or exponents in the US, are a short way of writing numbers. For example, 53 is just a short way of writing 5 multiplied by itself 3 times, so 53 = 5 x 5 x 5 = 125. 51 = 5 and 50 = 1, which makes sense as each power is 5 times bigger than the one before it.

Powers can also be negative (2-1 is the same as 1/2 ) and fractional (21/2 is the same as √2).

Prime factorization

Every whole number can be written as a unique product of prime factors. We can use factor trees to find the prime factors. So for example, 60 can be written as the product of its prime factors as 2 x 2 x 3 x 5.

Pythagoras’ theorem

Pythagoras Theorem says that if we find the area of the squares on the two shortest sides of a right-angled triangle (only) and add them together, it will be equal to the area of the square on the longest side of the triangle.

Reciprocal

The reciprocal of a fraction is just the upside down version of it: 2/3 and 3/2 are reciprocals.

If we multiply two reciprocals together we always get 1, for example 2/3 x 3/2 = 6 / 6 = 1.

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The reciprocal of a whole number n is 1/n. To see why, consider the whole number as a fraction n /1, then multiply them together: n /1×1/n=n/n=1.

Relatively prime

Two numbers that are relatively prime do not share any common prime factors. So 6 and 8 are not relatively prime (they have 2 in common) but 5 and 8 are.

Triangle numbers

The triangle numbers 1, 3, 6, 10, … are the ones we get by putting dots in triangles like this:

You can think of each term as the (partial) sum of the whole numbers:

1, 1 + 2, 1 + 2 + 3, …

The nth term of the triangle numbers (which is also the sum of the first n whole numbers) is:

n (n+1 )2

You can work out this formula as shown in the text, or by this little trick. Write out the sum of the first n numbers twice, one above the other like this:

1 + 2 + 3 + … + (n-1) + nn + (n-1) + … + 3 + 2 + 1

Adding each pair gives n lots of 1 + n, or n x (n+1). Then because this gives us twice what we need (we have just added two sets of the numbers), we just divide by two to get the formula.

This is a special case of the sum of an arithmetic series with first term 1 and last term n, so we could also work out the formula that way.

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Bibliography

Problem/puzzle books

The Inquisitive Problem Solver (Vaderlind, Guy, Larson)

Mathematics Galore (James Tanton)

Solve This: Math Activities for Students and Clubs (James Tanton)

Math Without Words (James Tanton)

What to Solve (Judita Cofman)

Linking Cubes and the Learning of Mathematics (Paul Andrews)

Lessons in Play (Michael Albert, Richard Nowakowski, David Wolfe)

Algorithmic Puzzles (Anay Levitin, Maria Levitin)

History of maths books

Euler: The Master of Us All (William Dunham)

The History of Mathematics (Jeff Suzuki)

Other interesting books and articles related to investigations in this book

Concrete Mathematics (Graham, Knuth, Patashnik)

The Book of Numbers (Conway, Guy)

The Strong Law of Small Numbers (Guy, article)

Proofs from The Book (Martin Aigner, Gunter Zeigler)

Proofs that Really Count: The Art of Combinatorial Proof (Arthur Benjamin, Jennifer Quinn)

On Numbers and Games (John Conway)

A Friendly Introduction to Number Theory (Joseph Silverman)

Proofs Without Words (Roger Nelsen)

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Winning Ways For Your Mathematical Plays (Elwyn Berlekamp, John Conway, Richard Guy)

Mathematics: The Man-Made Universe (Sherman Stein)

Ingenuity in Mathematics (Ross Honsberger)

Fair Game (Richard Guy)

The Enjoyment of Math (Hans Rademacher, Otto Toeplitz)

Your Move (David Silverman)

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