Chalkdust, Issue 10

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In this issue... 4

In conversation with Clifford Cocks We chat to the crypto chief

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Can computers prove theorems? Your Kevin Buzzard is in another castle

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Artificial music Carmen Cabrera-Arnau doesn’t play an instrument

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On √2 Yiannis Petridis is not a square

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42 54

Curiosities of linearly ordered sets Andrei Chekmasov is out of order Crocheting fractals Felix Stokes Felix Stokes

Felix Stokes

Felix Stokes

Felix Stokes

Felix Stokes

Secrets, surveys and statistics Paula Rowińska asks embarrassing questions

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50

4 1

59

They might not be giants Angela Brett might not be standing on their shoulders

3 8 17 22 24 31

Page 3 model

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Comic: The inverse homotopy by Tom Hockenhull

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How to make... ...tessellating shortbread

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The Goldwasser–Miacli cryptosystem by Zubair Junjunia

48 50

Crossnumber

53 56 58 60

Reviews

63 64

Chalkdust holiday snaps

What’s hot and what’s not Which mathematician are you? Dear Dirichlet A mathematical song The big argument: axiom of choice

Spotlight on: Pamela E Harris by Talithia Williams Puzzles Letters On the cover: Islamic geometry by Emily Maw Top ten: issues of Chalkdust autumn 2019

chalkdust Welcome to issue 10 of Chalkdust! For fans of the decimal system, this is a big deal. Personally, we’re most excited that the issue number no longer has a leading zero. But alas, it now has a trailing zero. The team Carmen Cabrera-Arnau Petru Constantinescu Thùy Dương “TD” Ðặng Eleanor Doman Sean Jamshidi Nikoleta Kalaydzhieva Jean Lagacé Emily Maw Johnny Nicholson Sam Porritt Tom Rivlin Sally Said Matthew Scroggs Belgin Seymenoğlu David Sheard Adam Townsend Albert Wood Cartoonist Tom Hockenhull Cover Adapted from watercolour and gouache painting by Samira Mian

d c a b l n e

chalkdustmagazine.com contact@chalkdustmagazine.com @chalkdustmag chalkdustmag chalkdustmag @chalkdustmag@mathstodon.xyz Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.

This issue, we are extremely proud to present a wide range of authors, including contributions from students and professors, professionals and hobbyists, from near and far. We believe that maths really is for everyone, and we are happy that our readership, authorship, and mothership reflects this. There are at least two running themes in issue 10. The first theme is cryptography: this issue features an interview with Clifford Cocks, who independently invented the RSA algorithm (see pages 4–7), and a brief introduction to probabilistic encryption (see pages 40– 41). Another recurring theme is music: you will find non-mathematical songs written by computers (see pages 18–21), and a mathematical song written by a non-computer (see page 24). You will also find a poem (see page 59), a heated argument (see page 31), and lots more (see page 1). In honour of this issue’s link to cryptography, we have decided to encrypt the final part of this editorial using a Vigenère cipher. Good luck! Dqfvb qdfmo rfawf fav odtb, qbfustimhhwt ouxxwfukbx ttwg. Mv dlgcg rdkh, mtahvtwzlb ch, cdcqvgih nu, crzeqgpapu mxbor, xgzzh. Kiesdwkjz rvuvid vzoiafo dpcfvk. Pod pzqh dkpvgw, bbfwadn murl, ecehmqgwvivg qr, imndliihr s, ooxci. Huw Evrasrhkv hvpu

Acknowledgements Big up to: all our authors for writing wonderful content; our sponsors for allowing us to continue making the magazine; Helen Wilson, Helen Higgins, Luciano Rila and everyone else at UCL’s Department of Mathematics; Matt Parker and Simon for their help with distribution; Matt Parker for his free advertising at TMiP; Tony Mann for inviting us to Greenwich Maths Time; Nazar Miheisi for signposting us to Cliff Cocks. ISSN 2059-3805 (Print). ISSN 2059-3813 (Online). Published by Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK. © Copyright for articles is retained by the original authors, and all other content is copyright Chalkdust Magazine 2019. All rights reserved. If you wish to reproduce any content, please contact us at Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK or email contact@chalkdustmagazine.com

chalkdustmagazine.com

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Eleanor Doman

If you model rabbits under ideal circumstances, you may ďŹ nd that the number of pairs of rabbits each month follows the Fibonacci sequence. In this case, ‘ideal circumstances’ is a euphemism for nonsense, as your assumptions would include blatant untruths such as ‘rabbits mate once a month every month except their ďŹ rst month alive’, ‘a pair of rabbits gives birth to exactly one pair of rabbits per month’, and ‘the hutch is inďŹ nitely big (and hence Starsky is very squashed)’. Fibonacci numbers, however, are not completely absent from nature. They accurately describe a vastly superior animal: the honeybee. Male bees (drones) come from unfertilised eggs, and so they only have one parent—the queen. Female bees (workers or queens) come from fertilised eggs and so have two parents—the current queen and a drone. If you follow a drone’s family tree backwards, you will see that a drone has:

♀ ♂ ♀ ♀ ♂ ♀ ♂ ♀ ♀ ♂ ♀ ♀ ♂ 13 great -grandparents ♀ ♂ ♀ ♀ ♂ ♀ ♂ ♀ 8 great -grandparents ♀ ♂ ♀ ♀ ♂ 5 great-great-grandparents 3 great- ♀ ♂ ♀ grandparents 2 grandparents ♀ ♂ 1 parent ♀ đ?&#x;Ś

đ?&#x;Ľ

the drone

♂

The number of ancestors of a male bee follows the Fibonacci sequence. Who would’ve expected that?!

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chalkdust

In conversation with

Clifford Cocks Clifford Cocks

Sean Jamshidi and Emily Maw

T

hroughout history, people have wanted to communicate in secret. But for a long time, the need for sender and recipient to agree on a way to encode their message (a ‘key’) meant that secure communication was costly, and mostly used by the military. But in the 1970s new mathematical ideas paved the way for public-key cryptography, a communication strategy that doesn’t rely on a mutually agreed key. If you’ve ever banked or shopped online then you’ve used public-key cryptography, most probably a type called the Diffie–Hellman protocol. (If you want to brush up on Diffie–Hellman, this is a great time to dig out your copy of Chalkdust issue 09 and read Axel Kerbec’s article ‘Hiding in plain sight’.) One of the lesser-known figures in the story of public-key cryptography is Clifford Cocks, a former chief mathematician at Britain’s GCHQ (the Government Communication Headquarters). Cliff’s relative anonymity is because, due to the secretive nature of his employer, his contribution was not made public for 24 years. We caught up with him via video call to find out what it felt like to have cracked the code, but kept it secret.

Finding the right factors In 1973, Cliff Cocks had just joined GCHQ in Cheltenham. He had completed his undergraduate degree at Cambridge and had begun a PhD at Oxford, working in number theory, before deciding that academia was not right for him. One afternoon Cocks’ mentor at GCHQ, Nick Patterson, introduced him to the concept of public-key cryptography as a “really cool problem, but one that nobody had really got anywhere on.” The idea had first been conceived a few years earlier by chalkdustmagazine.com

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chalkdust James Ellis, another GCHQ employee, and the concept was simple yet groundbreaking. What if the intended recipient of the secure message (say, Alice) has two keys, a public one that she gives out freely to anybody, and a private key that nobody else knows. The sender, Bob, can use the public key to encrypt his message, then send it on to Alice, who can decode it using her private key. Since one needs the private key to decode the message, only Alice can read it, but her and Bob never have to agree on what the private key is. As Cocks puts it, “the two sides can have a secure conversation starting with absolutely nothing in common.” The issue that Ellis had faced was, how could this ‘asymmetric’ encryption work in practice? Cocks was intrigued by the problem, and found himself turning it over in his head at home that evening. He quickly made progress. “Luckily, Patterson had described the problem to me in a very mathematical way... I knew that what I needed was a mathematical function that was easy to do one way but hard to undo. I’d been working in number theory, so factorising seemed like something that could work.” Overnight, he had laid the groundwork for a feasible method. The basic idea behind Cocks’ method is that the private key is two large prime numbers, and the public key is their product (which Alice can easily compute). To decrypt the message, one needs to spot the original primes just by looking at the product. If the primes are large enough, then this is very hard to do.

Flickr user Rhian, CC BY 2.0

Multiplying large primes is like mixing paint; once it’s done it’s hard to undo it or to tell which colours were used.

With great modesty, Cocks describes his discovery as a case of being “in the right place at the right time, and having the problem explained to me in the right way.” He suggests that his freshness at GCHQ was an advantage: “I hadn’t quite realised how many people had worked on it in the past and not really got anywhere.” Some of those people who hadn’t got anywhere thought that Ellis’ unconventional idea was “crazy”, and that there had to be “some reason why it wasn’t possible”, and it wasn’t until Cocks approached the problem in a more mathematical way that any progress was made. Once the work had been done, Cocks says he had no idea how significant it would prove. “In those days, cryptography was only really used by governments. Very few people were interested in it professionally outside that.” Of course, all of that changed with the internet and with the need to exchange encrypted information—in particular, information relating to online payments—between people that have never met. “I certainly did not foresee where technology would go in the future!”

Coming in from the cold GCHQ were impressed with Cocks’ idea and shared it with their intelligence allies but, due to the secrecy of the organisation, it was never made public. A couple of years later MIT engineers Ron Rivest, Adi Shamir and Leonard Adleman independently came up with the same solution, and published their work in the journal Communications of the Association for Computing Machinery. The algorithm is now known as RSA in their honour, and is still one of the most widely-used encryption methods today. Astonishingly, the Diffie–Hellman protocol was also discovered at GCHQ 5

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chalkdust in secret, by Malcolm Williamson, Cocks’ colleague and childhood friend. In fact, GCHQ decided not to implemented Cocks’ idea at that time. “They thought very seriously about how to do it, but came to the conclusion, quite correctly at the time, that it was too expensive.” Ellis had originally conceived of public-key cryptography as a way to save money on the cost of sharing keys between senders and recipients, which was fast becoming the stumbling block to secure communication. “If you think about what computers were capable of at the time, (public-key encryption) would have to have been implemented on special-purpose hardware, and even then it would have been very slow. Given that the aim was to save money, it wouldn’t have been feasible.” Fast forward to 1997, 24 years after Cocks first made the discovery. “I was going to give a talk in Cirencester, presenting some new idea about public-key cryptography. We thought inevitably there was going to be questions like ‘had I been involved in developing early versions of the algorithm?’” To avoid this speculation, it was decided that GCHQ would declassify Cocks’ work. After Cirencester, Cocks gave a talk to the British Society for the History of Mathematics. The story was picked up by the author Simon Singh, and it went on from there. But how did it feel to finally ‘go public’? “I wasn’t sure how it was going to go. There’s some reassurance about being in the shadows.” Perhaps Cocks’ announcement was made easier by the fact that, for many years, rumours had been circulating about GCHQ’s work. “A lot of academics who I’d worked with would ask me the question, and it was a bit awkward really! So it was more about clearing the air.”

Not-so-secret communication Although Cocks always maintained a close relationship with the academic world, he soon realised that there just wasn’t enough time to keep up-to-date with all of the relevant theoretical work. This is exemplified by the fact that he only learned of Rivest, Shamir and Adleman’s rediscovery of public-key encryption when he happened to read a feature on it in a copy of Scientific American one lunchtime. It’s for this reason—the need to bring in fresh ideas from academics with their finger on the pulse of current research—that he pushed hard to strengthen links between the intelligence community and academia during his time as chief mathematician, first sponsoring research positions at British universities and later setting up the Heilbronn Cocks with a copy of ChalkInstitute for Mathematical Research in Bristol, where academics dust issue 09. spend half their time pursuing projects directed by GCHQ. As somebody who made the transition from university to government, he was keen that the institute was compatible with academia and “that there was the right balance of career development and providing pointers on difficult problems.” There are, however, some important differences between working at GCHQ and a university. “GCHQ is much more collaborative, although academia is getting better. You don’t stand up in a chalkdustmagazine.com

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chalkdust seminar and say, ‘I’ve got this half-baked idea that I can’t make work’. But at GCHQ you want to get solutions, and you don’t really care if they’re yours, or a colleague’s.” And once you’ve solved the problem, there’s no chance of getting public recognition for it (at least, not for a few years). “The focus of what you do is internal, but it’s not like we’re going to solve the Riemann hypothesis and keep it to ourselves. It’s more about the applications.” So learning to give up the glory, and in particular to share work that isn’t finished, is something that academics need to be trained to do when they make the move into intelligence. But there are some positive aspects to the secretive life of a GCHQ employee. “You can’t take your work home with you, so there’s a separation between work and home life. Well, apart from those problems that stick in your head and you can’t get rid of them!” Does Cocks ever wonder what his life would have been like if he’d continued with his PhD at Oxford, working on elliptic curves under Bryan Birch? “It’s interesting to speculate. It certainly would have been a slower career.” Upon leaving Oxford, GCHQ seemed like a natural choice for a job, being one of the only places that recruited people to do “something approximating pure mathematics”. His decision to leave his PhD was prompted by the realisation that, in his field of number theory, it Public-key cryptography has would have taken a lot of work until he could start doing somemade online security possible. thing ‘useful’. “I didn’t want to spend three years just getting to the coal face.” Having come up with a solution to public-key encryption within six weeks of starting his new job, it’s fair to say that he had cracked that coal face wide open, and taken the first peek at a rich seam that is still being mined today. Sean Jamshidi and Emily Maw Sean and Emily are PhD students at University College London. Neither of them has solved any famous mathematical problems in secret.

a @sean_jamshidi a @emilyharrietmaw Four hexagons by Catriona Shearer

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?

There are four regular hegagons. The area of the blue hexagon is 12. What is the area of the red hexagon?

You can find Catriona and loads more of her puzzles on Twitter a @Cshearer41.

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Maths is a fickle world. Stay à la mode with our guide to the latest trends.

& WHAT’S

WHAT’S

HOT NOT Issue 10

HOT

We’re so excited about releasing issue 10!

Issue 3628800 NOT Wait another 1.8 million years for this one? No thanks.

Lean

Agree? Disagree? a @chalkdustmag b chalkdustmag l chalkdustmag f chalkdustmag

Manually proving theorems

See pages 9–16.

Yawn.

HOT

Disliking the food in the university canteen

NOT

Recycling jokes from issue 06

NOT

Proroguing the canteen to take back control of the menu This joke has been deemed null and void by Peter Dirichlet, president of the Chalkdust supreme humour court.

HOT

HOT

Writing new content is hard.

Sudoku

NOT

They’re all exactly the same. I say Su-no-ku.

Really big numbers Shut up, Graham.

Really big triangles

NOT HOT

Bonus points if they’re scalene. More free fashion advice online at d chalkdustmagazine.com

Pictures Sudoku: Héctor Rodríguez, CC BY-NC 2.0.

chalkdustmagazine.com

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chalkdust

Can computers prove theorems? Kevin Buzzard

H

ow do we prove that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś? At school, this might have been taught to you in the following way. You were given a box of little plastic cubes, two cubes were put in one of your hands and then two more cubes in the other, and you were challenged to count how many cubes you had in total.

But this doesn’t really prove that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś: it proves that đ?&#x;¤ plastic cubes + đ?&#x;¤ plastic cubes = đ?&#x;Ś plastic cubes. You could try it again with pencils and show that đ?&#x;¤ pencils + đ?&#x;¤ pencils = đ?&#x;Ś pencils, and after you’ve tried it with sufficiently many things you might become convinced that there is an underlying pattern. But is this a proof that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś? Proving it like this feels a bit like an experimental science—it works with cubes, it works with pencils, and this is evidence that it works in general. I think we are all pretty confident that, whatever the actual rules of maths are, they probably don’t mention pencils. But what are the rules of maths? Are there any rules at all, or do we all just have some inbuilt intuition as to what constitutes a valid mathematical argument? Before the 1900s, people worked intuitively, and there was broad agreement as to what constituted a correct argument. But as people began to do more complex and abstract mathematics, this approach became problematic, because people’s intuitions could differ. Ask a room full of teenagers whether đ?&#x;˘.đ?&#x;Ťđ?&#x;Ťđ?&#x;Ťđ?&#x;Ťđ?&#x;Ťđ?&#x;Ťđ?&#x;Ť... = đ?&#x;Ł, and you will get different opinions. This is because different people have different intuitions about what the real numbers actually are. Differences of opinion as to whether arguments were valid forced mathematicians into actually writing down an official rulebook: the axioms of maths. 9

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chalkdust So how do we prove đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś just using the axioms of maths and no plastic cubes? Come to think of it, how do we define 2, and how do we define +? The Italian mathematician Giuseppe Peano had some ideas about this in 1889, and his ideas are now called Peano’s axioms.

Peano’s axioms Peano wanted to define a number system called the natural numbers, which for him started at zero. So the question is how to find axioms that model the set {đ?&#x;˘, đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ, đ?&#x;Ś, ...}. Here is how Peano did it. He defined a constant zero (to be thought of as 0 in our mental model), and a successor function đ?‘† , which takes a natural number đ?‘› as input, and spits out another natural number đ?‘†(đ?‘›) (to be thought of as the number after đ?‘›). Here then are Peano’s axioms for defining the natural numbers. Axiom 1 Axiom 2 Axiom 3

Giuseppe Peano zero is a natural number. If � is a natural number, then its successor �(�) is also a natural number. That’s it.

In a computer language, these axioms might look something like this: inductive nat | zero : nat | S (n : nat) : nat We’ll say more about this computer language later; before we do, let’s play with these axioms for a bit, and see if we can make a few natural numbers. At the beginning, we have no natural numbers at all, so axiom 2 is useless to us. However we can use axiom 1 and this lets us make the natural number zero. Now axiom 2 comes into play, and we can use it to make a new natural number đ?‘†(zero). A good name for this new number would be one. We can similarly define two to be đ?‘†(one), otherwise written as two âˆś= đ?‘†(one), we can set three âˆś= đ?‘†(two) and so on. Here’s the computer language again, taking us up to four. definition definition definition definition

one := S(zero) two := S(one) three := S(two) four := S(three)

Each natural number can be expressed in a reduced form đ?‘†(đ?‘†(đ?‘†(...(đ?‘†(zero))...) and two natural numbers were defined to be equal by Peano if and only if their reduced forms were the same. This was how Peano ruled out things such as đ?&#x;˘ = đ?&#x;Ł.

Note that we’ve defined four to be �(three), but we can unravel the definition completely, and see that we could have defined four to be �(�(�(�(zero)))) (note that there are four occurrences of � in the definition of four). In fact, it is not hard to convince yourself (because of axiom 3) that the only way of making natural numbers is by starting with zero and then applying � some number of times; that’s all there is. chalkdustmagazine.com

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chalkdust We now have most of the players in our equation two + two = four; all that is missing is the definition of addition. Of course addition is not a number, it is a function.

DeďŹ ning functions Before a child learns to add, they learn to count. They are challenged to do things such as counting to ten and then to one hundred. They realise that they can in theory get to any number just by starting at zero and counting. This idea is what Peano axiomatised. But once a child’s counting is solid, they are ready to learn addition, and this is what we will define next. The addition function add takes two natural numbers đ?‘š and đ?‘› as input, and it outputs their sum add(đ?‘š, đ?‘›), a number more commonly written as đ?‘š + đ?‘›. Before we make the general definition, let’s step back, forget about axioms, go back to the usual notation for numbers, and think about what we know already about addition. Let’s see if we can isolate the ideas which capture the concept of addition perfectly. Let’s consider the special case where đ?‘› = đ?&#x;˘. What should đ?‘š +đ?&#x;˘ be? We are trying to define addition which agrees with our intuitive notation: of course đ?‘š + đ?&#x;˘ should be đ?‘š, so this case is easy.

Now here’s a more interesting question. Say we’re building mathematics from scratch, and we’re currently halfway through defining the addition function. The answers to some addition questions have been defined already, but others have not. Say đ?‘š is a number and we know that đ?‘š + đ?&#x;§đ?&#x;Ş = đ?&#x;Ťđ?&#x;¤, but we haven’t defined đ?‘š + đ?&#x;§đ?&#x;Ť yet. What should we define đ?‘š + đ?&#x;§đ?&#x;Ť to be? We could solve for đ?‘š and work everything out, but there’s a quicker way! We know that đ?‘š + đ?&#x;§đ?&#x;Ť is one more than đ?‘š + đ?&#x;§đ?&#x;Ş, so if đ?‘š + đ?&#x;§đ?&#x;Ş is 92 then đ?‘š + đ?&#x;§đ?&#x;Ť should be 93. Or as Peano might say, if đ?‘š + đ?&#x;§đ?&#x;Ş = đ?&#x;Ťđ?&#x;¤, then đ?‘š + đ?‘†(đ?&#x;§đ?&#x;Ş) = đ?‘†(đ?&#x;Ťđ?&#x;¤). And of course 58 is not special: if đ?‘š + đ?‘› = đ?‘Ą then đ?‘š + đ?‘†(đ?‘›) should equal đ?‘†(đ?‘Ą).

We have isolated two general principles of addition here: Rule 1 Rule 2

đ?‘š + đ?&#x;˘ = đ?‘š. If we have defined đ?‘š + đ?‘› already, then đ?‘š + đ?‘†(đ?‘›) equals đ?‘†(đ?‘š + đ?‘›).

In code it looks like this:

def add : nat → (nat → nat) | m zero := m | m (S(n)) := S(add m n) Our first rule tells us how to define đ?‘š + đ?&#x;˘ for every đ?‘š. Because đ?&#x;Ł = đ?‘†(đ?&#x;˘), our second rule now tells us how to define đ?‘š + đ?&#x;Ł for every đ?‘š. Because đ?&#x;¤ = đ?‘†(đ?&#x;Ł), we can use the second rule again to define đ?‘š + đ?&#x;¤ for every đ?‘š. This is looking promising! But can we prove that we have defined đ?‘š + đ?‘› for every value of đ?‘›? This looks like it should be a proof by induction. The first rule gives us the base case, that đ?‘š + đ?&#x;˘ is defined. The second rule gives us the inductive step: if đ?‘š + đ?‘› is defined, then so is đ?‘š + đ?‘†(đ?‘›).

But how do we know that induction works for Peano’s natural numbers? It’s because of Peano’s axiom 3. Every natural number is either zero, or the successor of a number that was born before it was, and that’s it. Hence if we define something for zero, and if we define it for �(�) given that it’s 11

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chalkdust defined for đ?‘›, then we have defined something for every natural number. We can conclude that we have now defined addition! Now let’s test it, and let’s see if we can prove that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś.

đ?&#x;¤+đ?&#x;¤=đ?&#x;Ś

Again let’s use usual mathematical notation for numbers in this section, remembering that they are really all defined in terms of zero and đ?‘† . We can now attempt to give a complete proof that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś. The problem with addition is that we only defined đ?‘š + đ?&#x;˘ and đ?‘š + đ?‘†(đ?‘›), so if we want to work out something like đ?‘š + đ?&#x;¤ then we have to rewrite đ?&#x;¤ as đ?‘†(đ?&#x;Ł) and then use the fact that đ?‘š + đ?‘†(đ?&#x;Ł) = đ?‘†(đ?‘š + đ?&#x;Ł) by definition. Here is a complete proof that đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś. 2 + 2 = = = = = = =

2 + S(1) S(2 + 1) S(2 + S(0)) S(S(2 + 0)) S(S(2)) S(3) 4

--------

by by by by by by by

definition definition definition definition definition definition definition

of of of of of of of

2 "+ S(n)" 1 "+ S(n)" "+ 0" 3 4

And here is how it looks on a computer (using the usual notation + for add): theorem two_add_two : two + two = four := begin refl end Wait—whatever does refl mean? And how come the computer’s proof is shorter than ours?

Proving theorems on a computer Lean is free, open source software being developed by a team led by Leonardo de Moura at Microsoft Research. Lean is a programming language whose commands correspond to the axioms of mathematics. Lean starts off by knowing how to do maths, but not knowing any maths at all. The code we saw above is valid Lean code. What is interesting is that after we taught Lean about natural numbers and addition, Lean figured out how to prove various things like đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś all by itself. Here refl is a tactic, and what it does is that it simpifies both sides as much The tactic select menu in a fictional game as it can and then checks that they have become the for the Sega Master System. same. If this works, the proof compiles and we’ve proved our theorem. If it doesn’t work, the tactic fails and we will need to try another tactic. Lean has a bunch of tactics built in, many of them far more powerful than refl. chalkdustmagazine.com

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chalkdust Lean is not a completely automatic theorem prover though—it often needs help from humans to prove theorems. Lean is an interactive theorem prover. Think about it like this. Lean is a framework which turns mathematical statements into levels of a computer game. Any maths theorem, from đ?&#x;¤ + đ?&#x;¤ = đ?&#x;Ś to Fermat’s Last Theorem, when formalised in Lean, becomes a level of the game. If you manage to use Lean’s tactics to prove a theorem, you have solved the level. My favourite computer game used to be the Zelda series but now my favourite computer game is Lean. Let me show you some more easy Lean levels. Let’s think about some more simple properties of addition, and see how we can prove them using Lean. By our definition of addition, we know that đ?‘› + đ?&#x;˘ = đ?‘› for every integer đ?‘›. But what about đ?&#x;˘ + đ?‘›? We might instinctively just say that this is obviously đ?‘› too, because this is immediately obvious using our “real worldâ€? model of addition. But this is not how the game works! The fact that đ?‘Ž + đ?‘? = đ?‘? + đ?‘Ž is a level in our game, but it’s quite a tricky one for beginners, and we have not got to it yet; it’s like the boss level for the Addition World level set in Lean. So let’s get back to đ?&#x;˘ + đ?‘›. Recall that the definition of đ?&#x;˘ + đ?‘› depends on whether đ?‘› is zero or a successor, and so we are going to have to prove đ?&#x;˘ + đ?‘› = đ?‘› by induction.

The final boss in Addition World.

Base case: đ?‘› = đ?&#x;˘. Here we have to prove đ?&#x;˘ + đ?&#x;˘ = đ?&#x;˘ and this is true by definition of + đ?&#x;˘, because đ?‘› + đ?&#x;˘ is defined to be đ?‘›.

Inductive step: let’s assume the hypothesis đ??ť âˆś đ?&#x;˘ + đ?‘‘ = đ?‘‘ and try and prove that đ?&#x;˘ + đ?‘†(đ?‘‘) = đ?‘†(đ?‘‘). In fact this is not too hard: đ?&#x;˘ + đ?‘†(đ?‘‘) = đ?‘†(đ?&#x;˘ + đ?‘‘) by definition of + and this equals đ?‘†(đ?‘‘) by the inductive hypothesis. We are done! In Lean this proof looks like this:

theorem zero_add (n : nat) : zero + n = n := begin induction n with d H, { -- base case refl -- true by definition }, { -- inductive step show S (zero + d) = S(d), rewrite H, -- H is the hypothesis that zero + d = d -- Lean finishes the proof automatically } end This is the last Lean proof which I will give here. The reason is that for proofs any more complex than this, it gets hard to follow them on paper. The way to understand these proofs best is to 13

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chalkdust look at them in Lean itself. If a proof were running in Lean on a computer then you could click anywhere on the proof and see Lean’s state (what it knows and what it is trying to prove). For example, just after the show tactic in the above proof, Lean’s state looks like this: 1 d H ⊢

goal : nat, : zero + d = d S (zero + d) = S d

It’s not hard to get Lean running on your computer— you don’t even have to install it, you can just use it through a web browser. For more details, go to my blog d xenaproject.wordpress.com and search for ‘Chalkdust’. There are links there that take you straight into Lean’s Addition World, which contains a few relatively simple theorems/levels about addition on Peano’s natural numbers.

Note that Lean writes S d not S(d), this is just some cool computer science thing that they do in functional programming languages, and Lean is a functional programming language.

If however you still prefer pen and paper, I will tell you that the next level in Addition World is to prove that addition is associative, ie that (đ?‘Ž +đ?‘?)+đ?‘? = đ?‘Ž +(đ?‘? +đ?‘?) for all natural numbers đ?‘Ž, đ?‘? and đ?‘? . Try proving it by induction on đ?‘? . Once you’ve done this, you could try proving addition is commutative, ie, that đ?‘Ž + đ?‘? = đ?‘? + đ?‘Ž. That level is the boss level for Addition World, so once you’ve done it you can move onto the next set of levels, called Multiplication World. How would you define multiplication? Can you prove đ?&#x;Ł Ă— đ?‘š = đ?‘š? What about đ?‘Ž Ă— (đ?‘? + đ?‘?) = đ?‘Ž Ă— đ?‘? + đ?‘Ž Ă— đ?‘? ? How would you define exponentiation? Can you prove (đ?‘Žđ?‘? )đ?‘? = đ?‘Žđ?‘?Ă—đ?‘? ? Note that exponentiation is not associative, in contrast to addition and multiplication. Doing these levels on paper can become a bit tricky; it’s much easier to do them within the software. There are several worlds for Peano’s natural numbers: there is an Inequality World for example, which has got some pretty tricky levels in. And then on top of that is a Diophantine Equation World, which has a bunch of levels about solving equations Multiplication World, featuring the notoriin natural numbers and contains such levels as Ferous exponentiation ghost. mat’s Last Theorem. Nobody has ever solved this level in Lean, or in any of the other computer proof verification systems either. Note that the level is extremely easy to make: theorem FLT (a b c n : N) (h : n > 2) : a^n + b^n = c^n → a * b = 0 := begin -- insert millions of lines of code here end It’s just very very very hard to solve. chalkdustmagazine.com

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Going further It’s all very well answering questions about the natural numbers. But what other kinds of pure mathematics can be done in this way? I believe that all of pure mathematics can be done in this way. Groups, rings, differential equations, complex manifolds, you name it. Lean’s maths library started in 2017 and is growing and growing. When I started learning about Lean in 2017 the maths library didn’t even have the complex numbers in it, but it did have the real numbers, and all the levels in Real Number World had been solved (computer scientists call this ‘making an API for the real numbers’) so I added the complex numbers myself: structure complex : Type := (re : R) (im : R) I now had to define addition, subtraction and multiplication and division of complex numbers, and prove basic things such as the fact that multiplication was commutative. All the results I needed about the real numbers were already proved however, so the task was not hard; Basic Complex Number World was a bunch of pretty easy levels. Lean now has a bunch of undergraduate level mathematics, including groups, fields, vector spaces, rings and modules, metric spaces and topological spaces, bilinear maps and so on. The library is growing really quickly. In 2018, I worked with a team of undergraduates at Imperial College London to define schemes in Lean. Schemes were discovered (or invented, depending on your philosophy) by Alexandre Grothendieck in the late 1950s and their discovery heralded the beginning of modern algebraic geometry. Schemes are normally taught at masters or beginning PhD level—but they’re just a definition really, like groups or the natural numbers. In 2011, the brilliant young German mathematician Peter Scholze discovered the notion of a perfectoid space. Scholze used perfectoid spaces to prove many new mathematical theorems, and in particular he made some great advances in the Langlands Philosophy, a series of conjectures that lie at the heart of modern number theory. Scholze was awarded a Fields medal, one of the highest honours in mathematics, in 2018. In 2019, Patrick Massot, Johan Commelin and I formalised the definition of a perfectoid space in Lean. It took well over 10,000 lines of computer code, a substantial part of which involved There are lots of levels that still need to be solving some very cunning levels in Topological Ring unlocked World, the computer game version of chapters 2 and 3 of Topologie Générale by Nicolas Bourbaki. Within two years Lean has gone from not knowing what the complex numbers are to knowing some very modern and fashionable mathematics, and one can ask what the future holds for this software. Perfectoid Space World is now unlocked in Lean, and the master puzzler Scholze has set some pretty fiendish levels in there. Unfortunately, we might have to unlock several more worlds before we are powerful enough to solve some of the serious levels in Perfectoid Space World. 15

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chalkdust Nobody really knows how far things can go. We mathematicians are making a database of theorems. Undergraduates from across the world are contributing. We are teaching Lean a standard undergraduate maths syllabus and then we will teach it more. It’s really good fun and anyone can join in. Occasionally we mathematicians find that we need a new tactic—for example proving that (đ?‘Ž+đ?‘?)đ?&#x;Ľ = đ?‘Žđ?&#x;Ľ + đ?&#x;Ľđ?‘Žđ?&#x;¤ đ?‘? + đ?&#x;Ľđ?‘Žđ?‘? đ?&#x;¤ + đ?‘? đ?&#x;Ľ just from the axioms of a ring is surprisingly difficult! But computer scientists are also involved. We asked the computer scientists if they could make us an “expand out the brackets and tidy upâ€? tactic, and they did. Any mathematical technique that can be explained as an algorithm can be turned into a tactic in Lean. Computer scientists are interested in writing AI programs that can beat humans at things. Computers can currently beat humans at chess, Go and StarCraft. Will computers one day be able to beat us at mathematics? Machine learning algorithms can learn many things, but they work best with a database. We mathematicians are making this database using Lean. Software like this can also be used for teaching. If students learnt the language which Lean uses, university lecturers could set homework requiring proofs in Lean, and students could write their solutions in Lean. One advantage of this for the students would be that they would know immediately if their proofs were correct or incorrect, giving them instant feedback. It would also be a neat new way for the students to learn mathematics. The advantages for lecturers like me is that I would not have to mark the work—the computer could mark it for me. However, if computers do get better than humans at proving theorems, I might be out of a job anyway. Kevin Buzzard Kevin is a professor of pure mathematics at Imperial College London. He blogs about Lean in his undergraduate-centred blog at d xenaproject.wordpress.com, and tweets about it at a @XenaProject. The Xena Project meets every Thursday evening during term time in the computer room of the maths department at Imperial College; bring a laptop. He is currently writing a book for undergraduate mathematicians about how to learn Lean. You can also find him on the Lean chat at d leanprover.zulipchat.com where you can pester him to finish the book.

Want more retro gaming? Be the master of your own universe in Reproduce or die, a variation of John Conway’s Life. Discover guns, gliders and hats at:

d chalkdustmagazine.com/game chalkdustmagazine.com

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Which mathematician are you?

START

YOU ARE

Pythagoras

โ Is 2 irrational?

NO

<latexit sha1_base64="DDXJc7RbexDc1e8H/lQN5x0VZTg=">AAACBXicdVBNS8NAEN3Ur1q/qh69LBbBiyGpFitFKXjxWMVWoQmy2W51cbOJuxOhhJ79BV71F3gTLx48+xMEr/4Pt2kFFX0w8Hhvhpl5QSy4Bsd5s3Jj4xOTU/npwszs3PxCcXGppaNEUdakkYjUaUA0E1yyJnAQ7DRWjISBYCfB5f7AP7lmSvNIHkMvZn5IziXvckrASG1PXylIvZpX658VS469U61Uyi7OyNbOiFScKnZtJ0Opvrv3bL1uvDfOih9eJ6JJyCRQQbRuu04MfkoUcCpYv+AlmsWEXpJz1jZUkpBpP81O7uM1o3RwN1KmJOBM/T6RklDrXhiYzpDAhf7tDcQ/Paa5hF/LoVv1Uy7jBJikw93dRGCI8CAS3OGKURA9QwhV3JyP6QVRhIIJrmBy+Xoe/09aZdvdtMuHJqAjNEQeraBVtI5ctI3q6AA1UBNRFKFbdIfurRvrwXq0noatOWs0s4x+wHr5BEldnP4=</latexit>

YOU ARE

YES

Georg Cantor

YOU ARE

How do you like your numbers?

UNCOUNTABLE YOU ARE

I PREFER

LET N

๐ บ How do you prove Fermatโ s Last Theorem?

YOU ARE

Andrew Wiles

YOU ARE

Gottfried Wilhelm Leibniz

FOR

p < 100 CANโ T DO IT

Did you invent calculus?

YES

PRIME

I PREFER SHAPES

Isaac Newton

Pierre de Fermat

YOU ARE

LOVE

YOU ARE

Do you like ?

ONLY SPHERICAL YOU ARE

NO

ร variste Galois

YOU ARE

Felix Klein

NOT BENDY ENOUGH

Why not?

Sophie Germain

Grace Chisholm

NO METRIC

How many jelly beans are in this jar?

1729

โ ยนโ โ โ

NOT SPIKY ENOUGH YOU ARE

Bernhard Riemann

YOU ARE

Benoit Mandelbrot

YOU ARE

Srinivasa Ramanujan

Pictures: Pythagorasโ theorem: Wikimedia user Petrus3743, CC BY-SA 4.0. Jellybeans: Flickr user cbgrfx123, CC BY-SA 2.0. Apple: Wikimedia user Richtom80, CC BY-SA 3.0

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Artificial music Carmen Cabrera-Arnau

O

ut of all the words in the English dictionary, art is possibly the one with the most debatable definition. In his 1897 book What Is Art?, Russian writer Leo Tolstoy argued that “art begins when a person, with the purpose of communicating to other people a feeling they once experienced, calls it up again within themself and expresses it by certain external signs”. An important aspect in Tolstoy’s argument is that of the artist’s sincerity—that is, the extent to which the artist has experienced the feeling that they are expressing—which is crucial in determining the appreciation of the work by others. Contrary to Tolstoy’s belief is the one popularised by the French writer Théophile Gautier in the early 19th century, summarised in the slogan l’art pour l’art—art for art’s sake. For Gautier, the intrinsic value of a work of art has to be completely detached from any sort of sentimental, social or moral context. New technologies add a layer of complexity to the old and neverending discussion about what should be considered art. What would the conversation between Tolstoy and Gautier be like after having been presented with one of Emmy’s musical compositions? Emmy, short for ‘Experiments in Music Intelligence’, was created in 1981 by David Cope, nowadays professor emeritus at the University of California, Santa Cruz. Cope, who was suffering from composer’s block, wanted to build software able to generate new material in line with his own pieces, using these pieces as the main input for the software. However, due to the lack of personal works, he started by taking the pieces of various classical composers as the input for his computer programs instead.

chalkdustmagazine.com

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chalkdust After spending some time perfecting Emmy, Cope was able to produce, in a matter of minutes, thousands of new instances of music in JS Bach’s style. This resulted in the 1993 release of Bach by Design, one of his several computer-generated music albums. Since Cope’s days, music-generating systems using artificial intelligence have experienced big advances. Nowadays, there are all sorts of user-friendly systems: IBM Watson Beat, Google Magenta’s NSynth Super, Jukedeck, Melodrive, Spotify’s Creator Technology Research Lab, Amper Music, and so on. Some music systems, like Amper, have explicitly been taught the rules of music theory. However, most AI music systems use artificial neural networks to generate output. The neural networks identify patterns from the multiple samples of source material they are fed with. These patterns are then used to create new music in the form of an audio file or a music A chorale harmonisation or a chorale is a muscore. While some systems will simply create sical piece traditionally intended to be sung a melody from a given note, others are able to by a congregation during a German Protesharmonise a given melody. tant service. It is often written for soprano, alto, tenor and bass. The soprano is the voice For a taste of what AI is capable of doing, you that holds the melody, which is usually a can have a look at the Google Doodle from 21 Lutheran hymn tune, while the other three March 2019, celebrating Bach’s 334th birthday. voices provide the harmony. Coconet is the machine learning model that makes this Doodle work. Trained with a relatively small dataset of 306 choral harmonisations by Bach, Coconet can harmonise a melody entered by the user in Bach’s contrapuntal style in a matter of seconds. The mechanisms behind the Doodle are explored in the following section.

Coconet in a nutshell Coconet’s task involves taking incomplete musical scores and filling them up with the missing material. For the result to be loyal to Bach’s style, Coconet needs to first be trained to know what is the ‘right’ style. This training is done by randomly erasing some notes from the original chorales composed by Bach and asking Coconet to reconstruct the erased notes. A rank is given to quantify the accuracy of Coconet’s version with respect to Bach’s. Coconet will then be encouraged to repeat high-ranked guesses in future reconstructions of incomplete music scores, while trying to avoid low-ranked guesses. So how is the music extracted from probability distributions? One could think naively that it is OK to just pick the pitch which corresponds to the highest probability assigned to the missing notes for each voice independently. However, Bach chorales are all about harmony and harmony is all about interactions between notes; the melodic lines of the different voices cannot be considered in isolation. To account for these interaction effects, there are several solutions. Perhaps the most obvious one would be to assign the highest probability pitches to one of the voices, and then feed Coconet with this new version of the incomplete chorale. The model would update the probability distributions 19

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chalkdust for the other voices. The process could then be iterated until all the voices are complete. Although it is simple, this solution is not ideal; very different results might be obtained depending on which voice is completed first. Coconet opts for a more robust solution. At first, all the pitches in the incomplete chorale are filled up simultaneously according to the highest probabilities for each of the individual voices. But this result is just taken as a draft. Then, some of the guesses are randomly erased and the new incomplete chorale is fed into Coconet again. New probability distributions are obtained for the new gaps. The process, called blocked Gibbs sampling, is repeated until the probability distributions given at consecutive iterations of the process are similar enough to always give the same pitch.

The diverse opinions about the final products are as interesting, if not more, as the mechanisms behind AI-generated music. The audience’s reaction to artificially generated music was spectacularly tested at the University of Oregon in 1997. There, the pianist Winifred Kerner performed three pieces: one written by her husband, the composer Steve Larson; another one written by Bach; and the last one, generated by Emmy. After her performance, the audience was asked to guess which was which. To Larson’s despair, the audience concluded that his composition had been created by Emmy and that Emmy’s work was genuine Bach. Larson was not the only one feeling uncomfortable about the fact that Emmy had been able to fool a whole audience. American professor of cognitive science Douglas Hofstadter, author of the 1979 Pulitzer prize-winning book Gödel, Escher, Bach, had argued a machine “would have to wander around the world on its own, fighting its way through the maze of life and feeling every moment of it” in order to produce anything similar to the masterpieces. In a 1997 article published by the New York Times, he claimed that the only comfort he could take from Larson’s experiment in front of the audience was that “Emmy doesn’t generate style on its own. It depends on mimicking prior composers”. The introduction of this sort of sophisticated and yet easy-to-use system not only opens a philosophical discussion on what should be called art, but it also brings in an ethical problem. In 2017, music-streaming service Spotify hired AI researcher François Pochet as the new director of the Spotify Creator Technology Research Lab. The hiring added even more weight to the accusation made by the magazine Music Business Worldwide that the platform had launched several playlists authored by fictional artists. These playlists, with around 500 million streams, were mood-themed with titles such as ‘peaceful piano’ or ‘ambient chill’: precisely the kind of atmospheric musical genres that AI is really good at generating. If this music had been created by Spotify’s AI, it would mean that they could have avoided paying royalties to the rights’ owners, as technically nobody would be the owner of this artificially created music. For the amount of streams that the playlists received, the cost would be in the range of $3m. In the end, Spotify declared that the music in the playlists was actually composed by real artists and that they were being paid the corresponding royalties. AI-generated music is controversial, but also exciting. AI is clever enough to generate short fragchalkdustmagazine.com

20

chalkdust ments of music in the style of Bach’s chorales. Indeed, despite the expressiveness in these pieces, the composition techniques used by the German genius to compose them tend to be rather algorithmic. It is also clever enough to create nice atmospheric music. However, AI still has a lot to learn in order to be able to produce a masterpiece in its own developed style, let alone the interpretation aspect. It will be a few years until the rise of a new Leonard Cohen. But AI is on the right track. As Pablo Picasso once put it: “Good artists borrow, great artists steal�... and this is precisely how machine learning works! Carmen Cabrera-Arnau Carmen Cabrera-Arnau is doing a PhD in applied mathematics at UCL with a focus on mathematical modelling of complex systems in urban environments. She enjoys maths outreach, eating cheese naan, and has been working for Chalkdust since 2017.

c c.cabrera-arnau@ucl.ac.uk Colourful arguments by Alaric Stephen and Alex Mayall It is impossible to arrange the seven tetrominoes (or Tetris pieces) to make a đ?&#x;Ś Ă— đ?&#x;Š rectangle. A common way to prove this is to colour the rectangle in a checkerboard pattern, and think about which colours each piece will cover. Pause here and think about how to do this before continuing.

A harder problem is to consider nine L tetrominoes. This time the goal is to arrange them into a đ?&#x;¨ Ă— đ?&#x;¨ square. Feel free to flip your L pieces over.

Can you use a colouring argument to prove that this is impossible to do? You can hear Alaric and Alex discuss puzzles like this on their superb podcast Odds and Evenings. Find out more at d oddsandevenings.com.

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chalkdust

Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the prof’s help? Contact c deardirichlet@chalkdustmagazine.com

Dear Dirichlet, Boy, am I in a world of woes! In the coming months, I have to go into surgery six times for a whole host of illness es. To its credit, the hospital has allowed me to arrange my own schedule. In your doctoral opinion, which pro ced ures should I undergo ďŹ rst?

— Under the weather, Cambridge â–

DIRICHLET SAYS: Bypass, Intestinal, Dental, Myotomy, Appendectomy,

Septoplasty: the usual order of operations. If you’re in the US, insurers will insist on Pacemakers, Excision, Myotomy and then D, A, S, which is unfortunate as you don’t need the ďŹ rst two things doing. Remember, only genius will get this right, and American healthcare is a disaster anyway.

Dear Dirichlet,

und in our pyjamas active. No more lounging aro ily fam the get to d ide dec I’ve rting by getting Maker streams on Twitch! I’m sta after đ?&#x;Ťam, watching Super Mario ngest is 3 ourhood after breakfast. The you ghb nei the und aro e bik to n the childre d. Know any ďŹ nd something appropriately size to ing ggl stru I’m and , old s month good retailers? gi, Dallas

— Lui

â–

DIRICHLET SAYS: No need to go shopping, just bring in the Floyd–

Warshall algorithm. Simply use it on all pairs of vertices (u,v) and then ďŹ nd the pair that minimises dist(u,v)+dist(v,u). Since this represents a path from u to v and then back to u, that will ďŹ nd you the smallest cycle. I notice you’ve included a donation of $3 in your letter and a lot of letter ‘R’s. Thanks for the continued support. Appreciate it. chalkdustmagazine.com

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Dear Dirichlet, I need to paint large letters on the side of my science communicatio n-themed narrow boat – “Yes! It’s LAVA!” etc . I’m finding getting the spacin g right between the characters a bit tricky though. The pairs “Ye” and “VA” are par ticu larly difficult. Do you know any guides which can help?

— Times New Showman, Edgbaston

■

DIRICHLET SAYS: I, too, have enjoyed many hours idling through the

beautiful canals of Birmingham on a barge with my badgers. With regard to your lettering, you may just find it easier to use a computerised typesetting system. Attract millions of pounds of funding and media interest by taking something we’ve been doing for years and renaming it ‘machine kerning’.

Dear Dirichlet,

my logging equiptty sure a colleague is stealing pre I’m and firm ber tim a in rk I wo all year but it’s shirt... They’ve been disappearing ment. Chainsaws, planes, flannel of range, thief recently by hiding just out the ch cat to d trie I il. Apr ce sin been worse ore? I know it e you heard of such a thing bef Hav r. iske wh a by m the sed but I mis sounds like a conspiracy. x

— Lumber Jack, The Bo

DIRICHLET SAYS: What you have there, Jack, is a devious toolbox plot. You say you noticed it starting after the first quartile of the year? Of course you did, but fret not, by the end of the third quartile, everything will quieten down. I’m sorry to hear your story but I will repeat what I’ve been telling my students forever: label your axes. ■

Dear Dirichlet, How many Platonic liquids are there?

■

— Sofia, Bulgaria

DIRICHLET SAYS: Two. More Dear Dirichlet, including two seasonal specials, online at d chalkdustmagazine.com 23

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George Boole (to the tune of Hey Jude by the Beatles)

Words by Tom Button a @tombutton

George Boole, don’t be so sad, You took an idea and made it better: A statement that is either true or false Can be represented as a letter. George Boole, don’t be afraid, You are remembered for your al-ge-bra. You thought everything was just black or white, Where they saw a horse you saw a zebra.

George Boole (1815–1864) was an English mathematician and the inventor of Boolean algebra, which deals with variables that can be either true or false.

And anytime you feel the pain, George Boole, refrain, You carry the world upon your numbers. You made a most fantastic tool, George Boole, it’s cool, For making computers unencumbered.

Boolean algebra is essential to all modern computers, as the operators that Boole applied to his truth variables behave in exactly the same way as the logic gates used to build computers.

Na-na-na, na na, na-na-na, na. George Boole, don’t be down, You’re regarded as a pacesetter. When switching circuits were first invented, There was no other theory better. So using one or using nought, George Boole, you taught Us operations like conjunction. Though electronics came after you, George Boole, you knew, We had what we needed for its construction. Na-na-na, na na, na-na-na, na, yeah. George Boole, don’t be afraid, You are remembered for your al-ge-bra. You thought that everything was just black or white, Where they saw a horse you saw a zebra. Zebra, zebra, zebra, zebra, yeah! Na, na, na, na-na-na na, na-na-na na, George Boole. (repeat the final line between 8 and ∞ times)

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chalkdust

On √2

Wikimedia commons user Hitoaki Koishikawa, CC BY-SA 3.0

Yiannis Petridis

O

ne of the first theorems lots of students see is that the square root of two is irrational (ie not a fraction). Therefore, we cannot restrict our attention to rational numbers only. Clearly √đ?&#x;¤ is a number we must have, as by Pythagoras’ theorem it represents the length of the hypotenuse of a right-angled isosceles triangle with vertical sides đ?&#x;Ł. What the theorem says is that √đ?&#x;¤ is never đ?‘Ľ/đ?‘Ś with đ?‘Ľ , đ?‘Ś integers. Or, to put it another way, đ?&#x;¤â‰

đ?‘Ľđ?&#x;¤ đ?‘Śđ?&#x;¤

⇔ đ?‘Ľ đ?&#x;¤ ≠đ?&#x;¤đ?‘Ś đ?&#x;¤ ,

that is, the square of an integer is never twice the square of another integer. However, they are both integers. The closest they can be apart is đ?&#x;Ł, ie đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł.

This is the simplest form of Pell’s equation: đ?‘Ľ đ?&#x;¤ − đ?‘ đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł. When đ?‘ = đ?&#x;¤, one can easily find a solution in the integers (đ?‘Ľ, đ?‘Ś) = (đ?&#x;Ł, đ?&#x;Ł). With a little more thinking we find the solutions (đ?‘Ľ, đ?‘Ś)

(đ?&#x;Ł, đ?&#x;Ł) (đ?&#x;Ľ, đ?&#x;¤) (đ?&#x;Š, đ?&#x;§) (đ?&#x;Łđ?&#x;Š, đ?&#x;Łđ?&#x;¤)

Why?

đ?&#x;Łđ?&#x;¤ − đ?&#x;¤ â‹… đ?&#x;Łđ?&#x;¤ đ?&#x;Ľđ?&#x;¤ − đ?&#x;¤ â‹… đ?&#x;¤đ?&#x;¤ đ?&#x;Šđ?&#x;¤ − đ?&#x;¤ â‹… đ?&#x;§đ?&#x;¤ đ?&#x;Łđ?&#x;Šđ?&#x;¤ − đ?&#x;¤ â‹… đ?&#x;Łđ?&#x;¤đ?&#x;¤

25

= −đ?&#x;Ł = +đ?&#x;Ł = −đ?&#x;Ł = +đ?&#x;Ł

autumn 2019

chalkdust The last one is more challenging. Why do we care? Writing đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł as đ?&#x;¤

đ?‘Ľ

đ?&#x;Ł ( ) = đ?&#x;¤ Âą đ?&#x;¤, đ?‘Ś đ?‘Ś

we see that finding solutions to Pell’s equation with larger denominator produces fractions đ?‘Ľ/đ?‘Ś with squares closer to đ?&#x;¤, ie better approximations of √đ?&#x;¤, since đ?&#x;Ł/đ?‘Ś đ?&#x;¤ will be smaller. For instance the pairs found above produce the approximations đ?&#x;Ł/đ?&#x;Ł = đ?&#x;Ł, đ?&#x;Ľ/đ?&#x;¤ = đ?&#x;Ł.đ?&#x;§, đ?&#x;Š/đ?&#x;§ = đ?&#x;Ł.đ?&#x;Ś, and đ?&#x;Łđ?&#x;Š/đ?&#x;Łđ?&#x;¤ = đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;¨đ?&#x;¨ for √đ?&#x;¤ = đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;Śđ?&#x;¤đ?&#x;Łđ?&#x;Ľđ?&#x;§đ?&#x;¨. We will see below that we can find infinitely many points with integer coordinates (đ?‘Ľ, đ?‘Ś) satisfying this Pell’s equation. It is desirable to be able to produce them systematically. We start by using the familiar identity đ?‘Žđ?&#x;¤ − đ?‘? đ?&#x;¤ = (đ?‘Ž + đ?‘?)(đ?‘Ž − đ?‘?) to rewrite the first solution (đ?‘Ľ, đ?‘Ś) = (đ?&#x;Ł, đ?&#x;Ł) as đ?&#x;Łđ?&#x;¤ − đ?&#x;¤ â‹… đ?&#x;Łđ?&#x;¤ = (đ?&#x;Ł + √đ?&#x;¤)(đ?&#x;Ł − √đ?&#x;¤) = −đ?&#x;Ł. This suggests that the number đ?&#x;Ł + √đ?&#x;¤ is important (we call it the fundamental unit). We then observe that (đ?&#x;Ł + √đ?&#x;¤)đ?&#x;¤ = (đ?&#x;Ł + √đ?&#x;¤)(đ?&#x;Ł + √đ?&#x;¤) = đ?&#x;Ľ + đ?&#x;¤âˆšđ?&#x;¤,

(đ?&#x;Ł + √đ?&#x;¤)đ?&#x;Ľ = (đ?&#x;Ľ + đ?&#x;¤âˆšđ?&#x;¤)(đ?&#x;Ł + √đ?&#x;¤) = đ?&#x;Š + đ?&#x;§âˆšđ?&#x;¤,

(đ?&#x;Ł + √đ?&#x;¤)đ?&#x;Ś = (đ?&#x;Š + đ?&#x;§âˆšđ?&#x;¤)(đ?&#x;Ł + √đ?&#x;¤) = đ?&#x;Łđ?&#x;Š + đ?&#x;Łđ?&#x;¤âˆšđ?&#x;¤.

There is a clear pattern that emerges: the coefficients of the expressions on the right are the pairs (đ?‘Ľ, đ?‘Ś) we found above. Let’s write (đ?&#x;Ł + √đ?&#x;¤)đ?‘› = đ?‘Ľđ?‘› + đ?‘Śđ?‘› √đ?&#x;¤,

with đ?‘Ľđ?‘› and đ?‘Śđ?‘› natural numbers (this is guaranteed by the binomial theorem if we expand and collect the terms that contain √đ?&#x;¤ and the terms that do not). Then it seems that đ?‘Ľđ?‘›đ?&#x;¤ − đ?&#x;¤đ?‘Śđ?‘›đ?&#x;¤ = Âąđ?&#x;Ł

with +đ?&#x;Ł when đ?‘› is even and −đ?&#x;Ł with đ?‘› odd. So to find solutions we have to expand (đ?&#x;Ł + √đ?&#x;¤)đ?‘› . Using the binomial theorem is not very practical. In fact, we try to find recursive formulae for the two sequences đ?‘Ľđ?‘› and đ?‘Śđ?‘› . We have đ?‘Ľđ?‘›+đ?&#x;Ł + đ?‘Śđ?‘›+đ?&#x;Ł √đ?&#x;¤ = (đ?&#x;Ł + √đ?&#x;¤)đ?‘›+đ?&#x;Ł

= (đ?‘Ľđ?‘› + đ?‘Śđ?‘› √đ?&#x;¤)(đ?&#x;Ł + √đ?&#x;¤)

= (đ?‘Ľđ?‘› + đ?&#x;¤đ?‘Śđ?‘› ) + (đ?‘Ľđ?‘› + đ?‘Śđ?‘› )√đ?&#x;¤.

We match coefficients to get the recurrences:

đ?‘Ľđ?‘›+đ?&#x;Ł = đ?‘Ľđ?‘› + đ?&#x;¤đ?‘Śđ?‘› ,

đ?‘Śđ?‘›+đ?&#x;Ł = đ?‘Ľđ?‘› + đ?‘Śđ?‘› .

Matching coefficients can be justified because we can rearrange to get √đ?&#x;¤ = chalkdustmagazine.com

đ?‘Ľđ?‘›+đ?&#x;Ł − đ?‘Ľđ?‘› − đ?&#x;¤đ?‘Śđ?‘› đ?‘Ľđ?‘› + đ?‘Śđ?‘› − đ?‘Śđ?‘›+đ?&#x;Ł

26

,

(f)

chalkdust with integer numerator and denominator. If đ?‘Śđ?‘›+đ?&#x;Ł is not equal to đ?‘Ľđ?‘› + đ?‘Śđ?‘› , we get that √đ?&#x;¤ is rational. We can apply (f) repeatedly to the pair (đ?&#x;Łđ?&#x;Š, đ?&#x;Łđ?&#x;¤) to get new solutions leading to better approximations of √đ?&#x;¤: (đ?‘Ľ, đ?‘Ś)

(đ?&#x;Śđ?&#x;Ł, đ?&#x;¤đ?&#x;Ť) (đ?&#x;Ťđ?&#x;Ť, đ?&#x;Šđ?&#x;˘) (đ?&#x;¤đ?&#x;Ľđ?&#x;Ť, đ?&#x;Łđ?&#x;¨đ?&#x;Ť)

Using (f), we see that

đ?‘Ľ/đ?‘Ś

đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;Ľđ?&#x;Š đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;Śđ?&#x;¤đ?&#x;Ş đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;Śđ?&#x;¤đ?&#x;˘

đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = (đ?‘Ľ + đ?&#x;¤đ?‘Ś )đ?&#x;¤ − đ?&#x;¤(đ?‘Ľ + đ?‘Ś )đ?&#x;¤ = −(đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ ). đ?‘Ľđ?‘›+đ?&#x;Ł đ?‘› đ?‘› đ?‘› đ?‘› đ?‘› đ?‘› đ?‘›+đ?&#x;Ł

(ff)

đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = −đ?&#x;Ł and vice versa. This explains the pattern we found above. If đ?‘Ľđ?‘›đ?&#x;¤ − đ?&#x;¤đ?‘Śđ?‘›đ?&#x;¤ = +đ?&#x;Ł, then đ?‘Ľđ?‘›+đ?&#x;Ł đ?‘›+đ?&#x;Ł In fact, one can actually consider also negative powers of đ?&#x;Ł + √đ?&#x;¤. Since (đ?&#x;Ł + √đ?&#x;¤)−đ?&#x;Ł = −(đ?&#x;Ł − √đ?&#x;¤), we can check that the solutions we get are the same up to the sign of đ?‘Ľđ?‘› and đ?‘Śđ?‘› .

An approximate history In ancient times, the numerators were called diameter or diagonal numbers and the denominators side numbers, because their ratio tends to the ratio of the hypotenuse divided by the vertical side of any right-angled isosceles triangle. The pair (đ?&#x;Š, đ?&#x;§) giving the approximation đ?&#x;Š/đ?&#x;§ for √đ?&#x;¤ was known to the Pythagoreans and Plato. Pell’s equation đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł and the recurrence formulae (f) appear in the work of Theon of Smyrna (On mathematics useful for the understanding of Plato, AD 130—about 1500 years before Pell) and must have been known to Plato and likely the Pythagoreans according to Commentary on Plato’s republic by Proclus. For a fuller history, see A history of Greek mathematics, vol. I by Thomas Heath (pages 91–93), and Greek mathematical works I by Ivor Thomas (pages 132– 139). In fact, Proclus refers to Euclid (Euclid II.10) for the proposition (đ?‘Ľ + đ?&#x;¤đ?‘Ś)đ?&#x;¤ + đ?‘Ľ đ?&#x;¤ = đ?&#x;¤(đ?‘Ľ + đ?‘Ś)đ?&#x;¤ + đ?&#x;¤đ?‘Ś đ?&#x;¤ ,

John Pell, after whom the equation đ?‘Ľ đ?&#x;¤ − đ?‘ đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł is named.

which is equivalent to (ff), and is exactly what we saw for the solutions (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ). Around AD 250, Diophantus considered integer solutions of the equation đ?‘Žđ?&#x;¤ đ?‘Ľ đ?&#x;¤ + đ?‘? = đ?‘Ś đ?&#x;¤

for various (đ?‘Ž, đ?‘?), eg đ?‘Ľ đ?&#x;¤ + đ?&#x;Ł = đ?‘Ś đ?&#x;¤ , đ?‘Ľ đ?&#x;¤ − đ?&#x;Ł = đ?‘Ś đ?&#x;¤ , đ?‘Ľ đ?&#x;¤ + đ?&#x;Łđ?&#x;¤ = đ?‘Ś đ?&#x;¤ . This is closely related to Pell’s equations. The great Indian mathematician Brahmagupta (born AD 598) discovered the identity (đ?‘Žđ?&#x;¤ − đ?‘ đ?‘? đ?&#x;¤ )(đ?‘Ľ đ?&#x;¤ − đ?‘ đ?‘Ś đ?&#x;¤ ) = (đ?‘Žđ?‘Ľ + đ?‘ đ?‘?đ?‘Ś)đ?&#x;¤ − đ?‘ (đ?‘Žđ?‘Ś + đ?‘?đ?‘Ľ)đ?&#x;¤ ,

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chalkdust which helps produce new solutions from old. The solution of the general Pell’s equation is due to Bhaskara II (12th century). Pell himself was an English mathematician alive in the 17th century, and is connected to the problem thanks to Euler, who mistakenly believed that a commentary Pell had written on the equation was in fact his own original work.

The geometric approach There is a geometric interpretation of the construction in (f), which associates the solutions we’ve found to points on a curve in the plane. The equations đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł represent two hyperbolae in the plane with asymptotes đ?‘Ś = Âąđ?‘Ľ/√đ?&#x;¤. They meet the axes at the points (Âąđ?&#x;Ł, đ?&#x;˘) and (đ?&#x;˘, Âąđ?&#x;Ł/√đ?&#x;¤). We search for points on these hyperbolae with coordinates in the natural numbers (we might as well concentrate on the first quadrant đ?‘Ľ, đ?‘Ś ⊞ đ?&#x;˘). The closest point to the origin is (đ?&#x;Ł, đ?&#x;˘). The next closest is (đ?&#x;Ł, đ?&#x;Ł). Starting from a point (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) on either hyperbola, we draw the tangent line to the hyperbola at that point. Implicit differentiation gives đ?&#x;¤đ?‘Ľ − đ?&#x;Śđ?‘Śđ?‘Ś ′ = đ?&#x;˘ â&#x;š đ?‘Ś ′ = đ?‘Ľ/(đ?&#x;¤đ?‘Ś), so that the slope of the tangent line is đ?‘Ľđ?‘› /(đ?&#x;¤đ?‘Śđ?‘› ). At the same time the line from (đ?‘Ľđ?‘›+đ?&#x;Ł , đ?‘Śđ?‘›+đ?&#x;Ł ) to (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) has slope đ?‘Ľđ?‘› đ?‘Śđ?‘›+đ?&#x;Ł − đ?‘Śđ?‘› = , đ?‘Ľđ?‘›+đ?&#x;Ł − đ?‘Ľđ?‘› đ?&#x;¤đ?‘Śđ?‘›

with the help of (f). We conclude that this line is the tangent line at (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ). To find the next solution, we draw the tangent line at (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) and find its point of intersection with the other hyperbola. So not only does the tangent line intersect the other hyperbola, but it intersects at a point with integer coordinates! This construction is called the tangent construction and predates the famous geometric construction of the group law on elliptic curves. The figure below shows the two hyperbolae and the tangent at (đ?&#x;Ł, đ?&#x;Ł). đ?&#x;Ľ

đ?‘Ś

đ?&#x;¤ đ?&#x;Ł đ?&#x;˘

đ?&#x;˘

đ?&#x;Ł

đ?&#x;¤ đ?‘Ľ

đ?&#x;Ľ

đ?&#x;Ś

We can also understand the sign of the remainder Âąđ?&#x;Ł using this geometric approach. The slope of the asymptote đ?‘Ś = đ?‘Ľ/√đ?&#x;¤ is đ?&#x;Ł/√đ?&#x;¤. If (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) is on the hyperbola đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = −đ?&#x;Ł, then đ?‘Śđ?‘› is relatively large compared to the asymptote, so đ?‘Śđ?‘› /đ?‘Ľđ?‘› > đ?&#x;Ł/√đ?&#x;¤ â&#x;ş đ?‘Ľđ?‘› /đ?‘Śđ?‘› < √đ?&#x;¤. The slope of the tangent line is smaller than the one of the asymptote. Similarly, if (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) is on the hyperbola đ?‘Ľ đ?&#x;¤ − đ?&#x;¤đ?‘Ś đ?&#x;¤ = đ?&#x;Ł, then đ?‘Śđ?‘› is relatively small compared to the asymptote, so đ?‘Śđ?‘› /đ?‘Ľđ?‘› < đ?&#x;Ł/√đ?&#x;¤ â&#x;ş đ?‘Ľđ?‘› /đ?‘Śđ?‘› > √đ?&#x;¤. The

chalkdustmagazine.com

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chalkdust slope of the tangent line is larger than the one of the asymptote. In both cases the tangent line at (đ?‘Ľđ?‘› , đ?‘Śđ?‘› ) will cross the asymptote and meet the other hyperbola at the point (đ?‘Ľđ?‘›+đ?&#x;Ł , đ?‘Śđ?‘›+đ?&#x;Ł ) with larger coordinates. As the tangent lines get squeezed between the hyperbolae, their slopes approach the slope of the asymptote. We see that đ?‘Ľđ?&#x;Ł đ?‘Ľđ?&#x;Ľ đ?‘Ľđ?&#x;§ đ?‘Ľđ?&#x;Ś đ?‘Ľđ?&#x;¤ < < < â‹Ż < √đ?&#x;¤ < â‹Ż < < . đ?‘Śđ?&#x;Ł đ?‘Śđ?&#x;Ľ đ?‘Śđ?&#x;§ đ?‘Śđ?&#x;Ś đ?‘Śđ?&#x;¤

This means that we have trapped √đ?&#x;¤ from above and below by the corresponding fractions. Using the fractions đ?&#x;¤đ?&#x;Ľđ?&#x;Ť/đ?&#x;Łđ?&#x;¨đ?&#x;Ť and đ?&#x;Ťđ?&#x;Ť/đ?&#x;Šđ?&#x;˘ we see that we found the first four decimal digits of √đ?&#x;¤ ≈ đ?&#x;Ł.đ?&#x;Śđ?&#x;Łđ?&#x;Śđ?&#x;¤.

Other forms of Pell’s equation

If one tries to work with Pell’s equation đ?‘Ľ đ?&#x;¤ − đ?&#x;Ľđ?‘Ś đ?&#x;¤ = Âąđ?&#x;Ł in order to approximate √đ?&#x;Ľ, then this can only be done with +đ?&#x;Ł (the reader may be interested to investigate why). The fundamental unit here is not đ?&#x;Ł + √đ?&#x;Ľ but đ?&#x;¤ + √đ?&#x;Ľ. This provides the smallest solution of the equation: (đ?&#x;¤, đ?&#x;Ł). The recurrence relations are different but equally easy to find: đ?‘Ľđ?‘›+đ?&#x;Ł = đ?&#x;¤đ?‘Ľđ?‘› + đ?&#x;Ľđ?‘Śđ?‘› , đ?‘Śđ?‘›+đ?&#x;Ł = đ?‘Ľđ?‘› + đ?&#x;¤đ?‘Śđ?‘› , producing from (đ?&#x;¤, đ?&#x;Ł) the solutions (đ?&#x;Š, đ?&#x;Ś), (đ?&#x;¤đ?&#x;¨, đ?&#x;Łđ?&#x;§) etc. Since đ?‘Ľđ?‘›đ?&#x;¤ is always greater than đ?&#x;Ľđ?‘Śđ?‘›đ?&#x;¤ , the method provides approximations of √đ?&#x;Ľ that are always larger than √đ?&#x;Ľ.

In both cases of Pell’s equation discussed so far (√đ?&#x;¤ and √đ?&#x;Ľ) it was easy to spot the smallest solution in the natural numbers. But this becomes increasingly hard: for đ?‘Ľ đ?&#x;¤ −đ?‘ đ?‘Ś đ?&#x;¤ = đ?&#x;Ł and đ?‘ = đ?&#x;Łđ?&#x;Ľ the smallest solution is (đ?&#x;¨đ?&#x;Śđ?&#x;Ť, đ?&#x;Łđ?&#x;Şđ?&#x;˘). Frenicle challenged Wallis to solve the equation with đ?‘ = đ?&#x;Ľđ?&#x;Łđ?&#x;Ľ. The smallest solution is (đ?&#x;Ľđ?&#x;¤đ?&#x;Łđ?&#x;Şđ?&#x;Şđ?&#x;Łđ?&#x;¤đ?&#x;˘đ?&#x;Şđ?&#x;¤đ?&#x;Ťđ?&#x;Łđ?&#x;Ľđ?&#x;Śđ?&#x;Şđ?&#x;Śđ?&#x;Ť, đ?&#x;Łđ?&#x;Şđ?&#x;Łđ?&#x;Ťđ?&#x;Ľđ?&#x;Şđ?&#x;˘đ?&#x;Łđ?&#x;§đ?&#x;Şđ?&#x;§đ?&#x;¨đ?&#x;Śđ?&#x;Łđ?&#x;¨đ?&#x;˘). The Archimedes cattle problem (see Greek mathematical works II, Ivor Thomas, page 202), which Gotthold Ephraim Lessing discovered as a poem of forty-four lines in a Greek manuscript, leads to Pell’s equation đ?‘Ľ đ?&#x;¤ −đ?&#x;Śđ?&#x;Łđ?&#x;˘đ?&#x;¤đ?&#x;Şđ?&#x;¨đ?&#x;Śđ?&#x;¤đ?&#x;Ľđ?&#x;¤đ?&#x;Šđ?&#x;Şđ?&#x;Śđ?&#x;¤đ?&#x;Śđ?‘Ś đ?&#x;¤ = đ?&#x;Ł. Here the smallest solution has 206,545 digits if written out explicitly. How does one proceed to find the smallest solution to Pell’s equation without guessing? There is a systematic technique using continued fractions. Let’s start with √đ?&#x;¤. We write √đ?&#x;¤ = (√đ?&#x;¤ + đ?&#x;Ł) − đ?&#x;Ł = đ?&#x;Ł + =đ?&#x;Ł+

using always that √đ?&#x;¤ + đ?&#x;Ł = đ?&#x;¤ + consider the fractions đ?&#x;Ł+

đ?&#x;Ł

đ?&#x;Ľ = , đ?&#x;¤ đ?&#x;¤

đ?&#x;¤+

đ?&#x;Ł . √đ?&#x;¤+đ?&#x;Ł

đ?&#x;Ł+

đ?&#x;Ł

đ?&#x;Ł

đ?&#x;¤+

đ?&#x;Ł

đ?&#x;Ł

√đ?&#x;¤+đ?&#x;Ł

đ?&#x;Ł

√đ?&#x;¤ + đ?&#x;Ł

=đ?&#x;Ł+

đ?&#x;¤+

=đ?&#x;Ł+ đ?&#x;Ł

đ?&#x;¤+

đ?&#x;Ł

đ?&#x;¤+

đ?&#x;Ł

đ?&#x;¤+

đ?&#x;Ł √đ?&#x;¤+đ?&#x;Ł

đ?&#x;Ł

đ?&#x;Ł

√đ?&#x;¤+đ?&#x;Ł

=â‹Ż

Ignoring the last fraction with the square root, we are led to

đ?&#x;¤ + đ?&#x;Ł/đ?&#x;¤

=đ?&#x;Ł+

đ?&#x;Ł

đ?&#x;Š = , đ?&#x;§/đ?&#x;¤ đ?&#x;§

đ?&#x;Ł+

đ?&#x;¤+

đ?&#x;Ł

đ?&#x;Ł

đ?&#x;¤+

đ?&#x;Ł

đ?&#x;¤

=

đ?&#x;Łđ?&#x;Š

đ?&#x;Łđ?&#x;¤

,

â‹Ż

The fractions we get are called the convergents đ?‘?đ?‘› /đ?‘žđ?‘› for the continued fraction of √đ?&#x;¤, which we write [đ?&#x;Ł; đ?&#x;¤], with the bar denoting that the pattern đ?&#x;¤ repeats itself at infinity. The convergents 29

autumn 2019

chalkdust đ?‘?đ?‘› /đ?‘žđ?‘› are the same rational approximations that we found above. This can be seen by writing recurrence relations for đ?‘?đ?‘› and đ?‘žđ?‘› : đ?‘?đ?‘›+đ?&#x;Ł = đ?‘?đ?‘› + đ?&#x;¤đ?‘žđ?‘› ,

đ?‘žđ?‘›+đ?&#x;Ł = đ?‘?đ?‘› + đ?‘žđ?‘› ,

which are the same as (f). It is true that the smallest solution of Pell’s equation (đ?‘Ľ, đ?‘Ś) will be among the pairs (đ?‘?đ?‘› , đ?‘žđ?‘› ), and, therefore, may be found by performing the continued fraction expansion and testing each successive convergent. From the calculation √đ?&#x;Ľ = đ?&#x;Ł + (√đ?&#x;Ľ − đ?&#x;Ł) = đ?&#x;Ł +

đ?&#x;¤

√đ?&#x;Ľ + đ?&#x;Ł

=đ?&#x;Ł+

đ?&#x;Ł

đ?&#x;Ł + (√đ?&#x;Ľ − đ?&#x;Ł)/đ?&#x;¤

=đ?&#x;Ł+

đ?&#x;Ł+

đ?&#x;Ł

đ?&#x;Ł đ?&#x;¤+(√đ?&#x;Ľâˆ’đ?&#x;Ł)

=â‹Ż

it is not too hard to see that √đ?&#x;Ľ = [đ?&#x;Ł; đ?&#x;Ł, đ?&#x;¤]. For đ?‘ = đ?&#x;Ľđ?&#x;Łđ?&#x;Ľ it is a lot more difficult: √đ?&#x;Ľđ?&#x;Łđ?&#x;Ľ = [đ?&#x;Łđ?&#x;Š; đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ś, đ?&#x;Łđ?&#x;Ł, đ?&#x;Ł, đ?&#x;Ł, đ?&#x;Ľ, đ?&#x;¤, đ?&#x;¤, đ?&#x;Ľ, đ?&#x;Ł, đ?&#x;Ł, đ?&#x;Łđ?&#x;Ł, đ?&#x;Ś, đ?&#x;¤, đ?&#x;Ł, đ?&#x;Ľđ?&#x;Ś]. The origins of continued fractions go back to Euclid and the Euclidean algorithm to find the greatest common divisor of two numbers. Rafael Bombelli in L’Algebra (1572) essentially proved that √đ?&#x;Łđ?&#x;Ľ has the infinite continued fraction √đ?&#x;Łđ?&#x;Ľ = đ?&#x;Ľ +

đ?&#x;¨+

đ?&#x;Ś

đ?&#x;¨+

đ?&#x;Ś

đ?&#x;Ś

đ?&#x;¨+â‹Ż

,

this being the first time that square roots and continued fractions are related. Even now there are đ?&#x;Ľ đ?&#x;¤. open questions on the continued fraction expansion of numbers like √ Yiannis Petridis Yiannis is a professor of mathematics at University College London. He works in analytic number theory, can write with both hands, and really likes polar bears.

Six months ago, we didn’t know... ...that 42 and 906 are sums of three cubes: đ?&#x;Śđ?&#x;¤ = (−đ?&#x;Şđ?&#x;˘đ?&#x;§đ?&#x;Ľđ?&#x;Şđ?&#x;Šđ?&#x;Ľđ?&#x;Şđ?&#x;Şđ?&#x;Łđ?&#x;¤đ?&#x;˘đ?&#x;Šđ?&#x;§đ?&#x;Ťđ?&#x;Šđ?&#x;Ś)đ?&#x;Ľ + đ?&#x;Şđ?&#x;˘đ?&#x;Śđ?&#x;Ľđ?&#x;§đ?&#x;Šđ?&#x;§đ?&#x;Şđ?&#x;Łđ?&#x;Śđ?&#x;§đ?&#x;Şđ?&#x;Łđ?&#x;Šđ?&#x;§đ?&#x;Łđ?&#x;§đ?&#x;Ľ + đ?&#x;Łđ?&#x;¤đ?&#x;¨đ?&#x;˘đ?&#x;¤đ?&#x;Łđ?&#x;¤đ?&#x;Ľđ?&#x;¤đ?&#x;Ťđ?&#x;Šđ?&#x;Ľđ?&#x;Ľđ?&#x;§đ?&#x;¨đ?&#x;Ľđ?&#x;Łđ?&#x;Ľ ,

đ?&#x;Ťđ?&#x;˘đ?&#x;¨ = (−đ?&#x;Šđ?&#x;Śđ?&#x;Ťđ?&#x;¤đ?&#x;Śđ?&#x;¤đ?&#x;§đ?&#x;Ťđ?&#x;Ľđ?&#x;Ťđ?&#x;§đ?&#x;¨đ?&#x;Łđ?&#x;˘đ?&#x;Ľđ?&#x;Ťđ?&#x;Š)đ?&#x;Ľ + đ?&#x;Šđ?&#x;¤đ?&#x;˘đ?&#x;§đ?&#x;Śđ?&#x;˘đ?&#x;Şđ?&#x;Ťđ?&#x;¨đ?&#x;Šđ?&#x;Ťđ?&#x;Ľđ?&#x;§đ?&#x;Ľđ?&#x;Ľđ?&#x;Šđ?&#x;Şđ?&#x;Ľ + đ?&#x;Ľđ?&#x;§đ?&#x;Ťđ?&#x;¨đ?&#x;Łđ?&#x;Ťđ?&#x;Šđ?&#x;Ťđ?&#x;¨đ?&#x;Łđ?&#x;§đ?&#x;Ľđ?&#x;§đ?&#x;¨đ?&#x;§đ?&#x;˘đ?&#x;Ľđ?&#x;Ľ .

114 is now the smallest positive integer for which it is unknown whether or not it can be written as the sum of three cubes. The only other numbers less than 1000 for which this is currently unknown are 165, 390, 579, 627, 633, 732, 921, and 975‌ ‌but work on this problem is progressing rapidly so this may have changed by now!

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30

đ?œƒ đ?œ™

Jean

LagacĂŠ

V

I’m staunchly pro-choice.

Johnny

Nicholson I say screw the axiom of choice!

Let anyone do with their sets as they please. And if “as they pleaseâ€? is choosing without discrimination an element out of every set in an inďŹ nite collection, then so be it! And I don’t want to Banach–Tarski my spheres in any other way. Let’s not go back to the 19th century, when people didn’t even know they had a choice, and keep on choosing, my friends!

I haven’t used the axiom of choice for two years and, while my proofs might take a bit longer than most, at least I can rest in the knowledge that my arguments are watertight. Would you believe, this liberal lot actually think that every vector space has a basis? For those of you who also can’t think of a basis for � as a ℚ-vector space, I invite you to join me in my rejection of the axiom of choice!

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Curiosities of linearly ordered sets Andrei Chekmasov

Y

ou might look at the title and ask yourself: what is a linearly ordered set? A good example of such a set would be counting the number of years that have passed since the Big Bang— the age of the universe. It will look something like this: Bang!

Now?

đ?&#x;˘ → đ?&#x;Ł → đ?&#x;¤ → đ?&#x;Ľ → â‹Ż → 13,772,000,000 → â‹Ż

More formally, a set đ?‘† is a collection of objects đ?‘Ľ , and if an object is in a set it is called an element of the set: đ?‘Ľ ∈ đ?‘† . Simply put, a linearly ordered set is an arrangement of the elements of a set one after another. If we assume that the universe will exist forever, our example represents a linear arrangement for the set N of non-negative integers. This specific order is called the natural order on N = Z+ âˆŞ {đ?&#x;˘}. Keep this example in mind while we define what a linear order is.

We begin by introducing a relation → on a set đ?‘† , which determines the arrangement of elements of đ?‘† ; for convenience we shall use the symbol → whenever there are elements đ?‘Ľ, đ?‘Ś ∈ đ?‘† , and đ?‘Ľ comes before (or to the left of) đ?‘Ś in the order of elements, writing đ?‘Ľ → đ?‘Ś .

We need the relation → to have some useful properties to become a linear order on � . First, the order needs to be consistent in some sense across the whole set, so we require the relation to have the transitivity property: if � → � and � → � , then � → � . Second, we want the order to apply to every element of the set � , so it is necessary for the relation to have the trichotomy property: for

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34

chalkdust all đ?‘Ľ, đ?‘Ś ∈ đ?‘† , exactly one of the following holds: đ?‘Ľâ†’đ?‘Ś

đ?‘Ľ=đ?‘Ś

�→�

Enforcing transitivity and trichotomy properties allows us to avoid ‘loops’ in the linear order like đ?‘Ľâ†’đ?‘Śâ†’đ?‘§â†’đ?‘Ľ

If → is a linear order on đ?‘† , then đ?‘† together with the relation → is called a linearly ordered set. Applying this definition to the natural order on N, we see that the relation → is represented by the more familiar ‘less than’ relation <N on the elements of N.

Dense orders Broadly speaking, there are two types of linear orders: dense and non-dense orders. A linear order on a set is dense if between any two elements of the set one can always find a third element of the same set. Otherwise, a linear order is non-dense. We challenge the reader to answer this question: is it possible to make a dense order on a set with a finite number of elements? An example of a dense linear order is well known to all of us. It is formed by the set of positive rational numbers, Q+ , and the relation <Q+ . With respect to <Q+ , between any two positive rational numbers, one can find a third positive rational number. For example, if đ?‘Ľ, đ?‘Ś ∈ Q+ and đ?‘Ľ <Q+ đ?‘Ś , then their average is also a rational number and is between them with respect to <Q+ : đ?‘Ľ <Q+

đ?‘Ľ +đ?‘Ś đ?&#x;¤

<Q + đ?‘Ś

You need an infinite set to have a dense order, but this does not mean that every linear order on an infinite set is dense! The natural order of non-negative integers is not dense, despite being infinite. For example, there is no integer between 2 and 3. Though we are limited to integers here we can still construct a dense linear order on the set N! Here is an example. On a straight line we start step đ?&#x;˘ by marking out đ?&#x;˘ and đ?&#x;Ł. Then at step đ?&#x;Ł we place đ?&#x;¤ at a midpoint between đ?&#x;˘ and đ?&#x;Ł on the straight line. We continue from left to right and at step đ?&#x;¤ find midpoints between đ?&#x;˘ and đ?&#x;¤, đ?&#x;¤ and đ?&#x;Ł and placing đ?&#x;Ľ and đ?&#x;Ś at those midpoints respectively. We continue like this forever. From step đ?&#x;Ł onwards, for each đ?‘– ∈ Z+ we fit the numbers đ?&#x;¤đ?‘–−đ?&#x;Ł + đ?&#x;Ł through to đ?&#x;¤đ?‘– into all the gaps between numbers in the previous step. The relative location of the first đ?&#x;Łđ?&#x;Š elements of N according to the linear order is shown in the figure overleaf. So is this linear order dense? At any step imagine two natural numbers đ?‘Ž and đ?‘? which have no numbers between them. On the next step we will put a number at the midpoint between đ?‘Ž and đ?‘? so that đ?‘Ž and đ?‘? are not ‘immediate’ neighbours any more. We built a dense order indeed! By applying this algorithm, we ‘packed’ all the integers inside the interval [đ?&#x;˘, đ?&#x;Ł] in order to construct a dense order on N. 35

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Step 0:

1

0

Step 1:

2

0

Step 2:

3

0

Step 3:

5 9

0

Step 4:

5

2

3 10

3

1

6 11

6

4 7

2 12

â‹Ž

8

4

7

13

2

1

14

15

4

1

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1

â‹Ž Constructing the dense order on the natural numbers.

Cantorian order Looking at how many more rationals there seem to be in comparison with the natural numbers (after all, there is an infinite number of rationals between đ?&#x;˘ and đ?&#x;Ł), you might expect that every linear order on Q+ is a dense order. Not at all! German mathematician Georg Cantor (1845–1918) showed that there is a linear order on Q+ that is not dense. Can you come up with one? Q+

Let us introduce the Cantorian order â&#x;ś on Q+ via a diagram:

â‹ą

đ?&#x;Ł đ?&#x;Ť

đ?&#x;Ł đ?&#x;Ş đ?&#x;¤ đ?&#x;Ť

đ?&#x;Ł đ?&#x;Š đ?&#x;¤ đ?&#x;Ş

đ?&#x;Ł đ?&#x;¨ đ?&#x;¤ đ?&#x;Š đ?&#x;Ľ đ?&#x;Ş

đ?&#x;Ł đ?&#x;§ đ?&#x;¤ đ?&#x;¨ đ?&#x;Ľ đ?&#x;Š

đ?&#x;Ł đ?&#x;Ś đ?&#x;¤ đ?&#x;§ đ?&#x;Ľ đ?&#x;¨ đ?&#x;Ś đ?&#x;Š

đ?&#x;Ł đ?&#x;Ľ đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ đ?&#x;§ đ?&#x;Ś đ?&#x;¨

đ?&#x;Ł đ?&#x;¤ đ?&#x;¤ đ?&#x;Ľ đ?&#x;Ľ đ?&#x;Ś đ?&#x;Ś đ?&#x;§ đ?&#x;§ đ?&#x;¨

đ?&#x;Ł đ?&#x;Ł

đ?&#x;¤ đ?&#x;¤ đ?&#x;Ľ đ?&#x;Ľ đ?&#x;Ś đ?&#x;Ś đ?&#x;§ đ?&#x;§

â‹Ž

đ?&#x;¤ đ?&#x;Ł đ?&#x;Ľ đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ đ?&#x;§ đ?&#x;Ś đ?&#x;¨ đ?&#x;§

đ?&#x;Ľ đ?&#x;Ł đ?&#x;Ś đ?&#x;¤ đ?&#x;§ đ?&#x;Ľ đ?&#x;¨ đ?&#x;Ś

đ?&#x;Ś đ?&#x;Ł đ?&#x;§ đ?&#x;¤ đ?&#x;¨ đ?&#x;Ľ đ?&#x;Š đ?&#x;Ś

đ?&#x;§ đ?&#x;Ł đ?&#x;¨ đ?&#x;¤ đ?&#x;Š đ?&#x;Ľ

đ?&#x;¨ đ?&#x;Ł đ?&#x;Š đ?&#x;¤ đ?&#x;Ş đ?&#x;Ľ

đ?&#x;Š đ?&#x;Ł đ?&#x;Ş đ?&#x;¤

đ?&#x;Ş đ?&#x;Ł đ?&#x;Ť đ?&#x;¤

đ?&#x;Ť đ?&#x;Ł

â‹ą

đ?&#x;Ś đ?&#x;§ đ?&#x;Ľ đ?&#x;Ť đ?&#x;Š đ?&#x;¨ đ?&#x;Ş đ?&#x;Ť formingđ?&#x;Ş a plane đ?&#x;Šof fractions đ?&#x;Ľ đ?&#x;¨ wheređ?&#x;§ numerators đ?&#x;Ś The order is created by increase as one goes diagođ?&#x;Ś đ?&#x;¨ đ?&#x;Ş đ?&#x;§ đ?&#x;Š đ?&#x;Ť nally right and denominators increase as one goes diagonally left. Then, by travelling along a path đ?&#x;Ť đ?&#x;Š đ?&#x;§ đ?&#x;Ş đ?&#x;¨ đ?&#x;Ś đ?&#x;§ back and forth, one creates a linear order đ?&#x;Ş which đ?&#x;Ťsimplify are skipped: for examđ?&#x;¨ on Q+đ?&#x;Š. Fractions đ?&#x;Š skippedđ?&#x;¨ (they are đ?&#x;Ť and sođ?&#x;Ş forth are đ?&#x;§ crossed out on the diagram). ple, đ?&#x;Ł/đ?&#x;¤ is present, but đ?&#x;¤/đ?&#x;Ś, đ?&#x;Ľ/đ?&#x;¨ đ?&#x;Ş đ?&#x;Ť đ?&#x;¨ đ?&#x;Š We can see that there are no elements in between adjacent fractions, yet this linear order contains đ?&#x;Š đ?&#x;Ť đ?&#x;¨ đ?&#x;Ş đ?&#x;Š whole đ?&#x;Şset Q+ . đ?&#x;Ť every possible fraction, thus covering the đ?&#x;Š đ?&#x;Ť đ?&#x;Ş đ?&#x;Ş đ?&#x;Ť chalkdustmagazine.com 36 đ?&#x;Ť đ?&#x;Ş đ?&#x;Ť đ?&#x;Ť

chalkdust Relative to the Cantorian order one can speak of the first positive rational number, the second positive rational number and so on. In other words, one can construct a one-to-one mapping from Z+ , the set of positive integers, to Q+ since two different positive rational numbers are always in two different positions in the Cantorian order. So there are the same number of positive rational numbers as there are positive integers! If you think of the apparent abundance of positive rational numbers relative to the apparent sparseness of positive integers on the number line, you will find this conclusion a bit mind-bending. Unsurprisingly, Cantor’s contemporaries were shocked by his findings, and many rejected his research. Cantor’s thinking was clearly ahead of his time!

A bigger picture Cantor had grander ideas than just defining a weird linear order. After all, at the time he was developing what is now known as modern set theory. One of the problems Cantor faced was how to define infinity; he thought of an elegant approach to grasp this ethereal concept. We’ll walk ourselves through it. To start off, we know that every positive integer has a successor—just add đ?&#x;Ł to it. Thus you can count positive integers. The number of elements in a set like Z+ is called the cardinality of the set. The infinite set whose elements can be counted is called countably infinite. Other sets that can be put into one-to-one mapping with positive integers are also countably infinite. Cantorian order proves that Q+ is countably infinite.

Georg Cantor

Countably infinite sets have some interesting properties. All of their subsets are either finite or countably infinite. For example, you can map one-to-one positive integers to positive even integers, which is a subset of Z+ . Another property is that if you combine two countably infinite sets, the resulting set would be countably infinite. Let’s combine the set N with the set of negative integers {‌ , −đ?&#x;Ľ, −đ?&#x;¤, −đ?&#x;Ł} and place the combined set in the following linear order đ?&#x;˘ → −đ?&#x;Ł → đ?&#x;Ł → −đ?&#x;¤ → đ?&#x;¤ → −đ?&#x;Ľ → đ?&#x;Ľ → ‌

In other words, odd terms are defined by (đ?‘– − đ?&#x;Ł)/đ?&#x;¤ and even terms are defined by −đ?‘–/đ?&#x;¤ where đ?‘– ∈ Z+ is the position of the number in the linear order. This linear order shows that the combined set is countably infinite. But there is more to infinite sets! Cantor showed that there are at least two types of infinity: countable as we have discussed above and uncountable. With uncountably infinite sets it is impossible to construct a one-to-one mapping between their elements and the positive integers. The set of real numbers R is uncountably infinite. In fact, its subset [đ?&#x;˘, đ?&#x;Ł) ⊂ R turns out to be uncountably infinite too. Cantor showed that there is no one-to-one mapping between Z+ and [đ?&#x;˘, đ?&#x;Ł).

So let’s illustrate Cantor’s idea by trying to construct some linear order for the real numbers from [đ?&#x;˘, đ?&#x;Ł) which is equivalent to the natural order on Z+ and see where this goes wrong. For our linear order we will be continuously generating random numbers đ?‘Ľ ∈ [đ?&#x;˘, đ?&#x;Ł) and placing them one after another, skipping numbers which were already generated. We might get a linear order similar to 37

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chalkdust this example:

đ?&#x;˘.đ?&#x;łđ?&#x;Ľđ?&#x;Ťâ€Ś → đ?&#x;˘.đ?&#x;˘đ?&#x;ąđ?&#x;§â€Ś → đ?&#x;˘.đ?&#x;Ľđ?&#x;Śđ?&#x;ľâ€Ś → â‹Ż → đ?&#x;˘.đ?&#x;¤đ?&#x;Şđ?&#x;Łâ€Śđ?&#x;łâ€Ś → â‹Ż

Now Cantor’s diagonal argument will help us to construct a mysterious real number from [đ?&#x;˘, đ?&#x;Ł) that does not fit into the linear order! In our approach we plan to construct the new real number so that it differs from every number in the linear order in at least one digit. The new number has its first digit different from the first digit of the first number in the linear order, its second digit different from the second digit of the second number in the linear order, and so on. We change digits by taking the original digit and adding đ?&#x;Ł to it if the digit is from đ?&#x;˘ to đ?&#x;Ş and substitute the digit đ?&#x;Ť with đ?&#x;˘. In other words, we take diagonal digits of elements from the original linear order and add đ?&#x;Ł to each of them or substitute đ?&#x;Ť with đ?&#x;˘ and construct the new real number from this array of digits. For our example the new real number would be đ?&#x;˘.đ?&#x;Şđ?&#x;¨đ?&#x;˘â€Śđ?&#x;Şâ€Ś

This new number cannot appear in the linear order: it differs from the first number in the linear order with its first digit, with the second number in the order with its second digit and so on. Yet it is a perfectly good real number! So there are more real numbers in [đ?&#x;˘, đ?&#x;Ł) than positive integers. We come to a conclusion that these sets do not have equal sizes. There are plenty more surprises in set theory! We have seen that there are at least two different sizes of infinity, but are there any infinities which are bigger than countable, but smaller than uncountable? For that question Cantor formulated his continuum hypothesis which reads: ‘There is no set whose cardinality is strictly between that of the positive integers and the real numbers.’ The continuum hypothesis can neither be proven nor disproven using the standard axioms of set theory; it stands independent of those axioms. In other words, you can create a set theory where the continuum hypothesis is true, or you can create a set theory where it is not. Cantor’s work creating set theory led to captivating results by other mathematicians, like Russell’s paradox—a famous logical contradiction which has profound implications for the foundations of mathematics. So dive in to learn more! Andrei Chekmasov Andrei Chekmasov is a 10th grader at Cypress Bay High School in Weston, Florida. At school, he enjoys studying calculus and computer science, as well as hanging out with friends. When not wrestling with schoolwork, Andrei enjoys building Lego models and competing in maths and coding competitions.

Six months ago, we didn’t know... ...that the Duffin–Schaeffer conjecture theorem is true. In July, James Maynard and Dimitris Koukoulopoulos announced a proof of this important result in metric number theory that was first conjectured in 1941. This results tells us how well most irrational numbers can be approximated by rational numbers in terms of the size of the denominator of the approximating rational. chalkdustmagazine.com

38

tessellating shortbread You will need 4 kg butter, 6 kg our, (đ?&#x;¤ + đ?œ€) kg sugar, an oven that has been preheated to 220°C, a rolling pin, an oven tray, a cooling rack, a knife or cookie cutter

Instructions

1

Combine 4 kg butter, 6 kg our and 2 kg sugar in a large bowl with your hands.

Roll out the dough on a at surface.

3

2

Cut the dough into equally shaped quadrilaterals.

Place quadrilaterals on the baking tray and bake for đ?&#x;˘.đ?&#x;Łđ?&#x;¨Ě‡ hours.

5

4

Place quadrilaterals on the cooling rack, sprinkle with đ?œ€ kg sugar, leave to cool, then tessellate and eat. Tube map platonic solids, FrĂśbel stars and slide rules: more How to make at d chalkdustmagazine.com 39

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G

–M 1982

Zubair Junjunia

Alice ďŹ rst generates a private & public key and sends the public key to the server. Bob retrieves the public key from the server & encrypts his message.

Bob sends the encrypted message and Alice uses the private key to decrypt it.

Introduction Goldwasser–Micali was the ďŹ rst public-key encryption scheme that is probabilistic: it yields dierent cipher texts when encrypting the same message. It was developed by ShaďŹ Goldwasser and Silvio Micali in 1982.

Quadratic residue Let đ?‘? be prime. A number đ?‘Ľ is called a quadratic residue modulo đ?‘? if there exists an integer đ?‘Ś  such that the dierence between đ?‘Ľ and đ?‘Ś đ?&#x;¤ is a multiple of đ?‘?, or in other words đ?‘Ľ = đ?‘Ś đ?&#x;¤ (mod đ?‘?). Otherwise, đ?‘Ľ is called quadratic non-residue modulo đ?‘?.

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Legendre & Jacobi +đ?&#x;Ł The Legendre symbol is deďŹ ned: ( đ?‘?đ?‘Ľ ) = { −đ?&#x;Ł

if đ?‘Ľ is a quadratic residue mod đ?‘?, if đ?‘Ľ is a quadratic non-residue mod đ?‘?.

The Jacobi symbol is deďŹ ned: ( đ?‘ đ?‘Ľ ) = ( đ?‘?đ?‘Ľ ) ( đ?‘?đ?‘Ľ ) ... ( đ?‘?đ?‘Ľ ), where đ?‘?đ?&#x;Ł đ?‘?đ?&#x;¤ ...đ?‘?đ?‘˜ is the prime đ?&#x;Ł

đ?&#x;¤

đ?‘˜

factorisation of đ?‘ . If ( đ?‘ đ?‘Ľ ) = −đ?&#x;Ł then đ?‘Ľ is a quadratic non-residue; but if it is đ?&#x;Ł, then đ?‘Ľ may or may not be a quadratic residue.

Generating a key Alice must ďŹ rst generate a private & public key Alice picks two primes, eg đ?‘? = đ?&#x;Łđ?&#x;Ł and đ?‘ž = đ?&#x;Łđ?&#x;Š, and calculates their product, đ?‘ = đ?&#x;Łđ?&#x;Şđ?&#x;Š. She must now pick a number đ?‘Ľ which is a quadratic non-residue modulo đ?‘ but has a Jacobi symbol of +đ?&#x;Ł. She does this by picking a random number and checking that it is a non-residue. In this example, she picks đ?‘Ľ = đ?&#x;Š as this is a non-residue mod đ?&#x;Łđ?&#x;Ł and mod đ?&#x;Łđ?&#x;Š. She has now generated the public key (đ?&#x;Łđ?&#x;Şđ?&#x;Š) and private key (đ?&#x;Š).

Encrypting Bob uses the public key to encrypt his message Bob now wants to use the public key (đ?‘ = đ?&#x;Łđ?&#x;Şđ?&#x;Š) to encrypt his message (in binary) and send it to Alice. As an example, suppose he is encrypting đ?‘? = đ?&#x;˘. He picks a random number between đ?&#x;Ł and đ?‘ that is coprime to đ?‘ , eg đ?‘Ś = đ?&#x;Ť. Bob calculates the cryptogram by calculating đ?‘? = đ?‘Ś đ?&#x;¤ đ?‘Ľ đ?‘? (mod đ?‘ ). In this example, Bob ďŹ nds đ?‘? = đ?&#x;Ťđ?&#x;¤ Ă— đ?&#x;Šđ?&#x;˘ = đ?&#x;Şđ?&#x;Ł (mod đ?&#x;Łđ?&#x;Şđ?&#x;Š) = đ?&#x;Şđ?&#x;Ł. Bob does this for every bit in the message and sends the cryptogram to Alice. Because Bob used a random number in his calculation, this cryptosystem is probabilistic: the next time the same algorithm is run, another number would be used resulting in a dierent cryptogram.

Decrypting Alice uses the private key to decrypt Bob’s message Alice receives the cryptogram and must now convert it back to original text. The ďŹ rst number she receives is đ?&#x;Şđ?&#x;Ł and she checks whether it is a quadratic residue modulo đ?&#x;Łđ?&#x;Ł and đ?&#x;Łđ?&#x;Š. It is for both. If one or more đ?‘Ľ had been included in the product when calculating đ?‘?, then đ?‘? would be a quadratic non-residue in either mod đ?&#x;Łđ?&#x;Ł or mod đ?&#x;Łđ?&#x;Š. đ?&#x;Şđ?&#x;Ł is a quadratic residue in both mods, so no copies of đ?‘Ľ were included in the product. From this, Alice can deduce that đ?‘? = đ?&#x;˘ so she has decrypted the ďŹ rst bit. Alice does this for every bit, then has decoded the message. 41

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chalkdust

Crocheting fractals Felix Stokes

E

ver since getting into the mathematical field of really cool shapes (I think some call it ‘topology’), I’ve been keen to bring as many of these shapes to life as possible. When you’re holding a shape in your hands and it still doesn’t make sense in your head, that’s when you know you’ve encountered one of the best. My method of choice for making these shapes is crochet. It’s essentially a way of connecting a series of loops of yarn to gradually build up a surface. Before I start, I have to mention the book Crocheting Adventures with Hyperbolic Planes by Daina Taimina: it’s what got me into mathematical crochet, and it goes into a lot of detail about geometry as well as providing plenty of patterns to try out. However, as much as I love them, hyperbolic surfaces aren’t going to be the topic of this article. The crocheting project I’m going to discuss instead is one which I developed myself, after my love of making cool shapes had been going on for a little while. At this point, I’d played around with various circle circumference functions, crocheted ‘integrals’ and ‘derivatives’ of the same function, and made Möbius strips, but what I really wanted to try and create was some kind of fractal. Choosing the fractal was easy enough: I wanted something that was made using a relatively simple rule, and didn’t get too squiggly to be difficult to make a decent approximation of. The Koch curve, for instance, despite being simple, gets very squiggly and has a lot of angles and turns. Likewise, the dragon curve, beautiful as it is, has nowhere for the yarn to attach to so we cannot create a continuous surface. In the end, I settled on Sierpinski’s triangle. It’s simple, it’s modular so I could potentially make smaller pieces and sew them together, and all the points are connected to each other. chalkdustmagazine.com

42

chalkdust The next question, then, was how? I decided to keep it simple: I wanted to make a long chain, and by connecting it to itself at certain points using slip-stitches, I wanted to trace out all the lines of the triangle. For non-crocheters, a ‘chain’ is exactly what you imagine: it’s just a long line of loops all linked to the next. A slip-stich is like adding an extra loop, except not only is the loop linked to the whole chain behind it, it’s also connected to the crochet at another point. I’ve drawn a few sets of diagrams below so you can picture what goes on. (1)

(2)

(3)

(4)

Creating a chain: (1) The yarn is looped around the hook, then pulled through the previous loop. Repeating this forms a chain, as shown in (2). Making a slip-stitch: (3) The hook is inserted into a stitch in the chain, then the yarn is looped around the hook. The yarn is then pulled through both stitches to form (4).

At this point, I’d figured out what I wanted to make and how. One thing I didn’t know, however, was if it was even possible. Can you create Sierpinski’s triangle by making a chain and attaching it to itself at certain points using slip-stitches? You can simplify the problem a little bit by removing the context of crochet from the equation, and just thinking about drawing a continuous line that connects to itself at certain points. In other words: is it possible to draw Sierpinski’s triangle in one continuous line? If so, then I’d be able to trace the path of the line with crochet, and create it. It might be fun to stop reading here and have a go at solving this problem yourself. As soon as I realised this was a problem about tracing a shape with one continuous line, my mind went straight to graph theory. Graph theory is another area of maths which I adore (my interests seem to be anything that isn’t in the A-level maths syllabus), so I was very excited to be using it to solve a genuine real-world problem I’d come up with. It’s the study of nodes and the connections between them, and is often used to figure out the best route between places (by modelling locations as nodes, and roads as the connections). I decided to start by modelling the basic first iteration of the triangle as a graph, or a series of nodes and connections, as you can see on the left on the next page. 43

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chalkdust Using the draw function on Excel (which has the handy property of allowing you to ‘replay’ your drawing, so you can watch the path being traced again), I had a quick go at tracing the lines, and found a solution, as you can see on the right:

Of course, a proof consisting of one example for one iteration wasn’t going to be sufficient, but at least I had some evidence that it was probably going to be possible. This was the point where I decided to properly bring out the graph theory, and apply an idea first proposed by Euler when he was solving the Seven Bridges of Königsberg problem. The Seven Bridges of Königsberg is a fairly famous problem. A series of seven bridges all cross the river at various points, and the question is whether it is possible to walk across all seven bridges, starting anywhere, but only crossing each bridge once. It turns out to be impossible, but the reasoning behind it is what is key to solving our Sierpinski’s triangle problem. The first step—as with our triangle—is to remove the problem from its context. We can model the top and bottom banks as nodes, as well as the ‘islands’ in the middle and on the right. The bridges become connections between the nodes.

The map simplified into a graph.

A map showing the seven bridges over the river at Königsberg.

Euler solved this problem by considering two kinds of nodes: a node with an even number of connections, and a node with an odd number of connections. Let’s start with the even node. If you were to start your walk from somewhere outside the node, then ‘entering’ the node would ‘cost’ you one connection and leaving it would ‘cost’ you one more. So passing through the node will cost you two connections. Because there are an even An ‘even’ node with four connumber of connections, and passing through the node uses up nections. an even number of connections too, you can ‘clear’ all the connections just by passing through the node as many times as you need. These kinds of even nodes chalkdustmagazine.com

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chalkdust are therefore redundant when your path starts outside the node. But what if you want to start your walk ‘inside’ the node? Again, because the number of connections are even, you have to leave the node as many times as you enter: starting outside the node means that if you entered the node, you have to leave it. Starting inside the node means that if you leave the node, you have to enter it again. In other words, if you start your walk on an even node, you have to finish it there. The next situation to consider is the odd node. This time, if you start your walk outside the node and pass through it as often as you can, you will always be left with one connection remaining. This also means that if you start your walk outside the node, you An ‘odd’ node with three conwill have to finish your walk inside the node. The same logic can nections. be used to show that if you start your walk inside the node, you have to finish it outside, and at a different node. This means that if you have an odd node, you either have to start your walk there or end it there. Let’s recap the rules we’ve just worked out: 1. For an even node, we can disregard it if we start our walk somewhere else. Otherwise, we have to both start and finish there. 2. For an odd node, we have to either start there or finish there. Now, clearly there’s only going to be one start point and one end point to our walk. That means we cannot have more than two odd nodes anywhere in our graph. It’s also impossible to have a graph with just one odd node. This leaves us with two situations where we know we’ll be able to trace a path that goes through all of our connections only once: either there are no odd nodes, or there are two of them. If we go back to the Seven Bridges of Königsberg problem and count the number of connections to each node, we can see that each of the four nodes has an odd number of connections. Therefore, it is impossible to walk across each bridge in Königsberg only once. An exciting discovery for the world of graph theory, perhaps, but rather disappointing for a visiting tourist. However, what we’re really interested in here is whether or not it’s possible to draw Sierpinski’s triangle in one continuous line. If we return to my little diagram and count the number of connections at each node, every single node has an even number of connections. We’ve already proven it’s possible to draw this shape in one line using an example, but now we can also prove it using graph theory. So, we’ve proven it’s possible to crochet any Sierpinski’s triangle, right? Not quite. To show it’s possible for any iteration of Sierpinski’s triangle, we’re going to use a trick whereby we show that if it’s possible for one iteration, it’s possible for the next one too. I’ve been The graph of Sierpinski’s triangle again. throwing around the term ‘iteration’ a lot, but I haven’t actually specified what I mean by it. In the case of this triangle, let’s say that each time you make an 45

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chalkdust iteration, you take three of your current triangles and glue them together to make a bigger one. In the above example, you can see that we took a regular triangle and attached three together at each corner. For the next iteration, we’ll take three copies of our triangle from the current iteration, and stick them together to make a bigger triangle again. You’ll notice that I’ve coloured in three pairs of nodes in the diagram on the right. These pairs of nodes are the only nodes whose number of connections will change after an iteration. You’ll also notice that I haven’t drawn any extra connections to connect the nodes of each pair. That’s because these nodes actually overlap when you make an iteration. They both ‘merge’ to create one of the corners of the new empty centre triangle. Regardless of what the rest of the triangle looks like, we know that the corners at the very edge of Sierpinski’s triangle will have The second iteration of Siertwo connections, because they’ll be corners of a triangle. So, pinski’s triangle. when we combine these corners into one node like on the diagram, we’ll end up with four connections. This means that each time we iterate our triangle, we will continue to have only even nodes, and we will therefore continue to have a shape which we can crochet. Since we’ve also shown that it’s possible to crochet the first iteration, we now know that it is possible to create any iteration after the first one.

The finished product!

Satisfied, I sat down with some yarn and a ridiculous number of stitch markers, and a few hours later I had managed to create a fourth-iteration Sierpinski’s triangle. Of course, this is just the fourth iteration, and it only comes out to about 25 cm in length. One of the many joys of combining maths with crochet is the number of different directions and adaptions you can easily make, allowing us to ask many different questions about crocheting fractals. How many iterations would you need to make a triangle that was as large as a person? Or a house? Or how about the whole of Russia? It might also be fun to think about whether it would be possible to create other shapes and patterns like this. For example, would it be possible to crochet a cube? If not, how might you adapt the design to make it possible?

If you want to crochet your own fractal, full instructions can be found at

d chalkdustmagazine.com.

Felix Stokes Felix is primarily a classics student at Oxford who loves knitting, crochet, and all things yarn. He’s also a huge fan of topology, making mathematical craft one of his favourite hobbies.

c felix.stokes@bnc.ox.ac.uk chalkdustmagazine.com

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Quiz Answer the following questions, adding up your points as you go. At the end, you will be given some life advice based on your score.

1. What is the best day of the year? a) Pi Approximation Day +5 points b) Pi Day +3 points +2 points c) Bonfire Night d) Halloween +0 points

3. What is the best area of maths? a) Numerical analysis +5 points b) Category theory +4 points +2 points c) Statistics d) Physics +0 points

2. Do you like maths? a) Yes, loads b) It’s not too bad c) A little d) What’s maths?

4. What is the best number? a) 11 b) i c) π d) τ

+5 points +3 points +2 points +0 points

Results 0 1 2–20

+5 points +3 points +2 points +0 points

You should not write an article for issue 11 of Chalkdust. This is impossible, you have cheated. You should write an article for issue 11 of Chalkdust and send it to

c contact@chalkdustmagazine.com Did you know... ...that not all conjectures turn out to be true? Karol Borsuk conjectured in 1932 that all finite regions of đ?‘›-dimensional space can be split into đ?‘› + đ?&#x;Ł smaller regions. This was eventually shown to be false by Jeff Kahn and Gil Kalai in 1993.

30 by Daniel Griller How many positive integers have exactly 30 positive factors, one of which is 30? You can find more puzzles by Daniel in his (highly recommended) book Elastic Numbers, on his blog d puzzlecritic.wordpress.com, and on Twitter a @puzzlecritic.

Did you know... ...that the area of the biggest hyperbolic triangle is finite? The area is Ď€, and corresponds to a triangle which has three 0° angles.

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#10 Set by Humbug 1

2

4

3 9

5

6

10

12

8

16

17

11

13 14

15 18

22

19

21

25

27

28

32

33

29

31 34 37

36

40

41 44

26

30

35

43

20

23

24

50

38

39

42 45

46

47 49

7

48

51

52

53

Rules Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0. Use of Python, OEIS, Wikipedia, etc is advised for some of the clues. To enter, send us the sum of the across clues via the form on our website (d chalkdustmagazine.com) by 2 February 2020. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 14 March 2020. One randomly-selected correct answer will win a ÂŁ100 Maths Gear goody bag, including non-transitive dice, a Festival of the Spoken Nerd DVD, a dodecaplex puzzle and much, much more. Three randomly-selected runners up will win a Chalkdust T-shirt. Maths Gear is a website that sells nerdy things worldwide. Find out more at d mathsgear.co.uk chalkdustmagazine.com

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Across

1 9 times this number is an anagram of this number. 3 All the digits of this number are equal. 10 Not equal to 32A or 40A. 11 A multiple of 8D. 12 The sum of 25A and 43A. 13 49A reversed. 14 Each digit of this number∗ is either one less or two less than the previous digit. 15 31D reversed. 18 5 more than a palindrome. 21 The highest common factor of 26A and 34A. 22 6 more than a palindrome. 23 Each digit of this number∗ is either one more or three less than the previous digit. 24 Not equal to 32A or 40A. 25 The difference between 31D and 15A. 26 The highest common factor of 21A and 34A. 27 Each digit of this number†is either the sum or the difference of the digits either side of it. 32 The highest common factor of 10A and 24A. 33 3 more than a palindrome. 34 The highest common factor of 21A and 26A. 35 Each digit of this number∗ is either one more or three less than the previous digit. 37 The difference between 49A and 13A. 40 The lowest common multiple of 10A and 24A. 41 4 more than a palindrome. 43 25A reversed. 45 7 more than a palindrome. 47 37A reversed. 48 The sum of 47A and 37A. 49 Each digit of this number∗ is (strictly) less than the previous digit. 51 The sum of the digits of 23A. 52 All the digits of this number are equal. 53 Equal to 1A.

(5) (9) (2) (3) (4) (3) (7) (3) (4) (2) (3) (9) (2) (3) (2) (15) (2) (3) (2) (9) (3) (2) (4) (3) (7) (3) (4) (3) (2) (9) (5) 49

Down

1 A multiple of 9D. 2 A multiple of 1D. 4 One of the digits of 33A appears more than once in this number. 5 The sum of this number, 23A and 30D is greater than 1,000,000,000. 6 The length of one of the other two sides of a triangle whose hypotenuse is 2665 units long and whose sides all have integer length. 7 A solution of đ?‘Ľ đ?&#x;¤ −(50D)đ?‘Ľ +(11A) = đ?&#x;˘. 8 A solution of đ?‘Ľ đ?&#x;¤ −(50D)đ?‘Ľ +(11A) = đ?&#x;˘. 9 A palindrome. 13 A prime number. 16 Each digit of this number†is either the sum or the difference of the digits either side of it. 17 The sum of this number’s digits is the square of its first digit. 19 2 more than a palindrome. 20 The sum of this number and 33A is greater than 10,000. 22 A non-prime number with only one (distinct) prime factor. 24 This number contains only two different digits. 27 A multiple of 13D. 28 Each digit of this number∗ is one less than the previous digit. 29 8 more than a palindrome. 30 A multiple of the final digit of 22A. 31 The sum of this number’s digits is 7. 36 A multiple of 9 that is less than 4D. 38 33A less than 39A. 39 Equal to 2D. 42 Equal to 18A. 44 1 more than a palindrome. 46 A factor of 14A. 49 The first digit of this number is one of the digits of 45A. 50 More than 8D. â€

(6) (6) (6) (8) (4)

(2) (2) (4) (2) (6) (5) (4) (4) (3) (6) (5) (4) (4) (8) (3) (6) (6) (6) (4) (4) (2) (2) (2)

∗ Except the first. Except the first and last.

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chalkdust

Spotlight on: Pamela E Harris

Talithia Williams

A

s far as Pamela E Harris knew when she was growing up, there were no Latina mathematicians. Through almost 20 years of schooling she had never met one. Then, one year shy of earning her PhD, she did. It meant a lot to know that she wouldn’t be the only one. The lack of role models who shared a similar heritage and background made Harris’ experience one of isolation. She says she feared that she wouldn’t be able to succeed as a mathematician. However, in 2012 when Harris attended a meeting of the Society for the Advancement of Chicanos/Hispanics and Native Americans in Science (Sacnas), it changed her life. She is now part of a large, supportive community that uplifts and helps each other become leaders in their respective fields.

The making of a mathematician Harris spent her childhood in Mexico and emigrated to California with her family when she was eight. Things were rough financially and, after a short return to Mexico, her family emigrated to Wisconsin. There, she attended Marquette University, and it’s here that she began to think seriously about becoming a mathematician. She says, “During my fourth year as an undergraduate student, my real analysis professor said, ‘When you go to graduate school…’. With this comment alone, she changed the course of my life. Her comment started me on the path to graduate school but, more importantly, her belief in my ability to succeed motivated me for years past the start of my graduate programme.” chalkdustmagazine.com

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chalkdust Harris attended the University of Wisconsin–Milwaukee where she earned a masters, then a PhD in mathematics. Her research interests became algebra and combinatorics. She explains her work in this way: “Consider the following combinatorial problem: In how many ways can the positive integer đ?‘› be written as a sum of positive integers (ignoring the order)?â€? For example, the number 3 can be written in the following three ways: 3, 2 + 1, 1 + 1 + 1. “Although this process is simple, determining a Flickr user MTSOfan, CC BY-NC-SA 2.0 formula for the partition function, which counts the number of Partitioning an integer involves integer partitions of đ?‘›, eluded generations of mathematicians dividing it up into smaller parts. and was only recently solved by Ken Ono, Jan Bruinier, Amanda Folsom, and Zach Kent in 2011. Their formula relied on the new and surprising discovery that partitions are fractal in nature.â€?

Finding formulae Now an assistant professor in the department of mathematics and statistics at Williams College in Massachusetts, Harris researches vector partition functions and graph theory: work that has been supported through awards from the National Science Foundation and the Center for Undergraduate Research in Mathematics. A vector partition function computes the number of ways that one can write a vector, say đ?’— , by summing given vectors {đ?’‚đ?&#x;Ł , đ?’‚đ?&#x;¤ , ‌} in such a way that the coefficient of each đ?’‚đ?‘– is a non-negative integer. For example, one could ask, how many ways are there to make ÂŁ5 from standard British coins? In this case the (one-dimensional) vector đ?’— is 5 and the set of given vectors đ?’‚đ?‘– is just the set of coin denominations: {đ?&#x;¤, đ?&#x;Ł, đ?&#x;˘.đ?&#x;§, đ?&#x;˘.đ?&#x;¤, đ?&#x;˘.đ?&#x;Ł, đ?&#x;˘.đ?&#x;˘đ?&#x;§, đ?&#x;˘.đ?&#x;˘đ?&#x;¤, đ?&#x;˘.đ?&#x;˘đ?&#x;Ł}. Harris says, “Vector partition functions have many interesting properties, but finding formulae for vector partition functions is also very difficult.â€? In particular, Harris has worked on a vector partition function known as Kostant’s partition function which is important for representation theory. Representation theory is a branch of mathematics that tries to solve problems about abstract algebraic objects by representing their elements as matrices, which are easier to work with. In the case where the abstract object is a Lie algebra (pronounced ‘Lee’) understanding the representation turns out to involve combinatorics and Kostant’s partition function.

Inspiring the next generation Fields medallist Artur Harris also enjoys working with undergraduates on mathematical Avila is one of the bestresearch. “I find that many undergraduate students do not know known Latin American what mathematical research is about, or how one does research. mathematicians. Working to help them understand how as a mathematician we can take a problem and generalise it further to find new results, is one of the most rewarding aspects of my job.�

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chalkdust Being Latina, an immigrant and the first in her Mathematics has taught me to be pafamily to graduate from university, Harris is tient, to work hard and to be resilient. firmly and actively dedicated to improving diI know most times I will fail to answer versity and retention rates among women and the questions I pose, but I do know that minorities in science, and in mathematics in along the way I will grow and develop particular. She travels widely—her favourite new insights. perk of being a mathematician—to share research findings and to co-organise research symposia and professional development sessions for the national conference of Sacnas. She was a Project NExT (New Experiences in Teaching) fellow from 2012 to 2013, and is an editor of the e-mentoring blog of the American Mathematical Society. Her work has created new research opportunities for underrepresented students that support and reinforce their identity as scientists. In 2016, she helped develop and create the website d lathisms.org, an online platform that features the extent of the research, teaching and mentoring contributions of Latinxs and Hispanics in the mathematical sciences.

Impact beyond mathematics Harris is grateful for the support of her community and her mentors, including that first analysis professor who gave her her early self-belief. “I have been very lucky to be surrounded by peers and mentors. They often remind me that as a Latina mathematician, my work has an impact outside of the walls of my institution and that I can make a difference in the mathematical community. Their support has been invaluable throughout my career, and I am grateful to have them in my corner. I certainly wouldn’t be where I am today without them.�

Lisa Jacobs

Harris (back centre) with her students.

Talithia Williams Talithia is an associate professor of mathematics at Harvey Mudd College. Her book Power in Numbers features the stories of more than 30 women in mathematics.

d talithiawilliams.com a @Dr_TalithiaW

l dr_talithiaw

Did you know... ...that in 1919, the Hungarian mathematician George PĂłlya conjectured that more than half of the natural numbers less than any given number have an odd number of prime factors? In 1958, Brian Haselgrove proved that PĂłlya was wrong by finding a number close to đ?&#x;Łđ?&#x;˘đ?&#x;Ľđ?&#x;¨ such that fewer than half of the natural numbers less than it have an odd number of prime factors.

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On this page, you can find out what we think of recent books, films, games, and anything else vaguely mathematical. Full reviews of many of the items featured here can be found at d chalkdustmagazine.com

The Maths of Life and Death Kit Yates A really enjoyable and entertaining journey through many topics in applied maths, including the maths used in medicine, the courtroom, and the media, plus a really interesting chapter looking at mathematical biology, Kit’s own area of research. Highly recommended for mathematicians, scientists, and everyone inbetween.

ggggh

Geometry Puzzles in Felt Tip

Mathigon.org Philipp Legner

Catriona Shearer

The perfect inductive-learning complement to school mathematics. Plenty of intuitive interactive exercises, which are adapted and personalised according to each user’s needs.

The perfect book for anyone who loves Catriona’s Twitter puzzles (a @CShearer41).

ggggg

ggggh

Inventing the mathematician Sara N Hottinger

Making Mathematics with Needlework

Sometimes challenging, but peppered with fascinating perspectives on a mathematician’s place in society.

Sarah-Marie Belcastro and Carolyn Yackel One of the best collections of creative mathematical ideas.

ggghi

ggggg

Greenwich Maths Time

Integers

Some of maths communication’s biggest hitters showcasing fun, interactive exhibits.

Overrated.

ggggi

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Secrets, surveys and statistics Flickr user Wessex Archaeology, CC BY-NC-SA 2.0

Paula Rowińska

H

ave you ever shoplifted? Have you ever cheated on your partner? Have you driven a car under the influence of alcohol or drugs?

If you ask such sensitive questions in a survey, don’t expect honest answers. Participants might worry that you wouldn’t be able to provide full anonymity and their embarrassing responses might end up in the wrong hands. They may skip more sensitive questions or, even worse, just blatantly lie, which would make your survey utterly useless. Luckily, there is a way to design fully anonymous surveys. All you need is money. No, I’m not asking you to bribe your respondents. A simple fair coin you keep in your pocket serves not only as the means to pay for services, but also as an excellent generator of events with some known probability (in this case with probability đ?&#x;Ł/đ?&#x;¤). In fact, a die or coloured balls in an urn would do, they’d just generate different probabilities.

Binary choices Let’s say that a question in your survey is a sensitive yes/no question, where ticking ‘yes’ would mean admitting to some embarrassing or even criminal activity. To minimise the incentive to lie, you might want to include the following extra instructions in the survey. Ask the participants to flip a coin before answering the question. If they get tails, they should respond truthfully. However, if they get heads, ask them to toss the coin again and record the results: ‘yes’ for heads and ‘no’ for tails. Since you won’t be able to distinguish between responses to the question and those chalkdustmagazine.com

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chalkdust who simply recorded the result of the coin flip, survey-takers can rest assured that they won’t be identified. Hopefully this will encourage honesty and give you some data to work with. But‌ how do you get any meaningful information from this coin-tossing exercise? Scientific studies usually require large samples, so let’s assume that you have managed to recruit a sufficiently large total number of subjects, which we will call đ?‘‡ . By the end of the study, you will have collected đ?‘‡ (total) answers, one from each participant: đ?‘Œ ‘yes’ responses and đ?‘ = đ?‘‡ − đ?‘Œ ‘no’ responses. On average, đ?‘‡ /đ?&#x;¤ of these will be truthful answers (since đ?‘‡ /đ?&#x;¤ people will get heads in the first coin flip), and đ?‘‡ /đ?&#x;¤ will be meaningless results of second coin tosses. This means that out of đ?‘‡ /đ?&#x;¤ people answering the actual question, đ?‘Œ − đ?‘‡ /đ?&#x;Ś confirmed and I prefer Ď„ charts. đ?‘ = đ?&#x;Ľđ?‘‡ /đ?&#x;Ś − đ?‘Œ denied the controversial statement. Now you know that on average (đ?‘Œ −đ?‘‡ /đ?&#x;Ś)/(đ?‘‡ /đ?&#x;¤) of the whole population (assuming that the sample was representative) have engaged in the embarrassing activity in question. Of course, if instead of a fair coin you decide to generate events with probabilities different to đ?&#x;Ł/đ?&#x;¤ using, for example, a biased coin or a die, you must adjust the numbers accordingly. Since the concept remains more or less the same, I’ll leave this as an exercise for the reader (just because I’ve always wanted to use this most-hated phrase of any PhD student).

A range of values Nice, but what if you want to know the number of times the respondent committed the crime as well as whether they had or not? Or in other words, what if the answer to your question is not just ‘yes’ or ‘no’ but a range of values? Don’t worry, maths will save your survey. Imagine you want to identify some quantity đ?‘‹ , which the respondents aren’t too happy to share. If they’re able to generate random numbers with known and positive mean đ?‘š and variance đ?‘Ł , you’re safe. You can ask them to generate one such number đ?‘? and, without disclosing it, answer the question with đ?‘Œ = đ?‘‹ đ?‘? . Again, you have no way of learning the value they’re so afraid of sharing. However, đ?‘‹ and đ?‘? are independent variables: the value of đ?‘‹ doesn’t influence the generated random number đ?‘? , and vice versa. Then the average of đ?‘Œ is just the product of averages of đ?‘‹ and đ?‘? , where the latter is equal to đ?‘š. Therefore, you can estimate the average value of đ?‘‹ as the average of responses đ?‘Œ divided by đ?‘š. You can even quantify the error of your estimation, but I’ll spare you the details. (If you’re interested, you can find them in the paper Scrambled randomized response methods for obtaining sensitive quantitative data by Eichhorn and Hayre.) So, please flip a coin. If you get heads, toss it again and give me ‘yes’ for heads and ‘no’ for tails. Otherwise, let me know: have you ever attempted to drink and derive? Paula RowiĹ„ska As a PhD student at Imperial College London, Paula uses maths and stats to study the impact of wind energy on electricity prices. The title of her TEDx talk ‘Let’s have a maths party!’ seems to summarise her two favourite activities.

d paularowinska.wordpress.com a @PaulaRowinska 55

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Puzzles

Looking for a fun puzzle but not got time to tackle the crossnumber? You’re on the right page.

Pieceword Arrange the pieces below in the grid to form a (mostly) mathematical crossword. One piece has been placed for you. The completed grid has order 2 rotational symmetry. A I R I P T I C

E D O P T

E R O T

T A N T S S

G E R A O

U

O E R M

Z E R D E P

T R I H M P E

A M T R I E D

I U S P

O R R A D I

C O M O V A C

A I R I P T I C

S G T O R U A

D D E A R

T Y I R M E A

M A T E T D

E S E P

O U N P

P L E E R E D

D E D I E N E

R A N G S

P O S Y U U M

A R C H L G R O

S L E T E

Arrange the digits I Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct.

+

T S U R

B

The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, đ?&#x;Ś + đ?&#x;Ľ Ă— đ?&#x;¤ is 14, not 10.

Ă— Ă— = 72

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+ +

Ă— Ă— = 63

+ + +

C H

Ă— Ă— = 80

= đ?&#x;Łđ?&#x;¤ = đ?&#x;Łđ?&#x;¨ = đ?&#x;Łđ?&#x;Š

chalkdust

Arrange the digits II Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, đ?&#x;Ś + đ?&#x;Ľ Ă— đ?&#x;¤ is 14, not 10.

+ + = 18

3

4

5

7

6 8

9

10 12

+ á

− + = 3

+ − Ă—

á Ă— = 24

= đ?&#x;Łđ?&#x;Š = đ?&#x;Łđ?&#x;¤ =đ?&#x;Ś

The only crossnumber by Humbug Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0.

1 2

+

Each clue gives one property that is only true of that clue: for example, 5A is a prime number, and none of the other numbers in the completed crossnumber are prime numbers.

11 13

Down 1 The only number that contains a 9. (2)

Across

4 The only cube number.

(2)

2 The only number that is the sum of (2) the square of the first đ?‘› primes (for some value of đ?‘›).

5 The only prime number.

(2)

3 The only even number.

(2)

7 The only multiple of 11.

(2)

4 The only multiple of 47.

(3)

8 The only square number.

(2)

9 The only factor of 612.

(2)

11 The only multiple of 7.

(2)

2 The only number whose digits add (3) to 18.

6 The only number whose digits are (3) all one more than the previous digit. 10 The only triangle number.

(2)

11 The only multiple of 13. (2) 12 The only number between 500 and (3) 13 The only number whose digits add (2) 560. up to 12. 57

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We love it when our readers write to us. Here are some of the best emails, tweets and letters we’ve been sent. Send your comments by email to c contact@chalkdustmagazine.com, on Twitter a @chalkdustmag, or by post to e Chalkdust Magazine, Department of Mathematics, University College London, Gower Street, London WC1E 6BT, UK.

Dear Chalkdust, Good afternoon. A friend has just introduced us to your fabulous magazine and my mathematically insatiable son has devoured issue 9. Rebecca, Cambridge

Dear Chalkdust , I’ve read the Matt Parker article, and the copy will stay in my work bag to read on the tube each day.

About to sit down and look at some of the puzzles in Chalkdust we got at #humblepi last night. Thanks a @mathsjem and a @standupmaths for organising!

a @WilkinsAmy

Samuel, London

Dear Chalkdust, I am an editor for Spektrum der Wissenschaft, which is a German popular science magazine affiliated with Scientific American. I really like your magazine and the work you do there. After reading the nice article of Tai-Danae Bradley about category theory, I decided that this would be also a nice topic for our magazine. Manon, Germany

My Monday was brightened up by the delivery of the latest Chalkdust! Copies available at the end of the maths corridor for any interested students (or teachers).

a @AquinasMath

They’ve arrived! I love dropping Chalkdust maths love bombs about a @UniofReading. Calvin, Reading a @C_J_Smith

Hello there! General Kenobi, Utapau chalkdustmagazine.com

58

They might not be giants “If I have seen farther,” the scientist said, “it’s not because I am a giant. “Great minds of the past have helped me get ahead; it’s their shoulders on which I’m reliant.”

Angela Brett

a @macaronique d angelastic.com m Cernoise

“Now listen to me!” said the great on whose shoulder the first one was glad to have stood. “I’m quite short of stature, it’s just that I’m older and those before me were so good.” And sure enough, this one was perched on the neck of a giantist of great renown who balanced in turn on another; by heck! It’s little guys all the way down. And some were thought giants, and some were thought midgets and some were thought nothing at all, but each would insist, “Those below were no idjits. It’s them that have made me so tall.” And scrambling around them their fans would aspire, to see something not seen before by climbing the tower of dwarves, ever higher for glimpses, or footholds, or more. Most could not scale to the summit in time, before their peak fitness would end. Some found it tough and abandoned the climb while some would, with vigour, descend: Aware that such heights were so taxing to reach, they helped to lift people and hopes, inventing new ladders and platforms to teach, securing and showing the ropes. “They might not be giants, but they must go far and that journey isn’t for me. I’ll boost them through science, raise them and the bar and profit from what they will see.” So said the teacher while lifting a child on shoulders so humble and stressed. The youth saw a vista that had them beguiled and bounded straight up to the crest.

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chalkdust

On the cover

Islamic geometry Emily Maw

F

rom the exquisite patterns of the Alhambra palace in Spain to a jigsaw puzzle on a rainy day, tessellations (tilings of the plane using shapes with no overlaps or gaps) are everywhere. They are sometimes used for practical reasons: providing durable and water-resistant surfaces, or for efficiencies of space (like hexagons in a honeycomb). And sometimes they are there for aesthetic reasons: tessellations are known to have been used in architecture since at least 4000BC when the Sumerians decorated walls with patterns of clay tiles.

Later, mosques were often decorated with tessellations— alongside calligraphy and arabesque plant forms—to avoid depicting people or animals that could become objects of worship. The repeating patterns are also said to be reminders of the infinite nature of God. Some of the earliest-known geometric forms in Islamic art can be found in the Great Mosque of Kairouan in Tunisia, which was established in the 7th century, but the artform soon spread across the Islamic world and diversified, developing regional characteristics. The significant intellectual contributions of Islamic mathematicians allowed the patterns to become much more complicated. Islamic designs are constructed on grids that require only a straight edge and compass to draw, building up from lines and circles that are repeated, overlapped chalkdustmagazine.com

60

Saeid Shakouri

The design that inspired the front cover of Chalkdust issue 10.

chalkdust and interlaced to form intricate patterns. The cover of this issue of Chalkdust features artwork by Samira Mian. The front cover is a pattern found on the shrine of Abdulazim Hasani, and the back cover is from a ‘jaali’ panel in the 14th century Al Rifa’i Mosque in Cairo. Samira researches historical patterns, and the constructs them by hand using efficient and mathematically-pleasing methods to obtain the correct proportions, before painting them in her own beautiful colour schemes. She used to be a maths teacher, but now runs Islamic geometry workshops—combining her creativity with her love of ruler and compass constructions. To create some of her patterns for yourself, check out m samiramian or l samira.mian.

Let’s tessellate

Samira Mian creates designs inspired by tessellations in Islamic art.

Tessellations aren’t just a beautiful application of ancient mathematics though; they involve lots of modern concepts and there are many problems that are still unsolved. The simplest tessellations are called ‘regular tilings’. These use only one regular polygon, and there are three of them: the plane can be tiled using the square, equilateral triangle, or regular hexagon. These tilings are also periodic: Contruction of part of the cover of this issue. Full instructions repeating patterns with translacan be found at d chalkdustmagazine.com tional symmetry in two directions. If you allow combinations of any shapes, periodic tilings can be categorised based on their symmetries into 17 different ‘wallpaper groups’. It is claimed that all 17 can be found in the Alhambra! Can the plane be tiled by repeating the same irregular shape? Conway came up with a sufficient criterion for deciding if a given shape tiles the plane, but no general rule has yet been found; although we now have a good understanding in the case of convex shapes (those where all interior angles are less than 180°). All triangles tile the plane (two stick together to make a parallelogram), and so does any hexagon in which each pair of opposite sides is parallel. In fact so do all quadrilaterals: two can be stuck together to make a hexagon. There are no tilings for convex đ?‘›-gons when đ?‘› is bigger than 6. But what about pentagons?

Regular pentagons can’t tile the plane because their internal angle (108°) doesn’t divide 360 (although they do tile other surfaces such as the sphere and the hyperbolic plane). In fact, it’s not known whether there is any other shape with order-5 rotational symmetry which tiles the plane. 61

autumn 2019

chalkdust But it is known that there are fifteen types of irregular, convex pentagon that do tile the plane. Four of them were discovered in the 1970s by Marjorie Rice, an amateur mathematician, after reading an article on tessellations by Martin Gardner in Scientific American, and the most recent one was only discovered in 2015 by Casey Mann, Jennifer McLoud-Mann, and David Von Derau. Michaël Rao gave a computer-assisted proof in 2017 that there are no more convex pentagons that work, so we now know all of the irregular, convex polygons that can tile the plane.

Tom Ruen, CC BY-SA 4.0

The 15th convex pentagon that tiles the plane.

Since they can all be made to tile periodically (though some can also be arranged to tile aperiodically) there does not exist a convex polygon that tiles only aperiodically—when a shifted copy of a tiling will never match the original. How about other combinations of shapes? Wang conjectured in 1961 that any set of tiles that tiles the plane can always be arranged to do so periodically. However, in 1966 his student Berger found a set of 20,426 tiles that only tile aperiodically. Robinson reduced it to six in 1971, and in 1974 Roger Penrose discovered an aperiodic set of two tiles. It is still not known if there is a single aperiodic tile (though it would have to be non-convex by our above observation).

You might think that an aperiodic pattern would be entirely without symmetry, but this is not the case! Penrose tilings have no translational symmetry, but they do have symmetries coming from the infinite repetition of any bounded patch of the tiling, and reflective and rotational (order 5 again!) symmetries of those patches. They are also ‘self-similar’, in that the same patterns occur at bigger and bigger scales (like fractals), meaning they can be created recursively using a process called inflation. Penrose tiling patterns have even been discovered in the structure of ‘quasicrystals’—substances with aperiodic order, including many aluminium alloys.

A periodic tiling on the Darb-e Imam shrine.

Although these discoveries are relatively recent, there is evidence to suggest that the properties of Penrose tiles were understood in the medieval Islamic world. In 2007 two physicists, Peter Lu and Paul Steinhardt, claimed that patterns on the 13th-century Darb-e Imam shrine in Iran showed a ‘nearly-perfect’ Penrose tiling that could not have been constructed without mathematical knowledge of the tiles’ properties. The same researchers also uncovered a 15th-century scroll that describes how to make self-similar patterns from the tiles, five centuries before their discovery in modern-day science. Emily Maw Emily is a PhD student at University College London. She works in symplectic geometry.

a @emilyharrietmaw

chalkdustmagazine.com

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Where has your copy of Chalkdust been on holiday? Let us know: c contact@chalkdustmagazine.com a @chalkdustmag l chalkdustmag 63

TOP TEN

This issue features the top ten issues of Chalkdust. To vote on the top ten pictures of scorpions, go to d chalkdustmagazine.com

At 10, it’s the tenth most egg-cellent issue of Chalkdust: issue 06.

At 9, it’s issue 02, the difficult second issue.

10

9

At 8, it’s John Lennon’s latest avant garde composition: issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09. issue 09.

6

4

2

8

At 6, it’s the issue that will explain why the vote to put this issue in sixth place is unfair: issue 03.

At 4, it’s the only issue printed on A5 paper (which was a mistake, just like the minidisc): issue 01.

At 5, it’s When I’m issue 04 by The Beatles.

5

At 3, it’s You’re my number three issue of Chalkdust by S Club issue 07.

3

After a whole year at number 1, it’s the issue that’s finally been knocked down to number 2: issue 08.

chalkdustmagazine.com

7

At 7, it’s Puff, the Magic Dragon Curves by Peter, issue 05 and Mary.

Topping the pops this issue, it’s a new entry and a brand new release: issue 10.

1

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