The odds of being crushed by a meteor are considerably lower (i.e. more likely) than those of winning the jackpot on the National Lottery.

— Quite Interesting (@qikipedia) January 11, 2018

In the account’s usual citationless factoid style, the Elves state that you’re more likely to be crushed by a meteor than to win the jackpot on the lottery.

The replies to this tweet were mainly along the lines of this one from my internet acquaintance Chris Mingay:

Should we not be getting almost weekly stories of people being crushed by a meteor then ?

— Chris Mingay (@GhostMutt) January 11, 2018

Yeah, why don’t we hear about people being squished by interplanetary rocks all the time? I’d tune in to that!

A couple of other helpful sorts have provided some extra data as context for this fact:

I asked on Twitter if any turbonerds keep a record of every jackpot ever, and of course they do: Peter Rowlett and Tim Stirrup both provided me with a link to Richard K. Lloyd’s comprehensive table, which reckons there have been 4749 winners, of which 3220 became millionaires.

4750 people have ever won the lottery (for a definition of ‘won’ that might not be the one we want, but it gives us an order of magnitude)

According to their website,4750 people have become millionaires since 1994 from UK lotto wins, so how many have been crushed?

— ste-b (@worldwarste) January 11, 2018

And only one person ever has been crushed by a meteor:

How can this possibly be true when only one person has ever been hit by a meteor in recorded history?

— Dan (@dev_meltus) January 11, 2018

I immediately hit the

(Hey, QI like to do it to their guests, so why can’t I?)

The statement sounds wildly incorrect on first inspection, so I reckon we’re not talking about the same kinds of odds.

It must be the case that:

- someone has worked out the odds of being killed by a meteor,
- someone has worked out the odds of winning the lottery, and
- someone has compared those two numbers.

I assume at least the first two someones were not QI Elves, and I reckon the third one probably wasn’t either. So, where did QI get their fact?

A search for “meteor lottery odds” got me this story on independent.co.uk published five days before QI’s tweet, so that’s probably their source. That links to “Review Journal”, a generically authoritative-sounding title, which turns out to be the Las Vegas Review Journal, who in 2015 published an article by someone affiliated with gobankingrates.com titled “20 things more likely to happen to you than winning the lottery”. That cheery listicle cites a 2008 article by Phil Plait on his Bad Astronomy blog where he cites Alan Harris’ answer to the Fermi question of working out your odds of being killed by a meteor, directly or indirectly. The “crushed” phrasing, which is a stronger statement than the one Harris looked at, seems to originate with the Las Vegas Review Journal. Maddeningly, Plait doesn’t give a citation for Alan Harris’s calculation and I can’t find a better source on Google, so the search stops here.

After all of that chasing, I’ve got a kind-of reputable source for the “1 in 700,000” odds of being killed by a meteor presented in the Independent article. That’s much better odds than the 1 in 45,057,474 chance of winning the lottery claimed by operators Camelot. We hear about people winning the lottery fairly often, so why isn’t “meteor squish” a journalistic cliché like “bus plunge”?

Well, the meteor figure is your *lifetime* odds of being killed, and the lottery figure is your odds of winning *each time you play*. That’s it – they’re measured in different units, effectively. Plait’s *Bad Astronomy* piece contained a good explanation of what the odds meant, but that got lost when the headline figure was spread in factoid form.

So we can’t compare the two numbers as stated – that’d be like me saying I’m taller than you are old. What can we do to get numbers for meteors and lotteries that we *can* compare?

One option is to assume both take place once – a meteor hits Earth, and you play the lottery. We know the odds of winning the lottery in one attempt, and one of Harris’s assumptions in his model was that an asteroid impact would kill everyone – so your probability of being killed is 1. No contest – you’re way more likely to be killed by an asteroid that hits Earth than for the lottery ticket you just bought to be a winner.

A more reasonable approach might be to look at your odds within a certain period of time. We’ve already got a figure of around 1 in 700,000 for being killed by a meteor in a 70-year lifespan, so we just need to get the corresponding figure – what are your odds of winning the National Lottery at least once in your lifetime?

Clearly, it depends on how often you play. My personal odds are zero – I’ve never so much as bought a scratchcard. Conversely, if you buy enough tickets, you can guarantee you win, a tactic executed to great success by Voltaire and later on some MIT students. Those strategies both relied on oversights in the rules of their respective lotteries to make them profitable, but if you’ve got a fortune to spare you could buy a National Lottery ticket corresponding to each combination of six balls and guarantee that exactly one of them will win.

For the sake of getting a reasonable number, let’s say you buy one ticket for each draw. There are two draws each week, so 104 draws each year. So your odds of winning the lottery at least once in 70 years is

At this point I wanted to use the fact that you can only play the lottery once you’re 16, and the life expectancy in the UK is 81.2 years, but I’ll stick with 70 years of playing so we can compare with the meteor number.

\[ 1 – \left( 1 – \frac{1}{45057474} \right)^{(104 \times 70)} \approx 1 \text{ in } 6190 \]

That’s a way, way lower number than the meteor number. So you’re vastly more likely to win the lottery in your lifetime than you are to be killed by a world-ending meteor – over 100 times more likely, in fact.

And if a meteor did kill everyone, you’d be unlikely to read about it in the news the next day.

]]>The results of the Royal Statistical Society’s Statistic of the Year competition, which we covered here when they announced a call for entries, have been announced. The winners include a UK Statistic of the Year (on the density of building in the UK – it’s lower than you think), an International Statistic of the Year (comparing the risk of a US citizen being killed by a terrorist or a lawnmower -guess which one is 34.5 times more dangerous), and five ‘highly commended’ entries.

This meant that stats from various different topics could be recognised, ranging from environmental issues, housing and teen pregnancy, to technological advances and the fundamentals of scientific research. Happily, the story was picked up by a couple of the papers, hopefully raising awareness of some of this (actually quite interesting and important) actual maths.

The full list of results, on the RSS website

Best statistics of 2017 are 69 and 0.1, at The Financial Times

The next London Mathematical Society Mary Cartwright Lecture, presented by Carola-Bibiane Schönlieb (University of Cambridge), will be held on Friday 2 March 2018 at De Morgan House in London. The lecture will be followed by a wine reception and dinner.

Organised annually by the LMS’s Women in Mathematics committee, the lecture is named for the British mathematician and chaos theorist. The event features two speakers – a warm-up from Andrea Bertozzi (UCLA) on Geometric graph-based methods for high dimensional data, followed by the Mary Cartrwight Lecture itself, given by Carola Bibiane-Schönlieb (University of Cambridge) on Model-based learning in imaging.

More information and how to register

We (Katie, Peter and Christian, and briefly Paul) were guests on one episode of Samuel Hansen’s Relatively Prime podcast just before Christmas – on one of our favourite hobby-horses, ridiculous made-up formulae in the press. We explained how the UK press does it better than anyone to Samuel’s US-based audience, then went on to tear a few of them apart. If you won’t be made sad by being reminded how it’s not Christmas any more, you can listen to the podcast through the RelPrime Website or on iTunes.

Formulaic Perfection, on RelPrime.com

]]>The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

]]>We now know 50 Mersenne primes! The latest indivisible mammoth, $2^{77,232,917}-1$, was discovered by Great Internet Mersenne Prime Search user Jonathan Pace on the 26th of December 2017. As well as being the biggest Mersenne prime ever known, it’s also the biggest prime of any sort discovered to date.

GIMPS works by distributing the job of checking candidate numbers for primality to computers running the software around the world. It took over six days of computing to prove that this number is prime, which has since been verified on four other systems.

Pace, a 51-year old Electrical Engineer from Tennessee, has been running the GIMPS software to look for primes for over 14 years, and has been rewarded with a \$3,000 prize. When a prime with over 100 million digits is found, the discoverer will earn a \$50,000 prize. That probably won’t be for quite a while: this new prime has $23{,}249{,}425$ decimal digits, just under a million more than the previous biggest prime, discovered in 2016.

If you’re really interested, the entire decimal representation of the number can be found in a 10MB ZIP file hosted at mersenne.org. Spoiler: it begins with a 4.

**More information: **press release at mersenne.org, home of the Great Internet Mersenne Prime Search.

This recording came about when I was working the *Codebreakers and Groundbreakers* exhibition for The Fitzwilliam Museum, Cambridge. The exhibition focuses on two codebreakers, the first being Michael Ventris, an architect and linguist who in 1952 deciphered the ancient Greek script known as Linear B. This breakthrough added about 500 years to our knowledge of ancient Greek history and is considered one of the great advances in classical scholarships.

The second codebreaker featured in the Fitzwilliam exhibition is Alan Turing, one of the leading British codebreakers at Bletchley Park who broke the German code machine Enigma during World War II. This gang of codebreakers was a mix of mathematicians, linguists, and puzzle-solvers. One of our aims for the exhibition is to show how this collaboration and sharing of skills made breaking Enigma possible.

The exhibition will run until the beginning of February 2018, and features unique Linear B clay tablets from the palace at Knossos, Crete; a rare U-boat Enigma machine and British TypeX machine both on loan from GCHQ; as well as various archival material from Michael Ventris and Alan Turing.

The exhibition also contains audio extracts from a BBC broadcast from Michael Ventris. So it was disappointing that we could not feature an equivalent audio extract from Alan Turing. However, the Turing archives at King’s College, Cambridge do contain Turing’s script for *Can Computer’s Think?* This script is available online via the Turing Digital Archive.

I was surprised how good Turing’s script was. Instead of the dense, academic language I feared, the script was a clear and simple explanation of the future of computing written for the layman – with perhaps the exception of the long desert island analogy, which I did not get on with at all. The second thing that surprised me was, for a script written in 1951, how current it all seemed. Turing is often described as ahead of his time, and the evidence is right here.

Turing talks about the computer’s ability to imitate any kind of calculating machine, a property known as universality, and considers if a machine will ever be able to imitate a brain. Turing then goes on to discuss whether a computer is capable of originality, or indeed free-will.

Turing then makes one firm prediction, that by the end of the 20th century computers would be able to answer questions in a manner indistinguishable from a human being – this is the famous Turing test. Turing’s prediction may have been a couple of decades early, but with the rise of digital assistants I would have to say he was completely right.

I was fascinated by Turing’s prediction of how unsettling it would be to design machines to look like people, an effect we now call the Uncanny Valley. Turing also saw no limits to what a computer would be able to achieve, and saw the future of programming to be closer to that of teaching, which is what we see today in the areas of Deep Learning and Neural Networks.

Throughout the lecture, Turing’s language is friendly and inclusive. He is also charmingly humble, admitting there are many other opinions and that these were just his own. I was also pleased to see Turing acknowledge the legacy of computing with a quotation from Ada Lovelace speaking about Charles Babbage’s Analytical Engine. I found the script so pleasing that I decided it would be nice to rerecord it so that new audiences would be able to hear Turing’s words as he intended.

Allan Jones has written more on this series of broadcasts by the BBC.

The *Codebreakers and Groundbreakers* exhibition is open at the Fitzwilliam Museum, Cambridge until Sunday the 4th of February 2018.

A transcript of the broadcast (PDF) has been submitted by Lewis Baxter.

]]>- Howard Groves, Member, Senior Mathematical Challenge Problems Group and Member, UK Mathematics Trust Challenges Sub Trust. OBE, for services to Education.
- Christl Donnelly FRS, Professor of Statistical Epidemiology, Imperial College London. CBE, for services to Epidemiology and the Control of Infectious Diseases.
- Ben Goldacre, Senior Clinical Research Fellow, Centre for Evidence-Based Medicine, University of Oxford and author of Bad Science. MBE, for services to Evidence in Policy.
- Andrew Morris, Professor of Medicine, Director of the Usher Institute of Population Health Sciences and Informatics, and Vice-Principal Data Science, University of Edinburgh. CBE, for services to Science in Scotland.
- Stephen Sparks, lately Professorial Research Fellow, University of Bristol and former chair of ACME. Knighthood, for services to Volcanology and Geology. (Via The Mathematical Association.)
- Bernard Silverman, lately Chief Scientific Adviser, Home Office and former President of the Royal Statistical Society (RSS). Knighthood, for public service and services to Science. (Via Hetan Shah.)
- John Curtice, Professor of Politics, University of Strathclyde and Senior Research Fellow, NatCen Social Research, and Honorary Fellow, RSS. Knighthood, for services to the Social Sciences and Politics. (Via Hetan Shah.)
- Diane Coyle, Professor of Economics, University of Manchester. CBE, for services to Economics and the Public Understanding of Economics. (Via Hetan Shah.)

Get the full list here. If you spot any others we should mention, please let us know in the comments.

]]>Exams have a nasty habit of sucking the joy out of a subject. My interest in proper literature was dulled by A-Level English, and I celebrated my way out of several GCSE papers – in subjects I’d picked because I enjoyed them – saying “I’ll never have to do that again.”

Geometry is a topic that generally suffers badly from this – but fortunately, Ed Southall and Vincent Pantaloni’s *Geometry Snacks* is here to set that right.

Geometric reasoning is joyful and beautiful. Memorising the difference between complementary and supplementary angles… rather less so. *Geometry Snacks* doesn’t shy away from the vocabulary of geometry, but its focus is squarely on the key ideas.

It’s a gorgeous little book (16cm by 16cm, 80 pages), with a stylised doughnut on the front to represent the bite-sized problems inside. It’s in five sections, covering “find the fraction” puzzles, angle-finding, proofs, areas and sangaku, lovely 17th- to 19th-century Japanese puzzles largely involving circles.

What makes it special for me is the clarity and elegance of the diagrams. Careful use of grey and red shading generally makes it obvious what they’re asking for; a lot of thought has gone into making the pictures pop (it’s not as colourful as Byrne’s *Elements*, but it’s probably clearer). I particularly like the device of using used and unused matchstick heads to define points!

The authors also take care to offer multiple solutions to problems, where appropriate. My one niggle with the book, though, is that the solution pages are more cluttered and harder to follow than the puzzles – in a perfect world, there would be a little more space for the answers. That said, I gather the publishers are set to release a special e-reader version of Geometry Snacks, in which some of the solutions will be animated. (The book mentions exploring the puzzles with GeoGebra, but it’s not clear whether that’s part of the e-reader version or just something recommended.)

Like all good puzzle books, *Geometry Snacks* goes from the straightforward (a parallelogram is inscribed in a square, with two opposite vertices at the midpoints of opposite sides of the square – what fraction is shaded?) to the much trickier (such as question 47, pictured). There’s something here for just about anyone with more than a passing familiarity with geometry – and especially for teachers looking for interesting extension problems.

It might be a little late to put it on your Christmas list (and certainly don’t let it replace *The Maths Behind*, that would be foolish), but *Geometry Snacks* is an excellent use for a book token, Amazon voucher or Christmas money.

Geometry Snacks is published by Tarquin and is available at the time of writing for £7.95.

]]>This year, he wanted to talk about a conjecture he’s recently investigated.

It’s just over an hour long. Sit down with a warm drink and enjoy some interesting recreational maths from the master.

]]>I guess I’m alone in this, since Matt Parker’s ‘guess the mean of the digits in all the entries’ competition received the most entries of any competition meaning he also won a prize.

Bored with ‘guess the mean of all the entries’ style competitions, I decided to come up with something that flips this concept on its head.

My competition was to come up with a function for mapping a set of $N$ integers onto a single value, and whoever’s function takes a pre-determined set of integers I’m thinking of closest to a particular predetermined value I’m thinking of is the winner. I was slightly sloppy in the way I defined the question (I used $N$ to refer to both the set of numbers, and the size of the set) but everyone responded beautifully – with only some people feeling the need to point out my error – in what became a virtual festival of making up your own notation.

I received 32 entries to the competition, not all of which were actually valid entries – several had the bright idea of using ‘the number you’re thinking of’ as a variable in their function, which is 100% cheating. Here are some of my favourites, in increasing order of amazingness, culminating with the winning entry (scroll down to the bottom if you just want to see that).

The set of numbers I was thinking of was in fact The Numbers, from the TV series Lost – in which a seemingly random set of integers play a mysterious yet crucial role in the plot. The numbers are 4, 8, 15, 16, 23 and 42. As well as being one character’s winning lottery numbers, and popping up in various other places, for a whole chunk of season 2 the numbers must be typed into a computer every 108 minutes, or else… something bad will happen.

I actually quite enjoyed it but I have notoriously bad taste in TV and films.

If you haven’t seen Lost, and you have a spare 122 hours of your life you’re happy to not get back, go for it.

My set of numbers is therefore as follows: 4, 8, 15, 16, 23 and 42 (in that order). The target number for the function was chosen to be zero, for reasons I can’t explain (mostly panic).

On to the entries!

Several examples fell into this category – including:

$f(x_0, x_1, \ldots, x_N) = N$

Simply counting the number of numbers (minus one). While we appreciate your lovely and obfuscating zero-indexing, this gives the answer $5$.

$f(x,y) = x^2 – y$

Since this function is only defined on two inputs, we took the first two, giving $ 4^2 – 8 = 8$.

$f(x_1, x_2, \ldots, x_N) = x_N^2 – 1$

Taking the last of the numbers and squaring it, then subtracting one. This gives $42^2 – 1 = 1763$.

Various

statistical functions, including thesquare of the mean, theupper quartile, thelower quartileand the ‘anti-mode‘ (the least common value, and if there’s more than one value equally uncommon, the mean of these – since the set is all distinct, this is just the mean LOL!).

These gave the values $240.25, 19.5, 11.5$ and $18$.

“

The first prime not a factor of any of the numbers”

This is $11$.

“

Return a random element of the set”

Since the target value isn’t in the set, this won’t work but it’s a nice try!

\[f: x_1 \ldots x_N \rightarrow \frac{e^{\min(x_i)} \cdot \pi^{\max(x_i)}}{N}\]

Almost like Euler’s identity in how it effortlessly combines mathematical constants, but sadly returns a result in the region of $7 \times 10^{21}$.

Many entries fell into this category:

$f(x_1, x_1, \ldots, x_N) = \frac{\displaystyle\sum_{i=1}^N (-1)^i i x_N}{N}$

If this was meant to be an alternating sum of multiples of successive entries, which is how I initially interpreted it, it also doesn’t sum to zero. And you should be more careful with notation, because this is exactly what was written.

An alternating sum of multiples of the largest entry, which given $i$ runs from $1$ to $6$ means you end up with a total of three times that entry, divided by 6 – giving $x_N/2 = 42/2 = 21$.

$f(n_1, n_1, \ldots, n_N) = \frac{\displaystyle\prod_{i=1}^N n_i}{N}$

The product of all the entries divided by the number of entries, which is $1,236,480$ (very much not zero).

The $\big(\displaystyle\prod_{i=1}^N x_i\big)^{th}$ prime number, modulo $\displaystyle\sum_{i=1}^N x_i$

This one returns 67.

$\frac{1}{M} = \frac{1}{x_1} +\frac{1}{x_2} + \cdots +\frac{1}{x_N}$

For reference, $M$ was the name I gave to my target value, and this was described as the ‘parallel resistance sum’. It works out to give $M = 1.7499207463\ldots$.

$f(x) =\displaystyle\sum_{k=1}^n \sin (kx_k)$

I like it, but it gives $-1.168\ldots$ – close, but no cigar.

Several entries took a chance on me, and hoped they could guess what number I was thinking of, defining their function as the constant function giving a single value regardless of the inputs. Entries of this form, in increasing order of size, included $\pi, 4, 7, 20$ and $37$, sadly none of which win.

Ask a stupid question, and you’ll get the following answers. We tried to calculate some of these in the ten minutes we had to judge the competition, but for some of them we merely established they were definitely not zero and gave up.

$f(x_1, x_2, \ldots, x_N) = x_1 + \frac{1}{x_2+ \frac{1}{\cdots + \frac{1}{x_{N-1} + \frac{1}{x_N}}}}$

A continued fraction. We calculated the answer but all that’s written on here is “4.something”. Not equal to zero.

“Arrange the numbers in order and multiply the difference between them in pairs.”

I enjoy a good bit of tedious busy work, and I got $3724$.

$f(n_1, n_2, \ldots, n_x) = (x+2) \cdot \sqrt{n_1 \cdot n_x \cdot\displaystyle\prod_{\textrm{all } x} n_x}$

Interesting notation switcheroo from what everyone else seems to be using. This was a pain to calculate too and my bit of paper says ‘TOO BIG’.

“The median of all the answers to this competition + 1”

Given that we had 32 different entries, some of which we couldn’t properly calculate, but there was literally no chance the median answer was $-1$, the only thing written on the paper is the word ‘NO’.

“Sum the number of letters of the positive integers, minus the number of letters of the negative integers”

It’s lovely! Our bit of paper says “about 40ish?” but if anyone wants to work it out, go for it.

“For each element of N, go to the Wikipedia page for N(i) and find the N(i)th letter on the page; typecast this letter into I32 and take the sum of these values”

A bit of Googling has allowed me to establish that this is a way of converting letters into numbers, but since I assume none of the standard letters convert to a negative number, this will definitely not equal zero.

$f(x_1, x_2, \ldots, x_N) = e^{e^{\displaystyle\sum_{i=1}^N x_i}}$

We had a go at calculating this in Google Sheets, which was our terrible calculating tool of choice under pressure, and the value was actually too big for it to handle. In some ways, this could be the least correct answer.

Given that I named the number I was thinking of $M$, I received a small selection of entries from people who didn’t want to play like normal people, at least one of which just specified $f(N)=M$ – this is OBVIOUSLY not what you were meant to do. Others tried to define a function in terms of $N$ and $M$, including the following:

\[f(x_1, x_2, \ldots, x_N) = \begin{cases}

\Bigg(\frac{\displaystyle\sum_{i=1}^N \displaystyle\sum_{j=1}^{x_i} x_i^j}{N!}\Bigg)^{\displaystyle\frac{1}{x_{\max{3,N}}}}&& \textrm{if }N\textrm{ is coprime with }M\\

6 && \textrm{otherwise}

\end{cases}\]

For the record, while typing this in LaTeX I’ve hit a record of four consecutive close braces. Sadly $6$, the number of entries in my list, is not coprime with $0$, so this gives the answer $6$ (and is disqualified anyway for using $M$ in the definition).

$f(N,M) = 0. \big(\displaystyle\sum_{i=1}^{N-2} x_i\big) + M$

This was presumably hoping to get as small a number as possible in the first part, then add it to M and hope nobody was closer. Turns out not only is this cheating, it was also not close enough, because we did actually get an exact winner, and it was glorious. See below.

$f(n_1, n_2, n_3, \ldots, n_N) = n_2 + n_3 + \ldots + n_N \mod n_1$

This function takes all but the first number, adds them together and then takes the result modulo the first number, aka remainder on division. In the case of The Numbers, the result is:

$f(4, 8, 15, 16, 23, 42) = (8 + 15 + 16 + 23 + 42) \mod 4 \equiv 0$

Amazingly, through pure chance, the sum of these five numbers (104) is a multiple of 4, and so the answer is zero! We have a winner. Congratulations to the winner, who forgot to write their name on the sheet, but was identified and awarded their prize – a wind-up robot, as per the rules of the Competition Competition, worth strictly less than £1.

]]>

In September 2017, John Duncan, Michael Mertens, and Ken Ono published a paper announcing a connection between the Pariah group known as the O’Nan group (after Michael O’Nan, who discovered it in 1976) and another modular form. Like Monstrous Moonshine, the new connection is through an infinite-dimensional shape which breaks up into finite-dimensional pieces. Also like Monstrous Moonshine, the modular form in question has a deep connection with elliptic curves. In this case, however, the connection is more subtle and leads through yet another set of important mathematical objects: the quadratic fields.

It’s also possible to have fields of things that aren’t numbers, which are useful in lots of other situations — see Section 4.5 of *The Mathematics of Secrets* for a cryptographic example.

What mathematicians call a field is a set of objects which are closed under addition, subtraction, multiplication, and division (except division by zero). The rational numbers form a field, and so do the real numbers and the complex numbers. The integers don’t form a field because they aren’t closed under division, and the positive real numbers don’t form a field because they aren’t closed under subtraction.

A common way to make a new field is to take a known field and enlarge it a bit. For example, if you start with the real numbers and enlarge them by including the number $i$ (the square root of $-1$), then you also have to include all of the imaginary numbers, which are multiples of $i$, and then all of the numbers which are real numbers plus imaginary numbers, which gets you the complex numbers. Or you could start with the rational numbers, include the square root of $2$, and then you have to include the numbers that are rational multiples of the square root of $2$, and then the numbers which are rational numbers plus the multiples of the square root of $2$. Then you get to stop, because if you multiply two of those numbers you get

\[ (a+b\sqrt{2})(c+d\sqrt{2}) = (ac + 2bd) + (ad+bc)\sqrt{2} \]

which is another number of the same form. Likewise, if you divide two numbers of this form, you can rationalize the denominator and get another number of the same form. We call the resulting field the rational numbers “adjoined with” the square root of $2$. Fields which are obtained by starting with the rational numbers and adjoining the square root of a rational number (positive or negative) are called quadratic fields.

Identifying a quadratic field is almost, but not quite, as easy as identifying the square root you are adjoining. For instance, consider adjoining the square root of $8$. The square root of $8$ is twice the square root of $2$, so if you adjoin the square root of $2$ you get the square root of $8$ for free. And since you can also divide by $2$, if you adjoin the square root of $8$ you get the square root of $2$ for free. So these two square roots give you the same field.

This is the same $b^2 – 4ac$ as in the quadratic formula.

For technical reasons, a quadratic field is identified by taking all of the integers whose square roots would give you that field, and picking out the integer $D$ with the smallest absolute value that can be written in the form $b^2 – 4ac$ for integers $a$, $b$ and $c$. This number $D$ is called the fundamental discriminant of the field. So, for example, $8$ is the fundamental discriminant of the quadratic field we’ve been talking about, not $2$, because $8 = 4^2 – 4 \times 2 \times 1$, but $2$ can’t be written in that form.

After addition, subtraction, multiplication, and division, one of the really important things you can do with rational numbers is factor their numerators and denominators into primes. In fact, you can do it uniquely, aside from the order of the factors. If you have number in a quadratic field, you can still factor it into primes, but the primes might not be unique. For example, in the rational numbers adjoined with the square root of negative 5 we have

\[ 6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5}) \]

where $2$, $5$, $1+\sqrt{-5}$, and $1-\sqrt{-5}$ are all primes. You’ll have to trust me on that last part, since it’s not always obvious which numbers in a quadratic field are prime. **Figures 1 and 2 **show some small primes in the rational numbers adjoined with the square roots of negative 1 and negative 3, respectively, plotted as points in the complex plane.

We express this by saying the rational numbers have unique factorization, but not all quadratic fields do. The question of which quadratic fields have unique factorization is an important open problem in general. For negative fundamental discriminants, we know that $D = -3, -4, -7, -8, -11, -19, -43, -67, -163$ are the only such quadratic fields; an equivalent form of this was conjectured by Gauss but fully acceptable proofs were not given until 1966 by Alan Baker and 1967 by Harold Stark. For positive fundamental discriminants, Gauss conjectured that there were infinitely many quadratic fields with unique factorization but this is still unproved.

Furthermore, Gauss identified a number, called the *class number*, which in some sense measures how far from unique factorization a field is. If the class number is $1$, the field has unique factorization, otherwise not. The rational numbers adjoined with the square root of negative $5$ ($D = -20$) have class number $2$, and therefore do not have unique factorization. Gauss also conjectured that the class number of a quadratic field went to infinity as its discriminant went to negative infinity; this was proved by Hans Heilbronn in 1934.

What about Moonshine? Duncan, Mertens, and Ono proved that the O’Nan group was associated with the modular form

\[ F(z) = e^{-8 \pi i z} + 2 + 26752 e^{6 \pi i z} + 143376 e^{8 \pi i z} + 8288256 e^{14 \pi i z} + \ldots \]

which has the property that the coefficient of $e^{2|D| \pi i z}$ is related to the class number of the field with fundamental discriminant $D \lt 0$. Furthermore, looking at elements of the O’Nan group sometimes gives us very specific relationships between the coefficients and the class number.

Mathematicians say a symmetry that gets you back where you started after you do it $n$ times is a “symmetry of order $n$”.

For example, the O’Nan group includes a symmetry which is like a 180 degree rotation, in that if you do it twice you get back to where you started. Using that symmetry, Duncan, Mertens, and Ono showed that for even $D \lt -8$, $16$ always divides $a(D) + 24h(D)$, where $a(D)$ is the coefficient of $e^{2|D|\pi i z}$ and $h(D)$ is the class number of the field with fundamental discriminant D. For the example $D = -20$ from above, $a(D) = 798588584512$ and $h(D) = 2$, and $16$ does in fact divide $798588584512 + 48$. Similarly, other elements of the O’Nan group show that $9$ always divides $a(D)+24h(D)$ if $D = 3k+2$ for some integer $k$ and that $5$ and $7$ always divide $a(D)+24h(D)$ under other similar conditions on $D$. And $11$ and $19$ divide $a(D)+24h(D)$ under (much) more complicated conditions related to points on an elliptic curve associated with each $D$, which brings us back nicely to the connection between Moonshine and elliptic curves.

Monstrous Moonshine showed that the Monster, and therefore the Happy Family, was related to modular forms and elliptic curves, as well as string theory. O’Nan Moonshine brings in two more sporadic groups, the O’Nan group and its subgroup the “first Janko group”. (**Figure 3** shows the connections between the sporadic groups. “M” is the Monster group, “O’N” is the O’Nan group, and “J_{1}” is the first Janko group.) It also connects the sporadic groups not just to modular forms and elliptic curves, but also to quadratic fields, primes, and class numbers. Furthermore, the modular form used in Monstrous Moonshine is “weight 0”, meaning that $k = 0$ in the definition of a modular form given in Part II. That ties this modular form very closely to elliptic curves.

Umbral Moonshine also uses weight $3/2$ modular forms.

The modular form in O’Nan Moonshine is “weight $3/2$”. Weight $3/2$ modular forms are less closely tied to elliptic curves, but are tied to yet more ideas in mathematical physics, like higher-dimensional generalizations of strings called “branes” and functions that might count the number of states that a black hole can be in.

That still leaves four more pariah groups, and the smart money predicts that Moonshine connections will be found for them, too. But will they come from weight $0$ modular forms, weight $3/2$ modular forms, or yet another type of modular form with yet more connections? Stay tuned! Maybe someday soon there will be a Part IV.

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