What mathematics is

Mark a few points on a circle and join every point to every other one with a straight line. These segments are called chords, and they cut the disc into pieces. How many pieces do you get?

Start small. One point: no chords, the disc is whole, one piece. Two points: one chord, two pieces. Three points make a triangle and four pieces: the triangle itself and three crusts around it. Four points cut the disc into eight pieces, five into sixteen. There is one condition: the points are placed so that no three chords pass through the same point inside the circle. Why it matters will become clear shortly.

One, two, four, eight, sixteen: twice as many each time. Before you read on, place your bet.

How many pieces are there with six points on the circle, if no three chords meet at one point?

Thirty-one. If you wrote 32, don't worry: that is exactly why this puzzle gets shown. Check it on the circle below by adding a sixth point.

Thirty-one

The circle below works like pencil and paper, except that it never loses count. It finds every crossing of the chords and every piece, even the tiny ones near the rim, and colours them so that neighbours don't blend together. You can drag the points around the circle, add them and remove them.

With six points there are 31 pieces. Move the points however you like: as long as no three chords meet at one point, there are exactly 31, and a thirty-second never appears. After that the gap only widens. Seven points give 57 pieces instead of 64, eight give 99 instead of 128.

At the tenth point the sequence brushes against a power of two once more: $256 = 2^8$. That's a coincidence; next come 386, 562, 794. The puzzle is known as Moser's circle problem, after the Canadian mathematician Leo Moser.

A trap inside the trap

The condition about three chords isn't pedantry. Put six points at the corners of a regular hexagon: the three long diagonals meet at the centre, and the count drops to 30. Where three chords would have crossed at three different points and fenced off a small triangle, there is now a single point and no triangle. Crossings that coincide always eat pieces (we'll prove this in the next section). So 31 is the largest possible number of pieces for six points, and that is what the puzzle asks about.

Eight points sit at the corners of a regular octagon, and all the chords are drawn. How many pieces are there?

$128$, as the doubling promises Exactly $99$, as in the table Fewer than $99$ More than $99$

All four long diagonals of a regular octagon meet at its centre, and at eight more points three chords meet. Every such coincidence eats pieces, and 88 are left. The 99 in the table is the largest possible count: you get it only when no three chords pass through one point.

Where 31 comes from

To see why there are 31 pieces and not 32, we don't need to count them one by one. We need to understand where pieces come from. Let's draw the chords one at a time and watch how many pieces each one adds.

The pattern jumps out at once: a chord that crosses $k$ chords already drawn adds $k + 1$ pieces. If that always happens, the number of pieces is

$$\text{pieces} = 1 + \text{chords} + \text{crossings}:$$

the original disc, plus one for each chord, plus one for each crossing. But after the business with 32 we won't take a pattern on trust. First we'll count the terms, and then we'll prove the formula always holds.

There are as many chords as there are pairs of points. Each of the $n$ points can be joined to $n - 1$ others, but that counts every chord twice, once from each end, so there are $\frac{n(n-1)}{2}$ chords. With six points that's $\frac{6 \cdot 5}{2} = 15$. The number of ways to choose $k$ things out of $n$ is written $\binom{n}{k}$ and read "$n$ choose $k$"; some books write $C(n, k)$ or ${}^nC_k$. So there are $\binom n2$ chords. We'll look at these numbers closely in the chapter on combinatorics.

The crossings are more interesting. It turns out that every crossing corresponds to exactly one set of four points on the circle: the ends of the two chords that meet there. So there are as many crossings as sets of four: $\binom n4 = \frac{n(n-1)(n-2)(n-3)}{24}$. We divide by 24 because the same four points can be picked in order in $4 \cdot 3 \cdot 2 \cdot 1 = 24$ ways. Why fours and crossings match up one to one, the proof will explain.

The original disc: one piece while there are no chords. One piece for each chord. There are as many chords as pairs of points: $\binom n2 = \frac{n(n-1)}{2}$. One piece for each crossing. There are as many crossings as sets of four points on the circle: $\binom n4 = \frac{n(n-1)(n-2)(n-3)}{24}$. For $n = 5$: $1 + 10 + 5 = 16$. For $n = 6$: $1 + 15 + 15 = 31$. For $n = 7$: $1 + 21 + 35 = 57$. The formula holds when no three chords pass through one point; otherwise there are fewer pieces.

Now let's state the result the way mathematicians do: as a theorem with a proof.

In this course theorems sit in coloured cards, and the proof comes straight after the card. Wherever an honest picture exists, the proof is shown as a drawing that builds up step by step along with the text: you can page back and forth through the steps and drag the points in the drawing to convince yourself that the argument doesn't depend on where they are.

The trick is not to count the pieces of the finished picture, but to build it one chord at a time and watch how many pieces each chord adds.

Before any chords are drawn, the disc is a single piece. That is the term $\p1{1}$ in the formula. Draw the chords one at a time, in any order. Suppose the next chord crosses $k$ of those already drawn. The crossings on it are all different, because no three chords pass through one point. These $k$ points split the new chord into $k + 1$ segments. Each segment runs across one of the existing pieces from edge to edge and cuts it into exactly two. Why not three? The line through a chord meets the disc in exactly that chord, so every piece consists of the points of the disc lying on one particular side of each line drawn so far. Such a piece is convex: with any two of its points it contains the segment joining them. And a segment running across a convex shape from edge to edge splits it into exactly two parts, one on each side of the line. So the new chord adds $1 + k$ pieces: one for itself and one more for each crossing it creates. Draw all the chords this way. The "one for itself" terms add up to the total number of chords. A chord is a pair of points, and $n$ points have $\binom n2$ pairs. That is the term $\p2{\binom n2}$. All the $k$ terms together add up to the number of crossings. Each crossing was counted exactly once: when we drew the second of the two chords that meet there. When we drew the first, the crossing didn't exist yet. It remains to count the crossings. Take four points in the order $P$, $Q$, $R$, $S$ around the circle. The chord $PR$ splits the disc in two, with $Q$ and $S$ on different sides, so the segment $QS$ must cross $PR$. But in the pairs $PQ$ and $RS$, and $PS$ and $QR$, both ends of one chord lie on the same side of the other, and the chords don't meet. So of the three ways to split four points into two pairs, only the diagonals of the quadrilateral cross: every set of four gives exactly one crossing. Conversely, every crossing gives one set of four, the ends of its two chords. There are as many crossings as sets of four: $\p3{\binom n4}$. Add up the original piece, the pieces for the chords and the pieces for the crossings: $R(n) = \p1{1} + \p2{\binom n2} + \p3{\binom n4}$. With six points, $1 + 15 + 15 = 31$.

The condition about three chords isn't in the theorem for decoration. Here is what happens when it's broken.

Let's rerun the proof of the theorem and see which term shrinks.

Three chords in general position cross pairwise at three different points and fence off a small triangle. Counting as in the proof: one piece, plus one for each chord and one for each crossing, $1 + 3 + 3 = 7$ in all. Slide the end $F$ of the third chord until the chord passes through the crossing of the other two. The triangle shrinks and finally disappears: all three chords now pass through one point. There are still three pairs of crossing chords, but only one crossing point. The third chord passes through a point that already exists and earns one unit for crossings, not two. That gives $1 + 3 + 2 = 6$ pieces, one fewer. The general case works the same way. Every step of the theorem's proof still holds, except that a crossing on the new chord is counted once, however many chords pass through it. In the theorem every pair of crossing chords brought its own unit, and there are always $\binom n4$ such pairs, one per set of four points. If $m \ge 3$ chords meet at a point, their $\binom m2$ pairs bring only $m - 1$ units between them: the first of those chords gets nothing for that point, each later one gets one. Since $m - 1 < \binom m2$ for $m \ge 3$, there are fewer pieces. At the centre of the regular hexagon $m = 3$, and $\binom 32 - 2 = 1$ piece is lost: $31 - 1 = 30$.

One puzzle remains: why did the first five answers match the powers of two so neatly? It isn't chance, and the explanation fits on one line.

Why the first five answers are powers of two

A power of two is the sum of a whole row of Pascal's triangle: $2^{n-1} = \binom{n-1}{0} + \binom{n-1}{1} + \dots + \binom{n-1}{n-1}$, since each of $n - 1$ things can be taken or left, and $\binom{n-1}{k}$ counts the choices of exactly $k$ of them (details in the chapter on combinatorics). Moser's answer is the sum of only the first five of these numbers. You can see this from Pascal's rule $\binom{n-1}{k-1} + \binom{n-1}{k} = \binom nk$:

$$\binom{n-1}{0} + \underbrace{\binom{n-1}{1} + \binom{n-1}{2}}_{\binom n2} + \underbrace{\binom{n-1}{3} + \binom{n-1}{4}}_{\binom n4} = 1 + \binom n2 + \binom n4.$$

While $n \le 5$ the row has no other numbers, and the match is perfect. At $n = 6$ the last term $\binom 55 = 1$ drops out, hence $31 = 32 - 1$.

The same sum answers a question that sounds completely different: into how many pieces do $n - 1$ three-dimensional "planes" in general position cut four-dimensional space? In ordinary space the trap springs sooner: four flat cuts divide a watermelon into at most $1 + 4 + 6 + 4 = 15$ pieces, not 16. The proof uses the same trick of watching how many pieces each new cut adds.

There are eight points on a circle, and no three chords pass through one point. How many crossings do the chords have inside the circle?

As many as there are sets of four points: $\binom 84 = \frac{8 \cdot 7 \cdot 6 \cdot 5}{24} = 70$. As a check: $1 + 28 + 70 = 99$ pieces, as in the table.

What is the smallest number of points for which there are more than a thousand pieces?

Use the formula: $R(12) = 1 + 66 + 495 = 562$, $R(13) = 1 + 78 + 715 = 794$, $R(14) = 1 + 91 + 1001 = 1093$. Answer: 14 points. Doubling would have passed a thousand already at the eleventh point: $2^{10} = 1024$.

Patterns that break

Moser's circle is not the only trap of its kind, and far from the most treacherous. Five agreements in a row is not much. Sometimes there are forty.

In 1772 Leonhard Euler noticed that $n^2 + n + 41$ gives prime numbers for $n = 0, 1, 2, \dots, 39$: 41, 43, 47, 53, 61, 71, 83, …, 1601. A prime is a number greater than one that is divisible only by one and by itself. Forty primes in a row make it easy to believe it will go on forever. But at $n = 40$ we get $40^2 + 40 + 41 = 40 \cdot 41 + 41 = 41^2 = 1681$. With hindsight the formula was doomed even without a search: at $n = 41$ all three terms are divisible by 41.

In 1640 Pierre de Fermat was sure that all numbers of the form $2^{2^n} + 1$ are prime: 3, 5, 17, 257, 65,537. The next one, $2^{32} + 1 = 4{,}294{,}967{,}297$, was hard to check by hand, and for nearly a century nobody did. In 1732 Euler found a divisor: $4{,}294{,}967{,}297 = 641 \cdot 6{,}700{,}417$. Since then, not one of the later numbers of this form that could be checked has turned out to be prime.

Or a very simple game: 31, 331, 3331, 33,331, 333,331, 3,333,331, 33,333,331. All seven numbers are prime. The eighth, $333{,}333{,}331$, equals $17 \cdot 19{,}607{,}843$.

Among our examples the record for patience belongs to a conjecture of the Hungarian mathematician George Pólya (1919). Break every number into prime factors and count them with repetition: $12 = 2 \cdot 2 \cdot 3$ has three factors, $10 = 2 \cdot 5$ has two, and 1 has none. Pólya conjectured that for every $N \ge 2$, at least half of the numbers from 1 to $N$ have an odd number of factors. In 1958 the British mathematician Haselgrove proved that the conjecture fails somewhere, without saying where. The smallest such $N$ was found in 1980 by the Japanese mathematician Minoru Tanaka: $906{,}150{,}257$. More than nine hundred million confirmations in a row, and the conjecture is still false.

Continue the sequence

"Continue the sequence" puzzles test your wits, but they have no single answer. A finite beginning doesn't determine what follows: the same first numbers belong to as many rules as you like. Here are six perfectly natural ones that all start the same way, $1, 2, 4, 8, 16$, and then go their separate ways. In the second tab you can choose the sixth number yourself, and there will be a formula that produces it.

The trick in the second tab is honest. Through any six points with different $x$-coordinates passes exactly one polynomial of degree at most five (why is explained in the chapter on polynomials), and it can be written down explicitly. So the answer "the sixth term is 17" is just as legitimate as "32"; you only have to show the rule. Incidentally, Moser's formula is a polynomial of degree four, and no other polynomial of degree at most four passes through the points $(1, 1)$, $(2, 2)$, $(3, 4)$, $(4, 8)$, $(5, 16)$. In that sense 31 is the most economical way to continue the sequence.

Practise catching patterns that break. The trainer offers a statement that holds for the first few $n$ and asks you to find the first $n$ where it fails. Almost all the statements are about prime numbers, which get a full treatment in chapter 3.

Why proof

If examples guarantee nothing, how do mathematicians ever know anything for sure? They prove it. Remember how we got Moser's formula: we didn't test it for six, seven and eight points; we understood where each piece comes from. That kind of argument works for every $n$ at once, for six points and for a million.

Every proof rests on something: on definitions, on theorems already proved, and at the very bottom on axioms, statements accepted without proof as the rules of the game (more on them in chapter 16). Here is the smallest possible model of such a building: one axiom of geometry and one consequence of it. In this course axioms sit in grey cards with a double border; theorems and propositions sit in coloured ones.

We argue by contradiction: suppose the statement is false and derive a contradiction with the axiom.

Suppose the distinct lines $a$ and $b$ have at least two common points; call them $A$ and $B$. The line $a$ passes through $A$ and $B$. The line $b$ also passes through $A$ and $B$. In the drawing it is bent: you can't draw a different straight line through $A$ and $B$, and that is exactly what the axiom says. Only one line passes through two distinct points, so $b$ coincides with $a$. So $a$ and $b$ are the same line, whereas we assumed they were distinct. A contradiction. So the assumption is false, and two distinct lines have at most one common point.

Three steps, but they contain everything any proof is made of: starting rules, a statement, and a chain in which every link rests on what has already been accepted. The "suppose the opposite" move will come up again and again; its most famous use is the proof that $\sqrt2$ cannot be written as a fraction (chapter 6).

What has been proved stays true forever. Around 300 BC Euclid proved that there are infinitely many primes, and his proof is still read today without a single correction; in chapter 3 it takes half a page. A new experiment can overturn a physical theory. It can't overturn a theorem, unless the proof contains a mistake.

The caveat "unless the proof contains a mistake" is not there for show. Proofs are written by people, and people make mistakes, so other mathematicians read and check them. In recent decades computers have increasingly checked them too, line by line; more on that in the last chapter.

The story of the circle is a mathematician's work in miniature. First we notice a pattern: 1, 2, 4, 8, 16. We make a conjecture: it doubles every time. We test it and find a counterexample, 31. We look for how things really work and prove a formula that always holds. Then we ask what happens if we break the condition (three chords through one point) and what happens in space (the watermelon). And along the way we notice that pairs and quadruples of points are counted by the same numbers $\binom nk$, which will turn up again in all sorts of different problems. When one observation explains many unlike problems at once, we speak of abstraction.

Sometimes a pattern is true, and even then a proof is needed. Add up the first odd numbers: $1 = 1^2$, $1 + 3 = 2^2$, $1 + 3 + 5 = 3^2$, $1 + 3 + 5 + 7 = 4^2$. This time there's no trap. In the chapter on sequences this is proved with a single picture, and you'll learn to prove such statements for all $n$ at once.

And sometimes a conjecture survives every test while a proof stays out of reach for centuries. In 1742 Christian Goldbach, in a letter to Euler, made a guess that today is stated like this: every even number greater than two is the sum of two primes. $4 = 2 + 2$, $10 = 3 + 7$, $100 = 3 + 97$.

Goldbach's conjecture has been checked by computer for all even numbers up to $4 \cdot 10^{18}$. What does that mean?

It is proved: that many checks are enough It holds up to $4 \cdot 10^{18}$, but in general it is still a conjecture It is almost certainly false, since no one has proved it yet It can no longer be refuted

A check is a lot of examples, and examples can't cover all numbers. Remember Pólya: there the first counterexample hid beyond nine hundred million. If Goldbach's conjecture has a counterexample, it is larger than $4 \cdot 10^{18}$. So far nobody has a proof, and nobody has a refutation either.

Why mathematics works

So far this all looks like a game: points on a circle, chords, prime numbers. Yet games like these grew into the calculations behind bridges, orbits and power grids. In 1960 the physicist Eugene Wigner wrote an article about it called "The Unreasonable Effectiveness of Mathematics in the Natural Sciences". There is less that's unreasonable here than it seems, and two stories show why.

In the 1840s astronomers' calculations didn't add up: Uranus moved across the sky not quite the way the law of universal gravitation required. Urbain Le Verrier supposed that an unknown planet was pulling on Uranus, and from the deviations he calculated where it had to be. On 23 September 1846 Johann Galle at the Berlin Observatory pointed a telescope at the predicted spot and found Neptune that same night, about a degree from where it was supposed to be. People say the planet was discovered "at the tip of a pen". Independently of Le Verrier, the Englishman John Couch Adams made a similar calculation.

The second story comes from engineering. Square roots of negative numbers appeared in the sixteenth century as an intermediate trick in the formula for cubic equations, and for a long time they were considered not real, "imaginary" (see chapter 15). Three centuries later, in the 1890s, Charles Steinmetz and other engineers began calculating alternating-current circuits with complex numbers, which turns differential equations into arithmetic. Electrical engineers still work this way, except that they write the imaginary unit as $j$, because $i$ is already taken by the current.

The secret, if there is one, is that mathematics studies structure: what follows from what. The law of gravitation describes the structure of the world. If it is true, then all of its consequences are true too, including those nobody has seen yet. A mathematician derives the consequences without looking through a telescope and tells you where to look.

Seven big questions

Seven questions run through the whole course. They sound simple, some almost childish, yet their answers became landmarks in the history of mathematics. Each question is asked here and answered in the chapter where the tools for the answer are finally at hand: the first answer comes in chapter two, the last in chapter fifty-nine. Pick a question, read the teaser and try a small experiment.

At school the sign rule is learned by heart: minus times minus makes plus. But what is a debt multiplied by a debt? The rule looks like a convention, and it seems we could have agreed on something else. Or could we?

Chapter 2. Zero and minus Chapter 15. Complex numbers

A circle and a square of the same area. Scissors cutting along straight lines can't turn the circle into the square, and ruler and compass can't construct such a square: in 1882 Lindemann proved that $\pi$ is transcendental. Yet in 1990 Miklós Laczkovich proved that cutting and reassembling is possible after all, if the pieces are ones no artist could draw, and there are about $10^{50}$ of them.

Chapter 6. The square root of two Chapter 20. Area and volume Chapter 43. Galois theory Chapter 54. Measure

Quadratic equations were solved as far back as Babylon; cubics and quartics fell in sixteenth-century Italy. Then for nearly three hundred years nobody could find a formula for the fifth degree. Not because people searched badly: there is no such formula, and this can be proved.

Chapter 14. Polynomials Chapter 43. Galois theory

There are infinitely many natural numbers, and infinitely many points on a segment, so it seems there is nothing to compare. In 1638 Galileo noticed something odd: there are as many squares as there are numbers, even though squares get rarer and rarer. And in 1874 Georg Cantor proved that infinities come in different sizes.

Chapter 52. Infinities

You agree on a secret key with your bank, and anyone at all can listen to the whole conversation. Yet the eavesdropper still doesn't learn the key. It all rests on operations that are easy to perform and, as far as anyone knows, practically impossible to undo.

Chapter 41. Remainders and ciphers Chapter 44. Elliptic curves

The equations that govern the air are known, and computers are fast. But in 1961 Edward Lorenz rounded one number in a calculation, $0.506127$, to $0.506$, and got completely different weather. Weather can be predicted about two weeks ahead at most, and weak computers are not the reason.

Chapter 31. Differential equations Chapter 59. Chaos

It seems that everything true in mathematics gets proved sooner or later, if you just try hard enough. In 1931 Kurt Gödel showed that this is not so: for any reasonable system of axioms for arithmetic there are true statements about numbers that it cannot prove, and new axioms won't fix that.

Chapter 56. Gödel and Turing

How the course works

The course has 61 chapters in ten parts: Introduction, Numbers, Algebra, Geometry, Calculus, Linear algebra, Structures, Chance & data, Foundations and Horizons. On the main page they are drawn as a metro map: each part is a line of its own, chapters are stations, and interchanges connect places where an idea from one part is at work in another.

Every chapter opens with a wall, a problem the tools so far can't solve, and builds a ladder over it. At the end, in the "Where next" section, a new wall appears, and the next chapter starts from it. You can read in order, or jump straight into any chapter: its header says what it builds on.

Every chapter has a level. "School" is for ages 11–15, "High school" for ages 16–18, "University" for the first two years, and "Beyond" for readers who aren't put off by abstraction. The course starts at school level and by the end reaches mathematics from the later years of university.

Inside a chapter you'll meet a few regular residents.

• Formulas taken apart, like Moser's formula above: tap a coloured part of the formula and its explanation lights up.
• Widgets: live drawings that answer "what happens if…". Drag, change, break things: nothing will be damaged.
• "Try it yourself": problems with an answer box. The checker accepts any correct form: $\frac12$, $0.5$ and $\frac24$ are all the same to it.
• Theorems and proofs: theorems in coloured cards, axioms in grey ones with a double border, and the proof comes right after the theorem, step by step, with the drawing building up along with the text.
• Sidebars marked History, Deeper, For students and Careful can be skipped without losing the thread.
• Terms, like counterexample and proof, are collected in a glossary with a short definition and a link to the chapter where each first appears.

Besides the chapters, the course has two tools. Trainers, like the counterexample hunt above, invent problems endlessly, accept an answer in any correct form and show a worked solution; they are all collected on the practice page. Here is another one, on Moser's formula:

Solvers take a problem of your own, say a quadratic equation or breaking a number into prime factors, and solve it step by step, explaining why each step is legitimate. And in the topic catalogue you can filter chapters by subject and level, for example everything that comes in handy for school-leaving exams.

Where next

We spent the whole chapter counting pieces of a circle, 1, 2, 4, 8, 16, 31, and never once wondered where the numbers themselves come from, or the way we write them. Leopold Kronecker, as Heinrich Weber recalled, used to say that God made the integers and all else is the work of man. Even if so, it was people who had to write them down. A shepherd can mark each sheep with a notch on a stick. But how do you write a number too big for any stick, and how do you add such numbers in columns? The answer took tens of thousands of years, from a notched bone to binary code. That is chapter 1.