Field

Number — page 2

The whole numbers, and how much structure they turn out to have.

The partition product's coefficients to q¹². A row of series coefficients computed by expanding a product, beside the same numbers obtained another way.

Every partition, hidden in a product

Multiply out one factor for each part size and the coefficient of q to the n is the number of partitions of n. Nothing is being approximated: the product is a bookkeeping device that does the counting by multiplying.

The product of (1 − qᵏ), and what survives at 12. The coefficients of the pentagonal product drawn as signed bars, with the partitions into distinct parts that Franklin's move leaves unpaired.

The terms that cancel almost everything

Multiply out the product of 1 − q, 1 − q², 1 − q³ and so on, and nearly every coefficient is zero. What survives is a single plus or minus one at 1, 2, 5, 7, 12, 15 — and the reason is a way of pairing partitions off so that each pair cancels.

p(n) to 60, against the Hardy–Ramanujan estimate. The number of partitions of each number up to sixty on a logarithmic scale, with the asymptotic estimate drawn over it and the ratio of the two tabulated.

The size of a number with no formula

There is no closed expression for the number of partitions of n. There is an expression for how large it is — with a square root in the exponent and a π in front — and it is accurate enough that rounding a few terms of its refinement gives the exact count.

Every fifth partition count divides, and the rank that says why. A row of partition counts with the ones in a congruence class marked, and a histogram of partitions sorted by rank.

Every fifth one divides

p(4) is 5, p(9) is 30, p(14) is 135, and every partition count at a number leaving four on division by five is divisible by five. Ramanujan read it off a table; the explanation is a way of splitting those partitions into five equal heaps.

The two supplements, and the residue classes that decide them. A table of odd primes with the Legendre symbols of minus one and two beside the residue of p modulo four and modulo eight.

The two supplements, and where the eight comes from

The main law relates two odd primes to each other and says nothing about −1 or about 2. Those two are settled separately, by their own counts, and the answers arrive modulo four and modulo eight — which is a clue about where the whole subject is really taking place.

Multiplication by 3 modulo 11, and the sign of the shuffle. Residues in two rows joined by strings showing where multiplication sends each one, with a strip beneath comparing the sign of the shuffle to the Legendre symbol for every multiplier.

The symbol is the sign of a shuffle

Multiplying every residue modulo p by a fixed number rearranges them. That rearrangement is a permutation, permutations have a sign, and the sign is exactly the Legendre symbol — so a question about squares becomes a question about crossings.

The Gauss sum for 13, added one root at a time. Partial sums of the p-th roots of unity signed by the Legendre symbol, drawn as a walk closing on a point at distance root p from the origin.

One sum, squared two ways

Add the p-th roots of unity, each taken with a plus or a minus according to whether its index is a square. The walk that results closes on a point at distance √p from the origin — and squaring that one number, evaluated two different ways, is the reciprocity law.

Where a congruence decides which primes a form represents, and where it does not. Rows of primes marked by whether each is represented by x squared plus n y squared, with the residue classes that decide it where such classes exist.

Which primes a form takes

A prime is the sum of two squares exactly when it is 1 modulo 4. Change the form slightly, to x² + 27y², and no congruence on p decides it at all — which is where the elementary subject ends and its successor begins.

The primes below 100,000, by remainder mod 4. A bar for each remainder on division by 4, showing how many primes below 100000 leave it. The 2 classes sharing no factor with 4 hold near-equal counts; the rest are empty or hold one prime.

Infinitely many of one kind

Euclid's argument produces a prime nobody had listed, and says nothing about what it looks like. Ask for infinitely many primes ending in 3, or leaving a remainder of 1 on division by 4, and the same construction has to be aimed — and for most targets nobody knows how to aim it.

Five families of primes, counted below 100,000. Five counting curves on logarithmic axes: every prime, the primes one more than a multiple of four, the twin pairs, the primes one more than a square, and the Mersenne primes. Two of the five families are known to be infinite and three are open questions.

Which infinitudes are proved

The primes never stop, and neither — apparently — do the twin pairs, the primes one more than a square, or the Mersenne primes. Three of those four statements are theorems and one is not, and counting the members of each family tells nobody which.

The primes 3 mod 4 against the primes 1 mod 4, out to 30,000. A plot of the difference between the counts of primes in two residue classes, against the bound, showing a persistent lead for one class and where it is lost.

Every class, and in equal shares

Euclid's argument aimed at a residue class reaches some classes and stalls at others. The theorem covering all of them is Dirichlet's, its proof abandons arithmetic entirely for analysis, and what it proves is stronger than infinitude — the classes are equal, though not at any point anybody has counted.

Two sums of reciprocals: 2.89 and climbing against 1.71 and level. Two curves against the logarithm of the bound: the sum of reciprocals of all primes, rising steadily, and the sum over the twin primes, flattening towards a limit.

The sieve that cannot finish

Sifting out the composites is the oldest method in the subject and it has a ceiling nobody has raised. The reciprocals of the twin primes add to a finite number, so no argument that measures thickness can reach them — and the inclusion–exclusion every sieve truncates goes wildly wrong before it goes right.

$The tree as words in two matrices. 6 nodes of the Stern–Brocot tree, each as the word of turns reaching it, the matrix that word multiplies out to, its two columns as fractions, and the mediant of those columns.$

Two matrices that generate the tree

A node of the Stern–Brocot tree is not really a fraction — it is the pair of fractions it lies between. Written as the columns of a matrix, the two turns of the tree become two multiplications, and the determinant that kept everything in lowest terms becomes a property of a product.

Every positive rational, in one sequence. The first 32 terms of Stern's diatomic sequence as bars, with the ratios of consecutive terms beneath. Every ratio is in lowest terms, no two agree, and each term counts the hyperbinary representations of its index.

Every rational in one sequence

The tree lists every positive fraction once and needs a tree to do it. One recursion on the whole numbers lists them in a single row — and each term of it counts something nobody was asking about, which is why the enumeration works.

$13 record approximations in 26 turns. The distance from π to each fraction the descent passes, against its denominator, on logarithmic axes. 13 of them beat every fraction with a smaller denominator.$

The fractions that beat every smaller one

Walking down the tree towards a number produces a sequence of fractions closing in on it. Most of them are steps along the way; a few are the best approximations there are — closer than every fraction with a smaller denominator — and which few is decided by where the turns change direction.

The Farey tessellation, and a line down to √2 − 1. Semicircles over the unit interval joining every pair of Farey neighbours with denominators up to 13, and a vertical line at √2 − 1. The 6 arcs it crosses are the intervals of the Stern–Brocot descent to √2 − 1, and their turns spell LRRLL.

The arcs a line crosses on its way to a number

Draw a semicircle over every pair of neighbouring fractions and the half-plane above the number line is cut into curved triangles that never overlap. A straight line dropped towards any number crosses those arcs one after another, and the arcs it crosses, and the side it leaves each triangle by, are exactly the steps of the Stern–Brocot descent towards that number.

$Minkowski's question-mark function. The graph of Minkowski's function ?(x) on the unit interval: continuous and increasing, sending each Stern–Brocot fraction to the binary fraction in the same position. It sends √2 − 1 to 2/5 and φ − 1 to 2/3.$

The function that sends fractions to binary

The Stern–Brocot tree and the tree of binary fractions have exactly the same shape, so there is a function that sends each fraction to the binary fraction in the same position. It is continuous and increasing, it turns every quadratic irrational into an ordinary fraction, and it does all of its rising on a set of numbers so thin that at almost every point its slope is nought.

Bhāskara's cyclic method on x² − 61y² = 1, step by step. A table of the cyclic method's rows: the helper m chosen at each step and the near miss a² − Db² = k it produces, ending at k = 1 with the fundamental solution.

A method that is allowed to miss

Bhāskara's cyclic method solves x² − Dy² = 1 by aiming at the wrong target. It keeps a pair a, b with a² − Db² = k for some small k, combines it with a helper chosen so that k can be divided out, and repeats until k is 1. For D = 61 it reaches the ten-digit fundamental solution in 13 steps, where walking the convergents of √61 takes 22 — and for every D up to 100 it is faster.

$The states the continued fraction of √61 can be in. A grid of whole-number pairs with the band of reduced states shaded, the pairs that qualify marked, and the cycle of states visited by the expansion of the square root numbered in order.$

Why the expansion has to repeat

The continued fraction of √61 runs 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 and then starts again. It must: each step's state is a pair of whole numbers trapped in a small band, and only 14 pairs fit. The expansion of √61 visits 11 of them in a cycle, the other 3 form a cycle of their own, and the period reads the same backwards before its last term, which is twice the first.

The powers of 2 modulo 13, as a ring of 12. The non-zero residues modulo 13 placed on a circle, with the successive powers of 2 joined by straight lines into a closed walk of 12 steps.

One residue whose powers are all of them

Fermat's theorem says every order divides p − 1. It does not say that anything has order exactly p − 1, which is a separate and stronger claim — and what forces it is a count of how many numbers share each divisor with p − 1.

The orders of the 8 units modulo 15. A strip of the units modulo 15 with each one's multiplicative order beneath it, the largest order marked at 4 against φ(15) = 8.

The exponent that is smaller than Euler's

Euler's theorem raises every unit to the count of the units and gets one. The smallest exponent that works for all of them at once is often much smaller — and a composite is invisible to Fermat's test exactly when that smaller number divides n − 1.

The two squares of 97, produced by division. A table of the division chain on 97 and a square root of minus one modulo it, with each row's quotient and remainder, the point at which the remainder falls below the square root marked, and the two squares that add to 97.

The two squares actually produced

Three proofs say a prime one more than a multiple of four is a sum of two squares, and not one of them hands over the squares. Running the Euclidean algorithm half-way does — and where to stop is the whole of the correctness argument.

$How rare a sum of two squares is. Two curves against the logarithm of the bound: the fraction of numbers below it that are sums of two squares, falling; and that count times the square root of the logarithm, divided by the bound, which is nearly constant.$

Almost no number is one

Sums of two squares look common — a sixth of all numbers up to a million are one. The fraction is falling to nothing, at a rate so slow that no computation will ever make it obvious, and the constant in front of it has been computed to fifty places and identified with nothing.

Where the aliquot sequence of every number up to 1000 goes. A grid of the starting values 2 to 1000 coloured by the fate of each aliquot sequence: 964 reach 1, 19 reach a perfect number, 3 enter a cycle, and 13 pass 10²² undecided.

The sum of the parts, taken again

Replace a number by the sum of its proper divisors and do it again. Most numbers fall to 1, a few land on a perfect number or a cycle, and some climb for hundreds of steps. Below a thousand there are twelve whose fate nobody knows — and the thing that keeps them climbing is a perfect number hiding in their factorisation.

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