Concept

Quadratic irrational

A number of the form (a + b√D)/c with whole numbers a, b, c and D not a perfect square. These are the irrational roots of quadratic equations with whole coefficients, and they are exactly the numbers whose continued fractions eventually repeat.

Named by 4 essays across one field — each of them below, with the objects they name alongside it.

Whole-number points on x² − 2y² = 1. The branch of the hyperbola x² − 2y² = 1 in the first quadrant, with the whole-number points on it marked and labelled, and the lattice drawn faintly behind.

One solution that makes all the others

The equation x² − 2y² = 1 has infinitely many whole-number solutions, and every one of them is a power of the smallest. The multiplication that produces them is what multiplying two numbers of the form a + b√2 comes to when the √2 terms are collected — so an equation about a hyperbola turns out to carry a group.

number · Pell
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.

number · Pell
How many digits Pell's first solution has, for D up to 1000. A scatter of the digit count of the smallest solution of Pell's equation against D, with the record-setting values ringed and labelled.

Sixty needs two digits and sixty-one needs ten

The smallest solution of x² − 60y² = 1 is x = 31. For x² − 61y² = 1 it is x = 1,766,319,049. The jump is not an accident of 61: the solution is exactly the product of one period's complete quotients, so its size is set by how long the continued fraction takes to come home — and for the cattle problem Archimedes is said to have posed, that product has 103,273 digits.

number · Pell
The first 600 terms of the continued fraction of ∛2. Partial quotients of ∛2 computed exactly: 1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3 …; the largest in the first 600 is 7451.

A cube root that looks like chance

The continued fraction of √61 repeats after eleven terms, and every square root's must. The cube root of 2 obeys no such law. Computed exactly, by a method of Lagrange's that never leaves the whole numbers, its terms run 1, 3, 1, 5, 1, 1, 4 … and throw up a 534 at the thirty-fifth place and a 7,451 at the 571st — and in every statistic anyone has measured, they look like the terms of a number picked at random.

number · Pell

Named alongside it

The objects these essays reach for when they reach for this one.

Continued fractionsPell equationFundamental solutionPeriodicityAlgebraic numberAlgorithmConjugateConvergentCubicGrowth rateHyperbolaLattice point

All concepts