The diagonal no unit measures

A golden rectangle eats itself: cut a square from it and a smaller golden rectangle is left. That one property is the whole of the golden ratio, and the rectangle is the tidy modern way to show it. It is not where the Greeks met the number. They met it in the regular pentagon, where it is the ratio of a diagonal to a side, and the pentagon has a version of the eating-itself property that is more striking than the rectangle’s because nothing has to be added to see it. Draw the five diagonals, and the smaller pentagon is already there.

That smaller pentagon carries an argument, and it may be the oldest proof in mathematics that a length is not a fraction of another. Kurt von Fritz argued in 1945 that the Pythagoreans discovered incommensurability not in the square, with its diagonal $\sqrt 2$, but in the pentagon — their emblem — by running Euclid’s subtraction on the diagonal and the side and watching it never stop. Whether or not the history happened that way, the argument is complete, and it is a drawing.

The star inside the pentagon #

Call the side $s$ and the diagonal $d$. The ratio $d/s$ can be found from similar triangles alone. Two diagonals from one vertex, together with the side opposite that vertex, make a tall isosceles triangle with apex angle $36°$ and legs $d$. One of the other diagonals cuts it into a smaller triangle of exactly the same shape — apex $36°$ — whose legs are $s$ and whose base is $d - s$. Same shape means same proportions:

$$\frac{d}{s} = \frac{s}{d - s}.$$

Cross-multiplying gives $d^2 - ds - s^2 = 0$, and dividing by $s^2$ gives $(d/s)^2 - (d/s) - 1 = 0$, whose positive root is $\varphi = (1 + \sqrt 5)/2 = 1.6180\ldots$ The figure measures the ratio on the drawn coordinates, not from the formula, in every pentagon it draws, and finds $1.618034$ each time.

The diagonal is cut by the two diagonals that cross it into three pieces. The two end pieces are each $s/\varphi$ and the middle piece is $s/\varphi^2$; together they make $s(2/\varphi + 1/\varphi^2) = s\varphi = d$. The middle piece is the side of the inner pentagon, and the end piece plus the middle piece — the stretch from a vertex to the far crossing — is exactly $s$, one full side. That coincidence is the whole proof.

Subtraction that never finishes #

Euclid’s way of comparing two lengths is to subtract the smaller from the larger, and keep going with the smaller and the remainder. If some unit measures both lengths exactly, the lengths are whole numbers of it, the remainders are too, and since whole numbers cannot decrease for ever the process must stop — at which point the last nonzero remainder is the largest common unit. It is Euclid’s algorithm, drawn as a tiling, applied to lengths.

Run it on the pentagon. The diagonal minus the side is $d - s = s/\varphi$, which is the inner pentagon’s diagonal. The side minus that is $s - s/\varphi = s/\varphi^2$, which is the inner pentagon’s side. Two subtractions have turned the pair (diagonal, side) of the big pentagon into the pair (diagonal, side) of the small one.

So the process can never stop: at every stage the pair is a pentagon’s diagonal and side again, in the ratio $\varphi$, and the next two subtractions produce the next pentagon in. The figure runs it for eight steps and finds the ratio $1.618034$ at every one.

That is the proof. If some unit measured both the diagonal and the side a whole number of times, every pair in the process would be whole numbers of that unit. The pairs shrink by $\varphi$ at every step without ever reaching zero. A sequence of positive whole numbers cannot decrease for ever. So there is no such unit — the diagonal and the side are incommensurable, and $\varphi$ is not a ratio of whole numbers. The argument is the same shape as the proof that folds a square into a smaller one to show $\sqrt 2$ is irrational: a supposed solution produces a smaller solution, forever. In the pentagon the smaller solution is not constructed by the argument; it is sitting in the picture.

The same descent, in whole numbers #

The picture’s argument can be translated into arithmetic, and the translation shows exactly what the smaller pentagon is. Suppose $\varphi = p/q$ with $p$ and $q$ whole numbers, $q$ as small as possible. The defining property $\varphi = 1 + 1/\varphi$ says $p/q = 1 + q/p$, which rearranges to $p^2 = pq + q^2$, and from that

$$\frac{q}{p - q} = \frac{p}{q}.$$

So $\varphi$ is also the ratio $q/(p - q)$ — and since $\varphi$ is between one and two, $p - q$ is a positive whole number smaller than $q$. That contradicts the choice of $q$ as the smallest possible denominator. There is no smallest, so there is none at all.

Read geometrically, $p$ and $q$ are the diagonal and side measured in a supposed common unit, $p - q$ is the diagonal minus the side, and the new fraction $q/(p - q)$ is the ratio of the next pair in the subtraction. The arithmetic proof is the pentagon picture with the pentagons taken away and only their measurements kept. It is also, line for line, the pattern of the classical proofs that square roots of non-squares are irrational: find, from a supposed fraction, a smaller fraction for the same number.

What incommensurability cost, and what repaired it #

The discovery that two lengths in a figure as simple as a pentagon have no common measure was not a curiosity to the people who made it. The Pythagorean view, as later writers describe it, was that all things are number — that every magnitude is some whole number of some unit — and geometry had been done on that assumption: two lengths were compared by finding a unit that measured both, and a proportion $a : b = c : d$ meant that the same whole numbers measured both pairs. The pentagon’s diagonal and side have no such unit, so for them a ratio, as then understood, did not exist.

The repair took most of a century and is Book V of Euclid’s Elements, attributed to Eudoxus of Cnidus. It defines equality of ratios without any common unit: $a : b$ equals $c : d$ if, for every pair of whole numbers $m$ and $n$, $ma$ exceeds, equals or falls short of $nb$ exactly when $mc$ exceeds, equals or falls short of $nd$. A ratio is known by which fractions it lies above and which it lies below. That is, almost word for word, how Richard Dedekind defined the real numbers in 1872 — a number is the cut it makes in the fractions — and Dedekind acknowledged the resemblance. The pentagon’s incommensurable diagonal is, by that route, one of the ancestors of the real number line.

How many times the smaller fits #

Euclid’s process records one more thing: how many times the smaller length fits into the larger before a remainder is left. Those counts are the digits of the ratio’s continued fraction, and they tell irrational ratios apart.

For the pentagon the side fits into the diagonal exactly once, and the remainder fits into the side exactly once, and so on for ever: $\varphi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}$. For the square, the side fits into the diagonal once and then the remainders fit twice, for ever: $\sqrt 2 = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cdots}}$. For $8/5$, which agrees with $\varphi$ to two decimal places, the process runs $1, 1, 1, 2$ and stops — because $8$ and $5$ have a common unit, namely $1$.

All ones is the extreme case in a precise sense: a continued fraction with small digits converges slowly, since each digit is how much a new approximation improves on the last, and ones are the smallest digits there are. So $\varphi$ is the number least well approximated by fractions, which is the same fact the golden rectangle showed as the slowest possible peeling of squares, and the reason it keeps turning up wherever a process wants to avoid resonance with any fraction. Made precise, it is Hurwitz’s theorem: every irrational number has infinitely many fractions $p/q$ within $1/(\sqrt 5\, q^2)$ of it, and the $\sqrt 5$ cannot be improved, because for $\varphi$ it is exactly right. The constant in the best general theorem about approximating irrationals is the pentagon’s.

The triangle that eats itself #

The pentagram’s points are triangles of the kind used above: apex $36°$, base angles $72°$. That triangle has its own version of the rectangle’s self-similarity.

Bisect one of the $72°$ base angles. The bisector meets the opposite side and cuts the triangle into two: a smaller triangle with angles $36°, 72°, 72°$ — the same shape, smaller by $\varphi$ — and a wide triangle with angles $36°, 36°, 108°$, sometimes called a golden gnomon. The bisector, the original base, and the piece of the far side it cuts off are all equal. Repeat on the smaller triangle and the cuts spiral inward for ever, exactly as the squares spiral inward in the golden rectangle.

The two triangles in that cut, the $36°$ and the $108°$ one, are exactly half of the two tiles in Penrose’s rhombus tiling, and the rule that cuts each into smaller copies is the rule that builds the tiling. That is where the next essay goes.

Every length is a power of φ #

Look back at the pentagram with all its nested pentagons, and list every length that appears: sides, diagonals, the pieces of diagonals between crossings. With the outer side as $1$, the outer diagonal is $\varphi$, the end pieces of each diagonal are $1/\varphi$, the middle pieces $1/\varphi^2$, the next pentagon’s diagonal $1/\varphi$ and its side $1/\varphi^2$, and so on inward. Every length in the figure is a power of $\varphi$ — positive, zero or negative — and nothing else ever occurs.

Since the pieces of a diagonal add up to the whole, and each piece is a power of $\varphi$, the figure is full of identities of the form $\varphi^{k} = \varphi^{k-1} + \varphi^{k-2}$: the end piece plus the middle piece is a side, $1/\varphi + 1/\varphi^2 = 1$; the side plus the end piece is the diagonal, $1 + 1/\varphi = \varphi$. Each is $\varphi^2 = \varphi + 1$ multiplied by a power of $\varphi$, and applied repeatedly it gives $\varphi^n = F_n\varphi + F_{n-1}$, with the Fibonacci numbers as coefficients: $\varphi^2 = \varphi + 1$, $\varphi^3 = 2\varphi + 1$, $\varphi^4 = 3\varphi + 2$, $\varphi^5 = 5\varphi + 3$. The Fibonacci numbers that the whirling squares produced as side lengths appear here as the bookkeeping of a single figure’s segments, and the reason is the same equation.

The pentagon’s equation #

The number $\varphi$ enters the pentagon through a quadratic equation, and the equation comes from the algebra of the fifth roots of one.

The vertices of a regular pentagon are the solutions of $z^5 = 1$, and apart from $z = 1$ they satisfy $z^4 + z^3 + z^2 + z + 1 = 0$. That equation is symmetric — reversing the coefficients leaves it unchanged — and dividing by $z^2$ and writing $w = z + 1/z$ collapses it to $w^2 + w - 1 = 0$. For a point on the unit circle, $z + 1/z = 2\cos\theta$, so $2\cos 72° = 1/\varphi$ and $2\cos 144° = -\varphi$. The pentagon’s $\varphi$ is a root of a quadratic because the fifth roots of unity pair up into a quadratic’s roots.

A quadratic with whole-number coefficients can be solved with one square root, and one square root can be constructed with compass and straightedge. So the regular pentagon is constructible, which Euclid did in Book IV of the Elements — and the reason it is constructible is the same reason $\varphi$ is irrational only mildly: it is a root of a degree-two equation, as simple as an irrational number can be.

Why no crystal has five-fold symmetry #

The same number has a consequence far from geometry, and it is the one that made the pentagon’s arithmetic famous in physics. A crystal is a lattice of atoms, repeating in every direction, and a rotation that is a symmetry of the crystal must carry the lattice onto itself. Written in the lattice’s own coordinates, such a rotation is a matrix of whole numbers, so its trace — the sum of its diagonal entries — is a whole number. But the trace of a rotation of the plane by an angle $\theta$ is $2\cos\theta$, whatever coordinates are used.

For a rotation by $72°$, the trace would be $2\cos 72° = 1/\varphi = 0.618\ldots$, which is not a whole number, and not even a fraction. So no lattice in the plane, and no crystal, has a symmetry of order five. The same argument allows only $2\cos\theta \in \{-2, -1, 0, 1, 2\}$, which is rotations of order one, two, three, four and six: the crystallographic restriction, and the reason floors are tiled with triangles, squares and hexagons and never with pentagons. The pentagon’s diagonal being incommensurable with its side and the pentagon being unable to tile a periodic pattern are, at bottom, the same fact about $\varphi$.

That restriction is exactly what made the discovery of quasicrystals in 1982, by Dan Shechtman, so startling: an alloy whose diffraction pattern had sharp ten-fold symmetry, which no lattice could produce. The resolution was that the atoms were ordered but not periodic — arranged in a pattern like Penrose’s tiling, whose five-fold symmetry is local and statistical rather than a symmetry of a repeating lattice. Shechtman received the Nobel Prize in Chemistry in 2011.

What the drawing cannot show #

It cannot show irrationality. Every drawing has finite precision, and a drawn ratio of $1.618034$ is equally consistent with $\varphi$ and with a fraction such as $1{,}618{,}034/1{,}000{,}000$. The descent argument is what rules the fraction out, and it is an argument about all steps of a process that the drawing shows only eight of. The figures confirm the ratio at every step drawn; the logic supplies the steps that are not.

It cannot show the history. Von Fritz’s reconstruction, that the pentagon rather than the square gave the Greeks their first incommensurable pair, is a hypothesis built from the importance of the pentagram to the Pythagoreans and from the form of Euclid’s algorithm. The ancient sources do not settle it.

And it cannot show the lattice argument. That the trace of a lattice symmetry is a whole number is a statement about matrices in integer coordinates, and no drawing of a pentagon shows a lattice failing to have five-fold symmetry; it shows only the number, $1/\varphi$, that the failure comes from.

Where the pentagon leads #

The rectangle’s self-similarity produced a spiral; the pentagon’s produces a star within a star, and the golden triangle’s produces a rule for cutting two shapes into smaller copies of themselves. That rule, applied to Penrose’s two rhombs, builds a tiling of the whole plane that has local five-fold symmetry everywhere, uses thick and thin tiles in the ratio $\varphi$, and — because $\varphi$ is irrational — never repeats. The pentagon cannot tile periodically; Penrose’s tiles, built from its triangles, tile the plane only non-periodically.

Sideways, the same number is the ratio of edge lengths that the icosahedron and dodecahedron are built from, which is why those two solids are the only regular polyhedra whose coordinates need a square root of five.

A star with a smaller star inside #

In a regular pentagon, the diagonal minus the side is the diagonal of the pentagon the pentagram encloses, and the side minus that is its side. So Euclid’s subtraction on diagonal and side reproduces a smaller pentagon every two steps and never stops, and since whole numbers cannot shrink for ever, no unit measures both: $d/s = \varphi$ is not a fraction. Its continued fraction is all ones, the slowest possible.

The pentagram’s points are $36°$–$72°$–$72°$ triangles that cut into smaller copies of themselves, the pentagon’s vertices are fifth roots of one whose symmetric equation reduces to $w^2 + w - 1 = 0$, and the rotation by $72°$ has trace $1/\varphi$, which is why no crystal has five-fold symmetry.

When a figure contains a smaller copy of itself, look for the process that produced it — if the process never ends, something in the figure is irrational.