The lengths dividers cannot reach

A compass does two jobs at once. It draws a circle, which can then be crossed with lines and other circles, and it carries a length: set its legs to a segment, lift it, and put it down somewhere else. Euclid’s own compass was not allowed to carry lengths — it collapsed when lifted — and his second proposition shows how to carry one anyway. This essay asks about the opposite instrument: a pair of dividers that carries lengths and draws no circles at all.

With a straightedge, the dividers are far stronger than a straightedge with a bare circle. They find midpoints, drop perpendiculars and draw parallels, so the line at infinity that a straightedge alone cannot recover is recovered at once. And yet they reach strictly less than a compass does, by a margin that no figure on the page reveals. David Hilbert described exactly what they reach in 1899, in the book that rebuilt Euclid’s geometry from axioms.

The figure below is what they add. A spiral of right triangles, each with a leg of length one standing square to the long side of the last: the long sides are $\sqrt 2, \sqrt 3, \ldots, \sqrt{17}$. Every triangle is one step of the only new operation the dividers provide.

A right angle without a circle #

A straightedge alone cannot draw a perpendicular. With dividers it can, and the construction is the whole reason the dividers are strong.

Mark off equal lengths on either side of a point $B$ of a line, then from the two marks carry a longer length inward until the two copies meet: the meeting point is the apex of an isosceles triangle, and the line from $B$ to the apex is square to the base. Carrying equal lengths is what the dividers do, and an isosceles triangle is what equal lengths make; no circle is ever drawn. From perpendiculars come parallels (two perpendiculars to one line), and from parallels come midpoints and every construction that needs only them.

Then the step in the figure. Given a length $x$ on a line, raise a perpendicular at its end, carry a unit up it, and carry the long side of the resulting right triangle back to the line. The new length is $\sqrt{1 + x^2}$, by Pythagoras. Starting from $x = \sqrt 2$ the figure lands on $\sqrt 3$, measured and checked.

More generally, the long side of any right triangle whose legs have been constructed can be carried anywhere: $\sqrt{a^2 + b^2} = a\sqrt{1 + (b/a)^2}$, and multiplying and dividing lengths are straightedge constructions once parallels are available. So the dividers can take the square root of any sum of squares of lengths already built. That is the whole of their new power.

Theodorus, and a field named after nobody #

Plato’s dialogue Theaetetus reports that Theodorus of Cyrene proved $\sqrt 3, \sqrt 5, \ldots, \sqrt{17}$ irrational one at a time, and stopped at seventeen. The spiral that now carries his name is a modern guess at how he drew them, and it happens to be exactly the construction above, repeated: the hero’s sixteen triangles turn through $351.2°$ in all, so the one for $\sqrt{18}$ would be the first to overlap the start.

Hilbert’s description of everything the dividers reach is in the same terms. Start from the number $1$, allow addition, subtraction, multiplication and division, and allow the one extra operation $x \mapsto \sqrt{1 + x^2}$. The numbers so obtained form a field, usually called the Pythagorean field, and Hilbert showed that the points a straightedge and dividers can construct from a unit segment are exactly those whose coordinates lie in it.

There is a quicker route to $\sqrt n$ than the spiral, and it uses a theorem from a different part of mathematics. Lagrange proved in 1770 that every whole number is a sum of four squares, $n = a^2 + b^2 + c^2 + d^2$. Then $\sqrt n$ is the hypotenuse of a right triangle whose legs are $\sqrt{a^2 + b^2 + c^2}$ and $d$, and $\sqrt{a^2 + b^2 + c^2}$ is the hypotenuse of one with legs $\sqrt{a^2 + b^2}$ and $c$, and so on: three right triangles, whatever the size of $n$. The spiral reaches $\sqrt{17}$ in sixteen steps; the four squares $17 = 16 + 1$ reach it in one. The identity that multiplies sums of squares is what lies behind Lagrange’s theorem, and here it bounds the cost of a construction.

That field contains a great deal. It contains $\sqrt 2 = \sqrt{1 + 1}$, $\sqrt 3 = \sqrt{1 + (\sqrt2)^2}$, every $\sqrt n$, the golden ratio $(1 + \sqrt 5)/2$, and nested roots like $\sqrt{2 + \sqrt 2} = 2\cos(\pi/8)$. Every one of them is a length a compass could also draw, since the compass reaches every number built from square roots. The question is what, of the compass’s reach, the dividers miss.

The step that only ever roots a positive number #

The answer turns on something that the step $\sqrt{1 + x^2}$ does automatically and a general square root does not.

A number built from square roots can often be read in more than one way. The number $1 + \sqrt 2$ is built from $\sqrt 2$, and nothing in the arithmetic of addition and multiplication distinguishes $\sqrt 2$ from $-\sqrt 2$: both square to two. Replacing one by the other throughout gives a second, equally consistent reading, in which $1 + \sqrt 2$ becomes $1 - \sqrt 2$. The different readings of a number are its conjugates — the other roots of the simplest polynomial equation with whole coefficients that it solves.

Now look at what the dividers’ step does. If $x$ is real under every reading, then $1 + x^2$ is at least one under every reading, and its square root is real under every reading too. Addition, subtraction, multiplication and division preserve the property as well. So every number the dividers reach is real under every reading — every one of its conjugates is a real number. A number with that property is called totally real.

The number $\sqrt{1 + \sqrt 2}$ is not. Under the second reading its radicand is $1 - \sqrt 2$, which is negative, and its square root is imaginary. So no sequence of divider steps can ever produce it, however long and however clever, although a compass produces it in one stroke.

The converse is the harder half of Hilbert’s theorem: every totally real number the compass can reach, the dividers reach too. The proof comes down to showing that a number positive under every reading can be written as a sum of squares of numbers already built, so that its square root is a hypotenuse the dividers can carry. That a totally positive number in a field of this kind is always a sum of squares is a theorem Carl Ludwig Siegel proved in general in 1921, answering a question of Hilbert’s own.

Reading reach off the conjugates #

With the theorem in hand, deciding whether the dividers can reach a length is a calculation about a polynomial, and never about a drawing.

The first five rows have every conjugate on the real line, and the dividers reach them all — including $2\cos(2\pi/17)$, whose polynomial has degree eight and all eight of its roots real. The next two have two conjugates each off the line. $\sqrt{1 + \sqrt2}$ solves $x^4 - 2x^2 - 1 = 0$, whose roots are $\pm\sqrt{1 + \sqrt 2}$ and $\pm i\sqrt{\sqrt 2 - 1}$. The fourth root of two solves $x^4 - 2 = 0$, with roots $\pm\sqrt[4]2$ and $\pm i\sqrt[4]2$. The compass reaches both; the dividers reach neither. The last row, the cube root of two, has degree three, so it is out of reach of the compass too, for the reason that makes doubling the cube impossible.

Notice what the table does not use. It never looks at a construction, a diagram or a length. The reach of an instrument, which sounds like a question about drawing, is settled by where the other roots of a polynomial lie — points that exist only in the complex plane and that no construction on the page ever touches.

Two nested roots, read four ways #

It is worth doing one comparison entirely by hand, because the two numbers look so alike.

Take $\alpha = \sqrt{2 + \sqrt 2}$ and $\beta = \sqrt{1 + \sqrt 2}$. Each is a square root of a square root, and each is a perfectly good length: $\alpha \approx 1.848$ and $\beta \approx 1.554$. Each has four readings, because there are two choices for the inner root and then two for the outer one.

For $\alpha$ the four readings are $\pm\sqrt{2 + \sqrt2}$ and $\pm\sqrt{2 - \sqrt2}$. Since $\sqrt 2 \approx 1.414$ is less than $2$, both radicands are positive, and all four readings are real numbers: $\pm 1.848$ and $\pm 0.765$. They are, in fact, $2\cos(\pi/8)$, $2\cos(3\pi/8)$ and their negatives — the kind of number the regular octagon is built from.

For $\beta$ the four readings are $\pm\sqrt{1 + \sqrt2}$ and $\pm\sqrt{1 - \sqrt 2}$, and $1 - \sqrt 2$ is negative. Two readings are real and two are imaginary.

Now run the dividers’ step backwards. If $\beta$ were a hypotenuse the dividers had built, it would be $\sqrt{1 + x^2}$ for some $x$ they had built earlier — or, more generally, the square root of a sum of squares of such numbers. Under every reading, a sum of squares of real numbers is positive, so every reading of $\beta$ would be real. Two of them are not. That is the whole impossibility proof, and it never mentions geometry.

The comparison also shows why the obstruction is invisible on the page. On the page, only the reading in which every root is taken positive exists: a segment of length $1.554$ is a segment like any other. The other readings live in the algebra of the number, and they decide what the page can hold.

A semicircle, where dividers never arrive #

The contrast with the compass is sharpest on the very number the dividers cannot reach.

A semicircle on a segment split into lengths $a$ and $b$ has, above the split, a height of $\sqrt{ab}$: the two right triangles on either side of that height are similar, which is the same argument that makes the altitude of a right triangle the mean of the pieces it cuts. With $a = 1$ and $b = 1 + \sqrt 2$, the height is $\sqrt{1 + \sqrt 2}$.

What the circle does that the dividers cannot is meet a line at a place not decided in advance. The height of the semicircle above the split is found by crossing the circle with the perpendicular, and where that crossing happens depends on the circle’s curvature, not only on lengths that have already been built. Dividers never cross a circle with anything; they only lay down lengths that already exist. Every point they reach is a hypotenuse of lengths already in hand, and hypotenuses are always square roots of sums of squares.

It is worth noticing which compass is doing the work. A compass stuck at one opening was shown to reach everything a full compass reaches, with a straightedge’s help. It cannot carry an arbitrary length at all. The dividers can carry any length and cannot draw a circle. Between the two, the fixed compass wins outright — drawing is what matters, and carrying is not.

Polygons are not what is lost #

It is tempting to guess that the dividers fail on the famous constructions, the regular polygons. They do not.

The polygons a compass can draw are the ones whose number of sides is a power of two times distinct Fermat primes, and every one of them is within the dividers’ reach. The coordinates of the corners are cosines and sines of $2\pi k/n$, and the conjugates of a cosine of this kind are other cosines of the same kind, all real. The same is true of the sines. So the triangle, the pentagon, the 17-gon, the 257-gon and the 65,537-gon can all be drawn without drawing a circle.

What the dividers lose is subtler: lengths like $\sqrt{1 + \sqrt 2}$ and $\sqrt[4]2$, and figures that need them — a square whose area is $\sqrt 2$, for instance, whose side is $\sqrt[4]2$. The dividers can build a square of area $2$, and of area $3$, and of area $2 + \sqrt 2$, but not of area $\sqrt 2$. Nothing about the shapes suggests the difference.

A plane where some circles never meet #

Hilbert did not ask the question as a puzzle about instruments. His axioms for geometry included one allowing any segment to be copied along any line from any point — the dividers, written as an axiom — and none guaranteeing that a line through the inside of a circle must cross it. He wanted to know how much of Euclid survived without that guarantee.

The Pythagorean field answers it with a model. Take as the plane all points whose coordinates lie in the field. Every axiom of incidence, order, congruence and parallels holds there. Now take the circle of radius $\sqrt 2$ about the origin and the vertical line $x = \sqrt{2 - \sqrt 2}$; both are in the field, since $2 - \sqrt 2$ is positive under both readings of $\sqrt 2$. In the ordinary plane they cross at the height $\sqrt{2 - (2 - \sqrt 2)} = \sqrt[4]{2}$ — the fourth root of two, which the table above showed is not in the field. In the Pythagorean plane the line passes through the inside of the circle and never meets it.

That is the method of models at work: a structure in which every other axiom holds and one statement fails proves that the statement does not follow from the others. Here the statement is that circles and lines meet when they should, and the structure is the reach of an instrument. The dividers are the geometry without the circle axiom, and the numbers they miss are the points where that geometry has holes.

What the conjugate strips cannot show #

The table computes conjugates as roots of stated polynomials, found by an iteration and checked by substitution. That the stated polynomial is really the simplest one each number solves — its minimal polynomial — is checked numerically rather than proved in the figure; for these eight numbers it is a standard fact, and for the degree-eight cosine it comes from the theory of the 17-gon rather than from any computation drawn here. A wrong polynomial with extra roots would put extra dots on the strip.

The spiral and the step are drawn, but a proof that some length is unreachable cannot be drawn at all, since it is a statement about every possible construction. The strip of conjugates is the certificate instead: one dot off the line settles the question for every construction at once, which is exactly the kind of argument that the cube that will not double made with degrees.

Where does this leave the instruments? #

Every classical pair of instruments has now been sorted into one of three levels. A straightedge alone, or with a bare circle, reaches only what incidences allow — the rational points of whatever is given, and their projective relatives. A straightedge with dividers reaches the totally real numbers built from square roots. And every instrument that can draw a circle — a full compass, a compass alone, a compass stuck at one opening, a straightedge with one centred circle — reaches all numbers built from square roots, totally real or not. Above that sit the instruments that solve cubics, the marked ruler and the conics, which reach further still.

What is still not understood is the finer question the price of a construction raised: not whether a point can be reached, but how many steps it takes, and whether a construction known to reach it is anywhere near the cheapest. For the dividers the question has an extra twist. Every compass construction of a totally real number can, by Hilbert’s theorem, be replaced by a divider construction, and how much longer the replacement must be is a question about cost of exactly the kind left unanswered for the compass itself.