Every step is a square root

The reachable points of a construction are wherever two drawn objects cross. There are three ways for that to happen, and all three come down to solving one equation.

Three intersections, two equations #

Take a plane with coordinates, and let $K$ be the set of numbers already reachable — closed under addition, subtraction, multiplication and division, which makes it a field. Every point built so far has both its coordinates in $K$.

A line through two such points has an equation $ax + by = c$ with $a$, $b$, $c$ all in $K$, because they are differences and products of the coordinates.

A circle centred at one such point through another has an equation $x^2 + y^2 + dx + ey + f = 0$, again with every coefficient in $K$, because the radius squared is a sum of squares of differences.

Now the three cases.

Line meets line. Two linear equations in two unknowns. Solving them uses only the four operations, so the crossing point has both coordinates in $K$ already. Nothing new is reached.

Line meets circle. Substitute the line into the circle and a quadratic in one unknown comes out. Its solutions are $\frac{-b ± \sqrt{b^2-4ac}}{2a}$ with $a$, $b$, $c$ in $K$ — so the new coordinates lie in $K$ together with one square root of an element of $K$, and nothing more.

Circle meets circle. Subtract the two circle equations. The $x^2$ and $y^2$ terms cancel, leaving a linear equation — the radical line of the two circles — and the problem collapses to the previous case.

So the third case is the second case, and the second case is the only one that reaches anything new. A construction step adjoins at most one square root.

What adjoining a square root costs #

The phrase “adjoins a square root” can be made exact, and making it exact is what turns a geometric fact into a countable one.

Suppose $K$ is a field and $r$ is in $K$ but $\sqrt r$ is not. Everything expressible using $K$ and $\sqrt r$ has the form $a + b\sqrt r$ with $a$ and $b$ in $K$ — because $(\sqrt r)^2$ is $r$, which is back in $K$, so no higher power of $\sqrt r$ survives. That set is closed under the four operations; the only one worth checking is division, and

$$\frac{1}{a+b\sqrt r} = \frac{a-b\sqrt r}{a^2-b^2r},$$

whose denominator is in $K$ and is not zero unless $\sqrt r$ was in $K$ after all.

The new field is therefore a two-dimensional space over the old one, with $1$ and $\sqrt r$ as a basis. That number — the dimension — is called the degree of the extension, and it is written $[K(\sqrt r) : K] = 2$.

The table beside the tower is not decoration. A claim that a set is a field of a stated size is a claim that products stay inside it, and that claim is finite: multiply every basis element by every other and look. Two elements, four products, all accounted for.

Degrees multiply #

The step that makes the whole argument work is one line of counting, and it is the reason a chain of doublings is a power of two rather than a sum.

If $L$ is a $d$-dimensional space over $K$, and $M$ is an $e$-dimensional space over $L$, then $M$ is a $de$-dimensional space over $K$. The basis is the obvious one: take each of the $e$ basis elements of $M$ over $L$, multiply it by each of the $d$ basis elements of $L$ over $K$, and the $de$ products are a basis.

Applied to a construction, this says that after $k$ steps the field reached has degree $2^{k}$ over the rationals at most — some steps adjoin a root already present and cost nothing, which is why it is at most.

The tower is the picture of that multiplication. Each rung is a doubling; the number at the top is the product of the rungs; and the table at each stage is the evidence that the stage is really the size the tower says it is.

A field is a vector space, which is where the counting comes from #

The word “dimension” is doing real work above and it is worth saying why it is available at all.

A field $L$ containing a smaller field $K$ can be regarded as a space of vectors with $K$ as the scalars: adding two elements of $L$ is vector addition, and multiplying an element of $L$ by an element of $K$ is scaling. Every axiom needed is one of the field axioms, so nothing has to be checked.

That single change of view is what makes the argument countable. A space over $K$ has a basis — a smallest set of elements from which everything is built, and built in exactly one way — and the number of elements in it is the dimension, which does not depend on which basis is chosen. All of that is the ordinary linear algebra of grids and coordinates, applied to an object that does not look like a grid at all.

So $\mathbb{Q}(\sqrt2)$ is a plane whose two axes are $1$ and $\sqrt2$, with rational coordinates. $\mathbb{Q}(\sqrt2,\sqrt3)$ is a four-dimensional space over $\mathbb{Q}$ with axes $1, \sqrt2, \sqrt3, \sqrt6$. The multiplication tables in the figures are that basis being tested: every product of two axis directions must be expressible in the axes, or the space is not closed and the dimension is a fiction.

This is the second time in the collection that a hard question has been made easy by finding the vector space it was hiding in — the first is the dot product as a shadow — and it will not be the last.

A worked chain, from two dots to a length #

The abstract statement is easier to trust after watching it happen once.

Start with the two dots, so the reachable field is $\mathbb{Q}$: everything with rational coordinates, and no drawing needed to get there.

Draw the two unit circles. They cross at height $\frac{\sqrt3}{2}$, which is not rational, so the field becomes $\mathbb{Q}(\sqrt3)$ — degree $2$.

Now use that height as a length and build $\sqrt{\frac{\sqrt3}{2}}$ with the semicircle construction. The number under the root is in $\mathbb{Q}(\sqrt3)$ and its square root is not, so the field becomes $\mathbb{Q}(\sqrt3, \sqrt{\sqrt3/2})$ — degree $4$ over the rationals.

Two steps, two doublings, and the number now reached satisfies a polynomial of degree four and none smaller. A third step would take it to eight, a fourth to sixteen. At no point can the count land on three, or six, or ten, because the only thing the machine ever does to the count is multiply it by two.

That is the entire content of the impossibility results, and everything after this is a matter of finding a number whose count is not a power of two.

Why “at most”, and not “exactly” #

A construction step need not double anything, and the cases where it does not are worth naming, because a proof that said “exactly” would be false.

A step may adjoin a square root of something that already has one in the field — drawing a circle whose radius squared is $4$ reaches $2$, which was there all along. A step may produce an intersection point whose coordinates were already available by another route, which happens constantly in real constructions and is why the second round of the closure is so redundant. And a step may be a line-line crossing, which adjoins nothing by the case analysis above.

So the degree of the field after $k$ steps is $2^j$ for some $j \le k$, and the theorem only ever needs the upper bound. That is a comfortable position to be in: the result is about what cannot be reached, so an argument that overestimates what can be reached is arguing on the safe side.

The theorem, and it is one sentence #

Putting the pieces together gives the result the rest of this ladder is built on.

If $x$ is constructible, then $[\mathbb{Q}(x) : \mathbb{Q}]$ is a power of two.

The argument: a construction reaching $x$ is a finite chain of steps, each adjoining at most one square root, ending in a field $F$ containing $x$ with $[F:\mathbb{Q}] = 2^k$. Now $\mathbb{Q}(x)$ sits inside $F$, and degrees multiply, so $[\mathbb{Q}(x):\mathbb{Q}]$ divides $2^k$. A divisor of a power of two is a power of two.

That is the whole proof, and it is worth noticing how little geometry is left in it. The only geometric input was the case analysis at the top of this page — three kinds of intersection, two kinds of equation. Everything after that is counting dimensions.

Degree, measured #

The degree of a single number over the rationals has a concrete meaning: it is the degree of the smallest polynomial equation with whole-number coefficients that the number satisfies, and that polynomial is unique up to scaling.

For $\sqrt2$ the polynomial is $x^2-2$, and the check that no smaller one exists is the check that $\sqrt2$ is not rational — which is the oldest impossibility proof there is and which the figure above performs by the crudest available method, trying every rational the rational root theorem allows.

That theorem is the workhorse of this entire field, so it is worth stating. If a polynomial with whole-number coefficients has a rational root $p/q$ in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient. There are finitely many such fractions, and testing them is division.

$\sqrt2+\sqrt3$ is a good example because it is not obviously a nested radical of the kind the previous essay described, and its degree is not obviously $4$. Squaring it twice and rearranging gives $x^4 - 10x^2 + 1 = 0$, and the table shows that this polynomial has no rational root, which rules out its factoring off a linear piece. It happens also not to factor into two quadratics over the rationals, so $4$ is the answer: the number lives in the top of the tower in the first figure and nowhere smaller.

What the theorem does not say #

Two cautions, both of which matter and one of which is a genuine gap.

The converse is false. Having degree a power of two does not make a number constructible. There are numbers of degree $4$ that no compass reaches, because the tower over them cannot be built two steps at a time — the relevant condition involves the whole symmetry group of the polynomial and not merely its degree. The theorem here is a one-way test, and one-way is enough for every impossibility in this ladder, because every one of them works by exhibiting a degree that is not a power of two.

The degree has to be found, not guessed. Writing down a polynomial the number satisfies is easy; showing it is the smallest is the work. For the three classical problems that work is small — a cubic with no rational root is irreducible, which is a special fact about degree three and four, since a factorisation would have to include a linear piece. For higher degrees it is genuinely harder and the rational root theorem alone is not enough.

The gap in the converse is worth one concrete example, because “there are numbers of degree four that are not constructible” is the kind of sentence that gets read as a technicality. Take a cubic with three real roots and no rational one — $x^3 - 3x - 1$ is the standard specimen, and it is exactly the equation that appears when trying to trisect $40°$. Its roots have degree three, so none of them is constructible. Now take a root of the quartic $x^4 - x - 1$: its degree is four, a power of two, and it is still not constructible, because the tower over it cannot be climbed in steps of two. What decides the matter is a count of the polynomial’s symmetries rather than a count of its degree, and that count is Galois’ contribution rather than Wantzel’s.

The distinction matters for reading this ladder honestly. Every impossibility here is proved by the easy direction, which is a genuine theorem and needs nothing further. No essay in it proves that anything is constructible by citing a degree — where a construction is claimed, a construction is drawn.

Where the doubling shows up as a picture #

It is worth returning to the drawing, because the algebra above has a geometric shadow that is easy to miss.

Every crossing on the axis of that figure has rational coordinates, and every crossing off it has a coordinate involving $\sqrt3$. That is the case analysis made visible: the line contributed nothing new and the circles contributed one square root.

It is also the reason the second round of the closure is so much bigger and yet so much less interesting. The points multiply, but the field grows only when a genuinely new square root appears, and most of the thousand-odd second-round intersections lie in fields already reached.

Where this ladder goes #

Everything needed for the impossibility proofs is now in place, and the proofs themselves are short.

To show a construction impossible, produce a number it would have to reach and show that number has degree three, or five, or anything that is not a power of two. Doubling the cube needs $\sqrt[3]2$, whose minimal polynomial is $x^3-2$. Trisecting sixty degrees needs a root of $8y^3-6y-1$. Both are cubics with no rational root, so both have degree three, and three does not divide any power of two.

The polygons are the same argument done carefully over all $n$ at once, and the answer turns out to depend on the size of a different count entirely. And squaring the circle is not this argument at all: $\pi$ has no degree, because it satisfies no polynomial equation with whole-number coefficients whatsoever, and proving that took another two thousand years and a completely different kind of mathematics.