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Turning a question about drawing into a question about degrees: each step extracts at most a square root, so a constructible number has degree a power of two — and three classical problems close at once.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to translate a construction into a tower of field extensions, say why each step has degree at most two, deduce that every constructible number has degree a power of two, apply that to doubling the cube, trisecting an angle and squaring the circle, say why the condition is necessary but not sufficient, and state which regular polygons are constructible.
The last four lessons built degrees, minimal polynomials, splitting fields and the tower law, all inside algebra. This lesson points them at three questions the Greeks asked and could not settle, and which stayed open for two thousand years. Nothing new is needed: the whole argument is the tower law, applied to a geometric process.
A point is constructible when it can be reached from $(0,0)$ and $(1,0)$ by finitely many intersections of lines through constructed points and circles centred at them. A quadratic tower is a chain $\mathbb{Q} = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_k$ with each step of degree $1$ or $2$. A number is transcendental when no polynomial over $\mathbb{Q}$ has it as a root. A Fermat prime has the form $2^{2^{m}} + 1$: the known ones are $3, 5, 17, 257, 65537$.
Setting up. Put the two given points at $(0,0)$ and $(1,0)$. A construction produces new points by intersecting:
In every case the new coordinates satisfy an equation of degree at most $2$ over the field generated so far. So each step enlarges the field by a degree of $1$ or $2$.
The theorem. If $\alpha$ is constructible then there is a tower
$$\mathbb{Q} = F_0 \subseteq F_1 \subseteq \cdots \subseteq F_k \ni \alpha, \qquad [F_{i+1} : F_i] \le 2,$$
so $[F_k : \mathbb{Q}] = 2^{j}$ for some $j$. By the tower law $[\mathbb{Q}(\alpha) : \mathbb{Q}]$ divides that, so
$$[\mathbb{Q}(\alpha) : \mathbb{Q}] \ \text{is a power of } 2.$$
The three classical problems, closed.
What a construction can do. The constructible numbers form a field closed under square roots: the four operations come from similar triangles, and $\sqrt{a}$ from the geometric mean in a semicircle on a diameter of length $a + 1$.
Necessary, not sufficient. Degree a power of $2$ does not guarantee constructibility. The sharp statement is that $\alpha$ is constructible exactly when the splitting field of its minimal polynomial has degree a power of $2$ over $\mathbb{Q}$ — a condition about the Galois group, which the next unit supplies. A quartic whose Galois group is $S_4$ gives a degree-$4$ number that cannot be drawn.
Regular polygons. A regular $n$-gon is constructible exactly when $n$ is a power of $2$ times distinct Fermat primes. So $3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20$ yes; $7, 9, 11, 13, 14, 18$ no. Gauss proved the $17$-gon constructible at nineteen, which is what decided him on mathematics.
Another way: picture
Think of the constructible points as everything reachable from two dots by a machine with exactly two moves. Each move can, at worst, extract one square root — and nothing else. So the reachable set is the numbers built from rationals by repeated square roots, and a cube root is simply not a shape the machine can make. The impossibility is a statement about the machine, not about the number.
Another way: steps
To decide whether something can be constructed: 1. Reduce the problem to constructing one length. 2. Find that number's minimal polynomial over $\mathbb{Q}$. 3. Take its degree. 4. If the degree is not a power of $2$, the construction is impossible — and that settles it. 5. If it is a power of $2$, the test is inconclusive: look at the splitting field, or find the construction.
The claim is about the general angle. Many particular angles trisect easily: $90^\circ$ into three $30^\circ$ angles is a standard construction, and $180^\circ$ is trivial. What is impossible is a construction that works for every angle, and one counterexample suffices.
The counterexample. $60^\circ$ is constructible — it is an angle of an equilateral triangle. Trisecting it would produce $20^\circ$, hence $\cos 20^\circ$. The triple-angle identity $\cos 3\theta = 4\cos^{3}\theta - 3\cos\theta$ with $\theta = 20^\circ$ gives
$$\tfrac12 = 4x^{3} - 3x, \qquad \text{that is} \qquad 8x^{3} - 6x - 1 = 0.$$
By the rational root test the candidates are $\pm 1, \pm \tfrac12, \pm\tfrac14, \pm\tfrac18$, and none is a root. A cubic with no rational root is irreducible over $\mathbb{Q}$, so $\cos 20^\circ$ has degree $3$ and is not constructible.
What changes with a different tool. A ruler with two marks on it trisects any angle — this is the neusis construction, known to Archimedes. So does folding a piece of paper. Adding a conic section to the toolkit also does it. The theorem is entirely about which instruments are allowed, and says nothing about $20^\circ$ being mysterious.
The regular polygons, for completeness. A regular $n$-gon needs $\cos(2\pi/n)$, whose degree is $\phi(n)/2$. That is a power of $2$ exactly when $n = 2^{a}p_1\cdots p_r$ with the $p_i$ distinct Fermat primes. The heptagon fails because $\phi(7)/2 = 3$; the $9$-gon fails because it would trisect $120^\circ$; the $17$-gon succeeds because $\phi(17)/2 = 8$.
A note on what was actually proved. Wantzel settled doubling the cube and trisection in 1837, using exactly the degree argument above. Squaring the circle waited until 1882 and needed a far harder theorem — transcendence — because the degree argument has nothing to bite on when there is no degree.
Thinking no angle can be trisected. Many can; what fails is a single method for all of them.
Thinking $\sqrt[3]{2}$ cannot be constructed by any means. It can — with a marked ruler, with paper folding, with conic sections. The theorem is about two specific instruments.
Treating the degree condition as sufficient. It is necessary only. A degree-$4$ number whose splitting field has degree $24$ is not constructible.
Assuming a longer construction could reach further. The argument bounds every construction of every length at once, which is why it ends the search rather than narrowing it.
*Confusing not constructible with irrational.* $\sqrt2$ is irrational and perfectly constructible.
Forgetting that squaring the circle needed a different theorem. No degree argument applies to $\pi$, because it has no degree; transcendence is a much stronger and much harder statement.
It needs $\cos(2\pi/5)$. From $\zeta_5$ satisfying $x^{4} + x^{3} + x^{2} + x + 1 = 0$, dividing by $x^{2}$ and writing $y = x + x^{-1} = 2\cos(2\pi/5)$ gives $y^{2} + y - 1 = 0$.
A quadratic in the cosine.
So $2\cos(2\pi/5) = \frac{-1 + \sqrt5}{2}$, of degree $2$ over $\mathbb{Q}$.
Degree two.
A power of $2$, and here the construction genuinely exists: a tower of one quadratic step is exactly one square root, which a compass extracts.
Constructible.
It needs $\cos(2\pi/7)$. The seventh cyclotomic polynomial has degree $\phi(7) = 6$, and the cosine generates the real subfield of index $2$.
Half of six.
So $[\mathbb{Q}(\cos(2\pi/7)) : \mathbb{Q}] = 3$.
Degree three.
Three is not a power of $2$, so the heptagon cannot be drawn with ruler and compasses — unlike the pentagon, and unlike the $17$-gon.
Impossible.
It needs a side of length $\sqrt[3]{5}$.
Name the number.
Its minimal polynomial is $x^{3} - 5$, irreducible by Eisenstein at $5$, so the degree is $3$.
Degree three.
Three is not a power of $2$, so it is not constructible. The same argument handles a cube of any volume that is not a perfect cube times a constructible cube — the number $2$ in the classical problem plays no special role.
Put in order the steps that turn a question about ruler and compasses into a question about degrees.
Number the steps in order (write the number in the box):
Is this constructible with ruler and compasses: a regular polygon with $17$ sides?
For each number, fill in its degree over $\mathbb{Q}$, and then the exponent that makes that degree a power of $2$ — writing $0$ when the degree is not a power of $2$ at all.
| Degree over the rationals | Exponent, or zero if not a power of two | |
|---|---|---|
| $\sqrt2$ | ||
| $\sqrt[3]{2}$ | ||
| $\cos 20^\circ$ | ||
| $\sqrt{1 + \sqrt3}$ | ||
| $\cos(2\pi/5)$ | ||
| $\cos(2\pi/17)$ |
Select every statement that is true of ruler-and-compass constructions.
This task has no paper form; do it on a device.
Match each classical construction problem to the number that decides it.
| $\sqrt[3]{2}$, of degree $3$ | $\cos 20^\circ$, of degree $3$ | $\sqrt{\pi}$, which has no degree at all | $\cos(2\pi/5)$, of degree $2$ | $\sqrt{2}$, of degree $2$ | |
|---|---|---|---|---|---|
| Doubling the cube | |||||
| Trisecting a $60^\circ$ angle | |||||
| Squaring the circle | |||||
| Drawing a regular pentagon |
Doubling a cube of side $1$ needs a segment of length $\sqrt[3]{2}$. What is $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}]$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put in order the steps that turn a question about ruler and compasses into a question about degrees.
Number the steps in order (write the number in the box):
You can decide whether a length is constructible by computing a degree, and say what the argument does not claim. Say in your own words why the impossibility is a statement about the instruments rather than about the number. Next: the group of symmetries of an extension, and the correspondence it sets up.
9. Your turn: can a cube of volume $5$ be constructed from a unit length?, step 3