Euclidean Constructibility

Definition 1. A real number $u$ is said to be constructible if we can construct a line segment of length $u$ using just a straightedge and a compass. By assumption, $1$ is constructible. We allow constructible numbers to be negative.

Problem 1. Show that the set $\frak C$ of constructible numbers forms a subfield of $\mathbb R$ that contains $\mathbb Q$ .

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Solution. We first show that $\mathbb Q \subseteq \frak C$ . Fix $p / q \in \mathbb Q^+$ with $\gcd(p, q) = 1$ without loss of generality. Given a point $(0, 0)$ , it is not hard to construct the point $(q, p)$ . The line that intersects both points has equation $y = px/q$ . In particular, it intersects the line $x = 1$ and the point $(1, p/q)$ . Therefore, the line segment connecting $(0,p/q)$ and $(1, p/q)$ have length $p/q$ , so that $p/q \in \frak C$ , as required.

To check that $\frak C$ forms a field, suppose $u, v \in \frak C$ . It suffices to check that $u+v \in \frak C$ , $uv \in \frak C$ , and $u/v \in \frak C$ .

For addition, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(v,0)$ . Use the parallel postulate to translate the line segment by $(0, 1)^{\mathrm T}$ , so that $s_2$ now passes through $(0, 1)$ and $(v, 1)$ . Use the parallel postulate to translate the line segment by $(u,-1)^{\mathrm T}$ . Then the line segment obtained by combining $s_1$ and $s_2$ has a total length of $u + v$ , as required.

For multiplication, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(0,v)$ . Let $s_3$ denote the line segment connecting $(0,0)$ and $(0,1)$ , and $s_4$ connect $(0,1)$ to $(u,0)$ . Using the parallel postulate, draw a line parallel to $s_4$ passing through $(0, v)$ . Extend $s_4$ to a line segment $s_5$ until it intersects said line. Let $\ell$ denote the length of $s_5$ . Using similar triangles, $\ell/u = v/1$ , so that $\ell = uv$ .

For division, let $s_1$ denote the line segment connecting $(0, 0)$ and $(u,0)$ , and $s_2$ denote the line segment connecting $(0,0)$ and $(0,v)$ . Let $s_3$ denote the line segment connecting $(0,0)$ and $(0,1)$ , and $s_4$ connect $(0,v)$ to $(u,0)$ . Using the parallel postulate, draw a line parallel to $s_4$ passing through $(0, 1)$ and intersecting $s_1$ . Denote the shorterned line segment by $s_5$ . Let $\ell$ denote the length of $s_5$ . Using similar triangles, $1/v = \ell/u$ , so that $\ell = u/v$ , as required.

Problem 2. Show that $u \in \frak C$ if and only if there exist real numbers $\alpha_1,\dots,\alpha_n$ such that $\alpha_i \in \mathbb F_i$ for each $i$ , and $u \in \mathbb F_{n+1},$ where

$\mathbb F_i := \mathbb Q(\sqrt{\alpha_1},\dots,\sqrt{\alpha_{i-1}}).$

In particular, we say that the angle $\theta$ is constructible if and only if $\cos(\theta)$ is constructible.

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Solution. Denote $\mathbb F_0 := \mathbb Q$ . Given $\mathbb F_i$ , iew numbers can be constructed via one of three methods:

Intersecting lines
Intersecting a line and a circle
Intersecting circles

Carrying out the first and third operations drawn using points in $\mathbb F_i^2$ produces that points belong to $\mathbb F_i^2$ , and thus constructs numbers that remain in $\mathbb F_i$ . Carrying out the second operation produces points that belong to $\mathbb F_{i+1} := \mathbb F_i(\sqrt{\alpha_i})$ for some $\alpha_i \in \mathbb F_i$ . Now, $u \in \frak C$ if and only if it can be constructed using a finite number of one of these three steps. This is equivalent to the existence of a chain

$\mathbb F_0 \subseteq \mathbb F_1 \subseteq \cdots \subseteq \mathbb F_n \subseteq \mathbb F_{n+1}$

such that $u \in \mathbb F_{n+1}$ , as required.

Problem 3. If $u \in \frak C$ , show that there exists a positive integer $r$ such that $[\mathbb Q(u) : \mathbb Q] = 2^r$ .

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Solution. Using Problem 2, write $\mathbb Q(u) = \mathbb F_{n+1}$ . As subfields, $[\mathbb F_{n+1} : \mathbb F_n] = 2$ . Therefore, by the tower rule,

$\begin{aligned} [\mathbb Q(u): \mathbb Q] &= [\mathbb F_{n+1} : \mathbb F_0] \\ &= [\mathbb F_{n+1} : \mathbb F_n] \cdot [\mathbb F_n : \mathbb F_{n-1}] \cdot \cdots \cdot [\mathbb F_1 : \mathbb F_0] \\ &= \underbrace{ 2 \cdot 2 \cdot \cdots \cdot 2 }_{n+1} \\ &= 2^{n+1}. \end{aligned}$

Set $r:=n+1$ and we are done.

Problem 4. Show that the angle $\pi/9$ is not constructible. Therefore, we cannot, in general, trisect the angle.

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Solution. Firstly, we recall that $\cos(\pi/3) = 1/2$ , so that the angle $\pi/3$ is constructible. Denote $u := \cos(\pi/9)$ . Using the triple angle identity,

$\cos(\pi/3) = 4 \cos^3(\pi/9) - 3 \cos(\pi/9).$

Substituting, $8u^3 - 6u - 1 = 0$ . Using the rational root test, the polynomial $f(x) = 8x^3 - 6x - 1$ is irreducible in $\mathbb Q[x]$ . By definition,