# 代數導論二 Week 10 (Part 1) - Classical Straightedge and Compass Construction

[TOC]

1. Doubling a cube: construct $a$ such that
$$
a^3 = 2
$$

2. Trisecting an angle

3. Squaring a Circle: exists construction $a$, such that
$$
a^2 = \pi
$$

> 用代數方法說明用叉歸做圖不可能做出上面這三件事情。

## Definition

:::warning

Given: two points on the plane (segment on the plane), straightedge, compass

1. Draw a line through two ==given== points
2. Draw a circle whose center is a ==found== point and with a distance between two points we have already found.

:::

Note that every lattice point on the plane can be identified under such construction.

## Definition：Contructible

:::warning

$a \in \Bbb R$ is called constructible if we can find a segment with length $|a|$ by (1) and (2)

:::

### Q1: What numbers are contructible

Observe that

1. $\Bbb Z \subseteq \Bbb R$ is contructible
2. if $a, b$ are constructible, then $a\pm b$, $ab$ are contructible, and $a/b$ with $b \neq 0$ are also contructible. So $\Bbb Q \subseteq \Bbb R$ are all construtible

> 用平行線可以做出相乘跟相除

3. Start from $(x, y) \in \Bbb Q$, the intersection of lines:

$$
\begin{cases}
a_1x + b_1y = c_1
\newline
a_2x + b_2y = c_2
\end{cases}
$$

where $a_i, b_i, c_i \in \Bbb Q$, is in $\Bbb Q$.

4. The intersection of a circle of line:

$$
\begin{cases}
a_1x + b_2y = c_1
\newline
(x - a_2)^2 + (y - b_2)^2 = c_2^2
\end{cases}
$$

is also contructible, which is a solution to quadratic equation of $x$ and $y$. In particular, square root of any contructible number is also constructible (子母性質).

### Proposition 1

:::danger

If $\alpha \in \Bbb R$ can be obtained from $\Bbb Q$ by $(1)$ and $(2)$, then

$$
[\Bbb Q(\alpha):\Bbb Q] = 2^k
$$

for some non-zero integer $k$。

:::

### Proposition 2

:::danger

All contructible numbers in $\Bbb R$ form a subfield of $\Bbb R$

:::

## Proposition 3

:::danger

Problem 1 is impossible to solve

:::

Suppose that, by contrdition, $\sqrt[3]{2}$ is contructible. Observe that $\sqrt[3]{2}$ is a silution of

$$
x^3 - 2
$$

which is an irreducible polynomial in $\Bbb Q$, then

$$
[Q(\sqrt[3]{2}):\Bbb Q] = 3
$$

By proposition $1$,

$$
3 \mid 2^k
$$

which leads to contradiction

## Proposition 4

:::danger

Problem 2 is impossible to solve

:::

> 可以三等份表示他的 $\cos$ 跟 $\sin$ 都是 *constructible*. 所以找出不 *constructible* 的例子就好了

By (1)(2), $\theta$ is contructible if and only if both $\cos \theta$ and $\sin \theta$ are contructible

Let

$$
\theta = 60^{\circ}
$$

then

$$
\cos \theta = 4 \cos^3 (\theta/3) - 3 \cos (\theta/3)
$$

So

$$
\frac {1}{2} = 4 \cos^3 (\theta/3) - 3 \cos (\theta/3)
$$

Multiply both side by $2$:

$$
1 = 8 \cos^3 (\theta/3) - 6 \cos (\theta/3)
$$

and let

$$
\alpha = 2 \cos \theta/3
$$

then $\alpha$ is a solution of

$$
x^3 - 3x - 1 = 0
$$

$\theta/3$ is contructible if and only if $\cos \theta/3$ and $\sin \theta/3$ are contructible if and only if $\alpha$ is contructible. However, since $x^3-3x - 1$ is an irreducible polynomial, so

$$
[\Bbb Q(\cos \theta/3):\Bbb Q] = 3
$$

which contraditcts to proposition 1

> 師：我們在 50 分鐘之內成功解決了古希臘三大難題