- A straight line may be drawn from any point to any other point.
- A finite straight line may be extended continuously in a straight line.
- A circle may be described with any center and radius.
- All right angles are equal to each other.
- Parallel postulate.
However, there are three problems that cannot be done with these tools:
- Squaring a circle: given a circle, construct a square with the same area.
- Doubling a cube: given a cube, construct a cube with double the volume.
- Trisection of angles: trisect an angle.
First, we need to establish 2D-coordinate system. In order to make sense a number $x$, we have to first be given (or define) the length of 1. In other words, we can assume we know a segment AB of length 1, and we want to construct numbers from it using coordinate systems ${(x,y)|x,y\in \mathbb{R}}$.
Definition. $\alpha \in \mathbb{R}$ is constructible if $(\alpha,0)$ is constructible.
Remark. $(\alpha,\beta)$ is constructible if and only if both $\alpha$ and $\beta$ are constructible.
What constructions can we do?
1. Basic Constructions
(1) Given a segment, we can divide it into n equal parts.
(2) Given a segment of length x, we can construct one of length $\sqrt{x}$ and $x^2$.
Proof.
What constructions can we do?
1. Basic Constructions
(1) Given a segment, we can divide it into n equal parts.
(2) Given a segment of length x, we can construct one of length $\sqrt{x}$ and $x^2$.
Therefore, given (0, 0) and (1, 0), we can construct all coordinates with integer entries and then by method (1), we can construct all rational numbers, hence $\mathbb{Q}^2$.
2. Constructions by Intersections
(3) Given A, B, C, D distinct points, $P=AB\cap CD$ if it exists.
(4) Given A, B, C, D distinct points, P is the intersection of circle $O = (|A|, |AB|)$ with line CD if it exists.
Lemma If A, B, C, D $\in K \subset \mathbb{R}$, where $K$ is a subfield, then in case (3), the point P has coordinates in $K$; in case (4), P has coordinates in $K(\sqrt{\alpha})$, where $\alpha >0, \alpha \in K$.
Proof.
In case (3), we have two lines, both of form $\alpha x +\beta y = \gamma$, where $\alpha, \beta, \gamma \in K$, then the solution of the two equations of lines also lie in $K$ by Cramer's Rule.
In case (4), we have one quadratic equation $(x-\alpha)+(y-\beta) = \gamma^2$ and another linear equation of a line, to solve the system of equations, we will end up with a quadratic polynomial of form $ax^2+bx+c=0$, with $a,b,c\in K$, hence the solutions like in $K(\sqrt{\Delta})$, where $\Delta = b^2-4ac\ge 0$ since the circle and line intersect.
Corollary Any constructible number $\alpha$ lies in a subfield $K_n\subset \mathbb{R}$, $\mathbb{Q} = K_0 \subset K_1\subset K_2\subset \dots \subset K_n$ and each $K_i = K_{i-1}(\sqrt{a_i})$ for some $a_i>0$ in $K_{i-1}$.
Proof. By induction on the moves to construct $\alpha$.
Theorem Any constructible real number is algebraic, and its degree over $\mathbb{Q}$ is a power of 2.
Proof. By last corollary, if $\alpha$ is constructible, $\alpha \in K_n$ for some $n$, with $[K_n:\mathbb{Q}]= 2^r$ for some $r\ge 0$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}]\ |\ 2^r$.
Remark. With more work (like the ones in basic constructions), one can show that the set of constructible numbers is a subfield of $\mathbb{R}$.
Now we are ready to show that the three problems are impossible to be solved by edge and compass:
(1) Squaring a circle:
If there were such a method, then we can square a circle of radius 1, so a square with area $\pi$, side length $\sqrt{\pi}$ would be constructible, hence $\pi$ is constructible, but $\pi$ is not algebraic, contradiction.
Doubling a cube of side length 1 amounts to constructing $^3\sqrt{2}$, the side length of the new cube. Because $x^3-2$ is irreducible, it's the minimal polynomial of $^3\sqrt{2}$ over $\mathbb{Q}$ hence $^3\sqrt{2}$ has degree 3 over $\mathbb{Q}$, which is not a power of 2, contradiction.
(3) Trisection of angles:
If we could trisect any angle, then we could trisect 60° angle, so $(\cos20°, \sin20°)$ is constructible, hence $2\cos 20^{\circ}$ is constructible. Note that $cos3x=4cos^3x-3cosx$, so $1/2=\cos 60° = 4\cos^3 20°-3\cos 20°$. Let $u=2\cos 20°$, then $u^3-u-1=0$. But $x^3-3x-1$ is irreducible over $\mathbb{Q}$ by rational root theorem, so $u$ has degree 3 over $\mathbb{Q}$, contradiction.
Remark. One can trisect any angle easily using a ruler and compass.
Constructing Regular n-gons
Corollary Any constructible number $\alpha$ lies in a subfield $K_n\subset \mathbb{R}$, $\mathbb{Q} = K_0 \subset K_1\subset K_2\subset \dots \subset K_n$ and each $K_i = K_{i-1}(\sqrt{a_i})$ for some $a_i>0$ in $K_{i-1}$.
Proof. By induction on the moves to construct $\alpha$.
Theorem Any constructible real number is algebraic, and its degree over $\mathbb{Q}$ is a power of 2.
Proof. By last corollary, if $\alpha$ is constructible, $\alpha \in K_n$ for some $n$, with $[K_n:\mathbb{Q}]= 2^r$ for some $r\ge 0$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}]\ |\ 2^r$.
Remark. With more work (like the ones in basic constructions), one can show that the set of constructible numbers is a subfield of $\mathbb{R}$.
Now we are ready to show that the three problems are impossible to be solved by edge and compass:
(1) Squaring a circle:
If there were such a method, then we can square a circle of radius 1, so a square with area $\pi$, side length $\sqrt{\pi}$ would be constructible, hence $\pi$ is constructible, but $\pi$ is not algebraic, contradiction.
(2) Doubling a cube:
(3) Trisection of angles:
If we could trisect any angle, then we could trisect 60° angle, so $(\cos20°, \sin20°)$ is constructible, hence $2\cos 20^{\circ}$ is constructible. Note that $cos3x=4cos^3x-3cosx$, so $1/2=\cos 60° = 4\cos^3 20°-3\cos 20°$. Let $u=2\cos 20°$, then $u^3-u-1=0$. But $x^3-3x-1$ is irreducible over $\mathbb{Q}$ by rational root theorem, so $u$ has degree 3 over $\mathbb{Q}$, contradiction.
Remark. One can trisect any angle easily using a ruler and compass.
Constructing Regular n-gons
Question: Can we construct a regular 7-gon? When is an n-gon constructible ($n\ge 3$)?
In general, a regular n-gon is constructible if and only if $\cos \frac{2\pi}{n}$ is constructible.
Answer:
Let $\zeta^7 = 1, \zeta \neq 1$. Then $\zeta$ is a root of $\frac{x^7-1}{x-1}=x^6+x^5+\dots+1$, which is a cyclotomic polynomial and irreducible by Eisenstein's Criterion. If a regular 7-gon is constructible, both $\sin\frac{2\pi}{7}$ and $\cos\frac{2\pi}{7}$ would be constructible, hence they lie in some extension $K$ of $\mathbb{Q}$ that has a degree $[K:\mathbb{Q}]=2^r$. Because $K\subset \mathbb{R}$, $K\subset K(i)$ and $[K(i):K]=2$. So $[K(i):\mathbb{Q}]=2^{r+1}$. $\zeta = \cos\frac{2\pi}{7}+\sin\frac{2\pi}{7}$ but $\zeta$ has degree 6, not a power of 2, contradiction.
Corollary. A p-gon cannot be constructible for $p$ prime if $p-1$ is not a power of 2. If a regular p-gon is constructible, then $p=2^r+1, r\ge 1$.
Corollary. Let $p$ be prime. If a regular p-gon is constructible, then $p$ is a Fermat prime.
Proof. Let $r=2^k x$, with $k$ the power of 2 in the prime factorization of $r$. Then $x$ is 1 or any other odd positive integer. Suppose $x\ge 3$ odd, then $2^{2^{k}t}+1=(2^{2^{k}})^t+1^t$ has a factor $2^{2^k}+1, k\ge 0$, contradiction.
Corollary. Let $p$ be prime. If a regular p-gon is constructible, then $p$ is a Fermat prime.
Proof. Let $r=2^k x$, with $k$ the power of 2 in the prime factorization of $r$. Then $x$ is 1 or any other odd positive integer. Suppose $x\ge 3$ odd, then $2^{2^{k}t}+1=(2^{2^{k}})^t+1^t$ has a factor $2^{2^k}+1, k\ge 0$, contradiction.
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