📂Abstract Algebra

Proof of the Three Classical Problems of Antiquity

Theorem ¹

The following three constructions are impossible:

• [1] Squaring the circle: Construct a circle with the same area as a given square.
• [2] Doubling the cube: Construct a cube with twice the volume of a given cube.
• [3] Trisecting the angle: Divide a given angle into three equal parts.

Disproof

It’s truly remarkable that these long-standing problems of geometry are solved by algebra. Essentially, the contrapositive of the following lemma is used.

Properties of constructible numbers: A number is said to be constructible if it can be obtained through a finite number of elementary operations and square roots, including $1$.

• (1): Constructible numbers are algebraic.
• (2): If $\gamma \not\in \mathbb{Q}$ is constructible, then there exists a finite sequence $\left\{ a_{i} \right\}_{i=1}^{n}$ satisfying $$\left[ \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} , a_{i} \right) : \mathbb{Q} \left( a_{1} , \cdots , a_{i-1} \right) \right] = 2 \\ \mathbb{Q} ( \gamma) = \mathbb{Q} \left( a_{1} , \cdots , a_{n} \right)$$ for some $r \in \mathbb{N}$ such that $$\left[ \mathbb{Q} \left( \gamma \right) : \mathbb{Q} \right] = 2^{r}$$

[1]

It is sufficient to show that a circle with area $\pi$ is a counterexample.

Algebraic and Transcendental Numbers: Let’s say the extension field of $F$ is $E$. For any non-constant $f(x) \in F [ x ]$, if there exists $\alpha \in E$ satisfying $f( \alpha ) = 0$, it is called algebraic over $F$, and if not, it is transcendental. If $\alpha \in \mathbb{C}$ is algebraic, it is called an algebraic number, and if it’s transcendental, a transcendental number.

To have a square of area $\pi$, the side length needs to be $\sqrt{\pi}$, but since $\pi$ is a transcendental number over $\mathbb{Q}$, it cannot be constructed according to the contrapositive of lemma (1). Therefore, its square root, $\sqrt{\pi}$, is also not constructible.

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[2]

It is sufficient to show that a cube with volume $1$ is a counterexample.

To have a cube with volume $2$, a side length of $\sqrt[3]{2}$ is necessary, $$2^{r} = \left[ \mathbb{Q} \left( \sqrt[3]{2} \right) : \mathbb{Q} \right] = 3$$ but since there’s no $r \in \mathbb{N}$ satisfying this equation, by the contrapositive of lemma (2), $\sqrt[3]{2}$ is not constructible.

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[3]

It’s sufficient to showcase an angle of size $60^{\circ}$ as a counterexample.

According to the triple angle formula of trigonometric functions, $$\cos 60^{\circ} = 4 \cos^{3} 20^{\circ} - 3 \cos 20^{\circ}$$ Considering $\displaystyle \cos 60^{\circ} = {{1} \over {2}}$ leads us to define $\displaystyle \alpha := \cos 20^{\circ}$, thus $$4 \alpha^3 - 3 \alpha = {{1} \over {2}} \implies 8 \alpha^3 - 6 \alpha - 1 = 0$$ meaning $\alpha$ is a root of the polynomial $( 8 x^3 - 6 x - 1 ) \in \mathbb{Q} [ x ]$. The only candidates for factors of this integer-coefficient polynomial are $$(8x \pm 1), (4x \pm 1), (2x \pm 1), (x \pm 1)$$ However, none of these yield zero when calculated. That $( 8 x^3 - 6 x - 1 )$ doesn’t factor into terms of $1$ implies it also doesn’t have terms of $2$. In summary, $$2^{r} = \left[ \mathbb{Q} \left( \alpha \right) : \mathbb{Q} \right] = 3$$ and no $r \in \mathbb{N}$ satisfying $2^r = 3$ exists. According to the contrapositive of lemma (2), $\cos 20^{\circ}$ is unconstructable, meaning we cannot trisect an angle given with size $60^{\circ}$.

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Common Knowledge

Especially, “Square the circle” was used in the Anglosphere to mean “doing the impossible” or “making sense”. It can be thought of in a similar vein to the Korean expression “try making soybean paste out of beans,” implying an attempt at an impossible task.