📂Abstract Algebra

Proof of the Three Impossible Construction Problems

Theorem¹

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 whose volume is twice that of a given cube.
• [3] Trisecting the angle: Trisect a given angle.

Disproof

Having long been a problem of geometry, it is truly marvelous that it is solved by algebra. Essentially, the contrapositive of the following lemma is used.

Properties of constructible numbers: A number that can be obtained from $1$ by a finite number of arithmetic operations and taking square roots is said to be constructible.

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

[1]

It suffices to show that a circle with area $\pi$ serves as a counterexample.

Algebraic and transcendental numbers: Let $E$ be an extension field of a field $F$. An $\alpha \in E$ satisfying $f( \alpha ) = 0$ for some non-constant $f(x) \in F [ x ]$ is said to be algebraic over $F$, and transcendental if it is not algebraic. When $F = \mathbb{Q}$ and $E = \mathbb{C}$, an $\alpha \in \mathbb{C}$ is called an algebraic number if it is algebraic, and a transcendental number if it is transcendental.

For a square to have area $\pi$, its side length must be $\sqrt{\pi}$; however, since $\pi$ is a transcendental number over $\mathbb{Q}$, it is non-constructible by the contrapositive of Lemma (1). Therefore its square root $\sqrt{\pi}$ is also non-constructible.

■

[2]

It suffices to show that a cube with volume $1$ serves as a counterexample.

For the cube to have volume $2$, the length of an edge must be $\sqrt[3]{2}$; however, no $r \in \mathbb{N}$ satisfies $$ 2^{r} = \left[ \mathbb{Q} \left( \sqrt[3]{2} \right) : \mathbb{Q} \right] = 3 $$ so by the contrapositive of Lemma (2), $\sqrt[3]{2}$ is non-constructible.

■

[3]

It suffices to show that an angle of $60^{\circ}$ serves as a counterexample.

By the triple angle formula for trigonometric functions, $$ \cos 60^{\circ} = 4 \cos^{3} 20^{\circ} - 3 \cos 20^{\circ} $$ Since $\displaystyle \cos 60^{\circ} = {{1} \over {2}}$, setting $\displaystyle \alpha := \cos 20^{\circ}$ gives $$ 4 \alpha^3 - 3 \alpha = {{1} \over {2}} \implies 8 \alpha^3 - 6 \alpha - 1 = 0 $$ That is, $\alpha$ is a zero of the polynomial function $( 8 x^3 - 6 x - 1 ) \in \mathbb{Q} [ x ]$. The only candidates for factors of this integer-coefficient polynomial function are $$ (8x \pm 1), (4x \pm 1), (2x \pm 1), (x \pm 1) $$ However, computing them directly shows that none of $$ \mp {{1} \over {8}} , \mp {{1} \over {4}} , \mp {{1} \over {2}} , \mp 1 $$ is a zero. That $( 8 x^3 - 6 x - 1 )$ does not factor with a linear term means it does not have a quadratic factor either. In summary, $$ 2^{r} = \left[ \mathbb{Q} \left( \alpha \right) : \mathbb{Q} \right] = 3 $$ and no $r \in \mathbb{N}$ satisfies $2^r = 3$. By the contrapositive of Lemma (2), $\cos 20^{\circ}$ is non-constructible, and thus an angle given as $60^{\circ}$ cannot be trisected.

■

Trivia

In particular, “Squre the circle” has been used in the English-speaking world to mean “to do the impossible” or as a retort like “talk sense”. In Korean terms, one may think of it as something like the expression “make meju out of red beans”.