3 Ancient Construction Problems
This chapter will employ the results to demonstrate the impossibility of trisecting the angle and doubling the cube. This formalisation is based on the work conducted during my project in Bonn during the Lean Course WiSe 23/24: https://github.com/Louis-Le-Grand/LeanCourse23Fork/tree/master/LeanCourse/Project
3.1 Doubling the cube
The doubling of the cube, also known as the Delian problem, represents an ancient geometric problem. The objective is to construct the edge of a second cube whose volume is double that of the first, using only a ruler and compass, given the edge of a cube. The construction of a second cube with double the volume of the original cube begins with a cube of volume \(a^3\), where \(a\) is the length of an edge. Thus, a cube with double the volume (\(2\cdot a^3\)) has an edge length of the cube root of two times the length of the original edge. If we now take the unit cube and reduce \(\mathcal{M}\), the problem is as follows: Let \(\mathcal{M} = \{ 0,1\} \).
(\(\sqrt[3]{2}\) is irrational) The third root of \(2\) is an irrational number.
The following theorem will be used without proof, as it is already available in MathLib:
For any \(x\in (\mathbb {R}\setminus \mathbb {Z})\) if there exist an \(n\in \mathbb {N}_{{\gt}0}\) and \(m\in \mathbb {Z}\) such that \(m = x^n\), then \(x\) is rational.
The fact that \((\sqrt[3]{2})^3=2\), allows us to deduce that the only remaining task is to prove that it is not an integer. This can be observed trough two relations.
\(P := X^3 - 2\) is irreducible over \(\mathbb {Q}\).
Since \(\mathbb {Q}\) is \(a\) subfield of \(\mathbb {C} [X]\), we know that
Suppose \(P\) is reducible, then
In particular it has a zero in \(\mathbb {Q}\), so there is a rational number \(a\) such that \(a^3 = 2\).
But we know that \(\zeta _3 \sqrt[3]{2}\) and \(\zeta _3^2 \sqrt[3]{2}\) are not real numbers and \(\sqrt[3]{2}\) is not rational 3.1. So \(P\) is irreducible over \(\mathbb {Q}\).
The cube can’t be doubled using a compass and straightedge.
By applying the corollary 2.31, it is sufficient to proof that no \(m\in \mathbb {N}\) exists such that
Since \(P\) is irreducible over \(\mathbb {Q}\) ??, monic and has \(\sqrt[3]{2}\) as a zero, we know that \([\mathbb {Q}(\sqrt[3]{2}):\mathbb {Q}] = 3\). And since
we can conclude that the cube can’t be doubled using a compass and straightedge.
3.2 Trisection of an angle
The trisection of an angle with a compass and ruler can be reduced to the following problem:
Let \(\mathcal{M} = \{ a, b, c\} \) with \(a, b, c\) not on a line and \(\alpha := \angle (b - a, c - a)\) be the resulting angle. Then \(\alpha \) can be trisected if and only if there is a point \(d\in \mathcal{M}_{\infty }\) such that \(\angle (b - a, d - a) = \alpha /3\). The use of a normed set \(\mathcal{M} = \{ 0,1,e^{i\alpha }\} \) leads to the following problem: Let \(\mathcal{M} = \{ 0,1,\exp (\textbf{i} \alpha )\} \).
In this context, since the numbers zero and one are rational numbers, it can be concluded that \(K_0\) is equal to
Given the corollary reference and the fact that \(e^{\imath\alpha /3}\) is a zero of \(X^3 - e^{\imath\alpha }\in \mathbb {Q}(e^{\imath\alpha },\overline{e^{\imath\alpha }})[X]\), the following is equivalent:
\(\exp (\textbf{i} \alpha /3) \notin \mathcal{M}_{\infty }\)
\(\text{degree}(\mu _{e^{\imath\alpha /3},K_0}) = 3\)
\(X^3 - e^{\imath\alpha }\) is irreducible over \(\mathbb {Q}(e^{\imath\alpha },\overline{e^{\imath\alpha }})\)
The following section will demonstrate that the angle of \(\frac{\pi }{3}=60°\) is not trisectable.
The degree of \(K_0 = \mathbb {Q}(e^{\imath\frac {\pi }{3}},\overline{e^{\imath\frac {\pi }{3}}})\) is equal to \(2\).
For all real numbers \(\alpha \), we have that
For \(\alpha = \pi / 3\) we get
Therefore \(\mathbb {Q}(e^{\imath\frac {\pi }{3}},\overline{e^{\imath\frac {\pi }{3}}})\) is in \(\mathbb {Q}(\imath\sqrt {3})\). And since \(\imath\sqrt {3}\) is a zero of \(X^2 + 3\), we know that the degree of \(K_0\) less then \(2\). To show that the degree is not \(1\), we apply the fact that \(\imath\sqrt {3} \notin \mathbb {Q}\).
The angle \(\pi / 3 = 60^{\circ }\) can’t be trisected using a compass and straightedge.
By utilising the aforementioned lemma 3.4 to apply the corresponding corollary 2.32, we can narrow our focus to the degree over \(\mathbb {Q}\). Now we use the fact that if \(x\in \mathcal{M}_{\infty }\), then \(x.real, x.imag \in \mathcal{M}_{\infty }\) 2.4. Thus we focus on \(\cos (\alpha /3)\), which the real part \(e^{\imath\frac {\pi }{3}}\) and a zero of
Suppose \(f\) is reducible over \(\mathbb {Q}\), then \(f\) has a rational zero \(a\), since \(f\) is of degree \(3\). According to the rational root theorem, a root rational root of \(f\) is of the form \(\pm \frac{p}{q}\) with \(p\) a factor of the constant term and \(q\) a factor of the leading coefficient. So the only possible rational zeros of \(f\) are
One can check that none of these numbers are a zero of \(f\). So \(f\) is irreducible over \(\mathbb {Q}\) and \(\cos (\alpha /3) \notin \mathcal{M}_{\infty }\). Therefore
So the angle \(\pi / 3 = 60^{\circ }\) can’t be trisected using a compass and straightedge.
A general angle can’t be trisected using a compass and straightedge.
Employ the previous lemma with the angle \(\pi / 3\).