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Section 8.7 What's Not Constructible

We are are now ready to solve the questions of antiquity about geometric constructions. Briefly, here is what we know already.

The constructible numbers are a field containing \(\Q\) and contained in \(\R\text{.}\) The field is closed under taking square roots of positive elements. Right away, this tells us that numbers like \(\sqrt{\frac{3 + \sqrt{5}}{\sqrt{7}}}\) are constructible.

We also know something about the converse: if a real number \(\alpha\) is constructible, then it exists in a field extension \(F_k\) which is “at the end” of a sequence of iterated extensions

\begin{equation*} {\mathbb Q} = F_0 \subset F_1 \subset \cdots \subset F_k \end{equation*}

such that \(F_i = F_{i-1}( \sqrt{ \alpha_i}\, )\) with \(\alpha_i \in F_i\text{.}\) In other words, the only way to get a constructible number is to find it inside a field extension you get by repeatedly adjoining square roots of elements in the previous field.

In particular, now that we know something about the degree of a field extension, we know that \([F_i:F_{i-1}] = 2\) in each of the iterated extensions, and by the tower rule ( Theorem 8.37) we know that if \(\alpha\) is constructible, then there exists an integer \(k \gt 0\) such that \([{\mathbb Q}(\alpha) : {\mathbb Q} ] = 2^k\text{.}\)

So how does this help prove that you cannot construct particular geometric figures? We will argue that if it were possible to construct such a figure, then a particular number would be constructible. To show that is impossible, we will argue about the degree of the field extension that contains that number.

Subsection 8.7.1 Doubling the Cube and Squaring the Circle

The first two constructions we will consider are both about constructing a shape with a given area. We will show that the length of the side of the shape is not constructible.

Doubling the cube is impossible.

Given the edge of the cube, it is impossible to construct with a straightedge and compass the edge of the cube that has twice the volume of the original cube. Let the original cube have an edge of length \(1\) and, therefore, a volume of \(1\text{.}\) If we could construct a cube having a volume of \(2\text{,}\) then this new cube would have an edge of length \(\sqrt[3]{2}\text{.}\) However, \(\sqrt[3]{2}\) is a zero of the irreducible polynomial \(x^3 -2\) over \({\mathbb Q}\text{;}\) hence,

\begin{equation*} [{\mathbb Q}(\sqrt[3]{2}\, ) : {\mathbb Q}] = 3 \end{equation*}

This is impossible, since \(3\) is not a power of \(2\text{.}\)

Squaring the circle.

Suppose that we have a circle of radius \(1\text{.}\) The area of the circle is \(\pi\text{;}\) therefore, we must be able to construct a square with side \(\sqrt{\pi}\text{.}\) This is impossible since \(\pi\) and consequently \(\sqrt{\pi}\) are both transcendental. Therefore, using a straightedge and compass, it is not possible to construct a square with the same area as the circle.

Subsection 8.7.2 Trisecting an Angle

The question is whether given any angle, it is possible to construct the angle which is one third of it. To prove this is impossible, we need only show that some angle is non-constructible while three times it is constructible.

Trisecting an arbitrary angle is impossible.

We will show that it is impossible to construct a \(20^\circ\) angle. Consequently, a \(60^{\circ}\) angle cannot be trisected. We first need to calculate the triple-angle formula for the cosine:

\begin{align*} \cos 3 \theta & = \cos( 2 \theta + \theta )\\ & = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta\\ & = ( 2 \cos^2 \theta - 1) \cos \theta - 2 \sin^2 \theta \cos \theta\\ & = ( 2 \cos^2 \theta - 1) \cos \theta - 2 (1- \cos^2 \theta) \cos \theta\\ & = 4 \cos^3 \theta - 3 \cos \theta\text{.} \end{align*}

The angle \(\theta\) can be constructed if and only if \(\alpha = \cos \theta\) is constructible. Let \(\theta = 20^{\circ}\text{.}\) Then \(\cos 3 \theta = \cos 60^\circ = 1/2\text{.}\) By the triple-angle formula for the cosine,

\begin{equation*} 4 \alpha^3 - 3 \alpha = \frac{1}{2}\text{.} \end{equation*}

Therefore, \(\alpha\) is a zero of \(8 x^3 - 6 x -1\text{.}\) This polynomial has no factors in \({\mathbb Z}[x]\text{,}\) and hence is irreducible over \({\mathbb Q}[x]\text{.}\) Thus, \([{\mathbb Q}( \alpha ) : {\mathbb Q }] = 3\text{.}\) Consequently, \(\alpha\) cannot be a constructible number.

Subsection 8.7.3 Historical Note

Algebraic number theory uses the tools of algebra to solve problems in number theory. Modern algebraic number theory began with Pierre de Fermat (1601–1665). Certainly we can find many positive integers that satisfy the equation \(x^2 + y^2 = z^2\text{;}\) Fermat conjectured that the equation \(x^n + y^n = z^n\) has no positive integer solutions for \(n \geq 3\text{.}\) He stated in the margin of his copy of the Latin translation of Diophantus' Arithmetica that he had found a marvelous proof of this theorem, but that the margin of the book was too narrow to contain it. Building on work of other mathematicians, it was Andrew Wiles who finally succeeded in proving Fermat's Last Theorem in the 1990s. Wiles's achievement was reported on the front page of the New York Times.

Attempts to prove Fermat's Last Theorem have led to important contributions to algebraic number theory by such notable mathematicians as Leonhard Euler (1707–1783). Significant advances in the understanding of Fermat's Last Theorem were made by Ernst Kummer (1810–1893). Kummer's student, Leopold Kronecker (1823–1891), became one of the leading algebraists of the nineteenth century. Kronecker's theory of ideals and his study of algebraic number theory added much to the understanding of fields.

David Hilbert (1862–1943) and Hermann Minkowski (1864–1909) were among the mathematicians who led the way in this subject at the beginning of the twentieth century. Hilbert and Minkowski were both mathematicians at Göttingen University in Germany. Göttingen was truly one the most important centers of mathematical research during the last two centuries. The large number of exceptional mathematicians who studied there included Gauss, Dirichlet, Riemann, Dedekind, Noether, and Weyl.

André Weil answered questions in number theory using algebraic geometry, a field of mathematics that studies geometry by studying commutative rings. From about 1955 to 1970, Alexander Grothendieck dominated the field of algebraic geometry. Pierre Deligne, a student of Grothendieck, solved several of Weil's number-theoretic conjectures. One of the most recent contributions to algebra and number theory is Gerd Falting's proof of the Mordell-Weil conjecture. This conjecture of Mordell and Weil essentially says that certain polynomials \(p(x, y)\) in \({\mathbb Z}[x,y]\) have only a finite number of integral solutions.

Exercises 8.7.4 Collected Homework


Prove, using field extensions, that a regular 9-gon is not constructible (with a compass and straight-edge).


First explain why constructing a regular 9-gon is equivalent to constructing a \(40^\circ\) angle.


Now prove that \(40^\circ\) is not constructible. Your proof should mimic the proof we did in class that \(20^\circ\) is not constructible. Make sure you are talking about field extensions and polynomials (do not just argue that it is impossible because if we could construct it, we would then be able to construct \(20^\circ\) which we know is impossible).


Prove that it is possible to construct a regular 20-gon.


Do not give the construction completely, just say how you know that the 20-gon is constructible. Note that if the central interior angle is constructible, then the 20-gon will be constructible.