Constructible numbers; Compass and Straightedge constructions

Simply put, a real number r is constructible if, starting form a line segment of unit length, a line segment of length |r| can be constructed with a compass and straightedge in a fintie number of steps. The study of constructible numbers is (elegantly) linked to 4 famous problems in Euclidean geometry: (1) Doubling the cube (2) Trisecting the angle (3) Squaring the circle (4) Constructing regular polygons. We will give the more detailed problem statements when looking at these problems in the second half of this post.

Here is a more rigourous definition of a constructible number. First, let us lay down the rules of the game:

  1. We start with two given points in the Euclidean plane, of unit length apart. These starting points are, by default, classified as constructed points.
  2. Given any two constructed points P and Q, we may construct a (infinite) line through them using the straightedge.
  3. Given any three constructed points A,B,C, we may construct a circle with centre at A and radius of length BC.
  4. The intersection points between any constructed lines and circles are considered constructed points.

A point, line, or circle is said to be constructible if they can be obtained in finitely many steps using the above rules.

Definition 1: A real number r is said to be constructible if there exists two constructible points of length |r| apart.

Before we proceed, a quick remark about our compass. The rules above assumes that we are given a fixed compass, as opposed to a collapsing compass which collapses whenever the compass is lifted from the page. In other words, a collapsing compass cannot be used to transfer distances directly, in which case rule #3 is replaced by the following statement:

3′. Given any two constructed points P,Q, we may construct a circle with centre at P and passing through Q.

However, it turns out that the use of a collapsing compass as opposed to a fixed compass is an unimportant restriction. Using a multi-step procedure, a distance can still be transferred with a collapsing compass. Interested readers may refer to Wikipedia article on the compass equivalence theorem. Henceforth, we will be using a fixed compass for simplicity.

Some basic constructions

Before we attempt to study the set of constructible numbers in detail, let us take a closer look at a couple of useful constructions that can be done with compass and straightedge. Acknowledgement: All the figures below are taken from [1].

Construction 2: Given a constructed line l and a constructed point p, we can construct a line perpendicular to l and passing through p.

Proof: Obviously integer lengths are constructible. By choosing a large enough integer radius, we can get a circle centered at p to intersect l at two points q_1, q_2. Drawing circles C_1, C_2 of sufficiently large equal radius centered at q_1, q_2, the two intersection points between C_1, C_2 yield the desired line. \square

Constructible1

Construction 3: Given a constructed line l and a constructed point p, we can construct a line parallel to l and passing through p.

Proof: First draw a line l' perpendicular to l passing through p, and then draw a line perpendicular to l' at p. \square

Constructible2

Corollary 4: The constructible numbers form a subfield of \mathbb{R}.

Proof: Given |P_1P_2|=p and |Q_1Q_2|=q where P_1,P_2,Q_1,Q_2 are constructed points, we need to show that p+q,pq,p^{-1} are constructible.

(p+q). Draw a circle centered at P_2 of radius q. Let R be the point of intersection between the circle and P_1P_2 extended (i.e. P_2 is between P_1 and R). Then |P_1R|=p+q.

(pq). Draw a line perpendicular to P_1P_2 at P_1, and use the compass to mark A on this line such that |AP_1|=1. Now, draw a line parallel to AP_1 passing through Q_1, and use the compass to mark B on this line such that |BQ_1|=q. Finally, by using the parallel line construction, find C such that AP_1P_2 is similar to BQ_1C. Then |Q_1C|=pq.

Constructible3

(p^{-1}). The idea is similar as above. First construct AP_1P_2 as before. This time, construct C such that |Q_1C|=1 before using the parallel line construction to find B to complete the similar triangle. In this case, |BQ_1|=p^{-1}. \square

Corollary 5: If a is a positive constructible number, then so is \sqrt{a}.

Proof: Let A_1,A_2 be constructed points such that |A_1A_2|=a and B be on A_1A_2 extended such that |A_2B|=1. By drawing circles of sufficiently large equal radius centered at A_1, B and drawing the common chord, we can find the midpoint C of A_1B.

Constructible4

Draw the circle of radius (a+1)/2 centered at C. Draw the line perpendicular to A_1B at A_2, letting P be the point of intersection with the circle. Then |A_2P|=\sqrt{a} (prove for similar triangles, for example). \square

Characterising the set of constructible numbers

Corollary 4 should give a hint that field theory would play a useful role in characterising the set of constructible numbers. Indeed, in this section we will assume some basic knowledge of field theory and describe a neat algebraic approach to understand constructible numbers.

For one direction of our characterisation, we have the following:

Theorem 6: Let \mathbb{Q}=F_0\subseteq F_1\subseteq \dots\subseteq F_n = K be a chain of subfields of \mathbb{R} such that [F_{i+1}:F_i]=2 for all i=0,\dots,n-1. Then every element of K is constructible.

Proof: When the characteristic \neq 2, a quadratic extension L/K is of the form L=K(\alpha) where we may assume \alpha^2\in K (this follows from the quadratic formula). Proceeding by induction along the chain, the base case i=0 follows from corollary 4 and the observation that integers are constructible. Now assuming that every element of F_{n-1} is constructible, write F_n = F_{n-1}(\alpha) where \alpha^2\in F_{n-1}. Then \alpha is constructible, by corollary 5. Applying corollary 4, we conclude that every element of F_n is constructible. \square

For the reverse direction, we first show the following.

Theorem 7: Suppose the two starting points have coordinates (0,0) and (1,0) and let P be a constructible point. Then there exists \mathbb{Q}=F_0\subseteq F_1\subseteq \dots\subseteq F_n = K a chain of subfields of \mathbb{R} with [F_{i+1}:F_i]=2 for all i=0,\dots,n-1, such that the coordinates of P are in K.

Proof: To attain the point P, the process requires drawing a finite sequence of lines and circles. If A,B are points with coordinates in a subfield F\subseteq \mathbb{R}, then the line AB has equation

ax+by+c=0 with a,b,c\in F

If a circle has centre C with coordinates in F and has radius an element of F, then the circle has equation

(x-a)^2+(y-b)^2=c^2 with a,b,c\in F

Proceeding in the spirit of induction, suppose that after drawing some number of lines and circles, all the constructed points have coordinates in a field L where L is the end of a chain of quadratic extensions. In drawing the next line or circle, we see that its equation has coefficients in L as described above. Let Q be a newly constructed point. We claim that the coordinates of Q are in some quadratic extension of L.

  • Case 1: Q is the intersection between two lines. Solving two linear equations with coefficients in L gives a point with coordinates in L.
  • Case 2: Q is the intersection between a line and circle. Using the line equation to write one variable in terms of the other, we obtain a quadratic equation in one unknown – the desired result follows.
  • Case 3: Q is the intersection between two circles. The difference between the two circle equations is a linear equation. The reasoning reduces to that of case 2.

As such, a finite number of quadratic extensions L\subseteq\dots\subseteq L' will ensure that all newly constructed points due to the new line or circle have coordinates in L'. \square

Corollary 8: Let \alpha be a constructible number. Then there exists \mathbb{Q}=F_0\subseteq F_1\subseteq \dots\subseteq F_n = K a chain of subfields of \mathbb{R} with [F_{i+1}:F_i]=2 for all i=0,\dots,n-1, such that \alpha \in K.

Proof: Suppose A,B are constructible points such that |AB|=\alpha. By drawing a circle of radius \alpha centered at the origin, we see that (\alpha,0) is a constructible point. Apply theorem 7. \square

4 famous problems in Euclidean geometry

The algebraic approach to understanding constructible numbers yields elegant solutions to many famous geometry problems which had defied centuries of attack. The key weapon to solving these problems is the following corollary.

Corollary 9: If \alpha is constructible, then \alpha is algebraic over \mathbb{Q} and moreover [\mathbb{Q}[\alpha]:\mathbb{Q}] is a power of 2.

Proof: Using the notation of corollary 8, [\mathbb{Q}[\alpha]:\mathbb{Q}] is finite and divides [K:\mathbb{Q}], which is a power of 2 by the multiplicative property of degree. \square

Without further ado, let us look at the four famous construction problems mentioned at the beginning of this post.

Problem 10: (Doubling the cube) Using only a compass and straightedge, is it possible to a construct the edge of a second cube whose volume is double that of the given first?

Solution: The problem is equivalent to asking whether \alpha= \sqrt[3]{2} is constructible. Clearly, \alpha satisfies X^3-2=0 which is irreducible (e.g. by Eisenstein’s criterion). As such, [\mathbb{Q}[\alpha]:\mathbb{Q}]=3 which is not a power of 2. The answer is no. \square

Problem 11: (Trisecting the angle) Using only a compass and straightedge, is it possible, given two lines meeting at an arbitrary angle 3\theta, to construct a two lines meeting at angle \theta?

Solution: Assume on the contrary that it is possible, in particular for the choice 3\theta=\pi/3.

Now, starting with the points (0,0) and (0,1), it is easy to construct an equilateral triangle of unit side length. By assumption, we can construct two lines l,l' that meet at P at angle \theta. By constructing Q on l such that |PQ|=1 and drawing the line perpendicular to l' passing through Q, we are able to construct the length \cos \theta.

From the triple angle formula \cos 3\theta = 4 \cos^3\theta - 3\cos\theta, we know that \cos \theta satisfies the polynomial 8X^3-6X-1, which is irreducible (e.g. using the rational root theorem). This yields the desired contradiction with corollary 9. \square

Problem 12: (Squaring the circle) Using only a compass and straightedge and given a circle, is it possible to construct a square with the same area?

Solution: Assume on the contrary that it is possible. Starting with the points (0,0) and (0,1), we are able to draw a circle with radius 1 and hence by assumption \sqrt{\pi} is constructible. By corollary 9, this implies that \sqrt{\pi} is algebraic and hence \pi is algebraic but \sqrt{\pi} is transcendental (see previous post on transcendental numbers). \square

Problem 13: (Constructing regular polygons) Using only a compass and straightedge and given two starting points, is it possible to construct a regular n-gon for any n?

Solution: We will focus on the case where n is an odd prime p and give a necessary condition for a regular p-gon to be constructible.

As the exterior angle to a regular p-gon is \frac{2\pi}{p}, we deduce that \cos \frac{2\pi}{p} is constructible. Let \zeta = e^{\frac{2\pi i}{p}}. Since \cos \frac{2\pi}{p} = (\zeta + \zeta^{-1})/2, we have the following field extension:

\mathbb{Q}\subseteq\mathbb{Q}[\cos \frac{2\pi}{p}]\subseteq\mathbb{Q}[\zeta]

It is well-known from basic field theory that the minimal polynomial of \zeta is X^{p-1}+\dots+1 and so [\mathbb{Q}[\zeta]:\mathbb{Q}]=p-1. At the same time, [\mathbb{Q}[\zeta]:\mathbb{Q}[\cos \frac{2\pi}{p}]]=2 because \zeta satisfies the equation

X^2 - 2\cos\frac{2\pi}{p} X + 1 = 0

and \zeta\notin\mathbb{Q}[\cos \frac{2\pi}{p}]. As such,

[\mathbb{Q}[\cos \frac{2\pi}{p}]:\mathbb{Q}]=\frac{p-1}{2}

By corollary 9, we conclude that \frac{p-1}{2} is a power of 2, i.e. p is of the form 2^r + 1. More specifically, r needs to be a power of 2 as well – if r has an odd factor t, then writing r=st, we can factorize p as

p=2^{st}+1 = (2^s + 1)((2^s)^{t-1}-(2^s)^{t-2}+\dots+1)

which is a contradiction. Therefore, p must be of the form p = 2^{2^k} + 1, i.e. p is a Fermat prime. \square

Remark: It turns out that the condition of p being a Fermat prime is also sufficient for the constructibility of the regular p-gon. This was shown by Gauss. More generally, a regular n-gon is constructible if and only if n is a product of a power of 2 and any number of distinct Fermat primes (including none). This is known as the Gauss-Wantzel theorem.

References

[1] Artin, M. (2010). Algebra. (2nd ed.). Prentice Hall.

[2] Milne, J.S. (2017). Fields and Galois Theory. (Version 4.53). http://www.jmilne.org/math/CourseNotes/FT.pdf

[3] Wikipedia. Constructible number. https://en.wikipedia.org/wiki/Constructible_number

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