LECTURE 5: STRAIGHTEDGE AND COMPASS

CONSTRUCTIONS

Constructible numbers. We begin with a rough definition which will be ex-plained in more detail below.

Definition (Constructible number - rough version). A real number α P R is con-structible if we can construct a line segment of length |α| in a finite number ofsteps using from a fixed line segment of unit length using only a straightedge andcompass.

We will give a more detailed description of what is meant by this presently.Let P Ď R2 be a finite a set of points in the plane. We are interested in the

following geometric figures:

‚ Straight lines which pass through a pair of distinct points x, y P P .‚ Circles centred at some point c P P of radius |x´y| for some pair of distinct

points x, y P P .

From these shapes we consider a new set of points P̄ which consists of the unionof P together with the intersections of the lines and circles of the above form. Wesay P̄ is the set of points which is constructible from P .

Example. Let P :“ tp0, 0q, p1, 0qu. Then the set of points which is constructiblefrom P is given by

P̄ “ tp0, 0q, p1, 0q, p´1, 0q, p0, 2q, p1{2,?

3{2q, p1{2,´?

3{2qu.

Indeed, from P we can form:

‚ The straight line ` corresponding to the x-axis.‚ Two circles C0, C1 of unit radius centred at p0, 0q, p1, 0q, respectively.

Note that

`X C0 “ tp´1, 0q, p1, 0qu, `X C1 “ tp0, 0q, p2, 0qu,

C0 X C1 “ tp1{2,?

3{2q, p1{2,´?

3{2qu.

Definition (Constructible number - precise version). Define a sequence P1, P2, . . .of finite subsets of R2 recursively as follows:

‚ P1 :“ tp0, 0q, p1, 0qu.‚ Supposing Pn has been defined for some n P N, let Pn`1 denote the set of

points which are constructible from Pn.

We say a number α P R is constructible if |α| is the distance between a pair ofpoints in

P :“8ď

n“1

Pn.

The set of all constructible numbers is denoted C.

Example. It is easy to verify that the following examples.

‚ From the above we see?

3{2 P C‚ N Ď C.‚ n

?3{2 P C for all n P N.

We note some basic facts about the set P.1

2 LECTURE 5: STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

Lemma. i) α P R is constructible if and only if p0,˘αq, p˘α, 0q P P.ii) If px, yq P P and α P C, then px˘ α, yq, px, y ˘ αq P P.

A key algebraic observation is the following.

Theorem. The set of constructible numbers C is a subfield of R.

Corollary. Q Ă C and QˆQ Ă P.

We make the following simple observations:

‚ The intersection of two lines is the simultaneous solution of a pair of linearequations.

‚ The points of intersection of a line and circle are the simultaneous solutionsof a linear and a quadratic equation.

‚ The points of intersection of two circles are the simultaneous solutions to apair of quadratic equations.

Thus, the co-ordinates of points in P (and therefore the constructible numbersC) are all obtainable from numbers in Q through a finite sequence of field operationsand taking square roots.

Theorem. Suppose γ P CzQ. Then there exist α1, . . . , αn P C with αn “ γ suchthat

rQpα1, . . . , αiq : Qpα1, . . . , αi´1qs “ 2

for i “ 1, . . . , n. In particular, rQpγq : Qs “ 2r for some r P N.

Proof. The key ideas in the proof are:

‚ The above discussion shows the existences of such a tower.‚ rQpγq : Qs “ 2r follows from the Tower Law (and uniqueness of prime

factorisation).

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Impossibility proofs.

Theorem. Doubling the cube is impossible. In particular, given a cube of volume1 it is not possible to construct with a straightedge and compass the length of a sideof a cube of volume 2.

Proof. The key ideas in the proof are:

‚ The theorem is equivalent to showing 3?

2 R C.‚ rQp 3

?2q : Qs “ 3.

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Theorem. Squaring the circle is impossible. In particular, given a circle of radius1 it is impossible to construct the length of a side of a square with the same area asthe circle.

Proof. The key ideas in the proof are:

‚ The theorem is equivalent to showing?π R C.

‚ Since π is transcendental,?π is transcendental.

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Theorem. Trisecting the angle is impossible. In particular, the angle π{3 cannotbe trisected using a ruler and compass.

Proof. The key ideas in the proof are:

‚ An angle θ can be constructed if and only if cos θ P C.‚ Since cosπ{3 “ 1{2 P Q, the problem is equivalent to showing α :“ cosπ{9 RC.

LECTURE 5: STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 3

‚ By the triple angle formula, cos 3θ “ 4 cos3 θ´ 3 cos θ so 8α3´ 6α2´ 1 “ 0.‚ fpxq :“ 8x3 ´ 6x2 ´ 1 is irreducible over Z - to see this assume there is a

linear factor pax ` bq. Then a|8 and b|1 so a P t1, 2, 4, 8u and b P t´1, 1u.This implies the set t˘1,˘1{2,˘1{4,˘1{8u contains a root of f , but onecan easily verify that this is not the case.

‚ irrpα,Qq “ f{8 and rQpαq : Qs “ 3.

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Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. Uni-

versity Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.

E-mail address: jehickman@uchicago.edu