LECTURE 5: STRAIGHTEDGE AND COMPASS CONSTRUCTIONSmath.uchicago.edu/~j.e.hickman/256 Lecture...
Transcript of LECTURE 5: STRAIGHTEDGE AND COMPASS CONSTRUCTIONSmath.uchicago.edu/~j.e.hickman/256 Lecture...
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: [email protected]