Download Lecture 5

LECTURE 5: STRAIGHTEDGE AND COMPASS
CONSTRUCTIONS
Constructible numbers. We begin with a rough definition which will be explained in more detail below.
Definition (Constructible number - rough version). A real number α P R is constructible if we can construct a line segment of length |α| in a finite number of
steps using from a fixed line segment of unit length using only a straightedge and
compass.
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 union
of P together with the intersections of the lines and circles of the above form. We
say 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 constructible
from 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 of
points in
8
ď
P :“
Pn .
n“1
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 Q ˆ Q Ă P.
We make the following simple observations:
‚ The intersection of two lines is the simultaneous solution of a pair of linear
equations.
‚ The points of intersection of a line and circle are the simultaneous solutions
of a linear and a quadratic equation.
‚ The points of intersection of two circles are the simultaneous solutions to a
pair of quadratic equations.
Thus, the co-ordinates of points in P (and therefore the constructible numbers
C) are all obtainable from numbers in Q through a finite sequence of field operations
and taking square roots.
Theorem. Suppose γ P CzQ. Then there exist α1 , . . . , αn P C with αn “ γ such
that
rQpα1 , . . . , αi q : Qpα1 , . . . , αi´1 qs “ 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).
Impossibility proofs.
Theorem. Doubling the cube is impossible. In particular, given a cube of volume
1 it is not possible to construct with a straightedge and compass the length of a side
of 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.
Theorem. Squaring the circle is impossible. In particular, given a circle of radius
1 it is impossible to construct the length of a side of a square with the same area as
the 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.
Theorem. Trisecting the angle is impossible. In particular, the angle π{3 cannot
be 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 cos π{3 “ 1{2 P Q, the problem is equivalent to showing α :“ cos π{9 R
C.
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.
‚ f pxq :“ 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 one
can easily verify that this is not the case.
‚ irrpα, Qq “ f {8 and rQpαq : Qs “ 3.
Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.
E-mail address: [email protected]