Geometrical Constructions and Algebra

by Lazarus Mudehwe

In this article, we want to discuss how we use algebraic methods to solve some geometrical construction problems that defeated the greatest mathematicians for over 2000 years. The only constructions we permit are those that can be made by means of a straight edge (a ruler without markings) and compasses. The problems are:

  1. doubling the cube i.e. construction of a cube whose volume is double that of a given cube
  2. trisecting the angle i.e. dividing a given angle into three equal angles
  3. construction of a regular heptagon

These famous problems were known to the Greeks well over 2000 years ago. But as an increasing number of people failed to solve them, it began to be suspected that they were impossible to solve. This was finally proved only only in the nineteenth century, by the methods we now present.

The proofs rely on some theory of equations - especially of cubic equations. The relevance to these construction problems of algebraic ideas like the solution of equations was first observed by François Viète in the sixteenth century.

Let us consider one of the simplest cubic equations, namely:

x3-1 = 0
Factorising, we obtain (x-1)(x2+x+1) = 0. Thus, x = 1  or  x2+x+1 = 0.

Now, if x2+x+1 = 0 then, using the quadratic equation formula, we have

x = (-1±   _____
Ö(12-4)
 
)/2 = (-1±   __
Ö(-3
 
)/2 = (-1±(Ö-1)Ö3)/2 = (-1±iÖ3)/2

Where i = Ö-1. Clearly, i is not a real number as the square of any real number is positive or zero. We call it an imaginary number. Thus the roots of the equation x3-1 = 0 are 

1,   -1/2+i(Ö3/2)  and  -1/2-i(Ö3/2).

Note that this equation has only one real root. In a similar way, we can show that the equation x3-2 = 0 has only one real root. We shall need this fact in showing that it is impossible to double the cube using a ruler and compasses only.

We need another fact about roots of a polynomial equations with integral coefficients. We will show that if a rational p/q, where p and q have no common factors except ±1, is a root of the polynomial equation

a0xn+a1xn-1+ . . . +an-1x+an = 0,  where  a0, a1, ...,an are  integers,
then  q must divide a0 and p must divide an. For, since p/q satisfies the equation, we can substitute it into the equation; multiplying by qn, we obtain
a0pn+a1pn-1q+ . . . +an-1pqn-1+anqn = 0.
On dividing by p and rearranging, we get
a0pn-1+a1pn-2q+ . . .+an-1qn-1 = -(anqn)/p
We note that the left hand side of the equation is an integer, implying that p divides an as p and q have no common factors except ±1. By a similar argument, we can show that q must divide a0. If a0 = 1 - in other words, if the equation is
xn+a1xn-1+ . . . +an-1x+an = 0,
then the only rational roots the equation can have are integral ones. Consequently, if such an equation does not possess integral roots, then it has no rational roots.

Now, given an arbitrary unit line segment we can construct all numbers that are obtained from unity by the rational algebraic operations of addition, subtraction, multiplication and division. This is the field of all rational numbers. Let us denote this field by F0. We further showed that this field can be extended by one operation of the compasses to include an irrational number - just take the square root of any non-square number k in F0, for example Ö2, Ö3, Ö11, etc. By further use of compasses and straight edge, we can construct any number obtained from Ök by the rational algebraic operations. These numbers can be shown to be of the form   a+bÖk   where a, b, k are in F0 but Ök is not in F0.

Do the following exercise for yourself. Let a, b, c, d be rational numbers. Show that (a+bÖ2)(c+dÖ2) and (a+bÖ2)/(c+dÖ2) can be expressed in the form p+qÖ2 where p and q are rational numbers.
(Hint: For the quotient, multiply numerator and denominator by c-dÖ2).

This field of numbers of the form p+qÖk, where p, q and k are rational numbers in F0, and where Ök is not in F0, is called an extension field. Let us denote this field by F1. With another operation of the compasses we can construct Ök where k is in F1 but Ök is not in F1. Furthermore, by use of ruler and compasses, we can construct numbers of the form   p+qÖk    where p,q and k are in F1 but Ök is not in F1. These numbers are in the extension field F2. This process can be continued indefinitely. Thus, starting with an arbitrary unit segment, we can construct the field of rational numbers F0 and extension fields F1, F2, F3, ..., Fn, ... such that F0 Ì F1 Ì F3 ... Ì Fn Ì ... . The set of all real numbers which belong to an extension field Fn, after some such finite sequence of operations, is called the set of constructible numbers. Note that this set is far smaller than the set of real numbers, many of which are not constructible, for example: e,  3Ö2,  7Ö8,  ln4 etc. Note also that every constructible number is  algebraic, in the sense that such a number satisfies a polynomial equation

a0xn+a1xn-1+...+an-1x+an = 0, where  a0, a1,..., an-1, an are  integers.
Thus the set of constructible numbers is a subset of the set of algebraic numbers. To see why, look at an example:
  æ
Ö
1+Ö5
 

is a typical constructible number, obtained by two square rooting operations. We'll show how it can be `unwrapped' to get the polynomial equation it satisfies.

Let us now tackle the problem of doubling the cube. It is enough for us to consider a cube whose sides are of unit length. Consequently, it will have unit volume. The problem is : Can we construct by ruler and compasses only the side of a cube whose volume is twice the given cube? If we let x be the length of the sides of the required cube, then x satisfies the equation

x3 = 2   or    x3-2 = 0.
Now from our discussion above, it is clear that the answer will be negative if we show that x is not constructible. We shall show this by the method of contradiction discussed by Prof. Fleischner in 2.1.

Thus, we suppose x IS constructible using a ruler and compasses only. Then x must lie in some extension field Fk. But x is not in F0, the field of rational numbers since it can easily be shown (by contradiction) that 3Ö2 is NOT a rational number. Thus x lies in some extension field Fk where k is positive. From the discussion above of extension fields, we saw that if x is in Fk, then x is in Fk+1 and so on. So let us take k to be the least positive number such that x lies in Fk. Thus x has the form

x = p+qÖw
where p, q, w are in Fk-1 but Öw is not in Fk-1  (if Öw  were in Fk-1, then x would be in Fk-1). We shall show that if x = p+qÖw is a solution of x3-2 = 0, then y = p-qÖw is also a solution of x3-2 = 0. Now, since x lies in the field   Fk,   x3  and  x3-2 also lie in the field  Fk. Thus x3-2 must have the form
x3 - 2 = a+ bÖw
Note that x3 = (p + qÖw)3 = p3+ 3p2(qÖw) + 3p(qÖw)2 + (Öw)3

= p3 + 3p2qÖw + 3pq2w + q3wÖw = (p3 + 3pq2w) + (3p2q + q3w)Öw Therefore, x3 - 2 = (p3 + 3pq2w - 2) + (3p2q + q3w)Öw = a + bÖw where a = p3 + 3pq2w- 2  and  b = 3p2q + q3w.

If we let y = p - qÖw, then, after expanding and substituting for a and b, we have

y3 - 2 = a - bÖw.
To show that y is also a root of the equation x3 - 2 = 0, we have to show that y3 - 2 = 0. In other words, we have to show that a and b are zero. Now, since x = p + qÖw is a solution of the equation, this implies that a + bÖw = 0. This means both a and b are zero. For if b were not zero, then Öw = -a/b. But -a/b is in Fk-1 and so Öw would be in Fk-1, contrary to our assumption. Therefore b must be zero. Now, if b is zero, a is zero. Consequently, a = b = 0.Thus y = p-qÖw is a root of the equation x3-2 = 0. We will further show that x ¹ y. Suppose they were equal. Then, p+qÖw = p-qÖw. Hence 2qÖw = 0. Now if either q or Öw is zero, then x = p. This means x is in Fk-1. This is contrary to the assumption that x is in Fk but not in Fk-1.

Now, our assumption that x is constructible has led us to two real roots of the equation x3 - 2 = 0. But, we saw that this equation has only one real root. We have thus reached a contradiction. Consequently, our hypothesis that x is constructible is false, and the cube cannot be doubled.

Next time we shall show how the problems of trisecting the angle and constructing a heptagon can be dealt with. - Lazarus Mudehwe

Reference:
R Courant and H Robbins  What is Mathematics   (Oxford University Press,   London, New York, Toronto, 1958).

File translated from TEX by TTH, version 1.50.

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