CONSTRUCTIBLE NUMBERS AND NUMBER FIELDS - IMPOSSIBILITY PROOFS AND ALGEBRA - GEOMETRICAL CONSTRUCTIONS. THE ALGEBRA OF NUMBER FIELDS - What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

CHAPTER III. GEOMETRICAL CONSTRUCTIONS. THE ALGEBRA OF NUMBER FIELDS

PART I. IMPOSSIBILITY PROOFS AND ALGEBRA

*§2. CONSTRUCTIBLE NUMBERS AND NUMBER FIELDS

1. General Theory

Our previous discussion indicates the general algebraic background of geometrical constructions. Every ruler and compass construction consists of a sequence of steps, each of which is one of the following: 1) connecting two points by a straight line, 2) finding the point of intersection of two lines, 3) drawing a circle with a given radius about a point, 4) finding the points of intersection of a circle with another circle or with a line. An element (point, line, or circle) is considered to be known if it was given at the outset or if it has been constructed in some previous step. For a theoretical analysis we may refer the whole construction to a coordinate system x, y (see p. 73). The given elements will then be represented by points or segments in the x, y plane. If only one segment is given at the outset, we may take this as the unit length, which fixes the point x = 1, y = 0. Sometimes there appear “arbitrary” elements: arbitrary lines are drawn, arbitrary points or radii are chosen. (An example of such an arbitrary element appears in constructing the midpoint of a segment; we draw two circles of equal but arbitrary radius from each endpoint of the segment, and join their intersections.) In such cases we may choose the element to be rational; i.e. arbitrary points may be chosen with rational coordinates x, y, arbitrary lines ax + by + c = 0 with rational coefficients a, b, c, arbitrary circles with centers having rational coördinates and with rational radii. We shall make such a choice of rational arbitrary elements throughout; if the elements are indeed arbitrary this restriction cannot affect the result of a construction.

For the sake of simplicity, we shall assume in the following discussion that only one element, the unit length 1, is given at the outset. Then according to §1 we can construct by ruler and compass all numbers that can be obtained from unity by the rational processes of addition, subtraction, multiplication and division, i.e. all the rational numbers r/s, where r and s are integers. The system of rational numbers is “closed” with respect to the rational operations; that is, the sum, difference, product, or quotient of any two rational numbers—excluding division by 0, as always—is again a rational number. Any set of numbers possessing this property of closure with respect to the four rational operations is called a number field.

Exercise: Show that every field contains all the rational numbers at least. (Hint: If a ≠ 0 is a number in the field F, then a/a = 1 belongs to F, and from 1 we can obtain any rational number by rational operations.)

Starting from the unit, we can thus construct the whole rational number field and hence all the rational points (i.e. points with both coordinates rational) in the x, y plane. We can reach new, irrational, numbers by using the compass to construct e.g. the number image which, as we know from Chapter II, §2, is not in the rational field. Having constructed image we may then, by the “rational” constructions of §1, find all numbers of the form

(1) image

where a, b are rational, and therefore are themselves constructible. We may likewise construct all numbers of the form

image

where a, b, c, d are rational. These numbers, however, may always be written in the form (1). For we have

image

where p, q are rational. (The denominator c2 – 2d2 cannot be zero, for if c2 – 2d2 = 0, then image = c/d, contrary to the fact that image is irrational.) Likewise

image

where r, s are rational. Hence all that we reach by the construction of image is the set of numbers of the form (1), with arbitrary rational a, b.

Exercises: From Image obtain the numbers

image

in the form (1).

These numbers (1) again form a field, as the preceding discussion shows. (That the sum and difference of two numbers of the form (1) are also of the form (1) is obvious.) This field is larger than the rational field, which is a part or subfield of it. But, of course, it is smaller than the field ofall real numbers. Let us call the rational field F0 and the new field of numbers of the form (1), F1. The constructibility of every number in the “extension field” F1 has been established. We may now extend the scope of our constructions, e.g. by taking a number of F1, say k = 1 + image and extracting the square root, thus obtaining the constructible number

image

and with it, according to §1, the field consisting of all the numbers

(2) image

where now p and q may be arbitrary numbers of F1, i.e. of the form image, with a, b in F0, i.e. rational.

Exercises: Represent

image

in the form (2).

All these numbers have been constructed on the assumption that only one segment was given at the outset. If two segments are given we may select one of them as the unit length. In terms of this unit suppose that the length of the other segment is α. Then we can construct the field Gconsisting of all numbers of the form

image

where the numbers a0, · · ·, am and b0,· · ·, bn are rational, and m and n are arbitrary positive integers.

Exercise: If two segments of lengths 1 and α are given, give actual constructions for 1 + α + α2, (1 + α)/(1 – α), α3.

Now let us assume more generally that we are able to construct all the numbers of some number field F. We shall show that the use of the ruler alone will never lead us out of the field F. The equation of the straight line through two points whose coördinates a1, b1 and a2, b2 are in F is (b1– b2)x + (a2– a1)y + (a1b2a2 b1) = 0 (see p. 491); its coefficients are rational expressions formed from numbers in F, and therefore, by definition of a field, are themselves in F. Moreover, if we have two lines, αx + βy – γ = 0 and α’x + β’y – γ’ = 0, with coefficients in F, then the coördinates of their point of intersection, found by solving these two simultaneous equations, are imageimage Since these are likewise numbers of F, it is clear that the use of the ruler alone cannot take us beyond the confines of the field F.

Exercises: The lines image, have coefficients in the field (1). Calculate the coordinates of their point of intersection, and verify that these have the form (1).— Join the points image and image by a line ax + by + c = 0, and verify that the coefficients are of the form (1).—Do the same with respect to the field (2) for the lines image, image, and the points image, respectively.

We can only break through the walls of F by using the compass. For this purpose we select an element k of F which is such that image is not in F. Then we can construct image and therefore all the numbers

(3) image,

where a and b are rational, or even arbitrary elements of F. The sum and the difference of two numbers image and image, their product, image, and their quotient,

image

are again of the form image with p and q in F. (The denominator c2kd2 cannot vanish unless c and d are both zero; for otherwise we would have image, a number in F, contrary to the assumption that image is not in F.) Hence the set of numbers of the form image forms a field F’. The field F’ contains the original field F, for we may, in particular, choose b = 0. F’ is called an extension field of F, and F a subfield of F’.

As an example, let F be the field image with rational a, b, and take image. Then the numbers of the extension field F’ are represented by image where p and q are in image, with rational a, b, a’, b’. Any number in F’ can be reduced to that form; for example

image

Exercise: Let F be the field image, where p and q are of the form image rational. Represent image in this form.

We have seen that if we start with any field F of constructible numbers containing the number k, then by use of the ruler and a single application of the compass we can construct image and hence any number of the form image, where a, b, are in F.

We now show, conversely, that by a single application of the compass we can obtain only numbers of this form. For what the compass does in a construction is to define points (or their coördinates) as points of intersection of a circle with a straight line, or of two circles. A circle with center image, image and radius r has the equation (ximage)2 + (yimage)2 = r2; hence, if image, image, r are in F, the equation of the circle can be written in the form

x2 + y2+ 2αx + 2βy + γ = 0,

with the coefficients α, β, γ in F. A straight line,

ax + by + c = 0,

joining any two points whose coördinates are in F, has coefficients a, b, c in F, as we have seen on page 129. By eliminating y from these simultaneous equations, we obtain for the x-coördinate of a point of intersection of the circle and line a quadratic equation of the form

Ax2 + Bx + C = 0,

with coefficients A, B, C in F (explicitly: A = a2 + b2, B = 2(ac + b2α – abβ), C = c2 – 2bcβ + b2γ). The solution is given by the formula

image

which is of the form image, with p, q, k in F. A similar formula holds for the y-coördinate of a point of intersection.

Again, if we have two circles,

x2 + y2 + 2αx + 2βy + γ =0,

x2 + y2 + 2α’x + 2β’y + γ’ = 0,

then by subtracting the second equation from the first we obtain the linear equation

2(α – α’)x + 2(β – β’)y + (γ – γ’) = 0,

which may be solved with the equation of the first circle as before. In either case, the construction yields the x- and y-coördinates of either one or two new points, and these new quantities are of the form image, with p, q, k in F. In particular, of course, image may itself belong to F, e.g., when k= 4. Then the construction does not yield anything essentially new, and we remain in F. But in general this will not be the case.

Exercises: Consider the circle with radius image about the origin, and the line joining the points image. Find the field F’ determined by the coordinates of the points of intersection of the circle and the line. Do the same with respect to the intersection of the given circle with the circle with radius image and center image.

Summarizing again: If certain quantities are given at the outset, then we can construct with a straightedge alone all the quantities in the field F generated by rational processes from the given quantities. Using the compass we can then extend the field F of constructible quantities to a wider extension field by selecting any number k of F, extracting the square root of k, and constructing the field F’ consisting of the numbers image, where a and b are in F. F is called a subfield of F’; all quantities in F are also contained in F’, since in the expression image we may choose b= 0. (It is assumed that image is a new number not lying in F, since otherwise the process of adjunction of image would not lead to anything new, and F’ would be identical with F.) We have shown that any step in a geometrical construction (drawing a line through two known points, drawing a circle with known center and radius, or marking the intersection of two known lines or circles) will either produce new quantities lying in the field already known to consist of constructible numbers, or, by the construction of a square root, will open up a new extension field of constructible numbers.

The totality of all constructible numbers can now be described with precision. We start with a given field F0, defined by whatever quantities are given at the outset, e.g. the field of rational numbers if only a single segment, chosen as the unit, is given. Next, by the adjunction of image, where k0 is in F0, but image is not, we construct an extension field F1 of constructible numbers, consisting of all numbers of the form image, where a0 and b0 may be any numbers of F0. Then F2, a new extension field of F1, is defined by the numbers image where a1 and b1are any numbers of F1, and k1 is some number of F1 whose square root does not lie in F1. Repeating this procedure, we shall reach a field Fn after n adjunctions of square roots. Constructible numbers are those and only those which can be reached by such a sequence of extension fields; that is, which lie in a field FB of the type described. The size of the number n of necessary extensions does not matter; in a way it measures the degree of complexity of the problem.

The following example may illustrate the process. We want to reach the number

image

Let F0 denote the rational field. Putting ko = 2, we obtain the field F1, which contains the number image. We now take image and k2 = 3. As a matter of fact, 3 is in the original field F0, and a fortiori in the field F2, so that it is perfectly permissible to take k2 = 3. We then take image, and finally image. The field F5 thus constructed contains the desired number, for image is also in F5, since image and image, and therefore their product, are in F3 and therefore also in F6

Exercises: Verify that, starting with the rational field, the side of the regular 2m-gon (see p. 124) is a constructible number, with n = m – 1. Determine the sequence of extension fields. Do the same for the numbers

image

2. All Constructible Numbers are Algebraic

If the initial field F0 is the rational field generated by a single segment, then all constructible numbers will be algebraic. (For the definition of algebraic numbers see p. 103). The numbers of the field F1 are roots of quadratic equations, those of F2 are roots of fourth degree equations, and, in general, the numbers of Fk are roots of equations of degree 2k, with rational coefficients. To show this for a field F2 we may first consider as an example image. We have image, a quadratic equation with coefficients in a field F1. By squaring, we finally obtain

(x2 − 1)2 = 2(2x + 1)2,

which is an equation of the fourth degree with rational coefficients.

In general, any number in a field F2 has the form

(4) image

where p, q, w are in a field F1, and hence have the form image, where a, b, c, d, e, f, s are rational. From (4) we have

x2 – 2px + p2 = q2w,

where all the coefficients are in a field F1, generated by image. Hence this equation may be rewritten in the form

image

where r, s, t, u, v are rational. By squaring both sides we obtain an equation of the fourth degree

(5) (x2 + ux + v)2 = s(rx + t)2

with rational coefficients, as stated.

Exercises: 1) Find the equations with rational coefficients for a) image.

2) Find by a similar method equations of the eighth degree for a) image

To prove the theorem in general for x in a field Fk with arbitrary k, we show by the procedure used above that x satisfies a quadratic equation with coefficients in a field Fk-1. Repeating the procedure, we find that x satisfies an equation of degree 22 = 4 with coefficients in a field Fk-2, etc.

Exercise: Complete the general proof by using mathematical induction to show that x satisfies an equation of degree 2t with coefficients in a field Fk-1, 0 < l ≤ k. This statement for l = k is the desired theorem.