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

PART I. IMPOSSIBILITY PROOFS AND ALGEBRA

*§3. THE UNSOLVABILITY OF THE THREE GREEK PROBLEMS

1. Doubling the Cube

Now we are well prepared to investigate the old problems of trisecting the angle, doubling the cube, and constructing the regular heptagon. We consider first the problem of doubling the cube. If the given cube has an edge of unit length, its volume will be the cubic unit; it is required that we find the edge x of a cube with twice this volume. The required edge x will therefore satisfy the simple cubic equation

(1) x³ – 2 = 0.

Our proof that this number x cannot be constructed by ruler and compass alone is indirect. We assume tentatively that a construction is possible. According to the preceding discussion this means that x lies in some field F_k obtained, as above, from the rational field by successive extensions through adjunction of square roots. As we shall show, this assumption leads to an absurd consequence.

We already know that x cannot lie in the rational field F₀, for is an irrational number (see Exercise 1, p. 60). Hence x can only lie in some extension field F_k, where k is a positive integer. We may as well assume that k is the least positive integer such that x lies in some F_k. It follows that x can be written in the form

where p, q, and w belong to some F_k-1, but does not. Now, by a simple but important type of algebraic reasoning, we shall show that if is a solution of the cubic equation (1), then is also a solution. Since x is in the field F_k, x³ and x³ – 2 are also inF_k, and we have

(2) 

where a and b are in F_k-1. By an easy calculation we can show that a = p³ + 3pq²w – 2, b =3p²q + q³w. If we put

then a substitution of − q for q in these expressions for a and b shows that

(2’) .

Now x was supposed to be a root of x³ – 2 = 0, hence

(3) 

This implies—and here is the key to the argument—that a and b must both be zero. If b were not zero, we would infer from (3) that . But then would be a number of the field F_k-1 in which a and b lie, contrary to our assumption. Hence b = 0, and it follows immediately from (3) that a= 0 also.

Now that we have shown that a = b = 0, we immediately infer from (2’) that is also a solution of the cubic equation (1), since y³ – 2 is equal to zero. Furthermore, y ≠ x, i.e. x – y ≠ 0; for, can only vanish if q = 0, and if this were so then x = p would lie in F_k-1, contrary to our assumption.

We have therefore shown that, if is a root of the cubic equation (1), then is a different root of this equation. This leads immediately to a contradiction. For there is only one real number x which is a cube root of 2, the other cube roots of 2 being imaginary (see p. 98); is obviously real, since p, q, and were real.

Thus our basic assumption has led to an absurdity, and hence is proved to be wrong; a solution of (1) cannot lie in a field F_k, so that doubling the cube by ruler and compass is impossible.

2. A Theorem on Cubic Equations

Our concluding algebraic argument was especially adapted to the particular problem at hand. If we want to dispose of the two other Greek problems, it is desirable to proceed on a more general basis. All three problems depend algebraically on cubic equations. It is a fundamental fact concerning the cubic equation

(4) z³ + az² + bz + c =0

that, if x₁, x₂, x₃ are the three roots of this equation, then

(5) x₁ + x₂ + x₃ = –a.†

Let us consider any cubic equation (4) where the coefficients a, b, c are rational numbers. It may be that one of the roots of the equation is rational; for example, the equation x³ – 1 = 0 has the rational root 1, while the two other roots, given by the quadratic equation x² + x + 1 = 0, are necessarily imaginary. But we can easily prove the general theorem: If a cubic equation with rational coefficients has no rational root, then none of its roots is constructible starting from the rational field F₀.

Again we give the proof by an indirect method. Suppose x were a constructible root of (4). Then x would lie in the last field F_k of some chain of extension fields, F₀, F₁, · · ·, F_k, as above. We may assume that k is the smallest integer such that a root of the cubic equation (4) lies in an extension field F_k. Certainly k must be greater than zero, since in the statement of the theorem it is assumed that no root x lies in the rational field F₀. Hence x can be written in the form

where p, q, w are in the preceding field, F_k-1, but is not. It follows, exactly as for the special equation, z³ – 2 = 0, of the preceding article, that another number of F_k,

will also be a root of the equation (4) As before, we see that q ≠ 0 and hence x ≠ y.

From (5) we know that the third root u of the equation (4) is given by u = –a – x – y. But since x + y = 2p, this means that

u = –a – 2p,

† The polynomial z³ + az² + bz + c may be factored into the product (z – x₁)(z – x₂)(z – x₃), where x₁, x₂, x₃ are the three roots of the equation (4) (see p. 101). Hence

z³ + az² + bz + c = z³ – (x₁+x₂+x₃)z² +(x₁x₂ + x₁x₃ + x₂x₃)z – x₁x₂x₃,

so that, since the coefficient of each power of z must be the same on both sides,

–a = x₁+x₂+x₃, b = x₁x₂ + x₁x₃ + x₂x₃, –c = x₁x₂x₃.

where has disappeared, so that u is a number in the field F_k−1. This contradicts the hypothesis that k is the smallest number such that some F_k-1 contains a root of (4). Hence the hypothesis is absurd, and no root of (4) can lie in such a field F_k. The general theorem is proved. On the basis of this theorem, a construction by ruler and compass alone is proved to be impossible if the algebraic equivalent of the problem is the solution of a cubic equation with no rational roots. This equivalence was at once obvious for the problem of doubling the cube, and will now be established for the other two Greek problems.

3. Trisecting the Angle

We shall now prove that the trisection of the angle by ruler and compass alone is in general impossible. Of course, there are angles, such as 90° and 180°, for which the trisection can be performed. What we have to show is that the trisection cannot be effected by a procedure valid forevery angle. For the proof, it is quite sufficient to exhibit only one angle that cannot be trisected, since a valid general method would have to cover every single example. Hence the non-existence of a general method will be proved if we can demonstrate, for example, that the angle 60° cannot be trisected by ruler and compass alone.

We can obtain an algebraic equivalent of this problem in different ways; the simplest is to consider an angle θ as given by its cosine: cosθ = g. Then the problem is equivalent to that of finding the quantity x = cos(θ/3). By a simple trigonometrical formula (see p. 97), the cosine of θ/3 is connected with that of θ by the equation

cos θ = g = 4 cos³ (θ/3) – 3 cos (θ/3).

In other words, the problem of trisecting the angle θ with cos θ = g amounts to constructing a solution of the cubic equation

(6) 4z³ – 3z – g = 0.

To show that this cannot in general be done, we take θ = 60°, so that . Equation (6) then becomes

(7) 8z³ – 6z = 1.

By virtue of the theorem proved in the preceding article, we need only show that this equation has no rational root. Let v = 2z. Then the equation becomes

(8) v³– 3v = 1.

If there were a rational number v = r/s satisfying this equation, where r and s are integers without a common factor > 1, we should have r³ – 3s²r = s³. From this it follows that s³ = r(r² – 3s²) is divisible by r, which means that r and s have a common factor unless r= ±1. Likewise, s² is a factor of r³ = s²(s + 3r), which means that r and s have a common factor unless s = ±1. Since we assumed that r and s had no common factor, we have shown that the only rational numbers which could possibly satisfy equation (8) are +1 or –1. By substituting +1 and –1 for v in equation (8) we see that neither value satisfies it. Hence (8), and consequently (7), has no rational root, and the impossibility of trisecting the angle is proved.

The theorem that the general angle cannot be trisected with ruler and compass alone is true only when the ruler is regarded as an instrument for drawing a straight line through any two given points and nothing else. In our general characterization of constructible numbers the use of the ruler was always limited to this operation only. By permitting other uses of the ruler the totality of possible constructions may be greatly extended. The following method for trisecting the angle, found in the works of Archimedes, is a good example.

Fig. 36. Archimedes’ trisection of an angle.

Let an arbitrary angle x be given, as in Fig. 36. Extend the base of the angle to the left, and swing a semicircle with O as center and arbitrary radius r. Mark two points A and B on the edge of the ruler such that AB = r. Keeping the point B on the semicircle, slide the ruler into the position where A lies on the extended base of the angle x, while the edge of the ruler passes through the intersection of the terminal side of the angle x with the semicircle about 0. With the ruler in this position draw a straight line, making an angle y with the extended base of the original angle x.

Exercise: Show that this construction actually yields y = x/3.

4. The Regular Heptagon

We shall now consider the problem of finding the side x of a regular heptagon inscribed in the unit circle. The simplest way to dispose of this problem is by means of complex numbers (see Ch. II, §5). We know that the vertices of the heptagon are given by the roots of the equation

(9) z⁷ – 1 = 0,

the coördinates x, y of the vertices being considered as the real and imaginary parts of complex numbers z = x + yi. One root of this equation is z = 1, and the others are the roots of the equation

(10) 

obtained from (9) by factoring out z – 1 (see p. 99). Dividing (10) by z³, we obtain the equation

(11) z³ + 1/z³ + z² + 1/z² + z + 1/z + 1 = 0.

By a simple algebraic transformation this may be written in the form

(12) (z + 1/z)³ – 3(z + 1/z) + (z + 1/z)² – 2 + (z + 1/z) + 1 = 0.

Denoting the quantity z + 1/z by y, we find from (12) that

(13) y³ + y² – 2y – 1 = 0.

We know that z, the seventh root of unity, is given by

(14) z = cos φ; + i sin φ,

where φ = 360°/7 is the angle subtended at the center of the circle by the edge of the regular heptagon; likewise we know from Exercise 2, page 97, that 1/z = cos φ – i sin φ, so that y= z + 1/z = 2 cos φ. If we can construct y, we can also construct cosφ and conversely. Hence, if we can prove that y is not constructible, we shall at the same time show that z, and therefore the heptagon, is not constructible. Thus, considering the theorem of Article 2, it remains merely to show that the equation (13) has no rational roots. This, too, is proved indirectly. Assume that (13) has a rational root r/s, where r and s are integers having no common factor. Then we have

(15) r³ + r²s – 2rs² – s³ = 0;

whence it is seen as above that r³ has the factor s, and s³ the factor r. Since r and s have no common factor, each must be ±1; therefore y can have only the possible values +1 and –1, if it is to be rational. On substituting these numbers in the equation, we see that neither of them satisfies it. Hence y, and therefore the edge of the regular heptagon, is not constructible.

5. Remarks on the Problem Of Squaring The Circle

We have been able to dispose of the problems of doubling the cube, trisecting the angle, and constructing the regular heptagon, by comparatively elementary methods. The problem of squaring the circle is much more difficult and requires the technique of advanced mathematical analysis. Since a circle with radius r has the area πr², the problem of constructing a square with area equal to that of a given circle whose radius is the unit length 1 amounts to the construction of a segment of length as the edge of the required square. This segment will be constructible if and only if the number π is constructible. In the light of our general characterization of constructible numbers, we could show the impossibility of squaring the circle by showing that the number π cannot be contained in any field F_k that can be reached by the successive adjunction of square roots to the rational field F₀. Since all the members of any such field are algebraic numbers, i.e. numbers that satisfy algebraic equations with integer coefficients, it will be sufficient if the number π can be shown to be not algebraic, i.e. to be transcendental (see p. 104).

The technique necessary for proving that π is a transcendental number was created by Charles Hermite (1822-1905), who proved the number e to be transcendental. By a slight extension of Hermite’s method F. Lindemann succeeded (1882) in proving the transcendence of π, and thus definitely settled the age-old question of squaring the circle. The proof is within the reach of the student of advanced analysis, but is beyond the scope of this book.