An Imaginary Tale: The Story of i [the square root of minus one] (Princeton Library Science Edition) [NOOK Book]

An Imaginary Tale

PRINCETON UNIVERSITY PRESS

Chapter One

1.1 The Cubic Equation

At the end of his 1494 book Summa de Arithmetica, Geometria, Proportioni et Proportionalita, summarizing all the knowledge of that time on arithmetic, algebra (including quadratic equations), and trigonometry, the Franciscan friar Luca Pacioli (circa 1445-1514) made a bold assertion. He declared that the solution of the cubic equation is "as impossible at the present state of science as the quadrature of the circle." The latter problem had been around in mathematics ever since the time of the Greek mathematician Hippocrates, circa 440 B.C. The quadrature of a circle, the construction by straightedge and compass alone of the square equal in area to the circle, had proven to be difficult, and when Pacioli wrote the quadrature problem was still unsolved. He clearly meant only to use it as a measure of the difficulty of solving the cubic, but actually the quadrature problem is a measure of the greatest difficulty, since it was shown in 1882 to be impossible.

Pacioli was wrong in his assertion, however, because within the next ten years the University of Bologna mathematician Scipione del Ferro (1465-1526) did, in fact,discover how to solve the so-called depressed cubic, a special case of the general cubic in which the second-degree term is missing. Because his solution to the depressed cubic is central to the first progress made toward understanding the square root of minus one, it is worth some effort in understanding just what del Ferro did.

The general cubic contains all the powers of the unknown, i.e.,

x³ + [a.sub.1]x² + [a.sub.2]x + [a.sub.3] = 0

where we can take the coefficient of the third-degree term to be unity without loss of any generality. If that coefficient is not one, then we just divide through the equation by the coefficient, which we can always do unless it is zero-but then the equation isn't really a cubic.

The cubic solved by del Ferro, on the other hand, has the general form of

x³ + px = q,

where p and q are non-negative. Just like Diophantus, sixteenth-century mathematicians, del Ferro included, avoided the appearance of negative coefficients in their equations. Solving this equation may seem to fall somewhat short of solving the general cubic, but with the discovery of one last ingenious trick del Ferro's solution is general. What del Ferro somehow stumbled upon is that solutions to the depressed cubic can be written as the sum of two terms, i.e., we can express the unknown x as x = u + v. Substituting this into the depressed cubic, expanding, and collecting terms, results in

u³ + v³ + (3uv + p)(u + v) = q.

This single, rather complicated-looking equation, can be rewritten as two individually less complicated statements:

3uv + p = 0

which then says

u³ + v³ = q.

How did del Ferro know to do this? The Polish-American mathematician Mark Kac (1914-84) answered this question with his famous distinction between the ordinary genius and the magician genius: "An ordinary genius is a fellow that you and I would be just as good as, if we were only many times better. There is no mystery as to how his mind works. Once we understand what he has done, we feel certain that we, too, could have done it. It is different with the magicians ... the working of their minds is for all intents and purposes incomprehensible. Even after we understand what they have done, the process by which they have done it is completely dark." Del Ferro's idea was of the magician class.

Solving the first equation for v in terms of p and u, and substituting into the second equation, we obtain

[u.sup.6] - qu³ - p³/27 = 0.

At first glance this sixth-degree equation may look like a huge step backward, but in fact it isn't. The equation is, indeed, of the sixth degree, but it is also quadratic in u³. So, using the solution formula for quadratics, well-known since Babylonian times, we have

u³ = q/2 ± [square root of q²/4 + p³/27].

or, using just the positive root,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Now, since v³ = q - u³, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Thus, a solution to the depressed cubic x³ + px = q is the fearsome-looking expression

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Alternatively, since [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], then the second term in this expression can have a - 1 factor taken through the outer radical to give the equivalent

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

You can find both forms stated in different books discussing cubics, but there is no reason to prefer one over the other.

Since p and q were taken by del Ferro to be positive, it is immediately obvious that these two (equivalent) expressions for x will always give a real result. In fact, although there are three solutions or roots to any cubic (see appendix A), it is not hard to show that there is always exactly one real, positive root and therefore two complex roots to del Ferro's cubic (see box 1.1).

Now, before continuing with the cubic let me say just a hit about the nature of complex numbers. A complex number is neither purely real nor purely imaginary, but rather is a composite of the two. That is, if a and b are both purely real, then a + b[square root of -1] is complex. The form used by mathematicians and nearly everybody else is a + ib (the great eighteenth-century Swiss mathematician Leonhard Euler, about whom much more is said in chapter 6, introduced the i symbol for [square root of -1] in 1777). This is written as a + jb by electrical engineers. The reason electrical engineers generally opt for j is that often occurs in theft problems when electric currents are involved, and the letter i is traditionally reserved for that quantity. Contrary to popular myth, however, I can assure you that most electrical engineers are not confused when they see an equation involving complex numbers written with i = [square root of -1] rather than with j. With that said, however, let me admit that in chapter 5 I, too, will use j rather than i for [square root of -1] when I show you a nice little electrical puzzle from the nineteenth century.

Complex numbers obey many of the obvious rules, e.g., (a + ib) (c + id) = ac + iad + ibc + i²bd = ac - bd + i(ad + bc). But you do have to be careful. For example, if a and b can both only be positive, then [square root of ab] = [square root of a] [square root of b]. But if we allow negative numbers, too, this rule fails, e.g., [square root of (-4)(-9)] = [square root of 36] = 6 [not equal to] [square root of -4] [square root of -9] = (2i)(3i) = 6i² = -6. Euler was confused on this very point in his 1770 Algebra.

One final, very important comment on the reals versus the complex. Complex numbers fail to have the ordering property of the reals. Ordering means that we can write statements like x > 0 or x < 0. Indeed, if x and y are both real, and if x > 0 and y > 0, then their product xy > 0. If we try to impose this behavior on complex numbers, however, then we get into trouble. An easy way to see this is by a counterexample. That is, let us suppose we can order the complex numbers. Then, in particular, it must be true that either i > 0 or i < 0. Suppose i > 0. Then, -1 = i · i > 0, which is clearly false. So we must suppose i < 0, which when we multiply through by -1 (which flips the sense of the inequality) says -i > 0. Then, - 1 = (-i) · (-i) > 0, just as before, and it is still clearly false. The conclusion is that the original assumption of ordering leads us into contradiction, and so that assumption must be false. Now back to cubics.

Once we have the real root to del Ferro's cubic, then finding the two complex roots is not difficult. Suppose we denote the real root given by del Ferro's equation by [r.sub.1]. Then we can factor the cubic as

(x - [r.sub.1])(x - [r.sub.2])(x - [r.sub.3]) = 0 = (x - [r.sub.1])[x² - x([r.sub.2] + [r.sub.3]) + [r.sub.2][r.sub.3]].

To find the two additional roots, [r.sub.2] and [r.sub.3], we can then apply the quadratic formula to

x² - x([r.sub.2] + [r.sub.3]) + [r.sub.2][r.sub.3] = 0.

For example, consider the case of x³ + 6x = 20, where we have p = 6 and q = 20. Substituting these values into the second version of del Ferro's formula gives

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Now, if you look at the original cubic long enough, perhaps you'll have the lucky thought that x = 2 works (8 + 12 = 20). So could that complicated-looking thing with all the radical signs that I just wrote actually be 2? Well, yes, it is. Run it through a hand calculator and you will see that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Then, to find the other two roots to f(x) = 0 = x³ + 6x - 20, we use the fact that one factor of f(x) is (x - 2) to find, with some long division, that

(x - 2)(x² + 2x+ 10) = x³ + 6x - 20.

Applying the quadratic formula to the quadratic factor quickly gives the two complex roots (solutions to the original cubic) of

[r.sub.2] = -1 + 3[square root of -1]

and

[r.sub.3] = -1 + 3[square root of -1]

1.2 Negative Attitudes about Negative Numbers

But this is all getting ahead of the story. Del Ferro and his fellow mathematicians did not, in fact, do any of the above sort of factoring to get the complex roots-the finding of a single, real, positive number for the solution of a cubic was all they were after. And, as long as mathematicians concerned themselves with del Ferro's original depressed cubic, then a single, real, positive root is all there is, and all was well. But what of such a cubic as x³ - 6x = 20, where now we have p = -6 < 0? Del Ferro would never have written such a cubic, of course, with its negative coefficient, but rather would have written x³ = 6x + 20 and would have considered this an entirely new problem. That is, he would have started over from the beginning to solve

x³ = px + q

with, again, both p and q non-negative. This is totally unnecessary, however, as at no place in the solution to x³ + px = q did he ever actually use the non-negativity of p and q. That is, such assumptions have no importance, and were explicitly made simply because of an unwarranted aversion by early mathematicians to negative numbers.

This suspicion of negative numbers seems so odd to scientists and engineers today, however, simply because they are used to them and have forgotten the turmoil they went through in their grade-school years. In fact, intelligent, nontechnical adults continue to experience this turmoil, as illustrated in the following wonderful couplet, often attributed to the poet W. H. Auden:

Minus times minus is plus. The reason for this we need not discuss.

The great English mathematician John Wallis (1616-1703), for example, whom you will meet in more detail later in the next chapter as the individual who made the first rational attempt to attach physical significance to [square root of -1], also made some incredible assertions concerning negative numbers. In his 1665 book Arithmetica Infinitorum, an influential book read with great interest by the young Isaac Newton, Wallis made the following argument. Since a ÷ 0, with a > 0, is positive infinity, and since a ÷ b, with b < 0, is a negative number, then this negative number must be greater than positive infinity because the denominator in the second case is less than the denominator in the first case (i.e., b < 0). This left Wallis with the astounding conclusion that a negative number is simultaneously both less than zero and greater than positive infinity, and so who can blame him for being wary of negative numbers? And, of course, he was not alone. Indeed, the great Euler himself thought Auden's concern sufficiently meritorious that he included a somewhat dubious "explanation" for why "minus times minus is plus" in his famous textbook Algebra (1770).

We are bolder today. Now we simply say, okay, p is negative (so what?) and plug right into the original del Ferro formula. That is, replacing the negative p with -p (where now p itself is non-negative) we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

as the solution to x³ = px + q, with p and q both non-negative. In particular, the formula tells us that the solution to x³ = 6x + 20 is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

which is indeed a solution to the cubic, as can be easily verified with a handheld calculator.

1.3 A Rash Challenge

The story of the cubic now takes a tortured, twisted path. As was the tradition in those days, del Ferro kept his solution secret. He did this because, unlike today's academic mathematicians who make their living publishing their results to earn first appointment to a junior professorship and later promotion and tenure, del Ferro and his colleagues were more like independently employed businessmen. They earned their livelihoods by challenging each other to public contests of problem solving, and the winner took all-prize money, maybe, certainly "glory," and with luck the support of an admiring and rich patron. One's chances of winning such contests were obviously enhanced by knowing how to solve problems that others could not, so secrecy was the style of the day.

In fact, del Ferro almost took the secret of how to solve depressed cubics to the grave, telling at most only a small number of close friends. As he lay dying he told one more, his student Antonio Mafia Fior. While Fior was not a particularly good mathematician, such knowledge was a formidable weapon and so, in 1535, he challenged a far better known and infinitely more able mathematician, Niccolo Fontana (1500-77). Fontana had come to Fior's attention because Fontana had recently announced that he could solve cubics of the general form x³ + px² = q.

Continues...

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Excerpted from An Imaginary Tale by Paul J. Nahin Copyright © 1998 by Princeton University Press
Excerpted by permission. All rights reserved. No part of this excerpt may be reproduced or reprinted without permission in writing from the publisher.

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