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COMPUTING SCIENCE

Foolproof

Mathematical proof is foolproof, it seems, only in the absence of fools

Brian Hayes

Wantzel's Theorem

The fact that trisection is impossible is common knowledge, but the reason it's impossible—the content of the proof—is not so widely known. Many authors mention it but few explain it. Even Underwood Dudley's splendid send-up of mathematical cranks, A Budget of Trisections, does not go through the proof step by step.

The origin and history of the proof are also somewhat shadowy. There is a lesson here for those who seek immortality by solving some trophy problem in mathematics. Trisection had been near the top of the most-wanted list for two millennia, and yet the author of the first published proof of impossibility has not earned a place in the pantheon.

That author was the French mathematician Pierre Laurent Wantzel (1814-1848), who is hardly a household name, even in mathematical households. His proof appeared in 1837. As far as I know, it has never been reprinted and has never been published in English translation. (I have posted a crude attempt at a translation on the American Scientist Web site.) Many citations of the paper give the wrong volume number, suggesting that even some of those who refer to the proof have not read it. And to pile on one further indignity, the paper itself gets the author's name wrong: He is listed as "M. L. Wantzel."

One reason for Wantzel's obscurity may be that his proof is almost unintelligible for modern readers. Later expositors offer more lucid accounts. Felix Klein, L. E. Dickson and Robert C. Yates all published versions of the proof; so did Willard Van Orman Quine, who wrote his version in response to a $100 challenge. For a thorough and accessible book-length exposition I highly recommend Abstract Algebra and Famous Impossibilities, by Arthur Jones, Sidney A. Morris and Kenneth R. Pearson.

As penance for my youthful career as a trisector, I would now like to try giving a brief sketch of the impossibility proof. The basic question is: What can you do with a straightedge and compass? You can draw lines and circles, obviously, but it turns out you can also do arithmetic. If the length of a line segment represents a number, then ruler-and-compass manipulations can add, subtract, multiply and divide such numbers, and also extract square roots. Suppose you are given a segment of length 1 to start with; what further numbers can you generate? All the integers are easy to reach; you can also get to any rational number (a ratio of integers). Square roots give access to certain irrationals; by taking square roots of square roots, you can also do fourth roots, eighth roots, and so on. But that's all you can do. There is no way to extract cube roots, fifth roots or any other roots not a power of 2—which is the crucial issue for trisection.

A purported trisection procedure is required to take an angle qand produce q/3. Since the procedure has to work with any angle, we can refute it by exhibiting just one angle that cannot be trisected. The standard example is 60 degrees. Suppose the vertex of a 60-degree angle is at the origin, and one side corresponds to the positive x axis. Then to trisect the angle you must draw a line inclined by 20 degrees to the x axis and passing through the origin.

To draw any line, all you need is two points lying on the line. In this case you already have one point, namely the origin. Thus the entire task of trisection reduces to finding one more point lying somewhere along the 20-degree line. Surely that must be easy! After all, there are infinitely many points on the line and you only need one of them. But the proof says it can't be done.

To see the source of the difficulty we can turn to trigonometry. If we knew the sine and cosine of 20 degrees, the problem would be solved; we could simply construct the point x=cos20, y=sin20. (Of course we need the exact values; approximations from a calculator or a trig table won't help.) We do know the sine and cosine of 60 degrees: The values are Ö—3/2 and 1?2. Both of these numbers can be constructed with ruler and compass. Furthermore, formulas relate the sine and cosine of any angle qto the corresponding values for q/3. The formulas yield the following equation (where for brevity the symbol ureplaces the expression cosq/3):

cosq=4u 3-3u.

For the 60-degree angle, with cosq=1?2,the equation becomes 8u 3-6u=1.Note that this is a cubic equation. That's the nub of the problem: No process of adding, subtracting, multiplying, dividing and taking square roots will ever solve the equation for the value of u. (The hard part of the proof, which I'm not brave enough to attempt here, shows that the cubic equation cannot be reduced to one of lower degree.)

This proof, with its excursions into trigonometry and algebra, would have been alien to Euclid, but the conclusion is easily translated back into the language of geometry: Not a single point along the 20-degree line (except the origin) can be reached from the 60-degree line by ruler-and-compass methods. There is something hauntingly counterintuitive about this fact. The two lines live on the same plane; they even intersect, and yet they don't communicate. You can't get there from here. Like Hobbes, I wouldn't believe it, except the proof compels belief. I wonder if my old friend Dmytro would be convinced.

© Brian Hayes




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