Wednesday, December 20, 2006


I like this, by Brian Hayes in American Scientist, via Arts and Letters Daily, regarding the proof that the trisection of angles is impossible. (Warning: not recommended for maths-phobes)

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

Socrates, drawing figures in the sand, undertakes to coach an untutored slave boy, helping him to prove a special case of the Pythagorean theorem. I paraphrase very loosely:
Socrates: Here is a square with sides of length 2 and area equal to 4. If we double the area, to 8 units, what will the length of a side be?

Boy: Umm, 4?

Socrates: Does 4 x 4 = 8?

Boy: Okay, maybe it's 3.

Socrates: Does 3 x 3 = 8?

Boy: I give up.

Socrates: Observe this line from corner to corner, which the erudite among us call a diagonal. If we erect a new square on the diagonal, note that one-half of the original square makes up one-fourth of the new square, and so the total area of the new square must be double that of the original square. Therefore the length of the diagonal is the length we were seeking, is it not?

Boy: Whatever...

A purported trisection procedure is required to take an angle ? and produce ?/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 ?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 ? to the corresponding values for ?/3. The formulas yield the following equation (where for brevity the symbol u replaces the expression cos?/3):

cos? = 4u 3 - 3u.

For the 60-degree angle, with cos? = 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."

No comments: