HAMPSHIRE COLLEGE: YOU CAN T GET THERE FROM HERE : WHY YOU CAN T TRISECT AN ANGLE, DOUBLE THE CUBE, OR SQUARE THE CIRCLE RAVI VAKIL Contents 1. Introduction 1 2. Impossibility proofs, and 2 2 3. Real fields 5 4. Recasting the classical problems 6 4.1. The first two cases 6 4.2. The third case: π 7 5. Conclusion 8 Here are important things to go on the board. 2 / Q. a, b, c, d Q, a + b 2 = c + d 2 a = b = c = d = 0, later changed to F. Two cubics 8x 3 6x 1 = 0, x 3 2 = 0. Three statements in final form. Sneaky Lemma. 1. Introduction Hi there. Let me start by thanking you all for the invitation to be here. Date: Friday, August 4, 2000. File hampshire.tex. 1
I came here today to tell you a great story. What I want to do is to prove for you the three classical problems of antiquity, the impossibility of trisecting an angle, doubling the cube, or squaring the circle. People hear about it pretty early, but only see the proof in an undergraduate course in Galois theory, which is pretty advanced. Galois theory is one of the coolest courses in undergrad, and the proof of these impossibilities is one of the coolest part of the Galois theory course. I m not going to tell you what Galois theory is, but if you understand and remember the argument I m going tell you, you ll already have an intuitive understanding for some of the ideas. (handout; questions) First of all, how many of you have heard of these problems? What are they? 1. Given an angle, can you trisect it using only ruler and compass? 2. Given a circle, can you construct a square of the same area using only ruler and compass? 3. Given a line segment of length 1, can you construct a segment of length 3 2? I m going to do something even easier. I ll show that you can t even trisect a 60 degree angle! Notice that we need to say with ruler and compass; if you are allowed other means, you can sometimes do these things. For example, Sarah-Marie told me that some of you know how to use origami folds to trisect an angle. Here s another way of trisecting an angle, using a marked straightedge ; this trick is due to Archimedes. In this figure, angle B is one third angle A. 2
B A 2. Impossibility proofs, and 2 Math departments around the country get hundreds of letters every year from people claiming to have done it. Some departments actually post these letters on a board for peoples amusement. What these people think is that no one has yet found a way to do these things, but that surely it is possible, and we just haven t been smart enough yet. But that s wrong. It just can t be done. It s not unreasonable for them to have this misconception. After all, how can you show something can t be done? That s why, when this impossibility was first shown, it was such a big deal. So to warm up, let me start with another classical impossibility proof, something most of you have undoubtedly seen, the fact that 2 is irrational. Give them this proof. Here s an immediate consequence. Suppose a + b 2 = c + d 2, where a, b, c, and d are rational. Then a = c and b = d. Reason: Otherwise 2 = (c a)/(b d). Just for future reference, here s another irrationality statement. cos 20 is a root of 8x 3 6x 1 = 0. How can we use this to show that cos 20 is irrational? The irrationality of 2 was a big deal, and it s hard to realize how big a deal it was, because the idea of the real numbers as some sort of continuum is so hard-wired into our brain. But here s how humankind thought of numbers at the time. There were the numbers we could immediately think of, the rationals, which had to do with, literally, number. So the ancient greeks vision of numbers was: Q. 3
But there were other things that looked vaguely like numbers as well: lengths. They were of a different flavor. For example, if you multiply two lengths, you get an area. But there was no obvious reason why lengths and numbers shouldn t be the same. Any reasonable person would expect lengths to be numbers. But: 2 is a length but not a number! So suddenly the ancient greeks vision of numbers was Q, and 2. All of a sudden the world got a lot weirder. It actually took a long time to make sense of a bigger set of numbers than Q that didn t look really ad hoc. That was the concept of real number R, and it was only in the last couple of centuries that it was put on a firm footing. Now in school, R is introduced to us as infinite decimals, with issues such as.999... glossed over, and the identification with lengths assumed. After the reals, of course, the idea of number continued to develop. Here are some other common sets of numbers: introduce complex numbers, and algebraic numbers. To prove this ancient result, I ll introduce a new set of numbers, that I ll call constructible numbers, that lie between Q and R. These are the real numbers that you can write down using rational numbers, the four arithmetic operations, and square roots. Everything you write down has to be real though. For example: (Do some.) Theorem. Let S be the set of points you could then construct using ruler and compass, starting with the points (0, 0) and (1, 0). Then S is precisely the set of points (x, y) where x and y are constructible. There are many parts to this proof, but they re all easy. First, let me convince you that you can construct any point (x, y) where x and y are constructible. It suffices to show that you can construct a line segment of length x and a line segment of length y. Explain. 4
So how can I get a line segment of some constructible length x? Answer: if I have line segments of length a and b, then I can construct line segments of length a+b, a b, ab, a/b if b 0, a and ra (r Q). I ll show you most of them. (Do it.) Then this is good enough. For example, to construct 1 4 + 39, I just... So I ve just shown you that you can construct all points with constructible coordinates. Now I have to show that you can t construct any more. I ll just do one part, and hopefully you ll get the idea. So how might you get new points? There are only so many things you could do. You could, for example, take a circle and intersect it with a line. The circle would be defined by choosing a center point and a radius point, and the line would be defined by choosing two points on it. All four points would have to be constructible. (Give them names.) Now if you work out the formula of the intersection, you ll realize that you ll get something involving all of these letters, and the four operations, and square roots. So the result is also constructible. Similarly, you can show that the intersection points of two constructible circles are also constructible, and the same with lines. Before goin on to the proof: pentagon is constructible, as sin 18 = ( 5 1)/4, so can construct 18 degree angles. 17-gon is constructible too! 3. Real fields The constructible numbers are an example of a real field. Definition. A real field F is a subset of R containing 0 and 1, closed under the four operations. In other words, if you add, subtract, multiply, or divide two elements of F, you ll get another element of F. The other idea I ll need is the idea of a quadratic extension of a real field. 5
Definition. Suppose F is a real field, and z > 0 is an element of F such that z isn t in F. Then the elements of the form a+b z (where a, b F ) also form a real field, and this is called a quadratic extension of F, and is denoted F ( z). Define the conjugate of a + b z to be a b z. As an example, Q( 2) is a quadratic extension of Q. In fact, if you understand Q( 2), then you basically understand any qudaratic extension F ( z). For example, it s a fact that if a, b, c, d F, and a + b z = c + d z, then a = b = c = d = 0; the proof is just the same as when F = Q and z = 2. Q( 2, 3) is a quadratic extension of Q( 2); we ll say its in a tower of quadratic extensions of Q. Then constructible numbers are precicsely those numbers which lie in some tower of quadratic extensions of Q. For example, 1/ 4 + 39 lies in... 4. Recasting the classical problems Now the three classical problems can be reduced to the following statements. 1. cos 20 isn t constructible. (Explain.) 2. 3 2 isn t constructible. 3. π isn t constructible. (Explain.) 4.1. The first two cases. Observe that the first two are roots of cubic polynomials with no rational roots: 8x 3 6x 1, x 3 2. So far we ve just spent our time with definitions. But all that s left is one sneaky lemma. Now pay attention to this, because I don t think this is in any book. 6
Sneaky lemma. Suppose F is a real field. Suppose you have a cubic with coefficients in F, and no root in F. Then it has no roots in a quadratic extension. Let me do one example. (1/8) = 0. Consider the polynomial x 3 (6/8)x Question: What are the sum of the roots of this polynomial? Answer: 0. You ve already shown me that there are no rational roots. I ll show that there are no roots in a quadratic extension of Q. Suppose a+b z is a root. Then b isn t 0, as there aren t any rational roots. That means when I stick it into the formula, I get zero. Imagine I did it. Then when I expand it, I get mess 1 + mess 2 z = 0. Then mess 1 and mess 2 must both be zero. What happens when I stick a b z into x 3 (6/8)x (1/8)? If you think about it, you ll realize that you ll get the same thing, except the sign in front of z will change; you ll get mess 1 mess 2 z = 0. Thus a b z is another root! So we ve found 2 roots already! And we can find the third because we know they sum to zero! What s the third? Ans: 2a. But that s rational! And we know that x 3 (3/4)x (1/8) = 0 has no rational roots, so we have a contradiction! Thus we ve shown that a + b z isn t a root of our cubic. Exactly the same argument proves the key lemma in general. Now we can finish the proof in Case 1. 8x 3 6x 1 = 0 has no roots in Q, and its coefficients are in Q, so the Sneaky Lemma tells us that it has no roots in a quadratic extension of F of Q. Now it has no roots in F, and its coefficients are in F, so the Sneaky Lemma tells us that it has no roots in a quadratic extension of G of F. 7
Similarly, by induction, it has no roots in any tower of quadratic extensions of Q. Thus it has no constructible roots! We ve proved the first case! The second case was to show that 3 2 isn t constructible. But by the identical argument, x 3 2 = 0 has no constructible roots, so we win again! 4.2. The third case: π. Finally, let s deal with π. Here I ll have to quote a result. You re probably aware that π is transcendental, and hence not algebraic. That s the one fact I won t prove for you, and it s hard. Fact. Constructible numbers are algebraic. That proves the third case! 5. Conclusion In short, I really hope you saw the general flavour of how that argument goes, and I hope that some of you followed it from beginning from end. This is one of the finest intellectual gems in human history, and it s remarkable that it s potentially comprehensible without years of training. I m going to stick around for dinner, so I hope some of you got stuck somewhere, and will have questions to ask me. And finally, I hope all of you found this as cool as I do. Thank you! 8