Letters to a Young Mathematician (8 page)

Dear Meg,

Please don’t try to trisect the angle. I’ll send you some interesting problems to work on if you want to flex your research muscles already; just stay clear of angle trisection. Why? Because you’ll be wasting your time.
Methods that go beyond the traditional unmarked ruler and compass are well known; methods that do not cannot possibly be correct. We know that because mathematics enjoys a privilege that is denied to most other walks of life. In mathematics, we can prove that something is impossible.

In most walks of life, “impossible” may mean anything from “I can’t be bothered” to “No one knows how to do it” to “Those in charge will never agree.” The science fiction writer Arthur C. Clarke famously wrote that “When an elderly and distinguished scientist declares something to be possible, he is very likely right. When he declares it to be impossible, he is almost certainly
wrong.” (Clarke was writing in 1963, when most scientists, especially elderly and distinguished ones, were almost certainly “he.”) But applied to mathematicians and mathematical theorems, Clarke’s statement is plain wrong. A mathematical proof of impossibility is a virtually unbreakable guarantee.

I say “virtually” because sometimes the impossible can become possible if the question is subtly changed. Then, of course, it’s not the same question. Archimedes knew how to trisect an angle using a
marked
ruler and compass.

My favorite simple impossible problem is a puzzle. Though it looks frivolous at first glance, it provides a lot of insight into logical inference in mathematics, and in particular into how we know that some tasks are impossible. The puzzle is this: given a chessboard with two diagonally opposite corners missing, can you cover it with thirty-one dominoes, each just the right size to cover two adjacent squares?

It must be understood that no “cheating” is allowed.
The dominoes must not overlap, or be cut up, or anything like that.

The first question to ask is reasonably straightforward and comes naturally to any mathematician: is the area an obstacle to success? The total area of the mutilated chessboard is 64 - 2 = 62 squares. The total area of the dominoes is 2 × 31 = 62 squares. So we have exactly the right number of dominoes to cover the chessboard. If we had
been given only thirty, then calculating the total area would immediately have proved that the task is impossible. But we’ve been given thirty-one, so the area is not an obstacle.

Meg, I know you’ve done a lot of math, but it’s just possible that you’ve never come across this puzzle. Puzzles do not feature prominently in university textbooks. Please try it. For the moment, don’t think about it; just cut out some cardboard dominoes and try to fit them together.

Done that? Did you get anywhere?

No. You tried and tried but nothing worked. You can see why if you count the black and white squares.

Every domino, no matter where it is placed, covers one black and one white square of the chessboard. So any arrangement of nonoverlapping dominoes must cover the same number of black squares as white. But the mutilated chessboard has thirty-two black squares and thirty white ones. No matter how you arrange the dominoes, two black squares (at least) must always stay empty.

If instead, two adjacent corners are removed—one black, one white—this argument fails. And in fact, the puzzle can then be completed. But the “parity” argument, counting the numbers of squares of the two colors and comparing them, rules out the version that I posed to begin with. The task is impossible . . . period.

The deeper message behind this puzzle applies throughout mathematics. When a problem leads you to
consider some huge number of possibilities—such as all the different ways the dominoes might be arranged— there is generally no practical way to deal with them one at a time. You must search for some common feature that does not change when you change the arrangement:
an
invariant
.

Here, the first invariant we tried was area. If you rearrange the dominoes, their total area remains the same. But that invariant doesn’t help here. So I resorted to a different invariant: the difference between the number of black squares and the number of white ones. This is always zero in any domino arrangement. So
no
arrangement in which the invariant is nonzero can be obtained by arranging dominoes according to our rules.

The proof leaves open the possibility that some arrangements with zero invariant might not be possible for other reasons. In fact, these do exist; maybe you can find some. The “area” invariant solves some puzzles but not others. So does the “parity” invariant, odd or even.
The same is true of most invariants.

Now we must move on, from puzzles to serious mathematical problems. Remarkably, and delightfully, similar ideas still apply.

Trisecting the angle is a case in point. We now know—we have known it since Gauss’s student Pierre Wantzel wrote down a proof in 1837—that it is impossible to trisect the angle using an unmarked straightedge and compass. That is, there is no geometric construction,
employing the traditional implements in the traditional manner, that will divide any given angle into three exactly equal parts.

There are zillions of approximate constructions. None will be exact. I can say this without the slightest fear of contradiction and without examining any proposed method. We
know
it must contain a mistake. We don’t know where or what the mistake is—and it may be very difficult to find—but we can be certain it’s there.

This, I know, sounds arrogant. It can be very annoying to any would-be trisector. “How can they possibly know this when they haven’t even looked at my proof?”

They know because such a construction has been proved to be impossible. If someone claimed they could run a mile in ten seconds, you wouldn’t have to watch them try it to know that there has to be a trick. Maybe they “run” using rocket assistance. Maybe their “mile” is not measured in the orthodox manner and is no longer than a bus. Maybe something is funny about their clock.
We don’t need to know which to smell a rat.

It’s like that in mathematics, but with a higher degree of certainty.

Very well: how do we know that angle trisection is impossible?

Although the problem concerned belongs to geometry, its resolution belongs to algebra. This is a standard ploy in mathematical research: try transforming your problem into something that is logically equivalent but
lies in a different area of mathematics. If you’re fortunate, the new area will permit the use of new techniques, casting new light on the problem. The idea of replacing geometry with algebra—or the reverse—goes back at least to René Descartes. In an appendix to his
Discours de
la Méthode
of 1637, with the title
La Géometrie
, he outlined the use of coordinates to turn geometric forms into algebraic equations, and back again. Today we call them Cartesian coordinates in his honor.

The idea will be familiar to you, Meg. Any point in the plane can be characterized by two numbers, distances measured in two directions at right angles to each other. Horizontal and vertical, or north/south and east/west. A line, circle, or other curve is just a collection of points, a set of pairs of numbers. Any statement about those lines and curves can be converted to a corresponding statement about numbers, and such statements belong to the realm of algebra. Thus the fact that a circle has a radius of one unit, when converted into algebra using Pythagoras’s theorem, becomes the fact that for any point on its circumference, the square of its horizontal coordinate, added to the square of its vertical coordinate, equals 1. In symbols,
x
²
+
y
²
= 1. This is the equation corresponding to that circle.

Every circle, every straight line, and every curve has a corresponding equation. And the points where, say, a circle meets a line are those pairs of numbers that satisfy both the equation for the circle and the equation for the
line. Instead of drawing lines and curves and finding their intersections points, we can just solve equations. More importantly, instead of
thinking
about drawing lines and curves and finding their intersections points, we can think about solving the corresponding equations. And this is how we can prove that angles cannot be trisected in the manner specified.

Here’s how it goes, stripped of technical details. Any geometrical construction begins with a collection of points and then constructs new points by one of three methods. Either we draw two lines through existing points and find out where those lines meet; or we draw one such line and find where it meets a circle centered on a known point and passing through another known point; or we draw two such circles and see where
they
meet. This small set of moves is a product of our tools: a straightedge makes only straight lines, and a compass makes only circles. Thus we build new points out of old ones, and we continue for a finite number of such moves and then stop.

This is another standard proof technique: break the problem down into the simplest possible parts.

It may appear that an angle does not fit this description. But an angle is specified by two lines that meet at a common point, and those two lines can be specified by the common point, another point on the first line, and another point on the second line. Three points suffice to define an angle. It takes only one further point to specify
an angle one-third the size. But locating that fourth point may in principle require constructing a lot of auxiliary points along the way, and the claim is that none of these will actually help.

To see why, we apply another standard proof technique: examine each of the simplest steps and try to find its essential features.

Geometrically, there are three distinct steps: two lines, line plus circle, two circles. But if we convert those steps into algebra, we find that the first one is equivalent to solving a linear equation and the other two are equivalent to solving a quadratic equation. In a linear equation we are told that some multiple of the unknown, plus some number, is zero. In a quadratic equation we are told that some multiple of the square of the unknown, plus some multiple of the unknown, plus some number, is zero.

Linear equations are “special cases” of quadratic ones: the relevant multiple of the square of the unknown is the multiple by zero. So all three steps boil down to solving quadratic equations.

Methods for solving quadratic equations were known to the Babylonians in 2000 BC, and the basic idea is that you can always do it using square roots. In short, we have replaced “constructible using unmarked straightedge and compass” by “expressible by a sequence of square roots (and other arithmetical operations such as addition and subtraction).” That characterizes all possible points that can arise from geometric constructions.

Suppose that some angle can be trisected using such a construction. Then the coordinates of the corresponding point—the one associated with an angle one-third the size—must be expressible by a sequence of square roots. Is that possible? Well, we know something about that new point, namely, that its coordinates are given by a cubic equation, one that also involves the cube of the unknown. This observation comes from trigonometry, where there is a standard formula relating the sine of an angle to the sine of three times that angle.

The whole shebang, then, reduces to a simple question: given a number that you know is the solution of a cubic equation, is it possible to express that number using only square roots? The intuition is that there is a mismatch here: no sequence of steps involving the number 2 should give rise to the number 3. A close examination of the properties of solutions of equations leads to an invariant, known as the
degree
. This has nothing to do with “degree” as a way to measure angles; it is a whole number specifying the type of equation that is being solved. Simple properties of the degree prove that the only time that you can solve a cubic equation using nothing more elaborate than square roots is when the cubic equation breaks down into a linear equation and a quadratic, or three linear ones.

However, a short calculation shows that with rare exceptions, the cubic equation associated with angle trisection is not like that. It does not break down. In particular,
this is the case when the initial angle is 60º, for example. Therefore the cubic equation cannot be solved
exactly
using only square roots. Indeed, if it could be solved in that manner, then the integer 3 would have to be an even number. But of course it isn’t.

I’ve left out the details—you can find them in many standard algebra texts if you want to see them—but I hope the story is clear. By transforming the geometry into algebra, we can reformulate angle trisection (indeed any construction) as an algebraic question: can some number associated with the desired construction be expressed using square roots? If we know something useful about the number concerned—here, that it is given by a cubic equation—then we may be able to answer the algebraic version of the question. In this case, the algebra rules out any possibility that such a construction exists, thanks to the invariant known as the degree.

It’s not a matter of being clever: however cunning you are, your purported construction will necessarily be inexact. It might be very accurate (sadly, most attempts aren’t; have a look at Underwood Dudley’s
A Budget of
Trisections
), but it cannot be exact. It’s not a matter of finding other methods for trisecting angles, either: those are already known, and that wasn’t the question. I always tell anyone who sends me an attempted trisection that I’m not concerned about their being wrong, and neither should they be. The problem is that if they are
right
,
then a direct consequence of their proof is the fact that 3 is an even number.