Games and Mathematics

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Games and Mathematics Page 20

by Wells, David


  Finally, if we understand its structure, it is often possible to transform a problem into a much easier problem, a theme that we shall illustrate by geometrical inversion.

  Pythagoras' theorem

  A good proof makes us wiser.

  [Manin 1979: 18]

  Figure 17.1 Empty space

  Figure 17.1 is a picture of some empty space. It is meant to be flat as it would be if the book were closed. Not much to look at, is there? So let's add some detail, a set of equidistant parallel lines (Figure 17.2).

  Figure 17.2 A sequence of equi-distant parallel lines

  This is still not obviously fascinating, so in Figure 17.3 we have added a second identical set of parallel lines, perpendicular to the first, creating a grid of squares which could, in theory, continue off-page as far as we choose.

  Figure 17.3 Two sets of equidistant parallel lines forming squares

  Squares are familiar objects, and there are many things we could do with them: what we actually do is to slide them so that each strip moves the same distance past its neighbours (Figure 17.4).

  Figure 17.4 Figure 17.3 with the squares shifted sideways

  Next, we slide the squares in the perpendicular direction, by the same distance, creating a pattern full of ‘square holes’ (Figure 17.5).

  Figure 17.5 A pattern of square holes

  This has a simple kind of symmetry. The squares march off the edge of the diagram in two directions at right-angles which we have highlighted by choosing some suitable points, the centres of the large squares, and joining them (Figure 17.6).

  Figure 17.6 9 Offset squares with their centres joined

  In a strange way we have got back to our first pattern of squares covering the plane, except these new squares are larger and tilted. In fact there is one large new square for each original-square-plus-square-hole. From this observation we deduce that the area of each large square equals the sum of the areas of a pair of smaller squares (Figure 17.7).

  Figure 17.7 Part of Figure 17–6 blown up

  Look again, and these five pieces can be fitted together to make either one large square or two smaller squares, so we have stumbled across a dissection which proves our point.

  We can also look at these squares in another way. If we draw the triangle in Figure 17.8, the three sizes of squares fit round it exactly.

  Figure 17.8 Dissection of large square into two smaller squares

  Aha! The theorem of Pythagoras which says that in any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides. By a few simple transformations plus, of course, looking we have discovered Pythagoras' theorem and a way of proving it. Or have we? On our way we made a rather basic assumption, as often happens when you are relying on experiment and observation. We assumed that we could cover a flat plane, represented by part of the page of this book, by a set of equidistant parallel lines, but this only works for a ‘flat’ surface and not, for example, on a sphere. We can indeed divide the surface of a sphere by two set of lines, lines of longitude and latitude, but the lines of longitude are not parallel to each other (they meet at the North and South poles) and the lines of latitude are not actually straight lines on the sphere, because particular lines of latitude are not the shortest distances between any two points on them. (The equator is the only exception.)

  It seems that instead of saying that on a plane surface, Pythagoras' theorem is true and we've proved it, it would be more accurate to say that if Pythagoras's theorem is true, all over a surface, then that surface is a plane! Instead of assuming that we know what a plane surface is – it's a kind of all-over-flat thing – we ought to admit that Pythagoras' theorem is a test for whether a surface is indeed a plane.

  Does this mean that we can't actually prove Pythagoras? That can't be right, because it's proved in school textbooks! Indeed, but only by assuming some alternative defining properties of a plane as found in Euclid. Using such basic properties, there are many proofs, so many that Elijah S. Loomis was able to publish 367 of them in his book The Pythagorean Proposition (1940). Since then many more have been discovered. Here is an old proof by Leonardo da Vinci (Figure 17.9).

  Figure 17.9 Leonardo's proof of Pythagoras

  Leonardo takes a standard figure, repeats the original triangle at the bottom, and joins the two top vertices to make a third copy of the original triangle. He then draws the two thin lines, which are perpendicular, creating four identical quadrilaterals: one pair makes two original triangles and the large square; the other makes two original triangles and the two smaller squares.

  But how illuminating is this proof? Drawing the two dotted lines looks like a trick which makes Pythagoras's theorem looks like a trick too, an idiosyncratic result that has no deeper significance. How misleading! The existence of hundreds of proofs of Pythagoras is rather a sign that it is very deep, though we can guarantee that few of them will explain why. They convince, but they do not illuminate.

  Some proofs confirm what we suspected, but do not show why in any deeper sense, because the question, ‘Why?’ is ambiguous. Are we asking for a logically convincing proof or an illuminating proof that highlights the structure of the situation? Many proofs are the former, the best proofs are both, and some arguments are illuminating but not entirely convincing so you look for ways to tighten them up.

  The artist John Constable wrote that, ‘We see nothing till we truly understand it.’ When you understand Pythagoras as a representing the deep structure of the geometrical plane, the diagram looks different and the theorem itself appears entirely obvious.

  Euclidean coordinate geometry

  Plane geometry is so rich because positions can vary independently in two dimensions. It was René Descartes (1596–1650) who explained that all the points of the plane could be represented by measurements along two independent axes, the coordinates of the point. By choosing a specific relationship between the coordinates, the point would move in a straight line, or in a circular motion, or along a parabola. Figure 17.10 shows the parabola 4y = x2.

  Figure 17.10 Parabola with points marked

  The focus is at (0, 1) and the directrix is the line y = −1. Any point on the parabola is equidistant from each. For example, the point, (1, 1/4) is 5/4 from each, as we find by Pythagoras (Figure 17.11).

  Figure 17.11 Parabola, points marked and focus-directrix property

  When the plane is covered by a grid of unit squares, Pythagoras' theorem becomes the statement that the square of the distance between two points (a,b) and (c,d) is (a−c)2+(b−d)2 and so the distance PQ = . This is neat but also rather complicated – it includes squares and square roots – which is surprising given how simple the theorem seemed at the end of the last chapter. It suggests that there may be occasions when we should try to avoid using Pythagoras, fundamental thought it is, when using coordinate geometry.

  That conclusion has an element of truth, which emphasises that the type of geometry you use ought to be chosen to fit the problem. Let's look at two examples. The first shows the coordinate approach working well, despite first impressions. The second show it working very well indeed on a problem that involves lengths but does not use Pythagoras.

  For our first example, let's find the circumcentre of a triangle, the point that is equidistant from all three vertices.

  Figure 17.12 Triangle, vertices marked and circumcentre

  If the circumcentre of the triangle in Figure 17.12 is (a,b) then by Pythagoras's theorem in coordinates we have the equations,

  and

  This is only superficially complicated: when we square each equation and simplify and we get the equations of two straight lines:

  These are the perpendicular bisectors of two of the sides (so we could have found them without using Pythagoras at all, Figure 17.13) and their intersection happens to lie on the third side because this triangle is right-angled.

  Figure 17.13 The same triangle with two perpendicular bisectors marked

  In
the next case Pythagoras simply can't be used, but we still get a result.

  The average of two points

  The coordinates of the mid-point of the line joining two points is the arithmetical ‘average’ of their coordinates, and also the centre of gravity of the two points, regarded as identical physical particles.

  Figure 17.14 Triangle, vertices and mid-points marked

  In Figure 17.14 the ‘average’ of A(2,3) and B(4,7) is (3,5) because 3 is the average of 2 and 4 and 5 is the average of 3 and 7. The three mid-points form a new triangle (Figure 17.15). We can now find the mid- points of the sides of that triangle, and then repeat, again and again. The second set of mid-points, by a simple calculation, is (4, 3), (5, 4) and (6, 3).

  Figure 17.15 The same with mid-points of mid-points marked

  We are clearly closing in on ‘a centre’ of the triangle, but which of the centres will it be, and what are its coordinates? By finding averages repeatedly, we only get what is called a linear function of the numbers we started with, meaning no squares, cubes, square roots, and so on. Also, our calculations are symmetrical, so the result must depend symmetrically on the original three points, but the only symmetrical linear function of A, B and C is their sum, A + B + C or a fraction or multiple of it. In fact, the limit of all these triangles is 1/3(A + B + C) which, calculating for each coordinate separately, is (16/3, 11/3). This is, as we might have guessed, the centre of gravity of the vertices of all three triangles.

  The skew quadrilateral

  The neat idea of ‘averaging’ two points offers an extra advantage. The mid-point of A and B is (A + B) in any number of dimensions, so if we draw a skew quadrilateral (Figure 17.16) – meaning that the four vertices and four edges do not all lie in one plane, then we can still conclude that the lines joining the mid-points of two pairs of sides will have a common mid-point, a fact which is certainly not immediately obvious.

  Figure 17.16 Skew quadrilateral with two pairs of mid-points joined

  Indeed it's not obvious that the mid-point lines even meet at all – most lines in three-dimensional space certainly do not. What is more, if we add two more edges to the skew quadrilateral, to make a tetrahedron (Figure 17.17) then we can conclude that all three lines joining the mid-points of the opposite edges of a tetrahedron, meet, and this is so in any number of dimensions.

  Figure 17.17 The same but all three pairs of mid-points joined

  We can go further. We need not limit ourselves to merely averaging two points, we can also calculate weighted averages (Figure 17.18).

  Figure 17.18 Line, with one of the one-third points marked X

  X is one third of the way from A to B, meaning it is twice as near to A as it is to B, so its coordinates will be the weighted average, (2A + B)/3 or (4,2). Suppose we take now take weighted averages of two pairs of the opposite sides of another skew quadrilateral, and join them (Figure 17.19).

  Figure 17.19 Skew quadrilateral with sides divided into 10 parts

  We have joined two pairs of points: (7A+3B)/10 to (7C+3D)/10 and (7C+3B)/10 to (7A+3D)/10. These two chords meet in space at their shared mid-point, (7A + 3B + 7C + 3D)/20.

  This may seem less than surprising, because there is a symmetry in our choice of the ends of the chords, but it so happens that two chords such as (2A+B)/3 to (2C+D)/3 and (3B+4C)/7 to (3D+4A)/7 will also meet in space, although they do not bisect each other. The fact is that ALL the chords in one set intersect ALL the chords in the other, forming a ruled surface which is a hyperbolic paraboloid (Figure 17.20).

  Figure 17.20 Skew quadrilateral forms hyperbolic paraboloid

  This surface has several curious properties, one more curious than most: since a straight line is the shortest distance between two points and this surface consists of straight lines, you might think that it would be the minimal surface spanning the original skew quadrilateral, but it isn't, as Hermann Schwarz discovered in 1865. However, if the skew quadrilateral is chosen so that its edges are four diagonals of a cube, then the ratio of the areas of the surfaces is very close to 1, about 1.0012 [Dalpé 1998: 6].

  Pythagoras is a deep and beautiful theorem but these weighted averages without Pythagoras are pretty smooth too, provided you only try to use them for appropriate problems. That's the secret! Every problem has a structure behind it, at least that's what mathematicians assume. Find the correct structure and the problem turns out to be relatively easy, choose an inappropriate structure and it becomes difficult – and fail to find any structure and it is impossible, or can only be solved by checking.

  Much mathematical research consists of looking for the structures hidden behind problem situations. This is often an extremely difficult challenge because almost by definition you cannot afford to think only in familiar terms. You have to think new, deep thoughts, for which you need that old standby, imagination.

  18 Hidden structure, common structure

  The primes and the lucky numbers

  Gauss as a young teenager, maybe 14 or 15, (he recorded his experience many years later) conjectured on the basis of experimental calculation that the number of primes less than n is approximately n/log n. Since the primes are so weird anyway, we can hardly be surprised that this is an unusual formula but we may still wonder if it occurs elsewhere because any such analogy might help us to understand the prime numbers better.

  Around about 1955 Stanislav Ulam (1909–1984) took an ingenious step in that direction by constructing a sequence by a process that is similar to the sieve of Erastosthenes, but different. The numbers that his technique generated, which he called the lucky numbers because they luckily escape being eliminated by the sieve, behave – surprise, surprise – much like the primes, in certain important respects.

  We start with the sequence of positive integers and cross out all the even numbers, leaving the odd. The second integer remaining is 3, so cross out every third number to leave, 1

  3

  7

  9

  1

  3

  15

  19

  21

  …

  The third integer remaining is 7, so next cross out every seventh number, starting with 19. The fourth number left is 9, so cross out every ninth number, starting with 27, and so on. Hence the sequence of lucky numbers starts, 1

  3

  7

  9

  13

  15

  21

  25

  31

  33

  37

  43

  49

  51

  63

  67

  69

  73

  75

  79

  87

  93

  99

  105

  111

  115

  127

  129

  133

  135

  141

  151

  …

  These are the lucky numbers and they share many properties with the primes. For a start, there are roughly similar numbers of each:

  primes

  lucky numbers

  < 100

  25

  23

  < 1000

  168

  153

  < 10,000

  1,229

  1,118

  < 100,000

  9,592

  8,772

  [Schneider 2002: Lucky Numbers].

  The primes and luckies also become scarcer as they get larger, at about the same rate, so the number of luckies less than n is also approximately n/log n.

  There are also twin lucky numbers matching the twin primes: the first few twin luckies are: 1–3, 7–9, 13–15, 31–33, 49–51, 67–69, 73–75…and there are 33 twin luckies under 1000 compared to 35 twin primes.

  The Lucky Goldbach Conjecture may well be true also. Computer checks by Walter Schneider up to 1010 show that every even number to that limit is the sum of two luckies.

  There are also roughly equal nu
mbers of luckies of the forms 4n + 1 and 4n + 3, as there are for primes, but there are none of the form 3n + 2, all of which are eliminated on the second stage of the Ulam sieve.

  Are there more sequences, also created by sieves, that share these same properties? Mathematicians would like to know, because whatever primes and luckies have in common cannot be at the core of the unique nature of prime numbers.

  Objects hidden behind a veil

  Here are two quadratic functions, x2 + 10 − 7x and x2 + 10 − 6x. The first has real zeros 2 and 5 (Figure 18.1) but a graph of the second does not cross the x-axis (Figure 18.2) and seems to have no geometrical roots.

 

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