Amazing Mathematics Day 29 th June 2007

1. Amazing Mathematics Day 29th June 2007 ‘Solving the Unsolvable?’ Dr David Fisher Department of Mathematics University of Surrey

2. Euclid’s ‘Elements’ Euclid of Alexandria (325 - 265 BC) wrote ‘The Elements’, collecting together the arithmetic and geometry known at the time Euclid’s work was a major influence on European mathematics, and formed the basis of school geometry until quite recently Classical Greek geometry was based on idealised concepts of points and lines. An important aspect was the construction of figures using only a straight-edge and compass, such as bisecting an angle

3. Geometrical constructions • Start with a unit length • Place units end-to-end to get any integer length, e.g. 3 units: • Rational lengths can be constructed using an unmarked ruler and compass only, e.g. to construct a line of length 5/3 : D OA = 1 OB = 3 OD = 5 C O A B Draw AC parallel to BD. Then OC = 5/3

4. Geometrical constructions • AB = x, BC = 1 • How long is BD? • D • A x B 1 C By similar triangles, BD2 = x BD =  x

5. Solving equations x + 2 = 4 can be solved for x in the natural numbers x + 4 = 2 can be solved for x in the integers 2x + 1 = 4 can be solved for x in the rational numbers x2 = 2 can be solved for x in the real numbers x2 + 2 = 0 cannot be solved for x in the real numbers The solutions of all but the last of these equations can be shown as points on a line, the ‘real number line’ : -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

6. Duplicating the cube Now consider the equationx3 = 2 x is the side of a cube whose volume is 2 cubic units 1 x From classical times, people tried to construct the cube root of 2 by straight-edge and compass only (‘duplicating the cube’) This was finally proved impossible in the 19th century So is there a number whose cube is 2?

7. Does 2 have a cube root? • 1.263 = 2.00376, which is quite close to 2 • We can approximate the cube root of 2 as closely as we like • An iterative method can be used: let x1 = 1 and, for n > 1, • This gives the sequence 1, 1.3333, 1.2353, 1.2679, …, • It seems to be converging to 1.259921… • If the sequence converges then, in the limit,

8. Quadratic equations • Classical Greek mathematicians could solve quadratics, but there was no algebraic formulation until about 100 AD • They didn’t believe in negative numbers, so x2 + ax= b was a different type of equation from x2 = ax + b • Solutions were geometrical, e.g. to solve x2 + 2ax = b: • b • ax

9. Quadratic equations • As we do believe in negative numbers, we just need to solve x2 + 2bx + c = 0 • Complete the square: (x + b)2 – b2 + c = 0 • (x + b)2 = b2 – c • x + b = ± (b2 – c) • x = –b ± (b2 – c) • We have solved the quadratic by radicals • Can higher-order equations be solved by radicals?

10. It depends what you mean by “solve” • A cubic equation : ax3 + bx2 + cx + d = 0 • Some cubics are easy : x3 – x = 0 has roots • –1, 0, 1 (where the graph cuts the x-axis) • A cubic equation can have up to three distinct real roots • We can easily find them approximately. A computer algebra program will solve x3 + x2 – 2x – 1 = 0 to give • x = –1.801937736, x = –0.4450418679, x = 1.246979604 • This gives no insight into where the solutions come from C.E.M. JOAD 1891 - 1953

11. Reducing a cubic • To solve the general quadratic we completed the square • Perhaps ‘completing the cube’ will help with the cubic • Solve x3 – 12x2 + 42x – 49 = 0 (*) • Note that (x – 4)3≡x3 – 12x2 + 48x – 64 • so equation (*) is (x – 4)3 – 6x + 15 = 0 • (x – 4)3 – 6(x – 4) – 9 = 0 • y3 – 6y – 9 = 0, where y = x – 4 • In this way we can always get rid of the ‘squared’ term

12. Getting over the next hurdle • Now we need to solve y3 – 6y – 9 = 0 (**) • Suppose we can’t spot a solution by inspection • Split y into two parts: write y = u + v, so • y3 = (u + v)3 = u3 + 3u2v + 3uv2 + v3 • Equation (**) is u3 + v3 – 9 + 3(u2v + uv2 – 2u – 2v) = 0 • Solve these two equations for uandv : u3 + v3 – 9 = 0, u2v + uv2 – 2u – 2v = 0 • This still involves cubes, so is it any easier?

13. We can solve cubics! u2v + uv2 – 2u – 2v = 0 gives (u + v)(uv – 2) = 0, so v = –uorv = 2/u v = –uis not consistent with u3 + v3 – 9 = 0, sov = 2/u u3 + v3 – 9 = 0then gives Multiply through by u3 to get u6–9u3 + 8 = 0 (u3 – 1)(u3 – 8) = 0 u3 = 1 or u3 = 8, so u = 1 or u = 2 y = u + 2/u, so y = 3, so x = 7

14. That wasn’t too bad ... • We can easily check that this is correct, but is it complete? • Equation (*) has only one real root, which we have found • Let’s try another one: solve x3 + 3x2 – 12x – 18 = 0 • Substitute x = y – 1 to get y3 – 15y – 4 = 0 • Put y = u + 5/u to get u6–4u3 + 125 = 0 • Put z = u3soz2–4z + 125 = 0 • (z – 2)2 + 121 = 0 z = 2 ± (–121)

15. Oh no, we can’t solve cubics yet! • So far everything has been routine algebra, but now we are stuck. (–121) is not a real number, yet a graph clearly shows that our equation has three real roots and at least one is a positive integer • This troubled people for a long time! Clearly it was ‘absurd’ to work with square roots of negative numbers, even once negative numbers were accepted • In 16th century Italy, competitions were held to see who could solve previously unsolved problems

16. The cubic formula • Niccolo Fontana discovered this formula for solving x3 + px = q where p and q are positive: • This formula does not seem to find three solutions, even when it’s clear that three exist • The quadratic formula works because 1 has two square roots, 1 and –1 • So perhaps 1 should have three cube roots! What are they?

17. Inventing a new number • The breakthrough comes if we keep going with our example • Suppose (–121) ‘exists’. Then it is 121(–1) or 11(–1) , so we just need (–1) to exist • (–1) is clearly not a ‘real’ number, but is it any more ‘unreal’ than –1 or 2 ? • Define the symbol i by the property i2 = –1 • This ‘imaginary’ i needs to behave itself when it’s combined with real numbers

18. Complex numbers • If a and b are real numbers, an expression of the form a + bi is called a complex number • Combine such expressions by normal algebra, but remember that i2 = – 1 • (2 + 3i) + (5 – 7i) = 7 – 4i • (2 + 3i)(5 – 7i) = 10 – 14i + 15i – 21i2 = 31 + i • It works! The complex numbers form a consistent arithmetical system which includes the real numbers

19. Continuing with the cubic We had z = 2 ± 11i where z = u3 Let u = a + bi , so (a + bi)3 = 2 ± 11i Multiplying out, and comparing real and imaginary parts, we find that a = 2, b = ± 1 i.e. (2 + i)3 = 2 + 11i and(2 – i)3 = 2 – 11i Notice that(2 + i)(2 – i)= 5 y = u + 5/uso y = (2 + i) + (2 – i)= 4 Finallyx = y– 1, sox = 3

20. Amazing Maths! • We have gone through complex numbers and come out with a real answer. Perseverance has paid off! • BUT we still only have one root of the cubic. Where are the other two roots? • The answer lies in a geometric way of looking at complex numbers • The expression a + bi contains two real numbers a and b • Plot (a, b) as a point on a graph (an `Argand diagram’) • i = 0 + 1i corresponds to the point (0, 1)

21. From algebra to geometry and back • Going from 1 to i(multiplying by i once) corresponds to a 90º rotation about (0, 0) • Going from 1 to –1(multiplying by i twice) corresponds to a 180º rotation about (0, 0) • Two successive 180º rotations about (0, 0) take us from 1 to 1, corresponding to the fact that (–1)2 = 1 • If multiplying by something three times in succession takes us from 1 to 1, that thing is a cube root of 1 • Going from 1 to 1 involves a 360º rotation about (0, 0). One third of this is a 120º rotation

22. Meet the cube roots of 1 The point 1/3 of the way round the unit circle from 1 to 1 is 1 60o As a complex number this is

23. 1 2 There are three cube roots of 1 We call this complex number  (omega)  =  It is easy to show that 3= 1 Also (2)3= 1 and 2= The three cube roots of 1 are 1,  and 2

24. Completing the first example Solving x3 – 12x2 + 42x – 49 = 0, we used the substitutions x = y + 4 and y = u + 2/u, and found u3 = 1 It now follows that u =1,  or 2 so y = 3, y =  + 22 or y =2 + 2 Thusx=7, x = orx = It IS possible to solve cubics by radicals, but it’s not always quite so easy!

25. Why stop there? • What about quartic equations? • They’re not too bad. By a suitable substitution we can always get rid of the ‘cubed’ term • Example: solve x4 – 2x2 + 8x – 8 = 0 • Suppose it has two quadratic factors • (x2 + kx + m)(x2 – kx + n) = 0 • Expand and compare coefficients to find k, m, n • (x2 + 2 x + 22)(x2 – 2x + 22) = 0 • Solve two quadratics to get the four roots

26. Again, why stop there? • Methods for solving quadratic, cubic and quartic equations by radicals were known by the 17th century AD • ‘By radicals’ means starting with the coefficients and using only addition, subtraction, multiplication, division and taking roots (square roots, cube roots, etc), including complex roots • Quintic (fifth degree) equations resisted all attempts to solve them by radicals • In 1824 the Norwegian mathematician Niels Abel proved that there is no general formula for solving a quintic by radicals • However, clearly SOME quintics are solvable by radicals, e.g. the solutions of x5 = 1 are just the five 5th roots of 1

27. Look for patterns! • To solve the quartic equation x4 – 2x2 + 8x – 8 = 0 we factorised it as (x2 + 2 x + 22)(x2 – 2x + 22) = 0 • The four solutions are • x = • There is some symmetry to these solutions • Roughly speaking, a symmetry of the roots of an equation is a way of swapping them round (a permutation) so that if an equation with integer coefficients is satisfied by some of the roots, it is still satisfied after the roots are permuted

28. A very simple example The quadratic equation x2 – 4x + 2 = 0 has roots x1 = 2 + 2andx2 = 2 – 2 Any polynomial equation satisfied by x1 is also satisfied by x2 Swapping x1 and x2 is a symmetry of the roots. Call it s Not swapping them is clearly also a symmetry. Call it n The table shows how n and s combine when one is followed by another

29. A brief introduction to group theory • In the early 19th century, mathematicians were starting to study groups of permutations • A group is not just a set of objects. It also has a structure • Sets with structure are what abstract algebraists study • A group has an operation defined on it (e.g. addition, multiplication, composition of functions) such that if you combine two elements of the group using this operation, you get an element of the group. Also: • There is an identity element (0 for +, 1 for x) • Every element has an inverse (–a for +, 1/a for x) • The operation is associative: a(bc) = (ab)c

30. To cut a long story short ... • The symmetries of the roots of an equation form a group • Does this group give us any information about the roots of the equation? • The answer turns out to be ‘yes’! • A polynomial equation is solvable by radicals if, and only if, the group of symmetries of its roots is a ‘solvable’ group • ‘Solvable’ means that the group splits up into smaller pieces in a particular way • The general quintic (and higher order) equation does not have a solvable group, so it is not solvable by radicals

31. Do you know this man? • Evariste Galois, born Paris 1811 • Not a great success at school! Grew bored and rebellious • Read books on Maths and tried to do original work, but it was disorganised and not appreciated • Got involved in republican politics • Challenged to a dual • Died 2nd June 1832, aged 20

32. Galois’s legacy • Galois’s work was eventually appreciated as a true work of genius, laying the foundations of modern pure mathematics • He explained the connection between symmetry groups and solvability of equations. This topic is studied at university under the name of ‘Galois Theory’ • In doing this he made big advances in group theory, which has since been used to analyse symmetry in geometrical figures, crystals, atomic particles, etc

33. Three impossible things • Work on solvability of equations led to some significant geometric results • Since classical times, people had been trying to carry out certain constructions by ruler and compass only: • Duplicating the cube • Squaring the circle • Trisecting an angle • It has been proved that all of these are impossible

34. Impossible constructions These can’t be done by ruler and compasses only: If you think you’ve found a way of doing any of these, you are wrong. But who knows what other amazing things you might discover by trying?

35. For information about Maths degrees at Surrey please visit our website www.maths.surrey.ac.uk