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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 Department of Mathematics Geometrical constructions • AB = x, BC = 1 • How long is BD? By similar triangles, D AB BD BD BC x BD BD 1 BD2 = x BD = x A x B Department of Mathematics 1 C 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 can be solved for x in the complex 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 Department of Mathematics 4 5 6 7 Duplicating the cube Now consider the equation x3 = 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 Department of Mathematics 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 a x Department of Mathematics 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? Department of Mathematics Cubic equations • ax3 + bx2 + cx + d = 0 • Some cubics are easy to solve. 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 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 Department of Mathematics 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 Department of Mathematics 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 u and v : u3 + v3 – 9 = 0, u2v + uv2 – 2u – 2v = 0 This still involves cubes, so is it any easier? Department of Mathematics We can solve cubics! u2v + uv2 – 2u – 2v = 0 gives (u + v)(uv – 2) = 0, so v = –u or v = 2/u v = –u is not consistent with u3 + v3 – 9 = 0, so v = 2/u u3 + v3 8 – 9 = 0 then gives u 3 9 0 u 3 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 Department of Mathematics 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 Put z = u3 so (z – 2)2 + 121 = 0 u6 – 4u3 + 125 = 0 z2 – 4z + 125 = 0 z = 2 ± (–121) = 2 ± 11i Department of Mathematics Continuing with the cubic z = 2 ± 11i where z = u3 Let u = a + bi , so (a + bi)3 = 2 ± 11i Now (2 + i)3 = 2 + 11i and (2 – i)3 = 2 – 11i Also (2 + i) (2 – i) = 5 y = u + 5/u so y = (2 + i) + (2 – i) = 4 Finally x = y – 1, so x = 3 To get a real solution we had to go via complex numbers. What about the other two real roots? Department of Mathematics The cubic formula • Niccolo Fontana discovered this formula for solving x3 + px = q where p and q are positive: x3 q p3 q 2 3 q p3 q 2 2 27 4 2 27 4 • 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? Department of Mathematics 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 Department of Mathematics The cube roots of 1 The point 1/3 of the way round the unit circle from 1 to 1 is cos 60 , sin 60 o o 1 3 , 2 2 As a complex number this is 1 3 i 2 2 Department of Mathematics 1 60o There are three cube roots of 1 We call this complex number (omega) 1 3 i = 2 2 It is easy to show that 3 = 1 Also (2)3 = 1 and 2 = 1 ω 1 2 The three cube roots of 1 are 1, and 2 Department of Mathematics 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 = + 22 or y = 2 + 2 Thus x = 7, x = 5 3 5 3 or x = i i 2 2 2 2 It IS possible to solve cubics by radicals, but it’s not always quite so easy! Department of Mathematics 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 + 22)(x2 – 2x + 22) = 0 Solve two quadratics to get the four roots Department of Mathematics 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 Department of Mathematics Look for patterns! • To solve the quartic equation x4 – 2x2 + 8x – 8 = 0 we factorised it as (x2 + 2 x + 22)(x2 – 2x + 22) = 0 • The four solutions are x= 2 28 2 , 2 2 2 8 2 2 • 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 the roots, it is still satisfied after the roots are permuted Department of Mathematics A very simple example The quadratic equation x2 – 4x + 2 = 0 has roots x1 = 2 + 2 and x2 = 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 Department of Mathematics n s n n s s s n 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 Department of Mathematics 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 Department of Mathematics 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 this was disorganised and not appreciated • Got involved in republican politics • Challenged to a duel • Died 2nd June 1832, aged 20 Department of Mathematics 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 now known by 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 • Work on solvability of equations led to some significant geometric results, e.g. impossible to trisect an angle by ruler and compass, or `square the circle’. Department of Mathematics Explore further • There is an article on Galois Theory on the NRICH website http://nrich.maths.org (Type ‘Galois’ in the search box) Department of Mathematics