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

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Department of Mathematics AB = x, BC = 1 How long is BD? D A x B 1 C Geometrical constructions BD 2 = x BD = x By similar triangles,

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Department of Mathematics 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 x 2 = 2 can be solved for x in the real numbers x 2 + 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’ : Solving equations -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

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Department of Mathematics Now consider the equation x 3 = 2 x is the side of a cube whose volume is 2 cubic units 1 x Duplicating the cube 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

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

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

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Department of Mathematics ax 3 + bx 2 + cx + d = 0 Some cubics are easy to solve. x 3 – 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 x 3 + x 2 – 2x – 1 = 0 to give x = –1.801937736, x = –0.4450418679, x = 1.246979604 This gives no insight into where the solutions come from Cubic equations

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Department of Mathematics To solve the general quadratic we completed the square Perhaps ‘completing the cube’ will help with the cubic Solve x 3 – 12x 2 + 42x – 49 = 0 (*) Note that (x – 4) 3 ≡ x 3 – 12x 2 + 48x – 64 so equation (*) is (x – 4) 3 – 6x + 15 = 0 (x – 4) 3 – 6(x – 4) – 9 = 0 y 3 – 6y – 9 = 0, where y = x – 4 In this way we can always get rid of the ‘squared’ term Reducing a cubic

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Department of Mathematics Now we need to solve y 3 – 6y – 9 = 0 (**) Suppose we can’t spot a solution by inspection Split y into two parts: write y = u + v, so y 3 = (u + v) 3 = u 3 + 3u 2 v + 3uv 2 + v 3 Equation (**) is u 3 + v 3 – 9 + 3(u 2 v + uv 2 – 2u – 2v) = 0 Solve these two equations for u and v : u 3 + v 3 – 9 = 0, u 2 v + uv 2 – 2u – 2v = 0 This still involves cubes, so is it any easier? Getting over the next hurdle

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Department of Mathematics u 2 v + uv 2 – 2u – 2v = 0 gives (u + v)(uv – 2) = 0, so v = –u or v = 2/u v = –u is not consistent with u 3 + v 3 – 9 = 0, so v = 2/u u 3 + v 3 – 9 = 0 then gives Multiply through by u 3 to get u 6 – 9u 3 + 8 = 0 (u 3 – 1)(u 3 – 8) = 0 u 3 = 1 or u 3 = 8, so u = 1 or u = 2 y = u + 2/u, so y = 3, so x = 7 We can solve cubics!

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Department of Mathematics 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 x 3 + 3x 2 – 12x – 18 = 0 Substitute x = y – 1 to get y 3 – 15y – 4 = 0 Put y = u + 5/u to get u 6 – 4u 3 + 125 = 0 Put z = u 3 so z 2 – 4z + 125 = 0 (z – 2) 2 + 121 = 0z = 2 ± (–121) = 2 ± 11i That wasn’t too bad...

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Department of Mathematics z = 2 ± 11i where z = u 3 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? Continuing with the cubic

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Department of Mathematics Niccolo Fontana discovered this formula for solving x 3 + px = q where p and q are positive: The cubic formula 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?

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Department of Mathematics Going from 1 to i (multiplying by i once) corresponds to a 90 º rotation about (0, 0) Going from 1 to – 1 (multiplying by i twice) corresponds to a 180 º rotation about (0, 0) Two successive 180 º 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 360 º rotation about (0, 0). One third of this is a 120 º rotation From algebra to geometry and back

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

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Department of Mathematics We call this complex number (omega) = There are three cube roots of 1 It is easy to show that 3 = 1 Also ( 2 ) 3 = 1 and 2 = 1 22 The three cube roots of 1 are 1, and 2

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

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Department of Mathematics What about quartic equations? They’re not too bad. By a suitable substitution we can always get rid of the ‘cubed’ term Example: solve x 4 – 2x 2 + 8x – 8 = 0 Suppose it has two quadratic factors (x 2 + kx + m)(x 2 – kx + n) = 0 Expand and compare coefficients to find k, m, n (x 2 + 2 x + 2 2)(x 2 – 2x + 2 2) = 0 Solve two quadratics to get the four roots Why stop there?

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Department of Mathematics 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 x 5 = 1 are just the five 5th roots of 1 Again, why stop there?

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Department of Mathematics To solve the quartic equation x 4 – 2x 2 + 8x – 8 = 0 we factorised it as (x 2 + 2 x + 2 2)(x 2 – 2x + 2 2) = 0 The four solutions are x = Look for patterns! 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

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Department of Mathematics The quadratic equation x 2 – 4x + 2 = 0 has roots x 1 = 2 + 2 and x 2 = 2 – 2 Any polynomial equation satisfied by x 1 is also satisfied by x 2 Swapping x 1 and x 2 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 A very simple example

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Department of Mathematics 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 A brief introduction to group theory

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Department of Mathematics 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 To cut a long story short...

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Department of Mathematics 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 Do you know this man?

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Department of Mathematics 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’. Galois’s legacy

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Department of Mathematics There is an article on Galois Theory on the NRICH website http://nrich.maths.org (Type ‘Galois’ in the search box) Explore further

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