12.2 Existence proofs: examples

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1 TOPIC 1: IMPOSSIBILITY AND EXISTENCE PROOFS Ref Sc american /gardner, 1.1 Mathematics is perhaps unique in that one may prove that certain situations are impossible. For example, it is not possible, using the usual ruler and compass constructions, to trisect an angle of arbitrary size. When such a statement is made, the layman may interpret this as meaning that no one has yet succeeded in doing it, that the method has, so far, not been discovered. Perhaps one day it will be? And so there are still those who hope to secure fame, and perhaps fortune, by being the first to 'crack' this problem which mathematicians, 'despite their best endeavours', have found impossible. What is not appreciated is that a solution is logically impossible, in the same way that none of the numbers in the range to 1000 is a perfect square - for is 100 and 101 is 1001, so there is 'no room' for a whole number whose square lies in this range. The activity sheet provides some accessible problems with which students may be expected to make some progress, perhaps with a little help. The discussion sheet Proofs of Impossibility deals with four classical impossibility problems, which, except for the first, are at a harder level. Some notes on the activity sheet problems are provided below. 1. Existence proofs: examples 1 The number of primes is infinite (Euclid) {This was discussed in the topic Valid Proofs } Proof Suppose not, and let,, 5, 7,...p L be a list of primes of which p L is the largest. Consider the number N = p L + 1. N is not divisible by any of,, 5, 7,...p L, since in each case, the remainder is 1. So either N is prime, or N is composite but with a prime factor not in our list. In either case we have shown the existence of a prime not in our list. Contradiction. Note:. + 1 = = = = 11 All these are prime, but the next in the sequence is an instance where the second possibility occurs: = 001 = The number of 4n+ primes is infinite.[ See Baxandall, et al, Proof in mathematics, 00, p67] a) If 4N+ is composite, it has at least one 4k+ factor Proof Suppose not. Then all its factors are of type 4k+1. But (4k+1)(4l+1) = 16 kl + 4l + 4k + 1 = 4Z+1, so no product of 4k+1 factors alone can produce a number of type 4N+. Contradiction. Note: Since (4k+)(4l+) is of form 4Z+1, there could be 1,, 5, 7,.. factors of type 4k+. b) There exist an infinite number of primes of type 4N+. Suppose not - let the list be, 7, 11, 19,, 1, 4,...p L, where p L is the largest prime of type 4k+. Consider N = 4( p L ) +. Note: we do not include, for then N would be M(); it makes no difference whether, 5, 1, 17,.. are included or not. Now N is not divisible by 7, 11, 19,,...p L (since in each case, the remainder is ). So, either N is prime (and is a 'new' 4N+ prime) or N is composite, and so by (a) above, has a prime factor of type 4k+ (applying (a) repeatedly, if necessary). In either case, we have found a prime of type 4k+ not already in our list. Contradiction. 1

2 (Note: 4( )+ = 5855 is clearly M(5), and so is not prime, but its other factor is 1171 which is a 4k+ prime - we use (a) above, which guarantees the existence of such a factor. One the other hand, = 11 is prime, as is = 41779) (See Book II for a proof that the number of 4k+1 primes is infinite.). (See Proof p67) Prove F i, F j (Fermat numbers) have no common factor if i j. Hint: express (F i -1) in terms of (F j -1), where i>j. Hence show the number of primes is infinite. Proof: Let F i j i -1 = x =, F j -1 = y =. Since i>j we have x>y; indeed, since i = j. (i-j), we have: i j i j j i j i j. x = = = ) = y, and, since i-j is even, i>j, we can write ( x = y k, (where k = i-j ). [Example: x= (i=5), y= 4 (j=), x = y 8 = y But they are both odd, hence there is no common factor. [The following version is an illustration of how a proof can be made more transparent by a suitable choice of notation]: Let F n = N + 1 where N = n Let F k = K + 1 where K = k Let K/N = Q = q, which is even if k>n (and so K>N) [In fact, Q is k-n, but we do not need this - only that Q is even] Then K = NQ = ( N ) Q (F k -1) = (F n -1) Q, where Q is even So F k - 1 = M(F n ) + 1 F k = M(F n ) +, so that F k, F n have no common factor except possibly. Another (easier) proof due to Polya: Let m = n. F n+1 - = m - 1 = ( m +1)( m -1) = F n (F n -) = F n F n-1 F n-..(f 1 -), with F 1 =5, so the only possible common factor is.] Now consider F 1, F, F,...F N and let each of these be expressed as a product of primes. Since they have no common factor, there must be at least N different primes involved. But N may be as large as we please. Note that all these proofs are non-constructive - they prove that certain prime numbers must exist, but they do not tell us explicitly what their values are. By way of contrast, the next example gives a constructive proof. 4. ('Proof'p67) Show that it is possible to find a set of 100 consecutive numbers, none of which is prime. Proof: The numbers 101! +, 101! +,...101! +101 are all composite. 5. The Handshakes theorem: In a group of 6 people there exists either a set of three, none of whom has shaken hands with the other two, or a set of three each of whom has shaken hands with the other two. Proof: Represent the 6 individuals by a set of 6 points. Join points P, Q by a red line if the individuals they represent have shaken hands, and by a blue line if they have not. Then there is either a blue triangle, or a red triangle, in the resulting configuration. - here, = i-j] So (F i -1) = (F j -1) k = M(F j ) +1, and any common factor of F i, F j is a factor of.

3 For consider any point P. There must emanate from P either red lines or blue lines (there are 5 in all). Suppose (without loss of generality) that there are red lines; the points at the other ends of each line form a triangle. If it is a triangle of blue lines, the theorem is true; if not, there is one red line in this triangle, which, with two of those emanating from P, makes a red triangle. So in either case, the theorem is true. (In fact, it can be proved that there are two such monochromatic triangles). [A similar theorem states that in a set of 18 people there is always a set of 4 all of whom have shaken hands with each other, or none of whom have]. 6. {For other existence proofs see Brouwer's Fixed Point theorem (Proof p64) and Gauss' Fundamental Theorem of Algebra (Proof p4). See also the Activities, Topic 7, on rational and irrational numbers.} 1. Proofs of Impossibility Activity Sheet 1. Cover a 4 4 chessboard with 8 dominoes, and 8 8 chessboard with dominoes. A 4 board has two opposite corners removed. Cover it with 5 dominoes? A 4 4 board has two opposite corners removed. Cover it with 7 dominoes? An 8 8 board with opposite corners removed cover with 1 dominoes? A 5 5 with opposite corners removed?. Which of these diagrams can be drawn without lifting the pencil from the paper and without tracing any arcs twice?. Three houses (1,, ) have to be joined up to three utilities (gas, water, electricity, denoted by A, B, C) in such a way that the trenches never cross (to avoid future mishaps). Is this possible? A B C 1 4. Are there whole numbers x, y such that x + y = 19? Are there whole numbers x, y, z such that x +y +z =? Are there whole numbers x, y such that x + y = 7? Are there whole numbers x, y such that x 4 + y 4 = 10? 5. Draw a map which cannot be coloured in colours. In colours. In 4 colours.

4 6. If m + 1 is prime, m>1, prove that m must be a power of. If m 1 is prime, prove that m must be prime 7. Proof that is irrational (See Proof, p64) B A 1 8. Geometric proof that is irrational see diagram (Fold AG on to AF) AB=1, AC =, CF=EF = 1 CE = ( -1)=-. So if AB, AC can be measured in terms of a common unit, CF, CE can be measured in terms of the same unit But the square on CF is smaller than that on AB. C - D F - 1 E G Similar constructions show (hexagon) and = ( 5-1)/ (pentagon) are irrational. 9. If a polynomial equation a 0 x n + a 1 x n a n-1 x + a n = 0 has a rational root x = p/q, then q is a factor of a 0 and p is a factor of a n. 10. e is irrational (Proof, p9) 11. Consider families of (primitive) Pythagorean triples eg. (,4,5), (5,1,1), (7,4,5) Difference 1 (8,15,17), (1,5,7) Difference. What are the possible values for (hypotenuse) (second-largest side)? 1. x n + y n = z n has no solution in integers, n> (Fermat, Wiles). Investigate negative integer n. 1.4 Proofs of Impossibility Discussion Sheet l. Königsberg bridges Each 'visit' to a junction uses up just of the paths going there. If every path is traced, each junction, except start and finish points, must have an even number of paths. 5. Classical problems of geometry. Each of these tasks is equivalent to solving an equation. a) Trisect angle: 4x -x = c (=0.5, say) b) Duplicate cube: x = c) Square circle:. with equal areas 4

5 Ruler and compass constructions can be shown equivalent to solution of quadratic equations. For (a), (b) (i) Expressions of the form may be shown to be roots of irreducible equations of degree m, for some whole number m. (ii) Both equations are irreducible, and of degree m For c): Very difficult theorem " is transcendental". ie. not the root of any algebraic equation (Lindemann, l88). So all three problems are impossible.. a) Formula for Quadratic equations x = (-B (B -4AC))/(A) b) Cubic: eg x -18x -5 = 0. Is there a similar formula? Cardan's method: Put x=u+v. Then x = (u+v) = u + v + uv(u+v) = (u +v ) + (uv)x. Write u +v = 5, uv = 6, so u v = l6. So u,v are the roots of t -5t +l6 = 0 ie. of (t-8)(t-7) = 0. u = 8, v = 7; u=, (, ), v=, (, ) where =l. For real x, x=+=5 is the only solution. c) Can we solve degree 4 in a similar way? (yes). Degree 5,6,7...? (NO) (Niels Abel; the proof uses what is now known as group theory). 4. Fermat's Last theorem: x n + y n = z n has no solutions in integers, n>. a) First consider the case n=. Pythagorean triads x + y = z. 'Primitive' if no common factor. x=p - q y=pq is a solution, (easily verified) z=p + q (where p,q have no common factor, and are not both odd), and, conversely, every solution is of this form. v Proof: Seek rational solutions of u + v = 1. Intersection of v=m(u+1) with this circle is a rational point iff u, m are rational. Intersections are where m (u+l) + u - 1 = 0, giving u=-l, or u=(1-m )/(l+m )= (p -q )/(p +q ) if m=p/q, rational. So x/z = (p -q )/(p +q ) and x:y:z as required. (If p,q have common factor, or are both odd, x,y (and z) have common factor). Table of results: p q x y z l l l l (b) We prove the impossibility in the case n=4, by proving: u 5

6 x 4 + y 4 = z has no solution (Note: this implies x 4 + y 4 = z 4 has no solution: case n=4). Proof: Suppose x 4 + y 4 = z has a solution. Then: x = p - q y = pq z = p + q, with p, q having no common factor, and one even. (This must be q, since x, mod 4). Let q=s, y =4ps p=z 1 and s=t. Note - p, s have no common factor, so each must be a perfect square. Then x =(z 1 ) - (t ) and so (x, t, z 1 ) are a Pythagorean triad Thus, for some l, m : z 1 =l +m x=l -m t =lm and this l=x 1 and m=y 1, since, as above, l, m have no common factor. Now z 1 = l + m = (x 1 ) + (y 1 ) = x y 14, which is of the same form as the original. But, z 1 < z. Contradiction [Method of Infinite Descent] (Since the first edition of Proof was written, in 1978, Fermat's Last theorem has been proved by Peter Wiles (199) The first public proof was faulty, but an amended version was given later. See Proof and Scientific American??? 1.5 Impossibility Proofs - Activities (Solutions) 1. Covering a chess board with dominoes. This may be done in many different ways. Illustrate some. If now two squares at opposite corners of the board are removed, is it still possible? [No, since each domino covers one black and one white square - the modified board has white and 0 black squares (or vice versa)] [A x4 board may be covered with or without the corner squares] Investigate the situation on a 7 7 board. [Not possible with or without the corner squares, since number of squares is odd!] (Similar considerations can be used to prove some solitaire problems insoluble see Averbach and Chein, p87 [Topic 1 refers], and also M Gardner, 'Further mathematical Diversions', p1ff). Euler's Königsberg bridges problem. An Eulerian graph has at most two odd vertices, and can be traced in this way. Gas/Water/Electricity problem Suppose a plane diagram can be drawn. There is a circuit 1AB1. C and must be both inside, or both outside, this circuit. There is also a route 1C which subdivides the circuit, with A on one part and B on the other, and adjacent to different regions bordered by 1C. cannot be in both these regions - yet A and B must both exist. Contradiction. A B C 1 6

7 4. Both problems are impossible. (i) 4k+ x + y Clearly x, y are not both even, and not both odd. Let x=m and y=n+1; then x = 4m and y = 4n +4n +1, so that x +y is of the form 4k+1. Contradiction (Alternatively, x = 0 or 1 mod 4, so that x + y = 0,1, mod 4) (ii) 8k +7 x + y + z, proved similarly: (x = 0, 1, or 4 mod 8, so RHS cannot be 7 mod 8, OR, alternatively, LHS = mod 4 so x, y z all odd, but (odd) = 1 mod 8, so impossible) Similarly, there are no whole numbers x, y such that x + y = 7k+ or that x 4 + y 4 = 5k+ 5. See diagrams for maps needing, 4 colours. By the 4-colour theorem (Proof p 68) any map in the plane can be coloured with 4 colours. 6. M +1 is prime only if M is a power of. {See Fermat numbers, 'Proof', p5.} Consider x +1 = (x+1)(x -x+1) x 5 +1 = (x+1)(x ) x 7 +1 = (x+1)(x ) (In each case f(-1) =0, so that (x+1) is a factor). In general, if f is odd, x f +1 has a factor x+1. Now let M have an odd factor f, M=fg, and write x= g. Then M +1 has a factor x+1 namely g +1 and so is not prime. Compare with the Mersenne numbers M - 1; these cannot be prime unless M is prime {and not necessarily then; Proof, p69}. If M=fg with f, g >1, then x= g gives M -1 = x f -1 = (x-1)(x f ) so that M -1 has a factor g -1 which is greater than 1 7. Proof that is irrational - see Proof p Geometric proof that is irrational - see diagram. AB = 1, AC =, CF = EF = -1, CE = ( -1) = - So if AB, AC can be measured in terms of a common unit, CF, CE can be measured in terms of the same unit. But the square on CF is smaller than that on AB. 9. Theorem If x=p/q is a root of a 0 x n +a 1 x n a n-1 x +a n =0 then q divides a 0 and p divides a n. (Assume p, q in lowest terms) - D Proof {uses UF theorem: if p divides ab and hcf(p,a) is 1 then p divides b} Substituting, and multiplying by q n, a 0 p n +a 1 p n-1 q +...a n-1 pq n-1 +a n q n =0 Every term except the last is divisible by p, so p divides a n q n. Thus p divides a n. Similarly q divides a 0 p n and so a 0. A B C B F - 1 E B A A 1 G Thus the only possible rational roots of x -=0 are +1, -1, +, -, none of which is a root - so is irrational. 7

8 Similarly the only possible rational roots of x -6=0 are 1,,,6 or their negatives, so that 6 is irrational. 10. e is irrational see Proof, p66-67 [Also, e, are transcendental Proof, p68] 11. Investigation shows possible values to be 1,, 8, 9, 5,... Using general solution of Pythagorean triad, x = p -q, y = pq, z = p +q, with p odd, q even, possible values for difference are: p + q -pq = (p-q) = (odd), ie. 1, 9, 5,... or p + q - (p - q ) = q = (even) ie., 8,, n=0: no solutions n=1: eg 1/ = 1/ + 1/6, 1/5 + 1/14 = 1/10 Generally, {x,y,z} = {c(c+k), k(c+k), ck} and multiples of these. n=: eg. 1/15 + 1/0 = 1/1. Generally, {x,y,z} = {ac,bc,ab} and multiples of these. (See Proof, App 9, problem 5, pp85 and 97) 8

Solutions to Practice Final

Solutions to Practice Final s to Practice Final 1. (a) What is φ(0 100 ) where φ is Euler s φ-function? (b) Find an integer x such that 140x 1 (mod 01). Hint: gcd(140, 01) = 7. (a) φ(0 100 ) = φ(4 100 5 100 ) = φ( 00 5 100 ) = (