A proof has to be rigorously checked before it is published, after which other mathematicians can use it to further develop the subject.

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1 Proof in mathematics is crucial to its development. When an idea is formulated or an observation is noticed it becomes necessary to prove what has been discovered. Then again, the observation may prove to be incorrect. In 177, Leonhard Euler, the Swiss mathematician, noted that n + n + 41 created prime numbers for n = 0,1,,3,..and as he tested this conjecture with successive values of n, he must have believed he had found a quadratic polynomial which produced primes for all the positive integers. Sadly, the polynomial fails when n = 40, where the value obtained turns out to be the square number Even if Euler or anyone since had continued to discover primes for a polynomial for all the positive integers up to say n = , the polynomial could still fail for n = or produce one of the infinite number of primes yet to be discovered! Task 1 1 = 1, an odd number, 3 = 9, an odd number, 5 = 5, an odd number. Prove that the square of every odd number is also an odd number Task A b B T 1 a c T 3 D b T c a C Use: area of triangle T 1 + area of triangle T + area of triangle T 3 = area of trapezium ABCD to prove Pythagoras theorem, a + b = c This proof was discovered by James Abram Garfield [ ], the 0 th President of the United States of America, in Garfield was assassinated in the year he became President. A proof has to be rigorously checked before it is published, after which other mathematicians can use it to further develop the subject Page 1 of 15

2 Task 3 a c Use Pythagoras theorem to prove: sin x + cos x = 1 x b Task 4 E A G B AB is parallel to CD Prove AG H = GH D C H D F The proof depends on two axioms [axioms are self-evident or assumed statements]: 1. if you pick any two distinct points on a straight line, the angle between those two points will be 180. if you take any two intersecting straight lines and shift one of the lines so it is in a different position, but still parallel to its original position, the angle between the two intersecting lines stays the same. Task 5 Use Task 4 to prove that the angles of a triangle total 180 o Task 6 Use sin x + cos x = 1 to prove tan x + 1 = sec x Page of 15

3 Task 7 Which calculation is correct? Task 8 What is wrong with this proof? Task 9 Let a = b a = ba a - b = ba - b (a b)(a + b) = b(a b) (a + b) = b a = a = 1 What is wrong with this proof? sin x + cos x = 1 cos x = 1 sin x cosx = 1 sin x 1 + cosx = sin x Let x = π 1 + cosπ = sin π 1 1 = = Page 3 of 15

4 Proof by contradiction One method of proof is that of proof by contradiction. You assume something to be true when you suspect it isn t and use this to create a contradiction, leading to the conclusion that the original statement is not true. Proving that is irrational is a famous proof by contradiction. Task 10 Assume is rational and write = a, where both a and b are positive integers with no b common factors to cancel down the rational. Explain the mathematics used for each step = a b = a b a = b a is even a is even [see Task 1!] a = m = (m) b b = 4m b = m b is even b is even both a and b are even Contradiction! a b is irrational Page 4 of 15

5 Task 11 Which is correct or are neither correct? T = T = (1 1) + (1 1) + (1 1) + (1 1) +... T = T = 0 T = T = 1 (1 1) (1 1) (1 1)... T = T = 1 There is an infinite set of solutions to Pythagoras theorem where all of a, b and c are positive integers. They are called Pythagorean triples: = = = 5 are three examples. Pierre de Fermat [ ] famously conjectured that no such triples existed for the following sequence of equations a 3 + b 3 = c 3, a 4 + b 4 = c 4, a 5 + b 5 = c 5,... Fermat stated that he had a proof but that the margin was too small to contain it. The mathematics which finally produced a proof for Andrew Wiles in 1993 was not developed until centuries after Fermat s death, making it extremely unlikely that Fermat had a proof and if he did, it was probably incorrect. Wiles, a professor of mathematics at Princeton university, spent 7 years in his study at his home working on his proof. After the proof was announced at the end of a series of 3 lectures, the checking process was begun and only a handful of mathematicians in the world had the skills necessary to undertake such a massive task. An error was found and despondent Wiles then began the extremely complex process to fix his proof. He was on the point of giving up when on September 19 th 1994, he realised how the proof could be fixed. Working then with a former pupil of his, Richard Taylor, the proof was finally completed and published in May The theorem, known as Fermat s last theorem, was so named as it was the last of the unproven theorem of Fermat s theorems to be completed. The formal statement of the theorem is the Diophantine equation x n + y n = z n has no integer solutions for n > and x, y and z 0. Even the greatest mathematicians can struggle for years over their mathematics Page 5 of 15

6 Task 1 Demonstrate that there as many even numbers as there are positive integers. A mathematical paradox is a statement which appears to contradict itself or be contrary to expectation! The Greek philosopher Zeno of Elea [490 BC 430 BC] produced a number of paradoxes. Here are two of them: The Dichotomy paradox: Before an object can travel a given distance 100m, it must travel a distance 50m. In order to travel 50m, it must travel 5m,.. Since this sequence goes on forever, it therefore appears that the total distance of 100m can never be travelled! The mathematics needed to solve this paradox was not available in Zeno s lifetime. It required the discovery of calculus and the proof that infinite geometric series can converge. Achilles and the tortoise paradox: A speedy Achilles is unable to catch a plodding tortoise which has been given a head start. During the time it takes Achilles to catch up to the tortoise s starting position, the tortoise has moved forward to a new position and when Achilles reaches this new position, the tortoise has moved forward again. This is obviously wrong since Achilles will clearly pass the tortoise quickly rather than never. The resolution is similar to that of the dichotomy paradox. Such paradoxes held up the development of Greek mathematics for a long time and it never really recovered. Task 13 Prove that the finite geometric series a + ar + ar + ar 3 + ar 4 + ar n 1 has sum a(1 rn ) 1 r Corollary [A follow up to a proof] If the geometric series a + ar + ar + ar 3 + ar 4 + is summed to infinity and 0 < r < 1 then its sum is a 1 r Page 6 of 15

Task 14 Discuss the philosopher Bertrand Russell s Antinomy [paradox] Let R be the set of all sets which are not members of themselves. Then R is neither a member of itself nor not a member of itself.

7 Task 14 Discuss the philosopher Bertrand Russell s Antinomy [paradox] Let R be the set of all sets which are not members of themselves. Then R is neither a member of itself nor not a member of itself. Discuss the Jourdain card paradox On one side of a card is the statement: The statement on the other side of this card is true On the other side of the card is the statement: The statement on the other side of this card is false The brilliant self-taught Indian mathematician Srinivasa Ramanujan [ ] was invited to Cambridge University to work with two of England s greatest pure mathematicians of the 0 th century - Godfrey Hardy and John Littlewood. It took Hardy and Littlewood a lot of persuasion before Ramanujan accepted his many conjectures and theorems needed rigorous mathematical proof. Ramanujan found life in Cambridge very difficult. He was a victim of racial abuse, had many health problems and the local cuisine did not satisfy his strict vegetarian diet. Despite all his difficulties Ramanujan contributed a lot to mathematical development, particularly in number theory and is recognised as one of the greatest mathematicians ever. The Man Who Knew Infinity is a film about his life in Cambridge. Ramanujan worked on Fermat s last theorem and produced an infinite family of near misses to the case when n = 3. Ramanujan s family of solutions are to the equations x 3 + y 3 = z 3 ± 1, where x, y, z are positive integers. If Ramanujan had found 3 positive integers satisfying x 3 + y 3 = z 3 then he would have shown that Fermat s last theorem was not true by the method of counter-example Page 7 of 15

8 Task 15 Find positive integers that satisfy x 3 + y 3 = z 3 ± 1 Task 16 Disprove by counter-example For real numbers a and b 1. If a > b then a > b. If a > b then a > b 3. 36n 810n is prime for n = 0, 1,,... oeis.org/a n is prime for n = 0, 1,,... oeis.org/a04189 Task 17 Disprove by counter-example that fg(x) = gf(x), for any pair of real functions f(x) and g(x). Task 18 Research and explain the steps for Euclid s proof of the existence of an infinity of primes. Task 19 If n is a positive integer, prove that Task = n+ n 1 n 1+ n n 1 Prove that for all values of realx, x 6x + 11 Task 1 Prove that the triangle with vertices A(1, 1), B(3, 3) and C(4, ) is right-angled without calculating the length of each side Task Prove that 0 < k < 1 5 if the quadratic equation kx + 5kx + 3 = 0 has no real roots Task 3 Explain why n 3 n is even and divisible by 3 for all positive integers n > An algebraic proof using the method of mathematical induction showing that n 3 n is divisible by 3 can be watched at: Mathematical induction is a powerful method which you will meet if you are studying further mathematics at A-level. Task 4 Prove that the triangle with vertices A(1, 1), B(3, 3) and C(5, ) is scalene Page 8 of 15

9 Tasks 5 31 Who wants to be a millionaire? Page 9 of 15

10 Teacher notes Task 1 If n is a positive integer, then n is even and therefore n + 1 or n 1 must be odd (n + 1) = 4n + 4n + 1 = (n + n) + 1, which is an even number plus 1 and therefore odd. Task Area of T 1 = ab Area of T = ab Area of T 3 = c Area of trapezium ABCD = (a+b) ab + ab + c = a + ab + b ab + c = a b + ab + c = a + b c = a + b Task 3 c = a + b 1 = a c + b c 1 = ( a c ) + ( b c ) 1 = cos x + sin x sin x + cos x = 1 Task 4 (a + b) = a +ab+b E A G B step 1: x step : 180 x by axiom 1 C F H D step 3: 180 x by axiom 1 step 4: x by axiom Task 5 Alternate angles are equal Task 4 a b c a b parallel a + b + c = 180 [axiom 1] Page 10 of 15

11 Task 6 sin x + cos x = 1 sin x cos x + 1 = 1 cos x tan x + 1 = sec x Task 7 They are both correct Calculator 1 is set to radian measure and calculator is set to degree measure Task 8 Line 5 is incorrect. Division by zero, (a b), from line 4 is not defined. Task 9 The penultimate line should read 1 1 = 1 ± 1 0 and it is the negative root which is needed to produce 0 = 0 and not 0 = Task 10 = a b = a b [Assuming is rational with a and b positive integers and a b in its lowest terms] [Squaring both sides] a = b [Multiplying both sides by b ] a is even a is even [see Task 1!] [It is a multiple of, therefore even] [If a were odd then a would also be odd [Task 1], therefore a must be even] a = m [a is a multiple of ] = (m) b [Squaring both sides] b = 4m [Multiplying both sides by b and squaring the bracket] b = m [Dividing both sides by ] b is even [A multiple of ] b is even [A consequence of Task 1 again] Page 11 of 15

12 both a and b are even [They would therefore cancel down by and a was in its lowest b terms] Contradiction! a b is irrational Task 11 T is either 1, for an odd number of terms, or zero, for an even number of terms. The series is infinite and its total oscillates between 0 and 1 Task The even numbers can be mapped 1 : 1 with the positive integers A way to solve puzzle two at Hilbert s Hotel. Puzzle 3 makes use of Euclid s proof of the existence of an infinite number of primes Task 13 Let T n = a + ar + ar + ar 3 + ar 4 + ar n 1 rt n = ar + ar + ar 3 + ar 4 + ar n T n rt n = a ar n T n (1 r) = a(1 r n ) T n = a(1 rn ) 1 r Corollary If 1 < r < 1, then r n approaches zero as n approaches infinity, r n 0 as n and T = a 1 r Task 14 Warning, these Paradoxes can damage your health! The Jourdain card paradox is best analysed with an actual card which you keep flipping and reading. The barber, a modification of Russell s Paradox Russell s Paradox using set notation Zeno s Dichotomy Paradox Page 1 of 15

13 Task = = and many more mathworld.wolfram.com/diophantineequation3rdpowers.html Task a = ( ), b = ( 3). a = ( 3), b = ( ) 3. When n = 45, 36n 810n = = , two distinct prime factors 4. When n = 6, n = = , three distinct prime factors Task 17 f(x) = x + 1 and g(x) = x, x 0 fg(x) = f(x ) = x + 1 gf(x) = g(x + 1) = (x + 1) = 4x + 4x + 1 x + 1 if x 0 Task 18 Task 19 1 n + n 1 n n 1 n n n 1 + n 1 1 n 1 n n 1 n = n n 1 n 1 n + n (n 1) (n 1) (n ) = n n n 1 n Task 0 = n n 1 + n 1 n = n 1 x 6x + 11 = (x 3) = (x 3) + (x 3) is a square number and therefore has a minimum value of 0, when x = 3 Therefore x 6x + 11 is always Page 13 of 15

14 Task 1 B Gradient = 1 1 = -1 C A Gradient = = 1 The product of the gradients of the line segments from B is -1 and so the line segment BA is perpendicular to the line segment BC and the triangle is right-angled at B Extension Prove that the gradients of perpendicular lines are negative reciprocals of each other Task A quadratic equation with no real roots has its discriminant < 0 b 4ac < 0 for the given quadratic (5k) 4(k)(3) < 0 5k 1k < 0 k(5k 1) < 0 0 < k < k Page 14 of 15

15 Task 3 n 3 n = n(n 1) = n(n 1)(n + 1) (n 1)n(n + 1) (n 1)n(n + 1) represent 3 consecutive integers and therefore must contain a multiple of 3 Example: 9, 10, 11 [multiple of 3 first] 8, 9, 10 [multiple of 3 in the middle] 7, 8, 9 [multiple of 3 at the end The products are: odd even odd or even odd even Both products therefore contain a multiple of and must be even Task 4 B 1 C 1 A By Pythagoras theorem 4 AC = 6, AB = 8 and BC = 5 and therefore, the triangle is scalene Tasks 5 31 Good luck! Page 15 of 15

Math 312, Lecture 1. Zinovy Reichstein. September 9, 2015 Math 312

Math 312, Lecture 1. Zinovy Reichstein. September 9, 2015 Math 312 Math 312, Lecture 1 Zinovy Reichstein September 9, 2015 Math 312 Number theory Number theory is a branch of mathematics Number theory Number theory is a branch of mathematics which studies the properties