SESSION 1: PROOF 1. What is a “proof” 1.1. An easy example. Roughly speaking, a mathematical proof is a rigorous demonstration that a mathematical statement is true. For an elementary example, consider the following mathematical statement: p Statement 1.1. If a and c are positive real numbers, then x = ± ac satisfies ax2 − c = 0. We first remark that in Statement 1.1, a was assumed to be positive; this is to make all the quantities in Statement 1.1 well defined. In particular a and c positive means that p a 6= 0 and ac > 0, so that ac and ac are well defined real numbers. It is not difficult to prove Statement 1.1: p Proof. If x = ± ac then r 2 c c 2 = . x = ± a a So, c ax2 − c = a − c = c − c = 0. a p So we have shown that x = ± ac satisfies ax2 − c = 0 as claimed. Statement 1.1 is not difficult to prove; the presentation is given to demonstrate how to write a clear, comprehensive, cohesive, rigorous argument. To underline why the above proof is well written, we provide a second “proof” of Statement 1.1 (which will later be criticised): “Proof” 1.2. c ax2 + c = x2 + = 0 ra c . =x⇒± a 1 2 SESSION 1: PROOF 1.2. How do I write a proof ? Note that the first proof may be read, and understood, independently of Statement 1.1, while “proof” 1.2 is completely obscure without the context of Statement 1.1! We now make the following remarks about writing good mathematical proofs: • Always clearly state your assumptions. • Always use English to clearly convey what you mean; good mathematical arguments should read like an essay. The first sentence in the proof of 1.1 reads “If x equals plus or minus the square root of c divided by a then x squared equals plus or minus the square root of c divided by a all squared which equals c divided by a”. Whenever you write a proof, you should read through your answer and check to see if it could be read aloud and still be understood. • Use English or symbols to correctly connect your mathematical statements (see below). • Always make a clear conclusion about what you have just shown. • Take pride in the mathematics you write; when you write a piece of mathematics, you are showing people exactly how you think. So, you should try to write as elegantly and beautifully as you can! We now discuss the following important mathematical symbols; “⇒”, “⇔”, “=”, and how they should be used. “=”: the equals sign is used to denote an equality between mathematical expressions. For example, 1+1=2, or sin2 θ + cos2 θ = 1. “⇒”: the implies sign is used to denote when one mathematical statement follows directly from another. For example, x = 3 ⇒ x2 = 9. Which is read “if x equals three then x squared equals nine” or “x equals three implies x squared equals nine”. “⇔”: the equivalence sign is used to denote the equivalence of two mathematical statements. If A and B are mathematical statements, then writing A ⇔ B means that both A ⇒ B and B ⇒ A hold. For example, x − 1 = 0 ⇔ x = 1. Note that it would be wrong to write x = 3 ⇔ x2 = 9. SESSION 1: PROOF 3 With the above remarks in mind, answer the following questions: Exercises: 1.1: Be as critical of “Proof” 1.2 as you can be; find as many mistakes/ambiguities/places where the argument can be improved as possible. 1.2: If a, b, c ∈ R with a 6= 0 and b2 − 4ac > 0, show that √ −b ± b2 − 4ac satisfies ax2 + bx + c = 0. x= 2a 1.3: (The factor theorem) Let p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial with real coefficients (i.e. ai ∈ R) and an 6= 0. Show that a real number r satisfies p(r) = 0 if and only if there exists a polynomial q(x) = bn−1 xn−1 + bn−2 xn−2 + · · · + b1 x + b0 with bn−1 6= 0 such that p(x) = (x − r)q(x). Hint: Use the polynomial remainder theorem (see Exercise 2.12). 1.4: (The rational root theorem) Let p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial with integer coefficients (i.e. ai ∈ Z) and an 6= 0. Show that if x = rs is a root of p(x) then r divides a0 and s divides an . 1.5: Factorise the following polynomials: • 5x3 − 9x2 + 3x + 1. • 2x4 − 11x3 − 26x2 + 29x + 42. • x5 − 2x4 − 3x3 + 6x2 + 2x − 4. 1.3. Why bother? We know from Exercise 1.1 that “Proof” 1.2 is far from perfect! However, you might argue that the person who wrote the answer more or less knew “why” p x = ± ac satisfied ax2 −c = 0; so why bother being so precise when formulating a proof? The trouble is that when you are not absolutely clear about what your assumptions are, or don’t properly justify implications, it is easy to run into trouble as the next example shows: Example 1.3. Let x = y so that y(x − y) = x(x − y) = x2 − xy = x2 − y 2 = (x + y)(x − y) = 2y(x − y) ⇒ y(x − y) = 2y(x − y) or 1 = 2. Okay, so if we accept that proofs are important, then we accept that proofs should be written carefully and precisely. However, we might ask why are proofs important? This 4 SESSION 1: PROOF is a big question, which I won’t pretend to answer; instead, I will outline a few reasons why I care about proof. • A proof of a mathematical statement is absolute; there is no exception to the rule. Such absolute statements are wonderful and do not exist in any other field of study! • Mathematics is EVERYWHERE! From music to the economy, mathematics is used to help us reason about the world we live in; if we weren’t able to apply theorems that we have proved to be true in mathematics, we would not know much about the world around us. • It is possible to know when you can not do better; for example, a cartographer wishing to buy as few colours as possible to colour his maps can be told by a mathematician that four different colours will be enough to colour his map (this is called the four colour theorem). Taking the opposite stand point, mathematicians have been able to tell chemists when certain crystal structures have no mathematical obstructions (which led Chemists to search for, and successfully produce, new crystal structures). There are many more examples of this nature. • Most importantly, a mathematical proof precisely captures an idea; these ideas are often beautiful and elegant (see Theorem 2.4). Exercises: 1.6: What is wrong with the following argument: (1) let x = 1 + · · · + 1 (x times). Multiplying both sides of Equality (1) by x we get: (2) let x2 = x + · · · + x. Differentiating both sides of the Equality (2) with respect to x we get; (3) 2x = 1 + · · · + 1 = x. Dividing both sides of Equality (3) by x, we conclude that 2=1. 1.7: What is wrong with the following argument: Let x = π+3 . One can check that; 2 (3 − x)2 = π2 4 − 3π 2 + 9 4 = (π − x)2 . From this we conclude that 3 − x = π − x and that π = 3. SESSION 1: PROOF 5 1.8: If you know about complex numbers, what’s wrong with the following argument: p √ √ √ 1 = 1 = (−1)(−1) = −1 −1 = i2 = −1. Hence, 1 = −1. 2. Different types of proof In Section 1 we provided a proof of Statement 1.1; this is an example of a “direct proof”; namely, a proof that is based on a series of inferences made from established facts or assumptions. In this section we will introduce three further techniques of proving mathematical statements; proof of the contrapositive statement, (dis)proof by counterexample, and proof by contradiction. 2.1. The contrapositive. Mathematical statements of the form “if A then B” are called conditional statements. For example, in Statement 1.1, we have “IF (a and c are positive p real numbers) THEN (x = ± ac satisfies ax2 − c = 0)”. The contrapositive statement of the conditional statement ‘if A then B” is ‘if not B then not A”. For example, the contrapositive of the statement “if my name is Frank then my name begins with an ‘F’” is “if my name does not begin with an ‘F’ then my name is not Frank”. Every conditional statement is equivalent to its contrapositive statement. Namely, (A ⇒ B) ⇔ ( not B ⇒ not A). This can be visualised by the following Venn diagram: “if x is in A then x is in B” implies that “if x is not in B then x is not in A”, and “if x is not in B then x is not in A” implies “if x is in A then x is in B”. Namely, “if x is in A then x is in B” and “if x is not in B then x is not in A” are logically equivalent. A B As a further example, the contrapositive of Statement 2.1 is Statment 2.2. Statement 2.1. If a, b are real numbers such that ab is irrational, then either a or b is irrational Statement 2.2. if a and b are rational numbers then ab is a rational number. 6 SESSION 1: PROOF Often it is considerably easier to prove the contrapositive of a mathematical statement; Statement 2.1 is difficult to prove directly, but the logically equivalent Statement 2.2 is easy to prove: Proof. If a and b are rational, then a = m n ab = and b = mx ny x y for integers m, n, x, y so that ∈ Q. Namely, if a and b are rational numbers then ab is a rational number. The contrapositive of this tells us that if a, b are real numbers such that ab is irrational, then either a or b is irrational. 2.2. Counterexamples. A counterexample is an exception to the rule; for example, an albino crow would be an exception to the rule that “all crows are black”. In mathematics, the existence of a counterexample to a mathematical statement means that the mathematical statement is false. For example, consider the following statement: There do not exist integers n, x, y, z such that xn + y n = z n . If we could find integers n, x, y, z such that xn + y n = z n , then n, x, y, z would be a counterexample to the above statement, and we would conclude that the above statement is false; namely there exist integers n, x, y, z such that xn + y n = z n . Indeed, n = x = y = 1, and z = 2 is a counterexample to the above statement 1. 2.3. Proof by contradiction. Suppose that we would like to prove a statement A. If we assume that A is false, and, from our assumption that A is false, directly prove a consequence that we know to be absurd, then our assumption that A was false must have been incorrect; namely, A must have been true. For example, suppose that we wanted to show that no human can do everything; we assume that there exists a person who can do everything. In particular (s)he can make a sandwich that is too big to be finished by any human. This means that there is a sandwich that (s)he is unable to finish, and so (s)he is unable to eat a sandwich; this contradicts our assumption that (s)he can do everything, and we conclude that our assumption that there exists a person who can do everything must have been false. To give a more mathematical example, we recall that a natural number p 6= 1 is said to be prime if the only natural numbers that divide p are 1 and p; we now give a famous example of a proof by contradiction. Theorem 2.3. There are infinitely many primes. 1For n > 2, and x, y, z > 0 there are no integer solutions to xn + y n = z n ; this is a famous theorem called Fermat’s Last Theorem (which is worth a google if you don’t already know about it). SESSION 1: PROOF 7 Proof. We assume that there are only finitely many prime numbers p1 , p2 , . . . , pn . We know that n > 0 because 2 is prime. We note that k = p1 p2 . . . pn + 1 is greater than each pi (so k 6= pi for any i) and that k divided by each pi has remainder 1. In particular, if p1 , p2 , . . . , pn is a complete list of primes then k’s only divisors are 1 and k meaning that k is a prime. This contradicts our assumption that {p1 , . . . , pn } was the set of all primes, and so the set of all primes is not finite. 2.4. Proof by mathematical induction. Suppose that we would like to prove a statement S(n) which is dependent upon a positive integer n. Suppose also, that we know that S(k) is true and that (%) if S(n) is true then S(n + 1) is true. In this case we can conclude that S(N ) is true for every positive integer N ≥ k. To see this, note (%) (%) (%) (%) S(k) =====⇒ S(k + 1) =======⇒ S(k + 2) =======⇒ . . . ========⇒ S(N ). with n=k with n=k+1 with n=k+2 with n=N −1 So, if S(k) is true and S(n) ⇒ S(n + 1) for n≥ k then S(N ) is true for all N ≥ k. The statement S(k) is called the base case, and the assumption that S(n) is true is called the inductive hypothesis. A proof by mathematical induction consists of: (1) Proving a base case. (2) Assuming the inductive hypothesis S(n) and using S(n) to prove S(n + 1). (3) Concluding that S(N ) holds for all N ≥ k. Induction is very good at establishing identities like: 1 + 3 + 5 + · · · + (2n − 1) = n2 n(n + 1)(2n + 1) 6 2 n (n + 1)2 1 + 2 3 + 3 3 + · · · + n3 = 4 However induction can be used to prove more versatile results: 1 + 2 2 + 3 2 + · · · + n2 = Theorem 2.4. Every positive integer N > 1 can be written as a product of prime numbers. Proof. We let S(n) be the statement that every positive integer greater than 1 and less than or equal to n can be written as a product of primes. We now note that 2 is a prime 8 SESSION 1: PROOF number and so S(2) holds. We assume the inductive hypothesis S(n); that is, we assume that every 1 < m ≤ n can be written as a product of primes. We will now show, assuming the inductive hypothesis, that S(n + 1) is true. If we have 1 < m ≤ n + 1 then either 1 < m ≤ n or m = n + 1. If 1 < m ≤ n then, by our inductive assumption, we that know m can be written as a product of primes. If m = n + 1 and n + 1 is prime then we are done. If m = n + 1 is not prime then m = pq for some 1 < p, q ≤ n. The inductive hypothesis tells us that p = p1 × · · · × pa and q = q1 × · · · × qb for some prime numbers p1 , . . . pa , q1 , . . . , qb . Hence m = n + 1 = pq = p1 × · · · × pa × q1 × · · · × qb is a product of prime numbers and so every integer 1 < m ≤ n + 1 can be written as a product of primes. By induction we see that S(N ) is true for all N > 1 and every positive integer is a product of prime numbers. Exercises: √ 2.1: Prove that if p is a prime number then p is an irrational number. √ Hint: assume that p = m/n where m and n are natural numbers with no common factors. 2.2: Prove that if n is a natural number such that n2 − 6n + 5 is even then n is odd. Hint: prove the contrapositive statement. 2.3: Disprove the following statement: for x a non-negative real number −x < x. 2.4: Prove or disprove: If f (x) is an increasing function on [0, ∞) then f (x) → ∞ as x → ∞. 2.5: Prove or disprove: if x, y are real numbers such that x + y > 1, 000, 000 then x ≥ 500, 000 or y ≥ 500, 000. 2.6: Prove or disprove: If n is a natural number and n2 is odd, then n is odd. √ √ 2.7: Prove or disprove: 2 + 3 is irrational. 2.8: Prove or disprove: every map can be coloured with three colours in such a way that no two adjacent countries have the same colour. 2.9: Fermat’s Last Theorem states that there do not exist positive integers x, y, z such that xn + y n = z n for any integer n > 2. Given that Fermat’s Last Theorem has been proved: (1) Prove or disprove: there do not exist integers x, y, z such that xn + y n = z n for any integer n ≥ 3. SESSION 1: PROOF 9 (2) Prove or disprove: there do not exist positive rational numbers x, y, z such that xn + y n = z n for any integer n ≥ 3. 2.10: The Fibonacci sequence {Fn } is defined by F1 = F2 = 1 and Fn = Fn−2 +Fn−1 . Show that F3n is even. 2.11: Show that if a square is removed from a grid with 2n × 2n squares, then the configuration can be tiled with L-shaped tiles consisting of three squares (of the form shown below). Conclude that 22n = 3k + 1 for some positive integer k. 2.12: (polynomial remainder theorem2) Prove the following statement: Given a polynomial p(x) of order n and a polynomial d(x) of order k, with k < n, show that there exists polynomials g(x) and r(x) with deg(r(x)) < deg(d(x)) such that p(x) = d(x)g(x) + r(x). 2.13: 3 Suppose that there are n lines in the plane with every line intersecting every other line and no three lines intersecting at a single point. Show that the number 2 . of regions bounded by the lines is n +n+2 2 2As a special case, the polynomial remainder theorem says that if d(x) = x − a then r(x) is a constant. 2 question: why do we guess the number n +n+2 in this question? 2 3Bonus 10 SESSION 1: PROOF 3. Some STEP style questions (1) (STEP 1 level) Let n be a positive integer greater than 999 whose last three digits are abc. Prove that if 4a + 2b + c is divisible by 8 then n is divisible by 8. (2) (STEP 1 level) (i) Prove that if a + 2b + 3c = 7x then a2 + b2 + c2 = (x − a)2 + (2x − b)2 + (3x − c)2 (ii) Prove that if 2a + 3b + 3c = 11x, then a2 + b2 + c2 = (2x − a)2 + (3x − b)2 + (3x − c)2 (iii) State and prove a general result of which (i) and (ii) are special cases. (3) (STEP 1 level) (i) By considering (1 + x + x2 + · · · + xn )(1 − x) show that, if x 6= 1, then 1 + x + x2 + · · · + xn = 1 − xn+1 1−x (ii) By differentiating both sides and setting x = −1 show that ( − n2 if n is even, 1 − 2 + 3 − 4 + · · · + (−1)n−1 n = n+1 if n is odd. 2 (4) (STEP 1 level) (i) Show that, if n is a positive integer such that (n − 3)3 + n3 = (n + 3)3 then n is even and n2 is a factor of 54. (ii) Deduce that no positive integer satisfies (n − 3)3 + n3 = (n + 3)3 . (5) (STEP 1 level) The right angled triangles with side lengths 6, 8, 10, and 5, 12, 13 both have the property that their area equals their perimeter. (i) Give a counterexample to show that the area of a right angled triangle does not necessarily equal the perimeter of the triangle. SESSION 1: PROOF 11 (ii) A triangle has sides x, y, and z with rational lengths where z is the longest side and m 4m − 8n 8n2 − 4mn + m2 x= , y= , z= n m − 4n n(m − 4n) where m and n are integers satisfying m > 4n. By writing z 2 − y 2 = (z − y)(z + y) or otherwise deduce that (a) x2 + y 2 = z 2 , (b) x + y + z = xy . 2 (iii) Hence deduce that there are infinitely many right angled triangles with rational side lengths have the property that the area equals the perimeter. (6) (STEP 1 level) Denote the k th derivative of a function h(x) by h(k) (x). Let f (x) and g(x) be n-times differentiable functions. Prove by mathematical induction that n X n (n−k) (n) (f g) (x) = f (x)g (k) (x) k k=0 (7) (STEP 2 level) Consider the quadratic equation p (∗) nx2 + 2x pn2 + q + rn + s where p > 0, p 6= r and n = 1, 2, 3, . . . . (i) For the case where p = 3, q = 50, r = 2, s = 15 find the set of values of n for which equation (∗) has no real roots. (ii) Prove that if p < r and 4q(p − r) > s2 , then (∗) has no real roots for any value of n. 2 (iii) If n = 1, p − r = 1 and q = s8 , show that (∗) has real roots if and only if √ √ s ≤ 4 − 2 2 or s ≥ 4 + 2 2. (8) (STEP 2 level) Let Sn (x) = ex 3 dn dxn 3 e−x . (i) Show that S2 (x) = 9x4 − 6x and find S3 (x). (ii) Prove by induction on n that Sn (x) is a polynomial. By means of your inductive argument determine the order of this polynomial and the coefficients of the highest power of x. d (iii) Show also that if dx (Sn ) = 0 for some value a of x, then Sn (a)Sn+1 (a) ≤ 0. 12 SESSION 1: PROOF (9) (STEP 2 level) Let n = 2k + 1 be a positive integer and let √ √ 1 for x ≥ 1. (x + x2 − 1)n + (x − x2 − 1)n y(x) = 2 Show that (i) y(x) = a0 x + a1 x3 + s2 x5 + · · · + ak x2k+1 (ii) a1 = (−1)k−1 2k(k + 1)(2k + 1) 3 (iii) (−1)k 2k(k − 1)(k + 1)(k + 2)(2k + 1) 15 Determine the value of a0 + a1 + · · · + ak . a2 =