Reductio ad Absurdum FUTURE REVISIONS OF THIS LICENSE

Chapter 1 Essential Techniques

1.1 Reductio ad Absurdum

In this section we will see examples of proofs by contradiction. That is, in trying to prove a premise, we assume that its negation is true and deduce incompatible statements from this. 1 Example Shew, without using a calculator, that 6 − √ 35 1 10 . Solution: Assume that 6 − √ 35 ≥ 1 10 . Then 6 − 1 10 ≥ √ 35 or 59 ≥ 10 √ 35. Squaring both sides we obtain 3481 ≥ 3500, which is clearly nonsense. Thus it must be the case that 6 − √ 35 1 10 . 2 Example Let a 1 , a 2 , . . . , a n be an arbitrary permutation of the numbers 1 , 2, . . . , n, where n is an odd number. Prove that the product a 1 − 1a 2 − 2 ···a n − n is even. Solution: First observe that the sum of an odd number of odd integers is odd. It is enough to prove that some difference a k − k is even. Assume contrariwise that all the differences a k − k are odd. Clearly S = a 1 − 1 + a 2 − 2 + ··· + a n − n = 0 , since the a k ’s are a reordering of 1 , 2, . . . , n. S is an odd number of summands of odd integers adding to the even integer 0. This is impossible. Our initial assumption that all the a k − k are odd is wrong, so one of these is even and hence the product is even. 3 Example Prove that √ 2 is irrational. Solution: For this proof, we will accept as fact that any positive integer greater than 1 can be factorised uniquely as the product of primes up to the order of the factors. Assume that √ 2 = a b , with positive integers a, b. This yields 2b 2 = a 2 . Now both a 2 and b 2 have an even number of prime factors. So 2b 2 has an odd numbers of primes in its factorisation and a 2 has an even number of primes in its factorisation. This is a contradiction. 4 Example Let a , b be real numbers and assume that for all numbers ε 0 the following inequality holds: a b + ε . 1 2 Chapter 1 Prove that a ≤ b. Solution: Assume contrariwise that a b. Hence a − b 2 0. Since the inequality a b + ε holds for every ε 0 in particular it holds for ε = a − b 2 . This implies that a b + a − b 2 or a b. Thus starting with the assumption that a b we reach the incompatible conclusion that a b. The original assumption must be wrong. We therefore conclude that a ≤ b. 5 Example Euclid Shew that there are infinitely many prime numbers. Solution: We need to assume for this proof that any integer greater than 1 is either a prime or a product of primes. The following beautiful proof goes back to Euclid. Assume that {p 1 , p 2 , . . . , p n } is a list that exhausts all the primes. Consider the number N = p 1 p 2 ··· p n + 1 . This is a positive integer, clearly greater than 1. Observe that none of the primes on the list {p 1 , p 2 , . . . , p n } divides N, since division by any of these primes leaves a remainder of 1. Since N is larger than any of the primes on this list, it is either a prime or divisible by a prime outside this list. Thus we have shewn that the assumption that any finite list of primes leads to the existence of a prime outside this list. This implies that the number of primes is infinite. 6 Example Let n 1 be a composite integer. Prove that n has a prime factor p ≤ √ n . Solution: Since n is composite, n can be written as n = ab where both a 1, b 1 are integers. Now, if both a √ n and b √ n then n = ab √ n √ n = n, a contradiction. Thus one of these factors must be ≤ √ n and a fortiori it must have a prime factor ≤ √ n. The result in example 6 can be used to test for primality. For example, to shew that 101 is prime, we compute T √ 101 U = 10. By the preceding problem, either 101 is prime or it is divisible by 2 , 3, 5, or 7 the primes smaller than 10. Since neither of these primes divides 101, we conclude that 101 is prime. 7 Example Prove that a sum of two squares of integers leaves remainder 0, 1 or 2 when divided by 4. Solution: An integer is either even of the form 2k or odd of the form 2k + 1. We have 2k 2 = 4k 2 , 2k + 1 2 = 4k 2 + k + 1 . Thus squares leave remainder 0 or 1 when divided by 4 and hence their sum leave remainder 0, 1, or 2. 8 Example Prove that 2003 is not the sum of two squares by proving that the sum of any two squares cannot leave remainder 3 upon division by 4. Solution: 2003 leaves remainder 3 upon division by 4. But we know from example 7 that sums of squares do not leave remainder 3 upon division by 4, so it is impossible to write 2003 as the sum of squares. 9 Example If a , b, c are odd integers, prove that ax 2 + bx + c = 0 does not have a rational number solution. Practice 3 Solution: Suppose p q is a rational solution to the equation. We may assume that p and q have no prime factors in common, so either p and q are both odd, or one is odd and the other even. Now a  p q ‹ 2 + b  p q ‹ + c = 0 = ⇒ ap 2 + bpq + cq 2 = 0 . If both p and p were odd, then ap 2 + bpq + cq 2 is also odd and hence 6= 0. Similarly if one of them is even and the other odd then either ap 2 + bpq or bpq + cq 2 is even and ap 2 + bpq + cq 2 is odd. This contradiction proves that the equation cannot have a rational root. Practice 10 Problem The product of 34 integers is equal to 1. Shew that their sum cannot be 0. 11 Problem Let a 1 , a 2 , . . . ,a 2000 be natural numbers such that 1 a 1 + 1 a 2 + ··· + 1 a 2000 = 1 . Prove that at least one of the a k ’s is even. Hint: Clear the denominators. 12 Problem Prove that log 2 3 is irrational. 13 Problem A palindrome is an integer whose decimal expansion is symmetric, e.g. 1 , 2, 11, 121, 15677651 but not 010, 0110 are palindromes. Prove that there is no posi- tive palindrome which is divisible by 10 . 14 Problem In △ABC, ∠A ∠B. Prove that BC AC. 15 Problem Let 0 α 1. Prove that √ α α . 16 Problem Let α = 0 .999 . . . where there are at least 2000 nines. Prove that the deci- mal expansion of √ α also starts with at least 2000 nines. 17 Problem Prove that a quadratic equation ax 2 + bx + c = 0 , a 6= 0 has at most two solutions. 18 Problem Prove that if ax 2 + bx + c = 0 has real solutions and if a 0, b 0, c 0 then both solutions must be negative.

1.2 Pigeonhole Principle