# Integers and Rational Numbers

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1 Chapter 6 Integers and Rational Numbers In this chapter, we will see constructions of the integers and the rational numbers; and we will see that our number system still has gaps (equations we can t solve) and needs to be extended further. The Study guide suggests that you collect a stock of examples. In the supplementary material we see that two famous numbers (the square root of 2, and the base of natural logarithms) are irrational. 6.1 Integers and rational numbers In this section we extend our number system, first to the integers, and then to the rational numbers, and describe some of the properties of these new numbers Why do we need to extend the number system? There are good practical reasons for needing a larger number system than the natural numbers. Natural numbers are ideal for counting that is what they are for! But as we become more sophisticated, we need new kinds of numbers. First, once we have a banking system, we need to face the fact that a customer might actually owe the bank money. I had 100 in the bank, but as a result of a sudden emergency, I had to withdraw 150, leaving me 50 in debt to the bank. I would say I am in the red, because this balance would be written in red in the bank s ledgers. But it would be better if, instead of two kinds of numbers (red and black), there was only one kind, which could be positive or negative. The other extension comes from the need for numbers in measurement. If I draw a square with side of length 1, and measure the diagonal, it comes out (as near as I can make it) to 1.414, while the circumference of a circle of diameter 1 turns out to be about I don t know whether these values are exact, but I 89

2 90 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS can t even express them with just natural numbers (the nearest natural numbers would be 1 and 3 respectively). From a mathematician s point of view, there is another reason: solving equations. Much of mathematics is concerned with exactly this. If we start with the natural numbers, we can solve the equation 3+x = 5 (the solution is x = 2, but we can t solve the equation 5 + x = 3. We need negative numbers for this. Similarly, we can solve the equation 3x = 6, but not the equation 3x = 7, unless we introduce fractions The integers There are two ways of getting from the natural numbers to the integers. Each has its drawbacks. We will consider them in term. First method: we know what we want! The integers consist of the natural numbers, zero, and the negatives of the natural numbers. We need a convenient way to distinguish the natural numbers from their negatives; we could write the negatives in red, but we will instead simply write the negative of n in the usual way as n. Thus, Z = N {0} { n : n N}. Already there is a problem, with a double use of : as a symbol for subtraction (as in 7 4 = 3), and as indicating the negative of a natural number. But worse is to come, if we want to give a definition of addition for integers. To define a + b, we need separate cases according as each of a and b is positive, zero, or negative. And it is worse than that. If a is positive and b is negative, we need separate rules for a + b depending on whether a is greater than, equal to, or less than b. For example, 5 + ( 3) = 2, 3 + ( 3) = 0, 3 + ( 5) = 2. This makes thirteen different cases that have to be specified. Imagine the problems that will arise when we have to show, for example, the associative law (a + b) + c = a + (b + c). The other important drawback of this method is that we assume that we already know what the integers look like, and formalise that. It would be better to construct them without using any such knowledge; this is what the second method will do. Although this method is terrible as a mathematical definition, you should certainly continue to think of the integers as positive or negative natural numbers or zero, just as you always have.

3 6.1. INTEGERS AND RATIONAL NUMBERS 91 Second method: put equivalence relations to work The reason we are extending our number system is to ensure that equations b + x = a always have a solution, no matter what a and b are. So we have to add new numbers so as to achieve this. This time, the method is much more complicated, but once the integers are constructed, verification of their properties is more straightforward; also, we do not begin with a preconception of what the resulting system looks like. The simplest solution is the most profligate. For every two natural numbers a and b, add a new number m(a,b) so that x = m(a,b) is a solution to this equation. The reason why we don t want to do this is that there will be far too much duplication. For example, the equations 5 + x = 3 and 7 + x = 5 should have the same solution. So instead we want to take the set of all equations of this form, and sort them into bags, each of which gives us a new number: equations in the same bag will have the same solution. Then we will label each bag with the integer which is the common solution to all the equations in that bag, and regard the bags as being the integers. When we use an integer in a sum, we just read the label on the bag; we don t have to worry about the contents. For example, all the equations 2 + x = 1, 3 + x = 2, 4 + x = 3,..., will correspond to the integer 1, as the picture shows x = x = x = x = x = x = x = x = x = x = x = x = Remember that partitions come from equivalence relations: we want to define an equivalence relation on the set of all the equations whose equivalence classes do what we want. So we do a bit of rough work. If the equations b + x = a and d + x = c are to have the same solution, then b + c = b + (d + x) = (b + x) + d = a + d, where we used a bit of manipulation involving the commutative and associative laws in there. (More precisely, we want to construct the integers in such a way that these laws continue to hold; this will force our choices.)

4 92 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS So we make a definition and prove a lemma about it. Definition Let S be the set of all equations of the form b+x = a, for natural numbers a and b. Define a relation R on the domain S by the rule (b + x = a) R (d + x = c) if and only if a + d = b + c. Lemma R is an equivalence relation on S. Proof We have three rules to check. Reflexive: a + b = b + a (by the commutative law), so (b + x = a) R (b + x = a). Symmetric: Suppose that (b + x = a) R (d + x = c). Then a + d = b + c. Hence c + b = d + a (by the commutative law again), so (d + x = c) R (b + x = a). Transitive: Suppose that (b + x = a) R (d + x = c) and (d + x = c) R ( f + x = e). Then a + d = b + c and c + f = d + e. Hence a + d + f = b + c + f = b + d + e, and the cancellation law gives a + f = b + e, so (b + x = a) R ( f + x = e). Now watch closely. The next bit is the real test of a mathematician. If you can understand what I am going to say next, you will have no trouble with anything else you meet in your degree. We have defined an equivalence relation on the set of equations of the form b + x = a so that two equations are equivalent if they have the same solution. So there is one solution for each equivalence class. Now, being brave, we say, take each equivalence class to be one of our new numbers: Definition An integer is an equivalence class of the relation R on the domain S just defined. To repeat: each equivalence class corresponds to a single integer; we take the integers to be the equivalence classes. We will now change our notation a bit and write the equivalence class R((b + x = a)) (consisting of all the equations related to this one) as [a,b]. Anyway, we have now defined the integers. We need to be able to add, multiply and order them. Remembering that the equivalence class [a,b] corresponds to solutions of the equation b + x = a, in other words, x = a b, we can figure out what the rules must be, by a short calculation:

5 6.1. INTEGERS AND RATIONAL NUMBERS 93 So (a) (a b) + (c d) = (a + c) (b + d), (b) (a b) (c d) = (ac + bd) (ad + bc), (c) (a b) < (c d) if and only if a + d < b + c. Definition We define addition and multiplication of integers by the rules [a,b] + [c,d] = [a + c,b + d], [a,b] [c,d] = [ac + bd,ad + bc], [a,b] < [c,d] a + d < b + c. There is some work which has to be done here. The definition of addition, for example, appears to depend on which choice of pair [a,b] we chose to represent a given integer. So we have to show that, if the equations b + x = a and b + x = a are equivalent (that is, if a + b = a + b), and if also d + x = c and d + x = c are equivalent (that is, if c + d = c + d), then the equations (b + d) + x = (a + c) and (b +d )+x = a +c are also equivalent. What we have to verify is that (a +c)+ (b + d ) = (a + c ) + (b + d). This can be shown by simple rearrangement. The point is, that the calculations use only properties of the natural numbers (which we already know to be true). Now it is just a case of laboriously verifying that the integers we have constructed satisfy properties like those of the natural numbers: the commutative, associative, distributive, and order laws, and the cancellation laws (except that we can t cancel 0 from multiplication: that is, if 0a = 0b, we are not allowed to conclude that a = b). But two more final jobs are also very important: Proposition Inside the integers, we can find a copy of the natural numbers. That is, the integer [a + n,a] (which contains all pairs (x + n,x)) behaves just like the natural number n: (a) [a + n,a] + [b + m,b] = [c + m + n,c]; (b) [a + n,a] [b + m,b] = [c + mn,c]; (c) [a + n,a] < [b + m,b] if and only if m < n. Proposition If e and f are any two integers, then the equation f +x = e has a unique integer solution.

6 94 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS Proof If e = [a,b] and f = [c,d], then the solution x is [a + d,b + c]. For [c,d] + [a + d,b + c] = [c + a + d,d + b + c] = [a,b]. This is because (c+a+d, d +b+c)r(a, b), since (c+a+d)+b = (d +b+c)+a. This has been a long and complicated ride. So let us summarise what we have done. Starting from the set N of natural numbers, we have constructed a new set of objects called integers such that (a) their addition, multiplication, and order relations satisfy all the usual rules ; (b) within them, we can find a subset whose addition, multiplication and order are exactly the same as those for the natural numbers; (c) equations of the form f + x = e have unique solutions in the new system. What does this mean in practice? What it certainly doesn t mean is that I want you to stop writing 23 and write [1, 24] or [100, 123] or something else instead. The integer 23 is equal to 1 24, or to , as it always was! The things we have constructed are the usual integers; but we have put them on a firm mathematical footing, based on our knowledge of the natural numbers. We have already taken this point of view in Chapter 3, where we saw that the set of integers is countably infinite Division and divisibility Now we turn to an important property of integers: divisibility and greatest common divisor. In this and the next section, it will be convenient to change the definition of natural numbers slightly, to allow 0 to be a natural number. You will see the reason for this soon. Division is the reverse of multiplication; but it cannot always be performed exactly in the integers. You can divide 6 by 2, since 6 = 2 3; but you cannot divide 7 by 2. And you can t divide anything by zero! Definition Let m and n be integers. We say that m divides n, or that n is divisible by m, if there is an integer k such that n = mk. We write m n to mean that m divides n.

7 6.1. INTEGERS AND RATIONAL NUMBERS 95 Note that this is a relation: if we put the inputs 2 and 6 into the black box for divisibility, it responds yes, but if we put 2 and 7 in, it responds no. Do not confuse m n with the fraction m/n, which is a number. Zero behaves in a slightly confusing way: Proposition Let n be an integer. Then (a) n 0; (b) 0 n holds if and only if n = 0. Proof (a) true since 0 = n 0; (b) if n = 0, then 0 n by (a); conversely, if 0 n then n = 0 k for some integer k, so necessarily n = 0. Although exact division is not always possible, there is a substitute. This is called the division algorithm, since to do the calculations you use the method for long division you learned in primary school. An algorithm is a constructive method or recipe for producing some specified result. This algorithm will be stated just for non-negative integers; it actually works for all integers, but the signs complicate things a bit. We will only use it when everything is non-negative. Theorem (The Division Algorithm) Let a and b be integers, with a 0 and b > 0. Then there exist unique natural numbers q and r such that (a) a = bq + r; (b) 0 r b 1. The numbers a, b, q, r are called the dividend, divisor, quotient and remainder respectively. Proof First we show that we can find q and r satisfying the two conditions. Let S be the set of positive integers k such that bk > a. Certainly S is non-empty since, for example, b(a + 1) > ba a, so a + 1 S. By (b) of Theorem 5.1.4, the set S has a smallest element. Let n be the smallest element. Since n 1, we can write n = q + 1, where q 0. Now bq a (since q is smaller than n, which was the smallest number for which bn > a; so certainly a = bq + r for some number r, where r 0. Finally, bq + r = a < bn = b(q + 1) = bq + b, so r < b, which implies that r b 1.

8 96 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS Now we have to prove that the quotient and remainder are unique. Accordingly, suppose that we have two quotient remainder pairs: that is, a = bq 1 + r 1 = bq 2 + r 2, where 0 r 1 b 1 and 0 r 2 b 1. If r 1 = r 2, then bq 1 = bq 2, and so q 1 = q 2. So suppose that r 1 and r 2 are unequal. One of them is larger; suppose that r 1 < r 2. Then bq 1 = a r 1 > a r 2 = bq 2, so q 1 > q 2. This means that q 1 = q 2 + c for some natural number c > 0. Then bq 2 + r 2 = bq 1 + r 1 = b(q 2 + c) + r 1 = bq 2 + bc + r 1, so that r 2 = bc + r 1. But then r 2 bc b, contrary to the assumption that r 2 b 1. So we can say: b a holds if and only if, when we divide a by b, the remainder is zero. This is why we added 0 to the natural numbers in this section. In this proof we used the cancellation laws for natural numbers: if a+c = b+c then a = b; and if ac = bc, then a = b. We stated them for the natural numbers not including zero; and indeed the second law fails if c = 0. So we have to check our proof to see that we didn t cancel 0. How do we find q and r in practice? Use the long division algorithm that you learned at school. The way it works is very similar to the proof we gave above. To divide a by b, you find (by trial division) the largest q such that bq a, and then put r = a bq. Of course, our base 10 representation of natural numbers allows us to get the quotient one digit at a time, which saves a lot of work! Greatest common divisor Definition Let a and b be two integers. The greatest common divisor or highest common factor of a and b, written gcd(a,b), is a natural number d such that (a) d a and d b; (b) if e is any integer such that e a and e b, then e d; (c) d 0. The last condition is included because, without it, we could not say whether the greatest common divisor of 4 and 6 is 2 or 2; each of them satisfies the first two conditions. We have to make a choice, so we choose 2 rather than 2.

9 6.1. INTEGERS AND RATIONAL NUMBERS 97 It is not obvious that a number satisfying these conditions even exists. But we can say for sure that there cannot be more than one! For suppose that d 1 and d 2 are both greatest common divisors of a and b. Then d 1 a and d 1 b; since d 2 is a greatest common divisor, it follows from part (b) of the definition that d 2 d 1. Similarly, d 1 d 2. Hence d 1 = d 2. It happens that gcd(0,0) = 0. Let us check this. We saw above that every natural number divides 0. So 0 0 and 0 0; and, if e 0 and e 0, then e 0. So by definition 0 is the greatest common divisor! If a and b are not both zero, say a 0, then the greatest common divisor cannot be larger than a, and it is (as the name suggests) the largest natural number which divides both a and b. (For if d 0 and e d, then e d, since d = e f for some natural number f.) If we had not included 0, we could have defined the greatest common divisor of two natural numbers to be the largest natural number dividing both of them; the definition is simpler, and there is no problem about uniqueness. But there is a reason for our strange choice, which will appear. We are going to describe Euclid s algorithm for finding the greatest common divisor of two integers. Changing signs doesn t affect the greatest common divisor, so we can assume that the integers are non-negative. Lemma (a) For every integer a, gcd(a,0) = a. (b) For any two integers a and b, if a = bq + r, then gcd(a,b) = gcd(b,r). Proof (a) is an exercise for you. (b) We use the general principle: if two pairs of integers have exactly the same divisors, then they have the same greatest common divisor. For, assuming that neither pair is (0, 0), the greatest common divisor is simply the largest number which divides both the numbers in the pair. So we need to show that, if a = bq + r, then (a,b) and (b,r) have the same set of divisors. Suppose that d a and d b. Then also d bq, so d r, since r = a bq. (In detail: d a and d b, so a = kd and b = ld for some natural numbers k and l. Then r = a bq = kd ldq = d(k lq), so d r.) In the other direction, if d b and d r, then d a, since a = bq + r. Euclid s Algorithm To find gcd(a,b) for two non-negative integers a and b: (a) if b = 0, then gcd(a,b) = a. (b) if b 0, apply the Division Algorithm to find q and r with a = bq + r and 0 r < b. Then calculate gcd(b,r); the result is equal to gcd(a,b).

10 98 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS Why does it work? We know that, if a = bq + r, then gcd(a,b) = gcd(b,r). Also, r < b. If r = 0, then the answer is a; otherwise we apply the algorithm again. The remainders get smaller at each step; we know that they can t go on getting smaller for ever, so after a finite number of steps the remainder reaches zero. It is easier to explain with an example. What is gcd(87,33)? So 87 = , 33 = , 21 = , 12 = , 9 = gcd(87,33) = gcd(33,21) = gcd(21,12) = gcd(12,9) = gcd(9,3) = gcd(3,0) = 3. The rule is: Continue dividing the previous divisor by the remainder until the remainder is zero. The last divisor used is the greatest common divisor. It is possible, though complicated, to give a proof that the algorithm works correctly; but hopefully this is clear to you. In the supplementary material we will see that Euclid s algorithm has another use as well The rational numbers The other solution to the problem that division is not always possible in the integers is to extend the number system to a larger one (the rational numbers) in which division is possible. Rather than say 7 divided by 2 gives quotient 3 and remainder 1, we will be able to say 7 divided by 2 is 7 2 = In this section, we will talk about how this extension is done. If I were to write out this section in full, it would look very similar to the construction of the integers, with only some differences in detail. Instead, I will just sketch the differences. We want to solve equations bx = a, where a and b are integers and b 0. The solution will eventually be the rational number a/b. Again there are two methods.

12 100 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS We re not finished yet! Mathematicians realised quite early on that this was not the end of the story for solving equations. We saw earlier that, empirically, if the side of a square is 1, then the diagonal is about But about is not good enough: what is it exactly? Pythagoras Theorem tells us that, if we take one of the two right-angled triangles into which the diagonal divides the square, then the square on the hypotenuse is equal to the sum of the squares on the other two sides. That is, the length of the hypotenuse is a number whose square is 2. Your calculator will tell you that = ; close to 2 but not exactly 2. If you try more places of decimals, you will get closer and closer to 2, but you will never get it exactly. This is because all the numbers you can type into your calculator are rational numbers (and rather special rational numbers at that, of the form / , say, where the denominator is a power of 10. Another famous theorem of Pythagoras tells us that there is no rational number whose square is equal to 2. We need one preliminary result. A natural number of the form 2k (for some natural number k) is even; one of the form 2k + 1 is odd. Every natural number is of one of these types: for if we apply the division algorithm to divide n by 2, we get n = 2k + r, where 0 r 1; that is, r = 0 or r = 1. Lemma The square of an even number is even, and the square of an odd number is odd. Proof If n is even, say n = 2k, then n 2 = 4k 2 = 2(2k 2 ); so n 2 is even. If n is odd, say n = 2k + 1, then n 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1; so n 2 is odd. Theorem There is no rational number x such that x 2 = 2. Proof First, an observation about rational numbers. Let m/n be a rational number, and suppose that m and n have a common factor d. This means that m = d p and n = dq for some integers p and q; so m/n = p/q. (To tie this in with what you learned in the last section, notice that mq = d pq = np, so the pairs (m,n) and (p,q) represent the same rational number.) Informally we say that a common factor can be cancelled from the numerator and denominator of a rational number without changing its value. So we can assume that all common factors have been cancelled, so that gcd(m,n) = 1. Now suppose that x = m/n satisfies x 2 = 2. That is, (m/n) 2 = 2, so that m 2 = 2n 2. This means that m 2 is even, so m is even (because the square of an odd number would be odd). Say m = 2k for some natural number k.

13 6.1. INTEGERS AND RATIONAL NUMBERS 101 Then we have 4k 2 = m 2 = 2n 2, so we can cancel a factor of 2 and obtain 2k 2 = n 2. Thus, n 2 is even, and so as before n is even. But now m and n are both even, so 2 is a common factor, contradicting our assumption that all the common factors have been cancelled. This contradiction shows that no such rational number x can exist. This provided a big problem for Pythagoras and his students. It is obvious that the length of the diagonal of a square has to be some number; but, if rational numbers are all that we have, there is no such number! As well as polynomial equations like x 2 2 = 0 that we can t solve in the rational numbers, there are other types of equations, for example (a) tanx = 1, which has a solution x = π/4, one-eighth of the circumference of a circle of unit radius; (b) logx = 1, which has a solution x = e, the base of natural logarithms. These numbers are both irrational. The proof for π is rather difficult. The proof for e is in the supplementary material, using a result from calculus. We see that we need to enlarge once again our stock of numbers. We do this in the next chapter Study skills 6: Make a stock of examples Later in the module we will discuss the uses of examples. But already you can see that they will be useful. You might remember from an earlier chapter that, although it often happens that, if p is prime, then the Mersenne number 2 p 1 is also prime, this is not always the case; it fails for p = 11, since = 2047 = If you remember this example, then you can immediately answer the question whether this general statement is true. In calculus it is particularly important to have a stock of examples at your disposal: a continuous function which is not differentiable, a convergent sequence of continuous functions whose limit is not continuous, and so on. But the principle applies in general to any branch of mathematics. Is a sum of rational numbers always rational? Yes, if it is a finite sum; but no, in general. Here is an example: 2 = = Here is a famous example which appears in many different parts of mathematics: n=1 1 n 2 = π2 + = 32 6.

14 102 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS If p is a prime number greater than 5, and you calculate 1/p as a decimal, it will recur, with period which divides p 1. Sometimes it is equal (for example, 1/7 = ); sometimes not (for example, 1/11 = ). You may find a use for these examples one day. Keep your own little stock, and add to it anything you find interesting. 6.2 Supplementary material More on Euclid s Algorithm Euclid s algorithm has the following consequence: Theorem Let a and b be natural numbers with gcd(a, b) = d. Then there are integers x and y such that ax + by = d. We will not prove this theorem; however, we will demonstrate how it works, using the example of Euclid s algorithm from the notes. But first a remark is in order. Suppose that a,b,d are integers such that (a) d a and d b; (b) d = ax + by for some integers x,y. Then d = gcd(a,b). For suppose that e a and e b. Then e ax and e by, so e ax + by = d. Now recall our example: 87 = , 33 = , 21 = , 12 = , 9 = We conclude that gcd(87,33) = 3. Now work up from the second last equation, expressing 3 in terms of what is above: 3 = = 12 ( ) 1 = = ( ) = = 33 2 ( ) 3 = , so gcd(87,33) = 3 = 87x + 33y, where x = 3 and y = 8. Again, it should be clear that this procedure will always work.

15 6.2. SUPPLEMENTARY MATERIAL The Fundamental Theorem of Arithmetic Theorem (Fundamental Theorem of Arithmetic) Every natural number greater than 1 can be factorised into primes; the factorisation is unique up to the order of the factors. Proof We saw earlier (Lemma 5.1.7) that every natural number greater than 1 has a prime factor. This is obviously the first step in proving that there is a factorisation into prime factors. (We leave the uniqueness until later.) So we will prove by Strong Induction the statement P(n): if n > 1, then n can be factorised into prime factors. So let us assume that P(m) holds for every natural number m < n, that is, every natural number m satisfying 1 < m < n can be factorised into prime factors. Now there are three possibilities for n: (a) n = 1. In this case there is nothing to prove, since P(1) is then trivially true. (b) n is prime. In this case we have our factorisation, with just one factor. (c) n is not prime. By Lemma 5.1.7, n has a prime factor p; and so n = pm, where 1 < m < n. (If m = 1, then p = n, contrary to the case assumption; and if m = n, then p = 1, contrary to the definition of a prime.) By the induction hypothesis, m = q 1 q 2 q r, where q 1,q 2,...,q r are primes; then n = pq 1 q 2 q r, where p,q 1,...,q r are primes. All in all, the inductive step is complete. The proof that the factorisation is unique, that is, that two factorisations of n into prime factors can differ only in the order of the factors, goes like this. We suppose that we have two factorisations of n into primes, say n = p 1 p 2 p r = q 1 q 2 q s. We have to show that r = s and that the qs are just the ps possibly in a different order. So we look first at p 1. If we could show that one of the primes on the right, say q i, was equal to p 1, then we could match p 1 with q i and cancel this factor. Then we could continue the process, matching p 2 with one of the remaining qs, and so on, until we had matched up the two factorisations. So the crucial thing is to show: If p is prime and p divides a 1 a 2...,a s, then p divides a i for some i.

16 104 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS Note that we allow the as to be arbitrary natural numbers here. If we then specialise to the case where they are primes, if p q i and q i is prime, then p = q i (since the only factors of q i are q i and 1, and p 1). We do this first for s = 2. Lemma Suppose that p is prime and p ab for some natural numbers a,b. Then either p a, or p b. Proof If p divides a, then our conclusion holds. So we may suppose that p does not divide a, and must conclude that then it divides b. If p does not divide a, then the greatest common divisor of p and a is 1 (since 1 and p are the only divisors of p). By our extended version of Euclid s algorithm (Theorem 6.2.1), we know that there are integers x and y such that xp + ya = 1. Multiplying by b, we see that xpb + yab = b. Now p obviously divides xpb; and by assumption, p divides ab, so p divides yab. Hence p divides the sum of these two numbers, which is xpb + yab = b, as required. Now it is easy to extend this to the product of more than two factors. Suppose that p is prime, and p divides a 1 a 2 a s. Then either p a 1 or p a 2 a s. In the first case, we have succeeded in showing that p divides one of the as. In the second case, either p a 2 or p a 3 a s. Continuing, we eventually conclude that p divides a i for some i Another proof of Pythagoras Theorem Pythagoras Theorem tells us that 2 is not a rational number. Here is a different proof, possibly closer to the one that Pythagoras found. Instead of 2, we will instead show that 1+ 2 is irrational. This is obviously equivalent. We will as usual argue by contradiction, and suppose that = m/n for some natural numbers m and n. Now the diagonal of a unit square clearly has length between 1 and 2, so lies between 2 and 3. This means that, when we apply the division algorithm to m and n, we obtain m = 2n + r, where 0 r m 1. And r cannot be zero, since this would mean that m = 2n, so = m/n = 2, which is obviously wrong.

17 6.2. SUPPLEMENTARY MATERIAL 105 Now n r = = = n m 2n n (1 + 2)n 2n = = m n. In the fourth line, we used the fact that ( 2 + 1)( 2 1) = 2 1 = 1. This means that, whatever fraction m/n we find for 1 + 2, we can find a fraction n/r with smaller numerator and denominator. This is impossible! Another way of saying it is that the equation m = 2n + r is the first step of Euclid s algorithm for finding the greatest common divisor of m and n. But the ratio n/r is the same as the ratio m/n. This will remain true at every step of the algorithm; so the algorithm will never terminate. This argument can also be expressed geometrically, more in line with the way that Pythagoras (and Euclid) probably thought. You can find this in Peter Cameron s Number Theory notes, on page The irrationality of e The number e, which is the basis of natural logarithms, is known to be irrational. The proof follows, using an expression for e which will be discussed further in the next chapter (and which also describes what we need for a sum like this to converge to a real number). For now, let us just assume that the expression below specifies a particular real number called e. The expression is: Proposition e = n=0 1 n!. We will need another sum, the geometric series: Proposition If 1 < r < 1, then ar n = a n=0 1 r.

18 106 CHAPTER 6. INTEGERS AND RATIONAL NUMBERS Given this, here is the proof that e is irrational. As you might expect by now, it is a proof by contradiction. Let us suppose that e is equal to the rational number p/q. Then q!e = p(q 1)!, since the q in the denominator cancels with the factor q in q!, and the remaining factors 1,2,...,q 1 have product (q 1)!. In particular, we see that q!e is an integer. But from the infinite series, we have q!e = q! n=0 We break this sum into two parts; first the terms from 0 to q, and then the remaining terms. For the first part, q! q n=0 1 n!. q 1 n! = q! n=0 n!. Since n q, the fraction q!/n! is an integer: it is equal to (n + 1) q, since the factors 1,...,n cancel). So the sum of all these terms is also an integer. For the second part, we have again to deal with fractions q!/n!, this time with n > q. Now the factors in the numerator cancel into the denominator, and we have q! n! = 1 (q + 1)(q + 2) n 1 (q + 1) n q, since we have replaced some of the factors in the denominator by smaller ones. So the second part of the sum is The last sum is q! n=q+1 n! < n=q+1 This is a geometric series, whose sum is 1 (q + 1) n q. 1 q (q + 1) 2 + 1/(q + 1) 1 1/(q + 1) = 1 q 1. So the integer q!e is the sum of two terms, of which the first one is an integer and the second is a positive number smaller than 1. This is impossible. The contradiction shows that e is irrational.

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