The mysterious and somewhat unreal set of real numbers

The mysterious and somewhat unreal set of real numbers 1 Introduction In the beginning were the natural numbers, 1, 2, 3,...They were soon followed by the rational numbers, so called because they are ratios of natural numbers. For the ancient Greeks (and others), mathematical objects were seen from a very geometric point of view; numbers were used to measure lengths, areas and volumes; they had no use (and probably no conception) of the notion of a negative number, or even 0. But with the hindsight provided by the millennia that passed since then, and thanks to the genius of some Indian mathematicians of the seventh or eighth century, we are (I hope) quite comfortable working with 0 and negative numbers. So here are the basic numbers, and the symbols we use to denote them. The set of natural numbers N = {1,2,3,4,...}. In this set we have two operations, + and ; if a,b N, then a + b and a b are in N. One usually writes a b or, even more commonly, ab for a b. As the number system expands, operations in the larger set when applied to elements of the smaller set have the same result as they had previously. By this I mean, for example: If a/b,c/darerationalnumbers, with a,b,c,d integers, we define a b + c d = ad+bc. If n,m are integers, then they are interpreted bd as fractions by n = n/1, m = m/1 and new n+m = n 1 + m 1 = n 1+1 m 1 1 = n+m 1 = old n+m. The set of integers Z = {0,±1,±2,±3,...,}. In this set we have the two operations + and. But now Z is what is called a group under +, meaning that there is an identity element 0 for +, and for every a Z there is a Z such that a+( a) = 0 = ( a)+a. Because a+b = b+a, Z with + is a commutative (also called abelian) group. The set Q of rational numbers. We can (somewhat informally) define Q as the set of all numbers of the form m/n where m,n Z and n 0, with the proviso that m n = m n if and only if mn = m n. In Q we again have the operations + and, but Q is a field under these operations, meaning that it is a commutative group under +, and that Q\{0} (i.e., Q without the 0 element) is a commutative group under, with identity element 1.

2 THE NUMBER LINE 2 It would be easy to assume that with Q we have all the numbers we could ever need. In some ways we do. We have little use for anything better than rational numbers in everyday life. If you measure the length of a table, for example, there is no such thing as absolute precision; once you get to the atomic level any further precision is meaningless. And rational numbers are quite sufficient to give the length to any desired meaningful precision. But mathematics, even applied mathematics, takes the real world and maps it onto an ideal world, in which all lengths (and areas, and volumes) are possible, works in this ideal world and, if it is applied mathematics, then interprets the results in real terms. The fact that mathematics works, that it helps to understand the world, is something that has been a matter of wonderment to many. It seems as if Plato and Galileo may have been right after all. Plato, in believing in the existence of an ideal world of which this world is a mere shadow; Galileo in that this world is governed by a blueprint written in mathematical terms. 2 The number line Let s say we only have Q. We can represent the points of Q on a straight line as follows. Select a point for 0 and a point to the right of 0 for 1. 0 1 Now one can add the rest of the points of Q, according to size. First we place the integers Z in the obvious way. Because the Greeks were firm believers in constructions using only a compass and straightedge (unmarked ruler), with center at 1, we put the other end of the compass at 0, draw a circle, and where this circle intersects again the line, we place the integer 2. Then with center at 2, other end at 1, we draw a circle intersecting the line at 1 and at another point which corresponds to 3. And so forth. Similarly we place the negative integers 1, 2,... going toward the left instead of to the right. If a Q, a 0, then a = m/n with m,n Z, n 0. We can always assume n > 0. Divide m by n; that is write m = kn+l, where k,l Z and 0 l < n. For example, if 0 m < n, then k = 0 and l = m. Thus a = m n = k + l n. To place the number a, divide the segment from k to k +1 into n equal parts (this can be done using only a compass!); the end points of the little segments so obtained correspond to the numbers k,k + 1 n,k + 2 n 1 n,...,k + n,k + 1, of which a is one. Here is a picture of the line with a few points explicitly marked. Grade, middle and high school kids would write this as k l and call it a mixed fraction n (or something like that), but I find this notation dangerous because it can be confused with k times m/n.

2 THE NUMBER LINE 3-4 -3-2 1 3 5 9 2 4 13 0 1 1 2 3 4 Now suppose we draw at the point 1 a line perpendicular to the number line, mark off a point at height 1. This can be done using only a compass and a straightedge. Join the point at height one, call it P, to the point 0: P 0 1 Next, with center at 0, placing the other end of the compass at P, draw a circle. This circle intersects the line at two points. Let Q be the intersection to the right of 1. P 0 1 Q If math is to be of any use in real life, Q should correspond to a number. But, as we know, this number would be the length of the segment 0P and this length, if it were a number, would be a number whose square is 2. To the horror of the Greeks, they could prove that there is no ratio of integers having a square equal to 2; the square root of 2 is irrational! If we do not expand our system of numbers past the rational level, we could not perform the most basic geometric

3 THE REAL NUMBERS 4 constructions in a meaningful way. And so, like it or not, the real numbers had to be invented. Or discovered 3 The real numbers The real numbers were being used long before there was any attempt to formalize them. There is a level at which mathematics and philosophymeet, and it is a level avoided by most mathematicians while doing mathematics. In mathematics, at least as we do it nowadays, what objects are is of little or no importance. The questions what is the number 1?, what is the nature of 3?, what is 7? may be of great interest from a philosophical point of view, but are totally irrelevant from a mathematical point of view. Mathematical objects are determined by their properties. Two objects having the same exact properties can be considered as being one and the same. Within reason, and some care. The way a mathematical theory is set up is by listing some basic properties (axioms) of the objects it works with, and then deducing everything else by logical arguments. Concerningthe real numbers, you all haveworkedalot with those numbers, and you have (I hope) a feeling for them. What I will do here is to state the defining properties of these numbers; everything (and by everything I mean everything) else having to do with them can be proved (i.e., obtained by logical arguments) from these properties. One can (and one should) ask oneself whether such objects really exist. Setting up a theory requires some conditions on the axioms, a very important one is consistency. Now if the theory involves only a few objects and axioms, consistency can be easy to determine. Otherwise, it can be quite hard. The standard way to show a theory is consistent is to build a model. One of the nicest and most elegant ways of building a model for the real numbers, though not necessarily the most intuitive way, is the way Richard Dedekind did it in the 19th century; his method is known as Dedekind cuts and is presented in Section 2 of Chapter 1 of our textbook. What this construction shows is that theobjectdefinedbythenexttheorem, andtobedenotedbyr, hasanexistence as solid as that of the natural numbers; if you believe in natural numbers (and standardlogic),youcanbelieveintherealnumbers. Onceyouaresatisfiedofthe consistency of R, the best thing to do is to forget the construction. Otherwise, it can be quite counterproductive. When you think of 2, for example, you should see it as a number more or less similar to any other number; in decimal terms it has an expansion 1.4142135623730950488016887242097... which goes on and on; you should not think of it as being a pair of sets of rational numbers (A,B), where A = {x Q : x > 0,x 2 < 2} {x Q : x 0}, B = {x Q : x > 0,x 2 > 2}. An interesting question is whether when mathematicians come up with new stuff, are they creating it or just discovering it? One should remember that the Greeks did NOT think in decimal terms.

3 THE REAL NUMBERS 5 I will state first a theorem and then a definition. The theorem may contain some unfamiliar terms. I ll explain them in detail after the statements. Theorem 1 Up to order preserving isomorphisms, there exists a unique ordered field that is complete in its order. Definition 1 The unique field mentioned in Theorem 1 is known as the field of real numbers and denoted by R. 3.1 The terms appearing in the theorem explained. Field. A field is a set F in which two operations have been defined, usually denoted by + and, where we write ab for a b, such that the following properties hold: 1. + is commutative; a+b = b+a for each a,b F. 2. + is associative; a+(b+c) = (a+b)+c for each a,b,c F. 3. There exists an element 0 F such that a+0 = 0+a = a for each a F. 4. For each a F, there exists a F such that a+( a) = 0. Comments: These four properties can be summarized by stating (F,+) is a commutative group. The element 0 is easily seen to be unique; for each a F it is easy to see that there is only one element that can work as a. Furthermore one sees easily that if a,b F and a+b = a,then b = 0. If a,b F, one writes a b for a+( b). One sees that x = a b is the unique solution of the equation b+x = a. 5. is commutative; ab = ba for each a,b F. 6. is associative; a(bc) = (ab)c for each a,b,c F. 7. There exists an element 1 F such that 1 0 and a 1 = 1 a = a for each a F. 8. For each a F, such that a 0, there exists a 1 F such that aa 1 = 1. Comments: These next four properties can be summarized by stating (F\{0}, ) is a commutative group. The element 1 is easily seen to be unique; for each a F, a 0, it is easy to see that there is only one element that can work as a 1. At this point there is no relation between + and One can see easily that if a,b F and ab = a, and if a 0, then b = 1. More generally, one sees that if a 0, then x = a 1 b is the unique solution of ax = b. What happens if a = 0 has to wait for the next and final property. If a,b F, a 0, one writes b/a for a 1 b. Thus the unique solution of ax = b if a 0 is usually written in the form b/a.

3 THE REAL NUMBERS 6 9. distributes with respect to +; a(b+c) = ab+ac for each a,b,c F. Comment: This property relates the two operations. Now one can prove that a 0 = 0 for each a F. In fact, let x = a 0. We have a+x = a 1+x = a 1+a 0 = a(1+0) = a 1 = a; since a +x = a, we conclude that x = 0. One can also prove other usual properties of these operations, such as: a( b) = ab = ( a)b a,b F, ( a)( b) = ab a,b F ab = ac,a 0 b = c There are zillions of different fields in existence. Because at the very least a field has to contain an element denoted by 0, an element denoted by 1, and 1 0, the minimum number of elements a field can have is two. There is a field of two elements. We may as well denote the elements by 0 and 1; the set F = {0,1} becomes a field defining etc. 0+0 = 0,0+1 = 1+0 = 1,1+1 = 0, 0 0 = 0,0 1 = 1 0 = 0,1 1 = 1. it is easy to see there is no other possible way of defining the operations. The only one in question might be why should 1+1 = 0? The alternative is 1 + 1 = 1, but then 1 = 0. One still has to see that the axioms are verified, a slightly boring but trivial exercise. Because there is only one way to define the operations, we can say that up to isomorphism there is only one field of two elements. If you want, instead of denoting the elements by 0,1 you denote them by a,b; then one of a,b has to play the role of 0, the other one of 1, and after that all is the same. It is customary to denote, as we did, the two identity elements (additive and multiplicative) of a field by 0, 1, respectively. Sometimes this could cause a confusion; for example, when one wants to distinguish between the 1 (or 0) of the field and the integer 1 (or 0). Or if there is more than one field in play. Then one can denote these identity elements by 0 F,1 F,respectively. Isomorphism. Suppose F,G are fields; I ll use the same symbols for the operationsin both F and G. An isomorphism fromf to Gis a one-to-one, onto map f : F G such that f(a+b) = f(a)+f(b), f(ab) = f(a)f(b) for each a,b F. It is easy to see that in these circumstances f(0 F ) = 0 G, f(1 F ) = 1 G and that the inverse f 1 is an isomorphism from G to

3 THE REAL NUMBERS 7 F. We say F is isomorphic to G, and write F = G, iff there exists an isomorphism f : F G. Since then f 1 is, as mentioned, an isomorphism in the opposite direction, we also have G = F so that the relation is symmetric and instead of saying F is isomorphic to G one can say F and G are isomorphic. Clearly, every field is isomorphic to itself (the identity map being an isomorphism) and it is also immediate that compositions of isomorphisms are isomorphisms, so that the relation is transitive. In brief, isomorphism is an equivalence relation for fields. Isomorphic fields are considered indistinguishable, as long as no other properties are involved. One can show, and you may have seen it alreadyin an algebracourse, that for each prime number p, each natural number n, there exists exactly one field (up to isomorphism) with p n elements. No other finite fields exist. So, is there a field of 7 elements? Yes, 7 is prime. Is there a field of 121 elements? Yes, 121 = 11 2 is the power of a prime. Is there a field of 6 elements? No, 6 is not the power of a prime. Ordered Field. We say F is an ordered field if it is a field (of course) and there is a relation, usually denoted by <, in F such that the following properties hold. 1. For each a,b,c F, if a < b and b < c, then a < c. (Transitivity) 2. For each a,b,c F, if a < b, then a+c < b+c. 3. For each a,b,c F, if a < b, and c > 0, then ac < bc. 4. For each a,b F one and exactly one of the three following possibilities holds: (a) a < b. (b) a = b. (c) b < a. (Trichotomy) If F is an ordered field and < is the relation, one defines: a > b if and only if b < a a b if and only if a < b or a = b a b if and only if b a To get a bit of the flavor of an ordered field, here is a simple lemma and its proof. Lemma 2 Assume F is an ordered field. If a,b F and if a b and b a, then a = b. Proof. Assume a b and b a. Because a b, then a < b or a = b. Whichever one of these two possibilities holds, b < a cannot hold; i.e.,

3 THE REAL NUMBERS 8 b a. But b a and b a is only possible if b = a. Not every field can be ordered. An important property of an ordered field is that all non-zero squares are positive. That is: If F is an ordered field, if a F, if a 0, then a 2 > 0. To get there we need to see first that if a > 0, then a < 0, and vice-versa. Let us be formal, state all this and more as a lemma, with proof. The fact that in any field (or additive group) ( a) = a is very easy and, if necessary, left as an exercise. Lemma 3 Assume F is an ordered field. 1. If a F, then a > 0 if and only if a < 0. 2. If a,b,c F, a < b and c < 0, then ac > bc. 3. If a F and if a 0, then and a 2 > 0. 4. If a F and a > 0, then 1/a > 0 5. If a,b F and 0 < a < b, then 0 < 1/b < 1/a. Proof. 1. Assume a > 0. By trichotomy, a is either = 0, > 0 or < 0. Now a = 0 implies a = 0, so a 0. Thus a > 0 or a < 0. Assume a > 0. Adding a to both sides we get ( a)+a > 0+a, which is 0 > a; contradicting a > 0. You see, I hope, how these proofs go; one shows that one of the three possible relations holds by showing that the other two cannot hold. For the converse, one proceeds similarly. Assuming a < 0,one sees that a 0, so either a > 0 or a < 0. If a < 0, add a to both sides to get a+( a) < 0+( a), hence 0 < a, a contradiction. 2. Assume a,b,c F, a < b and c < 0. Adding a to the first inequality, we get b a > 0. Since c < 0 we have that c > 0. In fact, the alternative is c < 0, which implies c > 0 by what we proved. Thus (b a)( c) > 0( c) = 0; i.e., bc+ac > 0. Add bc to both sides to get the desired conclusion. 3. Assume a F and a > 0. Then a > 0 or a < 0. If a > 0, we can multiply the inequality a > 0 by a to get a 2 = a a > a 0 = 0; i.e., a 2 > 0. If a < 0, then a > 0 and by what we proved a 2 = ( a)( a) = ( a) 2 > 0. 4. Assume a > 0. Since a 1 a = 1 0, we see that 1/a 0. Thus the alternative to 1/a > 0 is 1/a < 0. If 1/a < 0, multiplying by a we get 1 < 0, a contradiction. The result follows.

3 THE REAL NUMBERS 9 5. Assume 0 < a < b. By what we proved 1/a > 0,1/b > 0. We can thus multiply a < b by 1/a to get 1 < b/a. Next multiply this last inequality by 1/b to get 1/b < 1/a. An important though simple corollary is: Corollary 4 Assume F is an ordered field. Then 1 > 0. Proof. 1 0, thus 1 = 1 2 > 0. Afirstpeculiarityoforderedfieldsisthattheyareinfinite!. Infact, assume first F is anyfield. What I am about to dowill be less confusingif I denote for a moment the element 1 of the field by 1 F, use 1 to denote the natural number 1. If n N we can define n 1 F F by: 1 1 F = 1 F, 2 1 F = 1 F +1 F, 3 1 F = 1 F +1 F +1 F, and so forth; inductively, assuming n 1 F defined, one defines (n+1) 1 F = n 1 F +1 F. If the field is finite, one cannot possible have all of these field elements being distinct; there have to be n, m N, m < n such that n 1 F = m 1 F. Setting k = n m one has k N and k 1 F = 0 F, the zero element of the field: In a finite field, adding again and again the element 1 to itself, eventually produces the 0 element. There always is k N such that k 1 F = 0 F. One can show (it isn t too hard) that the smallest such positive integer k is prime, called the characteristic of the field; if p is the characteristic of F (i.e., p is the smallest positive integer such that p 1 F = 0 F ), then k 1 F = 0 if and only if k is a multiple of p. But things are very different if we have an ordered field. In this case, (I drop the subscript F) because 1 > 0, we have 1+1 > 1+0 = 1 > 0, since 1 + 1 > 0, we have 1 + 1 + 1 > 1 + 0 = 1 > 0; and so forth. Adding 1 any number of times always results in a positive element of the field; so n 1 m 1 if m < n; otherwise (n m) 1 = 0. The set {n 1 : n N} is infinite forcing the field F to be infinite since it contains an infinite set. We can identify 1 + 1 with the integer 2, 1 + 1 + 1 with the integer 3, and so forth; thus an ordered field F contains a replica of N; identifying the additive inverse of 1,2,3,... in F with the integers 1, 2, 3,..., respectively, we see that an ordered field contains a replica of Z; a perfect replica because one can see that operating on the images of the integers in the field corresponds exactly to operating on the integers. But wait!, there is more. If F is an ordered field, it contains a replica of the integers. As mentioned above, it is so perfect a replica that we simply identify the elements of F

3 THE REAL NUMBERS 10 with the corresponding integers. Because all that matters are properties, we can even say that Z is a subset of F. But if Z is in it, so are the multiplicative inverses of Z; in short, every ordered field F contains a perfect replica of Q, perfect in the sense that applying the field operations to the replica elements gives the same results as applying the operations of Q first and then replacing by the replicand. Order is also preserved. I will try to make all this precise as follows, using again the notation 1 F, 0 F, for the 1 and 0 elements of F, to minimize confusion: Theorem 5 Assume F is an ordered field. There exists a unique map Φ : Q F with the following properties: 1. Φ is one-to-one. 2. Φ(0) = 0 F and Φ(1) = 1 F. 3. If a,b Q, then Φ(a+b) = Φ(a)+Φ(b) and Φ(ab) = Φ(a)Φ(b) 4. If a Q, then Φ( a) = Φ(a). 5. If a Q, a 0, then Φ(1/a) = 1 F /Φ(a). 6. If a,b Q, a < b, then Φ(a) < Φ(b). I won t prove it in detail, just sketch a bit the proof. First of all, several properties are a bit redundant. Assume, for example that one has a map Φ : Q F that is simply not identically zero (i.e., there is a Q such that Φ(a) 0 F and which satisfies simply If a,b Q, then Φ(a + b) = Φ(a) + Φ(b) and Φ(ab) = Φ(a)Φ(b). It then is automatically one-to-one and satisfies all the other properties. In fact, in the first place Φ(0) = 0 F, since Φ(0) = Φ(0+0) = Φ(0)+Φ(0) and in any field, x+x = x implies x = 0 F. Next Φ(1) = 1 F. In fact, here is where the existence of a Q such that Φ(a) 0 F comes in. We have Φ(a) = Φ(a 1) = Φ(a)Φ(1), hence Φ(1) = 1 F since Φ(a) 0 F. We can next prove that x Q, x 0 implies Φ(x) 0 F. In fact, assuming Φ(x) = 0 F we would have the following nonsensical situation: 1 F = Φ(1) = Φ(x 1 x ) = Φ(x)Φ(1 x ) = 0 F Φ( 1 x ) = 0 F. Next we see that Φ( a) = Φ(a), and Φ(1/a) = 1 F /Φ(a) if a 0. Both proofs are similar. We have 0 F = Φ(0) = Φ(a+( a)) = Φ(a)+Φ( a), thus Φ(a) = Φ( a). Similarly, if a Q, a 0, 1 F = Φ(1) = Φ(a 1 a ) = Φ(a)Φ(1 a ),

3 THE REAL NUMBERS 11 thus Φ(a) 1 = Φ( 1 a ). To see Φ is one-to-one, suppose a,b Q, a b. Then a b 0 and Φ(a) Φ(b) = Φ(a b) 0; i.e., Φ(a) Φ(b). Let us see also that the conditions we assume on Φ; ie., that Φ satisfies 3 and does not map everything to 0 F, make it unique. That is quite immediate now. It is easy to see that if a = m/n Q, then the only choice for Φ(a) is m 1 F /n 1 F. Finally, this also shows that if a Q, a > 0, then Φ(a) > 0, and from this it is very easy to conclude that Φ(a) < Φ(b) if a < b. The only remaining question is whether such a Φ exists. Well, yes, simply define Φ(m/n) = m 1 F /n 1 F, and see that the propertieshold. An importantpropertytoverifyisthatifm,m,n,n Z, n 0 n and m/n = m /n, then m 1 F /n 1 F = m 1 F /n 1 F. That is quite immediately verified, but needs to be done to see the definition makes sense. Having seen that every ordered field contains a replica of Q, we see that Q is in a way the smallest ordered field. Order Isomorphism. Assume F, G are ordered fields. An order isomorphism from F to G is a map f : F G such that 1. f is an isomorphism. 2. If a,b F and a < b, then f(a) < f(b). (I am using the same notation for the relation in G as for the relation in F.) Completeness in the order As mentioned, there are zillions of different fields. Imposing an order relation cuts their number down significantly; for one, it eliminates all finite fields. But zillions still remain, all containing a replica of Q. The final property we have to consider, which reduces the number down to essentially only one, is completeness in the order. For this we need to know what it means to be bounded above, below, and bounded, in an ordered field. Actually, the concept makes sense in any ordered set; i.e., a set S in which one has defined a transitive relation <. But let s stick to fields. Assume until further notice that F is an ordered field. Let S F and let b F. We say that b is an upper bound of S iff x b for each x S. We say a subset S of F isbounded above iff it has an upper bound; i.e., there exists b F such that x b for all x F. Some comments may be in order. A set may or may not have an upper bound. For example, in Q, the set N does not have an upper bound. But if it has one upper bound, it has an infinity of them. In fact, if b is an upper bound of S, so are b+1,b+1+1, etc. (By b+1,b+1+1,..., I

3 THE REAL NUMBERS 12 mean, of course, b+1 F, b+1 F +1 F,...) Upper bounds may or may not belong to the set of which they are an upper bound. TwoimportantalbeittrivialsubsetsofanyfieldF arethe fieldf itselfand the empty set. Considerfirst the field F. Can it be bounded above? The answeris clearly no; if b F werean upper bound of F, since b+1 F we would need to have b+1 b, which is nonsense. The case of the empty set is a bit more subtle. A fundamental principle of logic is that if a premise of a reasoning is false, whatever conclusion one comes up with, the whole reasoning is right. For example, consider the following statements: If Boca Raton is the capital of the Unites States, then monkeys can fly. If all dogs are white, then all cats are black. If 2 < 1, then all equations have solutions. They are all perfectly true. So let us analyze what it means for a set to be bounded above. Assume S F; S is bounded above if there exists b F such that b is an upper bound of S. The defining property of being an upper bound is: b F is an upper bound of S if and only if the following statement is true: If x S, then x b. Well, assume now S =. Then x S; i.e., x is automatically false so any implication having this statement as a premise will be true. That is, let b be any element of F. Then the statement If x, then x b is true. Think about it. The conclusion is: Every element of F is an upper bound of ; the set of upper bounds of is F. The empty set is about as bounded above as any set can be! What goes up must come down or, what can be done above, can be done below. Similarly one defines: Let S F and let b F. We say that b is a lower bound of S iff x b for each x S. We say a subset S of F is bounded below iff it has a lower bound; i.e., there exists b F such that b x for all x F. Similar comments to those we made abovefor upper bounds hold for lower bounds. A set may or may not be bounded below but, if it is, it has an infinity of lower bounds. The empty set is bounded below; every element of F is a lower bound of. Finally, a subset of F is said to be bounded if it is both bounded above and below. A subset of F may or may not have a maximum (or minimum) element. I ll define both concepts at once, using the high road/low road method.

3 THE REAL NUMBERS 13 To get one definition take always the high road, for the other one, the low road. { } maximum Let S F. We say S has a iff there is b S such that { } minimum x b for each x S. b x Notice the big difference with the definition of upper/lower bound. The assumption now is b S. Looking at the definition of maximum, we see that if there is b S such that x b for each x S, then it is unique. In fact, if we also have c S such that x c for each x S, then we can take x = c in x b; x = b in x c, to get c b, b c, hence b = c. So if S has a maximum, we denote by maxs the one and only element of S such that x maxs for all x S and call it, naturally, the maximum of S. Similarly for a minimum; if S has a minimum, we denote by mins the one and only element of S such that mins x for all x S. Certain properties are quite immediate. For example, it is immediate that if S F and S contains an upper bound, then that upper bound is the maximum of S. Similarly, if b{ S and b is} a lower bound of S, maximum then b = mins. Thus a set has a if and only if it is { } { minimum } bounded above an upper bound and contains. In this case it bounded below a lower bound { } { } upper bound upper bound can contain exactly one and this is { } lower bound lower bound max the of the set. min Consider the set S = {1,1/2,1/3,...} Q. It is clear that S is bounded and maxs = 1. It isalsobounded below; by0andeverynegativerational. However, it does not have a minimum; if b S, then b = 1/n for some positive integer n and if m N and m > n, then 1/m S and 1/m < 1/n = b, so b cannot be the minimum. The following result is sort of obvious, but useful. Lemma 6 Let S be a non-empty finite subset of F. Then mins,maxs exist. Proof. It will be convenient to deal first with the case of a set of 1 or 2 elements. If S = {a,b}, where a,b F, then either a b or a > b. In the first case we obviously have mins = a and maxs = b; in the latter case

3 THE REAL NUMBERS 14 mins = b and maxs = a. Generally speaking, if a,b F, we can define max(a,b),min(a,b) by similarly max(a,b) = a if b a, otherwise max(a,b) = b; min(a,b) = a if a b, otherwise min(a,b) = b. Let n be the number of elements of S; since S, we have n N. Write S = {s 1,...,s n } and proceed by induction on n. If n = 1, it is clear that maxs = s 1 = mins. Assume the result proved for some n 1. Let S = {s 1,...,s n+1 } have n + 1 elements. Let T = {s 1,...,s n }; by the induction hypothesis, maxt,mint exist. It is now easy to see that max(maxt,s n+1 ) = maxs, min(mint,s n+1 ) = mins. We finally arrive at the threshold of the last property. Let S F. We say that S has a least upper bound or supremum iff there exists an element σ of S (and call σ a supremum of S) with the following properties: 1. σ is an upper bound of S. 2. If b is an upper bound of S, then σ b. In other words, a least upper bound is exactly what its name says it is, a least upper bound. Now least objects tend to be unique; that is why they are called least and not lesser. The same is true with least upper bounds. Suppose we could have two least upper bounds of a set S, call them σ and τ. Because τ is an upper bound and σ is all upper bounds, we haveσ τ. Switchingroles, becauseσ isan upperbound and τ is all upper bounds, we have τ σ. That τ = σ follows. Since, if there is a least upper bound there is only one, we can refer to it (when it exists) as the least upper bound of the set S; it is denoted by sups (sup for supremum). Similarly, we say that S has a greatest lower bound or infimum iff there exists a necessarily unique element σ of S, to be called the greatest lower bound or infimum of s and denoted by infs, such that: 1. σ is a lower bound of S. 2. If b is a lower bound of S, then b σ. Here is now the relevant definition. Assume that F is an ordered field. We say it is complete in the order iff every non-empty subset that is bounded above has a least upper bound.

4 WHAT DOES THIS MEAN FOR US? 15 Why does one not also require that every non-empty subset that is bounded below should have a greatest lower bound? The answer is that one gets it for free. In fact, assume sups exists for every non-empty subset that is bounded above. Assume now S and S is bounded below. Let T be the set of all lower bounds of S. Because S is bounded below, T is not empty. Because S is not empty, T is bounded above; in fact, let s S. Then t s if t T. Every element of S is an upper bound of T. Thus supt exists. It is easy to see that supt must be infs. In fact, let s S. Then, as mentioned, s is an upper bound of T, thus supt s. This shows that supt is a lower bound of S. Let b be a lower bound of S. Then b T and (since supt is an upper bound of T), we have b supt. Thus supt is every lower bound of S. It follows that supt = infs. An alternative proof can be obtained showing that defining S = { x : x S}, then infs = sup( S). And that is it! Theorem 1 should now be reasonably clear. The proof is not so easy. Two things need to be proved: There exists an ordered field that is complete in its order, and there is only one such field (up to unique order isomorphisms). That is, if F, G are ordered fields complete in their order, there exists a unique order isomorphism f : F G. The existence part of the proof can be accomplished, for example, by the method of Dedekind cuts (Chapter 1, 2 of our textbook). The uniqueness part is actually quite easy, a matter of standard tricks, and doing the obvious. 4 What does this mean for us? What Theorem 1 does for us is to distill the essence of the set of real numbers. Each one of us has an interpretation, an idea, of what the real numbers are; and that is fine, and should be maintained. But stripped of all decorations the real numbers are completely characterized by being a completely ordered field that is complete in its order. As an ordered field, it contains a replica of Q; we simply say that Q is a subfield of R. A good way of looking at it, perhaps, and one in which Q is more naturally a subset of R is as follows. The field Q is ordered but not complete in its order. Consider, for example, the following subset S of Q: S = {x Q : x 0 and x 2 < 2}. This set is clearly not empty, 0 S, so is 1. It is bounded above, by 2 for example. In fact, if x 2, then x 2 4 > 2, so nothing 2 can be in the set, making 2 an upper bound. Can this set have a sup in Q? One can show, I won t go into details here, that if σ = sups, then σ 2 = 2. Thus σ / Q (Incidentally, this also proves that 2 R). One can think of obtaining R from Q by adding suprema (the correct plural of supremum is suprema) to all bounded non empty sets that lack a supremum in Q. From a logical, constructive point of view, this is a bad definition, but it could be intuitively a better way of looking at it.

5 AN IMPORTANT QUIRK 16 There are holes in Q; filling them produces R. The main thing to keep in mind is that EVERY property of the real numbers should be deducible using at last instance ONLY the fact that R is an ordered field complete in its order. 5 An important quirk I began these notes talking about the appearance of irrational numbers. One consequence of their appearance for the Greeks was the difficulty in proving equality of proportions, a situation which was finally resolved by Eudoxus. His theory of proportions is developed in Book V of Euclid s Elements. Very, very roughly he sees a fact that was then used masterfully by both Euclid and Archimedes. Stated in modern terms it sort of says that the only way to prove that two real numbers are equal is to prove that they are not unequal. Let me be more specific. Say a,b R and we want to prove a = b. Here is a typical proof: Assume first a < b; derive a contradiction. Next, assume b < a; derive a contradiction. By trichotomy, a = b. If you get only one thing out of these notes let it be this one thing, repeated for emphasis: Essentially, to say that two real numbers are equal is to say that they are not unequal. It is almost impossible to prove a = b for real numbers, except by contradiction. Since the late nineteenth, early twentieth century, certain mathematical logicians do not accept some basic reasoning premises, most especially the so called principle of the excluded third which justifies proofs by contradiction. This leads to some curious results, to a strange mathematics. It should be mentioned that all of these mathematical logicians are above all mathematicians, and most of them also do classical mathematics when not being infused by the logical spirit. But we are ONLY doing mainstream mathematics here. A non-believer in proofs by contradiction may ask, for example, is 0.999999...= 1? From a mainstream point of view, we first have to define carefully what is meant by 0.999999... It is lim n 9 10 + 9 100 + + 9 10 n Using completeness in the order it is not hard at all to prove that this limit exists. Let us call it s. So is s = 1? There is NO direct way of proving this, that I know of. But one can show that s < 1 leads to a contradiction, and that s > 1 leads to a contradiction. Thus s = 1.