AXIOMS FOR THE INTEGERS

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1 AXIOMS FOR THE INTEGERS BRIAN OSSERMAN We describe the set of axioms for the integers which we will use in the class. The axioms are almost the same as what is presented in Appendix A of the textbook, with a couple of differences explained in Remark 1 below. Near the end, there are also a couple of examples of proof by contradiction. Although we have been careful when using commutativity and associativity in this note, you do not need to invoke them explicitly in your own proofs. The axioms. The integers, which we denote by Z, is a set, together with a nonempty subset P Z (which we call the positive integers), and two binary operations addition and multiplication, denoted by + and, satisfying the following properties: (Commutativity) For all integers a, b, we have a + b = b + a and a b = b a. (Associativity) For all integers a, b, c, we have a + (b + c) = (a + b) + c and a (b c) = (a b) c. (Distributivity) For all integers a, b, c, we have (a + b) c = a c + b c. (Identity) There exist integers 0 and 1, such that for all integers a, we have a + 0 = a and a 1 = a. (Additive inverses) For any integer a, there exists an integer a such that a + ( a) = 0. (Closure for P) If a, b are positive integers, then a + b and a b are positive integers. (Trichotomy) For every integer a, exactly one of the following three possibilities hold: either a is a positive integer, or a = 0, or a is a positive integer. (Well-ordering) Every nonempty subset of the positive integers has a smallest element. For the well-ordering property, we have not yet defined smallest, so we can rephrase it more formally as follows: if S is a nonempty set of positive integers, then S contains an element a such that for every b in S which is not equal to a, we have b a is a positive integer. Remark 1. There are a couple of differences from the axioms as presented in Appendix A of the textbook. First, the book lists closure as an axiom, but we don t need to state it separately, because when we say that + and are binary operations on the integers, this means precisely that they give rules which associate to any pair of integers a and b a new integer a + b (or a b). More significantly, the book lists the cancellation law as an axiom, but we will see in Corollary 11 that it actually follows from the other axioms. We use the conventional order of operations, with multiplications occurring prior to additions. Thus, in the distributive law, a c + b c means (a c) + (b c). As usual, we will write a b as an abbreviation for a + ( b). 1

2 Uniqueness observations. Notice first that 0 and 1 have been defined implicitly in the axioms, as the additive and multiplicative identities. However, in order to know that there is no ambiguity in defining them this way, we need to know: Lemma 2. 0 and 1 are uniquely defined by the property of being the additive and multiplicative identity, respectively. Proof. Suppose that 0, 0 are integers, and both are additive identities, so that for all integers a, we have a + 0 = a = a + 0. Then we want to show that we must have 0 = 0. But using also commutivity, we have 0 = = = 0, The same argument works for 1 with multiplication in place of addition. Similarly, we want to know that inverses are unique (otherwise, in the statement of trichotomy, we might have to worry about which a we are considering. Lemma 3. If a is any integer, then a is uniquely defined by the property that a + ( a) = 0. Proof. Suppose that b, b are integers and a + b = 0 = a + b. We want to see that b = b. But using commutivity and associativity, b = b + 0 = b + (a + b ) = (b + a) + b = b + (a + b) = b + 0 = b, It then makes sense to say the following: Proposition 4. 0 = 0. Proof = 0 by definition of 0, so by definition of additive inverse we see that 0 is the additive inverse of itself, which is to say that 0 = 0. Ordering and consequences. We will define ordering on the integers using the set of positive integers: Definition 5. If a, b are integers, we say a is greater than b, and write a > b, if a b is positive. We say a is smaller than b, and write a < b, if b a is positive. As usual, we write a b to mean that either a > b or a = b, and similarly with a b. Notice in particular that a > 0 if and only if a is positive. This is not our definition, but we see that it is the same using Proposition 4! We take for granted what are sometimes called properties of equality : for instance, if a = b, then for any c we have a c = b c. This is because a = b means that a and b are the same integers, and by definition multiplication is a well-defined operation on integers, so if we write a given integer two different ways, that doesn t affect what happens when we multiply it. However, inequalities are a different matter: because we have defined them in terms of the set of positive integers (about which in principle we know nothing beyond what is stated in the axioms), we ought to check that the familiar properties of inequalities still hold in the integers. For instance, we have: Proposition 6. Suppose that a, b, c are integers with a > b and c > 0. Then a c > b c. This turns out to require some intermediate steps! Lemma 7. If a is any integer, then a 0 = 0. 2

3 Proof. Using distributivity, we have a 0 = a (0 + 0) = a 0 + a 0. If we add (a 0) to both sides, we get 0 = (a 0) + (a 0) = (a 0 + a 0) + (a 0) = a 0 + (a 0 + (a 0)) = a = a 0, Lemma 8. If a is any integer, then ( a) = a. Proof. By definition of additive inverse, we want ( a) + a = 0, but a + ( a) = 0 by definition, so this follows from commutativity. Lemma 9. If a, b, c are any integers, then: (i) a = a ( 1); (ii) ( a) b = (a b) = a ( b); (iii) ( a) ( b) = ab; (iv) (a b) c = a c b c; Proof. For (i), we want a + a ( 1) = 0. But using distributivity and Lemma 7, For (ii), we apply (i) repeatedly to find a + a ( 1) = a 1 + a ( 1) = a (1 + ( 1)) = a 0 = 0. a ( b) = a (b ( 1)) = (a b) ( 1) = (a b). Using this (with a and b reversed) we also see that ( a) b = b ( a) = (b a) = (a b), proving (ii). For (iii), we use (ii) and Lemma 8 to see that ( a) ( b) = (a ( b)) = ( (a b)) = a b. Finally, for (iv) we can use (ii) and distributivity to get (a b) c = (a + ( b)) c = a c + ( b) c = a c + ( (b c)) = a c b c, Proof of Proposition 6. By definition, a c > b c is the same as saying a c b c is positive. But using Lemma 9 (iv) we have a c b c = (a b) c, and a > b and c > 0 mean that a b and c are positive, so by the closure axiom (a b) c is also positive, and we conclude that a c > b c, However, it s impractical to prove every property of inequality, and most of them are more straightforward: for instance, that a > b and b > c implies a > c follows easily from the closure of P, and a > b implies b > a is also easy from what we ve done. So while it s a good habit to think carefully about which properties of inequalities are being used and why they re true, we ll generally take them for granted. 3

4 Cancellation. As asserted above, the cancellation law can actually be deduced from our axioms. We are now ready to prove this. Having built up some familiarity with basic proofs, our arguments will use a couple of steps at a time now. Lemma 10. Suppose that a and b are integers, and a b = 0. Then either a = 0 or b = 0. Proof. We use trichotomy to consider all possible cases with a and b both nonzero. First, if a and b are both positive, then a b must also be positive, by closure. If a is positive and b is positive, then by Lemma 9 (ii) we have that a ( b) = (a b) is positive, so again a b is not zero. Similarly, if a and b are positive, then (a b) is again positive, and if a and b are positive, then ( (a b)) = a b is positive (using Lemma 8). In all of these cases, a b is not zero, and these are only cases for which neither a or b is zero, so we conclude the lemma. Corollary 11 (Cancellation law). Suppose that a, b, c are integers, and c 0. Then if a c = b c, it follows that a = b. Proof. Given that a c = b c, we have from Lemma 9 (iv) that 0 = a c b c = (a b) c. Since we assume that c 0, it follows from Lemma 10 that a b = 0, and hence that a = b, as desired. The smallest positive integer. We defined 1 to be the multiplicative identity, but it is also the smallest positive integer, as we now explain. Proposition is the smallest positive integer. Proof. By the well-ordering principle, we know that there is some smallest positive integer; let s call it a. First suppose that a < 1. By Proposition 6, we have a 2 < 1 a = a. But if a is positive, a 2 is also positive by closure, and this contradicts that a is the smallest positive integer. In order to rule out a > 1, we just need to know that 1 is positive. But if 1 were positive, then for any positive a we d have a = a ( 1) also positive, contradicting trichotomy. Also, if we have 1 = 0, then for any positive a we d have a = a 1 = a 0 = 0, again contradicting trichotomy (note that here we are using that P is assumed nonempty). We thus conclude by trichotomy that 1 is positive, and hence is the smallest positive integer. Although such specific definitions won t really come up for us during the class, we could now define 2 to be the smallest positive integer which isn t 1, and 3 to be the smallest positive integer which isn t 1 or 2, etc. More symbols. In this note, we ve tried to minimize use of symbols to make it friendlier. However, usually in math we use more symbols, including for in, for for all, and for there exists. We would thus write the closure axiom for positive integers as: for all a, b P, we have a + b P and a b P, or even more briefly, as a, b P, we have a + b P and a b P. More consequences of well-ordering. We ll begin by proving a couple of variants of the wellordering principle. Proposition 13. Any nonempty subset of the integers which is bounded below has a least element. By bounded below, we mean that there exists an integer which is less than or equal to every integer in the subset. Proof. Let S be a nonempty subset of Z, bounded below by some integer a. Then let T = {b a+1 : b S}. Then since b a for all b S, we see that T consists entirely of positive integers. By the well-ordering principle, T has a least element, say c. Then c + a 1 is the least element of S. 4

5 Proposition 14. Any nonempty subset of the integers which is bounded above has a greatest element. Proof. Let S be a nonempty subset of Z, belowed above by some integer a. Then let T = { b : b S}. Then T is bounded below by a, so T has a least element, say c. Then c is the greatest element of S. More substantively, we discuss proofs by induction, and why their validity follows from well ordering. A proof by induction works as follows: if you want to prove that a statement is true for all positive integers n, you first prove it for n = 1 (the base case ), and then show that if the statement is true for some n (the induction hypothesis ), then it also has to be true for n + 1 (the induction step ). One then concludes that it is true for all n. We can formalize this in the following theorem. Theorem 15 (Mathematical induction). Let S be a subset of P which contains 1, and which has the property that for any n, if n S then n + 1 S. Then S = P. In the statement, we picture that the set S is the set of n for which our desired statement is true. Proof. Let S be as above, and let T be the set of positive integers which are not contained in S. We thus want to show that T is empty. If T is not empty, then the well-ordering principle tells us that it contains a least element, say a. Then a 1, since we have assumed 1 S. Thus, by Proposition 12, we must have a > 1, so a 1 P. Since a is the least element of T, we conclude that a 1 S. But then by hypothesis, a S, which is a contradiction. What this is saying is that the fact that proofs by induction work follows from the well-ordering principle. This is saying that any proof by induction could instead be phrased as a proof using the well-ordering principle. In fact, it is also true that the well-ordering principle follows from induction, so that we could use either one in our axioms for the integers, and any proof using well-ordering could be stated in terms of induction. However, in a given proof it is often more natural or convenient to use one or the other. More often than not, we will find it more convenient to use the well-ordering principle directly in our arguments in this class, but we now give an example of a proof using induction. Proposition 16. For any n P, we have n k = n(n + 1). 2 Proof. We can rewrite the desired formula to avoid division as n 2 k = n(n + 1). The base case is n = 1. In this case, we just have 2 1 k = 2 = 1 2, so it is true. Now, suppose the statement holds for a particular n, so that for the induction step we want to show it is also true for n + 1. We calculate (using the induction hypothesis for the second equality) ( n+1 n ) 2 k = 2 k + 2(n + 1) = n(n + 1) + 2(n + 1) = (n + 2)(n + 1) = (n + 1)(n + 2). This proves the desired formula for n + 1, and by induction we conclude the statement of the proposition. 5

6 Note that induction can also be used to prove statements that hold for all integers n which are greater than or equal to a given integer c. In this case, we use n = c as the base case instead of n = 1. 6