The natural numbers As mentioned earlier in the course, the natural numbers can be constructed using the axioms of set theory. In this note we want to discuss the necessary details of this construction. Recall that Axiom VII, often called the axiom of infinity, guarantees the existence of at least one inductive set that is, a set X with the property that (i) X, and (ii) A {A} X whenever A X. Notice that such an inductive set X must contain the sets, { }, {, { }}, {, { }, {, { }}},. However, it may also contain a lot of other elements. In order to get rid of all conceivable extraneous elements in X we can form the smallest possible subset of X which is still inductive, called its inductive core: C(X) {I I X and I is and inductive set} {A X A I for every inductive set I with I X}. We would like to define the natural numbers N to be this inductive core C(X). For this definition to be meaningful, however, we need to check if one arrives at the same inductive core regardless of the inductive set one starts out with. To this end, suppose that Y is also an inductive set. We wish to show that C(X) C(Y ). First, observe that X Y is an inductive set, too. Also, since the sets involved in the intersection defining C(X Y ) are all those sets involved in the intersection defining C(X) plus possibly more, we conclude that C(X Y ) C(X). Conversely, let A C(X). In order to show that A C(X Y ), we let I be an arbitrary inductive subset of X Y and argue that A I. But that follows immediately upon observing that I X is an inductive subset of X and must therefore contain A. We have now shown that C(X Y ) C(X). Similarly, C(X Y ) C(Y ). Hence C(X) C(Y ). This little discussion enables us to make our definition: Definition. Let X be any inductive set. We define the set of natural numbers as N C(X). Next, let us give the elements of N shorter names. We shall write 0, 1, 2, 3, for the sets, { }, {, { }}, {, { }, {, { }}},. That is to say, we write 0 for, 1 for {0}, 2 for {0, 1}, 3 for {0, 1, 2}, and so on. We will also use the convenient abbreviation A + 1 A {A}. Hence, 1 0 + 1, 2 1 + 1, 3 2 + 1, and so on. This does not mean, however, that we already know how to add natural numbers. We will define addition and multiplication of natural numbers later. At this point, A + 1 simply stands for A {A}. Notice that the principle of mathematical induction follows immediately from the definition of N: Theorem [Mathematical induction]. Let a predicate P (n) be given. Suppose that P (0) is true and that P (n + 1) is true whenever P (n) is true. Then P (n) is true for all n N. 1
Proof. Let A {n N P (n) is true }. Then 0 A and n + 1 A whenever n A. Hence, A is an inductive set, so that N C(A) A N. That is A N. We now put an ordering on the natural numbers: Definition. For m, n N we define m < n to mean m n. For example, 3 < 5 since 3 5 {0, 1, 2, 3, 4}. Exercises. Use mathematical induction to prove the following statements for all m, n, k N. [Solutions are sketched at the end of this note.] 1. m < n + 1 if and only if either m < n or m n. 2. m n if and only if either m < n or m n. 3. If m < n, then m n. 4. Writing m n for m < n or m n, we get: (i) n n. (ii) If m n and n m, then m n. (iii) If m n and n k, then m k. (iv) If m < n and n < k, then m < k. 5. If m < n, then m + 1 < n + 1. 6. Exactly one of the following must be true: m < n, m n, or n < m. 7. If m + 1 n + 1, then m n. Theorem [Recursion Theorem]. Let A be a set, a A, and r : N A A any function. Then there exists a function f : N A such that (i) f(0) a, and (ii) f(n + 1) r(n, f(n)) for all n N. We say that the rule r together with the initial value a recursively define the sequence f. Proof. Put f {S N A (0, a) S n x[(n, x) S (n+1, r(n, x)) S]} N A. Then (i) (0, a) f and (ii) n x[(n, x) f (n + 1, r(n, x)) f]. Therefore, once we have shown that f is a function it will automatically have properties (i) and (ii) of the theorem. (A quick induction argument shows that the domain of f is N.) We 2
will prove by induction on m that f passes the vertical line test and is therefore a function: (m, x 1 ), (m, x 2 ) f x 1 x 2. Basic Step: Suppose there are (0, x 1 ), (0, x 2 ) f with x 1 x 2. Then either x 1 a or x 2 a. Say, x 1 a. Form the set S f \ {(0, x 1 )} N A. Observe that (i) (0, a) S, and (ii) that if (n, x) S, then (n, x) f so that (n + 1, r(n, x)) f. Hence, (n + 1, r(n, x)) S, because (n + 1, r(n, x)) (0, x 1 ) (since n + 1 0). So, by definition of f, f S. This contradicts the definition of S. Inductive Step: Suppose that (m + 1, x 1 ), (m + 1, x 2 ) f with x 1 x 2. If both x 1 r(m, z 1 ) for some (m, z 1 ) f and x 2 r(m, z 2 ) for some (m, z 2 ) f, then z 1 z 2 by inductive hypothesis, so that x 1 x 2 which is not the case. We can therefore assume, without loss of generality, that x 1 r(m, z) for all (m, z) f. Form the set S f \ {(m + 1, x 1 )} N A. As before, observe that (i) (0, a) S (since 0 m+1) and (ii) that if (n, x) S, then (n, x) f so that (n+1, r(n, x)) f. Hence, (n + 1, r(n, x)) S (otherwise m + 1 n + 1 and x 1 r(n, x) r(m, x) with (m, x) (n, x) f, which is not allowed). The definition of f implies now that f S, which in turn contradicts the definition of S. This completes the proof. We now use the recursion theorem to introduce addition and multiplication into our set of natural numbers. First we want to define what it means to add a natural number to a fixed m N. We do this by recursively defining a function f m : N N. We require that (i) f m (0) m; and (ii) f m (n + 1) f m (n) + 1. The recursion theorem guarantees the existence of such a function f m. We will of course write m + n for f m (n). (Notice that f m (1) f m (0) + 1 m + 1 so that there is no ambiguity in using m + 1 for both m {m} and f m (1).) In this more suggestive notation, the above two items read: (i) m + 0 m; and (ii) m + (n + 1) (m + n) + 1. Exercises. 8. Show that (m + n) + k m + (n + k) for all m, n, k N. 9. Show that m + n n + m for all m, n N. 3
We also define multiplication of a fixed m N by a natural number via the recursion theorem by way of a function g m : N N with the properties (i) g m (0) 0; and (ii) g m (n + 1) g m (n) + m. Writing m n for g m (n), this recursive definition can be written as (i) m 0 0; and (ii) m (n + 1) m n + m. Here, we make the usual convention that multiplication is carried out before addition, unless specified otherwise. Exercises. Prove the following for all m, n, k N. 10. m 1 1 m m. 11. (m + n) k m k + n k. 12. m n n m. 13. (m n) k m (n k). 14. If m < n, then m + k < n + k. 15. If m < n and k 0, then m k < n k. 16. If m k n k and k 0, then m n. 17. If m + k n + k then m n. 18. If m n 0, then either m 0 or n 0. 19. If m n, then there is exactly one d N such that m + d n. 4
Solutions. 1. This follows straight from the definition of n + 1 n {n} and the meaning of the symbol <. 2. We will prove, by induction on n, that m < n implies m n. The case n 0 is vacuously true. Now suppose that m < n + 1. Then either m < n or m n, by Exercise 1. Hence, by the inductive hypothesis, m n, so that m n {n} n + 1. Once more, we induct on n. If n 0, then m n. Inductively, suppose that m n + 1 n {n}. If n m, then m n, so that by inductive hypothesis either m n or m < n. Either way, this case implies that m n {n}, i.e. m < n+1. On the other hand, if n m, then n m by the first part of this proof. So, m n {n} n + 1. 3. We show by induction on n that n n for all n N. Clearly, 0 0, so that the statement is true for n 0. Next, assuming that n n, we wish to prove that n + 1 n + 1. Suppose to the contrary that n + 1 n + 1. Then n + 1 < n or n + 1 n by Exercise 1. If n + 1 < n, then n {n} n + 1 n by Exercise 2. This implies n n; a contradiction. The alternative, n + 1 n, implies the same contradiction. 4. By Exercise 2, m n if and only if m n. 5. We induct on n. The basic step is vacuously true. Now suppose that m < n+1. We need to show that m + 1 < (n + 1) + 1. By Exercise 1, either m n or m < n. In the first case, m + 1 n + 1 < (n + 1) + 1. In the second case, the inductive hypothesis implies that m + 1 < n + 1 < (n + 1) + 1. Hence, m + 1 < (n + 1) + 1 by Exercise 4 (iv). 6. By Exercises 2 and 3, at most one of the three alternatives is possible. To show that at least one of them occurs, we induct on n. If n 0, then n m. (A quick induction shows that 0 k for all k N.) Our inductive hypothesis is now that either m < n, m n, or n < m. We need to show that either m < n + 1, m n + 1, or n + 1 < m. If m n, then m < n + 1. If m < n, then m < n < n + 1. Lastly, if n < m, then n + 1 < m + 1 by Exercise 5. So, by Exercise 1, either n + 1 < m or n + 1 m. 7. Follows from Exercises 5 and 6. 8. We induct on k. If k 0, then (m+n)+0 m+n m+(n+0). Assuming that the statement holds for k, we compute (m + n) + (k + 1) def [(m + n) + k] + 1 I.H. [m + (n + k)] + 1 def m + [(n + k) + 1] def m + (n + (k + 1)). 9. (a) First we prove, by induction on m, that m + 0 0 + m. The basic step is a triviality: 0 + 0 0 + 0. Here is the inductive step: (m + 1) + 0 m + 1 (m + 0) + 1 I.H. (0 + m) + 1 #8 0 + (m + 1). 5
(b) Next we show, by induction on m, that m+1 1+m. We can use Part (a) for the basic step: 0+1 1+0 (although this also follows straight from the definitions). The inductive step is as follows: (m + 1) + 1 I.H. (1 + m) + 1 def 1 + (m + 1). (c) Now we prove, by induction on n, that m + n n + m. We did the basic step in Part (a). The inductive step is as follows: m + (n + 1) def (m + n) + 1 I.H. (n + m) + 1 def n + (m + 1) (b) n + (1 + m) #8 (n + 1) + m. 10. First notice that m 1 m (0 + 1) def m 0 + m def 0 + m #9 m + 0 m. So we only have to show that 1 m m also. One shows this by induction on m. The basic step follows from the definition: 1 0 0. The inductive step is as follows: 1 (m + 1) def 1 m + 1 I.H. m + 1. def 11. Induct on k. Here is the inductive step: (m + n) (k + 1) (m+n) k +(m+n) #10+I.H. m k +n k +m 1+n 1 #9+def m (k +1)+n (k +1). 12. Induct on n. Here is the inductive step: m (n + 1) def m n + m I.H.+#10 n m + 1 m #11 (n + 1) m. 13. Induct on k. Here is the induction step: (m n) (k + 1) def (m n) k + m n I.H. m (n k) + m n #11,12 m (n k + n) def m (n (k + 1)). 14. Induct on k, using Exercise #5. 15. Using the previous exercises, this is an easy induction on k. 16. The contrapositive of this statement follows immediately from Exercises #6 and #15. 17. In conjunction with Exercise #7, this is a straight forward induction on k. 18. We prove the contrapositive: suppose m 0 and n 0. Since n < 0 (i.e. n ) is impossible, then, by Exercise #6, 0 < n. Hence, by Exercises #15 and #12, 0 m 0 0 m < n m m n, so that m n 0 (again by Exercise #6). 19. We only have to show the existence of such a d, uniqueness follows from Exercise #17. This is most easily done with an induction on n. If n 0, then m 0 and we can take d 0. Now suppose that m (n + 1). We will show that there is a d N such that m + d n + 1. If m n + 1, we take d 0. If m < n + 1, then m n by Exercise #1. We then use the inductive hypothesis to get a c N with m + c n and take d c + 1. 6