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The natural numbers For a subset A of R, A is called inductive (or inductively closed) if for all x, x ∈ A implies x + 1 ∈ A. One inductive set is R itself. The set of natural numbers is defined to be the set N= \ {A : A is inductive with 1 ∈ A}. Theorem. N is an inductive subset of R containing 1. If A is any inductive set that contains 1, then A ⊃ N. Proof. First, there is an inductive subset of R containing 1, namely R itself, so N exists and is a subset of R. T From the definition of intersection, 1 ∈ {A : A is inductive and 1 ∈ A} = N. To see that N is inductive, let x ∈ N. Then, x ∈ A, for all inductive sets A with 1 ∈ A. Hence x + 1 ∈ A, for all inductive sets A with 1 ∈ A. Therefore, x+1∈ \ {A : A is inductive with 1 ∈ A} = N, by definition. Thus x ∈ N =⇒ x + 1 ∈ N, so N is inductive. For the second statement, let A be inductive with 1 belonging to it. If x ∈ N, then x ∈ A, by definition of intersection. So N ⊂ A. The end of the proof is a special case of the general fact that the intersection of a family of sets is always contained in each set in the family. The above immediately yields the Principle of Mathematical Induction: Theorem. (PMI) (1) If A is an inductive set of natural numbers including the element 1, then A = N. (2) If P (n) is a statement about natural numbers n, such that (i) P (1) is true, and (ii) whenever P (n) is true, P (n + 1) is true, then P (n) is true for all n ∈ N. Proof. (1) By hypothesis, A ⊂ N; Also A is inductive with 1 ∈ A, so N ⊂ A by the previous result. Thus A = N. (2) Let A = {n ∈ N : P (n) is true }. By hypothesis, 1 ∈ A and x ∈ A =⇒ x + 1 ∈ A. Thus A is inductive, so A = N, by part (1). That is, for all n ∈ N, P (n) is true. 13/9/2008 843 mam 17 18 THE NATURAL NUMBERS Theorem. For all n ∈ N, n ≥ 1. Proof. Here the statement to be proved by induction on n is: n ≥ 1. Now, (i) 1 ≥ 1 is true of course and, (ii) if n ≥ 1, then n + 1 ≥ 1 + 0 ≥ 1. Thus, by PMI n ≥ 1, for all n ∈ N. Corollary. 0 ∈ / N. Proof. If 0 ∈ N we would have 0 ≥ 1, by the previous theorem. But 0 < 1, so this would be a contradiction. So, if n = 1, its predecessor n − 1 is not a natural number, but this is the only such case. Theorem. If n ∈ N and n > 1, then n − 1 ∈ N. Proof. Since every natural number is greater than or equal to 1, saying n > 1 is the same as saying n 6= 1. Let A be {n ∈ N : n = 1 or n − 1 ∈ N}. Then A contains 1, by definition. Suppose now n ∈ A. Then n ∈ N; hence, (n + 1) − 1 = n ∈ N. Thus n ∈ A =⇒ n + 1 ∈ A, so A is inductive. Hence N = A, by PMI. Thus, for all n ∈ N, n = 1 or n − 1 ∈ N, In other words, n 6= 1 implies n − 1 ∈ N. The last step of the proof used the fact, from elementary logic, that the statement p ∨ q is equivalent to the statement (not p) =⇒ q. (∨ means “or” and =⇒ means “implies”.) Theorem. If n ∈ N and n < m ∈ N, then n + 1 ≤ m. Proof. We use induction on n: So let A = {n ∈ N : for all m ∈ N, if n < m then n + 1 ≤ m}. Let m ∈ N. If 1 < m, then m − 1 ∈ N, so 1 ≤ m − 1, hence 1 + 1 ≤ m. Since m was arbitrary, 1 ∈ A. Suppose n ∈ A, and let m ∈ N again be arbitrary. If n + 1 < m, then 1 < m, so m − 1 ∈ N and n < m − 1; hence, n + 1 ≤ m − 1, by the inductive hypothesis (that is, since n ∈ A). Thus, (n + 1) + 1 ≤ m. And this was true for arbitrary m, so A is inductive, and A = N, as required. In the above proof we used a general principle. If we have a statement to prove about two natural numbers n and m, we try to do induction on just one of them (in the above case, n). We get more power by making the inductive hypothesis strong, saying “for all m ∈ N”. The idea is that if you assume more, you can prove more. 13/9/2008 843 mam THE NATURAL NUMBERS 19 Corollary. If n ∈ N and n < a < n + 1, then a ∈ / N. In other words, there is no natural number between successive ones. Proof. Let n ∈ N and n < a < n + 1. If a ∈ N, then by the theorem n + 1 ≤ a, which contradicts a < n + 1. Well-ordering property. Each non-∅ subset of N has a least element. If A is a set of real numbers, to say that a0 is a least element of A means a0 ∈ A and for each a ∈ A, a0 ≤ a. Clearly, a set A can have no more than one least element (why?). Proof. Let M ⊂ N and let M be non-empty. Suppose M has no least element. We will show that this leads to a contradiction. Let B be the set of n ∈ N for which n ≤ m, for all m ∈ M . We claim that 1 ∈ B and B is inductive. First, 1 ∈ B. Indeed, 1 ≤ m, for all m ∈ M , since M is a set of natural numbers. Now let n ∈ B. Then n ≤ m, for all m ∈ M , But n ∈ / M , for otherwise n would be the least element of M which we are assuming doesn’t exist. Thus, n < m, for all m ∈ M , so n + 1 ≤ m, for all m ∈ M , by the previous result, that is n + 1 ∈ B. Thus, n ∈ B implies n + 1 ∈ B. This shows B is inductive, so contains all the elements of N. In other words, for all natural numbers n, n ≤ m, for all m ∈ M . Since M 6= ∅ it contains some element, say m0 . But m0 is a natural number so m0 ≤ m, for all m ∈ M . So m0 is the least member of M after all, a contradiction. 13/9/2008 843 mam 20 THE NATURAL NUMBERS There is a strengthening of the Principle of Mathematical induction which is often more convenient, known as the 2nd Principle of Mathematical Induction or the Principle of Complete Induction. As usual {1, . . . , n} means the set of all natural numbers k with 1 ≤ k ≤ n. Theorem. (2nd PMI) (1) Let B be a set of natural numbers such that 1 ∈ B and that for all n, {1, . . . , n} ⊂ B implies n + 1 ∈ B, then B = N. (2) If P (n) is a statement about natural numbers n such that (i) P (1) is true, and (ii) whenever P (k) is true for all natural numbers k ≤ n, P (n + 1) is also true, then P (n) is true for all n ∈ N. Proof. Let A be the set of all n ∈ N such that {1, . . . , n} ⊂ B. We will show that A is inductive. Since 1 ∈ B, {1} ⊂ B, so 1 ∈ A. Let n ∈ A. Then {1, . . . , n} ⊂ B, so n + 1 ∈ B. But then, {1, . . . , n + 1} ⊂ B so that n + 1 ∈ A. Thus, by the Principle of Mathematical Induction, A = N, so that for all n ∈ N, {1, . . . , n} ⊂ B, in particular, N ⊂ B. Since B ⊂ N , B = N. Theorem. N is closed under addition. The statement means that for all m, n ∈ N, m + n ∈ N. In the proof, we will use the general principle mentioned earlier: We will choose one of m and n and prove the statement with the other universally quantified. Proof. Let A = {n ∈ N : m + n ∈ N, for all m ∈ N}. Then 1 ∈ A, because this just says m + 1 ∈ N for all m ∈ N, which we already know. Now let n ∈ A. Then, for each m ∈ N, m + n ∈ N, so m + (n + 1) = (m + n) + 1 ∈ N, so that n + 1 ∈ A. Thus, A is inductive, so each n ∈ N belongs to A, which says that for all n ∈ N and for all m ∈ N, m + n ∈ N, as required. In the above proof, how did we know to add n + 1 to the m? Because, we assumed n ∈ A and we were trying to show n + 1 in A. For that, we needed to show n + 1 satisfied the definition of the members of the set A. Thus, we fixed one m and added n + 1 to it to see if m + (n + 1) still belonged to N. It did. We also followed a common custom of not bothering to write out the line: n ∈ A implies n + 1 ∈ A. 13/9/2008 843 mam THE NATURAL NUMBERS 21 Theorem. N is closed under multiplication. Proof. Let A = {n ∈ N : mn ∈ N, for all m ∈ N} Then, 1 ∈ A, since 1 is a multiplicative identity, so suppose n ∈ A and fix m ∈ N. Then, m(n + 1) = mn + m, by the distributive law. Since n ∈ A, mn ∈ N and thus, by closure under addition, mn + m ∈ N. This proves n + 1 ∈ A, so A is inductive. Therefore by PMI, for all n ∈ N and for all m ∈ N mn ∈ N. Definition. For n ∈ N and x ∈ R we define xn using the idea of induction: x1 = x, and assuming xn defined we let xn+1 = xn x. A definition of this type is called a recursive definition or a definition by recursion. One also lets x0 = 1. (In some books, 00 is left undefined, but these still tend to use it as though its definition were 00 = 1, for example in an expression such as P n k 0 k=0 x , when x is allowed to be 0. We take 0 = 1 always. We have to be very careful, though, not to give this expression properties that it doesn’t have.) 1. Sometimes the 2nd principle of induction is stated in only one step: If B is a set of natural numbers such that n ∈ N and k ∈ B, for all k ∈ N with k < n implies n ∈ B, then B = N. Show that in using this version one still has to prove 1 ∈ N. 2. If m, n ∈ N, with m > n, then m − n ∈ N. 3. The Binomial Theorem: Let a, b ∈ R (or in any field), for any n ∈ N, (a + b)n = ` ´ Pn `n´ k n−k n! . Here n is the binomial coefficient k!(n−k)! . It satisfies Pascal’s k=0 k a b `n+1´ `n´ ` n ´k Law: k+1 = k + k+1 . 4. The Bernoulli inequalities: For n ∈ N, if b > −1, (1+b)n ≥ 1+nb; if b < 1, (1−b)n ≥ 1−nb. P n−i bi−1 . 5. For each n ∈ N, an − bn = (a − b) n i=1 a 6. If S is a non-empty finite subset of R, then S has a maximum (that is, a largest element) and a minimum ( smallest element). (This can be proved by induction on the number of elements of S.) 7. Prove that if S is a non-empty finite subset of R, then S has a maximum, by induction on the number of elements of S. 13/9/2008 843 mam