[轉錄]Re: [問題] 請問1+1=2是如何證出來的

看板ILSH-97309作者kimo6414 (一生懸命)時間15年前 (2009/06/23 22:07)推噓4(4推 0噓 4→)

留言8則, 5人參與討論串1/1

※ [本文轉錄自 Math 看板] 作者: plover (+oo) 看板: DAIC 標題: Re: [問題] 請問1+1=2是如何證出來的時間: Sat Jan 11 17:27:47 2003 ※ 引述《bigjuto (用過的都說棒)》之銘言： : 是用皮亞諾公設嗎... : 該如何去證？ Author: Pinter We will proceed as follows: we define 0 = {}. In order to define "1," we must fix a set with exactly one element; thus 1 = {0}. Continuing in fashion, we define 2 = {0,1}, 3 = {0,1,2}, 4 = {0,1,2,3}, etc. The reader should note that 0 = {}, 1 = {{}}, 2 = {{},{{}}}, etc. Our natural numbers are constructions beginning with the empty set. The preceding definitions can be restarted, a little more precisely, as follows. If A is a set, we define the successor of A to be the set A^+, given by A^+ = A ∪ {A}. Thus, A^+ is obtained by adjoining to A exactly one new element, namely the element A. Now we define 0 = {}, 1 = 0^+, 2 = 1^+, 3 = 2^+, etc. 現在問題來了, 有一個 set 是包括所有 natural numbers 的嗎 ? (甚至問一個 class). 這邊先定義一個名詞, 接著在引 A9, 我們就可以造出一個 set 包括所有的 natural numbers. A set A is called a successor set if it has the following properties: i) {} [- A. ii) If X [- A, then X^+ [- A. It is clear that any successor set necessarily includes all the natural numbers. Motivated bt this observation, we introduce the following important axiom. A9 (Axiom of Infinity). There exist a successor set. As we have noted, every successor set includes all the natural numbers; thus it would make sense to define the "set of the natural numbera" to be the smallest successor set. Now it is easy to verify that any intersection of successor sets is a successor set; in particular, the intersection of all the successor sets is a successor set (it is obviously the smallest successor set). Thus, we are led naturally to the following definition. 6.1 Definition By the set of the natural numbers we mean the intersection of all the successor sets. The set of the natural numbers is designated by the symbol ω; every element of ω is called a natural number. 6.2 Theorem For each n [- ω, n^+≠0. Proof. By definition, n^+ = n ∪ {n}; thus n [- n^+ for each natural number n; but 0 is the empty set, hence 0 cannot be n^+ for any n. 6.3 Theorem (Mathematical Induction). Let X be a subset of ω; suppose X has the following properties: i) 0 [- X. ii) If n [- X, then n^+ [- X. Then X = ω. Proof. Conditions (i) and (ii) imply that X is a successor set. By 6.1 ω is a subset of every successor set; thus ω 包含於 X. But X 包含於 ω; so X = ω. 6.4 Lemma Let m and natural numbers; if m [- n^+, then m [- n or m = n. Proof. By definition, n^+ = n ∪ {n}; thus, if m [- n^+, then m [- n or m [- {n}; but {n} is a singleton, so m [- {n} iff m = n. 6.5 Definition A set A is called transitive if, for such x [- A, x 包含於 A. 6.6 Lemma Every natural number is a transitive set. Proof. Let X be the set of all the elements of ω which are transitive sets; we will prove, using mathematical induction (Theorem 6.3), that X = ω; it will follow that every natural number is a transitive set. i) 0 [- X, for if 0 were not a transitive set, this would mean that 存在 y [- 0 such that y is not a subset of 0; but this is absurd, since 0 = {}. ii) Now suppose that n [- X; we will show that n^+ is a transitive set; that is, assuming that n is a transitive set, we will show that n^+ is a transitive set. Let m [- n^+; by 6.4 m [- n or m = n. If m [- n, then (because n is transitive) m 包含於 n; but n 包含於 n^+, so m 包含於 n^+. If n = m, then (because n 包含於 n^+) m 包含於 n^+; thus in either case, m 包含於 n^+, so n^+ [- X. It folloes by 6.3 that X = ω. 6.7 Theorem Let n and m be natural numbers. If n^+ = m^+, then n = m. Proof. Suppose n^+ = m^+; now n [- n^+, hence n [- m^+; thus by 6.4 n [- m or n = m. By the very same argument, m [- n or m = n. If n = m, the theorem is proved. Now suppose n≠m; then n [- m and m [- n. Thus by 6.5 and 6.6, n 包含於 m and m 包含於 n, hence n = m. 6.8 Recursion Theorem Let A be a set, c a fixed element of A, and f a function from A to A. Then there exists a unique function γ: ω -> A such that I. γ(0) = c, and II. γ(n^+) = f(γ(n)), 對任意的 n [- ω. Proof. First, we will establish the existence of γ. It should be carefully noted that γ is a set of ordered pairs which is a function and satisfies Conditions I and II. More specifically, γ is a subset of ω╳A with the following four properties: 1) 對任意的 n [- ω, 存在 x [- A s.t. (n,x) [- γ. 2) If (n,x_1) [- γ and (n,x_2) [- γ, then x_1 = x_2. 3) (0,c) [- γ. 4) If (n,x) [- γ, then (n^+,f(x)) [- γ. Properties (1) and (2) express the fact that γ is a function from ω to A, while properties (3) and (4) are clearly equivalent to I and II. We will now construct a graph γ with these four properties. Let Λ = { G | G 包含於 ω╳A and G satisfies (3) and (4) }; Λ is nonempty, because ω╳A [- Λ. It is easy to see that any intersection of elements of Λ is an element of Λ; in particular, γ = ∩ G G[-Λ is an element of Λ. We proceed to show that γ is the function we require. By construction, γ satisfies (3) and (4), so it remains only to show that (1) and (2) hold. 1) It will be shown by induction that domγ = ω, which clearly implies (1). By (3), (0,c) [- γ; now suppose n [- domγ. Then 存在 x [- A 使得 (n,x) [-γ; by (4), then, (n^+,f(x)) [- γ, so n^+ [- domγ. Thus, by Theorem 6.3 domγ = ω. 2) Let N = { n [- ω | (n,x) [- γ for no more than one x [- A }. It will be shown by induction that N = ω. To prove that 0 [- N, we first assume the contrary; that is, we assume that (0,c) [- γ and (0,d) [- γ where c≠d. Let γ^* = γ - {(0,d)}; certainly γ^* satisfies (3); to show that γ^* satisfies (4), suppose that (n,x) [- γ^*. Then (n,x) [- γ, so (n^+,f(x)) [- γ; but n^+≠0 (Theorem 6.2), so (n^+,f(x))≠(0,d), and consequently (n^+,f(x)) [- γ^*. We conclude that γ^* satisfies (4), so γ^* [- Λ; but γ is the intersection of all elements of Λ, so γ 包含於 γ^*. This is impossible, hence 0 [- N. Next, we assume that n [- N and prove that n^+ [- N. To do so, we first assume the contrary -- that is, we suppose that (n,x) [- γ, (n^+,f(x)) [- γ, and (n^+,u) [- γ where u≠f(x). Let γ^。 = γ - {(n^+,u)}; γ^。 satisfies (3) because (n^+,u)≠(0,c) (indeed, n^+≠0 by Theorem 6.2). To show that γ^。 satisfies (4), suppose (m,v) [- γ^。; then (m,v) [- γ, so (m^+,f(v)) [- γ. Now we consider two cases, according as (a) m^+≠n^+ or (b) m^+ = n^+. a) m^+≠n^+. Then (m^+,f(v))≠(n^+,u), so (m^+,f(v)) [- γ^。. b) m^+ = n^+. Then m = n by 6.7, so (m,v) = (n,v); but n [- N, so (n,x) [- γ for no more than one x [- A; it follows that v = x, and so (m^+,f(v)) = (n^+,f(x)) [- γ^。. Thus, in either case (a) or (b), (m^+,f(v)) [- γ^。, thus, γ^。 satisfies Condition (4), so γ^。[- Λ. But γ is the intersection of all the elements of Λ, so γ 包含於 γ^。; this is impossible, so we conclude that n^+ [- N. Thus N = ω. Finally, we will prove that γ is unique. Let γ and γ' be functions, from ω to A which satisfy I and II. We will prove by induction that γ = γ'. Let M = { n [- ω | γ(n) = γ'(n) }. Now γ(0) = c = γ'(0), so 0 [- M; next, suppose that n [- M. Then γ(n^+) = f(γ(n)) = f(γ'(n)) = γ'(n^+), hence n^+ [- M. If m is a natural number, the recurion theorem guarantees the existence of a unique function γ_m: ω -> ω defined by the two Conditions I. γ_m(0)=m, II. γ_m(n^+) = [γ_m(n)]^+, 對任意的 n [- ω. Addition of natural numbers is now defined as follows: m + n = γ_m(n) for all m, n [- ω. 6.10 m + 0 = m, m + n^+ = (m + n)^+. 6.11 Lemma n^+ = 1 + n, where 1 is defined to be 0^+ Proof. This can be proven by induction on n. If n = 0, then we have 0^+ = 1 = 1 + 0 (this last equality follows from 6.10), hence the lemma holds for n = 0. Now, assuming the lemma is true for n, let us show that it holds for n^+: 1 + n^+ = (1 + n)^+ by 6.10 = (n^+)^+ by the hypothesis of induction. 把 n = 1 並且注意 2 = 1^+, 故 1 + 1 = 2. -- ※ 發信站: 批踢踢實業坊(ptt.csie.ntu.edu.tw) ◆ From: 140.112.247.33

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