作者plover (+oo)
看板DAIC
标题Re: [问题] 请问1+1=2是如何证出来的
时间Sat Jan 11 17:27:47 2003
※ 引述《bigjuto (用过的都说棒)》之铭言:
: 是用皮亚诺公设吗...
: 该如何去证?
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.
现在问题来了, 有一个 set 是包括所有 natural numbers 的吗 ? (甚至问
一个 class). 这边先定义一个名词, 接着在引 A9, 我们就可以造出一个 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)), 对任意的 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) 对任意的 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)]^+, 对任意的 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 并且注意 2 = 1^+, 故 1 + 1 = 2.
--
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180F:推 hhh1234321 : 腊月十七推一下 02/05 08:33
181F:推 johnnyttttt : 2015/02/13 朝圣 02/13 14:19
182F:推 waasabi : 2015/03/06 03/06 11:53
183F:推 chwa : 朝圣 2015/03/16 03/16 12:53
184F:推 iPolo3 : 工虾 跨拢抹 03/16 13:02
185F:推 FateOFP : 朝圣 2015/03/25 03/25 02:39
186F:推 ltameion : 朝圣 2015/03/25 03/25 03:09
187F:推 rataliepos : 2015/3/25 03/25 06:46
188F:推 gold97972000: 朝圣 2050/03/25 03/25 11:26
189F:推 aij : 朝圣 2100/03/25 03/25 12:50
190F:推 jameskey : 朝圣 2015/04/24 04/24 14:22
191F:推 jerry73204 : 朝圣 2015/05/09 看到神了 <(_ _)> 05/09 17:03
192F:推 jaytony : 朝圣 2015/05/09 已跪 05/26 16:17
193F:推 haocker : 朝圣 2015/07/25 闪尿了 07/25 01:07
194F:推 x710142857 : 朝圣 2015/08/20 08/20 22:39
195F:推 Desperato : 朝圣 2015/08/20 \ow o/ 08/20 22:41
196F:推 Luwan : 2015 8/29 太屌了... 08/29 09:53
197F:推 hhjkjk11 : 2015/9/5 09/05 21:51
198F:推 ray81712 : 朝圣 2015/9/5 夸拢某 09/05 21:54
199F:推 janna5566 : 朝圣RRR 看不懂QQ 2015/ 09/05 21:59
200F:推 opmew : 2015/09/05 09/05 22:05
201F:推 l2687316988 : 太扯啦 09/05 22:16
202F:推 leafhow : 朝圣 09/05 22:18
203F:推 SIRIUS : 朝圣 2015/09/05 09/05 22:28
204F:推 caffpetiy : 快推 不然人家以为我看不懂 09/05 23:09
205F:推 Curry5566 : 朝圣 2015/09/05 09/05 23:18
206F:推 kobe9527 : 朝圣 09/05 23:25
207F:推 tomoyari : 朝圣 09/06 01:09
209F:→ ChihYaoLin : 朝圣 09/15 08:54
210F:推 god829 : 朝圣2015/09/05 另外紧握小根根到底是什麽东西= = 09/15 09:12
211F:推 KANEISBEACH : 跟我想得差不多 给推 10/23 07:30
212F:推 hhh1234321 : 怪不得我的加法一直无法突破,懂了..... d(^_^)b 11/21 08:15
213F:推 LeeMY : 2016 04/11 04/11 02:13
214F:推 regen1999 : 2016/06/27 06/27 01:54
215F:推 evilture : 2016/07/15 我到底看了三小 07/15 22:30
216F:推 ckjonathan : 2016/08/05 08/05 16:33
217F:推 kevinyin9 : 2016/8/30 08/30 23:00
218F:推 josephcc : 2016/9/29 09/29 06:02
219F:推 ms0705718 : 来朝圣 2016/12/13 12/13 14:08
220F:推 Sidney0503 : 恩恩 跟我想的一样 2017/ 01/09 09:40
221F:推 FuwafuwaCAT : 太深奥了 01/10 01:32
222F:推 twbbsbbs : 大狮必推! 03/07 11:19
223F:推 sustainer123: 朝圣 2017/4/25 04/25 21:35
224F:推 a83a83cjcj : 朝圣 2017/05/09 XD 05/09 12:03
225F:推 Leafypc : 2017/09/01 朝圣 plover大大生日快乐! 09/01 12:55
226F:推 canucksteve : 朝圣 2017/09/12 09/13 21:27
227F:推 dionysus522 : 朝圣 2017/10/23 10/23 10:10
228F:推 ck0987515477: 2017/11/12 朝圣 11/12 15:19
229F:推 sos976431 : 2018 1 10 推 01/10 12:09
230F:推 AsllaPiscu : 朝圣推 01/14 22:41
231F:推 gfhnrtjpoiuy: 朝圣 2018/01/14 01/14 22:42
232F:→ Anonym5566 : 朝圣 2018.01.14 01/14 22:45
233F:推 SteveNeko : 2018/01/14 朝圣 01/14 22:46
234F:推 LiptonTea : 2018/01/22 朝圣 01/22 02:35
235F:推 Gauss : 2018/1/22 01/22 14:58
236F:推 mark10133 : 2018/02/03朝圣 02/03 20:29
237F:推 GKki2012 : [-是指 "belong to"吗 ∈ 03/12 09:44
238F:推 Gauss : 20180406 04/06 15:39
239F:推 Felix30810 : 20180603朝圣 06/03 00:19
240F:推 zxuanjia : 2018/06/03朝圣 06/03 22:34
241F:推 Aionaon : 20180604 朝圣 06/04 13:55
242F:推 anderson0815: 2018/9/10 朝圣 09/10 18:24
243F:推 Jetinacn : 2018.9.25 朝圣 09/25 01:33
244F:推 tom282f3 : 107.09.25 09/25 21:23
245F:推 tom282f3 : 紧握小根根 09/25 21:25
246F:推 joj4211 : 2018.10.15朝圣数学的起源 10/15 15:00
247F:推 waazxc77548 : 高二生瑟瑟发抖 10/17 22:12
248F:推 xikless : 2018.12.02 12/02 01:47
249F:推 gcobs0834 : 2018.12.5 12/05 15:42
250F:推 qqaatw : 2018/12/5 朝圣 12/05 15:53
251F:推 delvinnew200: ...朝圣 12/08 19:26
252F:推 d880126d : 2019.1.3 朝圣 01/03 14:35
253F:推 a12349221 : 朝圣…… 03/17 00:04
254F:推 steven56138 : 2019/9/25 考古朝圣 09/25 12:22
255F:推 Azraelx : 2020.06.28 朝圣 06/28 19:47
256F:推 jiexyz : 20211127朝圣 11/27 04:06
257F:推 Qkirito : 2021/12/17朝圣 这次比较看的懂罗 12/17 01:37
258F:推 oskens : 看了李永乐的证明终於弄懂这文章在干啥了 朝圣 03/13 19:55
259F:推 boriszhang : 朝圣 2023/12/07 12/07 12:48