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推文 x0
[学习] 大家知道1+1=2原理吗?我证明给你看!
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 包含于 X. But X 包含于 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 包含于 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 包含于 n;
but n 包含于 n^+, so m 包含于 n^+. If n = m, then (because n
包含于 n^+) m 包含于 n^+; thus in either case, m 包含于 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 包含于 m and m 包含于 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 包含于 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 包含于 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 包含于 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
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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献花 x2 回到顶端 [楼 主] From:台湾亚太线上 | Posted:2006-02-12 16:47 |
albee543
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actually yo should say in one dimentional space, 1 + 1 = 2
and no one in earch yet proofed why 1 + 1 = 2
if you can do that probably u can get a nobel price

or

by defination, 1 + 1 = 2
basically you cannot proof it, its a defination.
I heard someone proofed 2 + 1 = 3, used like 1000 pages


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献花 x0 回到顶端 [1 楼] From:台湾亚太线上 | Posted:2006-02-12 16:49 |
circlemap
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我是比较倾向于 1+1 = 2 乃是定义
凡定义就不被证明
所以 3 = (1+1) +1
4 = ((1+1)+1)+1

以后继者的观点来定义后于1的整数当然也行


献花 x0 回到顶端 [2 楼] From:台湾中华电信 | Posted:2006-02-12 18:58 |
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大学以下的我建议你们以定义来看就好^o^


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献花 x0 回到顶端 [3 楼] From:台湾亚太线上 | Posted:2006-02-12 20:08 |
chenweiyeh
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挖~有看没有懂~
实在是太深澳了\


献花 x0 回到顶端 [4 楼] From:台湾中华电信 | Posted:2006-02-12 20:31 |
夷希微
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天啊~都是英文 ^^" 表情
想看也看不懂说 :P 表情
1+1=2 的定理只在数学领域范围中成立而已~
在其它领域里就不成立了 ^_^
例如在电子逻辑的领域里是 1+1=1 的 表情 表情
其它领域例:1 只手 + 1 只手 = 1 双手
还有: 1 包面粉 + 1 杯水 = 1 团面团
另还有: 1 只公鸡 + 1 只母鸡 = 数不清的鸡
.....
嘻嘻~我是来闹场的~酸 表情 表情


视之不见,名曰夷;
听之不闻,名曰希;
搏之不得,名曰微。
此三者不可致诘,故混而为一。

[截自老子道德经第十四章]
献花 x2 回到顶端 [5 楼] From:韩国Hanaro电信 | Posted:2006-02-12 23:41 |
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下面是引用夷希微于2006-02-12 23:41发表的 :
天啊~都是英文 ^^" 表情
想看也看不懂说 :P 表情
1+1=2 的定理只在数学领域范围中成立而已~
在其它领域里就不成立了 ^_^
例如在电子逻辑的领域里是 1+1=1 的 表情 表情
.......
哈哈哈... 我原本很正经的看待每一段话的解释...
但是我不得不佩服你.. 尤其当我看到你最后一段的时候.. 我大笑了 表情


献花 x0 回到顶端 [6 楼] From:台湾中华电信 | Posted:2006-02-13 00:30 |
albee543
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下面是引用夷希微于2006-02-12 23:41发表的 :
天啊~都是英文 ^^" 表情
想看也看不懂说 :P 表情
1+1=2 的定理只在数学领域范围中成立而已~
在其它领域里就不成立了 ^_^
例如在电子逻辑的领域里是 1+1=1 的 表情 表情
.......

哈哈!!分明就来扎场的
不知道那是什么理论 表情


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献花 x0 回到顶端 [7 楼] From:台湾亚太线上 | Posted:2006-02-13 00:40 |
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下面是引用夷希微于2006-02-12 23:41发表的 :
天啊~都是英文 ^^" 表情
想看也看不懂说 :P 表情
1+1=2 的定理只在数学领域范围中成立而已~
在其它领域里就不成立了 ^_^
例如在电子逻辑的领域里是 1+1=1 的 表情 表情
.......
天阿~~大大你该不会是传说中的kuso达人呢? 表情


点下面的连结就是对我最大的回馈,并不花你半毛钱
http://bbs.mychat.to/index.php?u=200012
请支持小弟开版(有关社区管理或社区生活上的相关问题回覆)
http://bbs.mychat.to/read.php?tid=595610
献花 x0 回到顶端 [8 楼] From:台湾中华电信 | Posted:2006-02-13 02:56 |
aa 手机 会员卡 葫芦墩家族
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以前当兵时,有一个学长他是念数学研究所的,他问我 0 的定义
我说 0 就是 0 啊,还要怎么定义,他就拿了一本数学的书,好厚的一本啊(大概5公分厚)
他说那本书就是在说明 0 的定义,看了都傻了.. 表情


献花 x0 回到顶端 [9 楼] From:台湾新世纪资通 | Posted:2006-02-13 16:54 |

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