10

## 『偽鐵人賽』IT技術--三十天讓你瞭解數字系統

### 13 則留言

0

iT邦高手 1 級 ‧ 2011-10-23 13:17:37

0
iT邦好手 1 級 ‧ 2011-10-23 20:50:06

billchung提到：

billchung提到：
1和0在電腦科學中構成最基本的運算單元。

billchung提到：

kradark iT邦好手 1 級‧ 2011-10-23 20:52:36 檢舉

(哈哈 開個微笑不要介意....)

http://www.tngs.tn.edu.tw/teaching/math/research/1%2B1=2.pdf
1+1=2 的證明，請問高微的元素在哪？

http://www.wretch.cc/blog/NBabbage/6888733

kradark iT邦好手 1 級‧ 2011-10-23 21:29:26 檢舉

1 + 1 = 2
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 ? ](<strong>現在問題來了, 有一個 set 是包括所有 natural numbers 的嗎 ? </strong>)(甚至問

kradark iT邦好手 1 級‧ 2011-10-23 21:30:25 檢舉

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.

kradark iT邦好手 1 級‧ 2011-10-23 21:31:13 檢舉

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.

kradark iT邦好手 1 級‧ 2011-10-23 21:31:35 檢舉

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.

kradark iT邦好手 1 級‧ 2011-10-23 21:31:49 檢舉

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

n^+ [- X. It folloes by 6.3 that X = ω.

kradark iT邦好手 1 級‧ 2011-10-23 21:32:08 檢舉

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

kradark iT邦好手 1 級‧ 2011-10-23 21:32:31 檢舉

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.
kradark iT邦好手 1 級‧ 2011-10-23 21:32:48 檢舉

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.

kradark iT邦好手 1 級‧ 2011-10-23 21:34:23 檢舉

[恕刪]

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.

kradark iT邦好手 1 級‧ 2011-10-23 21:46:06 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 21:54:23 檢舉

simon581923提到：

0

iT邦超人 1 級 ‧ 2011-10-23 22:22:19

kradark iT邦好手 1 級‧ 2011-10-23 22:27:35 檢舉

billchung提到：

kradark iT邦好手 1 級‧ 2011-10-23 22:36:13 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:59:59 檢舉

0
billchung
iT邦新手 3 級 ‧ 2011-10-23 22:22:37

kradark iT邦好手 1 級‧ 2011-10-23 22:31:35 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:41:00 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:41:44 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:44:24 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:44:54 檢舉

0

iT邦新手 4 級 ‧ 2011-10-23 22:25:27

0

iT邦超人 1 級 ‧ 2011-10-23 22:26:40

1+1為什麼一定要等於2??

...那...爭論哲學...可以嗎??

billchung提到：

kradark iT邦好手 1 級‧ 2011-10-23 22:40:23 檢舉

0

iT邦超人 1 級 ‧ 2011-10-23 22:31:13

studyazure提到：
30天的流水帳

kradark iT邦好手 1 級‧ 2011-10-23 22:33:44 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:47:10 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 22:57:13 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 23:03:12 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 23:06:37 檢舉

kradark iT邦好手 1 級‧ 2011-10-23 23:08:03 檢舉

studyazure提到：

kradark iT邦好手 1 級‧ 2011-10-23 23:08:30 檢舉

billchung提到：

To Simon大:

0
billchung
iT邦新手 3 級 ‧ 2011-10-23 22:31:36

0

iT邦大師 1 級 ‧ 2011-10-23 23:08:47

=.=|||

tecksin提到：

wiselou: 有學問! 給你一個讚!

billchung提到：

billchung提到：

billchung提到：

wiselou提到：

0
billchung
iT邦新手 3 級 ‧ 2011-10-24 00:11:44

0
billchung
iT邦新手 3 級 ‧ 2011-10-24 01:44:48

To Simon: 你在回文中提到 『這樣子說, 真的很不妥, 有點咒人家的意思...』所以照這個方式推理，媒體在咒每個哀悼Jobs的人，就是這樣的意思。

0

iT邦超人 1 級 ‧ 2011-10-24 09:07:40

billchung提到：

Bill是寫什麼呢? 對他的這篇文'如喪考妣'...不倫不類至極, 也就說, 他認為卡大對他文的哀痛至極, 是這樣子嗎? 卡大只是提出一點不同的看法, 就要被這樣子'屈辱', 是否太過份?

Bill又引用佛家語...此心未明，如喪考妣，此心既明，更如喪考妣.

To Simon: 你對於忽略自己做過的事功力倒是很高。

0
gkkangel
iT邦好手 1 級 ‧ 2012-10-16 16:24:30