自行對號入座...
樓上的不介意擠一下吧。
(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
我只看到代數學和集合論，高微在哪？

現在你咕一個高中數學的網頁，來問高微的元素在哪？
拍謝啦!!! 請您自己過目囉，我也不知道對不對，
至少證1+1=2是高微課程，我唸數學系時，學長、同學、教授從沒有人懷疑過。

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
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.

[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.

把 n = 1 並且注意 2 = 1^+, 故 1 + 1 = 2.

我只能說, 當年沒轉去應數系是對的....因為大一時微積分都幾乎滿分, 應數系的助教和教授一直想要我轉去應數系...

學IT的應該不用懂這麼嚴謹的數學吧。
我先自鞭，我沒修過高等微積分喔，大二就從數學系烙跑了，

請參考在下劣作: 資訊學院的30門課-微積分(上)calculus
資訊學院的30門課-微積分(下) calculus

simon581923提到：
沒轉去應數系是對的

一定是對的，我大二時，兩個微積分被當檔修，一個自己不修，一個轉去會計系，一個轉去資訊系，六個人中只有一個室友修高等微積分，當他花兩個多禮拜關房間看高微時，我們在客廳坐一排看鎖碼台，然後他只考2,30幾分，還說已經算很高了。

我也有點後悔當時沒有『勇氣』從零介紹到二十九，起碼轉大富翁可以賺一堆東西。

明知道得獎無望的人，還能夠持續無私的發文，就是值得鼓勵的。

這就是一種參與、一種對社群的認同。

希望大家能夠給予發文大大一點點回應，就算是擠出幾個字也好。

billchung提到：
勇氣

真的...需要有勇氣才敢PO那些文...

剩三篇，我想惡搞從0介紹到2耶，
發文大大會不會鞭我啊？

不會，你如果可以每一篇還縮到十個字以內，我應該會非常開心。

非也，非也，寫錯就要承認，
我還要感激分享過程中，幫我辛苦校稿的各位大大，
不然我自己錯了，我都不知道，再次感激，
重點是分享的過程，大家有沒有成長。

老實說，這篇能有什麼營養？我自己都看不出來有啥營養。

那copy要不要承認？
不是指這篇文，有人的文被我發現疑似抄自維基，但那位仁兄也沒承認啊~

非啥非呢？就說這篇文的解釋我說了算，如果要是非，也只能是我才 "是"，眾人皆"非"，這樣懂嗎？

為甚麼你說了算？

你也可以砍文？
寫錯為甚麼可以你說了算?

彼此彼此。

因為這篇文是我發的。當然解釋就在我囉

更何況，我也沒說他對，因為要用中文寫作技巧解釋。

如果有興趣，你也可以自己寫一篇廢文...

抱頭痛哭，我寫了20幾篇廢文，原來是一堂課就可以上完的。
唉，對不起恩師，對不起站方，對不起好多人。

To kradark:
請不要在我的版上如喪考妣，這會觸霉頭的，你可以在自己的版隨便痛哭，但在這邊請保持微笑。

警告billchung人身攻擊
已抓圖

請公開道歉 請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉請公開道歉

隨便你..

隨便你..

要不要我也幫你抓一張

哪裡人身攻擊了？

如喪考妣=悲痛至極

Reference:
http://140.111.34.46/chengyu/mandarin/fulu/dict/cyd/0/cyd00779.htm

我很想知道哪裡有人身攻擊...

若你被這樣講，你若覺得不是人身攻擊????
這已經不在討論下去了。

謝謝您

就說隨便你了，你也公開發文了，還在這幹嘛呢？

如果是我，我會先去看那句成話是否真的構成毀謗。
否則，很容易留下笑柄...

畢竟真告下去，沒告成反而被告誣告，還不是自己倒楣？

再回一次，隨便你 again

studyazure提到：
告下去

這可能還需要大名鼎鼎的billchung公開資料
在下沒這個能耐 我只是小兵一枚

先去睡了 各為晚安

我為什麼要公開資料呢？

剛沒看到15號回應，你也晚安

billchung提到：
請不要在我的版上如喪考妣

這樣子說, 真的很不妥, 有點咒人家的意思...

你可以說..請不要在我的版上加油添醋, 或...請不要在我的版上搬弄是非, 但...這句成語..最好不要亂用, 因為那是非常不好的情況...

版大摸著自己的良心自問, 如果有人奉送你這旬成語, 你做何感想??

將心比心, 不希望別人對你講的, 就不對別人講...這是做人處世最基本的道理.

有嗎 ？應該沒有吧？

有嗎 ？應該沒有吧？

這個場合適合搬沙發嗎？

板凳好一點，沙發太佔位置了

搬盾牌實際點....

=.=|||

好像很多人都被『考妣』的台語音給誤導了...
考妣...指的是父母的意思...跟台語的...靠北...邊走...意義不一樣...
所以...

tecksin提到：
適合搬沙發嗎？

搬總裁出來當擋箭牌....

wiselou: 有學問! 給你一個讚!

billchung提到：
有學問

不敢當...只是年少時期...看的閒書多了些...
這個學問啊...要搬鐵大出來了...
鐵大.....大..大..大...（echo）

看起來所有哀悼Jobs的人都應該去找媒體算帳

中時的標題沒錯啊! '如喪考妣'真的只有一種場景會用啊對Jobs幹嘛要'如喪考妣'??可是, 我見很多PO文對Jobs...唉懷念就好了嘛!!真的, 我不太懂現在年輕人在想什麼??