聽說相當困難, 要證明1+1=2非常複雜
是嗎?
可否簡單解說一下原則...
2005-05-22 06:26:33 · 10 個解答 · 發問者 Anonymous in 科學 ➔ 數學
其實這是很難的
相信我, 1+1"本來"就等於二
你怎麼知道?
2005-05-22 06:31:16 · update #1
基本上,這只是定義及假設的問題。
何謂「證明」?沒有公理假設及邏輯系統法則,就沒有所謂證明。現代數學家大部份都接受集合論為起點,其中有九大公理〔包括極有爭議性的選擇公理〕。
在上列式子中,出現了「1」「2」、「+」、「=」四個符號。
在回答你的問題前,必須先作一些數學系統上的假設,並且明確定義出以上四個符號的意義,才會導出結論,不然只是各說各的,每人談的都是自己的世界。
我們就從集合論的觀點去定義以上四個符號,請注意這四個符號並非最基本的符號,是需要定義出來的。
篇幅及精力所限〔我自己寫的回答,要自己打字的啊,可憐我吧〕,我只是粗略地說明一下,詳情請自行參讀一些集合論的書,很多都有談及自然數系統的建構過程。
集合論最基本的關係符號,事實上就是以邏輯關係來說明,有包含〔subset〕、交集〔intersection〕及聯集〔union〕。首先,是等號〔=〕的定義,兩個集合 A 及 B,如果互相包含,我們就稱他們為相等,以 A = B 表示。
然後,我們開始定義空集合。然後開始建構自然數系統,方便起見,空集合就是稱為 0,令它為第一個自然數。然後以數學歸納法的方式定義所有自然數:假設已經定義好 n 了,把 n 這個集合視為 {n} 這個集合的一個元素,取 n 與 {n} 的聯集,稱為 n + 1。聰明的讀者,應該知道,這其實就是在以「包集合」的方式來數東西,因此,以上的例子中,1 是被定義為 0 與 {0} 的聯集,而 2 是被定義為 1 與 {1} 的聯集。而從這兒,我們也順便定義了什麼是數 n 次,可稱 1 為 0 的第一個後繼者,2 為 0 的第 2 個後繼者等等。
數學歸納法,事實上也定義了何謂「+」,令 m 及 n 為兩自然數,m + n 的意思就是 m 的第 n 個後繼者。
於是式子的問題其實是:1 的第一個後繼者是否為 2?答案當然是肯定的啦。
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沒錯,一加一的確等於二,但我可以很明確的告訴你,一加一有時候或許並不會得到二這個答案,這並不是某小學生算錯或胡說八道,而是有它的道理在的。但在深究這個道理之前,我們倒是可以先來談談一個我們已知的事實:為什麼一加一等於二?
將時間推到幾千年前,一個不見數學蹤跡,一個人們不曉何謂算數記數的年代。一天,小明很高興地從湖邊補到一條魚,他高興地大喊:」晚上有一條魚可以吃了!」回到山洞的家中,看見爸爸也補了一條魚回來,他再度高興地大喊:」晚上有一條魚和一條魚可以吃了!」沒多久小明的弟弟回來,手邊也拿回一隻活蹦亂跳的魚兒,他又高興地大喊:」晚上有一條魚和一條魚和一條魚可以吃了!」中午,媽媽回來了,他大喊」待會兒有一條魚和一條魚和一條魚和一條魚可以吃了!」但小明有些不開心,他開始覺得這樣很笨。黃昏一刻,爺爺,奶奶,阿姨,小舅,叔叔,姑姑都回來山洞,也都各帶了一條魚回來,小明看見後就發瘋似地飛奔出去,他翹家了。
晚上小明想了很久,他決定把一條魚用一種符號代替,一條魚和一條魚用另一種符號代替......,直到可以代替全家所有的魚為止。隔天他回到家,驕傲地跟大家說明他了不起的,出類拔萃的見解:我有一條魚,爸爸有一條魚,總共就有兩條魚。(也就是1+1=2)
或許這就是數學發展的濫觴,它提醒了我們:無論多麼高深的數學理論,多麼艱澀的數值計算,其實都源自於生活,源自於大自然。
我們還要來看看1+1=2這個式子,將它分解成左邊的1+1與右邊的2這兩個部分,可以這麼假設,左邊發生在右邊之前,於是我們這麼說:小明的一條魚與爸爸的一條魚擺在一起,是兩條魚。這隱含了一個重要的意義:一條魚永遠是一條魚,不管是在水裡,天空中,甚至在爸爸的肚子裡(被分解而不是消失)。(註一)
但是挑戰來了,兩萬年後小明的後代,搭乘太空船來到天鵝座X-1星(註二),並朝X-1的方向丟了一條魚,他一路觀察魚飛往X-1的軌跡, 在靠近X-1一定距離(註三)的某個瞬間,魚兒突然消失不見,非咀嚼後吞下肚子那樣被分解,而是徹頭徹尾地消失在X-1,這個人類最早發現的黑洞前。 嚴謹一點的說法是,這條魚所對應的波函數的一部分,在進入黑洞後就不見蹤影,而完全無法進行觀測,導致下面一個結論,能區別這條魚之所以為魚的資訊,已經消失大半。相當於這條魚已經不再是一條魚了,而它究竟變成什麼,是在黑洞外的觀測者永遠無法知道的。(註四)
於是站在黑洞旁的我們,還能堅持1+1=2嗎?
依舊不以為然?唔...也罷,讓我們先來瞧瞧1+1是如何等同於3的。
有黑洞這樣不斷吸入物質的天體,相對地,我們也可以推論出一個不斷吐出物質的天體,由於有與黑洞相反的性質,故我們稱之為」白洞」。假設小明的後代有一天和女友去看電影,他們一人帶了一塊錢在身上,但兩張電影票總共要三塊錢,當他們倆雙手一攤時,發覺加起來竟有三塊錢,原因就在於旁邊的白洞吐了一塊錢給他們!當然運氣不好時,獲得的也有可能是一坨大便就是。
終究這是個可笑的例子,但它所象徵的意義是不可言喻的。
又根據觀察數據指出,宇宙的肇始源自於大霹靂。我們可以這樣想像,在一百多億年以前的某個瞬間,整個宇宙從"無"急速膨脹,逐漸演化成現今的規模。這件事透露了什麼樣的訊息?
行文至此,勉力向諸位說明了為何1+1=3,但我真正想說的不僅如此,而是想要揭示一個」沒有硬性規定,沒有強制束縛,沒有各式制限,一切都是可能的」的世界,這個世界就在我們四週,但很多人察覺不到,而更多人害怕改變。我所能做的,只是為諸位揭開這扇門,而是否願意走進這個世界並且接納它,就不在在下的能力範圍之內了。
註一:相當於亞里斯多德的同一律:在思考過程中,某物的意義是保持不變的
註二:天鵝座X-1為一雙星系統,主星為一超巨星,伴星為質量在太陽5~10倍的黑洞,本文的天鵝座X-1專指其伴星。
註三:此距離為事件視界,或謂事象地平面。
註四:上述的」魚消失不見」事件,其實觀測者永遠無法見得,而會看見前往黑洞的物體行進速度愈來愈慢,直到停留在事件視界上。但為簡化敘述,我們假設觀測者能夠觀測此事件
2005-05-22 06:34:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋
過了一段時間就會有人問,然後又有人抄我寫的解答,算了吧,不要再問了。
2005-05-26 18:32:55 補充:
要抄我請註明:
http://tw.knowledge.yahoo.com/question/?qid=1004123000037
2005-05-25 11:45:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋
我覺得第二位的答案很有趣也蠻容易懂的~~^^
2005-05-24 15:42:41 · answer #3 · answered by Anonymous · 0⤊ 0⤋
以下三個公設可以證明 1+1=2 .
一. next(1)=2, next(2)=3, next(3)=4 ...(自然數的定義)
二. prev(next(n))=n, if n 是自然數 (前後關係的定義)
三. m + n = next(m)+prev(n) if n1
or m + n = next(m) if n=1 其中 m,n 皆為自然數 (加法的定義)
由以上三式得 1+1=next(1)=2 .
2005-05-24 11:56:29 · answer #4 · answered by Anonymous · 0⤊ 0⤋
這不是很難!
這只是要證明出公設的公式而已。
1+1=2是由一位叫〝皮亞諾〞的數學家,
所寫出來的公設,一種大家都承認而且認定的一種公式。
而把皮亞諾的公式簡化後,
可以得到
N*=N+1 (N* 在數學上稱為後繼元素)
1*=1+1
1*=2
這只是一種概念,並不會很難的!!
2005-05-23 18:58:58 · answer #5 · answered by Anonymous · 0⤊ 0⤋
不對哦...1+1=2
正確的做法是
1+1-1=2-1
是要兩邊-1才是正確的數學
移動是跳動的數學
在自然數1,2,3,4,5,6,7
1+1絕對=2
怎講..自然的東西..一個人+一個人
就是2個人...一個小孩+一個大人
還是2個人...
2005-05-23 03:32:53 · answer #6 · answered by Anonymous · 0⤊ 0⤋
1+1=2 ?
必須先知道幾件事?
1. 1,2是屬於哪一個集合(數系)?
2. 1,2是什麼樣的元素?
3.加法是如何定義的?
4.此加法運算具有何種性質?
這些就是 代數學 中
群 group----> 群 已能解釋以上問題
環 ring
體 field
的概念
請自行參閱 代數學 的書
我推薦 台大 莫中堅的近世代數 與 康明昌的代數學
ps.我以前數學老師也這麼說 (1+1=2 是某人博士論文)
這是以訛傳訛
2005-05-22 15:44:38 · answer #7 · answered by KK 2 · 0⤊ 0⤋
我前陣子看到了署名 Pinter 的一段證明,僅供分享交流,如下:
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.
2005-05-22 06:51:23 · answer #8 · answered by 渡邊源 5 · 0⤊ 0⤋
1+1=2
1個+1個=2個
像上面講的
1=2-1
1+1≠2
有其他可能就是...
電腦程式
2005-05-22 06:40:37 · answer #9 · answered by Anonymous · 0⤊ 0⤋
幹麻證明
1+1本來就等於2
如果把+1移到後面去
就變成1=2-1
還是相等
所以1+1就是等於2
2005-05-22 06:28:24 · answer #10 · answered by Anonymous · 0⤊ 0⤋