請問1+1=2要怎麼證明?(15分)

因為1+1=2,這種基本中的基本

根本不知道從何證明起= =

所以還請各位數學好的大大來幫我解決這問題囉

9 Answers

Rating
  • Anonymous
    2 decades ago
    Favorite Answer

    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.

    2005-01-07 23:08:06 補充:

    之前也有人問過

    但不知什麼原因被移除了...

    2005-01-08 07:52:01 補充:

    To狂 草香

    我覺得你的證法有漏洞耶`

    題目是證明1+1=2

    但你的證明過程中

    用到2-1=1

    if已知2-1=1

    那1+1=2啦`

    Source(s): 台大ptt-<<Math>>
  • 8 years ago

    ...

    .

    .

    .

    .

    .

    .

  • Anonymous
    2 decades ago

    完全看不懂....= =

  • 2 decades ago

    一團麵粉加一團麵粉~~還是等於一團大麵粉ㄚ!!???

  • How do you think about the answers? You can sign in to vote the answer.
  • 2 decades ago

    移項法則:

    等號左邊等於等號右邊

    將等號兩邊化為最簡數

    移項時,變動到等號兩邊時

    必須變號

      1(+1)=2      符號跟著後面的數字大家都知道吧!

    → 1=2—1        把(+1)移到=右邊 P.S需變號

    → 1=1          則 1=1      (↑+號變—號)

                     1+1=2

    2005-01-08 13:52:42 補充:

    痾...抱歉..我是用移項..所以會變成那樣啊!!!

    Source(s): 自己
  • 2 decades ago

    數字是印度人發明的 如果你想要知道正確的答案 你應該要去問印度人

    不過以我的知識 我可以告訴你:

    人類在很早很早以前就懂得算數 而證明就是

    一顆蘋果和另一顆蘋果擺再一起 一定會有兩顆蘋果

    這是永遠不會改變的

  • 2 decades ago

    要證明1+1等於2 得先探討阿拉伯數字 為什麼1的後面接的是2呢?

  • 2 decades ago

    因為以前原始人不是都有打獵

    ㄊ們用1顆小石子代表1隻獵物

    1隻獵物加1隻獵物

    等於

    1顆小石子加1顆小石子

    等於2之列物

    ㄊ們獵物太多用小時子代替

    1~...

    Source(s):
  • Anonymous
    2 decades ago

    1代表1個單位,2代表兩個單位,所以1+1等於兩個1,不用ㄑ探討為什麼要叫做2,這只是一個代表,所以你高興的話再你心中也可以把零當成是任何的數,畢竟這只是世人通用的語言跟符號罷了

Still have questions? Get your answers by asking now.