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看到如何证明1+1=2这一问题被网友提到过几次,google 了一下,找到个证明如下:http://answers.yahoo.com/question/index?qid=20080622015630AABM5dF

本文发表在 rolia.net 枫下论坛The proof starts from the Peano Postulates, which define the natural numbers N. N is the smallest set satisfying these postulates:

P1. 1 is in N.
P2. If x is in N, then its "successor" x' is in N.
P3. There is no x such that x' = 1.
P4. If x isn't 1, then there is a y in N such that y' = x.
P5. If S is a subset of N, 1 is in S, and the implication
(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:
Def: Let a and b be in N. If b = 1, then define a + b = a'
(using P1 and P2). If b isn't 1, then let c' = b, with c in N
(using P4), and define a + b = (a + c)'.

Then you have to define 2:
Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.
Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which
replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
definition of addition to this:
Def: Let a and b be in N. If b = 0, then define a + b = a.
If b isn't 0, then let c' = b, with c in N, and define
a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the
Theorem above is a little different:

Proof: Use the second part of the definition of + first:
1 + 1 = (1 + 0)'
Now use the first part of the definition of + on the sum in
parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

更多精彩文章及讨论,请光临枫下论坛 rolia.net
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Replies, comments and Discussions:

  • 工作学习 / 科技领域杂谈 / 看到如何证明1+1=2这一问题被网友提到过几次,google 了一下,找到个证明如下:http://answers.yahoo.com/question/index?qid=20080622015630AABM5dF
    本文发表在 rolia.net 枫下论坛The proof starts from the Peano Postulates, which define the natural numbers N. N is the smallest set satisfying these postulates:

    P1. 1 is in N.
    P2. If x is in N, then its "successor" x' is in N.
    P3. There is no x such that x' = 1.
    P4. If x isn't 1, then there is a y in N such that y' = x.
    P5. If S is a subset of N, 1 is in S, and the implication
    (x in S => x' in S) holds, then S = N.

    Then you have to define addition recursively:
    Def: Let a and b be in N. If b = 1, then define a + b = a'
    (using P1 and P2). If b isn't 1, then let c' = b, with c in N
    (using P4), and define a + b = (a + c)'.

    Then you have to define 2:
    Def: 2 = 1'

    2 is in N by P1, P2, and the definition of 2.

    Theorem: 1 + 1 = 2

    Proof: Use the first part of the definition of + with a = b = 1.
    Then 1 + 1 = 1' = 2 Q.E.D.

    Note: There is an alternate formulation of the Peano Postulates which
    replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the
    definition of addition to this:
    Def: Let a and b be in N. If b = 0, then define a + b = a.
    If b isn't 0, then let c' = b, with c in N, and define
    a + b = (a + c)'.

    You also have to define 1 = 0', and 2 = 1'. Then the proof of the
    Theorem above is a little different:

    Proof: Use the second part of the definition of + first:
    1 + 1 = (1 + 0)'
    Now use the first part of the definition of + on the sum in
    parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

    更多精彩文章及讨论,请光临枫下论坛 rolia.net
    • PF. Question: are all of the postulates required?
      • 这问题问得好。
      • Nope. P3,P4,P5 are not necessary at all, none of them is necessary for the “proof”. :-(
    • 哈哈,这哪是证明啊,只不过是对2作为1的后继元素之定义的一个表达方式而已。
      • 你说的对,这不能算做证明。
    • 所以说整数系在加法下是完备的。