f*i
2 楼
1+1可以等于0,比如 F2,中,呵呵
M*c
3 楼
坐等答复
w*a
4 楼
就是这么定义的，不为啥
f*t
5 楼
公理，无须证明。
d*x
6 楼
这基本是公理的范畴
http://zh.wikipedia.org/wiki/%E7%9A%AE%E4%BA%9A%E8%AF%BA%E5%85%
注意2只是一个符号，用以表示1的后继。
【在 C**********r 的大作中提到】
: rt
http://zh.wikipedia.org/wiki/%E7%9A%AE%E4%BA%9A%E8%AF%BA%E5%85%
注意2只是一个符号，用以表示1的后继。
【在 C**********r 的大作中提到】
: rt
I*t
7 楼
其实你可以“证明”1+1=2的
http://mathforum.org/library/drmath/view/51551.html
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.
http://mathforum.org/library/drmath/view/51551.html
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.
C*r
8 楼
numbers
so, the key is the definition of 2 as 1', like phoenix mentioned?
【在 I*********t 的大作中提到】
: 其实你可以“证明”1+1=2的
: http://mathforum.org/library/drmath/view/51551.html
: 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.
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