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.
相关阅读
[合集] 为什么毛利兰对康黑的造谣视而不见[合集] 现在气氛多好vesper8这种市井上海人渣女人 (转载)[合集] Survivor benefit 的官方解释关于外F——6个常见问题的科普 （原创）不管WSN喜欢不喜欢，WF会愈演愈烈怎样与黑人斗争及我的成功经验你们纠结外内发的有意思吗？多一种选择罢了吃不到葡萄说葡萄酸索男们恨外F，终究是因为自己没老婆外f的老公是不是应该叫做外公？男的还有替外f说话的？三条法律外的理由：为什么Zimmerman无罪对付外F还得外F[合集] 靠，找到尹榆的ID了[合集] 我给每个黄瓜种子底下埋了一个煮鸡蛋她是我的红烧肉汉语中文八级考题荤评中国好声音第一集其实索男和外F是一类人