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標題:
why does 1+1=2 ?
發問:
What is the reason for 1+1=2? Please kindly answer in English with lots and lots of explanation. I know it is an university topic but I'm only a form.3 student but I want to learn about it. Please!
最佳解答:
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.
其他解答:
why does 1+1=2 ?
發問:
What is the reason for 1+1=2? Please kindly answer in English with lots and lots of explanation. I know it is an university topic but I'm only a form.3 student but I want to learn about it. Please!
最佳解答:
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.
其他解答:
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Oh so easy 1pig and 1 pig is how many? ^o^|||||You may want to know Alfred North Whitehead and Bertrand Russell in their "The Principia Mathematica" had used more than 360 pages to prove 1+1=2.
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