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- 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.
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