IS there a Proof for 1+1= 2? yes, and it is easy if you pay attention because math is the language of the universe. let me speak math with you. 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 isnt 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 isnt 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 isnt 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. so, nice talking to you.
Posted on: Sun, 14 Sep 2014 16:57:06 +0000
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