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Math Help - associative law

  1. #1
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    associative law

    Prove for any natural numbers  a,b,c , we have  (a+b)+c = a+(b+c) . So fix  a and  b and induct on  c . For  c = 0 ,  a+b = a+b . Suppose inductively that  (a+b)+c = a+(b+c) . We have to prove that  (a+b) + c++ = a+(b+c++) . So  (a+b)+c++ = (a+b+c)++ . Also  a+(b+c++) = (a+b+c)++ .  \blacksquare

    Is this correct?
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    Prove for any natural numbers  a,b,c , we have  (a+b)+c = a+(b+c) . So fix  a and  b and induct on  c . For  c = 0 ,  a+b = a+b . Suppose inductively that  (a+b)+c = a+(b+c) . We have to prove that  (a+b) + c++ = a+(b+c++) . So  (a+b)+c++ = (a+b+c)++ . Also  a+(b+c++) = (a+b+c)++ .  \blacksquare
    Remember how we define addition. We define +(a,0) = a and we denote it by a+0 = a. And +(a,b+1) = +(a,b) + 1. We denote a+(b+1) = (a+b)+1. Now use induction. Say (a+b)+c = (a+b)+c is true. Then (a+b)+(c+1) = ((a+b)+c)+1 = (a+(b+c))+1 = a + ((b+c)+1) = a+(b+(c+1)).
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    did I do that?
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  4. #4
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    Quote Originally Posted by particlejohn View Post
    did I do that?
    The answer is “we don’t know”.
    I am glad that TPH answered the way he did. I was about do the almost the same thing as he has done. But I would have used different notation. The fact is that there are almost as many notations used in Peano arithmetic as there are textbooks that address the topic. So the reason we don’t is we have not seen the notation you are using.
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