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Looking for a paper

  1. Dec 11, 2005 #1
    I was asked to find a paper with proofs on the basis of arthimetics. For example proof that the product of a serie of factors is the same no matter what is the order the're read, and other arthimetic rules.
     
  2. jcsd
  3. Dec 11, 2005 #2
    what you're talking about is associativity. i think any book on abstract algebra would have what you're looking for.
     
  4. Dec 12, 2005 #3

    matt grime

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    And they're axioms not theorems
     
  5. Dec 12, 2005 #4

    shmoe

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    You know, I don't think I could name one. You'll usually get something like "using induction and the associativity law, we can unambiguously write multiplications without brackets," but I can't recall ever seeing it worked out in full gory detail.
     
  6. Dec 12, 2005 #5

    JasonRox

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    Find a book that starts with the Peano Axioms and that should do it.

    You even have to prove that 1 + x = x + 1!

    It can be done.
     
  7. Dec 12, 2005 #6

    shmoe

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    I know it can be done, I just can't recall ever seeing it in an abstract algebra book. It's usually waffled over though and with good reason-it's not really difficult or enlightening, just messy.
     
  8. Dec 12, 2005 #7

    JasonRox

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    Yeah, an Abstract Algebra text certainly would never have it.

    I don't see why it should though. You talk about associativity, but that's if groups have that property, which is what you must show yourself.
     
  9. Dec 12, 2005 #8

    shmoe

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    groups are all associative, it's part of the definition of a group.
     
  10. Dec 12, 2005 #9
    I might be remembering wrong, but doesn't Spivak's Calculus start with some of this stuff?
     
  11. Dec 12, 2005 #10

    JasonRox

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    DAMN! Big mistake.

    I meant to say something else. We check if sets under a given binary operation is associative, which is one of the steps to showing that a set is a group under a given binary operation.

    That makes me laugh because it's so so so wrong what I said. :rofl:
     
  12. Dec 12, 2005 #11

    JasonRox

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    He does, but I think his proof of associativity is just the same as any Abstract Algebra text.

    Well, you can prove that (a + b) + c = a + ( b + c), then it is a direct consequence that it will be true for n elements.
     
  13. Dec 12, 2005 #12
    Can someone simply proove it right here?
     
  14. Dec 12, 2005 #13

    JasonRox

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  15. Dec 12, 2005 #14
    As a matter of fact I did but I didn't find anything relevant. Thanks alot, but this is only concerning sums, not multiplications/divisions, not what I was looking for.
     
    Last edited: Dec 12, 2005
  16. Dec 12, 2005 #15

    JasonRox

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    You are welcome. :biggrin:
     
  17. Dec 12, 2005 #16

    HallsofIvy

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  18. Dec 12, 2005 #17
    HallsofIvy, I lack the proper knowledge to understand your paper. Is the proof of these axioms absolutly require basis in Number Theory?
     
  19. Dec 12, 2005 #18

    JasonRox

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    Axioms don't require proofs, only Theorems.

    That is the basics for Number Theory, and you don't need much mathematics for it.

    I had a link to a free basic Number Theory book, but I guess I lost it. It's easier to read, and doesn't use quantifiers, which is probably what you don't understand in HallsofIvy's text.

    Regardless, HallsofIvy wrote a nice monograph, and maybe one day I'll one too. Maybe a different topic though.
     
  20. Dec 13, 2005 #19

    HallsofIvy

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    Well, you asked about basic properties of numbers, didn't you? There are other ways of approaching the natural numbers but they essentially boil down to the same proofs.

    The essential property of the natural numbers, also called the "counting numbers" is- counting! Which, in technical terms is just "induction": the fact that given any natural number there is a "next" number and that if we start at 1 and to each "next" number in turn, we get all natural numbers.

    All proofs of basic properties of natural numbers reduce to proofs by induction. You must remember that anytime you ask for proofs of the fundamentals of anything, you are getting into deep water!

    Here is, briefly, the proof of the associative law for addition:
    The set of natural numbers, N, is first defined as a set of objects together with a function f (the "successor function) such that:
    a) There exist a member of N, called "1" such that f is a one-to-one, onto, function from N to N\{1}. That is, for every natural number, n, f(n), the "successor" of n, is a natural number and every natural number except 1 is the successor of some other unique natural number.
    b) If U is a set of natural numbers such that
    (i) 1 is in U
    (ii) Whenever n is in U, f(n) is also in U
    then U is the set of all natural numbers: U= N.
    (The principal of induction- the natural numbers are precisely those numbers we get by starting at 1 and going to the "next" number (the successor).)

    Addition is defined by: For any natural number, n, n+ 1= f(n). If m is not 1, then m is the successor of sum natural number, i: m= f(i). We define n+ m= n+ f(i)= f(n+i).
    One would use induction to show that this is "well defined" (that the sum of two natural numbers is a unique natural number)- that was "theorem 1".

    Theorem 2 (associative law for addition):
    For all a,b,c ε N, (a+b)+c = a+ (b+c)

    For a,b ε N let Uab= {c | (a+b)+ c= a+ (b+c)}
    1) (a+b)+ 1= f(a+b)= a+f(b)= a+ (b+1)
    Therefore 1 ε Uab.
    2) Assume c ε Uab.
    Then (a+b)+ f(c)= f((a+b)+ c)= f(a+ (b+c)) since cε Uab
    f(a+ (b+c))= a+ f(b+c)= a+ (b+ f(c))
    Therefore f(c) ε Uab.
    Therefore Uab = N.
     
    Last edited: Dec 13, 2005
  21. Dec 13, 2005 #20
    maybe i'm too fast for my own good. iit isn't just associativity?? :confused:
     
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