1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

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

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Homework Helper
    Gold Member

  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

    User Avatar
    Homework Helper
    Gold Member

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

    HallsofIvy

    User Avatar
    Science Advisor

    Here's a link to a monograph I wrote myself several years ago. It details proofs of basic properties of: Natural numbers, Integers, Rational numbers, and Real numbers. The natural numbers are the first chapter:
    http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/$file/NUMBERS.pdf [Broken]
     
    Last edited by a moderator: May 2, 2017
  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

    User Avatar
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor

    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 by a moderator: Dec 13, 2005
  21. Dec 13, 2005 #20
    maybe i'm too fast for my own good. iit isn't just associativity?? :confused:
     
  22. Dec 13, 2005 #21

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    I had thought it was associativity for a general number of elements, what lets you write a product unambiguously as abcd without brackets for example. This is what I don't think I've seen in an abstract algebra book, just mentioned or left as an exercise. Which is fair, once you're at that stage you should be capable of muddling through the details.

    Werg22, you can prove this sort of thing by induction on the number of elements in the product. I don't know the simplest way to proceed, but you could assume that every product with n terms is equal no matter how the parentheses are arranged. Then for n+1 terms, break it up into smaller products and go from there. Here you'd try to prove that no matter what you start with, you could reduce it to the same arrangement of parantheses, say multiply from left to right e.g. for 4, reduce to ((ab)c)d.
     
  23. Dec 13, 2005 #22

    HallsofIvy

    User Avatar
    Science Advisor

    I can't speak for how fast you are but I did say "Here is, briefly, the proof of the associative law for addition".

    Yes, it is the associative law- that's why I called it that!

    Of course, once you have (a+ b)+ c= a+ (b+ c) the more general form follows easily.
     
  24. Dec 13, 2005 #23

    shmoe

    User Avatar
    Science Advisor
    Homework Helper

    So I don't seem totally off my rocker, Werg22 had said "For example proof that the product of a serie of factors is the same no matter what is the order the're read", which I understood to be associativity for a general number of parenthesis.
     
  25. Dec 13, 2005 #24
    I found a way to proove it somehow by induction.

    It is easy to proove those identities:

    1. [tex]ab=ba[/tex]

    2.[tex]abc=(ab)c=a(bc)[/tex]

    Then we can proove that for this special case (3 elements) the order dosen't matter. Thus for 4 elements;

    [tex]abcd = (abc)d [/tex] By definition .

    Now we can proove tha not matter what the order in the parenthesis, the result remains unchanged. The proove the general case, we have to proove that you can place any of the element to any wished position without chaging the result. By property 1, d can be the first element and the last. It is easy to see that any number can be put first, and since there is 3 remaining elements, and we already prooved it can be arranged a wished, then any combination can be formed with those elements without changing the result. Then we prooved this for 4 elements, and we proove it for 5 the same way.

    Q.E.D.?
     
    Last edited: Dec 13, 2005
  26. Dec 13, 2005 #25

    JasonRox

    User Avatar
    Homework Helper
    Gold Member

    What you described as a proof of associativity for 4 elements isn't a proof for associativity. You basically just described commutativity.

    The element d being on the left or right has nothing to do with associativity.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook