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2 very basic questions

  1. Aug 13, 2013 #1
    Hi

    I am trying to learn how the number systems was created, and there are two very basic thing I don't get.

    first question:


    My book describes and proves that addition is well-defined for integers Z.

    that is if, z2=z3, then
    z1+z2 = z1+z3

    It also does the same for rational number, it gives a proof that

    if q2 = q3, then
    q1+q2=q1+q3


    However one thing that puzzles me is that I can not find a proof that it holds for naturlar numbers.

    That is if:

    n2= n3

    then:
    n1+n2 = n1+n3

    I know this is probably very basic, can I assume it is correct, or should it also be proved?

    They define the natural numbers as cardinal numbers of sets. And proves many laws like m+n=n+m etc. for natural numbers, but not the one I asked above.


    second question:
    This question is probably very stupid, but since it seems like everything should be proved at this basic level, why can I assume that if a = b, then b = a, is this how = is defined?
     
  2. jcsd
  3. Aug 13, 2013 #2

    mfb

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    All natural numbers are integers (and all integers are rational numbers).
    If something is true for all integers, it is true for all natural numbers.

    It is a part of its definition.
     
  4. Aug 13, 2013 #3

    phinds

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    Well, natural numbers are a subset of integers so wouldn't it follow that if it's true for integers it HAS to be true for natural numbers?

    EDIT: I see mfb beat me to it :smile:
     
    Last edited: Aug 13, 2013
  5. Aug 13, 2013 #4
    The problem with what you are saying with, N beeing a subset of Z, therefore if addition is well defined in Z then it is in N, is that for me it seemed like they used that addition in N was well defined when they proved it in Z.

    the proof for Z is like this:
    [m,n], [p,q] and [r,s] are integers. Where m,n,p,q,r,s is in N.
    then we have that [p,q] = [r,s], so p+s = q+r

    then they prove that [m+p,n+q] = [m+r,n+s]

    they do it like this:

    (m+p) + (n+s) = m+n+p+s = m+n+q+r = (n+q) + (m+r),
    so by the definition of Z [m+p,n+q] = [m+r,n+s]

    but when we write :
    m+n+p+s = m+n+q+r
    Aren't we using that addition is well defined for N?

    EDIT:

    Because if m+n = a, p+s=b, and q+r = c, then we have since p+s=q+r, then b=c
    so we use that a+b = a+c, if b=c
     
    Last edited: Aug 13, 2013
  6. Aug 13, 2013 #5

    SteamKing

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    'derfor'?
     
  7. Aug 13, 2013 #6

    phinds

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    That's Upper Slobovian for "therefore" :smile:
     
  8. Aug 13, 2013 #7
    Sorry for using a word in a foreign language, it is fixed now.
     
  9. Aug 13, 2013 #8
    I tried making a proof, is this proof valid:

    First my book defines addition in W(whole numbers) as. If, m,n[itex]\in[/itex] W, and A and B are sets suck that m = #(A) and n = #(B) and A [itex]\cap[/itex] B = [itex]\phi[/itex], then m+n = #(A[itex]\cup[/itex]B)
    #is the cardinal number

    then I want to prove that if b=c, then a+b=a+c
    To prove this I assume that
    a = #(A) for a set A, and b = #(B) for a set B, I also assume that A[itex]\cap[/itex]B = [itex]\phi[/itex]
    I also assume that c = #(C) for a set C

    Now since A and B are disjoint I get directly from the definition that a+b = #(A[itex]\cup[/itex]B)
    I also have that c=#(C)=#(B)=b

    Now instead of saying that c=#(C), I can just use that c = #(B), and since this it all it takes to use the definition of addition(?, can I just choose to use the other set), then I have that since A and B are disjoint
    a+c = #(A[itex]\cup[/itex]B)
    and because of this
    a+b = #(A[itex]\cup[/itex]B) = a+c
     
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