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Natural numbers Z?

  1. Mar 27, 2007 #1
    Do people usually prove that Z is an abelian group under (normal) addition or is it the definition of the natural numbers Z?
     
  2. jcsd
  3. Mar 27, 2007 #2

    radou

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    What exactly is your question? (Z, +) is an abelian group. I don't believe the integers are defined that way. They only "fit into this given definition".
     
  4. Mar 27, 2007 #3

    HallsofIvy

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    First, the "natural numbers" are NOT Z. The natural numbers include only the positive integers (some texts include 0) while Z is all integers. Obviously the natural numbers does NOT form a group since they do not have additive inverses. Typically, Z is defined in terms of natural numbers (say, as equivalence classes of pairs of natural numbers) and then the fact that they form an abelian group is proved.
     
  5. Apr 1, 2007 #4
    That + is abelian follows from

    (x + y)z = xz + yz
    x(y + z) = xy + xz (both proved by Peano i think)

    Given a, b in Z.

    (a + b)(1 + 1) = a(1 + 1) + b(1 + 1) = a + a + b + b,
    (a + b)(1 + 1) = (a + b)1 + (a + b)1 = a + b + a + b


    this implicates

    a + b + a + b = a + a + b + b

    so

    a + b = b + a

    so + is abelian
     
    Last edited: Apr 1, 2007
  6. Apr 1, 2007 #5

    radou

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    This shows that it's not necessary to assume that, in a ring with addition, addition is commutative, since it follows from the other ring axioms.

    The distributive law is an axiom, btw, there's nothing to prove.
     
  7. Apr 1, 2007 #6

    HallsofIvy

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    An axiom in what system? Certainly the distributive law is part of the definition of "ring" and so a axiom in that sense. The fact that the distributive law is true for the natural numbers, integers, rational numbers, etc. can be proved.
     
  8. Apr 1, 2007 #7

    radou

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    Yes, that's what I meant. I assumed Ultraworld was referring to a ring, since there's multiplication and addition in his post.
     
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