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- Thread starter pivoxa15
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- #2

radou

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What exactly is your question? (Z, +) is an abelian group. I don't believe the integers are

- #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.

- #4

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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

(x + y)z = xz + yz

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

Given a, b in

(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

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- #5

radou

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That + is abelian follows from

(x + y)z = xz + yz

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

Given a, b inZ.

(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

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.

- #6

HallsofIvy

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- #7

radou

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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.

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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