# Natural numbers Z?

## Main Question or Discussion Point

Do people usually prove that Z is an abelian group under (normal) addition or is it the definition of the natural numbers Z?

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Homework Helper
Do people usually prove that Z is an abelian group under (normal) addition or is it the definition of the natural numbers Z?
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".

HallsofIvy
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Do people usually prove that Z is an abelian group under (normal) addition or is it the definition of the natural numbers Z?
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.

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:
Homework Helper
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
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.

HallsofIvy