MandysNotes

Wednesday, 12 February 2014 02:03

Sets, Rings, and Groups of Numbers.

By  Gideon
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Sets, Rings, and Groups of Numbers

The set $$\mathbb{N}$$, of $$\mathbf{Natural Numbers}$$, is the set of counting numbers starting from zero.
$\mathbb{N} = \{ 0, 1, 2, 3, ...\}$
We can add any two natural numbers, $$n$$, and $$m$$, together and the result, $$p = n+ m$$, is also a natural number. However, addition is a one-way-street'' when it comes to natural numbers: we can put two natural numbers together, but we cannot pull them apart again.

The problem is that for any natural number $$n$$, the additive inverse of $$n$$, that is $$(-1)n$$, is not a natural number. Natural numbers have addition but no subtraction.

In order to rectify this situation we need to add to the set $$\mathbb{N}$$, the set $$(-1) \mathbb{N}$$, of all natural numbers multiplied by $$(-1)$$.

The set $$\mathbb{Z}$$, called the set of $$\mathbf{Integers}$$ is equal to $$\mathbb{N} \cup (-1) \mathbb{N}$$, where the symbol $$\cup$$'' (union) means the addition of one whole set to another.

In other words,
$\mathbb{Z} = \{ ..., -3, -2, -1, 0, 1, 2, 3, ...\}.$

The integers have some nice algebraic properties with respect to addition:

• There exists a unique number, $$0$$, in $$\mathbb{Z}$$, such that for any other number $$k$$ in $$\mathbb{Z}$$, $$k+0 = k$$. Zero is called the $$\mathbf{Additive Identity}$$ of $$\mathbb{Z}.$$
•   If $$k$$ and $$l$$ are any two number in $$\mathbb{Z}$$, then $$m= k + l$$ is also a number in $$\mathbb{Z}.$$
• For any $$k$$ in the set $$\mathbb{Z}$$, there is another number $$k'$$ such that $$k+k'=0$$. I.e., $$k' = -k.$$

A set with addition, such as $$\mathbb{Z}$$, that satisfies the above three criteria is called an $$\mathbf{Additive Group}.$$