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 \(\mathbf{Additive Group}.\)



Read 1862 times Last modified on Wednesday, 12 February 2014 02:37
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