Difference between revisions of "rational number"

Latest revision as of 19:44, 30 September 2008

The rational numbers are the set of all fractions:

Q = {0, 1, -1, 1/2, -1/2, 2, -2, ...}

They are an Abelian group under addition, and, if {0} is removed from the set, they form an Abelian group under multiplication as well. Thus, the rational numbers form a field.

To make this concrete, we can construct a model for the rational numbers. First, for our purposes here, let us define

I = integers = { ... , -2, -1, 0, 1, 2, ... }

P = Positive integers = { 1, 2, 3, 4, ... }

Let the "<", "+", "x" operations on P be defined by restriction of the same operations on I.

Next, we form a set of ordered pairs of integers:

<math> \mathbb{Q}_0 \equiv \{(x,y) | x \in \mathbb{I}, y \in \mathbb{P} \}</math>

Define an equivalence relation on the set:

<math>(x,y) \sim (z,w)\ \equiv\ \exists k \in \mathbb{I} \left ( x = kz \and y = kw \right )</math>

This just says that all "equivalent" fractions, which differ only by a common factor in the numerator and denominator, are actually equal. We then form the set

<math>\mathbb{Q} = \mathbb{Q}_0 / \sim</math>

This is similar to discarding everything from the set except the "reduced fractions" -- those for which the numerator and denominator have no common divisor.

If we use [q] to represent the equivalence class of q, where <math>q \in \mathbb{Q}_0</math>, then by using the relation "<" defined on the integers, we can define "<" in Q as the relation:

<math>[(x,y)] < [(z,w)]\ \equiv\ xw < yz</math>

Multiplication is also easy to define:

<math>[(x,y)] \cdot [(z,w)] \equiv [(xz,yw)]</math>

And addition is nearly as simple:

<math>[(x,y)] + [(z,w)] \equiv [(xw + zy, yw)]</math>

It is straightforward to prove the axioms which describe the rational numbers within this model, thus showing that it is, indeed, a model for the rationals.

Finally, as an aside, we exhibit a very small theorem. Define the set of all reduced fractions:

<math>F = \{(x,y) \in \mathbb{Q}_0\ | \ \not\exists k \in \mathbb{I} \ (k \neq x \and x = ky) \}</math>

Then given any element <math>(x,y) \in \mathbb{Q}_0</math>, we will show that there exists a reduced fraction, <math>f \in F</math>, such that [f] = [(x,y)]. Consider the set of all denominators for elements of <math>\mathbb{Q}_0</math> which are equivalent to (x,y):

<math>D = \{w \in \mathbb{P} \ |\ \exists z \in \mathbb{I} \left ([(z,w)] = [(x,y)] \right )\}</math>

Then <math>D \subset \mathbb{P}</math>. But P is well ordered so D must have a smallest element. Call that element W, and the corresponding numerator Z. Then (Z,W) must be a reduced fraction which is equal to (x,y).

The rational numbers, in addition to being important in and of themselves, are one of the building blocks we use to construct the real numbers.