We previously defined integers in terms of natural numbers. Similarly we can define rational numbers in terms of integers.
\(\forall ab \in \mathbb{I} (\neg (b=0)\rightarrow \exists c (b.c=a))\)
A rational is an ordered pair of integers.
\(\{\{a\},\{a,b\}\}\)
So that:
\(\{\{a\},\{a,b\}\}=\dfrac{a}{b}\)
Integers can be shown as rational numbers using:
\((i,1)\)
Integers can then be turned into rational numbers:
\(\mathbb{Q}=\dfrac{a}{1}\)
\(a=\dfrac{a_1}{a_2}\)
\(b=\dfrac{b_1}{b_2}\)
\(c=\dfrac{c_1}{c_2}\)
There are an infinite number of ways to write any rational number, as with integers. \(\dfrac{1}{2}\) can be written as \(\dfrac{1}{2}\), \(\dfrac{-2}{-4}\) etc.
The class of these terms form an equivalence class.
We can show these are equal:
\(\dfrac{a}{b}=\{\{a\},\{a,b\}\}\)
\(\dfrac{ca}{cb}=\{\{a\},\{a,b\}\}\)
\(\dfrac{ca}{cb}=\{\{ca\},\{ca,cb\}\}\)
\(\{\{a\},\{a,b\}\}=\{\{ca\},\{ca,cb\}\}\)
Then we can define addition as:
\((a,b)+(c,d)=(a.d+b.c,b.d)\)
\(a+b=c\)
\(c_1=a_1b_2+a_2b_1\)
\(c_1=a_2b_2\)
\(a-b=c\)
\(c_1=a_1b_2-a_2b_1\)
\(c_1=a_2b_2\)
Similarly, multiplication can be defined as:
\((a,b).(c,d)=(a.c, b.d)\)
\(ab=c\)
\(c_1=a_1b_1\)
\(c_1=a_2b_2\)
\(\dfrac{a}{b}=c\)
\(c_1=a_1b_2\)
\(c_1=a_2b_1\)
We can see rational numbers as cartesian products of integers. That is:
\(\mathbb{Q}=Z.Z\)
We can order the rational numbers like so:
\(\{\dfrac{1}{1},\dfrac{2}{1},\dfrac{1}{2},\dfrac{1}{3},\dfrac{2}{2}\dfrac{3}{1}...\}\)
These can be mapped from natural numbers, so there is a bijunctive function.
So:
\(|\mathbb{Q} |=|\mathbb{Z}.\mathbb{Z} |=|\mathbb{N} |=\aleph_0\)
As: \(|\mathbb{Z}.\mathbb{Z} |=|\mathbb{Z}|^2\)
\(|\mathbb{N}|^n=\mathbb{N}\)
\(\dfrac{A}{B}+\dfrac{C}{D}=\dfrac{AD+BC}{BD}\)
\(\dfrac{A}{B}\dfrac{C}{D}=\dfrac{AC}{BD}\)
\(C+\dfrac{A}{B}=\dfrac{BC+A}{B}\)
\(C\dfrac{A}{B}=\dfrac{AC}{B}\)
\(\dfrac{A+B}{C}=\dfrac{A}{C}+\dfrac{B}{C}\)
\(\dfrac{A}{B}=\dfrac{AC}{BC}\)
We have: \(\dfrac{1}{A.B}\)
We want this in the form of:
\(\dfrac{a}{A}+\dfrac{b}{B}\)
First, lets define \(M\) as the mean of these two numbers, and define \(\delta=M-B\). Then:
\(\dfrac{1}{AB}=\dfrac{1}{(M+\delta)(M-\delta)}=\dfrac{a}{M+\delta}+\dfrac{b}{M-\delta}\)
We can rearrange the latter two to find:
\(1=a(M-\delta)+b(M+\delta)\)
Now we need to find values of \(a\) and \(b\) to choose.
Let’s examine \(a\).
\(a=\dfrac{1-b(M+\delta)}{M-\delta}\)
\(a=-\dfrac{bM+b\delta -1}{M-\delta}\)
\(a=-\dfrac{bM+b\delta -1}{M-\delta}\)
For this to divide neatly we need both the numerator to be a constant multiplier of the denominator. This means the ratio the multiplier for the left hand side of the denominator is equal to the right:
\(\dfrac{bM}{M}=\dfrac{b\delta -1}{-\delta}\)
\(b=\dfrac{b\delta -1}{-\delta}\)
\(b=\dfrac{1}{2\delta}\)
We can do the same for \(a\).
\(a=-\dfrac{1}{2\delta}\)
We can plug these back into our original formula:
\(\dfrac{1}{(M+\delta)(M-\delta)}=\dfrac{-\dfrac{1}{2\delta}}{M+\delta}+\dfrac{\dfrac{1}{2\delta}}{M-\delta}\)
\(\dfrac{1}{(M+\delta)(M-\delta)}=\dfrac{1}{2\delta}[\dfrac{1}{M-\delta}-\dfrac{1}{M+\delta}]\)
For any pair of rationals, there is another rational between them:
\(a=\dfrac{p}{q}\)
\(b=\dfrac{m}{n}\)
Where \(b>a\).
We define a new rational:
\(c=\dfrac{a+b}{2}\)
\(c=\dfrac{pn+qm}{2qn}\)
This is a rational number.
We can write:
\(a=\dfrac{2pn}{2qn}\)
\(b=\dfrac{2qm}{2qn}\)
As \(b>a\) we know \(2qm>2pn\)
So: \(a < c < b\)