robin/Robin_PHD
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

tidying

Browse Source
This commit is contained in:
Robin P. Clark 2015-06-29 14:33:31 +01:00
parent 39fbe109ff
commit d6c80b1709
1 changed files with 76 additions and 45 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

119
papers/fermat/fermat.tex
View File

@ -23,10 +23,20 @@
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newcommand{\pf}{prime~factor}
\newcommand{\pfs}{prime~factors}
\def\layersep{1.8cm}
\linespread{1.0}
\title{flt primes}
\begin{document}
% numbers at outer edges
\pagenumbering{arabic} % Arabic page numbers hereafter
@ -39,10 +49,13 @@
\paragraph{Keywords:} fermat; prime;
%\small
\abstract{ % \em
} % abstract
\abstract{
Viewing integers as collections of prime numbers and
using properties of prime numbers under addition and multiplication
this paper offers a proof of Fermats last theorem for
all positive integers $> 2$.
}
\section{Introduction}
Fermat's Last Theorem
@ -53,9 +66,9 @@ equation $a^n + b^n = c^n$ for any integer value of n greater than two.
\section{Breaking positive integers into constituent products of bags of primes}
Any positive integer can be represented as a collection (or bag) of prime numbers multiplied together.
A function $bpf()$ or `bag of prime factors' is defined to represent this.
A function $bpf()$ or `bag of {\pfs}' is defined to represent this.
\begin{equation}
\prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n
a^n + b^n = \prod{bpf(a)}^n + \prod{bpf(b)}^n = c^n
\end{equation}
%The function $bpf()$ will always contain 1.
@ -66,7 +79,7 @@ A `Set' in mathematics is a collection of objects that may have only one of each
A `bag' is similar to a Set, except that it may have duplicates.
The number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$.
Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~prime~factors:
Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~{\pfs}:
\begin{equation}
\label{eqn:primesexpanded0}
\prod{cbpf(a,b)}^n \prod{ubpf(a)^n} + \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n \; ,
@ -83,8 +96,9 @@ this can be re-written as:
\section{Properties of numbers viewed as products of bags of prime factors}
\subsection{ Primes are guaranteed not preserved in addition }
\subsection{Conditions where some Primes are guaranteed not preserved in addition}
%
% ADDITION DESTROY UNCOMMON PRIME FACTORS
%
This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$
are are those common in $a$ and $b$.
@ -100,11 +114,14 @@ If a prime is added to another prime number the result
cannot be a prime number, simply because all prime numbers above two are odd;
the result of the addition must even and therefore have at least a prime factor of two.
%
Prime numbers are unique. Adding to them, or adding other prime numbers to them, takes that unique
property away.
%
Thus only common prime factors in $a$ and $b$ are preserved
as a result of equation~\ref{eqn:primesexpanded1}.
%
\subsection{conditions for having a integer root}
\subsection{Conditions for having a integer root}
To have an integer root $n$ all prime numbers that comprise the number to be rooted must be at least
to the power of $n$.
@ -130,39 +147,48 @@ that number must be at the power of n or greater.
% Adding two prime numbers at any power greater than 1
% and then taking a root means getting an irrational number.
\subsection{}
\begin{equation}
\label{eqn:primesexpanded21}
\prod{cbpf(a,b)}^n \big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) = c^n
\end{equation}
\begin{equation}
a^n + b^n = \prod{bpf(c)^n}
\end{equation}
%assuming its true $c^n$ must be $ 2 \prod{cpf(a,b)}^n \prod{upf(a)^n} \prod{upf(b)^n} $
Extending this concept, taking a number as a bag of prime factors and then taking it to the
power of $n$, means taking the number of individual primes in the bag of prime factors and multiplying that number by $n$.
For instance the number 306, as a bag of prime factors is $\{2,3,3,17\}$ i.e. $306=\prod \{2,3,3,17\}$.
Cubing; $306^3$ gives 28652616: as a bag of prime factors 28652616 is $\{2,2,2,3,3,3,3,3,3,17,17,17\}$.
Viewing the result of the cubing in terms of bags of primes numbers,
\begin{itemize}
\item 306 has 3 twice as a prime factor, $306^3$ has 3 six times as a prime factor:
\item 306 has 2 once as a prime factor, $306^3$ has 2 3 times as a prime factor:
\item 306 has 17 once as a prime factor, $306^3$ has 17 3 times as a prime factor.
\end{itemize}
% For instance the number
% \begin{equation}
% \label{eqn:primesexpanded21}
% \prod{cbpf(a,b)}^n \big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) = c^n
% \end{equation}
%
% \begin{equation}
% 2 \prod{cpf(a,b)}^n = \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}}
% a^n + b^n = \prod{bpf(c)^n}
% \end{equation}
%
%
% %assuming its true $c^n$ must be $ 2 \prod{cpf(a,b)}^n \prod{upf(a)^n} \prod{upf(b)^n} $
%
%
% %
% % \begin{equation}
% % 2 \prod{cpf(a,b)}^n = \frac{\prod{bpf(c)^n}}{\prod{upf(a)^n} \prod{upf(b)^n}}
% % \end{equation}
%
%
% That means that ${ubpf(a)^n}$ and ${ubpf(b)^n}$ multiply the prime numbers in $cbpf(a,b)$
% $n$ times each.
% These are a component of $c^n$.
%
%
% \begin{equation}
% \label{eqn:primesexpanded22}
% \prod{cbpf(a,b)}^n = \frac{c^n}{\big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) }
% \end{equation}
That means that ${ubpf(a)^n}$ and ${ubpf(b)^n}$ multiply the prime numbers in $cbpf(a,b)$
$n$ times each.
These are a component of $c^n$.
\begin{equation}
\label{eqn:primesexpanded22}
\prod{cbpf(a,b)}^n = \frac{c^n}{\big( \prod{ubpf(a)^n} + \prod{ubpf(b)^n} \big) }
\end{equation}
\section{contradiction}
\section{Proof by Contradiction.}
%
For $a^n + b^n = c^n$ to be true for whole numbers $ > 2$, the highest prime factors on both sides of the equation must be equal.
%
@ -177,6 +203,10 @@ both $a$ and $b$ must contain the highest prime in $c$.
If $a$ and $b$ are whole numbers they either create a result with
the highest prime more than once, or it is destroyed by addition.
For $a^n + b^n = c^n$, for the highest prime, this means $a+b=1$.
This means that where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
This concept can be extended to numbers where there are duplicate highest primes.
%% Simple case where only one of highest prime factor in c^n
% describe contradiction for simple case:
@ -188,25 +218,26 @@ Taking the value $c$ as the product of a bag of prime numbers, it must have
a largest prime in $c$ (to a power $t$ which is one or more), i.e. $p^t$.
%
When $c$ is taken to the power $n$, $c^n$, that
means this prime factor becomes $p^{t+n}$.
means this prime factor becomes $p^{tn}$.
%
Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(t+n)}$, for that prime in the result.
Therefore, for that highest prime in $c$, $a^n + b^n$ must add up to $p^{(tn)}$, for that prime in the result.
%
Because prime numbers are by definition indivisible by other
whole numbers, the only way to get a prime number taken to
a power $p^t$ by addition is to add proportions that add up to one $p^t$.
%
This means both a and b must contain this prime factor {\em in some proportion}
so that $a p^{t+n} + b p^{t+n} = p^{t+n} $ satisfy the highest prime in $c$.
In order for this to be true $a$ and $b$ must both be fractions of a whole number.
so that $a p^{tn} + b p^{tn} = p^{tn} $ satisfy the highest prime in $c$.
%
In order for this to be true $a$ and $b$ must both be fractions of a whole number:
again this means $a+b$ must equal 1.
Thus where $a$ and $b$ are $ > 2$; $a^n + b^n \neq c^n$ for whole numbers.
\subsection{trivial case}
Take the trivial case where $c$ has the prime number 7:
Take the trivial case where $n=2$ and $c$ has the prime number 7 as one of its prime~factors:
%
$$ a^n + b^n = 7^n = 49 $$
$$ a^n + b^n = 7^n = 49 \; . $$
%
In order to get the prime factor 7 in the result both a and b must have the prime number 7 in them.
That is the numbers $a$ and $b$ must both have the number 7 as a common prime factor