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Robin P. Clark 2015-06-29 11:42:58 +01:00
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22
papers/fermat/fermat.tex
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@ -64,9 +64,9 @@ The numbers $a$ and $b$ may have common and uncommon prime factors; these can b
three 'bags', those only in $a$; $ubpf(a)$, those only in $b$; $ubpf(b)$ and those common to both; $cbpf(a,b)$. three 'bags', those only in $a$; $ubpf(a)$, those only in $b$; $ubpf(b)$ and those common to both; $cbpf(a,b)$.
A `Set' in mathematics is a collection of objects that may have only one of each type of element. A `Set' in mathematics is a collection of objects that may have only one of each type of element.
A `bag' is similar to a Set, except that it may have duplicates. A `bag' is similar to a Set, except that it may have duplicates.
Thus the number $32$ is represented as the product of a bag of prime numbers thus: $\prod \{2,2,2,2,2\}$ i.e. $2^5 = 32$. 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 bag of prime factors: Viewing the addition of $a^n +b^n$ as products of bags of common and uncommon~prime~factors:
\begin{equation} \begin{equation}
\label{eqn:primesexpanded0} \label{eqn:primesexpanded0}
\prod{cbpf(a,b)}^n \prod{ubpf(a)^n} + \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n \; , \prod{cbpf(a,b)}^n \prod{ubpf(a)^n} + \prod{cbpf(a,b)}^n \prod{ubpf(b)^n} = c^n \; ,
@ -84,15 +84,14 @@ this can be re-written as:
\section{Properties of numbers viewed as products of bags of prime factors} \section{Properties of numbers viewed as products of bags of prime factors}
\subsection{ Primes are guaranteed not preserved in addition } \subsection{ Primes are guaranteed not preserved in addition }
%
Only common prime factors are guaranteed preserved as a result of equation~\ref{eqn:primesexpanded1}.
% %
This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$ This means for $a^n+b^n$ the only prime factors guaranteed to be in $c^n$
are are those common in $a$ and $b$. are are those common in $a$ and $b$.
% %
Adding numbers creates a dissolving of the prime factors in the result. Adding numbers creates a `dissolving' of prime factors in the result:
Addition of primes causes the highest prime factors to become that is addition of numbers causes the uncommon prime factors to become
lost, but increases the number of smaller prime factors. lost, but increases the number other prime factors.
% %
Consider $43 +21 = 64$. These primes add up to a result with Consider $43 +21 = 64$. These primes add up to a result with
a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^6$. a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^6$.
@ -100,7 +99,10 @@ a bag of six twos i.e. $bpf(64) = \{2,2,2,2,2,2\}$ or more conventionally $64=2^
If a prime is added to another prime number the result 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; 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. the result of the addition must even and therefore have at least a prime factor of two.
%
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}
@ -197,7 +199,7 @@ 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} 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 $c$. 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 fractions of a whole number. In order for this to be true $a$ and $b$ must both be fractions of a whole number.
\subsection{trivial case} \subsection{trivial case}
@ -209,7 +211,7 @@ $$ a^n + b^n = 7^n = 49 $$
In order to get the prime factor 7 in the result both a and b must have the prime number 7 in them. 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 That is the numbers $a$ and $b$ must both have the number 7 as a common prime factor
to get seven as a prime factor in the result. to get seven as a prime factor in the result.
Any other number will not give a 7 in the bag of prime numbers result. Any other number will not give a 7 in the bag of prime numbers representation of the result.