A proof of the Collatz problem... I hope it is !!!
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roupam
Jun 17, 2009, 8:19:14 AM6/17/09
to True But Unproven - the Collatz Conjecture
Once again... the phoenix tries to rise from its ashes...
I have included the LaTeX source below...
Compile it to PDF and then read the paper...
I have shortened many things...
But in case you think I am missing something please do let me know...
Thanks
Roupam
---(Source starts below)---
\documentclass[12pt]{amsart}
\usepackage{geometry} % See geometry.pdf to learn the
layout options. There are lots.
\geometry{letterpaper} % ... or a4paper or a5paper
or ...
%\geometry{landscape} % Activate for for rotated page
geometry
%\usepackage[parfill]{parskip} % Activate to begin paragraphs with
an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/
`basename #1 .tif`.png}
%\date{} % Activate to
display a given date or no date
\begin{document}
\begin{center}
\bigskip
{\Huge A proof of the Collatz problem}
\bigskip
{\Large Roupam Ghosh}
$\emph{roupam...@gmail.com}$
\bigskip
\textit{Abstract}
\end{center}
We prove the Collatz conjecture, by trying to complete an incomplete
sequence, and eliminating out the contradictions. We also check and
eliminate out the possibility of an infinite loop. In the end we show
that every integer has a finite stopping time, which in other words
solves the Collatz conjecture.
\bigskip
\section{The Collatz Problem}
Consider the function
\begin{equation*}
f(x) =
\begin{cases}
\frac{x}{2} & \text{if $x$ is even}\\
3x+1 & \text{if $x$ is odd}
\end{cases}
\end{equation*}
\bigskip
The Collatz conjecture states that there exists an integer $d (\ge 0)$
corresponding to $x$ such that
$f^d(x) = 1$ for all natural numbers $x$
\\
where $f^d(x) = f(f(f(... $d times ...$(f(f(f(x))))))) $
\\
\section{Considering only odd integers}
Now if we consider only odd integers of the Collatz sequence, and
modify the function as follows
Consider the function for odd integers
\begin{equation*}
T(x) = \frac{(3x+1)}{2^k}
\end{equation*}
where k is the highest power of 2 that divides $3x+1$
\\
\\
\emph{If we can prove that for all odd integers the stopping time
corresponding to T(x) is finite, then the Collatz problem is proved.}
\newpage
\section{Considering an incomplete sequence}
\paragraph{Consider a given incomplete sequence}
$a_1, a_2, ... , a_m$ , where $a_k = T^k(a_1)$, and $a_1$ is the
minimum among $a_1, ... , a_m$.
\newline
\newline
Then we shall have
$2^{S_m} = (3 + \frac{1}{a_1})(3 + \frac{1}{a_1}) ... (3 + \frac{1}
{a_1}) \frac{a_1}{a_{m+1}}$
\newline
But since $a_1$ is mimimum, hence, we shall,\\
\begin{equation}
2^{S_m} < (3 + \frac{1}{a_1})^m\frac{a_1}{a_{m+1}}
\end{equation}
\bigskip
Now lets suppose
\begin{equation}
a_{m+1} < a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{c_m}{2^{S_m}} < a_1
\end{equation}
\newline
\newline
Now, for $m > 3, c_m > 3^m$, hence we have
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{3^m}{2^{S_m}} < a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}} (a_1 + 1) < a_1
\end{equation}
\begin{equation}
3^m(\frac{1+a_1}{a_1}) < 2^{S_m}
\end{equation}
From (1) and (6) we have
\begin{equation}
3^m(\frac{1+a_1}{a_1}) < (3 + \frac{1}{a_1})^m\frac{a_1}{a_{m+1}}
\end{equation}
From (7) we get after rearranging the terms
\begin{equation}
3^m(a^2_1 - a_1 a_{m+1} - a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac
{^mC_m}{a^{m-2}_1} > 0
\end{equation}
Here by (2) we have the equation
\begin{equation}
L.H.S > 3^m(- a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac{^mC_m}{a^
{m-2}_1}
\end{equation}
Which reduces to the equation (10) if we assume (11) to be true
\begin{equation}
(ma_1 - 3a_{m+1}) + \frac{2^m - m - 1}{3^{m-1} a^{m-2}_1} > 0
\end{equation}
\begin{equation}
3^m(- a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac{^mC_m}{a^{m-2}_1} \\
> (ma_1 - 3a_{m+1}) + \frac{2^m - m - 1}{3^{m-1} a^{m-2}_1}\\
>0
\end{equation}
Now for $m > 3$
\begin{equation}
\frac{2^m - m - 1}{3^{m-1} a^{m-2}_1} < 1
\end{equation}
Hence,
\begin{equation}
(ma_1 - 3a_{m+1}) > 0
\end{equation}
which is true since $m > 3$ and $a_1 > a_{m+1}$
\\
\\
\subsection{Statement 1}
Hence, we get from (1) and (2) and assumption (11), the equation (12)
which is true. If (1), (2) or (11) was false, (12) would also have
been false. Hence, (2) as well as (12) is true. So we can state that,
for any arbitrary number $a_1$ there exists $m > 3$ such that $a_{m+1}
< a_1$. Which , obviously, gives a hint that all integers have a
finite stopping time.
\newline
\newline
\paragraph{Now consider, if for all $m \ge 1$}
\begin{equation}
a_{m+1} > a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{c_m}{2^{S_m}} > a_1
\end{equation}
But, for $m \ge 1$
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{m4^{m-1}}{2^{S_m}} \ge \frac{3^m}{2^
{S_m}}a_1 + \frac{c_m}{2^{S_m}} ,\\
\end{equation}
\newline
Therefore,
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{m4^{m-1}}{2^{S_m}} > a_1
\end{equation}
Implies,
\begin{equation}
a_1 < \frac{m4^{m-1}}{2^{S_m} - 3^m}
\end{equation}
Which in turn, implies,
\begin{equation}
a_1 < m4^{m-1}\\
\end{equation}
Since $2^{S_m} - 3^m \ge 1$ for all $m \ge 1$
\\
But here, $a_1$ is arbitraily chosen, and doesn't depend on the value
of m
Hence, (19) does not hold true for all m. So, our assumption $a_{m+1}
> a_m$ for all $m \ge 1$, must be wrong
\\
\\
\\
This gives us an indication that any infinite trajectory cannot have a
minimum $a_1$.
Which is clearly impossible. Hence, infinite trajectories do not
exist. Considering the fact that infinite trajectories do not exist,
we may say from \textbf{Statement 1} that $m > 3$ has a finite value.
All of the above sums up to the fact, that every number has a finite
stopping time.
\subsection{Q.E.D}
Does this solve the Collatz problem??? You decide... :)!!!
\end{document}
I have included the LaTeX source below...
Compile it to PDF and then read the paper...
I have shortened many things...
But in case you think I am missing something please do let me know...
Thanks
Roupam
---(Source starts below)---
\documentclass[12pt]{amsart}
\usepackage{geometry} % See geometry.pdf to learn the
layout options. There are lots.
\geometry{letterpaper} % ... or a4paper or a5paper
or ...
%\geometry{landscape} % Activate for for rotated page
geometry
%\usepackage[parfill]{parskip} % Activate to begin paragraphs with
an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/
`basename #1 .tif`.png}
%\date{} % Activate to
display a given date or no date
\begin{document}
\begin{center}
\bigskip
{\Huge A proof of the Collatz problem}
\bigskip
{\Large Roupam Ghosh}
$\emph{roupam...@gmail.com}$
\bigskip
\textit{Abstract}
\end{center}
We prove the Collatz conjecture, by trying to complete an incomplete
sequence, and eliminating out the contradictions. We also check and
eliminate out the possibility of an infinite loop. In the end we show
that every integer has a finite stopping time, which in other words
solves the Collatz conjecture.
\bigskip
\section{The Collatz Problem}
Consider the function
\begin{equation*}
f(x) =
\begin{cases}
\frac{x}{2} & \text{if $x$ is even}\\
3x+1 & \text{if $x$ is odd}
\end{cases}
\end{equation*}
\bigskip
The Collatz conjecture states that there exists an integer $d (\ge 0)$
corresponding to $x$ such that
$f^d(x) = 1$ for all natural numbers $x$
\\
where $f^d(x) = f(f(f(... $d times ...$(f(f(f(x))))))) $
\\
\section{Considering only odd integers}
Now if we consider only odd integers of the Collatz sequence, and
modify the function as follows
Consider the function for odd integers
\begin{equation*}
T(x) = \frac{(3x+1)}{2^k}
\end{equation*}
where k is the highest power of 2 that divides $3x+1$
\\
\\
\emph{If we can prove that for all odd integers the stopping time
corresponding to T(x) is finite, then the Collatz problem is proved.}
\newpage
\section{Considering an incomplete sequence}
\paragraph{Consider a given incomplete sequence}
$a_1, a_2, ... , a_m$ , where $a_k = T^k(a_1)$, and $a_1$ is the
minimum among $a_1, ... , a_m$.
\newline
\newline
Then we shall have
$2^{S_m} = (3 + \frac{1}{a_1})(3 + \frac{1}{a_1}) ... (3 + \frac{1}
{a_1}) \frac{a_1}{a_{m+1}}$
\newline
But since $a_1$ is mimimum, hence, we shall,\\
\begin{equation}
2^{S_m} < (3 + \frac{1}{a_1})^m\frac{a_1}{a_{m+1}}
\end{equation}
\bigskip
Now lets suppose
\begin{equation}
a_{m+1} < a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{c_m}{2^{S_m}} < a_1
\end{equation}
\newline
\newline
Now, for $m > 3, c_m > 3^m$, hence we have
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{3^m}{2^{S_m}} < a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}} (a_1 + 1) < a_1
\end{equation}
\begin{equation}
3^m(\frac{1+a_1}{a_1}) < 2^{S_m}
\end{equation}
From (1) and (6) we have
\begin{equation}
3^m(\frac{1+a_1}{a_1}) < (3 + \frac{1}{a_1})^m\frac{a_1}{a_{m+1}}
\end{equation}
From (7) we get after rearranging the terms
\begin{equation}
3^m(a^2_1 - a_1 a_{m+1} - a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac
{^mC_m}{a^{m-2}_1} > 0
\end{equation}
Here by (2) we have the equation
\begin{equation}
L.H.S > 3^m(- a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac{^mC_m}{a^
{m-2}_1}
\end{equation}
Which reduces to the equation (10) if we assume (11) to be true
\begin{equation}
(ma_1 - 3a_{m+1}) + \frac{2^m - m - 1}{3^{m-1} a^{m-2}_1} > 0
\end{equation}
\begin{equation}
3^m(- a_{m+1}) + ^mC_1 3^{m-1}a_1 + ... + \frac{^mC_m}{a^{m-2}_1} \\
> (ma_1 - 3a_{m+1}) + \frac{2^m - m - 1}{3^{m-1} a^{m-2}_1}\\
>0
\end{equation}
Now for $m > 3$
\begin{equation}
\frac{2^m - m - 1}{3^{m-1} a^{m-2}_1} < 1
\end{equation}
Hence,
\begin{equation}
(ma_1 - 3a_{m+1}) > 0
\end{equation}
which is true since $m > 3$ and $a_1 > a_{m+1}$
\\
\\
\subsection{Statement 1}
Hence, we get from (1) and (2) and assumption (11), the equation (12)
which is true. If (1), (2) or (11) was false, (12) would also have
been false. Hence, (2) as well as (12) is true. So we can state that,
for any arbitrary number $a_1$ there exists $m > 3$ such that $a_{m+1}
< a_1$. Which , obviously, gives a hint that all integers have a
finite stopping time.
\newline
\newline
\paragraph{Now consider, if for all $m \ge 1$}
\begin{equation}
a_{m+1} > a_1
\end{equation}
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{c_m}{2^{S_m}} > a_1
\end{equation}
But, for $m \ge 1$
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{m4^{m-1}}{2^{S_m}} \ge \frac{3^m}{2^
{S_m}}a_1 + \frac{c_m}{2^{S_m}} ,\\
\end{equation}
\newline
Therefore,
\begin{equation}
\frac{3^m}{2^{S_m}}a_1 + \frac{m4^{m-1}}{2^{S_m}} > a_1
\end{equation}
Implies,
\begin{equation}
a_1 < \frac{m4^{m-1}}{2^{S_m} - 3^m}
\end{equation}
Which in turn, implies,
\begin{equation}
a_1 < m4^{m-1}\\
\end{equation}
Since $2^{S_m} - 3^m \ge 1$ for all $m \ge 1$
\\
But here, $a_1$ is arbitraily chosen, and doesn't depend on the value
of m
Hence, (19) does not hold true for all m. So, our assumption $a_{m+1}
> a_m$ for all $m \ge 1$, must be wrong
\\
\\
\\
This gives us an indication that any infinite trajectory cannot have a
minimum $a_1$.
Which is clearly impossible. Hence, infinite trajectories do not
exist. Considering the fact that infinite trajectories do not exist,
we may say from \textbf{Statement 1} that $m > 3$ has a finite value.
All of the above sums up to the fact, that every number has a finite
stopping time.
\subsection{Q.E.D}
Does this solve the Collatz problem??? You decide... :)!!!
\end{document}
roupam
Jun 17, 2009, 2:37:33 PM6/17/09
to True But Unproven - the Collatz Conjecture
Download my paper in pdf format from here...
http://www.filedropper.com/mypaper
No need to compile the LaTeX source...
Thanks
Roupam
http://www.filedropper.com/mypaper
No need to compile the LaTeX source...
Thanks
Roupam
Mensanator
Jun 17, 2009, 11:48:19 PM6/17/09
to True But Unproven - the Collatz Conjecture
On Jun 17, 1:37�pm, roupam <roupam.gh...@gmail.com> wrote:
> Download my paper in pdf format from here...http://www.filedropper.com/mypaper
>
> No need to compile the LaTeX source...
Good idea. You're going to have enough trouble
> No need to compile the LaTeX source...
even to get someone to look at a .pdf, let alone
compiling Latex.
By the way, you've got it all ass-backwards again.
There is no such thing as a FINITE sequence,
all sequences are infinite, it's just a
matter of whether or not the sequences contain
DISTINCT numbers and how many different
non-DISTINCT sequences there are. When you
reach the exact opposite conclusion, there's
certainly no point in reading further.
You really ought to learn what the actual
problem is (I told you before that that
"stopping at 1" nonsense is a red herring).
If you make no attempt to learn anything,
then yes, "crank" is the correct term.
>
> Thanks
> Roupam
roupam ghosh
Jun 18, 2009, 12:42:54 AM6/18/09
to TrueBut...@googlegroups.com
Good idea. You're going to have enough trouble
even to get someone to look at a .pdf, let alone
compiling Latex.
Thanks... Its very encouraging... I've had much trouble already...
By the way, you've got it all ass-backwards again.
There is no such thing as a FINITE sequence,
all sequences are infinite, it's just a
matter of whether or not the sequences contain
DISTINCT numbers and how many different
non-DISTINCT sequences there are. When you
reach the exact opposite conclusion, there's
certainly no point in reading further.
Isnt that obvious!!!
Don't you think I understand that???
I think others are not understanding what I have to say...
If you make no attempt to learn anything,
then yes, "crank" is the correct term.
Thanks... So I am a crank... and I should stop doing maths because I am not trying to learn anything... This is really very constructive criticism...
Anyways, i dont want to alienate myself from people who are truly devoted to the Collatz problem... So just wanted to tell that, a good samaritan, has POLITELY pointed me towards an error...
Mensanator
Jun 18, 2009, 1:39:31 PM6/18/09
to True But Unproven - the Collatz Conjecture
On Jun 17, 11:42 pm, roupam ghosh <roupam.gh...@gmail.com> wrote:
> > Good idea. You're going to have enough trouble
> > even to get someone to look at a .pdf, let alone
> > compiling Latex.
>
> Thanks... Its very encouraging... I've had much trouble already...
>
> By the way, you've got it all ass-backwards again.
>
>
>
> > There is no such thing as a FINITE sequence,
> > all sequences are infinite, it's just a
> > matter of whether or not the sequences contain
> > DISTINCT numbers and how many different
> > non-DISTINCT sequences there are. When you
> > reach the exact opposite conclusion, there's
> > certainly no point in reading further.
>
> Isnt that obvious!!!
> Don't you think I understand that???
finite sequences? When you make a proof, you need
to show your work. Maybe what you meant is "finite
up to the point where it becomes an infinite
non-distinct sequence", but you didn't say that.
What are we supposed to do, read your mind?
> I think others are not understanding what I have to say...
>
> > If you make no attempt to learn anything,
> > then yes, "crank" is the correct term.
>
> Thanks... So I am a crank... and I should stop doing maths
and claiming it's a proof. You should NOT claim something
is a proof if you're not done checking it.
> because I am not
> trying to learn anything... This is really very constructive criticism...
but I see little benefit to doing so.
>
> Anyways, i dont want to alienate myself from people who are truly devoted to
> the Collatz problem...
the few who care. You've had more replies from me than all of
sci.math combined. I could block you from posting to this group,
but I am still giving you the benefit of a doubt.
> So just wanted to tell that, a good samaritan, has
> POLITELY pointed me towards an error...
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