My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree27Thanks
Reply
 
LinkBack Thread Tools Display Modes
March 30th, 2017, 11:03 PM   #21
Senior Member
 
Joined: Mar 2017
From: .

Posts: 208
Thanks: 2

Math Focus: Number theory
Quote:
Originally Posted by Mariga View Post
Hey guys.. I have the proof to the Conjecture. Is it safe to submit it here? I have had it since 2012 but have trust issues. Also I am quite sure about it.
Here it is..

Proof to Collatz conjecture.

Consider any positive integer n from which the sequence is formed. n has a probability of 0.5 of either being odd or even. If even, we divide it by two. If odd, we multiply it by three, add one and then divide the result by two since the resulting number must be even. This is same as multiplying n by 1.5 and adding 0.5. The resulting integer, say m, will hence either be n/2 or 1.5n+0.5 and also has a probability of 0.5 of being either even or odd. The 0.5 that is added has considerable effect on outcome only if n is 1 (which can explain the repeating cycle when the sequence reaches 1). In this case we will neglect it.
Since n has equal chances of being odd or even, it therefore has equal chances of being divided by 2 or being multiplied by 1.5, and so is m and the rest of the outcomes. The factor by which the outcomes are divided by is greater than the factor they are multiplied with and hence, the sequence will converge.
Mariga is offline  
 
March 31st, 2017, 08:01 AM   #22
Senior Member
 
Joined: Aug 2012

Posts: 1,373
Thanks: 322

I'm stealing this.
Thanks from topsquark and Joppy
Maschke is online now  
March 31st, 2017, 08:34 AM   #23
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,778
Thanks: 2195

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Mariga View Post
Proof to Collatz conjecture.
The probability that die roll results in an odd number is 0.5, and the probability that that it results in an even number is 0.5. Each of the even numbers 2, 4 and 6 is greater than one of the odd numbers 1, 3 or 5.

Your logic claims that if we add odd numbers and subtract evens from a running total we are guaranteed to reach negative numbers at some time.

But there are infinite sequences of rolls that do not give us negative numbers (e.g. any sequence containing only odd numbers). Therefore, the probabilistic argument fails. It is thus not robust enough to prove Collatz either.
Thanks from Joppy
v8archie is online now  
March 31st, 2017, 11:11 PM   #24
Senior Member
 
Joined: Mar 2017
From: .

Posts: 208
Thanks: 2

Math Focus: Number theory
if you choose any integer randomly, it has an equal chance of either being even or odd
Mariga is offline  
April 1st, 2017, 06:38 AM   #25
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,778
Thanks: 2195

Math Focus: Mainly analysis and algebra
Yes. But there are still infinite sequences of integers that are all odd, that could be chosen randomly.
Thanks from Joppy
v8archie is online now  
April 2nd, 2017, 10:38 AM   #26
Senior Member
 
Joined: Mar 2017
From: .

Posts: 208
Thanks: 2

Math Focus: Number theory
say we randomly pick 1000 integers. there would be around 500 even and 500 odd ones. same as if we flip a coin. In fact the larger the sample space, the more accurate it is. I this case, our sample is infinite coz it consists of all positive integers. It is not possible to have only odd numbers recurring. It is same as saying flipping a coin let's say only a million times and having a million tails. That's impossible.
Mariga is offline  
April 2nd, 2017, 11:11 AM   #27
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: CA

Posts: 1,238
Thanks: 637

Quote:
Originally Posted by Mariga View Post
say we randomly pick 1000 integers. there would be around 500 even and 500 odd ones. same as if we flip a coin. In fact the larger the sample space, the more accurate it is. I this case, our sample is infinite coz it consists of all positive integers. It is not possible to have only odd numbers recurring. It is same as saying flipping a coin let's say only a million times and having a million tails. That's impossible.
it's not impossible

it simply has probability $\left(\dfrac 1 2\right)^{1000000}$

quite unlikely yes but it is not 0 probability
romsek is offline  
April 2nd, 2017, 11:33 AM   #28
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,778
Thanks: 2195

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by Mariga View Post
It is same as saying flipping a coin let's say only a million times and having a million tails. That's impossible.
What is the greatest number of consecutive heads that it is possible to have? Suppose that it is $N$ and that we have just achieved that number of consecutive heads. How does the coin know that on the next toss, the probability of a tail has to be 100% instead of 50%? What mechanism enacts this change? Suppose I was using two coins instead of one. Can I now get $2N$ heads before a tail? Or do the coins somehow know how many coins I'm tossing and are able to adjust their probabilities appropriately? What if myself and a colleague decide to do this experiment in different rooms at the same time, pooling our results. How do the two coins communicate their current state to each other?

I'm sure, with a little thought you can come up with more situations that need addressing. Perhaps, once you've claimed the prize for your Collatz proof, you can solve these related problems.

Last edited by skipjack; April 2nd, 2017 at 11:35 AM.
v8archie is online now  
April 3rd, 2017, 09:33 AM   #29
Senior Member
 
Joined: Mar 2017
From: .

Posts: 208
Thanks: 2

Math Focus: Number theory
Quote:
Originally Posted by romsek View Post
it's not impossible

it simply has probability $\left(\dfrac 1 2\right)^{1000000}$

quite unlikely yes but it is not 0 probability
Romsek that's incorrect. It is not under binomial. Take a coin and flip it 50 times. count the number of heads and tails. You will notice it is around 25 give or take a few heads and 25 give or take a few tails. eg maybe 22 heads and 28 tails. it even gets better for 100 flips.. way better for 1000 flips and so forth.
Mariga is offline  
April 3rd, 2017, 05:48 PM   #30
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 6,778
Thanks: 2195

Math Focus: Mainly analysis and algebra
What's even better? I think you are wrong anyway. But that needn't worry you as long as you understand why you are so likely to get something close to a 50/50 split. Hint: it's to do with the "probability space". The important thing about the probability space is that is includes plenty of sequences where only 10% or fewer of the flips come up heads.
v8archie is online now  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
collatz, conjecture, proof



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
On the Collatz Conjecture JwClaassen Number Theory 0 March 18th, 2017 08:47 AM
collatz conjecture isaac Number Theory 6 March 15th, 2016 01:12 AM
About Collatz conjecture vlagluz Number Theory 10 November 5th, 2014 12:27 AM
The marvellous proof of the collatz conjecture lwgula Number Theory 2 October 30th, 2014 02:02 PM
Collatz conjecture & More (Please Help) Aika Number Theory 6 April 29th, 2012 06:34 AM





Copyright © 2017 My Math Forum. All rights reserved.