My Math Forum Proof to Collatz conjecture.

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March 30th, 2017, 11:03 PM   #21
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 Originally Posted by Mariga 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.

 March 31st, 2017, 08:01 AM #22 Senior Member   Joined: Aug 2012 Posts: 1,571 Thanks: 379 I'm stealing this. Thanks from topsquark and Joppy
March 31st, 2017, 08:34 AM   #23
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 Originally Posted by Mariga 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.

 March 31st, 2017, 11:11 PM #24 Senior Member   Joined: Mar 2017 From: . Posts: 262 Thanks: 4 Math Focus: Number theory if you choose any integer randomly, it has an equal chance of either being even or odd
 April 1st, 2017, 06:38 AM #25 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2295 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
 April 2nd, 2017, 10:38 AM #26 Senior Member   Joined: Mar 2017 From: . Posts: 262 Thanks: 4 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.
April 2nd, 2017, 11:11 AM   #27
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 Originally Posted by Mariga 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

April 2nd, 2017, 11:33 AM   #28
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 Originally Posted by Mariga 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.

April 3rd, 2017, 09:33 AM   #29
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 Originally Posted by romsek 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.

 April 3rd, 2017, 05:48 PM #30 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,973 Thanks: 2295 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.

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