My Math Forum  

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

Number Theory Number Theory Math Forum


Thanks Tree28Thanks
Reply
 
LinkBack Thread Tools Display Modes
April 9th, 2017, 05:22 PM   #101
Senior Member
 
Joined: Aug 2012

Posts: 1,888
Thanks: 525

Quote:
Originally Posted by Mariga View Post
The first 10 positive odd integers are 1,3,5,7,9,11,13,15,17,19. If we multiply each by 3, add one and divide by two, we get these results 2,5,8,11,14,17,20,23,26,29. Results are also half odd and half even meaning for any odd number, when we multiply it by 3, add 1 and then divide by 2, chances are 50/50.. either odd or even.
No, it's not probabilistic. It's true than half are odd and half are even, but it's not random, any more than if I asked you what are the odds that $5$ is even.
Maschke is online now  
 
April 9th, 2017, 05:29 PM   #102
Senior Member
 
Joined: Aug 2012

Posts: 1,888
Thanks: 525

Quote:
Originally Posted by Mariga View Post
So what does all that mean? it simply means, if we have a number, lets say 69, the number is odd, and so we will have to multiply it by 3, add one and then divide by two.. the number we would get, has 50/50 chance of being either even or odd.
No that's not true. $3 \times 69 + 1 = 208 = 2^4 \times 13$. It's deterministic, not random. Flipping a coin is random. Determining if half of $208$ is even or odd is deterministic.

Quote:
Originally Posted by Mariga View Post
If it is even, we will divide the number by 2 to get the next outcome. the outcome will still be 50/50.. either odd or even. If the number is odd, still a 50/50 chance the result will be even and so forth.
Since that's the case for each and every outcome, that it is either odd of even, and we are dividing by 2 if it is even or multiplying by 1.5 if odd, then it means the sequence will converge.
That last bit comes out of nowhere, and it's not true. Why must the sequence converge? It may be that on average the sequences converge, but on average, prime numbers are odd.

Last edited by Maschke; April 9th, 2017 at 05:42 PM.
Maschke is online now  
April 9th, 2017, 09:15 PM   #103
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Quote:
Originally Posted by Maschke View Post
We should perhaps try to drill down one point at a time.

It is true that 3n+1 is always even for odd n. But if you think that's because of a probabilistic argument, then you must think that since the odds of a prime number being odd are 1, all prime numbers are odd.

3n+1 is not even because of a probabilistic approach. It is even because of the formula itself. We very well know that if we multiply an odd number by another odd number we get an odd number and when we add an odd number and another odd number, we always get an even number.
When we phrase that in terms of probability, we will say the probability of getting an even number after multiplying an odd number by an odd number is zero. The statement is very true. for any number.
When it comes to the example you have used on prime numbers, it doesn't relate at all to what we are doing here and that is an assumption that is not based on any fact and it doesn't hold.
Mariga is offline  
April 9th, 2017, 09:45 PM   #104
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Quote:
Originally Posted by Maschke View Post
But we are entirely sure. If the even number is divisible by 4 then half of it is even. If it's divisible by 8 then half of half of it is even. Etc. There is nothing probabilistic. if you show me a number I'll tell you exactly how many times it's divisible by 2. It's not random or unknown. Every number $n = 2^k m$ where $m$ is odd and $k$ is maximal. You just look at the number's prime factorization. If $m = 1$ then $n$ is a power of $2$.
I'll still ask you to take another example.. take this scenario.. A gun is pointed at your head and you are given half a second to give answers, meaning you don't have time to calculate the values. If the half second passes, you are killed. If your answer is wrong, you are killed.
Take number 47, the guy is not interested in the value of any outcome but is interested in whether the numbers are even or odd. So he asks you 3n+1.. odd or even? to rescue yourself, if you are an accomplished mathematician, you would say even even without the need to calculate the value. What if he asks you
$\frac{3n+1}{2}$ whether the number is odd or even.. you don't have the time to calculate the value.. would you be sure of your answer?
what if he asks you a number like 4326.. when you divide it by two.. will you get an even number or an odd number?
How would you rescue your life? by plain guess work.
Is there any difference between such a scenario with a scenario where you are asked to guess heads or tail in a coin flip?
Mariga is offline  
April 9th, 2017, 10:10 PM   #105
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Quote:
Originally Posted by Maschke View Post
No, it's not probabilistic. It's true than half are odd and half are even, but it's not random, any more than if I asked you what are the odds that $5$ is even.
Do u agree with the example i have used for even numbers? if you do then apply the same for this.
If we continue on with the chain, the results for consecutive odd numbers will be alternating even and odd numbers. this shows that if I choose any random odd number, i multiply it by 3,add 1 then divide by 2, the number is we get is as likely to be odd as it is to be even.
I don't know how to better explain this point if you still don't get it but take your time.
or let's take this example.. take all odd integers that are below say 1000. We have shown that for each consecutive odd numbers the outcomes will be alternating and so the even outcomes will be 500 and odd outcomes will be 500. So if I pick any random odd number below 1000, chances are, the outcome will be among the 500 odd numbers or the 500 even numbers.
Mariga is offline  
April 9th, 2017, 10:18 PM   #106
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Quote:
Originally Posted by Maschke View Post
No that's not true. $3 \times 69 + 1 = 208 = 2^4 \times 13$. It's deterministic, not random. Flipping a coin is random. Determining if half of $208$ is even or odd is deterministic.
Refer to 104.
Mariga is offline  
April 9th, 2017, 10:37 PM   #107
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Quote:
Originally Posted by Maschke View Post
That last bit comes out of nowhere, and it's not true. Why must the sequence converge? It may be that on average the sequences converge, but on average, prime numbers are odd.
If the numbers have same chances of being multiplied by 1.5 as they have of being divided by 2, then the sequence will converge.

If you still have a problem understanding all those points then also asking someone who understands it better may be helpful because the way I explain them may still not align with the way you view the problem. This discussion would go on and on because we are seeing the problem differently yet each of us has a point.
From the angle I have shown where we are not calculating the values, I have clearly shown how it is not deterministic. We will never know a cat is in which box unless we open a box. We can't know whether the next number is odd or even unless we calculate it.
Also we have exhausted on the point of infinitary probability by showing that it is impossible to have any number from which the sequence infinitely rises. Such a number from which such outcomes would be obtained exists only in theory.

Last edited by Mariga; April 9th, 2017 at 10:40 PM. Reason: To grammatically correct a sentence.
Mariga is offline  
April 10th, 2017, 03:06 AM   #108
Global Moderator
 
Joined: Dec 2006

Posts: 18,956
Thanks: 1603

What does your method tell us about the corresponding problem where the multiplier 3 is replaced by a larger odd number?
skipjack is offline  
April 10th, 2017, 03:39 AM   #109
Banned Camp
 
Joined: Mar 2017
From: .

Posts: 338
Thanks: 8

Math Focus: Number theory
Unless we encounter a cycle, the sequence should infinitely diverge.
we are either dividing by 2 or multiplying by 2.5
Mariga is offline  
April 10th, 2017, 07:27 AM   #110
Senior Member
 
Joined: Jun 2015
From: England

Posts: 822
Thanks: 243

Quote:
Originally Posted by Mariga View Post
The numbers are not meaningless... also probability doesn't depend on occurrence, but occurrence depends on probability.
Also sorry I don't understand your point.
The proof itself is clearly stated and if anything in it is incorrect, refer to it for easier understanding and if not properly phrased, we can edit it.
I detect a certain contempt and offhandness in this and your subsequent reply to my posts.

You don't understand it so it must be wrong or dismissed?

As far as I can see you are using what is known as the classical or Laplace definition of probability.

This is known to be at best incomplete and at worst flawed.

Have you heard of the Bertrand Paradox?

https://en.wikipedia.org/wiki/Bertra..._(probability)

I quote the key phrase from my link

Quote:
an example to show that probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined.
So, exactly as I said the XX or mechanism is a vital part of probability.

The statement

The probability is 0.5 is meaningless since it possesses no mechanism or outcome.

The statement the probability of a fair coin toss being heads is 0.5 is meaningful, although that is no guarantee of its veracity.

I think the problem with probability in this proof lies with the fact that it is an applied concept that has borrowed from construct(s) in pure maths that are not (necessarily) addressed by the application.

Here are some problems.

1) Proving the Conjecture requires you to prove that there must exist a finite n such that the necessary convergence occurs.

(This is what Masche means by terminating)

But you are offering the convergence of an infinite sequence as part of your proof.
Yet the convergence of an infinite sequence is unaffected by the prior convergence of any finite number of terms.

So with my fair coin if I always start with a head the probability at that first toss will be 1, yet the infinite sequence will still converge to 0.5.

Last edited by skipjack; April 10th, 2017 at 08:03 AM.
studiot 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 4th, 2014 11:27 PM
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 © 2018 My Math Forum. All rights reserved.