November 3rd, 2014, 04:58 AM  #1 
Member Joined: Apr 2013 Posts: 56 Thanks: 1  About Collatz conjecture
Collatz conjecture is just a small part of a big group of unsolved Number Theory problems. Problem 1. S is a subset of odd natural numbers which satisfies the following conditions: 1) 1 belongs to S 2) If n belongs to S then 2n + 1 belongs to S 3) If 3n belongs to S then n belongs to S Prove of disprove: S is the set of all odd natural numbers. I posted this problem in the Number Theory List, and the biggest world Number theory profs can not solve it. You can make up many similar unsolved problems, for example, change 3 into 5, etc. This shows that, really, we know nothing about natural numbers... 
November 3rd, 2014, 07:02 AM  #2  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra 
There is a significant difference between there being a lot that we don't know about something and there being little that we do know about it. Quote:
Last edited by v8archie; November 3rd, 2014 at 07:13 AM.  
November 3rd, 2014, 12:06 PM  #3 
Member Joined: Apr 2013 Posts: 56 Thanks: 1 
You should be able to state the hypothesis correctly. Of course, I do not mean the set of odd numbers in this case. For example, 1) 1 belongs S 2) If n belongs S then 3n + 1 belongs S 3) If 2n belongs S then n belongs S. Hypothesis: Is S the set of all natural numbers not divisible by 3? Try to prove or disprove. Last edited by vlagluz; November 3rd, 2014 at 12:28 PM. 
November 4th, 2014, 03:48 AM  #4  
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  Quote:
 
November 4th, 2014, 05:25 AM  #5 
Member Joined: Apr 2013 Posts: 56 Thanks: 1 
Oh, my God! I never said that these problems are EQUIVALENT to Collatz conjecture. They just belong to the same class of the problems, which nature is a big mystery. Collatz conjecture 1) 1 belongs to S 2) If n belongs to S then 2n belongs to S 3) If 3n + 1 belongs to S then n belongs to S. Hypothesis: Is S the set of all natural numbers? It is a modification of mathematical induction. To prove that a statement P(n) is true for all natural values of n, prove that: 1) P(1) is true 2) If P{k) is true then P(2k) is true 3) If P(3k + 1) is true then P(k) is true. Then, according to Collatz conjecture, the statement would be true for all natural values of n. Now, flag into your hands, and forward!!!  to attack the fortress which name is Collatz conjecture!!! Good luck!!! Last edited by vlagluz; November 4th, 2014 at 05:55 AM. 
November 4th, 2014, 06:39 AM  #6 
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26 
Just so I understand, since your explanation isn't entirely clear: Are you saying that if you could prove this then the Collatz conjecture would be proved or not? 
November 4th, 2014, 07:40 AM  #7 
Member Joined: Apr 2013 Posts: 56 Thanks: 1 
Investigate this question yourself

November 4th, 2014, 07:52 AM  #8 
Senior Member Joined: Mar 2012 Posts: 572 Thanks: 26  I have and (assuming you are referring to my question) I would argue 'not'. Let's say that S is the set of all natural numbers. Let's divide S into S1 and S2. S1 is the set of all numbers that comply with the Collatz conjecture (eg they have a chain that leads to 1) S2 is the set of all numbers that don't comply with the Collatz conjecture. 1) 1 doesn't belong to S2 However 2) If n belongs to S2 then 2n belongs to S2 3) If 3n + 1 belongs to S2 then n belongs to S2. Last edited by Hedge; November 4th, 2014 at 07:55 AM. 
November 4th, 2014, 10:18 AM  #9 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Right. I went to a talk last month where Conway discussed this class of problems and explained why he thinks most instances which cannot be solved easily cannot be solved at all (i.e., are formally undecidable).

November 4th, 2014, 05:50 PM  #10  
Member Joined: Apr 2013 Posts: 56 Thanks: 1  Quote:
Then What? Didn't get your arguments  

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