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 November 3rd, 2014, 05:58 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, 08:02 AM   #2
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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.

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The case with 5 is, I think false, since $9 \times 5^m \ne 2n+1$ where $n$ is odd and $m$ is natural. This is easily proved, because$$5 (4a+1) = 20a + 5 = 4(5a + 1) + 1 =4b + 1$$

Last edited by v8archie; November 3rd, 2014 at 08:13 AM.

 November 3rd, 2014, 01:06 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 01:28 PM.
November 4th, 2014, 04:48 AM   #4
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Quote:
 Originally Posted by vlagluz 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.
I'm not sure this does equate to the Collatz conjecture, though you may be able to correct me - how do the conditions above rule out a loop of large numbers that doesn't reach 1? Or (less plausibly) a chain that rises infinitely and never reaches 1? So far as I can see all the numbers in such a loop or infinite chain would satisfy these conditions.

 November 4th, 2014, 06:25 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 06:55 AM.
November 4th, 2014, 07:39 AM   #6
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Just so I understand, since your explanation isn't entirely clear:

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 Originally Posted by vlagluz Oh, my God! 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?
Are you saying that if you could prove this then the Collatz conjecture would be proved or not?

 November 4th, 2014, 08:40 AM #7 Member   Joined: Apr 2013 Posts: 56 Thanks: 1 Investigate this question yourself
November 4th, 2014, 08:52 AM   #8
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 Originally Posted by vlagluz Investigate this question yourself
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 08:55 AM.

November 4th, 2014, 11:18 AM   #9
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Quote:
 Originally Posted by vlagluz They just belong to the same class of the problems, which nature is a big mystery.
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, 06:50 PM   #10
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Quote:
 Originally Posted by Hedge 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.

Then What? Didn't get your arguments

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