User talk:Senojesse

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Cheers! prime mover (talk) 15:34, 7 January 2021 (UTC)

Edits on Collatz conjecture
Please do not discuss the question in the page itself. Please do it in the talk page. HINT: See the "Discussion" link at the top of the Collatz Conjecture/Supposed Proof page.

You will need to learn the protocol and conventions around the use of a wiki before you are going to be able to contribute meaningfully on.

And I'm sorry, but your analysis of the Collatz conjecture is flawed. Backing up your statements by "It is a fact!" is not a proof. --prime mover (talk) 06:20, 20 January 2021 (UTC)


 * If you would do us the courtesy of writing your contributions in $\LaTeX$, then this would be greatly appreciated.


 * It would also do you well to make yourself familiar with how proofs are structured.


 * You may also wish to consider taking a course on number theory, so as to give you the chance of understanding some of the concepts involved.


 * Bottom line: there is nothing in the structure of a Collatz sequence that suggests that any number form (be it $4 n + 1$ or $4 n + 3$ or divisible by $4$ or divisible by $2$ or whatever) is more common than any other form. Just because (and this is still only a hypothesis) it appears as though all the numbers are eventually "covered" in the set of ALL Collatz Sequences does not mean that every number appears equally often in the sum collection of all those Collatz sequences..

What does this mean?


 * It means that (from what I can make out -- your argument is bletheringly incoherent) the premises that you start with do not hold.--prime mover (talk) 06:40, 1 February 2021 (UTC)


 * Mind you, your whole argument is made much more difficult to follow because of your inability of refusal to mark your code up so as to make it easily readable. Please make the effort to do this. The world has moved on a little since the days of typewriters. --prime mover (talk) 08:43, 30 January 2021 (UTC):

Mind you, your rules are made much more difficult to follow because your "Help" and "FAQ" sections do not answer some of my questions.


 * What questions are those?


 * Suppose I wanted to use the Latex symbol for "set"? There is no way that I can

search the answer to this question. I have searched numerous subjects which have no responses.

I missed a critical tutorial when I registered and have been unable to recall it. Would someone please help get restarted? Senojesse (talk) 21:13, 8 February 2021 (UTC)

Everybody else who contributes here has no problem learning how to use this site. Some users refuse to use our house style, for whatever reasons, but nobody is so incapable of learning as not to be able to work out how to use a MediaWiki editor to present their contributions.


 * I mean, you don't even seem to have grasped the basic fact of how to indent lines in a discussion page so as to indicate the flow of discourse. --prime mover (talk) 06:40, 1 February 2021 (UTC)


 * Here is a link to an article which may help to explain why your probabilistic approach is invalid: https://www.quantamagazine.org/why-mathematicians-still-cant-solve-the-collatz-conjecture-20200922/

This not a "probabilistic" approach. My approach is the proof by contradiction.

BUT You do not seem to know how the calculation of a Collatz sequence works.

let me educate you.


 * Yep, feel free to try. I'm waiting for you to start.--prime mover (talk) 06:40, 1 February 2021 (UTC)

If 3*N+1 is 0MOD2 the odd term is increased by only 50%.


 * What is that supposed to mean?--prime mover (talk) 06:40, 1 February 2021 (UTC)

This the only time the CS increases. My argument is that it decreases as often as it increases. Usually by only 25%, frequently by 65; not uncommonly by 80%, and in longer sequences by 90%.


 * "It decreases as often as it increases. Usually..." THIS IS A PROBABILISTIC APPROACH.--prime mover (talk) 06:40, 1 February 2021 (UTC)

Definitions: Usually 25% of the terms :: I never used the phrase "of the terms". I meant the value of the current odd term as a percentage of the penultimate odd term.

Frequently 12.5% of the terms Not uncommonly 6.25% of the terms In longer sequences 3.125% of the terms
 * It is obvious that you challenge these numbers. 3*n +1 would need to be

0MOD2 in > 60% terms to prevent collapse to 1.


 * Of course I challenge these numbers. You haven't proved that they are correct. You have asserted, but you have not proved.--prime mover (talk) 06:40, 1 February 2021 (UTC)


 * You refuse to understand or accept my premises. Simply (3*n+1)2 = 1.5*n

3*n+1)/4 = 0.75*n, (3*n+1)8 = 0.375*n, (3*n+1)/32 = 0.1875*n

Another essential fact that Proofwiki cannot accept is the categorization of an odd number as expressible as 4k+1 or 4k+3.

let n = 4*k+1. Then 3*n+1 = 12*k+4 which is divisible by 4 let n = 4*k+3 Then 3*n+3 = 12*k+10 which is never divisible by 4

an infinitely long Collatz Sequence existed; then the 4*k+3 numbers were significantly more numerous than the 4*k+1.


 * Which your proof fails to do. --prime mover ([[User tN


 * My proof is based upon the FACT that number of 4*k+3 terms would need

to be almost as twice as many as the number of 4*k+1 terms; to sustain minimal growth.
 * Minimal growth is loosely defined as the last odd term is greater than

the initial seed


 * I'd like to see where your proof is that the Collatz Sequence doesn't go into a loop, say, $a_n \to a_{n+1} \to a_{n+2} \to \dotsb \to a_{n+m} \to a_n$ etc. There's more to just proving it doesn't go up for ever.

(talk) 06:40, 1 February 2021 (UTC)

Senojesse (talk) 02:20, 9 February 2021 (UTC)


 * Please, do not use an observation for a proof. "Usually", "frequently" and "uncommonly" are not big enough. You do realize, that since we are talking about all natural numbers and subsets theirof, there are as many numbers divisible by 2, 4, 8, 16 and so on? There are actually as many numbers divisible by 2 as there are divisible by $2^{googol}$. This is a weird property of an infinitely countable set: as long as there is a 1-to-1 correspondence, the sizes are the same. Your argument at strongest can be a use to prove this conjecture up to some $N < \infty$, but definitely not the full theorem. Basically, suppose you stop your analysis at some absurdly large number. I will ask you then what about numbers larger by one more order of magnitude? As long as I can ask such a question later, there is no proof. Recall that we have calculated nontrivial zeros of Riemann zeta function to very high powers yet nobody is claiming that this prooves anything.--Julius (talk) 06:22, 1 February 2021 (UTC)