What is the Collatz Problem and how can it be solved?

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The discussion centers on the Collatz problem and the implications of fixing the variable k within its mathematical framework. Participants debate whether k should be considered fixed or variable, with arguments suggesting that treating k as fixed leads to contradictions in the proof structure. The concept of decidability is also scrutinized, with claims that the Collatz problem is undecidable due to its reliance on the axioms of infinity and the inherent symmetry of the Binary Tree. The conversation highlights the complexity of proving the Collatz conjecture and the necessity of clarifying terms like "out of range" and "fixed" in mathematical discourse. Ultimately, the participants emphasize the need for rigorous definitions and logical consistency in mathematical proofs related to the Collatz problem.
  • #391
Matt,
1) And we agreed the th first column is (1010101010... ) the second (110011001100..) and so on - the nth is 2^n ones, 2^n 0s, looping again and again?

2) remember how we showed you that that implies that every row has only a finite number of zeroes in it? remember how that implies the string ...01010101) with an infinite number of 0s in it is not on the list?

1) Yes I agree that this is the redundant way to show how the columns are constructed, but:

2) You showed nothing because the three representations below are one and only one thing:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> 1
 ...,1,1,1,0 <--> 2
 ...,1,1,0,1 <--> 3 
 ...,1,1,0,0 <--> 4 
 ...,1,0,1,1 <--> 5 
 ...,1,0,1,0 <--> 6 
 ...,1,0,0,1 <--> 7 
 ...,1,0,0,0 <--> 8 
 ...,0,1,1,1 <--> 9 
 ...,0,1,1,0 <--> 10
 ...,0,1,0,1 <--> 11
 ...,0,1,0,0 <--> 12
 ...,0,0,1,1 <--> 13
 ...,0,0,1,0 <--> 14
 ...,0,0,0,1 <--> 15
 ...,0,0,0,0 <--> 16
 ...
Now let us make a little redundancy diet:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
...  [b]1[/b]-1-1-1 <--> 1
     \  \ \0 <--> 2
      \  0-1 <--> 3 
       \  \0 <--> 4 
       [b]0[/b]-[b]1[/b]-1 <--> 5 
        \ \[b]0[/b] <--> 6 
         0-1 <--> 7 
          \0 <--> 8 
 ... [b]0[/b]-[b]1[/b]-1-1 <--> 9 
     \  \ \0 <--> 10
      \  [b]0[/b]-[b]1[/b] <--> 11
       \  \0 <--> 12
       0-1-1 <--> 13
        \ \0 <--> 14
         0-1 <--> 15
          \0 <--> 16
 ...
and we get:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
          /1 <--> 1
         1 
        / \0 <--> 2
       1   
       /\ /1 <--> 3 
      /  0
     /    \0 <--> 4 
 ... [b]1[/b]    
     \    /1 <--> 5 
      \  [b]1[/b] 
       \/ \[b]0[/b] <--> 6
       [b]0[/b]  
        \ /1 <--> 7
         0
          \0 <--> 8
          
          /1 <--> 9 
         1
        / \0 <--> 10
       [b]1[/b]  
       /\ /[b]1[/b] <--> 11
      /  [b]0[/b] 
     /    \0 <--> 12
 ... [b]0[/b]    
     \    /1 <--> 13
      \  1
       \/ \0 <--> 14
       0  
        \ /1 <--> 15
         0
          \0 <--> 16
 ...
 
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  • #392
So we have absolutely nailed the construction of the array.

Now, are you claiming the the cardinality of the rows is 2^aleph-0 because for every element in the power set of N, there is a row which corresponds to the indicator function of that element in the power set? Yes, or No is again all that is required. Note this is equivlaent to saying that the set of rows has card 2^aleph-0 because the rows form a 'complete' list of every string of 0s and 1s, ie *ANY* possible string of 0s and 1s occurs as one of the rows.
 
  • #393
Matt,

If we use your terminology, then the answer is yes.
 
  • #394
But we agreed that we are using the cantor view point on cardinality, didn't we?

So, let z be some arbitrary element in P(N), it must if your conjecture is true be one of the rows. Let the number of the row (they're enumerated by N) be R.

Are we ok so far? Yes or no?
 
  • #395
Matt,
(they're enumerated by N)
No, they're enumerated.

This is what you don't understand, the list of unique notations that related to R is longer then the list of unique notations that related to N.

In your termenology the complete R list is longer than the complete N list.
 
Last edited:
  • #396
So the rows are not of cardinality aleph-0 as you've been claiming all through this, then?
 
  • #397
Matt,

You simply don't read what I write.

So here it is again:

1) Both R list and N list are enumerable.

2) |R|>|N|
 
  • #398
So, when I asked you if the rows could be enumerated by N, and you said no, were you lying or mistaken? because that's what enumerable means in this context. so we ask again, are the set of rows in bijective correspondence with N as you claim they are? So picking an element z in P(N) it occurs at row r for some r in N (this r was the R last time, i didn't mean R as in real number, sorry if that's the confusion), that is what you mean by the rows are enumerable, which they are by the construction we've agreed on. And you are claiming that each element of P(N) corresponds to some row.
 
  • #399
Incidentally, the word enumerable is the same as countable. So, remembering that we are usgin real mathematics here, your last post states there is a bijection between R and N, and that there isn't a bijection between R and N.
 
  • #400
Matt,
are the set of rows in bijective correspondence with N as you claim they are?
You know what? let us play your game.

1) I constructed a list with unique sequences of 0 1 combinations each, which its length has 2^aleph0 magnitude.

2) I also showed a bijection from N to P(N).


Conclusion:

The transfinite univereses do not exist.
 
Last edited:
  • #401
But we are playing my game, we agreed to use the proper definitions of all the words.

So there is a bijectin from N to P(N) because there is a row in your diagram for each element of the power set - the element being given by the indicator function described by each row. that is the bijection isn't it?
 
  • #402
Matt,

We have found that |P(N)|>=|N|

Therefore the tansfinite cardinality does not exist.
 
  • #403
Please answer my question as to how the bijection you are claiming between N and P(N) arises.

here it is


So [in your opinion] there is a bijection from N to P(N) because there is a row in your diagram for each element of the power set - the element being given by the indicator function described by each row. that is the bijection isn't it?

again all that is required is a simple yes or no answer, no more.
 
  • #404
Matt,

There is no YES/NO answer here.

Conclusions:


1) |R| > |N| (R list is longer then N list).

2) There is no necessary connection between N and being enumerable.

3) The words “complete” or “all” are meaningless when they are related to a collection of infinitely many elements, because they are leading to contradiction (|R| >= |N|).

Instead, the word "any is used.

Maybe this will help:

Code: 
<--------------------Arithmetic magnitude

 {...,3,2,1,0} = Z*
     2 2 2 2
     ^ ^ ^ ^         Geometric magnitude
     | | | |                  | 
     v v v v                  |    
{...,1,1,1,[b]1[/b]} <--> 1          | 
 ...,1,1,[b]1[/b],                   |
 ...,1,[b]1[/b] <---- Cantor’s       |
 ...,[b]1[/b],1, ,    Diagonal       |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...                          |
 ...,0,0,0,0 <--> 2           | 
 ...,0,0,0, ----.             |
 ...,0,0, ,     |             |
 ...,0,0, ,     | Not Covered |
 ...,0, , ,     |--  By N     |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ... -----------‘             |
 ...,1,1,1,1 <--> 3           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
...                           |
 ...,0,0,0,0 <--> 4           | 
 ...,0,0,0,                   |
 ...,0,0, ,                   |
 ...,0,0, ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          |
 ...,1,1,1,1 <--> 5           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
                              |
 ...                          V
 
Last edited:
  • #405
I would like to remind you that you are operating in the world of conventional mathematics and therefore all of the reasons you just gave are irrelevant, not to say misleading and erroneous.

You have stated there is a bijection between the set of rows and P(N), yet you cannot give this bijection or offer an existence proof. As your claims are supposedly constructive this is obviously a big error on your part. I can predict you will accuse me of just playing with words and misunderstanding the 'true' picture of what maths is, but I think we've seen enough of your position to see it is indefensible; you certainly haven't been able to defend it.

In particular enumerable and countable are defined by bijections with the set of Natural numbers, irrespective of what your opinion is on the matter.

Just because one set appears 'bigger than another' is not important. You think that |Q| equals |N| in whatever meaning you give to those symbols, yet clearly one is 'smaller' than the other, in the sense that a list of Q elements contains a list of N elements as a sublist, to put it in your terms.

After all this do you think that you perhaps ought to learn the meaning of any of the terms you use, because you evidently haven't even after somewhere in the region of 600 posts on it to my knowledge.
 
  • #406
Matt,

There exist a countable list of all(=your termenology) R members represented by infinitely long combinations of 01 sequences, in front of your eyes.

Prove that this list does not exist, but before that take a look again on:
Code: 
<--------------------Arithmetic magnitude

 {...,3,2,1,0} = Z*
     2 2 2 2
     ^ ^ ^ ^         Geometric magnitude
     | | | |                  | 
     v v v v                  |    
{...,1,1,1,[b]1[/b]} <--> 1          | 
 ...,1,1,[b]1[/b],                   |
 ...,1,[b]1[/b] <---- Cantor’s       |
 ...,[b]1[/b],1, ,    Diagonal       |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...                          |
 ...,0,0,0,0 <--> 2           | 
 ...,0,0,0, ----.             |
 ...,0,0, ,     |             |
 ...,0,0, ,     | Not Covered |
 ...,0, , ,     |--  By [b]N[/b]     |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ... -----------‘             |
 ...,1,1,1,1 <--> 3           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
...                           |
 ...,0,0,0,0 <--> 4           | 
 ...,0,0,0,                   |
 ...,0,0, ,                   |
 ...,0,0, ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          |
 ...,1,1,1,1 <--> 5           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
                              |
 ...                          V
As you can see, Cantor's diagonal does not cover this list and also
N members do not cover it, therefore there is no bijection
from |N| to |P(N)| (|P(N)|>|N|).

Therefore the "bijection" that you see between R and N is only an illusion, because each
n <--> ...000/...111 capture only one slice form each 0 or 1 bulk of geometric magnitude, as you can clearly see below:
Code: 
 <---Arithmetic magnitude 

 {...,3,2,1,0} = Z*
     2 2 2 2  
     ^ ^ ^ ^   
     | | | |   
     v v v v  
{...,[b]1-1-1-1[/b]} <--> 1  Geometric magnitude(based on the 
 ...,1,1,1,[b]0[/b]  <--> 2          |          thin notations)          
 ...,1,1,[b]0[/b]/                   |
 ...,1,1/0,                   |
 ...,1,[b]0[/b], ,                   |
 ...,1/0, ,                   |
 ...,1|0, ,                   |
 ...,1|0, ,                   |
 ...,[b]0[/b]/ , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          V

As for Q and N, both have an arithmetic magnitude.
 
Last edited:
  • #407
So now you admit there is no bijection from N to P(N)? So why did you claim that aleph-0 equals 2^aleph-0? Have you completely recanted your assertion, or are you going to claim that you've never insisted that they are the same? becaue you've repeatedly insisted that the reals are countable that implies there is a bijection from N to P(N).

Or are you going to simultaneously claim that there is a set of cardinality 2^aleph-0 that is countable, yet not in bijection with N, contradicting the definition of countable?
 
  • #408
Matt,


I said that there is a bijection from |N| to |P(N)| when I play your game, and by your game the conclusion is that the transfinite cardinality cannot exist because we get |N|=<|P(N)|.

But if we want to understand what is going on, we have no choice but to say bye bye to your game and move to another game, where R is enumerable but |R|>|N|.

This result, which is impossible trough your game, can be simply and clearly explained and demonstrated by my game.


Now we are in my game, so this time please read all that is below and try to understand it:

There exist a countable list of all(=your terminology) R members represented by infinitely long combinations of 01 sequences, in front of your eyes.

Prove that this list does not exist, but before that take a look again on:
Code: 
<--------------------Arithmetic magnitude

 {...,3,2,1,0} = Z*
     2 2 2 2
     ^ ^ ^ ^         Geometric magnitude
     | | | |                  | 
     v v v v                  |    
{...,1,1,1,[b]1[/b]} <--> 1          | 
 ...,1,1,[b]1[/b],                   |
 ...,1,[b]1[/b] <---- Cantor’s       |
 ...,[b]1[/b],1, ,    Diagonal       |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...                          |
 ...,0,0,0,0 <--> 2           | 
 ...,0,0,0, ----.             |
 ...,0,0, ,     |             |
 ...,0,0, ,     | Not Covered |
 ...,0, , ,     |--  By [b]N[/b]     |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ...,0, , ,     |             |
 ... -----------‘             |
 ...,1,1,1,1 <--> 3           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
...                           |
 ...,0,0,0,0 <--> 4           | 
 ...,0,0,0,                   |
 ...,0,0, ,                   |
 ...,0,0, ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          |
 ...,1,1,1,1 <--> 5           | 
 ...,1,1,1,                   |
 ...,1,1, ,                   |
 ...,1,1, ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
 ...,1, , ,                   |
                              |
 ...                          V
As you can see, Cantor's diagonal does not cover this list and also
N members do not cover it, therefore there is no bijection
from |N| to |P(N)| (|P(N)|>|N|).

Therefore the "bijection" that you see between R and N is only an illusion, because each
n <--> ...000/...111 capture only one slice form each 0 or 1 bulk of geometric magnitude, as you can clearly see below:
Code: 
 <---Arithmetic magnitude 

 {...,3,2,1,0} = Z*
     2 2 2 2  
     ^ ^ ^ ^   
     | | | |   
     v v v v  
{...,[b]1-1-1-1[/b]} <--> 1  Geometric magnitude(based on the 
 ...,1,1,1,[b]0[/b]  <--> 2          |          thin notations)          
 ...,1,1,[b]0[/b]/                   |
 ...,1,1/0,                   |
 ...,1,[b]0[/b], ,                   |
 ...,1/0, ,                   |
 ...,1|0, ,                   |
 ...,1|0, ,                   |
 ...,[b]0[/b]/ , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...,0, , ,                   |
 ...                          V

As for Q and N, both have an arithmetic magnitude.



Conclusions:


1) |R| > |N| (R list is longer then N list).

2) There is no necessary connection between N and being enumerable.

3) The words “complete” or “all” are meaningless when they are related to a collection of infinitely many elements, because they are leading to contradiction (|R| >= |N|).

Instead, the word "any" is used.
 
Last edited:
  • #409
I am perfectly open to new ideas, always have been, however, you have repeatedly said that proper mathematics is wrong and its error means that Cantor is wrong, and that there is only one kind of infinity. It doesn't matter what your definitions and issues are here, only that you must then prove something is wrong in Cantor's argument. Your method is to then redefine all of the terms to mean something else. Thus you are being a complete moron in claiming he is wrong, because you are deliberately misinterpreting what he said.


I don't particularly care for your opinion on your new mathematics, I don't see anything interesting in it, nor new. My only position has been that you do not understand tht which you defame and that we cane easliy prove Cantor is correct because you evidently don't have a clue what's going on. Shall we agree on that? I don't care about your opinion, and you know squat about mathematics?

There is no bijection in my mathematical world from N to P(N). So why do you keep saying that? It's always been your contention that there must be, but you've never been able to prove it, and now you're claiming you never wanted to...? Pillock.
 
  • #410
Matt,

Ok, prove by your system that my matrix does not have the complete 01 combinations.

...0101 or ...1010 is in the list, for example:

Let us take again our set:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
[b]{[/b]...,1,1,1,1[b]}[/b]<--> 1
 ...,1,1,1,0 <--> 2
 ...,1,1,0,1 <--> 3 
 ...,1,1,0,0 <--> 4 
 ...,1,0,1,1 <--> 5 
 ...,1,0,1,0 <--> 6 
 ...,1,0,0,1 <--> 7 
 ...,1,0,0,0 <--> 8 
 ...,0,1,1,1 <--> 9 
 ...,0,1,1,0 <--> 10
 ...,0,1,0,1 <--> 11
 ...,0,1,0,0 <--> 12
 ...,0,0,1,1 <--> 13
 ...,0,0,1,0 <--> 14
 ...,0,0,0,1 <--> 15
 ...,0,0,0,0 <--> 16
 ...
Now let us make a little redundancy diet:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
...  [b]1[/b]-1-1-1 <--> 1
     \  \ \0 <--> 2
      \  0-1 <--> 3 
       \  \0 <--> 4 
       [b]0[/b]-[b]1[/b]-1 <--> 5 
        \ \[b]0[/b] <--> 6 
         0-1 <--> 7 
          \0 <--> 8 
 ... [b]0[/b]-[b]1[/b]-1-1 <--> 9 
     \  \ \0 <--> 10
      \  [b]0[/b]-[b]1[/b] <--> 11
       \  \0 <--> 12
       0-1-1 <--> 13
        \ \0 <--> 14
         0-1 <--> 15
          \0 <--> 16
 ...
and we get:
Code: 
 {...,3,2,1,0}=Z*
     2 2 2 2
     ^ ^ ^ ^
     | | | |
     v v v v
          /1 <--> 1
         1 
        / \0 <--> 2
       1   
       /\ /1 <--> 3 
      /  0
     /    \0 <--> 4 
 ... [b]1[/b]    
     \    /1 <--> 5 
      \  [b]1[/b] 
       \/ \[b]0[/b] <--> 6
       [b]0[/b]  
        \ /1 <--> 7
         0
          \0 <--> 8
          
          /1 <--> 9 
         1
        / \0 <--> 10
       [b]1[/b]  
       /\ /[b]1[/b] <--> 11
      /  [b]0[/b] 
     /    \0 <--> 12
 ... [b]0[/b]    
     \    /1 <--> 13
      \  1
       \/ \0 <--> 14
       0  
        \ /1 <--> 15
         0
          \0 <--> 16
 ...
you have repeatedly said that proper mathematics is wrong and its error means that Cantor is wrong, and that there is only one kind of infinity.
My system is reacher then Cantor's transfinite universes bacause:

1) By my system aleph0+1 > aleph0 , 2^aleph0 < 3^aleph0

2) By Cantor's system aleph0+1 = aleph0 , 2^aleph0 = 3^aleph0


By the way, when we move from the 01 matrix representation to the Binary Tree representation, the meaning of the word magnitude become clearer, because several sequential 1 or 0 notations of each column in the matrix, are compressed to a single notation, which its magnitude equivalent to the quantity of the notations that it represents.
 
Last edited:
  • #411
My new motto will be don't feed the trolls.

My new motto will be don't feed the trolls.


sory i did not answer you matt
i am to busy

but way you call Organic a Troll ?

Don't you see he is just a monkey ...

And aren't we all the monkey of Euclid ?


Moshek
 

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