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Boolean Logic cannot deal with infinitely many objects

  1. Dec 10, 2003 #1
    I realized that my last thread (named by "permutations") can’t be understood by professional mathematicians.

    Because I don’t know how to write my idea in the common formal way, I am going to do it in a non-formal way, but I will do my best to write it in the clearest way.

    So here it is:

    Let us check these lists.

    P(2) = {{},{0},{1},{0,1}} = 2^2 = 4

    and also can be represented as:

    00
    01
    10
    11


    P(3) = {{},{0},{1},{2},{0,1},{0,2},{1,2},{0,1,2}} = 2^3 = 8

    and also can be represented as:

    000
    001
    010
    011
    100
    101
    110
    111

    Let us call any full 01 list, combinations list.

    Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

    000
    001
    010
    011
    100
    101
    110
    111

    We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

    001
    011
    010
    000
    101
    100
    111
    110

    The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

    The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

    In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

    The number of the combinations, which are out of the range of Cantor's diagonal is:

    2^n - n

    Every column, which belongs to some combinations list is a sequence of 01 notations, based on some periodic frequency changes, for example:

    the right column of number 3 combinations list, is based on 2^0(=1).

    Therefore the periodic frequency changes are 1, and the result in this case is:
    01010101.

    The result of the middle column is based on 2^1(=2), therefore the sequence is:
    00110011.

    The result of the left column is based on 2^2(=4), therefore the sequence is:
    00001111.

    and we get the full combinations list of number 3:

    000
    001
    010
    011
    100
    101
    110
    111

    We can get a combinations list of infinitely many places, by using the ZF Axiom of infinity induction, on the left side of our combinations list, by using the induction on the power_value of each column, for example:

    2^0, 2^1, 2^2, 2^3, ...

    In this stage we have proven, by induction, that Cantor's diagonal cannot cover any full 01 combinations list, finite or infinite.

    Therefore its result is not a new combination (that has to be added to the list).

    Because Cantor's diagonal cannot cover the full 01 combinations list (of aleph0 places for each combination) we can conclude that 2^aleph0 > aleph0.

    But, because no diagonal's result is a new combination (and therefore not added to the list) each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

    ...000 <--> 1
    ...001 <--> 2
    ...010 <--> 3
    ...011 <--> 4
    ...100 <--> 5
    ...101 <--> 6
    ...110 <--> 7
    ...111 <--> 8
    ...

    Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

    (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects in infinitely many magnitudes.

    One can say that at least the sequence ...111 is not in the list, for example:

    ...000 <--> 1
    ...001 <--> 2
    ...010 <--> 3
    ...011 <--> 4
    ...100 <--> 5
    ...101 <--> 6
    ...110 <--> 7
    ...111 <--> 8
    ...

    Let us examine the infinite from another point of view.

    When we have ...111 AND ...000 in an ordered combinations list, it means that the list is complete.

    But this is the whole point, infinitely many objects cannot be completed, otherwise they are finitely many objects.

    Therefore ...111 AND ...000 are not in the list of infinity many objects.

    In other words [...000, ...111) XOR (...000, ...111] .

    There are 2 possible structural types of infinitely many 01 notations:

    (?...0]
    (?...1]

    We know how some infinitely long combination starts, but its left side is
    unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

    Therefore we can find a meaningful missing result by Cantor's Diagonalization method, only in a finite combinations list.


    For more details please look at:

    http://www.geocities.com/complementarytheory/RiemannsBall.pdf


    Organic
     
    Last edited: Dec 13, 2003
  2. jcsd
  3. Dec 10, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    As I said before, Cantor's Diagonalization method requires that you have as many digits in each number as you have numbers (the way it is normaly applied, this is a countable list of numbers each having a countable number of digits).

    In addition, the "input" for Cantor's Diagonalization method is the list of numbers not any one number.

    It is not a matter of not knowing how to "write" mathematics. You've never taken the time to understand the mathematics itself.
     
  4. Dec 10, 2003 #3
    Dear HallsofIvy,


    Now i know that you don't really want to understand what you read.

    And to understand what i wrote you have to do 2 basic things:

    1) You have to read it from the first word until the last word.

    2) After you read all of it, you have to check if you understand it.

    3) After you understand it, than and only than, please write a detailed reply.

    Thank you.



    Organic
     
    Last edited: Dec 11, 2003
  5. Dec 10, 2003 #4

    russ_watters

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    Staff: Mentor

    Frankly, Organic, the only thing that would keep me from laughing at the mess you made with that last thread would be my respect for you as a person (until proven otherwise, I assume everyone is worthy of my respect). Posts like this one diminish that. You won't get very far here unless you drop your attitude. And you will get even further if you first learn some real math before trying to invent your own new math.
     
  6. Dec 11, 2003 #5
    Dear russ_watters,


    First you have to show that you understand what i wrote in this thread, and you can do in this way:

    1) You have to read it from the first word until its last word.

    2) After you read all of it, you have to check if you understand it.

    3) After you understand it, than and only than, please write your detailed reply.


    If you can't follow these 3 steps, then you did not show to the persons that read your last raply, that you have any meaningful thing to say about what i wrote.


    Yours,


    Organic
     
    Last edited: Dec 11, 2003
  7. Dec 11, 2003 #6

    NateTG

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    Science Advisor
    Homework Helper

    Perhaps you should try to understand Cantor's diagonal argument before you start claiming that it is invalid.

    Usually the thing you call 'combinations list' in your post is referred to as [tex]2^A[/tex].

    Cantor's argument is that [tex]|2^A| \neq |A|[/tex]. This should painfully clear if [tex]A[/tex] is a finite set.

    A particularly choice abuse is:
    Your example pairings are either not well defined, or for example, if the ...'s are all zeros will never map any sequence that contains an infinite number of 1's.
     
    Last edited: Dec 11, 2003
  8. Dec 11, 2003 #7
    I've read all of it.

    I've understood what you are trying to say / prove. However, I do not agree with it.

    Your reasoning is flawed. Cantor's argument ONLY works for a list of n numbers, each containing exactly n digits. Since your list does not follow this requirement, you can't use Cantor.
     
  9. Dec 11, 2003 #8
    Re: Combinations

    and I also didn't understand what do you mean by this......

     
  10. Dec 11, 2003 #9
    To NateTG,

    Please show in what i wrote how do you come to the conclusion that what you call A, is finite.
     
  11. Dec 11, 2003 #10
    To suyver,

    You are right, Cantor's diagonal is limited to aleph0^2(=aleph0), and this is exactly the reason why he can find infinitely many new numbers, which are not in this aleph0^2(=aleph0) list.

    The complete list of all R numbers is 2^aleph0.

    I proved, by using the ZF axiom of infinity induction, that 2^aleph0 > aleph0.

    Then i proved that 2^aleph0 = aleph0 by this:

    Each infinitely long sequence of 01 notations can be mapped with some natural number, for example:

    ...000 <--> 1
    ...001 <--> 2
    ...010 <--> 3
    ...011 <--> 4
    ...100 <--> 5
    ...101 <--> 6
    ...110 <--> 7
    ...111 <--> 8
    ...

    Therefore we can conclude that 2^aleph0 = aleph0, and we come to contradiction.

    (2^aleph0 >= aleph0) = {}, and we have a proof saying that Boolean Logic cannot deal with infinitely many objects.
     
    Last edited: Dec 11, 2003
  12. Dec 11, 2003 #11
  13. Dec 11, 2003 #12
    To Guybrush Threepwood,

    You wrote:
    Here it is again:

    Now, let us use Cantor's Diagonalization method on some finitely long combinations list, for example, the combinations list of number 3:

    000
    001
    010
    011
    100
    101
    110
    111

    We can change the order of the rows, and then use Cantor's Diagonalization method, for example:

    001
    011
    010
    000
    101
    100
    111
    110

    The input for Cantor's Diagonalization method in the first example is 000 and the output is 111.

    The input for Cantor's Diagonalization method in the second example is 010 and the output is 101.

    In both examples we find that the result is already in the combinations list, and this combination, which is already in the list, is one of the combinations that Cantor's Diagonal does not cover.

    The number of the combinations, which are out of the range of Cantor's diagonal is:

    2^n - n
     
  14. Dec 11, 2003 #13
    Organic, you don't have to repaste the original message.

    If the combination is already on the list, what do you mean is not covered?

    And also the argument HallsOfIvy made about the incorrect use of the diagonalization method in this case stands.
     
  15. Dec 11, 2003 #14
    I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases.
     
  16. Dec 11, 2003 #15
    Please read my answer to To suyver.
     
  17. Dec 11, 2003 #16

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Now, that's just bad elementary mathematics!
     
  18. Dec 11, 2003 #17
    I ask again how?? I provided a link to the ZF set of axioms earlier. I just don't see it....

    no, you forgot at least the all '1' sequence.....
     
  19. Dec 11, 2003 #18
    Dear HallsofIvy,

    Ok, show in detailed reply why it is bad elementary mathematics.
     
  20. Dec 11, 2003 #19
    Let us examine the infinite from another point of view.

    When we have ...11111 AND ...00000 in an ordered combinations list, it means that the list is complete.

    But this is the whole point, infinitely many objects cannot be completed, otherwise they are finite.

    Therefore ...11111 AND ...00000 are not in the list of infinity many objects.

    In other words [...000, ...111) XOR (...000, ...111] .

    There are 2 possible structural types of infinitely many 01 notations:

    (?...0]
    (?...1]

    We know how some infinitely long combination starts, but its opposite side is
    unknown (can be 0 XOR 1) and this missing information is essential to the existence of the induction.

    Therefore we can find meaningful missing result by Cantor's diagonal method, only in a finite combinations list.

    For more details please look at:

    http://www.geocities.com/complementarytheory/RiemannsBall.pdf
     
    Last edited: Dec 11, 2003
  21. Dec 11, 2003 #20

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The statement "I proved it for n, i proved it for n+1, therefore i proved it for infinitely many cases." doesn't make any sense.
    If you mean "n" as a single specific integer, then "I proved it for n, i proved it for n+1" only proves it for those two particular numbers.
    If you meant n to be any natural number, then after "I proved it for n" it isn't necessary to prove it for n+1!
    If you meant this as "proof by induction", then it is "Assuming the statement is true for n, prove it is true for n+1" and is only part of what is necessary for a proof by induction.
     
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