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A formal solution to Hilbert's 1st and 6th problems

  1. Sep 8, 2003 #1
    Dear researcher,




    A formal solution to Hilbert's 1st and 6th problems
    ---------------------------------------------------

    A and B are sets.

    q and p are members.

    Option 1: q and p are members of A, but then q is not equal to p .

    Option 2: q is a member of A , p is a member of B .

    D = Discreteness = q XOR p = a localized element = {.}

    C = Continuum = q to p correspondence = a non-localized element = {.___.}

    In the Common Math 0^0 is not well defined, because each member is D.

    Let us say that power 0 is the simplest level of existence of some set's content.

    Because there are no Ds in C, its base value = 0, but because it exists (unlike the emptiness), its cardinality = 0^0 = 1.

    There are now 3 kinds of cardinality:

    |{}| = 0 = the cardinality of the Empty set.

    |{._.}| = 0^0 = 1 = the cardinality of C.

    |{.}| = 1^0 = 1 = the cardinality of D.


    Any point is a D element. Any line a C element.

    It means that there is a XOR ratio between LINES to POINTS.


    XOR ratio between LINES to POINTS
    ---------------------------------

    0(LINE) 0(POINT) -> 0-(No information) -> no conclusion.

    0(LINE) 1(POINT) -> 1-(Clear Particle-like information) -> conclusions on points.

    1(LINE) 0(POINT) -> 1-(Clear Wave-like information) -> conclusions on lines.

    1(LINE) 1(POINT) -> 0-(No clear information) -> no conclusion.



    Some explanation:
    -------------------
    D = Discreteness = q XOR p = a localized element = {.}

    C = Continuum = q to p correspondence = a non-localized element = {._.}

    By the above definitions, for the first time in modern mathematics, there is clear and sharp distinction between the Continuum and the Discreteness concepts, not by their Quantity Concept, but by their Structural concept.

    By real analysis the Continuum is "infinitely many elements with no gaps between them".

    By defining the correspondence itself as a legitimate member, I redefine the original lexicographical meaning, back to the Continuum concept, and change the perception of Continuum and Discreteness concepts in Modern Mathematics.


    Then, in the detailed menuscript, I clearly show that the Structure concept has more interesting information than the Quntity consept in Mathematics, in general.


    For more detailed information, please see:

    http://www.geocities.com/complementarytheory/CATpage.html

    I know that it is hard to understand, because I have changed the most abvious paradigm, which says that Math is first of all, to deal with Quantities.

    By my new theory of numbers, that follows this opening on Hibert's 1st and 6th problems, I clearly show that Math first of all is, to deal with Structures that are built from associations between oppiste conceptsts.




    Sincerely yours,

    Doron Shadmi
     
    Last edited by a moderator: Sep 11, 2003
  2. jcsd
  3. Sep 8, 2003 #2

    HallsofIvy

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    "q = a , but if q = {} then q = |a| ."

    What does this mean? You said that a was a member of set A. Are you assuming that A is a set of sets? What is |a|?

    You seem to be seriously confused about the distinction between sets and members of sets.
    As usual you are terms in non-standard ways without defining them. I am still wondering what an "XOR ratio" is.
     
  4. Sep 8, 2003 #3
    HallsofIvy

    Thank you for your reply.

    I am not assuming that A OR B are sets of sets. all what I do is not to conclode {} as a member, but its cardinality, which is 0.

    Capital A or capital B are sets.

    The lowercase letters: a,b,q and p are members.

    |a| = the cardinal of a

    XOR is an exclusive OR.
     
    Last edited by a moderator: Sep 8, 2003
  5. Sep 8, 2003 #4

    HallsofIvy

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    If the members of A are not sets then what do you mean by "q = a , but if q = {} then q = |a|".

    I KNOW what XOR means. I even know what a ratio is! I do not know what "XOR ratio" means.
     
  6. Sep 8, 2003 #5
    Dear HallsofIvy,

    Again, thank you for your reply.

    A or B can be any kind of set, which means 3 options:

    1) Set of sets.
    2) Set of members that are not sets + members that are sets.
    3) Set of members that are not sets.

    Code (Text):

    {0,1,2,3,...}={[B]{ }[/B],[B]{[/B]{ }[B]}[/B],[B]{[/B]{ },{{ }}[B]}[/B],[B]{[/B]{ },{{ }},{{ },{{ }}}[B]}[/B],[B]{[/B]...  
                   |0| |-1-| |----2----| |----------3----------| |--4
                    |    ^        ^                 ^               ^    
                    |____|        |                 |               |
                      |           |                 |               |
                      |___________|                 |               |
                            |                       |               |
                            |_______________________|               |
                                        |                           |  
                                        |___________________________|
                   
                   
     
    From the above example you can learn that any number (but not 0) has an internal structure that is built by Empty set recursion.

    Through this point of view, I can check 1 to 1 correspondance between members that are sets.

    More than that: 0.101101... and |{}|.|{{}}| |{}| |{{}}| |{{}}| |{}| |{{}}| ... , are the same.


    Please give the same meaning to "XOR" and to "XOR ratio".

    In my language, Hebrew, we add the word "ratio"(in Hebrew it is "YACHAS") before the logical condition. So, maybe I wrongly translated it to English. Please tell me if I can omit it.

    Another important thing is that I have added some explanation to my 1st message in this thread, and I hope it will help you to understand the meaning of my work.

    Yours,

    Doron
     
    Last edited by a moderator: Sep 9, 2003
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