Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question on a<b

  1. Sep 25, 2003 #1
    Let us say that a and b are real numbers, and we have
    [a,b] = {x: a <= x <= b}.

    Can x be a 1 to 1 correspondence (map) between any two real numbers included in [a,b] ?


    Organic
     
    Last edited: Sep 25, 2003
  2. jcsd
  3. Sep 25, 2003 #2

    ahrkron

    User Avatar
    Staff Emeritus
    Gold Member

    You probably need to elaborate a little.

    In the expression you show, x is a dummy variable. It stands for anything.

    If you want it to be a 1-1 correspondence, then you also have to define a way to compare such a thing with a real number, since you are supposed to look for those x objects that satisfy a <= x <= b (and you already stated that a and b are real numbers).

    Also, if you do define a way to decide if a real number is "<=" than a 1-1 correspondence, you should change the notation you use for the set of 1-1corresps, since [a,b] already stands for a set of real numbers.
     
  4. Sep 25, 2003 #3
    Hallo ahrkron,

    Since x can be a 1 to 1 correspondence between any two real numbers, is the set of all x maps is complete (have the power of the continuum)?


    Organic
     
  5. Sep 25, 2003 #4

    ahrkron

    User Avatar
    Staff Emeritus
    Gold Member

    Hi Organic,

    You misunderstood my reply.

    Your first post is flawed. It does not make sense as stated. You need to define properly the objects and relations you are using. Otherwise, no further comclusions can be drawn from it.
     
  6. Sep 25, 2003 #5
    Hi ahrkron,

    May be you can help me to write it in a formal way.


    The question is:


    Theorem: 2^aleph0 < c


    Proof:

    Let A be the set of all negative real numbers included in (-1,0).

    Let B be the set of all positive real numbers included in (0,1).

    Let M be the set of maps (1 to 1 and onto) between any two single real numbers of
    A and B sets.

    Therefore |M| = 2^aleph0.

    (0,1) = {x: 0 < x < 1 }, where x is a 1-1 correspondence between any two real numbers included in (0,1), and any x element has no more than 1 real number as a common element with some other x element.

    Let T be the set of all x (1-1 correspondence) elements included in (0,1).

    Therefore |T| = |M| = 2^aleph0.

    B is a totally ordered set, therefore we can find x element between any two different real numbers included in (0,1).

    Any x element must be > 0 and cannot include in it any real number.

    Therefore 2^aleph0 < c (does not have the power of the continuum).

    Q.E.D


    A structural model of the above:
    Code (Text):

                set      set       set  
                 A        M         B
                 |        |         |
                 |        |         |
                 v        v         v
                 !__________________!<---- set
                 !__________________!<----  T members
                 !__________________!
                 !__________________!   Any point is some real number
                 !__________________!   (A or B members).
                 !__________________!  
                 !__________________!   Any line is some 1-1 correspondence    
                 !__________________!   (M or T members).
                 !__________________!
                 !__________________!
     




    Am I right ?



    Organic
     
    Last edited: Oct 2, 2003
  7. Sep 26, 2003 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Doron, is there a reason for not posting this under the name you used before? The mathematician formerly known as "Doron Shadmi"?

    The point of the first response by Organic, which you still haven't addressed is that it makes no sense to assert that x is a NUMBER, as you do in say 0<= x<= 1, and then state that x is a 1-1 correspondence: "(0,1) = {x: 0 < x < 1 }, where x is a 1-1 correspondence between any two real numbers included in (0,1)."

    x is either a number or a 1-1 correspondence between numbers. It can't be both.
     
  8. Sep 26, 2003 #7
    Hi,

    x is not a NUMBER but an element that exists between any two different real numbers included in (0,1).



    Organic
     
  9. Sep 26, 2003 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    So, once again, you are not talking about "normal" mathematics and are not defining your terms. In "normal" mathematics, inequality with numbers is only defined FOR numbers. WHAT KIND of "element" exists between any two numbers but is not a number itself? What in the world do you mean by "element"?
     
  10. Sep 26, 2003 #9
    Hi,

    Any x is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers.

    No single "normal" real number has this property (to be a connector between some two other different "normal" real numbers).

    Please do not forget that negative, irrational, complex and transfinite numbers where not "normal" elements, when they first discovered.

    I think the important thing is to find if those "non-normal" elements
    can be useful for the development of Mathematics Language, by opening new areas for research.

    I think if we associate these "non-normal" elements with, so called, "normal" numbers, we shall obtain a new point of view on the abstract concept of a NUMBER.




    Organic
     
    Last edited: Sep 26, 2003
  11. Sep 26, 2003 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, they surely won't be useful if you don't DEFINE them.
     
  12. Sep 26, 2003 #11
    Please open the attached pdf file.
     

    Attached Files:

    • cat.pdf
      cat.pdf
      File size:
      89 KB
      Views:
      69
  13. Sep 26, 2003 #12

    ahrkron

    User Avatar
    Staff Emeritus
    Gold Member

    This is far, far out.
    To TD it goes.
     
  14. Oct 2, 2003 #13
    "Any x is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers."

    what is your PRECISE definition of "<"? so, when you say 0<x, where x is a connector, what precisely does that mean? certainly, this isn't the same < when we say 0<1. it would seem that your < must be an EXTENSION of the binary relation < as defined for real numbers. if it's not an extension, then i would think the whole thing falls apart, for then the two <'s have utterly different meanings even when confined to the concept of real numbers.

    let's pretend i don't have adobe. what's a connector? if it's a bijection, what is the domain and range? and how is this bijection inserted into the continum in such a way that it makes sense to say it is between two real numbers?

    this is not automatically bizarre because in nonstandard analysis, there are infinitely many hyperreals between two real numbers. each real number has a halo of hyperreals that are infinitesimally close to it.

    i'd like to point out that he never said {x &epsilon; R: 0<x<1}, he just said {x: 0<x<1}; that could be a set of hyperreals or what not... clearly, more precision would be nice.

    cheers,
    phoenix
     
    Last edited: Oct 2, 2003
  15. Oct 2, 2003 #14
    Hi phoenixthoth,

    First, thank you for your reply.

    p and q are real numbers.

    If p < q then
    [p, q] = {x : p <= x <= q} or
    (p, q] = {x : p < x <= q} or
    [p, q) = {x : p <= x < q} or
    (p, q) = {x : p < x < q} .

    A single-simultaneous-connection is any single real number included in p, q
    ( = D = Discreteness = a localized element = {.} ).

    Double-simultaneous-connection is a connection between any two different real numbers included in p, q , where any connection has exactly 1 D as a common element with some other connection ( = C = Continuum = a non-localized element = {.___.} ).

    Therefore, x is . XOR .___.

    Any C is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers (D elements).

    No single "normal" real number (a D element) has this property, to be a connector between some two different "normal" real numbers (D elements).

    Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.


    For more detailes please look at:


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


    Please look also in this thread:

    https://www.physicsforums.com/showthread.php?s=&threadid=6427


    I'll be glad to get your remarks and insights.






    Yours,

    Organic
     
    Last edited: Oct 6, 2003
  16. Oct 2, 2003 #15

    russ_watters

    User Avatar

    Staff: Mentor

    This may help. Its his first post in his other thread.
    Today I went to the bank and tried to define a new definition of "exchange rate" but for some reason they weren't buying. Hmm, I don't understand why not...
     
  17. Oct 3, 2003 #16
    russ_watters, no, it doesn't help. your comments remind me of those who didn't like the idea of irrational, transcendental, hyperreal, or complex numbers. gauss, as far as in know, invented a new kind of number and they were used in his PhD thesis. i say it doesn't help because it neither proves or disproves the existence of connector elements in such a way that they are naturally embeddable within R.

    organic, a few things:

    i'm still in need of a more precise definition of what .___. or "connector" means. i know you say it's a 1-1 correspondance (ie, bijection) element between two real numbers. one way to express functions is with notation like this (as we all know): f(x) = stuff. but another way is to represent them as sets of ordered pairs such that no two first-coordinates match (ie, no two different ordered pairs are present with the same first coordinate). so when you say x is a bijection between real numbers, a and b, let's say, that means that x is representable as the following set of ordered pairs (ie, function): { (a, b) }, the singleton containing the ordered pair (a, b). ok, fair enough. x = { (a, b) }.

    so far, x is not any more of exotic of a number than the complex number (a,b) or a + b i. but exoticness is irrelevant as long as there is consitency. some people, like david hilbert, define mathematical existance as consistency. (and if physical existance also equals consistency, then, bingo, mathematical existance = physical existence. ha ha ha.)

    here's the big question that will improve your theory for as it stands right now, there is a big gap. and here it is.

    suppose y is a real number (to say "normal" real number is redundant) and x is a connector, x = { (a, b) }, where a and b are two real numbers. for x to be a connector, this implies a != b. (note that != is standard notation for unequal.)

    this is the big question you must answer:

    how can you extend the binary relation < so that a statement like

    x < y

    makes sense?

    what do i mean by "make sense?"

    well, at first glance, statements like { (1,2) } < 14 don't make sense.

    here's how you might fix this. when a = b, then whatever your extended definition of < is should reduce to the definition of < for real numbers for when a = b, x is identified with the real number a. "identified" means that there is a bijection between the set of sets of the form { (a, a) } and the set of real numbers; this bijection is given by
    f( { (a, a) } ) = a.
    this is what i mean by saying that your definition of < meant for connectors and reals alike should be an extension of what < means for two real numbers.

    doing this will, i think, embed the set of connectors within the set of real numbers, which is what i think you want to do.

    then, once you have a definition of <, you can then talk about a connector being between two real numbers.

    the first big theorem for you, once you have appropriately precise definitions, is an existence theorem. you need to prove the following, which is NOT clear to me:

    let a < b be real numbers. there exists a connector x such that
    a < x < b. (where < has been precisely defined to apply to connectors and reals.)

    do that, and you've actually got your own pet theory on a new kind of number.

    to explore a bit further, find the cardinality of the set of all connectors between a and b.

    next, find applications!

    the main thing about nonstandard analysis, what makes it useful, is that STANDARD analysis results can be obtained by working exclusively with hyperreals. i'm talking about nonstandard analysis for they also talk about there being things between real numbers. godel, well known for his incompletness thoerem, said that nonstandard analysis will be the analysis of the future. well, he's no analyst! go back to the corner, godel, and study your logic!

    cheers,
    phoenix
     
    Last edited: Oct 3, 2003
  18. Oct 5, 2003 #17
    Hi phoenixthoth,

    First, thank you for your positive attitude.

    Well, I have ideas, but I am not good in formal definitions.

    So, if you can help me in this point, I'll be thankful to you.

    In general I look on Math as a form of comnunication that can be developed through team work between persons, therefore I do not understand the aggression of some persons when I show them my ideas in a wrong "formal" way.

    I think this forum will be a better place through positive attitude to each other's ideas.

    OK..., more to the point.

    I'll write my definitions again, and then I'll try to explain my ideas
    to you the best I can (also sorry about my poor English).


    p and q are real numbers.

    If p < q then
    [p, q] = {x : p <= x <= q} or
    (p, q] = {x : p < x <= q} or
    [p, q) = {x : p <= x < q} or
    (p, q) = {x : p < x < q} .

    A single-simultaneous-connection is any single real number included in p, q
    ( = D = Discreteness = a localized element = {.} ).

    Double-simultaneous-connection is a connection between any two different real numbers included in p, q , where any connection has exactly 1 D as a common element with some other connection ( = C = Continuum = a non-localized element = {.___.} ).

    Therefore, x is . XOR .___.

    Any C is not a "normal" real number but a connector (a 1-1 correspondence element) between any two different "normal" real numbers (D elements).

    No single "normal" real number (a D element) has this property, to be a connector between some two different "normal" real numbers (D elements).

    Between any two different arbitrary close Ds there is at least one C, and only C has the power of the continuum.

    Now I'll try to explain the above:

    By p<q I mean that there exists some interval > 0 between any two different real numbers, which means that p or q never reach each other by definition (both of them are singletons).

    You wrote:
    I'll try to answer (hope I understand you):

    a=p , b=q


    Let us write the connctor between a and b as aCb.

    So, x = aCb.

    By x < y we mean that y is bigger than aCb range (I know that the word "range" or "domain" already have their meaning in Math languauge, so maybe instead of "range" I use "path-place").

    If I draw it, it will look like this:
    Code (Text):

     a      b    y(a [b][i]D[/i][/b] element)
     .______.    .

    If y = b[b][i]C[/i][/b]c ,where c > b then
     a      b    c
     .______.____.
     
    If a = b then x is a D element (a "normal" real number).

    As you see "<" "=" ">" works for both D and C elements.

    D = {.} = a localized element.

    C = {.__.} = a non-localized element (Double-simultaneous D places).

    By (p, q) = {x : p < x < q} I mean that some C exists between a and b (aCb) where a > p and b < q, therefore x = aCb.

    Hope I am understood.

    Please tell me what do you think.

    Thank you.


    Organic
     
    Last edited: Oct 6, 2003
  19. Oct 5, 2003 #18
    Hi Organic.
    from your file:
    Huh... really? However, you are not talking about "normal" mathematics. If I get you right, you are talking about symmetry, well, but if a set really has a symmetry - it has topology,
    if a set has topology - it has signature, and if it has signature - it must have more than one element (may be uncertain nature)!
    Now, just tell us - how do you construct the symmetry on empty set without signature in such a way that it makes sense to say it is a real symmetry? These (and many others) contradictions doesn't allow considering your math's invention...
     
  20. Oct 5, 2003 #19
    Hi Anton A. Ermolenko,

    Why you are so aggresive, please take Math as some form of communication where persons share each other's ideas.

    I'll be glad to get any remarks from you.


    More to the point, I connect between simplicity and symmetry concepts.

    The more you are simple the more you are symmetric.

    The {} content is the most simple, therefore the most symmetric.

    Please write to me on any part of my work, where you find contradictions.

    You can find my work here:

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

    Thank you.



    Organic
     
    Last edited: Oct 5, 2003
  21. Oct 5, 2003 #20
    if there are any terms in the following post not clear, consult http://www.mathworld.com .

    if x = { (a, b) }, which is the bijection with domain {a} and range {b}, is a connector between the real numbers a and b, and y is some other real number, then are you saying that < means the following:

    x < y if b < y?

    thanks.

    cheers,
    phoenix
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A question on a<b
  1. B-field phase question (Replies: 1)

  2. A relative to B (Replies: 1)

  3. B field question? (Replies: 5)

  4. B field question. (Replies: 5)

  5. Force and B field (Replies: 5)

Loading...