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Circle's area and the CH problem

  1. Oct 8, 2003 #1
    |Q|<|R|=c

    r=1

    Polygon's area = pi r^2 iff r has some unchanged value in any arbitrary direction, or in another words, the polygon's perimeter has 2^aleph0 points.

    Let S1 = the area of this closed element.



    Let us say that we have a polygon, which is made of aleph0 points.
    So, its area must be less than pi r^2 because we have at least 1 r length in any arbitrary direction which is < 1.

    Let S2 = the area of this closed element.


    S1-S2 = x > 0


    My qeustion is: can we find x value and use it to conclude something on the CH problem ?


    Organic
     
    Last edited: Oct 8, 2003
  2. jcsd
  3. Oct 8, 2003 #2

    HallsofIvy

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    Ok, you are using the term "polygon" where "r has some unchanged value in any arbitrary direction" to mean "circle"?

    In my opinion, it would be better to SAY "circle". At least then we wouldn't have to worry about what you meant by "r".

    No, we can't say that. A line segment, or sequence of line segments, or curve CAN'T have only aleph0 points. Any line segment, curve or other continuous one-dimensional figure has 2^aleph0= c points.

    No, any one-dimensional curve must have c points and that has no relationship whatsoever to its size.

    Any figure with aleph0 points must have measure (lenght, area, volume, whatever) 0.
     
  4. Oct 8, 2003 #3
    Hi HallsofIvy,

    Thank you for your reply.

    I think you missed the point.

    I did not use the word curve or circle, but polygon.

    To make it clearer, I am talking about a polygon in 3 basic states.

    1) A polygon which is made of 2^aleph0 vertices.

    2) A polygon which is made of aleph0 vertices.

    3) A polygon which is made of some n>2 vertices.


    Case(1) has the area of pi r^2(=a).

    Case(3) has area c, which is < a .

    Case(2) has area b, which is < a and > c.

    Therefore we get c < b < a.

    My question is aboust b < a.



    Organic
     
    Last edited: Oct 9, 2003
  5. Oct 8, 2003 #4

    Hurkyl

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    Polygons, by definition have only a finite number of vertices. They can have neither aleph0 nor c vertices.

    And for good reason; one runs into severe technical difficulties if one tries to define a polygon with an infinite number of vertices. In general, we aren't even guaranteed a single pair of adjacent vertices; how can we define what the edges of the polygon are?


    There is a general rule one must take to heart when one deals with the infinite: the infinite does not behave like the finite. You really should study some analysis and/or set theory to understand how to properly work with infinity and infintiessimals.



    postscript: in nonstandard analysis, one can define a "hyperpolygon" to have n vertices where n is a hyperinteger (which may be nonfinite). However, a hyperpolygon with a nonfinite number of vertices must have uncountably many vertices; it cannot have aleph0 vertices. Also, the area of a regular hyperpolygon with nonfinite vertices is not &pi;r2 (though it will be infinitessimally close).
     
  6. Oct 9, 2003 #5
    Hi Hurkyl,

    If I understand you, then I put my argument as an infinitessimal question:

    Let Rf be the magnitude level of roughness of some closed 1 dim. element's perimeter.

    L1 = some circle's priameter is 1 unit.

    Therefore:

    RfA = L1/2^aleph0 = circle's roughness magnitude.


    RfB = L1/aleph0 = a closed 1 dim. element's roughness magnitude.

    n>2
    RfC = L1/n = polygon's roughness.



    Let S2 be the area of a closed 1 dim. element with RfA magnitude .

    Let S1 be the area of a closed 1 dim. element with RfB magnitude,
    where the area of the 1 dim. element is close as possible(approaches) to the L1 circle's area.


    S2-S1 = x > 0


    My qeustion is: can we find x value and use it to conclude something on the CH problem ?





    Organic
     
    Last edited: Oct 9, 2003
  7. Oct 9, 2003 #6

    Hurkyl

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    (a) Infinitessimals don't exist1, so it makes even less sense to reformulate your question to use infinitessimals.

    (b) Your definition of "the magnitude level of roughness" makes no sense whatsoever.

    (c) Not all cardinal numbers are natural numbers. You cannot divide natural numbers by transfinite cardinals (such as aleph0) unless you explicitly define the meaning of such division.

    (d) Among polygons inscribed in a given circle, their areas can get as close as desired to that of the circle, but cannot obtain it. There is no "closest possible" polygon, because no matter how close a polygon's area is to that of the circle, there always exists another polygon whose area is closer.

    (e) Just so you're aware, the continuum hypothesis has been proven independant of the axioms of ZFC. One ramification of this fact is that if you can prove the CH within ZFC, you have proven ZFC to be inconsistent.

    (f) I'm not sure why you think this example would shed any light on whether there exists a set with cardinality between aleph0 and c.


    1: I mean in standard analysis, of which the Euclidean plane is part.
     
  8. Oct 9, 2003 #7
    Hi Hurkyl,
    My definitions of roughness magnitude give them their existance and sence.
    Please show me in a detailed way why "it makes no sence whatsoever".
    In this stage I am talking about 2^aleph0 and aleph0 , so I dont understand what do you mean by "Not all cardinal numbers are natural numbers".

    Again, my definitions of roughness magnitude explicitly define the meaning of such division.

    Please show me why my definitions are meaningless.


    I know about cohen's forcing method (1963) and godel.
    "closest as possible" related to RfB element, which is not a ploygon(please look again at my definitions). by "closest as possible" I mean that the distance between any 2 aleph0 points = L1/aleph0


    I'll write my definitions again:

    Let Rf be the magnitude level of roughness of some closed 1 dim. element's perimeter.

    L1 = some circle's priameter is 1 unit.

    Therefore:

    RfA = L1/2^aleph0 = circle's roughness magnitude.


    RfB = L1/aleph0 = a closed 1 dim. element's roughness magnitude.

    n>2
    RfC = L1/n = polygon's roughness.

    y = the size of the radius of L1 circle

    The magnitude of 2^aleph0 is greater than the the magnitude of aleph0.

    In the case of RfB there are aleph0 radii with r=y, but there are 2^aleph0 radii with r<y.

    aleph0 < 2^aleph0, therfore r<y.



    Let S2 be the area of a closed 1 dim. element with RfA magnitude .

    Let S1 be the area of a closed 1 dim. element with RfB magnitude, where the disnace between any 2 aleph0 points = L1/aleph0.

    In case of RfA we know that S2 = 3.14... r^2 (where r=y).

    But in case of RfB r<y, therefore S1 = 3.14... r^2 (where r<y).



    Therefore S2-S1 = x > 0




    Organic.
     
    Last edited: Oct 9, 2003
  9. Oct 9, 2003 #8

    Hurkyl

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    (a) You haven't defined "roughness magnitude"; you stated a symbol, said it is "roughness magnitude" (plus a few other undefined nouns and adjectives), and went on to state conclusions.

    (b) Definition != existance. e.g. If I define goglomp to be a $50 debt you owe me, I think you'd agree that a goglomp does not exist simply because I've defined it. :smile:


    My mistake; you have not given a definition, so it was incorrect to say "your definition does not make sense".


    aleph0 and 2^aleph0 are cardinal numbers. You are trying to divide something by them. No such operation is defined.


    Because your "definitions" do not convey any information about what your concepts are, do not convey any information about how to use the concepts, and do not provide a single statement that can be used in a proof.
     
  10. Oct 9, 2003 #9
    OK dear hurkyl,

    You helped me to define Equation Tree.

    Please check if you can find a formal definition for the ideas that is written below.

    Please read all of it, and then please check if there is a way to define Rf.

    Thank you.


    Organic



    I'll write my ideas again:

    Let Rf be the magnitude level of roughness of some closed 1 dim. element's perimeter.

    L1 = some circle's priameter is 1 unit.

    Therefore:

    RfA = L1/2^aleph0 = circle's roughness magnitude.


    RfB = L1/aleph0 = a closed 1 dim. element's roughness magnitude.

    n>2
    RfC = L1/n = polygon's roughness.

    y = the size of the radius of L1 circle

    The magnitude of 2^aleph0 is greater than the the magnitude of aleph0.

    In the case of RfB there are aleph0 radii with r=y, but there are 2^aleph0 radii with r<y.

    aleph0 < 2^aleph0, therfore r<y.



    Let S2 be the area of a closed 1 dim. element with RfA magnitude .

    Let S1 be the area of a closed 1 dim. element with RfB magnitude, where the disnace between any 2 aleph0 points = L1/aleph0.

    In case of RfA we know that S2 = 3.14... r^2 (where r=y).

    But in case of RfB r<y, therefore S1 = 3.14... r^2 (where r<y).



    Therefore S2-S1 = x > 0
     
  11. Oct 9, 2003 #10

    Hurkyl

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    First, let's pick apart "closed 1 dim. element"

    I presume "dim" is an abbreviation for "dimensional".

    You say "element", but you don't say of what set this is supposed to be an element. It makes sense to say things like "Let A be an element of the power set of S", "Let r be an element of the real numbers" or "Let p be a prime element of the natural numbers"... though we usually say things like "Let A be in the power set of S" or "Let p be a prime natural number". This is my nitpicky English side speaking, but the phrase "is an element of" means the same thing as "is in", so it's usually better to use "is in" because it's grammatically simpler.

    If I had to guess, I think what you mean is "closed curve". A closed curve is, essentially, a one-dimensional geometric figure that has no endpoints.


    Second, you talk about polygons. Well, what polygons? You give no additional specifications about the polygons in which you are interested. I am guessing you are interested in regular polygons inscribed in some given circle, but you don't say anything to this effect anywhere in your posts.


    Is "L1" supposed to be a statement, a number, or a geometrical object? Based on your later usage, it seems that you want L1 to be a number, or maybe a geometrical object, but this quoted passage is defining L1 to be a statement.

    The basic mathematical grammar associated with "=" is:

    <object> = <object>

    So you are defining L1 to be the statement "some circle's perimeter is 1 unit". Is this what you really want?


    As for roughness magnitude, I'm not really sure what you mean by it. If you can't define it explicitly, maybe giving an explicit example of how to use it or compute it. e.g.

    Draw a circle of radius 1.
    In this circle, inscribe an equaliteral triangle.
    The RF of the triangle is _______

    (fill in the blank)


    You use the word "Therefore", implying that these 3 statements have been logically proven using previous statements... are these really supposed to be definitions?


    If my guess as to the meaning of "closed 1 dim. element" is right, this definition does not work because circles and polygons are all closed curves.

    Is "L1/X" (for various X) supposed to be division, or is this a symbol you're defining? If so, from what set does "L1" come? From what set does "X" come? How do we perform the division?

    Reflecting on your posts, I *think* this means that you want to place X points along the perimeter of the circle, then construct a figure by drawing edges between adjacent points... but there is a severe technical problem; when X is transfinite, you generally won't even have a single pair of points you can call adjacent.

    And, incidentally, the only meaningful definition I can imagine off hand for this would imply that your "L1/aleph" (if I'm guessing correctly what this means) really is a circle, and so is "L1/aleph0".
     
  12. Oct 10, 2003 #11
    Dear Hurkyl,


    Thank you very much for your reply.

    Dim. and dimensional is the same.

    L1 is a circle with a perimeter's length 1.

    / is "divid by"

    n>2

    RFC is the roughness of some polygon which is closed by L1 circle and calculated as L1/n.

    A circle is not a polygon, therefore we are talking about its roughness-magnitude, which is notated as RFA and calculated by L1/2^aleph0 (=0)

    A 1 dimensional closed geometrical object is not a circle and it is not a polygon. Let us call it SC for semi-circle.

    RFB is the roughness-magnitude of some SC which is closed by L1 circle and calculated as L1/aleph0 (>0).

    I think that SC must exist if there is cirlce.

    Organic
     
    Last edited: Oct 10, 2003
  13. Oct 10, 2003 #12

    HallsofIvy

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    Do you mean the polygon is inscribed in the circle (i.e. all its vertices are ON the circle)? If so then this is only defined for polygons inscribed in a circle or radius 1/(2 &pi;). I assume, although you did not say it, that n is the number of vertices.


    Hurkyl has told you repeatedly that you cannot divide a real number by an infinite cardinal. It is possible to use this as a "shorthand" for "limit of L1/x, as x goes to infinity, is 0" but in that case the "infinity" involved is NOT 2^aleph0.

    Where to start? You specifically defined "roughness magnitude" only for (a limited set of) polygons (and gave an invalid definition for a circle) and then asserted that SC is neither a polygon nor a circle. Roughness magnitude is not defined for SC. Once again, you cannot divide a real number by an infinite cardinal. Even as "shorthand" for a limit, I cannot imagine why you think this result would be larger than 0.
     
  14. Oct 10, 2003 #13
    Dear HallsofIvy,


    Thank you very much for your reply too.

    You help me to be more accurate.

    So, I will write it again according to your remarks.


    L1 is a circle with a perimeter's length 1.

    / is "divided by"

    n>2

    RFC is the roughness of some polygon which is closed by L1 circle (all its vertices are ON the circle) and calculated as L1/n.

    A circle is not a polygon, therefore we are talking about its roughness-magnitude, which is notated as RFA and calculated by L1/2^aleph0 (=0)

    A 1 dimensional closed geometrical object is not a circle and it is not a polygon. Let us call it SC for semi-circle.

    RFB is the roughness-magnitude of a SC which is closed by L1 circle (all its points are ON the circle) and calculated as L1/aleph0 (>0).

    I think that SC must exist if the circle exists as a geometrical object.

    More to the point:

    Let us say that [oo] = 2^aleph0 = c (has the power of the continuum).

    Therefore:

    1/[oo] = 0

    1/0 = [oo]

    The first known transfinite cardinals are aleph0 and 2^aleph0.

    2^aleph0 = c (has the power of the continuum) therefore RFA = 1/2^aleph0 = 0 (there are no "holes").

    aleph0 < c (does not have the power of the continuum) therefore RFB = 1/aleph0 > 0 (there are "holes").

    y = the size of the radius of L1 circle

    The magnitude of 2^aleph0 is greater than the the magnitude of aleph0.

    In the case of RfB there are aleph0 radii with r=y, but there are 2^aleph0 radii with r<y.

    aleph0 < 2^aleph0, therfore r<y.

    S2 is the area of a RFA geometrical object (a circle).

    S1 is the area of a RFB geometrical object (a SC).

    In case of RfA we know that S2 = 3.14... r^2 (where r=y).

    But in case of RfB r<y, therefore S1 = 3.14... r^2 (where r<y).

    S2 - S1 = x where x > 0.

    A question: can we use x to ask meaningful questions on the CH problem ?


    Organic
     
    Last edited: Oct 10, 2003
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