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Math Help - Archimedes Paradox

  1. #16
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    Re: Archimedes Paradox

    Reference: Deveno post #13. Deveno, you are assuming the field is Archimedian and then proving Archimedes law based on that assumption.
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  2. #17
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    Re: Archimedes Paradox

    Quote Originally Posted by topsquark View Post
    I'm at a loss. The only Archimedes Paradox I have found deals with an floating object and the bouyant force involved?

    -Dan
    Thanks for confirming the originality of my proposal. I also did a search but could find nothing on Archimede's Math Paradox. Can I quote you? I'm sure your opinion carries a lot more weight than mine.
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  3. #18
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    Re: Archimedes Paradox

    Quote Originally Posted by Hartlw View Post
    Reference: Deveno post #13. Deveno, you are assuming the field is Archimedian and then proving Archimedes law based on that assumption.
    i'm assuming no such thing. in fact, it's unclear to me what you are trying to get at. your prior posts have made reference to things such as "infinite number". what is the context, what kinds of things are we talking about?
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  4. #19
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    Re: Archimedes Paradox

    Quote Originally Posted by Deveno View Post
    i'm assuming no such thing. in fact, it's unclear to me what you are trying to get at. your prior posts have made reference to things such as "infinite number". what is the context, what kinds of things are we talking about?
    Let r be the number of points in the unit interval [0,1].

    "Cantor's Theorem: The set of all points in the unit interval [0,1] is uncountable."

    Shilov, Elementary Real and Complex Analysis. Dover. Theorem 2.41, pg 33
    (Thanks Cantor)

    EDIT: I see a problem here. Is r a real number, ie, does it qualify for the Archimedan "postulate?" Anyhow, I previously made it abundantly clear by what I meant. You might not agree, but what I meant was very clearly spelled out. For example, I can give you a number which violates Archimedes postulate: n. Then you give me n+1. Then I give you n+2. For each n, we can each give n+1. By induction, we both win.

    The whole point, and the problem is, that both sides of Archimedes Postulate are unbounded.

    You begin your previous post with the assumption of an Archimedean field. I don't know how to get there from here to quote it. OK I'll look it up.

    You might like my post "Archimedes and Pythagoras" in the math forum.
    Last edited by Hartlw; January 5th 2013 at 11:02 AM.
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  5. #20
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    Re: Archimedes Paradox

    Deveno, if your team can use the argument that the left side is unbounded, then I can use the argument that the right side is unbounded. You might respond, "but you have to choose first, you don't get a second try, and you can't use infinity." In that case, it should be made clear in the postulate. Damn. But that's the same as saying the postulate only applies for r finite. I give up for now.

    Think I'll go out for a troll, whoops, er, stroll.
    Last edited by Hartlw; January 5th 2013 at 11:32 AM.
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  6. #21
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    Re: Archimedes Paradox

    this is what i mean by "unclear":

    your original post says:

    Quote Originally Posted by Hartlw View Post
    For any number r, n exists st n > r.
    Let r be any member of the set of all integers.
    Then there is an integer greater than the set of all integers.
    my first question is: what do you mean by "number"? what is our context? not everything is true "for all numbers", and some things are true for a number n "as one kind of number" and false for n as "another kind of number". for example:

    as integers, 3 is not divisible by 2. as rational numbers, 3 IS divisible by 2 (3 = 2(3/2)).

    the statement:

    "for all r, there exists n such that n > r"

    is not true in all contexts. for example, if r is a polynomial with integer coefficients (that is, an element of Z[x]), we can order first by deg(r), and then order secondly by the leading coefficient. in this ordering, x > n, for all integers n, so this ordering is not archimedean. if you insist on having a field, we can consider instead the field of all rational function in x, Z(x), and define:

    p(x)/q(x) > 0 iff the leading coefficient of p(x) > 0 in Z, and for a(x) = p(x)/q(x), b(x) = r(x)/(s(x), a(x) > b(x) iff a(x) - b(x) > 0.

    it's clear that for any integer n, the polynomial x - n > 0 (the leading coefficient is 1 > 0), so x > n.

    now, this is probably not what you meant by "number", but still: what DID you mean? in any kind of reasonable debate, defining terms is crucial...perhaps one party may well misunderstand the other's point if this is NOT done.

    in the USUAL "number systems", an element u is not of unbounded magnitude. all real numbers are finite. all rational numbers are finite. all integers are finite. number systems that include "infinite elements" act rather bizarrely, and i would not trust my intuition to reason about them.
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  7. #22
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    Re: Archimedes Paradox Math

    Quote Originally Posted by Hartlw View Post
    Any number r can be finite or infinite. If r is qualified as finite (it isnít), Archimedes Postulate is correct, otherwise it is incorrect.

    The trouble with Deveno's argument is that he is, in effect, assuming r is finite.

    I didn't simply say "Archimedes postulate is incorrect" (a paradox) and throw it out to see what happens. I repeatedly prove it and no one proves me wrong..
    In the real number system, which is what we are talking about if we are talking about Archimedes' Postulate, all numbers are finite. It is not Deveno who is assuming that!
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  8. #23
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    Re: Archimedes Paradox Math

    Quote Originally Posted by HallsofIvy View Post
    In the real number system, which is what we are talking about if we are talking about Archimedes' Postulate, all numbers are finite. It is not Deveno who is assuming that!
    Integers "are" real numbers. The integers are not finite.
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  9. #24
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    Re: Archimedes Paradox

    the set of integers is not a finite set. any particular integer is a finite number.
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    Irrational Numbers

    Irrational numbers are defined by limits. This requires that integer N exist st:
    N>1/e for any e>0. e canít be irrational because that requires definition of irrational.
    let e = 1/Ní
    Therefore N exists st N>Ní for any (all) integers Ní (Archimedes Paradox).

    Note in this case you have to specify N for any Ní I come up with. If you beat Ní, Iím allowed to come up with another Ní (say N'+1).
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  11. #26
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    Re: Irrational Numbers

    OK, So letís try a modified Archimedes Postulate: N>r, r finite (note this is a definition of r finite).

    Then:
    (N>r, r finite) iff (r=N, then n= N+1 exists), ie,
    (N>r, r finite) iff (every integer has a successor).

    At this point we have reached rock bottom: Peanosí Axiom #II (every integer has a auccessor). Axiom #I is (1 is an integer) (Landau).

    Note you reach the same paradox with Peanoís axiom #II: I give you n, you give me n+1, then I give you n+2 (which I can do because of axiom #2), you give me n+3, Ö But you have to stop somewhere.

    So technically, you should replace Archimedes modified Postulate with Peanoís Postulate #II, but Archimedes got there first, so Peanoís postulate #II should be Archimedes modified Postulate.

    Finally, I suppose, you could repeat the above argument with Archimedes Postulate as it stands, but you can see why Iím not anxious to go there, although you could still argue that itís a paradox in the same sense that Peanoís Axiom #II is a paradox. Hmmmm, I like that, Peanos Axiom #2 should be called Peanos Postulate (it really is if you think about it). Another peer post?

    Finally, itís possible to state that (ignoring the paradoxís):

    (Archimedes (modified?) Postulate) iff (Peano Axiom #II) iff (the sqrt2 exists).
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  12. #27
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    Re: Archimedes Paradox

    If I can only define a real number to within a finite degree of accuracy, then a^2 = 2 has a rational solution to within a finite degree of accuracy. Why bother?

    Edit: the rational numbers and real numbers both satatisfy the same field properties. So just go merrily along assuming a^2 = 2 has a rational solution, until you get to a point where you have to calculate a value to sqrt2 and then do a rational approximation. You won't do any better by defining it with a cut or series.

    Edit: pi is a rational number to any degree of accuracy. A series converges means it can be given by a rational number to any degree of accuracy. If a series converges to a "real" number, then the "real" number can only be given by a rational number to any degree of accuracy.
    Last edited by Hartlw; January 8th 2013 at 08:33 AM.
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  13. #28
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    Re: Archimedes Paradox

    Or, referring to my last post, if you have an infinite number of integers available to define a real number, then if you allow an infinite number of integers in the numerator and denominator of a rational number, a ^2=2 has a rational solution (the usual proof assumes p and q finite) in the same sense that it has a real solution.
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  14. #29
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    Re: Archimedes Paradox

    as far as fields, go, if all you are going to do is algebra, there isn't the need for all the real numbers.

    but if you want a number x0 that represents where the graph of f(x) = x2 -2 cross the x-axis, you'll have to look for something besides a rational.

    in other words: we can't solve all polynomial equations with rational numbers.

    if one insists that there are rational numbers that are "close enough" (within some rational epsilon), that's fine...up to a point. if epsilon is large enough, and many calculations are performed using it, the error can be significant.

    in particular, approximating the length of curves becomes problematic, because if we use line-segments to approximate these lengths, say if we have N line-segments, and there is a possible error of 1/N, we can be WAY off.

    the property that R has that Q does not, is the "least upper bound property", which is to say, we can consider any cauchy series as converging to a number.

    now, for practical purposes (i want to know how long my hose is, to the nearest "some kind of unit of length") rational numbers serve just fine, in any laboratory, all measurements made ARE rational. the trouble is with THEORY. if we want to study moving objects subjected to forces (such as a rocket bearing a payload of a global communications satellite), the assumption of continuity greatly simplifies calculation. instead of analyzing a large number of finite differences (over small discrete time-intervals), we can simply differentiate. this can be done entirely symbolically, without evaluating the original function or the derivative at a particular point.

    but a derivative is a limit, and to assert we have a well-defined function is to assert that all limit points of the domain exist. the rationals aren't "large enough" for this to be true.

    i think the paradox you're searching for is more along the lines of this one:

    1) any real number is definable as the limit of SOME sequence of rationals
    2) the definition of such a sequence amounts to an enumeration of successive approximations
    3) the list of all such known enumerations is a countable union of a countable set, hence countable
    4) the real numbers are uncountable

    therefore

    5) some real numbers are not definable

    which turns out to be true. so if SOME real numbers are undefinable, how can ALL of them (as a set) be?

    this problem comes up because the power set of a countably infinite set is uncountable. so it's "so big" there's no way we can EVER list all of its elements (though showing a particular thing is in this set is relatively easy). so how do we know "what's in it"?

    **********

    i'd like to point out that the existence of a field containing Q that has a solution to x2 - 2 can be created without reference to the archimedean property of the rationals. here is how you do it:

    create the set QxQ. identify the elements of Q with the pairs (q,0) where q is in Q. define:

    (q,r) + (q',r') = (q+q',r+r')
    (q,r)*(q',r') = (qq'+2rr',qr'+q'r)

    note that (0,1)*(0,1) = (0+2,0+0) = (2,0) = 2.

    it is tedious to verify that (QxQ,+,*) is a field, but it can be done. the additive identity is (0,0):

    (q,r) + (0,0) = (q,r)

    the multiplicative identity is (1,0):

    (q,r)*(1,0) = (q+0,0+r) = (q,r)

    and the inverse of (q,r), where neither is 0 is:

    (q,r)-1 = (q/(q2-2r2), -r/(q2-2r2))

    since (q,r)*(q/(q2-2r2), -r/(q2-2r2)) = ((q2-2r2)/(q2-2r2), (-qr+qr)/(q2-2r2)) = (1,0).

    (quick note: why is q2-2r2 never 0?)

    the set {(q,0):q in Q} does indeed "behave well" with respect to our new operations:

    (q,0) + (q',0) = (q+q',0)
    (q,0)*(q',0) = (qq' + 2*0, q*0 + q'*0) = (qq',0)

    so, as you can see we don't "need" the archimedean property of Q (or R) to have the √2 exist.

    *************

    as far as the Peano axioms go, it is true that what makes the natural numbers archimedean is the stipulation that s(n) ≠ n (that is, that s be injective, together with 0 not in s(N)).

    for if |N| = k, for some finite number k, then for every number z in N, we have s(z) in N, and we have k-1 such s(z) that are not 0 (these are all distinct since s is injective).

    thus we have k-1 z such that s(z) ≠ 0, which means our last element has no choice but to map to 0, which is forbidden.

    this is a fancy way of saying: we can always count up to k, and then count one more. no one knows if this is actually "true", but our mathematics is based on the hypothesis that it is.
    Thanks from MarkFL and topsquark
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  15. #30
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    Re: Archimedes Paradox

    You can replace any real number with a rational number if you allow an infinite denominator in lowest terms. A field allows n+1 for any n so it allows infinite numbers. OK, if you want to call a rational number with infinite denominator a real number, that's semantics. Real numbers are a subset of the rational numbers, the ones with infinite (allowable ) denominator.

    But that's not why I called:

    Euclids Algorithm:
    For any N1 & N2, N exists st N1 = NN2 + p, 0 < p < N2.

    N = N1/N2 Ė p/N1 > N1/N2, Archimedes Postulate (Theorem)

    Archimedes Postulate:
    For any N1 & N2, N exists st N > N1/N2

    Let Ní be smallest such N. Then Ní-1 < N1/N2.
    N1/N2 < Ní < N1/N2 +1
    N1 < NíN2 < N1 + N2
    NíN2 = N1 + N2 - p, 0 < p < N2
    (Ní-1)N2 = N1 Ė p
    N1 = NN2 + p, Euclidean Algorithm (Theorem)

    Archimedes Postulate iff Euclids Algorithm.


    EDIT: Replace "For any" by "Given." I removed (relented) my last sentence. I now buy: Given any number r, N exists st N>r. "Given"
    instead of "for" makes a big difference.
    EDIT: Nah, its still a paradox. Limits require beating 1/e for ANY e.
    Last edited by Hartlw; January 9th 2013 at 12:46 PM.
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