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Math Help - Archimedes Paradox

  1. #31
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    Re: Archimedes Paradox

    Reference your last post Deveno. Your definitions are inconsistent- please correct. QXQ = ((q1,0) , (q2,0))? What does that have to do with (q,r)? Which member of your "field" is equal to sqrt2?

    EDIT: Also, you have come up with the statement that (undefined symbol)^2 = 2
    If you have a real number on the right you have to have a real number on the left. If your "field' contains that real number, what is it?
    Last edited by Hartlw; January 9th 2013 at 11:44 AM.
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  2. #32
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    Re: Archimedes Paradox

    Deveno, why would you go to such horrific lengths to come up with some monstrosity (your post #29) you claim is a field which has a “member” such that “member”^2 = 2 and is not Archimedean (a pure field). It’s a total waste since fields are isomorphic.

    Here’s how its done (reference my post #28):
    The field of rationals is complete because infinite denominators are included in the field. Thereforee x^2=2 has a rational solution. (No Archimedean "Postulate")
    Last edited by Hartlw; January 9th 2013 at 12:59 PM. Reason: enter post #'s
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  3. #33
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    Re: Archimedes Paradox

    the problem here is your mis-matching "structure types" and you don't even know it.

    Euclid's algorithm works in ANY field, simply choose, for b ≠ 0, q = a/b, and for b = 0, q = 0. then we have b = qa + 0 (the explicit "euclidean d function" for a field is: d(x) = 1, for all x in F*).

    but fields, in general, need NOT be archimedean: you are over-generalizing from one special case. in this special case, it is true.

    but realize: the archimedean postulate for a field is a statement concerning various things:

    1. a field, and a sub-ring of "integers" in this field
    2. an order on the field, compatible with an order on the integers

    the euclidean algorithm only works for those rings where we have a "degree" function d:R*-->N with: for any a in R, and b in R*, we have q and r such that: b = aq + r, and either:

    a) r = 0
    b) d(r) < d(a)

    for R = Z, the usual d-function is d(n) = |n|.
    for R = F[x], polynomial with coefficients in F, d(p(x)) = deg(p).
    for R = Z[i], the gaussian integers: d(a+ib) = a2+b2

    the last example is particularly relevant here: even though Z[i] is euclidean, it does not possess a total order that respects the field operations. that is Q[i] (the rational complex numbers) isn't even an ordered field, much less an archimedean ordered field.

    another way to put this is that "+" and "*" (multiplication) and "<" don't always mean what you think they must mean. the integers, and the rationals are "special" while the terms "euclidean" and "archimedean" have meanings that extend far beyond their applications to Z and Q.

    it is possible to devise fields for which the archimedean property fails in a spectacular way: FINITE fields. here is one example:

    F = {0,1,a,a+1}

    0+0 = 0
    0+1 = 1
    0+a = a
    0+(a+1) = a+1

    1+1 = 0
    1+a = a+1
    1+(a+1) = a

    a+a = 0
    a+(a+1) = 1

    (a+1)+(a+1) = 0

    0*0 = 0
    0*1 = 0
    0*a = 0
    0*(a+1) = 0

    1*1 = 1
    1*a = a
    1*(a+1) = a+1

    a*a = a+1
    a*(a+1) = 1

    (a+1)*(a+1) = 1 which the sums and products not listed given by commutativity of addition and multiplication.
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  4. #34
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    Re: Archimedes Paradox

    Quote Originally Posted by Hartlw View Post
    Deveno, why would you go to such horrific lengths to come up with some monstrosity (your post #29) you claim is a field which has a “member” such that “member”^2 = 2 and is not Archimedean (a pure field). It’s a total waste since fields are isomorphic.

    Here’s how its done (reference my post #28):
    The field of rationals is complete because infinite denominators are included in the field. Thereforee x^2=2 has a rational solution. (No Archimedean "Postulate")
    "infinite denominators" aren't rational numbers. there is a DIFFERENCE between "an infinite number" and "an infinite number of numbers" that is:

    \lim_{n \to \infty} n \neq \lim_{n \to \infty} \{a_1,...,a_n\}

    it is not clear how you are going to, in your "proposed" system, evaluate:

    \frac{-1}{1}\cdot \frac{-1}{1} \cdot \frac{-1}{1} \cdots = ??

    even if you try to limit yourself to creating "super-rational" numbers in [0,1], you have a problem of UNIQUENESS:

    \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdots =? \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdots

    what criterion, pray tell, do you intend to use to tell if the LHS is greater than, equal to, or less than the RHS?

    as far as my "monstrosity" goes: Q and Qx{0} are isomorphic as fields, any statement P(q) in the rationals has a corresponding statement P((q,0)) which is also true. this is NO DIFFERENT than when we identify the INTEGER n with the RATIONAL n/1. we have a ring-isomorphism n-->n/1.

    in FACT, the rational numbers are an equivalence relation on ZxZ* we say (a,b) ~ (c,d) if and only if ad = bc (in Z). we represent the equivalence class as a/b.

    for example: 1/2 ~ 2/4 (that is, the pair (1,2) ~ (2,4)) because 4*1 = 2*2). we often write a/b as if this determined a unique pair, when it fact it doesn't: a/b = (ac)/(bc) for ANY non-zero integer c.

    ************
    allowing for "infinite combinations" of things is QUITE different than allowing for FINITE combinations of things. infinite sums, and infinite products don't always "converge" (however you define THAT to mean in your system of "infinite rationals"). i'm pretty sure you don't see the problem with what you've done: MOST of the "infinite denominator" rationals are going to get very small, very fast. the only way to "balance" that is to allow "infinite numerators", too, in which case you have a LOT of:

    ∞/∞ situations to handle, which are going to be VERY difficult to reconcile.

    of course, the "easy ones" to deal with are the ones in which all but a finite number of factors top and bottom are 1, but these are just the ordinary rational numbers.

    infinite SUMS are far easier to deal with than infinite PRODUCTS, which is why we use cauchy SERIES of rational numbers to define real numbers. even trying to add just TWO of your "infinite denominator" rationals is going to be prohibitively difficult, how do you imagine it will be done?

    (what goes wrong with the formula: a/b + c/d = (ad + bc)/(bd) ?)
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  5. #35
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    Re: Archimedes Paradox

    The topic of discussion is obviously integers, rational numbers, and real numbers.
    I have no doubt that by arbitrarily defining "*" and "+" for a field and then applying it to an algorithm and postulate intended for integers, one can come up with strange things; if you had simply said so, I would have believed you.

    Perhaps there should be a separate thread for different ways to define "+" and "*" for a field and the effect on Euclids Algorithm and Archimedes Postulate. I have no objection, nor interest, in this type of excercise, unless it is used to intimidate or insult people. And clearly off topic.

    I admit I brought up infinite decimals (rationals) as a real field, which is kind of off-topic unless one wants to get around Archimedes Paradox. I weakened on this when I discovered the number of primes was infinite, so you could show even an infinte decimal (fraction) couldn't equal sqrt2.

    Finally, I note there has as yet been no repudiation of my initial post, arbitrary field definitions not withstanding. At this point it looks like it isn't going to happen, so perhaps we need a new thread, "Is Archimedes "postulate" necessary for a real field?


    EDIT: Come to think of it, infinite decimals require Euclids Algorithm, which I showed was equivalent to Archimedes "Postulate," so that's certainly not getting around it.
    Last edited by Hartlw; January 9th 2013 at 11:06 PM.
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  6. #36
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    Re: Archimedes Paradox

    How about we conclude that N > r for all r not because integer N is unbounded (so is r), but because of a primal property of integers that every integer has a larger integer. Works for me. (Archimedes Postulate is OK)
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  7. #37
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    Re: Archimedes Paradox

    by definition, the real numbers are the unique complete ordered archimedean field (up to isomorphism). so limiting your discussion to them is sort of like saying: "all numbers are even, since i'm limiting my discussion to the multiples of 2".

    a lot of ideas that are distinct in other structures happen to be the same for the real numbers, this says more about the real numbers than the ideas themselves. just sayin'.
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  8. #38
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    Re: Archimedes Paradox

    Deveno, regarding your last post, maybe I can learn something. For z, are you saying d acts as an ordering function?

    But Archimedes Postulate says something about the elements of the structure, not a derived function, and the elements of Z aren't ordered.

    What is an example of an ordered structure where the elements themselves are ordered?
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  9. #39
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    Re: Archimedes Paradox

    I could see it if you took Archimedes Postulate to be a statement about a property of elements called distance, where distance for the real numbers is the number ltself (ok, magnitude).

    EDIT But there would have to be a 1:1 correspondence between the elements and d, which isn't the case for Z
    Last edited by Hartlw; January 11th 2013 at 08:11 AM.
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  10. #40
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    Re: Archimedes Paradox

    Quote Originally Posted by Hartlw View Post
    How about we conclude that N > r for all r not because integer N is unbounded (so is r), but because of a primal property of integers that every integer has a larger integer. Works for me. (Archimedes Postulate is OK)
    I assume that "N" here is the of positive integers but what is "r"? The set of real numbers (normally "R")? If so what order ">" are you using? The normal order on number sets is "subset" and then it is true that "N< R", not the other way around. Another commonly used order is "A< B if and only if cardinality of A is less than cardinality of B" but then we would still have "N< R", not the other way around.
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  11. #41
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    Re: Archimedes Paradox

    HallsofIvy: N is an integer and r is a real number. I was assuming a general familiarity with Archemides Postulate. If that is not the case, I suggest you google it or consult a textbook on algebra.
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  12. #42
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    Re: Archimedes Paradox

    hartlw: Stop insulting the natives. If I am correct (and please correct me if I'm wrong Halls), Halls was asking if we were dealing with r as a real number or the set of real numbers. It's a matter of clarification, not lack of learning.

    Actually, even if I am misinterpreting what Halls was asking about, I looked back over this whole conversation. I readily admit that I could be wrong, but I have two possibilities I would like to put out for consideration.
    1. I'm not sure you are working with the set \mathbb{N}. I think you may be talking about the set \matbb{N} \cup \{ \infty \}, which is a whole different beast. My reason for this thought is that you were talking at one point about the set of real numbers [0,1]. The real line in its usual ordering is isomorphic to (0, 1) in its usual ordering. That is to say there is no largest nor most negative number in \mathbb{R} or (0, 1). Unless you are comparing it to the set \mathbb{R} \cup \{ \infty \} \cup \{ -\infty \} , which is isomophic to \mathbb{N} \cup \{ \infty \} . etc.

    2. Is it possible that you are not talking about integers here, but cardinal numbers?

    Neither of these possibilities come from Archimedes Property but since there are a number of semantic points being made about it, it seemed a possibility as to where the conversation has lead.

    -Dan
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  13. #43
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    Re: Archimedes Paradox

    I thought it was clear throughout the discussion that we were talking Archimedes Classic Math Postulate (Principle, Theorem) as stated in my first post or, for example, Birkhoff and McLean, and as responded to by most participants. This thread was started by suggesting it was a paradox. But I relented and posted:

    "How about we conclude that N > r for all r not because integer N is unbounded (so is r), but because of a primal property of integers that every integer has a larger integer. Works for me. (Archimedes Postulate is OK) "

    If you wish to start (why at this point) a new topic about some abstract version, I suggest a new thread appropriateley titled for those interested.

    EDIT To repeat: Archimedes Postulate: For any real number r an integer N exists st N > r. What is your version?
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    Re: Archimedes Paradox

    Quote Originally Posted by Hartlw View Post
    I thought it was clear throughout the discussion that we were talking Archimedes Classic Math Postulate (Principle, Theorem) as stated in my first post or, for example, Birkhoff and McLean, and as responded to by most participants. This thread was started by suggesting it was a paradox. But I relented and posted:

    "How about we conclude that N > r for all r not because integer N is unbounded (so is r), but because of a primal property of integers that every integer has a larger integer. Works for me. (Archimedes Postulate is OK) "

    If you wish to start (why at this point) a new topic about some abstract version, I suggest a new thread appropriateley titled for those interested.

    EDIT To repeat: Archimedes Postulate: For any real number r an integer N exists st N > r. What is your version?
    The same as yours. I was just commenting on what I thought was the general trend the thread was going to. If I was wrong, my apologies.

    -Dan

    PS The actual problem has long since been beaten into the ground. It seemed to me that the discussion became a question of which set and which ordering a set must have to have to be Archimeden (Archimedian?) Just going with the flow.

    -Dan
    Last edited by topsquark; January 11th 2013 at 12:14 PM.
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  15. #45
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    Re: Archimedes Paradox

    personally, i think there is a "sub-text" confusion between "number" (meaning an element of some algebraic structure, typically some sub-structure of the real numbers) and "number" (meaning cardinality). so, for example:

    numbers (meaning #1) are finite.

    the number of numbers (meaning #2) is infinite.

    there's a certain ontological tension, here....can we consider a number (meaning #2) as a number (meaning #1)?

    the answer is "sort of"....we CAN do it, but "it changes the rules of the game", and some of the statements we made about numbers (original meaning #1) are no longer true. for example, if we add the cardinal number \aleph_0 to the natural numbers, we lose the archimedean property, because now we HAVE an N > r, for all r ≠ N, and adding 1 to r doesn't change this. in this "extended number system", the "biggest element" (N) no longer behaves well with regard to the euclidean algorithm:

    if we are to write:

    N = qa + r

    what are our possible choices for q?

    if q ≠ N, then qa + r < N, so the only reasonable choice is q = N, in which case we must have:

    N = Na + r

    but here we have a problem: any choice of r will do (we lose the uniqueness of r).

    so, in this sense, adding N to our set of "finite numbers" breaks the archimedean property and the euclidean algorithm "at the same time" (in essentially similar ways).

    a euclidean "d-function" is not quite an order, usually it's only a "partial order". for example, in the gaussian integers Z[i], any integer pairs (a,b) that lie on a circle of given radius in the complex plane, are "all the same size"

    the archimedean postulate DOES say something about an order. it's important to realize that when considering an ordered structure, we can go 2 ways:

    1) we can generalize the scope of the "algebraic operations" (such as expanding the integers to include "fractions")

    2) we can relax the order to a partial order, which maintains the original order as a "sub-order".

    we can also do some combination of (1) & (2), which can lead to different ways the euclidean algorithm and archimedean property might diverge in applicability. often, we have "compatibility restrictions", such as:

    a ≤ b implies a+c ≤ b+c

    a ≤ b and 0 ≤ c implies ac ≤ bc (in other words + and * "respect" the order).

    the interaction of these different aspects of our structure can be fairly involved: for example, dividing polynomials requires more arithmetic than dividing integers, and certain things we "take for granted" can be lost along the way:

    the INTEGER 2 (for example) is "indivisible" (prime) in the "regular integers" but factors over Z[i] as 2 = (1+i)(1-i). so things which seem "intuitively true" in our "everyday experience" can suddenly not be true if we consider "more general examples of structures". the concept of "number" can be broadened to include a lot of "strange things", and not all of these act like our intuition says they should.

    "totally (strict) ordered sets" are rather restrictive: if we start with "one thing" (let's call it A) and some manner of "adding A" (for the moment let's just call this AA), we wind up with expressions like this:

    A
    AA
    AAA
    .......
    AAAAAAAAAAAAAAAAAAAAAAAAAA

    etc.

    which we can make "bigger" by tacking on an extra "A" at the end. any such system winds up looking a lot like the natural numbers with:

    A = 1
    AA = 2

    etc.

    and even though it's not obvious at the outset that the natural numbers are in fact an archimdean structure, and that the euclidean algorithm DOES work, the recursive nature of their definition can be leveraged to show that:

    1) the natural numbers are well-ordered
    2) there is no largest natural number
    3) the euclidean division algorithm always terminates <--we have to add 0 to the natural numbers for this to work. no big deal.

    in a sense, the natural numbers are "minimal" with respect to properties (1) - (3) above: in categorical terms, this is often phrased as: the integers are an INITIAL object in the class of all ordered commutative rings. if we are going to have +, * AND <, the integers are "the basic place to start". just this information:

    we can add
    we can multiply (and ab = ba, which seems "intuitively clear" but isn't always so, nxn matrices give an easy "exception")
    we can always decide for two different things a,b "which is bigger"

    (and surely we need all these things to state the euclidean algorithm, although we could do with just the order for archimedes, if we have "a distinguished subset") leads us to consider the integers as "the prototypical example".

    in a nutshell, these things pretty much sum up "the basis of most of mathematics" from antiquity to about 1600 or so. the field of number theory (which pretty much takes the above as the only given) which does not, in general require the use of the machinery of calculus (although in the last century, analytic methods have been used to explore deep connections between "algebra" and "analysis", particularly in the behavior of algebraic function spaces) is still a thriving and relatively unexplored field. we do not even understand the integers very well, and we've been at it for at least 2,000 years. the "mysteries of the prime numbers" are so deep, and factoring so hard, that its often used as a basis for encryption software (where it is assumed "hackers" will have access to high-speed computing equipment).

    cantor's insightful investigations of non-finite entities notwithstanding, we know even LESS about the realms beyond the finite, where logic dissolves into utter madness. for every level of complexity we think we "get", there lie several more beyond, that are hard to even talk about, much less reason about. for example, Hilbert spaces are usually "very big", and yet they appear to govern what happens "at the very small" (which is to say quantum state analysis uses Hilbert spaces as a model). we are well beyond "counting", at this stage, into "general principles of behavior", the examples of which have a complexity belied by the simplicity of the notation. we have "uncountable variants" of uncountable objects built upon uncountable copies of uncountable things. and we use these notions to study "actual particles", the number of which is surely finite.

    from this perspective...the entire contents of this thread amount to: "eh? integers? boring...."
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