mm-1859 === I have p(r,n) the succession defined as: p(r,n) = 0 for all n < r If we build the table: r|n 1 2 3 4 ...... 1 1 1 1 1 ...... 2 0 1 2 3 ...... 3 0 0 1 3 ...... 4 0 we could make the conjecture that p(r,n) = { { 0 otherwise where S_k(n) = sum_{i=1}^n i^k I tried to prove that by induction, but I arrive to contradiction: this equality is true For n =r+1 no problem, but for the induction step, I arrive that (1) S_{r-1}(n-r) + S_{r-2}(n-(r-1)) = S_{r-1}(n+1-r) has to happen. But clearly (1) does not happen. Can you help me in detecting the error? Xan. posting-account=SAlCkQwAAADOQfb8GuNFkWcQtC01OYCg This is exactly the Pascal's Triangle, which consists of Binomial Coefficients. (Binomial Coefficients B(n,k) are usually defined for k<0 See: http://mathworld.wolfram.com/BinomialCoefficient.html http://mathworld.wolfram.com/PascalsTriangle.html Martin This looks like Pascal's triangle. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ Yes. It's Pascal triangle turned in left. Without this hint I will no see that I'm wrong at the begginning. Xan. And there is no smaller value for p,q ! (fundamental solution of Pell equation a^2 - 13b^2 = 1) for sqrt(15), a^2 - 15b^2 = 1 has fundamental solution 4/1(obvious) Some horrible values, worse than sqrt(13) : sqrt(61) ? 1766319049 / 226153980 is the smallest value ! next huge values are for sqrt(109) 158070671986249 / 15140424455100 These values were asked by Fermat to Frenicle in 1657 saying : [61 and 109] being not too big numbers so you don't have too much work ;-) For general method look for Pell equation and PQa algorithm. -- philippe (chephip at free dot fr) posting-account=YEgZ2gwAAABJXWwDrJ38p9qyq9A1Zi2G work I did'nt know that this problem is the same as the solution of Pell equations. Do you know the solution for sqrt(19) and sqrt(22)? Ludovicus [...] (if starting from a/b the limit is sqrt(L) = sqrt(a^2 -1) / b) this means that L = (a^2 - 1)/b^2 or a^2 - L*b^2 = 1 Apply PQa to 19 and 22 ... This is not impossible by hand : PQa ends after just 6 steps for these values. (170,39) solution of x^2 - 19y^2 = 1 (197,42) solution of x^2 - 22y^2 = 1 However there are lots of software, and on line solvers to do computations for you. My own site has one in Javascript at http://chephip.free.fr/ie/exepqa.html and describes the PQa algorithm, but it is in French... Good descriptions of the Pell equation solving by John P. Robertson or Keith Matthews (Google for Pell equation) Some more general solver is at Dario Alpern's site http://www.alpertron.com.ar/QUAD.HTM for a generic quadratic solver Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 -- philippe (chephip at free dot fr) Brounker found this one by hand: xx - 313yy = 1 x = 32188120829134849 y = 1819380158564160 and he said in a letter to Fermat that it had taken him an hour or two to work out. Ah, the good old days... In several freak cases of the Pell equation xx - Dyy = 1, x is 49 mod 200, and D is 1 or 49 mod 60. I have no idea why. LH Groan! So Aussies met L'Hopital's rule *before* meeting the series for the trig functions. The clever country indeed! :-( Anyway, I'd advise you to learn the common power series anyway -- they are extremely useful! for all x, e^x = 1 + x + x^2/2 + x^3/6 + ... + x^n/n! + ... , sin(x) = x - x^3/6 + x^5/120 - ... + (-1)^n x^(2n+1)/(2n+1)! + ... , cos(x) = 1 - x^2/2 + x^4/24 - ... + (-1)^n x^(2n)/(2n)! + ... , and for |x| < 1 (1+x)^a = 1 + ax + a(a-1)x^2/2 + ... + a(a-1)(a-2)...(a-n+1)x^n/n! + ... [the binomial series], log(1+x) = x - x^2/2 + x^3/3 - ... - (-1)^n x^n/n + .... -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ X-RFC2646: Original + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + I think that's typical in the United States, too. Most of the undergraduate calculus books I've looked at present L'Hopital's rule and improper integrals in a chapter preceding the chapter on infinite series and power series. Aristotle Polonium + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + if S is a (n .81~m)-matrix and a (n .81~ n)-matrix and tr(S'HS)=Sum from i = 1 to n of (SS')_ij H_ij _ij means subscript ij thankx You need to sum over j as well, and one of your ij shuld be ji. LD posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY Now I can at least give a more general exposition on surrogate factoring by focusing on a single quadratic, in the *hope* that maybe something can be resolved, though now there are posters--surprise, surprise--who are clearly working to simply disrupt discussion with their posts. Focusing on the single quadratics yz^2 - Az + j^2 = 0 where j is some non-zero integer that you choose, it's clear that if you have some integer M in mind, with j coprime to M, there is no reason to think that quadratic or its solutions are at all related to the factors of M. But I can introduce a relationship in a very easy way by using T = M^2 - j^2, and adding T to both sides which gives yz^2 - Az + M^2 = T and from there I can solve to get Az = f_1 b_2 + f_2 b_1 - 2M^2 where f_1 f_2 = T and b_1 b_2 = M^2 and I've introduced a relationship between factors of M, and T, and a root of the previous quadratic times A, from that quadratic. So what does Az have to do with roots of M? Nothing. However, f_1, f_2, b_1 and b_2 can be *rationals* so that equation is actually talking about finding rationals such that you get the relationship. That's why I've talked about this method relying on the entire set of rationals, as it does. Basically if any rational factors of T and M exist, such that Az = f_1 b_2 + f_2 b_1 - 2M^2 then you can get those solutions out, as it's easy enough to solve for one of them, like you have b_2 f_1 = (Az + 2M^2 +/- sqrt((Az + 2M^2)^2 - 4M^2T))/2 found simply enough by using the substitution b_1 = M^2/b_2 and here you get a difference of squares, as somehow Az has been related to M, where T clearly is important. Now imagine you already know M's factors, and with them pick some rationals to get Az from Az = f_1 b_2 + f_2 b_1 - 2M^2 then you know that will fit into the solution for b_2 f_1, and it raises the question of, well what if you don't know those factors, can you just check Az values to try and get them? That's where surrogate factoring comes into it, as that's the idea of it, to get those factors of M, from using other factorizations. It turns out you can delve more deeply into the question of those rational factors, as you can simply consider b_1 = (g_1 d_2)/d_1 and b_2 = (g_2 d_1)/d_2 where g_1, g_2, d_1, and d_2 are integers, as notice then b_1 b_2 = ((g_1 d_2)/d_1)((g_2 d_1)/d_2) = g_1 g_2 so I can have g_1 and g_2 be integers, and consider ALL possible rational factors of M. Making the substitutions for b_1 and b_2 and simplifying a bit gives Az d_1 d_2 = f_1 g_1 d_2^2 + f_2 g_2 d_1^2 - 2M^2 d_1 d_2 so considering d_1 and d_2 you get a conic section. That is, assuming you have all the other values and wish to know if there exists integers d_1 and d_2, the curve of the solution for all possible complex values for d_1 and d_2 is a conic. It turns out that whether it's elliptic or hyperbolic depends on the sign of T. If T is positive, then it's elliptic. If T is negative then it's hyperbolic. The gist of it then is that somehow I've related this quadratic with an arbitrary integer yz^2 - Az + j^2 = 0 with conics, in order to relate to the factorization of M. If you can solve either of the conics, then you can factor M. James Harris Still factoring only Quadratics? Factor this z ^ 2 = a* y ^ 2*x + b* x*y + c * x ^ 2 + d* y * z + e * z * x + f X-RFC2646: Original Why don't you post your messages *after* you have come up with something interesting to say? [snip general exposition on surrogate factoring] ...long post with obvious errors and no useful results: Surprise, surprise, Gomer! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com posting-account=82j3EgwAAACYxp9hHOcWy78r2IGkmc3t Closer. Hey magic. A minus sign has changed to a plus sign. And without James Harris having to say a thing! I think what you want is Az = 2M^2 - (f_1 b_2 + f_2 b_1) - William Hughes X-Enigmail-Version: 0.90.0.0 X-Enigmail-Supports: pgp-inline, pgp-mime 100 LOL points I don't agree. The idea of area is freely used in secondary and even primary education, long before it can be rigorously justified (rather like the Fundamental Theorem of Arithmetic) and it seems to me that one of the first tasks of a university maths course should be to shore up the basic idea of area (and volume). -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland Why? Doing too much non-rigorously makes rigor hard to learn later. Area can be better justified if the rigorous properties are stated, at least. This one can be justified as soon as it is used IF the properties of the integers, including induction, are taught early, as they should be. As it is less basic than you think, why? The Euclid approach to area was to approximate the figure by a union of figures whose areas could be obtained by elementary methods, such as unions of rectangles, and to use (without any rigor, but with good intuition) that congruent figures had equal areas. The problem is harder than it looks. This can be extended to all sets in two dimensions, but not to all in three, and this is not by any means elementary; some form of the Axiom of Choice is needed for both of these. Length and other continuous measures are not the basic ones; counting is much more basic, and length is defined in terms of counting. It turns out, with analysis, that length can be defined in arbitrary Euclidean spaces, but above that, it gets much trickier. Archimedes did not have this when he came up with reasonable arguments about the upper bound for the circumference of a circle, or the area of a sphere. Similarly, the Riemann integral is reduced to the limit of integration on finite measure spaces. That is what is basic. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 We are probably talking at cross-purposes. My 8-year old grand-daughter is taught about area and volume (with much splilt water) and unique factorisation has also been mentioned in her class at school. She thinks there is a largest number, so induction would be wasted on her. Even in secondary school, I don't think pupils would have sufficient mathematical maturity to understand a rigorous definition of area, while on the other hand they are perfectly capable of talking accurately on the subject, essentially regarding area as a physical entity. So it seems to me, as I said, that one of the first tasks of a university mathematical course should be to set these ideas on a firm footing. If and when that has been done, it would seem to me to provide the simplest introduction to integration. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland posting-account=84XyagwAAAAn211ho6L11j1JK1UhtY3o [cut] the area of the figure by a known area less as well as a known area more. The style of proof would be the following: Theorem. Area of circle is pi * r * r. Proof: Let A be the area of the circle. Case 1. Suppose A < pi * r * r. .... .... Thus, I can find an inscribed regular polygon of area A_n within A such that A < A_n < pi * r * r. Contradiction. Case 2. Suppose pi * r * r < A. .... .... Thus, I can find a circumscribed regular polygon of area A_m around A such that pi * r * r < A_m < A. Contradiction. Hence, A = pi * r * r. QED Note that a regular polygon is convexed. See below. I believe that this style of proof is more rigorous than you seem to indicate. I don't recall off-hand any area proofs given by Euclid (e.g, the area of a circle). However, I remember a few bits on how Archimedes proved surface areas of the sphere and similar surfaces. In particular, I remembered one of the axioms given *explicitly* by Archimedes: namely, that the approximating figure below had to be convex in order to guarantee that the area of the approximating figure is less than the area of the given figure. This axiom provides the basis for the contradiction given above. Thus, their proofs were rigourous (from axioms) and not intuitive. Note that if the approximating figure is not convex with respect to the given figure, then the well-known stair case approximation to the diagonal of a square would show that the diagonal of a square would be 2 and not the square root of 2. -- Bill Hale I've sifted through them all, and again, and again. And then went to check a few books. And I finally think I got it! To be honest, now it seems relativly 'simple'. As a concept like vector space seems more like something to accept than to get. Or in other words, while for example 2 and 3 dimensional vector space can be visualized. It seems best not to dance around those for too long and just accept the generalisation vector space offers and accept that you can in fact work in an n-dimensional space without being able to really summon a visualisation in front of you. On the other hand, getting into the concept of vector space just raised a whole lot more questions! Which I am sifting through now (For example, I'm trying to unravel affine space as we speak!) There is one question I would like to ask though: If,for example, I define a vector space V consisting of all the 2x2 matrices of reals. Then my vector space V is defined over the field F of reals? (and not matrices) I would assume so, as matrices aren't a field. in V would be in fact scalar multiplication of matrices. Not matrix multiplication, I presume. I'm just asking so I'm very sure. Books alone don't mention a lot of things, unfortunatly! posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs accept able Yes. Many mathematical problems have more than 3 variables, and vector space concepts can be hugely useful in understanding the nature of those problems. But it's still often helpful to illustrate some n-space geometrical concept in 3-space to help your intuition along. Yes. Think of your matrices as 4-element column vectors, for vector space purposes. The space of 4-vectors is isomorphic to the space of 2x2 matrices. multiplication Multiplication of a matrix by a scalar, e.g. 3A. Right. - Randy posting-account=BjC-YAwAAADQ91Zm3XkS3aGs3XlaqZ4X Actually, i just thought, may be You don't know, what a field is. One example of a field is the set of real numbers with addition and multiplication and some rules, like distributivity, that there is a Zero and an One. And in difference to the vectors, they are called scalars (as they scale for example the size of vectorarrows) And a vectorspace over a field has nothing to do with over, it just means, one needs beside the set of vectors also a set of scalars, and a lot of rules. Have success Hero To my mind you are just adding an extra layer of confusion. All the machines under discussion are theoretical concepts. The essential point in my view is that while you might be able to carry out a computation more efficiently with a quantum computer the actual computations you can carry out are exactly the same. I don't understand this. By the algorithmic entropy of a Turing machine T I would mean the length of the shortest string t such that T(s) = U(ts) for every string s, where U is a chosen universal machine. This has nothing to do with the use to which you might want to put T. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland tim@birdsnest.maths.tcd.ie says... Nonsense. Physical electronic and quantum computers have been built, and the former have been developed to a scale at which they can do useful work. So far, physical quantum computers are rudimentary, but they do exist. If you don't mean by 'quantum computer' a physical computer, just what are you suggesting the term has been reserved for? I am not convinced that that is an essential point. If, for example, it turns out that a quantum computer can carry out certain operations more efficiently, surely the real world difference may be that a physical quantum computer can do what a physical electronic computer cannot? If the 'essential mathematics' concerning them cannot consider the difference, so much the worse for it - it has lost its 'essentiality' and needs to be replaced by a mathematical model that connects with relevant issues. I thought it had been made clear that the abstract quantum computers [with or without their programs] under consideration can also be described by finite strings. Which from the above seems to be the total meaning of your reference to algorithmic entropy. So I don't understand what you meant when you said they have infinite algorithmic entropy. - Gerry Quinn There is a perfectly good mathematical theory of computability; and there is a perfectly good mathematical theory of the efficiency of algorithms. But they are different theories. In my understanding, quantum computing has no bearing on computability; but it does have a bearing on algorithmic efficiency, eg through Shor's algorithm. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland f(3*x)+1 ) As an application of 'Local equalities' I would like your solving of the above 'thing'. Some people asked me in a recent past to 'disguise' somehow presentations of problems. Amiti.8es,Alain. posting-account=OT2XzgwAAACfOom0T1e3nydGQHkWwi3r Here's what I'd do: First, plug in x = 0 and you'll have a quadratic in f(0). Solving this gives f(0) = -1 or f(0) = -2/3; however, if f(0) = -1 then the denominator on the LHS is zero when x = 0, so for solutions which are defined at the origin we need f(0) = -2/3. Now multiply this out so you have a quadratic in f(3x) on one side and one in f(x) on the other. Differentiate each side and again plug in x = 0. All the constant terms cancel out, so you get that f'(0) = 0 (if my calculations are correct). Differentiate that, and I get f''(0) = 10/13 (although there were a bunch of numbers in there and I'm not sure I added them right.) Proceeding in this manner, you can generate the Maclaurin series coefficients for the solution. Of course, it'd probably be easier with mathematica instead of pencil and paper. posting-account=uJhfTw0AAACZ85X1hg4ZQuYw9kXQVPPG words.... lol damn - and I was all set to follow-up by asking what the intuitive word for *connected* was... ;) X-RFC2646: Original What do you mean by irrational fraction type number? An irrational fraction is a contradiction in terms. posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY Yup. Both are algebraic numbers, but the roots of the second quadratic are not algebraic integers, while one of the roots of the first is. degree except the I do use the terminology properly, but people like you seem intent on refusing to acknowledge that there's a problem with it. For instance, with 2x^2 + 3x + 1 only one of the roots is an algebraic integer, while both are algebraic numbers, and with 2x^2 + 5x + 1 neither of the roots are algebraic integers, but both are algebraic numbers. Mathematicians have traditionally leapt to the conclusion that finding out neither root is an algebraic integer with the second one, is significant. But I can prove that the second example may be like the first, except with irrationals, where one root is like the integer root before, and the other is like the root that is a fraction, but both are irrational. Describe that with current terminology please, or quit complaining, as it's not my fault, but the fault of math society if less awkward terminology does not even exist. James Harris posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs irrational. I'd be interested in this proof. But before that, I'd be interested in a statement of what it is you are proving. What does it mean to be like the integer root or like a fraction? Like in what way? - Randy posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn except and I will presume to answer on behalf of the Great Sphinx. Like the integer root means: an algebraic integer. Like a fraction means: an algebraic number which is not an algebraic integer. It is worth noting that, given any algebraic integer A, there exists another algebraic integer B and a rational integer n, such that A = B/n, i.e., every algebraic number is a quotient of an algebraic integer and an ordinary integer. For a polynomial like 2 x^2 + 5 x + 1, Harris appears to be saying he can find a ring R which contains the roots of this polynomial and in which the only units are 1 and -1, and one of roots is an integer of some kind in this ring, perhaps even an algebraic integer, and other root is a quotient, i.e., an algebraic number which is not an algebraic integer. In fact, if 2 x^2 + 5 x + 1 is factored in the form (a x + 1)*(b x + 1), neither a nor b can be an algebraic integer, so this polynomial may not be adequate to illustrate what he wants. The real question is, why does he want a ring which has the property that he describes? It is clear now that this is not a matter of his thinking that the mathematics of algebraic number theory is *incorrect* - it is instead a matter of personal taste, i.e., his feeling that polynomials should factor in a different way in some alternative ring versus how they necessarily factor in the ring of algebraic numbers. He wants more similarity between reducible and irreducible polynomials than is allowed for within the ring of algebraic integers. Why he wants this is not clear. Well, actually it is clear. The tail is wagging the dog. He wants it because otherwise there is no point to his paper Advanced Polynomial Factorization. And then, lurking behind APF, is the mighty Hammer itself with which to destroy the Evil Mathematical Establishment. If the results of APF hold, then Harris believes he will have a complete (and short) proof of Fermat's Last Theorem. That is the Hammer. It is a house of cards. Knowing that his objection to the way factorization is done with existing algebraic number theory is just a matter of preference means that there is really no rigorous basis at all behind APF, and the Hammer is just a derivative delusion. Do not expect him to recognize this anytime soon. Nora B. Nntp-Posting-Host: apps.cwi.nl ... ... ... Can you do the same with 2x^2 + 6x + 1 ? If not, what is the difference between the two quadratics you gave and this quadratic? (Note: if you adjoin one of the roots to a ring that contains 1, 1/2 will also be in that ring.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn coefficients. coefficients. equation with posters terminology, type algebraic finding Significant or not, it is simply a provable fact. I think you know this. irrational. So you try to invent another ring which is larger than the algebraic integers, but contains a few numbers which are not algebraic integers. The only units in this larger ring are 1 and -1. However it is easy to show that it is not possible to construct such a ring for the polynomial P(x) = 2 x^2 + 5 x + 1. Here is why. The two roots of P(x) are r1 = (-5 + sqrt(17))/4 and r2 = (-5 - sqrt(17))/4. Now suppose you adjoin one of those roots, say r1, to the ring A of algebraic integers: i.e., consider A[r1]. Note that 1/r1 = r2/(r1*r2) = -r2 / (1/2) = -2 r2. But 2 r2 = (-5 - sqrt(17))/2, which is an algebraic integer. Therefore r1 has an inverse in A[r1]; it is a unit, but it is not +1 or -1. So if you are going to find a ring with the properties you want for this polynomial, it is not going to be a ring which contains the algebraic integers. But the real question is: what do you think is wrong with accepted theory? That is, why is it bad that both roots of (2 x^2 + 5 x + 1) are NOT algebraic integers, while one root of (2 x^2 + 3 x + 1) is a fraction (-1/2) and the other is an integer (-1) ? Why is that a problem? Other than the fact Advanced Polynomial Factorization ? Is there some underlying symmetry here which you think must be achieved? It is clear now that you now accept what algebraic number theory and Galois theory both say about this. But for some reason, apparently an esthetic reason, you think things should be otherwise. Because of this, whatever your reason is, you think a big chunk of the math accepted by algebraists for over 100 years, is somehow wrong - not wrong exactly, but just not what you want it to be or what you think it SHOULD be. Can you provide more of an explanation for why you think current theory is somehow deficient ? Is it more than esthetics, or what ? Nora B. as days. My association with the Department is that of an alumnus. [.snip.] I don't think this is an accurate restatement. What is being asked is not that the ring contain no units other than 1 and -1 (this is already false in all number fields except for most of the imaginary are units are 1 and -1. This is trivally equivalent to asking that R intersect Q be equal to Z. Now, there are many, many, many such rings. Bill Dubuque and others have given explicit examples. Pick any algebraic number which is not an algebraic number and such that no power of whom is rational, call it r, and take A[1/r]. That ring has that property. The real difficulty lies precisely in that there are many, many, such rings, and that while Zorn's Lemma guarantees the existence of maximal such rings, there are many such maximal rings; talking about the ring with these properties is nonsense. Your ring A[r1] is probably one such ring. So is the ring A[r2]. But there is no ring which contains BOTH r1 and r2 and satisfies the condition of having intersection with Q equal to Z. Because if a ring contains both r1 and r2, then it contains r1*r2 = (25-17)/16 = 1/2. So A[r1] is contained in some object ring, and A[r2] is contained in another, but there is no object ring that contains both. IF you further ask, as it was at one point implicitly required, that Q and has at least one root in R, then it has all its roots in R), subring of the algebraic numbers with that property AND with the property that it intersects Q at Z. But that ring is... the ring of all algebraic integers. [.snip.] -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu How do I extend my wisdom with such infinite irrational powers? With my weak rational powers I cannot seem to devise a principle of choice to pick an algebraic number that is not an algebraic number. That must be a surreal transcendentally transcendental number that will exist once Harris emerges from the 19'th century and begins to study 20'th century authors such as Conway and Rubel. Or perhaps I need to imbibe more of the Zima spirit, especially those containing bugs (including worms), see --Bill Dubuque posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn algebraic such contains which Arturo, You're right - my mistake, I should have looked up what he said previously. As often happens it is a little hard to discern what Harris is thinking. With the example P(x) = 2 x^2 + 5 x + 1, one has the impression that he would like for one of the roots to be an algebraic integer, and the other to be an algebraic number which is not an algebraic integer. This would parallel the reducible case. Letting r1 and r2 be the roots, one clearly has r1 * r2 = 1/2, or (2r1) r2 = 1. Thus the first term (2r1) is an algebraic integer, and the second term (r2) is an algebraic number which can be written in the form r2 = s2/2, where s2 is an algebraic integer. Thus r2 is a kind of fraction. (Of course that is true for all algebraic integers). So what Harris wants is a ring that contains r2 and in which the only rational units are 1 and -1. Presumably this ring also contains (2r1). Thus the minimal such ring is A(2r1, r2) = A(r2) (since 2r1 is in A already). neither of these is a unit in the ring of algebraic integers. maximal Agreed. Right. I don't think Harris accepted the condition of being closed under conjugates. The main point in all this is, why is Harris so insistent that there be a similarity between a reducible polynomial, like 2 x^2 + 3 x + 1, and an irreducible one like 2 x^2 + 5 x + 1? The former has one root which is an integer and the other which is a fraction. The latter, Harris thinks, SHOULD have one root which is an algebraic integer and the other of which is an algebraic fraction (i.e., an algebraic number, which of course can be written as a quotient of an algebraic integer and an ordinary integer) ? Why does he think that is so important? He quit claiming long ago that his previously announced main result was what he said it was at first - that is, he recognizes that neither of the roots of 2 x^2 + 5 x + 1 are algebraic integers, and he no longer claims that if Q(x) = 65 x^3 - 12 x + 1 is factored in the form (a x + 1)*(b x + 1)*(c x + 1), then one of a, b, or c must be coprime to 5. But he still thinks there is something wrong with the algebraic integers because roots of irreducible polynomials act differently from roots of reducible polyomials. It's no longer a matter of his thinking algebraic number theory or Galois theory are *wrong* - it is more a matter of his just plain *not liking* what these theories imply about factorizations. The fact that he now refuses to stick his neck out and actually state some definitions and theorems does not help if we want to understand his (so-called) thinking. One other ingredient in all this. I don't think Harris has ever understood, or at least has ever accepted, the fact that his argument based on constant terms are constant is wrong. This adds to the confusion. If that argument were correct, then since it leads to a contradiction of known theorems, mathematics would be inconsistent. Harris knows this. So he has this problem of things not factoring the way he wants and facts being inconsistent with the constant terms are constant argument, and he somewhat illogically resolves these two by saying the algebraic integers do not have the properties that they SHOULD have. More briefly: what Harris thinks is wrong with algebraic number theory is that it doesn't give the results that he wants it to. Nora B. days. My association with the Department is that of an alumnus. [.snip.] Oh, boy, this is a real mess, isn't it. That should be: There are many, many, many examples of rings that properly contain the algebraic integers and whose intersection with Q is just equal to Z. Pick any algebraic integer r such that no power of it is rational, call it r, and take A[1/r]. Or take any algebraic number which is not an algebraic integer, and such that no power of it is rational, call it s, and take A[s]. [.snip.] As I recall, there was never any real discussion. Some of his directly replied to those who pointed out this condition would limit him to subrings of the algebraic integers, but since he dropped all the arguments surrounding these assertions (at least publically) it is not clear if he now avoids those requirements or not. Because his original argument arranged things so that two of the coefficients were noncoprime to p and one was. Then he ran into trouble with units, i.e., whether or not p would be a unit in that ring. So he switched to algebraic integers (where the units would not be a problem, since no integer other than 1 and -1 would be invertible in the ring), and tried to pass his argument through by claiming that it would necessarily be the case that two of the coefficients were noncoprime to p and one was. The proof that his arguments were wrong involved showing that certain things were not algebraic integers. To show that, their minimal polynomial was produced. These polynomials were nonmonic, primitive, irreducible. Since he could follow all the easy algebra except for the final conclusion (if r is a root of a nonmonic, primitive, irreducible polynomial with integer coefficients, then r is not an algebraic integer) he attacked that for over a year, refusing to look over proofs of the basic fact. Finally, backed into a wall, he looked over the proof, and had no choice but to agree to it, it being completely elementary (well, he did try to deny it several times, claiming errors in it, but eventually he came around). And so the only way he could continue to push his original argument was to argue that there is something wrong with algebraic integers, what, nobody knows. Throughout his denials about roots of nonmonic primitive irreducibles, a lot of verbiage was thrown around about rational, irrational, integer-like, fractions, and lots of examples like the above and attempts at counterexamples to the proposition with reducible polynomials. All the flailing simply came because there was no knowledge or understanding, just a very deep and powerful desire to have the original argument go through. I think these are just the successors/inheritors of that flailing. One wonders if there is even a memory of where this all came from on his part... [.snip.] -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu ...hmmm. Am I reading this correctly? Could you give an example of such a number?... -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com days. My association with the Department is that of an alumnus. Sorry about that; I meant any algebraic number which is not an algebraic INTEGER, and such that no power of it is a rational. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu That's obviously because the second polynomial is irreducible and first polynomial is reducible. Reducible, in this case, means that the first polynomial can be expressed as the product of a monic polynomial with integer coefficients and a non-monic polynomial with integer coefficients. The product will contain all roots -- both the algebraic integer roots and the non-algebraic integer roots. Big deal, and so what? ...because it is reducible to the product of (2x + 1) which is an irreducible non-monic polynomial with integer coefficients, and (x + 1) which is a monic polynomial with integer coefficients. Hence, there will be one root which is an algebraic integer and one which is not. QED. ...because it is irreducible. Please cite any mathematician who leapt to the conclusion that there is some special significance to this obvious finding, beyond what is already implied by the definitions.If you do not provide such a citation, it is clearly because you *cannot* do so. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ... The basic difference between the two polynomials is of course that one is reducible over the integers and the other not... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Notice also that 2x^2+3x+1 factors over the integers, while 2x^2+5x+1 does not. irrational fractional type is not *standard* terminology. And all we claim is that in both cases, both are algebraic numbers. Since all algebraic integers are algebraic numbers, there is no conflict. In what way have we leapt to that conclusion? Significant in what respect? sqrt(2) is an irrational algebraic integer. What do you mean by except with irrationals? awkward is often associated with precise. When simple terminology is adequate, we use it. -- Will Twentyman email: wtwentyman at copper dot net Is it, in fact? But that does not at all affect my argument. IF a bijection for finite n is valid to show that the proof (bijection) holds for the infinite set N then an argument valid for finite intervals (n,n+1] is valid for the whole set of those intervals My method *is* vaid for the whole of R. Can you give an element Can you give an x e R which does not define a finite interval (-x,x)? The bijection works for *all* R ontains finite elements x only. Your methods works for *all* intervals defined The intervals (n, n+1] are elements of a set of intervals which intervals. And there are finite n only, as you stated above. Surrender! There is no reasonable argument in favour of tansfinity. Regars, WM No. There is none. So it works for all intervals defined by a finite 'x'. R is not one of those intervals, and so the conclusion is not valid for R. Yup, so it is valid for all finite intervals, but not for R itself. Wrong. R is not a set of intervals. It is a set of real numbers. So (n, n+1] is *not* an element of R. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ Just as I suspected. Professor Muck really cannot distinguish between elements and subsets of sets. And he is teaching mathematics :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ Actually i do not know what he is teaching (physics?). But certainly not mathematics :-)) -- use mail for mail not nonail Fair comment: what I should have said was as far as I am aware he is contracted to teach mathematics :-(. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ finite N, and so works for *all* elements of N (because N contains finite elements only). This is not an argument, if we have to distinguish between sets and elements. Nntp-Posting-Host: hera.cwi.nl ... Sorry I should have written: The bijection works for *all* finite *elements* of N, and so works for *all* elements of N. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ I'm also strikingly handsome and I plan to be wealthy someday soon. I guess you really could tell me the odds on that one.... Consider the two problems stated below. Problem 1: A field u satisfies Laplace equation in a 2-D region Omega and satisfies Zero Dirichlet BC on the external boundary Gamma_Ext and non-zero Neumann boundary condition on an internal hole Gamma_Hole (assume however that the NET flux on Gamma_Hole is zero ... equilibrating flux). Now remove the external boundary Gamma_Ext and create an infinite region Omega* but with a hole. Problem 2: Consider the Exterior Neumann problem where a field v satisfies the Laplace equation in the infinite region Omega* and satisfies the above Neumann boundary condition (same flux as before) Define a functional F(w; Gamma_hole) = Integral(|w|^2) over the boundary Gamma_Hole Is the following conjecture true? Conjecture: F(u; Gamma_hole) <= F(v; Gamma_hole) If so, is there a proof? If not, is there a counter example? Note: 1. I believe the conjecture can be proven when Omega is an annular region through Fourier analysis 2. Numerical experiments suggest that the conjecture might be true for arbitrary regions (even in 3-D) Never call a dame chick, They *hate* that. Really? My wench never said nothing. Never call a spade a spade.... -- Patterns of floral functions R = a + b*sin(p*A/q)^M are given on web page http://www.b192907.com/mathgraph for a = 1 and b = 1 or a = 1 and b = 2, etc. Where p = 1,3,5,7,9,11; q = 1, 2,3,4; q = 1,2,3,4 and M = 1,2,3 are used. The variable A is from 0 to 2*p*pi. Similarly, star patterns of R = a + b*sec(p*A/q)^M in the web site. These patterns are also available on application program on the web page. 10500 patterns are published in the book of pattern mathematics by Dr. K. G. Shih. It is my hope that you can introduce the web page to your friends. My mailing address : Dr. Keh-Gong Shih 11 Lantana Terrace Dartmouth, NS, Canada. B2X 3L3 posting-account=lg_NlA0AAACBfQRWSK9R13m3Ur1lV4OT The inverse of a hyperbola shows all four legs (which diverge to infinity in Cartesian coordinates) converge at the point in the center of the inverse map. Is this point-infinity at the center of the inverse actually: 1) a point, possibly in hyperspace, where the legs actually do meet? 2) an instance of a pseudometric, where distances between separated objects can be zero? 3) not really a mapping at all, but a failure of the map at a limit value, i.e. the mapping is not one-to-one at infinity 4) really should map to multiple points in 3-space (hyperspace to the 2D object), but get mapped to the same point when the thing is squished into 2D? 5) something else, which I haven't thought of? Thanx! =[ d PS: my suspicion / best guess: pseudometric PPS, and I hate to have to say this, but: If you consider answering this beneath your dignity, then just shut the up about it and let someone else answer it. I'm sick and tired of lectures about asking stupid questions. Take the standard hyperbola x^2 - y^2 =1 or parameterised with t in [ 0, 2 pi [ { x(t) = 1/cos(t) { y(t) = tan(t) Perhaps with inverse, you mean that every point (x,y) gets transformed to ( x/r^2, y/r^2 ) where r^2 = x^2+y^2, such that a point at distance r, gets transformed to a point at distance 1/r with the same t (angle in polar coordinates). In that case the inverted hyperbola would be an eight figure, parameterised as { x(t) = cos(t) / (1+sin^2(t)) { y(t) = sin(t) cos(t) / (1+sin^2(t)) satisfying These are 4 ways to reach the central point, corresponding to the four directions of the asymptotes. Dirk Vdm It may be that you are talking about the one-point compactification of the Argand plane. If so, the point is denoted oo (i.e. infinity symbol). Since you talk of the legs of a hyperbola, I suspect that you have in mind the _graph_ of a function such as But what do you mean by the inverse of a hyperbola? I'm not aware of graphs having inverses. The _function_ that is graphed may have an inverse, but then the term is ambiguous. The inverse of may be i.e. the multiplicative inverse of f(x), or it may be where f^{-1} is such that f^{-1}(f(x)) = f(f^{-1}(x)) = x. In both senses of inverse only some f's have inverses. Any subset of the plane has an inverse in the sense of the geometry of inversion; or rather, it has one inverse for each circle, namely, its image by inversion with respect to that circle. In that context, the word inverse, unqualified, usually means image of inversion in the unit circle, that being the map (defined on the plane together with a single point at infinity) which takes (0,0) to infinity, infinity to (0,0), and a point (x,y) other than (0,0) to (x/r^2,y^r^2), where r^2=x^2+y^2. The inverse of an algebraic curve P(x,y)=0, where P is a polynomial of degree exactly d, is the algebraic curve (r^2d)(P(x/r^2,y/r^2). The inverse of the hyperbola where xy-1=0 is thus the curve of degree 4 where xy-(x^2+y^2)^2=0, a lemniscate if I'm not mistaken. Near (0,0), this curve looks like the curve defined by the leading term of its polynomial, namely, xy; so two branches cross there, for a total of 4 semi-branches--the images by inversion of the four legs of the original hyperbola. Lee Rudolph Yes of course! My apologies to the op for going off on a tangent (or perhaps asymptote). I now fancy he must mean inverse in the sense you explain. posting-account=lg_NlA0AAACBfQRWSK9R13m3Ur1lV4OT Yes, Lee and Dirk were correct, I was referring to the circular inversion (inversion with respect to a circle, specifically the unit circle centered at O), and yes, the hyperbola does map to a Lemniscate, or sideways figure 8, (but only if the circle you're taking the inverse with respect to is centered at the origin). Here's a picture of the situation. The green thing is the unit circle and the red thing is the inverse: http://www-groups.dcs.st-and.ac.uk/~history/Curvepics/Hyperbola/H yperbola5.gif I'm already aware that the ends of the legs of the hyperbola map to the single point that represents infinity in the inverse, namely the point at the center of the figure 8, where all four lines cross. What I'm really curious about is whether, in this or any other circular inverse, the point in the center is actually the image under the mapping of a single point where all infinite values have the same physical coordinates (sometimes called the point at infinity), OR whether the four infinite values of the hyperbola (the ends of the legs) do NOT really intersect, but only appear to in the circular inverse, possibly for one of the reasons I guessed at in my original post. The fact that the point at infinity is also the place where parallel lines meet (in Euclidian space), when they don't REALLY meet at all, is what leads me to doubt that the ends of the hyperbola really meet too. See: http://mathworld.wolfram.com/PointatInfi nity.html See, someone told me that, mathematically, the two-sheeted hyperboloid really only has one sheet, but that in 3D we can't see the plane where the two halves touch, revealing the hyperboloid to be a single, simple, higher-dimensional solid of genus 1, rather than two cone-like things facing each other. This is important (to me) because it's part of the description of what's really going on in spacetime when an object is accelerated in 3-space, so I didn't want to just dumbly BELIEVE it, I wanted to. verify it for myself. I don't DISBELIEVE what someone smarter than me says -- I just want to really understand it, that's all. But since the equation of a hyperbola involves infinite-valued coordinates at those points, I can't just plug a pair of X's and Y's into the equation and see if they map to the same point on the hyperbola. So I figured the neat-o circular inverse function was a cool way to tell. Thing is, It's ambiguous to me whether it really does or not. If anyone has another suggestion of how I can prove that the two halves of the hyperbola are really connected (in a higher-dimensional space), that would be just as good, particularly if the proof involved equations rather than just looking at a pretty picture. =[ d His what? Is that a misprint for execute me? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ On Tue, 22 Mar 2005 15:02:22 -0600, Jon Slaughter It sure looks that way to me. You've said over and over that sqrt(2) is infinite. Now you say you're _not_ talking about an infinite real number? Choose one: (i) sqrt(2) is not a real number (ii) when you say sqrt(2) is infinite you don't mean that sqrt(2) is infinite. Uh, the distinction is exactly what _you_ seem to be missing, based on what you've been saying. ************************ David C. Ullrich On Tue, 22 Mar 2005 15:02:22 -0600, Jon Slaughter Which interpretation of infinity do you have in mind? So, when you say that sqrt(2) is infinite you do _not_ mean that some real number (the positive one which when squared equals 2) is infinite? I'm glad to hear it. You _may_ mean that 1.414213562373.... has infinitely many digits. You're right, _if_ that's what you do mean. Or you may mean something else. My guess: something to do with the way sqrt(2) _may_ be defined or constructed (I'm using constructed in a loose sense). Are you going to tell us what you do mean? To expand on my guess: if you define sqrt(2) to be a Dedekind cut, then, yes, infinite sets are involved. Similarly with the other popular ways of defining real numbers from rationals: Cauchy sequences and nested intervals. But if you define _any_ real number to be a Dedekind cut (etc) then infinite sets are involved. Also sqrt(2) need not be defined by reference to infinite things or infinite processes: Robin Chapman has given an example. Principal! -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Elegance is an algorithm Iain M. Banks, _The Algebraist_ Adam Hartshorne ... Sorry a liitle correction the formula should have been more clearly Any basic textbook on differential geometry will tell you the basics about the principle curvature. In particular they are the two zeros of a certain quadratic function, and that quadratic might be the basis of the code you have for differentiating them. Hope this helps. LH It's interesting to see two different ways of doing this as Borel summation. Since we need a series of the form z*f(z), and ours isn't, you did z*F(z) and when you put in z=1 we get F(1), part without the 1. Mine comes out as: 3 / /[-1 3] 1 - - int|hypergeom|[--, -], [2], t| exp(-t), t = 0 .. infinity| 4 [2 2] / / // / (1/2) / (1/2) / (1/2) 3 || 4 t - 1/ 1 + t / EllipticKt / 1 - - int||- --------------------------------------------- 4 3 t Pi / (1/2) (-4 + 8 t) EllipticEt /| | + ----------------------------| exp(-t), t = 0 .. infinity| 3 t Pi / / -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Hello. Is there a convenient way to check if all roots of a polynomial are less than 1 in absolute value? Something like the Hurwitz criterion would be very nice (it allows to say if all roots have a negative real part by simply calculating a set of determinants composed up of the coefficients). Estimations of a number of complex roots within a given region are also welcome (to start with, one may use Rouche's theorem). The Schur-Cohn algorithm. its Schur transform Tp(z) = sum_{k=0}^{n-1} (conjugate(a_0) a_k - conjugate(a_{n-k}) a_n) z^k a polynomial of degree n-1 (we consider the degree as n even though some of the leading coefficients may be 0). Let T^k p(z) be obtained by iterating this transform for k=2,3,...,n. Let gamma_k = T^k p(0). Note that all gamma_k is real for k=1,2,...,n. Theorem: If p is a polynomial of degree n, all zeros of p lie outside Of course you can transform all zeros outside to all zeros inside using the reciprocal polynomial. The number of zeros in the disk can also be counted using these gamma_k (under the assumption that the gamma_k are all nonzero). Namely, if k_1 < k_2 < ... < k_m are those k's for which gamma_k < 0, the number of zeros in the unit disk is sum_{j=1}^m (-1)^(j-1) (n+1-k_j). Reference: P. Henrici, Applied and Computational Complex Analysis vol. 1, Wiley 1974, section 6.8. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada Yes! - Rouche's theorem or the Cauchy index and Sturm sequences are possibilities. If you already know about the Hurwitz criterion, why don't you use the biholomorphic map to transform the unit disc to the negative half plane? You will also find the following books useful, I hope: The Geometry of the Zeros of a Polynomial in a Complex Variable by Morris Marden (1949) [The second, extended edition is called Geometry of polynomials.] Analytic Theory of Polynomials by Q.I.Rahman und G.Schmeisser (2002). Shame on me, I don't remember that map; is it rational? If it's not, then applying it to a polynomial will make it not a polynomial anymore, so I won't be able to apply the Hurwitz criterion. I'll look for these books, thank you. posting-account=YKijbA0AAAB7TZmds574IxyIgO_eqQ9W I have following problem that I am trying to solve and really appreciate any help: I need the solution of the equation in discerete functions domain where n is an integer: j(n)*[p(n).x(n)] = p(n).[j(n)*x(n)] where: * stands for convolution, . multiplication j(n): bessel function with argument d. (besselj(n,d) in matlab notation) p(n): a known function which is even in n x(n): the unknown function to be solved. To put it another way: Let the operator J be convolution with j(n) and operator P be multiplication by p(n), then (JP-PJ)x(n)=0 is the equation. Indeed finding an expression for the commutator of J and P, (JP-PJ) will help me even more. Going in to fourier domain does not help much since multiplication and convolution is interchanged and the type of the equation remains the same. Necmi Bugdayci. As I have stated before, I do not think that the cute epsilon-delta definition should be used. However, there are somewhat less cute, but equivalent, approaches which can be used quite well. BUT, do not just state the definition of limit and expect the student to understand the concept; concepts are not learned that way. A lot of what Euler does, and having the student do examples until the concept is understood, but not many more, is needed. This holds for any other concept, including the conceptS of integers. Also, keep using limits, continuity, etc., instead of making the rest of the course computation. I agree about the epsilon-delta argument as usually done, but not about the rest. It is not necessary to be complete, but it is necessary to be rigorous. Of course, it helps if the students have had a real Euclid course, instead of the garbage usually taught. Also, the college algebra course needs to spend time on induction, unless this is done earlier. Induction belongs in elementary school; it is an essential property of the integers. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 At least for me, is this normal? It's working OK for me, accessing it via The last couple of posts were actually from yesterday (March 22). There was e.g. one by Ali Taghavi on the subject (Fredholm) Index of Functional operatores and one by Mark Sapir on recursively enumerabe sets. Do you see those? Don't be surprised if there's sometimes a gap of 24 or even 48 hours between sci.math.research postings. The moderators have day jobs too, and sometimes might only moderate once or twice in a day. And the volume of postings is not extremely high, so it could just be that nobody has posted for a while. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada X-Abuse-Notes: Abuse reports must be submited via the usenetabuse.com portal listed above. X-Abuse-Notes2: Reports sent via any other method will not be processed. If you make a post to a moderated newsgroup (like sci.math.research), it will not appear immediately, but that is normal and not a sign that something is down. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Star function : R = a + b*sec(p*A/q)^M www.b192907.com/mathgraph Author : Dr. K. G. Shih posting-account=tg5b7Q0AAACFjo4N-uoZ90NzHuc-dWPu I'm thinking about a variation of the traveling salesmen problem called the Euclidean traveling salesmen problem. In it the distances between the cities are a set of straight, coplanar lines. I'm also only concerned with a set of complete cities (so there is a road between every city) Given a set of such distances my fist task might be to check that distances meet the geometric requirements for a Euclidean TSP. If there are n cities there are n(n-1)/2 roads. But, how much work will it take to find out if a set of distances between cities is Euclidean? If we have only one city no check is needed. With two cities no check is needed ... but with three we need to check that triangle inequality is not violated. The fourth city will require some kind of verification using the distance formula. There will be two possible locations for the 4th city. The addition of a 5th city adds 4 new edges and each must be checked. Here is a table of what will happen: Vertices ---edges---checks 1-------------0----------0 2-------------1----------0 3-------------3----------1 4-------------6----------2 5-------------10----------6 6-------------15----------10 7-------------21----------15 8-------------28----------21 9-------------36----------28 At this point I want to stop and ask if my thinking so far makes sense? The progression of these numbers seems to be the same as the number of edges, but for some reason it doesn't start out the same way. Should I be concerned about that? Depending upon how you count, this could be considered to be multiple inequality checks; but we'll ignore that and call it one. One edge needs no checking. A second edge needs only to be checked for the triangle inequality. At that point, there are in general two solutions. If the first three cities are collinear, there will be only one solution. The third edge needs to be tested for equality with (either of) the solution(s). Call it one check anyway. Equality checks are annoying to do in floating point arithmetic, just as a warning in case you weren't aware of it. You need to allow for round-off error, which can be surprisingly large in some cases. This can apply to the edge cases of inequality checks as well. One of the edges doesn't need any checks. A second edge needs to be checked only for the triangle inequality. A third edge needs to be checked for equality (probably twice). The fourth edge needs to be checked for equality (possibly twice). Cities Checks 1 0 2 0 3 1 (1 inequality) 4 3 (2 inequalities, 1 equality) 5 6 (3 inequalities, 3 equalities) 6 10 (4 inequalities, 6 equalities) ...etc - Tim Google for Kent Paul Dolan. He has many years of experience with the Traveling Salesman problem. Alternatively, you may want to email him. Watch out: He posts to the newsgroup talk.bizarre. -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable which doesn't actually exist due to contradiction with the definition of a probability distribution, - Tim posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8 Hi Tim, So, is probability density function, in this case not a real function or a nonstandard function, more acceptible for that definition? You're correct, I guess, in terms of the context of a standard model, but not necessarily the implied nonstandard model's context. So, no. When I, or anyone else, say uniform probability distribution over the integers, you might understand why that means a) the probability of any of the integers being selected is independent and equal to the probability of any others, for the natural integers m, n P(n) = P(m), and b) the sum of P(n) for n in N = 1. In considering concepts like probability zero, or infinitesimal probability, versus impossibility, part of the notion thus addressed is that those probability zero events are not impossible, thus that their probability is qualitatively and quantitatively greater than zero and positive, and, basically, infinitesimal. About a uniform probability distribution over the integers, consider any contiguous sequence (m, m+1, ..., m+n) for natural m and n. As n diverges, the probability that one of the n elements of that sequence, a contiguous subset of the naturals or a subset of the naturals representing a contiguous sequence of naturals, is an even number is one half. Select a number at random from 1-100, the probability that it's even is one half. Transferringly, select a number at random from 1-oo, the probability that it's even is one half. If you accept that there is a uniform probability distribution over the reals, then the conditional probability that a number is an even integer given that it is an integer is one half. Flip a coin: you're 1/n the way there of sampling a real number. Flip another coin, you're 1/n the way there of sampling a second real number, and 2/n of the way there of sampling the previous. Tim, a different Tim, Tim Mellor, and I were discussing this sampling of the reals numbers and stuff last year, if for some bizaare reason you want to read more of my opinion about the real numbers and all of mathematical logic research real-valued sample on sci.logic. I've also considered some other notions of the powerset coded as binary. What do you think about that, nonstandard probability, infiniftesimal probability, distributions over the reals? There's probably a forty year old operations research journal or finance or something, Bayesian statistics I guess, that covers so much of that stuff that it would take some years to reformulate the vocabulary to understand it, I spent a week or two just trying to read a history of statistics book in two volumes and they're thick, but by the same token, flip a coin. Do it again. Ross If you want to choose different definitions, sure. Just make sure to carefully and rigorously explain those nonstandard definitions. Also make sure to prove any theorems you might want to use, because theorems based on standard definitions can't just be assumed to work with nonstandard definitions. You can equally well do the same thing for prime and square to talk about the properties of square prime numbers. Probability distributions require a lot more properties than that! As it happens, your listed properties contradict the other properties implicit in the definition of probability distributions. Have you checked that the subsets of N satisfying this implicit condition form a sigma-algebra? (Obviously not, because they don't) There isn't. There are uniform probability distributions over some subsets (e.g. intervals) of the reals, but not over the reals as a whole. Likewise there are uniform probability distributions over subsets of integers, but not over the whole set. Since you seem to be talking about Bayesian prior distributions, you are probably already aware that they satisfy many of the same rules as probability distributions, but they are not in general probability distributions themselves. There are several important results that hold true for probability distributions that do not hold true for such entities. Confusing them is very likely to mislead you greatly. - Tim posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs This is a red herring. The people you refer to are mathematicians who might object to Cantor's theory on mathematical grounds, who might have alternate axiomizations of set theory. But people like that start out by understanding what they are attacking. The Usenet crackpots come from the viewpoint that if I can't understand it, it is by definition a bad theory. These are exactly the same people who attack relativity, they have just happened to become obsessed with a different target. The two sets of posts are remarkably similar in style and titling (Cantor wuz wrong! Einstein wuz wrong! Cantor/Einstein committed fraud! Disproof of relativity/diagonal argument. Who does that refer to? Many undergraduates and high school students are exposed to Cantor's ideas every day, and absorb them successfully. It's not that they're inaccessible. It's that they require you to suspend your intuition in favor of deductive logic. But that is precisely where all the anti-Cantorians begin jumping up and down. Well, some of them. Others are obsessed with the diagonal proof of uncountability of the reals, and think they can modify it to show it is wrong. So as soon as you say there are sets which are in one-to- one correspondence with their proper subsets as a true statement, you are no longer talking about the same people I'm talking about. - Randy When you start a new thread, it might be nice if you told us what you're talking about, or included a reference to the other thread (saying the other thread doesn't help, because there are hundreds of threads). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada Robert Israel, The topic about what we are discussing is pretty clearly stated when I started the Thread, and that is precisely the subject : (Requesting for) Non-trivial Examples of Singularities (See Msg 0 of this Thread : dt 3/18/05 : http://mathforum.org/kb/message.jspa?messageID=3701927&tstart=0) I had also posted the same query in alt.math.undergrad, and that is what I meant by other thread - to which Ali Taghavi has replied, and I referred to his reply. I agree that I could have stated more explicitly about the other Thread, but since practically, almost no tech discs have taken place at all in either, I just mentioned in the passing so that Ali Taghavi realises that I have seen his reply. It would be more productive to get actual examples for the subject matter of this thread ; let us change gear to more technical matters pl. .. Those (myself included) reading this in sci.math wouldn't see the postings in mathforum and alt.math.undergrad, so especially when you don't include a specific reference or link to those postings we can hardly be expected to know about them. Since now you did include a link to the Mathforum posting, I'll copy some of that for the benefit of sci.math readers. Removable |Singularity. |I would like to know of 3 or 4 more non-trivial examples of Removable |Singularities (something concrete like sin(z)/z preferred ; if the examples are |very abstract, then they need more explanations for most of us to comprehend.) Roll your own: take any analytic functions g(z) and h(z) and any point a removable singularity at z=a. |2) Many books give (z-a)^(1/2) as a Branch Singularity Example, for eg with |say, a = 3. |Can we generally also say that ANY (z-a)^(1/n), n being a positive even |integer, is an example of Branch Singularity at z = +/-a ? As long as 1/n is not an integer, (z-a)^(1/n) will have a branch point at z = a (not at z=-a though). There will be finitely many branches if n is rational, infinitely many if n is irrational. |2-b) If a = 0 above, will f(z) = z^(1/n), again n being a positive even |integer, be called an Isolated Singularity ? No, because no branch of f(z) can be analytic in {z: 0 < |z| < epsilon}: you'll always have a branch cut. |3) We know that if f(z) = {u + iv} is analytic in a domain, let us call the |domain as REG, then, u and v are harmonious in REG. Is the reverse true ie, |if u and v are harmonious in REG, can we say that f(z) = {u + iv} is analytic |in REG ie, does the above meet Necessary Condition also ? The word is harmonic, not harmonious. The answer is no. They have to satisfy the Cauchy-Riemann equations. |If not, I would like to know of 3 or 4 non-trivial, examples where it is only |sufficient but not necessary. (Again, it would be highly appreciated and |understood better, the less abstract these non-trivial examples are.) Take any pair of non-constant harmonic functions u and v. If by some miracle your u + i v is analytic, replace u by 2 u. |4) f(z) being analytic is only a sufficient condition to imply that f(z) is |continuous, but not the reverse. Most books give f(z) = conjugate (z) as an |example. |I would like to know of 3 or 4 more non-trivial, examples where f(z) being |analytic is only sufficient to say that f(z) is continuous but not |necessary. Again, the less abstract these examples are, the better it is to |convey the examples and comprehend them. Take any polynomial in z and conjugate(z) that is not just a polynomial in z. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada X-Spam-This: SpamCopies@YahooGroups.Com (Sorry for tardy reply but the mouse on my computer stopped working so I had no way to do hardly anything online until today.) That's correct but irrelevant. I never made any such claim. My claim was that a machine might sit computing for a very long time, churning out successive digits as time goes on, so that the longer the machine is allowed to run the more digits are generated. Mathematically we say that there's an infinite sequence of digits, where the machine computes an initial segment that grows with time to include every element of the digit sequence eventually. (For each digit of infinite sequence, that digit will eventually be reached.) If so, the infinite sequence is computable. Alternately, a machine might be given as input a number which specifies how many digits are desired and the machine the generates exactly that many digits. The restriction is that given two different inputs m < n, the digits generated from m exactly match the first m digits generated from n. The union of all those overlapping initial sequences defines the mathematical infinite sequence. Likewise that infinite sequence is computable. Alternately, a machine might be given as input the number of digits of precision to use internally, and then generate whatever number of digits of output are forced by those internal computations. Two restrictions: (1) As internal precision gets arbitrary large, output Initial-segment match as above. Likewise union of overlapping initial segments is mathematical infinite sequence which is computable. Horsefeathers! All of the above three definitions of computable infinite sequence are equivalent, each of which is equivalent to the definition you give below: That is yet another definition, equivalent to the three I gave. Do you disagree that given any machine satisfying any of the four definitions (i.e. computing single digits or initial segments or growing unbounded initial segment), it's possible to build a machine that emulates the first machine internally to produce an overall machine satisfying any of the remaining three definitions? Actually not. The third definition, namely determine beforehand the internal precision, then run it and get some smaller number of output digits, with internal precision increased signficantly every few years yielding corresponding significant increase in output sequence length. Only an idiot would spend large amounts of computer time just to compute a single million(-some-odd)th digit and discard all the previous digits, then go through even more work to compute a single million(-some-odd-plus-one)th digit again discarding all previous digits, etc. Note also that there is no known way to generate the nth digit of pi without in fact generating all or most of the earlier digits at the same time, or at least generating all the information necessary to generate those earlier digits. In fact usually the calculations are done in binary in the first place, generating all or most of the bits in an approximation to pi, then that approximation is used to generate the decimal digits corresponding to it. If you really only want a single digit, I suppose you could generate the binary approximation and then use a finite-state modular algorithm to directly generate just the single desired digit without actually generating all the earlier digits. posting-account=L1sZYA0AAADNLE8xURzoFV6m50rxazp3 [snipped various TM definitions of computable numbers, which are basically ok] generate and the You need to catch up with new developments in the last 10 years. Google for BBP algorithm (Bailey, Borwein and Plouffe) which generates the nth hexadecimal digit (hence, trivially, binary digit) of Pi (and other polylogarithmic transcendental numbers) directly and efficiently, without having to compute all preceding bits. This method has been used to compute the trillionth bit or so of Pi, whereas Pi's full expansion has only been explored up to 50 billion digits or so (by Prof. Kanada, in Japan). More recently, Plouffe found a similar way to compute the nth decimal digit of Pi also. This method is actually less efficient (in terms of computation time) than the best traditional methods (which compute all preceding digits), but it does have the advantage that it needs very little space to hold intermediary results (unlike traditional methods where the entire sequence of digits has to be available at some point). Michel. If anybody stands for the traditional in geometry, it's Euclid. Euclid did have what he called a definition for points, though it doesn't make much sense as a definition for us moderns. Elements, Book 1, Definition 1: A point is that which has no part. (see e.g. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada But I never opposed that. I simply said that I felt no need to conform to definitions that I agreed with, and never said that a mathematical definition is what I was after. why it was long as everywhere. as Again, for the thousandth time, the problem is number of elements, not infinite. One of them is I believed that some of mathematicians HERE did because when Tony wanted to claim that the set of integers had more elements than the set of even integers they insisted that they had the same number of elements on the basis that number of elements just meant cardinality. Which is the point I entered the discussion. Good. Now solve the problem by telling us what you mean by number of elements. In particular when do disjoint sets have the same number of elements if they ever do. Bob Kolker other access to mind. not These are all the things we have to figure out by examining inner experience -- the best we can. You commented in another post that you wanted to re-read Kant. Looking at his philosophy of mind might really help you. You might want to read some stuff on the topic written by Andy Brook. He's a prof here and has some really interesting ideas on these sorts of questions, and he's a materialist nice introductory text with Rob Stainton, but it might be too basic for you. Kant was very bright and Very Wrong. There is no such thing as a synthetic a priori judgement. All synthetic judgements are a posteriori. He was dead wrong about Euclidean Geomtry which he consider apodactic being based on necessary synthetic a priori judgements. The existence of non-Euclidean geometry indicates Kant was mistaken. Bob Kolker Did a quick search on Kant's Ding-an-sich, and confirmed my understanding: it's the reality out there that we cannot know directly but only means of our senses. Kant showed that what we know of the external world, ie our perceptions of it, is not the thing-in-itself, which he defined as the thing apart from any relations between it and other things, including the relation of being known. In this respect, he corrected earlier notions that what we could know the world directly. If I understand him properly, he realised that we use our senses just as we use our instruments in order to know, and that our method of knowing something determines what we know of it. However, it appears he was inconsistent in his use of the term; and also used the concept to decide what was knowable (ie, what things we could know as our perceptions of them), and what was not, basing his choices on the state of scientific capabilities of the time. At this point, I realised that googling isn't good enough - all this method yields is summaries and quotations, which aren't enough. So, will have to read more. IIRC, he introduced the notion of a prior synthetic postulates in order to solve the problem of how our perceptions of things could nevertheless give us true knowledge of the Dinger-an-sich. Our perceptions give us only the Erscheinung of things, and not the things themselves, and the only check on the truth of the Erscheinungen are themselves Erscheinungen, so there has to be some guarantee that (or at least good grounds) for accepting some Erscheinungen as somehow primary and therefore the bases of deciding the truth of others. IMO, this is a very clever variation of Descartes' method of doubt, but like Descartes, Kant made a mistake. (Erscheining translates as appearance, which in English as well as in German may also mean illusion, but that's not what Kant meant: he use the term in Plato's sense. One could take his philosophy as a revision of Platonism.) Anyhow, that's where I'm at. at And this relates to his philosophy of mind ... how, precisely? Space and time do not live in our heads, they live Out There. Kant glorified the subjective. Bob Kolker On Wed, 23 Mar 2005 20:30:23 -0500, robert j. kolker Spatial dimensionality lives in our heads, Bob. Space is Out There. We move about in it. Spatiality (an abstraction from Space) is in our heads. Time is Out There. We perceive chenge in things external to us. Temporality (an abstraction from Time) is in our heads. Bob Kolker On Thu, 24 Mar 2005 11:37:57 -0500, robert j. kolker Yoda say: progress Bob makes when barking like a dog he doesn't. Is time really out their? Or only change? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Actually, change is the primary and time is the derived concept. Which leads to a problem, that mentioned by Heraklitus. To see a change is to compare changless things. Change has its associated problems. Suppose we did not have any memory at all. Could we be aware of change? Probably not. Memory which straddles Out There (impressions of things Out There are the content of memory) and In Here is what we need to be aware of change at all. If we had no memory we would be timeless. There would only be Now. In any case the elements of time (interval and order) have connections to thengs external to us. There is an objective aspect to it. But there is a made up aspect too (what Kant would call an a priori synthetic). How do we know when two time intervals are equal? We can't pick them up and lay them next to each other. Bob Kolker On Thu, 24 Mar 2005 14:04:12 -0500, robert j. kolker All we really know, Bob, are differences or contradictions and we have to work out whatever else we can know in terms of them and not what we might like to imagine is out there. Out whose? -- Giuseppe Oblomov Bilotta E la storia dell'umanit.88, babbo? Ma niente: prima si fanno delle cazzate, poi si studia che cazzate si sono fatte (Altan) (And what about the history of the human race, dad? Oh, nothing special: first they make some foolish things, then you study what foolish things have been made) He meant Out There. Bob Kolker Oh, I know. Just poking fun at them, since they seemed so attentive to my typos, I thought they would rather watch out for theirs. Or are they practising not what they preach? -- Giuseppe Oblomov Bilotta Axiom I of the Giuseppe Bilotta theory of IT: Anything is better than MS On Thu, 24 Mar 2005 22:34:03 +0100, Giuseppe Bilotta Hell I don't mind practicing typos, just preaching typos. My original objection only came after three consecutive typos which made your point completely indecipherable, in reaction to which you began to preach typos as some sort of gospel. Ok, I'll reread some Kant. Real Soon. :-) Notice, Dave, how Albert doesn't address the issue, but instead insults you. That's his techqniue when you say something sensible that he doesn't agree with, because to agree with it would be to admit that he was wrong, or worse, made a mistake (maybe even -- horrors! - a silly mistake.) Of course your statement is a definition. It shows how to use infinite in specified context. But Albert doesn't like specific contexts, and even less does he like specific examples, because that would mean he has to admit that his vague notions about what a word means might actually be nonsense. See how he attacked me for providing an example of how to use accurate and precise -- he said I was obviously incapable of constructing a general definition of these terms. But he was unwilling to provide one himself. Then he attacked me for a comment in which I assumed that he used metaphysical in the sense given in my dictionary, and he called me a ing idiot. I replied once more in a civilised fashion, and he insulted me again. There's no point. He suffers from some pathology of the ego, and lashes out in rage whenever he is caught making some kind of mistake. He thinks of debating and argument in terms of winning and losing: note his unseemly crowing like a demented cockerel whenever he thinks he has won a debating point. The last thing he wants to do is learn anything from other people. Many years ago, during one of the pomo invasions of sci.physics, somebody brought up the terms Apolonian argument and Dionisian argument (supposedly these terms exist since antiquity). In a nutshell, an Apolonian argument is a tool, used to arrive at some conclusion. A Dionisian argument, on the other hand, is its own purpose, those who engage in such don't care about learning what's right or wrong, what's true of false, the sole goal is winning (though it is a quite silly notion of winning). The only worthwhile reason to respond to practitioners of Dionisian argument is when you're not really talking to them, but through them, i.e. when your goal is not to convince your opponent (a waste of time) but to educate bystanders. If that's what you're doing in this argument, fine. Else, I suggest you read my sig.:-) Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same [...] Else, I suggest you read my sig.:-) Ta, will heed. Yes I concur except that the goal is neither to convince opponents (as if there could be opponents to universal truth in science) nor to educate bystanders but to discover what is universally and necessarily true and why, something alien to mathematikers and physikers alike. So why did you post here at all in the first place? -- Giuseppe Oblomov Bilotta I weep for our generation -- Charlie Brown On Thu, 24 Mar 2005 00:09:54 +0100, Giuseppe Bilotta I post where ignorance is rife and pretentious is not. You have brought both ignorance and pretentiousness to the NG. Well done, Young Zick. Obi-wan says: You will be the death of me yet, Young Zick. Yoda says: Ignorant and Stupid is Zick. Dumber still get he cannot. Bob Kolker On Thu, 24 Mar 2005 11:31:46 -0500, robert j. kolker Now you're back to barking like a dog again instead of thinking, Bob. May the force be without you, Bob, unless you get back to thinking. It seems to me I recollect as how Obi-wan discouraged Luke from thinking. Terminally bad advice when you have to use words instead of light sabers. Oh I dunno, Bob. I might surprize you yet. We already know how dumb you can get. What we don't know and would like to find out is how undumb you might become. Yes, maybe Dave will kiss you boo-boo, Wolf. Hide there behind his skirts until he notices you. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy On Wed, 23 Mar 2005 11:13:48 -0500, Wolf Kirchmeir Not exactly, Wolf. Albert suffers from a desire for universal and not parochial knowledge, a malaise you and Bob don't suffer. It was pretty obvious right from the start that Albert Wagner and Lester Zick are just trolling around (I'm still not sure about Tony; he's either a much more refined troll, or a truly interested, but blatantly uninformed, chap; I'll give him the benefit of doubt and keep answering his posts more seriously). But I must say that taking their bait and feeding it back is an exceedingly funny (in a grotesque way, possibly) way to spend the time when taking a break from serious work :) Maybe we should cut all the groups out from the follow-up chain, though. -- Giuseppe Oblomov Bilotta Can't you see It all makes perfect sense Expressed in dollar and cents Pounds shillings and pence (Roger Waters) On Wed, 23 Mar 2005 17:57:48 +0100, Giuseppe Bilotta Sure. There are much easier ways of trolling around than defining mathmatical concepts for mathematikers they can't quite seem to define for themselves. Unless of course you're really stupid enough to think that your usual spelling for Tolstoy is really usual. Sounds attractive, but you'll just get them discussing the universal tautological properties of the empty set. -- Richard Herring If one could convince them to test the tautological properties feature. /BAH Subtract a hundred and four for e-mail. Sure as hell can't expect mathematikers to discuss the universal properties of anything, Red. On Wed, 23 Mar 2005 09:56:05 +0100, Giuseppe Bilotta The only universal definition for circle I can think of offhand is a curve with the ratio of pi/2 between two points. A curve is defined universally as a line which doesn't occur between two points. On Tue, 22 Mar 2005 23:16:04 -0500, Wolf Kirchmeir There is a distinction however where rigorous definition is specifically used to mean undefined. The undefined terms you refer to above are merely incidentally undefined and not defined to mean undefined, which is what infinite is taken to mean. Common usage of the domain in which it is to be used (whatever that means) is irrelevant to explicit lack of definition. So what are you saying here, Dave, that you don't know what infinite means or cardinality means but you propose to define them anyway in terms of how they're applied when you don't know what they are to begin with? I think you might just be better off assuming a defintion which can be applied in the manner you propose than claiming your usage defines them. Usage can define lots of things.You can illustrate redness in red things quite adequately without claiming to define red. In a post on a collateral thread Jesse admits the concept of infinity has no apparent prior definition, from which it is easy enough to conclude that infinity means lack of definition or undefined. That does not indicate how the lack of definition can be used, but it does define infinity in universal terms. Talk to your grade school teacher who either did or should have warned you about the universal sufficiency of ostensible usage definitions. Of course this just assumes you have a class of objects to begin with when you can't even say what the criterion for object selection is. And all I'm saying is that said-when definitions are inherently particular in nature and not universal because they rely on your own classification criteria. Hell, you don't even say is-when so the definition can be traced back to you. You just use the said-when version of definition so you can always claim you're just citing authority, but the problem is that you're the one supposedly doing the definition and the authority you cite isn't here to be cross examined on his definition. one-to-one but not onto or admit that this is just how the concept, whatever it means, is used in set theory. we have a definition? And I can't believe you really think this is a serious basis of mathematical definition. On Wed, 23 Mar 2005 10:02:17 +0100, Giuseppe Bilotta My defition of infinite is more than one typo in a single word. Dave, when you have finally made a passing grade in Jr. High English, then you should check out Ludwig Wittgenstein. You are /way/ over your head in attempting to lecture anyone on proper sentence construction. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy The short and sweet version is: a qualifier of nouns. No. That /was/ your turn. But thank you for an excellent example of nonsense. It's a shame that you know so little of English grammar and syntax. What does 'itself' refer to? 'Infinite'? If so then this 'definition' is recursive and therefore meaningless. I have already pointed out that you have provided no definition of the word 'infinite', but rather just used it as an adjective without prior definition. Why would you say that. It seems that I am the one insisting on a prior definition of terms. You are content to simply point and grunt like a chimp and assume that what you point at constitutes a definition. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy In the above context infinite is a property of sets. An adjective describing some (but not all) sets. In this context infinite means of infinite cardinality. So a set has infinite cardinality if and only if it can be mapped one to one onto a proper subset of itself. Bob Kolker On Wed, 23 Mar 2005 10:28:39 -0500, robert j. kolker Which is just a derivative consequence of its lack of definition, Bob. I tried to warn you not to attempt this, Bob. But you had to try anyway, didn't you? Infinite cannot be defined as an infinite property of anything. In _MIND AND NATURE: A Necessary Unity_ Gregory Bateson describes this faulty logic: --------------------------------------- A common form of empty explanation is the appeal to what I have called dormitive principles, borrowing the word dormitive from Moli.8fre. There is a coda in dog Latin to Moli.8freÍs Le Malade Imaginaire, and in this coda, we see on the stage a medieval oral doctoral examination. The examiners ask the candidate why opium puts people to sleep. The candidate triumphantly answers, Because, learned doctors, it contains a dormitive principle. We can imagine the candidate spending the rest of his life fractionating opium in a biochemistry lab and successively identifying in which fraction the so-called dormitive principle remained. A better answer to the doctorsÍ question would involve, not the opium alone, but a relationship between the opium and the people. In other words, the dormitive explanation actually falsifies the true facts of the case but what is, I believe, important is that dormitive explanations still permit abduction. Having enunciated a generality that opium contains a dormitive principle, it is then possible to use this type of phrasing for a very large number of other phenomena. We can say, for example, that adrenalin contains an enlivening principle and reserpine a tranquilizing principle. This will give us, albeit inaccurately and epistemologically unacceptably, handles with which to grab at a very large number of phenomena that appear to be formally comparable. And, indeed, they are formally comparable to this extent, that invoking a principle inside one component is in fact the error that is made in every one of these cases. ------------------------------------- -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Albert Wagner says... The definition of infinite has nothing in common with what Bateson is talking about. There is nothing circular about saying A set S is infinite if there exists a bijection between S and a proper subset of S. Bijection is defined without any mention of infinity. Proper subset is defined without any mention of infinity. Therefore, defining defining infinite in terms of bijection and proper subset is not circular. -- Daryl McCullough Ithaca, NY On 23 Mar 2005 08:55:46 -0800, stevendaryl3016@yahoo.com (Daryl It's not so much circular, Daryl, as not a definition. It's just a list of consequences for infinity when applied to set operations. It's more an illustration of what infinity means in a particular instance by list of ostensible consequences. What is a definition? -- Giuseppe Oblomov Bilotta Can't you see It all makes perfect sense Expressed in dollar and cents Pounds shillings and pence (Roger Waters) On Wed, 23 Mar 2005 21:49:16 +0100, Giuseppe Bilotta A list of predicates stated in relation to one another. In the case of the term infinite the only predicate is specific lack of definition. Things defined as infinite lack some specific character in certain respects which have certain implications for operations or other circumstances specified in terms of that specific character.. And this would be a definition of definition? Pretty much contradictory. (And I'm not even going into what list, predicate, relation mean ...) -- Giuseppe Oblomov Bilotta I'm never quite so stupid as when I'm being smart --Linus van Pelt On Thu, 24 Mar 2005 13:04:09 +0100, Giuseppe Bilotta So you're more of the definition is when something is defined . . . school of thought? Lester Zick says... My definition is a definition in the same sense as the following definitions there does not exist integers y and etc. These are definitions, in the sense that they explain the meaning of are already assumed to be understood. The definition of infinite has exactly that character: proper subset of x such that there exists a bijection between x and y. -- Daryl McCullough Ithaca, NY On 23 Mar 2005 11:58:51 -0800, stevendaryl3016@yahoo.com (Daryl Well I agree, Daryl, except what you seem to be defining is x under various conditions. Just saying you're defining what is on the left of what's on the left overall that's being defined in general terms. So the most immediate term. And in your example that means x eveness, x primacy, x infinite etc. Which is completely different from defining eveness, primacy, and infinite in general terms. This is exactly why said-when and is-when ostensible listings aren't general definitions. The best you can say is that you're describing set infinity but not infinity itself in general terms. For that you have to say infinite is or infinite means . . . And the only answer is undefined. This reminds me of a grade school teacher I had who couldn't see that 4.499 rounded to 4 and not 5 because she thought you began at the rightmost fractional digit and rounded successively up instead of at the leftmost fractional digit. She just didn't get it. (And no I'm not still in her class as many here would like to pretend.) Lester Zick says... No. x is a variable. What is being defined is the *predicate* x is even, or x is a perfect square, etc. -- Daryl McCullough Ithaca, NY On 23 Mar 2005 16:17:02 -0800, stevendaryl3016@yahoo.com (Daryl Saying x is a variable doesn't mean you aren't defining it. What you're defining is x eveness, x perfect squareness, etc. You could just as easily say that x eveness or x perfect squareness is a variable that you're defining. If you want to define eveness just say eveness is . . . Yes, it does. No, that's not what is being defined. The point of a (free) variable in a definition is that you can substitute any constant term (of the right type) for the variable to get a true statement. So in you can substitute any constant term of type natural for x and get a true statement: etc. -- Daryl McCullough Ithaca, NY On 24 Mar 2005 07:49:37 -0800, stevendaryl3016@yahoo.com (Daryl So if you aren't defining x, what makes you think you are defining even, prime, etc? Lester Zick says... Because if you know what multiple and integer and 2 mean, then you know what it means for an integer to be a multiple of 2. If you know what it means for an integer to be a multiple of 2, then you know what it means for an integer to be even, since x is a multiple of 2 and x is even mean the same thing, for all x. -- Daryl McCullough Ithaca, NY On 24 Mar 2005 09:59:56 -0800, stevendaryl3016@yahoo.com (Daryl So why even bother with x? Just go right to the subject of definition. Lester Zick says... In my experience, the use of free variables in a technical definition is much clearer, once you get used to that style. It's a replacement for the use of pronouns, which can be confusing. -- Daryl McCullough Ithaca, NY On 24 Mar 2005 11:34:01 -0800, stevendaryl3016@yahoo.com (Daryl The problem then is whether it is a technical definition or not. Certainly what you're defining is not necessarily technical because you're drawing on a specialized subset of meaning. If you take a term which is used to indicate uncountable and apply it to counting, a conflict results. x is the dummy variable of a schema. It is eliminated when a particular value is substituted for it. Bob Kolker On Thu, 24 Mar 2005 14:36:22 -0500, robert j. kolker Or we could just skip right to the subject of definition which for infinite certainly doesn't include matching or not matching. Negative integers too? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Albert Wagner says... Yes, -2 is a multiple of 2, since it is equal to 2 * -1. -- Daryl McCullough Ithaca, NY OK. Chalk up another mathematical definition that is different from common usage. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Yes and zero too. Divisibility by 2 is all that matters. Bob Kolker Daryl's phrase was 'multiple of 2', which by my understanding could not be negative. I would have preferred 'divisibility by 2' to keep signs straight. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy What is -4*2 ? Bob Kolker X-AUTHid: wagners5 : Lester Zick says... : Because if you know what multiple and integer and 2 mean, then : you know what it means for an integer to be a multiple of 2. If you : know what it means for an integer to be a multiple of 2, then you : know what it means for an integer to be even, since x is a multiple : of 2 and x is even mean the same thing, for all x. The philosophs seem to be claiming that you cannot define an adjective. After all, an adjective must describe some object, and you apparently must define the object before you can define the adjective. However most adjectives can be applied to many, many different objects and their meaning does not depend on the exact type of object to which it is applied. Defining the adjective does not require you to define the objects to which it applies first. For example, x is empty if x contains nothing. This defines a meaning of x is empty, assuming of course that x contains nothing has previously been defined. It applies to any x to the extent that x contains nothing is meaningful. Is a rock empty? Is a rock full? Of course if you do not like the if notation you can simply remove the x is and the if empty : contains nothing even : a multiple of two self-conscious : aware of oneself as an individual or of one's own being, actions, or thoughts. infinite : able to be put in a one-to-one correspondence with a proper subset of itself Clearly these adjectives do no apply to all objects. For example, asking if a rock is even is the same as asking if a rock is a multiple of two. Only integers can be multiples of two, and rocks are not integers. Likewise only sets have proper subsets, and a rock is not a set, so rocks are not infinite either. Stephen And here I am still waiting for the definition of real mathematicians. Or you could just skip right to definition of the adjective. Or we might just vouchsafe that empty means containing nothing. Okay, better. The question is whether the definition vouchsafed is self consistent. Your definition for infinite produces contradiction between proper subset and contained in as noted by Bob earlier today. Well one might argue that rocks can be a multiple of two. But leaving that aside we have a conflict between sets and proper subsets when infinity is assigned to the countability of sets and subsets. X-AUTHid: wagners5 : Okay, better. The question is whether the definition vouchsafed is : self consistent. Your definition for infinite produces contradiction : between proper subset and contained in as noted by Bob earlier today. This makes no sense. There is nothing in the above definition of infinite about contained. There is no contradiction. The definition of contained makes no reference to bijections and the definition of bijection makes no reference to contained. There are sets that can be put in a one-to-one correspondence with proper subsets of themselves. The natural numbers are an example. What contradiction do you think this leads to? Stephen Whether proper subsets defined in finite instances can remain proper subsets in infinite instances according to one and the same definition for proper subsets. A template for true statements is not a definition. (BTW. Is -4 even?) -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Albert Wagner says... Yes, it is, at least for extentional predicates (which are the ones that are most important for mathematics and science). Yes. -- Daryl McCullough Ithaca, NY Yes. It is divisible by 2. -4 = -2*2. Bob Kolker Yes. I thought so. Then Daryl's definition above is incomplete. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy X-AUTHid: wagners5 : I thought so. Then Daryl's definition above is incomplete. No, his definition made no mention of natural numbers. There is no mention of natural numbers there. It applies to all possible values for x. It clearly applies to -4. -4 is a multiple of 2, therefore -4 is even. Stephen Apparently, mathematicians define multiple in absolute terms. In common usage there is no way that -4 is a multiple of 2. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy [...] Oh, yes there is, Albert. In common usage, -20 degrees is twice as cold as -10 degrees. I know it's physically nonsense, but that's the way people talk. I've heard' em say so myself. And the way people talk is common usage, no? If your balance sheet showed -$1,000,000 last year, and -$2,000,000 this year, you lost twice as much this year. Most people will say that, won't they? Not on the Kelvin scale. But soft! I merely just with you. Bob Kolker Oh, so you're one of those who doesn't consider negative numbers as numbers? No surprise you're unable to accept infinites in that same class. -- Giuseppe Oblomov Bilotta I weep for our generation -- Charlie Brown Historically negative numbers used to be called impossible or fantastical numbers. We still have a vestige of that. Sqrt(-1) is still referred to as imaginary although it is no more or less real (in the ontological sense) than so-called real numbers. I means what is real about a number that needs an infinite number of decimal digits to define it? Bob Kolker Daryl was using type natural, ie, natural numbers in his example. Natural numbers do not include the negative numbers. Some people exclude zero from the natural numbers (that's what I was taught in Austria), but that convention is apparently dying out. Integers refers to positive and negative numbers and zero. Rationals (any number that can be represented as n/m) includes integers, since any integer n can be represented as n/1. Etc. So, Albert, what is a definition? The base notion seems to be that a definition somehow points to the referent of the term. That's why I disagree that X is Y is the only acceptable form of a definition, since it is ambiguous, and needs to be unpacked, as one of my phil profs used to say. The form X refers to Y is usually OK, but it sometimes is equivalent to X means Y, which I exclude, since that merely asserts that two terms have the same meaning, and does not indicate what they refer to, so that at least one of the terms must be defined. And so on. IOW, definition is not at all a clear term, whether applied to the process or to the result. Often, the import of a definition is unclear unless and until one has examples of the term's usage, inclduying whatver inferenbcs can be drawn by using it. Thus, one could argue that a text on number theory, taken as a whole, is a definition of number. Or that Kant's Critique of Pure Reason is a defintion of Ding-an-sich (and other terms.) Etc. When Keats said Beauty is truth, truth beauty, which type of definition did he think he was making? One of my favourite definitions is A hole is to dig. Would that all definitions were that clear. X-AUTHid: wagners5 : Lester Zick says... : My definition is a definition in the same sense as the : following definitions Oh c'mon Daryl, can't you see that the obvious implication is that x is also even and that your definition says nothing more than Really. Haven't you learned anything yet. :) :) Stephen Nope. Not a definition. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Albert Wagner says... Yes, it is. -- Daryl McCullough Ithaca, NY Discussion, linux) I would ask him what he means by definition, but the prospect seems too damn circular for my tastes. -- Jesse F. Hughes Now 'pure math' makes sense as well as clearly it's a peacock game, where some of you see it as a way to show you as being highly intelligent and thus more desirable to women. -- James S. Harris -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy On Wed, 23 Mar 2005 10:45:38 -0600, Albert Wagner Albert, I have to sincerely congratulate you on this contribution to the description of a common failing among putative mathematicians. I'd just like to note though that your own Infinite cannot be defined as an infinite property of anything expresses the problem much better than the citation. Bateson's commentary has the advantage of putting the problem in clear historical relief however that reminds me of the meander of scientific progress throughout western history. And it also puts me in mind of the problem most people have with the idea of mind as a kind of dormitive principle but one which I think can be resolved I am not attempting to perform mathematical operations on the concept 'infinite' as if it were a number; You are. I refuse to accept your burden of proof. Nice sidestep, but it fails. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy You are the one sidestepping. I didn't ask you to perform operations on it. I asked you to give a definition. So the one sidestepping is you. (And I'm not even remarking how people (like you) have refused to acknowledge negatives and irrationals as numbers, not to talk about the imaginary *shiver* numbers.) -- Giuseppe Oblomov Bilotta Can't you see It all makes perfect sense Expressed in dollar and cents Pounds shillings and pence (Roger Waters) On Wed, 23 Mar 2005 17:52:57 +0100, Giuseppe Bilotta The definition for infinity is that it is undefined. Statement contradicted by any dictionary. -- Giuseppe Oblomov Bilotta Can't you see It all makes perfect sense Expressed in dollar and cents Pounds shillings and pence (Roger Waters) On Wed, 23 Mar 2005 21:50:07 +0100, Giuseppe Bilotta Lots of statements are contradicted by any dictionary, sport, just like your claimed most usual roman letter spelling for Tolstoy. ... Congratulations. Why don't you write a dictionary, while you're at it, instead of books noone will read? -- Giuseppe Oblomov Bilotta Hic manebimus optime On Thu, 24 Mar 2005 00:09:56 +0100, Giuseppe Bilotta Who is this guy noone? Is he defined or infinite? Ask him when he comes to ask you for an autograph on his copy. -- Giuseppe Oblomov Bilotta E la storia dell'umanit.88, babbo? Ma niente: prima si fanno delle cazzate, poi si studia che cazzate si sono fatte (Altan) (And what about the history of the human race, dad? Oh, nothing special: first they make some foolish things, then you study what foolish things have been made) X-AUTHid: wagners5 : Statement contradicted by any dictionary. My dictionary lists 7 definitions for infinity. Seems like a strange thing to do for a word that is undefined. Most of them refer to infinite. For infinite it has 9 definitions, none of which are undefined. The closest is unbounded or unlimited; perfect : God's infinite mercy The 5th definition is: 5. Math. a. not finite. b. (of a set) having elements that can be put into one-to-one correspondence with a subset that is not the given set This is the 1994 Webster's New Universal Unabridged Dictionary Stephen 5a strikes a little close to home however unless you prefer to say infinite is not when something is finite because finite doesn't mean defined so infinite can't mean undefined. Lester Zick says... That's an awkward way to put it, but yes---finite does not mean defined and infinite does not mean undefined. sin(1/0) is undefined, but it isn't infinite. -- Daryl McCullough Ithaca, NY And the limit of 1/x as x approaches zero from the positive direction is infinite and is defined. While 1/x is undefined when x equals 0. -- bz please pardon my infinite ignorance, the set-of-things-I-do-not-know is an infinite set. bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap In this case the term undefined means no predicate is defined whereas in the case of ordinary application of the term infinite only means countability lacks definition but not other predicates. But the term infinite itself still means undefined for any predicate to which it is applied. So why don't dictionaries say so? Is there some Cabal in action that wants to hide the Truth? (Of course, we'd better gliss over the fact that just because *you* can't grasp the concept of countability in case of infinity, it doesn't mean it's lacking. Just like people unable to accept the concept of square root for negative numbers only showed a limitation of theirs, not a lack of said square roots). -- Giuseppe Oblomov Bilotta I'm never quite so stupid as when I'm being smart --Linus van Pelt On Thu, 24 Mar 2005 22:34:04 +0100, Giuseppe Bilotta Probably because mathematikers haven't explained it to them. What I can't grasp in the case of infinity is finity which apparently you have no problem grasping. It's what makes you a mathematiker. Excuse me, why should mathematiker tell dictionary compilers about *your* definition of infinite as undefined? What, the ability to grasp more things than you? -- Giuseppe Oblomov Bilotta I weep for our generation -- Charlie Brown No. 1/x cannot approach a finite real number as a limit. Given any magnitude R x can be chosen with a small enough magnitued so that |1/x| compactification of the Real number space. Bob Kolker On 24 Mar 2005 07:34:57 -0800, stevendaryl3016@yahoo.com (Daryl Daryl, I addressed this early on. This is an awkward way to put it because it isn't a definition. There are many predicates in a given instance. In the case of set theory sets counting is one predicate. The term infinite applies to counting wherein you have undefined countability. This doesn't affect definition of other predicates but does mean that certain contradictions are possible where counting is the critical concept. And this makes it look like a concept defined in other respects is defined when it may not actually be. What remain defined are other aspects of the definition. I don't know how else to say it. Lester Zick says... I thought you were saying that you were using undefined and infinite as synomous. If you agree that they are not synonymous, then on what basis are you using the word undefined? -- Daryl McCullough Ithaca, NY On 24 Mar 2005 12:36:04 -0800, stevendaryl3016@yahoo.com (Daryl Well the basis is somewhat hyperbolic, Daryl, to emphasize what the term actually does mean when applied to predicates in isolation. I've never denied there is an epistemological problem generated by selective application of the term infinite or undefined to predicate complexes. The idea is to get people to understand what the idea implies in basic terms and then move on to its application in complex contexts. A lot of people want to simply ignore the general meaning of infinite so as to pretend there is no lack of definition whatsoever. A set is finite if and only if 1. It is empty or to one correspondence with {1, ..., n} A set if infinite if it is not finte. Bob Kolker On Thu, 24 Mar 2005 10:38:35 -0500, robert j. kolker Well Bob considering the perspicacity you demonstrated earlier in your previous analysis of set properties and conflicts among them in the case of infinities, I would really rather discuss these kinds of topics in those terms instead of bantering back and forth about this. Good work, Bob. Now, how would you generalize finite to apply to things other than sets? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Any other thing than sets? On quantities I could lay a definition of finite. As a generality, finiteness/infiniteness does not extend everywhere. There are no generally infinite or finite things. For some things the concept simply does not apply. Bob Kolker No, not *any* other thing, other finite(or infinite) things. Sorry, an example is useful only /after/ definition. One or the other does. All particular finite things can have their finiteness generally defined. All particular infinite things can have their infiniteness generally defined. Finite, infinite or indeterminate? What would you consider to be indeterminate? Would 'not finite' then refer to the pair infinite|indeterminate? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy [...] I was under the impression that one starts with examples. So an example that captures the meaning of the definition one has derived from examples is just fine with me. What's more, without an example, I can't tell what a general definition might refer to. Most words have several meanings, and all meanings are fuzzy round the edges, so to speak. What are your examples of, if no definition yet exists? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy How can you have a definition, if you have no examples to test it against? On Thu, 24 Mar 2005 16:48:03 -0500, Wolf Kirchmeir Well, Wolf, I have a definition for universal predicate but no examples to test it against. There are actually only two instances where no exemplary empirical test is possible: where the predicate is universal and where there is no predicate. On Thu, 24 Mar 2005 13:58:16 -0500, Wolf Kirchmeir One starts with examples as the basis of inference but then generalizes definitions in universal terms. If not we would be satisfied with simple lists of examples instead of definitions. Young children are when they learn to speak. The will do their own generalizing without any help from adults. Young children are born generalizers. Bob Kolker On Thu, 24 Mar 2005 15:09:03 -0500, robert j. kolker Adults are also born generalizers, Bob. It gets bred out of them somewhere along the way. You are talking about yourself again? Bob Kolker This is because the 8th definition of infinity is undefined, as is the 10th definition of infinite. Why would you expect to find the undefined in a book of definitions? -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Oho. So is the 256th definition of finite, of definition etc. Great work, Watson. -- Giuseppe Oblomov Bilotta I'm never quite so stupid as when I'm being smart --Linus van Pelt Ooooooo! *Shiver* numbers. Well, that settles it. Without acknowledging *shiver* numbers, how could I possibly define 'infinite'? LOL. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy [...] In Bilotta's sentence, Shiver is an apposition to the phrase imaginary numbers, not a an adjective qualifying numbers. wonder why he injected *shiver*. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy You're welcome. Because the idea of imaginary numbers (think about it: numbers whose square was negative!) seemed appallingly absurd to many. Just like the idea of negative numbers. Or of irrational numbers. In just the same way, you refuse the idea of transfinite (this is the proper term, actually) numbers. -- Giuseppe Oblomov Bilotta I'm never quite so stupid as when I'm being smart --Linus van Pelt On Wed, 23 Mar 2005 22:53:07 +0100, Giuseppe Bilotta You sure for a change? Actually you're always sure most especially when you're wrong. I wonder if the term transfinite might mean transcending definition? Just a thought since you're so sure the proper term is not infinite. Kinda makes me wonder though if infinite is not the proper term why everyone seems to have a definition for it. Transfinite numbers are the cardinals onto which you can map infinite sets. -- Giuseppe Oblomov Bilotta Hic manebimus optime iD8DBQFCQu0uvmGe70vHPUMRAqsxAJwPlgFcZTx8mTVARw/9X2r2OJRUjgCfTidL wTXfZ8DLz40hZyxoFJJ1ygE= =FHAi What about the transfinite ordinals? One of the cool thing about set theory is that I've never heard my professor say number during the lessons :) -- Giuseppe Oblomov Bilotta I'm never quite so stupid as when I'm being smart --Linus van Pelt Yeah! Really cool. A mathematiker that doesn't like the sound of the word 'number'. LOL. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy Hey, it just occurred to me that 'number' in English also means more numb! Appropriate, to say the least. -- Giuseppe Oblomov Bilotta They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. Benjamin Franklin Er, well, not exactly. Number has two pronunciatioans: with and withou English. Their technical name is homograph. Words that are pronounced the same but spelled differently are homophones. Poorly trained language arts teachers call them both homonyms, which is very confusing to students. Other examples: invalid, minute, wound, wind, bow, etc. Some pronunciation differences distinguish between noun and verb, as in house, record, etc. But the visual, spelling pun works. Oh, yeah, you're right. Is the mathematical number the one with the b? (Eye can sea the see allright ;)) -- Giuseppe Oblomov Bilotta They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. Benjamin Franklin this isn't a language group, so I'll refrain from pontificating further. :-) Eenglish spelling is so dum! Right? Rite? Ryt? Reyet? Bob Kolker On Thu, 24 Mar 2005 13:04:08 +0100, Giuseppe Bilotta Yeah, well, I guess that answers someones question. It's just a little tough figuring out whose. Much less who cares. Why don't you try answering a few more questions no one cares about? You seem to have a penchant for transcendental irrelevancies. Very funny, and one more sidestep. I'll give give you the wriggly easy way out: it's spelt Tolstoij, in Latin letters, usually. -- Giuseppe Oblomov Bilotta Can't you see It all makes perfect sense Expressed in dollar and cents Pounds shillings and pence (Roger Waters) Yes, but not in English. Sorry about that. Russian names are transliterated differently in different languages. It is a universal statement that reduces to a contradiction of a tautology. Haven't you been reading? Or are you a mathematiker? Bob Kolker That is the one question which Wagner and Zick repeatedly refuse to answer. They will not even attempt definitions or clarifications. All they produce are vain repetitions of nonsense and snotty insults. Bob Kolker On Wed, 23 Mar 2005 08:31:37 -0500, robert j. kolker See, Bob, this is classic Bob where you complain about Wagner and Zick and their refusal to answer questions or attempt definitions and clarifications and insult and demean them but carefully edit just enough of the original comment so Wagner and Zick cannot tell what term is referred to that you claim they didn't or maliciously won't define that justifies your contumely. Cite relevant text or shut up. I have stated repeatedly that 'infinity' is undefinable and therefore undefined. -- I know that most men, including those at ease with problems of the greatest complexity, can seldom accept even the simplest and most obvious truth if it be such as would oblige them to admit the falsity of conclusions which they have delighted in explaining to colleagues, which they have proudly taught to others, and which they have woven, thread by thread, into the fabric of their lives. - -- Tolstoy And others have defined the concept. Infinity. That which pertains to the concept of infinite. Infinite is a property possesed by some sets and not others. Infinity is also used (in some contexts) to refer to unboundedness where arbitrarily large distance between objects is possible. In projective geometry there are also points at infinity (look it up using Google). In the theory of real variables you have integrals of functions where the range of integration is from minus infinity to plus infinity. So you are quite wrong. Others have defined infinite in specific contexts and the definitions are meaningful in these contexts. Bob Kolker