mm-889 Subject: Kronecker symbol (a/b), periodic in b for Žxed a? I have been reading Henri Cohen¹s A Course in Computational Algebraic Number Theory and I have a question (at least one). On page 28 Theorem 1.4.9 he states: (4) a != 0 being Žxed (positive or negative), the Kronecker symbol (a/b) is periodic in b of period |a| if a = 0 or 1 (mod 4), otherwise it is periodic of period 4|a|. As a counter example of this, let a = 3. The period of (3/b) should be 12, but (3/2) = -1 and (3/14) = +1. I checked this using his GP/PARI CALCULATOR Version 2.2.8. Since (3/2) != (3/14) the period is not 12 as the theorem requires. Is the theorem false? What do you think? -Harry === Subject: Re: Kronecker symbol (a/b), periodic in b for Žxed a? |4) a != 0 being Žxed (positive or negative), the Kronecker symbol (a/b) is |periodic in b of period |a| if a = 0 or 1 (mod 4), |otherwise it is periodic of period 4|a|. | |As a counter example of this, let a = 3. The period of (3/b) should be 12, |but (3/2) = -1 and (3/14) = +1. Some authors leave (n/2) undeŽned when n is of the form 4k+3. Cohen and some others deŽne (8k+3/2)=-1, (8k+7/2)=+1. Under the other deŽnition, (3/b) is undeŽned if b is even, and by quadratic reciprocity, (3/b) = (b/3)(-1)^{(b-1)/2}, which is periodic (where it is deŽned) with period 12. I haven¹t checked, but it appears as though this claim may be generally true for the other deŽnition, and got slightly mistated here. Keith Ramsay === Subject: Re: Kronecker symbol (a/b), periodic in b for Žxed a? X-RFC2646: Format=Flowed; Original > |4) a != 0 being Žxed (positive or negative), the Kronecker symbol (a/b) is > |periodic in b of period |a| if a = 0 or 1 (mod 4), > |otherwise it is periodic of period 4|a|. > |As a counter example of this, let a = 3. The period of (3/b) should be 12, > |but (3/2) = -1 and (3/14) = +1. > Some authors leave (n/2) undeŽned when n is of the form 4k+3. > Cohen and some others deŽne (8k+3/2)=-1, (8k+7/2)=+1. Under > the other deŽnition, (3/b) is undeŽned if b is even, and by quadratic > reciprocity, (3/b) = (b/3)(-1)^{(b-1)/2}, which is periodic (where it > is deŽned) with period 12. > I haven¹t checked, but it appears as though this claim may be > generally true for the other deŽnition, and got slightly mistated > here. > Keith Ramsay You¹re right. Here is Henri¹s answer: === Subject: Re: Kronecker symbol (a/b), periodic in b for Žxed a? You are right, as stated the theorem is false. In case a is not congruent to 0 or 1 mod 4, then the symbol is periodic of period dividing 4|a| when one restricts to ODD numbers b. With even numbers, as you mention, it is false. Best, Henri Cohen === Subject: Casimir Force is not a Free Zero Point Virtual Photon Vacuuum Force I understand your math and the physics perfectly well. I don¹t understand why you are using the wrong math in this example and limiting yourself to a 1 dimensional problem when in fact 3 are required to model reality. Because, my model is true any dimension. Also the piston in the cylinder is the simplest situation and it is effectively 1D by symmetry since the important parameter is the movement of the piston in the cylinder. The same calculation works in all dimensions because the space-dimensions are LINEARLY INDEPENDENT. The key conclusion that, contrary to Hal Puthoff¹s frequent claims to science journalists, FREE virtual zero point photons without any j.A coupling to electrons and protons make exactly ZERO contribution to the measured Casimir force, which is not at all a direct zero point force, but is, rather, an electrostatic Van Der Waals force between real charges placed outside the vacuum in spatially separated charge neutral conŽgurations. The free virtual photon pressure is CONSTANT in the case of zero vacuum coherence and it obeys w = -1 with a positive ZPE density and an equal in magnitude but opposite in sign NEGATIVE PRESSURE. I prove this explicitly below using the standard continuum spherically symmetric Lorentz-invariant spectral density of Želd oscillators, which is an APPROXIMATION to the more exact discrete sum in a Žnite cavity problem. Contrary to QED, in GR you cannot subtract off even these constant free virtual photon ZPE densities, which have STRONG direct space-time warping power when not suppressed by MACRO-QUANTUM VACUUM COHERENCE out of which Einstein¹s gravity and the inertia of ordinary matter emerges. Note Hal Puthoff¹s category confusion here. I am NOT claiming that the Casimir force is in anyway related to the origin of gravity and inertia as Hal has suggested in numerous pop interviews with the media. Look the idea is very simple. Suppose you have a Žnite 3D box cavity, independent x,y,z axes with a little box of variable size X, Y, Z inside a big box of Žxed size Lx, Ly, Lz. The counting of the LINEARLY INDEPENDENT standing wave modes along each direction works the same way. Count the number of independent modes along each space direction exactly as I did. Assume a common short-wave cutoff a for simplicity. The EXACT total number of modes is then inside the little box N = NxNyNz = (1/8((X/a)(X/a + 1)(Y/a)(Y/a + 1)(Z/a)(Z/a + 1) But there is only 1 virtual photon per independent mode, and their virtual zero point energies add linearly even though the modes in different independent directions multiply. That is the Hamiltonian for a sum of independent Želd oscillators is H = Sum over all independent modes of hck(a*kak + 1/2) where we sandwich the LINEAR ENERGY OPERATOR between the VACUUM STATE |0> that is a product of the vacuum states |0>k one for each independent mode. Standing wave cavity modes in different directions are independent as well as those along the same direction with different nx, ny, nz and k<0|a*kak|0>k = 0 using 2nd quantization in Fock occupation number space in the traditional textbook way. a* creates a real photon, a destroys a real photon. The wave number for each independent standing wave cavity mode Želd oscillator, that in the vacuum has NO real photons is knx = nxpi/X where nx = 1 to X/a, similarly for y, z DO NOT CONFUSE THESE MODE INTEGERS WITH THE EIGENVALUES OF a*kak for the number of REAL PHOTONS in each independent cavity mode Želd oscillator k that is a standing wave obeying the cavity boundary conditions. Note that Nx = SUM of ALL the MODE INTEGERS nx from 1 to X/a (in sense of nearest least integer to X/a). Similarly for ny & nz. Each independent mode of wave number knx = nxpi/X contributes a zero point energy hckni/2 with a Žxed coefŽcient hcpi/2X, so that the key quantity is sum of nx from 1 to (X/a) which is (1/2)(X/a)(X/a + 1). Similarly for y and z and you simply ADD the separate linearly independent zero point virtual photon energies for EACH independent oscillator! So my EXACT calculation obviously works in 1D, 2D, 3D ... ND. This is simple and obvious. Now you¹re not just saying that Hal is wrong, you¹re trying to tell me that Planck, Casimir, Milonni and everything else QED has to say about the ZPF is wrong without any justiŽcation other than your desire to prove you¹re right and everyone else is wrong. I know better than to believe such nonsense. I never said Planck was wrong. Show me where Planck did such a calculation? Show what is wrong with the above simple exact DISCRETE counting? The standard CONTINUUM SPHERICALLY SYMMETRIC argument is: Density of oscillator states in 3D is 4pik^2dk per polarization. k^2 = kx^2 + ky^2 + kz^2 = (nxpi/X)^2 + (nypi/Y)^2 + (nzpi/Z)^2 FOR GLOBALLY FLAT MOMENTUM SPACE WITH EUCLIDEAN CONTINUUM GEOMETRY USING PYTHAGOREAN THEOREM. But this discrete sum is approximated by a continuous integral below and spherical symmetry is assumed. These two approximations introduce mathematical artifacts LIKE LORENTZ INVARIANCE! The total zero point energy density is then Integral (hc/2)4pik^3dk = 4pi(hc/8)k(MAX)^4 k(MAX) = 2pi/a a = short wave cut off Total zero point energy ZPE is then ZPE = 4pi(hc/8)(2pi/a)^4V > 0 in a spatial volume V. Note LINEAR DEPENDENCE OF TOTAL VIRTUAL PHOTON ZPE ON VOLUME V. Note that even in this calculation, the PRESSURE is - d(ZPE)/dV = - 4pi(hc/8)(2pi/a)^4 is CONSTANT independent of changes in volume V. AND IT IS NEGATIVE and it is - the ZPE DENSITY! That is w = -1! Therefore even in this continuum spherical symmetric model one gets the same effective result I get with my more exact purely discretum model with NO SPHERICAL SYMMETRY IMPOSED ADHOC! There is NO net ZPE pressure differential on any wall. You have exactly the same negative free virtual photon pressures on each side of the wall NO MATTER HOW YOU CALCULATE IT! If the formula used to derive the ZPF mode pressure presented by Milonni is wrong, then so is the spectral energy density of the ZPF and the Lorentz invariance of that density. No it¹s not as I just showed. To the extent that V --> inŽnity and a --> 0 which is what you need for globally žat continuum of special relativity with EFFECTIVE SPATIAL SPHERICAL SYMMETRY (no Žnite cavity walls of arbitrary shape), you get same essential result! That same essential result is that there is NO contribution to the Casimir force from purely free virtual zero point photons! The Casimir force requires the j.A coupling to properly compute as a Van Der Waals force between real charges. This is NOT a pure vacuum problem! That cannot be so. It has been veriŽed by more than 1 kind of experiment. QED is the best tested theory there ever was. IMO, test after test it has proven itself correct. It is even more battle tested and proven in day to day life than GR ever was. Not true! The domain of validity of QED is much more limited than anyone realized even though the accuracy of the QED measurements is very high and even though renormalization methods work very well - too well, although not even Richard Feynman understood why! Feynman told me that directly BTW! main_engineering -----Original Message----- completely. I count the number of modes exactly. There is one FREE virtual photon per mode (AKA Želd oscillator). Therefore, the calculation is exact and simple. === Subject: Final Step of the Prine Number Theorem, I can not understand by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8G1pHT13271; Could you turn to the page 366 of Introduction to the Theory of Numbers by Hardy and Wright? I posted a message on the proof some time ago, but I could understand that part. (It was indeed a matter of understanding the deŽnition of limsup.) But this part is unclear (to me). 1. Is O(1/eta_0) = o(1) because eta_0 is in [zeta, zeta + delta - alpha], and eta_0 -> oo as zeta -> oo? 2. The book says, By (22.15.8), V(eta) decreases steadily... where V(eta) increases. But then, at the next sentence, I am suddenly confused. Hence, in our interval, either V(eta_0) = 0 for some eta_0 or V(eta) changes sign at most once. How could this be true??? The most unsubtle part is, why at most once??? It¹s been almost two years since I bought this book, and Žnally, I am going to understand what I wanted to understand most (WITH YOUR to my school, I have it now.) Please help me understand this stuff. erdos fan === Subject: Re: Final Step of the Prine Number Theorem, I can not understand > 2. The book says, By (22.15.8), V(eta) decreases steadily... > where V(eta) increases. But then, at the next sentence, I am > suddenly confused. Hence, in our interval, either V(eta_0) = 0 > for some eta_0 or V(eta) changes sign at most once. How could this > be true??? The most unsubtle part is, why at most once??? V is piecewise continuous with jump discontinuities. In between its discontinuities it is decreasing. Its jump discontunities are increasing. We are supposing that V is nonzero on I. It cannot change sign except at its discontinuities. If it changes sign at x say, immediately to the right of x it is positive and decreasing. It decreases and remains positive until the next jump where it increases again etc. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: metric units by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8G1pHN13280; what metric units would you use to measure a tree? your shoes? a pin? the water in the swimming pool? water in a bath tub? the size of planets the ammount in an injection/? the thickness of a hair? the area of a stamp? === Subject: Re: metric units > what metric units would you use to measure .... The news group misc.metric-system might be a better place for this question. Ken Pledger. X-mailer: xrn 9.02 === Subject: Re: metric units Mail-To-News-Contact: abuse@dizum.com >what metric units would you use to measure >a tree? Ms (mega-seconds) >your shoes? kg (kilo-grams) >a pin? mhos >the water in the swimming pool? Kelvins >water in a bath tub? kg (kilo-grams) >the size of planets decimal degrees >the ammount in an injection/? Assuming that this is a current injection, Amperes >the thickness of a hair? meters >the area of a stamp? square meters -- Michael F. Stemper #include If we aren¹t supposed to eat animals, why are they made from meat? === Subject: Re: metric units > what metric units would you use to measure > a tree? Seconds.* > your shoes? Theres (nearly) always two, but that doesn¹t need any units. > a pin? kilogrammes* > the water in the swimming pool? degrees kelvin > water in a bath tub? still degrees kelvin > the size of planets# Ok, you got me: this is a fairly sensible question. That would have to be meters. Or meters cubed (you could use litres instead). Probably not meters squared.* > the ammount in an injection/? moles > the thickness of a hair? meters* > the area of a stamp? meters squared* Note 1, for the answers marked *, you might wish to use a preŽx, see http://www.unc.edu/~rowlett/units/preŽxes.html === Subject: Re: metric units >> what metric units would you use to measure >> the ammount in an injection/? >moles >> the thickness of a hair? >meters* Wait, have I got the line breaks wrong? Surely the hair is thickest right at the moles? === Subject: Re: metric units > what metric units would you use to measure > the ammount in an injection/? >>moles > the thickness of a hair? >>meters* >Wait, have I got the line breaks wrong? Surely the hair is thickest >right at the moles? Anyway, for measuring thickness of hairs a real engineer would use the eponymous unit RCH or BCH, appropriate choice to be determined by Želd conditions. Lee Rudolph === Subject: Re: metric units > what metric units would you use to measure > a tree? [Assuming height: meters] > your shoes? [centimeters] > a pin? [Assuming thickness: millimeters] > the water in the swimming pool? [cubic meters] > water in a bath tub? [liters] > the size of planets [kilometers] > the ammount in an injection/? [milliliters] > the thickness of a hair? [micrometers] > the area of a stamp? [square millimeters] === Subject: Help with Limit problem needing some assistance in answering. Here¹s the problem: f(x) = ((1/2) - ((x^2)/24)) g(x) = (1-cos(x))/(x^2) h(x) = (1/2) Find the limit as x -> 0 of (1-cos(x))/(x^2) using the Sandwich Theorem and f(x)< g(x) < h(x) I know that by the Sandwich Theorem, limit as x approaches 0 of f(x) = limit as x approaches 0 of h(x) = (1/2), therefore lim as x approaches 0 of g(x) is (1/2). My question is this: How come I do not get (1/2) if I multiply f, g, and h by x^2? If I do this and then take the limit of f and h as x approaches 0, I get the limit = 0. Any shedding of light would be appreciated. Will === Subject: Re: Help with Limit problem === Subject: Help with Limit problem If you¹re going to post same question is more than one newsgroup, then cross post to both newsgroups in a single post. It is easier for you and for us when you do so. Ie >f(x) = ((1/2) - ((x^2)/24)) >g(x) = (1-cos(x))/(x^2) >h(x) = (1/2) >Find the limit as x -> 0 of (1-cos(x))/(x^2) using the Sandwich >Theorem and f(x)< g(x) < h(x) >I know that by the Sandwich Theorem, limit as x approaches 0 of f(x) >= limit as x approaches 0 of h(x) = (1/2), therefore lim as x >approaches 0 of g(x) is (1/2). >My question is this: >How come I do not get (1/2) if I multiply f, g, and h by x^2? >If I do this and then take the limit of f and h as x approaches 0, I >get the limit = 0. No, you¹re taking the limit of x^2 f(x) and x^2 h(x) and the limit you get is the limit of x^2 g(x) = 1 - cos x and as lim(x->0) cos x = 1 directly without sandwich theorem is lim(x->0) (1 - cos x) = 0 Exercise: from lim(x->0) (1 - cos x)/x^2 = 1/2 show lim(x->0) (sin x)/x = 1 ---- === Subject: Re: Help with Limit problem > needing some assistance in answering. > Here¹s the problem: > f(x) = ((1/2) - ((x^2)/24)) > g(x) = (1-cos(x))/(x^2) > h(x) = (1/2) > Find the limit as x -> 0 of (1-cos(x))/(x^2) using the Sandwich Theorem and > f(x)< g(x) < h(x) > I know that by the Sandwich Theorem, limit as x approaches 0 of f(x) = limit > as x approaches 0 of h(x) = (1/2), therefore lim as x approaches 0 of g(x) > is (1/2). > My question is this: > How come I do not get (1/2) if I multiply f, g, and h by x^2? If you change the deŽnition of the function you should expect the values it takes to change, no? > If I do this and then take the limit of f and h as x approaches 0, I get the > limit = 0. Well, yes. Giving these new functions names, F(x) = x^2 . f(x) = x^2/2 - x^4/24 G(x) = x^2 . g(x) = 1 - cos(x) H(x) = x^2 . h(x) = x^2/2 All are deŽned at x=0 so we don¹t even have to take limits: F(0) = G(0) = H(0) = 0. What¹s the problem? -- Larry Lard Replies to group please === Subject: Re: Help with Limit problem > Here¹s the problem: > f(x) = ((1/2) - ((x^2)/24)) > g(x) = (1-cos(x))/(x^2) > h(x) = (1/2) > Find the limit as x -> 0 of (1-cos(x))/(x^2) using the Sandwich Theorem and > f(x)< g(x) < h(x) > I know that by the Sandwich Theorem, limit as x approaches 0 of f(x) = limit > as x approaches 0 of h(x) = (1/2), therefore lim as x approaches 0 of g(x) > is (1/2). > My question is this: > How come I do not get (1/2) if I multiply f, g, and h by x^2? > If I do this and then take the limit of f and h as x approaches 0, I get the > limit = 0. > Any shedding of light would be appreciated. Why should you get 1/2? Suppose that f(x) = g(x) = h(x) = 1. Then the limits of f(x), g(x), and h(x) at 0 are all equal to 1. But if you multiply all of them by x, the limits become 0. What¹s strange about that? Jose Carlos Santos === Subject: Re: the reported decline in wages there was an astonishing story in the NYTimes, I think, about the sudden rise in the use of offshore tax-shelters, the kind of places where the companies have no other business to speak of. mentioned was Bermuda, which is a British dominionate, I think. not mentioned was Accenture, the company that was Arthur Anderson, now in Bermuda, after the Enron Žacso. much of the personnel of Enron, itself, was absorbed into Warburg GBH, or what ever it is. > I am talking about it. Kerry is talking about it too. === Subject: Re: the reported decline in wages > there was an astonishing story in the NYTimes, I think, > about the sudden rise in the use of offshore tax-shelters, > the kind of places where the companies have no other business > to speak of. mentioned was Bermuda, > which is a British dominionate, I think. > not mentioned was Accenture, > the company that was Arthur Anderson, now in Bermuda, > after the Enron Žacso. > much of the personnel of Enron, itself, was absorbed > into Warburg GBH, or what ever it is. This is because our tax structure penalizes companies that do business in other countries. Not only do they have to pay taxes in the countries where they make sales, they have to pay US taxes on that same foreign income. This kind of tax penalty does not exist in most other countries, so most foreign companies have a signiŽcant competitive advantage in that regard. So OF COURSE US companies are going to reincorporate overseas - they have to to stay competitive. Our government all but forces companies to take this step to remain competitive, and now the government is trying to assault these companies for taking this step that they, the government, forced these businesses to take in the Žrst place! It¹s all a big sham and another step on the road to socialism which both parties are taking us down. Bush won¹t stop the march of socialism in this country, and he appears into to help it. Kerry, of course, would be the champion of it were he to be elected. *sigh* === Subject: Re: the reported decline in wages the alleged double taxation was also addressed China was named as one of the newer havens. what we really need is a tariff on oil, while it is still possible to do it, a hand¹s breadth from (or into) Hubbert¹s peak, > much of the personnel of Enron, itself, was absorbed > into Warburg GBH, or what ever it is. > This is because our tax structure penalizes companies that do > business in other countries. Not only do they have to pay taxes > in the countries where they make sales, they have to pay US taxes > on that same foreign income. This kind of tax penalty does not > exist in most other countries, so most foreign companies have a > signiŽcant competitive advantage in that regard. So OF COURSE > US companies are going to reincorporate overseas - they have to > to stay competitive. > Our government all but forces companies to take this step to > remain competitive, and now the government is trying to assault > these companies for taking this step that they, the government, > forced these businesses to take in the Žrst place! --Give Earth a Trickier Dick Cheeny -- out of ofŽce, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rwgrayprojects.com/synergetics/plates/Žgs/plate02. html === Subject: Re: the reported decline in wages more talk? There Is No ŒUpswing¹ in the Swing States (PDF on http://larouchepub.com) Iraq: Moral Authority Is Greater Than Military Might Neo-Cons Knee Deep in Caucasus Provocations CIA Nominee Goss¹s Lies Threaten National Security (Part 2) > I blame the worker. --Chair Man George! http://tarpley.net === Subject: theory of geometric algebras tell me === Subject: Re: theory of geometric algebras > tell me I¹ve been working on a geometric algebra based on A.N. Whitehead¹s book A Treatise on Universal Algebra. It¹s an algebra of projective geometry. You can download a text from http://homepage.ntlworld.com/stebla/Whitehead.html . === Subject: Re: theory of geometric algebras > tell me Sometimes geometric algebra refers to Clifford algebra. === Subject: Re: theory of geometric algebras > tell me Try looking it up on Google, you will undoubtedly Žnd at least something. Dave === Subject: Re: Simple math question. Help! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8G3KQE21289; >>You want to set up a paddock for horses. It is required that each horse >have >>10 metres squared for each horse. There are 15 horses. she wants a >square >>paddock. What will the dimensions of the paddock be? How much fencing >will >>you need? >> >>If you answer, show the detail please. >> When someone does your homework for you, do you given them credit in >> your writeup, or do you appropriate their work and lie to your >> professor by claiming it is yours? >> -- >> It¹s not denial. I¹m just very selective about >> what I accept as reality. >> --- Calvin (Calvin and Hobbes) >> Arturo Magidin >> magidin@math.berkeley.edu >10^2 times 15 horses = 150 meter^2 >Square root of 150 = side >Side * side = 10 * 15 neter >4 *side = outside fence only length exact >You wantum fence between every horse? or not? >Stuckum on this HW analysis. >It is only about 3.2 meter length per horse, hope they are small. Not that small. 10 m x 10 m x 15 horses = 1500 sq. m phil === Subject: Re: Tapped areas in Math research? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8G3KQf21285; >> I am at a point where I need to start choosing a >> research topic for my thesis. >... >> I would appreciate your comments in this respect >> and/or comments on how to choose a topic ( I am aware >> that it is ultimately my decision, but I would appreciate >> your comments just to have some input) >Surely the interests and talents of the mathematicians in whose >presence you will be working (for instance, whoever ends up >being your advisor in a real or formal sense) are most >important? If you are such an outlier that you can do >research on a topic about which none of the people in your >immediate intellectual community have either interest >or knowledge, then you don¹t need (and probably wouldn¹t >beneŽt from) the advice of sci.math; otherwise, you ought >to be talking to *them* and Žnding out what *they* know. >> or accounts of >> how you chose your topic. >I stumbled onto it. After spending my Žrst three years at >graduate school disregarding what I¹ve just suggested to you >(I had got it into my head that it would be interesting to >study local moduli of Lie algebras, via a topological and >algebro-geometric analysis of the variety of their structure >constants; and so it would have been, but no one thereabouts >at the time had the slightest interest in the idea), while-- >luckily--also soaking up a lot of atmosphere and some knowledge >about differential topology, by hanging out with people who >were doing that. During the summer between my third and fourth >years, Dennis Sullivan convened a problem seminar that was >attended by local topologists and singularists. One of the >problems got me thinking about the topological structure of >4-manifolds in general, and complex algebraic surfaces in >particular. I spent a lot of time drawing pictures and >playing with equations. One day I realized that by cobbling >together a couple of things I¹d learned from reading Milnor¹s books, >I could prove something interesting about hypersurfaces in CP^3; >and that was my thesis. (It got outdated pretty quickly. But by >then I was thinking about other, different-though-related, things.) I guess one lesson I can gather from your comment is to read up on things that seem interesting even if I do not understand them all that well at the beginning, and playing around with such material. Would you agree? . I have not been able to do that up till now, since I have had to prepare for the qualiŽers which I Žnally got off my back, so that now I am able to do some more exploring. >Lee Rudolph === Subject: Re: Tapped areas in Math research? >> > I am at a point where I need to start choosing a > research topic for my thesis. > >>... >> > I would appreciate your comments in this respect > and/or comments on how to choose a topic ( I am aware > that it is ultimately my decision, but I would appreciate > your comments just to have some input) > >>Surely the interests and talents of the mathematicians in whose >>presence you will be working (for instance, whoever ends up >>being your advisor in a real or formal sense) are most >>important? If you are such an outlier that you can do >>research on a topic about which none of the people in your >>immediate intellectual community have either interest >>or knowledge, then you don¹t need (and probably wouldn¹t >>beneŽt from) the advice of sci.math; otherwise, you ought >>to be talking to *them* and Žnding out what *they* know. >> > >or accounts of > how you chose your topic. > >>I stumbled onto it. After spending my Žrst three years at >>graduate school disregarding what I¹ve just suggested to you >>(I had got it into my head that it would be interesting to >>study local moduli of Lie algebras, via a topological and >>algebro-geometric analysis of the variety of their structure >>constants; and so it would have been, but no one thereabouts >>at the time had the slightest interest in the idea), while-- >>luckily--also soaking up a lot of atmosphere and some knowledge >>about differential topology, by hanging out with people who >>were doing that. During the summer between my third and fourth >>years, Dennis Sullivan convened a problem seminar that was >>attended by local topologists and singularists. One of the >>problems got me thinking about the topological structure of >>4-manifolds in general, and complex algebraic surfaces in >>particular. I spent a lot of time drawing pictures and >>playing with equations. One day I realized that by cobbling >>together a couple of things I¹d learned from reading Milnor¹s books, >>I could prove something interesting about hypersurfaces in CP^3; >>and that was my thesis. (It got outdated pretty quickly. But by >>then I was thinking about other, different-though-related, things.) >> > I guess one lesson I can gather from your comment is to read > up on things that seem interesting even if I do not understand them > all that well at the beginning, and playing around with such > material. Would you agree? . I have not been able to do that > up till now, since I have had to prepare for the qualiŽers > which I Žnally got off my back, so that now I am able to do some > more exploring. Does your department have a graduate seminar where interested students and faculty get together to read papers in some active area and discuss unsolved problems? That¹s the best way to get started in mathematics research. If you don¹t have such a seminar, consider starting one. === Subject: Analysis question If f:[0,1]--> R is Lebesgue measurable and fg is in L^2 for every g in L^2. Is it necessary true that f is in L^{inŽnity}? === Subject: Re: Analysis question > If f:[0,1]--> R is Lebesgue measurable and fg is in L^2 for every g >in L^2. Is it necessary true that f is in L^{inŽnity}? Yes. Robin Chapman has already given a simple argument starting from nothing. Here¹s a slightly slick way to prove it. (I actually prefer an argument like the one RC gave - I point out the following because sometimes an argument like the following is much easier to come up with than a more direct/elementary proof): DeŽne T:L^2 -> L^2 by Tf = gf. It¹s easy to see from the Closed Graph Theorem that T is bounded (for the details you can use the fact that convergence in norm implies almost everywhere convergence of a subsequence.) But if {x : |g(x)| > A} has positive measure it¹s easy to see the norm of T must be at least A. So there exists A such that this set has measure zero, qed. Exercise: Contemplate how it is that both arguments depend crucially on the completeness of L^2. ************************ David C. Ullrich === Subject: Re: Analysis question > If f:[0,1]--> R is Lebesgue measurable and fg is in L^2 for every g >in L^2. Is it necessary true that f is in L^{inŽnity}? > Yes. Robin Chapman has already given a simple argument starting > from nothing. Here¹s a slightly slick way to prove it. (I actually > prefer an argument like the one RC gave - I point out the following > because sometimes an argument like the following is much easier > to come up with than a more direct/elementary proof): > DeŽne T:L^2 -> L^2 by Tf = gf. It¹s easy to see from the Closed > Graph Theorem that T is bounded (for the details you can use > the fact that convergence in norm implies almost everywhere > convergence of a subsequence.) But if {x : |g(x)| > A} has > positive measure it¹s easy to see the norm of T must be at > least A. So there exists A such that this set has measure zero, > qed. > Exercise: Contemplate how it is that both arguments depend > crucially on the completeness of L^2. I don¹t see how the from nothing approach relies on the completeness of L^2. Just use Fatou¹s Lemma or monotone convergence. === Subject: Re: Analysis question >> If f:[0,1]--> R is Lebesgue measurable and fg is in L^2 for every g >>in L^2. Is it necessary true that f is in L^{inŽnity}? >> Yes. Robin Chapman has already given a simple argument starting >> from nothing. Here¹s a slightly slick way to prove it. (I actually >> prefer an argument like the one RC gave - I point out the following >> because sometimes an argument like the following is much easier >> to come up with than a more direct/elementary proof): >> DeŽne T:L^2 -> L^2 by Tf = gf. It¹s easy to see from the Closed >> Graph Theorem that T is bounded (for the details you can use >> the fact that convergence in norm implies almost everywhere >> convergence of a subsequence.) But if {x : |g(x)| > A} has >> positive measure it¹s easy to see the norm of T must be at >> least A. So there exists A such that this set has measure zero, >> qed. >> Exercise: Contemplate how it is that both arguments depend >> crucially on the completeness of L^2. >I don¹t see how the from nothing approach relies on the completeness of >L^2. Just use Fatou¹s Lemma or monotone convergence. Hmm. I could argue that that¹s more or less the same thing. But that would be silly, since it¹s not. I could argue that it¹s _kinda_ the same thing... never mind. ************************ David C. Ullrich === Subject: Re: Analysis question > If f:[0,1]--> R is Lebesgue measurable and fg is in L^2 for every g > in L^2. Is it necessary true that f is in L^{inŽnity}? Yes. Suppose that f is not essentially bounded above: {x: f(x) > n} has positive measure for all n. Then one can easily construct a sequence of disjoint measurable sets A_1, A_2, ... such that each A_n has positive measure a_n and f(x) > n for all x in A_n. Let b_1, b_2, ... be a sequence of positive numbers and deŽne g to be the sum over n of b_n times the characteristic function of A_n. Then integral |g|^2 = sum b_n^2 a_n and integral |fg|^2 = sum n^2 b_n^2 a_n. If we take b_n^2 = 1/(a_n n^2) then g is L^2 but fg isn¹t. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: HOW DO YOU THINK ABOUT REFORM FIZEAU EXPERIMENT? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8G3Wwx22294; THE REFORM FIZEAU EXPERIMENT Supposing that Fizeau experiment is run in a model which satisfy the following conditions: - The vacuum laboratory for containing all of experimental apparatuses is put in a airplane or spacecraft which is moving at v0 uniform speeds, it means the experimental will be run in a laboratory which is moving at v0 uniform speeds. - A part of spreading way of the light (AB distance) is the water medium (n = 1,3) which is moving at u0 uniform speed expecting to the laboratory (airplane or spacecraft). - The water medium and the laboratory (airplane or spacecraft) travel in the same direction. Some questions: How is it about the achievement result? Are there any difference when comparing with a similar experiment which is run in the rest laboratory on the face of the earth? Are achievement values of experiment result varied with the change of v0 speed of the laboratory? I think that: If achievement values donāt vary with v0 speed of the laboratory (airplane or spacecraft), the Fizeau experiment interpretation of Einsteinās Relativity will be right. On the contrary, we will not be able to interpret a such experiment by the speed addition of Einsteinās Relativity. So,how can we interpret it in such case? We can argue that when once achievement values change with v0 speed of the laboratory (airplane or spacecraft), u speed value geting from the result of calculation must be recognized as an absolute quantity of the measurement in an absolute frame. === Subject: Re: HOW DO YOU THINK ABOUT REFORM FIZEAU EXPERIMENT? > THE REFORM FIZEAU EXPERIMENT > Supposing that Fizeau experiment is run in a model which satisfy the following conditions: > - The vacuum laboratory for containing all of experimental apparatuses is put in a airplane or spacecraft which is moving at v0 uniform speeds, it means the experimental will be run in a laboratory which is moving at v0 uniform speeds. > - A part of spreading way of the light (AB distance) is the water medium (n = 1,3) which is moving at u0 uniform speed expecting to the laboratory (airplane or spacecraft). > - The water medium and the laboratory (airplane or spacecraft) travel in the same direction. > Some questions: > How is it about the achievement result? > Are there any difference when comparing with a similar experiment which is run in the rest laboratory on the face of the earth? > Are achievement values of experiment result varied with the change of v0 speed of the laboratory? > I think that: > If achievement values don?t vary with v0 speed of the laboratory (airplane or spacecraft), the Fizeau experiment interpretation of Einstein?s Relativity will be right. On the contrary, we will not be able to interpret a such experiment by the speed addition of Einstein?s Relativity. So,how can we interpret it in such case? > We can argue that when once achievement values change with v0 speed of the laboratory (airplane or spacecraft), u speed value geting from the result of calculation must be recognized as an absolute quantity of the measurement in an absolute frame. i hate stupid foreigners === Subject: Re: The real numbers, and general comments >... That thread was started by Ross Finlayson, >not myself. > Here is a minor point of attribution: Uncountable sets in CZF? was started by Agamemnon. Sorry. It still wasn¹t me, though. Why does your name appear in quotes in the subject line? It shouldn¹t, should it? Andrew Usher === Subject: Re: The real numbers, and general comments > I think I have realised why the real numbers do not have a good > deŽnition. > > I have also observed that this trend does not seem to have affected > and thus properly think about the value of their work. Modern math has > made this impossible; I know that they would reply that this is > because mathematics is more intellectually sophisticated. But, as Alan > Sokal so memorably demonstrated, certain Želds of the humanities are > intellectually vacuous despite (or because of?) their esoteric jargon. > Could this not be partially true of mathematics today? > Well, that makes your task rather simple. Write a medium-length > and submit it to a mathematics journal - just as Alan Sokal did, of > course. When it is accepted, let us all know. Would that then mean that math IS an experimental science? :) John === Subject: Re: The real numbers, and general comments >> N = {0, 1, 2, ...} >> >> is understood by all of us. >> >> >> But is it a Žnite deŽnition? I mean it only gives the Žrst 3. >> >> Since the rationals are countable, they must be allowable, and since >> sequences involve a countable number of terms they must be acceptable, so >> are cauchy sequences of rationals not acceptable. > > It gives the rule to construct all the other terms, therefore it¹s a > Žnite deŽnition. > > And it is a way of deŽning the real numbers contradicting your original > claim, I think. I don¹t have the original post anymore and can¹t be > I deŽned N, not R. Pay atterntion or don¹t bother replying. > Andrew Usher Erm, from the format of your reply you seemed to be agreeing that my deŽnitions were Žnite. Why don¹t you pay attention? Or deŽne your terms properly. === Subject: Re: The real numbers, and general comments > It gives the rule to construct all the other terms, therefore it¹s a > Žnite deŽnition. > > And it is a way of deŽning the real numbers contradicting your original > claim, I think. I don¹t have the original post anymore and can¹t be > > I deŽned N, not R. Pay atterntion or don¹t bother replying. > Erm, from the format of your reply you seemed to be agreeing that my > deŽnitions were Žnite. Why don¹t you pay attention? Or deŽne your > terms properly. I said a deŽnition was Žnite if it gives the rule to make all of the set. Since no rule is possible giving all Dedekind cuts, or all Cauchy sequences, R does not have a Žnite deŽnition. Andrew Usher === Subject: Re: The real numbers, and general comments > By deŽnition no computable sequence can have a non-computable limit, > and no computable set can have a non-computable supremum, so what do > we really need completeness for? Chaitin¹s Omega is the limit of a computable sequence of rational numbers. === Subject: Re: The real numbers, and general comments > By deŽnition no computable sequence can have a non-computable limit, > and no computable set can have a non-computable supremum, so what do > we really need completeness for? > Chaitin¹s Omega is the limit of a computable sequence of rational > numbers. No, read one of the standard summaries on Chaitin¹s constant. We can compute only the Žrst N terms (N depending on implementation), beyond we come to an undecidable machine (probably Collatz-like). Think about it in basic terms: if a sequence is computable, so is its limit, for that is the very deŽnition of computability! Andrew Usher === Subject: Re: The real numbers, and general comments > By deŽnition no computable sequence can have a non-computable limit, > and no computable set can have a non-computable supremum, so what do > we really need completeness for? > > Chaitin¹s Omega is the limit of a computable sequence of rational > numbers. > No, read one of the standard summaries on Chaitin¹s constant. We can > compute only the Žrst N terms (N depending on implementation), beyond > we come to an undecidable machine (probably Collatz-like). Think about > it in basic terms: if a sequence is computable, so is its limit, for > that is the very deŽnition of computability! Here¹s the kind of thing I was thinking of. Let a_n be the rational number n binary digits long whose kth bit is 1 iff the kth Turing machine halts in n steps or less. If the kth machine doesn¹t halt, then all a_n will have the kth bit 0. If the kth machine does halt, then all a_n for sufŽciently large n will have the kth bit 1. This sequence is certainly computable. The algorithm for deciding if a_n = x consists of running Turing machines 1 through n for n steps each, and comparing the result. You could write a computer program to do it. Note that this is not the same as saying that you could write a computer program to output the Žrst n bits of Chaitin¹s Omega; that¹s impossible. === Subject: Re: The real numbers, and general comments > No, read one of the standard summaries on Chaitin¹s constant. We can > compute only the Žrst N terms (N depending on implementation), beyond > we come to an undecidable machine (probably Collatz-like). Think about > it in basic terms: if a sequence is computable, so is its limit, for > that is the very deŽnition of computability! > Here¹s the kind of thing I was thinking of. Let a_n be the rational number > n binary digits long whose kth bit is 1 iff the kth Turing machine halts > in n steps or less. If the kth machine doesn¹t halt, then all a_n will > have the kth bit 0. If the kth machine does halt, then all a_n for > sufŽciently large n will have the kth bit 1. > This sequence is certainly computable. The algorithm for deciding if > a_n = x consists of running Turing machines 1 through n for n steps > each, and comparing the result. You could write a computer program to do > it. Note that this is not the same as saying that you could write a > computer program to output the Žrst n bits of Chaitin¹s Omega; that¹s > impossible. I think you might have a counterexample here. You must prove, though, that the average time-to-halt of halting programs increase in the enumeration order, othwise this sequence will not converge to Omega. Andrew Usher === Subject: Re: The real numbers, and general comments > No, read one of the standard summaries on Chaitin¹s constant. We can > compute only the Žrst N terms (N depending on implementation), beyond > we come to an undecidable machine (probably Collatz-like). Think about > it in basic terms: if a sequence is computable, so is its limit, for > that is the very deŽnition of computability! > > Here¹s the kind of thing I was thinking of. Let a_n be the rational number > n binary digits long whose kth bit is 1 iff the kth Turing machine halts > in n steps or less. If the kth machine doesn¹t halt, then all a_n will > have the kth bit 0. If the kth machine does halt, then all a_n for > sufŽciently large n will have the kth bit 1. > > This sequence is certainly computable. The algorithm for deciding if > a_n = x consists of running Turing machines 1 through n for n steps > each, and comparing the result. You could write a computer program to do > it. Note that this is not the same as saying that you could write a > computer program to output the Žrst n bits of Chaitin¹s Omega; that¹s > impossible. > I think you might have a counterexample here. You must prove, though, > that the average time-to-halt of halting programs increase in the > enumeration order, othwise this sequence will not converge to Omega. So you agree that this sequence is computable, and the apparent limit is not, so if it converges then it¹s a counterexample, right? Okay, let¹s prove it converges. If the kth Turing machine halts, let H(k) be the number of steps it takes to halt. If it doesn¹t halt, let H(k)=1. So for all n>=H(k), the kth bit of a_n equals the kth bit of Omega. (I¹m not claiming we can compute H(k), just that it exists) Now let f(k)=max{H(1),H(2),H(3),...,H(k)}. So for all n>=f(k), bits 1 through k of a_n equal bits 1 through k of Omega. In other words, for each positive integer k, there exists a positive integer f(k) such that for all n>=f(k), |Omega - a_n| < 1/2^k. So {a_n} converges to Omega, QED. === Subject: Re: The real numbers, and general comments > I think you might have a counterexample here. You must prove, though, > that the average time-to-halt of halting programs increase in the > enumeration order, othwise this sequence will not converge to Omega. > So you agree that this sequence is computable, and the apparent limit is > not, so if it converges then it¹s a counterexample, right? Okay, let¹s > prove it converges. > If the kth Turing machine halts, let H(k) be the number of steps it takes > to halt. If it doesn¹t halt, let H(k)=1. So for all n>=H(k), the kth bit > of a_n equals the kth bit of Omega. (I¹m not claiming we can compute > H(k), just that it exists) > Now let f(k)=max{H(1),H(2),H(3),...,H(k)}. So for all n>=f(k), bits 1 > through k of a_n equal bits 1 through k of Omega. > In other words, for each positive integer k, there exists a positive > integer f(k) such that for all n>=f(k), |Omega - a_n| < 1/2^k. > So {a_n} converges to Omega, QED. Yes, I forgot an elementary property of inŽnite sequences. I was wrong, then, that _every_ computable sequence has a computable limit. But almost all do. I wonder, though, if you could change the deŽnition of computability to Œthe limit of a computable sequence¹. Would that then make my statements true? Andrew Usher === Subject: Re: The real numbers, and general comments |I wonder, though, if you could change the deŽnition of computability |to Œthe limit of a computable sequence¹. Would that then make my |statements true? I don¹t know which statements you have in mind, and I don¹t feel like reviewing the thread right now. In this larger Želd, the ordering loses some of the computational properties it has for the computable reals. If r and s are reals computed by given programs, and r |I wonder, though, if you could change the deŽnition of computability > |to Œthe limit of a computable sequence¹. Would that then make my > |statements true? > In this larger Želd, the ordering loses some of the computational > properties it has for the computable reals. If r and s are reals > computed by given programs, and r that r of computable sequences of rationals. (A limit of a computable > sequence of computable reals is always a limit of a computable > sequence of rationals.) One can¹t guarantee being able to give an > integer as an upper or lower bound on the limit. I see. However, it is already true that we don¹t know how far we need to go to Žnd out rs. > They can be added, subtracted, multiplied, divided and so on, though. > It makes sense to deŽne a computable sequence r_m of these limiting > reals to be a sequence for which there exists a computable function > a(m,n) from pairs of positive integers (m,n) to the rationals, where > lim{n} a(m,n) = r_m for each m. There¹s no guarantee that if r_m has > a limit, that the limit can be given as the limit of a computable > sequence of rationals. So the problem of not being able to compute > limits is only sort of postponed a little. Actually, this isn¹t right because I just deŽned all such limits to be computable. Therefore sequences of limits are also, and so inŽnitely. DeŽne S_0 to be the Œnormal¹ computables, S_1 the Žrst step of admitting limits, and so on. S_omega is the union of all these, and it will still be countable because we are adding countably many sequences at each of countably many steps. Now: Is S_omega computably-complete? Andrew Usher === Subject: Re: The real numbers, and general comments > Yes, I forgot an elementary property of inŽnite sequences. I was > wrong, then, that _every_ computable sequence has a computable limit. > But almost all do. I think this almost all can only be true in the sense of almost all that you will come across in actual practice. After all, the set of computable sequences with computable limits and the set of computable sequences with non-computable limits are both countably inŽnite sets. It¹s just so much easier to give examples of the former. > I wonder, though, if you could change the deŽnition of computability > to Œthe limit of a computable sequence¹. Would that then make my > statements true? I think you¹ll Žnd it would make them tautological. :) As has already been implied in this thread, you really need to change a lot more of your axioms and deŽnitions to get a completely computable theory of mathematics. For example, I think the counterexample I gave no longer works if you use a deŽnition of convergence based on computability. In fact, I¹m pretty sure that sequence isn¹t even computably Cauchy. I was a little surprised that you accepted my proof, as I thought you might already be using such restrictive rules. I won¹t speculate on whether or not the computable numbers satisfy a computable completeness condition. Mainstream mathematics seems to choose its foundations with an inclusive heuristic, accepting the existence of more objects rather than fewer. There are those who research the consequences of more restrictive systems, such as constructivists, and they certainly produce interesting results. However, most mathematicians use lots of non-constructive methods, and I believe they¹re aware of the fact. If you¹re uneasy about accepting such methods, it might help to remember the example of the Axiom of Choice. This axiom asserts the existence of many many objects without any speciŽed process for Žnding them. But AC is consistent with and independent of ZF. So ZFC, with so much extra non-constructiveness in it, contains no real new inconsistencies (although ZF is, admittedly, far from constructive). If you¹re interested in the consequences of assuming only computable things exist, you might look at what others have done in this area. I know nothing about it, but I suspect it¹s been researched before. === Subject: Re: The real numbers, and general comments >I think you might have a counterexample here. You must prove, though, >that the average time-to-halt of halting programs increase in the >enumeration order, othwise this sequence will not converge to Omega. >>So you agree that this sequence is computable, and the apparent limit is >>not, so if it converges then it¹s a counterexample, right? Okay, let¹s >>prove it converges. >>If the kth Turing machine halts, let H(k) be the number of steps it takes >>to halt. If it doesn¹t halt, let H(k)=1. So for all n>=H(k), the kth bit >>of a_n equals the kth bit of Omega. (I¹m not claiming we can compute >>H(k), just that it exists) >>Now let f(k)=max{H(1),H(2),H(3),...,H(k)}. So for all n>=f(k), bits 1 >>through k of a_n equal bits 1 through k of Omega. >>In other words, for each positive integer k, there exists a positive >>integer f(k) such that for all n>=f(k), |Omega - a_n| < 1/2^k. >>So {a_n} converges to Omega, QED. > Yes, I forgot an elementary property of inŽnite sequences. I was > wrong, then, that _every_ computable sequence has a computable limit. > But almost all do. That last statement is very dangerous to make. When dealing with standard analysis, it turns out that almost all numbers are not computable. Almost all functions are discontinuous. They may not be the numbers/functions we use most often, but that is far from the same thing. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments |> Chaitin¹s Omega is the limit of a computable sequence of rational |> numbers. | |No, read one of the standard summaries on Chaitin¹s constant. We can |compute only the Žrst N terms (N depending on implementation), beyond |we come to an undecidable machine (probably Collatz-like). Think about |it in basic terms: if a sequence is computable, so is its limit, for |that is the very deŽnition of computability! No, it isn¹t. It¹s computable if the rate of convergence is itself computable, and not necessarily otherwise. One form of the standard deŽnition of r¹s being a computable real is that there be a sequence a1,a2,a3,... of rationals such that |r-a_n|<1/n for each n>0. Chaitin¹s Omega is the least upper bound of a bounded, computable, increasing sequence of rational numbers. Roughly, if Chaitin¹s Omega is the probability of a program halting, we can let a_n be the probability that a program is among the Žrst n in some enumeration and halts within n steps. One can¹t use the convergence of a_n to Omega to compute Omega, simply because one has no way of knowing how long it will take before a_n is within some epsilon>0 of Omega. Eventually, enough of the programs that halt will have halted to make the approximation agree with Omega in its Žrst 2N bits. But the only way to know that we¹ve gotten there is to prove that enough of the rest of the programs are not going to halt. Redoing real analysis without the least upper bound principle is a very serious revision. Constructive real analysis is one way to go about it (since the least upper bound principle isn¹t constructive), and it would be difŽcult to say categorically what other approaches one might take. I don¹t think there¹s any simple route. Keith Ramsay === Subject: Re: The real numbers, and general comments > |No, read one of the standard summaries on Chaitin¹s constant. We can > |compute only the Žrst N terms (N depending on implementation), beyond > |we come to an undecidable machine (probably Collatz-like). Think about > |it in basic terms: if a sequence is computable, so is its limit, for > |that is the very deŽnition of computability! > No, it isn¹t. It¹s computable if the rate of convergence is itself > computable, and not necessarily otherwise. > One form of the standard deŽnition of r¹s being a computable real > is that there be a sequence a1,a2,a3,... of rationals such that > |r-a_n|<1/n for each n>0. > Chaitin¹s Omega is the least upper bound of a bounded, computable, > increasing sequence of rational numbers. Roughly, if Chaitin¹s Omega > is the probability of a program halting, we can let a_n be the > probability that a program is among the Žrst n in some enumeration > and halts within n steps. Ah, this way won¹t work; it is not monotonic, so you can¹t prove it converges. A correct way was given by Nathan in the other reply. > One can¹t use the convergence of a_n to Omega to compute Omega, > simply because one has no way of knowing how long it will take before > a_n is within some epsilon>0 of Omega. Eventually, enough of the > programs that halt will have halted to make the approximation agree > with Omega in its Žrst 2N bits. But the only way to know that we¹ve > gotten there is to prove that enough of the rest of the programs are > not going to halt. If it does converge, of course the limit must be Omega. I see now the problem - I had my own mental deŽnition of computability that would actually make Omega computable (I do know the standard deŽnition.). It was Œcan get the number exactly in countably inŽnite steps¹. Running all the programs Œto inŽnity¹ is countably many steps, and would determine the halting constant. Andrew Usher === Subject: Re: The real numbers, and general comments |> Chaitin¹s Omega is the least upper bound of a bounded, computable, |> increasing sequence of rational numbers. Roughly, if Chaitin¹s Omega |> is the probability of a program halting, we can let a_n be the |> probability that a program is among the Žrst n in some enumeration |> and halts within n steps. | |Ah, this way won¹t work; it is not monotonic, so you can¹t prove it |converges. A correct way was given by Nathan in the other reply. At what point is it not monotonic? If a program is among the Žrst n, then it¹s also among the Žrst n+1. If a program halts within n steps, then it also halts within n+1 steps. I didn¹t say it stops on step n, merely within n steps. Keith Ramsay === Subject: Re: The real numbers, and general comments >R is normally conceptualised as an inŽnite geometric line. I say that >what we call a number is a point on this line, and so many points as >we can locate can be called numbers. But, it is not a valid inference >> >from this to state that the line is made up of numbers! >> >>No, it¹s an axiom that each of the points *corresponds* to a distinct >>number. There is no inference, just an axiom... an assumption. >Ihis Œaxiom¹ is actually a redeŽnition of Œnumber¹. This is justiŽed >based on the completeness axiom, isn¹t it? Basically, yes. But here¹s the point, it is an axiom. It is a given assumption. How you interpret it is your own business. >By deŽnition no computable sequence can have a non-computable limit, >and no computable set can have a non-computable supremum, so what do >we really need completeness for? >>Dealing with the points on the real line that are non-computable. > Circular argument, isn¹t it? You are justifying the Œaxiom¹ that gives > us non-computable numbers by using the concept of non-computable > numbers. The existence of numbers that are not computable is a result of our deŽnition of computability (Turing Machines, Partial Recursive Functions, etc). There is nothing about associating each number (non-computable or not) with a point on the real number line that has anything to do with whether that particular number is computable. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments >Ihis Œaxiom¹ is actually a redeŽnition of Œnumber¹. This is justiŽed >based on the completeness axiom, isn¹t it? > Basically, yes. But here¹s the point, it is an axiom. It is a given > assumption. How you interpret it is your own business. Do you think axioms should have no justiŽcation? May I then take Œ1 = 0¹ as an axiom, and have this arithmetic be as good as yours? >By deŽnition no computable sequence can have a non-computable limit, >and no computable set can have a non-computable supremum, so what do >we really need completeness for? >> >>Dealing with the points on the real line that are non-computable. > > Circular argument, isn¹t it? You are justifying the Œaxiom¹ that gives > us non-computable numbers by using the concept of non-computable > numbers. > The existence of numbers that are not computable is a result of our > deŽnition of computability (Turing Machines, Partial Recursive > Functions, etc). There is nothing about associating each number > (non-computable or not) with a point on the real number line that has > anything to do with whether that particular number is computable. No, those methods only establish sequences that are non-computable. Again, it¹s another thing entirely to declare them Œnumbers¹. Andrew Usher === Subject: Re: The real numbers, and general comments >Ihis Œaxiom¹ is actually a redeŽnition of Œnumber¹. This is justiŽed >based on the completeness axiom, isn¹t it? >>Basically, yes. But here¹s the point, it is an axiom. It is a given >>assumption. How you interpret it is your own business. > Do you think axioms should have no justiŽcation? May I then take Œ1 = > 0¹ as an axiom, and have this arithmetic be as good as yours? The only usual justiŽcation for a set of axioms is that they are consistent, which is demonstrated by the existence of a model. Other than that, you can choose whatever axioms you want. For example, the axiom Œ1=0¹ is perfectly good, as long as you don¹t intend to append it to the axioms of Peano Arithmatic. >By deŽnition no computable sequence can have a non-computable limit, >and no computable set can have a non-computable supremum, so what do >we really need completeness for? >> >>Dealing with the points on the real line that are non-computable. >Circular argument, isn¹t it? You are justifying the Œaxiom¹ that gives >us non-computable numbers by using the concept of non-computable >numbers. >>The existence of numbers that are not computable is a result of our >>deŽnition of computability (Turing Machines, Partial Recursive >>Functions, etc). There is nothing about associating each number >>(non-computable or not) with a point on the real number line that has >>anything to do with whether that particular number is computable. > No, those methods only establish sequences that are non-computable. > Again, it¹s another thing entirely to declare them Œnumbers¹. Perhaps this is when it¹s time to agree to disagree. You appear to have a different set of axioms that you are using for your foundation. Until you state what they are clearly, there isn¹t much else I can say. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments >Ihis Œaxiom¹ is actually a redeŽnition of Œnumber¹. This is justiŽed >based on the completeness axiom, isn¹t it? > > Basically, yes. But here¹s the point, it is an axiom. It is a given > assumption. How you interpret it is your own business. > Do you think axioms should have no justiŽcation? Axioms must be self-consistent. They need have no justiŽcation other than that. > May I then take Œ1 = 0¹ as an axiom, and have this arithmetic > be as good as yours? Possibly. Depends on what meaning you¹re going to assign to 1 and 0 before taking that axiom, what other properties they will have. More likely you¹ll end up with an inconsistent system, something provable from your other axioms that is not consistent with this one. - Randy === Subject: Re: The real numbers, and general comments >>But that *is* a Žnite deŽnition. I deŽned them, I didn¹t tell you >>how to enumerate them, since that is impossible. Also, I didn¹t know >>that you believe quantifying over all subsets of an inŽnite set is >>illegitimate. If that is the case, then you are working with a >>completely different set theory from ZFC, or even ZF. >In another thread (ŒUncountable sets in CZF?¹) we are discussing the >Loewenheim-Skolen paradox, which shows that ZFC is incomplete. Given >that, why should ZFC have any preferred status? >>Only because it¹s the one I¹m most familiar with. What are the axioms >>of CZF? More to the point, how is it different from ZFC? > I don¹t know what CZF is. That thread was started by Ross Finlayson, > not myself. > I was asking if the incompleness of ZFC should cause us to reject it > as the true set theory. Is a complete set theory possible, in the > sense that no additional axioms can falsify any of its theorems? I strongly suspect the answer is no, though I¹d have to do some checking. Also, even if CZF is complete, does it accurately describe what you would normally thing of as set theory? Does it correspond with your understanding of how sets work? >>There¹s a simple problem with that: Let f be the following function: >>f(x) = 0 when x is irrational, f(x) = 1 when x is rational. > >This is not piecewise continuous. (Note the restriction above.) >> >>Good point, however not all functions in real analysis are piecewise >>continuous. >Can you give an example of a useful function that isn¹t? (I don¹t >count the delta Œfunction¹, as that isn¹t even a valid function.) >>Useful for what? The above is useful for demonstrating that a function >>can have countably inŽnitely many non-zero values, yet have an integral >>= 0. > Pathology for the sake of pathology is not what I consider Œuseful¹. > Interesting, yes, but not useful. Pathology for the sake of illustrating unusual or unexpected properties that run contrary to what most people would expect is useful. There is a tendancy to assume that all functions continuous almost everywhere that the function is normally deŽned, simply because that is all most people encounter. In this case, it is worthwhile to keep people from making wrong assumptions. Also, it is possible to take the position that, since continuous functions make up a very small percentage of the functions from R to R, that continuous functions are the pathological ones and should be ignored. I know that would make calculus students happy. >Though I¹d rather deŽne continuity to be uniform continuity over an >interval, following the naive Œcan be drawn without lifting the >pencil¹. An open interval can be deŽned as the union of a sequence of >closed intervals, hence we can still say that > f(x) = 1/x >is continous on its entire domain, and similarly all algebraic >functions. >>Fair enough. > Do you agree with my general point here, that analysis can be done > over any countable dense set, and for all piecewise continuous > functions, will give the same results? It depends on how you deŽne the domain of the function. If you deŽne the domain of f(x) = 1/x as R - {0}, then it is continuous on its entire domain. If you deŽne the domain to be R, then f(x) is discontinuous at 0. You can take the Žrst position, most people seem to understand the second to be the case. >>If R is not just numbers, what are the other things? >Uncomputable sequences would be a good description. >>It seems to me that it would be easier to use a new deŽnition of R in >>which you restrict yourself to computable numbers. Of course, that >>would result in R being countable. > I¹m not sure if there¹s a good, short deŽnition of the computables > that doesn¹t mention any speciŽc computing machine. My deŽnition is > the union of all deŽnable archimedean ordered Želds. In this case, > it can only be gotten as the sup of an inŽnite sequence: > Q < F1 < F2 < F3 < ... to inŽnity = S > where the Fn are archimedean ordered Želds (Not all such sequences > will give all of S, and any that does will be uncomputable.). By that deŽnition, it appears that you are saying R is uncountable. Since it is known that the computable numbers are countable, you then appear to be saying that R has non-computable members. Since R is normally described as being numbers, wouldn¹t that suggest that there are non-computable numbers? >>I don¹t see a problem with calling an uncomputable sequence a number, >>however. > You can¹t calculate with it. In a practical sense, you can¹t calculate with any number that is not algebraic. If you are working with decimals, you can¹t calculate with any number that isn¹t rational. Regardless, I don¹t view not being able to calculate with a number as a problem. >I hope you aren¹t just being obtuse here; the deŽnition of parallel >of course is changed in the obvious fashion. I said that no actual >theorems of either geometry are disproved, and that is clearly true. >>That is because each geometry is consistent. That does not mean they >>have the same results. If you are changing deŽnitions to accomodate >>the axioms (which is common), then you can still state different results >>by replacing parallel in one system with the deŽnition from the the >>other system. >>In one system, there are lines that do not intersect (Euclidean). In >>the other, every pair of lines intersects (projective). These are still >>different results. > Given that the two geometries are consistent with each other, what is > the problem with unifying them as one? They are not consistent with each other. Each is consistent with itself. If you try to take all the axioms/deŽnitions together, you will get an inconsistent system. At that point, it becomes useless. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments > I don¹t know what CZF is. That thread was started by Ross Finlayson, > not myself. > I was asking if the incompleness of ZFC should cause us to reject it > as the true set theory. Is a complete set theory possible, in the > sense that no additional axioms can falsify any of its theorems? > I strongly suspect the answer is no, though I¹d have to do some > checking. Also, even if CZF is complete, does it accurately describe > what you would normally thing of as set theory? Does it correspond with > your understanding of how sets work? I just said I know nothing about CZF. >Can you give an example of a useful function that isn¹t? (I don¹t >count the delta Œfunction¹, as that isn¹t even a valid function.) >> >>Useful for what? The above is useful for demonstrating that a function >>can have countably inŽnitely many non-zero values, yet have an integral >>= 0. > > Pathology for the sake of pathology is not what I consider Œuseful¹. > Interesting, yes, but not useful. > Pathology for the sake of illustrating unusual or unexpected properties > that run contrary to what most people would expect is useful. There is > a tendancy to assume that all functions continuous almost everywhere > that the function is normally deŽned, simply because that is all most > people encounter. In this case, it is worthwhile to keep people from > making wrong assumptions. Also, it is possible to take the position > that, since continuous functions make up a very small percentage of the > functions from R to R, that continuous functions are the pathological > ones and should be ignored. I know that would make calculus students happy. No one even thought that these things should be called Œfunctions¹ for the Žrst 200 years of analysis. That should tell you how useful they are. > Do you agree with my general point here, that analysis can be done > over any countable dense set, and for all piecewise continuous > functions, will give the same results? > It depends on how you deŽne the domain of the function. If you deŽne > the domain of f(x) = 1/x as R - {0}, then it is continuous on its entire > domain. If you deŽne the domain to be R, then f(x) is discontinuous at > 0. You can take the Žrst position, most people seem to understand the > second to be the case. This isn¹t an answer to my question. But, yes, I do deŽne the domain the second way; since 1/x is not deŽned at 0, it isn¹t part of the domain. >>If R is not just numbers, what are the other things? > >Uncomputable sequences would be a good description. >> >>It seems to me that it would be easier to use a new deŽnition of R in >>which you restrict yourself to computable numbers. Of course, that >>would result in R being countable. > > I¹m not sure if there¹s a good, short deŽnition of the computables > that doesn¹t mention any speciŽc computing machine. My deŽnition is > the union of all deŽnable archimedean ordered Želds. In this case, > it can only be gotten as the sup of an inŽnite sequence: > > Q < F1 < F2 < F3 < ... to inŽnity = S > > where the Fn are archimedean ordered Želds (Not all such sequences > will give all of S, and any that does will be uncomputable.). > By that deŽnition, it appears that you are saying R is uncountable. > Since it is known that the computable numbers are countable, you then > appear to be saying that R has non-computable members. Since R is > normally described as being numbers, wouldn¹t that suggest that there > are non-computable numbers? I did not mention R or Œuncountable¹ above, so I don¹t see where you are coming from here. I choose the letter S for the computables because that letter is not used for any of the normal sets, and because of its generality in my view. >>I don¹t see a problem with calling an uncomputable sequence a number, >>however. > > You can¹t calculate with it. > In a practical sense, you can¹t calculate with any number that is not > algebraic. If you are working with decimals, you can¹t calculate with > any number that isn¹t rational. Regardless, I don¹t view not being able > to calculate with a number as a problem. What do you mean by Œcalculate¹ here? Rational numbers can be calculated with exactly for arithmetic operations. Algebraic irrationals, though, can only be done through symbolic expressions, which is the same as for transcendentals. I say computable numbers are the set calculable over because you can evaluate any expression involing them; yes, I know this is essentially the deŽnition of computable, but it works. >>In one system, there are lines that do not intersect (Euclidean). In >>the other, every pair of lines intersects (projective). These are still >>different results. > > Given that the two geometries are consistent with each other, what is > the problem with unifying them as one? > They are not consistent with each other. Each is consistent with > itself. If you try to take all the axioms/deŽnitions together, you > will get an inconsistent system. At that point, it becomes useless. You have not shown any inconsistency. Euclidean, afŽne, and projective are just different types of theorems in my geometry - they differ in the types of symmetry, and the types of invariants they can employ. Andrew Usher === Subject: Re: The real numbers, and general comments |>Also, it is possible to take the position |> that, since continuous functions make up a very small percentage of the |> functions from R to R, that continuous functions are the pathological |> ones and should be ignored. I know that would make calculus students |happy. | |No one even thought that these things should be called Œfunctions¹ for |the Žrst 200 years of analysis. That should tell you how useful they |are. They had not settled upon a common notion of function, which is rather different from having happily assumed that they were all continuous, by a deŽnition of continuous that also only became standard later. Moreover, more than 90% of the real analysis that has been done, has been done since then. So no, it really *doesn¹t* tell you how useful discontinuous functions are. Keith Ramsay === Subject: Re: The real numbers, and general comments > |>Also, it is possible to take the position > |> that, since continuous functions make up a very small percentage of the > |> functions from R to R, that continuous functions are the pathological > |> ones and should be ignored. I know that would make calculus students > |happy. > |No one even thought that these things should be called Œfunctions¹ for > |the Žrst 200 years of analysis. That should tell you how useful they > |are. > They had not settled upon a common notion of function, which > is rather different from having happily assumed that they were all > continuous, by a deŽnition of continuous that also only became > standard later. I think they had a good understanding of Œcontinuous¹; it means the curve doesn¹t have any breaks in it. > Moreover, more than 90% of the real analysis that has been done, > has been done since then. So no, it really *doesn¹t* tell you how > useful discontinuous functions are. It¹s essentially impossible to say anything about a general function. Only when you assume at least continuity do you get some theorems. Andrew Usher === Subject: Re: The real numbers, and general comments >> |>Also, it is possible to take the position >> |> that, since continuous functions make up a very small percentage of the >> |> functions from R to R, that continuous functions are the pathological >> |> ones and should be ignored. I know that would make calculus students >> |happy. >> | >> |No one even thought that these things should be called Œfunctions¹ for >> |the Žrst 200 years of analysis. That should tell you how useful they >> |are. >> They had not settled upon a common notion of function, which >> is rather different from having happily assumed that they were all >> continuous, by a deŽnition of continuous that also only became >> standard later. > I think they had a good understanding of Œcontinuous¹; it means the > curve doesn¹t have any breaks in it. No it doesn¹t. >> Moreover, more than 90% of the real analysis that has been done, >> has been done since then. So no, it really *doesn¹t* tell you how >> useful discontinuous functions are. very useful, try learning some integration stuff. > It¹s essentially impossible to say anything about a general function. > Only when you assume at least continuity do you get some theorems. > Andrew Usher does the symbol L^2 mean anything? === Subject: Re: The real numbers, and general comments > It¹s essentially impossible to say anything about a general function. > Only when you assume at least continuity do you get some theorems. Look up the Lebesgue Dominated Convergence Theorem. -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: The real numbers, and general comments > I don¹t know what CZF is. That thread was started by Ross Finlayson, > not myself. > I was asking if the incompleness of ZFC should cause us to reject it > as the true set theory. Is a complete set theory possible, in the > sense that no additional axioms can falsify any of its theorems? > I strongly suspect the answer is no, though I¹d have to do some > checking. Also, even if CZF is complete, does it accurately describe > what you would normally thing of as set theory? Does it correspond with > your understanding of how sets work? I just said I know nothing about CZF. >Can you give an example of a useful function that isn¹t? (I don¹t >count the delta Œfunction¹, as that isn¹t even a valid function.) >> >>Useful for what? The above is useful for demonstrating that a function >>can have countably inŽnitely many non-zero values, yet have an integral >>= 0. > > Pathology for the sake of pathology is not what I consider Œuseful¹. > Interesting, yes, but not useful. > Pathology for the sake of illustrating unusual or unexpected properties > that run contrary to what most people would expect is useful. There is > a tendancy to assume that all functions continuous almost everywhere > that the function is normally deŽned, simply because that is all most > people encounter. In this case, it is worthwhile to keep people from > making wrong assumptions. Also, it is possible to take the position > that, since continuous functions make up a very small percentage of the > functions from R to R, that continuous functions are the pathological > ones and should be ignored. I know that would make calculus students happy. No one even thought that these things should be called Œfunctions¹ for the Žrst 200 years of analysis. That should tell you how useful they are. > Do you agree with my general point here, that analysis can be done > over any countable dense set, and for all piecewise continuous > functions, will give the same results? > It depends on how you deŽne the domain of the function. If you deŽne > the domain of f(x) = 1/x as R - {0}, then it is continuous on its entire > domain. If you deŽne the domain to be R, then f(x) is discontinuous at > 0. You can take the Žrst position, most people seem to understand the > second to be the case. This isn¹t an answer to my question. But, yes, I do deŽne the domain the second way; since 1/x is not deŽned at 0, it isn¹t part of the domain. >>If R is not just numbers, what are the other things? > >Uncomputable sequences would be a good description. >> >>It seems to me that it would be easier to use a new deŽnition of R in >>which you restrict yourself to computable numbers. Of course, that >>would result in R being countable. > > I¹m not sure if there¹s a good, short deŽnition of the computables > that doesn¹t mention any speciŽc computing machine. My deŽnition is > the union of all deŽnable archimedean ordered Želds. In this case, > it can only be gotten as the sup of an inŽnite sequence: > > Q < F1 < F2 < F3 < ... to inŽnity = S > > where the Fn are archimedean ordered Želds (Not all such sequences > will give all of S, and any that does will be uncomputable.). > By that deŽnition, it appears that you are saying R is uncountable. > Since it is known that the computable numbers are countable, you then > appear to be saying that R has non-computable members. Since R is > normally described as being numbers, wouldn¹t that suggest that there > are non-computable numbers? I did not mention R or Œuncountable¹ above, so I don¹t see where you are coming from here. I choose the letter S for the computables because that letter is not used for any of the normal sets, and because of its generality in my view. >>I don¹t see a problem with calling an uncomputable sequence a number, >>however. > > You can¹t calculate with it. > In a practical sense, you can¹t calculate with any number that is not > algebraic. If you are working with decimals, you can¹t calculate with > any number that isn¹t rational. Regardless, I don¹t view not being able > to calculate with a number as a problem. What do you mean by Œcalculate¹ here? Rational numbers can be calculated with exactly for arithmetic operations. Algebraic irrationals, though, can only be done through symbolic expressions, which is the same as for transcendentals. I say computable numbers are the set calculable over because you can evaluate any expression involing them; yes, I know this is essentially the deŽnition of computable, but it works. >>In one system, there are lines that do not intersect (Euclidean). In >>the other, every pair of lines intersects (projective). These are still >>different results. > > Given that the two geometries are consistent with each other, what is > the problem with unifying them as one? > They are not consistent with each other. Each is consistent with > itself. If you try to take all the axioms/deŽnitions together, you > will get an inconsistent system. At that point, it becomes useless. You have not shown any inconsistency. Euclidean, afŽne, and projective are just different types of theorems in my geometry - they differ in the types of symmetry, and the types of invariants they can employ. Andrew Usher === Subject: Re: The real numbers, and general comments >Can you give an example of a useful function that isn¹t? (I don¹t >count the delta Œfunction¹, as that isn¹t even a valid function.) >> >>Useful for what? The above is useful for demonstrating that a function >>can have countably inŽnitely many non-zero values, yet have an integral >>= 0. >Pathology for the sake of pathology is not what I consider Œuseful¹. >Interesting, yes, but not useful. >>Pathology for the sake of illustrating unusual or unexpected properties >>that run contrary to what most people would expect is useful. There is >>a tendancy to assume that all functions continuous almost everywhere >>that the function is normally deŽned, simply because that is all most >>people encounter. In this case, it is worthwhile to keep people from >>making wrong assumptions. Also, it is possible to take the position >>that, since continuous functions make up a very small percentage of the >>functions from R to R, that continuous functions are the pathological >>ones and should be ignored. I know that would make calculus students happy. > No one even thought that these things should be called Œfunctions¹ for > the Žrst 200 years of analysis. That should tell you how useful they > are. By that logic, using letters for variables isn¹t very useful either. The ancient Greeks did Žne without variables, yet we teach students to use them almost all the time. How recently something was discovered/named has little bearing on its usefulness. >Do you agree with my general point here, that analysis can be done >over any countable dense set, and for all piecewise continuous >functions, will give the same results? >>It depends on how you deŽne the domain of the function. If you deŽne >>the domain of f(x) = 1/x as R - {0}, then it is continuous on its entire >>domain. If you deŽne the domain to be R, then f(x) is discontinuous at >>0. You can take the Žrst position, most people seem to understand the >>second to be the case. > This isn¹t an answer to my question. But, yes, I do deŽne the domain > the second way; since 1/x is not deŽned at 0, it isn¹t part of the > domain. I want to be careful about answering these types of questions because I¹m rusty on my analysis. I don¹t immediately see a problem with doing analysis on countable dense sets. I¹m far from certain what types of results you will get, however. >>If R is not just numbers, what are the other things? > >Uncomputable sequences would be a good description. >> >>It seems to me that it would be easier to use a new deŽnition of R in >>which you restrict yourself to computable numbers. Of course, that >>would result in R being countable. >I¹m not sure if there¹s a good, short deŽnition of the computables >that doesn¹t mention any speciŽc computing machine. My deŽnition is >the union of all deŽnable archimedean ordered Želds. In this case, >it can only be gotten as the sup of an inŽnite sequence: > Q < F1 < F2 < F3 < ... to inŽnity = S >where the Fn are archimedean ordered Želds (Not all such sequences >will give all of S, and any that does will be uncomputable.). >>By that deŽnition, it appears that you are saying R is uncountable. >>Since it is known that the computable numbers are countable, you then >>appear to be saying that R has non-computable members. Since R is >>normally described as being numbers, wouldn¹t that suggest that there >>are non-computable numbers? > I did not mention R or Œuncountable¹ above, so I don¹t see where you > are coming from here. I choose the letter S for the computables > because that letter is not used for any of the normal sets, and > because of its generality in my view. I¹m trying to understand what you think R is. You responded with a set S, which you called the computable numbers. Somehow, S itself is not computable, which suggests it is also uncountable. Now, if S is what you think of as the real numbers, you clearly have a different understanding of them than I do. If not, I still don¹t know what you mean by the real numbers. I¹m getting the feeling that you would prefer to work with some non-standard analysis, but I have no idea what axioms you wish to work from. I also don¹t know how much you¹ve studied it. >>I don¹t see a problem with calling an uncomputable sequence a number, >>however. >You can¹t calculate with it. >>In a practical sense, you can¹t calculate with any number that is not >>algebraic. If you are working with decimals, you can¹t calculate with >>any number that isn¹t rational. Regardless, I don¹t view not being able >>to calculate with a number as a problem. > What do you mean by Œcalculate¹ here? Rational numbers can be > calculated with exactly for arithmetic operations. Algebraic > irrationals, though, can only be done through symbolic expressions, > which is the same as for transcendentals. I mean any form of symbol manipulation that can be done with a Žnite number of symbols. Using fraction bars and radicals is perfectly Žne with me. However it¹s represented, however, there should be a Žnite method for accessing any given decimal digit of any stage of the result. > I say computable numbers are the set calculable over because you can > evaluate any expression involing them; yes, I know this is essentially > the deŽnition of computable, but it works. Since a computable number has a Žnite representation (the algorithm to produce it to arbitrary precision), that works for me as well. >>In one system, there are lines that do not intersect (Euclidean). In >>the other, every pair of lines intersects (projective). These are still >>different results. >Given that the two geometries are consistent with each other, what is >the problem with unifying them as one? >>They are not consistent with each other. Each is consistent with >>itself. If you try to take all the axioms/deŽnitions together, you >>will get an inconsistent system. At that point, it becomes useless. > You have not shown any inconsistency. Euclidean, afŽne, and > projective are just different types of theorems in my geometry - they > differ in the types of symmetry, and the types of invariants they can > employ. Theorem 1: There exists two lines that do not intersect. Theorem 2: There do no exist two lines that do not intersect. Theorem 1 is an axiom in Euclidean geometry. Theorem 2 is a theorem/axiom of projective geometry. Theorem 1 is true if and only if Theorem 2 is false. This means you cannot have a model the union of Euclidean and projective geometries. They are inconsistent because they produce mutually exclusive theorems. If you disagree with my claim that they are inconsistent, please show me where my mistake is. More useful would be to spell out the deŽnitions/axioms/invariants of your geometry. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: The real numbers, and general comments >>Pathology for the sake of illustrating unusual or unexpected properties >>that run contrary to what most people would expect is useful. There is >>a tendancy to assume that all functions continuous almost everywhere >>that the function is normally deŽned, simply because that is all most >>people encounter. In this case, it is worthwhile to keep people from >>making wrong assumptions. Also, it is possible to take the position >>that, since continuous functions make up a very small percentage of the >>functions from R to R, that continuous functions are the pathological >>ones and should be ignored. I know that would make calculus students happy. > > No one even thought that these things should be called Œfunctions¹ for > the Žrst 200 years of analysis. That should tell you how useful they > are. > By that logic, using letters for variables isn¹t very useful either. > The ancient Greeks did Žne without variables, yet we teach students to > use them almost all the time. This is a different kind of thing. No one can diagree with improvements in notation because they režect what was already being done (here, algebraic manipulation). On the other hand, discontinous functions weren¹t Œbeing done¹ until some rigorists used them as pathological examples. > I want to be careful about answering these types of questions because > I¹m rusty on my analysis. I don¹t immediately see a problem with doing > analysis on countable dense sets. I¹m far from certain what types of > results you will get, however. OK. My restriction to piecewise-continuous functions is because they only have a countable amount of information., so that one will not lose any by going to a countable set. >I¹m not sure if there¹s a good, short deŽnition of the computables >that doesn¹t mention any speciŽc computing machine. My deŽnition is >the union of all deŽnable archimedean ordered Želds. In this case, >it can only be gotten as the sup of an inŽnite sequence: > > Q < F1 < F2 < F3 < ... to inŽnity = S > >where the Fn are archimedean ordered Želds (Not all such sequences >will give all of S, and any that does will be uncomputable.). >> >>By that deŽnition, it appears that you are saying R is uncountable. >>Since it is known that the computable numbers are countable, you then >>appear to be saying that R has non-computable members. Since R is >>normally described as being numbers, wouldn¹t that suggest that there >>are non-computable numbers? > > I did not mention R or Œuncountable¹ above, so I don¹t see where you > are coming from here. I choose the letter S for the computables > because that letter is not used for any of the normal sets, and > because of its generality in my view. > I¹m trying to understand what you think R is. You responded with a set > S, which you called the computable numbers. Somehow, S itself is not > computable, which suggests it is also uncountable. Now, if S is what > you think of as the real numbers, you clearly have a different > understanding of them than I do. If not, I still don¹t know what you > mean by the real numbers. It¹s a theorem that S is non-computable. Assume it is. Then you can diagonalise over it to get another number, which is computable. Contradiction. One could say it is Œcomputably uncountable¹. When I say R, or the real numbers, I mean the same thing you do. > I¹m getting the feeling that you would prefer to work with some > non-standard analysis, but I have no idea what axioms you wish to work > from. I also don¹t know how much you¹ve studied it. I would like to study it, yes. > I say computable numbers are the set calculable over because you can > evaluate any expression involing them; yes, I know this is essentially > the deŽnition of computable, but it works. > Since a computable number has a Žnite representation (the algorithm to > produce it to arbitrary precision), that works for me as well. Right. > You have not shown any inconsistency. Euclidean, afŽne, and > projective are just different types of theorems in my geometry - they > differ in the types of symmetry, and the types of invariants they can > employ. > Theorem 1: There exists two lines that do not intersect. > Theorem 2: There do no exist two lines that do not intersect. > Theorem 1 is an axiom in Euclidean geometry. > Theorem 2 is a theorem/axiom of projective geometry. > Theorem 1 is true if and only if Theorem 2 is false. > This means you cannot have a model the union of Euclidean and projective > geometries. They are inconsistent because they produce mutually > exclusive theorems. If you disagree with my claim that they are > inconsistent, please show me where my mistake is. > More useful would be to spell out the deŽnitions/axioms/invariants of > your geometry. You haven¹t stated the theorems correctly. Th. 1: there exist two lines that do not intersect _on the Euclidean plane_. Th. 2: there do not exist two lines that do not intersect _on the projective plane_. These are consistent with one another. (The Euclidean/afŽne plane does not include the ideal line.) Andrew Usher === Subject: Re: The real numbers, and general comments >>Pathology for the sake of illustrating unusual or unexpected properties >>that run contrary to what most people would expect is useful. There is >>a tendancy to assume that all functions continuous almost everywhere >>that the function is normally deŽned, simply because that is all most >>people encounter. In this case, it is worthwhile to keep people from >>making wrong assumptions. Also, it is possible to take the position >>that, since continuous functions make up a very small percentage of the >>functions from R to R, that continuous functions are the pathological >>ones and should be ignored. I know that would make calculus students happy. >No one even thought that these things should be called Œfunctions¹ for >the Žrst 200 years of analysis. That should tell you how useful they >are. >>By that logic, using letters for variables isn¹t very useful either. >>The ancient Greeks did Žne without variables, yet we teach students to >>use them almost all the time. > This is a different kind of thing. No one can diagree with > improvements in notation because they režect what was already being > done (here, algebraic manipulation). On the other hand, discontinous > functions weren¹t Œbeing done¹ until some rigorists used them as > pathological examples. How is new terminology (functions) different from new symbols? Both represent a new way of thinking about an old concept. When you upgrade/formalize your concept, you sometimes discover new possibilities. >You have not shown any inconsistency. Euclidean, afŽne, and >projective are just different types of theorems in my geometry - they >differ in the types of symmetry, and the types of invariants they can >employ. >>Theorem 1: There exists two lines that do not intersect. >>Theorem 2: There do no exist two lines that do not intersect. >>Theorem 1 is an axiom in Euclidean geometry. >>Theorem 2 is a theorem/axiom of projective geometry. >>Theorem 1 is true if and only if Theorem 2 is false. >>This means you cannot have a model the union of Euclidean and projective >>geometries. They are inconsistent because they produce mutually >>exclusive theorems. If you disagree with my claim that they are >>inconsistent, please show me where my mistake is. >>More useful would be to spell out the deŽnitions/axioms/invariants of >>your geometry. > You haven¹t stated the theorems correctly. > Th. 1: there exist two lines that do not intersect _on the Euclidean > plane_. > Th. 2: there do not exist two lines that do not intersect _on the > projective plane_. > These are consistent with one another. (The Euclidean/afŽne plane > does not include the ideal line.) This is where laying down your terms becomes important. Based on what you had said, I had know way of knowing that you had two different planes in mind. Until you state the deŽnitions, theorems, axioms, etc, how is someone to know you have a new geometry in mind that encompasses both of the old ones? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Academy123.com - Solutions to Algebra Textbook Problems I am proud to announce that Academy123.com has launched: http://www.academy123.com We are offering Algebra homework help for middle school and high school students. You can actually hear a teachers voice and see the handwriting ink strokes (like an invisible teacher) for each and every problem in the most popular Algebra I textbooks. Please take a look! === Subject: Re: Important, best mathematics universities > wish to be a pure & applied mathematics major (well, majors), I was > hoping for the suggestions of the participants of this forum for any > suggestions on where they believe the best mathematics education is > offered. > The colleges I already plan on applying deŽnitely are MIT, Columbia, > Yale, and Berkeley. If you have any other suggestions, or helpful > comments about what you know about these schools, they are greatly > -Bobby (an avid reader of the newsgroup) Go to http://www.phds.org/rankings/ and enter your requirments. I was surprised that my #1 was UC Santa Barbara, ranked above the usual places, MIT, CIT, Harvard, Princeton, Berkeley, Yale. (What about Stanford--I don¹t recall seeing it). Does anyone know anything about UCSB. I had not thought of it in that light. I don¹t even recall if I searched for Physics or Math--physics I think, yes. For math, you will have to do it yourself. Van === Subject: Re: Important, best mathematics universities Ecole Normale Sup.8erieure rue d¹Ulm, France: You can try to get in there: http://www.ens.fr/international/UK/index.htm A good thing to know. Students don¹t pay anything: they¹re actually paid by the french government. Paul. (I wish I were there :p) __________________________________ That¹s not exhaustive... but here are a part of their wall of Fame: Mathematicians: Bourbaki¹s group Henri Cartan Alain Connes (m.8edaille Fields 1982) .83variste Galois Laurent Lafforgue (Fields medal 2002) Laurent Schwartz (Fields medal 1950) Jean-Pierre Serre (Fields medal 1954) Ren.8e Thom (Fields medal 1958) Jean-Christophe Yoccoz (Fields medal 1994) Physicians: Claude Cohen-Tannoudji (prix Nobel 1997) Pierre-Gilles de Gennes (prix Nobel 1991) Alfred Kastler (prix Nobel 1965) Gabriel Lippmann (prix Nobel 1908) Louis N.8eel (prix Nobel 1970) Jean Perrin (prix Nobel 1926) ______________________ === Subject: Math Brain Teaser Number Game http://www.vsisystems.com/numberboy.asp === Subject: Re: Math Brain Teaser Number Game > http://www.vsisystems.com/numberboy.asp (2x+4)/2-x = 2 And the only real question is: why limit the choice of x to a number between 1 and 25? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Best Softwares & Cracks HERE => Download NOW Hottest and best full Software Titles & Cracks For Sale. 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logarithm. Is there something in a beacht tent what can be related to logarithms. I understand that the frame is made of triangles so that¹s goniometric but can anybody help me with logarithms? === Subject: Re: beach tent and logarithm > Does anybody know the relation between a beach tent and the math function > logarithm. > Is there something in a beacht tent what can be related to logarithms. I > understand that > the frame is made of triangles so that¹s goniometric but can anybody help me > with logarithms? I don¹t believe that it has anything to do with logs. I don¹t know about any triangles, but its gravity that makes the shape. A line with gravity makes a catenary (see CRC for a picture, or a physics text). The tent is gravity acting over a surface. It sounds like the soap bubble problem, which is a great problem, I recomment it. I don¹t know if the shape has a name or not. Do a search for the soap bubble problem. Van === Subject: Time delation. A real concept or a mistake of physics? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GC6NW32463; Please try to come in http://ftp.ninhthuanpt.com.vn/ntacademy/xemnoidung.asp?TID= 270&PN=1 === Subject: Re: Time delation. A real concept or a mistake of physics? Whatever it is (apart from a spelling mistake) it¹s a physical not a mathematical concept and so is off-topic here. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: beach tent and logarithm I think I was wrong about soap bubbles--surface tension is too important. I am sure this problem has been worked on though-- gravity on 2D rather than 1D, as for catenary is still how I would do it. === Subject: Re: beach tent and logarithm I don¹t believe that it has anything to do with logs. I don¹t know about any triangles, but it is gravity that makes the shape. A line with gravity makes a catenary (see CRC for a picture, or a physics text). The tent is gravity acting over a surface. Van === Subject: Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GDZcp08750; The question is in the subject : Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? It seems to me that it may be related to the distribution of sin(n) in [-1,1] but I don¹t know much about such topics. DETOURRE Eric eric.detourre@laposte.net === Subject: Re: Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? > The question is in the subject : > Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? > It seems to me that it may be related to the distribution of sin(n) in [-1,1] > but I don¹t know much about such topics. Permit me to ask two stupid questions: what is the function E? And is it really supposed to be the exponent of the -1? If by E(2*sin(n)) you mean exp(2*sin(n)) then I don¹t know what (-1) to this power is supposed to BE. Some sort of principal value? If so, this looks hard. --Ron Bruck === Subject: Re: Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? > The question is in the subject : > Is the inŽnite series sum (-1)^E(2*sin(n))/n convergent ? > It seems to me that it may be related to the distribution of sin(n) in [-1,1] > but I don¹t know much about such topics. What is E? Are you summing [(-1)^E][(2*sin(n))/n]? or [(-1)^E(2*sin(n))]/n? or (-1)^[E(2*sin(n))/n]? or what? -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: Euclidean geometry problem by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GDZbw08704; 2 CIRCLES 1 RADIUS 2CM FIT INSIDE A RECTANGLE 5CM X 6CM BOTH CIRCLES TOUCH EACH SIDE OF THE RECTANGLE AND ALSO EACH OTHER WHAT IS THE RADIUS OF THE 2ND CIRCLE === Subject: Re: Euclidean geometry problem ETAtAhRD21ZILaJ+qJBg13/ nWo5Zuk72HAIVALWapZo4sfNxM8shisqT4cmgRD5n I get only a 2nd degree equation. 20 lines to solution 15 10 5 Let A, B, C, D be the vertices of the rectangle in rotational order, with AB = 6 and BC = 5. Let P_1 be the enter of a circle having the given 2-cm radius and put that circle inside the rectangle, žush against AB and BC. The second circle is then žush against CD and DA, with radius r and center P_2. The distance form P_1 to each of the tangent sides o Circle 1 is 2, while the corresponding distance on Circle 2 is r. Therefore the projections of P_1 P_2 onto AB and BC are, respectively, 4-r and 3-r, and from the Pythagorean Theorem we have: (4-r)^2 + (3-r)^2 = (P_1 P_2)^2 Now if the circles are externally tangent to each other as well as the sides of the rectangle, then P_1 P_2 is simply r+2 because the line segment between the centers passes through the point of tangency. So r^2 - 18r + 21 = 0 giving r = 9-sqrt(60). The other root of the quadratic equation is too large to Žt into the rectangle. It corresponds to a circle that encloses B and is tangent to EXTENSIONS of CD and DA beyond the rectangle. --OL === Subject: Re: Euclidean geometry problem >2 CIRCLES 1 RADIUS 2CM FIT INSIDE A RECTANGLE 5CM X 6CM >BOTH CIRCLES TOUCH EACH SIDE OF THE RECTANGLE AND ALSO EACH OTHER WHAT IS THE RADIUS OF THE 2ND CIRCLE MAPLE OR MATHEMATICA EXPERTS --- I THINK THIS LEADS TO AN 8TH DEGREE EQUATION. (x-r)^2 + (y-r)^2 = r^2 (with r=unknown.) (x-3)^2 + (y-4)^2 = 4 (radius 2 circle.) The last is because the radius 2 circle at the far corner must have center (3,4) to touch both edges which are the lines x=5 and y=6. These equations tell us that (x,y) is on both circles, but it does not tell us that the two circles are tangent to one another. The line connecting the two centers in this case must also contain (x,y), and the slope of this line must be the same measured any way. (x-r)/(y-r) = (x-3)/(y-4) = (r-3)/(r-4) so (x-r)(r-4) = (y-r)(r-3) CAN SOMEBODY WHO IS GOOD AT USING MAPLE PLEASE CHECK THIS OUT? IS THERE A SOLUTION IN RADICALS? The three equations are: (x-r)^2 + (y-r)^2 = r^2 (x-3)^2 + (y-4)^2 = 4 (x-r)(r-4) = (y-r)(r-3) === Subject: Re: Euclidean geometry problem >2 CIRCLES 1 RADIUS 2CM FIT INSIDE A RECTANGLE 5CM X 6CM >BOTH CIRCLES TOUCH EACH SIDE OF THE RECTANGLE AND ALSO EACH OTHER WHAT IS THE RADIUS OF THE 2ND CIRCLE > (x-r)^2 + (y-r)^2 = r^2 (with r=unknown.) > (x-3)^2 + (y-4)^2 = 4 (radius 2 circle.) Solving the above for x and y, Mathematica gives a quantity under radical sign:- 3969 + 7938 r - 5967 r^2 + 2172 r^3 - 399 r^4 + 34 r^5 - r^6, which should vanish for touching circles.Apart from the required solution r = 1.25404, r = 3,(thrice repeated), 7 and 16.746 are other roots. r = 3 is an internal contacting circle, but wonder to what other conŽgurations radii 7 and 16.746 corrospond. === Subject: Re: Euclidean geometry problem ETAsAhRMbQo4tx009mVAPIGkRU2GbQitNgIUV1CCyFs74U3Y72Zi8trZpb9IAc w= The larger values of r correspond to circles that are tangent to EXTENSIONS of the sides of the rectangle. r = 7 corresponds to internal contact, with the larger circle warrping around the given one. r = 16.7+ (actually 9+sqrt(60)) corresponds to external contact with the given circle; this circle wraps around a vertex of the rectangle and is tangent to the extensions of two sides opposite that vertex. The circle required in the original problem has r = 9-sqrt(60) and this is the 1.25+ root. See my Žrst posting in this thread for the derivation of the radical roots. --OL === Subject: Re: Euclidean geometry problem <414a110e.22862558@news.east.earthlink.net> > CAN SOMEBODY WHO IS GOOD AT USING MAPLE PLEASE CHECK > THIS OUT? IS THERE A SOLUTION IN RADICALS? > The three equations are: > (x-r)^2 + (y-r)^2 = r^2 > (x-3)^2 + (y-4)^2 = 4 > (x-r)(r-4) = (y-r)(r-3) Four solutions with Maple 9.5: x y r 3 6 3 7/5 14/5 7 (195+20*s)/d (234+24*s)/d 9+2*s (195-20*s)/d (234-24*s)/d 9-2*s where s=sqrt(15) and d=61. === Subject: Re: Euclidean geometry problem ETAtAhQrlOC2QcjjCFti06cM8VzMk9NAYAIVAJabsvmekRP1egaFHg4K9qgk/ /5i There are only four solutions because the system derived by Jeffrey H is only of devree 4, not 8. The difference etween the curcle equations is linear, hence the reduction of degtree. The quartic equation for r is actually (r^2-10r+21)*(r^2-18r+21) = 0 where the factor r^2-10r+21 corresponds to the smaller circle being inside rather than outside the larger one. We know from my solution that r^2-18r+21 is the proper factor for putting the smaller circle outside the larger one. --OL === Subject: Re: Euclidean geometry problem ETAuAhUAqGqDZ1vhLvG+2ScLx1evDed95dUCFQDIhPhKt+kIMuyGf+ rY2xQvrb7YFA== nt t say that the difference between the Žrst tw equations is inear in x and y. (Œm not used to r being an unknown in this kid of system.) We also do have a linear quation in r, so certain it is that the system ends up being quartic. --OL === Subject: Re: rule of signs by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GDgEp09308; >to solve for the roots of a polynomial equation is by using the >descartes rule of signs. how about if dealing with continuous >function, is descartes rule of signs still applicable or is their any >methods to be used. I¹m not sure I understand your question. DesCarte¹s rule of signs does not solve polynomial equations- it only puts an upper bound on the number of positive or negative real roots it has: If there are n changes in sign when the polynomial is written in increasing or decreasing order of powers, then the number of positive roots is n, or n minus some even number. To Žnd the same for negative roots, replace x with -x. In particular, DesCarte¹s rule, applied to x^4- x^3+ x^2- x+ 1= 0, tells us, since there are 4 changes in sign, there COULD be 4 or 2 or 0 positive roots but tells us nothing about what those roots could be. (Replacing x with -x we get x^4+ x^3+ x^2+ x+ 1 which has NO sign changes and so no positive roots so the original equation has no negative roots.) As far as continuous functions are concerned, well, of course, polynomials ARE continuous functions! If you mean non-polynomial functions, then, since you CAN¹T put the function in increasing or decreasing order of powers it¹s not clear HOW DesCartes rule would apply. === Subject: Inverse of space? Hi all, Is there something like inverse of space. Can anybody suggest me a Papu. === Subject: Re: Demonstrating that 0.999... = 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GENmE13532; >>In case you¹re not convinced that these questions need >>to be answered: Regard the string ...111 as a view of >>a number. Now manipulate that number. Call it x: >> x = ....111 >> 10x = ...1110 >>Subtract those two equations and you get 9x = -1. >>So by regarding an inŽnite string as a view of a number >>and manipulating that string we¹ve shown that ...111 = -1/9. >>Presumably you don¹t believe that. And of course the error >>in the proof is that ...111 is not a view of a number; >>the manipulations are not valid. >But that isn¹t the same as what OP was doing. His step speciŽcaly avoided doing >anything that would lead to the question: But what happens at Œthe end¹? pardon me what end!!!! we have an inŽnite number of 9¹s what end?????? >Your example, on the other hand, very much would prompt the question: But what >happens to the leftmost one when you multiply by 10?. >I thought that OP had demonstrated quite clearly the validity of the >manipulations being made. That was, surely, the point of the method. === Subject: Re: Demonstrating that 0.999... = 1 >>But that isn¹t the same as what OP was doing. His step speciŽcaly avoided doing >>anything that would lead to the question: But what happens at Œthe end¹? >pardon me what end!!!! >we have an inŽnite number of 9¹s what end?????? Exactly ! === Subject: Re: I wanna be a big mathematician! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GENlL13504; >Hi everybody, >My level of mathematics is standard and I am getting older :=( >so I would like to know which books should I start with in order to be a big mathematician! >For calculus? mathematic analysis? any other Želd? If I wanted to be a big mathematician I would start by eating at McDonalds 3 times a day..... :-) === Subject: Re: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GENld13496; >Please try to come in http://ftp.ninhthuanpt.com.vn/ntacademy/xemnoidung.asp?TID= 270&PN=1 In addition to being off topic, the entire website is in Vietnamese! Sorry, that¹s one of the many languages I can¹t read. === Subject: Re: Final Step of the Prine Number Theorem, I can not understand by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GENlZ13500; >V is piecewise continuous with jump discontinuities. >In between its discontinuities it is decreasing. Its jump >discontunities are increasing. We are supposing that V is nonzero >on I. This is the main point I can not understand. How do we know that V is nonzero on I? Is the supposition valid? What if there are more than one discontinuities in I? >It cannot change sign except at its discontinuities. >If it changes sign at x say, immediately to the right of x >it is positive and decreasing. It decreases and remains positive >until the next jump where it increases again etc. >-- >Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 >Francis Wheen, _How Mumbo-Jumbo Conquered the World_ erdos fan === Subject: Re: Final Step of the Prine Number Theorem, I can not understand >>V is piecewise continuous with jump discontinuities. >>In between its discontinuities it is decreasing. Its jump >>discontunities are increasing. We are supposing that V is nonzero >>on I. > This is the main point I can not understand. How do we know that > V is nonzero on I? We don¹t. That¹s why Wright splits into cases according to whether V vanishes on the interval or not. Theorem 436 deals with the case where V vanishes. > Is the supposition valid? What if there are > more than one discontinuities in I? Either V vanishes, or it changes sign once or not at all. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Why are there more Mersenne primes than Fermat primes??? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GElXS15413; >I¹ve been looking around, and everyone says that there are so MANY >Mersenne primes (compared to a Žnite set), but almost NO Fermat >primes! >I just don¹t understand this! >Just because Fermat numbers grow faster, They don¹t just grow faster, they grow a LOT faster. This is important. >that means that there should >be less primes in the sequence? But why? These heuristic arguments >don¹t really have much sense, no offense intended... None taken. But you need to say *why* you believe they don¹t make sense. Do you understand the Prime Number Theorem? Do you understand that it applies uniformly? Take a large integer X. The probability that an integer randomly drawn from a set of integers in a close interval around X is prime is 1/log(X). For Mersenne primes, sum to inŽnity of 1/log(M_p) diverges. We thus expect inŽnitely many. For Fermat primes, sum to inŽnity of 1/log(F_n) converges quite quickly. We thus expect Žnitely many. Why is this a mystery? Consider the Fermat numbers less than M. Call this F(M). Consider the Mersenne numbers less than M. Call this Me(M). It is easy to show that limit m-->oo F(m)/Me(M) is 0. The Fermat numbers form a very very thin set within the integers. The Mersenne numbers merely form a very thin set. There just aren¹t ENOUGH Fermat numbers within the integers to have inŽnitely many of them be prime on probabilistic grounds. >I mean, the sequence of primes is inŽnite, so what are chances that >there are no sub-sequences that grow faster than e^(n^2) ??? One must be careful about deŽning the proper measures here, but one would expect that almost all integer sub-sequences such that 1/log(n) converges have only Žnitely many primes. >Not to mention the existance of Mill¹s constant(s). >I know that from actual computational expirience, it seems that >Elliptic Divisibility Sequences have only Žnitely many primes, but if >I had to actually guess, I would guess that there are inŽnitely many >sequences who have inŽnitely many primes. You are correct. There are. But the set of such sequences has density 0 relative to the set of all possible sequences. [again, being careful about measure]. It is even easy to construct sets with the same density as Fermat numbers that DO have inŽnitely many primes. But such sets are extremely rare among the set of all possible such sets. example: Let FP(n) be the smallest prime > 2^(2^n). Here, EVERY element is prime. This set has the same asymptotic density as Fermat numbers. But such sets are RARE. === Subject: Re: Demonstrating that 0.999... = 1 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GF9al18062; Why do so many people feel that it is axiomatic that 0.999... is not equal to 1? It seems that the fundamental problem is that these people object to the idea of one real number having more than one decimal representation. Why is this so hard to accept? Decimals are not the same thing as reals, so why would you assume that there would automatically be some isomorphic equivalence between them? This just seems like a really strong assumption to make. Matthew >I know how frequently the topic comes up here, although I¹ve not taken much >notice of it (So excuse me if this method has been suggested before). >I was trying to explain this to someone who wasn¹t at all convinced, and I >thought I¹d post the step by step method I used to Žnally convince her that it >was in fact the case. >It involves asking the doubter a series of questions: >1) Do you accept that if you divide 1 by 9, you will get an answer consisting of >a decimal point followed by an inŽnitely long string of 1¹s >2) Do you accept that string is known as 0.111... ? >3) Do yo accept that if you multiply a number by a single digit, and there are >no carries, then the answer is simply each digit in the original string replaced >by that digit multipled by the selected single digit >4) Do you therefore accept that 0.111... x 9 = 0.999... >5) Do you accept that x / 9 * 9 = x >If you accept all the above, you must accept that 0.999... = 1 >I found the advantage of this approach was that it avoids the doubter having to >worry about what happens at inŽnity :), and also avoids the need to even >touch on limits. I always feel that a lot of the proofs of this are very >over-complicated, and this only increases the Œsuspicion¹ in those who are not >convinced. >I haven¹t yet found another Œvictim¹ to try it out on. >OK, žame away! >PRL === Subject: Re: Demonstrating that 0.999... = 1 > Why do so many people feel that it is axiomatic that 0.999... is not equal to 1? IMO, because they are trained that if two decimals x and y are not equal, then there is a leftmost digit at which they differ; e.g.: x = 0.136278934177... y = 0.136274184567... ^ The conceptual problem is that they do not realize that the converse of this statement is not true: if x and y are decimals which differ at some leftmost digit, then they are not equal. Counterexample: 0.999... = 1.0 Since they have not been taught that decimals are limits, this fails to make sense to them; in almost every case, the converse _is_ true. === Subject: analytic...¹ hello.....doctor~ Žnd analytic function such that f(1/n) = e^(-n) for all positive integer n. ---------------------------------------- um.....i think..... f(z) = e^(-1/z). but this is not analytic at point z=0. even if i deŽned to f(0) = 0, it¹s not analytic at point z=0. help me, please~ thank you very much for your advice. === Subject: Re: analytic...¹ >Žnd analytic function such that f(1/n) = e^(-n) for all positive integer n. >f(z) = e^(-1/z). >but this is not analytic at point z=0. If you want an answer that is analytic at z=0, you¹re out of luck. Hint: if f has a zero of order m at z=0, what can you say about |f(1/n)| n^m as n -> inŽnity? Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: analytic...¹ > Žnd analytic function such that f(1/n) = e^(-n) for all positive integer n. > ---------------------------------------- > um.....i think..... > f(z) = e^(-1/z). > but this is not analytic at point z=0. > even if i deŽned to f(0) = 0, it¹s not analytic at point z=0. > help me, please~ Help us, please: Where is f supposed to be analytic? Is it Žnd all or Žnd a? You often leave out important details like this. === Subject: Re: analytic...¹ > Žnd analytic function such that f(1/n) = e^(-n) for all positive integer n. > ---------------------------------------- > um.....i think..... > f(z) = e^(-1/z). > but this is not analytic at point z=0. > even if i deŽned to f(0) = 0, it¹s not analytic at point z=0. > help me, please~ > Help us, please: Where is f supposed to be analytic? Is it Žnd all or > Žnd a? You often leave out important details like this. oh...i am sorry. but it¹s not my mistake. i think that original problem is imperfect. thank you very much for your indication. === Subject: Lines in R^2 Someone told me that the set of all lines in R2 has a natural structure as a differentiable manifold. Can anyone tell me how I can see this? I know the set of all lines passing through the origin is just the RP1. But how about those which do not pass through the origin? === Subject: derivative... hello.....doctor~ derivative of sin(1/x) is lim [sin{(1/x)+h} - sin(1/x)] / h h->0 i want to deduce (-1/(x^2))*cos(1/x) from this. is it possible calculation ? but i don¹t know the method. thank you very much for your advice. === Subject: Re: derivative... > hello.....doctor~ > derivative of sin(1/x) is > lim [sin{(1/x)+h} - sin(1/x)] / h > h->0 No, it¹s lim_h->0 [sin{(1/(x+h)} - sin(1/x)] / h > i want to deduce (-1/(x^2))*cos(1/x) from this. > is it possible calculation ? Why not use the chain rule? === Subject: Re: derivative... > hello.....doctor~ > derivative of sin(1/x) is > lim [sin{(1/x)+h} - sin(1/x)] / h > h->0 oh....i am sorry. mistatke. lim [sin{(1/(x+h)} - sin(1/x)] / h h->0 > i want to deduce (-1/(x^2))*cos(1/x) from this. > is it possible calculation ? > but i don¹t know the method. > thank you very much for your advice. === Subject: Re: derivative... >>hello.....doctor~ >>derivative of sin(1/x) is >>lim [sin{(1/x)+h} - sin(1/x)] / h >>h->0 > oh....i am sorry. mistatke. > lim [sin{(1/(x+h)} - sin(1/x)] / h > h->0 >>i want to deduce (-1/(x^2))*cos(1/x) from this. >>is it possible calculation ? >>but i don¹t know the method. >>thank you very much for your advice. First, use this trigononometric identity: sin(A) - sin(B) = 2 cos((A+B)/2) sin((A-B)/2) to simplify [sin{(1/(x+h)} - sin(1/x)] into this form: 2 cos( [ 1/(x+h) + 1/x ]/2 ) sin( [ 1/(x+h) - 1/x ]/2 ) or: 2 cos( (2x+h)/(2x(x+h)) ) sin( -h / (2 x (x+h)) ) Next, notice that the cosine factor approaches cos(1/x) as h --> 0. About the sine factor: lim_{h-->0} ( sin( -h/(2x(x+h)) )/h ) You could use the fact that sin(x)/x --> 1 as x --> 0 to do this: As (-h/(2x(x+h)) --> 0, sin( -h /(2x(x+h))) / (-h/(2x(x+h))) --> 1 so, sin( -h/(2x(x+h)) ) / h ------------------------ ---> 1 (-h/(2x(x+h))) / h so, the two limits lim_{h-->0} { sin( -h/(2x(x+h)) )/h } and lim_{h-->0} { (-h/(2x(x+h))) / h } are the same (i.e., they either both exist and have the same value, or they both fail to exist). The latter limit, however, is simple to compute, since the fraction simpliŽes: (-h/(2x(x+h))) / h = -h/(2x(x+h)h) = -1/(2x(x+h)) and this goes to -1/(2x^2) as h --> 0. Finally, we put it all together: lim_{h-->0} ( [sin{(1/(x+h)} - sin(1/x)] / h ) = lim_{h->0}( 2 cos( (2x+h)/(2x(x+h)) ) sin( -h / (2 x (x+h)) )/h } = 2 lim_{h->0}cos( (2x+h)/(2x(x+h)) ) * lim_{h->0} { (1/h)*sin( -h / (2 x (x+h)) ) } We¹ve calculated the individual limits, so we put everything where it belongs: = 2 cos(1/x) (-1/(2x^2)) = -cos(1/x)/x^2 === Subject: Re: Raatikainen¹s critique of Chaitin > The part we don¹t agree is case 1: > 1) There is an inŽnite sequence of stronger and stronger theories > the characteristic constant of which are equal > The characteristic constant of a theory T is determined by the smallest > index e, s.t. for no m does T prove > Ad<=e ( {d}(0) is undeŽned / {d}(0)=/= m ) > where {k} is the machine with index k. [The term characteristic constant is not being used anywhere, could we please stick with the terms in the original work?] Let¹s make it very clear that we have been referring to the same number because your above formula does not make sense to me. [Is Ad=H(d)? Why <=? What is {k}(0)?] All along, Raatikainen and you have been talking about Theorem LB, which states something else. I am going to quote Theorem LB (referred to as Chaitin¹s Theorem in Raatikainen¹s paper) in its entirety from third edition of AIT monograph, page 198, and ask you a question to clarify it. Question: Is the characteristic constant you have been talking about Œn¹ or ŒO(1)¹ in Œn <= H(s) + O(1)¹ in the formulation of Chaitin¹s weakness incompleteness result, namely Theorem LB of AIT? According to Raatikainen¹s paper, it ought to be Œn¹. (This is explained in the very Žrst page) Theorem LB [Chaitin (1974a,1974b,1975a,1982b)] Consider a formal theory all of whose theorems are assumed to be true. Within such a formal theory a speciŽc string cannot be proven to have information content more than O(1) greater than the information content of the axioms of the theory. I.e., if H(s) >= n is a theorem only if it is true, then it is a theorem only if n <= H(axioms) + O(1). Conversely, there are formal theories whose axioms have information content n+O(1) in which it is possible to establish all true propositions of the form H(s) >= n and of the form H(s) = k with k < n. I am asking this so that we can avoid further misunderstandings. Now, what is it that you guys don¹t like about n <= H(axioms) + O(1) In addition, do you know what the invariance theorem is, and how it relates to this small theorem? Do you understand that we are dealing with only self-delimiting programs (Levin) so that subadditivity of algorithmic information holds in the proof? Do you understand that O(1) depends on only the rules of inference, and the universal computer chosen, and is thus covered by invariance theorem, as well as H(axioms)? Do you appreciate the consequences of your intuitive denial of the invariance theorem, admittedly the simplest fact in Kolmogorov complexity? Do you understand that the indexing arguments are in fact denying the universality of algorithmic complexity, and not just Theorem LB? Either you deny every theorem, or you don¹t deny anything, in my opinion. And furthermore why is it said that algorithmic complexity is asymptotic? If you can give determinate answers to these questions, then we will have enough information to judge how sensible Raatikainen was in his critique of Theorem LB, and its stronger cousins, because these questions are not answered by himself in the paper. (And hence, why I thought he didn¹t know about these facts) I can quote the proof as well, should these explanations about the proof be insufŽcient. -- Eray Ozkural === Subject: Re: Raatikainen¹s critique of Chaitin >The part we don¹t agree is case 1: > 1) There is an inŽnite sequence of stronger and stronger theories > the characteristic constant of which are equal >>The characteristic constant of a theory T is determined by the smallest >>index e, s.t. for no m does T prove >> Ad<=e ( {d}(0) is undeŽned / {d}(0)=/= m ) >>where {k} is the machine with index k. > [The term characteristic constant is not being used anywhere, could > we please stick with the terms in the original work?] As I said earlier, the characteristic constant of a theory T is the smallest c, s.t. T does not prove that the complexity of any string is greater than c. > Let¹s make it very clear that we have been referring to the same > number because your above formula does not make sense to me. Then read it again. Using a different notation we might write The characteristic constant of a theory T is the index of smallest Turing machine e, s.t. for no m does T prove for all d <= e( the machine with index d does not halt on empty input or if it does, its output is not m) You can insert a Universal Turing machine of your choice in the above formulation, if you wish. Chaitin¹s incompleteness theorem shows that such a constant exists for every theory T, i.e. every axiomatisable theory can prove only Žnitely many strings to be random. Furthermore, he gives an upper bound on the characteristic constant which depends on the complexity of the set of axioms of T. Now, Chaitin wishes to draw from this the morale that the complexity of the axioms somehow determine how complex things we can prove from them as witnessed by his stated goals: I would like to measure the the power of a set of axioms and rules of inference. I would like to be able to say that if one has ten pounds of axioms and a twenty pound theorem, then that theorem cannot be derived from those axioms. Chaitin has certainly established this if one interpretes Œweigth¹ of a set of axioms as its algorithmic complexity and the Œweight¹ of a theorem as m if it¹s of form K(c)>m and as, say, 0 if it¹s not of this form. However, I suspect he had something more interesting in mind. What Raatikainen¹s, Solovay¹s, van Lambangen¹s and other¹s examples show is that while Chaitin¹s upper bound certainly holds, it does not necessarily tell us anything particularly interesting about the theory. By choosing a suitable Universal Turing machine (i.e. an acceptable indexing) we can show that theories which would be ordinarily considered very strong, e.g. ZFC, have characteristic constant 0, i.e. do not prove any string to be random. We can also show that there are inŽnite sequences of stronger and stronger and more complex and complex theories which prove exactly the same strings to be random, or equivalently, have the same characteristic constant. See e.g. my construction in the previous post. > Now, what is it that you guys don¹t like about > n <= H(axioms) + O(1) Nothing. It just is not as ludicurously signiŽcant as Chaitin and his followers would have us believe. > In addition, do you know what the invariance theorem is, and how it > relates to this small theorem? Yes. The invariance theorem simply states that the complexity of a given string varies by an additive constant when going from one Universal Turing machine to another, or alternatively, by using a different acceptable indexing. Nothing in Raatikainen¹s results contradicts this. > Do you appreciate the consequences of your intuitive denial of the > invariance theorem, admittedly the simplest fact in Kolmogorov > complexity? Could you elaborate on this? I don¹t see where I have Œintuitively denied¹ the invariance theorem. It seems you do not understand what Raatikainen¹s arguments are about. The invariance theorem does hold for acceptable indexings, so I don¹t see how one would have to deny it in any sense to accept Raatikainen¹s results. > Do you understand that the indexing arguments are in > fact denying the universality of algorithmic complexity, and not just > Theorem LB? Since I don¹t understand what you mean by Œuniversality of algorithmic complexity¹ I can¹t tell. We don¹t even need these Œindexing arguments¹ to make Raatikainen¹s point. There are basicly two ways to establish that the characteristic constant of a theory T is not a good indicator of its strength in any reasonable sense: Žx a theory T and vary the indexing or Žx an indexing and vary T. If you have qualms about the former, then concentrate on the latter. -- Aatu Koskensilta (aatu.koskensilta@xortec.Ž) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Raatikainen¹s critique of Chaitin Hi Aatu, > Since I don¹t understand what you mean by Œuniversality of algorithmic > complexity¹ I can¹t tell. We don¹t even need these Œindexing arguments¹ > to make Raatikainen¹s point. There are basicly two ways to establish > that the characteristic constant of a theory T is not a good indicator > of its strength in any reasonable sense: Žx a theory T and vary the > indexing or Žx an indexing and vary T. If you have qualms about the > former, then concentrate on the latter. Hmm. Okay, let¹s concentrate on the latter, then. Let¹s put that argument in a really simple and clear form. Is there such a thing in Raatikainen¹s paper? Which section do you think does that? I wish to talk very precisely from now on. [Although that does not necessarily mean bad formalism, just enough to be precise] Let¹s Žx a universal computer U, this is nice, because it does not contradict with the basic way of studying asymptotic program size complexity. [Let¹s forget the former case for a moment as you suggest, because I do have qualms about it and Raatikainen¹s subsection devoted to showing that it is sensible is not convincing, IMO] So, everything we say is about asymptotic complexity. Behavior of big numbers. Now you are suggesting that we start with an FAS T, and its axiom string A, and vary this theory in which way? I think you would like to Žx the inference rules as well like Chaitin, so we are going to change the axiom string A. Now, how exactly do you suggest we vary it? -- Eray Ozkural PS: David Bernier suggested such a construction based on PA + individual halting problems, but it did not seem to contradict with Chaitin¹s opinion. === Subject: Re: Raatikainen¹s critique of Chaitin > Now you are suggesting that we start with an FAS T, and its axiom > string A, and vary this theory in which way? I think you would like to > Žx the inference rules as well like Chaitin, so we are going to > change the axiom string A. > Now, how exactly do you suggest we vary it? On second thought, I remembered that I already gave a better example for this than some construction, formalizing maximally informative. This one is the one we already agree on. But of course you don¹t see why indexing arguments don¹t work. And that is the only thing that Raatikainen brings into discussion. I¹ll explain later... -- Eray === Subject: Re: Raatikainen¹s critique of Chaitin >> But the statement 18500956786009609206809234576723987575921479192754659183918374 982983798734 9 >> = 18500956786009609206809234576723987575921479192754659183918374 9829837987349 >> is not so short. So it would seem that a short statement (low entropy) >> can have a long statement (high entropy) as a theorem. > Sorry, I didn¹t read your posting carefully enough. > I¹ll have to think about your example. I have a little spare time before the start of the next academic year, when I am giving a half-course on Algorithmic Information Theory. I¹ve given the course before, but haven¹t discussed the application of the theory to Godel¹s Theorem (or unprovability). However, I thought now I would try to get to the bottom of it, and to that end I am looking through some of Chaitin¹s works. This is basically an account of my progress to date. I start from a statement in the Abstract of Chaitin¹s paper Godel¹s theorem and information, Int J Th Phys 22 (1982) pp 941-954, reprinted in Information, Randomness & Incompleteness: Papers on Algorithmic Information Theory, ed 2 (1982). Godel¹s theorem may be demonstrated using arguments having an information-theoretic žavor. In such an approach it is possible to argue that if a theorem contains more information than a given set of axioms, then it is impossible for the theorem to be derived from the axioms. Now this seems a pretty clear assertion to me, subject only to a deŽnition of what one means by (1) the information in a given set of axioms, and (2) the information in a theorem. On the Žrst, I was wrong in an earlier posting in suggesting that it meant the entropy of the axioms (together presumably with the rules of inference) considered simply as strings. In fact Chaitin¹s deŽnition is slightly more subtle. He deŽnes a formal system as a recursively enumerable set of propositions. This means that there exists a Turing machine T which will output the propositions in the theory as T(0), T(1), T(2), ... when the numbers 0, 1, 2 are fed in (encoding n say as n 0¹s followed by a 1). Now Chaitin deŽnes the entropy (I will use that word rather than information) of the system as the entropy H(T) of T, which we may take as the length of the shortest string t which will cause the chosen universal machine U to emulate T, ie U(ts) = T(s) for all strings s. [Alternatively, we could lay down how a Turing machine T is to be speciŽed by a string t¹, and deŽne H(T) to be H(t¹). The two deŽnitions would be equivalent, I believe, up to a constant depending only on U. But let us stick to the Žrst, which is simpler to state.] This seems to me a perfectly clear and proper deŽnition, since the provable propositions in any formal system will certainly form such a recursively enumerable set. That brings us to the second question: how do we deŽne the entropy of a theorem (or proposition) P? The obvious deŽnition is the entropy H(p) of the string p expressing P in the formal system, and I will assume for the moment that that is what Chaitin means. Having given a precise meaning to Chaitin¹s assertion, I am searching for a proof of it in Chaitin¹s work, and I must admit that this is proving much more difŽcult than I expected. [As it happens, I¹m also reading Truesdell¹s The tragicomical history of thermodynamics, 1822-1854, and I can¹t help applying some of his remarks to Chaitin. In no other discipline had the same equations been published over and over again so many times by different authors in different ill-deŽned notations and therefore claimed as his own by each; in no other has a singly author seen Žt to publish essentially the same ideas over and over again within a period of twenty years; and nowhere else is the ratio of talk and excuse to reason and result so high.] But to return to the task in hand, I¹m looking at Chaitin¹s book Information-theoretic incompleteness (1992, reprinted 1998) in which one would surely expect to Žnd the sought-for proof. Unfortunately, we come up against one of the difŽculties in Chaitin¹s work: his passion for LISP, in terms of which many of his deŽnitions are set. Another is his passion for the number Omega. But I Žnd a statement very similar to the one we started with in the paper Informational-theoretic incompleteness included in the above work, to appear (then) in Applied Mathematics and Computation. We show that no formal system of complexity n can exhibit a speciŽc object with complexity greater than n+c. Now I have to tease out exactly what this means. Could the use of LISP, or what I assume is equivalent, the lambda calculus, actually lead to a different deŽnition of the entropy of a theorem? I leave my research at that point. Any comments gratefully received. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > Godel¹s theorem may be demonstrated using arguments > having an information-theoretic žavor. > In such an approach it is possible to argue that > if a theorem contains more information than a given set of axioms, > then it is impossible for the theorem to be derived from the axioms. > Now this seems a pretty clear assertion to me, Why? It is possible to argue any silliness. The statement that a theorem cannot have higher complexity than the axioms of the theory is trivially false on any obvious interpretation. > We show that no formal system of complexity n > can exhibit a speciŽc object with complexity greater than n+c. > Now I have to tease out exactly what this means. No problem here. It¹s just a formulation of Chaitin¹s version of the incompleteness theorem. It does nothing to show that if a theorem contains more information than a given set of axioms... etc. === Subject: Re: Raatikainen¹s critique of Chaitin >> Godel¹s theorem may be demonstrated using arguments >> having an information-theoretic žavor. >> In such an approach it is possible to argue that >> if a theorem contains more information than a given set of axioms, >> then it is impossible for the theorem to be derived from the axioms. >> Now this seems a pretty clear assertion to me, > Why? It is possible to argue any silliness. The statement that a > theorem cannot have higher complexity than the axioms of the theory > is trivially false on any obvious interpretation. You are basically saying that Chaitin is being fraudulent. I don¹t believe that. If a mathematician says It is possible to argue that Wiles¹ work shows the Riemann Hypothesis to be true (I am of course taking a ridiculous example) this means in my view that the author claims that he has such an argument. A lawyer might argue that all he is saying is that _someone_ _might_ have such an argument. But I don¹t think that is acceptable in mathematics. In any case, the preceding sentence Godel¹s theorem may be demonstrated ... clearly means that he is going to demonstrate it. >> We show that no formal system of complexity n >> can exhibit a speciŽc object with complexity greater than n+c. >> Now I have to tease out exactly what this means. > No problem here. It¹s just a formulation of Chaitin¹s version of the > incompleteness theorem. It does nothing to show that if a theorem > contains more information than a given set of axioms... etc. Well, obviously one has to ask if a speciŽc object can include a theorem. I shall continue my reading in the Žrm belief that Chaitin means what he says, or at least what I think he says. My view of Chaitin¹s work is that he made one great discovery, but only one, and that he spent a lot of time dressing up that idea in various guises. But the same could be said, IMHO, of Turing. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > [ snip ] >Well, obviously one has to ask if a speciŽc object >can include a theorem. >I shall continue my reading in the Žrm belief >that Chaitin means what he says, >or at least what I think he says. >[ snip ] Don¹t know how much time you have for this. Chaitin has authored many papers. Have you tried to ask him directly? I would be interested in seeing his answer. === Subject: Re: Raatikainen¹s critique of Chaitin >>I shall continue my reading in the Žrm belief >>that Chaitin means what he says, >>or at least what I think he says. >>[ snip ] > Don¹t know how much time you have for this. Chaitin has authored many > papers. Actually, the informational content of his papers is remarkably small, since he keeps saying the same thing over and over again. I feel a Turing machine could compress them to a few pages. > Have you tried to ask him directly? I would be interested in seeing his > answer. I did actually have a short correspondence with him some years ago. He was extremely polite, and even sent me a complete set of his papers (in 4 very thick books); but he didn¹t answer the small query I put to him, probably because he was too busy to read my missive thoroughly. My view of Chaitin seems to differ from others in this correspondence, who seem to fall into two classes - those who believe he has done nothing, and those who believe he has revolutionised mathematics. I believe he introduced one important concept - the entropy H(s) of a string - and showed that it behaved nicely. But I don¹t believe that this idea has huge philosophical implications. However, it does seem to me a very important idea, since I cannot see any other sense in which one can interpret the term information as eg information coming out of a black hole or information in a DNA sample. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > [ snip ] >My view of Chaitin seems to differ from others in this correspondence, >who seem to fall into two classes - those who believe he has done nothing, >and those who believe he has revolutionised mathematics. >I believe he introduced one important concept - >the entropy H(s) of a string - >and showed that it behaved nicely. >But I don¹t believe that this idea has huge philosophical implications. >However, it does seem to me a very important idea, >since I cannot see any other sense in which one can interpret >the term information as eg information coming out of a black hole >or information in a DNA sample. OK, since you studied Chaitin¹s contributions, perhaps you can tell us: How does his information complexity differ from Kolmogorov¹s? I Žgure you are probably in best position to answer this. === Subject: Re: Raatikainen¹s critique of Chaitin > OK, since you studied Chaitin¹s contributions, perhaps you can tell us: > How does his information complexity differ from Kolmogorov¹s? I think the two are exactly equivalent. As I understand it, Kolmogorov and Chaitin developed the theory independently at the same time. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin >>OK, since you studied Chaitin¹s contributions, perhaps you can tell us: >>How does his information complexity differ from Kolmogorov¹s? > I think the two are exactly equivalent. > As I understand it, Kolmogorov and Chaitin developed the theory > independently at the same time. And the third person to whom independent discovery of this concept is attributed is Ray Solomonoff, although he was mainly concerned with formalising inductive reasoning rather than complexity per se. Solomonoff¹s papers A Formal Theory of Inductive Inference part I and A Preliminary Report on a General Theory of Inductive Inference are available at his homepage http://world.std.com/~rjs/. -- Aatu Koskensilta (aatu.koskensilta@xortec.Ž) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Raatikainen¹s critique of Chaitin > > >>OK, since you studied Chaitin¹s contributions, perhaps you can tell us: >>How does his information complexity differ from Kolmogorov¹s? > > I think the two are exactly equivalent. > As I understand it, Kolmogorov and Chaitin developed the theory > independently at the same time. > And the third person to whom independent discovery of this concept is > attributed is Ray Solomonoff, although he was mainly concerned with > formalising inductive reasoning rather than complexity per se. > Solomonoff¹s papers A Formal Theory of Inductive Inference part I and > A Preliminary Report on a General Theory of Inductive Inference are > available at his homepage http://world.std.com/~rjs/. Aatu is right, hence why I referred to Kolmogorov/Chaitin/Solomonoff complexity at one time in the discussions. The coolest name is Kolmogorov, so when I want nice sounding statements I say Kolmogorov complexity, perhaps the better terms are simply program size, programmatic or algorithmic complexity, which are discoverer neutral. Too bad I¹m not Russian, I could have a name like Kolmogorov¹s. :/ -- Eray Ozkural === Subject: Re: Raatikainen¹s critique of Chaitin > My view of Chaitin seems to differ from others in this correspondence, > who seem to fall into two classes - those who believe he has done nothing, > and those who believe he has revolutionised mathematics. I don¹t think any sensible correspondent here has advocated either of these extremes. > I believe he introduced one important concept - > the entropy H(s) of a string - > and showed that it behaved nicely. > But I don¹t believe that this idea has huge philosophical implications. Just like most of us (with a couple of noisy exceptions). > However, it does seem to me a very important idea, > since I cannot see any other sense in which one can interpret > the term information as eg information coming out of a black hole > or information in a DNA sample. So, does Hawking or do geneticists actually use Chaitin¹s deŽnition of entropy? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Raatikainen¹s critique of Chaitin >> My view of Chaitin seems to differ from others in this correspondence, >> who seem to fall into two classes - those who believe he has done >> nothing, and those who believe he has revolutionised mathematics. > I don¹t think any sensible correspondent here has advocated > either of these extremes. Really? I would have thought the two most proliŽc contributors to this thread - Torkel Franzen and Eray Ozkural - Žt into these two categories fairly snugly. >> However, it does seem to me a very important idea, >> since I cannot see any other sense in which one can interpret >> the term information as eg information coming out of a black hole >> or information in a DNA sample. > So, does Hawking or do geneticists actually use Chaitin¹s deŽnition > of entropy? As I understand it, the concept of entropy has gone through 4 phases: thermodynamics, statistical mechanics, Shannon¹s Information Theory and Chaitin/Kolmogorov theory. There have been constant arguments at all stages about the equivalence or non-equivalence of these concepts. I have certainly read references to the entropy of a black hole which explicitly referred to Kolmogorov complexity, eg in the collection of papers Complexity, Entropy and the Physics of Information ed Zurek, 1990, Santa Fe Institute. The fact that Hawking uses the term information implies to my mind that he is using the term in the sense of Information Theory, which leaves the choice between Shannon and Chaitin. I don¹t see how one can apply Shannon¹s theory directly to deŽne the entropy of a black hole, since there does not appear to be the necessary ensemble. So I have to assume that he is using the term in the sense of Chaitin and Kolmogorov. I need not add that I am not a physicist, and am more than willing to be corrected. As to genetics, I have certainly seen references to the information content of DNA in terms of the code¹s compressibility, which is equivalent to Chaitin¹s deŽnition. However, I haven¹t understood most of the references to entropy in genetics that I have seen, and am not completely convinced that they make sense (eg references to the concentration of entropy at a particular gene). But I know even less about genetics than I do about physics. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > > My view of Chaitin seems to differ from others in this correspondence, > who seem to fall into two classes - those who believe he has done > nothing, and those who believe he has revolutionised mathematics. >> I don¹t think any sensible correspondent here has advocated >> either of these extremes. > Really? > I would have thought the two most proliŽc contributors to this thread - > Torkel Franzen and Eray Ozkural - > Žt into these two categories fairly snugly. Please read my sentence carefully. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Raatikainen¹s critique of Chaitin > Really? Why do you insist that Chaitin hasn¹t done anything? I suggest that you consider his actual work, which is really quite accessible - for example his incompleteness theorem. === Subject: Re: Raatikainen¹s critique of Chaitin >> Really? > Why do you insist that Chaitin hasn¹t done anything? That¹s a very odd remark to address to me, as I have repeatedly said that I believe Chaitin¹s Algorithmic Information Theory is an important mathematical development. You are the person who has constantly belittled Chaitin¹s work. If in fact you have had a change of heart I am very glad to hear of it. The Really? above expressed surprise at Robin Chapman¹s statement, and had nothing to do with my opinion of Chaitin¹s work. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > That¹s a very odd remark to address to me, > as I have repeatedly said that I believe > Chaitin¹s Algorithmic Information Theory > is an important mathematical development. Really? Haven¹t you in fact gone on and on about how Chaitin has done nothing? === Subject: Re: Raatikainen¹s critique of Chaitin >>I shall continue my reading in the Žrm belief >>that Chaitin means what he says, >>or at least what I think he says. >> >>[ snip ] >> > Don¹t know how much time you have for this. Chaitin has authored many > papers. > Actually, the informational content of his papers is remarkably small, > since he keeps saying the same thing over and over again. > I feel a Turing machine could compress them to a few pages. > Have you tried to ask him directly? I would be interested in seeing his > answer. > I did actually have a short correspondence with him some years ago. > He was extremely polite, and even sent me a complete set of his papers > (in 4 very thick books); > but he didn¹t answer the small query I put to him, > probably because he was too busy to read my missive thoroughly. > My view of Chaitin seems to differ from others in this correspondence, > who seem to fall into two classes - those who believe he has done nothing, > and those who believe he has revolutionised mathematics. > I believe he introduced one important concept - > the entropy H(s) of a string - > and showed that it behaved nicely. > But I don¹t believe that this idea has huge philosophical implications. > However, it does seem to me a very important idea, > since I cannot see any other sense in which one can interpret > the term information as eg information coming out of a black hole > or information in a DNA sample. But, nobody claimed that Chaitin¹s Theories concern nom-existent information in DNA Samples. Since for one thing DNA samples aren¹t even samples, they are drawings as the mathe-ma-stupid are reminded often, And they equally obviously concern nothing with the non-existent information out of Black Holes. Since Black Holes are not even Information *collecting* *events*. As the G-g, Geo Einstone IDIOTS are reminded semi-daily. Because a system *Collapses*, implies nothing about *information* in or out of the system. Except in Israel, where it¹s always been true, even the Sanai snakes believe it¹s true, === Subject: Kolmogorov Complexity and physical information (Was Re: Raatikainen¹s critique of Chaitin) > My view of Chaitin seems to differ from others in this correspondence, > who seem to fall into two classes - those who believe he has done nothing, > and those who believe he has revolutionised mathematics. > I believe he introduced one important concept - > the entropy H(s) of a string - > and showed that it behaved nicely. > But I don¹t believe that this idea has huge philosophical implications. > However, it does seem to me a very important idea, > since I cannot see any other sense in which one can interpret > the term information as eg information coming out of a black hole > or information in a DNA sample. I don¹t think these examples of physical information are incommensurable with Kolmogorov/Chaitin complexity. For one thing, in a RUCA, the algorithmic information corresponds to the minimum thermodynamical work required to get a job done... So, if digital physics is right, then algorithmic information can become an appropriate measure of information coming out of a black hole, and for similar reasons of information in a DNA sample since the biology of an organism will again have to be a discrete computer, however, the biological information might be losing some of the properties of these nice theoretical cellular automata. These ideas are analogous to the identity of code length of random sources (classical information theoretic entropy) to entropy in thermodynamics. And indeed, if you would read the introductory chapter of Cover and Thomas¹s excellent Elements of Information Theory book, you would see that Kolmogorov complexity is seen as _more_ fundamental than Shannon information, simply because Shannon information can be formally derived from Kolmogorov complexity. (I can scan and send you the chapter for criticism if you would like) However, I do not assume, either, that algorithmic complexity is the ultimate measure of complexity. I have set forth a small research agenda for myself, to try, experimentally and theoretically, alternative complexity measures. I believe that there should be a mathematical theory that can formalize our entire scientiŽc conception of knowledge, e.g. computational ontology, and that is what I will be looking at. A promising research in that direction is algorithmic statistics of Paul Vitanyi. My present philosophical analysis suggests that while the ideas of coding and machine are central to information measures, there is no singular such measure, and the undecidability of Kolmogorov complexity suggests that it is an over-idealization for some of the mundane kinds of information, here in the real world. The variety of measures is best exempliŽed by Schmidhuber¹s Generalized Kolmogorov Complexity. Can we truly say that Schmidhuber information is the real information content of a string, rather than Kolmogorov complexity? Mathematically, it would seem so. But philosophically, I sense plurality. -- Eray Ozkural === Subject: Re: Kolmogorov Complexity and physical information (Was Re: Raatikainen¹s critique of Chaitin) > And indeed, if you would read the introductory chapter > of Cover and Thomas¹s excellent Elements of Information Theory book, > you would see that Kolmogorov complexity is seen as _more_ fundamental > than Shannon information, simply because Shannon information can be > formally derived from Kolmogorov complexity. (I can scan and send you > the chapter for criticism if you would like) If this offer is for me, I think I have access to this book, thank you very much. I hope what you say is true, and I will certainly look at Thomas & Cover to see how they put it. > I believe that there should be a > mathematical theory that can formalize our entire scientiŽc > conception of knowledge, e.g. computational ontology, and that is > what I will be looking at. A promising research in that direction is > algorithmic statistics of Paul Vitanyi. I suppose quantum information theory, which seems quite active, is another vista. However, I Žnd it difŽcult enough to understand Chaitin without straying into further Želds ... -- 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 === Subject: Re: Kolmogorov Complexity and physical information (Was Re: Raatikainen¹s critique of Chaitin) > And indeed, if you would read the introductory chapter > of Cover and Thomas¹s excellent Elements of Information Theory book, > you would see that Kolmogorov complexity is seen as _more_ fundamental > than Shannon information, simply because Shannon information can be > formally derived from Kolmogorov complexity. (I can scan and send you > the chapter for criticism if you would like) > If this offer is for me, I think I have access to this book, > thank you very much. > I hope what you say is true, and I will certainly look at Thomas & Cover > to see how they put it. Ok. Here is the quote from Chapter 1, in somewhat mangled form by the conversion procedure, but mostly readable: Computer Science (Kolmogorov Complexity). Kolmogorov, Chaitin and Solomonoff put forth the idea that the complexity of a string of data can be deŽned by the length of the shortest binary program for computing the string. Thus the complexity is the minimal description length. This deŽnition of complexity turns out to be universal, that is, computer independent, and is of fundamental importance. Thus Kolmogorov complexity lays the foundation for the theory of descriptive complexity. Gratifyingly, the Kolmogorov complexity K is approximately equal to the Shannon entropy H if the sequence is drawn at random that has entropy H. So the tie-in between from a distribution theory and Kolmogorov complexity is perfect. Indeed, we information than Shannon consider Kolmogorov complexity to be more fundamental entropy. It is the ultimate data compression and leads to a logically consistent procedure for inference. relationship between algorithmic There is a pleasing complementary complexity. One can think about compucomplexity and computational tational complexity (time complexity) and Kolmogorov complexity (program length or descriptive complexity) as two axes corresponding to time and program length. Kolmogorov complexity program running focuses on minimizing along the second axis, and computational complexity focuses on minimizing along the Žrst axis. Little work has been done on the simultaneous minimization of the two. Note that Cover&Thomas make the point I made very clearly. There is no room for doubt, this was one of the most inžuential things I have ever read, and thus I remember it exactly... (since signiŽcant emotional states cause memorization in the brain) > I believe that there should be a > mathematical theory that can formalize our entire scientiŽc > conception of knowledge, e.g. computational ontology, and that is > what I will be looking at. A promising research in that direction is > algorithmic statistics of Paul Vitanyi. > I suppose quantum information theory, which seems quite active, > is another vista. Is another vista, but not as fundamental as Kolmogorov complexity, although since quantum codes are *shorter* than Shannon codes, quantum info theorists argue that well, this is the ultimate data compression. Nice stuff for EEE people, but computer scientists are unfortunately a little far from that venue at the present. I still don¹t understand those operators too well :) What is truly intriguing is that the quantum computers are a non-deterministic machine that can provide O(n^2) speedup for certain problems, but not O(2^n) speedup as it currently seems. This means that this is essentially a parallel machine of some sort for us theorists, and requires novel algorithm design! > However, I Žnd it difŽcult enough to understand Chaitin > without straying into further Želds ... I believe that Chaitin¹s work must not be criticized from an intellectual vacuum, like the way Raatikainen did. The followers of digital philosophy movement seek an explanation from from many sciences. -- Eray Ozkural === Subject: Re: Kolmogorov Complexity and physical information >> However, I Žnd it difŽcult enough to understand Chaitin >> without straying into further Želds ... > I believe that Chaitin¹s work must not be criticized from an > intellectual vacuum, like the way Raatikainen did. An ad hominem attack, as ever :-( -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Kolmogorov Complexity and physical information > >> However, I Žnd it difŽcult enough to understand Chaitin >> without straying into further Želds ... > > I believe that Chaitin¹s work must not be criticized from an > intellectual vacuum, like the way Raatikainen did. > An ad hominem attack, as ever :-( Not at all. I am only saying that some arguments against Chaitin¹s philosophy are from a constrained point of view, which neglects the relevance of descriptive complexity to physics, biology, artiŽcial intelligence and complexity sciences in general. At any rate, it was unnecessary for me to mention Raatikainen any further. It is not on topic in this thread... -- Eray Ozkural === Subject: Re: Kolmogorov Complexity and physical information >> Eray are there integers with an inŽnite number of digits? Ozkural >> > However, I Žnd it difŽcult enough to understand Chaitin > without straying into further Želds ... >> >> I believe that Chaitin¹s work must not be criticized from an >> intellectual vacuum, like the way Raatikainen did. >> An ad hominem attack, as ever :-( > Not at all. Exactly. Describing Raatikainen¹s intellectual position as a vacuum is simply abuse. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Kolmogorov Complexity and physical information > Exactly. Describing Raatikainen¹s intellectual position as a vacuum > is simply abuse. He is criticizing Chaitin¹s philosophy from some singular point of view. I shall not elaborate further on this thread. It¹s off topic. The topic is the relation of Kolmogorov complexity to physical information. Either stick with the topic or change it. === Subject: Re: Raatikainen¹s critique of Chaitin > You are basically saying that Chaitin is being fraudulent. I have no idea what Chaitin was thinking in the quoted passage. > Well, obviously one has to ask if a speciŽc object > can include a theorem. It doesn¹t matter if it¹s a theorem or anything else: for any theory T there is an n such that T does not prove, for any k, theorem or not, that k has complexity greater than n. === Subject: Re: Raatikainen¹s critique of Chaitin >> Well, obviously one has to ask if a speciŽc object >> can include a theorem. > It doesn¹t matter if it¹s a theorem or anything else: for any theory > T there is an n such that T does not prove, for any k, theorem or not, > that k has complexity greater than n. I¹m not sure what you are saying. Chaitin¹s statement was: We show that no formal system of complexity n can exhibit a speciŽc object with complexity greater than n+c. Are you saying that this is obvious to you? I don¹t think one could make a statement about any k, as you do, without specifying in some way what set or class k belongs to. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > Chaitin¹s statement was: > We show that no formal system of complexity n > can exhibit a speciŽc object with complexity greater than n+c. > Are you saying that this is obvious to you? No, I¹m saying that it¹s a formulation of Chaitin¹s incompleteness theorem: for any consistent theory T, there is an m such that for no k does T prove that k has complexity greater than m. Note that can exhibit does not mean has as a theorem: trivially, a theory with axiom (x)(x=x) has theorems of unbounded complexity. Can exhibit an object with complexity greater than m means can prove of an object that it has complexity greater than m. > I don¹t think one could make a statement about any k, as you do, > without specifying in some way what set or class k belongs to. k is just a bit string, or a number if you like. You will recall that Kolmogorov complexity is deŽned for any bit string. === Subject: Re: Raatikainen¹s critique of Chaitin >> Chaitin¹s statement was: >> We show that no formal system of complexity n >> can exhibit a speciŽc object with complexity greater than n+c. >> Are you saying that this is obvious to you? > No, I¹m saying that it¹s a formulation of Chaitin¹s incompleteness > theorem: for any consistent theory T, there is an m such that for > no k does T prove that k has complexity greater than m. Note that > can exhibit does not mean has as a theorem: trivially, a theory > with axiom (x)(x=x) has theorems of unbounded complexity. Can exhibit an > object with complexity greater than m means can prove of an > object that it has complexity greater than m. I don¹t think one can refer to Chaitin¹s incompleteness _theorem_, since (as has been pointed out here) he explicitly and repeatedly states that he has proved several incompleteness theorems. What you say is certainly equivalent to the Žrst of his theorems. I¹m not entirely sure that it is what Chaitin meant in the above quotation. You are saying, if I understand you correctly, that exhibit an object with complexity greater than n means give a formal proof within the theory that the object has complexity greater than n. The trouble with this interpretation, as I see it, is that he has described a formal system as any recursively enumerable set of propositions; and I don¹t see why such a set should contain any statement of the kind you refer to. -- 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 === Subject: Re: Raatikainen¹s critique of Chaitin > I¹m not entirely sure that it is what Chaitin meant in the above > quotation. So come up with some other meaning. > The trouble with this interpretation, as I see it, > is that he has described a formal system > as any recursively enumerable set of propositions; > and I don¹t see why such a set should contain any statement > of the kind you refer to. If it doesn¹t contain any statement of the form the complexity of k is greater than m at all, the theorem applies vacuously. === Subject: Re: Raatikainen¹s critique of Chaitin > > And if you are going to criticize a mathematician¹s *philosophy*, > don¹t call his work footnotes to X. Nothing entitles you to such ad > hominem arguments. > > But you feel entitled to imply that Raatikainen is incompetent and žat > out state that he does not understand basic mathematical arguments and > concepts? I pretty much think almost all of my objections are philosophical, little mathematics is involved anyway. But there may be a conceptual confusion, I cannot yet tell how much of it is due to Raatikainen, how much due to Chaitin, and how much due to others like us, so many posts in this thread, more than half of them are serious žaming... :/ I will try to reply to your careful explanations of the one signiŽcant point, about truth of arithmetic statements, and the characteristic constant of a theory. I haven¹t missed your informative posts. Maybe, then, I can. > By the way, have you found Hintikka¹s insidious comments yet? I am > genuinely interested in what he has to say. > Hintikka is a fool, and I will demonstrate why he is so, the quote > will certainly be worth the several months of wait I have unwillingly > cost you. I am sorry for having used these silly words, I thought you were implying that no such insidious comments existed. They do, and I will show them. I am not anybody¹s lawyer and I am not on a propaganda mission. *That* would be foolish. -- Eray === Subject: Re: Raatikainen¹s critique of Chaitin Discussion, linux) > I pretty much think almost all of my objections are philosophical, > little mathematics is involved anyway. I half-agree with you here, but I don¹t see any reason to disparage philosophy like that. -- Who knows, maybe that may be the only way to settle this crap. It¹s not like it¹d be that hard for me to go back and get a math degree. I can penetrate the math social group and then Žnish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: Raatikainen¹s critique of Chaitin > I pretty much think almost all of my objections are philosophical, > little mathematics is involved anyway. > I half-agree with you here, but I don¹t see any reason to disparage > philosophy like that. I appreciate your position. -- Eray === Subject: Re: Raatikainen¹s critique of Chaitin > Correction. I should have said any FAS that is as powerful as PA or > more powerful than PA. > No, you should have removed the unnecessary assumption that the > system has only a Žnite number of axioms. It has never been the suggestion that the number of axioms, hence the axiom string is Žnite. Instead, the suggestion has been that the Kolmogorov complexity of the axiom schema is Žnite, all along, regardless of the length of the axiom string. Once more, you have skipped upon the exact deŽnition to arrive at a conclusion you would prefer. Take care, -- Eray Ozkural === Subject: Re: Raatikainen¹s critique of Chaitin > It has never been the suggestion that the number of axioms, hence the > axiom string is Žnite. Instead, the suggestion has been that the > Kolmogorov complexity of the axiom schema is Žnite, all along, An effectively axiomatized theory need not have any axiom schemata. As for the idea of an axiom schema of inŽnite Kolmogorov complexity, this would seem to be related to your earlier contemplation of integers with an inŽnite number of digits. === Subject: Re: Raatikainen¹s critique of Chaitin > It has never been the suggestion that the number of axioms, hence the > axiom string is Žnite. Instead, the suggestion has been that the > Kolmogorov complexity of the axiom schema is Žnite, all along, > An effectively axiomatized theory need not have any axiom schemata. Daryl and I seem to agree on that we need it (Daryl pointed in an earlier discussion that there are many valid axiom schema). Schmidhuber also allows for axiom schema in his AI work. Some axiom strings are inŽnite, this has no effect on the theorems, it doesn¹t matter. Chaitin insists that axiom strings should be Žnite in AIT, which I do not Žnd myself in clear agreement with. But, perhaps you are right, insisting on a Žnite axiom string, we can handle axiom schema as higher order statements: deŽne program P as: i <- 2 while true output i = i i <- i * i Each theorem output by program P is correct. So that c.e. sets of axioms could be handled easily. As far as I can tell, AIT does not prevent us from having such axioms as above. Hence, we can think of the compressed axiom schema... > As for the idea of an axiom schema of inŽnite Kolmogorov > complexity, this would seem to be related to your earlier > contemplation of integers with an inŽnite number of digits. Hmmm. So you perhaps think that is a foolish idea. To be able to formally deŽne maximum informative I have to actually take that step, so that you can talk about K(x|Omega). I haven¹t found a better way to do it, if you have an idea, then please tell me. -- Eray Ozkural === Subject: Re: derivative... > hello.....doctor~ > derivative of sin(1/x) is > lim [sin{(1/x)+h} - sin(1/x)] / h > h->0 No. Let f(x) = sin(1/x). f(x+h) = sin(1/(x + h)). Or the chain rule with y(x) = 1/x, f(x) = sin(y(x)), df/dx = (df/dy) (dy/dx) ; dy/dx = y¹ = -1/x^2 sin(y+h) - sin(y) = sin(y) cos(h) + sin(h) cos(y) - sin(y) cos(h) --> 1 - h^2/2, etc. === Subject: Square root of non-symmetric positive deŽnite It is standard that a symmetric/hermitian positive semideŽnite matrix admits a square root. How about a non-symmetric/non-hermitian positive semideŽnite matrix? I¹m pretty sure there is still a square root, but I have no proof. DeŽnition: A is pos. semideŽnite (real case) iff x¹ A x >= 0 for all real vectors x. Note that when A is pos semideŽnite but not symmetric, its eigenvalues are not necessarily positive. Mimo -- Should work better if you insert 2718 after my name. :-) === Subject: Re: Square root of non-symmetric positive deŽnite >It is standard that a symmetric/hermitian positive semideŽnite matrix >admits a square root. >How about a non-symmetric/non-hermitian positive semideŽnite matrix? I¹m >pretty sure there is still a square root, but I have no proof. >DeŽnition: A is pos. semideŽnite (real case) iff x¹ A x >= 0 for all real >vectors x. A matrix with no square root will have a vector v with A^2 v = 0 but A v <> 0 (see e.g. my sci.math posting from December 1992 on the subject Re: Square root of a matrix, Now let w(t) = v - t A v for real scalar t. We have A w(t) = A v and w(t)¹ A w(t) = (v¹ - t (A v)¹) A v = v¹ A v - t ||A v||^2 But this < 0 for sufŽciently large t, so A is not positive semideŽnite. In the complex case, replace Œ by * (the Hermitian conjugate). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: Square root of non-symmetric positive deŽnite Am 16.09.04 18:30 schrieb Mimosa: > It is standard that a symmetric/hermitian positive semideŽnite matrix > admits a square root. > How about a non-symmetric/non-hermitian positive semideŽnite matrix? I¹m > pretty sure there is still a square root, but I have no proof. > DeŽnition: A is pos. semideŽnite (real case) iff x¹ A x >= 0 for all real > vectors x. > Note that when A is pos semideŽnite but not symmetric, its eigenvalues are > not necessarily positive. Yes, but if it is positive semideŽnite, then the eigenvalues are greater/equal zero (that¹s the deŽnition of pos. semideŽnite). Then you just diagonalize A by D = Q * A * inv(Q) Žnd D^(1/2) and then it should simply be C = inv(Q) * D^(1/2) * Q Gottfried Helms === Subject: Re: Square root of non-symmetric positive deŽnite > Am 16.09.04 18:30 schrieb Mimosa: >> It is standard that a symmetric/hermitian positive semideŽnite matrix >> admits a square root. >> How about a non-symmetric/non-hermitian positive semideŽnite matrix? I¹m >> pretty sure there is still a square root, but I have no proof. >> DeŽnition: A is pos. semideŽnite (real case) iff x¹ A x >= 0 for all >> real vectors x. >> Note that when A is pos semideŽnite but not symmetric, its eigenvalues >> are not necessarily positive. > Yes, but if it is positive semideŽnite, then the eigenvalues are > greater/equal zero (that¹s the deŽnition of pos. semideŽnite). The deŽnition of pos. semideŽnite in the case of non-symmetric matrices is the one I gave above. It doesn¹t imply that the eigenvalues are >=0. As an example A:=[[1,1],[-1,0]] is pos. semideŽnite but its eigenvalues are complex. Mimo -- Should work better if you insert 2718 after my name. :-) === Subject: Re: Square root of non-symmetric positive deŽnite Am 16.09.04 19:26 schrieb Mimosa: >>Yes, but if it is positive semideŽnite, then the eigenvalues are >>greater/equal zero (that¹s the deŽnition of pos. semideŽnite). > The deŽnition of pos. semideŽnite in the case of non-symmetric matrices is > the one I gave above. It doesn¹t imply that the eigenvalues are >=0. I remembered the statement that I made from my book, but in fact it was in the chapter of hermitean matrices. Sorry. So I don¹t have a solution for your question right at hand. Gottfried Helms === Subject: Still wasting your time on that silly numerology crap, Daryl? > 18 19 20 21 22 23 24 25 26 27 28 29 <-George¹s positions > Y A N N A C O U L I A S > 25 1 14 14 1 3 15 21 12 9 1 19 = 135 <-value > George was the 14th person to provide stats at this year¹s > Saskatoon Fringe Festival. And I bet the rest of the attendees looked pretty mainstream and well-adjusted compared to you, ya ing kook. Go back to the looney ward so the Hindus can torture you, and we can all have a good laugh... @:O) === Subject: Re: Still wasting your time on that silly numerology crap, Daryl? > 18 19 20 21 22 23 24 25 26 27 28 29 <-George¹s positions > Y A N N A C O U L I A S > 25 1 14 14 1 3 15 21 12 9 1 19 = 135 <-value > George was the 14th person to provide stats at this year¹s > Saskatoon Fringe Festival. > And I bet the rest of the attendees looked pretty mainstream and > well-adjusted compared to you, ya ing kook. Go back to the looney ward > so the Hindus can torture you, and we can all have a good laugh... @:O) fuffy === Subject: Re: Still wasting your time on that silly numerology crap, Daryl? Still Daryl¹s bitch I see. Time to stop obsessing and move on. === Subject: Re: Still wasting your time on that silly numerology crap, Daryl? > Still Daryl¹s bitch I see. Time to stop obsessing and move on. He is James Takayama. First James was incessently harrassing Lester Chow, I came to Lester Chow¹s defense and now James Takayama is incessently harrassing me. James Takayama claims to be a police ofŽcer in Hawaii. -Daryl S. Kabatoff === Subject: Re: John Edmund Milburn - May 7th 1975 > M I L B U R N > 13 9 12 2 21 18 14 = 89 > John (the son born in 1975) was bartendering at Hose and Hydrant, > he asked me to do his stats and then never gave me anything for my > work. Smart guy.... > He said that he would buy me a beer the next time I came in but > he never did, and anyway my work is worth more than a beer. Obviously very few people agree with that assessment. People with half a brain who aren¹t complete and total wacknuts know when it¹s time to try something new. Apparently, you¹re ing stupid enough to keep doing the same old, same old, and piss and moan when you get the same results. Get real, Daryl - NOBODY GIVES A ABOUT YOUR STUPID-ASS NUMEROLOGICAL RANTINGS. Sure, they will throw a loonie your way once in a while, either as žeeting entertainment, a laugh at your expense, or to get you off your back - but they don¹t take your seriously. The problem is that YOU do - and you get all angry and pissy when others don¹t. Get a life, before you turn 70 and the current generation of Usenet posters¹s kids have to put up with the same old crap... > John makes his money in the Broadway commercial district, it¹s the > street with the penis poster poles. In 1988 I was arrested and > repeatedly tortured for daring to call these penis poster poles and > the obelisks on church roofs (and at The Vatican, The Whitehouse and > at Saskatoon City Hall) the dinks that they are. You don¹t learn, do you? > Collectively you > people spent millions of dollars having me repeatedly arrested and > tortured, my interest in math was used as an excuse to do this to me, > then when I meet with your kids and show them evidence that their > names are a gift from God, they are so cheap and ignorant that they > can¹t even offer to buy me a cookie for my work. I bust my arse for 17 > years trying to show you people who your God is, and you people reward > me with torture and utter poverty. No, you spent 17 years wasting your life on TOTALLY USELESS CRAP, yet it¹s everyone else¹s fault that you¹re too ing STUPID to Žgure that out... > I begged for years for assistance > to get out of the country and away from my persecutors, and you people > can¹t even pay me minimum wage (or in this case a single penny) so > that I could get away on my own dime. Your so-called work isn¹t even WORTH minimum wage! Try žipping burgers - you will make more, and at least provide SOME semblance of a useful service to your fellow citizens... === Subject: Probabilty Question Noel at probability.net suggested you might be able to help with the following question (it is also posted on his Probability Yahoo Group): Scenario: In a competitive situation, say a race, election or quiz, there are 5 candidates (C1, C2, C3, C4 and C5) who (by assessment and/or statistical analysis) each have a probability of winning of: C1=0.47 C2=0.25 C3=0.14 C4=0.10 C5=0.04 Question: Is it possible to expand that data to re-present the information as the probability for each candidate Žnishing 1st, 2nd, 3rd, 4th or 5th. It seems possible if you assume that the probabilty of winning fairly represents each candidate¹s relative performance to each other but I¹m having a problem thinking it through. Obviously the probability for C1 to win is 0.47, that¹s given, so the probability of C1 Žnishing 2nd, 3rd, 4th or 5th is 0.53. The task is how do we split the 0.53 value between Žnishing 2nd, 3rd, 4th and 5th and then do likewise for the other candidates. All input gratefully received. David Caple === Subject: Re: Probabilty Question >Scenario: >In a competitive situation, say a race, election or quiz, there are 5 >candidates (C1, C2, C3, C4 and C5) who (by assessment and/or >statistical analysis) each have a probability of winning of: >C1=0.47 >C2=0.25 >C3=0.14 >C4=0.10 >C5=0.04 >Question: >Is it possible to expand that data to re-present the information as >the probability for each candidate Žnishing 1st, 2nd, 3rd, 4th or >5th. In general, no. The assessment algorithm will produce 120 probabilities, corresponding to the 120 possible rankings. You can¹t determine anything about these numbers (other than that they add in groups of 24 to the values given) without assuming something about the assessment algorithm. Mike Guy === === Subject: Re: The real numbers, and general comments by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GHmUh32104; >>... That thread was started by Ross Finlayson, >>not myself. >> Here is a minor point of attribution: Uncountable sets in CZF? >> was started by Agamemnon. > Sorry. It still wasn¹t me, though. > Why does your name appear in quotes in the subject line? It > shouldn¹t, should it? > Andrew Usher I posted it through Math Forum: http://www.mathforum.org/discuss/sci.math/ Postings made through that web site do not always propagate. The one you mention took several days to appear. Math Forum is convenient to read current postings, as they are presented within an hour or so of their postings, faster than other news post browsing websites, its composition facilities leave much to be desired, and while posts often do get sent when submitted through Math Forum, sometimes they do not. It is most often used by anonymous and one-time posters. Math Forum also archives a variety of other mathematical dis- cussions. http://www.mathforum.org/discussions/ Hey I searched for you on the Internet, I think you are a Math Ph.D. from Ireland. Ross F. === Subject: Re: analytic...¹ by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GHmV732123; >hello.....doctor~ >Žnd analytic function such that f(1/n) = e^(-n) for all positive integer n. >---------------------------------------- >um.....i think..... >f(z) = e^(-1/z). >but this is not analytic at point z=0. >even if i deŽned to f(0) = 0, it¹s not analytic at point z=0. Someone correct me if I am wrong.... This is perhaps a matter of deŽnition. Does analytic here mean analytic over the whole plane? Or does it mean, analytic anywhere the function is deŽned? If the latter, just take f(z) = exp(-1/z) for Re(z) > 0, i.e. deŽne f only over the right half plane. Or, since the question asked about positive integers, does analytic mean real analytic? Does f need to be deŽned over the entire plane? === Subject: Set theory - VERY basic question: Why do we need elements when we have sets??? Why do we need the concept of element in set theory? It seems redundant with the concept of set. Elements seem to be a sub-class of sets. Couldn¹t we simply do this: - Sets are containers of other sets. - DeŽne the empty set {} - A is a subset of B if all subsets of A are also a subset of B. Then replace all mentions of element with set and is element of with is subset of and get the same results with a simpler set of axioms. I¹m not a mathematician, just curious... Alain === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŽth X-Treme: C&C,DWS at 12:27 PM, alainlavoie@msn.com (JAL) said: >Why do we need the concept of element in set theory? Who says that we do? There are set theories, e.g., ZFC, in which sets are all there is. Other set theories have a concept of atoms, which are not built from other sets. >Then replace all mentions of element with set and is element >of with is subset of Those are very different concepts. {1} is a subset of {1,2}, but it is not an element of it. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? >Why do we need the concept of element in set theory? It seems >redundant >with the concept of set. Elements seem to be a sub-class of sets. >Couldn¹t we simply do this: >- Sets are containers of other sets. >- DeŽne the empty set {} >- A is a subset of B if all subsets of A are also a subset of B. >Then replace all mentions of element with set and is element of >with is subset of and get the same results with a simpler set of >axioms. >I¹m not a mathematician, just curious... In ZF and NBG, all elements are sets. However, is an element of does not mean the same thing as is a subset of, and your third condition is at best unclear, and in practice seems useless. In all usual models, subset is deŽned in terms of elements. It is possible to have models with urelemente, elements which are not sets. These are useful for counterexamples, but set theory is not greatly changed by allowing them. -- 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 === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Long ago there was a book written along those lines. Set Theory without Elements or maybe Topology without Elements ?? My memory of it is a bit uncertain. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? >>Why do we need the concept of element in set theory? It seems >>redundant >>with the concept of set. >> >Long ago there was a book written along those lines. >Set Theory without Elements >or maybe Topology without Elements ?? >My memory of it is a bit uncertain. Also known as pointless topology. === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? >>Why do we need the concept of element in set theory? It seems >>redundant >>with the concept of set. >> >> >Long ago there was a book written along those lines. >Set Theory without Elements >or maybe Topology without Elements ?? >My memory of it is a bit uncertain. > Also known as pointless topology. I¹m thinking of writing a book titled, Math without Numbers. Will it be sellable if it turns out empty? ...tonyC === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > I¹m thinking of writing a book titled, Math without > Numbers. Mathematics without Numbers, Geoffrey Hellman, 1993, Oxford University Press. === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > I¹m thinking of writing a book titled, Math without > Numbers. > Mathematics without Numbers, Geoffrey Hellman, 1993, Oxford > University Press. ...tonyC === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have s... |Why do we need the concept of element in set theory? It seems redundant |with the concept of set. Elements seem to be a sub-class of sets. The notion of set used in set theory is to be distinguished from the notion of an aggregate composed of parts. There is a theory of the relationship between part and whole that has a closer resemblance to what you describe. (Usually there is no special notion of element; you don¹t need to assume that there are parts that can¹t be further divided into parts, although you could.) But think of membership in a set as being a different sort of relationship. In mathematics, one eventually deals with a lot of sets whose members are themselves sets. In those cases we tend to really need the concept of set and not the concept of an aggregate of parts. Say for example we had some points p1,...,p10 in the plane. For each subset of {p1,...,p10}, we can ask whether the subset is collinear or not. This property is naturally associated with a set, the set of subsets of {p1,...,p10} that are collinear. This could be something like {{p1,p2,p3},{p3,p4,p5},...}. There are 1024 subsets of {p1,...,p10}, and for each one there is a separate question, whether the subset is collinear or not. This kind of information is too much to be captured by a simple relationship of part to whole. If we just took the union of the subsets, we¹d just get {p1,...,p10} again, which isn¹t what we want. We could say something similar about the set of committees in the U.S. congress. They are overlapping, some are contained inside others, and so on. In order to consider the set of the committees, we are treating the committees to some degree as being individuals that are distinct from each other, even if they are overlapping. In the theory of voting, one sometimes considers the sets of voters who could, if they agreed, ensure that a given candidate wins. That¹s another set (a.k.a. family) of sets of individuals where the distinction between being a member or not isn¹t a matter of being included, but is designated based on some property that each set may or may not have. Keith Ramsay === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? Correction > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Elements seem to be a sub-class of sets. > Couldn¹t we simply do this: > - Sets are containers of other sets. > - DeŽne the empty set {} > - A is a subset of B if all subsets of A are also a subset of B. > Then replace all mentions of element with set and is element of > with is subset of and get the same results with a simpler set of > axioms. I have adopted another approach in my DC Proof software to hopefully simplify matters. I have a Set predicate that designates a variable as a set. Designating an object as a set allows you to invoke a number of set axioms for that object -- selecting a subset from it, and inferring the existence of its power set, etc. To illustrate, I introduce the natural numbers in a premise giving the equivalent of the Peano¹s axioms as follows: Set(n) & 1 e n & ALL(a):[a e n => next(a) e n] & ALL(a):[a e n => ~next(a)=1] & ALL(a):ALL(b):[a e n & b e n & next(a)=next(b) => a=b] & ALL(a):[Set(a) & 1 e a & ALL(b):[b e n & b e a => next(b) e a] => ALL(b):[b e n => b e a]] Here, for example, I designate n as a set, but not the number 1. I haven¹t found any need for 1 to be a set. And it would be impossible to prove that 1 is a set in my system, starting from this premise. Something to think about, at any rate. Dan Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Elements seem to be a sub-class of sets. > Couldn¹t we simply do this: > - Sets are containers of other sets. > - DeŽne the empty set {} > - A is a subset of B if all subsets of A are also a subset of B. > Then replace all mentions of element with set and is element of > with is subset of and get the same results with a simpler set of > axioms. I have adopted another approach in my DC Proof software to hopefully simplify matters. I have a Set predicate that designates a variable as a set. Designating an object to be a set allows you to invoke a number of set axioms on that object -- selecting a subset from it, and inferring the existence of its power set, etc. To illustrate, I introduce the natural numbers in a premise giving the equivalent of the Peano¹s axioms as follows: Set(n) & 1 e n & ALL(a):[a e n => next(a) e n] & ALL(a):[a e n => ~next(a)=1] & ALL(a):ALL(b):[a e n & b e n & next(a)=next(b) => a=b] & ALL(a):[Set(a) & 1 e a & ALL(b):[b@n & b e a => next(b) e a] => ALL(b):[b@n => b e a]] Here, for example, I designate n as a set, but not the number 1. I haven¹t found any need for 1 to be a set. And it would be impossible to prove that 1 is a set in my system starting from this premise. Something to think about, at any rate. Dan Download DC Proof 1.0 at http://www.dcproof.com === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Elements seem to be a sub-class of sets. > Couldn¹t we simply do this: > - Sets are containers of other sets. > - DeŽne the empty set {} > - A is a subset of B if all subsets of A are also a subset of B. > Then replace all mentions of element with set and is element of > with is subset of and get the same results with a simpler set of > axioms. > I¹m not a mathematician, just curious... > Alain DeŽne S as the set of all sets that contain themselves. Think about your third axiom and see what happens. A. === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? Discussion, linux) >> Why do we need the concept of element in set theory? It seems >> redundant >> with the concept of set. Elements seem to be a sub-class of sets. >> Couldn¹t we simply do this: >> - Sets are containers of other sets. >> - DeŽne the empty set {} >> - A is a subset of B if all subsets of A are also a subset of B. >> Then replace all mentions of element with set and is element of >> with is subset of and get the same results with a simpler set of >> axioms. >> I¹m not a mathematician, just curious... >> Alain > DeŽne S as the set of all sets that contain themselves. Think about > your third axiom and see what happens. I give up. What happens? And since he hasn¹t said explicitly what sets exist, how do we know your S exists? -- Who knows, maybe that may be the only way to settle this crap. It¹s not like it¹d be that hard for me to go back and get a math degree. I can penetrate the math social group and then Žnish the takedown from inside. -- James S. Harris contemplates a new strategy. === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Elements seem to be a sub-class of sets. No, not a sub-class. It¹s the very same class. Every element is a set, and every set is an element of some other set. > Couldn¹t we simply do this: > - Sets are containers of other sets. > - DeŽne the empty set {} > - A is a subset of B if all subsets of A are also a subset of B. > Then replace all mentions of element with set and is element of > with is subset of and get the same results with a simpler set of > axioms. Except that in ZF, none of the axioms actually mentions sets at all. It¹s implicitly assumed that everything that exists according to the axioms is a set. Therefore, there is no redundancy. Element of is the fundamental concept. > I¹m not a mathematician, just curious... > Alain -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. X-mailer: xrn 9.02 === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? Mail-To-News-Contact: abuse@dizum.com >> with the concept of set. Elements seem to be a sub-class of sets. >No, not a sub-class. It¹s the very same class. Every element is a set, >and every set is an element of some other set. >Except that in ZF, none of the axioms actually mentions sets at all. >It¹s implicitly assumed that everything that exists according to the >axioms is a set. Therefore, there is no redundancy. Element of is the >fundamental concept. I thought that I had a reasonably good layman¹s understanding of sets, but this statement blind-sided me. A set can only be a collection of sets? So it¹s turtles all the way down then? I¹d always thought that { Communism, my bicycle, the word mosquito } was a perfectly valid set. -- Michael F. Stemper #include Indians scattered on dawn¹s highway bleeding; Ghosts crowd the young child¹s fragile eggshell mind. === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > with the concept of set. Elements seem to be a sub-class of sets. >>No, not a sub-class. It¹s the very same class. Every element is a set, >>and every set is an element of some other set. >>Except that in ZF, none of the axioms actually mentions sets at all. >>It¹s implicitly assumed that everything that exists according to the >>axioms is a set. Therefore, there is no redundancy. Element of is the >>fundamental concept. > I thought that I had a reasonably good layman¹s understanding of sets, but > this statement blind-sided me. A set can only be a collection of sets? So > it¹s turtles all the way down then? > I¹d always thought that { Communism, my bicycle, the word mosquito } > was a perfectly valid set. Well some people allow non-sets to be elements of sets (called ur-elements) but the point is that one can do all of standard mathematics in ZF (or usually ZFC) without ur-elements at all: every set being a set of sets. See, say, Halmos¹s Naive Set Theory. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? > Why do we need the concept of element in set theory? It seems > redundant > with the concept of set. Elements seem to be a sub-class of sets. > No, not a sub-class. It¹s the very same class. Every element is a set, > and every set is an element of some other set. What about urelements? (Does anyone still use those things?) > Couldn¹t we simply do this: > - Sets are containers of other sets. > - DeŽne the empty set {} > - A is a subset of B if all subsets of A are also a subset of B. > Then replace all mentions of element with set and is element of > with is subset of and get the same results with a simpler set of > axioms. > Except that in ZF, none of the axioms actually mentions sets at all. > It¹s implicitly assumed that everything that exists according to the > axioms is a set. Therefore, there is no redundancy. Element of is the > fundamental concept. > I¹m not a mathematician, just curious... > Alain === Subject: Re: Set theory - VERY basic question: Why do we need elements when we have sets??? >> Why do we need the concept of element in set theory? It seems >> redundant >> with the concept of set. Elements seem to be a sub-class of sets. >> No, not a sub-class. It¹s the very same class. Every element is a set, >> and every set is an element of some other set. > What about urelements? (Does anyone still use those things?) I was only talking about ZF, though I didn¹t mention that. (And right, ZF doesn¹t have proper classes, but the meaning is clear.) -- Dave Seaman Judge Yohn¹s mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > Your little group of censorious little cretins žunked out in my > estimation. James has been intemperate and hot headed. What is NOT > missing is his heart and his basic sense of being a MAN, rather than a > spineless whelp like many of you who worship orthodoxy without > understanding it. > I have had my disagreements with James on certain issues, but he is > more a MAN in the sense of what counts than any of you whelps. And > with that, I go back to my work. > I despise you for picking on him . I LOATHE cowards, and that is what > most of you are. > Your group žunked out in my estimation. While you are at it, grow > some balls if you can and attack some REAL basic problems in > mathematics, rather than act like censorious little cretins who only > know how to critique, rather than create something new. > Andrew Beckwith, PhD Unfortunately JSH¹s work as posted on sci.math has been only too well understood by several people on this newsgroup. It seems he himself does not understand mathematics very well, though. As for you, Dr. Beckwith, you seem to place great stock in deserving respect because you have a PhD. Many disreputable people have earned PhD degrees from reputable universities, and there are plenty of diploma mills which will grant any degree one wants if they are paid. Your posting above is ample demonstration of your own character žaws. === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > Your little group of censorious little cretins žunked out in my > estimation. James has been intemperate and hot headed. What is NOT > missing is his heart and his basic sense of being a MAN, rather than > a > spineless whelp like many of you who worship orthodoxy without > understanding it. > I have had my disagreements with James on certain issues, but he is > more a MAN in the sense of what counts than any of you whelps. And > with that, I go back to my work. > I despise you for picking on him . I LOATHE cowards, and that is what > most of you are. > Yes. It takes a lot of courage to type nasty things on your keyboard, > locked safely away in your house. > Your group žunked out in my estimation. While you are at it, grow > some balls if you can and attack some REAL basic problems in > mathematics, rather than act like censorious little cretins who only > know how to critique, rather than create something new. > Andrew Beckwith, PhD > There are a few people here who have at least tried to be of some > help to people who come here with questions. I¹m not clear on what > you are doing here, except calling people names. > I fear most of us will be forgotten completely in a couple hundred > years or so. It would be nice to imagine that there is a Gauss or > Galileo here, but I fear its not so, we just have to do our best > with our meager gifts. witten? hawking? === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > I fear most of us will be forgotten completely in a couple hundred > years or so. It would be nice to imagine that there is a Gauss or > Galileo here, but I fear its not so, we just have to do our best > with our meager gifts. > witten? hawking? Though they may lurk here (somehow I doubt it) I don¹t think either one is a contributor to sci.math. - Randy === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > I fear most of us will be forgotten completely in a couple hundred > years or so. It would be nice to imagine that there is a Gauss or > Galileo here, but I fear its not so, we just have to do our best > with our meager gifts. > > witten? hawking? > Though they may lurk here (somehow I doubt it) I don¹t > think either one is a contributor to sci.math. > - Randy i meant here as in.. in this day and age.. as for sci.math itself... the only person who probably comes close is james harris.. he¹s so brilliant he¹s laughed at in his time. same thing happened with many geniuses of antiquity. one implies the other right? === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > I fear most of us will be forgotten completely in a couple hundred > years or so. It would be nice to imagine that there is a Gauss or > Galileo here, but I fear its not so, we just have to do our best > with our meager gifts. > > witten? hawking? > > Though they may lurk here (somehow I doubt it) I don¹t > think either one is a contributor to sci.math. > > i meant here as in.. in this day and age.. as for sci.math > itself... the only person who probably comes close is james harris.. > he¹s so brilliant he¹s laughed at in his time. same thing happened > with many geniuses of antiquity. one implies the other right? Being ridiculed is no guarantee that you are a genius ahead of your time. True, they laughed at Darwin, the Wright Brothers, and Einstein, but they laughed at Bozo the Clown, too. Carl Sagan === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > > I fear most of us will be forgotten completely in a couple hundred > > years or so. It would be nice to imagine that there is a Gauss or > > Galileo here, but I fear its not so, we just have to do our best > > with our meager gifts. > > witten? hawking? > > Though they may lurk here (somehow I doubt it) I don¹t > think either one is a contributor to sci.math. > > i meant here as in.. in this day and age.. as for sci.math > itself... the only person who probably comes close is james harris.. > he¹s so brilliant he¹s laughed at in his time. same thing happened > with many geniuses of antiquity. one implies the other right? > Being ridiculed is no guarantee that you are a genius ahead > of your time. True, they laughed at Darwin, the Wright > Brothers, and Einstein, but they laughed at Bozo the Clown, too. > Carl Sagan harris is the Œbozo the clown¹ of sci.math === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you X-CompuServe-Customer: Yes X-Coriate: interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Pose: George Cox X-Punge: Micro$oft X-Sanguinate: The MVS Guy X-Terminate: SPA(GIS) X-Tinguish: Mark GrifŽth X-Treme: C&C,DWS at 09:14 PM, rwill9955@hotmail.com (andy) said: >Your little group of censorious little cretins žunked out in my >estimation. James has been intemperate and hot headed. What is NOT >missing is his heart and his basic sense of being a MAN, A *MAN* who wants to call someone a coward does it to his face. FOAD. >Andrew Beckwith, PhD In what, education? *PLONK* -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: STRANGE PROPERTY OF THE GOLDEN RATIO by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8GJWpW09332; I recall hearing that the positive integers are partitioned by the two sets deŽned as below, and wondered if anyone could refer me to a proof (I understand one may be somewhere on Sci.Math). Letting G(x) denote the greatest integer less than or equal to x and r be the golden ratio [ r = (1+ SQRT(5))/2 ], deŽne A = { G(nr) : n = 1, 2, 3, ... } B = { G(nr^2) : n = 1, 2, 3, ... }. An elementary density argument shows a related uniqueness result: if sets deŽned in this manner partition the set of positive integers, then r must be the golden ratio; but the converse is giving me a problem. When r is the golden ratio, we have r^2 = r+1; so that if we characterize A = {a(n) : n = 1, 2, 3, ... }, then B = {a(n) + n : n = 1, 2, 3, ... }, where the elements of both sets are enumerated in increasing order. This characterization leads to an easy inductive, in parallel consruction a unique pair of partitioning sets. What I have not been able to show that it deŽnes the same sets as the ones deŽned in the second paragraph; but I do wonder if this observation may be involved in any proof. The unresolved issues are the disjointness and ontoness as consequences of the greatest integer forms of their elements. What is so fascinating to me is that though both sets are somewhat chaotic, they are so complementary! === Subject: Re: STRANGE PROPERTY OF THE GOLDEN RATIO >I recall hearing that the positive integers are partitioned by the >two sets deŽned as below, and wondered if anyone could refer me to a >proof (I understand one may be somewhere on Sci.Math). >Letting G(x) denote the greatest integer less than or equal to x and >r be the golden ratio [ r = (1+ SQRT(5))/2 ], deŽne > A = { G(nr) : n = 1, 2, 3, ... } > B = { G(nr^2) : n = 1, 2, 3, ... }. For more interesting observations about the golden ratio, see http://www.goldenmuseum.com/ --Keith Lewis klewis {at} mitre.org The above may not (yet) represent the opinions of my employer. === Subject: Re: STRANGE PROPERTY OF THE GOLDEN RATIO > I recall hearing that the positive integers are partitioned by the > two sets deŽned as below, and wondered if anyone could refer me to a > proof (I understand one may be somewhere on Sci.Math). > Letting G(x) denote the greatest integer less than or equal to x and > r be the golden ratio [ r = (1+ SQRT(5))/2 ], deŽne > A = { G(nr) : n = 1, 2, 3, ... } > B = { G(nr^2) : n = 1, 2, 3, ... }. > An elementary density argument shows a related uniqueness result: if > sets deŽned in this manner partition the set of positive integers, > then r must be the golden ratio; but the converse is giving me a > problem. Any two irrationals u and v so that 1/u + 1/v = 1 give two sequences A = { G(nu) } and B = { G(nv) } with the same property that all integers are once and only once in {A, B}. See Beatty Sequence (Wolfram) > When r is the golden ratio, we have > r^2 = r+1; dividing by r^2 is 1/r + 1/r^2 = 1 > so that if we characterize > A = {a(n) : n = 1, 2, 3, ... }, > then > B = {a(n) + n : n = 1, 2, 3, ... }, See Wythoff Array and Wythoff game > where the elements of both sets are enumerated in increasing order. > This characterization leads to an easy inductive, in parallel > consruction a unique pair of partitioning sets. -- philippe (chephip at free dot fr) === Subject: Question about closed and compact sets Consider 2 sets, both subsets of the set of all rationals Q: A = {p: 2< p^2 <3, p rational} B = {p: a < p < b, for a,p,b rational} 1.-It appears to me that A is closed subset of Q, but not B. Is this correct? I have a text that gives B as an example of a closed subset of Q. 2.- How does one go about proving that A is a compact subset of B. All help appreciated. === Subject: Re: Question about closed and compact sets > Consider 2 sets, both subsets of the set of all rationals Q: > A = {p: 2< p^2 <3, p rational} > B = {p: a < p < b, for a,p,b rational} > 1.-It appears to me that A is closed subset of Q, but not B. Is this > correct? I have a text that gives B as an example of a closed subset > of Q. B is not closed in Q. No open neighborhood of the point a in Q misses the subset B, so the complement of B in Q cannot be open. Since B¹s complement fails to be open, B fails to be closed. > 2.- How does one go about proving that A is a compact subset of B. A is not compact. Consider the open covering of A by the intervals I_n = (Ln , Rn) = { q in Q | Ln < q < Rn } and where: Ln is the least number of the form N/10^n strictly greater than sqrt(2) Rn is the greatest number of the form N/10^n strictly less than sqrt(3) as follows: L1 = 1.5, L2 = 1.42, L3 = 1.415, ... (decreasing to sqrt(2)) R1 = 1.7, R2 = 1.73, R3 = 1.732, ... (increasing to sqrt(3)) That is, A > ... > I_(n+1) > I_n > I_(n-1) > ... > I_1 and A = union_n(I_n). Notice that this covering has no Žnite subcover. Thus A cannot be compact. > All help appreciated. Dale === Subject: Re: Question about closed and compact sets > Consider 2 sets, both subsets of the set of all rationals Q: > > A = {p: 2< p^2 <3, p rational} > A is not compact. Consider the open covering of A by the intervals > I_n = (Ln , Rn) = { q in Q | Ln < q < Rn } > and where: > Ln is the least number of the form N/10^n > strictly greater than sqrt(2) > Rn is the greatest number of the form N/10^n > strictly less than sqrt(3) > as follows: > L1 = 1.5, L2 = 1.42, L3 = 1.415, ... (decreasing to sqrt(2)) > R1 = 1.7, R2 = 1.73, R3 = 1.732, ... (increasing to sqrt(3)) > That is, A > ... > I_(n+1) > I_n > I_(n-1) > ... > I_1 > and A = union_n(I_n). > Notice that this covering has no Žnite subcover. Thus A cannot be > compact. Why not just consider the open cover (-oo, sqrt(3) - 1/n), n = 1, 2, ... ? === Subject: Re: Question about closed and compact sets > > Consider 2 sets, both subsets of the set of all rationals Q: > > A = {p: 2< p^2 <3, p rational} > A is not compact. Consider the open covering of A by the intervals > > I_n = (Ln , Rn) = { q in Q | Ln < q < Rn } > > and where: > Ln is the least number of the form N/10^n > strictly greater than sqrt(2) > Rn is the greatest number of the form N/10^n > strictly less than sqrt(3) > > as follows: > L1 = 1.5, L2 = 1.42, L3 = 1.415, ... (decreasing to sqrt(2)) > R1 = 1.7, R2 = 1.73, R3 = 1.732, ... (increasing to sqrt(3)) > > That is, A > ... > I_(n+1) > I_n > I_(n-1) > ... > I_1 > > and A = union_n(I_n). > > Notice that this covering has no Žnite subcover. Thus A cannot be > compact. > Why not just consider the open cover (-oo, sqrt(3) - 1/n), n = 1, 2, ... ? === Subject: Re: Question about closed and compact sets days. My association with the Department is that of an alumnus. >Consider 2 sets, both subsets of the set of all rationals Q: >A = {p: 2< p^2 <3, p rational} >B = {p: a < p < b, for a,p,b rational} >1.-It appears to me that A is closed subset of Q, but not B. Is the topology of Q the induced topology (from the real line, usual topology, of course)? > Is this correct? A subset X of Q will be closed in the induced topology if and only if there exists a closed set C (of the real line) such that C intersect Q is equal to X. Clearly, A = [sqrt(2),sqrt(3)] intersect Q. So A is indeed a closed subset of Q. I don¹t think B is a closed subset in the induced topology, assuming of course that a I have a text that gives B as an example of a closed subset >of Q. Are you positive? It would be if a and b were irrationals, for example. >2.- How does one go about proving that A is a compact subset of B. Well, Žrst, it is unclear that A is a subset of B. What are a and b? Are you sure it is a compact subset? -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Question about closed and compact sets >Consider 2 sets, both subsets of the set of all rationals Q: >A = {p: 2< p^2 <3, p rational} >B = {p: a < p < b, for a,p,b rational} >1.-It appears to me that A is closed subset of Q, but not B. > Is the topology of Q the induced topology (from the real line, usual > topology, of course)? > Is this correct? > A subset X of Q will be closed in the induced topology if and only if > there exists a closed set C (of the real line) such that C intersect Q > is equal to X. > Clearly, A = [sqrt(2),sqrt(3)] intersect Q. So A is indeed a closed > subset of Q. > I don¹t think B is a closed subset in the induced topology, assuming > of course that a Q at B. Then C contains a sequence of elements converging to a (for > each positive integer n, let p(n) be a rational which satisŽes > a < p(n) < min (a+1/n, b). This must exist, and p(n) converges to a; > since C is closed, that means that a lies in C; but then C intersect Q > would contain a, which contradicts the assumption that it is equal to > B. > I have a text that gives B as an example of a closed subset >of Q. > Are you positive? It would be if a and b were irrationals, for example. You are correct, as usual. Example is for a and b irrationals. One of these days my posts will be error-free! >2.- How does one go about proving that A is a compact subset of B. > Well, Žrst, it is unclear that A is a subset of B. What are a and b? > Are you sure it is a compact subset? Again, I mis-posted. Example is for A is subset of Q (the rationals), === Subject: Re: Question about closed and compact sets days. My association with the Department is that of an alumnus. >>Consider 2 sets, both subsets of the set of all rationals Q: >> >>A = {p: 2< p^2 <3, p rational} What I posted was slightly incorrect: A is equal to ([sqrt(2),sqrt(3)]intersect Q) UNION ([-sqrt(3),-sqrt(2)] intersect Q), the union of two relatively closed sets. >>B = {p: a < p < b, for a,p,b rational} [...] >> I have a text that gives B as an example of a closed subset >>of Q. >> Are you positive? It would be if a and b were irrationals, for example. >You are correct, as usual. Example is for a and b irrationals. One >of these days my posts will be error-free! Then the same argument as given before works: if B = { p in Q: a< p < b, a and b irrationals} then B = [a,b] intersect Q, and therefore is closed in the induced topology. >>2.- How does one go about proving that A is a compact subset of B. >> Well, Žrst, it is unclear that A is a subset of B. What are a and b? >> Are you sure it is a compact subset? >Again, I mis-posted. Example is for A is subset of Q (the rationals), A is not a compact subset of Q. The family of relative open sets {U_n}, n in N, where U_n = [ (-inŽnity,sqrt(3)-1/n) intersect Q ] covers A: any rational p satisfying p^2<3 lies in one of these intervals. But clearly it has no Žnite subcover. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Interesting Posers [...] > 4. A corner is a Žgure composed of 3 unit squares, obtained by > removing any one unit square from a 2 by 2 block square. > A 5 by 7 chess board is to be covered by such corners in such a way > that each unit square on the chess board is covered by the same number > of corners. > Is it possible ? The only rectangles that do not admit a uniform multicovering by corners are 1xn, 3x(2n+1), 5x5, and 5x7. Proof: 1xn is clearly impossible. For the other impossible cases, note they are all (2m+1)x(2n+1) rectangles. Assign weights to the squares of the rectangle in the following pattern. -2 1 -2 1 -2 1 -2 1 1 1 1 1 1 1 -2 1 -2 1 -2 1 -2 1 1 1 1 1 1 1 -2 1 -2 1 -2 1 -2 The weight of any corner is 0 or 3, so the weight of any multicovering by corners is nonnegative. However, since the multicovering is uniform, the weight of the multicovering must be a multiple of the total weight of the rectangle, which is (m-1)(n-1) - 3. This is negative when m=1 (3x(2n+1)), when m=n=2 (5x5), and when m=2,n=3 (5x7), so these rectangles do not admit a uniform multicovering. To demonstrate that all the other cases can be multicovered, Žrst _ _ _ _ _ note that the 2x3 brick | |_ | and the 2x2 block |_|_| admit uniform |_ _|_| |_|_| multicoverings, the latter by superimposing four corners. These can be combined to form a multicover of any (2m)x(2n+1) rectangle. The diagrams and _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ | _| |_ | _| | |_ | _| _|_ | |_| |_ _|_|_| | |_ _|_|_| |_| _|_| |_ _| | |_ _| |_ |_ |_ _|_|_ | | |_ _|_ _|_ | | |_| |_| _| | |_| |_ _|_ |_ |_| |_ _|_ _|_|_ _|_ _| | _| |_| |_| | |_|_ _|_ _|_ _| depict 1-coverings (tilings) of a 5x9 rectangle and a 7x7 square minus a block. The latter can be combined with a multicovered block for a multicovering of the 7x7 square. Any larger (2m+1)x(2n+1) square can be formed by laminating (2n)xm rectangles to these, completing the proof. Closer examination of this proof will reveal that for rectangles that admit multicoverings, the ones whose area is a multiple of 3 admit 1-coverings (tilings), while the others admit 3-coverings. The questions listed in the original post are indeed interesting posers. What is their source? Dan Hoey Hoey@AIC.NRL.Navy.Mil === Subject: My paper, and the cheaters I¹ve made some pointed criticisms of math society but if you actually pay attention you¹ll realize that the evidence against math society is nastier than I go into detail about usually. Supposedly a correct math result is what¹s important, but despite my having a quite correct paper, which was to be published, sci.math¹ers managed to get it yanked IN ONE DAY with some hostile emails. The paper has been by a lot of editors and mathematicians and to date, no major error has been found, which I say because there was one minor error that actually got pointed out by a sci.math poster. So the lie is that if you have a correct result and send it to a journal, then that¹s what¹s important, but the reality is that the sci.math¹ers managed to censor my paper with a few bogus emails...lying to an editor who didn¹t follow fair procedures, and yanked my paper. That¹s the REAL math world. If you actually look at my paper, which you can see at http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid =26 the Žrst thing you should notice is that it is SHORT. The paper was at a math journal for OVER NINE MONTHS and I was told it was peer reviewed. the electronic journal¹s website, and sci.math¹ers went after it--after ripping on the journal and its editors. They are HATEFUL AND VICIOUS PEOPLE who so far have gotten away with cheating the system, but where are mathematicians now? Do mathematicians actually care about their vaunted peer review system? Apparently not if it has to do with my paper. These people lie. They clearly behave politically. They clearly act looking out for some agenda other than the truth, and I don¹t know why any of you trust them. Just look at http://rattler.cameron.edu/swjpam/vol2-03.html and see where my paper used to be at a journal that¹s STILL PUBLISHING, and consider just how the evidence properly should be evaluated. My paper is quite correct, and in fact mostly uses basic algebra, but math people lie about it, and get away with lying because the MATH SYSTEM IS CORRUPT. So yes, you are sheep and cows who follow along because people tell you what to think and you refuse to actually think for yourselves. That¹s why Andrew Wiles¹s work is accepted, despite a dramatic logical žaw, and the demonstrated inability of posters to make even a basic defense of it by answering my null test, but you people are sheep! You follow because it makes you feel good. Actually thinking for yourselves would be too hard, and it might hurt your feelings if you realized how easily mathematicians can lie to you. Yes, I am angry but justiŽably so, as there¹s just no way around the reality that bull worked with my paper. I FOLLOWED THE RULES, my paper was put out there, and PEOPLE WHO CHEATED, got it yanked in a day with some freaking emails. THAT is the math world. THAT is the vaunted modern math system. THAT Is the reality of a political world where the truth is less important to these people than the consequences of the truth. They don¹t care about the math, and if you support them, or just sit by while they piss on the system, then you don¹t either. History is your judge. This generation has earned the contempt of future mathematicians, and they will deservedly ignore your mathematical work. After all, you CANNOT BE TRUSTED, so why, when that is clear, should people in the future bother to dig through the crap to try and Žnd a decent result here and there? YOU ARE DESERVED OF CONTEMPT AS A SOCIETY! You people are working for nothing, except money, and what little trivialities society gives you now, but I guess for you mathematicians paying your mortgages, and getting your kids through college IS what it¹s all about. But you will be spat upon by real mathematicians in the future, who will cite you with contempt. James Harris === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > So the lie is that if you have a correct result and send it to a > journal, then that¹s what¹s important, but the reality is that the > sci.math¹ers managed to censor my paper with a few bogus > emails...lying to an editor who didn¹t follow fair procedures, and > yanked my paper. > That¹s the REAL math world. > If you actually look at my paper, which you can see at > http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid= 26 > the Žrst thing you should notice is that it is SHORT. The Žrst thing I notice is your co-author has a PhD in physics and came here abusing us for how we were disrespectful of your veteran status. I don¹t see how he is either a qualiŽed co-author or a person who does his research. He acts more like a buddy sticking up for his friend without discovering the facts. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: My paper, and the cheaters did you ever attempt to acertain the reason for the withdrawal (sic; rejection) of your paper, other than ther mere fact of a miniature mass-movement -- I wouldn¹t call it a society, exactly -- that you¹ve induced with your social experiment? since referees aren¹t perfect, aren¹t they allowed to change their minds, such as if they got a good critique of a žaw in the paper -- isn¹t that normal? otherwise, it¹s going to be just as corrupt as any thing in Universe, more or less. > http://rattler.cameron.edu/swjpam/vol2-03.html > YOU ARE DESERVED OF CONTEMPT AS A SOCIETY! --Give Earth a Trickier Dick Cheeny -- out of ofŽce, after gigayears! http://tarpley.net/bush12.htm http://www.benfranklinbooks.com/ http://members.tripod.com/~american_almanac http://www.wlym.com/pdf/iclc/howthenation.pdf http://www.rwgrayprojects.com/synergetics/plates/Žgs/plate02. html === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > So the lie is that if you have a correct result and send it to a > journal, then that¹s what¹s important, but the reality is that the > sci.math¹ers managed to censor my paper with a few bogus > emails...lying to an editor who didn¹t follow fair procedures, and > yanked my paper. > That¹s the REAL math world. > If you actually look at my paper, which you can see at > http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid= 26 > the Žrst thing you should notice is that it is SHORT. Actually, the Žrst thing to notice is that the result is WRONG. All the complaining in the world won¹t change that--it was WRONG before and it¹s still WRONG now. You¹re just making yourself look foolish. Again. Rick === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > So the lie is that if you have a correct result and send it to a > journal, then that¹s what¹s important, but the reality is that the > sci.math¹ers managed to censor my paper with a few bogus > emails...lying to an editor who didn¹t follow fair procedures, and > yanked my paper. > That¹s the REAL math world. > If you actually look at my paper, which you can see at > http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid= 26 > the Žrst thing you should notice is that it is SHORT. > The paper was at a math journal for OVER NINE MONTHS and I was told it > was peer reviewed. > the electronic journal¹s website, and sci.math¹ers went after > it--after ripping on the journal and its editors. > They are HATEFUL AND VICIOUS PEOPLE who so far have gotten away with > cheating the system, but where are mathematicians now? > Do mathematicians actually care about their vaunted peer review > system? > Apparently not if it has to do with my paper. > These people lie. They clearly behave politically. They clearly act > looking out for some agenda other than the truth, and I don¹t know why > any of you trust them. > Just look at > http://rattler.cameron.edu/swjpam/vol2-03.html > and see where my paper used to be at a journal that¹s STILL > PUBLISHING, and consider just how the evidence properly should be > evaluated. > My paper is quite correct, and in fact mostly uses basic algebra, but > math people lie about it, and get away with lying because the MATH > SYSTEM IS CORRUPT. > So yes, you are sheep and cows who follow along because people tell > you what to think and you refuse to actually think for yourselves. > That¹s why Andrew Wiles¹s work is accepted, despite a dramatic logical > žaw, and the demonstrated inability of posters to make even a basic > defense of it by answering my null test, but you people are sheep! > You follow because it makes you feel good. > Actually thinking for yourselves would be too hard, and it might hurt > your feelings if you realized how easily mathematicians can lie to > you. > Yes, I am angry but justiŽably so, as there¹s just no way around the > reality that bull worked with my paper. > I FOLLOWED THE RULES, my paper was put out there, and PEOPLE WHO > CHEATED, got it yanked in a day with some freaking emails. > THAT is the math world. > THAT is the vaunted modern math system. > THAT Is the reality of a political world where the truth is less > important to these people than the consequences of the truth. > They don¹t care about the math, and if you support them, or just sit > by while they piss on the system, then you don¹t either. > History is your judge. This generation has earned the contempt of > future mathematicians, and they will deservedly ignore your > mathematical work. > After all, you CANNOT BE TRUSTED, so why, when that is clear, should > people in the future bother to dig through the crap to try and Žnd a > decent result here and there? > YOU ARE DESERVED OF CONTEMPT AS A SOCIETY! > You people are working for nothing, except money, and what little > trivialities society gives you now, but I guess for you > mathematicians paying your mortgages, and getting your kids through > college IS what it¹s all about. > But you will be spat upon by real mathematicians in the future, who > will cite you with contempt. > James Harris Math content: 0% Social content: 100% === Subject: Re: My paper, and the cheaters ... > Math content: 0% > Social content: 100% Antisocial content: 100% === Subject: Re: My paper, and the cheaters Discussion, linux) > That¹s why Andrew Wiles¹s work is accepted, despite a dramatic > logical žaw, and the demonstrated inability of posters to make even > a basic defense of it by answering my null test, but you people are > sheep! You still haven¹t shown how the null test works. Let¹s look at two arguments, one valid and one not. Here¹s the valid one. ,----[ Proof ] | 1. 0 < 1 Ax. | 2. 0 + 1 = 1 Ax. | 3. 0 < 0 + 1 Subst. | 5. (A x)(A y)( x < y -> x + 1 < y + 1 ) Ax. | 6. (A y)( n < y -> n + 1 < y + 1) UE | 7. n < n + 1 -> n + 1 < (n + 1) + 1 UE | 8. (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) UI | 9. 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) | &I | 10. ( 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) ) -> | (A x)( x < x + 1 ) Ax. | 11. (A x)( x < x + 1 ) MP `---- Here¹s one that¹s invalid. ,---- | 1. (E x)P(x) (Assumption) | ,---- | 2. | P(y) (E-elim) | 3. | (A x)P(x) (A-intro, but this is an invalid application of the | | rule.) | `---- | 4. (E x)P(x) -> (A x)P(x) (-> intro) `---- Could you please apply your test to both trivial arguments? Show which step is the keystone step of the Žrst argument and show how the null test proves the second argument is invalid. I¹ll give you a hint. For the Žrst, assume NOT (A x)( x < x + 1 ) and show which line that assumption contradicts. For the second, show that the assumption of NOT (E x)P(x) -> (A x)P(x) doesn¹t contradict any of the four lines of the proof. point, the null test will become a standard tool of proof veriŽcation. [Followups set to sci.math] -- Analysis/ editors have evaluated the paper, they accepted it for publication and they have the copyright of its contents - and thus they are responsible for its correctness,¹ she [said]. === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. Please do not ever again claim to be an honest person. Your proof contradicts known mathematics. At least 4 counterproofs have been given. Dale Hall¹s counterproof can be checked with simple arithmetic. The exact spot in your proof where you make your major error has been pointed out repeatedly. You know all this and yet you continue to deny it. Below is one of the simplest counterproofs. Feel free to point out errors (assuming you are still interested in math as opposed to social studies). This is, incidentally, for your future reference, what real proof is supposed to look like. James Harris has written a paper called Advanced Polynomial Factorization in which he claims the following: if the polynomial 65*x^3 - 12*x + 1 is factored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1), where a1, a2 and a3 are algebraic integers, then exactly two of a1, a2, and a3 are divisible in the algebraic integers by sqrt(5), and the third one is coprime to 5. It should be noted that -a1, -a2, and -a3 are roots of x^3 + 12*x^2 - 65. That Harris¹s claim is false can be seen from the following: Assume m(x) is an irreducible monic polynomial with integer coefŽcients and algebraic integers a and b are two roots of m(x). Theorem. If p is a nonzero rational integer and a is divisible by sqrt(p) in the algebraic integers, then b is also divisible by sqrt(p). Proof: ------------------------------------------------------------- ----------- First, a couple of facts about irreducible polynomials. A polynomial with rational coefŽcients is said to be irreducible over the rationals if it cannot be written as a product of two other polynomials each of degree at least 1, with rational coefŽcients. If r is a root of an irreducible polynomial m(x), and r is also a root of another polynomial s(x), then m(x) divides s(x), i.e., there exists another polynomial t(x) such that s(x) = m(x) * t(x). This shows that m(x) has the minimal degree of any polynomial with rational coefŽcients for which r is a root; it is therefore called the minimum polynomial for r. ------------------------------------------------------------- ----------- Now assume p is a nonzero integer and sqrt(p) divides a in the algebraic integers. That is, a = sqrt(p)*u, where u is an algebraic integer. Note that a^2 = p * u^2. Let f(x) be the minimum monic polynomial for u^2. Thus f(a^2/p) = 0. DeŽne a new polynomial g(x) by g(x) = f(x/p). Finally deŽne h(x) = g(x^2). Note that h(x) is a polynomial with rational coefŽcients. Since h(a) = g(a^2) = f(a^2/p) = 0, the polynomial h(x) is divisible by m(x). Therefore h(b) = 0 also. That is, f(b^2/p) = 0. Now, since u^2 is an algebraic integer, we can assume the polynomial f(x)has integer coefŽcients and is monic. Therefore b^2/p is an algebraic integer, which implies b / sqrt(p) is also an algebraic integer. One thus concludes that b is also divisible by sqrt(p) in the algebraic integers, QED. ************************************************************* ******* Here is how this applies to the polynomial x^3 + 12*x^2 - 65. This polynomial is monic and irreducible with integer coefŽcients. By the theorem above, if one of the roots is divisible by sqrt(5), then they all are. This is not possible because the product of the roots is 65 = 5 * 13. Therefore Harris¹s claim that ANY of the roots are divisible by sqrt(5) is false. Nora B. [snip usual rant] === Subject: Re: My paper, and the cheaters This page might be of help, to James and others: http://mathforum.org/dr.math/faq/faq.cubic.equations.html It only discusses brute-force methods for solving cubic and quartic equations [not 65*x^3 - 12*x + 1 speciŽcally], but such BFMs are useful for weeding out numerical counterexamples. === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > > Please do not ever again claim to be an honest person. > Your proof contradicts known mathematics. At least 4 > counterproofs have been given. Dale Hall¹s counterproof > can be checked with simple arithmetic. The exact spot > in your proof where you make your major error has been pointed > out repeatedly. You know all this and yet you continue to > deny it. That¹s not true. In fact, I don¹t disagree with much of what you say. And I¹d like you to admit that in a post, so that I can move on to what I actually *DO* say. First though, I need you to read what I have below and acnowledge that you see the agreement on a key point. > Below is one of the simplest counterproofs. Feel free > to point out errors (assuming you are still interested in > math as opposed to social studies). This is, incidentally, > for your future reference, what real proof is supposed to > look like. > James Harris has written a paper called Advanced Polynomial > Factorization in which he claims the following: if the > polynomial > 65*x^3 - 12*x + 1 > is factored in the form > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > where a1, a2 and a3 are algebraic integers, then exactly > two of a1, a2, and a3 are divisible in the algebraic integers > by sqrt(5), and the third one is coprime to 5. > It should be noted that -a1, -a2, and -a3 are roots of > x^3 + 12*x^2 - 65. > That Harris¹s claim is false can be seen from the following: > Assume m(x) is an irreducible monic polynomial with integer coefŽcients > and algebraic integers a and b are two roots of m(x). > Theorem. If p is a nonzero rational integer and a is divisible by > sqrt(p) in the algebraic integers, then b is also divisible > by sqrt(p). That last should say, in the algebraic integers. Theorem accepted. Proof not needed here so I deleted it out to focus on what¹s important. It is in fact true that if Œa¹ is divisible by sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) ***in the ring of algebraic integers***. Concede agreement Nora Baron and then I¹ll explain the rest. > This polynomial is monic and irreducible with integer coefŽcients. > By the theorem above, if one of the roots is divisible by sqrt(5), > then they all are. This is not possible because the product of the > roots is 65 = 5 * 13. Therefore Harris¹s claim that ANY of the > roots are divisible by sqrt(5) is false. The poster has shifted from the true claim--that the result applies to algebraic integers--to a far vaguer and more inclusive claim, basically that it applies in general. Repeatedly, posters keep making claims true in the ring of algebraic integers, I accept those claims in that ring, and they then act like I don¹t!!! I repeat, the result Nora Baron gave is correct *in the ring of algebraic integers* and I now challenge that poster and others in that camp to Žnally and for once concede that I am not arguing against that point. Then I¹d like, with the permission of Nora Baron to explain exactly what my proof actually says. But Žrst I want the poster Nora Baron to accept that I am not challenging on this speciŽc point. That¹s the Žrst step. It¹s up to that poster to respond. James Harris === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > > > Please do not ever again claim to be an honest person. > Your proof contradicts known mathematics. At least 4 > counterproofs have been given. Dale Hall¹s counterproof > can be checked with simple arithmetic. The exact spot > in your proof where you make your major error has been pointed > out repeatedly. You know all this and yet you continue to > deny it. > That¹s not true. > In fact, I don¹t disagree with much of what you say. > And I¹d like you to admit that in a post, so that I can move on to > what I actually *DO* say. > First though, I need you to read what I have below and acnowledge that > you see the agreement on a key point. What??? You want people to agree to your own statement that you understand and accept a proof? Why don¹t you just say you accept it and go on? Why the game-playing? > > Below is one of the simplest counterproofs. Feel free > to point out errors (assuming you are still interested in > math as opposed to social studies). This is, incidentally, > for your future reference, what real proof is supposed to > look like. > > > James Harris has written a paper called Advanced Polynomial > Factorization in which he claims the following: if the > polynomial > > 65*x^3 - 12*x + 1 > > is factored in the form > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > > where a1, a2 and a3 are algebraic integers, then exactly > two of a1, a2, and a3 are divisible in the algebraic integers > by sqrt(5), and the third one is coprime to 5. > > It should be noted that -a1, -a2, and -a3 are roots of > > x^3 + 12*x^2 - 65. > > That Harris¹s claim is false can be seen from the following: > > > Assume m(x) is an irreducible monic polynomial with integer coefŽcients > and algebraic integers a and b are two roots of m(x). > > Theorem. If p is a nonzero rational integer and a is divisible by > sqrt(p) in the algebraic integers, then b is also divisible > by sqrt(p). > That last should say, in the algebraic integers. > Theorem accepted. Proof not needed here so I deleted it out to focus > on what¹s important. It is in fact true that if Œa¹ is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. Seems obvious that¹s what she¹s saying. > Concede agreement Nora Baron and then I¹ll explain the rest. She did I believe. > > This polynomial is monic and irreducible with integer coefŽcients. > By the theorem above, if one of the roots is divisible by sqrt(5), > then they all are. This is not possible because the product of the > roots is 65 = 5 * 13. Therefore Harris¹s claim that ANY of the > roots are divisible by sqrt(5) is false. > The poster has shifted from the true claim--that the result applies to > algebraic integers--to a far vaguer and more inclusive claim, > basically that it applies in general. I don¹t think that was her intention. I think she showed that a/sqrt(5) is not an algebraic integer. Look at her proof. She isn¹t claiming more than that. > Repeatedly, posters keep making claims true in the ring of algebraic > integers, I accept those claims in that ring, and they then act like I > don¹t!!! > I repeat, the result Nora Baron gave is correct *in the ring of > algebraic integers* and I now challenge that poster and others in that > camp to Žnally and for once concede that I am not arguing against > that point. You want the people in that camp to agree that you agree with them ? > Then I¹d like, with the permission of Nora Baron to explain exactly > what my proof actually says. You need PERMISSION to do this ??? > But Žrst I want the poster Nora Baron to accept that I am not > challenging on this speciŽc point. > That¹s the Žrst step. > It¹s up to that poster to respond. Like I say, she did. Incidentally, what did they say about APF at the journal to which you submitted it earlier in the summer? Andrzej > James Harris === Subject: Re: My paper, and the cheaters >> Theorem. If p is a nonzero rational integer and a is divisible by >> sqrt(p) in the algebraic integers then so are all conjugates b of a. > Theorem accepted. Proof not needed here so I deleted it out to focus > on what¹s important. It is in fact true that if Œa¹ is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. You can¹t avoid this theorem by shifting to your elusive object ring J of JSH integers because the theorem holds in far greater generality. Indeed I prove below that it holds for any set of algebraic numbers closed under conjugates, squares, and sqrts. Recall, as I¹ve already mentioned [1], that your ring J would be too weak for your intended applications if it were not closed under conjugates. And surely you want J closed under sqrts, since these are the simplest nontrivial algebraic integers. Hence you¹re forced to accept the theorem in J. Below is the promised proof, which is essentially the same proof I gave in a prior post here. First we require the following Lemma, which shows that if a, b are conjugate algebraic numbers then the same holds true of their images under any polynomial g(X) in Q[X] because the minimum poly f of g(a) is also a minimum poly for g(b). Thus g(a) and g(b) are conjugates, being roots of irreducible f(X). Hence a conjugate-closed set J contains g(a) iff J contains g(b). LEMMA Let a, b be conjugate algebraic numbers, i.e. any two roots of an irreducible polynomial m(X) in Q[X]. Let J be any conjugate-closed set of algebraic numbers. Then g(a) in J <=> g(b) in J, for all g(X) in Q[X]. PROOF g(a) in J => f(g(a)) = 0 for irred f(X) in Q[X] => m(X)|f(g(X)) in Q[X] => f(g(b)) = 0 => g(b) in J via J conjugate-closed QED THEOREM With the same hypotheses as the above Lemma if J is square,sqrt-closed then for all integers n sqrt(n)|a in J <=> sqrt(n)|b in J PROOF a/sqrt(n) in J <=> a^2/n in J via J closed under square,sqrt <=> b^2/n in J via Lemma with g(X) = X^2/n <=> b/sqrt(n) in J via J closed under square,sqrt QED --Bill Dubuque === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > > > Please do not ever again claim to be an honest person. > Your proof contradicts known mathematics. At least 4 > counterproofs have been given. Dale Hall¹s counterproof > can be checked with simple arithmetic. The exact spot > in your proof where you make your major error has been pointed > out repeatedly. You know all this and yet you continue to > deny it. > That¹s not true. > In fact, I don¹t disagree with much of what you say. > And I¹d like you to admit that in a post, so that I can move on to > what I actually *DO* say. > First though, I need you to read what I have below and acnowledge that > you see the agreement on a key point. > > Below is one of the simplest counterproofs. Feel free > to point out errors (assuming you are still interested in > math as opposed to social studies). This is, incidentally, > for your future reference, what real proof is supposed to > look like. > > > James Harris has written a paper called Advanced Polynomial > Factorization in which he claims the following: if the > polynomial > > 65*x^3 - 12*x + 1 > > is factored in the form > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > > where a1, a2 and a3 are algebraic integers, then exactly > two of a1, a2, and a3 are divisible in the algebraic integers > by sqrt(5), and the third one is coprime to 5. > > It should be noted that -a1, -a2, and -a3 are roots of > > x^3 + 12*x^2 - 65. > > That Harris¹s claim is false can be seen from the following: > > > Assume m(x) is an irreducible monic polynomial with integer coefŽcients > and algebraic integers a and b are two roots of m(x). > > Theorem. If p is a nonzero rational integer and a is divisible by > sqrt(p) in the algebraic integers, then b is also divisible > by sqrt(p). > That last should say, in the algebraic integers. Agreed. > Theorem accepted. Proof not needed here so I deleted it out to focus > on what¹s important. It is in fact true that if Œa¹ is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. > Concede agreement Nora Baron and then I¹ll explain the rest. Of course I concede. That¹s what I proved. The question now becomes, what did you claim to be true in Advanced Polynomial Factorization. Just below is the full text of APF exactly as it was submitted [Begin APF] Electronic Journal: Southwest Journal of Pure and Applied Mathematics Internet: http://rattler.cameron.edu/swjpam.html ISSN 1083 ADVANCED POLYNOMIAL FACTORIZATION James Harris Eail Address: jstevh@msn.com ABSTRACT. Algebraic method for determining distribution of factors within a polynomial factorization, which breaks through what was seen as a barrier from overinterpretations of Galois Theory. A.M.S. (MOS) Subject ClassiŽcation Codes. 11R04,11R09 KEY WORDS AND PHRASES. Polynomial factorization, Galois theory, Factorization lemma, Ring of algebraic integers ADVANCED POLYNOMIAL FACTORIZATION APPROACHED. Determining the distribution of factors within irrational algebraic integers has long been considered impossible as it is not possible to do using Galois Theory. However a simple technique through the introduction of more variables makes it possible. To highlight the standard belief consider the algebraic integer roots of x^2 + x - 5. While you know that the algebraic integer factors are themselves factors of 5, can either not have non unit factors of 5? How do you know? In looking to consider distribution of algebraic integer factors within a factorization I¹ll be using a more complicated example than x^2 + x - 5. This paper will show, using basic algebraic methods, that given the factorization, in the ring of algebraic integers, 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5. First I¹ll need a simple lemma to generalize beyond factors of a polynomial that are themselves polynomials. FACTORIZATION LEMMA: Given a factor g of a polynomial P(x), further deŽned as a factor for all x, which means that the value of g for a value Œa¹ of x is a factor of P(a), within the ring of algebraic integers, there exists r and c such that g = r + c where r=0, or varies as x varies, and c is a factor of the constant term P(0) and is itself constant. Let x=0, then g must be a factor of P(0), so at that point c = g. If when x does not equal 0, g=c, r=0. If when x does not equal 0, g!=c there must exist r which varies with x. That is, r=g. [] As an example consider sqrt(x + 1) which is a non polynomial factor of x+1, and while there are an inŽnity of irrational solutions consider the rational solution at x=35. Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5, and c=1. But for different values of x, g and r will vary, while c will not. PRIMARY ARGUMENT. Given 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) in the ring of algebraic integers. Let P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2 + u^3 f) Here f is a non unit, non zero algebraic integer coprime to 3 and x, and u a non unit, non zero algebraic integer coprime to f. Note P(m) has a factor that is f^2 . That expression comes from expanding (v^3 + 1) x^3 - 3 vxy^2 + y^3, using the substitutions v = -1 + m f^2, and y = uf, where additional variables provide an additional degree of freedom. Now consider the factorization P(m) = (a1 x + uf)(a2 x + uf)(a3 x + uf) where multiplying out shows that a1 a2 a3 = m^3 f^6 - 3 m^2 f^4 + 3m f^2 = f^2 (m^3 f^4 - 3 m^2 f^2 + 3m) so a1 a2 a3 = m f^2 (m^2 f^4 - 3 m f^2 + 3). Therefore, at least one of the a¹s cannot be coprime to m, and at least one of the a¹s must equal 0 when m=0. (Note: The a¹s are roots of a monic polynomial with algebraic integer coefŽcients so they are algebraic integers.) Notice that the constant term P(0) is given by P(0) = f^2 (3x u^2 + u^3 f) and also that P(0)/f^2 = 3x u^2 + u^3 f, which is coprime to f. Then I have the factor of P(m), g1, where g1 = a1 x + uf, where here I also have that a1 is not coprime to m. g1 = c = uf meaning f is a factor of the constant term. Therefore, exactly two of the a¹s equal 0, when m=0, to get the factor f^2 in the constant term P(0), while one must not equal 0, or f^3 would be the factor. Now as noted before in general P(m) has a factor that is f^2, and separating that factor off, gives a constant term coprime to f; therefore, given g1 = a1 x + uf where with m = 0, g1 gives a factor of f it must have that same factor in general, proving that two of the a¹s have a factor that is f . Therefore, one factor is coprime to f. Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn¹t change the a¹s, I have (m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3(-1 + mf^2)xu^2 + u^3 = 65x^3 - 12x + 1 which may be more easily seen from using v = -1 + mf^2 = 4, y=1 with (v^3 + 1) x^3 - 3vxy^2 + y^3 . Therefore, with the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5, which shows where some of the algebraic integer factors distribute despite the factors being irrational. [End of APF] ------------------------------------------------------------- --------- It is clear from this text that you were working in the ring of algebraic integers. For example, you say: This paper will show, using basic algebraic methods, that given the factorization, in the ring of algebraic integers, 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5. Isn¹t that EXACTLY what you said in the paper that you submitted, and about which you are now complaining that mathematicians have lied, etc. ? Over and over again, throughout APF, you refer to factorization in the algebraic integers. There is no hint that you might be thinking of any other ring. If in fact you were thinking the conclusion applied to some other ring, would you not have stated that? The paper makes absolutely no sense on any level without such a statement. What is the reader to think if suddenly, without warningat the end, you are supposedly talking about factorization and coprimeness in some other ring? It is thus utterly clear that that is what you *intended*. Now I think you want to rewrite history and claim that the whole theorem was not about algebraic integers, but about numbers in some other ring which is not even deŽned, not even MENTIONED, in the paper! > > This polynomial is monic and irreducible with integer coefŽcients. > By the theorem above, if one of the roots is divisible by sqrt(5), > then they all are. This is not possible because the product of the > roots is 65 = 5 * 13. Therefore Harris¹s claim that ANY of the > roots are divisible by sqrt(5) is false. > The poster has shifted from the true claim--that the result applies to > algebraic integers--to a far vaguer and more inclusive claim, > basically that it applies in general. No, absolutely not. When I said divisible above, I meant divisible in the ring of algebraic integers. Just as you did in APF when you said one of the a¹s is coprime to 5. Never in that paper did you suggest you might be doing arithmetic in any other ring. > Repeatedly, posters keep making claims true in the ring of algebraic > integers, I accept those claims in that ring, and they then act like I > don¹t!!! READ YOUR OWN PAPER. Again, I quote: This paper will show, using basic algebraic methods, that given the factorization, in the ring of algebraic integers, 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5. Now, when you say ... factorization, in the ring of algebraic integers, any sensible person would assume that you mean a1, a2, and a3 are algebraic integers. When you say one of the a¹s is coprime to 5, any sensible person would assume you meant coprime to 5 in the algebraic integers. That is CLEARLY what your paper is about. Any reader of the journal in which it was published (for a few hours) would assume that. The paper makes no sense otherwise. > I repeat, the result Nora Baron gave is correct *in the ring of > algebraic integers* and I now challenge that poster and others in that > camp to Žnally and for once concede that I am not arguing against > that point. Certainly you appear to be accepting what I proved. But then you must logically also accept that it contradicts what you claim to have proved in APF. > Then I¹d like, with the permission of Nora Baron to explain exactly > what my proof actually says. Go for it. > But Žrst I want the poster Nora Baron to accept that I am not > challenging on this speciŽc point. > That¹s the Žrst step. > It¹s up to that poster to respond. Consider it done. Your turn. Nora B. Postscript: I suspect you now want to invoke the ring of objects. In the past you have said that the ring of algebraic integers is somehow incomplete and that that is what leads to the problem with APF. You have said that your ring of objects is bigger than the ring of algebraic integers. That is, the ring of objects *contains* the ring of algebraic integers. Given this, note that: (1) Any factorization of a polynomial in the ring of algebraic integers is also a factorzation in the ring of objects, and (2) There does exist a factorization of 65*x^3 - 12*x + 1 of the form you claimed, (a1*x + 1)*(a2*x + 1)*(a3*x + 1) where each of a1, a2, and a3 is an algebraic integer. This follows from the fact that if 65*x^3 - 12*x + 1 is factored as above, then -a1, -a2, and -a3 are roots of y^3 - 12*y^2 - 65 which by deŽnition means they are algebraic integers. That the latter polynomial does have a factorization is guaranteed by the Fundamental Theorem of Algebra. Now put (1) and (2) together, and you conclude that there is a factorization of the polynomial 65*x^3 - 12*x + 1 of the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1) where a1, a2, and a3 are OBJECTS. None of a1, a2, and a3 are coprime to 5 in the ring of algebraic integers. Are any of them coprime to 5 in the ring of OBJECTS? Take a1, for example. It is not coprime to 5 in the ring of algebraic integers. Therefore there exist nonunit algebraic integers s, t, and w such that a1 = s * w 5 = t * w. Of course w is the common factor of a1 and 5 in the algebraic integers. But since s, t, and w are algebraic integers, they are also OBJECTS. Therefore w is also a common factor of a1 and 5 in the ring of OBJECTS. That is, a1 and 5 are also not coprime in the ring of OBJECTS. The same holds true for a2 and a3. This contradicts your statement in APF, *even if you were talking about coprimeness in the ring of Objects.* Again, I must point out: your statements in APF are crystal clear. You were NOT talking about any ring of objects. The term objects was not even mentioned. You were doing all your algebra in the ring of algebraic integers. The paper was incoherent unless that is the ring in which your conclusions applied. > James Harris === Subject: Re: My paper, and the cheaters > I¹ve made some pointed criticisms of math society but if you actually > pay attention you¹ll realize that the evidence against math society is > nastier than I go into detail about usually. > > Supposedly a correct math result is what¹s important, but despite my > having a quite correct paper, which was to be published, sci.math¹ers > managed to get it yanked IN ONE DAY with some hostile emails. > > The paper has been by a lot of editors and mathematicians and to date, > no major error has been found, which I say because there was one minor > error that actually got pointed out by a sci.math poster. > > > Please do not ever again claim to be an honest person. > Your proof contradicts known mathematics. At least 4 > counterproofs have been given. Dale Hall¹s counterproof > can be checked with simple arithmetic. The exact spot > in your proof where you make your major error has been pointed > out repeatedly. You know all this and yet you continue to > deny it. > > That¹s not true. > > In fact, I don¹t disagree with much of what you say. > > And I¹d like you to admit that in a post, so that I can move on to > what I actually *DO* say. > > First though, I need you to read what I have below and acnowledge that > you see the agreement on a key point. > > > > Below is one of the simplest counterproofs. Feel free > to point out errors (assuming you are still interested in > math as opposed to social studies). This is, incidentally, > for your future reference, what real proof is supposed to > look like. > > > James Harris has written a paper called Advanced Polynomial > Factorization in which he claims the following: if the > polynomial > > 65*x^3 - 12*x + 1 > > is factored in the form > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > > where a1, a2 and a3 are algebraic integers, then exactly > two of a1, a2, and a3 are divisible in the algebraic integers > by sqrt(5), and the third one is coprime to 5. > > It should be noted that -a1, -a2, and -a3 are roots of > > x^3 + 12*x^2 - 65. > > That Harris¹s claim is false can be seen from the following: > > > Assume m(x) is an irreducible monic polynomial with integer coefŽcients > and algebraic integers a and b are two roots of m(x). > > Theorem. If p is a nonzero rational integer and a is divisible by > sqrt(p) in the algebraic integers, then b is also divisible > by sqrt(p). > > That last should say, in the algebraic integers. > > Agreed. > Theorem accepted. Proof not needed here so I deleted it out to focus > on what¹s important. It is in fact true that if Œa¹ is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. > > Concede agreement Nora Baron and then I¹ll explain the rest. > Of course I concede. That¹s what I proved. > The question now becomes, what did you claim to be true in > Advanced Polynomial Factorization. > Just below is the full text of APF exactly as it was submitted That¹s unnecessary. No sense piling on a lot of information for readers, eh? Now then at issue is the factorization 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) and we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor of sqrt(5) in the ring of algebraic integers. Do you understand that? If so then more progress can be made. Now you should be able to understand that and no longer make posts accusing me of being dishonest and then making statements that I don¹t actually disagree with, as if I do. I¹ll repeat it so that there¹s no room for confusion: Given the factorization 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor of sqrt(5) in the ring of algebraic integers. Now is that settled? James Harris === Subject: Re: My paper, and the cheaters > I¹ll repeat it so that there¹s no room for confusion: > Given the factorization > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor of > sqrt(5) in the ring of algebraic integers. > Now is that settled? > James Harris That means that your paper is incorrect. === Subject: Re: My paper, and the cheaters > > I¹ve made some pointed criticisms of math society but if you actually > > pay attention you¹ll realize that the evidence against math society is > > nastier than I go into detail about usually. > > > > Supposedly a correct math result is what¹s important, but despite my > > having a quite correct paper, which was to be published, sci.math¹ers > > managed to get it yanked IN ONE DAY with some hostile emails. > > > > The paper has been by a lot of editors and mathematicians and to date, > > no major error has been found, which I say because there was one minor > > error that actually got pointed out by a sci.math poster. > > > > Please do not ever again claim to be an honest person. > Your proof contradicts known mathematics. At least 4 > counterproofs have been given. Dale Hall¹s counterproof > can be checked with simple arithmetic. The exact spot > in your proof where you make your major error has been pointed > out repeatedly. You know all this and yet you continue to > deny it. > > That¹s not true. > > In fact, I don¹t disagree with much of what you say. > > And I¹d like you to admit that in a post, so that I can move on to > what I actually *DO* say. > > First though, I need you to read what I have below and acnowledge that > you see the agreement on a key point. > > > > Below is one of the simplest counterproofs. Feel free > to point out errors (assuming you are still interested in > math as opposed to social studies). This is, incidentally, > for your future reference, what real proof is supposed to > look like. > > > James Harris has written a paper called Advanced Polynomial > Factorization in which he claims the following: if the > polynomial > > 65*x^3 - 12*x + 1 > > is factored in the form > > (a1*x + 1)*(a2*x + 1)*(a3*x + 1), > > where a1, a2 and a3 are algebraic integers, then exactly > two of a1, a2, and a3 are divisible in the algebraic integers > by sqrt(5), and the third one is coprime to 5. > > It should be noted that -a1, -a2, and -a3 are roots of > > x^3 + 12*x^2 - 65. > > That Harris¹s claim is false can be seen from the following: > > > Assume m(x) is an irreducible monic polynomial with integer coefŽcients > and algebraic integers a and b are two roots of m(x). > > Theorem. If p is a nonzero rational integer and a is divisible by > sqrt(p) in the algebraic integers, then b is also divisible > by sqrt(p). > > That last should say, in the algebraic integers. > > > Agreed. > > > Theorem accepted. Proof not needed here so I deleted it out to focus > on what¹s important. It is in fact true that if Œa¹ is divisible by > sqrt(p) in the algebraic integers then b is also divisible by sqrt(p) > ***in the ring of algebraic integers***. > > Concede agreement Nora Baron and then I¹ll explain the rest. > > > Of course I concede. That¹s what I proved. > > The question now becomes, what did you claim to be true in > Advanced Polynomial Factorization. > > Just below is the full text of APF exactly as it was submitted > > That¹s unnecessary. No sense piling on a lot of information for > readers, eh? Right. Just a little bit of information will sufŽce. Here are a couple of quotes from ŒAPF¹ : ------------------------------------------------------------- ------------- [Quote from beginning of Œproof¹] This paper will show, using basic algebraic methods, that given the factorization, in the ring of algebraic integers, 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5. [End of Quote #1] ------------------------------------------------------------- ------------- [Quote from the end of the paper] Therefore, with the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one of the a¹s is coprime to 5, which shows where some of the algebraic integer factors distribute despite the factors being irrational. [End of Quote #2] ------------------------------------------------------------- ------------- I¹m quoting these to show what you were claiming in APF. You were talking about factorizations *in the algebraic integers* and coprimeness *in the algebraic integers*. This could not be clearer or less ambiguous. If that isn¹t what you meant, you certainly chose a bizarre way, even a *deceptive* way, to state your results. > Now then at issue is the factorization > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > and we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor > of sqrt(5) in the ring of algebraic integers. Yes. > Do you understand that? I understand that you are now accepting this as a proven fact, yes. > If so then more progress can be made. > Now you should be able to understand that and no longer make posts > accusing me of being dishonest and then making statements that I don¹t > actually disagree with, as if I do. How long ago did you accept the proofs that the claimed main results of APF are false? Was it before or after your most recent submission of the results of the paper to a journal? Have you withdrawn that submission? > I¹ll repeat it so that there¹s no room for confusion: > Given the factorization > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > we ARE IN AGREEMENT that neither a_1, a_2, nor a_3 have a factor of > sqrt(5) in the ring of algebraic integers. > Now is that settled? I agree that you now appear to have agreed with this. Let me just summarize where that leaves us. First, note that -a_1, -a_2, and -a_3 are roots of the polynomial y^3 + 12*y^2 - 65 All roots of such a polynomial are algebraic integers. That follows from the deŽnition of algebraic integers. I believe you also now accept the fact that none of a_1, a_2, and a_3 are coprime to 5 in the algebraic integers. Thus there is a factorization of 65*x^3 - 12*x + 1 of the form (a_1*x + 1)*(a_2*x + 1)*(a_3*x + 1) such that a_1, a_2, and a_3 are algebraic integers, and none of them are coprime to 5 in the algebraic integers. Agreed? Clearly, if you review the two quotes from APF above, this fact blatantly contradicts your main claims in that paper. Therefore the main conclusion of APF is incorrect. Agreed? If so, our disagreements are over. The fact above does not imply that there is anything wrong or deŽcient with the algebraic integers. The only thing Œwrong¹ about it from your point of view is that it doesn¹t agree with what you want to be true. Now, you may be saying, I didn¹t mean exactly what I said in APF. APF was a Žrst step toward showing that there is a basic problem with core mathematics. Or some such. If so, then you were submitting APF as a stand-alone paper, and one in which you *knew* the main result to be false as it was worded. Why would you do such a thing? Would that be an honest thing¹ to do? You have already admitted to a minor error in APF. Are you now saying you knew all along that the whole thing was false or that it contradicted known mathematics, and you submitted to a new journal anyway ? Perhaps you are now saying, The conclusion about coprimeness in APF did not mean coprimeness in the algebraic integers. I was talking about coprimeness in some other ring. No, that is not a tenable statement! If you had meant some other ring, you would have speciŽed it. You would not have submitted and re-submitted a paper several times with that kind of key condition left out. Don¹t try to rewrite history. It is really clear that your original intention in APF was to prove a theorem that was about factorizations *in the algebraic integers*; the main conclusion was clearly intended to be a statement about coprimeness *in the algebraic integers*. That is the only way to interpret the paper that makes any sense. Don¹t try now to weasel out of what you originally and clearly thought was true by Žnally admitting that all our counterproofs are valid, and trying to invent some imaginary explanation based on a ring that you cannot even deŽne and that was never even hinted at in the original paper. You have a *lot* of explaining to do. Try to keep it honest. Nora B. > James Harris === Subject: Re: My paper, and the cheaters days. My association with the Department is that of an alumnus. [.snip.] >Postscript: > I suspect you now want to invoke the ring of objects. >In the past you have said that the ring of algebraic >integers is somehow incomplete and that that is what >leads to the problem with APF. You have said that your >ring of objects is bigger than the ring of algebraic >integers. That is, the ring of objects *contains* the >ring of algebraic integers. > Given this, note that: > (1) Any factorization of a polynomial in the ring > of algebraic integers is also a factorzation in > the ring of objects, and > (2) There does exist a factorization of > 65*x^3 - 12*x + 1 > of the form you claimed, > (a1*x + 1)*(a2*x + 1)*(a3*x + 1) > where each of a1, a2, and a3 is an algebraic integer. > This follows from the fact that if 65*x^3 - 12*x + 1 > is factored as above, then -a1, -a2, and -a3 are > roots of > y^3 - 12*y^2 - 65 > which by deŽnition means they are algebraic integers. > That the latter polynomial does have a factorization is > guaranteed by the Fundamental Theorem of Algebra. > Now put (1) and (2) together, and you >conclude that there is a factorization of the polynomial > 65*x^3 - 12*x + 1 >of the form > (a1*x + 1)*(a2*x + 1)*(a3*x + 1) >where a1, a2, and a3 are OBJECTS. None of a1, >a2, and a3 are coprime to 5 in the ring of >algebraic integers. Are any of them coprime to 5 >in the ring of OBJECTS? > Take a1, for example. It is not coprime to 5 >in the ring of algebraic integers. Therefore there >exist nonunit algebraic integers s, t, and w such that > a1 = s * w > 5 = t * w. > Of course w is the common factor of a1 and 5 in the >algebraic integers. > But since s, t, and w are algebraic integers, they are >also OBJECTS. Therefore w is also a common factor of >a1 and 5 in the ring of OBJECTS. That is, a1 and 5 are >also not coprime in the ring of OBJECTS. You forgot one key thing. An element w may be a non-unit in the ring of algebraic integers, yet be a unit in the ring of Objects. In that case, it is perfectly possible for a1 and 5 to be non-coprime in the algebraic integers yet coprime in a larger ring. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My paper, and the cheaters > This polynomial is monic and irreducible with integer coefŽcients. > By the theorem above, if one of the roots is divisible by sqrt(5), > then they all are. This is not possible because the product of the > roots is 65 = 5 * 13. Therefore Harris¹s claim that ANY of the > roots are divisible by sqrt(5) is false. > The poster has shifted from the true claim--that the result applies to > algebraic integers--to a far vaguer and more inclusive claim, > basically that it applies in general. No, she hasn¹t. [rest deleted] === Subject: Re: My paper, and the cheaters >I¹ve made some pointed criticisms of math society but if you actually >pay attention you¹ll realize that the evidence against math society is >nastier than I go into detail about usually. >Supposedly a correct math result is what¹s important, but despite my >having a quite correct paper, which was to be published, sci.math¹ers >managed to get it yanked IN ONE DAY with some hostile emails. >The paper has been by a lot of editors and mathematicians and to date, >no major error has been found, which I say because there was one minor >error that actually got pointed out by a sci.math poster. >> Please do not ever again claim to be an honest person. >>Your proof contradicts known mathematics. At least 4 >>counterproofs have been given. Dale Hall¹s counterproof >>can be checked with simple arithmetic. The exact spot >>in your proof where you make your major error has been pointed >>out repeatedly. You know all this and yet you continue to >>deny it. > That¹s not true. It is certainly true that I have provided a refutation of the main result of your paper, in a way that can be checked (i.e., either veriŽed or refuted, and everyone who has done the work has found my result veriŽed), using simple, if tedious, arithmetic. It is certainly true that you continue to deny the truth of what I have done, and refuse to provide any evidence at all of your counterclaim that I have made some unwarranted assumption. My result can be veriŽed using standard arithmetic. If there were a hidden or unwarranted assumption, then arithmetic would need to have made that assumption, since that¹s all that is required to do the job. ... stuff deleted ... > It¹s up to that poster to respond. contains an incorrect conclusion, and therefore is žawed. > James Harris Dale === Subject: Re: My paper, and the cheaters > >> >I¹ve made some pointed criticisms of math society but if you actually >pay attention you¹ll realize that the evidence against math society is >nastier than I go into detail about usually. > >Supposedly a correct math result is what¹s important, but despite my >having a quite correct paper, which was to be published, sci.math¹ers >managed to get it yanked IN ONE DAY with some hostile emails. > >The paper has been by a lot of editors and mathematicians and to date, >no major error has been found, which I say because there was one minor >error that actually got pointed out by a sci.math poster. > >> >> Please do not ever again claim to be an honest person. >>Your proof contradicts known mathematics. At least 4 >>counterproofs have been given. Dale Hall¹s counterproof >>can be checked with simple arithmetic. The exact spot >>in your proof where you make your major error has been pointed >>out repeatedly. You know all this and yet you continue to >>deny it. > > > That¹s not true. > > It is certainly true that I have provided a refutation of the > main result of your paper, in a way that can be checked (i.e., > either veriŽed or refuted, and everyone who has done the work > has found my result veriŽed), using simple, if tedious, arithmetic. No, you have not. I¹m not interested in games here. Acknowledge that I accept the result from the poster Nora Baron, and then I¹ll go into further detail. For the rest of you, I need posters like W. Dale Hall, Arturo Magidin, and Nora Baron to acknowledge what I accept as they repeatedly come back and claim that I¹m denying it, and then use that lie to push their case. First step is to get them in posts to acknowledge what I accept so that they can¹t come back later and claim I don¹t. I¹m shutting down their options. > It is certainly true that you continue to deny the truth of what > I have done, and refuse to provide any evidence at all of your > counterclaim that I have made some unwarranted assumption. You are lying. I¹ve repeatedly explained *exactly* what you did. You just keep coming back as if I haven¹t. Now then, acknowledge that ALL of your results are *in the ring of algebraic integers* and I¹ll point out that I am not arguing with that point. You know that then I¹ll come back and explain rather easily why in the ring of algebraic integers is not enough, which is why it¹s important for me to get people like you to acknowledge basic facts upfront. > My result can be veriŽed using standard arithmetic. If there > were a hidden or unwarranted assumption, then arithmetic would > need to have made that assumption, since that¹s all that is > required to do the job. Yes, what you give is correct IN THE RING OF ALGEBRAIC INTEGERS, and I do not argue with that point! Acknowledge the facts here and then I can move on to the next mathematical issue. > ... stuff deleted ... > It¹s up to that poster to respond. > > contains an incorrect conclusion, and therefore is žawed. I¹m not evading anything. If my work is in fact žawed, as you claim, then how can acknowledging that I accept certain points change anything? Where is Nora Baron now? Why did you answer for that poster, repeat your claims, and act like I¹m dodging? I¹m not dodging or evading. I don¹t have to evade. Now then, acknowledge that I accept your result in the ring of algebraic integers. James Harris === Subject: Re: My paper, and the cheaters >> >> >I¹ve made some pointed criticisms of math society but if you actually >pay attention you¹ll realize that the evidence against math society is >nastier than I go into detail about usually. > >Supposedly a correct math result is what¹s important, but despite my >having a quite correct paper, which was to be published, sci.math¹ers >managed to get it yanked IN ONE DAY with some hostile emails. > >The paper has been by a lot of editors and mathematicians and to date, >no major error has been found, which I say because there was one minor >error that actually got pointed out by a sci.math poster. > >> >> Please do not ever again claim to be an honest person. >>Your proof contradicts known mathematics. At least 4 >>counterproofs have been given. Dale Hall¹s counterproof >>can be checked with simple arithmetic. The exact spot >>in your proof where you make your major error has been pointed >>out repeatedly. You know all this and yet you continue to >>deny it. >That¹s not true. >>It is certainly true that I have provided a refutation of the >>main result of your paper, in a way that can be checked (i.e., >>either veriŽed or refuted, and everyone who has done the work >>has found my result veriŽed), using simple, if tedious, arithmetic. > No, you have not. > I¹m not interested in games here. Acknowledge that I accept the > result from the poster Nora Baron, and then I¹ll go into further > detail. > For the rest of you, I need posters like W. Dale Hall, Arturo Magidin, > and Nora Baron to acknowledge what I accept as they repeatedly come > back and claim that I¹m denying it, and then use that lie to push > their case. > First step is to get them in posts to acknowledge what I accept so > that they can¹t come back later and claim I don¹t. > I¹m shutting down their options. >>It is certainly true that you continue to deny the truth of what >>I have done, and refuse to provide any evidence at all of your >>counterclaim that I have made some unwarranted assumption. > You are lying. > I¹ve repeatedly explained *exactly* what you did. > You just keep coming back as if I haven¹t. > Now then, acknowledge that ALL of your results are *in the ring of > algebraic integers* and I¹ll point out that I am not arguing with that > point. > You know that then I¹ll come back and explain rather easily why in > the ring of algebraic integers is not enough, which is why it¹s > important for me to get people like you to acknowledge basic facts > upfront. >>My result can be veriŽed using standard arithmetic. If there >>were a hidden or unwarranted assumption, then arithmetic would >>need to have made that assumption, since that¹s all that is >>required to do the job. > Yes, what you give is correct IN THE RING OF ALGEBRAIC INTEGERS, and I > do not argue with that point! Let¹s take the statements that I claim deny this: First, from the introduction: This paper will show, using basic algebraic methods, that given the factorization, in the ring of algebraic integers, 65x^3 .89Ćā 12x+ 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x+ 1) one of the a.89„śs is coprime to 5. and at the end of the paper: Therefore, with the factorization 65x^3 .89Ćā 12x+ 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x+ 1) one of the a.89„śs is coprime to 5, which shows where some of the algebraic integer factors distribute despite the factors being irrational. Did you see the phrase in the ring of algebraic integers in the Žrst passage? Are you now repudiating that phrase? I mean, my argument showed that there are algebraic integers that divide both 5 and each of the a¹s (a separate one for each a_i). Did I suggest that I was discussing anything *other* than the ring of algebraic integers? If you think so, then point out where. > Acknowledge the facts here and then I can move on to the next > mathematical issue. >> ... stuff deleted ... >It¹s up to that poster to respond. >>contains an incorrect conclusion, and therefore is žawed. > I¹m not evading anything. > If my work is in fact žawed, as you claim, then how can acknowledging > that I accept certain points change anything? > Where is Nora Baron now? Why did you answer for that poster, repeat > your claims, and act like I¹m dodging? > I¹m not dodging or evading. I don¹t have to evade. > Now then, acknowledge that I accept your result in the ring of > algebraic integers. Therefore, you will retract your paper, or revise it by removing the statement that any of the a¹s (in the factorization in the above excerpt of your paper from the URL: http://www.ne-plus-ultra.net/index.php?option=content&task= view&id=46&Itemid =26 that you recently promoted as being mathematically correct. If you have changed your mind, and accept that it is in error, that¹s Žne. If you can accept the fact that your argument leads to the žawed conclusion, as I stated, then I can and will certainly acknowledge your admission of error. > James Harris Dale === Subject: Re: My paper, and the cheaters > Below is one of the simplest counterproofs [...] The exact same proof was posted to sci.math August 4th. There I presented a clearer way to structure it, see For convenience I repeat it below. The proof is clearer if one Žrst abstracts out a slightly more general Lemma as below. This clariŽes the essence of the matter, namely that conjugates a, b are algebraically indistinguisable, so g(a) is an Algebraic Integer iff g(b) is too, g(X) in Q[X]. Then the rest of the proof is immediate as follows: a/sqrt(p) in AI with AI := Algebraic Integers <=> a^2/p in AI via AI closed under square,sqrt <=> b^2/p in AI via Lemma below with g(X) = X^2/p <=> b/sqrt(p) in AI via AI closed under square,sqrt LEMMA Let a, b be conjugate algebraic numbers, i.e. any two roots of an irreducible polynomial m(X) in Q[X]. Then g(a) in AI <=> g(b) in AI, for all g(X) in Q[X] PROOF g(a) in AI => f(g(a)) = 0 for monic f(X) in Z[X] => m(X)|f(g(X)) in Q[X] => f(g(b)) = 0 => g(b) in AI QED --Bill Dubuque === Subject: Re: My paper, and the cheaters > > Below is one of the simplest counterproofs [...] > The exact same proof was posted to sci.math August 4th. > There I presented a clearer way to structure it, see > For convenience I repeat it below. > The proof is clearer if one Žrst abstracts out a slightly more > general Lemma as below. This clariŽes the essence of the matter, > namely that conjugates a, b are algebraically indistinguisable, > so g(a) is an Algebraic Integer iff g(b) is too, g(X) in Q[X]. > Then the rest of the proof is immediate as follows: > a/sqrt(p) in AI with AI := Algebraic Integers > <=> a^2/p in AI via AI closed under square,sqrt > <=> b^2/p in AI via Lemma below with g(X) = X^2/p > <=> b/sqrt(p) in AI via AI closed under square,sqrt > LEMMA Let a, b be conjugate algebraic numbers, i.e. > any two roots of an irreducible polynomial m(X) in Q[X]. > Then g(a) in AI <=> g(b) in AI, for all g(X) in Q[X] > PROOF g(a) in AI => f(g(a)) = 0 for monic f(X) in Z[X] > => m(X)|f(g(X)) in Q[X] > => f(g(b)) = 0 > => g(b) in AI QED posted back in August. Perhaps it can be extended as Arturo suggested in another reply to your post. Of course the extension to other rational powers of p is trivial. Nora B. > --Bill Dubuque === Subject: Re: My paper, and the cheaters days. My association with the Department is that of an alumnus. >> Below is one of the simplest counterproofs [...] >The exact same proof was posted to sci.math August 4th. >There I presented a clearer way to structure it, see >For convenience I repeat it below. As long as we have your attention... Nora and I discussed briežy over e-mail the possibility of extending the argument to show (*) If a and b are conjugate algebraic integers, and r is an algebraic integer that divides a (in the ring of all algebraic integers), then there is a conjugate of r that divides b. This is of course very easy with some simple Galois Theory; but one of the good points of the arguments you and Nora presented is that there is no need to invoke any Galois Theory at all. Neither of us could quite get it to work, though to be honest we did not think too hard or long on it. Is there a similar argument using divisibility of polynomials (or some such) that you can think of that will establish (*) without having to invoke Galois Theory? -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My paper, and the cheaters > > Below is one of the simplest counterproofs [...] >> The exact same proof was posted to sci.math August 4th. >> There I presented a clearer way to structure it, see >> For convenience I repeat it below. > As long as we have your attention... > Nora and I discussed briežy over e-mail the possibility of extending > the argument to show > (*) If a and b are conjugate algebraic integers, and r is an > algebraic integer that divides a (in the ring of all algebraic > integers), then there is a conjugate of r that divides b. > This is of course very easy with some simple Galois Theory; but one of > the good points of the arguments you and Nora presented is that there > is no need to invoke any Galois Theory at all. > Neither of us could quite get it to work, though to be honest we did > not think too hard or long on it. Is there a similar argument using > divisibility of polynomials (or some such) that you can think of that > will establish (*) without having to invoke Galois Theory? Off the top of my head I suspect that one may constructivize the Galois-theoretic argument in the usual manner, i.e. by elimination, say by the use of resultants. However it is probably unlikely that the general case would yield anything as explicit as this simple case. --Bill Dubuque === Subject: Re: My paper, and the cheaters days. My association with the Department is that of an alumnus. >The proof is clearer if one Žrst abstracts out a slightly more >general Lemma as below. This clariŽes the essence of the matter, >namely that conjugates a, b are algebraically indistinguisable, I suspect that it is that statement (that a and b are algebraically indistinguishable) that has led to many confusions on the parts of the original poster. Consider for example his statement, when told that Newton polygons could be used to study the roots, that I had been lying when I said the same thing: It is also, I think, the source of the cryptic claim on the alleged over interpretation of Galois theory in the manuscript. Not that it is isn¹t a bang-on, dead-on-target, statement, mind. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My paper, and the cheaters > So the lie is that if you have a correct result and send it to a > journal, then that¹s what¹s important, but the reality is that the > sci.math¹ers managed to censor my paper with a few bogus > emails...lying to an editor who didn¹t follow fair procedures, and > yanked my paper. > That¹s the REAL math world. Hmmm... haven¹t those very people said that the editor made an error in yanking your paper? How does that go with censoring? The editor should not have pulled the paper after having accepted it. He should have published the comments. > My paper is quite correct, and in fact mostly uses basic algebra, but > math people lie about it, Had your work been published along with the mathematical critiques, people could judge both issues for themselves. Just as they do here, where both appear. - Randy === Subject: Re: My paper, and the cheaters > The editor should not have pulled the paper after having > accepted it. He should have published the comments. I suggested as much in a letter to the editor of the journal at the time when this controversy erupted, with the addendum that if the critics had a point, the onus would be on James to answer their objections in a rejoinder to that same journal. === Subject: Re: My paper, and the cheaters >I¹ve made some pointed criticisms of math society but if you actually >pay attention you¹ll realize that the evidence against math society is >nastier than I go into detail about usually. >Supposedly a correct math result is what¹s important, but despite my >having a quite correct paper, which was to be published, sci.math¹ers >managed to get it yanked IN ONE DAY with some hostile emails. You should really retaliate. Send some hostile emails to the journals where we publish _our_ papers and get them yannked. Guffaw. >The paper has been by a lot of editors and mathematicians and to date, >no major error has been found, Hard to decide whether you¹re lying, deluded, or you actually don¹t understand the simple counterexamples that have been posted. (No, the counterexamples don¹t depend on that false fact you claim elsewhere they depend on.) >which I say because there was one minor >error that actually got pointed out by a sci.math poster. >[...] >But you will be spat upon by real mathematicians in the future, who >will cite you with contempt. Hard to understand why you feel so unloved around here. >James Harris ************************ David C. Ullrich === Subject: Re: My paper, and the cheaters days. My association with the Department is that of an alumnus. >You should really retaliate. Send some hostile emails to the >journals where we publish _our_ papers and get them yannked. Hey! Your arm¹s better! Glad to see it. -- It¹s not denial. I¹m just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My paper, and the cheaters >>You should really retaliate. Send some hostile emails to the >>journals where we publish _our_ papers and get them yannked. >Hey! Your arm¹s better! Glad to see it. Yeah. Not entirely better, but it works Žne for lightweight things, so we suspect the remaining problem is just that the muscles are weak from weeks of disuse. Funny thing is I¹m typing a lot better than I used to. I was afraid the opposite would happen cuz the left hand would forget where keys are, but it seems the weeks of hunt&peck have given both hands a _better_ idea of where keys are. I don¹t quite understand this. ************************ David C. Ullrich === Subject: Re: My paper, and the cheaters >You should really retaliate. Send some hostile emails to the >journals where we publish _our_ papers and get them yannked. >>Hey! Your arm¹s better! Glad to see it. > Yeah. Not entirely better, but it works Žne for lightweight > things, so we suspect the remaining problem is just that the > muscles are weak from weeks of disuse. > Funny thing is I¹m typing a lot better than I used to. > I was afraid the opposite would happen cuz the left hand > would forget where keys are, but it seems the weeks > of hunt&peck have given both hands a _better_ idea of > where keys are. I don¹t quite understand this. Sounds odd, but it¹s nice to see you back in form. Perhaps you got out of some bad habits? -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: My paper, and the cheaters days. My association with the Department is that of an alumnus.