mm-2321 === Subject: Re: alternating group >What is the third one (M)? > >Let P = N intersect A_n >P has index of 4, hence S_n is the union of P, aP, bP, cP. >Suppose P/aP = N and P/bP = A_n. >What is the third normal subgroup? P/cP? (it does not seem to be a >subgroup) Yes, that's right, the third normal subgroup is P/cP. Why don't you try and prove that for yourself - it is not difficult! Derek Holt. >> Suppose N is your normal subgroup and is not A_n. Then N intersect A_n >> has index 4 in S_n, and S_n is the union of three normal subgroups of >> index 2, N, A_n, and another one, say M. Either M or N must contain >> a transposition, giving a contradiction. >> >> Derek Holt. > === Subject: Rsa Labs offline ? Hi all, does anyone know if something's wrong with RSA labs site (www.rsasecurity.com) ? I don't seem to be able to reach them ... Dave === Subject: Re: Rsa Labs offline ? > Hi all, > does anyone know if something's wrong with RSA labs site > (www.rsasecurity.com) ? I don't seem to be able to reach them ... Works fine. Sometimes a server may be down for few minutes to few hours, no need to panic ;-) Computers are not an exact science ! === Subject: Re: division by zero and complex nymbers > As said at http://mathworld.wolfram.com/RiemannSurface.html > > Riemann surfaces are one way of representing multiple-valued functions; > another is branch cuts. > > These are two solutions to define rigorously what was nonsensically called > multiple-valued functions. This term is only an abuse, and is still used > by _some_ mathematicians for the sake of custom. But it is still > nonsensical. If one wants to be completely rigorous then one cannot speak of sqrt existing as a function at all then. In this case it's a function on its Riemann surface, when we speak of sqrt defined on the complex plane we then think of sqrt(pi^-1(z)) where pi is the projection from the riemann surface to C, and thus pi^-1(z) has two values. It doesn't help clarity to keep speaking of pi^-1. And you were not being very rigorous either since you wanted sqrt(-1) = i, but then you have not specified which branch you take. Assuming the principal branch for sqrt over R^+ makes sense given the geometry, but not so in C. It does make a lot of sense to use this abuse of notation in this case. If you want to specify it rigorously then I suppose you could say: sqrt is not a function on C{0} as it has two values at every point, but on any given simply connected subdomain D of C{0} one can produce two holomorphic functions which correspond to sqrt. So people rather abuse the notation since everyone knows what it means when we say that a holomorphic function is multiple valued. It really, REALLY makes no sense to me to just say sqrt(-1) = i and pretend that we are being rigorous. Jiri === Subject: Re: division by zero and complex nymbers >> As said at http://mathworld.wolfram.com/RiemannSurface.html >> >> Riemann surfaces are one way of representing multiple-valued functions; >> another is branch cuts. >> >> These are two solutions to define rigorously what was nonsensically called >> multiple-valued functions. This term is only an abuse, and is still used >> by _some_ mathematicians for the sake of custom. But it is still >> nonsensical. > > If one wants to be completely rigorous then one cannot speak of sqrt > existing as a function at all then. It can be define as a function C -> C, but it's not continuous then. That's not a problem, discintunous functions exist, in usual mathematics ;-) > In this case it's a function on > its Riemann surface, when we speak of sqrt defined on the complex plane > we then think of sqrt(pi^-1(z)) where pi is the projection from the > riemann surface to C, and thus pi^-1(z) has two values. It doesn't > help clarity to keep speaking of pi^-1. > And you were not being very > rigorous either since you wanted sqrt(-1) = i, but then you have not > specified which branch you take. Assuming the principal branch for > sqrt over R^+ makes sense given the geometry, but not so in C. I said: You should never write sqrt(-1) either, unless you have previously defined it as a function C -> C (and then, either sqrt(-1)=i, either sqrt(-1)=-i). I never said it's sufficient to let sqrt(-1)=i to define a branch. > It does make a lot of sense to use this abuse of notation in this > case. A function has one value. Nothing meaningful by saying otherwise. > If you want to specify it rigorously then I suppose you could > say: sqrt is not a function on C{0} as it has two values at every > point, Yes > but on any given simply connected subdomain D of C{0} one can > produce two holomorphic functions which correspond to sqrt. Yes > So people > rather abuse the notation since everyone knows what it means when we > say that a holomorphic function is multiple valued. That's precisely to make sure people know what they are talking about, that this abuse should be avoided whenever possible. > It really, REALLY makes no sense to me to just say sqrt(-1) = i and > pretend that we are being rigorous. I said [...] if you have previously defined it as a function, but you can take a sentence out of context to support your argumentation. That's just as nonsensical as the concept of multi-valued function... === Subject: Re: Is this expression viewable on sci.math? >Just two examples that, in case they work, will make it >much more easy to explain what my mind is confused by. >Is this readable for a standard user here? I know i can >see it but that might be due to all that stuff i have >installed on my computer. >> xy:(x=y) >and >> /32 >> f(x,y) dydx >> 0 - >(Disclaimer: Since i'm going to ask some heavier questions >that more or less require either 1. ability to show formulas >as above, or 2. few pages of explaination that i have no >desire to write nor make anybody else read - then the >above IS INDEED on-topic even if it doesn't look like it.) I can read something, more than what got in as I quoted it. But I suggest you use what is usually called math TeX, which is TeX notation with, for example, $ left out. Just about anything more than ASCII will give different results on different newsreaders. -- 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: Numerical quadrature for uneven spacing? I'm trying to calculate the integral of a function given values at some points. The points are not evenly spaced, so a direct use of Simpson's Rule is not possible. I could use Trapezoid Rule, but I'd prefer the lower error that Simpson's Rule gives. I tried deriving the Lagrange polynomial for three arbitrary points and then integrating it, and this seems to have worked quite well - but I'd like to confirm that I'm correct. If the points are x1,f1, x2,f2, and x3,f3, and I take h1=x2-x1 and h2=x3-x2, then I get I=(1/6)*(h1+h2)*(3(f1+f3) + ((h1+h2)^2/(h1*h2))*(f2-f1-(f3-f1)*h1/(h1+h2))) (this is hard to read - here's a jpg instead: http://members.shaw.ca/russford/integrate.jpg) Does this look right? I know that if I put h1=h2=h, I get Simpson's rule. === Subject: Re: Numerical quadrature for uneven spacing? I'm curious: how did this problem arise (is it homewrok? If so, what are you studying?). What languages do you know (could you handle a FORTRAN program, or would it have to be in C or something else)? Michael === Subject: Re: Numerical quadrature for uneven spacing? Hi Michael, This problem is for my MSc work in Structural Engineering. I'm trying to compute a line integral over a curve of the products of some functions. I have no control over the selection of knots, and the spacing of the knots along the curve will often be uneven. I've already implemented it in C and I'm just trying to confirm that it's not fundamentally flawed before I write. === Subject: Re: Numerical quadrature for uneven spacing? >I'm trying to calculate the integral of a function given values at some >points. The points are not evenly spaced, so a direct use of Simpson's >Rule is not possible. I could use Trapezoid Rule, but I'd prefer the >lower error that Simpson's Rule gives. > >I tried deriving the Lagrange polynomial for three arbitrary points and >then integrating it, and this seems to have worked quite well - but I'd >like to confirm that I'm correct. > >If the points are x1,f1, x2,f2, and x3,f3, and I take h1=x2-x1 and >h2=x3-x2, then I get > >I=(1/6)*(h1+h2)*(3(f1+f3) + >((h1+h2)^2/(h1*h2))*(f2-f1-(f3-f1)*h1/(h1+h2))) > >(this is hard to read - here's a jpg instead: >http://members.shaw.ca/russford/integrate.jpg) > >Does this look right? I know that if I put h1=h2=h, I get Simpson's >rule. I verified your computation. It is correct. However I think the following rearrangement of it is more expressive: I = C [ h2 (2h1 - h2) f1 + (h1 + h2)^2 f2 + h1 (2h2 - h1) f3 ] where C = (h1 + h2)/(6 h1 h2). -- Rouben Rostamian === Subject: Re: Numerical quadrature for uneven spacing? === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> At one time it was probably the best they could do. Now we can do better. Now we can measure differences in the isotopic composition of the purest water - conductivity grade H2O. A *standard* cannot be subject to uncontrolled variabilities that make measurable differences. The continuing refinements in metrology drive the need for improvements in the quality of our *standards*. Tom Davidson Richmond, VA === Subject: Re: Why the metric committee chose the kilogram as the standard mass <42bf5402$0$62503$892e7fe2@authen.white.readfreenews.net> > At one time it was probably the best they could do. > > Now we can do better. Now we can measure differences in the isotopic > composition of the purest water - conductivity grade H2O. A > *standard* cannot be subject to uncontrolled variabilities that make > measurable differences. > > The continuing refinements in metrology drive the need for improvements > in the quality of our *standards*. > Still there will always be manufacturing tolerences; due to human abilities even by the most skilled professionals. Don > Tom Davidson > Richmond, VA === Subject: Re: Why the metric committee chose the kilogram as the standard mass > > But this doesn't imply that mass is derived from weight/gravity. > Similarly, the speed of light doesn't derive from length/time, as Seto claims. === Subject: Re: Why the metric committee chose the kilogram as the standard mass > > > > But this doesn't imply that mass is derived from weight/gravity. > > > It doesn't have to imply: Mass _is_ the mathematical ratio of a body's weight (w), divided by the acceleration (g) at which it will free fall, anyplace, anywhere, and is a constant for any given body. Don > Similarly, the speed of light doesn't derive from length/time, as > Seto claims. === Subject: Re: Help me master mathematics! <4EFve.1032264$w62.31182@bgtnsc05-news.ops.worldnet.att.net> Don't change the subject. We're talking about math INVENTED in the past 10-20 years. Mark can't think of a single thing of use that mathematicians invented over the past 20 years. Math research has become largely irrelevant, just living off the reputation of ancient works of Newton and Liebnitz and Bernoulli. MD, PhD === Subject: Re: Help me master mathematics! > Don't change the subject. We're talking about math INVENTED in the > past 10-20 years. Mark can't think of > a single thing of use that mathematicians invented over the past 20 > years. Wavelets were largely developped during the last 20 years (ok, they were _invented_ much earlier, but their use was not discovered). >Math research has become largely irrelevant, just living off > the reputation of ancient works of Newton and Liebnitz and Bernoulli. > > MD, PhD > === Subject: Re: Help me master mathematics! > Mark can't think of > a single thing of use that mathematicians invented over the past 20 > years. That's not true. Mark can think of at least fifteen such things. === Subject: Re: Help me master mathematics! > Mark can't think of > a single thing of use that mathematicians invented over the past 20 > years. That's not true. Mark can thing of at least fifteen such things. === Subject: Re: Euclidean Geometry in Schools >In message , Jean-Claude >>> In message , Jean-Claude >>> ... >>>> every majorized subset of R has a sup. >>> ... >>> What is a majorized subset? >>A included-in R s.t. there exists M in R s.t. >>for all x in A, x < M. >>Maybe the word majorized is not correct. >>has an upper bound would be better ? >Definitely; just learning the meaning of sup and inf (which are Latin >abbreviations) stresses the brains of the poor students to breaking >point. You mean they do not know the meanings of superior and inferior? Well, I wouldn't be surprised. -- 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: Euclidean Geometry in Schools >>>> Construction is so messy. Let's instead teach axioms. >>>> The reals are a complete Archimedean ordered field. >>> What exactly does complete mean here? >> Every Cauchy series converges. >So you have a metric on the field? >How do you define a metric without the reals? Try using a least upper bound for bounded sets. However, you can define a metric-type uniformity just using the rational numbers; completion in it is equivalent to completion in a real-valued metric. -- 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: Euclidean Geometry in Schools >> The teachers seem unable to unlearn the idea that numbers are >> strings of decimal digits, to be manipulated. They cannot >> understand that the representation by decimal digits is a >> derived representation and a convenience, not at all basic. >And, of course, they should believe expert opinions over what they >were taught in their formative years and how that presentation has >served them adequately throughout their lives. If you assume knowing how to calculate without knowing what it means served them adequately, you are correct about that point. As far as making a case for maintaining what was taught in their formative years, this is exactly the case the creationists are making in attempting to block teaching evolution, or in earlier times, changing from the earth being at the center of the solar system, or changing from alchemy to chemistry. The start of the new math program was when the daughter of a mathematician asked her father about word problems, and found that, while her ability to carry out computations was excellent, she had no idea of what she was doing. The ones who could use mathematics as a universal processing language were the ones who could escape from their primary indoctrination, and it was for precisely this reason that the program was set up to start from the beginning. >The burden of proof lies with jackasses such as yourself and has not >been met outside of peer review committees whose expertise is >self-proclaimed and staunchly protected by university communities that >act out of a political self-interest having little to do with mathematics >education. The educationists would not have even accepted the program provisionally had it not been tested with tens of thousands of school children. >When one actually investigates the historical developments behind your >reference, one is left with the distinct impression of an agenda rather than >a research program. Of course it was an agenda, an agenda to teach mathematical concepts instead of getting the children into a situation from which it would be hard to get the concepts later. The difficulties of the teachers adequately demonstrate that this was, and is, the case. None of the mathematics was new. The use of set algebra was a full century old, and the cardinal number concept 75 years old, at the time. In large part, that agenda was dictated by reaction >to Immanuel Kant's critical philosophy because it proposed that >mathematics was autonomous from logic. It is unfortunate that his >work had been so badly misinterpreted. The synthetic construction >of his philosophy associated mathematics with spatial and temporal intuition >and this became a focus for his detractors. However, Kant's treatment >of causality led him to characterize the relationship between mathematics >and >logic in terms of a rule-based schematism. It is perfectly compatible with >the computational aspects influencing modern mathematics. Mathematics has always been conceptual; the Greeks managed to reduce the subject of geometry to a few axioms (too few, as they made lots of assumptions, but to them axioms were self evident truths) and then derived logically. >What Kant actually said concerning the language of mathematicians >was that geometry constituted the ostensive language of mathematicians >and algebra constituted the symbolic language of mathematicians. Algebra, which is much more recent than geometry, constitutes a powerful symbolic language, which applies to everything. Algebra as a language belongs with beginning reading, and should not be considered as only applying to mathematical objects. It is a very SIMPLE language, and its simplicity is why it is so universal. Every time it got extended, it became simpler. >Naturally, I am presuming you to be familliar with the relationship of >polynomials and the representations of numbers via a place-wise >arrangement of digits. So, maybe... just maybe..., the stupid idiots >who teach mathematics to children and teenagers aren't convinced >that they should give up the number line just because a vocal >plurality of logicists know what a number is. I do not know of anyone who advocated giving up the number line. >As for derived representations what the heck are you talking about? >Cantor's >real numbers are derived representations as are Dedekind cuts. Yet >neither he nor >Dedekind was stupid enough to completely discount the importance of the >correspondence with geometric visualization. The problem they had been >addressing >was the presupposition of a limit's existence. Apparently, rational numbers >had some >intrinsically self-evident existence... Apparently, pairs of integers had >some intrinsically >self-evident existence... Apparently, pairs of natural numbers had some >intrinsically >self-evident existence... Apparently, classes defined by the group (groop) >structure of >transverse designs had some intrinsically self-evident existence... >Apparently, the >class of objects that are self-contradictory had an intrinsically >self-evident existence from >which all other mathematical existence follows. At this stage, the question is about the integers, and then fractions. Limit should come with infinite decimals or earlier, but do not try to do it all at once. The problem starts earlier. Dedekind claimed to be able to say what numbers were in his paper, Was sind und wie sollen die Zahlen Peano came up with an abstract characterization of the integers, I personally did not like the cardinal approach, but preferred that of Peano, which is formally more complicated, but actually easier. It is harder to prove some theorems, but in it, addition becomes equivalent to continuing to count. >Oh yeah!!! I can see how that is just going to help all of those kids >balance their >first bank statement because they now know what a self-appointed group of >recluses >notorious for poor communication skills believe about the concept of >number The idea that one can learn all the shortcuts and theorems from the definitions is nonsense. But one still does not need to memorize the tables and blindly apply them. I have not memorized the tables for the other bases which I sometimes use, but that does not mean that I cannot do the operations, and one acclaimed method of multiplication does not have the multiplication tables memorized. ................... >Oh yeah (Number Two)!!! Let's all tell the teachers that the words are >meaningless but that >we should teach these meaningless symbols without the pictures anyway. Who said that? There were picture in all of the new math texts. Two sets had the same number of elements if one could match the elements; in the early texts, this was done by drawing lines. >Of course, maybe I have just gotten all of this a little bit wrong. You have a lot of it wrong. >Why don't you tell me exactly what a number is? That is the mistake Dedekind fell into, and which Peano partly avoided, and which I will avoid. There is the notion of cardinal number, the notion of ordinal number, the notion of point on a line, the notion of quantity. These are different intuitions, and the fact that the integers behave the same way in all of them is what is needed. I would use the Peano approach, and start by representing the integers as a string of tick marks. In other words, counting. I will not go into the details here, but I believe the cardinal meaning needs to be taught with it. With the ordinal approach, the sum of two integers is given by counting up from the first until the second is added in; with the cardinal approach, the sum is given by taking two disjoint sets with the appropriate numbers of elements and taking their union. The second seems easier, but it is not. -- 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: sexiest math PhD babe, Enigma <05vte.2$45.931@news.uchicago.edu> Mark rented and watched Enigma, the movie. It sucks, although Mark would not mind shagging the slighly chubby Kate Winslet in glasses. === Subject: Re: sexiest math PhD babe <05vte.2$45.931@news.uchicago.edu> <3i6k10Fk6ak8U1@individual.net> <87fyv4fb0o.fsf@phiwumbda.org> Losers. === Subject: Re: Gaussian Kernel Integral >Hi all, >I am facing doubts and difficulty in calculating the integral value of >a gaussian kernel function. I am quite puzzled because it would seem to >be a classical one (inverse of a lognormal variable), though I >investigated diverse technics (holomorphic analysis, numerical Taylor >expansion) without success. >I would be really pleased if any could put me into the right track. I >would be eternally in debts (!?). Any specialised books references >welcome. >The integral form is as following : integration on whole real axis R of >exp(-u*u/2)/(1+exp(u)). I can do this using a property of the exponential function. We have 1/(1+t) + 1/(1+1/t)) = 1. The rest is straightforward. -- 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: Gaussian Kernel Integral <42c01c67$0$648$edfadb0f@dread16.news.tele.dk> it extends to exp(-u*u/2)/(1+exp(b * u)), for b real variable, as well. Any idea to calculate exp(-u*u/2)/(1+ c * exp(b * u)), with c real variable === Subject: Re: Gaussian Kernel Integral > > it extends to exp(-u*u/2)/(1+exp(b * u)), for b real variable, as well. > > Any idea to calculate exp(-u*u/2)/(1+ c * exp(b * u)), with c real > variable > Maple doesn't do the case c=2 (and b=1). Why do you think it can be written in simpler form? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Comments: This message did not originate from the Sender address above. It was remailed automatically by anonymizing remailer software. Please report problems or inappropriate use to the remailer administrator at . === Subject: Occult Pentagrams in Washingtons architecture Mail-To-News-Contact: abuse@dizum.com What does a Pentagram mean geometrically? Is it related to a Platonic solid? Have heard there are occult demonic interpretations. www.theforbiddenknowledge.com/chapter3 === Subject: Book on Introduction to lie groups and lie algebras I am looking for a book for selfstudy on lie goups and lie algebras but most of the books in library are advanced or atleast are intended for graduate level. What I am looking for is a book with many example to make me understand the subject and requests all of you to point me to a suitable TEXT BOOK. I am a physics student THANKS. fareed === Subject: Re: Book on Introduction to lie groups and lie algebras Fulton and Harris (1991) Representation Theory: A First Course (Springer) fulfils all the criteria you've specified. It's introductory, it's packed with examples and very clear explanations and though it claims to be a graduate textbook it should be comprehensible to most undergraduates. Jonny > > I am looking for a book for selfstudy on lie goups and lie algebras but most > of the books in library are advanced > or atleast are intended for graduate level. What I am looking for is a book > with many example to make me understand the subject and requests all of you > to point me to a suitable TEXT BOOK. I am a physics student > THANKS. > > fareed === Subject: Re: Request comment on Probability calculation scenario to positive integers. Your approach is similar to what I had been thinking about. Admittedly, a little is lost if it assumed that all 4 integers are different. Assume C1, C2, r1, and r2 are all positive and all different. The integers, of course, are well ordered, thus one of the 4 must be the smallest. If Abs(C1-C2) < r1+r2 then the 'circles' will intereset. Suppose C1, by happenstance, is the smallest then if C1. I might add, perhaps a true non-sequitor, that if one allows '1' as a prime, in a sense similar to it being declared 'quadrat frei' (square free) then Goldbach's conjecture can be restated as Every integer greater than 1 is equidistant from two primes. with what might be viewed as an hallucination === Subject: Re: Request comment on Probability calculation scenario It's kind of hard to tell what you're saying. If you are only selecting one integer for the coordinate of C1 (as opposed to two), then circles have nothing to do with it; you are just asking whether abs(C1 - C2) <= abs(r1) + abs(r2) Or did you intend the x and y coordinates of C1 and C2 to be arbitrarily selected ? === Subject: Re: Request comment on Probability calculation scenario Sorry, perhaps I should have been more specific. The two integers (positive) are selected on the x-axis in the xy plane so that you end up with two circles whose centers are located on the x-axis with their corresponding radii. I avoided using coordinate pairs to place the circle centers anywhere in the xy plane for ancillary reasons. === Subject: geometrical object What is the correct English name for a geometrical object in (general) N-dimensional Cartesian space, defined as a Cartesian product: [A1,B1] x [A2,B2] x ... x [AN,BN] where [Ai,Bi] are intervals along any i-th axes? I assume that the lengths of the intervals are not necessarily equal, so that the object IS NOT an N-dimensional hypercube. L.B. *-------------------------------------------------------------------* | Dr. Leslaw Bieniasz, | | Institute of Physical Chemistry of the Polish Academy of Sciences,| | Department of Electrochemical Oxidation of Gaseous Fuels, | | ul. Zagrody 13, 30-318 Cracow, Poland. | | tel./fax: +48 (12) 266-03-41 | | E-mail: nbbienia@cyf-kr.edu.pl | *-------------------------------------------------------------------* | Interested in Computational Electrochemistry? | | Visit my web site: http://www.cyf-kr.edu.pl/~nbbienia | *-------------------------------------------------------------------* === Subject: Re: geometrical object Hypercuboid? > > What is the correct English name for a geometrical object > in (general) N-dimensional Cartesian space, defined as a > Cartesian product: > > [A1,B1] x [A2,B2] x ... x [AN,BN] > > where [Ai,Bi] are intervals along any i-th axes? > I assume that the lengths of the intervals are not necessarily > equal, so that the object IS NOT an N-dimensional hypercube. > > L.B. > > *-------------------------------------------------------------------* > | Dr. Leslaw Bieniasz, | > | Institute of Physical Chemistry of the Polish Academy of Sciences,| > | Department of Electrochemical Oxidation of Gaseous Fuels, | > | ul. Zagrody 13, 30-318 Cracow, Poland. | > | tel./fax: +48 (12) 266-03-41 | > | E-mail: nbbienia@cyf-kr.edu.pl | > *-------------------------------------------------------------------* > | Interested in Computational Electrochemistry? | > | Visit my web site: http://www.cyf-kr.edu.pl/~nbbienia | > *-------------------------------------------------------------------* === Subject: Re: geometrical object Ì Leslaw Bieniasz ó.8d.98.87.8b.8c .97.99.95 .92.86.94.9d.92.87 > > > > What is the correct English name for a geometrical object > in (general) N-dimensional Cartesian space, defined as a > Cartesian product: > > [A1,B1] x [A2,B2] x ... x [AN,BN] > > where [Ai,Bi] are intervals along any i-th axes? > I assume that the lengths of the intervals are not necessarily > equal, so that the object IS NOT an N-dimensional hypercube. For N=2 you'll get a rectangle. For N=3, a parallepiped. So the answer seems to be an N-dimensional hyperparallepiped. Alternatively, you may call it a hyperprism, although prism is sometimes reserved for triangular shapes. > L.B. > > *-------------------------------------------------------------------* > | Dr. Leslaw Bieniasz, | -- I. N. Galidakis http://users.forthnet.gr/ath/jgal/ Eventually, _everything_ is understandable === Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? My understanding is that Mr. Prof. Andrew Wiles proved Fermat Last theorem for n>3. Does he had a original proof for n=3? If not, why he got Wolfsskehl Prize? If not ,that means I am the first person,the first time ,to give a more complet proof of FLT for n=prime >2.That is where the dispute is possible. We have to read Mr. Fermat' statement very precise. GHEORGHE GHIATA === Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? days. My association with the Department is that of an alumnus. > >Here's a critique of part of your posted FLT proof: > >> HERE it is Fermat's Proof of FLT >> Theorem: >> he EQUATION: >> X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. >> >> OBSERVATION: >> X,Y ,Z relative Prime numbers > >A word of warning (which you might be aware of, but I'll say it >anyway): This means that there is no factor d that divides evenly into >X, Y, and Z, i.e., >gcd (X,Y,Z) = 1. > >It's still possible to have >gcd (X,Y) > 1, gcd (X,Z) > 1, and gcd (Y,Z) > 1. >For instance, if X = 6, Y = 10, and Z = 15. Not if you are also assuming X^n + Y^n = Z^n. More generally, if a^n + b^m = c^q, with a, b, c, m, n, q positive integers, then any prime that divides two of a, b, and c will necessarily divide all three. === Subject: Re: To Mr. Andrew Wiles:Do You Agree Fermat 's Proof Of FLT ever existed? days. My association with the Department is that of an alumnus. > >Here's a critique of part of your posted FLT proof: > >> HERE it is Fermat's Proof of FLT >> Theorem: >> he EQUATION: >> X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. >> >> OBSERVATION: >> X,Y ,Z relative Prime numbers > >A word of warning (which you might be aware of, but I'll say it >anyway): This means that there is no factor d that divides evenly into >X, Y, and Z, i.e., >gcd (X,Y,Z) = 1. > >It's still possible to have >gcd (X,Y) > 1, gcd (X,Z) > 1, and gcd (Y,Z) > 1. >For instance, if X = 6, Y = 10, and Z = 15. Not if X, Y, and Z satisfy an equation where (a power of) one of them is a sum of (powers of) the other two. If a^m + b^n = c^q, a, b, c, m, n, q positive integers, then any common factor of two of a, b, and c will necessarily be a common factor of the third. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Orlow cardinality question > >> TO's claim, for example, that the standard construction of the set of > >> natural numbers requires the existence of what he calls 'infinite > >> natural numbers has been shown, in a variety of ways, by a variety of > >> people, to be false. > > The standard construction does not prohibit infinite values, and the > > requirement that the set be infinite demands it. > > There is no such 'requirement'. We just note that there is no end to > the set, as opposed to all finite sets we know, and we therefore call > it 'infinite'. It's just a name. You imagine that this 'infinite' is > suddenly some number, but that really is just something you pull out > of your hat (or something). > > > Jan > > You may not be able to identify an end to it, but I can't identify an end to the possible values its elements can have either. Sure, this sounds like a situation of infinity, and I choose to see both as infinite. The artificial part of the standard position is the requirement that all naturals be finite, which is not stated in the definition of the set, and only based on a flawed inductive proof and nothing else, while it is easily proven that the set which is considered infinite cannot possibly only contain finite numbers. This condition of finitude is pointless and illogical, and simply wrong. -- Smiles, Tony === Subject: Re: Orlow cardinality question Virgil said: > > > > > Virgil said: > > > > > > > > > > > > Such an assumption is your idea that there cannot be an infinite > > > > > > > set > > > > > > > of finite elements, for example. Or that the limit of a sequence > > > > > > > must > > > > > > > somehow become a part of that sequence. > > > > > > > > > > > > But I did not ASSUME there cannot be an infinite number of finite > > > > > > naturals. I > > > > > > proved it mathematically > > > > > > > > > > Those proofs require assming the equivalent of of twhat they purport to > > > > > prove. > > > > > > > > > That's simply not true. Respond to them specifically, without snipping > > > > and > > > > otherwise amngling them. > > > > > > OK, I will amngle not. > > I bet you don't even know what amngling is! ;) > > I guess it's a sort of relf-seferential term. > > > > > But anyway, three is too many. If it were we who > > > were the idiots, and you were trying to explain something, you might > > > reasonably try three different explanations, but if you claim you have > > > proofs in the normal mathematical sense, one is enough. If the proof is > > > valid, the existence or otherwise of other proofs is irrelevant. So > > > *you* choose one of your proofs, and I (prolly others too; why _do_ > > > cranks attract such attention? why is it all just so tantalisingly > > > fascinating?) will point out the first error, without mangling or > > > anything else. Then if you wish to you can revise it and try again, or > > > just expand the claim that we show is false. > > > > Would that that were sufficient. But, since they all get rejected for reasons > > that don't pan out, I have produced three. If three proofs from different > > areas > > all agree, that does bolster the argument. So, you want to hack at one at a > > time? Okay. Try the information theory one: > > > > > > > > Given a set of symbols with size S, we can produce a set of all strings using > > those symbols that have length L, and the size of this set will be S^L. > > Digital > > number systems fall into this category, with S being the number base, which > > is > > always finite (2 for binary, 10 for decimal). > > > If we want to have an infinite > > set of digital strings, therefore, S^L needs to be infinite, but S is finite, > > so L, the length of the strings, needs to be infinite to have an infinite set > > of such strings. > > In the above TO makes the assumption that a set of finite strings cannot > be infinite. The set S = {x, x, xxx, ...} is made up only of > finite strings, each string being 1 character longer than its > predecessor. Unless TO can prove that such a set has a longest string in > it, one can construct the Function on S which appends an character to > each member and thus injects S into a proper subset of itself. Thus the > set S of finite strings is Cantor_infinite, even if it were TO_finite. > > Thus TO's assumption that an infinite set of strings must contain an > infinite string is shown to be false, at least by the Cantor standard of > finite versus infinite for sets. > > > > > > > > > So, where in this argument do you see an error? Can S^L ever be infinite and > > produce an infinite number of strings, without infinite S and/or L? Keep in > > mind the relation to trees. S is the number of branchings at each node. L is > > the number of levels in the tree, or branchings in each path. > > TO assumes, possibly because it holds for any finite set of strings, one > can find a longest string, allowing TO to identify an L, that the same > must hold for infinite sets of strings, which is esssentially assuming > his result, that a set of all finite strings must be finite. > > We claim that the number of members in an ordered set is finite if and > only if every non-empty subset of it, including the set itself, has a > first member and last member relative to that ordering. > > > We observe that all of the sets {1}, {1,2}, {1,2,3}, ...,{1,2,3,...,n}, > which have obvious first an last members, are finite by this definition, > and we call each of their last members a finite natural. > > We claim that the union of all such sets is a set N which has no last > member, and is, by the above definition, not a finite set. > > And unless someone can prove that such a union contains a last or > largest member, it satisfies the Cantor definition of not-finite, too. > Well, obviously none of this satisfies me, as I have said, since the Cantor definition takes back seat to actual axioms of arithmetic and value. Is it your position that if a set has a first and last member, that it is finite, or only that a finite set must have a first and last member? -- Smiles, Tony === Subject: Re: Orlow cardinality question <85mzpqta8t.fsf@lola.goethe.zz> Given that logical arguments don't work with tony, I'll keep it short: > So, you agree that S^L is the size of the set of L-length strings constructible > from a smymbol set of size S, yes > and that, for that size to be infinite for a > regular digital number system, L must be infinite? Doesn't that mean you can't > have an infinite set of such digital whole numbers, which are strings, without > having either infinitely long strings, or an infinite set of symbols? Doesn't > this describe the binary or digital numbers, and doesn't that imply that an > infinite set of whole numbers must contain some such infinitely long members? no > If not, why not? see previous post. Jiri === Subject: Re: Orlow cardinality question > > P1. 1 is in N. > > P2. If x is in N, then its successor x' is in N. > > P3. There is no x in N such that x' = 1. > > P4. If y in N isn't 1, then there is a x in N such that x' = y. > > Interesting, not a postulate I see everywhere, check mathworld for > instance. I think this can be proved by the other postulates, but perhaps not as stated. I think it comes as a result of P6 which means that if you had other things not being a successor of another element, then you could take a smaller subset of N and still have it obey all properties. > > P5. If x and y are in N and x' = y', then x = y. > > P6. If S is a subset of N, 1 is in S, and the implication > > (x in S ==> x' in S) holds, then S = N. > > > > Note that N is, by definition, the SMALLEST such set, and since > > excluding any infinite naturals leaves us with a smaller set which > > satisfies all of these postulates, there can be no infinite naturals > > in N. > > I see nothing in the postulates that states that it should be the smallest > such set. I wonder how the Peano axioms as quoted in mathworld could > support only finite numbers as being numbers. I think that P4 above is > crucial. P6 guarantees that N is the smallest such set. If you can find a smaller set with x in S => x' in S, then P6 says that set is N, so that's why it's the smallest. Jiri === Subject: Re: Orlow cardinality question > stephen@nomail.com said: > > >> >> Not according to you. > > >> >> > > >> >> Recall the definition of the naturals? If p is a natural number, then > > >> >> so is p+1. > > >> >> > > >> >> Take the set of all natural numbers, N. You say that the size of the > > >> >> set, which we will also call N, is equal to the biggest number in N. > > >> >> You have some sort of idea with induction that you say proves this. > > >> >> > > >> >> However, because of the definition of the natural numbers, N+1 is also > > >> >> in N. And, according to Bigulosity, N+1 > N, and therefore, N is no > > >> >> longer the largest element in N. Even worse, we can keep doing this > > >> >> infinitely often, and get N+N, or 2*N. > > >> >> > > >> >> So, one of your 'theorems' has to be wrong. It is now time to choose > > >> >> your weakest link :-) > > >> >> > > >> >> > > >> >> Jan > > >> >> > > >> >> > > >> > You seem to misunderstand. N is taken as the unit discrete infinity, declared > > >> > as the size of the set of integers, and used for comparing other sets to that > > >> > standard set. It is also equal to the length of the standard infinite line, as > > >> > the maximum value in the reals is only 1 (0.111111....) greater than the > > >> > maximum value of the integers. There is nothing in my set of rules that says > > >> > one cannot add to this number, or multiply it, or perform any arithmetic > > >> > operation on it. You see, I am not hung up on the largest integer question, > > >> > which is unanswerable except in the most arbitrary ways. > > >> > > >> *Sigh* all right, I'll dig up the reference... > > >> > > >> Your words: > > >> By induction, the set size is ALWAYS the same as the maximal > > >> number. > > >> > > >> So if you call the size of N (the set of naturals) N, then N should > > >> also be the maximal element of N. However, by definition, N+1 is also > > >> in N, and since N+1>N, N is no longer the maximum element of N. > > > Uh huh. That's a problem isn't it? Well if you declare the size to be N then > > > you can always add another element, and get a set of size N+1 can't you? The > > > same problem exists for both maximal element AND size for the finite naturals, > > > which Cantor has resolved by falsely caliming the finite naturals constitute > > > an infinite set. > > >> > > >> Let us call the size of N 's', for a while, to avoid clutter. Again, > > >> in short: > > >> > > >> |N| = s => s is the maximum element of N => s in N => s+1 in N > > >> => s is not the maximum element of N (since s+1 > s). > > >> > > >> This is a contradiction, plain and simple. Surely even you won't deny > > >> this? > > > I never claimed there was a maximal element to the set of finite naturals as a > > > whole, just as I never claimed it has any specific size. > > > > You seem to be claiming it right now!! > > > > > However, if the > > > maximal element or upper bound is finite, then the same is true of the set > > > size, as I have shown in three different ways, and if the set size is infinite, > > > then the maximal element or upper bound is also infinite. > > > > But you have not shown that the maximal element or upper bound is finite!! > > In fact you apparently are denying that there is a maximal element. > > Now you have introduced 'upper bound' as well. Do the finite > > numbers have a finite upper bound? > > > > Look, if you claim that > > 1. if the maximal element or upper bound is finite, then the same is > > true of the set size > > you cannot conclude that the set size is finite until you prove either > > a. the maximal element is finite > > or > > b. the upper bound is finite. > > > > You apparently now agree that a is false, because there is > > no maximal element. I do not know what b means to you. > > Do you think there is a finite upper bound on the finite numbers? > > What is this upper bound? If the upper bound is B, what is B+1? > Let's go back to a. If every element in the set is finite, is it not safe to > say that the maximal element, even if unknown, is also finite? Safe? Dunno. Are you buying or selling? Is it safe to assert that a maximal element simply must exist? If so, why? Must it? How can it? However unknown, however twilit, however evanescent, if a maximal integer n exists, one axiom says that n+1 is also an integer; does this not contradict its being maximal? How do you _define_ 'maximal'? > > So your definition of finite sets include sets with no end or bound. > > There is no upper bound on how large a finite number can be. There > > is no largest, or last, finite number. They are endless. You > > apparently agree to that but still claim the set is not infinite. > Finite whole numbers don't go on forever, but only for a finite number of > iterations. There can be no identifiable end, but they are defined as having an > end. Really? There's an Orlovian axiom that just says The integer stop eventually, is there? Do you claim this bit of definition is also to be found in normal mathematics? Brian Chandler http://imaginatorium.org === Subject: Re: Orlow cardinality question >> stephen@nomail.com said: >> > Look, if you claim that >> > 1. if the maximal element or upper bound is finite, then the same is >> > true of the set size >> > you cannot conclude that the set size is finite until you prove either >> > a. the maximal element is finite >> > or >> > b. the upper bound is finite. >> > >> > You apparently now agree that a is false, because there is >> > no maximal element. I do not know what b means to you. >> > Do you think there is a finite upper bound on the finite numbers? >> > What is this upper bound? If the upper bound is B, what is B+1? >> Let's go back to a. If every element in the set is finite, is it not safe to >> say that the maximal element, even if unknown, is also finite? > Safe? Dunno. Are you buying or selling? > Is it safe to assert that a maximal element simply must exist? If so, > why? > Must it? How can it? However unknown, however twilit, however > evanescent, if a maximal integer n exists, one axiom says that n+1 is > also an integer; does this not contradict its being maximal? How do you > _define_ 'maximal'? It is the largest element except for all those elements larger than it. Can't you follow Tony's simple and intuitive arguments? :) >> > So your definition of finite sets include sets with no end or bound. >> > There is no upper bound on how large a finite number can be. There >> > is no largest, or last, finite number. They are endless. You >> > apparently agree to that but still claim the set is not infinite. >> Finite whole numbers don't go on forever, but only for a finite number of >> iterations. There can be no identifiable end, but they are defined as having an >> end. > Really? There's an Orlovian axiom that just says The integer stop > eventually, is there? Do you claim this bit of definition is also to > be found in normal mathematics? I love the circular reasoning evident in his response. Finite whole numbers don't go on forever, but only for a finite number of iterations. This roughly translates as Finite whole numbers don't go on forever because finite numbers don't go on forever. Finite numbers have to end because they are finite, that is Tony's mantra, and he will repeat it endlessly, which according to his twisted logic, means he will have to stop at some point when he reaches the end. :) Stephen === Subject: Re: Orlow cardinality question > Finite numbers have to end because they are finite, that > is Tony's mantra, and he will repeat it endlessly, which > according to his twisted logic, means he will have to stop > at some point when he reaches the end. :) Unfortunately, you won't know it when he does, because (according to Tony) the end is not identifiable :-) Jan === Subject: Re: Orlow cardinality question >> Finite numbers have to end because they are finite, that >> is Tony's mantra, and he will repeat it endlessly, which >> according to his twisted logic, means he will have to stop >> at some point when he reaches the end. :) > Unfortunately, you won't know it when he does, because (according to > Tony) the end is not identifiable :-) Yes, it will unidentifiably endless end. I can't imagine why anyone would argue with that. It is so intuitive even a 9 year old would understand it. :) Stephen === Subject: Re: Orlow cardinality question <87k6ko3i5t.fsf@phiwumbda.org> Discussion, linux) > stephen@nomail.com said: >> >> Here is a function that is a bijection from the set of pairs of natural >> numbers to the set of natural numbers. >> Given (x,y) map it toz=2^x(2y+1)-1. >> The inverse functions exist but are not so pretty. >> >> Stephen >> > Yes it's true not all mapping functions are simple formulas. What > you have here is really a mapping of a 2D space (the xy plane) to a > third dimension, through a function on two variables. There is no > exact inverse function in this case, but three complementary > functions which define x, y and z in terms of each other. Exact inverse function? What does that mean? -- Jesse F. Hughes I have put all the information that you need at [a Yahoo! group] where you'll notice a significantly better signal to noise ratio, as I'm just about the only person posting. -- James S. Harris on noise A first possible Hystorical reconstruction: Well,as I have shown on this disscusion group the origin of Pell's equation can be traced to the studies by trial and error of special Pytogora's triangles a^2+b^2=(b+1)^2 probably from 500.b.c to Archimedes time.Archimedes challenged his contemporaries to resolve a Pell's equation possible around 225 b.c.discovering very possible the Fermat's algorythm shown on this disscusion group. He knew how to calculate the summe of finite and inffinite of geometrical series.Before him this topic goes back to at least1200 b.c.as a papyrus shows ,and Euclid at least 100 years before Archimedes gives a clumsy formula to calculate this summe). Archimedes'formula is the same like the formula we used today. That shows that he knew that : X^n+Y^n=(X+Y)[X^(n-1)-Y*X^(n-2)+(Y^2)*X^(n-3).........+Y^(n-1)] For sure he knew how to expand (X+Y)^2 (it goes back at least to Euclid time). What we can detect from his work is that where in his time some problems which the mathematiciens before him tried to find long due answers to, and Archimede studied them and was succsefull about many of them. What kind the questions or problems? Well,to name 3: 1)Is there any solutions for any D(square free)to the Pell's equation? 2)Is there anyway an answer to what is the summ of infinite geometrical series? 3)Can the Pytagora's triangle formula be extended to powers greater than 2? I claim that Archimedes knew how to expand the bynomial (X+Y)^n and knew at least that the coeficient of X^(n-1)term and Y^(n-1) term is equal to n.This guy Was Smart(what a platitude!). He resolved much harder problems than this.Let's recognise! So he atacked The 3d problem Or question(Fermat last Theorem) and come out with Fermatlast theorem. What I Am saying is the in the time of Fermat where more of Archimedes's works manuscripts than today. And we know that Fermat had a collection of Archimedes works and in one of them (which did not survive )Fermat met the Archimedes*s challenges about Pell'Equation and Last theorem at least.And he went on and proved them. george ghiata Transform Y^2=D*X^2+1 into Pytogora triangle a^2+b^2=c^2 Here it is; Y^2=D*X^2+1 Case 1: D=2*C-1 a=Y-X b=X*(C*X-Y) c=X*(C*X-Y)+1 a^2+b^2=c^2 Case 2: D=C-1=even number X must be=2*K a=Y-X b=X*(C*K-Y) c=X*(C*K-Y)+1 a^2+b^2=c^2 Observation:Y and X are INTEGERS:We can invent a new problem,no? Create by george ghiata Here it is another one: Here is the theorem: If A,B,C are natural numbers then it takes less than 15 arithmetical operations( summ,difference,multiplication and division) tofind X,Y,Z natural numbers such that A*X+B*Y=C*Z; Let's take A , B , C then write : 2*A*B*C + Q - B^2 = C^2*(1 +2*A*B*C) Let's take x=A*B*C +Q and y =A*B*C - B^2 Now x+y is divisible by C^2 Now x - y = Q + B^2; and Q - B^2 =C*S Let's take :[ y*(Q +B^2) + 2*Q*B^2] and [ - x*(Q+B^2)+2*Q*B^2] If we add-up the two above expressions we get a result divisible by C^2 If we substract the two expressions we get a result divisible by C^2 Therefore [ y*(Q+B^2) + 2*Q*B^2] is divisible by C^2 If we substitute y with its value we get that [ A*B*C*(Q +B^2) - B^2*(Q +B^2) + 2*Q*B^2] is divisible by C^2 and B Therefore EQ1= A*B*C*(Q+B^2) + B^2*(Q - B^2) = C^2*B*Z Therefore A*(Q+B^2) + B*S=C*Z Therefore A*X + B*Y = C*Z If we count the number of arithmetical operations to get this result we find to be 11(eleven).QED-or may be 15 any way. Created by Gheorghe Ghiata HEllo Here is an experiment: symbol for ballbearing -> O symbol for solid board -> --------------- symbol for solenoid -> [] symbol for imaginary line = = = = -------------O----------- -------------O----------- Above we see two solid boards with the implented ballbearings and the solenoids between them Now we fix a rectangular magnet[tinny bar -like] in the ballbearing in the upper board (hangindown betwwen solenoids but above the imaginary double line shown inthe figure) Now we fix a round-flat magnet in the ballbearing in the lower board sitting upright between solenoids but below the imaginary line shown in the figure. Now we get closer and closer the two magnets by moving the boards verticaly toward each other. They will start to rotate inducing electric curent in the solenoids.I think that will be a selfreliant and cheap way to get electricity(thousands of small magnets can be used) Please describe how many things will be changed. Created by George ghiata My name is Gheorghe Ghiata- A second possible historical reconstruction Archimedes had a method to resolve Pell's equation Fermat Redescovere it and proved by infinite descent method that allways works. That method is shown first herein the text. Archimedes Proved FLT for n=3 by infinite Descend method and Fermat rediscovered it. and the formula: Cnk=n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)/1*2*3*$.......*k which Was published by italian Math.Cardano in 1575 and Fermat's friend Merssene republish it in France in 1631.Archimedes probaly did not know the formula for large n's . That is why Fermat challenged his contemporaries FLT proof for n=3 The proof was the Same for any other odd prime n. The elementary Proof Of FLT by infinite Descend Method for any n=odd prime>1 is shown in this text First: About Pell's Equation I want to simplify my presentation of the method for which I have a proof by infinite descend methodthat It works allways to solve Pell's EQuation I presented it in writing to University Of victoria B.C CANADA and some of it before that to another professors.(In fact a sketch of it I sent 3-4 yaers ago to Mr.Prof Andrew Wiles!) You are going to see what Euler and many other mathematiciens miss to find. I never seen any where this :in no elemenetary theory of numbers books or advanced ones. Well ,it is only elementary high school algebra! Here is a simple exemple: Y^2=D*X^2+1 Let's choose D=29. Let's choose a<(29)^(1/2)>a+1 a=5. Now we write: 5^2-29=(1)*(-4) b^2-29=-4*k We choose b as being the largest number <(29)^(1/2)such that b+5 is divisible by factor 4(in this case) and we continuu the algorithm following this rule. So we have: 5^2-29=(1)*(-4) I 3^2-29=-4*5 II 2^2-29=-5*5 III 3^2-29=-5*4 5^2-29=-4*1 Now we know that (a^2+s*b^2)*(c^2+s*d^2)=(a*c-s*b*d)^2+s*(a*d+b*c)^2 So we multiply I*II=a^2-29*b^2=5*(4)^2 The proof shows that 4^2 simplifies and we get: I'=c^2-29*d^2=5 Now we multiply (I')*III= e^2-29*f^2=-5*5^2 After we divide by 5^2 we continuu this algorithm to the end and get x^2-29*y^2=-1 Now we multiply this identity by itself and get X^2-29*Y^2=1 The proof is elementary high school algebra technique. How in the world Euler could miss to find this algorythm and he never was able to proof Pell's equation ? More than that.It took for him 7 years to proof that a prime=4*k+1 can be represented as a^2+b^2 and to get a method how to find this representation. Whereas this algorithm which i called Fermat's-Archimedes Algorythm shows how to do it: As you see we got above the solution x^2+1=29*y^2 We can get from above a^2+1=29*b where a <29 and apply Fermat's-Archimedes algorythm c^2+1=b*e where c+a=b* ........ ........ x^2+1=y*1 Now we do the multiplication algorythm and get the representation X^2+Y^2=29 As you see we get y*1 as the last product. Well is needed only high school algebra to show that if we have x^2+s=D*b where s >0 then in the last product y*v we have 1= < v=< (s+1)/2 When s <0 there is a similar result but but different. In this way Fermat proved all his statements about the representations of diferrent primes .I proved them too. Try the one about, that all the primes 4*k+3 of X^2+5*Y^2 have as the last digit 3 or 7 Created by Gheorghe Ghiata Professor Harold Edwards in his book Fermat Last Theorem' Pell's eq. always have a infinity of solutios by proprely applyed infinit descend method.Mr Prof.says that at this time(1977) we do not see how Fermat could had aplyed that method for Pell's equation. Well, I rediscovered this Algorythm(may be somedoyelse too)and I named it Fermat Algorythm(Ishould had called it Archimedes algorythm) and I rediscovered the infinite descend method which proves that the algorythm always works(It is a high schoolAlgebra level). I would show now the begining of the proof by infinite descend method which Fermat talked about and which I send to Mr. Prof.Andrew Wiles in Dec.2002.The whole proof had be sent later to Mr.Prof.K Sound from University Of Michigan and Prof.David Leeming from University Victoria B.C. Canada Here It Is : Let's take: X^2=A*Y^2+1 Let''take : (R0)^2-A=-(k0)*(k1) where (R0)^22 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n The end of OBSERVATION CASE1 For simplicity of presentation lets take X*Y*Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=p^n Y=B+P=-B+W-Q=p*(z2) B=d*p Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k*q*p*u=b*u Therefore b is divisible by q*p Lets take : X^n-X +Y^n-Y = Z^n-Z+Z-X-Y X+Y-Z=B is divisible By n Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z If we follow the same procedure with EQ2 we get the same results: 1) [(z1)-1] is divisible by n 2) (W^n-P^n)/(W-P)=(z1) 3) (z1) can not be equal to 1 If we follow the same procedure with EQ3 we get the same results: 1) [(z2)-1)] is divisible by n 2) (W^n-Q^n)/(W-Q)=(z2) 3)(z2) can not be equal 1 Now we take: EQ4 : X^n-Q^n+Y^n-P^n -Z^n+W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+(W-Q-P) The EQ4 can be writen as : EQ4: n*m*B=(n^2)*J+2*B Therefore B is divisible by n^2 Therefore [(T/z)-1]=[z^(n-1)-1] is divisible by n^2 Since (z-1) is divisible by n we get from above that (z-1) is divisible by n^2 Since z+b=u^(n-1) and since b is divisible by n^2 we get that [ u^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z1)-1] is divisible by n^2 2) [q^(n-1)-1] is divisible by n^2 The same way we get that 1) [(z2)-1] is divisible by n^2 2) {p^(n-1)-1} is divisible by n^2 With this new data we get that EQ4 can be written : n*B*m={n^3)*J'+2*B Therefore we started a Infinite descend. Case2: If X*Y*Z is divisible by n the proof is almost the same. Let's Take Z divisible by n. X+Y=2*B+Q+P=(u^n)*n^(n-1) Z=B+Q+P=n*u*z Therefore 2*B+Q+P=2*b*u*n+q^n +p^n=(u^n)*n^(n-1) Therefore from above and OBSERVATION we get that q^n+p^n is divisible by n^2. Therefore 2*b*n*u is divisible by n^2 . Therefore u is divisible by n. Using EQ2 and EQ3 we follow the same proof as in the case X*Y*Z not divisible by n and get to the stage of the proof in which we get that B is divisible by n^3. Since B=b*n*u we get that u is divisible by n^2. So as we continuu the infinite descent we will get u divisible by n^(m-1) when we get B divisible by n^m.That is how we get the infinite descent in the the case 2. Therefore Fermat Last Theorem is True. Created By GHEORGHE GHIATA- 06-06/05 Copy of this text has been sent to Mr.Andrew Wiles and Mr.Andrew Granville To MR. PROF.Andrew Wiles My goal has been greater than to proof Fermat Last theorem (like you.) This has been a EXPERIMENT which I described in one of the e-mail sent They might still have it in the Archive of Www.Mathforum.org Shortly it has to do whith proving that there is UNIVERSAL MEMORY. A budhist saying says: IF you try to find a solution to any kind of problem ,and you strugle, you better try to REMEMBER IT. AND all my proofs reflect that. And in the end I REMEMBEReD the PROOF of FLT. MY REGARDS, gheorghe ghiata 5/23/05 Hello. Here is another proof of FLT : The ARCHIMEDES PROOF OF FERMAT LAST THEOREM (No need for him to know formula of Cnk) Theorem: he EQUATION: X^n+Y^n=Z^n where X,Y,Z are relative prime Integers ,n=prime>2 is impossible. OBSERVATION: X,Y ,Z relative Prime numbers Let's say that X^n+Y^n=Z^n Lets take: X+Y=W and R=X^(n-1)-(X^(n-2))*Y +(X^(n-3))*Y^2.........+Y^(n-1)= = W^(n-1)-(Cn1)*[W^(n-2)]*Y+(Cn2)*[W^(n-3]*Y^2- -(Cn3)*[W^(n-4)]*Y^ +(Cn4)*[W^(n-5)]*Y^4-............ ........-(Cn2)*[W^1)*Y^(n-2 ) +(Cn1)*Y^(n-1) where Cnk=[n*(n-1)*(n-2)*(n-3)*.........*(n-k+1)]/2*3*4*.... .*k When Z is not divisible by n we see that (X+Y)=W and R do not have any common divisor . Therefore they are relative prime and must be: R=z^n and W=X+Y=u^n and Z=u*z When Z is divisible by n,that is Z=u*n*z we see that W and R have as common divisor only n Therfore W=(u^n)*n^(n-1) and R=n*z^n End of OBSERVATION Lets take Z not divisible by n When Z*X*Y is divisible by n the proof is almost the same. PROOF: Let's take : X^n+Y^n=Z^n X+Y- Z=B X=B+Q Y=B+P Therefore : EQ1: (B+Q)^n+(B+P)^n=(B+Q+P)^n X+Y=2*B+Q+P=u^n=W Z=B+Q+P=u*z B=b*u and Q+P=u*s and - z=-u^(n-1) +b and z-b=s 2*b+s=u^(n-1) b+s=z We see that we can write : X= - B+W-P and Q= -2*B +W-P Y= - B +W-Q and P= - 2*B+W-Q Z= - B+W Therefore we get : EQ2: (-B+W)^n +(-B-P)^n=(-B+W-P)^n Q=-2*B+W-P=q^n X=B+Q= -B+W-P=q*(z1) Therefore we get : B=c*q EQ3: (-B +W)^n +(-B-Q)^n=(-B+W-Q)^n P=-2*B+W-Q=m*p^n Y=B+P=-B+W-Q=r*p*(z2) WHen Y is not divisible by n then r=1 and m=1 When Y is divisible by n the r=n and m=n^(n-1) B=d*p*r Since X,Y, Z are relative prime we have u,q,p relative prime. Therefore B=k* r*q*p*u=b*u Therefore b is divisible by q*p*r Let's take: EQ1 : (B+Q)^n+(B+P)^n=(B+Q+P)^n EQ1: (2*B+Q+P)*T=(B+Q+P)^n where T=z^n can not be equal to 1 . Therefore z is not equal to 1 T= [B^(n-1)+(c1)*B^(n-2)+(c2)*B^(n-3)+....(ct)*B+(Q^n+P^n)/(Q+P)]=z^n and (c1),(c2).....(ct) are integers coeficients and (ct)= {n*Q^(n-1)+n*P^(n-1)-2*[Q^n+P^n]/(Q+P)}/(Q+P)=(Q+P)*V T can not be equal to 1 Therefore z must be not equal 1 Since X^n-X+Y^n-Y is divisible by n we get that (T-1)=z^n-1 is divisible by n. Therefore (z-1) is divisible by n. Since B and Z have as common divisior only u We can write : T= Z*B*F(q,p)+[Q^n+P^n]/(Q+P)=z^n=[-b+u^(n-1)]^n where F(q,p) is a function=F of (q,p) When we develop the parantheses of z^n We see that z^n=G(q,p) where G(q,p) is a Function =G of (q,p) of (q,p) Let's multiply T by (Q+P): T*(Q+P)=(Q+P)*Z*B*F(q,p) +(Q^n+P^n)=(Q+P)*z^n=(u*s)*z^n T*(Q+P)= (Q+P)*Z*B*F(q,p) +(u*s)*(z*M)=(Q+P)*z^n Since z*u=Z we can divide by Z and get: T*(Q+P)/Z= (Q+P)*B*F(q,p) +s*M= s*z^(n-1) Since s=z-b we can write : T*(Q+P)/Z= (Q+P)*B*F(q,p)+s*M=z^n-b*z^(n-1) Since b is divisible by q*p we that s*M=A(q,p) where A(q,p) is a function =A of (q,p) Therefore: (Q^n+P^n)= (Q+P)*[Q^(n-1)-P*Q^(n-2) +........P^(n-1)] =Z*A(q,p)=(B+Q+P)*A(q,p) whereZ=B+Q+P=D(q,p) is a function D of (q,p) But we know that [Q^n+P^n] is a reducible polynom of (q,p) only one way to two factors. Therefore the only solution is [Q^n+P^n]/(Q+P) =z Now if we substitute in T {Q^n+P^n]/(Q+P) with z we see that z=f(q,p) Now we have F= (Q^n+P^n)=Q+P)*z=s*Z=(z-b)*(B+Q+P) We know that z=f(q,p) and b is divisible by q*p. But Q^n+P^n is a polynomial of (q,p) which can be reducible only one way as the product of two polynomial factors of (q,p) ,but F shows that there is another way of decomposition in two polinomial factors of (q,p). That is impossible . Therefore Fermat Last Theorem is true. Created by GHEORGHE GHIATA-06/06/05 To AStrophysicists: When 5-D univers started burning from inside its center outwards, is that what I call Big Bang.And out of that continuu burning'of 5-D univers our 4-D univers, the inside circular uneven surface appeared and continuu to expand . Therefore our univers is a interface between two 5-D universes,sandwiched betwwen them. george ghiata THE END* *Author's note: This is the end of the real story which have taken place over a period of 4 years VIA electronic messages and postal service If you read all about it and you do not believe it ,then you can look I have to say TANKS to every PARTICIPANT. You might ask who the stage DIRECTOR has been. Here it is my answer: In a village, an unusual animal appeared and the people never seen before this kind of animal.So they decide to go and ask the sage of village.The sage told them to wait for the answer because he needs to meditate to find it.So,he fell in deep meditation and after awhile he start to cry and cry.Then at one moment he burst laughing and laugh and laugh.This made people very curios about these reactions and when the sage woke -up from the meditation they asked him why he cried so much His answer was :Well,when I see so many IGNORANTpeople who do not know what this strange animal is all about, I was suffering so much that I could not hold my tears. Why did you burst laughing then? the people asked. Well,I REALIZED that I did not know the Answer either. Created by GHeorghe ghiata Now Iam saying Good-by and I withdraw in my Ivory tower murmuring verses from the poem Lucifer by romanian poet Mihai Eminescu GHEORGHE GHIATA-06/06/05 _______________________________________________________________ About Mr. Prof.ANDREW WILES'PROOF OF FLT Plain Text Reply My understanding is that Mr. Prof. Andrew Wiles proved Fermat Last theorem for n>3. Does he had a original proof for n=3? If not, why he got Wolfsskehl Prize? If not ,that means I am the first person,the first time ,to give a more complet proof of FLT for n=prime >2.That is where the dispute is possible. We have to read Mr. Femat' statement very precise. GHEORGHE GHIATA GHEORGHE GHIATA (cod Nr.050322) _________________________________________________________________ Comments: This message did not originate from the Sender address above. It was remailed automatically by anonymizing remailer software. Please report problems or inappropriate use to the remailer administrator at . === Subject: Nukes Mail-To-News-Contact: abuse@dizum.com 5000 nukes are now targeted at the people in Europe and Usa. Russia and US has 97 of all nuclear weapons and the wars in Iraq and now Iran (Yes it has started) have increased tensions. The leaders does not have to fear. They just run and hide in their underground cities if a nuke exchange is triggered. But what about you and your family? Please sign this petition and pass it on to as many as possible and as fast as possible. It can easily travel around the globe this week if people care. Listen to your heart and you know what is right. Evil is just the absence of good. www.petitiononline.com/prop1/petition.html