Given two arrays: P 3 x n and p 2 x m where m << n and given P_sub 3 x k and p_sub 2 x k where k < m s.t. these two arrays have correponding columns using P_sub and p_sub we can recover the transformation (A,t) between them i.e. p_sub = A * P_sub + t the question is: using A,t and p (2d points) how can I find the corresponding 3d points P1 (subset of P) ? Using P1 = A^-1 * (p - t) is not correct, true ? ==== this is originally informatic problem, but it's a math challenge too. I cant figure out what are the secret numbers they want me to search for such that the conversation can be possible ( see the problem). Problem C Secret Numbers Input File: C.DAT Program Source File: C.PAS or C.C or C.CPP Two natural numbers a and b are chosen (1 > this is originally informatic problem, but it's a math challenge too. > I cant figure out what are the secret numbers they want me to search > for such that the conversation can be possible ( see the problem). > > > Problem C > Secret Numbers > > Input File: C.DAT > Program Source File: C.PAS or C.C or C.CPP > > Two natural numbers a and b are chosen (1 multiple of a and b (a*b), and person S is told the sum of a and b > (a+b). The discussion between M and S goes like this: > > M: I do not know the numbers a and b. This implies that the product m=a*b has (counting multiplicity) at least three prime factors (and m is not the cube of prime). > S: I do not know them either, but I knew you would not know them. This implies that the sum s=a+b can't be represented as the sum of two primes or as the sum of a prime and its square (otherwise S would not know that M could not know ...). Every even number in the range given can be represented as the sum of two odd primes, and a prime plus its square is always even, so s can't be even. Also, s can't be of the form p+2, p an odd prime. > M: Now I know the numbers! So there must be only one factorization of m which produces a sum which is odd and not p+2 with p prime. > S: Now I know them, too! So there must be only one way of getting s as sum for which the product of the summands can be factored in only one way in which the sum of the factors is odd, not of the form p+2 with p prime. ==== >M: I do not know the numbers a and b. > > This implies that the product m=a*b has (counting multiplicity) > at least three prime factors (and m is not the cube of prime). > >S: I do not know them either, but I knew you would not know them. > > This implies that the sum s=a+b can't be represented as the sum of > two primes or as the sum of a prime and its square (otherwise S would > not know that M could not know ...). Every even number in the range > given can be represented as the sum of two odd primes, and a prime > plus its square is always even, so s can't be even. Also, s can't > be of the form p+2, p an odd prime. Can you give me a reference about this point? just what is an odd prime? prime that is not 2? thanks > >M: Now I know the numbers! > > So there must be only one factorization of m which produces a sum > which is odd and not p+2 with p prime. > >S: Now I know them, too! > > So there must be only one way of getting s as sum for which the > product of the summands can be factored in only one way in which > the sum of the factors is odd, not of the form p+2 with p prime. ==== [ snip ] >> Every even number in the range >> given can be represented as the sum of two odd primes, and a prime >> plus its square is always even, so s can't be even. Also, s can't >> be of the form p+2, p an odd prime. > > Can you give me a reference about this point? Google for Goldbach+conjecture; all smallish even numbers can be represented as a+b with a, b odd primes. > just what is an odd prime? > prime that is not 2? Yes, a number that is odd and prime. ==== In sci.math, Dave Seaman : >> In sci.math, The Ghost In The Machine >><10l301-l7p.ln1@lexi2.athghost7038suus.net>: > In sci.math, Ziga Habjan > : >> test >> > > [1] Prove Fermat's last theorem. > (there exist infinitely many primes of the form 2^(2^n) + 1) > >> Oops. Someone already did and I transcribed the wrong thing anyway. :-) >> Aargh. > > > [2] Prove Goldbach's conjecture. > (any even number > 2 is the sum of two primes) > > [3] Prove that the number of real points in the line segment [0,1] (C) > is equivalent to the cardinality of the set of all subsets of the > natural numbers (aleph-1). > >:-) > > In light of what you said under [1], I thought perhaps [3] was > deliberate. But just in case it wasn't: the cardinalities of R and of > P(N) are both equal to c = 2^aleph_0. The hypothesis that c = aleph_1, > the cardinality of the set of all countable ordinals, is the continuum > hypothesis. > Well, [3] was deliberate in terms of my knowledge, although I'm not entirely certain what aleph_1 = 2^aleph_0 means; as far as I'm concerned, if card(N) = aleph_0, then the set of all subsets of N (including itself, as it turns out) could be termed 2^N, with cardinality aleph_1. At least, such is my understanding of the alephs. I could be wrong. Also, could you explain the notation P(N)? I'm not familiar with it. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== |Well, [3] was deliberate in terms of my knowledge, although I'm not |entirely certain what aleph_1 = 2^aleph_0 means; as far as I'm |concerned, if card(N) = aleph_0, Correct. |then the set of all subsets of N |(including itself, as it turns out) could be termed 2^N, Correct. In general X^Y is used to represent the set of functions from Y to X. (This meshes with the meaning m^n has for nonnegative integers m and n. If X has m elements, and Y has n elements, the number of functions from Y to X is m^n.) Since 2 is used to represent a two-element set such as {0,1}, 2^N stands for the set of functions from N to {0,1}. These can be identified with subsets of N by associating the function f(n) = {1 if n is in S {0 if n is not in S with the set S. (This uses the law of excluded middle, in the form of either n is in S or n is not in S. Rarely will anyone mention where they use the law of excluded middle, but I sometimes do because of my interest in constructive mathematics, where the law of excluded middle is not used.) ;-) The same exponential notation is used for cardinalities. If |X| and |Y| are the cardinalities of X and Y, then |X|^|Y| stands for |X^Y|. So 2^aleph_0 stands for the cardinality of 2^N. That cardinality is also known as the continuum. |with |cardinality aleph_1. At least, such is my understanding of the alephs. Not necessarily correct. By definition, aleph_1 is the cardinality of the smallest uncountable ordinal. It's a cardinality > aleph_0, with no cardinalities between them. Assuming the Axiom of Choice the cardinals are linearly ordered by <, so we can call aleph_1 the first cardinal greater than aleph_0. (Usually the axiom of choice is just assumed. It's not mentioned very often that it's being assumed. This is again something I mention because it's the other principle not used in constructive mathematics.) ;-) The equation aleph_1 = 2^aleph_0 is known as the continuum hypothesis. Goedel proved that it's consistent with the standard axiom system for set theory (the Zermelo-Fraenkel axioms plus the axiom of choice, ZFC), assuming that ZFC is consistent to begin with. Cohen proved that aleph_1 < 2^aleph_0 is also consistent with ZFC, assuming again that ZFC is consistent. Some people think that the continuum hypothesis is analogous to the parallel axiom in geometry. They think that there's no such thing as the correct answer to whether it's true. Some people think that there is a correct answer. The last I read, it was alleged that among the set theorists who think there's a real answer, more of them think that the continuum hypothesis is false than think it's true. I tend to suspect it is false as well. Why should there be a one-to-one correspondence between the smallest uncountable ordinal and the subsets of N? But that's just speculative opinion. Goedel for a time suspected that 2^aleph_0 was aleph_2. |I could be wrong. | |Also, could you explain the notation P(N)? I'm not familiar with it. If X is a set, P(X) is often used to denote the set of all subsets of X. The correspondence between subsets of X and functions from X to {0,1} can be expressed as |P(X)| = |2^X| or |P(X)| = 2^|X|. Keith Ramsay ==== >A Google search for nonabelianity yields 2 hits. I haven't yet tried >abelianity or (non)abelianism. Abelianity gets 65, wow! One of them is k-abelianity, another one is i-abelianity. It's fun to make up words like this. My favorite nonexistent word I've come up with this way so far is abelianiyat (to parallel afghaniyat). Nonabelianiyat lives! Keith Ramsay ==== >Yes, at m=0. We agree that a_3 is coprime to f when m=0. >> And I'm sensing that you must still think that there's some variable >> dependency on m, or there wouldn't be further discussion. > Well, yes I do. The a's are clearly dependent on m and f. Their cubic is >described above. So I don't really know what you mean, I don't think. The m and >f are almost independent - they can be anything as long as f is coprime to 3 and >m. Some progress may have been made if you accept that you're trying to > get a variable dependency for how f^2 divides off, though it's a > constant. > I don't think I am. I might be wrong though. I don't think it matters at all how f^2 divides off, you end up with in the same place. > Possibly you're confused because the a's are dependent on m and f, but > f^2 is a constant factor of P(m), and it is not. I agree that f is not dependent on m. Is that what you mean? Also m and f are completely independent in general, while I introduce > specific restrictions for special purposes at particular points. > No problem. a_3 is demonstrably not coprime to f when f is not coprime to 3, and m is not coprime to f. I'm happy to agree that they're independent, and it's the specific restrictions that bear looking at more closely. > Again, the math can't read minds, so it's setup to handle the general > case, where m and f are completely independent, and it doesn't bother > to shift because of my choices, as the mathematical logic is rigid. > Are you still trying to claim that you are not? >> If you're not then I can just check at m=0, confident that I've >> covered when m does not equal 0, right? >> If not, why do you believe so? > I'm sure this is not the case. a is a function of m. > Trivial analogy: >Say a_3 = 3 + m (1 + f) >Imagine f=5 so a_3 is coprime to f when m=0. >Now, when m=7, also coprime to 5, a_3 is not coprime to 5. > I know your a_3's aren't as above, but I honestly don't accept that checking at >m=0 is >sufficient, unless I can see a proof of it. Well your own example should show you why. Imagine the possibility > that you had some expression where if f=5, your a_3 had a factor of 5 > for ALL m, but if f didn't equal 5 it equaled 1 at m=0. Do you believe that is possible? I think you're alluding to the w_1w_2w_3 construction below. So I'll leave it till then. You didn't yet show the proof of how checking at m=0 is sufficient to prove that a_3 is coprime to f for all m. If you follow mathematical logic that should finish your objections. The mathematical logic hasn't been presented yet. > > In fact the b's are never forced into a field, but they are forced out >> of the ring of algebraic integers. > OK. It's important that it's recoginsed that they are often not algebraic >integers. It is important as it shows a problem with the definition of algebraic > integers, as it's not as inclusive as it should be. > Your assumption that the field of algebraic numbers is required, which >> you've repeated several times, but have been unable to prove. >As long as we're mindful that the b's are not always algebraic integers. I have said so myself, and in fact that is why there's a problem with > the ring of algebraic integers. It may seem esoteric to readers on sci.physics and sci.skeptic, but > mathematics requires zero errors, and what I've managed to show with > some fascinatingly basic algebra is an error created by that > definition of algebraic integers as roots of monic polynomials with > integer coefficients. That definition leaves gaps by not including certain numbers that > should be included which leads to fascinating contradiction like that > ***in the ring of algebraic integers*** you can have abc = 5, where a, > b and c are coprime to 5. That coprime just means they don't share non-unit factors, i.e. not > factors of 1, with 5, but they multiply together to give 5, and a, b > and c are each algebraic integers. Mathematicians missed this little thing for over a hundred years, but > I can prove there's a problem in the ring with a short argument using > basic algebra, which comes at the end of this post. > Clearly, you've seized on one idea, and you keep holding on to it, > despite my efforts to get you to follow the math. >>I'm trying to follow the math. I keep getting stuck here though: >>You claim that because >>b_1 b_2 a_3 = m^3 f^4 - 3m^2 f^2 + 3m >>and RHS is coprime to f (which it is as m and f are coprime) >>then a_3 is coprime to f. >>This is only true for integer m and f when m=0. >> Which indicates that you believe that it might be different when m >> does not equal 0, which forces a dependency on m. > Yes, the values of the a's and b's depend on m. The RHS is coprime to f, >of course. I'm hoping that you don't mean something as trivial as that. It is >agreed that b_1b_2a_3 is coprime to f. It's not agreed that a_3 is coprime to f >for all valid m and f. However, disagreeing there requires that you go against mathematical > logic. > Which mathematical logic? I've shown you example of how it can be that b1*b2*a3 is coprime to f does not mean that a3 is coprime to f Recall? b1*b2*a3 = 21, coprime to 5. But a3 = 25. This happens because b1 and b2 are not algebraic integers. In your work, b1 and b2 are not in general algebraic integers, so you need to provide a proof that a_3 is coprime to f. So far it is just an assertion. It might even be true, but it is a gap in your proof. If you can prove that a_3 is coprime to f when m==0, you'll get there as well. So far, however, these proofs don't exist. > In this case it's hopefully easily seen by considering that if f=3, > ALL of the a's have a constant factor that is 3, plus if m isn't > coprime to 3 they can have additional factors in common with m and 3. That shows factors of f can't jump around when f is coprime to 3 as > you apparently continue to wish, as in fact that would be forcing a > dependency on f^2 which does not exist, as f^2 is a constant with > regard to m. > Now it's quite simple, admit that you believe there's a variable >> dependency on m, and then I can show there is none. >I think this would be extremely valuable. But you shouldn't feel any pressure to >convince me of anything. >The choice, and the floor is yours if you want. I believe there's a variable >dependency on m.The a's and b's are variables here, right? They're the things I >think are dependent on m. Well you're suddenly sounding passive. The question here is, can you > follow mathematical logic? Or will you hang on to some belief, > possibly for personal comfort *against* mathematical logic? I find that question intriguing. I'm just waiting to see the mathematical logic. If I see it, and it's valid, and I reject it, then you won't be intrigued for long, you'd be entitled to be dismissive of me. > However, if you continue to maintain that there is no dependency on m, >> but then try to give emphasis on the case when m=0, then that >> fundamental contradiction in your thinking means that you cannot be >> following the math. >>Except at m=0 when b_1 and b_2 are =0 and are algebraic integers. >> Well, try m=1 with f=sqrt(2), and welcome to a more complicated > mathematical world than you might have realized. >>I'm sure it will work out fine with those values, so I won't try. >>How about YOU try when f=5, and m is your favorite non-zero integer coprime > to >>5. >> Why? > No reason. I'm guessing though that a_3 is not coprime to 5. There is no need to guess. It's mathematics, and it's possible to > prove that it is. No guessing needed. > wrong assumption, which is that the constant f^2 divides off in some >> way as a function of m or variable dependent on m, which you have >> repeatedly demonstrated by citing the m=0 case as if it is a special >> case, but when I say that's your assumption, you claim it is not. > In my mind, it doesn't matter how the f^2 divides off. If you resolve the a's >and b's to their cubic definitions, the steps you took to get there disappear in >any case. >I haven't actually said that m=0 is a special case at any point. Although >secretly I think that solving for m=0 isn't sufficient to show that the result >applies for all m. Not so secretly, obviously. You apparently have seized on the idea that m=0 is a special case, and > simply choose to ignore counterexamples like m=1, with f=sqrt(2), > though there apparently you may think that f not being an integer > makes a difference. I thought f was an integer, but may have been wrong. No problem if it's extended. The problems I have are: a_3 is a function of m and f. You prove that a_3 is coprime to f when m=0, and assert that therefore it is coprime to f for all m. I question this assertion. b_1*b_2*a_3 is coprime to f.You assert that a_3 is coprime to f when b_1 and b_2 are not algebraic integers. I question this assertion. m=1 f=sqrt(2) is not a counterexample. b_1 and b_2 are algebraic integers in that case. And indeed a_3 is coprime to f, as mathematical logic would have it. However, there is no need to guess, or go by hunches, as if you follow > the math, the conclusion is clear. >> Looking at other examples will not help you, but trying to focus you >> onto the basic contradiction in your position--the claim that there is >> no dependency on m, when you keep pointing at m=0 as if there is--just >> might. >> And if you refuse to ever acknowledge something that obvious, then I >> don't see where there's much likelihood that you will ever follow the >> math. >That's all good. Perhaps there is something you can do to enlighten me. I know >the a's and b's depend on m; but I know you know that as well, it makes me think >that I'm just misunderstanding what you mean by variable dependency. > So if I'm merely misunderstanding, I'll stop for now. > Phil Nicholson. Well, the argument which settles things is *luckily* short and rather > direct, so I'll give it here. Some may think it's exactly what > they've seen before, as I've been posting it a lot of places, but I've > seen need to put in minor corrections. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. > Strictly speaking you've proven that w_1 w_2 b_3 = 3 and w_1 w_2 w_3 = f > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Strictly speaking w_1 w_2 w_3 = 3 in this case. You've just chosen to make different substitutions for w_1 w_2 in the two separate examples. Which is fine, it neither proves nor disproves anything. That is, the w's are now all constant with regard to m and have the > same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, > f=sqrt(2), but that's a special case as generally they are not, which > shows a problem with the ring of algebraic integers. I didn't ever suggest that the w's varied. I was hoping to see you prove how you could verify that a_3 was coprime to f at m== 0. Maybe next time? Oh, I guess that's a 3rd problem I have then. I disagree with your assertion that there's a problem with the ring of algebraic integers. Perhaps when we see that proof that a_3 must be coprime to f, we can consider this assertion. I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where 1 is the only > member that is both a unit, i.e. factor of 1, and an integer, where no > non-unit member is a factor of any two integers that are coprime. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. > James Harris ==== of course, that could be classical codependency. > why don't you try working on another problem, > can you find evidence of even one, > who has stuck with your prevarications, > after his (or her) initial locquaciousness? --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== Barbier's theorem is that all curves of constant width of width w have the same perimeter, pi * w. http://www.cut-the-knot.org/ctk/Barbier.shtml gives a proof without calculus. Is there a simple proof that uses calculus? http://mathworld.wolfram.com/CurveofConstantWidth.html ==== >,, e denotes Napier's constant and let [.] > be the integral part. Prove or disprove that Integral_{t=0 to t=infty}e^{-t}(t-1)^{n} dt = [ n!/e + 1/2] . -- Spammers: I don't want a small digital camera to post photos of a large, low weight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI ==== > ,, e denotes Napier's constant and let [.] > be the integral part. Prove or disprove that > > Integral_{t=0 to t=infty}e^{-t}(t-1)^{n} dt = [ n!/e + 1/2] . Of course, this fails if n = 0. In this case, we have: Integral_{t=0 to t=infty}e^{-t}(t-1)^{0} dt = [ 0!/e + 1/2] Since we can split the integral and limit towards n=0, we don't need to worry about the 0^0 that could result. Integral_{t=0 to t=infty}e^{-t} dt = [ 1/e + 1/2] The left side is 1, the right is 0. Looking beyond this, we know that n must be assumed to be an integer. (The left side is continuous outerwise, and the right discrete.) int[t=0..inf] e^-t (t-1)^n dt This looks remarkably like the gamma function... In fact, we have: int[t=0..inf] e^-t t^n dt = n! (This relation is easily proven by induction. n=0 is the base case. It conveniently happens that one application of integration by parts gives precisely the desired recurrence relation.) int[t=0..inf] e^-t (t-1)^n dt Split the range. int[t=0..1] e^-t (t-1)^n dt + int[t=1..inf] e^-t (t-1)^n dt int[t=0..1] e^-t (t-1)^n dt + int[t=0..inf] e^-(t+1) t^n dt int[t=0..1] e^-t (t-1)^n dt + 1/e int[t=0..inf] e^-t t^n dt int[t=0..1] e^-t (t-1)^n dt + n!/e The integral is certainly less than 1 for any n >= 0. What is more, it alternates in sign and diminishes. For n>=1, the first term is less than 1/2. For this reason, if you can show that the original integral is always integral in value, the proof is complete, except for its failure at 0. int[t=0..inf] e^-t (t-1)^n dt Since n is assumed to be an integer, we have: (t-1)^n = t^n - C(n,1) t^n + C(n,2) t^2 - ... + (-1)^x int[t=0..inf] e^-t [t^n - C(n,1) t^(n-1) + C(n,2) t^(n-2) - ... + (-1)^x] dt int[t=0..inf] e^-t [t^n] dt - int[t=0..inf] e^-t [C(n,1) t^(n-1)] dt + int[t=0..inf] e^-t [C(n,2) t^(n-2)] dt - ... + int[t=0..inf] e^-t [(-1)^x] dt n! - C(n,1) (n-1)! + C(n,2) (n-2)! - ... + 1 Thus, it is just the sum and difference of integers. It follows that it is itself an integer. The theorem holds, for integers n > 0. ==== David Bernier >>Suppose we want to know what the image of pi is under >>some field automorphism phi of the real numbers. > Luckily R has very few field automorphisms. :-) > Yes. We might soon run out of letters otherwise. > This made me wonder what alphabets are used for symbols in > contemporary mathematics literature written in English: - Roman (with i,j,k,u,v,w) > - Greek > - Hebrew (aleph, beth) > - Cyrillic?? (The Tate-Shafarevich group??) > - what about the Weierstrass P function? I think this P comes from an old German typeface, as does the R produced by the TeX symbol Re. The inertia of notation is rather odd, isn't it? E.g. people still write s=sigma+it when talking about the zeta function (but no other subject) -- the same notation introduced by Dirichlet in his 1837 paper on arithmetic progressions. LH ==== Nitpicking... > Let's get back to real math. >> You have said, applying your methods in Advanced Polynomial >>Factorization, that if you factor the polynomial >> P(x) = 65*x^3 - 12*x + 1 >>in the form >> P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1), >>where a1, a2, and a3 are algebraic integers, then two of >>the a's are divisible by sqrt(5) in the algebraic >>integers. Right? > > Nope. > >> Say a1 is divisible by sqrt(5). Let a1 = sqrt(5) * c1, >>where c1 is an algebraic integer. >> Note that -1/a1 is a root of P(x). Therefore >> P(-1/(sqrt(5)*c1)) = 0. >>This implies >> -65*(1/(5*sqrt(5)*c1^3)) - 12*(-1/(sqrt(5)*c1)) + 1 = 0. >>Multiply through by 1/(5*sqrt*5)*c1^3). You get >> -65 + 12*5*c1^2 + 5*sqrt(5)*c1^3 = 0. >>Divide out 5, move things around: >> sqrt(5)*c1^3 = -12*c1^2 + 13. >>Square both sides: >> 5*c1^6 = 144*c1^4 - 312*c1^2 + 169. >>Rewrite this as >> 5*c1^6 - 144*c1^4 + 312*c1^2 - 169 = 0. >>Use your favorite piece of software to show that this is >>a non-monic and ***irreducible*** polynomial in c1. >> Then apply a well-known theorem from algebraic number >>theory: >> THEOREM: If r is a root of a non-monic polynomial >> with integer coefficients, ***irreducible*** over >> the rationals, the r cannot be an algebraic integer. >>and conclude that c1 cannot be an algebraic integer. You must also include the hypothesis that the polynomial is >primitive. Since nonzero constants are units in Q[x], they are not >considered nontrivial factors, so the hypothesis must be explicitly >included. > > That is correct. And it is true that c1 cannot be an algebraic > integer. > > That has not been under debate. > > It's also NOT under debate as to whether or not given > > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > > any a's exist, within the ring of algebraic integers, such that > sqrt(5) is a factor of them in that ring. > I see that the current version of APF does not make this claim, though I believe you have said exactly that in the past. Here is what the current version of APF *does* claim: Let f = prime > 3, m = integer coprime to f, v = -1 + m*f^2, and u = integer coprime to f, and P(x) = (v^3 + 1)*x^3 + 3*v*x*u^2*f^2 + u^3*f^3. Then P(x)/f^2 may be factored in the form [1] P(x)/f^2 = (a1*x + u)*(a2*x + u)*(a3*x + u*f), where a1, a2, and a3 are algebraic integers. -------------------------------------------------------- Let f = 5, m = 1, u = 1. Then P(x)/f^2 = 553*x^3 + 72*x + 5. If this is factored in form [1], it will look like P(x)/f^2 = (a1*x + 1)*(a2*x + 1)*(a3*x + 5). This means that -1/a1 is a root of 553*x^3 + 72*x + 5 = 0. That is, 553*(-1/a1^3) + 72*(-1/a1) + 5 = 0, or 5*a1^3 -72*a1^2 - 553 = 0. But this last expression is a non-monic, irreducible, primitive polynomial in a1. Therefore a1 cannot be an algebraic integer. Therefore the conclusion of APF is false. > The problem is that neither a_1, a_2, nor a_3 have ANY non-unit > factors in common with 5 in the ring of algebraic integers. > That would certainly be a problem, given that their product is 65. So it looks like both of us arrive at a contradiction. We draw different conclusions from it, apparently. I conclude that your claim is false and that therefore there is necessarily an error in your proof [and I have described where that error is and what it is at length]. You conclude that there is something wrong with the ring of algebraic integers, perhaps that it is incomplete. What that means is not clear, at least not to me. It could mean that the a.i.'s do not really form a ring - perhaps that they are not closed under addition and multiplication or that the distributive law does not hold, etc.. This however is a very old theorem and is not in doubt. The main question here is, if you arrive at a contradiction, why do you assume the problem must be somewhere other than in your own proof? Why, in view of your atrocious track record over 8 years, do you now assume that you are infallible? Have you had a message from God, or what? Nora B. > > James Harris ==== >>I thought Nora was talking about something other than a letter saying >>This guy's a crank. It was my assumption that by specifying what those >>counterarguments are she meant that the editor would be provided with >>valid mathematical reasons why James' paper is incorrect. The editor >>(or reviewer) would be able to read both James' submission and the >>counterarguments and make his/her own decision. > That's what I assumed she meant as well, probably because that's > what she said. It's a terrible idea, for the reasons Randy suggests, > and also not necessary, for reasons he suggests. OK, fine. It just seems to me that as long as the information is correct, then the source, or how it was obtained, or even the motives of the person providing it are unimportant. But I've had no experience with the review process and am happy to defer to those who do. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== [.snip.] >>theory: >> THEOREM: If r is a root of a non-monic polynomial >> with integer coefficients, ***irreducible*** over >> the rationals, the r cannot be an algebraic integer. >>and conclude that c1 cannot be an algebraic integer. >> >> You must also include the hypothesis that the polynomial is >> primitive. Since nonzero constants are units in Q[x], they are not >> considered nontrivial factors, so the hypothesis must be explicitly >> included. That is correct. And it is true that c1 cannot be an algebraic >integer. That has not been under debate. Which is why I said I was nitpicking: pointing out a minor error that is well understood. >It's also NOT under debate as to whether or not given 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) any a's exist, within the ring of algebraic integers, such that >sqrt(5) is a factor of them in that ring. This is rather confused. You have a not at the beginning, a whether or not after that, and a qualifier any for the a's. It's pretty close to nonsense. What you are really saying, presumably, is: Given a1, a2, a3 algebraic integers such that 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) [as a polynomial identity], then none of a_1,a_2,a_3 are multiples (in the ring of algebraic integers) of sqrt(5). This is also true, and has been established. >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. And that's false. I am pretty sure that Dale produced explicit common factors; but in any case, your claim here is certainly false, since their product is not coprime to 65. Lemma. Let R be the ring of all algebraic integers, and let a, b, c be any elements of R. If a and b are coprime to c, then a*b is coprime to c. Proof. We use the characterization of coprime valid for commutative rings with 1: a and b are coprime in R if and only if there exist x and y in R such that ax+by = 1. Since a and c are coprime by assumption, there exist n and m in R such that an+cm = 1. Since b and c are coprime by assumption, there exist r and s in R such that br+cs = 1. Multiplying both together, we have 1 = (an+cm)(br+cs) = abrn + acns + cbmr + c^2*ms = ab(rn) + c(ans + bmr + cms). Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and ab*x + c*y = 1. Therefore, ab and y are coprime. QED So, assume you were correct and neither a_1, a_2, nor a_3 have ANY non-unit factors in common with 5 in the ring of algebraic integers. Then, by the lemma, neither does a1*a_2; and applying the lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which clearly has 5 as a nonunit common factor with 5. This contradicts the assumption that none of a_1, a_2, a_3 have common non-unit factors with 5 in the ring of algebraic integers. Therefore, your assertion is false. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== In sci.math, James Harris <3c65f87.0308061055.c2ca0c2@posting.google.com>: >> In sci.math, James Harris >> <3c65f87.0308050808.61ebf91d@posting.google.com>: >>It occurred to me that some of you may be hampered in understanding >>certain math arguments of mine because >> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 + u^3 f) >>has that constant factor of f^2. >>Normally when considering factorizations, you separate off constant >>factors, as otherwise you don't have a unique factorization even with >>polynomial factors. >>For instance >> >> 4(x^2 + 2x + 1) = (2x + 2)(2x + 2) = (x+1)(4x + 4) >>along with an infinity of other factorizations, but typically you'd >>just have >> 4(x^2 + 2x + 1) = 4(x+1)(x+1). >> >> I suspect there are a number of ways of dealing with >> this issue; I'd probably state that the factorization >> would require that all non-trivial polynomials have >> coefficients with gcd 1. (4 is a coefficient of the >> trivial polynomial 4 * x^0 and would have to be treated >> as a special case, but one can also chop up 4 into its >> constituent prime factors if need be.) > > It's easy enough just to separate the 4 to the side as I did above, > but things become more complicated with an expression like > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > as can be seen by the *months* of discussion I've gone through, though > now there should be progress as I've nailed down a false assumption > that others must be having, which is the belief the f^2 can divide > from the factors > > (a_1 x + uf), (a_2 x + uf), and (a_3 x + uf) > > as a function of m. But f^2 *can* divide from those factors. If f is prime, one merely needs to have either exactly one a be divisible by f^2 (in which case u and f have to relate somehow) or exactly two a's be divisible by f. If f is a nonprime additional possibilities ensue, depending on u. I'd have to work out the gloppy details. This subproblem gets a little complicated, at first blush, but it's not unmanageable. > > It's the kind of weird false assumption that can just hang out there > if no one puts it forward directly, and I think that mathematicians > would not make it. > > After all, f^2 is a constant factor of P(m), why would it have an m > dependency? > > Luckily, I can easily show that it does not for those who get really > stuck on the false assumption. > >>Besides all that the expression I use is rather imposing, and it has a >>lot of symbols, so I thought I'd remind you of a few things. >>1. You *can* look at an actual example with m=1, f=sqrt(2), as then >>all that complexity drops away and you have >> P(1) = 2x^3 - 3x + 1 >>which actually does reduce over Q. >> >> P(1) = 2*x^3 - 6*u^2*x + 2*sqrt(2)*u^3 >> >> One has to set u to be 1/sqrt(2) as well. > > Oh yeah, I left out several steps, like using y=uf. Notice > > f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) = > > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - > > 3(-1+mf^2 )x u^2 f^2 + u^3 f^3. > > Now using y=uf, I have > > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - > > 3(-1+mf^2 )xy^2 + y^3. > > So, if you factor to get something like > > (a_1 x + y)(a_2 x + y)(a_3 x + y) > > the a's are independent of y, so I can let y=1, so I have > > > (m^3 f^4 - 3m^2 f^2 + 3m)f^2 x^3 - > > 3(-1+mf^2 )x + 1 > > and with m=1, f=sqrt(2) that is > > 2x^3 - 3x + 1. > > If you prefer to keep y, you have > > 2x^3 - 3y^2 + y^3. > > If you really *must* keep y=uf, so that you need y=sqrt(2)u, then you > may do so. > > The expression is still, of course, reducible over Q. That it is. > > >>Some of you may have realized that you can consider m=1(mod sqrt(2)) >>to blow apart several assertions made by some posters. >> >> P(1 + k(sqrt(2)) = >> ((2 * sqrt(2)*k^3 + 6*k^2 + 3*sqrt(2)*k + 1)*f^6 >> + (-6*k^2 - 6*sqrt(2)*k - 3)*f^4 + (3*sqrt(2)*k + 3)*f^2)*x^3 >> + ((-3*sqrt(2)*k - 3)*u^2*f^4 + 3*u^2*f^2)*x + u^3*f^3 >> >> for any integer (or, for that matter, non-integer) k. >> This is not reducible over Q except when k = 0, even >> if f is equal to 2^(1/4). >> >> (This expression courtesy of Pari GP, which may explain its >> slight oddity, but I'm not about to bust my brains out >> to clean it up except for replacing 2.8284271247... with 2 * sqrt(2), >> etc.) > > And the important point is that it is only reducible for k=0. > > That shreds the objections where posters have claimed that > reducibility over Q is actually controlling whther or not two of the > a's have a factor that is f, with > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > And it seems that what they were actually depending on was the > possibility of confusion where people falsely assumed that f^2 could > divide out from > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > as a *function* or variable dependent on m, despite it being constant. > > Luckily that strange and false assumption can be easily refuted by > letting f=3, or letting f have any non unit factor in common with 3, > in case someone thinks that f=3 exactly makes a difference. It doesn't. If one equates your P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) the a's by necessity have to have certain properties, especially if one's assuming the a's are all rational. However, since you've defined P(m) as a product of f^2 with something else, one can always compute, say, Q(m) = P(m) / f^2, fairly trivially. The divisor is not dependent on m. (Whether this is useful is not clear.) > > I suggest to readers that the math experts never made the strange > assumption, but might have surmised that others could fall prey to it. > >>2. A requirement I give is that f be coprime to 3, but letting f=3, >>you get that *each* of the a's in the factorization >> 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - >> 3(-1+m3^2 )x u^2 + u^3 3) = >> (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u) >>has a non-unit factor in common with 3, which is a radical factor of >>3, and there's no reason to believe it varies with m, or that it cares >>if the polynomial is irreducible over Q. That actually destroys >>several claims made about using Galois Theory where reducibility over >>rationals is an issue. >>What I want you to understand is that for trained mathematicians, >>these are not issues. However, when it comes to confusing people >>about even relatively basic mathematics, who would be better at it >>than mathematicians? >> >> Non-mathematicians, in some cases. Training tends to wear a groove >> in some people's minds. :-) > > My work is *basic* algebra. It's hard to believe that discussions > could have gone on for so many months with mathematicians, i.e. math > experts by definition, unaware of the truth. > > On the other hand, admitting the truth has a definite social > consequence. > > Given the improbability that math experts were in fact lost on strange > and false math assumptions, where they might have seen a clear benefit > to obscuring the truth by various means, it's more reasonable to > suppose that they acted on social motivations. > >>They need to confuse you here for *social* reasons. >> >> Mathematics is in part a social science; all sciences are, by >> virtue of peer review. > > That is true. I am, however, not a mathematician. I'm an admitted > discoverer for profit, who has made extraordinary math finds. > > Mathematicians may see a social benefit to obscuring my finds from the > world to among other things, preserve their current social structure > and control over mathematics itself. > > Power corrupts after all. And consider how much power mathematicians > have now when it comes to saying what is true in mathematics. Be careful, or you'll have to browse this website: http://zapatopi.net/afdb.html :-) > >>Notice that with f=3, the constant term P(0) = u^2(3x + 3u) = >>3u^2(x+u), so it *still* has a factor that is 3, and that's why I >>always have the condition that f be coprime to 3. >>Here, however, I'm hoping it'll help to point out why that requirement >>is there, and what happens if you ignore it. >>Well then, what are some posters trying to convince you about >> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - >> 3(-1+mf^2 )x u^2 + u^3 f)? >>They're trying to convince you that there is a mathematical limitation >>based on reducibility over Q that determines how f^2 can divide >>through when you have the factorization >> P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) >> >> I think there's some confusion. P(m)'s 0 term is in fact u^3*f^3 >> when multiplied out. > > Nope. Setting m=0, gives P(0) = 3xu^2 + u^3 f = u^2(3x + uf). I did not say P(0). I said P(m)'s 0 term, which probably needs to be clarified to P(m)'s x^0 term -- the constant. If one assumes P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) then the x^0 term is by necessity u^3f^3 -- which turns out to in fact be the case as you've defined P(m) in a certain way. The x^3 term is by necessity a_1a_2a_3. (Since I don't know what the a's are I can't go much further although I can equate the product thereof to the term in front of x^3, if I wished to, and work out the rest of the terms to establish 3 equations in 3 unknowns relating the a's.) Apologies if that wasn't clear. As you can see the f^2 factor waltzes in again; one has to be careful if one drops it from the intermediate computations. This is fine. As for your computation of P(0) -- you've simply left out the f^2 term, from the looks of it. Call that an oopsie. :-) > > What's fascinating about it is that you can see echoes of the > factorization > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > as it's clear that at m=0, two and only two of the a's equal 0, as > that's the only way to get that u^2 in u^2(3x+uf). P(m)'s x^3 term is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3. P(m)'s x^2 term is always zero. P(0)'s x^3 term and x^2 terms are therefore both zero. Therefore exactly two of the a's are 0, as P(0) is a linear equation in x, not a cubic one. I wouldn't really expect any other result in that case. :-) > > It's actually rather fascinating. Which just makes it that much > clearer that mathematicians have been avoiding interesting *math* > instead choosing to focus on obscuring recognition of its validity, > either directly in posts attacking my me or my work, or indirectly by > ignoring my work. > >>and the first question that should come to you is, how could a >>constant factor be constrained by reducibility over rationals? >>Now that question is resolvable, but I know the answer is in my favor, >>so mathematicians are avoiding even letting you know that IS the >>question, and instead those who post work to confuse. >>Now given that I know I have a short proof of Fermat's Last Theorem, >>and that mathematicians have been avoiding dealing with reality, while >>some posters have gotten away with *deliberately* confusing people, >>why would I quit talking about my proof of FLT? >>If you'd found a short proof of Fermat's Last Theorem, would you quit >>talking about it? >> >> Depends on how many demonstrable errors there were in the proof. >> My short perusal through your webpages suggests that you might >> want to clarify your thinking and/or show your work a bit more, as >> you leap from equation to equation without grinding it out in some >> cases. I'd have to look to be more specific at this point. > > Being specific is important, otherwise your comments can't be put into > context. > > Giving what I've seen I'm not willing to just be trusting. I like the way Reagan put it: Trust but verify. :-) > > And the great thing about mathematics is that I don't have to be. True, but one does have to be a little more careful at times. Euclid made at least one error in some of his proofs, and apparently the diagonizalization proof of Cantor proving the uncountability of the reals needs shoring up as well. > > FYI my website is http://groups.msn.com/AmateurMath > > so the proof is out there. > >> You might profit by studying Andrew Wiles' proof as well. I don't >> know if it's on the Web. > > Why? Mostly because he proved Fermat's last theorem. It may come down to whose proof is simpler, but this sort of thing occasionally happens in mathematics: two people, working independently, discover a very similar proof, method, or identity. Check out the history of solving x^3 + ax^2 + bx + c, for example, in http://mathworld.wolfram.com/CubicEquation.html (along with the actual solution :-) ). > > > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, James Harris <3c65f87.0308080723.3d28b8b8@posting.google.com>: > > Not even close. First, in this case, f factors out of your > polynomial 3 times, not 2 times as in the cases you were > considering (f <> 3). This is a special case of no interest > to you or me. It is irrelevant. The Galois argument does not > apply here. No claims based on that argument are 'destroyed'. > Second, only one of the proofs that you are wrong in the cases > where f <> 3 is dependent on Galois Theory. The other proofs are > based on an elementary theorem from algebraic number theory, > which you have previously accepted. All of the proofs *do* > require irreducibility of P(x)/f^2. >>Well, it IS the case that for f=sqrt(2) only *two* of the a's have a >>factor that is sqrt(2), so your claim that it is otherwise is false. >> >> In your main applications, as in your proof of FLT, >> f is an integer. I assumed that here. > > Your assumption is irrelevant to that fact, as you have tried to > confuse people by working to convince that m=0 is a special case. Unfortunately, if one is solving for the a's in P(m) = (a_1x + uf)(a_2x + uf)(a_3x + uf), m=0 *is* a special case as exactly two of the a's become zero. For m != 0, none of the a's are zero. 0 introduces problems in factorization, as you may well appreciate. [snip for brevity] > Previous to that I said that your position requires that people > believe that given > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f) = > > (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) > > that some f^2 divides off as some function of m, or variable dependent > on m. I'm leaving this bit in merely to define P(m), for those who may wander in later. :-) > > Splitting that sentence up with your own comments and then posting as > if I'd said something else is clearly dishonest. > > > James Harris -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > > [.snip.] > >>theory: >> >> THEOREM: If r is a root of a non-monic polynomial >> with integer coefficients, ***irreducible*** over >> the rationals, the r cannot be an algebraic integer. >> >>and conclude that c1 cannot be an algebraic integer. >> >> You must also include the hypothesis that the polynomial is >> primitive. Since nonzero constants are units in Q[x], they are not >> considered nontrivial factors, so the hypothesis must be explicitly >> included. > >That is correct. And it is true that c1 cannot be an algebraic >integer. > >That has not been under debate. > > Which is why I said I was nitpicking: pointing out a minor error > that is well understood. It is true that c1 cannot be an algebraic integer. >It's also NOT under debate as to whether or not given > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > >any a's exist, within the ring of algebraic integers, such that >sqrt(5) is a factor of them in that ring. > > This is rather confused. You have a not at the beginning, a whether > or not after that, and a qualifier any for the a's. It's pretty > close to nonsense. What you are really saying, presumably, is: > > Given a1, a2, a3 algebraic integers such that > > 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1) > > [as a polynomial identity], then none of a_1,a_2,a_3 are multiples > (in the ring of algebraic integers) of sqrt(5). It is true that neither a_1, a_2, nor a_3 has sqrt(5) as a factor ***in the ring of algebraic integers***. > This is also true, and has been established. Yup as I've stated. >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. > > And that's false. I am pretty sure that Dale produced explicit common > factors; but in any case, your claim here is certainly false, since > their product is not coprime to 65. I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit factors in common with 5 in the ring of algebraic integers. > Lemma. Let R be the ring of all algebraic integers, and let a, b, c be > any elements of R. If a and b are coprime to c, then a*b is coprime to > c. > > Proof. We use the characterization of coprime valid for commutative > rings with 1: a and b are coprime in R if and only if there exist x > and y in R such that ax+by = 1. By that definition only *one* of the a's is coprime to 5, but none of them has a factor in common with 5 either. The ring of algebraic integers is really screwed up. For those who don't understand, consider that in the ring of evens, which does not have 1, you can't use that definition of coprime that Arturo Magidin gives, though it is, interestingly enough, true that in fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is not in the ring. However, rather than use dueling definitions or argue about definitions I can simply switch to saying that 2 does not share non-unit factors in the ring of evens with 6. > Since a and c are coprime by assumption, there exist n and m in R such > that an+cm = 1. Since b and c are coprime by assumption, there exist r > and s in R such that br+cs = 1. > > Multiplying both together, we have > > 1 = (an+cm)(br+cs) > = abrn + acns + cbmr + c^2*ms > = ab(rn) + c(ans + bmr + cms). > > Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and > ab*x + c*y = 1. Therefore, ab and y are coprime. QED > > So, assume you were correct and neither a_1, a_2, nor a_3 have ANY > non-unit factors in common with 5 in the ring of algebraic > integers. Then, by the lemma, neither does a1*a_2; and applying the > lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which > clearly has 5 as a nonunit common factor with 5. This contradicts the > assumption that none of a_1, a_2, a_3 have common non-unit factors > with 5 in the ring of algebraic integers. Therefore, your assertion is > false. Well by your definition of coprime NONE of the a's have a factor in common with 5, in the ring of algebraic integers, and you cannot prove that any of them do. Now if you don't want to call that coprime fine. It doesn't change the situation. What I can do is show that with a very quick argument using basic algebra as I've done. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. That is, the w's are now all constant with regard to m and have the same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, and using m=1, f=sqrt(5), gives 65x^3 - 12xy^2 + y^3 which results in b's that are NOT algebraic integers. I've found the Ring of Objects which includes the ring of algebraic integers, and does not have this problem, as the b's are all included in it. The Ring of Objects is the set of all numbers where -1 and 1 are the only members that are both a unit, i.e. factor of 1, and an integer, where no non-unit member is a factor of any two integers that are coprime. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. James Harris ==== >> ...start with a dodecagon (12-gon) [...] make a path that moves >> from vertex to vertex [...] visiting each vertex exactly one time. >> And the path returns to the starting-point. But in this puzzle, >> consecutive vertexes MAY be connected by a segment. >> So, I give a list of nonnegative integers below. As the path is drawn >> (as opposed to after the path is completed), the n_th segment crosses >> a(n) previously drawn segments, where a(n) is the n_th term of the >> integer-list. >> The path starts at 12. And the first segment goes from 12 to 8. >> The list {a(n)}: 0, 0, 1, 0, 2, 2, 0, 3, 2, 2, 5, 2 ... >17 of the 10! paths that start off {12, 8} match that crossing- >count sequence, so solutions seem fairly rare - about 1 per 213000 >paths - but on the other hand, there are perhaps O((n-3)!) possible >crossing-count sequences, so 17 could instead be an unusually high >prevalence, and this case no more rare than thousands of others. >I plan to look at this more and find out, next weekend. >-jiw ... > I wonder which sequences (if any), for the 12-gon, produce one > solution, but are interesting. > If you, or anyone, happens to find such a sequence, then it would be > interesting to post it to sci.math and rec.puzzles as a challenge for > us all. ... For the 12-gon there are 893597 different sequences that begin 12,8, and only 315027 of them belong to unique paths; I don't know if any of those paths are interesting. :) For example, the sequential-crossing-counts list 0 0 0 1 1 1 1 0 6 7 8 2 belongs to a unique path. On the other hand, the list 0 0 0 0 1 2 2 3 4 4 5 4 belongs to 164 different paths, and no list belongs to more than that. Occ.Count #Occ.Count Extension An example count sequence 1 315027 315027 0 0 0 1 1 1 1 0 6 7 8 2 2 179162 358324 0 0 0 1 1 1 1 0 6 0 7 7 3 102670 308010 0 0 0 1 1 1 1 0 0 0 1 7 4 71484 285936 0 0 0 1 1 1 0 0 0 1 1 1 The first column is how many paths a pattern belongs to. The second column is how many patterns belong to that number of paths. The third column is the product of the first two, or sum of products when the first column is a range. The last group of columns shows an example of a count sequence that belongs to the number of paths given in column 1. (In the file, there is also an Eg# column that shows program code numbers of the example count sequences. Also, the file shows similar data for 5, 6, ... 11-gons as well as 12-gons.) -jiw ==== |Both of these theorems would be referred to as the T theorem as if there was |a unique theorem. But these are not the same sentences. Is the T theorem |supposed to refer to one of the similar ways of writing the theorem? No, it refers to the family of expressions that are considered obviously enough equivalent to each other. A similar custom is observed with definitions; if two are obviously equivalent, they're considered definitions of the same concept. I suppose it could be called a family resemblance concept. Others here have referred to logical equivalence, but by itself logical equivalence is too weak a relationship. Two theorems can be logically equivalent in a way that's not obvious enough for them to count as the same fact expressed two ways. I think there's a subjective element to it. Two statements that are considered separate facts in an elementary discussion may be deemed the same fact on a higher level of sophistication. Stabs have been made at rigorously defining a concept of obvious equivalence, but so far as I know they haven't been very successful! This is done sometimes by philosophers for a theory of knowledge. Some of them would like to say that if a person knows a certain statement is true, then they also know that inessential variations on the statement are true. But the idea seems to have serious pitfalls. None of this appears to be a problem in practice, in mathematics. One can always switch to a stricter way of talking in which one refers specifically to particular expressions of the theorem, without worrying about whether the equivalence with another expression is obvious or not. Keith Ramsay ==== >But I suspect the conjecture may imply Bertrand's postulate, i.e. if >A_{1,1} >= 2, A_{1,j} = 1 otherwise, A_{i,j+1} = |A_{i,j} - A_{i+1,j}|, >and A_{i,1} is strictly increasing, then A_{i+1,1} < 2 A_{i,1} > 2 3 5 7 9 15 33 1 2 2 2 6 18 1 0 0 4 12 1 0 4 8 1 4 4 1 0 1 I think this can be extended infinitely to the right and down. Keith Ramsay ==== >I absolutely agree that the conjecture has more to do with the growth >rate than the arithmetic properties of the elements, but what does it >say exactly? Does it say more, or less, than Bertrand's Postulate? > > More, I think. There are sequences that satisfy a Bertrand's Postulate > that won't satisfy the conjecture, e.g, I think, this one: > > 4 5 9 11 19 23 39 47 79 > 1 4 2 8 4 16 8 32 > 3 2 6 4 12 8 24 > 1 4 2 8 4 16 > 3 2 6 4 12 > 1 4 2 8 > 3 2 6 > 1 4 > 3 > > (where the second line seems to be EIS sequence A076736) > Still, to play The Bill's Advocate for a second, could it be that the following is true? (*) If a_0 = 2, a_i is odd for i>0, and a_i < a_(i+1) < 2 a_i, then the left diagonal of the absolute-difference table is always 1. I think this is the point of the claim that the conjecture is a weak statement about the primes, and the example above gets by because it starts with a 4. I see two options. If (*) is true, then the statement on primes follows by Bertrand's Postulate and so has no added arithemtic value. But then the essence is to prove (*), which is perhaps an interesting combinatorial, prime-free, statement, and appears to be non-trivial. If (*) is false, then there is indeed extra content to the statement on the primes beyond Bertrand's Postulate, in which case we go back to my original question of what is it that the statement is saying about the primes, exactly. I couldn't decide one way or another regarding (*), but I may have missed a simple example. - EM ==== Snip--- > Here is more which I have just tested. > > Pursuing this further I tried forcing errors to prove my point in the > above post. > > I found that switching certain primes that are together in the > sequence, trying one switch at a time, either had no effect on the 4 > patterns or had a similar effect like the 277,53 switch. Where prime > 53 is placed just before 277 to create a temporary end to the 4 left > diagonal delta patterns. > > In this column reverse In this column doing the > same > the smaller number to be first reversal will not have any > effect > to create a temporary end to the on the 4 left diagonal delta > patterns. > the 4 left diagonal delta patterns. > > [5,2] reversal drops 1st diagonal pattern. [7,3] reversal change > has > [13,5] like (277,53) switch. no effect on > patterns > [23,7] ditto [43,11] ditto > [53,13] ditto > [73,17] ditto [83,19] ditto > [103,23] ditto > [139,29] ditto > [151,31] ditto > [181,37] ditto > [199,41] ditto [223,43] ditto > [241,47] ditto > [277,53] ditto > > Etc. As these adjoined primes above appear in the sequence to create an error, use the values in the left column only, shift the smaller right prime to the left and the larger left prime to the right. One set at a time for each trial. The above paragraph is a partial correction of the paragraph below. > The column on the left above are those certain adjoining primes in the > sequence when switched produce a temporary end to the 4 left diagonal > delta patterns then after a number of delta rows will return to those > same 4 patterns and continue. Whereas the above column on the right > the pairs when swiched never have any affect on the 4 patterns. Always > shift the smaller left prime to the right and the larger right prime > to the left to create a possible error. > > I believe this to be an important discovery because this sequence > should be in a certain order and if some certain pairs of primes in > the sequence are reversed the 4 left diagonal delta patterns start out > ok but then are temporarily ended when more delta rows are generated. > Then with more terms iterated creating more delta rows the 4 left > delta diagonal patterns eventually reappears and the same 4 patterns > continue on. > > When a known forced error in a certain position caused by switching > two different primes that are next to each other in the sequence will > end the 4 patterns at a certain point. Then at a certain point of more > delta rows will resume this pattern. How can these 4 patterns begin > and then end for a short duration and then restart again and continue > with the same 4 patterns? > What is going on here? > > Can someone duplicate this with another sequence if you only reverse a > certain two consecutive terms and at one point end the 4 left diagonal > patterns then restart them again after more iterations and delta rows? > > It will help in the understanding of this sequence if the delta rows > are created to show the 4 patterns and to create errors to see first > hand what happens. > > I hope I explained clearly of what is going on! > If not,any questions or replies welcome. > > > Dan Dan ==== [snip] > You can do it in O(n log(n)), as long as at most O(n log(n)) pairs are > within eps of each other. > > myproc:= proc(e, n, eps) > # e is a numeric array with index 0 to n-1 > local L,i,j,k,T,delta; > L:= sort([$0..n-1],(i,j) -> (e[i] < e[j])); > T:= Array(1..n); > for i from 1 to n-1 do > for j from i+1 to n while e[L[j]] < e[L[i]]+eps do > delta:= abs(L[j]-L[i]); > T[delta]:= T[delta]+1 > od > od; > for k from 1 to n do > if T[k] = n-k then > printf(Found period of %dn,k); > return > fi > od; > printf(No period found, maybe increase eps); > end; Robert, many thanks, but I forgot to mention that the e entries are in general complex. I am also getting a strange error in the sort subproc for real entries, like: >myproc(e,10,1e-2); Error, (in myproc) sort: 2nd argument must be a boolean valued function. I suspect this probably has to do with the fact that I am using Maple V release 4? > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. ==== lim n->oo (1+1/x)n! / (n! n^(n/x) (n/x)) = 1 Watch as x goes to n. ((1+1/n)n)! / ( n! n^(n/n) (n/n)) ((1+1/n)n)! / ( n! n^1 1) (n+1)! / (n! n) lim ((n+1) / n) = 1 How about ((1+1/x)n)! = n! n^(n/x) (n/x) for x =1? lim (2n)! / (n! n^n n) = 1 The Handbook of Mathematical Functions has an identity for Gamma(2z). gamma(2z) = sqrt(2Pi) 2^(2z-1/2) gamma(z) gamma(z + 1/2) . (2n)! = gamma(2n+1) = 2n gamma(2n) . 2n gamma(2n) / ( n! n^n n ) = 1 . 2 gamma(2n) / (n! n^n ) = 1 . gamma(2n) = (n! n^n ) / 2 . sqrt(2Pi) 2^(2n-1/2) gamma(n) gamma(n+1/2) = (n! n^n ) / 2 . 2sqrt(2Pi) 2^(2n -1/2) gamma(n) gamma(n+1/2) = gamma(n+1) n^n . sqrt(Pi) 2^(2n+1) gamma(n) gamma(n+1/2) / (gamma(n+1) n^n ) = 1 . sqrt(Pi) 2^(2n+1) gamma(n+1/2) / n^(n+1) = 1 The function gamma(n+ 1/2) is sqrt(Pi) 1*3*5*7*...*(2n-1) / 2^n. lim ( 2Pi 2^n) (1*3*5*7*...*(2n-1)) / n^(n+1)) = 1 That looks funny. Do you know any expressions g(n) such that the infinite product of the expression 2n-1, that is f(n) = II_n=1^oo (2n-1), besides f(n)=g(n), that their quotient lim ( f(n) / g(n) ) = 1? Also for x=1: lim n->oo (2n)! / (n! n^n n) = 1 The expression has the term n^n which finds a place in Stirling's equation: lim n->oo n! e^n / n^n sqrt(2Pi) sqrt(n) = 1 n^n = n! e^n / sqrt(2Pi) sqrt(n) (2n)! sqrt(2Pi) sqrt(n) / (n! n! e^n) = 1 (2n)! sqrt(2Pi) sqrt(n) / ( n!^2 e^n) = 1 We have that; lim y->oo sqrt(y Pi/2) y! / (y/2)!^2 2^y = 1 y!/ (y/2)!^2 = 2^y / sqrt(y Pi/2) Setting n=y/2, y = 2n y! sqrt(2Pi) sqrt(y/2) / (y/2)!^2) e^(y/2) = 1 y! / (y/2)!^2 = e^(y/2) / sqrt(y/2) 2^y / sqrt(y Pi/2) = e^(y/2) / sqrt(y/2) 2^2n / sqrt(2n Pi/2) = e^n / sqrt(n) 2^2n / sqrt(n) sqrt(Pi) = e^n / sqrt(n) 2^2n / sqrt(Pi) = e^n lim n-> oo 2^2n / e^n = sqrt(Pi) lim n->oo 2^4n / e^2n = Pi Consider again: lim n->oo (1+1/x)n! / (n! n^(n/x) (n/x)) = 1 Let x be 1/n, approaching zero from the positive side. That isn't an integer. ((1+n)n)! / (n! n^(n^2) n^2) = 1 (n^2 + n)! / (n! n^(n^2) n^2) = 1 Ross ==== >> >>I think it's interesting that there is an expression for (5n/4)! in >>terms of (n!), (n/4), and n^(n/4). >>(5n/4)! = n! (n/4) n^(n/4) >> >> What's the point? Have you ever posted anything anywhere anytime >> that wasn't total bollocks? > > > Why do you care? Shouldn't you be glad that someone does? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen <3c6b9c1e.0308050103.3a81fb73@posting.google.com> <3c6b9c1e.0308051131.170b39f3@posting.google.com> <3c6b9c1e.0308061313.208699b3@posting.google.com> <3c6b9c1e.0308081453.4451c994@posting.google.com> ==== : : :I know (5/4)! / (n! (n/4) n^(n/4)) does not equal one for any finite :integer n. You had (5n/4)! on top before - and you then in fact did say they were equal. :What I have surmised is that the limit as n diverges of :that expression is equal to one. That's not what you said, so it was hardly to be inferred - but anyway the ratio does not converge to 1. Indeed, if you do mean (5/4)! on the top (by which one presumes you mean Gamma(9/4) then clearly the limit is 0. On the other hand, if you really mean (5n/4)! on the top then the expression diverges - it grows exponentially. I might as well snip the rest of the discussion to preserve my sanity. ==== I know that ( kg = k B U ) for geodesic curvature formula. ( operation of k B U inner product between character ) now, my question is how to derive to ( kg = k B U ) kg : geodesic curvature k:curvature B : unit binormal vector U : unit normal vector please, sir. thanks to work out my problem. sorry, I have committed an error about ASCII. ==== > I know that ( kg = k B U ) for geodesic curvature formula. > > ( operation of k B U inner product between character ) > > now, my question is how to derive to ( kg = k B U ) > > kg : geodesic curvature > k:curvature > B : unit binormal vector > U : unit normal vector > > please, sir. > thanks to work out my problem. > > sorry, I have committed an error about ASCII. Just a point about notation: if B and U are vectors, for the inner product of B with U, it's perfectly ok to type: B.U David Bernier ==== oh....that's good idea. ==== >I was thinking of generation by string concatenation, where >a^3=b^2=c^2=d^2=1 and five re-write rules are needed to create S4. Two >permutations, two tau matrices, or two gamma matrices also create S4. >You say A5 needs 3 generators, and he replied > I didn't say that A5 needs 3 generators. I said that A5^20 (the direct product > of 20 copies of A5) needs 3 generators. A5^19 is still a 2-generator group. 1) I apologise for a careless miss-reading. 2) You expose an error in my (self-taught, not programmed = not understood) account of groups and loops. Is relators the accepted name for my string generators? 3) Do you imply that all groups smaller than A5^20 have 2 generators? Roger (still learning at 74) Beresford. ==== >I was thinking of generation by string concatenation, where >>a^3=b^2=c^2=d^2=1 and five re-write rules are needed to create S4. Two >>permutations, two tau matrices, or two gamma matrices also create S4. >>You say A5 needs 3 generators, and he replied > I didn't say that A5 needs 3 generators. I said that A5^20 (the direct product >> of 20 copies of A5) needs 3 generators. A5^19 is still a 2-generator group. 1) I apologise for a careless miss-reading. >2) You expose an error in my (self-taught, not programmed = not >understood) account of groups and loops. Is relators the accepted >name for my string generators? If a group G is generated by elements x1, x2, ..., then a relator for G in these generators is a string or word in the symbols xi and xi^-1 which evaluates to the identity in G. For example, S3 is generated by x=(1,2,3) and y=(1,2), and (xy)^2 = xyxy is a relator for S3 in these generators. >3) Do you imply that all groups smaller than A5^20 have 2 generators? No, of course not! There is a group of order 8 (Z2 x Z2 x Z2) which requires 3 generators. But A5^n is a 2-generator group for all n < 20. (And A5^n can be generated by 3 generatorsiff n <= 1668.) I was just offering A5^19 and A5^20 as examples for which it could be difficult to decide whether they are 2-generator or 3-generator groups without specialised knowledge. There may well be smaller equally difficult examples. Derek Holt. ==== > If a group G is generated by elements x1, x2, ..., then a relator for G in > these generators is a string or word in the symbols xi and xi^-1 which > evaluates to the identity in G. > For example, S3 is generated by x=(1,2,3) and y=(1,2), and (xy)^2 = xyxy > is a relator for S3 in these generators. Is the term relator rather than relation now accepted usage? I see no etymological reason for preferring relator, which presumably should be a person or object that relates to something, eg Derek Holt is my relator in this matter. -- Timothy Murphy tel: +353-86-233 6090 ==== > If a group G is generated by elements x1, x2, ..., then a relator for G in >> these generators is a string or word in the symbols xi and xi^-1 which >> evaluates to the identity in G. >> For example, S3 is generated by x=(1,2,3) and y=(1,2), and (xy)^2 = xyxy >> is a relator for S3 in these generators. Is the term relator rather than relation now accepted usage? >I see no etymological reason for preferring relator, >which presumably should be a person or object that relates to something, >eg Derek Holt is my relator in this matter. The distinction that's common among combinatorial group theorists of a topological bent is that a relation is an equation between two words, while a relator is a word (by intent, one which will be put into an equation with the identity on the other side). For groups, this seems really to be a matter of taste; but for other structures (like non-cancellation semigroups), obviously not every relation can necessarily be replaced by an equivalent relation which sets a word equal to the identity (in fact, there may be no identity in the structure being considered!). was using the = of explication, not missaying (xy)^2 = xyxy is a relation (which would be silly in the context of groups; but not that of non-associative algebra, for instance). Sometimes I yearn for a standard typographical device to distinguish the = of explication (or derivation) from the = of equality, just as := and =: have come to be (fairly) standard notations for the = of assignment (or definition). Lee Rudolph ==== > I am looking for a problem referenced > by Martin Gardner, Chapter 11 in his > collection entitled The 2nd Scientific > American Book of Mathematical > Puzzles & Diversions, >Simon and Schuster, 1961. To quote Gardner precisely: Lewis Carroll was fond of inventing quaint and enormously complicated problems of this sort. Eight are to be found in the appendix of his _Symbolic Logic_. One monstrous Carrollian problem (involving 13 variables and 12 premises from which one is to deduce that no magistrates are snuff-takers) was fed to and IBM 704 computer by John G. Kemeny, chairman of the mathematics department at Dartmouth College. This doesn't really imply that the magistrate/snuff-taker problem is in Symbolic Logic. John Robertson ==== >> [.snip.] >>>Well I did find a problem with the definition of the object ring that >>>I'd given, and I've updated it. >>There have been at least two changes in recent memory, one sort of >>announced, one done in silence. And then there was another change in >>the past 36 hours, presumably what you are refering to here. >> >> And yet another unannounced change has now occured. >> >> Yesterday, the definition at >> >> http://www.msnusers.com/AmateurMath/objectmathematic.msnw >> >> was: >> >> >>The Object Ring is the set of all numbers where any member that is a >>unit, i.e. factor of 1, and its multiplicative inverse are units in >>all possible commutative rings in which either and all integers are >>members, and where no non-unit member a is a factor of any two >>integers that are coprime. >> >> >> The Object Ring is the set of all numbers where 1 is the only member >> that is both a unit, i.e. factor of 1, and an integer, where no >> non-unit member is a factor of any two integers that are coprime. >> >> You are still being sloppy in saying set of all numbers. I suspect >> that you mean to restrict yourself to complex numbers, if not >> ALGEBRAIC numbers, and to give this set the inherited structure. If >> this is the case, then since -1 is both a unit and an integer in any >> subring of the complex numbers, it looks like you have nothing, yet >> again. Hey, you're right. Good catch. I'll update the page. remind you: >> Now, assuming you meant to say 1 and -1 are the only elements which >> are both units and integers, then you still must prove that Object >> ring under this definition specifies a unique such object. >> Presumably, you want to say largest subring of the complex numbers >> such that..., because otherwise, the integers are The Object Ring, >> but so is any subring of the ring of all algebraic integers. It would >> be of paramount importance to make sure that it defines a unique >> thing, if you are going to call refer to it by using the singular >> >> I am also pretty certain that this definition includes way too many >> things that you do not want. But it is obvious that once again all you >> are doing is trying to fix, by fiat, the problems that plagued your >> original proof of two years ago. >> >> I must, however, confess that I am flabbergasted at your brilliance: >> here we have what, by your own account, is the key, central, germain, >> touchstone, concept of your approach. And even though you have been >> able to change the definition in significant ways over the past 8 >> months, yet your proof is so solid that changing this key definition >> does not require you to change even a single word of the rest of your >> developement to take into account these changes. Truly, a work of >> genius. >James Harris ************************ David C. Ullrich ==== >> [.snip.] >>>Well I did find a problem with the definition of the object ring that >>>I'd given, and I've updated it. >>There have been at least two changes in recent memory, one sort of >>announced, one done in silence. And then there was another change in >>the past 36 hours, presumably what you are refering to here. >> >> And yet another unannounced change has now occured. >> >> Yesterday, the definition at >> >> http://www.msnusers.com/AmateurMath/objectmathematic.msnw >> >> was: >> >> >>The Object Ring is the set of all numbers where any member that is a >>unit, i.e. factor of 1, and its multiplicative inverse are units in >>all possible commutative rings in which either and all integers are >>members, and where no non-unit member a is a factor of any two >>integers that are coprime. >> >> >> The Object Ring is the set of all numbers where 1 is the only member >> that is both a unit, i.e. factor of 1, and an integer, where no >> non-unit member is a factor of any two integers that are coprime. >> >> You are still being sloppy in saying set of all numbers. I suspect >> that you mean to restrict yourself to complex numbers, if not >> ALGEBRAIC numbers, and to give this set the inherited structure. If >> this is the case, then since -1 is both a unit and an integer in any >> subring of the complex numbers, it looks like you have nothing, yet >> again. Hey, you're right. Good catch. I'll update the page. Yeah, I noticed you said a poster had pointed out the rather silly error. I also notice that you continue to state all numbers without specifying whether you are considering only algebraic numbers, or in fact all numbers. If you are considering all numbers, then what you have need not even have operations defined: it would include actual polynomials in any number of incompatible variables, and some elements of positive characteristic. If you are going to update the page, why not bother to fix the definition COMPLETELY? Current definition states: The Object Ring is the set of all numbers where -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. Coprime where? In the final product, or in the ring of integers? It's possible that two integers are coprime in a larger ring without being coprime in the integers. The final clause is empty if you mean coprime in the integers: in ANY subring of the complex numbers, an element which divides two integers which are relatively prime in the ring of integers must be a unit. Putting it in the definition only obscures the latter. Is your object ring unique? Is it a ring? You are using the singular complex numbers which satisfies this condition: in fact, at least EVERY subring of the algebraic integers does, and more besides, like Z[pi], for example. And you only define it as a ->set<-, but you also claim it is a ring. Since you are not specifying that it is a subring of the complex numbers, what are the ring operations? Does your ring contain multiplicative inverses for algebraic integers which are NOT integers? Let f(x) be any monic irreducible cubic with integer coefficients, |f(0)|>1, and assume moreover that its discriminant is not a square in Q (so that the extension given by a single root is not normal). Does your ring contain 1/r? It need not cause any integer other than 1 and -1 to become invertible, but it would still possibly cause problems with your congruences. And let me repeat what I said before, which you removed without addressing or acknowledging: Now, assuming you meant to say 1 and -1 are the only elements which are both units and integers, then you still must prove that Object ring under this definition specifies a unique such object. Presumably, you want to say largest subring of the complex numbers such that..., because otherwise, the integers are The Object Ring, but so is any subring of the ring of all algebraic integers. It would be of paramount importance to make sure that it defines a unique thing, if you are going to call refer to it by using the singular I am also pretty certain that this definition includes way too many things that you do not want. But it is obvious that once again all you are doing is trying to fix, by fiat, the problems that plagued your original proof of two years ago. I must, however, confess that I am flabbergasted at your brilliance: here we have what, by your own account, is the key, central, germain, touchstone, concept of your approach. And even though you have been able to change the definition in significant ways over the past 8 months, yet your proof is so solid that changing this key definition does not require you to change even a single word of the rest of your developement to take into account these changes. Truly, a work of genius. ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== In sci.math, James Harris <3c65f87.0308081016.460d766c@posting.google.com>: >> In sci.physics, James Harris >> <3c65f87.0308061529.3dc1cce5@posting.google.com>: > > >However, you still seem to not understand what a mathematical proof >>is. >>It is a perfect argument that begins with a truth and proceeds by >>logical steps to a conclusion which then must be true. >>So it's impossible to find an error in a proof. >> >> Attempt at counterexample: I claim to prove that 1 = 2. > > Which shows that like the poster I was answering before you fail to > understand what a mathematical proof is. A mathematical proof is a > *perfect* argument, so no counterexample exists. A mathematical proof is a sequence of steps. If those steps are performed correctly and all assumptions are accounted for, then it's a reasonably good proof. Perfect? I don't know how to measure perfect. Some interesting things happen when one changes the assumptions though; the classical one is arguably Lobachevksy attempting to find an absurdity by replacing the parallel postulate (an axiom), and instead developing an entirely new geometric form, hyperbolic geometry. > Possibly you've been > programmed by social conventions where claims of proof are called > proofs. > > But it's like if I say I have proof that you are a dog. > > My *saying* I have proof does not create a proof. > > So if a person says they have proof you're a dog, does that prove that > a proof can be in error? It proves people can be in error when claiming proofs. > > No, it's just that they're in error, and do not have proof you're a > dog. > > If they did have proof you're a dog, then you'd be a dog. > >> Let a = b = 1. >> >> Then a^2 = ab. >> >> a^2 - b^2 = ab - b^2. >> >> (a+b)(a-b) = b(a-b) >> >> Dividing by a-b we get > > And given that a=b that's an attempt at dividing by zero in the > classic example. Exactly. > > This example only seems to work by *human* error as human beings see > 'a' and they see 'b' and think, different things, despite them being > defined to be the same at the beginning. > > Doing the substitution a=b, ignoring the 1 for the moment gives > > a^2 = a^2 > > a^2 - a^2 = a^2 - a^2 > > (a+a)(a-a) = a(a-a) > > Dividing by a-a would be an error, as a-a=0. > > Now using the full substitution of a=b=1, you have > > 1^2 = 1^2 > > 1^2 - 1^2 = 1^2 - 1^2 > > (1+1)(1-1) = 1(1-1) > > and dividing by 1-1 would be an error as it equals 0. Exactly. It's not a proof, merely a claim at one. I claim that you claim to have a proof of Fermat's Last Theorem. This is one reason why peer review is so important; while it doesn't totally eliminate error, it at least allows for more eyeballs to check for errors in the proof. The author, presumably, then corrects those errors and republishes, or abandons the effort. (I would think abandonment would be extremely rare, unless the author, say, dies or something. :-) ) > >> a+b = b. >> >> 1 = 2. >> >> QED. >> >> This is of course a claim of a proof only, and the error is >> (hopefully) easily spotted. Many other claims have far more >> obscure errors. > > Given a claim of proof, you can test it by determining if the argument > begins with a truth, and proceeds by logical steps to a conclusion > which then must be true. > > Unfortunately many people say proof when they mean claim of proof, > so a lot of people believe that a math proof can be wrong, but they > wouldn't believe that proof in any other context can be wrong, as then > they realize it simply wasn't proof. > > If you have proof that someone committed a crime, then you have proof. Crime commissions are in the legal realm. Of course one can set up interesting logical problems a la Sherlock Holmes, if one wishes. The legal realm merely requires proof beyond a reasonable doubt. Reason is used in both proofs and criminology. > > If it's not proof, then it's not proof. > > That when math is stuck next to proof some people suddenly think > something changes is problematic, and may be why some can accept the > possibility of error in a math proof. I've found at least two errors in your proof submission. The latter one is fatal; I can't work around it. The former's effect on your proof is unclear. Please fix. :-) > >>However, a would-be discoverer *can* make errors in describing a >>proof, or think they see a proof where none exists, and potentially >>that can be found out by starting at the beginning of the proof, and >>proceeding through it checking each step to make certain that it is a >>logical one. >> >> I submit you have a claim. Has it at least been peer-reviewed? :-) > > That's an interesting question and the answer is, I don't know. I > have sent my work to math journals, and a paper is currently at a math > journal, and I'm waiting to hear from them. Well, we'll see; presumably they are reviewing it. :-) > > Have any of the other journals I've sent papers to actually > peer-reviewed? > > I don't know. > > What is important to remember though, is that math society is a > society, and I've already outlined the weird notion that a proof can > be in error, where people actually believe that a *math proof* can be > in error, when what they should realize is that a claim of proof can > be in error. > > Math proofs are perfect, just like any other proof that actually is a > proof. > > People, on the other hand, can say proof when in fact they don't > have a proof. And how do we know a proof is perfect? > > Just like someone can say you are a dog, claim they have proof, but it > be nonsense. > > Hopefully I've cleared that issue up, and I've gone on about it > because it was being questioned!!! Questioning is part of the peer review process, methinks. Obviously in Usenet the questioning is highly informal (and occasionally peppered with insults, especially if alt.syntax.tactical gets involved :-), or irrelevancies). I don't have a clue as to how the more formal mathematical or scientific peer review process works. I don't even know how the PhD peer review process works, although I suspect one has to stand and defend his papers against the attacks from his peers, presumably in a forum of, maybe 6 to 12 peers. > > Now here's a math proof. Those who doubt that fact can believe it's a > claim of proof, but it's verified to be a proof by tracing the > argument out. > > In this case, I begin with an expression. The expression exists, so > that is the truth from which you start. > > Consider, in the ring of algebraic integers, I'm assuming here that you are using a definition similar to http://mathworld.wolfram.com/AlgebraicInteger.html which defines an algebraic integer r (of degree N, if r satisfies no lower equation) as a solution to a polynomial x^N + a_{N-1}x^{N-1} + ... + a_1x + a_0 = 0. This gets a little weird, as algebraic integers of degree > 2 may not always be factorable in an elegant fashion. We shall proceed ... carefully. :-) > > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f). > > That is, I have the identity which defines P(m) in terms of various > symbols, and it's all in the ring of algebraic integers, which means > that the symbols can only represent numbers that are algebraic > integers. > > Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization > > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > > where w_1 w_2 w_3 = f, and > > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), > > and at m=0 > > P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), > > so two of the b's must equal 0, which means > > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > > which is > > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) > > proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves > b_3 = 3. Be careful here. You have proven that P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) but you have *not* constrained u. If u is 0 things get ridiculous and relatively uninteresting. If u is not 0 one can compute P(0)/(u^2f^2) = (b_3 w_1 w_2 x + uf) = 3x + uf and you've actually proven that b_3 w_1 w_2 = 3, if x != 0 (which is also relatively uninteresting). Such potholes are easily avoided of course (usually), but your proof's real problem here is you leapt to the wrong conclusion as you left out b_3. Since (3 - 2sqrt(2)) * (3 + 2sqrt(2)) = 1 over the algebraic integers, one also has to be careful about other conclusions regarding this product as well. I'm not sure there's a smallest algebraic integer > 0. (It's easy to prove the set of algebraic integers clusters towards 1+ by considering the equation y^n + 2; as n -> oo the primary root y = (2^(1/n)) tends to 1, and one gets an infinite subset. By replacing y = (x + w) where w is any integer and grinding out the resultant equation, one can show that the set of algebraic integers clusters around any integer, including 0+. Therefore there isn't a smallest positive algebraic integer.) I don't know if this is a fatal flaw, but it is a problem. > > Now that was a lot of steps, but each was a logical one. > > First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are defined > by the factorization > > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > > then I set m=0, and used the definition of P(m) to get P(0). > > That told me that at m=0 two of the b's are 0, because then > > P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), > > where the u^2 couldn't get there unless two of the b's are 0. > > Then using that result I get from > > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > > that > > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) > > and multiplying through by w_1 w_2 I have > > P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) > > which with > > P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), > > tells me that w_1 w_2 = 1, when m=0. Again, where did b_3 go? > > Essentially objections to how f^2 divides off now come down to > claiming that the w's are functions of m, but consider that w_1 w_2 = > 1, when m=0, if f is coprime to 3. > > Now I'm focusing on what has been revealed to be an area of confusion. > Apparently some people believe that when I divide off f^2 that it can > divide off as a *function* of m, so that m=0 might be a special case. It is possible to define f(x) = K, where K is an arbitrary constant. Usually such functions are relatively uninteresting. Therefore I fail to see why this is even a problem, let alone why people would object thereto. Of course your definition P(m) would more properly be defined P(m,f,u,x), in certain contexts. In the computer engineering realm you've basically defined a function/algorithmic procedure P(m) with one parameter and three globals, one of which you're attempting apparently to solve for (x). This isn't a real big problem in mathematical circles, though. > I'm now starting the argument to address that belief by noting again > that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when f > doesn't have 3 as a factor. > > But that was an arbitrary choice, so let f=3. > > That is, I *said* f is coprime to 3 but in considering this > possibility it's worth it to relax that restriction and now consider > what would happen if it equals 3. > > Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. > > Seeing that is as simple as looking at > > P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > > 3(-1+mf^2 )x u^2 + u^3 f > > with f=3 as then you have > > P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 - > > 3(-1+m3^2 )x u^2 + 3u^3 > > so *every* coefficient has a factor that is 3, as you can tell by > looking. > > So with > > P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > > each of the b's and each of the w's has a factor that is 3^{1/3}, If we grind out this mess with f = 3, we get P(m)/9 = (81 m^3 - 27 m^2 + 3m)x^3 - 3(-1+9m)xu^2 + 3u^3 so we can conclude that b_1 b_2 b_3 = (81 m^3 - 27 m^2 + 3m). and w_1 w_2 w_3 = 3u^3. We *cannot* conclude that all b's have an algebraic integer factor 3^{1/3} without additional information; for all I know b_1 = (81 m^3 - 27 m^2 + 3m) and b_2 and b_3 are 1 -- unlikely, admittedly, but theoretically possible. Ditto for the w's. You might as well divide P(m)/27 and compute b'_1, etc. and w'_1, etc., as well. Whether this is useful is not clear to me. This *is* a fatal flaw. [rest snipped, as it requires reanalysis] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== In sci.math, Randy Poe <585ab5d8.0308081312.74fc47e1@posting.google.com>: >>In sci.physics, James Harris >><3c65f87.0308061529.3dc1cce5@posting.google.com>: >> >> > However, you still seem to not understand what a mathematical proof > is. > > It is a perfect argument that begins with a truth and proceeds by > logical steps to a conclusion which then must be true. > > So it's impossible to find an error in a proof. >>Attempt at counterexample: I claim to prove that 1 = 2. >> >> Which shows that like the poster I was answering before you fail to >> understand what a mathematical proof is. A mathematical proof is a >> *perfect* argument, so no counterexample exists. > > Way too wide an opening there James. The obvious question is > why do counterexamples to your proofs abound? Could it > mean (gasp) your proof is less than perfect? Please describe one of these counterexamples; I'm mildly curious. Admittedly, my previous post details at least two flaws; these presumably can lead to some interesting counterexamples. Or one can postulate f = u = 0 and generate some uninteresting ones. :-) I've also noted that (3 - 2*sqrt(2)) * (3 + 2*sqrt(2)) = 1 is an interesting product of algebraic integers as well. Obviously this can lead to some weird problems, as one can prove positive integers have unique factorizations, but algebraic integers do not: 1 = 1 * 1 = (3 - 2*sqrt(2)) * (3 + 2*sqrt(2)) = (4 - sqrt(15)) * (4 + sqrt(15)) = ... = (n - sqrt(n^2 - 1)) * (n + sqrt(n^2 - 1)) etc. Mr. Harris does jump to some interesting conclusions, though; how does a * b * c = 3 in the algebraic integers yield the requirement that each of a, b, and c has a factor of 3^(1/3) (an algebraic integer of degree 3)? This particular one puzzles me. > > - Randy -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > [.snip.] > The Object Ring is the set of all numbers where 1 is the only member > that is both a unit, i.e. factor of 1, and an integer, where no > non-unit member is a factor of any two integers that are coprime. > set of all numbers rational? real? algebraic? algebraic integers? complex? Are the integers included (implied elsewhere on his webpage but never explicitly shown)? This is the best I can translate his definition. Let S be the object ring. 1 e S, -1 (not e) S If n =/= 1 and n e Z then if n e S then 1/n (not e) S If A,B e Z and (A,B) = 1 and there does not exist x e S | xn = 1, if there exists a, b e S | an = A and bn = B then n (not e) S So the question is: what sort of structure does this set have? As defined, it is obviously NOT a ring (-1) is excluded, so is the set closed under addition and multiplication? Is it commutative? -Tralfaz ==== > > Does your ring contain multiplicative inverses for algebraic integers > which are NOT integers? Let f(x) be any monic irreducible cubic > with integer coefficients, |f(0)|>1, and assume moreover that its > discriminant is not a square in Q (so that the extension given by a > single root is not normal). Does your ring contain 1/r? It need not > cause any integer other than 1 and -1 to become invertible, but it > would still possibly cause problems with your congruences. There are also examples in quadratic fields, e.g. see below >>Jim Propp asks: > > Does there exist an algebraic number w that is NOT an algebraic integer, > but that nevertheless has the property that the only rational numbers > in Z[w] are the rational integers? >> w = 1/(3 + sqrt(2)). >> w is integral at P = (3 - sqrt(2)), but not at P' = (3 + sqrt(2)). >> it follows that Z[w] is integral everywhere, excepting at P', >> and hence Q / Z[w] is integral everywhere, including at 7. > > This is a nice argument that is worth explaining in more detail. > Among other things, it illustrates the power of localization. > > The Key Point is v_p(x) < 0 for some prime p in Q > => v_P(x) < 0 for *every* prime P lying over p. > > Let F = Q(w) and let J be the ring of integers in F. We want to > have v_p(x) >= 0 for every x in Z[w] and every rational prime p. > We claim to achieve this, it suffices to choose, for each rational > prime p, one prime P in J lying over p, and ensure that v_P(w) >= 0 > for all these P. For then v_P(x) >= 0 for all x in Z[w], and the > above Key Point will force v_p(x) >= 0 should x happen to be in Q. > > On the other hand, we don't want to make v_P(w) >= 0 for *all* > primes P in J, for then w would be an algebraic integer. So now > it's clear what we need: we need to choose a prime p that splits > in J and choose w such that v_P(w) < 0 for some prime lying > over p and v_P'(w) >= 0 for some other prime P' over the same p. > To find the simplest example, take the UFD: F = Q(r), r^2 = 2, > and note that 7 splits into (3+r)(3-r) in J. Then it is > easy to check that w = 1/(3+r) satisfies v_P(w) >= 0 for > all primes P in J except P = (3+r)J, so it does the trick. See also other posts in the thread containing the above post http://groups.google.com/groups?selm=74k4r1%24sa0%40schubert.mit.edu -Bill Dubuque ==== Hey, > f(x) = 2 ^ ([log(x * 8) / log (2)] - 1) / 4 yeah this works - also got this one working (for PHP) $newvalue = pow(2,floor(log(16*$value)/log(2))-2) / 4; > int k, z, i ; > > k = (log((float) x) + 0.0000001) / log(2.0) - 1; > > z = 1 ; > for (i = 0; i < k; i++) > z = 2 * z ; > > Then z should be the numerator of the expression on the right. Nikolai Onken ==== Well true, but I ment it more like f(3/8) = 1/4 so I was looking for a function which returns the right values which would be for example: f(x) = 2^( floor(ln(16*x)/ln(2)) - 2 ) / 4 Nikolai ==== First of all I want to say I am no mathematician. So anyone who is nice enough to respond to this, please explain in layman's terms (if you start using mathmatical jargon beyond the basic stuff, please define it first or I probably won't know what you're talking about). I have recently become very interested in solving what I have learned is called a Diophantine equation, that is, an equation where the solutions are just integers. After researching this on the internet, I have also found that my equation is very similar to an equation called the Pell equation, but not exactly. The Diophantine equation I want to know how to solve is: x^2 - y^2 = N Now, from reading about the Pell equation, which is x^2 - Dy^2 = 1, it is always required that D is not a perfect square. Well, in my case the D is a perfect square. But then the 1 is not a 1, but an N, meaning, I want to give this Diophantine equation solver a positive integer N and it give me the possible pairs (x,y) that satisfy that equation. By the way, I only want values of x and y that are >= 0. For example, if N=7, then the possible solution pairs are just one, namely (4,3). Has anyone studied this? Any internet sites I could be directed to? It seems like such a simple obvious thing that I would think someone would have looked into it. At the webpage http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html I found this: The more complicated equation x^2 - Dy^2 = c can also be solved for certain values of c and D, but the procedure is more complicated (Chrystal 1961). However, if a single solution is known, other solutions can be found using the standard technique for the Pell equation. Since I always know one solution, from this I would think all the solutions can be found using this Pell equation technique which I wish someone would explain to me. (Hmm, after studying that page more thoroughly, I now believe that technique will not work for D equals 1, sigh) ==== > The Diophantine equation I want to know how to solve is: > > x^2 - y^2 = N > > Now, from reading about the Pell equation, which is x^2 - Dy^2 = 1, it > is > always required that D is not a perfect square. Well, in my case the D is > a > perfect square. But then the 1 is not a 1, but an N, meaning, I want to > give this Diophantine equation solver a positive integer N and it give me > the > possible pairs (x,y) that satisfy that equation. By the way, I only want > values of x and y that are >= 0. For example, if N=7, then the possible > solution pairs are just one, namely (4,3). Has anyone studied this? Any > internet sites I could be directed to? It seems like such a simple > obvious thing that I would think someone would have looked into it. Your equation is much simpler than Pell's equation, and you are getting into unnecessary complication by considering the latter. Your equation is solved by simple factorisation: (x - y)(x + y) = N. So x - y = u, x + y = v where N = uv. Note that u and v are both even, or both odd. So N must be divisible by 4 if it is even (which is the only condition for the equation to have a solution). -- Timothy Murphy tel: +353-86-233 6090 ==== > The Diophantine equation I want to know how to solve is: > > x^2 - y^2 = N > > By the way, I only want values of x and y that are >= 0. For example, > if N=7, then the possible solution pairs are just one, namely (4,3) This solution is far easier than you suspect. Rewrite it as (1) (1) (x+y)(x-y) = N and then you can see that solutions for (x+y) and (x-y) the two factors of N, depend upon the prime factorization of N. For N prime, which is the case in your example we have N = 1*7. This gives (x+y)=7 and (x-y)=1 I trust you can see why it is NOT the other way around. We then have two simultaneous equations in x,y to solve. Or (2) x+y = 7 x-y = 1 ------- 2x = 8 (by adding) x = 4 and 4-y = 1 4-1 = y 3 = y which is the pair (4,3) which you found. If N is composite, then things get much more interesting, since there are pairs of values, depending upon the factorization. For example with 24 we have 24 = 1*24 2*12 3*8 4*6 and solutions of (x,y) are: x=25/2, y=23/2 x=7 y=5 x=11/2 y=5/2 x=5 y=1 Some are rational, others are integer, so multiple solutions exist. I hope this helps. *SPECIAL NOTE: We don't have to stay with integer factorizations of N, say 24 = 3/2 * 16 then (x,y) = (35/4, 29/4) is a solution. This will work for any 2 rational numbers whose product is N. ==== >The Diophantine equation I want to know how to solve is: >x^2 - y^2 = N Other posters have shown you how to solve this equation. Write N=ab in all possible ways (including those with a or b negative), solve x-y=a, x+y=b for x, y, and keep the solutions in positive integers x, y. For a discussion applicable to any x^2 - Dy^2 = N, when D is a square, see the section Delta > 0 is a square in the file Solving the equation ax^2 + bxy + cy^2 + dx + ey + f = 0 at http://members.aol.com/_ht_a/jpr2718/ For your equation, Delta = 4. >At the webpage http://mathworld.wolfram.com/DiophantineEquation2ndPowers.html >I found this: The more complicated equation x^2 - Dy^2 = c can also be >solved >for certain values of c and D, but the procedure is more complicated >(Chrystal 1961). Matthews and Mollin have recently found a method (actually known to Lagrange, but forgotten), that is hardly more complicated than the methods for solving the x^2 - Dy^2 =+-1 equation, for D not a square. See Solving the generalized Pell equation, and other files, at the website above. In addition to the Matthews method, which is probably the best, this file summarizes the method given in Chrystal, also due to Lagrange, and called Lagrange's system of reductions. Both of the two files cited discuss ways to generate all solutions to any of these equations. John Robertson ==== > > Your equation is much simpler than Pell's equation, > and you are getting into unnecessary complication by considering the latter. > Your equation is solved by simple factorisation: > > (x - y)(x + y) = N. > > So > x - y = u, x + y = v > where > N = uv. > Note that u and v are both even, or both odd. > So N must be divisible by 4 if it is even > (which is the only condition for the equation to have a solution). Oh, I forgot to mention. I am only concerned with odd N. Now I don't understand your response (which is the only condition for the equation to have a solution). I just showed you the solution to N = 7 ('solution' meaning values for x and y that satisfy the equation). What I am after is the solution(s) for any odd N. How is that determined? Can the Pell equation solution be used? ==== > >An interesting class of integral domains are the rings F[x] >of polynomials over a finite field F, >and the finite algebraic extensions of these. >These rings behave very much like number rings; >almost every result for number rings holds also for these >function-rings. In particular, they have finite ideal class groups. >two ideals I,J in R are said to be in the same ideal class >if aI = bJ for some non-zero a,b in R. >It is easy to show that the ideal classes form an abelian semigroup. Indeed a monoid. I notice IJ as given above isn't always the ideal intersection of I and J but an ideal smaller than the intersection. >The condition you give is the condition that the ideal classes >should form a group. >I believe that a necessary and sufficient condition for this >is that R should be a Dedekind domain, >ie ever ideal in R is uniquely expressible as a product of prime >ideals. That may be possible. Likely then R has only finite many ideals. Now as R itself is an ideal, it'd be a unique product of ideals making all the ideals of R, products of prime ideals. The important thing is a useful collection of rings for which UFD <-> PID to give intuitive reason why the ideal class group size is a measure how far a ring is from being a UFD. >An alternative way (neater in my opinion) to view this >is to consider fractional ideals, ie subsets of the >quotient-field k of R of the form xI, where I is an ideal in R >and x is in k (ie x = a/b with a,b in R). >Then we set I^{-1} = {x in k: xI < R}; >and the condition you give is equivalent to I I^{-1} = R. But 1 in I^-1, so where's the ideal of R? If I^-1 has finite number of denominators, then it makes some small sense. If I^-1 is finite, I^-1 = { n1/d1,.. nj/dj }, then the ideal I^-1 = (n1,.. nk) and I I^-1 = (d1*..*dk) ??? What do I do about 1 in I^-1 ? ---- ==== > >An alternative way (neater in my opinion) to view this > >is to consider fractional ideals, ie subsets of the > >quotient-field k of R of the form xI, where I is an ideal in R > >and x is in k (ie x = a/b with a,b in R). > >Then we set I^{-1} = {x in k: xI < R}; > >and the condition you give is equivalent to I I^{-1} = R. > > But 1 in I^-1, so where's the ideal of R? If I^-1 has finite > number of denominators, then it makes some small sense. > If I^-1 is finite, I^-1 = { n1/d1,.. nj/dj }, then the ideal > I^-1 = (n1,.. nk) and I I^-1 = (d1*..*dk) ??? > What do I do about 1 in I^-1 ? I^{-1} is a fractional ideal, so it can contain 1. This just means that I^{-1} contains R. The ideal (or fractional ideal) I is said to be invertible if I*I^{-1} = R. This is just your condition. The fractional ideals form a group if and only if every ideal is invertible. Integral domains with this property are called Dedekind domains, and they include number rings (rings of integers in number fields) but also other rings (eg finite algebraic extensions of the ring F[x] of polynomials over a field F). The principal fractional ideals xR (where x is in k) form a subgroup of the group of fractional ideals; and the ideal class group is the quotient-group. -- Timothy Murphy tel: +353-86-233 6090 ==== Buddhism is the only religion that is compatible with science and mathematics. It is a way of thinking which oroginated in India in the is paying the price for this till today. If India was Buddhist, the Industrial revolution would have happened in India in 1000 AD. ==== > Buddhism is the only religion that is compatible with science and > mathematics. It is a way of thinking which oroginated in India in the > is paying the price for this till today. If India was Buddhist, the > Industrial revolution would have happened in India in 1000 AD. Horseshit. The industrial revolution occurred in Christian nations and Jews and Christians have won lots of Nobel prizes in science. I think that the only Nobel prize ever won by a Buddhist was Tenzin Gyatso's Nobel PEACE prize. George ==== > Horseshit. The industrial revolution occurred in Christian > nations and Jews and Christians have won lots of Nobel > prizes in science. I think that the only Nobel prize ever > won by a Buddhist was Tenzin Gyatso's Nobel PEACE > prize. You may have a point. Who is that historian who traced the industrial revolution to the protestant mind-set? (I should know but I'm having a senior moment) Anyway, the idea of India remaining Buddhist makes an interesting what-if. ==== Buddhism is the only religion that is compatible with science and >mathematics. It is a way of thinking which oroginated in India in the >is paying the price for this till today. If India was Buddhist, the >Industrial revolution would have happened in India in 1000 AD. Horseshit. The industrial revolution occurred in Christian > nations and Jews and Christians have won lots of Nobel > prizes in science. I think that the only Nobel prize ever > won by a Buddhist was Tenzin Gyatso's Nobel PEACE > prize. > holy Marx!! even commisar amrtya sen won a nobel. nobel can rhyme with gobel when you want it to. In my highly educated opinion, which will come to be shared soon among all nobel laureates, India's problem is that it is deep in mohamadism, and surrounded by mohamadism. It is, however, good to see hindus have started to break out of the historic morass since early 90s, finding their true hindu spirit of mookti (ie freedom), and finding, in the words of India's Prime minister Vajpayee, natural allies in the western world. > George ==== Why do bengalis spell certain sanscrit words like muslims and arabs??? Deb >Horseshit. The industrial revolution occurred in Christian >nations and Jews and Christians have won lots of Nobel >prizes in science. I think that the only Nobel prize ever >won by a Buddhist was Tenzin Gyatso's Nobel PEACE >prize. You may have a point. Who is that historian who traced the industrial > revolution to the protestant mind-set? (I should know but I'm having a > senior moment) Anyway, the idea of India remaining Buddhist makes an interesting > what-if. ==== >> Horseshit. The industrial revolution occurred in Christian >> nations and Jews and Christians have won lots of Nobel >> prizes in science. I think that the only Nobel prize ever >> won by a Buddhist was Tenzin Gyatso's Nobel PEACE >> prize. You may have a point. Who is that historian who traced the industrial >revolution to the protestant mind-set? (I should know but I'm having a >senior moment) Weber. Gareth ==== >>Could someone explain this to me... >>1) if a>2) if a<=b then a>My text says this is right, but has no explaination for it. A proof >>or something might be nice... >> > Steve, I can sort of grasp the either/or of the matter, but if in >case 2) >the possibility of a=b is not met, as in 2<=3, why is the statement if >2<=3 then 2<3 false? Have I muddled something up? >Josh > > Josh, > > In this context, to say 2) is true means that it needs to be true for > *any possible* values of a and b. Sure, 2<=3 and 2<3 are both true, > but 2) is not true *in general*, which is intended here. Get it? > > In this, you might stumble over if 3<2 then 3<=2, which is regarded > as a true conditional since the premise is false. I can only suggest > that such conditionals are best thought of as *vacuously true*, true > for purposes like stamping 1) true for all possible values of a and > b. But all we really care about are the values of a and b where the > premise a<=b is really true. > > Bob Bob, Josh ==== N^n injects into N? How is this possible? Does this mean that Z^n also >injects into Z? >This seems ridiculous to me, but I will have to think about it more after I >read all of the >responses. > > Yes there is an injection Z^n -> Z. Even Z^n -> N. And even > (as I said earlier) Q^n -> N. All assuming n is a natural > number, i.e. finite. > > A complete enumeration of Z^2 is easy to visualize. Start at > (0,0). Around that point there is a square of eight points; > traverse them counterclockwise starting at the x-axis: > (1,0), (1,1), (0,1), (-1,1), (-1,0), (-1,-1), (0,-1), (1,-1). > Around this square there is a larger square of 16 points which > we traverse by the same rule: > (2,0), (2,1), (2,2), (1,2), (0,2), (-1,2), (-2,2), (-2, 1), > (-2,0), (-2,-1), (-2,-2), (-1,-2), (0,-2), (1,-2), (2,-2), > (2,-1) > thus you see we have a bijection between all points in this > subset of Z^2 and the first 25 elements of N -- and of course > we can continue in the same vein to ever larger squares, and > thereby cover all of Z^2 in sequence. Adding dimensions only > makes our rule more complicated (and our progress outward > slower) but the basic idea still works. > > My example is clearly a bijection, but with Cantor-Bernstein > all we really need to show is an injection; hence the other > example you were given -- f(a,b) = 2^a * 3^b -- is also a > satisfactory answer for the case N^2 -> N. I ran across this nice bijection between N^2 and N in Hungerford: (m, n) |-> 2^(m-1) * (2n -1) In other words you decompose a natural number into the product of the highest power of 2 that divides it, and an odd number. So (1,n) gives you all the odd numbers; (2,n) gives 2 times the odd numbers; (3,n) gives a4 times all the odd numbers; and so forth. This is in the beginning of the book in the section on cardinal arithmetic. In that section he proves that for any cardinal k*k = k but the proof invokes Zorn's lemma and is nonconstructive. ==== > You're being remarkably dense Nora Baron. > > Think about it. > > Math is not a popularity contest. It's not a fashion show. The truth > matters. Never mind that. Just answer her question. Jan Bielawski ==== In his paper Advanced Polynomial Factorization, James Harris considered the polynomial P(x) = 65*x^3 - 12*x + 1. Assume this polynomial is factored in the form P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1), where a1, a2, and a3 are algebraic integers. James Harris in a recent post in the sci.math thread Constant factors and polynomials has said: ... neither a1, a2, nor a3 have ANY nonunit factors in common with 5 in the ring of algebraic integers. It is not clear how he arrives at this, and it disagrees sharply with several proofs of the fact that ALL of a1, a2, and a3 have nonunit factors in common with 5. But the interesting thing here is to see where he goes with his conclusion. Since a1*a2*a3 = 65 = 5 * 13, it is clearly the case that at least one of a1, a2 or a3 must have a nonunit factor in common with 5. Hence a contradiction. Most people would very reasonably deduce from such a contradiction that they have made a mistake. They would conclude that the quoted text above must be incorrect, and there must be an error in the logic which led to it. They would check their argument until they found an error. Not, however, James Harris. He concludes that there is some kind of flaw in the ring of algebraic integers that has been overlooked by mathematicians since the time of Gauss, Kummer, and Dedekind. Harris disbelieves mathematical arguments that have been presented here which lead to contradictions of his claims. He calls us liars for presenting them. He says we have not disproved his claims and we have not found any errors in his proofs. Then he arrives (somehow!) at a contradiction of his own, as noted above. Perhaps he says to himself: So yes, I have a contradiction. Of course it is TOTALLY IMPOSSIBLE that I have any errors in my math or logic. After all, I have written a sequence of words and equations and stuff that I call a 'proof', and I cannot see anything wrong with it. By definition if you call something a proof, it must be perfect and correct. Proofs cannot duel. The math doesn't care if you say it is wrong. A proof is perfect. True, I have been in this situation many times before and have been proved wrong every time. True, the contradiction implies that 150 years of basic mathematics checked by thousands of people is incorrect. True, I have not been able to find any errors in the various independent counterarguments presented by other sci.math posters. Still, the one possibility that I must eliminate immediately is that I have made a mistake. True, people have pointed to explicit parts of my argument which they claim are incorrect. However, I have DISPROVED those arguments by noting that certain factors (the w's) cannot be capable of recognizing that the polynomial they are factoring is irreducible. True, this is not exactly a rigorous mathematical argument, but I am absolutely certain it is correct because I MADE IT UP AND I AM SMARTER THAN EVERYBODY ELSE. The prime directive here is that I CANNOT BE WRONG. I AM INFALLIBLE AND MY PROOF CANNOT HAVE ERRORS. I can dismiss the contradictions of my claims by other people as being due to their obvious lying and and jealousy and stupidity. I cannot do the same with the contradiction that I arrived at myself because I AM INFALLIBLE. I AM INCAPABLE OF ERROR (THIS TIME). Therefore there must be a basic problem with mathematics. No, I do not conclude that mathematics is inconsistent (though perhaps I should ...). Instead I conclude that the ring of algebraic integers is INCOMPLETE. Don't ask me exactly what that means. I don't know. Just take my word for it. After all, I AM INFALLIBLE. Nora B. ==== The real JSH posts from msn.com. This poster is a fake. ==== > > If I want to solve a puzzle such as the one below, is there a mathematical > way of doing it? I'm guessing it could be done by algebra or something > similar? As you may have guessed I'm not a maths expert by a long shot... > 8) > > if Andrew, Bob, Charles, Dave have 4963 between then, and Andrew has 598 > more than Bob, and Bob has 415 more than Charles, and Charles has twice > as much as Dave, how much does Dave have? > > [1] A + B + C + D = 4963 > > [A] A - B = 598 > > [B] B - C = 415 > > [C] C - 2D = 0 > > What number is D? > > Now it took me about 60 seconds to enter the formular in Excel and type > numbers into it until I got the answer of 505, but that is kind of > cheating - what is the proper way to solve these puzzles? Others have posted solutions but haven't highlighted the key idea. Because the above system of equations has a triangular form, one may easily successively eliminate each variable as follows. By [A] eliminate A in [1] obtaining [2] in variables {B,C,D} By [B] eliminate B in [2] obtaining [3] in variables {C,D} By [C] eliminate C in [3] obtaining [4] in variable {D} Now solve for D the linear equation [4]. Notice how the triangular form ensures that once you have eliminated a variable, it will never be reintroduced by a later step. For example, because equation [C] involves no variables preceding C (i.e. neither A nor B), such variables are not reintroduced when using [C] to eliminate C in [3]. In general, whenever one performs elimination it is usually useful to first check to see if the structure of the system of equations lends itself to any optimizations such as above. -Bill Dubuque ==== > The object of puzzles is, of course, to find an > elegant solution (otherwise, they aren't recreation, they're work). But > a puzzle that requires more than purely minimal computation (in the eyes > of the beholder, of course) isn't very interesting. (Play is what you > get to do, work is what you have to do.) > how about this one? A tree is placed every 60 yards from a given point, a pile of gravel every 40 yards. Apart from the first point, how far out does the tree and gravel apear in the same spot? (answer : on the 8th tree, 520 yards) Answered by writing out the distances manually, but how to solve it mathematically??? Is there a website that teaches maths/algebra with these types of questions? ==== The object of puzzles is, of course, to find an >elegant solution (otherwise, they aren't recreation, they're work). But >a puzzle that requires more than purely minimal computation (in the eyes >of the beholder, of course) isn't very interesting. (Play is what you >get to do, work is what you have to do.) > > how about this one? A tree is placed every 60 yards from a given point, a pile of gravel every > 40 yards. Apart from the first point, how far out does the tree and gravel > apear in the same spot? (answer : on the 8th tree, 520 yards) Answered by writing out the distances manually, but how to solve it > mathematically??? I think that's wrong. The third tree is 120 yards from the first tree and there will also be a pile of gravel there. That's because the least common multiple of 40 and 60 is 120. Google on least common multiple and greatest common divisor. Make the prime factorization of both 40 = 2^3.5 60 = 2^2.3.5 and keep the highest exponents for LCM = 2^3.3.5 = 120 and the lowest exponents for GCD = 2^2.5 = 20 As you can see, in general LCM(x).GCD(x) = x, or in this case 20 times 120 = 40 times 60 = 2400 ==== > The object of puzzles is, of course, to find an >> elegant solution (otherwise, they aren't recreation, they're work). But >> a puzzle that requires more than purely minimal computation (in the eyes >> of the beholder, of course) isn't very interesting. (Play is what you >> get to do, work is what you have to do.) > how about this one? > A tree is placed every 60 yards from a given point, a pile of gravel every >40 yards. Apart from the first point, how far out does the tree and > gravel >apear in the same spot? (answer : on the 8th tree, 520 yards) > Answered by writing out the distances manually, but how to solve it >mathematically??? I think that's wrong. The third tree is 120 yards from the first tree and > there will also be a pile of gravel there. That's because the least common > multiple of 40 and 60 is 120. Google on least common multiple and > greatest common divisor. my mistake, it should have read 65 for the trees... ==== > The object of puzzles is, of course, to find an >> elegant solution (otherwise, they aren't recreation, they're work). But >> a puzzle that requires more than purely minimal computation (in the eyes >> of the beholder, of course) isn't very interesting. (Play is what you >> get to do, work is what you have to do.) > how about this one? > A tree is placed every 60 yards from a given point, a pile of gravel every >40 yards. Apart from the first point, how far out does the tree and > gravel >apear in the same spot? (answer : on the 8th tree, 520 yards) > Answered by writing out the distances manually, but how to solve it >mathematically??? I think that's wrong. The third tree is 120 yards from the first tree and > there will also be a pile of gravel there. That's because the least common > multiple of 40 and 60 is 120. Google on least common multiple and > greatest common divisor. Make the prime factorization of both 40 = 2^3.5 > 60 = 2^2.3.5 and keep the highest exponents for LCM = 2^3.3.5 = 120 > and the lowest exponents for GCD = 2^2.5 = 20 As you can see, in general LCM(x).GCD(x) = x, or in this case 20 times 120 = > 40 times 60 = 2400 This should of course be LCM(x,y).GCD(x,y) = xy, was a bit hasty ==== >The object of puzzles is, of course, to find an >>elegant solution (otherwise, they aren't recreation, they're work). > But >>a puzzle that requires more than purely minimal computation (in the > eyes >>of the beholder, of course) isn't very interesting. (Play is what you >>get to do, work is what you have to do.) >> how about this one? >> A tree is placed every 60 yards from a given point, a pile of gravel > every >> 40 yards. Apart from the first point, how far out does the tree and >gravel >> apear in the same spot? (answer : on the 8th tree, 520 yards) >> Answered by writing out the distances manually, but how to solve it >> mathematically??? > I think that's wrong. The third tree is 120 yards from the first tree and >there will also be a pile of gravel there. That's because the least common >multiple of 40 and 60 is 120. Google on least common multiple and >greatest common divisor. my mistake, it should have read 65 for the trees... Then 40 = 2^3.5 65 = 5.13 LCM(40,65) = 2^3.5.13 = 520 GCD(40,65) = 5 ==== let R is ring with unity let R = C[0,1] , that is all continuous function with interval [0,1] let M = {f in C[0,1] | f(a) = 0 , a is exist in [0,1]} show, M is maximal ideal of R ----------------------------------- M not equal to R if N is ideal of R such that M in N in R if N not equal to R , M in N Any f in N, f(a) = 0 => f in M (***) hence N in M hence N = M therefore M is maximal ---------------------------------- (***) section i think that this section was wrong. if f(x) =1 is in N, this section is wrong how do you think about (***) ?? correct ? or incorrect?? please, point out an error~sir~ thanks to read ==== > let R is ring with unity > let R = C[0,1] , that is all continuous function with interval [0,1] > let M = {f in C[0,1] | f(a) = 0 , a is exist in [0,1]} > Let f(x) = x and g(x) = 1-x, then both f and g in M but f+g = 1 is not in M, so M not ideal. I think you mean for a in [0,1] let M_a = { f in C[0,1] | f(a) = 0 } > show, M is maximal ideal of R > show M = M_a is maximal ideal of R Assume some g not in M and show the ideal generated from M and g is R. So g(a) /= 0. Now let f be any continuous function. Thus h(x) = f(x) - f(a)g(x)/g(a) is in M and f(a)g(x)/g(a) is in the ideal generated by g. Hence f(x) = h(x) - f(a)g(x)/g(a) is in the ideal generated by g and M. > M not equal to R > Ok. > if N is ideal of R such that M in N in R > if N not equal to R , M in N > Huh? Instead of 'in' do you mean M contained in N ? > Any f in N, f(a) = 0 => f in M (***) > hence N in M > hence N = M > therefore M is maximal > Makes no sense. (***) section > i think that this section was wrong. > if f(x) =1 is in N, this section is wrong > how do you think about (***) ?? > correct ? or incorrect?? > ==== >let R is ring with unity let R = C[0,1] , that is all continuous function with interval [0,1] let M = {f in C[0,1] | f(a) = 0 , a is exist in [0,1]} _That_ set is not an ideal! You meant Let a be in [0,1], and let M = {f in C[0,1] | f(a) = 0}. (it makes a big difference whether you say what a is inside or outside the {}.) >show, M is maximal ideal of R ----------------------------------- M not equal to R if N is ideal of R such that M in N in R if N not equal to R , M in N Any f in N, f(a) = 0 => f in M (***) hence N in M hence N = M therefore M is maximal ---------------------------------- (***) section i think that this section was wrong. if f(x) =1 is in N, this section is wrong how do you think about (***) ?? correct ? or incorrect?? please, point out an error~sir~ Certainly (***) is incorrect. It's not true that f(a) = 0 for any f in N. You're trying to prove that N = M. That's impossible. Instead try to prove that N = M _or_ N = R. (Suppose that N is _not_ equal to M. Then show N = R.) >thanks to read > ************************ David C. Ullrich ==== thank you. sir ==== thank you. sir ==== Refers to : http://www.ics.uci.edu/~eppstein/junkyard/teabag.html http://mathforum.org/discuss/sci.math/a/m/509288/509622 undeformed square bag length a. At this point of time, a neither a theoretical final value, nor one supported by analysis could be conclusively arrived at. It value is approximately a^3/5 . The above volume figure can be computed by FEA if this can be modeled in finite element analysis with geometric non-linearity, with large in-plane stiffness matrix coefficients compared to small in-plane coefficients as applicable to a flexible/inextensible bag. I do hope someone would respond for analysis, especially as it has inter-disciplinary significance. G.L.Narasimham Ex Design Head and Advisor, Composites, Vikram Sarabahi Space Center,India ==== In sci.math, michael : > > >> You know what happens after I make a post like this one? > > Sales of all products containing caffeine increase tenfold? That reminds me. I need another cup of coffee/cocoa mixture. Be right back... > >> Just remember, I've been looking at posts on this newsgroup for years, >> and I've seen quite a few people come and go during that time. > > Then why are you posting this to alt.fiction.original, when it's of no > interest to us whatsoever? Most likely because JSH is a bit sloppy regarding newsgroup postings. :-) Someone else is suggesting that JSH is sloppy in other areas as well. :-) (Not me. I *know* JSH's math is a bit on the careless side, not because of his equation manipulations but because he leaps to conclusions that need to be carefully tackled instead.) > > next post from jstevh@msn.com telling us that this is a forgery. > > Go play in the traffic. > Now now, this *is* sci.math (among others); at least phrase it as a creative math problem: (1) A person decides to attempt to cross an 8-lane freeway, for some reason which shall remain unspecified by the math problem (why does Billy throw the ball to Jane anyway?). The person can walk at 3 mi/hr = 4.4 ft/sec. Assuming a uniform population of sedans of 16 feet in length and 6 feet in width and exactly in the middle of each lane, that the lanes are 12 feet wide, that every sedan is following the speed limit and the 2-second rule [*], that the sedans are otherwise randomly distributed, that the person, once he starts to cross, blithely walks at a uniform velocity straight across the highway, as opposed to doing something more intelligent (like running zig-zags), and that the sedans don't brake before hitting him, what is the probability that he'll be struck? (2) Same as (1), except the sedans are using a different rule, the car-length every 10 mph rule. This rule is obviously not quite as safe but I happen to live in a metro area so know even this relaxed (?!) rule is broken routinely at speed. (3) Same as (1) except the traffic is bumper-to-bumper stop-and-go. (This one should be easy. Of course being struck at 2 mph isn't quite as deadly as 65 mph unless one's head gets stuck under a wheel or something.) (4) Same as (1), except we now assume a mix of cars: 8 parts sedan, 2 parts SUV (length 16 1/2 feet, width 6 1/2 feet). (5) Same as (1) and (3), except we assume the mix 7 parts sedan, 2 parts SUV, and 1 part semi-tractor trailer (length 64 feet (48 foot trailer, 16 foot cab), width 8 feet [+]). Since truckers are (hopefully) more knowledgable we increase their following distance to 4 seconds. (6)-(10). Assume the sedans and SUV's can see a distance of 1,000 feet and brake within a distance of 300 feet (of which up to 50 feet can be reaction time). The SUVs take 350 feet since they're heavier. The trucks can brake within 500 feet. Assume also that the cars and SUVs don't skid out of control while braking and the trucks don't jackknife. Assume perfect visibility to the vehicle's right (e.g., no trees in the way or blind spots in the vehicle). (11)-(15) Assume in (6)-(10) a visibility of 200 feet by either nighttime conditions, an obscuring hill, or fog. (16)-(30). Now assume the person can only walk at 1.5 mph = 2.2 ft/sec. (Good luck. He'll need it.) [*] The 2-second rule is a common one, and basically stipulates that the rear of the preceding car, assuming you and he are traveling at the same speed, shall be 2 seconds in front of your front bumper. Admittedly this rule has some interesting ramifications with respect to highway capacity, as it indicates that no matter how much one increases the speed the capacity of the highway is largely constant, assuming everyone, erm, rigorously follows this rule... [+] a cargo container apparently has dimensions 40' by 96; a trailer can be as big as 48' x 96 as mentioned -- maybe even longer. I'm not sure how big a truck cab is but I'm assuming it's similarly sized to a sedan or SUV except that there's no area for the kids; it's mostly motor. :-) Since 96 = 8' we're in fairly good shape here; also, the truck cab partially occupies the trailer space because of the hitch, making dimensions a little weird. But then, this *is* a hypothetical math problem anyway... :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== I may have no idea about your problem, but I _can_ line your columns up for you, when viewed in a fixed-width font such as Courier: | C H A I R S | 0 1 2 3 4 5 6 7 8 | | 0 1 1 1 1 1 1 1 1 1 | 1 1 2 3 4 5 6 7 8 9 |P 2 1 3 7 13 21 31 43 57 73 |E 3 1 4 13 34 73 136 229 358 529 |O 4 1 5 21 73 209 501 1045 1961 3393 |P 5 1 6 31 136 501 1546 4051 9276 19081 |L 6 1 7 43 229 1045 4051 13327 37633 93289 |E 7 1 8 57 358 1961 9276 37633 130922 394353 -- Clive Tooth http://www.clivetooth.dk ==== >Hey guys, you were fantastic last time I ran into a problem, and so I >thought I'd post something here that I've been confused about for quite a >while. I'm trying to study Order Statistics, and if you have U1...Un iid uniform on >(0,1), and work out their joint pdf you get f(u(1),.....u(n)) = n! where >u(1) <= ... <= u(n). Why is this? How do you prove that it is n!? I mean I understand that all >orderings are equally as likely, and so there are n! combinations, but how >does this translate to the pdf? This is covered in one of the books by Ross, which I do not have at hand at this location. So I'll have to see if I can reconstruct the argument from memory. First, instead of the density, let's look (non-rigorously) at the probability of being in a small interval around (u(1), ..., u(n)). That probability is f(u(1), ..., u(n)) * du(1) du(2) ... du(n). The du's add nothing except to put it in terms of an actual probability instead of a density. Let the original rvs be called x1, x2, ..., xn. Now the probability that the order statistics lie in this interval is the same as that one of the x's take the value u(1), another take u(2), etc. It doesn't matter what the order is. The probability that (x1, x2, ... , xn) lies in the n-dimensional interval (u(1), u(2), ... u(n))+(du(1), du(2), ...du(n)), i.e., that x1 takes a value near u(1), that x2 takes a value near u(2), etc., is just du(1) du(2) ... du(n). But there are n! identical events (the reorderings of the x's) that will give rise to exactly the same order statistics. So the probability that the order statistics fall near (u(1), u(2), ..., u(n)) is n! du(1) du(2) ... du(n). Now compare to what I said the probability was in terms of f(u(1), ...u(n)) and you can see, dropping the du's, that the density is n!. The probability is of course 0 that the order statistics don't satisfy u(1) <= u(2) <=... <= u(n). - Randy ==== > I can't tell you if Peter Lynds is correct, but I believe he is on to > something here. A problem I've had for a while now is that with our > current understanding of time and physical matter, something must have > come from nothing. Think about it, current theory is that it all > started with a big bang. Where did the matter come from? Some say it > was a contraction of a previous universe and maybe that is so, but if > you follow it back to the beginning, you have to conclude that > something came from nothing or that the matter always existed. > > Neither concept can be understood within the current framework of > physics or philosophy. Now, if our concept of time is incorrect then > maybe we can start to understand more about our origins. > > As human beings, we will have a very hard time grasping these concepts > of time and motion. We are hard wired to see things in a particular > way and everything we see reinforces these beliefs. > > Good luck Peter. I read both of Lynds's papers. The only question is whether he's sincere but naive, or a deliberate troll in the tradition of the Sokol hoax. There is no intellectual content. He doesn't understand calculus or real analysis. In one section he says that it's impossible to assign a velocity to a moving object. But in the previous paragraph he gives an example of a train moving at 100 km/hr, contradicting his own theory. He defines velocity as delta-position over delta-time, demonstrating an ignorance of freshman calculus. All he is really doing is describing for us his own thoughts on encountering the Zeno paradoxes. He adds nothing new. What he calls uncertainty is really just measurement error. ==== http://digilander.libero.it/fraterno/zenone.htm ==== I don't understand what the sign || means around a variable.. for example: |x| < x+1 < 5 -Paul ==== I don't understand what the sign || means around a variable.. for example: |x| < x+1 < 5 > -Paul |x| means the absolute value of x. It is the same as x when x is positive and -x when x is negative So, |3| = 3 |34.12435| = 34.12435 |-4| = 4 |0| = 0 |-2.34| = 2.34 ==== |x| ----- it is the modulus of x (or, the absolute value of x). i.e. |5| = 5 |-2| = 2 |3+4i| = 5 Michael Leung Paul ???????:3F34EE8C.3060906@ppp.com... I don't understand what the sign || means around a variable.. for example: |x| < x+1 < 5 > -Paul ==== >I have been told, that he was not all that >careful about the distiction between manifolds and, what we now call, >algebraic varieties. As I recall in some languages other than English, varieties are called algebraic manifolds. Not that one doesn't need to be careful about the distinction, but in a sense they're very similar. Keith Ramsay ==== A mathematical proof begins with a truth, and proceeds by logical steps to a conclusion which then must be true. I've pulled a detailed exposition of a short argument that quickly shows a problem with algebraic integers. It starts after the reference. Now here's a math proof. Those who doubt that fact can believe it's a claim of proof, but it's verified to be a proof by tracing the argument out. In this case, I begin with an expression. The expression exists, so that is the truth from which you start. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). That is, I have the identity which defines P(m) in terms of various symbols, and it's all in the ring of algebraic integers, which means that the symbols can only represent numbers that are algebraic integers. Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Now that was a lot of steps, but each was a logical one. First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are defined by the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) then I set m=0, and used the definition of P(m) to get P(0). That told me that at m=0 two of the b's are 0, because then P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), where the u^2 couldn't get there unless two of the b's are 0. Then using that result I get from P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) that P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) and multiplying through by w_1 w_2 I have P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) which with P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), tells me that w_1 w_2 = 1, when m=0. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. Now I'm focusing on what has been revealed to be an area of confusion. Apparently some people believe that when I divide off f^2 that it can divide off as a *function* of m, so that m=0 might be a special case. I'm now starting the argument to address that belief by noting again that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when f doesn't have 3 as a factor. But that was an arbitrary choice, so let f=3. That is, I *said* f is coprime to 3 but in considering this possibility it's worth it to relax that restriction and now consider what would happen if it equals 3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Seeing that is as simple as looking at P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f with f=3 as then you have P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3 so *every* coefficient has a factor that is 3, as you can tell by looking. So with P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) each of the b's and each of the w's has a factor that is 3^{1/3}, while the b's can have additional factors in common with 3, the w's cannot, as when 3 is separated out, notice you have P(m)/3^2 = 3((m^3 3^3 - 3m^2 3 + m) x^3 - (-1+m3^2 )x u^2 + u^3). But before at m=0, they were coprime to f, now they are not when f=3, as they are constant. Clearly, they are constant in both cases with respect to m, without regard to the value of f. Which makes sense as f^2 is not a function of m, and it is what is being divided off. That is, if they were functions of m, so that w_1 w_2 = 1 at m=0 as a *function* of m, then it wouldn't matter if f had a factor of 3 or not, you'd STILL get w_1 w_2 = 1 at m=0, without regard to the value of f. But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0, which only works if the w's are independent of m, which they are. It makes sense that they are anyway, as f^2 isn't a function of m, but I've seen that for some people the idea can take hold after seeing m=0 highlighted. But if the w's were functions of m, then w_1 w_2 would equal 1, without regard to the value of f, but it does not. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. And here I've packed in a lot of information as well. First, with f coprime to 3, I now know that the factorization is P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) as the w's are constant with respect to m, so I can just check at m=0, which revealed that w_1 w_2 = 1. Now that doesn't necessarily force w_1 and w_2 to each equal 1, but even if they were factors of 1, i.e. unit factors, that would only change b_1 and b_2. So I have my factorization without regard to m in terms of where the f goes, and then I point out that you can actually check my work using m=1, f=sqrt(2), as then you get a polynomial which you can factor rather simply. So you can actually get the values for the b's and check them, and see that they are all algebraic integers, and all are coprime to 2. However, usually, for f values that are coprime to 3, you don't get b's that are algebraic integers, which shows a problem with the ring of algebraic integers. Now the nice thing about a mathematical proof is that if someone disagrees they have to find some misstep. Unfortunately, people can *say* that proof is not a proof, even when it is, just like if you tried to say you were human, and not a dog, someone might dispute any proof you might give, claiming it false. James Harris ==== > A mathematical proof begins with a truth, and proceeds by logical step. But you never use this method. ==== >A mathematical proof begins with a truth, and proceeds by logical >steps to a conclusion which then must be true. I've pulled a detailed exposition of a short argument that quickly >shows a problem with algebraic integers. It starts after the >reference. >Now here's a math proof. Those who doubt that fact can believe it's a >claim of proof, but it's verified to be a proof by tracing the >argument out. > Not! See below. >In this case, I begin with an expression. The expression exists, so >that is the truth from which you start. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). That is, I have the identity which defines P(m) in terms of various >symbols, and it's all in the ring of algebraic integers, which means >that the symbols can only represent numbers that are algebraic >integers. Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves >b_3 = 3. > When m = 0. >Now that was a lot of steps, but each was a logical one. > So far, so good. >First I introduced b_1, b_2, b_3, w_1, w_2, and w_3, which are defined >by the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) then I set m=0, and used the definition of P(m) to get P(0). That told me that at m=0 two of the b's are 0, because then P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), where the u^2 couldn't get there unless two of the b's are 0. Then using that result I get from P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) that P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) and multiplying through by w_1 w_2 I have P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) which with P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), tells me that w_1 w_2 = 1, when m=0. > Yep, all just fine when m = 0. No problemo so far. >Essentially objections to how f^2 divides off now come down to >claiming that the w's are functions of m, but consider that w_1 w_2 = >1, when m=0, if f is coprime to 3. Now I'm focusing on what has been revealed to be an area of confusion. > Apparently some people believe that when I divide off f^2 that it can >divide off as a *function* of m, so that m=0 might be a special case. >I'm now starting the argument to address that belief by noting again >that w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when f >doesn't have 3 as a factor. But that was an arbitrary choice, so let f=3. > f = 3 is irrelevant to what you want. There is no reason to consider it. As you will see below, it is a red herring and it does not show what you want. >That is, I *said* f is coprime to 3 but in considering this >possibility it's worth it to relax that restriction and now consider >what would happen if it equals 3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. Seeing that is as simple as looking at P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f with f=3 as then you have P(m)/3^2 = (m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3 so *every* coefficient has a factor that is 3, as you can tell by >looking. > Let's look at this in detail when m = 1 and u = 1. Then P(m)/3^2 = (81 - 27 + 3)*x^3 - 3*8*x + 3 = 3*(19*x^3 - 8*x + 1). >So with P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) each of the b's and each of the w's has a factor that is 3^{1/3}, >while the b's can have additional factors in common with 3, the w's >cannot, as when 3 is separated out, notice you have P(m)/3^2 = 3((m^3 3^3 - 3m^2 3 + m) x^3 - (-1+m3^2 )x u^2 + u^3). > As above when m = 1, u = 1, this is 3*(19*x^3 - 8*x + 1). The polynomial inside the parentheses can be factored in the form [1] (19*x^3 - 8*x + 1) = (b1*x + 1)*(b2*x + 1)*(b3*x + 1), where b1, b2, and b3 are the negatives of the roots of the associated polynomial u^3 + 8*u^2 - 19. Since the roots of the latter polynomial are algebraic integers, one concludes b1, b2, and b3 are algebraic integers also. Now: how might you distribute the 3 in the expression 3*(b1*x + 1)*(b2*x + 1)*(b3*x + 1) ? Answer: LOTS of ways! There is no unique way. Here are several: 1. (b1*x + 1)*(b2*x + 1)*(3*b3*x + 3) 2. (sqrt(3)*b1*x + sqrt(3))*(sqrt(3)*b2*x + sqrt(3)*(b3*x + 1) 3. (3^{1/3}*b1*x + 3^{1/3})*(3^{2/3}*b2*x + 3^{2/3})*(b3*x + 1) 4. (3^{1/5}*b1*x + 3^{1/5})*(3^{3/5}*b2*x + 3^{3/5}) *(3^{1/5}*b3*x + 3{1/5}) In fact INFINITELY many ways. Any way you want to split 3 as a product of three numbers gives a factorization. And in all the examples just given (and in infinitely many others) note that (1) the coefficients of the x's are algebraic integers, and (2) the w terms are also algebraic integers. Proving what, exactly, you ask? #### Proving that the f = 3 case tells you NOTHING useful about the necessary values of the w's. There is no unique way to write them down. This is a special, exceptional case in which too many of the of the f terms can be factored out. >But before at m=0, they were coprime to f, now they are not when f=3, >as they are constant. Clearly, they are constant in both cases with >respect to m, without regard to the value of f. Which makes sense as >f^2 is not a function of m, and it is what is being divided off. > But in this case you can divide off *another* factor of f, and that is what leads to the nonuniqueness shown above, which wrecks your argument. This does not happen when f is a prime bigger than 3 and m is coprime to f. f = 3 is a special case of no interest or relevance to your main argument. It is a nuisance distraction. It proves nothing. >That is, if they were functions of m, so that w_1 w_2 = 1 at m=0 as a >*function* of m, then it wouldn't matter if f had a factor of 3 or >not, you'd STILL get w_1 w_2 = 1 at m=0, without regard to the value >of f. But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0, which only >works if the w's are independent of m, which they are. > See above. In the special case with f = 3, the w's are not uniquely determined. You cannot draw conclusions from it about how the f terms are distributed among the w's or the linear factors. This case is a red herring, and it is of no interest in your general argument, where you require that f is a prime > 3. Worst of all, it does not imply what you want. >It makes sense that they are anyway, as f^2 isn't a function of m, but >I've seen that for some people the idea can take hold after seeing m=0 >highlighted. But if the w's were functions of m, then w_1 w_2 would equal 1, >without regard to the value of f, but it does not. > w1 = w2 = 1 and w3 = 3 is just one of the infinite range of possibilities. It is Example 1 above. >Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, >f=sqrt(2), but that's a special case as generally they are not, which >shows a problem with the ring of algebraic integers. > In this special case, P(x) does not even have rational coefficients. It too is of no interest or value for your main argument. >And here I've packed in a lot of information as well. > Not enough, clearly. >First, with f coprime to 3, I now know that the factorization is P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf) as the w's are constant with respect to m, so I can just check at m=0, >which revealed that w_1 w_2 = 1. Now that doesn't necessarily force >w_1 and w_2 to each equal 1, but even if they were factors of 1, i.e. >unit factors, that would only change b_1 and b_2. So I have my factorization without regard to m in terms of where the f >goes, and then I point out that you can actually check my work using >m=1, f=sqrt(2), as then you get a polynomial which you can factor >rather simply. So you can actually get the values for the b's and >check them, and see that they are all algebraic integers, and all are >coprime to 2. However, usually, for f values that are coprime to 3, you don't get >b's that are algebraic integers, which shows a problem with the ring >of algebraic integers. > Wrong! You *can* get algebraic integers, but *** not with the properties you want ***, and there is no problem with the ring of algebraic integers. Here is how things work when f = 5, m = 1, u = 1, v = -1 + m*f^2, and P(x) = (v^3 + 1)*x^3 - 3*v*x*(u*f)^2 + (u*f)^3: P(x)/f^2 = P(x)/25 = 553*x^3 - 72*x + 5. By the Magidin-McKinnon theorem (essentially proved earlier by someone else [P. M. Cohn?]), this can be factored in the form 553*x^3 - 72*x + 5 = (b1*x + w1)*(b2*x + w2)*(b3*x + w3), where b1, b2, b3, and w1, w2, w3 are algebraic integers. You can show using elementary Galois theory that EACH of w1, w2, and w3 is not coprime to f = 5. Thus: a factorization of the desired form DOES exist, but it does NOT have one of the properties that you desperately want. No problem with the ring of algebraic integers, and no valid proof for you. You lose on two counts. Too bad! >Now the nice thing about a mathematical proof is that if someone >disagrees they have to find some misstep. > See above at #### ! The misstep in the current argument has been found. >Unfortunately, people can *say* that proof is not a proof, even when >it is, just like if you tried to say you were human, and not a dog, >someone might dispute any proof you might give, claiming it false. > In this case, you have tried to use an irrelevant red-herring argument to show what you want. Unfortunately, in the special case you selected, the number f (= 3) does factors *** non-uniquely *** through the linear terms of your polynomial factorization, and you end up being able to conclude: *** N O T H I N G *** about the cases in which you are interested. But you have made progress. Do you realize how long it took us to get through to you that there is actually a nontrivial problem with generalizing from m = 0 to m <> 0 ? Do you realize how many incorrect arguments you have already burned through (including the present one) in trying to handle that problem? Do you realize that all of this is a waste of time, because your main claims have already been shown to be false and cannot be fixed by twiddling with the details? Nora B. James Harris ==== >> |CUSPIDAL: (1) Belonging to the apex (of a cone). >> | (2) Having, relating to, or of the nature of, a cusp. >> >> In the study of modular forms, cuspidal has a technical meaning. >> Maybe it can be covered by (2), but it's kind of a stretch! [...] >Isn't a cuspidal modular form (or sometime I hear cusp forms, which I >think is synonymous) related to a Riemann surface with cusp points? And >isn't much information about the form given by examining these cusps? That's why I said maybe it can be covered by (2), yes. Relating to is a catchall for such uses. Keith Ramsay ==== what would be the best way to prove rational number + rational number = rational number and rational number + irrational = irrational This is what I have: 1) since all rational numbers can be expressed as a ratio of two integers, n=p/q then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers i.e. 1*4+2*3/4*3=10/12 2) using the above, with irrationals, there is no ratio such that n=p/q i.e n=sqrt(x) so that p/q + sqrt(x) ... that is where I begin to get really informal. anyone wanna let me know how I could improve on this? would induction or an indirect proof be of any use? josh ==== > what would be the best way to prove > > rational number + rational number = rational number > and > rational number + irrational = irrational > > This is what I have: > > 1) > since all rational numbers can be expressed as a ratio of two integers, > n=p/q > then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers Well that does irrational. > 2) using the above, with irrationals, there is no ratio such that n=p/q > i.e n=sqrt(x) > so that p/q + sqrt(x) ... Hmmm. Now every positive real number has the form sqrt(x) for some x but even if sqrt(x) is irrational, x may still be irrational. Now just suppose we had a rational plus an irrational equalling a rational. Say a + b = c where a and c are rational but b is irrational. What can we say about b? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > what would be the best way to prove rational number + rational number = rational number > and > rational number + irrational = irrational > Assume r irrational and sum rational. Then for some integers a,b,u,v a/b + r = u/v Produce a contradiction. > This is what I have: 1) > since all rational numbers can be expressed as a ratio of two integers, n=p/q > then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers > Correct. > i.e. 1*4+2*3/4*3=10/12 > Wrong and hard to read for cramming all white space out of equation. > 2) using the above, with irrationals, there is no ratio such that n=p/q > i.e n=sqrt(x) > so that p/q + sqrt(x) ... > that is where I begin to get really informal. > Huh? ==== >what would be the best way to prove rational number + rational number = rational number >and >rational number + irrational = irrational This is what I have: 1) >since all rational numbers can be expressed as a ratio of two integers, n=p/q >then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers i.e. 1*4+2*3/4*3=10/12 That is the right idea. You should also verify that your denominator q1q2 is nonzero. 2) using the above, with irrationals, there is no ratio such that n=p/q >i.e n=sqrt(x) >so that p/q + sqrt(x) ... >that is where I begin to get really informal. anyone wanna let me know how I could improve on this? >would induction or an indirect proof be of any use? >josh Not all irrational numbers have the form sqrt(x) where x is rational. The definition of irrational involves a negation: A real number x is irrational if x is not rational, or A real number x is irrational if it cannot be expressed as p/q where p, q are integers and q <> 0. [There are some more positive ways to express it: A real number x is irrational if whenever p = q*x with p, q integers, we have p = q = 0.] You want to prove rational + irrational is always irrational. That is, if (x is rational) and (y is irrational) then (x+y is irrational), I'll assume you know that the sum of two real numbes is real, so x + y is surely real. Now you want to show (x, y assumed real) if (x is rational) and (y is not rational) then (x+y is not rational). The conclusion is a negation, so an indirect proof suggests itself. Assume x, y are real x is rational y is not rational x+y is rational Try to get a contradiction. When you work out your full strategy, see how 1) applies. You mention induction, which is applicable primarily for propositions involving integers. This argument is about rationals and irrationals. It's conceivable that since rationals are defined in terms of integers, that you may need to invoke induction (perhaps the absolute values of their denominators). But try the indirect proof technique first. -- Spammers: I don't want a small digital camera to post photos of a large, low weight, penis on a re-financed Nigerian domain site. Peter-Lawrence.Montgomery@cwi.nl Home: San Rafael, California Microsoft Research and CWI ==== >what would be the best way to prove [...] >rational number + irrational = irrational > What is rational minus rational? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== > > rational number + rational number = rational number > and > rational number + irrational = irrational > > This is what I have: > > 1) > since all rational numbers can be expressed as a ratio of two integers, n=p/q > then p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integers > > i.e. 1*4+2*3/4*3=10/12 > > 2) using the above, with irrationals, there is no ratio such that n=p/q > Suppose that rational + irrational = rational, then rational - rational = irrational which contracdicts what you proved in (1), so it must be the case that rational + irrational = irrational. i.e n=sqrt(x) > so that p/q + sqrt(x) ... > that is where I begin to get really informal. > > anyone wanna let me know how I could improve on this? > would induction or an indirect proof be of any use? > josh ==== For any regular polygon, given the enclosed area A and the number of sides n, is it possible to calculate the other major dimensions (inscribed circle radius/diameter, circumscribed cirle radius/diameter, length of one side)? I am looking for a general formula that can be applied to polygons of any n. ==== > For any regular polygon, given the enclosed area A and the number of > sides n, is it possible to calculate the other major dimensions > (inscribed circle radius/diameter, circumscribed cirle > radius/diameter, length of one side)? I am looking for a general formula that can be applied to polygons of > any n. Let the side of the polygon be 2*L. Let the circumradius be R0. Let the inradius be R1. We have A=n*L^2*cotan(pi/n) So, L=sqrt(A*tan(pi/n)/n) Also R0=L*cosec(pi/n) R1=L*cotan(pi/n) -- Clive Tooth http://www.clivetooth.dk ==== Now, if I can impose once again, is there a way to find the inside and outside radii of a circular ring, given only the area and the wall thickness? b = inside radius c = outside radius w = wall thickness = b - a A = area = Pi(c^2 - b^2) Restating the question, given 'A' and 'w', can one calculate 'b' and 'c'? For any regular polygon, given the enclosed area A and the number of >sides n, is it possible to calculate the other major dimensions >(inscribed circle radius/diameter, circumscribed cirle >radius/diameter, length of one side)? > I am looking for a general formula that can be applied to polygons of >any n. Let the side of the polygon be 2*L. > Let the circumradius be R0. > Let the inradius be R1. We have A=n*L^2*cotan(pi/n) So, L=sqrt(A*tan(pi/n)/n) Also > R0=L*cosec(pi/n) > R1=L*cotan(pi/n) -- > Clive Tooth > http://www.clivetooth.dk ==== Now, if I can impose once again, is there a way to find the inside and > outside radii of a circular ring, given only the area and the wall > thickness? b = inside radius > c = outside radius > w = wall thickness = b - a A = area = Pi(c^2 - b^2) Restating the question, given 'A' and 'w', can one calculate 'b' and 'c'? Is this homework? -- Clive Tooth http://www.clivetooth.dk ==== >> Is this homework? << I kinda wish it was, but I've been out of school for many more years than I care to admit and my algebra/trig/geometry is quite rusty. am trying to figure out if 'A' and 'w' are enough info to enable one to find 'b' and 'c'. Now, if I can impose once again, is there a way to find the inside and >outside radii of a circular ring, given only the area and the wall >thickness? > b = inside radius > c = outside radius > w = wall thickness = b - a > A = area = Pi(c^2 - b^2) > Restating the question, given 'A' and 'w', can one calculate 'b' and 'c'? Is this homework? -- > Clive Tooth > http://www.clivetooth.dk ==== >> Is this homework? << I kinda wish it was, but I've been out of school for many more years than I > care to admit and my algebra/trig/geometry is quite rusty. am trying to figure out if 'A' and 'w' are enough info to enable one to find > 'b' and 'c'. > Now, if I can impose once again, is there a way to find the inside and >> outside radii of a circular ring, given only the area and the wall >> thickness? >> b = inside radius >> c = outside radius >> w = wall thickness = b - a >> A = area = Pi(c^2 - b^2) >> Restating the question, given 'A' and 'w', can one calculate 'b' and > 'c'? Ok. A = pi(c^2-b^2) w = c-b [not b-a as you give above] So, b = c-w Thus, A = pi*(c^2-b^2) = pi*(c^2-(c-w)^2) = pi*(c^2-c^2+2*c*w-w^2) = pi*(2*c*w-w^2) = pi*w*(2*c-w) A/(pi*w) = 2*c-w So, c = (A/(pi*w)+w)/2 and b = c-w = (A/(pi*w)-w)/2 -- Clive Tooth http://www.clivetooth.dk ==== >> Is this homework? << > I kinda wish it was, but I've been out of school for many more years than > I >care to admit and my algebra/trig/geometry is quite rusty. > I >am trying to figure out if 'A' and 'w' are enough info to enable one to > find >'b' and 'c'. > > message >>Now, if I can impose once again, is there a way to find the inside and >>outside radii of a circular ring, given only the area and the wall >>thickness? >> b = inside radius >> c = outside radius >> w = wall thickness = b - a >> A = area = Pi(c^2 - b^2) >>Restating the question, given 'A' and 'w', can one calculate 'b' and >'c'? Ok. A = pi(c^2-b^2) > w = c-b [not b-a as you give above] So, > b = c-w Thus, > A = pi*(c^2-b^2) > = pi*(c^2-(c-w)^2) > = pi*(c^2-c^2+2*c*w-w^2) > = pi*(2*c*w-w^2) > = pi*w*(2*c-w) A/(pi*w) = 2*c-w So, > c = (A/(pi*w)+w)/2 and > b = c-w > = (A/(pi*w)-w)/2 -- > Clive Tooth > http://www.clivetooth.dk ==== >I say I have a proof. The math should be trivial for mathematicians. >The work is available online 24 hours a day around the world. Why is there still a debate? Because you're too dense to understand the objections. (Or too obstinate to agree that clearly explained objections are correct.) I mean for heaven's sake, the Proof evidently still uses the notion of objects, and the definition of object is _still_ incoherent. (No, adding the -1 doesn't suffice to fix it - that was just the only one of Arturo's objections that you understood.) You should answer Bernier's question: Is Pi an Object? Is 2^sqrt(3) an Object? (And explain how the yes or no follows from the definition. Actually you don't even have to _find_ the answer - just explain how a yes or no _would_ follow from the definition if you could do various calculations.) You also don't seem to have noticed a point Arturo made (he's not the only person who's noticed this aspect of things): the definition of object has been _changing_ a lot lately. But nonethless the Proof, _using_ the notion of object, has _not_ changed! How can the same proof be correct in two different versions, if it uses objects and the meaning of the word object keeps changing? This makes it clear to the meanest intelligence that whatever it is you have it's _not_ a proof based on deductions from definitions. >Because the truth is that I'm right. Mathematics is being taught that >is false, and it has been taught for quite some time. Mathematicians >claim that they don't have any errors in core mathematics, but here >is one. Also if they admit the error then they have to acknowledge >me, then my proof of Fermat's Last Theorem and my prime counting work >should come out as well. And yes, the Hammer has arrived and is in full swing. I have the >momentum I've been looking for, so it's time to change the >establishment, for the betterment of all. You really do sound very wacky when you talk this way. Honest. >And someone brought up your current crop of great mathematicians >which included Ribet, Wiles, Taylor, Frey, and some other guy, and I'm >now speaking directly to them--You should be ashamed of yourselves, >and you should have known the day of reckoning was coming soon. I've been looking for a simple solution using elementary methods, as a >hobby, for almost seven years. Despite having started from scratch, I >think I made a little progress and I'm talking about it. Over the span of time I've been pursuing my little hobby, I've created >a lot of enemies on this newsgroup by jumping to my desired conclusion >and talking about it, only to find out later I was wrong. At times, >I've also questioned the morals or competency of those enemies And when it turned out that their objections were correct, I mean when _you_ finally agreed that their objections were correct, there's never even a _hint_ of apology for calling them liars when in fact they were just telling what even you finally acknowledged to be the truth. And no matter how many times the cycle repeats, the next time someone disagrees with you they're immediately called liars. _That's_ why people despise you - it's not because of your mistaken ideas about mathematics. >(especially when they were calling me names, questioning my sanity, or >otherwise being obnoxious). So in the meantime the debate continues. Some of you now know that >mathematicians are worse than not being quite what you might have >thought they were. But the disillusionment may soon get worse. They are people who in not admitting they are wrong are apparently >willing to continue to teach false mathematics to students who trust >them because that's an inevitable consequence of ignoring my work. They ignore that paper; then they'll be teaching false mathematics. I'm waiting for them to do it, so hopefully the federal authorities >can pounce on them for fraud. But I'm warning like this post because >I don't think mathematicians believe that they are subject to the >rules of society. I think they'll read this post and think they can get away with it. >It turns out that destructive ideas, what I call hostile memes, can >take over the human mind. They are like viruses and can remove the >ability to think rationally. People under the influence of hostile memes can behave as if >possessed. They do odd things like attack countries that are from all appearances >actually trying to comply with the international mood. They also do interesting things like proclaim that they are experts >about diseases which are also called mysterious. More interestingly to me people under the influence of hostile memes >can start a war claiming they are trying to help and free people they >are attacking!!! These hostile memes can be the tip of the iceberg for groups of ideas >that in their totality are more sentient than homo sapiens sapiens. They like you though, and have endless fun playing with you, and some >of you call them demons or devils. You all depend on me shutting up, so that people won't know the truth. Insults, including talk of racial slurs, and continual references >back to the rest of sci.math with the claim that no one believes me The claim that no one believes you? Name _one_ person who does. >are apparently efforts to get me to quiet down by using intimidation >before the world finds out that there are mathematicians who will not >only will lie about important mathematics, but who seem to live in >their own little world where they make up their own rules. They are immortal. And they have been around for longer than you >have, and will be here after you're gone. However, they play by >rules, unlike many people. So I put it out there so that when they're facing the public, you know >the truth. If they whine about their importance to society, Jesus. Not one of your critics has _ever_ said anything about his importance to society. You're projecting again - the only person in all this who exhibits that sort of megalomania is you. >as if >that means they should be able to get away with betraying it, think of >the young people they were willing to teach false mathematics to, and >consider their contempt for those young minds, and the future they >represent. I'm curious about how some of you would react if you found out that >indeed I was right, and that for all these months there's been a short >proof of Fermat's Last Theorem known, but resisted by mathematicians. Would you care? Would it matter to you if they were confused or deliberately hiding >the truth? Do you think it'd matter to you if it turned out it was just a few >people who've been posting here or if a bigger number of >mathematicians than you supposed knew the truth but kept quiet? If you're a mathematician, do you think it'd have any impact on you >personally? Professionally? If you're not a mathematician, do you think it'd have any impact on >your trust of things mathematicians say or have said? Some of you may know that I also recently found what I've called the >functional definition of the prime counting function. Do you see any significance in my using the term functional? If mathematicians have been avoiding an important bit of work in prime >number theory do you think they would be doing so because they >*believe* it's unimportant, or shockingly important? If you find out that it is important work, but a large number of >mathematicians deliberately ignored it even though it was brought to >their attention in private communications, would you be more or less >likely to trust mathematicians specifically about prime numbers? What if you found out that I had information that proved my case >conclusively but was instead waiting to see if mathematicians would >act in a way that showed they would lie for their own interests. Do you think I would have justification for witholding this >information to see if they'd tell the truth? Would you feel better if I held this information until they told the >truth, waited a while and then produced it whether they told the truth >or not, or would you just as soon I shut-up whether I'm right or not >because you're just sick of me, and you couldn't care less how >important the math I've discovered is? Do you believe that if I did have important mathematical work that I >could just send it to a math journal as you feel confident that a >journal would consider it and report the information to the world if >it were correct? If you find out that even journals failed in this case, would you find >yourself more or less likely to trust pronouncements made in journals >in the future? How about science journals versus math journals? Would you consider a very large failure to tell the truth in the math >field when looking at result in other fields? If, if, if, if, if, if, if... none of these things has happened except in your imagination. >If I tell you now to buy futures in the natural gas market, and that's it, >am I not making an assertion about my expertise? In the regular world, you'd probably have context to help you evaluate my >true expertise but this is a newsgroup on the INTERNET, and it's a far more >difficult proposition. So, in the past I've told you NOT to just trust me but to check the math, >and I've often provided math for you to check. A while back I was doing a search using google at www.google.com, >where I was using my name, and various words like prime, prime >counting, and prime counting function, when I noticed something odd >using just prime counting, which was that links to some of my posts >were coming up as high as number 4 in a list of over 100,000 search >results. It turned out that only MathWorld was beating me out when it came to >the subject of counting primes. I found that fascinating, and contemplated it. I can understand that you'd be perturbed at the idea that you should >question Galois Theory (or better yet your own work which you claim >depends on it) as that is probably an idea that gets a very emotional >reaction from you. Uh, yes, when you suggest that we should question Galois theory you do indeed get an emotional reaction from mathematicians. But not the emotion you seem to think - the emotion this elicits is a sort of hilarity. >However the choice is clear, given that polynomials *are* reducible, >and the simplicity of my argument where ultimately reliance is on the >distributive principle. my prime counting work. I mentioned this interesting oddity to one of >them, and it stopped. Well, I should say the behavior *changed*, as >now you can just go to google and type in prime counting, and you >will find it more difficult to see what I've actually said, while a >link to a flame page against me now gets top billing (all still >amazing high in the search list). (I'd appreciate verification from someone else, as I'm not certain >that google doesn't have some cookie or something set so that my own >name is being used in searches when I do it. Um, you might want to >hurry though, as I'm still wondering about the speed of the last >change, which may have been a coincidence, but after this post, things >may change again.) Oh yeah, another leading mathematician told me that one out of five >graduate students who do work in the area find something like my prime >not been worth publishing either. Yeah. It's truly amazing, that the same thing can be discovered so many times by so many people and still be suppressed. Maybe there's a simpler explanation: maybe it's actually no big deal. >So here you have my claim that I've found a *short* proof of Fermat's >Last Theorem, where the methods used involve factoring polynomials >into non-polynomial factors, which you can't find in all of >mathematics outside of my work, meanwhile I face a lot of hostility >over my work, from people who can't show an error within the work >itself. So the gist of it is that the short FLT Proof, which can be found at >my website for my math discoveries, which is >http://www.msnusers.com/AmateurMath is currently unwanted and unloved, >except by me of course, as I think it's really neat!!! Along with it is a prime counting function, which you can't find in >any established math reference or even online if it's not connected to >me (last time I checked), and it's unwanted as well. So you have all these math bigwigs ganging up on my *short* math >results, and strutting around the newsgroup, and I do admit that I'd >like them cut down to size. But it looks like it'll take me some time. However, now that I have the Hammer and am getting a feel for its heft >and weight, it might finally be giant thumping time!!! I talk of the FLT Proof as Thor's Hammer because, yes I'm a mythology >buff. Is that the reason? Huh. All these years I'd assumed that the reason you refer to your Proof as Thor's Hammer is that you enjoy sounding like a complete idiot in front of the entire planet. >And it is massively incredible to have such a thing as your own >discovery of a short FLT Proof, which is indestructible, and quite >powerful, but also light and sublime. I rely heavily on its power to get me through these dark days. Still, I guess it's really not mine, but it does make me feel quite >powerful, And that's it, in a nutshell as it were - all this is about how it makes you feel. >like Thor, while my pitiful mortal frame wields it for a >time. James Harris ************************ David C. Ullrich ==== > >I say I have a proof. The math should be trivial for mathematicians. >The work is available online 24 hours a day around the world. > >Why is there still a debate? > > Because you're too dense to understand the objections. (Or too > obstinate to agree that clearly explained objections are correct.) > > I mean for heaven's sake, the Proof evidently still uses the > notion of objects, and the definition of object is _still_ > incoherent. (No, adding the -1 doesn't suffice to fix it - that > was just the only one of Arturo's objections that you understood.) > > You should answer Bernier's question: Is Pi an Object? Is > 2^sqrt(3) an Object? (And explain how the yes or no follows from > the definition. Actually you don't even have to _find_ the > answer - just explain how a yes or no _would_ follow from the > definition if you could do various calculations.) > > You also don't seem to have noticed a point Arturo made (he's > not the only person who's noticed this aspect of things): the > definition of object has been _changing_ a lot lately. But > nonethless the Proof, _using_ the notion of object, has > _not_ changed! How can the same proof be correct in > two different versions, if it uses objects and the meaning > of the word object keeps changing? > > This makes it clear to the meanest intelligence that whatever > it is you have it's _not_ a proof based on deductions from > definitions. > >Because the truth is that I'm right. Mathematics is being taught that >is false, and it has been taught for quite some time. Mathematicians >claim that they don't have any errors in core mathematics, but here >is one. Also if they admit the error then they have to acknowledge >me, then my proof of Fermat's Last Theorem and my prime counting work >should come out as well. > >And yes, the Hammer has arrived and is in full swing. I have the >momentum I've been looking for, so it's time to change the >establishment, for the betterment of all. > > You really do sound very wacky when you talk this way. Honest. > > [most of text snipped] David, The current theory is that the person to whom you are responding (jstevh@yahoo.com, NOT jstevh@msn.com) is not exactly James Harris, but rather someone who has collected Harris quotations over a long period of time and is now patching them together and posting them essentially as troll bait. That is why the wording, tone, etc., sound so much like the real Harris: they ARE the real Harris. You are of course right that as usual he sounds somewhat wacky, though the real Harris has not mentioned The Hammer for a while. Nora B. > >Still, I guess it's really not mine, but it does make me feel quite >powerful, > > And that's it, in a nutshell as it were - all this is about how > it makes you feel. > >like Thor, while my pitiful mortal frame wields it for a >time. > >James Harris > > ************************ > > David C. Ullrich ==== I found the definition that I was looking for, in a book review (I don't recommend the book): A brain is necessary to consciousness, but is it sufficient? of course, Herr Doktor-Professor Ullrich's *crux mathematicorum* also applies to this iff. why don't you find some truly recreational math to do, for a while, instead of blowing your brains out on a proof that's been done by all accounts, if not very simply.... which reminds me of one of your silly statements about that, that Andy W. didn't admit taht Gauss would've seen the gist, and fixed the error at the git-go. even if so; so, What? there is no notion, that I've heard, that Wiles has added anything elementary to the world, such as Gauss did in spades. > _not_ changed! How can the same proof be correct in > two different versions, if it uses objects and the meaning > of the word object keeps changing? --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== > > It should be instructive to readers of these various newsgroups to > note that the objections waged against my short proof of FLT are in > fact tired, treadworn, and well refuted old news. > Speaking of tired, treadworn, and well refuted... your FTL falls into that catagory. Speed of Light http://scienceworld.wolfram.com/physics/SpeedofLight.html Special Relativity http://scienceworld.wolfram.com/physics/SpecialRelativity.html Oh, and thanks for registering at http://www.google.com/search?q=%22James+Harris%22+site%3Awww.crank.net http://www.crank.net/harris.html ==== > > It should be instructive to readers of these various newsgroups to >note that the objections waged against my short proof of FLT are in >fact tired, treadworn, and well refuted old news. > > Speaking of tired, treadworn, and well refuted... your FTL falls into > that catagory. is FLT (Fermat's Last Theorem). He is presenting a short proof. After all, Fermat said he had a nice proof but there was not enough room in the margin to write it. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net ==== > > It should be instructive to readers of these various newsgroups to >> note that the objections waged against my short proof of FLT are in >> fact tired, treadworn, and well refuted old news. > Speaking of tired, treadworn, and well refuted... your FTL falls into >that catagory. > > is FLT (Fermat's Last Theorem). He is presenting a short proof. After > all, Fermat said he had a nice proof but there was not enough room in > the margin to write it. > > Chuck ==== >I am stuck on Pool of Radiance. I seem to have run out of commissions >and so I don't know what to do. The last commission given to me was >rescueing the heir of the House of B. The documentation seems to >imply there are more commissions, but they havent been offered >(something involving hobgoblins?, and something about a princess?). >What should I do? Attack the castle? Ah, well there's your problem. When you rescued the heir you should have gotten a sword. If you sold it you could easily become stuck. And that's why FTL is impossible. ==== > >I am stuck on Pool of Radiance. I seem to have run out of > commissions >and so I don't know what to do. The last commission given to me was >rescueing the heir of the House of B. The documentation seems to >imply there are more commissions, but they havent been offered >(something involving hobgoblins?, and something about a princess?). >What should I do? Attack the castle? > > Ah, well there's your problem. When you rescued the heir you should have > gotten a sword. If you sold it you could easily become stuck. > > And that's why FTL is impossible. Sigh! It is a morning of dyslexia. The thread is about a short proof of FLT and not about FTL. I think I will mention this to my exwife. She is a member of DAM (Mothers Against Dyslexia). Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net ==== > It should be instructive to readers of these various newsgroups to > note that the objections waged against my short proof of FLT are in > fact tired, treadworn, and well refuted old news. > > For example, I quote from a post from 1989: > > http://groups.google.com/groups?selm=19961226224000.RAA29848%40ladder01.news. aol.com&oe=UTF-8&output=gplain > >I am stuck on Pool of Radiance. I seem to have run out of > commissions >and so I don't know what to do. The last commission given to me was >rescueing the heir of the House of B. The documentation seems to >imply there are more commissions, but they havent been offered >(something involving hobgoblins?, and something about a princess?). >What should I do? Attack the castle? > > The relevance of this post to my short proof of FLT is self-evident. > Yeah; the princess was rescued during the assault on the kobold fortress. (As a troll and a wild bore, you were obviously fighting on the side of the kobolds.) ==== [...] I urge people to check the from address before replying to these things. jsteh@yahoo.com is NOT the James Harris well-known for claiming a proof of Fermat. The real James Harris posts from an msn address. Keith Ramsay ==== Thoughts for the day: 1. We don't normally wake up and find ourselves in Rawalpindi, Pakistan. But what if the flash mob phenomenon makes this more likely? 2. What if flash mobs make expectant mothers give birth to completely unrelated children? supermarket sales in Kiev, make money fast in Urdu, .... 4. What if this started a chain reaction of causality, and suddenly great masses of people completely lose control over their lives? ==== I'm looking for an (possibly efficient) algorithm to solve systems of quadratic equalities and inequalities where the number of quadratic terms (i.e. x*y or x^y) is small. Losely speaking, I have a lot of linear constraints, but sometimes there could be some products. Marco ==== > I'm looking for an (possibly efficient) algorithm to solve systems of > quadratic equalities and inequalities where the number of quadratic > terms (i.e. x*y or x^y) is small. > Losely speaking, I have a lot of linear constraints, but sometimes > there could be some products. > > Marco you better ask in the newsgroups sci.math.num-analysis sci.op-research Check out http://plato.asu.edu/guide.html for QP solvers. They allow an additional objective function which you don't have, so you just set it equal to zero, say. There are also solvers specifically for feasibility problems like yours. One example is FILTRANE which is part of GALAHAD (see under QP in above source). Hans Mittelmann ==== Mathematica can solve systems of polynomial inequalities. ==== > I'm looking for an (possibly efficient) algorithm to solve systems of > quadratic equalities and inequalities where the number of quadratic > terms (i.e. x*y or x^y) is small. > Losely speaking, I have a lot of linear constraints, but sometimes > there could be some products. The buzzphrase you are looking for is Quadratic Programming try typing it into , or even into Google. -- Gordon D. Pusch perl -e '$_ = gdpusch@NO.xnet.SPAM.comn; s/NO.//; s/SPAM.//; print;' ==== Let z = sum ( i = 0,1,2,3,... ) 1/(2^(f(i))) where f(x) = x iff the Continuum Hypothesis is true f(x) = x+1 iff the Continuum Hypothesis is false Is z well defined? Is z a number? Is z some sort of weird Shroedinger's Cat (sp.)? Is z complete gibberish? ==== * Doug B. > Let z = sum ( i = 0,1,2,3,... ) 1/(2^(f(i))) > > where > > f(x) = x iff the Continuum Hypothesis is true > f(x) = x+1 iff the Continuum Hypothesis is false > > Is z well defined? Is z a number? Is z some sort of weird > Shroedinger's Cat (sp.)? Is z complete gibberish? It depends on your mathematical philosophical position: An intuitionist will say it is not a number. A platonist and a formalist think it is a number. -- Jon Haugsand ==== > Let z = sum ( i = 0,1,2,3,... ) 1/(2^(f(i))) > > where > > f(x) = x iff the Continuum Hypothesis is true > f(x) = x+1 iff the Continuum Hypothesis is false > > Is z well defined? Is z a number? Is z some sort of weird > Shroedinger's Cat (sp.)? Is z complete gibberish? A simpler example is Let x = 1 iff Continuum Hypothesis is true Let x = 0 iff Continuum Hypothesis is false. I have seen Theorems that go something like this: There exists an object with certain properties if the Continuum Hypothesis is true, and there does not exist an object with the same properties if Martin's Axiom is true. I believe that Martin's Axiom is some other unprovable axiom, that is inconsistent with the Continuum Hypothesis, and also inconsistent with the Axiom of Choice. The object with certain properties was something that one would not initially have thought of as obscure set theory. For example, the existence of so called derivatives on certain kinds of Banach algebras. (Sorry it was a long time ago, so I am not even quite sure what the definition of derivative is.) Anyway, such proofs must essentially start with constructions like the ones described above. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen ==== In sci.math, Jani Yusef : > Does anyone know of any tools to convert tex documents to xml? Maybe > some other structured document format? I know there are a few tex2html > tools out there but these actually lose the math in generating > aethetically pleasing markup. There are always issues in conversions. However, there is a MathML extension to either HTML or XHTML (I forget which) which may solve part of your problem. I don't know if there's a tool to do an automatic conversion from one to the other, though. http://www.w3.org/Math/ for details on the spec. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >> I just received a calculus textbook I ordered for self study. >> Unfortunately, I didn't get the CD containing the worked out solutions >> to the old numbered exercises in the textbook which I had thought was >> included in the CD attached to the textbook. > > Would that be the ODD numbered exercises? Yes, that would be ODD numbered exercises. ==== > I just received a calculus textbook I ordered for self study. > Unfortunately, I didn't get the CD containing the worked out solutions > to the old numbered exercises in the textbook which I had thought was > included in the CD attached to the textbook. > > This incident brings me to the observation, which I suspect is common > in college math textbooks, of requiring students to pay for the > additional solutions manual instead of either (i) including the worked > out solutions in the back of the textbook or (ii) including the > solutions in the CDROM that already comes with each textbook. In the > case of my textbook, the solutions are already in the back of the > textbook. It would only require maybe 20 additional pages to include > all the worked out examples, instead of just the solutions since the > font type used is already small. In the case of the CDROM which also > comes with my textbook, there is more than sufficient space remaining > on the CDROM to include the solutions manual in PDF format (the CDROM > for the textbook had 3 MB used). > > The point that I am trying to make is that for very little additional > cost to the publisher, the worked out solutions could be included as a > few additional pages in the text or for no additional cost on the > accompanying CDROM. Instead, the publisher charges a rediculous price > for the solutions manual (in my case, over $50) most of which is > additional profit. > > It makes me angry that after spending over $120 for a textbook, I need > to spend over another $50 to pay for the solutions manual when the > publisher could have added the solutions manual to the textbook for > virtually no additonal cost (i.e. included in the attached CDROM). I > think that the publishers are abusing their situation and therefore > students should behave accordingly towards publishers when the > opportunity arises. It seems that many some readers of my post inferred that I was complaining about the price of the textbook. My complaint was for not including the worked out solutions to the odd numbered questions either together with the textbook or on the CDROM which is already included with the textbook. I agree that academic textbooks are expensive and even more so if you live in Canada. Since Canadian universities and colleges use the same textbooks as in the US, the price of textbooks are even more expensive because of the lower exchange rate for Canadian currency. For a discrete mathematics course that I had taken recently, the course used Rosen's Discrete Mathematics and Its Applications which costs $135 CDN plus the solutions manual for $55 and after tax, the amount was over $200 (the US cost after currency conversion would have been even more). Fortunately, some publishers charge a lower price for Canadian purchasers otherwise textbooks would routinely cost over $150 CDN for each textbook. Unfortunately, higher level textbooks are even more expensive relative to the size of the books and publishers are less likely to offer differential pricing to Canadian purchasers. A introductory proof text that I had recently looked at was a small 400 page hardcover text that cost $133 CDN plus tax. I am not knowledgeable about the academic book publishing business so I do not have an informed opinion as to whether the textbook costs are reasonable or not. It looks like I will be shopping for secondhand textbooks. ==== , > > >The point of my original post was that textbook publishers are >charging significant fees for solution manuals when they could have >included them with the text for very little or no additional cost. I >think the publishers are taking advantage of the situation and I think >this is wrong. > > Speaking of textbook publishers charging outrageous amounts of money, has > anyone noticed that there is now a 5th Edition of Stewart's _Calculus_? > Looking through both copies, I have so far found nothing more than minor > changes (mostly in the problems). With so many schools requiring this > text > for calc classes, James Stewart seems to be a one-man industry. I > half-wonder if they just release a new edition every time sales slow down > from a glut of used copies. > > -Davis > I'm pretty sure that's true. Most books, if they become somewhat successful, then to go into new editions every three or so years. Don ==== >>> Counterexamples to FLT can certainly be imagined >>I don't think so. I think one could imagine things that seem like >>counter examples of FLT though. >> Then maybe how you can explain how Ribet proved his theorem? >> I was under the impression that he started by imagining a counterexample >> to FLT, and from that hypothesis he reached the conclusion that there was >> also a counterexample to the Taniyama-Shimura conjecture. Are you saying >> what he did is impossible? Then why did Wiles get so excited when he >> heard the news? > Well these things are beyond my knowledge. I assumed FLT was proven. > If counter example means something that proves FLT is false, and yet experts > say FLT is true, then I assume no such counter example can exist, but as I > said > this is beyond me so there is no point in me trying to answer.. I simply > don't know (one way or the other). When Ribet proved his theorem, the news opened the door for Wiles to start trying to prove FLT by proving the Taniyama-Shimura conjecture. Seven years later, Wiles emerged with the proof (but it was more like nine years before the proof was accepted, because the first attempt had a gap). So, at the time when Ribet started imagining that a counterexample to FLT might exist, nobody knew whether one actually existed or not. Are you claiming that the mere existence of a proof makes it impossible to -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. ==== >>>I've not yet seen a careful definition that >>>establishes exactly what existence means in >>>this context (yet some people talk about it >>>often), and more importantly why it should >>>matter to me. In the most general sense, all >>>mathematical objects exist because they >>>can be imagined. Right? >> I suggest you read carefully the last quoted sentence in that paragraph >> above. Do you deny having written that? >No, but I do recall distancing myself from mathematical objects, don't >you? I had actually overlooked the possibility that you considered anything >related >to that phrase after I invested so much energy into renouncing it early in >our exchange, if that was your original beef, you must be over it now, no? I thought the important words in that sentence were exist because they > can be imagined, not mathematical objects. Especially in light of > your recent reaffirmation of the contrapositive of that statement, when > you claimed that things that don't exist are things that can't be > imagined. Sorry, but your St. Anselm act is not working. Oh, I have an act now. >Shouldn't it have been obvious to you, or any rational person, that >after all these efforts to renounce that term, that my original claim >was more or less retracted? If you don't think so - I'll retract it now. >Done. Does your retraction also apply to these statements from your previous > post? My retraction can applied to everything I've written in this thread, because it doesn't really matter. It's all informal, so who cares? It was foolish of me to involve myself in this thread because generally, as an observer I notice the vast majority of such threads (informal, philosophical type threads) are of no value, and I try to avoid them... but ya know... every so often... I get an urge. >I don't think my most general notion of exists is as controversial >as you make it out to be. Certainly, things which have no physical >presence, can't be said to exist if they can't be thought of. If one can not think of something, and one can not physically contact... You are still arguing, even as recently as a few hours ago, as if > existence is indistinguishable from being imaginable. It isn't. Yea well, I don't see it that way, sorry. Adios. -- > Dave Seaman > Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. > ==== |[...] it is easy to |see Caratheodory implies that an uncountable totally ordered set X |cannot be written as a countable union of countable sets, and that |apparently cannot be proved in ZF (e.g. for X = w_1). No, not for w_1, but for the continuum. Remember that w_1 is the smallest uncountable ordinal; the well-ordering of it gets us over the need for AC. Cohen proved that if a model of ZF exists then there exists a model in which the continuum is a countable union of countable sets. But it's possible to prove in ZF that the union of a countable family of countable subsets of w_1 is countable. If S_1, S_2, ... is a sequence of countable subsets of w_1, then they have least upper bounds a_1, a_2, ... in w_1, and the sequence a_1, a_2,... in turn has a least upper bound a. At this point we can tell that the union is not all of w_1. But that also shows that it's countable: by the definition of w_1, the segment below a is a countable set, and the union of the S_i being an infinite subset of a countable set is countable. Keith Ramsay ==== [snip] > Whatever I think of 1, 2, 3, N, P(N) is irrelevant, and whatever > I think of mathematical objects also is irrelevant... because that > never was an issue. What was an issue was: What does Herman's > student mean by mathematical object? I still want to know, but > I don't think I will.... Oh well. Since my name is mentioned quite often in this sub-thread, i've been looking for a good moment to jump in. :-) I wasn't thinking of any particular 'definition' of 'existence', and i don't think it is needed. The point was that, starting with such assumptions (whether accepting or rejecting infinite sets), has influence on what *properties* we'll give infinite sets, in the end. I think that, even without defining existence, i can explicitize what i meant with 'the student doesn't think infinite sets exist', namely by a comparison: In what sense do 'infinitesimals' exist, in classical math? Well: intuitively, we can talk about them, but 'officially' they don't exist, i.e. they are not elements of R. They are just an intuitive idea, and sentences containing them must first be 'coded' into something more formal and exact. Likewise, it should be a pretty trivial observation that sentences about N and P(N) can be likewise coded into sentences that officially only know about finite sets. What is less known, is that that leads to a form of mathematics that can give totally different answers to many mathematical questions. ==== > > |[...] it is easy to > |see Caratheodory implies that an uncountable totally ordered set X > |cannot be written as a countable union of countable sets, and that > |apparently cannot be proved in ZF (e.g. for X = w_1). > > No, not for w_1, but for the continuum. Remember that w_1 is the smallest > uncountable ordinal; the well-ordering of it gets us over the need for AC. > > Cohen proved that if a model of ZF exists then there exists a model > in which the continuum is a countable union of countable sets. > > But it's possible to prove in ZF that the union of a countable family of > countable subsets of w_1 is countable. If S_1, S_2, ... is a sequence of > countable subsets of w_1, then they have least upper bounds a_1, a_2, ... > in w_1, and the sequence a_1, a_2,... in turn has a least upper bound a. How do we know that the sup of a_1,a_2,... is < w_1? For that matter how do we even know a_1,a_2... are not equal to w_1? Certainly you can reduce the statement to showing that a countable set of countable ordinals is bounded above by a countable ordinal. In fact you can even replace countable set of countable ordinals with strictly increasing sequence of countable ordinals without too much difficulty but I still don't see how to prove that the sup is strictly less than w_1. And I've been told you can't prove it without AC, though of course I can't justify that. I didn't actually know for certain that the same held for the continuum. That's even worse - it means there's no hope of constructing a sensible measure on R in ZF (i.e. a measure for which we can prove the singletons are null and R itself isn't). Michael ==== >> >> |[...] it is easy to >> |see Caratheodory implies that an uncountable totally ordered set X >> |cannot be written as a countable union of countable sets, and that >> |apparently cannot be proved in ZF (e.g. for X = w_1). >> >> No, not for w_1, but for the continuum. Remember that w_1 is the smallest >> uncountable ordinal; the well-ordering of it gets us over the need for AC. >> >> Cohen proved that if a model of ZF exists then there exists a model >> in which the continuum is a countable union of countable sets. >> >> But it's possible to prove in ZF that the union of a countable family of >> countable subsets of w_1 is countable. If S_1, S_2, ... is a sequence of >> countable subsets of w_1, then they have least upper bounds a_1, a_2, ... >> in w_1, and the sequence a_1, a_2,... in turn has a least upper bound a. How do we know that the sup of a_1,a_2,... is < w_1? For that matter >how do we even know a_1,a_2... are not equal to w_1? Certainly you can reduce the statement to showing that a countable set >of countable ordinals is bounded above by a countable ordinal. In fact >you can even replace countable set of countable ordinals with >strictly increasing sequence of countable ordinals without too much >difficulty but I still don't see how to prove that the sup is strictly >less than w_1. And I've been told you can't prove it without AC, >though of course I can't justify that. I don't see how either. I started a reply explaining that the point was that by well-ordering we could _define_ a bijection from a to N (or an injection into N or whatever, doesn't matter) for every a < w1. But none of my explanations _quite_ worked; I always found myself choosing a sequence at some point. >I didn't actually know for certain that the same held for the >continuum. That's even worse - it means there's no hope of >constructing a sensible measure on R in ZF (i.e. a measure for which >we can prove the singletons are null and R itself isn't). Michael ************************ David C. Ullrich ==== > A little story. > A few years ago, i was attending a lecture. the audience consisted of > about 30 or 40 professional mathematicians of the kind you describe: > from all kinds of 'normal' mathematical subjects, not especially interested > in set theory or foundations. During the lecture, there was some talk of AD, and how strange it was > that AD is 'not true' (most people accept ZFC as true, and ZFC + AD > is inconsistent, so a consequence is, they think AD is not true). > There was some noise yeah, yeah, it's strange, but the shrugging of > shoulders > continued, nonetheless. For those who don't know AD, the following variant of AD is already > contradicting ZFC: Given a set A subset of N^N, and two players 1, and 2 playing a game. > 1 chooses a natural number, then 2 chooses one, then 1 chooses one, etc. > After infinitely many steps, the result is an element of N^N. > If this element is in A, 1 wins, otherwise, 2 wins. > The axiom says: either 1 has a winning strategy or 2 has. > (Either 1 can manipulate the end-result into A, or 1 cannot, that is: > 2 can prevent the en-result to be in A.) Now, if you think a little bit about this principle, it is just as > plausible a principle as AC, if not much more plausible. Finite > games are decided, why aren't infinitary games decided? It is really > strange. But nobody seemed to care. It is an interesting story, and I had never heard of AD before. But I sure don't share your claim that this principle is just as plausible as AC or more so. On the contrary, AC (in certain forms) is just obviously so, according to my intuitions. I'm thinking here of the infinite product of non-empty sets is non-empty version. The situation with AC and AD are similar. For both, we have facts that hold for finite cases and we wish to assert them for infinite cases. Maybe if I was more familiar with games, I would agree with you that AD is superficially plausible. But the fact that finite games are decided seems much more esoteric than the fact that finite products of non-empty sets are non-empty. Extending the latter to infinite sets seems more reasonable than the former. Again, maybe it's just a matter of my background. I'm not sure. -- Jesse F. Hughes I have written many words to sci.math, some of them are not even meaningless. --Ross Finlayson ==== > [snip] > Well, we do have someone who doesn't believe infinite sets exist, and goes >> so far as to say sets don't exist at all! That person is me. >> I don't believe in the Platonic realm or whatever, but I do assign greater >> degrees of realness to mathematical objects. Sets have my lowest >> realness rating. But you do talk and think about them, right? Perhaps only in disguised form. > And when you do, you probably think about them *as if* they are fixed, > completed totalities? In that case, you are assuming lots of properties > about them.... And, indeed, most mathematicians are not aware that they do exactly that: > assuming lots of things silently. They don't even care that they do. But what can be done about it? Sigh. Why must anything be done about it at all? Let the mathematicians do mathematics and let the philosophers of mathematics fuss about with what it all means and who's doing it right. It's worked well enough up to now. Not everyone cares to get into esoteric discussions of whether mathematical objects really exist, or whether Julius Caesar is a number. Apparently, even if one is apathetic about these issues, he can do good mathematics (Petry's claims notwithstanding). Others can discuss the deep (or silly) philosophical issues surrounding mathematics all they wish. It's not likely to have much impact on what it is that the majority of mathematicians do. I should point out: I don't assume (of course) that the sets of mathematicians and philosophers of mathematics are disjoint, but I do suppose that they are distinct. Also, I consider myself a member of the intersection, although my background in philosophy of mathematics is not deep. -- I've ... contacted [some of the...] highest I.Q.'s in the country... I've even helped the FBI out a few times... I've met at least one governor..., a senator... and I've had some really good seats at sports games. My experiences are not your experiences. --JSH != you ==== [snip] > > _Pace_ Wigner, what's unreasonable isn't that correct mathematics > correctly understood and applied can be effective in understanding > the world, it's that Cargo Cult Mathematics seems to actually > work sometimes (whereas Cargo Cult Science hardly ever does). > > Lee Rudolph what does cargo cult mathematics mean? Kevin ==== > I must respectfully disagree with the seeming consensus of the > mathematicians whose analyses I scanned. The seeming paradox has little to > do with the specific distributions assumed for the source of the pairs. It > is resolved in all cases by inclusion of the boundary cases in the > calculation of expected gain over any finite series. I haven't thought it through completely, but it would seem that a winning > strategy would be to switch in every case in which you observe a value less > than the maximum value you have previously observed, otherwise to stick with > your first choice. Of course if you are told a maximum bound ahead of time there is no problem. > Always switch unless you see that value. > knowing the bound IS having knowledge of the distribution! Herc ==== The Valley is still green for IITians Times News Network Mumbai - Minutes after receiving the President of India Gold Medal at the 41st Convocation of the Indian Institute of Technology, Mumbai, on Friday, Digvijay Raorane proudly announced that he's headed to the University of California for a master's degree in Nano technology. If research conditions improve here, I might consider heading back, said the 22-year-old mechanical engineer. Raorane's statement seemed an echo of what IITian Rajiv Gupta, chairperson of Rohm and Haas Company, USA, said a little earlier. We left the country because of its lack of infrastructure and efficiency, Gupta, who was the chief guest for the function, said. But such drawbacks should not deter us from our responsibility of uplifting the community to build a stronger India, he added. Quoting Mahatma Gandhi, Gupta urged the young graduates to be part of the change you want to see in the world. Adding that great ideas were empty unless one can transform them into something tangible, the 1967 IIT graduate said, One can imagine possibilities, but then you have to innovate reality. Chairman and chief mentor of Infosys Technologies Narayan Murthy was awarded the Degree of Doctor of Science (Honoris Causa) by chairman and managing director of Baja Auto Rahul Bajaj. Handing over the citation, Bajaj, who is also the chairman of the Board of Governors of IIT, Mumbai, threw in a crushing hug for Murthy, who he described as a leader, visionary and philanthropist. Accepting the citation Murthy said, Our aspirations are our possibilities. So you young graduates must always aim high. Mrthy's wife Sudha and daughter Akshata aslo accompanied him to the function. As many as 1,247 degrees were awarded to successful candidates at various levels at the function. The Institute Gold Medal was awarded to Premal Shah and the Dr Shankar Dayal Sharma Gold Medal for all-round excellence went to Nitin Dewan. Both the bright sparks have secured jobs with consultancy firms in Delhi. Listing the achievements of the institute over the past year, IIT director Ashok Misra said, We have received Rs 23.7 crores through the sponsored research programmes. This represents a 48 per cent increase over the previous year. To increase industry-academia partnerships the institute is now setting up the Society for Innovation and Entrepreneurship. Read the complete news at: http://www.timesofindia.com News Plus http://www.mantra.com/newsplus Jai Maharaj http://www.mantra.com/jai Om Shanti Shubhanu Nama Samvatsare Dakshinaya Nartana Ritau Kark Mase Shukl Pakshe Shukr Vasara Yuktayam Mool Nakshatr Vaidhruti-Vishakumbh Yog Bav-Balav Karan Dvadashi Yam Tithau http://www.mantra.com/holocaust http://www.hindu.org http://www.hindunet.org The truth about Islam and Muslims http://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for the educational purposes of research and open discussion. The contents of this post may not have been authored by, and do not necessarily represent the opinion of the poster. The contents are protected by copyright law and the exemption for fair use of copyrighted works. considered or answered if it does not contain your full legal name, are not necessarily those of the poster. ==== I can't believe what suckers we have been as Indian citizens for these past decades .... providing a top-notch good-as-any-other education to these bastards who simply hop on the first bloody flight to the west and flee. If it were up to me every %!#$@ admitted IIT student would pay the Rs equivalent of $40K/yr for tuition/room/board unless they agreed to be contractually bound to working in India for at least X number of years. If you want a good education - pay for it you filthy bloodsuckers. If you want to head out west .. then feel free to go out there and spend $140K on your fucking undergrad degree. Don't count on the Indian govt. to help you get there for free. And before any of you friggin trolls start bitching ... I am an Indian ... and I did choose the West too ... but I didnt leech a quality undergraduate education off of the Indian taxpayer - I moved to the US and worked my way through college .. shelled out that $140K for a undegrad degree. I don't see why the !@##! the government doesnt get its act together and charge you what that education is worth. -Bum > The Valley is still green for IITians Times News Network Mumbai - Minutes after receiving the President of India > Gold Medal at the 41st Convocation of the Indian > Institute of Technology, Mumbai, on Friday, Digvijay > Raorane proudly announced that he's headed to the > University of California for a master's degree in Nano > technology. If research conditions improve here, I might consider > heading back, said the 22-year-old mechanical engineer. Raorane's statement seemed an echo of what IITian Rajiv > Gupta, chairperson of Rohm and Haas Company, USA, said a > little earlier. We left the country because of its lack of > infrastructure and efficiency, Gupta, who was the chief > guest for the function, said. But such drawbacks should not deter us from our > responsibility of uplifting the community to build a > stronger India, he added. Quoting Mahatma Gandhi, Gupta urged the young graduates > to be part of the change you want to see in the world. > Adding that great ideas were empty unless one can > transform them into something tangible, the 1967 IIT > graduate said, One can imagine possibilities, but then > you have to innovate reality. Chairman and chief mentor of Infosys Technologies Narayan > Murthy was awarded the Degree of Doctor of Science > (Honoris Causa) by chairman and managing director of Baja > Auto Rahul Bajaj. Handing over the citation, Bajaj, who is also the > chairman of the Board of Governors of IIT, Mumbai, threw > in a crushing hug for Murthy, who he described as a > leader, visionary and philanthropist. Accepting the citation Murthy said, Our aspirations are > our possibilities. So you young graduates must always aim > high. Mrthy's wife Sudha and daughter Akshata aslo accompanied > him to the function. As many as 1,247 degrees were > awarded to successful candidates at various levels at the > function. The Institute Gold Medal was awarded to Premal Shah and > the Dr Shankar Dayal Sharma Gold Medal for all-round > excellence went to Nitin Dewan. Both the bright sparks > have secured jobs with consultancy firms in Delhi. Listing the achievements of the institute over the past > year, IIT director Ashok Misra said, We have received Rs > 23.7 crores through the sponsored research programmes. > This represents a 48 per cent increase over the previous > year. To increase industry-academia partnerships the institute > is now setting up the Society for Innovation and > Entrepreneurship. Read the complete news at: > http://www.timesofindia.com News Plus > http://www.mantra.com/newsplus Jai Maharaj > http://www.mantra.com/jai > Om Shanti > Shubhanu Nama Samvatsare Dakshinaya Nartana Ritau > Kark Mase Shukl Pakshe Shukr Vasara Yuktayam > Mool Nakshatr Vaidhruti-Vishakumbh Yog > Bav-Balav Karan Dvadashi Yam Tithau http://www.mantra.com/holocaust http://www.hindu.org > http://www.hindunet.org The truth about Islam and Muslims > http://www.flex.com/~jai/satyamevajayate o Not for commercial use. Solely to be fairly used for the > educational purposes of research and open discussion. The contents of > this post may not have been authored by, and do not necessarily represent > the opinion of the poster. The contents are protected by copyright law > and the exemption for fair use of copyrighted works. > considered or answered if it does not contain your full legal name, > are not necessarily those of the poster. ==== Apparently Bum could not get into IIT. Bum barfed, farted, whined and revealed his jealousy thus: > I can't believe what suckers we have been as Indian citizens for these past > decades .... providing a top-notch good-as-any-other education to these > bastards who simply hop on the first bloody flight to the west and flee. If > it were up to me every %!#$@ admitted IIT student would pay the Rs > equivalent of $40K/yr for tuition/room/board unless they agreed to be > contractually bound to working in India for at least X number of years. > > If you want a good education - pay for it you filthy bloodsuckers. If you > want to head out west .. then feel free to go out there and spend $140K on > your fucking undergrad degree. Don't count on the Indian govt. to help you > get there for free. > > And before any of you friggin trolls start bitching ... I am an Indian ... > and I did choose the West too ... but I didnt leech a quality undergraduate > education off of the Indian taxpayer - I moved to the US and worked my way > through college .. shelled out that $140K for a undegrad degree. I don't see > why the !@##! the government doesnt get its act together and charge you what > that education is worth. > > -Bum > ==== (This can't be exactly right, because the most famous conformal >field theory with the Monster as symmetries is not supersymmetric, >and its partition function is the j-function, which is a modular >function of weight 0, not a modular form of weight 24. So, my >brain must have been a bit fried by the time we got to this really >far-out stuff.) You might be interested in: BEAUTY AND THE BEAST: SUPERCONFORMAL SYMMETRY IN A MONSTER MODULE. By Lance J. Dixon (Princeton U.), P. Ginsparg (Harvard U.), Jeffrey A. Harvey (Princeton U.). HUTP-88-A013, PUPT-1088, Apr 1988. 30pp. Published in Commun.Math.Phys.119:221-241,1988 There's a scanned version on line. Note that they are working in lightcone gauge so c=24. Aaron -- Aaron Bergman ==== >> Still waiting for you to refute Nora Baron, James. What's the holdup? >> Toby What you missed it? I actually replied to Nora Baron in the thread >Constant factors and polynomials and basically shredded > > in your dreams that >poster's arguments. I just checked and Nora Baron has not replied. Yup, Nora Baron has given some math whoppers. Don't believe me? >Then believe the math. > > Good! Let's see these whoppers! They are available to readers in the Google archives by use of Google Groups. >Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). >Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and > > *** Make a note of this for reference later: > > w_1 w_2 w_3 = f. Ok. > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves >b_3 = 3. Essentially objections to how f^2 divides off now come down to >claiming that the w's are functions of m, but consider that w_1 w_2 = >1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. > *** Note this also for reference later. Ok. > >Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) > Above you said: > > > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) > > > Clearly in the expression you just gave, > > P(m)/f^2 = (b_1 x + u)(b_2 x + u)(b_3 x + uf), > > > you intend that w_1 = 1, w_2 = 1, and w_3 = f. Right? Yup, when f is coprime to 3. > But above you say w_1 w_2 = 3^{2/3}. If w_3 = f and f = 3, > you get > > w_1 w_2 w_3 = 3^{5/3}, > > which is clearly wrong, given that you know the constant > term of P(x)/f^2 is u^3*f. As you noted above, > > w_1 w_2 w_3 = f = 3. > > This just looks totally confused. And this is the second > time you have posted this. Maybe you should think about > fixing these whoppers. Well the real dependency is on the *coefficients*, which is why it matters whether or not f is coprime to 3. Readers should now realize that Nora Baron is either clueless or possibly in that sad state of hanging on to beliefs that have been refuted. Some readers have been led astray into believing that reducibility over Q is what matters when it comes to distribution of factors in a factorization, when in fact it's the coefficients that matter, as I show above. Here Nora Baron is reduced to ignoring the mathematical facts, which is that the factorization varies based on whether or not f is coprime to 3, and not on the value of m. That shows you the difference when mathematics is about society. If Nora Baron weren't defending a social position, why would the poster make such a post? I say it's the fault of the sci.math newsgroup for siding so often with posters for social reasons, instead of mathematical ones. It corrupts the mathematics allowing errors to sneak in, and destroy the value of the work. James Harris ==== To James Harris (and company) Why don't you show your FLT proof (post it in their Math Forum) to some very enthusiastic Math amateurs/would-be professional mathematicians in http://eng.mathdb.org/ These fellows are quite keen in elementary mathematics (being mostly composed of past IMO contestants). Math Observer ==== Hey James, I gave an HONEST suggestion. I thought that your elementary proof of the FLT could perhaps be propagated here in HK as well (with so many young math enthusiasts around!). Math Observer PS: I'm an HK Math Observer (and I don't know the guys you mentioned!). PPS: I am simply interested in making the amateurs here do something useful (discuss and scrutinize your elementary proof (?) using elementary methods). ==== > To James Harris (and company) > > Why don't you show your FLT proof (post it in their Math Forum) to > some very enthusiastic Math amateurs/would-be professional > mathematicians in > > http://eng.mathdb.org/ > > These fellows are quite keen in elementary mathematics (being mostly > composed of past IMO contestants). > > Math Observer Hey, I went to the link and tried to make a post, but just got some error. How about you do me a favor and post the following there and post back if there are any interesting comments? Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must equal 1, if f is coprime to 3, which leaves b_3 = 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m. That is, the w's are now all constant with regard to m and have the same value no matter what the value of m is. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. I've found the Ring of Objects which includes the ring of algebraic integers, and does not have this problem, as the b's are all included in it. The Ring of Objects is the set of all numbers where -1 and 1 are the only members that are both a unit, i.e. factor of 1, and an integer, where no non-unit member is a factor of any two integers that are coprime. That definition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also find information on my other math research. James Harris ==== I just read under the button, Math Lab and, although I'd be chary of introducing constructions without a pair of compasses & paper (and a flat or spherical table) as a demo, at the least, I have to agree with the general idea. I've never used any such software, at any rate. (actually, this was the *bete noir* of my own 1-year geometry course in public school: no constructions; totally leftbrained .-) > http://eng.mathdb.org/ --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== [snip] C'mon, James. Give us just *one* number which should be in the ring of algebraic integers but is 'left out'. The algebraic integers are defined as the roots of monic polynomials with integer coefficients. What's missing? You have repeatedly made the claim that there are such numbers -- name one! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== that really sounds like Son of Harris and Plutonium, I must say!... they maintain about the same level of grammatical finesse, as well; eh? > Mathematics isn't supposed to be this way. Some might have thought > there was some dignity in the discipline, some decorum. --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== In sci.math, |-|erc [naked woman snipped] You are a weird one, sir. Far better porn is available at http://www.persiankitty.com, for example. (18 and older only. Legal restrictions on viewing porn may apply in some areas. Some images may be offensive to some viewers. The poster disclaims all knowledge of and responsibility for any acts, lewd or otherwise, resulting from viewing of pictures on this site. The poster has no affiliation with that site or sites referred therefrom. Offer void where prohibited. May cause warts, hairy palms, and disgusting bodily emissions.) Followups to the obvious place. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== Some countable ordinals have a finite desciption in English (abbreviated hereon to FDIE). For example 'the first countable ordinal which is greater than any finite ordinal' is a FDIE of a countable ordinal. So is 'the first countable ordinal which is greater than any finite ordinal raised to the power of the first countable ordinal which is greater than any finite ordinal'. However, not all countable ordinals do have a FDIE. The reason for this is because the number of FDIEs is countable (quite straightforwardly, the set of finite combinations of a finite set of symbols is countable); On the other hand, the set of countable ordinals is UNcountable. This raises the question of whether the FIRST countable ordinal which has no FDIE has a FDIE. If it DOES, this would contradict the fact that it's the first countable ordinal which does NOT have a FDIE. If it DOESN'T, this would contradict the fact that I've just give a FDIE of it, namely: 'The first countable ordinal which does not have a FDIE'. ==== >Some countable ordinals have a finite desciption in English >(abbreviated hereon to FDIE). > [...] >This raises the question of whether the FIRST countable ordinal which >has no FDIE has a FDIE. > Sounds to me like the proof that each natural number is an interesting number... -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== |>Some countable ordinals have a finite desciption in English |>(abbreviated hereon to FDIE). The restriction to countability isn't strictly necessary, of course. |[...] | |>This raises the question of whether the FIRST countable ordinal which |>has no FDIE has a FDIE. | |Sounds to me like the proof that each natural number is an interesting |number... I agree. I think the possibility of describing a mathematical object like an ordinal (by way of a well-ordering on the natural numbers, say) is intimately related to its holding an interest for us, enabling it to be distinguished from among the closely related objects by means of a finite amount of pointing out. I would call it a strengthening of the paradox of the first uninteresting natural number, because while one can imagine that the natural numbers are all at least microscopically interesting, it's not so easy to imagine from a realist perspective that ALL ordinals (going up to ones of arbitrarily large cardinality) continue to have some positive interest to them. The paradox is that these concepts are slippery enough not to be able to reflect back on themselves in such a way. Keith Ramsay ==== > Nonsense, just take the probability of an event to be > its Lebesgue measure. Try some post-1850 mathematics > for a change. Game over. Aw, Wade, you're right but also that's kind of harsh, considering that a significant percentage of posts have no math content whatsoever, posts in this thread being among them. By the way, have you found a job yet? ==== > Nonsense, just take the probability of an event to be >> its Lebesgue measure. Try some post-1850 mathematics >> for a change. Game over. Aw, Wade, you're right but also that's kind of harsh, That's kind of harsh? I wonder why there would be harsh replies to your posts in this thread? >considering that a significant percentage of posts have >no math content whatsoever, posts in this thread being >among them. What Dale said: You should make up your mind. Making posts that _appear_ to be attempts at math and then just disclaiming them this way is a little... can't quite come up with the right word, lemme just repeat what he said: >> In short, since you have no academic stake in the discussion, >> you are free to toss out as many nonsensical statements as >> you like, and no one should call you to explain your reasoning >> for any of them. >> >> Note that I don't intend this as any form of castigation, and >> I apologize if that sort of tone comes across: such would be >> unintentional. However, civility suggests that people engaged >> in a discussion deserve to know who is actually engaged in the >> topic, and who is merely using the discussion as a handy source >> of amusement. > ************************ David C. Ullrich ==== In sci.math, Nat Silver : > http://mr-31238.mr.valuehost.co.uk/assets/Flash/psychic.swf > Clever, but one notices after awhile that all of the multiples of 9 have exactly the same symbol. Makes one go hmm.... even if one has not seen a variant of this game before (which I for one have). -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== For continuous data f(x) on [a,b], we can calculate the mean or average value as A = 1/(a-b) int{a,b} f(x) dx My question is can we calculate the variance of a continuous function as 1/(a-b) int{a-b} (f(x) - A)^2 dx For N discrete data points, we can divide by either N or N-1 to get biased or unbiased estimator, but is there this distinction in continuous case? ==== > math is my main passion, however unfortunately i dunno whether i have > the deductive reasoning needed to go all the way to phd. ahh life, > God seldom matches what you like most with what you're most talented > at. anyway, im practically entirely self-taught and am just now after > many years getting to a point where college is an option financially. > but having very little college classtime im utterly unfamiliar with > what the general requirements for a 4-year degree are. i was always > under the impression a 4 year degree required some extremely high > maths, but recently had a brief opportunity to talk to a coworker who > has one, and from what i gathered she didn't do much beyond your basic > intro theory of algebra/theory of calculus!.. i asked what about > galois theory and she didnt even know what it was... so now im > thinking maybe i have a chance for a 4 year after all (in the subject > where my heart lies) (in lieu of a math degree i was thinking of going for something like > phd in computer programming with a pure math minor, but after that > conversation now i dunno WHAT to do... computer programming is where > i seem to have the most innate talent, though my heart is already > stolen by maths...) Here are is the Two different Degree Plans offered by University of Colorado, having about Average standing in Math among Large State Universities. *********** Mathematics Plan I Required Courses Semester Hours Calculus 1, 2, and 3 MATH 3000 Introduction to Abstract Mathematics or MATH 3200 Introduction to Topology MATH 3130 Introduction to Linear Algebra MATH 3140 Abstract Algebra MATH 4310 Introduction to Analysis A two-semester upper-division sequence approved by the Department of Mathematics and upper-division math electives 12 Mathematics Plan II Required Courses Semester Hours Calculus 1, 2, and 3 12- MATH 3130 Introduction to Linear Algebra MATH 4430 Ordinary Differential Equations MATH 4650 Intermediate Numerical Analysis One of the following courses: MATH 4510 Introduction to Probability, MATH 4470 Introduction to Partial Differential Equations, MATH 4450 Introduction to Complex Variables, MATH 4330 Fourier Analysis, or MATH 4120 Introduction to Operations Research A two-semester upper-division sequence approved by the Department of Mathematics and upper-division math electives ***** Comment by RJ P Upper Division Math Electives Include Topology. Set Theory. Statistics, Number Theory and Many other Choices. Also Inter-plan electives are sometimes chosen. It looks like Plan 1. would fit your interests more. Coupled with a few Computer Science Courses in Programming and Data/Control Structures, your prospect of Technical Employment would be competitive among Entry-level jobs in a serious Info Tech career. Skills rather than intensive Mathematics, but the Colleges of Education and State Certifications have strengthened the requirements from the Old days where anyone who cot a C in Calculus and took Intro to Abstract Math for Secondary Teachers, and Statistics for Educators ( Watered-down theoretical content) could have enough credits to be certified to teach Math. Don't get me started on who's Actually teaching Math to our youngsters!! Good Luck!! Bob Pease ==== > In reading the book The Man Who Loved Only Numbers about Paul Erdos, by > Paul Hoffman, I learned that Erdos presented a elementary proof that for all > natural number n, exists a prime number between n and 2n. Can anyone tell > me where can I find Erdos' proof? _Proofs from the Book_ by Martin Aigner and Gunter Ziegler (Springer, 1998). Look up Bertrand's Postulate in the book. ==== > > > In reading the book The Man Who Loved Only Numbers about Paul Erdos, by > Paul Hoffman, I learned that Erdos presented a elementary proof that for all > natural number n, exists a prime number between n and 2n. > > Can anyone tell me where can I find Erdos' proof? Prof. Robin Chapman has written a simple proof of Bertrands's postulate. http://www.maths.ex.ac.uk/~rjc/etc/bertrand.pdf