mm-56 === Given two arrays: P 3 x n and p 2 x m where m << nand given P_sub 3 x k and p_sub 2 x k where k < ms.t. these two arrays have correponding columnsusing P_sub and p_sub we can recover the transformation (A,t) between them i.e.p_sub = A * P_sub + tthe question is: using A,t and p (2d points)how can I ?d 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 ?ure out what are the secret numbers they want me to searchfor such that the conversation can be possible ( see the problem).Problem C Secret NumbersInput File: C.DATProgram Source File: C.PAS or C.C or C.CPPTwo natural numbers a and b are chosen (1 this is originally informatic problem, but it's a math challenge too.> I cant ?ure 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 oftwo primes or as the sum of a prime and its square (otherwise S wouldnot know that M could not know ...). Every even number in the rangegiven can be represented as the sum of two odd primes, and a primeplus its square is always even, so s can't be even. Also, s can'tbe 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 sumwhich 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 theproduct of the summands can be factored in only one way in whichthe 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 numberscan 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 in?itely 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 notentirely certain what aleph_1 = 2^aleph_0 means; as far as I'mconcerned, if card(N) = aleph_0, then the set of all subsets of N(including itself, as it turns out) could be termed 2^N, withcardinality 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.netIt'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 fromY to X. (This meshes with the meaning m^n has for nonnegative integersm and n. If X has m elements, and Y has n elements, the number offunctions from Y to X is m^n.) Since 2 is used to represent a two-elementset such as {0,1}, 2^N stands for the set of functions from N to {0,1}.These can be identi?d with subsets of N by associating the function f(n) = {1 if n is in S {0 if n is not in Swith the set S. (This uses the law of excluded middle, in the form ofeither n is in S or n is not in S. Rarely will anyone mention where theyuse the law of excluded middle, but I sometimes do because of my interestin 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|. So2^aleph_0 stands for the cardinality of 2^N. That cardinality is alsoknown as the continuum.|with|cardinality aleph_1. At least, such is my understanding of the alephs.Not necessarily correct. By de?ition, aleph_1 is the cardinality of thesmallest uncountable ordinal. It's a cardinality > aleph_0, with nocardinalities between them. Assuming the Axiom of Choice the cardinals arelinearly ordered by <, so we can call aleph_1 the ?st cardinal greaterthan aleph_0. (Usually the axiom of choice is just assumed. It's notmentioned very often that it's being assumed. This is again something Imention because it's the other principle not used in constructivemathematics.) ;-)The equation aleph_1 = 2^aleph_0 is known as the continuum hypothesis.Goedel proved that it's consistent with the standard axiom system forset theory (the Zermelo-Fraenkel axioms plus the axiom of choice, ZFC),assuming that ZFC is consistent to begin with. Cohen proved thataleph_1 < 2^aleph_0 is also consistent with ZFC, assuming again thatZFC is consistent.Some people think that the continuum hypothesis is analogous to theparallel axiom in geometry. They think that there's no such thing as thecorrect answer to whether it's true. Some people think that there isa correct answer. The last I read, it was alleged that among the settheorists who think there's a real answer, more of them think that thecontinuum hypothesis is false than think it's true. I tend to suspect itis false as well. Why should there be a one-to-one correspondence betweenthe smallest uncountable ordinal and the subsets of N? But that's justspeculative 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} canbe 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 oneis i-abelianity. It's fun to make up words like this. My favoritenonexistent word I've come up with this way so far isabelianiyat (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 mand>f are almost independent - they can be anything as long as f is coprime to 3and>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 allhow 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> speci? 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, andit's the speci? 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, con?ent 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 checkingat>m=0 is>suf?ient, 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 ittill then.You didn't yet show the proof of how checking at m=0 is suf?ient to prove thata_3 is coprime to f for all m. If you follow mathematical logic that should ?ish your objections.The mathematical logic hasn't been presented yet.> > In fact the b's are never forced into a ?ld, 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 de?ition of algebraic> integers, as it's not as inclusive as it should be. > Your assumption that the ?ld 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> de?ition of algebraic integers as roots of monic polynomials with> integer coef?ients. That de?ition 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. Itis>agreed that b_1b_2a_3 is coprime to f. It's not agreed that a_3 is coprimeto 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 bethat b1*b2*a3 is coprime to f does not mean that a3 is coprime to fRecall? b1*b2*a3 = 21, coprime to 5. But a3 = 25. This happens becauseb1 and b2 are not algebraic integers.In your work, b1 and b2 are not in general algebraic integers, so you need toprovide 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 anypressure to>convince me of anything.>The choice, and the ?s 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 thethings 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 ?d that question intriguing.I'm just waiting to see the mathematical logic. If I see it, and it's valid, andI reject it, then you won't be intrigued for long, you'd be entitled to bedismissive 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 ?e with those values, so I won't try.>>How about YOU try when f=5, and m is your favorite non-zero integercoprime> 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 thea's>and b's to their cubic de?itions, the steps you took to get theredisappear 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 suf?ient to show that theresult>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'sextended.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 thisassertion.b_1*b_2*a_3 is coprime to f.You assert that a_3 is coprime to f when b_1 andb_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 inthat 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. Iknow>the a's and b's depend on m; but I know you know that as well, it makes methink>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 thatw_1 w_2 b_3 = 3 andw_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 twoseparate examples. Which is ?e, 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 howyou 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 assertionthat there's a problem with the ring of algebraic integers. Perhaps when we seethat 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 de?ition and more is linked to from my primary website http://groups.msn.com/AmateurMath where you can also ?d 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 ?d 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.shtmlgives 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, lowweight, penis on a re-?anced 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 needto 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 dtThis 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 partsgives precisely the desired recurrence relation.)int[t=0..inf] e^-t (t-1)^n dtSplit the range.int[t=0..1] e^-t (t-1)^n dt + int[t=1..inf] e^-t (t-1)^n dtint[t=0..1] e^-t (t-1)^n dt + int[t=0..inf] e^-(t+1) t^n dtint[t=0..1] e^-t (t-1)^n dt + 1/e int[t=0..inf] e^-t t^n dtint[t=0..1] e^-t (t-1)^n dt + n!/eThe integral is certainly less than 1 for any n >= 0. What is more,it alternates in sign and diminishes. For n>=1, the ?st term isless than 1/2. For this reason, if you can show that the originalintegral is always integral in value, the proof is complete, exceptfor its failure at 0.int[t=0..inf] e^-t (t-1)^n dtSince 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)^xint[t=0..inf] e^-t [t^n - C(n,1) t^(n-1) + C(n,2) t^(n-2) - ... +(-1)^x] dtint[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] dtn! - C(n,1) (n-1)! + C(n,2) (n-2)! - ... + 1Thus, it is just the sum and difference of integers. It follows thatit 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 ?ld automorphism phi of the real numbers.> Luckily R has very few ?ld 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 bythe TeX symbol Re.The inertia of notation is rather odd, isn't it? E.g. people still writes=sigma+it when talking about the zeta function (but no other subject) --the same notation introduced by Dirichlet in his 1837 paper on arithmeticprogressions.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 coef?ients, ***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 thisclaim, 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 cannotbe an algebraic integer. Therefore the conclusionof 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 productis 65. So it looks like both of us arrive at a contradiction. Wedraw different conclusions from it, apparently. I concludethat your claim is false and that therefore there is necessarilyan 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. Whatthat 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 theyare not closed under addition and multiplication or that thedistributive law does not hold, etc.. This however is a veryold 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 yourown proof? Why, in view of your atrocious track record over8 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, ?e. 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 theperson providing it are unimportant. But I've had no experience withthe review process and am happy to defer to those who do.-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@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 coef?ients, ***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 errorthat 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 whetheror not after that, and a quali?r any for the a's. It's prettyclose 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 commonfactors; but in any case, your claim here is certainly false, sincetheir product is not coprime to 65.Lemma. Let R be the ring of all algebraic integers, and let a, b, c beany elements of R. If a and b are coprime to c, then a*b is coprime toc.Proof. We use the characterization of coprime valid for commutativerings with 1: a and b are coprime in R if and only if there exist xand y in R such that ax+by = 1. Since a and c are coprime by assumption, there exist n and m in R suchthat an+cm = 1. Since b and c are coprime by assumption, there exist rand s in R such that br+cs = 1.Multiplying both together, we have1 = (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, andab*x + c*y = 1. Therefore, ab and y are coprime. QEDSo, assume you were correct and neither a_1, a_2, nor a_3 have ANYnon-unit factors in common with 5 in the ring of algebraicintegers. Then, by the lemma, neither does a1*a_2; and applying thelemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, whichclearly has 5 as a nonunit common factor with 5. This contradicts theassumption that none of a_1, a_2, a_3 have common non-unit factorswith 5 in the ring of algebraic integers. Therefore, your assertion isfalse. === = === ==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 ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite 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 Magidinmagidin@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 in?ity 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>> coef?ients with gcd 1. (4 is a coef?ient 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 divisibleby 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 nonprimeadditional possibilities ensue, depending on u. I'd haveto work out the gloppy details.This subproblem gets a little complicated, at ?st blush, butit'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, especiallyif one's assuming the a's are all rational. However, since you'vede?ed P(m) as a product of f^2 with something else, one canalways compute, say, Q(m) = P(m) / f^2, fairly trivially. Thedivisor is not dependent on m. (Whether this is useful isnot 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 de?ition, unaware of the truth.> On the other hand, admitting the truth has a de?ite 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 bene?> 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 pro?, who has made extraordinary math ?ds.> Mathematicians may see a social bene? to obscuring my ?ds 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 needsto be clari?d 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 turnsout to in fact be the case as you've de?ed P(m) in acertain 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 muchfurther although I can equate the product thereof to theterm in front of x^3, if I wished to, and work out therest of the terms to establish 3 equations in 3 unknownsrelating the a's.)Apologies if that wasn't clear. As you can see the f^2 factorwaltzes in again; one has to be careful if one drops it fromthe intermediate computations. This is ?e.As for your computation of P(0) -- you've simply left out thef^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) isa 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 ?st 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 speci? at this point.> Being speci? 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, andapparently the diagonizalization proof of Cantor provingthe 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 pro? 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 comedown to whose proof is simpler, but this sort of thingoccasionally happens in mathematics: two people, workingindependently, discover a very similar proof, method,or identity.Check out the history of solving x^3 + ax^2 + bx + c,for example, inhttp://mathworld.wolfram.com/CubicEquation.html(along with the actual solution :-) ).> James Harris-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === 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 inP(m) = (a_1x + uf)(a_2x + uf)(a_3x + uf),m=0 *is* a special case as exactly two of the a'sbecome 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 de?e P(m), for those who maywander 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.netIt's still legal to go .sigless. === > [.snip.]>>theory:> THEOREM: If r is a root of a non-monic polynomial>> with integer coef?ients, ***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 quali?r 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-unitfactors 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 de?ition only *one* of the a's is coprime to 5, but none ofthem 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 de?ition of coprime thatArturo Magidin gives, though it is, interestingly enough, true that infact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 isnot in the ring.However, rather than use dueling de?itions or argue aboutde?itions I can simply switch to saying that 2 does not sharenon-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 de?ition of coprime NONE of the a's have a factor incommon with 5, in the ring of algebraic integers, and you cannot provethat any of them do.Now if you don't want to call that coprime ?e. It doesn't changethe situation.What I can do is show that with a very quick argument using basicalgebra 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 leavesb_3 = 3.Essentially objections to how f^2 divides off now come down toclaiming 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 thesame 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, andusing m=1, f=sqrt(5), gives 65x^3 - 12xy^2 + y^3which results in b's that are NOT algebraic integers.I've found the Ring of Objects which includes the ring of algebraicintegers, and does not have this problem, as the b's are all includedin it.The Ring of Objects is the set of all numbers where -1 and 1 are theonly 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 arecoprime.That de?ition and more is linked to from my primary website http://groups.msn.com/AmateurMathwhere you can also ?d 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 ?st 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 ?d 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 ?d 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 anyof those paths are interesting. :) For example, thesequential-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 4belongs 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 1The ?st column is how many paths a pattern belongs to.The second column is how many patterns belong to that number ofpaths. The third column is the product of the ?st two, orsum of products when the ?st column is a range. The lastgroup of columns shows an example of a count sequence that belongs to the number of paths given in column 1.(In the ?e, there is also an Eg# column that shows programcode numbers of the example count sequences. Also, the ?e 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 obviouslyenough equivalent to each other. A similar custom is observed withde?itions; if two are obviously equivalent, they're consideredde?itions of the same concept. I suppose it could be called a familyresemblance concept.Others here have referred to logical equivalence, but by itself logicalequivalence is too weak a relationship. Two theorems can be logicallyequivalent in a way that's not obvious enough for them to count as thesame fact expressed two ways.I think there's a subjective element to it. Two statements that areconsidered separate facts in an elementary discussion may be deemed thesame fact on a higher level of sophistication.Stabs have been made at rigorously de?ing a concept of obviousequivalence, but so far as I know they haven't been very successful!This is done sometimes by philosophers for a theory of knowledge. Some ofthem 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 canalways switch to a stricter way of talking in which one refers speci?allyto particular expressions of the theorem, without worrying about whetherthe 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 15 331 2 2 2 6 181 0 0 4 121 0 4 81 4 41 01I think this can be extended in?itely to the rightand 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 thatthe following is true?(*) If a_0 = 2, a_i is odd for i>0, and a_i < a_(i+1) < 2 a_i, thenthe left diagonal of the absolute-difference table is always 1.I think this is the point of the claim that the conjecture is a weakstatement about the primes, and the example above gets by because itstarts with a 4.I see two options. If (*) is true, then the statement on primesfollows by Bertrand's Postulate and so has no added arithemtic value.But then the essence is to prove (*), which is perhaps an interestingcombinatorial, prime-free, statement, and appears to be non-trivial.If (*) is false, then there is indeed extra content to the statementon the primes beyond Bertrand's Postulate, in which case we go back tomy original question of what is it that the statement is saying aboutthe primes, exactly.I couldn't decide one way or another regarding (*), but I may havemissed 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 ?st 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 anerror, use the values in the left column only, shift the smaller rightprime to the left and the larger left prime to the right. One set at atime 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 ?st> hand what happens.> I hope I explained clearly of what is going on!> If not,any questions or replies welcome.> > DanDan === [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> ? od;> printf(No period found, maybe increase eps);> end;Robert,many thanks, but I forgot to mention that the e entries are in generalcomplex.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 Vrelease 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-- Ioannishttp://users.forthnet.gr/ath/jgal/____________________ _______________________Eventually, _everything_ is understandable. === lim n->oo (1+1/x)n! / (n! n^(n/x) (n/x)) = 1Watch 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) = 1How about ((1+1/x)n)! = n! n^(n/x) (n/x) for x =1?lim (2n)! / (n! n^n n) = 1The 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) = 1The 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)) = 1That looks funny. Do you know any expressions g(n) such that thein?ite 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) = 1The expression has the term n^n which ?ds a place in Stirling'sequation:lim n->oo n! e^n / n^n sqrt(2Pi) sqrt(n) = 1n^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) = 1We have that;lim y->oo sqrt(y Pi/2) y! / (y/2)!^2 2^y = 1y!/ (y/2)!^2 = 2^y / sqrt(y Pi/2)Setting n=y/2, y = 2ny! sqrt(2Pi) sqrt(y/2) / (y/2)!^2) e^(y/2) = 1y! / (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^nlim n-> oo 2^2n / e^n = sqrt(Pi)lim n->oo 2^4n / e^2n = PiConsider again:lim n->oo (1+1/x)n! / (n! n^(n/x) (n/x)) = 1Let x be 1/n, approaching zero from the positive side. That isn't aninteger.((1+n)n)! / (n! n^(n^2) n^2) = 1(n^2 + n)! / (n! n^(n^2) n^2) = 1Ross === >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 ?ite:integer n.You had (5n/4)! on top before - and you then in fact did say they wereequal.: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 theratio does not converge to 1. Indeed, if you do mean (5/4)! on the top (bywhich one presumes you mean Gamma(9/4) then clearly the limit is 0. On theother hand, if you really mean (5n/4)! on the top then the expressiondiverges - 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 curvaturek:curvatureB : unit binormal vectorU : unit normal vectorplease, 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.UDavid 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 ?e 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 = notunderstood) account of groups and loops. Is relators the acceptedname 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 ?e 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 inthese generators is a string or word in the symbols xi and xi^-1 whichevaluates to the identity in G.For example, S3 is generated by x=(1,2,3) and y=(1,2), and (xy)^2 = xyxyis 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 requires3 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 bedif?ult to decide whether they are 2-generator or 3-generator groupswithout specialised knowledge.There may well be smaller equally dif?ult 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 theoristsof a topological bent is that a relation is an equation betweentwo words, while a relator is a word (by intent, one which willbe put into an equation with the identity on the other side).For groups, this seems really to be a matter of taste; but forother structures (like non-cancellation semigroups), obviously not every relation can necessarily be replaced by an equivalent relationwhich sets a word equal to the identity (in fact, there may be noidentity in the structure being considered!).was using the = of explication, not missaying (xy)^2 = xyxy isa relation (which would be silly in the context of groups; butnot that of non-associative algebra, for instance). Sometimes I yearn for a standard typographical device to distinguishthe = of explication (or derivation) from the = of equality,just as := and =: have come to be (fairly) standard notationsfor the = of assignment (or de?ition).Lee Rudolph === > I am looking for a problem referenced> by Martin Gardner, Chapter 11 in his> collection entitled The 2nd Scienti?> American Book of Mathematical> Puzzles & Diversions,>Simon and Schuster, 1961.To quote Gardner precisely:Lewis Carroll was fond of inventing quaint and enormously complicated problemsof this sort. Eight are to be found in the appendix of his _Symbolic Logic_. One monstrous Carrollian problem (involving 13 variables and 12 premises fromwhich one is to deduce that no magistrates are snuff-takers) was fed to and IBM704 computer by John G. Kemeny, chairman of the mathematics department atDartmouth College.This doesn't really imply that the magistrate/snuff-taker problem is inSymbolic Logic.John Robertson === >> [.snip.]>Well I did ?d a problem with the de?ition 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 de?ition 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 de?ition speci?s 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 de?es a unique>> thing, if you are going to call refer to it by using the singular> I am also pretty certain that this de?ition includes way too many>> things that you do not want. But it is obvious that once again all you>> are doing is trying to ?, by ?t, the problems that plagued your>> original proof of two years ago.> I must, however, confess that I am ?gasted 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 de?ition in signi?ant ways over the past 8>> months, yet your proof is so solid that changing this key de?ition>> 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 ?d a problem with the de?ition 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 de?ition 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 sillyerror. I also notice that you continue to state all numbers withoutspecifying whether you are considering only algebraic numbers, or infact all numbers. If you are considering all numbers, then whatyou have need not even have operations de?ed: it would includeactual polynomials in any number of incompatible variables, and someelements of positive characteristic. If you are going to update thepage, why not bother to ? the de?ition COMPLETELY?Current de?ition states:The Object Ring is the set of all numbers where -1 and 1 are the onlymembers that are both a unit and an integer, where no non-unit memberis a factor of any two integers that are coprime. Coprime where? In the ?al product, or in the ring of integers?It's possible that two integers are coprime in a larger ring withoutbeing coprime in the integers.The ?al clause is empty if you mean coprime in the integers: inANY subring of the complex numbers, an element which divides twointegers which are relatively prime in the ring of integers must be aunit. Putting it in the de?ition only obscures the latter.Is your object ring unique? Is it a ring? You are using the singularcomplex numbers which satis?s this condition: in fact, at leastEVERY subring of the algebraic integers does, and more besides, likeZ[pi], for example. And you only de?e it as a ->set<-, but you alsoclaim it is a ring. Since you are not specifying that it is a subringof the complex numbers, what are the ring operations?Does your ring contain multiplicative inverses for algebraic integerswhich are NOT integers? Let f(x) be any monic irreducible cubicwith integer coef?ients, |f(0)|>1, and assume moreover that itsdiscriminant is not a square in Q (so that the extension given by asingle root is not normal). Does your ring contain 1/r? It need notcause any integer other than 1 and -1 to become invertible, but itwould still possibly cause problems with your congruences.And let me repeat what I said before, which you removed withoutaddressing or acknowledging:Now, assuming you meant to say 1 and -1 are the only elements whichare both units and integers, then you still must prove that Objectring under this de?ition speci?s a unique such object.Presumably, you want to say largest subring of the complex numberssuch that..., because otherwise, the integers are The Object Ring,but so is any subring of the ring of all algebraic integers. It wouldbe of paramount importance to make sure that it de?es a uniquething, if you are going to call refer to it by using the singularI am also pretty certain that this de?ition includes way too manythings that you do not want. But it is obvious that once again all youare doing is trying to ?, by ?t, the problems that plagued youroriginal proof of two years ago.I must, however, confess that I am ?gasted 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 beenable to change the de?ition in signi?ant ways over the past 8months, yet your proof is so solid that changing this key de?itiondoes not require you to change even a single word of the rest of yourdevelopement to take into account these changes. Truly, a work ofgenius. === === ===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 ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite 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 Magidinmagidin@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 ?d 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 areperformed correctly and all assumptions are accounted for,then it's a reasonably good proof. Perfect? I don't know howto measure perfect.Some interesting things happen when one changes the assumptionsthough; the classical one is arguably Lobachevksy attempting to?d an absurdity by replacing the parallel postulate (an axiom),and instead developing an entirely new geometric form, hyperbolicgeometry.> 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> de?ed 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 itdoesn't totally eliminate error, it at least allows for moreeyeballs to check for errors in the proof. The author,presumably, then corrects those errors and republishes, orabandons the effort. (I would think abandonment would be extremelyrare, 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 canset 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'seffect on your proof is unclear.Please ?. :-)>>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 (andoccasionally peppered with insults, especially if alt.syntax.tacticalgets involved :-), or irrelevancies). I don't have a clue as tohow the more formal mathematical or scienti? peer review processworks. I don't even know how the PhD peer review process works,although I suspect one has to stand and defend his papers againstthe attacks from his peers, presumably in a forum of, maybe 6 to 12peers.> Now here's a math proof. Those who doubt that fact can believe it's a> claim of proof, but it's veri?d 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 de?ition similar tohttp://mathworld.wolfram.com/AlgebraicInteger.htmlwhich de?es an algebraic integer r (of degree N, if r satis?sno lower equation) as a solution to a polynomialx^N + a_{N-1}x^{N-1} + ... + a_1x + a_0 = 0.This gets a little weird, as algebraic integers of degree > 2may 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 de?es 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 thatP(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 ridiculousand relatively uninteresting. If u is not 0 one can computeP(0)/(u^2f^2) = (b_3 w_1 w_2 x + uf) = 3x + ufand you've actually proven that b_3 w_1 w_2 = 3, if x != 0 (whichis also relatively uninteresting).Such potholes are easily avoided of course (usually), butyour proof's real problem here is you leapt to the wrong conclusionas you left out b_3.Since (3 - 2sqrt(2)) * (3 + 2sqrt(2)) = 1 over the algebraicintegers, one also has to be careful about other conclusionsregarding this product as well. I'm not sure there's asmallest algebraic integer > 0. (It's easy to prove theset of algebraic integers clusters towards 1+ by consideringthe equation y^n + 2; as n -> oo the primary root y = (2^(1/n))tends to 1, and one gets an in?ite subset. By replacingy = (x + w) where w is any integer and grinding out theresultant equation, one can show that the set of algebraicintegers clusters around any integer, including 0+. Thereforethere isn't a smallest positive algebraic integer.)I don't know if this is a fatal ?ut 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 de?ed> 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 de?ition 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 de?e f(x) = K, where K is an arbitrary constant.Usually such functions are relatively uninteresting. ThereforeI fail to see why this is even a problem, let alone why peoplewould object thereto.Of course your de?ition P(m) would more properly be de?edP(m,f,u,x), in certain contexts. In the computer engineeringrealm you've basically de?ed a function/algorithmic procedureP(m) with one parameter and three globals, one of which you'reattempting apparently to solve for (x). This isn't a realbig 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* coef?ient 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 getP(m)/9 = (81 m^3 - 27 m^2 + 3m)x^3 - 3(-1+9m)xu^2 + 3u^3so 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 factor3^{1/3} without additional information; for all I knowb_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. andw'_1, etc., as well. Whether this is useful is not clear to me.This *is* a fatal ?est snipped, as it requires reanalysis]-- #191, ewill3@earthlink.netIt'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 ?d 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 ?thesepresumably can lead to some interesting counterexamples. Orone can postulate f = u = 0 and generate some uninteresting ones. :-)I've also noted that (3 - 2*sqrt(2)) * (3 + 2*sqrt(2)) = 1is an interesting product of algebraic integers as well.Obviously this can lead to some weird problems, as onecan 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 yieldthe requirement that each of a, b, and c has a factorof 3^(1/3) (an algebraic integer of degree 3)? This particularone puzzles me.> - Randy-- #191, ewill3@earthlink.netIt'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 butnever explicitly shown)?This is the best I can translate his de?ition.Let S be the object ring.1 e S, -1 (not e) SIf n =/= 1 and n e Z then if n e S then 1/n (not e) SIf A,B e Z and (A,B) = 1 and there does not exist x e S | xn = 1, if thereexists a, b e S | an = A and bn = B then n (not e) SSo the question is: what sort of structure does this set have? As de?ed,it is obviously NOT a ring (-1) is excluded, so is the set closed underaddition 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 coef?ients, |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 ?lds, 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 suf?es 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 ?d 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) satis?s 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 posthttp://groups.google.com/groups?selm=74k4r1%24sa0% 40schubert.mit.edu-Bill Dubuque === Hey,> f(x) = 2 ^ ([log(x * 8) / log (2)] - 1) / 4yeah this works - also got this one working (for PHP)$newvalue = pow(2,?og(16*$value)/log(2))-2) / 4;> int k, z, i ;> k = (log((?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 likef(3/8) = 1/4so I was looking for a function which returns the right values whichwould be for example:f(x) = 2^( ?n(16*x)/ln(2)) - 2 ) / 4Nikolai === First of all I want to say I am no mathematician. So anyone who is niceenough to respond to this, please explain in layman's terms (if you startusing mathmatical jargon beyond the basic stuff, please de?e it ?st or Iprobably won't know what you're talking about). I have recently become very interested in solving what I have learned iscalled a Diophantine equation, that is, an equation where the solutions arejust integers. After researching this on the internet, I have also found thatmy equation is very similar to an equation called the Pell equation, but notexactly. 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 isalways required that D is not a perfect square. Well, in my case the D is aperfect square. But then the 1 is not a 1, but an N, meaning, I want to givethis Diophantine equation solver a positive integer N and it give me thepossible pairs (x,y) that satisfy that equation. By the way, I only wantvalues of x and y that are >= 0. For example, if N=7, then the possiblesolution pairs are just one, namely (4,3). Has anyone studied this? Anyinternet sites I could be directed to? It seems like such a simple obviousthing that I would think someone would have looked into it.At the webpage http://mathworld.wolfram.com/ DiophantineEquation2ndPowers.htmlI found this: The more complicated equation x^2 - Dy^2 = c can also be solvedfor certain values of c and D, but the procedure is more complicated (Chrystal1961). However, if a single solution is known, other solutions can be foundusing the standard technique for the Pell equation.Since I always know one solution, from this I would think all the solutionscan be found using this Pell equation technique which I wish someone wouldexplain to me. (Hmm, after studying that page more thoroughly, I now believethat 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 = vwhere 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) = Nand 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 wayaround.We then have two simultaneous equations in x,y to solve. Or(2) x+y = 7 x-y = 1 ------- 2x = 8 (by adding) x = 4and 4-y = 1 4-1 = y 3 = ywhich is the pair (4,3) which you found.If N is composite, then things get much more interesting, since there are pairsof values, depending upon the factorization.For example with 24 we have 24 = 1*24 2*12 3*8 4*6and solutions of (x,y) are: x=25/2, y=23/2 x=7 y=5 x=11/2 y=5/2 x=5 y=1Some 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 * 16then (x,y) = (35/4, 29/4) is a solution. This will work for any 2 rationalnumbers whose product is N. === >The Diophantine equation I want to know how to solve is:>x^2 - y^2 = NOther posters have shown you how to solve this equation. Write N=ab in allpossible 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 discussionapplicable to any x^2 - Dy^2 = N, when D is a square, see the section Delta >0 is a square in the ?e 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 solvingthe x^2 - Dy^2 =+-1 equation, for D not a square. See Solving the generalizedPell equation, and other ?es, at the website above. In addition to theMatthews method, which is probably the best, this ?e summarizes the methodgiven in Chrystal, also due to Lagrange, and called Lagrange's system ofreductions.Both of the two ?es cited discuss ways to generate all solutions to any ofthese 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 Idon't understand your response (which is the only condition forthe equation to have a solution). I just showed you the solutionto N = 7 (?solution' meaning values for x and y that satisfy theequation). What I am after is the solution(s) for any odd N. Howis that determined? Can the Pell equation solution be used? === >An interesting class of integral domains are the rings F[x] >of polynomials over a ?ite ?ld F, >and the ?ite 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 ?ite 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 idealintersection 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 suf?ient 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 ?ite many ideals.Now as R itself is an ideal, it'd be a unique product of ideals makingall the ideals of R, products of prime ideals.The important thing is a useful collection of rings for whichUFD <-> PID to give intuitive reason why the ideal class groupsize 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-?ld 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 ?itenumber of denominators, then it makes some small sense.If I^-1 is ?ite, 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-?ld 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 ?ite> number of denominators, then it makes some small sense.> If I^-1 is ?ite, 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 invertibleif 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 ?lds)but also other rings (eg ?ite algebraic extensionsof the ring F[x] of polynomials over a ?ld 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 andmathematics. It is a way of thinking which oroginated in India in theis paying the price for this till today. If India was Buddhist, theIndustrial 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.Horse. The industrial revolution occurred in Christiannations and Jews and Christians have won lots of Nobelprizes in science. I think that the only Nobel prize everwon by a Buddhist was Tenzin Gyatso's Nobel PEACEprize.George === > Horse. 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 industrialrevolution to the protestant mind-set? (I should know but I'm having asenior moment)Anyway, the idea of India remaining Buddhist makes an interestingwhat-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. Horse. 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 gobelwhen you want it to.In my highly educated opinion, which will come to be shared soon among allnobel laureates, India's problem is that it is deep in mohamadism, andsurrounded by mohamadism. It is, however, good to see hindus have startedto break out of the historic morass since early 90s, ?ding their truehindu spirit of mookti (ie freedom), and ?ding, in the words ofIndia's Prime minister Vajpayee, natural allies in the western world.> George === Why do bengalis spell certain sanscrit words like muslims and arabs???Deb>Horse. 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. === >> Horse. 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.> BobBob, 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. ?ite.> 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 ?st 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 threadConstant 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 itdisagrees sharply with several proofs of the fact that ALL of a1, a2, and a3 have nonunit factors incommon with 5. But the interesting thing here is to see where hegoes with his conclusion. Since a1*a2*a3 = 65 = 5 * 13,it is clearly the case that at least one ofa1, a2 or a3 must have a nonunit factor incommon 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 anerror. Not, however, James Harris. He concludes thatthere is some kind of ? the ring of algebraicintegers that has been overlooked by mathematicianssince the time of Gauss, Kummer, and Dedekind. Harris disbelieves mathematical arguments that have been presented here which lead to contradictionsof his claims. He calls us liars for presenting them.He says we have not disproved his claims and we havenot found any errors in his proofs. Then he arrives (somehow!) at a contradiction ofhis 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 de?ition 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 ?d 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 alater step. For example, because equation [C] involves novariables preceding C (i.e. neither A nor B), such variablesare not reintroduced when using [C] to eliminate C in [3].In general, whenever one performs elimination it is usuallyuseful to ?st check to see if the structure of the systemof equations lends itself to any optimizations such as above.-Bill Dubuque === > The object of puzzles is, of course, to ?d 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 every40 yards. Apart from the ?st point, how far out does the tree and gravelapear in the same spot? (answer : on the 8th tree, 520 yards)Answered by writing out the distances manually, but how to solve itmathematically???Is there a website that teaches maths/algebra with these types of questions? === The object of puzzles is, of course, to ?d 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 ?st point, how far out does the tree andgravel> 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 ?st tree andthere will also be a pile of gravel there. That's because the least commonmultiple of 40 and 60 is 120. Google on least common multiple andgreatest common divisor.Make the prime factorization of both40 = 2^3.560 = 2^2.3.5and keep the highest exponents for LCM = 2^3.3.5 = 120and the lowest exponents for GCD = 2^2.5 = 20As 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 ?d an>> elegant solution (otherwise, they aren't recreation, they're work).But>> a puzzle that requires more than purely minimal computation (in theeyes>> 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 gravelevery>40 yards. Apart from the ?st 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 ?st 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 ?d an>> elegant solution (otherwise, they aren't recreation, they're work).But>> a puzzle that requires more than purely minimal computation (in theeyes>> 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 gravelevery>40 yards. Apart from the ?st 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 ?st 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 = 2400This should of course be LCM(x,y).GCD(x,y) = xy, was a bit hasty === >The object of puzzles is, of course, to ?d 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 whatyou>>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 ?st 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 ?st treeand>there will also be a pile of gravel there. That's because the leastcommon>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...Then40 = 2^3.565 = 5.13LCM(40,65) = 2^3.5.13 = 520GCD(40,65) = 5 === let R is ring with unitylet 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 Rif N is ideal of R such that M in N in Rif N not equal to R , M in NAny f in N, f(a) = 0 => f in M (***)hence N in Mhence N = Mtherefore M is maximal----------------------------------(***) sectioni think that this section was wrong.if f(x) =1 is in N, this section is wronghow 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 Mbut 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 RAssume 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 Mand 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 unitylet 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 meantLet 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 isinside or outside the {}.)>show, M is maximal ideal of R-----------------------------------M not equal to Rif N is ideal of R such that M in N in Rif N not equal to R , M in NAny f in N, f(a) = 0 => f in M (***)hence N in Mhence N = Mtherefore M is maximal----------------------------------(***) sectioni think that this section was wrong.if f(x) =1 is in N, this section is wronghow do you think about (***) ??correct ? or incorrect??please, point out an error~sir~Certainly (***) is incorrect. It's not truethat f(a) = 0 for any f in N.You're trying to prove that N = M. That'simpossible. 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.htmlhttp:// mathforum.org/discuss/sci.math/a/m/509288/509622undeformed square bag length a. At this point of time, a neither atheoretical ?al value, nor one supported by analysis could beconclusively arrived at. It value is approximately a^3/5 .The above volume ?ure can be computed by FEA if this can be modeledin ?ite element analysis with geometric non-linearity, with largein-plane stiffness matrix coef?ients compared to small in-planecoef?ients as applicable to a ?e/inextensible bag.I do hope someone would respond for analysis, especially as it hasinter-disciplinary signi?ance.G.L.NarasimhamEx 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.?tion.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 areasas well. :-) (Not me. I *know* JSH's math is a bit on thecareless side, not because of his equation manipulations butbecause he leaps to conclusions that need to be carefullytackled instead.)> next post from jstevh@msn.com telling us that this is a forgery.> Go play in the traf?.> Now now, this *is* sci.math (among others); at least phraseit as a creative math problem:(1) A person decides to attempt to cross an 8-lane freeway,for some reason which shall remain unspeci?d by the mathproblem (why does Billy throw the ball to Jane anyway?).The person can walk at 3 mi/hr = 4.4 ft/sec. Assuming auniform population of sedans of 16 feet in length and 6feet in width and exactly in the middle of each lane, thatthe lanes are 12 feet wide, that every sedan is followingthe speed limit and the 2-second rule [*], that the sedansare otherwise randomly distributed, that the person,once he starts to cross, blithely walks at a uniformvelocity straight across the highway, as opposed to doingsomething more intelligent (like running zig-zags), andthat the sedans don't brake before hitting him, what isthe 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 obviouslynot quite as safe but I happen to live in a metro area soknow even this relaxed (?!) rule is broken routinely at speed.(3) Same as (1) except the traf? is bumper-to-bumper stop-and-go.(This one should be easy. Of course being struck at2 mph isn't quite as deadly as 65 mph unless one's headgets 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 partssedan, 2 parts SUV, and 1 part semi-tractor trailer (length64 feet (48 foot trailer, 16 foot cab), width 8 feet [+]).Since truckers are (hopefully) more knowledgable weincrease their following distance to 4 seconds.(6)-(10). Assume the sedans and SUV's can see a distanceof 1,000 feet and brake within a distance of 300 feet (ofwhich up to 50 feet can be reaction time). The SUVs take350 feet since they're heavier. The trucks can brakewithin 500 feet. Assume also that the cars and SUVsdon't skid out of control while braking and the trucksdon't jackknife. Assume perfect visibility to the vehicle'sright (e.g., no trees in the way or blind spots in thevehicle).(11)-(15) Assume in (6)-(10) a visibility of 200 feetby 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 stipulatesthat the rear of the preceding car, assuming you and he aretraveling at the same speed, shall be 2 seconds in front ofyour front bumper. Admittedly this rule has some interestingrami?ations with respect to highway capacity, as it indicatesthat no matter how much one increases the speed the capacity ofthe highway is largely constant, assuming everyone, erm, rigorouslyfollows 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 butI'm assuming it's similarly sized to a sedan or SUV exceptthat there's no area for the kids; it's mostly motor. :-)Since 96 = 8' we're in fairly good shape here; also, thetruck cab partially occupies the trailer space becauseof the hitch, making dimensions a little weird. But then,this *is* a hypothetical math problem anyway... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === I may have no idea about your problem, but I _can_ line your columns up foryou, when viewed in a ?ed-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 Toothhttp://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 athand at this location. So I'll have to see if I can reconstruct theargument from memory.First, instead of the density, let's look (non-rigorously) at theprobability 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 actualprobability instead of a density.Let the original rvs be called x1, x2, ..., xn. Now the probabilitythat 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'tmatter what the order is. The probability that (x1, x2, ... , xn) liesin 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 avalue near u(2), etc., is just du(1) du(2) ... du(n).But there are n! identical events (the reorderings of the x's) thatwill give rise to exactly the same order statistics. So theprobability 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 satisfyu(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 de?es 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 positiveand -x when x is negativeSo,|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| = 5Michael LeungPaul ???????: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 calledalgebraic manifolds. Not that one doesn't need to be careful aboutthe distinction, but in a sense they're very similar.Keith Ramsay === A mathematical proof begins with a truth, and proceeds by logicalsteps to a conclusion which then must be true.I've pulled a detailed exposition of a short argument that quicklyshows a problem with algebraic integers. It starts after thereference.Now here's a math proof. Those who doubt that fact can believe it's aclaim of proof, but it's veri?d to be a proof by tracing theargument out.In this case, I begin with an expression. The expression exists, sothat 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 de?es P(m) in terms of varioussymbols, and it's all in the ring of algebraic integers, which meansthat the symbols can only represent numbers that are algebraicintegers.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 leavesb_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 de?edby 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 de?ition 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 toclaiming 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 candivide 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 againthat w_1 w_2 = 1, when m=0, if f is coprime to 3. That is, when fdoesn'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 thispossibility it's worth it to relax that restriction and now considerwhat 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 fwith 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^3so *every* coef?ient has a factor that is 3, as you can tell bylooking.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'scannot, 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 withrespect to m, without regard to the value of f. Which makes sense asf^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 ornot, you'd STILL get w_1 w_2 = 1 at m=0, without regard to the valueof f.But in fact, if f=3, you have w_1 w_2 = 3^{2/3} at m=0, which onlyworks 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, butI've seen that for some people the idea can take hold after seeing m=0highlighted.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, whichshows 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 forcew_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 fgoes, and then I point out that you can actually check my work usingm=1, f=sqrt(2), as then you get a polynomial which you can factorrather simply. So you can actually get the values for the b's andcheck them, and see that they are all algebraic integers, and all arecoprime to 2.However, usually, for f values that are coprime to 3, you don't getb's that are algebraic integers, which shows a problem with the ringof algebraic integers.Now the nice thing about a mathematical proof is that if someonedisagrees they have to ?d some misstep.Unfortunately, people can *say* that proof is not a proof, even whenit 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 logicalstep. 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 veri?d 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 de?es 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 de?ed>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 de?ition 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 ?e 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 isno reason to consider it. As you will see below, itis 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 fwith 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^3so *every* coef?ient 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 factoredin 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 rootsof the associated polynomial u^3 + 8*u^2 - 19.Since the roots of the latter polynomial are algebraicintegers, 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 areseveral: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 aproduct of three numbers gives a factorization. And in all theexamples just given (and in in?itely many others) note that (1) thecoef?ients of the x's are algebraic integers, and (2) the w termsare also algebraic integers. Proving what, exactly, you ask?#### Proving that the f = 3 case tells you NOTHING useful about thenecessary values of the w's. There is no unique way to write themdown. This is a special, exceptional case in which too many of theof 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, whichwrecks 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'sare not uniquely determined. You cannot draw conclusionsfrom 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. Worstof 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 in?ite rangeof 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 haverational coef?ients. It too is of no interestor 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 withthe properties you want ***, and there is no problem withthe ring of algebraic integers. Here is how things work whenf = 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 ofw1, w2, and w3 is not coprime to f = 5. Thus: a factorization of the desired form DOES exist, but itdoes NOT have one of the properties that you desperately want.No problem with the ring of algebraic integers, and no validproof 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 ?d 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) doesfactors *** non-uniquely *** through the linear terms of yourpolynomial factorization, and you end up being able toconclude: *** 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 ittook us to get through to you that there is actually anontrivial problem with generalizing from m = 0 tom <> 0 ? Do you realize how many incorrect argumentsyou have already burned through (including the presentone) in trying to handle that problem? Do you realizethat all of this is a waste of time, because your main claims have already been shown to be false andcannot be ?ed 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 tois a catchall for such uses.Keith Ramsay === what would be the best way to proverational number + rational number = rational numberandrational number + irrational = irrationalThis is what I have:1)since all rational numbers can be expressed as a ratio of two integers, n=p/qthen p1/q1 + p2/q2 = (p1q2+p2q1)/q1q2 which is still a ratio of integersi.e. 1*4+2*3/4*3=10/122) using the above, with irrationals, there is no ratio such that n=p/qi.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 integersWell 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 xbut even if sqrt(x) is irrational, x may still be irrational.Now just suppose we had a rational plus an irrational equallinga rational. Say a + b = c where a and c are rational but bis 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/vProduce 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 proverational number + rational number = rational number>and>rational number + irrational = irrationalThis 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 integersi.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 de?ition 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 isirrational 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 rationalTry to get a contradiction. When you work out your full strategy, see how 1) applies. You mention induction, which is applicable primarily forpropositions involving integers. This argument is aboutrationals and irrationals. It's conceivable that since rationalsare de?ed in terms of integers, that you may need to invokeinduction (perhaps the absolute values of their denominators).But try the indirect proof technique ?st.-- Spammers: I don't want a small digital camera to post photos of a large, lowweight, penis on a re-?anced 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 ofsides n, is it possible to calculate the other major dimensions(inscribed circle radius/diameter, circumscribed cirleradius/diameter, length of one side)?I am looking for a general formula that can be applied to polygons ofany 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 Toothhttp://www.clivetooth.dk === Now, if I can impose once again, is there a way to ?d the inside andoutside radii of a circular ring, given only the area and the wallthickness? 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 ?d 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 Toothhttp://www.clivetooth.dk === >> Is this homework? <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 thanI> care to admit and my algebra/trig/geometry is quite rusty. am trying to ?ure out if ?A' and ?w' are enough info to enable one to?d> ?b' and ?c'.> Now, if I can impose once again, is there a way to ?d 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-wThus,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-wSo,c = (A/(pi*w)+w)/2andb = c-w = (A/(pi*w)-w)/2-- Clive Toothhttp://www.clivetooth.dk === >> Is this homework? <<> I kinda wish it was, but I've been out of school for many more yearsthan> I>care to admit and my algebra/trig/geometry is quite rusty.>I>am trying to ?ure out if ?A' and ?w' are enough info to enable one to> ?d>'b' and ?c'.>message>>Now, if I can impose once again, is there a way to ?d the insideand>>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 tooobstinate to agree that clearly explained objections are correct.)I mean for heaven's sake, the Proof evidently still uses thenotion of objects, and the de?ition of object is _still_incoherent. (No, adding the -1 doesn't suf?e to ? it - thatwas 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 fromthe de?ition. Actually you don't even have to _?d_ theanswer - just explain how a yes or no _would_ follow from thede?ition if you could do various calculations.)You also don't seem to have noticed a point Arturo made (he'snot the only person who's noticed this aspect of things): thede?ition of object has been _changing_ a lot lately. Butnonethless the Proof, _using_ the notion of object, has_not_ changed! How can the same proof be correct intwo different versions, if it uses objects and the meaningof the word object keeps changing?This makes it clear to the meanest intelligence that whateverit is you have it's _not_ a proof based on deductions fromde?itions.>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 ?d out later I was wrong. At times,>I've also questioned the morals or competency of those enemiesAnd when it turned out that their objections were correct, I meanwhen _you_ ?ally agreed that their objections were correct, there'snever even a _hint_ of apology for calling them liars when in factthey were just telling what even you ?ally acknowledged to bethe truth. And no matter how many times the cycle repeats, thenext time someone disagrees with you they're immediately calledliars._That's_ why people despise you - it's not because of yourmistaken 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 in? 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 in? 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 meThe claim that no one believes you? Name _one_ person whodoes.>are apparently efforts to get me to quiet down by using intimidation>before the world ?ds 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 hisimportance to society. You're projecting again - the only person inall 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 de?ition of the prime counting function.Do you see any signi?ance 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 ?d 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 speci?ally 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 justi?ation 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 con?ent that a>journal would consider it and report the information to the world if>it were correct?If you ?d out that even journals failed in this case, would you ?d>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>?ld when looking at result in other ?lds?If, if, if, if, if, if, if... none of these things has happened exceptin 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>dif?ult 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