mm-3869 === Subject: Re: Abstract Algebra > I'm preparing for my upcoming final and I'll have a few questions. > I was hoping that you can check my proof. > Prove that the ideal < x2 +1> is prime in Z[x] but not maximal in Z[x]. > Proof: x^2+1= (x +i)(x-i) =>no roots in Z. So x^2+1 is irreducible in Z[x]. > Since Z[x] is a U.F.D., we get that x^2 +1 is prime. This implies > Z[x]/ is an integral domain. is not maximal since x+1 is a > non-unit in Z[x]/. Asserting that x + 1 is not a unit isn't proving that x + 1 is not a unit. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: David Petry advocates gas chambers for humanists? wanted to show someone his more idiotic mathematical rantings. >Unfortunately I ran accross the following post in which david argues >for building gas chambers and killing people. (Do note the subject of >that message too) > Okay, as much as I've suspected that David Petry is a little > weird in his militant anti-Cantorianism, I had never suspected > him of being *this* much of a crackpot. People should be aware > of this before engaging him > A snippet from David Petry's 1996 post to misc.activism.militia: >The Humanism problem is the most outstanding problem facing America >today. If we have a revolution, but fail to solve the Humanism >problem, the revolution will have been for naught. >Humanism is a disease which has corrupted the soul of our society. We >simply must be willing to do anything and everything necessary to wipe >Humanism off the face of the earth. If it means killing Humanists, so >be it. If it means building gas chambers, so be it. If it leads to the >death of two or three innocent people for every Humanist killed, well, >that's not too high a price to pay for freedom. problem: If society would simply recognize that Humanism is a religion, then we could solve the Humanism problem simply by applying the laws which keep religion separate from the government and out of the schools. conclusions. === Subject: Re: David Petry advocates gas chambers for humanists? wanted to show someone his more idiotic mathematical rantings. >Unfortunately I ran accross the following post in which david argues >for building gas chambers and killing people. (Do note the subject of >that message too) Okay, as much as I've suspected that David Petry is a little > weird in his militant anti-Cantorianism, I had never suspected > him of being *this* much of a crackpot. People should be aware > of this before engaging him > A snippet from David Petry's 1996 post to misc.activism.militia: >The Humanism problem is the most outstanding problem facing America >today. If we have a revolution, but fail to solve the Humanism >problem, the revolution will have been for naught. >Humanism is a disease which has corrupted the soul of our society. We >simply must be willing to do anything and everything necessary to wipe >Humanism off the face of the earth. If it means killing Humanists, so >be it. If it means building gas chambers, so be it. If it leads to the >death of two or three innocent people for every Humanist killed, well, >that's not too high a price to pay for freedom. > problem: > If society would simply recognize that Humanism is a religion, then we > could solve the Humanism problem simply by applying the laws which > keep religion separate from the government and out of the schools. So the absence of religion is itself a religion? So all science, mathematics, history, art, language, etc. should be kept out of schools? What, exactly, do you plan to teach? > conclusions. If crackpot notions were thought out by the originator, then no one would have to jump to conclusions. === Subject: Re: David Petry advocates gas chambers for humanists? then no one would have to jump to conclusions. And, seeing just how _much_ Petry hates scientists, helps explain why he has dedicated the last 16 years of his life to disrupting their efforts, as with the current thread. FWIW xanthian. === Subject: Re: Why are we sent in this world? > Please answer this question. To ignore David Petry's postings. === Subject: Re: Applied Trig Have a look at the diagram at http://homepages.ihug.co.nz/~acgrant/untitled.JPG I have just done a picture for bank at the moment, and I think it is a case of adding the angles??? HTH -Al === Subject: Re: Applied Trig > Have a look at the diagram at > http://homepages.ihug.co.nz/~acgrant/untitled.JPG > I have just done a picture for bank at the moment, and I think it is a > case of adding the angles??? > HTH > -Al What you might try is this. Orient yourself with some x-y-z axes, with the x-y plane horizontal and the z axis vertical. Take a pencil, representing the laser, and point it along the x-axis. No go through the four rotations I listed in my earlier post, in that order, and see if you think the pencil ends up pointing where it ought to for that combination of azimuth, elevation, bank and pitch. If it is then I think your best bet is to compute the rotations I described (which is relatively easy to do); if not then I've misunderstood the setup. === Subject: Z/2Z How do I show that the ring of all functions from any infinite set X to Z/2Z is not Noetherian? === Subject: Re: Z/2Z > How do I show that the ring of all functions from any infinite set X to > Z/2Z is not Noetherian? Let J_0 be the zero ideal, that is the ideal generated by the function that maps every element of X to zero. For any x_1 in X, let J_1 be the ideal generated by the function that maps x_1 to 1, and everything else to zero. So then J_0 is strictly contained in J_1. For any x_2 different from x_1, define J_2 to be... Hopefully you see where this is going and can finish the rest. Mike === Subject: Trigonometry eguation Please, help me to solve the equation [provided it is possible] without using arcus function [because I know the result is in this case x=arcsin(sqrt(2)/4): sin(x)+cos(x)=0.5 I need to get the exact result [eg for equation: cos(x)=0; the exact result is x=PI/2+k*PI ] Is it possible to solve without arcus functions? === Subject: Re: Trigonometry eguation > Please, help me to solve the equation [provided it is possible] without > using arcus function [because I know the result is in this case > x=arcsin(sqrt(2)/4): > sin(x)+cos(x)=0.5 > I need to get the exact result [eg for equation: cos(x)=0; the exact > result is x=PI/2+k*PI ] > Is it possible to solve without arcus functions? Square both sides and then use identities to get it into something you can solve. It should go something like this sin^2(x)+2sin(x)cos(x)+cos^2(x)=1/4 1+sin2x=1/4 sin2x = -3/4 Now you finish. Dave === Subject: Re: Trigonometry eguation As you may already have tried, it is easy to get a quadratic equation in either sin(x) or cos(x) and solve for the sin or cos. However to get x itself you need the inverse trig function. === Subject: The Numbers Game The DaVinci Code movie is set to open on 05/19/06. That combines: 19 and 95 (I was 19 in 95, and it was a wonderfully magical year for me), and 56 (the birth year of the woman who inspired my poetry book). The Number Game goes on. Ilya Shambat. === Subject: Re: Question <4bj65mF11g6aoU1@individual.net> <4bjlvvF11vjjuU1@individual.net> <4bjv3oF120sraU1@individual.net> <4bkb11F11te9gU1@individual.net> <4bkmsbF11l70rU1@individual.net> <4blqbbF11n2mpU1@individual.net> Every real can be represented by a convergent sequence of rationals in >> the Cauchy model and the representation can be chosen in a variety of >> ways, including the one in which 0.999... is regarded as the limit >> (which can be proved to exist) of the sequence, 0.9, 0.99, 0.999, .... > I agree! and disagree . > 0.9999..... is not the limit of the sequence 0.9 ,0.99,0.999,.... , it > is the upper bound of that sequence. > Number 1.000..... is the limit of that sequnce. > But by then you should define the world limit in a clear manner. > If what you mean by lim ( 0.9 ,0.99,0.999,.... ) is the smallest REAL > number that is bigger than all terms belonging to the sequnce 0.9 > ,0.99,0.999,.... . , then: lim ( 0.9 ,0.99,0.999,.... ) = > 1.000......... no doubt. > But if what you mean by lim ( 0.9 ,0.99,0.999,.... ) is the smalles > ordinal number that is bigger than all terms belonging to the sequnce > 0.9 ,0.99,0.999,.... then: lim ( 0.9 ,0.99,0.999,.... ) = > 0.9999.......1 since that ordinal is smaller than one and yet bigger > than all the terms in the sequence 0.9 ,0.99,0.999,.... . Your 0.999...1 is not a real number. All of the other numbers can be represented as either nonterminating decimal fractions or as sums of infinite series, e.g., 0.999... = sum{i = 1 to oo} 9 x 10^-i = 1 1.000... = 1 + sum{i = 1 to oo} 0 x 10^-i = 1 Your 0.999...1 cannot be represented by either notation, so it is not a real number. === Subject: Re: Calculus XOR Probability > stephen@nomail.com said: > What, exactly, IS that flaw, besides being different from the standard > pseudotheory? Which one? TO has so many. > > David Tribble came up with a rather devastating argument I thought, > but Tony seems to refuse to undestand it. It may be the case > that he does not understand it, or he may recognize that it > undercuts his logic, so he refuses to understand it. The > basic argument is that if you take some finite number k, > and increment it m times, you generate m new finite numbers. > Now given Tony's general belief that there is no real difference > between finite and infinite numbers, it does not seem that it > should matter if m is finite or not. Therefore if you increment > k m times, and m is an infinite number, why do you not generate > an infinite number of finite numbers? How many finite numbers > do you generate? > You think that's a devastating argument? Maybe I don't get the point. I tried > to answer David, but he didn't seem to understand what I was saying, but > that's > fairly common. The reason you do not generate an infinite number of finite > numbers is that, as soon as you have generated an infinite number of > naturals, > you have applied an infinite number of increments to your starting value, > thereby increasing it by an infinite amount, and causing it to be infinite in > value. TO seems to think that a process that only generates finite numbers must somehow also generate infinite numbers, but never explains why satisfactorily. > And, you ask how many finite naturals are generated? Well, I've already said > that without a definitie value range, measuring the set precisely is > impossible, so the closest answer I can give to that is aleph_0. In other words, infinitely many finite numbers are generated! > >> Of course your stairstep example was meant to be exactly >> that, but Tony does not see it that way, especially since >> he has now decided that points have a direction, although >> he did admit that some points have multiple directions. > > I wonder what direction the isolated point (0,0) is supposed to have. > All directions? No directions? An uncountable number of directions? > > I sometimes have a hard time believing that Tony can actually > be serious about points having a direction, but given the > amount of energy and emotion he has invested into this, > I suppose it is possible he will grasp any straw he can. > He seemed to concede that the point at the origin has > a vertical and horizontal direction, but why failed > to recognize that that must be true of all points, and > of all directions. > It depends on the space it's in. Suppose this point is in the plane space of Euclid's elements, all by its lonesome. Towards what other point(s) does this point point? === Subject: Re: Calculus XOR Probability > And, if you think the direction of the approximating elements doesn't > matter in measuring the curve, then you're simply wrong, and need to > look back at the basics of how calculus works. The measure of the length of a curve is standardly defined to be the least upper bound, if there is one, of the polygonal lengths, with polygons or broken lines having a finite number of segments. In the case of an actual staircase curve, the longest polygonal approximation is the staircase itself. But for the limit function all polygons lie strictly along the diagonal. Thus the lengths of the staircases are strictly greater that the length of the limit diagonal as the step size goes to zero. === Subject: Re: Calculus XOR Probability > cbrown@cbrownsystems.com said: > To be honest, my response was blah blah blah. That is essentially TO's only response to anything. === Subject: Re: Calculus XOR Probability > Virgil said: > > Virgil said: > > Mike Kelly said: > > > I'm sorry but I don't know what externally infinite means, maybe > you could explain. > > One that achieves infinity through infinite range, rather than > through infinite density, because it adds new elements outside of its > range, rather than within it. Does that make sense? > > NO! An open real interval ( or an open rational interval, as far as > that > goes) is externally infinite in the TO sense but of finite range. > > If you can provide a model of y -> x sure. > > Given any open real (or rational) interval, I and any y in I then there > are x and z both in I such that x < y < x. Thus y -> x < y < z follows > in TO's peculiar notation. > Where did I say this was restricted to some particular interval. External > infinities produce new element OUTSIDE the current interval. This merely creates a larger interval. One can always do this in such a way that the union of all such intervals is still a finite interval, unless one requires, for example, that there be a metric on the space and that no distances between successive points be less that some positive amount. > So externally infinite and bounded sets of rational and reals are > quite possible. So what is the point of worrying about external or > internal infiniteness? > They follow from distinct axiomatic statements regarding order. Not at all. One can have finite ordered sets which are neither internally nor externally infinite. > The naturals aren't generated by some iterative process that ends. > They are defined by the Peano axioms and, hey, there they are, all > infinitely many of them. > > What a dandy statement, chock full of new information! > > New to TO, perhaps, but things don't seem to stay with him long, so > what > was old yesterday is always new to him today. > > > Except the finite values don't cover an infinite range. > > An infinite set of finite values cover as much range as there is. > Is [0,1] all there is? In [0,1], yes! > > TO hasn't even got the naturals straight and now wants to tackle the > hyperreals? > Nothing is straight, not even the number line. TO's notion of the number line may be bent, but his vision of anything mathematical is too strabismic to be relied on. > > There are only finite 'points' in the naturals, the integers, the > rationals, the reals, or even the complexes. > In the standard world, sure. So what? We live in that standard world. > No, your ordinal-driven argument starting with 0 is bogus. > > It is no more bogus than starting with 1, as both equally satisfy the > Peano axioms. TO's bogus arguments fail when one uses the 0 startup, > which is the only reason he has for objecting to it. One could equally > well start with 1-,or -2, and so on, as all equally satisfy the Peano > axioms, but there are better reasons for starting at 0 that fora any > other starting point. > Stating a constant equality between element position and value is a > lot more relaible than saying each element is the set of all that > came before it, which really means everything up to its predecessor, > and then claiming there is no predecessor to some thing that's > supposed to be the set of all naturals. That thing is the union of all vN naturals which is guaranteed to exist by another axiom. > I go with well-stated > equalities before vague notions of more than all the rest- ness. TO's well-stated equalities are all ultimately based on the very axiom system which gives us those provable results that TO chooses to ignore and contradict when stating his well-stated equalities. . > > You can't start with nothing, TO, and get anywhere. If you start with > the standard well understood numbers, you have also assumed basic set > theory with Peano et al. > > Fine. It's the transfinite stuff that's garbage. If you accept the axioms, as you say you do, you must accept their consequences as well. Either that or provide mathematically valid disproofs based only on those axioms, and logic, of those 'consequences' you wish to reject. === Subject: Re: Calculus XOR Probability > Virgil said: > Of course IFR depends on the definition of quantity and arithmetic > formulas. > Consider those already defined as usual. > > Defining them as usual requires the mechanism of the very set theory > that TO objects to, o that TO is already assuming all that he wants to > reject. > The usual definitions for arithmetic operations have nothing to do with > transfinite set theory, though that may be artificially woven into some > axioms. The same axioms that give us transfinite set theory are needed to give us the naturals, and from them the other number systems on which TO is depending. TO can't get to where he wants to go without the same foundation that provides the structures he is trying vainly to avoid. > > That's all quite well mapped out already in standard math. Just consider > f > and > g to define a bijection as mutually inverse functions. > > Defining them as usual requires the mechanism of the very set theory > that TO objects to, so that TO is already assuming all that he wants to > reject. > > One can compare sets over a finite domain, or an infinite domain. > > But not over an unknown domain. > Which is exactly my point about needing definite domains and ranges, and not > vague definitions like all finites. A domain is no more than a set, and the set of von Neumann naturals, or the resulting sets of integers or of rationals, or of reals, are still sets. Without a specific set as domain and another as codomain for a giv en function, there is no way to tell whether it is invertible or not. > If f(g(x))=g(f(x))=x for all relevant x, as TO claims, then they > have to be bijections from some set to itself. > No, they're mutual injections between the naturals and some other sequential > set, together defining a bijection. The equation stated, f(g(x))=g(f(x))=x requires that each element of the domain of f is also in its codomain, each element in the domain if g is also in its codomain and each element in the domain of either is in the domain of the other. So it either domain is restricted to the naturals, both are. > They absolutely have the same cardinality, that's true, but when you > observe > them as sets of points on the real line, if one function always has more > elements than another within any given value range, then it seems > perfectly > intuitive and logical to say that it has more elements overall. > > Then TO is saying that , for N being the standard set of naturals and S > being {1/(n+1): n in N}, there are more members in S than in N, as this > exactly meets his standards. > > The most egeregious error in this regard is the equivalence > derived between the set of naturals and the set of rationals, > when there are clearly an infinite number of the latter per unit > on the number line, and only one of the former. So, while what > you say is true of cardinality, the conflation of that to size > of the set is absolutely unfounded and at the heart of your > counterintuitive results. > > Does TO insist that the cardinality of the rationals is greater > than that of the naturals? > > No, but the set of rationals is infinitely larger than the set of > naturals, your diagonalized bijection notwithstanding. Larger in what sense? One finite set is larger that another if and only if there is an injection but no surjection from the smaller to the larger. TO is always going on about what is true for finite sets must carry over to infinite ones, but balks at the most important carryover of all. That rule rule (that one set is larger than another if and only if there is an injection but no surjection from the smaller to the larger) need not be constrained to finite sets. . > > No, that's what you get when you spew pure negativity, rather > than saying anything substantive. It's not a particular mapping > or function, but a rule regarding the relationship between set > size and mapping function. > > Such rules are mappings. > Perhaps you could consider it a mapping from the mapping function > used to used to define the bijection and the value range to some > expression of the set size. I think of it as a formulaic > relationship. In other words, a mapping. === Subject: Re: Calculus XOR Probability > Virgil said: > For example, given two ordered sets, such as real closed intervals, > which are order isomorphic, their power sets are not ordered, or > even orderable, in any natural way, but are still clearly > bijectable. > > > Of course their power sets are orderable. > > But for uncountable ordered sets, not in any natural way based only on > the ordering of the base set. > That depends on the order you're considering. If you define a linear > sequential > order, then the power set can also be ordered. Linear sequential orderings, at least as mathematics defines them, are limited to countable sets. > > Any real interval can be ordered ala the H-riffics > > Any real interval is already ordered. TO's delusions are not needed to > do it. > But not as a discrete sequence. Linear sequential orderings, at least as mathematics defines them, are limited to countable sets. What goes on outside of mathematics is irrelevant here. === Subject: Re: Calculus XOR Probability > imaginatorium@despammed.com said: > OK, Tony - here's a hint (suggestion) for starting your > axiomatisation programme. The first thing we all want to know is the > basis for this declare. A set has no maximum value, so we declare > one, then everything's all right. Just how much can one declare away > like this? How about considering the second even prime. There isn't > one, but if we declare that there is, what happens? > If you want to axiomatize it as a rule, then there is logical implication > involved. So, given the premise that there is a definite value range, then we > can say that the measure of a quantitative set over that range is given by > IFR. > For lack of a definite value range, no measure can really be taken, and sets > cannot be compared. Maybe not by TO, but everyone else can compare them. > > And can you give us any idea where this declared maximal element for > the standard naturals is? Would it be sort of larger than any of the > standard naturals as the set exists before you start messing it up? > What if we prove something about a set with no maximal element, then we > declare one anyway, do we have to do everything all over again, just in > case bits got messed up? (They rather obviously will: we might have a > set for which it's actually hard work to prove there is no maximal > element - then you come along and declare one? > You're being ridiculous. I never said every set HAS to have a maximal > element. > I said that a usable MEASURE of a set requires it. Except that cardinality is eminently usable on all sorts of sets without maximal elements. That TO does not like to use cardinality in no way interferes with its proven usability Nor, despite all his ranting has TO /proved/ that there is anything wrong with cardinality. In fact, I am not aware that TO has ever proved anything in his reams of claims. > > You are obviously not entirely stupid, but I'm left gasping trying to > imagine quite _how_ you misunderstand what set theory is if you can > swallow the stuff above. > Can you understand that any axiomatic rule is a statement of logical > implication, and that the premise that there is a definite value range, > finite > or infinite, leads to the conclusion that there is a precise way to compare > sets which is not possible without it? I think you can. But making such an assumption that a value range exists for sets for which it has been several times validly proven that it cannot exist only guarantees that TO's system is self contradictory. === Subject: Group of matrices Show that th group of matrices with determinant=1 is an algebraic set in A^(n^2). === Subject: Re: Transfinite Ordinal Multiplication <300420061449312346%anniel@nym.alias.net.invalid> In Introduction to mathematical philosophy [by Bertrand Russell] >> n.Omega means Omega summed >> by itself n times, while Omeage.n means n summed by itself Omega times. >> small irrelevant question: why do they do it opposite from the way >> mathematicians do it? > you mean philosophers? I don't know really. Bertrand Russell was both a philosopher and a mathematician. That's why he could write about the philosophy of mathematics with some authority. === Subject: Re: Prime Buddies >On the surface it appears to be a probabilistic argument that there are >*likely* to be infinitely many twin prime pairs. Is that accurate? > Yes, on the surface it appears to be a probabilistic argument, no it > isn't actually one. > That's actually true: There is no probabilistic argument. > There is no calculation of the expected density of twin primes. > Erm, what about the Twin-prime and Hardy-Littlewood constants? I was commenting on the quoted posts about whether Winer has a probabilistic argument. He doesn't. My There only referred to his paper. I know others have made a plausible estimate for twin prime counts. Another quote from Winer's proof: Any non-zero probability (even a decreasing one) event given enough chances will eventually occur. Since we can take as many chances as we want (the number line is of infinite size), we will eventually get another prime twin or prime singleton. Thus the set of prime twins (and primes) is infinite because for any prime twin (or prime singleton) we can get the next one. Since he doesn't make twin estimates or use the prime number theorem or any heuristics at all, he could have used the same argument about Fermat primes, but they have a finite heuristic expectation. -- Jens Kruse Andersen === Subject: Re: affine algebraic set <4bph3bF12ie9vU1@news.dfncis.de Show that if the field k is not algebraically closed, > then any affine algebraic set V in A^n can be represented as V=Z(f) > where f is a single polynomial. > It seems as if Homework Panic Season has started. > I am afraid that you will not get much attention in this NG > unless you provide more information on what you have already > done (even if you tried and failed). > A hint to get you started on this one: try to prove by induction > on n, that there is a homogeneous polynomial h in k[x_1,...,x_n] > that vanishes _only_ at (0,0,...,0). > Now if V is given as Z(f_1,...,f_n) the desired polynomial is just > h( f_1, ..., f_n ). > In the above induktive proof you will find that the case n = 1 is obvious, > the induction step from n-1 to n is straightforward for n > 2 > and that the case n = 2 is where the actual work has to be done. > Marc I started as follows: When k = IR, if V = Z(f_1, ..., f_n), then V = Z(F(f_1, ..., f_n)), where F(x_1, ..., x_n) = (x_1)^2 + ... + (x_n)^2. Then for non-closed k we need to pick F such that ... And I don't know how to pick F. === Subject: Black String/Hole I was talking to one of my ta's a while back and he was telling me of some research that a few professors were working on that dealt with using hyperbolic PDEs to study the Black String. I was just wondering what kind of significance studying the black string/black holes can have. What is the purpose of studying something like this? === Subject: Re: How to prove that a special matrix is invertible. >> as a rational function. As while integrating a rational function >> using partial fractions, we can substitute any numerical >> value for X if the denominator doesn't vanish there. >> Assuming r1, r2, r3 are distinct, multiply the identity by X - r1: >> c1 + c2*(X-r1)/(X-r2) + c3*(X-r1)/(X-r3) = 0*(X-r1) = 0. >> Take the limit as X -> c1 to conclude c1 = 0. >Can you take X-->r1? (I guess you meant r1). I think X can take only >special values in the original problem these values are wN^(2*i)*we^2. >In this case we can take r1 only when epsilon=0. >I just want to make sure that I don't miss anything. >Toygar. Yes, it should be X --> r1 rather than X --> c1. Let C (= complex numbers) be a field. The field of rational functions n variable X over C is defined in terms of equivalence classes [just as, for example, 4/6 = 10/15 when we define ordinary fractions.] As rational functions, X^2 - 1 ------- - 3X and 1 + X/(-1/2) X - 1 are in the same equivalence class, since each simplifies to 1 - 2X when we apply standard rules from algebra. They are considered the same. Note, however, that one of these is undefined at X = 1. A polynomial of degree d defined over a field has at most d roots. If you can find d+1 distinct roots, you must have the zero polynomial. That is, your polynomial must vanish everywhere, not just at those d+1 roots you found. This is a powerful observation. For your problem, the obvious way to add the fractions in the g(X) definition leads to a rational function whose numerator has degree <= K-1 and whose denominator has degree K. We know K different values for which the rational function is defined and evaluates to zero. Therefore its numerator polynomial, after summing the fractions, must be identically zero, including at r1, r2, r3. I'll let someone else make the clarifications next time. -- VP Cheney Burr-ed his gun as a bird flew past The nation responds burr as we await bird flu shots and fight a real cold war. pmontgom@cwi.nl Microsoft Research and CWI Home: Bellevue, WA === Subject: Infinite field Let k be an infinite field and let V=Z(f) be a hypersurface in A^n. where f(x_1,...,x_n) is a non-zero, non-constant polynomial. Show that there are infinitely many points in A^n that are not in V. I started by Fix any point (a_1, ..., a_n) in k^n. The polynomial p(x_1) = f(x_1, a_2, ..., a_n) has at most only finitely many zeros. === Subject: Re: JSH: Factoring problem, trivial? ... > where k = 8*m in this specific formulation. Since that equality is > the only thing the final N = sqrt(1+8*mT) is rational guarantees, > it's got no way to force how the factors of T distribute between N+1 > and N-1. But, while possible, it's probably uncommon for a two-factor > T to have both of them end up in the same one. When it happens, you > can search for another m. > [Dik T. Winter] >> It appears to me that it is just as likely as with other methods that >> try to express something as the difference between two squares. So >> I would think that the probability of it occurring is 1/2. > That's because you're just thinking ;-) > [also Dik] > Without using only special values of m, there are 381 possible values of > m such that 8.m.219 + 1 is a square with m <= 1000000. Of these T=219 > splits over n-1 and n+1 in 191 cases. Just as expected... > But note that at m=12 (the smallest such m), N = 145 < T-1. Given that > N^2-1 = k*T for some natural k, both factors of T have to end up in N-1 and > N+1, but whenever N < T-1 they can't both end up in the same one (else that > one would be >= T, but N-1 < N+1 < T under the assumption N < T-1). So > searching smallest first isn't random at least to the extent that if any N > < T-1 exists, it will find the smallest such N first, and the method > succeeds then (and those bounds aren't sharp, just suggestive). > Of course the special properties of m make it a bit more difficult. > I don't bother writing up testing results for James anymore, but I will for > you :-) > I tried factoring all composites of the form T = p*q where p and q are odd > primes less than 100, p <= q, and T mod 4 = 3 (as noted earlier, when T mod > 4 = 1 the method as given fails at once, since the required modular inverse > doesn't exist then). There are 143 such composites. In each case, I tried > all m's of the special form starting from the smallest such positive m and > increasing until gcd(T, N-1) and gcd(T, N+1) revealed p and q. The overall > stats were appalling, but the gcds worked about 75% of the time: > number of 1+8*m*T quantities checked: 5986985 > of those, the number that were squares: 190 > of those, the number that revealed p & q: 143 > So in only 47 of the 190 cases tried did the gcd yield trivial factors. > As an aside, with T=219 you needed to check 280 integers for being > square using values of m that have a special form. > Right. For a real hoot, try factoring 71 * 73 = 5183 this way -- it takes > way over a million(!) tries to hit N=13045683. It may have atypical behavior, factoring better with LARGE numbers that have large primes than with small ones. Also the complaints about the modular inverse not always existing are silly as you just pick an n_1 where you do have a modular inverse. You can pick n_1 to be any natural that you choose. I just suggested floor(T/2) off the top of my head. This method may bizarrely get more efficient as T increases in size and as the gap between p_1 and p_2 increases. Also there is a real question about what is the best value for n_1, but you can play with it at will making it as large or as small as you like, as long as you can find a modular inverse, and it's a natural number greater than 1. If you play with n_1 = 2 or anything that small though, you're really stupid. Time is a wasting. If this thing does behave atypically getting ever more efficient as T increases in size, then while you people are jabbering, others might be doing a lot better. James Harris