. mm-2519 === Subject: Re: My Work--Objective Review > How can they ask that when I gave P(x)=x+1 in the ring of integers as > an example to try and help them understand? > Are you confused by the profound and extraordinary possibilities of > P(x)=x+1 in the ring of integers? I certainly am. It's very confusing to write P(x) in the ring of integers. Taken literally, that ought to mean that P(x) is an integer; but it's not - you could think of it as a function from the set of integers to itself, perhaps, but that's not the same as being an integer. Surely what you mean to say is P(x) in the ring of polynomials with integer coefficients. If you can't be bothered to write out the ring of polynomials with integer coefficients, I'm sure we'd be happy with Z[x]. === Subject: Re: My Work--Objective Review > I have NOTHING to hide about my work. It's available 24 hours a day, > 7 days a week at > http://groups.msn.com/AmateurMath > and included the paper Advanced Polynomial Factorization, which is > sweeping in its simplicity, while being the most important math paper > of the year because it highlights an error in taught mathematics--a > problem in core. > Let's have a look at it. The main claim, it appears, > is that if P(x) is the polynomial > P(x) = 65*x^3 - 12*x + 1 > and P(x) is factored in the form > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1) > where a1, a2, and a3 are algebraic integers, then > at least one of a1, a2 or a3 is coprime to 5. > Is this correct ? [snip analysis of claim] > Therefore if one of a1, a2, or a3 is coprime > to 5, they all are. > But a1*a2*a3 = 65. Thus at least one of a1, > a2, and a3 is NOT coprime to 5. Thus a contradiction. > The claim in the Advanced Polynomial Factorization > paper is therefore wrong. > If any of you can find an error then present it. > See above. > Then can someone who > is objective and invested in either side push forward any claims of > error that seem cogent? > If so, then I'll address THAT person's post. > Go for it. > I predict that either (1) James will ignore your post altogether, in spite of the fact that he actively solicited such responses, or (2) he will completely fail to grasp what you posted and, as an alternative to constructive dialog, will accuse you of lying, cheating, blowing smoke, 'blocking' his contribution, etc. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: My Work--Objective Review Take a dose of your own advice and LEARN MATHEMATICS! I am well versed in mathematics, as is everyone else is that reads your stuff so if so many people say it's wrong, get a clue. David Moran === Subject: Re: My Work--Objective Review James! Stop - bloody - whining! If you think the mathematic arguments by Ullrich et al. are wrong, please say where you think there are wrong. That's all. I don't want to read these rants that have nothing to do with mathematics, and I don't think anyone else wants to either. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ You could take his life and... - Mirja Tolsa === Subject: Re: My Work--Objective Review Visiting Assistant Professor at the University of Montana. [.snip.] >How can they ask that when I gave P(x)=x+1 in the ring of integers as >an example to try and help them understand? You call yourself a writer; the sentence above does not parse, let alone make mathematical sense. Mathematical formulas are part of the language. They are read out loud How can they ask that[,] when I gave capital P of x is equal to x plus one in the ring of integers as an example to try and help them understand? >Are you confused by the profound and extraordinary possibilities of >P(x)=x+1 in the ring of integers? Again, this does not parse. Setting that aside, you are still confusing polynomials, functions, and their values. Polynomials, qua polynomials, are not functions, though they can be used to define functions. However, their properties as functions do not always translate to properties as polynomials: the constant function 2 from the integers to the integers divides the function x|->(x^2+x) in the ring of all functions from the integers to the integers, but the polynomial 2 does not divide the polynomial x^2+x in the ring of all polynomials with integer coefficients. And neither the polynomial function nor the function are ->integers<-. As to your preference of examples by way of explanation, when will it get through that when people ask for a ->definition<-, and example will ->never<- do by itself? [.snip.] >Why don't you show just how well you understand Sawyer by explaining >with > P(x) = x + 1 >in the ring of integers? Because if it is a polynomial, it is not an integer (except for constant polynomials, which sometimes are considered to be integers as well). If it is an integer, then at best you can think of it as a constant polynomial. But you are clearly not doing either. So P(x) is ->not<- in the ring of integers. At least not as the words are understood by the entire world bar James Harris. >> So you need either: to explain which way you are thinking of your >> polynomials, or to explain why Sawyer's carefully delineated >> distinction doesn't exist. >More than once I've talked about polynomials as a family of values. Which is incorrect. Perhaps you should talk about the polynomial function associated to the polynomial instead... [.snip.] >>But if every polynomial divides every other polynomial, >> isn't it rather hard to have an irreducible polynomial? >You don't know much mathematics do you? >Irreducibility is typically expressed over Q, or the field of >rationals, though it can also be about Z, the integers. And I learned >that from arguing with people about it. You don't know much mathematics, do you? Irreducibility of a polynomial is ALWAYS expressed relative to a particular ring of polynomials. Only when it is understood from context can this be dropped. Irreducible over Q means that the polynomial is irreducible, when considered as an element of the ring of polynomials with rational coefficients, Q[x]. [.snip.] >Even with current math teaching I doubt anyone would talk of >irreducibility over the ring of algebraic integers. You are incorrect; it makes perfect sense to talk about irreducibility over the ring of algebraic integers. It means that the polynomial, when considered as an element of A[x], where A is the ring of all algebraic integers, is irreducible. As it happens, the only irreducible polynomials are the linear polynomials which are primitive; that is, polynomials ax+b, where a and b are algebraic integers, a nonzero, and a and b are coprime (there exist r,s algebraic integers such that ar+bs=1). [.snip.] >Well you can't help. You betrayed that you don't know even basic >mathematics. You are being very careful with your rare exotic flower, aren't you? >Oh yeah, it'd help if you know mathematics. It would help is ->YOU<- knew mathematics. But you do not. Certainly, not the kind of mathematics you are trying to talk about. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My Work--Objective Review > I have NOTHING to hide about my work. It's available 24 hours a day, > 7 days a week at > > http://groups.msn.com/AmateurMath > > and included the paper Advanced Polynomial Factorization, which is > sweeping in its simplicity, while being the most important math paper > of the year because it highlights an error in taught mathematics--a > problem in core. > > Let's have a look at it. The main claim, it appears, > is that if P(x) is the polynomial > P(x) = 65*x^3 - 12*x + 1 > and P(x) is factored in the form > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1) > where a1, a2, and a3 are algebraic integers, then > at least one of a1, a2 or a3 is coprime to 5. > Is this correct ? > [snip analysis of claim] > Therefore if one of a1, a2, or a3 is coprime > to 5, they all are. > But a1*a2*a3 = 65. Thus at least one of a1, > a2, and a3 is NOT coprime to 5. Thus a contradiction. > The claim in the Advanced Polynomial Factorization > paper is therefore wrong. > If any of you can find an error then present it. > See above. > Then can someone who > is objective and invested in either side push forward any claims of > error that seem cogent? > > If so, then I'll address THAT person's post. > Go for it. > > I predict that either (1) James will ignore your post altogether, in spite of the fact that he actively > solicited such responses, or (2) he will completely fail to grasp what you posted and, as an alternative to > constructive dialog, will accuse you of lying, cheating, blowing smoke, 'blocking' his contribution, etc. > -- Yes, looks like prediction (1) is on target, but of course prediction (2) might come true eventually. You realize that neither of these predictions puts you up there in a class with Jeanne Dixon or other renowned prophets. A third possibility is that he may say, You did not find an error in my paper. Therefore if you are right and I am right, there is a contradiction in 'core' mathematics, which is what I said in the first place. The problems with that argument are (1) sight unseen, it is far more likely that there is an error in his argument than that there is a heretofore over- looked contradiction in 'core' math, whatever that is - if you see a strange light in a dark wooded area, you don't assume as a first explanation that it is extraterrestrials out hunting for mushrooms; and (2) his paper, particularly the latter part, is quite unreadable. A fourth possibility is that my argument is wrong and he will find an error in it. I don't totally rule that out. However I agree that the chance that he even understands my argument is pretty slim. Nora B. > There are two things you must never attempt to prove: the unprovable -- and the obvious. > -- > Democracy: The triumph of popularity over principle. === Subject: Re: My Work--Objective Review > Their names are quite familiar to me and to many of you who have kept > up with the discussions that have raged about my mathematical work: > > Ullrich, Magidin, Wayne Brown, Christian Bau, etc. > > A group uniform in that they are unrelentingly negative, often take > cheapshots, like Ullrich, a math professor at a state university, who > loves the word idiot or likening me to a monkey with a typewriter. > > > I have NOTHING to hide about my work. It's available 24 hours a day, > 7 days a week at > > http://groups.msn.com/AmateurMath > > and included the paper Advanced Polynomial Factorization, which is > sweeping in its simplicity, while being the most important math paper > of the year because it highlights an error in taught mathematics--a > problem in core. > > Let's have a look at it. The main claim, it appears, > is that if P(x) is the polynomial > P(x) = 65*x^3 - 12*x + 1 > and P(x) is factored in the form > P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1) > where a1, a2, and a3 are algebraic integers, then > at least one of a1, a2 or a3 is coprime to 5. > Is this correct ? Yup. > Let x = 1/u. Then P(x) = 0 implies > P(x) = 65/u^3 - 12/u + 1 = 0. > Multiplying through by u^3 gives > u^3 - 12*u^2 + 65 = 0. > Call this polynomial Q(u). Let r1, r2, and r3 be > roots of Q. Note that a1, a2 and a3 happen to be > the negatives of r1, r2, and r3 - that is, a1 = -r1, > a2 = -r2, a3 = -r3 [the order does not matter]. > The polynomial Q(u) is irreducible. Let H be the > field of algebraic numbers; clearly r1, r2, and > r3 are in H. Let F12 be an automorphism of H such > that F12(r1) = r2, and F12 leaves fixed the subfield > of rational numbers. Such exists because of the > irreducibility of Q. Note that F12(a1) = a2, since > a1 = -r1 and a2 = -r2. > Without loss of generality, suppose a1 is coprime > to 5 in the ring of algebraic integers. By definition > this means that there exist algebraic integers r and s > such that Notice the call to the definition. > r*a1 + s*5 = 1. > Now apply F12 to both sides: > (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). > Note that F12(5) = 5 and F12(1) = 1 because > the automorphism F12 leaves fixed the subfield > of rational numbers. Also F12(a1) = a2. > Let r' = F12(r) and s' = F12(s). Note that > both r' and s' are algebraic integers. Thus eqn (1) > reduces to > r'*a2 + s'*5 = 1. > That is, a2 is also coprime to 5. > Similarly one shows that a3 is coprime to 5. Which looks like a good example to show those who wondered how what I've shown highlights an error in *taught* mathematics. It's basically an abuse of Galois Theory. > Therefore if one of a1, a2, or a3 is coprime > to 5, they all are. > But a1*a2*a3 = 65. Thus at least one of a1, > a2, and a3 is NOT coprime to 5. Thus a contradiction. > The claim in the Advanced Polynomial Factorization > paper is therefore wrong. Now I've explained more than once that I prove that x has y as a factor, but x/y is not an algebraic integer, which is a contradiction. So *obviously* there's a problem somewhere with what mathematicians are doing if they think they can prove something that's false. Finally, notice the attempt to disprove a proof with *another* argument, which is claimed to be a proof. It's like trying to fight a proof with a proof, but proofs can't fight each other. Now given my original post, which called for an objective review of the actual paper, which is available 24 hours a day, one might understand why I see Nora as a yet another jackass. === Subject: Re: My Work--Objective Review > Now given my original post, which called for an objective review of > the actual paper, which is available 24 hours a day, one might > understand why I see Nora as a yet another jackass. Blovating jackass! Bob Kolker === Subject: Re: My Work--Objective Review > Right. Well, since you complain a lot about people who (as far as I > can see) are genuinely asking you to explain _exactly_ what you mean > by a polynomial, I've given the following couple of pages of a > (smelly-paper) book, which are also available 24/7 (I hope; I use > pair.com hosting): > How can they ask that when I gave P(x)=x+1 in the ring of integers as > an example to try and help them understand? I think the point is that they all know that polynomial refers to something with a combination of letters and numbers, with plus signs in the middle, and bits raised to different powers. > Are you confused by the profound and extraordinary possibilities of > P(x)=x+1 in the ring of integers? Yes, I suppose, because like everyone else I don't know what you mean by in the ring of integers. > The scope of x, pp 34-35 of Sawyer's A concrete approach to > abstract algebra: > http://imaginatorium.org/private/sawyer.gif > (136K, 800x600) > > In this extract, Sawyer describes two ways of looking at a polynomial; > depending on whether you view the polynomial as a formal object or > whether you view it as simply a generalised form of arithmetic (i.e. > you are always thinking of just its evaluation), you will get > different answers to quite basic questions like Are these two > polynomials equal? or Is this polynomial a factor of that one? > Hmmm...I think you're trying to blow smoke. I'm paraphrasing Sawyer: are you saying he's blowing smoke? (You could at least have accused me of blowing smoke rings - that might have been mildly funny.) > Why don't you show just how well you understand Sawyer by explaining > with > P(x) = x + 1 > in the ring of integers? > So you need either: to explain which way you are thinking of your > polynomials, or to explain why Sawyer's carefully delineated > distinction doesn't exist. > More than once I've talked about polynomials as a family of values. OK: this sounds a little like the idea of a polynomial function: for each integer, n say, there's another integer m, got by evaluating this P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the 'bright pupils' tend to see introductory algebra. > But at times I also speak of them as objects in their own right. OK: this sounds a bit like the formal view of a polynomial. How Sawyer suggests the 'bright pupils' tend to see introductory algebra. (**Of course**, he's not saying that one view is right or wrong, or weak or powerful, just that it turns out that different beginning students grapple with this differently.) Well, forgive me for choosing a different example, but I'd like to take > Um, actually, have I understood your latest claim correctly? You say > that the algebraic integers are incomplete, because you have x and y, > such that y is a factor of x, but x/y isn't an algebraic integer? So > in your new system of completing the ring of algebraic integers (is > this related to Objects, by the way? I'll called them CAIs for > completed algebraic integers for now) - is it the case that for any > CAIs, r and s, that r/s is also a CAI? So consider two general > polynomials P(x) and Q(x) in the ring of CAIs: since you say that > the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is > also a CAI? > No. >But if every polynomial divides every other polynomial, > isn't it rather hard to have an irreducible polynomial? > You don't know much mathematics do you? > Irreducibility is typically expressed over Q, or the field of > rationals, though it can also be about Z, the integers. And I learned > that from arguing with people about it. > Even with current math teaching I doubt anyone would talk of > irreducibility over the ring of algebraic integers. > Damn. So you're just another jackass. > > If any of you can find an error then present it. Then can someone who > is objective and invested in either side push forward any claims of > error that seem cogent? > > Well, I can't understand what an object is. Do you know anyone except > yourself who can? Perhaps a good tack to take would just be a humble > appeal to the silent masses for some good soul to step forward and > explain objects so that at least people like me could understand them > (even if the Nasty Mathematician Mafia say they can't). > Well you can't help. You betrayed that you don't know even basic > mathematics. > Oh well, is there anyone out there who can be objective? > Oh yeah, it'd help if you know mathematics. > === Subject: Re: My Work--Objective Review [OOOPS!! Sorry, previous version was misdigitalisation...] > Right. Well, since you complain a lot about people who (as far as I > can see) are genuinely asking you to explain _exactly_ what you mean > by a polynomial, I've given the following couple of pages of a > (smelly-paper) book, which are also available 24/7 (I hope; I use > pair.com hosting): > How can they ask that when I gave P(x)=x+1 in the ring of integers as > an example to try and help them understand? I think the point is that they all know that polynomial refers to something with a combination of letters and numbers, with plus signs in the middle, and bits raised to different powers. > Are you confused by the profound and extraordinary possibilities of > P(x)=x+1 in the ring of integers? Yes, I suppose, because like everyone else I don't know what you mean by in the ring of integers. > The scope of x, pp 34-35 of Sawyer's A concrete approach to > abstract algebra: > http://imaginatorium.org/private/sawyer.gif > (136K, 800x600) > > In this extract, Sawyer describes two ways of looking at a polynomial; > depending on whether you view the polynomial as a formal object or > whether you view it as simply a generalised form of arithmetic (i.e. > you are always thinking of just its evaluation), you will get > different answers to quite basic questions like Are these two > polynomials equal? or Is this polynomial a factor of that one? > Hmmm...I think you're trying to blow smoke. I'm paraphrasing Sawyer: are you saying he's blowing smoke? (You could at least have accused me of blowing smoke rings - that might have been mildly funny.) > Why don't you show just how well you understand Sawyer by explaining > with > P(x) = x + 1 > in the ring of integers? > So you need either: to explain which way you are thinking of your > polynomials, or to explain why Sawyer's carefully delineated > distinction doesn't exist. > More than once I've talked about polynomials as a family of values. OK: this sounds a little like the idea of a polynomial function: for each integer, n say, there's another integer m, got by evaluating this P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the 'bright pupils' tend to see introductory algebra. > But at times I also speak of them as objects in their own right. OK: this sounds a bit like the formal view of a polynomial. How Sawyer suggests the 'bright pupils' tend to see introductory algebra. (**Of course**, he's not saying that one view is right or wrong, or weak or powerful, just that it turns out that different beginning students grapple with this differently. It seems you do have some idea of the distinction he's making.) Well, forgive me for choosing a different example, but I'd like to take: Q(x) = 2x+1 When you consider this in the ring of integers, does it have any roots? > Um, actually, have I understood your latest claim correctly? You say > that the algebraic integers are incomplete, because you have x and y, > such that y is a factor of x, but x/y isn't an algebraic integer? So > in your new system of completing the ring of algebraic integers (is > this related to Objects, by the way? I'll called them CAIs for > completed algebraic integers for now) - is it the case that for any > CAIs, r and s, that r/s is also a CAI? So consider two general > polynomials P(x) and Q(x) in the ring of CAIs: since you say that > the polynomials are also CAIs, doesn't it follow that P(x)/Q(x) is > also a CAI? > No. Why ever not? Is there a false step in my outline proof? Can you show me where it is? After all, we know proofs are irrefutable... [snip: Arturo Magidin dealt with this] > Damn. So you're just another jackass. Aha!! I'll have my point back, please. Brian Chandler ---------------- geo://Sano.Japan.Planet_3 http://imaginatorium.org === Subject: Re: My Work--Objective Review Visiting Assistant Professor at the University of Montana. [.snip.] >> Without loss of generality, suppose a1 is coprime >> to 5 in the ring of algebraic integers. By definition >> this means that there exist algebraic integers r and s >> such that >Notice the call to the definition. Definitions are shorthand. In computer science terms, something you claim to be familiar with, coprime is a macro; saying by definition just means that you are expanding the macro. Saying a1 and 5 are coprime means there exists r and s such that r*a1+s*5=1. Just like saying a is even means, ->by definition of even<-, that there exist an integer k such that a=2*k. [.snip.] >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. >It's basically an abuse of Galois Theory. It is an abuse of Galois Theory that someone who admits he does not know anything about it will nonetheless insist he can spot when it is being used correctly and when it is not. [.snip.] >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a contradiction. Which means you do not know what the word factor means. That's all. You do not understand what factor means. What you write above is exactly the same as saying that you have found and even integer n such that n/2 is not an integer. It is an abuse of terminlogy. >So *obviously* there's a problem somewhere Yes: the problem is that you are an ignoramus who refuses to learn what the words he uses mean. DEF. Let R be a commutative ring. Let x and y be elements of R. We say y is a factor of x (in R) if and only if there exist z in R such that y*z = x. If the ring R is understood from context, we can omit (in R). DEF. Let R be a commutative ring. An element x in R is called a unity if and only if for all y in R, x*y=y*x = y. If so, we denote x by 1, and we say that R is a commutative ring with 1. DEF. Let R be a commutative ring. We say that R is an INTEGRAL DOMAIN if and only if it has a 1 for every x,y in R, if x*y=0, then x=0 or y=0. DEF. Let R be an integral domain. The FIELD OF FRACTIONS of R is constructed as follows: (1) Let k be the collection of all formal symbols x/y, with x,y in R, y nonzero. (2) Define an equivalence relation on k, by letting (x/y)~(z/w) if and only if x*w=z*y in R. (3) Let K be the set whose elements are the equivalence classes of elements of k under ~. Denote the class of x/y by [x/y]. (4) Define an addition and a multiplication in K by: [x/y] + [z/w] = [(x*w+z*y)/y*w] [x/y] * [z/w] = [(x*z)/(y*w)]. THEOREM. The field of fractions of an integral domain is a field; the 0 is the class [0/1]. The one is [1/1]. R is a subring of K by the map that identifies the element r in R with the element [r/1] of K. THEOREM. Let R be an integral domain, and let x,y be elements of R. Then y is a factor of x in R if and only if the element [x/y] of K lies in the image of R. We express this by writing x/y lies in R. Therefore, what you have claimed is that there is an algebraic integer which is not an algebraic integer. And that is nonsense on the same level as saying that you can find an even number which is not a multiple of 2. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My Work--Objective Review > Without loss of generality, suppose a1 is coprime > to 5 in the ring of algebraic integers. By definition > this means that there exist algebraic integers r and s > such that > Notice the call to the definition. Uh, yes. In verifying whether a1 is coprime to 5, we see whether it meets the definition of coprime to 5. Do you really think that there's something wrong with that? - Randy === Subject: Re: My Work--Objective Review neutral, disinterested, but mathematically compentent third party willing to review one's work. Nice job. >Their names are quite familiar to me and to many of you who have kept >up with the discussions that have raged about my mathematical work: >Ullrich, Magidin, Wayne Brown, Christian Bau, etc. >A group uniform in that they are unrelentingly negative, often take >cheapshots, like Ullrich, a math professor at a state university, who >loves the word idiot or likening me to a monkey with a typewriter. Did he do that? That's not right. You're more like a monkey with a computer hooked up to the internet. A monkey with only a typewriter is far more limited in where he can fling his crap. >and included the paper Advanced Polynomial Factorization, which is >sweeping in its simplicity, while being the most important math paper >of the year because it highlights an error in taught mathematics--a >problem in core. What were some of the runners-up for most important math paper of the year? Was it hard to judge, objectively, between those in the field of core mathematics and those in, say, algebraic topology? >>Without loss of generality, suppose a1 is coprime >>to 5 in the ring of algebraic integers. By definition >>this means that there exist algebraic integers r and s >>such that > Notice the call to the definition. >> r*a1 + s*5 = 1. Oh yeah, look at that! When Nora uses the word coprime she takes it to mean what its definition says it means. Whereas you use the word more creatively, like an artist really. This powerful technique allows you to prove, well, anything you want! But when will the stodgy mathematical community come to accept your avant-garde vision? >>Note that F12(5) = 5 and F12(1) = 1 because >>the automorphism F12 leaves fixed the subfield >>of rational numbers. Also F12(a1) = a2. >>Let r' = F12(r) and s' = F12(s). Note that >>both r' and s' are algebraic integers. Thus eqn (1) >>reduces to >> r'*a2 + s'*5 = 1. >>That is, a2 is also coprime to 5. >>Similarly one shows that a3 is coprime to 5. > Which looks like a good example to show those who wondered how what > I've shown highlights an error in *taught* mathematics. > It's basically an abuse of Galois Theory. Is any use of Galois Theory an abuse of it, or just those that debunk your arguments? Please don't let the fact that you don't know the first thing about Galois Theory stop you from giving a ruling! We need to know ASAP whether to stop stop teaching it. > Finally, notice the attempt to disprove a proof with *another* > argument, which is claimed to be a proof. > It's like trying to fight a proof with a proof, but proofs can't fight > each other. Even proofs that use Galois theory? Okay, denote Nora's proof N, and let N^* be [must stop fingers from typing . . . losing control . . .] the dual proof [aarrghhh!] > Now given my original post, which called for an objective review of > the actual paper, which is available 24 hours a day, one might > understand why I see Nora as a yet another jackass. A jackass on the TV show Jackass recently attached a bungie cord to his underwear, then jumped out of a tree in pursuit of the ultimate wedgie. It was generally agreed that he found it. To call you a jackass would be an insult to that guy, James. === Subject: Re: My Work--Objective Review Putting aside, for a moment, the urge to comment about your histrionics (which often reach epic proportions) and concentrating on the scientific content of your work, it is clear that you are not yet qualified to enroll in algebra 101. Your character flaws consistently foil any attempt on your part to devise a rational argument suitable for presentation to others, leaving you with the double stigma of being a lousy mathematician and a worse human being. If you really do have a degree in physics, which I doubt, I think you have compelling grounds to demand your money back. There is no evidence of scientific thought present in your flawed arguments or your passionate, nay maniacal, defense of them. On a positive note, you can probably still get a job as a crash dummy. -- Words mean exactly what I want them to mean! -- the caterpillar in Alice in Wonderland and, more recently, . -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: My Work--Objective Review >> Their names are quite familiar to me and to many of you who have kept >> up with the discussions that have raged about my mathematical work: >> >> Ullrich, Magidin, Wayne Brown, Christian Bau, etc. >> >> A group uniform in that they are unrelentingly negative, often take >> cheapshots, like Ullrich, a math professor at a state university, who >> loves the word idiot or likening me to a monkey with a typewriter. >> >> >> I have NOTHING to hide about my work. It's available 24 hours a day, >> 7 days a week at >> >> http://groups.msn.com/AmateurMath< /a> >> >> and included the paper Advanced Polynomial Factorization, which is >> sweeping in its simplicity, while being the most important math paper >> of the year because it highlights an error in taught mathematics--a >> problem in core. >> >> Let's have a look at it. The main claim, it appears, >> is that if P(x) is the polynomial >> P(x) = 65*x^3 - 12*x + 1 >> and P(x) is factored in the form >> P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1) >> where a1, a2, and a3 are algebraic integers, then >> at least one of a1, a2 or a3 is coprime to 5. >> Is this correct ? >Yup. >> Let x = 1/u. Then P(x) = 0 implies >> P(x) = 65/u^3 - 12/u + 1 = 0. >> Multiplying through by u^3 gives >> u^3 - 12*u^2 + 65 = 0. >> Call this polynomial Q(u). Let r1, r2, and r3 be >> roots of Q. Note that a1, a2 and a3 happen to be >> the negatives of r1, r2, and r3 - that is, a1 = -r1, >> a2 = -r2, a3 = -r3 [the order does not matter]. >> The polynomial Q(u) is irreducible. Let H be the >> field of algebraic numbers; clearly r1, r2, and >> r3 are in H. Let F12 be an automorphism of H such >> that F12(r1) = r2, and F12 leaves fixed the subfield >> of rational numbers. Such exists because of the >> irreducibility of Q. Reference for this: http://www.math.niu.edu/~beachy/aaol/galois.html especially Proposition 8.6.2 on that page. Or see the textbook, Abstract Algebra, by John Beachy and William Blair. An *excellent* book. >> Note that F12(a1) = a2, since >> a1 = -r1 and a2 = -r2. >> Without loss of generality, suppose a1 is coprime >> to 5 in the ring of algebraic integers. By definition >> this means that there exist algebraic integers r and s >> such that >Notice the call to the definition. You have a problem with my using a standard definition ??? >> r*a1 + s*5 = 1. >> Now apply F12 to both sides: >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). >> Note that F12(5) = 5 and F12(1) = 1 because >> the automorphism F12 leaves fixed the subfield >> of rational numbers. Also F12(a1) = a2. >> Let r' = F12(r) and s' = F12(s). Note that >> both r' and s' are algebraic integers. Thus eqn (1) >> reduces to >> r'*a2 + s'*5 = 1. >> That is, a2 is also coprime to 5. >> Similarly one shows that a3 is coprime to 5. >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. >It's basically an abuse of Galois Theory. What's the abuse? >> Therefore if one of a1, a2, or a3 is coprime >> to 5, they all are. >> But a1*a2*a3 = 65. Thus at least one of a1, >> a2, and a3 is NOT coprime to 5. Thus a contradiction. >> The claim in the Advanced Polynomial Factorization >> paper is therefore wrong. > >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a contradiction. >So *obviously* there's a problem somewhere with what mathematicians >are doing if they think they can prove something that's false. I would say obviously there is a problem somewhere, yes. >Finally, notice the attempt to disprove a proof with *another* >argument, which is claimed to be a proof. You claimed something was true. I gave an argument to show it wasn't. Four possibilities: 1. I have an error in my argument. You however did not point out any error. 2. You have an error in your argument. I did not point it out in my post, but it has been pointed out many times previously. Your argument depends on the form that a factorization has in a degenerate, singular case (when m = 0). You argue implicitly that that same form of factorization must hold in the nonsingular case in which you are actually interested. It does not. This has been pointed out innumerable times. Arturo and others have tried many times to get you to understand it, but they have failed. I will probably fail as well. 3. Both of us have errors in our arguments. 4. Neither of us has an error in our arguments. There is a fundamental contradiction in mathematics. I think we should just put this idea aside. It is the least likely of the possibilities and it does not obviously lead anywhere. It's also possible that the fact I used about automorphisms, which comes from Galois theory, is wrong. That too seems unlikely since Galois theory has been around for over 160 years and looked at by many very smart people, and this fact is encountered at the most basic level. I think that too should be put aside. >It's like trying to fight a proof with a proof, but proofs can't fight >each other. We both present arguments. You say yours is a proof. I say mine is. We are not both right. In the end we have to try to convince other people and each other. >Now given my original post, which called for an objective review of >the actual paper, which is available 24 hours a day, one might >understand why I see Nora as a yet another jackass. comprehensive review, but I addressed your main point and I showed it was wrong. I was objective. I did not address Lemma 1.1 because in my view it is trivial: given the hypotheses, and noting that g is actually a function of x, you then let the constant c = g(0) [you did do that part], and then you define r(x) = g(x) - c. End of proof. But again, your central error is assuming that the form of a factorization is the same in a singular case as it is in the general case. It is always dangerous to generalize based on degenerate cases or singularities. You can probably guess why. Nora B. > === Subject: Re: My Work--Objective Review >> Their names are quite familiar to me and to many of you who have kept >> up with the discussions that have raged about my mathematical work: >> >> Ullrich, Magidin, Wayne Brown, Christian Bau, etc. >> >> A group uniform in that they are unrelentingly negative, often take >> cheapshots, like Ullrich, a math professor at a state university, who >> loves the word idiot or likening me to a monkey with a typewriter. >> >And yes, lately I myself have taken to calling people jackass. >It's nasty but maybe there's another route to resolution as I thought >I'd come back and go after Nora's argument a little more directly. You mean you actually thought that calling people jackass was a route to resolution? Maybe you should give mathematics a rest and lend your services to the diplomatic corps. There are tons of wars simply dying to get started -- YOU could be one that gets them going! >> Now apply F12 to both sides: >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). >> Note that F12(5) = 5 and F12(1) = 1 because >> the automorphism F12 leaves fixed the subfield >> of rational numbers. > >Prove it. [S]he just did! Is it the word automorphism you don't understand? Just look it up, then! Or do you disbelieve the existence of such an automorphism? What, exactly, is it that you want proven? >If Nora isn't being a jackass, then sorry for calling you a jackass >Nora. >But if your argument is right, you have shown Galois Theory to be >false. As has been pointed out before, there are at least four possibilities here: 1) Nora's argument is wrong; 2) Your argument (or alleged proof) is wrong; 3) Both arguments are wrong; 4) Galois Theory is wrong. Galois Theory has never been proven wrong; you have (almost) never been proven un-wrong. >So you see, it doesn't matter, so I don't mind. >In the end, mathematicians have to face my proof, and not try to run >away from it, or disprove it with another proof because you see, >proofs don't duel. So the proof that your proof is correct is the fact that it is a proof? Therefore, any disproof of your proof is nessesarily a disproof of itself and anything that has been used to derive the disproof, right? Is there ANY way one can convince you that your alleged proof is actually wrong? -- Thomas Wasell | A pencil with no point needs no eraser. wasell@bahnhof.se | === Subject: Re: My Work--Objective Review <3EEF5326.3F8AC63F@btinternet.com> > or likening me to a monkey with a typewriter. > This I find harder to explain. Perhaps David once met a monkey with a > typewriter and found it to be an unpleasant experience? >Eh, no. This was an allusion to the knowledge that a million monkeys >at typewriters will at some time produce a work of Shakespeare. Even >I (non-native English speaker) understood it. Sorry Dirk, but you missed George's very droll British sense of humour. I thought George's remark very funny, but maybe that just says something about me! BraileTrail -- === Subject: Re: My Work--Objective Review [...] >>In the end, mathematicians have to face my proof, and not try to run >>away from it, or disprove it with another proof because you see, >>proofs don't duel. >So the proof that your proof is correct is the fact that it is a proof? That's correct. No matter _how_ many times he eventually has to say oh by the way, that was wrong, sorry, it's still true that the next time he thinks he's proved something the fact that it's a proof is what proves it must be correct, and that anyone who disputes its correctness is a lying incompetent fool who should be fired (recently abbreviated to jackass.) Truly remarkable. >Therefore, any disproof of your proof is nessesarily a disproof of itself >and anything that has been used to derive the disproof, right? Is there ANY >way one can convince you that your alleged proof is actually wrong? ************************ David C. Ullrich === Subject: Re: My Work--Objective Review Nora: >> Without loss of generality, suppose a1 is coprime >> to 5 in the ring of algebraic integers. By definition >> this means that there exist algebraic integers r and s >> such that JSH: >Notice the call to the definition. Nora: > You have a problem with my using a > standard definition ??? This is only a wild hypothesis, but here goes. Suppose JSH thinks a proof is rather like a computer program. Then he thinks of bits like for loops, assignments, ifs, and whatnot as pure logic. But notice his exact phrasing above - a call to the definition. Suppose he thought that a definition was rather like a #define in C: can be used for unhygienic purposes, and is generally deprecated. Wouldn't this explain a number of his wilder claims? (Well I can't see how to account for the abuse of Galois theory bit, but I think we have to go one step at a time.) Brian Chandler ---------------- geo://Sano.Japan.Planet_3 http://imaginatorium.org/ JSH Jackass rating: 1 (so far) === Subject: Re: My Work--Objective Review >Notice the call to the definition. > You have a problem with my using a > standard definition ??? I cruised through and saw definition and made an assumption. It was a mistake. I later posted in a second reply noting that fact. However, I notice that the second reply didn't get much attention. >> >> r*a1 + s*5 = 1. >> >> Now apply F12 to both sides: >> >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). >> >> Note that F12(5) = 5 and F12(1) = 1 because >> the automorphism F12 leaves fixed the subfield >> of rational numbers. Also F12(a1) = a2. As requested in my second reply, you need to give more detail here. Also as noted repeatedly, proofs don't duel. If I'm wrong then there's an error in my proof. Trying to attack one proof by claiming it contradicts with another is useless. You can cast doubt that way, but to prove a proof false, you have to deal with the actual argument within it. >> >> Let r' = F12(r) and s' = F12(s). Note that >> both r' and s' are algebraic integers. Thus eqn (1) >> reduces to >> >> r'*a2 + s'*5 = 1. >> >> That is, a2 is also coprime to 5. >> >> Similarly one shows that a3 is coprime to 5. >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. >It's basically an abuse of Galois Theory. > What's the abuse? My belief has been that posters like yourself have been abusing Galois Theory as I don't believe it's wrong. I think you're cheating, which is why you wish to go to some other argument rather than deal with the one I've presented. So I see it as an abuse. Alternatively, Galois Theory IS wrong. Or you may *believe* I'm wrong, but if I'm wrong there would be a way to show it using my work. That is, if mathematicians are experts, which by definition they are, and I'm not a mathematician, but claim to have a proof, it hardly makes sense for mathematicians to use an alternative argument, as I'll point out the potential for abuse. It'd seem more rational for mathematicians to save their energy and attack my work itself. >> Therefore if one of a1, a2, or a3 is coprime >> to 5, they all are. >> >> But a1*a2*a3 = 65. Thus at least one of a1, >> a2, and a3 is NOT coprime to 5. Thus a contradiction. >> >> The claim in the Advanced Polynomial Factorization >> paper is therefore wrong. > >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a > contradiction. >So *obviously* there's a problem somewhere with what mathematicians >are doing if they think they can prove something that's false. > I would say obviously there is a problem somewhere, > yes. Then it hardly makes sense for you to spend time and effort not addressing my central claim which is that I have a proof. After all, if you found an error in that proof, you could make a VERY short post, and it'd be over instantly--no more discussion needed. CERTAINLY if I refused to acknowledge the truth, then I might continue arguing, but at least other posters would know there was a discovered error which they could view themselves to make their own decision. Now then are you a mathematician? If so, then you are a math expert by definition. Then if there is an error in my paper you should be able to point it out, and quit wasting so much time trying to fight my proof with claims of other proofs of your own. >Finally, notice the attempt to disprove a proof with *another* >argument, which is claimed to be a proof. > You claimed something was true. I gave an argument > to show it wasn't. You fought a conclusion of one argument, which I claim is a proof, with another argument which you claim is a proof. I challenge you to find an error in my argument; you challenge me to find an error in yours. It's a waste of time. I'm not a mathematician. My argument is simpler. So now there are dueling claims. You claim you gave an argument--I presume you'd claim it is correct--proving me wrong, I say you're wrong. Isn't anyone else bothered by the implication that mathematics is a mess? What's with all the debate? Are all mathematicians such losers that they can't handle a claim from an admitted non-mathematician, about a paper that's on display 24 hours a day? ARE YOU ALL LOSERS??!!! Yup, I'm frustrated. > Four possibilities: > 1. I have an error in my argument. You however did > not point out any error. I'm not a mathematician. Why in the hell should I necessarily be able to find an error in your argument? Dueling claims continue. > 2. You have an error in your argument. I did not > point it out in my post, but it has been pointed > out many times previously. Your argument depends > on the form that a factorization has in a degenerate, > singular case (when m = 0). You argue implicitly > that that same form of factorization must hold in the > nonsingular case in which you are actually interested. > It does not. This has been pointed out innumerable times. > Arturo and others have tried many times to get > you to understand it, but they have failed. I will > probably fail as well. If that's true then you can reference actual statements in the paper. What I see is an unsupported statement, and worse, you have an appeal to authority by mentioning Arturo Magidin. It seems to me that you are unsure of your own statement, and possibly wish to convince others without presenting actual evidence. > 3. Both of us have errors in our arguments. > 4. Neither of us has an error in our arguments. There > is a fundamental contradiction in mathematics. I > think we should just put this idea aside. It > is the least likely of the possibilities and it > does not obviously lead anywhere. There is no fundamental contradiction in mathematics. I have presented the most logical possibility before which is that mathematicians are lying. > It's also possible that the fact I used about automorphisms, > which comes from Galois theory, is wrong. That too seems > unlikely since Galois theory has been around for over 160 > years and looked at by many very smart people, and this > fact is encountered at the most basic level. I think that > too should be put aside. I don't care how long it's been around, if it's false it's false. If it's true it's true. And that condition is independent of time. However, I should point out that Galois Theory is a field theory. My work is not over fields. Still either you screwed up that F12 bit above, or Galois Theory is wrong. I'm interested in other replies that address whether you screwed up Nora. >It's like trying to fight a proof with a proof, but proofs can't > fight >each other. > We both present arguments. You say yours is a proof. > I say mine is. We are not both right. In the end > we have to try to convince other people and each other. That's stupid. A proof begins with a truth and proceeds by logical steps to a conclusion which then must be true. So all that's necessary is to start at the beginning and trace out each step. That IS possible with my paper, and I've done it which is why I get to have fun calling people like you out, and also calling you a jackass. You'll have a harder time tracing out your argument, but it might help you out to try, if you have the ability Nora. Yup, are you smart enough Nora? Screw convincing people. People are often stupid enough to be convinced of just about anything. === Subject: Re: My Work--Objective Review > [.snip.] >> Without loss of generality, suppose a1 is coprime >> to 5 in the ring of algebraic integers. By definition >> this means that there exist algebraic integers r and s >> such that >Notice the call to the definition. > Definitions are shorthand. In computer science terms, something you > claim to be familiar with, coprime is a macro; saying by > definition just means that you are expanding the macro. Saying a1 > and 5 are coprime means there exists r and s such that r*a1+s*5=1. Other readers will remember that I replied *again* noting that my issue with the call to the definition here didn't matter. However that post where I also asked about a specific claim in the author's argument hasn't gotten nearly the same number of replies from what I'm seeing now in Google Groups. So for those of you who wonder if I'm just blowing smoke, notice how posters, including mathematicians like Magidin here who got his Ph.d from Berkeley, pile on when they think they see an opening. Yup, as the original post brings out, they're jackasses. > Just like saying a is even means, ->by definition of even<-, that > there exist an integer k such that a=2*k. > [.snip.] >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. >It's basically an abuse of Galois Theory. > It is an abuse of Galois Theory that someone who admits he does not > know anything about it will nonetheless insist he can spot when it is > being used correctly and when it is not. > [.snip.] Readers who look back might also notice that Magidin deleted out all parts of the argument that Nora presented which I left in my post. Readers in looking over a post of Nora's that I replied to before this one will see where she refers to Arturo in an appeal to authority, and yes, that's Arturo Magidin, who here mentions nothing that she said, and doesn't mention her at all. Mathematicians are so fascinating as social creatures. >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a contradiction. > Which means you do not know what the word factor means. > That's all. You do not understand what factor means. What you write > above is exactly the same as saying that you have found and even > integer n such that n/2 is not an integer. It is an abuse of > terminlogy. That's false. It turns out that the *definition* of an algebraic integer as the ROOT of a monic polynomial with integer coefficients is what allows the ring to be incomplete. Now any rational person looking over that definition for an algebraic integer would be at a loss to see how the definition of factor applies. The appeal to a definition here is the kind of crap that I thought I was jumping on before with Nora. >So *obviously* there's a problem somewhere > Yes: the problem is that you are an ignoramus who refuses to learn > what the words he uses mean. Well, you're a jackass. > DEF. Let R be a commutative ring. Let x and y be elements of R. We say > y is a factor of x (in R) if and only if there exist z in R such > that y*z = x. If the ring R is understood from context, we can omit > (in R). > DEF. Let R be a commutative ring. An element x in R is called a unity > if and only if for all y in R, x*y=y*x = y. If so, we denote x by 1, > and we say that R is a commutative ring with 1. > DEF. Let R be a commutative ring. We say that R is an INTEGRAL DOMAIN > if and only if it has a 1 for every x,y in R, if x*y=0, then x=0 or y=0. > DEF. Let R be an integral domain. The FIELD OF FRACTIONS of R is > constructed as follows: > (1) Let k be the collection of all formal symbols x/y, with x,y > in R, y nonzero. > (2) Define an equivalence relation on k, by letting (x/y)~(z/w) if > and only if x*w=z*y in R. > (3) Let K be the set whose elements are the equivalence classes of > elements of k under ~. Denote the class of x/y by [x/y]. > (4) Define an addition and a multiplication in K by: > [x/y] + [z/w] = [(x*w+z*y)/y*w] > [x/y] * [z/w] = [(x*z)/(y*w)]. > THEOREM. The field of fractions of an integral domain is a field; the > 0 is the class [0/1]. The one is [1/1]. R is a subring of K by the map > that identifies the element r in R with the element [r/1] of K. > THEOREM. Let R be an integral domain, and let x,y be elements of > R. Then y is a factor of x in R if and only if the element [x/y] of K > lies in the image of R. We express this by writing x/y lies in R. > Therefore, what you have claimed is that there is an algebraic integer > which is not an algebraic integer. And that is nonsense on the same > level as saying that you can find an even number which is not a > multiple of 2. Ok, so let's say readers wish to believe that I've been refuted. Why did Magidin go through the trouble? And what can I say here? That's he's full of it? That he's boldly lying yet again? But when you look over what he presented and think to yourself, wow, that looks like what I figure math stuff should look like, what am I to do? Even if I take him apart piece by piece if you *believe* it doesn't matter. I feel sorry for many of you because your faith does not constitute knowledge. Just remember Jim Jones convinced quite a few people that sucking down poisoned kool-aid was a great idea! You're wired to believe. You're wired to pick an authority and just believe. What can mere logic do against that wiring in your heads? === Subject: Re: My Work--Objective Review > That's false. It turns out that the *definition* of an algebraic > integer as the ROOT of a monic polynomial with integer coefficients is > what allows the ring to be incomplete. > Now any rational person looking over that definition for an algebraic > integer would be at a loss to see how the definition of factor > applies. No, a rational person would say this: Let c factor into a*b in the ring of algebraic integers. Since algebraic integers are roots of monic polynomials with integer coefficients, I conclude that a and b are roots of monic polynomials with integer coefficients. What loss did you think a rational person would be at? There is NO OTHER MEANING of factor! - Randy === Subject: Re: My Work--Objective Review [cut] >> r*a1 + s*5 = 1. >> >> Now apply F12 to both sides: >> >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). >> >> Note that F12(5) = 5 and F12(1) = 1 because >> the automorphism F12 leaves fixed the subfield >> of rational numbers. Also F12(a1) = a2. > As requested in my second reply, you need to give more detail here. Which statement(s) do you need more justification for: a) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1) b) F12(5) = 5 c) F12(1) = 1 d) F12(a1) = a2 -- Bill Hale === Subject: Re: My Work--Objective Review Visiting Assistant Professor at the University of Montana. Newsgroups trimmed. >> [.snip.] >> >>> Without loss of generality, suppose a1 is coprime >>> to 5 in the ring of algebraic integers. By definition >>> this means that there exist algebraic integers r and s >>> such that >> >>Notice the call to the definition. >> Definitions are shorthand. In computer science terms, something you >> claim to be familiar with, coprime is a macro; saying by >> definition just means that you are expanding the macro. Saying a1 >> and 5 are coprime means there exists r and s such that r*a1+s*5=1. >Other readers will remember that I replied *again* noting that my >issue with the call to the definition here didn't matter. Will you ->EVER<- learn about distribution in usenet? That message did not make it to my server until after I had posted this. Your innuendos notwithstanding. [.snip.] >> Which means you do not know what the word factor means. >> That's all. You do not understand what factor means. What you write >> above is exactly the same as saying that you have found and even >> integer n such that n/2 is not an integer. It is an abuse of >> terminlogy. >That's false. It turns out that the *definition* of an algebraic >integer as the ROOT of a monic polynomial with integer coefficients is >what allows the ring to be incomplete. Please provide a definition of incomplete. Because, if we use the words definition, algebraic integer, and factor in their usual sense, then what you have said is nonsense. >Now any rational person looking over that definition for an algebraic >integer would be at a loss to see how the definition of factor >applies. DEFINITION. Let c be a complex number. Then c is an algebraic integer if and only if there exists a monic polynomial f(x) with integer coefficients such that c is a root of f(x). DEFINITION. Let R be any commutative ring. Let x and y be elements of R. Then y is a factor of x in R if and only if there exists z in R such that y*z = x. THEOREM. The collection of all algebraic integers form an integral domain, contained in the field of complex numbers. Which of the above do you disagree with? Remember: definitions cannot be right or wrong; they are only shorthand. Now, you claim that: (a) You have algebraic integer y and x; (b) that y is a factor of x in the ring of algebraic integers; (c) that there does not exist a z in the ring of algebraic integers such that y*z = x. Therefore, you are speaking nonsense. [.snip.] >>So *obviously* there's a problem somewhere >> Yes: the problem is that you are an ignoramus who refuses to learn >> what the words he uses mean. >Well, you're a jackass. I see you do not contest my statement above. For the record, your appreciation of whether or not I am, as you claim, a jackass is immaterial. >> DEF. Let R be a commutative ring. Let x and y be elements of R. We say >> y is a factor of x (in R) if and only if there exist z in R such >> that y*z = x. If the ring R is understood from context, we can omit >> (in R). >> DEF. Let R be a commutative ring. An element x in R is called a unity >> if and only if for all y in R, x*y=y*x = y. If so, we denote x by 1, >> and we say that R is a commutative ring with 1. >> DEF. Let R be a commutative ring. We say that R is an INTEGRAL DOMAIN >> if and only if it has a 1 for every x,y in R, if x*y=0, then x=0 or y=0. >> DEF. Let R be an integral domain. The FIELD OF FRACTIONS of R is >> constructed as follows: >> (1) Let k be the collection of all formal symbols x/y, with x,y >> in R, y nonzero. >> (2) Define an equivalence relation on k, by letting (x/y)~(z/w) if >> and only if x*w=z*y in R. >> (3) Let K be the set whose elements are the equivalence classes of >> elements of k under ~. Denote the class of x/y by [x/y]. >> (4) Define an addition and a multiplication in K by: >> [x/y] + [z/w] = [(x*w+z*y)/y*w] >> [x/y] * [z/w] = [(x*z)/(y*w)]. >> THEOREM. The field of fractions of an integral domain is a field; the >> 0 is the class [0/1]. The one is [1/1]. R is a subring of K by the map >> that identifies the element r in R with the element [r/1] of K. >> THEOREM. Let R be an integral domain, and let x,y be elements of >> R. Then y is a factor of x in R if and only if the element [x/y] of K >> lies in the image of R. We express this by writing x/y lies in R. >> Therefore, what you have claimed is that there is an algebraic integer >> which is not an algebraic integer. And that is nonsense on the same >> level as saying that you can find an even number which is not a >> multiple of 2. >Ok, so let's say readers wish to believe that I've been refuted. Why do you not comment on anything of the above? Because you do not understand it. Despite being mathematics. >Why did Magidin go through the trouble? Read the .sig. >And what can I say here? That's he's full of it? That he's boldly >lying yet again? You could say that. That seems to be your only defense when faced with undeniable evidence of your errors. >But when you look over what he presented and think to yourself, wow, >that looks like what I figure math stuff should look like, what am I >to do? Read it, try to understand it, and if there is something you do not understand, ask. Nobody is asking you to take it on faith. We only ask that you take the time to LEARN the stuff you try to pontificate aboue. >Even if I take him apart piece by piece if you *believe* it doesn't >matter. Because you never do take apart piece by piece, except in your imagination. Your complaints are always either wrong, nonsense, or based on ignorance and misunderstanding. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan Arturo Magidin magidin@math.berkeley.edu === Subject: Re: My Work--Objective Review > That's false. It turns out that the *definition* of an algebraic > integer as the ROOT of a monic polynomial with integer coefficients is > what allows the ring to be incomplete. > > Now any rational person looking over that definition for an algebraic > integer would be at a loss to see how the definition of factor > applies. > No, a rational person would say this: > Let c factor into a*b in the ring of algebraic integers. > Since algebraic integers are roots of monic polynomials with > integer coefficients, I conclude that a and b are roots of > monic polynomials with integer coefficients. Sure, *if* it so factors. You're running in a circle. As I've stated more than once and as anyone can verify who looks over my paper, I prove that a number I'll call x has a number I'll call y as a factor, where they both are algebraic integers, but it turns out that x/y is NOT an algebraic integer. Here's a test of your ability to think rationally Randy Poe, can you see how nothing you have said prevents that possibility? If you say that x factors into ab in the ring of algebraic integers, then of course x/a or x/b is an algebraic integer. But I didn't say that x factors into y times some algebraic integer. I said that x *provably* has y as a factor in the ring of algebraic integers, but that leads to a contradiction as x/y is not in the ring, so the ring isn't complete. That is, x/y is *supposed* to be in the ring, but provably is not. For those who find themselves totally confused, imagine I say that rargles are a complete ring, and I give you x = ab in the ring of rargles. Now then I also give some definition of the ring which I'll leave up in the air, and one of you, who is a real mathematician, comes along and says, hey your definition works sometimes, but there are times that your ring of rargles allows me to prove that x has a factor of y within the ring, but x/y is not in the ring, which is a contradiction. So then, there erupts a huge fight, let's say, because so many people love the rargle definition, but the real mathematician refuses to accept that nonsense and sticks to his guns. Ultimately, people look and realize that the definition for the rargle ring left an opening which lead to the contradiction that he pointed out. > What loss did you think a rational person would be at? There > is NO OTHER MEANING of factor! > - Randy But people like Randy went insane as they repeated over and over again that there is NO OTHER MEANING OF FACTOR! Logic is not for everyone. Math is not for everyone because it doesn't care if your mental wiring isn't up to the task of handling a mathematical argument. Now I can explain carefully how it is that x has y as a factor when x/y does not, and where that all comes from, especially considering that division isn't even defined for rings, but what's the point? First I need to have rational people who can handle a logical argument. Not people whose brains can't handle the facts. === Subject: Re: My Work--Objective Review nevermind. let's start another thread in this item! seriously, bright pupil is sort of an oxymoron, considering that Jimmy's eyes are always dilated. > I'm paraphrasing Sawyer: are you saying he's blowing smoke? (You > could at least have accused me of blowing smoke rings - that might > OK: this sounds a little like the idea of a polynomial function: for > each integer, n say, there's another integer m, got by evaluating this > P(x) at x=n. Of course m is just n+1. This is how Sawyer suggests the > 'bright pupils' tend to see introductory algebra. > Q(x) = 2x+1 > When you consider this in the ring of integers, does it have any > roots? > [snip: Arturo Magidin dealt with this] > Damn. So you're just another jackass. > Aha!! I'll have my point back, please. --les ducs d'Enron! http://quincy4board.homestead.com/ Funny.html (schoolboard stuffin') === Subject: Re: My Work--Objective Review >As I've stated more than once and as anyone can verify who looks over >my paper, I prove that a number I'll call x has a number I'll call y >as a factor, When you state that x has y as a factor, what do you mean? Please be precise. I'm not looking for examples, I'm looking for a definition. I want a way to take two items, x and y, and determine if y is a factor of x. Alan -- Defendit numerus === Subject: Re: My Work--Objective Review ye gadzooks -- I think we've dyscovered the generalized Turing Machine that always halts, no matter what the input! > That's correct. No matter _how_ many times he eventually has to > say oh by the way, that was wrong, sorry, it's still true that the > next time he thinks he's proved something the fact that it's a proof > is what proves it must be correct, and that anyone who disputes > its correctness is a lying incompetent fool who should be fired > (recently abbreviated to jackass.) >Therefore, any disproof of your proof is nessesarily a disproof of itself >and anything that has been used to derive the disproof, right? Is there ANY >way one can convince you that your alleged proof is actually wrong? --les ducs d'Enron! http://quincy4board.homestead.com/ Funny.html (schoolboard stuffin') === Subject: Re: My Work--Objective Review > > >Notice the call to the definition. > > > > You have a problem with my using a > standard definition ??? > I cruised through and saw definition and made an assumption. > It was a mistake. I later posted in a second reply noting that fact. > However, I notice that the second reply didn't get much attention. >> >> r*a1 + s*5 = 1. >> >> Now apply F12 to both sides: >> >> (1) F12(r)*F12(a1) + F12(s)*F12(5) = F12(1). >> >> Note that F12(5) = 5 and F12(1) = 1 because >> the automorphism F12 leaves fixed the subfield >> of rational numbers. Also F12(a1) = a2. > As requested in my second reply, you need to give more detail here. Keep reading - I gave a reference to a website and a book that explain this very clearly. Oh, I see you deleted this reference. It was: http://www.math.niu.edu/~beachy/aaol/galois.html See esp. section 8.6.2. > Also as noted repeatedly, proofs don't duel. > If I'm wrong then there's an error in my proof. > Trying to attack one proof by claiming it contradicts with another is > useless. > You can cast doubt that way, but to prove a proof false, you have to > deal with the actual argument within it. Actually, I don't agree with this. True, it is good to identify where a person has made an error when they claim to have proved something. But if you provide a proof that what they are claiming is wrong, you have headed off not only their current proposed argument, but also all future arguments in support of that same claim. In any case, I did both. I did say explicitly where your error occurs and why it is wrong - see below. >> >> Let r' = F12(r) and s' = F12(s). Note that >> both r' and s' are algebraic integers. Thus eqn (1) >> reduces to >> >> r'*a2 + s'*5 = 1. >> >> That is, a2 is also coprime to 5. >> >> Similarly one shows that a3 is coprime to 5. > >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. > >It's basically an abuse of Galois Theory. > > > What's the abuse? > My belief has been that posters like yourself have been abusing Galois > Theory as I don't believe it's wrong. I think you're cheating, which > is why you wish to go to some other argument rather than deal with the > one I've presented. > So I see it as an abuse. I still don't see the abuse. If I'm cheating, please tell me where. > Alternatively, Galois Theory IS wrong. > Or you may *believe* I'm wrong, but if I'm wrong there would be a way > to show it using my work. Yes - see below. > That is, if mathematicians are experts, which by definition they are, > and I'm not a mathematician, but claim to have a proof, it hardly > makes sense for mathematicians to use an alternative argument, as I'll > point out the potential for abuse. > It'd seem more rational for mathematicians to save their energy and > attack my work itself. I think both approaches should be taken - it's one thing to point to a specific place in your argument and say, You have not justified this, to which you might answer It's obvious and anyone can see it and it doesn't need to be justified. That can get frustrating. It's of value also to say, Here is an argument that shows your argument, *whatever it is*, must be wrong. I chose to take the latter course initially. >> Therefore if one of a1, a2, or a3 is coprime >> to 5, they all are. >> >> But a1*a2*a3 = 65. Thus at least one of a1, >> a2, and a3 is NOT coprime to 5. Thus a contradiction. >> >> The claim in the Advanced Polynomial Factorization >> paper is therefore wrong. > > > >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a > contradiction. > >So *obviously* there's a problem somewhere with what mathematicians >are doing if they think they can prove something that's false. > > > > I would say obviously there is a problem somewhere, > yes. > Then it hardly makes sense for you to spend time and effort not > addressing my central claim which is that I have a proof. > After all, if you found an error in that proof, you could make a VERY > short post, and it'd be over instantly--no more discussion needed. > CERTAINLY if I refused to acknowledge the truth, then I might continue > arguing, Yes - I think that is what will happen - anyway, keep reading - > but at least other posters would know there was a discovered > error which they could view themselves to make their own decision. > Now then are you a mathematician? Yep. > If so, then you are a math expert by definition. Then if there is an > error in my paper you should be able to point it out, and quit wasting > so much time trying to fight my proof with claims of other proofs of > your own. Keep reading. >Finally, notice the attempt to disprove a proof with *another* >argument, which is claimed to be a proof. > > > You claimed something was true. I gave an argument > to show it wasn't. > You fought a conclusion of one argument, which I claim is a proof, > with another argument which you claim is a proof. > I challenge you to find an error in my argument; you challenge me to > find an error in yours. I didn't challenge you. I just pointed out that you hadn't. Anyway, see below. > It's a waste of time. I have nothing to lose in this. You are the one who wants to be recognized for great discoveries. You cannot afford to ignore proofs that you are wrong. > I'm not a mathematician. My argument is simpler. > So now there are dueling claims. You claim you gave an argument--I > presume you'd claim it is correct--proving me wrong, I say you're > wrong. > Isn't anyone else bothered by the implication that mathematics is a > mess? > What's with all the debate? > Are all mathematicians such losers that they can't handle a claim from > an admitted non-mathematician, about a paper that's on display 24 > hours a day? > ARE YOU ALL LOSERS??!!! > Yup, I'm frustrated. > Four possibilities: > > 1. I have an error in my argument. You however did > not point out any error. > I'm not a mathematician. Why in the hell should I necessarily be able > to find an error in your argument? It's a simple, short argument. You know enough math to evaluate it. You are perfectly capable of finding an error in it if there is one. Might be worth noting that none of the other posters here have ventured to say whether my argument is right or wrong. It could easily be wrong, eh? > Dueling claims continue. I don't get this emphasis on duelling claims. Why are you stressing that so much? It's standard procedure if one guy says he has a proof, and another guy says he is wrong, the second guy either finds a counterexample (which often involves a proof of its own) or a counter-proof. > 2. You have an error in your argument. I did not > point it out in my post, but it has been pointed > out many times previously. Your argument depends > on the form that a factorization has in a degenerate, > singular case (when m = 0). You argue implicitly > that that same form of factorization must hold in the > nonsingular case in which you are actually interested. > It does not. This has been pointed out innumerable times. > Arturo and others have tried many times to get > you to understand it, but they have failed. I will > probably fail as well. > If that's true then you can reference actual statements in the paper. You should be able to tell EXACTLY what I am talking about from my reference to a degenerate case (where your polynomial, which is usually of degree 3, becomes a linear function, i.e., a polynomial of degree 1) and my reference to m = 0. What I have given would be sufficient for any normal person to understand what is wrong. Therefore I refuse to give you more help than that at this point. You have the choice of (1) actually doing some work to figure out what I am talking about, or (2) resigning yourself to the the fact that I have specified exactly where your error is and that, separately, I have a counterargument which proves you are wrong. I have led the horse to water. If the horse wishes to die of thirst, there is not much I can do about it. Well, theoretically, there is one alternative. You could give a detailed proof of why properties of a factorization in a degenerate case imply that those same properties hold in nondegenerate cases. That is, you can fill in the main gap in your argument. Or try to. And, theoretically, there is another alternative. You could find an explicit error in MY argument. Do that and I will go away. Finally, if after studying it for a while you still don't get what I am saying, let me know. > What I see is an unsupported statement, and worse, you have an appeal > to authority by mentioning Arturo Magidin. By no means! I mentioned Arturo just to jog your memory. 'Authority' here is not an issue. My statements stand strictly on their own. For all I know, Arturo thinks my argument is B.S.. > It seems to me that you are unsure of your own statement, and possibly > wish to convince others without presenting actual evidence. Not at all. I thought what I said was more than sufficient. Apparently not. See above. > 3. Both of us have errors in our arguments. > > 4. Neither of us has an error in our arguments. There > is a fundamental contradiction in mathematics. I > think we should just put this idea aside. It > is the least likely of the possibilities and it > does not obviously lead anywhere. > There is no fundamental contradiction in mathematics. > I have presented the most logical possibility before which is that > mathematicians are lying. Doesn't seem all that likely that ALL of us would lie. In any case, *I'm* definitely not lying. > It's also possible that the fact I used about automorphisms, > which comes from Galois theory, is wrong. That too seems > unlikely since Galois theory has been around for over 160 > years and looked at by many very smart people, and this > fact is encountered at the most basic level. I think that > too should be put aside. > I don't care how long it's been around, if it's false it's false. > If it's true it's true. > And that condition is independent of time. > However, I should point out that Galois Theory is a field theory. All I used here was a little introductory theorem about field automorphisms. A key feature is that automorphisms exist which transform one root of an irreducible polynomial into another root, they will leave the set of algebraic integers invariant. That is not really a field property, but it's relevant and useful here. > My work is not over fields. > Still either you screwed up that F12 bit above, or Galois Theory is > wrong. Or your argument is wrong. How have you excluded that possibility? > I'm interested in other replies that address whether you screwed up > Nora. You are? Well, people are perfectly free to make such replies. >It's like trying to fight a proof with a proof, but proofs can't > fight >each other. > > > We both present arguments. You say yours is a proof. > I say mine is. We are not both right. In the end > we have to try to convince other people and each other. > > That's stupid. Is it? If so, why do you spend so much time trying to convince everyone here that your arguments are correct? > A proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. A correct proof does that. An incorrect proof either leaves out a step or has an erroneous step. An argument is not a proof just because its author says it is a proof. > So all that's necessary is to start at the beginning and trace out > each step. > That IS possible with my paper, and I've done it which is why I get to > have fun calling people like you out, and also calling you a jackass. > You'll have a harder time tracing out your argument, but it might help > you out to try, if you have the ability Nora. See above. > Yup, are you smart enough Nora? Nah. I'm just a jackass. Why do you think anyone would be convinced by what a jackass would say? Nora B. > Screw convincing people. People are often stupid enough to be > convinced of just about anything. > === Subject: Re: My Work--Objective Review >> That's false. It turns out that the *definition* of an algebraic >> integer as the ROOT of a monic polynomial with integer coefficients is >> what allows the ring to be incomplete. >> >> Now any rational person looking over that definition for an algebraic >> integer would be at a loss to see how the definition of factor >> applies. >> No, a rational person would say this: >> Let c factor into a*b in the ring of algebraic integers. >> Since algebraic integers are roots of monic polynomials with >> integer coefficients, I conclude that a and b are roots of >> monic polynomials with integer coefficients. >Sure, *if* it so factors. If it doesn't, neither a nor b is a factor of c. >You're running in a circle. >As I've stated more than once and as anyone can verify who looks over >my paper, I prove that a number I'll call x has a number I'll call y >as a factor, The expression y is a factor in the algebraic integers means x/y is an algebraic integer. > where they both are algebraic integers, but it turns out >that x/y is NOT an algebraic integer. Then y is not a factor of x. Once again, the term y is a factor of x in the algebraic integers means x/y is an algebraic integer. That's what it means to every but you. What does it mean to you? In particular, what does it mean for y NOT to be a factor of x in the algebraic integers? Just tell me that. If I have two arbitrary algebraic integers, what test to I apply to tell if y is a factor of x? I'm going to hammer on that question for awhile to see if you answer it. >Here's a test of your ability to think rationally Randy Poe, can you >see how nothing you have said prevents that possibility? No I can't say that, since x/y is an algebraic integer and y is a factor of x in the algebraic integers are the same statement. >If you say that x factors into ab in the ring of algebraic integers, >then of course x/a or x/b is an algebraic integer. Yep. >But I didn't say that x factors into y times some algebraic integer. Then y is not a factor of x in the algebraic integers. See how that works? >I said that x *provably* has y as a factor in the ring of algebraic >integers, but that leads to a contradiction as x/y is not in the ring, >so the ring isn't complete. That is, x/y is *supposed* to be in the >ring, but provably is not. OK, I think I see what you're saying. You think you have a proof that x/y is an algebraic integer. And another proof that it isn't. Contradiction. So what is your line of argument that x/y is an algebraic integer? - Randy === Subject: Re: My Work--Objective Review >> That's false. It turns out that the *definition* of an algebraic >> integer as the ROOT of a monic polynomial with integer coefficients is >> what allows the ring to be incomplete. >> >> Now any rational person looking over that definition for an algebraic >> integer would be at a loss to see how the definition of factor >> applies. >> No, a rational person would say this: >> Let c factor into a*b in the ring of algebraic integers. >> Since algebraic integers are roots of monic polynomials with >> integer coefficients, I conclude that a and b are roots of >> monic polynomials with integer coefficients. >Sure, *if* it so factors. >You're running in a circle. >As I've stated more than once and as anyone can verify who looks over >my paper, I prove that a number I'll call x has a number I'll call y >as a factor, where they both are algebraic integers, but it turns out >that x/y is NOT an algebraic integer. Yes, you've stated this more than once. As has been stated in reply many times, this is sheer nonsense, unless you clarify that you mean factor in some sense other than factor in the algebraic integers, and explain what you _do_ mean. [...] >First I need to have rational people who can handle a logical >argument. >Not people whose brains can't handle the facts. It appears that there's not a single mathematician on the planet who can handle a rational argument. Very curious. (Hmm, maybe there's another explanation...) > ************************ David C. Ullrich === Subject: Re: My Work--Objective Review >>So *obviously* there's a problem somewhere >> Yes: the problem is that you are an ignoramus who refuses to learn >> what the words he uses mean. > Well, you're a jackass. The problem is that you have *no* evidence to support what you're saying a period of *years* that prove what he's saying about you. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: My Work--Objective Review > Newsgroups trimmed. >>Other readers will remember that I replied *again* noting that my >>issue with the call to the definition here didn't matter. > Will you ->EVER<- learn about distribution in usenet? That message did > not make it to my server until after I had posted this. Your innuendos > notwithstanding. James is completely ignorant of USENET. That's the trouble with Google Groups; it gives idiots access to USENET without requiring them to have even the minimum knowledge about it that even the most clueless users used to have. >>Why did Magidin go through the trouble? > Read the .sig. Oops, that requires USENET (or at least email) knowledge. You'd better explain to him what a .sig is. (It won't do any good, of course, as he never learns from explanations or definitions.) -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: My Work--Objective Review ... stuff deleted ... > As requested in my second reply, you need to give more detail here. > Also as noted repeatedly, proofs don't duel. > If I'm wrong then there's an error in my proof. > Trying to attack one proof by claiming it contradicts with another is > useless. > You can cast doubt that way, but to prove a proof false, you have to > deal with the actual argument within it. >>Let r' = F12(r) and s' = F12(s). Note that >>both r' and s' are algebraic integers. Thus eqn (1) >>reduces to >> >> r'*a2 + s'*5 = 1. >> >>That is, a2 is also coprime to 5. >> >>Similarly one shows that a3 is coprime to 5. >Which looks like a good example to show those who wondered how what >I've shown highlights an error in *taught* mathematics. >It's basically an abuse of Galois Theory. >> What's the abuse? > My belief has been that posters like yourself have been abusing Galois > Theory as I don't believe it's wrong. I think you're cheating, which > is why you wish to go to some other argument rather than deal with the > one I've presented. > So I see it as an abuse. > Alternatively, Galois Theory IS wrong. > Or you may *believe* I'm wrong, but if I'm wrong there would be a way > to show it using my work. > That is, if mathematicians are experts, which by definition they are, > and I'm not a mathematician, but claim to have a proof, it hardly > makes sense for mathematicians to use an alternative argument, as I'll > point out the potential for abuse. > It'd seem more rational for mathematicians to save their energy and > attack my work itself. >>Therefore if one of a1, a2, or a3 is coprime >>to 5, they all are. >> >>But a1*a2*a3 = 65. Thus at least one of a1, >>a2, and a3 is NOT coprime to 5. Thus a contradiction. >> >>The claim in the Advanced Polynomial Factorization >>paper is therefore wrong. > >Now I've explained more than once that I prove that x has y as a >factor, but x/y is not an algebraic integer, which is a >> contradiction. >So *obviously* there's a problem somewhere with what mathematicians >are doing if they think they can prove something that's false. >>I would say obviously there is a problem somewhere, >>yes. > Then it hardly makes sense for you to spend time and effort not > addressing my central claim which is that I have a proof. > After all, if you found an error in that proof, you could make a VERY > short post, and it'd be over instantly--no more discussion needed. > CERTAINLY if I refused to acknowledge the truth, then I might continue > arguing, but at least other posters would know there was a discovered > error which they could view themselves to make their own decision. How about this: Let z be any root of the polynomial: p(x) = x^3 - 12 x^2 + 65, and define these polynomials: q(x) = 8 x^2 - 76 x -185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 Then the following are trivial, if tedious, to show: 5 = q(z)*r(z). z = r(z)*s(z). Finally, r(z) has minimal polynomial MinPoly(r(z)) = x^3 - 969 x^2 + 315 x + 5 This shows that 1. 5 and z [remember z is *any* root of p(x)] have a mutual factor, r(z). 2. r(z) is an algebraic integer. 3. The remaining factor of 5 is another algebraic integer q(z). 4. The remaining factor of z is another algebraic integer s(z). 5. r(z) is NOT a unit. produce the common factor between z and 5 is in accord with what one expects from Galois theory; the common minimal polynomial for the (varying) factor of 5 is similarly in accord with what one would expect from Galois theory. Since the a's in the factorization of 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) are precisely equal to (-1) times the roots of p(x) = x^3 - 12 x^2 + 65 = (x - r1)(x - r2)(x - r3), this result about the roots of p(x) directly shows that JSH is wrong. > Now then are you a mathematician? > If so, then you are a math expert by definition. Then if there is an > error in my paper you should be able to point it out, and quit wasting > so much time trying to fight my proof with claims of other proofs of > your own. The above shows that the so-called Primary argument in your paper is in error. I may spend the time and effort to isolate the first error in that section, but I will make no promises at this time. >Finally, notice the attempt to disprove a proof with *another* >argument, which is claimed to be a proof. >> You claimed something was true. I gave an argument >> to show it wasn't. > You fought a conclusion of one argument, which I claim is a proof, > with another argument which you claim is a proof. > I challenge you to find an error in my argument; you challenge me to > find an error in yours. > It's a waste of time. > I'm not a mathematician. My argument is simpler. I have shown *directly* that your conclusion is in error. I may look closer to find the initial error. It is not my responsibility to wipe your rumpus every time you make poopy, though, so I may just leave you all stinky. > So now there are dueling claims. You claim you gave an argument--I > presume you'd claim it is correct--proving me wrong, I say you're > wrong. But I've shown *directly* that your conclusion is incorrect. Is that any sort of thing you should pay attention to? > Isn't anyone else bothered by the implication that mathematics is a > mess? Doesn't it bother you to be just plain wrong? AGAIN???? > What's with all the debate? > Are all mathematicians such losers that they can't handle a claim from > an admitted non-mathematician, about a paper that's on display 24 > hours a day? Are you such a loser that you need to harangue, taunt, cast aspersions, and make a nuisance of yourself? All for reasons that wouldn't be acceptable if you were in fact correct, yet you are apparently NEVER correct. > ARE YOU ALL LOSERS??!!! > Yup, I'm frustrated. >> Four possibilities: >> 1. I have an error in my argument. You however did >> not point out any error. > I'm not a mathematician. Why in the hell should I necessarily be able > to find an error in your argument? > Dueling claims continue. I have shown you to be wrong. Irrefutably. Do the arithmetic. The above numbers are r(z) are common factors of 5 and z (where z ranges over the roots of the polynomial p(x)), so they are also common factors between 5 and your ai's . Each ai has a factor in common with 5. Theses factor are (1) integral over Q, (2) not units. >> 2. You have an error in your argument. I did not >> point it out in my post, but it has been pointed >> out many times previously. Your argument depends >> on the form that a factorization has in a degenerate, >> singular case (when m = 0). You argue implicitly >> that that same form of factorization must hold in the >> nonsingular case in which you are actually interested. >> It does not. This has been pointed out innumerable times. >> Arturo and others have tried many times to get >> you to understand it, but they have failed. I will >> probably fail as well. > If that's true then you can reference actual statements in the paper. The actual statement is this: Therefore, with the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) one fo the a'2 is coprime to 5. > What I see is an unsupported statement, and worse, you have an appeal > to authority by mentioning Arturo Magidin. What I see is a pattern of obsession with the evil Arturo Magidin, and an inability of JSH to distinguish between poo and poonola. > It seems to me that you are unsure of your own statement, and possibly > wish to convince others without presenting actual evidence. It seems to me that you should go back and find out where you are wrong, and perhaps learn to verify a proof *before* calling people names. >> 3. Both of us have errors in our arguments. >> 4. Neither of us has an error in our arguments. There >> is a fundamental contradiction in mathematics. I >> think we should just put this idea aside. It >> is the least likely of the possibilities and it >> does not obviously lead anywhere. > There is no fundamental contradiction in mathematics. > I have presented the most logical possibility before which is that > mathematicians are lying. Oops, in the presence of evidence, the most logical possibility is the one that's in agreement with the evidence. To date, all the evidence we have is your track record, as compared to the track record of the contributors of sci.math in demolishing your false claims. These track records suggest something quite different from what you're suggesting as being most likely. Could it be that you're blinded by your obsessive need to show up those fargin iceholes? >> It's also possible that the fact I used about automorphisms, >>which comes from Galois theory, is wrong. That too seems >>unlikely since Galois theory has been around for over 160 >>years and looked at by many very smart people, and this >>fact is encountered at the most basic level. I think that >>too should be put aside. > I don't care how long it's been around, if it's false it's false. > If it's true it's true. Galois theory has been tested FAR more than (1) Relativity, (2) Quantum mechanics, (3) Object math [snrk] > And that condition is independent of time. > However, I should point out that Galois Theory is a field theory. Again, you betray abysmal ignorance. You must imagine that results derived from field considerations cannot be used to make statements about subrings of that field. That's pretty much an indictment of the level at which you operate, and that level will place a strict limit on how far you can proceed, except by sheer accident. > My work is not over fields. > Still either you screwed up that F12 bit above, or Galois Theory is > wrong. Or, as I have just shown *directly*, you are just plain wrong. Couldn't get your brain wrapped around that concept, now, could you? > I'm interested in other replies that address whether you screwed up > Nora. Nora is correct. JSH is incorrect. >It's like trying to fight a proof with a proof, but proofs can't >> fight >each other. >> We both present arguments. You say yours is a proof. >>I say mine is. We are not both right. In the end >>we have to try to convince other people and each other. > > That's stupid. > A proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. Oh, but, er, didn't I just show you how wrong you were? Your conclusion IS WRONG! > So all that's necessary is to start at the beginning and trace out > each step. Too bad you screwed up, isn't it? > That IS possible with my paper, and I've done it which is why I get to > have fun calling people like you out, and also calling you a jackass. But your arguments lead you to incorrect conclusions. Why on earth would you continue with such poor results? Doesn't it embarass you to make a fool of yourself with such amazing regularity? > You'll have a harder time tracing out your argument, but it might help > you out to try, if you have the ability Nora. No, standard mathematics is well-enough constructed, and comes with sufficient infrastructure (by means of an array of lemmas & theorems, standard examples and techniques) that allow one not only to produce proofs but also to isolate errors. > Yup, are you smart enough Nora? It's all about how *smart* someone is, isn't it? The world isn't always just about how *smart* people are, but sometimes it's about producing correct results. No matter how *smart* you think you are, as long as you continue to be wrong while insisting you're right, as long as you continue to insult those who *are* right, you're going to be chasing your own tail. How long do you intend to keep this stupid game going? Lots of us use this as a form of perverse entertainment; my guess is that you are also using it in that way, believing somehow that you're fooling someone. If you truly believe you're correct, you'll do something to make that manifest, such as getting yourself some education in the fields where you pretend to have ability. > Screw convincing people. People are often stupid enough to be > convinced of just about anything. It's clear you can be convinced of anything you scribble down. > Dale === Subject: need a function Hi. I'm trying to find a function satisfying the conditions below: Define S:= {(x,y) in R : 0 Define S:= {(x,y) in R : 0 Define T:= {x in R : 0 Problem: Use the fact that every real number has a decimal expansion > to produce an injective function that maps S into T. Think about how you can take two infinite decimal expansions and turn them into one infinite decimal expansion. (If you've heard of Hilbert's Hotel, that might give you an idea.) -- word to the appropriate symbol to email me. Joshua P. Bowman === Subject: Re: need a function > Hi. I'm trying to find a function satisfying the conditions below: > Define S:= {(x,y) in R : 0 Define T:= {x in R : 0 Problem: Use the fact that every real number has a decimal expansion > to produce an injective function that maps S into T. For the reals which can have terminating decimal representations, require them to be so represented, then for x with decimal representation .abcdefg... and y with decimal representation .ABCDEFG... map(x,y) onto z with decimal representation .aAbBcCdDeEfFgG... === Subject: Re: need a function >Hi. I'm trying to find a function satisfying the conditions below: >Define S:= {(x,y) in R : 0Define T:= {x in R : 0Problem: Use the fact that every real number has a decimal expansion >to produce an injective function that maps S into T. The usual solution is a bit of a trick, hard to hint at without simply giving it away. Given (x,y), take the decimal expansions of x and y and do a riffle shuffle. Brian === Subject: Need help !! I don't know proper method to prove following : Let G is a finite group of order 39. Prove that : If G is a nonabelian group, then G has 13 Sylow 3-subgroups. _____________________________________________________________ By Sylow theorem, n_3=1 or 13 , where n_3 means # of Sylow 3-subgroups. I know if G is an Abelian Group, then n_3=1. But, I'm not sure whether the converse is established or not. If the converse(i,e, n_3=1 then G is Abelian) is established, then this problem can be proved by contraposition. Except this contrapositive method, Is there any other method for proving this ? Or, If you know the method for showing converse, please post reply. === Subject: Re: Need help !! > I don't know proper method to prove following : > Let G is a finite group of order 39. > Prove that : If G is a nonabelian group, then G has > 13 Sylow 3-subgroups. > _____________________________________________________________ > By Sylow theorem, n_3=1 or 13 > , where n_3 means # of Sylow 3-subgroups. > I know if G is an Abelian Group, then n_3=1. > But, I'm not sure whether the converse is established or not. > If the converse(i,e, n_3=1 then G is Abelian) is established, then > this problem can be proved by contraposition. > Except this contrapositive method, > Is there any other method for proving this ? First n_13 = 1. Let S_13 be the unique (therefore normal) Sylow 13-subgroup. If n_3 = 1, Let S_3 be the unique Sylow 3-subgroup. Note both S_13 and S_3 are normal and they are disjoint so G contains a subgroup isomorphic to S_13 x S_3 (namely (S_13)(S_3)). But | S_13 x S_3 | = 39. That would say G is isomorphic to S_13 x S_3 and thus was abelian. -- Paul Sperry Columbia, SC (USA) === Subject: Obsolete Ring Theory Definition? Hello All, I have found two definitions for the term 'primary ideal'. I was wondering if they are equivalent, or since one is from older sources, if one of them is obsolete. The first version was found in Grillet's _Algebra_ and Hazewinkel's Soviet _Math Encyclopedia_. It states that the primary ideal of a commutative ring R is a subset of the ring such that for any two elements in the ring whose product is in the ideal, either the second factor in the product is a member of the ideal or a positive power of the first factor is a member of the ideal. In symbols, if a,b are members of R and a*b is a member of the ideal I of R, then either b is in I or a^n is in I for some n in Z+. The second definition appears in two earlier books on ideals one was written by Northcott, the other by Burton. It describes a primary ideal as one in which the presence of the product of two elements in the ideal and the fact that the first element is not a member of the ideal, always implies that some positive power of the second factor is an element of the ideal. That is if a*b is in I and a is not a member of I, then b^n is in I for some n in Z+. Evidently the second definition requires that one of the factors of the product not be included in the primary ideal, whilst the first definiton does not make this stipulation. My question are 1. Are these two definitions equivalent? 2. Is the second definition not really used anymore? LS Thomas === Subject: Re: Obsolete Ring Theory Definition? >Hello All, > I have found two definitions for the term 'primary ideal'. I was >wondering if they are equivalent, or since one is from older sources, >if one of them is obsolete. >The first version was found in Grillet's _Algebra_ and Hazewinkel's >Soviet _Math Encyclopedia_. It states that the primary ideal of a >commutative ring R is a subset of the ring such that for any two >elements in the ring whose product is in the ideal, either the second >factor in the product is a member of the ideal or a positive power of >the first factor is a member of the ideal. In symbols, if a,b are >members of R and a*b is a member of the ideal I of R, then either b is >in I or a^n is in I for some n in Z+. >The second definition appears in two earlier books on ideals one was >written by Northcott, the other by Burton. It describes a primary >ideal as one in which the presence of the product of two elements in >the ideal and the fact that the first element is not a member of the >ideal, always implies that some positive power of the second factor is >an element of the ideal. That is if a*b is in I and a is not a member >of I, then b^n is in I for some n in Z+. >Evidently the second definition requires that one of the factors of >the product not be included in the primary ideal, whilst the first >definiton does not make this stipulation. >My question are >1. Are these two definitions equivalent? Yes, trivially. The equivalence follows from the fact that A --> B v C is logically equivalent to A & ~B --> C. >2. Is the second definition not really used anymore? It's simply a clumsier restatement of the first; there's no real difference. [...] Brian === Subject: Re: Obsolete Ring Theory Definition? > Hello All, > I have found two definitions for the term 'primary ideal'. I was > wondering if they are equivalent, or since one is from older sources, > if one of them is obsolete. > The first version was found in Grillet's _Algebra_ and Hazewinkel's > Soviet _Math Encyclopedia_. It states that the primary ideal of a > commutative ring R is a subset of the ring such that for any two > elements in the ring whose product is in the ideal, either the second > factor in the product is a member of the ideal or a positive power of > the first factor is a member of the ideal. In symbols, if a,b are > members of R and a*b is a member of the ideal I of R, then either b is > in I or a^n is in I for some n in Z+. > The second definition appears in two earlier books on ideals one was > written by Northcott, the other by Burton. It describes a primary > ideal as one in which the presence of the product of two elements in > the ideal and the fact that the first element is not a member of the > ideal, always implies that some positive power of the second factor is > an element of the ideal. That is if a*b is in I and a is not a member > of I, then b^n is in I for some n in Z+. > Evidently the second definition requires that one of the factors of > the product not be included in the primary ideal, whilst the first > definiton does not make this stipulation. > My question are > 1. Are these two definitions equivalent? Yes. The second is phrased a little oddly but what it _means_ is that _if_ the first element is not in the ideal then some power of the second _is_ in the ideal (the ring is commutative). You can read the second as saying if a*b is in I and a is not in I then b^n is in I... -- Paul Sperry Columbia, SC (USA) === Subject: Re: Online Petition Request--my prime counting function > Wolfram's company Wolfram Research has a website that supposedly > presents mathematics online for users interested in learning. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ It's of no interest to you then. > ... > MAKE Wolfram Research own up to the responsibility they took by > *supposedly* presenting known mathematics!!! Wolfram Research are doing Eric Weisstein and the the rest of us a big favour. In that regard they have _no_ responsibility, but common sense dictates that they'll not willfully publish rubbish.... GC ... === Subject: Re: Online Petition Request--my prime counting function > Wolfram's company Wolfram Research has a website that supposedly > presents mathematics online for users interested in learning. > However I just went to > http://mathworld.wolfram.com/PrimeCountingFunction.html > and noted that I've STILL been snubbed when it comes to my prime > counting function, which is not listed, though it works. Your prime counting function is listed about at the middle of that page under the name of its inventor, Legendre. You are two hundred years late. === Subject: Re: Online Petition Request--my prime counting function > And if Stephen Wolfram thinks he can ignore me, then he's got another > thing coming. Not that he even knows about me, as if it matter given > that the poor man thinks he's explained everything with his book. Oh > well. I think you outclass SW slightly in arrogance and grossly in ignorance. Dirk Vdm === Subject: Re: Online Petition Request--my prime counting function >Wolfram's company Wolfram Research has a website that supposedly >presents mathematics online for users interested in learning. >However I just went to > http://mathworld.wolfram.com/PrimeCountingFunction.html >and noted that I've STILL been snubbed when it comes to my prime >counting function, which is not listed, though it works. You really have no idea what an idiot you sound like when you talk this way. Hints: 1. It's been explained to you many times by many people who nobody cares much about your Prime Counting Function. 2. Even if it _were_ something of mathematical interest, it doesn't follow that it's supposed to be presented at that site. Find a library. Look at the dozens of math journals. Each month every one of them contains dozens of examples of actual new mathematics that _are_ of interest to at least a few people. That stuff isn't all on wolfram's site. 3. Even if your thing _was_ appropriate for inclusion, they're under no obligation to include it. [...] >And if Stephen Wolfram thinks he can ignore me, then he's got another >thing coming. Nope. He _can_ ignore you. >Not that he even knows about me, as if it matter given >that the poor man thinks he's explained everything with his book. Oh >well. > ************************ David C. Ullrich === Subject: Re: Online Petition Request--my prime counting function >>Wolfram's company Wolfram Research has a website that supposedly >>presents mathematics online for users interested in learning. >>However I just went to >> http://mathworld.wolfram.com/PrimeCountingFunction.html >>and noted that I've STILL been snubbed when it comes to my prime >>counting function, which is not listed, though it works. >You really have no idea what an idiot you sound like when you >talk this way. Hints: >1. It's been explained to you many times by many people who >nobody cares much about your Prime Counting Function. >2. Even if it _were_ something of mathematical interest, it >doesn't follow that it's supposed to be presented at that >site. Find a library. Look at the dozens of math journals. >Each month every one of them contains dozens of examples >of actual new mathematics that _are_ of interest to at least >a few people. That stuff isn't all on wolfram's site. >3. Even if your thing _was_ appropriate for inclusion, >they're under no obligation to include it. Somehow I forgot the most obvious reason this is hilarious: 4. Even if none of the above applied, and Wolfram was in fact doing an evil thing by suppressing the Truth about the Prime Counting Function, the idea that you should ask for help from your dear friends on sci.math is hilarious. >[...] >>And if Stephen Wolfram thinks he can ignore me, then he's got another >>thing coming. >Nope. He _can_ ignore you. >>Not that he even knows about me, as if it matter given >>that the poor man thinks he's explained everything with his book. Oh >>well. >> >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: Online Petition Request--my prime counting function > Wolfram's company Wolfram Research has a website that supposedly > presents mathematics online for users interested in learning. > > However I just went to > > http://mathworld.wolfram.com/PrimeCountingFunction.html > > and noted that I've STILL been snubbed when it comes to my prime > counting function, which is not listed, though it works. > Your prime counting function is listed about at the middle of that page > under the name of its inventor, Legendre. You are two hundred years > late. So, James is a PLAGIARIST! === Subject: Re: Online Petition Request--my prime counting function > Somehow I forgot the most obvious reason this is > hilarious: > 4. Even if none of the above applied, and Wolfram was > in fact doing an evil thing by suppressing the Truth > about the Prime Counting Function, the idea that you > should ask for help from your dear friends on sci.math > is hilarious. so they can enjoy the joke (although it may just annoy them). I also think I'll suggest that they include a notice on their Prime Counting page that some poseur named is trying to steal credit for Legendre's work... -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Online Petition Request--my prime counting function > Wolfram's company Wolfram Research has a website that supposedly > presents mathematics online for users interested in learning. I don't think one can learn anything from Wolfram Research web pages. They are sometimes useful as a reference for a mathematician (even that is tenuous). I'm not sure what's the use for them to tell you the truth. === Subject: PGF's Got a queue m/m/4 Arrivals at 30 per hour central queue if no free assistant service time exponential mean 4 mins Now I have to show that the probability generating function is: 1/23 ( 3 + 6s + 6s^2 + 4s^3 + (4s^4/2-s)) I undestand all this so far but cant see how they get the 2-s in the last term. Based on the above queue information I have worked out that the traffic intensity (rho = lambda/n x epsilon) is 0.5. I have been told that the probability distribution has form Px = 1/k n^x/x! rho^x for 0<=xn I have calculated that k = (8(3 + 9rho + 12rho^2 + 8rho^3))/24(1-rho) and so p0 = 1/k with rho = 0.5 giving 3/23 Using the above (substitutuing into Px) I have worked out that p1=6/23 p2=6/23 p3=4/23 rest = 4/23 So now I have to prove that the PGF for the equilibrium queue size is Pi(s) = 1/23 ( 3 + 6s + 6s^2 + 4s^3 + (4s^4/2-s)) I have plugged in the values for P0..rest above (3/23, 6/23 etc), but cannot see where the 2-s comes from. I then have to evaluate the mean equilibium queue size, to do this I differentiated the PGF above and set s = 1? Any more help to see how they get the 2-s in the last term of the pgf would be most appreciated. Debbie === Subject: Presentations of a cyclic group Suppose you have a finite abelian cyclic group . Is there a canonical way (or any way, really) to express this alternatively in terms of two generators and two relations as: Generators = {g,h} Relations = {ag=0, bh=cg} where a, b, and c are integers. ? If so, this would be very useful to me, and I'd appreciate it if anyone could post with a way to do this (if it indeed can be done). Is there any kind of _Mathematica_ or Maple command that does this automatically, perhaps? (say once you enter the generator of the cyclic group and its order) Or perhaps if it can be done in certain special cases, maybe someone could point to those? === Subject: Re: Presentations of a cyclic group > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with generator (1,1) as prototype of theorem. That should work unless order of group is power of prime. === Subject: Re: Presentations of a cyclic group > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. It's not clear to me exactly what you're given in this problem. If by this you mean we are given n, and told only that the group G = , where g and h are unspecified elements of G, is isomorphic to Z_n; then it's not clear that a non-trivial presentation of the given form exists. By non-trivial here, I mean a presentation where neither = G nor = G; clearly, is always going to give a representation of Z_a, but that's probably not what you want. Consider the cyclic group Z_6. We must select g and h from {2,3,4}, since otherwise one of g or h will generate Z_6 on its own; so either {g,h} = {2,3} or {g,h} = {3,4}. If we take g = 2 or 4, then the only possible presentation of your given form is ; if we take g = 3, then . But these are both equivalent to . There is a homomorphism from this group onto D_3 (dihedral group) with presentation , and onto Z_6 with presentation ; so it would seem there is no non-trivial representation of Z_6 of your form. > If so, this would be very useful to me, and I'd appreciate it if > anyone could post with a way to do this (if it indeed can be done). Is > there any kind of _Mathematica_ or Maple command that does this > automatically, perhaps? (say once you enter the generator of the > cyclic group and its order) > Or perhaps if it can be done in certain special cases, maybe someone > could point to those? === Subject: Re: Presentations of a cyclic group > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > It's not clear to me exactly what you're given in this problem. See my other post where I try to work this out in general, starting from the hints I gave in my first post. In working thru this, the reason I disallowed the order of the group to be a power of a prime is elucidated. More comments below. > If by this you mean we are given n, and told only that the group G = > , where g and h are unspecified elements of G, is isomorphic to > Z_n; then it's not clear that a non-trivial presentation of the given > form exists. > By non-trivial here, I mean a presentation > where neither = G nor = G; clearly, is > always going to give a representation of Z_a, but that's probably not > what you want. > Consider the cyclic group Z_6. We must select g and h from {2,3,4}, > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. > If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. And h = 2 or h = 4. Don't take h = 2, take h = 4, as in accordance to my work in the other post. Thus Z_6 = = <-1>, does it not? > But these are both equivalent to . There is a > homomorphism from this group onto D_3 (dihedral group) with > presentation , and onto Z_6 with > presentation ; so it would seem > there is no non-trivial representation of Z_6 of your form. === Subject: Re: Presentations of a cyclic group > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > I suggest using the example Z/6 as subdirect product of Z/2 + Z/3, with > generator (1,1) as prototype of theorem. That should work unless order of > group is power of prime. Let G be a finite cyclic group with order n. Assume n > 1 isn't a power of a prime. Thus some cofinite j,k > 1 with n = rs Let (1,0) and (0,1) be generators of Z/r & Z/s. r(1,0) = (0,0) = s(0,1) If I've done this right, G = <(1,1)> = <(1,0) + (0,1)> is a cyclic group of order n with generator (1,1) Oh, if you want it in the - form then use (1,1) = (1,0) - (0,s-1) Whence (0,s-1) is a generator of Z/s and r(1,0) = (0,0) = s(0,s-1) When n the order of G is a power of a prime, then be contented with trivial solution, g = 0, h = 1, G = <0-1> = <-1>, 1g = 0 = nh. === Subject: Re: Presentations of a cyclic group > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > Consider the cyclic group Z_6. We must select g and h from {2,3,4}, > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. > If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. > And h = 2 or h = 4. Don't take h = 2, take h = 4, > as in accordance to my work in the other post. > Thus Z_6 = = <-1>, does it not? Yes, it works in the sense that does indeed generate Z_6. But, as I read it, the poster wants a presentation of the group of the form: G = such that G is isomorphic to Z_n for a given n. In the case n = 6, no such non-trivial (in the sense of my post) presentation is possible, unless we _require_ that the group be Abelian (i.e., unless there is an additional, _implict_ relation g+h = h+g). --------------------------------------------------- C Brown Systems Designs Multimedia Environments for Museums and Theme Parks --------------------------------------------------- === Subject: Re: Presentations of a cyclic group <3F1C7264.303BBB4D@cbrownsystems.com> > Suppose you have a finite abelian cyclic group . Is there a > canonical way (or any way, really) to express this alternatively in > terms of two generators and two relations as: > > Generators = {g,h} > Relations = {ag=0, bh=cg} where a, b, and c are integers. > Consider the cyclic group Z_6. We must select g and h from {2,3,4}, > since otherwise one of g or h will generate Z_6 on its own; so either > {g,h} = {2,3} or {g,h} = {3,4}. > > If we take g = 2 or 4, then the only possible presentation of your > given form is ; if we take g = 3, then 2g = 0, 2g = 3h>. > And h = 2 or h = 4. Don't take h = 2, take h = 4, > as in accordance to my work in the other post. > Thus Z_6 = = <-1>, does it not? > Yes, it works in the sense that does indeed generate Z_6. > But, as I read it, the poster wants a presentation of the group of the > form: > G = > such that G is isomorphic to Z_n for a given n. In the case n = 6, no > such non-trivial (in the sense of my post) presentation is possible, > unless we _require_ that the group be Abelian (i.e., unless there is an > additional, _implict_ relation g+h = h+g). Indeed, either a technical oversight or a catagorical presumption of Abelian groups, by the OP. === Subject: primes: a piece of the puzzle? to indicate summation and inf means infinity. See note at end for more information about notation etc. Theorem 1. a 1 If p(x) = lim E ------------------------------- a->inf n=2 ( 1 + (1/2) sin(pi x / n) ) ^ a then the equation p(x) = 1 has exactly the set of prime numbers as its solution set. This means taht p(x) = 1 if and only if x is a prime number. That is, the only values of x that solve p(x) = 1 are those x that are prime numbers, and these values always solve it. PROOF 1. sin(x) = 0 if and only if x is an integral multiple of pi. 2. sin(pi x / n) = 0 if and only if n|x; that is, if and only if (x / n) is an integer. This follows from (1): sin(pi u) = 0 if and only if u is an integer, and if u = x / n then this is true only if x / n is an integer, something that is true if and only if x is an integral multiple of n. The latter means that x has n as a factor, e.g. n|x (n divides x). For example, 3|12 is true because 12 = (2)(2)(3) has 3 as a factor; likewise, sin(pi 12 / 3) = sin(pi 4) = 0. 3. lim u ^ -a = 0 if and only if u is not 1; it is equal to 1 if u is 1. a->inf This is because if u is 1, u ^ -a is 1 for all a and is hence 1 also in the limitting case. On the other hand, if u is not 1, as a goes to infinity, u ^ a goes to infinity, so its reciprical, u ^ -a, goes to zero, and is hence 0 in the limiting case. 4. y(x, n) = lim (1 + sin(pi x / n)) ^ -a a->inf = 1 if x has n as a factor (meaning x is an integral multiple of pi), and = 0 otherwise. In (2) we found that sin(pi x / n) = 0 if and only if x has n as a factor. Thus, if x has n as a factor, equation (4) reduces to (1 + 0) ^ -a which, according to step (3), we have found is equal to 1 in the limit as a goes to infinity. If, on the other hand, x does not have n as a factor, then sin(pi x / n) must not be 0, according to step (2). Therefore, 1 + sin(pi x / n) must not be 1 and hence, the limit of (1 + sin(pi x / n)) ^ -a as a goes to infinity must not be 1; in fact, according to step (3), it must be 0. 5. There is an equation for finding an expected prime from the prime number density distribution; this equation does not work, however. In the next theorem I discuss finding a suitable replacement that actually works; this consists of an infinite number of terms, the first N of which, for a finite number N, may be found. To facilitate this, and to facilitate looking for patterns in the term as N approaches infinity, something I have not done because I feel it would probably require a new non-analytic system (for the currently existing methods of analysis), I have changed equation (4) to the following: 1 y(x, n) = lim ----------------------- a->inf (1 + sin(pi x / n)) ^ a = 1 if x has n as a factor (meaning x is an integral multiple of pi), and = 0 otherwise. This of course follows from (4). The intent of using this form is that the behavior of the expansion of (1 + u) ^ a into unfactored polynomial form is well known. 6. The definition of a prime number is: an integer greater than or equal to 2 that has no integers greater than or equal to 2 as factors besides itself. This is equivelent to saying: if you count the number of factors a number has, it is equal to 1 if and only if the number in question is a prime number. Here, number of factors means the number of unique integers greater than or equal to 2 that the number is divisible by. Thus 6 has 3 factors, because it is divisible by 2, 3, and 6. 9 has 2 factors, because it is divisible by only 3 and 9. Equation (5) defines y(x, n) such that y(x, n) = 1 if x has n as a factor, and y(x, n) = 0 if x does not have n as a factor; that is, it indicates a boolean value of 1 for yes and 0 for no. A prime number is defined as a number that has no for the answer for numbers 2, 3, 4, ..., p-1 for some prime number p. That means that the answer is no for the number 2 and for the number 3, and the number 4, etc. This and of boolean values can be accomplished in several ways. In the second theorem I will accomplish it via a system of equations. Here, I accomplish it via a summation: the sum of all the values y(x, n) for values of n equal to 2, 3, 4, ..., p-1 for a prime number p, is 0 only if y(x, n) was 0 for each value of n in question, e.g. for the number 2 and for 3, etc. Clearly y(x, n) is 0 for all n > x, because e.g. 7 does not have 8 as a factor, nor does it have 9 as a factor, or 10, or any number greater than 8. Hence, the summation of y(x, n) for values of n equal to 2, 3, 4, ... to infinity is equal to a finite number for any finite number x. Furthermore, if x is prime, this summation is equal to precisely 1, because in that case, only y(x, x) = 1; all the other y(x, n) values equal 0. Now the idea of the theorem is that we can use the former summation of y(x, n) to define some equation p(x) such that p(x) = 1 if and only if x is a prime number. The theorem takes care of multiple infinities. There are two infinities: first, in each value of y(x, n) which has a limit as a -> infinity, and second, in the summation, which is for the values n as n goes to infinity. It is customary in elementry calculus to use a limits for infinite summations, and it is well known that multiple infinities converge (e.g. the limit as a goes to infinity of 2 ^ -a is the same as the limit as a goes to infinity of 3 ^ -a), which concludes this proof. Theorem 2. Define q(x) as n - x q(x, n) = lim ------------------------------- a -> inf ( 1 + (1/2) sin(pi x / n) ) ^ a Then the (simultaneous) system of equations q(x, 2) = 0 q(x, 3) = 0 q(x, 4) = 0 q(x, 5) = 0 . . . has as its solution set the set of prime numbers, meaning the following: if the system of equations could somehow be solved for all values of x, then the values of x that would be determined, would be precisely all the prime numbers. To find all the prime numbers no greater than 100, one does the following: write the system of equations q(x, 2) = 0; q(x, 3) = 0; q(x, 4) = 0; ... q(x, 100) = 0 and then solve it; the solutions are the prime numbers. As it is required that a approach infinity, a manual (e.g. numerical, via a computer) solution requires some sort of understanding of how large a needs to be before all the x's thus obtained can simply be e.g. rounded to the nearest integer and then used. I have not attempted to do this. The former theorem is better suited for proving primality; that theorem tells the number of factors, as defined in the proof of the theorem, that a given number has. But even it requires taking the limit as a goes to infinity, as well; the present theorem could also prove primality by plugging in a prime number into each equatio nand then testing the equation to see if it is equal to 0. Although this is a mathematical method, it isn't particularly useful for numerical analysis, because directly testing division on a computer via division (and checking for a remainder) has the same effect. However, theorem 2 may provide hints about the future of the theory behind computers, something I don't discuss here. The use of theorem 2 is that unlike theorem 1 it provides a way to solve for x. Theorem 1 can't be solved for x. Theorem 2 is of much importance because it provides some clues about what form a function f(m) that returns the m'th prime number, may take. I don't discuss finding the former function f(m) for all m. But this theorem sheds some light on finding some function F(m, n) which is equal to f(m) if m is less than or equal to n, and is undefined otherwise. No one knows how to find f(m) which is the limit of f(m, n) as n goes to infinity. I doubt an analytical technique can be found to do this using only one level of infinity; there will probably be multiple infinities; it could turn out that they may be combined in notation only. The result may involve fractals and the complex plane, and a new theory about computers and algorithms; I don't discuss this here, but the assumption that some notation for which f(m) exists may be found, presents insight about why finding F(m, n) is so important - even to national security (cryptography implications may well exist - I do not discuss this, and my opinion about the future scientific calculator consisting of a 100 page book containing a table of numbers, and a piece of paper and pencil). Now F(m, n) is not so easy to state explictly because the m'th prime is somewhat nonmathematical (however I have succeded in turning the original nonmathematical problem into a mathematical one so it may be possible to state F(m, n) explictly and in a computationally efficient way). However, the spirit of F(m, n) is already here in theorem 2. Just write the given system of equations out: q(x, 2) = 0, q(x, 3) = 0, ..., q(x, n) = 0. This can be represented as a matrix, almost. Each q(x, n) is not quite a polynomial. But for every finite value of a, it is equal to (n - x) divided by a polynomial of order a. If q(x, n) could be represented as a polynomial for every finite a, then, for every finite a, there exists some matrix representing the system of equations for q(x, 2) = 0, q(x, 3) = 0, ..., q(x, n) = 0. The solution to the system of equations can be found by performing elementry matrix operations on the matrix, or through Cramer's rule which is probably the best choice. Now further work lies in the following. Represent q(x, n) as a polynomial for each finite a. Then, express each prime number using cramer's rule, for each finite a. Finally, let the system of equations go from 2 through a (instead of n). The result is a series of equations, representing each prime number, e.g. F(m, a) where m is a subscript, not an argument. By taking the limit as a approaches infinity, one thus obtains f(m) where m is a subscript, e.g. the m'th prime. The question to ask is whether the former will be straightforward or will require more renormalization work. PROOF I only note that q(x, n) is the product of (n - x) and the y(x, n) of theorem 1. Using theorem 1 and the fact that n - x = 0 if and only if n = x, it follows that q(x, n) = 0 if and only if EITHER x is has n as a factor, OR x equals n. Consequently, q(x, n) = 0 is true for all integers n greater than or equal to 2, IF AND ONLY IF x is prime. UNRELATED ASIDE NOTE pertaining to theorem 1: step (3) was the missing piece that took some time to come up with. Its predecessor was the equation f(x) = g(g(g( ... g(x) ... ))) where g(x) = x + sin(x). Presumably this can be normalized (using only one level of infinity and not using infinite recursion (composition)) using some new system like fractals, but how? Here, f(1) = pi, and f(x) can be used in place of (3) when slightly modified, but this leads to multiple levels of infinity in the final prime number system. I am embarrased to say the former f(x), which represents years of (hobbyist) research into both calculating pi and prime numbers, was made obsolete by remembering that in elementry calculus courses it is assumed that the limit as a goes to infinity of u ^ -a is 0. I don't remember courses noting this is not true for u = 1, but clearly this is the case. The f(x) approach required some kind of renormalization; I never bothered to attempt to devise some kind of Renormalization Theory. Instead I found a way around the problem. In the original reasoning, it was noted that u can be positive or negative; if negative, u ^ a becomes positive in the limit as a goes to infinity since this happens for all a greater than or equal to 2. This holds for its reciprical too. --- Note: This is just a quick note about prime numbers, posted for your information and review. The equations in this post presumes fixed-width font is used. I have tried to make this brief, but honesty and integrety (and rambling) has kept me from leaving out critical pieces of the puzzle, because I have the perception that someone might possibly find some use for the work in question. inf is used for (positive) infinity, and E is used in place of the greek letter sigma, to indicate summation. As always, a^b indicates a to the power of b and a/b indicates a divided by b. Unfortunately this is more than 200 lines long (on my screen) and even then I've left a lot out... please let me know what you think and if you see any problems! Also have you ever heard of anything like this being done before? Perhaps I have simply retraced someone else's footsteps; I don't know what Riemann's hypothesis is or what the zeta function is, for example. === Subject: Re: primes: a piece of the puzzle? [...] > 3. lim u ^ -a = 0 if and only if u is not 1; it is equal to 1 if u > is 1. > a->inf > This is because if u is 1, u ^ -a is 1 for all a and is hence 1 also > in the limitting case. On the other hand, if u is not 1, as a goes to > infinity, u ^ a goes to infinity, so its reciprical, u ^ -a, goes to > zero, and is hence 0 in the limiting case. lim( (1/2)^(-a), a -> infinity) = ? [...] -- Paul Sperry Columbia, SC (USA) === Subject: Re: primes: a piece of the puzzle? [...] > lim( (1/2)^(-a), a -> infinity) = ? > [...] you very much for finding this flaw!!! See below to see how I corrected it. I originally didn't have the 1 + in the denominator due to a different mistake, so I figured the denominator, which was sin(pi x / n) ^ a at the time, would never be negative since we take the limit as a goes to infinity. But the moment I changed this to the following, the error appeared: a 1 p(x) = lim E ------------------------------- a->inf n=2 ( 1 + (1/2) sin(pi x / n) ) ^ a We can fix the problem by making it sin^2 instead of just sin. The range of sin^2 is [0, 1] rather than [-1, 1] so the former problem of the denominator reducing to ( 1 - (1/2) ) ^a no longer arises if we do this: a 1 p(x) = lim E ----------------------------- a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a It appears that this fixes the problem. The (1/2) wasn't needed anyways but was put there so that ^ -a could be used instead of ^ a; this property still holds since the denominator will never be zero even without the (1/2) since 1 + sin^2 will never be less than 1. Originally I was hesitant about using sin^2 because sin has a nice series expansion, but the fact that the sin had to be moved into the denominator together with the fact that sin^2 probably has a nice expansion too, is good reason to make the corrected equation above the new one. any more mistakes! Actually, take a look at the following page: http://willow.schlanger.name/math/primes.html where I have made the former correction and also cleaned the whole thing up and corrected a typographical mistake. ** Please let me know what you think and if you find any (other) -- Willow Schlanger === Subject: Re: primes: a piece of the puzzle? > a 1 > p(x) = lim E ----------------------------- > a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a lim sin(pix/n) = 0 as n -> infinity, so: lim (1+sin((pix/n))^2) = 1, n->infinity even with the exponentiation, (*)^a, the terms do not go to zero, independent of x, so the series diverges === Subject: Re: primes: a piece of the puzzle? > a 1 > p(x) = lim E ----------------------------- > a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a > lim sin(pix/n) = 0 as n -> infinity, > so: lim (1+sin((pix/n))^2) = 1, n->infinity > even with the exponentiation, (*)^a, the > terms do not go to zero, independent > of x, so the series diverges p(x) = sum(n=2..oo) lim(k->oo) 1/(1 + sin^2 pi.x/n)^k is easy fix since eventually sin pi.x/n /= 0 which assures the sum is finite. Verbose M.Tao needs to clarify the ranges of x and a. Reals, positive numbers, integers or what? Likely we can presume a is positive integer. Even so, he should clarify if limit is continuous limit or discrete sequence limit. Perhaps M.Tao would be tempted to contemplate integral(2,oo) lim(k->oo) 1/(1 + sin^2 pi.x/n)^k if limit and integral can be interchanged if integrand can be integrated if integral form is useful for the purpose at hand. === Subject: Re: primes: a piece of the puzzle? > a 1 > p(x) = lim E ----------------------------- > a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a > lim sin(pix/n) = 0 as n -> infinity, > so: lim (1+sin((pix/n))^2) = 1, n->infinity > even with the exponentiation, (*)^a, the > terms do not go to zero, independent > of x, so the series diverges > p(x) = sum(n=2..oo) lim(k->oo) 1/(1 + sin^2 pi.x/n)^k > is easy fix since eventually sin pi.x/n /= 0 > which assures the sum is finite. Comments made below to M.Tao are acutally for Willow.S. M.Tao may thank himself for such misplaced remarks as he removed all indications of who said what. > Verbose M.Tao needs to clarify the ranges of x and a. > Reals, positive numbers, integers or what? > Likely we can presume a is positive integer. > Even so, he should clarify if limit is continuous limit or > discrete sequence limit. > Perhaps M.Tao would be tempted to contemplate > integral(2,oo) lim(k->oo) 1/(1 + sin^2 pi.x/n)^k > if limit and integral can be interchanged > if integrand can be integrated > if integral form is useful for the purpose at hand. === Subject: Re: primes: a piece of the puzzle? This is in response to the problem pointed out by Modular Tao. It appears that the equation from Theorem 1 should be changed to this: x 1 p(x) = lim E ----------------------------- a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a Here, the upper limit of the summation has been changed from a (which approaches infinity) to x, a finite value. I still am not convinced this change is necessary, but it's clear the change fixes the problem pointed out by Modular Tao, that the limit as n -> infinity of sin(pi x / n) is 0. Indeed, the original equation is, if he's right (and I think he is), identical to this: inf p(x) = E y(n, x) n=2 1 y(n, x) = lim ----------------------------- a->inf ( 1 + (sin(pi x / n))^2 ) ^ a Now y(n, x) = 1 if x / n is an integer, and y(n, x) = 0 otherwise. But the limit as n -> infinity of x / n is 0, so lim n->infinity y(n, x) = 1. Modular Tao is of the opinion that this makes it so that the series does not converge if we take the summation from n=2 to infinity as I have done. Making the upper limit x corrects the problem, and the p(x) = 1 still is true if and only if x is a prime number. William Elliot notes that p(x) might possibly be able to use an integral instead of a summation. This is indeed correct. This is because for any integer x > 1, y(n, x) = 0 if n is not also an integer. So, research into whether or not the summation needs to have a finite number as its upper limit, he suggests, could be done by looking into these things: > Perhaps M.Tao would be tempted to contemplate > integral(2,oo) lim(k->oo) 1/(1 + sin2 pi.x/n)^k > if limit and integral can be interchanged > if integrand can be integrated > if integral form is useful for the purpose at hand. and later > Comments made below to M.Tao are acutally for Willow.S. > M.Tao may thank himself for such misplaced remarks > as he removed all indications of who said what. In discussions with me Modular Tao was most helpful; I think he just clipped out who said what for consiseness. Now indeed an integral can be used for the purpose at hand, but I am having trouble with the first two points - mainly whether or not the integrand can be integrated with infinity as its upper limit, trying both places for the limit (after and before the limit). I'll probably need help with determining whether or not the integral, but I think Modular Tao ultimately is probably correct that the upper limit needs to be finite. Can anyone explain to me why switching the order of the limits (for a and n, as below) could make any difference? N 1 p(x) = lim lim E ----------------------------- a->inf N->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a My view is that the order of the limits can't possible affect the ultimate result. I pointed out to M. Tao that (sin(pi x / n))^2 > 0 for all n > x, but he noted this is true only for a finite n, and in the limit as n -> infinity, the value is 0. The story so far can be seen here: http://willow.schlanger.name/math/primes.html If anyone can help me try to prove that the integral that has infinity as its upper limit can not be integrated, that would be most appreciated. If that can be shown, then it is also shown that the summation must have a finite upper-limit, which sucks but is acceptable nevertheless. Theorem 2 isn't affected, except for the case where an infinite number of equations is written down. Willow Schlanger === Subject: Re: primes: a piece of the puzzle? OK, I'm now convinced the integral doesn't exist (in integeral form, something suggested by William Elliot). In response to the flaw pointed out by Modular Tao, here's the latest version: Theorem 1. x 1 p(x) = lim E ----------------------------- a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a then p(x) = 1 if and only if x is a prime number. This means that the equation p(x) = 1 has exactly the set of prime numbers as its solution set. That is, the only values of x that solve p(x) = 1 are those x that are prime numbers, and these values always solve it. PROOF - See original post. Theorem 2. n - x If y(n, x) = lim ----------------------------- a->inf ( 1 + (sin(pi x / n))^2 ) ^ a then the (simultaneous) system of equations y(2, x) = 0 y(3, x) = 0 y(4, x) = 0 . . . has the set of prime numbers as its solution set. This means that if the former system of equations were to be somehow solved for all values of x, the values obtained would be all the prime numbers (of which there is an infinite number). PROOF - See original post. Rather than looking at the original post, please look at the following page for the proofs: http://willow.schlanger.name/math/primes.html That page has updated and slightly corrected the proofs. --- It looks like for Theorem 2 I'm going to have to add the restriction that there must be a finite number of equations, because lim n->inf y(n, x) = 1 for any finite x. This is very, very, very disappointing. The original idea was that since sin has an infinite series expansion, we can approximate each equation as a polynomial with a finite order, with that order approaching infinity. Now a system of N equations, each of which is a polynomial of order M, for M < N, can be solved for x. So why can't we solve all the equations of Theorem 2 to find all the prime numbers? Well, each y(n, x) has a sin function in it, which has an infinite series expansion, e.g. a polynomial with an infinite number of terms. But we have only a finite number of equations - unfortunately - because lim n->inf y(n, x) does not exist (due to the n - x in the numerator), so only a finite number of equations may exist. On the other hand, I don't understand why there must be a finite number of equations and have lingering doubts... after all, if y(q, x) exists for some finite integer q, then y(q + 1, x) exists too; here, q + 1 is also a finite integer. So why should lim q-> inf y(q, x) not exist? It is confusing. Now surely there must be a way to approximate prime numbers by using a finite number of terms for the series expansion of sin and then solving the resulting equations, where the number of equations is some finite number N, e.g. N = 1000, where we wish to determine all the prime numbers less than or equal to N. Approximating prime numbers is something that can be done by relying on expected primes; it could turn out that this method is a first approximation, analogous to using only one term for a series expansion. I believe the former method of solving equations from Theorem 2 can be used to get a better approximation. Now it appears the approximation can't increase in accuracy without limit, but is this necessary? What if, upon solving the system of equations, we get e.g. 1.9999, 2.99999, 4.99999, 6.99999 etc.? If we can show that using any more terms in the expansion for sin will not be enough for those numbers to increase beyond 2, 3, 5, and 7, then we will have a good enough solution. This is a summary of the discussion so far. Feel free to read the page at the address given above. Please let me know what your thoughts are... Willow === Subject: Re: primes: a piece of the puzzle? > Theorem 1. > x 1 > p(x) = lim E ----------------------------- > a->inf n=2 ( 1 + (sin(pi x / n))^2 ) ^ a > then p(x) = 1 if and only if x is a prime number. What's the scope of x and a? By common convention x and a are usually reals. So that's how your statement reads. I you want x and a to be intgers, then the common convention is n,m,i,j,k. Less confusing and more encouraging to other readers dropping in the middle of this thread is to use Sum in place of E. You could switch the limit and sum as the sum is finite. Further you could p(x) = sum(x=2..oo) lim a->inf ... as the sum actually is finite. > This means that the equation p(x) = 1 has exactly the set of prime > numbers as its solution set. That is, the only values of x that solve > p(x) = 1 are those x that are prime numbers, and these values always > solve it. > PROOF - See original post. Bah! What part of OP? Some of it was rejected. If you want us to review your work, then give us a complete copy without having to go chasing around to find all the parts and puzzle which ones to use, which to discard and where and how to use them. > Theorem 2. > n - x > If y(n, x) = lim ----------------------------- > a->inf ( 1 + (sin(pi x / n))^2 ) ^ a > then the (simultaneous) system of equations > y(2, x) = 0 > y(3, x) = 0 > has the set of prime numbers as its solution set. This means that if > the former system of equations were to be somehow solved for all > values of x, the values obtained would be all the prime numbers (of > which there is an infinite number). > PROOF - See original post. > Rather than looking at the original post, please look at the following > page for the proofs: > http://willow.schlanger.name/math/primes.html Bah, then I have to cut and paste and juggle text around just to make comments on your work. Besides jpg stuff doesn't show up for everybody as well as you think it might. > It looks like for Theorem 2 I'm going to have to add the restriction > that there must be a finite number of equations, because lim n->inf > y(n, x) = 1 for any finite x. This is very, very, very disappointing. What you mean isn't clear and it's even more confusing as you haven't appears y(x) = p(x) - 1 = 0 is, over the integers at least, a function that has all the primes and only primes for solutions. > The original idea was that since sin has an infinite series expansion, > we can approximate each equation as a polynomial with a finite order, > with that order approaching infinity. > Now a system of N equations, each of which is a polynomial of order M, > for M < N, can be solved for x. > So why can't we solve all the equations of Theorem 2 to find all the > prime numbers? Well, each y(n, x) has a sin function in it, which > has an infinite series expansion, e.g. a polynomial with an infinite > number of terms. But we have only a finite number of equations - > unfortunately - because lim n->inf y(n, x) does not exist (due to the > n - x in the numerator), so only a finite number of equations may > exist. > On the other hand, I don't understand why there must be a finite > number of equations and have lingering doubts... after all, if y(q, x) > exists for some finite integer q, then y(q + 1, x) exists too; here, q > + 1 is also a finite integer. So why should lim q-> inf y(q, x) not > exist? It is confusing. > Now surely there must be a way to approximate prime numbers by using a > finite number of terms for the series expansion of sin and then > solving the resulting equations, where the number of equations is some > finite number N, e.g. N = 1000, where we wish to determine all the > prime numbers less than or equal to N. Such an algorithm would be unusably slow; recall you also have to take limit. === Subject: probability question I had statistics in college but that was long ago. I think my dog ate the book some years back. Anyway, here's my question: Suppose I have 1 million things I need to check. And suppose I take a random sample of 1,000 from the million and check those. If the 1,000 are all correct, what can I say about the other 999,000? How do we express that? We are x percent certain that y are correct -- something like that. Now, what if I sample one percent of the million (10,000) and they're all correct? Can you give me a relevant formula? Alan --Alan Dechert Most people love a good story more than they love the truth. http://www.go2zero.com/adechert/ === Subject: Re: probability question > I had statistics in college but that was long ago. I think > my dog ate the book some years back. Has the dog graduated? === Subject: Re: probability question >> I had statistics in college but that was long ago. I think >> my dog ate the book some years back. >Has the dog graduated? Okay, forget that part. Seriously, I will be grateful for help on this question. Alan === Subject: Re: probability question G'day This is a guess. The numbers were too big on your example and my calc kept on displaying a math error so to reduce the problem. If I have 10 computers and know there are 2 defective (20% defective). After inspecting 3 computers what is the probability of finding no defective computers? If the above is what you are after it is a hypergeometric distribution and the formula is: p(x) = ( ^{n1}C_{x} * ^{n2}C_{r-x} ) / ( ^{n1+n2}C_{r} ) where n1=no of successes (faulty computers0 n2=number of failures r=sample size p(0)= ( ^{2}C_{1} * ^{8}C_{2} ) / ( ^{10}C_{3} ) = 46% >I had statistics in college but that was long ago. I think my dog ate the book >some years back. Anyway, here's my question: Suppose I have 1 million things >I need to check. And suppose I take a random sample of 1,000 from the million >and check those. If the 1,000 are all correct, what can I say about the other >999,000? How do we express that? We are x percent certain that y are correct >-- something like that. >Now, what if I sample one percent of the million (10,000) and they're all >correct? >Can you give me a relevant formula? >Alan >--Alan Dechert >Most people love a good story more than they love the truth. >http://www.go2zero.com/adechert/ -- Craig & Fiona Tipping, Trawalla, Australia If it looks like a duck and quacks like a duck it might be a space alien, but is probably just a duck - Victorian Skeptics === Subject: Problem related to confidence interval A diet food company has 120 salesman. Monthly sales of each salesman is approximately normally distributed with a mean sales amount of $53,000 and a standard deviation of $15,000. A random sample of 10 salesman is randomly selected. Suppose that the sample mean sales amount turn out to be $63,000, construct a 95% confidence interval for the true mean among the 120 salesman. Solution: I calculated the mean and standard deviation of the sample mean sales to be $53,000 and $4384.67 respectively. bar X = 63000, n = 120, s.d.= 15,000 95% confidence = (1-a) where a = 0.05; hence a/2 = 0.025 ,so Zvalue(.025) = 1.96 = bar X +/- Z(a/2) (sigma/sqroot(n)) = 63,000 +/- (1.96)*15000/sqrt(120) = 63,000 +/- 2,683.8405 LCL = 60316.1595 UCL = 65683.8405 Is the solution make sense? === Subject: problems with using induction to prove generalized DeMorgan Identities (I use LateX notation below, so it may help to know that in LateX, bigcup is the union symbol, and bigcap is the intersection, in is the member of symbol, and _ and ^ denote subscripts and superscrips, respectively) Using induction, and one of the basic 2 set DeMorgan identities, one may prove the corresponding identity for any finite number of sets. e.g. (bigcup_{iin I}A_i)^c = bigcap_{iin I}A_i^c where I = {1,2,...n} is an index for the sets, A_i. Alternatively, one may prove the above indentity directly from the definitions of union and intersection for an indexed family of sets. My question: What about an infinite number of sets? i.e. Are both of the above mentioned methods of proof still valid to prove the identity below? (bigcup_{n=1}^{infty}A_n)^c = bigcap_{n=1}^{infty}A_n^c I read a text that made a small mention that the induction argument is invalid for the Generalized DeMorgan identity above. I cannot figure out why that may be the case. (In other words, I need a counterexample.) In addition, would the more direct proof, using definitions of union and intersection for indexed families of sets be valid for BOTH finite and infinite index sets? If so, why is this form of reasoning more valid than using induction? === Subject: Re: problems with using induction to prove generalized DeMorgan Identities One of the common ways to show equality among sets is the following: x is in (UA_i)^c iff x is not in UA_i iff x is not in A_i, for every i iff x is in (A_i)^c, for each i iff x is in the intersection of (A_i)^c. Hope this helps, Brian > (I use LateX notation below, so it may help to know that in LateX, > bigcup is the union symbol, and bigcap is the intersection, in > is the member of symbol, and _ and ^ denote subscripts and > superscrips, respectively) > Using induction, and one of the basic 2 set DeMorgan identities, one > may prove the corresponding identity for any finite number of sets. > e.g. > (bigcup_{iin I}A_i)^c = bigcap_{iin I}A_i^c > where I = {1,2,...n} is an index for the sets, A_i. > Alternatively, one may prove the above indentity directly from the > definitions of union and intersection for an indexed family of sets. > My question: What about an infinite number of sets? i.e. Are both of > the above mentioned methods of proof still valid to prove the identity > below? > (bigcup_{n=1}^{infty}A_n)^c = bigcap_{n=1}^{infty}A_n^c > I read a text that made a small mention that the induction argument is > invalid for the Generalized DeMorgan identity above. I cannot figure > out why that may be the case. (In other words, I need a > counterexample.) In addition, would the more direct proof, using > definitions of union and intersection for indexed families of sets be > valid for BOTH finite and infinite index sets? If so, why is this > form of reasoning more valid than using induction? === Subject: Re: problems with using induction to prove generalized DeMorgan Identities > (I use LateX notation below, so it may help to know that in LateX, > bigcup is the union symbol, and bigcap is the intersection, in > is the member of symbol, and _ and ^ denote subscripts and > superscrips, respectively) > Using induction, and one of the basic 2 set DeMorgan identities, one > may prove the corresponding identity for any finite number of sets. > e.g. > (bigcup_{iin I}A_i)^c = bigcap_{iin I}A_i^c > where I = {1,2,...n} is an index for the sets, A_i. > Alternatively, one may prove the above indentity directly from the > definitions of union and intersection for an indexed family of sets. > My question: What about an infinite number of sets? i.e. Are both of > the above mentioned methods of proof still valid to prove the identity > below? > (bigcup_{n=1}^{infty}A_n)^c = bigcap_{n=1}^{infty}A_n^c > I read a text that made a small mention that the induction argument is > invalid for the Generalized DeMorgan identity above. I cannot figure > out why that may be the case. (In other words, I need a > counterexample.) Well, you won't get a counterexample since DeMorgan's Laws are true for any family of sets. However, (finitie) induction works like this: For positive integer n let P(n) be the statement For any family of subsets, A_1, ... , A_n of a set U, (A_1 / ... / A_n)^c = A_1^c / ... / A_n^c). If, say, P(2) is true and P(k+1) is true whenever P(k) is true then P(n) is true for all integers >= 2. Note that P(n) is referring to _finite_ unions and intersections which is why (finite) induction will not speak to the issue of infinite unions. > In addition, would the more direct proof, using > definitions of union and intersection for indexed families of sets be > valid for BOTH finite and infinite index sets? Yes. > If so, why is this form of reasoning more valid than using > induction? In the finite case, it isn't - valid is valid. In the infinite case, finite induction won't work because it doesn't apply. I don't know if you are familiar with transfinite induction but that, I think, _would_ work for the infinite case. However, I suspect that a necessary step would require an element-wise argument so you might as well do the whole thing element-wise. -- Paul Sperry Columbia, SC (USA) === Subject: prob with derivation of Laplace xfrm of time shifted function hi: i'm reviewing Laplace xfrms and ran into this. it's from an electronics text and i can't find a similar xfrm in a tbl of elementary Laplace xfrms to check it's accuracy. any pointers to a good table on the Web would be appreciated, but let me try to explain the prob so you can correct me if i'm wrong. i'll use the following notation: INT the integral symbol f(t-T) function of t, f(t) shifted T units to the right U(t) unit step function where U(t) = 0 for t<=0 and 1 for t>0 U(t-T) shift it right T units f(t-T)U(t-T) this causes the time shifted function to begin at time T, i.e., it is zero for t < T e epsilon - the natural log base oo infinity (kinda cute lookin' in this news reader :-) ) F(s) Laplace xfrm of f(t) defined next. F(s) = INT f(t) e^(-st) dt ; integrated from 0 to oo anyway, after a bunch of blah blah on shifted functions and stating that the limits of integration will be omitted for now, the text says **** F(s) = INT f(t-T)U(t-T) e^(-s(t-T)) dt Or, since e^(-s(t-T)) = e^(-st) e^(sT), the last equation can be written in the form e^(-sT) F(s) = INT f(t-T)U(t-T) e^(-st) dt because e^(sT) is held fixed during the integration. ***** and they go on to finish deriving the xfrm pair and use it in examples and problems with answers. spacecraft fail over little errors like this. see how when they moved e^(sT) outside the integral they changed it to e^(-sT) ? that's what's screwing me up. can't remember a thing in the past that would make me do this. as the text says, it's held fixed. a constant. any help/correction on this is greatly appreciated. mike c. === Subject: Re: prob with derivation of Laplace xfrm of time shifted function Cc: mcolasono@earthlink.net > hi: > i'm reviewing Laplace xfrms and ran into this. it's from an electronics > text and i can't find a similar xfrm in a tbl of elementary Laplace > xfrms to check it's accuracy. any pointers to a good table on the Web > would be appreciated, but let me try to explain the prob so you can > correct me if i'm wrong. > i'll use the following notation: > INT the integral symbol > f(t-T) function of t, f(t) shifted T units to the right > U(t) unit step function where U(t) = 0 for t<=0 and 1 for t>0 > U(t-T) shift it right T units > f(t-T)U(t-T) this causes the time shifted function to begin at time T, > i.e., it is zero for t < T > e epsilon - the natural log base > oo infinity (kinda cute lookin' in this news reader :-) ) > F(s) Laplace xfrm of f(t) defined next. > F(s) = INT f(t) e^(-st) dt ; integrated from 0 to oo To try to avoid confusing things, I'll use the generic x(t) and X(s) instead of f/F. You'll see. The definition of the unilateral L-transform is: X(s) = L[x(t)] = INT x(t) e^(-st) dt, from t=0 to t=+oo In our case, x(t) = f(t-T)u(t-T), so X(s) = INT f(t-T)u(t-T) e^(-st) dt from t=0 to t=+oo Do a good ol' change of variables: z = t - T; t = z + T dz = dt Note that the limits on z are: z_low = t_low - T = 0 - T = -T z_high = t_high - T = +oo - T = +oo so, X(s) = INT f(z)u(z) e^[-s(z+T)] dz from z=-T to +oo but the presence of u(z) means the integrand vanishes for z<0, and so we can say the limits of integration are z=0 to +oo without changing the value of the integral (and we can also stop writing u(z) in the integrand, since its effect is now reflected in the limits of integration) X(s) = INT f(z) e^[-s(z+T)] dz from z=0 to +oo Distribute the power and expand it: X(s) = INT f(z) e^(-sz) e^(-sT) dz from z=0 to +oo e^(-sT) is a constant relative to z, so pull it out of the integral X(s) = e^(-sT) INT f(z) e^(-sz) dz from z=0 to +oo But the integrand is F(s) by definition! X(s) = e^(-sT) F(s) and since X(s) = L[x(t)] = L[f(t-T)u(t-T)], L[f(t-T)u(t-T)] = e^(-sT) L[f(t)] -- Rich Carreiro rlcarr@animato.arlington.ma.us === Subject: Re: prob with derivation of Laplace xfrm of time shifted function substitution in the blah blah part i mentioned. it's a salient point. see if this ( the blah blah part leading up to my OP) is confusing enough: they used small tau instead of z in the substitution, i'll stick with z. the text reads: ********* Note that the function f(t)U(t-T) has the value zero for all values of time less than t=T. Now let us consider a function, f(z)U(z). By basic definition of eq.3, [unilateral L-transform] the Laplace tansform of the function is, F(s) = INT f(z)U(z) e^(-sz) dz from z=0 to z->+oo eq.19 where, U(z) = 1 for all z>0. Suppose that z is time measured from the instant t=T as in [figure of function multiplied by shifted Unit step function -- funny, i just noticed that they don't shift the actual function in the fig]. Then, z=t-T, and also dz=dT, because T is constant in any given problem. Thus, using teh relationship, z=t-T and dz=dt, (and omitting the limits of integration for the time being), eq.19 becomes, F(s) = INT f(t-T)U(t-T) e^(-s(t-T)) dt ********** which is the eq with which i started the quote in the OP. see, you did a good job explaining this by sticking with the variable substitution, where the text jumps back to the original setup. being a non-mathematician, and having not done calculus in years, how harshly should i be chastized for not seeing this? should i beat myself wih a stick? :-) mike c. rlcarr@animato.arlington.ma.us says... > hi: > > i'm reviewing Laplace xfrms and ran into this. it's from an electronics > text and i can't find a similar xfrm in a tbl of elementary Laplace > xfrms to check it's accuracy. any pointers to a good table on the Web > would be appreciated, but let me try to explain the prob so you can > correct me if i'm wrong. > > i'll use the following notation: > > > INT the integral symbol > f(t-T) function of t, f(t) shifted T units to the right > U(t) unit step function where U(t) = 0 for t<=0 and 1 for t>0 > U(t-T) shift it right T units > f(t-T)U(t-T) this causes the time shifted function to begin at time T, > i.e., it is zero for t < T > > e epsilon - the natural log base > > oo infinity (kinda cute lookin' in this news reader :-) ) > > F(s) Laplace xfrm of f(t) defined next. > > F(s) = INT f(t) e^(-st) dt ; integrated from 0 to oo > To try to avoid confusing things, I'll use the generic > x(t) and X(s) instead of f/F. You'll see. > The definition of the unilateral L-transform is: > X(s) = L[x(t)] = INT x(t) e^(-st) dt, from t=0 to t=+oo > In our case, x(t) = f(t-T)u(t-T), so > X(s) = INT f(t-T)u(t-T) e^(-st) dt from t=0 to t=+oo > Do a good ol' change of variables: > z = t - T; t = z + T > dz = dt > Note that the limits on z are: > z_low = t_low - T = 0 - T = -T > z_high = t_high - T = +oo - T = +oo > so, > X(s) = INT f(z)u(z) e^[-s(z+T)] dz from z=-T to +oo > but the presence of u(z) means the integrand > vanishes for z<0, and so we can say the limits > of integration are z=0 to +oo without changing > the value of the integral (and we can also > stop writing u(z) in the integrand, since its > effect is now reflected in the limits > of integration) > X(s) = INT f(z) e^[-s(z+T)] dz from z=0 to +oo > Distribute the power and expand it: > > X(s) = INT f(z) e^(-sz) e^(-sT) dz from z=0 to +oo > e^(-sT) is a constant relative to z, so pull it > out of the integral > X(s) = e^(-sT) INT f(z) e^(-sz) dz from z=0 to +oo > But the integrand is F(s) by definition! > X(s) = e^(-sT) F(s) > and since X(s) = L[x(t)] = L[f(t-T)u(t-T)], > L[f(t-T)u(t-T)] = e^(-sT) L[f(t)] === Subject: question about elementary functions I was told that they are polynomial functions, trig functions(sine, cosine, tan etc), exponential function(e^x). Are there any other functions that are called elementary functions? Is the term elementary equivalent to the term analytical ? I have been wondering to why it is possible to differentiate sine or cosine or exponetial functions infinitely many times. Is that because those functions can be represented as a infinite series? If so, if we can find some function that can't be expressed as any infinite series, is it impossible to differenciate those functions infinetly many times? Is it possible to represent any kind of functions, using infinite series? === Subject: Re: question about elementary functions > I was told that they are polynomial functions, trig functions(sine, > cosine, tan etc), exponential function(e^x). Are there any other > functions that are called elementary functions? Is the term elementary > equivalent to the term analytical ? x+y, xy, x^y, sqr, log, sin etc, arcsin etc. No. > I have been wondering to why it is possible to differentiate sine or > cosine or exponetial functions infinitely many times. Is that because > those functions can be represented as a infinite series? If so, if we > can find some function that can't be expressed as any infinite series, > is it impossible to differenciate those functions infinetly many times? > Is it possible to represent any kind of functions, using infinite series? Contemplate upon |x| = sqr x^2 around x = 0. === Subject: Re: question about elementary functions > I was told that they are polynomial functions, trig functions(sine, > cosine, tan etc), exponential function(e^x). Are there any other > functions that are called elementary functions? Rational functions (polynomial fractions), inverse trig functions, logarithms and exponentials (any base, not just e). That's usually it. There isn't a canonical list, so you're free to toss in another or toss out some if you want. > Is the term elementary equivalent to the term analytical ? No. Analytic has a specific meaning -- the function equals it's power series (on an open set). > I have been wondering to why it is possible to differentiate sine or > cosine or exponetial functions infinitely many times. Is that because > those functions can be represented as a infinite series? No. f(x) = e^(-1/x^2) (and 1 at x = 0) can be differentiated infinitely many times. It can be represented as an infinite series, but it's a pretty bad representation. The MacLaurin series exists, but it doesn't equal the function (except, of course, at x = 0). > If so, if we can find some function that can't be expressed as any > infinite series, is it impossible to differenciate those functions > infinetly many times? Is it possible to represent any kind of > functions, using infinite series? It probably is, depending on how you interpret represent. As an undergraduate, I only met convergent series, and they were expected to converge to the functional values. However, there are other ways to interpret represent. The most widely used method is asymptotic convergence. An asymptotic series representing a function might even diverge! You run into asymptotic representations a lot with low Reynolds number (high viscosity) fluid flow. Jon Miller === Subject: Re: question about elementary functions Also hyperbolic functions (sinh, cosh, etc.) and their inverses. === Subject: Re: question about elementary functions > Also hyperbolic functions (sinh, cosh, etc.) and their inverses. Oops...these are really just special kinds of exponential and logarithmic functions, respectively. -- Darrell === Subject: question about uniformely convergence When I was reading my book, I've got an idea that if the series converges uniformely to some function, then we can use that infinite series to approximate the function. So there is a use for uniformely convergent series, but what about those series that are convergent to some function, but not uniformely? Are they useless? I also would like to know if there is any way to know how to come up with those infinite series that are convergent to a given function. For example, if the given function was: f(x)= 1/(1-x) , how could we have guessed that the corresponding infinite series was s(x)=1+x^2+x^3+.. I understand that it is easy to write down s(x), and then realize that when we simplify the right hand side of the infinite series, it happens that it equals to 1/(1-x). But I am interested in the other way arround, without assuming that f(x) is an analytical function or f(x) can be represented as power series. I also would like to know if there are many convergent infinite series for a given function. Is it only one infinite series for one function? === Subject: Re: question about uniformely convergence > When I was reading my book, I've got an idea that if the series > converges uniformely to some function, then we can use that infinite > series to approximate the function. So there is a use for uniformely Even when it's not uniform but it won't be as good an approximation. The hooha about uniform convergence is the limit function is continuous. > convergent series, but what about those series that are convergent to > some function, but not uniformely? Are they useless? I also would like > to know if there is any way to know how to come up with those infinite > series that are convergent to a given function. For example, if the > given function was: f(x)= 1/(1-x) , how could we have guessed that the > corresponding infinite series was s(x)=1+x^2+x^3+.. If not by Talyor's series, then by talent. > I understand that it is easy to write down s(x), and then realize that > when we simplify the right hand side of the infinite series, it happens > that it equals to 1/(1-x). But I am interested in the other way arround, > without assuming that f(x) is an analytical function or f(x) can be > represented as power series. I also would like to know if there are many > convergent infinite series for a given function. Is it only one > infinite series for one function? e^x has uncountably many. For all real a, e^x = sum e^a (x-a)^n / n! === Subject: Re: question about uniformely convergence > For example, if the > given function was: f(x)= 1/(1-x) , how could we have guessed that the > corresponding infinite series was s(x)=1+x^2+x^3+.. Depends on the function, I guess. Since this one's rational why not do long division: 1+x+x^2+x^3+... _________________ 1-x |1 1-x ----- x x-x^2 ------ x^2 x^2-x^3 -------- x^3 x^3-x^4 -------- ... ...but this does not address the interval of convergence. There is another method, among others, that does, involving geometric series. Notice how the form of f closely resembles that of the sum of a geometric series: Sum(n=0,oo] ar^n = a/(1-r), |r|<1 Let a=1, r=x: Sum(n=0,oo) x^n = 1/(1-x), |x|<1 1 + x + x^2 + x^3 +... = 1/(1-x) whenever -1 e^x has uncountably many. For all real a, > e^x = sum e^a (x-a)^n / n! It seems to me that sum e^a (x-a)^n / n! is just sum x^n / n! is disguise. (i.e., doesn't your series, when you expand the (x-a) terms and collect like terms, simply equal sum x^n / n! ?) Travis === Subject: Question on Quaternion & Quotient Group Exer. Hello All, I have a couple of questions about a generic group theory problem. Although I am familiar with the method for proving that a group is abelian or has a certain order; I am still unclear as to what elements of a quotient group look like to the extent that I am unsure what form the group elements involved in this exercise have. The question is stated as follows: Q is the quaternion group {+/- (1,i,j,k)} of order 8, H = Q x Q, A = {(1,1),(-1,-1)}, G= H/A and Z(G) is the center of G. I. Prove: |Z(G)|=2 i.e. that the center of G has order 2 II. Show: G/Z(G) is abelian. I am assuming that the group operation throughout is multiplication since that is what the sketchy solution seems to use. My questions are 1. What form do the elements of G have? Is it {hA: for all h in H} or {aH : for all a in A} ? and 2. What form do the elements of G/Z(G) have? I know Z(G)= {x in G| xg=gx for all g in G}, but that is about all. LS Thomas === Subject: Re: Question on Quaternion & Quotient Group Exer. > The question is stated as follows: Q is the quaternion group {+/- > (1,i,j,k)} of order 8, H = Q x Q, A = {(1,1),(-1,-1)}, G= H/A and Z(G) > is the center of G. > I. Prove: |Z(G)|=2 i.e. that the center of G has order 2 > II. Show: G/Z(G) is abelian. > 1. What form do the elements of G have? Is it {hA: for all h in H} or > {aH : for all a in A} ? The former. > 2. What form do the elements of G/Z(G) have? > I know Z(G)= {x in G| xg=gx for all g in G}, but that is about all. === Subject: Re: Question on Quaternion & Quotient Group Exer. > Hello All, > I have a couple of questions about a generic group theory problem. > Although I am familiar with the method for proving that a group is > abelian or has a certain order; I am still unclear as to what elements > of a quotient group look like to the extent that I am unsure what > form the group elements involved in this exercise have. > The question is stated as follows: Q is the quaternion group {+/- > (1,i,j,k)} of order 8, H = Q x Q, A = {(1,1),(-1,-1)}, G= H/A and Z(G) > is the center of G. > I. Prove: |Z(G)|=2 i.e. that the center of G has order 2 > II. Show: G/Z(G) is abelian. > I am assuming that the group operation throughout is multiplication > since that is what the sketchy solution seems to use. > My questions are > 1. What form do the elements of G have? Is it {hA: for all h in H} or > {aH : for all a in A} ? > and > 2. What form do the elements of G/Z(G) have? > I know Z(G)= {x in G| xg=gx for all g in G}, but that is about all. > LS Thomas OK. There may well be prettier ways but I'm going to beat the poor thing to death. Suppose (a,b)A is in the center and y in Q is arbitrary. [(a,b)A][(1,y)A] = (a,by)A = { (a,by), (-a,-by) } and [(1,y)A][(a,b)A] = (a,yb)A = { (a,yb), (-a, -yb) }. These two must be equal and since a =/= -a, we get yb = by for all y. That says b = +/-1. Similarly a = +/-1. Now (1,1)A = (-1,-1)A = A and (1,-1)A = (-1,1)A so Z(G) = {A, (1,-1)A} Now, the notation is going to get messy. I'm going to let Z = Z(G). An arbitrary element of G/Z is ((a,b)A)Z = {(a,b)AA, (a,b)A(1,-1)A } = {(a,b)A,(a,-b)A } = { {(a,b), (-a,-b)}, {(a,-b),(-a,b)} }. Consider ((a,b)A)Z((c,d)A)Z = ((a,b)A(c,d)A)Z = ((ac,bd)A)Z = { {ac,bd),(-ac,-bd)},{(ac,-bd),(-ac,bd)} } (*) And ((c,d)A)Z((a,b)A)Z = ... = { {(ca,db),(-ca,-db)}, {(ca,-db),(-ca,db} } (**) Now the question is does (*) = (**) ? If ac = -ca and db = -bd, the answer is yes. If ac = ca [for example a = c] then if bd = db the first set of (*) is the first set of (**) and the second set of (*) is the second set of (**). If bd = -db the first set of (*) is the second set of (**) and the second set of (*) is the first set of (**). And etc. -- Paul Sperry Columbia, SC (USA) === Subject: quotient groups Let G = (finite, abelian, 2-generated by x and y) where order(x) = a and order(y) = b are the only relations imposed (so G is isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = where f, g in Z. Now, given a group element n in G, let N be the coset of H in G containing n. That is, an uppercase letter is simply the coset containing the element of G that is the lowercase letter. Suppose the following facts are true: (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in Z }. (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in H). How do I present G/H in the form G/H = where r, s, t in Z ??? === Subject: Re: quotient groups > Let G = (finite, abelian, 2-generated by x and y) where > order(x) = a and order(y) = b are the only relations imposed (so G is > isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = > where f, g in Z. > Now, given a group element n in G, let N be the coset of H in G > containing n. That is, an uppercase letter is simply the coset > containing the element of G that is the lowercase letter. Suppose the > following facts are true: > (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in > Z }. > (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in > H). > How do I present G/H in the form > G/H = > where r, s, t in Z ??? Let's see if I've got this straight. G is abelian, G = + with o(x) = a, o(y) = b and / = {0}. H is a cyclic subgroup of G and G/H is generated by (x - y) + H. (I guess that's possible - I haven't checked.) (2) Is no new information.The first equation follows from G/H = <(x - y) + H>. The second equation just says y = y. So, it boils down to G = + , H is a cyclic subgroup and G/H = <(x - y) + H> What I don't follow is what you want. Clearly G/H is generated by x + H and y + H. a*(x + H) = H = b*(y + H) = 0 -- Paul Sperry Columbia, SC (USA) === Subject: Re: quotient groups > Let G = (finite, abelian, 2-generated by x and y) where > order(x) = a and order(y) = b are the only relations imposed (so G is > isomorphic to Z_a x Z_b). Let H be a cyclic subgroup of G where H = > where f, g in Z. Simply write X = x+H, Y = y+H or just use x+H, y+H. > (1) G/H is cyclic with G/H = , i.e. G/H = { c(x - y) + H | c in > Z }. > (2) x = w(x - y) + h, y = (w - 1)(x - y) + h (for some w in Z, h in > H). > How do I present G/H in the form > G/H = > where r, s, t in Z ??? This is rather complex, however it does address some issues I raised in a similar post a few days ago. Your problem has eerie resemblance to the problem posed by Heather in 'Presentations of a cyclic group'. I refer you to my response in her thread as possible suggestion. Not having gone thru the details of your presentation, I'm jumping to conclusion in making the referral. If I've time later, I'll go thru the details; as for now I'm sensing some confusion in your presentation or lack of understanding upon my behalf for not having slugged thru the details. === Subject: Radius of convergence issues Hello all. I've come to a brick wall in an early analysis class. The problem is as follows: Let {a_n} be a bounded sequence. Let R be the radius of convergence of sum((a_n)x^n). Prove that R >= 1. Now, I go about it like this: Since a_n is bounded, |a_n|<=M for all n. So |(a_n)x^n|<=M|x^n|. But sum(M|x|^n) coverges for all |x|<1 and thus, R=1. So, I can show that R=1 but not that, possibly, R>1. Am I fundamentally not understanding the idea of the radius of convergence? Any and all help is a lot of appreciated. Ben Scott === Subject: Re: Radius of convergence issues > Let {a_n} be a bounded sequence. Let R be the radius of convergence of > sum((a_n)x^n). Prove that R >= 1. > Now, I go about it like this: > Since a_n is bounded, |a_n|<=M for all n. So |(a_n)x^n|<=M|x^n|. But > sum(M|x|^n) coverges for all |x|<1 and thus, R=1. > So, I can show that R=1 but not that, possibly, R>1. Am I fundamentally not > understanding the idea of the radius of convergence? > Any and all help is a lot of appreciated. Wake up, R >= 1 iff (R > 1 _or_ R = 1) === Subject: Re: Radius of convergence issues > I've come to a brick wall in an early analysis class. Me too of some sort, please delete my other post. > Let {a_n} be a bounded sequence. Let R be the radius of convergence of > sum((a_n)x^n). Prove that R >= 1. > Now, I go about it like this: > Since a_n is bounded, |a_n|<=M for all n. So |(a_n)x^n|<=M|x^n|. But > sum(M|x|^n) coverges for all |x|<1 and thus, R=1. > So, I can show that R=1 but not that, possibly, R>1. No, if a_n = 1/2^n, then you'll see R = 2. > Am I fundamentally not > understanding the idea of the radius of convergence? R is the smallest number with for all x, (|x| < R implies sum a_n x^n converges) You've shown not R < 1, thus 1 <= R. === Subject: Re: Radius of convergence issues > I've come to a brick wall in an early analysis class. > Me too of some sort, please delete my other post. Yicks, it's just not my day. > Let {a_n} be a bounded sequence. Let R be the radius of convergence of > sum((a_n)x^n). Prove that R >= 1. > Now, I go about it like this: > Since a_n is bounded, |a_n|<=M for all n. So |(a_n)x^n|<=M|x^n|. But > sum(M|x|^n) coverges for all |x|<1 and thus, R=1. > So, I can show that R=1 but not that, possibly, R>1. > No, if a_n = 1/2^n, then you'll see R = 2. > Am I fundamentally not > understanding the idea of the radius of convergence? > R is the smallest number with > for all x, (|x| < R implies sum a_n x^n converges) R is the _largest_ number with for all x, (|x| < R implies sum a_n x^n converges) > You've shown not R < 1, thus 1 <= R. === Subject: Re: Radius of convergence issues > I've come to a brick wall in an early analysis class. > Me too of some sort, please delete my other post. > Yicks, it's just not my day. > Let {a_n} be a bounded sequence. Let R be the radius of convergence of > sum((a_n)x^n). Prove that R >= 1. > > Now, I go about it like this: > > Since a_n is bounded, |a_n|<=M for all n. So |(a_n)x^n|<=M|x^n|. But > sum(M|x|^n) coverges for all |x|<1 and thus, R=1. > > So, I can show that R=1 but not that, possibly, R>1. > No, if a_n = 1/2^n, then you'll see R = 2. > Am I fundamentally not > understanding the idea of the radius of convergence? > > R is the smallest number with > for all x, (|x| < R implies sum a_n x^n converges) > R is the _largest_ number with > for all x, (|x| < R implies sum a_n x^n converges) > You've shown not R < 1, thus 1 <= R. help! Ben Scott === Subject: real analysis proofs I'd like to know if my proofs make sense logically since I have a lot of doubts about them. They aren't supposed to be formal so please disregard any formalities. Subscript is denoted by () for example sup (k) is sup subscript k Suppose that s (n) is bounded Prove that any open interval containing lim sup s (n) there are infinitely many n with s (n) inside that interval Sup(k)s(n)>=sup(k+1)s(n) since {s(n)|n>k+1} is a subset of {s(n)|n>k}. Since the sequence is bounded and sup (k) is bounded monotone non-increasing subsequence it will converge (This is one of the parts I'm not sure about if it is a bounded monotone non-increasing subsequence). Therefore choose epsilon>0 then there exists n>N implies |sup s(n)-s| lim sup s(n) then for some open interval containing x, there are infinitely many n such that s(n) does not lie in that interval Since by definition of a limit there is only a finite number of s (n) existing outside epsilon. Choose x>epsilon. Let epsilon>0, x>s, then let epsilon=x-s then n>N implies |sup s (n)-s| I'd like to know if my proofs make sense logically since I have a lot of > doubts about them. They aren't supposed to be formal so please disregard any > formalities. > Subscript is denoted by () for example sup (k) is sup subscript k > Suppose that s (n) is bounded > Prove that any open interval containing lim sup s (n) there are infinitely > many n with s (n) inside that interval I'm having an awful time with your notation so I'm going to rewrite as I go. Let A(n) = sup{s(k) : k > n} so that limsup(s(n)) = lim(A(n),n->oo). > Sup(k)s(n)>=sup(k+1)s(n) since {s(n)|n>k+1} is a subset of {s(n)|n>k}. > Since the sequence is bounded and sup (k) is bounded monotone non-increasing > subsequence it will converge (This is one of the parts I'm not sure about if > it is a bounded monotone non-increasing subsequence). OK. A(n) is non-increasing and bounded, since s(n) is bounded, and thus converges. The limsup exists and we can set limsup(s(n)) = S. > Therefore choose > epsilon>0 then there exists n>N implies |sup s(n)-s|< epsilon. |A(n) - S| < epsilon. OK, but you're going backwards - start with an open interval containing S > Therefore > there is an infinite number of s (n) in the interval (epsilon-s, epsilon+s) Because .... The rest needs some work. > If x> lim sup s(n) then for some open interval containing x, there are > infinitely many n such that s(n) does not lie in that interval > Since by definition of a limit there is only a finite number of s (n) > existing outside epsilon. Choose x>epsilon. Let epsilon>0, x>s, then let > epsilon=x-s then n>N implies |sup s (n)-s| (n)|-|s| I'm also supposed to prove this, I would like to know if it is possible to > do using a Cauchy sequence, please avoid any suggestions on how to solve it. > Suppose that A is a closed interval of the form [a,b], S is a non-empty set > of real numbers, and S is a subset of A prove that supS belongs to A and > infS belongs to A _I_ would not use Cauchy sequences. -- Paul Sperry Columbia, SC (USA) === Subject: Series for one from e^(-x^2) Hello again, Yet another issue in an analysis class. The problem is thus: Show that 1 = sum((-1)^(n+1) (2n+1) / ((2^n) (n!))) by considering d^2/dx^2(exp(-x^2)). So far I've showed that d^2/dx^2(exp(-x^2)) = -2 exp(-x^2) + 4x^2 exp(-x^2) = sum((-1)^(n+1) (4n-2) / (n-1)! x^(2n)) which means -exp(-x^2)+2x^2 exp(-x^2) = sum((-1)^(n+1) (2n-1) / (n-1)! x^(2n)). And that's about as far as I get. I just can't seem to get the one sum in Ben Scott === Subject: Re: Series for one from e^(-x^2) === Subject: Series for one from e^(-x^2) >Show that 1 = sum((-1)^(n+1) (2n+1) / ((2^n) (n!))) by considering >d^2/dx^2(exp(-x^2)). >So far I've showed that >d^2/dx^2(exp(-x^2)) = -2 exp(-x^2) + 4x^2 exp(-x^2) = sum((-1)^(n+1) >(4n-2) / (n-1)! x^(2n)) >which means >-exp(-x^2)+2x^2 exp(-x^2) = sum((-1)^(n+1) (2n-1) / (n-1)! x^(2n)). -2 e^(-x^2) + 4x^2 e^(-x^2) = -2 sum (-1)^n x^2n /n! + 4xx sum (-1)^n x^2n / n! = -2 - 2 sum (-1)^(n+1) x^2(n+1) /(n+1)! + 4xx sum (-1)^n x^2n /n! = -2 + 2 xx sum (-1)^n x^2n /(n+1)! + 4xx sum (-1)^n x^2n /n! = -2 + 2 xx sum (-1)^n x^2n (1/(n+1)! + 2/n!) = -2 + 2 xx sum (-1)^n (2n + 3)x^2n /(n+1)! = -2 - 2 xx sum (-1)^(n+1) (2n + 3)x^2n /(n+1)! = -2 sum (-1)^n (2n + 1)x^2n /n! >And that's about as far as I get. I just can't seem to get the one >sum in the form of the other. Once again, any help is much Did that do it? It'd help were you to present us your calculations. Pay attention to details, do one step at a time, lest you invite error. BTW, when you write 'sum', where does n start from? ---- === Subject: Re: Series for one from e^(-x^2) === > Subject: Series for one from e^(-x^2) >Show that 1 = sum((-1)^(n+1) (2n+1) / ((2^n) (n!))) by considering >d^2/dx^2(exp(-x^2)). >So far I've showed that >d^2/dx^2(exp(-x^2)) = -2 exp(-x^2) + 4x^2 exp(-x^2) = sum((-1)^(n+1) >(4n-2) / (n-1)! x^(2n)) >which means >-exp(-x^2)+2x^2 exp(-x^2) = sum((-1)^(n+1) (2n-1) / (n-1)! x^(2n)). > -2 e^(-x^2) + 4x^2 e^(-x^2) > = -2 sum (-1)^n x^2n /n! + 4xx sum (-1)^n x^2n / n! > = -2 - 2 sum (-1)^(n+1) x^2(n+1) /(n+1)! + 4xx sum (-1)^n x^2n /n! > = -2 + 2 xx sum (-1)^n x^2n /(n+1)! + 4xx sum (-1)^n x^2n /n! > = -2 + 2 xx sum (-1)^n x^2n (1/(n+1)! + 2/n!) > = -2 + 2 xx sum (-1)^n (2n + 3)x^2n /(n+1)! > = -2 - 2 xx sum (-1)^(n+1) (2n + 3)x^2n /(n+1)! > = -2 sum (-1)^n (2n + 1)x^2n /n! >And that's about as far as I get. I just can't seem to get the one >sum in the form of the other. Once again, any help is much > Did that do it? It'd help were you to present us your calculations. > Pay attention to details, do one step at a time, lest you invite error. > BTW, when you write 'sum', where does n start from? > ---- The sums go from n=0 to infinity. I think my fundamental mistep was not equating the series of -2 e^(-x^2) + 4x^2 e^(-x^2) directly; I differentiated the power series for e^(-x^2) twice. Although the result is the same, certain steps become a bit hazy when done from that angle. Ben Scott === Subject: Re: Series for one from e^(-x^2) >Show that 1 = sum((-1)^(n+1) (2n+1) / ((2^n) (n!))) by considering >d^2/dx^2(exp(-x^2)). >d^2/dx^2(exp(-x^2)) = -2 exp(-x^2) + 4x^2 exp(-x^2) = sum((-1)^(n+1) >(4n-2) / (n-1)! x^(2n)) >-exp(-x^2)+2x^2 exp(-x^2) = sum((-1)^(n+1) (2n-1) / (n-1)! x^(2n)). > -2 e^(-x^2) + 4x^2 e^(-x^2) > = -2 sum (-1)^n x^2n /n! + 4xx sum (-1)^n x^2n / n! > = -2 - 2 sum (-1)^(n+1) x^2(n+1) /(n+1)! + 4xx sum (-1)^n x^2n /n! > = -2 + 2 xx sum (-1)^n x^2n /(n+1)! + 4xx sum (-1)^n x^2n /n! > = -2 + 2 xx sum (-1)^n x^2n (1/(n+1)! + 2/n!) > = -2 + 2 xx sum (-1)^n (2n + 3)x^2n /(n+1)! > = -2 - 2 xx sum (-1)^(n+1) (2n + 3)x^2n /(n+1)! > = -2 sum (-1)^n (2n + 1)x^2n /n! > BTW, when you write 'sum', where does n start from? > The sums go from n=0 to infinity. What about your first sum? :-) === Subject: set theory proof I'm trying to figure out how to prove the following (I'll write some symbols in LateX notation): If A_1 supseteq A_2 supseteq A_3 supseteq A_4 ... are all finite, nonempty sets of real numbers, then the intersection, bigcap_{n=1}^{infty} A_n is finite and nonempty. In other words, A_{n+1} is a subset of A_n, and A_1 is a superset for all A_n. === Subject: Re: set theory proof Visiting Assistant Professor at the University of Montana. >I'm trying to figure out how to prove the following (I'll write some >symbols in LateX notation): >If A_1 supseteq A_2 supseteq A_3 supseteq A_4 ... are all finite, >nonempty sets of real numbers, then the intersection, >bigcap_{n=1}^{infty} A_n is finite and nonempty. >In other words, A_{n+1} is a subset of A_n, and A_1 is a superset for >all A_n. Since they are finite, write A_1 = {x_1,....,x_m}. Suppose the intersection is empty: then for each i, there must exist a j which depends on i (say j(i) ) such that x_i is not in A_j. Let J = max {j(1),...,j(m)}. What can you say about A_J? It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: show that sqrt{3} + sqrt{2} is algebraic I'm trying to find integer coefficients for a polynomial which has sqrt{3} + sqrt{2} as a root. Any ideas? === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? Start with setting x = (the root you want). Since this expression only has square roots, it should be easy to manipulate (moving terms around, squaring) this expression into an equation between a polynomial in x and zero. -- word to the appropriate symbol to email me. Joshua P. Bowman === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic >I'm trying to find integer coefficients for a polynomial which has >sqrt{3} + sqrt{2} as a root. Any ideas? Let x = sqrt(3) + sqrt(2). Then x^2 = 5 + 2*sqrt(6), so sqrt(6) = (x^2 - 5)/2. Can you take it from there? Brian === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? > x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) x = sqr 3 + sqr 2 x^2 = 5 + 2 sqr 6 (x^2 - 5)^2 = 24 Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3? === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? Start with x = sqrt(2)+sqrt(3) Rearrange such that one side becomes simpler when squared. Square both sides. Rearrange such that one side becomes simpler when squared. Square both sides. Repeat until there's nothing left to do. Note - you'll be doubling the number of roots to the equation with each square, but that can't be avoided. Phil === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic > I'm trying to find integer coefficients for a polynomial which has > sqrt{3} + sqrt{2} as a root. Any ideas? See pages 5 and 6 of my notes on algebraic number theory for a method of solving all such problems (and an example slightly harder than this). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: show that sqrt{3} + sqrt{2} is algebraic >> I'm trying to find integer coefficients for a polynomial which has >> sqrt{3} + sqrt{2} as a root. Any ideas? >> x^4 - 10*x^2 +1 has roots the 4 roots +/-sqrt(2) +/-sqrt(3) > x = sqr 3 + sqr 2 > x^2 = 5 + 2 sqr 6 > (x^2 - 5)^2 = 24 > Is there an integer polynomial in Z[x] with roots sqr 2, sqr 3? That depends... if you require the polynomial to have sqrt 2 and sqrt 3 as _the_ roots, then the answer is no. Clearly one polynomial with these roots is f(x) = (x-sqrt(2))(x-sqrt(3)) = x^2 - (sqrt(2)+sqrt(3))x + sqrt(6). Any other polynomial with exactly sqrt(2) and sqrt(3) as roots is a constant times this polynomial. Now If you require the polynomial to belong to Z[x], then you can only multiply f by integers; but since sqrt(6) is irrational, no polynomial in Z[x] is an integer multiple of f. If you only require that sqrt(2) and sqrt(3) are roots of the polynomial, then the answer is clearly yes. Try g(x) = (x-sqrt(2))(x+sqrt(2))(x-sqrt(3))(x+sqrt(3)) /Rasmus --