> I think I've figured out a way to show basically all of you, including > people who think they don't know any math that mathematicians have > been lying about my work. It's so trivial you *should* wonder why > they thought they could get away with it. Okay. (I've always *wanted* to see all those lying mathematicians shown up, ;) but maybe there's something to it.) Let's see this easy proof of mathematician lies. > Here goes. My paper Advanced Polynomial Factorization depends on considering a > factor of a polynomials that I call g. ITYM factor of a polynomial, singular. These sorts of syntactic issues can be very important in any hard science. Or math, too. :) > (Paper linked to at http://groups.msn.com/AmateurMath as usual.) I'm not going to read a whole paper. This is supposed to be an *easy* proof. (I haven't visited the link.) > And in my paper I start by showing that I can write that as g = r + c where either r=0, or r changes as the polynomial's value changes, > while c does not. So, like if the original polynomial was F(x) = x^2 - 1, then g could be (x-1) or (x+1), and r=x and c=-1 (or +1). Makes sense, although it's kind of a funny way to write it. > Now you can consider all factors of a given polynomial using g's, with > something like g_1...g_k = P(x) where you have k factors. (x-1)(x+1) = x^2 - 1. Yep. Gotcha. It raises a slight problem if the original polynomial is, say, 2x^2-2, or something, since then r doesn't equal x; it equals some factor times x. But all right. > For instance, for P(x)=x^2 + 2x + 1, g_1 = x+1, g_2 = x+1, gives you 2 factors. Those are polynomial factors, but I'm generalizing in a simple way to > say that for the factors g, in general, you have an element I call r, > which changes as the independent variable changes, and you have > another element I call c, which does not. Generalizing *beyond* polynomials? To what? Integers? :) > For my example up above it's easy, as with g_1 = x + 1, x varies as x > varies, while 1 does not. True dat. > Now that's enough that the proof in the paper is straightforward, but > posters have argued with me anyway, with some trying to argue over the > definition of polynomial, amazingly enough. However, consider that the g's have an important feature, which is > that when x=0, I have g_1...g_k = P(0). Obviously. > For instance, with my simple example, with P(x) = x^2 + 2x + 1, with > x=0, P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1. You see, P(0) gives the constant term, so at x equal 0, the g's must > multiply to give the constant term. Yep. > So then, maybe you still want to believe the mathematicians and > question that I can write g = r + c. Why would anyone dispute it? > Well consider that substituting gives me g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives r_1...r_k +...+c_1...c_k = P(x), which is r_1...r_k +...+P(0) = P(x), Geez, that's confusing. Must be the lack of whitespace. I'll stick to Perl, if you don't mind. Anyway, it seems like some sort of example of why the above bit of text is right, and I already agree with that. So I'll skip it. > which means that if you believe the mathematicians then they've > convinced you to doubt algebra itself, as then you must believe that > everything to the left of P(0) above can *maybe* be constant, but also > *maybe* vary as x varies. So why would mathematicians argue against such a simple result? blah, blah, blah... I wanna see the proof! So I skip the polemicizing. > Two reasons I suggest. First because they wish to disagree with me. > Second because they probably believe that they can get away with it. That is, MOST of you will doubt algebra itself rather than consider > that mathematicians, whom you probably don't even personally know, > would lie. So where does this lead? Whew. Okay, where *does* it lead? > Well the polynomial I show in the paper is P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf which seems to be just complicated enough to give mathematicians room > to lie. You can say *that* again (but without the political innuendos, please). Lessee... P(m) = ((mf-1)^3 + 1) * x^3 - (3*(mf-1)*y^2) * x + (y^3) Looks reasonable, although I don't see the point of all those extra variables. I assume they *are* variables, right? > For instance, you may be saying, HEY, what's with the 'm' when you > had 'x' before??!!! Wait... you still *have* 'x'! Is 'm' supposed to be the independent variable here? That makes the algebra yuckier. > Well, there's no rule that says that you have to use the letter x as > the variable label for a polynomial. Also, there are historical > reasons for my usage as it goes back to my work with FLT where x, y > and z are used with x^p + y^p = z^p. (A little egotistical. You're not supposed to quote *yourself* as a historical reason.) > Finally, the weirder thing is that one poster in particular got a lot > of mileage out of questioning my finding the constant term with an > expression like the above by using m=0, as that gives me P(0) = 3xy^2 + y^3 and he got a lot of mileage for YEARS (before I had discovered proof I > want to add and after) by emphasizing that two of the ROOTS of such an > expression considered as a polynomial with respect to x are not > defined at that point. Huh? Nothing wrong with a polynomial that only has one root. It's just linear in x, one might say. > Well that's easy enough to see as the original expression is (v^3+1)x^3 - 3vxy^2 + y^3 which if you *wish* to see it as a polynomial with respect to x, is of > degree 3, but when v=-1, it's of degree 1, so if you solve for the > roots, you'll get funky stuff. Now when I was finding the proof of FLT...remember the process took > some years... (Danger, Will Robinson! Danger!) > at times I'd talk of polynomials with respect to other > than m, but I refined my discourse as my understanding improved. However, people arguing with me did not. You may *choose* to believe that they did not because they don't know > enough mathematics to follow, but we're talking about actual > mathematicians here. (Danger!) > What's more rational? I say it's more rational to suppose that they *did* figure out that it > worked as described, but also noticed that as long as they disagreed, > no one seemed to call them on making false statements, except me, and > they knew my credibility wasn't so great. For most of you, there's probably the belief that there's some funky > higher math involved that your pitiful brain (Hoo boy. Let's just ride this one out.) > can't follow or you > don't know about, as you may suppose that mathematicians either > wouldn't lie, or they wouldn't lie in such a dumb way where I could > catch them so easily. But consider what's in the balance: 1. I discovered a proof of Fermat's Last Theorem that's more > available to people in general than most of what mathematicians have > been producing lately. Available how? I've never seen any proof of FLT, although I know that Wiles' proof is probably written up somewhere. You'd think a neat FLT proof would make big waves. (Oh, never mind.) > 2. Worse I did so having said I'd find it years ago, and having spent > years looking for it posting a lot of my ideas, and getting in insult > battles with posters, quite a few who happened to be mathematicians. 3. Then to add insult to injury, I keep questioning mathematicians in > terms of their ethics, and maybe *extremely importantly* I doubt that > Wiles found a proof of Fermat's Last Theorem. What do you *think* he proved, then? > And those are just highlights as there's more but I think that kind of > gives my point. For mathematicians the situation could be considered one of the worst > possible disasters they can imagine. EXCEPT, it looks like all they have to do is either stay quiet, or > *claim* I'm wrong. They could also claim you're right. That just about exhausts the possibilities, there. Unfortunately, nobody thinks that you know any math, which means you're almost certainly *not* right. Hmm. > Many of you simply believe them, and question algebra itself, which is > rather sad. I'd think at least some of you valued your educations. Oh, I do. My education *taught* me algebra. Did yours? Seriously, though. Stop making these oblique general insults and give me that easy proof you promised. > Others of you may figure it doesn't matter, maybe because Western > civilization seems to be based on lying anyway, and maybe you figure I > should just grow up, accept that everybody lies and move on. And I > don't have to talk about Enron or pedophile Catholic priests or things > in that vein. blah blah blah... > I mean, look at George W. Bush and Iraq. If people can be *killed* > over lies, without consequences to the liars, then what's with some > freaking stupid math? Good point. Mathematicians *are* a part of society after all. Why should they > tell the truth now? It'd be like Bush owning up. They can just sit > tight, and be quiet, like so many American citizens or they can out > and out lie, like so many other patriots. blah blah blah... > After all, that's so easy, now isn't it? Which is why you need to understand why I talk about mathematicians > potentially being prosecuted. Liars don't just stop because the gig > is up, as then, they wouldn't necessarily be liars, then eh? blah blah blah... It'd be against their *true* natures. James Harris > Wait! *WHAT*!?! Where's that proof you promised? I waded through that whole diatribe for *nothing*?! Dangit, Harris, you said you had an easy proof that mathematicians lied, and all I get is a discussion of how you choose to express factors of polynomials with 'g' and 'r' and 'c' instead of normal-people letters? Where's the beef? -Arthur ==== If anybody has any time, I wouldn't mind you guys trying this: http://www.chriscentral.com .... it's a message board which supports entering equations/math in standard notation...so you can put maths in your posts.Give it a go please post any problems you find either there or here. Chris. ==== I've long thought that was a major opportunity for improvement in USENET, but unlike me, you took some action! Good job! For your next project, why don't you find out what impossible feat would be required to add something like this to USENET itself. I used to think this would be anti-democratic, as part of the beauty of USENET is that you could be a full and equal citizen writing in from the African bush on a KAYPRO over a 1200 BAUD modem. Well, it's a nice fantasy: but since it has nothing to do with reality, may as well up the technical standard. I'm afraid your Java applet overwhelms my old machine, but I'm getting a more up to date one any day ... ==== Dear Edward Green: ... > For your next > project, why don't you find out what impossible feat would be required > to add something like this to USENET itself. I used to think this > would be anti-democratic, as part of the beauty of USENET is that you > could be a full and equal citizen writing in from the African bush on > a KAYPRO over a 1200 BAUD modem. Well, it's a nice fantasy: but since > it has nothing to do with reality, may as well up the technical > standard. A lot of web pages use imbedded graphics for the formulae. Usenet accepts HTML posts, which can have imbedded graphucs. It is considered bad manners to post this way, however. David A. Smith ==== >Usenet accepts HTML posts, which can have imbedded graphucs. It is >considered bad manners to post this way, however. In fact, both graphics and HTML are forbidden by Usenet standards (outside of groups with binaries in their names). This is not just theory: many news servers enforce this rule by silently graphics and such will be drastically limiting their audience. If you need HTML and graphics, put up a Web page and then post the URL here. But unless your equations or figures are quite complex, it's probably better to do the best you can in plain text. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/ Walrus meat as a diet is less repulsive than seal. ==== Actually, all considering, it wouldn't be very hard. A standard known as MathML allows for the encoding of math in web pages. The images are made by my server to make sure everyone can see them, BUT, you could download a plugin that lets you view mathml properly, and subsequently you could put mathml into usenet without too much difficulty. I welcome you to use my site where it helps :-) Chris http://www.chriscentral.com Message Board with Equation Support >Usenet accepts HTML posts, which can have imbedded graphucs. It is >considered bad manners to post this way, however. > > In fact, both graphics and HTML are forbidden by Usenet standards > (outside of groups with binaries in their names). This is not > just theory: many news servers enforce this rule by silently > graphics and such will be drastically limiting their audience. > > If you need HTML and graphics, put up a Web page and then post the > URL here. But unless your equations or figures are quite complex, > it's probably better to do the best you can in plain text. ==== > If anybody has any time... Yes, you realy have to have time. It takes minutes (760kb/s) until the java applet WebEQ Editor is loaded. Than it takes time to type your formula with the not very user friendly WebEQ Editor and as a result you will see a link to an image in your text, but you can not correct the math anymore. By the way WebEQ Editor has a very limited font set. It only uses the symbol font. There are several existing message boards that use LaTeX-like input code for equations, that will than be rendered on the fly by the server. I think that is a much better solution than using a fat java UI, with lots of limitations, like WebEQ Editor. Bernhard > > If anybody has any time, I wouldn't mind you guys trying this: > http://www.chriscentral.com .... it's a message board which supports > entering equations/math in standard notation...so you can put maths in your > posts.Give it a go please post any problems you find either there or here. > > Chris. ==== >Dear Edward Green: ... >> For your next >> project, why don't you find out what impossible feat would be required >> to add something like this to USENET itself. I used to think this >> would be anti-democratic, as part of the beauty of USENET is that you >> could be a full and equal citizen writing in from the African bush on >> a KAYPRO over a 1200 BAUD modem. Well, it's a nice fantasy: but since >> it has nothing to do with reality, may as well up the technical >> standard. A lot of web pages use imbedded graphics for the formulae. >Usenet accepts HTML posts, which can have imbedded graphucs. It is >considered bad manners to post this way, however. Any time you deviate from defined Usenet standards, be aware that some people will be using software that can't view it, and some servers simply will not pass the message along. -- Is that plutonium on your gums? Shut up and kiss me! -- Marge and Homer Simpson ==== In sci.physics, Chris Muktar <73d480e3.0306202246.2ac1705d@posting.google.com>: > Actually, all considering, it wouldn't be very hard. A standard known > as MathML allows for the encoding of math in web pages. The images are > made by my server to make sure everyone can see them, BUT, you could > download a plugin that lets you view mathml properly, and subsequently > you could put mathml into usenet without too much difficulty. With about the same results as one gets when putting HTML into Usenet now: a general response of eewwwwwwwwwwww, gross. :-P I use SLRN. It doesn't understand HTML *or* base64-encoding. I suspect a number of others are in the same boat. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > > I've long thought that was a major opportunity for improvement in > USENET, but unlike me, you took some action! Good job! For your next > project, why don't you find out what impossible feat would be required > to add something like this to USENET itself. I used to think this Use latex (http://www.tex.ac.uk) in your posts, then if someone feels the need to work out what $frac{partialvec{B}}{partial t}&=&0$ means they can always run it through their latex programme. Latex is free for Unix and Windows and any other system you care to think about. -- Frodo Morris http://users.ox.ac.uk/~wadh1342 All your bast are belong to us AKA Graham Lee, Wadham College SpectrumSofts currently on show at URL/speccy/: Speccy@Home SETI Client Also the home of iloveyou.bas, the first PC virus ported to the ZX82!!! ==== >Dear Edward Green: ... >> For your next >> project, why don't you find out what impossible feat would be required >> to add something like this to USENET itself. I used to think this >> would be anti-democratic, as part of the beauty of USENET is that you >> could be a full and equal citizen writing in from the African bush on >> a KAYPRO over a 1200 BAUD modem. Well, it's a nice fantasy: but since >> it has nothing to do with reality, may as well up the technical >> standard. A lot of web pages use imbedded graphics for the formulae. >Usenet accepts HTML posts, which can have imbedded graphucs. It is >considered bad manners to post this way, however. > > Any time you deviate from defined Usenet standards, be aware that some > people will be using software that can't view it, and some servers simply > will not pass the message along. Noted. Maybe that's why I added impossible. I suspect or know it wouldn't be terribly technically difficult for others, if not for me, to implement something like this, but standard is the bugaboo: you would need a critical mass of people on the same page before trying to change things. It seems to me what's really wanted (for some hyptothetical NeoNet(SM)) is a standard of embedding (compressed) binary images in text, plus, say, index card size touch screen peripheral writing tablets -- you are writing your post, you define a box with a few key strokes or mouse clicks, and then put your equations or sketches inside it, by hand, with the writing tablet. Transparent, and no need to standardize on some math typsetting language. This would mimic the way people communicate using a blackboard or a pad of paper. Given the existence of so-called binary groups, bandwidth really wouldn't really be an issue: such a group would use data/post comparable to an ordinary binary. The issues are inventing, promulgating and propagating a technical standard. Interesting, the touch sensitive writing tablet seems like a natural peripheral, but doesn't seem to be widely known or used. Of course it might be problematic competing for disk space with images of moose genetalia and all the other material whose promulgation must be maintained to preserve the right for free speech. impossible after all: given a standard, a NeoGroup(SM) could fit neatly inside existing networks and hardware, blah, blah, blah, and have about the same overhead as a binary group. In fact, maybe it could be distributed as a binary with some embedded coding which would simply tell end machines with the right software how to read the posts. So one wouldn't have to mess with the Sacred Structure of Usenet much after all ... a new standard could sort of insinuate itself inside the old, waiting for converts. I think such a thing would be a big hit on all the science groups: text + handwritten equations and sketches would be an order of magnitude more powerful and fluid (not to mention luminous ;-) as a medium than text + ascii math or text + web page references. Any Usenet savvy programmers out there want to help me develop this for love? ==== ...By the way WebEQ Editor has a very limited font set. > It only uses the symbol font. That has never been true, but it is certainly not true for the current version of WebEQ, which supports all the characters in the MathML 2.0 Recommendation -- and then some. Bob Mathews bobm@dessci.com Director of Training 830-990-9699 http://www.dessci.com/free.asp?free=news Design Science, Inc. -- How Science Communicates MathType, WebEQ, MathPlayer, MathFlow, Equation Editor, TeXaide ==== ...By the way WebEQ Editor has a very limited font set. > It only uses the symbol font. And I replied: > That has never been true, but it is certainly not true for the > current version of WebEQ, which supports all the characters in > the MathML 2.0 Recommendation -- and then some. Actually that's not correct either. Currently WebEQ supports around 400 characters, with more planned. That's a great deal more than using only the symbol font, but less than the full MathML 2.0 spec. Sorry for the confusion. Bob Mathews bobm@dessci.com Director of Training 830-990-9699 http://www.dessci.com/free.asp?free=news Design Science, Inc. -- How Science Communicates MathType, WebEQ, MathPlayer, MathFlow, Equation Editor, TeXaide ==== Here's a look at the methods highlighted in my paper Advanced Polynomial Factorization where I bring down the number of symbols. Notice if you can follow to the conclusion you're looking at an *error* in taught mathematics. If you can't follow when the only symbol left is m, then I don't know how you could ever follow. The ring is algebraic integers, though at one point you're pushed out of the ring by the flaw in taught mathematics. I'm curious to know if any of you can find that point. Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). However, now I'll let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) and P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5). Now for P(0), I have P(0) = 25 (3(2) + 5), which is that constant term I talk about a lot. And now notice that P(0)/25 = 3(2) + 5, which is coprime to 5(yes, f is coprime to 3, x, and u which is given in the paper). Here's where it gets interesting as while the key variable is m, I notice that focusing on those x's it looks like another cubic, so I consider the factorization P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) and specifically focus on the factor 2 a_1 x + 5, which I'll call g. I notice that if a_1 is 0, when m=0, as at least one of the a's must be, then at that point g=5, and has a factor that is 5. However, P(m)/25 has a factorization as well, but its constant term is P(0)/25 = 3(2) + 5 which is coprime to 5, and considering g I see that factor of 5 that is visible with g = a_1 x + 5 has to go away, and in going away it must go through a_1. That is, I'm considering P(m)/25 = 8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5 and I have to do it somewhat obliquely because I don't know what the factors of the a's are right off just from looking at P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) though because P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) I know that the a's have some factors in common with 5, which *should* be trivial. Now I hope that putting in some numbers helped. James Harris ==== > Here's a look at the methods highlighted in my paper Advanced > Polynomial Factorization where I bring down the number of symbols. > Notice if you can follow to the conclusion you're looking at an > *error* in taught mathematics. > No. Your conclusion is incorrect. You continue to insist that in the polynomial factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) with a1,a2,a3 algebraic integers, at least one of the a's is coprime to 5. More recently, you have insisted that *all* of the a's are coprime to 5. I show that none of them is coprime to 5, that there is, for each ai, a factor r(-ai) which is (1) an algebraic integer, and (2) a factor of 5. If you insist that all three are units, then you are compelled to claim that 5 is a unit in the ring of algebraic integers. What is the error in the following computation? It is a *DIRECT* calculation that shows that your claim regarding the a's is incorrect. Do you retract that claim? If not, why not? If you do not retract the claim, you must disagree with my assertion that is easily verified. You must disavow the truth. Are you doing that? Why are you disavowing the truth? De truth been berry berry good to me. I'll repeat, for the record, that I have produced, for EACH of the a's, two factorizations: 5 = q(-ai)*r(-ai) ai = r(-ai)*s(-ai) That is, there are three a's. There are three factorizations, and there are three common factors (one for each ai, a factor in common between ai and 5). For the sake of completeness, here are the polynomials q, r, and s: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 A quick bit of arithmetic will show the following: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x, where p(x) = x^3 - 12 x^2 + 65. Note that this immediately shows that for z = any root of p(x), q(z)*r(z) = 5, r(z)*s(z) = z. Despite your ranting, you do not appear to be up to the task of computing the values of q,r,and s for the roots, so I'll do that here (courtesy of DOE Macsyma [version: Maxima 5.9.0]), for one of them. Recall that I have given these results in terms of roots of the polynomial: p(x) = x^3 - 12 x^2 + 65 Those roots are all (-1) times the coefficients ai that you have made these foolish claims about. Here are the roots: Let u = (63 + i*sqrt(12415))/2, ubar = (63 - i*sqrt(12415))/2, zeta = (-1 + i*sqrt(3))/2, zetabar = (-1 - i*sqrt(3))/2. Then the roots of p(x) can be given as follows: x1 = u^(1/3) + ubar^(1/3) + 4 x2 = zeta*u^(1/3) + zetabar*ubar^(1/3) + 4 x3 = zetabar*u^(1/3) + zeta*ubar^(1/3) + 4 where the (...)^(1/3) above are the values with argument = 1/3 times the argument of (...). Note that this convention forces the three roots to be real (as they must be, for this polynomial). The roots are approximately: x1 ~ 11.50930 x2 ~ -2.14375 x3 ~ 2.63445 The factorizations claimed above are as follows, in the case of the root x1 = -a1: q1 = q(x1) = 8(63 + i sqrt(12415))^(2/3)/2^(2/3) + 8 (63 - i sqrt(12415))^(2/3)/2^(2/3) + 256/2^(2/3) - 12 (63 + i sqrt(12415))^(1/3)/2^(1/3) - 12 (63 - i sqrt(12415))^(1/3)/2^(1/3) - 361 r1 = q(x1) = 8(63 + i sqrt(12415)^(2/3)/2^(2/3) + 8(63 - i sqrt(12415)^(2/3)/2^(2/3) + 256/2^(2/3) + 60 (63 + i sqrt(12415))^(1/3)/2^(1/3) + 60 (63 - i sqrt(12415))^(1/3)/2^(1/3) + 67 s1 = s(x1) = 4(63 + i sqrt(12415)^(2/3)/2^(2/3) + 4(63 - i sqrt(12415)^(2/3)/2^(2/3) + 128/2^(2/3) + 5 (63 + i sqrt(12415))^(1/3)/2^(1/3) + 5 (63 - i sqrt(12415))^(1/3)/2^(1/3) - 188 While verification that q1*r1 = 5 and r1*s1 = x1 is tedious, the above identities: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x suffice to produce the desired result (given that p(x1) = 0). > If you can't follow when the only symbol left is m, then I don't know > how you could ever follow. It doesn't matter what you're saying, because your methodology is flawed, since it gives incorrect results (namely: all a's coprime to 5). ... irrelevant nonsense deleted ... > I know that the a's have some factors in common with 5, which *should* > be trivial. > Please address the question, James Harris. > Now I hope that putting in some numbers helped. > Ditto, James Harris. Answer the question. What is the flaw in my DIRECT calculation (no complicated logic involved, only arithmetic) that shows you're wrong? > > James Harris Dale. ==== > Here's a look at the methods highlighted in my paper Advanced > Polynomial Factorization where I bring down the number of symbols. > Notice if you can follow to the conclusion you're looking at an > *error* in taught mathematics. > > If you can't follow when the only symbol left is m, then I don't know > how you could ever follow. I was in a bit of a rush earlier. That should be leaving only m and the a's, as you have a_1, a_2 and a_3 below. > The ring is algebraic integers, though at one point you're pushed out > of the ring by the flaw in taught mathematics. I'm curious to know if > any of you can find that point. > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > However, now I'll let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > and > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5). Should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now for P(0), I have P(0) = 25 (3(2) + 5), which is that constant term > I talk about a lot. And now notice that P(0)/25 = 3(2) + 5, which is > coprime to 5(yes, f is coprime to 3, x, and u which is given in the > paper). > > Here's where it gets interesting as while the key variable is m, I > notice that focusing on those x's it looks like another cubic, so I > consider the factorization > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) Again should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > and specifically focus on the factor 2 a_1 x + 5, which I'll call g. That should be 2 a_1 + 5. > I notice that if a_1 is 0, when m=0, as at least one of the a's must > be, then at that point g=5, and has a factor that is 5. > > However, P(m)/25 has a factorization as well, but its constant term > is > > P(0)/25 = 3(2) + 5 > > which is coprime to 5, and considering g I see that factor of 5 that > is visible with > > g = a_1 x + 5 Should be g = 2 a_1 + 5. > has to go away, and in going away it must go through a_1. > > That is, I'm considering > > P(m)/25 = 8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5 > > and I have to do it somewhat obliquely because I don't know what the > factors of the a's are right off just from looking at > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) Should be P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > though because > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > I know that the a's have some factors in common with 5, which *should* > be trivial. > > Now I hope that putting in some numbers helped. But I'm wary. It shouldn't have been necessary for me to put in numbers, and I'm wary that some poster is going to come along, throw up a bunch of technical b.s and people will just believe yet again. Mathematicians CAN lie to you. There's no rule written into the laws of physics that prevents them from LYING TO YOU. James Harris ==== > Here's a look at the methods highlighted in my paper Advanced > Polynomial Factorization where I bring down the number of symbols. > Notice if you can follow to the conclusion you're looking at an > *error* in taught mathematics. > > If you can't follow when the only symbol left is m, then I don't know > how you could ever follow. > > The ring is algebraic integers, though at one point you're pushed out > of the ring by the flaw in taught mathematics. I'm curious to know if > any of you can find that point. > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > However, now I'll let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > and > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5). > > Now for P(0), I have P(0) = 25 (3(2) + 5), which is that constant term > I talk about a lot. And now notice that P(0)/25 = 3(2) + 5, which is > coprime to 5(yes, f is coprime to 3, x, and u which is given in the > paper). > > Here's where it gets interesting as while the key variable is m, I > notice that focusing on those x's it looks like another cubic, so I > consider the factorization > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) > > and specifically focus on the factor 2 a_1 x + 5, which I'll call g. > > I notice that if a_1 is 0, when m=0, as at least one of the a's must > be, then at that point g=5, and has a factor that is 5. > Note for the discussion below: you are saying explicitly here that when m = 0, a_1 = 0; and that you are saying that therefore * 5 is a factor of a_1 *. > However, P(m)/25 has a factorization as well, but its constant term > is > > P(0)/25 = 3(2) + 5 > > which is coprime to 5, and considering g I see that factor of 5 that > is visible with > > g = a_1 x + 5 > > has to go away, and in going away it must go through a_1. > As you have said above, when m = 0, a_1 = 0. Of course 0 is a multiple of 5, albeit in a rather trivial sense. And right here is the key to the error in your reasoning. You implicitly argue that since a_1 is a multiple of 5 when m = 0, it must also be such when m <> 0. The key thing here is, *** a_1 is not a constant! ***. It is an algebraic integer *variable* whose values depend on the integer m. The fact that a_1 is 0 when m = 0 tells you absolutely nothing about what its values may be, or how they are related to the number 5, when m is not equal to zero. You have made a giant leap here, based on an intuitive hunch - no proof, no underlying mathematical principle. And as demonstrated by both me and W. Dale Hall, your hunch here is dead wrong. See below. > That is, I'm considering > > P(m)/25 = 8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5 > > and I have to do it somewhat obliquely because I don't know what the > factors of the a's are right off just from looking at > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 x + 5) > > though because > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > I know that the a's have some factors in common with 5, which *should* > be trivial. > Correct. In fact all of the a's share factors with 5. > Now I hope that putting in some numbers helped. > > > James Harris Putting in numbers just makes it harder in this case for you to see the forest for the trees. It obscures the fact that a_1, a_2, and a_3 are FUNCTIONS of m. If a_1 were a constant with respect to m, everything would work out. But obviously it cannot be. When m = 0, a_1 = 0. When m <> 0, a_1 also cannot be equal to 0. The key thing here is, you CANNOT use the fact that a_1 = 0 when m = 0 to deduce properties of a_1 when m <> 0. Think of it this way. When m = 0, a_1 = 0, so not only is it divisible by 5, it is also divisible by 7, 11, 137, 10^9 + 1, etc.. Does that imply that when m <> 0, a_1 is also divisible by 7, 11, etc. ??? Of course not! But that is exactly where your logic leads. You have started a new thread here rather than respond to my proof and comments, and Dale Hall's proof, in other threads. Dale has forcefully reminded you of his proof that the main conclusion of APF is wrong, and that you can verify this for yourself with simple computations. Dale specifically challenges you to find an error in his argument. Similarly I remind you of my independent proof, reproduced below, to which you have not provided a valid objection: ==================================================================== > > The claim is made in Advanced Polynomial Factorization that if > > P(x) = 65*x^3 - 12*x + 1 > > 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, and a3 must be coprime to 5 in the algebraic integers. > > Thus in the following, assume that one of the ai, > say a1, is, as claimed, coprime to 5 in the algebraic integers. > That is, there exist algebraic integers s and t such that > > s*a1 + t*5 = 1. [1] > > Let x = 1/u. Then P(x) = 0 implies that > > P(x) = 65/u^3 -12/u + 1 = 0, or > > Q(u) = u^3 - 12*u^2 + 65 = 0. > > Let r1, r2, and r3 be the roots of this > equation. All of r1, r2, and r3 are algebraic > integers. Note that r1, r2, and r3 happen to > be the negatives a1, a2, and a3 in some order - > say, a1 = -r1, a2 = -r2, a3 = -r3. > > Note that Q(u) is irreducible over the rationals. > 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. Note also that > F12(a1) = a2, since a1 = -r1 and a2 = -r2. > > {That such an automorphism F12 exists is well-known, > and is described in: > > http://www.math.niu.edu/~beachy/aaol/galois.html, > > see especially Proposition 8.6.2 on that page. Or > see the excellent textbook, Abstract Algebra, by > John Beachy and William D. Blair.} > > > Now apply the automorphism F12 to both sides of > equation [1]: > > > F12(s)*F12(a1) + F12(t)*F12(5) = F12(1). > > But F12(1) = 1, F12(5) = 5, and F12(a1) = a2. Thus > we obtain > > s' * a2 + t' * 5 = 1, [2] > > where s' = F12(s) and F12(t) = t'. Note that the > automorphism F12 preserves the property of being > an algebraic integer; that is, s' and t', like > s and t, are algebraic integers. Therefore > equation [2] implies that a1 is coprime to 5 > also. > > Similarly one shows that a3 is coprime to 5. > > Thus, from the assumption that one of a1, a2, > or a3 is coprime to 5, we deduce that all three > of them must be coprime to 5. > > But a1 * a2 * a3 = 65. Therefore at least one > of a1, a2, or a3 is NOT coprime to 5. Therefore > a contradiction. The source of the contradiction > was the initial assumption that one of a1, a2, or > a3 is coprime to 5. Therefore NONE of a1, a2, and > a3 are coprime to 5, contradicting the main result > of Advanced Polynomial Factorization. > > ======================================================================= Thus the main claim in APF is absolutely false. There are two different, independent proofs of its falsity. I think it may be the case that essentially your same APF argument is employed in your claimed proof of Fermat's Last Theorem. This clearly should be the end of this discussion. There is no hope now of retrieving what you wanted from APF. In starting this thread, you have only driven more nails into the coffin: you have made it even more clear how your thinking went astray. If you venture to add anything to this you will simply be beating a horse that died on you some time ago. Finally, I believe you submitted APF to one of the journals of the American Mathematical Society, and that it was rejected. I wonder if you might be willing to share with us what the editors or reviewers gave as their reasons for rejection? Nora Baron ==== [snip analysis and refutation of James' latest failed attempt to salvage his proof] > Finally, I believe you submitted APF to one of the journals of the > American Mathematical Society, and that it was rejected. I wonder > if you might be willing to share with us what the editors or > reviewers gave as their reasons for rejection? Nora Baron If he does answer your above question, which I doubt, he is most likely to respond that his manuscript was rejected because the reviewers were biased members of a vast conspiracy dedicated to repelling invasions by 'outsiders'. James Harris believes he is infallible and his proofs irrefutable. I wonder if anyone has ever seen a more spectacular example of self-imposed blindness before? -- 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 ==== must ... start ... another ... item! (I have no hands, but I must type .-) > Here's a look at the methods highlighted in my paper Advanced > Polynomial Factorization where I bring down the number of symbols. > Notice if you can follow to the conclusion you're looking at an > *error* in taught mathematics. > > > No. Your conclusion is incorrect. You continue to insist that in the > polynomial factorization > > 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) > > with a1,a2,a3 algebraic integers, at least one of the a's is coprime to > 5. More recently, you have insisted that *all* of the a's are coprime to > 5. I show that none of them is coprime to 5, that there is, for each ai, > a factor r(-ai) which is (1) an algebraic integer, and (2) a factor of > 5. If you insist that all three are units, then you are compelled to > claim that 5 is a unit in the ring of algebraic integers. --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... For a 1000-year anglo-american hegemony? HEY, JIMMY; LET'S US and SU FIGHT -then-PM of England & Zbiggy http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX ==== Well I'm going to try and break it down even more to try and see if y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, which seem to confuse people. The ring is algebraic integers, and let me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be written like r+c, where r=0, or r varies as the polynomial variable varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), which equals 0, when m=0, so at least one of the a's must equal 0, when m=0. And to get that factor that is 25, you must have two a's that go to 0, when m=0. (Note: Some posters have gotten a lot of mileage out of calling that a degenerate case, but they were just fooling you into forgetting your basic algebra and what you know about polynomials. I think they did so deliberately as the math isn't complicated.) Now comes the question of what happens when m is NOT 0, and answering that question requires looking again at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11) and noticing that the constant term is 25(11), but you can divide off that 25 to get P(m)/25 which gives you a constant term that's 11. And 11 and 5 are coprime. That's very important. In fact, that's the *key* fact which should stick in your mind. So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = (5000m^3 - 600 m^2 - 126m + 11) I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor is a factor of the constant term, and in fact, it'd have a factor of the constant term that is 5. So why would you think that the factor of the constant term would move or change when m changes? Well it can't. Given that with 2 a_1 + 5, that 5 in there is a factor of the constant term of 25(5000m^3 - 600 m^2 - 126m + 11) you *still* have a factor of the constant term when 25 is separated off. But now your constant term is 11. That forces all the factors of 5 to go away from 2 a_1 + 5. Now you may wish for there to be someway for some factors to remain, but what actuall happens is you get (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of 5. But let's let m=0 again and see what happens now, as that gives (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11 and you'll notice I have 1 where I had 5 before, so it can all work. That is, the a's that went to 0 before will go to 0 again. Some posters have claimed otherwise which I've called voodoo math. Of course, if you think about it, you'll understand that they probably knew the truth and just lied to you as they didn't think you could figure it out, and probably didn't expect me to simplify in this way. Those people aren't your friends. You may trust them. You may admire them, but they clearly don't think much of you. So go back now and read posts from Arturo Magidin, Keith Ramsay, and Nora Baron among others, and ask yourself: What do they think of me, really? James Harris ==== [snip] > So go back now and read posts from Arturo Magidin, Keith Ramsay, and > Nora Baron among others, and ask yourself: What do they think of me, > really? Harrigance - the guy on the balcony: http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance1.jpg http://users.pandora.be/vdmoortel/dirk/Stuff/Arrogance2.jpg Dirk Vdm ==== [snip techno-babble] > So go back now and read posts from Arturo Magidin, Keith Ramsay, and > Nora Baron among others, and ask yourself: What do they think of me, > really? I can't speak for them, but I think you are an arrogant imbecile with an attention span of about a nanosecond, and a serious disorder of the cognitive faculties. -- 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 ==== [.snip.] >So go back now and read posts from Arturo Magidin, Keith Ramsay, and >Nora Baron among others, and ask yourself: What do they think of me, >really? What does it matter? Your arguments fall on their own. Now, go back and read posts from James Harris, and ask yourself: what does he think of me, really? And why? ====================================================================== 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 ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: > > Let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get > > P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and now you can see what the polynomial P(m) looks like without so > many symbols. > > Now from before where I had x, I *still* have that > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 - 75m^2 + 3m) > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims to the effect that you think these functions are continuous, so they certainly should take on non-algebraic integer values. Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It would make your work easier to read and much clearer. -- Will Twentyman ==== [snip] > Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It > would make your work easier to read and much clearer. ... and much easier to debunk, so he won't do it. Dirk Vdm ==== > > [snip] > > >>Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. > > > ... and much easier to debunk, so he won't do it. > Maybe if he did it he could find his own mistakes and not bother us with them. Or maybe it has something to do with his Object Math. I looked at his website but couldn't make much sense of it. Lack of clear definitions with examples, perhaps. -- Will Twentyman ==== >> >> [snip] >> >> >Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >would make your work easier to read and much clearer. >> >> >> ... and much easier to debunk, so he won't do it. >> Maybe if he did it he could find his own mistakes and not bother us with >them. Or maybe it has something to do with his Object Math. I looked >at his website but couldn't make much sense of it. Lack of clear >definitions with examples, perhaps. The shortcomings of his definition of object have been pointed out numerous times. While denying any problems repeatedly, he has nonetheless made subtle changes to it; yet, it still remains a problem. Since it is supposed to be the basis of his whole enterprise, that is enough to establish that it tumbles down. At one point, I tried to explain exactly why his definition of object was faulty, and why even if one were to give it its most liberal reading, the definition is such that it only includes integers. He never replied. http://groups.google.com/groups?selm=b8p0q2%241ahu%241%40agate.berkeley.edu James is not interested in finding his own mistakes. Never has been. In fact, he will avoid confronting them as long as humanly possible, if not longer. ====================================================================== 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 ==== [snip] >>Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. > ... and much easier to debunk, so he won't do it. > Maybe if he did it he could find his own mistakes and not bother us with > them. Or maybe it has something to do with his Object Math. I looked > at his website but couldn't make much sense of it. Lack of clear > definitions with examples, perhaps. IMO the only thing someone who suffers from this kind of illness is interested in, is attention - it doesn't matter if it is positive or negative. He's getting plenty of it, so I guess we are all very kind to James :-) Dirk Vdm ==== > > >[snip] >Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. >... and much easier to debunk, so he won't do it. >Maybe if he did it he could find his own mistakes and not bother us with >>them. Or maybe it has something to do with his Object Math. I looked >>at his website but couldn't make much sense of it. Lack of clear >>definitions with examples, perhaps. > > > The shortcomings of his definition of object have been pointed out > numerous times. While denying any problems repeatedly, he has > nonetheless made subtle changes to it; yet, it still remains a > problem. Since it is supposed to be the basis of his whole enterprise, > that is enough to establish that it tumbles down. > > At one point, I tried to explain exactly why his definition of object > was faulty, and why even if one were to give it its most liberal > reading, the definition is such that it only includes integers. He > never replied. > > http://groups.google.com/groups?selm=b8p0q2%241ahu%241%40agate.berkeley.edu > > James is not interested in finding his own mistakes. Never has > been. In fact, he will avoid confronting them as long as humanly > possible, if not longer. tendency to start new threads and simply reassert his claims is something I've become all too familiar with in the past month. I still haven't figured out why he wants all of us to join MSN's Amateur Math to just post direct links there. Right now I wonder how long it will be before I tire of reading his and the practice serves no purpose. Do you think there's a chance of getting him to play around with Set Theory or Computability Theory? -- Will Twentyman ==== > > >[snip] >Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. >... and much easier to debunk, so he won't do it. >Maybe if he did it he could find his own mistakes and not bother us with >>them. Or maybe it has something to do with his Object Math. I looked >>at his website but couldn't make much sense of it. Lack of clear >>definitions with examples, perhaps. > > > IMO the only thing someone who suffers from this kind of > illness is interested in, is attention - it doesn't matter if it is > positive or negative. He's getting plenty of it, so I guess we > are all very kind to James :-) > > Dirk Vdm If only he would reciprocate and give each of us as much attention as we give him. Answers to objections might be nice. Minus the generalized attacks on character would be nicer. -- Will Twentyman ==== >[snip] >Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. >... and much easier to debunk, so he won't do it. >Maybe if he did it he could find his own mistakes and not bother us with >>them. Or maybe it has something to do with his Object Math. I looked >>at his website but couldn't make much sense of it. Lack of clear >>definitions with examples, perhaps. > IMO the only thing someone who suffers from this kind of > illness is interested in, is attention - it doesn't matter if it is > positive or negative. He's getting plenty of it, so I guess we > are all very kind to James :-) Dirk Vdm If only he would reciprocate and give each of us as much attention as we > give him. Answers to objections might be nice. Minus the generalized > attacks on character would be nicer. That would be nice indeed. It would also be nice if mental disease would be effectively curable with proper medication. But don't let me put you off - enjoy it while you can :-) Dirk Vdm ==== >Well I'm going to try and break it down even more to try and see if >y'all will accept the mathematics: [...] So go back now and read posts from Arturo Magidin, Keith Ramsay, and >Nora Baron among others, and ask yourself: What do they think of me, >really? Fascinating argument: I know, I'll behave like a complete and total ass. Then people will hate me, and then when they say I'm wrong about the math I can explain that it's just because they hate me. Too bad nobody's stupid enough to buy it. >James Harris ************************ David C. Ullrich ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: > > Let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get > > P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and now you can see what the polynomial P(m) looks like without so > many symbols. > > Now from before where I had x, I *still* have that > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 - 75m^2 + 3m) That's like saying, if I write y=mx+b, that you think it should be y(x)=mx+b. > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > > Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims to the effect > that you think these functions are continuous, so they certainly should > take on non-algebraic integer values. The a's are roots of a monic polynomial with integer coefficients, given that m is an integer. Now if you wish to make m rational then you can explore continuity. > Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It > would make your work easier to read and much clearer. Why oh why do people ever insist on writing y=mx+b instead of y(x)=mx+b? It seems to me that you need to begin a crusade to make certain that such a horror is no longer committed. You can be a paragon of mathematical style on a quest to save the ignorant masses from this travesty of ever writing such a thing as y=mx+b as instead you make certain that people write y(x) = mx + b!!! James Harris ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so > many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). This is greatly simplified. I'm with you so far. Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), Well, no, I don't notice that. But I don't think it matters. which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > OK. So you've declared that two of the a's are 0 at m=0. Therefore, the third a=3 at m=0. I'll simplify if you don't mind; makes it easier for me: a_3 = b_3 + 3 (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(11). OK. Still with you. And all of a_1, a_2, b_3 are 0 at m=0. > (Note: Some posters have gotten a lot of mileage out of calling that a > degenerate case, but they were just fooling you into forgetting your > basic algebra and what you know about polynomials. I think they did > so deliberately as the math isn't complicated.) Now comes the question of what happens when m is NOT 0, and answering > that question requires looking again at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11) and noticing that the constant term is 25(11), but you can divide off > that 25 to get P(m)/25 which gives you a constant term that's 11. And > 11 and 5 are coprime. That's very important. In fact, that's the > *key* fact which should stick in your mind. So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = (5000m^3 - 600 m^2 - 126m + 11) > I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor > is a factor of the constant term, and in fact, it'd have a factor of > the constant term that is 5. So why would you think that the factor > of the constant term would move or change when m changes? Well it can't. I agree with you. I think I'm following. 5 is a factor of 5 for any value of m. Given that with 2 a_1 + 5, that 5 in there is a factor of the constant > term of 25(5000m^3 - 600 m^2 - 126m + 11) you *still* have a factor of the constant term when 25 is separated > off. But now your constant term is 11. That forces all the factors of 5 to go away from 2 a_1 + 5. Oops. Think you just lost me. You've shown that 5 is a factor of 2a_1 + 5 when a_1 is zero; not for a_1 non-zero. Could you clarify what you mean here? Now you may wish for there to be someway for some factors to remain, > but what actuall happens is you get (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of > 5. > Well, not quite. They independently proved that *none* of the a's were coprime to 5 for a completely different polynomial. I can't see the relevance. > But let's let m=0 again and see what happens now, as that gives (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11 and you'll notice I have 1 where I had 5 before, so it can all work. That is, the a's that went to 0 before will go to 0 again. Some posters have claimed otherwise which I've called voodoo math. You're being far too polite. All your doing is dividing by 5, and asserting that if a is zero then a/5 is zero. If people are claiming otherwise they're complete liars. Of course, if you think about it, you'll understand that they probably > knew the truth and just lied to you as they didn't think you could > figure it out, and probably didn't expect me to simplify in this way. > Yep, they're liars. > Those people aren't your friends. You may trust them. You may admire > them, but they clearly don't think much of you. > Yep, they're not my friends. > So go back now and read posts from Arturo Magidin, Keith Ramsay, and > Nora Baron among others, and ask yourself: What do they think of me, > really? > It's not my place to opine what they think of you. I haven't noticed any of them claiming that a/5 is not zero if a=0 however. But. There's no conclusion here. What are you saying? Given: (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? How about (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) = 25(5000m^3 - 600 m^2 - 126m + 11) By exactly the same reasoning, some of the c's are coprime to 7? I think there's something missing in all this. James Harris Phil Nicholson ==== > Maybe if he did it he could find his own mistakes and not bother us with them. What you call bothering, he proudly calls modern problem solving techniques. Take a look - http://groups.google.com/groups?selm=3c65f87.0304261407.7bb46534@posting.goo gle.com ==== > That would be nice indeed. It would also be nice if mental > disease would be effectively curable with proper medication. > But don't let me put you off - enjoy it while you can :-) I enjoy the entertainment JSH provides as much as anyone, but ... I have a nagging feeling that this is a modern intellectual version of the medieval court enjoying the antics of the fool, who was a cripple or a hunchback or otherwise handicapped. Gib ==== > That would be nice indeed. It would also be nice if mental > disease would be effectively curable with proper medication. > But don't let me put you off - enjoy it while you can :-) I enjoy the entertainment JSH provides as much as anyone, but ... I have > a nagging feeling that this is a modern intellectual version of the > medieval court enjoying the antics of the fool, who was a cripple or a > hunchback or otherwise handicapped. ... or the village idiot providing the necessary entertainment on a hot Sunday afternoon. Of course, that's very obvious. Dirk Vdm ==== > > >> That would be nice indeed. It would also be nice if mental >> disease would be effectively curable with proper medication. >> But don't let me put you off - enjoy it while you can :-) > > > I enjoy the entertainment JSH provides as much as anyone, but ... I have > a nagging feeling that this is a modern intellectual version of the > medieval court enjoying the antics of the fool, who was a cripple or a > hunchback or otherwise handicapped. That does seem to be the case. The difference is: in medieval courts, the fool was aware of his situation, as well as the true purpose of his being there. I think similar analogies could be made to people throwing rocks in the school yard. There are times when I have to wonder if we wouldn't perform a better service by responding only to his math and ignoring/deleting everything else that comes from his mouth. Maybe then he would learn to focus on the math as well. -- Will Twentyman ==== > > >>Maybe if he did it he could find his own mistakes and not bother us with them. > > > What you call bothering, he proudly calls modern problem solving techniques. Take a look - > http://groups.google.com/groups?selm=3c65f87.0304261407.7bb46534@posting.goog le.com > > > disturbing. -- Will Twentyman ==== > >Well I'm going to try and break it down even more to try and see if >y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and >looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, >which seem to confuse people. The ring is algebraic integers, and let >me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still >finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so >many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be >written like r+c, where r=0, or r varies as the polynomial variable >varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), >>I think this should be a_1(m) a_2(m) a_3(m) = 25(625 m^3 - 75m^2 + 3m) > > > That's like saying, if I write y=mx+b, that you think it should be > y(x)=mx+b. You say this as if y(x)=mx+b is uncommon. When teaching about linear functions in algebra, it's usually written as y=mx+b. In calculus, I've seen both. This is especially true when switching rapidly back and forth between y as a function of x and y evaluated at a particular value of x. >which equals 0, when m=0, so at least one of the a's must equal 0, >when m=0. And to get that factor that is 25, you must have two a's that go to 0, >when m=0. >>Why can't a_2(0) = 5, a_3(0) = -2/3? You have made claims to the effect >>that you think these functions are continuous, so they certainly should >>take on non-algebraic integer values. > > > The a's are roots of a monic polynomial with integer coefficients, > given that m is an integer. I missed that m is an integer. What does that have to do with the possible values of a_i(0) though? You've said the a's are roots of a monic polynomial, but also that they don't need to be polynomials themselves. As a result, I see no reason for assuming anything about the codomain of those functions. >>Also, why do you INSIST on not using the notation a_i(m) or a_i(0)? It >>would make your work easier to read and much clearer. > > > Why oh why do people ever insist on writing y=mx+b instead of > y(x)=mx+b? Because it's a shortcut AND then don't use the symbol y to refer to both mx+b and m*0+b with little warning as to which it will be in the next line. > It seems to me that you need to begin a crusade to make certain that > such a horror is no longer committed. I'll take that under advisement. Would you care to assist me? > You can be a paragon of mathematical style on a quest to save the > ignorant masses from this travesty of ever writing such a thing as > y=mx+b as instead you make certain that people write y(x) = mx + b!!! Can I use you as a model of the increased clarity that happens? Maybe a before and after snapshot. -- Will Twentyman ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so > many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). > > This is greatly simplified. I'm with you so far. Good. Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > Well, no, I don't notice that. But I don't think it matters. It's very important as it shows part of the relationship between the a's and m, which is fully defined by three expressions, but I'm focusing on one. Here the point is that at *least* one of the a's is equal to 0, when m=0. Remember the a's come from the factorization of P(m), where P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f), and I've put in some numbers for f, x, and u, as f=5, x=2, and u=1. The factorization is P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). Note to readers: I'm leaving everything in from the previous post, which makes things kind of bulky, but I'm hoping that in this case you'll be better served by having all the information readily available. Though I do fear there may be a bit of overload for some of you. which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. And to get that factor that is 25, you must have two a's that go to 0, > when m=0. OK. So you've declared that two of the a's are 0 at m=0. No that's required by the given expressions. > Therefore, the third a=3 at m=0. I'll simplify if you don't mind; makes it > easier for me: > a_3 = b_3 + 3 Whatever works for you is fine with me, as long as it works, and is mathematically correct. That looks ok as you're identifying part of the constant term for the factor 2a_3 + 5 as the full value is 11 for its constant term is 11. Here you have b_3 as part of the varying portion. > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = 25(11). > > OK. Still with you. And all of a_1, a_2, b_3 are 0 at m=0. Yup. > (Note: Some posters have gotten a lot of mileage out of calling that a > degenerate case, but they were just fooling you into forgetting your > basic algebra and what you know about polynomials. I think they did > so deliberately as the math isn't complicated.) Now comes the question of what happens when m is NOT 0, and answering > that question requires looking again at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11) and noticing that the constant term is 25(11), but you can divide off > that 25 to get P(m)/25 which gives you a constant term that's 11. And > 11 and 5 are coprime. That's very important. In fact, that's the > *key* fact which should stick in your mind. So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = (5000m^3 - 600 m^2 - 126m + 11) > I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor > is a factor of the constant term, and in fact, it'd have a factor of > the constant term that is 5. So why would you think that the factor > of the constant term would move or change when m changes? Well it can't. > > I agree with you. I think I'm following. 5 is a factor of 5 for any value of > m. Ok, let see what's next. Given that with 2 a_1 + 5, that 5 in there is a factor of the constant > term of 25(5000m^3 - 600 m^2 - 126m + 11) you *still* have a factor of the constant term when 25 is separated > off. But now your constant term is 11. That forces all the factors of 5 to go away from 2 a_1 + 5. > > Oops. Think you just lost me. You've shown that 5 is a factor of 2a_1 + 5 > when a_1 is zero; not for a_1 > non-zero. Could you clarify what you mean here? Hmmm...how about this explanation? Remember that the polynomial is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and you see that factor 25. Now imagine that I have the polynomial Q(m), where P(m) = 25 Q(m) so you have as a factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 and the question is, how does that factor of 25 divide out? Well, checking at Q(0), gives 11, so how can those factors of 25 divide through Q(m) in such a way as to give 11, at m=0? There's only one way, which gives you Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). If that bothers you, remember that at m=0, two of the a's equal 0. If it still bothers you, try and get everything to work some other way. And yes, it can be shown rigorously, but right now there's the problem of getting that feel for the math. Then you can go to the paper Advanced Polynomial Factorization. Now you may wish for there to be someway for some factors to remain, > but what actuall happens is you get (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of > 5. > Well, not quite. They independently proved that *none* of the a's were > coprime to 5 for a > completely different polynomial. I can't see the relevance. Nope. It turns out that they're making their claim for the polynomial I've given you, but in the past there have been *symbols* where I've now given numbers. > But let's let m=0 again and see what happens now, as that gives (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11 and you'll notice I have 1 where I had 5 before, so it can all work. That is, the a's that went to 0 before will go to 0 again. Some posters have claimed otherwise which I've called voodoo math. > > You're being far too polite. All your doing is dividing by 5, and asserting > that if a is zero then a/5 is > zero. If people are claiming otherwise they're complete liars. Hmmm...looks like I screwed up in my statement, as what I should have said is that they've made that claim when m does NOT equal 0. My apologies. The voodoo math has come in when m doesn't equal 0, as if suddenly constant factors shift around. Of course, if you think about it, you'll understand that they probably > knew the truth and just lied to you as they didn't think you could > figure it out, and probably didn't expect me to simplify in this way. Yep, they're liars. Well that is true as can be seen by the replies from some of them in several threads I created yesterday, like this one. My simplification removes areas for doubts, and if they weren't liars, they'd simply acknowledge the truth at this point. > Those people aren't your friends. You may trust them. You may admire > them, but they clearly don't think much of you. Yep, they're not my friends. Hmmm...so much agreement puts me somewhat at a loss for words. > So go back now and read posts from Arturo Magidin, Keith Ramsay, and > Nora Baron among others, and ask yourself: What do they think of me, > really? It's not my place to opine what they think of you. I haven't noticed any of > them claiming that a/5 is not zero > if a=0 however. That is true. > But. There's no conclusion here. > What are you saying? > > Given: > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) = > 25(5000m^3 - 600 m^2 - 126m + 11) > > That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, which is why the ring has problems. > How about > (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) = > 25(5000m^3 - 600 m^2 - 126m + 11) > > By exactly the same reasoning, some of the c's are coprime to 7? No. > I think there's something missing in all this. That's ok. My suggestion is to consider my reply and see if you have that same feeling. Remember what I'm doing is considering a polynomial factorization, with specific reference to its constant term *because* I'm using that as my anchor given that I'm using non-polynomial factors of that polynomial. That is, see the constant term as a lighthouse in the darkness. Or better yet, see the math here as requiring that you fly by your instruments, as otherwise you may spiral into the ground, figuratively speaking. James Harris ==== Fair bit of snippage. I've got some work to do now to follow this up properly. a's and m, which is fully defined by three expressions, but I'm > focusing on one. Here the point is that at *least* one of the a's is equal to 0, when > m=0. Which was this. And agreed. when m=0. OK. So you've declared that two of the a's are 0 at m=0. No that's required by the given expressions. > OK. I'm guessing that you're right. But I'll check this and get back if necessary. > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of > 5. > Well, not quite. They independently proved that *none* of the a's were > coprime to 5 for a > completely different polynomial. I can't see the relevance. Nope. It turns out that they're making their claim for the polynomial > I've given you, but in the past there have been *symbols* where I've > now given numbers. I don't recall any claims being made by others. I did see proofs made by others that for the polynomial 65x^3 -12x +1, none of the a's are coprime to 5. The polynomial under discussion here is 5000m^3 -600m^2 -126m +11. 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, > which is why the ring has problems. Surely not. What about when m=5? What are the a's in that case? How about > (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) = > 25(5000m^3 - 600 m^2 - 126m + 11) By exactly the same reasoning, some of the c's are coprime to 7? No. I believe you'd be right. James Harris ==== > Fair bit of snippage. I've got some work to do now to follow this up > properly. > > > Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), Well, no, I don't notice that. But I don't think it matters. > > Sorry, poorly worded on my part. I *DID* notice that a_1 a_2 a_3 *DOESN'T* > in fact equal > your main point. Yeah, you're correct. That should be a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m). It's very important as it shows part of the relationship between the > a's and m, which is fully defined by three expressions, but I'm > focusing on one. Here the point is that at *least* one of the a's is equal to 0, when > m=0. > > Which was this. And agreed. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. OK. So you've declared that two of the a's are 0 at m=0. No that's required by the given expressions. OK. I'm guessing that you're right. But I'll check this and get back if > necessary. Well I guess you're thinking that I just picked the possibility that two of the a's are 0, and maybe it's possible for, say, only one of the a's to equal 0. But if you do that then you can only get a factor of 5 that is 5, for the result. > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of > 5. > Well, not quite. They independently proved that *none* of the a's were > coprime to 5 for a > completely different polynomial. I can't see the relevance. Nope. It turns out that they're making their claim for the polynomial > I've given you, but in the past there have been *symbols* where I've > now given numbers. > > I don't recall any claims being made by others. I did see proofs made by > others that > for the polynomial 65x^3 -12x +1, none of the a's are coprime to 5. > The polynomial under discussion here is 5000m^3 -600m^2 -126m +11. Then you haven't looked closely. The claim is against the polynomial factorization (v^3+1)x^3 - 3v xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and readers can get from there to 25(5000m^3 -600m^2 -126m +11) easily enough by plugging in f=5, x=2, y=5. > 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, > which is why the ring has problems. > > Surely not. What about when m=5? What are the a's in that case? The problem is with the ring of algebraic integers as I've explained. Consider the following which you deleted out from my previous post: Hmmm...how about this explanation? Remember that the polynomial is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and you see that factor 25. Now imagine that I have the polynomial Q(m), where P(m) = 25 Q(m) so you have as a factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 and the question is, how does that factor of 25 divide out? Well, checking at Q(0), gives 11, so how can those factors of 25 divide through Q(m) in such a way as to give 11, at m=0? There's only one way, which gives you Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). If that bothers you, remember that at m=0, two of the a's equal 0. If it still bothers you, try and get everything to work some other way. How about > (2 c_1 + 7)(2 c_2 + 7)(2 c_3 + 7) = > 25(5000m^3 - 600 m^2 - 126m + 11) By exactly the same reasoning, some of the c's are coprime to 7? No. > I believe you'd be right. The math should be simple now that so many symbols have been replaced with numbers. The issue is an error in taught mathematics revealed by what I've shown here. My position is that mathematicians should be diligent in teaching the truth, and should acknowledge an error, so that it is no longer taught. James Harris ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: > > Let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get > > P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and now you can see what the polynomial P(m) looks like without so > many symbols. > > Now from before where I had x, I *still* have that > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > Interesting. When the x's were left unspecified, the form of the factorization was P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5). Now that you have substituted in x = 2, it is no longer a factorization of a polynomial in x. It is just a factorization as a product of three numbers. So 2*a_1 + 5, for example, is just an algebraic integer. The factorization is [1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5). So then what you are asserting below is that if you factor the number P(m) as in [1], then one of a1, a2, or a3 must be coprime to 5. Right? Let's take m = 1. Then P(m) = 25 * 4285. This can be factored as 5 * 5 * 4285, which yields a1 = 0, a2 = 0, and a3 = 2140. None of these is coprime to 5. End of story. Don't like a1 = a2 = 0 ? Other things work too - e.g., a1 = -5, a2 = -5, a3 = 2140. None coprime to 5, as before. Read on, however - > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > > (Note: Some posters have gotten a lot of mileage out of calling that a > degenerate case, but they were just fooling you into forgetting your > basic algebra and what you know about polynomials. I think they did > so deliberately as the math isn't complicated.) > > Now comes the question of what happens when m is NOT 0, and answering > that question requires looking again at > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11) > > and noticing that the constant term is 25(11), but you can divide off > that 25 to get P(m)/25 which gives you a constant term that's 11. And > 11 and 5 are coprime. That's very important. In fact, that's the > *key* fact which should stick in your mind. > > So now looking at > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = > > (5000m^3 - 600 m^2 - 126m + 11) > > > I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor > is a factor of the constant term, and in fact, it'd have a factor of > the constant term that is 5. So why would you think that the factor > of the constant term would move or change when m changes? > > Well it can't. > > Given that with 2 a_1 + 5, that 5 in there is a factor of the constant > term of > > 25(5000m^3 - 600 m^2 - 126m + 11) > > you *still* have a factor of the constant term when 25 is separated > off. > > But now your constant term is 11. > > That forces all the factors of 5 to go away from 2 a_1 + 5. > > Now you may wish for there to be someway for some factors to remain, > but what actuall happens is you get > > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = > > (5000m^3 - 600 m^2 - 126m + 11) > > while posters have argued that *all* the a's would have some factor of > 5. > Yep, sure. See above. With actual numbers! > But let's let m=0 again and see what happens now, as that gives > > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = 11 > > and you'll notice I have 1 where I had 5 before, so it can all work. > > That is, the a's that went to 0 before will go to 0 again. > > Some posters have claimed otherwise which I've called voodoo math. > > Of course, if you think about it, you'll understand that they probably > knew the truth and just lied to you as they didn't think you could > figure it out, and probably didn't expect me to simplify in this way. > > Those people aren't your friends. You may trust them. You may admire > them, but they clearly don't think much of you. > > So go back now and read posts from Arturo Magidin, Keith Ramsay, and > Nora Baron among others, and ask yourself: What do they think of me, > really? > > > James Harris > It's almost funny until you realize why there was so much frenetic > energy, and why these people are so desperate to hide the math that > they continually delete it out: > > The mathematics is rather simple, the algebra is basic, but the > conclusion is dramatic. > > So let's say you were facing people who needed to keep people from > looking closely, and every time you tried to simplify and show > details, they'd jump in and hide details and make things more > convoluted? > > Well you might do what I've done today and stretch them out. > What this means is: start a lot of new threads in the hope that people won't notice that the problems in the old ones were not answered. Mr. Harris imagines that he is playing to a huge audience of silent lurkers, perhaps even math undergrads. I am sure they will be impressed by this clever evasive maneuver! See below for what happens when people look closely and show details ... be sure to note where I have hidden details and made things more convoluted ... where I was so desperate to hide the math that I continually deleted it out ... > What I'm doing is not very complicated as I'm using the extra symbols > to factor a polynomial into non-polynomial factors. > > Now that's a fascinating idea for factoring polynomials that I guess > is new to the math world. > > And yes, bad apples can take advantage when there's something new. > > What is a surprise though is that mathematical society can be taken in > by people like Arturo Magidin or Nora Baron, when they're posting so > desperately, so quickly when stretched out, that they can barely get > to the math, or say things that are clearly false. > > But what if they never really believed in mathematics? > > What if for them it is a fashion show? > > Remember mathematics can be about appearance for some people. They > might have gotten a lot of mileage out of being able to put down math > that looked good. > > That is, you may have people who are cons in your midst who are used > to playing a game upon other mathematicians, and now they're > trapped--stretched out--by a lot of threads and math where I did the > sudden move of putting in numbers where once there were symbols. > > It seems to me that in mathematics there is a certain respect that is > to be given to ideas, logic, and mathematical truth. Sure being a > rather large society it's not surprising that corrupt people can slip > in, and maybe get some stature, or get through a Ph.d program. > > But in society when the corrupt people reveal themselves clearly by > finally pushing beyond obvious limits--like denying basic algebra--it > becomes time to clean house. Look over those threads, look at their > desperation, and their contempt for algebra and your algebra > knowledge, then please ask yourselves how it cannot now be that time. > > > James Harris I looked a little more closely at Advanced Polynomial Factorization. It turns out that, in addition to basic conceptual problems, it also has careless mistakes in algebra. Start with P(m) = f^2 *((m^3*f^4 - 3*m^2*f^2 + 3*m)*x^3 - 3*(-1 + m*f^2)*x*u^2 + u^3*f), as you give early in Section 2 of APF. You note that P(m) has a factor of f^2. Nothing wrong with this particular formula. However, on the last page, when you substitute in m = 1, f = sqrt(5), and u = 1, you do NOT obtain 65*x^3 - 12*x + 1. Instead you get 65*x^3 - 60*x + 5*sqrt(5). Now if this is factored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1), it obviously doesn't work: 1*1*1 is not equal to 5*sqrt(5). Back on the second page, in a parenthetical note at the bottom, you say: Note: the a's are roots of a monic polynomial with algebraic integer coefficients so they are algebraic integers. Not true at all. Note that the polynomial P(m), as given above, if considered as a polynomial in x, does not have constant term equal to 1 or -1. The constant term with respect to x is u^3*f^3. Thus the polynomial of which the a's are roots is not monic. I am not sure what you actually intended. This is carelessly written, even independent of the unsoundness of the ideas. Parts of your posts from yesterday on this topic also contained similar algebraic errors. I think it was still your intention in APF however to show that if 65*x^3 - 12*x + 1 were factored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1), where a1, a2, and a3 are algebraic integers, then one of a1, a2, or a3 is coprime to 5. No matter how you cut it, that is still false, as shown by the proofs given in other threads by me and W. Dale Hall. Yes, you have started other several new threads today, but you have explicitly avoided a direct response to either me or Dale Hall on this. Even if you are unable to perform the computations that Dale proposes and are incapable of understanding the proof that I presented, you cannot ignore the simple algebraic errors in APF. Something there has to change. Nora B. ==== > Fair bit of snippage. I've got some work to do now to follow this up > properly. And again. > 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, > which is why the ring has problems. > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. It's actually fascinating as it's a bizarre result, and I think you realize that I'm correct as you *again* deleted out the following wonderful simplification which shows clearly that I'm right. The problem is with the ring of algebraic integers as I've explained. Consider the following which you deleted yet again from my previous post: Hmmm...how about this explanation? Remember that the polynomial is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and you see that factor 25. Now imagine that I have the polynomial Q(m), where P(m) = 25 Q(m) so you have as a factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 and the question is, how does that factor of 25 divide out? Well, checking at Q(0), gives 11, so how can those factors of 25 divide through Q(m) in such a way as to give 11, at m=0? There's only one way, which gives you Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). If that bothers you, remember that at m=0, two of the a's equal 0. If it still bothers you, try and get everything to work some other way. Readers who want some sense of what I'm up against should read the various posts in this thread from people trying to escape that obvious conclusion. Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. That's trivial enough and has been admitted, but posters seem to want the constants to move around if m doesn't equal 0, which is what I call voodoo math. James Harris ==== a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. Yuck. Unfortunately in my haste I neglected to refute the claim, as it is in fact NOT true that none of the a's are coprime to 5, as in fact the fascinating as it's a bizarre result is that they ALL are coprime to 5 in the ring of algebraic integers. > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), which gives the reader a chance to see P(m) with only the m left as a symbol > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? I want readers to consider how many posters have apparently attempted to make them believe that the factor of f^2, here 25, was in some sense welded into the expression, when in fact it's a factor of P(m) such that I can write P(m) = 25 Q(m) as I've done here, or P(m) = f^2 Q(m), in general. > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. And remember that posters in trying to refute have continuously called m=0 a degenerate case when in fact they need you to ignore the obvious. That is, given that f^2 is this factor that's multipled times P(m) as it is, then of course, it can be separated off, and it's not so complicated and extraordinary that you need Galois Theory or any of all that extra technicality. > If it still bothers you, try and get everything to work some other > way. Indeed. > Readers who want some sense of what I'm up against should read the > various posts in this thread from people trying to escape that obvious > conclusion. > > Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. > > That's trivial enough and has been admitted, but posters seem to want > the constants to move around if m doesn't equal 0, which is what I > call voodoo math. But what's at stake is the *belief* that mathematicians could not have taught erroneous mathematics for over a hundred years. And in fact the flawed mathematics is STILL in the textbooks. It is rather weak-minded that mathematicians would continue to fight over this, but I'm unfortunately familiar with how weak people often are. The truth can be a bitter pill, expecially for fakes fighting to preserve the lie. James Harris ==== [snip] > The truth can be a bitter pill, expecially for fakes fighting to > preserve the lie. James Harris Well, enjoy the pill! You've earned it. The argument you have given is erroneous. It leads to a false conclusion. The problem is not any contradiction in math, or incompleteness of any ring, but in the errors you made in your argument. You are now flogging a dead horse. Please mount the beast and ride off into the wilderness where you belong. -- 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 ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. Perhaps you mean that none of the a's are coprime to 5? >Which is trivially obvious and uninteresting after all. >Sorry to have bothered you. >>It's actually fascinating as it's a bizarre result, and I think you >>realize that I'm correct as you *again* deleted out the following >>wonderful simplification which shows clearly that I'm right. > > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > Pretty much like the a's in the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) are all coprime to 5? Why should your argument work in the one case, and not in the other? Oh, in case you forgot: I have shown explicit factors that are shared by 5, and *EACH* of the a's in this example. As a result, 5 CANNOT be coprime to any of these a's. Why do you flee from the case you should be fixing? > >>The problem is with the ring of algebraic integers as I've explained. >> You have not proven a thing. >>Consider the following which you deleted yet again from my previous >>post: >>Hmmm...how about this explanation? >>Remember that the polynomial is >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > > which gives the reader a chance to see P(m) with only the m left as > a symbol > > >>and you see that factor 25. Now imagine that I have the polynomial >>Q(m), where >> P(m) = 25 Q(m) So Q(m) = 5000 m^3 - 600 m^2 - 126 m + 11? >>so you have as a factorization >> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 Rather, a factorization: P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)? What result guarantees that such a factorization is possible? >>and the question is, how does that factor of 25 divide out? > What happened to the extra dollar? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > I'm sorry, I was thinking about your other case (I mentioned it above) where you said all the a's were coprime to 5, yet I *showed* you the common factors. I was wondering why you wouldn't even address that failure of your method, rather than moving to another (most likely incorrect) case? Why wouldn't a person address what can be computed directly, rather than running to another case where the computations are almost surely more complicated? Is it the very intractibility of the new problem that makes you safe? If no one can tell you're wrong, does that make you right? What if you could be shown wrong with Galois theory? I see in your paper that you call such applications *overinterpretations* of Galois theory, although you have admitted that you know less about Galois theory than virtually any human, & you prove that point by pointing out (as though itwere some flaw) that Galois theory typically deals with fields, where you're working with rings. > >>Well, checking at Q(0), gives 11, so how can those factors of 25 >>divide through Q(m) in such a way as to give 11, at m=0? >>There's only one way, which gives you >> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). >>If that bothers you, remember that at m=0, two of the a's equal 0. > > > And remember that posters in trying to refute have continuously called > m=0 a degenerate case when in fact they need you to ignore the > obvious. > The case is degenerate in that the degree of the polynomial drops from 3 to 1 when m=0. Frequently, calculations change in degenerate cases. > That is, given that f^2 is this factor that's multipled times P(m) as > it is, then of course, it can be separated off, and it's not so > complicated and extraordinary that you need Galois Theory or any of > all that extra technicality. > Do you mean divided? Why say separated? Is it necessary to use your own private language? > >>If it still bothers you, try and get everything to work some other >>way. > > > Indeed. > > >>Readers who want some sense of what I'm up against should read the >>various posts in this thread from people trying to escape that obvious >>conclusion. >>Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. >> What do you mean? My PC works when m=0, and it has nothing whatsoever to do with those a's. My car works, and it is totally unaware of the values of m and the a's. I work, for that matter, and I haven't worried about your m and a's, not ever. >>That's trivial enough and has been admitted, but posters seem to want >>the constants to move around if m doesn't equal 0, which is what I >>call voodoo math. > What has been admitted? > > But what's at stake is the *belief* that mathematicians could not have > taught erroneous mathematics for over a hundred years. > Could you please address your failure to fix the claim you had about the above polynomial factorization? Please make correct assertions before you claim that mathematicians have taught erroneous mathematics. > And in fact the flawed mathematics is STILL in the textbooks. > > It is rather weak-minded that mathematicians would continue to fight > over this, but I'm unfortunately familiar with how weak people often > are. You *are* familiar with how weak a *certain person* often is. You are the source of dishonesty, error, and cynicism in these threads, yet you make claims of being entirely honest and forthright. Doesn't your practice of hypocrisy cause you any discomfort? What do your parents think of your hypocrisy? Are they hypocrites as well? > > The truth can be a bitter pill, expecially for fakes fighting to > preserve the lie. > Fighting to preserve the lie. That's how JSH refers to a person pointing out that he [JSH] is wrong; pointing this fact out for days on end, and with algebra that is as simple and direct as anyone could imagine. Fakes. That's who can show what JSH says doesn't exist, and who can point those non-existent things out in a simple fashion. > > James Harris Perhaps our esteemed Mister Harris is warning everyone about himself. Dale. ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > which gives the reader a chance to see P(m) with only the m left as > a symbol > Yes! I like this example. See below. > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions of m. When m = 0, two of them are 0. However when m is not zero, none of them are zero. So how do their values when m = 0 determine what they will be when m <> 0 ? > And remember that posters in trying to refute have continuously called > m=0 a degenerate case when in fact they need you to ignore the > obvious. > > That is, given that f^2 is this factor that's multipled times P(m) as > it is, then of course, it can be separated off, and it's not so > complicated and extraordinary that you need Galois Theory or any of > all that extra technicality. > > If it still bothers you, try and get everything to work some other > way. > > Indeed. > > Readers who want some sense of what I'm up against should read the > various posts in this thread from people trying to escape that obvious > conclusion. > > Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. > > That's trivial enough and has been admitted, but posters seem to want > the constants to move around if m doesn't equal 0, which is what I > call voodoo math. > > But what's at stake is the *belief* that mathematicians could not have > taught erroneous mathematics for over a hundred years. > Let's go back to your original polynomial, P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and assume it is factored *as you propose* in the form [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). Now instead of m = 0, let's try m = 1: P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. This can be factored in the form [*] by letting a_1 = -5, a_2 = -5, and a_3 = 2140. Note that EACH ONE of these is divisible by 5: NONE ARE COPRIME TO 5. Clearly m = 0 is a special case. It is different in an essential way from m = 1 and other nonzero values of m. Now, what's your explanation? Nora B. > And in fact the flawed mathematics is STILL in the textbooks. > > It is rather weak-minded that mathematicians would continue to fight > over this, but I'm unfortunately familiar with how weak people often > are. > > The truth can be a bitter pill, expecially for fakes fighting to > preserve the lie. > > > James Harris ==== > But what's at stake is the *belief* that mathematicians could not have > taught erroneous mathematics for over a hundred years. No, it's not at stake. It's not even a serious consideration for anyone but 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 ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > which gives the reader a chance to see P(m) with only the m left as > a symbol > > > Yes! I like this example. See below. Fascinating. > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. > > > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions > of m. When m = 0, two of them are 0. However when m is not zero, > none of them are zero. So how do their values when m = 0 > determine what they will be when m <> 0 ? Well you have P(m) = 25 Q(m) and that factor 25, which has been the focus of all the arguing. Now considering Q(m) it turns out that looking at the factorization Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 the a's are in the way, and it's not clear how you divide out. Now you *know* that the 25 divides through some way, and it's clear that whatever way that is would be true for any m, just like with 2(x^2 + 2x + 1) = (x+1)(2x+2) when you divide off 2, you do it for all x. For readers, people like Nora Baron have basically been arguing that you can have something like Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2) where h_1 h_2 h_3 = 5, and each is not a unit. But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25, when in fact Q(0)=11. Given that rigorous fact they claim that m=0 is a some kind of special case and try to cast doubt on the result, which is an attack on algebra that I call voodoo math. Surprisingly they've been quite successful from what I've gathered in convincing people, which makes you wonder about people's understanding of algebra. Because from algebra, it turns out that you can just set m=0, to figure out how that 25 divides out, as I've done. > And remember that posters in trying to refute have continuously called > m=0 a degenerate case when in fact they need you to ignore the > obvious. > > That is, given that f^2 is this factor that's multipled times P(m) as > it is, then of course, it can be separated off, and it's not so > complicated and extraordinary that you need Galois Theory or any of > all that extra technicality. > > If it still bothers you, try and get everything to work some other > way. > > Indeed. > > Readers who want some sense of what I'm up against should read the > various posts in this thread from people trying to escape that obvious > conclusion. > > Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. > > That's trivial enough and has been admitted, but posters seem to want > the constants to move around if m doesn't equal 0, which is what I > call voodoo math. > > But what's at stake is the *belief* that mathematicians could not have > taught erroneous mathematics for over a hundred years. > > > Let's go back to your original polynomial, > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and assume it is factored *as you propose* in the form > > [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now instead of m = 0, let's try m = 1: Ok. > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. > > This can be factored in the form [*] by letting > > a_1 = -5, a_2 = -5, and a_3 = 2140. Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that seem to be just a tad bit strange to you? > Note that EACH ONE of these is divisible by 5: Ok, you *pick* values from the a's as if they're not defined by a cubic, which they are, and your picked values are supposed to prove something? > NONE ARE COPRIME TO 5. So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? Remember you simply tossed those numbers out there, but there *is* a way to calculate the a's so your assertion can be tested, as they are roots to a cubic. Now if you can do that Nora Baron then clearly I made some kind of mistake and there's no more room for me to argue. So why don't you go back to the cubic which defines the a's, stick in all the values and see if the a's come out as you have above, and then it's over. > Clearly m = 0 is a special case. It is different > in an essential way from m = 1 and other nonzero > values of m. Clearly you haven't proven your assertion. > Now, what's your explanation? > > Nora B. Well I've explained above, and again, of course, you can't just *pick* values for the a's for a particular m, as their values are set rigorously. What I'm curious about are the people who believe Nora Baron had a valid point. Speak up, and don't be shy as I suspect you may believe she still has a valid point and I need to understand why she's so effective in convincing people. Come on, speak up, do you think I answered her objections? Do you believe they were valid in the first place? James Harris ==== > > >>and you see that factor 25. Now imagine that I have the polynomial >>Q(m), where >> P(m) = 25 Q(m) >>so you have as a factorization >> Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 >>and the question is, how does that factor of 25 divide out? I want readers to consider how many posters have apparently attempted >to make them believe that the factor of f^2, here 25, was in some >sense welded into the expression, when in fact it's a factor of P(m) >such that I can write P(m) = 25 Q(m) as I've done here, or P(m) = f^2 Q(m), in general. >>Well, checking at Q(0), gives 11, so how can those factors of 25 >>divide through Q(m) in such a way as to give 11, at m=0? >>There's only one way, which gives you >> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). >>If that bothers you, remember that at m=0, two of the a's equal 0. > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions >>of m. When m = 0, two of them are 0. However when m is not zero, >>none of them are zero. So how do their values when m = 0 >>determine what they will be when m <> 0 ? > > > Well you have P(m) = 25 Q(m) and that factor 25, which has been the > focus of all the arguing. Now considering Q(m) it turns out that > looking at the factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > the a's are in the way, and it's not clear how you divide out. > > Now you *know* that the 25 divides through some way, and it's clear > that whatever way that is would be true for any m, just like with > > 2(x^2 + 2x + 1) = (x+1)(2x+2) > > when you divide off 2, you do it for all x. > > For readers, people like Nora Baron have basically been arguing that > you can have something like > > Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2) > > where h_1 h_2 h_3 = 5, and each is not a unit. > > But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25, when in > fact Q(0)=11. > > Given that rigorous fact they claim that m=0 is a some kind of special > case and try to cast doubt on the result, which is an attack on > algebra that I call voodoo math. > > Surprisingly they've been quite successful from what I've gathered in > convincing people, which makes you wonder about people's understanding > of algebra. > > Because from algebra, it turns out that you can just set m=0, to > figure out how that 25 divides out, as I've done. > > >And remember that posters in trying to refute have continuously called >m=0 a degenerate case when in fact they need you to ignore the >obvious. That is, given that f^2 is this factor that's multipled times P(m) as >it is, then of course, it can be separated off, and it's not so >complicated and extraordinary that you need Galois Theory or any of >all that extra technicality. >>If it still bothers you, try and get everything to work some other >>way. Indeed. >>Readers who want some sense of what I'm up against should read the >>various posts in this thread from people trying to escape that obvious >>conclusion. >>Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. >>That's trivial enough and has been admitted, but posters seem to want >>the constants to move around if m doesn't equal 0, which is what I >>call voodoo math. But what's at stake is the *belief* that mathematicians could not have >taught erroneous mathematics for over a hundred years. > Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >>and assume it is factored *as you propose* in the form >>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >> Now instead of m = 0, let's try m = 1: > > > Ok. > > >> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. >> This can be factored in the form [*] by letting >> a_1 = -5, a_2 = -5, and a_3 = 2140. > > > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? You haven't provided a way to define what the a_i(m) are, so of course she can simply select values that work. >> Note that EACH ONE of these is divisible by 5: > > Ok, you *pick* values from the a's as if they're not defined by a > cubic, which they are, and your picked values are supposed to prove > something? They AREN'T defined by a cubic. They are constrained by a cubic. If you want them to be defined by a cubic, you'll have to say more about how they are supposed to be constructed. >> NONE ARE COPRIME TO 5. > > > So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > Remember you simply tossed those numbers out there, but there *is* a > way to calculate the a's so your assertion can be tested, as they are > roots to a cubic. Not that you've provided. If you have a way in mind, please provide it to us and include it in your paper. You've only claimed this things exist and have certain properties. If you want to change them, so be it. But that may cause more problems for you. > Now if you can do that Nora Baron then clearly I made some kind of > mistake and there's no more room for me to argue. room for you to argue. > So why don't you go back to the cubic which defines the a's, stick in > all the values and see if the a's come out as you have above, and then > it's over. For each value of m, she can define a_i(m) to be whatever three values makes both sides equal. You've given nothing more specific about the values to require more. >> Now, what's your explanation? >> Nora B. > > > Well I've explained above, and again, of course, you can't just *pick* > values for the a's for a particular m, as their values are set > rigorously. What I'm curious about are the people who believe Nora > Baron had a valid point. Where have you established how to rigorously compute them. What are they? > Speak up, and don't be shy as I suspect you may believe she still has > a valid point and I need to understand why she's so effective in > convincing people. Because you have not stated all of the conditions you want to exist. Once you do that, you will lose the results you are trying to achieve. > Come on, speak up, do you think I answered her objections? Do you > believe they were valid in the first place? I think you didn't understand her objections. -- Will Twentyman ==== the only answer is, JSH is always-and-only every where dense. > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and assume it is factored *as you propose* in the form > > [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now instead of m = 0, let's try m = 1: > > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. > > This can be factored in the form [*] by letting > > a_1 = -5, a_2 = -5, and a_3 = 2140. > > Note that EACH ONE of these is divisible by 5: > > NONE ARE COPRIME TO 5. > > Clearly m = 0 is a special case. It is different > in an essential way from m = 1 and other nonzero > values of m. > > Now, what's your explanation? --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > which gives the reader a chance to see P(m) with only the m left as > a symbol > > > Yes! I like this example. See below. > > Fascinating. > > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. > > > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions > of m. When m = 0, two of them are 0. However when m is not zero, > none of them are zero. So how do their values when m = 0 > determine what they will be when m <> 0 ? > > Well you have P(m) = 25 Q(m) and that factor 25, which has been the > focus of all the arguing. Now considering Q(m) it turns out that > looking at the factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > the a's are in the way, and it's not clear how you divide out. > > Now you *know* that the 25 divides through some way, and it's clear > that whatever way that is would be true for any m, just like with > > 2(x^2 + 2x + 1) = (x+1)(2x+2) > > when you divide off 2, you do it for all x. > > For readers, people like Nora Baron have basically been arguing that > you can have something like > > Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2) > > where h_1 h_2 h_3 = 5, and each is not a unit. > > But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25, when in > fact Q(0)=11. > > Given that rigorous fact they claim that m=0 is a some kind of special > case and try to cast doubt on the result, which is an attack on > algebra that I call voodoo math. > > Surprisingly they've been quite successful from what I've gathered in > convincing people, which makes you wonder about people's understanding > of algebra. > > Because from algebra, it turns out that you can just set m=0, to > figure out how that 25 divides out, as I've done. > No voodoo at all. What you would like to have here is that the same a_1, a_2, and a_3 work when m = 0 as when m <> 0. But you know, and everyone else knows, that that is totally impossible. Why? Because when m = 0, two of the a's must be zero. When m <> 0, NONE of the a's can be zero. It's real simple. The a's are functions of m. And I don't mean CONSTANT functions. Got that? Different m's give different a's. You believe that because a_1 is divisible by 5 when m = 0, it must be divisible by 5 for other values of m also. It is some kind of hunch that you have, based on intuition. It seems like a nice parallel construction. You think it's obvious. It isn't. In fact it is false. Your intuition this time is wrong. Your hunch has been disproved. What the a's are when m = 0 is a special case. It does NOT determine what they are when m <> 0. Sure, when m = 0, you can assume that a_1 and a_2 are zero. Can you can say that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of course you can - they are both 0! After all, ANY number divides 0. For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are algebraic integers also, since both are 0. Does that mean that 137 must divide two of the a's, when m <> 0 ???? By your logic, YES! Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS OF M. WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT BE ASSUMED TO BE TRUE WHEN M <> 0. ***** Why is this so hard for you to understand? Why do you keep trying to act like I am pulling some kind of trick when I say this, or pretending that I am lying? > And remember that posters in trying to refute have continuously called > m=0 a degenerate case when in fact they need you to ignore the > obvious. > > That is, given that f^2 is this factor that's multipled times P(m) as > it is, then of course, it can be separated off, and it's not so > complicated and extraordinary that you need Galois Theory or any of > all that extra technicality. > > If it still bothers you, try and get everything to work some other > way. > > Indeed. > > Readers who want some sense of what I'm up against should read the > various posts in this thread from people trying to escape that obvious > conclusion. > > Above the *only* thing that works at m=0, is a_1=a_2=0, while a_3 = 3. > > That's trivial enough and has been admitted, but posters seem to want > the constants to move around if m doesn't equal 0, which is what I > call voodoo math. > > But what's at stake is the *belief* that mathematicians could not have > taught erroneous mathematics for over a hundred years. > > > Let's go back to your original polynomial, > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and assume it is factored *as you propose* in the form > > [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now instead of m = 0, let's try m = 1: > > Ok. > > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. > > This can be factored in the form [*] by letting > > a_1 = -5, a_2 = -5, and a_3 = 2140. > > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? > Nope. It was your idea, not mine, to replace the x's with a constant value of 2. When you did that you were no longer factoring a polynomial. You are just factoring an ordinary number. There are no hidden rules that say it must be factored as a polynomial. You can't have it both ways. You want it factored as a polynomial, leave the x's in there (and see below). You want it factored as an ordinary number (remember? VERY much simplified) then one must assume that any factorization is legitimate, and you get my example above. So let's go back to assuming you want it factored as a polynomial. Of course you know that both I and W. Dale Hall have proofs that that is not going to work. You have not provided a valid objection to my proof, and you have not provided any response whatsoever to what Dale did. I don't see how you can. It's a simple computation. You do the arithmetic, and Dale is either right or he is wrong. You don't have to understand ANYTHING. There is no wool to be pulled over anyone's eyes. Try it and see. See also below. > Note that EACH ONE of these is divisible by 5: > > Ok, you *pick* values from the a's as if they're not defined by a > cubic, which they are, and your picked values are supposed to prove > something? > Yes. It proves what you said was wrong. You said that one of the a's had to be coprime to 5. Clearly, totally, unambiguously false. > NONE ARE COPRIME TO 5. > > So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > Remember you simply tossed those numbers out there, but there *is* a > way to calculate the a's so your assertion can be tested, as they are > roots to a cubic. > They *would* be roots of a cubic in the variable x, but you replaced x by 2. At that point there is no longer a polynomial. The restriction that they be roots of a cubic is gone. It was your idea, not mine, to oversimplify things and make the substitution. Remember: if you want to go back to assuming that a_1, a_2, and a_3 are roots of a cubic in the variable x, to continue to maintain what you think is true, you are going to have to find an error in the proof I have presented (and is appended below). > Now if you can do that Nora Baron then clearly I made some kind of > mistake and there's no more room for me to argue. > Yep. You should not have substituted x = 2. You want it factored as a polynomial, you had better leave it a polynomial and not try to oversimplify. You have shown confusion previously over what the difference between a polynomial and the evaluation of that polynomial when you choose a particular value for x. The moral here is, this little example should show you pretty convincingly that when people tried to get you to understand that difference, they were not just playing with semantics. There is another more general moral here, and you have run afoul of this one many a time in the past also: Examples are not a substitute for proofs. And do NOT oversimplify when you are trying to do mathematics. > So why don't you go back to the cubic which defines the a's, stick in > all the values and see if the a's come out as you have above, and then > it's over. > I will do better than that. I will reproduce my proof that your claim that one of the a's is coprime to 5 is wrong, but I will do it in more generality. That way I will be heading off any examples you might try to come up with in the future. What you want is a consequence of the following claim, which you have stated elsewhere: CLAIM: It is possible to find a monic polynomial of degree 3, with integer coefficients, irreducible over the rationals. and with roots a1, a2, and a3, such that at least one of a1, a2, or a3 is coprime, in the ring of algebraic integers, to a prime factor of the constant term of the polynomial. Right? You have claimed as recently as yesterday. DISPROOF OF CLAIM: Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by definition, a1, a2, and a3 are algebraic integers. The claim is that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory***, there exists an automorphism F12 of the field of algebraic numbers such that: 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(a1) = a2. Thus one obtains: r'*u2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts the CLAIM made above. Please feel free to point out any errors in the proof I just gave. ***: Reference on automorphisms: http://www.math.niu.edu/~beachy/aaol/galois.html or the excellent textbook Abstract Algebra, by John Beachy and William D. Blair > Clearly m = 0 is a special case. It is different > in an essential way from m = 1 and other nonzero > values of m. > > Clearly you haven't proven your assertion. > > Now, what's your explanation? > > Nora B. > > Well I've explained above, and again, of course, you can't just *pick* > values for the a's for a particular m, as their values are set > rigorously. If you can just *pick* x = 2, changing the problem from factoring a polynomial to factoring an ordinary number, certainly I can *pick* any factorization that works. Lesson here: be careful what you *pick*. > What I'm curious about are the people who believe Nora > Baron had a valid point. > > Speak up, and don't be shy as I suspect you may believe she still has > a valid point and I need to understand why she's so effective in > convincing people. > > Come on, speak up, do you think I answered her objections? Do you > believe they were valid in the first place? > On this point I agree. Speak up, lurkers! Nora B. > > James Harris ==== Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >>and assume it is factored *as you propose* in the form >>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >> Now instead of m = 0, let's try m = 1: > > > Ok. > > >> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. >> This can be factored in the form [*] by letting >> a_1 = -5, a_2 = -5, and a_3 = 2140. > > > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? > > You haven't provided a way to define what the a_i(m) are, so of course > she can simply select values that work. That's false as the a's are given by the factorization (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf. For my example I used x=2, f=5, u=1. And now letting m=1, to test your claim I have 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) so the a's are the roots of the cubic a^3 + 72a^2 - 13825 = 0. And checking with a=-5, gives -12150, and not 0. So Nora Baron's claim and yours is refuted. >> Note that EACH ONE of these is divisible by 5: That was Nora Baron's comment for her picked values for the a's. > Ok, you *pick* values from the a's as if they're not defined by a > cubic, which they are, and your picked values are supposed to prove > something? > > They AREN'T defined by a cubic. They are constrained by a cubic. If > you want them to be defined by a cubic, you'll have to say more about > how they are supposed to be constructed. That statement is false as I've shown by giving the cubic. >> NONE ARE COPRIME TO 5. > > > So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? That was my question to Nora Baron. > Remember you simply tossed those numbers out there, but there *is* a > way to calculate the a's so your assertion can be tested, as they are > roots to a cubic. > > Not that you've provided. If you have a way in mind, please provide it > to us and include it in your paper. You've only claimed this things > exist and have certain properties. If you want to change them, so be > it. But that may cause more problems for you. I simplify and people like you come in and try to confuse, and sadly, I think that the readership of sci.math wants to be confused by you. > Now if you can do that Nora Baron then clearly I made some kind of > mistake and there's no more room for me to argue. > > room for you to argue. Well now that I've given the cubic, I challenge you to admit *you* made a mistake. > So why don't you go back to the cubic which defines the a's, stick in > all the values and see if the a's come out as you have above, and then > it's over. > > For each value of m, she can define a_i(m) to be whatever three values > makes both sides equal. You've given nothing more specific about the > values to require more. That is false as I've repeatedly given the expression defining the a's. >> Now, what's your explanation? >> Nora B. > > > Well I've explained above, and again, of course, you can't just *pick* > values for the a's for a particular m, as their values are set > rigorously. What I'm curious about are the people who believe Nora > Baron had a valid point. > > Where have you established how to rigorously compute them. What are they? For m=1, f=5, the a's are the roots of the cubic a^3 + 72a^2 - 13825 = 0. > Speak up, and don't be shy as I suspect you may believe she still has > a valid point and I need to understand why she's so effective in > convincing people. > > Because you have not stated all of the conditions you want to exist. > Once you do that, you will lose the results you are trying to achieve. I've stated those conditions repeatedly. Are you now claiming that you have never seen the factorization (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)? > Come on, speak up, do you think I answered her objections? Do you > believe they were valid in the first place? > > I think you didn't understand her objections. I think you're lying. James Harris ==== [.snip.] >> What I'm curious about are the people who believe Nora >> Baron had a valid point. >> >> Speak up, and don't be shy as I suspect you may believe she still has >> a valid point and I need to understand why she's so effective in >> convincing people. >> >> Come on, speak up, do you think I answered her objections? Do you >> believe they were valid in the first place? >> On this point I agree. Speak up, lurkers! Hmmm... Within the last 24 Hours, James incorrectly claimed that Keith Ramsay was doing an appeal to the gallery, and correctly noted that appealing to the gallery is a logical fallacy. Perhaps James does not realize that what he is doing here is a ->classic<- example of appealing to the gallery? After all, he does not know what appeal to authority, ad hominem, and any of the other logical fallacies he claims others are committing are... ====================================================================== 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 ==== > > > > > Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >>and assume it is factored *as you propose* in the form >>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >> Now instead of m = 0, let's try m = 1: >Ok. > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. >> This can be factored in the form [*] by letting >> a_1 = -5, a_2 = -5, and a_3 = 2140. >Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that >seem to be just a tad bit strange to you? >>You haven't provided a way to define what the a_i(m) are, so of course >>she can simply select values that work. > > > That's false as the a's are given by the factorization > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf. > > For my example I used x=2, f=5, u=1. And now letting m=1, to test > your claim I have > > 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > so the a's are the roots of the cubic > > a^3 + 72a^2 - 13825 = 0. > > And checking with a=-5, gives -12150, and not 0. > > So Nora Baron's claim and yours is refuted. I thought you said the a's vary with m. If they are roots of the cubic, then they are constants. Note that in your paper you use the a's in two different ways. Which way are they used in THIS case? I believe Dale Hall is the one who showed that if you view the a's as constants, they still don't have the properties you want, but I could easily be mistaken. >> Note that EACH ONE of these is divisible by 5: > > > That was Nora Baron's comment for her picked values for the a's. > > >Ok, you *pick* values from the a's as if they're not defined by a >cubic, which they are, and your picked values are supposed to prove >something? >>They AREN'T defined by a cubic. They are constrained by a cubic. If >>you want them to be defined by a cubic, you'll have to say more about >>how they are supposed to be constructed. > > > That statement is false as I've shown by giving the cubic. What you've shown is that you are inconsistent in the use of your notation. Perhaps if you used a_1(m) when dealing with functions and a_1 when dealing with constants this confusion wouldn't arise. >> NONE ARE COPRIME TO 5. >So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > > That was my question to Nora Baron. > > >Remember you simply tossed those numbers out there, but there *is* a >way to calculate the a's so your assertion can be tested, as they are >roots to a cubic. >>Not that you've provided. If you have a way in mind, please provide it >>to us and include it in your paper. You've only claimed this things >>exist and have certain properties. If you want to change them, so be >>it. But that may cause more problems for you. > > > I simplify and people like you come in and try to confuse, and sadly, > I think that the readership of sci.math wants to be confused by you. I didn't use the same symbols to represent two different types of quantities. >Now if you can do that Nora Baron then clearly I made some kind of >mistake and there's no more room for me to argue. >>room for you to argue. > > > Well now that I've given the cubic, I challenge you to admit *you* > made a mistake. Challenge declined. You want the a's to both vary with m and be constants. You'll have to make up your mind which it is. >So why don't you go back to the cubic which defines the a's, stick in >all the values and see if the a's come out as you have above, and then >it's over. >>For each value of m, she can define a_i(m) to be whatever three values >>makes both sides equal. You've given nothing more specific about the >>values to require more. > > > That is false as I've repeatedly given the expression defining the > a's. If the a's are constants. If they are, then your expression above fails to be true when you vary m. >> Now, what's your explanation? >> Nora B. >Well I've explained above, and again, of course, you can't just *pick* >values for the a's for a particular m, as their values are set >rigorously. What I'm curious about are the people who believe Nora >Baron had a valid point. >>Where have you established how to rigorously compute them. What are they? > > > For m=1, f=5, the a's are the roots of the cubic > > a^3 + 72a^2 - 13825 = 0. There's nothing in the nature of the problem that (to me) makes such a definition either obvious or necessary. >Speak up, and don't be shy as I suspect you may believe she still has >a valid point and I need to understand why she's so effective in >convincing people. >>Because you have not stated all of the conditions you want to exist. >>Once you do that, you will lose the results you are trying to achieve. > > > I've stated those conditions repeatedly. Are you now claiming that > you have never seen the factorization > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)? Is THIS where it's defined??? Ok... and do you want the choice of your a_i(m) to hold for all x and y? Your simplification does not make that obvious. Worse, there is no good way of assigning the values generated for each choice of m to each a_i(m). Note: when you CHOSE values for x and y, you changed the nature of the problem being looked at. The observations made by myself and (I believe) Nora are based on the special case of x=2, y=5. The irony is, you've been defending a leap from specific to general in your arguments, but won't let us look at the specific while ignoring the general. Have you considered keeping everything in one thread so it's easier to follow things? >Come on, speak up, do you think I answered her objections? Do you >believe they were valid in the first place? >>I think you didn't understand her objections. > > > I think you're lying. That is your perogative. > James Harris -- Will Twentyman ==== > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? For strangeness, nothing equals JSH's attempts at mathematics. ==== > I think you didn't understand her objections. > > I think you're lying. > > > James Harris We doubt that you are thinking and we know you're lying! ==== > > > > > Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >>and assume it is factored *as you propose* in the form >>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >> Now instead of m = 0, let's try m = 1: >Ok. > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. >> This can be factored in the form [*] by letting >> a_1 = -5, a_2 = -5, and a_3 = 2140. >Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that >seem to be just a tad bit strange to you? >>You haven't provided a way to define what the a_i(m) are, so of course >>she can simply select values that work. > > > That's false as the a's are given by the factorization > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf. > > For my example I used x=2, f=5, u=1. And now letting m=1, to test > your claim I have > > 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > so the a's are the roots of the cubic > > a^3 + 72a^2 - 13825 = 0. > > And checking with a=-5, gives -12150, and not 0. > > So Nora Baron's claim and yours is refuted. > > I thought you said the a's vary with m. If they are roots of the cubic, > then they are constants. Note that in your paper you use the a's in two > different ways. Which way are they used in THIS case? I believe Dale > Hall is the one who showed that if you view the a's as constants, they > still don't have the properties you want, but I could easily be mistaken. Are you being deliberately obtuse? Nora Baron's assertion is with m=1. >> Note that EACH ONE of these is divisible by 5: > > > That was Nora Baron's comment for her picked values for the a's. > > >Ok, you *pick* values from the a's as if they're not defined by a >cubic, which they are, and your picked values are supposed to prove >something? >>They AREN'T defined by a cubic. They are constrained by a cubic. If >>you want them to be defined by a cubic, you'll have to say more about >>how they are supposed to be constructed. > > > That statement is false as I've shown by giving the cubic. > > What you've shown is that you are inconsistent in the use of your > notation. Perhaps if you used a_1(m) when dealing with functions and > a_1 when dealing with constants this confusion wouldn't arise. The following should not be new to you: (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf. And what I did to *simplify* was to let x=2, f=5, u=1. Then Nora Baron made an assertion about the a's for m=1, and I gave the defining cubic. >> NONE ARE COPRIME TO 5. >So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > > That was my question to Nora Baron. > > >Remember you simply tossed those numbers out there, but there *is* a >way to calculate the a's so your assertion can be tested, as they are >roots to a cubic. >>Not that you've provided. If you have a way in mind, please provide it >>to us and include it in your paper. You've only claimed this things >>exist and have certain properties. If you want to change them, so be >>it. But that may cause more problems for you. > > > I simplify and people like you come in and try to confuse, and sadly, > I think that the readership of sci.math wants to be confused by you. > > I didn't use the same symbols to represent two different types of > quantities. If I have y=mx+b, and then a little later I have y=x+1, and then later talk of with x=1, y=2, is that using the same symbols to represent different types of quantities? If you're having trouble keeping up, then you can work through it *yourself* putting in whatever symbols help you. However, I've yet to be convinced that others would be more helped by my adding in extra than hurt by the additional level of complexity. For instance, if I have P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x + y)(a_2(m) x + y)(a_3(m) x + y) does that help more than it hurts? And it's even busier with the FLT argument where there are even MORE symbols. Are there people who will now get even more confused? What kind of arguments might I face then from people who want to drag me into arguments about functions? >Now if you can do that Nora Baron then clearly I made some kind of >mistake and there's no more room for me to argue. >>room for you to argue. > > > Well now that I've given the cubic, I challenge you to admit *you* > made a mistake. > > Challenge declined. You want the a's to both vary with m and be > constants. You'll have to make up your mind which it is. You lack intellectual honesty. >So why don't you go back to the cubic which defines the a's, stick in >all the values and see if the a's come out as you have above, and then >it's over. >>For each value of m, she can define a_i(m) to be whatever three values >>makes both sides equal. You've given nothing more specific about the >>values to require more. > > > That is false as I've repeatedly given the expression defining the > a's. > > If the a's are constants. If they are, then your expression above fails > to be true when you vary m. You can't be that lost. I simply find it hard to believe that you don't know more as I think you do but are attempting to confuse others. >> Now, what's your explanation? >> Nora B. >Well I've explained above, and again, of course, you can't just *pick* >values for the a's for a particular m, as their values are set >rigorously. What I'm curious about are the people who believe Nora >Baron had a valid point. >>Where have you established how to rigorously compute them. What are they? > > > For m=1, f=5, the a's are the roots of the cubic > > a^3 + 72a^2 - 13825 = 0. > > There's nothing in the nature of the problem that (to me) makes such a > definition either obvious or necessary. You mean like expressions that actually define the a's? You seem to think you don't need to bother with the math! >Speak up, and don't be shy as I suspect you may believe she still has >a valid point and I need to understand why she's so effective in >convincing people. >>Because you have not stated all of the conditions you want to exist. >>Once you do that, you will lose the results you are trying to achieve. > > > I've stated those conditions repeatedly. Are you now claiming that > you have never seen the factorization > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)? > > Is THIS where it's defined??? Ok... and do you want the choice of your > a_i(m) to hold for all x and y? Your simplification does not make > that obvious. Worse, there is no good way of assigning the values > generated for each choice of m to each a_i(m). That is b.s. as you can easily enough solve for the a's, even with the expression in that form by creating a cubic as I have done, and then solving it. There'll be a LOT of symbols, but it can be done. > Note: when you CHOSE values for x and y, you changed the nature of the > problem being looked at. The observations made by myself and (I > believe) Nora are based on the special case of x=2, y=5. The irony is, > you've been defending a leap from specific to general in your arguments, > but won't let us look at the specific while ignoring the general. Actually my work has always covered a family of values, which is why I can stick in values as I did in this thread. Previously, with FLT, of course, I couldn't stick in values for x, y and z. > Have you considered keeping everything in one thread so it's easier to > follow things? The threads become HUGE with hundreds of posts, and then it's harder to follow things. >Come on, speak up, do you think I answered her objections? Do you >believe they were valid in the first place? >>I think you didn't understand her objections. > > > I think you're lying. > > That is your perogative. You've either been making a horrendous number of mistakes in replying to me while continually ignoring information that has been given, or you're lying. James Harris ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > which gives the reader a chance to see P(m) with only the m left as > a symbol > > > Yes! I like this example. See below. > > Fascinating. > > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. > > > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions > of m. When m = 0, two of them are 0. However when m is not zero, > none of them are zero. So how do their values when m = 0 > determine what they will be when m <> 0 ? > > Well you have P(m) = 25 Q(m) and that factor 25, which has been the > focus of all the arguing. Now considering Q(m) it turns out that > looking at the factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > the a's are in the way, and it's not clear how you divide out. > > Now you *know* that the 25 divides through some way, and it's clear > that whatever way that is would be true for any m, just like with > > 2(x^2 + 2x + 1) = (x+1)(2x+2) > > when you divide off 2, you do it for all x. > > For readers, people like Nora Baron have basically been arguing that > you can have something like > > Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2) > > where h_1 h_2 h_3 = 5, and each is not a unit. > > But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25, when in > fact Q(0)=11. > > Given that rigorous fact they claim that m=0 is a some kind of special > case and try to cast doubt on the result, which is an attack on > algebra that I call voodoo math. > > Surprisingly they've been quite successful from what I've gathered in > convincing people, which makes you wonder about people's understanding > of algebra. > > Because from algebra, it turns out that you can just set m=0, to > figure out how that 25 divides out, as I've done. > > > > No voodoo at all. What you would like to have here is that > the same a_1, a_2, and a_3 work when m = 0 as when m <> 0. > But you know, and everyone else knows, that that is totally > impossible. Why? Because when m = 0, two of the a's must be > zero. When m <> 0, NONE of the a's can be zero. That everyone else knows is an appeal to the gallery which is a logical fallacy. The a's are defined by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y), where v=-1+mf^2, and I guessed that with so many symbols people were confused, so I stuck in some values to lessen the symbol load. > It's real simple. > > The a's are functions of m. And I don't mean CONSTANT functions. > > Got that? > > Different m's give different a's. > > You believe that because a_1 is divisible by 5 when m = 0, > it must be divisible by 5 for other values of m also. It is > some kind of hunch that you have, based on intuition. It seems > like a nice parallel construction. You think it's obvious. > It isn't. In fact it is false. Your intuition this time is > wrong. Your hunch has been disproved. Here you claim that I'm acting on a hunch about the a's. But, when I used x=2, f=5, y=5, I have P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and noticing that 25 I used P(m) = 25 Q(m), so I have Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 where the question is, how does that 25 divide through? But since I know that Q(0)=11, I know that it must be Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). At Q(0) that is Q(0) = (2 (0) + 1)(2 (0) + 1)(2 (3) + 5) = 11. So in fact it's algebra and not a hunch. It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can see how the 2 divides out, but if you couldn't see, and couldn't guess, like if you had P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), with a_1 a_2 = 2x^2, a_1 + a_2 = 4x, you could have P(x) = 2Q(x), and set x=0, with Q(0) to get 1, showing that Q(x) = (a_1 + 1)(a_2/2 + 1), is correct. Of course, you can also just see that is the only way that works. But regardless there is only ONE way that will work, and it doesn't change as x changes, and it doesn't change above as m changes. Assaulting algebra is a rather disturbing step taken by posters continuing to argue with me. What is also troubling is how uncaring the newsgroup seems to be about their denial of basic algebra. But it is telling. James Harris ==== [snip] > Assaulting algebra is a rather disturbing step taken by posters > continuing to argue with me. What is also troubling is how uncaring > the newsgroup seems to be about their denial of basic algebra. But it > is telling. No one is assaulting algebra. They are denying that you have constructed a valid proof. And they are correct. You are wrong. What is telling here is your monumental incompetence. -- 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 ==== >>Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >>and assume it is factored *as you propose* in the form >>[*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). >>Now instead of m = 0, let's try m = 1: >> P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. >>This can be factored in the form [*] by letting >> a_1 = -5, a_2 = -5, and a_3 = 2140. >>>>>Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that >>>seem to be just a tad bit strange to you? >>You haven't provided a way to define what the a_i(m) are, so of course >>she can simply select values that work. >That's false as the a's are given by the factorization (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf. For my example I used x=2, f=5, u=1. And now letting m=1, to test >your claim I have 13825x^3 - 72xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) so the a's are the roots of the cubic a^3 + 72a^2 - 13825 = 0. And checking with a=-5, gives -12150, and not 0. So Nora Baron's claim and yours is refuted. >>I thought you said the a's vary with m. If they are roots of the cubic, >>then they are constants. Note that in your paper you use the a's in two >>different ways. Which way are they used in THIS case? I believe Dale >>Hall is the one who showed that if you view the a's as constants, they >>still don't have the properties you want, but I could easily be mistaken. > > > Are you being deliberately obtuse? Nora Baron's assertion is with > m=1. Correct. She was DEFINING the values of the a_i(1). The discussion had dealt with what the a_i(0) were, but she simply showed that when m changes from 0 to 1, it is possible for the a_i(m) to change from having the properties you wanted to ones that you did NOT want. Nothing obtuse. Now, do you understand what a function is? Do you perhaps see the importance of defining the characteristics of these functions at all values of m if you want them to have certain properties? >>Note that EACH ONE of these is divisible by 5: >That was Nora Baron's comment for her picked values for the a's. > >>>Ok, you *pick* values from the a's as if they're not defined by a >>>cubic, which they are, and your picked values are supposed to prove >>>something? >>They AREN'T defined by a cubic. They are constrained by a cubic. If >>you want them to be defined by a cubic, you'll have to say more about >>how they are supposed to be constructed. >That statement is false as I've shown by giving the cubic. >>What you've shown is that you are inconsistent in the use of your >>notation. Perhaps if you used a_1(m) when dealing with functions and >>a_1 when dealing with constants this confusion wouldn't arise. > > > > The following should not be new to you: > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf. > > And what I did to *simplify* was to let x=2, f=5, u=1. Then Nora > Baron made an assertion about the a's for m=1, and I gave the defining > cubic. When you set x=2,f=5,u=1, then you change the nature of the problem. If you would prefer to think of the a_i as varying over m,x,f,u, we can do that but then Nora is still able to choose the values she did. If you don't want people defining the a_i's in inconvenient ways, you will have to specify what is legal. Don't be surprised if the specifications you provide limit you to cases that don't give the broad result you want to achieve. >>NONE ARE COPRIME TO 5. >>>>>So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? >That was my question to Nora Baron. >>Remember you simply tossed those numbers out there, but there *is* a >>>way to calculate the a's so your assertion can be tested, as they are >>>roots to a cubic. >>Not that you've provided. If you have a way in mind, please provide it >>to us and include it in your paper. You've only claimed this things >>exist and have certain properties. If you want to change them, so be >>it. But that may cause more problems for you. >I simplify and people like you come in and try to confuse, and sadly, >I think that the readership of sci.math wants to be confused by you. >>I didn't use the same symbols to represent two different types of >>quantities. > > > If I have y=mx+b, and then a little later I have y=x+1, and then later > talk of with x=1, y=2, is that using the same symbols to represent > different types of quantities? I have no problem with this. > If you're having trouble keeping up, then you can work through it > *yourself* putting in whatever symbols help you. I do. That's why you keeping seeing my posts with a_1(m). > However, I've yet to be convinced that others would be more helped by > my adding in extra than hurt by the additional level of complexity. It's not for the others, it's to clearly define what are the independent variables for the functions a_i. It's so that you know what you're talking about and can clearly communicate it to us. > For instance, if I have > > P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1(m) x + y)(a_2(m) x + y)(a_3(m) > x + y) > > does that help more than it hurts? It helps a great deal. How does it hurt? > And it's even busier with the FLT argument where there are even MORE > symbols. > > Are there people who will now get even more confused? Unlikely. It may be busier but it is now precise. You have clearly indicated that the a_i depend only on m, and that when you change x or y, the a_i will not change. Of course, y and v are both defined in terms of f, so there's probably a little more going on. Your introductions of v and y is hiding a relationship. > What kind of arguments might I face then from people who want to drag > me into arguments about functions? Ones that you can either clearly defend against or concede to as everyone will be discussing the same things. Right now there are no definitions so people are interpreting the same symbols differently. >>>Now if you can do that Nora Baron then clearly I made some kind of >>>mistake and there's no more room for me to argue. >>room for you to argue. >Well now that I've given the cubic, I challenge you to admit *you* >made a mistake. >>Challenge declined. You want the a's to both vary with m and be >>constants. You'll have to make up your mind which it is. > > You lack intellectual honesty. You are the one that is being dragged into defining your symbols. >>>So why don't you go back to the cubic which defines the a's, stick in >>>all the values and see if the a's come out as you have above, and then >>>it's over. >>For each value of m, she can define a_i(m) to be whatever three values >>makes both sides equal. You've given nothing more specific about the >>values to require more. >That is false as I've repeatedly given the expression defining the >a's. >>If the a's are constants. If they are, then your expression above fails >>to be true when you vary m. > > You can't be that lost. I simply find it hard to believe that you > don't know more as I think you do but are attempting to confuse > others. If the a's vary with m, then there must be a way to define them for each choice of m. You have NOT done that. You have given what looks like a definition but would lead to the a's being constants. Perhaps if you go back and POST your paper HERE we can see if you've changed things enough to address the issues that you feel are valid. I'm working with a copy that is approximately a month old. >>Now, what's your explanation? >>Nora B. >>>>>Well I've explained above, and again, of course, you can't just *pick* >>>values for the a's for a particular m, as their values are set >>>rigorously. What I'm curious about are the people who believe Nora >>>Baron had a valid point. >>Where have you established how to rigorously compute them. What are they? >For m=1, f=5, the a's are the roots of the cubic a^3 + 72a^2 - 13825 = 0. >>There's nothing in the nature of the problem that (to me) makes such a >>definition either obvious or necessary. > > > You mean like expressions that actually define the a's? > > You seem to think you don't need to bother with the math! Ok, I just noticed the below... That connection would have been helpful. Why do you berate me for not connecting something below with that above? >>>Speak up, and don't be shy as I suspect you may believe she still has >>>a valid point and I need to understand why she's so effective in >>>convincing people. >>Because you have not stated all of the conditions you want to exist. >>Once you do that, you will lose the results you are trying to achieve. >I've stated those conditions repeatedly. Are you now claiming that >you have never seen the factorization (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)? >>Is THIS where it's defined??? Ok... and do you want the choice of your >>a_i(m) to hold for all x and y? Your simplification does not make >>that obvious. Worse, there is no good way of assigning the values >>generated for each choice of m to each a_i(m). > > That is b.s. as you can easily enough solve for the a's, even with the > expression in that form by creating a cubic as I have done, and then > solving it. > > There'll be a LOT of symbols, but it can be done. Here's the funny thing: Nora did it. It worked. You just don't like it. If she failed to do it correctly, point out the *specific* flaw. >>Note: when you CHOSE values for x and y, you changed the nature of the >>problem being looked at. The observations made by myself and (I >>believe) Nora are based on the special case of x=2, y=5. The irony is, >>you've been defending a leap from specific to general in your arguments, >>but won't let us look at the specific while ignoring the general. > > Actually my work has always covered a family of values, which is why I > can stick in values as I did in this thread. Previously, with FLT, of > course, I couldn't stick in values for x, y and z. What family of values? It might help if you informed people about that in advance. >>Have you considered keeping everything in one thread so it's easier to >>follow things? > > > The threads become HUGE with hundreds of posts, and then it's harder > to follow things. It's easier than trying to keep track of 20 short threads. I can mark a thread for future reference. It's harder to mark a lot of threads, and harder to follow the logic between the threads. A concession in one thread is not apparent in another thread. > James Harris -- Will Twentyman ==== > >> [deleted] > > It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can see how > the 2 divides out, but if you couldn't see, and couldn't guess, like > if you had > > P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), > > with a_1 a_2 = 2x^2, a_1 + a_2 = 4x, Correction 1: 2 a_1(x) + a_2(x) = 4x. Correction 2: only if a_1(x) = x, a_2(x) = 2x. If you want to allow anything else, your assertion is incorrect. And you consistently want to allow the a_i(x) to be non-polynomials. -- Will Twentyman ==== > > > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) > > For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. > > Perhaps you mean that none of the a's are coprime to 5? > Which is trivially obvious and uninteresting after all. > Sorry to have bothered you. > > It's actually fascinating as it's a bizarre result, and I think you > realize that I'm correct as you *again* deleted out the following > wonderful simplification which shows clearly that I'm right. > > Yuck. Unfortunately in my haste I neglected to refute the claim, as > it is in fact NOT true that none of the a's are coprime to 5, as in > fact the fascinating as it's a bizarre result is that they ALL are > coprime to 5 in the ring of algebraic integers. > > The problem is with the ring of algebraic integers as I've explained. > > Consider the following which you deleted yet again from my previous > post: > > Hmmm...how about this explanation? > > Remember that the polynomial is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > which gives the reader a chance to see P(m) with only the m left as > a symbol > > > Yes! I like this example. See below. > > Fascinating. > > and you see that factor 25. Now imagine that I have the polynomial > Q(m), where > > P(m) = 25 Q(m) > > so you have as a factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > and the question is, how does that factor of 25 divide out? > > I want readers to consider how many posters have apparently attempted > to make them believe that the factor of f^2, here 25, was in some > sense welded into the expression, when in fact it's a factor of P(m) > such that I can write > > P(m) = 25 Q(m) > > as I've done here, or P(m) = f^2 Q(m), in general. > > Well, checking at Q(0), gives 11, so how can those factors of 25 > divide through Q(m) in such a way as to give 11, at m=0? > > There's only one way, which gives you > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > If that bothers you, remember that at m=0, two of the a's equal 0. > > > Proving what, exactly? Clearly a_1, a_2 and a_3 are functions > of m. When m = 0, two of them are 0. However when m is not zero, > none of them are zero. So how do their values when m = 0 > determine what they will be when m <> 0 ? > > Well you have P(m) = 25 Q(m) and that factor 25, which has been the > focus of all the arguing. Now considering Q(m) it turns out that > looking at the factorization > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > the a's are in the way, and it's not clear how you divide out. > > Now you *know* that the 25 divides through some way, and it's clear > that whatever way that is would be true for any m, just like with > > 2(x^2 + 2x + 1) = (x+1)(2x+2) > > when you divide off 2, you do it for all x. > > For readers, people like Nora Baron have basically been arguing that > you can have something like > > Q(m)=(2 a_1/h_1 + h_2 h_3)(2 a_2/h_2 + h_1 h_3)(2 a_3/h_3 + h_1 h_2) > > where h_1 h_2 h_3 = 5, and each is not a unit. > > But notice then that setting m=0, gives (h_1 h_2 h_3)^2 = 25, when in > fact Q(0)=11. > > Given that rigorous fact they claim that m=0 is a some kind of special > case and try to cast doubt on the result, which is an attack on > algebra that I call voodoo math. > > Surprisingly they've been quite successful from what I've gathered in > convincing people, which makes you wonder about people's understanding > of algebra. > > Because from algebra, it turns out that you can just set m=0, to > figure out how that 25 divides out, as I've done. > > > > No voodoo at all. What you would like to have here is that > the same a_1, a_2, and a_3 work when m = 0 as when m <> 0. > But you know, and everyone else knows, that that is totally > impossible. Why? Because when m = 0, two of the a's must be > zero. When m <> 0, NONE of the a's can be zero. > > That everyone else knows is an appeal to the gallery which is a > logical fallacy. > Just a statement of fact, and anyway backed up immediately by a specific justification. As for an appeal to the gallery: that seems like an odd complaint coming from you, with your frequent indictments of mathematicians as corrupt liars who want to deceive the innocent naive silent readers of sci.math, alt.math, and alt.math.undergrad. > The a's are defined by > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y), > > where v=-1+mf^2, and I guessed that with so many symbols people were > confused, so I stuck in some values to lessen the symbol load. > You oversimplified things and paid the price. > It's real simple. > > The a's are functions of m. And I don't mean CONSTANT functions. > > Got that? > > Different m's give different a's. > > You believe that because a_1 is divisible by 5 when m = 0, > it must be divisible by 5 for other values of m also. It is > some kind of hunch that you have, based on intuition. It seems > like a nice parallel construction. You think it's obvious. > It isn't. In fact it is false. Your intuition this time is > wrong. Your hunch has been disproved. > > Here you claim that I'm acting on a hunch about the a's. > > But, when I used x=2, f=5, y=5, I have > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and noticing that 25 I used P(m) = 25 Q(m), so I have > > Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 > > where the question is, how does that 25 divide through? > > But since I know that Q(0)=11, I know that it must be > > Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). > > At Q(0) that is > > Q(0) = (2 (0) + 1)(2 (0) + 1)(2 (3) + 5) = 11. > > So in fact it's algebra and not a hunch. > What you have here is that when m = 0, a_1 = a_2 = 0, and therefore a_1 and a_2 are divisible by 5, and 2*a_1 / 5 and 2*a_2 / 5 are both algebraic integers (because both are 0, and conclude that a_1 and a_2 must be divisible by 5 when m takes on nonzero values also. You do not prove this. You do not cite any mathematical principle that justifies it. It is clearly just your hunch. It looks like a nice pattern, so it must be true. Of course a_1 and a_2 are actually dependent on m. When m = 0, they are both 0 and both divisible by 5, but how does that tell you anything about either of them when m <> 0 ? You have not provided a justification. You simply make the statement. And you deleted out the rest of my post which dealt with this. Why? Further, you have not found a valid objection to my proof that your central claim related to this is incorrect. Nor have you responded at all to Dale Hall's separate proof. To show that Dale is wrong, all you would have to do would be carry out a simple computation. You don't have to understand anything about Galois theory or Gauss' lemma or anything else from the algebraic theory of numbers. But you haven't done anything at all. Why? Since you deleted my proof that your central claim is wrong, I am appending it again below. Feel free to point out any errors in it, or post a question if there are parts of it that you don't understand. ================================================================== JSH CLAIM: It is possible to find a 3rd degree polynomial with integer coefficients, monic and irreducible over the rationals, such that if a1, a2, and a3 are the roots, then at least one of a1, a2, or a3 is coprime (in the algebraic integers) to a prime factor of the constant term of the polynomial. DISPROOF OF CLAIM: Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by definition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory, there exists an automorphism F12 of the field of algebraic numbers with the following properties *** : 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts your claim which was quoted above. Again, please feel free to point out any errors in the proof I just gave. *** Reference on automorphisms: http://www.math.niu.edu/~beachy/aaol/galois.html See especially Proposition 8.6.2 on that page. Or see the excellent textbook, Abstract Algebra, by John Beachy and William D. Blair.} ================================================================== > It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can see how > the 2 divides out, but if you couldn't see, and couldn't guess, like > if you had > > P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), > > with a_1 a_2 = 2x^2, a_1 + a_2 = 4x, > > you could have P(x) = 2Q(x), and set x=0, with Q(0) to get 1, showing > that > > Q(x) = (a_1 + 1)(a_2/2 + 1), is correct. > > Of course, you can also just see that is the only way that works. > > But regardless there is only ONE way that will work, and it doesn't > change as x changes, and it doesn't change above as m changes. > Oversimplified examples are NOT a substitute for proofs. I would think you would have learned that by now, if nothing else. Nora B. > Assaulting algebra is a rather disturbing step taken by posters > continuing to argue with me. What is also troubling is how uncaring > the newsgroup seems to be about their denial of basic algebra. But it > is telling. > > > James Harris ==== sorry for violating my promise, today; tomorrow, it will hoepfully remain in full force. anyway, I had to say that your statement is peculiar, because an appeal to the gallery can be countered by anyone who happens to be *in* the gallery (*I* don't know that e.g.); the important part, apparently, was that *you* should perhaps know it, given your extensive 10-year mission to prove la premiere theoreme de Fermat. That everyone else knows is an appeal to the gallery which is a logical fallacy. The a's are defined by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y), where v=-1+mf^2, and I guessed that with so many symbols people were confused, so I stuck in some values to lessen the symbol load. --UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?... La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto: (FOSSILISATION [McCainanites?] (TM/sic))/ BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm. Http://www.tarpley.net/bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 ==== Previously I answered Nora Baron's claim that I was acting on a hunch as I refuted that claim mathematically, and then I deleted out the rest as I wanted to shorten the length of posts. This time I'm deleting down to just past her hunch assertion. Nora Baron's comments follow. > What the a's are when m = 0 is a special case. It does NOT > determine what they are when m <> 0. Sure, when m = 0, > you can assume that a_1 and a_2 are zero. Can you can say > that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of course > you can - they are both 0! After all, ANY number divides 0. > > For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are > algebraic integers also, since both are 0. Does that mean that 137 must > divide two of the a's, when m <> 0 ???? By your logic, YES! > > Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS OF M. > WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT BE ASSUMED > TO BE TRUE WHEN M <> 0. ***** I'm leaving that in as part of the setup without saying much more here, as I handle everything below. What's fascinating is how far she goes in trying to push against algebra, as I'll show below. > Why is this so hard for you to understand? Why do you keep trying to > act like I am pulling some kind of trick when I say this, or pretending > that I am lying? Well either you're lying or you think you've refuted algebra, as I show below. I'd think that you'd prefer that people think you're lying. Nora Baron's comments follow. > > Let's go back to your original polynomial, > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and assume it is factored *as you propose* in the form > > [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now instead of m = 0, let's try m = 1: > > Ok. > > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. > > This can be factored in the form [*] by letting > > a_1 = -5, a_2 = -5, and a_3 = 2140. > > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? > > > > Nope. It was your idea, not mine, to replace the x's > with a constant value of 2. When you did that you were > no longer factoring a polynomial. You are just factoring > an ordinary number. There are no hidden rules that say it > must be factored as a polynomial. > > You can't have it both ways. You want it factored as > a polynomial, leave the x's in there (and see below). You > want it factored as an ordinary number (remember? VERY much > simplified) then one must assume that any factorization is > legitimate, and you get my example above. The factorization exists for all x so why would it go away for a particular x? > So let's go back to assuming you want it factored as a > polynomial. Of course you know that both I and W. Dale > Hall have proofs that that is not going to work. You have > not provided a valid objection to my proof, and you have > not provided any response whatsoever to what Dale did. I > don't see how you can. It's a simple computation. You > do the arithmetic, and Dale is either right or he is wrong. > You don't have to understand ANYTHING. There is no wool > to be pulled over anyone's eyes. Try it and see. > > See also below. Well you're fighting algebra. Guess what? The algebra wins. See below... > Note that EACH ONE of these is divisible by 5: > > Ok, you *pick* values from the a's as if they're not defined by a > cubic, which they are, and your picked values are supposed to prove > something? > > > Yes. It proves what you said was wrong. You said that one > of the a's had to be coprime to 5. Clearly, totally, > unambiguously false. It can only be false if you wish to refute algebra Nora Baron. See below... > NONE ARE COPRIME TO 5. > > So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > Remember you simply tossed those numbers out there, but there *is* a > way to calculate the a's so your assertion can be tested, as they are > roots to a cubic. > > > > They *would* be roots of a cubic in the variable x, but you > replaced x by 2. At that point there is no longer a polynomial. > The restriction that they be roots of a cubic is gone. It was > your idea, not mine, to oversimplify things and make the > substitution. Let me ask you a question, are you claiming that given P(x) = (x+1)(x+2) that if I stick in actual values for x, the factorization is gone? Yes, if I have P(2)=12, it is true that you just see a number, but notice that P(2) = 3(4). The expression I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and now you are arguing that the factorization goes away simply because I set x=2? Um, that's not algebra Nora Baron; that's more voodoo math. What I've done is put in actual numbers for x, f, and u, as I set x=2, f=5, and u=1, which gave me a polynomial. But the factorization *still* exists, and your wish that it doesn't--no matter how desperate--won't change that reality. > Remember: if you want to go back to assuming that a_1, a_2, and > a_3 are roots of a cubic in the variable x, to continue to > maintain what you think is true, you are going to have to find > an error in the proof I have presented (and is appended > below). Why? > Now if you can do that Nora Baron then clearly I made some kind of > mistake and there's no more room for me to argue. > > > > Yep. You should not have substituted x = 2. You want it > factored as a polynomial, you had better leave it a polynomial > and not try to oversimplify. You have shown confusion previously > over what the difference between a polynomial and the evaluation > of that polynomial when you choose a particular value for x. So you think that factorizations go away if you stick in values? > The moral here is, this little example should show you pretty > convincingly that when people tried to get you to understand that > difference, they were not just playing with semantics. > > There is another more general moral here, and you have run > afoul of this one many a time in the past also: Examples are > not a substitute for proofs. And do NOT oversimplify when you > are trying to do mathematics. It doesn't seem that I can oversimplify for you Nora Baron. > So why don't you go back to the cubic which defines the a's, stick in > all the values and see if the a's come out as you have above, and then > it's over. > > > > I will do better than that. I will reproduce my proof that your > claim that one of the a's is coprime to 5 is wrong, but I will do it in > more generality. That way I will be heading off any examples > you might try to come up with in the future. What you want is > a consequence of the following claim, which you have stated > elsewhere: > > CLAIM: It is possible to find a monic polynomial of degree > 3, with integer coefficients, irreducible over the rationals. > and with roots a1, a2, and a3, such that at least one of > a1, a2, or a3 is coprime, in the ring of algebraic > integers, to a prime factor of the constant term of > the polynomial. > > Right? You have claimed as recently as yesterday. > > > DISPROOF OF CLAIM: And Nora Baron rattles on a bit more. Where from before her primary position is that given a factorization I lose the factorization by actually putting in a value for x, as I used x=2. The key expression is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and I put in actual numbers for several values as a simplification to remove room for bogus objections like Nora Baron's here. Yet Nora Baron in one reply simply put in some picked values for the a's for m=1 in what she thought of as a refutation, and now seems to think that my saying x=2 takes away the factorization above. That's hardly even voodoo mathematics. I think it's more simply: nonsensical. James Harris ==== > >> What the a's are when m = 0 is a special case. It does NOT >>determine what they are when m <> 0. Sure, when m = 0, >>you can assume that a_1 and a_2 are zero. Can you can say >>that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of course >>you can - they are both 0! After all, ANY number divides 0. >> For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are >>algebraic integers also, since both are 0. Does that mean that 137 must >>divide two of the a's, when m <> 0 ???? By your logic, YES! >> Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS OF M. >>WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT BE ASSUMED >>TO BE TRUE WHEN M <> 0. ***** >> Why is this so hard for you to understand? Why do you keep trying to >>act like I am pulling some kind of trick when I say this, or pretending >>that I am lying? > > > Well either you're lying or you think you've refuted algebra, as I > show below. Or you don't understand. > I'd think that you'd prefer that people think you're lying. > > Nora Baron's comments follow. > > >> Let's go back to your original polynomial, >> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), >> Please note: you are factoring 25(5000m^3 - 600m^2 - 126m + 11) Where is x? Answer: nowhere. Now stop talking about it, or stop talking about this polynomial. [deleted] >Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that >seem to be just a tad bit strange to you? > Nope. It was your idea, not mine, to replace the x's >>with a constant value of 2. When you did that you were >>no longer factoring a polynomial. You are just factoring >>an ordinary number. There are no hidden rules that say it >>must be factored as a polynomial. >> You can't have it both ways. You want it factored as >>a polynomial, leave the x's in there (and see below). You >>want it factored as an ordinary number (remember? VERY much >>simplified) then one must assume that any factorization is >>legitimate, and you get my example above. > > > The factorization exists for all x so why would it go away for a > particular x? Because you got rid of x. If you see an x up there, get glasses. >> So let's go back to assuming you want it factored as a >>polynomial. Of course you know that both I and W. Dale >>Hall have proofs that that is not going to work. You have >>not provided a valid objection to my proof, and you have >>not provided any response whatsoever to what Dale did. I >>don't see how you can. It's a simple computation. You >>do the arithmetic, and Dale is either right or he is wrong. >>You don't have to understand ANYTHING. There is no wool >>to be pulled over anyone's eyes. Try it and see. >> See also below. > > > Well you're fighting algebra. Guess what? The algebra wins. > > See below... > > >> Note that EACH ONE of these is divisible by 5: Ok, you *pick* values from the a's as if they're not defined by a >cubic, which they are, and your picked values are supposed to prove >something? > Yes. It proves what you said was wrong. You said that one >>of the a's had to be coprime to 5. Clearly, totally, >>unambiguously false. > > > It can only be false if you wish to refute algebra Nora Baron. See > below... > > >> NONE ARE COPRIME TO 5. So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? Remember you simply tossed those numbers out there, but there *is* a >way to calculate the a's so your assertion can be tested, as they are >roots to a cubic. > They *would* be roots of a cubic in the variable x, but you >>replaced x by 2. At that point there is no longer a polynomial. >>The restriction that they be roots of a cubic is gone. It was >>your idea, not mine, to oversimplify things and make the >>substitution. > > > Let me ask you a question, are you claiming that given > > P(x) = (x+1)(x+2) > > that if I stick in actual values for x, the factorization is gone? > > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). Notice: P(2)=2(6)=1(12) The uniqueness of the factorization into 2 factors is gone. > > The expression I use is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? Does this look anything like the polynomial above? Does it have the same number of variables? > Um, that's not algebra Nora Baron; that's more voodoo math. You substituted for x but want to keep x. That's voodoo alright. > What I've done is put in actual numbers for x, f, and u, as I set x=2, > f=5, and u=1, which gave me a polynomial. But the factorization > *still* exists, and your wish that it doesn't--no matter how > desperate--won't change that reality. The factorization still exists, but not uniquely. Perhaps I've been wrong all along and you meant a_i(m,x,f,u). If this is the case, what she did *still* works. She defined a_i(1,2,5,1). She just didn't do it the way you want. >> Remember: if you want to go back to assuming that a_1, a_2, and >>a_3 are roots of a cubic in the variable x, to continue to >>maintain what you think is true, you are going to have to find >>an error in the proof I have presented (and is appended >>below). > > Why? Because otherwise your paper is wrong. >Now if you can do that Nora Baron then clearly I made some kind of >mistake and there's no more room for me to argue. > Yep. You should not have substituted x = 2. You want it >>factored as a polynomial, you had better leave it a polynomial >>and not try to oversimplify. You have shown confusion previously >>over what the difference between a polynomial and the evaluation >>of that polynomial when you choose a particular value for x. > > > So you think that factorizations go away if you stick in values? See above with P(2)=12. >So why don't you go back to the cubic which defines the a's, stick in >all the values and see if the a's come out as you have above, and then >it's over. > I will do better than that. I will reproduce my proof that your >>claim that one of the a's is coprime to 5 is wrong, but I will do it in >>more generality. That way I will be heading off any examples >>you might try to come up with in the future. What you want is >>a consequence of the following claim, which you have stated >>elsewhere: >> CLAIM: It is possible to find a monic polynomial of degree >> 3, with integer coefficients, irreducible over the rationals. >> and with roots a1, a2, and a3, such that at least one of >> a1, a2, or a3 is coprime, in the ring of algebraic >> integers, to a prime factor of the constant term of >> the polynomial. >> Right? You have claimed as recently as yesterday. >> DISPROOF OF CLAIM: > > > At this point, how are you dealing with her proof? I would think that of all people, you would know better than to delete material and then respond to it. You were very upset when I did that by accident, yet you make sure everyone knows you deleted it. > And Nora Baron rattles on a bit more. Where from before her primary > position is that given a factorization I lose the factorization by > actually putting in a value for x, as I used x=2. The key expression > is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and I put in actual numbers for several > values as a simplification to remove room for bogus objections like > Nora Baron's here. Yet Nora Baron in one reply simply put in some > picked values for the a's for m=1 in what she thought of as a > refutation, and now seems to think that my saying x=2 takes away the > factorization above. That's hardly even voodoo mathematics. > > I think it's more simply: nonsensical. > > > James Harris This does not refute her argument. It does not even deal with it. There is something you are welcome to own, however. voodoo mathematics. I'm willing to let you have it. It's all yours. Whatever it is. -- Will Twentyman ==== > > > [deleted] > > > It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can see how > the 2 divides out, but if you couldn't see, and couldn't guess, like > if you had > > P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), > > with a_1 a_2 = 2x^2, a_1 + a_2 = 4x, > > Correction 1: 2 a_1(x) + a_2(x) = 4x. > Correction 2: only if a_1(x) = x, a_2(x) = 2x. > > If you want to allow anything else, your assertion is incorrect. And > you consistently want to allow the a_i(x) to be non-polynomials. Hmmm...which might lead a *rational* reader to suppose you're trying to claim that it doesn't work with non-polynomial factors. How about this? p(x) = sqrt(2(x^2+2x+1)) = sqrt((a_1+1)(a_2+2)) with a_1 a_2 = 2x^2, 2a_1 + a_2 = 4x? Mathematicians, what a crew. But you know what? They take themselves SO seriously, even when they're fighting basic algebra. So why would *mathematicians* fight algebra? Because they're actually like English professors...really stuck-up English professors...who believe they're owed a certain amount of respect. But you see, English professors can pooh-pooh a literary work, so mathematicians acting like really stuck-up English professors think they can pooh-pooh valid mathematics, which inevitably leads to them attacking the foundations of mathematics because math is funny that way. James Harris ==== > ... stuff deleted ... >> So let's go back to assuming you want it factored as a >>polynomial. Of course you know that both I and W. Dale >>Hall have proofs that that is not going to work. You have >>not provided a valid objection to my proof, and you have >>not provided any response whatsoever to what Dale did. I >>don't see how you can. It's a simple computation. You >>do the arithmetic, and Dale is either right or he is wrong. >>You don't have to understand ANYTHING. There is no wool >>to be pulled over anyone's eyes. Try it and see. >> See also below. > > > Well you're fighting algebra. Guess what? The algebra wins. > > See below... > ... stuff deleted ... > > It can only be false if you wish to refute algebra Nora Baron. See > below... > ... stuff deleted ... > The expression I use is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > > Um, that's not algebra Nora Baron; that's more voodoo math. > > What I've done is put in actual numbers for x, f, and u, as I set x=2, > f=5, and u=1, which gave me a polynomial. But the factorization > *still* exists, and your wish that it doesn't--no matter how > desperate--won't change that reality. > You obtained the polynomial 65 x^3 - 12 x + 1 by substituting v=4, y=1 into that expression, correct? You claim that this factorization: 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) with the a's being algebraic integers, has the a's being coprime to 5, correct? What you say is wrong, and repeating it doesn't change the facts. I've lost count of how many times I've posted this particular argument, but I don't mind, since I can easily I have proposed that 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 have the property that, for any root z of the polynomial p(x)= x^3 - 12 x^2 + 65, we have q(z)*r(z) = 5 r(z)*s(z) = z. I've shown how this can be verified, by doing the following multiplications (courtesy of DOE Macsyma): First, here are the products that I'm making claims about: q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325 r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680 Next, a couple of products of p(x) = x^3 - 12 x^2 + 65 with polynomials of degree 1: (64 x + 128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320 (32 x + 72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680 Finally, we compare the results and see this: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x, Note that, for any value xo that makes p(xo) = 0, that same value xo will make q(xo)r(xo) = 5, so r(xo) is a factor of 5. That value of xo also makes r(xo)*s(xo) = xo, so r(xo) is a factor of xo. In short, r(xo) becomes a factor of *both* xo and 5. Since r(x) is a polynomial with integral coefficients, r(xo) is an algebraic integer whenever xo is. It is similarly simple to demonstrate that, whenever xo is a root of p(x), then r(xo) is a root of the polynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5: First, expand the polynomial mpr(r(x)): mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5 = (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2 + 315 (8 x^2 - 4 x - 45) + 5 = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 Next, multiply these two polynomials: p(x) = x^3 - 12 x^2 + 65, and w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808 to get this: p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 Notice the equality mpr(r(x)) = p(x)*w(x). That means for every value of x, the polynomial you get by computing r(x), then evaluating mpr(x) at that value, is equal to the product of p(x) and w(x). If xo is a root of p(x), you have p(xo) = 0, so p(xo)*w(xo) = 0, and therefore r(xo) is a root of mpr(x). Note that there are three such roots of mpr(x), and (taking the three roots x1,x2,x3 of p(x)), three values r1 = r(x1) ~ 968.67481 r2 = r(x2) ~ -0.01517 r3 = r(x3) ~ 0.34036 correspond to the three real roots of mpr(x). As such, their product must be -5. Your earlier claim that the a's must be coprime to 5 implies that the r's must be units (since they're common factors of ai with 5, in the ring of algebraic integers). If you multiply units, even you must realize that the product is again a unit. Therefore, -5 (and equivalently, 5 itself) must be a unit in the ring of algebraic integers. Do you agree or not? If so, then say so. I'll make it easy; here's a form for you to fill out and post to your full array of newsgroups: I, James S. Harris, affirm my belief that, in the ring of algebraic integers, the (rational) integer 5 [five] is a unit. James S. Harris. If not, then show me (hey, don't worry about me, show your public!) where my error is. I've done all the multiplication; you can verify or refute all this very easily, given the ability to multiply or expand polynomial expressions. Show how highly you value algebra: Do some. It's easy: ordinary polynomials, ordinary polynomial multiplication, nothing up my sleeves, no salesman will call. You don't even need to solve any equations. Just multiply, combine terms, show me wrong! Prove me wrong. Your mama says you can't. ... stuff deleted ... > > And Nora Baron rattles on a bit more. Where from before her primary > position is that given a factorization I lose the factorization by > actually putting in a value for x, as I used x=2. The key expression > is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and I put in actual numbers for several > values as a simplification to remove room for bogus objections like > Nora Baron's here. Yet Nora Baron in one reply simply put in some > picked values for the a's for m=1 in what she thought of as a > refutation, and now seems to think that my saying x=2 takes away the > factorization above. That's hardly even voodoo mathematics. > > I think it's more simply: nonsensical. > I haven't bothered with your evaluation at x=2, and don't care about whether you think anyone is doing anything. What I do notice is the level of cowardice it must take to sit there and claim that people are, what was it, fighting algebra? You don't even fight. The algebra I've presented is simple enough that you shouldn't evade it. Well, I know that if I had made an actual mistake (as contrasted to the typos that Wayne Brown picked up), you would be all over it like flies on a pile of poo, but if someone shows up with the actual goods, you're Brave Sir Robin: Brave Sir Robin ran away. Bravely ran away, away! When danger reared its ugly head, He bravely turned his tail and fled. Yes, brave Sir Robin turned about And gallantly he chickened out. Bravely taking to his feet He beat a very brave retreat, Bravest of the brave, Sir Robin! He is packing it in and packing it up And sneaking away and buggering up And chickening out and pissing off home, Yes, bravely he is throwing in the sponge... You might think I'll give up on this. > > James Harris Dale. ==== > > >[deleted] >It's like with 2(x^2+2x+1) = (x+1)(2x+2), where here you can see how >the 2 divides out, but if you couldn't see, and couldn't guess, like >if you had P(x)= 2(x^2+2x+1) = (a_1 + 1)(a_2 + 2), with a_1 a_2 = 2x^2, a_1 + a_2 = 4x, >>Correction 1: 2 a_1(x) + a_2(x) = 4x. > > > > >>Correction 2: only if a_1(x) = x, a_2(x) = 2x. >>If you want to allow anything else, your assertion is incorrect. And >>you consistently want to allow the a_i(x) to be non-polynomials. > > > Hmmm...which might lead a *rational* reader to suppose you're trying > to claim that it doesn't work with non-polynomial factors. > > How about this? > > p(x) = sqrt(2(x^2+2x+1)) = sqrt((a_1+1)(a_2+2)) > > with a_1 a_2 = 2x^2, 2a_1 + a_2 = 4x? > > Mathematicians, what a crew. But you know what? They take themselves > SO seriously, even when they're fighting basic algebra. > > So why would *mathematicians* fight algebra? Because they're actually > like English professors...really stuck-up English professors...who > believe they're owed a certain amount of respect. > > But you see, English professors can pooh-pooh a literary work, so > mathematicians acting like really stuck-up English professors think > they can pooh-pooh valid mathematics, which inevitably leads to them > attacking the foundations of mathematics because math is funny that > way. > > > James Harris Let's stick with the original problem. P(x)=2(x^2 + 2x + 1) P(x)= (a_1(x) + 1)(a_2(x) + 2) a_1(x)=sqrt(x) a_2(x)=2x sqrt(x) - 2x + 6 sqrt(x) - 8 + 8/(sqrt(x)+1) These are valid factorizations in the algebraic functions. a_1(x) a_2(x) = 2x^2 - 2x sqrt(x) + 6x - 8 sqrt(x) + [8 sqrt(x)]/[sqrt(x)+1] Therefor, a_1(x) a_2(x) =/= 2x^2. Put bluntly: I *am* claiming that non-polynomial factors don't have the properties you want them to have. Simply put, non-polynomial factorizations do not behave like polynomial factorizations. -- Will Twentyman ==== > > Previously I answered Nora Baron's claim that I was acting on a hunch > as I refuted that claim mathematically, and then I deleted out the > rest as I wanted to shorten the length of posts. > > This time I'm deleting down to just past her hunch assertion. > > Nora Baron's comments follow. > Some of! > What the a's are when m = 0 is a special case. It does NOT > determine what they are when m <> 0. Sure, when m = 0, > you can assume that a_1 and a_2 are zero. Can you can say > that 2*a_1/5 and 2*a_2/5 are both algebraic integers? Of course > you can - they are both 0! After all, ANY number divides 0. > > For example, 137 divides 0. Both 2*a_1/137 and 2*a_2/137 are > algebraic integers also, since both are 0. Does that mean that 137 must > divide two of the a's, when m <> 0 ???? By your logic, YES! > > Again: ***** THE A's ARE NOT CONSTANTS. THEY ARE FUNCTIONS OF M. > WHATEVER WEIRD PROPERTIES THEY MAY HAVE WHEN M = 0, CANNOT BE ASSUMED > TO BE TRUE WHEN M <> 0. ***** > > I'm leaving that in as part of the setup without saying much more > here, as I handle everything below. What's fascinating is how far she > goes in trying to push against algebra, as I'll show below. > > Why is this so hard for you to understand? Why do you keep trying to > act like I am pulling some kind of trick when I say this, or pretending > that I am lying? > > Well either you're lying or you think you've refuted algebra, as I > show below. > > I'd think that you'd prefer that people think you're lying. > People should certainly judge for themselves, unless you consider that playing to the gallery. > Nora Baron's comments follow. > Again? > > Let's go back to your original polynomial, > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and assume it is factored *as you propose* in the form > > [*] P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > Now instead of m = 0, let's try m = 1: > > Ok. > > P(m) = 25 * (5000 - 600 - 126 + 11) = 25 * 4285. > > This can be factored in the form [*] by letting > > a_1 = -5, a_2 = -5, and a_3 = 2140. > > Um, Nora Baron, now you willy-nilly give the a's values. Doesn't that > seem to be just a tad bit strange to you? > > > > Nope. It was your idea, not mine, to replace the x's > with a constant value of 2. When you did that you were > no longer factoring a polynomial. You are just factoring > an ordinary number. There are no hidden rules that say it > must be factored as a polynomial. > > You can't have it both ways. You want it factored as > a polynomial, leave the x's in there (and see below). You > want it factored as an ordinary number (remember? VERY much > simplified) then one must assume that any factorization is > legitimate, and you get my example above. > > The factorization exists for all x so why would it go away for a > particular x? > My point in showing what happened when you substituted 2 for x was in part that it completely changes the problem. You started with statements about factoring a polynomial. You evaluated the polynomial at x = 2. The evaluation is simply an integer. Your proposed factorization was the factorization of an ordinary integer. If you wanted us to continue regarding it as a polynomial factorization, you should have left it as a polynomial. This point is admittedly peripheral to the main argument here, that is, statements you have made regarding factorizations of polynomials. However I thought it was worth making anyway, to demonstrate how polynomials and their evaluations are different things. Sorry you didn't get it. > So let's go back to assuming you want it factored as a > polynomial. Of course you know that both I and W. Dale > Hall have proofs that that is not going to work. You have > not provided a valid objection to my proof, and you have > not provided any response whatsoever to what Dale did. I > don't see how you can. It's a simple computation. You > do the arithmetic, and Dale is either right or he is wrong. > You don't have to understand ANYTHING. There is no wool > to be pulled over anyone's eyes. Try it and see. > > See also below. > > Well you're fighting algebra. Guess what? The algebra wins. > > See below... > > Note that EACH ONE of these is divisible by 5: > > Ok, you *pick* values from the a's as if they're not defined by a > cubic, which they are, and your picked values are supposed to prove > something? > > > Yes. It proves what you said was wrong. You said that one > of the a's had to be coprime to 5. Clearly, totally, > unambiguously false. > > It can only be false if you wish to refute algebra Nora Baron. See > below... > Nope. You made a statement about a factorization of an ordinary integer. You said it could not be of a certain form. I showed directly and unambiguously that your statement was wrong. You did NOT say that, after you had evaluated the polynomial and obtained an ordinary integer, that you still wanted the factorization to be that associated with the polynomial. You were, literally, wrong. You tried to get an oversimplified example to prove your point. I gave a perfectly valid counterexample. Now you want to go back and say, No, I meant only polynomial-type factorizations! To which I say, Well, then, why didn't you leave it in the form of a polynomial in the first place? It doesn't make it any simpler to replace 'x' with '2' if you are just going to consider the '2' as a polynomial variable, rather than a constant. Why did you evaluate to an ordinary integer anyway if you are not going to take advantage of it? What did you expect to gain? Again, this is peripheral to the main argument. If you don't get my point here and want to think of this as a minor distraction, it's fine with me, as long as you address the *real* problem as discussed below. > NONE ARE COPRIME TO 5. > > So are your claiming that at m=1, a_1 = -5, a_2 = -5, and a_3 = 2140? > > Remember you simply tossed those numbers out there, but there *is* a > way to calculate the a's so your assertion can be tested, as they are > roots to a cubic. > > > > They *would* be roots of a cubic in the variable x, but you > replaced x by 2. At that point there is no longer a polynomial. > The restriction that they be roots of a cubic is gone. It was > your idea, not mine, to oversimplify things and make the > substitution. > > Let me ask you a question, are you claiming that given > > P(x) = (x+1)(x+2) > > that if I stick in actual values for x, the factorization is gone? > No, it's just the opposite. If you stick in actual values for x, you may get factorizations which are not consistent with the polynomial factorization. For example, you let x = 2. P(x) = 12. The factorization that is consistent with the polynomial factorization that you just gave is 12 = 3 * 4 = (2 + 1)*(2 + 2). There are, however, other factorizations: for example 2 * 6 or 1 * 12. These are NOT consistent with the polynomial factorization. > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). > See above. > The expression I use is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > Not at all. You don't lose any of the polynomial factorizations when you evaluate. You gain some *new* ones. That is exactly what happened in what I posted. > Um, that's not algebra Nora Baron; that's more voodoo math. > Nonsense. It's just boring arithmetic. > What I've done is put in actual numbers for x, f, and u, as I set x=2, > f=5, and u=1, which gave me a polynomial. But the factorization > *still* exists, and your wish that it doesn't--no matter how > desperate--won't change that reality. > You got it backwards. You don't lose any of the polynomial factorizations. You just gain some other ones. That is why your oversimplified example didn't make any sense. It most certainly did not prove your point. It went exactly the other direction: it showed that factorizations existed which were exactly of the form you said could not occur. You would have been better off leaving it as a polynomial in x. > > Remember: if you want to go back to assuming that a_1, a_2, and > a_3 are roots of a cubic in the variable x, to continue to > maintain what you think is true, you are going to have to find > an error in the proof I have presented (and is appended > below). > > Why? > A truly strange question. I have pointed out an explicit place in your argument where you have made a mistake. I have pointed out exactly what went wrong with your reasoning. You still apparently do not understand it. That is unfortunate. But in addition to identifying your error [which basically boils down to assuming without a shred of proof that the form of a factorization is the same when m <> 0 as it is in the degenerate case, when m = 0], I have a proof that your main conclusion *** cannot be correct ***. It is not a long or difficult proof. Most people would react by trying to find an error in my proof. You have failed to do so. It stands. I have found your error. You have not found mine, if I have one. It's that simple. Worse yet: W. Dale Hall also has a proof, totally inde- pendent from mine, that one of your main claims is false. In his case, all you have to do to check whether he is right or not is carry out a little computation. You have not done it, or if you have, you are being mighty quiet about it. Dale's proof stands also. Yes, most people would not be happy to have two independent unrefuted proofs that their beloved work was wrong, hanging out there unrefuted. They would try find where the proofs went wrong. That's why. > Now if you can do that Nora Baron then clearly I made some kind of > mistake and there's no more room for me to argue. > > > > Yep. You should not have substituted x = 2. You want it > factored as a polynomial, you had better leave it a polynomial > and not try to oversimplify. You have shown confusion previously > over what the difference between a polynomial and the evaluation > of that polynomial when you choose a particular value for x. > > So you think that factorizations go away if you stick in values? > No, no, no, again: you don't lose factorizations which work for the underlying polynomial when you stick in values. You gain new ones. > The moral here is, this little example should show you pretty > convincingly that when people tried to get you to understand that > difference, they were not just playing with semantics. > > There is another more general moral here, and you have run > afoul of this one many a time in the past also: Examples are > not a substitute for proofs. And do NOT oversimplify when you > are trying to do mathematics. > > It doesn't seem that I can oversimplify for you Nora Baron. > No, on the contrary. As has often happened with your simplistic little toy examples, you went too far. It backfired on you. > So why don't you go back to the cubic which defines the a's, stick in > all the values and see if the a's come out as you have above, and then > it's over. > > > > I will do better than that. I will reproduce my proof that your > claim that one of the a's is coprime to 5 is wrong, but I will do it in > more generality. That way I will be heading off any examples > you might try to come up with in the future. What you want is > a consequence of the following claim, which you have stated > elsewhere: > > CLAIM: It is possible to find a monic polynomial of degree > 3, with integer coefficients, irreducible over the rationals. > and with roots a1, a2, and a3, such that at least one of > a1, a2, or a3 is coprime, in the ring of algebraic > integers, to a prime factor of the constant term of > the polynomial. > > Right? You have claimed as recently as yesterday. > > > DISPROOF OF CLAIM: > > And Nora Baron rattles on a bit more. Where from before her primary > position is that given a factorization I lose the factorization by > actually putting in a value for x, as I used x=2. The key expression > is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and I put in actual numbers for several > values as a simplification to remove room for bogus objections like > Nora Baron's here. Yet Nora Baron in one reply simply put in some > picked values for the a's for m=1 in what she thought of as a > refutation, and now seems to think that my saying x=2 takes away the > factorization above. That's hardly even voodoo mathematics. > > I think it's more simply: nonsensical. > > > James Harris ==== [...] | DISPROOF OF CLAIM | | Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are | integers, and assume Q(x) is irreducible over the rationals. | Assume that c = p * v, where p is a prime and v is an integer. | Let a1, a2, and a3 be the roots of Q(x). Note that by | definition, since Q(x) is monic, a1, a2, and a3 are algebraic | integers. | | Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime | to p. It seems to me that it should be possible to prove this without relying upon the existence of the automorphism of the algebraic numbers you used. Keith Ramsay ==== >| DISPROOF OF CLAIM >| >| Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are >| integers, and assume Q(x) is irreducible over the rationals. >| Assume that c = p * v, where p is a prime and v is an integer. >| Let a1, a2, and a3 be the roots of Q(x). Note that by >| definition, since Q(x) is monic, a1, a2, and a3 are algebraic >| integers. >| >| Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime >| to p. > > It seems to me that it should be possible to prove this without > relying upon the existence of the automorphism of the algebraic > numbers you used. > > Keith Ramsay > It seems to me that you are both wasting your time if you hope to educate JSH - I have been following the many JSH initiated threads for maybe all of 2 months or so now and I have not yet seen a instance of him refuting an apparently sensible and logical argument that is contrary to his position. I can't follow the mathematics involved but the standard of discussion involved is simultaneously entertaining and educational: I definitely learn more as I attempt to follow the sensible discussion and I am most certainly amused by the mental acrobactics that JHS is displaying for public entertainment ... all this and I don't have to worry about paying for my degree (*laughs loudly*) and it comes complete with a free pass to watch the monkey in the zoo! As I have said previously, I seriously hope that you (Keith) and Nora (and Will Twentyman and Arturo Magidin and David Ulrich - sorry if I don't have all the names perfectly correct) will continue to respond seriously and sensibly to JSH as this is fantastic learning stuff for me ... as you have to make your argument and discussion more and more simple in an attempt to show JSH the many ways in which you believe him to be incorrect I learn more and more. My latest instruction, thanks very much to the efforts of Keith and Nora, has been in the area of mathematical proof - I always believed that proofs belonged solely to the realm of geometry (showing one triangle to be the same as the other) and now I have seen how proof pervades even (especially ?) the most esoteric math (not that I believe, looking at other threads in this NG, that this stuff is particularly esoteric). I wonder, in the spirit of JSH, whether that last sentence is syntactically and gramatically correct ? I'm not sure if I have set the NG stuff correctly - if I've done this right then only sci.math will see me ... I don't understand why JSH feels a need to post to sci.*, alt.*, *.fr and so on ... if it's mathematics then it belongs here doesn't it ? (unless it's specifically related to undergraduate studies, the definition of which varies by country) Ivan McDonagh. ==== Let me ask you a question, are you claiming that given > > P(x) = (x+1)(x+2) > > that if I stick in actual values for x, the factorization is gone? > > > No, it's just the opposite. If you stick in actual values > for x, you may get factorizations which are not consistent > with the polynomial factorization. For example, you > let x = 2. P(x) = 12. The factorization that is consistent > with the polynomial factorization that you just gave is > > 12 = 3 * 4 = (2 + 1)*(2 + 2). > > There are, however, other factorizations: for example 2 * 6 > or 1 * 12. These are NOT consistent with the polynomial > factorization. That is true. Hopefully there's some progress being made Nora Baron!!! > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). > > > See above. > > The expression I use is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > > > Not at all. You don't lose any of the polynomial factorizations > when you evaluate. You gain some *new* ones. That is exactly what > happened in what I posted. Well let's consider what you actually *did* which was to put in values for a_1, a_2, and a_3, as if you could just pick them at will. However, above you admit that the factorization still remains even when I put in a value for x. And in fact, the x's and the value of the a's are independent anyway, as can be seen from (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) so your position that my picking a value for x, as I picked x=2, affected the a's is NOT algebra. That is, there is no rational way you could have supposed that my picking an actual value for x would affect the a's in such a way that you thought you could just pick values for the a's as you did in your post. So I have (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and to simplify in order to lessen the ability of posters to confuse because of the symbol load, I set f=5, u=1, and x=2. Of those choices *only* the selection of f=5, affected the value of the a's, not x, Nora Baron, so your focus on x is bogus. James Harris ==== > > > > Let me ask you a question, are you claiming that given P(x) = (x+1)(x+2) that if I stick in actual values for x, the factorization is gone? > No, it's just the opposite. If you stick in actual values >>for x, you may get factorizations which are not consistent >>with the polynomial factorization. For example, you >>let x = 2. P(x) = 12. The factorization that is consistent >>with the polynomial factorization that you just gave is >> 12 = 3 * 4 = (2 + 1)*(2 + 2). >>There are, however, other factorizations: for example 2 * 6 >>or 1 * 12. These are NOT consistent with the polynomial >>factorization. > > > That is true. Hopefully there's some progress being made Nora > Baron!!! > Ah good. You understand that simplifying can introduce extraneous solutions that do not help you analyze your original problem. > >Yes, if I have P(2)=12, it is true that you just see a number, but >notice that P(2) = 3(4). > See above. >The expression I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and now you are arguing that the >factorization goes away simply because I set x=2? > Not at all. You don't lose any of the polynomial factorizations >>when you evaluate. You gain some *new* ones. That is exactly what >>happened in what I posted. > > > Well let's consider what you actually *did* which was to put in values > for a_1, a_2, and a_3, as if you could just pick them at will. Or perhaps you don't. > > However, above you admit that the factorization still remains even > when I put in a value for x. It remains, but not uniquely. You appear to have missed the entire point. Put in a value for x and you get *extra* *valid* factorizations. If you don't want them, don't plug a value in for x. > And in fact, the x's and the value of > the a's are independent anyway, as can be seen from > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) Nora pretty clearly discredited that statement in her post. > > so your position that my picking a value for x, as I picked x=2, > affected the a's is NOT algebra. That is, there is no rational way > you could have supposed that my picking an actual value for x would > affect the a's in such a way that you thought you could just pick > values for the a's as you did in your post. What part of her algebra is incorrect? At what point did she do something where the right side doesn't equal the left side? If she did something that is not algebra, there is a mistake. Where is it? > > So I have > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and to simplify in order to lessen the > ability of posters to confuse because of the symbol load, I set f=5, > u=1, and x=2. > > Of those choices *only* the selection of f=5, affected the value of > the a's, not x, Nora Baron, so your focus on x is bogus. Assigning x a value changed it to a completely different problem with MORE factorizations. One of them was inconvenient. I guess you didn't understand after all. If you don't like the results, point out the specific error in her work. Otherwise, you are wasting bandwidth. -- Will Twentyman ==== > [...] > | DISPROOF OF CLAIM > | > | Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are > | integers, and assume Q(x) is irreducible over the rationals. > | Assume that c = p * v, where p is a prime and v is an integer. > | Let a1, a2, and a3 be the roots of Q(x). Note that by > | definition, since Q(x) is monic, a1, a2, and a3 are algebraic > | integers. > | > | Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime > | to p. > > It seems to me that it should be possible to prove this without relying > upon the existence of the automorphism of the algebraic numbers you used. > > Keith Ramsay You are right about that. W. Dale Hall has produced an independent proof for a special case that is simpler to verify. However on balance I think the automorphism argument is the shortest and simplest to understand. Nora B. ==== > >>[...] >>| DISPROOF OF CLAIM >>| >>| Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are >>| integers, and assume Q(x) is irreducible over the rationals. >>| Assume that c = p * v, where p is a prime and v is an integer. >>| Let a1, a2, and a3 be the roots of Q(x). Note that by >>| definition, since Q(x) is monic, a1, a2, and a3 are algebraic >>| integers. >>| >>| Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime >>| to p. >>It seems to me that it should be possible to prove this without relying >>upon the existence of the automorphism of the algebraic numbers you used. >>Keith Ramsay > > > You are right about that. W. Dale Hall has produced an independent > proof for a special case that is simpler to verify. However on > balance I think the automorphism argument is the shortest and > simplest to understand. > > Nora B. I agree that the Galois theory argument is more informative, shortest, and [given even a glimmer of understanding of what Galois theory is all about], simplest to understand. Further, it has the advantage of admitting some amount of generalization, which my approach foregoes entirely. That said, I think there is a major hurdle in getting JSH to accept the fact that Galois theory is correct, and applicable in the context of the factorization of polynomials. He seemingly has a huge bug up in an unmentionable orifice, about the fact that Galois theory canonically deals with fields and field extensions, not realizing that the ring of integers of a number field enjoys some properties that arbitrary rings do not. I had vainly hoped that JSH would [irrespective of any acknowledgement of the hated source] take a look at the verification of those common factors that I gave, via direct polynomial multiplications, verify those multiplications for himself to see that I wasn't lying, and come to his senses about his argument. Well, I *did* strongly suspect that my errand was in vain, but I wanted to give him the benefit of the doubt, and failing that, give him enough opportunity to demonstrate his unwillingness to face reality. I have apparently been granted that second wish rather than the first. Dale ==== > > > Let me ask you a question, are you claiming that given > > P(x) = (x+1)(x+2) > > that if I stick in actual values for x, the factorization is gone? > > > No, it's just the opposite. If you stick in actual values > for x, you may get factorizations which are not consistent > with the polynomial factorization. For example, you > let x = 2. P(x) = 12. The factorization that is consistent > with the polynomial factorization that you just gave is > > 12 = 3 * 4 = (2 + 1)*(2 + 2). > > There are, however, other factorizations: for example 2 * 6 > or 1 * 12. These are NOT consistent with the polynomial > factorization. > > That is true. Hopefully there's some progress being made Nora > Baron!!! > This seems a little ironic. You were claiming previously that by evaluating the polynomial, one would lose the factor- ization associated with it. I pointed out that you had it exactly backwards - that by evaluating the polynomial, in general you introduce some factorizations that are different from the polynomial factorization. Of course you still have the polynomial factorization as well. Also of course I knew all this. So when you say there's some progress being made, it is evidently on your end, not on mine. Also as I pointed out previously, in showing the factorization of P(2), I was making the point that it does you no good whatsoever to try to simplify by substituting in an actual value for x, when you have no intention of considering any other factorization than the polynomial factorization. It seemed to me that you did not understand that little subtlety, but now perhaps you are getting it. It is not the central point anyway, so I don't really care on this one. > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). > > > See above. > > The expression I use is > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > > > Not at all. You don't lose any of the polynomial factorizations > when you evaluate. You gain some *new* ones. That is exactly what > happened in what I posted. > > Well let's consider what you actually *did* which was to put in values > for a_1, a_2, and a_3, as if you could just pick them at will. > No, I didn't pick them at will. I did retain exactly the *form* of the factorization, (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5), and of course I factored in such a way that a1, a2, and a3 were all divisible by 5. Since when you let x = 2, it was a perfectly valid factorization of an ordinary integer, not of a polynomial. You said it could not be factored in this form and I proved that it could. Again, this is a minor side-issue. If you still don't get, don't worry about it. The main things you should worry about are what you deleted. I will say them again in a different way that you may possibly be able to understand: 1. You are considering a degree 3 polynomial P(x) which is also a function of m. When m = 0, that polynomial becomes of first degree in x. You note that if the factorization is of the form (a1*x + 5)*(a2*x + 5)*(a3*x + 5), then when m = 0, to retain this form for the factorization, two of the a's, say, a1 and a2, must be zero. Of course zero is divisible by 5, so you can say that when m = 0, a1 and a2 are multiples of 5. Up to this point everything is OK. Then you make the great leap. You conclude that not only are a1 and a2 multiples of 5 when m = 0, they must be multiples of 5 for other values of m also. It goes without saying here that for different values of m, the values of a1 and a2 are different. You may consider them as functions of m, and it would make sense to denote them as a1(m) and a2(m). So you are saying: a1(m) and a2(m) are divisible by 5 when m = 0 therefore a1(m) and a2(m) are divisible by 5 for ALL OTHER values of m. Right? So where's the problem? Why is everyone being so obtuse about this? Because you do not have the slightest hint, not the slightest shred, not the faintest wisp of justfication for the word therefore. It is pure hunch, pure intuition, pure guess. And pure error. It is a false conclusion. You do not cite any general theorem or mathematical principle that justifies this. You just say it. You think it is obvious and everyone else should just endorse it on the dotted line. Saying it is enough, right? You say it, and as with so many other things you have said, everyone else should just shut up and believe it. No further proof needed, really. It's not enough. It's false. There is another slightly interesting issue here, related to the degeneracy and singularity that occurs when m = 0. I will post something on this later. 2. How do I know it's false? Because W. Dale Hall and I have given separate, independent proofs that your main claim is false. Since you have YET AGAIN deleted out the section of my post that contained my proof, and since you have not previously found any valid objection to it, I am going to give you yet another chance and reproduce it here. I am sure your many fans out there are beginning to be embarrassed by your failure to cope with this, and they will appreciate the fact that you are being given another shot at it. Right, fans ? Not to mention the many future math historians, when they starting writing your biography: ============================================================== JSH CLAIM: It is possible to find a 3rd degree polynomial with integer coefficients, monic and irreducible over the rationals, such that, if a1, a2, and a3 are the three roots, then at least one of a1, a2 or a3 is coprime in the algebraic integers to a prime integer which divides the constant term of the polynomial. DISPROOF OF CLAIM: Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by definition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory, there exists an automorphism F12 of the field of algebraic numbers such that: 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts your claim noted above. Please feel free to point out any errors or gaps in the proof I just gave. ============================================================== > However, above you admit that the factorization still remains even > when I put in a value for x. Yes, of course the original polynomial factorization is still there. I never said it wasn't. I just pointed out that when you evaluate, you get other factorizations also - factorizations which clearly, obviously violate your claims. > And in fact, the x's and the value of > the a's are independent anyway, as can be seen from > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > so your position that my picking a value for x, as I picked x=2, > affected the a's is NOT algebra. That is, there is no rational way > you could have supposed that my picking an actual value for x would > affect the a's in such a way that you thought you could just pick > values for the a's as you did in your post. > Wrong. The original a's were derived from the roots of a polynomial in x. When you chose x = 2, you were abandoning the polynomial. What you had was no longer a polynomial in x. When you talked about factoring it, you were factoring an ordinary number. You didn't say, OK, now I have an ordinary number, but I still want to factor it as if it were a polynomial in x. That would have been a silly statement to make, and it would have implied that your simplification was really quite pointless. The only sensible interpretation is that after you evaluated the polynomial, you were thinking about factoring an ordinary number. Now you are trying weasel out of that because you see it was obviously false. AGAIN: this is a minor side issue. If you STILL really don't get it, stop worrying about it. Your real big-time problems are the two I listed above. To summarize: A. I and Dale Hall have found separate proofs that your central claims are incorrect. Unless you can find an error in both our proofs, it really doesn't matter much what you say. There is an error in your proofs and an error in your thinking. B. I have found an explicit place in your argument where your thinking is incorrect. I have described it in excruciating detail: see above. So far you have either been unable to understand it or unwilling to understand it. It is conceivable that I am wrong and that you could convince me that your argument actually makes sense. Of course you would also need to show that my proofs and Dale's are incorrect, because they say your CONCLUSION is wrong, regardless of how you got there. However so far you have not made even a feeble attempt on either front. Basically you just say, If it's true when m = 0, it must be true for all m. Period, end of argument. I got news for you. That ain't a proof. Nora B. > So I have > > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and to simplify in order to lessen the > ability of posters to confuse because of the symbol load, I set f=5, > u=1, and x=2. > > Of those choices *only* the selection of f=5, affected the value of > the a's, not x, Nora Baron, so your focus on x is bogus. > > > James Harris ==== Why not just take a simpler approach with James? Just ignore him. Nobody really believes him anyway (with the *possible* exception of himself). Why bother entertaining him? (If nothing else, it gives him the misleading appearance of credibility.) ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: > > Ok I have > > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > > That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: > > Let x=2, f=5, u=1, so that I have > > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > > which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get > > P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > > and now you can see what the polynomial P(m) looks like without so > many symbols. > > Now from before where I had x, I *still* have that > > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > > > > Interesting. When the x's were left unspecified, > the form of the factorization was > > P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5). > > Now that you have substituted in x = 2, it is no > longer a factorization of a polynomial in x. It is > just a factorization as a product of three numbers. However the a's in general are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so x has *nothing* to do with the value of the a's. > So 2*a_1 + 5, for example, is just an algebraic > integer. The factorization is > > [1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5). > > > So then what you are asserting below is that if you factor > the number P(m) as in [1], then one of a1, a2, or > a3 must be coprime to 5. Right? Yup, one of them is clearly coprime to 5. The coprimeness result leads to the conclusion that they all must be coprime to 5 in the ring of algebraic integers, which is what's wacky, and shows a problem with the ring. > Let's take m = 1. Then > > P(m) = 25 * 4285. > > This can be factored as > > 5 * 5 * 4285, > > which yields a1 = 0, a2 = 0, and a3 = 2140. > > None of these is coprime to 5. End of story. That is not correct as you can't just pick values for the a's in that way. > Don't like a1 = a2 = 0 ? Other things work > too - e.g., a1 = -5, a2 = -5, a3 = 2140. None > coprime to 5, as before. > > Read on, however - Ok. > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > > (Note: Some posters have gotten a lot of mileage out of calling that a > degenerate case, but they were just fooling you into forgetting your > basic algebra and what you know about polynomials. I think they did > so deliberately as the math isn't complicated.) > > Now comes the question of what happens when m is NOT 0, and answering > that question requires looking again at > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11) > > and noticing that the constant term is 25(11), but you can divide off > that 25 to get P(m)/25 which gives you a constant term that's 11. And > 11 and 5 are coprime. That's very important. In fact, that's the > *key* fact which should stick in your mind. > > So now looking at > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = > > (5000m^3 - 600 m^2 - 126m + 11) > > > I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor > is a factor of the constant term, and in fact, it'd have a factor of > the constant term that is 5. So why would you think that the factor > of the constant term would move or change when m changes? > > Well it can't. > > Given that with 2 a_1 + 5, that 5 in there is a factor of the constant > term of > > 25(5000m^3 - 600 m^2 - 126m + 11) > > you *still* have a factor of the constant term when 25 is separated > off. > > But now your constant term is 11. > > That forces all the factors of 5 to go away from 2 a_1 + 5. > > Now you may wish for there to be someway for some factors to remain, > but what actuall happens is you get > > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = > > (5000m^3 - 600 m^2 - 126m + 11) > > while posters have argued that *all* the a's would have some factor of > 5. > > > Yep, sure. See above. With actual numbers! Wrong as I pointed out above as Nora Baron apparently never realized that the a's are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so she can't just pick, though you can see she tried. And in fact for m=1, where she actually picked values for the a's the cubic is a^3 + 72a - 13825 = 0 and none of her picks work. James Harris ==== > Fair bit of snippage. I've got some work to do now to follow this up > properly. > And again. > 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, > which is why the ring has problems. > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. Perhaps you mean that none of the a's are coprime to 5? Which is trivially obvious and uninteresting after all. Sorry to have bothered you. > James Harris Phil Nicholson. ==== > >Well I'm going to try and break it down even more to try and see if >y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and >looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, >which seem to confuse people. The ring is algebraic integers, and let >me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still >finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so >many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > Interesting. When the x's were left unspecified, >>the form of the factorization was >> P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5). >>Now that you have substituted in x = 2, it is no >>longer a factorization of a polynomial in x. It is >>just a factorization as a product of three numbers. > > > However the a's in general are given by > > a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, > > so x has *nothing* to do with the value of the a's. You keep saying this as if it's obvious... Let's look at it. P(m)= f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) distribute in f^2 P(m) = (m^3 f^6 - 3m^2 f^4 + 3mf^2) x^3 - 3(-1+mf^2)xu^2 f^2 + u^3 f^3 using v=-1+mf^2, v^3+1 is the coefficient on x^3 P(m) = (v^3+1) x^3 - 3vxu^2 f^2 + u^3 f^3 This can be viewed as a polynomial in terms of x, or v, or x and v. Viewed as a polynomial in terms of x and reversing it's coefficients gives the expression you have listed. Here's the catch, keeping x as an unknown RESTRICTS the possible values of the a's. Watch what happens when you plug in f=5, u=1 P(m) = (v^3+1) x^3 - 75 vx + 125 where now v=-1+25m. Now plug in x=2 P(m) = 8(v^3+1) - 600 v + 125 P(m) = 8v^3 - 600 v + 133 where v=-1+25m This can *not* be viewed as a polynomial in terms of x. It can only be the a's as you have suggested. When you set x=2, you lose information about the a's, and introduce additional values. In this example you can clearly see the difference between having x as a variable, and x=2. You fundamentally change the nature of the problem. Worse, when x=2, P(m) is no longer a monic polynomial under any available interpretation. The roots of this polynomial need not be algebraic integers. > > >>So 2*a_1 + 5, for example, is just an algebraic >>integer. The factorization is >>[1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5). >>So then what you are asserting below is that if you factor >>the number P(m) as in [1], then one of a1, a2, or >>a3 must be coprime to 5. Right? > > > Yup, one of them is clearly coprime to 5. The coprimeness result > leads to the conclusion that they all must be coprime to 5 in the ring > of algebraic integers, which is what's wacky, and shows a problem with > the ring. > > >>Let's take m = 1. Then >> P(m) = 25 * 4285. >>This can be factored as >> 5 * 5 * 4285, >>which yields a1 = 0, a2 = 0, and a3 = 2140. >>None of these is coprime to 5. End of story. > > > That is not correct as you can't just pick values for the a's in that > way. You lost information when you chose x, and the available assignments of the a's grew. Inconvenient, but true. >>Don't like a1 = a2 = 0 ? Other things work >>too - e.g., a1 = -5, a2 = -5, a3 = 2140. None >>coprime to 5, as before. >>Read on, however - > > > Ok. > > >So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be >written like r+c, where r=0, or r varies as the polynomial variable >varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), which equals 0, when m=0, so at least one of the a's must equal 0, >when m=0. And to get that factor that is 25, you must have two a's that go to 0, >when m=0. (Note: Some posters have gotten a lot of mileage out of calling that a >degenerate case, but they were just fooling you into forgetting your >basic algebra and what you know about polynomials. I think they did >so deliberately as the math isn't complicated.) Now comes the question of what happens when m is NOT 0, and answering >that question requires looking again at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11) and noticing that the constant term is 25(11), but you can divide off >that 25 to get P(m)/25 which gives you a constant term that's 11. And >11 and 5 are coprime. That's very important. In fact, that's the >*key* fact which should stick in your mind. So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = (5000m^3 - 600 m^2 - 126m + 11) >I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor >is a factor of the constant term, and in fact, it'd have a factor of >the constant term that is 5. So why would you think that the factor >of the constant term would move or change when m changes? Well it can't. Given that with 2 a_1 + 5, that 5 in there is a factor of the constant >term of 25(5000m^3 - 600 m^2 - 126m + 11) you *still* have a factor of the constant term when 25 is separated >off. But now your constant term is 11. That forces all the factors of 5 to go away from 2 a_1 + 5. Now you may wish for there to be someway for some factors to remain, >but what actuall happens is you get (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of >5. > Yep, sure. See above. With actual numbers! > > > Wrong as I pointed out above as Nora Baron apparently never realized > that the a's are given by > > a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, > > so she can't just pick, though you can see she tried. And in fact for > m=1, where she actually picked values for the a's the cubic is > > a^3 + 72a - 13825 = 0 > > and none of her picks work. > > > James Harris Your assertion about the a's only holds true when x is kept as a variable. See my argument above. Perhaps if you clearly defined which variables the a's are dependent on we could clear up a lot of this mess. The only thing you've made clear is that they should depend on m. Do they also depend on f? u? x? If so, you must be careful which letters you substitute values in for. It appears that you do NOT wish m to be a function of x. This means that you cannot include a simplification that includes choosing values for x. If you do, you have *over*simplified and changed the problem. -- Will Twentyman ==== [snip] > I had vainly hoped that JSH would [irrespective of any acknowledgement > of the hated source] take a look at the verification of those common > factors that I gave, via direct polynomial multiplications, verify > those multiplications for himself to see that I wasn't lying, and > come to his senses about his argument. I'm afraid that your hope was truly in vain. In the few instances where James tried to follow up a simple multiplication of a few binomials, he constistently got the exponents and the signs wrong. He is hopelessly sloppy in his algebra. I think if you want to gain some ground (no promises) you will have to post the exact values of the numbers a1, a2 and a3, and then derive the expressions which prove they are not coprime to 5 in the ring of algebraic integers so that they are public record. I don't think James is disposed to do this or has the ability. It is simpler to for him just to assert that you must be wrong, since his proof is irrefutable. In a previous post he even said not to bother citing errors in his proof because if any existed *he* would let everyone know about it. > Well, I *did* strongly suspect that my errand was in vain, but I > wanted to give him the benefit of the doubt, and failing that, give > him enough opportunity to demonstrate his unwillingness to face reality. I have apparently been granted that second wish rather than the first. Dale -- 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 ==== Your (and Will's) patience is admirable, but Quixotic. JSH has shown time and time again that he either fails to understand the nature of mathematical proof, or simply does not feel constrained by it. Observing your attempts to enlighten him reminds me of a Simpsons episode where Homer is talking to the dog, but from the dog's end it is just meaningless blah blah blah interspersed with occurrences of the only word the dog understands, his own name. Gib ==== > > Your (and Will's) patience is admirable, but Quixotic. JSH has shown > time and time again that he either fails to understand the nature of > mathematical proof, or simply does not feel constrained by it. Observing > your attempts to enlighten him reminds me of a Simpsons episode where > Homer is talking to the dog, but from the dog's end it is just > meaningless blah blah blah interspersed with occurrences of the only > word the dog understands, his own name. > > Gib > Being an educator, I like to maintain the belief that all people are capable of learning. James has shown a capacity for thinking about math, so with my (perhaps false but still comforting) belief and his professed willingness to learn, I am willing to explain as long as he listens. I also find it fascinating to observe the various ways in which he avoids listening and the various mental gymnastics he displays in defending his version of truth. Finally, it's good for me. I've knocked a lot of rust off the mental gears by following the arguments and presenting my own. -- Will Twentyman ==== | Why not just take a simpler approach with James? Just ignore him. | Nobody really believes him anyway (with the *possible* exception of | himself). Why bother entertaining him? (If nothing else, it gives | him the misleading appearance of credibility.) By now this suggestion has been made many times, and there hasn't been anything like the kind of general cooperation that it would take to end the discussion. If the only point were to prevent people from being taken in, I agree just letting him attempt to persuade people would be as effective as arguing with him. But there are other reasons why people keep at it. For one thing, sometimes it's interesting to try to convince someone who's exceptionally resistant to being convinced, and see how it it that they manage to hang on to their opinions anyway. I think a number of us just find it irritating to see someone making claims we know are wrong and not being corrected. I won't claim that's a rational irritation. Some people find it entertaining in other ways. The main thing that would convince me to stop would be if I thought it would be better for his own well-being not to have me as a distraction from his other issues. It seems possible to me that he'd prefer I stopped discussing his proofs with him too, but I don't know. Keith Ramsay ==== > | Why not just take a simpler approach with James? Just ignore him. > | Nobody really believes him anyway (with the *possible* exception of > | himself). Why bother entertaining him? (If nothing else, it gives > | him the misleading appearance of credibility.) > > By now this suggestion has been made many times, and there hasn't > been anything like the kind of general cooperation that it would > take to end the discussion. If the only point were to prevent > people from being taken in, I agree just letting him attempt to > persuade people would be as effective as arguing with him. But > there are other reasons why people keep at it. > > For one thing, sometimes it's interesting to try to convince someone > who's exceptionally resistant to being convinced, and see how it it > that they manage to hang on to their opinions anyway. I think a number > of us just find it irritating to see someone making claims we know > are wrong and not being corrected. I won't claim that's a rational > irritation. Some people find it entertaining in other ways. > > The main thing that would convince me to stop would be if I thought > it would be better for his own well-being not to have me as a > distraction from his other issues. It seems possible to me that he'd > prefer I stopped discussing his proofs with him too, but I don't know. > > Keith Ramsay Perhaps I am a little behind on this discussion, but why can't we just give James a polynomial and see if he can make his technique work? Working from a polynomial you made yourself around your method (which if I checked correctly cannot be factored with integers anyway) is quite different than trying to apply it in practice. It also seems that - though a bit of work - the traditional p/q approach and sign examination are pretty simple. They are also fairly straightforward to implement programmatically. (I wonder if James has seen this?) ==== > > Your (and Will's) patience is admirable, but Quixotic. JSH has shown > time and time again that he either fails to understand the nature of > mathematical proof, or simply does not feel constrained by it. > Observing your attempts to enlighten him reminds me of a Simpsons > episode where Homer is talking to the dog, but from the dog's end it is > just meaningless blah blah blah interspersed with occurrences of the > only word the dog understands, his own name. > > Gib JSH is intelligent enough to do basic algebra pretty much through the quadratic equation and has learned random facts past that - e.g., he now knows what algebraic integers are - but he has not absorbed what it means to construct a complete rigorous proof. In the present controversy the problem is that he has an overpowering intuition that there must be a formula connecting the roots of a polynomial equation with the constant term - in fact he is right, in the sense that for polynomials through 4th degree, there are formulas involving radicals that specify the roots in terms of the coefficients. He believes that such formulas extend to degenerate cases. He then finds a formula-type relationship for a degenerate case, and then assumes that this formula must hold in nondegenerate cases as well. He cannot conceive that in nondegenerate cases, the formula generalizes in any but the obvious way. Ideally what we would do to prove this is incorrect is actually write down the roots of his polynomial and show how each root shares algebraic integer factors with prime divisors of the constant term. Incredibly enough, this is not completely simple even in the case of a quadratic. In the case he is interested in, the cubic, the formulas for the roots are quite complicated and showing directly that they include algebraic integer factors of the constant term is a horrible mess. It is much easier to prove it indirectly. That is what I have done using automorphisms and what W. Dale Hall has done in a quite different The automorphism argument is, I think, a nice example of how sometimes having some theoretical superstructure can create a shorter, more easily understood path to the answer than a direct frontal assault. Sometimes the long way round the mountain is actually the shorter than trying to drill your way through it. JSH essentially refuses to look at either of our arguments. That is the other problem. Although not stupid, he has truly enormous ego investment in his argument. He does not see that it has a flaw or gap. He thinks that bit about generalizing from the degenerate case is the natural and obvious thing to do, because he has that simple formula (a_1/5). He would rather conclude that there is something wrong with the definition of algebraic integers or there is a flaw in Galois theory than look really critically and rigorously at his own proof. However I actually don't think this can continue forever. I think we will eventually find a way through his armor. It has happened before. Simpson's dog: my recollection of that is that it came from a Far Side cartoon. In one panel the dog's owner is saying something like No, Ginger! You must not bark in the house, Ginger! Do you hear me, Ginger?, and the dog hears Blah, Ginger! Blah blah blah, Ginger! Blah blah blah, Ginger! I wonder which came first - Far Side, or the Simpson's version? In any case, yes, JSH has reinvented this also. Nora B. ==== certainly, but can he Complete the Square? it's not really a property of tetragona per se, but the diagram will help, either way. (I prefer the lunes proof of the pythagoreean th.; and, I finally realized what the spatial analog is, a couple o'weeks, ago .-) > JSH is intelligent enough to do basic algebra pretty much through > the quadratic equation and has learned random facts past that - e.g., --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX ==== > (A lot of the usual stuff snipped.) > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 = 0 and conclude that at least one of the a_i's, say a_1, is zero. Putting this into your equation following the line Now setting m=0 gives me will then give you 5(2 a_2 +5)(2 a_3 + 5) = 25(11), and I can divide both sides by 5. Now, why couldn't it be that a_2 and a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by 5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you arrive at your statement, And to get that factor that is 25, you must -- Mark Thornquist ==== > > (A lot of the usual stuff snipped.) > > So > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > > 25(5000m^3 - 600 m^2 - 126m + 11). > > Now setting m=0 gives me > > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > > Notice that > > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > > Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 > = 0 and conclude that at least one of the a_i's, say a_1, is zero. > Putting this into your equation following the line Now setting m=0 > gives me will then give you > > 5(2 a_2 +5)(2 a_3 + 5) = 25(11), > > and I can divide both sides by 5. Now, why couldn't it be that a_2 and > a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by > 5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you > arrive at your statement, And to get that factor that is 25, you must Sure. I'm considering non-polynomial factors of P(m), and I have from my lemma that any such factor can be written as r+c, where r=0, or varies with m, while c remains constant and is a factor of the constant term. So with g_1 = (2 a_1 + 5), and since one of the a's MUST equal 0, when m=0, selecting a_1 as the one gives me g_1 = 5, so c=5, and of course r = g_1 - c, and as m varies, r varies, while, of course, at m=0, it also equals 0. Now then I notice that P(m)/25 has a constant term that is coprime to 5, as P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11. So when that 25 goes, then a factor has to come out of g_1 as well. That is, looking at P(0)/25 = 11, I have that the constant term is coprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out. It's like P(m) = g_1 g_2 g_3, and I checked at P(0), to find that at that point g_1=5. But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5 goes. But when m doesn't equal 0, that only handles one of the 5's as 25 has two factors where each is 5. Therefore, ONE other of the a's must go to 0, when m=0, and it also has a factor that is 5 which separates out. That's why the lemma is KEY, and is the linchpin of the argument. I've also used P(m) = 25 Q(m) to explain, which can help you by letting you consider factors of Q(m). For instance, if you have h_1 = 2a_1/5 + 1 as a factor of Q(m), you can check at m=0, to find that h_1=1. But let's say that h_1 = 2a_1/s + 5/s where s is some non unit factor of 5, but 5/s is not coprime to 5. Then at m=0, you'd have h_1 = 5/s, which contradicts with Q(0)=11. Now you MAY wish to forget that a_1 = 0 from before with P(m) but doing so is not logical. If someone wishes any portion of what I just showed you expanded on, then please point out which section. I'm quite willing to explain in detail, but I can't read your mind. You need to tell me where you're not sure, so that I can give you more details at that point. James Harris ==== > >>(A lot of the usual stuff snipped.) >So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be >written like r+c, where r=0, or r varies as the polynomial variable >varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), How do you figure? a_1, a_2, a_3 are non-polynomial functions of m. They don't have to be anything so simple as this. which equals 0, when m=0, so at least one of the a's must equal 0, >when m=0. And to get that factor that is 25, you must have two a's that go to 0, >when m=0. >Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 >>= 0 and conclude that at least one of the a_i's, say a_1, is zero. >>Putting this into your equation following the line Now setting m=0 >>gives me will then give you >>5(2 a_2 +5)(2 a_3 + 5) = 25(11), >>and I can divide both sides by 5. Now, why couldn't it be that a_2 and >>a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by >>5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you >>arrive at your statement, And to get that factor that is 25, you must > > > Sure. I'm considering non-polynomial factors of P(m), and I have from > my lemma that any such factor can be written as r+c, where r=0, or > varies with m, while c remains constant and is a factor of the > constant term. > > So with g_1 = (2 a_1 + 5), and since one of the a's MUST equal 0, > when m=0, selecting a_1 as the one gives me See objection above regarding this. Note: you have never addressed the objections to your assumption that non-polynomial factors behave like polynomial factors. > > g_1 = 5, > > so c=5, and of course r = g_1 - c, and as m varies, r varies, while, > of course, at m=0, it also equals 0. > > Now then I notice that P(m)/25 has a constant term that is coprime to > 5, as > > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so > > P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11. > > So when that 25 goes, then a factor has to come out of g_1 as well. > > That is, looking at P(0)/25 = 11, I have that the constant term is > coprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out. > > It's like > > P(m) = g_1 g_2 g_3, and I checked at P(0), to find that at that point > g_1=5. > > But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5 > goes. > > But when m doesn't equal 0, that only handles one of the 5's as 25 has > two factors where each is 5. Therefore, ONE other of the a's must go > to 0, when m=0, and it also has a factor that is 5 which separates > out. Why? I'll suggest an alternative: a_2 = -2, a_3 = 25. Along with a_1=0 you end up with: g_1=5, g_2=1, g_3=55 > > That's why the lemma is KEY, and is the linchpin of the argument. > > I've also used P(m) = 25 Q(m) to explain, which can help you by > letting you consider factors of Q(m). For instance, if you have > > h_1 = 2a_1/5 + 1 > > as a factor of Q(m), you can check at m=0, to find that h_1=1. > > But let's say that > > h_1 = 2a_1/s + 5/s > > where s is some non unit factor of 5, but 5/s is not coprime to 5. > > Then at m=0, you'd have h_1 = 5/s, which contradicts with Q(0)=11. Not if h_2 = s/5 and h_3 = 11 > > Now you MAY wish to forget that a_1 = 0 from before with P(m) but > doing so is not logical. > > If someone wishes any portion of what I just showed you expanded on, > then please point out which section. I'm quite willing to explain in > detail, but I can't read your mind. You need to tell me where you're > not sure, so that I can give you more details at that point. Ok, I'd love to see why non-polynomial a_i must have the product you claimed. I've provided counter-examples elsewhere and you have yet to address them. > James Harris -- Will Twentyman ==== Good morning, Is the symmetric algebra generated by diagonal tensors? Tern_ ==== Let E is a extension field of a field F. Prove that: If u in E is algebraic over F then, u^2 is algebraic over F my pf>Let [F(u^2):F]=n, then {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F. so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over F. therefore, c_0,c_1,...,c_n are not all zero, where c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0. so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x), then f(u^2)=/=0 and deg f(x) >=1 . hence, u^2 is algebraic over F. ---------------------------------------------------------- I wonder whether this proof is really correct or not. If not, please show me right way. In addition, How can I prove that F(u)=F(u^2) ? If anyone know this, please post reply. ==== in message : > Let E is a extension field of a field F. > Prove that: If u in E is algebraic over F > then, u^2 is algebraic over F [...] > In addition, How can I prove that F(u)=F(u^2) ? You can't, because it's not true in general. Consider F = Q, u = sqrt(2). -- Jim Heckman ==== If E is finite extension field of a field F and [F(u):F] is odd, then Is it true, in general ? If then, Can you give me some hint for proving ... in message : Let E is a extension field of a field F. > Prove that: If u in E is algebraic over F > then, u^2 is algebraic over F [...] In addition, How can I prove that F(u)=F(u^2) ? You can't, because it's not true in general. > Consider F = Q, u = sqrt(2). -- > Jim Heckman ==== >Let E is a extension field of a field F. >Prove that: If u in E is algebraic over F > then, u^2 is algebraic over F my pf>Let [F(u^2):F]=n, > Right here, you've assumed what you want to prove. You have to prove that [F(u^2):F] is finite. What do you want to do to prove that u^2 is algebraic? I get the idea that you're trying to use a theorem (if E>K>F and E algebraic, then K algebraic and [E:F]=[E:K][K:F]). Not knowing what book you're using (if you are using a book), I don't know if the point of the exercise is to reinforce the theorem (and make you check the assumptions), or if you're just supposed to use the definition of algebraic to see if u^2 is algebraic (which is what I do assume in the absence of any other information). That is, this exercise is setting up that nice theorem on the degrees of extensions. > then > {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F. > Why? (disclaimer below signature) > so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over F. > therefore, c_0,c_1,...,c_n are not all zero, where > c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0. > so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x), then > f(u^2)=/=0 and deg f(x) >=1 . hence, u^2 is algebraic over F. >---------------------------------------------------------- >I wonder whether this proof is really correct or not. >If not, please show me right way. In addition, How can I prove that F(u)=F(u^2) ? > > The basic idea here would be to to find [F(u):F(u^2)]. You've really only got two choices here. But once you've done the first part (u^2 is algebraic), you know that [F(u):F] = [F(u):F(u^2)][F(u^2):F] (if you haven't proven this yet, now is a good time, because it's a very convenient tool), which tells you an awful lot. Jon Miller ==== > Let E is a extension field of a field F. > Prove that: If u in E is algebraic over F > then, u^2 is algebraic over F my pf>Let [F(u^2):F]=n, then do you know such an n exists? this is pretty much what you're trying to prove... > {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F. > so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over F. > therefore, c_0,c_1,...,c_n are not all zero, where > c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0. > so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x), then > f(u^2)=/=0 and deg f(x) >=1 . > hence, u^2 is algebraic over F. this doesn't work as your starting point is wrong. But it looks as though you've already shown that u is algebraic over F iff [F(u):F] is finite. (And in fact that [F(u):F] is the degree of the minimal polynomial of u over F, but we don't need this for what follows). If so, you can make your approach work in a slightly modified form... You need to start from what is given... u is algebraic over F, so F(u):F is of finite dimension? Also u^2 is in F(u), and so powers of u^2 are all in F(u). But there's a limit to how many independent powers of u^2 there can be in F(u):F (why?), which is where your approach above can naturally come in... HTH Mike. > ---------------------------------------------------------- > I wonder whether this proof is really correct or not. > If not, please show me right way. In addition, How can I prove that F(u)=F(u^2) ? If anyone know this, please post reply. ==== [one top-posting fixed for free, next one will not be answered] in message : in message : Let E is a extension field of a field F. > Prove that: If u in E is algebraic over F > then, u^2 is algebraic over F [...] In addition, How can I prove that F(u)=F(u^2) ? You can't, because it's not true in general. > Consider F = Q, u = sqrt(2). If E is finite extension field of a field F and [F(u):F] is odd, > then Is it true, in general ? Yes. > If then, Can you give me some hint for proving ... Another poster already has. Consider that [F(u):F] = [F(u):F(u^2)] [F(u^2):F]. -- Jim Heckman ==== T Y S S E L A N D 20 25 19 19 5 12 1 14 4 = 119 Angie was playing piano on the NW corner of 21st Street and 4th Ave., I stopped to listen and to collect her stats. She is from Saskatoon, has produced a CD a few years ago and was working toward another. 189 Angie 27 2 61 58/307 1471 Angie 36 Grace 34 Tysseland 119 Angie was born in 61 (Exodus 11), her last name adds to 119. Her last 11 letters add to 128 (Numbers 11). Her first 7 letters add to 61 (Exodus 11 and is her year of birth). She has 11 consonants, they add to 137 (the 3x11th prime) and are in positions adding to 118 (the 8x11th non-prime). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Angie was born on day 58 (the 7 primes up to 17). Her first name adds to 36, or 7 plus the 7th prime (17) plus the 7th non-prime (12), or simply 7+7p+7np. Her middle name adds to 17+17, her given names add together for 70. Her last name adds to 7x17, or 7 times the 7th prime. Her common name adds to the 155 verses of Ephesians, Bible Book 49 (7x7 and is the 17+17th non-prime). Her last two names add together for 153 (1 through 17 and is the 117th non-prime, it's the terminating chapter of Numbers and is the number of fish in the net in John 21). Her full name adds to 189 (the first 17 primes minus the first 17 non-primes). The 7 different letters in her given names add to 57. The consonants in her given names add to 49 (7x7 and is the 17+17th non-prime). In her given names, her consonants exceed her vowels by 28 (1 through 7). She has 12 (7th non-prime) letters in all, her 7+7 missing letters exceed her 12 (7th non-prime) represented letters by 77. She has 7 unrepeated letters, they add to 91 (Leviticus 1, the first of the chapters to contain the length of 17 verses). Her consonants exceed her vowels by 85 (5x17 and is 17p+17np, Ruth with 85 verses is the 17th shortest Book in the Bible). Her odd valued letters add to 107 (Leviticus 17), it's the 28th (1 through 7) prime. Her primes and squares add together for the 108 verses of Bible Book 59 (the 17th prime), it is the 17th prime short of the 167 verses of Book 17). Her initials add to 28 (1 through 7). She was born on the 27th day of the month, corresponding to Daniel with 357 (7x17+7x17+7x17) verses. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Angie was born on day 58 (the 19th Book of the New Testament). She has 19 letters and a last name adding to 119. In her given names, her consonants exceed her vowels by 28 (19th non-prime). The odd valued letters in her given names add to 38 (19+19). The 7 different letters in her given names add to 57 (19+19+19). In all her unrepresented letters exceed her represented letters by 77 (the primes up to 19). Her consonants add with their positions for 255 (First Samuel 19). Her squares add to 38 (19+19). Her odd valued letters add to 107 (the 19th prime in non-prime position). Her initials add to 28 (the 19th non-prime). We meet on the 172nd day of the year (Deuteronomy 19), we meet with 193 days remaining in the year. Her birthday was 114 (6x19 and is 19+19p+q9np) days ago. Tysseland (119) and I had our birthdays were an average of 119 days ago. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Angie's consonants add to 137, vowels to 52 (37th non-prime). Her represented letters add to 137. Her names add to the 25th, 23rd and 89th non-primes, together for 137. In her given names, her unrepresented letters exceed her represented letters by 237. She was born with 307 days remaining in the year. Angie's consonants add to the 33rd prime, vowels to the 37th non-prime, together for 70, pretty as her given names add together for 70. Angie's last name adds to 119, her common name adds to 155 (the 119th non-prime). Angie Tysseland was born on the 27th. The consonants in her given names are together in positions adding to 27. In her given names, her consonants plus their positions together exceed her vowels plus their positions by 27. The primes in her given names add to 27. In her full name, her unrepresented letters exceed her represented letters by 77 (Exodus 27). Her odd valued letters are in positions adding to 117 (Leviticus 27). Her primes and squares together add with their positions for 108 (4x27). Her full name adds to 189 (7x27). Old Testament Book 27 and New Testament Book 27 are the major Books of end-times prophecy. Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== T Y S S L A N D 20 25 19 19 12 1 14 4 = 119 Angie was playing piano on the NW corner of 21st Street and 4th Ave., I stopped to listen and to collect her stats. She is from Saskatoon, has produced a CD a few years ago and was working toward another. 189 Angie 27 2 61 58/307 1471 Angie 36 Grace 34 Tyssland 119 Angie was born in 61 (Exodus 11), her last name adds to 119. Her last 11 letters add to 128 (Numbers 11). Her first 7 letters add to 61 (Exodus 11 and is her year of birth). Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 <-7th-> 12 -- -- 58 50 Angie was born on day 58 (the 7 primes up to 17). Her first name adds to 36, or 7 plus the 7th prime (17) plus the 7th non-prime (12), or simply 7+7p+7np. Her middle name adds to 17+17, her given names add together for 70. Her last name adds to 7x17, or 7 times the 7th prime. Her common name adds to the 155 verses of Ephesians, Bible Book 49 (7x7 and is the 17+17th non-prime). Her last two names add together for 153 (1 through 17 and is the 117th non-prime, it's the terminating chapter of Numbers and is the number of fish in the net in John 21). Her full name adds to 189 (the first 17 primes minus the first 17 non-primes). The 7 different letters in her given names add to 57. The consonants in her given names add to 49 (7x7 and is the 17+17th non-prime). In her given names, her consonants exceed her vowels by 28 (1 through 7). She has 12 (7th non-prime) letters in all, her 7+7 missing letters exceed her 12 (7th non-prime) represented letters by 77. She has 7 unrepeated letters, they add to 91 (Leviticus 1, the first of the chapters to contain the length of 17 verses). Her consonants exceed her vowels by 85 (5x17 and is 17p+17np, Ruth with 85 verses is the 17th shortest Book in the Bible). Her odd valued letters add to 107 (Leviticus 17), it's the 28th (1 through 7) prime. Her primes and squares add together for the 108 verses of Bible Book 59 (the 17th prime), it is the 17th prime short of the 167 verses of Book 17). Her initials add to 28 (1 through 7). She was born on the 27th day of the month, corresponding to Daniel with 357 (7x17+7x17+7x17) verses. Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Angie was born on day 58 (the 19th Book of the New Testament). She has 19 letters and a last name adding to 119. In her given names, her consonants exceed her vowels by 28 (19th non-prime). The odd valued letters in her given names add to 38 (19+19). The 7 different letters in her given names add to 57 (19+19+19). In all her unrepresented letters exceed her represented letters by 77 (the primes up to 19). Her squares add to 38 (19+19). Her odd valued letters add to 107 (the 19th prime in non-prime position). Her initials add to 28 (the 19th non-prime). We meet on the 172nd day of the year (Deuteronomy 19), we meet with 193 days remaining in the year. Her birthday was 114 (6x19 and is 19+19p+q9np) days ago. Tyssland (119) and I had our birthdays were an average of 119 days ago. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Angie's consonants add to 137, vowels to 52 (37th non-prime). Her represented letters add to 137. Her names add to the 25th, 23rd and 89th non-primes, together for 137. In her given names, her unrepresented letters exceed her represented letters by 237. She was born with 307 days remaining in the year. Angie's consonants add to the 33rd prime, vowels to the 37th non-prime, together for 70, pretty as her given names add together for 70. Angie's last name adds to 119, her common name adds to 155 (the 119th non-prime). Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! ==== > > T Y S S L A N D > 20 25 19 19 12 1 14 4 = 119 Actually it is... T Y S S E L A N D 20 25 19 19 5 12 1 14 4 = 119 ==== S L U S A R C Z Y K 19 12 21 19 1 18 3 26 25 11 = 155 260 Ania 23 12 82 357/8 9440 Ania 25 Elzbieta 80 Slusarczyk 155 Ania is from Krakow Poland, she provided stats in the afternoon at the Extra Food Store on Broadway Ave., it's the street with the penis poster poles. Ania was born on day 357, it's the number of verses in Daniel, the Book is in part about 666. Her last two names add together for 235, the 184th prime (1097) and the 184th non-prime (235) averages 666. The consonants in her given names add to 74 (a factor of 666). Her odd valued letters add to 130 and also her even valued letters add to 130, corresponding to Numbers 13 (the 6th prime). Her names add to 25, 80 and 155, these are the 16th, 58th and the 119th non-primes, together for 193 (Bible Book 6 chapter 6). Her first and last names average 90 (66th non-prime), Exodus contains 1213 verses (66+66+66th prime) and brings the Bible up to 90 chapters (66th non-prime). Her name adds to the 260 chapters of the New Testament, New Testament chapter 260 is Revelation 22, it's the terminating chapter of Bible Book 66. Her first name adds to 25 and her initials add to 25. Her 357th day of birth exceeds her 260 valued name by 97 (25th prime). She was born in the 12th month (Second Kings with 25 chapters). I am 25.84 years older than Ania. 389 <-77th prime 104 <-77th non-prime 77 <-77 --- 570 The Four 57's Genesis 41 -> 41 Leviticus 14 -> 104 Judges 9 -> 220 <-I dreamt of 220 roofs blown John 11 -> 1008 off homes in the Dakotas ---- 1373 <-220th prime Chapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime <-together for 57-> Major Books of End-Times Prophecy (Daniel and Revelation are in part about 666 while Isaiah contains 66 chapters): Daniel - 357 verses Revelation - 404 verses <-57 plus the 57th prime plus the 57th non-prime Isaiah - 1292 verses <-an average of 19.575757... verses per chapter Ania was born on day 357, her first name adds to 25 and her initials add to 25 (Bible chapter 57 contains 25 verses and also Bible Book 57 contains 25 verses, pretty as 57 and 25 are the 41st and 16th non-primes, together for 57). Her given names add together for 105, it's the 78th non-prime while 78 in turn is the 57th non-prime (105 is the 57th non-prime in non-prime position). Her vowels add to 78 (57th non-prime). Her consonants exceed her vowels by 104 (Leviticus 14 with 57 verses), prettier as 77 plus the 77th prime (389) plus the 77th non-prime (104) adds to 570. She was born on the 23rd, corresponding to Isaiah (the Book contains an average of 19.575757... verses per chapter). I meet Ania and she provides the stats on the 57+57+57th day of the year. Ania's day, month and year of birth adds to the 117 verses of Bible Book 22 and she has 22 letters. Ania was born on day 357, her day and month of birth adds together for 35. Her last two names add together for 235. Ania was born 349 days closer to the end of the year than to the beginning of the year, she was born 49 weeks (and 6 days) closer to the end of the year than to the beginning of the year. Her last name adds to the 155 verses of Ephesians, Bible Book 49. Non-Primes Non In Prime Primes Positions 1 1 2 4 <- 4 3 6 <- 6 4 8 5 9 <- 9 6 10 7 12 <- 12 8 14 9 15 10 16 11 18 <- 18 12 20 13 21 <- 21 14 22 15 24 16 25 17 26 <- 26 <-7th non-prime in prime position The nubile sweety was born on day 357 (7x17+7x17+7x17), she was born exactly 17+17+17 weeks into the year. Her name adds to the 260 chapters of the New Testament, or 10 times the 17th non-prime (26). She was born in the 12th month and has 12 letters in her given names (7th non-prime). Her last name adds to the 155 verses of Bible Book 49 (7x7 and is the 17+17th non-prime). She might take my 62 valued last name and end up with a name adding to the 167 verses of Bible Book 17, it is the 7 primes in prime positions up to the 17th prime. Esther becomes Queen in Book 17 and Q is the 17th letter of the alphabet. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 23 10 29 11 31 <- 31 12 37 13 41 <- 41 14 43 15 47 16 53 17 59 <- 59 --- 167 Esther Book 17 After showing Ania evidence that her name was a gift from God, she rewarded my work by requesting my copy of the record of our meeting. But it was my paper, my pen, my calculator and my work, so I told her that she was not welcomed to it. Starting in 1988 I was repeatedly arrested and tortured in psychiatric facilities in an effort to make me shut up about the penis poster poles on Broadway Ave. and to make me shut up about the Egyptian penises at The Vatican, Whitehouse, in front of Saskatoon City Hall and on the roofs of the churches, and to shut up about the mathematical harmony between people's names, birthdays and numbers in the Bible. Year after year they tortured me, year after year I begged people in the community for assistance to flee the country, but they are so cheap and so ignorant that they can't even offer me a cookie for my work. Protestants and Catholics lobbied my abusive parents to shut me up, then Protestants and Catholics sat on psychiatric appeal panel hearings and repeatedly gave When I complained to people about the way these Christians and that I was there because I was supposed to be there. And so it is with Ania, she is here on the usenet because she is supposed to be here on the usenet. And likely even if she did give me $2 for my work, she and her friends and relatives annually and collectively spend thousands upon thousands of dollars turning pagan fertility symbols (evergreen trees, they stay green all year) into decorated idols. In December, their churches would have a decorated tree at the entrance, another next to the pulpit, and yet another downstairs in the lunch room, maybe Ania would attend church once a year to please her mom, and it would be in late December, and she would comment on how beautiful the trees are, and will give them money. I said that the penis poster poles on Broadway Avenue were representations of penises like the Egyptian obelisks on church roofs, I get tortured year after year, and year after year the Broadway merchants add additional representations of penises along their street. Then Ania comes to Penis Street and I show her some gems, and she rewards my work by wanting my work and our record of our meeting. You recent eastern European, Asian and Latin American immigrants have done much to bring organized crime and violence to Canada and disrupt whatever peace I am able to find here. While back in your home land you hear that millions of dollars have been spent on torturing me in Canada, and your only concern is to come to Canada so you can make some of this money too. Your traditions are in conflict with God's Commandments, and I am on my knees begging God to honor Exodus 20:5 and Hosea 4:6 as promises, and terminate the lives of Ania and her siblinks. If I hear of the deaths of any of the people in her family I will cheer, it will be in accordance to Scripture (Psalm 137:9), and I will post these stats again. Fifteen years of this work and you people reward me with torture, you are an ignorant and compassionless piece of crap and I am on my knees begging people are so cheap and ignorant that you don't even have the decency evidence of who your God is, and that He provides you with your very name. They tortured me here for years, nobody cared then and nobody cares now, all you care about is raising money so you can fly off to India in the search of truth. Go to hell!!! Daryl Shawn Kabatoff Box 7134 Saskatoon Saskatchewan Canada S7K 4J1 Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!