| <- e1*e2 = e2*e1 = e3 e1*e3 = e3*e1 = -e2 e2*e3 = e3*e2 = -e1 OL === Subject: Re: More properties of mutated quaternions Originator: bergv@math.uiuc.edu (Maarten Bergvelt) Not to burst your bubble, but your mutaated quaternions do not satisfy associativity: (e1*e2)*e3 = -e3*e3 = +1 e1*(e2*e3) = e1*e1 = -1 This complicates multiplication of three or more elements. Quaternions with zero divisors are not new. Try this multiplication table: 1*ek = ek*1 = ek for k = 1, 2, 3 ek*ek = -1 for k = 1, 2, 3 e1*e2 = e2*e1 = e3 e1*e3 = e3*e1 = -e2 e2*e3 = e3*e2 = -e1 Multiplication is commutative and associative. Zero divisors include 1 (+/-) e3 and e1 (+/-) e2, and there exists an idempotent pair (1 (+/-) e3)/2 along with the trivial pair of and 1. OL === Subject: JSH: Strange will to be wrong I think one of the biggest puzzles for me as I wait and hope that the latest very simple explanation of a problem in number theory might finally get some traction is this strange will to be wrong that many of you clearly have. One would think that some of you actually would like to use correct mathematics, and not waste time and effort with mathematical ideas proven to be wrong, and that at least some of you would actually care about learning powerful mathematical techniques that DO WORK versus wasting time on flawed ones that have been proven not to work. But I guess most of you tell yourselves that's not what you're doing so you can do it, which is how this situation has continued for so many years. And then you must feel pride at learning useless crap that doesn't work, but maybe impresses people because so many people are a part of the error? So this time I updated a technique that I first thought up late last year as now I start with the factorization 2975x^2 - 15x + 2 = 2(f(x) + 1)*(g(x) + 1) where the change was to leverage two prime numbers so that I can step through the derivation with 7 and 17 versus just 7 as that didn't work with you as I got some of the same people on sci.math who would come over to other newsgroups mind you as well, tossing up specious objections which sounded intelligent, I guess, and that crap worked like it has worked so many times before. By using two prime numbers though I remove the ability of those people to make seemingly intelligent objections but people remember I have not only explained and explained as I've come up with different ways to try and get mathematicians to behave like mathematicians and accept mathematical proof, I've even had the result published in a peer reviewed mathematical journal!!! Usenet crushed the process as sci.math'ers managed to convince the editors to pull my paper, and later the entire journal died. A an entire mathematical journal died here yet still the error has stayed in place along with the denial in this will to be wrong. That takes a will to be wrong for that to happen and years to go by while I labor to try and find some way to explain that doesn't allow you to tell yourselves lies about the math so you can keep doing wrong mathematics. Wrong mathematics is EASIER. There is a thrill in feeling you are correct because you are using flawed ideas that allow you to convince yourself that you have proven something you haven't. But now we have a situation where a lot of people around the world are choosing easy, doing wrong mathematics, on a huge scale. So yes, I can understand if say Andrew Wiles wouldn't want to accept this result as it takes away his research, so no, he did not prove Fermat's Last Theorem. And Ribet might want to hide from this because it takes away his, and I'm not saying either of them are doing so, but I'm giving them as dramatic examples to explain the why of the very human behavior here. But wrong is wrong. You people can keep doing wrong math to prove things all you want and you are not doing anything of value no matter how many of you get together to live in the error. Wrong mathematical ideas are easier because they let you prove anything, like the classic examples that boil down to divide by zero errors. So yeah, do the wrong math, and use the ring of algebraic integers wrong, without understanding its quirks and real mathematical properties, and you can think you proved Fermat's Last Theorem when you didn't. Wrong math is easier, but it's still wrong. My hope is that some of you start appreciating mathematics. Because of the last few years Usenet has been doing the opposite, using group processes to make this situation much harder in a fight to be wrong. And in a fight to keep bringing other people into the error as pity the poor students, who could have been learning the truth years ago, but instead are currently wasting time and mental energy on bogus mathematical ideas that are appealing because they are wrong, and in mathematics, wrong is easier. It is so, so much harder to be right in mathematics. James Harris === Subject: JSH: Galois Theory, so what's wrong? I am sure there will be a lot of confusion about the significance of the result that I have showing a problem with use of the ring of algebraic integers. It is such a huge problem in number theory that it's hard to grasp the full impact, but I can maybe help at least with Galois Theory. The result shows that Galois Theory tells you nothing more about non- rationals than it does about rationals. That is the succinct way to explain the impact there, and why I say it does not say Galois Theory is wrong, exactly, but it greatly limits its usefulness to number theorists as to taking it away for the most part as a meaningful tool. And I want to emphasize that mathematically it just never was. People just can make mistakes, and as time goes on those mistakes can be found and the truth learned. It is a process that has gone on for as long as there have been people. We live. We learn. James Harris === Subject: Re: JSH: Galois Theory, so what's wrong? Nice to see you back. Sorry you are recycling the same old material. === Subject: Re: JSH: Galois Theory, so what's wrong? Since it's such a huge problem in number theory, you should have no difficulty coming up with an example of an *actual theorem* that you believe you have shown to be false. So please do. Really? Then what theorem of Galois theory concerning non-rationals is false? In that case, presumably there is some theorem that number theorists claim to have proved using Galois theory which is incorrect. What is it? You certainly don't seem to learn. -Rotwang === Subject: Re: Pseudo-Education Software Fails The Test I think any generalization about technology is not useful. If the product being used has been tested with accepted methods AND if it addresses instructional purposes important to the course of which it is a part AND if it is properly implemented as part of a program AND if the teachers understand how to use it AND if the students use it as intended, then the outcomes may or may not reveal to you something about learning styles. But these ANDs rarely meet up with one another in the known PS universe. It's because no teacher in a school can expect to be given the time, money, equipment or clout to implement anything, regardless of what one is told he is supposed to accomplish. The time, money, equipment and clout go to people who don't know you, don't know what you're doing or what you need, but who experience some sense of importance and accomplishment when they drop shrinkwrapped packages on some lowly, much-hassled teacher's wobbly schoolroom desk. TSK May those who damn us be damned. alhuriyehNOBOTS@NOBOTSyahoo.com ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** http://www.usenet.com === Subject: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ? I need to factor [ x^3 - 5x^2 - 8x + 4 ] in order to solve a recurrence for an algorithms class, but I seem to be stuck. The closest I've been able to get is (x - 1)(x - 2)(x - 2), which yields [ x^3 - 5x^2 + 8x - 4 ] , but as you can see, the signs on the 3rd and last term are off. The particular method for solving the recurrence strictly calls for factoring of this polynomial, but I'm starting to think this polynomial can't be factored. Anyway, before I say this to my professor, I want to make sure there is no other way to factor this.... Ed === Subject: Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ? No nice roots. There are a number of online polynomial root finders. Here's one: === Subject: Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ? Hint: If the roots are r1, r2, and r3, what is their product? What does this say about possible rational roots? Hint: If the polynomial is reducible then it must have at least one linear factor and correspondingly at least one rational root. Hint: Since the polynomial is monic, if it has a rational root, then one can conclude something further about that root. Hint: If you find one root, (say) r, then you can divide by (x-r) to get a quadratic. I assume you know how to factor a quadratic. === Subject: Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ? Must it be factored into three monomials with integer coefficients? === Subject: Re: Is there any way to factor this x^3 - 5x^2 - 8x + 4 ? Hmm... I belive so because I need to plug-in both the roots and the coefficients into the general solution for the recurrence I'm trying to solve (which is of the form [ a_n = C_1r_1^n + C_2r_2^n + C_3r_3^n ] (note I used the underscore to indicate a subscript). === Subject: Re: How do you solve this simple equation? really need a refresher. === Subject: Re: SOLVE IT I am getting a different answer: 100! * (-7)/12 -Vishvas Vasuki | |
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