mm-436 Subject: Re: JSH: Non-uniqueness of factorization > That's why I hate posters like Dik Winter and Nora Baron because > they're so adept at hiding things. > > Like notice how they used the example of > x(x+1)/2 > always being an integer in the ring of integers?Yup, but apparently you have not understood why that was brought up. Itwas brought to show that when a polynomial over a ring is divisible bya constant in that ring, there is not necessarily a polynomial factorisationwhere one of the terms is always divisible by that factor. Something youthought at that time apparently. And a non-polynomial factorisation canbe pretty erratic and haphazard. Like in this case, define: w1(x) = gcd(x, 2) w2(x) = 2 / w1(x)than we can factorise as: [x / w1(x)][(x + 1)/w2(x)]just about as haphazard as with the polynomial of Rick Decker. > That's because the ring of integers is a unique factorization domain, > so you can factor *every* integer as 2j(j+1) or j(2j+1) with j an > integer in the ring of integers.No and no. That is not the reason. > Now how many of you know that but listened to those posters?How many times have I told you that the algebraic integers do not forma unique factorisation domain? Rather, I would say that the algebraicintegers do not even form a factorisation domain, as (as I have writtenmany times before) there are no primes not indecomposibles in thealgebraic integers. > That's what's so frustrating!!!What is frustrating is that you have no idea what is going on but remainbragging about your understanding. And when refutations come up you donot understand them, or read them, or follow them; you just say it isnonsense and go off at a tangent. Actually, my pages on your attemptfor FLT when n=4 are very illustrative. > Dik Winter and Nora Baron are the two most capable, most able at > little tricks like that, where they hide their arguments in stuff that > you should catch, like that x(x+1)/2 is always true in integers > because integers have uniqueness of factorization.No. > But algebraic integers do NOT!As I have written many times. > The ideas of Dik Winter and Nora Baron do NOT pass the smell-test.Eh? > Ultimately they're relying on confusion, and that you don't pay > careful attention so that they can sneak in little tricks like > x(x+1)/2 in integers. > > Without uniqueness of factorizaton Dik Winter can't give a reason why > for any particular w_1(x) and w_2(x), and notice he doesn't even > try!!!Eh, note the w_1(x) and w_2(x) I defined above? > Why should he?I think I have posted the factorisation earlier, but do not remember. > As long as I'm supposedly just some nut, and you're just willing to > attack my ideas without wondering about the underlying mathematics, as > long as a dumb device like x(x+1)/2 in the ring of integers get's past > you, why should he worry? > > Meanwhile he keeps up his own personal attacks with a webpage harping > on an old abandoned argument of mine from YEARS ago on x^4 + y^4 = > z^4.As far as I know, you *never* have proved that simple case to satisfaction. > The guy is SCUM. And he's fooled you repeatedly, and that's why IOh, perhaps I am a SCUM. So what? When a scum comes up with a validrefutation, you have to consider it, not just put it aside.dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Non-uniqueness of factorization > Can someone please give me step by step details on how to ignore posts by > and replies to his posts? I just don't have the willpower to > not read his bull.contains jstevh@msn.com in either header or body will be the best.(Until he changes address of course.)dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Financial Equation: Please help me to formulate an equation for this problem.> Hello to any Computer Science Majors in here:> I have probably a simple question that I need help on. I need to> formulate an equation that if given the monthly interest rate on a> credit card, and the starting bce, and monthly payment would tell> you how long it would take you to pay off the credit card. My> professor came up with this equation that doesn't seem to be working:> m=months left, i=monthly interest rate, b=current bce, and P is> monthly payment.> m=[-log(1- i * b/P) / log (1 + i)> As well he gave this equation which would estimate the payment necess> to pay off loan in a certain number of months (e.g, use -12 for -m if> paying in one year):> P= i * b / 1 (1 + i) ^m> Please if you can help with any of this, I would appreciate it. I need> to be able to use this equation in C++ program. Let A_n be the (principal) amount owed just after the n'th payment (with A_0 as the initial amount)Let p be the amount of each paymentLet i be the rate of interest for one payment periodThen the amounts owed immediately after each payment must satisfy the recurrence equation A_{n+1} = A_n + i*A_n - pI.e., new principal = principal + interest - paymentThis recurrence relation can be solved as A_n = A_0*(1+i)^n - p*(1 + (i+i) +^(i+i)2 + ... + (i+i)^(n-1))Which simplifies toA_n = A_0*(1+i)^n - p*((1+i)^n -1)/i.This is the basic periodic payment equation on which all other periodic payment calculations are based.For your problem, take A_n = 0 (all paid off), A_0 = b (current bce) i = i p = Pand solve for n:(1+i)^n*[A_0 - p/i] = A_n - p/i(1+i)^n = (A_n - p/i)/(A_0 - p/i) = (i*A_n - p)/(i*A_0 - P)Then take logs on both sides. === Subject: Re: Need Help> f:(-1,1)->R, f is continuous on x=0> and f(x)=f(x^2) for all x on (-1,1)> Show that f(x)=f(0) for all x on (-1,1)> I try to solve this problem for several times,> But, I cannot think any good idea.> If anyone can help me, please post reply.Note that for any x in (-1,1), the sequence x, x^2 , x^4, x^8, ..., converges to zero, which is a point of continuity of f(x). === Subject: Re: JSH: Chaotic math? > No, it will not. As shown above, for w = 3 the degree will be 6. For > w = 4 the degree will be larger than 25, but I do not know because I do > not know the class number of Q(sqrt(-551)), but it is related to that. > > I have a web page that calculates the class number for negative > discriminants. The web page is at:Yup, so it is 26. > Rather than using ideals, I use quadratic forms. You need to > enter 4 times the value of the square root that you are > interested in. Of course, this only works for negative values.I am wondering. When I enter -551 I get 26, just as if I enter 2204.I think it is based on the discriminant rather than the value underthe root. (And when that value is of the form 4.n+1 the discriminantis the same, as is the case with -551.) > For example, the class number of Q(sqrt(-167)) is obtained > by entering in -668 and getting the result h(-668) = 11.h(-167) also gives 11.On the other hand h(-165) gives 0, as the discriminant is 660. > > The class number of Q(sqrt(-551)) is obtained by entering > in -2204 and getting the result h(-2204) = 26.Or by entering -551.Yes, it is confusing.dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Chaotic math? > Indeed. Do you want to know what has been found upto now:(corrected spelling error). > x degree w_1(x) > -3 6 (1 + sqrt(-38))^(1/3) > -2 10 (-128 + 3.sqrt(-47))^(1/5) > -1 1 7 > 0 1 7 > 1 2 sqrt(7) > 2 22 (44444 - 111.sqrt(-167))^(1/11) > 3 6 [(-25 + 3.sqrt(-83))/2]^(1/3) > Have a look about how Keith Ramsay found his result. But let us give > an example. Set x = -4, we have: a_1(-4) = (-5 + sqrt(-311))/2. The > class number here is 19, so a_1(-4)^19 and 7^19 have a common factor > in Z(sqrt(-311)). So we find: > a_1(-4)^19 = (- 3279828388182942545 - 110677566089868539.sqrt(-311))/2 > and > 7^19 = 11398895185373143. > The next step is to find integer or half-integer values p and q such > that p^2 + 311.q^2 = 11398895185373143. You can automate that. > But this takes quite some effort... p = 99831352and q = 2146257will do just that. So are there r and s such that (+-p +- q.sqrt(-311))(r + s.sqrt-311)) = a_1(-4)^19?To be continued.dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Algebra> graph the function h(x)=5x-6$ gnuplot} set terminal ascii} set output 'foo.txt'} plot [-0.2, 0.2] 5*x - 6} exit$ cat foo.txt-5| | / | / | / | / | /-6| / | / | / | / | / | /-7+--------------------- | | |-0.2 0 0.2$I hope this answers your question.P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. === I am an undergrad in numerical analysis and would like to read as muchcurrent research projects or even older stuff. Is there some free source ofsuch information on the internet? === >I am an undergrad in numerical analysis and would like to read as much>current research projects or even older stuff. Is there some free source of>such information on the internet?Start with your university library -- the math library, if it isorganized by departments.Many libraries display new issues of journals in a special place. Thatis a great way to get started browsing. Ask your librarian about online access. It is a complex issue, butmuch/most online access depends on what your univ subscribes to. Askthem. If you want to look up something specific, or to try to browse aspecific journal that you have discovered online, go ahead and try it;you may or may not have access.Many journals will send anyone (subscribed or not) email alerts (oftenwith table of contents) that new issues have been published.But start with your local library, either by asking the librarian orperhaps checking their web page -- cuz they are the ones who have theanswer that is relevant to you. (Example... The home page for the UnivCalif Berkeley Biology library has a link to Electronic Journals,which lists and links to all journals to which we have access.(However, the list is not always accurate, so it is good to check anyparticular item of interest.)noise, try one of the science-oriented search engines, such as scirus.I do not know how good their coverage of math is, but you can check.bob === >I am an undergrad in numerical analysis and would like to read as much>current research projects or even older stuff. Is there some free source of>such information on the internet?If you've already tried Google, you should say so and let us know more specifically what you're looking for.Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: interesting observation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LDK1d18255;when u calculate average speed, it is the weighted harmonic mean ofthe speed during different legs of the journey.hope that solves mattersvibhanshuhttp ://www.geocities.com/vibhanshuhttp:// www.geocities.com/vibhanshu It turns out you can prove that there's an error in core with rather> basic math, using a quadratic:I don't even bother reading your proofs any more, because I already knowthey are wrong.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--.--http://www.crbond.com === Subject: Re: Core error proof, simpler, shorter> It turns out you can prove that there's an error in core with rather> basic math, using a quadratic:> Let> P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.> Then> a = 1/b and b = 1/a> and to highlight the simple case let (8a + b) = 9, and substituting> with> a = 1/b gives,> 8(1/b) + b = 9, so 8 + b^2 = 9b, so> b^2 - 9b + 8 = (b-8)(b-1) = 0.> But why two solutions? Obviously, you can just have a=1/8, and get> (x + 8(1/8))(x + 8) = (x + 8)(x + 1)> so it's trivially easy what's going on with those two solutions.> But let's make it interesting by considering 8a + b = 17, as then you> have> 8/b + b = 17, so b^2 - 17b + 8 = 0,> but here it's really interesting as checking with the quadratic> formula, I have> b = (17 +/- sqrt(257))/2> and mathematicians teach that *now* something significant has happened> as that is irrational, and now consider what happens if we now solve> for 'a'.> Using now b=1/a,> 8a + 1/a = 17, so 8a^2 - 17a + 1 = 0.> BUT 8a^2 - 17a + 1 is a primitive non-monic polynomial irreducible> over Q!> That means that 'a' *cannot* be an algebraic integer, as algebraic> integers cannot be roots of primitive non-monic polynomials> irreducible over Q!> But notice that 'b' IS an algebraic integer as it's the root of a> monic polynomial with integer coefficient.> Solving 8a^2 - 17a + 1 = 0 with the quadratic formula gives> a = (17 +/- sqrt(257))/16> but it's clear that *only* one of those roots can be like 1/8 before,> while the other is like 8 from before, but *neither* is an algebraic> integer!> So what's the core error?> The assumptions of some mathematicians would mean that> P(x) = (x+8a)(x+b),> is impossible in an ring where 'a' and 'b' have properties like> integers, because it's impossible in the ring of algebraic integers,> if ab = 1 and (8a + b) is an integer, when the result is a polynomial> irreducible over Q that has 'a' as a root!> See the odd illogic that has been behind posters arguing with me?> Clearly there isn't anything special about that case besides our> inability to *look* at the roots because of the irreducibility, unlike> with integers!> It's an integer prejudice, which is kind of funny now, but also silly.> It turns out that there *has* to be a ring beyond algebraic integers,> which I call the object ring, where there's no problem.> Now you can fight the core error for all you're worth, and you'll just> be silly.> The error has been in mathematics long enough.> James, there is NO core error in mathematics. The only error is in yourunderstanding. You have to understand math before you can judge it; clearlyyou don't because even I can see you're wrong. C'mon admit it James, youdon't care about the math. It's obvious you don't.David Moran === Subject: Re: OT: Uni Lecturers strike: Calling students to defend their Rights.>Apologies for the OT post, but it is an important issue to me If you post to an appropriate newsgroupThis is an appropriate newsgroup, I am an undergraduate in the UK and thisaffects me directly. === Subject: Re: OT: Uni Lecturers strike: Calling students to defend their Rights.>> If you post to an appropriate newsgroup>This is an appropriate newsgroup, I am an undergraduate in the UK and this>affects me directly.So do the poll tax and EU regulations, but that doesn't mean this is the right venue to discuss them.Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: OT: Uni Lecturers strike: Calling students to defend their Rights.> The AUT recently proposed and confirmed strike action. This means final year> students may not graduate, or will do with unclassified degrees.> The NUS support this action, using a poll of 0.03% of students to support> their stance. This union should represent and defend its members rights. It> is not doing so.> We need to email the NUS AUT and Vice chancellors of your universities to> tell them that you condemn the damage of students interests for lecturers> gain. Note I support their cause myself, but not the action to solve the> problem.> If not an honours year student please support your colleagues and help us to> let everyone know that this is unacceptable.> I have created a 'petition'/form for you to sign and email / send to> (probably better) the NUS, AUT and Vice chancellors.> Find it here:> http://tinyurl.com/39hvwBad address> in word format or, here:> http://tinyurl.com/3grmoAlso bad address> in text format.> Addresses are:> or http://tinyurl.com/2zcj9Good address> and find your vice chancellors address on the relevant university website.> Also if you could email me or reply to this post if you do plan to use the> form so I can see how many people are helping! Cheers.Why are the petition sites bad addresses?Where is explaination of the story?Your's is too short to known anything. === Subject: Re: OT: Uni Lecturers strike: Calling students to defend their Rights.Today Benjamin, MISSINGes0u1136TERMINATOR, and UNEXPECTED_DATA_AFTERes0u113...:> Apologies for the OT post, but it is an important issue to me, and hopefully> many other students reside here and can help!> The AUT recently proposed and confirmed strike action. This means final year> students may not graduate, or will do with unclassified degrees.> The NUS support this action, using a poll of 0.03% of students to support> their stance. This union should represent and defend its members rights. It> is not doing so.This has never been the case. As far as I can tell, the NUS is a forum forpolitically amibitious students to get on the public stage when theyshould be concentrating on their studies.> We need to email the NUS AUT and Vice chancellors of your universities to> tell them that you condemn the damage of students interests for lecturers> gain. Note I support their cause myself, but not the action to solve the> problem.So, if the NUS called on students to boycott lectures in protest at top-upfees, *you* would be attending lectures as usual, right?> If not an honours year student please support your colleagues and help us to> let everyone know that this is unacceptable.If your university is any use, it will have an exam results appealsprocedure which should be receptive to the case that your result does notreflect your true abilities due to the fact that questions were set onmaterial which was not lectured due to whatever reason.[1]This situation is no different from when time is lost due to the lecturerbeing ill or attending conferences.This assumes, of course, that lecturers are not fair-minded and will notexclude from the examination material which they were unable to lecturefor whatever reason.There remains the problem of not being taught the omitted material, butyou should be able to teach yourself, and if your leaving furthereducation anyway then it's not going to affect your result if you don'tlearn it.[1] And there should be an external examiner who is required toapprove any questions on the exam and should ensure that the questions arenot unfair.P.A.C. SmithThe vast majority of Iraqis want to live in a peaceful, free world.And we will find these people and we will bring them to justice. === Subject: JSH: My fear, consider thisIt's so sad that I've tried to come up with simple examples for solong, and posters like Nora Baron, Dik Winter and Arturo Magidinhave *successfully* come back with their own posts and kept winning atconvincing you all that I was wrong!It's been so frustrating that I'm terrified that they will just getaway with it again, so here's another angle as I desperately try yetagain to get someone to care about what's mathematically correctknowing the kind of people who are out there to come right back andpush incorrect math.Previously I noted that I could useP(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.And I considered 8a + b = 17, as then you have8/b + b = 17, so b^2 - 17b + 8 = 0.Now imagine *any* non-unit factor f in the ring of algebraic integers,like 1+i that might be a factor of b, and let b = fz, and substituteand you getf^2 z^2 - 17fz + 8 = 0, sof z^2 - 17z + 8/f = 0and notice you STILL have that f on the front.I don't want to hear that it isn't applicable because f isn't aninteger, as if you will have to get a polynomial reducible over Q ifyou pick the right f, as that's just bogus.The problem is that 17. For it to work, you need to have somethingeven!I'm so damned tired. I can't be sure if anyone will listen to me. Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will comeback like they have before, now won't they?They'll come back and post something stupid, and wrong, and just plainevil, and you'll go along like you've done before.It's so evil, so frustrating. Nothing matters in mathematics,mathematicians don't give a damn about the truth.NOBODY ING CARES ABOUT THETRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! JSH === Subject: Re: JSH: My fear, consider this> It's so sad that I've tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong!alt.math.undergrad? You pathetic heap of parrot droppings. Yourignorance, incompetence, and psychosis are not of interest to theworld at large. Quite the contrary. You are not even an interestinglaughingstock.Hey stooopid loud troll James Always in error, never in doubt!Harris, put up or shut up. , King of the Primes! Whereare your sceptor and crown, delusional , your regalclothes? Is a $10,000 prize no questions asked too small to justifyyour submission of two little prime numbers? Or are you a psychoticimpotent gelding?http://www.rsasecurity.com/rsalabs/challenges/ factoring/faq.htmlhttp://www.rsasecurity.com/rsalabs/ challenges/factoring/numbers.htmlhttp://www.crank.net/ harris.html It's not every braying jackass that gets a whole page at crank.net--Uncle Alhttp://www.mazepath.com/uncleal/qz.pdfhttp:// www.mazepath.com/uncleal/eotvos.htm (Do something naughty to physics) === Subject: Re: JSH: My fear, consider this> It's so sad that I've tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong!Sad for you, perhaps, but hilarious for us to watch your weaseling when confronted by undeniable evidence of your errors.Keep it up. It's as much fun as watching the inmates at Bedlam. === Subject: Re: JSH: My fear, consider this> It's so sad that I've tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong!Especially since you have been wrong on every occasion? Just Googleof which are admissions of having been incorrect. Here's the mostrecent (top hit in above Google search, sorted by date): === Subject: JSH: Chaotic math? It all boils down to a system of equations: w_1(x) w_2(x) = 7 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) a^2 - (x - 1)a + 7(x^2 + x) Now then solve for w_1(x) or w_2(x). YES, I WAS WRONG with a specific assertion of mine that I made repeatedly for some time, in that I didn't believe you could always find w's in the ring of algebraic integers, but now what? ... stuff deleted ...barely 2 days later with a revisionist version, in which you were beingstymied by people winning at convincing sci.math that you were wrong?> It's been so frustrating that I'm terrified that they will just get> away with it again, so here's another angle as I desperately try yet> again to get someone to care about what's mathematically correct> knowing the kind of people who are out there to come right back and> push incorrect math.Yes, being correct does allow someone to get away with it.Life is unfair in that way. It allows those who produce the *correct*results to *get away with it*.Meanwhile, the one who has repeatedly made incorrect assertions, dueto a flawed method, is desperately trying to get folks to care aboutwhat? What's mathematically correct.Wait! Weren't you *mathematically INCORRECT* about that factoringbusiness? Didn't all the evil sci.math folks *get away with* beingcorrect? So where was the *incorrect math* in those responses? Huh?> Previously I noted that I could use> P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.> And I considered 8a + b = 17, as then you have> 8/b + b = 17, so b^2 - 17b + 8 = 0.> Now imagine *any* non-unit factor f in the ring of algebraic integers,> like 1+i that might be a factor of b, and let b = fz, and substitute> and you get> f^2 z^2 - 17fz + 8 = 0, so> f z^2 - 17z + 8/f = 0> and notice you STILL have that f on the front.> I don't want to hear that it isn't applicable because f isn't an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as that's just bogus.> The problem is that 17. For it to work, you need to have something> even!> I'm so damned tired. I can't be sure if anyone will listen to me. > Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come> back like they have before, now won't they?What are you trying to say? That people who know algebra can unravel themess you weave, and turn it into a valid mathematical question that theythen answer? Or is it that those answers show you're wrong?Is that what's sticking in yer craw, champ?> They'll come back and post something stupid, and wrong, and just plain> evil, and you'll go along like you've done before.Yes, the answers that show you're in over your head are just that: stupid, and wrong, and just plain evil.> It's so evil, so frustrating. Nothing matters in mathematics,> mathematicians don't give a damn about the truth.Your notion of truth is what's screwed. Despite your ranting, itis you who insists that he has a special place in the mathematicaluniverse, who has a privileged understanding of all of algebra, andwho ALONE can save the human race from its destiny. You are the oneengaging in magical thinking, and you are the one who tries to intuitalgebraic reality.YOU. Not the sci.math crowd.> NOBODY ING CARES ABOUT THE> TRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Apparently not, not the way you perceive it, because youare seriously deluded. Your method of derivation is justplain flawed, *because* it does not hew to logic, based onthe properties of the system you're dealing with. The evidenceof that is unmistakable: your method produces flawed results.It doesn't matter how much faith you have in it, or howwell you think you understand it; as long as your methodleads you to make incorrect assertions, it will be wrong.How long has it been that you have been giving your undyingallegiance to a system of thought that paints you as a fool?> JSH === Subject: Re: My fear, consider this< snip > I'm so damned tired. I can't be sure if anyone will listen to me.They won't. You have zero credibility. No, That's not quitetrue. Your credibility is purely imaginary -- it's all in your mind.-- Bob Day === Subject: Re: JSH: My fear, consider this>[...]>I'm so damned tired. That's so sad.>I can't be sure if anyone will listen to me. Being unsure about that indicates remarkable stupidity.Anyone with the intelligence of a brick would be certainby now that nobody was going to listen.>Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come>back like they have before, now won't they?Yup. And they'll be _right_.>They'll come back and post something stupid, and wrong, and just plain>evil, and you'll go along like you've done before.>It's so evil, so frustrating. Nothing matters in mathematics,>mathematicians don't give a damn about the truth.>NOBODY ING CARES ABOUT THE>TRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Giggle. You don't seem to be keeping count, but every oneelse is. It's happened 1,342 times that you've complainedabout the fact that nobody cares about the truth. And1,341 of those times even _you_ eventually agrees thatthe rotten liars were right.No, James, going back to your recent post about why you'rebetter than us: You're not. One of many reasons is thatyou are _so_ incredibly slow to catch on to things. Imean rats in laboratories learn a lot faster than you do.>JSH === Subject: Re: JSH: My fear, consider thisThis thread was originated by . Therefore its contents are false.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--.--http://www.crbond.com === Subject: Re: JSH: My fear, consider this> It's so sad that I've tried to come up with simple examples for so> long, and posters like Nora Baron, Dik Winter and Arturo Magidin> have *successfully* come back with their own posts and kept winning at> convincing you all that I was wrong!Because they were right. You are a crackpot.> It's been so frustrating that I'm terrified that they will just get> away with it again, so here's another angle as I desperately try yet> again to get someone to care about what's mathematically correct> knowing the kind of people who are out there to come right back and> push incorrect math...and never a doubt in your mind that you could be wrong.> Previously I noted that I could use> P(x) = (x+8a)(x+b), and ab = 1, so P(x) = x^2 + (8a + b)x + 8.> And I considered 8a + b = 17, as then you have> 8/b + b = 17, so b^2 - 17b + 8 = 0.> Now imagine *any* non-unit factor f in the ring of algebraic integers,> like 1+i that might be a factor of b, and let b = fz, and substitute> and you get> f^2 z^2 - 17fz + 8 = 0, so> f z^2 - 17z + 8/f = 0> and notice you STILL have that f on the front.> I don't want to hear that it isn't applicable because f isn't an> integer, as if you will have to get a polynomial reducible over Q if> you pick the right f, as that's just bogus.> The problem is that 17. For it to work, you need to have something> even!> I'm so damned tired. I can't be sure if anyone will listen to me. > Arturo Magidin or Dik Winter or Nora Baron or Rick Decker will come> back like they have before, now won't they?> They'll come back and post something stupid, and wrong, and just plain> evil, and you'll go along like you've done before.> It's so evil, so frustrating. Nothing matters in mathematics,> mathematicians don't give a damn about the truth.> NOBODY ING CARES ABOUT THE> TRUTH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! You're right. We're only here to laugh at crackpots. === Subject: Linear Algebra - this seems like a mistake in my textbookHi All:My textbook is showing how to calculate the adjoint of a 4x4 matrix. Here's their example:A =1 -3 0 -23 -12 -2 -6-2 10 2 5-1 6 1 3They then (without showing their work or steps) go on to sayt that thecofactor (A21) is -1No matter what I do, I keep getting (-7)(-1) = 7 for the cofactor ofA21. The -7 is the determinant of M21 and the -1 is the (-1)^i+j wherei = 2, j = 1. A few of their other cofactor answers for that matrix don't match upeither to what I'm getting. I know that I'm stupid but even mycalculator comes up with -7 as the determinant of the minor for (2,1)Can someone else tell me what they get as the cofactor for (2,1)? I'dlike to see what I'm doing wrong. Here's what the book has as theanswer for the adjoint of the above matrix:0 -1 0 -2-1 1 2 -20 -1 -3 32 -2 -3 2I'm simply not getting that same answer.thanks!! === Subject: Re: Linear Algebra - this seems like a mistake in my textbook> Hi All:> My textbook is showing how to calculate the adjoint of a 4x4 matrix. > Here's their example:> A => 1 -3 0 -2> 3 -12 -2 -6> -2 10 2 5> -1 6 1 3> They then (without showing their work or steps) go on to sayt that the> cofactor (A21) is -1> No matter what I do, I keep getting (-7)(-1) = 7 for the cofactor of> A21. The -7 is the determinant of M21 and the -1 is the (-1)^i+j where> i = 2, j = 1. > A few of their other cofactor answers for that matrix don't match up> either to what I'm getting. I know that I'm stupid but even my> calculator comes up with -7 as the determinant of the minor for (2,1)> Can someone else tell me what they get as the cofactor for (2,1)? I'd> like to see what I'm doing wrong. Here's what the book has as the> answer for the adjoint of the above matrix:> 0 -1 0 -2> -1 1 2 -2> 0 -1 -3 3> 2 -2 -3 2> I'm simply not getting that same answer.> thanks!!I don't know what you are doing wrong, but my answers agree with the book answers.How are you evaluating the 3 by 3 minors?The cofactor of the row 2 column 1 element of your original matrix should be the negative of the determinant of 3 -2 -6-2 2 5-1 1 3 === Subject: Re: Linear Algebra - this seems like a mistake in my textbook> Hi All:> My textbook is showing how to calculate the adjoint of a 4x4 matrix. > Here's their example:> A => 1 -3 0 -2> 3 -12 -2 -6> -2 10 2 5> -1 6 1 3> They then (without showing their work or steps) go on to sayt that the> cofactor (A21) is -1> No matter what I do, I keep getting (-7)(-1) = 7 for the cofactor of> A21. The -7 is the determinant of M21 and the -1 is the (-1)^i+j where> i = 2, j = 1. > A few of their other cofactor answers for that matrix don't match up> either to what I'm getting. I know that I'm stupid but even my> calculator comes up with -7 as the determinant of the minor for (2,1)> Can someone else tell me what they get as the cofactor for (2,1)? I'd> like to see what I'm doing wrong. Here's what the book has as the> answer for the adjoint of the above matrix:> 0 -1 0 -2> -1 1 2 -2> 0 -1 -3 3> 2 -2 -3 2> I'm simply not getting that same answer.> thanks!!Your text is correct. Did you forget that the adjoint matrix is the_transpose_ of the cofactor matrix? That is the i-j entry of theadjoint is (-1)^(i+j)*det(M(j,i)).Paul SperryColumbia, SC (USA) === Subject: Re: Linear Algebra - this seems like a mistake in my textbook> Your text is correct. Did you forget that the adjoint matrix is the> _transpose_ of the cofactor matrix? That is the i-j entry of the> adjoint is (-1)^(i+j)*det(M(j,i)).problem before I even attempt to perform the transpose to find theadjoint. I am having the problem with the determinant of the M21 and finding A21> 1 -3 0 -2> 3 -12 -2 -6> -2 10 2 5> -1 6 1 3Correct me if I'm wrong, but isn't |M21| = det of-3 0 -210 2 5 6 1 3which is -3 |2 5| - 0 |10 5| + -2|10 2| |1 3| |6 3| |6 1|And going thru that, I get -7. the cofactor I get is (-1)(-7) = 7.The book says that A21 = -1I just don't see how they get that. I keep getting 7 as the cofactorfor A21. I got the cofactors correct for row 1 though. thanks... === Subject: Re: Linear Algebra - this seems like a mistake in my textbook>Your text is correct. Did you forget that the adjoint matrix is the>_transpose_ of the cofactor matrix? That is the i-j entry of the>adjoint is (-1)^(i+j)*det(M(j,i)).> > problem before I even attempt to perform the transpose to find the> adjoint. > I am having the problem with the determinant of the M21 and finding A21>1 -3 0 -2>3 -12 -2 -6>-2 10 2 5>-1 6 1 3> Correct me if I'm wrong, but isn't |M21| = > det of> -3 0 -2> 10 2 5 > 6 1 3Yes> which is -3 |2 5| - 0 |10 5| + -2|10 2|> |1 3| |6 3| |6 1|Yes> And going thru that, I get -7. > the cofactor I get is (-1)(-7) = 7.No. You get -3*(2*3 - 5*1) -2*(10*1 - 2*6) =-3*1 -2*(-2) = 1I'll bet you did -3 - 4 instead of -3 + 4.> The book says that A21 = -1> I just don't see how they get that. I keep getting 7 as the cofactor> for A21. I got the cofactors correct for row 1 though. > thanks...Paul SperryColumbia, SC (USA) === Subject: Try out this one folks!!!!!!!! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1LLaMJ23796;This apuzzle .there r 3 men and 3 cannibals on an piece of land separated fromanother by a stream.They need to go the other part of land separated from the 1st by the river.Now..there's a boat which they can use but it can accomodate only 2 ata time.Also,during the placement of men and cannibals..if there r more numberof cannibals at a given place then the men ..then the men r eaten up.get the cannibals and the men to the other side in least possibleways. === Subject: Re: JSH: Apology to Ramsay, why I post> I have no faith in intelligence tests. You, perhaps, do.>Let me put it this way. If you compared your IQ to mine, >you would not like the result. That holds for a great many>other posters on sci.math as well.> I resent the insinuation. It may well be that if I were to> compare my IQ to yours I would like the result. Oh, you> didn't mean that? The problem is your use of that. > That is a dangerous thing to do. I thought you meant that, when in> fact you didn't mean that but instead meant that. That can lead to> confusion.> -William Hughes Yes, I thought about that later & said to myself: *that* didn'tcome out quite right. What I meant was that if Harris compared his IQ to that of a great many other sci.math posters (you included, Mr. Hughes), he would not like the result. On the otherhand you might be delighted to compare yours to mine. Harris was identified at an early age as 'gifted'. This appearsto have given him the impression that he is Special and maybe thathe doesn't need to work hard to learn things like other people do.It's probably a good thing to identify bright people early in life, but here is an example of where it may have done irreparableharm. Marilyn Vos Savant is another example. No real intellectual accomplishments to her name - a silly general-advice column inParade magazine, based on absolutely nothing more than her worldrecord IQ (now surpassed, I believe, by a black girl). Ms. VosSavant made the mistake of thinking she too did not have to workto understand math - she actually published a little book claiming thatWiles' proof was incorrect because it used non-Euclidean geometry!She had absolutely no concept of what she was talking about. Itwas an embarrassment. I don't think she has ever acknowledged how far off base she was. Mathematicians as a group have the image of inarticulate cold-blooded arithmetic whizzes with a feeble grasp on reality, no girl friends (or boy friends), total nerds and dimwits exceptfor this one thing they do well. When I was in grad school there*was* a subset like that. They were *not* the top students. The bestones resembled Andrew Wiles: well-rounded, incredibly intelligentpeople, well-socialized, interesting, etc.. [I claim no such properties for myself.] Some of us are, to be sure, obsessive -there are both good and bad things about that. Doing well on IQ tests is not important. You do or you don't.In general I would bet sci.math groupies do quite well. It doesn'tmatter. IQ is a highly artificial construct, dreamed up by some psychologists who thought that multiple-choice tests could capturewhatever quality it was that Newton, Fermat, Kant, Aristotle, Confucius, Darwin, Tolstoy, El Greco, and other people with real accomplishments had - even though those accomplishments had NOTHING in common with,say, visualizing little 3D objects or having a large vocabulary. Accomplishment takes talent, but more than that, it takes hard work, a lot of learning, intensity, some kind of inner drive,concentration, curiosity, etc.. Wiles' case is interesting for another reason. His great work was not done when he was a teenager or in his 20s. There is a myth that math people are washed up by age 30.That may not be true any more. Marston Morse did good work in his 80s. A lot of the easy math has been mined out; the hard stuff that is left requires a huge amount of erudition. That is Harris's most serious error: thinking that because he is Special he can take a short cut right to the finish line without having to learn or understand much of anything. I'm not saying this out of jealousy or desire to protectthe Establishment. Just look at his track record. He makes a lifetime'sworth of serious conceptual errors every month or so ... without eventrying. In the end, not one single realized accomplishment, althoughI do think he has set some kind of high-water mark for being the most incredibly successful and annoying jerk on the Internet up to the present time. The Isaac Newton, the pioneering Daniel Boone of crank/trolls! This alone might eventually get him a spot on Letterman. Nora B. === Subject: Re: Power of a construction > Focusing on > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are > roots of > > a^2 - (x - 1)a + 7(x^2 + x) > > was always just half the story. The rest was always what does it look > like when you actually divide that 7 from both sides. > > Some posters have basically put forward that it's a mystery wrapped up > in an enigma, choosing to use unknowns w_1(x) and w_2(x),Not exactly a mystery and not exactly an enigma, I have provided expressionsfor w_1(x) and w_2(x)... How to calculate them is a different story. > in an enigma, choosing to use unknowns w_1(x) and w_2(x), so you'd > have > > (5b_1(x) + w_1(x))(5b_2(x) + w_2(x)) = 25x^2 + 30x + 2 > > but what does that tell you?It tells us that you reversed the function of w_1(x) and w_2(x).Using b_1(x) = a_1(x)/w_1(x) a more proper expression would be: > (5b_1(x) + w_2(x))(5b_2(x) + w_1(x)) = 25x^2 + 30x + 2 > It's actually worse than uninformative as it gives a false > implication, since it appears that the w's here are in some way > factors of 2,Why? That would be the case if b_1(x) and or b_2(x) are zero forsome x, but at most one of them is zero for x = 0 or x = -1. > but w_1(x) w_2(x) = 7, and the idea is that they're > algebraic integers. > > Now then, what's the difference with > > (5b_1(x)/sqrt(7) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2?Methinks you mean 7 rather than sqrt(7). But b_1(x)/7 is provablynot an algebraic integer. On the other hand, b_1(x)/w_1(x) *is*an algebraic integer for all x with a proper definition of w_1(x).Here a table (based in part on work by Keith Ramsay and Rick Decker):x w_1(x)-5 [(257 + 3.sqrt(-131))/2]^(1/5)-4 (99831352 + 2146257.sqrt(-311))^(1/19)-3 (1 + 3.sqrt(-38))^(1/3) (due to Rick)-2 (- 128 + 3.sqrt(-47))^(1/5)-1 7 (due to James) 0 7 (due to James) 1 sqrt(7) (due to Rick) 2 (44444 - 111.sqrt(-167))^(1/11) (due to Keith) 3 [(- 25 + 3.sqrt(-83))/2]^(1/3) 4 (194324 + 10359.sqrt(-551))^(1/13) 5 (11 + 9.sqrt(-206))^(1/5) > Well it works for one thing, and does so in such a way that b_1(x) and > b_2(x) can be functions in a ring where -1 and 1 are the only integer > units.You never have given a definition of such a ring so that one coulddetermine whether a particular algebraic number is in that ring or not.More proper, as all of the numbers above are algebraic integers, and asyour ring contains the algebraic integers, all of the numbers above arein your ring. So even in your ring both factors are not coprime to 7,unless you say that quite a few algebraic integers are units in yourring... > Notice that with my mathematics I can divide off that 7, with yours > you cannot do so without confusion.The confusion is only on your end. > So is it really *my* mathematics then, or isn't it actually the > mathematics, while some of you are holding on to something awkward, > misleading and confusing? Think about it.I expanded the table above using basic mathematical theory. Apparentlythat basic theory leads to results, so I do not know why you think itis misleading. It is indeed awkward, and it may be confusing, but notmisleading, because it gives results.dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Power of a construction> Look at> (5b_1(x)/sqrt(7) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> and consider.> Focusing on > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are> roots of> a^2 - (x - 1)a + 7(x^2 + x)> was always just half the story. The rest was always what does it look> like when you actually divide that 7 from both sides.> Some posters have basically put forward that it's a mystery wrapped up> in an enigma, A tired, trite phrase [misquoted], and here of course utterly untrue. We specified w_1(x) and w_2(x) as follows: w_1(x) = GCD((5 a_1(x) + 7, 7). w_2(x) = (5 a_2(x) + 7)/f_1(x),where f_1(x) = 7/w_1(x). There is no mystery or enigma about them, although it must be said that actually computing them is difficult. We have shown how they canbe computed for various values of x. There is a deterministicalgorithm with well-defined answers. There are computer programswhich implement the algorithm. You want a *formula*. There is no formula. There is an algorithm.Just as your prime counting function is an algorithm. You cannotwrite down a *formula*, but you can write a well-defined algorithmwhich does the computation. Just as with your algorithm, you canwrite a computer program that implements it. Your saying it is a mystery is deliberately misleading propaganda.On the other hand, perhaps to you is *is* a mystery, since evidentlyyou cannot understand or accept it.> choosing to use unknowns w_1(x) and w_2(x), They are not unknown. They have explicit definitions andthey are computable by a well-defined algorithm. See above.> so you'd> have> (5b_1(x) + w_1(x))(5b_2(x) + w_2(x)) = 25x^2 + 30x + 2> but what does that tell you? Using the definitions I gave above, it tells you that there isa solution for which all the coefficients on the left arealgebraic integers. And it tells you more than that. Thereis not simply an existence proof. Behind it there is analgorithm, due essentially to Dedekind, for computing w_1(x)and w_2(x). Keith Ramsay used that algorithm to solve forw_1(2) and w_2(2). Since then, Dik has shown the answer forw_1(3) and w_2(3). You have already seen Decker's result for x = 1. Decker also found the result for x = -3. I'm not saying it's EASY to compute. Usually it's not. For manyvalues of x, you would not want to do it by hand. But it'sPOSSIBLE to compute w_1(x) and w_2(x) for any integer x.> It's actually worse than uninformative as it gives a false> implication, since it appears that the w's here are in some way> factors of 2, It appears that way to you. No one else claims that. Clearlythey're not.> but w_1(x) w_2(x) = 7, and the idea is that they're> algebraic integers. That's part of the idea.> Now then, what's the difference with> (5b_1(x)/sqrt(7) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2?> Well it works for one thing, Not really. It works in the ring of algebraic *numbers*, but that isof course trivial. It does NOT work in the ring of algebraic*integers*, which is what you want. It is the wrongfactorization.> and does so in such a way that b_1(x) and> b_2(x) can be functions in a ring where -1 and 1 are the only integer> units.> Notice that with my mathematics I can divide off that 7, with yours> you cannot do so without confusion. No, there's no confusion at all. If you divide the firstfactor by 7, you do not get algebraic integer coefficients. What is confusing about that? It just tells you that you have chosen the wrong factorization. There is another CORRECTfactorization with w_1(x) and w_2(x) as defined above whichgives you A.I. coefficients everywhere. It depends on what you want. If you want an easy but incorrect factorization, you should stick with yours. If youwant a harder, but correct factorization, you should usew_1(x) and w_2(x) as defined above. Which do you want?Isn't correct more important than easy? In either case, do NOT try to tell people, as you have over and over again, that yours is the ONLY POSSIBLE factorization. Yours is a factorization which basicallydoes not work. You cannot conclude that no other factorizationworks either. You have seen that that is false with examples*** which you said repeatedly were impossible ***. Just out of curiosity, how many more examples do you need?> So is it really *my* mathematics then, or isn't it actually the> mathematics, while some of you are holding on to something awkward,> misleading and confusing? The choices you are outlining are: 1. Simple but wrong [yours] 2. More complex but correct [ours] *Not* awkward, misleading and confusing. Nora B.> Think about it.> > Decker Quadratic Source Information> ---------------------> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.> If you wish to see his original post here are some headers which also> show that he posts from Hamilton College: === > Subject: Re: Mathematical consistency, courage> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: Power of a construction> Look at> (5b_1(x)/sqrt(7) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2> and consider.> Focusing on > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where the a's are> roots of> a^2 - (x - 1)a + 7(x^2 + x)> was always just half the story. The rest was always what does it look> like when you actually divide that 7 from both sides.> Some posters have basically put forward that it's a mystery wrapped up> in an enigma, choosing to use unknowns w_1(x) and w_2(x), so you'd> have> (5b_1(x) + w_1(x))(5b_2(x) + w_2(x)) = 25x^2 + 30x + 2> but what does that tell you?It tells me that you either don't follow what people say with anycare, or that this is more mendacious posting on your behalf.The w's given to you before are not the w's you use here. > It's actually worse than uninformative as it gives a false> implication, since it appears that the w's here are in some way> factors of 2, but w_1(x) w_2(x) = 7, and the idea is that they're> algebraic integers.You see, that's just wrong. No one has said the w's given to you byDik orginally (I believe) were divisors of 2 in the algebraicintegers.> Now then, what's the difference with> (5b_1(x)/sqrt(7) + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2?> Well it works for one thing, and does so in such a way that b_1(x) and> b_2(x) can be functions in a ring where -1 and 1 are the only integer> units.> Notice that with my mathematics I can divide off that 7, with yours> you cannot do so without confusion.> So is it really *my* mathematics then, or isn't it actually the> mathematics, while some of you are holding on to something awkward,> misleading and confusing?Your mathematics, for want of a better word, is deeply flawed, as youcannot even tell us what this Object Ring is. Nor can you seem toaccept, despite the plethora of evidence, that you're wrong in yourassertions about the algebraic integers.> Think about it.> Decker Quadratic Source Information> ---------------------> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.> If you wish to see his original post here are some headers which also> show that he posts from Hamilton College: === > Subject: Re: Mathematical consistency, courage> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).