mm-186 === Subject: Re: the anticlassicalist }{ ii: the spectre =-=-=-=-=-=-: :Now, with this long history of logical analyses of language,I always find it strange:: Well, it is not strange.: You are ßaunting ignorance here.It is strange, since it is unnecessary.I am ßaunting ignorance there.that there are comments like:: :Attention being devoted here entirely to the classicaltwo-valued theories of truth-functions, quantifiers, andidentity, with syntax, semantics, and pragmatics builtupon their basis, there will be no concern with alternativeforms of logic, so-called three-valued (or, more generally,n-valued) logics, modal logics, intuitionist logics, andthe like. The view is that whatever is valuable in thesealternatives can be achieved more readily within theclassical framework by suitable extensions.: :in the introduction to Semiotics and linguistic structureby R. M. Martin, where such a representation is notfaithful to actual usage or expression.:: Actual usage or expression is just irrelevant. Unless you are alinguist: whose object of study is how real people use natural language, natural: language IS JUST STUPID, because PEOPLE are just stupid.There is a formalisation to logic. We can formalise many forms ofreasoning. There is no necessity to choose one logical form over anotherexcept for representing the model faithfully.So unfornutately, your comment doesn't help resolve my ignorance.Now, I just recently posed some questions on thenewsgroups concerning nonclassical logic, and certainlinguists and physicists were actually quiteconfrontational about the idea of educating about thesenonclassical logics. Expressibility was always profferedas a reason, even though no one can even claim thatboolean structures possess a uniqueness in this statement is incredibly stupid. Lambda calculus is: BIGGER AND BADDER than classical boolean anything; it is MORE: complex. SO OF COURSE it is rich enough to express: whatever. That is not even the question. The question: is whether you can achieve similar expressiveness with: SIMPLER machinery (like classical boolean algebra).: There is, after all, a set of first-order axioms defining: lambda-calculus.So it is complexity, then? Because I know an orthomodular logic to describequantum propositions that is simpler than any embedding could be intoBoolean. I know many models whose Heyting structure is far more simplisticthan the corresponding Boolean embedding.And since Heyting algebras have a potential universality hinted by theCurry-Howard isomorphism, why is it so necessary to fall back on theclassical approach. I still do not see from where this desire arises atforcing the ontology of a model...-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: === anticlassicalist }{ ii: the spectre continues> galathaea with this long history of logical analyses of language,> I always find it strange>> Well, it is not strange.> You are ßaunting ignorance here.>> It is strange, since it is unnecessary.>> I am ßaunting ignorance there.>> that there are comments like:>> Attention being devoted here entirely to the classical> two-valued theories of truth-functions, quantifiers, and> identity, with syntax, semantics, and pragmatics built> upon their basis, there will be no concern with alternative> forms of logic, so-called three-valued (or, more generally,> n-valued) logics, modal logics, intuitionist logics, and> the like. The view is that whatever is valuable in these> alternatives can be achieved more readily within the> classical framework by suitable extensions.>> in the introduction to Semiotics and linguistic structure> by R. M. Martin, where such a representation is not> faithful to actual usage or expression.>> Actual usage or expression is just irrelevant. Unless you are a> linguist> whose object of study is how real people use natural language, natural> language IS JUST STUPID, because PEOPLE are just stupid.>> There is a formalisation to logic. We can formalise many forms of> reasoning. There is no necessity to choose one logical form over another> except for representing the model faithfully.>> So unfornutately, your comment doesn't help resolve my ignorance.>> Now, I just recently posed some questions on the> newsgroups concerning nonclassical logic, and certain> linguists and physicists were actually quite> confrontational about the idea of educating about these> nonclassical logics. Expressibility was always proffered> as a reason, even though no one can even claim that> boolean structures possess a The parenthetical statement is incredibly stupid. Lambda calculus is> BIGGER AND BADDER than classical boolean anything; it is MORE> complex. SO OF COURSE it is rich enough to express> whatever. That is not even the question. The question> is whether you can achieve similar expressiveness with> SIMPLER machinery (like classical boolean algebra).> There is, after all, a set of first-order axioms defining> lambda-calculus.>> So it is complexity, then? Because I know an orthomodular logic to describe> quantum propositions that is simpler than any embedding could be into> Boolean.Just FYI:If by Ôsimpler', you mean Ôfewer defining properties', then that's notthe same as the complexity that was probably meant, here. I know many models whose Heyting structure is far more simplistic> than the corresponding Boolean embedding.Can you name them? Heyting algebras are always infinite, afaik.> And since Heyting algebras have a potential universality hinted by the> Curry-Howard isomorphism, why is it so necessary to fall back on the> classical approach. I still do not see from where this desire arises at> forcing the ontology of a model...Perhaps the following may help (and perhaps not :-) ).A Heyting algebra is a mathematical structure of some kind.It's defined as an infinite set, together with some operators andrelations, that satisfies certain (first-order logic) conditions.So even to understand the notion of a Heyting algebra, mostpeople require a good intuitive picture of the Tarski semanticsof first order (classical) predicate logic.Constructivism has so many variants that, in order to studythem well, most systems are defined and studied using classicalmeans and classical thinking habits. In more vague terms:'reasoning on the meta level is still classical'.You may find that awful, but that's the way it is. One reasonfor this is that there is no reason to prefer one form ofconstructivism over another. You like Heyting algebras,someone else likes some other system. And unlikethe several formalizations of classical logic, these variantsare far from being mutually equivalent or translatable.Using classical definitions of the lot allows us at leastto understand all variants at the same time, and to makemutual comparisons. There is plenty of room for moreresearch and other points of view, but as something thatis supposed to give students a good basis for furtherresearch, classical logic is still essential, constructivisman extra. That may serve as an explanation why not manyhave responded very enthousiastic to your pamßet. === ii: the spectre continues[...]: Just FYI:: If by Ôsimpler', you mean Ôfewer defining properties', then that's not: the same as the complexity that was probably meant, here.You're absolutely right. I intended simpler here in terms of definingaxioms.[...]: Heyting algebras are always infinite, afaik.The axiom collection is finite for general Heyting algebras. There is aninfinite hierarchy of expressions which evaluate to truth values which canbe shown to be different, but the collection of expressions in booleanlogics is also infinite.In other words, both have ways to build sentences in an infinte number ofways but are both finitely presentable.[...]: Perhaps the following may help (and perhaps not :-) ).:: A Heyting algebra is a mathematical structure of some kind.: It's defined as an infinite set, together with some operators and: relations, that satisfies certain (first-order logic) conditions.: So even to understand the notion of a Heyting algebra, most: people require a good intuitive picture of the Tarski semantics: of first order (classical) predicate logic.Although you may decide to model the logic this way, understand that it is amodel of a model, and there are other models of Heyting structures that arepossible. And even then, I am not talking of models of models. I'vealready mentioned that this extra indirection allows alot of metalogics.What I am talking about is the original model itself, built with the logicof something like a Heyting algebra. I can model that in anotherHeyting-based model if I desire, or I can use a first order classical model,or indeed anything else that has enough expressibility to build the definingrelations for a Heyting structure.Again, the defining relations for Heyting algebras seem in many ways to be_prior_ to those of the Boolean algebras, and I see no reason to look forembeddings with more complicated structure or the indirection of models onmodels. And there are definitely many propositions that may fail to haveboolean truth values (does program x halt?, is statement y provable insystem z?, did electron e go through the right slit?, etc.), where themetamodel does not change the fact that the basic model must still have anextended logic.: Constructivism has so many variants that, in order to study: them well, most systems are defined and studied using classical: means and classical thinking habits. In more vague terms:: Ôreasoning on the meta level is still classical'.: You may find that awful, but that's the way it is. One reason: for this is that there is no reason to prefer one form of: constructivism over another. You like Heyting algebras,: someone else likes some other system. And unlike: the several formalizations of classical logic, these variants: are far from being mutually equivalent or translatable.Heyting algebras seem to cover the spectrum for constructive theories, atleast from what I've seen (I would like references otherwise, though). I doknow the distinctions between the Russian school, Brouwer's ideas, and such,but that is why I focus on the common Heyting form of the collection ofconstructive axioms. This seems intricately related to the view ofconstruction in proof and programming found in the Curry-Howard isomorphism.: Using classical definitions of the lot allows us at least: to understand all variants at the same time, and to make: mutual comparisons. There is plenty of room for more: research and other points of view, but as something that: is supposed to give students a good basis for further: research, classical logic is still essential, constructivism: an extra. That may serve as an explanation why not many: have responded very enthousiastic to your pamßet. :-)But my whole point is that classical logic is _not_ essential at theselevels. It has its place in certain realms of mathematics, but the reasonsfor our metamodels being classic is historical, not of any fundamentalrequirement. When you keep saying that classical logics appear in the metalevel, you have to realise that the choice is not unique, and does notchange the fact that the immediate logic of what is being modeled is veryoften not Boolean. We could have standardised Heyting algebras as the logicto express our metamodels in, and would have found it completely amenable.And very often in these models, it is important to understand the rules ofpropositions on the immediate level, and how they differ from classicalterms of quantum logic in hopes of making this much more clear, but I wantedto respond to you more immediately to underline these points.-- -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea: === James Harris>Come on, reply to my posts!!! Come why don't you say something else,>huh?> [infantile tantrum snipped]> As always, you are talking about yourself. You live in dread of losingyour> audience on usenet. Pathetic but true. If one of your worthless threadsgoes> unanswered for even 12 hours, you panic, and put up three more within one> hour. If you behave like this and you expect to be admired, or even> tolerated, then you are insane.> Bad news, Harris: I will not respond to you again.>>Evidently, JSH thrives on attention. Sad that a === Re: principal ideals in F[x,y]Arturo's proof is certainly correct, but I think that after the line> h(x,y)*f(x,y) = x and h(x,y)*g(x,y) = y.it's easier to simply plug in some values. Putting x=0 into the firstidentity gives h(0,y)*f(0,y)=0, so either h(0,y)=0 or f(0,y)=0.Now if f(0,y)=0, then f(x,y)=x*F(x,y) for some polynomial F, and henceh(x,y)*x*F(x,y)=x. This implies that h is a constant. But h(0,0)=0,since h is in the ideal, so h(x,y)=0. Contradiction. Hence f(0,y) isnot 0. This proves that h(0,y)=0, so h(x,y)=x*H(x,y) for somepolynomial H.Substituting into the second equation gives x*H(x,y)*g(x,y)=y, andputting x=0 into this identity yields the absurdity 0 = y. Hence theideal is not principal.BTW, a similar argument shows that if p is a prime number, then theideal { f(y) in Z[y] such that f(0) = 0 (mod p) }is not a principal ideal. (Notice that this is the ideal generated byp and y, so it's very similar to the ideal of F[x,y] generated by xand y.)Joe Silverman> >If F is a field, and I is the ideal {fin F[x,y] : f(0,0)=0} in> >F[x,y], then why is I not principal?> >> >I'm trying to show by contradiction using the fact that> >deg(fg)=deg(f)+deg(f). hasn't worked so principal, say I=(h(x,y)), then there must be polynomials> f(x,y) and g(x,y) such that> > h(x,y)*f(x,y) = x and h(x,y)*g(x,y) = y.> > of any monomial in h(x,y) is 1; that is, h(x,y) must be of the form> h(x,y)=a+bx+cy.> > The same is true for both f and g. Say f(x,y) = r+sx+ty.> > Then ct=0, (bt+cs)=0, ra=0, (at+rc)=0, and as+br=1.> > If g(x,y)=u+vx+wy, then> > ua = 0, vb=0, cw=0, (bw+cv)=0, (av+bu)=0, and (aw+uc)=1.> > So either a=0, or both r and u are 0. > > If a is zero, then either b=0 or u=0 (from av+bu)=0; but you cannot> have both a and b zero, since as+br=1. So a=0 implies u=0. But then> aw+uc=0, and it is supposed to be 1. So a is not zero.> > Therefore, r=u=0. Since a cannot be zero and at+rc=0, then t=0. Which> means that cs=0; If s=0, then as+br=0, which is impossible. So> c=0. Since bw+cv=0, then b=0 or w=0. But w cannot be zero (since> aw+uc=1), so b=0. Now we have a=t=u=c=b=0. But then as+br=1 is> impossible. > > Thus, no such h exists. (Other, simpler arguments, are of course> possible). Therefore, I is not principal.> > > -- > I accept as reality.> --- Calvin (Calvin and Hobbes)> === Circle>For b, what does a C3v (invariant under 120 deg rotation) >symmetric arrangement yield? (Place the outermost edge of>the squares tangent to the circle.)> > Here's a similar situation that does better than the 80.4%> with part (a). Use this 120 deg symmetry, but place the diagonals> of the square along the radii. The optimum placement of the> squares (they can be shifted along the radii) covers 83.0% of> the circle's area. This is the best that I've found, but I don't> know if it's the best possible.To be a little more precise, the area is 2.60808705... I have an exactform, but it's pretty messy. This happens when the diagonal of eachsquare extends .13980931 past the center of the circle === Circle>> For b, what does a C3v (invariant under 120 deg rotation) >> symmetric arrangement yield? (Place the outermost edge of>> the squares tangent to the circle.)>>Here's a similar situation that does better than the 80.4%>with part (a). Use this 120 deg symmetry, but place the diagonals>of the square along the radii. The optimum placement of the>squares (they can be shifted along the radii) covers 83.0% of>the circle's area. This is the best that I've found, but I don't>know if it's the best possible.> > To be a little more precise, the area is 2.60808705... I have an exact> form, but it's pretty messy. This happens when the diagonal of each> square extends .13980931 past the center of the circle (along the> diameter).Oh, that's a nice configuration to consider!I checked your result, and I agree with your area Ac = 2.60808705,although the extension of the square beyond the center of the circle isa little off: it should be rc = 0.13977317. (I would guess that youobtained the value of Ac to single precision, which yields only half theprecision for rc).For general r (distance diameter extends beyond the circle's center),the area A of the resulting figure can be found as follows:Let u = sqrt(2) - r, v = sqrt(2 - u^2), d = (u - v)/2, andp = (1 + u v)/2, then A = 3 (p - sqrt(p) d - (3 + sqrt(3)) r^2/2 + Arcsin(d)).This area is maximized when r equals the critical value rc: rc = (19 - 5 sqrt(3) - sqrt(176 - 86 sqrt(3))) / (26 sqrt(2)).The formula for Ac = A(rc) simplifies greatly (as one would expect).Let dc = rc (3 + sqrt(3))/2. Then Ac = 3 (1 - sqrt(2) dc + Arcsin(dc)).To summarize the results so far, the best result posted for case (a)(no overlap) is(a) pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8) = 2.57003584 = 81.8067816% of the circle,and the best result posted for case (b) (overlap allowed) is(b) Ac (see above) = 2.60808705 = 83.0179893% of the circle.-Jim === does a C3v (invariant under 120 deg rotation) >>symmetric arrangement yield? (Place the outermost edge of>>the squares tangent to the circle.)>> >> Here's a similar situation that does better than the 80.4%>> with part (a). Use this 120 deg symmetry, but place the diagonals>> of the square along the radii. The optimum placement of the>> squares (they can be shifted along the radii) covers 83.0% of>> the circle's area. This is the best that I've found, but I don't>> know if it's the best possible.>> To summarize the results so far, the best result posted for case (a)> (no overlap) is> > (a) pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8) => 2.57003584 = 81.8067816% of the circle,I have an improvement for case (b) (over Jake's 83.0% configuration).It again has C3 symmetry, but not D3 symmetry (as Jake's and Hauke'sconfigurations do).Let one corner of a square be at (-r,0), and the opposite corner beat (-r + sqrt(2) cos(phi), sqrt(2) sin(phi)) -- i.e., the diagonalconnecting (-r,0) to the opposite corner is oriented at an angle phi.Let the other two squares be the image of this one under rotationsby 2 pi/3 and 4 pi/3, respectively. (Note that for phi = 0, thisreduces to Jake's configuration.)The optimal case has r = rc and phi = phic, where rc = 0.192008425, and phic = 0.678933044.The area Ac achieved by this configuration is Ac = 2.61034477 = 83.0898548% of the circle.The value 2.60808705 for Jake's configuration is a local maximum fora constrained 1-parameter problem, and the value above is a localmaximum for a constrained 2-parameter problem. But of greater interestis whether the values are local maxima for the unconstrained problem,which has 9 parameters (3 for each square).I looked into this. Both constrained maxima are indeed critical pointsfor the unconstrained problem (the gradient of the area function is 0for each). To assess whether the critical points are local maxima, weform the Hessian matrix and look at its eigenvalues. One eigenvaluemust be 0 (corresponding to the eigenvector that rotates the entiresystem). For Jake's configuration, there are 5 negative, and 3 positivepositive eigenvalues, so it is a saddle rather than a local maximum.For the new configuration I've given, all 8 non-zero eigenvalues of theHessian are negative. The configuration is indeed a local maximum ofthe unconstrained problem, and hence a viable candidate for the titleof global maximum.Note: if you graph the configuration it may look like some of the cornersof the squares lie on the circle, but they do not. They are at a distance0.99803 from the center of the === Michelson-Morley-Miller,> with Michelson being the pioneer of interferomteric methods> for this. D.C.Miller's refinement highlighted the anomolies> that MM had foundThey found none. Miller's results could not be duplicated by anyone butMiller with his apparatus.> but this seems to have been dysmissed, because> of a comment that Eintstein made,It was dismissed because it could not be repeated. It was dismissedbecause it was not verfiable.> on one of his brief forays to his office at CalTech.> there is no vacuum, per se, anyway, unless> you rely on the formalism of atoms as zero-dimensional points> (and === the Bell curveWhat i m trying to get, is very interesting fundamentally. got an interesting insight, by working with the Traveling Sales Manproblem, which is known to be NP-Hard.It seems that distribution of paths are gaussian. And based on thisassumption, it is possible to construct an Suboptimal alogorithmfinder that finds a solution in O(N^2).The solution however, heuristically seems to be acccurate, to onlyabout sqrt(N!)/N! = 1/sqrt(N!)1/sqrt(N!) of all total paths are missed by this algorithm.Ofcourse there many other assumptions, the algorithm makes. One isthat costs are symmetric and that costs are unique.Now , if N-> infty, 1/sqrt(N!) would goto 0. In a way, may bequantum computers, === approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id sci.mathSmall mistake below, at [*****].NB>> I let it get personal, and made an earlier post with a silly sign>> mistake. Then I just got really pissed, and made some posts. But>> here's what I feel is the proper approach, with a focus on an implied>> dependency that Dik Winter never bothered to handle.>> >> Consider the Decker quadratic example (reference at bottom).>> >> 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).>> >> Dik Winter has repeatedly asserted that there exists varying algebraic>> integer functions w_1(x) and w_2(x), such that>> >> w_1(x) w_2(x) = 7>> >> and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and>> w_2(x) as factors, respectively.>> >> Your description is accurate. I agree with Dik on this. We have>both given explicit definitions of w_1(x) and w_2(x). See below.>> Now then, introducing f_1(x) and f_2(x) as the other factors of the>> a's, you have>> >> w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, >> >> so>> >> a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5.>> >> Now it matters to use the fact that the a's are roots of>> >> a^2 - (x - 1)a + 7(x^2 + x),>> >> as solving that quadratic, and picking a_1(x) for the positive sign>> root gives>> >> a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so>> >> (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2>> >> so>> >> w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so>> >> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x),>> >> which implies a relationship between f_1(x) and w_1(x), but>> >> f_1(x) f_2(x) = 25x^2 + 30x + 2>> >> Correct. First, a_1(x) is a root of>>[*] a^2 - (x - 1)*a + 7*(x^2 + x).>>Thus a_1(x) is clearly dependent on x. Note however that it>is also dependent on 7: if you replace 7 by, say, 17, then>then a_1(x) is not a root of [*] but instead is a root of>> a^2 - (x - 3)*a + 17.[*****} This should have been a^2 - (x - 3) + 17*(x^2 + x).>> Now, back to a_1(x) when 7 is used: w_1(x) is also a function>of both x and 7, because when 7 is used, you get>> w_1(x) = GCD(a_1(x), 7).>>and when 17 is used, you would get >> w_1(x) = GCD(a_1(x), 17).>> Note that when you replace 7 by 17, a_1(x) also changes;>therefore w_1(x) does also.> What you call f_1(x) is:>> f_1(x) = (5 a_1(x) + 7)/w_1(x),>>which is ALSO clearly dependent on x, but also clearly dependent>on 7: again, if you replace 7 by 17, you would get a different>value for a_1(x), w_1(x), AND f_1(x). If you want all these>functions to be distinguished for situations other than 7, you >really should write them as a_1(x, 7), a_1(x, 17), f_1(x, 7), >f_1(x, 17), etc..>> and it is arbitrary that 7 was the multiple before as it could have>> been 11 or 13 or any of an infinity of other numbers, and w_1(x) and>> w_2(x) are themselves not determined, so it's a spurious appearance of>> dependency.>> >> No - it's not spurious at all. If you replace 7 by something >else, everything changes. For example, with 17, the original>assumption would be>>[**] (5 a_1(x) + 17)*(5 a_2(x) + 17) = 17*(25*x^2 + 30*x + 2).>>There is no reason to think that the same a_1(x) will work for>17 as for 7; the factorizations are quite different. [For one>thing, the CONSTANT terms are different.] Similarly>for w_1(x) and f_1(x): they will be different for 17 than they>are for 7, etc.. > That's the little detail that Dik Winter never bothered to address.>> >> Why should he? He was considering only the 7 factorization.>You appear to think that a_1(x), etc., will not change when you >replace 7 by other numbers. Clearly from the left side of>[**] above, that is not going to be true.>> Now then, if f_1(x) is in fact independent of w_1(x), then how do you>> account for the appearance of a dependency in>> >> f_1(x) is NOT independent of w_1(x). You *defined* it as>> f_1(x) = (5 a_1(x) + 7)/w_1(x).>> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)?>> >> If you try to push the issue that it isn't so required consider what>> happens if you try to divide through by w_1(x), as then you have>> >> f_1(x) = >> >> (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x))>> >> and the problem is that all of the w's need to go away. So assuming>> that a_1(x) has w_1(x) as a factor means, introducing g(x), that you>> have>> >> a_1(x) = w_1(x) g(x), so>> >> f_1(x) = (5g(x)/2+ 2 w_2(x))>> >> which indicates that f_1(x) is dependent on w_2(x).>> >> It is, of course. w_1(x) and w_2(x) are highly dependent>also since their product is 7.>> However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them as>> an infinity of functions will work, > Not for fixed values like 7. There are a lot of constraints. >For example, for 7, a_1(x) is a root of a^2 - (x - 1) + 7*(x^2 + x),>so a_1(x) is restricted to only being one of two possible numbers.>Then since the right definition of w_1(x) is>> w_1(x) = GCD(a_1(x), 7), >>w_1(x) is determined once you have chosen a_1(x). Why? Because>the GCD function is well-defined (to within units) in the ring>of algebraic integers.>> while f_1(x) is independent of>> w_1(x) and w_2(x) since>> >> f_1(x) f_2(x) = 25x^2 + 30x + 2.>> >> The point is you can multiply some polynomial like 25x^2 + 30x + 2, by>> anything you choose. There's no way that it's locking into it that>> functions that are factors of 7 are required.>> >> Not right - see above. If you replace 7 by something else, like 17,>all of a_1(x), w_1(x), and f_1(x) change. They are all linked >together by the equation>> f_1(x) = (5 a_1(x) + p)/w_1(x),>>where p could be 7 or 17, etc.>> The simplest answer is that w_1(x) either equals 7, or it equals 1,>> and checking with the first gives>> >> f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/14 + 2)/w_1(x)>> That simplest answer however does not give algebraic integer>coefficients as you well know, and that is what you wanted in >the first place. It's simple, but it is the wrong factorization.>Things are not always simple!>> >> implying that 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x))) has a factor that>> is 14, which is false in the ring of algebraic integers, so Dik Winter>> and his cohorts falsely seize on that to hide the issue with their own>> claims.>> >> They've gone so far that Keith Ramsay even claimed to have posted a>> solution for the w's, but how did he pick w's from infinity?>> >> He didn't. There is a well-defined algorithm for computing>the w's, starting with the number (1 + sqrt(-167))/2. If you>had replaced 7 with 17, the starting number would have been>(-1 + sqrt(-407))/2 and the result would be different.>> Given that his polynomial is of degree 22, it's quite possible that he>> just chose a really big polynomial!>> >> Actually I think it had degree 11. I think he used a program>to do the computation, and I think it finds the lowest-degree>polynomial that works. I believe it computes powers of an>ideal, ^M, until it finds an M where that ideal>is principal. It is not a matter of human choice, just a >programmable algorithm.>> Remember the issue here is f_1(x) and it's *lack* of a dependency with>> w_1(x).>> >> See above. They are quite dependent, and both change>if you replace 7 by other numbers.>> The problem for Dik Winter and his buddies is that you have two>> independent equations:>> >> w_1(x) w_2(x) = 7>> >> and>> >> f_1(x) f_2(x) = 25x^2 + 30x + 2>> >> To try and dispute my results, posters like Dik Winter or Nora Baron>> simply skip past mathematical consequences of their claims, like the>> independence between those equations.>> >> We focussed on the 7 case because that was Decker's original>example and your own main interest. Other cases are similar>*but not the same*. We never claimed they were.>> I'm still pissed. > Too bad. >> In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about >this. We are not evil conspirators trying to rob you of your>rightful credit. We are just ordinary people trying to show you>where you are making mistakes.>> Nora B.>> But there's not a lot I can do when people like Dik>> Winter can so easily get away with basic problems in their claims, on>> a newsgroup that doesn't seem to give a damn about mathematical truth.>> >> >> James === long time ago, J.J. Sylvester posed the problem:if I have arbitrarily many 5 cent or 17 cent stamps,what is the largest denomination I cannot make?In general, if we have p and q cent stamps, it turnsout the answer is pq-p-q (granted p and q are coprime).I have derived a solutionto the problem, but I'd like to teach this to myundergraduates, some of whom have a limited background.So my question is: is there a very nice & friendlyproof of this fact? (For example, which avoids anynonobvious facts from number theory.)In case you're interested, the class I'm teaching islinear algebra, but as you can see I like to givepuzzles to the class which are not === long time ago, J.J. Sylvester posed the problem:>if I have arbitrarily many 5 cent or 17 cent stamps,>what is the largest denomination I cannot make?>>In general, if we have p and q cent stamps, it turns>out the answer is pq-p-q (granted p and q are coprime).>I have derived a solution>to the problem, but I'd like to teach this to my>undergraduates, some of whom have a limited background.>So my question is: is there a very nice & friendly>proof of this fact? (For example, which avoids any>nonobvious facts from number theory.)I answered this question recently. Take a look .Rob Johnson take out the trash before === someone tell me how to do this problem? Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmetic> sequence. I looked in the back and saw that the answer was (-3/2), but I am not> sure how they got that.> If they must be *consecutive* terms, as listed, for some arithmetic> sequence, then the differences between consecutive terms must all be> equal and (2x+1) - (x+3) = (5x+3) - (2x+1). If they need not be consecutive, for any arithmetic sequence, the> problem has no unique solution.But, notice what the real question is... The answer given in the === in message > Can someone tell me how to do this problem?> > Find x so that x+3, 2x+1, and 5x+3 are terms of an arithmetic> sequence.> > I looked in the back and saw that the answer was (-3/2), but I am not> sure how they got that.> > If they must be *consecutive* terms, as listed, for some arithmetic> sequence, then the differences between consecutive terms must all be> equal and (2x+1) - (x+3) = (5x+3) - (2x+1). If they need not be consecutive, for any arithmetic sequence, the> problem has no unique solution.> > But, notice what the real question is... The answer given in the back of the> book appears to be wrong.> > You're right, -3/2 is === first is in douhnuts but not in pizza>my second is in bubbles but not in toothpaste>my third is in kiss but not in branston pickle ??The connection to alt.MATH.undergrad escapes me.Perhaps rec.puzzles would be a better choice. -- 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 === approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id can give me some formulas on how to get thepercentage of the amount : my target is 850 and I get only 730 how much is the percentagetotal of 730 out of 850.... please ... i need only the formula to getthe percent of 730 ... === approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id Please if somebody can give me some formulas on how to get the>percentage of the amount :> my target is 850 and I get only 730 how much is the percentage>total of 730 out of 850.... please ... i need only the formula to get>the percent and, since per centMEANS per hundred this is === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, with following problem I want to solve this equationsin(3x)+5i=0where i=sqrt(-1).How can I solve this equation for x.Does somebody have answer for this === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, stuck with following problem I want to solve this equation>sin(3x)+5i=0>where i=sqrt(-1).>How can I solve this equation for x.>Does somebody have answer for this >Pratap What's wrong with the method you learned when you were 12 years old?If sin(3x)+ a= 0 (a is ANY number) then sin(3x)= -a so 3x= arcsin(-a)=-arcsin(a) and x= -arcsin(a)/3. In this case, a= 5i sox= -arcsin(5i)/3. How do you find arcsin(5i)? Same way you would the arcsin of anynumber: use a calculator. Many modern calculators will do operationswith complex numbers. If you don't have one you will need to find aformula for that arcsin in terms of real numbers. Do you know that sin(x)= (e^ix- e^(-ix))/2i ? with sin x= 5i, thiswould be e^(ix)- e^(-ix)= (5i)(2i)= -10. Can you solve that (for === approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id inventions or discoveries?They are simply stepping stones to true enlightenment. When one day>we UNconver (neither discover nor invent) the ultimate single >equation which explains everything!:) And we find out that the solution to that === find out that the solution to that equation is 47!Surely you === Goooooooogled and there is some history on this question but it's> five years ago now ( I knew Usenet had gone downhill) so: are mathematical> theorems inventions or discoveries?> > cheers> > dd> > Good proofs are inventions, but the theorems === truthConstant factors of polynomials used to be irrelevant multiples thatyou just got rid of, until I came along.Now suddenly, you're told that constants are really products offunctions of x, and must be treated as such with polynomials. So whatchanged?It's as easy as looking at that example from Rick Decker, where he'sactually *supporting* the idea of a constant--7--being the product offunctions of x.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).If you check, it, yes, it all works out, but what about that 7. Ifyou're trained at all in mathematics then you should have learned todiscard it and focus on whether or not 25x^2 + 30x + 2 is reducibleover rationals, or on how to get its roots, but I changed the focuswith my research.So why decide that 7 is now a product of functions of x? Because thetruth hurts.The truth is that mathematicians have a definition that doesn't dowhat they think it does, and you can either accept the truth, or youcan fight for what is now a religion.Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, andthese are *varying* functions that are the factors of (5a_1(x) + 7)and (5a_2(x) + 7), in the ring of algebraic integers.They're fighting for functions so that they can stay in the ring ofalgebraic integers.Trouble is that I can show in multiple ways that what they're sayingis false, besides the fact that it's silly.You're *taught* to discard constant factors of polynomials like 7*because* they are independent of x. Dik Winter and Rick Decker, andpeople like them don't care about mathematics. They don't care howmany ways I prove them wrong.They just keep coming back with their silly claims.So why w_1(x) and w_2(x)? They can't tell you, except to say because,because otherwise their beliefs about mathematicians and the ring ofalgebraic integers are wrong, because then they'd have to admit thatI'm right, because then they'd have to accept mathematical truth.Can they derive w_1(x) and w_2(x)?No.Instead you have one of their cohorts Keith Ramsay throwing out adegree 22 polynomial claiming to have found it some kind of way, yetif you look atw_1(x) w_2(x) = 7(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) anda^2 - (x - 1)a + 7(x^2 + x)there is no way to derive the w's as varying functions of x.The only thing that actually makes sense is something likew_1(x) = 7, w_2(x) = 1.I can point out the obvious, like why 7? Why not 13? Why can't youjust have something likew_1(x) w_2(x) = 13(5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) anda^2 - (x - 1)a + 13(x^2 + x)?Mathematically there has to be a reason, right?But not if you're Dik Winter or Rick Decker promoting what is really a*religion* and not mathematics.But most of you go along with them, you have to be going along as thestory is too big, too much important mathematics here formathematicians to try and claim that constants like 7 are no longer sosimple. You have to be complicit in their deceit, and an activeparticipant in betraying mathematics, or you would speak up.That's why I'm better than you are.I can handle the truth.Years ago when I was working out various ideas that I thought provedFermat's Last Theorem, dealing with a lot of hostile posters, callingme names, insulting me, I found out I was wrong with an argument thatI'd held on to for months. I went from thinking I was hero to feelinglike a zero.But I came out and admitted I was wrong because I realized that inmathematics there is no gain from holding on to false beliefs.That's why I'm a better person than you. If you don't learn thatlesson I'm a better person than you can ever be.Not because I made some math discoveries, not because of my personalcharacteristics, like I curse and get absusive myself, and hey, I'mselfish, but I can handle the truth, so no matter who you are, whatyou've done, or what other people tell you about yourself or what youtell yourself about yourself, I'm a better person than you are whileyou can't.I'm better than you because I actually believe in mathematics, inmathematical truth, even when it hurts.So you keep fighting along side Dik Winter and Rick Decker for what isnow your mathematical religion, fighing for lies, fighting for that 7that's now a product of functions!But no matter how many people stand next to you, and congratulate you,or just silently support you, you will still be wrong becausemathematics is supposed to be more than a vote of confidence.Mathematics is about the truth.Can you ever handle truth? Will any of you start growing as people? Will any of you learn to better than you were?James HarrisDecker Quadratic Source Information---------------------Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he posts from Hamilton === College:Subject: Re: Mathematical consistency, courageDecker 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 + === apps.cwi.nl > Constant factors of polynomials used to be irrelevant multiples that > you just got rid of, until I came along.They are not really irrelevant, but you can indeed ignore them on occasion. > Now suddenly, you're told that constants are really products of > functions of x, and must be treated as such with polynomials. So what > changed?Well, perhaps because in your situation the functions are not polynomials?You say it so clear non-polynomial factorisation. But indeed, if youhave R(x).S(x) = p.T(x), with R, S and T polynomials with integercoefficients and p some number (not necessarily prime), you have that R isdivisible by a factor of p, and S divisible by the other factor. But thisworks only in this situation. If R and S are arbitrary functions thisis no longer necessarily true. You can have the following too: Q(x) = gcd(2, x) R(x) = Q(x).x S(x) = 2.x/Q(x) T(x) = 2.x^2Now Q, R, S and T are functions from integers to integers. T is divisibleby 2 as such a function, but neither R nor S are. But R(x)S(x) = T(x). > It's as easy as looking at that example from Rick Decker, where he's > actually *supporting* the idea of a constant--7--being the product of > functions of x.Note that this is needed *only* when you wish to stay in the algebraicintegers. > (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). > > If you check, it, yes, it all works out, but what about that 7. If > you're trained at all in mathematics then you should have learned to > discard it and focus on whether or not 25x^2 + 30x + 2 is reducible > over rationals, or on how to get its roots, but I changed the focus > with my research.Eh, no. *If* you are trained in mathematics you do *not* discard that 7.It is immediately clear that 25x^2 + 30x + 2 is not reducible over therationals, on the other hand you have that 7(25x^2 + 30x + 2) is theproduct of two algebraic integers for each algebraic integer x, and thatproduct is divisible by 7. > So why decide that 7 is now a product of functions of x? Because the > truth hurts.No, because if the factorisation is independent of x you would havethat the two factor are divisible by the factors in that factorisation,*which is easily shown to be false*. So you go looking for ways todistribute the factors of 7 dependent on x to get the proper divisibility. > The truth is that mathematicians have a definition that doesn't do > what they think it does, and you can either accept the truth, or you > can fight for what is now a religion.Perhaps it does not do what we think it does do, but in this case itdoes do exactly what we think it does do. > Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, and > these are *varying* functions that are the factors of (5a_1(x) + 7) > and (5a_2(x) + 7), in the ring of algebraic integers. > > They're fighting for functions so that they can stay in the ring of > algebraic integers.Well, you say we are forced out of the algebraic integers, we think thatis false and give functions that show it is false and that you can stayin the algebraic integers. > Trouble is that I can show in multiple ways that what they're saying > is false, besides the fact that it's silly.Silly is in the eye of the beholder. > You're *taught* to discard constant factors of polynomials like 7 > *because* they are independent of x. Dik Winter and Rick Decker, and > people like them don't care about mathematics. They don't care how > many ways I prove them wrong.We care about mathematics. If this is the rule you are taught, goback to college and ask your money back. > They just keep coming back with their silly claims. > > So why w_1(x) and w_2(x)? They can't tell you, except to say because, > because otherwise their beliefs about mathematicians and the ring of > algebraic integers are wrong, because then they'd have to admit that > I'm right, because then they'd have to accept mathematical truth. > > Can they derive w_1(x) and w_2(x)? > > No.I did quite a few times for your perusal, for your cubic and for thisquadratic, but you just ignored them. I will do it for a last time:Given: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) Define: w_1(x) = gcd(5a_1(x) + 7, 7) w_2(x) = 7/w_1(x)Then (5a_1(x) + 7) is (in the algebraic integers) divisible by w_1(x) for all x. (5a_2(x) + 7) is (in the algebraic integers) divisible by w_2(x) for all x (*). And w_1(x)w_2(x) = 7 for all x.(*) This is a bit more difficult to show. Let's have some short namings: f1 = 5a_1(x) + 7 f2 = 5a_2(x) + 7 The claim is equivalent to 7/gcd(f1,7) | f2, where | means divides. We have gcd(f1*f2,7) = 7 | gcd(f1,7)gcd(f2,7), and as gcd(f1,7) | 7, 7/gcd(f1,7) | gcd(f2,7) | f2. QED. > Instead you have one of their cohorts Keith Ramsay throwing out a > degree 22 polynomial claiming to have found it some kind of way,You already apologised to him. He did for x = 2. Rick Decker did forx = 1 and x = -3. Are you also apologising to him? > yet > if you look at > > w_1(x) w_2(x) = 7 > > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > and > > a^2 - (x - 1)a + 7(x^2 + x) > > there is no way to derive the w's as varying functions of x.What about the derivation above? > The only thing that actually makes sense is something like > > w_1(x) = 7, w_2(x) = 1.That is just what you think, but what you think is not always the truth. > I can point out the obvious, like why 7? Why not 13? Why can't you > just have something like > > w_1(x) w_2(x) = 13 > > (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) > > and > > a^2 - (x - 1)a + 13(x^2 + x)?You can. Use my definitions of the w's above, change some 7's to 13's,and look what happens. > Mathematically there has to be a reason, right?Yup, there is. In short, Galois theory has shown that your assumptionthat of a peculiar cubic only two factors where co-prime to some primeis false in the algebraic integers. That is a truth that refuses tosink in apparently. > I can handle the truth.Except that for 6 weeks you could not handle the truth of Keith Ramsay'sfacts.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn === amsterdam, nederland; http://www.cwi.nl/~dik/Subject: Re: surrounded by the walls of Mathematics. Andthose walls have to be built and guarded by men with brains. Who's gonna doit? You? You, James Harris? Mathematicians have a greater responsibilitythan you can possibly fathom. You weep over the rebuttal of your proofsand you curse Mathematicians. You have that luxury. You have the luxury ofnot knowing what they know: That your proofs' rebuttal, while tragic toyou, stems from fact. And their existence, while grotesque andincomprehensible to you, enforces those facts. You don't want the truth. Because deep down, in places you don't talkabout at parties, you want them on that wall. You need them there.They use words like Proof, Conjecture, Lemma...They use these words as thebackbone to a life spent defending something. You use Ôem as a punchline.They have neither the time nor the inclination to explain themselves to aman who rises and sleeps under the blanket of the very theoretical frameworkthey provide, then questions the rigorous manner in which they provide it.I'd prefer you just said thank you and went on your way. Otherwise, Isuggest you prove your idea within the confines of that framework, and standa post.You want the truth?You can't handle the truth!moth(sincerest appologies to the script writers of a few good men for butcheringJack's speech in this === the truth> Constant factors of polynomials used to be irrelevant multiples that> you just got rid of, until I came along.> > Now suddenly, you're told that constants are really products of> functions of x, and must be treated as such with polynomials. So what> changed?> > It's as easy as looking at that example from Rick Decker, where he's> actually *supporting* the idea of a constant--7--being the product of> functions of x.Are you saying that such things as sin(x)^2 + cos(x)^2 = 1 are not allowed becasue the right hand side is a constant?To have f_1(x)*f_2(x) = 7, one only needs to have f_2(x) = 7/f_1(x).Are you saying that such equations are imposssible?> > 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).> > If you check, it, yes, it all works out, but what about that 7. If> you're trained at all in mathematics then you should have learned to> discard it and focus on whether or not 25x^2 + 30x + 2 is reducible> over rationals, or on how to get its roots, but I changed the focus> with my research.So JSH, who is, by his own admission, not trained at all in mathematics, is now empowered to decide how those who are so trained may think.Next he will be explaining brain surgery to brain surgeons.> > So why decide that 7 is now a product of functions of x? Because the> truth hurts.Let w_1(x) = 2 + sin(x) and w_2(x) = 7/(2+sin(x)). What is their product, Jimmy boy, or is that one too tough for you to figure out? > > The truth is that mathematicians have a definition that doesn't do> what they think it does, and you can either accept the truth, or you> can fight for what is now a religion.The truth is that mathematicians have oodles of definitions that JSH doesn't understand and doesn't like. Definitions don't DO anything except one to say complicated things more brießy.> > Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, See example above> and> these are *varying* functions that are the factors of (5a_1(x) + 7)> and (5a_2(x) + 7), in the ring of algebraic integers.> > They're fighting for functions so that they can stay in the ring of> algebraic integers.It is a fight that they have long since won. All that is left is for the loser to regain consciousness after his extended delirium.> > Trouble is that I can show in multiple ways that what they're saying> is false, besides the fact that it's silly.There is certainly a good deal of silliness involved, but it is not in what THEY are saying.> > You're *taught* to discard constant factors of polynomials like 7> *because* they are independent of x. Dik Winter and Rick Decker, and> people like them don't care about mathematics. They don't care how> many ways I prove them wrong.As long as the number of times remains at its current value of zero, they certainly don't care.Except that, when you stop, you will be depriving a good many people of a good deal of innocent amusement.> > They just keep coming back with their silly claims.And then proving them. Nasty of them, isn't it.> > So why w_1(x) and w_2(x)? They can't tell you, except to say because,> because otherwise their beliefs about mathematicians and the ring of> algebraic integers are wrong, because then they'd have to admit that> I'm right, because then they'd have to accept mathematical truth.> > Can they derive w_1(x) and w_2(x)?> > No.They can, for one x at a time and with a good deal of labor, but choose not to, as it is a bootless task to prove individul cases when, as here, the general rule has already been established.> > Instead you have one of their cohorts Keith Ramsay throwing out a> degree 22 polynomial claiming to have found it some kind of way, A way which found a common non-unit algebraic integer factor for 7 and (1_sqrt(-167))/2, and which can be easly verified ( I have verified it myself). yet> if you look at> > w_1(x) w_2(x) = 7> > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > > and> > a^2 - (x - 1)a + 7(x^2 + x)> > there is no way to derive the w's as varying functions of x.> > The only thing that actually makes sense is something like> > w_1(x) = 7, w_2(x) = 1.Which has been shown not to make sense for some values of x other than 0.> > I can point out the obvious, like why 7? Why not 13? Why can't you> just have something like> > w_1(x) w_2(x) = 13> > (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) > > and> > a^2 - (x - 1)a + 13(x^2 + x)?> > Mathematically there has to be a reason, right?Wrong. You can have it, but not with the same functions for which the product equals 7, since 7 and 13 are not equal.> > But not if you're Dik Winter or Rick Decker promoting what is really a> *religion* and not mathematics.> > But most of you go along with them, you have to be going along as the> story is too big, too much important mathematics here for> mathematicians to try and claim that constants like 7 are no longer so> simple. You have to be complicit in their deceit, and an active> participant in betraying mathematics, or you would speak up.> > That's why I'm better than you are.> > I can handle the truth.Present evidence contradicts that claim.> > I'm better than you because I actually believe in mathematics, in> mathematical truth, even when it hurts.If you really beleive in what you're saying, then it doesn't yet hurt enough. I am thinking of the mule trainer and a 2 by 4,> > So you keep fighting along side Dik Winter and Rick Decker for what is> now your mathematical religion, fighing for lies, fighting for that 7> that's now a product of functions!There are infinitely many ways of expressing any constant as a product of functions. > Mathematics is about the truth.And JSH is about self-aggrandizement.> > Can you ever handle truth? Try us with === truthX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>[...]That's why I'm better than you are.Actually you're not better than any of us, except at trolling.>I can handle the truth.It really disgusts me when you talk about the Truth this way.I pointed out years ago, and it's still true, that you're the_only_ person I've _ever_ seen here say that if what he justsaid was wrong we shouldn't bother saying so, becausehe didn't want to know.>Years ago when I was working out various ideas that I thought proved>Fermat's Last Theorem, dealing with a lot of hostile posters, calling>me names, insulting me, I found out I was wrong with an argument that>I'd held on to for months. I went from thinking I was hero to feeling>like a zero.But I came out and admitted I was wrong because I realized that in>mathematics there is no gain from holding on to false beliefs.That's why I'm a better person than you. If you don't learn that>lesson I'm a better person than you can ever be.No, actually this is why you're _not_ better than us. That bitabout how you thought you were right but found out you werewrong applies to _all_ of your discoveries, except the currentone. No matter _how_ many times you finally realize you werewrong about something you still show _no_ doubt about whatever the current discovery is.It's truly remarkable how _much_ worse you are than therest of us in this regard. You talk a lot about how you learnfrom your mistakes, but those statements are as absurdas the things you say about mathematics.>Not because I made some math discoveries, not because of my personal>characteristics, like I curse and get absusive myself, and hey, I'm>selfish, but I can handle the truth, so no matter who you are, what>you've done, or what other people tell you about yourself or what you>tell yourself about yourself, I'm a better person than you are while>you can't.I'm better than you because I actually believe in mathematics, in>mathematical truth, even when it hurts.So you keep fighting along side Dik Winter and Rick Decker for what is>now your mathematical religion, fighing for lies, fighting for that 7>that's now a product of functions!But no matter how many people stand next to you, and congratulate you,>or just silently support you, you will still be wrong because>mathematics is supposed to be more than a vote of confidence.Mathematics is about the truth.Can you ever handle truth? Will any of you start growing as people? >Will any of you learn to better than you were?ing lying slanderous moron.>James HarrisDecker 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, courageDecker 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).************************David C. === polynomials used to be irrelevant multiples that> you just got rid of, until I came along.They were? Please provide a of polynomials like 7> *because* they are independent of x. Dik Winter and Rick Decker, and> people like them don't care about mathematics. They don't care how> many ways I prove them wrong.Once is enough. Unfortunately, they have proven *you* are.Hahahahahaha.... It is to laugh! James, you crack me up!> I can handle the truth.You can handle the Ôsplitting field' between your rear cheeks. The truthis elsewhere.> Years ago when I was working out various ideas that I thought proved> Fermat's Last Theorem, dealing with a lot of hostile posters, calling> me names, insulting me, I found out I was wrong with an argument that> I'd held on to for months. I went from thinking I was hero to feeling> like a zero.During those months everyone (but you) knew you were wrong. You were azero the whole time.> But I came out and admitted I was wrong because I realized that in> mathematics there is no gain from holding on to false beliefs.Everyone else already knew that.> That's why I'm a better person than you. If you don't learn that> lesson I'm a better person than you can ever be.Better person? You are a liar and a crank. You have no standing to compareyourself to your critics, most of whom are not only far bettermathematicians than you, but also far better human beings. Your consistenttrack record of proving yourself to be an arrogant, obnoxious blowhardplaces you in the intellectual and moral dumpster.[snip continued self aggrandizing rant.]> James I am the Magnificent, W.O.O.O. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over === factors of polynomials used to be irrelevant multiples that> you just got rid of, until I came along. Now suddenly, you're told that constants are really products of> functions of x, and must be treated as such with polynomials. So what> changed? It's as easy as looking at that example from Rick Decker, where he's> actually *supporting* the idea of a constant--7--being the product of> functions of x. 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). If you check, it, yes, it all works out, but what about that 7. If> you're trained at all in mathematics then you should have learned to> discard it and focus on whether or not 25x^2 + 30x + 2 is reducible> over rationals, or on how to get its roots, but I changed the focus> with my research. So why decide that 7 is now a product of functions of x? Because the> truth hurts. The truth is that mathematicians have a definition that doesn't do> what they think it does, and you can either accept the truth, or you> can fight for what is now a religion. Decker and a guy named Dik Winter say that w_1(x) w_2(x) = 7, and> these are *varying* functions that are the factors of (5a_1(x) + 7)> and (5a_2(x) + 7), in the ring of algebraic integers. They're fighting for functions so that they can stay in the ring of> algebraic integers. Trouble is that I can show in multiple ways that what they're saying> is false, besides the fact that it's silly. You're *taught* to discard constant factors of polynomials like 7> *because* they are independent of x. Dik Winter and Rick Decker, and> people like them don't care about mathematics. They don't care how> many ways I prove them wrong. They just keep coming back with their silly claims. So why w_1(x) and w_2(x)? They can't tell you, except to say because,> because otherwise their beliefs about mathematicians and the ring of> algebraic integers are wrong, because then they'd have to admit that> I'm right, because then they'd have to accept mathematical truth. Can they derive w_1(x) and w_2(x)? No. Instead you have one of their cohorts Keith Ramsay throwing out a> degree 22 polynomial claiming to have found it some kind of way, yet> if you look at w_1(x) w_2(x) = 7 (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) and a^2 - (x - 1)a + 7(x^2 + x) there is no way to derive the w's as varying functions of x. The only thing that actually makes sense is something like w_1(x) = 7, w_2(x) = 1. I can point out the obvious, like why 7? Why not 13? Why can't you> just have something like w_1(x) w_2(x) = 13 (5a_1(x) + 13)(5a_2(x) + 13) = 13(25x^2 + 30x + 2) and a^2 - (x - 1)a + 13(x^2 + x)? Mathematically there has to be a reason, right? But not if you're Dik Winter or Rick Decker promoting what is really a> *religion* and not mathematics. But most of you go along with them, you have to be going along as the> story is too big, too much important mathematics here for> mathematicians to try and claim that constants like 7 are no longer so> simple. You have to be complicit in their deceit, and an active> participant in betraying mathematics, or you would speak up. That's why I'm better than you are. I can handle the truth. Years ago when I was working out various ideas that I thought proved> Fermat's Last Theorem, dealing with a lot of hostile posters, calling> me names, insulting me, I found out I was wrong with an argument that> I'd held on to for months. I went from thinking I was hero to feeling> like a zero. But I came out and admitted I was wrong because I realized that in> mathematics there is no gain from holding on to false beliefs. That's why I'm a better person than you. If you don't learn that> lesson I'm a better person than you can ever be. Not because I made some math discoveries, not because of my personal> characteristics, like I curse and get absusive myself, and hey, I'm> selfish, but I can handle the truth, so no matter who you are, what> you've done, or what other people tell you about yourself or what you> tell yourself about yourself, I'm a better person than you are while> you can't. I'm better than you because I actually believe in mathematics, in> mathematical truth, even when it hurts. So you keep fighting along side Dik Winter and Rick Decker for what is> now your mathematical religion, fighing for lies, fighting for that 7> that's now a product of functions! But no matter how many people stand next to you, and congratulate you,> or just silently support you, you will still be wrong because> mathematics is supposed to be more than a vote of confidence. Mathematics is about the truth. Can you ever handle truth? Will any of you start growing as people?> Will any of you learn to better than you were?> James Harris 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).James, shut up. Quit putting people down and STICK TO THE MATH. You are nobetter than us and vice versa.David === I want to do a least square approximation with the simple basisfunctions 1,t,t^2,...,t^(N-1). I choose a time grid ast=[0,1/(M-1),2/(M-1),...1]. This gives me a MxN matrix, say A. The basisfunctions phi_n(t)=t^n are polynomials and therefore I use the ususalintegral inner product in (0,1).The inner product becomes smaller and smaller when m,n->M but of course itis not zero for any n,m. Does this prove that the columns becomes morelinear dependent as n,m->M?What if I evaluate the matrix for the gridpoints and then use the sum ofsquares dot product to measure ortogonality between === so important to these people?You are important for the same === Re: JSH: Why am I so important? ... earth-shatteringly incisive analysis deleted ...> > Clearly I'm VERY IMPORTANT to these people!!!> > Oh well, I'll just keep posting, and I guess these people will keep> replying.> > > James HarrisYou mispelled impotent. I know that those words are pronounced (poe-nownst) the same way in Georgia (Joe-ja). People of yoursensitive disposition use the abbreviation ED (for ErectileDysfunction). Impotent seems, so, er, direct (pardon the implicitrhyme with erect).Here's the URL you want: === just made a couple of posts, one of which had an obvious sign error. Now then, probably quite a few posters will gleefully leap upon those> posts, as if it's such a big deal. Why? Why do I see webpages scattered across the Internet dedicated to> pushing negatives about me? Why am I so important to these people? Given the fact that I've been posting for years, and making mistakes> for years, it stands to reason that posters who so happily reply to me> as if it matters are strangely deluded. What changes? It's a bizarre drama, where I'm this guy, who posts a lot and has a> blog. For some odd reason, there are all these other people, like Dik> Winter, Erik Max Francis and Rick Decker who have these webpages,> which always find some way to insult me. Now sure, maybe they get their jollies doing that, but why keep at it> for YEARS. Look at Decker's webpage, and notice the date. Look at Dik Winter's webpages and notice the dates to which he's> referring. Clearly I'm VERY IMPORTANT to these people!!! Oh well, I'll just keep posting, and I guess these people will keep> replying.> James HarrisJames, QUIT COMPLAINING. If you don't want replies from certain people,don't post. You act like you own usenet and these threads. NEWSFLASH: youdon't.David === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, 2x+1, 5x+3 is an arith. sequence, we have that1. x+3=(2x+1)+y and2. x+3=(5x+3)+2y,where y is the term of the arith. sequence.Solving the linear system of equations (e.g. equation === research and politics, algebraic integersNow the way I see it, math research is just about what follows frommathematics, but I think you can now see that it's also aboutpolitics.I mean, look at your math professors. Mostly guys of a certain typewho have various outlooks on life, but they have a certain slant totheir mathematical views. It's just human nature.Now then look at this example from a post by Rick Decker, a professorat Hamilton College:(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). That doesn't fit into your professors' worldview.Decker was trying to imitate my work where I have cubics, carefullyderived, but I like his example, which I think he kind of guessed at,as it's a quadratic, and can STILL tell you that the math scene haschanged.Now then, you can go along with posters trying to tell you it hasn'tor decide that if you don't hear from your professors it hasn't, butjust go to your textbooks, look diligently, and try to find an examplelike Decker's quadratic.There's that 7 times a polynomial and you will see in your textbooksinstructions on how to handle such a factor--divide it off.You're getting a quick growing up lesson in mathematics and I'mtalking to you this way because maybe you're not thoroughlyindoctrinated in the *politics* of mathematics yet.If Gauss were around or Dedekind or Euler I could intrigue them withmy cubics, and be celebrated. But your professors aren't Gauss orDedekind or Euler or people who are in any way even close. They'reßawed as mathematicians, rather small in the mathematical arena, nowfaced with something too big for them to handle.My hope is that one of you IS something like a Gauss or a Dedekind oran Euler and you are less concerned with politics and how the truthwill impact the social scene of mathematics--like scare all thoseprofessors--than you are in what's correct.If you've seen the arguments over that example, you may think it'ssettled in some simple way, like you have functions w_1(x) and w_2(x),where w_1(x) w_2(x) = 7, and they vary in *some* way as x changes.Some posters are relying on basics like with x^2 + bx + 8, you canpick various integer b's such that the factors of the roots vary, buthere you have(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). It's a special beastie. If you rely on simple, then you don't havewhat it takes to be a significant mathematician.If you imagine you have what it takes to be a mathematician, test theassertion.Then you have 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 derive w_1(x), and w_2(x).Or, if you wish, find a professor who can.Here's the opportunity of a lifetime, or potentially the waste of alife, as some of you will choose wrongly, and trust, making apolitical decision.Later, when the truth comes out, you will lose faith, and face asituation where you can't go back in time and do something different.On that day, some of you will walk out of === Dik Winter's claims revisited, dependency issue solution for the w's, but how did he pick w's from infinity?> > Given that his polynomial is of degree 22, it's quite possible that he> just chose a really big polynomial!> Well I read over what Keith Ramsay has and it looks ok to me, though Ididn't check all his numbers.So I was wrong here, no === deal.Being wrong isn't a big deal. But consistently accusing those who discover your errors of being liars *is*a big deal. It demonstrates clearly that you have NO integrity.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over === claims revisited, dependency issue> jstevh@msn.com (James that Keith Ramsay even claimed to have posted a> solution for the w's, but how did he pick w's from infinity?> > Given that his polynomial is of degree 22, it's quite possible that he> just chose a really big polynomial!> > > Well I read over what Keith Ramsay has and it looks ok to me, though I> didn't check all his numbers.> > So I was wrong here, no big deal.> > > James HarrisThat makes a perfect score of none right in all these protracted disputes and a humongous number of times wromg.Doesn't this record indicate that your forte is not in === revisited, dependency issue 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).> > Dik Winter has repeatedly asserted that there exists varying algebraic> integer functions w_1(x) and w_2(x), such that> > w_1(x) w_2(x) = 7> > and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and> w_2(x) as factors, respectively.> Right. You've seen what happens when x=1 and x=2. Let's try adifferent one, where things are simple enough to verify byhand.Suppose we take x = -3. Then we have (5a_1(-3) + 7)(5a_2(-3) + 7) = 7(25(9) + 30(-3) + 2) = 7(137)where the a's satisfy a^2 + 4a + 7((-3)^2 + (-3)) = a^2 + 4a + 42.We find that a_1 = -2 + sqrt(-38) and a_2 = -2 - sqrt(-38)It's easy enough to verify that the a's aren't divisible by7 or by sqrt(7).However, a_1 does share a factor in common with 7,namely w_1 = (1 + 3sqrt(-38))^{1/3}We can verify that1. w_1^3 divides 7^3, since (1 + 3sqrt(-38))(1 - 3sqrt(-38)) = 343 = 7^3 So w_1 = (1 + 3sqrt(-38))^{1/3} divides 7.2. w_1^3 divides a_1^3 = 220 - 26sqrt(-38), since (1 + sqrt(-38)(-8 - 2sqrt(-38)) = 220 - 26 sqrt(-38) So w_1 = (1 + 3sqrt(-38))^{1/3} divides a_1Thus w_1 is a common divisor of a_1 and 7. (In fact,it's a gcd of a_1 and 7.)In a similar way, we find that w_2 = (1 - 3sqrt(-38))^{1/3}is a common divisor of a_2 and 7.In this case, with (5a_1 + 7)(5a_2 + 7) = 7(137)We see that the factor 7 on the right splits as w_1 * w_2 = 7and that f_1 = (5a_1 + 7)/w_1 = 5(-8 - 2sqrt(-38))^{1/3} + (1 - 3sqrt(-38))^{1/3}and f_2 = (5a_1 + 7)/w_1 = 5(-8 + 2sqrt(-38))^{1/3} + (1 + 3sqrt(-38))^{1/3} f_1 * f_2 = 137exactly as expected. Likewise, it's easy to show thatw_1, w_2, f_1, and f_2 are all algebraic integersand none of them are units.I hasten to add that this behavior is what happens foralmost all values of x, namely that in (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)with the a's satisfying a^2 -(x - 1)a + 7(x^2 + x)we will be able to find algebraic integers w_1(x) and w_2(x)with w_1(x) * w_2(x) = 7 and w_i(x) dividing a_i(x) so thatwith f_i(x) = (5a_1(x) + 7)/w_i(x)we will have f_1(x) * f_2(x) = 25x^2 + 30x + 2as long as 7 doesn't divide 25x^2 + 30x + 2.With more or less difficulty, one could do the sameconstruction for most integers x, just as I did for x=1and Keith did for x=2. In most cases we'll find that 7splits into two nonunit factors, distributed between(5a_1+7) and (5a_2+7). In general, as with w_i(-3) = (1 +/- 3sqrt(-38))^{1/3}it won't be immediately obvious that w_1(x) and w_2(x)are divisors of 7 and of a_i(x), in the sense thatyou can't look at them and immediately notice thatthey === claims revisited, dependency issueX-DMCA-Notifications: -0800, norabaron@hotmail.com (Nora Baron)>>[...]>> I'm still pissed. > Too bad. In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about >this.Don't tell him that. He's a lot more fun to read when he's inraving lunatic mode.Oops, I guess that wasn't a very nice thing to say. I for onehave decided that the idea that we're supposed to be niceto everyone, regardless of how they behave, is just silly.> We are not evil conspirators trying to rob you of your>rightful credit. We are just ordinary people trying to show you>where you are making mistakes. Nora B.************************David C. === norabaron@hotmail.com (Nora Baron)> >>[...]>> I'm still pissed. > Too bad. In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about >this.> > Don't tell him that. He's a lot more fun to read when he's in> raving lunatic mode.> > Oops, I guess that wasn't a very nice thing to say. I for one> have decided that the idea that we're supposed to be nice> to everyone, regardless of how they behave, is just silly.> Some professor you turned out to be Ullrich.You're a joke and a disgrace to your profession.Oh yeah, off.James === dependency issueX-DMCA-Notifications: 16:44:40 -0800, norabaron@hotmail.com (Nora Baron)>> >[...]> I'm still pissed. >> Too bad. >> In fact, seeing the number of threads you have started today>>on this, and their content, it looks to me like you are frantic,>>perhaps even dangerously upset. You need to calm down about >>this.>> >> Don't tell him that. He's a lot more fun to read when he's in>> raving lunatic mode.>> >> Oops, I guess that wasn't a very nice thing to say. I for one>> have decided that the idea that we're supposed to be nice>> to everyone, regardless of how they behave, is just silly.>> Some professor you turned out to be Ullrich.You're a joke and a disgrace to your profession.Actually being polite to people who behave like raving lunaticsbut insist they understand mathematics much better than_every_ professional mathematician on the planet is notpart of the job description.If people had been refusing to help you learn math you'dhave a valid complaint about what sort of professors theywere. But you've shown over and over for years that youhave no interest in learning any math.>Oh yeah, off.>James Harris************************David C. === message>>[...]> >> I'm still pissed.> > Too bad.> > In fact, seeing the number of threads you have started today>on this, and their content, it looks to me like you are frantic,>perhaps even dangerously upset. You need to calm down about>this. Don't tell him that. He's a lot more fun to read when he's in> raving lunatic mode. Oops, I guess that wasn't a very nice thing to say. I for one> have decided that the idea that we're supposed to be nice> to everyone, regardless of how they behave, is just silly.> Some professor you turned out to be Ullrich. You're a joke and a disgrace to your profession. Oh yeah, off.> James HarrisSome man you are. You should either be ashamed, embarrassed, or both. I'venever seen a man === Winter's claims revisited, dependency issue> ... stuff deleted ...> > >>Given that his polynomial is of degree 22, it's quite possible that he>>just chose a really big polynomial!> > > Actually I think it had degree 11. I think he used a program> to do the computation, and I think it finds the lowest-degree> polynomial that works. I believe it computes powers of an> ideal, ^M, until it finds an M where that ideal> is principal. It is not a matter of human choice, just a > programmable algorithm.> I thought the gcd was of degree 11 over the field Q(sqrt(-167)), butwhen you put it in terms of rational integers, the degree had to be 22.It turned out that way in KASH, which I used. The class number of thesplitting field of (whatever polynomial it was) was 11, so it tookraising the ideal to the 11th power to force it to beprincipal, as you pointed out. However, this is still relative tothe (integers of the) splitting field of that quadratic. ... the rest deleted ...Except for this:> In fact, seeing the number of threads you have started today> on this, and their content, it looks to me like you are frantic,> perhaps even dangerously upset. You need to calm down about > this. We are not evil conspirators trying to rob you of your> rightful credit. We are just ordinary people trying to show you> where you are making mistakes.> > Nora B.> !labaC lacitamehtaM livE terceS eht tuoba no tel ot ton tseBOr, ixnay on the onspiracycay!> > >>But there's not a lot I can do when people like Dik>>Winter can so easily get away with basic problems in their claims, on>>a newsgroup that doesn't seem to give a damn about mathematical truth.>>As Dan Quayle said, How terrible it is to lose your mind. Or not to have a mind at === Winter's claims revisited, dependency issueNntp-Posting-Host: apps.cwi.nl > 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). > > Dik Winter has repeatedly asserted that there exists varying algebraic > integer functions w_1(x) and w_2(x), such that > w_1(x) w_2(x) = 7Yup, right, they were defined in terms of the gcd function on algebraicintegers. > and that the factors (5a_1(x) + 7) and (5a_2(x) + 7) have w_1(x) and > w_2(x) as factors, respectively.It is a property of the gcd function that gcd(5a_1(x) + 7, 7) is afactor of both (5a_1(x) + 7) and of 7. I constructed such w_1(x)and w_2(x) such that also w_1(x)w_2(x) = 7. You never pointed toan error in those definitions. > Now then, introducing f_1(x) and f_2(x) as the other factors of the > a's, you have > w_1(x) f_1(x) = 5a_1(x) + 7, and w_2(x) f_2(x) = 5a_2(x) + 7, Indeed. > so > a_1(x) = (w_1(x) f_1(x) - 7)/5, and a_2(x) = (w_2(x) f_2(x) - 7)/5. > > Now it matters to use the fact that the a's are roots of > a^2 - (x - 1)a + 7(x^2 + x), > as solving that quadratic, and picking a_1(x) for the positive sign > root gives > > a_1(x) = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, so > > (w_1(x) f_1(x) - 7)/5 = ((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 > > so > > w_1(x) f_1(x) - 14 = 5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2, soGetting sloppy again? 14 should be 7. > f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x), > which implies a relationship between f_1(x) and w_1(x), but > f_1(x) f_2(x) = 25x^2 + 30x + 2But there *is* such a relation: w_1(x)f_1(x) = (5a_1(x) + 7), > and it is arbitrary that 7 was the multiple before as it could have > been 11 or 13 or any of an infinity of other numbers, and w_1(x) and > w_2(x) are themselves not determined, so it's a spurious appearance of > dependency.I do not understand what you are writing here at all. If you meanthat w_1(x).w_2(x) = 7 could just as easily have been 11 or 13, youare talking nonsense. They were *defined* such that their productis 7. And if I provide definitions in full detail, I do not expectthat somebody says they are not determined. Go over the definitionsand play state which part is incomplete, does not define an algebraicinteger, or what. > That's the little detail that Dik Winter never bothered to address.There is one little detail that you never bothered to address. Thereare complete definitions of w_1(x) and w_2(x). > Now then, if f_1(x) is in fact independent of w_1(x), then how do you > account for the appearance of a dependency in > f_1(x) = (5((x-1)+sqrt((x-1)^2 + 28(x^2 + x)))/2 + 14)/w_1(x)?f_1(x) is *not* independent of w_1(x), you have just defined it suchthat w_1(x)f_1(x) = (5a_1(x) + 7). So where do you find theindenpendence? And when you replace 14 by 7 above, it just statesthe same. > If you try to push the issue that it isn't so required consider what > happens if you try to divide through by w_1(x), as then you have > > f_1(x) = > > (5((x-1)/w_1(x)+sqrt((x-1)^2 + 28(x^2 + x)))/(2w_1(x) + 2w_2(x))You have lost me here. > and the problem is that all of the w's need to go away. So assuming > that a_1(x) has w_1(x) as a factor means, introducing g(x), that you > haveA strong assumption you make here, and it is not valid. a_1(x) does*not* have w_1(x) as a factor, it is (5a_1(x) + 7) that has w_1(x)as a factor. > which indicates that f_1(x) is dependent on w_2(x).Whacky, but indeed, it is so, because w_1(x) is dependent on w_2(x)and the reverse. They are connected through the relation w_1(x)w_2(x) = 7. > However, as I pointed out w_1(x) w_2(x) = 7 does NOT determine them as > an infinity of functions will work, while f_1(x) is independent of > w_1(x) and w_2(x) since > > f_1(x) f_2(x) = 25x^2 + 30x + 2.This does not make them independent on w_1(x) and w_2(x). You have*defined* f_1(x) as the function such that w_1(x)f_1(x) = (5a_1(x) + 7)so you yourself has made it dependent on w_1(x). Similar withf_2(x) and w_2(x). > The point is you can multiply some polynomial like 25x^2 + 30x + 2, by > anything you choose. There's no way that it's locking into it that > functions that are factors of 7 are required.*You* required it by having the factorisation (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2). > They've gone so far that Keith Ramsay even claimed to have posted a > solution for the w's, but how did he pick w's from infinity? > > Given that his polynomial is of degree 22, it's quite possible that he > just chose a really big polynomial!Ah, I understand, you did not check what he you have two > independent equations: > > w_1(x) w_2(x) = 7 > > and > > f_1(x) f_2(x) = 25x^2 + 30x + 2 > > To try and dispute my results, posters like Dik Winter or Nora Baron > simply skip past mathematical consequences of their claims, like the > independence between those equations.Try to read my definitions of w_1(x) and w_2(x) and see that they force w_1(x)f_1(x)w_2(x)f_2(x) = (5a_1(x) + 7)(5a_2(x) + 7) = = 7(25x^2 + 30x + 2).And as w_1(x), f_1(x), w_2(x) and f_2(x) are (with my definitions) allalgebraic integer functions I see no problem. > I'm still pissed. But there's not a lot I can do when people like Dik > Winter can so easily get away with basic problems in their claims, on > a newsgroup that doesn't seem to give a damn about mathematical truth.It is clear through your remark on Keith Ramsay's solution that youreally do not read is correct or not, you just assume.Neither (I assume) did you check my definitions of w_1(x) and w_2(x).-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; === revisited, dependency issue Adjunct Assistant Professor at the University of Montana. [.snip.]>A strong assumption you make here, and it is not valid. a_1(x) does>*not* have w_1(x) as a factor, it is (5a_1(x) + 7) that has w_1(x)>as a factor.Actually, in this context it is valid: w_1(x) divides 7, and divides5a_1(x)+7, so it divides 5a_1(x); since w_1(x) divides 7, it iscoprime to 5, and therefore divides a_1(x).-- accept as reality. --- Calvin (Calvin and === claims revisited, dependency issueThere was a small mistake in my previous reply to this:where I said a^2 - (x - 1)*a + 17it should have been a^2 - (x - 1)*a + 17*(x^2 + x).Nora B.[rest === simpler to just post to all the newsgroups that I posted beforean apology for questioning Keith Ramsay's honesty. It seems he did infact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and7, so I was wrong.So I apologize to Keith Ramsay for questioning his honesty here.As for the rest of the obsessive repliers, no apologies to thatdogmatic crew!They're still at it, replying, and replying and replying.Still now you can see why Usenet IS useful to me, as I can have peoplecheck things for me, like Keith Ramsay === post > It's simpler to just post to all the newsgroups that I posted before > an apology for questioning Keith Ramsay's honesty. It seems he did in > fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and > 7, so I was wrong.Yup. > So I apologize to Keith Ramsay for questioning his honesty here.Why not apologise to those who did repost that result for your perusal? > As for the rest of the obsessive repliers, no apologies to that > dogmatic crew! > > They're still at it, replying, and replying and replying. > > Still now you can see why Usenet IS useful to me, as I can have people > check things for me, like Keith Ramsay did.But if people check things they are abused...What happens is that if somebody posts a counterexample or something,you just state it is nonsense, without ever checking. This example hasbeen posted in total 10 times by 4 different posters (Keith Ramsay,Nora Baron, Rick Decker and me) in a period of 6 weeks (the first oneon 7 January). And finally after 6 weeks it took you only one falsestart to check that the result was true indeed... If you had doneyour checking 6 weeks ago Usenet would have been useful for you andit would have spared us a large number of threads.Now when are you going to check my definitions of w_1(x) and w_2(x)?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; === simpler to just post to all the newsgroups that I posted before> an apology for questioning Keith Ramsay's honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.Why is the non-unit algebraic integer factor of (1 + sqrt(-167))/2 and 7important ?Likewise, is there any significance to the non-unit irrational === to all the newsgroups that I posted before> an apology for questioning Keith Ramsay's honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.> > Why is the non-unit algebraic integer factor of (1 + sqrt(-167))/2 and 7> important ?It is not all that important in the great scheme of things, but it isimportant in the ongoing discussion of Mr. Harris's ideas, because hehad previously stated that there was no such factor. > Likewise, is there any significance to the non-unit irrational === honesty here. As for the rest of the obsessive repliers, no apologies to that> dogmatic crew! They're still at it, replying, and replying and replying.And you'll still be at it, too, lying and lying and lying.> James Often in error, but never in doubt. Harris--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over === to Ramsay, why I postX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>It's simpler to just post to all the newsgroups that I posted before>an apology for questioning Keith Ramsay's honesty. It seems he did in>fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and>7, so I was wrong.So I apologize to Keith Ramsay for questioning his honesty here.As for the rest of the obsessive repliers, no apologies to that>dogmatic crew!In particular none to the people that you've accused of various sorts of dishonesty when they pointed out that he'd given thisexample? Huh.ing moronic asshole.>They're still at it, replying, and replying and replying.Still now you can see why Usenet IS useful to me, as I can have people>check things for me, like Keith Ramsay did.>James Harris************************David C. === postOriginator: a@shell3.shore.net (a)>Still now you can see why Usenet IS useful to me, as I can have people>check things for me, like Keith Ramsay did.Of course. That's a symptom of your Narcissistic Personality Disorder: 6. is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own === Re: JSH: Apology to Ramsay, why I post> It's simpler to just post to all the newsgroups that I posted before> an apology for questioning Keith Ramsay's honesty. It seems he did in> fact post a non-unit algebraic integer factor of (1+sqrt(-167))/2 and> 7, so I was wrong.> > So I apologize to Keith Ramsay for questioning his honesty here.> > As for the rest of the obsessive repliers, no apologies to that> dogmatic crew!> > They're still at it, replying, and replying and replying.> > Still now you can see why Usenet IS useful to me, as I can have people> check things for me, like Keith Ramsay did.> > > James HarrisKeith Ramsay merely contributed futher proof of JSH's inanity.There is no evidence that JSh will === Algebraic integers check proof> Well I kept fiddling at it, and now have on my blog the complete proof> that there's a problem with the ring of algebraic integers by checking> what happens with the roots of x^2 - x + 42 and y^2 + by - 7, with an> algebraic integer b.> > It's all at> > http://mathforprofit.blogspot.com/> Nope. I was wrong so I deleted off that blog entry.Sorry about the mistake there, but === support1.mathforum.org (8.11.6/8.11.6/The Math Forum, wondering if anyone knew how to slove this story problem. Here it is below.A truck driver drives 55mph on a 200 mile trip to pick up a truckload. On the drivers way home the driver drives 40mph with the load. Find the drivers average speed round === Significant 2 factor ANOVA result with a non-sig post hoc mult undergraduate psychology student and am having difficulty withunderstandingwhether a significant 2 factor ANOVA result can have a non-significant posthoc multiple t-test? I will try to elaborate.The two factors are: 1) Emotion (which had three levels: negative, neutraland positive) and 2) Life period (which had two levels: under 16 years ofage and over 16 years of age).What I'm interested in is seeing if there is any difference between thelevels within the first factor.The results from my 2 factor ANOVA test are as followsF = 11.303, df = 2, p = 0.000Which given a significance level of 5% is significant.However, using a post-hoc t-test on all possible combinations of factor 1and factor 2 (6 combinations) adjusting the significance level to 0.00833(to keep the same level of probability of a type I error occuring). Thosetests give the following results. None of them are significant to the 5%level.Before 16Negative and neutralt = -3.166, df = 11, p = 0.009Neutral and positivet = 2.313, df = 11, p = 0.041Negative and positivet = -2.701, df = 11, p=0.021After 16Negative and neutralt = -2.909, df = 13, p = 0.12Neutral and positivet = 1.437, df = 13, p = 0.174Negative and positivet = -1.266, df = 13, p=0.228What assumptions can I make about differences between factor 1 given thedifferences in significance levels? Is it possible to have one test sayingthere is a significant difference and a post-hoc t-test suggesting there areno tHatDudeUK....My Raw data is below to see if there are any obvious mistakes I have madeabove, although can be ignored...Mean Reaction Times (ms) before 16:P Negative Neutral Positive1 3034.30 3048.30 3953.602 5045.60 7315.00 6462.603 7165.30 7112.40 6667.704 5243.40 6674.90 5998.005 3913.70 4466.40 4968.206 3634.30 7144.40 4265.007 4717.50 5871.90 4789.608 6331.80 4686.80 5325.509 3144.10 4987.70 3938.9010 3649.20 6039.70 4196.3011 5347.40 6831.40 6331.7012 7340.30 9434.20 8128.90Mean Reaction Times (ms) after 16:P Negative Neutral Positive1 2210.30 2508.70 2587.802 1596.40 2869.80 2738.003 4769.80 6976.90 6479.304 3194.60 5409.60 4182.005 4834.20 5766.20 5256.006 4276.10 5256.50 4735.607 5909.60 5303.80 6427.308 3194.60 5409.60 4182.009 2851.40 2440.30 3546.1010 4732.80 6014.70 3277.8011 6978.00 8316.00 7374.7012 3036.40 2610.80 3971.6013 3194.60 5409.60 1482.0014 7455.70 6442.50 6659.50