mm-138 => Please look at my thread in:> > http://www.physicsforums.com/showthread.php?s=&threadid= 6539DonOt bother. It merely links to a passowrd protected pdf 'le.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) =>suppose (i) that f is C^infty and (ii) that the Taylor series has>positive (or in'nite) radius of convergence at any point. Can E be>that large in this case?> > No. Suppose f is C^in'nity and its Taylor series has positive (or > in'nite) radius of convergence everywhere. For any positive integer > N let E_N = {x: for all n, |f^(n)(x)| <= N^(n+1) n!}. Then E_N, being> an intersection of closed sets, is closed. The condition on the > radius of convergence implies that union_N E_N = R. By Baire Category, > some E_N has nonempty interior. But f is analytic on any open interval > contained in E_N.IOm sorry to keep adding new posts to this thread, but I have one morequestion: suppose (ii) above is replaced by the relaxed condition thatthe Taylor series has a null radius of convergence at most at isolatedpoints. If x is an accumulation point of E, then the radius ofconvergence in x is 0, right?PS: for those who didnOt read the rest of the thread, E is the set ofpoints where f (smooth) fails to be analytic.Michele-- DOOh, Google doesnOt a provide a .sig! => IOm sorry to keep adding new posts to this thread, but I have one more> question: suppose (ii) above is replaced by the relaxed condition that> the Taylor series has a null radius of convergence at most at isolated> points. If x is an accumulation point of E, then the radius of> convergence in x is 0, right?No. If a < b, you can de'ne f(x) = exp(-1/[(x-a)(b-x)]) on (a,b), f = 0 on R (a,b). Then f is C^oo(R) with all derivatives 0 on R (a,b). For each n, choose f_n as above with respect to (1/(n+1), 1/n), n = 1,2, ... There exist constants c_n > 0 such that sum_(n=1,oo) c_n*f_n converges in C^oo(R) to some f. For this f, E = {1/n : n = 1,2,...} U {0}, 0 is an accumulation point of E, and the radius of convergence of the Taylor series is oo at each point of E, including 0. =>IOm sorry to keep adding new posts to this thread, but I have one more>question: suppose (ii) above is replaced by the relaxed condition that>the Taylor series has a null radius of convergence at most at isolated>points. If x is an accumulation point of E, then the radius of>convergence in x is 0, right?>PS: for those who didnOt read the rest of the thread, E is the set of>points where f (smooth) fails to be analytic.No. For example, let g be a function thatOs C^in'nity on R andanalytic nowhere, and let f(x) = exp(-1/x^2) g(x) for x <> 0, 0 forx = 0. Then f(x) = O(x^n) as x -> 0 for every n, so the Taylor seriesof f at x=0 is 0, which has in'nite radius of convergence. But 's nowhere analytic.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => Luigi that I read, the addition and subtraction formulas (sine> and cosine) come all you beginning from the cos(x+y).>> then the cos(x+y) formula has proved through some geometric> considerations.>> The problem that I have is this:>> this proof is valid only if x and y are inclusive between zero and> 45 (so that x+y is inclusive between 0 and 90).>> I know that every angle can be reduced to the 'rst octant, but to> prove that the cos(x+y) formula is true for every x and every y, I> would need to analyze various cases...>> they are some about 'fty!>> 0 <= X <=45 0 <= Y <=45> 0 <= x <=45 45 0 <= x <=45 45 90> 0 <= x <=45 90 ...>> Is there a quicker way?>this is probably not what you want, but if you have some complexanalysis under your belt and this equation:e^(ix) = cos x + i*sin x has been proved to your satisfaction, then you have a method forrigorously deriving those trig identities for all angles and withvirtually no work,just takee^i(x+y) = e^(ix) * e^(iy), and as a bonus you get both the sin andthe cos formulas at the same time.this is how i derive them on tests, i donOt bother trying to memorizethem.sorry if this is totally unhelpful.justin =>>Luigi escribi.97 en subtraction formulas (sine>and cosine) come all you beginning from the cos(x+y).>>then the cos(x+y) formula has proved through some geometric>considerations.>>[...]> >this is probably not what you want, but if you have some complex>analysis under your belt and this equation:>e^(ix) = cos x + i*sin x >has been proved to your satisfaction, then you have a method for> [...]While weOre on the subject, the addition formula can be proved directly from the fact that sine and cosine are the solutions to the differential equation y + y = 0 satisfying the appropriate initial conditions. Of course, Luigi was looking for a geometric proof.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu => This geomatric proof is valid for any angle:> http://www.geocities.com/scroussette/addtrig.htmlthank you very much...your help has been fundamental to meLuigi =>> this proof is valid only if x and y are inclusive between zero and> 45 (so that x+y is inclusive between 0 and 90).> If the sum of two non-negative numbers is no more than 90 it DOES NOT follow> that the two numbers are less than 90!! Consider 70 + 20 or 89.999 + .001>Sorry I intended (0<=x<=45) ^ (0<=y<=45) => [0<=(x+y)<=90]not (0<=x<=45) ^ (0<=y<=45) <> [0<=(x+y)<=90]Luigi =>a sub 0 mod c = a^0 mod c and>a sub 1 mod c = a^1 mod c and>a sub 2 mod c = = ( a^1 * a ) mod c and>a sub 0 = a^0 and>a sub 1 = a^1 but, aa sub 2 mod c <= a^2 and 2a sub n mod c <= a^n, that is, itOs possible that>a sub n mod c # a^n so for either a or b, 1<=a1 implies>either>a sub 3 mod c # a^3 or>b sub 3 mod c # b^3 so>0<=n<=2>ThatOs about right. Let me add a bita sub n = ( (a sub n-1* a ) mod ca sub 0 = a^0 mod c so a.0 + b.0 = c implies a solution (suf'ces),while a<=b sqrt(5), but1,1 < sqrt(2) but n<=c so this is the only onea sub 3 <= a^n but ifa,b < cuberoot(c), then a.3 + b.3 = c suf'ces, and any such solution would be a prize, but while1,1 < cuberoot(3) BUT1,2 <> cuberoot (3) and for n <= c this is seems to bemore than a coincidence, that isn <= c implies n <= 2 because2,2 < fourthroot4) and1,3 <> fourthroot (4) and sofor n >2, for c = cmin = n anda <= b < c = nE a,b,c | b > nthroot(c) in other wordsn>2 implies E b < c <= n | b>nthroot(c) and a.n + b.n = c does not suf'ce 3>n implies E b < c <= n | bc.Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Af'davit and Description at http://www.riaa.org/I will be signing an amended Af'davit soon. =Aside from knowing and understanding that P^n is a vector space, which mighthelp, I canOt help.P^1 is the space of all 'rst order polynomialsP^n is the space of all nth order polynomials.Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Af'davit and Description at http://www.riaa.org/I will be signing an amended Af'davit soon. => >> They couldnOt quite get the platinum-irridium artifact to weigh exactly>> 9.866 65 newtons - nohow - so they gave up and adopted it by proclamation:>> Now theyOre looking to rede'ne that adopted kilogram:>>I believe itOs 9.80665 N. (I may be wrong). m. ;-) > > You are wrong, but in a different way than you think you might be.> > That value has nothing whatsoever to do with the de'nition of a> kilogram as a unit of mass.> > Furthermore, if you stick to SI units, any standard acceleration of> free fall is worthless as tits on a boar.I thought that free fall meant that acceleration was zero?For taking F = ma, if F = 0, m > 0, sysengr => > >> They couldnOt quite get the platinum-irridium artifact to weigh exactly>> 9.866 65 newtons - nohow - so they gave up and adopted it by proclamation:>> Now theyOre looking to rede'ne that adopted kilogram:>>I believe itOs 9.80665 N. (I may be wrong). m. ;-) > > You are wrong, but in a different way than you think you might be.> > That value has nothing whatsoever to do with the de'nition of a> kilogram as a unit of mass.> > Furthermore, if you stick to SI units, any standard acceleration of> free fall is worthless as tits on a boar.> > I thought that free fall meant that acceleration was zero?No. An object falling ballistically in gravity is infree fall, as is a satellite in orbit. One onlinedictionary gives this: The ideal falling motion of something subject only to a gravitational 'eld. - Randy => > >> They couldnOt quite get the platinum-irridium artifact to weigh exactly>> 9.866 65 newtons - nohow - so they gave up and adopted it by proclamation:>> Now theyOre looking to rede'ne that adopted kilogram:>>I believe itOs 9.80665 N. (I may be wrong). m. ;-) > > You are wrong, but in a different way than you think you might be.> > That value has nothing whatsoever to do with the de'nition of a> kilogram as a unit of mass.> > Furthermore, if you stick to SI units, any standard acceleration of> free fall is worthless as tits on a boar.> > I thought that free fall meant that acceleration was zero?> > No. An object falling ballistically in gravity is in> free fall, as is a satellite in orbit. One online> dictionary gives this: The ideal falling motion of > something subject only to a gravitational 'eld.The above does not contradict acceleration being zero for an object infree fall.-- Johan =Suppose I have a real 3x3 rotation matrix, R. So, R is orthogonal andROR=I, where I is a 3x3 identity matrix.I need to compute the singular value decomposition of (R - I), and I wouldlike to do it symbolically (as opposed to numerically).Does the fact that R is orthogonal help with this problem, or will theanswer be ugly (symbolically)?Anybody already know of an answer (published or otherwise)?John =>Suppose I have a real 3x3 rotation matrix, R. So, R is orthogonal and>ROR=I, where I is a 3x3 identity matrix.>I need to compute the singular value decomposition of (R - I), and I would>like to do it symbolically (as opposed to numerically).>Does the fact that R is orthogonal help with this problem, or will the>answer be ugly (symbolically)?rotation, and it will work out pretty nicely.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =Robert, YouOre killing me. IOm old and tired. Do you really expect me to work atthis if you already know the answer? ;-) Any more (detailed) hints?John>Suppose I have a real 3x3 rotation matrix, R. So, R is orthogonal and>ROR=I, where I is a 3x3 identity matrix.>>I need to compute the singular value decomposition of (R - I), and Iwould>like to do it symbolically (as opposed to numerically).>>Does the fact that R is orthogonal help with this problem, or will the>answer be ugly (symbolically)?>> rotation, and it will work out pretty nicely.>> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2 = - tired of spending all your free time trying to 'nd the right job? - have you met your goals of sending a certain number of resumes each week? - are you ready fr success in your career search? We have answers to these questions for you, come and visit us atwww.KareerSearch.com! =S^c is not open <=> Ae>0,Ey in X:y in (U(x,e) n S)you say that itOs not true but then it means that the following isalso not trueS^c is open <=> Ee>0,Ay in X:y in U(x,e) => y in S^cBut this is a de'nition of open set or is it?No it isnOt, but if you insist on writing it like that it would be:S^c is open <=> Ax in S^c, Ee>0: U(x,e) contained in S^cDo you mean that the following is correct?S^c is open <=> Ax in S^c, Ee>0, Ay in X: y in U(x,e) => y in set of facts and the set of parts intechnological order.For the set of parts, in technological order, from Finite Mathematics withBusiness Applications, there exists a vector, a, and a square lower lefttriangular quantity matrix, Q with zero diagonal q.nn, such that:The elements of a are in technological order, in other words, an item a.i doesnot appear on the list until all of the parts and subassemblies that must gointo it have already appeard in elements a.1 through a.(i-1), andq.ij is the number of units of item a.j directly needed to assemble one itema.i., that is if we had on hand q.i1 units of item a1, q.i2 units of iterm a2,etc., and q.in units of a.n then we could immediately assemble one unit of a.iand have no parts left over. a.i is listed once in terms of any or no parts a.1through a.i on row i with q.j1 through q.j(i-1) the parts needed, if any, orotherwise zero.Stage requirements exists on Q. Q^k = 0 for some k<=n. The entries of Q^1 givethe one-stage requirments for a.imax and its subassemblies. The entries of Q^2are the two stage requirements for a.imax and etc. The entries of Q^(n<-imax)are the n-stage requirements for a.imax, that is the 'nal product, and anyneed subassemblies. Therefore since the zero diagonal, Q^2 has a zero diagonaljust below the zero diagonal, and kmin is the order of Q, not the rank orsize.I believe but have no example that this matrix is as useful for baking a cakeor sheet of n cookies per sheet as it is for proving a theorem with numberedsingle stage axioms and other postulates, corrolaries, and sub-proofs. The bookgives a clear example of the assembly of A tripod mast constructed from half-legs assembled from rods and bolts, legsassembled from half-legs and bolts, and a plate and some bolts.Now for a factory taking orders for individual items of manufacture such as oneVCR with a free tape and manual, and three VCR tapes, each with a free label,it is easy to extend Q to this theorem:x = x1 + x2 + ... x.imax is the unknown production vector, how many of eachitem to pull before assembly of the order begins. The elements can be whollyseparated, such as a VCR and a CD player, or inherently linked as above,including orders for spare parts if offered. The production vector times theprice vector can be the price of the order if nothing is double billed, andthatOs just a restriction on the price vector for a particular catalog.But E d = d1, d2, ..., dn, the outside demand vector, or the order.How are x, Q, and d, related?x = xQ +dd = x*(I-Q) and if I-Q has an inversex = (d*(I-q)^-1)They show the inverse exists.(I-Q)^-1 = I + Q + Q^2 + ... Q^(k-1), k<=imax the 'rst k for which Q^k = 0The lower left triangular structure of Q makes this particularly easy.By precomputing (I-Q)^(-1) per catalog, any order d from that catalog producesx = d * ((I - Q)^(-1))by one sole matrix multiplication, very fast in particular in Mathcad.Perhaps this will produce by analogy a more subtle proof of God. I think youmay have over-linearized the problem At least you didnOt attempt to prove thereis a Minkowski space inside the psyche, resulting from EinsteinOs GeneralRelativity, as a crackpot did in talk.origins a few years ago! I 'nd youranalysis intruiging.However, your analysis seems to imply there is a 'rst reason for everything,or did I misunderstand it?While thatOs not necessarily true, you may not have implied it. I just seemsthat if E xm]Reason^n(x)=god,in the same sense that one can sometimes writelim[x->c]f(x)=inf in calculus.E) OBVIOUS CORALLARIES- Every x in F has a god-limit which can be thoughtof a distance from god.Suppose that x has god-limit m,then for each y,there will be an n satisfying 0 <= n < m such that x<=Reason^n(x) and y<=Reason^n(x),Very roughly speaking,a common reason for x and any fact y can always be foundbetween x and its god-limit.Yours,Doug Goncz (at aol dot com)Replikon Research Read the RIAA Clean Slate Program Af'davit and Description at http://www.riaa.org/I will be signing an amended Af'davit soon. =I am suprised that anyone reads that stuff from so far back.I had psychological problems at the time and was a crank evenif I was unaware of it at the time. Its over with andhopefully IOll never return to crankdom. But nowthat were talking about all that stuff I posted long agoIOll mention that, for the time being, IOve tried tostay away from topics like this, although I donOtwant to renounce my old views completely. I thinkthat reality needs a prime uni'er (conciousness),and I guess it could also be construed as the reasonfor everything. I donOt even know if math couldapproach a topic like this because math talks aboutthings that donOt really exist. Everything inthe phenomenal world never really *is* because itundergoes change, and math speaks of absolutely existingentities. Indeed, thatOs probably not a failing of math.Its a failing of intellect itself when we divide the worldinto beings. Even talking about the contents of a thought is meaningless if there are *no* *individual* *thoughts*.At any rate either conciousness moves in time whilewhat is perceived is static, or what is perceivedchanges while conciousness remains (paradoxically) 'xed.claims that conciousness (i.e. me and you) is absolutely 'xedand I consider that to be a likely proposition.(i.e. the prime uni'er = you, now and forever).I apologise to the readers of sci.math for havingstarted this crankery long ago. I am cross posting this postto alt.philosophy and I would appreciateit if all replies to this post were posted there(and not to sci.math), since this subject isnot mathematical in nature.Sorry Doug for not reading and digesting your own reply.It really isnOt up my alley right now, and I do notmean to imply that your ideas are uninterestingor not worthy of consideration. =X is the number of aces in a set of 13 cards drawn.I would say its Poisson distribution because, foreach card drawn not being ace, the probability thatthe next card is an ace increases.On the other hand the probability of getting more thanfour aces is zero, no matter how many cards we willdraw. Any suggestions?Please note that Im not looking for an actual answerbut the resoning why it should or should not be aPoisson (or any other) distribution.-- KindlyKonrad-- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy => X is the number of aces in a set of 13 cards drawn.> .... You may 'nd the sci.stat.math news group a better place to ask. Ken Pledger. =>X is the number of aces in a set of 13 cards drawn.>I would say its Poisson distribution because, for>each card drawn not being ace, the probability that>the next card is an ace increases.Why would that indicate itOs Poisson?>On the other hand the probability of getting more than>four aces is zero, no matter how many cards we will>draw. Any suggestions?Then how could it be Poisson?| Please note that Im not looking for an actual answer| but the resoning why it should or should not be a| Poisson (or any other) distribution.You stated the reason, you just donOt realize that you did.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =>> I would say its Poisson distribution because, for>> each card drawn not being ace, the probability that>> the next card is an ace increases.>> Why would that indicate itOs Poisson?No calls to a larm central for 5 minutes gives lower chancethat a call will occure within next minute than 10 minuteswithout a call. ThatOs how i 'gure. Where do i step wrong?> ...the probability of getting more than four aces is zero...> Then how could it be Poisson?According to my teacher, the number of spelling errors ona page can be Poisson distributed, even if the chance ofthem being, say 1.000.000, is zero. ThatOs how. :)> You stated the reason, you just donOt realize that you did.If not Poisson, then which one?-- KindlyKonrad-- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy => I would say its Poisson distribution because, for> each card drawn not being ace, the probability that> the next card is an ace increases.>> Why would that indicate itOs Poisson?>No calls to a larm central for 5 minutes gives lower chance>that a call will occure within next minute than 10 minutes>without a call. ThatOs how i 'gure. Where do i step wrong?YouOre wrong to think that this makes it a Poisson distribution.If youOre talking about a Poisson process (where the number ofoccurrences of something over any given time interval is a random variable with Poisson distribution), then the numbers ofoccurrences over disjoint intervals are supposed to be independent.So if having no calls over the last 10 minutes increases the probability of a call in the next minute, this is _not_ a Poissonprocess and the total number of calls will probably _not_ havea Poisson distribution.> ...the probability of getting more than four aces is zero...>> Then how could it be Poisson?>According to my teacher, the number of spelling errors on>a page can be Poisson distributed, even if the chance of>them being, say 1.000.000, is zero. ThatOs how. :)Well, either your teacher is confused or you have misunderstoodwhat (s)he said. The Poisson distribution with parameter lambdasays Pr(X=x) = exp(-lambda) lambda^x/x! for _all_ nonnegative integers x. If Pr(X=1000000) = 0, itOs not Poisson. Now it mightbe _approximated_ by a Poisson distribution, but thatOs not what you asked about.>If not Poisson, then which one?Hypergeometric.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =-- KindlyKonrad-- places;their souls be chased by demons in Gehenna from one room toanother for all eternity and more.Sleep - thing used by ineffective people as a substitute for coffeeAmbition - a poor excuse for not having enough sence to be lazy = which tools can do tests of randomness? which one is the best? and rank data available! => > which tools can do tests of randomness? http://csrc.nist.gov/rng/ =[...]|> However, Professor McKenzie began claiming my work is algebraic|> geometry which is out of his 'eld. So I explained the entire thing|> over more than an hour, having driven up four hours from Atlanta|> metro, and he tells me itOs out of his 'eld.[...]In other words, you talk stuff that is orthogonal to real mathematics,not even wrong, and you were wasting the guyOs time and so he sent youon.IOd have done the same... much like the Indians who 'gured out theycould get rid of the Spaniards by telling them the next village had lotsof gold.If he tells you youOre wrong or talking nonsense, youOll argue. If hetells you it isnOt his 'eld and he doesnOt know much about it, but youcan go talk to that guy over there who might be interested, he is ridof you.IOll only tell you this once...-- cu,Brucedrift wave turbulence: http://www.rzg.mpg.de/~bds/ => > [...]> > |> However, Professor McKenzie began claiming my work is algebraic> |> geometry which is out of his 'eld. So I explained the entire thing> |> over more than an hour, having driven up four hours from Atlanta> |> metro, and he tells me itOs out of his 'eld.> > [...]> > In other words, you talk stuff that is orthogonal to real mathematics,> not even wrong, and you were wasting the guyOs time and so he sent you> on.Which de'es the fact that I explained it point-by-point. Theprofessor challenged me repeatedly on points and I handled allchallenges.The discussion was over an hour in two sessions.What amazes me is how clearly some of you are dedicated to falsebeliefs. Given the evidence that matters you instead turn to *social*pronouncements as if simply *saying* something changes the truth.ItOs like something is broken in your heads and your rationalfunctions are impaired or absent. Fascinating.James Harris => > [.snip.]> >Well, rationally, one would suppose that in explaining a math argument>a person has to go step-by-step, which reveals all the logical links. You would think so. IOve explained step by step why your claims about> the roots of x^3+3x-2 are false. Did you look at it?> > [.snip.]No. >However, youOve given the information that you know of him, and he IS>your senior.> > Surely thatOs irrelevant? I thought it wasnOt about who had what> degree, but about the math?> > (Yes, he is my better, by far. HeOs a world-renowned expert in> Universal Algebra, he solved the Tarski Problem a few years back, he> invented Tame Congruence Theory, and has a signi'cant number of> accomplishments under his bel)And he liked me, and I must admit that the more I think about it themore I wonder if I didnOt make a mistake taking the path I have.But then IOve spent so much time wasted with a liar like you Magidin.Maybe mathematicians in general arenOt so bad, but you are.And I think back now on what I gave him. >He dismissed the objection youOve used so often, and in fact, thatOs>not surprising as it never made any sense to think that f divided off>P(m) as some function of m or dependent on m, as itOs just not>mathematics.> > Apparently, he dismissed what you think or what you told him was my> objection. Your track record is clear: you have great dif'culty in> correctly paraphrasing other peopleOs objections.Bull. He trashed your objection. ThatOs it.> We have no way of knowing if what you told him was an accurate> representation of my view, other than your say so. And you have been> wrong on this subject pretty much every time youOve tried to state> what my objection is.> > But, tell you what: state my objection in full context, and provide> a link to a post where I made it.> > ItOs not that f divided off P(m) as some function of m or dependent> on m because I have never used the words divided off, so thatOs> your paraphrase. And it depends very much on just what the heck f and> P(m) are supposed to be.> > So, go ahead. Provide the complete statement, weOll see how accurate> you were in reporting it.Come on Magidin, thatOs only necessary for all those losers who donOtreally know mathematics.IOve hung out with Professor McKenzie. I know where I stand.>IOve called it voodoo math. Yes. YouOve also said that saying that something is a parameter is> rejecting algebra. Did this professor also agree with you on that> point?> > [.snip.]The professor was quite gracious in many ways, and IOve been ratherobnoxious.I hate you Magidin. YouOve twisted things, lied, and challenged me atevery point even when IOve been right.You jealous bastard. So what if IOm a better mathematician. Is thatany reason to trash everything you were taught? >>In any case, you say elsewhere that he blew you off. You>>seem to be implying that he agreed your work was correct ->>given that itOs very hard to 'gure out what you mean by saying>>he blew you off...>> >> Well, thereOs another point. One of the problems with his Advanced>> Polynomial Factorization is that it is not clear what it is he is>> saying; and as many have pointed out over several months, he could be>> saying true things, or he could be saying false things. However, as>> many others have pointed out as well, the true things he could be>> saying are not applicable to the situation he wants to apply them to,>> and the things which are applicable are not true. So it is entirely>> possible (even likely) that someone could say that what James says in>> Advanced Polynomial Factorization is correct; that does not mean>> that it is what James thinks he is saying is correct.>>A math professor senior to Magidin when presented with the objection>Magidin has given for months, quickly dismissed it.> > Non sequitur.> > By your own standards, seniority is irrelevant, but let that be as it> may. I have absolutely no problems admitting that Ralph McKenzie is> way smarter than I am.> > However, there is not an iota of evidence that what you presented to> this professor was an accurate report of my objection.I presented your objection which relies on the assertion that factorsof f can distribute as functions of m, or dependent on m, and theprofessor dismissed it.IOve talked to you too much Magidin. YouOve convinced me that othermathematicians are corrupt and Professor McKenzie at my own alma materwas nice to me and I responded in anger when he didnOt give me exactlywhat I wanted: vindication.>Rather than simply tell the truth, Magidin now backs away from his>original position, > > What was my original position?That my work is wrong.>I guess now claiming that Professor McKenzie didnOt>understand what I was discussing.> > No, I did not make that claim. What I said, as shoudl be clear from a> simple reading of the above, is that your Advanced Polynomial> Factorization is unclear, and many statements are ambiguous. That> certain interpretations of those statements lead to absolutely correct> statements, which are not applicable in the situation you are trying> to apply them (your FLT argument); and that certain interpretations of> the ambiguous statements lead to absolutely false statements. Which interpretation did you present?I explained the argument point-by-point you sick asshole.Professor McKenzie was skeptical but I answered his objections.You sick twisted bastard, IOve argued with you for years, and youstill refuse to accept the math.>The conclusion of the paper Advanced Polynomial Factorization is that>for the factorization>> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)>>where the aOs are all algebraic integers, one of them is coprime to 5.> > And that conclusion is false. WeOve gone over it, in detail. But here> wrong. (Original calculations done by Dale Hall):> > 1. Let > > q1 = 8 (a_1)^2 - 76 (a_1) - 185> r1 = 8 (a_1)^2 - 4 (a_1) - 45> s1 = 4 (a_1)^2 - 37 (a_1) - 104> > Since a_1 is an algebraic integer, each of q1, r1, s1 are algebraic> integers. 2. Likewise, let> > q2 = 8 (a_2)^2 - 76 (a_2) - 185> r2 = 8 (a_2)^2 - 4 (a_2) - 45> s2 = 4 (a_2)^2 - 37 (a_2) - 104> > and> > q3 = 8 (a_3)^2 - 76 (a_3) - 185> r3 = 8 (a_3)^2 - 4 (a_3) - 45> s3 = 4 (a_3)^2 - 37 (a_3) - 104> > Each of q2, r2, s2, q3, r3, s3 are algebraic integers.> > 3. We have that> > q1*r1 = [8(a_1)^2 - 76(a_1) - 185][8(a_1)^2 - 4(a_1)-45]> = 64(a_1)^4 - 32(a_1)^3 - 360(a_1)^2> -608(a_1)^3 + 304(a_1)^2 + 3420(a_1)> -1480(a_1)^2 + 740(a_1) + 8325> > = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325> > > Since (a_1)^3 - 12(a_1)^2 + 65 = 0, we have that> > (a_1)^3 = 12(a_1)^2 - 65 (a_1)^4 = 12(a_1)^3 - 65(a_1)> = 12( 12(a_1)^2 - 65) - 65(a_1)> = 144(a_1)^2 - 780 - 65(a_1)> = 144(a_1)^2 - 65(a_1) - 780,> > so> > q1*r1 = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325> = 64 [144(a_1)^2 - 65(a_1) - 780] > - 640 [12(a_1)^2 - 65]> - 1536(a_1)^2 + 4160(a_1) + 8325> = 9216(a_1)^2 - 4160(a_1) - 49920 - 7680(a_1)^2 + 41600> -1536(a_1)^2 + 4160(a_1) + 8325> = 41600+8325-49920> = 5.> > Since (a_2)^3 - 12(a_2)^2 + 65 = 0 and (a_3)^3 - 12(a_3)^2 + 65 = 0,> we also have> > q2*r2 = 5.> q3*r3 = 5.> > > > 4. Using the same de'nitions, we have:> > r1*s1 = ( 8(a_1)^2 - 4(a_1) - 45) * (4(a_1)^2 - 37(a_1) - 104)> = 32(a_1)^4 - 296(a_1)^3 - 832(a_1)^2> - 16(a_1)^3 + 148(a_1)^2 + 416(a_1)> - 180(a_1)^2 +1665(a_1) + 4680> > = 32(a_1)^4 - 312(a_1)^3 - 864(a_1)^2 + 2081(a_1) + 4680> = 32( 144(a_1)^2 - 65(a_1) - 780) - 312( 12(a_1)^2 - 65)> - 864(a_1)^2 + 2081(a_1) + 4680 = 4608(a_1)^2 - 2080(a_1) - 24960 - 3744(a_1)^2 + 20280> - 864(a_1)^2 + 2081(a_1) + 4680> = (a_1) + 20280 + 4680 - 24960> = a_1> > And so we also have> > r2*s2 = a_2> r3*s3 = a_3.> > 5. Since r1, s1, q1 are algebraic integers, r1*q1 = 5 and r1*s1 = a_1,> it follows that r1 is a common algebraic integer factor of a_1 and> 5.> > 6. Since r2, s2, q2 are algebraic integers, r2*q2 = 5 and r2*s2 = a_2,> it follows that r2 is a common algebraic integer factor of a_2 and> 5.> > 7. Since r3, s2, q3 are algebraic integers, r3*q3 = 5 and r3*q3 = a_3,> it follows that r3 is a common algebraic integer factor of a_3 and> 5.> > 8. We claim that r1, r2, and r3 are roots of the polynomial:> > f(x) = x^3 - 969 x^2 + 315 x + 5.> > To verify this, plug in the value of r1, and use the following> identities:> > > (a_1)^3 = 12(a_1)^2 - 65.> (a_1)^4 = 144(a_1)^2 - 65(a_1) - 780.> > (a_1)^5 = (a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2> = 12(a_1)^4 - 65(a_1)^2> = 12(144(a_1)^2 - 65(a_1)-780) - 65(a_1)^2> = 1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2> = 1663(a_1)^2 - 780(a_1) - 9360.> > (a_1)^6 = (a_1)^4 (a_1)^2> = (144(a_1)^2 - 65(a_1) - 780) (a_1)^2> = 144(a_1)^4 - 65(a_1)^3 - 780(a_1)^2> = 144 (144(a_1)^2 - 65(a_1) - 780) > - 65(12(a_1)^2 - 65)> - 780(a_1)^2> = 20736(a_1)^2 - 9360(a_1) - 112320> -780(a_1)^2 + 4225> -780(a_1)^2> = 19176(a_1)^2 - 9360(a_1) - 108095.> > Same for r2 and r3, replacing a_1 for a_2 and a_3, respectively> (omitted for space).> > 9. f(x) is monic, primitive, and irreducible over Q. For the latter,> the polynomial is reducible over Q if and only if it has a root> over Q, since it is degree 3. The only possible rational roots, by> the p/q test, are 1, -1, 5, and -5, and> f(1) = -648> f(-1)=-1280> f(5) = -22520> f(-5)=-25920.> > > 10. r1 is an algebraic integer unit if and only if 1/r1 (its> multiplicative inverse) is also an algebraic integer. But 1/r1 is> a root of the polynomial we obtain from > f(x) = x^3 - 969 x^2 + 315 x + 5. > by plugging in 1/x, setting equal to 0, and solving, that is, 1/r1> is a root of> > g(x) = 5x^3 + 315x^2 - 969x + 1> > which is a primitive, non-monic, irreducible polynomial over> Q. Therefore, none of its roots are algebraic integers. So 1/r1, a> root, is not an algebraic integer. So r1 is not an algebraic> integer unit.Which is the break in the logical chain you anti-mathematician.What is wrong with you Magidin?Now why donOt you 'll in the gap?My point is that an algebraic integer can be coprime to ALL primeinteger and yet its multiplicative inverse can NOT be an algebraicinteger.IOve proven my point repeatedly, and given the *math* but you refuseto accept it, so how can you be a mathematician?Consider http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759where *basic* algebra refutes you. ItOs *algebra* that refutes youMagidin, but you refuse to accept the math.You have hounded me for years, convinced people of falsehoods, andnow, even when repudiated by Professor McKenzie, whom you know, youSTILL attack the math.WHAT IS WRONG WITH YOU??!!!James Harris =Uh oh...Your worst fears con'rmed... they really are lying to you.Of course this brings up a sticky issue: who to believe? Thise that agreewith you, or those that donOt.As for Madigan, of this affair. At the very least, theyOll take away his slide ruleand perhaps his library card. Maybe you could write to the NYT, I mean whoknows what other things heOs lied about. Is he really Madigan (maybe heOsSuddam trying to undermine America)? Does he _really_ come from Berkley? Andof course the list of questions could go on and on and on and on....Thanb you for alerting us to this grave danger! => the NYT, I mean who knows what other things heOs lied about. Is he> really Madigan (maybe heOs Suddam trying to undermine America)? Does> he _really_ come from Berkley? And of course the list of questionsYou mispelt Amercia.-- = [.snip.]>> And that conclusion is false. WeOve gone over it, in detail. But here>> wrong. (Original calculations done by Dale Hall):>> >> 1. Let >> >> q1 = 8 (a_1)^2 - 76 (a_1) - 185>> r1 = 8 (a_1)^2 - 4 (a_1) - 45>> s1 = 4 (a_1)^2 - 37 (a_1) - 104>> >> Since a_1 is an algebraic integer, each of q1, r1, s1 are algebraic>> integers.>> >> 2. Likewise, let>> >> q2 = 8 (a_2)^2 - 76 (a_2) - 185>> r2 = 8 (a_2)^2 - 4 (a_2) - 45>> s2 = 4 (a_2)^2 - 37 (a_2) - 104>> >> and>> >> q3 = 8 (a_3)^2 - 76 (a_3) - 185>> r3 = 8 (a_3)^2 - 4 (a_3) - 45>> s3 = 4 (a_3)^2 - 37 (a_3) - 104>> Each of q2, r2, s2, q3, r3, s3 are algebraic integers.>> >> 3. We have that>> >> q1*r1 = [8(a_1)^2 - 76(a_1) - 185][8(a_1)^2 - 4(a_1)-45]>> = 64(a_1)^4 - 32(a_1)^3 - 360(a_1)^2>> -608(a_1)^3 + 304(a_1)^2 + 3420(a_1)>> -1480(a_1)^2 + 740(a_1) + 8325>> >> = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> >> >> Since (a_1)^3 - 12(a_1)^2 + 65 = 0, we have that>> >> (a_1)^3 = 12(a_1)^2 - 65>> >> (a_1)^4 = 12(a_1)^3 - 65(a_1)>> = 12( 12(a_1)^2 - 65) - 65(a_1)>> = 144(a_1)^2 - 780 - 65(a_1)>> = 144(a_1)^2 - 65(a_1) - 780,>> >> so>> >> q1*r1 = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> = 64 [144(a_1)^2 - 65(a_1) - 780] >> - 640 [12(a_1)^2 - 65]>> - 1536(a_1)^2 + 4160(a_1) + 8325>> = 9216(a_1)^2 - 4160(a_1) - 49920 - 7680(a_1)^2 + 41600>> -1536(a_1)^2 + 4160(a_1) + 8325>> = 41600+8325-49920>> = 5.>> >> Since (a_2)^3 - 12(a_2)^2 + 65 = 0 and (a_3)^3 - 12(a_3)^2 + 65 = 0,>> we also have>> >> q2*r2 = 5.>> q3*r3 = 5.>> >> >> >> 4. Using the same de'nitions, we have:>> >> r1*s1 = ( 8(a_1)^2 - 4(a_1) - 45) * (4(a_1)^2 - 37(a_1) - 104)>> = 32(a_1)^4 - 296(a_1)^3 - 832(a_1)^2>> - 16(a_1)^3 + 148(a_1)^2 + 416(a_1)>> - 180(a_1)^2 +1665(a_1) + 4680>> >> = 32(a_1)^4 - 312(a_1)^3 - 864(a_1)^2 + 2081(a_1) + 4680>> = 32( 144(a_1)^2 - 65(a_1) - 780) - 312( 12(a_1)^2 - 65)>> - 864(a_1)^2 + 2081(a_1) + 4680>> >> = 4608(a_1)^2 - 2080(a_1) - 24960 - 3744(a_1)^2 + 20280>> - 864(a_1)^2 + 2081(a_1) + 4680>> = (a_1) + 20280 + 4680 - 24960>> = a_1>> >> And so we also have>> >> r2*s2 = a_2>> r3*s3 = a_3.>> >> 5. Since r1, s1, q1 are algebraic integers, r1*q1 = 5 and r1*s1 = a_1,>> it follows that r1 is a common algebraic integer factor of a_1 and>> 5.>> >> 6. Since r2, s2, q2 are algebraic integers, r2*q2 = 5 and r2*s2 = a_2,>> it follows that r2 is a common algebraic integer factor of a_2 and>> 5.>> >> 7. Since r3, s2, q3 are algebraic integers, r3*q3 = 5 and r3*q3 = a_3,>> it follows that r3 is a common algebraic integer factor of a_3 and>> 5.>> >> 8. We claim that r1, r2, and r3 are roots of the polynomial:>> >> f(x) = x^3 - 969 x^2 + 315 x + 5.>> >> To verify this, plug in the value of r1, and use the following>> identities:>> >> >> (a_1)^3 = 12(a_1)^2 - 65.>> (a_1)^4 = 144(a_1)^2 - 65(a_1) - 780.>> >> (a_1)^5 = (a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2>> = 12(a_1)^4 - 65(a_1)^2>> = 12(144(a_1)^2 - 65(a_1)-780) - 65(a_1)^2>> = 1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2>> = 1663(a_1)^2 - 780(a_1) - 9360.>> >> (a_1)^6 = (a_1)^4 (a_1)^2>> = (144(a_1)^2 - 65(a_1) - 780) (a_1)^2>> = 144(a_1)^4 - 65(a_1)^3 - 780(a_1)^2>> = 144 (144(a_1)^2 - 65(a_1) - 780) >> - 65(12(a_1)^2 - 65)>> - 780(a_1)^2>> = 20736(a_1)^2 - 9360(a_1) - 112320>> -780(a_1)^2 + 4225>> -780(a_1)^2>> = 19176(a_1)^2 - 9360(a_1) - 108095.>> Same for r2 and r3, replacing a_1 for a_2 and a_3, respectively>> (omitted for space).>> >> 9. f(x) is monic, primitive, and irreducible over Q. For the latter,>> the polynomial is reducible over Q if and only if it has a root>> over Q, since it is degree 3. The only possible rational roots, by>> the p/q test, are 1, -1, 5, and -5, and>> f(1) = -648>> f(-1)=-1280>> f(5) = -22520>> f(-5)=-25920.>> >> >> 10. r1 is an algebraic integer unit if and only if 1/r1 (its>> multiplicative inverse) is also an algebraic integer. But 1/r1 is>> a root of the polynomial we obtain from >> f(x) = x^3 - 969 x^2 + 315 x + 5. >> by plugging in 1/x, setting equal to 0, and solving, that is, 1/r1>> is a root of>> >> g(x) = 5x^3 + 315x^2 - 969x + 1>> >> which is a primitive, non-monic, irreducible polynomial over>> Q. Therefore, none of its roots are algebraic integers. So 1/r1, a>> root, is not an algebraic integer. So r1 is not an algebraic>> integer unit.>>Which is the break in the logical chain you anti-mathematician.There is no break. r1 is an algebraic integer unit if and only if (a)it is an algebraic integer; and (b) its multiplicative inverse is alsoan algebraic integer. You tried using this result with your failedpolynomial x^3 -3x + 2, remember?We know r1 is an algebraic integer. And weOve seen that itsmultiplicative inverse is not an algebraic integer. Therefore, r1 isnot an algebraic integer unit.>What is wrong with you Magidin?>>Now why donOt you 'll in the gap?BecausetI can do nothing about the empty space between your ears.>My point is that an algebraic integer can be coprime to ALL prime>integer and yet its multiplicative inverse can NOT be an algebraic>integer.Your point is wrong, and is proven wrong above. IOve producedEXPLICITLY a non-unit common factor of a1 and 5. Your claim is wrong. = reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = [.snip.]>> Apparently, he dismissed what you think or what you told him was my>> objection. Your track record is clear: you have great dif'culty in>> correctly paraphrasing other peopleOs objections.>>Bull. He trashed your objection. ThatOs it.So you claim. Looks, from your report, that he trashed what youreported to be my objection. Let me guess, you said What he isbasically saying is... and then you simpli'ed the situation so hecould see the gist of what you said my claim was? [.snip.]>> So, go ahead. Provide the complete statement, weOll see how accurate>> you were in reporting it.>>Come on Magidin, thatOs only necessary for all those losers who donOt>really know mathematics.Let me quote you on this.>>Facts not in evidence.>>For those who donOt know, and in case I messed up with my usage, that>>means that the poster has said something that requires outside>>veri'cation. That is, to 'nd out if heOs telling the truth, you>>have to go do research.>ItOs a nasty little trick because who is going to go do all that>research to it a nasty little trick when you make claims not in evidence? Minewere available to check by anyone who wanted to do so. Your claims areuncheckable. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu>> [...]>>But then IOve spent so much time wasted with a liar like you Magidin.>>Maybe mathematicians in general arenOt so bad, but you are.>And I think back now on what I gave him.>[...]>>The professor was quite gracious in many ways, and IOve been rather>obnoxious.>>I hate you Magidin. YouOve twisted things, lied, and challenged me at>every point even when IOve been right.>>You jealous bastard. So what if IOm a better mathematician. Is that>any reason to trash everything you were taught?> >> [...]>>IOve talked to you too much Magidin. YouOve convinced me that other>mathematicians are corrupt and Professor McKenzie at my own alma mater>was nice to me and I responded in anger when he didnOt give me exactly>what I wanted: vindication.> [...]>>I explained the argument point-by-point you sick asshole.>>Professor McKenzie was skeptical but I answered his objections.>>You sick twisted bastard, IOve argued with you for years, and you>still refuse to accept the math.>[...]>>You have hounded me for years, convinced people of falsehoods, and>now, even when repudiated by Professor McKenzie, whom you know, you>STILL attack the math.>>WHAT IS WRONG WITH YOU??!!!>James HarrisDavid C. Ullrich**************************As far as IOm concerend youOre trying to wait until I die, so I 'guremaybe you should die instead. How about that, eh? WouldnOt that be abetter twist?You refuse to follow the math, so the great Powers that controlreality and *speak* in mathematics decide to kill you instead of me.So what do you think about that, eh? Oh, canOt hear Them talking?Well, I guess thatOs because you donOt really understand Mathematics,the true language, which is THE language.TheyOre talking about you now, and They agree with my assessment, andwill not penalize me as They allowed the others like Galois and Abelto be penalized.They will kill you instead.James Harris speaking on Weird factorization, genius => >> ... stuff deleted ...> >The conclusion of the paper Advanced Polynomial Factorization is that>for the factorization>> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)>>where the aOs are all algebraic integers, one of them is coprime to 5.>>And that conclusion is false. WeOve gone over it, in detail. But here>>wrong. (Original calculations done by Dale Hall):>>1. Let >> q1 = 8 (a_1)^2 - 76 (a_1) - 185>> r1 = 8 (a_1)^2 - 4 (a_1) - 45>> s1 = 4 (a_1)^2 - 37 (a_1) - 104>> Since a_1 is an algebraic integer, each of q1, r1, s1 are algebraic>> integers.>>2. Likewise, let>> q2 = 8 (a_2)^2 - 76 (a_2) - 185>> r2 = 8 (a_2)^2 - 4 (a_2) - 45>> s2 = 4 (a_2)^2 - 37 (a_2) - 104>>and>> q3 = 8 (a_3)^2 - 76 (a_3) - 185>> r3 = 8 (a_3)^2 - 4 (a_3) - 45>> s3 = 4 (a_3)^2 - 37 (a_3) - 104>> Each of q2, r2, s2, q3, r3, s3 are algebraic integers.>>3. We have that>> q1*r1 = [8(a_1)^2 - 76(a_1) - 185][8(a_1)^2 - 4(a_1)-45]>> = 64(a_1)^4 - 32(a_1)^3 - 360(a_1)^2>> -608(a_1)^3 + 304(a_1)^2 + 3420(a_1)>> -1480(a_1)^2 + 740(a_1) + 8325>> = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>>Since (a_1)^3 - 12(a_1)^2 + 65 = 0, we have that>> (a_1)^3 = 12(a_1)^2 - 65>> (a_1)^4 = 12(a_1)^3 - 65(a_1)>> = 12( 12(a_1)^2 - 65) - 65(a_1)>> = 144(a_1)^2 - 780 - 65(a_1)>> = 144(a_1)^2 - 65(a_1) - 780,>>so>> q1*r1 = 64(a_1)^4 - 640(a_1)^3 - 1536(a_1)^2 + 4160(a_1) + 8325>> = 64 [144(a_1)^2 - 65(a_1) - 780] >> - 640 [12(a_1)^2 - 65]>> - 1536(a_1)^2 + 4160(a_1) + 8325>> = 9216(a_1)^2 - 4160(a_1) - 49920 - 7680(a_1)^2 + 41600>> -1536(a_1)^2 + 4160(a_1) + 8325>> = 41600+8325-49920>> = 5.>>Since (a_2)^3 - 12(a_2)^2 + 65 = 0 and (a_3)^3 - 12(a_3)^2 + 65 = 0,>>we also have>>q2*r2 = 5.>>q3*r3 = 5.>>4. Using the same de'nitions, we have:>> r1*s1 = ( 8(a_1)^2 - 4(a_1) - 45) * (4(a_1)^2 - 37(a_1) - 104)>> = 32(a_1)^4 - 296(a_1)^3 - 832(a_1)^2>> - 16(a_1)^3 + 148(a_1)^2 + 416(a_1)>> - 180(a_1)^2 +1665(a_1) + 4680>> = 32(a_1)^4 - 312(a_1)^3 - 864(a_1)^2 + 2081(a_1) + 4680>> = 32( 144(a_1)^2 - 65(a_1) - 780) - 312( 12(a_1)^2 - 65)>> - 864(a_1)^2 + 2081(a_1) + 4680>> >> = 4608(a_1)^2 - 2080(a_1) - 24960 - 3744(a_1)^2 + 20280>> - 864(a_1)^2 + 2081(a_1) + 4680>> = (a_1) + 20280 + 4680 - 24960>> = a_1>> And so we also have>> r2*s2 = a_2>> r3*s3 = a_3.>>5. Since r1, s1, q1 are algebraic integers, r1*q1 = 5 and r1*s1 = a_1,>> it follows that r1 is a common algebraic integer factor of a_1 and>> 5.>>6. Since r2, s2, q2 are algebraic integers, r2*q2 = 5 and r2*s2 = a_2,>> it follows that r2 is a common algebraic integer factor of a_2 and>> 5.>>7. Since r3, s2, q3 are algebraic integers, r3*q3 = 5 and r3*q3 = a_3,>> it follows that r3 is a common algebraic integer factor of a_3 and>> 5.>>8. We claim that r1, r2, and r3 are roots of the polynomial:>> >> f(x) = x^3 - 969 x^2 + 315 x + 5.>> To verify this, plug in the value of r1, and use the following>> identities:>> (a_1)^3 = 12(a_1)^2 - 65.>> (a_1)^4 = 144(a_1)^2 - 65(a_1) - 780.>> (a_1)^5 = (a_1)^3(a_1)^2 = (12(a_1)^2-65)(a_1)^2>> = 12(a_1)^4 - 65(a_1)^2>> = 12(144(a_1)^2 - 65(a_1)-780) - 65(a_1)^2>> = 1728(a_1)^2 - 780(a_1) - 9360 - 65(a_1)^2>> = 1663(a_1)^2 - 780(a_1) - 9360.>> (a_1)^6 = (a_1)^4 (a_1)^2>> = (144(a_1)^2 - 65(a_1) - 780) (a_1)^2>> = 144(a_1)^4 - 65(a_1)^3 - 780(a_1)^2>> = 144 (144(a_1)^2 - 65(a_1) - 780) >> - 65(12(a_1)^2 - 65)>> - 780(a_1)^2>> = 20736(a_1)^2 - 9360(a_1) - 112320>> -780(a_1)^2 + 4225>> -780(a_1)^2>> = 19176(a_1)^2 - 9360(a_1) - 108095.>> Same for r2 and r3, replacing a_1 for a_2 and a_3, respectively>> (omitted for space).>>9. f(x) is monic, primitive, and irreducible over Q. For the latter,>> the polynomial is reducible over Q if and only if it has a root>> over Q, since it is degree 3. The only possible rational roots, by>> the p/q test, are 1, -1, 5, and -5, and>> f(1) = -648>> f(-1)=-1280>> f(5) = -22520>> f(-5)=-25920.>> >>10. r1 is an algebraic integer unit if and only if 1/r1 (its>> multiplicative inverse) is also an algebraic integer. But 1/r1 is>> a root of the polynomial we obtain from >> f(x) = x^3 - 969 x^2 + 315 x + 5. >> by plugging in 1/x, setting equal to 0, and solving, that is, 1/r1>> is a root of>> g(x) = 5x^3 + 315x^2 - 969x + 1>> which is a primitive, non-monic, irreducible polynomial over>> Q. Therefore, none of its roots are algebraic integers. So 1/r1, a>> root, is not an algebraic integer. So r1 is not an algebraic>> integer unit.> > > Which is the break in the logical chain you anti-mathematician.> Do you claim that this step is incorrect?Why?I suggest, since you apparently have a friendly, trustingrelationship with Professor McKenzie, that you cut and pastethe text (steps 1 - 17, together with your preceding paragraphof the trouble youOre having with some of the readers here, andask him for a judgment on the correctness (or error, since youbelieve this argument to be §awed) of the argument. I maintainthat the argument is correct, as has every other reader who hasexamined the steps.This is a perfectly simple task, and if youOre correct, wouldbuy you a modicum of credibility. ... stuff deleted ...> Now why donOt you 'll in the gap? My point is that an algebraic integer can be coprime to ALL prime> integer and yet its multiplicative inverse can NOT be an algebraic> integer.> Either this statement is demonstrably wrong, or I fail to understandit. Taken literally, youOre suggesting that there is an algebraicinteger r, such that r is coprime to r^2. You must realize that thiscannot be true: Let us suppose that r is coprime to r^2 in the ring of algebraic integers. Let m,n be algebraic integers, for which mr + nr^2 = 1. Factoring, one has r(m + nr) = 1, giving (m + nr), the multiplicative inverse of r, as an algebraic integer. Thus, r must be a unit, for its multiplicative inverse is an algebraic integer.> IOve proven my point repeatedly, and given the *math* but you refuse> to accept it, so how can you be a mathematician?> You will no doubt claim that the above argument is §awed, as well.What is the error in that argument? ... stuff deleted ...> You have hounded me for years, convinced people of falsehoods, and> now, even when repudiated by Professor McKenzie, whom you know, you> STILL attack the math.> If you wish to count Professor McKenzie as a proponent of the claimthat the above argument, as well as MagidinOs re-phrasing of my earlierargument, you are doing him a disservice by not allowing him to expresshis independent opinion (which, by your accounting, will surely supportyour position).It canOt be made any simpler. Let Professor McKenzie see the arguments.Let him speak for himself. I have been tempted to address him myself,but have chosen not to do so out of respect for your relationship withhim. If you also have respect for that relationship, you will not asserthis position, but will invite him to weigh in on it himself.invite him to respond to you directly. Be sure to obtain his permissionto quote his response to those arguments, and report back to sci.math.> > James HarrisDale. =Just 'xing a poorly-formed sentence. My fault for posting too lateat night, no doubt.> ... mumble ...> If you wish to count Professor McKenzie as a proponent of the claim> that the above argument, as well as MagidinOs re-phrasing of my earlier> argument, ... (insert) ... are incorrect ... > ... , you are doing him a disservice by not allowing him to express> his independent opinion (which, by your accounting, will surely support> your position).> ... and so forth ...> > Dale.> Sorry for the sloppy wording.Sorrier to have failed to follow it myself.Dale. => But then IOve spent so much time wasted with a liar like you Magidin.> Prof. Magadin has spent countless hours of his time carefully and meticulously understanding, explaining, and 'nding the errors in your arguments. He is unfailingly polite and always sticks to the math. I donOt understand your anger towards him. You should thank him. =>> Tee-hee.>Nothing makes a pig happier than a roll in the wallow, as this tittering porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg& width=314&height=400# => I thought I was set. With a high I.Q. group set to publish my paper> on factoring polynomials into non-polynomial factors I 'gured that> now certainly I could push my agenda to fame and fortune. I was> wrong.If a society is founded on the idea of not being accountable to others because of superior intelligence, it is not likely to be a source of respected papers.> So here I am back again, humbled yet again, as thereOs no escaping it,> there is no way IOll get anywhere pissing off mathematicians.You need to remember this throughout the week, at least, before weOll believe that *you* believe that.> So here are some concessions.> [concessions deleted]> > So quit the lying you dark evil people as IOm not out here claiming to> have a proof of FermatOs Last Theorem and IOm not out here claiming to> have found THE prime counting function.Umm, how are insults supposed to constitute not pissing off mathematicians?> > Take down all those webpages attacking me, and quit with the posts> calling me a crank. IOm 'nally tired of being called a crank.IOm not one of them, but I suspect theyOll be taken down when you *prove* that you are willing to listen to people who are trying to explain the math to you.> > I want to go legit.> > Um, there is that little problem with algebraic integers to discuss> though; however, IOm open-minded and willing to The proofs are waiting there.> > LetOs get back to it folks. No FLT. NO ING PRIME COUNTING!!!I got that the 'rst time. No need to shout.> > But 'nally I want some straight answers on the ring of algebraic> integers.Ok, how about starting with what you understand the de'nitions of ring and algebraic integers to be.> And thereOs no website of mine, so no way to claim that IOm NOT> dropping FLT and THE prime counting you evil bastards.Again, I thought you didnOt want to piss of mathematicians. What part of this is not intended to be at least irritating?-- Will Twentyman =well, what has the IBF ... I mean,the Homeland Security out't found, thus far? your terminology of correct math is unde'ned, althoughmany of us would say that Liebniz criteriumof iff is what you really mean; is it?... some folks say,bijective binary operation, or some thing, an implicationthat can be read either way, dependingupon how one uses the wording for necessity & suf'ciency. the same lack of de'nition applies, it seems,to your veri'cation or proof of the lying liarsO lies,who are so tedious in their belittling of your efforts,even ham-handed ... but that doesnOt apply to Magidin,nor to Nora Baboobda Baron. I mean,they seem fairly honest (and nice, butthatOs beside the point, with an ingracious character,like Yusef) to me. what have you proven about them,again?is prof. Mackenzie going to deliver his judgement,soon?well, IOm not coming back to this til *at least* after Oct.7;after all, itOs polls from AOL-TimeWarner-HBO-CNN that I haveto deal with, plus Warren BuffetOs hulking puppet,while we account the actual polling-place ordeal. at least,heOs not taking steroids any more!... see my sig. (I mean,youOve never replied to me, before, so,thereOs not really any reason ... but,IOm hoping to see some action from those high school wonders .-) > I could meet each and every such challenge and show how each step> followed logically, and therefore correctly. > ItOs not about believing any of you but about your willingness or> unwillingness to accept correct mathematics. What I veri'ed is my > What my meeting with Professor McKenzie also con'rmed for me is that> a professional mathematician wouldnOt accept their primary objection,> which con'rms to me that they were deliberately lying, as itOs> nonsensical to believe that factors of f vary as functions or as> dependents on m, which Professor McKenzie didnOt even seriously> consider as a possibility. > What I have clearly veri'ed is that certain posters who *are*> mathematicians, like Magidin, have been lying about the mathematics,> and lying repeatedly with objections that a professional mathematician> quickly rejected.--les ducs de Buffet!http://larouchepub.com =Hey,Could I ask for some homework help? Apologies beforehand if this isthe wrong group; I saw alt.algebra.help but this isnOt algebra...d = deltaSo, I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, thenfor all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then wehave f(x) > N.Obviously, the problem revolves around choosing appropriate N. Anysuggestions would be welcomed.Also, does anyone have a book recommendation for a 4th semestercalculus class? We did the basic 3 semesters of calculus (plug & chugw/ Thomas), and now are using Calculus by Spivak as a 1 semesterreview of all calculus focused on proofs. IOm not sure that I likethe book very much and am interested in a companion book to see adifferent approach to the various mathematics. Any recommendationswould be welcome.-earl- => Hey,> > Could I ask for some homework help? Apologies beforehand if this is> the wrong group; I saw alt.algebra.help but this isnOt algebra...> > d = delta> > So, I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.> > Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we> have f(x) > N.> > Obviously, the problem revolves around choosing appropriate N. Any> suggestions would be welcomed.No. You are not allowed to choose N. You must de'ne d in an appropriate way *for each N*. You will, in general, be looking for d as a function of N.The proof is similar to epsilon-delta proofs where you found delta as a function of epsilon.-- Will Twentyman => I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.> > Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we> have f(x) > N.The de'nition of lim (x->a) f(x) = +inf is that for each N > 0 there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then f(x) > N. ThereOs a difference between this and the above.Wishful thinking: if 0 < |x-3| < d, then 1/|x-3| > 1/d, which implies 1/(x-3)^2 > 1/d^2. If we only knew 1/d^2 > N ... hmmm .... => I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.>> >> Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then>> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we>> have f(x) > N.>>The de'nition of lim (x->a) f(x) = +inf is that for each N > 0 there is >a d > 0 s.t., for all x, if 0 < |x-a| < d, then f(x) > N. ThereOs a >difference between this and the above.??? IOve never been much of a proofreader. The only differenceI can see is for all N changed to for each N > 0 in yourversion. Is that the difference youOre referring to? _That_difference doesnOt make any difference...>Wishful thinking: if 0 < |x-3| < d, then 1/|x-3| > 1/d, which implies >1/(x-3)^2 > 1/d^2. If we only knew 1/d^2 > N ... hmmm ....Yeah, itOs a toughie. Maybe d = 0.0000001 works.************************David C. Ullrich =>> Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then>> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we>> have f(x) > N.>>The de'nition of lim (x->a) f(x) = +inf is that for each N > 0 there is >a d > 0 s.t., for all x, if 0 < |x-a| < d, then f(x) > N. ThereOs a >difference between this and the above.> > ??? IOve never been much of a proofreader. The only difference> I can see is for all N changed to for each N > 0 in your> version. Is that the difference youOre referring to? _That_> difference doesnOt make any difference...The de'nition is an iff, not an if, then. Is there no difference between for each N and for all N above? Maybe youOre right. I thought for each indicated better that d depends on N, but maybe not. Better than either might be for each N > 0 there corresponds a d > 0. (Meanwhile the OP was busy trying to choose an appropriate N, which I missed completely.) => Hey,>> Could I ask for some homework help? Apologies beforehand if this is> the wrong group; I saw alt.algebra.help but this isnOt algebra...>> d = delta>> So, I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.>> Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we> have f(x) > N.>> Obviously, the problem revolves around choosing appropriate N. Any> suggestions would be welcomed.Nononononono. Your enemy gives you N. It is up to you to pick d to beatthe enemy.So, you want 1/((x-3)^2)>N if |x-3|some functiong(d). So you wantg(d)>N. Now, you have to solve this inequality for d.Jon Miller => Hey,>> Could I ask for some homework help? Apologies beforehand if this is> the wrong group; I saw alt.algebra.help but this isnOt algebra...>> d = delta>> So, I want to prove lim (x approaches 3) of f(x) = 1/((x-3)^2) = +inf.>> Where the de'nition of limit is that, if lim (x->a) f(x) = +inf, then> for all N, there is a d > 0 s.t., for all x, if 0 < |x-a| < d, then we> have f(x) > N.>> Obviously, the problem revolves around choosing appropriate N. Any> suggestions would be welcomed.>> Nononononono. Your enemy gives you N. It is up to you to pick d to beat> the enemy.ThatOs way to think about it - a challenge!The enemy picture is real nice :-)>> So, you want 1/((x-3)^2)>N if |x-3| knowing only that> |x-3|some function> g(d). So you want> g(d)>N. Now, you have to solve this inequality for d.>> Jon MillerIOd write it down like this:Given an arbitrary positive N (by the enemy),then we have: 1/(x-3)^2 > N<==> (x-3)^2 < 1/N<==> |x-3| < 1/sqrt(N)<== (reading this arrow as as soon as) |x-3| < d and d = 1/sqrt(N)So for each N we have found a d such that |x-3| < d ==> 1/(x-3)^2 >findirk Vdm <3f73c8f2$1$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft = at 03:42 PM, Bob Pease different approaches.Some rigorous, some just using handwaving.>What happened is that a lot of folks insisted that the approach that>their Calc I book used was the only approach, even when quoted>chapter and verse from other texts.There are a lot of different Calculus texts. Some use soundapproaches, some donOt. IOm not aware of a Calc I book that de'nesthe term differential in a rigorous fashion.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to = Shmuel (Seymour J.) Metz different Calculus texts. Some use sound> approaches, some donOt. IOm not aware of a Calc I book that de'nes> the term differential in a rigorous fashion.IIRC, at least one edition of Calculus by Tom Apostol does. tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft = at 06:10 PM, raf@tiki-lounge.com (Ross A. Finlayson) said:>I know that bijections exist among in'nite sets. Q can be>considered greater than N for a variety of reasons.Not for reasons that make any Mathematical sense.>it emits a random rational number. De'ne random rational number.>more than half, What does that mean?>Who says that there are uncountable sets that are countable>unions of countable sets? Somebody who is misinformed.>About 2^x, itOs just what it is, 2^x. ThatOs not an answer, and has no relevance to Mathematics.> Consider 3^x. Say you have a>trinary logic with tree values, 0, 1, and 2.ThatOs cool when x is an integer. What about whn x is a set?>Form each sequence of>length p of values from {0, 1, 2}, the count of permutations is 3^x.What permutations? How is p related to x?>That leads back to (1+iota) ^ x. What is Iota?>When x is in'nite, I prefer to consider it a unit scalar in'nity,What does that mean?>using in'nitesimal analysis.What is in'nitesimal analysis?>ItOs called integrating x dx from zero to one.No. Integration has nothing to do with anything called in'nitesimalanalysis.>Virgil, I donOt accept the antidiagonal argument. It doesnOt hold>true under the examination that I have given it,You havenOt given it an analysis.>dual representation of some rationals,Irrelevant; you can de'ne the mapping to only use one of them.>I consider in'nite binary sequences with rational densities of>ones. Meaningless unless and until you de'ne density.>Are the irrationals everywhere discontinuous?Meaningless; only functions are continuous or discontinuous.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft = at 02:13 PM, raf@tiki-lounge.com (Ross A. Finlayson) said:>This is where the antidiagonal argument fails due to dual >representation of some rationals, No. You just de'ne the mapping to only use on of the representations.>IOve been thinking some how this affects my logical system. >Basically IOm still working on 'nding a way to actually show that>all in'nite sets are equivalent,ItOs easy; just pick an inconsistent axiom set.>propensityPlease de'ne.You seem to be freely inventing new terms without ever de'ning them,and using existing terms in ways not consistent with their acceptedde'nitions. ThatOs not Mathematics, just rhetoric.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to = Shmuel (Seymour J.) Metz raf@tiki-lounge.com (Ross A. Finlayson) said:> >This is where the antidiagonal argument fails due to dual >representation of some rationals, > > No. You just de'ne the mapping to only use one of the representations.Actually, most variations on the Cantor diagonal proof de'ne an antidiagonal in a way that prevents it from having more than one representation.In base 10, for example, a real whose (non-terminating) decimal representation does not contain digit 9 or digit 0 has a unique decimal representation.Ross gets up tight about the fact that you canOt do this in base two without, in effect, switching to base four by using pairs of binary digits instead of single digits. =Half of the irrationals means that, for example, of the irrationals onthe interval [0,2] that half of them are in [0,1] and the other hal'n [0,2], and that of the irrationals on [0,3] that a third of themare in [0,1], etcetera.N+1 s the set of each element of N union {N}, id est, N U {N}. P(N)is a superset of N U {N}.About the in'nite summation, in some respects the limit concept iseasily discarded, quali'edly, its main, or rather, one of its mainjusti'cations is the use for where the actual result would beunde'ned, for example the limit of x/x as x goes to zero equals one. When you sum 2^(-n) for for each and every non-negative 'nite integern, the sum is actually two.Relative density means density of one thing in terms of the density ofanother. for example, the density of the integers in the integers istwice relative that of the even integers in the integers.About semi-algebraic, extra-, trans-, or x- transcendentals,thatOs actually about those things. For example 2^ (sqrt 2) istranscendental, from what I hear, where 2 and sqrt 2 are eachalgebraic. Where does that leave pi^e?WhatOs continuity? I think itOs a construction on a point settopology.About the alternation of rationals and irrationals in the reals, itOssomething along the lines of saying hereOs a rational, itOs neighborsare irrationals, their neighbors are rationals, for in'nitely manyneighbors in either direction the value is inde'nite, and any otherrational or irrational selected (named except via inde'nite neighborrelation to the de'nite point) is in'nitely far away on the sequenceof neighbors.Integration was seen by its founders as an in'nite summation o'n'nitely small areas and was called in'nitesimal analysis. Foralmost all intents and purposes it is that.Have a good one, Ross => > Half of the irrationals means that, for example, of the irrationals on> the interval [0,2] that half of them are in [0,1] and the other half> in [0,2], And here I thought that both halves of [0,2] were in [0,2]. Just shows how wrong I can be, I guess!> N+1 s the set of each element of N union {N}, id est, N U {N}. If N+1 is N union {N}, then what are N+2, and N+3, etc.?> P(N) is a superset of N U {N}.Wrong!1, which is a member of N U {N}, is also a member of many members of P(N) but is not itself a member of P(N), so P(N) is NOT a superset of N U {N}!> About the in'nite summation, in some respects the limit concept is> easily discarded, Not by those who know how to use it. It may be elided, but not discarded.> WhatOs continuity? I think itOs a construction on a point set> topology.In what context are you wanting to use continuity? The most common one is in the context of real functions, which do not seem relevant here. About the alternation of rationals and irrationals in the reals, itOs> something along the lines of saying hereOs a rational, itOs neighbors> are irrationals, their neighbors are rationals, for in'nitely many> neighbors in either direction the value is inde'nite, and any other> rational or irrational selected (named except via inde'nite neighbor> relation to the de'nite point) is in'nitely far away on the sequence> of neighbors.Within the ordered 'eld of reals, no number has a next door neighbor on either side, so what do you mean by neighbors of a real number? And what do you mean by in'nitely far away on the sequence of neighbors? No two real numbers can be in'nitely far away from each other.Irrelevancies snipped. => > > > Half of the irrationals means that, for example, of the irrationals on> the interval [0,2] that half of them are in [0,1] and the other half> in [0,2], > > And here I thought that both halves of [0,2] were in [0,2]. Just > shows how wrong I can be, I guess!> > > N+1 s the set of each element of N union {N}, id est, N U {N}. > > If N+1 is N union {N}, then what are N+2, and N+3, etc.?> > P(N) is a superset of N U {N}.> > Wrong!> > 1, which is a member of N U {N}, is also a member of many members of > P(N) but is not itself a member of P(N), so P(N) is NOT a superset > of N U {N}!> > About the in'nite summation, in some respects the limit concept is> easily discarded, > > Not by those who know how to use it. It may be elided, but not > discarded.> > > WhatOs continuity? I think itOs a construction on a point set> topology.> > In what context are you wanting to use continuity? The most common > one is in the context of real functions, which do not seem relevant > here. > > About the alternation of rationals and irrationals in the reals, itOs> something along the lines of saying hereOs a rational, itOs neighbors> are irrationals, their neighbors are rationals, for in'nitely many> neighbors in either direction the value is inde'nite, and any other> rational or irrational selected (named except via inde'nite neighbor> relation to the de'nite point) is in'nitely far away on the sequence> of neighbors.> > Within the ordered 'eld of reals, no number has a next door > neighbor on either side, so what do you mean by neighbors of a > real number? > > And what do you mean by in'nitely far away on the sequence of > neighbors? No two real numbers can be in'nitely far away from each > other.> Virgil, N U {N} is a subset of P(N). Represent each natural as an ordinal. Then, for example, the subset {0} is an element of P(N), {0} = 1. Theempty set, zero, is an element of P(N), as is each singleton {n} for nin N, as is N. Thus P(N) contains each element 0, 1, 2, ..., as wellas N. Thus N U {N} is a subset of P(N).N+1 U {N+1} is N+2, etcetera, the order type of N is N.The limit concept can be discarded when its result is the same as i't were considered discarded. Calculate the area of a square: itOseasy, square the length of a side.About [0,2], that was actually a typing error. Half of theirrationals on [0,2] are between zero and one, the other half arebetween one and two. The function to map between them is f(x)=x+1.Consider the function f(x)=1 if x is irrational and f(x)=0 if x isrational. Is it everywhere discontinuous?About immediate neighbor points, IOm talking about just that: theimmediate neighbors of a point on the continuous reals. Their realdifference is in'nitesimal.Ross =:> :> P(N) is a superset of N U {N}.:> :> Wrong!:> :> 1, which is a member of N U {N}, is also a member of many members of :> P(N) but is not itself a member of P(N), so P(N) is NOT a superset :> of N U {N}!:> : Virgil, : N U {N} is a subset of P(N). Represent each natural as an ordinal. : Then, for example, the subset {0} is an element of P(N), {0} = 1. The: empty set, zero, is an element of P(N), as is each singleton {n} for n: in N, as is N. Thus P(N) contains each element 0, 1, 2, ..., as well: as N. Thus N U {N} is a subset of P(N).Virgil is correct. P(N) does not contain the element 0. Every element of P(N) is a set of integers. 0 is not a set. { 0 } is an element of P(N),but 0 != { 0 }.Stephen =Suppose M k-subsets are independently drawnfrom a set P={1,2, ....., n} (that is, drawing k elementsfrom P to form a subset and there are M of this.).WhatOs the probability that these M subsets cover theoriginal set P, that is, the union of these M subsets = P itself ? =>Suppose M k-subsets are independently drawn>from a set P={1,2, ....., n} (that is, drawing k elements>from P to form a subset and there are M of this.).>WhatOs the probability that these M subsets cover the>original set P, that is, the union of these M subsets = P itself ?You already asked a question equivalent to this. Note that theunion is P if and only if the intersection of the complements is empty.See my answer from Sept. 25 in the thread subset counting problem.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right? .`. .-./ _=_ .-. { (,(oYo),) }} {{ | |} } { { ()/ }} {{ }O-=-O{ } } { { }._:_.{ }} {{ } -:- { } } {_{ }` `{ _} (((() (/))))> > > James Harris =[...]|> Basically he was blowing me off, but not because I was wrong. |> Remember, IOd gone over each and every point step-by-step. It seems|> to me that clearly he *knew* he could simply reject my work because|> heOs a mathematician, and thatOs what a mathematician can do.[...]Learn the concept, not even wrong ...-- cu,Brucedrift wave turbulence: http://www.rzg.mpg.de/~bds/ =>> >>Obviously IOm rather gleeful at having managed to talk over my work>>which reveals that wacky problem with algebraic integers with an>>actual, real-live mathematician, and not a possibly fake one that just>>posts a lot on Usenet!!!>>Yes I used chalk and went over it all on the chalkboard and even>>talked over some of the math history like talking about Dedekind.>>What I veri'ed is that oddity that you CAN talk to a mathematician,>>give a correct math argument, go over each and every point>>step-by-step, yet have that mathematician simply dismiss you. >> >> Fascinating. He did in fact dismiss your work, but nonetheless>> this 'lls you with glee.>>However, the imporant point is that he couldnOt 'nd an error.So you claim. If so all that proves is that heOs not very good at'nding errors. (Or more likely, that heOs not very good attrying to translate things that make no sense as written intothings that make sense so that one _can_ say what theproblem is.)>There are people who *dismiss* the idea that man landed on the moon.>>The key issue here is mathematical correctness. My point is that I>went through the math point-by-point and the professor could not 'nd>an error.> >>In this>>case the professor claimed it was out of his area, and gee, guess>>what? According to him no one else at Vanderbilt University had the>>necessary expertise.>>Basically he was blowing me off, but not because I was wrong. >> >> Of course not. No matter how many people, usenet posters,>> mathematicians you visit in person, no matter _how_ many of>> them say youOre wrong, itOs simply not possible that the reason>> theyOre all saying that is that youOre _wrong_.>>But youOre now lying David Ullrich as my point is that Professor>McKenzie did NOT say that I was wrong.>>It is such an obvious falsehood that I feel con'dent in calling you>out here as a liar.>>The issue is mathematical correctness.>>And your implication that itOs so rejected is false, as in fact a key>paper of mine is to be published.>Giggle. >>Remember, IOd gone over each and every point step-by-step. It seems>>to me that clearly he *knew* he could simply reject my work because>>heOs a mathematician, and thatOs what a mathematician can do.>>However, now at least I know that I can explain my work, refute all>>objections, and still face mathematicians willing to just ignore it>>like that professor, or lie about it on Usenet.>>And hey, you all knew that, right? You knew that people on Usenet>>like lie all the time and that even mathematicians who post will lie.>>Well IOll give that professor one bit of praise, at least he didnOt>>claim my work was incorrect. And he dismissed the objection that>>certain lying mathematician posters have used repeatedly, I guess>>because they know that most of you are too stupid to catch them. The>>professor trashed it in an instance.>> >> Answer the question. YouOve been asked at least four times by now:>> Did he refute a _verbatim_ _quote_ of MagidinOs objections, or your>> paraphrase of them?>>He refuted the assertion that f or factors in common with f can divide>off dependent on m or as functions of m.>>He quickly dismissed it, which con'rmed for me that an experienced>mathematician wouldnOt take it seriously, even for a moment,>con'rming that Magidin was, as I 'gured, lying. Unless you wish to>claim that Magidin is incompetent.>Too bad other mathematicians who have posted a lot on Usenet werenOt>>smart enough at least not to lie about my work, but hey, they know>>they can lie to most of you, now canOt they?>>And I donOt really think itOs that youOre all too stupid to catch the>>lies, as I think you just *want* to believe. After all, itOs a>>simpler world where IOm just a nut, instead of a major discoverer who>>found an over hundred year old error in core mathematics, right?>>That quick dismissal of a key objection con'rmed my suspicion of a>high *tolerance* of readers to lies from certain sources.>>Apparently, many of you decided to accept false mathematics from>posters like Magidin; however, in a different context--off Usenet--an>experienced mathematician quickly rejected the same objection which>apparently satis'ed many of you who are on sci.math for MONTHS.>>Magidin gave you what you wanted, where you clearly wanted to hear>that I was wrong, and the mathematical truth didnOt matter to you, as>you so readily accepted the lies.>>The point is that the math didnOt matter to readers on the sci.math>newsgroup, or MagidinOs false claims would have been dismissed as>quickly there, as in that professorOs of'ce.>>The math didnOt matter to you.>James Harris************************David C. Ullrich => >>Obviously IOm rather gleeful at having managed to talk over my work>>which reveals that wacky problem with algebraic integers with an>>actual, real-live mathematician, and not a possibly fake one that just>>posts a lot on Usenet!!!>>Yes I used chalk and went over it all on the chalkboard and even>>talked over some of the math history like talking about Dedekind.>>What I veri'ed is that oddity that you CAN talk to a mathematician,>>give a correct math argument, go over each and every point>>step-by-step, yet have that mathematician simply dismiss you. >> >> Fascinating. He did in fact dismiss your work, but nonetheless>> this 'lls you with glee.>>However, the imporant point is that he couldnOt 'nd an error.> > So you claim. If so all that proves is that heOs not very good at> 'nding errors. (Or more likely, that heOs not very good at> trying to translate things that make no sense as written into> things that make sense so that one _can_ say what the> problem is.)Unlike David Ullrich, a math professor at Oklahoma State University,Professor McKenzie is a *distinguished* math professor, who doesnOtspend a lot of time posting on the sci.math newsgroup, as probably hehas better things to do with his time.HereOs a link to VanderbiltOs page on him: http://sitemason.vanderbilt.edu/site/czXIYMDavid Ullrich questioning his judgement is a rather pathetic displayas professors arenOt supposed to engage in such attacks.Then again, how many professors spend as much time as Ullrich postingon newsgroups like sci.math?For those who wonder how complicated the math actually is, or how easyit is to get confused about it, hereOs a link to a website in HongKong where they allow use of LaTeX, so itOs prettier:http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg= 759James Harris =>> > >Obviously IOm rather gleeful at having managed to talk over my work>which reveals that wacky problem with algebraic integers with an>actual, real-live mathematician, and not a possibly fake one that just>posts a lot on Usenet!!!>>Yes I used chalk and went over it all on the chalkboard and even>talked over some of the math history like talking about Dedekind.>>What I veri'ed is that oddity that you CAN talk to a mathematician,>give a correct math argument, go over each and every point>step-by-step, yet have that mathematician simply dismiss you. > > Fascinating. He did in fact dismiss your work, but nonetheless> this 'lls you with glee.>>However, the imporant point is that he couldnOt 'nd an error.>> >> So you claim. If so all that proves is that heOs not very good at>> 'nding errors. (Or more likely, that heOs not very good at>> trying to translate things that make no sense as written into>> things that make sense so that one _can_ say what the>> problem is.)>>Unlike David Ullrich, a math professor at Oklahoma State University,>Professor McKenzie is a *distinguished* math professor, who doesnOt>spend a lot of time posting on the sci.math newsgroup, as probably he>has better things to do with his time.>>HereOs a link to VanderbiltOs page on him:http://sitemason.vanderbilt.edu/site/czXIYM>>David Ullrich questioning his judgement is a rather pathetic display>as professors arenOt supposed to engage in such attacks.WasnOt questioning his judgement. You seem to have missed thewords So you claim. If so....>Then again, how many professors spend as much time as Ullrich posting>on newsgroups like sci.math?>>For those who wonder how complicated the math actually is, or how easy>it is to get confused about it, hereOs a link to a website in Hong>Kong where they allow use of LaTeX, so itOs prettier:>>http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg =759>James Harris************************David C. Ullrich =>> For those who wonder how complicated the math actually is, or how easy> it is to get confused about it, hereOs a link to a website in Hong> Kong where they allow use of LaTeX, so itOs prettier:>> http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759>IOm a little too lazy to work through all the algebra on that page, butplease help on this question: Do all those assumes and arbitrays causeany problems? =James Harris scribbled the followingon sci.math:> Unlike David Ullrich, a math professor at Oklahoma State University,> Professor McKenzie is a *distinguished* math professor, who doesnOt> spend a lot of time posting on the sci.math newsgroup, as probably he> has better things to do with his time.So spending time on sci.math answering peopleOs questions makesUllrich non-distinguished? Considering the rather tremendous amountof work Ullrich went through to help me learn general (non-metric)topology, I would have to disagree with that.-- /-- Joona Palaste (palaste@cc.helsinki.') | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |-- Finland rules! /ItOs not survival of the fattest, itOs survival of the 'ttest. - Ludvig von Drake =>James Harris scribbled the following>on sci.math:>> Unlike David Ullrich, a math professor at Oklahoma State University,>> Professor McKenzie is a *distinguished* math professor, who doesnOt>> spend a lot of time posting on the sci.math newsgroup, as probably he>> has better things to do with his time.>>So spending time on sci.math answering peopleOs questions makes>Ullrich non-distinguished? No, itOs just that James made a typo. It should have beenDistinguished Professor in mathematics. Prof. McKenzieOs title inVanderbilt is Distinguished Professor (as opposed to OLecturerO,OAdunct ProfessorO, OAssistant ProfessorO, OAssociate ProfessorO,OEmeritus ProfessorO, OFull ProfessorO, OUniversity ProfessorO, etc.)Glancing at the page in Oklahoma State University, I cannot seeProf. UllrichOs title, but IOm guessing from previous posts that itwould be Professor or Full Professor.Tsk, tsk, tsk, James. And you call yourself a writer. According to theChicago Manual of Style, TITLES AND OFFICES 7.17 [...] A title used alone, in place of a personal name, is capitalized in such contexts as toasts or formal introductions[.] Titles used in place of names in directa ddress are capitalized[.] 7.22 Among professional titles, names academic professorships and fellowships are usually capitalized wherever they appear[.]So Prof. McKenzieOs title, Distinguished Professor, should have beencapitalized, not written as *distinguished* math professor.I assume, of course, that your use of the word distinguished derivesfrom his title; I doubt you had any other basis for making thatassertion. = expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = Magidinmagidin@math.berkeley.edu =>>James Harris scribbled the following>>on sci.math:> Unlike David Ullrich, a math professor at Oklahoma State University,> Professor McKenzie is a *distinguished* math professor, who doesnOt> spend a lot of time posting on the sci.math newsgroup, as probably he> has better things to do with his time.>>So spending time on sci.math answering peopleOs questions makes>>Ullrich non-distinguished? >>No, itOs just that James made a typo. It should have been>Distinguished Professor in mathematics. Prof. McKenzieOs title in>Vanderbilt is Distinguished Professor (as opposed to OLecturerO,>OAdunct ProfessorO, OAssistant ProfessorO, OAssociate ProfessorO,>OEmeritus ProfessorO, OFull ProfessorO, OUniversity ProfessorO, etc.)>>Glancing at the page in Oklahoma State University, I cannot see>Prof. UllrichOs title, but IOm guessing from previous posts that it>would be Professor or Full Professor.My actual title would take too much space to type. You can justcall me Sir.************************David C. Ullrich => => Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A manOs capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of 'gures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the inde'nite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> =I doubt de Morgan was as uselessly obsessed with Smith as you guys are with Harris. =>>I doubt de Morgan was as uselessly obsessed with Smith as you guys are with >Harris.You might want to check out the Budget of Paradoxes before making sucha statement.Considering the amount of space devoted to that individual, and thecost and work involved in de MorganOs replies to him (publishing arefutation to a pamphlet was certainly something that required morework than the amount I need to spend in replying to one of JamesOspost), I would not sharing. = [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu>>I doubt de Morgan was as uselessly obsessed with Smith as you guys are with >Harris.> > You might want to check out the Budget of Paradoxes before making such> a statement.> > Considering the amount of space devoted to that individual, and the> cost and work involved in de MorganOs replies to him (publishing a> refutation to a pamphlet was certainly something that required more> work than the amount I need to spend in replying to one of JamesOs> post), I would not be so sure that this is the case.Fine, have it your way: de Morgan was as as uselessly obsessed as you guys are.> ... In such work as > this... it is idle to look for the calm reasoning of a scholar;> we do not 'nd it, and there is little or no advantage in> attacking the obvious inconsistencies and absurdities that abound.> -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in> _The Shakespearan Ciphers Examined_, by William F.> Friedman and Elizebeth S. FriedmanNow that I can agree with. =Arturo Magidin scribbled the followingon sci.math:>>James Harris scribbled the following>>on sci.math:> Unlike David Ullrich, a math professor at Oklahoma State University,> Professor McKenzie is a *distinguished* math professor, who doesnOt> spend a lot of time posting on the sci.math newsgroup, as probably he> has better things to do with his time.>>So spending time on sci.math answering peopleOs questions makes>>Ullrich non-distinguished? > No, itOs just that James made a typo. It should have been> Distinguished Professor in mathematics. Prof. McKenzieOs title in> Vanderbilt is Distinguished Professor (as opposed to OLecturerO,> OAdunct ProfessorO, OAssistant ProfessorO, OAssociate ProfessorO,> OEmeritus ProfessorO, OFull ProfessorO, OUniversity ProfessorO, etc.)> Glancing at the page in Oklahoma State University, I cannot see> Prof. UllrichOs title, but IOm guessing from previous posts that it> would be Professor or Full Professor.> Tsk, tsk, tsk, James. And you call yourself a writer. According to the> Chicago Manual of Style, (snip)I think itOs more likely that James Harris made that typo *onpurpose*, to make it *look like* McKenzie was indeed more distinguishedthan Ullrich. Because Ullrich has said §at-out that James Harris iscompletely *wrong*, and McKenzie hasnOt. James Harris must have takenpersonal issue with this and developed a hatred towards Ullrich, whichhasnOt (yet) been the case with McKenzie.-- /-- Joona Palaste (palaste@cc.helsinki.') | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |-- Finland rules! /You have moved your mouse, for these changes to take effect you must shut downand restart your computer. Do you want to restart your computer now? - Karri Kalpio =>Arturo Magidin scribbled the following>on sci.math:>James Harris scribbled the following>on sci.math:>> Unlike David Ullrich, a math professor at Oklahoma State University,>> Professor McKenzie is a *distinguished* math professor, who doesnOt>> spend a lot of time posting on the sci.math newsgroup, as probably he>> has better things to do with his time.>>So spending time on sci.math answering peopleOs questions makes>Ullrich non-distinguished? > No, itOs just that James made a typo. It should have been>> Distinguished Professor in mathematics. Prof. McKenzieOs title in>> Vanderbilt is Distinguished Professor (as opposed to OLecturerO,>> OAdunct ProfessorO, OAssistant ProfessorO, OAssociate ProfessorO,>> OEmeritus ProfessorO, OFull ProfessorO, OUniversity ProfessorO, etc.)> Glancing at the page in Oklahoma State University, I cannot see>> Prof. UllrichOs title, but IOm guessing from previous posts that it>> would be Professor or Full Professor.> Tsk, tsk, tsk, James. And you call yourself a writer. According to the>> Chicago Manual of Style, >>(snip)>>I think itOs more likely that James Harris made that typo *on>purpose*, to make it *look like* McKenzie was indeed more distinguished>than Ullrich. Well, yes. The only reason James had for thinking that Prof. McKenziewas distinguished was his title at Vanderbilt. In fact, IOll wagerthat he cannot name a single accomplishment of Prof. McKenzieOs thathe does not copy from my post in which I mention a few of them (andthere are quite a few others); I doubt he can even being to explainthe ones that he could copy from that post. He also has no idea ofwhat Prof. Ullrich may or may not have accomplished as a professional,so his comparison must be based on job titles, not on mathematicalstatus. In fact, Prof. McKenzie ->is<- a very distinguished mathematician.But JamesOs description of him as a *distinguished* math professormust have been, in fact, a reference to his title, not to hisprominence; James was ignorant as to Prof. McKenzieOs prominence. Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = Giggle. >Nothing makes a pig happier than a roll in the wallow, as this giggling porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg& width=314&height=400# => >> .... After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right?>> Yes, right. Or at least, nearly so. IOd replace nut with silly> person who could learn some [not all that dif'cult] mathematics, but> chooses not to.>> Which de'es the fact that I *explained* my mathematical argument> point-by-point in person to an actual math professor. That professor> did NOT 'nd any error, but instead claimed my work was out of his> area.>> However, given that information, you choose to claim that I am a> silly person who needs to learn mathematics, which is an odd> response.>> Oh, in case any of you are wondering, he did challenge me throughout> the discussion on many points. ItOs just that the mathematics is> rather basic and easy, so even a well-trained mathematician canOt> successfully challenge a *single* point of it with mathematics.> > Nonsense. It has not only been successfully challenged, repeatedly, but successfully refuted, repeatedly. You are in denial to the> point of complete fabrication.Bull. You wish.The math is in my favor, while youOre deluded.IOve explained it to a distinguished math professor so losers onUsenet have no power over me.Others can see http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759James Harris => .... After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right?>> Yes, right. Or at least, nearly so. IOd replace nut with silly> person who could learn some [not all that dif'cult] mathematics, but> chooses not to.>> Which de'es the fact that I *explained* my mathematical argument> point-by-point in person to an actual math professor. That professor> did NOT 'nd any error, but instead claimed my work was out of his> area.>> However, given that information, you choose to claim that I am a> silly person who needs to learn mathematics, which is an odd> response.>> Oh, in case any of you are wondering, he did challenge me throughout> the discussion on many points. ItOs just that the mathematics is> rather basic and easy, so even a well-trained mathematician canOt> successfully challenge a *single* point of it with mathematics.OK, James Harris. Here is a challenge I posted repeatedly that has been 100% successful. It is simply this: Give us *one* number thatOshouldO be in the ring of algebraic integers, but which is not in the ring of algebraic integers. You claim to have proven that thereare such numbers, but you have never produced one -- not a single one. If you want to continue with the delusion that no one has eversuccessfully challenged a *single* point of your argument, here is your chance. If you do NOT respond to this post with a speci'cnumber (numeric value) which should be in the ring of algebraic integers, but which is not, I claim you are a liar and that you cannotproduce the results you claim.YouOre on....--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com => > .... After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right?>> Yes, right. Or at least, nearly so. IOd replace nut with silly> person who could learn some [not all that dif'cult] mathematics, but> chooses not to.>> Which de'es the fact that I *explained* my mathematical argument> point-by-point in person to an actual math professor. That professor> did NOT 'nd any error, but instead claimed my work was out of his> area.>> However, given that information, you choose to claim that I am a> silly person who needs to learn mathematics, which is an odd> response.>> Oh, in case any of you are wondering, he did challenge me throughout> the discussion on many points. ItOs just that the mathematics is> rather basic and easy, so even a well-trained mathematician canOt> successfully challenge a *single* point of it with mathematics.> > OK, James Harris. Here is a challenge I posted repeatedly that has been 100% successful. It is simply this: Give us *one* number that> OshouldO be in the ring of algebraic integers, but which is not in the ring of algebraic integers. You claim to have proven that there> are such numbers, but you have never produced one -- not a single one. If you want to continue with the delusion that no one has ever> successfully challenged a *single* point of your argument, here is your chance. If you do NOT respond to this post with a speci'c> number (numeric value) which should be in the ring of algebraic integers, but which is not, I claim you are a liar and that you cannot> produce the results you claim.> > YouOre on....Well in my paper Advanced Polynomial Factorization I prove that withthe factorization 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),where the aOs are algebraic integer, one of the aOs is coprime to 5.One approach to the argument can be found (it looks pretty) at thefollowing link:http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg= 759Well, it turns out that *two* of the aOs should have a factor that issqrt(5) in the ring of algebraic integers, but donOt because theirother factor is not an algebraic integer, which leads to a simple wayto show one of these numbers presented by a poster on sci.math monthsago.The aOs are roots of a^3 + 12a^2 - 65, so knowing that two have afactor that is sqrt(5) use a = sqrt(5)b, which gives you 5sqrt(5)b^3 + 12(5)b^2 - 65 = 0 and dividing off 5 gives you sqrt(5)b^3 + 12b^2 - 13 = 0 so *two* of the roots for b are the requested numbers while the thirdis not.For those confused by the impact of using a substitution likea=sqrt(5)b knowing that only two of the aOs have sqrt(5) as a factor,hereOs an example where you can look at the impact.Consider (x+2)(x+2)(x+1) = x^3 + 5x^2 + 8x + 4=0, with x=2y, so youhave 8y^3 + 4(5)y^2 + 8(2)y + 4 = 0, and dividing off 4 gives you 2y^3 + 5y^2 + 2y + 1 = 0.However, while that is reducible over rationals, sqrt(5)b^3 + 12b^2 - 13is not, meaning that while you can solve for the roots, you canOt tell*which* two of them are the desired numbers.James Harris posted repeatedly that has been 100% successful. It is simply this: Give us *one* number that> OshouldO be in the ring of algebraic integers, but which is not in the ring of algebraic integers. You claim to have proven that there> are such numbers, but you have never produced one -- not a single one. If you want to continue with the delusion that no one has ever> successfully challenged a *single* point of your argument, here is your chance. If you do NOT respond to this post with a speci'c> number (numeric value) which should be in the ring of algebraic integers, but which is not, I claim you are a liar and that you cannot> produce the results you claim.>> YouOre on....>> Well in my paper Advanced Polynomial Factorization I prove that with> the factorization 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),> where the aOs are algebraic integer, one of the aOs is coprime to 5.>> One approach to the argument can be found (it looks pretty) at the> following link:>> http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759>> Well, it turns out that *two* of the aOs should have a factor that is> sqrt(5) in the ring of algebraic integers, but donOt because their> other factor is not an algebraic integer, which leads to a simple way> to show one of these numbers presented by a poster on sci.math months> ago.>> The aOs are roots of a^3 + 12a^2 - 65, so knowing that two have a> factor that is sqrt(5) use a = sqrt(5)b, which gives you>> 5sqrt(5)b^3 + 12(5)b^2 - 65 = 0>> and dividing off 5 gives you>> sqrt(5)b^3 + 12b^2 - 13 = 0>> so *two* of the roots for b are the requested numbers while the third> is not.>> For those confused by the impact of using a substitution like> a=sqrt(5)b knowing that only two of the aOs have sqrt(5) as a factor,> hereOs an example where you can look at the impact.>> Consider (x+2)(x+2)(x+1) = x^3 + 5x^2 + 8x + 4=0, with x=2y, so you> have>> 8y^3 + 4(5)y^2 + 8(2)y + 4 = 0,>> and dividing off 4 gives you>> 2y^3 + 5y^2 + 2y + 1 = 0.>> However, while that is reducible over rationals,>> sqrt(5)b^3 + 12b^2 - 13>> is not, meaning that while you can solve for the roots, you canOt tell> *which* two of them are the desired numbers.>> James HarrisYou have, once again, failed to meet the challenge. The challenge is still 100% successful. You failed to produce ONE SINGLE NUMBER whichshould be in the ring of algebraic integers but which is not.Score another point for me. Deduct one for you.Next time you decide to answer a challenge, why not address the challenge posted, which was (to refresh your failing memory) to produce onenumber, just one numerical value, which should be in the ring of algebraic integers but which is not. Your wandering monologue above does notaddress the challenge.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com => .... After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right?>> Yes, right. Or at least, nearly so. IOd replace nut with silly> person who could learn some [not all that dif'cult] mathematics, but> chooses not to.>> Which de'es the fact that I *explained* my mathematical argument> point-by-point in person to an actual math professor. That professor> did NOT 'nd any error, but instead claimed my work was out of his> area.>> However, given that information, you choose to claim that I am a> silly person who needs to learn mathematics, which is an odd> response.>> Oh, in case any of you are wondering, he did challenge me throughout> the discussion on many points. ItOs just that the mathematics is> rather basic and easy, so even a well-trained mathematician canOt> successfully challenge a *single* point of it with mathematics.>> Nonsense. It has not only been successfully challenged, repeatedly, but successfully refuted, repeatedly. You are in denial to the> point of complete fabrication.>> Bull. You wish.>> The math is in my favor, while youOre deluded.I am deluded that you have been successfully challenged repeatedly? I rest my case on who is deluded here.> IOve explained it to a distinguished math professor so losers on> Usenet have no power over me.>> Others can see http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759>> James Harrisa trained mathematician can successfully challenge a *single* point of your argument. This is patently false and absurd. Thisnewsgroup abounds with speci'c challenges to your argument which you have failed to refute. In most cases, as in the recent post byMagidin, you simply refuse to acknowledge the challenges. DonOt you think other readers can see through you?--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com => > If any wish to dispute that assessment, I welcome them or any of those> who have disputed my argument in the past to come forward here with a> *mathematical* objection and IOll explain--yet again--why itOs wrong> or doesnOt apply.> > Easy. You posted this about 6 hours after Magidin asked which of 17> points were wrong. I see you havenOt answered. Please do so. The message> > Several people have already commented about you not answering.> > V.ThatOs funny. I post from Google which means that several hours cango by before I can reply through Google.It actually helps to keep down the posting volume.It looks like an update occurred so I should come across the message.Still I note for readers that you, like a coward, backed away from mychallenge.James Harris => Obviously IOm rather gleeful at having managed to talk over my work> which reveals that wacky problem with algebraic integers with an> actual, real-live mathematician, and not a possibly fake one that just> posts a lot on Usenet!!!> > Yes I used chalk and went over it all on the chalkboard and even> talked over some of the math history like talking about Dedekind.> > What I veri'ed is that oddity that you CAN talk to a mathematician,> give a correct math argument, go over each and every point> step-by-step, yet have that mathematician simply dismiss you. In this> case the professor claimed it was out of his area, and gee, guess> what? According to him no one else at Vanderbilt University had the> necessary expertise.If the person says this isnOt his area of specialty, it *might* mean that heOs either too rusty or never learned enough to consider himself a quali'ed judge of the validity of your work. The amount of math that is out there is so vast that no one can pick up an arbitrary piece of work and be assured of understanding it. IOve noticed that a number of the regulars on sci.math only respond to certain topics. There are others that they simply never respond to. Your paper may be one where your real-live mathematician would never have responded if you werenOt face to face because he didnOt trust himself to give completely correct information.> > Basically he was blowing me off, but not because I was wrong. > Remember, IOd gone over each and every point step-by-step. It seems> to me that clearly he *knew* he could simply reject my work because> heOs a mathematician, and thatOs what a mathematician can do.Or maybe he wasnOt interested, or maybe itOs something he doesnOt know enough about without some study, or...> > However, now at least I know that I can explain my work, refute all> objections, and still face mathematicians willing to just ignore it> like that professor, or lie about it on Usenet.> > And hey, you all knew that, right? You knew that people on Usenet> like lie all the time and that even mathematicians who post will lie.I wasnOt aware of that, though I was aware you repeatedly claim that.> > Well IOll give that professor one bit of praise, at least he didnOt> claim my work was incorrect. And he dismissed the objection that> certain lying mathematician posters have used repeatedly, I guess> because they know that most of you are too stupid to catch them. The> professor trashed it in an instance.Perhaps youOd provide the details of his rebuttal?> > Too bad other mathematicians who have posted a lot on Usenet werenOt> smart enough at least not to lie about my work, but hey, they know> they can lie to most of you, now canOt they?> > And I donOt really think itOs that youOre all too stupid to catch the> lies, as I think you just *want* to believe. After all, itOs a> simpler world where IOm just a nut, instead of a major discoverer who> found an over hundred year old error in core mathematics, right?James, as much as I hate to admit it, I think you are better at math than at character analysis. Please stick to that, at least in sci.math.-- Will Twentyman => > Obviously IOm rather gleeful at having managed to talk over my work> which reveals that wacky problem with algebraic integers with an> actual, real-live mathematician, and not a possibly fake one that just> posts a lot on Usenet!!!> > Yes I used chalk and went over it all on the chalkboard and even> talked over some of the math history like talking about Dedekind.> > What I veri'ed is that oddity that you CAN talk to a mathematician,> give a correct math argument, go over each and every point> step-by-step, yet have that mathematician simply dismiss you. In this> case the professor claimed it was out of his area, and gee, guess> what? According to him no one else at Vanderbilt University had the> necessary expertise.> > If the person says this isnOt his area of specialty, it *might* mean > that heOs either too rusty or never learned enough to consider himself a > quali'ed judge of the validity of your work. The amount of math that > is out there is so vast that no one can pick up an arbitrary piece of > work and be assured of understanding it. IOve noticed that a number of > the regulars on sci.math only respond to certain topics. There are > others that they simply never respond to. Your paper may be one where > your real-live mathematician would never have responded if you werenOt > face to face because he didnOt trust himself to give completely correct > information.ItOs weird watching posters go through gyrations, so hereOs more ofthe story.IOd been sending my paper Advanced Polynomial Factorization around tomath journals and mostly got blank negatives in reply. Like theAmerican Journal of Mathematics the editor for one journal was at my almamater Vanderbilt University.So I sent it to him and made *certain* to mention that IOm an alumnus.Well based on his 'rst helpful response I realized that the paperneeded some serious work, so I made a lot of changes, but didnOt hearfrom him.Time went by, and IOd sent him what I thought of as a stable version,Mazur at Harvard, and then he replied saying heOd been busy.Then he said that he needed a face-to-face meeting, and thenemphasized that when I suggested a phone call, and a few weeks later ISo in fact the professor has had my work in mind for several months,and itOs really nonsensical to believe that by the time IOd explainedwhat is actually a rather basic and simple argument to him that hewouldnOt have seen an error.ItOs also telling that he shot down Arturo MagidinOs objection soquickly.Now Professor McKenzie was actually a rather nice guy, and we had adecent talk though I was frustrated by his refusal to properlyacknowledge my work. And his saying *then* that it was out of hisarea was such an obvious cop-out that it ticked me off.Remember, IOve found a hundred year old error that has sat in anesoteric branch of mathematics. YouOd think I could get *some*recognition for my 'nd.Obviously my hope was that my alma mater would back me up, but I had aserious disappointment that day. I had four hours to think about thehorror of my situation on the drive back from Nashville. Four hoursto wonder what to do when mathematicians try to blow off importantmathematics.Yup, Professor McKenzie was in many ways a nice guy, but as far as IOmconcerned he failed in the area that really counted.James Harrs McKenzie was in many ways a nice guy, but as far as IOm> concerned he failed in the area that really counted.>> James HarrsOn the contrary. He succeeded admirably in determining that the work you presented wasincomprehensible crap, and that you were uneducable. Too bad you wasted his time.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com => > Obviously IOm rather gleeful at having managed to talk over my work> which reveals that wacky problem with algebraic integers with an> actual, real-live mathematician, and not a possibly fake one that just> posts a lot on Usenet!!!As near as I can tell by the recent spate of posts, Jamesis claiming that the fact that he talked about his workout loud proves that itOs valid.> Basically he was blowing me off, but not because I was wrong. > Remember, IOd gone over each and every point step-by-step. It seems> to me that clearly he *knew* he could simply reject my work because> heOs a mathematician, and thatOs what a mathematician can do.> > Or maybe he wasnOt interested, or maybe itOs something he doesnOt know > enough about without some study, or...... or maybe he would have said exactly the same thingsto somebody who covered his chalkboard with a proof thatthe world was created last Tuesday, that the moon was madeof cheese, and that you could trisect an angle and squarethe circle using nothing but a pencil and a piece ofmoldy bread.> Well IOll give that professor one bit of praise, at least he didnOt> claim my work was incorrect. And he dismissed the objection that> certain lying mathematician posters have used repeatedly, I guess> because they know that most of you are too stupid to catch them. The> professor trashed it in an instance.> > Perhaps youOd provide the details of his rebuttal?I would also like to hear James paraphrase of the objectionthat certain lying mathematician posters have used repeatedly.I cynically suspect it bears no resemblance to anythinganybody but James has actually said. - Randy => >Obviously IOm rather gleeful at having managed to talk over my work>which reveals that wacky problem with algebraic integers with an>actual, real-live mathematician, and not a possibly fake one that just>posts a lot on Usenet!!!>>Yes I used chalk and went over it all on the chalkboard and even>talked over some of the math history like talking about Dedekind.>>What I veri'ed is that oddity that you CAN talk to a mathematician,>give a correct math argument, go over each and every point>step-by-step, yet have that mathematician simply dismiss you. > > Fascinating. He did in fact dismiss your work, but nonetheless> this 'lls you with glee.> > However, the imporant point is that he couldnOt 'nd an error. Since there is no such thing as an error in *Number Theory*, you probably should be thankful he even let you keep your computer.> There are people who *dismiss* the idea that man landed on the moon. They should, since it wasnOt an idea.> The key issue here is mathematical correctness. My point is that I> went through the math point-by-point and the professor could not 'nd> an error. Then you sould post your proof to someplace other than alt.math.undegrad, Since the only thing anybody does in undergrad is give money to moron *physicists*.> >In this>case the professor claimed it was out of his area, and gee, guess>what? According to him no one else at Vanderbilt University had the>necessary expertise.>>Basically he was blowing me off, but not because I was wrong. > > Of course not. No matter how many people, usenet posters,> mathematicians you visit in person, no matter _how_ many of> them say youOre wrong, itOs simply not possible that the reason> theyOre all saying that is that youOre _wrong_.> > But youOre now lying David Ullrich as my point is that Professor> McKenzie did NOT say that I was wrong.> It is such an obvious falsehood that I feel con'dent in calling you> out here as a liar.> > The issue is mathematical correctness. ThatOs impossible. Since mathematical correctness is isomorphic to gravity correctness, neither of which exist. =:> this 'lls you with glee.: However, the imporant point is that he couldnOt 'nd an error.Does this mean that he accepted your logic at each step? Orcould he 'nd nothing lucid enough to call an error? => IOve seen, for JJ, sets have no properties, set mappings are a> pipe dream, in'nity is ineffable, and nonesense is a virtue.Ha Ha. Lovely typo that nonesense, given its ambiguity between nonsenseand nonessence. ;)Since some of JJ remarks seem loosely Wittgensteinian, perhaps he could agreeto both. The early Wittgenstein of the _Tractatus Logico-Philosophicus_regarded his own position as by its own lights nonsense, but as a ladder thatyou can kick away once youOve reached the top; whereas the later Wittgensteinof the _Philosophical Investigations_ regarded the quest for Socraticde'nitions (de'ning characteristeristics or essential properties) asgenerally wrongheaded and plumped instead for a notion of familyresemblances.Karl => It turns out that the de'nition for algebraic integers where they are>> roots of *monic* polynomials leaves off certain numbers that should be>> included, which is how I found a way to show some wacky things.>> >> ItOs easy enough to explain as consider>> >> x^3 - 3x + 2>> >> which as it turns out has a root that is coprime to 2. That is, it>> doesnOt share any non-unit factors with 2 in the ring of algebraic>> integers.>>The roots of this cubic are three algebraic integers, and their product>For this reason, I am surprised by the statement that one of them is>coprime to 2.> > The three roots are 1 (twice) and -2; 1 is coprime to 2, in the> trivial way. The reason this is not a problem and not something wacky> is that the polynomial is reducible over Q:> > x^3-3x+2 = (x-1)(x-1)(x+2).> > James later stated that he should have written x^3+3x-2, which is> irreducible over Q. In that case, none of the roots are units. all of> them are factors of 2 and of themselves, and therefore you would be> well to be surprised at the statement that one of them is coprime to> 2.ThatOs actually a useful statement to show how the wacky error withalgebraic integers can lead to a false proof for those of you whothink it doesnOt matter to have this particular error in core.Here Magidin IS correct in that none of the roots can be units, butthatOs because *in the ring of algebraic integers* itOs provably truethat any unit is a root of a monic polynomial with integercoef'cients and a last coef'cient of 1 or -1.But that does NOT prove that none of the roots are coprime to 2.Here Magidin is using one thing, but switching to another, as if youjust believe that to be coprime to 2, one of the roots has to be aunit, then you accept his claim as true. Notice he says well to besurprised.To catch Magidin here you have to force him to *prove*non-coprimeness, and he cannot. Various posters have tried thisparticular trick, and IOd catch them on it, and theyOd start going incircles.James Harris =>It turns out that the de'nition for algebraic integers where they are>>roots of *monic* polynomials leaves off certain numbers that should be>>included, which is how I found a way to show some wacky things.>>ItOs easy enough to explain as consider>> x^3 - 3x + 2>>which as it turns out has a root that is coprime to 2. That is, it>>doesnOt share any non-unit factors with 2 in the ring of algebraic>>integers.>>The roots of this cubic are three algebraic integers, and their product>For this reason, I am surprised by the statement that one of them is>coprime to 2.>>The three roots are 1 (twice) and -2; 1 is coprime to 2, in the>>trivial way. The reason this is not a problem and not something wacky>>is that the polynomial is reducible over Q:>>x^3-3x+2 = (x-1)(x-1)(x+2).>>James later stated that he should have written x^3+3x-2, which is>>irreducible over Q. In that case, none of the roots are units. all of>>them are factors of 2 and of themselves, and therefore you would be>>well to be surprised at the statement that one of them is coprime to>>2.> > > ThatOs actually a useful statement to show how the wacky error with> algebraic integers can lead to a false proof for those of you who> think it doesnOt matter to have this particular error in core.> > Here Magidin IS correct in that none of the roots can be units, but> thatOs because *in the ring of algebraic integers* itOs provably true> that any unit is a root of a monic polynomial with integer> coef'cients and a last coef'cient of 1 or -1.> > But that does NOT prove that none of the roots are coprime to 2.> > Here Magidin is using one thing, but switching to another, as if you> just believe that to be coprime to 2, one of the roots has to be a> unit, then you accept his claim as true. Notice he says well to be> surprised.> > To catch Magidin here you have to force him to *prove*> non-coprimeness, and he cannot. Various posters have tried this> particular trick, and IOd catch them on it, and theyOd start going in> circles.> Okay, IOll take a stab at it:x^3+3x-2 factors as (x-a)(x-b)(x-c) where a,b,c are roots.Multiplying this out, we get that:-a-b-c=0ab+ac+bc=3abc=2Looking at the last equation, we see three factors of 2: a,b,c since2=a(bc),2=b(ac),2=c(ab).These factors are each factors of themselves: a=1*a, b=1*b, c=1*cSo, a is a factor of a and 2,b is a factor of b and 2,c is a factor of c and 2.Since you admit that a,b,c are not units, and you have de'ned two numbers to be coprime if they have no non-unit factors in common,a is not coprime to 2, because they have a common non-unit factor: ab is not coprime to 2, because they have a common non-unit factor: bc is not coprime to 2, because they have a common non-unit factor: cTherefor, *none of the roots are coprime to 2*. QED.Now, having said that, I expect you will 'nd an objection somewhere.-- Will Twentyman => James later stated that he should have written x^3+3x-2, which is>>irreducible over Q. In that case, none of the roots are units. all of>>them are factors of 2 and of themselves, and therefore you would be>>well to be surprised at the statement that one of them is coprime to>>2.>>> > Here Magidin IS correct in that none of the roots can be units, but> thatOs because *in the ring of algebraic integers* itOs provably true> that any unit is a root of a monic polynomial with integer> coef'cients and a last coef'cient of 1 or -1.> > But that does NOT prove that none of the roots are coprime to 2.Of course not, nor did he claim that as a proof. However itOs trivialto see that none of the roots are coprime to 2. Denote the roots by r1,r2, and r3, so thatx^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)Then r1*r2*r3 = 2, so r1 divides 2 in the ring of algebraic integers(in fact 2 / r1 = r2*r3). Thus r1 and 2 share a common nonunit factor,namely r1. The same holds for r2 and r3, so none of the roots arecoprime to 2.Rick = > James later stated that he should have written x^3+3x-2, which is>>irreducible over Q. In that case, none of the roots are units. all of>>them are factors of 2 and of themselves, and therefore you would be>>well to be surprised at the statement that one of them is coprime to>>2.> Here Magidin IS correct in that none of the roots can be units, but> thatOs because *in the ring of algebraic integers* itOs provably true> that any unit is a root of a monic polynomial with integer> coef'cients and a last coef'cient of 1 or -1.> > But that does NOT prove that none of the roots are coprime to 2.> > > Of course not, nor did he claim that as a proof. However itOs trivial> to see that none of the roots are coprime to 2. Denote the roots by r1,> r2, and r3, so that> > x^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)> > Then r1*r2*r3 = 2, so r1 divides 2 in the ring of algebraic integers> (in fact 2 / r1 = r2*r3). Thus r1 and 2 share a common nonunit factor,> namely r1. The same holds for r2 and r3, so none of the roots are> coprime to 2.That does not prove that there does not exist algebraic integers OaOand ObO such that r_1 a + 2b = 1 in the ring of algebraic integers,and in fact for one of the rOs it is true that they *do* exist as bothyou and Magidin are wrong.Again, the trick is to try and use the result that it is not a unit,when in fact, my point is that the ring of algebraic integers isscrewed up in that it is not a unit.Mistakes in core mathematics can be great fun to play with, so posterslike Rick Decker can run themselves in knots, if they donOt follow themath.Remember an error in core is a big deal. People can prove all kindsof things with such an error, which is why it should be handled.James Harris =>> >> James later stated that he should have written x^3+3x-2, which is>irreducible over Q. In that case, none of the roots are units. all of>them are factors of 2 and of themselves, and therefore you would be>well to be surprised at the statement that one of them is coprime to>2.> >> Here Magidin IS correct in that none of the roots can be units, but>> thatOs because *in the ring of algebraic integers* itOs provably true>> that any unit is a root of a monic polynomial with integer>> coef'cients and a last coef'cient of 1 or -1.>> >> But that does NOT prove that none of the roots are coprime to 2.>> >> >> Of course not, nor did he claim that as a proof. However itOs trivial>> to see that none of the roots are coprime to 2. Denote the roots by r1,>> r2, and r3, so that>> >> x^3 + 3x - 2 = (x - r1)(x - r2)(x - r3)>> >> Then r1*r2*r3 = 2, so r1 divides 2 in the ring of algebraic integers>> (in fact 2 / r1 = r2*r3). Thus r1 and 2 share a common nonunit factor,>> namely r1. The same holds for r2 and r3, so none of the roots are>> coprime to 2.>>That does not prove that there does not exist algebraic integers OaO>and ObO such that r_1 a + 2b = 1 in the ring of algebraic integers,Yes, it does. For rewriting 2 as r1*(r2*r3), we have that if such an aand b existed, then:1 = r_1*a + = r_1*a + r1*r2*r3*b = r_1*(a+r2*r3*b)Since r2, r3, and b are algebraic integers, so is r2*r3*b. Since a isalso an algebraic integer by assumption, so is a+r2*r2*b.So if a and b existed, then there would be an algebraic integer x(namely, x=a+r2*r3*b) such that r_1*x = 1.Therefore, r_1 would be an algebraic integer unit.But the ONLY complex number x with the property that r_1*x = 1 isx=1/r_1.Therefore, you are claiming that a+r2*r3*b = 1/r1.But we know that a+r2*r3*b is an algebraic integer. And we know that1/r1 is NOT an algebraic integer.This is a contradiciton. The contradiction arises from assuming that aand b exist. Therefore, no such a and no such b exist.The same argument, word for word, exchanging r1 and r2, shows the samefor r2; and exchanging r1 and r3, shows it for r3.But this is all a red herring.>and in fact for one of the rOs it is true that they *do* exist as both>you and Magidin are wrong.Please state explicilty(a) which r;(b) what is the value of a; and(c) what is the value of b.Otherwise, your claim is unfounded. But this is all a red herring.>Red herring.We used YOUR de'nition of coprime. It is using YOUR de'nition ofcoprime that you conclude that one of r1, r2, or r3 is coprime to2. But using THAT de'nition, it turns out that NONE of r1, r2, or r3are coprime to 2. So your claim is wrong.>Again, the trick is to try and use the result that it is not a unit,>when in fact, my point is that the ring of algebraic integers is>screwed up in that it is not a unit.So: the trick is to use that it is not a unit, when in fact it is nota unit.Why is it a trick to use something that is true?>>Mistakes in core mathematics can be great fun to play with, so posters>like Rick Decker can run themselves in knots, if they donOt follow the>math.You seem to be tying yourself in knots. You claim it is a trick (or anerror) to use a true fact. Why? = [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu mathematics can be great fun to play with, so posters> like Rick Decker can run themselves in knots, if they donOt follow the> math.>> Remember an error in core is a big deal. People can prove all kinds> of things with such an error, which is why it should be handled.>> James HarrisThe error is in your argument, not in core mathematics. The only big deal here is the magnitude of your egoand its reciprocal: your vanishingly small intelligence.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com = [.snip.]>> x^3-3x+2 = (x-1)(x-1)(x+2).>> >> James later stated that he should have written x^3+3x-2, which is>> irreducible over Q. In that case, none of the roots are units. all of>> them are factors of 2 and of themselves, and therefore you would be>> well to be surprised at the statement that one of them is coprime to>> 2.>>ThatOs actually a useful statement to show how the wacky error with>algebraic integers can lead to a false proof for those of you who>think it doesnOt matter to have this particular error in core.>>Here Magidin IS correct in that none of the roots can be units, but>thatOs because *in the ring of algebraic integers* itOs provably true>that any unit is a root of a monic polynomial with integer>coef'cients and a last coef'cient of 1 or -1.Excellent! Last time I claimed (and proved) this, you stated this 241%40agate.berkeley.eduTHEOREM. Let f(x) be a monic polynomial with integer coef'cients,which is irreducible over Q. Then a root of f(x) is an algebraicinteger unit if and only if the constant term of f(x) is 1 or -1.Proof. The de'nition of algebraic integer unit is an algebraicinteger u such that 1/u is also an algebraic integer.Let r be a root. If the constant term is 1 or -1, then r divides 1 or-1 (in the ring of algebraic integers), so it is a unit.Conversely, suppose that r is an algebraic integer unit. Lets=1/r. Then s is an algebraic integer. Writef(x) = x^n + ... + a_0.0 = s^nf(r) = s^n(r^n+...+a_0) = 1 + a_{n-1}s + ... + a_0s^n.So s is a root of g(y) = a_0x^n+...+1; since g(y) is the polynomial weget by taking f(1/y) and multiplying through by y^n, it follows fromirreducibility of f(x) that g(x) is irreducible.Since s is the root of the irreducible primitive polynomial g(y) withinteger coef'cients, it follows that the leading coef'cient of g(y)is either 1 or -1. So a_0=1 or a_0=-1. Therefore, if r is an algebraicinteger unit, then the constant term of f(x) is either 1 or -1. QEDYour reply it:-- Begin Insert --> THEOREM. Let f(x) be a monic polynomial with integer coef'cients,> which is irreducible over Q. Then a root of f(x) is an algebraic> integer unit if and only if the constant term of f(x) is 1 or -1.> > Proof. The de'nition of algebraic integer unit is an algebraic> integer u such that 1/u is also an algebraic integer.> > Let r be a root. If the constant term is 1 or -1, then r divides 1> or> -1 (in the ring of algebraic integers), so it is a unit.> > Conversely, suppose that r is an algebraic integer unit. Let> s=1/r. Then s is an algebraic integer. WriteAs I guessed you use the *de'nition* which means your argument IScircular as it has to be because what you have canOt be a theorem,when IOve proven it false. -- End Insert --So you now agree that the theorem is correct. Good.>But that does NOT prove that none of the roots are coprime to 2.Nobody said that by itself proved it. What PROVES that none of theroots are coprime to 2 is:1. If there is a non-unit common factor of x and y, then x and y are not coprime.2. r1 is not a unit; r2 is not a unit; r3 is not a unit.3. r1 is a factor of r1 and of 2. r2 is a factor of r2 and 2. r3 is a factor of r3 and 2.4. Therefore, r1 is a non-unit common factor of r1 and 2. (2 and 3)5. Therefore, r1 and 2 have a non-unit common factor. (4)6. Therefore, r1 and 2 are not coprime. (1 and 4)7. r2 is a non-unit common factor of r2 and 2 (2 and 3).8. Therefore, r2 and 2 have a non-unit common factor (7)9. Therefore, r2 is not coprime to 2. (1 and 8)10. r3 is a non-unit common factor of r3 and 2 (2 and 3)11. Therefore, r3 and 2 have a non-unit common factor (10)12 Therefore, r3 and 2 are not coprime (1 and 11).13. Therefore, none of the roots are coprime to 2 (6, 9, and 12).>Here Magidin is using one thing, but switching to another, as if you>just believe that to be coprime to 2, one of the roots has to be a>unit, then you accept his claim as true. That would be an incorrect statement as to fact. I never assume thatif a root is coprime to 2 then it is not a unit. I never used what you CLAIM I am using. I dare you to point out whichof the steps above uses what you claim I am using.>To catch Magidin here you have to force him to *prove*>non-coprimeness, and he cannot. I just did.>Various posters have tried this>particular trick, and IOd catch them on it, and theyOd start going in>circles.Actually, you are. You keep claiming that we are using the converseof if x and y have a non-unit common factor then they are notcoprime, but we do not. Also, as long as weOre at it:YouOve said that the DEFINITION of being coprime isx and y are coprime means that any common factor of x and y is aunit.Is that not the same as stating:1. If x and y are coprime then any common factor of x and y is a unit;and2. If every common factor of x and y is a unit, then x and y are coprime.?Why not? If the de'nition does not mean both things, then it is nota de'nition, it is just a ->property<- of being coprime.If it is the same, then surely their contrapositives are valid aswell:x and y are not coprime means that there is a common factor of xand y which is not a unitand that means BOTHI. If x and y are not coprime, then there is a common factor of x and y which is not a unit.ANDII. If there is a common factor of x and y which is not a unit, then x and y are not coprime.But that is immaterial. You admit that II is true, and that is theonly thing I am using. = [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu> It turns out that the de'nition for algebraic integers where they are> roots of *monic* polynomials leaves off certain numbers that should be> included, which is how I found a way to show some wacky things.> > ItOs easy enough to explain as consider> > x^3 - 3x + 2> > which as it turns out has a root that is coprime to 2. True enough. I make its roots x=1 (twice) and x=-2; 1 is comprime to 2in the Algebraic Integers.> Well, that *should* make it a unit i.e. factor of 1, but turn the> polynomial around using x=1/y and set it to 0 and you get> > 1/y^3 - 3/y + 2 = 0> > which is> > 2y^3 - 3y^2 + 1 = 0Which has a repeated root of y=1 and a root of y=-1/2.> which would force a root to be a unit as well, but thereOs a quite> correct theorem proving that the root of a non-monic primitive can NOT> be an algebraic integer.Bollocks. Absolute crap. I think the correct formulation of thattheorem states that such a primitive must have at least one root thatisnOt an algebraic integer, not that all its roots must be. Considerthe set of polynomial equations of the form 2x^2-(2n+1)x+n=0, where nis an integer. Then they are all non-monic primitives, but they allhave the algebraic integer solution x=n.> Maybe, after all, IOm looking for a 'ght.It took you that long to work that out?Jack Rudd => >Oh yeah, youOre right. That should be x^3 + 3x - 2.> > Okay, this one is irreducible over Q: if it were reducible it would> have a root, and the only possible rational roots are 1, -1, 2, and> -2. Checking each shows that it does not have rational roots. Well here we go again. I noticed a poster trying to make a big dealabout the lateness of my reply but hey, I reply through Google and ittakes it a while to put up messages.The math proving my point is basically high school stuff, so itOseasy, which makes it fascinating to me that so many of you *trust*Magidin. Now I understand, I think, why he keeps lying, as whatchoice does he have?But why do any of you believe him and his voodoo math? > So, do you agree or disagree with each of the following?> > 1. If r1, r2, r3 are the three roots of x^3+3x-2, then > x^3+3x-2 = (x-r1)(x-r2)(x-r3).Agree. > 2. Therefore, r1*r2*r3*(-1) = 2.Agree.> 3. Therefore, r1 is a factor of 2 in the ring of all algebraic> integers, since (-1)*r2*r3 is an algebraic integer, and> r1* ( (-1)*r2*r3 ) = 2.Trivially, yes, so I agree.> 4. Likewise, r2 is a factor of 2 in the ring of all algebraic> integers, since (-1)*r1*r3 is an algebraic integer, and> r2* ( (-1)*r1*r3 ) = 2.Again, trivially, yes, so I agree.> 5. Likewise, r3 is a factor of 2 in the ring of all algebraic> integers, since (-1)*r2*r3 is an algebraic integer, and> r3* ( (-1)*r2*r3 ) = 2.Yet again, trivially, yes, so I agree.> 6. In the ring of all algebraic integers, if x and y are algebraic> integers, then x is a common factor of x and x*y.Agree.> 7. In the ring of all algebraic integers, if x and y are algebraic> integers and x is not an algebraic integer unit, then x is a> non-unit common factor of x and x*y.Hmmm...what does being a unit have to do with anything?If I have xy = 2, where y=2, isnOt x STILL a factor of 2?Then again, yeah, sounds ok to me.Agree.> 8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and> 2; r3 is a common factor of r3 and 2.Agree. > 9. r1 is not a unit in the ring of algebraic integers, because 1/r1> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,> irreducible, primitive polynomial with integer coef'cients, and> therefore its roots are not algebraic integers.Agree.> and 9).Agree.> 11. r2 is not a unit in the ring of algebraic integers. Because 1/r2> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,> irreducible, primitive polynomial with integer coef'cients, and> therefore its roots are not algebraic integers.Agree.> and 11).Agree.> 13. r3 is not a unit in the ring of algebraic integers. Because 1/r3> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,> irreducible, primitive polynomial with integer coef'cients, and> therefore its roots are not algebraic integers.Agree.> and 13).Agree.> 15. In the ring of all algebraic integers, x and y are not coprime if> and only if there exists a non-unit common factor of x and y (in> the ring of all algebraic integers).> > [This is what you use as a de'nition; it is equivalent to the> standard de'nition for the ring of all algebraic integers, so it does> not matter in ->that<- context]Agree.> 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprimeDisagree.Actually IOm glad you presented it that way Magidin as it shows howyouOve fooled so many by getting them to take an illogical step.Here it IS true that x and y are NOT coprime if there exists anon-unit factor common to both x and y, but the converse, is NOT truein the ring of algebraic integers as it is screwy.You have a gap.For those readers confused consider that in the ring of *evens* 2 and6 donOt share non-unit factors, but by MagidinOs de'nition they areNOT coprime.Why?Because 3 is not even, but 2(3) = 6, so by MagidinOs de'nition ofcoprime they are NOT coprime. Now the problem with algebraic integersis more complicated, but that example should help you to see thatMagidin is lying to you.ItOs trickery.Further, since his position has been repudiated by Professor McKenziewho shot down his key objection he is against the discipline ofmathematics as are his supporters.Again see http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759James Harris =[cut]> 15. In the ring of all algebraic integers, x and y are not coprime if> and only if there exists a non-unit common factor of x and y (in> the ring of all algebraic integers).> > [This is what you use as a de'nition; it is equivalent to the> standard de'nition for the ring of all algebraic integers, so it does> not matter in ->that<- context]> > Agree.Actually, you say below that you do not agree with this.> > 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime> > Disagree.> > Actually IOm glad you presented it that way Magidin as it shows how> youOve fooled so many by getting them to take an illogical step.> > Here it IS true that x and y are NOT coprime if there exists a> non-unit factor common to both x and y, but the converse, is NOT true> in the ring of algebraic integers as it is screwy.Step 15 is the de'nition that is being used for the term coprime.For a de'nition, the converse is always true. It does not requirea proof: there is no gap.Otherwise, when someone says Suppose r1 is not coprime to 2, wewould not be able to know what was meant by that statement. Thestatement Suppose r1 is not coprime to 2 can be unwind usingthe converse of the de'nition of coprime into a statementinvolving non-unit factors common to r1 and 2, thereby eliminatingthe term coprime. That is what de'nitions do: they give namesto new relationships involving previous de'ned or primitive terms.> You have a gap.> > For those readers confused consider that in the ring of *evens* 2 and> 6 donOt share non-unit factors, but by MagidinOs de'nition they are> NOT coprime.No, by MagidinOs de'nition, they are coprime.> Why?> > Because 3 is not even, but 2(3) = 6, so by MagidinOs de'nition of> coprime they are NOT coprime.But, 3 is not an even integer. So, you have not exhibited a non-uniteven integer that is a common factor of 2 and 6. You have not shownthat 2 and 6 are not coprime.-- Bill Hale =-I do not claim, that James Harris made any errors in his Advancedpolynomial factorization paper, because it is really hard to 'gureout what he really means, so itOs even harder to decide whether it iscorrect.One reason it maybe hard to convince him that he is making mistakesall over is that the main statement of his paper is actually true.concept to grasp that a proof of a true statement may be itsel'ncorrect.So this is just to let you all know that what he claims in that paperis actually a triviality:It is actually not obvious 'rst what his main statement is, becauseJames does not believe in labeling things as theorem or proof.Nonetheless the following is a quote from his paper:This paper will show, using basic algebraic methods, that given thefactorization, in the ring of algebraic integers, 65x^3-12x + 1 =(a1*x + 1)(a2*x + 1)(a3*x + 1) one of the aOs is coprime to 5And here is why this is trivial. In other words, here is a proof:(this is no longer a quotation)Since (a1*x + 1)(a2*x + 1)(a3*x + 1)=a1*a2*a3*x^3+...+(a1+a2+a3)*x+1,one obtains, that a1+a2+a3=-12, and therefore if none of the aOs iscoprime to 5, then 12 is not coprime to 5 either. But then 1=5*5-2*12is also not coprime to 5. This is a contradiction.Complete =>It is actually not obvious 'rst what his main statement is, because>James does not believe in labeling things as theorem or proof.>Nonetheless the following is a quote from his paper:>>This paper will show, using basic algebraic methods, that given the>factorization, in the ring of algebraic integers, 65x^3-12x + 1 =>(a1*x + 1)(a2*x + 1)(a3*x + 1) one of the aOs is coprime to 5>>And here is why this is trivial. In other words, here is a proof:>(this is no longer a quotation)>Since (a1*x + 1)(a2*x + 1)(a3*x + 1)=a1*a2*a3*x^3+...+(a1+a2+a3)*x+1,>one obtains, that a1+a2+a3=-12, and therefore if none of the aOs is>coprime to 5, then 12 is not coprime to 5 either.6 is not coprime to 30.10 is not coprime to 30.15 is not coprime to 30.Therefore, 6+10+15 = 31 is not coprime to 30. ItOs not denial. IOm just very selective about what I accept as reality. Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu [.personal attack removed.]>> So, do you agree or disagree with each of the following?>> >> 1. If r1, r2, r3 are the three roots of x^3+3x-2, then >> x^3+3x-2 = (x-r1)(x-r2)(x-r3).>>Agree.> >> 2. Therefore, r1*r2*r3*(-1) = 2.>>Agree.> 3. Therefore, r1 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r2*r3 is an algebraic integer, and>> r1* ( (-1)*r2*r3 ) = 2.>>Trivially, yes, so I agree.> 4. Likewise, r2 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r1*r3 is an algebraic integer, and>> r2* ( (-1)*r1*r3 ) = 2.>>Again, trivially, yes, so I agree.> 5. Likewise, r3 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r2*r3 is an algebraic integer, and>> r3* ( (-1)*r2*r3 ) = 2.>>Yet again, trivially, yes, so I agree.> 6. In the ring of all algebraic integers, if x and y are algebraic>> integers, then x is a common factor of x and x*y.>>Agree.> 7. In the ring of all algebraic integers, if x and y are algebraic>> integers and x is not an algebraic integer unit, then x is a>> non-unit common factor of x and x*y.>Hmmm...what does being a unit have to do with anything?The last statement is x is a NON-UNIT common factor of x andx*y. Clearly, we need x to be not a unit for that to be true.>If I have xy = 2, where y=2, isnOt x STILL a factor of 2?Yes, but it is not a non-unit factor of 2, because in that case x=1,so x is a unit.>Then again, yeah, sounds ok to me.>>Agree.> 8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and>> 2; r3 is a common factor of r3 and 2.>>Agree.> >> 9. r1 is not a unit in the ring of algebraic integers, because 1/r1>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.>>Agree.> and 9).>>Agree.> 11. r2 is not a unit in the ring of algebraic integers. Because 1/r2>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.>>Agree.> and 11).>>Agree.> 13. r3 is not a unit in the ring of algebraic integers. Because 1/r3>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.>>Agree.> and 13).>>Agree.> 15. In the ring of all algebraic integers, x and y are not coprime if>> and only if there exists a non-unit common factor of x and y (in>> the ring of all algebraic integers).>> >> [This is what you use as a de'nition; it is equivalent to the>> standard de'nition for the ring of all algebraic integers, so it does>> not matter in ->that<- context]>>Agree.> 16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime>>Disagree.Why?By 10, which you agree to, r1 is a non-unit common factor of r1 and2. Therefore, r1 and 2 have a non-unit common factor (namely, r1). By15 that means that r1 and 2 are not coprime (since there exists anon-unit common factor of x and y, namely r1).Same thing with r2 and 2, and r3 and 2.The 3 is a typo.>Actually IOm glad you presented it that way Magidin as it shows how>youOve fooled so many by getting them to take an illogical step.What step is illogical and y?>Here it IS true that x and y are NOT coprime if there exists a>non-unit factor common to both x and y, but the converse, is NOT true>in the ring of algebraic integers as it is screwy.Point the zeroth. so in fact you are contesting point 15, not pointreally read more carefully. Point 15 was:>> 15. In the ring of all algebraic integers, x and y are not coprime if>> and only if there exists a non-unit common factor of x and y (in>> the ring of all algebraic integers).So you are saying that this is not true. It is not if and only if,you claim that only only if holds. Fine.Replace 15 with:15O. In the ring of all algebraic integers, if there exists a non-unit common factor of x and y (in the ring of all algebraic integers) then x and y are not coprime.That one you claim to agree to.Point the 'rst: I never used the converse: there exists a non-unitfactor common to both r1 and 2 (point 10), and so I conclude that r1and 2 are not coprime. There exists a non-unit factor common to bothr2 and 2 (point 12), so I conclude that r2 and 2 are notcoprime. There exists a non-unit factor common to both r3 and 2 (point14), so I conclude that r3 and 2 are not coprime.Where did I supposedly use the converse? The converse would be If r1and 2 are not coprime, then there exists a non-unit factor common toboth r1 and 2. I never invoked that.Point the second: Actually, the converse IS true in the ring ofalgebraic integers. I already gave you a 40agate.berkeley.eduDedekindOs Theorem on the 'niteness of the class number.>You have a gap.Nope. I never used the converse.>For those readers confused consider that in the ring of *evens* 2 and>6 donOt share non-unit factors, but by MagidinOs de'nition they are>NOT coprime.This is a red herring. We are working in the ring of all algebraicintegers, and I did NOT use the converse of the statement, I only usedthe statement in its direct form.You have not correctly pointed any gap. You INSINUATE that I use theconverse of a result, when I used the result itself.Please read again points 15 and 16 carefully. Replace point 15 bypoint 15O, and you will see that point 16 follows from using ONLYpoint 15O, not the converse. == ==Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = >Oh yeah, youOre right. That should be x^3 + 3x - 2.>>Okay, this one is irreducible over Q: if it were reducible it would>>have a root, and the only possible rational roots are 1, -1, 2, and>>-2. Checking each shows that it does not have rational roots. > > > Well here we go again. I noticed a poster trying to make a big deal> about the lateness of my reply but hey, I reply through Google and it> takes it a while to put up messages.> > The math proving my point is basically high school stuff, so itOs> easy, which makes it fascinating to me that so many of you *trust*> Magidin. Now I understand, I think, why he keeps lying, as what> choice does he have?> > But why do any of you believe him and his voodoo math?> Why do you continually claim that people are lying? Is your worldso delicate that there is no possibility that you are in error?I, for one, 'nd MagidinOs arguments to be carefully formulated,with none of the sham informality that your arguments cloak themselvesin, and when I study them carefully, I 'nd them to be correct. Thetimes IOve done the same with regard to your arguments, I 'nd themfestooned with holes, in fact I 'nd them comprising more hole thansubstance.>So, do you agree or disagree with each of the following?>>1. If r1, r2, r3 are the three roots of x^3+3x-2, then >> x^3+3x-2 = (x-r1)(x-r2)(x-r3).> > > Agree. >>2. Therefore, r1*r2*r3*(-1) = 2.> > > Agree.> > >>3. Therefore, r1 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r2*r3 is an algebraic integer, and>> r1* ( (-1)*r2*r3 ) = 2.> > > Trivially, yes, so I agree.> > >>4. Likewise, r2 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r1*r3 is an algebraic integer, and>> r2* ( (-1)*r1*r3 ) = 2.> > > Again, trivially, yes, so I agree.> >5. Likewise, r3 is a factor of 2 in the ring of all algebraic>> integers, since (-1)*r2*r3 is an algebraic integer, and>> r3* ( (-1)*r2*r3 ) = 2.> > > Yet again, trivially, yes, so I agree.> > >>6. In the ring of all algebraic integers, if x and y are algebraic>> integers, then x is a common factor of x and x*y.> > > Agree.> > >>7. In the ring of all algebraic integers, if x and y are algebraic>> integers and x is not an algebraic integer unit, then x is a>> non-unit common factor of x and x*y.> > > > Hmmm...what does being a unit have to do with anything?> > If I have xy = 2, where y=2, isnOt x STILL a factor of 2?> > Then again, yeah, sounds ok to me.> > Agree.> > >>8. r1 is a common factor of r1 and 2; r2 is a common factor of r2 and>> 2; r3 is a common factor of r3 and 2.> > > Agree.> > >>9. r1 is not a unit in the ring of algebraic integers, because 1/r1>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.> > > Agree.> > >> and 9).> > > Agree.> > >>11. r2 is not a unit in the ring of algebraic integers. Because 1/r2>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.> > > Agree.> > >> and 11).> > > Agree.> > >>13. r3 is not a unit in the ring of algebraic integers. Because 1/r3>> satis'es the polynomial 2x^2 - 3x + 1, which is non-monic,>> irreducible, primitive polynomial with integer coef'cients, and>> therefore its roots are not algebraic integers.> > > Agree.> > >> and 13).> > > Agree.> >15. In the ring of all algebraic integers, x and y are not coprime if>> and only if there exists a non-unit common factor of x and y (in>> the ring of all algebraic integers).>>[This is what you use as a de'nition; it is equivalent to the>>standard de'nition for the ring of all algebraic integers, so it does>>not matter in ->that<- context]> > > Agree.> > >>16. r1 is not coprime to 2. r2 is not coprime to 2. r3 is not coprime> > > Disagree.> > Actually IOm glad you presented it that way Magidin as it shows how> youOve fooled so many by getting them to take an illogical step.> > Here it IS true that x and y are NOT coprime if there exists a> non-unit factor common to both x and y, but the converse, is NOT true> in the ring of algebraic integers as it is screwy.> You state: x and y are NOT coprime if there exists a non-unit factor common to x and yMagidin has shown, and you have agreed, to the following: r1 is a non-unit factor common to r1 and 2. (step 10) r2 is a non-unit factor common to r2 and 2. (step 12) r3 is a non-unit factor common to r3 and 2. (step 14).Yet, in step 16, modulo a typo (Magidin surely intended to typer3 is not coprime to 2), you disagree with the combined statement.Or, perhaps you didnOt read his statement as containing the clear typo-graphical error (typing 2 when he clearly [referencing step 14] meantto type 3).> You have a gap.> > For those readers confused consider that in the ring of *evens* 2 and> 6 donOt share non-unit factors, but by MagidinOs de'nition they are> NOT coprime.> No. Magidin uses the de'nition that r and s are coprime in the ring Rif the minimal ideal in R containing r and s is R itself. If the ringcontains a multiplicative identity, this is equivalent to the existenceof members m and n for which mr + ns = 1.Given the fact that 2Z doesnOt have a multiplicative identity, it isnecessary to deal with the previous form of the de'nition (involvingideals, rather than a decomposition of 1 into the prospective coprimeelements).Given that de'nition, and the ring 2Z, the members 2 and 6 arecoprime, since the minimal ideal containing 2 and 6 is 2Z. How do wesee this? Let I be an ideal (an additive subgroup of the ring, which is closed under multiplication by ring elements) of 2Z, containing both 2 and 6. Since I is closed under multiplication by elements of 2Z, and 2 is already in I, then -2*2 = -4 is also in I. Since 6 is also in I, and since I is an additive subgroup of 2Z, we then know that -4 + 6 = 2 is in I (I know, I already knew that, but we continue). Multiplying by ring elements, we obtain all other elements of 2Z. Actually, the minimal ideal of 2Z containing 2 is already 2Z. Thus, 2 is coprime to every other element, including 6, of the ring 2Z.> Why?> > Because 3 is not even, but 2(3) = 6, so by MagidinOs de'nition of> coprime they are NOT coprime. Now the problem with algebraic integers> is more complicated, but that example should help you to see that> Magidin is lying to you.> No, youOre mis-representing his position. Part of honest argumentationinvolves attempting to attack the position that the person *actually*holds, rather than purposely misreading it and attacking that mistakenposition.> No, you only produce gaps by taking advantage of typographical orother inconsequential errors. When you attempt to address his *actual*arguments, you consistently fail.> Further, since his position has been repudiated by Professor McKenzie> who shot down his key objection he is against the discipline of> mathematics as are his supporters.> Hey, if you want to keep holding Professor McKenzie up as a person whosupports your claim, you should at least offer Professor McKenzie thecourtesy of speaking for himself. It is a sign of profound disrespectto this person who has befriended you, without any apparent recompenseto himself, to act otherwise.> Again see http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > > James HarrisDale. =Some corrections: [.snip.]>No. Magidin uses the de'nition that r and s are coprime in the ring R>if the minimal ideal in R containing r and s is R itself.This is not the de'nition I was using, either. That would becomaximal in a ring with 1, and in rings without 1 (or in theabsence of the Axiom of Choice) it need not be equivalent to coprime.IOll give the de'nitions in a second, but let me just point out thatJamesOs objection is moot. I used only HIS de'nition in theargument I gave (which is really just copying some results that you,Dale, had gotten and posted, with a few more multiplications ofpolynomials made explicitly.Anyway, the de'nitions:Let R be a ring, which may or may not have a multiplicativeidentity. An ideal I of R is a prime ideal if and only if it is notequal to R, and whenever x and y are elements of R such thatx*y is in I, then either x is in I or y is in I.An ideal J of R is a maximal ideal if and only if it is an ideal,properly contained in R, and there does not exist any ideal I whichproperly contains J and is properly contained in R.Two ideals A and B are coprime if and only if there does not exist aprime ideal containing both. Two elements r and s of R are coprime if and only if the principalideals they generate, (r) and (s), are coprime.Two ideals A and B are coprime if and only if there does not exist amaximal ideal containing both.Two elements r and s of R are comaximal if and only if the principalideals they generate, (r) and (s), are comaximal. For rings with 1,this is equivalent to the minimal ideal in R that contains r and sbeing R itself.If the ring is commutative and has a 1, then every maximal idealis prime. So coprime -> comaximal.If the ring has a 1, and you assume the Axiom of Choice (ZornOsLemma), then every proper ideal is contained in a maximal ideal, socomaximal->coprime.So, if your ring is commutative, has a 1, and you assume the Axiom ofChoice, then coprime is equivalent to comaximal. Your de'nition isthe de'nition of comaximal.> If the ring>contains a multiplicative identity, this is equivalent to the existence>of members m and n for which>> mr + ns = 1.This is true for comaximal.>Given the fact that 2Z doesnOt have a multiplicative identity, it is>necessary to deal with the previous form of the de'nition (involving>ideals, rather than a decomposition of 1 into the prospective coprime>elements).>>Given that de'nition, and the ring 2Z, the members 2 and 6 are>coprime, since the minimal ideal containing 2 and 6 is 2Z. How do we>see this?>> Let I be an ideal (an additive subgroup of the ring, which is> closed under multiplication by ring elements) of 2Z, containing> both 2 and 6.>> Since I is closed under multiplication by elements of 2Z, and> 2 is already in I, then -2*2 = -4 is also in I. Since 6 is> also in I, and since I is an additive subgroup of 2Z, we then> know that -4 + 6 = 2 is in I (I know, I already knew that, but> we continue). Multiplying by ring elements, we obtain all other> elements of 2Z.Ehr, no. Multiplying by ring elements once you have 2 gives you allthe multiples of 4. However, an ideal is also closed under differences,since you have 2 you have 0=2-2; and you have -4 so you have4=0-(-4); and you have 6 so you have 4-6 = -2.It is also closed under sums, so you have 2+2, 2+2+2, 2+2+2+2,....,,so you get all positive multiples of 2; and you also have -2, -2+(-2),-2+(-2)+(-2),.... so you get all negative multiples of 2, so you getall of 2Z.> Actually, the minimal ideal of 2Z containing> 2 is already 2Z.Yes. The minimal ideal containing 2n is {2n*k : k an integer}. And theideal generated by 2n is {4n*k: k an integer}.Using the de'nition on principal ideals, note that (2)={4k: k is aninteger} and (6) = {12k: k is an integer}. The former contains thelatter, but (2) is not prime: 2*2 is in (2), but 2 is not. Therefore,if there were a prime ideal containing both, then it would properlycontain {4k: k an integer}. That means that there is an even integer awith a = 2 (mod 4) in the ideal. But that means that a-2=4k, so a-4k =2; therefore, the ideal contains 2, and as you show above, that meansthat the ideal is all of 2Z, and hence is not prime. Thus, (2) and (6)are coprime, so 2 and 6 are coprime.They are not, however, comaximal: (2) = {4k: k is an integer} is amaximal ideal, as just shown above. And it contains (6); so (2) and(6) are both contained in the maximal ideal (2), hence they are notcomaximal.James is using the de'nition a and b are coprime in R if and only ifevery common divisor of a and b in R is a unit. For this de'nitionto even ->begin<- to make sense, you need to be able to talk aboutunits in R. The only way you can even begin to talk about units in Ris if R has a multiplicative identity. So JamesOs de'nition does noteven ->begin<- to make sense in 2Z, making each and every mention of2Z nothing but a red herring.It is easy to verify that if a and b are coprime under the de'nitionabove involving prime ideals (assuming the Axiom of Choice and that Ris commutative and with 1), then they are coprime under thisde'nition. The two are not equivalent in general, as the old exampleZ[sqrt(-5)] shows, since 2 and 1+sqrt(-5) are coprime under theOcommon divisorO de'nition, but not under the de'nition as ideals.However, in the case of the ring of all algebraic integers, one canshow that the two de'nitions are equivalent (again, assuming theAxiom of Choice to be able to get the OcoprimeO as linear combinationproperty). It follows from the Finiteness of the Class Number. Iposted a proof some time ago, glossing only over the last 40agate.berkeley.edu =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = realized that itOs simpler to refer people to a Hong Kong website> where they allow posting with LaTeX so it looks prettier anyway.> > http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759Do you want the responses posted there or here? It might be easier to simply copy the relevant material here so everyone can easily look at it in context.> > The math IS rather basic and I still think itOs mostly at high school> level, so itOs amazing that so many of you keep 'ghting it to the> extent that I have to keep pushing and pushing.So are the counter-arguments.> > But itOs not amazing if mathematicians are willing to lie about> mathematics for *social* reasons, or then again, you might be ashamed> at such an error, while I think itOs fascinating.How about sticking to math instead of social analysis?-- Will Twentyman =sorry, but I just canOt stop it.so, given the high school level of your latest treatise,which I donOt dare to look at -- hey, IOve made *some* progress --that is probably the audience that would be so unencumberedby the hypothetically evil math, as to be able to vet your hypothesesin a way that you might concur with. well?however, how can you prove a proof,just by the tautology of it be math, QED? some time ago, on http://listserv.acsu.buffalo.edu/on its geodesic list, a guy was belaboring his effortsto prove the Kepler conjecture about 12 balls around one;recently, I realized taht he was merely digesting,down to identical arbitrary factors, a bookthat went through HaleOs alleged proof. now,having perused taht book at a bookstore, I realized, why:the book shows taht the proof was not complete,in spite of itOs having been called, prooven. since youOve never said taht youOve ever proven any thing*but* le derniere theorem de Fermat, as it is speciously (orpopularly) called, you might have to explain why it is that,Class, weOre just going to jump directly into The Math; after all,it is mathematics! even if you were wrong,theyOd still learn some thing; eouldnOt they?so, you dystrust professional math folks, as wellas anyone who deigns to take them seriouslyin a sci.math forum. so,seek your proof, else where, amongst those who are receptive(no matter how high their IQs .-)well? > Hmmm...are mathematicians as a group evil? I question; you decide.> > I realized that itOs simpler to refer people to a Hong Kong website> where they allow posting with LaTeX so it looks prettier anyway.> > http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > The math IS rather basic and I still think itOs mostly at high school> level, so itOs amazing that so many of you keep 'ghting it to the> extent that I have to keep pushing and pushing. > It IS math after all. ItOs not like IOm making things up.--les ducs de Buffet!http://larouchepub.com =Hope I have the spelling correct on palimpset.Just a few comments on that TV show. I am not at all surprized that Archimedes 1st had the Calculus longbefore Liebniz or Newton and to think that if Archimedes work had beenmade the bible of ancient times instead of the gooey men of deserts asthe bulwark of society. As the program suggests that we would be 100years more advanced. Nay, I would say 1,000 years more advancedbecause collective-religion is nothing but a drag and resistance onhumanity. If Archimedes 1st work and perspective had been made theworld religion then humanity will have colonized Moon, Mars, Europa bycolonies. Organized religion has always been a human retardant.How is it that Volume is the dual of Motion, vis a vis, Integral dualof Derivative.We know that Energy is the dual of Time in physics and we know thatMomentumis the dual of Position in physics. So, somehow, Volume must be energyor volume must be momentum!!!!!I know that Archimedes 1st did not have a decimal number system, butcurious as to how he derived 10^69 grains of sand to cover the Cosmosand what sort of Presumptions he made to give the Cosmos a containerto hold that sand????The TV show was wrong in getting too many mathematicians to talk forit leaves the impression that Archimedes 1st was mostly aMathematician when that is patently false. Archimedes 1st was foremosta Engineer and Physicists and lastly and only minorly was he amathematician. For in fact, virtually all of his mathematics sprungforth from his work on engineering and physics.Archimedes 1st is the worldOs 'rst Physicist for he systematizedin point.An accurate history of Archimedes 1st would be 50 pages of Physics and30 pages of Engineering and the last 20 pages of a 100 page historywould be mathematics.The historical account of the scene of his death is 'cticious. Whenyou have a man so smart and brilliant although 75 years old, he is notgoing to be doodling mathematics after having defended Syracuse fromthe Romans. Archimedes 1st was killed after trying §ee the city alongwith many others. He was just too old and could not §ee fast enough.And NOVA TV should not be stating history as true history when thosefacts were simply not known as to the scene of his death.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies =I found a description of ArchimedesO methods for large numbers and his SandReckoner problem hl=en&lr=&ie=UTF-8&oe=UTF-8&selm=200007152124.OAA27104% 40math-cl-n03.ucr.edu&rnum=4IOm not sure about the suggestion that Archimedes works replacing religion.I believe this was tried already with Pythagoras:http://www.math.tamu.edu/~don.allen/history/pythag/ pythag.htmlWhat really annoyed me about the program is that they suggest Archimedes hadcalculus and dealt with the in'nitesimal, but stops there. The closet IOvehttp://news-service.stanford.edu/news/november6/archimede s -116.html onReviel Netz, but it too stops just short: The proof in question is too complex to explain here, but suf'ce it to say that its shakes up the historical view of pre- calculus to its very foundations. And this is something Netz particularly enjoys.Does anyone here know the details of this? =Looking for math textbooks (not workbooks) equivalent to the actualclassroom textbooks used in grades 3 through 12. Request internetreferences (keywords to search, websites to check out, ...) orspeci'c publishers (et al) to contact. Money relevant, so 'rstchoice is downloading equivalent texts in pdf format (et al) off o'nternet. => Looking for math textbooks (not workbooks) equivalent to the actual> classroom textbooks used in grades 3 through 12.Meaning you are looking for free stuff on the Internet, right? ThatOs noteasy, but letOs see what other readers come up with. There is an onlinecourse, called ALEKShttp://www.aleks.com/that seems all right and is not too expensive, and the best elementaryschool textbook series available in English (in the opinion of many) is theSingapore Primary Mathematics series, available in the United States throughhttp://www.singaporemath.com/> Request internet> references (keywords to search, websites to check out, ...) or> speci'c publishers (et al) to contact. Money relevant, so 'rst> choice is downloading equivalent texts in pdf format (et al) off of> internet.A good source of freedbies is the Web site of Professor Hung-hsi Wu ofBerkeleyhttp://math.berkeley.edu/~wu/There is MUCH available at the algebra level. Use Google to search foralgebra review 'letype:pdf or beginning algebra 'letype:pdf orwhatever keywords you are looking for. You could try prealgebra orprecalculus in those searches too. Geometry is much more poorlyrepresented online. There is a not half bad prealgebra textbook, all free,and very mathematical, athttp://ms.yuba.cc.ca.us/~jb2/Hope this helps!-- Karl M. Bunday Christ has set us free. Galatians 5:1Learn in Freedom (TM) http://learninfreedom.org/ => >He even shot down the objection that Magidin and others have used for>so long dismissing it almost on sight.> > He did? Did he shoot down my (single?) objection? Did he read ->my<-> words, or was it something you, ehr, explained to him?> > Because you have a track record of just not getting what the objection> is. It is interesting that I do not see anything in your> correspondences below with Ralph McKenzie or anyone else that even> mentions me, though.Professor McKenzie dismissed your objection claiming that the aOs canhave varying factors of f dependent on m.For background and pretty math as they allow LaTeX, readers should seehttp://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg= 759Notice how *short* the math argument is. MagidinOs objection would bethat the gOs you see on that page can have varying factors of fdependent on the value of m, as he seems hellbent on attacking algebraitself and the result that only *two* of the gOs have f as a factorwhile one is coprime, a result independent of m, when f is coprime to3, x and u.Magidin is an anti-mathematician attacking algebra itself, and evenquestioning the dismissal of his bogus objection by ProfessorMcKenzie, a man senior to him whom he claims to have seen at Berkeley.He is a proven liar and the proof is at that link. Magidin wants youto believe in voodoo math, where because he says so, f^2 divides of'n different ways from P(m) depending on the value of m, but you see,thatOs not mathematics.Magidin has proven himself to be against mathematicians, now againstProfessor McKenzieOs dismissal of a bogus objection, against algebra,with his bogus objection, and against you, by continuing to try andconfuse and convince people of falsehoods.James Harris => >He even shot down the objection that Magidin and others have used for>so long dismissing it almost on sight.> > He did? Did he shoot down my (single?) objection? Did he read ->my<-> words, or was it something you, ehr, explained to him?> > Because you have a track record of just not getting what the objection> is. It is interesting that I do not see anything in your> correspondences below with Ralph McKenzie or anyone else that even> mentions me, though.> > Professor McKenzie dismissed your objection claiming that the aOs can> have varying factors of f dependent on m.> > For background and pretty math as they allow LaTeX, readers should see> > http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > Notice how *short* the math argument is. MagidinOs objection would be> that the gOs you see on that page can have varying factors of f> dependent on the value of m, as he seems hellbent on attacking algebra> itself and the result that only *two* of the gOs have f as a factor> while one is coprime, a result independent of m, when f is coprime to> 3, x and u.> > Magidin is an anti-mathematician attacking algebra itself, and even> questioning the dismissal of his bogus objection by Professor> McKenzie, a man senior to him whom he claims to have seen at Berkeley.> > He is a proven liar and the proof is at that link. Magidin wants you> to believe in voodoo math, where because he says so, f^2 divides off> in different ways from P(m) depending on the value of m, but you see,> thatOs not mathematics.> > Magidin has proven himself to be against mathematicians, now against> Professor McKenzieOs dismissal of a bogus objection, against algebra,> with his bogus objection, and against you, by continuing to try and> confuse and convince people of falsehoods.> > > James Harris Advanced Polynomial Factorization, which I believe you discussed with Prof. McKenzie, says that if the polynomial P(x) = 65*x^3 - 12*x + 1is factored in the form P(x) = (a1*x + 1)*(a2*x + 1)*(a3*x + 1),where a1, a2, and a3 are algebraic integers, then at leastone of a1, a2, and a3 is coprime to 5. Right? Here is a theorem from algebraic number theory: Theorem: If Q(x) is a primitive irreducible monic polynomial with integer coef'cients and constant term b0, and q is any prime dividing b0, then any root of Q(x) is an algebraic integer which is not coprime to q [that is, any root has a nonunit factor in common with q].Some questions -1. Did Prof. McKenzie comment on whether your result about 65^x^3 - 12*x + 1 is correct or not?2. Do you think the Theorem above is true?3. Would you like to see a proof of it?4. If the theorem is true, does it have anything to do with the result stated above from Advanced Polynomial Factorization ? Andrzej => >He even shot down the objection that Magidin and others have used for>so long dismissing it almost on sight.>>He did? Did he shoot down my (single?) objection? Did he read ->my<->>words, or was it something you, ehr, explained to him?>>Because you have a track record of just not getting what the objection>>is. It is interesting that I do not see anything in your>>correspondences below with Ralph McKenzie or anyone else that even>>mentions me, though.> > > Professor McKenzie dismissed your objection claiming that the aOs can> have varying factors of f dependent on m.> > For background and pretty math as they allow LaTeX, readers should see> > http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > Notice how *short* the math argument is. Notice that you have responses waiting. >MagidinOs objection would be> that the gOs you see on that page can have varying factors of f> dependent on the value of m, as he seems hellbent on attacking algebra> itself and the result that only *two* of the gOs have f as a factor> while one is coprime, a result independent of m, when f is coprime to> 3, x and u.HeOs also right. You demonstrate behavior when m=0. This behavior need not carry over to when m<>0. Many people have told you this. The objections will not stop coming. Perhaps you should send McKenzie a link to this site and see what his response is to my observations there.[attacks on Magidin snipped]-- Will Twentyman =>> >>He even shot down the objection that Magidin and others have used for>>so long dismissing it almost on sight.>> >> He did? Did he shoot down my (single?) objection? Did he read ->my<->> words, or was it something you, ehr, explained to him?>> >> Because you have a track record of just not getting what the objection>> is. It is interesting that I do not see anything in your>> correspondences below with Ralph McKenzie or anyone else that even>> mentions me, though.>>Professor McKenzie dismissed your objection claiming that the aOs can>have varying factors of f dependent on m.Stating facts not in evidence.Did you show him the polynomial you are using, and did you show him mywords? Or did you use one of your simpli'ed examples where thecomplications that create your mistake donOt exist?Like when you use 2x^2+4x+4 to prove that what you are doing is valid? [.snip.]>Notice how *short* the math argument is. MagidinOs objection would be>that the gOs you see on that page can have varying factors of f>dependent on the value of m,Oh, well, if you can tell what my objections will be, I certainlydonOt need to post them any more.Too bad you almost never understand what I say, so you cannot betrusted to know what my objection would be.>He is a proven liar and the proof is at that link.The link proves nothing, least of all anything about my honesty.> Magidin wants you>to believe in voodoo math, where because he says so, f^2 divides off>in different ways from P(m) depending on the value of m, but you see,>thatOs not mathematics.So you claim. Your claim is unfounded.>Magidin has proven himself to be against mathematicians, now against>Professor McKenzieOs dismissal of a bogus objection, against algebra,>with his bogus objection, and against you, by continuing to try and>confuse and convince people of falsehoods.Assumes facts that are not in evidence and cannot be substantiated:you are not a trusted source of reports of other peopleOsstatements. You cannot be trusted to have given an accurate report ofmy objections, and you cannot be trusted to be giving an accuratereport of the response you got.Your assertions are therefore worthless. = such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manOs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan = >>He even shot down the objection that Magidin and others have used for>>so long dismissing it almost on sight.>> >> He did? Did he shoot down my (single?) objection? Did he read ->my<->> words, or was it something you, ehr, explained to him?>> >> Because you have a track record of just not getting what the objection>> is. It is interesting that I do not see anything in your>> correspondences below with Ralph McKenzie or anyone else that even>> mentions me, though.>>Professor McKenzie dismissed your objection claiming that the aOs can>have varying factors of f dependent on m.> > Stating facts not in evidence.> > Did you show him the polynomial you are using, and did you show him my> words? Or did you use one of your simpli'ed examples where the> complications that create your mistake donOt exist?> > Like when you use 2x^2+4x+4 to prove that what you are doing is valid?> > [.snip.]> >Notice how *short* the math argument is. MagidinOs objection would be>that the gOs you see on that page can have varying factors of f>dependent on the value of m,> > Oh, well, if you can tell what my objections will be, I certainly> donOt need to post them any more.> > Too bad you almost never understand what I say, so you cannot be> trusted to know what my objection would be.> >He is a proven liar and the proof is at that link.> > The link proves nothing, least of all anything about my honesty.http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg= 759That link shows the rather simple and basic argument proving my point. > Magidin wants you>to believe in voodoo math, where because he says so, f^2 divides off>in different ways from P(m) depending on the value of m, but you see,>thatOs not mathematics.> > So you claim. Your claim is unfounded.Well then, are you *now* saying that f^2 just divides off withoutregard to what value m is?IOll use whatOs athttp://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759as a reference here.Are you then accepting that two of the gOs should have a factor of f,as follows from whatOs on that page?I say should because the problem with the ring of algebraic integersis that they do not *in the ring of algebraic integers*. >Magidin has proven himself to be against mathematicians, now against>Professor McKenzieOs dismissal of a bogus objection, against algebra,>with his bogus objection, and against you, by continuing to try and>confuse and convince people of falsehoods.> > > Assumes facts that are not in evidence and cannot be substantiated:> you are not a trusted source of reports of other peopleOs> statements. You cannot be trusted to have given an accurate report of> my objections, and you cannot be trusted to be giving an accurate> report of the response you got.> > Your assertions are therefore worthless.Then are you now accepting the result athttp://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759which shows that two of the gOs *should* have a factor that is f?James Harris => >He even shot down the objection that Magidin and others have used for>so long dismissing it almost on sight.> > He did? Did he shoot down my (single?) objection? Did he read ->my<-> words, or was it something you, ehr, explained to him?> > Because you have a track record of just not getting what the objection> is. It is interesting that I do not see anything in your> correspondences below with Ralph McKenzie or anyone else that even> mentions me, though.>>Professor McKenzie dismissed your objection claiming that the aOs can>>have varying factors of f dependent on m.>> >> Stating facts not in evidence.>> >> Did you show him the polynomial you are using, and did you show him my>> words? Or did you use one of your simpli'ed examples where the>> complications that create your mistake donOt exist?>> >> Like when you use 2x^2+4x+4 to prove that what you are doing is valid?>> >> [.snip.]>> >>Notice how *short* the math argument is. MagidinOs objection would be>>that the gOs you see on that page can have varying factors of f>>dependent on the value of m,>> >> Oh, well, if you can tell what my objections will be, I certainly>> donOt need to post them any more.>> >> Too bad you almost never understand what I say, so you cannot be>> trusted to know what my objection would be.>> >>He is a proven liar and the proof is at that link.>> >> The link proves nothing, least of all anything about my honesty.>>http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg= 759That links proves nothing least of all anything about my honesty.>That link shows the rather simple and basic argument proving my point.That link provides the same tired old argument that has been shot downrepeatedly on this newsgroup.>> Magidin wants you>>to believe in voodoo math, where because he says so, f^2 divides off>>in different ways from P(m) depending on the value of m, but you see,>>thatOs not mathematics.>> >> So you claim. Your claim is unfounded.>>Well then, are you *now* saying that f^2 just divides off without>regard to what value m is?I still donOt know just what it is you think divides off means. I amsaying that I have NEVER said f^2 divides off in different ways fromP(m) depending on the value of m.That is nothing but your paraphrase of your understanding of what Imay or may not have said.I have no way of knowing what it was you presented to Prof. McKenzie,either in terms of what I claim or what you claim. For all I know, youwent ahead and used your stupid 2x^2+4x+2 example and claimed I saidsomething about it. Fact is: the values of the aOs depends on m. Whenever, for a speci'cvalue of m, coprime to f, m different from 3, the resulting polynomialis irreducible over Q of degree 3, then none of the aOs are coprime tof.None of your examples satisfy the conditions. In the speci'c case of65x^3 - 12x + 1 (f=u=1, m=5), we have explicitly shown that none of the aOs arecoprime to 5, despite your claims that (a) one of them is; and (b)that our explicit calculations are wrong.>IOll use whatOs at>>http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759>> as a reference here.>>Are you then accepting that two of the gOs should have a factor of f,>as follows from whatOs on that page?By your standards, I could just call you a dishonest git and refuse tolook at anything that you do not post here and now. ThatOs what you doall the time.But letOs look at it.You have the polynomialP(m,x) = f^2( (m^3f^4 -3m^2f^2 +3m)x^3 -3 (-1+mf^2)xu^2 +u^3f )and you factor it asP(m,x) = (a1x+uf)(a2x+uf)(a3x+uf)claiming that a1, a2, a3 can be chosen to be algebraic integers.The aOs are given as the roots ofg(y) = y^3 + 3(-1+mf^2)y^2 - f^2(m^3f^4 -3m^2f^2+3m)[a typo in your reference; the second term lacks the y^2. If you aregoing to refer people there, you should specify upon reference thatthere is an error and what it is]So the aOs are given as functions of m. To verify, we check:We need a1*a2*a3 = f^2(m^3f^4 - 3m^2f^2 + 3m), which is what you getfrom g(y). Good.We need uf(a1*a2 + a1*a3 + a2*a3) to be equal to 0, which is whatyou get from g(y) once your error is corrected. GoodYou need u^2f^2(a1+a2+a3) = -3(-1+mf^2)u^2f^2, so you need a1+a2+a3 =-3(-1+mf^2), which is what you get from g(y). Good.So you de'ne g_1, g_2, and g_3, which are function of m with valuesin the polynomials with algebraic integer coef'cients, asg_1(m) = a_1(m)*x + ufg_2(m) = a_2(m)*x + ufg_3(m) = a_3(m)*x + ufThe you consider P(x,m=0) = f^2(3xu^2 + u^3f)which is a polynomial of degree 1 in x, andfor which we getg_1(0) = ufg_2(0) = ufg_3(0) = 3x+u'ndicating that a_1(0) = 0, a_2(0)=0, and a_3(0)=3.No problem there, as long as you realize that g_1, g_2, and g_3 arefunctiong_i : {values of m} -> {Linear polynomials in X with coef'cients in A[x]}.Then you write:But P(0)/f^2 = 3xu^2 + uf, as f divides off only two of the gOs,while the with the third it is blocked, as long as it is coprime to 3and to x, so assume it is, and assume as well that f is coprime to u.A little late in the game to be throwing assumptions, but what theheck.Then it follows from the constant term that g_1 and g_2 each have afactor that is f.This statement is imprecise as written. If you meanthen it follows that g_1(m) and g_2(m), for each such value of m, aredivisible by f, then I disagree with that statement. It does notfollow from what you have.If you mean Then it follows that g_1(0) and g_2(0) are each multiplesof f, then that statement is both correct and trivial: g_1(0)=u*f,g_2(0)=u*f.All you have done is repeat the same unjusti'ed claim you have beenrepeating ad nauseam for over a year. That because g_1(0) is amultiple of f, then it follows that g_1(m) is ALWAYS a multiple off. You have not proven this to be true. That because g_2(0) is amultiple of f, then it follows that g_2(m) is ALWAYS a multiple off. You have not proven this to be true. That because a_1(0) is notcoprime to f, a_2(0) is not coprime to f, and a_3(0) IS coprime to f,then it follows that a_1(m) and a_2(m) are always not coprime to f,and a_3(m) is always coprime to f. You have not proven this to betrue.So, once again, you waste everyoneOs time by simply repeating the sameset of (a) unjusti'ed assertions; that (b) have been proven wrongexplicitly with the polynomial 65x^3 - 12x + 1.Why do the gOs depends on m? ItOs obvious! You haveg_i(m) = a_i(m)x + ufand you have de'ned a_i(m) to be a root ofy^3 + 3(-1+mf^2)y^2 - f^2(m^3f^4 -3m^2f^2+3m)!Are you trying to claim that the values of a_i donOt change as mchanges?Note also that there is no constant term of g_i with respect to mbecause g_i is NOT a polynomial in m. The value of a_i(m) is de'nedin terms of cubic roots of expressions of m, and the values of thecubic roots cannot be evaluated on their own: you mus choose them insuch a way as to make the choices compatible with a_1, a_2, and a_3(using CardanoOs formulas).It is entirely possible for an algebraic expression involving m to becoprime to some number when m=0, but not coprime to it for othervalues of m.>I say should because the problem with the ring of algebraic integers>is that they do not *in the ring of algebraic integers*.Nonsense. [.snip.]>>Magidin has proven himself to be against mathematicians, now against>>Professor McKenzieOs dismissal of a bogus objection, against algebra,>>with his bogus objection, and against you, by continuing to try and>>confuse and convince people of falsehoods.>> >> >> Assumes facts that are not in evidence and cannot be substantiated:>> you are not a trusted source of reports of other peopleOs>> statements. You cannot be trusted to have given an accurate report of>> my objections, and you cannot be trusted to be giving an accurate>> report of the response you got.>> >> Your assertions are therefore worthless.>>Then are you now accepting the result at>>http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759>> which shows that two of the gOs *should* have a factor that is f?Non sequitur. I do not know what it was you told Prof. McKenzie wasmy claim, but you are not to be trusted to accurately report mywords. Unless you provided him with a transcript of what you said andwhat I replied, your claims about what Prof. McKenzie may or may nothave said about what you think I said are worthless.This above is yet another piece of evidence in my favor: You aremisrepresenting my claims about this matter. Since you are doing sowhile having my precise words before your very eyes, it is clear thatwe cannot assume that you can accurately report my claims at all, letalone to Prof. McKenzie from memory. [Gabriele Rossetti] has left a vast body of writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu>> Because you have a track record of just not getting what the objection>> is. It is interesting that I do not see anything in your>> correspondences below with Ralph McKenzie or anyone else that even>> mentions me, though.> Professor McKenzie dismissed your objection claiming that the aOs can> have varying factors of f dependent on m.What were the *exact words* you used in describing the objection?Because unless you quoted Prof. MagidinOs argument word-for-word,it wasnOt his objection that was dismissed, it was your distortedmisunderstanding of it that was dismissed.-- Wayne Brown | When your tailOs in a crack, you improvisefwbrown@bellsouth.net | if youOre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock => > Maybe, you should stick to nursing in the Army Reserves James.> > LurchI wasnOt a nurse, though I did act as a nurse a few times.You ing loser though, if you want to attack me as an Army veteran,then I can assure you itOs a battle you will lose.The US Army has proven itself time and time again, and I dare you toreply to me on that playing 'eld.I think youOre a ing coward. I dare you to attack me as an Armyveteran.Come on head. Come at me for my military service. Do your best.James HarrisUS ARMY VETERAN => Come on ********. Come at me for my military service. Do your best.You must have an interesting service record. I can just imagine yourevaluations: Spends most of his time hiding under his bunk, cryingfor his Mommy.-- Wayne Brown | When your tailOs in a crack, you improvisefwbrown@bellsouth.net | if youOre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =Wayne Brown scribbled the followingon sci.math:>> Come on ********. Come at me for my military service. Do your best.> You must have an interesting service record. I can just imagine your> evaluations: Spends most of his time hiding under his bunk, crying> for his Mommy.Why do you care about James HarrisOs military service record? He mightbe the greatest war hero ever in the history of the USA, but thatdoesnOt make his *proof* any more *right*. WeOre on sci.math here,not in soc.culture.military.-- /-- Joona Palaste (palaste@cc.helsinki.') | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |-- Finland rules! /When a man talks dirty to a woman, thatOs sexual harassment. When a woman talksdirty to a man, thatOs 14.99 per minute + local telephone charges! - Ruben Stiller => Wayne Brown scribbled the following> on sci.math:>> Come on ********. Come at me for my military service. Do your best.> > You must have an interesting service record. I can just imagine your> evaluations: Spends most of his time hiding under his bunk, crying> for his Mommy.> > Why do you care about James HarrisOs military service record? In this country, we have something called the Constitution. It is thedocument that protects the rights and freedoms of the citizens.It is so important that it is part of the PresidentOs Oath of Of'ce:Presidential Oath of Of'ceI, name, do solemnly swear (or af'rm) that I will faithfully executethe of'ce of President of the United States, and I will to the bestof my ability, preserve, protect, and defend the Constitution of theUnited States. Since the President is the Commander-In-Chief of the US Armed Forces,the military is the means by which the Constitution is defended. Forany member of the military to advocate the overthrow of theConstitution by a foreign power (as Harris has done) is treason.Thus, James (Traitor) HarrisO military service record is as laughableas his physics degree. > He might> be the greatest war hero ever in the history of the USA, but that> doesnOt make his *proof* any more *right*. WeOre on sci.math here,> not in soc.culture.military. => Wayne Brown scribbled the following> on sci.math:> Come on ********. Come at me for my military service. Do your best.>> You must have an interesting service record. I can just imagine your>> evaluations: Spends most of his time hiding under his bunk, crying>> for his Mommy.> Why do you care about James HarrisOs military service record? He might> be the greatest war hero ever in the history of the USA, but that> doesnOt make his *proof* any more *right*. WeOre on sci.math here,> not in soc.culture.military.I donOt care about it. He brought it up, and IOm just pointing out thathe probably was as incompetent in that area as in any other. I suspecthe is incompetent in *every* area of his life.-- Wayne Brown | When your tailOs in a crack, you improvisefwbrown@bellsouth.net | if youOre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =Wayne Brown scribbled the followingon sci.math:>> Wayne Brown scribbled the following>> on sci.math:>> Come on ********. Come at me for my military service. Do your best.> You must have an interesting service record. I can just imagine your> evaluations: Spends most of his time hiding under his bunk, crying> for his Mommy.>> Why do you care about James HarrisOs military service record? He might>> be the greatest war hero ever in the history of the USA, but that>> doesnOt make his *proof* any more *right*. WeOre on sci.math here,>> not in soc.culture.military.> I donOt care about it. He brought it up, and IOm just pointing out that> he probably was as incompetent in that area as in any other. I suspect> he is incompetent in *every* area of his life.Why did you have to point it out? James HarrisOs competence in militaryservice matters exactly *zilch* in this debate. I thought we wereridiculing his proof attempts, not himself.-- /-- Joona Palaste (palaste@cc.helsinki.') | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |-- Finland rules! /Normal is what everyone else is, and youOre not. - Dr. Tolian Soran => Wayne Brown scribbled the following> on sci.math:> Wayne Brown scribbled the following> on sci.math:> Come on ********. Come at me for my military service. Do your best.>> You must have an interesting service record. I can just imagine your>> evaluations: Spends most of his time hiding under his bunk, crying>> for his Mommy.> Why do you care about James HarrisOs military service record? He might> be the greatest war hero ever in the history of the USA, but that> doesnOt make his *proof* any more *right*. WeOre on sci.math here,> not in soc.culture.military.>> I donOt care about it. He brought it up, and IOm just pointing out that>> he probably was as incompetent in that area as in any other. I suspect>> he is incompetent in *every* area of his life.> Why did you have to point it out? James HarrisOs competence in military> service matters exactly *zilch* in this debate. I thought we were> ridiculing his proof attempts, not himself.clearly was losing control of himself and I wanted to see just how farout of control he would go. Unfortunately, this side discussion maywell distract him enough to lose the opportunity. :-(Besides, I leave ridiculing his proof attempts to others moremathematically quali'ed than myself.-- Wayne Brown | When your tailOs in a crack, you improvisefwbrown@bellsouth.net | if youOre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =It turns out that thereOs a group of rather decent people who seem tobe primarily based in Hong Kong who have a math website:http://eng.mathdb.org/They have a great feature in that you can use LaTeX to post whichgives prettier postings:See my work at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759But maybe more importantly they have a top down posting format, wherelater posters get tossed onto the tail, which has foiled a lot ofsci.math posters who for YEARS used a rather frustrating technique onme when I posted.IOd make a post then theyOd pile on with a lot of replies which woulddilute the original message, where IOd have to hope that people readthe original post, knowing that many readers, like Google Groups, putup the *last* post by automatically showing it 'rst, which means thatyahooOs looking to disrupt get equal or higher billing.Entire threads can be hijacked where they run off on a completelydifferent subject than in the original post.However if you go tohttp://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759you will come to my work 'rst and have to scroll down to see thepostings of others.Notice how quiet that thread is.James Harris => It turns out that thereOs a group of rather decent people who seem to> be primarily based in Hong Kong who have a math website:> > http://eng.mathdb.org/> > They have a great feature in that you can use LaTeX to post which> gives prettier postings:> > See my work at http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > But maybe more importantly they have a top down posting format, where> later posters get tossed onto the tail, which has foiled a lot of> sci.math posters who for YEARS used a rather frustrating technique on> me when I posted.> > IOd make a post then theyOd pile on with a lot of replies which would> dilute the original message, where IOd have to hope that people read> the original post, knowing that many readers, like Google Groups, put> up the *last* post by automatically showing it 'rst, which means that> yahooOs looking to disrupt get equal or higher billing.> > Entire threads can be hijacked where they run off on a completely> different subject than in the original post.> > However if you go to> > http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=759> > you will come to my work 'rst and have to scroll down to see the> postings of others.> > Notice how quiet that thread is.> > > James HarrisPerhaps because far fewer people look there than the entire thread. You can then read the thread in chunks of 10 partial posts at a time, with the ability to go into entire replies as necessary.-- Will Twentyman => Notice how quiet that thread is.Yep, itOs quiet because most of us havenOt followed you over there, andthe people who already were there apparently donOt feel youOve postedanything worth talking about.-- Wayne Brown | When your tailOs in a crack, you improvisefwbrown@bellsouth.net | if youOre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock => Notice how quiet that thread is.>>Yep, itOs quiet because most of us havenOt followed you over there, and>the people who already were there apparently donOt feel youOve posted>anything worth talking about.Actually I was a little disappointed to see just now that there weresome replies from others- was a lot more poignant a day or so ago when all there was was the OP and then 4 replies, all from James.************************David C. Ullrich =|many readers, like Google Groups, put up the *last* post by|automatically showing it 'rst,i probably agree more serious defect is that, at least at one time, it wasvirtually impossible to search for other occurrences of certain kindsof data in a thread beyond the 'rst (which perhaps was actually thelast) occurrence found. i donOt know whether that defect was ever'xed. there are a number of other useful than it could be.anyone who tries to argue that the things that i consider defects areactually good features is an evil moron.-- = > > |many readers, like Google Groups, put up the *last* post by > |automatically showing it 'rst, > > > know, I would think that Google Groups is not a reader but anews reader I use (and the newsreaders used by all people I know)-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =James Harris has stated several times in this newsgroup that there havebeen no successful challenges to his work. However, I have successfullychallenged him recently on at least two issues which deserve ahistorical recap, so that the record is not falsi'ed by his claim thatthere have been none.I. James claimed repeatedly that it is possible for abc = 5, where OaO,ObO and OcO are non-unit algebraic integers, and where one or more ofthe terms on the left side is coprime to 5. I showed, also repeatedly,that dividing both sides by OaO, ObO or OcO results in an algebraicinteger consisting of the product of two algebraic integers on the left,and a quotient, which must therefore be the same algebraic integer, onthe right. Hence, each of the terms on the left divides 5 in the ring ofalgebraic integers. Hence NONE of the terms on the left is coprime to to5, only assuming that all of them are algebraic integers and none ofthem are unity.This is more than a challenge to his claim, it is a conclusiverole this claim holds in establishing his claim that the ring ofalgebraic integers is §awed is lost.II. James repeatedly claimed that there are numbers which should be inthe ring of algebraic integers, but which are not. I have, alsorepeatedly, challenged him to produce one such number. Any one. Onlyone. Just one number (numeric value) which should be in the ring ofalgebraic numbers but which is not. He has never produced such a number.This challenge will continue to go unanswered by him because THERE ARENO SUCH NUMBERS. The ring of algebraic integers is bona 'de ring.Nothing is missing.There you go James. Two successful challenges. Refute that!--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com => II. James repeatedly claimed that there are numbers which should be in> the ring of algebraic integers, but which are not. I have, also> repeatedly, challenged him to produce one such number. Any one. Only> one. Just one number (numeric value) which should be in the ring of> algebraic numbers but which is not. He has never produced such a number.> This challenge will continue to go unanswered by him because THERE ARE> NO SUCH NUMBERS. The ring of algebraic integers is bona 'de ring.> Nothing is missing.Please, I implore, to all who percieve that I am sophmoric, to forgiveme. However, perhaps I am mistaken, but arenOt the only things thatshould be in something, already de'ned to be in that thing? Forinstance, how could I try to argue another number into the Set ofRational Numbers, or Positive Integers, or perfect squares, that isnOtalready in there? => II. James repeatedly claimed that there are numbers which should be in> the ring of algebraic integers, but which are not. I have, also> repeatedly, challenged him to produce one such number. Any one. Only> one. Just one number (numeric value) which should be in the ring of> algebraic numbers but which is not. He has never produced such a number.> This challenge will continue to go unanswered by him because THERE ARE> NO SUCH NUMBERS. The ring of algebraic integers is bona 'de ring.> Nothing is missing.>> Please, I implore, to all who percieve that I am sophmoric, to forgive> me. However, perhaps I am mistaken, but arenOt the only things that> should be in something, already de'ned to be in that thing? For> instance, how could I try to argue another number into the Set of> Rational Numbers, or Positive Integers, or perfect squares, that isnOt> already in there?Beats me. Ask James.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =It asks to identify the c-level surface of: f(x,y,z) = 9x^2 - 4y^2 + 36z^2where i) c < 0, ii) c = 0, iii) c > 0 -c /36 = x^2/2^2 - y^2/3^2 + z^2None of the weird surface equations in my book match it. The best Icould 'nd was a hyperboloid of 2 sheets.For ii, I ended up with an elliptic cone.And for iii, since the only change is -c to +c, I made it ahyperboloid of 1 sheet.Am I correct?And another question asks:V(x,y) = k/[(r^2 - x^2 - y^2)^(1/2)]k and r are positive constants. What are the level curves of V.Would it become a circle with something like this:r^2 - x^2 - y^2 = k^2/c^2x^2 + y^2 = r^2 - k^2/c^2 =>It asks to identify the c-level surface of:> f(x,y,z) = 9x^2 - 4y^2 + 36z^2>>where i) c < 0, ii) c = 0, iii) c > 0>> -c /36 = x^2/2^2 - y^2/3^2 + z^2>>None of the weird surface equations in my book match it. What are you looking for in the way of matching? Coef'cientby coef'cient identity?>The best I>could 'nd was a hyperboloid of 2 sheets.If you explain what measure of goodness you are using tosay that this is The best you could 'nd, then probablysome knowledgeable person here will be able to say if youOreright or wrong. (Does your textbook, refered to above as It,spell out what it would *mean* to identify the c-levelsurface? Does it give a general de'nition of hyperboloidof 2 sheets, elliptic cone, or hyperboloid of 1 sheet?I wouldnOt be surprised if the answers to both questions are no.)>For ii, I ended up with an elliptic cone.>>And for iii, since the only change is -c to +c, I made it a>hyperboloid of 1 sheet.>>Am I correct?>>And another question asks:>>V(x,y) = k/[(r^2 - x^2 - y^2)^(1/2)]>>k and r are positive constants. What are the level curves of V.>>Would it become a circle with something like this:>>r^2 - x^2 - y^2 = k^2/c^2>x^2 + y^2 = r^2 - k^2/c^2Sure.Lee RudolphHereOs a process that de'nes a structure - IOd like to know whether ornot itOs a fractal, and if so, how to calculate its dimension.The domain of the structure is the unit square 0 <= x < 1, 0 <=y <1.First add the lines x=1/3 and x=2/3Then add the following lines with y< 1/2: x=1/9, x=2/9, x=4/9, x=5/9, x=7/9, x=8/9WhatOs in the unit square now looks like this: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |At the next stage add lines with y<1/4, and x=1/27, 2/27, 4/27...26/27, getting this: | | | | | | | | | | | | | | | | | | | | | | | ||||||||||||||||||||||||||||||||||||||||||||||||| At each successive stage you add the lines y< 1/2^n, x =a/b where a is all the integers up to 3^n not equal to 0 mod 3, and b = 3^nDoes the resulting 'gure (the limit as n->oo) have fractal dimension? Can we work out its area?Sue => HereOs a process that de'nes a structure - IOd like to know whether or> not itOs a fractal, and if so, how to calculate its dimension. The domain of the structure is the unit square 0 <= x < 1, 0 <=y <1.> > First add the lines x=1/3 and x=2/3> > Then add the following lines with y< 1/2: x=1/9, x=2/9, x=4/9, x=5/9, > x=7/9, x=8/9> > WhatOs in the unit square now looks like this:> > > | |> | |> | |> | |> | | | | | | | |> | | | | | | | |> | | | | | | | |> | | | | | | | |> > At the next stage add lines with y<1/4, and x=1/27, 2/27, 4/27...26/27, > getting this:> > | |> | |> | |> | |> | | | | | | | |> | | | | | | | |> ||||||||||||||||||||||||> |||||||||||||||||||||||| > > At each successive stage you add the lines y< 1/2^n, x =a/b where a is all > the integers up to 3^n not equal to 0 mod 3, and b = 3^n Does the resulting 'gure (the limit as n->oo) have fractal dimension? Can > we work out its area?Ok IOve realised, I think, that the 'gure will not have any area, since it will be too sparse - there will only be lines with rational fractal? It seems self-similar, in that at every scale it exhibits two things then one thing twice as big. =>> HereOs a process that de'nes a structure - IOd like to know whether or>> not itOs a fractal, and if so, how to calculate its dimension. >> The domain of the structure is the unit square 0 <= x < 1, 0 <=y <1. >> First add the lines x=1/3 and x=2/3 >> Then add the following lines with y< 1/2: x=1/9, x=2/9, x=4/9, x=5/9, >> x=7/9, x=8/9... >> At the next stage add lines with y<1/4, and x=1/27, 2/27, 4/27...26/27, >> getting this:...>> At each successive stage you add the lines y< 1/2^n, x =a/b where a is all >> the integers up to 3^n not equal to 0 mod 3, and b = 3^n >> Does the resulting 'gure (the limit as n->oo) have fractal dimension? Can >> we work out its area?>Ok IOve realised, I think, that the 'gure will not have any area, since >it will be too sparse - there will only be lines with rational >x-coordinates, and no irrational x-coordinates.>But is it a fractal? It seems self-similar, in that at every scale it >exhibits two things then one thing twice as big.It is not fractal in the sense of non-integer Hausdorff dimension: this object obviously has sigma-'nite 1-dimensional Hausdorff measure, and so its Hausdorff dimension is 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => > It is not fractal in the sense of non-integer Hausdorff dimension: > this object obviously has sigma-'nite 1-dimensional Hausdorff measure, > and so its Hausdorff dimension is 1.> It is fractal in the sense of being self-similar. (Or, moreprecisely in this case, self-af'ne.)These two answers show how fractal can be an imprecise term.-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ =In fact, if f is an L^1 function, you can say thatlim ||u||=0as t gets large. Try that one as an exercise.BDG>>Hmmm...I see why making the change of variables makes sense, but now I>canOt proceed from>>||u|| = sup (sum) |1/sqrt(4 pi t) INT(-oo,oo)>e^(-y^2/(4t))[f(x_{i+1}-y)-f(x_i-y)] dy|.>>If f was in L^1, it seems like I could use some sort of Cauchy-Schwarz>inequality in the integral, but itOs still a mystery how to end up>comparing ||u|| to ||f|| rather than ||INT(-oo,oo) f||...>> Well IOll just do it then - make certain to mention my name when> you hand it in:>> |u(x_1, t) - u(x_2, t)| + ...>> <= 1/sqrt(4 pi t) INT(-oo,oo) e^(-y^2/(4t))> (|f(x_1-y)-f(x_2-y)| + ...) dy>> <= 1/sqrt(4 pi t) INT(-oo,oo) e^(-y^2/(4t)) ||f|| dy>> = ||f|| .>De'ne the total variation norm || || of a function on the line to be>>||f|| = sup( sum over all j |f(x_{j+1})-f(x_j)| ),>> _where_ x_1 < x_2 < ... < x_n.>>How do you show that, if u(x,t) is the solution to the heat equation>>u_t=u_{xx}, u(x,0)=f(x), then u satis'es>>||u(.,t)|| <= ||f|| ?>>I know that u(x,t) = 1/sqrt(4 pi t) int(-oo,oo) exp( -(x-y)^2/(4t) )>>f(y) dy, but with the x in the exponent I havenOt been able to make>>||u(.,t)|| look like anything I can start writing inequalities>>with?any suggestions?>> Make a change of variables, to show that>> u(x,t) =>> 1/sqrt(4 pi t) int(-oo,oo) exp( -y^2/(4t) ) f(x-y) dy .>> Use the fact that>> 1/sqrt(4 pi t) int(-oo,oo) exp( -y^2/(4t) ) dy = 1.>> ************************>> David C. Ullrich>> ************************>> David C. Ullrich =>In fact, if f is an L^1 function, you can say that>>lim ||u||=0>>as t gets large. Try that one as an exercise.Well thatOs easy - dunno if itOs in fact, because itdoesnOt give the inequality in the OP (the inequalitythe OP asked about is not true unless f has boundedvariation):If ||.|| is the total variation norm and * is convolutionthen in general ||F*G|| <= ||F||_1 ||G||, as the argumentbelow shows. To get the inequality the OP wanted youlet G = f and F = g_t, where g_t(x) = 1/sqrt(4 pi t) e^(-x^2/(4t)),while to get the inequality you mention you let F = fand G = g_t (and note that ||g_t|| -> 0.)>BDG>>Hmmm...I see why making the change of variables makes sense, but now I>>canOt proceed from>>||u|| = sup (sum) |1/sqrt(4 pi t) INT(-oo,oo)>>e^(-y^2/(4t))[f(x_{i+1}-y)-f(x_i-y)] dy|.>>If f was in L^1, it seems like I could use some sort of Cauchy-Schwarz>>inequality in the integral, but itOs still a mystery how to end up>>comparing ||u|| to ||f|| rather than ||INT(-oo,oo) f||...>> Well IOll just do it then - make certain to mention my name when>> you hand it in:>> |u(x_1, t) - u(x_2, t)| + ...>> <= 1/sqrt(4 pi t) INT(-oo,oo) e^(-y^2/(4t))>> (|f(x_1-y)-f(x_2-y)| + ...) dy>> <= 1/sqrt(4 pi t) INT(-oo,oo) e^(-y^2/(4t)) ||f|| dy>> = ||f|| .>>De'ne the total variation norm || || of a function on the line to be>>||f|| = sup( sum over all j |f(x_{j+1})-f(x_j)| ),>> _where_ x_1 < x_2 < ... < x_n.>How do you show that, if u(x,t) is the solution to the heat equation>u_t=u_{xx}, u(x,0)=f(x), then u satis'es>>||u(.,t)|| <= ||f|| ?>>I know that u(x,t) = 1/sqrt(4 pi t) int(-oo,oo) exp( -(x-y)^2/(4t) )>f(y) dy, but with the x in the exponent I havenOt been able to make>||u(.,t)|| look like anything I can start writing inequalities>with?any suggestions?>> Make a change of variables, to show that>> u(x,t) => 1/sqrt(4 pi t) int(-oo,oo) exp( -y^2/(4t) ) f(x-y) dy .> Use the fact that>> 1/sqrt(4 pi t) int(-oo,oo) exp( -y^2/(4t) ) dy = 1.> ************************>> David C. Ullrich>> ************************>> David C. Ullrich>************************David C. Ullrich => http://www.behavior.org/journals_BP/2000/Place.pdf> > There is quite a large literature on social conformity which could be > referred to, but whilst of some heuristic value (in the helpful, > illustrative sense) I really 'nd that social psychological literature > relatively trivial compared to the rather detailed and powerful work > which has been done for decades in what I rather loosely refer to in > this newsgroup as behaviour analysis and learning theory.'nd that very conforming to hear again.Longley Citation Statistics [8 long years and counting]-- --folk psychology [derogatorily] in over 1000 postsQuine [glowingly] in over 1800 postssilly [regarding othersO ideas] over 400 timesSkinner [in awe] in over 800 postsbehaviorism [as accepted dogma] in over 500 posts[pernicious] mentalism over 100 times[pernicious] cognitivism over 250 timesintensional [as wrongish] over 1600 timesextensional [as correctish] over 1850 timesfrag.html [self-promotion] over 300 times => > In the meantime the OEIS entry> http://www.research.att.com/projects/OEIS?Anum=A085000 has> been updated twice. There are two minor glitches in the updated page:> The estimate given by Tim Paulden for a(6) was> a(6) is at least 1862125166. and not 1865999570 (which is the value> IOve found a little bit later with my program). The example given for> the> 3X3 matrix says:> The following 3 X 3 matrix is one of 72 whose determinant is 412:> There are only 36 matrices with det(A_3)=412, the other 36 matrices have> det(A_3)=-412.Both things have been corrected.Hugo =How do I calculate the lowest common multiple of three numbers.a) if the numbers are integersb) if the numbers are fractions eg 11.5 14 and 16.5-- _______________________ /_____________________(_) | | |__________________(_) Peter_Morris_1 at | |/____________________ btinternet dot com |_____________________(_) => How do I calculate the lowest common multiple of three numbers.> > a) if the numbers are integersLCM(x,y,z) = LCM(LCM(x,y),z)> > b) if the numbers are fractions eg 11.5 14 and 16.5Depends on what you mean by LCM in this case. What do you consider to be multiples of a fraction? Do you mean only integer multiples?If so, then convert all your values to rationals with a least common denominator, 'nd the LCM of the new numerators over the least common denominator.> > --> _______________________> /_____________________(_)> | | |__________________(_) Peter_Morris_1 at> | |/____________________ btinternet dot com> |_____________________(_)> > > =>> How do I calculate the lowest common multiple of three numbers.>> a) if the numbers are integers>> LCM(x,y,z) = LCM(LCM(x,y),z)>> b) if the numbers are fractions eg 11.5 14 and 16.5>> Depends on what you mean by LCM in this case. What do you consider> to be multiples of a fraction? Do you mean only integer multiples?> If so, then convert all your values to rationals with a least common> denominator, 'nd the LCM of the new numerators over the least> common denominator.I want a number P such that :P= 11.5xP= 14yP=16.5zwhere x,y,z are all integers.or, to put it another wayStarting at time 0 events x, y & z occur.event x repeats every 11.5 seconds.event y repeats every 14 secondsevent z repeats every 16.5 seconds.How long will it be before all 3 events occur simultaneously again?give the answer, and show the working, please.> --> _______________________> /_____________________(_)> | | |__________________(_) Peter_Morris_1 at> | |/____________________ btinternet dot com> |_____________________(_)> => I want a number P such that :> > P= 11.5x> P= 14y> P=16.5z> > where x,y,z are all integers.> > or, to put it another way> > Starting at time 0 events x, y & z occur.> > event x repeats every 11.5 seconds.> event y repeats every 14 seconds> event z repeats every 16.5 seconds.> > How long will it be before all 3 events occur simultaneously again?> give the answer, and show the working, please.How about this: First let P be the lcm of the numerators of your fractions. Then calculate the corresponding x, y, and z. Then divide P by the gcd of x, y, and z, and use that quotient as P.-- => How do I calculate the lowest common multiple of three numbers.>> a) if the numbers are integers>> LCM(x,y,z) = LCM(LCM(x,y),z)>> b) if the numbers are fractions eg 11.5 14 and 16.5>> Depends on what you mean by LCM in this case. What do you consider>> to be multiples of a fraction? Do you mean only integer multiples?>> If so, then convert all your values to rationals with a least common>> denominator, 'nd the LCM of the new numerators over the least>> common denominator.>>I want a number P such that :>>P= 11.5x>P= 14y>P=16.5z>>where x,y,z are all integers.>>or, to put it another way>>Starting at time 0 events x, y & z occur.>>event x repeats every 11.5 seconds.>event y repeats every 14 seconds>event z repeats every 16.5 seconds.>>How long will it be before all 3 events occur simultaneously again?>give the answer, and show the working, please.>I would do the following: lcm( lcm(11.5*10,14*10), 16.5*10)/10 so thatall 3 numbers are integers so you can use the lcm algorithm.I get an answer of lcm=10626 and x=924 y=759 z=644. Is it really thelowest, I donOt know :)[ lcm(a,b) can be calculated by a*b/gcd(a,b) where gcd is for exampleEuclidOs greatest common divisor algorithm. ] =>How do I calculate the lowest common multiple of three numbers.>>a) if the numbers are integersThe lcm of a, b, and c is the lcm of the lcm of a and b, and c.[a,b,c] = [ [a,b],c]where [x,y] = lowest common multiple of x and y.The lcm can be calculated by using the gcd:[x,y] = |xy|/(x,y)where |xy| is the absolute value of xy, and (x,y)= greatest commondivisor of x and y.The greatest common divisor of x and y can be found by using theEuclidean algorithm.>b) if the numbers are fractions eg 11.5 14 and 16.5Then it makes, a priori, no sense to talk about lowest commonmultiple. If you allow fractions, then any fraction is a commonmultiple of all of them.So you must want something different. I am guessing the following willaccomplish what you want:Write each as a reduced fraction,a r p-, -, -b s qFind the lowest common multiple of b, s, and q (letOs call it x), andrewrite each fraction as a fraction with that denominator:c t o-, -, -x x xNow 'nd the smallest common multiple of c, t, and o. Call it y. Theny/x is the number you are looking for (IOm guessing). = selective about what I accept as reality. Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu> Ullrich of the world, smart people with good credentials,> who got on the naive side of the question, are so intent to remain> there.> > Because the Ullrichs of the world essentially make it very dif'cult for> any crackpot to pass through this (particular) forum. Or if you want it> more bluntly put, in order to pass through those various Ullrichs, you> have to 'netune your math in advance, so as to not only make sense, but> to be quite elegant as well.> > As such, in general, they reduce the noise to ratio volume.> > I am not saying you are a crackpot (because IOve only browsed very> quickly through this thread, so I wouldnOt know anyway), just offering a> viable explanation for the infamous so called Ullrich 'lter in> sci.math.> > ItOs even caught me, once, claiming (indirectly) that I was a crackpot> who was studying the analytic continuation of the x^y function. Of> course the 'lter missed the extra ^. So there. EldonIOm an airplane driver, not a mathematician, have no reputation touphold and donOt care if they call me a crackpot. Ullrich calls me adumb , from him, thatOs okay.I refuse to believe they are confused by the non elegance of my math.IOve caught them on the naive side of a counter intuitive question andthey donOt wish to admit it.Dr. Holt has admitted that I have a correct solution, he isnOtconvinced that itOs unique. Dr. Ullrich hasnOt, as of yet, recognizedmy solution.Then, Dr. Holt, said that my correct solution, is not as plausible asDr. UllrichOs interpretation, which can be made because he can assumethat the problem statement isnOt true. He has deferred to a panel ofexperts to decide which assumptions can be made.Dr. Ullrich has refused to argue the case because he has a majorityopinion somewhere that the words at least one is mean somethingother than what they say.I have knocked out the Oassuming the statement isnOt trueO argument.So far, no capitulation from either doctor. They donOt seem to havecomplete agreement between themselves.IOm the dumb ?Eldon:) =ADD the following to the list:The FCT for graphs is fundamentally different from the FCT for maps.Maps have regions which have sides; graphs do not have regions.Maps are always planar; graphs are seldom planar!Maps are two dimensional; graphs are one dimensional! > SOME THINGS YOU NEED TO KNOW ABOUT THE FOUR COLOR THEOREM!!!> > DEFINITIONS: > CHI(G) is the chromatic number of graph G. > If CHI(G) <= k, then graph G is k-colorable> If CHI(G) = k, then graph G is k-chroma.> > A graph is planar if it contains no subgraph homeomorphic to> K5 or K3,3. > > Every maximal planar graph [MPG] has exactly 3n-6 edges.> > If every MPG is 4-colorable; then every planar graph is 4-colorable.> > All MPGOs are generated by the intersection of two triangulations of an > n-sided polygon.> > Every triangulation of every convex polygon is 3-colorable.> > All 4-partite graphs, both planar and non-planar, are 4-colorable. =>WhatOs written below is essentially a rant because poor old Larry has>great dif'culties with objectivity. This is a common problem in people>who donOt have any formal training in psychology, and perhaps>understandably so.>The dilemma is this - if a discussion is so abstract that people have>dif'culty seeing what it really refers to, many untrained folk will>just regard the material as academic. If, on the other hand, it is>presented in a much more direct way, this can be quite disconcerting and>OpersonalO as readers see fragments of their own behaviour being>examined and publicly exposed whether it really is addressing their>personal behaviour directly or not.>[...]David, I am not hanging out in the AI newsgroups much,however I would like to respond a bit:1) Do not assume that everyone else or anyone else has no formaltraining in Psychology.2) Do not assume that anyone or everyone will respond to directbehavioural descriptions by personalizing the topic.In particular, you are making assumption 1 about peopleabout whom I strongly believe it is incorrect, and youappear to be making assumption 2 to de§ect criticismrather than as a legitimate objective debating point.-george william herbertgherbert@retro.com =>>WhatOs written below is essentially a rant because poor old Larry has>>great dif'culties with objectivity. This is a common problem in people>>who donOt have any formal training in psychology, and perhaps>>understandably so.>>The dilemma is this - if a discussion is so abstract that people have>>dif'culty seeing what it really refers to, many untrained folk will>>just regard the material as academic. If, on the other hand, it is>>presented in a much more direct way, this can be quite disconcerting and>>OpersonalO as readers see fragments of their own behaviour being>>examined and publicly exposed whether it really is addressing their>>personal behaviour directly or not.>>[...]>>David, I am not hanging out in the AI newsgroups much,>however I would like to respond a bit:>>1) Do not assume that everyone else or anyone else has no formal>training in Psychology.I donOt.>>2) Do not assume that anyone or everyone will respond to direct>behavioural descriptions by personalizing the topic.>I donOt.>In particular, you are making assumption 1 about people>about whom I strongly believe it is incorrect, and you>appear to be making assumption 2 to de§ect criticism>rather than as a legitimate objective debating point.>No - these are *your* assumptions.Have you bothered to read anything IOve referenced?-- David Longley => Sizemore and I have independently tried to short circuit this type of > behaviour in c.a.p, not only in Michaels, but in Ozkural and one or two > others who post to the newsgroup. Actually, Longley has engaged in 1000s [yes, thousands] of suchinterchanges over the past 8 years, with 100s of different people onc.a.p.ItOs simply his pattern to harass people who talk about standard AItopics like cognition and the mind.Longley Citation Statistics [8 years and counting]-- folk psychology [derogatorily] in over 1000 postsQuine [glowingly] in over 1800 postssilly [regarding othersO ideas] over 400 timesSkinner [in awe] in over 800 postsbehaviorism [as accepted dogma] in over 500 posts[pernicious] mentalism over 100 times[pernicious] cognitivism over 250 timesintensional [as wrongish] over 1600 timesextensional [as correctish] over 1850 times =Message 2 in thread Read the rest of this message... (707 more lines) =Mr. Michaels: As you have correctly perceived, Longley is not a guy withquirky andfuturistic ideas - a premature mindset in Gunter Stentian terms -but simply someone clinging to 80 year old ideas which the vastmajority of people in science moved past decades ago. It is very clearthat those ideas are not going to help solve the AI problems oftomorrow - which is what c.a.p is meant to discuss in the 'rst place.Dr. Sizemore: As I have pointed out repeatedly, the majority of postersaround here do not have any clue as to what constitutes valid argumentation.Here we see the common fallacy: behaviorism is a minority position,therefore behaviorism is wrong.>> Mr. Longley has the problem that he is quite possibly> entirely correct, in the context of the 1950Os and the> psychology of mind as understood then, but is not making> a meaningful contribution in the context of the 2000Os> and the software problem of an AI mind.>> He is exactly a broken clock.>> As you have correctly perceived, Longley is not a guy with quirky and> futuristic ideas - a premature mindset in Gunter Stentian terms -> but simply someone clinging to 80 year old ideas which the vast> majority of people in science moved past decades ago. It is very clear> that those ideas are not going to help solve the AI problems of> tomorrow - which is what c.a.p is meant to discuss in the 'rst place.> => Mr. LongleyOs problem is _not_ that he has an innovative> viewpoint which is having trouble being accepted by a> is in the position of those who doubted Einstein, trying to> cling to an outmoded mindset which has been overtaken by> events. He is 'ghting a battle to prevent progress on any> other terms than his, he is utterly condemned to failure,> and meanwhile his chattering, drivel, and repetitive> negativism waste the time of people who could be making> positive contributions to the develoment of AI using modern> knowledge not much connected to the detailed understanding> of carbon-based intelligence.>> The following shows the extent of these contributions:>> Longley Citation Statistics> > folk psychology [derogatorily] in over 1000 posts> Quine [glowingly] in over 1800 posts> silly [regarding othersO ideas] over 400 times> Skinner [in awe] in over 800 posts> behaviorism [as accepted dogma] in over 500 posts> [pernicious] mentalism over 100 times> [pernicious] cognitivism over 250 times> intensional [as wrongish] over 1600 times> extensional [as correctish] over 1850 times> => He tells people _doing_ AI, that they are doing it entirely> wrong, despite that he has no contributions of his own, in> software on silicon, which he can present as a superior> alternative,>> This point has been made countless times to no effect. =>And what if you (and others) have both misunderstood what behaviourism >is and wonOt be corrected?Do you understand what behaviourism is?Do you understand what its underlying assumptions andrelevant limitations are?You are displaying a number of symptoms of OHammer SyndromeO.-george william herbertgherbert@retro.com =GWH: Do you understand what behaviourism is?Do you understand what its underlying assumptions andrelevant limitations are?You are displaying a number of symptoms of OHammer SyndromeO.GS: Suppose you tell us what behaviorism is, what the underlying assumptionsare, and its limitations. Wanna bet that I can show you to be wrong?>And what if you (and others) have both misunderstood what behaviourism>is and wonOt be corrected?>> Do you understand what behaviourism is?>> Do you understand what its underlying assumptions and> relevant limitations are?>> You are displaying a number of symptoms of OHammer SyndromeO.> -george william herbert> gherbert@retro.com> => Mr. Michaels: As you have correctly perceived, Longley is not a guy with> quirky and> futuristic ideas - a premature mindset in Gunter Stentian terms -> but simply someone clinging to 80 year old ideas which the vast> majority of people in science moved past decades ago. It is very clear> that those ideas are not going to help solve the AI problems of> tomorrow - which is what c.a.p is meant to discuss in the 'rst place.> > Dr. Sizemore: As I have pointed out repeatedly, the majority of posters> around here do not have any clue as to what constitutes valid argumentation.> Here we see the common fallacy: behaviorism is a minority position,> therefore behaviorism is wrong.> No, the argument was behaviorism is a miniscule position, thereforebehaviorism is irrelevant. =Ad populum.. Exactly my point. And you revel in it. Amazing.> Mr. Michaels: As you have correctly perceived, Longley is not a guy with> quirky and> futuristic ideas - a premature mindset in Gunter Stentian terms -> but simply someone clinging to 80 year old ideas which the vast> majority of people in science moved past decades ago. It is very clear> that those ideas are not going to help solve the AI problems of> tomorrow - which is what c.a.p is meant to discuss in the 'rst place.>> Dr. Sizemore: As I have pointed out repeatedly, the majority of posters> around here do not have any clue as to what constitutes validargumentation.> Here we see the common fallacy: behaviorism is a minority position,> therefore behaviorism is wrong.> No, the argument was behaviorism is a miniscule position, therefore> behaviorism is irrelevant. => Ad populum.. Exactly my point. And you revel in it. Amazing.>> No, the argument was behaviorism is a miniscule position, therefore>> behaviorism is irrelevant.The question here isnOt about popularity, itOs about relative WORTH ascompared to other concepts.bp <947208bc88d54dd60b374e12b098cec3@news.teranews.com> =>> Mr. Michaels: As you have correctly perceived, Longley is not a guy with>> quirky and>> futuristic ideas - a premature mindset in Gunter Stentian terms ->> but simply someone clinging to 80 year old ideas which the vast>> majority of people in science moved past decades ago. It is very clear>> that those ideas are not going to help solve the AI problems of>> tomorrow - which is what c.a.p is meant to discuss in the 'rst place.>> Dr. Sizemore: As I have pointed out repeatedly, the majority of posters>> around here do not have any clue as to what constitutes valid argumentation.>> Here we see the common fallacy: behaviorism is a minority position,>> therefore behaviorism is wrong.>>No, the argument was behaviorism is a miniscule position, therefore>behaviorism is irrelevant.And what if you (and others) have both misunderstood what behaviourism is and wonOt be corrected?-- David Longley =This was posted in a new thread, but apparently you didnOt see it.Again, avoiding -istics and -isms.So Mr. Longley. I have read your material. I reviewed some of the history ofthis. And hereOs what I think:When looked at from a qualitative point of view, thereOs no doubt about thequality of your use of language. You are a superb writer, in that you havegood command of the english language to forward your own purpose.However...it seems to me that in your drive to forward your purpose, youtend to ignore ALL evidence that even remotely casts doubt on your owntheories/postulations, even when many others in the community forward themin various states of conversational mode, everything from polite toviolent.As to the quality of your CONTENT...IOm simply a layman in this area. Itdoes, however, seem to me that you rely far too much on terminology thatyouOve designated as applying correctly to one context, and one contextonly. Deviation from this self-dictated norm seems to result in tirades ofnearly incomprehensible ire.As for quantity...The sheer quantity of posts brimming with invective, ire,condescension, insult, and otherwise trivialization of all who disagree isfrightening in its immensity. I believe the term Low signal-to-noiseratio would apply.Secondly for quantity: It would seem that the quantity of people whoOveattempted in various methods to help, cajole, drive, force, or otherwisein§uence your theories would be enough of a indicator that a reasonableperson would, in the face of such evidence, cause a degree of introspection(something with you have innumerable times blamed uncountable people of,apparently).YouOll notice I say it would seem a lot. ThatOs precisely what happens.The semblance of these things, while perhaps not the objective reality ofthe situation, surely do count towards the vector sum of the opinionsarrayed against you, and most certainly some of the blame does lie on you.So thatOs what I think.bp <3LOdnYcq9oVxiOmiU-KYuA@metrocast.net> <3f74bc38_1@news.iprimus.com.au> <3F78CDB7.7EFD621B@sonic.net> =WhatOs written below is essentially a rant because poor old Larry hasgreat dif'culties with objectivity. This is a common problem in peoplewho donOt have any formal training in psychology, and perhapsunderstandably so.The dilemma is this - if a discussion is so abstract that people havedif'culty seeing what it really refers to, many untrained folk willjust regard the material as academic. If, on the other hand, it ispresented in a much more direct way, this can be quite disconcerting andOpersonalO as readers see fragments of their own behaviour beingexamined and publicly exposed whether it really is addressing theirpersonal behaviour directly or not.In reality, the behaviours are shared by millions - the fact that Larry,Eray, Neil, Patty, Kent etc etc share those behaviours is incidental. Inorder to bring the vicissitudes of the intensional heuristics into boldrelief it is useful to highlight some of the more obvious logicalfailures of intensionality in the newsgroup.Where this tends to become problematic is where some individuals justrespond to whatOs been said disparagingly. Both Glen and I have indeedreprimanded others for some types of behaviour - these are invariablyresponses to quite egregious posts from people who have arrogantlyasserted that they understand what has been explained to them, that itis trivially obvious, or something like that - and yet then go on toshow very clearly that they havenOt understood a thing!! - Fragmentsis very much about all of this - hence the title and the centrality ofreferential opacity and LeibnizOs Law.I suggest that the individuals who are currently moaning and complainingtake Otime-outO, re-read the papers I have referenced (instead ofcounting words) - particularly Fragments (and perhaps the papers onCognitive Skills), and that they give some serious thought to how andwhy cognitivism may not be what they think it is. oOo>FD621B@sonic.net>...>> There are 3 or 4 contrary cases. One that still bothers me was a> fairly good piece of creative sarcasm he did. All the right>inferences> pulled out, best way (I thought) to refute the argument addressed.>I> just 'gure he must have stayed up all night with that one, or just> possibly he remembered it from a book. The others could be 100:1> accidents considering his volume. OTOH, I always require a theory>to> 't all the facts, so really canOt let that one piece go. That and>one> other post by themselves are the reason it took me so long to>question> my beliefs about Longley (I was his greatest defender for a while).>>It is the contrary cases, I think, that can give you the best>>information about what hypotheses 't or donOt 't. One of the>>best writers I have ever known would wake up in the morning with>>one of two different brains; when he was OupO, he was brilliant.>>He had fantastic skills of synthesis and could pull together>>words in a way that was pure poetry and, at the same time, capital-T>>True. Of course, heOd also write a lot of capital-C Crap when he>>was OupO; it was hit-or-miss.>>when he was OdownO, he was incapable of performing at that>>level, but could analyze and dissect arguments, do derivations and>>technical writing, and edit the Crap out of what he had written when>>he was OupO. He was generally unhappy in his OdownO state, and had>>little or no emotional attachment to anything. Altogether, he was>>brilliant, but bipolar. When he took his medication, he was, all>>told, a happier person, but unable to untangle arguments or>synthesize>>or create at anything like his unmedicated levels. Sometimes heOd>>change meds and a bit of his OmagicO would reappear for a month or>>three, but then it always faded again. Finally, he decided heOd>>rather be brilliant than happy and gave up the meds entirely. As>>he said, Without my crutch, I fall down a lot. But with it, I can>>never §y.>>Had he routinely taken his medication, I doubt that heOd have>>or, years later, committed suicide. :-|>>My point in bringing this up, of course, is that brilliance is>>sometimes unstable. Longley in his usual case sounds a lot like my>>departed friend when he was on his meds; itOs all there, but itOs>>not 'ring together or being synthesized correctly, and the mind>>winds up falling back on well-rehearsed tracks that it can maintain>>without the active need for the facilities which are blunted. The>>existence of a few contrary posts, where things seem to be coming>>together, may indicate moments when brain chemistry, for whatever>>reason, was different and the blunted facilities were sharp again.>>considered before. This makes a great deal of sense to me re: Longley>because it solves two problems. First are the contrary cases, which I>agree are as important as you say. Those cases are not a matter of>degree like changes in mental capacity thru the day or day-to-day. At>least I have variously observed (more-or-less) his quality of analysis>to be very stable around §oor level. Then, in about 1 in 100-150>cases, itOs like another person at a second level. Damn, if I had been>thinking about this I would have noticed if it happened in multiple>posts on the same day. Well, another plus for de'ning the problem. (I>should have speculated more wildly?) Never seen anything remotely brilliant, no more than average>analytical with superior memory (BTW, I just grant superior memory>because it is irrelevant to my case and others have remarked on it,>but *IOve* never seen it demonstrated.) Infrequently better is not>the same as infrequently worse. Medication seems the most obvious>explanation for infrequently better. It goes without saying that we>are speculating and there are lots of alternative possibilities. Most>obvious only means most obvious, blah, blah, blah.>>>Second, Longley was in the UK prison system for a number of years with>promotions eventually to Principal Psychologist. Other than the>possibility of nepotism, the Longley that was promoted there is highly>unlikely to be the Longley we see today, especially considering the>visible and delicate nature of such a position. Having owned and>operated three business, and making part of my living now as an>organizational consultant, I feel as unquali'ed as anyone to judge>this. In fact, the Longley of today would not be hired for an>administrative position in the 'rst place in any organization I am>familiar with, except for a menial job or perhaps sales job (which>wouldnOt last long). Exceptions to every rule. Of course I am>factoring out newsgroup posting personality attributes like>insulting, etc., and giving him the bene't of any doubts. Nobody>would hire Larry either, including me. >He has called his prison program a failure, though useful and makes>various (I have argued invalid which argument still stands) scienti'c>claims about it, and that he blames others (mainstream or cognitive>psychologists). So we would have to postulate some event happening>that required continuing medication between the time of the failure>and the posting behavior, which I think was a short time, year or two,>so I have no idea of how probable it is. But it does explain the>contrary cases better than anything else I can think of at the moment.>>I can state from experience that when people fail in some grand>business endeavor, having invested other people in their efforts>more-or-less, and then blame that failure on others, the justi'cation>could be paralyzing and affect them greatly their whole lives,>especially if the failure is criticized by someone they highly respect>(wife, father, peers, etc.), and (guessing) a circle of friends />general public in a highly visible job if still located in the>same area. The ones who properly blame themselves try, try again and>almost always do succeed. Of course in reality is is partly otherOs>fault, partly bad luck, and partly their fault. But blaming the 'rst>two gets one nowhere and is a total waste of time. I have one case of>a person who was just passed over by (he thought) an inferior>subordinate in a factory which socially paralyzed him from then on ->the conversation always got to that issue, incessantly repeated.>> DonOt know about radical behaviorism, the other disruptive part>of> his dogma. I am unquali'ed. But itOs not the problem that started> this discussion - it was the language dogma, speci'cally how it> supposedly invalidates logical argumentation. In light of the>above,> it is no surprise that Longley wants to do that.>>There was a time when everyone thought everything about people was>>innate. Skinner demonstrated that it wasnOt true, and the radical>>form of its negation became an equally absolute doctrine, that kids>>were born tabula rasa and that no cognitive experience could be>>universal or even valid. Radical behaviorism held that people>>were automatons capable of being conditioned to any response or>>behavior, and if there is such a thing as subjective experience or>>experienced emotion, it isnOt relevant. It is now generally believed>>that this wasnOt right either. The truth, as generally understood>>nowadays, lies somewhere between, at an inconvenient shade of gray>>which requires actual consideration and understanding.>>I agree with that as I am informed (little), otherwise take it at You have been studying and practicing behavior for as many decades>as you are old, longer than the experience required to be an expert in>most professions.>>2. The quality of your study depends on what kind of person you are.>Are you scienti'cally trained? Are you an objective observer? Do you>notice objective tests of your ideas and actions outside of yourself?>Are you methodical, and can you apply statistics to your samples in a>mature and pragmatic way? How successful have you been at getting>people around you to behave in a preferred way?>>3. While you are on the subway studying behavior or at work or>manipulating your parents or guiding your children, psychologists are>in a lab studying rats. I de'nitely consider you the winner here (but>I support them because *someday* they will catch up and surpass and>thatOs the only way to do that).>>4. Your samples are anecdotal so that your statistics can only be>your judgment aid. Behavioral scientists are guessing human behavior>into numbers, then translating resulting numbers back into behavior,>disallowing judgment. Thus (I say) their methodology is as §awed as>yours. And the results are usually not being independently con'rmed>in real-time tests, but all I read in their literature is inferences>that indicate certain conjectures with support of this approach.>>But, being allowed to use judgment, you constantly predict future>behavior (real-time), note the results, and make corrections with a>brain in'nitely superior to a multiple regression calculator in this>application and many others (I have plenty of samples of CEOOs who>throw out the (quite statistically sophisticated) computer sales>forecast and 'nd the staring at the wall method much superior>according to actual future sales - real-time tests). The computer>printout is only for the bank and bankers have told me they ignore>it too! (But they want to see that you can do it - experienced CEOOs>multiply everything by 2 because they know the bank will divide>everything by 2, so much for statistics (these people *have* to use>judgment and objective tests or they go out of business).>>5. The psychologists have one advantage. When discussing behavior with>each other, they have special words within that task so that they can>more easily and precisely understand each other. When you and I>discuss behavior, we have to use common words. Thus, we are in second>place here, but if we are very good communicators we can come so close>that the difference becomes trivial in most applications, IMO. This is>no con'rmation of LongleyOs ambiguous language dogma, as everybody>has known about special words within a task forever.>>6. The proof is in the pudding. Factoring out clinical psychology,>name some behavioral area as applicable and important as your own>behavioral interests where psychologists do a better job of predicting>than you as proven in independent scienti'cally and statistically>sound real-time tests. If you run a business, can a psychologist make>up a better employment test than you? Well, that depends a lot on you.>In my case, they de'nitely cannot in the business with which I am>intimately familiar. IOve made them up and have objective tests of>high success. You and I are not biased wrt race, gender, etc. and>testing thousands of applicants (remaining justi'cations of>psychological testing after all the mumbo jumbo is factored out).>>Net result: If any psychologist wants to argue human behavior with>you, IOd give you 50-50 odds. The only reason I give them even odds is>because some (old, wise) psychologists throw most of that away in>practice anyway, and use a lifetime of judgment, just as the old>preacher, or general, other leader, or scienti'cally trained grandma.>And these others having the best informed wisdom usually outperform>the psychologists, and thatOs why they are in charge of things, and>why psychologists (especially those who quote philosophy books and>have a narrow language dogma) are not in charge of things.>>I started posting in c.a.p. with an open mind, especially anxious to>'nd out what was new in psychology since my last review in the 80Os.>That is mainly because my current organizational genius wants to>retire, and it was suggested to me that the 'eld of psychology might>offer a replacement. Not only was there nothing new worth talking>about, my review in the 80Os now seems foolishly>optimistic. Certainly reading the c.a.p. group has formed a strong>negative bias. Anyway, I will go back to former proven organizational>track records in military at the 04 - 05 level 'rst, then business,>then retired, and drag my cajoles thru 50 feet of broken glass to hear>them fart over the phone, and try to think of perks like golf and>concerts for their wives and §ying airplanes to interest them in the>occasional walk-thru because they wonOt work for love nor money, but>you canOt live without their psychological and organizational>experience/judgment/analytical talents.>>If oneOs facilities for consideration and understanding are blunted,>>I suppose that the ambiguity of it may cause a certain level of>>dif'culty and may motivate a retreat to one of the simpler extreme>>positions. However, I am not a psychologist, nor do I play one on>>TV.>>I loved that episode where Newhart has the dream and heOs back in bed>with Susan Pleshette in the old apartment. I think heOs great because>he makes it look easy. I also think Newhart would make a great>psychoanalyst in real life (again factoring out clinical psychology).>Also Suzanne of course, or any old, wise, patient and intelligent>person (patient lets me out ).> Whether anything like this is happening in LongleyOs case is>>not affected by my speculations, and bluntly I donOt know whether>>the supposition is even valid, let alone whether it applies to him.>>Well, it goes without saying that guessers and gamblers like myself>are only going to get so far with newsgroup diagnosis of in'nitely>complex brains. That never froze me in place when I had to make a>decision before, and it wonOt now. But for once I have more than 5>minutes to decide, which does make it interesting and educational.> I would like to hear about any of your creative housekeeping> techniques if explaining wouldnOt interfere and if you are so> inclined.>>Compassion is a good start. On days when compassion is too much>>of a reach, I 'nd that a kill'le can be an adequate substitute.>>One thing that will cut almost all kooks out of any usenet>>discussion, along with threads in which they are active, with>>very little effect on any other conversation, is to simply>>followup-to lines. It is actually quite rare for a non-kook>>to start a crossposted thread or for a kook to start a thread>>without crossposting. Many kooks, perhaps even a majority, will>>not even _respond_ to any thread without crossposting their>>responses.>>Longley is a special case. He usually sticks to c.a.p. He jumps into>threads. He wastes the time of posters unfamiliar with him. He>(inadvertently) spreads an insidious dogma about language reliability.>To some, like myself, it is too serious to ignore. He is especially>disruptive because he pits other posters against each other (new>posters go thru a stage of sympathy and defense), creating and>reinforcing factions, and overtly tries to do that (again) today by>calling for a poll for and against, a potential §ame war over him>which I expect he would enjoy. I imagine in his dreams last night he>had a few of the new posters and one radical behaviorist in the group>listing my and other antagonistOs faults, me and them returning with a>list of their faults, intermingled with his octal sequence of ritual>words, with him jumping in as the professor approving,>reprimanding, getting the upper hand (words he has used in this>context) with his unruly students, who happen all to be educated>adults who just want a friendly, interesting discussion of arti'cial>intelligence and who have no interest in enrolling in the discredited>Longley Philosophy 001 course which is not his 'eld anyway.>>I donOt know anything for sure. I engage and sometimes win ('nd out>IOm incorrect and thus learn something) while my counterpart loses>(teaches me something and learns nothing).>>Larry>> Bear-- David Longley => The dilemma is this - Both Glen and I have indeed> reprimanded others for some types of behaviour - Fragments> is very much about all of this.The part I miss is where you became the be-all, end-all authority thatconsidered himself so authoritative that you earned the right to*reprimand* ANYONE for having an opinion different than yours, as if youwere some pedantic schoolmaster in 18th century London.bp <7T01LKEmHqc$EwjH@longley.demon.co.uk> <3f73a0d0_1@news.iprimus.com.au> <3LOdnYcq9oVxiOmiU-KYuA@metrocast.net> <3f74bc38_1@news.iprimus.com.au> =>WhatOs written below is essentially a rant because poor old Larry has>great dif'culties with objectivity. This is a common problem in people>who donOt have any formal training in psychology, and perhaps>understandably so.*mmmph**snort**gggg** whaaaaahaaaahaaaaaaahaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa a*youOve never*heee*been*heeeeeeeeeeeeeeeeeeeeeeeeeeeee*to*mmmph* sci.psychology.psychotherapy*heee*have you?-- astri*hahahahhahahahaha**phew* => >WhatOs written below is essentially a rant because poor old Larry has>great dif'culties with objectivity. This is a common problem in people>who donOt have any formal training in psychology, and perhaps>understandably so.> > *mmmph*> *snort*> *gggg*> * whaaaaahaaaahaaaaaaahaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa a*> > youOve never> > *heee*> > been> > *heeeeeeeeeeeeeeeeeeeeeeeeeeeee*> > to> > *mmmph*> > sci.psychology.psychotherapy> > *heee*> > have you?> > -- astri> *hahahahhahahahaha*> *phew*Well, it made me laugh.But then almost everything makes me laugh.Larry => -- astri> *hahahahhahahahaha*> *phew*Remind yourself you didnOt want to read this,and scolded me for bringing it here.These cutups are hilarious, like marionettesworked by asylum residents. The lips move,sounds come out, they dance their ritual jigs,but the heads are still made all of wood, thedances betray no meaning, and the sounds tellless than wind moving through the trees.xanthian.And the best part is, they perform for free.-- =>Rickert: If LongleyOs views are compatible with credible scientists in>psychology related 'elds, why is he not posting in the>sci.psychology newsgroups?>GS: Perhaps you need remedial reading lessons, Rickert. Behaviorists are in>the minority, here and elsewhere.Dogmatic behaviorists are in a minority -- as they deserve to be.However, there are plenty sensible non-dogmatic behaviorists in AIareas.>Rickert: Longley regularly takes *interpretations* of experimental results,>and declares them to be evidence. He seems to not even be aware that>the interpretations were made on the assumption of the folk>psychological views that Longley decries.>GS: He reinterprets the results of experiments conducted in the mentalist>tradition.Fine. There is no problem with that.The problem comes when he asserts that his interepretation is theevidence, and when he insists that anyone who disagrees with hisinterpretation is ignoring the evidence.> The interaction between mentalistic concepts and the experimental>procedures are not as straightforward as you are making them out to be.Presumably that utterance is a mechanical response to stimuli youreceived. We can ignore it. Any as you are making them out to bein your statement is pure 'ction.>Rickert: For example, he cites work of Kahneman and Tversky. I donOt have>any>problem with their experimental work. It is sound research. But>what Longley asserts, is that this provides evidence of>irrationality. But he can only claim this based on>folk-psychological accounts of rationality. If he wants to reject>folk psychology, he should equally reject those accounts of>rationality.>GS: As I have told you many times (and you have apparently ignored) one must>mention folk-psychological terms sometimes in order to point to the>BEHAVIORAL PHENOMENA said to require them. The phenomena are real enough,>but the mentalistic accounts are rubbish.It is up to Longley to provide a behaviorist account of rationality(if he can). But he doesnOt. He takes the mentalist account, andasserts that it is behaviorist.>Rickert: He regularly cites Quine. But QuineOs accounts of science (as in>folk-psychological assumptions. Likewise, much of QuineOs writing on>language is derivative of folk-psychological assumptions.>Strictly speaking, radical behaviorism also is heavily dependent on>folk psychological assumptions. If you removed all such assumptions>from radical behaviorism, little of use would remain.>GS: On the contrary, radical behaviorism is useful precisely because it>treats the phenomena said to require mentalistic notions without pointing to>such entities as causes.Then I guess we should say that physics, chemistry, geology are allmentalistic. For they certainly deal with causes. <2eb3320723dad25fefbb4e1a8bf9b934@news.teranews.com> <_qreb.23585$ev2.4804540@newssrv26.news.prodigy.com> => Rickert: He regularly cites Quine. But QuineOs accounts of science (as in>>folk-psychological assumptions. Likewise, much of QuineOs writing on>>language is derivative of folk-psychological assumptions.This really does reveal how little you have understood. You confuse your own misunderstandings with what I have written and as you slowly start to grasp what IOve been saying you have the audacity and arrogance to think itOs your own idea!. read Fragments properly. Why do you think experimental psychology studies folk psychology? What is the relationship between Anomalous Monism and Eliminative Materialism?IOve made this point about your muddled, solipsistic thinking time and time again but like some of the other ignoramuses now frequenting this newsgroup the point doesnOt sink in because you donOt listen and you wonOt read! Either you have inadequate understanding of what behaviour science is all about or you are guilty of perhaps one of the most egregious sins one can 'nd in an academic - time will tell.>Strictly speaking, radical behaviorism also is heavily dependent on>>folk psychological assumptions. If you removed all such assumptions>>from radical behaviorism, little of use would remain.>GS: On the contrary, radical behaviorism is useful precisely because it>>treats the phenomena said to require mentalistic notions without pointing to>>such entities as causes.>>Then I guess we should say that physics, chemistry, geology are all>mentalistic. For they certainly deal with causes.>He said entities !-- David Longley <2eb3320723dad25fefbb4e1a8bf9b934@news.teranews.com> <_qreb.23585$ev2.4804540@newssrv26.news.prodigy.com> => Rickert: If LongleyOs views are compatible with credible scientists in>>psychology related 'elds, why is he not posting in the>>sci.psychology newsgroups?>GS: Perhaps you need remedial reading lessons, Rickert. Behaviorists are in>>the minority, here and elsewhere.>>Dogmatic behaviorists are in a minority -- as they deserve to be.>However, there are plenty sensible non-dogmatic behaviorists in AI>areas.>Rickert: Longley regularly takes *interpretations* of experimental results,>>and declares them to be evidence. He seems to not even be aware that>>the interpretations were made on the assumption of the folk>>psychological views that Longley decries.>GS: He reinterprets the results of experiments conducted in the mentalist>>tradition.>>Fine. There is no problem with that.>>The problem comes when he asserts that his interepretation is the>evidence, and when he insists that anyone who disagrees with his>interpretation is ignoring the evidence.> The interaction between mentalistic concepts and the experimental>>procedures are not as straightforward as you are making them out to be.>>Presumably that utterance is a mechanical response to stimuli you>received. We can ignore it. Any as you are making them out to be>in your statement is pure 'ction.>Rickert: For example, he cites work of Kahneman and Tversky. I donOt have>>any>>problem with their experimental work. It is sound research. But>>what Longley asserts, is that this provides evidence of>>irrationality. But he can only claim this based on>>folk-psychological accounts of rationality. If he wants to reject>>folk psychology, he should equally reject those accounts of>>rationality.>GS: As I have told you many times (and you have apparently ignored) one must>>mention folk-psychological terms sometimes in order to point to the>>BEHAVIORAL PHENOMENA said to require them. The phenomena are real enough,>>but the mentalistic accounts are rubbish.>>It is up to Longley to provide a behaviorist account of rationality>(if he can). But he doesnOt. He takes the mentalist account, and>asserts that it is behaviorist.>Are you really this stupid? Go and re-read what I have written in Fragments - *carefully*.-- David Longley =My favorite fragments:You said this: >The two dogmas listed above, are, it will be argued, a consequence of a>simple, but quite radical misinterpretation of the data which inspired>this revolution and which has subsequently adversely changed the direction>of much of contemporary research in experimental psychology. The most important item in that sentence is where you saidmisinterpretation. This means itOs your OPINION. NOT fact. Opinionsdiffer. Are you going to call me stupid now?You would prefer human behaviour to be as regular as clockwork, a result ofpurely quantitative processes, and you say so, constantly. You explicitlyignore all qualitative factors as irrelevant (or should Isay...intensional?) and repeatedly expound on the glories of thequantitative, pure numbers, in which you apparently hope to reducebehaviour to mere formulae.Hence, you said: >IF such_and_such_behaviour THEN food_pops_out_at_X >IF so_and_so_behaviour THEN NOT food_pops_out_at_X.as applied to a study of the behaviour of rats. Rats are not intelligent.They lack the cognitive abilities you so despise. They lack, dare I say,INUTITION.For the sheer fun of it, letOs apply that to you, using your own rules.IF Longley_Behaviour_Is_Challenged THEN Accuse_challenger_of_irrationalityIF Longley_Is_Agreed_with THEN Praise_person_in_agreementI think that works for me. How do you feel about it?bp =[another copy, including trailing replicated copiesof the orginals to which he was responding and whichhe had already cut and pasted into the main body ofhis discourse]Having no clue as to the public norms of Usenet,Mr. Sizemore manages to post most of what he hasto say in quadruplicate.Sigh.xanthian.-- =This is all you can muster? Somehow I 'gured. Anyway, when most of thepeople around here deserve some respect, IOll take the time to trim posts.BTW, it is Dr. Sizemore to lamers like you.> [another copy, including trailing replicated copies> of the orginals to which he was responding and which> he had already cut and pasted into the main body of> his discourse]>> Having no clue as to the public norms of Usenet,> Mr. Sizemore manages to post most of what he has> to say in quadruplicate.>> Sigh.>> xanthian.> -- => [another copy, including trailing replicated copies> of the orginals to which he was responding and which> he had already cut and pasted into the main body of> his discourse]Style notwithstanding, youOd think some common-sense editing would come intoplay...oh wait...that would be folk psychology! Evil!bp => The German words ACHT (8) and EINS (1) represent numbers and have> their letters in alphabetical order. I can think of only> one de'nite English example, excluding one-letter answers (e, i)> (but have two other English examples if we get pushy).> Roman numerals such as CCIV (204) are not acceptable,> but a letter may be duplicated if all occurrences are adjacent.> DonOt forget that some numbers have multiple names, > such as FOURTH or QUARTER for 1/4.> > For readers whose native language is not English, > how many examples can you 'nd in your language?LetOs try some more exotic languages. (Accuracy not guaranteed sincenone of these languages is my native tongue.)Chinese (Mandarin)Well in its own script the question cannot be asked since it does notrepresent the sound and there is one character for a whole word. Butthere is a standard romanization called pin yin. In that:0 ling No1 yi No2 er Yes3 san No4 si No5 wu No6 liu No7 qi No8 ba No9 jiu No10 shi NoNo further words are required to get up to 99 but all of 11 to 99include shi and hence do not qualify.100 bai No1000 qian No10000 wan NoI forget any higher than that but these words get you up to 99,999. So Chinese manages 2 out of 100,000. Ahead of Finnish but nonethelessa low score.All these words should have tone marks but I have no idea how to writethose in a newsgroup.Tagalog (Filipino)1 isa No2 dalawa No3 tatlo No4 apat No5 lima No6 anim No7 pito No8 walo No9 siyam No10 sampu NoAll of 11 to 19 start labi and hence are no20, 30, up to 90 are compounds of 2 to 9 and hence are no100 daan no1,000 libu noI donOt know any bigger numbers. They are never used. If the subjectwas money, Spanish numbers would be used. If the subject wastechnical then English would be used.So another zero langauge. MalayI would need to check but I would expect a similar success rate toTagalog. They are related languages are both are very fond of theletter a which rules out most words.ThaiI would like to do this but it has its own alphabet and I would haveto revise it. It has loads of letters, with lots of redundancy. Thealphabet starts with a letter which is between English g and k. It isthen followed by four ks.JapaneseAs for Chinese there is the script issue but again there is a standardromanization called romanji. I would need to refresh my memory. ichi, ni, san.J =I am offering free basic website monitoring, looking for some peopleinterested in trying it out.http://www.mjcsoftware.com/main.asp?page=servicesMike Curry =The unfortunate reality that *professional* mathematicians have beenworking to instill false mathematics into many of you has made my joba lot more dif'cult. However, my hope is that this latest expositionwill work to foil these anti-mathematicians.ConsiderP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)where as I demonstrate with P(x)=11x+123=11^2 + 11x + 2 the specialgrouping is to factor into non-polynomial factors. In my example Iused x but now I use m where the letter change shouldnOt matter but Inote it in case some 'nd it confusing.That factorization isP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)so the aOs are roots of the cubic a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)and I have three factorsg_1=(a_1 x + uf), g_2=(a_2 x + uf),g_3=(a_3 x + uf)(notice the *symmetry* here)where P(0) gives the constant term, of course, and P(0)=f^2(3x u^2 + u^3 f), so two of the aOs go to 0 when m=0 which is also seen from the cubicde'ning the aOs.Then arbitrarily picking a_1 and a_2 as the ones that go to 0 at m=0you haveg_1=uf, g_2=uf, g_3=3x+uf.(Notice now though that the symmetry is broken. ItOs that brokensymmetry which helps show why Galois Theory canOt be used to attack myargument.)But P(0)/f^2=3x u^2 + u^3 f as f divides off only two of the gOs whilewith the third it is blocked, as long as it is coprime to 3 and x, soassume it is, and assume as well that f is coprime to u.Then it follows from the constant terms that g_1 and g_2 each have afactor that is f.Many have disputed that conclusion, so IOll elaborate further.Now consider that asa_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) = f^2 m(m^2 f^4 - 3m f^2 + 3)and as they go to 0 when m=0, two of the aOs have a factor of m, andconsider the possibility that factor is sqrt(m).But a_3 does NOT go to 0 when m=0, but consider the possibility thatat m=0, a_3 - 3 has a factor that is m.Those choices are consistent witha_1 a_2 a_3 = f^2 m(m^2 f^4 - 3m f^2 + 3) so now let a_1 = sqrt(m)h_1(m), a_2 = sqrt(m) h_2(m), and a_3 = mh_3(m) + 3where I introduce the hOs as functions of m.Then from g_1=(a_1 x + uf), g_2=(a_2 x + uf),g_3=(a_3 x + uf)I haveg_1=(sqrt(m)h_1(m) x + uf), g_2=(sqrt(m)h_2(m) x + uf),g_3=((mh_3(m) + 3) x + uf)which works for m=0.Now then P(m) = g_1 g_2 g_3, but P(m) has a factor that is f^2 asP(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and dividing off f^2 gives a result with a constant term that iscoprime to f, which requires that g_1 and g_2 have a factor that is f,while g_3 does not.Notice here that the non-polynomial nature of the factorization is notterribly complicated with that possibility.So then, is it possible that both a_1 and a_2 go to 0 when m=0 withouthaving a factor that is sqrt(m)?The answer is, it doesnOt matter as the point is that *both* must have*some* factor of m, so the primary result would remain as the point isthat the *constant* terms with respect to m, are indeed constant.There is NO way for f to divide off of P(m) given that a_1 and a_2must have some factor of m, without dividing through them.Now there have been posters who have diligently worked to confuse onthis issue.My assessment is that they rely on you not being able to *physically*see the factorization, while the rules of algebra are actually easyhere.My hope is that looking atg_1=(sqrt(m)h_1(m) x + uf), g_2=(sqrt(m)h_2(m) x + uf),g_3=((mh_3(m) + 3) x + uf)will help at least some of you realize that yes, actual, realmathematicians, who I doubt believe what they themselves are saying,as the math is so easy, are lying to you quite deliberately, knowingthat you trust them.Notice that the broken symmetry is key to the result.If you doubt me, try and 'nd a way to divide f^2 off without goingthrough sqrt(m) h_1(m) and sqrt(m)h_2(m) given that the constant termof P(m)/f^2 is coprimre to f.YouOve been lied to by people who probably donOt care about you atall.They just want you to believe false math.Given the rapidity with which the primary objection IOve seen §oatedfor months was dismissed by a Professor McKenzie at VanderbiltUniversity, when I mentioned it to him in a face-to-face meeting, Ithink it safe to conclude that posters like Robin Chapman, Dave Rusin,and Robert Israel never accepted the claims of Arturo Magidin. Ofcourse, I donOt know that they de'nitely read his claims, but if theydid, they didnOt correct him.James Harris => The unfortunate reality that *professional* mathematicians have been> working to instill false mathematics into many of you has made my job> a lot more dif'cult. My professors failed utterly to instill any mathematics at all into me. ThatOs why I hang out here watching the parade. However, my hope is that this latest exposition> will work to foil these anti-mathematicians.Curses! Foiled again!> Consider> > P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)> > where as I demonstrate with P(x)=11x+123=11^2 + 11x + 2 the special> grouping is to factor into non-polynomial factors. In my example I> used x but now I use m where the letter change shouldnOt matter but I> note it in case some 'nd it confusing.> This is already too complicated for me to follow. Can you state in a simple manner exactly what you think the problem with the algebraic integers is? That there is a number that should be in them but isnOt? And why?Just explain in words where youOre going with all this and it would make it easier for a lot of us to follow you. =>The unfortunate reality that *professional* mathematicians have been>working to instill false mathematics into many of you has made my job>a lot more dif'cult. However, my hope is that this latest exposition>will work to foil these anti-mathematicians.>>Consider>>P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)>>where as I demonstrate with P(x)=11x+123=11^2 + 11x + 2 the special>grouping is to factor into non-polynomial factors. In my example I>used x but now I use m where the letter change shouldnOt matter but I>note it in case some 'nd it confusing.>>That factorization is>>P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)P(m,x). >so the aOs are roots of the cubic >>a^3 + 3(-1+mf^2)a^2-f^2(m^3 f^4 - 3m^2 f^2 + 3m)>>and I have three factors>>g_1=(a_1 x + uf), g_2=(a_2 x + uf),g_3=(a_3 x + uf)>>(notice the *symmetry* here)>>where P(0) gives the constant term, of course, and >>P(0)=f^2(3x u^2 + u^3 f), P(m=0,x).>so two of the aOs go to 0 when m=0 which is also seen from the cubic>de'ning the aOs.When m=0.>Then arbitrarily picking a_1 and a_2 as the ones that go to 0 at m=0>you have>>g_1=uf, g_2=uf, g_3=3x+uf.g_1 is a function that may or may not depend on m, its values beingpolynomials with algebraic integer coef'cients in the variable x; g_2is a function that may or may not depend on m, its values beingpolynomials with algebraic integer coef'cients in the variable x; g_3is a function that may or may not depend on m, its values beingpolynomials with algebraic integer coef'cients in the variable x; (infact, they ->do<- depend on m, but let that be for now). Sog_1(0) = uf;g_2(0) = uf;g_3(0) = 3x+uf.>(Notice now though that the symmetry is broken. ItOs that broken>symmetry which helps show why Galois Theory canOt be used to attack my>argument.)>But P(0)/f^2=3x u^2 + u^3 f as f divides off only two of the gOs while>with the third it is blocked, as long as it is coprime to 3 and x, so>assume it is, and assume as well that f is coprime to u.P(m=0,x) >Then it follows from the constant terms that g_1 and g_2 each have a>factor that is f.There is no constant terms of g_1 and g_2. The constant term ofg_1(0) and the constant term of g_2(0) are multiples of f, yes:g_1(0)=uf, and g_2(0)=uf.But you have NOT shown that g_1, g_2, or g_3, as functions of m thatthey are, have a well de'ned constant term. For all we know, g_1(1)is a polynomial with a DIFFERENT constant term from g_1(0); same forg_2 and g_3. You have your big hunking gap here.You would have to establish this.>Many have disputed that conclusion, so IOll elaborate further.>Now consider that as>>a_1 a_2 a_3 = f^2(m^3 f^4 - 3m^2 f^2 + 3m) = >> f^2 m(m^2 f^4 - 3m f^2 + 3)>>and as they go to 0 when m=0, two of the aOs have a factor of m, and>consider the possibility that factor is sqrt(m).This does not follow. Why do we not consider other possibilities?Maybe the are something else? What about other possibilities? You havenot shown that the factor cannot be something else.>But a_3 does NOT go to 0 when m=0, but consider the possibility that>at m=0, a_3 - 3 has a factor that is m.Why should we?What if a3 has a factor of sqrt(m+1)+1? ThatOs a factor of m, since(sqrt(m+1)-1)*(sqrt(m+1)+1) = m+1-1 = m, and it does not go to 0 when m=0.>Those choices are consistent with>>a_1 a_2 a_3 = f^2 m(m^2 f^4 - 3m f^2 + 3) Being consistent does not mean it is true. If we writex^2 + 2ax + a^2 = (x+a)^2then when x=0, a=2 is consistent with 4=2*2, but it is not consistentwith 4=1*4.>so now let >>a_1 = sqrt(m)h_1(m), a_2 = sqrt(m) h_2(m), and a_3 = mh_3(m) + 3>>where I introduce the hOs as functions of m.What sort of values do the hOs take? We can always do what you dohere, provided we allow h_1, h_2, h_3 to take arbitrary complex numbervalues. So this proves nothing.>Then from >>g_1=(a_1 x + uf), g_2=(a_2 x + uf),g_3=(a_3 x + uf)>>I have>>g_1=(sqrt(m)h_1(m) x + uf), >>g_2=(sqrt(m)h_2(m) x + uf),>>g_3=((mh_3(m) + 3) x + uf)>>which works for m=0.You have not established that the case m=0 will yield informationabout other cases.>Now then P(m) = g_1 g_2 g_3, but P(m) has a factor that is f^2 as>>P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)>>and dividing off f^2 gives a result with a constant term that is>coprime to f, which requires that g_1 and g_2 have a factor that is f,>while g_3 does not.g_1(0), g_2(0) are multiples of f. That tells you nothing about thevalues of g_1, g_2, and g_3 at other values of x.What do the hOs have to do with this anyway?>So then, is it possible that both a_1 and a_2 go to 0 when m=0 without>having a factor that is sqrt(m)?Yes.>The answer is, it doesnOt matter as the point is that *both* must have>*some* factor of m, a_1(0)=0, a_2(0)=0. >so the primary result would remain as the point is>that the *constant* terms with respect to m, are indeed constant.a_1(m), a_2(m), a_3(m) are not POLYNOMIALS; they have no constantterm. They are functions whose domain are the values of m, and whoserange are the algebraic integers.What does it mean to talk about the constant terms of an arbitraryfunction? >There is NO way for f to divide off of P(m) given that a_1 and a_2>must have some factor of m, without dividing through them.I have no idea what divide off and divide through is supposed tomean here.>Now there have been posters who have diligently worked to confuse on>this issue.>>My assessment is that they rely on you not being able to *physically*>see the factorization, while the rules of algebra are actually easy>here.Weird. We dealt with 65x^3 - 12x + 1 directly and explicitly. But youconfuse the issue claiming that a certain assumption was made, when itwas not.That leads inexorably to the conclusion that one of the following mustbe true:(1) The explicit factorizations for 65x^3 - 12x + 1 are wrong. That is possible, but seems unlikely. Everything is basic, plain, algebra, which you have veri'ed and admitted. The only objections are from you, and your objection right now is basically that even though you accept A, B, and C separately as true, you do not accept that A, B, and C are true holds.(2) Your arguments in Advanced Polynomial Factorization are wrong. That is possible; I donOt know, as I have not seen the latest version. The last time you posted your absolutely no doubt about it correct version, you had to post at least two corrections, and never answered further problems with the statements. Most of them are so confused as to being not even wrong.(3) Your arguments in Advanced Polynomial Factorization are correct, but they do not apply to this situation. That is even more likely, since your arguments in APF are about polynomials in one variable, while here you are dealing with polynomials in multiple variables.(4) Your arguments in Advanced Polynomial Factorization are correct, they DO apply in this situation, but the conclusion is not what you claim it is. Seems reasonable as well: above you concluded something for the case m=0 and claim that it follows for arbitrary m, donOt see why.(5) Your arguments in APF are correct, they DO apply, the conclusion is what you claim it is, and the explicit factorization and calculations are also correct. Then Mathematics is inconsistent. This is extremely, highly, unlikely. I am not willing to consider this as a viable possibility.>My hope is that looking at>>g_1=(sqrt(m)h_1(m) x + uf), >>g_2=(sqrt(m)h_2(m) x + uf),>>g_3=((mh_3(m) + 3) x + uf)>You have said nothing about h1, h2, h3. Can you PROVE that theirvalues are algebraic integers for every value of m? If you cannotprove that, then what you are doing is useless. [.snip.]>Given the rapidity with which the primary objection IOve seen §oated>for months was dismissed by a Professor McKenzie at Vanderbilt>University, when I mentioned it to him in a face-to-face meeting,There is no evidence to substantiate this claim other than your sayso. You have in the past been unable to accurately present theobjections of others. There is no reason to believe that youaccurately presented the primary objection to Prof. McKenzie, andtherefore there is no reason to believe your claim above. = writings... in which he has attempted to prove the truth of his unorthodox interpre- tation of medieval literature. They present a formidable record of unsystematic research in which we see an enthusiast plunging farther and farther and farther from the logic of facts and good sense until truth is lost in the dreadful nightmare of an idee 'xe. There is no real evolution of the Theory although it grows and expands until it embraces ever wider horizons. The numerous inaccuracies of deduction, mis-statements of historical fact, and self-contradictions...have caused critics to turn away from them in disgust... [...] It is impossible to read far... without realizing that we have to deal with a work of faith and imagination rather than of reasoning. There is an appearance of reason, for the author is set on proving by logic the truth of what he already believes by intuition. The truth is plain to him and he cannot comprehend why others do not immediately accept it, but as they desire demonstration he has multiplied his proofs. It is the redundancy and confusion of a prophet expounding by a familiar method the truth revealed to his own simple soul in a §ash of inspiration... In such work as this... it is idle to look for the calm reasoning of a scholar; we do not 'nd it, and there is little or no advantage in attacking the obvious inconsistencies and absurdities that abound. -- E.R. Vincent, _Gabriele Rossetti in England_, quoted in _The Shakespearan Ciphers Examined_, by William F. Friedman and Elizebeth S. Friedman =Arturo Magidinmagidin@math.berkeley.edu> 1. Mathematics is the science in which we make something out of> nothing.> > 2. All of man-made Mathematics consists of an abstraction from> physical processes.> > 3. Since Mathematics in general needs nothing to be created, then the> human mind seems to be limited in that it can only consider> possibilities that are analogous to physical processes.> > Charlie Volkstorf> Cambridge, MA> http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ 20021008.1/1> http://www.arxiv.org/html/cs.lo/0003071Deep thoughts indeed ... pulled out from the depth of your ass, I presume. => Deep?> Texan and Alaskan are whizzing off of a bridge.> Texan sez> This heah watahOs cold!!>> Alaskan sez Yeah, Deep,too!!>> RJ P>> I donOt see any intellectual content in the above, but perhaps you can> further enlighten me. You seem to be attempting to abstract from> physical processes (reading my discoveries and a driving accident), an> example of # 2 above.>> Charlie Volkstorf> Cambridge, MA>> Slack out!!> > My point was supposed to be that the intellectual content of the original> post was muddy, at best, and so the introduction of a> deep joke might add to the spirit ( fermenti??) of the thread.> > Dr. S.Dr. Seuss??Mayhaps you mean that the meaning of the original post is muddy(regardless of the intellectual content), but in any case, IOd be gladto clarify the matter if youOd like. (See the other responses that Ihave already provided.)Hopefully my thoughts are deeper than his cold water.1. Math needs nothing to be created.2. Man creates Math only from physical observations.3. Therefore, are we limited in our ability to conceive of the Maththat waits to be discovered?BTW: Compare my de'nition of Math to the popular ones that talk aboutthe science of numbers or patterns, which are at the wrong levelof abstraction! (Numbers are only part of Math and all sciencesanalyze patterns.) Math is so general that itOs gotta have a simple,no constants/literals (e.g. numbers) de'nition.Charlie VolkstorfCambridge, MA => 1. Mathematics is the science in which we make something out of> nothing.> > 2. All of man-made Mathematics consists of an abstraction from> physical processes.> > 3. Since Mathematics in general needs nothing to be created, then the> human mind seems to be limited in that it can only consider> possibilities that are analogous to physical processes.> the aim> for mathematicians has to be to abstract still further: start from> quite simple abstractions of physical reality (like the natural or> real numbers as in another post) and discover what lies behind.Yes. Abstracting from physical reality produces Mathematics. Abstracting from Mathematics produces metamathematics. Andabstracting from metamathematics produces metametamathematics. etc.Now, we all know what Mathematics and metamathematics look like. Butwhat does metametamathematics consist of?Charlie VolkstorfCambridge, MA => >1. Mathematics is the science in which we make something out of>nothing.>>2. All of man-made Mathematics consists of an abstraction from>physical processes.>>3. Since Mathematics in general needs nothing to be created, then the>human mind seems to be limited in that it can only consider>possibilities that are analogous to physical processes.> > >>the aim>>for mathematicians has to be to abstract still further: start from>>quite simple abstractions of physical reality (like the natural or>>real numbers as in another post) and discover what lies behind.> Yes. Abstracting from physical reality produces Mathematics. > Abstracting from Mathematics produces metamathematics. And> abstracting from metamathematics produces metametamathematics. etc.> > Now, we all know what Mathematics and metamathematics look like. But> what does metametamathematics consist of?What is (or was, at least) referred to as metamathematics is a part of mathematics, so metametamathematics is just a particular portion of metamathematics.-- Aatu Koskensilta (aatu.koskensilta@xortec.')Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus =I know some people here must use MikTex to create Tex and LaTexdocumentson windows, so maybe you can help me. I have just downloaded andinstalledit, and I need to get started learning by creating some trivial littletest documents.Oddly enough, there seems to be nothing to help me do this on theMikTexsite... and from what I could see, full blown TeX manuals seem to burymewith too many =For every natural n, let a_n= (1^p+...n^p)/(n^(p+1)), p>=0. IOm almostquite sure that a_n --> 1/(p+1), what is easy to prove if p isinteger. But so far I couldnOt give a proof that this is true forevery p>=0.ItOs not hard to see that, for every n, 1/n<= a_n<= 1. We also seethat, if we 'x n and let p goes to in'nity, then we approach 1/n.This suggests that, if we de'ne f_n(p) = (1^p+...n^p)/(n^(p+1)) andf(p) =1/(p+1), then f_n --> f uniformly, but IOm not sure.Any suggestions are welcome.Artur => For every natural n, let a_n= (1^p+...n^p)/(n^(p+1)), p>=0. IOm almost> quite sure that a_n --> 1/(p+1), what is easy to prove if p is> integer. But so far I couldnOt give a proof that this is true for> every p>=0.Rewrite as [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n). These Riemann sums -> int_[0,1] x^p dx = 1/(p+1) as n -> oo. (The result is also valid for -1 < p < 0.) =>> For every natural n, let a_n= (1^p+...n^p)/(n^(p+1)), p>=0. IOm almost> quite sure that a_n --> 1/(p+1), what is easy to prove if p is> integer. But so far I couldnOt give a proof that this is true for> every p>=0.>> Rewrite as [(1/n)^p + (2/n)^p + ... + (n/n)^p]*(1/n). These Riemann sums ->> int_[0,1] x^p dx = 1/(p+1) as n -> oo. (The result is also valid for -1 < p> < 0.)> Right. And a numerical analyst would supply material for errorestimate using Trapezoidal Rule: 1^p + ... + n^p = (n^(p+1) - 1) / (p+1) + (1 + n^p) / 2 + p * (n^(p-1) - 1) * T / 12where 0 <= T <= 1 , and T depends on n and p.Then just divide by n^(p+1) .So, for p>=0, the leading term in the error is 1/(2*n). => Lim (t -> 0) 1/(4 pi t) INTEGRAL( -oo , oo ) exp[ -(x-y)^2 / (4t) ]> f(y) dy = [ F_l(x) + F_r(x) ] / 2By the way, shouldnOt that be 1/sqrt(4 pi t) out in front? =for an exam. IOm stuck with the following. Any ideasLet S be the set of integers, a , b and c rational numbers.De'ne f: S->S by f(s)=a*s^2 + bs + c.Determine necessary and suf'cient conditions for a, b and csuch that f de'nes a function in S.It isnOt necessary that a and b and c are integers,for example f(s)=(1/2)*s*(s+1) = (1/2)*s^2 + (1/2)*s isalways and integer when s is an integer.Let S be the set of integers of the form 2^m*3^n with m>=0 and n>=0and T the set of positive integers. Prove thereOs a bijectionbetween S and T. => > What do you think?> > IOd rather use a real programming language where other important details > than the arithmetic have been well thought out, that has been designed > to make code easily readable, and with which I can easily use > arbitrary-bitlength integers without having to declare how long they > are. Like, say, Python.> > Why not Perl? Why is Python around if Perl can do things just 'ne? > IOm doing the same thing, making something slightly different for> different needs. IOll be adding functions, mainly for recursion, but> not structures or pointers. So IOll keep to an overall theme of> predetermination of data size.> > As for readable code, the grammar doesnOt allow declarations mixed> with statements--something I always thought contriubted to> unreadability. As for arbitrary-bitlength integers, it would be easy> to implement as I use them anyway in the Java implementation. But it> wouldnOt 't in with the language goals. Feel free to declare widths> all the way up to (2^31)-1, it wonOt allocate the whole thing.> > There are some problems for which 32 bits are insuf'cient. The following is> a run from my Python program to locate and test the ith, jth generation > Type 11212 Mersenne Hailstone. The 1st 9th generation (j=8) has over > 129 million bits. ItOs not enough simply to calculate it, I run it through> the inverse Collatz function to verify that the path contains j+1 repetitions> of the 11212 pattern before reaching the terminating 0 (it may run past the> j+1 repetition before reaching 0, but that will not be a 11212 sequence). 129 million is still less than 2^31. Say you want mersenne prime2**CD7D25h-1. Just do in CD7D25 $a;$a:2**CD7D25-1;.The program runs uses java bigintegers, which are arbitrary lengthintegers, so the calculations should be similar. => I will derive the proof AFTER the test results,> use only your common sense to answer..... ITOS EASY> So what is the relationship between the author and post content of the> 'rst three responses to your post, and how does this demonstrate> paranormal activity? Here they are below:>> So if I get one of my kids to post under my name- because my name is> obvious (and only one kid, always the same one), you will be able to> guess *his* or *her* name, right? not the surname, but the 'rst and> middle names?>> Guess which one of those 3 names had the above as their 1st reply to me?>> (without using Google).>> At the start of the thread that you instigated you said this was EASY.Not like a skeptic to renig is it? I said the orginal 3 questions were easy.YouOre off on a tangent then going back on your word.> Two days later and youOre stil unable to concoct a connection using> your bogus word-Numerology?> IOve Googled the thread, both in a.f.a-b and here in s.s., these are> the 'rst replies from each of the posters on this thread. PLease> post a message ID contradicting this.Just renigging on !Its from a different ng.>> Its intermitent, I this year and>> Sounds like data mining to me.No, 50 posts in 60 days is signi'cant, because there were only 100 total. If I roll a 6 ona dice 50 out of 100 times its signi'cant to billions to one against chance, that is exactlythe odds I am paralleling.>> found 50 with a good connection. Only 6 or 7 of those could you guess the name from 100> date of Proof of God one year exactly before, and I also predicted the effect would work>> So a 3% success rate from the total 200, or 6% from the 100 you data> mined and only 12% success rate from the 50 where you *were* able to> 'nd a tenuous connection? How many respondents made up this list,> 100 distinct individuals? You say yourself that the claim doesnOt> hold for repeated posts to you, so I must assume so.50 out of 100 total posts were signi'cant, (could be guessed with 10 options)THAT 50%6% could be guessed with over 100 options.44 good, 6 excellent out of 100 OK, read it properly>> The 2 month 50 sample is a bit hard, you only need 25 out of 50 right to prove 9,999 in 10,000odds>> I suggest you brush up on your statistics - in the meantime stay away> from racetracks and casinos.>Get a quali'cation in statistics before you make half dozen errors in a single post again. http://www.automeasure.com/chance.htmlTable II. Range of Number of Successes Attributable to Random Chance, for Final Testing.Chance of number of successes being outside the range shown: less than 1 in 10,000.Probability in % Probability as a fraction 5 Tries 10 15 20 30 40 5025.0 1/4 - 0-8 0-11 0-13 0-17 0-21 1-25That is : for 1 in 10,000 error, 50 tries, 4 options, *expected range* is 1 to 25.IOW : with 4 options you only need to score half of them right (with enough data of 50 questions).> there is a correlation but people wonOt admit to scoring mid range on a test,> this leaves me 2> options,> the 'nd a really good 20 or 30> from over 3 years of posting.>> You forgot the third option that doesnOt relate to datamining to> ensure success, and that is that there is no correlation.>YouOre welcome to demonstrate that but your speculation is §awed and futile.> 02 02 20021 > Randi will test you when you properly apply to be tested. Sign up here:>> http://www.randi.org/research/challenge.html>> This is by Rich Shewmaker>> any idea why?>> of course you have, hE ricH our king>> Uh? Where does hE ricH our king come from in the above post, and how> does it relate to the posters name? I *really* cannot see the> connection your making here.Eric H. ricH richI post the rich showmaker example asking for the connection,you reply regarding the connections with the name THATOS THE ANSWER - HE RICH!>> And are you saying that that this phenomenon is so tenous that the> best(?) example you can come up with is one 18 month old post amongst> the 100s in the past few years.>6 months old.The reason IOm using this example is people have already spotted it from a list in sci.math,its 1000 to 1 post in itself, and I got 2 other very good ones all on the same day.And that day was the day after the shuttle explosion, which I predicted here.and I predicted the effect would happen the day before in alt.magic.Since like all skeptics you renig in going along with a simple analysis> So if I get one of my kids to post under my name- because my name is> obvious (and only one kid, always the same one), you will be able to> guess *his* or *her* name, right? not the surname, but the 'rst and> middle names?>> Carl R. Osterwald> Wally Anglesea> Irony AlertThis paragraph was WallyOs 'rst reply, that why it COULD NOT WORK this thread, simple hey?HeOs asking me to try the effect on his kid, angel see.With a little thought, you could have guessed the name!Herc = demonstrated his kookiness whenGet a quali'cation in statistics before you make half dozen errors in a single post again.CAre to tell us what *Your* quali'cations are, kook? Uh? Where does hE ricH our king come from in the above post, and how>> does it relate to the posters name? I *really* cannot see the>> connection your making here.>>Eric H. ricH rich>>I post the rich showmaker example asking for the connection,>you reply regarding the connections with the name THATOS THE ANSWER - HE RICH!So now, you see the mental gymnastics the Herc takes, and why he is akook.HIs claims are just his own wild fantasies, just like he thinks he isthe real Truman, and we are all living in his show.Since like all skeptics you renig in going along with a simple analysis> So if I get one of my kids to post under my name- because my name is>> obvious (and only one kid, always the same one), you will be able to>> guess *his* or *her* name, right? not the surname, but the 'rst and>> middle names?>> Carl R. Osterwald>> Wally Anglesea>> Irony Alert>>This paragraph was WallyOs 'rst reply, that why it COULD NOT WORK this thread, simple hey?>>HeOs asking me to try the effect on his kid, angel see.So you see the lunacy in his claim? The mental gymnastics he goesthrough are merely signs that he needs medication, and now you knowwhy he got a restraining order placed on him.Furthermore heOs never accepted my challenge. IOm quite happy toallow one of my kids to post 3 news messages. It will be up to him toguess that childs full name. Since he know he doesnOt stand a chance,he runs off, backp[eddalling at a million miles an hour.>>With a little thought, you could have guessed the name!But you are afraid to try to guess the names of one of my kids, huh?-- Find out about AustraliaOs most dangerous Doomsday Cult:http://users.bigpond.net.au/wanglese/pebble.htmYou canOt fool me, itOs turtles all the way down. => Not like a skeptic to renig is it? I said the orginal 3 questions wereeasy.> YouOre off on a tangent then going back on your word.You never answered any of my questions. This means you are not psychic.Now tell me who it is that reneged on demonstrating a psychic ability.You then asked us questions. If you didnOt already know the answers thenyou are not psychic.Face it, quack. You are not a psychic. Grow up, move out of mommyOsbasement, and get a life of your own.-- Felony case 02-CR-0617 9/1/03: Oregon Department ofJustice V. Raymond Ronald Karczewski, Defendant.The defendantOs name is NOT copyrighted. =Today I found a nice approx to the circumf of the ellipse x^2/a^2 + y^2/b^2 = 1, namely: Pi*(a+b)*(19*b^2+26*a*b+19*a^2)/((3*a+5*b)*(5*a+3*b))This is quite accurate and too pretty not to be already known, perhaps even well known.So IOm wondering, does anyone recognize it? Does it have a name? Who 'rst discovered it? In general, what as a Pade approximant to the Ellipic integral that gives the exact circumference.Jim BuddenhagenTo reply copy jbuddenh@REMOVEtexas.net to address bar and edit out REMOVE => Today I found a nice approx to the circumf of the ellipse> x^2/a^2 + y^2/b^2 = 1, namely:>> Pi*(a+b)*(19*b^2+26*a*b+19*a^2)/((3*a+5*b)*(5*a+3*b))>> This is quite accurate and too pretty not to be already known,> perhaps even well known.>> So IOm wondering, does anyone recognize it? Does it have a name?> Who 'rst discovered it? In P.S. I found it as a Pade approximant to the Ellipic integral that gives> the exact circumference.Yes, it is a previously known approximation. (For that matter, so are allof the other low-order Pade approximants.) Your approximation is creditedto E. S. Selmer (1975).BTW, the coef'cient of Pi(a+b) can also be written nicely as 1 + 4(a-b)^2/((3a+5b)(5a+3b)). David Cantrell =Is there a space X with a universal cover X~ so that the 'ber of some pointin X is uncountable.Tyler SmithUniversity of Illinois at Urbana Champaign =>Is there a space X with a universal cover X~ so that the 'ber of some point>in X is uncountable.How about the plane with points with both coordinates rational removed?Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 => Referring to Al and me , You say :> When are the wedding bells , Jeff ? > Any day now ... Al is going to be my bitch .At long last, the other shoeOs 'nally dropped. The wife problems, when all is said and done, actually stem from ... her having the wrong body parts. =Steve:>IOve been looking at a vector based representation of a second order polynomial. >I wondered if it is possible to represent higher order polynomials in the same way?>>I cant see a straight-forward extension to 3rd order for example, if there is one.>This is the seond order representation:>>f(x) = xOAx+bOx+c>Since no one else has answered, as a guess, will tensor notation help?-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu => > [...]> > YOknow, some people donOt like having their names in subject lines. I know I wouldnOt.> > Also, some people might not like it if you discuss them in public> without them knowing it.> > If I were this guy at your alma mater and I graciously spent time> discussing stuff with you, then found out you were plastering my name> all over usenet, IOd be mighty ticked off.the problem with professor so-and-so is that he is not very bright, atleast socially speaking... it should have taken less than a minute tocatch onto the fact that james harris is a high-caliber crackpot. forthat reason professor so-and-so probably deserves the ridicule thecrackpot has subjected him to...> V. => Since that number-grid puzzle I posted a week or so ago> (http://mathforum.org/discuss/sci.math/t/539121)> was too hard for us all, apparently, I will just go ahead and post> this easier number-grid puzzle below.> > Fill a 4-by-4 grid with 16 integers, each integer >= 2 and not> necessarily distinct, such that, if a(k,j) is an element in the grid,> > (each index taken mod-4, ie. the grid wraps around vertically and> horizontally) a(k,j) = the highest-prime-divisor of a(k-1,j) +> the highest-prime-divisor of a(k-2,j)> > AND a(k,j) = > > the lowest-prime-divisor of a(k,j-1) +> the lowest-prime-divisor of a(k,j-2).> > > Find EVERY set of 16 integers which satis'es this.> > Leroy Quet> > Clue: I used a result conjectured by me recently, and proved by> others, to eliminate a case or cases.Another clue: With many many of these 'nd all such solutionsmath-puzzles having only one or two solutions anyway, I will admitthat there is only one solution, IF I did my math correctly.Leroy Quet => offended you.Please stay self-identical!PH appreciate you taking your time out to answer such simpletonquestions.my signature containing my reply adress. thatOs just my way of lettingpeople know that there is a reply adress.I really appreciate your input. that helps me tremendously. gratefully, sean > > --snip-->> Then I advise you to concentrate on the steady states. Ideally, you>> should be able to 'nd the steady states for each set of parameters.>> Then classify them according to their stability using linearisation>> (you know how to do this?). This should split your parameter space>> into several regions, each with a seperate type of steady states. It>> would look something like this:>> >> Region 1: one attractive steady state,>> Region 2: three steady states, one of which attractive>> Region 3: three steady states, none attractive> > You asked the right there Dr. Ulm, And No.. I have to admit I donOt.> IOm still tryin gto understand the idea of stability of dynamical> system. It seems liek if a variable settles into a steady state or> equilibrium then it is said to be stable?> > (Sorry for the late reply; IOm terribly busy at the moment)> > Steady state analysis is (in theory) easy. If you have an equation> > yO(t) = F(y)> > where y is an n dimensional vector valued function, then the steady > states are the solutions of the equation> > F(y) = 0.> > Let y0 be such a solution. If the function F is differentiable> at y0 then you can examine the Taylor expansion of F:> > F(y) = F(y0) + FO(y0).(y-y0) + Rest> > Note, that FO is a n x n matrix. If you ignore the rest above,> then your ODE will be a linear one which however should resemble> the behaviour of the original ODE if the solution is near y0.> The linear system will be stable (i.e. all solutions will converge> to y0 ) if all eigenvalues of FO(y0) have negative real part.> > So, steady state analysis consists of 'nding all solutions of> the equation F(y)=0 and then examining the eigenvalues of FO> at those solutions (in particular you would have to 'nd out> how many of those eigenvalues do have negative real part).> This can give a very good picture of the behaviour of solutions> for many ODEs.> > I have no idea how to classify the steady states using> linearisation... Perhaps you woull be kind to suggest some manuals or> books?>> It was some time since I last had to use steady state analysis.> I really cant remember any good book on the subject. Sorry. However,> sci.math is very good at 'nding such books. So I suggest you start> a new thread asking for literature on the subject.> > --snip--> > PS. Please note the reply to address is sean_incali01@yahoo.com> > if the subject of discussion gets too much off-topic. Maybe someone> else will someday have a similar problem, and then she can > Michael.