mm-9 === We are working on software that will speak mathematics embedded in a webpage. Our main goal is to make math in web pages accessible to visuallyimpaired readers for which we've recently been awarded an NSF grant. Afterdemonstrating an early version of this software to teachers that happened \\tohave normal sight, they claimed that they could see an additional use inteaching normally sighted students how math is spoken. This would of coursebe most useful in situations where a teacher is not present. Does anyoneknow of any work done on this subject within the educational community? Arethere any opinions on the worth of this sort of thing? Any comments arewelcome.Paul ToppingDesign Science, Inc.www.dessci.com-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === We are working on software that will speak mathematics embedded in a web> page. Our main goal is to make math in web pages accessible to visually> impaired readers for which we've recently been awarded an NSF grant. > After demonstrating an early version of this software to teachers that > [had] normal sight, they claimed that they could see an additional use > in teaching normally sighted students how math is spoken. Does > anyone know of any work done on this subject within the educational > community?Not I, but see (and its sub-page ).> Are there any opinions on the worth of this sort of thing? Any comments > are welcome.I think it's worthwhile. You might also ask in other NGs, such as, perhaps, comp.infosystems.www.browsers.misc or sci.math. Hth.AM, Math, Wash. U. St. Louis I've been erasing too much UBE.msh210@math.wustl.edu Of a reply, then, if you have been cheated,-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === >I need to determine a percentage for a SALES application.>>$168,000.00. They grew in sales by a total of $28,231.00. Do I take>sales) that gives me .8319 and say that they grew by 17%????>>When I take the lowest number and multply by 20% I get $27,953.00.>>How does the business world determine a true percentage for this type>of scenario??I would caution you to not be lured that there is some \\true\\ oruniquely correct answer. The key point is to decide what you mean, andcalculate the % to agree with what you mean.They started at $140, and went to $170 (simplifying the numbers). Sothey increased by $30 -- from the $140 where they had been. So if youwant the % they increased from where they had been...On the other hand, someone might look at this and say, wow last yearwe were <> lower than we are this year. That would lead to adifferent calculation.So the key is to think about what you want.bob-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I would use the calculator to check your work only. You must learn todo these things by hand just as you learned how to do other math \\byhand\\ like adding, subtracting, multi-place multiplaction and longdivision. If you become dependent on the calculator, you really willnot learn the process very well, no matter how well you understand theconcept. It does take time and patience, but that's what homework isabout. Math courses demand a lot of time, some of which is devoted totechnical tedium, just like learning music requires you to play scalesand do exercises that are not as much fun as playing melodies.Unless your book is giving you exercises with very large matrices, Iwould strive to learn to do them by hand.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === 87,160 formulas and 10,828 graphics about mathematical functionsare now available free at The Wolfram Functions Site, animportant new resource for mathematicians, scientists, engineers,and students at:http://functions.wolfram.comIn the applications of mathematics to science and engineering--aswell as in pure mathematics itself--there are several hundred so--called \\special functions\\ that have been intensively used for acentury or more. These special functions--with names like Besselfunctions, hypergeometric functions, and totient functions--de?e focal points of mathematical knowledge. The WolframFunctions Site provides in a readily accessible way the largestcollection, by far, of such knowledge ever assembled.Several widely used handbooks of mathematical functions have beenpublished, the largest of which contained about 15,000 formulas,meticulously compiled from thousands of technical papers.Traditional handbooks have also included only handfuls ofgraphics illustrating the properties of functions. WithMathematica, a huge number of new visualizations of functionshave become possible. The Wolfram Functions Site assembles over10,000 of these, with many more being planned.A major tour de force of reference website construction, TheWolfram Functions Site contains over 30 gigabytes of data.Material in The Wolfram Functions Site can be downloaded inseveral standard formats, including Mathematica InputForm andStandardForm, MathML, and PDF. Formulas can be copied from thesite and immediately used as input to a computer system. For easeof citation, each formula has been assigned a unique permanent ID.While having already far surpassed previous knowledge bases formathematical functions, continued growth is projected for TheWolfram Functions Site, with new searching capabilities, externalcontributions, and new classes of graphics and information. Visitthe site at:http://functions.wolfram.com-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Some of you here may be interested in the solutions to the ReaderInvestigations (high school level) from the February issue of theAMESA KZN Mathematics Journal, which have been published in the Novhttp://mzone.mweb.co.za/residents/profmd/homepage4.html under the linkReader Investigations from KZN Mathematics Journal.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === >> thats what i'm trying to ?d out!!!!!!!!> A quick Google on \\lb pounds\\ led me to:> http://www.unc.edu/~rowlett/units/dictP.html> You'll ?d your answer there.> MTEven better, type into the Google box \\454g in lb\\ and see what you get!Stephen D. T. FroggattHead of Mathshttp://www.oakspark.redbridge.sch.uk-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > thats what i'm trying to ?d out!!!!!!!!Latin - librum/libri \\pound\\ came from libra/librae \\scale(s)\\ upon \\whichthe weights were placed. Financially the English pound has the same originand the Pound sign is just an elaborate L.Stephen === Stephen D. T. FroggattHead of Mathshttp://www.oakspark.redbridge.sch.uk === === =-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I am given the following date for the number of times a population ofsix families dined out during the previous month.4 6 9 4 5 7a. Compute the mean, median and the mode for this date.b. Compute the range of this datac. Compute the variance and standard deviationd. Assume that these data represent a sample rather than a population. Compute the variance and standard deviation. Discuss thedifference between the value computed here and in part b.Any help you can give me will be greatly appreciated in this matter.I only need help on d, because it is puzzling me so bad I have been upfor almost ?e hours trying to ?ure d out before I pull my hairout.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > I am given the following date for the number of times a population of> six families dined out during the previous month.> 4 6 9 4 5 7> a. Compute the mean, median and the mode for this date.> b. Compute the range of this data> c. Compute the variance and standard deviation> d. Assume that these data represent a sample rather than a > population. Compute the variance and standard deviation. Discuss the> difference between the value computed here and in part b.The difference between a POPULATION VARIANCE and a SAMPLE VARIANCEshould be well-described in your textbook. (You did look in the book,didn't you?)The reason that there are two different formulas for variance is thatwhen you have a small sample, the variance of the sample is smallerthan the variance of the whole population, just because you have asmall sample. (Consider a sample of size 1---its variance is always0, no matter what population it is taken from.)There is a standard correction for sample size that gives you a \\bestestimate\\ of the variance of the population from a sample. Rather than give it to you here, I'm going to ask that you go back toyour book and re-read the section on sample variance.-- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karpluslife member (LAB, Adventure Cycling, American Youth Hostels)Effective Cycling Instructor #218-ck (lapsed)Professor of Computer Engineering, University of California, Santa CruzUndergraduate and Graduate Director, BioinformaticsAf?iations for identi?ation only.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > I am given the following data for the number of times a population of> six families dined out during the previous month.>> 4 6 9 4 5 7>> a. Compute the mean, median and the mode for this date.The mean is just average, the median is the middle and the mode is the mostcommon number> b. Compute the range of this dataWhat is the min and max of the data?> c. Compute the variance and standard deviationVariance is a little more complicated...add the following....probability adata point times each data point squaredSo for yours six data points...the probility is (1/6) times (4^2 + 6^2 + \\...+ 7^2)then (omega)^2 = (above result) - (mean)^2 = VarianceSD = sqrt of the variance = (omega)> d. Assume that these data represent a sample rather than a> population. Compute the variance and standard deviation. Discuss the> difference between the value computed here and in part b.> Any help you can give me will be greatly appreciated in this matter.Not sure what the difference is in part d.Good Luck.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === how do you determine the porportion of closing stock prices that areat least 50?How do you write a short statement that describes the stock pricedata?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Need more infomation here....the proportion of what (number of prices)?> how do you determine the porportion of closing stock prices that are> at least 50?> How do you write a short statement that describes the stock price> data?>-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Download beta 1 - www.master-graph.com/mgraph20b1.exe (only 477KB).All who will help me to improve this program will get a freeregistration key.You can do the following:-?d bugs-feedback about you impression by the program-advice me to change or add some new features-translate it to the other language (seewww.master-graph.com/instructions/interface.html)Feel free to contact with me - roman@master-graph.comSincerely,Roman.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > For a while now I've been issuing a clarion call about a problem in a> somewhat esoteric branch of mathematics, where relatively basic> algebra reveals a strange problem that has been in the discipline for> over a hundred years.> Recently objections to my work have centered on a particular> factorization where the actual objections can help readers see why it> is necessary that I go outside the math community for help:> My reply is not copied to sci.skeptic or sci.physics. If youwant to forward it to them, ?e. If you do so you might mentionto them that you are starting this new thread on a topic that was still being debated in other threads, that you had failed to answer speci? challenges in those other threads, and thatyou are essentially running away from questions about your proof that you cannot answer.> The key factorization I use is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where v=-1+mf^2, and the ring is something called algebraic integers,> which is just the set of numbers that are roots of monic polynomials> with integer coef?ient, like x^2 + 2x + 2, where the lead> coef?ient is 1, so it is monic, and its coef?ients are 1, 2, and> 2, which are, of course, integers.> Posters for some time on the math newsgroups have managed to object by> casting doubt on the possibility that x and y are constant while only> m varies.> > Yet that is refuted by considering that the a's are the roots of> a^3 + 3va^2 - (v^3+1)=0, where remember v=-1+mf^2,> Yes ... so in particular, the constant term, v^3 + 1, is divisible by m * f^2. > so it's hard to understand how anyone could object by worrying about x> and y.> Why do they object? Because you have claimed that at least one of the three rootsa1, a2, and a3 of that polynomial is coprime to f (if f is prime). The reason we object to that is that we have proofs that it isfalse. You have not provided a valid objection to those proofs. In fact you keep starting new threads in order to avoid facing upto what the proofs say. You answer other less important parts ofthose posts but you evade questions about the proofs by me and W. Dale Hall. So for your convenience I will provide my proof again here, and perhaps this time you can point out an error in my argument: Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by de?ition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. *** Assume a1 is coprime to p. *** By standard theory, there exists an automorphism F12 of the ?ld of algebraic numbers such that: 1. F12 leaves the sub?ld of rational numbers ?ed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, \ which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts what you have claimed: that at least one of a1, a2, or a3 is coprime to p. Your response?> Good question but it is relevant that they are> trying to cast doubt on my ability to use a value at m=0, where two of> the a's equal 0, and one equals 3, with the expression> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where things get just complicated enough for confusion.> To try and lessen that confusion I put in numbers, choosing f=5, x=2,> y=5, so you have> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),> and the factorization> > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).> So all I did was stick in values for x, y and f, which as m varies I> have the polynomial P(m), which has the value shown and factors as> shown, which follows from> Personally I think it just confuses things to stick inx = 2, since the only factorizations you are interested inconsidering here are those associated with the original polynomial as a polynomial in x. It is better to leaveit as a polynomial in the indeterminate x.> P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y).> So now I move to P(m) = 25 Q(m), and notice that Q(0)=11, and now I> have that> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25> where there's a question about how the 25 divides out.> So to answer that question I let m=0, which is the move mathematicians> have been ?hting, which gives me Q(0) = 11.> There's only one way that can happen which is to have> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> as at m=0, you have a_1=a_2=0, while a_3 = 3. > Yep. In this case you can factor 5 out of both a1 and a2: a1 = 0 = 0 * 5, and a2 = 0 = 0 * 5.> So that tells me how to divide out that 25, and there shouldn't be any> arguing.> Nope. No arguing when m = 0.> But there has been a LOT of arguing as, you see, in this esoteric ring> of algebraic integers, a_1/5 is provably not an algebraic integer for> m not equal to 0.> Remember that the a's are the roots of> a^3 + 3va^2 - (v^3+1)=0, where v=-1+mf^2,> so at m=0, v=-1, so you have a^2(a+3) = 0.> Since I'm just ?uring out how that 25 divides out, it's not a> surprise that> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> should work for all m, and not just m=0. And it's that result which> mathematicians have been ?hting.> Yes, that is the key all right. You say should work, asif it were obvious. That phrase is highly revealing of whatis going on in your head. It should work, but you are notgiving any real reason. There really IS no reason other than the fact that you cannot think of any other way to factor 5*5 out of a1, a2, and a3. Yes, when m = 0 and only one of a1, a2, or a3 is nonzero, you are right - there is essentially no other way. There is a speci? and obvious reason for that. But that speci? and obvious reason is no longer present when m <> 0. Why? Because the key to it when m = 0 is that 2 of the 3 a's must be zero. But when m <> 0, NONE of them are zero. The keything is missing when m <> 0. The logic no longer applies. That means that the only reason you are left with tojustify this when m <> 0 is your feeling that somehow it should be parallel to, or similar to, what happens when m = 0. But *your feeling* is not a proof. It is an intuitive guess. It has enormous appeal to you simply because you cannotthink of any other way things could happen. But - to coin a phrase - *the math doesn't care what you think or feel*. It doesn't care about your unproven intuition.It may very well ?d another way to factor 25 out of the product of three factors. In fact that is exactly what happens. Algebraic integerfactors of 5 are divisors of EACH of a1, a2 and a3 when m <> 0.*See my proof above.*> So what's the big deal?> Well, if the 25 divides out as I've shown above then you're forced out> of the ring of algebraic integers, which leads to the bizarre result> that> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.> This last bit is new and rather puzzling. I don't see how you arrive at this, even using your own faulty logic. If things are as you described above, a1/5 and a2/5 would be algebraic integers, and you would have to conclude that a1 and a2 were NOT coprime to 5. What are you thinking? Nora B.[delete blather directed toward sci.skeptic and sci.physics] === OK. I'm still following this. Having made some attempts to work out formyself what a1, a2, a3 are,I think you have but a single error. This is despite a lot of erroneousstuff thrown at you, such as I choosea_1=-5, a_2=-5,The a's are not arbitrary. I ?ally follow you. They depend on m and f, butnot x.> For a while now I've been issuing a clarion call about a problem in a> somewhat esoteric branch of mathematics, where relatively basic> algebra reveals a strange problem that has been in the discipline for> over a hundred years.>> Recently objections to my work have centered on a particular> factorization where the actual objections can help readers see why it> is necessary that I go outside the math community for help:>> The key factorization I use is>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>> where v=-1+mf^2, and the ring is something called algebraic integers,> which is just the set of numbers that are roots of monic polynomials> with integer coef?ient, like x^2 + 2x + 2, where the lead> coef?ient is 1, so it is monic, and its coef?ients are 1, 2, and> 2, which are, of course, integers.>> Posters for some time on the math newsgroups have managed to object by> casting doubt on the possibility that x and y are constant while only> m varies.>> Yet that is refuted by considering that the a's are the roots of>> a^3 + 3va^2 - (v^3+1)=0, where remember v=-1+mf^2,>> so it's hard to understand how anyone could object by worrying about x> and y.>> Why do they object? Good question but it is relevant that they are> trying to cast doubt on my ability to use a value at m=0, where two of> the a's equal 0, and one equals 3, with the expression>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>> where things get just complicated enough for confusion.>> To try and lessen that confusion I put in numbers, choosing f=5, x=2,> y=5, so you have>> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),>> and the factorization>> P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).>> So all I did was stick in values for x, y and f, which as m varies I> have the polynomial P(m), which has the value shown and factors as> shown, which follows from>> P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y).>> So now I move to P(m) = 25 Q(m), and notice that Q(0)=11, and now I> have that>> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>> where there's a question about how the 25 divides out.>> So to answer that question I let m=0, which is the move mathematicians> have been ?hting, which gives me Q(0) = 11.>> There's only one way that can happen which is to have>> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>> as at m=0, you have a_1=a_2=0, while a_3 = 3.>> So that tells me how to divide out that 25, and there shouldn't be any> arguing.>> But there has been a LOT of arguing as, you see, in this esoteric ring> of algebraic integers, a_1/5 is provably not an algebraic integer for> m not equal to 0.>> Remember that the a's are the roots of>> a^3 + 3va^2 - (v^3+1)=0, where v=-1+mf^2,>> so at m=0, v=-1, so you have a^2(a+3) = 0.>> Since I'm just ?uring out how that 25 divides out, \ it's not a> surprise that>> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>> should work for all m, and not just m=0. And it's that result which> mathematicians have been ?hting.>> So what's the big deal?For everything above, I concur with it all.>> Well, if the 25 divides out as I've shown above then \ you're forced out> of the ring of algebraic integers, which leads to the bizarre result> that>I agree with the ?st half of that sentence. With Q(m) you've stepped outof the algebraic integers.a_1/5 and a_2/5 are not algebraic integers.> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.OK. I will paraphrase, tell me if this is what you mean or not.(When f is 5), for certain m (at least) it looks like a_1 (and a_2 and a_3)might be NOT coprime to5 (in the algebraic integers)However a_1/5 is not an algebraic integer when m does not equal 0.Therefore a_1 is coprime to 5 when m does not equal 0.Is that your argument?Because that's incorrect. a_1/5 not an algebraic integer does NOT imply a_1coprime to 5.If that's not your argument, just what is your argument?Phil.>> James Harris === Revision 1Reason for revision: Fix error a^2(a+3)=0, add in b's. ___JSH------------------------------------For a while now I've been issuing a clarion call about a problem in asomewhat esoteric branch of mathematics, where relatively basicalgebra reveals a strange problem that has been in the discipline forover a hundred years.Recently objections to my work have centered on a particularfactorization where the actual objections can help readers see why itis necessary that I go outside the math community for help:The key factorization I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),where v=-1+mf^2, and the ring is something called algebraic integers,which is just the set of numbers that are roots of monic polynomialswith integer coef?ient, like x^2 + 2x + 2, where the leadcoef?ient is 1, so it is monic, and its coef?ients are 1, 2, and2, which are, of course, integers.Posters for some time on the math newsgroups have managed to object bycasting doubt on the possibility that x and y are constant while onlym varies.Yet that is refuted by considering that the a's are the roots of a^3 + 3va^2 - (v^3+1)=0, where remember v=-1+mf^2,so it's hard to understand how anyone could object by worrying about xand y.Why do they object? Good question but it is relevant that they aretrying to cast doubt on my ability to use a value at m=0, where two ofthe a's equal 0, and one equals 3, with the expression (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),where things get just complicated enough for confusion.To try and lessen that confusion I put in numbers, choosing f=5, x=2,y=5, so you have P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),and the factorization P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).So all I did was stick in values for x, y and f, which as m varies Ihave the polynomial P(m), which has the value shown and factors asshown, which follows from P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y).So now I move to P(m) = 25 Q(m), and notice that Q(0)=11, and now Ihave that P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25where there's a question about how the 25 divides out.So to answer that question I let m=0, which is the move mathematicianshave been ?hting, which gives me Q(0) = 11.There's only one way that can happen which is to have P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)as at m=0, you have a_1=a_2=0, while a_3 = 3.That tells me that for two of the a's--arbitrarily selected as a_1 anda_2--I have a factor that is 5, so I can introduce b_1 and b_2 where a_1 = 5 b_1, and a_2 = 5 b_2, so I have Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).So that tells me how to divide out that 25, and there shouldn't be anyarguing.But there has been a LOT of arguing as, you see, in this esoteric ringof algebraic integers, a_1/5 is provably not an algebraic integer form not equal to 0.Remember that the a's are the roots of a^3 + 3va^2 - (v^3+1)=0, where v=-1+mf^2,so at m=0, v=-1, so you have a^2(a-3) = 0.Since I'm just ?uring out how that 25 divides out, it's not asurprise that Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5)should work for all m, and not just m=0. And it's that result whichmathematicians have been ?hting.So what's the big deal?Well, if the 25 divides out as I've shown above then you're forced outof the ring of algebraic integers, which leads to the bizarre resultthat a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.That is, b_1 and b_2 are not algebraic integers when m does not equal0, but the 25 can only divide out one way, so something is wrong.Here coprime means that they don't share non unit algebraic integerfactors with 5, for instance, like 1+2i as (1+2i)(1-2i)=5, and a unitfactor is just a factor of 1.It's important to remember that algebraic integers do not form a ?ldwhere you can have fractions, but are a special kind of number that'slike integers, which is why they're \ called algebraic integers.Well the problem comes from the de?ition which is over a hundredyears old.Mathematicians from what I've read pride themselves on having avoidedmajor upheavals from errors in their ?ld, and while it may seemsomewhat odd to many of you, what I have is a major upheaval.Well if you're a physicist or skeptic, why should you care?Consider that if mathematicians have a problem that they avoid, evenif it seems trivial to you, and they continue to teach the ?athematics, then a rot has entered the discipline. But physicistsdepend on mathematics, so what if the rot spreads?How many of you *really* think you have the expertise to always checkthe mathematical tools you use with an eye out for esoteric andstrange errors?Don't you depend on mathematicians to keep things in order?Sure, eventually if some physical theory depended on bogus math, itmight occur to someone that the math was ? but might they not?st question the theory? And what if some physicist questioned themath but felt the weight of the mathematical establishment behind themath, like what I'm facing now?Yup, you guessed it. Others would probably think the physicist was anut, and trust the mathematicians.And I *have* written a paper which proves that there's a problem, andit's available on the web \ at my websitehttp://groups.msn.com/AmateurMath where it's a pdf, which is onlyavailable to those who join that group, but I have the mathematicsalso athttp://groups.msn.com/AmateurMath/ algebraoffactorizations.msnwwhere you can look at it without joining the group.And, yes, I've sent the paper to math journals, and no, it hasn't beenaccepted by them. The reasons have ranged from it's too short, toclaims of various errors.If you just *trust* the mathematicians and *believe* I must be wrongfor that reason, then go back and read through this post to see whatyou're believing in, as the algebra is basic.And that goes back to my point as mathematicians shouldn't ?ht thetruth here, but in ?hting it they highlight the problem I mention asso far I have been blocked, which is why I need to go outside the mathcommunity.Still this post is also going to the newsgroup sci.math so you canalso see any responses that might come from there to see if *any*actually make mathematical sense.James Harris === >Another good quote from that work, suitable for James (given his>propensity to talk about Truth and Mathematics), would be:>He believed in a door. He must ?d that door. The door was> a way to ... to ....> The Door was The Way.> Good.> Capital letters were always the best way of dealing with> things you didn't have a good answer to.> === === ===>No, my arguments were based on no hypothesis at all.> === === ===>Arturo Magidin>magidin@math.berkeley.edu> Wasn't there a guy in Yellow Submarine who would walk around with a towel> around his face, because he believed that if *he* couldn't see someone, they> couldn't see him, either???Beast of Traal, which thinks that if you can't see it, it can't see you (butdespite this seemed to be a very successful predator, presumably due to thelarge number of people who don't know where their towel is). Possibly thisis what you're thinking of?-- Dave TaylorBut I don't know \ how to teach - I'm a Professor![Futurama] === > a_1 = 5 b_1, and a_2 = 5 b_2, so I have>> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).I have no problem is this. Despite there being quite a few steps not shownto your audience, your algebra appears to be correct.> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.Prove it.>> That is, b_1 and b_2 are not algebraic integers when m does not equal> 0, but the 25 can only divide out one way, so something is wrong.It doesn't follow from this that a_1, a_2 are coprime to 5.The only thing wrong is your conclusion above.Prove it. Or would you rather I proved that they're not coprime to 5?> James HarrisPhil Nicholson === [snip]Here is James Harris' recipe for continuing to ?s dead horse, tomount and ride it, and to enter it into the Kentucky Derby:1) When another poster presents a valid refutation of his argument, ignoreit,2) Misrepresent the other posters criticism so that it appears to have atotally different character,3) Attack the misrepresented position furiously, with disparaging sideremarks,4) Announce, triumphantly, that his proof has again survived allcriticisms,5) Accuse the critic of conspiracies to overthrow basic algebra.James ?ds many ways to explain the existence of his critics, except thecorrect explanation -- that he is wrong.--No matter how low a charlatan sets the bar, he will manage to slitherunder it.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Bug-blatter> Beast of Traal, which thinks that if you can't see it, it can't \ see you(but> despite this seemed to be a very successful predator, presumably due tothe> large number of people who don't know where their towel is). Possibly this> is what you're thinking of? === > What gets to be a little scary is that it's starting to seem like it's> _literally_ all \ the mathematicians on the planet engaged in this> conspiracy to suppress the Truth. I mean when it was just> sci.math, well that's a close-knit group of cronies, or so I> imagine it appears to some people, so they could be all> in it together. But you've been sending the paper to journals> in Germany, Russia, China... they _all_ seem to Know that> Your paper is Not To Be Published.Well sure. Unfortunately, he's cross-posting this to sci.physicshoping for non-conspirators, but as everybody knows all thephysicists in the world and everybody who's ever passed acollege physics course are engaged in the Great Conspiracy toPerpetuate the Absurd Relativity Hoax to Ensure TheirOwn Job Security.>which is why I need to go outside the math>community.> Makes a lot of sense. The people on sci.physics and sci.skeptic> have been very receptive to your ideas, right?Alas... see above. Conspiracies are just thick on theground out here in Usenet land. - Randy === >>>>And, yes, I've sent the paper to math journals, and no, it hasn't been>>accepted by them. The reasons have ranged from it's too short, to>>claims of various errors.>>>Truly fascinating. People on sci.math studiously post explicit>proofs that the claims are false. Some of them you ignore, some>you totally misunderstand. Journals claim there are various>errors. You _know_ that _many_ of the things you've claimed>about math in the past have been wrong - these were things>that you were just as certain of as you're certain right now>that your paper is correct. But in spite of all this the idea that>you might be simply wrong doesn't come up.>> This reminded me of the following passage from Dirk Gently's Holistic> Detective Agency, an underappreciated work of the late, great Douglas> Adams:>> The Monk currently believed that the valley and everything in the> valley and around it, including the Monk itself and the Monk's horse,> was a uniform shade of pale pink. This made for a certain dif?ulty in> distinguishing any one thing from any other thing, and therefore made> doing anything anything or going anywhere impossible, or at least> dif?ult and dangerous. Hence the immobility of the Monk and the> boredom of the horse, which had had to put up with a lot of silly things> in its time but was secretly in the opinion that this was one of the> silliest.>> How long did the Monk believe these things?>> Well, as far as the Monk was concerned, forever. The faith which moves> mountains, or at least believes them against all the available evidence> to be pink, was a solid and abiding faith, a great rock against which> the world could hurl whatever it would yet it would not be shaken. In> practice the horse knew twenty-four hours was usually about its lot.>> Give James time, and he will decide that (perhaps) he was wrong. The> only question is what will replace his current belief/delusion. Maybe it> will be something innocuous, like pink mountains, which will take him> far from sci.math :-) Or possibly he'll come up with a short proof of> the four colour problem.>> Mark Atherton>I _love_ that book!! The Electric Monk is one of my favorite Douglascharacters. That, and the couch stuck in the stairwell.Pete K. === > a_1 = 5 b_1, and a_2 = 5 b_2, so I have>> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).> I have no problem is this. Despite there being quite a few steps not shown> to your audience, your algebra appears to be correct.I am willing to elaborate on any step. > a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.> Prove it.>>That is, b_1 and b_2 are not algebraic integers when m does not equal>0, but the 25 can only divide out one way, so something is wrong.> It doesn't follow from this that a_1, a_2 are coprime to 5.> The only thing wrong is your conclusion above.> Prove it. Or would you rather I proved that they're not coprime to 5?The math can only give one answer. Here I'll elaborate on what thatanswer is.The key factorization I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),which gives me a^3 + 3v a^2 - (v^3+1) = 0, where a_1, a_2, and a_3 are its roots,and I also have that v=-1+mf^2.So now I let m=1, and then I have 13825x^3 -72 xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),so I have a^3 + 72 a^2 - 13825 = 0, and the three roots are the a's.Now using a = 5b means that the equation will give me three results tomatch with the three a's, where those results are a_1/5, a_2/5, anda_3/5.That substitution gives me 125b^3 + 72 (25)b^2 - 13825 = 0, and dividing off 25 gives 5b^3 + 72 b^2 - 5530 = 0, which is what's called irreducible over Q.A reducible polynomial over Q is just one that factors in rationals,for instance x^2 + 2x + 1 is reducible to (x+1)^2, and Q is just math slang for rational numbers.Since 5b^3 + 72 b^2 - 5530 is not reducible over rationals it's calledirreducible over Q, and also because the lead coef?ient is 5 and not1 or -1, it's not monic, so it's fully called a non-monic irreducibleover Q.However there is math proof that an algebraic integer cannot be theroot of such a non-monic irreducible over Q, so none of its roots arealgebraic integers.Remember algebraic integers are de?ed to be roots of a monicpolynomial with integer coef?ients.One of the a's was already shown to be coprime to 5, and now giventhat b_1 and b_2 are not algebraic integers they can't be calledfactors of the a's in the ring of algebraic integers, so in fact allof the a's have now been shown to be coprime to 5, which is the resultthat shows a problem with the ring of algebraic integers.James Harris === > Revision 1> Reason for revision: Fix error a^2(a+3)=0, add in b's. ___JSH> ------------------------------------> For a while now I've been issuing a clarion call about a problem in a> somewhat esoteric branch of mathematics, where relatively basic> algebra reveals a strange problem that has been in the discipline for> over a hundred years.> Recently objections to my work have centered on a particular> factorization where the actual objections can help readers see why it> is necessary that I go outside the math community for help:> The key factorization I use is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where v=-1+mf^2, and the ring is something called algebraic integers,> which is just the set of numbers that are roots of monic polynomials> with integer coef?ient, like x^2 + 2x + 2, where the lead> coef?ient is 1, so it is monic, and its coef?ients are 1, 2, and> 2, which are, of course, integers.> Posters for some time on the math newsgroups have managed to object by> casting doubt on the possibility that x and y are constant while only> m varies.> Yet that is refuted by considering that the a's are the roots of> a^3 + 3va^2 - (v^3+1)=0, where remember v=-1+mf^2,> This should be a^3 - 3*v*a^2 + (v^3 + 1) = 0.> so it's hard to understand how anyone could object by worrying about x> and y.> Why do they object? Good question but it is relevant that they are> trying to cast doubt on my ability to use a value at m=0, where two of> the a's equal 0, and one equals 3, with the expression> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where things get just complicated enough for confusion.> To try and lessen that confusion I put in numbers, choosing f=5, x=2,> y=5, so you have> P(m) = 25(5000m^3 - 600 m^2 - 126m + 11),> and the factorization> > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5).> The substitution of 2 for x doesn't do you anygood at all, since you are continuing to consideronly factorizations as a polynomial in the variable x.The simpli?ation here is strictly illusional, UNLESSyou actually want people to take it literally andconsider factorizations of the NUMBER P(2), as opposed to the POLYNOMIAL P(x). But you don't, sincethat leads to immediate disaster (as opposed to slightly deferred disaster) for your claims.> So all I did was stick in values for x, y and f, which as m varies I> have the polynomial P(m), which has the value shown and factors as> shown, which follows from> P(m)= (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y).> So now I move to P(m) = 25 Q(m), and notice that Q(0)=11, and now I> have that> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25> where there's a question about how the 25 divides out.> So to answer that question I let m=0, which is the move mathematicians> have been ?hting, That is absolutely not true. No one objects to your lettingm = 0. It is what you do afterward that causes the problem.>which gives me Q(0) = 11.> There's only one way that can happen when m = 0> which is to have> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> as at m=0, you have a_1=a_2=0, while a_3 = 3.> That tells me that for two of the a's--arbitrarily selected as a_1 and> a_2--I have a factor that is 5, so I can introduce b_1 and b_2 where> a_1 = 5 b_1, and a_2 = 5 b_2, so I have> Yes - in fact, you can write b_1 = b_2 = 0 if you want. 5 isa factor of a_1 and a_2 in a somewhat unusual way - both are zero.In fact, anything is a factor of zero. What you have just writtenwould be equally true if you replaced 5 by 107: a_1 = 107 * _1 and a_2 = 107 * b_2. Can we therefore assume that a_1 and a_2 are divisible by 107 when m <> 0 ? Why not? > Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).> So that tells me how to divide out that 25, and there shouldn't be any> arguing.> No arguing at all, as long as you stay with m = 0.> But there has been a LOT of arguing as, you see, in this esoteric ring> of algebraic integers, a_1/5 is provably not an algebraic integer for> m not equal to 0.> Remember that the a's are the roots of> a^3 + 3va^2 - (v^3+1)=0, where v=-1+mf^2,> Right. Which means a1, a2, and a3 are functions of m.> so at m=0, v=-1, so you have a^2(a-3) = 0.> Since I'm just ?uring out how that 25 divides out, it's not a> surprise that> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5)> should work for all m, and not just m=0. Not a surprise ??? [***] should work for all m ??? Why on earth can you not see that this, the absolute heart of your argument, is nothing but wishful thinking ? THIS IS NOT A PROOF. THIS IS LIP-FLAPPING. THIS ISWISHFUL THINKING. NO MATHEMATICIAN WILL EVER ACCEPT THISAS A LOGICAL ARGUMENT.> And it's that result which> mathematicians have been ?hting.> So what's the big deal?> > Well, if the 25 divides out as I've shown above then \ you're forced out> of the ring of algebraic integers, which leads to the bizarre result> that> {$} a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.> Of course this directly con?with the fact that a1*a2*a3 = multiple of 5. I honestly do not know how you have concluded {$}, and it issomething new that you have not stated previously. By whateverlogic you deduce it, however, it should scream out one message toyou, loud and clear: YOU REASONING IS INCORRECT. THERE IS AN ERRORSOMEWHERE IN WHAT YOU HAVE DONE. I have speci?d above [***] one place where you have, at an absolute minimum, a great gaping hole in your argument. Thatalone would raise extreme doubt in the mind of any rationalmathematician. Compound that with the fact that there are TWOextant proofs that your main conclusion is wrong, and doubtbecomes certainty: HOWEVER YOU GOT HERE, YOU HAVE MADE A SERIOUS MISTAKE !!! If I were you, I would quit worrying about the x = 2 thingand start seriously considering what I and others have saidregarding your leap from m = 0 to m <> 0. You must admit thatsomething there is desperately wrong. And it is not Galoistheory. It isn't even rocket science. Nora B. [most guff directed at sci.physics, sci.skeptic deleted]> If you just *trust* the mathematicians and *believe* I must be wrong> for that reason, then go back and read through this post to see what> you're believing in, as the algebra is basic.> And that goes back to my point as mathematicians shouldn't ?ht the> truth here, but in ?hting it they highlight the problem I mention as> so far I have been blocked, which is why I need to go outside the math> community.> Still this post is also going to the newsgroup sci.math so you can> also see any responses that might come from there to see if *any*> actually make mathematical sense.> James Harris === >>>>>>> a_1 = 5 b_1, and a_2 = 5 b_2, so I have>>>>>> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).>>>>I have no problem is this. Despite there being quite a few steps not shown>>to your audience, your algebra appears to be correct.> I am willing to elaborate on any step.>>> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.>>>>Prove it.>>>>>>>That is, b_1 and b_2 are not algebraic integers when m does not equal>>>0, but the 25 can only divide out one way, so something is wrong.>>>>It doesn't follow from this that a_1, a_2 are coprime to 5.>>The only thing wrong is your conclusion above.>>>>Prove it. Or would you rather I proved that they're not coprime to 5?> The math can only give one answer. Here I'll elaborate on what that> answer is.> The key factorization I use is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> which gives me > a^3 + 3v a^2 - (v^3+1) = 0, where a_1, a_2, and a_3 are its roots,> and I also have that v=-1+mf^2.> Now using f=5, leads me to> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)and a_i's are dependent on f.> and from before in my previous post I have that > a_1 = 5 b_1, a_2 = 5 b_2, so> 25 b_1 b_2 a_3 = 25(625 m^3 - 75 m^2 + 3m)> so> b_1 b_2 a_3 = (625 m^3 - 75 m^2 + 3m)> which tells me that if m is coprime to 5, then b_1 b_2 a_3 must be as> well.I don't see how this follows.> So now I let m=1, and then I havea_i's dependent on f and m.> 13825x^3 -72 xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> so I have> a^3 + 72 a^2 - 13825 = 0, and the three roots are the a's.> Now using a = 5b means that the equation will give me three results to> match with the three a's, where those results are a_1/5, a_2/5, and> a_3/5.> That substitution gives me> 125b^3 + 72 (25)b^2 - 13825 = 0, and dividing off 25 gives> 5b^3 + 72 b^2 - 5530 = 0, which is what's called irreducible over Q.> A reducible polynomial over Q is just one that factors in rationals,> for instance> x^2 + 2x + 1 is reducible to (x+1)^2, > and Q is just math slang for rational numbers.> Since 5b^3 + 72 b^2 - 5530 is not reducible over rationals it's called> irreducible over Q, and also because the lead coef?ient is 5 and not> 1 or -1, it's not monic, so it's fully called a non-monic irreducible> over Q.> However there is math proof that an algebraic integer cannot be the> root of such a non-monic irreducible over Q, so none of its roots are> algebraic integers.> Remember algebraic integers are de?ed to be roots of a monic> polynomial with integer coef?ients.> One of the a's was already shown to be coprime to 5, and now given> that b_1 and b_2 are not algebraic integers they can't be called> factors of the a's in the ring of algebraic integers, so in fact all> of the a's have now been shown to be coprime to 5, which is the result> that shows a problem with the ring of algebraic integers.You have (possibly) shown that the b_1(m=1,f=5), b_2(1,5), a_3(1,5) are coprime to 5. How does that imply that a_1(1,5) a_2(1,5) are coprime to 5? Recall, a_i(1,5) = 5*b_i(1,5). It seems to me that a_1(1,5) stands a reasonable chance of not being coprime to 5, even though b_1(1,5) is not.For example: 1 is coprime to 5 in the integers, since (6)1+(-1)5=1That does not imply that 5 is coprime to 5 in the integers.> James Harris-- Will Twentyman === >>> a_1 = 5 b_1, and a_2 = 5 b_2, so I have>>> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).>>I have no problem is this. Despite there being quite a few steps notshown>to your audience, your algebra appears to be correct.>> I am willing to elaborate on any step.>>> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.>Prove it.>>>> That is, b_1 and b_2 are not algebraic integers when m does not equal>> 0, but the 25 can only divide out one way, so something is wrong.>It doesn't follow from this that a_1, a_2 are coprime to 5.>The only thing wrong is your conclusion above.>>Prove it. Or would you rather I proved that they're not coprime to 5?>> The math can only give one answer. Here I'll elaborate on what that> answer is.>> The key factorization I use is>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>> which gives me>> a^3 + 3v a^2 - (v^3+1) = 0, where a_1, a_2, and a_3 are its roots,>> and I also have that v=-1+mf^2.>>> Now using f=5, leads me to>> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)>> and from before in my previous post I have that>> a_1 = 5 b_1, a_2 = 5 b_2, so>> 25 b_1 b_2 a_3 = 25(625 m^3 - 75 m^2 + 3m)>> so>> b_1 b_2 a_3 = (625 m^3 - 75 m^2 + 3m)>> which tells me that if m is coprime to 5, then b_1 b_2 a_3 must be as> well.It's a bit dangerous using the word coprime in reference to b_1 and b_2,seeingas they're not algebraic integers. Some will attempt to hang you for it.But the algebra is ?e.>> So now I let m=1, and then I have>> 13825x^3 -72 xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>> so I have>> a^3 + 72 a^2 - 13825 = 0, and the three roots are the a's.>> Now using a = 5b means that the equation will give me three results to> match with the three a's, where those results are a_1/5, a_2/5, and> a_3/5.>> That substitution gives me>> 125b^3 + 72 (25)b^2 - 13825 = 0, and dividing off 25 gives>> 5b^3 + 72 b^2 - 5530 = 0, which is what's called irreducible over Q.>> A reducible polynomial over Q is just one that factors in rationals,> for instance>> x^2 + 2x + 1 is reducible to (x+1)^2,>> and Q is just math slang for rational numbers.>> Since 5b^3 + 72 b^2 - 5530 is not reducible over rationals it's called> irreducible over Q, and also because the lead coef?ient is 5 and not> 1 or -1, it's not monic, so it's fully called a non-monic \ irreducible> over Q.>> However there is math proof that an algebraic integer cannot be the> root of such a non-monic irreducible over Q, so none of its roots are> algebraic integers.>> Remember algebraic integers are de?ed to be roots of a monic> polynomial with integer coef?ients.Yep no problem with any of that.>> One of the a's was already shown to be coprime to 5, and now given> that b_1 and b_2 are not algebraic integers they can't be called> factors of the \ a's in the ring of algebraic integers, so in fact all> of the a's have now been shown to be coprime to 5, which is the result> that shows a problem with the ring of algebraic integers.The problem is the statement so in fact all of the a's have now been shownto be coprime to 5It's clear that b_1 and b_2 aren't algebraic integers, but nowhere have youshown that the a's arecoprime to 5. In fact it's easy to show the opposite.Apart from that, everything appears to be correct.I'm still waiting to see the proof.>>> James HarrisPhil Nicholson. === >>>>>>And, yes, I've sent the paper to math journals, and no, it hasn't been>>>accepted by them. The reasons have ranged from it's too short, to>>>claims of various errors.>>>> Truly fascinating. People on sci.math studiously post explicit>> proofs that the claims are false. Some of them you ignore, some>> you totally misunderstand. Journals claim there are various>> errors. You _know_ that _many_ of the things you've claimed>> about math in the past have been wrong - these were things>> that you were just as certain of as you're certain right now>> that your paper is correct. But in spite of all this the idea that>> you might be simply wrong doesn't come up.>>This reminded me of the following passage from Dirk Gently's Holistic>Detective Agency, an underappreciated work of the late, great Douglas>Adams:>>The Monk currently believed that the valley and everything in the>valley and around it, including the Monk itself and the Monk's horse,>was a uniform shade of pale pink. This made for a certain dif?ulty in>distinguishing any one thing from any other thing, and therefore made>doing anything anything or going anywhere impossible, or at least>dif?ult and dangerous. Hence the immobility of the Monk and the>boredom of the horse, which had had to put up with a lot of silly things>in its time but was secretly in the opinion that this was one of the>silliest.>>How long did the Monk believe these things?>>Well, as far as the Monk was concerned, forever. The faith which moves>mountains, or at least believes them against all the available evidence>to be pink, was a solid and abiding faith, a great rock against which>the world could hurl whatever it would yet it would not be shaken. In>practice the horse knew twenty-four hours was usually about its lot.>>Give James time, and he will decide that (perhaps) he was wrong. The>only question is what will replace his current belief/delusion. Maybe it>will be something innocuous, like pink mountains, which will take him>far from sci.math :-) Or possibly he'll come up with a short proof of>the four colour problem.>>Mark Atherton>> I _love_ that book!! The Electric Monk is one of my favorite Douglas> characters. That, and the couch stuck in the stairwell.>> Pete K.>>I liked the ?h that you put in your ear and it translates every language.I remember listening to the HGTTG on NPR when I was a kid. They also had alive-action show of it on the local PBS station, too. Zaphiod Beeblebrox wasthe best!! === >One of the a's was already shown to be coprime to 5, and now given>that b_1 and b_2 are not algebraic integers they can't be called>factors of the a's in the ring of algebraic integers, so in fact all>of the a's have now been shown to be coprime to 5, which is the result>that shows a problem with the ring of algebraic integers.> The problem is the statement so in fact all of the a's have now been shown> to be coprime to 5> > It's clear that b_1 and b_2 aren't algebraic \ integers, but nowhere have you> shown that the a's are> coprime to 5. In fact it's easy to show the opposite.It's impossible to \ prove that the a's are not coprime to 5 in the ringof algebraic integers, when m is coprime to 5.The math only works one way.Remember from my earlier post that I had P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25with the question of how that 25 divided out.Checking at Q(0) proved that it divided out like P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)so I let a_1 = 5 b_1, and a_2 = 5 b_2.The check at m=0, involved noticing that Q(0)=11, so a_1=a_2=0, atm=0, while a_3=3.Since the 25 divides out only one way--independent of m--by settingm=0, as I did, I ?ured out how it divides for all m, which is apoint that others have disputed.However, the alternative is that the 25 divide out one way for m=0,and another way for some other m, which can't be true mathematically. > Apart from that, everything appears to be correct.> I'm still waiting to see the proof.The key point in my exposition has to do with the proof that Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)is the proper way to divide out the 25, which means that two of thea's *should* have a factor that is 5, but provably do not in the ringof algebraic integers, which reveals a problem with the ring. P(m) = 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11),and my smart idea was to ?ure out how to factor that intonon-polynomial factors, which reveals the problem with algebraicintegers, as I've shown.So, the central point is that the 25 divides out *one* way for all m,so I can check at m=0, to ?d out what that one way is.James Harris === >>> a_1 = 5 b_1, and a_2 = 5 b_2, so I have>>> Q(m) = (2 b_1 + 1)(2 b_2 + 1)(2 a_3 + 5).>>I have no problem is this. Despite there being quite a few steps not shown>to your audience, your algebra appears to be correct.> I am willing to elaborate on any step.>> a_1, a_2 and a_3 are each coprime to 5 when m does not equal 0.>Prove it.>>>> That is, b_1 and b_2 are not algebraic integers when m does not equal>> 0, but the 25 can only divide out one way, so something is wrong.>It doesn't follow from this that a_1, a_2 are coprime to 5.>The only thing wrong is your conclusion above.>>Prove it. Or would you rather I proved that they're not coprime to 5?> The math can only give one answer. Here I'll elaborate on what that> answer is.> The key factorization I use is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> which gives me > a^3 + 3v a^2 - (v^3+1) = 0, where a_1, a_2, and a_3 are its roots,> and I also have that v=-1+mf^2.> Now using f=5, leads me to> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)> and from before in my previous post I have that > a_1 = 5 b_1, a_2 = 5 b_2, so> If you are saying here that b_1 and b_2are algebraic integers, you have shown this only for m = 0. For m <> 0 you have claimed itto be true but you do not have a proof, and in fact it is known to be false. Therefore in the following we must assume that b_1 and b_2 are algebraic *numbers*, not algebraic*integers*. > 25 b_1 b_2 a_3 = 25(625 m^3 - 75 m^2 + 3m)> so> b_1 b_2 a_3 = (625 m^3 - 75 m^2 + 3m)> which tells me that if m is coprime to 5, then b_1 b_2 a_3 must be as> well.> The right side is indeed clearly an integer coprime to 5. Thereforethe left side is also, even though b_1 and b_2 may not bealgebraic integers.> So now I let m=1, and then I have> 13825x^3 -72 xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> > so I have> a^3 + 72 a^2 - 13825 = 0, Note that this should be a^3 - 72*a^2 + 13825 = 0, andyou need to set y = 1.> and the three roots are the a's.> Now using a = 5b means that the equation will give me three results to> match with the three a's, where those results are a_1/5, a_2/5, and> a_3/5.> > That substitution gives me> 125b^3 + 72 (25)b^2 - 13825 = 0, and dividing off 25 gives> 5b^3 + 72 b^2 - 5530 = 0, which is what's called irreducible over Q.> A reducible polynomial over Q is just one that factors in rationals,> for instance> x^2 + 2x + 1 is reducible to (x+1)^2, > and Q is just math slang for rational numbers.> Since 5b^3 + 72 b^2 - 5530 is not reducible over rationals it's called> irreducible over Q, and also because the lead coef?ient is 5 and not> 1 or -1, it's not monic, so it's fully called a \ non-monic irreducible> over Q.> However there is math proof that an algebraic integer cannot be the> root of such a non-monic irreducible over Q, so none of its roots are> algebraic integers.> Interesting idea. You have actually learned some math.> Remember algebraic integers are de?ed to be roots of a monic> polynomial with integer coef?ients.> One of the a's was already shown to be coprime to 5, No - that previous part of your argument is wrong.It is true for m = 0, but you do not have a proof for m <> 0.> and now given> that b_1 and b_2 are not algebraic integers they can't be called> factors of the a's in the ring of algebraic integers, so in fact all> of the a's have now been shown to be coprime to 5, No, this conclusion does not follow. You are saying, for example, that a_1 = 5 * b_1,and b_1 is not an algebraic integer. Therefore a_1 must be coprime to 5. This logic is faulty. Suppose for example b_1 = 1/sqrt(5). This is an algebraic number but not an algebraic integer. However, a_1 = 5 * (1/sqrt(5)) = sqrt(5),which IS an algebraic integer, and IS NOT coprime to 5 inthe ring of algebraic integers. More generally, the following is NOT a theorem: | If a = p * b, | | where p is a prime and a is an algebraic integer | and b is an algebraic number which is NOT an | algebraic integer, then a and p are coprime.> which is the result> that shows a problem with the ring of algebraic integers.> I concede that you did have the germ of an idea here, but it does not work out to show what you wanted. There arestill two major mistakes. Nora B.> James Harris === >> >>> One of the a's was already shown to be coprime to 5, and now given>> that b_1 and b_2 are not algebraic integers they can't be called>> factors of the a's in the ring of algebraic integers, so in fact all>> of the a's have now been shown to be coprime to 5, which is the result>> that shows a problem with the ring of algebraic integers.>>The problem is the statement so in fact all of the a's have now beenshown>to be coprime to 5>>It's clear \ that b_1 and b_2 aren't algebraic integers, but nowhere haveyou>shown that the a's are>coprime to 5. In fact it's easy to \ show the opposite.>> It's impossible to prove that the a's are \ not coprime to 5 in the ring> of algebraic integers, when m is coprime to 5.Nah. It's really easy to prove. Watch: a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) =5*5*(integer coprime to 5) There are plenty of algebraic integer factors of 5 on the RHS There has to be just as many on the LHS Therefore at least one of those a's has algebraic integer factors whichare factors of 5.QED>> The math only works one way.>> Remember from my earlier post that I had>> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>> with the question of how that 25 divided out.>> Checking at Q(0) proved that it divided out like>> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>> so I let a_1 = 5 b_1, and a_2 = 5 b_2.>> The check at m=0, involved noticing that Q(0)=11, so a_1=a_2=0, at> m=0, while a_3=3.>> Since the 25 divides out only one way--independent of m--by setting> m=0, as I did, I ?ured out how it divides for all m, which is a> point that others have disputed.>> However, the alternative is that the 25 divide out one way for m=0,> and another way for some other m, which can't be true mathematically.>>Apart from that, everything appears to be correct.>>I'm still waiting to see the proof.>> The key point in my exposition has to do with the proof that>> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>> is the proper way to divide out the 25, which means that two of the> a's *should* have a factor that is 5, but provably do not in the ring> of algebraic integers, which reveals a problem with the ring.Well, the single problem this time is the statement which means that two ofthe a's *should*have a factor that is 5As you correctly say they provably do not.Why should they? Are you saying you have a problem with a_1/5, a_2/5 notbeing algebraic integers?>> P(m) = 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11),>> and my smart idea was to ?ure out how to factor that into> non-polynomial factors, which reveals the problem with algebraic> integers, as I've shown.>> So, the central point is that the 25 divides out *one* way for all m,> so I can check at m=0, to ?d out what that one way is.Truth is, you can divide out the 25 any way you want. Different ways willjust give different answers.As you've explained, Q(m) is not a monic polynomial, it's factors aren'talgebraic integers. You won'tbreak anything, so I'm happy for \ you to want to divide out the 25 in the wayyou've shown.>>> James Harris === >>>>>One of the a's was already shown to be coprime to 5, and now given>>that b_1 and b_2 are not algebraic integers they can't be called>>factors of the \ a's in the ring of algebraic integers, so in fact all>>of the a's have now been shown to be coprime to 5, which is the result>>that shows a problem with the ring of algebraic integers.>>> The problem is the statement so in fact all of the a's have now been> shown>> to be coprime to 5>>> \ It's clear that b_1 and b_2 aren't algebraic integers, but nowhere have> you>> shown that the a's are>> coprime to 5. In fact it's easy to show the opposite.>>It's impossible to \ prove that the a's are not coprime to 5 in the ring>of algebraic integers, when m is coprime to 5.> Nah. It's really easy to prove. Watch:> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)> =5*5*(integer coprime to 5)> There are plenty of algebraic integer factors of 5 on the RHS> There has to be just as many on the LHS> Therefore at least one of those a's has algebraic integer factors which> are factors of 5.> QEDWell it would seem so, but remember the ring is algebraic integers,where the de?ition of an algebraic integer is root of a monicpolynomial with integer coef?ients. I've found a problem in thering, where the mathematical reality de?s commonsense. >>The math only works one way.>>Remember from my earlier post that I had>> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>>with the question of how that 25 divided out.>>Checking at Q(0) proved that it divided out like>> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>so I let a_1 = 5 b_1, and a_2 = 5 b_2.>>The check at m=0, involved noticing that Q(0)=11, so a_1=a_2=0, at>m=0, while a_3=3.>>Since the 25 divides out only one way--independent of m--by setting>m=0, as I did, I ?ured out how it divides for all m, which is a>point that others have disputed.>>However, the alternative is that the 25 divide out one way for m=0,>and another way for some other m, which can't be true mathematically.>>> Apart from that, everything appears to be correct.>>> I'm still waiting to see the proof.>>The key point in my exposition has to do with the proof that>> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>is the proper way to divide out the 25, which means that two of the>a's *should* have a factor that is 5, but provably do not in the ring>of algebraic integers, which reveals a problem with the ring.> Well, the single problem this time is the statement which means that two of> the a's *should*> have a factor that is 5> As you correctly say they provably do not.> Why should they? Are you saying you have a problem with a_1/5, a_2/5 not> being algebraic integers?Possibly you believe that a_1/5 or a_2/5 can in some sense be afraction, but that is impossible as I showed above, as the 25 onlydivides out *one* way, and it is 5000m^3 - 600 m^2 - 126m + 11that is being factored. I simply discovered a way to factor it intonon-polynomial factors which reveals the problem with the ring ofalgebraic integers.>> P(m) = 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11),>>and my smart idea was to ?ure out how to factor that into>non-polynomial factors, which reveals the problem with algebraic>integers, as I've shown.>>So, the central point is that the 25 divides out *one* way for all m,>so I can check at m=0, to ?d out what that one way is.> Truth is, you can divide out the 25 any way you want. Different ways will> just give different answers.> As you've explained, Q(m) is not a monic polynomial, it's factors aren't> algebraic \ integers. You won't> break anything, so I'm happy for you to \ want to divide out the 25 in the way> you've shown.Let me consider your assertion about dividing out the 25, withsomething simpler like 2x^2 + 4x + 2 = (x+1)(2x+2)and, yes, you may *believe* that you can divide that 2 out any way youwant, but there's only one *sensible* way to divide it out.Possibly you're confused by the 5 in (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)but remember the polynomial is 5000m^3 -600 m^2 -126m +11 = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5). Remember the linchpin of my case is that given 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11) where 25(5000m^3 -600 m^2 - 126m + 11) =(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)the 25 divides out *one* way, so I can ?d that one way byconsidering m=0.Also remember, I used a simple method to factor the polynomial intonon-polynomial factors as I used the expression (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),where v=-1+mf^2, where I have the polynomial above using f=5, y=5,x=2, and the expression is unique in that it can be factored in morethan one way, which is why there are so many symbols.That's the point of my use of that expression. It can be factoredmore than one way.James Harris === ... > Now using a = 5b means that the equation will give me three results to > match with the three a's, where those results are a_1/5, a_2/5, and > a_3/5.... > One of the a's was already shown to be coprime to 5, and now given > that b_1 and b_2 are not algebraic integers they can't be called > factors of the a's in the ring of algebraic integers, so in fact all > of the a's have now been shown to be coprime to 5, which is the result > that shows a problem with the ring of algebraic integers.The latter conclusion is wrong. If a_1/5 = b1 is not an algebraic integer,this does *not* mean that a_1 is coprime to 5. It *only* means that 5 isnot a divisor of a_1.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [snip]> ... , but remember the ring is algebraic integers,> where the de?ition of an algebraic integer is root of a monic> polynomial with integer coef?ients. I've found a problem in the> ring, where the mathematical reality de?s commonsense.No you haven't. \ You've arrived at a false conclusion which proves there is an error in your argument.Thereis no such thing as an incomplete ring. Either the algebraic integers form a ring, in which case eachelement satis?s the requirements imposed by the de?ition of a ring, or else they do not form a ring.--There are degrees of certainty. They move from possible to plausible to convincing to conclusive. Alwaysknow which weight is appropriate to the subject at hand.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > > >>>One of the a's was already shown to be coprime to 5, and now given>>>that b_1 and b_2 are not algebraic integers they can't be called>>>factors of the a's in the ring of \ algebraic integers, so in fact all>>>of the a's have now been shown to be coprime to 5, which is the result>>>that shows a problem with the ring of algebraic integers.>>>>The problem is the statement so in fact all of the a's have now been shown>>to be coprime to 5>>>>It's clear that b_1 and b_2 aren't algebraic integers, but nowhere have you>>shown that the a's are>>coprime to 5. In fact it's easy to show the opposite.> > It's impossible to prove that the a's are not \ coprime to 5 in the ring> of algebraic integers, when m is coprime to 5.> > The math only works one way.> Remember from my earlier post that I had> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25> with the question of how that 25 divided out.> > Checking at Q(0) proved that it divided out like> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> so I let a_1 = 5 b_1, and a_2 = 5 b_2.> The check at m=0, involved noticing that Q(0)=11, so a_1=a_2=0, at> m=0, while a_3=3.> Since the 25 divides out only one way--independent of m--by setting> m=0, as I did, I ?ured out how it divides for all m, which is a> point that others have disputed.Wait, 25 divides out on way independent of m... as long as m=0??? You can't use results for m=0 to determine anything about the general case without a LOT more work. What you have shown is that 25 divides out a particular way when m=0... but not that this way works for other values of m.> However, the alternative is that the 25 divide out one way for m=0,> and another way for some other m, which can't be true mathematically.> Of course it can be true.Consider a(x) * b(x) where a(x) = 2x+5, b(x) = 20x+5.Now a(0) = 5, b(0) = 5, a(1) = 7, b(1) = 25.For both x=0, x=1, a(x)b(x) is divisible by 5. In the case x=0, both a(x) and b(x) are divisible by 5. In the case x=1, a(x) is NOT divisble by 5, but b(x) is now divisible by 25.Clear?>>Apart from that, everything appears to be correct.>>>>I'm still waiting to see the proof.> The key point in my exposition has to do with the proof that> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> is the proper way to divide out the 25, which means that two of the> a's *should* have a factor that is 5, but provably do not in the ring> of algebraic integers, which reveals a problem with the ring.> P(m) = 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11),> > and my smart idea was to ?ure out how to factor that into> non-polynomial factors, which reveals the problem with algebraic> integers, as I've shown.How does a non-monic polynomial tell you anything about solutions to a monic-polynomial? You are dealing with polynomials that don't relate to your claims. This is HIGHLY suspicious.> So, the central point is that the 25 divides out *one* way for all m,> so I can check at m=0, to ?d out what that one way is.> > James Harris-- Will Twentyman === > >>>>>>>>>>>>>>>>>>>>One of the a's was already shown to be coprime to 5, and now given>>>>>that b_1 and b_2 are not algebraic integers they can't be called>>>>>factors of the \ a's in the ring of algebraic integers, so in fact all>>>>>of the a's have now been shown to be coprime to 5, which is the result>>>>>that shows a problem with the ring of algebraic integers.>>>>>>>>The problem is the statement so in fact all of the a's have now been>>>> shown>>>>>>to be coprime to 5>>>>>>>>It's clear that b_1 and b_2 aren't algebraic integers, but nowhere have>>>> you>>>>>>shown that the a's are>>>>coprime to 5. In fact it's easy to show the opposite.>>>>>>It's impossible to prove that the a's \ are not coprime to 5 in the ring>>>of algebraic integers, when m is coprime to 5.>>>>Nah. It's really easy to prove. Watch:>> a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)>> =5*5*(integer coprime to 5)>> There are plenty of algebraic integer factors of 5 on the RHS>> There has to be just as many on the LHS>> Therefore at least one of those a's has algebraic integer factors which>>are factors of 5.>>QED> Well it would seem so, but remember the ring is algebraic integers,> where the de?ition of an algebraic integer is root of a monic> polynomial with integer coef?ients. I've found a problem in the> ring, where the mathematical reality de?s commonsense.> The polynomial is non-monic. What makes you think you are dealing with algebraic integers?>>>The math only works one way.>>>>>>Remember from my earlier post that I had>>>>>> P(m)/25 = Q(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25>>>>>>with the question of how that 25 divided out.>>>>>>Checking at Q(0) proved that it divided out like>>>>>> P(m)/25 = Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>>>>>so I let a_1 = 5 b_1, and a_2 = 5 b_2.>>>>>>The check at m=0, involved noticing that Q(0)=11, so a_1=a_2=0, at>>>m=0, while a_3=3.>>>>>>Since the 25 divides out only one way--independent of m--by setting>>>m=0, as I did, I ?ured out how it divides for all m, which is a>>>point that others have disputed.>>>>>>However, the alternative is that the 25 divide out one way for m=0,>>>and another way for some other m, which can't be true mathematically.>>>>>>>>>>Apart from that, everything appears to be correct.>>>>>>>>I'm still waiting to see the proof.>>>>>>The key point in my exposition has to do with the proof that>>>>>> Q(m) = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>>>>>is the proper way to divide out the 25, which means that two of the>>>a's *should* have a factor that is 5, but provably do not in the ring>>>of algebraic integers, which reveals a problem with the ring.>>>>Well, the single problem this time is the statement which means that two of>>the a's *should*>>have a factor that is 5>>As you correctly say they provably do not.>>Why should they? Are you saying you have a problem with a_1/5, a_2/5 not>>being algebraic integers?> Possibly you believe that a_1/5 or a_2/5 can in some sense be a> fraction, but that is impossible as I showed above, as the 25 only> divides out *one* way, and it is> 5000m^3 - 600 m^2 - 126m + 11> that is being factored. I simply discovered a way to factor it into> non-polynomial factors which reveals the problem with the ring of> algebraic integers.It only divides out ONE way, when m=0.>>> P(m) = 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11),>>>>>>and my smart idea was to ?ure out how to factor that into>>>non-polynomial factors, which reveals the problem with algebraic>>>integers, as I've shown.>>>>>>So, the central point is that the 25 divides out *one* way for all m,>>>so I can check at m=0, to ?d out what that one way is.>>>>Truth is, you can divide out the 25 any way you want. Different ways will>>just give different answers.>>As you've explained, Q(m) is not a monic polynomial, it's factors aren't>>algebraic integers. \ You won't>>break anything, so I'm happy for you to want to \ divide out the 25 in the way>>you've shown.> Let me consider your assertion about dividing out the 25, with> something simpler like> 2x^2 + 4x + 2 = (x+1)(2x+2)> and, yes, you may *believe* that you can divide that 2 out any way you> want, but there's only one *sensible* way to divide it out.> Possibly you're confused by the 5 in> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)> but remember the polynomial is> 5000m^3 -600 m^2 -126m +11 = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).> > Remember the linchpin of my case is that given> 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11) where> 25(5000m^3 -600 m^2 - 126m + 11) =(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)> the 25 divides out *one* way, so I can ?d that one way by> considering m=0.No you can't. You can ?d out how it divides out with m=0. That says nothing about m<>0.> Also remember, I used a simple method to factor the polynomial into> non-polynomial factors as I used the expression> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where v=-1+mf^2, where I have the polynomial above using f=5, y=5,> x=2, and the expression is unique in that it can be factored in more> than one way, which is why there are so many symbols.> That's the point of my use of that expression. It can be factored> more than one way.> James Harris-- Will Twentyman === >Possibly you believe that a_1/5 or a_2/5 can in some sense be a>fraction, but that is impossible as I showed above, as the 25 only>divides out *one* way, and it is>> 5000m^3 - 600 m^2 - 126m + 11>>that is being factored. I simply discovered a way to factor it into>non-polynomial factors which reveals the problem with the ring of>algebraic integers.> It only divides out ONE way, when m=0.Of course, some of you are smart enough to realize that since f=5 wasan arbitrary choice, I can choose some other value to see how many ofyou caught the mathematicians lying to you.Remember the expression is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),where v=-1+mf^2, and while before I used f=5, this time I'll usef=sqrt(2), m=1, which gives me, v=1, so I have 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),which IS reducible over Q, as you have 2x^3 - 3xy^2 + y^3 = (x - y)(2x^2 + 2xy - y^2),so you have a_3 = 1, and a_1 = -1-sqrt(3), a_2 = -1+sqrt(3).And guess what? Look through this thread and you can see people wholied to you, certain that they could get away with it.And why shouldn't they be certain? All they have to do is cast doubton me, and depend on your trust of mathematicians.Some of you may be wondering how you can tell that a_1 and a_2 have afactor of sqrt(2).Well, let a_1 = sqrt(2)b_1, so you have sqrt(2) b_1 = -1-sqrt(3), and adding 1 to both sides gives sqrt(2) b_1 + 1 = -sqrt(3), and squaring both sides gives 2b_1^2 + 2sqrt(2)b_1 + 1 = 3, which is 2b_1^2 + 2sqrt(2)b_1 - 2 = 0, and ?ally dividing off 2 gives b_1^2 + sqrt(2)b_1 -1 = 0, which is b_1^2 - 1 = -sqrt(2) b_1, and squaring and collecting gives b_1^4 - 4 b_1^2 + 1 = 0,so b_1 IS an algebraic integer, as that is a monic polynomial withinteger coef?ients.So what gives? Didn't I say the problem was that b_1 and b_2 aren'talgebraic integers? Yeah, for f=5 and a lot of other f's. It justturns out that f=sqrt(2) is a special case which is why I picked it toshow you that you'd been lied to by posters arguing with me.So why would they lie?Because the methods I've outlined here do more than provide anembarrassment by showing an over hundred year old error in taughtmathematics, they are also key methods used in a short proof ofFermat's Last Theorem that I discovered, so literally, millions ofdollars are at stake.Now if you liked being a patsy to mathematicians, who knew they couldlie to you with impunity, then sit back and laugh at the joke--whichis on you.Then remember that it's more than just some esoteric problem in abackwater of number theory that they're ?hting to hide, but also ashort proof of \ Fermat's Last Theorem by an unwanted outsider.I just showed you how easily mathematicians lie to you.Read more below with your eyes open this time. >Let me consider your assertion about dividing out the 25, with>something simpler like>> 2x^2 + 4x + 2 = (x+1)(2x+2)>>and, yes, you may *believe* that you can divide that 2 out any way you>want, but there's only one *sensible* way to divide it out.>>Possibly you're confused by the 5 in>> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>but remember the polynomial is> 5000m^3 -600 m^2 -126m +11 = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).>Remember the linchpin of my case is that given>> 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11) where>> 25(5000m^3 -600 m^2 - 126m + 11) =(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)>>the 25 divides out *one* way, so I can ?d that one way by>considering m=0.> No you can't. You can ?d out how it divides out with m=0. That says > nothing about m<>0.And isn't that really all a mathematician has to do, dispute anassertion without even bothering to back up what they say withmathematics?You *trust* them.>>Also remember, I used a simple method to factor the polynomial into>non-polynomial factors as I used the expression>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>>where v=-1+mf^2, where I have the polynomial above using f=5, y=5,>x=2, and the expression is unique in that it can be factored in more>than one way, which is why there are so many symbols.>>That's the point of my use of that expression. It can be factored>more than one way.The math is basic algebra. Now consider the power of mathematiciansto lie about basic algebra, beyond the question of how they couldpossibly do such a thing, consider their demonstrated ability here toalmost get away with it.Millions of dollars and prestige around the world is at stake. I'mone man forced to stand stand against the mathematical world becausethey're lying about my discoveries, and I now know they have one holdcard--your trust.Yup, if you follow the math and logic, then the truth comes out, andmaybe you start wondering where mathematicians learned their lessons.How do they know they can lie in such a situation?So what's in it for you?Well why should they lie and get away with it? Besides, don't youfeel even a little bit stupid for being taken in with basic *algebra*?And worse come to worse, since my story should be worth severalmillion dollars, I'll consider some kind of monetary compensation forsomeone who has CLEARLY and to a GREAT DEGREE helped my cause and beenONLY POSITIVE, as I'm wary of the people who know the truth who'vebeen lying trying to do an about face. I said *consider* and I don'tlike it, but I'm facing the reality of the real power of the mathworld.These people have the power to lie when it comes to math, and theyclearly know it.James Harris === >>[...]>>And worse come to worse, since my story should be worth several>million dollars, I'll consider some kind of monetary compensation for>someone who has CLEARLY and to a GREAT DEGREE helped my cause and been>ONLY POSITIVE, as I'm wary of the people who know the truth who've>been lying trying to do an about face. Fascinating, but not new - you've offered to pay people to agree you're right before. Didn't work \ then, so I don't see why you'dthink it would work now, since \ you're not doing anything different.If you want someone to take you up on this you need to be muchmore speci? - say exactly how much you're willing to pay andspecify _exactly_ what sort of support would earn payment.Someone would have to be a fool to fall for an offer speci?das fuzzily as what you say above.Oh yeah, you also need to somehow give a _guarantee_ thatthe supporter will be paid. Not that anyone has any reasonto think that you would lie, but everyone knows that there area lot of bad people out there, so people tend not to truststrangers. Maybe you could deposit a few thousand dollarsin advance with a trusted third party and publish the detailsof the contract signed by the third party, where he agreesto give the money to someone who meets certain clearlyspeci?d conditions.>I said *consider* and I don't>like it, but I'm facing the reality of the real \ power of the math>world.I'll _consider_ paying you $10,000 if you agree never topost anything related to mathematics on usenet.>These people have the power to lie when it comes to math, and they>clearly know it.>>>James Harris************************David C. Ullrich === > >> Possibly you believe that a_1/5 or a_2/5 can in some sense be a>> fraction, but that is impossible as I showed above, as the 25 only>> divides out *one* way, and it is>>> 5000m^3 - 600 m^2 - 126m + 11>>> that is being factored. I simply discovered a way to factor it into>> non-polynomial factors which reveals the problem with the ring of>> algebraic integers.>>It only divides out ONE way, when m=0.> > Of course, some of you are smart enough to realize that since f=5 was> an arbitrary choice, I can choose some other value to see how many of> you caught the mathematicians lying to you.> Remember the expression is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where v=-1+mf^2, and while before I used f=5, this time I'll use> f=sqrt(2), m=1, which gives me, v=1, so I have> 2x^3 - 3xy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> which IS reducible over Q, as you have> 2x^3 - 3xy^2 + y^3 = (x - y)(2x^2 + 2xy - y^2),> so you have a_3 = 1, and a_1 = -1-sqrt(3), a_2 = -1+sqrt(3).> And guess what? Look through this thread and you can see people who> lied to you, certain that they could get away with it.> And why shouldn't they be certain? All they have to do is cast doubt> on me, and depend on your trust of mathematicians.> Some of you may be wondering how you can tell that a_1 and a_2 have a> factor of sqrt(2).> Well, let a_1 = sqrt(2)b_1, so you have > sqrt(2) b_1 = -1-sqrt(3), and adding 1 to both sides gives> sqrt(2) b_1 + 1 = -sqrt(3), and squaring both sides gives> 2b_1^2 + 2sqrt(2)b_1 + 1 = 3, which is> 2b_1^2 + 2sqrt(2)b_1 - 2 = 0, and ?ally dividing off 2 gives> b_1^2 + sqrt(2)b_1 -1 = 0, which is> b_1^2 - 1 = -sqrt(2) b_1, and squaring and collecting gives> b_1^4 - 4 b_1^2 + 1 = 0,> so b_1 IS an algebraic integer, as that is a monic polynomial with> integer coef?ients.> JSH's underlying claim here is that if P(x) = (v^3 + 1)*x^3 - 3*v*x*y^2 + y^3,and v = -1 + m*f^2, and a1, a2, and a3 are algebraic integerssuch that P(x) = (a1*x + y)*(a2*x + y)*(a3*x + y),then exactly two of a1, a2, and a3 are divisible in the algebraicintegers by f, while a3 is coprime to f. JSH has claimed a great length that this is true when m = 1and f = 5. JSH's opponents say that is false. JSH now says, look, I can show it explicitly for f = sqrt(2).Therefore my opponents are lying. Actually what JSH's opponents say is that if P(x) = (v^3 + 1)*x^3 -3*v*x + 1is a polynomial with integer coef?ients as shown,and is irreducible over the rationals, and if P(x) isfactored in the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1),and q is any prime factor of A, then EACH of a1, a2,and a3 is an algebraic integer which is not coprime to f.(again here, v = -1 + m*f^2). It does appear that JSH and his opponents do not agreeon this. Are the opponents lying about the case where f = 5? No. There is a key important difference. The opponentsassume P(x) is IRREDUCIBLE, and that is the case when m = 1 and f = 5 or f = sqrt(5). However P(x) is NOTirreducible when f = sqrt(2), as JSH has noted above. There is no contradiction. The opponents are not lying.> So what gives? Didn't I say the problem was that b_1 and b_2 aren't> algebraic integers? Yeah, for f=5 and a lot of other f's. It just> turns out that f=sqrt(2) is a special case Exactly. It is one case where P(x) turns out not tobe irreducible. But it is not suf?ient for your purposesto focus only on reducible P(x). In general, for variousvalues of m and f, you cannot expect that P(x) is irreducible.> which is why I picked it to> show you that you'd been lied to by posters arguing with me.> So why would they lie?> They did not lie. Their proofs apply when P(x) is *irreducible*,and that is has been explicitly stated over and over again.Two such cases that you have considered at length are m = 1 andf = sqrt(5) or f = 5. P(x) is irreducible in both of these.> Because the methods I've outlined here do more than provide an> embarrassment by showing an over hundred year old error in taught> mathematics, they are also key methods used in a short proof of> Fermat's Last Theorem that I discovered, so literally, millions of> dollars are at stake.> Now if you liked being a patsy to mathematicians, who knew they could> lie to you with impunity, then sit back and laugh at the joke--which> is on you.> Then remember that it's more than just some esoteric problem in a> backwater of number theory that they're ?hting to hide, but also a> short proof of Fermat's Last Theorem by an unwanted outsider.> I just showed you how easily mathematicians lie to you.> Read more below with your eyes open this time.>> Let me consider your assertion about dividing out the 25, with>> something simpler like>>> 2x^2 + 4x + 2 = (x+1)(2x+2)>>> and, yes, you may *believe* that you can divide that 2 out any way you>> want, but there's only one *sensible* way to divide it out.>>> Possibly you're confused by the 5 in>>> (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5)>>> but remember the polynomial is>>> 5000m^3 -600 m^2 -126m +11 = (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5).>>> Remember the linchpin of my case is that given>>> 25 Q(m) = 25(5000m^3 - 600 m^2 - 126m + 11) where>>> 25(5000m^3 -600 m^2 - 126m + 11) =(2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)>>> the 25 divides out *one* way, so I can ?d that one way by>> considering m=0.>>No you can't. You can ?d out how it divides out with m=0. That says >nothing about m<>0.> And isn't that really all a mathematician has to do, dispute an> assertion without even bothering to back up what they say with> mathematics?> You *trust* them.> You shouldn't, and you don't have to. All you have to do isread our proofs, which speak for themselves. If you simplytake our word for it without reading the math, JSH is right.>>> Also remember, I used a simple method to factor the polynomial into>> non-polynomial factors as I used the expression>>> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),>>> where v=-1+mf^2, where I have the polynomial above using f=5, y=5,>> x=2, and the expression is unique in that it can be factored in more>> than one way, which is why there are so many symbols.>>> That's the point of my use of that expression. It can be factored>> more than one way.> The math is basic algebra. Most of it is. But the step where you generalize from m = 0to m <> 0 is not algebra, but an intuitive leap. You are sayingwhen m = 0, a1/5 is 0, and is therefore an algebraic integer.Therefore (you say) a1/5 must be an algebraic integer when m <> 0.Would be true if a1 were a constant, but it is not. It can beregarded as a function of m. Its being divisible by 5 when m = 0says nothing about its divisibility for other values of m. No, it is not algebra and it is not basic. In fact it is false.There are two explicit disproofs of it (mine and that of W. DaleHall). JSH has not posted a valid objection to either one. Maybe this will all be clearer in the quadratic case than it is in the cubic case. Here is a parallel of the JSH argument for P(x)a quadratic: Let P(x) = 5*m*x^2 + (m + 1)*x*y + y^2. Suppose P(x) is factored in the form (a1*x + y)*(a2*x + y). Let y = 5. Then P(x) = 5*m*x^2 + 5*(m + 1)*x + 25,and the factorization is then (a1*x + 5)*(a2*x + 5). Now let m = 0: P(x) = 5 * x + 25 = 5*(x + 5). This means that the factorization must be a1*x + 5 = 5, and a2*x + 5 = x + 5,where a1 = 0 and a2 = 1. Therefore when m = 0, the factorization canbe written in the form 5 * ((a1/5)*x + 1) * (a2*x + 5),where a1/5 is an algebraic integer. Note thatwhen m = 0, a2 = 1, which is coprime to 5. Does this work? Note that if I divide P(x) by 5, (1/5)*P(x) = m*x^2 + (m + 1)*x + 5,which must equal ((a1/5)*x + 1)*(a2*x + 5). Note that (1/5)*P(x) = 6 when m = 0, andof course in that case a1/5 = 0 and a2 = 1,so it does seem like everything checks out.If this form for the factorization is rightand a1/5 is an algebraic integer, it must bethe case that a2 is coprime to 5. For example,if m = 1 and x = 1, (1/5)*P(x) = 8, which is coprime to 5. If a2 were not coprimeto 5, this could not happen. Right?Looks like everything is pretty consistentwith what JSH says. Right?.............. But consider again (1/5)*P(x) = m*x^2 + (m + 1)*x + 5again. Let m = 1. It is: (1/5)*P(x) = Q(x) = x^2 + 2*x + 5. This is irreducible over the rationals. It factors as Q(x) = (x - r1)*(x - r2)where r1 = (-2 + sqrt(-16)))/2 = -1 + 2*iand r2 = (-2 - sqrt(-16)))/2 = -1 - 2*i.Q(x) is monic. Both r1 and r2 are algebraic integers.As it happens, both r1 and r2 are divisors of5 in the algebraic numbers. In fact, r1 = 5/(-1 - 2*i) and r2 = 5/(-1 + 2*i).This is pretty obvious actually when you notethat r1*r2 = 5.Now if the factorization of Q(x) is writtenas ((a1/5)*x + 1)*(a2*x + 5),then it must be the case that: -1/(a1/5) = r1 and -5/a2 = r2.(or vice versa). This implies a1 = -5/r1 and a2 = -5/r2.But -5/r1 = -r2, and -5/r2 = -r1.Therefore a1 = -r2 = 1 + 2*i, and a2 = -r1 = 1 - 2*i.And BOTH of these numbers are divisors of 5in the algebraic integers. Note that it is NOT true that a1/5 is an algebraic integer. It is (1 + 2*i)/5. OK, JSH can say: yes, all this is just arithmeticand it doesn't prove anything about MY argumentbecause I was talking about an irreducible cubic.*Not good enough*. JSH will have to ?d an explicitstep in the argument above where his argument forthe irreducible cubic breaks down when it is applied to the quadratic. Most of us already knowwhere that step is, eh?> Now consider the power of mathematicians> to lie about basic algebra, beyond the question of how they could> possibly do such a thing, consider their demonstrated ability here to> almost get away with it.> Millions of dollars and prestige around the world is at stake. I'm> one man forced to stand stand against the mathematical world because> they're lying about my discoveries, and I now know they have one hold> card--your trust.> Yup, if you follow the math and logic, then the truth comes out, and> maybe you start wondering where mathematicians learned their lessons.> > How do they know they can lie in such a situation?> So what's in it for you?> Well why should they lie and get away with it? Besides, don't you> feel even a little bit stupid for being taken in with basic *algebra*?> And worse come to worse, since my story should be worth several> million dollars, I'll consider some kind of monetary compensation for> someone who has CLEARLY and to a GREAT DEGREE helped my cause and been> ONLY POSITIVE, as I'm wary of the people who know the truth who've> been lying trying to do an about face. I said *consider* and I don't> like it, but I'm facing the reality of the real power of the math> world.> Looks like Nora B. will not get a cut.> These people have the power to lie when it comes to math, and they> clearly know it.> Point out where in the above post I have lied,or where in my proof (posted elsewhere) that the main result of Advanced Polynomial Factorization is wrong. Nora B.> James Harris === > For a while now I've been issuing a clarion call about a problem in a> somewhat esoteric branch of mathematics, where relatively basic> algebra reveals a strange problem that has been in the discipline for> over a hundred years.> Recently objections to my work have centered on a particular> factorization where the actual objections can help readers see why it> is necessary that I go outside the math community for help:> The key factorization I use is> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where v=-1+mf^2, and the ring is something called algebraic integers,> which is just the set of numbers that are roots of monic polynomials> with integer coef?ient, like x^2 + 2x + 2, where the lead> coef?ient is 1, so it is monic, and its coef?ients are 1, 2, and> 2, which are, of course, integers.> Posters for some time on the math newsgroups have managed to object by> casting doubt on the possibility that x and y are constant while only> m varies.> Yet that is refuted by considering that the a's are the roots of> a^3 + 3va^2 - (v^3+1)=0, where remember v=-1+mf^2,> so it's hard to understand how anyone could object by worrying about x> and y.> Why do they object? Good question but it is relevant that they are> trying to cast doubt on my ability to use a value at m=0, where two of> the a's equal 0, and one equals 3, with the expression> (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),> where things get just complicated enough for confusion.> OK. Let's let v = -1 + m*f^2, f = sqrt(5), and y = 5.In this case the polynomial is R(x) = 65*x^3 - 300*x + 125 = 5 *(13*x^3 - 60*x + 25). If this is factored as above, R(x) = (a1*x + 5)*(a2*x + 5)*(a3*x + 5),then JSH claims two of the a's are divisible by5 in the algebraic integers, and the third oneis coprime to 5. Thus assume that a3 is coprime to 5. Let x = u / 13. The equation R(x) = 0 implies that[1] u^3 - 60*13*u + 25*13^2 = 0. This may be written as u*(u^2 - 60*13) = -25*13^2. Let d be any root of [1]. Thus d*(d^2 - 60*13) = -25*13^2. Now if we assume [*] d is coprime to 5, this implies that d^2 - 60*13 is NOT coprime to 5. However, since 60 is a multiple of 5, d^2 alsomust not be coprime to 5. This implies, d itself must not be coprime to 5. Thiscontradicts our assumption [*] and we conclude:Each root of [1] is an algebraic integer which isnot coprime to 5. Now let r be any root of R(x). This means 5*(13*r^3 - 60*r + 25) = 0.Since this is assumed to be factored in the form (a1*x + 5)*(a2*x + 5)*(a3*x + 5),we can assume r = -5/a3. We have shown above that d = 13*r is an algebraic integerwhich is not coprime to 5. Therefore [because the algbraic integers are a Bezout domain] there exist algebraic integers c, c', and e such that 13*r = c * e and 5 = c' * e,and c and c' \ are coprime and nonunits.Therefore -5/a3 = -(c'*e)/a3 = 13*c*e, and a3 = -c' / (13*c), or 13*a3*c = -c'.Since 13 and c \ are coprime to c', a3 must NOT becoprime to c'. \ c' is a nonunit divisor of 5.Therefore a3 is not coprime to 5.And thus *another* proof that JSH's claim aboutthe roots of P(x) is incorrect. Nora Baron[snip] === > For a while now I've been issuing a clarion call about a problem in a> somewhat esoteric branch of mathematics, where relatively basic> algebra reveals a strange problem that has been in the discipline for> over a hundred years.> Recently objections to my work have centered on a particular> factorization where the actual objections can help readers see why it> is necessary that I go outside the math community for help:Did you try the psychiatric community for help with your problems?They offer the best likelihood of effective solutions! === > There was a misprint in the following section, at *****:> Actually what JSH's opponents say is that if> P(x) = (v^3 + 1)*x^3 -3*v*x + 1> is a polynomial with integer coef?ients as shown,> and is irreducible over the rationals, and if P(x) is> factored in the form> (a1*x + 1)*(a2*x + 1)*(a3*x + 1),>***** and q is any prime factor of A, then EACH of a1, a2,> and a3 is an algebraic integer which is not coprime to f.> (again here, v = -1 + m*f^2).> Here is what it should have said -> Actually what JSH's opponents say is that if>> P(x) = (v^3 + 1)*x^3 -3*v*x + 1>> is a polynomial with integer coef?ients as shown,> and is irreducible over the rationals, and if P(x) is> factored in the form>> (a1*x + 1)*(a2*x + 1)*(a3*x + 1),>> then EACH of a1, a2, and a3 is an algebraic integer which> is not coprime to f.(again here, v = -1 + m*f^2). Nora B. === There were some errors in the following section -> Therefore > -5/a3 = -(c'*e)/a3 = 13*c*e, and> a3 = -c' / (13*c), or> \ 13*a3*c = -c'.>>> Since 13 and c are coprime to c', a3 must \ NOT be> coprime to c'. c' is a nonunit divisor of 5.> \ Therefore a3 is not coprime to 5. What this should have been was -5/a3 = r, and therefore -13*5/a3 = 13*r = c * e, which yields -13 * c' * e / a3 = c * e, or -13 * c' = a3 * c.Now if c' is a nonunit, the conclusion followsas before. The fact that c and c' are coprime does not imply that either is not a unit. So faras I can tell it is possible for c' to be a unit. Thus I do not see a way to rescue this proof. === === ==> And thus *another* proof that JSH's claim about> the roots of P(x) is incorrect.> No, this proof does not quite seem to work. Howeverthe other proofs by me and W. Dale Hall still apply. Nora B. > Nora Baron> [snip] === I'm trying without great success, to prove to myself why James Harris' a_3 isnotcoprime to 5.Focussing here on Nora Baron's proof.>> Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are> integers and c = p*v, where p is a prime and v is another> integer. Q(x) is clearly monic. Assume Q(x) is> irreducible. Let a1, a2, and a3 be roots of Q(x). Note> that by de?ition, a1, a2, and a3 are algebraic> integers.>> You are claiming that at least one of a1, a2, or a3 is> coprime to p. *** Assume a1 is coprime to p. ***>> By standard theory, there exists an automorphism F12 of> the ?ld of algebraic numbers such that:>> 1. F12 leaves the sub?ld of rational numbers ?ed,> i.e., if q is rational, F12(q) = q.>> 2. F12(a1) = a2.>> 3. If t is an algebraic integer, F12(t) is also an> algebraic integer.>> Now since a1 is relatively prime to p, there exist> algebraic integers r and s such that>> [1] r*a1 + s*p = 1.>> Now apply the automorphism F12 to both sides of [1]:>> F12(r)*F12(a1) + F12(s)*F12(p) = F12(1).>> By the properties above, F12(p) = p and F12(1) = 1.> Moreover, r' = F12(r) is an algebraic integer, and> s' = F12(s) is an algebraic integer. Finally,> F12(a1) = a2. Thus one obtains:>> r'*a2 + s'*p = 1,>> which says: a2 and p \ are coprime in the algebraic> integers.>> Similarly one shows that a3 and p are coprime.>> Therefore if one of a1, a2, or a3 is coprime> to p, then they all are.>> But a1 * a2 * a3 = p * v. That is, p divides> the product of a1, a2, and a3. Therefore p cannot> be coprime to each of a1, a2, and a3.>> Putting all this together, one concludes that> NONE of a1, a2, or a3 can be coprime to p.>> This directly contradicts what you have claimed:> that at least one of a1, a2, or a3 is coprime to p.>> Your response?>OK, let Q(x) = x^3 - 3x^2 - 25x + 75One of the prime factors of 75 is 3, so let p=3So your proof tells me that none of the roots of my Q(x) are coprime to 3.But the roots are 3,5, -5.So either your proof is a dud or my understanding is wrong.Could you clarify for me? === @ I'm trying without great success, to prove to myself why James Harris' a_3 is@ not@ coprime to 5.@ Focussing here on Nora Baron's proof.@ @ >@ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are@ > integers and c = p*v, where p is a prime and v is another@ > integer. Q(x) is clearly monic. Assume Q(x) is@ > irreducible.[cut] @ OK, let Q(x) = x^3 - 3x^2 - 25x + 75@ One of the prime factors of 75 is 3, so let p=3@ @ So your proof tells me that none of the roots of my Q(x) are coprime to 3.@ @ But the roots are 3,5, -5.@ So either your proof is a dud or my understanding is wrong.@ @ Could you clarify for me?Your Q(x) is not irreducible.You have Q(x) = (x-3)(x-5)(x+5).-- Bill Hale === >> @ I'm trying without great success, to prove to myself why James Harris' a_3is> @ not> @ coprime to 5.> @ Focussing here on Nora Baron's proof.> @> @ >> @ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are> @ > integers and c = p*v, where p is a prime and v is another> @ > integer. Q(x) is clearly monic. Assume Q(x) is> @ > irreducible.> [cut]> @ OK, let Q(x) = x^3 - 3x^2 - 25x + 75> @ One of the prime factors of 75 is 3, so let p=3> @> @ So your proof tells me that none of the roots of my Q(x) are coprime to 3.> @> @ But the roots are 3,5, -5.> @ So either your proof is a dud or my understanding is wrong.> @> @ Could you clarify for me?>> Your Q(x) is not irreducible.>> You have Q(x) = (x-3)(x-5)(x+5).>> -- Bill HaleAhh, my understanding is wrong.Phil. === >>@ I'm trying without great success, to prove to myself why James Harris' a_3> is>@ not>@ coprime to 5.>@ Focussing here on Nora Baron's proof.>@>@ >>@ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are>@ > integers and c = p*v, where p is a prime and v is another>@ > integer. Q(x) is clearly monic. Assume Q(x) is>@ > irreducible.>[cut]>@ OK, let Q(x) = x^3 - 3x^2 - 25x + 75>@ One of the prime factors of 75 is 3, so let p=3>@>@ So your proof tells me that none of the roots of my Q(x) are coprime to 3.>@>@ But the roots are 3,5, -5.>@ So either your proof is a dud or my understanding is wrong.>@>@ Could you clarify for me?>>Your Q(x) is not irreducible.>>You have Q(x) = (x-3)(x-5)(x+5).>>-- Bill Hale> Ahh, my understanding is wrong.> Phil.Don't be too quick to believe them as despite irreducibility there's away to test whether or not Nora Baron knows what she's talking about,which is to have her do an ACTUAL automorphism. That is, you can plugin a value into a formula and it spits out an actual number. You cantest her by getting her to try and give the automorphism she claims todepend on for her argument, and then get her to put in a number.Remember, there ARE experts who can look at Nora Baron's work and tellyou what they think, but where are they? For instance, Arturo Magidinhas in the past made effort to point out he's an expert in GaloisTheory.James Harris === >I'm trying without great success, to prove to myself why James Harris' a_3 is>not>coprime to 5.>Focussing here on Nora Baron's proof.>>>>> Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are>> integers and c = p*v, where p is a prime and v is another>> integer. Q(x) is clearly monic. Assume Q(x) is>> irreducible. Let a1, a2, and a3 be roots of Q(x). Note>> that by de?ition, a1, a2, and a3 are algebraic>> integers.>>>> You are claiming that at least one of a1, a2, or a3 is>> coprime to p. *** Assume a1 is coprime to p. ***>>>> By standard theory, there exists an automorphism F12 of>> the ?ld of algebraic numbers such that:>>>> 1. F12 leaves the sub?ld of rational numbers ?ed,>> i.e., if q is rational, F12(q) = q.>>>> 2. F12(a1) = a2.>>>> 3. If t is an algebraic integer, F12(t) is also an>> algebraic integer.>>>> Now since a1 is relatively prime to p, there exist>> algebraic integers r and s such that>>>> [1] r*a1 + s*p = 1.>>>> Now apply the automorphism F12 to both sides of [1]:>>>> F12(r)*F12(a1) + F12(s)*F12(p) = F12(1).>>>> By the properties above, F12(p) = p and F12(1) = 1.>> Moreover, r' = F12(r) is an algebraic integer, and>> s' = F12(s) is an algebraic integer. Finally,>> F12(a1) = a2. Thus one obtains:>>>> r'*a2 + s'*p = 1,>>>> which says: a2 and \ p are coprime in the algebraic>> integers.>>>> Similarly one shows that a3 and p are coprime.>>>> Therefore if one of a1, a2, or a3 is coprime>> to p, then they all are.>>>> But a1 * a2 * a3 = p * v. That is, p divides>> the product of a1, a2, and a3. Therefore p cannot>> be coprime to each of a1, a2, and a3.>>>> Putting all this together, one concludes that>> NONE of a1, a2, or a3 can be coprime to p.>>>> This directly contradicts what you have claimed:>> that at least one of a1, a2, or a3 is coprime to p.>>>> Your response?>>>>OK, let Q(x) = x^3 - 3x^2 - 25x + 75>One of the prime factors of 75 is 3, so let p=3>>So your proof tells me that none of the roots of my Q(x) are coprime to 3.>>But the roots are 3,5, -5.>So either your proof is a dud or my understanding is wrong.Look at lines 3 and 4:>> Assume Q(x) is>> irreducible. Let a1, a2, and a3 be roots of Q(x). You see where it says assume Q(x) is irreducible?Your Q(x) is NOT irreducible: it factors asQ(x) = (x-3)(x-5)(x+5).The key to Nora's argument, which was pointed out to James almost twoyears ago, is the irreducibility of Q(x) (over the rationals). WHENQ(x) is irreducible of degree 3, his conclusions cannot follow. Andfor most values of his parameters, Q(x) is irreducible. === == === =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === === === =Arturo Magidinmagidin@math.berkeley.edu === >>>>@ I'm trying without great success, to prove to myself why James Harris' a_3>> is>>@ not>>@ coprime to 5.>>@ Focussing here on Nora Baron's proof.>>@>>@ >>>@ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are>>@ > integers and c = p*v, where p is a prime and v is another>>@ > integer. Q(x) is clearly monic. Assume Q(x) is>>@ > irreducible.>>[cut]>>@ OK, let Q(x) = x^3 - 3x^2 - 25x + 75>>@ One of the prime factors of 75 is 3, so let p=3>>@>>@ So your proof tells me that none of the roots of my Q(x) are coprime to 3.>>@>>@ But the roots are 3,5, -5.>>@ So either your proof is a dud or my understanding is wrong.>>@>>@ Could you clarify for me?>>>>Your Q(x) is not irreducible.>>>>You have Q(x) = (x-3)(x-5)(x+5).>>>>-- Bill Hale>>> Ahh, my understanding is wrong.>>> Phil.>>Don't be too quick to believe them as despite irreducibility there's a>way to test whether or not Nora Baron knows what she's talking about,>which is to have her do an ACTUAL automorphism.The existence of the automorphism is not in question, and she hasgiven everything that is needed to actually calculate it. Just becauseyou are too ignorant to know it does not mean that she has not done so.> That is, you can plug>in a value into a formula and it spits out an actual number. You can>test her by getting her to try and give the automorphism she claims to>depend on for her argument, and then get her to put in a number.>>Remember, there ARE experts who can look at Nora Baron's work and tell>you what they think, but where are they? Where do you think they should be? Making posts saying yeah, that'sright? Nora's argument \ is essentially the very same one I posted overa year ago to disprove your nonsense claims.> For instance, Arturo Magidin>has in the past made effort to point out he's an expert in Galois>Theory.Provide a citation, liar.I have ->never<- claimed or insinuated that I am an expert in GaloisTheory. But I guess that, from the position of utter ignorance(which is, after all, where ->you<- are) anyone who knows the basicsmust look like an expert.No, I am not an expert in Galois theory. I know some Galois Theory,certainly a bit beyond what is covered in a typical ?st semestergraduate course in Algebra, but not much more beyond that. For therecord, the amount of Galois Theory needed in the disproof of yourclaims is what is technically refered to as piddling. === === ===Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === === === =Arturo Magidinmagidin@math.berkeley.edu === >>>>>>@ I'm trying without great success, to prove to myself why James Harris' a_3>>> is>>>@ not>>>@ coprime to 5.>>>@ Focussing here on Nora Baron's proof.>>>@>>>@ >>>>@ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are>>>@ > integers and c = p*v, where p is a prime and v is another>>>@ > integer. Q(x) is clearly monic. Assume Q(x) is>>>@ > irreducible.>>>[cut]>>>@ OK, let Q(x) = x^3 - 3x^2 - 25x + 75>>>@ One of the prime factors of 75 is 3, so let p=3>>>@>>>@ So your proof tells me that none of the roots of my Q(x) are coprime to 3.>>>@>>>@ But the roots are 3,5, -5.>>>@ So either your proof is a dud or my understanding is wrong.>>>@>>>@ Could you clarify for me?>>>>>>Your Q(x) is not irreducible.>>>>>>You have Q(x) = (x-3)(x-5)(x+5).>>>>>>-- Bill Hale>>>>> Ahh, my understanding is wrong.>>>>> Phil.>>>>Don't be too quick to believe them as despite irreducibility there's a>>way to test whether or not Nora Baron knows what she's talking about,>>which is to have her do an ACTUAL automorphism.>>The existence of the automorphism is not in question, and she has>given everything that is needed to actually calculate it. Just because>you are too ignorant to know it does not mean that she has not done so.>>> That is, you can plug>>in a value into a formula and it spits out an actual number. You can>>test her by getting her to try and give the automorphism she claims to>>depend on for her argument, and then get her to put in a number.>>>>Remember, there ARE experts who can look at Nora Baron's work and tell>>you what they think, but where are they? >>Where do you think they should be? Making posts saying yeah, that's>right?Yeah, that's right.> \ Nora's argument is essentially the very same one I posted over>a year ago to disprove your nonsense claims.>>> For instance, Arturo Magidin>>has in the past made effort to point out he's an expert in Galois>>Theory.>>Provide a citation, liar.>>I have ->never<- claimed or insinuated that I am an expert in Galois>Theory. But I guess that, from the position of utter ignorance>(which is, after all, where ->you<- are) anyone who knows the basics>must look like an expert.>>No, I am not an expert in Galois theory. I know some Galois Theory,>certainly a bit beyond what is covered in a typical ?st semester>graduate course in Algebra, but not much more beyond that. For the>record, the amount of Galois Theory needed in the disproof of your>claims is what is technically refered to as piddling.What's funny is that my recollection is that the _last_ time Jamesaccused you of being an expert in Galois theory you said moreor less exactly what you say here. Memory problems, I guess.> === == === =>Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A man's capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of ?ures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the inde?ite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> === == === =>>Arturo Magidin>magidin@math.berkeley.edu>>>************************ David C. Ullrich === [.snip.]>>> For instance, Arturo Magidin>>>has in the past made effort to point out he's an expert in Galois>>>Theory.>>>>Provide a citation, liar.>>>>I have ->never<- claimed or insinuated that I am an expert in Galois>>Theory. But I guess that, from the position of utter ignorance>>(which is, after all, where ->you<- are) anyone who knows the basics>>must look like an expert.>>>>No, I am not an expert in Galois theory. I know some Galois Theory,>>certainly a bit beyond what is covered in a typical ?st semester>>graduate course in Algebra, but not much more beyond that. For the>>record, the amount of Galois Theory needed in the disproof of your>>claims is what is technically refered to as piddling.>>What's funny is that my recollection is that the _last_ time James>accused you of being an expert in Galois theory you said more>or less exactly what you say here. Memory problems, I guess.You know, I had completely forgotten about that; I had to do a googlesearch to ?d the post...http://groups.google.com/groups?selm=apbvds%242ok3% 241%40agate.berkeley.edu >No I was addressing your statement that you see no progress and noted >that would not be unexpected if you're actually an expert in the >?ld. \ I'm sorry, but you are simply wrong here. First, I am not an expert in Galois Theory; the Galois Theory I've been using is baby Galois Theory; the most basic; the stuff I learned as an undergraduate, which is taught to undergraduates.Hmmm... Amazing, that I can keep my lies straight even withoutremembering them... Must be like what Richard Schiff's character saidin an episode of _The West Wing_: I told him [Presidential CandidateJosiah Bartlet] to tell the truth, if for no other reason because itis easier to remember. === === ===Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === === === =Arturo Magidinmagidin@math.berkeley.edu===> >>> @ I'm trying without great success, to prove to myself why James Harris' a_3> is>> @ not>> @ coprime to 5.>> @ Focussing here on Nora Baron's proof.>> @>> @ >>> @ > Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are>> @ > integers and c = p*v, where p is a prime and v is another>> @ > integer. Q(x) is clearly monic. Assume Q(x) is>> @ > irreducible.>> [cut]>> @ OK, let Q(x) = x^3 - 3x^2 - 25x + 75>> @ One of the prime factors of 75 is 3, so let p=3>> @>> @ So your proof tells me that none of the roots of my Q(x) are coprime to 3.>> @>> @ But the roots are 3,5, -5.>> @ So either your proof is a dud or my understanding is wrong.>> @>> @ Could you clarify for me?>>> Your Q(x) is not irreducible.>>> You have Q(x) = (x-3)(x-5)(x+5).>>> -- Bill Hale>>Ahh, my understanding is wrong.>>Phil.> Don't be too quick to believe them as despite irreducibility there's a> way to test whether or not Nora Baron knows what she's talking about,> which is to have her do an ACTUAL automorphism. That is, you can plug> in a value into a formula and it spits out an actual number. You can> test her by getting her to try and give the automorphism she claims to> depend on for her argument, and then get her to put in a number.> Actually this is a good question. Essentially you need ?st topermute two of the roots, say, a1 and a2. This is simple enough.You use that permutation in an obvious way to construct an isomorphism of Q(a1) and Q(a2) within the splitting ?ld for the polynomial. That too is simple and straightforward. Then you extend that isomorphism to an automorphism F12 of the splitting ?ld itself, using the cited theorem in the Beachy-Blair book. This is all relatively straightforward. But that is not enough. Other algebraic integers are involved in the expression of coprimeness of a1 and p (p = 5 in the examples discussed here): r*a1 + s*5 = 1,i.e., r and s are algebraic integers which may not be in the splitting ?ld of the polynomial. You need to extend the automorphism of the splitting ?ld to a larger ?ld thatincludes r and s, as well as r' and s' mentioned later on, That such an \ extension exists is assured by Theorem 17.30of Algebra - A Graduate Course, by I. M. Isaacs ofU. Wisconsin [Brooks/Cole Publ. Co. Paci? Grove, CA,1994]. It is notable that Isaacs invokes Zorn's Lemma forthis. I am not sure that is absolutely necessary for a?ite extension of the splitting ?ld as would be associated with adding in r, s, r', and s'. Nora B.> Remember, \ there ARE experts who can look at Nora Baron's work and tell> you what they think, but where are they? For instance, Arturo Magidin> has in the past made effort to point out he's an expert in Galois> Theory.> James Harris === ... but not in?itessimal? > record, the amount of Galois Theory needed in the disproof of your> claims is what is technically refered to as piddling.--Dec.2000 ?WAND' Chairman Paul O'Neill, reelectedto Board. Newsish?http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanac === >[snip] >Remember, there ARE experts who can look at Nora Baron's work and tell>you what they think, but where are they?>> Where do you think they should be? Making posts saying yeah, that's> right?>[snip]It is my impression that some posters to sci.math, Robert Israel forexample, are rarely followed up. Somebody asks a question, Robert providesthe answer. Nobody bothers to say yeah, that's right :)-- Clive Toothhttp://www.clivetooth.dk === >> [.snip.]>>>>> For instance, Arturo Magidin>>>>has in the past made effort to point out he's an expert in Galois>>>>Theory.>>>>>>Provide a citation, liar.>>>>>>I have ->never<- claimed or insinuated that I am an expert in Galois>>>Theory. But I guess that, from the position of utter ignorance>>>(which is, after all, where ->you<- are) anyone who knows the basics>>>must look like an expert.>>>>>>No, I am not an expert in Galois theory. I know some Galois Theory,>>>certainly a bit beyond what is covered in a typical ?st semester>>>graduate course in Algebra, but not much more beyond that. For the>>>record, the amount of Galois Theory needed in the disproof of your>>>claims is what is technically refered to as piddling.>>>>What's funny is that my recollection is that the _last_ time James>>accused you of being an expert in Galois theory you said more>>or less exactly what you say here. Memory problems, I guess.>>>You know, I had completely forgotten about that; I had to do a google>search to ?d the post...>>http://groups.google.com/groups?selm=apbvds%242ok3% 241%40agate.berkeley.edu>>No I was addressing your statement that you see no progress and noted>that would not be unexpected if you're actually an expert in the>?ld.>> I'm sorry, but you are simply wrong here. First, I am not an expert> in Galois Theory; the Galois Theory I've been using is baby Galois> Theory; the most basic; the stuff I learned as an undergraduate, which> is taught to undergraduates.>>>Hmmm... Amazing, that I can keep my lies straight even without>remembering them... Indeed.>Must be like what Richard Schiff's character said>in an episode of _The West Wing_: I told him [Presidential Candidate>Josiah Bartlet] to tell the truth, if for no other reason because it>is easier to remember.Heh-heh. Although of course the idea of advising someone to tellthe truth because it's easier to keep things straight that way doesnot originate with The West Wing (don't recall who I have in mindhere, but it was no doubt not original with him either.)> === === ===>Why do you take so much trouble to expose such a reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A man's capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of ?ures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the inde?ite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> === == === =>>Arturo Magidin>magidin@math.berkeley.edu************************ David C. Ullrich === >>>[snip]>>Remember, there ARE experts who can look at Nora Baron's work and tell>>you what they think, but where are they?>>>> Where do you think they should be? Making posts saying yeah, that's>> right?>>[snip]>>It is my impression that some posters to sci.math, Robert Israel for>example, are rarely followed up. Somebody asks a question, Robert provides>the answer. Nobody bothers to say yeah, that's right :)Yeah, that's right.************************David C. Ullrich === [.snip.]>>Hmmm... Amazing, that I can keep my lies straight even without>>remembering them... >>Indeed.>>>Must be like what Richard Schiff's character said>>in an episode of _The West Wing_: I told him [Presidential Candidate>>Josiah Bartlet] to tell the truth, if for no other reason because it>>is easier to remember.>>Heh-heh. Although of course the idea of advising someone to tell>the truth because it's easier to keep things straight that way does>not originate with The West Wing (don't recall who I have in mind>here, but it was no doubt not original with him either.)Oh, it is almost certainly something old; my guess is you would beable to track it down to some Parliamentary debate at some pointbefore losing it in the mists of time...It does raise an interesting follow-up, though, in that James seems tocontinually have to withdraw and/or amend his recollections of thepast. While we seem to have no trouble keeping our lies straight,even without remembering, James seems to have trouble keeping thetruth straight, even when he does google searches ahead of time... === === === =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === === === =Arturo Magidinmagidin@math.berkeley.edu===> > ...>Must be like what Richard Schiff's character said> >in an episode of _The West Wing_: I told him [Presidential Candidate>Josiah Bartlet] to tell the truth, if for no other reason because it>is easier to remember.> Heh-heh. Although of course the idea of advising someone to tell> the truth because it's easier to keep things straight that way does> not originate with The West Wing (don't recall who I have in mind> here, but it was no doubt not original with him either.)>Well, Richard Whately, Archbishop of Dublin 1831-1863 said Honesty is the best policy; but he who is governed by that maxim is not an honest man.Hence (perhaps) Karl Popper's reworking: Honesty is better than policy.GC === > ...> It is my impression that some posters to sci.math, Robert Israel for> example, are rarely followed up. Somebody asks a question, Robert provides> the answer. Nobody bothers to say yeah, that's right :)> Wow, I have something in common with Robert Israel :-(GC === >>>> ...>>>Must be like what Richard Schiff's character said>> >in an episode of _The West Wing_: I told him [Presidential Candidate>>Josiah Bartlet] to tell the truth, if for no other reason because it>>is easier to remember.>>> Heh-heh. Although of course the idea of advising someone to tell>> the truth because it's easier to keep things straight that way does>> not originate with The West Wing (don't recall who I have in mind>> here, but it was no doubt not original with him either.)>>>>>>Well, Richard Whately, Archbishop of Dublin 1831-1863 said>> Honesty is the best policy; but he who is governed> by that maxim is not an honest man.>>Hence (perhaps) Karl Popper's reworking:>> Honesty is better than policy.>>>GCWell, a quick google found it attributed to Quintillion MendacemOportet Esse Memorem .. usually loosely translated ?a liar needs agood memory'.Larry(this space unintentially left blank ..... === > What exactly is a meter to the fourth power?I don't know, it's only a math question.> How does square meters per cubic second mean absorbed energy?I don't know, it's only a \ math question.> Better you should ask: does anyone care?I hope so. A couple do. Please, help me.Becki>Please de?e what you are looking for more clearly.>>In particular, explain why you believe it's not the case that an>in?ite number of fundamental constants could be generated.>>-Jonathan>>>Given seven (7) SI System base values; >>>>1) luminous intensity = 1.9720204(06) x 10^-45 cd >>2) Einstein time = 1.3511888(38) x 10^-43 s>>3) wavelength of gravity = 4.0507622(30) x 10^-35 m >>4) uni?d substance = 1.6605402(10) x 10^-27 kmol >>5) gravitational mass = 5.4563086(06) x 10^-8 kg>>6) electric current = 1.1857538(29) x 10^24 A >>7) Celsius temperature = 3.5518470(84) x 10^32 K >>>>And two (2) SI System base angles;>>>8) relative permeability = 6.8517994(75) x 10^1 sr >>9) inverse ?e-structure = 1.3703598(95) x 10^2 rad >> >>How many primary fundamental physical>>constants could possibly be generated?>>>Does anyone know?>>Because the physical world is 3, perhaps 4, dimensions; andstudying the following 133, the units never exceed the 4th powerof any base unit. So to rephrase the question, how many uniquecombinations of the above 7 SI System base units with the 2 SISystem base angles can be generated to the 4th dimension?Becki[?st 133 known primary physical properties below]001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg 002) volume of gravity = 6.6467639(49) x 10^-104 m^3 003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg 004) luminous ef?acy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 005) current volume = 1.3838179(77) x 10^-93 m^2/A 006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s 007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s 008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 009) gravitational ?y = 1.0031222(77) x 10^-70 m-s/kg 010) area of gravity = 1.6408674(64) x 10^-69 m^2011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s013) Samuel constant = 3.4161915(66) x 10^-59 m/A014) electric moment = 6.4900394(48) x 10^-54 A-s-m 015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 019) luminous intensity = 1.9720204(06) x 10^-45 cd 020) Einstein time = 1.3511888(38) x 10^-43 s 021) luminous ?1.3511888(38) x 10^-43 cd-sr 022) gravitational moment = 2.2102208(82) x 10^-42 kg-m 023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 024) absorption-emission = 2.4763790(61) x 10^-36 s/kg 025) wavelength of gravity = 4.0507622(30) x 10^-35 m 026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s 028) relative expansion = 2.8154365(22) x 10^-33 /K 029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 031) uni?d substance = 1.6605402(10) x 10^-27 kmol 032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s 033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2035) inverse electric current = 8.4334536(86) x 10^-25 /A 036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K 038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg 041) elementary charge = 1.6021773(38) x 10^-19 A-s 042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2043) ?st radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s045) speci? heat = 2.5303881(55) x 10^-16 m^2/s^2-K 046) magnetic ?4.1356692(24) x 10^-15 kg-m^2/A-s^2 047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg 049) permittivity of vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m 051) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 052) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 053) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 054) radiant distribution constant = 3.335640952... x 10^-9 s/m 055) gravitational mass constant = 5.4563086(06) x 10^-8 kg 056) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr 057) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 058) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 059) ?e-structure constant = 7.2973530(80) x 10^-3 /rad 060) second radiation constant = 1.4387688(01) x 10^-2 m-K 061) dielectric constant = 1.4594706(16) x 10^-2 /sr 062) spin half constant = 0.500000000 sr/rad 063) spin two constant = 2.000000000 rad/sr 064) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s 065) relative permeability = 6.8517994(75) x 10^1 sr 066) inverse ?e-structure = 1.3703598(95) x 10^2 rad 067) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K 068) spin angle constant = 9.3894312(09) x 10^3 sr-rad 069) Joel constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m070) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 071) Amos constant = 1.8997879(14) x 10^5 A^2-s072) inverse gravitational mass = 1.8327409(10) x 10^7 /kg 073) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol 074) speed of light in vacuum = 2.99792458 x 10^8 m/s 075) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 076) Obadiah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 077) Jonah constant = 5.6954208(84) x 10^13 A^2-s078) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 079) electric displacement = 3.9552490(31) x 10^15 A-s/m080) speci? energy = 8.9875517(87) x 10^16 m^2/s^2081) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 082) Micah constant = 1.4701479(23) x 10^18 kg-m^3/s^3083) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 084) molar mass constant = 3.2858635(84) x 10^19 kg/kmol 085) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 086) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K 087) Nahum constant = 7.2489467(08) x 10^22 A-s/kg-m088) electric current constant = 1.1857538(29) x 10^24 A 089) luminance constant = 1.2018157(77) x 10^24 cd/m^2 090) Habakkuk constant = 2.6944002(42) x 10^25 m^3/s^3091) luminous ?nsity = 8.2346007(05) x 10^25 cd-sr/m^2 092) Zephaniah constant = 4.4073925(95) x 10^26 kg-m^4/s^4093) Avogadro constant = 6.0221366(15) x 10^26 /kmol094) gravitational ?ld = 1.3469831(84) x 10^27 kg/m 095) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 096) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 097) Celcius temperature = 3.5518470(84) x 10^32 K 098) Haggai constant = 8.0776087(15) x 10^33 m^4/s^4099) gravity wave number = 2.4686711(86) x 10^34 /m 100) mass ?te constant = 4.0381539(96) x 10^35 kg/s 101) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol 102) Malachi constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3103) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 104) frequency of gravity = 7.4008900(30) x 10^42 /s 105) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 106) inverse luminous intensity = 5.0709414(41) x 10^44 /cd 107) angular velocity = 1.0141882(88) x 10^45 rad/s108) Zechariah constant = 8.5954663(19) x 10^46 A^2-s/kg-m 109) Matthew constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4110) Mark constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3111) electric ?nsity = 9.7642093(17) x 10^49 A-s/m^2 112) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr 113) gravity ?ld strength = 2.2187310(14) x 10^51 m/s^2 114) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 115) magnetic ?nsity = 2.5204163(73) x 10^54 kg/A-s^2 116) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K 117) Luke constant = 1.7895265(87) x 10^57 A-s/kg-m^2118) magnetic ?ld strength = 2.9272363(12) x 10^58 A/m 119) absorbed energy = 6.6515882(43) x 10^59 m^2/s^3120) John constant = 1.0880403(11) x 10^61 kg-m^3/s^4121) surface density constant = 3.3252585(75) x 10^61 kg/m^2 122) electric ?ld strength = 7.5560181(98) x 10^62 kg-m/A-s^3 123) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s 124) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 125) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 126) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 127) angular acceleration = 7.5058959(93) x 10^87 rad/s^2128) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K 129) current density constant = 7.2263839(39) x 10^92 A/m^2 130) gravitational density = 8.2089700(31) x 10^95 kg/m^3 131) energy density = 7.3778543(28) x 10^112 kg/m-s^ 2 132) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr133) irradiance constant = 2.2118250(84) x 10^121 kg/s^3etc., etc., etc. === >>What exactly is a meter to the fourth power?> > I don't know, it's only a math question.>>How does square meters per cubic second mean absorbed energy?> I don't know, it's only a math question.>>Better you \ should ask: does anyone care?> I hope so. A couple do. Please, help me.> Becki>>Please de?e what you are looking for more clearly.>>>>In particular, explain why you believe it's not the case that an>>in?ite number of fundamental constants could be generated.>>>>-Jonathan>>>>>>>Given seven (7) SI System base values; >>>>>>1) luminous intensity = 1.9720204(06) x 10^-45 cd >>>2) Einstein time = 1.3511888(38) x 10^-43 s>>>3) wavelength of gravity = 4.0507622(30) x 10^-35 m >>>4) uni?d substance = 1.6605402(10) x 10^-27 kmol >>>5) gravitational mass = 5.4563086(06) x 10^-8 kg>>>6) electric current = 1.1857538(29) x 10^24 A >>>7) Celsius temperature = 3.5518470(84) x 10^32 K >>>>>>And two (2) SI System base angles;>>>>>8) relative permeability = 6.8517994(75) x 10^1 sr >>>9) inverse ?e-structure = 1.3703598(95) x 10^2 rad >>>>>>How many primary fundamental physical>>>constants could possibly be generated?>>>>>>Does anyone know?>>>> Because the physical world is 3, perhaps 4, dimensions; and> studying the following 133, the units never exceed the 4th power> of any base unit. So to rephrase the question, how many unique> combinations of the above 7 SI System base units with the 2 SI> System base angles can be generated to the 4th dimension?> Becki> As I recall, there are theories that go up to 8 or 10 dimensions. The last one I heard was a Hausdorf Space cubed.Now which of those dimensions would be spacial, and which temporal is beyond me.-- Will Twentyman === > As I recall, there are theories that go up to 8 or 10 dimensions.The question is only up to four (4) dimensions.How many unique combinations would be possible?1) luminous intensity = 1.9720204(06) x 10^-45 cd 2) Einstein time = 1.3511888(38) x 10^-43 s3) wavelength of gravity = 4.0507622(30) x 10^-35 m 4) uni?d substance = 1.6605402(10) x 10^-27 kmol 5) gravitational mass = 5.4563086(06) x 10^-8 kg6) electric current = 1.1857538(29) x 10^24 A 7) Celsius temperature = 3.5518470(84) x 10^32 K 8) relative permeability = 6.8517994(75) x 10^1 sr 9) inverse ?e-structure = 1.3703598(95) x 10^2 rad The following is only a guide.Becki--First 144 Primary Fundamental Physical Constants 001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg 002) volume of gravity = 6.6467639(49) x 10^-104 m^3 003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg 004) luminous ef?acy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 005) current volume = 1.3838179(77) x 10^-93 m^2/A 006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s 007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s 008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 009) gravitational ?y = 1.0031222(77) x 10^-70 m-s/kg 010) area of gravity = 1.6408674(64) x 10^-69 m^2011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s013) Samuel constant = 3.4161915(66) x 10^-59 m/A014) electric moment = 6.4900394(48) x 10^-54 A-s-m 015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 019) luminous intensity = 1.9720204(06) x 10^-45 cd 020) Einstein time = 1.3511888(38) x 10^-43 s 021) luminous ?1.3511888(38) x 10^-43 cd-sr 022) gravitational moment = 2.2102208(82) x 10^-42 kg-m 023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 024) absorption-emission = 2.4763790(61) x 10^-36 s/kg 025) wavelength of gravity = 4.0507622(30) x 10^-35 m 026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s 028) relative expansion = 2.8154365(22) x 10^-33 /K 029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 031) uni?d substance = 1.6605402(10) x 10^-27 kmol 032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s 033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2035) inverse electric current = 8.4334536(86) x 10^-25 /A 036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K 038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg 041) elementary charge = 1.6021773(38) x 10^-19 A-s 042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2043) ?st radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s045) speci? heat = 2.5303881(55) x 10^-16 m^2/s^2-K 046) magnetic ?4.1356692(24) x 10^-15 kg-m^2/A-s^2 047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg 049) permittivity vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m 051) Joel constant = 4.8682699(55) x 10^-11 cd/m052) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 053) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 054) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 055) radiant distribution constant = 3.335640952... x 10^-9 s/m 056) gravitational mass constant = 5.4563086(06) x 10^-8 kg 057) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr 058) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 059) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 060) Amos constant = 1.9689919(12) x 10^-3 A-kmol061) ?e-structure constant = 7.2973530(80) x 10^-3 /rad 062) second radiation constant = 1.4387688(01) x 10^-2 m-K 063) dielectric constant = 1.4594706(16) x 10^-2 /sr 064) Obadiah constant = 1.4594706(16) x 10^-2 cd/s065) spin half constant = 5.000000000 x 10^-1 sr/rad 066) length fraction = 1.000000000 m/m067) mass fraction = 1.000000000 kg/kg068) time fraction = 1.000000000 s/s069) current fraction = 1.000000000 amp/amp070) temperature fraction = 1.000000000 K/K071) spin two constant = 2.000000000 rad/sr 072) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s 073) Jonah constant = 6.8517994(74) x 10^1 s/cd074) relative permeability = 6.8517994(75) x 10^1 sr 075) inverse ?e-structure = 1.3703598(95) x 10^2 rad 076) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K 077) spin angle constant = 9.3894312(09) x 10^3 sr-rad 078) Micah constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m079) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 080) Nahum constant = 1.8997879(14) x 10^5 A^2-s081) Habakkuk constant = 5.8979849(03) x 10^5 kmol-K082) inverse gravitational mass = 1.8327409(10) x 10^7 /kg 083) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol 084) speed of light in vacuum = 2.99792458 x 10^8 m/s 085) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 086) Zephaniah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 087) Haggai constant = 2.0541178(06) x 10^10 m/cd088) Zechariah constant = 5.6954208(84) x 10^13 A^2-s089) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 090) electric displacement = 3.9552490(31) x 10^15 A-s/m091) absorbed dose = 8.9875517(87) x 10^16 m^2/s^2092) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 093) Malachi constant = 1.4701479(23) x 10^18 kg-m^3/s^3094) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 095) molar mass constant = 3.2858635(84) x 10^19 kg/kmol 096) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 097) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K 098) Matthew constant = 7.2489467(08) x 10^22 A-s/kg-m099) electric current constant = 1.1857538(29) x 10^24 A 100) luminance constant = 1.2018157(77) x 10^24 cd/m^2 101) Mark constant = 2.6944002(42) x 10^25 m^3/s^3102) luminous ?nsity = 8.2346007(05) x 10^25 cd-sr/m^2 103) Luke constant = 4.4073925(95) x 10^26 kg-m^4/s^4104) Avogadro constant = 6.0221366(15) x 10^26 /kmol105) gravitational ?ld = 1.3469831(84) x 10^27 kg/m 106) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 107) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 108) Celcius temperature = 3.5518470(84) x 10^32 K 109) John constant = 8.0776087(15) x 10^33 m^4/s^4110) gravity wave number = 2.4686711(86) x 10^34 /m 111) mass ?te constant = 4.0381539(96) x 10^35 kg/s 112) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol 113) Timothy constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3114) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 115) frequency of gravity = 7.4008900(30) x 10^42 /s 116) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 117) inverse luminous intensity = 5.0709414(41) x 10^44 /cd 118) angular velocity = 1.0141882(88) x 10^45 rad/s119) Titus constant = 8.5954663(19) x 10^46 A^2-s/kg-m 120) Philemon constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4121) James constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3122) electric ?nsity = 9.7642093(17) x 10^49 A-s/m^2 123) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr 124) gravity ?ld strength = 2.2187310(14) x 10^51 m/s^2 125) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 126) magnetic ?nsity = 2.5204163(73) x 10^54 kg/A-s^2 127) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K 128) Peter constant = 1.7895265(87) x 10^57 A-s/kg-m^2129) magnetic ?ld strength = 2.9272363(12) x 10^58 A/m 130) absorbed dose rate = 6.6515882(43) x 10^59 m^2/s^3131) Jude constant = 1.0880403(11) x 10^61 kg-m^3/s^4132) surface density constant = 3.3252585(75) x 10^61 kg/m^2 133) electric ?ld strength = 7.5560181(98) x 10^62 kg-m/A-s^3 134) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s 135) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 136) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 137) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 138) angular acceleration = 7.5058959(93) x 10^87 rad/s^2139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K 140) current density constant = 7.2263839(39) x 10^92 A/m^2 141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 142) energy density = 7.3778543(28) x 10^112 kg/m-s^ 2 143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3 === >As I recall, there are theories that go up to 8 or 10 dimensions.> > The question is only up to four (4) dimensions.> How many unique combinations would be possible?Ok, you got 9 fundamental units, any of which can appearin the numerator or the denominator up to a power of 4.A unit cannot appear in _both_ numerator and denominatorbecause the smaller power would cancel.Labeling the constants a..i and looking at just the aconstant, there are 9 ways (out of 25 possible) that aappears in the fraction (na is the power of a in thenumerator and da is the power of a in the denominator):na da0 00 10 20 30 41 02 03 04 0And since you have 9 constants, there are9^9 or 387,420,489 combinations.A typical example is combination #24,376:na nb nc nd ne nf ng nh ni da db dc dd de df dg dh di0 4 0 2 0 0 0 0 0 3 0 3 0 3 0 0 0 0which isa^0 b^4 c^0 d^2 e^0 f^0 g^0 h^0 i^0-----------------------------------a^3 b^0 c^3 d^0 e^3 f^0 g^0 h^0 i^0or s^4 kmol^2-------------cd^3 m^3 kg^3> 1) luminous intensity = 1.9720204(06) x 10^-45 cd > 2) Einstein time = 1.3511888(38) x 10^-43 s> 3) wavelength of gravity = 4.0507622(30) x 10^-35 m > 4) uni?d substance = 1.6605402(10) x 10^-27 kmol > 5) gravitational mass = 5.4563086(06) x 10^-8 kg> 6) electric current = 1.1857538(29) x 10^24 A > 7) Celsius temperature = 3.5518470(84) x 10^32 K > 8) relative permeability = 6.8517994(75) x 10^1 sr > 9) inverse ?e-structure = 1.3703598(95) x 10^2 rad > The following is only a guide.> Becki> --> First 144 Primary Fundamental Physical Constants Good luck coming up with names for the other 387,420,345 constants.> 001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg > 002) volume of gravity = 6.6467639(49) x 10^-104 m^3 > 003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg > 004) luminous ef?acy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 > 005) current volume = 1.3838179(77) x 10^-93 m^2/A > 006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s > 007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s > 008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 > 009) gravitational ?y = 1.0031222(77) x 10^-70 m-s/kg > 010) area of gravity = 1.6408674(64) x 10^-69 m^2> 011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m> 012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s> 013) Samuel constant = 3.4161915(66) x 10^-59 m/A> 014) electric moment = 6.4900394(48) x 10^-54 A-s-m > 015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2> 016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 > 017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s> 018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 > 019) luminous intensity = 1.9720204(06) x 10^-45 cd > 020) Einstein time = 1.3511888(38) x 10^-43 s > 021) luminous ?1.3511888(38) x 10^-43 cd-sr > 022) gravitational moment = 2.2102208(82) x 10^-42 kg-m > 023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 > 024) absorption-emission = 2.4763790(61) x 10^-36 s/kg > 025) wavelength of gravity = 4.0507622(30) x 10^-35 m > 026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4> 027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s > 028) relative expansion = 2.8154365(22) x 10^-33 /K > 029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 > 030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 > 031) uni?d substance = 1.6605402(10) x 10^-27 kmol > 032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s > 033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3> 034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2> 035) inverse electric current = 8.4334536(86) x 10^-25 /A > 036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s> 037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K > 038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 > 039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m> 040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg > 041) elementary charge = 1.6021773(38) x 10^-19 A-s > 042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2> 043) ?st radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 > 044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s> 045) speci? heat = 2.5303881(55) x 10^-16 m^2/s^2-K > 046) magnetic ?.135 6692(24) x 10^-15 kg-m^2/A-s^2 > 047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 > 048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg > 049) permittivity vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 > 050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m > 051) Joel constant = 4.8682699(55) x 10^-11 cd/m> 052) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 > 053) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 > 054) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 > 055) radiant distribution constant = 3.335640952... x 10^-9 s/m > 056) gravitational mass constant = 5.4563086(06) x 10^-8 kg > 057) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr > 058) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 > 059) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 > 060) Amos constant = 1.9689919(12) x 10^-3 A-kmol> 061) ?e-structure constant = 7.2973530(80) x 10^-3 /rad > 062) second radiation constant = 1.4387688(01) x 10^-2 m-K > 063) dielectric constant = 1.4594706(16) x 10^-2 /sr > 064) Obadiah constant = 1.4594706(16) x 10^-2 cd/s> 065) spin half constant = 5.000000000? x 10^-1 sr/rad > 066) length fraction = 1.000000000? m/m> 067) mass fraction = 1.000000000? kg/kg> 068) time fraction = 1.000000000? s/s> 069) current fraction = 1.000000000? amp/amp> 070) temperature fraction = 1.000000000? K/K> 071) spin two constant = 2.000000000? rad/sr > 072) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s > 073) Jonah constant = 6.8517994(74) x 10^1 s/cd> 074) relative permeability = 6.8517994(75) x 10^1 sr > 075) inverse ?e-structure = 1.3703598(95) x 10^2 rad > 076) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K > 077) spin angle constant = 9.3894312(09) x 10^3 sr-rad > 078) Micah constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m> 079) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 > 080) Nahum constant = 1.8997879(14) x 10^5 A^2-s> 081) Habakkuk constant = 5.8979849(03) x 10^5 kmol-K> 082) inverse gravitational mass = 1.8327409(10) x 10^7 /kg > 083) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol > 084) speed of light in vacuum = 2.99792458 x 10^8 m/s > 085) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 > 086) Zephaniah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 > 087) Haggai constant = 2.0541178(06) x 10^10 m/cd> 088) Zechariah constant = 5.6954208(84) x 10^13 A^2-s> 089) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 > 090) electric displacement = 3.9552490(31) x 10^15 A-s/m> 091) absorbed dose = 8.9875517(87) x 10^16 m^2/s^2> 092) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 > 093) Malachi constant = 1.4701479(23) x 10^18 kg-m^3/s^3> 094) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 > 095) molar mass constant = 3.2858635(84) x 10^19 kg/kmol > 096) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 > 097) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K > 098) Matthew constant = 7.2489467(08) x 10^22 A-s/kg-m> 099) electric current constant = 1.1857538(29) x 10^24 A > 100) luminance constant = 1.2018157(77) x 10^24 cd/m^2 > 101) Mark constant = 2.6944002(42) x 10^25 m^3/s^3> 102) luminous ?nsity = 8.2346007(05) x 10^25 cd-sr/m^2 > 103) Luke constant = 4.4073925(95) x 10^26 kg-m^4/s^4> 104) Avogadro constant = 6.0221366(15) x 10^26 /kmol> 105) gravitational ?ld = 1.3469831(84) x 10^27 kg/m > 106) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 > 107) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 > 108) Celcius temperature = 3.5518470(84) x 10^32 K > 109) John constant = 8.0776087(15) x 10^33 m^4/s^4> 110) gravity wave number = 2.4686711(86) x 10^34 /m > 111) mass ?te constant = 4.0381539(96) x 10^35 kg/s > 112) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol > 113) Timothy constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3> 114) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 > 115) frequency of gravity = 7.4008900(30) x 10^42 /s > 116) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 > 117) inverse luminous intensity = 5.0709414(41) x 10^44 /cd > 118) angular velocity = 1.0141882(88) x 10^45 rad/s> 119) Titus constant = 8.5954663(19) x 10^46 A^2-s/kg-m > 120) Philemon constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4> 121) James constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3> 122) electric ?nsity = 9.7642093(17) x 10^49 A-s/m^2 > 123) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr > 124) gravity ?ld strength = 2.2187310(14) x 10^51 m/s^2 > 125) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 > 126) magnetic ?nsity = 2.5204163(73) x 10^54 kg/A-s^2 > 127) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K > 128) Peter constant = 1.7895265(87) x 10^57 A-s/kg-m^2> 129) magnetic ?ld strength = 2.9272363(12) x 10^58 A/m > 130) absorbed dose rate = 6.6515882(43) x 10^59 m^2/s^3> 131) Jude constant = 1.0880403(11) x 10^61 kg-m^3/s^4> 132) surface density constant = 3.3252585(75) x 10^61 kg/m^2 > 133) electric ?ld strength = 7.5560181(98) x 10^62 kg-m/A-s^3 > 134) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s > 135) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 > 136) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 > 137) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 > 138) angular acceleration = 7.5058959(93) x 10^87 rad/s^2> 139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K > 140) current density constant = 7.2263839(39) x 10^92 A/m^2 > 141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 > 142) energy density = 7.3778543(28) x 10^112 kg/m-s^ 2 > 143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr> 144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3 === The following is a suggestion whichBecki--Instead of using biblical names for the next 387,420,345 constants(there are not enough names), why not use AT&T, Kingston, Verizon,Surely there are at least 144,000 unique names for the new constantsthere. Some of them might even be physicists' names. Then if youneed more names, you can always resort to their yellow pages. I ampositive you will ?d companies favorable to having a physicalproperty, and its corresponding constant, named after them.White Pages-- First 144 Primary Fundamental Physical Constants 001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg 002) volume of gravity = 6.6467639(49) x 10^-104 m^3 003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg 004) luminous ef?acy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 005) current volume = 1.3838179(77) x 10^-93 m^2/A 006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s 007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s 008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 009) gravitational ?y = 1.0031222(77) x 10^-70 m-s/kg 010) area of gravity = 1.6408674(64) x 10^-69 m^2011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s013) Samuel constant = 3.4161915(66) x 10^-59 m/A014) electric moment = 6.4900394(48) x 10^-54 A-s-m 015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 019) luminous intensity = 1.9720204(06) x 10^-45 cd 020) Einstein time = 1.3511888(38) x 10^-43 s 021) luminous ?1.3511888(38) x 10^-43 cd-sr 022) gravitational moment = 2.2102208(82) x 10^-42 kg-m 023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 024) absorption-emission = 2.4763790(61) x 10^-36 s/kg 025) wavelength of gravity = 4.0507622(30) x 10^-35 m 026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s 028) relative expansion = 2.8154365(22) x 10^-33 /K 029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 031) uni?d substance = 1.6605402(10) x 10^-27 kmol 032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s 033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2035) inverse electric current = 8.4334536(86) x 10^-25 /A 036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K 038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg 041) elementary charge = 1.6021773(38) x 10^-19 A-s 042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2043) ?st radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s045) speci? heat = 2.5303881(55) x 10^-16 m^2/s^2-K 046) magnetic ?4.1356692(24) x 10^-15 kg-m^2/A-s^2 047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg 049) permittivity of vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m 051) Joel constant = 4.8682699(55) x 10^-11 cd/m052) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 053) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 054) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 055) radiant distribution constant = 3.335640952... x 10^-9 s/m 056) gravitational mass constant = 5.4563086(06) x 10^-8 kg 057) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr 058) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 059) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 060) Amos constant = 1.9689919(12) x 10^-3 A-kmol061) ?e-structure constant = 7.2973530(80) x 10^-3 /rad 062) second radiation constant = 1.4387688(01) x 10^-2 m-K 063) dielectric constant = 1.4594706(16) x 10^-2 /sr 064) Obadiah constant = 1.4594706(16) x 10^-2 cd/s065) spin half constant = 5.000000000 x 10^-1 sr/rad 066) length fraction = 1.000000000 m/m067) mass fraction = 1.000000000 kg/kg068) time fraction = 1.000000000 s/s069) current fraction = 1.000000000 amp/amp070) temperature fraction = 1.000000000 K/K071) spin two constant = 2.000000000 rad/sr 072) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s 073) Jonah constant = 6.8517994(74) x 10^1 s/cd074) relative permeability = 6.8517994(75) x 10^1 sr 075) inverse ?e-structure = 1.3703598(95) x 10^2 rad 076) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K 077) spin angle constant = 9.3894312(09) x 10^3 sr-rad 078) Micah constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m079) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 080) Nahum constant = 1.8997879(14) x 10^5 A^2-s081) Habakkuk constant = 5.8979849(03) x 10^5 kmol-K082) inverse gravitational mass = 1.8327409(10) x 10^7 /kg 083) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol 084) speed of light in vacuum = 2.99792458 x 10^8 m/s 085) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 086) Zephaniah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 087) Haggai constant = 2.0541178(06) x 10^10 m/cd088) Zechariah constant = 5.6954208(84) x 10^13 A^2-s089) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 090) electric displacement = 3.9552490(31) x 10^15 A-s/m091) absorbed dose = 8.9875517(87) x 10^16 m^2/s^2092) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 093) Malachi constant = 1.4701479(23) x 10^18 kg-m^3/s^3094) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 095) molar mass constant = 3.2858635(84) x 10^19 kg/kmol 096) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 097) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K 098) Matthew constant = 7.2489467(08) x 10^22 A-s/kg-m099) electric current constant = 1.1857538(29) x 10^24 A 100) luminance constant = 1.2018157(77) x 10^24 cd/m^2 101) Mark constant = 2.6944002(42) x 10^25 m^3/s^3102) luminous ?nsity = 8.2346007(05) x 10^25 cd-sr/m^2 103) Luke constant = 4.4073925(95) x 10^26 kg-m^4/s^4104) Avogadro constant = 6.0221366(15) x 10^26 /kmol105) gravitational ?ld = 1.3469831(84) x 10^27 kg/m 106) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 107) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 108) Celcius temperature = 3.5518470(84) x 10^32 K 109) John constant = 8.0776087(15) x 10^33 m^4/s^4110) gravity wave number = 2.4686711(86) x 10^34 /m 111) mass ?te constant = 4.0381539(96) x 10^35 kg/s 112) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol 113) Timothy constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3114) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 115) frequency of gravity = 7.4008900(30) x 10^42 /s 116) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 117) inverse luminous intensity = 5.0709414(41) x 10^44 /cd 118) angular velocity = 1.0141882(88) x 10^45 rad/s119) Titus constant = 8.5954663(19) x 10^46 A^2-s/kg-m 120) Philemon constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4121) James constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3122) electric ?nsity = 9.7642093(17) x 10^49 A-s/m^2 123) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr 124) gravity ?ld strength = 2.2187310(14) x 10^51 m/s^2 125) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 126) magnetic ?nsity = 2.5204163(73) x 10^54 kg/A-s^2 127) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K 128) Peter constant = 1.7895265(87) x 10^57 A-s/kg-m^2129) magnetic ?ld strength = 2.9272363(12) x 10^58 A/m 130) absorbed dose rate = 6.6515882(43) x 10^59 m^2/s^3131) Jude constant = 1.0880403(11) x 10^61 kg-m^3/s^4132) surface density constant = 3.3252585(75) x 10^61 kg/m^2 133) electric ?ld strength = 7.5560181(98) x 10^62 kg-m/A-s^3 134) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s 135) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 136) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 137) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 138) angular acceleration = 7.5058959(93) x 10^87 rad/s^2139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K 140) current density constant = 7.2263839(39) x 10^92 A/m^2 141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 142) energy density = 7.3778543(28) x 10^112 kg/m-s^2 143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3>> As I recall, there are theories that go up to 8 or 10 dimensions.>>The question is only up to four (4) dimensions.>How many unique combinations would be possible?>>> Ok, you got 9 fundamental units, any of which can appear> in the numerator or the denominator up to a power of 4.> A unit cannot appear in _both_ numerator and denominator> because the smaller power would cancel.> Labeling the constants a..i and looking at just the a> constant, there are 9 ways (out of 25 possible) that a> appears in the fraction (na is the power of a in the> numerator and da is the power of a in the denominator):> na da> 0 0> 0 1> 0 2> 0 3> 0 4> 1 0> 2 0> 3 0> 4 0> And since you have 9 constants, there are> 9^9 or 387,420,489 combinations.> A typical example is combination #24,376:> na nb nc nd ne nf ng nh ni da db dc dd de df dg dh di> 0 4 0 2 0 0 0 0 0 3 0 3 0 3 0 0 0 0> which is> a^0 b^4 c^0 d^2 e^0 f^0 g^0 h^0 i^0> -----------------------------------> a^3 b^0 c^3 d^0 e^3 f^0 g^0 h^0 i^0>> or> s^4 kmol^2> -------------> cd^3 m^3 kg^3>>>1) luminous intensity = 1.9720204(06) x 10^-45 cd >2) Einstein time = 1.3511888(38) x 10^-43 s>3) wavelength of gravity = 4.0507622(30) x 10^-35 m >4) uni?d substance = 1.6605402(10) x 10^-27 kmol >5) gravitational mass = 5.4563086(06) x 10^-8 kg>6) electric current = 1.1857538(29) x 10^24 A >7) Celsius temperature = 3.5518470(84) x 10^32 K >8) relative permeability = 6.8517994(75) x 10^1 sr >9) inverse ?e-structure = 1.3703598(95) x 10^2 rad >>> Good luck coming up with names for the other 387,420,345 constants. === >Message-id: <4da5bf5c.0307170826.35a9aa19@posting.google.com>>>>The following is a suggestion whichI hope you name one after me.The Mensanator Constant. I like the sound of that.>>>Becki>>-->>>Instead of using biblical names for the next 387,420,345 constants>(there are not enough names), why not use AT&T, Kingston, Verizon,>Surely there are at least 144,000 unique names for the new constants>there. Some of them might even be physicists' names. Then if you>need more names, you can always resort to their yellow pages. I am>positive you will ?d companies favorable to having a physical>property, and its corresponding constant, named after them.>>White Pages>>-- >>First 144 Primary Fundamental Physical Constants >>001) radiant volume = 1.3554076(23) x 10^-113 m-s^2/kg >002) volume of gravity = 6.6467639(49) x 10^-104 m^3 >003) gravitational volume = 1.2181796(21) x 10^-96 m^3/kg >004) luminous ef?acy = 3.7229891(12) x 10^-96 cd-sr-s^3/kg-m^2 >005) current volume = 1.3838179(77) x 10^-93 m^2/A >006) luminous energy = 1.8257112(76) x 10^-86 cd-sr-s >007) charge volume = 4.1485819(27) x 10^-85 m^3/A-s >008) moment of inertia = 8.9530792(67) x 10^-77 kg-m^2 >009) gravitational ?y = 1.0031222(77) x 10^-70 m-s/kg >010) area of gravity = 1.6408674(64) x 10^-69 m^2>011) Joshua constant= 1.3234483(74) x 10^-63 A-s^3/kg-m>012) Ruth constant = 4.9191969(04) x 10^-61 m^3/s>013) Samuel constant = 3.4161915(66) x 10^-59 m/A>014) electric moment = 6.4900394(48) x 10^-54 A-s-m >015) Ezra constant = 1.6752612(69) x 10^-49 kg-m^3/A-s^2>016) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 >017) Nehemiah constant = 1.1634040(12) x 10^-47 kg-m/A^2-s>018) magnetic moment = 1.9456648(79) x 10^-45 A-m^2 >019) luminous intensity = 1.9720204(06) x 10^-45 cd >020) Einstein time = 1.3511888(38) x 10^-43 s >021) luminous ?1.3511888(38) x 10^-43 cd-sr >022) gravitational moment = 2.2102208(82) x 10^-42 kg-m >023) self-mutual inductance = 3.4877974(86) x 10^-39 kg-m^2/A^2-s^2 >024) absorption-emission = 2.4763790(61) x 10^-36 s/kg >025) wavelength of gravity = 4.0507622(30) x 10^-35 m >026) Esther constant = 1.2379901(47) x 10^-34 s^4/m^4>027) Planck constant = 6.6260755(09) x 10^-34 kg-m^2/s >028) relative expansion = 2.8154365(22) x 10^-33 /K >029) electric resistivity = 1.0456153(81) x 10^-30 kg-m^3/A^2-s^3 >030) Job constant = 3.2671588(69) x 10^-29 A-s^3/kg-m^2 >031) uni?d substance = 1.6605402(10) x 10^-27 kmol >032) kinematic viscosity = 1.2143879(66) x 10^-26 m^2/s >033) Solomon constant = 3.7114010(91) x 10^-26 s^3/m^3>034) Isaiah constant = 1.9864474(64) x 10^-25 kg-m^3/s^2>035) inverse electric current = 8.4334536(86) x 10^-25 /A >036) Jeremiah constant = 1.3795107(62) x 10^-23 kg-m/A-s>037) heat capacity constant = 1.3806578(67) x 10^-23 kg-m^2/s^2-K >038) thermal resistance = 9.7865580(66) x 10^-21 s^3-K/kg-m^2 >039) Ezekiel constant = 9.7946958(79) x 10^-21 A-s^2/kg-m>040) gravitational molality = 3.0433399(76) x 10^-20 kmol/kg >041) elementary charge = 1.6021773(38) x 10^-19 A-s >042) Daniel constant = 1.1126500(56) x 10^-17 s^2/m^2>043) ?st radiation = 5.9552196(79) x 10^-17 kg-m^4/s^3 >044) Hosea constant = 2.5282858(10) x 10^-16 m/A-s>045) speci? heat = 2.5303881(55) x 10^-16 m^2/s^2-K >046) magnetic ?4.1356692(24) x 10^-15 kg-m^2/A-s^2 >047) electric permittivity = 1.2922426(95) x 10^-13 A^2-s^4/kg-m^3 >048) magnetic exposure = 2.9363759(53) x 10^-12 A-s/kg >049) permittivity of vacuum = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 >050) magnetic pole strength = 4.8032068(25) x 10^-11 A-m >051) Joel constant = 4.8682699(55) x 10^-11 cd/m>052) Newton constant = 6.6723563(41) x 10^-11 m^3/kg-s^2 >053) S-B primary constant = 1.3897405(80) x 10^-10 kg/s^3-K^4 >054) density of states = 2.0391992(76) x 10^-10 s^2/kg-m^2 >055) radiant distribution constant = 3.335640952... x 10^-9 s/m >056) gravitational mass constant = 5.4563086(06) x 10^-8 kg >057) permeability of vacuum = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr >058) electric conductance = 3.8740461(38) x 10^-5 A^2-s^3/kg-m^2 >059) magnetic permeability = 8.6102251(57) x 10^-5 kg-m/A^2-s^2 >060) Amos constant = 1.9689919(12) x 10^-3 A-kmol>061) ?e-structure constant = 7.2973530(80) x 10^-3 /rad >062) second radiation constant = 1.4387688(01) x 10^-2 m-K >063) dielectric constant = 1.4594706(16) x 10^-2 /sr >064) Obadiah constant = 1.4594706(16) x 10^-2 cd/s>065) spin half constant = 5.000000000.89 x 10^-1 sr/rad >066) length fraction = 1.000000000.89 m/m>067) mass fraction = 1.000000000.89 kg/kg>068) time fraction = 1.000000000.89 s/s>069) current fraction = 1.000000000.89 amp/amp>070) temperature fraction = 1.000000000.89 K/K>071) spin two constant = 2.000000000.89 rad/sr >072) gravitational momentum = 1.6357601(69) x 10^1 kg-m/s >073) Jonah constant = 6.8517994(74) x 10^1 s/cd>074) relative permeability = 6.8517994(75) x 10^1 sr >075) inverse ?e-structure = 1.3703598(95) x 10^2 rad >076) molar heat capacity = 8.3145102(91) x 10^3 kg-m^2/s^2-kmol-K >077) spin angle constant = 9.3894312(09) x 10^3 sr-rad >078) Micah constant = 1.1614098(14) x 10^4 A^2-s^2/kg-m>079) electric resistance = 2.5812805(64) x 10^4 kg-m^2/A^2-s^3 >080) Nahum constant = 1.8997879(14) x 10^5 A^2-s>081) Habakkuk constant = 5.8979849(03) x 10^5 kmol-K>082) inverse gravitational mass = 1.8327409(10) x 10^7 /kg >083) Faraday constant = 9.6485308(14) x 10^7 A-s/kmol >084) speed of light in vacuum = 2.99792458 x 10^8 m/s >085) gravitational energy = 4.9038856(17) x 10^9 kg-m^2/s^2 >086) Zephaniah constant = 1.4987209(15) x 10^10 kg-s^2/m^3 >087) Haggai constant = 2.0541178(06) x 10^10 m/cd>088) Zechariah constant = 5.6954208(84) x 10^13 A^2-s>089) Josephson primary = 2.4179883(50) x 10^14 A-s^2/kg-m^2 >090) electric displacement = 3.9552490(31) x 10^15 A-s/m>091) absorbed dose = 8.9875517(87) x 10^16 m^2/s^2>092) luminous density = 2.7467671(33) x 10^17 cd-sr-s/m^3 >093) Malachi constant = 1.4701479(23) x 10^18 kg-m^3/s^3>094) gravity displacement = 4.4930522(70) x 10^18 kg-s/m^2 >095) molar mass constant = 3.2858635(84) x 10^19 kg/kmol >096) magnetic potential = 1.0209607(45) x 10^20 kg-m/A-s^2 >097) thermal conductance = 1.0218097(04) x 10^20 kg-m^2/s^3-K >098) Matthew constant = 7.2489467(08) x 10^22 A-s/kg-m>099) electric current constant = 1.1857538(29) x 10^24 A >100) luminance constant = 1.2018157(77) x 10^24 cd/m^2 >101) Mark constant = 2.6944002(42) x 10^25 m^3/s^3>102) luminous ?nsity = 8.2346007(05) x 10^25 cd-sr/m^2 >103) Luke constant = 4.4073925(95) x 10^26 kg-m^4/s^4>104) Avogadro constant = 6.0221366(15) x 10^26 /kmol>105) gravitational ?ld = 1.3469831(84) x 10^27 kg/m >106) electric potential = 3.0607633(12) x 10^28 kg-m^2/A-s^3 >107) electric conductivity = 9.5637460(76) x 10^29 A^2-s^3/kg-m^3 >108) Celcius temperature = 3.5518470(84) x 10^32 K >109) John constant = 8.0776087(15) x 10^33 m^4/s^4>110) gravity wave number = 2.4686711(86) x 10^34 /m >111) mass ?te constant = 4.0381539(96) x 10^35 kg/s >112) molar energy = 2.9531869(13) x 10^36 kg-m^2/s^2-kmol >113) Timothy constant = 6.3723329(52) x 10^38 kg-m/A^2-s^3>114) surface concentration = 1.0119892(35) x 10^42 kmol/m^2 >115) frequency of gravity = 7.4008900(30) x 10^42 /s >116) gravitational force = 1.2106081(12) x 10^44 kg-m/s^2 >117) inverse luminous intensity = 5.0709414(41) x 10^44 /cd >118) angular velocity = 1.0141882(88) x 10^45 rad/s>119) Titus constant = 8.5954663(19) x 10^46 A^2-s/kg-m >120) Philemon constant = 1.9103773(59) x 10^47 kg-m^2/A^2-s^4>121) James constant = 5.9692181(67) x 10^48 A-s^2/kg-m^3>122) electric ?nsity = 9.7642093(17) x 10^49 A-s/m^2 >123) radiant intensity = 5.2968739(54) x 10^50 kg-m^2/s^3-sr >124) gravity ?ld strength = 2.2187310(14) x 10^51 m/s^2 >125) gravitational power = 3.6293118(17) x 10^52 kg-m^2/s^3 >126) magnetic ?nsity = 2.5204163(73) x 10^54 kg/A-s^2 >127) thermal conductivity = 2.5225121(74) x 10^54 kg-m/s^3-K >128) Peter constant = 1.7895265(87) x 10^57 A-s/kg-m^2>129) magnetic ?ld strength = 2.9272363(12) x 10^58 A/m >130) absorbed dose rate = 6.6515882(43) x 10^59 m^2/s^3>131) Jude constant = 1.0880403(11) x 10^61 kg-m^3/s^4>132) surface density constant = 3.3252585(75) x 10^61 kg/m^2 >133) electric ?ld strength = 7.5560181(98) x 10^62 kg-m/A-s^3 >134) dynamic viscosity = 9.9688744(17) x 10^69 kg/m-s >135) molar concentration = 2.4982686(65) x 10^76 kmol/m^3 >136) surface tension constant = 2.9885933(65) x 10^78 kg/s^2 >137) electric charge density = 2.4104622(20) x 10^84 A-s/m^3 >138) angular acceleration = 7.5058959(93) x 10^87 rad/s^2>139) thermal transfer constant = 6.2272531(21) x 10^88 kg/s^3-K >140) current density constant = 7.2263839(39) x 10^92 A/m^2 >141) gravitational density = 8.2089700(31) x 10^95 kg/m^3 >142) energy density = 7.3778543(28) x 10^112 kg/m-s^2 >143) radiance constant = 3.2280937(18) x 10^119 kg/s^3-sr>144) irradiance constant = 2.2118250(84) x 10^121 kg/s^3>>>> As I recall, there are theories that go up to 8 or 10 dimensions.>>>>The question is only up to four (4) dimensions.>>How many unique combinations would be possible?>>>>>> Ok, you got 9 fundamental units, any of which can appear>> in the numerator or the denominator up to a power of 4.>> A unit cannot appear in _both_ numerator and denominator>> because the smaller power would cancel.>>> Labeling the constants a..i and looking at just the a>> constant, there are 9 ways (out of 25 possible) that a>> appears in the fraction (na is the power of a in the>> numerator and da is the power of a in the denominator):>>> na da>> 0 0>> 0 1>> 0 2>> 0 3>> 0 4>> 1 0>> 2 0>> 3 0>> 4 0>> >> And since you have 9 constants, there are>>> 9^9 or 387,420,489 combinations.>>> A typical example is combination #24,376:>>> na nb nc nd ne nf ng nh ni da db dc dd de df dg dh di>> 0 4 0 2 0 0 0 0 0 3 0 3 0 3 0 0 0 0>>> which is>>> a^0 b^4 c^0 d^2 e^0 f^0 g^0 h^0 i^0>> ----------------------------------->> a^3 b^0 c^3 d^0 e^3 f^0 g^0 h^0 i^0>>>>> or>>> s^4 kmol^2>> ------------->> cd^3 m^3 kg^3>>>>>>1) luminous intensity = 1.9720204(06) x 10^-45 cd >>2) Einstein time = 1.3511888(38) x 10^-43 s>>3) wavelength of gravity = 4.0507622(30) x 10^-35 m >>4) uni?d substance = 1.6605402(10) x 10^-27 kmol >>5) gravitational mass = 5.4563086(06) x 10^-8 kg>>6) electric current = 1.1857538(29) x 10^24 A >>7) Celsius temperature = 3.5518470(84) x 10^32 K >>8) relative permeability = 6.8517994(75) x 10^1 sr >>9) inverse ?e-structure = 1.3703598(95) x 10^2 rad >>>>>> Good luck coming up with names for the other 387,420,345 constants.--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === =While sur?g the Web, I stumbled upon thefollowing site: http://www.mathpages.com/The *.com made me wonder who mightbe af?iated to this site.and so on, I thought it might be worth mentioningit in this NG.Would anyone care to share their views/reviewsof this site?David Bernier_________Then assuredly the world was made, not in time, but simultaneously with time. --St. Augustine === > While sur?g the Web, I stumbled upon the> following site:> http://www.mathpages.com/Oooh, thanks for that!> The *.com made me wonder who might> be af?iated to this site.Unless it is hidden in the HTML source, you areapparently doomed to keep wondering; I couldn't ?da trace of ownership claim.> and so on, I thought it might be worth mentioning> it in this NG.> Would anyone care to share their views/reviews> of this site?I am most struck by how _sane_ and _readable_ the siteis. The parts I found weren't overwhelmingly technical,either, a relief for someone with my modest math skills.The reader will gain a lot of historical perspectivefairly painlessly.xanthian.-- === David Bernier> While sur?g the Web, I stumbled upon the> following site:>> http://www.mathpages.com/>> The *.com made me wonder who might> be af?iated to this site.Indeed. He has a lot to say but is not trying to fatten his CV or sell usany book on which he makes a commission. Most unusual :)>> Would anyone care to share their views/reviews> of this site?I've seen some but not all of these interesting bits and pieces before. IMOit's a good site for any teacher who is looking for enrichment material.Larry === >While sur?g the Web, I stumbled upon the>following site:>> http://www.mathpages.com/>>The *.com made me wonder who might>be af?iated to this site.http://www.coolwhois.com/?d=mathpages.comis a way to answer that question. (Who oh why do so many authors put no contact information on their sites?)-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com/You ?d yourself amusing, Blackadder.I try not to ?the face of public opinion. === > While sur?g the Web, I stumbled upon the> following site:> http://www.mathpages.com/> The *.com made me wonder who might> be af?iated to this site.This was set up by Kevin S. Brown. It was started asa conributor in the early to mid nineties. He I believeholds the dubious honor of being the ?st person on sci.mathto attempt to explain math to JSH.> and so on, I thought it might be worth mentioning> it in this NG.> Would anyone care to share their views/reviews> of this site? ===