mm-148 It's NOT the same story if instead of being Tom Sawyer the main> character is called Tim Silversmith.> Mathematics works one way, but literature works another.> Character names are *part* of the story and aren't arbitrary like> variable names.>> It makes a difference what name a writer picks for a character.>> I do 'nd it fascinating that some of you don't get it.>> Since you, presumably, get it, then please explain to us what> the difference would be. How would calling the main character Tim> Silversmith change the story? Give speci'c examples from speci'c> passages, please.>> I notice James walked away from this one. Gee, what a> surprise.Well, this is a rare instance where James is absolutely right. Literatureis not mathematics, and names *are* signi'cant. Tim Silversmith, forexample, doesn't at all convey the character that Tom Sawyer does.Certain combinations of words or syllables add a subtle color tocharacter's name. Altering the characters' names will substantiallychange a literary work. =>> It's NOT the same story if instead of being Tom Sawyer the main> character is called Tim Silversmith.> Mathematics works one way, but literature works another.> Character names are *part* of the story and aren't arbitrary like> variable names. It makes a difference what name a writer picks for a character.> I do 'nd it fascinating that some of you don't get it.>> Since you, presumably, get it, then please explain to us what> the difference would be. How would calling the main character Tim> Silversmith change the story? Give speci'c examples from speci'c> passages, please.>> I notice James walked away from this one. Gee, what a> surprise.> Well, this is a rare instance where James is absolutely right. Literature> is not mathematics, and names *are* signi'cant. Tim Silversmith, for> example, doesn't at all convey the character that Tom Sawyer does.> Certain combinations of words or syllables add a subtle color to> character's name. Altering the characters' names will substantially> change a literary work.Is it a new work of literature? Can James release his versionof The adventures of Tim Silversmith and not be accused ofplagiarism? Can I take any existing novel, change the namesof the characters and claim I have a new work of literature?Why is it that publisher's won't accept that claim? - Randy => Don't beat me for my fumblings...But feel free to correct my meanderings and> if neccessary, direct me to a more appropriate forum.Probably sci.physics or even better, sci.physics.relativity.But I don't know how to say this without coming across asbeating you: I have no clue what question you're tryingto ask.> I was pondering the whole speed of light issue, and was wondering about the> argument of time being an extra dimension, all that...What is the speed of light issue? What is the argumentabout time being an extra dimension?> It seemed to me that this was more of a ratio effect than anything else, What is? What effect does this refer to?> i.e. as velocity in space increases, velocity in time decreased,What does velocity in time mean to you?> with a> velocity in space of 0 resulting in a velocity in time of X (being the> max)...hence with a speed of c in space, the resultant velocity in time> would then be 0.> Is this even remotely plausible?I don't know what this is, but so far as I can make it out,no. It doesn't 't with by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h87MbYQ20293; =>Don't beat me for my fumblings...But feel free to correct my meanderings and>if neccessary, direct me to a more appropriate forum.>>I was pondering the whole speed of light issue, and was wondering about the>argument of time being an extra dimension, all that...>>It seemed to me that this was more of a ratio effect than anything else, >i.e. as velocity in space increases, velocity in time decreased, with a>velocity in space of 0 resulting in a velocity in time of X (being the>max)...hence with a speed of c in space, the resultant velocity in time>would then be 0.>>Is this even remotely plausible?>>bp Well, 'rst, you are in the wrong forum. This has nothing to do with mathematics. Try sci.Physics. Secondly, how do you conclude that as velocity in space increases, velocity in time decreased? What exactly do you mean by velocity in time? Seems to me we all progress approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88FbWU09780; =>I'm trying to embed a large key (used for cryptography) into a>relatively simple polynomial over a prime 'eld. What I need is a>simple algorithm that will let me convert the key into a polynomial, and>then later convert the polynomial back into the key. I'm wondering if>anybody has any you convert a number of length n into a poly of degree n-1, modulo p eg 372 to 3x^2 + 7x + 2 mod 10 say.to get back to the original number you just read off the coef'cients.As this solution is trivial i assume i have misunderstood the question. could you clarify what you are interested in asking?cem salp =(revised)> Hey all,>> I have a question, if I have two seperate formulas that model two> phenomena and the two phenomena are related by cause and effect, is it> possible to model the cause and effect relationship between the two> formulas without leaving any question to interpretation as to which is> the cause, and which is the effect?>How about the following:If, for all x, P(x) causes Q(x), then(1) there exists an x such that P(x), and(2) there exists an x such that not Q(x), and(3) for all x, P(x) implies Q(x).The converse is not true, however.Dan =(revised)> Hey all,>> I have a question, if I have two seperate formulas that model two> phenomena and the two phenomena are related by cause and effect, is it> possible to model the cause and effect relationship between the two> formulas without leaving any question to interpretation as to which is> the cause, and which is the effect?>How about the following:If, for all x, P(x) causes Q(x), then(1) there exists an x such that P(x), and(2) there exists an x such that not Q(x), and(3) for all x, P(x)=> Q(x).The converse is not true, however.Dan => Hey all,>> I have a question, if I have two seperate formulas that model two> phenomena and the two phenomena are related by cause and effect, is it> possible to model the cause and effect relationship between the two> formulas without leaving any question to interpretation as to which is> the cause, and which is the effect?>How about the following:If, for all x, P(x) causes Q(x), then (1) there exists an x such P(x), and(2) for all x, P(x)=> Q(x).The converse is not true, however.Dan =When working with complex numbers, is there any reason that we cannot takeeven roots of negative integers, other than the square root of -1? Or is itjust some sort of de'ned convention? Or does no such restriction exist, andin some space (-1)^(.25) has meaning?--riverman =>When working with complex numbers, is there any reason that we cannot take>even roots of negative integers, other than the square root of -1? Or is it>just some sort of de'ned convention? Or does no such restriction exist, and>in some space (-1)^(.25) has meaning?>>--riverman>>to provide. =>>When working with complex numbers, is there any reason that we cannottake>even roots of negative integers, other than the square root of -1? Or isit>just some sort of de'ned convention? Or does no such restriction exist,and>in some space (-1)^(.25) has meaning?>--riverman to provide.Well, I guess I'll have to live with missing your excellent solution set atthe expense of such a sanctimoniouspost.--riverman => When working with complex numbers, is there any reason that we cannot take> even roots of negative integers, other than the square root of -1? Or is it> just some sort of de'ned convention? Or does no such restriction exist, and> in some space (-1)^(.25) has meaning?> --riverman> If x^(1/(2*n)) = y, then y^(2*n) = x, but for every real y, y^(2*n) = (y^n)^2 = (y^2)^n is non-negative.Thus even-indexed roots of negative reals (integers or not) cannot be real. =Virgil scribbled the following:>> When working with complex numbers, is there any reason that we cannot take>> even roots of negative integers, other than the square root of -1? Or is it>> just some sort of de'ned convention? Or does no such restriction exist, and>> in some space (-1)^(.25) has meaning?> If x^(1/(2*n)) = y, then y^(2*n) = x, > but for every real y, y^(2*n) = (y^n)^2 = (y^2)^n is non-negative.> Thus even-indexed roots of negative reals (integers or not) cannot > be real.Whoever said they had to be real? Do you think the square root of -1is real? If yes, then which real is it? If not, then why does it holda special place among all the other imaginary numbers?-- /-- Joona Palaste (palaste@cc.helsinki.') ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/War! Huh! Good God, y'all! What is it good for? We asked Mayor Quimby. - Kent Brockman => Virgil scribbled the following:>> When working with complex numbers, is there any reason that we cannot take> even roots of negative integers, other than the square root of -1? Or is >> it>> just some sort of de'ned convention? Or does no such restriction exist, >> and>> in some space (-1)^(.25) has meaning?> If x^(1/(2*n)) = y, then y^(2*n) = x, > but for every real y, y^(2*n) = (y^n)^2 = (y^2)^n is non-negative.> Thus even-indexed roots of negative reals (integers or not) cannot > be real.> Whoever said they had to be real? Do you think the square root of -1> is real? If yes, then which real is it? If not, then why does it hold> a special place among all the other imaginary numbers?Sorry, didn't read it carefully enough.Over the complex number system, every non-zero complex has n nth roots.And the negative of the 2*m th power of an integer has a (2*m)th root which is an integer multiple of i: [-(m*(2*n))]^(1/(2*n)) = (m*i or -m*i ) if restricted to purely imaginary roots => When working with complex numbers, is there any reason that we cannot take> even roots of negative integers, other than the square root of -1? Or is it> just some sort of de'ned convention? Or does no such restriction exist, and> in some space (-1)^(.25) has meaning? When working with complex numbers, you may take any roots (even or odd or fractional) of any complex numbers (including negative integers, or numbers with both real and imaginary parts.) (-1)^(1/4) is like the sqrt of +i and -i which are the +/- combinations of 1/sqrt(2) and i/sqrt(2). (try it youself: take 1/sqrt(2)+ i/sqrt(2) and raise that to the 4th power. Similarly, try it with a minus sign on the real part, or on the imaginary part, or on both parts. You should get -1 in all cases.) Just as there are two square roots of any number, there are 4 fourth roots. Plotting these points on a the complex plane will reveal that the n roots are always uniformly distributed around the unit circle |z|=1.J =>> When working with complex numbers, is there any reason that we cannottake> even roots of negative integers, other than the square root of -1? Or isit> just some sort of de'ned convention? Or does no such restriction exist,and> in some space (-1)^(.25) has meaning?>> When working with complex numbers, you may take any roots (even or odd> or fractional) of any complex numbers (including negative integers, or> numbers with both real and imaginary parts.)>> (-1)^(1/4) is like the sqrt of +i and -i which are the +/- combinations> of 1/sqrt(2) and i/sqrt(2). (try it youself: take 1/sqrt(2)+ i/sqrt(2) and> raise that to the 4th power. Similarly, try it with a minus sign on the> real part, or on the imaginary part, or on both parts. You should get -1> in all cases.)> Just as there are two square roots of any number, there are 4 fourth> roots. Plotting these points on a the complex plane will reveal that the n> roots are always uniformly distributed around the unit circle |z|=1.Wow, well explained and I get it. Stranger, even, is that my teaching textsfor my HS classes make a real point of saying ïno even roots for integers'except those texts that go into complex numbers, who allow sqrt(-1), butstill maintian ïno even roots for negative integers'. Even whenteaching complex roots to Precalculus students, the texts avoid negativebases to even powers, but I know the kids would understand it, since we 'ndcomplex roots to higher order quadratic functions (x^4+x^2+x, etc) all thetime with the quadratic formula. And I'm more suprised that I never cameacross it in 3+ years of Calculus, and all the other undergrad math classesI took. Where does this get addressed?But while we're on it (and negative bases to reducible rational powers? For example, I'vebeen teaching the advanced algebra kids that (-16)^(2/4) does not exist, asit is not well-de'ned since: (-16)^(2/4) should equal (-16)^(1/2) = 4i, howeversqrt(sqrt[(-16)^2]) = sqrt(16) and -sqrt(16) = 4 and 4i, etc.--riverman = > When working with complex numbers, is there any reason that we > cannot >> take > even roots of negative integers, other than the square root of > -1? Or is >> it > just some sort of de'ned convention? Or does no such restriction > exist, >> and > in some space (-1)^(.25) has meaning? When working with complex numbers, you may take any roots (even or >> odd or fractional) of any complex numbers (including negative >> integers, or numbers with both real and imaginary parts.) (-1)^(1/4) is like the sqrt of +i and -i which are the +/- >> combinations of 1/sqrt(2) and i/sqrt(2). (try it youself: take >> 1/sqrt(2)+ i/sqrt(2) and raise that to the 4th power. Similarly, >> try it with a minus sign on the real part, or on the imaginary >> part, or on both parts. You should get -1 in all cases.) Just as >> there are two square roots of any number, there are 4 fourth roots. >> Plotting these points on a the complex plane will reveal that the >> n roots are always uniformly distributed around the unit circle >> |z|=1. > Wow, well explained and I get it. Stranger, even, is that my teaching > texts for my HS classes make a real point of saying ïno even roots > for integers' except those texts that go into complex numbers, who > allow sqrt(-1), but still maintian ïno even roots for > negative integers'. Even when teaching complex roots to Precalculus > students, the texts avoid negative bases to even powers, but I know > the kids would understand it, since we 'nd complex roots to higher > order quadratic functions (x^4+x^2+x, etc) all the time with the > quadratic formula. And I'm more suprised that I never came across it > in 3+ years of Calculus, and all the other undergrad math classes I > took. Where does this get addressed? > [ snip ] >Well, there is some sense in doing what your textbooks do, possibly ---depending on exactly how they do it. The point is that once you startconsidering general roots, you have to remember there is always morethan one, and it is easy generate confusion when using expressions suchas sqrt(x) in equations.Even writing i = sqrt(-1) is not really correct, strictly speaking,because -1 has two quadratic roots in complex numbers, and so you haveto keep in mind what exactly this equation is supposed to mean. It means-1 has two square roots and one of them is i.As long as we stay within real numbers, all is 'ne, because we use theconvention that for example sqrt(4)=2, and not sqrt(4)=-2. So there isno ambiguity, and we can write formulas such as the well-known one for the solutions of the quadratic equation.But once we allow roots to be complex numbers, extra care is needed tounderstand what exactly we mean by expressions like3 + sqrt(-4) + sqrt(-16)and what the rules are for manipulating them. Does this expression standfor a single value, or do we take it to mean all possible combinationsof the values of the roots in it?N =Not sure I have understood you but(-1)^(.25) = sqrt(i) = (1+i)/sqrt(2)also other roots with - instead of +> When working with complex numbers, is there any reason that we cannot take> even roots of negative integers, other than the square root of -1? Or isit> just some sort of de'ned convention? Or does no such restriction exist,and> in some space (-1)^(.25) has meaning?>> --riverman>> =riverman complex numbers, is there any reason that we cannot take> even roots of negative integers, other than the square root of -1? Or is it> just some sort of de'ned convention? Or does no such restriction exist, and> in some space (-1)^(.25) has meaning?I don't understand your question. Of course you can take even roots ofnegative integers. For example sqrt(sqrt(-1)) is a well-de'ned complexnumber. Its value is equal to sqrt(1/2) + i*sqrt(1/2).-- /-- Joona Palaste (palaste@cc.helsinki.') ---------------------------| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.'/~palaste W++ B OP+ |----------------------------------------- Finland rules! ------------/Immanuel Kant but Genghis Khan. - The Of'cial Graf'tist's Handbook =Please excuse a non-mathematician for posting a question which is possiblyfrequently asked...As a diversion from counting sheep one sleepless night an idea occurred tome. Is there a number n such that n is the product of two primes a and b,but also the product of two different primes, d and e. Intuition (hah!)tells me that there isn't, but I don't have the foggiest notion of how toprove it, or the maths ability for that matter.I can no longer use it as a getting-to-sleep aid, so can anyone offer aneasy to understand proof (or rebuttal)?Ta muchly,Steve =Despite what some posters have said, it is neither obvious nor completelytrivial. Some of the proofs above simply assume the thereom to prove it.The simplest, easiest proper proof I could 'nd in 2 minutes searching is:http://www.cut-the-knot.org/blue/Euclid.shtmlHTHPeter Webb> Please excuse a non-mathematician for posting a question which is possibly> frequently asked...>> As a diversion from counting sheep one sleepless night an idea occurred to> me. Is there a number n such that n is the product of two primes a and b,> but also the product of two different primes, d and e. Intuition (hah!)> tells me that there isn't, but I don't have the foggiest notion of how to> prove it, or the maths ability for that matter.>> I can no longer use it as a getting-to-sleep aid, so can anyone offer an> easy to understand proof (or rebuttal)?>> Ta muchly,>> Steve>> => me. Is there a number n such that n is the product of two primes a and b,> but also the product of two different primes, d and e. Intuition (hah!)> tells me that there isn't, but I don't have the foggiest notion of how to> prove it, or the maths ability for that matter.>Working with an example:Knowing only that 9 and 5 are factors of 45, it is not obvious that 3 and 15are also factors of 45 unless it is obvious that 9 factors into 3 and 3.Then there is (3 * 3) * 5 or 3 * (3 * 5) which is 3 * 15 .So additional factors were found by factoring 9 but since 5 is prime andsince 3 is prime, there are no more factors to be found... => Working with an example:>> Knowing only that 9 and 5 are factors of 45, it is not obvious that 3 and15> are also factors of 45 unless it is obvious that 9 factors into 3 and 3.> Then there is (3 * 3) * 5 or 3 * (3 * 5) which is 3 * 15 .>> So additional factors were found by factoring 9 but since 5 is prime and> since 3 is prime, there are no more factors to be found...>...So if a number factors into a pair of primes then there are no additionalfactors. =No. Factorization into primes is unique.I will give a hint. Show that if p is primeand p divides a*b then either p divides aOR p divides p. => Please excuse a non-mathematician for posting a question which is possibly> frequently asked...> As a diversion from counting sheep one sleepless night an idea occurred to> me. Is there a number n such that n is the product of two primes a and b,> but also the product of two different primes, d and e. Intuition (hah!)> tells me that there isn't, but I don't have the foggiest notion of how to> prove it, or the maths ability for that matter.> I can no longer use it as a getting-to-sleep aid, so can anyone offer an> easy to understand proof (or rebuttal)?> Ta muchly,> Steve> Old theorem:Unique factorization into primes! => Please excuse a non-mathematician for posting a question which is>> possibly frequently asked...>> As a diversion from counting sheep one sleepless night an idea>> occurred to me. Is there a number n such that n is the product of>> two primes a and b, but also the product of two different primes, d>> and e. Intuition (hah!) tells me that there isn't, but I don't have>> the foggiest notion of how to prove it, or the maths ability for>> that matter.>> I can no longer use it as a getting-to-sleep aid, so can anyone>> offer an easy to understand proof (or rebuttal)?>> Ta muchly,>> Steve>> Old theorem:> Unique factorization into primes!Can you supply a proof? I think this calls for a proof by contradiction, butI don't know how :-(TIASteven(not a mathematician either) =Another way to ask your question is: Is is possible to factorize a number nin two different ways? The answer is obviously: NO. => Another way to ask your question is: Is is possible to factorize a number n> in two different ways? The answer is obviously: NO.12 = 3*4 = 2*6.I think you need to say ïcompletely' factorize. =>Another way to ask your question is: Is is possible to factorize a number n>in two different ways? The answer is obviously: NO.Actually, the way you've phrased the question, it's not only not obvious,it's not even true. 18 = 2*9 = 3*6, for instance. Doug =Doug Norris skrev i melding>>Another way to ask your question is: Is is possible to factorize anumber n>in two different ways? The answer is obviously: NO.>> Actually, the way you've phrased the question, it's not only not obvious,> it's not even true. 18 = 2*9 = 3*6, for instance.Yes. I ment factorizing to primes.> Doug>> =Martin>> Yes. I ment factorizing to primes.> Is there a number n such that n is the product of two primes a and b, but also the product of two different primes, d and e.Your question was posed perfectly. I love it and don'thave a clue, how to prove ab = de impossible forprimes a,b,d,e and {a,b} different from {d,e}.I *know* it's impossible. But I don't know how to proveit. So there are three possibilities for me: (a) Thinking it over (nighttime again ...) (b) Waiting for a good answer by e.g. Dave Rusin :-) (c) Reading a book.I'll try (a) 'rst, starting with let a*b be the smallest...and walking happily towards ... contradiction; q.e.d. orsomething like that.Couldn't you have asked for Are there in'nitely many primes?This is in my standard repertoire :-)I surely have to add the other one!Rainer Rosenthalr.rosenthal@web.de =Here is a short proof: http://www.math.uiuc.edu/~dan/ShortProofs/uf.html =>Martin>> Yes. I ment factorizing to primes.You're confusing Martin with the original poster. Doug => Martin>> Yes. I ment factorizing to primes.> Is there a number n such that n is the> product of two primes a and b, but also> the product of two different primes, d and e.>> Your question was posed perfectly. I love it and don't> have a clue, how to prove ab = de impossible for> primes a,b,d,e and {a,b} different from {d,e}.>> I *know* it's impossible. But I don't know how to prove> it. So there are three possibilities for me:>> (a) Thinking it over (nighttime again ...)> (b) Waiting for a good answer by e.g. Dave Rusin :-)> (c) Reading a book.>Just _any_ book? Like Lotte in Weimar? or better go for Tirant lo Blanc? =bjftq4$fqj$1@nntp.itservices.ubc.ca...>'nd all n oisitive integer (d(n))^3 = 4n>> I assume you mean positive integers, and d(n) is the number of positive> divisors of n.>> According to Maple, the solutions for n < 10000 are 2, 128 and 2000.>> If n = prod_{i=1}^k p_i^e_i where p_i are primes,> d(n) = prod_{i=1}^k (e_i + 1)>> Since 4n is a cube, taking p_1 = 2, e_1 = 1 mod 3.> And since d(n) is even, some e_i is odd.>> Write n = A B where A is the product of the p_i^e_i where> p_i^e_i < (e_i+1)^3, and B is the product of the others.> Then A can only have primes 2,3,5,7, with exponents <= 10 for 2,> 4 for 3, 2 for 5, 1 for 7. So there are 'nitely many possibilities> for A. And given A, there are only 'nitely many possibilities for> B since there are only 'nitely many (p_i, e_i) pairs with e_i >= 1> for which p_i^e_i/(e_i+1)^3 is less than a given bound. So there are> only 'nitely many solutions. An exhaustive search should be feasible> (although I haven't tried it).>> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> =David C. Ullrich a .8ecrit dans le message de>>f:R+--->R+*, C^ 1, f ï(x)>f(x) for any x in R+>>An = int_[0;n] exp(t)/f ï(t) dt converge>>Show that Bn = int_[0;n] exp(t)/f(t) dt converge>> This is equivalent to the question in the other thread,> integral 2 (not that I see how to do either one):>> The condition f ï(x)>f(x) says that exp(-x) f(x) is> strictly increasing, so f(x) = g(x) exp(x), where> g is strictly increasing.>> So we are given that the integral of dt/(g(t) + g'(t))> is bounded and we have to show that the integral> of dt/g(t) is bounded.> (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88Hxam20727; =>> I have a problem: when sending a function (such as sin(x)) to a proc, when I try to use it, I get sin(x)(0) instead of my desired >> sin(0). Can anybody help me?>>Can you give an example of what you are trying to do?sure:b:=proc(n,f,x) local k,res; res:=0; for k from 0 to n do res:=res+(n!/(k!*(n-k)!))*(x^k)*((1-x)^(n-k))*f(k/n); end do; return res;end;I need to use this for different functions: sin, cos, x^... and also derivaties and integrals of functions(btw, I couldn't use sigma because of the something^0=0 even if something=0..)>>Note that sin(x) is not a function; sin is a function, but sin(x)>is an expression representing the value of the sin function at a point x.>Perhaps what you need to do is specify the function sin as an argument>to your proc instead of the expression sin(x). as I said I need to use derivaties and integrals which prevents me from just sending sin.Without seeing actual>code, it's dif'cult to say more than that.>-- >Dave Seaman>Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.><http://www.commoncouragepress.com/index.cfm? action=book&bookid=228> => I have a problem: when sending a function (such as sin(x)) to a proc, when I try to use it, I get sin(x)(0) instead of my desired > sin(0). Can anybody help me?>>Can you give an example of what you are trying to do?> sure:> b:=proc(n,f,x)> local k,res;> res:=0;> for k from 0 to n do> res:=res+(n!/(k!*(n-k)!))*(x^k)*((1-x)^(n-k))*f(k/n);> end do;> return res;> end;So your proc expects f to be a function and not an expression, as Isuspected.> I need to use this for different functions: sin, cos, x^... and also derivaties and integrals of functions> (btw, I couldn't use sigma because of the something^0=0 even if something=0..)>>Note that sin(x) is not a function; sin is a function, but sin(x)>>is an expression representing the value of the sin function at a point x.>>Perhaps what you need to do is specify the function sin as an argument>>to your proc instead of the expression sin(x). > as I said I need to use derivaties and integrals which prevents me from just sending sin.Why? If f is a function, then its derivative is D(f). In particular,D(sin) = cos. I am not an expert on Maple, but two ways of expressing anantiderivative of a function f in function form are: x -> integrate(f(x),x);or unapply(integrate(f(x),x),x);Perhaps there is a less esoteric method that I am unaware of. Eitherway, substituting a function such as sin for f in the expressionsabove seems to do what you expect.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. =>>Note that sin(x) is not a function; sin is a function, but sin(x)>>is an expression representing the value of the sin function at a point x.>Perhaps what you need to do is specify the function sin as an argument>>to your proc instead of the expression sin(x). There is some possibility for confusion with this terminology becauseunfortunately Maple and its documentation refers to the expressionsin(x) as a function. I believe this is nonstandard usage. `sin` byitself is a procedure or an operator. > If f is a function, then its derivative is D(f). In particular,> D(sin) = cos. I am not an expert on Maple, but two ways of expressing an> antiderivative of a function f in function form are:> x -> integrate(f(x),x);> or> unapply(integrate(f(x),x),x); > Perhaps there is a less esoteric method that I am unaware of.No, that's it. And the unapply form is much preferable. =Let A >1 realFind all f such thatf : R+---> R continous such that for any x in R+f(x) = int_[0;Ax]f(t)dtthank you m.o. => Let A >1 real> Find all f such that> f : R+---> R continous such that for any x in R+> f(x) = int_[0;Ax]f(t)dt> thank you m.o.You could try differentiating both sides wrt x.This gives f'(x) = A f(Ax), maybe this is easier to deal with. =>Let A >1 real>Find all f such that>f : R+---> R continous such that for any x in R+>f(x) = int_[0;Ax]f(t)dtf(0) = 0 and f'(x) = A f(Ax).Then f''(x) = A d/dx f(Ax) = A^2 f'(Ax) = A^3 f(A^2 x)and f^(n)(x) = A^(n(n+1)/2) f(A^n x).In particular, f is C^in'nity and all its derivatives at 0 are 0. So the only possible analytic solution is f(x) = 0. I don't know if there are non-analytic solutions.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 =>One of the problems with coming up with something brilliant in>>mathematics is that other mathematicians can get jealous, lie to other>>people about how brilliant your work is, and wait to hope until you>>die, then you get to be some tragic 'gure or something.One of the problems of being human is that I make mistakes. I try to 'x thempromptly.>Please refer to>>http://users.aol.com/DGoncz/Publications/ Factorization.gifThis was a mistake, has been 'xed, and shows that James's factorization isconsistent.What the factorization is good for, I wish I knew.Yours,Doug Goncz (at aol dot com)Replikon Research Replikon Research researches replikons, which are self-reproducingcon'gurations of non-living matter in environments that support replication,analogous to approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h8643d426760; =>>One of the problems with coming up with something brilliant in>>mathematics is that other mathematicians can get jealous, lie to other>>people about how brilliant your work is, and wait to hope until you>>die, then you get to be some tragic 'gure or something.>Here's a weird factorization to show you yahoos how to factor a>>polynomial with placemarkers into non-polynomial factors. What's with all the yahoo references. Have you been reading Jonathan>> Swift lately or something?> P(x) = 11^2 + 11x + 2>>where x is, oh, let's let x be an integer. Sigh. If x is an integer, then this is not a polynomial, it is an>> integer. If this is a polynomial, then x is not an integer, it is a>> polynomial variable.>>Well readers of sci.math you must know by now that Arturo Magidin was>ring-leader and chief obfuscator for YEARS, and he's still working at>confusing you.>>Now do I care whether or not you want x to be a freaking variable or>whatever?>>Nope. Now if Arturo Magidin weren't pissing in his pants scared that>some of you now know what he's been doing to your minds--screwing with>them--then he wouldn't have to act like it's slapping Mother Theresa>or something, and could just work to clear it up.>>I'm telling you, this guy is balanced to evil. Arturo Magidin has to>be one of the most evil people currently alive on the planet, and he>seems to LOVE screwing with your heads.> If you mean that you want to think of the polynomial P(x) as a>> function from the integers to themselves, then say so; that means you>> are considering it as an element of Z^Z, or perhaps of some other>> larger ring R^Z (of functions from the integers to R).>>Whatever.>>Well, I just use the 11's as placemarkers, so I can *act* like it's>>the polynomial y^2 + y + 2, to get the factorization Surely you mean y^2 + yx + 2.>>Oh yeah, you're right. That was my bad, my mistake. Good catch.> You mean, presumably, that P is the result of evaluating the>> expression y^2+yx+2 at y=11. Now, is x a ->'xed<- integer, or is x>> some integer? That is, is x just an unknown but 'xed value, or is it>> a variable which takes integer values, or what? If it is a variable, then you can either consider y^2+yx+2 as an>> element of the ring of polynomials with coef'cients in Z^Z, or as an>> element of Z^{ZxZ}, the ring of functions on TWO integer-valued>> variables which takes values in the integers. Which one do you mean?>>Hey, quit trying to confuse people about y!!!!!>>It's the *polynomial* P(x) = 11x + 123, so whatever makes it a>polynomial.>>I'm telling you other readers this guy is freaking evil, but good at>being evil. Yup, he confused the hell out of a lot of you for YEARS,>and I have to give him credit for being so effective.>> P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)>>which is an example of a non-polynomial factorization. This expression is certainly not taking place in Z^Z, since the>> factors 11+ (x+sqrt(x^2-8)/2) and 11+(x-sqrt(x^2-8)/2) are not elements of>> Z^Z; even when x is an integer, they do not always take integer>> values. So in what ring is this factorization supposed to take>> place?>>Now I'd say that given x being an integer the expressions would each>evaluate to algebraic integers.>>Ok evil, confusing dude, what do you say?>> It can take place in R^Z (real-valued functions of an integer>> variable, x); or in C^Z (complex-valued functions of an integer>> variable); or in A^Z, the ring of all algebraic integer valued>> functions of integer variable. Which one? If it takes place in a ring R, it will also hold in any>> larger ring which contains a copy of R.>>I thought you said polynomials couldn't be in a ring.>>Well, let's say it's R, what then?>Now at 'rst I didn't 'gure something like that was so new, which was>>just another mistake. If you are clear, then it is both correct and not new. If you are not>> clear and you bandy terms like factor without any concern for their>> context, then it is new, but not correct.>>See what I mean readers?>>It seems to me that *mathematically* given something like>> P(x) = 11x + 123 =>> (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)>>that you can't say that *maybe* it's correct, but *maybe* it's not>correct, but Arturo Magidin 'nds a way.>>I'm telling you, he's balanced to evil. It takes a person so balanced>to screw with other people with mathematics that way, as mathematics>is supposed to be the pure language.>>So evil attempts to control the language and control your minds>through corrupting it.>>You may not have realized it, but Arturo Magidin is an agent of the>Dark Powers.>> Later I 'gured that people would just follow>>math rules, and eventually get it, which was another mistake.But from people like Mark Steinberger, Andrew Granville and Barry>>Mazur to poster here on sci.math, I've seen a refusal to JUST FOLLOW>>THE MATH!!! Without specifying WHERE the factorization takes place, you are not>> doing math. There is no math to follow. Factor and factorization>> are context dependent statements. You are trying to use them>> content-free. That's like the old question, How is a duck different>> from?, and the answer One of its legs is the same. [.snip.]>>So now some of you can see the battle between Good and Evil revealed>once again before human eyes.>>My job is to 'ght the corruption of mathematics and bring it closer>to Mathematics, the pure language of the Powers, what some of you call>gods and those in Christian mythology call archangels.>>Arturo Magidin is an agent of evil, and as such he works to confuse>you and push you away from the pure language into darkness and>confusion.>>I'm revealing these things to you because Judgement has been rendered,>and Death incarnated, and She walks the earth as a human female,>collecting human souls. To escape Her, you must learn to be *more*>than human, and as a 'rst step you must get closer to Mathematics,>the pure language.>>Then you must learn to truly listen and hear the music of the>Universe.>>But, continue to listen to Arturo Magidin, and your soul will be>taken, ripped from your body, and soon thereafter, your body will die.>>The 'rst wave is going now, and the entire process will take now a>bit under 200 years.>James HarrisWell readers of sci.math you must know by now that Arturo Magidin wasring-leader and chief obfuscator for YEARS, and he's still working atconfusing you.Now do I care whether or not you want x to be a freaking variable orwhatever? Readers of sci.math must know by now that Aruro Magidin must know one hell of lot more mathematics than you do and is trying to keep you honest (which he probably knows by now is a losing battle). The fact that you do NOT care whether x is a freaking variable or not is exacly why what you say is meaningless. =I have just started a math class and I am having a terrible time proving thefollowing:The quadratic mean of two integers a and b equals sqrt((a^2 + b^2) / 2).By computing the arithmetic and quadratic means of different pairs ofpositive integers, formulate a conjecture about their relative sizes andprove your conjecture.So far I have that when a=1 and b=2, their geometric mean (GM) is sqrt(5/2),which is approximately 1.58, not proved by my examples, because, no matter howmany pairs of values I examine, their will always be more pairs that Ihave not tested.Am I on the right track? Any further help appreciated. => I have just started a math class and I am having a terrible time provingthe> following:> The quadratic mean of two integers a and b equals sqrt((a^2 + b^2) / 2).> By computing the arithmetic and quadratic means of different pairs of> positive integers, formulate a conjecture about their relative sizes and> prove your conjecture.>>Just => I have just started a math class and I am having a terrible time proving the> following:> The quadratic mean of two integers a and b equals sqrt((a^2 + b^2) / 2).> By computing the arithmetic and quadratic means of different pairs of> positive integers, formulate a conjecture about their relative sizes and> prove your conjecture.> So far I have that when a=1 and b=2, their geometric mean (GM) is sqrt(5/2),> which is which> is 1.5. Trying some other pairs I get:-> a=1 b=1 GM=1 examples, because, no matter how> many pairs of values I examine, their will always be more pairs that I> have not tested.> Am I on the right track? Any further help appreciated. > Firstly, the geometric mean is *not* the quadratic mean. Which, incidentally, I have never come across before. The geometric mean of a_1, a_2,...,a_n is equal to (a_1*a_2*...*a_n)^(1/n) - in other words given n numbers it is the nth root of all the numbers multiplied together.Now to the quadratic mean...First state your conjecture. Is it perhaps expression for QM and then a >= sign and then an expression where you get.a .pdf document with the details.Mark Atherton--usepackage{amsmath}begin{document}Conjecture: for all real $a$ and $b$ the quadratic mean is greater than the arithmetic mean.begin{align*}&& sqrt{frac{a^2+b^2}{2}}&geqfrac{a+b}{2}&iff & frac{a^2+b^2}{2}&geqleft({frac{a+b^2}{2}}right)^2 &iff & frac{a^2+b^2}{2}&geqfrac{a^2+2ab+b^2}{4}&iff & frac{a^2+b^2}{2}&geqfrac{a^2+b^2}{4}+frac{2ab}{4}& iff & frac{a^2+b^2}{4}&geqfrac{2ab}{4}&iff & a^2+b^2&geq 2ab&iff & a^2-2ab+b^2&geq 0&iff & a^2-2ab+b^2&geq 0&iff & (a-b)^2&geq 0end{align*}Any real number squared is greater than or equal to zero.end{document} => I have just started a math class and I am having a terrible time proving the> following:> The quadratic mean of two integers a and b equals sqrt((a^2 + b^2) / 2).> By computing the arithmetic and quadratic means of different pairs of> positive integers, formulate a conjecture about their relative sizes and> prove your conjecture.> So far I have that when a=1 and b=2, their geometric mean (GM) is sqrt(5/2),> which is which> is 1.5. Trying some other pairs I get:->> a=1 b=1 GM=1 examples, because, no matter how> many pairs of values I examine, their will always be more pairs that I> have not tested.>> Am I on the right track? Any further help appreciated.>Its all in the maths lingo :formulate a conjecture about their relative sizes = which one is bigger?Herc => I have just started a math class and I am having a terrible time proving the> following:> The quadratic mean of two integers a and b equals sqrt((a^2 + b^2) / 2).> By computing the arithmetic and quadratic means of different pairs of> positive integers, formulate a conjecture about their relative sizes and> prove your conjecture.> So far I have that when a=1 and b=2, their geometric mean (GM) is sqrt(5/2),> which is approximately 1.58, and their arithmetic not proved by my examples,What conjecture? You have not stated any conjecture!> because, no matter how> many pairs of values I examine, their will always be more pairs that I> have not tested.You don't need to worry about the pairs that have not been testedto make a conjecture.For example, considering the example of 2, I can conjecture thatall primes are even. That is a valid conjecture.> Am I on the right track? Any further help appreciated.First, you need to make your conjecture.Next, you need to try to prove the conjecture.Before proving your conjecture, you might want to try a fewmore critical cases to see if your conjecture holds up. But,this is not necessary.-- support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88ILRZ22723; =>>Can you give an example of what you are trying to do?>here's something that works, but only for -cos(x), -sin(x) and x^int.cos(x) and sin(x) it doesn't like...b:=proc(n,f,x) local k,res.tmp; res:=0; for k from 0 to n do assign(x,k/n); tmp:=evalf(f(x)); res:=res+(n!/(k!*(n-k)!))*(X^k)*((1-X)^(n-k))*tmp; unassign(x); end do; return res;end;riddle me this...Annabelle => here's something that works, but only for -cos(x), -sin(x) and x^int.> cos(x) and sin(x) it doesn't like...> b:=proc(n,f,x)> local k,res.tmp;> res:=0;> for k from 0 to n do> assign(x,k/n);> tmp:=evalf(f(x));> res:=res+(n!/(k!*(n-k)!))*(X^k)*((1-X)^(n-k))*tmp;> unassign(x);> end do;> return res;> end;I think you have some typos there. That procedure doesn't do anythingfor me. (Error, (in assign) invalid arguments). Can you carefullypost the correct procedure? Also, post examples of how you would liketo call it, and what you expect the output to be.n!/(k!*(n-k)!) is the same as binomial(n,k)-cos(x) can be expressed as a procedure as x-> -cos(x), or simply-cos. =>Can you give an example of what you are trying to do?> here's something that works, but only for -cos(x), -sin(x) and x^int.> cos(x) and sin(x) it doesn't like...> b:=proc(n,f,x)> local k,res.tmp;> res:=0;> for k from 0 to n do> assign(x,k/n);> tmp:=evalf(f(x));> res:=res+(n!/(k!*(n-k)!))*(X^k)*((1-X)^(n-k))*tmp;> unassign(x);> end do;> return res;> end;> riddle me this...Did you mean to have both lowercase x's and capital X's in your proc?In any case, you didn't take my suggestion of letting f be the functionsin instead of the expression sin(x), etc.Note followups.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.> So pq = rs>> Now because r is prime and s is prime it must be> true that p divides either r or s>It *is* true.But the question was: how can it be proved! You mayuse the de'nition of primality for r and s, i.e.they don't have factors besides 1 and themselves.But where do you get your conclusion from? It isjust the question posed by the OP.The OP and I are looking for a logical explanation ofthe following fact: Given there are a blocks of b elements each: || 1 block 2 block 3 block a---------------------------- etc. ---------- 123..b 123..b 123..b 123..bThe blocks cannot be divided into equal parts exceptfor the trivial case of size 1 for the parts. (b isprime). And you cannot arrange the blocks in packagesof equal size (except for size 1), because a is prime. you cannot make it d blocks with e elements: || 1 block 2 block 3 block d---------------------------- etc. ---------- 123..e 123..e 123..e 123..ewhere d and e are primes.Good nightRainer Rosenthalr.rosenthal@web.de => So pq = rs>> Now because r is prime and s is prime it must be>> true that p divides either r or s>>It *is* true.>But the question was: how can it be proved! You may>use the de'nition of primality for r and s, i.e.>they don't have factors besides 1 and themselves.You don't need the primality of r and s. If p is prime, and p dividesrs, for positive integers r and s, then p divides r or s.One way of proving this is 'rst to prove:Lemma - For any positive integers x and y, there exist (notnecessarily positive) integers a and b such that ax + by = gcd(x,y).Assuming the lemma for the moment, if p divides rs, with p prime, andp does not divide r, then p and r must be coprime, so by the lemma thereexist a and b with ap + br = 1, and hence aps + brs = s.Now p divides the LHS of this last equation, so p divides s.So we have proved that p divides r or s.Of course we still have to prove the lemma. One way of doing this isby induction on max(x,y). If x = y, then gcd(x,y) = x = 1.x + 0.y.Otherwise, say x > y, in which case gcd(x,y) = gcd(x-y,y), andby inductive hypothesis there exist a, b with a(x-y) + by = gcd(x,y),so ax + (b-a)y = gcd(x,y) and we are support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88IQWr23112; =>> What does F(a,b;c;d) mean, where a,b,c,d are numbers?>> Is this function in Mathematica?>>Hypergeometric function.>>-->Julien Santini> =(snip of my badly formulated question)> I have to say that I don't understand the question.OK, I try to formulate it better, below...> I'm reasonably> conversant in Morse Theory, but this question doesn't seem to be about> that.OK.> I also am not familiar with double logarithmic cartesian> coordinates is supposed to mean (I can guess that it consists of> cartesian coordinates in which the x and y axes are placed on a log> scale).Right!> Maybe a bit more detail about what you want will elicit some response.Here I try again:First: I have not 'nd - yet - any Morse potential function in the doublelogarithmic cartesian coordination system. (i.e. my poor referencereasearch, mea culpa) :-)Second: Many Morse like potential function modi'cations have differentnames in normal cartesian xy-coordination system depending on the 'rstresearcher - unfortunately. :-(Here comes the function in the ln-ln (double logarithmic) cartesian system:ln(y)= - ((V.o)((1-e^((ln(x)-ln(x.o))/d))^2)-V.o), where .o, i.e.dot orefers subindex.This function has - for me - unknown name as plotten in the xy cartesiancoordination system instead of the double logarithmic cartesian system.See - for example data - below.Thus, what is the name of the function in the normal xy cartesiancoordination system? (Not - in the double logarithmic cartesian system.)I try to 'nd original references. The right function name helps - I assume.:-)You may consider my problem this way:Plot this data in the excel sheet graphics:y 1 0,9653 0,9287 0,9049 0,8301 0,7938 0,7571 0,809 0,8962 1,009 1,147 1,316 1,518 1,762 2,04 2,38 2,77 3,22 4,37 5,9 7,94 10,44x0 0,001 0,005 0,01 0,05 0,1 0,5 1 1,5 2 2,5 3 3,5 4 4,5 5 5,5 6 7 8 9 10Modify axes (x and y) into ln-ln form ( by using: select axis, format axis,select scale, logarithmic scale) and you see the above mentioned Morsepotential function that satis'es the equation written above.The original function, in the xy-cartesian coordination system, is unknownfor me. I would like to know the name of the original function in the x,ycartesian coordination system, though it has Morse potential function inthe double logarithmic cartesian coordination system. I assume - but I donot know - it has some spesi'c name and I would like to 'nd references inthe approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88JJ2427046; =Does anybody have a code for Bernstein polynomial operator?Annabelle => Does anybody have a code for Bernstein polynomial operator?>> Annabelle>Code: 007 =When looking for a random prime in some interval, often one will pick arandomodd integer n (in the interval) and use a primality test to test eachelement in thesequence n, n+2, n+4, ...., n + 2*k, for some 'xed k, until the 'rstprobable primeis found. (you will most likely 'rst apply a sieving procedure to thesequence and testthe elements in the resulting sequence, but I don't care about thathere...).For k = in'nity, one is guaranteed to 'nd a prime in the sequence sincethere isan in'nity of primes. Can anything interesting be said for 'nite k?Such as for all n, for k > X(n), there is at least one prime in thesequence with probabilityY(n,k), for some formulas X, Y, ideally with Y(n,k) = 1.Or maybe something in the lines of for k > X, for all n, there is at least one prime in the sequence withprobability Y(n),but that seems less likely.The fact that for x >= 114, we have x / ln(x) < pi(x) < 5*x / 4*ln(x)where pi(x) is the number of primes smaller or equal to x,can be used to calculate a lower bound on the number of primes in [n, n +2*k],but I don`t get a nice formula with that.Any references would be appreciated!--Anton => For k = in'nity, one is guaranteed to 'nd a prime in the sequence> since there is an in'nity of primes. Can anything interesting be> said for 'nite k? Such as for all n, for k > X(n), there is at> least one prime in the sequence with probability Y(n,k), for some> formulas X, Y, ideally with Y(n,k) = 1. Or maybe something in the> lines of for k > X, for all n, there is at least one prime in the> sequence with probability Y(n), but that seems less likely.I remember a theorem that for any n > 1, there's always a prime betweenn and 2n. =>> For k = in'nity, one is guaranteed to 'nd a prime in the sequence>> since there is an in'nity of primes. Can anything interesting be>> said for 'nite k? Such as for all n, for k > X(n), there is at>> least one prime in the sequence with probability Y(n,k), for some>> formulas X, Y, ideally with Y(n,k) = 1. Or maybe something in the>> lines of for k > X, for all n, there is at least one prime in the>> sequence with probability Y(n), but that seems less likely.> I remember a theorem that for any n > 1, there's always a prime between> n and 2n.There's a prime between 1 and 2, for suf'ciently generousinterpretations of between. Chebyshev said it, And I'll say it again. There is always a prime Between n and 2n. - P. Erdos-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. =>> For k = in'nity, one is guaranteed to 'nd a prime in the sequence>> since there is an in'nity of primes. Can anything interesting be>> said for 'nite k? Such as for all n, for k > X(n), there is at>> least one prime in the sequence with probability Y(n,k), for some>> formulas X, Y, ideally with Y(n,k) = 1. Or maybe something in the>> lines of for k > X, for all n, there is at least one prime in the>> sequence with probability Y(n), but that seems less likely.>> I remember a theorem that for any n > 1, there's always a prime> between n and 2n.I thought it was a conjecture (by Bertrand, I had to look it up), but you'reright: it was proven by Chebychev. => When looking for a random prime in some interval, often one will pick> a random> odd integer n (in the interval) and use a primality test to test each> element in the> sequence n, n+2, n+4, ...., n + 2*k, for some 'xed k, until the 'rst> probable prime> is found. (you will most likely 'rst apply a sieving procedure to> the sequence and test> the elements in the resulting sequence, but I don't care about that> here...).>> For k = in'nity, one is guaranteed to 'nd a prime in the sequence> since there is> an in'nity of primes. Can anything interesting be said for 'nite k?> Such as for all n, for k > X(n), there is at least one prime in the> sequence with probability> Y(n,k), for some formulas X, Y, ideally with Y(n,k) = 1.> Or maybe something in the lines of> for k > X, for all n, there is at least one prime in the sequence> with probability Y(n),> but that seems less likely.>> The fact that for x >= 114, we have> x / ln(x) < pi(x) < 5*x / 4*ln(x)> where pi(x) is the number of primes smaller or equal to x,> can be used to calculate a lower bound on the number of primes in [n,> n + 2*k],> but I don`t get a nice formula with that.>> Any references would be appreciated!>> --AntonIIRC, there's a conjecture that there is at least 1 prime between n and 2*nfor n > 1. =>> IIRC, there's a conjecture that there is at least 1 prime between n and2*n> for n > 1.> It is open whether or not for each n>1 if there is a prime between n^2 and(n+1)^2. =I have a freeware calculator that I have spent about 3 months updatingand now have up on my website for download. I need user feedback onusability, any bugs, suggested features, etc. It does expression evaluation, programable with VB Scriptinglanguage, does 2D (polar, cartesian and parametric) and 3D plotting,Matrix manipulation, complex number math, 1-variable equation solving,simultaneous equation solving (2x2 to 8x8), Integer math to 32000places, 2D statistics with graphing and regression curve 'tting, unitconversions and geometric area/volume calculations. For documentingcalculations, it has a built in RTF editor, image editor, formulaillustrator and schematic capture module. The design started out as an aid for engineers but has evolved moreinto a learning tool for math students. The link is: www.dazyweblabs.com/dazycalc/index.htmlVicDWL =Nice, why not go commercial?Displaying #1 - 100 (of 189 matching domains) 1 - 112calcio.com (DELETED on 10-28-02) 2 - 2calchemy.com (ON HOLD since 08-04-03) 3 - 2calcutta.com (DELETED on 06-15-03) 4 - 4calc.com (DELETED on 06-22-03) 5 - acalcutta.com (DELETED on 12-05-02) 6 - anycalc.com (DELETED on 08-11-03) 7 - area-calc.com (ON HOLD since 07-11-03) 8 - b2bcalc.com (DELETED on 05-15-03) 9 - bcalculo.com (DELETED on 07-02-03) 10 - bizcalc.com (ON HOLD since 06-22-03) 11 - blendcalc.com (DELETED on 07-17-03) 12 - bvcalcs.com (DELETED on 07-08-03) 13 - cal-calc.com (DELETED on 06-22-03) 14 - calc-this.com (DELETED on 06-19-03) 15 - calc24.com (DELETED on 04-17-03) 16 - calcabo.com (DELETED on 06-07-03) 17 - calcadult.com (DELETED on 08-04-03) 18 - calcaliv.com (DELETED on 12-25-02) 19 - calcarewc.com (DELETED on 07-06-03) 20 - calcarpin.com (ON HOLD since 09-07-03) 21 - calcasino.com (ON HOLD since 09-06-03) 22 - calcaster.com (DELETED on 09-07-03) 23 - calcastle.com (DELETED on 12-23-02) 24 - calcat2.com (DELETED on 01-28-03) 25 - calccio.com (DELETED on 08-21-03) 26 - calcer.com (DELETED on 07-22-03) 27 - calcex.com (DELETED on 11-11-02) 28 - calchai.com (ON HOLD since 08-22-03) 29 - calchice.com (DELETED on 10-09-02) 30 - calchina.com (DELETED on 08-16-03) 31 - calchow.com (DELETED on 10-25-02) 32 - calcina.com (DELETED on 12-25-02) 33 - calcinit.com (DELETED on 10-17-02) 34 - calcinite.com (DELETED on 10-25-02) 35 - calcio-39.com (DELETED on 06-06-03) 36 - calcio365.com (DELETED on 05-09-03) 37 - calcio39.com (DELETED on 06-06-03) 38 - calciod.com (DELETED on 05-02-03) 39 - calciofan.com (DELETED on 06-15-03) 40 - calciomio.com (DELETED on 03-07-03) 41 - calciook.com (DELETED on 05-16-03) 42 - calciosms.com (DELETED on 04-29-03) 43 - calciplus.com (ON HOLD since 07-23-03) 44 - calcitexl.com (ON HOLD since 04-23-03) 45 - calclick.com (ON HOLD since 07-01-03) 46 - calclinic.com (DELETED on 06-10-03) 47 - calcma.com (ON HOLD since 08-15-03) 48 - calcmm.com (ON HOLD since 06-22-03) 49 - calco-net.com (ON HOLD since 06-23-03) 50 - calcoil.com (DELETED on 05-20-03) 51 - calcolare.com (DELETED on 11-29-02) 52 - calcomets.com (DELETED on 12-18-02) 53 - calconc.com (ON HOLD since 09-04-03) 54 - calconinc.com (ON HOLD since 09-05-03) 55 - calcoo.com (DELETED on 10-10-02) 56 - calcool.com (ON HOLD since 07-12-03) 57 - calcoop.com (DELETED on 06-01-03) 58 - calcooper.com (DELETED on 03-21-03) 59 - calcotltd.com (DELETED on 08-05-03) 60 - calcotspa.com (ON HOLD since 07-12-03) 61 - calcou.com (DELETED on 10-10-02) 62 - calcougs.com (ON HOLD since 08-13-03) 63 - calcouple.com (DELETED on 06-07-03) 64 - calcousa.com (ON HOLD since 07-05-03) 65 - calcpainc.com (DELETED on 01-23-03) 66 - calcpics.com (DELETED on 06-27-03) 67 - calcserv.com (ON HOLD since 08-18-03) 68 - calcskins.com (DELETED on 04-13-03) 69 - calcsmart.com (DELETED on 04-20-03) 70 - calcstop.com (ON HOLD since 08-16-03) 71 - calctsa.com (DELETED on 07-28-03) 72 - calcul-ef.com (DELETED on 08-22-03) 73 - calculess.com (DELETED on 08-18-03) 74 - calculia.com (DELETED on 11-11-02) 75 - calculman.com (DELETED on 06-16-03) 76 - calculusu.com (DELETED on 11-07-02) 77 - calcumate.com (DELETED on 04-21-03) 78 - calcupa.com (ON HOLD since 08-30-03) 79 - calcurate.com (DELETED on 08-17-03) 80 - calcusoft.com (DELETED on 03-11-03) 81 - calcutter.com (DELETED on 12-29-02) 82 - catercalc.com (DELETED on 11-30-02) 83 - cecalcs.com (DELETED on 11-23-02) 84 - chemocalc.com (DELETED on 11-15-02) 85 - corecalc.com (DELETED on 12-02-02) 86 - e-calcium.com (DELETED on 03-06-03) 87 - e-calcul.com (DELETED on 12-24-02) 88 - e-calculs.com (DELETED on 12-24-02) 89 - ecalca.com (DELETED on 11-05-02) 90 - ecalcx.com (DELETED on 05-20-03) 91 - elcalcio.com (ON HOLD since 04-15-03) 92 - ener-calc.com (DELETED on 05-30-03) 93 - enrcalcs.com (DELETED on 06-01-03) 94 - equicalc.com (DELETED on 06-05-03) 95 - escalc.com (DELETED on 04-26-03) 96 - eucalc.com (DELETED on 11-13-02) 97 - excalco.com (DELETED on 10-12-02) 98 - ez-calc.com (ON HOLD since 07-13-03) 99 - fabricalc.com (DELETED on 07-04-03) 100 - farmcalc.com (ON HOLD since 07-09-03) 101 - 'scalco.com (DELETED on 11-29-02) 102 - 'scalcom.com (DELETED on 11-29-02) 103 - §ashcalc.com (DELETED on 01-16-03) 104 - §ocalc.com (DELETED on 04-10-03) 105 - §ycalc.com (DELETED on 01-16-03) 106 - focalcord.com (DELETED on 11-28-02) 107 - furncalc.com (DELETED on 12-18-02) 108 - gemcalc.com (DELETED on 08-30-03) 109 - gfrcalc.com (DELETED on 05-26-03) 110 - go2calc.com (DELETED on 06-16-03) 111 - go4calcio.com (DELETED on 12-26-02) 112 - gpcalc.com (DELETED on 08-18-03) 113 - grascalce.com (DELETED on 06-27-03) 114 - handcalc.com (DELETED on 01-20-03) 115 - hedgecalc.com (DELETED on 06-06-03) 116 - hvacalcs.com (ON HOLD since 08-02-03) 117 - icalcio.com (DELETED on 06-19-03) 118 - incalca.com (DELETED on 01-25-03) 119 - inetcalc.com (DELETED on 08-11-03) 120 - inscalc.com (DELETED on 03-24-03) 121 - italcalce.com (DELETED on 05-26-03) 122 - ixtacalco.com (DELETED on 09-07-03) 123 - kamcalc.com (DELETED on 10-30-02) 124 - kcalc.com (ON HOLD since 08-06-03) 125 - kcalcard.com (DELETED on 08-13-03) 126 - leasecalc.com (ON HOLD since 08-06-03) 127 - livecalc.com (ON HOLD since 06-22-03) 128 - loadcalcs.com (ON HOLD since 08-02-03) 129 - localcad.com (DELETED on 11-27-02) 130 - localchix.com (DELETED on 09-05-03) 131 - localcm.com (DELETED on 10-21-02) 132 - localco.com (DELETED on 03-26-03) 133 - localcon.com (DELETED on 12-06-02) 134 - localcube.com (DELETED on 04-07-03) 135 - lpapcalc.com (DELETED on 12-11-02) 136 - m-calc.com (DELETED on 03-23-03) 137 - magicalcd.com (DELETED on 04-07-03) 138 - mcalchemy.com (DELETED on 06-20-03) 139 - medicalcu.com (DELETED on 09-03-03) 140 - miocalcio.com (DELETED on 04-27-03) 141 - mmsscalc.com (ON HOLD since 08-13-03) 142 - msdmcalc.com (DELETED on 07-16-03) 143 - mycalcs.com (DELETED on 06-18-03) 144 - namcalc.com (DELETED on 10-30-02) 145 - oilcalc.com (ON HOLD since 09-02-03) 146 - oncalc.com (DELETED on 06-24-03) 147 - opticalcn.com (DELETED on 01-05-03) 148 - orocalcum.com (ON HOLD since 08-26-03) 149 - palmcalc.com (DELETED on 05-30-03) 150 - pascalco.com (DELETED on 01-01-03) 151 - pcalchemy.com (ON HOLD since 08-18-03) 152 - quicalcio.com (DELETED on 03-11-03) 153 - racalcorp.com (DELETED on 04-07-03) 154 - rascalcam.com (DELETED on 06-06-03) 155 - rascalcar.com (DELETED on 04-14-03) 156 - routecalc.com (DELETED on 06-01-03) 157 - safecalc.com (DELETED on 11-28-02) 158 - scorecalc.com (DELETED on 10-28-02) 159 - socalcat.com (DELETED on 07-23-03) 160 - socalcert.com (ON HOLD since 08-07-03) 161 - socalclan.com (DELETED on 05-18-03) 162 - socalcn.com (DELETED on 04-03-03) 163 - socalcolo.com (DELETED on 07-04-03) 164 - socalcpug.com (ON HOLD since 08-13-03) 165 - softcalc.com (ON HOLD since 08-30-03) 166 - solarcalc.com (DELETED on 05-03-03) 167 - sscalc.com (DELETED on 06-13-03) 168 - stacalco.com (DELETED on 12-22-02) 169 - statucalc.com (DELETED on 10-14-02) 170 - stimcalc.com (DELETED on 06-14-03) 171 - tecalcorp.com (DELETED on 08-20-03) 172 - techcalc.com (DELETED on 08-03-03) 173 - tekcalc.com (DELETED on 08-03-03) 174 - thecalc.com (ON HOLD since 09-04-03) 175 - ticalc83.com (DELETED on 12-19-02) 176 - truckcalc.com (DELETED on 06-24-03) 177 - twmedcalc.com (DELETED on 07-26-03) 178 - unicalc.com (DELETED on 08-05-03) 179 - uscalcium.com (DELETED on 01-22-03) 180 - usedcalcs.com (DELETED on 04-30-03) 181 - visicalc.com (ON HOLD since 01-27-03) 182 - vitacalc.com (DELETED on 06-14-03) 183 - vocalca.com (DELETED on 11-08-02) 184 - vocalcafe.com (DELETED on 10-25-02) 185 - w3calc.com (DELETED on 08-31-03) 186 - wapcalcio.com (ON HOLD since 05-28-03) 187 - wcalc.com (DELETED on 04-11-03) 188 - wtbcalc.com (DELETED on 07-16-03) 189 - zrcalce.com (ON HOLD since 08-23-03)Displaying #1 - 71 (of 71 matching domains) 1 - alloycalculator.com (DELETED on 11-05-02) 2 - auto-calculator.com (ON HOLD since 08-04-03) 3 - btucalculator.com (DELETED on 08-13-03) 4 - calculator-games.com (DELETED on 11-11-02) 5 - calculatorcom.com (ON HOLD since 08-05-03) 6 - calculatorgames.com (ON HOLD since 11-04-02) 7 - calculatorh.com (DELETED on 05-26-03) 8 - calculatorhouse.com (DELETED on 01-05-03) 9 - calculatorltd.com (DELETED on 07-17-03) 10 - calculatorp.com (DELETED on 06-13-03) 11 - calculators4less.com (DELETED on 10-30-02) 12 - calculators4u.com (DELETED on 12-18-02) 13 - calculatorsales.com (DELETED on 08-01-03) 14 - calculatorscales.com (DELETED on 06-28-03) 15 - calculatorscom.com (ON HOLD since 08-05-03) 16 - calculatorskit.com (DELETED on 06-11-03) 17 - calculatorsltd.com (DELETED on 06-30-03) 18 - calculatorsset.com (DELETED on 06-13-03) 19 - calculatorstreet.com (ON HOLD since 09-05-03) 20 - calculatorsurfer.com (ON HOLD since 09-05-03) 21 - calculatortools.com (ON HOLD since 09-05-03) 22 - calculatorwatch.com (ON HOLD since 09-05-03) 23 - calculatorworks.com (ON HOLD since 09-05-03) 24 - chemicalculator.com (DELETED on 11-18-02) 25 - chinacalculator.com (ON HOLD since 09-05-03) 26 - dcalculator.com (DELETED on 05-13-03) 27 - dieselcalculator.com (DELETED on 04-21-03) 28 - dincalculator.com (ON HOLD since 08-06-03) 29 - e-bizcalculator.com (DELETED on 11-16-02) 30 - fuzzy-calculator.com (DELETED on 07-01-03) 31 - fuzzycalculator.com (DELETED on 07-01-03) 32 - gascalculator.com (DELETED on 04-21-03) 33 - getcalculators.com (DELETED on 01-25-03) 34 - iolcalculator.com (DELETED on 01-14-03) 35 - iq-calculator.com (DELETED on 10-05-02) 36 - ixcalculator.com (DELETED on 07-17-03) 37 - laoncalculators.com (ON HOLD since 08-11-03) 38 - ldcalculator.com (DELETED on 11-05-02) 39 - lifecalculator.com (DELETED on 06-26-03) 40 - lonecalculators.com (ON HOLD since 08-11-03) 41 - love-calculator.com (ON HOLD since 02-27-02) 42 - love-calculators.com (ON HOLD since 08-22-03) 43 - lovelcalculator.com (DELETED on 10-24-02) 44 - lpgcalculator.com (DELETED on 04-21-03) 45 - m-calculator.com (DELETED on 03-23-03) 46 - magiccalculator.com (ON HOLD since 08-05-03) 47 - micalculator.com (DELETED on 04-06-03) 48 - movecalculator.com (DELETED on 03-15-03) 49 - netcalculators.com (DELETED on 10-26-02) 50 - petrolcalculator.com (DELETED on 04-21-03) 51 - pointscalculator.com (DELETED on 08-09-03) 52 - realcalculators.com (DELETED on 04-28-03) 53 - refundcalculator.com (DELETED on 12-08-02) 54 - riskcalculator.com (ON HOLD since 09-04-03) 55 - robotcalculator.com (DELETED on 05-19-03) 56 - rpncalculator.com (DELETED on 07-13-03) 57 - seacalculator.com (DELETED on 08-22-03) 58 - shop-calculator.com (DELETED on 04-28-03) 59 - shop-calculators.com (DELETED on 04-28-03) 60 - shopcalculator.com (DELETED on 12-15-02) 61 - shopcalculators.com (DELETED on 04-28-03) 62 - stock-calculator.com (DELETED on 12-31-02) 63 - suncalculator.com (ON HOLD since 08-12-03) 64 - threadcalculator.com (DELETED on 10-05-02) 65 - ti83calculator.com (DELETED on 03-29-03) 66 - tradecalculator.com (DELETED on 01-04-03) 67 - vcalculator.com (DELETED on 05-20-03) 68 - vpncalculator.com (DELETED on 12-27-02) 69 - wapscalculator.com (DELETED on 08-12-03) 70 - yourcalculator.com (DELETED on 03-31-03) 71 - zakatcalculator.com (DELETED on 08-04-03)Picks of the bunch:anycalccalcsmartcalcusoft 172 - techcalc.com (DELETED on 08-03-03) 173 - tekcalc.com (DELETED on 08-03-03) 174 - thecalc.com (ON HOLD since 09-04-03)unicalcvisicalc 8 - calculatorhouse.com (DELETED on 01-05-03)13 - calculatorsales.com (DELETED on 08-01-0321 - calculatortools.com (ON HOLD since 09-05-03) 46 - magiccalculator.com (ON HOLD since 08-05-03)70 - yourcalculator.com (DELETED on 03-31-03)I'd go thecalc.comgreat store name, thewatch.com sold for $600 a few months ago toa watch shop owner, good deal both ways.amazingly SciCalc.com is available I would get it if I was you and you'll have100s of times the amount of downloads and a recognised product.$9 at www.godaddy.comor $15 at www.namesecure.com with no ads redirection (better if you don't have DNS on yourwebsite).Herc => I have a freeware calculator that I have spent about 3 months updating> and now have up on my website for download. I need user feedback on> usability, any bugs, suggested features, etc.>> It does expression evaluation, programable with VB Scripting> language, does 2D (polar, cartesian and parametric) and 3D plotting,> Matrix manipulation, complex number math, 1-variable equation solving,> simultaneous equation solving (2x2 to 8x8), Integer math to 32000> places, 2D statistics with graphing and regression curve 'tting, unit> conversions and geometric area/volume calculations. For documenting> calculations, it has a built in RTF editor, image editor, formula> illustrator and schematic capture module.>> The design started out as an aid for engineers but has evolved more> into a learning tool for math students. The link is:>> www.dazyweblabs.com/dazycalc/index.html> Vic> DWLBetter use (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88KbFF00722; =>
> > Both of you are
tripping all over semantics.>>Another hit and run poster who
does more damage for not following the>argument of a thread.>> Would this argument continue on if you replace 00:00:00 and
the priceless 1>> second with 0 meters, 0 centimeters, 0
millimeters and the priceless 1>> centimeter?>>You forget that
00:00:00 is actually 24:00:00 or the 86 400th>second,however 0
meters:0 centimeters:0 millemeters mean exactly>0,nothing
,nought,nichts.> It is the EXACT same
argument.>>Funny,funny,funny,can you please capitalise the
whole sentence you I>can enjoy it more.>> The odometer in
your car works>> just like a clock, it measures forward
progression.>That's right,it is a good measure of distance
traveled,exactly how a>clock relates to rotation of the Earth
based on a human geometric>construct and if you forget that a
clock is based on a cycle and>circle you probably mistake it
for a measure of a quantity rather than>what it really is - a
PROPORTION,as a bonus you might discover the>difference
between a stopwatch registering 0 and a clock
registering>0.>> How do your arguments hold up there
Oriel36 when you look at how clocks were>> used in China?>>I
wish you were around in 1905 to point this out to a dreamy
kid> Was the water clock based on geometry and the
progression of>> stars - NO.>>Good idea !,it such a silly
premise to base geometry and motion of the>cosmos on
clocks. Put down your Plato and come back to the real
world.>>I never left the real world,it is a place where there
is no time>dilation, no lenght contraction, no twin paradoxes,
no multiverses, no>branes,just an intricate geometrical system
left by our ancestors who>were pretty much like us except that
suddenly for 100 years those who>inherited their insights
became plain dumb and silly.Relativity is not>offensive,it is
an insult to intelligence of mankind.>
=#include #include // solution of differential equation y''+sin(y)=0double JacobiSN(double z, double m){ long double f[40], p[20]; long double z2; f[3]=6; f[5]=120; f[7]=5040; f[9]=39916800; f[11]=6227020800; f[13]=f[11]*156; f[15]=f[13]*210; p[1]=1+m; p[2]=1+(14+m)*m; p[3]=1+(135+(135+m)*m)*m; p[4]=1+(1228+(5478+(1228+m)*m)*m)*m; p[5]=1+(11069+(165826+(165826+(11069+m)*m)*m)*m)*m; p[6]=1+(99642+(4494351+(13180268+(4494351+(99642+m)*m)*m)*m)*m )*m; p[7]=1+(896803+(116294673+(834687179+(834687179+(116294673+( 896803+m) *m)*m)*m)*m)*m)*m; z2 = pow(z,2); return z*(1-z2*(p[1]/f[3]-z2*(p[2]/f[5]-z2*(p[3]/f[7]-z2*(p[4]/f[9]- z2* (p[5]/f[11]-z2*(p[6]/f[13]-z2*p[7]/f[15]))))))); }void main(){ int i; double T = 6.534345170850724,y;// FILE *out; // out = fopen(JacobiSN.txt,wt); for(i=0;i<= 10;i++) { // fprintf(out,%e %en,T/40*i,JacobiSN(T/40*i, sqrt(0.5))); printf(%e %en,T/40*i,JacobiSN(T/40*i,sqrt(.5))); } // it should be equal to 1.000000, but it it rather very different// what is wrong: I seems not to be bad convergence of the serie} =N A D R O W S K I14 1 4 18 15 23 19 11 9 = 114 Oh dear. Megan makes her appearance in today's SaskatoonStar-Phoenix (sun-sunbird) obituaries.161+ DamonDamon 47 Nadrowski 114144+ KathyKathy 65 Lusby 79251 Megan 16 2 85 47/318 10226Megan 40 Kathryn 97 Nadrowski 114155+ KalleKalle 41 Nadrowski 114165+ MichaelMichael 51 Nadrowski 114 Megan was born on the 16th. Her 'rst and last names add togetherfor the 154 verses of Bible Book 25 Lamentations (the 16th non-prime).She has 16 letters in her last two names. She is missing 16 lettersfrom her given names. The primes, squares and cubes in her given namesadd together for 69, or 16 plus the 16th prime (53). Her middle nameadds to the 25th prime while 25 in turn is the 16th non-prime. Hergiven names add to the 28th non-prime and to the 25th prime, togetherfor 53 (the 16th prime). Her 251 valued name averages with her 47thday of birth for the 149 verses of Galatians, Bible Book 48(16+16+16). Megan lived for 18.53 years (53 is the 16th prime). Megandied 196 days after her birthday, there are 196 verses in TheTimothies (New Testament Books 15 and 16). Megan died on August 31stwhen she collided head-on with a vehicle driven by a 48 (16+16+16)year old driver on hiway 16 near North Battleford. That 48 (16+16+16)year old driver was Neil (40) Francis (70) Howarth (93), his name addsand Neil were 18 and 48 years old, together for 66 (Exodus 16), theirnames differ in value by 48 (16+16+16) and average 227 (Judges 16). Megan was born a multiple of 17 days into the century and she wasalso born a multiple of 59 (the 17th prime) days into the century(31093=17x31x59). Her name adds to 251 (the 'rst 17 non-primes). Her'rst 17 letters add to 189 (the 'rst 17 primes minus the 'rst 17non-primes). She was born in 85 (5x17 and is the 17th prime plus the17th non-prime), Ruth with 85 verses is the 17th shortest Book in theBible.Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 59 <-17th-> 26 --- --- 440 251 Dad's 'rst name adds to 47, mom's 'rst name adds to 65 (47thnon-prime). Megan was born on day 47, her last two names add togetherfor 211 (47th prime). Megan and her siblinks have 'rst and lastnames adding together for 474. Megan was born on and died on days ofthe month adding to 47. Daniel and Revelation are the major Books ofend-times prophecy, they differ in length by 47 verses.Primes 2 61 149 3 67 151 5 71 157 7 73 163 11 79 167 13 83 173 17 89 179 19 97 181 23 101 191 29 103 193 31 107 197 37 109 199 41 113 211 <-47th 43 127 223 47 131 227 53 137 229 59 139 233 Megan's last name adds to 114 (57+57). Her given names differ invalue by 57. Her vowels add to 57. Her 57+57 valued last name adds to285% (5x57%) of her 40 valued 'rst name. Her 'rst name adds to 40,Bible Book 40 is Matthew, Daniel is the 27th Book of the Old Testamentwith 357 verses while Matthew is the 'rst of the 27 Books of the NewTestament with 357+357+357 verses. Megan was goink out with somefeller who's last name adds to 114 (57+57). She died on day 243(66.57% into the year). There are 357 verses in Daniel while the 404verses of Revelation is 57 plus the 57th prime (269) plus the 57thnon-prime (78), these are the major Books of end-times prophecy (theyare in part about 666). The kids have 'rst names adding together for132 (66+66). Megan's last two names add together for 211 (theterminating chapter of Bible Book 6). Megan died 3.6660 years into thecentury (1339/365.25=3.6660). Her parents divorced and now have lastnames adding together for 193 (Bible Book 6 chapter 6), I guess theywere unable to see God's role in bringing the family together. Megan was a U of S student, she hoped to study law. Aftercriticizing churches I was tortured for years at the U of S, peopledidn't care while they were torturing me, and people continue to notcare now that they stopped arresting and torturing me. The lawyersappointed to me were powerless against the Mental Health Act, oncecharged under the Mental Health Act in Canada, people lose theirrights for eternity, the lawyers only went through motions at appealpanel hearings so that they could make a few dollars. All I can do iscry out to God against you people and request that He honors Exodus20:5 and Hosea 4:6 and terminates the lives of your children. And soMegan's death is good news and I rejoice with utter glee with the newsof her death. I lost summer after summer after summer after summerafter summer after summer after summer to psychiatric torture, and thesubsequent summers were effectively lost as well for I remain franticto §ee this country, and you people are so cheap and ignorant and soGod-damned compassionless, that you can't even assist me to get awayfor one single winter. The parents are having Megan cremated as is sheboth Reverend and Father to conduct the funeral service,Reverend means worthy of worship while we are instructed in theBible to call nobody Father but our Father in Heaven. But it simplydothn't matter what the Bible says, for the parents have traditionsthat supercede God's Commandments and all common sense. One down, twoto go, you people are the of the earth and I rejoice at the deathof your daughter!!! Go put a few extra blinkin' lights onto your paganevergreen tree in honor of your smitten progeny, you ignorantarseholes.Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!Exodus 20:5, Hosea 4:6 - And He took away your daughter!!! => H A Y M A N> 8 1 25 13 1 14 = 62 I met Mark at Timothies in the Canada Building on 21st Street, he> is a lawyer. Mark's sister has never been married, probababbly he is> actively seeking a husband for the nubile sweety.> 62+ Dad 27 4 41 117/248 +5775> 62+ Mom 24 10 45 297/68 +4134> 175 Andrea 10 5 68 131/235 4100> Andrea 43 Lynne 70 Hayman 62> 184 Mark 24 4 70 114/251 4814> Mark 43 William 79 Hayman 62> Primes Non-Primes Numbers> 2 1 1> 3 4 2> 5 6 3> 7 8 4> 11 9 5> 13 10 6> 17 12 7> 19 14 8> 23 15 9> 29 16 10> 31 18 11> 37 20 12> 41 21 13> 43 <-14th-> 22 <-14th-> 14> --- --- ---> 281 176 105> The parents were born in years averaging 43 (14th prime) and in> months adding to 14. The parents were born on days of the year adding> to 414. The parents were together born 7x14 days closer to the end of> their years than to the beginning of their years. Dad was born 131> days closer to the beginning of the year than to the end of the year> (Numbers 14). Andrea adds to 43 (14th prime), she was born on day 131> (Numbers 14). Mark adds to 43 (14th prime), he was born on day 114.> Andrea (43) and Mark (43) were born on days and in months adding to> 43. Andrea's middle name adds to 70 (5x14). Mark's middle name adds to> 79 (14+14p+14np and is the 14th prime in non-prime position). Both> kids have common names adding to 105 (1 through 14 and is the 'rst 14> primes minus the 'rst 14 non-primes). Andrea's full name adds to 175,> corresponding to Deuteronomy 22 (14th non-prime). Mark's given names> add together for 122 (22 is the 14th non-prime) and differ in value by> the 36 chapters of Bible Book 14, the 36 chapters of Bible Book 14 is> 14 plus the 14th prime (22). And Book 14 contains 822 verses, pretty> as 22 is the 14th non-prime, 22 exceeds 8 by 14 while 14 is the 8th> non-prime. Mark's last two names add together for 141. Mark's 'rst> two letters add to 14. The kids have middle names adding together for> 141. The kids have initials adding together for 65 (14p+14np). Mark's> vowels add to 22 (14th non-prime), and his consonants exceed his> vowels by 140. Andrea's 131st day of birth is 114.91% of Mark's 114th> day of birth. Mark's full name adds to 105.14% of his sister's full> name, keeping in mind that 105 is 1 through 14 and is the 'rst 14> primes minus the 'rst 14 non-primes. The kids were born in years> adding to 138 (the 105th non-prime). Mark was born 1.48 years after> his parent's birthdays. The family was born on days 27, 24, 10 and 24,> together these Bible Books contain 112 (8x14) chapters. They were born> in 41, 45, 68 and 70, these are the 13th prime, 31st, 49th and 51st> non-primes, together for 144. They were born in years averaging 56> (4x14). I am 4814 days older than Mark. The kids are separated by 714> (14x51) days, prettier as the parents were born on days of the month> adding to 51. The family was born on days of the century adding to> 82473, it is 225.79 years (225 is Judges 14 while 79 is 14+14p+14np).> Their name ends with the 14th letter of the alphabet.> Non-Primes> 1 57 110> 4 58 111> 6 60 112> 8 62 114> 9 63 115> 10 64 116> 12 65 117> 14 66 118> 15 68 119> 16 69 120> 18 70 121> 20 72 122> 21 74 123> 22 75 124> 24 76 125> 25 77 126> 26 78 128> 27 80 129> 28 81 130> 30 82 132> 32 84 133> 33 85 134> 34 86 135> 35 87 136> 36 88 138 <-105th> 38 90 140> 39 91 141> 40 92 142> 42 93 143> 44 94 144> 45 95 145> 46 96 146> 48 98 147> 49 99 148> 50 100 150> 51 102 152> 52 104 153> 54 105 154> 55 106 155> 56 108 156> The parents were born in years adding to 86 while the kids were> born with 486 days remaining in their years. The kids have 'rst names> adding together for 86. Mom was born with 68 days remaining in the> year and 'rst gave birth in 68.> The kids have given names adding to 113 and to 122, these are the> 30th prime and the 92nd non-prime, together for 122, and the kids> generally have their birthdays on days of the year averaging 122.> Mom was born in 45, the kids were born on days of the year adding> to 245. The family was born on days 27, 24, 10 and 24, these Bible> Books contain an average of 945 verses.> Primes > In Prime> Primes Positions> 1 2> 2 3 <- 3> 3 5 <- 5> 4 7> 5 11 <- 11> 6 13> 7 17 <- 17> 8 19> 9 23> 10 29> 11 31 <- 31> 12 37> 13 41 <- 41> ---> 108> I violated God's Sabbath and bought a coffee, and met Mark. Dad was> born in 41 while Gospel Mark is Bible Book 41. The parents are> separated by 1641 days, I am 4100 days older than the sister. The> short version of Gospel Mark contains 666 verses while 41 is the 6th> prime in prime position. The kids were born on days 10 and 24, the> former is 41.666...% of the latter. The family was born on days and in> months adding together for 108 (the primes in prime positions up to> 41). Mark's name adds to 184... the 184th prime (1097) and the 184th> primes, together for 36 (6x6 and 1 through 36 adds to 666). The last> name adds to 62, keeping in mind that chapter 1062 closes New> Testament Book 6, it is 666 plus 6x66 and is the 658 verses of Book 6> plus the 404 verses of Book 66. The family was together born 143 days> closer to the beginning of their years than to the end of their years,> chapter 666 brings Ecclesiastes up to 143 verses. The kids were born> on days 10 (6th non-prime) and 24 (6+6+6+6), Old Testament Books 6 and> 10 (6th non-prime) both contain 24 (6+6+6+6) chapters. Dad was born on> the 27th, Old Testament Book 27 and New Testament Book 27 are the> major Books of end-times prophecy, they are both in part about 666.> The family was born on days and in months and years adding to 332> (166+166).> 1-50 - Genesis> 51-90 - Exodus> 91-117 - Leviticus> 118-153 - Numbers> 154-187 - Deuteronomy> 188-211 - Joshua> 930-957 - Matthew> 958-973 - Mark> 974-997 - Luke> 998-1018 - John> 1019-1046 - Acts> 1047-1062 - Romans> 188 <-the opening chapter of Book 6 is 6x6x6 short> of the 404 verses of Bible Book 66, it is the> 6th prime squared (13x13) short of the 357> verses of Daniel (also in part about 666)> 193 <-Book 6 chapter 6 is the 44th prime,> while 44 is in turn 66.666...% of 66> 211 <-the terminating chapter of Book 6 is> approximately 66.6% of the 66th prime (317)> 357 <-the opening chapter of Book 6 plus the 6th> prime squared is the 357 verses of Daniel> (in part about 666)> 404 <-the 6th prime squared (13x13) plus the 6th> prime squared (13x13) plus 66 adds to the> 404 verses of Bible Book 66> 1062 <-666 plus 6x66 is a combination of the 658> verses of Bible Book 6 plus the 404 verses> of Bible Book 66, and is the terminating> chapter of New Testament Book 6> 1070 <-666 plus the 404 verses of Book 66 is the> 1070 verses of Job (Book 6+6+6)> 1213 <-Exodus terminates at chapter 90 (66th non-> prime) with 1213 verses (the 198th or the> 66+66+66th prime)> 1292 <-the 658 verses of Book 6 plus twice the> 66th prime (317) is the 1292 verses of> Isaiah (the Book contains 66 chapters)> The parents were together born 169 (13x13 or the 6th prime squared)> days further into their years than the kids. The parents were born on> days 27 and 24, together these Bible Books contain 13x13+13x13 more> verses more than kid's Books 10 and 24. Mark is Bible Book 41 (the> 13th prime), he and his sister were born in years adding to 138. The> family was born in years adding to 224 (Judges 13).> Daryl Shawn Kabatoff> Box 7134> Saskatoon Saskatchewan> Canada> S7K 4J1> Isaiah 45:4, Ephesians 3:15 - God gives you your name!!!And what's the probababbly of that? = Sire, there is no royal road to geometry.But the _is_ the Harris road.Gib =Suppose you have a population that can be partitioned by region,party, race, religion, gender, shoe size, and so forth. And you wanta small congress to proportionally represent this population -- notjust by region or party, but by all of these partitionings,simultaneously. You can't get perfectly proportional representationfor all partitionings, but you can come close. Just random selectionwill come close, but you could probably do better with effort.What's the name of this problem? What are the usual algorithms forsolving it? => Suppose you have a population that can be partitioned by region,> party, race, religion, gender, shoe size, and so forth. And you want> a small congress to proportionally represent this population -- not> just by region or party, but by all of these partitionings,> simultaneously. You can't get perfectly proportional representation> for all partitionings, but you can come close. Just random selection> will come close, but you could probably do better with effort.> What's the name of this problem? Politics.What are the usual algorithms for> solving it?Get a whole boatload of cash and learn how spread it around effectively to get the most for your particular group. => 2. Consider any plane, containing c=2^aleph 0 points.> Cover the plane with disjoint closed sets all with non zero area,> similiar to countries on a map.> One way is to let the cover consist of a single set (the entire plane).Isn't this the only way? A path-connected space cannot be the union oft disjoint nonempty closed sets for any countable t > 1. And, as youpointed out, the non-zero area requirement means the covering must becountable. =>> 2. Consider any plane, containing c=2^aleph 0 points.>> Cover the plane with disjoint closed sets all with non zero area,>> similiar to countries on a map. One way is to let the cover consist of a single set (the entire plane).> Isn't this the only way? A path-connected space cannot be the union of> t disjoint nonempty closed sets for any countable t > 1. And, as you> pointed out, the non-zero area requirement means the covering must be> countable.Let C_1 be a closed subset of the plane that approaches within distance 1of each point (for example, a set of isolated points with appropriatespacing will work).Let C_2 be a closed subset of R^2C_1 such that each point of R^2 lieswithin distance 1/2 of (C_0 union C_1).Inductively, let K_n = union_{k=1}^n C_k and choose C_(n+1) to be aclosed subset of R^2K_n such that each point of R^2 lies within distance1/(n+1) of K_(n+1).Then the C_n are a countable collection of disjoint closed sets thatcover R^2. In particular, K_n is closed for each n, and therefore anypoint p in the complement of K_n must lie at some distance epsilon > 0from K_n. Choosing n' > 1/epsilon, we conclude that p lies in K_n' andtherefore is in the union of all the C_n.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. => Well, what does just empirical and experiential mean anyway? Anytime> we talk about anything in the word we are applying a mental construct. So> I don't understand what your point is. If you're saying that Zeno's> paradoxes relies on the fact that certain models (or as you call them> mental constructs) are paradoxical when applied in certain ways, then I> agree and I never said otherwise. My basic point all along has been that> Zeno's paradoxes are not just a mathematical puzzle, but rather questions> about the more general metaphysical nature of the universe.Math deals with arguments and manipulations of a very limited naturein the larger scheme of things. With the standard math models ofreality I seen no real problem with Zeno's Paradox. But in my opinionZeno is arguing that the standard math models do not explain realityas we experience it, despite the usefulness of the standard Euclidean(Newtonian models). When one hears the standard Zeno argument of therunner and the hare, the naive mind rebels at the thought of endlesslyexamining smaller and smaller space and time intervals.As far as we can visualize, space and time can be dividedinde'nitely, butthe passage of time as we experience it can not be, dispite the commondiscriptions of time standing still and lives passing in front ofone's eyes. But why not if experience is based on time and space. Amodern explaination might be based on the common size and charge ofall electrons, but Zeno knew none of that. He was arguing on commonexperience.Now we do not know what Zeno's whole argument was, essentiallyeverything we know about them was from detractors (Plato? and maybesome others who thought they could answer them. But I think both heand democrates had very good arguments for the limitations of thestandard continous models.With other arguments, Zeno argued against descrete models.Any detractors of Zeno should ask themselves if they could 'nd betterarguments for the necessity of Quantum physics based upon Macroobservations.As for metaphysics, thats beyond my league, I just try to understandarguments. => I disagree with your perspective on limits. A major impetus for> developing the calculus was physical applications (for Newton there was no> difference between physics and calculus). And there is a de'nite> physical intuition behind the calculus. The concept of limit (as in> standard analysis) did not iron out the problems with using this physical> intuition; I am mostly speaking with regard to in'nitesimals.Any physical intuition behind in'nitesimals?> I don't see the limit as a very elegant solution. It does work, in some> sense, but I think Abraham Robinson's nonstandard analysis works a lot> better and is a better formalization of the intuition behind the calculus.I totally disagree. In'nitesimals are unintuitive and have nophysical interpretation. Limits are simple and make perfect physicaland mathematical sense. -- Robin Chapman, approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h88LR1s04701; =>>Martin>> Yes. I ment factorizing to primes.> Is there a number n such that n is the> product of two primes a and b, but also> the product of two different primes, d and e.> No, he (Martin) did not write that. That was written by geardown, the original poster.>Your question was posed perfectly. I love it and don't>have a clue, how to prove ab = de impossible for>primes a,b,d,e and {a,b} different from {d,e}.>>I *know* it's impossible. But I don't know how to prove>it. So there are three possibilities for me:>> (a) Thinking it over (nighttime again ...)> (b) Waiting for a good answer by e.g. Dave Rusin :-)> (c) Reading a book.>>I'll try (a) 'rst, starting with let a*b be the smallest...>and walking happily towards ... contradiction; q.e.d. or>something like that.>>Couldn't you have asked for Are there in'nitely many primes?>This is in my standard repertoire :-)>I surely have to add the other one!>>Rainer Rosenthal>r.rosenthal@web.de>> I believe that uniqueness of prime factorization goes back to Euclid, as does in'nitely any primes. =Timothy Murphy a dit :>> For SLn(C) and GLn(C), you have to consider their fundamental group>> : PI(SLn(C)) is isomorphic to the trivial group and PI(GLn(C)) is>> isomorphic to (Z, +).> Isn't it simpler just to observe that they are of different> dimensions? (Admittedly there is a little machinery involved in> de'ning dimension.) The theory of topological dimensions is a little harder that the>> little bit of homotopy here involved.> Is it?> How do you establish the fundamental group of GL(n,C)?> In any case, it is much easier to see informally> that the dimensions are different> than that the fundamental groups are different,> and the argument applies at once to most classical groups;> so I would suggest that if one is interested> in the question whether spaces are homeomorphic> it is certainly worth investing the time> to see that locally euclidean spaces of different dimensions> cannot be homeomorphic.> Homotopy is still interesting.You can easily see that a space with a hole cannot be homeomorphic to a space without a hole (ie. simply connected), and you can explain this to a 5-year-old child.Actually, the simpler to show is that PI(GLn(C)) is not the trivial group (ie. contains more than one element).-- Alexandre CharitopoulosEm6 / Eb7(5b) / Dm7 / Db7(5b, 9b) / Cmaj7 => dont know it.> Can anyone approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h832q6R28473; => dont know it.>> Can anyone help me ?>>Sometimes 1+1=3. They call it synergy.>See:http://www.rwgrayprojects.com/synergetics/s01/'gs /f0801.html Cliff Nelson =Simple problem that's making my hair fall off!The optimization in question is to reduce the number of multiplications inan expressioni.e.:x = a*b*c-a*d+b*c (5 multiplications)i = b*cx = a*(i-d)+i (optimized, 2 multiplications, same result)Following the same concept, I want to do the same thing with these twoexpressions:x = a*b*c-a*d+b*cy = a*b*c+a*c-b*d (8 multiplications total)So far, this is what I got:i = b*cx = a*(i-d) + iy = a*(i+c) - b*d (4 multiplications)My professor says that it can be done with just 3 multiplications.... Idon't see how. Any hints?Padu => .... > Following the same concept, I want to do the same thing with these two> expressions:> x = a*b*c-a*d+b*c> y = a*b*c+a*c-b*d (8 multiplications total)> So far, this is what I got:> i = b*c> x = a*(i-d) + i> y = a*(i+c) - b*d (4 multiplications)> My professor says that it can be done with just 3 multiplications.... I> don't see how. Any hints?... Here's a pretty broad hint. You've got x with twomultiplications. Now think about x + (y - x). Ken Pledger. => : Moreover, I am well aware of the> : adjustment dif'culties involved here.> : But the problem is something that you just do not> : understand>> You are LYING, mitch.I could be in error, George. Lying implies intention where there isnone.>> There is a VAST universe of things that I do not understand.> I majored in Philosophy as an undergrad but my goal was always> to be a computer programmer and all my further coursework has> been in Computer Science.I don't see the word ïmathematics' here.> The reason why the newsgroup is an> appropriate place for me to be is that I get a chance to learn> about topics that I did not have the luxury of studying at the> graduate level. I fully and freely concede the breadth of my> ignorance. But my ignorance IS NOT ANY sort of problem here,> because *I* am NOT the one claiming to know things that I don't> know! YOU ARE!I am sure you believe that, George.That is why I have taken the time to transcribe de'nitions from mysource texts and have been completely open about which texts I am using.If I am not using de'nitions correctly, then I am deserving ofcriticism. But, that is not what has been going on here.I *do* ask people to make an imaginative leap (or ten), and, that ishard. But, that is where I need veri'cation. I am not a Turingmachine asking if my new tape reader is cute enough to be part of thein crowd.>> If something that you just do not understand is ANYbody's> problem, it is YOURS, NOT mine! And the infuriating thing> about it is that while the things I don't understand are> HARD, the things YOU don't understand (like why it is idiotic> for you to be talking about your having problems with the> identity relation in a context such as 'rst-order ZFC) are EASY!>Formal identity between duplicate symbol strings is easy.Substitutivity in classical models is easy.A rigid notion of truth for models of set theory is not easy. Itdepends on constant bindings under interpretation mappings and variablebindings under satisfaction mappings. It is not trivial.In fact, it is not really possible by virtue of the underlying algebraicstructure. However, what is possible is a determination of a canonicalmodel to serve as basis for extensions. The constructible universesatis'es that role because it is a consistent global labeling ratherthan an extension obtained from collapsing a multiplicative set into anideal. The question boils down to understanding the concept of analgebra of labels.I had cause to go back to a document I prepared in 1994. I proposed thefollowing axiom in it:AxAy (x proper subset y -> x in (Goedel Closure(TransitiveClosure({y}))))I forgot about it. But, it gives you V=L and there are good reasons fordoing it.>> You have been asked a lot of simple direct questions that> you have completely refused to engage.That is inaccurate. There are issues involving different backgroundshere. Resolving those issues requires patience.> UNTIL you have engaged> them, NOTHING you have to say about more complicated things> that I or anyone else allegedly doesn't understand is going> to get parsed. You have well and truly earned that level> of dismissal.Perhaps two quotes from Whitehead are most appropriate here:Philosophy destroys its usefulness when it indulgesin brilliant feats of explaining away. It is then trespassingwith the wrong equipment upon the 'eld of particularsciences.The use of such a matrix is to argue from it boldly andwith rigid logic. The scheme should therefore be statedwith the utmost precision and de'niteness, to allow ofsuch argumentation. The conclusion of the argumentshould then be confronted with circumstances to whichit should apply.Between the two of us, George--who engages in brilliant feats ofexplaining away and who has actually presented a precise and de'nitescheme--albeit one that incorporates the circular de'nition ofpredicates (as opposed to *objects*).:-)mitch =[non-math deleted]> Oh yeah, other posters were getting away with teaching some wacky> crap, and I can show that with a simple expression:> Consider> P(x) = 11^2 + 11x + 2> where you have a special grouping simply to allow the non-polynomial> factorization, as you can see that P(x) is also> P(x) = 11x + 123> so I've just used the 11's as placemarkers, so I can *act* like it's> the polynomial y^2 + xy + 2, to get the factorization> P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)> but the polynomial is STILL P(x)= 11x + 123. These posters worked to> convince that the factorization, with more complicated expressions,> changed the polynomial in some way. Apparently plenty of sci.math> readers and especially alt.math.undergrad readers, who may be more> susceptible, were convinced by them.> This is perfectly ok. I've seen no objections to this. I've seen objections to how you *talk* about this. Next?[non-math deleted]-- Will Twentyman =[snip social commentary]> Oh yeah, other posters were getting away with teaching some wacky> crap, and I can show that with a simple expression:> Consider> P(x) = 11^2 + 11x + 2> where you have a special grouping simply to allow the non-polynomial> factorization, as you can see that P(x) is also> P(x) = 11x + 123> so I've just used the 11's as placemarkers, so I can *act* like it's> the polynomial y^2 + xy + 2, to get the factorization> P(x) = (11 + (x+sqrt(x^2-8))/2)(11 + (x-sqrt(x^2-8))/2)I've said in other threads there's a much less confusingway to say all this. You're not doing anything special here,and there's an easier road to get there which doesn'tsound like this wacky 11 is a variable thing.It goes like this: Let P(x) = 11 x + 123.Consider another function Q_x(y) = y^2 + xy + 2, where x appears as the linear coef'cient, a parameter.(Hence the notation Q_x). Clearly Q_x(11) = P(x) forall x.Q_x(y) can be factored by the quadratic theorem usinga = y^2, b = x, c = 2:Q_x(y) = (y - (-x+sqrt(x^2-8))/2)(y - (-x-sqrt(x^2-8))/2)and this holds for all complex x, y. Therefore, P(x) = Q_x(11) = (11 - (-x+sqrt(x^2-8))/2)(11 - (-x-sqrt(x^2-8))/2)> but the polynomial is STILL P(x)= 11x + 123.Yes. That's obvious. As usual, you've focused on the elementaryopening steps and ignored all the actual objections.> These posters worked to> convince that the factorization, with more complicated expressions,> changed the polynomial in some way.Nobody has done this. - Randy => Now I can 'nally not only refute posters arguing with me, which is> something that I've actually been doing for months, but I can 'nally> show you what bogus crap they've been teaching you, as I've found> something simple enough that it seems to get the message across.> You have succeeded in fooling yourself. You have also been evading an> Advanced Polynomial Factorization.Those are interesting charges. Let's see what this poster managed toactually come up with *mathematically*. > In the 'nal section of that paper, you conclude the follwing:> Therefore, with the factorization> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)> one of the a's is coprime to 5, which shows where some of the> algebraic integer factors distribute despite the factors> being irrational.> This being the conclusion of the section entitled Primary Argument, a> person might come to the opinion that its success was the apex of your> achievement, a real indicator of the power of your method.> I have shown the conclusion to be false.A mathematical proof is irrefutable. > Just in case you have forgotten, allow me to refresh your memory.> you respond to my request, after I had *repeatedly* pointed out that> the a's *all* have common factors with 5 in the ring of algebraic> integers: (> is me):> ... stuff deleted ...>> What? Are you saying that the factorization I've posted and> provided all the necessary calculation to verify that, is> somehow inadequate?>> Please explain. Show me the error.> That is, show how one of these is not correct:>> 1. 5 = q(a)*r(a)> 2. a = r(a)*s(a)> 3. r(a) is an algebraic integer> 4. r(a) is NOT a unit.> You have not given proof that r(a) cannot be coprime to 5.> ... stuff deleted ...> However, you have not proven that r cannot be coprime to 5, as> that can't be proven, as in fact, for one of the three a's, r IS> coprime to 5.> For the other two a's, r does not share non-unit factors in> common with 5 ***in the ring of algebraic integers***.> However, recently, you have seen at least a slight amount of the truth,> as you recently posted factor of z.>> Can you agree or disagree with that statement ...>> Well, if I have abc=p, than a is a factor of p.>> Can you agree or disgree with that statement James Harris. > (I'd use triple quotes if I could 'nd them!)>> I agree and I'm glad you brought that up as I've been making a> rather simple mistake with my example abc=p.> I've said that neither ïa', ïb', nor ïc' share non-unit factors> with p, when in fact they share *themselves* so I was wrong.> They also share factors of themselves as well, so in fact, they> each share an in'nity of non-unit factors with p, since they> can be simply decomposed, like sqrt(a), so it's even worse.> I've been thinking about a certain decomposition and screwed up> explaining it.> James HarrisYup. Notice readers how posters try to take *anything* and run withit, as if the mathematics changes.If I make statements that are wrong mathematically, they are justwrong.> In short, you now must be admitting the following. Let a be any of the> coef'cients to which you refer in the above-cited statement from your> paper:> 1. Since 5 = q(a) r(a), r(a) shares a non-unit factor with 5,> namely r(a)> 2. Since a = r(a) s(a), r(a) shares a non-unit factor with a,> namely r(a).> 3. Since r(a) is at once a shared non-unit factor between r(a)> and 5, and also a shared non-unit factor between r(a) and a,> r(a) is a shared non-unit factor between a and 5.> 4. a and 5 cannot be coprime in the ring of algebraic integers.> Your continued assertions that the paper Advanced Polynomial> Factorization is correct, and the *speci'c* assertion that its method> is correct, are untrue. You must accept that the method is §awed.Nope.Actually it is possible that ïa' is coprime to 5, which means thatboth r(a) and s(a) are coprime, which leaves q(a) NOT coprime to 5,but strangely enough *in the ring of algebraic integers* it does nothave 5 as a factor, though it does in a more inclusive ring.So in fact your construction cannot disprove my paper, which is howmathematics works, as you can't disprove a proof. If you are to maintain the claim that you are correct, you must do one> of the following two things:> 1. Refute at least one of the statements 1-4 above> or> 2. Withdraw your paper *and* any claims of its correctness.Mathematics isn't a debate, and math proofs don't duel.Now I've refuted your objection so what should you do?> What's sad to me is that I'm seeing *increased* hostile postings as> some of you seem intent on telling yourselves and the newsgroup what> you want to hear as if math isn't enough for you.> any hostile statement in it, so your view must be based on somethingWell it'd be a positive sign if you've decided to actually follow themath.> Remember, math is about truth. All of the math that any of you learn> had to be discovered by someone, and if it were just politics or what> made people feel good, what would anyone know today?> You may wish to heed your own advice. If math is about truth (and> actually, it's about logical consistency), then the truth (or what I> would suggest, logical correctness) of my argument should tell you what> you must do.Your conclusion does not follow from the mathematical argument you'vegiven.It turns out that you've neglected the possibility that ïa' is coprimeto 5, and in fact, cannot prove that it is not, as it IS possible thatit is coprime to 5 (readers should note that there are three possiblevalues for ïa').So your *assumption* that ïa' is not coprime to 5 is false, and is alogical §aw, which takes away the conclusion you desire. > My task was 'nding out what followed from mathematical logic.> Yet you have assiduously avoided addressing a key point: your method> is §awed to the extent that it enables you to deduce false statements.> Please explain.Actually, that's what you're doing. Math proofs don't duel.If anyone could 'nd an error in my paper there wouldn't be any ofthis side commentary with supposed alternate proofs.What looking for alternate proofs allows you to do is 'nd room tofool yourself, like where you have ignored the possibility that ïa' iscoprime to 5.> Many of you apparently thought that you could use me to feel better> about psychological scars in§icted upon you as children by bullies> that unfortunately you learned to want to *be*, which is classic in> the psychology.> Sorry, not mathematics. You are imposing your own preferences on the way> mathematics operates, and misinterpreting your own mistakes as being> evidence that others are mistaken. Put the pieces together correctly,> and you'll learn something.I am noting the horri'c behavior where many posters have gone out oftheir way to make hateful statements, and the newsgroup has not budgedor has actually cheered them on.That kind of behavior is telling.> That is, while you hated the bullies, and wished that society would> protect you, rather than learn that their behavior was wrong, you> learned to hate yourselves and wish for the power. Seeing me as weak> and set upon by the crowd, many of you became the bullies you'd always> wanted to be.> Can you name *one* person who has threatened your livelihood, threatened> your life, or informed the FBI about your misadventures?> Just one?Relevance?> But actually I was searching for the mathematics, as I *believe* in> mathematics, and know that it doesn't matter what you believe, or how> strong you think your crowd is, as mathematical truth is independent> of such things.> Mathematics depends on proof, and your method of proof is §awed. Thus,> your search for the mathematics has been futile. If you mean to be> searching for the mathematics, then you'll correct your methods.Well, your argument is §awed mathematically, as you made anassumption that ïa' can't be coprime to 5.Based on your false assumption you are quite arrogant now and thinkyou have cause to criticize me, when in fact, you are wrong.> So as one man, I can stand against the entire math society, if> necessary.> Yeah, right. Stand all you like. Insult all you like. Rant and rave> and plead and whine, just like you have done for going on 8 years, and> claim it's just the corrupt mathematical establishment. Until you> actually *change* what it is you're doing, you will get nowhere.What I can show is that mathematicians try to run math society like ademocracy, where what people *want* to believe is allowed to be calledtruth against actual mathematical truth.But then none of you are actually mathematicians, now are you?Math is not a democracy.> Many of you were broken all those years ago by bullies who pushed> themselves into you, and became a part of how you react.> I think I detect a bit of projection here. And there.Well there usually is *some* reason for lashing out, and having seenso many of you lash out, like you W. Dale Hall that it seems logicalto infer a reason, unless you're just innately evil, and like tryingto hurt others.> I suggest you go to true power and switch to mathematics. It can be> enough for you.> Perhaps you have too much of a romantic view of how things work. Your> idealized notions don't really help, not as long as you insist you're> correct and ignore the real demonstrations that you're not.> ... irrelevant stuff deleted ...Like your §awed attempt where you falsely assume that ïa' can't becoprime to 5?Again I remind you that proofs don't duel.You can't 'nd a correct argument to disprove a proof.> It's actually kind of sad, but then again, many of you were bullies.> I'll ask again: can you name ONE person who has threatened either your> livelihood or your life? Can you recall that you have done both?I've noted that society pays mathematicians to do a job, andreasonably mathematicians can expect to be held accountable if theydidn't do their job.Also I noted that mathematicians may be waiting and hoping that likepast math discoverers, like Galois and Abel, that I might dietragically before they have to acknowledge my work, and said thatinstead *they* should die.> Has anyone informed the FBI about your activities? Many people (well,> not really, but I'll imagine for a moment) could 'nd such an act to> be an implied threat.Mathematicians are a signi'cant part of the national securityapparatus of the United States and it is VERY important to 'nd outthat they are willing to lie.Mathematicians who can lie now about my work may actually be workingor 'nd cause to work for foreign governments against the interests ofthe United States.It is the job of the FBI to protect the citizens of the United Statesfrom enemies both foreign and domestic.It is my duty as a citizen to bring the FBI's attention to lyingmathematicians.> So 'nding out you were actually believing in some bogus b.s. is kind> of just desserts, so those bullies who beat themselves into you as> children, have beaten you yet again, which is also very sad. Just *who* is the bully here? The person who provides proof of your> errors, or the one who threatens lawsuits, congressional action, murder> at the hands of the US Army, or investigation by the FBI?In the United States under the Constitution you are innocent until*proven* guilty.Agencies of the United States should not be feared if you are notguilty of a crime, and you should be con'dent that you would beexonerated if an investigation took place.James Harris =Now I can 'nally not only refute posters arguing with me, which is>something that I've actually been doing for months, but I can 'nally>show you what bogus crap they've been teaching you, as I've found>something simple enough that it seems to get the message across.>You have succeeded in fooling yourself. You have also been evading an>>Advanced Polynomial Factorization.> Those are interesting charges. Let's see what this poster managed to> actually come up with *mathematically*.>>In the 'nal section of that paper, you conclude the follwing:>> Therefore, with the factorization>> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>> one of the a's is coprime to 5, which shows where some of the>> algebraic integer factors distribute despite the factors>> being irrational.>>This being the conclusion of the section entitled Primary Argument, a>>person might come to the opinion that its success was the apex of your>>achievement, a real indicator of the power of your method.>>I have shown the conclusion to be false.> A mathematical proof is irrefutable.>>Just in case you have forgotten, allow me to refresh your memory.>> you respond to my request, after I had *repeatedly* pointed out that>> the a's *all* have common factors with 5 in the ring of algebraic>> integers: (> is me):>> ... stuff deleted ...> What? Are you saying that the factorization I've posted and>> provided all the necessary calculation to verify that, is>> somehow inadequate?>> Please explain. Show me the error.>> That is, show how one of these is not correct:>> 1. 5 = q(a)*r(a)>> 2. a = r(a)*s(a)>> 3. r(a) is an algebraic integer>> 4. r(a) is NOT a unit.>> You have not given proof that r(a) cannot be coprime to 5.>> ... stuff deleted ...>> However, you have not proven that r cannot be coprime to 5, as>> that can't be proven, as in fact, for one of the three a's, r IS>> coprime to 5.>> For the other two a's, r does not share non-unit factors in>> common with 5 ***in the ring of algebraic integers***.>>However, recently, you have seen at least a slight amount of the > If I have z=xy, then x is a factor of z.>> Can you agree or disagree with that statement ...>> Well, if I have abc=p, than a is a factor of p.>> Can you agree or disgree with that statement James Harris. >> (I'd use triple quotes if I could 'nd them!)>> I agree and I'm glad you brought that up as I've been making a>> rather simple mistake with my example abc=p.>> I've said that neither ïa', ïb', nor ïc' share non-unit factors>> with p, when in fact they share *themselves* so I was wrong.>> They also share factors of themselves as well, so in fact, they>> each share an in'nity of non-unit factors with p, since they>> can be simply decomposed, like sqrt(a), so it's even worse.>> I've been thinking about a certain decomposition and screwed up>> explaining it.>> James Harris> Yup. Notice readers how posters try to take *anything* and run with> it, as if the mathematics changes.> If I make statements that are wrong mathematically, they are just> wrong.>>In short, you now must be admitting the following. Let a be any of the>>coef'cients to which you refer in the above-cited statement from your>>paper:>> 1. Since 5 = q(a) r(a), r(a) shares a non-unit factor with 5,>> namely r(a)>> 2. Since a = r(a) s(a), r(a) shares a non-unit factor with a,>> namely r(a).>> 3. Since r(a) is at once a shared non-unit factor between r(a)>> and 5, and also a shared non-unit factor between r(a) and a,>> r(a) is a shared non-unit factor between a and 5.>> 4. a and 5 cannot be coprime in the ring of algebraic integers.>>Your continued assertions that the paper Advanced Polynomial>>Factorization is correct, and the *speci'c* assertion that its method>>is correct, are untrue. You must accept that the method is §awed.> Nope.> Actually it is possible that ïa' is coprime to 5, which means that> both r(a) and s(a) are coprime, which leaves q(a) NOT coprime to 5,> but strangely enough *in the ring of algebraic integers* it does not> have 5 as a factor, though it does in a more inclusive ring.> So in fact your construction cannot disprove my paper, which is how> mathematics works, as you can't disprove a proof.You missed the point. We don't care if r(a) and s(a) are coprime. What matters is r(a) * s(a) = a.q(a) is not coprime to 5 because q(a) * r(a) = 5.Please read the above again and realize 1) r(a) is a factor of a, 2) r(a) is a factor of 5.Note: there were no assumptions made as to whether 5 and a are coprime. It is the conclusion of the argument, not a premise.[rest deleted]-- Will Twentyman =Now I can 'nally not only refute posters arguing with me, which is>something that I've actually been doing for months, but I can 'nally>show you what bogus crap they've been teaching you, as I've found>something simple enough that it seems to get the message across.>You have succeeded in fooling yourself. You have also been evading an>>Advanced Polynomial Factorization.> Those are interesting charges. Let's see what this poster managed to> actually come up with *mathematically*.>>In the 'nal section of that paper, you conclude the follwing:>> Therefore, with the factorization>> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>> one of the a's is coprime to 5, which shows where some of the>> algebraic integer factors distribute despite the factors>> being irrational.>>This being the conclusion of the section entitled Primary Argument, a>>person might come to the opinion that its success was the apex of your>>achievement, a real indicator of the power of your method.>>I have shown the conclusion to be false.> A mathematical proof is irrefutable.>>Just in case you have forgotten, allow me to refresh your memory.>> you respond to my request, after I had *repeatedly* pointed out that>> the a's *all* have common factors with 5 in the ring of algebraic>> integers: (> is me):>> ... stuff deleted ...> What? Are you saying that the factorization I've posted and>> provided all the necessary calculation to verify that, is>> somehow inadequate?>> Please explain. Show me the error.>> That is, show how one of these is not correct:>> 1. 5 = q(a)*r(a)>> 2. a = r(a)*s(a)>> 3. r(a) is an algebraic integer>> 4. r(a) is NOT a unit.>> You have not given proof that r(a) cannot be coprime to 5.>> ... stuff deleted ...>> However, you have not proven that r cannot be coprime to 5, as>> that can't be proven, as in fact, for one of the three a's, r IS>> coprime to 5.>> For the other two a's, r does not share non-unit factors in>> common with 5 ***in the ring of algebraic integers***.>>However, recently, you have seen at least a slight amount of the > If I have z=xy, then x is a factor of z.>> Can you agree or disagree with that statement ...>> Well, if I have abc=p, than a is a factor of p.>> Can you agree or disgree with that statement James Harris. >> (I'd use triple quotes if I could 'nd them!)>> I agree and I'm glad you brought that up as I've been making a>> rather simple mistake with my example abc=p.>> I've said that neither ïa', ïb', nor ïc' share non-unit factors>> with p, when in fact they share *themselves* so I was wrong.>> They also share factors of themselves as well, so in fact, they>> each share an in'nity of non-unit factors with p, since they>> can be simply decomposed, like sqrt(a), so it's even worse.>> I've been thinking about a certain decomposition and screwed up>> explaining it.>> James Harris> Yup. Notice readers how posters try to take *anything* and run with> it, as if the mathematics changes.> If I make statements that are wrong mathematically, they are just> wrong.>>In short, you now must be admitting the following. Let a be any of the>>coef'cients to which you refer in the above-cited statement from your>>paper:>> 1. Since 5 = q(a) r(a), r(a) shares a non-unit factor with 5,>> namely r(a)>> 2. Since a = r(a) s(a), r(a) shares a non-unit factor with a,>> namely r(a).>> 3. Since r(a) is at once a shared non-unit factor between r(a)>> and 5, and also a shared non-unit factor between r(a) and a,>> r(a) is a shared non-unit factor between a and 5.>> 4. a and 5 cannot be coprime in the ring of algebraic integers.>>Your continued assertions that the paper Advanced Polynomial>>Factorization is correct, and the *speci'c* assertion that its method>>is correct, are untrue. You must accept that the method is §awed.> Nope.> Actually it is possible that ïa' is coprime to 5, which means that> both r(a) and s(a) are coprime, which leaves q(a) NOT coprime to 5,> but strangely enough *in the ring of algebraic integers* it does not> have 5 as a factor, though it does in a more inclusive ring.> It doesn't matter whether r(a) and s(a) are coprime, nor does itmatter whether q(a) is coprime to 5. What does matter is that r(a)is a common non-unit factor of both a and 5. You do not deny this,and yet you persist.So, you are now claiming that, despite the fact that the number r(a)is a non-unit divisor of *both* a and 5, somehow there is no commonnon-unit divisor of both a and 5?It's astonishing the lengths you'll go to to prop up that non-proof.Perhaps you could de'ne coprime? > So in fact your construction cannot disprove my paper, which is how> mathematics works, as you can't disprove a proof.> Saying it's a proof, even believing it's a proof, doesn't make it so.Your argument is invalid, and that can be seen *because* it producesan incorrect conclusion.Your only counterargument is this: my argument is correct because it's a proof.That is not substantially different from this argument: your argument is not a proof because it is not correct.If you accept the form of the 'rst argument, then you must acceptthe form of the second.Further, the debate is *precisely* whether you have a proof. It is nota valid argument to assume the conclusion.>>If you are to maintain the claim that you are correct, you must do one>>of the following two things:>> 1. Refute at least one of the statements 1-4 above>>or>> 2. Withdraw your paper *and* any claims of its correctness.> Mathematics isn't a debate, and math proofs don't duel.> Now I've refuted your objection so what should you do?> Your claim of refutation is not accepted. It's surprising thatyou would imagine anyone could accept it.I should point out that you have decided that the de'nition ofcoprime isn't important at all. Your continued claim that theexistence of a non-unit common factor doesn't yield a non-unitcommon factor is laughable. Please continue.What's sad to me is that I'm seeing *increased* hostile postings as>some of you seem intent on telling yourselves and the newsgroup what>you want to hear as if math isn't enough for you.>any hostile statement in it, so your view must be based on something> Well it'd be a positive sign if you've decided to actually follow the> math.> You can say what you'd like, but to make the claim that I haven'tfollowed the math is patently absurd.>Remember, math is about truth. All of the math that any of you learn>had to be discovered by someone, and if it were just politics or what>made people feel good, what would anyone know today?>You may wish to heed your own advice. If math is about truth (and>>actually, it's about logical consistency), then the truth (or what I>>would suggest, logical correctness) of my argument should tell you what>>you must do.> Your conclusion does not follow from the mathematical argument you've> given.Really?Let's be precise here. Your de'nition of coprime is this: Two members a and b of the ring R are coprime iff whenever z is a common factor of a and b (that is: a = xz and b = yz for x,y,z in R), then z is a unit of R.If you have a disagreement with this phrasing of your de'nition, thenplease say so, and then provide the de'nition you would use.Given *that* de'nition, I have shown this: 5 = qr, a = rs,with q,r,s in the ring of algebraic integers.Note the form, which I'll place in juxtaposition to that of thede'nition: DEFINITION MY RESULT Hypothesis: Hypothesis: a,b coprime a,5 coprime R a commutative ring A the ring of algebraic integers a,b in R a, 5 in A x,y,z in R q,r,s in A a = xz a = sr b = yz 5 = qr ==> z a unit ==> r a unit.Note that the left column follows the de'nition I gave (if you have ade'nition that is at variance with this, then please give it). Theright column follows that pattern with the speci'cs of this case.Note, also, that the conclusion that the *de'nition* requires doesNOT hold in the right column. I have shown that the number r is not aunit.Thus, by the de'nition, a and 5 cannot be coprime in the ring ofall algebraic integers.Yet you claim they are.> It turns out that you've neglected the possibility that ïa' is coprime> to 5, and in fact, cannot prove that it is not, as it IS possible that> it is coprime to 5 (readers should note that there are three possible> values for ïa').> I have three possible values for a. For each a (note that I have writtenhave given the polynomials that yield these values).> So your *assumption* that ïa' is not coprime to 5 is false, and is a> logical §aw, which takes away the conclusion you desire.> Your ignorance is simply too glaring for this to be an accident. I haveNEVER assumed a and 5 are not coprime. It is the conclusion of these factorizations: 5 = qr, a = rsin the ring of algebraic integers, with r not a unit. Pray tell, whatlogic tells you that I have *assumed* a and 5 not to be coprime?I'm certain that you haven't fount it in any argument I've written.>My task was 'nding out what followed from mathematical logic.>Yet you have assiduously avoided addressing a key point: your method>>is §awed to the extent that it enables you to deduce false statements.>>Please explain.> Actually, that's what you're doing. Math proofs don't duel.> That statement has no meaning. Of course math proofs don't duel. Theyhave no hands to hold the foil or epee (whichever style of fencing youprefer). If your meaning is that one cannot use a mathematical argumentto refute another mathematical argument, then you are simply incorrect,and again show how deep your ignorance runs.> If anyone could 'nd an error in my paper there wouldn't be any of> this side commentary with supposed alternate proofs.> I showed you the error. You have decided to misinterpret everything,including your own de'nition of coprime, the nature of hypothesisand conclusion, and> What looking for alternate proofs allows you to do is 'nd room to> fool yourself, like where you have ignored the possibility that ïa' is> coprime to 5.> Duh. You read the conclusion of an argument, and read it as anassumption. Real smart, Mister Former Gifted Child. Please go backand read *what was written* and return when you have properlyread that argument.>Many of you apparently thought that you could use me to feel better>about psychological scars in§icted upon you as children by bullies>that unfortunately you learned to want to *be*, which is classic in>the psychology.>Sorry, not mathematics. You are imposing your own preferences on the way>>mathematics operates, and misinterpreting your own mistakes as being>>evidence that others are mistaken. Put the pieces together correctly,>>and you'll learn something.> I am noting the horri'c behavior where many posters have gone out of> their way to make hateful statements, and the newsgroup has not budged> or has actually cheered them on.> Perhaps you should see the other side of the coin: you are a person whoabuses the notion of debate, who deliberately misreads what people say,who refuses to crack open a book that might be somewhat enlightening.Further, you take every correction as attack, worthy of retaliation.> That kind of behavior is telling.> And yours isn't? I think the readers of these threads pretty much haveyou pegged.>That is, while you hated the bullies, and wished that society would>protect you, rather than learn that their behavior was wrong, you>learned to hate yourselves and wish for the power. Seeing me as weak>and set upon by the crowd, many of you became the bullies you'd always>wanted to be.>Can you name *one* person who has threatened your livelihood, threatened>>your life, or informed the FBI about your misadventures?>>Just one?> Relevance?> No, I didn't think you could.As far as your question, note that you are making the assertion thatmany of your critics are bullies. I am simply asking you to do betterthan to make some wild claim: substantiate it with particulars.So, I asked you the details of what you might have suggested what othersdid. However, I do not agree that sci.math has bullied you, because Ihave witnessed your unprovoked aggression and your general abuse of thenewsgroup. Bullying occurs in the absence of any provocation, and thenegative writing that has been directed at you has been almost entirelyin response to your provocation.The items in that above list were chosen from things I've seen YOU do.>But actually I was searching for the mathematics, as I *believe* in>mathematics, and know that it doesn't matter what you believe, or how>strong you think your crowd is, as mathematical truth is independent>of such things.>Mathematics depends on proof, and your method of proof is §awed. Thus,>>your search for the mathematics has been futile. If you mean to be>>searching for the mathematics, then you'll correct your methods.> Well, your argument is §awed mathematically, as you made an> assumption that ïa' can't be coprime to 5.> Based on your false assumption you are quite arrogant now and think> you have cause to criticize me, when in fact, you are wrong.So as one man, I can stand against the entire math society, if>necessary.>Yeah, right. Stand all you like. Insult all you like. Rant and rave>>and plead and whine, just like you have done for going on 8 years, and>>claim it's just the corrupt mathematical establishment. Until you>>actually *change* what it is you're doing, you will get nowhere.> What I can show is that mathematicians try to run math society like a> democracy, where what people *want* to believe is allowed to be called> truth against actual mathematical truth.> But then none of you are actually mathematicians, now are you?> There are ample pieces of evidence on line to verify whether many of usare mathematicians.> Math is not a democracy.> So? Are you claiming to be the King of Mathematics?>Many of you were broken all those years ago by bullies who pushed>themselves into you, and became a part of how you react.>I think I detect a bit of projection here. And there.> Well there usually is *some* reason for lashing out, and having seen> so many of you lash out, like you W. Dale Hall that it seems logical> to infer a reason, unless you're just innately evil, and like trying> to hurt others.> Really? Lashing out? I am simply treating you as the punk you are.Behave civilly, and you will get civil treatment out of me. I cannotspeak for others, but I trust that (given time for the residual echoesof the discord you've sown to die down) eventually, you will be treatedas a normal person. You apparently don't see any reason to behave in acivil fashion, preferring instead to bask in the glow of the nastinessthat you generate.>I suggest you go to true power and switch to mathematics. It can be>enough for you.>Perhaps you have too much of a romantic view of how things work. Your>>idealized notions don't really help, not as long as you insist you're>>correct and ignore the real demonstrations that you're not.>> ... irrelevant stuff deleted ...> Like your §awed attempt where you falsely assume that ïa' can't be> coprime to 5?> Show me the assumption in what I've written. Convince me that I'veassumed rather than concluded, and I'll retract what I've written.I have *not* assumed that a can't be coprime to 5. I have known itfrom other information, but I never used that fact. Instead, Ifound the factorizations and *concluded* that a cannot be coprimeto 5.> Again I remind you that proofs don't duel.> You can't 'nd a correct argument to disprove a proof.> The problem is that you don't have a proof.Haven't I been saying that all along?>It's actually kind of sad, but then again, many of you were bullies.>I'll ask again: can you name ONE person who has threatened either your>>livelihood or your life? Can you recall that you have done both?> I've noted that society pays mathematicians to do a job, and> reasonably mathematicians can expect to be held accountable if they> didn't do their job.> *DOES* society pay mathematicians? I have actually looked at mypaychecks, and have never seen the society written anywhere on them.*Society* is an aggregate, not a de'nite entity. *Society* pays no one,hires no one, 'res no one. My boss will let me go if I don't do my job,but won't yours?I note that no one has written in threatening terms regarding yourlivelihood. You rationalize to cover the fact that you *have* writtenin such terms regarding mathematicians.> Also I noted that mathematicians may be waiting and hoping that like> past math discoverers, like Galois and Abel, that I might die> tragically before they have to acknowledge my work, and said that> instead *they* should die.> I don't care about your life. I trust that you'll live like a normalperson. It's you who has the tragic/heroic 'gure complex, not us.However, your remark that others should die is misplaced.>>Has anyone informed the FBI about your activities? Many people (well,>>not really, but I'll imagine for a moment) could 'nd such an act to>>be an implied threat. Mathematicians are a signi'cant part of the national security> apparatus of the United States and it is VERY important to 'nd out> that they are willing to lie.> No, the national security apparatus is not the amorphous everyone whohas an important job to do melange that you are imagining. Rather, itcomprises the various security and intelligence agencies of the USFederal Government. However, there is a side issue: it is your claimalone that supports the assertion [we] are willing to lie.What happens when you go to the Feds about Death stalking the earth?What sort of reaction do you get when you tell them that Nature hasgiven you a special role, or that you have been testing mankind, andthat mankind has failed? Those are remarks that are equal in importance,don't you think?> Mathematicians who can lie now about my work may actually be working> or 'nd cause to work for foreign governments against the interests of> the United States.> Keep making those claims. The more detailed, the more speci'c, and themore grandiose, the better. Be sure to keep detailed notes and forwardthem to the FBI.> It is the job of the FBI to protect the citizens of the United States> from enemies both foreign and domestic.> It is my duty as a citizen to bring the FBI's attention to lying> mathematicians.> Go right ahead. However, you cannot claim to be the innocent recipientof nasty behavior. You have a full role as a participant (instigatorwould be a more correct term) in this mud'ght.So 'nding out you were actually believing in some bogus b.s. is kind>of just desserts, so those bullies who beat themselves into you as>children, have beaten you yet again, which is also very sad.>Just *who* is the bully here? The person who provides proof of your>>errors, or the one who threatens lawsuits, congressional action, murder>>at the hands of the US Army, or investigation by the FBI?> In the United States under the Constitution you are innocent until> *proven* guilty.> Agencies of the United States should not be feared if you are not> guilty of a crime, and you should be con'dent that you would be> exonerated if an investigation took place.> Oh, golly, did you think anyone was *actually* threatened? All I'msaying is that you have to assume that your actions are taken seriously(even if there is no chance on earth that such is the case), wheneveryou make a verbal attack, call the cops, threaten Congressional action,inform the FBI, whatever.Even if you're not taken seriously, your acts are seen as indicative ofyour intentions. Hostile intentions beget hostile reactions (I'm sureyour mother told you about these things).> James HarrisDaleScore: JSH rationalization of own goo'ness: +1 JSH refutation of wdh argument: -10 =>Now I can 'nally not only refute posters arguing with me, which issomething that I've actually been doing for months, but I can 'nally>show you what bogus crap they've been teaching you, as I've found>something simple enough that it seems to get the message across.You have succeeded in fooling yourself. You have also been evading an>>Advanced Polynomial Factorization.> Those are interesting charges. Let's see what this poster managed to> actually come up with *mathematically*.>>In the 'nal section of that paper, you conclude the follwing:>> Therefore, with the factorization>> 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>> one of the a's is coprime to 5, which shows where some of the>> algebraic integer factors distribute despite the factors>> being irrational.>This being the conclusion of the section entitled Primary Argument, a>>person might come to the opinion that its success was the apex of your>>achievement, a real indicator of the power of your method.>>I have shown the conclusion to be false.> A mathematical proof is irrefutable.>>Just in case you have forgotten, allow me to refresh your memory.>> you respond to my request, after I had *repeatedly* pointed out that>> the a's *all* have common factors with 5 in the ring of algebraic> integers: (> is me):>> ... stuff deleted ... What? Are you saying that the factorization I've posted and> provided all the necessary calculation to verify that, is>> somehow inadequate?> Please explain. Show me the error.> That is, show how one of these is not correct:> 1. 5 = q(a)*r(a)>> 2. a = r(a)*s(a)>> 3. r(a) is an algebraic integer>> 4. r(a) is NOT a unit.>> You have not given proof that r(a) cannot be coprime to 5.>> ... stuff deleted ...>> However, you have not proven that r cannot be coprime to 5, as>> that can't be proven, as in fact, for one of the three a's, r IS>> coprime to 5.>> For the other two a's, r does not share non-unit factors in>> common with 5 ***in the ring of algebraic integers***.>>However, recently, you have seen at least a slight amount of the truth,>>as you If I have z=xy, then x is a factor of z.> Can you agree or disagree with that statement ...> Well, if I have abc=p, than a is a factor of p. Can you agree or disgree with that statement James Harris. >> (I'd use triple quotes if I could 'nd them!) >> I agree and I'm glad you brought that up as I've been making a>> rather simple mistake with my example abc=p.> I've said that neither ïa', ïb', nor ïc' share non-unit factors>> with p, when in fact they share *themselves* so I was wrong.>> They also share factors of themselves as well, so in fact, they>> each share an in'nity of non-unit factors with p, since they>> can be simply decomposed, like sqrt(a), so it's even worse.>> I've been thinking about a certain decomposition and screwed up>> explaining it.>> James Harris> Yup. Notice readers how posters try to take *anything* and run with> it, as if the mathematics changes.> If I make statements that are wrong mathematically, they are just> wrong.>>In short, you now must be admitting the following. Let a be any of the>>coef'cients to which you refer in the above-cited statement from your>>paper:>> 1. Since 5 = q(a) r(a), r(a) shares a non-unit factor with 5,>> namely r(a)>> 2. Since a = r(a) s(a), r(a) shares a non-unit factor with a,> namely r(a).>> 3. Since r(a) is at once a shared non-unit factor between r(a)>> and 5, and also a shared non-unit factor between r(a) and a,>> r(a) is a shared non-unit factor between a and 5.>> 4. a and 5 cannot be coprime in the ring of algebraic integers.>>Your continued assertions that the paper Advanced Polynomial>Factorization is correct, and the *speci'c* assertion that its method>>is correct, are untrue. You must accept that the method is §awed.> Nope.> Actually it is possible that ïa' is coprime to 5, which means that> both r(a) and s(a) are coprime, which leaves q(a) NOT coprime to 5,> but strangely enough *in the ring of algebraic integers* it does not> have 5 as a factor, though it does in a more inclusive ring.> It doesn't matter whether r(a) and s(a) are coprime, nor does it> matter whether q(a) is coprime to 5. What does matter is that r(a)> is a common non-unit factor of both a and 5. You do not deny this,> and yet you persist.What is said is that it is possible that ïa' is coprime to 5, as arer(a) and s(a), which leaves only q(a) NOT coprime to 5, but oddlyenough it doesn't have a non-unit factor in common with 5 *in the ringof algebraic integers*, which leaves you with nothing. > So, you are now claiming that, despite the fact that the number r(a)> is a non-unit divisor of *both* a and 5, somehow there is no common> non-unit divisor of both a and 5?The ring of algebraic integers is §awed, as I've proven.You are trying to stand on a broken leg.> It's astonishing the lengths you'll go to to prop up that non-proof.> Perhaps you could de'ne coprime?I am using the de'nition that there exists f and g, in the ring ofalgebraic integers, such that af+5g=1.You cannot refute that possibility, so should withdraw your claims andconcede to the mathematical truth.> So in fact your construction cannot disprove my paper, which is how> mathematics works, as you can't disprove a proof.> Saying it's a proof, even believing it's a proof, doesn't make it so.> Your argument is invalid, and that can be seen *because* it produces> an incorrect conclusion.That is a false statement. Now I've shown you the §aw in yourreasoning, you can follow the math, or show yourself to not berational.> Your only counterargument is this:> my argument is correct because it's a proof.> That is not substantially different from this argument:> your argument is not a proof because it is not correct.> If you accept the form of the 'rst argument, then you must accept> the form of the second.> Further, the debate is *precisely* whether you have a proof. It is not> a valid argument to assume the conclusion.That is irrelevant to the fact that I've shot down your claims byshowing you have an assumption you *cannot* prove, so quit trying tochange the subject.Your claims have been dashed. You should follow the math. >>If you are to maintain the claim that you are correct, you must do one>>of the following two things:> 1. Refute at least one of the statements 1-4 above>>or> 2. Withdraw your paper *and* any claims of its correctness.> Mathematics isn't a debate, and math proofs don't duel.> Now I've refuted your objection so what should you do?> Your claim of refutation is not accepted. It's surprising that> you would imagine anyone could accept it.> I should point out that you have decided that the de'nition of> coprime isn't important at all. Your continued claim that the> existence of a non-unit common factor doesn't yield a non-unit> common factor is laughable. Please continue.I didn't mention anything about units, as I noted that you aredepending on the assumption that your ïa' cannot be coprime to 5.I have made a *speci'c* objection, so stop squirming. >What's sad to me is that I'm seeing *increased* hostile postings as>some of you seem intent on telling yourselves and the newsgroup what>you want to hear as if math isn't enough for you.any hostile statement in it, so your view must be based on something Well it'd be a positive sign if you've decided to actually follow the> math.> You can say what you'd like, but to make the claim that I haven't> followed the math is patently absurd.I have made a speci'c objection noting that your claims depend on theassumption that your ïa' can't be coprime to 5, but you cannot provethat assumption.Stop whining.>Remember, math is about truth. All of the math that any of you learn>had to be discovered by someone, and if it were just politics or whatmade people feel good, what would anyone know today?>>You may wish to heed your own advice. If math is about truth (and>>actually, it's about logical consistency), then the truth (or what I>>would suggest, logical correctness) of my argument should tell you what>>you must do.> Your conclusion does not follow from the mathematical argument you've> given.> Really?> Let's be precise here. Your de'nition of coprime is this:> Two members a and b of the ring R are coprime iff whenever> z is a common factor of a and b (that is: a = xz and b = yz> for x,y,z in R), then z is a unit of R.> If you have a disagreement with this phrasing of your de'nition, then> please say so, and then provide the de'nition you would use.I am using the de'nition that there exists f and g, in the ring ofalgebraic integers, such that af+5g=1.> Given *that* de'nition, I have shown this:> 5 = qr, a = rs,> with q,r,s in the ring of algebraic integers.And again, it is possible that ïa' is coprime to 5, as is ïr'. > Note the form, which I'll place in juxtaposition to that of the> de'nition:> DEFINITION MY RESULT> Hypothesis: Hypothesis:> a,b coprime a,5 coprime> R a commutative ring A the ring of algebraic integers> a,b in R a, 5 in A> x,y,z in R q,r,s in A> a = xz a = sr> b = yz 5 = qr ==> z a unit ==> r a unit.Now you are trying to rely on the unit de'nition, when the very pointof my work is that the ring of algebraic integers is screwed up, whichmeans that though ïa' is coprime to 5 it is NOT a unit in the ring ofalgebraic integers.> Note that the left column follows the de'nition I gave (if you have a> de'nition that is at variance with this, then please give it). The> right column follows that pattern with the speci'cs of this case.> Note, also, that the conclusion that the *de'nition* requires does> NOT hold in the right column. I have shown that the number r is not a> unit.Yet because the ring of algebraic integers is screwed up, it can STILLbe coprime to 5.You cannot prove that it cannot be.> Thus, by the de'nition, a and 5 cannot be coprime in the ring of> all algebraic integers.That is a false statement.It is true that ïa' is not a unit. However, the unit de'nitionassumes that if ïa' is a unit there exists an algebraic integer ïb'such that ab=1.However, the ring is §awed so that no such ïb' exists, while ïa' iscoprime to 5.> Yet you claim they are.You are chasing your tail.> It turns out that you've neglected the possibility that ïa' is coprime> to 5, and in fact, cannot prove that it is not, as it IS possible that> it is coprime to 5 (readers should note that there are three possible> values for ïa').> I have three possible values for a. For each a (note that I have written> have given the polynomials that yield these values).And I have looked those polynomials over and the possibility that ïa'is coprime to 5 is not eliminated by anything that you've given.> So your *assumption* that ïa' is not coprime to 5 is false, and is a> logical §aw, which takes away the conclusion you desire.> Your ignorance is simply too glaring for this to be an accident. I have> NEVER assumed a and 5 are not coprime. It is the conclusion of these > factorizations:> 5 = qr, a = rs> in the ring of algebraic integers, with r not a unit. Pray tell, what> logic tells you that I have *assumed* a and 5 not to be coprime?> I'm certain that you haven't fount it in any argument I've written.The ring of algebraic integers is screwed up, so that r can be coprimeto 5, yet not be a unit *in the ring of algebraic integers*.>My task was 'nding out what followed from mathematical logic.Yet you have assiduously avoided addressing a key point: your method>>is §awed to the extent that it enables you to deduce false statements.>>Please explain.> Actually, that's what you're doing. Math proofs don't duel.> That statement has no meaning. Of course math proofs don't duel. They> have no hands to hold the foil or epee (whichever style of fencing you> prefer). If your meaning is that one cannot use a mathematical argument> to refute another mathematical argument, then you are simply incorrect,> and again show how deep your ignorance runs.You cannot 'nd a proof that refutes a proof. Proofs do NOT duel.As I've given a proof and can go step-by-step through it showing thatit is a proof, it is futile to look for a counterexample.> If anyone could 'nd an error in my paper there wouldn't be any of> this side commentary with supposed alternate proofs.> I showed you the error. You have decided to misinterpret everything,> including your own de'nition of coprime, the nature of hypothesis> and conclusion, andI am using the de'nition that there exists f and g, in the ring ofalgebraic integers, such that af+5g=1.> What looking for alternate proofs allows you to do is 'nd room to> fool yourself, like where you have ignored the possibility that ïa' is> coprime to 5.> Duh. You read the conclusion of an argument, and read it as an> assumption. Real smart, Mister Former Gifted Child. Please go back> and read *what was written* and return when you have properly> read that argument.I have made a speci'c objection to your claims, and thereby have shotthem down.Your position is not supported by the math.You can refuse to follow mathematical logic, if you wish, but it willnot change the mathematical truth. >Many of you apparently thought that you could use me to feel better>about psychological scars in§icted upon you as children by bullies>that unfortunately you learned to want to *be*, which is classic in>the psychology.Sorry, not mathematics. You are imposing your own preferences on the way>>mathematics operates, and misinterpreting your own mistakes as being>evidence that others are mistaken. Put the pieces together correctly,>>and you'll learn something.> I am noting the horri'c behavior where many posters have gone out of> their way to make hateful statements, and the newsgroup has not budged> or has actually cheered them on.> Perhaps you should see the other side of the coin: you are a person who> abuses the notion of debate, who deliberately misreads what people say,> who refuses to crack open a book that might be somewhat enlightening.> Further, you take every correction as attack, worthy of retaliation.If you refuse to follow the math, I treat you accordingly.> That kind of behavior is telling.> And yours isn't? I think the readers of these threads pretty much have> you pegged.Mathematics is not a democracy.>That is, while you hated the bullies, and wished that society wouldprotect you, rather than learn that their behavior was wrong, you>learned to hate yourselves and wish for the power. Seeing me as weak>and set upon by the crowd, many of you became the bullies you'd always>wanted to be.>>Can you name *one* person who has threatened your livelihood, threatened>>your life, or informed the FBI about your misadventures?>>Just one?> Relevance?> No, I didn't think you could.> As far as your question, note that you are making the assertion that> many of your critics are bullies. I am simply asking you to do better> than to make some wild claim: substantiate it with particulars.> So, I asked you the details of what you might have suggested what others> did. However, I do not agree that sci.math has bullied you, because I> have witnessed your unprovoked aggression and your general abuse of the> newsgroup. Bullying occurs in the absence of any provocation, and the> negative writing that has been directed at you has been almost entirely> in response to your provocation.> The items in that above list were chosen from things I've seen YOU do.Then why don't you give such an instance.>But actually I was searching for the mathematics, as I *believe* in>mathematics, and know that it doesn't matter what you believe, or how>strong you think your crowd is, as mathematical truth is independent>of such things.>>Mathematics depends on proof, and your method of proof is §awed. Thus,>>your search for the mathematics has been futile. If you mean to be>>searching for the mathematics, then you'll correct your methods.> Well, your argument is §awed mathematically, as you made an> assumption that ïa' can't be coprime to 5.> Based on your false assumption you are quite arrogant now and think> you have cause to criticize me, when in fact, you are wrong.So as one man, I can stand against the entire math society, if>necessary.Yeah, right. Stand all you like. Insult all you like. Rant and rave>>and plead and whine, just like you have done for going on 8 years, and>>claim it's just the corrupt mathematical establishment. Until you>actually *change* what it is you're doing, you will get nowhere.> What I can show is that mathematicians try to run math society like a> democracy, where what people *want* to believe is allowed to be called> truth against actual mathematical truth.> But then none of you are actually mathematicians, now are you?> There are ample pieces of evidence on line to verify whether many of us> are mathematicians.Mathematicians follow the math.> Math is not a democracy.> So? Are you claiming to be the King of Mathematics?Math has no king.>Many of you were broken all those years ago by bullies who pushed>themselves into you, and became a part of how you react.I think I detect a bit of projection here. And there.> Well there usually is *some* reason for lashing out, and having seen> so many of you lash out, like you W. Dale Hall that it seems logical> to infer a reason, unless you're just innately evil, and like trying> to hurt others.> Really? Lashing out? I am simply treating you as the punk you are.> Behave civilly, and you will get civil treatment out of me. I cannot> speak for others, but I trust that (given time for the residual echoes> of the discord you've sown to die down) eventually, you will be treated> as a normal person. You apparently don't see any reason to behave in a> civil fashion, preferring instead to bask in the glow of the nastiness> that you generate.If you do not follow the math, you will be treated accordingly.My suggestion is that those who would call themselves mathematicians,behave like mathematicians.>I suggest you go to true power and switch to mathematics. It can be>enough for you.Perhaps you have too much of a romantic view of how things work. Your>>idealized notions don't really help, not as long as you insist you're>correct and ignore the real demonstrations that you're not.> ... irrelevant stuff deleted ...> Like your §awed attempt where you falsely assume that ïa' can't be> coprime to 5?> Show me the assumption in what I've written. Convince me that I've> assumed rather than concluded, and I'll retract what I've written.> I have *not* assumed that a can't be coprime to 5. I have known it> from other information, but I never used that fact. Instead, I> found the factorizations and *concluded* that a cannot be coprime> to 5.Then prove it.> Again I remind you that proofs don't duel.> You can't 'nd a correct argument to disprove a proof.> The problem is that you don't have a proof.> Haven't I been saying that all along?It doesn't matter what you *say* as people can say anything.A mathematical proof begins with a truth and proceeds by logical stepsto a conclusion which then must be true.You cannot present a mathematical proof to support your claims.I can.>It's actually kind of sad, but then again, many of you were bullies.I'll ask again: can you name ONE person who has threatened either your>>livelihood or your life? Can you recall that you have done both?> I've noted that society pays mathematicians to do a job, and> reasonably mathematicians can expect to be held accountable if they> didn't do their job.> *DOES* society pay mathematicians? I have actually looked at my> paychecks, and have never seen the society written anywhere on them.> *Society* is an aggregate, not a de'nite entity. *Society* pays no one,> hires no one, 'res no one. My boss will let me go if I don't do my job,> but won't yours?> I note that no one has written in threatening terms regarding your> livelihood. You rationalize to cover the fact that you *have* written> in such terms regarding mathematicians.Society has a *reasonable* expectation that mathematicians will tellthe truth about mathematics.If they do not, then it's just as reasonable to expect that they willbe held accountable.> Also I noted that mathematicians may be waiting and hoping that like> past math discoverers, like Galois and Abel, that I might die> tragically before they have to acknowledge my work, and said that> instead *they* should die.> I don't care about your life. I trust that you'll live like a normal> person. It's you who has the tragic/heroic 'gure complex, not us.> However, your remark that others should die is misplaced.Those wishing I should die, should die themselves instead.>Has anyone informed the FBI about your activities? Many people (well,>>not really, but I'll imagine for a moment) could 'nd such an act to>>be an implied threat.> Mathematicians are a signi'cant part of the national security apparatus of the United States and it is VERY important to 'nd out> that they are willing to lie.> No, the national security apparatus is not the amorphous everyone who> has an important job to do melange that you are imagining. Rather, it> comprises the various security and intelligence agencies of the US> Federal Government. However, there is a side issue: it is your claim> alone that supports the assertion [we] are willing to lie.> What happens when you go to the Feds about Death stalking the earth?> What sort of reaction do you get when you tell them that Nature has> given you a special role, or that you have been testing mankind, and> that mankind has failed? Those are remarks that are equal in importance,> don't you think?The FBI has no jurisdiction over Death.> Mathematicians who can lie now about my work may actually be working> or 'nd cause to work for foreign governments against the interests of> the United States.> Keep making those claims. The more detailed, the more speci'c, and the> more grandiose, the better. Be sure to keep detailed notes and forward> them to the FBI.That's not necessary, as the record is being made here, and stored byGoogle.> It is the job of the FBI to protect the citizens of the United States> from enemies both foreign and domestic.> It is my duty as a citizen to bring the FBI's attention to lying> mathematicians.> Go right ahead. However, you cannot claim to be the innocent recipient> of nasty behavior. You have a full role as a participant (instigator> would be a more correct term) in this mud'ght.I present my mathematical work, and try to explain it, which at timesinvolves upending false claims.>So 'nding out you were actually believing in some bogus b.s. is kind>of just desserts, so those bullies who beat themselves into you aschildren, have beaten you yet again, which is also very sad.Just *who* is the bully here? The person who provides proof of your>>errors, or the one who threatens lawsuits, congressional action, murder>>at the hands of the US Army, or investigation by the FBI?> In the United States under the Constitution you are innocent until> *proven* guilty.> Agencies of the United States should not be feared if you are not> guilty of a crime, and you should be con'dent that you would be> exonerated if an investigation took place.> Oh, golly, did you think anyone was *actually* threatened? All I'm> saying is that you have to assume that your actions are taken seriously> (even if there is no chance on earth that such is the case), whenever> you make a verbal attack, call the cops, threaten Congressional action,> inform the FBI, whatever.> Even if you're not taken seriously, your acts are seen as indicative of> your intentions. Hostile intentions beget hostile reactions (I'm sure> your mother told you about these things).If you do not follow the math, you will be treated accordingly.> James Harris> Dale> Score:> JSH rationalization of own goo'ness: +1> JSH refutation of wdh argument: -10Your mathematical argument has been shown to be §awed as you assumethat a number you call in some cases ïa' and in others ïr' cannot becoprime to 5.That is a *speci'c* charge against your claims.Talking doesn't change mathematical truth.James Harris = My apologies for this top-posting segment, but I felt it improved the reader's ability to follow discussion. Some time ago, I summarized the four points below. You have not attempted to refute any of them, so I assume that you accept them as being correct. For reference, we have q(a) = 8 a^2 + 76 a - 185 r(a) = 8 a^2 + 4 a - 45 s(a) = 4 a^2 + 37 a - 104 minpoly(r(a)) = x^3 - 969 x^2 + 315 x + 5 The following properties hold for any of the coef'cients ïa' of the factorization 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>> 1. 5 = q(a)*r(a)>> 2. a = r(a)*s(a)>> 3. r(a) is an algebraic integer>> 4. r(a) is NOT a unit.>>This is a summary of an argument I am promoting:>>In short, you now must be admitting the following. Let a be any of the>>coef'cients to which you refer in the above-cited statement from your>>paper:>> 1. Since 5 = q(a) r(a), r(a) shares a non-unit factor with 5,>> namely r(a)>> 2. Since a = r(a) s(a), r(a) shares a non-unit factor with a,>> namely r(a).>> 3. Since r(a) is at once a shared non-unit factor between r(a)>> and 5, and also a shared non-unit factor between r(a) and a,>> r(a) is a shared non-unit factor between a and 5.>> 4. a and 5 cannot be coprime in the ring of algebraic integers.>>Your continued assertions that the paper Advanced Polynomial>>Factorization is correct, and the *speci'c* assertion that its method>>is correct, are untrue. You must accept that the method is §awed.>Nope.>>Actually it is possible that ïa' is coprime to 5, which means that>both r(a) and s(a) are coprime, which leaves q(a) NOT coprime to 5,>but strangely enough *in the ring of algebraic integers* it does not>have 5 as a factor, though it does in a more inclusive ring.>It doesn't matter whether r(a) and s(a) are coprime, nor does it>>matter whether q(a) is coprime to 5. What does matter is that r(a)>>is a common non-unit factor of both a and 5. You do not deny this,>>and yet you persist.> What is said is that it is possible that ïa' is coprime to 5, as are> r(a) and s(a), which leaves only q(a) NOT coprime to 5, but oddly> enough it doesn't have a non-unit factor in common with 5 *in the ring> of algebraic integers*, which leaves you with nothing.> No, you have already decided to accept the fact that a non-unit divisoris never coprime to the number it divides. If you care to retract thatstatement, you are certainly free to do so. However, it would be amathematical error.>>So, you are now claiming that, despite the fact that the number r(a)>>is a non-unit divisor of *both* a and 5, somehow there is no common>>non-unit divisor of both a and 5?> The ring of algebraic integers is §awed, as I've proven.> I asked a simple question. An answer such as ïyes' or ïno' would havebeen appropriate. You could have said as much, but you decided to bringyour desired conclusion into the discussion. That is an invalidresponse, as it assumes what you intend to prove.> You are trying to stand on a broken leg.> Nonsense. I've had a broken leg. I've stood on it, in fact I endedup having to walk a mile on it to get help. This is nothing like that.>>It's astonishing the lengths you'll go to to prop up that non-proof.>>Perhaps you could de'ne coprime?> I am using the de'nition that there exists f and g, in the ring of> algebraic integers, such that af+5g=1.> Fine. That's the correct de'nition. I'll modify the argument Igave last time.NOTICE THAT THERE IS NO ASSUMPTION THAT ïa' AND 5 are coprime: >BEGINNING OF ARGUMENT<<<<<<<<<<<<<<<<<<<<END OF ARGUMENT<<<<<<<<<<<<<<<<<<<<<<<<<<<,I have started with an accepted set of premisses (the given data, together with the fact that r is not a unit), and the de'nition ofthe property of being coprime, and a brief sequence of algebraicmanipulations, to produce the following: r (sf + qg) = 1.This contradicts the fact that r is not a unit.You have maintained that I *assumed* that 5 and a could not becoprime, yet you will never 'nd that assumption, either explicitlyor implicitly made in the above argument.Please either verify that or shut the hell up about what I have orhaven't assumed.> You cannot refute that possibility, so should withdraw your claims and> concede to the mathematical truth.> What possibility? I have just proven that 5 and a are not coprime, usingthe de'nition you provided, the facts (1-4) that you have alreadyaccepted, and basic arithmetic.>So in fact your construction cannot disprove my paper, which is how>mathematics works, as you can't disprove a proof.>Saying it's a proof, even believing it's a proof, doesn't make it so.>>Your argument is invalid, and that can be seen *because* it produces>>an incorrect conclusion.> That is a false statement. Now I've shown you the §aw in your> reasoning, you can follow the math, or show yourself to not be> rational.>>Your only counterargument is this:>> my argument is correct because it's a proof.>>That is not substantially different from this argument:>> your argument is not a proof because it is not correct.>>If you accept the form of the 'rst argument, then you must accept>>the form of the second.>>Further, the debate is *precisely* whether you have a proof. It is not>>a valid argument to assume the conclusion.> That is irrelevant to the fact that I've shot down your claims by> showing you have an assumption you *cannot* prove, so quit trying to> change the subject.> What assumption have I made? Point it out by citing something in theargument (that is, between the lines [above] that I've marked like this: >BEGINNING OF ARGUMENT<<<<<<<<<<<<<<<<<<<<END OF ARGUMENT<<<<<<<<<<<<<<<<<<<<<<<<<<<,please identify the offending statement or statements).> Your claims have been dashed. You should follow the math.> Again, please indicate where I made the assumption you're alleging.I claim that I have not made any such assumption, so as the personmaking a positive charge that I *did* make such an assumption, itis really up to you to point out where I made that assumption.Just 'nd the assumption. ... stuff deleted ...> I have made a *speci'c* objection, so stop squirming.> So, I'm asking politely. Would you please indicate, within the argumentitself, *where* I made the assumption you are objecting to?And again you make the same charge:> I have made a speci'c objection noting that your claims depend on the> assumption that your ïa' can't be coprime to 5, but you cannot prove> that assumption.> Stop whining.> Please please tell me where I made that assumption. I can't see it.Apparently you can see it, and you are being overly coy by not showingwhere (that is, in *which statement*) the assumption lies?> I am using the de'nition that there exists f and g, in the ring of> algebraic integers, such that af+5g=1.>Given *that* de'nition, I have shown this:>> 5 = qr, a = rs,>>with q,r,s in the ring of algebraic integers.> And again, it is possible that ïa' is coprime to 5, as is ïr'.> I'm sorry, but you have already acceded the point that 5 sharesnon-unit factors with r. I have already shown that if there arenon-unit factors in common between two numbers, then they cannotbe coprime (argument is pretty much identical to the one I gaveabove relating ïa' and 5).>>Note the form, which I'll place in juxtaposition to that of the>>de'nition:>> DEFINITION MY RESULT>> Hypothesis: Hypothesis:>> a,b coprime a,5 coprime>> R a commutative ring A the ring of algebraic integers>> a,b in R a, 5 in A>> x,y,z in R q,r,s in A>> a = xz a = sr>> b = yz 5 = qr>> ==> z a unit ==> r a unit.> Now you are trying to rely on the unit de'nition, when the very point> of my work is that the ring of algebraic integers is screwed up, which> means that though ïa' is coprime to 5 it is NOT a unit in the ring of> algebraic integers.> What do you mean rely on the unit de'nition? Your point may well bethat the ring of algebraic integers is screwed up, but simply statingthe point over and over does not establish that point, does it?>>Note that the left column follows the de'nition I gave (if you have a>>de'nition that is at variance with this, then please give it). The>>right column follows that pattern with the speci'cs of this case.>>Note, also, that the conclusion that the *de'nition* requires does>>NOT hold in the right column. I have shown that the number r is not a>>unit.> Yet because the ring of algebraic integers is screwed up, it can STILL> be coprime to 5.> Oh, you mean that since the ring of algebraic integers is screwed up,then no method of proof is valid? I have given the proof for your oldde'nition of coprime, for the real de'nition of coprime, and ateach stage, the only information I've used is either (1) from thegiven data, which does NOT include any assumption about whether aand 5 are coprime, or (2) algebraic manipulations, valid in anycommutative ring, or (3) de'nition. All three are valid sources forstatements in proofs, and none makes use of the assumption that youhave claimed several times I'm making.Prove it.> You cannot prove that it cannot be.> And yet I've done just that! Ain't it amazin'?>>Thus, by the de'nition, a and 5 cannot be coprime in the ring of>>all algebraic integers. That is a false statement.Sez you. You can't read a simple> It is true that ïa' is not a unit. However, the unit de'nition> assumes that if ïa' is a unit there exists an algebraic integer ïb'> such that ab=1.> That is not an assumption. That is the de'nition of unit. I haveshown that if ïa' and 5 are coprime in the ring of algebraic integers,that there IS an algebraic integer which is the multiplicative inverseof their common factor r. I have given the expression above for sucha multiplicative inverse.> However, the ring is §awed so that no such ïb' exists, while ïa' is> coprime to 5.> You are confused. ïa' being coprime to 5 is not at all related towhether ïa' is a unit. Of course, ïa' is not a unit. Where the notionof units comes in is in looking at the common factor r(a) that isshared by ïa' and 5.I do not claim, nor do I care to claim, that ïa' should be a unit.>>Yet you claim they are.> You are chasing your tail.> Sez you. You make a speci'c complaint about an argument, and shouldeither put up or shut up. Show me where I made the *assumption* that'a' and 5 are coprime.>It turns out that you've neglected the possibility that ïa' is coprime>to 5, and in fact, cannot prove that it is not, as it IS possible that>it is coprime to 5 (readers should note that there are three possible>values for ïa').>I have three possible values for a. For each a (note that I have written>>have given the polynomials that yield these values). And I have looked those polynomials over and the possibility that ïa'> is coprime to 5 is not eliminated by anything that you've given.> Really? Even knowing that they share a non-unit as a common factor?You really don't want to admit to the truth, do you?>So your *assumption* that ïa' is not coprime to 5 is false, and is a>logical §aw, which takes away the conclusion you desire.>>Your ignorance is simply too glaring for this to be an accident. I have>>NEVER assumed a and 5 are not coprime. It is the conclusion of these >>factorizations:>> 5 = qr, a = rs>>in the ring of algebraic integers, with r not a unit. Pray tell, what>>logic tells you that I have *assumed* a and 5 not to be coprime?>>I'm certain that you haven't fount it in any argument I've written.> The ring of algebraic integers is screwed up, so that r can be coprime> to 5, yet not be a unit *in the ring of algebraic integers*.> I got it. Here you go. Take your favorite coef'cient fromthat factorization above, and these numbers: r(a) = 8 a^2 + 4 a - 45 5.Two algebraic integers. Show they're coprime, using the de'nitionyou're now using. That is, show me the algebraic integers u and v forwhich u*r(a) + v*5 = 1.I'll give you actual credit for something if you can do that.After all, I gave my argument that no such numbers could exist. Showthat I was wrong by *exhibiting* u and v. ... stuff deleted ...> You cannot 'nd a proof that refutes a proof. Proofs do NOT duel.> No, you are making the claim (in that remark) that you *do* have aproof. I have found a counterexample to your conclusion. Get offthe sloganeering and listen for once.> As I've given a proof and can go step-by-step through it showing that> it is a proof, it is futile to look for a counterexample.> Isn't it a bit confusing that you have to make these exotic claimsfor the ring of algebraic integers, claims like a number k can divide a number m, with k coprime to mwhile everyone else has lived with algebraic integers for decades,'nding no problem proving all sorts of things? ... stuff deleted ...>>Duh. You read the conclusion of an argument, and read it as an>>assumption. Real smart, Mister Former Gifted Child. Please go back>>and read *what was written* and return when you have properly>>read that argument.> I have made a speci'c objection to your claims, and thereby have shot> them down.> You claimed, without any evidence, that I made an assumption that, infact, I did *not* make. Provide evidence that I assumed what you claim,and I'll listen. Until that time, I'll make the assumption that you arejust full of poo.> Your position is not supported by the math.> I gave you the math, buddy, and you failed to read it. You insinuatethat I make assumptions that I know full well I haven't made. Show thatI did what you claim.> You can refuse to follow mathematical logic, if you wish, but it will> not change the mathematical truth.> So you say. I am not a logician. I didn't see any mention of Lowenheim- Skolem, nor of ZFC, nor Forcing. There weren't any large cardinals, nomodels, no axiom schemata. Nope, not mathematical logic.You may have meant the ordinary logic associated with proofs? I do thaton a semi-regular basis, so your observation is §awed. No big surprise,given your talent for misuse of language. ... stuff deleted ...>>Perhaps you should see the other side of the coin: you are a person who>>abuses the notion of debate, who deliberately misreads what people say,>>who refuses to crack open a book that might be somewhat enlightening.>>Further, you take every correction as attack, worthy of retaliation.> If you refuse to follow the math, I treat you accordingly.> Ah, now I am refusing to follow the math. How come you couldn'tidentify the hypotheses of my argument, and seem stalled on yourclaim that I've made an assumption that is actually derived as aresult? Is it I who won't follow the math, or you who can't?>That kind of behavior is telling.>And yours isn't? I think the readers of these threads pretty much have>>you pegged.> Mathematics is not a democracy.> Oh, the King of Math.>That is, while you hated the bullies, and wished that society would>protect you, rather than learn that their behavior was wrong, you>learned to hate yourselves and wish for the power. Seeing me as weak>and set upon by the crowd, many of you became the bullies you'd always>wanted to be.>Can you name *one* person who has threatened your livelihood, threatened>>your life, or informed the FBI about your misadventures?>>Just one?>Relevance?>No, I didn't think you could.>>As far as your question, note that you are making the assertion that>>many of your critics are bullies. I am simply asking you to do better>>than to make some wild claim: substantiate it with particulars.>>So, I asked you the details of what you might have suggested what others>>did. However, I do not agree that sci.math has bullied you, because I>>have witnessed your unprovoked aggression and your general abuse of the>>newsgroup. Bullying occurs in the absence of any provocation, and the>>negative writing that has been directed at you has been almost entirely>>in response to your provocation.>>The items in that above list were chosen from things I've seen YOU do.> Then why don't you give such an instance.How about these:(apologies for the length, folks, but I couldn't resist) ... stuff deleted ... Oh well, new plan as now I get to use the Hammer with full force, which is probably what was intended. Now, no feeling sympathy for mathematicians who start marching with signs like Will work for food in the future, as I've given them lots of opportunities to play right. I will not show mercy going forward. I was trained as a soldier in the United States Army after all. And when the US Army plays a game, we play to win. Yup, you guessed it. If worse comes to worse, I *will* turn to the Army to help me with mathematicians. And then mathematicians don't think the NSA or CIA can save your asses, as generals LIKE me. And I think I know the CIA and NSA better than any mathematician. When push comes to shove, they'll throw you out with the garbage. They'd personally shoot you themselves, if it were necessary. If I have to sic Army generals on you, I will be really pissed. James HarrisWhat is this, if not an explicit threat? Note the implied death threat? alt.writing So speaking to you mathematicians: Do you really think that *everyone* will side with you for prestige reasons forever? And what about when the party's over? Will you all gang up on Magidin and try to say it's all just his fault? I think many of you are quietly sitting around hoping that he keeps getting away with it, but Fred is a computer guy. Computer people tend to be less interested in accepting *deliberate* falsehoods for the system. So maybe Fred and the other computer people will decide to keep quiet, or maybe this time they'll get you for telling falshoods or sitting by while they're being told. And then what do you lose, eh? Millions of dollars in the US alone. You guys are betting your entire futures on a game you can't win. I have a Royal Flush, and you've put everything into the pot. Call. ... stuff deleted .. If I go to Congress and say they shouldn't fund ANY math research in the United States anymore as there's almost no way to tell when it's junk if mathematicians decide to sit quietly or worse, lie, then I remind that physicists and chemists can teach math, probably better than any of you anyway, what will you say then? Maybe you can get Magidin to argue for you then, eh? Oh don't be too surprised, he might be testifying before Congress as it is anyway. You people have been holding me up when I have a short proof of Fermat's Last Theorem!!! Quit betting against mathematical truth. Supposedly mathematicians stood for it. James HarrisNo doubt you don't think that's a threat against peoples' livelihoods.And who can forget the imaginary talk before the full session ofCongress: ... stuff deleted... Honored representatives of this great nation. I regret to tell you that given the time frames it stretches the limits of reason to accept that mathematicians did not know of the importance of my work during times while they sat by. There is also no credible explanation for their refusal to give that work its due in a timely manner except that they willfully, and irresponsibly sought to deny knowledge to the world based on either the belief that they could succeed, or in a willful denial of the truth based on what I conclude is a basic contempt for truth itself. I fear they felt that they had a right to do nothing, which ignores the basic professional ethics by which most professionals operate, including lawyers, as many of you are, who are duty bound to uphold the truth above all else. ... stuff deleted ... It is my recommendation that an independent body be put into place for a thorough review of all federal funds dispensed to colleges and universities that might be used for mathematical research with an emphasis on determining the veracity and importance of the work it is funding. My personal feeling is that over 90% of current mathematical research is suspect, and that a signi'cant reduction in funding may be the only way to weed out the bad apples, as those doing truly valuable work will be able to prove its worth, when asked to do so. I emphatically ask that we do so. I have no doubt that lobbyists for the mathematical community will claim foul, and try to scare this body and this country with claims of doom and gloom, or try to make you believe that our future will be negatively impacted by making mathematicians accountable, but I want to remind you of what has just taken place. A hundred years of research by some the 'nest minds in history failed to 'nd the mathematical work, which brought me worldwide attention, *despite* the best efforts of our modern mathematicians to deny the value of that work. Yes, that's the shocking thing here, that people so trusted and honored could so callously betray that trust. As the person who discovered that work, I am here to tell you that every doom and gloom claim made by the mathematical lobby, actually applies if we do NOT make necessary changes. If we do not hold mathematicians accountable. I want to emphasize that it is my strong belief that if I had been brought through the system of mathematical training currently taught, I'd never have made these discoveries. It is therefore imperative that we act NOW both to rescue the current generation of creative minds, and insure the future of *fruitful* mathematical research. And I am serious as this represents a position that I DO want presented to Congress, should anything happen to me. ... stuff deleted ... James Harrisor this: alt.writing ... stuff deleted ... Those of you who step out now to promote or defend bogus mathematical arguments do not have the right to not expect accountability. Whether you realize it or not, your *public* statements can be used against you in various forums. Your posts are PUBLIC statements. ... stuff deleted ... It also seems to me that you must be aware that proper recognition has monetary value for me, so malice aforethought can be assumed. I am also making it clear to you that I deem your statement and the defense of them by at least one other mathematican to be evidence of a severe problem within the mathematical community, which at least in the United States, I feel should be a matter that is considered by Congress in order to determine proper distribution for funding of mathematical research in the United States. ... stuff deleted ... Again I remind all posters that your posts are public statements that can be used against you. What is happening is not a game. It has a signi'cance to people worldwide and you have no protection based on your ignorance of international law. The truth matters. If I have to prove that by bankrupting some universities, then so be it. James HarrisI had not seen this one before, but it's certainly in the category: I have posted a correct proof of FLT. I contend that any webpage declaring that my proof is incorrect is false and libelous. Any such webpage should be updated as quickly as possible. I will allow a reasonable amount of time as determined by me for that update. Failure to comply may make the authors of that webpage liable for 'nancial damages which I contend I can prove are in the hundreds of thousands (maybe millions) of dollars. James HarrisI don't want to spend too long in Google, but you should get the drift:you *regularly* engage in abusive and threatening behavior. No doubtyou get some in response, but you cannot make a successful claim ofbeing bullied. ... stuff deleted ...> Mathematicians follow the math.>Math is not a democracy.More slogans. You should get some signs printed up.>So? Are you claiming to be the King of Mathematics? Math has no king.>Many of you were broken all those years ago by bullies who pushed>themselves into you, and became a part of how you react.>I think I detect a bit of projection here. And there.>Well there usually is *some* reason for lashing out, and having seen>so many of you lash out, like you W. Dale Hall that it seems logical>to infer a reason, unless you're just innately evil, and like trying>to hurt others.>Really? Lashing out? I am simply treating you as the punk you are.>>Behave civilly, and you will get civil treatment out of me. I cannot>>speak for others, but I trust that (given time for the residual echoes>>of the discord you've sown to die down) eventually, you will be treated>>as a normal person. You apparently don't see any reason to behave in a>>civil fashion, preferring instead to bask in the glow of the nastiness>>that you generate.> If you do not follow the math, you will be treated accordingly.> My suggestion is that those who would call themselves mathematicians,> behave like mathematicians.> However, by presenting proofs that you happen to 'nd unpleasant, youmake unfounded charges and engage in prevarication.>I suggest you go to true power and switch to mathematics. It can be>enough for you.>Perhaps you have too much of a romantic view of how things work. Your>>idealized notions don't really help, not as long as you insist you're>>correct and ignore the real demonstrations that you're not.>> ... irrelevant stuff deleted ...>Like your §awed attempt where you falsely assume that ïa' can't be>coprime to 5?>Show me the assumption in what I've written. Convince me that I've>>assumed rather than concluded, and I'll retract what I've written.>>I have *not* assumed that a can't be coprime to 5. I have known it>>from other information, but I never used that fact. Instead, I>>found the factorizations and *concluded* that a cannot be coprime>>to 5.> Then prove it.> I'll note that when asked directly (see the above: show me theassumption in what I've written), you choose to evade. That'spretty much the way you operate, make unfounded charges, andthen don't respond to the request for substantiation.No wonder you used to be a Republican. Kinda like Bush, Shrub,Gingrich, and oh who's your other pal in Georgia, that redneckprick with the axe handle? Lester Maddox, that's who. Your kindof folks. They haven't changed (those who are living), and you'rein perfect company. Hog Heaven.As far as your suggestion that I ïprove it' in respect to my remarkabout concluding rather than assuming the non-coprimeness of ïa'By the way, as the person who has asserted on several occasionsthat I made this assumption, it is really up to you to point it out. ... stuff deleted...>>The problem is that you don't have a proof.>>Haven't I been saying that all along? It doesn't matter what you *say* as people can say anything. A mathematical proof begins with a truth and proceeds by logical steps> to a conclusion which then must be true.> The must be true is only the case *if* the logical steps are airtight.Your problems with terminology and de'nition, not to mention the actualsteps in the proof, suggest that the whole mess could be rife witherror.Your further insistence, that the validity of an argument cannot betested by the presence of counterexamples to its conclusions, andthat is your primary argument here, is entirely bogus. Many potentialtheorems are shot down, not by the immediate detection of logical errorwithin the proof, but by the development of counterexamples.The logic is this: it is simpler to analyze a speci'c instanceof an argument than to dig through the full detail of a generalargument. The existence of a *single* counterexample invalidates theargument. That is, you're fond of saying its results are correct because it's a proof,while the contrapositive, it's not a proof because its result is incorrectis equally valid.> You cannot present a mathematical proof to support your claims.> I can.> Really? Then show the two algebraic integers f and g that show that ïa'and 5 are coprime in the ring of algebraic integers. Just show them, andproduce the sum: a f + 5 g = 1.That can't be so hard, now, can it? I mean, you have given a correctde'nition of coprimeness in a ring, you think you know what you're doing? Then do it. I don't believe your multi-variable polynomialargument, because I can prove that ïa' and 5 are not coprime, notfor *any* of the a's. However, you want to play that game, I'vejust told you how: Produce the numbers. Show the evidenceCan't do it? Won't do it? Wassamadda, your mamma won't let youplay with big numbers?Show the numbers. .. stuff deleted ...> Those wishing I should die, should die themselves instead.> ??? Has anyone expressed the sincere hope you'd die? I recall a fewthem as being weak attempts at humor.appears to be what you're talking about. I responded: >THIS IS WITH GREAT REGRETS THAT WE LET YOU KNOW THAT PERTTI >>LOUNESTO HAS DIED WHEN SWIMMING IN CRETE ON JUNE 21. >> I guess the 'rst question I would ask is whether somehow > we can get James Harris to take up swimming. > Dave. > Hey, While I understand the frustration of dealing with JSH, that's all it is. Nothing to get exorcised about. As for mortality, Jim Morrison said it best: No one here gets out alive. Let's have a little perspective. Dale.I read it as gallows humor, and inappropriate at that. You may not haveseen it, and so much the better, but I don't think *anyone* seriouslywishes you ill. Many are frustrated by your prideful ignorance and yourappalling rudeness to people who, quite frankly, have better things todo than to teach you something you could have learned dozens of timesover in the period you have been §ailing about here.>>Has anyone informed the FBI about your activities? Many people (well,>>not really, but I'll imagine for a moment) could 'nd such an act to>>be an implied threat.>Mathematicians are a signi'cant part of the national security>apparatus of the United States and it is VERY important to 'nd out>that they are willing to lie.>No, the national security apparatus is not the amorphous everyone who>>has an important job to do melange that you are imagining. Rather, it>>comprises the various security and intelligence agencies of the US>>Federal Government. However, there is a side issue: it is your claim>>alone that supports the assertion [we] are willing to lie.>>What happens when you go to the Feds about Death stalking the earth?>>What sort of reaction do you get when you tell them that Nature has>>given you a special role, or that you have been testing mankind, and>>that mankind has failed? Those are remarks that are equal in importance,>>don't you think?> The FBI has no jurisdiction over Death.> Nor do you.Do you actually think the FBI has jurisdiction over whether I believethat you have done any correct mathematics? That's what you're talkingabout when you say people are lying about math, you know. You justwant people to buy your argument.Any claim that people think it's correct, but are just lying about it is plain batty. Mathematicians simply do not lie about what they considercorrect. They may make errors of one sort or another, but in over 35years in dealing regularly with mathematicians, I have not once, NOTONCE, found one who lied about what he or she thought was correct ornot. Some may be more or less conservative when it comes to stuff theydon't have the insight to analyze at a §ash.No one lies about mathematics. Well, no one who is a mathematician,and your continuous cries of liar! are, to these ears, utterlyoffensive. And utterly uncalled for.>Mathematicians who can lie now about my work may actually be working>or 'nd cause to work for foreign governments against the interests of>the United States.>Keep making those claims. The more detailed, the more speci'c, and the>>more grandiose, the better. Be sure to keep detailed notes and forward>>them to the FBI.> That's not necessary, as the record is being made here, and stored by> Google.> Oh, no. You would be much better served to maintain meticulous records.>It is the job of the FBI to protect the citizens of the United States>from enemies both foreign and domestic.>>It is my duty as a citizen to bring the FBI's attention to lying>mathematicians.>Go right ahead. However, you cannot claim to be the innocent recipient>>of nasty behavior. You have a full role as a participant (instigator>>would be a more correct term) in this mud'ght.> I present my mathematical work, and try to explain it, which at times> involves upending false claims.> Your claim about ïa' and 5 is such a false claim, yet you 'ght toothand nail to defend it. In fact, virtually all your mathematical claims(maybe every single one, but I am not sure I have seen *all* yourmathematical claims) have proven to be false whenever they varied fromconventional mathematical teaching. Doesn't that suggest even a bitthat you need to exercise better judgment?>So 'nding out you were actually believing in some bogus b.s. is kind>of just desserts, so those bullies who beat themselves into you as>children, have beaten you yet again, which is also very sad.>Just *who* is the bully here? The person who provides proof of your>>errors, or the one who threatens lawsuits, congressional action, murder>>at the hands of the US Army, or investigation by the FBI?>In the United States under the Constitution you are innocent until>*proven* guilty.>>Agencies of the United States should not be feared if you are not>guilty of a crime, and you should be con'dent that you would be>exonerated if an investigation took place.>Oh, golly, did you think anyone was *actually* threatened? All I'm>>saying is that you have to assume that your actions are taken seriously>>(even if there is no chance on earth that such is the case), whenever>>you make a verbal attack, call the cops, threaten Congressional action,>>inform the FBI, whatever.>>Even if you're not taken seriously, your acts are seen as indicative of>>your intentions. Hostile intentions beget hostile reactions (I'm sure>>your mother told you about these things).> If you do not follow the math, you will be treated accordingly.> Let's hear it for SLOGANS!!!! boooooo.James Harris>>Dale>>Score:>> JSH rationalization of own goo'ness: +1>> JSH refutation of wdh argument: -10> Your mathematical argument has been shown to be §awed as you assume> that a number you call in some cases ïa' and in others ïr' cannot be> coprime to 5.> The number ïa' is any of the coef'cients of the factorization: 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)Your claim is that one of these a's is coprime to 5.My number ïa' is an arbitrary choice from the set {a1, a2, a3}.My numbers q(a), r(a), s(a) [or more simply q,r,s] are givenas these values (depending on a): q(a) = 8 a^2 + 76 a - 185 r(a) = 8 a^2 + 4 a - 45 s(a) = 4 a^2 + 37 a - 104You show your confusion by imagining that they are the same.In fact, ïr' is a divisor of ïa', and also is a divisor of 5.That fact proves that ïa' cannot ever be coprime to 5, sinceI also show that ïr' is never a unit (given the ïa' valuesas speci'ed earlier). Further, the argument that establishesthese facts is independent of any assumption whatever on ïa',save the fact that ïa' is one of the coef'cients of theabove factorization, as already stated. > That is a *speci'c* charge against your claims.> And a false charge. Please 'nd the place in which Ihave made such an assumption. Just do that, will you?I have already asked, and what did you say? Nothing.I have already claimed that your charge is unfounded,and your response? Prove it.That is not a legitimate answer, and you know it.> Talking doesn't change mathematical truth.> So? I have not made any unwarranted assumptions (well,it appears to be risky assuming that you'll readcorrectly, and respond genuinely). My arguments havebeen correct, as can be af'rmed by the fact that myalgebraic betters have not jumped in to correct anyblunders.If you care to continue this debate, you will respondwith the speci'c place where I assume that (1) ïa' is not coprime to 5 [and the concluding statement of an argument does *not* count as assumption: it counts as conclusion!],or (2) r(a) is not coprime to 5,although I do prove that one quickly, as it's a divisorof 5. It is a triviality to observe that for a non-unitdivisor D of a ring element N, D is never coprime to N.That result is valid in any ring with unit in which thenotion of coprime makes sense. It may be true morebroadly than that.> James HarrisDale. => ... a bunch of stuff, including the following un'nished thought:> I showed you the error. You have decided to misinterpret everything,> including your own de'nition of coprime, the nature of hypothesis> and conclusion, and> ... and probably a bunch of other things, none of which have come to mind, so I'll leave it at:You have decided to misinterpret everything, including your own de'nition of coprime and the nature of hypothesis and conclusion.Sorry for the confused ending of that sentence.> Dale> Score:> JSH rationalization of own goo'ness: +1> JSH refutation of wdh argument: -10> Dale =>>[...] Sorry, not mathematics. You are imposing your own preferences on the way>> mathematics operates, and misinterpreting your own mistakes as being>> evidence that others are mistaken. Put the pieces together correctly,>> and you'll learn something.>>I am noting the horri'c behavior where many posters have gone out of>their way to make hateful statements, and the newsgroup has not budged>or has actually cheered them on.>>That kind of behavior is telling.What's telling is the way you interpret simple statements about the math as hateful. Considering the way you tend to call people _evil_and explain that they're going to be _killed_ on account of the thingsthey say about math it's also hilarious.>[...]>>Also I noted that mathematicians may be waiting and hoping that like>past math discoverers, like Galois and Abel, that I might die>tragically before they have to acknowledge my work, Yes, you noted, on the basis of absolutely _no_ evidence, thatmathematicians may be waiting for this.>and said that>instead *they* should die.No, you said that they _will_ be _killed_. And killed by entitiesthat only you can hear.>> Has anyone informed the FBI about your activities? Many people (well,>> not really, but I'll imagine for a moment) could 'nd such an act to>> be an implied threat.>>Mathematicians are a signi'cant part of the national security>apparatus of the United States and it is VERY important to 'nd out>that they are willing to lie.>>Mathematicians who can lie now about my work may actually be working>or 'nd cause to work for foreign governments against the interests of>the United States.>>It is the job of the FBI to protect the citizens of the United States>from enemies both foreign and domestic.>>It is my duty as a citizen to bring the FBI's attention to lying>mathematicians.This kind of raving is why people... oh, never mind.>> So 'nding out you were actually believing in some bogus b.s. is kind>> of just desserts, so those bullies who beat themselves into you as>> children, have beaten you yet again, which is also very sad.> Just *who* is the bully here? The person who provides proof of your>> errors, or the one who threatens lawsuits, congressional action, murder>> at the hands of the US Army, or investigation by the FBI?>>In the United States under the Constitution you are innocent until>*proven* guilty.>>Agencies of the United States should not be feared if you are not>guilty of a crime, and you should be con'dent that you would be>exonerated if an investigation took place.>James HarrisDavid C. Ullrich**************************As far as I'm concerend you're trying to wait until I die, so I 'guremaybe you should die instead. How about that, eh? Wouldn't that be abetter twist?You refuse to follow the math, so the great Powers that controlreality and *speak* in mathematics decide to kill you instead of me.So what do you think about that, eh? Oh, can't hear Them talking?Well, I guess that's because you don't really understand Mathematics,the true language, which is THE language.They're talking about you now, and They agree with my assessment, andwill not penalize me as They allowed the others like Galois and Abelto be penalized.They will kill you instead.James Harris speaking on Weird factorization, genius =>Based on your false assumption you are quite arrogant now and think>you have cause to criticize me, when in fact, you are wrong.Talk about arrogant? What about the way you demean everyone on the group? Youshould be grateful people are willing to help you. A physics degree doesn'tmean you know everything about math.>Mathematicians are a signi'cant part of the national security>apparatus of the United States and it is VERY important to 'nd out>that they are willing to lie.>>Mathematicians who can lie now about my work may actually be working>or 'nd cause to work for foreign governments against the interests of>the United States.>>It is the job of the FBI to protect the citizens of the United States>from enemies both foreign and domestic.>>It is my duty as a citizen to bring the FBI's attention to lying>mathematicians.Give me a break. FBI has other things to do than go after mathematicians whohave NOT committed any crime.>In the United States under the Constitution you are innocent until>*proven* guilty.>>Agencies of the United States should not be feared if you are not>guilty of a crime, and you should be con'dent that you would be>exonerated if an investigation took place.James, don't you think that if one mathematician was exonerated, all of themwould be? After all, they all practice the same mathematics. Think about it.Not logical, is it?David Moran =On and off for a month now I have been trying to tackle a couple ofproblem inv olving a rubix cube.The 'rst problem is I would like to know how many differentarrangements there are.The problem might seem straightforward, however if you are used tocompleting the cube you will realise added dif'culties.I am absolutely sure that some seemingly possible arrangements areimpossible to acheive (similar to the famous 16-15 problem).One such impossible arrangement would occur if you took a completedcube and switched the coloured stickers on a cube with shows only twofaces.The second problem is to prove that some arrangements are impossible. I would appreciate any views. I have tried to apply orbit-stabilisertheorems to count, but it doesnt seem to work. => On and off for a month now I have been trying to tackle a couple of> problem inv olving a rubix cube.> The 'rst problem is I would like to know how many different> arrangements there are.> The problem might seem straightforward, however if you are used to> completing the cube you will realise added dif'culties.> I am absolutely sure that some seemingly possible arrangements are> impossible to acheive (similar to the famous 16-15 problem).> One such impossible arrangement would occur if you took a completed> cube and switched the coloured stickers on a cube with shows only two> faces.> The second problem is to prove that some arrangements are impossible. > I would appreciate any views. I have tried to apply orbit-stabiliser> theorems to count, but it doesnt seem to work.(in the ï80's I think). There are lots of cube sites on the web where they discuss the theory and practice of cubing. I understand Rubik's cube is making a comeback. Next, hula hoops. =I would like to estimate the following sum from below:sum_{i=1}^{m}{frac{1}{n-x_{i}}}, where each x_{i} is an integer between 1 and n-2. Ideally, the estimate would be in terms of n and the sum of thex_{i}'s. A natural approach is to apply Jensen's inequality - I triedthat but, unfortunately, the bound derived from it wasn't strongenough for my purposes.I'd most eager to hear other suggestions or ideas.TIA,Felix. =The appended appeared on another newsgroup. Anyone know whether there's any truth to the story? *************************************************** <------ ------->[Heard through the grapevine a while ago. Quite possibly true.]******************************************************** *******-- =Now, as you must know, with the standard digital-clock readout, sometimes represent other times if the clock happens to be lyingupside-down.(12:01 <> 10:21, for example)(But, even if we have a 4-digit display for all times {in which case, 01:21 <> 12:10 is a valid rotation},then 03:10 is NOT equal to 01:30 after §ipping, because 3 is not anynumber-digit after being rotated 180-degrees.)So, we have a 12-hour digital clock with the standard readout.ie. the digits are formed from some combination of 7 line-segments,each segment being lit or un-lit, the segments arranged as: --| | --| | --There is also a colon separating the minutes and the hours, as istypical.Now, we take a diplayed time, that if §ipped 180-degrees producesanother valid time where the colon is in the same place.(example: 10:10 <> 01:01 is valid, but 02:10 <> 12:00 is NOT validbecause the colon has moved.)So, between every time and its rotation, there is an absolutedifference in the two times (many differences, actually). But here, Iam interested in only the two positive differences between 0 and 12hours.Now these differences themselves can be other valid times whendisplayed on a digital-clock and rotated 180-degrees.And, in turn, we can take the absolute difference between the previousdifference and its rotation, continuing.What, then, is the time that has the most times which, by a chain ofdifferences and rotations, leads to this particular time?By searching by-hand and brute-force (which is what I suggest that youdo, since a computer-search is not worthwhile, this puzzle being tooeasy to justify that), I came up with a time which has a total of 14times which lead eventually to this time.But this is actually 7 pairs equal-by-rotation, and these pairs existwithin the branches of the tree, not necessarily at the leaves.(I get 5 pairs which are the leaves of the tree.)(And, yes, leading 0's are allowed, the colon is unmoved, and all ofthe other conditions implied above.)There could be a better answer. But I will post mine in the next fewdays, unless I do not feel the need to.There are several other puzzles that can be made using the basic ideaabove.You can ask about, say, using 24-hour clocks, allowing the moving ofthe colon (putting 0's wherever needed on the ends of the times),allowing differences greater than 12 hours, asking about othersymmetries (such as re§ections), or (oh, yeah) having the puzzle onan ANALOG (hands on a circlular face) clock.Even though this puzzle is not mathematical, in any purest sense, Iwill cross-post it to sci.math anyway, since some on that group may'nd this somewhat interesting anyway.Leroy Quet