mm-328 http://www.megafoundation.org/Ultranet/UltranetApp.html === Subject: Re: My research, publication announcement There's more to my work than just arguing on Usenet, so I'd like to point out that my paper Advanced Polynomial Factorization is sla to be published: See http://www.megasociety.net/NoesisHighlights.html The Mega Foundation is an organization of high IQ people, and I'm glad to be associa with them. To learn further about the organization you can use Google, or see: Why a group like the Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htm I hope at least some of you will appreciate that often the most important ideas in history have to get past people limi by their lack of imagination and their prejudices, who act against scientific progress. What I want you to see is that there's more to me than Usenet, so that you can begin to understand that the revolution I'm giving you a chance to be a part of is bigger than the small-minded people who continually throughout history work to halt progress. Thank you for your time and attention. http://mathforprofit.blogspot.com/ === Subject: Re: My research, publication announcement There's more to my work than just arguing on Usenet, so I'd like to point out that my paper Advanced Polynomial Factorization is sla to be published: See http://www.megasociety.net/NoesisHighlights.html The Mega Foundation is an organization of high IQ people, and I'm glad to be associa with them. To learn further about the organization you can use Google, or see: Why a group like the Mega Foundation? http://www.ultrahiq.org/Mega/WhyMega.htm I hope at least some of you will appreciate that often the most important ideas in history have to get past people limi by their lack of imagination and their prejudices, who act against scientific progress. What I want you to see is that there's more to me than Usenet, so that you can begin to understand that the revolution I'm giving you a chance to be a part of is bigger than the small-minded people who continually throughout history work to halt progress. http://mathforprofit.blogspot.com/ === Subject: Re: My research, publication announcement > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is sla > to be published: See http://www.megasociety.net/NoesisHighlights.html The Mega Foundation is an organization of high IQ people, and I'm glad > to be associa with them. To learn further about the organization > you can use Google, or see: Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm I hope at least some of you will appreciate that often the most > important ideas in history have to get past people limi by their > lack of imagination and their prejudices, who act against scientific > progress. What I want you to see is that there's more to me than Usenet, so that > you can begin to understand that the revolution I'm giving you a > chance to be a part of is bigger than the small-minded people who > continually throughout history work to halt progress. Thank you for your time and attention. http://mathforprofit.blogspot.com/ Dammit, he's back on top again. Somoeone will have to reply to push him back down again. === Subject: Re: My research, publication announcement >There's more to my work than just arguing on Usenet, so I'd like to >point out that my paper Advanced Polynomial Factorization is sla >to be published: Does the Usenet Posting Guide you're working on say anything about how to keep your browser from posting multiple copies of the same post? === === Subject: Re: My research, publication announcement http://www.polymath-systems.com/intel/hiqsocs/megasoc/noes152 /publish.html http://www.polymath-systems.com/intel/hiqsocs/megasoc/noes151 /cease&de.html === Subject: Re: My research, publication announcement > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is sla > to be published: See http://www.megasociety.net/NoesisHighlights.html Your work is interesting, but highly ßawed. I for one can pick out some of the ßaws without too much trouble; others have done more detailed analysis. At best, it appears that some minor corrections are needed; at worst, major surgery is the only option. I am not sure how your magic polynomial fits into your larger proof, either. === Subject: Re: My research, publication announcement > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is sla > to be published: See http://www.megasociety.net/NoesisHighlights.html The Mega Foundation is an organization of high IQ people, and I'm glad > to be associa with them. To learn further about the organization > you can use Google, or see: > http://mathforprofit.blogspot.com/ === http://mathforprofit.blogspot.com/ === r time and attention. http://mathforprofit.blogspot.com/ http://larouchepub.com/ === Subject: Re: My research, publication announcement > http://mathforprofit.blogspot.com/ Oooh. Well, I'm going to have to attack this. Sorry James, but this has the same problems as your other proofs. 1. Let P(x) = 5^3*7^6*x^3 - 3*5^3*7^4*x^2 - 2^3*3^2*5*7^2*x - 2*7^2*11 . 2. P(x) = 7^2*(5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11). P(x) = (5 * a_1(x) + 7) * (5 * a_2(x) + 7) * (5 * a_3(x) + 7) The astute reader may notice that 7^3 = 343 cannot be the last term so at least one of the a_i(x) must contain a constant addend of some sort. I will ignore this bizarre rewrite and instead note that P(x) = (5 * b_1(x) + 7) * (5 * b_2(x) + 7) * (5 * b_3(x) - 22) probably works better, for various reasons. This is the form you eventually get to in step 4, by different methods. We can now attempt to work out the forms for the b_i. We already know the following: [1] The product b_1(x) * b_2(x) * b_3(x) must contain an x^3 term. In fact, this x^3 term must be 7^6 * x^3. [2] If we assume P(x) = 7^6 * 5^3 * (x - x_1) * (x - x_2) * (x - x_3) for some algebraic numbers x_1, x_2, and x_3, then we can conclude that, for some permutation of x_i, b_1(x) = -7*x/x_1 b_2(x) = -7*x/x_2 b_3(x) = 22*x/x_3 is a valid solution. However, other solutions are possible, such as the rather trivial b_1(x) = P(x)/5 - 7/5 b_2(x) = -6/5 b_3(x) = 23/5 or something like b_1(x) = -14*x/x_1 - 7/10 b_2(x) = -7*x/x_1 b_3(x) = -11*x/x_1 + 11/5 In light of this observation I cannot draw any other conclusions at this time without additional constraints. [3] If we assume, as you did in an earlier proof attempt, that P(x) = (5 * c_1 * x + 7) * (5 * c_2 *x + 7) * (5 * c_3 * x - 22) where the c_i do not depend on x, then we can indeed state that, for some permutation of the x_i, c_1 = -7/x_1 c_2 = -7/x_2 c_3 = 22/x_3 However, are any of these algebraic integers? At this point I can state that, if the x_i satisfy P(x)/49 = (5^3*7^4*x^3 - 3*5^3*7^2*x^2 - 2^3*3^2*5*x - 2*11) = 0 then it must be the case that the values y_i = 1/x_i satisfy P(1/y)*y^3/49 = 2*11*y^3 + 2^3*3^2*5*y^2 + 3*5^3*7^2*y - 5^3*7^4 If we further assume y = z/7, then z = 7 * y = 7/x, and these satisfy the equation 2*11*z^3 + 2^3*3^2*5*7*z^2 + 3*5^3*7^4*z - 5^3*7^7 = 0 Substitute w = -z and we simply ßip signs: 2*11*w^3 - 2^3*3^2*5*7*w^2 + 3*5^3*7^4*w + 5^3*7^7 = 0 This polynomial is irreducible. If one looks at v = 22/x, then x = 22/v and we get -v^3 * P(22/v)/49 = 2*11*v^3 + 2^4*3^2*5*11*v^2 + 2^2*3*5*7^2*11^2 + 2^3*5^3*7^4*11^3 We note the common factor 2*11 and divide again, getting: -v^3 * P(22/v)/(2*7^2*11) = v^3 + 2^3*3^2*5*v^2 + 2*3*5*7^2*11 + 2^2*5^3*7^4*11^2 so yes, c_3 is an algebraic integer -- but that's the only one that can be. [4] This is not an attack on your person, but on your proof. However, if you continue to call those in the mathematical establishment idiots, why are you so surprised at the shipments of high-explosive vitriol coming on your doorstep? (This is assuming there *is* a Mathematical Establishment, admitly; presumably, there is a well-established procedure for peer review of mathematical results, which is about as close to an establishment as I personally can see at this time -- since I've not attemp to publish peer-reviewed papers in scientific journals I'm unlikely to know the details of this process.) Concentrate on the math. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: My research, publication announcement > See http://www.megasociety.net/NoesisHighlights.html Ye gods, what horrid typesetting. Not only that; I note that they carry a book review by William Dembski, one of the leading lights of the ID movement; people who think that evolution can not account for life on earth and that there has to have been an Intelligent Designer. You're in good company, James. V. -- email: lastname at cs utk edu homepage: cs utk edu tilde lastname === Subject: Re: My research, publication announcement In sci.physics, Victor Eijkhout <1g45l67.14bvddq1b6pkjbN%see.sig@for.addy>: > See http://www.megasociety.net/NoesisHighlights.html Ye gods, what horrid typesetting. Not only that; I note that they carry a book review by William Dembski, > one of the leading lights of the ID movement; people who think that > evolution can not account for life on earth and that there has to have > been an Intelligent Designer. You're in good company, James. V. === What does it cost to become a member? What does it cost to get a paper published? -- === See http://www.megasociety.net/NoesisHighlights.html -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: I can't stand it anymore amanda replied: [...] > http://www.japantoday.com/e/?content=news&cat=9&id=278105 http://tibet.dharmakara.net/ictracism.html > Long time ago - it was soon after 9/11 - I happened to go into an > egpytian group thinking I would learn more about the ancient egyptian > culture. Instead, some Indian Hindu hate mongers were trashing the > Muslims (mainly Pakistanis) pickign on their religion and treatment on > women. So I borught up suttee, caste system, and treatment of women > in their society. Caste systen has been abolished but has it really > gone? I told them to clean up their hosue before picking on others. I > didn't use the same user name. I found it. http://groups.google.com/groups?q=amanda+suttee&hl=en&lr=&ie= UTF-8&selm=9b4c 6065.0303110536.7bb32a3f%40posting.google.com&rnum=1 > All thanks to Islamic inßuence. > It's always someone else's fault, isn't it? Just shows how ill-informed persone you really are. I suggest you read history of your religion of Ôpeace' aka Islam, and what it did when when it ransacked ßourishing cultures and civilization. Islamic doctrine is like Nazisms, which it more than matches in fanaticism and fascism, especially when it come to lying and distorting history. In other few hundred years Nazis will also say that Hitler had a profoundly positive and spiritual inßuence on barbaric jews. Just as shameless retards are denying Islamic butchery and barbarity of over the past millenia. [...] > Rare..amusing. Forogt suttee too? Suttee was common among Rajput wives of warriors who did not want to be captured by invading Muslims. The number of women who commit suttee among 0.8 billion Hindus is around one every few decade. It voluntary and even then against the law.Now compare this with brutal, but legally enforced murder of thousands of Muslim women for committing adultery. The latter is sanctioned by Allah, and Pakistan, Afghanistan, Iran, Saudis implement it. Limb chopping being other favorite pastime of pedoe prophets' retarded followers. [...] > What percentage of Muslims women are covered in black tent? t In Islamic paradises like Afghanistan, Saudi Arabia, Iran it is close 100%. If the women dont obey they are beaten mercilessly. > married at like cattle at any age > How old were the Hindu girls when sold by their parents to rapists? t You are thinking of many cases of Indian Muslims who sell their preteen daughters to dirty old Arabs, especially in Hyderabad. They are rescued by Hindu police. [...] > An ignorant hatemonger calling one of the greatest human being in > history a pedoe doesn't count. Get that into your tick head. Just like Hitler is called the greatest human being in history by Nazis. Pol Pot by his own follwer. Just because he butchered people and had temporal lobe epilepsy with paranoid psychosis and delusions, does not mean he was great. Today he would be in lunatic asylum, where he belonged. There would have been no islamic mass murderers like Bin Ladens, Gaznavis and Timmurs. Regarding his pedoia; truth always hurts. Did he not marry Ayesha when she was 6 year old and he 54 yr old, and had sex with her when she was only 9 yr old? Did he not marry young wife of his own adop son? Did he not marry a Jewish women he captured after slaughtering her entire family and 700 plus surrendered jewish men? He even gave instruction on how to divide booty from loot. BTW, it looks like bride burning is today's problem. What say you about this? No one knows how suttee star, or for that matter where or when. It's not unique to India--widow suicide is known to have occurred among the Egyptians, Chinese, Vikings, and others. Some say its origin on the subcontinent dates back 5,500 yeawhile others believe it arrived much later, around 1 AD. I've heard Indians deny there's anything specifically Hindu about it, in that it doesn't figure in Hinduism's core texts. Today it's most closely associa with remote villages domina by the Rajputs. Suttee is different from bride burning, in which a newly married Indian woman is burned to death by her in-laws for failing to meet demands for a larger dowry, the traditional gift given to the couple by the bride's parents. Thousands of such murders have been repor. In contrast, suttee is a voluntary act, theoretically at least, meant to atone for the couple's sins and ensure their reunion in the afterlife. But horrified Indian feminists say that in practice the suttee victim often had little choice. Sometimes family membeincluding other women, browbeat her into it; sometimes she was bound or hopped up on drugs. Much of the time even that wasn't enough. It's said music was played at high volume during suttee so no one could hear the widow's screams. Seems you're as down on Hindu culture as you are on Muslim culture. I'm having a difficult time reconciling the claims about the frequency of suttee with the last accounts of how it occurred in practice. Nontheless, Hindu culture hardly seems blemish free and I disagree that you have to be perfect to point out faults you see in others. BTW, I agree with you about islam. >As a minority in the country I was born (my parent were born and >raised thare too) who suffered discrimination and were not considered >as full-citizens, you must be joking accusing me as a racist. >>What does that have to do with whether you are racist or not? > Because it was due to race that we were discrimina. The issue is whether *you* are racist, not others. It is very clear that you think you are saying that since you were discrimina against you cannot be racist. >>I don't think you really understand the concept. You were the one who didn't understand what I said. I do understand it, you still don't. >>It's like saying you are black, you can't be racist. >I got news for ya, if you can say that, you are. > I got news for you. You underestima me in assuming that I think > black cannot be racists. But you think that since *you* were discrimina against, that you cannot be racist. What's the difference? >Beside, as a person whose blood covers basically almost all races of >the world (not through Caucasoid and Negroid per se but through >Indo-Aryan which you probably have no clue about), including semite >(not jewish), you gotta be kidding to accuse me as a racist. >>You've gotta be kidding if you think this denies that you can be >>a racist. >>It's clear that you will profit from any pro-Asian racism. What are you talking about? >>Think Affirmative Racism. Think Ôdiversity'. >I was/am going with the second one, WHICH WAS MY WHOLE POINT OF >criticizing the use of the term Asian in IQ test, duh... >>That is was not inven by an Asian. That was your objection, how >>dare a non-Asian even imply that they can say anything about an Asian. My objection was that I as an Asian did not need to run and take the > IQ test to find out my intelligence level when, I repeat WHEN *that* > guy told me that I must not have done well blah..blah..blah and ehnce > against the IQ tests. I disagree. Your objection was, and I quote... Don't you realize that us Asians don't need the tests inven by non-Asians to find out whether we are intellgent or not? Your objection was that the teste were inven by non-Asians. Presumably had the tests been inven by Asians this would be a non-issue, as the tests would be fine. >>It's almost certain that you have. The most stupid thing I have ever heard. What I have had was getting >pulled over by the red-neck police >>Racism. Red neck is red neck. A racist term is a racist term. Simple as that. > Simple as that. Here in CA, I approached a > police car once to ask for some direction. The first things he said, > with a very pleasant manner was what can I do for you?. In > Houston, it would be like What's your problem?. Another time, as > the police blocked the route to my place, I asked the police standing > there what I should do and he was really nice. In Houston, it would be > like I am busy; get lost. Seems you are comparing your experiences in CA with your prejudices about Houston. > I even told that (that most of them are red neck) to a police officer > once. It was 11 pm at night, not on the highway but he happened to be > really closed by and I didn't see his car because I wan to get home > which was not too far and did not slow down as he expec. I was the > only car around there. He told you I am busy; get lost? > I told him to just give me the ticket and spare the lecture. He > asked me why I talked like that. I told him that I haven't met a > police officer who did not give me a ticket. I told him that most are > red necks. He just laughed. May be he didn't get mad because I was not > far from the campus and he knew that I was a student. (I did tell him > that I was tired and didn't see him when he asked why I didn't slow > down after seeing him). Or may be I looked cute to him (he wasn't very > old) when i said to him that most of them are red necks. Who knows? As > he gave me the ticket, he said that he would not have given me the > ticket if I didn't act the way I did. I replied - silently - yeah, > right! So you have no issues with stereotyping Americans. Odd that you have issues with stereotyping Asians. Why is this? >in Houston for driving a beat-up >car in my student days because I look a bit Hispanic or South >Americans. >>Odd, I was always pulled over by the cops when I had my old 1965 >>Mustang convertible. Racism obviously, right? In my case, there were other cars speeding but they didn't get > stopped. I'm not asking about your case. >>Rich Note: I will not be wasting any more time defending groundless >accusations. >>You've already proven that they are not groundless. Whatever. Ahh, the superior Asian IQ rears it's ugly head. Rich >>Rich > === Subject: Re: I can't stand it anymore I found it. http://groups.google.com/groups?q=amanda+suttee&hl=en&lr=&ie= UTF- > 8&selm=9b4c6065.0303110536.7bb32a3f%40posting.google.com&rnum =1 I took a look and you got the posts written by I also noticed that you copied the stuff written by the guy whom I referred to as Indian Hindu hate monger. All he did was coming up with crap. I am sorry that you took him seriously. You should have got some books written by unbiased historians before you decided to trash Islam and its Prophet, will you? There is not much difference in Islam and Christianity when it comes to crap, crpas which were made by humans. For the story of Muhammed, I strongly suggest that you get a few books written by real historians. BTW, I agree with you about islam. Which part did you agree with? Are you sure that what you read was written by me? By the way, here is something you may want to read about Muhammed. He was Caesar and Pope in one; but he was Pope without Pope's pretensions, Caesar without the legions of Caesar: without a standing army, without a bodyguard, without a palace, without a fixed revenue; if ever any man had the right to say that he ruled by the right divine, it was Mohammed, for he had all the power without its instruments and without its supports. ---- ----- MAHATMA GANDHI, speaking on the character of Muhammad, (pbuh) says in (YOUNG INDIA): I wan to know the best of one who holds today's undispu sway over the hearts of millions of mankind....I became more than convinced that it was not the sword that won a place for Islam in those days in the scheme of life. It was the rigid simplicity, the utter self-effacement of the Prophet, the scrupulous regard for his pledges, his intense devotion to this friends and followehis intrepidity, his fearlessness, his absolute trust in God and in his own mission. These and not the sword carried everything before them and surmoun every obstacle. When I closed the 2nd volume (of the Prophet's biography), I was sorry there was not more for me to read of the great life. -------- My choice of Muhammad to lead the list of the world's most inßuential persons may surprise some readers and may be questioned by othebut he was the only man in history who was supremely successful on both the religious and secular level. Michael H. Hart The 100: A Ranking of the Most Inßuential Persons in History. New York: Hart Publishing Company, Inc. 1978, p. 33. ---- ------- If greatness of purpose, smallness of means, and astounding results are the three criteria of human genius, who could dare to compare any great man in modern history with Muhammad? The most famous men crea arms, laws and empires only. They founded, if anything at all, no more than material powers which often crumbled away before their eyes. This man moved not only armies, legislations, empires, peoples and dynasties, but millions of men in one-third of the then inhabi world; and more than that, he moved the altathe gods, the religions, the ideas, the beliefs and souls. . . his forbearance in victory, his ambition, which was entirely devo to one idea and in no manner striving for an empire; his endless prayehis mystic conversations with God, his death and his triumph after death; all these attest not to an imposture but to a firm conviction which gave him the power to restore a dogma. This dogma was twofold, the unity of God and the immateriality of God; the former telling what God is, the latter telling what God is not; the one overthrowing false gods with the sword, the other starting an idea with words. osopher, orator, apostle, legislator, warrior, conqueror of ideas, restorer of rational dogmas, of a cult without images; the founder of twenty terrestrial empires and of one spiritual empire, that is Muhammad. As regards all standards by which human greatness may be measured, we may well ask, is there any man greater than he? Lamartine, HISTOIRE DE LA TURQUIE, Paris, 1854, Vol. II, pp. 276-277. ---- ------- What nonMuslim say about Muhhamed http://cyberistan.org/islamic/quote1.html Tolerance in Islan by Pickthall, an English convert http://cyberistan.org/islamic/toleran1.html I apologized those at sci.math,sci.physics,sci.chem; this is my last post on this thread. === Subject: Re: I can't stand it anymore > I found it. > http://groups.google.com/groups?q=amanda+suttee&hl=en&lr=&ie= UTF- > 8&selm=9b4c6065.0303110536.7bb32a3f%40posting.google.com&rnum =1 I took a look and you got the posts written by I also noticed that you copied the stuff written by the guy whom I > referred to as Indian Hindu hate monger. All he did was coming up > with crap. I am sorry that you took him seriously. You should have got some books written by unbiased historians before > you decided to trash Islam and its Prophet, will you? There is not much difference in Islam and Christianity when it comes > to crap, crpas which were made by humans. For the story of Muhammed, I strongly suggest that you get a few books > written by real historians. > BTW, I agree with you about islam. Which part did you agree with? Are you sure that what you read was > written by me? > By the way, here is something you may want to read about Muhammed. He was Caesar and Pope in one; but he was Pope without Pope's > pretensions, Caesar without the legions of Caesar: without a standing > army, without a bodyguard, without a palace, without a fixed revenue; > if ever any man had the right to say that he ruled by the right > divine, it was Mohammed, for he had all the power without its > instruments and without its supports. ----- ---- MAHATMA GANDHI, speaking on the character of Muhammad, (pbuh) says in > (YOUNG > INDIA): I wan to know the best of one who holds today's undispu > sway > over the hearts of millions of mankind....I became more than convinced > that > it was not the sword that won a place for Islam in those days in the > scheme > of life. It was the rigid simplicity, the utter self-effacement of the > Prophet, the scrupulous regard for his pledges, his intense devotion > to this > friends and followehis intrepidity, his fearlessness, his absolute > trust > in God and in his own mission. These and not the sword carried > everything > before them and surmoun every obstacle. When I closed the 2nd > volume (of > the Prophet's biography), I was sorry there was not more for me to > read of > the great life. -------- My choice of Muhammad to lead the list of the world's most inßuential > persons may surprise some readers and may be questioned by othebut > he > was the only man in history who was supremely successful on both the > religious and secular level. Michael H. Hart > The 100: A Ranking of the Most Inßuential Persons in History. New > York: > Hart Publishing Company, Inc. 1978, p. 33. > ----- ------ If greatness of purpose, smallness of means, and astounding results > are the > three criteria of human genius, who could dare to compare any great > man in > modern history with Muhammad? The most famous men crea arms, laws > and > empires only. They founded, if anything at all, no more than material > powers > which often crumbled away before their eyes. This man moved not only > armies, > legislations, empires, peoples and dynasties, but millions of men in > one-third of the then inhabi world; and more than that, he moved > the > altathe gods, the religions, the ideas, the beliefs and souls. . . > his > forbearance in victory, his ambition, which was entirely devo to > one idea > and in no manner striving for an empire; his endless prayehis > mystic > conversations with God, his death and his triumph after death; all > these > attest not to an imposture but to a firm conviction which gave him the > power > to restore a dogma. This dogma was twofold, the unity of God and the > immateriality of God; the former telling what God is, the latter > telling > what God is not; the one overthrowing false gods with the sword, the > other > starting an idea with words. osopher, orator, apostle, legislator, warrior, conqueror of > ideas, > restorer of rational dogmas, of a cult without images; the founder of > twenty > terrestrial empires and of one spiritual empire, that is Muhammad. As > regards all standards by which human greatness may be measured, we may > well > ask, is there any man greater than he? > Lamartine, HISTOIRE DE LA TURQUIE, Paris, 1854, Vol. II, pp. 276-277. > ----- ------ What nonMuslim say about Muhhamed > http://cyberistan.org/islamic/quote1.html > Tolerance in Islan by Pickthall, an English convert > http://cyberistan.org/islamic/toleran1.html > I apologized those at sci.math,sci.physics,sci.chem; this is my last > post on this thread. I went to the link I provided (http://cyberistan.org/islamic/quote1.html) and as I read the quotes there, I have decided to post the following since Rich has copied and pos the crap written by an Indian hate monger who claimed to be an ex-Muslim. Th fact that Rich so easily copied his words and pos here made me realize that even in these science ngs, there are those who know nothing about Islam and Muhammed and yet ready to accept any negative statements from someone like that guy without a doubt. This is my last post as I do not want to make this thread any longer especially when it has completely gone out of context. Rich , If you have anything to say to me about my Asian Racism, etc., feel free to email me. Below is more about using the sword and the great man you referred to as a Phedoe, after rading the hatemongers' posts. Keep in mind that thoes hatemongers has no historians writing Biography about them but Muhammed has many. Note the name K. S. Ramakrishna Rao. It is an Indian Hindu name. - - - - - - - - - - - - - - - - - - - - - - - - - - - - K. S. Ramakrishna Rao in ÔMohammed: The Prophet of Islam,' 1989 My problem to write this monograph is easier, because we are not generally fed now on that (distor) kind of history and much time need not be spent on pointing out our misrepresentations of Islam. The theory of Islam and sword, for instance, is not heard now in any quarter worth the name. The principle of Islam that there is no compulsion in religion is well known. ---- --- Lawrence E. Browne in ëThe Prospects of Islam,' 1944 Incidentally these well-established facts dispose of the idea so widely fostered in Christian writings that the Muslims, wherever they went, forced people to accept Islam at the point of the sword. ----- James Michener in ëIslam: The Misunderstood Religion,' Reader's Digest, May 1955, pp. 68-70. No other religion in history spread so rapidly as Islam. The West has widely believed that this surge of religion was made possible by the sword. But no modern scholar accepts this idea, and the Qur'an is explicit in the support of the freedom of conscience. Like almost every major prophet before him, Muhammad áá. (I cut the rest). -- W. Montgomery Watt in ÔMuhammad at Mecca,' Oxford, 1953. His readiness to undergo persecution for his beliefs, the high moral character of the men who believed in him and looked up to him as a leader, and the greatness of his ultimate achievement - all argue his fundamental integrity. To suppose Muhammad an impostor raises more problems that it solves. Moreover, none of the great figures of history is so poorly apprecia in the West as Muhammad.... Thus, not merely must we credit Muhammad with essential honesty and integrity of purpose, if we are to understand him at all; if we are to correct the errors we have inheri from the past, we must not forget the conclusive proof is a much stricter requirement than a show of plausibility, and in a matter such as this only to be attained with difficulty. ------ Dr. William Draper in ÔHistory of Intellectual Development of Europe' Four years after the death of Justinian, A.D. 569, was born in Mecca, in Arabia, the man who, of all men, has exercised the greatest inßuence upon the human race... To be the religious head of many empires, to guide the daily life of one-third of the human race, may perhaps justify the title of a Messenger of God. --------- === Subject: Re: I can't stand it anymore > amanda replied: [...] >as they have been told by the 19th century European racists >>Is there an Asian country which is not racist? >>Whya re you talking abput cpuntries. Why not individuals? >>Cause it's not just individuals, it's a cultural thing. >>Why don't you talk about individual Americans? > You are the one who accused all Asian countries as racists. It was an observation. I stand by it. All? Even Thialand? > You do not understand how Western culture (as they see from the > movies) scare the parents in those countries. These governments feel obliga to deter the bad part of western culture. I'd say that it often goes much farther than this. Just like with every other group. > In doing so, > sometimes they go extreme. When the movie Flash Dance came out, any kid who twist their body would be taken away by the police, at least > for a moment. My cousin, about 13, was really good at it. his father > never knew that he could do that. If the father knew, he would > probably scold the son. >>Just look at the nice >>things high Japanese govt officials have said about Americans. >who believed that they were superior to all other humans. >>As do all Asian races currently. >>Currently is the keyword but you are wrong to generalize all Asians. >>So you say. > Nice that you are using what Mlaysia said as the ruler. Never mind > that Malaysian government a Muslim government. Show me any Muslims > who thinks western culture is suitable to their children. Actually, Malaysia is not your typical Muslim country, it's as good as > it gets. But Malaysia is about a quarter chinese. Chinese represent nearly a quarter of the population, and > Indians are some 8%, but provide 20% of the lawyers and 15% > of doctors. Any racially inßammatory news tends to be self > censored in the major media and there is a conscious > effort among all ethnic groups to work together. Free elections > have been held since 1969 and the standard of living, in 1960 > on the level of Haiti, now surpasses much of Eastern Europe > and Latin America. Education represents over 20% of the > national budget and tens of thousands of students are sent > abroad. Well, there was a time where there is a limit on how many properties Chinese can own. How about Japan? I never mentioned which particular Asian group are racist in my opinion and which one are not. Only that I do not agree that all Asians are racists. By the way Japan has a lof of foreigners from other Asian countries over staying there and work. If Japanese goovernment do not want *them* stay there illegally (to work), why don't they punish the employers who hire them? It's double standard. http://www.japantoday.com/e/?content=news&cat=9&id=278105 Kanagawa governor apologizes for calling foreigners Ôthieves' for calling foreigners sneaky thieves while campaigning over > the weekend for a candidate running in Sunday's House of > Representatives general election. My strong desire to maintain security led me to make the false, > inappropriate remarks. I apologize and take them back, the > governor said. There are some foreigners who come to Japan > with student visas but end up overstaying illegally and commit > serious crimes. I want to control such crimes. The Kanagawa prefectural government received more than 100 > emails after the speech both criticizing and supporting the > remarks, Matsuzawa said. (Kyodo News) Or China? Chinese in general is one group that I now *claim* who think they are superior to all races. http://tibet.dharmakara.net/ictracism.html Racism: China's Secret Scourge ICT Report Refutes Beijing's Denial of Racism in China Washington, D. C. - International Campaign for Tibet (ICT) > will release a report at the UN World Conference Against > Racism (WCAR) documenting the origin and nature of racism > against Tibetans and how the Chinese government perpetuates > racist attitudes and policies. The 60- page report, entitled Jampa: The Story of Racism > in Tibet, describes how racist language and concepts permeate > China's constitution, laws and policy and how this has > contribu to the racism and discrimination Tibetans face > today. It is the first comprehensive analysis of this > phenomenon, a subject that has not been widely addressed by > scholahuman rights groups and others who generally focus > on more conventional human rights violations in Tibet. While highlighting racism in the west, China has effectively > suppressed racism as a domestic issue. This is their shameful > secret, said Tsering Jampa, Director of International Campaign > for Tibet- Europe. In the months leading up to the World Conference on Racism, > China has portrayed racism as a Western phenomenon that does > not exist in China. In a February 2001 submission to the UN, > China sta that all ethnic groups are living in harmony > in China. The Chinese government's denial that racism is a significant > problem in China is a policy which prevents Tibetans and others > from addressing racism in meaningful, constructive ways, said > John Ackerly, President of ICT. The title of the report, Jampa, refers to the protagonist > of a ubiquitous 1963 Communist Party propaganda film depicting > Tibetans as a backward people who can only be uplif by the > civilizing force of the Chinese. All Tibetans live under the shadow of this film, said Tsering > Jampa. The Chinese government has used it to denigrate Tibetan > culture and justify its occupation of Tibet. At the conference ICT will urge the government of China to > acknowledge the extent of the problem and to remove derogatory, > chauvinist or paternalistic language from laws and policy > statements. ICT is also urging Chinese non-governmental > organizations based in the west to work with Tibetan groups > on educational programs and initiatives to help combat this > long-standing problem. Although China tried to block the accreditation of Tibetan > human rights groups to the World Conference against Racism > a vote by UN member counties approved accreditation for > ICT and one other Tibetan organization. ICT has invi also Xiao Qiang, Director of Human Rights > in China, whose organization was not accredi, to join > its delegation to the conference. But then, you probably don't care. Your problem is that you assume too much. But you use probably and so I will give you a break. >Asian racism is probably not a > problem you face. Again, you assume TOO much. >> http://news.bbc.co.uk/1/hi/entertainment/showbiz/2582833.stm >> ÔAnti-Asian' Brad Pitt advert banned >> A car advert featuring Brad Pitt has been banned in Malaysia >> because it is an insult to Asians, according to reports >> in the country. [...] >My saying that Asian do not need a test inven by someone else >>See? You're into Ôus' and Ôthem'. Obviously for you Ôus' is Asians, an > d >>'them' is everyone else. > Again, another silly accustaions that I am into us and them. It's what you said. You took it out of context, i.e I didn't not say Us were superior to them. It was merely as reference to groups > Beside, we both know who inven IQ tests. Tell me who. C'mmon. > So don't make it look like I meant everyone else when Is aid them. That's what them means in this context. The context was when the person from that group accused me of not doing well on the IQ test and so my using them* refers to the group he belong to and not everyone else. >>And you find Ôthem' wanting in the same fashion >>as Zainuddin Maidin above. > Pretty bold accusation. You think you know but you do not know how > Asians see western culture very well. I think that they see it much as Ghandi did. Do you prefer that they say with enthusiasm, we should adopt it? Reporter to Mahatma Ghandi: Mr. Ghandi, what do you think > of Western Civilization? > Ghandi: I think it would be a good idea. >If you can't understand that, that's your problem. >>That's the problem. I see what you actually write and object to, and >>it's no different from the stuff others post. You find things wanting >>just because they are not Asian in origin. You can't see it, I underst > and >>that. > Whatever? I do not know how you concluded that I belive the Asians are superior to anyone else. You seem to be upset that I implied that > Asians are superior to you. May eb you had have some experience with > soem Asians acting like that. Like it's a rare and unexpec thing? So you did have some experience. And I got picked on because of that? But it depends on which Asians you met and where. In US, most Asians would act snobbish. > Heck..I have met Chinese who claimed to be superior to anyone else. You talk like there are Chinese who don't believe this. Yes there are. I mean there are truly Intelligent Chinese. > In fact, one time, when this friend of mine said that America is doing well because of its Chinese population. Immediately after I smiled > and replied I don't think so, I realized that that moment was the > beginning of the end of our friendship. Interesting. > And she wasn't the first Chinese I met who put down on Amercans > (talking about white Americans). Everytime, I have defended from the > Americans' side. For what? To be accused by people like you that I > think white American are inferior to Asians. Very nice! Oddly, I've not seen anything from you that looks like a defense of > white Americans. Why should I be defending for them when I didn't say that they are inferior. Oh .. I see. To you defending white Americans would be to say that they are superior. >>Why don't you also complain about Asian racism? >> Why should I complain about it here? >>Because I'm asking you to, just to show that you care about racism, no > t >>just Asians. But I wasn't talking about racism. I was talking about the miuse of term Asian. > If I didn't care about racism, I would not have star this thread. As I said, there is a difference between objecting to racism against > *anyone* and racism against the group you happen to belong to. Few will > actually do the first, no matter what they claim. So you think because I am Asian, I won't point out racism against among Asian group? Why are you insistent on that? Like I said, I was after the term Asian. I didn't initiate this thread to talk about Asian racism. > Why are you clueless about that? Just said it again. Maybe you'll catch it this time. Maybe not. Again ... I was after the term Asian. I didn't initiate this thread to talk about Asian racism. >I have complained many times about Indian caste system somewhere else > . >>Elsewhere? What are your complaints? > Long time ago - it was soon after 9/11 - I happened to go into an > egpytian group thinking I would learn more about the ancient egyptian > culture. Instead, some Indian Hindu hate mongers were trashing the > Muslims (mainly Pakistanis) pickign on their religion and treatment on women. So I borught up suttee, caste system, and treatment of women > in their society. Caste systen has been abolished but has it really > gone? I told them to clean up their hosue before picking on others. I didn't use the same user name. I found it. http://groups.google.com/groups?q=amanda+suttee&hl=en&lr=&ie= UTF- > 8&selm=9b4c6065.0303110536.7bb32a3f%40posting.google.com&rnum =1 > All thanks to Islamic inßuence. > It's always someone else's fault, isn't it? Just shows how ill-informed persone you really are. I suggest you > read history of your religion of Ôpeace' aka Islam, What made you say that my religion is Islam? >and what it did > when when it ransacked ßourishing cultures and civilization. > Islamic doctrine is like Nazisms, which it more than matches in > fanaticism and fascism, especially when it come to lying and > distorting history. Not surprised. There are many variations (of books) on that subject. It's up to the personw which view to hold. I know which one you chose. By the way, tell me which Muslim armies went to what is now Malaysia and Indenoesia and ransacked and forced conversion? In other few hundred years Nazis will also say that Hitler had a > profoundly > positive and spiritual inßuence on barbaric jews. Just as shameless > retards > are denying Islamic butchery and barbarity of over the past millenia. [...] > Rare..amusing. Forogt suttee too? Suttee was common among Rajput wives of warriors who did not want to > be captured by invading Muslims. Yeah..right. You are so misinformed. Ever watched the movie Far Pavillion? Based on a true story. Was she being chased byt he Muslims she wan suttee? > The number of women who commit suttee > among 0.8 billion Hindus is around one every few decade. The fact is that it exis. >It voluntary > and even then against the law. Namely, it was voluntary. >Now compare this with brutal, but > legally enforced murder of thousands of Muslim women for committing > adultery. The latter is sanctioned by Allah, No it was not. It's human who did that. > and Pakistan, > Afghanistan, Iran, Saudis implement it. Pakistan, Afhganistan, And Saudi Arabia are not the place where most Muslims are. You seem to be very ill-informed about Muslim and Islam. > Limb chopping being other favorite pastime of pedoe prophets' > retarded followers. Oh..completely misinformed about him too. I suggest you get better sources for you info. [...] > What percentage of Muslims women are covered in black tent? In Islamic paradises like Afghanistan, Saudi Arabia, Iran it is close > 100%. > If the women dont obey they are beaten mercilessly. > married at like cattle at any age > How old were the Hindu girls when sold by their parents to rapists? You are thinking of many cases of Indian Muslims who sell their > preteen daughters to dirty old Arabs, especially in Hyderabad. The keyword is Indian Muslims. They have carried on their pre-Islamic culture into their current lives. > They > are rescued by Hindu police. Where were that Hindu police when the Hiundus were setting fires on Muslims, including babies and pregnanat women, using gasoline. It happend just not too long ago. Were you out of this planet at the time? [...] > An ignorant hatemonger calling one of the greatest human being in > history a pedoe doesn't count. Get that into your tick head. Just like Hitler is called the greatest human being in history by > Nazis. Pol Pot by his own follwer. Just because he butchered people > and had temporal lobe epilepsy with paranoid psychosis and delusions, > does not mean he was great. Today he would be in lunatic asylum, where > he belonged. There would have been no islamic mass murderers like Bin > Ladens, Gaznavis and Timmurs. Regarding his pedoia; truth always hurts. Were you raised as Baptist. If so, it is very understanding that you ahve been given misinformation. > Did he not marry Ayesha > when she was 6 year old and he 54 yr old, and had sex with her when > she was only 9 yr old? He married here to make peace with that group. He did not consummate the marraige until she was 13, which was normal for the culture of the time. >Did he not marry young wife of his own adop > son? Adop son is not his son. >Did he not marry a Jewish women he captured after slaughtering > her entire family and 700 plus surrendered jewish men? No he did not. Find books not written by the Baptists and Christaain propagandists, will you? He even gave instruction on how to divide booty from loot. Step in the time and see things form the perspective of the time, will you? ALso, do not forget how he pardoned the enemies and made peace with them. BTW, it looks like bride burning is today's problem. What say you about this? Oh did you just get back form outer space? No one knows how suttee star, or for that matter where or when. > It's not unique to India--widow suicide is known to have occurred > among the Egyptians, Chinese, Vikings, and others. Some say its > origin on the subcontinent dates back 5,500 yeawhile others > believe it arrived much later, around 1 AD. I've heard Indians > deny there's anything specifically Hindu about it, in that it > doesn't figure in Hinduism's core texts. Today it's most closely > associa with remote villages domina by the Rajputs. It was the British who outlawed the suttee. India was the only country who practised it as long as they did. Suttee is different from bride burning, in which a newly married > Indian woman is burned to death by her in-laws for failing to meet > demands for a larger dowry, the traditional gift given to the > couple by the bride's parents. Thousands of such murders have been > repor. Why are you telling me this as if I don't know? > In contrast, suttee is a voluntary act, theoretically at > least, Yes. theorectical ONLY. >meant to atone for the couple's sins and ensure their > reunion in the afterlife. Failed to consider the politics behind it. > But horrified Indian feminists say > that in practice the suttee victim often had little choice. They were not horrified faminists. They are humane human. > Sometimes family membeincluding other women, browbeat her > into it; sometimes she was bound or hopped up on drugs. Much > of the time even that wasn't enough. It's said music was > played at high volume during suttee so no one could hear the > widow's screams. Seems you're as down on Hindu culture as you > are on Muslim culture. No..I am down on the abusive practises of people (human). I'm having a difficult time reconciling the claims about the frequency of suttee with the last accounts of how it occurred in practice. The last official one was when the British outlawed it but it continued to happen .. > Nontheless, Hindu culture hardly seems blemish free and I disagree that you have to > be perfect to point out faults you see in others. BTW, I agree with you about islam. No surprise there. You are missing out a lot on Islam. Now, do not get confused thinking that I am praising the Arabs. >As a minority in the country I was born (my parent were born and >raised thare too) who suffered discrimination and were not considered as full-citizens, you must be joking accusing me as a racist. >>What does that have to do with whether you are racist or not? > Because it was due to race that we were discrimina. The issue is whether *you* are racist, not others. The issue which I came to understand PERFECTLY now is that you want to me to say Asians (including I am) are racists. >It is very clear > that you think you are saying that since you were discrimina against > you cannot be racist. You twis what I said. I replied that it was due to race that we were discrimin. Now you are using it to attack me. >>I don't think you really understand the concept. > You were the one who didn't understand what I said. I do understand it, you still don't. >>It's like saying you are black, you can't be racist. >I got news for ya, if you can say that, you are. > I got news for you. You underestima me in assuming that I think > black cannot be racists. But you think that since *you* were discrimina against, that you > cannot be racist. What's the difference? I repeat..You twis what I said. I replied that it was due to race that we were discrimin. Now you are using it to attack me. >Beside, as a person whose blood covers basically almost all races of >the world (not through Caucasoid and Negroid per se but through >Indo-Aryan which you probably have no clue about), including semite >(not jewish), you gotta be kidding to accuse me as a racist. >>You've gotta be kidding if you think this denies that you can be >>a racist. >>It's clear that you will profit from any pro-Asian racism. >> What are you talking about? >>Think Affirmative Racism. Think Ôdiversity'. I was/am going with the second one, WHICH WAS MY WHOLE POINT OF >criticizing the use of the term Asian in IQ test, duh... >>That is was not inven by an Asian. That was your objection, how >>dare a non-Asian even imply that they can say anything about an Asian. My objection was that I as an Asian did not need to run and take the > IQ test to find out my intelligence level when, I repeat WHEN *that* > guy told me that I must not have done well blah..blah..blah and ehnce > against the IQ tests. I disagree. Your objection was, and I quote... Don't you realize that us Asians don't need the tests inven > by non-Asians to find out whether we are intellgent or not? Your objection was that the teste were inven by non-Asians. Presumab > ly > had the tests been inven by Asians this would be a non-issue, as the > tests would be fine. Presumably is the keyword. >>It's almost certain that you have. >>The most stupid thing I have ever heard. What I have had was getting pulled over by the red-neck police >>Racism. > Red neck is red neck. A racist term is a racist term. Simple as that. Oh..now you want me to be politcally correct? > Simple as that. Here in CA, I approached a > police car once to ask for some direction. The first things he said, > with a very pleasant manner was what can I do for you?. In > Houston, it would be like What's your problem?. Another time, as > the police blocked the route to my place, I asked the police standing > there what I should do and he was really nice. In Houston, it would be like I am busy; get lost. Seems you are comparing your experiences in CA with your prejudices > about Houston. I am comparing with my experiences with the red necks from Houston. > I even told that (that most of them are red neck) to a police officer > once. It was 11 pm at night, not on the highway but he happened to be really closed by and I didn't see his car because I wan to get home which was not too far and did not slow down as he expec. I was the > only car around there. He told you I am busy; get lost? No he did not. One did give that kind of impression once. (In words, it would be I am busy; get lost. ) When he saw my disappointment on my face, he immediately said I am sorry. By then, I already got my car window rolling up. > I told him to just give me the ticket and spare the lecture. He > asked me why I talked like that. I told him that I haven't met a > police officer who did not give me a ticket. I told him that most are > red necks. He just laughed. May be he didn't get mad because I was not far from the campus and he knew that I was a student. (I did tell him > that I was tired and didn't see him when he asked why I didn't slow > down after seeing him). Or may be I looked cute to him (he wasn't very old) when i said to him that most of them are red necks. Who knows? As he gave me the ticket, he said that he would not have given me the > ticket if I didn't act the way I did. I replied - silently - yeah, > right! So you have no issues with stereotyping Americans. No... not Americans. Only red necks, i.e red necks are (many) individuals. >Odd that you have > issues with stereotyping Asians. Why is this? I do not stereotype any group based on race or ethnicity. >in Houston for driving a beat-up >car in my student days because I look a bit Hispanic or South >Americans. >>Odd, I was always pulled over by the cops when I had my old 1965 >>Mustang convertible. Racism obviously, right? > In my case, there were other cars speeding but they didn't get > stopped. I'm not asking about your case. When you got pulled over, it was for your old Buick? When I got pulled over, it was for speeding but he didn't stop the SUV next to me. Why? because it was drven by a white? >>Rich >>Note: I will not be wasting any more time defending groundless >accusations. >>You've already proven that they are not groundless. > Whatever. Ahh, the superior Asian IQ rears it's ugly head. I will not be wasting any more time defending groundless accusations. Rich >>Rich > Good to know that I was right that you had some experience with some Asians acting superior to you. I guess it hurts since you expec that you would be exempt form it for being white, huh? Sorry for that experience though. You may be surprised that I almost got that experience too (I may have gotten it somewhere without knowing it) until that Chinese girl realized that way back in my ethnic group, there was Mongoloid mix. Her comment was this: You are smart because of the Mongolid blood. Well..I just wan to remind you that initia this thread to express my frustration about the term used in IQ tests in grouping people. So, here is the list of my suggestions: - Get over your hurt witht he Chinese and stop expecting me to say whites are superior to Asians (inclduign Chinese) . - Stop accusing me that I think Asians are superior to white. - Stop expecting me to say that all Asians are racists. - Stop assuming that my being Asian has benefit me. And - Do not twist my sentences. === Subject: Re: I can't stand it anymore > amanda replied: > [...] >as they have been told by the 19th century European racists >>Is there an Asian country which is not racist? >>Whya re you talking abput cpuntries. Why not individuals? >>Cause it's not just individuals, it's a cultural thing. >>Why don't you talk about individual Americans? > You are the one who accused all Asian countries as racists. > It was an observation. I stand by it. All? Even Thialand? Do you mean Thailand? http://www.eslcafe.com/jobinfo/asia/sefer.cgi?display: 1059593233-26315.txt http://www.eslcafe.com/jobinfo/asia/sefer.cgi?display: 1050508179-60192.txt http://www.eslcafe.com/jobinfo/asia/sefer.cgi?display: 978838545-13729.txt I'm betting the last guy is white. I've been to China a couple of times. People trea me well but I overheard someone saying that she couldn't imagine how someone could date Western men because they were hairy animals. People stared openly at me when I walked around a small industrial city a couple hours outside of Guangzhou and I was goofed on openly by a cab driver in Jilin provence. I shot a handgun at targets at an amusement park (much different than western amusement parks) I visi. A lot of people crowded around to watch. Everyone is racist to some extent. I've been called white boy within the past week while walking around in my neighborhood. Specifically, the phrase was I'm going to kick your ass white boy. A black guy yelled it out the window of a passing car. I wondered if he was one of the two guys I fought off when I was mugged (attemp anyway, I didn't have any cash) last year. An ex-gf of mine grew (from my Montreal days) grew up in Quebec City. She was of Chinese descent. Her mother was from the mountains and was tall; her father was from Hong Kong and was short. She's about 5'9. The Quebec City chinese community is predominantly smaller from what she says. They ostracized her because she was large. They told her she was part white. She's gotten crap from whites also. Once while trying to get back into Canada after visiting Niagra Falls, NY a border gaurd kept asking her for her passport. She showed him her drivers license and talked to him in French but he wouldn't believe she was Canadian. My mother is anti-semetic. She was raised Catholic but is of Jewish descent. Go figure. I've discovered (admit to myself) that I've got some deep sea racial bias. Ah well. === Subject: Re: I can't stand it anymore Amanda, If you are interes in learning more about the history and this topic called The Mismeasure of Man. Since the inception of standardized IQ tests almost 100 years ago, there has been a steady current of scientific opposition who sees the limitations and misuses of these tests. It's an interesting topic. === Subject: Re: I can't stand it anymore > Amanda, If you are interes in learning more about the history and > this topic called The Mismeasure of Man. Since the inception of > standardized IQ tests almost 100 years ago, there has been a steady > current of scientific opposition who sees the limitations and misuses > of these tests. It's an interesting topic. I was a ware of scientific opposition (on limitations and misuse) by hearing about it here and there. I was/am interes in finding out the limitations but never got around to really bother to find info on it. Thanks for the info. I will definitely get Gould's book. === Subject: Re: I can't stand it anymore > Amanda, If you are interes in learning more about the history and > this topic called The Mismeasure of Man. Since the inception of > standardized IQ tests almost 100 years ago, there has been a steady > current of scientific opposition who sees the limitations and misuses > of these tests. It's an interesting topic. I was a ware of scientific opposition (on limitations and misuse) by > hearing about it here and there. I was/am interes in finding out > the limitations but never got around to really bother to find info on > it. You're worked up enough about it to c post your whining complaints about the evils of IQ tests (or was it the evils of just talking about IQ tests) to three news groups, but you never bothered to find info on it beforehand. Amazing. === Subject: Re: I can't stand it anymore > Amanda, If you are interes in learning more about the history and > this topic called The Mismeasure of Man. Since the inception of > standardized IQ tests almost 100 years ago, there has been a steady > current of scientific opposition who sees the limitations and misuses > of these tests. It's an interesting topic. I was a ware of scientific opposition (on limitations and misuse) by > hearing about it here and there. I was/am interes in finding out > the limitations but never got around to really bother to find info on > it. You're worked up enough about it to c post your > whining complaints about the evils of IQ tests (or > was it the evils of just talking about IQ tests) to > three news groups, but you never bothered to find info > on it beforehand. Amazing. Are you TOO DUMB not to understand that not bothering to find out in detail (referring to do as a research work) did not mean I knew nothing about the limitations, especially after I said I was aware of the ..? === Subject: Re: I can't stand it anymore Better yet, why not refrain from making them? === Subject: Re: I can't stand it anymore >Better yet, why not refrain from making them? I had to sit on my hands to not say that. It was a useful post. For some strange reason, a high IQ rating without the training to use it appears to be desired. /BAH Subtract a hundred and four for e-mail. === Subject: Logarithm of Stirling numbers of the second kind I have a problem in which I need to calculate the logarithm of Stirling numbers of the second kind for large n (in the range 1 to thousands and preferably millions) and m in the same range. Due to the large range for n and m I prefer not to work with the Stirling numbers of the second kind directly, as my working precision is not that great. Are there any methods that calculate the logarithm of Stirling numbers of the second kind, or good approximations? An approximation with errors (of the natural logarithm of the Stirling number of the second kind) up to +-10 would still be useful for me, although the more accurate the better of course. John Reidar Mathiassen === Subject: An urgent question for William Elliot about a reference. Dear Mr. Elliot: Thanks for helping me proving that the edge connectivity of a complete graph equals n-1. I need to write a reference for that proof. Was it yours so I write your name as a reference or there was another reference like a book for example? Please let me know as soon as you can. Thanks. Subject: Re: An urgent question for William Elliot about a reference. === > Dear Mr. Elliot: Thanks for helping me proving that the edge > connectivity of a complete graph equals n-1. > I need to write a reference for that proof. Was it yours so I write > your name as a reference or there was another reference like a book > for example? > Please let me know as soon as you can. Thanks. > You're welcome and thanks for the compliment. For me, edge connectivity is a new concept which I've not see except at sci.math. Proof came of my efforts and not from a text. Uncomplica as it was, is it not already know? Surely the theorem is well known. For why you ask? Are you writing a thesis? === Subject: [Set Theory] A set has a disjoint copy of itself ? I am looking for a proof that any set has a disjoint copy of itself, i.e. that given a set A, there exists a set B with the same cardinality as A, and such that A/B=0. I am hoping this is true with ZF or ZFC. Noel. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? > I am looking for a proof that any set has a disjoint copy of itself, > i.e. that given a set A, there exists a set B with the same > cardinality as A, and such that A/B=0. I am hoping this is true with > ZF or ZFC. If you don't mind using the Axiom of Foundation (Regularity), you could just take B = {{a,A}: a in A}. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? > If you don't mind using the Axiom of Foundation (Regularity), you > could just take B = {{a,A}: a in A}. It took me a while to work that one out For all a in A, a in {a,A}/A. Hence from the axiom of foundation, I obtain that {a,A}/a=0 for all a in A. It follows that if {a,A} in A for some a, I have {{a,A},A}/{a,A}=0 which is a contradiction. So B/A=0. Thanks very much. Noel. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? Adjunct Assistant Professor at the University of Montana. I am looking for a proof that any set has a disjoint copy of itself, >i.e. that given a set A, there exists a set B with the same >cardinality as A, and such that A/B=0. I am hoping this is true with >ZF or ZFC. Just paint it a color. Given any set A, there exists a set B such that B is not an element of A. Take Ax{B}. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? Adjunct Assistant Professor at the University of Montana. >>I am looking for a proof that any set has a disjoint copy of itself, >>i.e. that given a set A, there exists a set B with the same >>cardinality as A, and such that A/B=0. I am hoping this is true with >>ZF or ZFC. Just paint it a color. Given any set A, there exists a set B such >that B is not an element of A. Take Ax{B}. Yuck. This doesn't quite work, as has been poin out elsewhere. But yes, there is a way to do it in ZF, by choosing an appropriate set B and taking A x {B}. B can be construc explicitly, so there is no need to invoke Choice. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? I am looking for a proof that any set has a disjoint copy of itself, >i.e. that given a set A, there exists a set B with the same >cardinality as A, and such that A/B=0. I am hoping this is true with >ZF or ZFC. Noel. B = {0} x A -- === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? >I am looking for a proof that any set has a disjoint copy of itself, >>i.e. that given a set A, there exists a set B with the same >>cardinality as A, and such that A/B=0. I am hoping this is true with >>ZF or ZFC. >>Noel. >B = {0} x A That is B = {(0,a), a in A}. What if A = {1, (0,1)} ? -- === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? >I am looking for a proof that any set has a disjoint copy of itself, >i.e. that given a set A, there exists a set B with the same >cardinality as A, and such that A/B=0. I am hoping this is true with >ZF or ZFC. >>Noel. B = {0} x A > >That is B = {(0,a), a in A}. What if A = {1, (0,1)} ? > Oops. You're right. Must be more careful. I trust Herman's response. -- === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? I am looking for a proof that any set has a disjoint copy of itself, >> i.e. that given a set A, there exists a set B with the same >> cardinality as A, and such that A/B=0. I am hoping this is true with >> ZF or ZFC. >> Noel. >> B = {0} x A > That is B = {(0,a), a in A}. >> What if >> A = {1, (0,1)} ? >Oops. You're right. Must be more careful. I trust Herman's response. Or let y be the least ordinal such that for all a in A, (y,a) is not in A. Then set B = {y} x A. If you include the axiom of foundation in ZF, can't we conclude that {A} x A is disjoint from A, or must we be a bit more careful? How about {a U {A} : a in A}? -- === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? >> I am looking for a proof that any set has a disjoint copy of itself, > i.e. that given a set A, there exists a set B with the same > cardinality as A, and such that A/B=0. I am hoping this is true with > ZF or ZFC. >> Noel. > B = {0} x A > That is B = {(0,a), a in A}. >> What if >> A = {1, (0,1)} ? > Oops. You're right. Must be more careful. I trust Herman's response. Or let y be the least ordinal such that for all a in A, (y,a) is > not in A. Then set B = {y} x A. If you include the axiom of foundation in ZF, can't we conclude that > {A} x A is disjoint from A, or must we be a bit more careful? How > about {a U {A} : a in A}? One doesn't need anything as gruesome as foundation. Going back to your original construction: although {1} x A may not be disjoint from A, {1} x A and {2} x A are two disjoint sets each with the cardinality of A. Extending this ideal let B be a set of cardinality > card(A) (e.g., B = P(A)). Then B x A has cardinality > card)A) and B x A contains subsets {b} x A (for b in B) each of cardinality card(A). There must be some b in B for which {b} x A is disjoint from A. -- === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? >I am looking for a proof that any set has a disjoint copy of itself, >i.e. that given a set A, there exists a set B with the same >cardinality as A, and such that A/B=0. I am hoping this is true with >ZF or ZFC. Let Q_0 = A, Q_{n=1} = U(Q_n), the union of the elements of Q_n, and let R be the union of all the Q's. Then if c is any set not in R, {: x in A} will do. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? > Let Q_0 = A, Q_{n=1} = U(Q_n), the union of the elements of > Q_n, and let R be the union of all the Q's. Then if c is > any set not in R, {: x in A} will do. Thank you very much. Just to make sur I understand correctly, assuming the ordered pair is defined as ={{c},{c,x}}, the proof would work just as well by taking R= Q_2 and c not in R ? Noel. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? I am looking for a proof that any set has a disjoint copy of itself, >i.e. that given a set A, there exists a set B with the same >cardinality as A, and such that A/B=0. I am hoping this is true with >ZF or ZFC. Let Q_0 = A, Q_{n=1} = U(Q_n), the union of the elements of > Q_n, and let R be the union of all the Q's. Then if c is > any set not in R, {: x in A} will do. Here is a slightly simpler version of your construction, which does not use the Axiom of Replacement. Let U be the set of all elements of elements of A (your Q_1). Let c be any set not in U, e.g., let c = {x in U: x is not an element of x). Let B = {{a,c}: a in A}. Then B is disjoint from A, and you get a bijection f:A --> B by defining f(a) = {a,c}. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? Here is a slightly simpler version of your construction, which does > not use the Axiom of Replacement. Let U be the set of all elements of elements of A (your Q_1). Let c be > any set not in U, e.g., let c = {x in U: x is not an element of x). > Let B = {{a,c}: a in A}. Then B is disjoint from A, and you get a > bijection f:A --> B by defining f(a) = {a,c}. This is great. Thank you very much. Noel. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? I am looking for a proof that any set has a disjoint copy of itself, > i.e. that given a set A, there exists a set B with the same > cardinality as A, and such that A/B=0. I am hoping this is true with > ZF or ZFC. Sure, if there are no ordinal numbers in A, then take the the the set {0,1,2,...} that has the same cardinality. If there is an ordinal number. Do the same thing, but start counting from the successor of A's highest ordinal. Noel. === Subject: Re: [Set Theory] A set has a disjoint copy of itself ? === > Do the same thing, but start counting from the successor of A's highest > ordinal. Which is a problem, if A contains no highest cardinal.... On the other hand, if A really is a set, then the ordinals in A are bounded, so you can just start off at any upper bound... I'm not sure if this is better, but here's (IMHO easier) another way to see it: If A has cardinality k, take the next cardinal number, let's call it k+1 (which is in particular a set). Then k+1A has cardinality k+1, so take any subset of k+1A of cardinality k. That's disjoint to A and has the same cardinality -> voila :) Cheers ipp === Subject: Re: Riemann's Zeta Function by H. M. Edwards > It seems to me that proving the functional equation for the Riemann Zeta > function is easier without using complex analysis at all. The proof that > I referenced earlier basically does it as follows: > Let Psi(t) = sum from n=1 to infinity of exp(-pi n^2 t). > Let R(s) = Integral from t=0 to infinity of Psi(t) t^{s/2} dt/t > Let Q(s) = 1/s + Integral from t=0 to 1 of Psi(t) t^{s/2} dt/t > Then you can prove (using Fourier transforms) that > R(s) = Q(s) + Q(1-s) > It immediately follows that R(s) = R(1-s). On the other hand, integrating > the power series for Psi(t) term by term gives > R(s) = sum from n=1 to infinity of pi^{-s/2} n^{-s} Gamma(s/2) > = pi^{-s/2) Zeta(s) Gamma(s/2) > So from R(s) = R(1-s), you get > pi^{-s/2) Zeta(s) Gamma(s/2) = pi^{-(1-s)/2) Zeta(1-s) Gamma((1-s)/2) > Of course, now that I think about it, this proof is a little > suspicious, because it uses a definition of the Zeta function > that is only valid for s > 1 (the power series representation), > while the result necessarily involves Zeta(s) for s < 1. This is Riemann's second proof. Nothing suspicious about it! > The point is, that Q(s) is an entire function of s and so > the formula R(s) = pi^{-s/2) zeta(s) Gamma(s/2) > defines zeta as a meromorphic function on C. As a matter of fact, I am studying the second proof. It's on page 15. This one I can follow much easier. The only line I don't follow is the following (where Theta(x)= Sum_1^inf exp(-n^2pi x)) -Int_inf^1 Theta(1/x)x^(-s/2)dx/x is made to equal Int_1^inf [x^(1/2)Theta(x)+x^(1/2)/2-1/2]x^(-s/2) dx/x I get the change of signs, but I don't understand the expansion in the brackets []. === Subject: Re: Riemann's Zeta Function by H. M. Edwards > It seems to me that proving the functional equation for the Riemann Zeta > function is easier without using complex analysis at all. The proof that > I referenced earlier basically does it as follows: > Let Psi(t) = sum from n=1 to infinity of exp(-pi n^2 t). > Let R(s) = Integral from t=0 to infinity of Psi(t) t^{s/2} dt/t > Let Q(s) = 1/s + Integral from t=0 to 1 of Psi(t) t^{s/2} dt/t > Then you can prove (using Fourier transforms) that > R(s) = Q(s) + Q(1-s) > It immediately follows that R(s) = R(1-s). On the other hand, integrating > the power series for Psi(t) term by term gives > R(s) = sum from n=1 to infinity of pi^{-s/2} n^{-s} Gamma(s/2) > = pi^{-s/2) Zeta(s) Gamma(s/2) > So from R(s) = R(1-s), you get > pi^{-s/2) Zeta(s) Gamma(s/2) = pi^{-(1-s)/2) Zeta(1-s) Gamma((1-s)/2) > Of course, now that I think about it, this proof is a little > suspicious, because it uses a definition of the Zeta function > that is only valid for s > 1 (the power series representation), > while the result necessarily involves Zeta(s) for s < 1. > This is Riemann's second proof. Nothing suspicious about it! > The point is, that Q(s) is an entire function of s and so > the formula R(s) = pi^{-s/2) zeta(s) Gamma(s/2) > defines zeta as a meromorphic function on C. As a matter of fact, I am studying the second proof. It's on page 15. > This one I can follow much easier. The only line I don't follow is the > following (where Theta(x)= Sum_1^inf exp((-n^2)pi x)) -Int_inf^1 (Theta(1/x))x^(-s/2)dx/x is made to equal > Int_1^inf [x^(1/2)(Theta(x))+(x^(1/2)/2)-(1/2)]x^(-s/2) dx/x I get the change of signs, but I don't understand the expansion in the > brackets. > [ ]. (I fixed the Ughs) === Subject: Re: Riemann's Zeta Function by H. M. Edwards PaulHjelmstad says... >(where Theta(x)= Sum_1^inf exp(-n^2pi x)) -Int_inf^1 Theta(1/x)x^(-s/2)dx/x is made to equal >Int_1^inf [x^(1/2)Theta(x)+x^(1/2)/2-1/2]x^(-s/2) dx/x I get the change of signs, but I don't understand the expansion in the >brackets >[]. This uses the fact 1. Theta(1/x) = [x^{1/2} Theta(x) + x^{1/2} - 1/2] The proof of this fact is via Fourier transforms. In general, if f(t) is a suitably well-behaved function, and we define the transform F(k) as F(k) = integral from t=-infinity to +infinity of f(t) exp(-i 2 pi k t) dt then 2. sum from n = -infinity to +infinity f(n) = sum from n = -infinity to +infinity F(n) So, let f_x(t) = exp(-pi t^2 x) (treating x as a parameter). Then 3. integral from t=-infinity to +infinity of exp(-pi t^2 x) exp(-i 2 pi k t) dt = x^{-1/2} exp(-pi k^2/x) So F_x(k) = x^{-1/2} exp(-pi k^2/x). Therefore, F_x(n) = x^{-1/2} f_{1/x}(n). Therefore, 4. sum from n = -infinity to +infinity f_x(n) = x^{-1/2} sum from n = -infinity to +infinity f_{1/x}(n) Since f_x(n) = f_x(-n), and f_x(0) = 1, we can write this as 5. 1 + 2 * sum from n=1 to infinity of f_x(n) = x^{-1/2} [1 + 2 * sum from n=1 to infinity of f_{1/x}(n)] In terms of Theta, this becomes: 6. 1 + 2 Theta(x) = x^{-1/2} [1 + 2 Theta(1/x) ] which when rearranged produces equation 1. The proof of equation 2 involves the use of both Fourier series and Fourier integrals. -- Daryl McCullough Ithaca, NY === Subject: Re: A Ôbasic' topology question about interiors > Of interest perhaps are these formulas: When U subset A subspace S > cl_A (U) = A / cl_S (U) > int_A (U) = A / int_S (SA / U) Thank you! These are most helpful formulae. You're welcome. Now prove them. Hint: > Prove the first and for the second use. > int_A(U) = A - cl_A (A-U) It appears that these work even if U is not strictly contained in A. Is that correct? Thanks for all your help. Ben Scott === Subject: Roots of a simple third-degree equation I know there must be a simple way to calculate the roots of a third-degree equation where the term in square is missing, like this one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois === Subject: Re: Roots of a simple third-degree equation Thanks very much all! Much apprecia! And thanks in particular to Sekhmet. Most equations of the third degree I have to solve are in the same form as the one pos. Dividing the equation to get a second degree equation is the easiest way to find all the roots. Have a good day. F. === Subject: Re: Roots of a simple third-degree equation In sci.math, Fran?ois <4e5874f3.0311111104.f3b1774@posting.google.com>: I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois A rather simple mechanism detailed in http://mathworld.wolfram.com/CubicEquation.html should breeze one through the general case, unless one happens to spot a root right away and can simply divide. -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Roots of a simple third-degree equation > In sci.math, Fran?ois > : > I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: > -r^3 + 3r + 2 = 0 > How does one do it? > Thanks in advance. > Fran.8dois A rather simple mechanism detailed in http://mathworld.wolfram.com/CubicEquation.html should breeze one through the general case, unless one happens > to spot a root right away and can simply divide. Starting around equation (22) on that page should show you how to do it (the general cubic is first transformed into a cubic with no quadratic term). r^3 - 3r - 2 = 0 Consider: (r - B)(r^2 + Br + B^2 + C) = (r^3 - Br^2) + (Br^2 - B^2r) + (B^2r - B^3) + (Cr - BC) = r^3 + Cr - (B^3 + BC) Which tells you that you can factor your cubic into this form, with C = -3 and B^3 + BC = 2, or B^3 - 3B = 2. === Subject: Re: Roots of a simple third-degree equation As several posters have poin out, if you are just looking for integer (or even rational) roots, it is not too hard. In general, though, it's no easier to solve a cubic with no quadratic term than it is to solve the general case. The reason is as follows: ax^3 + bx^2 + cx + d = 0 Substitute x = y - (b/3a). If you multiply it all out, you'll find that the new polynomial in y has no quadratic term. So if you can find a root y=r of this new equation, then x = r-(b/3a) is a root of the original equation. The method for solving Ay^3 + By + C = 0 was discovered (inven?) in Italy a very long time ago. It's usually called Cardano's formula, although it appears that Cardano actually stole it from someone else. You can find the formula (and it's derivation) at http://mathworld.wolfram.com/CubicEquation.html JoeS -------- I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois === Subject: Re: Roots of a simple third-degree equation I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois If you don't have a square term, jump to the standard form: ====================================== x^3 + a x^2 + b x + c = 0 Substitute x = z - a/3 giving with p = -a^2 / 3 + b q = 2 a^3 / 27 - a b / 3 + c the Standard form: z^3 + p z + q = 0 Calculate discriminant: D = (q/2)^2 + (p/3)^3 1) if D < 0 calculate t such that cos(t) = q/2 / sqrt( -(p/3)^3 ) calculate z1 = -2 sqrt( -p/3 ) cos( t/3 ) z2 = -2 sqrt( -p/3 ) cos( 2/3 pi + t/3 ) z3 = -2 sqrt( -p/3 ) cos( 2/3 pi - t/3 ) 2) if D = 0 (special case of next case) calculate z1 = -2 (q/2)^(1/3) z2 = z3 = (q/2)^(1/3) 3) if D >= 0 calculate u = ( -q/2 + sqrt(D) )^(1/3) v = ( -q/2 - sqrt(D) )^(1/3) calculate z1 = u + v z2 = - (u+v)/2 + (u-v)/2 sqrt(3) i z2 = - (u+v)/2 - (u-v)/2 sqrt(3) i Finally substitute: x1 = z1 - a/3 x2 = z1 - a/3 x3 = z1 - a/3 ====================================== for your example, -r^3 + 3 r + 2 = 0 directly go to the standard form: z^3 - 3 z - 2 = 0 so p = -3 and q = -2 discriminant D = 1 - 1 = 0 D = 0 z1 = -2 (-1)^(1/3) = 2 z2 = z3 = (-1)^(1/3) = -1 Two solutions: -1 and 2 hth Dirk Vdm === Subject: Re: Roots of a simple third-degree equation I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois If one or more of the roots are rational numbeas in the above example, there are a variety of simple methods, as outlined in other responses to your question. If none of the roots are rational, things get more difficult. The following will work in all cases, but is only needed for the tough ones: First, divide through by the coefficient of the cubic term to get the standard form x^3 + a*x + b = 0. Let x1, x2 and x3 represent the 3 roots, and it may be shown that x1 + x2 + x3 = 0, x1*x2 + x2*x3 + x3*x1 = a, and x1*x2*x3 = -b If one assumes that x1 = u + v, x2 = u*w + v/w, x3 = u/w + v*w, where w is either complex root of x^3 = 1, so w^2 + w + 1 = 0, then 3*u*v = =a and u^3 + v^3 = -b. Let r be any square root of (a/3)^3 + (b/2)^2, possibly complex, then take u as any cube root of (-b/2 + r) and v = -a/(3*u), and you have your solutions. For your example of -x^3 + 3*x + 2 = 0, you have a = -3 and b = -2, r = 0, u = v = 1, x1 = 1 + 1 = 2 , x2 = x3 = w + 1/w = -1 === Subject: Re: Roots of a simple third-degree equation U find by guess the root r = -1 Them -r^3 + 3r + 2 could be divided for r + 1 to guive -r^2 + r + 2 with roots 2 and -1 then u have a double root -1 and a single root 2 -r^3 + 3r + 2 = - (r-2) (r+1)^2 I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? Thanks in advance. Fran.8dois === Subject: Re: Roots of a simple third-degree equation I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? > Well, there are formulas for the solutions of polynomial equations of degree <= 5. In this particular case, 2 is easily seen to be a root. So you can divide the polynomium by (x-2) and solve the resulting second-degree equation. /David === Subject: Re: Roots of a simple third-degree equation David Rasmussen escribi.97 en el mensaje > I know there must be a simple way to calculate the roots of a > third-degree equation where the term in square is missing, like this > one: -r^3 + 3r + 2 = 0 How does one do it? > Well, there are formulas for the solutions of polynomial equations of > degree <= 5. David meant Ôof degree < 5' > In this particular case, 2 is easily seen to be a root. So you can > divide the polynomium by (x-2) and solve the resulting second-degree > equation. To the O.P.: The rational roots of a polinomial equation with integer cofficients, expressed as a reduced fraction, have a numerator that divides the independent term, and a denominator that divides the leading coefficient. Then, if the leading coefficient is +/- 1, all the rational roots are integers. So, your equation only can has 1, -1, 2 or -2 as rational roots. If the coefficients of the equation are rationals no integeyou allways can multiply the equation by the minimum common multiple of the denominatoin order to get an equivalent equation with integer coefficients. -- Best Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Sum [n in Set] (1/n) X-No-Confirm: yes X-Face: &UxwkUqGT`JINvVBu`q*X?=.dG|i0BLYLaPh|1J}h<>P)0E I recall seeing a theorem saying Sum [n in Set] (1/n) converges iff (some criterion on Set) where Set is a subset of the positive integers. I seem to recall it brought in measure theory. Does anyone know a reference? Or at least the statement? (I'm actually only interes in the statement, not the proof.) Tia, msh210@math.wustl.edu Of a reply, then, if you have been chea, http://math.wustl.edu/~msh210/ Likely your mail's by mistake been dele. === Subject: Re: Sum [n in Set] (1/n) X-No-Confirm: yes X-Face: &UxwkUqGT`JINvVBu`q*X?=.dG|i0BLYLaPh|1J}h<>P)0E >I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) >where Set is a subset of the positive integers. I seem to recall it >brought in measure theory. I believe the set of polynomials { t^n, n in Set } spans a dense > subset of the vector space of continuous functions on [0,1] iff > Sum [n in Set] (1/n) diverges. Could that (or something like that) > be what you meant? Yes! > There is also a conjecture of Erdos which I think states that > the sum diverges iff Set contains arbitrarily long arithmetic > progressions. Ah, that I'd never heard. Thanks for both! In the future, you can post or mail; doing both is unnecessary. msh210@math.wustl.edu Of a reply, then, if you have been chea, http://math.wustl.edu/~msh210/ Likely your mail's by mistake been dele. === Subject: Re: Sum [n in Set] (1/n) >>I recall seeing a theorem saying >> Sum [n in Set] (1/n) converges iff (some criterion on Set) >> There is also a conjecture of Erdos which I think states that >> the sum diverges iff Set contains arbitrarily long arithmetic >> progressions. That should be => , not iff . Obviously we could have arbitrarily long arithmetic progressions and still have convergence, e.g. if Set = { 10^k + m , 0 < m < k }. I have a notes saying Erdos offered $3000 for a proof of => . You can still collect. >In the future, you can post or mail; doing both is unnecessary. Maybe, but one of these days I'll learn that what IS necessary is to remember to ßag my email so that it doesn't look like it was (also) a newsgroup post ... dave === Subject: Re: Sum [n in Set] (1/n) X-DMCA-Notifications: http://www.giganews.com/info/dmca.html >I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) >> There is also a conjecture of Erdos which I think states that > the sum diverges iff Set contains arbitrarily long arithmetic > progressions. That should be => , not iff . Obviously we could have arbitrarily >long arithmetic progressions and still have convergence, e.g. if >Set = { 10^k + m , 0 < m < k }. I have a notes saying Erdos offered >$3000 for a proof of => . You can still collect. >>In the future, you can post or mail; doing both is unnecessary. Maybe, but one of these days I'll learn that what IS necessary is to >remember to ßag my email so that it doesn't look like it was (also) a >newsgroup post ... Well, I for one was pleased to see his reply to your email (although puzzled where the post from you was) because I didn't know about the Erdos thing... >dave === Subject: Re: Sum [n in Set] (1/n) Content-transfer-encoding: 8bit > I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) > where Set is a subset of the positive integers. I seem to recall it > brought in measure theory. Does anyone know a reference? Or at least the > statement? (I'm actually only interes in the statement, not the > proof.) Perhaps that the span of {x^s : s in S} is dense in C[0,1] iff sum 1/s is infinite? I forget whose theorem it is. The proof is in Rudin's Real and Complex Analysis. (And I may have some details wrong. Perhaps it's the ALGEBRA genera by these which is dense iff...) --Ron Bruck === Subject: Re: Sum [n in Set] (1/n) X-DMCA-Notifications: http://www.giganews.com/info/dmca.html > I recall seeing a theorem saying >> Sum [n in Set] (1/n) converges iff (some criterion on Set) >> where Set is a subset of the positive integers. I seem to recall it >> brought in measure theory. Does anyone know a reference? Or at least the >> statement? (I'm actually only interes in the statement, not the >> proof.) Perhaps that the span of {x^s : s in S} is dense in C[0,1] iff sum >1/s is infinite? Oh, I forgot about that - that could well be what he was wondering about. >I forget whose theorem it is. The proof is in >Rudin's Real and Complex Analysis. (And I may have some details wrong. >Perhaps it's the ALGEBRA genera by these which is dense iff...) No, you got it right. Sketch of proof of the easier direction: Say the sum of 1/n for n in S is infinite. Suppose mu is a measure on [0,1] that annihilates all the t^n for n in S. Define f(z) = int_0^1 t^(i/z) d mu(t). Show f is a bounded holomorphic function in the upper half-plane; now f(i/n) = 0 for n in S, so the Blaschke condition shows that f = 0. Hence f(i/n) = 0 for all n, hence mu = 0. >--Ron Bruck === Subject: Re: Sum [n in Set] (1/n) X-DMCA-Notifications: http://www.giganews.com/info/dmca.html >I recall seeing a theorem saying > Sum [n in Set] (1/n) converges iff (some criterion on Set) >where Set is a subset of the positive integers. I seem to recall it >brought in measure theory. Does anyone know a reference? Or at least the >statement? (I'm actually only interes in the statement, not the >proof.) Well, there are infinitely many such statements, hard to know which one you're referring to. Hmm. Say S is the set, and d(n) is card(S intersect {1,2,...n}) / n. Then I think it's clear that the sum_{n in S} 1/n converges if and only if sum(d(2^n)) converges. >Tia, msh210@math.wustl.edu Of a reply, then, if you have been chea, >http://math.wustl.edu/~msh210/ Likely your mail's by mistake been dele. === Subject: partial deriviatives! hi! Kinda basic question... if you have a function F(q(t),t) and you have the derivative d/dt( d/dq[F(q(t),t) ) then it should be d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] right? Or am i mistaken...?? /M === Subject: Re: partial deriviatives! > Kinda basic question... if you have a function F(q(t),t) > and you have the derivative d/dt( d/dq[F(q(t),t) ) then it should be d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] right? > Or am i mistaken...?? I think mistaken. The notation is a mess. Start with F(q,r) as a function of two variables. Then G(q,r) = [dF/dq](q,r) is also function of two variables. You want dG(q(t),t)/dt, which equals [dG/dq](q(t),t)*dq/dt + [dG/dr](q(t),t)*1. Recalling what G is, the answer is [d^2F/dq^2](q(t),t)*dq/dt + [d^2F/drdq](q(t),t). === Subject: Re: partial deriviatives! > hi! Kinda basic question... if you have a function F(q(t),t) > and you have the derivative d/dt( d/dq[F(q(t),t) ) then it should be d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] right? > Or am i mistaken...?? /M Assuming that you are asking for the total time derivative (not the partial time derivative), then the answer is frac{partial^2 F}{partial q^2} frac{dq}{dt} + frac{partial^2 F}{partial t partial q}. Kostas. === Subject: Vacuum Propellers Jack, I saw this paper while surfing the web and thought it looked similar to your work: http://arxiv.org/abs/gr-qc/9612022 Yes. There are some similarities but more important differences. Modanese's paper complements mine as is shown in detail in my paper in Progress in Quantum Physics Research (Nova Scientific Publishers - a copy is in http://qedcorp.com/APS/ ) 1. Modanese is dealing with a rotating real superconducting ground state of a huge number of real on mass shell electron pair bound states all in dealing with a virtual superconducting vacuum state of a huge number of off-mass-shell bound electron-positrons near the -mc^2 Fermi edge inside the 2mc^2 Dirac energy gap. These two macro-quantum condensates can couple together in a kind of Josephson virtual-real junction. I have the equation in paper I ci. 2. I derive both Einstein's gravity and the unified exotic vacuum dark energy/matter field from the vacuum condensate as an emergent process in the sense of Sakharov and Anderson. Modanese's paper is not fundamental like that. He assumes Einstein's gravity to start with. PS I correc an important typo in a slide in http://qedcorp.com/APS/StatGate.mov about dark matter with negative zero point energy density and POSITIVE pressure. The typo said negative pressure. My Vigier IV paper from Paris in final form is at http://qedcorp.com/APS/EmergentGravity.doc or same with .pdf not .doc Possible quantum gravity effects in a charged Bose condensate under variable e.m. field Authors: G. Modanese, J. Schnurer Comments: 26 pages, PDF, 7 figures. Final journal version Report-no: UTF-391/96 Journal-ref: Phys.Essays 14 (2001) 93-105 In the weak field approximation to quantum gravity, a local positive cosmological term mu^2(x) corresponds to a local negative squared mass term in the Lagrangian and may thus induce instability and local pinning of the gravitational field. Such a term can be produced by the coupling to an external Bose condensate. In the functional integral, the local pinning acts as a constraint on the field configurations. We discuss this model in detail and apply it to a phenomenological analysis of recent experimental results. Best wishes, Mark === Subject: Re: Vacuum Propellers yor titl does bad worded. them wirling thingies insid vacuums isnt Ôpropellers' they is Ôimpellers' === Subject: Marketing shift, core issues I've been thinking about my problems with getting any kind of admission that my math arguments showing the core error in mathematics are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics from modern research on human psychology to see if I can't break through the logjam. It seemed to me that it'd be a good idea to warn you ahead of time, as I'd prefer it that the math by itself would be enough, but that's not the way it works. I guess. At least now I won't have to feel guilty if I decide to use tactics, as I've given fair warning. http://mathforprofit.blogspot.com/ === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. > Way to go James, that is an excellent idea!! I will borrow from it and copy your techniques in my war against the evil Cantorians. With subtle, unethical psychological manipulation behind me, I know that now I can open the ßoodgates fully and do away once and for all with the collective bag of nonsensical gibberish that is Cantor. Would you like to join forces with me so that we can fight Cantor together? Your friend, Nathaniel Deeth Age 11 === Subject: Re: Marketing shift, core issues >Would you like to join forces with me so that we can fight Cantor >together? I'm sure if he was alive he'd be rolling in his grave. === Subject: Re: Marketing shift, core issues >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics >from modern research on human psychology to see if I can't break >through the logjam. It seemed to me that it'd be a good idea to warn you ahead of time, as >I'd prefer it that the math by itself would be enough, but that's not >the way it works. I guess. At least now I won't have to feel guilty if I decide to use tactics, >as I've given fair warning. And if this doesn't work, you can start pestering Congress to get your theorem passed into law. -- Let us learn to dream, gentlemen, then perhaps we shall find the truth... But let us beware of publishing our dreams before they have been put to the proof by the waking understanding. -- Friedrich August Kekul.8e === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. Most of the good marketing books will tell you that marketing works best when you have a salable product. You ought to try it! Subject: Re: Marketing shift, core issues === > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. Psychology won't make your arguments correct. It also isn't likely to make anyone forget that they are wrong. -- === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. > I just wan to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. Psychology won't make your arguments correct. It also isn't likely to > make anyone forget that they are wrong. It made forget that he is wrong. See http://www.mentalhealth.com/dis/p20-pe07.html for a description of Narcissistic Personality Disorder. === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. You are more liable to break the logjam by playing Russian roulette solitaire, as the logjam is all in your imagination. It seemed to me that it'd be a good idea to warn you ahead of time, as > I'd prefer it that the math by itself would be enough, but that's not > the way it works. I guess. At least now I won't have to feel guilty if I decide to use tactics, > as I've given fair warning. As JSH's past performances lead one to believe that he would not feel guilty even for using an Abomb on the mathematical world..... === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. > Just so that we all understand you this once, do you mean that you've decided to give up on math and become a bookseller or do you mean that you've given up on the math and are hoping to use modern marketing techniques to push your self-described proof on a gullible public? Either way, best of luck. Rick p.s. I guess you missed my earlier post, so I'm reprinting it here. Let's take a simpler example and see how your argument works. Let Q(x) = 7(25x^2 + 30x + 2) [1] = 7(x^2 + x)(5^2) + 7(x - 1)(5) + 7^2 and write Q(x) = (5a_1(x) + 7)(5a_2(x) + 7) [2] Now we mirror your construction and observe that we may take a_1(x) and a_2(x) to be roots of r(x) = a^2 - (x - 1)a + 7(x^2 + x) [3] We see that Q(0) = 14 = (7)(2), and r(0) = a^2 + a has roots a_1(0) = 0, a_2(0) = -1. We see that this assignment is consistent with your factorization [2], since Q(0) = (5(0) + 7)(5(-1) + 7) = (7)(2) [4] and we see that we can rewrite [4] as Q(0)/7 = (5(a_1(0)/7) + 1)(5a_2(0) + 7) = (1)(2) or, emphasizing the constant term in [1] we may write, letting b_2(0) = a_2(0) + 1, Q(0)/7 = (5(a_1(0)/7) + 1)(5b_2(0) + 2) = (1)(2) Now, however, you would have us write each factorization over the algebraic integers as Q(x)/7 = (5(a_1(x)/7) + 1)(5b_2(x) + 2) [5] regardless of the (rational integer) value of x. This constraint in [5] does indeed work in some cases. For example, if we take x = -1 we see that Q(-1) = -21 = (7)(-3) r(-1) = a^2 + 2a and we see that we may take a_1(-1) = 0, a_2(-1) = -2 and that doing so gives us the factorization Q(-1) = (5(0) + 7)(5(-2) + 7) and we see that Q(-1)/7 = (5(a_1(-1)/7) + 1)(5a_2(-1) + 7) = (1)(-3) The problem is that your restriction in [5] depends strongly on the polynomial r(x) used to define the terms a_1(x) and a_2(x). For x = 0, -1 the factorization in [5] worked because of the way r(x) splits into particular linear factors, but consider the case when x = 1. Now we have Q(1) = 399 = (7)(57) = (7)(3 * 19) r(1) = a^2 + 14 which gives us a_1(1) = sqrt(-14), a_2(1) = -sqrt(-14) and so we have the factorization Q(1) = (5 sqrt(-14) + 7)(-5 sqrt(-14) + 7) which is indeed equal to 399, as expec. However, it's immediately obvious that sqrt(-14) isn't divisible by 7 in the algebraic integeso the factorization in [5] does not hold. Instead of distributing the value 7 as 7 in the first factor and 1 in the second, we have the distribution sqrt(7) in both factoso we have the factorization in algebraic integers Q(1)/7 = [(5 sqrt(-14) + 7)/sqrt(7)] * [(-5 sqrt(-14) + 7)/sqrt(7)] In fact, your constraint on the way 7 (or 49, in the example you have been using) is split among the terms in your nonpolynomial factorization is valid only in a small number of cases for x. R. === Subject: Re: Marketing shift, core issues X-DMCA-Notifications: http://www.giganews.com/info/dmca.html >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. You know, in case you're curious, this sounds really really stupid. If your results were correct you'd be able to convince people of them by explaining the proofs carefully. But in fact they're wrong, people continually explain what the errors are, and tactics from marketing books are not going to change that. >I just wan to warn readers that I may be employing various tactics >from modern research on human psychology to see if I can't break >through the logjam. It seemed to me that it'd be a good idea to warn you ahead of time, as >I'd prefer it that the math by itself would be enough, but that's not >the way it works. I guess. At least now I won't have to feel guilty if I decide to use tactics, >as I've given fair warning. http://mathforprofit.blogspot.com/ === Subject: Re: Marketing shift, core issues >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. You know, in case you're curious, this sounds really really stupid. > If your results were correct you'd be able to convince people of > them by explaining the proofs carefully. But in fact they're wrong, > people continually explain what the errors are, and tactics from > marketing books are not going to change that. If his results were correct, the same people who point out errors to him with saintly patience would instead have provided all these explanations. === Subject: Re: Marketing shift, core issues >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. > You know, in case you're curious, this sounds really really stupid. > If your results were correct you'd be able to convince people of > them by explaining the proofs carefully. But in fact they're wrong, > people continually explain what the errors are, and tactics from > marketing books are not going to change that. If his results were correct, the same people who point out errors to him > with saintly patience would instead have provided all these explanations. Having spent some time now reading marketing tactics, I can clearly see that both Ullrich and Bau are selling a viewpoint to the readership. Here the assertion is that if I were right then people would necessarily agree with me!!! Is that true in your experience? === Subject: Re: Marketing shift, core issues > Having spent some time now reading marketing tactics, I can clearly > see that both Ullrich and Bau are selling a viewpoint to the > readership. But they have such a much better mousetrap! Here the assertion is that if I were right then people would > necessarily agree with me!!! While truth will out need not be true in areas outside of mathematics, it really does work in mathematics! Is that true in your experience? With regard to mathematical statements, yes! > === Subject: Re: Marketing shift, core issues >I've been thinking about my problems with getting any kind of >admission that my math arguments showing the core error in mathematics >are correct, so I've gone to marketing books. > You know, in case you're curious, this sounds really really stupid. > If your results were correct you'd be able to convince people of > them by explaining the proofs carefully. But in fact they're wrong, > people continually explain what the errors are, and tactics from > marketing books are not going to change that. > If his results were correct, the same people who point out errors to him > with saintly patience would instead have provided all these explanations. Having spent some time now reading marketing tactics, I can clearly > see that both Ullrich and Bau are selling a viewpoint to the > readership. Here the assertion is that if I were right then people would > necessarily agree with me!!! In mathematics? Yes, if accompanied by a cogent argument. I believe Ramanujan initially met with some resistance due to his unfamiliarity with standard notation and style. That doesn't mean you're Ramanujan. You've had ample chance to explicate your proofs, to answer specific questions where people couldn't understand what the web page was saying. Those chances all dead-ended. === Subject: Re: Marketing shift, core issues X-DMCA-Notifications: http://www.giganews.com/info/dmca.html >>I've been thinking about my problems with getting any kind of >>admission that my math arguments showing the core error in mathematics >>are correct, so I've gone to marketing books. You know, in case you're curious, this sounds really really stupid. >> If your results were correct you'd be able to convince people of >> them by explaining the proofs carefully. But in fact they're wrong, >> people continually explain what the errors are, and tactics from >> marketing books are not going to change that. If his results were correct, the same people who point out errors to him >> with saintly patience would instead have provided all these explanations. Having spent some time now reading marketing tactics, I can clearly >see that both Ullrich and Bau are selling a viewpoint to the >readership. Here the assertion is that if I were right then people would >necessarily agree with me!!! Is that true in your experience? It's certainly true in _my_ experience that when I'm right about something and have a valid proof that I'm right then competent mathematicians agree I'm right, after I've explained the proof, yes. That includes cases where they were certain at first I was wrong, by the way. > === Subject: Re: Marketing shift, core issues >>I've been thinking about my problems with getting any kind of >>admission that my math arguments showing the core error in mathematics >>are correct, so I've gone to marketing books. > You know, in case you're curious, this sounds really really stupid. >> If your results were correct you'd be able to convince people of >> them by explaining the proofs carefully. But in fact they're wrong, >> people continually explain what the errors are, and tactics from >> marketing books are not going to change that. > If his results were correct, the same people who point out errors to him >> with saintly patience would instead have provided all these explanations. Having spent some time now reading marketing tactics, I can clearly >see that both Ullrich and Bau are selling a viewpoint to the >readership. Here the assertion is that if I were right then people would >necessarily agree with me!!! Is that true in your experience? It's certainly true in _my_ experience that when I'm right about > something and have a valid proof that I'm right then competent > mathematicians agree I'm right, after I've explained the proof, > yes. That includes cases where they were certain at first I was > wrong, by the way. Don't you agree, David, that from C1-C4--indeed, from C3,C4 alone--any *competent* mathematician could have deduced Ex~(x=x)? C1 AxAy[x=y -> Az(z in x <-> z in y)] C2 AxAy[Az(z in x <-> z in y) -> Az(x in z <-> y in z)] C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A)Classification C4 AxAy[Az(z in x <-> z in y) -> {Et(x in t & y in t) <-> x=y}] (Weak Extensionality) someone will point out the error.? Did your Homies correct you? Or did they *defend* your mistake-- and you for having made it? --John === Subject: Re: Marketing shift, core issues I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. [snip] You are thus not merely incorrigbly stooopid, you are also an out and out fraud and admit as much (little). http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Marketing shift, core issues I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. > Many people have poin out the the errors in mathematics are yours. Have you any idea why crank dot net devotes a whole set of web pages on you? http://www.crank.net/harris.html Example: For the past seven yea has been a regular fixture on the Usenet newsgroup, sci.math. James, an amateur mathematician, has produced scores, if not hundreds of self-proclaimed elementary Ôproofs' of Fermat's Last Theorem, none of which, to date, have withstood serious scrutiny. Due to James's unfamiliarity with mathematical exposition, the work available on his main and Ôarea one' web sites and his contributions to the newsgroup have often been difficult to understand and have been augmen and clarified by many others over the years. This page is my attempt to gather together what has been said by and to Mr. Harris about FLT and present it in a style which meets at least minimal standards for clarity. === Subject: Re: Marketing shift, core issues > I've been thinking about my problems with getting any kind of > admission that my math arguments showing the core error in mathematics > are correct, so I've gone to marketing books. I just wan to warn readers that I may be employing various tactics > from modern research on human psychology to see if I can't break > through the logjam. It seemed to me that it'd be a good idea to warn you ahead of time, as > I'd prefer it that the math by itself would be enough, but that's not > the way it works. I guess. At least now I won't have to feel guilty if I decide to use tactics, > as I've given fair warning. Let me guess what this means: your algebra hasn't convinced anyone, so you are going to try to trick people into believing you. === Subject: Basic factorization ideas If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a problem realizing that only two of the c's have 7 as a factor. But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 22. I want to emphasize that point as notice there's only *one* way to divide through by 49 if 7 is not a factor of 22. Usually you can *see* the other factors of 7, but I want you to abstract, and generalize. Please pay careful attention to that example. You may see people who reply claiming that the word polynomial has some significance, as if it's a mystical thing which refutes basic logic, so if something isn't polynomial it no longer behaves logically. Now then, in my advanced factorization work, I just use functions of x that are a lot more complica than f_1(x) = c_1 x, and unfortunately there are people who can use an unfamiliar leap in complexity to confuse others. Some of you have learned various advanced math topics, now imagine if in your classrooms there were some hecklers who continually hollered out at your teacher, or otherwise disrup the class? What if when there were difficult concepts those hecklers would try to confuse everyone as they sought to discredit the mathematics? If you find that hard to imagine, imagine me in your class with you questioning the professor and calling him names. How much would you have learned? I need those of you interes in mathematics to focus on the basics, so that you can understand the advanced. http://mathforprofit.blogspot.com/ === Subject: Re: Basic factorization ideas In sci.physics, <3c65f87.0311111302.799cf8e3@posting.google.com>: > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. Oh, look, he's changed polys on us! But OK, let's check this one. Recall that, if P(x) = 49 * (x - x_1) * (x - x_2) * (x - x_3), then P(x) = (c_1 * x + 7) * (c_2 * x + 7) * ( c_3 * x + 1) implies that c_1 = -7 / x_1, c_2 = -7 / x_2, c_3 = -1 / x_3, for some permutation of the x_i. Again, c_1 * c_2 * c_3 = 49, as required, since x_1 * x_2 * x_3 = -1. Therefore, c_1 and c_2 both satisfy the equation c^3 * P(-7 / c)/49 = c^3 - 21*c^2 + 245*c - 343 and c_3 of course satisfies c^3 * P(-1 / c)/49 = c^3 - 3*c^2 + 5*c - 1 so it turns out that this time, James, you got it more or less right. (And c_3 is even a unit, to boot. Come to think of it, so are the x_i, which is probably one big reason why this particular case actually works.) Also, it is obvious that c_1 and c_2 have factors of 7, as well, in the ring of algebraic integers. But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 22. OK, stupid question. Where did the 22 come from? But yes, for this particular example, your odd math actually does work, as you are dealing with x_i which are in fact units. Now change a_0 to 3 and get back to us. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Basic factorization ideas > In sci.physics, > : > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. Oh, look, he's changed polys on us! But OK, let's check > this one. Recall that, if P(x) = 49 * (x - x_1) * (x - x_2) * (x - x_3), > then P(x) = (c_1 * x + 7) * (c_2 * x + 7) * ( c_3 * x + 1) > implies that c_1 = -7 / x_1, c_2 = -7 / x_2, c_3 = -1 / x_3, > for some permutation of the x_i. Again, c_1 * c_2 * c_3 = 49, > as required, since x_1 * x_2 * x_3 = -1. Therefore, c_1 and c_2 both satisfy the equation c^3 * P(-7 / c)/49 = c^3 - 21*c^2 + 245*c - 343 and c_3 of course satisfies c^3 * P(-1 / c)/49 = c^3 - 3*c^2 + 5*c - 1 so it turns out that this time, James, you got it more > or less right. (And c_3 is even a unit, to boot. > Come to think of it, so are the x_i, which is probably > one big reason why this particular case actually works.) Also, it is obvious that c_1 and c_2 have factors of 7, > as well, in the ring of algebraic integers. > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 22. OK, stupid question. Where did the 22 come from? Oh, I've been used to arguing with people about (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 b_3(x) + 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) where the a's are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) and b_3(x) = a_3(x) - 3, which is a substitution made because a_3(0) = 3, so that I can isolate constant terms. My point is that certain rules apply as even with my example above, I just have rather basic functions of x. > But yes, for this particular example, your odd math actually does > work, as you are dealing with x_i which are in fact units. So you believe that math is quirky? > Now change a_0 to 3 and get back to us. [rest snipped] My thinking is that some of you don't believe in the same mathematics that most people do, but instead have your own ideas about applicability of rules in mathematics, and apparently my work attracts a certain type of person who posts loudly about it. That is, I think that readers will find that posters are often challenging some rather basic math, using rules that vary with their moods. My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Basic factorization ideas > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. > If c_1 and c_2 (and hence c_3) are ordinary integethen yes, c_1 and c_2 must be divisible by 7, and c_3 must be coprime to 7 (does this hold for algebraic integers as well?) but the proof of this depends strongly on the fact that the LHS is composed of polynomial factors. > But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Yes, but this is dangerous. Someone who was not paying close attention might assume that the f_i were arbitrary functions and get the idea that things that are true of polynomials are also true of arbitrary functions. > Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 22. I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > Yes, if the f_i are defined as above. If the f_i are arbitary functions this statement is wrong. - William Hughes === Subject: Re: Basic factorization ideas > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. If c_1 and c_2 (and hence c_3) are ordinary integethen > yes, c_1 and c_2 must be divisible by 7, and c_3 must be > coprime to 7 (does this hold for algebraic integers as well?) > but the proof of this depends strongly on the fact that the LHS > is composed of polynomial factors. Why? That is, why does it matter if the factors are polynomial factors? What about (c_1 sqrt(x) + 7)(c_2 sqrt(x) + 7)( c_3 sqrt(x) + 1) = 49(x^{3/2} + 5x + 3sqrt(x) + 1)? > But, of course, you're looking at *functions* of x, as you have > f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, > so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). > Yes, but this is dangerous. Someone who was not paying close > attention might assume that the f_i were arbitrary functions > and get the idea that things that are true of polynomials > are also true of arbitrary functions. But why does it matter if they're polynomials? > Notice that dividing both sides by 49 gives > (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 > as long as you're in a ring where 7 is not a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Yes, if the f_i are defined as above. If the f_i are arbitary > functions this statement is wrong. Well I have to ask why again, and if readers think I'm just repealy asking for no good reason, consider that William Hughes has very *definitely* marked out a position without giving a reason. Rather than assume I can figure out what he was thinking when he pos, I'd just as soon ask him. My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Basic factorization ideas ... > If c_1 and c_2 (and hence c_3) are ordinary integethen > yes, c_1 and c_2 must be divisible by 7, and c_3 must be > coprime to 7 (does this hold for algebraic integers as well?) > but the proof of this depends strongly on the fact that the LHS > is composed of polynomial factors. Why? That is, why does it matter if the factors are polynomial > factors? What about (c_1 sqrt(x) + 7)(c_2 sqrt(x) + 7)( c_3 sqrt(x) + 1) = > 49(x^{3/2} + 5x + 3sqrt(x) + 1)? Now you work with polynomials in sqrt(x), so you still have polynomial factors in that ring. -- === Subject: Re: Basic factorization ideas ... > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. If c_1 and c_2 (and hence c_3) are ordinary integethen > yes, c_1 and c_2 must be divisible by 7, and c_3 must be > coprime to 7 (does this hold for algebraic integers as well?) No. For some x, (c3 x + 1) is actually *not* coprime to 7. The reason is that for those x, (x^3 + 5x^2 + 3x + 1) is divisible by 7, and that additional 7 has to go somewhere (it is distribu among the three factors). -- === Subject: [JSH] Crank Net!: Re: Basic factorization ideas === > I need those of you interes in mathematics to focus on the basics, > so that you can understand the advanced. http://www.crank.net/harris.html === Subject: Re: Basic factorization ideas > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. A bit of tricking? Your marketing? Let r1, r2 and r3 be the roots of the cubic in some order, than we have: c1 = 7/r1, c2 = 7/r2, c3 = 1/r3 all algebraic integers. So indeed, c1 and c2 are divisible by 7 in the algebraic integers. > But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 1) = 49(x^3 + 5 x^2 + 3x + 1). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 1) = x^3 + 5 x^2 + 3x + 1 as long as you're in a ring where 7 is not a factor of 22. What has 7 being a factor or not of 22 to do with it? This is also true when 7 is a factor of 22. > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Oh, you mean that when 7 is a factor of 22 than there are other ways to do the division, as in that case 7 is a unit in the ring. Moreover, there may be values of x such that (f3(x) + 1) is divisible by 7, I think. If this is the case there are other ways to distribute the 7's. ... > Now then, in my advanced factorization work, I just use functions of x > that are a lot more complica than f_1(x) = c_1 x, and unfortunately > there are people who can use an unfamiliar leap in complexity to > confuse others. Unfortunately, the functions here are trivially divisible by 7 in the ring of algebraic integers. In your advance factorisation work they are *not*. But you think that is a ßaw... But indeed, that your functions are much more complica *makes* a difference. c1 x *is* divisible by 7 because c1 is divisible by 7, regardless of the value of x (c1 is a constant), so (c1 x + 7) is trivially divisible by 7. (c3 x + 1) is *not* divisible by 7, unless x has a specific value as I mentioned above, which might or might not be possible, but that is irrelevant, because the first two factors are trivially divisible by 7. In your (more complica case) you *know* that a1(x) and a2(x) are *not* divisible by 7 for all x (you say that is a ßaw). It has been shown that (b3(x) + 22) is divisible by factors of 7 for almost all x. And those factors compensate for the lack of factors in a1(x) and a2(x). -- === Subject: Re: Basic factorization ideas > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. ... > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Oh, you mean that when 7 is a factor of 22 than there are other ways > to do the division, as in that case 7 is a unit in the ring. > Moreover, there may be values of x such that (f3(x) + 1) is divisible by > 7, I think. If this is the case there are other ways to distribute the > 7's. Lo and behold, there are indeed (not 7, but factors of 7)! When x = 7y + 2 for some integer y, (c3 x + 1) is *not* coprime to 7. This example is too easy to crack. Define: w3(x) = gcd(c3 x + 1, 7) (this one is not always 1 (*)), w2(x) = 7/w3(x). We now could divide the first factor by 7, the second by w2(x) and the third by w3(x) and remain in the algebraic integers. So, again, the possible ways to divide 49 through may very well depend on the value of x. ---- (*) Proof: We have (c3 x + 1) where c3 is 1/r, with r the negative of one of the roots of x^3 + 5 x^2 + 3 x + 1 Now (x/r + 1) = (r + x)/r, so the first is not coprime to 7 when (r + x) is not coprime to 7. Set x = 7y + 2. (r + x) is not coprime to 7 when (r + 7y + 2) is not coprime to 7, and that is when r + 2 is not coprime to 7. You may verify that (r + 2) is a root of y^3 - 11 y^2 + 35 y - 35. So it is an algebraic integer and not coprime to 7. -- === Subject: Re: Basic factorization ideas > If you saw > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) > with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. > ... > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. > Oh, you mean that when 7 is a factor of 22 than there are other ways > to do the division, as in that case 7 is a unit in the ring. > Moreover, there may be values of x such that (f3(x) + 1) is divisible by > 7, I think. If this is the case there are other ways to distribute the > 7's. Lo and behold, there are indeed (not 7, but factors of 7)! When x = 7y + 2 > for some integer y, (c3 x + 1) is *not* coprime to 7. This example is too > easy to crack. Define: > w3(x) = gcd(c3 x + 1, 7) > (this one is not always 1 (*)), > w2(x) = 7/w3(x). Well, yes, but what about (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = x^3 + 5x^2 + 3x + 1? Here the 49 has been divided through, and doesn't that change things? > We now could divide the first factor by 7, the second by w2(x) and > the third by w3(x) and remain in the algebraic integers. So, again, > the possible ways to divide 49 through may very well depend on the > value of x. > ---- But that now seems problematic as in mathematics, how do you determine time? Isn't there just the factorization (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = x^3 + 5x^2 + 3x + 1 without reference to *time* itself? Your example is irrelevant above as though c_3 x + 1 may have 7 as a factor, so what? The question is how to divide 49 from both sides of (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) and I think there's only one way to do so, and stay in the ring of algebraic integers. That is, one way to keep 1 from having 7 as a factor. My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Basic factorization ideas > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a > problem realizing that only two of the c's have 7 as a factor. > ... > I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Oh, you mean that when 7 is a factor of 22 than there are other ways > to do the division, as in that case 7 is a unit in the ring. > Moreover, there may be values of x such that (f3(x) + 1) is divisible by > 7, I think. If this is the case there are other ways to distribute the > 7's. Lo and behold, there are indeed (not 7, but factors of 7)! When x = 7y + 2 > for some integer y, (c3 x + 1) is *not* coprime to 7. This example is too > easy to crack. Define: > w3(x) = gcd(c3 x + 1, 7) > (this one is not always 1 (*)), > w2(x) = 7/w3(x). Well, yes, but what about > (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = > x^3 + 5x^2 + 3x + 1? > Here the 49 has been divided through, and doesn't that change things? It does not change a thing. For some values of x all three factors on the left are not coprime to 7. > We now could divide the first factor by 7, the second by w2(x) and > the third by w3(x) and remain in the algebraic integers. So, again, > the possible ways to divide 49 through may very well depend on the > value of x. But that now seems problematic as in mathematics, how do you > determine time? By looking at the clock. Otherwise I have no idea what you mean. > Isn't there just the factorization > (c_1/7 x + 1)(c_2/7 x + 1)( c_3 x + 1) = > x^3 + 5x^2 + 3x + 1 > without reference to *time* itself? I do not know what you mean with time. But indeed, that is one possible factorisation, but there are other factorisations, as I just showed above. > Your example is irrelevant above as though c_3 x + 1 may have 7 as a > factor, so what? The question is how to divide 49 from both sides of > (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = > 49(x^3 + 5x^2 + 3x + 1) > and I think there's only one way to do so, and stay in the ring of > algebraic integers. That is, one way to keep 1 from having 7 as a > factor. w3(x) = gcd(c3 x + 1), 7) (this is not identically equal to 1.) w2(x) = 7/w3(x), [(c1 x + 7)/7] [(c2 x + 7)/w2(x)] [(c3 x + 1)/w3(x)] Now, *which* of the three factors is *not* an algebraic integer for all x? And *how* do you find that 7 is a factor of 1? -- === Subject: Re: Basic factorization ideas >If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a >problem realizing that only two of the c's have 7 as a factor. Personally I have a problem with the fact that I can think of no values for c_1, c_2 and c_3 such that c_1, c_2 and c_3 are algebraic integers and the left and the right side of the equation are equal. Could you help me out and give me the values of c_1, c_2 and c_3? === Subject: Re: Basic factorization ideas If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a >problem realizing that only two of the c's have 7 as a factor. Personally I have a problem with the fact that I can think of no > values for c_1, c_2 and c_3 such that c_1, c_2 and c_3 are algebraic > integers and the left and the right side of the equation are equal. James has tricked you. In this version the roots of the cubic are units. Part of his marketing strategy, I think. -- === Subject: Re: Basic factorization ideas In sci.math, Dik T. Winter : >If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) with the c's algebraic integeI think few of you would have a >problem realizing that only two of the c's have 7 as a factor. > Personally I have a problem with the fact that I can think of no > values for c_1, c_2 and c_3 such that c_1, c_2 and c_3 are algebraic > integers and the left and the right side of the equation are equal. James has tricked you. In this version the roots of the cubic are > units. Part of his marketing strategy, I think. I'll admit I wish I knew...but I did note the change. The more general cubic: (c_1 * x + 7) * (c_2 * x + 7) * (c_3 * x + a_0) = 49 * (x^3 + a_2 * x^2 + a_1 * x + a_0) where x^3 + a_2 * x^2 + a_1 * x + a_0 is irreducible, might be an interesting proof attempt, were it to work ... which, judging from his earlier example, probably wouldn't. Is James pulling these specific examples out of thin air or is there a broader (mathematical) substructure from whence these are coming which ultimately leads to his Proof Of Something Significant? If the latter, where? -- #191, ewill3@earthlink.net It's still legal to go .sigless. === Subject: Re: Basic factorization ideas > In sci.math, Dik T. Winter > ... > James has tricked you. In this version the roots of the cubic are > units. Part of his marketing strategy, I think. I'll admit I wish I knew...but I did note the change. The more > general cubic: (c_1 * x + 7) * (c_2 * x + 7) * (c_3 * x + a_0) > = 49 * (x^3 + a_2 * x^2 + a_1 * x + a_0) where x^3 + a_2 * x^2 + a_1 * x + a_0 is irreducible, might > be an interesting proof attempt, were it to work ... which, > judging from his earlier example, probably wouldn't. No, indeed it certainly will not work if a0 != 1 and a0 also not a power of 7 and the polynomial irreducible. If a0 is a power of 7 there are possibilities. > Is James pulling these specific examples out of thin air or is > there a broader (mathematical) substructure from whence these > are coming which ultimately leads to his Proof Of Something > Significant? Apparently out of thin air. He tries a few polynomials which he can handle, sees that it works, and then makes it a general principle. If the polynmial does not work (like the unchanged version with 2 as constant) he silently shifts pollynomial. -- === Subject: Re: Basic factorization ideas > In sci.math, Dik T. Winter > : there a broader (mathematical) substructure from whence these > are coming which ultimately leads to his Proof Of Something > Significant? > As you might be able to guess from his personality (disorder) Harris must maintain the belief that he's working on (or has finished) a Work of Monumantal Importance. For the past seven years or so (it may be eight by now), he's been banging away at an elementary proof of Fermat's Last Theorem. The this cubic MUST have two of its factors divisible by a prime examples he's been pestering us with have to do with a step in his proof for the case of exponent 3. If you want to look at a de-obfusca treatment of an old version of his proof, just ask. Rick === Subject: Re: Basic factorization ideas >James has tricked you. In this version the roots of the cubic are >units. Part of his marketing strategy, I think. I don't completely understand what you are saying. Please read the following and tell me where I go wrong. The whole Ôdiscusion' is about algebraic integers. An algebraic integer is any value of x that satisfies: x^n + a_(n-1)*x^(n-1) + ... + a_1*x + a_0 = 0 (with the a's being integers) For example: sqrt(2) is an algebraic integer as it is the solution to x^2-2=0. The algebraic integers form a ring, which amongst other things means that the addition of two algebraic integers is an algebraic integer as well (the same goes for multiplication of two algebraic integers). Another property is that for an algebraic integer a, -a is also an algebraic integer. Now look at this: P(x) = 49*(x^3 + 5*x^2 + 3*x + 1) If the value of x is an algebraic integer, the value of P(x) for that x will be an algebraic integer as well. This is true, because 1,3,5 and 49 are algebraic integeso with x an algebraic integer as well, the result must be an algebraic integer. This can also be seen if P(x) is factored: P(x) = 49*(x+r_1)*(x+r_2)*(x+r_3) r_1, r_2 and r_3 are (obviously) algebraic integeso if the value of x is an algebraic integer, the value of P(x) is also an algebraic integer. Now Mr. Harris claims that P(x) can be rewritten to: P(x) = (c_1*x+7)*(c_2*x+7)*(c_3*x+1) in such a way that c_1, c_2 and c_3 are algebraic integers. This is what I can come up with: P(x) = 49*(x+r_1)*(x+r_2)*(x+r_3) P(x) = 7*7*r_1*r_2*r_3*(x/r_1+1)*(x/r_2+1)*(x/r_3+1) P(x) = r_1*r_2*r_3*(7*x/r_1+7)*(7*x/r_2+7)*(x/r_3+1) P(x) =(7*x/r_1+7)*(7*x/r_2+7)*(r_1*r_2*x+r_1*r_2*r_3) P(x) =((7/r_1)*x+7)*((7/r_2)*x+7)*(r_1*r_2*x+r_1*r_2*r_3) So: c_1 = 7/r_1, c_2 = 7/r_2, c_3 = r1*r2 and r1*r2*r3 = 1 c_3 is obviously an algebraic integer as it is the product of two algebraic integers. c_1 and c_2 can also be written like the product of algebraic integers: c_1 = 7/r_1 = 7*r_2*r_3/r_1*r_2*r_3 = 7*r_2*r_3 c_2 = 7/r_2 = 7*r_1*r_3/r_1*r_2*r_3 = 7*r_1*r_3 (@Dik: I guess this is what you tried to tell me...) So P(x) = (7*r_2*r_3*x + 7)*(7*r_1*r_3*x + 7)*(r1*r2*x + 1) By dividing P(x) by 49 we get: P(x)/49 = (r_2*r_3*x+1)*(r_1*r_3*x+1)*(r1*r2*x+1) The Ôfactors' of P(x)/49 (between the brackets) are obviously algebraic integers if x is an algebraic integer. However, we could just as easily have divided P(x) like this: P(x) = (((r_2*r_3)/7)*x + (1/7))*(7*r_1*r_3*x + 7)*(r1*r2*x + 1) or: P(x) = (7*r_2*r_3*x + 7)*(r_1*r_3*x + 1)*(((r1*r2)/7)*x + 1/7) In these cases the Ôfactors' don't have to be algebraic integers (but they can be, for instance for x=7 :) ). Now did I get that right? If so, why isn't there a 22 in my story? === Subject: Re: Basic factorization ideas James has tricked you. In this version the roots of the cubic are >units. Part of his marketing strategy, I think. I don't completely understand what you are saying. Please read the > following and tell me where I go wrong. My comment was about (c1 x + 7)(c2 x + 7)(c3 x + 1) = 49(x^3 + 5x^2 + 3x + 1) where you said you did not see algebraic integers. I assumed you had seen his previous attempt with the last 1 changed to 2, where indeed the c's are not algebraic integers. And I thought you based your comment on that. Probably my fault. > The whole Ôdiscusion' is about algebraic integers. An algebraic > integer is any value of x that satisfies: x^n + a_(n-1)*x^(n-1) + ... + a_1*x + a_0 = 0 (with the a's being > integers) Yup. > For example: sqrt(2) is an algebraic integer as it is the solution to > x^2-2=0. The algebraic integers form a ring, which amongst other things means > that the addition of two algebraic integers is an algebraic integer as > well (the same goes for multiplication of two algebraic integers). > Another property is that for an algebraic integer a, -a is also an > algebraic integer. You may note that the algebraic integers form a ring. I.e. adding algebaric integers gives an algebraic integer, as does multiplying. ... > Now Mr. Harris claims that P(x) can be rewritten to: > P(x) = (c_1*x+7)*(c_2*x+7)*(c_3*x+1) > in such a way that c_1, c_2 and c_3 are algebraic integers. This is what I can come up with: > P(x) = 49*(x+r_1)*(x+r_2)*(x+r_3) > P(x) = 7*7*r_1*r_2*r_3*(x/r_1+1)*(x/r_2+1)*(x/r_3+1) > P(x) = r_1*r_2*r_3*(7*x/r_1+7)*(7*x/r_2+7)*(x/r_3+1) > P(x) =(7*x/r_1+7)*(7*x/r_2+7)*(r_1*r_2*x+r_1*r_2*r_3) > P(x) =((7/r_1)*x+7)*((7/r_2)*x+7)*(r_1*r_2*x+r_1*r_2*r_3) So: > c_1 = 7/r_1, c_2 = 7/r_2, c_3 = r1*r2 and r1*r2*r3 = 1 Here we have it. A number in a ring is a unit in that ring if the multiplicative inverse is also in a ring. As you say, r2*r3 is the multiplicative inverse of r1, so r1 is a unit. That was the meaning of my comment. In general, when an algebraic integer is the root of a polynomial with integer coefficients, and the constant term of the polynomial is 1, the roots are units. The inverse also does hold. > (@Dik: I guess this is what you tried to tell me...) Indeed. > However, we could just as easily have divided P(x) like this: P(x) = (((r_2*r_3)/7)*x + (1/7))*(7*r_1*r_3*x + 7)*(r1*r2*x + 1) Indeed not, as you no in your follow-up. > If so, why isn't there a 22 in my story? 22 is one of the 22 red herrings James likes to throw around. -- === Subject: Re: Basic factorization ideas >In these cases the Ôfactors' don't have to be algebraic integers (but >they can be, for instance for x=7 :) ). Oops... this is obviously not true. :) I forgot the +(1/7)... silly me. === Subject: Re: Basic factorization ideas Crank Information http://www.crank.net/harris.html http://www.crank.net/usenet.html http://www.google.com/search?q=harris+site%3Awww.crank.net http://www.google.com/search?q=%22james+harris%22+site% 3Ausers.pandora.be === Subject: Re: Vedic Mathematics --- Myth and Reality > For instance, 2000 years down the road, the people of Punjabi > descent may claim that Punjab made great strides in physics > and chemistry during 1950-2050 and they can show numerous > physics and chemistry books by Punjabi authors to support > their claim. But as we know, nearly all the physics and > chemistry contained in these so-called Punjabi books > has come from the West. > While one does not know will happen after 2000 yeaone thinks one > knows what happened 2000 years ago, when one is guided by one's trust > in the people he considers worthy of respect, So claims don't need to be verified by evidence? Statements made under oath are taken as evidence in law courts. It is on such visual/oral/aural (non-literary, that is) evidence that people are convic for murder. Heard you heard of the term eye witness? By no means is literary evidence the only sort of evidence. Also, the genuineness of the literary evidence is a major issue. So, just as one takes as genuine evidence the word of a witness who cannot be proved false, one takes as genuine evidence the statements of people in positions of respect. Of course, there could be genuine mistakes - a witness can think he is speaking the truth, when actually he is not. Like, most physicists genuinely believe in Einstein's Theory of Relativity. They are not liaonly deluded, according to the one who can prove that theory baseless. > and the various objects > around one that are of relevance. Such as, a book I picked up at a > reputable shop, that has been the object of vilification. Why is it vilification to ask if the math algorithms that > are being marke as Vedic mathematics were really > developed by the Vedic people in the Vedic era? Nobody is just asking that. They are saying that the writer of the book is a fraud. This was the tone of the first post, and some followups. To show that the writer is a fraud, they have to prove that those algorithms were developed elsewhere, and then they have to show where the algorithms came from, who outside India developed them, where they were used, etc. That they have not done - those people who are attacking the writer of the book, I mean. When one cannot show that, one has to consider it very plausible that they were developed in ancient India, and followed down the ages. The book describes the multiplication algorithm in a single descriptive Sanskrit word, which goes into the core of the actual process. As I have shown, the ancient Indians really understood the place value underlying the decimal system, and used it very efficiently. Look, I have explained the one-line multiplication. But they have also done one-line division, one-line square root, etc. If there are existing methods elsewhere relating to one-line division or one-line square-root evaluation, then give the references. Or else, accept that the whole thing is really new for the world outside India. Certainly, I have never come ac this method earlier, and I have written multiprecision arithmetic software routines (consulting Knuth) myself using bases of 10 and higher (for faster computation). If I knew about all this Vedic arithmetic, I could have done a better job, and more easily too. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality > While one does not know will happen after 2000 yeaone thinks one > knows what happened 2000 years ago, when one is guided by one's trust > in the people he considers worthy of respect, > So claims don't need to be verified by evidence? So, just as one takes as genuine evidence the word of a witness > who cannot be proved false, one takes as genuine evidence the > statements of people in positions of respect. Values that are based on respect/reverance for personalities fall within the boundaries of temples/mosques/churches. Historical research doesn't depend on such beliefs. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality > My father taught me in my childhood days, 12 * > 34 > -------- > 48 > 36 > -------- > 408 this is the way children are still taught in at least Australian primary schools, and in most Indian schools as well. As you can see, you need three lines to get this result (48, 36 and 408). In Vedic arithmetic, you get the representation of the result in just one line, no matter what the length of the factors. Now, in India there probably still exist a lot of people who never did this long and rather silly-looking Western method, they always did it the ancient Vedic way, which was so much better. (I suspect that while they got the zero and other Indian arithmetic stuff into the West, they never learnt to use them the way the ancient Indians did.) Now, they are only telling us such methods by publishing them in a book. I have studied only the multiplication, but they claim they can do one-line division, square-root, etc. as well. Basically, I don't see why we should call them frauds (as the TIFR person indica) simply because these brilliant methods work out simpler and faster, and are altogether better. > cALCULATION IS > 12 * > 34 > -------- > 4*1 | 4*2 > 3*1 | 3*2 | > ------ > 3 | 10 | 8 = 408 Perfectly correct, however, computers convert them into binary numbers (1s and 0s) and then add and multiply using shifting techniques that are very efficient. However, for really long numbethey have to get a number of integers for the representation,as they use the same basic algorithm that you have demonstra. And there has to be a lot of linking between these numbers. With Vedic arithmetic, you can define a number as an arrays of integeand you do the job simply and also quickly, mainly because the internal computation does not involve extensive array processing. > So 12*34=508 Oh really? Again, never mind! To give a mathematical (and suitably, one-line!) definition of Vedic multiplication: The multiplicative product of two numbers A and B represen as a sequence of characters of length m and n respectively, of base X, and of value A(m-1)*X^(m-1) + A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) + B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) + C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by sum of the combinations all the A(i) and B(j) satisfying i+j=k, where i, j and k range from 0 to m, n and m+n-1 respectively. In lay language, this means gathering all the units, tens, hundreds, etc. in just one go, by finding out all the possible combinations for same. For instance, to find the units, you can only multiply the last digits of the two numbers. Let anyone try to multiply say 12345 by 67809 using this method, in one line, as an exercise! (If you give up, I'll do it!). At present, I am a bit busy, and have was too much time already! :) :) With best wishes to all, Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality > My father taught me in my childhood days, 12 * > 34 > -------- > 48 > 36 > -------- > 408 this is the way children are still taught in at least > Australian primary schools, and in most Indian schools as well. As > you can see, you need three lines to get this result (48, 36 and 408). > In Vedic arithmetic, you get the representation of the result in just > one line, no matter what the length of the factors. Now, in India I can also do it in one line: 12 * 34 -------- 48 36 408 But seriously, this vedic method is no different as our method. Our method is basically: 2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408 See, exactly the same as your method. Wilbert === Subject: Re: Vedic Mathematics --- Myth and Reality > My father taught me in my childhood days, 12 * > 34 > -------- > 48 > 36 > -------- > 408 > this is the way children are still taught in at least > Australian primary schools, and in most Indian schools as well. As > you can see, you need three lines to get this result (48, 36 and 408). > In Vedic arithmetic, you get the representation of the result in just > one line, no matter what the length of the factors. Now, in India I can also do it in one line: 12 * > 34 > -------- > 48 36 408 But seriously, this vedic method is no different as our method. > Our method is basically: 2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408 See, exactly the same as your method. Not at all! > Wilbert No, you don't see the difference. You have written 48 as 12*4, whereas in the Vedic method you take only one coefficient at one time. Like, you cannot take more than one coefficient, as you have done. Try doing what you have done for say 12345*567809! That is the very important difference, and that which will make it much simpler for computer calculation. I know it is very difficult to comprehend the simple and the brilliant, especially when the mind has been narrowly and closely formed. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality >> See, exactly the same as your method. >Not at all! It's 4 things to combine, no matter how it's done. Whether doing them 2 at a time in your head (or even by rote memory) or not is irrelevant. It's still 4. This is 3 products (1*3, 2*4 and 3*8), not 4. 12 * 34 = 308 + 210 - 110 = 408. >Try doing what you have done for say 12345*567809! It's 30 pairs either way. If you think there's anything substantially different, try doing it with: 23423499912809809123889728971349817238974409123841290827092347 832 x 34923423098230941820948120928340928029384029384223094823023432 423 ------- I know a quick way of doing this without doing ANY multiplication whatsoever, and yet requiring about the same number of additions that are required to combine the rows of numbers anyhow. Try matching that with that ingenious Vedic method of yours. [Hint: c.f., dynamic programming.] === Subject: Re: Vedic Mathematics --- Myth and Reality There was an Indian mathematican called Shakuntala who toured the west beating the computers at multiplying two large numbers - must have used vedic methods, === Subject: Re: Vedic Mathematics --- Myth and Reality > My father taught me in my childhood days, 12 * > 34 > -------- > 48 > 36 > -------- > 408 this is the way children are still taught in at least > Australian primary schools, and in most Indian schools as well. As > you can see, you need three lines to get this result (48, 36 and 408). > In Vedic arithmetic, you get the representation of the result in just > one line, no matter what the length of the factors. Now, in India I can also do it in one line: 12 * > 34 > -------- > 48 36 408 But seriously, this vedic method is no different as our method. > Our method is basically: 2*4 + 10*4 + 2*30 + 10*30 = 48 + 360 = 408 See, exactly the same as your method. Not at all! > Wilbert No, you don't see the difference. You have written 48 as 12*4, > whereas in the Vedic method you take only one coefficient at one time. coefficient at the time. > Like, you cannot take more than one coefficient, as you have done. > Try doing what you have done for say 12345*567809! That is the very > important difference, and that which will make it much simpler for > computer calculation. Our way (including the steps you make in your mind): 10000*9 + 2000*9 + 300*9 + 40*9 + 5*9 + 10000*800 + 2000*800 + ... Precisely the same as your method, only the ordering of adding is different. Big deal. > I know it is very difficult to comprehend the simple and the > brilliant, especially when the mind has been narrowly and closely > formed. Adding numbers in a different order, yes very brilliant. Wilbert === Subject: Re: Vedic Mathematics --- Myth and Reality ... > To give a mathematical (and suitably, one-line!) definition of Vedic > multiplication: The multiplicative product of two numbers A and B > represen as a sequence of characters of length m and n > respectively, of base X, and of value A(m-1)*X^(m-1) + > A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) + > B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) + > C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by sum of > the combinations all the A(i) and B(j) satisfying i+j=k, where i, j > and k range from 0 to m, n and m+n-1 respectively. If that's one line I'm a Dutchman. It looks like 8 lines to me. This doesn't look like it differs from the standard definition of long multiplication. What's Vedic about it? -- Unpatched IE vulnerability: protocol control chars Description: Circumventing content filters Reference: http://badwebmasters.net/advisory/012/ Exploit: http://badwebmasters.net/advisory/012/test2.asp === Subject: Re: Vedic Mathematics --- Myth and Reality > ... > To give a mathematical (and suitably, one-line!) definition of Vedic > multiplication: The multiplicative product of two numbers A and B > represen as a sequence of characters of length m and n > respectively, of base X, and of value A(m-1)*X^(m-1) + > A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) + > B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) + > C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by sum of > the combinations all the A(i) and B(j) satisfying i+j=k, where i, j > and k range from 0 to m, n and m+n-1 respectively. If that's one line I'm a Dutchman. It looks like 8 lines to me. Well, it is one long sentence, which could have been written in one long line. Try explaining the current Western method so brießy, if you can! > This doesn't look like it differs from the standard definition > of long multiplication. Try to be a bit more clever, please. Also, look at my explanation to the other chap, that I just pos. > What's Vedic about it? People who inven the method consider themselves Vedic. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality > ... > To give a mathematical (and suitably, one-line!) definition of Vedic > multiplication: The multiplicative product of two numbers A and B > represen as a sequence of characters of length m and n > respectively, of base X, and of value A(m-1)*X^(m-1) + > A(m-2)*X^(m-2)+...+A(0)*X^0 and B(n-1)*X^(n-1) + > B(n-2)*X^(n-2)...+B(0)*X^0 will be C(m+n-1)*X^(m+n-1) + > C(m+n-2)*X^(m+n-2) +...+ C(0)*X^0, where any C(k) is found by sum of > the combinations all the A(i) and B(j) satisfying i+j=k, where i, j > and k range from 0 to m, n and m+n-1 respectively. > If that's one line I'm a Dutchman. It looks like 8 lines to me. Well, it is one long sentence, which could have been written in one > long line. Try explaining the current Western method so brießy, if > you can! The current western method? If it's small, do it by inspection, else use a calculator or computer. Much shorter, and applicable to a wider range of problems too. > This doesn't look like it differs from the standard definition > of long multiplication. Try to be a bit more clever, please. Sheesh, so says someone who thinks that reeordering _the same freaking number of operations_ is clever. For the kinds of calculations I do, your technique is _useless_. Utterly useless. It's not clever, it's a chocolate teapot. Sorry to burst your bubble, but you, or your vedic cronies, have taken a technique which requires theta(m*n) single-digit multiplications, and theta(2*m*n) single-digit accumulations, and turned it into a method that requires theta(m*n) single- digit multiplications and theta(m*n) two-digit accumulations -- with exactly the same multiplicative constants (where m and n are the lengths of the numbers. That's _exactly_ the same workload. I'm fully aware of the Ôwestern', Ôrussian peasant', Ôfrench peasant' and Ôvedic' methods for multiplication, and on the whole I find the Ôwestern' one the easiest to use if I've got a piece of paper and a pen handy, and I always have a piece of paper and a pen handy. The Ôvedic' method is basically the Ôfrench peasant' method with JIT multiplications (and thus not requiring the paper scratch area for the calculations). And it's _exactly_ the definition of multiplication as used in multiplication by convolution, which is the most common method of doing big number multiplication nowadays. However, the multiplication by convolution has an asymptotically much lower Big-Oh than the naive counting-on-fingers technique you describe above. > Also, look at my explanation to > the other chap, that I just pos. > What's Vedic about it? People who inven the method consider themselves Vedic. And the people who propagate it I consider vedic-fanboy drones. You're as bad as Mac-zealots and linux-weenies. If people are too lazy to find the technique for problem solution that suits them best then that's _their_ problem. Stop proselytising. -- Unpatched IE vulnerability: Security zone transfer Description: Automatically opening IE + Executing attachments Published: March 22nd 2002 Reference: http://security.greymagic.com/adv/gm002-ie/ === Subject: Re: Vedic Mathematics --- Myth and Reality >Now, in India there probably still exist a lot of people who never did >this long and rather silly-looking Western method, they always did it the >ancient Vedic way, which was so much better. Ironically, multiplication is far easier -- almost combinatoric in fact -- in Roman numerals. And with a minor adjustment, the system can be made positional (even without a zero!), with the algorithm suitable extended. It's not widely known, but a 0 is not a requirement for a positional number system -- not even one (in fact) capable of representing 0(!) or even negative numbers: something the Hindu-Arabic system can't do without extaneous markers (the - symbol). The Roman system is, in fact, nearly already that kind of system; making it (in conjunction with the above-mentioned adjustment) both combinatorial and algebraic. The generalized IV rule is rendered as such: For sequences of M,D,C,L,X,V,I of the form S = z0 a1 z1 a2 z2 ... an zn with z0 > z1 > z2 > ... > zn each single digits; ai (i=1,...,n) each a string of 0,1,2, or more digits all of value less than zi, the value of S is z0 + z1 + z2 + ... + zn - the sum of the values of a1,a2,...,an The abovementioned adjustments are (1) Everything under a bar counts 1000-fold (2) Everything after a comma counts 1000-fold (3) Everything after a dash counts 10-fold. (4) Sequences are interpre only between dashes, commas and under bars. That makes the Roman system a superset of a positional 0-less system, with an interesting combinatorial addition and multiplication algorithm. >In lay language, this means gathering all the units, tens, hundreds, >etc. in just one go, by finding out all the possible combinations for >same. It's the same thing as the modern grade-school method done in a different order. Western methods include methods based on the fast Fourier transform, which can multiply 2 N digit numbers using on the order of N ln(N) multiplications, instead of N^2. Even with a simple modification of the grade school technique, you can get away with on the order of I think somewhere around N^{8/5} multiplcations. I generally go 36*49 = 1254 + 90*13 - 660 = 1764, or 57*28 = 1056 + 120*10 - 660 = 1596, or 73*39 = 2127 + 100*12 - 480 = 2847. === Subject: Re: Vedic Mathematics --- Myth and Reality >Now, in India there probably still exist a lot of people who never did >this long and rather silly-looking Western method, they always did it the >ancient Vedic way, which was so much better. Ironically, multiplication is far easier -- almost combinatoric in fact -- > in Roman numerals. And with a minor adjustment, the system can be > made positional (even without a zero!), with the algorithm suitable > extended. Try writing 123987739383883938989874 in Roman numerals. > It's not widely known, but a 0 is not a requirement for a positional > number system -- not even one (in fact) capable of representing 0(!) > or even negative numbers: something the Hindu-Arabic system can't > do without extaneous markers (the - symbol). They used an abacus to give a visual representation of a large number. > The Roman system is, in fact, nearly already that kind of system; > making it (in conjunction with the above-mentioned adjustment) both > combinatorial and algebraic. The generalized IV rule is rendered as such: > For sequences of M,D,C,L,X,V,I of the form > S = z0 a1 z1 a2 z2 ... an zn > with z0 > z1 > z2 > ... > zn each single digits; > ai (i=1,...,n) each a string of 0,1,2, or more digits > all of value less than zi, > the value of S is > z0 + z1 + z2 + ... + zn - the sum of the values of a1,a2,...,an Did the Romans have any knowledge of algebra? > The abovementioned adjustments are > (1) Everything under a bar counts 1000-fold > (2) Everything after a comma counts 1000-fold > (3) Everything after a dash counts 10-fold. > (4) Sequences are interpre only between dashes, commas and under bars. That makes the Roman system a superset of a positional 0-less system, > with an interesting combinatorial addition and multiplication algorithm. >In lay language, this means gathering all the units, tens, hundreds, >etc. in just one go, by finding out all the possible combinations for >same. It's the same thing as the modern grade-school method done in a different > order. Tell us about the grade-school method. What it is, and how it is the same as the method given. Only one coefficient is being taken at a time, in this method, for the computation. This method may well have been rediscovered lately, and used (after all the book mentioned was published in 1965, and the methods described there are far older), but certainly this method is not taught in schools, as we find out from other posters in this thread. My main point is to show that the writer of that book is not a fraud. His methods can be very useful, as opposed to current methods taught in schools. If the Vedic method is already used, fine! > Western methods include methods based on the fast Fourier transform, > which can multiply 2 N digit numbers using on the order of N ln(N) > multiplications, instead of N^2. Okay, so multiply 12345 by 67809 using FFT, right here. Show us how it is done more simply than what I have done below. You have to do a lot better if you want to debunk this method, convincingly. > Even with a simple modification of > the grade school technique, you can get away with on the order of I think > somewhere around N^{8/5} multiplcations. I generally go 36*49 = 1254 + 90*13 - 660 = 1764, > or 57*28 = 1056 + 120*10 - 660 = 1596, > or 73*39 = 2127 + 100*12 - 480 = 2847. I don't know what you are trying to prove, here!! You certainly have not succeeded in multiplying 12345 by 67809 by *any* method! *)*) Oh, by the way, looks like I have not done my one-line multiplication of 12345 by 67809. Not that anyone seemed really interes! The one line answer is (6)(19)(40)(61)(91)(85)(67)(36)(45) or 837102105. With a little bit of practice, one get this quicker. Following the method that I had rigorously defined in the earlier post, (but which has been dele here), and of course using base X as 10 (our familiar decimal system) : (45) = 5*9 (36) = 4*9+5*0 (67) = 3*9+8*5+4*0 (85) = 2*9+7*5+3*0+4*8 (91) = 1*9+6*5+2*0+7*4+3*8 (61) = 4*6+1*0+2*8+3*7 (40) = 1*8+6*3+2*7 (19) = 1*7+2*6 (6) = 1*6 As is evident, you always need only to take only one digit at a time. The whole algorithm is also very well sui for parallel processing. Especially, when fixed lengths are involved. Good that someone pos this to sci.math. Hopefully, some among you will check out my mathematical derivation for unlimi energy, in my homesite, and see if you can find any ßaws. http://www.users.bigpond.com/adda1234/index.htm Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality >> Ironically, multiplication is far easier -- almost combinatoric in fact -- >> in Roman numerals. And with a minor adjustment, the system can be >> made positional (even without a zero!), with the algorithm suitable >> extended. >Try writing 123987739383883938989874 in [the above-mentioned >modification of the] Roman numerals [this reply is in reference to]. CXXIII,XIIIM,DCCXIL,CXXDIII,CXXMIII,XIILM,XIM,CXXXMIV 123987739383883938989874 >> It's not widely known,... >> do without extraneous markers (the - symbol). >They used an abacus to give a visual representaiton of a large number. Irrelevant to what this is in reply to. > Did the Romans have any knowledge of algebra? Likewise, irrelevant. [Actually, the theory of fields and of Euclidean geometry are equivalent, each does precisely what the other does. So, even the Greeks knew (and used) algebra -- and even (as recently discovered) Calculus, via Archimedes.] >Tell us about the grade-school method. O(N^2) pairs for two N-digit numbers. >What it is, and how it is the same as the method given. O(N^2) pairs for two N-digit numbers. Everything else is irrelevant. >Okay, so multiply 12345 by 67809 using FFT, right here. Wrong question. The correct question is: Try multiplying 34912340291341230948213049821340982134098213490812309482139401 23434234- 23423423423412309482310948231094812309481230498123094812039481 23342342- 94230923423342342342391423049128340921384012398401239840123942 34234234 by 23423421341234092138402193480219348021394801239840213984023198 40123433- 23421341234213423141230498123049821341234213412342134231423142 34234233- 98509582340985203495834029580298345094385093480349850349850349 58454434 by any method *BUT* the well-known FFT-based method. search.yahoo.com/search?p=FFT-Based Multiplcation >Show us how it is done more simply than what I have done below. O(N^2) vs. O(N ln(N)) operations, for 2 N digit numbers. >You have to do a lot better if you want to debunk this method, >convincingly. No. It's O(N^2). >> I generally go 36*49 = 1254 + 90*13 - 660 = 1764, >> or 57*28 = 1056 + 120*10 - 660 = 1596, >> or 73*39 = 2127 + 100*12 - 480 = 2847. >I don't know what you are trying to prove, here!! Of course not. The method is O(N^log_2(3)) and is fairly well-known. > You certainly have not succeeded in multiplying 12345 by 67809 It works quite well there too. >(45) = 5*9 >(36) = 4*9+5*0 >(67) = 3*9+8*5+4*0 >(85) = 2*9+7*5+3*0+4*8 >(91) = 1*9+6*5+2*0+7*4+3*8 >(61) = 4*6+1*0+2*8+3*7 >(40) = 1*8+6*3+2*7 >(19) = 1*7+2*6 >(6) = 1*6 25 = 5^2 multiplications; O(N^2). >As is evident, you always need only to take only one digit at a time. O(N^2) operations required. Where they're done (in your head or on paper) is irrelevant. N^2 is N^2. >The whole algorithm is also very well sui for parallel processing. FFT-based methods, that is. An O(N^2) method is very poorly sui due to its inefficiency and the large degree of intermixing. An FFT-based method does not require ANY cing at all. Other parallel methods exists which don't require any multiplication whatsoever and operate in O(N) time with O(N) space. === Subject: Re: Vedic Mathematics --- Myth and Reality > My main point is to show that the writer of that book is not a fraud. > His methods can be very useful, as opposed to current methods taught > in schools. If current methods are indeed useless, how did people build the Golden Gate Bridge or the Queen Mary? > If the Vedic method is already used, fine! >Western methods include methods based on the fast Fourier transform, >>which can multiply 2 N digit numbers using on the order of N ln(N) >>multiplications, instead of N^2. === Subject: Re: Vedic Mathematics --- Myth and Reality My main point is to show that the writer of that book is not a fraud. > His methods can be very useful, as opposed to current methods taught > in schools. If current methods are indeed useless, how did people build the Golden > Gate Bridge or the Queen Mary? I did not expect this sort of response from you, Ranjit. This sort I expect from the type of Mr Singh & Co. I never mentioned in any post that the current methods are useless; I said that this Vedic method is better. That did *not* imply that the current methods are *useless*. You are twisting my words, the same way Mr Singh does. As for your question, people built the Golden Gate Bridge and the Queen Mary the same way the Egyptian built pyramids. It is generally understood that the ancient Egyptians did not know trigonometry, or any modern mathematics. You don't have to know the best and most correct theories, and possess the latest toys, to build anything. Whatever that works, and is known to succeed, leads to sound engineering. Often, with a trial-and-error learning process. The Egyptians had plumblines and spirit-levels, and rulers (the same as modern builders of skyscrapers), and were very ingenious with whatever else they had. Similarly, they did not need supercomputers to build The Golden Gate Bridge, or Queen Mary. They probably did their work with log tables and slide rules, the way I did in my engineering studies. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality > My main point is to show that the writer of that book is not a fraud. >> His methods can be very useful, as opposed to current methods taught >> in schools. > If current methods are indeed useless, how did people build the Golden > Gate Bridge or the Queen Mary? Don't expect a sensible answer, Arindam is a nutjob ... -- ------ Got to get behind the mule in the morning and plow === Subject: Re: Vedic Mathematics --- Myth and Reality >Subject: Re: Vedic Mathematics --- Myth and Reality >Message-id: >Now, in India there probably still exist a lot of people who never did >>this long and rather silly-looking Western method, they always did it the >>ancient Vedic way, which was so much better. Ironically, multiplication is far easier -- almost combinatoric in fact -- >in Roman numerals. And with a minor adjustment, the system can be >made positional (even without a zero!), with the algorithm suitable >extended. It's not widely known, Ok. >but a 0 is not a requirement for a positional number system Are you sure about that? You're not using 0 and pretending it's not there are you? >-- not even one (in fact) capable of representing 0(!) >or even negative numbers: something the Hindu-Arabic system can't >do without extaneous markers (the - symbol). The Roman system is, in fact, nearly already that kind of system; >making it (in conjunction with the above-mentioned adjustment) both >combinatorial and algebraic. The generalized IV rule is rendered as such: > For sequences of M,D,C,L,X,V,I of the form > S = z0 a1 z1 a2 z2 ... an zn > with z0 > z1 > z2 > ... > zn each single digits; > ai (i=1,...,n) each a string of 0,1,2, or more digits > all of value less than zi, > the value of S is > z0 + z1 + z2 + ... + zn - the sum of the values of a1,a2,...,an The abovementioned adjustments are >(1) Everything under a bar counts 1000-fold >(2) Everything after a comma counts 1000-fold Was that supposed to be 100-fold? >(3) Everything after a dash counts 10-fold. >(4) Sequences are interpre only between dashes, commas and under bars. That makes the Roman system a superset of a positional 0-less system, >with an interesting combinatorial addition and multiplication algorithm. I'm not following this. Can you actually spell out a few examples of this system? It seems to me that M, D, C, L and X are not needed, but maybe I just don't understand your system. Why is M needed if _I represents 1000? >In lay language, this means gathering all the units, tens, hundreds, >>etc. in just one go, by finding out all the possible combinations for >>same. It's the same thing as the modern grade-school method done in a different >order. Western methods include methods based on the fast Fourier transform, >which can multiply 2 N digit numbers using on the order of N ln(N) >multiplications, instead of N^2. Even with a simple modification of >the grade school technique, you can get away with on the order of I think >somewhere around N^{8/5} multiplcations. I generally go 36*49 = 1254 + 90*13 - 660 = 1764, >or 57*28 = 1056 + 120*10 - 660 = 1596, >or 73*39 = 2127 + 100*12 - 480 = 2847. -- === Subject: Re: Vedic Mathematics --- Myth and Reality >My father taught me in my childhood days, >> 12 * >> 34 >> -------- >> 48 >> 36 >> -------- >> 408 > > this is the way children are still taught in at least >Australian primary schools, and in most Indian schools as well. As >you can see, you need three lines to get this result (48, 36 and 408). > In Vedic arithmetic, you get the representation of the result in just >one line, no matter what the length of the factors. > [...] As I recall from my teens, there is a book by Trachtenberg on methods of quick arithmetic. These include long mulitplication where the calculations are all done on one line. -- === Subject: Re: Vedic Mathematics --- Myth and Reality >My father taught me in my childhood days, >> 12 * >> 34 >> -------- >> 48 >> 36 >> -------- >> 408 > > this is the way children are still taught in at least >Australian primary schools, and in most Indian schools as well. As >you can see, you need three lines to get this result (48, 36 and 408). > In Vedic arithmetic, you get the representation of the result in just >one line, no matter what the length of the factors. [...] As I recall from my teens, there is a book by Trachtenberg on methods of > quick arithmetic. These include long mulitplication where the > calculations are all done on one line. Very likely. After all, this is an old method. In the book on Vedic Mathematics, they also do one-line division, square-root etc. Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality [snip] > As I recall from my teens, there is a book by Trachtenberg on methods of > quick arithmetic. These include long mulitplication where the > calculations are all done on one line. Very likely. After all, this is an old method. In the book on Vedic > Mathematics, they also do one-line division, square-root etc. Do you know how old exactly? And... do you have a reference for that book? (Online somewhere?) The Ôone-register' method of multiplication is indeed very nice. But... it is arithmetic, rather than mathematics. That may also explain why the math teaching world is not very fond of tricks like this. Mathematics teachers are desperately fighting the general prejudice that mathematics is Ôjust' arithmetic. Herman Jurjus Arindam Banerjee. === Subject: Re: Vedic Mathematics --- Myth and Reality >[snip] >>As I recall from my teens, there is a book by Trachtenberg on methods of >quick arithmetic. These include long mulitplication where the >calculations are all done on one line. Very likely. After all, this is an old method. In the book on Vedic >>Mathematics, they also do one-line division, square-root etc. > >Do you know how old exactly? >And... do you have a reference for that book? (Online somewhere?) The Ôone-register' method of multiplication is indeed very nice. >But... it is arithmetic, rather than mathematics. That may also explain >why the math teaching world is not very fond of tricks like this. >Mathematics teachers are desperately fighting the general prejudice >that mathematics is Ôjust' arithmetic. I doubt this had anything to do with Vedic mathematics (though I may just be unaware of the sources). It was a system partly inven (or perhaps merely collec) by Jacow Trachtenberg in the twentieth century. The standard reference in English is The Trachtenberg Speed System of Basic Mathematics by Cutler and McShane, though this was transla and adap from Trachtenberg's original German (I think). You can find it at Amazon. The descriptions available on the web (search for Trachtenberg and arithmetic) mainly talk about tricks for multiplying by single digits, 11, and 12, which is the first part of the book. Later chapters discuss techniques of long addition, multiplication, and division. -- === Subject: Re: Vedic Mathematics --- Myth and Reality >Do you know how old exactly? >And... do you have a reference for that book? (Online somewhere?) The Ôone-register' method of multiplication is indeed very nice. >But... it is arithmetic, rather than mathematics. That may also explain >why the math teaching world is not very fond of tricks like this. >Mathematics teachers are desperately fighting the general prejudice >that mathematics is Ôjust' arithmetic. > I doubt this had anything to do with Vedic mathematics (though I may > just be unaware of the sources). It was a system partly inven (or > perhaps merely collec) by Jacow Trachtenberg in the twentieth > century. The standard reference in English is The Trachtenberg Speed > System of Basic Mathematics by Cutler > original German (I think). You can find it at Amazon. The descriptions available on the web (search for Trachtenberg and > arithmetic) mainly talk about tricks for multiplying by single digits, > 11, and 12, which is the first part of the book. Later chapters discuss > techniques of long addition, multiplication, and division. Thanks. But actually, i really *am* curious about what else there is in Vedic maths, not just the single-register multiplication. In the meantime i've found online sources on vedic mathematics. It appears to be (re)discovered not before 1910 or so. Convolution is older. Nevertheless, for historic reasons it could be worth a look. However... all they ever show is their Ôcwise-horizontal'. If you want to know the rest, they make you buy books. Another typical line: sorry, i'm no expert, so i'm unable to tell more; but it's all there, in our books / courses. Not very trustworthy... Herman Jurjus -- === Subject: Mike Turner in NY Times Science 11/11/03 I think we are so confused that we should keep an open mind to tinkering with gravity, said Dr. Michael Turner, a cosmologist at the University of Chicago. previously Q to Ed Witten: How can the cosmological constant be so close to zero but not zero? Ed Witten's answer: I really don't know. It's very perplexing that astronomical observations seem to show that there is a cosmological constant. It's definitely the most troublesome, for my interests, definitely the most troublesome, observation in physics in my lifetime. In my career that is. My answer to the same question is at http://qedcorp.com/APS/StarGate1.mov and http://qedcorp.com/APS/EmergentGravity.doc or same with .pdf extension for Acrobat General relativity predic the bending of light, the expansion of the universe and black holes, and has served as the foundation for modern cosmology, but theorists have never presumed that it would be the last word on gravity. For one thing, it is mathematically incompatible with the quantum laws very small distances or very high energies corresponding to the first moments after the Big Bang, where space and time become discontinuous, general relativity has to be merged with quantum theory, a project that has enged the present generation of physicists. But recently some experts have been wondering out loud if it is time to rewrite Einstein's version of the law as it applies to the other end of the length scale, to very long distances. The motivation comes from the predominance of what is sometimes called the dark sector in the universe. According to what has recently become a highly celebra standard model, ordinary atoms make up only 5 percent of the stuff of the cosmos. Some kind of mysterious dark matter, perhaps consisting of while the rest .84 a whopping 70 percent .84 consists of something even more mysterious, known as dark energy. Obviously a theory that leaves 95 percent of the universe unexplained is less than a complete triumph. Neither dark energy nor dark matter has been observed or detec directly. Each has been inferred from its gravitational effects on the tiny fraction of stuff we can see. As a result, some scientists have sugges that what astronomers have discovered in the last 20 years is their own ignorance of gravity. In particular, the discovery, five years ago, that the expansion of the universe is apparently accelerating, under the inßuence of that dark energy, has occasioned a re-evaluation of the old certainties. The simplest explanation for dark energy is something called the cosmological constant, first inven by Einstein, a cosmic repulsion caused by the energy residing in empty space. But attempts to calculate this energy have resul in numbers 1060 bigger than what astronomers have measured .84 so large that the universe would have blown apart before atoms or galaxies could have formed .84 causing theorists to throw up their hands. I think we are so confused that we should keep an open mind to tinkering with gravity, said Dr. Michael Turner, a cosmologist at the University of Chicago. As a result of all this, physics literature has become peppered with suggestions of ways to change gravity. This fall, given a choice of explanations for dark energy during cosmology workshop at the Kavli Institute for Theoretical Physics in Santa Barbara, Calif., 20 of the 44 participants vo for some variation of Einstein was wrong. Some of these proposals take their cue from the science-fiction-sounding string theory, the putative theory of everything, which holds out the possibility that our universe might be a 4-dimensional membrane (or brane) in an 11-dimensional space. nature in string theory would be stuck to the brane, like the nap on a rug. But the strings responsible for transmitting gravity would be able to drift away or leak into the meta-space surrounding the brane as they traveled along it from distant objects, according to a theory set forth in 2000 by Dr. Gia Dvali, Dr. Gregory Gabadadze and Dr. Massimo Porrati of New York University. The effect, they say, would be to make distant galaxies appear as if they were accelerating as they moved away from us. Also in a stringy vein is Cardassian expansion, named after the villainous race on Star Trek, and dreamed up by Dr. Katherine Freese and Dr. Matthew Lewis of the University of Michigan. According to their theory, the universe accelerates as a result of other branes tugging on our own. One can get an accelerating universe without having any dark energy, Dr. Freese said. Other theorists are going back and modifying general relativity possible equations that would carry out his ideas. But more complica equations might be necessary. That was the approach taken by Dr. Turner and his colleagues, Dr. Sean Carroll and Dr. Vikram Duvvuri of Chicago, and Dr. Mark Trodden of Syracuse. The result was a universe that would speed up as it got bigger and emptier. That might sound crazy, Dr. Turner said, but not any crazier than the idea 80 years ago that the universe would be expanding. The model raises as many questions as it answebut it and others like it are still worth pursuing, Dr. Carroll said. Something funny is going on when the universe gets to be 10 billion years old, he said, and none of our current ideas is standing up and declaring itself to be the right answer, so we have to be bold. === Subject: Re: Mike Turner in NY Times Science 11/11/03 I think we are so confused that we should keep an open mind to > tinkering with gravity, said Dr. Michael Turner, a cosmologist at the > University of Chicago. > previously Q to Ed Witten: How can the cosmological constant be so close to zero > but not zero? Close to zero? WTF does close mean? Is there a maximum value we can compare it to to see how close it is? -- Denis Loubet dloubet@io.com http://www.io.com/~dloubet === Subject: Re: JSH: $100,000 US offer, Abel Prize > One thing I've been fascina by as I've considered replies to my > posts is a loose group coordination between posteas some try to > post with math, and others just post various jibes, but all keep > focused on pushing the false notion that my rather basic argument > showing a problem with the definition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in > the math community understand that NONE of you can overcome that kind > of coordina effort, and even if you post your own version of my own > argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in > core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with > ONE person or one group that collaborates to produce the check by > machine from the Abel Prize, assuming, of course that I win it. This offer is rescinded. That is, I have changed my mind and am no longer offering to give $100,000 US or promise of any sum of money to anyone for helping me with my current research, including doing a computer check, or any other help whatsoever. I repeat there is no monetary offer on the table for helping me with recgnition for my work. Thank you for your time and attention. http://mathforprofit.blogspot.com/ === Subject: Re: JSH: $100,000 US offer, Abel Prize >Message-id: <3c65f87.0311111357.7fb8ad39@posting.google.com >> One thing I've been fascina by as I've considered replies to my >> posts is a loose group coordination between posteas some try to >> post with math, and others just post various jibes, but all keep >> focused on pushing the false notion that my rather basic argument >> showing a problem with the definition of algebraic integers is wrong. Now I've tried to use money before, but failed as I think people in >> the math community understand that NONE of you can overcome that kind >> of coordina effort, and even if you post your own version of my own >> argument, the rest of math society will shred you. However, the shortness and simplicity of the proof of the problem in >> core offers another solution--machine proof checking. That means I can make the offer of sharing at least $100,000 US with >> ONE person or one group that collaborates to produce the check by >> machine from the Abel Prize, assuming, of course that I win it. This offer is rescinded. Damn! >That is, I have changed my mind and am no >longer offering to give $100,000 US or promise of any sum of money to >anyone for helping me with my current research, including doing a >computer check, or any other help whatsoever. I repeat there is no monetary offer on the table for helping me with >recgnition for my work. Thank you for your time and attention. http://mathforprofit.blogspot.com/ -- === Subject: Re: Classification error, algebraic integer issue > Despite the history of science and physics in particular, it seems to > me that many of you believe that mathematics as a field is > invulnerable to a simple error. However I *have* found a > classification error that is over a hundred years old, as > mathematicians failed to realize there were numbers that had slipped > between the cracks of their classification scheme. > Basically, the ring of algebraic integers is too small to include all > the numbers they believe it includes, and it's relatively easy to > show. [Very bad math DELE ...] which proves that f must have 2 itself as a factor for g to be an >> algebraic integer. > OOPS! I've concluded that nothing in the post proves that statement. It turns out that based on what's given, maybe it's true, but it's not proven. OOPS indeed! Wait just a minute on this. We need to remind you of some things you said about this previously --- which proves that f must have 2 itself as a factor for g to be an >> algebraic integer. So simply beautiful that it handles the question of whether or not >people arguing with me are correct, in such a short space. > OOPS! Looks like the people arguing with you WERE correct, right? Just for future reference: Were they lying? Were they cranks? Were they conspiring to suppress your startling discovery, even though they secretly knew your proof was right ? Is that what happened ? And will you remember this the next time a disagreement with your b.s. comes up, and you accuse us of lying ??? Possibly I moved too quickly here, as notice that the coefficient >prove that two of the roots share factors with 2, while just *one* >does not, as the last two are even, while the second is not. > OOPS! Possibly you *did* move too quickly here - but you have often been quick to conclude what you *want* rather than what is correct. Easy mathematics, with such a clear result is quite satisfying. > OOPS! Easy mathematics, yes, but also *wrong* mathematics. *No* clear result, and *not* quite satisfying. Looks like you gloa too soon! >> So c_1 and c_2 while not algebraic integers cannot be written as a >> ratio of non-unit coprime algebraic integers either. > OOPS! This whopper contradicts a famous well-known theorem! >Which is fascinating in and of itself, but remember, it's the same >thing for algebraic integers themselves, which should give you some >perspective on why these numbers belong in the same category. OOPS! Not so fascinating after all! Has it ever occurred to you that you are over-using that word? You are not, after all, the brainy second-in-command on the original Starship Enterprise, eh? It was a simple classification error made before any of you were born >as it's over a hundred years old, where I'm the one man who stepped >forward to point it out and fix it. > OOPS! Apparently the one man who stepped forward here ought to step back and think a little harder ... >That puts me in a singular place in human history. > Hard to disagree with this one! You continue to build your reputation as a world-class boob! >> These numbers represent an unexplored frontier of mathematics, and an >> opportunity for many of you to make an impact like you never could >> have dreamed of before in the field. >> OOOPS! These numbers represent the unexplored frontier of, uh, ... the empty set! Might be kind of hard to get a PhD thesis out of that! >> It's fresh ground. > OOOPS again! It's fresh ... what ? The point here: try, try, try very hard to remember this little fiasco, which you reproduced over the last few days several times in several different threads, the next time you claim your critics are liaincompetents, cranks, or whatever. WE'RE NOT!!! >http:// mathforprofit.blogspot.com / === Subject: Re: Classification error, algebraic integer issue > It's fresh ground. OOOPS again! It's fresh ... what ? Fresh ground coffee. Hey, sometimes people just forget a when they type. After all, didn't someone just say JSH is doing marketing research? He's selling _coffee_ fergoodnesssake. No wonder we didn't recognize it as _mathematics_! dave === Subject: Semdirect product of categories? Semidirect product is a common construction in group theory. But it is obvious that if one relaxes the required assumptions, it can be generalized to monoids. Now it occurs to me that another simple generalization step can bring the concept to a categorical formulation: let A,B be categories and let (A,A) (the monoid of functors A->A) be regarded as a category itself. Then choose a functor sigma:B->(A,A) and define (a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1). It is straightforward to check that the definition is well-posed. But now the question is: is it of any utility? Is it taken into account somewhere? Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Re: Semdirect product of categories? |Semidirect product is a common construction in group theory. But it is |obvious that if one relaxes the required assumptions, it can be |generalized to monoids. | |Now it occurs to me that another simple generalization step can bring |the concept to a categorical formulation: let A,B be categories and |let (A,A) (the monoid of functors A->A) be regarded as a category |itself. Then choose a functor sigma:B->(A,A) and define | |(a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) | |for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1). | |It is straightforward to check that the definition is well-posed. But |now the question is: is it of any utility? Is it taken into account |somewhere? actually i'm not sure i understand your notation well enough to see whether what you're describing actually works, but i'm a bit skeptical about it because you don't seem to be giving the most straightforward generalization of the semi-direct product construction to a context involving arbitrary categories. the most straightforward such generalization is some version of the homotopy colimit construction. you start with a category c and a functor (or in some versions a pseudo-functor) f from c to the category of categories. the homotopy colimit of f is the category where an object is a pair (x,y) with x an object in c and y an object in f(x), and a morphism from (x,y) to (z,w) is a pair (m,n) with m:x->z in c and n:f(m)(y)->w in f(z), with composition of morphisms defined in a semi-obvious way. semi-direct product of monoids is then the special case where c has a unique object x and f(x) has a unique object y. one version of the homotopy colimit construction is called the grothendieck construction and has some pretty important uses in mathematics. the name homotopy colimit that i used above is perhaps not very standard terminology, but considering the many different version of the construction which have been studied, i'm not sure whether standard terminology really exists in this case. some names for other versions (or in some cases maybe the same version) of the construction that you might want to look up include 2-colimit, pseudo-colimit, lax colimit, and so forth. by the way the most general version of the construction is general enough to encompass not just semi-direct products of groups but in fact arbitrary group extensions. the extra twisting data needed to describe an arbitrary group extension can be considered as part of the data forming a pseudo-functor. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Semdirect product of categories? |Now it occurs to me that another simple generalization step can bring >the concept to a categorical formulation: let A,B be categories and >let (A,A) (the monoid of functors A->A) be regarded as a category >itself. Then choose a functor sigma:B->(A,A) and define (a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) for all a2,b2,a1,b1 s.t. Dom(a2)=sigma(b2)Cod(a1), Dom(b2)=Cod(b1). actually i'm not sure i understand your notation well enough to see >whether what you're describing actually works, but i'm a bit skeptical >about it because you don't seem to be giving the most straightforward >generalization of the semi-direct product construction to a context >involving arbitrary categories. Well, I didn't know the construction mentioned hereafter, and in this respect my one is indeed extremely naive. But I don't see how it could not work, and what is not clear: sigma is a functor B->(A,A), so sigma(b) is a functor A->A for all morphisms b. Its action on objects is obviously trivial. Note that sigma(e)=1_A (the identical functor) for all identities e of B. The notation (a2,b2)(a1,b1):=(a2sigma(b2)a1,b2b1) is a straightforward generalization of that used for groups/monoids. The only difference being that the obvious compatibility relation for the composition of the given morphisms is required: a1:X->Y, a2:sigma(b2)Y->Z, b1:S->T, b2:T->U. Verification that both the associative and the identity properties hold is merely a mechanical task. >the most straightforward such generalization is some version of the >homotopy colimit construction. you start with a category c and a >functor (or in some versions a pseudo-functor) f from c to the >category of categories. the homotopy colimit of f is the category >where an object is a pair (x,y) with x an object in c and y an object >in f(x), and a morphism from (x,y) to (z,w) is a pair (m,n) with >m:x->z in c and n:f(m)(y)->w in f(z), with composition of morphisms >defined in a semi-obvious way. semi-direct product of monoids is then the special case where c has a >unique object x and f(x) has a unique object y. Then if I'm not mistaken (or misunderstanding), my construction is the special case when c is arbitrary and f(x) has a unique object A (in the notation above). In other words it is the same construction with the difference that the codomain of f is restric to a subcategory of the category of categories, namely the category of functors A->A. Michele -- > Comments should say _why_ something is being done. Oh? My comments always say what _really_ should have happened. :) - Tore Aursand on comp.lang.perl.misc === Subject: Call for Participation , NFSNET is a distribu computing project devo to pushing factorization limits. It allows users to run client software which sends data back to a central site. The data is collec and eventually used to produce a factorization. However, the data is sent via an HTTP connection, which means that clients must be on-line. I am working on trying to do some slightly smaller numbebut are on the ÔMOST-WAN' list for factorizations. I could use some help, as I have very few machines. I will provide executables (even source if people want to look at it) and data via email. People will then run the code OFF-LINE then email the results back. The current project is to factor 2^653+1, which is the smallest unfactored number of the form 2^n+1 and is in fact THE ÔMost-Wan' factorization. We will then proceed with other numbers of the form 2^n+1. I solicit your help. Please email me at rsilverman@draper.com. I am posting from my AOL account because I do not have newsgroup access at work. The method for doing this work is the way NFSNET used to work, before it was automa. But this older method allows off-line processing. Bob Silverman You can lead a horse's ass to knowledge, but you can't make him think. === Subject: Re: Call for Participation > , NFSNET is a distribu computing project devo to pushing factorization > limits. It allows users to run client software which sends data back to a > central > site. The data is collec and eventually used to produce a factorization. However, the data is sent via an HTTP connection, which means that clients > must be on-line. I am working on trying to do some slightly smaller numbebut are on the > ÔMOST-WAN' list for factorizations. I could use some help, as I have very > few machines. I will provide executables (even source if people want to look at it) and > data via email. People will then run the code OFF-LINE then email the results back. Bob, I wish you luck with this, but wouldn't it be easier to try to bolt on off-line-ness onto NFSnet? Either way, I heartily recommend that those who are sitting around with PCs doing nothing useful volunteer to help out with this. -- Unpatched IE vulnerability: NavigateAndFind file proxy Description: c-domain scripting, cookie/data/identity theft, command execution Reference: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ PU-Content.HTM Exploit: http://safecenter.net/liudieyu/NAFfileJPU/NAFfileJ PU-MyPage.htm === Subject: Re: partial deriviatives! >> if you have a function F(q(t),t) >> and you have the derivative >> d/dt( d/dq[F(q(t),t) ) >> then it should be >> d^2/dq^2*dq/dt[F(q(t),t)] + d^2/dtdq[F(q(t),t)] >> right? >> Or am i mistaken...?? >> /M Assuming that you are asking for the total time derivative (not the partial >time derivative), then the answer is [shnip] but the real question is... is the d/dq a total q derivative? So (with the notation for partial derivative as pF/px): let q(t) be invertable d/dq[ F(q,t(q)) ] = (pF/pq) + (pF/pt)(dt/dq) let pF/pq == G(q(t),t) let pF/pt == H(q(t),t) d/dt[ G(q(t),t) + H(q(t),t) (dt/dq) ] = pG/pt + (pG/pq)(dq/dt) + pH/pt (dt/dq) + (pH/pq)(dq/dt)(dt/dq) = pG/pt + pH/pq + (pG/pq)(dq/dt) +(pH/pt)(dt/dq) == (p^2F/ptpq) + (p^2F/pqpt) + (p^2F/pq^2)(dq/dt) + (p^F/pt^2)(dt/dq) = ((p^2F/pq^2)(dq/dt) + (p^2F/pqpt)) + (p^F/pt^2)(dt/dq) + (p^2F/ptpq) ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ where this is the answer if only the OP d/dq means partial adam === Subject: Re: Four Color Graphs > Every complete 4-partite graph is four-colorable. The Mathworld > descriptions of Complete Graph, Complete k-Partite Graph, > k-Partite Graph are essentially appropriate to the following > discussion. The complete graph Kn is of course n-partite. In Mathworld, the > complete k-partite graph (Ck) is deno Kp,q, ... ,r; where p+q+ ... > +r = n. Here, a slightly different nomenclature is adop; ie, Ck = > (P1,P2,P3, ... Pk). > To minimize confusion, let Pi represent partition Ôi' and pi represent > the nummer of vertices in partition Ôi'. So we can also write that Ck = {p1, p2, p3, ... , pk} Then, C4 = {p1, p2, p3, p4}. The number of edges (Ec4) in C4 is, Ec4 = p1*(p2+p3+p4) + P2*(p3+p4) + p3*p4 > For example, let p1 = p2 = p3 = p4 = n/4; then Ec4 = n/4*(n/4+n/4+n/4) + n/4*(n/4+n/4) + n/4*n/4 = > [6*(n/4*n/4)} > Ec4 = .375*(n^2) Let n = 16, then pi = 4; and Ec4 = .375*(16*16) = 96. To check > Ec4 = 4*(4+4+4) + 4*(4+4) + 4*4 = 48 + 32 + 16 = 96. If we let Ec4 = INT(0.375*(n^2)), then the formula works for all > values of n when p1 ~ p2 ~ p3 ~ p4. For example; for n = 14, C4 = {4, > 4, 3, 3}. Without wrting out the calculations in detail; Ec4 = 4*10 + 4*6 + 9 = 40 + 24 + 9 = 73 > Ec4 = INT(.375*(14^14)) = INT(.375*(196)) = 73 At the other end of the spectrum, let p1= (n-3), p2 = p3 = p4 = 1. > Then Ec4 = (n-3)*(1+1+1) + 1*(1+1) + 1*1 = 3*n -6 We propose that 3*n-6 <= Ec4 <= 0.375*(n^2) for all n and all possible > partitionings thereof. Let n =8. The following partitionings are possible; {2,2,2,2), {3.2.2,1}, {3,3,1,1}. {4,2,1,1} & {5,1,1,1}. Howwever, no planar graph may have more than INT(n/2) vertices in any one partition. So the partitioning {5,1,1,1} is not applicable. Then Partitioning max edges max diagonals {2,2,2,2} 24 16 {3,2,2,1} 23 15 {3,3,1,1} 22 14 {4,2,1,1} 21 13 Note that any planar graph may be depic as an n-sided polygon with up to (2*n-6) diagonals. If n = 8, then 2n-6 = 10. Therefore, the partitioning {2,2,2,2) can be reduced to Binomial(16,10) = 8,008 different 4-color graphs with (3n-6)= 18 edges. But there are only 2,772 planar graphs with 8 vertices and 18 edges. (see HYPOTHESIS below) The total number of configurations due to all four partitionings is 12,298. It is impossible to determine the number of duplicate configurations. Further, there are Binomial(20,16) = 4,845 complete 4-partite graphs with n=8 and {2,2,2,2} partitioning. So it is conservatively safe to say that there are > 8,008 4-colorable graphs with 8 vertices and 18 edges. Of which, only 2,772 may be planar. HYPOTHESIS, The number of maximal planar graphs is [C_(n)*C_(n-1)]/2; where C_(n) and C_(n-1) are the Catalan numbers for (n) and (n-1). Specifically; C_(8) = 132, C_(7) = 42. Afterthought. A possible counter example to the FCT would be a complete 4-partite graph with less than 3n-6 edges; if one exis. As n increases, the ratio of all 4-color graphs to 4-color planar graphs increases rapidly. === Subject: JSH: Deprogramming needed? (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math discoveries? After all, I've been at this since April 1995 having spent a lot of time and effort, with literally thousands of posts along with all kinds of other activities, websites, and email to mathematicians all over the world. But, what if I'm wrong? What if all the time and energy I've inves in my work has made it difficult for me to see, along with the *harsh* and unforgiving hostility from several people who seem to have made it their mission to make me miserable and find pleasure in mocking or trying to humiliate me, have made it extremely difficult for me to see the truth? Think about the crushing sense of shame and misery if indeed I find out that the logical connections I so carefully and impatiently discovered over the years are simply not really there, but are a need induced delusion. It seems to me that marketing ploy though these statements may be, dealing with people who've made it their business to try and make me miserable, only to at times claim they're trying to help me is just too much. Aren't there any *other* people who suppose they are rational, who can follow a logical argument, who might comment? Why is it always the same people? Or people imitating them in hostile and mocking displays of animosity or anger? Aren't there any rational people who can trace out the steps in the work I've presen, who might step forward at this time, and demonstrate an ability to just be objective? I don't want any replies of people offering, but if you might consider it, I just want you to think about it. Sure I've been reading this marketing book, but it seems to me that still *someone* out there might be the right person, as of course, I don't believe I'm wrong. I've traced out every step in the arguments I have, and I think that irrational people have been dominating the discussion using group effects to hide the truth. Think about it. I'll come back to the subject later. http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Deprogramming needed? Originator: grubb@lola >Aren't there any *other* people who suppose they are rational, who can >follow a logical argument, who might comment? Sure. The first question I have is what you mean by the Ôconstant term' of a non-polynomial. In particular, why do you assume that the constant term of 5a_1(x) +7 is 7? Why do you assume that contant terms for non-polynomials act the same way as for polynomials? You have not proved this anywhere, so that constitutes a gap in your argument. Next, why do you assume that 49 being a factor of P(x) means that 7 has to be a factor of one of the terms in step 5? This is not proved, so constitutes a gap in your proof. At this point, I see 2 significant gaps in your logic. That is enough for me to ignore the whole argument. >Why is it always the same people? Or people imitating them in hostile >and mocking displays of animosity or anger? My guess is that nobody else cares. The people that are commenting on your argument are generally respec because of other things they post about. They have shown themselves able to give proofs of their own and to comment intelligently on arguments that others give. Given the qualifications of those arguing against you, you would have to give arguments that don't have glaring ßaws to convince anyone else that you are right. Unfortunately, your argument *does* have glaring ßaws. In particular, you assume that Ôconstant terms' for non-polynomials work the same as they do for polynomials over the integers. This has not been proved. The lack of a proof is a glaring ßaw. Now you have the following choices. Either admit that your argument is incomplete and supply the missing details OR admit that your argument is incomplete and admit defeat OR not admit you argument is incomplete and admit that you are simply a troll. --Dan Grubb === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? > You have been before, many times. You know the consequences. At this point, it is not even clear what you are claiming. At one time you were saying that in the factorization of your cubic polynomial P(x), at least two of the coefficients were divisible in the algebraic integers by f, a prime number > 3. That has been proven false. You appear now to have backed off on that claim. Perhaps you are claiming that at least two of the a's are divisible by f in the ring of objects. That is of highly dubious value because the ring of objects is not well-defined, and there is no way to tell whether a given number is or isn't in that ring, and basic theorems about that ring are completely lacking. Bottom line, no one cares about the ring of objects. This avenue cannot lead to a proof that there is a core error or a proof of Fermat's Last Theorem. So the real question here now is not whether you are wrong. The question is, what are you claiming to be true? > After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. But, what if I'm wrong? > Then all that time was was. > What if all the time and energy I've inves in my work has made it > difficult for me to see, along with the *harsh* and unforgiving > hostility from several people who seem to have made it their mission > to make me miserable and find pleasure in mocking or trying to > humiliate me, have made it extremely difficult for me to see the > truth? > A highly compound question. Putting in all the time and energy is no guarantee of anything. The world does not owe you a living. As for the hostility: you have earned at least some of it. As for making it extremely difficult for you to see the truth: that is exactly what some of us have been trying to do. You have resis it with every atom in your body. > Think about the crushing sense of shame and misery if indeed I find > out that the logical connections I so carefully and impatiently > discovered over the years are simply not really there, but are a need > induced delusion. > Are you asking for sympathy or mercy, or what? If we honestly think you are wrong (we do!), we are not going to say you are right just because you will feel bad otherwise. You're an adult after all. > It seems to me that marketing ploy though these statements may be, > dealing with people who've made it their business to try and make me > miserable, only to at times claim they're trying to help me is just > too much. > If your marketing ploy here is just to beg for mercy, it simply doesn't work with mathematicians. What they want is just sound mathematics. In spite of your lead-off sentences here, I don't think you accept any possibility that you are wrong. We see it completely oppositely. If you continue to claim that your arguments show a core error or a proof of FLT, there is no possibility that you are right. We have found examples that contradict your claims - examples which you have not refu in any way - and we have poin to exact spots in your argument that your errors occur. What more do you need? > Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? Why is it always the same people? Or people imitating them in hostile > and mocking displays of animosity or anger? > We're the most persistent. > Aren't there any rational people who can trace out the steps in the > work I've presen, who might step forward at this time, and > demonstrate an ability to just be objective? > No. You should have no question by now that we HAVE traced out the steps in your argument, and we find them lacking. They do not constitute a proof of anything. > I don't want any replies of people offering, but if you might consider > it, I just want you to think about it. Sure I've been reading this > marketing book, but it seems to me that still *someone* out there > might be the right person, as of course, I don't believe I'm wrong. > I've traced out every step in the arguments I have, and I think that > irrational people have been dominating the discussion using group > effects to hide the truth. > Marketing strategy is a terrible guide if you want vindication as a mathematician. It is a waste of time. It's not a matter of salesmanship or presentation. It is entirely a matter of having or not having a sound mathematical argument. You don't. > Think about it. I'll come back to the subject later. http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Deprogramming needed? as S. Se Baroness says, you can always go back to necessity & sufficiency, that is to say, the use of those words in simple English wordproblemmas. don't know it htat's possible, after what they did to you at Vanderbilt U. you can continue to hide behind Devlin's skirts of Sophistry, if you;re prepared to define yourself as a left- or right-wing prover. or, just show yourself that you can prove (say) the pythagorean theorem, and the numbertheoretic application to triples -- the trivially *true* case of solutions to FLT.... you might say that the PT is obvious or trivial, but it's not the same as the earth is demonstrably round, which is easy to say without any knowledge of trigonometry or navigation or astronomy, since most of us have experienced jet-induced time-zone differentials; eh?... clearly, Devlin cannot prove it for you; that'd blow his cover!... I mean, how can you offer insight into the apparently infinite cases where there are no solutions, if you don't know how there are *any* solutions, at all?... what, as they say, is the point? I want to thank another poster (GG?) for stating that your run-around with algebraic integers is crucial to the case for N=3; I'd never have known that! now, I officially quit; let others deal with the nostrums of marketing, as they already do from X hours of TV per day!... or they don't; how else would Captain Arnie have become the elec, here in the Sunshine State? so, your evil plan could actually work, I guess. welcome to the googolplex; there's a sucker born every nanosecond! > So the real question here now is not whether you are wrong. The > question is, what are you claiming to be true? > http://mathforprofit.blogspot.com/ --Trickier Dick Cheney's cover-up! === Subject: Re: JSH: Deprogramming needed? X-DMCA-Notifications: http://www.giganews.com/info/dmca.html >(Marketing ploy warning) So you warn us in advance that this post is not going to contain logical arguments explaining why you're right, but it's going to try to persuade for non-logical reasons instead. Thanks for the warning, but it wasn't needed - this is clear you're right by whining about how awful it would be if you were wrong. If anything this will have the opposite effect: Your whining about how awful it would be if you were wrong will not change the opinion of anyone competent to evaluate your work, but it _could_ convince people who know no math that you must be wrong, because if you were right you wouldn't have to use marketing ploys. Duh. >What if indeed I *am* wrong, and I don't have these great math >discoveries? After all, I've been at this since April 1995 having spent a lot of >time and effort, with literally thousands of posts along with all >kinds of other activities, websites, and email to mathematicians all >over the world. But, what if I'm wrong? What if all the time and energy I've inves in my work has made it >difficult for me to see, along with the *harsh* and unforgiving >hostility That hostility is not because you're wrong. You need to tell yourself that the reason people are so hostile is because they can't stand the possibility that you might be right. But that's just one of your many self-aggrandizing delusions - in fact the hostility stems from the fact that you're so hostile to others. You don't believe that. But you must agree that you _are_ hostile to others. For example you've complained to my employer about me being mean to you, but you've also called me a sing piece of dogs - you remember that, right? As evidence in support of the idea that people would not be so nasty to you if you were not nasty to them I offer the case of Finlayson. Look up some of his posts through the years. Essentially everything he says is wrong, usually hilariously obviously incoherently wrong. But he doesn't get the same sort of hostile reaction that you do - people explain he's all wrong without calling him nasty names. >from several people who seem to have made it their mission >to make me miserable and find pleasure in mocking or trying to >humiliate me, have made it extremely difficult for me to see the >truth? Think about the crushing sense of shame and misery if indeed I find >out that the logical connections I so carefully and impatiently >discovered over the years are simply not really there, but are a need >induced delusion. It seems to me that marketing ploy though these statements may be, >dealing with people who've made it their business to try and make me >miserable, only to at times claim they're trying to help me is just >too much. Aren't there any *other* people who suppose they are rational, who can >follow a logical argument, who might comment? Are you ever going to give up, and just decide that you're the only rational person in the universe? I mean that's what the evidence seems to indicate: There are no rational people on sci.math, the mathematicians you harass via email are not rational, the mathematicians you visit in person are not rational, the editors of journals you send papers to are not rational... Now, the conclusion a _rational_ person would draw from this was that maybe he was actually _wrong_, that it's just not rational to assume that _all_ those people share the same irrationality. Of course concluding that is beyond you, because it would destroy the picture of you as supreme genius that you cling to do desperately to give your otherwise empty life some meaning. Fine. But I'd think that at some point even you would be rational enough to just give up and conclude that you were the only rational person on the planet, instead of repeating these pleas for someone rational to step forward. You've been begging for that for yeaand it never happens. It's pathetic, actually. >Why is it always the same people? Or people imitating them in hostile >and mocking displays of animosity or anger? Aren't there any rational people who can trace out the steps in the >work I've presen, who might step forward at this time, and >demonstrate an ability to just be objective? I don't want any replies of people offering, but if you might consider >it, I just want you to think about it. Sure I've been reading this >marketing book, but it seems to me that still *someone* out there >might be the right person, as of course, I don't believe I'm wrong. >I've traced out every step in the arguments I have, and I think that >irrational people have been dominating the discussion using group >effects to hide the truth. Think about it. I'll come back to the subject later. http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Deprogramming needed? Discussion, linux) Cancel-Lock: sha1:NoXKuQcKQ1ac0dRnXMe5om8FiOc= > Are you ever going to give up, and just decide that you're the only > rational person in the universe? I mean that's what the evidence > seems to indicate: There are no rational people on sci.math, the > mathematicians you harass via email are not rational, the > mathematicians you visit in person are not rational, the editors > of journals you send papers to are not rational... It's not such a new conclusion. Here's one of my long-time favorite .sigs. [I]t's the damndest thing. There's something wrong with every last one of you, and I *never* thought that was a possibility. But now I feel it's the only reasonable conclusion. --JSH sees some sorta light -- Jesse Hughes You see 300 of something, anything, and you go `[Man], that's a lot of stuff.' -- Jim Bigler, quo in the Pittsburgh Post-Gazette. Subject: Re: JSH: Deprogramming needed? === > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? You are. After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. Effort doesn't indicate accuracy. I could spend 8 years throwing a football, running laps, etc, and believe that I would make a great quarterback. Unfortunately, with low stamina, lousy aim, and a fairly light frame, I'd be more likely to get killed. My effort and belief won't get me into the NFL. But, what if I'm wrong? Then you're wrong. Very simple. What if all the time and energy I've inves in my work has made it > difficult for me to see, along with the *harsh* and unforgiving > hostility from several people who seem to have made it their mission > to make me miserable and find pleasure in mocking or trying to > humiliate me, have made it extremely difficult for me to see the > truth? Back up. The hostility doesn't exists except as a reaction to rudeness or hostility. Disagreement does not mean hostile. Also, if you are miserable, that is a result of your choices. You may choose to be miserable or not, but we cannot make you so, especially since we can't even make you pay attention to what we say. Think about the crushing sense of shame and misery if indeed I find > out that the logical connections I so carefully and impatiently > discovered over the years are simply not really there, but are a need > induced delusion. There is no shame in being wrong. Most of us have experienced it before. It seems to me that marketing ploy though these statements may be, > dealing with people who've made it their business to try and make me > miserable, only to at times claim they're trying to help me is just > too much. These are a marketing ploy only in that they are a blatant play for sympathy. If you were to admit you've was 8 years of your life, you would have it. As you haven't made the admission yet, there is a different priority. Also, the unfortunate consequences of it being a waste of your time does not make you right. Aren't there any *other* people who suppose they are rational, who can > follow a logical argument, who might comment? Tons. But with the specialized topic, they will all be people with at least a bachelor's degree in math. This makes looking in certain directions unlikely to produce intelligent comment because of the required level of knowledge to intelligently look for ßaws. Why is it always the same people? Or people imitating them in hostile > and mocking displays of animosity or anger? Because you've tapped the well. Aren't there any rational people who can trace out the steps in the > work I've presen, who might step forward at this time, and > demonstrate an ability to just be objective? You are either assuming your work is simpler than it is, or you are not as clear as you believe you are. I don't want any replies of people offering, but if you might consider > it, I just want you to think about it. Sure I've been reading this > marketing book, but it seems to me that still *someone* out there > might be the right person, as of course, I don't believe I'm wrong. > I've traced out every step in the arguments I have, and I think that > irrational people have been dominating the discussion using group > effects to hide the truth. Try an experiment. Try to work with the idea that perhaps you *are* wrong and understand the details of the objections. If you *are* right, you have nothing to lose. If you are wrong, perhaps you will learn why. Think about it. I'll come back to the subject later. You have pos a plea for agreement with your argument based on sympathy. This is not the basis for a mathematical proof. If you can't come up with better advice from your marketing book, perhaps you should ditch it. -- === Subject: Re: Deprogramming needed? boundary=----=_NextPart_000_009C_01C3A8CD.A1B79170 --------- charset=iso-8859-1 > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. Since April 1995 and not right yet? === Subject: Re: JSH: Deprogramming needed? > What if all the time and energy I've inves in my work has made it > difficult for me to see, along with the *harsh* and unforgiving > hostility from several people who seem to have made it their mission > to make me miserable and find pleasure in mocking or trying to > humiliate me, have made it extremely difficult for me to see the > truth? Think about the crushing sense of shame and misery if indeed I find > out that the logical connections I so carefully and impatiently > discovered over the years are simply not really there, but are a need > induced delusion. Fact is: You have written your own kick me sign and attached it firmly to your back. Your own behavior makes everyone want to kick you. Something good comes from it occasionaly - for example the fast prime counting function on my web site (which by the way beats yours speed wise by a factor of about THOUSAND. It was a pleasure for me to kick your teeth out). Get medical help. === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? > Then you should probably stop posting to sci.math and find something more productive to do with your time. [rest dele] === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) > What if indeed I *am* wrong, and I don't have these great math > discoveries? Then you should probably stop posting to sci.math and find something > more productive to do with your time. [rest dele] Why? I've spent a lot of time talking about mathematics, so *clearly* I have some interest in the subject, and sci.math is a *public* forum for people with interest, many varying interests, in the subject of mathematics, so why would you try to convince me to stop posting? It seems to me that some of you just want to hurt people, and for some of you I'm just a target for your abuse, as I doubt any of you don't realize the nature of Usenet. So aren't there *any* reasonable people who might reply instead? If hateful people were judging their success on discouraging you from posting, would you give them victory? Why can't decent people on the newsgroups work at all to deßect these hostile people? My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) > What if indeed I *am* wrong, and I don't have these great math > discoveries? > Then you should probably stop posting to sci.math and find something > more productive to do with your time. > [rest dele] Why? I've spent a lot of time talking about mathematics, so *clearly* > I have some interest in the subject, and sci.math is a *public* forum > for people with interest, many varying interests, in the subject of > mathematics, so why would you try to convince me to stop posting? > Well, I meant stop posting about your alleged mathematical discoveries. I didn't realize you had any interest in posting about anything else. It was just an answer to a hypothetical question anyway. No need to get all rufßed. > It seems to me that some of you just want to hurt people, and for some > of you I'm just a target for your abuse, as I doubt any of you don't > realize the nature of Usenet. So aren't there *any* reasonable people who might reply instead? If hateful people were judging their success on discouraging you from > posting, would you give them victory? Why can't decent people on the > newsgroups work at all to deßect these hostile people? > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Deprogramming needed? > I've traced out every step in the arguments I have Are you a good tracker of steps? Do you still think that Z[1/2] contain 1/3? === Subject: Re: JSH: Deprogramming needed? you've already got your market-man, Devlin. you'll just have to decide with him, whether you are an L- or R-winger ... I guess. first, though, ask him if he can prove the pyhtagorean theorem, and generate the triples; perhaps, he can prove that it is Devlin-unsolvable. thank you! > (Marketing ploy warning) > I don't want any replies of people offering, but if you might consider > it, I just want you to think about it. Sure I've been reading this > marketing book, but it seems to me that still *someone* out there > might be the right person, as of course, I don't believe I'm wrong. > http://mathforprofit.blogspot.com/ --ils duces d'Enron! === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? After all, I've been at this since April 1995 having spent a lot of > time and effort, with literally thousands of posts along with all > kinds of other activities, websites, and email to mathematicians all > over the world. But, what if I'm wrong? That just could be why you are having so much trouble selling your product. === Subject: Re: JSH: Deprogramming needed? > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? We get more random numbers (from your post): ed6af0c92fa24f0810455fd94f4ccdaa David Bernier === Subject: Re: Deprogramming needed? X-DMCA-Notifications: http://www.giganews.com/info/dmca.html > (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? > << (Marketing ploy warning) What if indeed I *am* wrong, and I don't have these great math > discoveries? > << consider the state of mind of the speaker... After > years of people specifically detailing your mental > constructs and pointing out your errors in logic you > still have a doubt of your pass or fail status in life.... > Paul R. Mays So do you think I'm less or *more* likely to face the truth because of your post Mr. Mays? Aren't you trying to say that I'm a failure in LIFE in your comments? But aren't I a success in posting? If you push the notion that I'm a failure in life, why shouldn't I continue in an area where I'm clearly a success? Aren't you really working to *keep* me posting, possibly in fact, working to get me to post far more?