mm-64 === Some of you were probably surprised to learn that I did indeed ?d away to count prime numbers by integrating a partial differenceequation. Some of you probably STILL doubt that no one else inrecorded history has managed such a feat because you need to believein mathematicians.But my point is that mathematicians have gone rogue and act againstthe needs of society by de facto censorship of information that theydon't think makes them look good, like the information about mypartial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,there's the passive act of refusing to acknowledge the discoveryitself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,sqrt(y-1))],S(x,1) = 0.And p(x, y) = ?) - S(x, y) - 1, and you get S as the sum of dSfrom dS(x,2) to dS(x,y).That's it. That's the knowledge which mathematicians have purviewover, in terms of the expectation from society that importantinformation of a mathematical nature will be acknowledged bymathematicians.Note that it's a *discrete* function, so for you programmers thatmeans you need to use int's or long's or some discrete variable type. Also, if you wish to implement it, please sum from dS(x,2) up to andINCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, didyou ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will seethe method I've just shown you, unless maybe they're out on Usenetreading my posts, because the mathematical establishment thinks it canignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.But you see, what bene? do they see to their society by allowingthat someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let theworld be convinced I'm just a crank, most of them passively justsitting by, and keeping quiet about my results--after all, that'squite effective, eh?Then they have de facto censorship because people BELIEVE theywouldn't do such a thing if my work were important!!!So you have a standstill with me pushing my research, and a fewmathematicians actively ?hting its acceptance on Usenet, while mostjust do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor ErnieCroot, giving him more information about my prime counting researchthan I've posted here. He replied back *once*, and seemed friendlyenough. I answered him and awaited further replies. After some weeksI sent a query to follow-up, and here is his reply:Dear James,No, I haven't gotten around to looking at it. I'll let you knowwhen I do.Best,ERnie> Professor Croot:Just checking to see if you still have any interest in my ?d of a way to > count prime numbers by integrating a partial difference equation, as I > haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me > to you because I'd like to check in with him as well.James Harris> -Intellectual laziness is about deciding> ahead of time what you wish to believe,> and daring God to be different.> http://lostincomment.blogspot.com/Will I ever hear back from Professor Croot? Well, consider theevidence:I've given something new, a partial difference equation integrationfor counting prime numbers, a ?st in recorded human history.Professor Croot has had some time to consider my work, but now begsoff, claiming not to have looked at it.It turns out that he's a ?st year professor and I was referred tohim by another professor at Georgia Tech who *asked* him to look overmy work.That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians atuniversities have a lot of power when it comes to acceptance of mywork, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like DavidUllrich, a tenured math professor at Oklahoma State University, toTELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate tocount prime numbers which NO ONE ELSE in recorded human history hasmanaged.I have other mathematical research, but as long as mathematiciansstick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign mywork, lie and generally act like asses, knowing that others will justsit, and wait, waiting for mathematicians in the mainstream to letthem know that it's important.To a large extent I now censor my *own* work in talking about it, as Ifocus on things that are hard for people to lie about, and hope forthe best.Right now, locked inside of me is information that could be lost tohumanity because I'm the genius maligned, trapped by a system thatlets mathematicians get away with hurting the society that feeds andclothes them, by de facto censorship.I know things, important things, that you may never know aboutnumbers, and mathematics.Mathematicians are no longer part of decent society, but are now roguehaving taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial differenceequation.Check for yourself.James HarrisMy math discoveries, found for pro?http://mathforpro?.blogspot.com/ === > Some of you were probably surprised to learn that I did indeed ?d a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians. Numerous others have created prime counting functions. Take a look athttp://mathworld.wolfram.com/PrimeCountingFunction.html> But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],S(x,1) = 0.And p(x, y) = ?) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y).I did a little checking, and this is just a rehash of Legendre'sFormula.http://mathworld.wolfram.com/ LegendresFormula.html> That's it. That's the knowledge which mathematicians have purview> over, in terms of the expectation from society that important> information of a mathematical nature will be acknowledged by> mathematicians.Note that it's a *discrete* function, so for you programmers that> means you need to use int's or long's or some discrete variable type.> Also, if you wish to implement it, please sum from dS(x,2) up to and> INCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, did> you ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will see> the method I've just shown you, unless maybe they're out on Usenet> reading my posts, because the mathematical establishment thinks it can> ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.My 6 year old niece can contact mathematicians all over the world. The real question is how many of them replied.> But you see, what bene? do they see to their society by allowing> that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the> world be convinced I'm just a crank, most of them passively just> sitting by, and keeping quiet about my results--after all, that's> quite effective, eh?Then they have de facto censorship because people BELIEVE they> wouldn't do such a thing if my work were important!!!> So you have a standstill with me pushing my research, and a few> mathematicians actively ?hting its acceptance on Usenet, while most> just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie> Croot, giving him more information about my prime counting research> than I've posted here. He replied back *once*, and seemed friendly> enough. I answered him and awaited further replies. After some weeks> I sent a query to follow-up, and here is his reply:> Dear James,No, I haven't gotten around to looking at it. I'll let you know> when I do.Best,ERnie>Professor Croot:Just checking to see if you still have any interest in my ?d of a way to >count prime numbers by integrating a partial difference equation, as I >haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me >to you because I'd like to check in with him as well.James Harris>-Intellectual laziness is about deciding>ahead of time what you wish to believe,>and daring God to be different.>http://lostincomment.blogspot.com/Will I ever hear back from Professor Croot? Well, consider the> evidence:I've given something new, a partial difference equation integration> for counting prime numbers, a ?st in recorded human history.Wrong there. Difference equations are not integrated.> Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.If he did look at it, he would probably just pile it on top of all theother letters from amatures claiming to have a breakthrough whenactually the result is either wrong or previously known.> It turns out that he's a ?st year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.> That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like David> Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.In this case, it is useless. There are faster prime counting functionsout there.> Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.Again, partial difference equation are NOT integrated.> I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my> work, lie and generally act like asses, knowing that others will just> sit, and wait, waiting for mathematicians in the mainstream to let> them know that it's important.To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.Go for the ultimate self-censorship: stop talking, writing andposting.> Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.> I know things, important things, that you may never know about> numbers, and mathematics.Mathematicians are no longer part of decent society, but are now rogue> having taken their own path into darkness. Don't believe me?I don't believe you.> Check my instructions for integrating that partial difference> equation.Check for yourself.> James HarrisMy math discoveries, found for pro?> http://mathforpro?.blogspot.com/ === > Some of you were probably surprised to learn that I did indeed ?d a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians.But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.After all, it's very compact, as here are the instructions, yet again:dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))],S(x,1) = 0.And p(x, y) = ?) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y).That's it. That's the knowledge which mathematicians have purview> over, in terms of the expectation from society that important> information of a mathematical nature will be acknowledged by> mathematicians.Note that it's a *discrete* function, so for you programmers that> means you need to use int's or long's or some discrete variable type.> Also, if you wish to implement it, please sum from dS(x,2) up to and> INCLUDING dS(x,y).Now if you're a programmer or have been taught as a programmer, did> you ever get an assignment to count prime numbers?Now then, think about kids currently in school who I doubt will see> the method I've just shown you, unless maybe they're out on Usenet> reading my posts, because the mathematical establishment thinks it can> ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.But you see, what bene? do they see to their society by allowing> that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the> world be convinced I'm just a crank, most of them passively just> sitting by, and keeping quiet about my results--after all, that's> quite effective, eh?Then they have de facto censorship because people BELIEVE they> wouldn't do such a thing if my work were important!!!So you have a standstill with me pushing my research, and a few> mathematicians actively ?hting its acceptance on Usenet, while most> just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie> Croot, giving him more information about my prime counting research> than I've posted here. He replied back *once*, and seemed friendly> enough. I answered him and awaited further replies. After some weeks> I sent a query to follow-up, and here is his reply:> Dear James,No, I haven't gotten around to looking at it. I'll let you know> when I do.Best,ERnie>Professor Croot:Just checking to see if you still have any interest in my ?d of a way to >count prime numbers by integrating a partial difference equation, as I >haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me >to you because I'd like to check in with him as well.James Harris>-Intellectual laziness is about deciding>ahead of time what you wish to believe,>and daring God to be different.>http://lostincomment.blogspot.com/Will I ever hear back from Professor Croot? Well, consider the> evidence:I've given something new, a partial difference equation integration> for counting prime numbers, a ?st in recorded human history.Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.It turns out that he's a ?st year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.> That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like David> Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my> work, lie and generally act like asses, knowing that others will just> sit, and wait, waiting for mathematicians in the mainstream to let> them know that it's important.To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.I know things, important things, that you may never know about> numbers, and mathematics.Mathematicians are no longer part of decent society, but are now rogue> having taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial difference> equation.Check for yourself.> James HarrisMy math discoveries, found for pro?> http://mathforpro?.blogspot.com/A sociologist of science with whom I am cordial has taken up the topicof who does what to whom and why, in NG's such as sci.math, sci.logic,and sci.physics. This is an area that has not been explored in thedetail that it deserves to be, perhaps because even sociologists arenot entirely comfortable with what these NGs show about those, many ofthem professors, who in?s much harm as possible onseekers-after-knowledge such as yourself in these groups.--John === >[...]What Jesse said:>A sociologist of science with whom I am cordial has taken up the topic>of who does what to whom and why, in NG's such as sci.math, sci.logic,>and sci.physics. That's interesting.> This is an area that has not been explored in the>detail that it deserves to be, perhaps because even sociologists are>not entirely comfortable with what these NGs show about those, many of>them professors, who in?s much harm as possible on>seekers-after-knowledge such as yourself in these groups.But suggesting that Harris is a seeker-after-knowledge is hilarious.He has repeatedly said he's not interested in learning any math.He's the only person I've ever seen state on sci.math that if whathe just said was wrong we shouldn't bother saying so because hedidn't want to know. Seeker after knowledge? Right.Does your conjecture about why this interesting topic has notbeen studied come from your sociologist friend, or is it justyour own surmising?>--JohnDavid C. Ullrich === > A sociologist of science with whom I am cordial has taken up the topic> of who does what to whom and why, in NG's such as sci.math, sci.logic,> and sci.physics. This is an area that has not been explored in the> detail that it deserves to be, perhaps because even sociologists are> not entirely comfortable with what these NGs show about those, many of> them professors, who in?s much harm as possible on> seekers-after-knowledge such as yourself in these groups.Such a study does sound interesting. Despite your usual attempt tohint at dark motives of professionals and casting James I'm in it forthe vast sums of money and chicks Harris as a seeker-after-knowledge.-- All intelligent men are cowards. The Chinese are the world's worst?hters because they are an intelligent race[...] An average Chinesechild knows what the European gray-haired statesmen do not know, thatby ?hting one gets killed or maimed. -- Lin Yutang === Some of you were probably surprised to learn that I did indeed ?d a> way to count prime numbers by integrating a partial difference> equation.[snip]What you found was somebody else's work that you could notcompetently plagiarize.http://www.crank.net/harris.html It's not every braying jackass that gets a whole page at crank.netThe NSA is all hot and bothered about prime numbers for crackingtrapdoor encryptions. Of course, idiot James Harris, if you screw theNSA you disappear into indetreminate offshore Federal detention whereeven a bleeding heart ACLU Liberal lawyer can't ?d you. Go ahead,make their day.-- Uncle Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals)Quis custodiet ipsos custodes? The Net! === >Some of you were probably surprised to learn that I did indeed ?d a>way to count prime numbers by integrating a partial difference>equation. Some of you probably STILL doubt that no one else in>recorded history has managed such a feat because you need to believe>in mathematicians.But my point is that mathematicians have gone rogue and act against>the needs of society by de facto censorship of information that they>don't think makes them look good, like the information about my>partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,>there's the passive act of refusing to acknowledge the discovery>itself.Um, those acts are not censorship. The only person around here I'veever seen attempt any actual censorship is _you_, complaining to myemployer because you wanted him to get me to shut up.>After all, it's very compact, as here are the instructions, yet again:[Yet another repetition of the supposedly censored algorithm snipped]Now then, think about kids currently in school who I doubt will see>the method I've just shown you, unless maybe they're out on Usenet>reading my posts, because the mathematical establishment thinks it can>ignore my results.Have I contacted mathematicians? Yes.I've contacted mathematicians all over the world.And not _one_ of them has the decency to acknowledge the _obvious_importance of your work. Does that really seem plausible, thatthere could be such _universal_ agreement on a lie? Universalagreement on the truth seems much more plausible to me.>But you see, what bene? do they see to their society by allowing>that someone NOT a mathematician found such a result?Worse, I have other math knowledge. But mathematicians can let the>world be convinced I'm just a crank, most of them passively just>sitting by, and keeping quiet about my results--after all, that's>quite effective, eh?Then they have de facto censorship because people BELIEVE they>wouldn't do such a thing if my work were important!!!Do people really believe that? Huh.>So you have a standstill with me pushing my research, and a few>mathematicians actively ?hting its acceptance on Usenet, while most>just do their best to ignore it.For instance, I contacted Georgia Tech and talked to a Professor Ernie>Croot, Why? You're not going to stop until you've contacted everymathematician on the planet?>giving him more information about my prime counting research>than I've posted here. He replied back *once*, and seemed friendly>enough. I answered him and awaited further replies. After some weeks>I sent a query to follow-up, and here is his reply:>Dear James,No, I haven't gotten around to looking at it. I'll let you know>when I do.Best,ERnie> Professor Croot:Just checking to see if you still have any interest in my ?d of a way to > count prime numbers by integrating a partial difference equation, as I > haven't heard from you since my last reply.If you've lost interest can you refer me back to the professor who sent me > to you because I'd like to check in with him as well.James Harris> -Intellectual laziness is about deciding> ahead of time what you wish to believe,> and daring God to be different.> http://lostincomment.blogspot.com/Will I ever hear back from Professor Croot? Well, consider the>evidence:I've given something new, a partial difference equation integration>for counting prime numbers, a ?st in recorded human history.Professor Croot has had some time to consider my work, but now begs>off, claiming not to have looked at it.It turns out that he's a ?st year professor and I was referred to>him by another professor at Georgia Tech who *asked* him to look over>my work.>Possibly he intends to look at it. Possibly he's trying to bepolite: Suppose for the sake of argument that he's recognizedyour crackpottery nature somehow. Would you really be happierif he came out and said so?>That's a professor at Georgia Tech.So I'm an *independent* researcher, which means that mathematicians at>universities have a lot of power when it comes to acceptance of my>work, Actually they don't, because the mathematical world does not decidethese things the way you think they do. If Barry Mazur and AndrewWiles published a paper saying your proof of FLT was correct itmight make an impression at ?st, but _very_ soon people wouldstart saying no it was nonsense and wonder what they were smoking.>but may see little point in helping me.That leaves me Usenet, where there are mathematicians, like David>Ullrich, a tenured math professor at Oklahoma State University, to>TELL people that my work is useless or wrong.Yet, I found a partial difference equation that you can integrate to>count prime numbers which NO ONE ELSE in recorded human history has>managed.I have other mathematical research, but as long as mathematicians>stick to their guns, who gets to hear it?Sure I can talk about it on Usenet, and watch as posters malign my>work, lie and generally act like asses, Like asses and also like ing dog.>knowing that others will just>sit, and wait, waiting for mathematicians in the mainstream to let>them know that it's important.To a large extent I now censor my *own* work in talking about it, as I>focus on things that are hard for people to lie about, and hope for>the best.Right now, locked inside of me is information that could be lost to>humanity because I'm the genius maligned, trapped by a system that>lets mathematicians get away with hurting the society that feeds and>clothes them, by de facto censorship.I know things, important things, that you may never know about>numbers, and mathematics.You know things that you've never told us about? That's hard tobelieve.>Mathematicians are no longer part of decent society, but are now rogue>having taken their own path into darkness. Don't believe me?Check my instructions for integrating that partial difference>equation.Check for yourself.Check _what_? Nobody denies that it gives the right answer,the only dispute is about importance and originality. Howis checking going to settle that?>James HarrisMy math discoveries, found for pro?>http://mathforpro?.blogspot.com/David C. Ullrich === > To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best.>And we appreciate that you do that! Now if you could only put the primecounting differebce equation, polynomial factorization and FLT back into thebox too, that would be superb!> Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.>Could you please keep all of those locked up?When you die, then maybe, just maybe your INCREDIBLE genius will be sharedwith the world and all will know just how tortured you were by theconspirators!> I know things, important things, that you may never know about> numbers, and mathematics.>Do you know that you actually have more than nine of these symptoms? Thatalso deals with numbers, including 9 which is the perfect square 3^2!***Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionarystatement)A pervasive pattern of grandiosity (in fantasy or behavior), need foradmiration, and lack of empathy, beginning by early adulthood and present ina variety of contexts, as indicated by ?e (or more) of the following:(1) has a grandiose sense of self-importance (e.g., exaggerates achievementsand talents, expects to be recognized as superior without commensurateachievements)(2) is preoccupied with fantasies of unlimited success, power, brilliance,beauty, or ideal love(3) believes that he or she is special and unique and can only beunderstood by, or should associate with, other special or high-status people(or institutions)(4) requires excessive admiration(5) has a sense of entitlement, i.e., unreasonable expectations ofespecially favorable treatment or automatic compliance with his or herexpectations(6) is interpersonally exploitative, i.e., takes advantage of others toachieve his or her own ends(7) lacks empathy: is unwilling to recognize or identify with the feelingsand needs of others(8) is often envious of others or believes that others are envious of him orher(9) shows arrogant, haughty behaviors or attitudesReprinted with permission from the Diagnostic and Statistical Manual ofMental Disorders, fourth Edition. Copyright 1994 American PsychiatricAssociation*** === Some of you were probably surprised to learn that I did indeed ?d a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians.But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation.What do I mean by de facto censorship?Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself.No, censorship is when you are stopped from expressing your views. Whatkind of delusions of grandeur do you have to suffer from to think thatpeople are obliged to acknowledge your discovery?-- G.C. === > Some of you were probably surprised to learn that I did indeed ?d a> way to count prime numbers by integrating a partial difference> equation. Some of you probably STILL doubt that no one else in> recorded history has managed such a feat because you need to believe> in mathematicians. But my point is that mathematicians have gone rogue and act against> the needs of society by de facto censorship of information that they> don't think makes them look good, like the information about my> partial difference equation. What do I mean by de facto censorship? Well, besides the active activity, like webpages labeling me a crank,> there's the passive act of refusing to acknowledge the discovery> itself. After all, it's very compact, as here are the instructions, yet again: dS(x,y) = [p(x/y, y-1) - p(y-1, sqrt(y-1))][ p(y, sqrt(y)) - p(y-1,> sqrt(y-1))], S(x,1) = 0. And p(x, y) = ?) - S(x, y) - 1, and you get S as the sum of dS> from dS(x,2) to dS(x,y). That's it. That's the knowledge which mathematicians have purview> over, in terms of the expectation from society that important> information of a mathematical nature will be acknowledged by> mathematicians. Note that it's a *discrete* function, so for you programmers that> means you need to use int's or long's or some discrete variable type. Also, if you wish to implement it, please sum from dS(x,2) up to and> INCLUDING dS(x,y). Now if you're a programmer or have been taught as a programmer, did> you ever get an assignment to count prime numbers? Now then, think about kids currently in school who I doubt will see> the method I've just shown you, unless maybe they're out on Usenet> reading my posts, because the mathematical establishment thinks it can> ignore my results. Have I contacted mathematicians? Yes. I've contacted mathematicians all over the world. But you see, what bene? do they see to their society by allowing> that someone NOT a mathematician found such a result? Worse, I have other math knowledge. But mathematicians can let the> world be convinced I'm just a crank, most of them passively just> sitting by, and keeping quiet about my results--after all, that's> quite effective, eh? Then they have de facto censorship because people BELIEVE they> wouldn't do such a thing if my work were important!!! So you have a standstill with me pushing my research, and a few> mathematicians actively ?hting its acceptance on Usenet, while most> just do their best to ignore it. For instance, I contacted Georgia Tech and talked to a Professor Ernie> Croot, giving him more information about my prime counting research> than I've posted here. He replied back *once*, and seemed friendly> enough. I answered him and awaited further replies. After some weeks> I sent a query to follow-up, and here is his reply:Did it ever occur to you that he may be too busy to reply? That doesn't meanhe's running from your work.David Moran> Dear James, No, I haven't gotten around to looking at it. I'll let you know> when I do. Best, ERnieProfessor Croot:> Just checking to see if you still have any interest in my ?d of a wayto>count prime numbers by integrating a partial difference equation, as I>haven't heard from you since my last reply.> If you've lost interest can you refer me back to the professor who sentme>to you because I'd like to check in with him as well.> James Harris>-> Intellectual laziness is about deciding>ahead of time what you wish to believe,>and daring God to be different.>http://lostincomment.blogspot.com/ Will I ever hear back from Professor Croot? Well, consider the> evidence: I've given something new, a partial difference equation integration> for counting prime numbers, a ?st in recorded human history. Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it. It turns out that he's a ?st year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.> That's a professor at Georgia Tech. So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me. That leaves me Usenet, where there are mathematicians, like David> Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong. Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed. I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it? Sure I can talk about it on Usenet, and watch as posters malign my> work, lie and generally act like asses, knowing that others will just> sit, and wait, waiting for mathematicians in the mainstream to let> them know that it's important. To a large extent I now censor my *own* work in talking about it, as I> focus on things that are hard for people to lie about, and hope for> the best. Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship. I know things, important things, that you may never know about> numbers, and mathematics. Mathematicians are no longer part of decent society, but are now rogue> having taken their own path into darkness. Don't believe me? Check my instructions for integrating that partial difference> equation. Check for yourself.> James Harris My math discoveries, found for pro?> http://mathforpro?.blogspot.com/ === Right now, locked inside of me is information that could be lost to> humanity because I'm the genius maligned, trapped by a system that> lets mathematicians get away with hurting the society that feeds and> clothes them, by de facto censorship.>If you can't beat ?em ,j oin e'm. That is to say, get a PhD in math. For agenius like you that should be easy. Since you like math so much, why not doit full time?MB === Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it.> Clearly he's part of the conspiracy. === [snip]> I've given something new, a partial difference equation integration> for counting prime numbers,Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus. Integration refers to ?ding anti-derivatives.> a ?st in recorded human history.I suppose it would be -- except that you haven't shown how to *integrate* anypartial difference equations.> Professor Croot has had some time to consider my work, but now begs> off, claiming not to have looked at it. It turns out that he's a ?st year professor and I was referred to> him by another professor at Georgia Tech who *asked* him to look over> my work.>You would dare to say anything that af?med your own magni?ence.> That's a professor at Georgia Tech. So I'm an *independent* researcher, which means that mathematicians at> universities have a lot of power when it comes to acceptance of my> work, but may see little point in helping me. That leaves me Usenet, where there are mathematicians, like David> Ullrich, a tenured math professor at Oklahoma State University, to> TELL people that my work is useless or wrong.Also there are non-mathematicians who tell people that your work is useless andwrong.> Yet, I found a partial difference equation that you can integrate to> count prime numbers which NO ONE ELSE in recorded human history has> managed.Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus.> I have other mathematical research, but as long as mathematicians> stick to their guns, who gets to hear it?You post your results repeatedly and incessantly. This thread is acounter-example to your claim. *Everyone* gets to hear it -- over and over andover and over...> Check my instructions for integrating that partial difference> equation.Partial difference equations are *not* integrated. Their solutions are foundusing the sum calculus.> Check for yourself.Did that. No integration found.> James Harris My math discoveries, found for pro?> http://mathforpro?.blogspot.com/--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I am not sure if I have to ask this in the tex newsgroup, but I think the math> people know more about this type of problem. I wanted to know if there is a> package I can use to plot elliptic curves in LaTeX, can one do it in gnuplot? I> may just be able to use plot command and square root of x^3+ax+b and then the> negative of it, then stick them together.. but that's messy, probably theres a> better way (I mean.. what if I have an algebraic curve de?ed by f(x,y)=0 and> I can't express y in terms of x very simply? is there no way to do it with> gnuplot? or maybe other free softwares, or latex packages). As a last resort, I might just travel to university just to use their> for any suggestions. Sincerely,> Jose Capcognuplot can do such plots and it can also output LaTeX code. Also, Postscript plotscan be imported using PSTricks.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === [slightly edited for clarity]>gnuplot can do such plots and it can also output LaTeX code. ^^^^Last time I used gnuplot it didn't have any facility to plot implicitfunctions. But I admit it's been a while...>Also, Postscript plots can be imported using PSTricks.(Encapsultated) Postscript plots can be imported as such (withgraphic{s,x}). PSfrag can be usefulMichele-- > 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 === >gnuplot can do such plots and it can also output LaTeX code. Also, Postscript plots>can be imported using PSTricks.>could you give me an example code? given a polynomial f(x,y) .. I want to plot the points that solve f(x,y)=0 ... say f(x,y)=y^2+y-x^3+x Sincerely,Jose Capco === > could you give me an example code? given a polynomial f(x,y) .. I want to plot > the points that solve f(x,y)=0 ... say f(x,y)=y^2+y-x^3+x I tried to get splot to work with f(x,y) = y^2 + y - x^3 + x, but it refused, soI completed the square for y^2, and created two real functions of y, calledf(x) and g(x) depending upon the +/- value of the square root functionHere's the gnuplot codegnuplot> f(x)=-0.5+sqrt(x*x*x-x+0.25)gnuplot> g(x)=-0.5-sqrt(x*x*x-x+0.25)gnuplot> h(x)=0gnuplot> set xrange [-2:3]gnuplot> plot f(x),g(x),h(x)crude, but a start.and yes, you can set up output for LaTex:gnuplot> set terminal latexTerminal type set to ?latex'Options are ?(document speci? font)'gnuplot> set output my elliptic curvegnuplot> replotto get back to your terminalgnuplot> set terminal X11 (or whatever)gnuplot> set output === >gnuplot can do such plots and it can also output LaTeX code. Also, Postscript> plots>can be imported using PSTricks.> could you give me an example code? given a polynomial f(x,y) .. I want to plot> the points that solve f(x,y)=0 ... say f(x,y)=y^2+y-x^3+x Sincerely,> Jose Capcognuplot has excellent documentation, demos and example ?es -- including plottingresponse. Do a search.--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I have some questions about the de?ition of the Riemann Stieltjesintegral of a real bounded function f over an interval [a,b], withrespect to a function g, also bounded on [a,b].According to one de?ition (like we see in Bartle's and Apostol'sbooks), the integral is related to the concept of tagged partition. Wesay the integral of f with respect to g is I if, for every eps>0,there's a partition Pe of [a,b], such that, if P is any re?ement ofPe, then |S(P,f,g) - I| 0, there's a delta>0 such that, for every partition withmesh I have some questions about the de?ition of the Riemann Stieltjes> integral of a real bounded function f over an interval [a,b], with> respect to a function g, also bounded on [a,b]. According to one de?ition (like we see in Bartle's and Apostol's> books), the integral is related to the concept of tagged partition. We> say the integral of f with respect to g is I if, for every eps>0,> there's a partition Pe of [a,b], such that, if P is any re?ement of> Pe, then |S(P,f,g) - I| Riemann-Stieltjes sum of P, f and g over [a,b]. In this de?ition,> the sum is related to tag points chosen in each of the intervals of P,> and inequality (1) must be satis?d regardless the choice of the tag> points. The only requirement about g is that it is bounded on [a,b].This is equivalent to the next de?ition _if_ g is of bounded variation,because any such g is the sum two functions, one monotonic up and the othermonotonic down.> Other de?ition, like we ?d in Rudin's book, involves the concepts> of upper and lower sums and there are no tag points. The function g,> however, is required to be monotonically increasing on [a,b]. If P is> a partition of [a,b], the Riemann-Stieltjes sum is de?ed by taking> the in?um (lower sum) or the supremum (upper sum ) of f on each> interval of P. If the in?um of the upper sums equals the supremum of> the lower sums, both taken over all possible partitions, themn this> common number is said to be the integral of f with respect to g. Isn't> the fact the g must be increasing a loss of generality when compared> to the ?st de?ition? If we restrain our attention to the Riemann> integral, it doesn't matterm because then we always hace g(x) =x, but> in the general case it seems to be a loss of generality.For more on this (plus some of the Henstock business, by another name) Isuggest Avner Friedman, _Foundations of Modern Analysis_, in the exercisesfor section 2.11.LH === I have found a link that might help, although I worry it is not advancedenough to answer your question.It is at:http://mathworld.wolfram.com/StieltjesIntegral.htmlI hope this helps!Kavon I have some questions about the de?ition of the Riemann Stieltjes> integral of a real bounded function f over an interval [a,b], with> respect to a function g, also bounded on [a,b]. According to one de?ition (like we see in Bartle's and Apostol's> books), the integral is related to the concept of tagged partition. We> say the integral of f with respect to g is I if, for every eps>0,> there's a partition Pe of [a,b], such that, if P is any re?ement of> Pe, then |S(P,f,g) - I| Riemann-Stieltjes sum of P, f and g over [a,b]. In this de?ition,> the sum is related to tag points chosen in each of the intervals of P,> and inequality (1) must be satis?d regardless the choice of the tag> points. The only requirement about g is that it is bounded on [a,b]. Other de?ition, like we ?d in Rudin's book, involves the concepts> of upper and lower sums and there are no tag points. The function g,> however, is required to be monotonically increasing on [a,b]. If P is> a partition of [a,b], the Riemann-Stieltjes sum is de?ed by taking> the in?um (lower sum) or the supremum (upper sum ) of f on each> interval of P. If the in?um of the upper sums equals the supremum of> the lower sums, both taken over all possible partitions, themn this> common number is said to be the integral of f with respect to g. Isn't> the fact the g must be increasing a loss of generality when compared> to the ?st de?ition? If we restrain our attention to the Riemann> integral, it doesn't matterm because then we always hace g(x) =x, but> in the general case it seems to be a loss of generality. There's a 3rd de?ition you can ?d in a recent Bartle's book, which> relies on the concept of mesh (or norm ) of a partition (de?ed as> the lenght of the largest interval of P). It's similar to the 1st> de?ition, but instead of re?ements, Bartle says that, for every> eps>0, there's a delta>0 such that, for every partition with> mesh the usual eps-delta de?ition of limit of a function, but it renders> the proof of some theorems very cumbersome. Bartle still de?es the Generalized Riemann Integral (seems this> concept is due to Henstock) by introducing a gauge function which is> simply a positive real function de?ed on [a,b]. And he says a> function is Lebesgue integrable if |f| is Generalized Riemann> integrable. But he doesn't de?e Lebesgue integral. In the case of> generalized Riemann integral, there's no function g and f need not be> bounded. I think the ?st and the second de?ition, if g is monotonically> incresing, are equivalent and lead to the same value for I, but I'm> not sure if the Henstock integral or the integral based on meshes are> equivalent to them. I'm a bit confused, could anyone please clarify such points?> Artur === If I remember correctly, Leonidas Alaoglu was born in Canada (Red Deer,Alberta).>That being said: However, if you search Alaoglu Greek on Google, you don't>get any pages referring to Greeks aside from, possibly, Leonidas Alaoglu. >On the other hand, if you search Alaoglu Turkish on Google, you get>several pages referring to present-day Turks with that surname. Moreover,>the word ala also has a meaning in Turkish (i.e. very good, excellent).>Yet, clearly, Leonidas is a Greek name and not a Turkish name.>So, here's what I'm wondering: Is it possible that the mathematician >Leonidas Alaoglu had a Greek mother and a Turkish father?Unlikely -- marriages between Greeks (read Orthodox Christians) and Turks > (read Muslims) in the Ottoman Empire were very uncommon, moreover a Turkish> man marrying a Christian woman back then (if not today as well) would have > no incentive at all (to put it mildly) to give a non-Muslim/Turkish name to > his son ... while a possibility you left out (Greek father, Turkish mother) > would be even less likely for reasons I am asking you to guess :-)[Now look at me: Greek national, father's last name Turkish, father's ?st > name Greek (Christos), father's parents both Greeks (Georgios, Antonia)...] baloglouAToswego.edu-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === >If I remember correctly, Leonidas Alaoglu was born in Canada (Red Deer,>Alberta).Perhaps, I do not know much about his life. But my guess is that his parents were Greek immigrants from Asia Minor (Anatolia/Turkey) -- whereGreek refers *not* to ?citizenship' but to language and/or religion... baloglouAToswego.edu === But is this the best bound?>I read somewhere that sqrt(n) log n/(8pi) is an exact bound, nothing>better, if the RH is true. Is this true?Assume RH, assume x >= 2657> Then |pi(x)-li(x)| <= sqrt(x)log(x)/(8Pi)> This is a result of Shoenfeld 1976, Math. Comp. 30.> === > But I was wondering something about the Riemann Hypothesis: What if> it were proven to be false? I know that it's an important resultfor> number theory, but what would the consequences be, if any, if it> turned out to be false?in de.sci.mathematik, there is a thread of that kind, where I>learned the following astonishing fact:>The following 18 numbers >1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>are the only known ones not of the form > pq + qr + rp with p,q,r > 0Astonishing indeed!What is also interesting about these 18 integers is, add 1 to eachinteger and you have 18 primes.Also an interesting fact is that 210,330 and 462, the last 3 integersin this list, where each are @ a different level in the Collatz treeand each have this same property. Mainly (n-1) == 2(mod)3. Also allthere subsequent higher level integers have this property thus nobranching occurs in higher levels above these integers.Could there be a possible relation to the Collatz Conjecture?Dan >According to Borwein and Choi, there are are no others and>there are certainly no more than 19 of them.>BUT: if the Riemann Hypothesis fails, there could be another>number not of this form. And it must be > 10^11 then.>(1) Sequence A025052 in the njas archive> http://www.research.att.com/~njas/sequences/> from 20.06.2001 00:15>related sequence A006093. ( See the thread pq+p+q ...)>Rainer === >Message-id: But I was wondering something about the Riemann Hypothesis: What if> it were proven to be false? I know that it's an important result>for> number theory, but what would the consequences be, if any, if it> turned out to be false?>in de.sci.mathematik, there is a thread of that kind, where I>learned the following astonishing fact:>The following 18 numbers >1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>are the only known ones not of the form > pq + qr + rp with p,q,r > 0Astonishing indeed!What is also interesting about these 18 integers is, add 1 to each>integer and you have 18 primes.Also an interesting fact is that 210,330 and 462, the last 3 integers>in this list, where each are @ a different level in the Collatz tree>and each have this same property. Mainly (n-1) == 2(mod)3. Isn't that also true for 6, 18, 30, 42, 78 and 102? If (n-1) == 2(mod3), thenn==0(mod3), so the numbers you selected are just those divisible by 3.So what's the story with 1, 2, 4, 10, 22, 58, 70, 130 and 190? Don't youhave to explain why some numbers on the list have that property while others don't?>Also all>there subsequent higher level integers have this property thus no>branching occurs in higher levels above these integers.That's true on any number ==0(mod3). What distinguishes this small group from the in?ite lsit of other numbers with that property?Could there be a possible relation to the Collatz Conjecture?What relation? That some of them are divisible by 3? I think you need more than that. Does that particular list of numbers show up in theCollatz Conjecture somwhere? As an example, the list of knowninteger crossover points is {1, -1, -5, -17}. And sure enough, one numberon the list matches. But that's just coincidence. If _every_ number matched,then we should start looking for a relationship.Dan >According to Borwein and Choi, there are are no others and>there are certainly no more than 19 of them.>BUT: if the Riemann Hypothesis fails, there could be another>number not of this form. And it must be > 10^11 then.>(1) Sequence A025052 in the njas archive> http://www.research.att.com/~njas/sequences/> from 20.06.2001 00:15>related sequence A006093. ( See the thread pq+p+q ...)>Rainer--MensanatorAce of Clubs === >Message-id: But I was wondering something about the Riemann Hypothesis: What if> it were proven to be false? I know that it's an important result> for> number theory, but what would the consequences be, if any, if it> turned out to be false?>in de.sci.mathematik, there is a thread of that kind, where Ilearned the following astonishing fact:>The following 18 numbers >1 2 4 6 10 18 22 30 42 58 70 78 102 130 190 210 330 462>are the only known ones not of the form pq + qr + rp with p,q,r > 0>Astonishing indeed!>What is also interesting about these 18 integers is, add 1 to each>integer and you have 18 primes.>Also an interesting fact is that 210,330 and 462, the last 3 integers>in this list, where each are @ a different level in the Collatz treeand each have this same property. Mainly (n-1) == 2(mod)3. Isn't that also true for 6, 18, 30, 42, 78 and 102? If (n-1) == 2(mod3), then> n==0(mod3), so the numbers you selected are just those divisible by 3.> So what's the story with 1, 2, 4, 10, 22, 58, 70, 130 and 190? Don't you> have to explain why some numbers on the list have that property while > others don't?>Also all>there subsequent higher level integers have this property thus no>branching occurs in higher levels above these integers.That's true on any number ==0(mod3). What distinguishes this small > group from the in?ite lsit of other numbers with that property? >Could there be a possible relation to the Collatz Conjecture?What relation? That some of them are divisible by 3? I think you need > more than that. Does that particular list of numbers show up in the> Collatz Conjecture somwhere? As an example, the list of known> integer crossover points is {1, -1, -5, -17}. And sure enough, one number> on the list matches. But that's just coincidence. If _every_ number matched,> then we should start looking for a relationship.You are right!I have to get back into the Collatz conjecture -- getting rusty. Dan >Dan According to Borwein and Choi, there are are no others andthere are certainly no more than 19 of them.>BUT: if the Riemann Hypothesis fails, there could be another>number not of this form. And it must be > 10^11 then.>(1) Sequence A025052 in the njas archive> http://www.research.att.com/~njas/sequences/> from 20.06.2001 00:15>related sequence A006093. ( See the thread pq+p+q ...)>Rainer === > Craig Feinstein schrieb>Assume RH, assume x >= 2657>Then |pi(x)-li(x)| <= sqrt(x)log(x)/(8Pi)> But is this the best bound?Well, the best bound I know. If you ?da better bound, please inform us! ;-)> But is this the best bound?IMHO there is no reason known which excludesthe possibility of the existence of N and Bsuch that the following is valid: Assume x >= NThen |pi(x)-li(x)| <= B < sqrt(x)log(x)/(8Pi)Approved: newgroups-request@isc.orgArchive-Name: misc.metric-system === LAST CALL FOR VOTES (of 2) unmoderated group misc.metric-system[ Note: This document is multiposted in 3 copies because too many large service providers implement crossposting limits and would otherwise drop it. - n.a.n moderation team ]misc.metric-system The International System of Units.This vote is being conducted by a neutral third party. Questions aboutthe proposed group should be directed to the proponent.Proponent: Markus Kuhn Votetaker: Bill Aten RATIONALE: misc.metric-systemUnits of measurement and related standards affect many aspects of ourdaily lives. The global standardization of a single consistentInternational System of Units was a major breakthrough for humancivilization and signi?antly simpli?d communication, learning,work and trade all over the planet.The introduction of the metric system still faces delays in someareas. Notable examples are consumer communication and traf?regulations in the United States and United Kingdom, as well as partsof the aeronautical and typographic industry. It is therefore nosurprise that discussions about the metric system ?p regularlyin many different newsgroups. In particular the slow progress withmetrication in the United States promises to fuel such debates formany years to come.A dedicated newsgroup will focus expertise and will provide a mediumfor professionals and hobbyists to ?d advice and suggestions onmetric product standards and conventions. None of the newsgroups inwhich metric-system issues ?p frequently is particularly suitedfor this topic by charter and readership. The popularity of thethat there is a signi?ant number of people interested in the topic.Considering the important role that units of measurement play ineveryone's life, this promises to become a quite lively newsgroup.The name of the proposed group has been the subject of some debate.The present proposal is motivated by these considerations: - Although discussions about the metric system focus much on its slow progress in a small number of countries, the topic is inherently international in nature and discussions tend to bene? very signi?antly from world-wide participation. Therefore, placing the group under us.* or uk.* would be inappropriate. - The metric system affects many ?lds, including consumer communication and road traf?. Discussions about the metric system range from basic science and applied engineering considerations to economic, social, psychological, legal, public policy and media aspects. This excludes sci.* and leaves misc.* as the most appropriate hierarchy. - The metric system is the only system of units used in almost every region and ?ld of application. Imperial and U.S. Customary units are usually discussed in relation to the metric system, which is within the scope of the proposed group. Other unit systems have very limited applications and are better discussed in specialized science or history groups. The proposed group is far more likely to have specialized children rather than equivalent siblings, which speaks against an entire misc.measurement.* hierarchy and justi?s a place directly under misc. - The term metric system remains the most well known and most easily recognized English language term for what is more formally called the International System of Units (SI). This speaks against group names such as *.si or *.metric.The proposed charter has equally been the subject of some debate. Thepresent proposal is motivated by these considerations: - It refers equally to both the of?ial modern name International System of Units (SI) and the colloquial English term metric system. This is in the interest of rapid recognition by both readers and search-engine users. Any further distinction between these terms is deliberately left to explanatory periodic postings. - It is broad enough to cover discussions about different historic variants of the metric system (e.g., CGS, MKS, various European customary units) as well as contemporary units that compete with the SI (e.g., inch, pound, Fahrenheit, calorie). - It is narrow enough to exclude topics that are not related to metric units (e.g., the history and rede?ition of calendars). - It covers product standards and conventions that are not part of any of?ial de?ition of the metric system, but that are closely related to metric units (e.g., metric clothing sizes, paper formats, engineering components, traf? regulations). These can be considerably more complex topics than the metric system itself, leading to discussions of particular interest to consumers and practitioners. - It leaves room for the possible later creation of a separate misc.metric-system.advocacy group for those with a particular interest in political activities related to metrication. - It was written with the expectation that the topic is unlikely to attract large non-plain-text postings, commercial advertising or unsocial behaviour in any particular way, leaving these issues to common sense and USENET etiquette. - It is brief.CHARTER: misc.metric-systemThis newsgroup is for discussion about the International System ofUnits (SI) or metric system, including its use in scienti?,technical, and consumer applications, its history and de?ition, andits adoption in ?lds and regions where other units of measurementare still prevalent (metrication). Included within its scope arerelated global standards and conventions, for example metric productspeci?ations and consumer-product labelling practice.END CHARTER.HOW TO VOTE:Extract the ballot from the CFV by deleting everything before and afterthe BEGINNING OF BALLOT and END OF BALLOT lines. 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The individuals listed below will needcorrect this problem. m.collado [at] aaron.ls.?upm.es Manuel Collado-- Bill Aten, UVV === Why don't you just post the program? === > Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy. It just does the job. ___JSH-----public class DefPrimeCounter { /** Creates a new instance of Main */ public DefPrimeCounter() { } /** * @param args the command line arguments */ public static void main(String[] args) { int x = Integer.parseInt(args[0]); DefPrimeCounter Counter = new DefPrimeCounter(); System.out.println(Counter.p(x, (int)Math.sqrt(x))); } public int S(int x, int y){ if (y==1) return 0; int max = y+1; int sum=0, j; for (j=2; jWhy don't you just post the program? Ok. Here's a straight-forward Java implementation. Nothing fancy.> It just does the job. ___JSH -----> public class DefPrimeCounter { /** Creates a new instance of Main */> public DefPrimeCounter() {> } /**> * @param args the command line arguments> */> public static void main(String[] args) { int x = Integer.parseInt(args[0]); DefPrimeCounter Counter = new DefPrimeCounter(); System.out.println(Counter.p(x, (int)Math.sqrt(x)));> } public int S(int x, int y){ if (y==1) return 0; int max = y+1;> int sum=0, j; for (j=2; j (int)Math.sqrt(j)) - p(j-1, (int)Math.sqrt(j-1))); }> return sum; } public int p(int x, int y){ return x - S(x,y) - 1; } }Notice also that this program does not perform any integration. It simply forms a ?ite sum.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I haven't programmed this algorithm yet. But here are a few questionsfor the newsgroup.1) Aside from the complaints about terminology (e.g. incorrectly usingthe term integration to describe a discrete summation), does thisformula work?2) If it does indeed work for certain input values, does it fail forothers?3) Is this formula just a restatement of something we already knowfrom number theory? === > I haven't programmed this algorithm yet. But here are a few questions> for the newsgroup.1) Aside from the complaints about terminology (e.g. incorrectly using> the term integration to describe a discrete summation), does this> formula work?2) If it does indeed work for certain input values, does it fail for> others?Correctly implemented, it works ?e, albeit slowly.Plenty of people have successfully tested one versionor another of James' recursion.3) Is this formula just a restatement of something we already know> from number theory?Legendre's inclusion-exclusion formula, ca. 1790 I believe. - Randy === > I haven't programmed this algorithm yet. But here are a few questions> for the newsgroup. 1) Aside from the complaints about terminology (e.g. incorrectly using> the term integration to describe a discrete summation), does this> formula work?James has coded the algorithm for the original case involving a discretesummation and it does produce correct results in that case.> 2) If it does indeed work for certain input values, does it fail for> others?He has claimed that replacing the ?1's in the original equation with'delta y' converts the difference equation to a partial differentialequation -- but this formulation is *not* a partial differential equationand it does *not* work as a method of counting primes.> 3) Is this formula just a restatement of something we already know> from number theory?Yes.--There are two things you must never attempt to prove: the unprovable --and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy. > It just does the job. ___J seems to work, at least for a few input values:# primes < 100 = 25# primes < 1000 = 169# primes < 1000000 = 78498C code follows:#include #include #include #de?e STATS 0#de?e CACHE 1#if CACHE == 1#de?e I_CACHE_SIZE 1000#de?e XDIVI_CACHE_SIZE 10000long term_cache[I_CACHE_SIZE];char term_found[I_CACHE_SIZE] = {0};long xtra_cache[XDIVI_CACHE_SIZE][I_CACHE_SIZE];char xtra_found[XDIVI_CACHE_SIZE][I_CACHE_SIZE] = {{0}};#if STATS == 1long i_max = 0;long xdivi_max = 0;long i_out_of_range = 0;long xdivi_out_of_range = 0;#endifstatic long S (long x, long y){ long i; long sum; long prev_term; long this_term; if (y < 2) return 0; sum = this_term = 0; for (i = 2; i < y+1; ++i) { long xdivi; long xtra_term; prev_term = this_term; xdivi = x/i;#if STATS == 1 if (i > i_max) i_max = i; if (xdivi > xdivi_max) xdivi_max = xdivi; if (i >= I_CACHE_SIZE) ++xtra_i_out_of_range; if (xdivi >= XDIVI_CACHE_SIZE) ++xdivi_out_of_range;#endif if (xdivi < XDIVI_CACHE_SIZE && i < I_CACHE_SIZE) { if (xtra_found[xdivi][i] == 0) { xtra_cache[xdivi][i] = (xdivi-1) - S(xdivi,i-1); xtra_found[xdivi][i] = 1; } xtra_term = xtra_cache[xdivi][i]; } else { xtra_term = (xdivi-1) - S(xdivi,i-1); } if (i < I_CACHE_SIZE) { if (term_found[i] == 0) { term_cache[i] = (i-1) - S(i,sqrt(i)); term_found[i] = 1; } this_term = term_cache[i]; } else { this_term = (i-1) - S(i,sqrt(i)); } sum += (xtra_term - prev_term) * (this_term - prev_term); } return sum;} #else /* CACHE != 1 *//* no caching */static long S (long x, long y){ long i; long sum; long prev_term; long this_term; if (y < 2) return 0; sum = this_term = 0; for (i = 2; i < y+1; ++i) { long xdivi; long xtra_term; prev_term = this_term; xdivi = x/i; xtra_term = (xdivi-1) - S(xdivi,i-1); this_term = (i-1) - S(i,sqrt(i)); sum += (xtra_term - prev_term) * (this_term - prev_term); } return sum;} #endif /* CACHE == 1 */int main (int argc, char * argv[]){ long n,r; if (argc != 2) { printf(usage: %s 0>n, argv[0]); exit(EXIT_FAILURE); } n = strtol(argv[1], NULL, 10); if (n <= 0) { printf(usage: %s 0>n, argv[0]); exit(EXIT_FAILURE); } r = (n-1) - S(n,sqrt(n)); printf(n%ldnn,r);#if CACHE == 1#if STATS == 1 printf(i_max:%ld xdivi_max:%ldnn, i_max, xdivi_max); printf(i_out_of_range:%ld xdivi_out_of_range:%ldn, i_out_of_range, xdivi_out_of_range);#endif#endif return EXIT_SUCCESS;} === > Why don't you just post the program?Ok. Here's a straight-forward Java implementation. Nothing fancy. >It just does the job. ___J seems to work, at least for a few input values:# primes < 100 = 25> # primes < 1000 = 169That should be 168. That is, there are 168 primes up to and including 1000.James HarrisMy math discoveries, found for pro?http://mathforpro?.blogspot.com/ === > Why don't you just post the program?>Ok. Here's a straight-forward Java implementation. Nothing fancy. >It just does the job. ___J seems to work, at least for a few input values:# primes < 100 = 25> # primes < 1000 = 169That should be 168. That is, there are 168 primes up to and including 1000.You are right. I mistyped. === > Which integers can have squares that end with four identical digits?>[snip Jyrki's solution] 10/10 for part 1 Jyrki, now generalise to other bases :-)>Here are some questions for other bases:For base 3: I ?d that the square of [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3is [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,1, 1]_3This square ends in 12 ones. Is there a limit to the number of 1's a squarecan end in for base b = 3.For base 5: I ?d:The square of [1, 3, 4, 4, 0, 3, 3, 1]_5is [3, 1, 0, 1, 4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1]_5and the square of [3, 2, 4, 3, 1, 2, 1, 2]_5is [2, 2, 4, 1, 2, 3, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4]_5Can the number of 1's and 4's be arbitrary for base 5?--Edwin Clark === >Here are some questions for other bases:>For base 3: I ?d that the square of> [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,>1, 1]_3>This square ends in 12 ones. Is there a limit to the number of 1's a square>can end in for base b = 3.No.Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1) so one of these == 0 mod 3^(n+1). Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >Here are some questions for other bases:>For base 3: I ?d that the square of> [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,>1, 1]_3>This square ends in 12 ones. Is there a limit to the number of 1's a square>can end in for base b = 3.No.Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but > not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then> 2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and> 2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1) > so one of these == 0 mod 3^(n+1). More generally, for any odd b, if (b-1) x^2 + d == 0 mod b^n but not modb^(n+1), where gcd(d,b) = 1, then gcd(x,b) = 1 and (b-1) (x + y b^n)^2 - d == (b-1) x^2 - d + 2 (b-1) b^n x y mod b^(n+1)which is 0 for the appropriate value of y mod b. So if d is a quadratic residue mod b with gcd(d,b)=1 there are squares ending in arbitrarily many d's in base b. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > More generally, for any odd b, if (b-1) x^2 + d == 0 mod b^n but not mod> b^(n+1), where gcd(d,b) = 1, then gcd(x,b) = 1 and > (b-1) (x + y b^n)^2 - d == (b-1) x^2 - d + 2 (b-1) b^n x y mod b^(n+1)> which is 0 for the appropriate value of y mod b. So if d is a quadratic > residue mod b with gcd(d,b)=1 there are squares ending in arbitrarily > many d's in base b.What about higher powers (instead of squares)?-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === > More generally, for any odd b, if (b-1) x^2 + d == 0 mod b^n but not mod> b^(n+1), where gcd(d,b) = 1, then gcd(x,b) = 1 and > (b-1) (x + y b^n)^2 - d == (b-1) x^2 - d + 2 (b-1) b^n x y mod b^(n+1)> which is 0 for the appropriate value of y mod b. So if d is a quadratic > residue mod b with gcd(d,b)=1 there are squares ending in arbitrarily > many d's in base b.>What about higher powers (instead of squares)?Sure, why not? A similar proof will work for k'th powers if gcd(b,k) = 1.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >Here are some questions for other bases: >For base 3: I ?d that the square of [1, 1, 1, 0, 2, 1, 2, 2, 1, 2, 1]_3>is> [2, 0, 1, 1, 2, 2, 0, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1,1,>1, 1]_3 >This square ends in 12 ones. Is there a limit to the number of 1's asquare>can end in for base b = 3. No. Suppose 2 x^2 + 1 == 0 mod 3^n (so x^2 ends in n 1's in base 3) but> not mod 3^(n+1), where n > 1. Of course x is not divisible by 3. Then> 2 (x+3^n)^2 - 1 = 2 x^2 - 1 + 3^n x mod 3^(n+1) and> 2 (x-3^n)^2 - 1 = 2 x^2 - 1 - 3^n x mod 3^(n+1)> so one of these == 0 mod 3^(n+1).>After a little effort in deciphering your proof I agree. Very nice!But shouldn't the 3rd and 4th lines of your proof be> 2 (x+3^n)^2 + 1 = 2 x^2 + 1 + 4*3^n x mod 3^(n+1) and> 2 (x-3^n)^2 + 1 = 2 x^2 + 1 - 4*3^n x mod 3^(n+1) === >Which integers can have squares that end with four identical digits? I believe that all integers that end in two zeroes have squares that endin four identical digits.If x^2 is an integer that ends in 4 identical digits aaaa thenx^2 is congruent to aaaa modulo 10^4. In this case the generalizedLegendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square modulo10^4 then it will be -1. Maple's procedure numtheory[quadres] (c,d)implements L(c/d) and the following calculation shows that a squarecan end in 4 identical digits only when they are 0's:> for i from 0 to 9 do> a:=i+i*10+i*10^2+i*10^3:> print(a,numtheory[quadres](a,10^4));> od: 0, 1 1111, -1 2222, -1 3333, -1 4444, -1 5555, -1 6666, -1 7777, -1 8888, -1 9999, -1 === > If x^2 is an integer that ends in 4 identical digits aaaa then> x^2 is congruent to aaaa modulo 10^4. In this case the generalized> Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square modulo> 10^4 then it will be -1. That's not quite accurate.L(a/n) = 1 if a is a square mod n,but the converse is not true in general,eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === If x^2 is an integer that ends in 4 identical digits aaaa then>x^2 is congruent to aaaa modulo 10^4. In this case the generalized>Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a squaremodulo>10^4 then it will be -1. That's not quite accurate.> L(a/n) = 1 if a is a square mod n,> but the converse is not true in general,> eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.>I think you are confusing the jacobi function with the generalized Legendresymbol (called quadres in Maple):> jacobi(2,15); 1> quadres(2,15); -1Here are the descriptions from Maple the help for these two functions aswell as for the Legendre symbol.The function quadres will compute a generalized Legendre symbol L(a/b) of aand b, which is de?ed to be 1 if a is a quadratic residue (mod b) and -1if a is a quadratic non-residue (mod b) . The number a is a quadraticresidue of b if it has a square root (mod b); i.e., an integer c exists suchthat c^2 is congruent to a (mod b).The function jacobi will compute the Jacobi symbol J(a/b) of a and b. If thefactorization of b is p1^k1 * ... * ps^ks, then jacobi(a, b) = legendre(a,p1)^k1 * ... * legendre(a, ps)^ks , where legendre(a,p) is the Legendresymbol of a and p.The legendre(a, p) function computes the Legendre symbol L(a/p) of a and p(a prime), which is de?ed to be 1 if a is a quadratic residue (mod p), -1if a is a quadratic non-residue (mod p), and 0 if a is congruent to 0 (modp). The number a is a quadratic residue of p if it is not a multiple of pand has a square root (mod p), that is, there is an integer c such that c^2is congruent to a (mod p). The number a is a quadratic non-residue of p i? is not a multiple of p and does not have a square root (mod p).Edwin === >If x^2 is an integer that ends in 4 identical digits aaaa then>x^2 is congruent to aaaa modulo 10^4. In this case the generalized>Legendre symbol L(aaaa/10^4) will be 1 and if aaaa is not a square> modulo>10^4 then it will be -1.> That's not quite accurate.> L(a/n) = 1 if a is a square mod n,> but the converse is not true in general,> eg L(2/15) = L(2/3) L(2/5) = -1.-1 = 1.> I think you are confusing the jacobi function with the generalized> Legendre symbol (called quadres in Maple):I was indeed.I never saw the term generalised Legendre symbol used in this sense.though I have seen the term applied to the Jacobi symbol, eg by Hecke.Do you have a reference for this usage?I just looked in a few number theory books I have,and none of them used the term.It seems to me both misleading and useless --useless because there is actually no way of calculating itas there is with the Jacobi symbol,except by computing the Legendre symbol for each factor p(probably using the Jacobi symbol!).-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === >I think you are confusing the jacobi function with the generalized>Legendre symbol (called quadres in Maple): I was indeed.> I never saw the term generalised Legendre symbol used in this sense.> though I have seen the term applied to the Jacobi symbol, eg by Hecke. Do you have a reference for this usage?Not except for Maple. I was afraid someone might ask about this. I alsotried unsuccessfully to ?d it in my number theory texts.> I just looked in a few number theory books I have,> and none of them used the term. It seems to me both misleading and useless --> useless because there is actually no way of calculating it> as there is with the Jacobi symbol,> except by computing the Legendre symbol for each factor p> (probably using the Jacobi symbol!).>Well, it was useful for this problem. :-) Here's the Maple code for quadres:> showstat(numtheory:-quadres);numtheory:-quadres := proc(a, p)local fac, ig, n, pf, s; 1 if nargs <> 2 then 2 error invalid arguments end if; 3 n := modp(a,p); 4 if type(n,integer) and issqr(n) then 5 return 1 end if; 6 if not (type(n,integer) and type(p,integer)) then 7 return ?procname(args)' end if; 8 if p < 6 then 9 return -1 end if; 10 if isprime(p) then 11 return mods((?power')(n,1/2*p-1/2),p) end if; 12 ig := igcd(n,p); 13 if ig <> 1 and igcd(denom(n/ig^2),p/ig) = 1 then 14 return numtheory:-quadres(n/ig^2,p/ig) end if; 15 pf := ifactor(p); 16 if type(pf,`^`) then 17 return legendre_pow(n,pf) end if; 18 for fac in pf do 19 if type(fac,`^`) then 20 s := legendre_pow(n,fac) else 21 s := numtheory:-quadres(n,op(fac)) end if; 22 if s = -1 then 23 return -1 end if end do; 24 1end proc---Edwin === In the code for the Maple procedure quadres that I previously posted therewas a procedure legendre_pow that is not normally accessible. However,Robert Israel and Alec Mihailovs showed me two ways to get it. I haveattached the method by Alec to the end of this message. --Edwin> Here's the Maple code for quadres:showstat(numtheory:-quadres); numtheory:-quadres := proc(a, p)> local fac, ig, n, pf, s;> 1 if nargs <> 2 then> 2 error invalid arguments> end if;> 3 n := modp(a,p);> 4 if type(n,integer) and issqr(n) then> 5 return 1> end if;> 6 if not (type(n,integer) and type(p,integer)) then> 7 return ?procname(args)'> end if;> 8 if p < 6 then> 9 return -1> end if;> 10 if isprime(p) then> 11 return mods((?power')(n,1/2*p-1/2),p)> end if;> 12 ig := igcd(n,p);> 13 if ig <> 1 and igcd(denom(n/ig^2),p/ig) = 1 then> 14 return numtheory:-quadres(n/ig^2,p/ig)> end if;> 15 pf := ifactor(p);> 16 if type(pf,`^`) then> 17 return legendre_pow(n,pf)> end if;> 18 for fac in pf do> 19 if type(fac,`^`) then> 20 s := legendre_pow(n,fac)> else> 21 s := numtheory:-quadres(n,op(fac))> end if;> 22 if s = -1 then> 23 return -1> end if> end do;> 24 1> end proc>Here's the code for legendre_pow following Alec's method:kernelopts(opaquemodules=false):showstat(numtheory:- legendre_pow);numtheory:-legendre_pow := proc(nn, pow)local b, n, r; 1 r := op(2,pow); 2 b := op(op(1,pow)); 3 n := modp(nn,b^r); 4 if n <> 0 then 5 while igcd(n,b) <> 1 do 6 if not type(n/b^2,integer) then 7 return -1 else 8 n := n/b^2 end if end do end if; 9 if has({0, 1, 4, 9, 16},n) then 10 return 1 end if; 11 if b = 2 then 12 if modp(n,2^min(3,r)) = 1 then 13 return 1 else 14 return -1 end if end if; 15 quadres(n,b)end procX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === at 08:31 PM, hot-girl said:>which of de?iton is right??Both. One de?es a compact space, the other defrines a compact subsetof a topological space. A subset of a topological space is compact if? is a compact space when given the subspace topology. It's a matterof taste which you start with.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === Earlier today I noticed that it is very easy to derive Ceva's theorem from Menelaus' theorem: with the Cevians AD, BE, CF of triangle ABC meeting at G, apply Menelaus' theorem to triangles ABD and ACD and multiply the outcomes;to be precise, (AF/FB)*(BC/CD)*(DG/GA) = -1 and (AG/GD)*(DB/BC)*(CE/EA) = -1lead to (AF/FB)*(BD/DC)*(CE/EA) = +1 via DB/CD = BD/DC, DG/GD = AG/GA = -1, and, *of course*, (-1)*(-1) = +1 :-)The emphasis on the multiplication above alludes to the eccentic observationthat, just as +1 cannot generate -1, Ceva's theorem does not seem to be capable of proving Menelaus' theorem: in view of the well established andcelebrated duality, I ?d this a bit surprising; moreover, and in view ofthe simplicity of the novel (?!) derivation above, one has to wonder why the two theorems are separated by no less than 16 centuries... Comments, anyone?[I would like to dedicate this post to the memory of my father ChristosBaloglou (1919-2002), who passed away a year ago this week, one and half years after publishing Scattered Drops of Geometry (in Greek).] baloglouAToswego.edu === George Baloglou> Earlier today I noticed that it is very easy to derive Ceva's theorem from> Menelaus' theorem: with the Cevians AD, BE, CF of triangle ABC meeting atG,> apply Menelaus' theorem to triangles ABD and ACD and multiply theoutcomes;> to be precise, (AF/FB)*(BC/CD)*(DG/GA) = -1 and (AG/GD)*(DB/BC)*(CE/EA)= -1> lead to (AF/FB)*(BD/DC)*(CE/EA) = +1 via DB/CD = BD/DC, DG/GD = AG/GA= -1,> and, *of course*, (-1)*(-1) = +1 :-) The emphasis on the multiplication above alludes to the eccenticobservation> that, just as +1 cannot generate -1, ...Nice. Eccentric, but I don't mind :)> ... Ceva's theorem does not seem to be capable of proving Menelaus'stheorem...Not exactly. This page goes into it:http://www.cut-the-knot.org/Generalization/ Menelaus.shtmlIt seems odd that the ancient theorem in the pair (Ceva/Menelaus) is the onewhich involves negative numbers.The above website, which is high-school level in the main, doesn't emphasizethat Ceva and Menelaus belong properly to af?e geometry, meaning that wedon't do any comparison of distances in two independant directions.Ceva and Menelaus can be thought of as statements about determinants inhomogeneous coordinates (a 19th Century invention). In that vein, Routh gavea formula for the area of the triangle formed by any three cevians,concurrent or not. The story is in Coxeter's celebrated book _Introductionto Geometry_.Thx for this interesting note, GB.LH === >George Baloglou>[snip]> ... Ceva's theorem does not seem to be capable of proving Menelaus's>theorem...>Not exactly. This page goes into it:>http://www.cut-the-knot.org/Generalization/ Menelaus.shtmlIndeed! I had done my web search before posting, by the way, and I had beenat the sister page http://www.cut-the-knot.org/Generalization/ceva.shtml, where a variety of proofs did not include the one I posted yesterday; butI see now that there is a remark toward the bottom of the page, following anumber of applications, that does allude to the equivalence between the two theorems...>[snip]>Thx for this interesting note, GB. baloglouAToswego.edu === >Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a ?ed number of calculations?)No.>I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?No. There is no such thing as an optimal method for a ?ednumber of calculations, unless you severely restrict the set of possible functions to be integrated. Each method will give 0 error for some functions, not for others.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a ?ed number of calculations?)No.>I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?No. There is no such thing as an optimal method for a ?ed> number of calculations, unless you severely restrict the set > of possible functions to be integrated. Each method will give > 0 error for some functions, not for others.Yes, this was clear to me (as stated in the aboveparagraph you quoted).I was hoping to ?d out if in general, or in acertain class of functions, or say, in a certainapproximate percentage of the cases it might betrue that Simpson's rule would beat higher orderpolynomials in terms of accuracy per number ofcalculations.I assume that even that is hard to ?d out?I guess I'll do some more reading on this, andmaybe some experimentation (what the hell, itwill be fun anyway! :-))Carlos-- === >Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a ?ed number of calculations?) > No. >I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true?> No. There is no such thing as an optimal method for a ?ed> number of calculations, unless you severely restrict the set > of possible functions to be integrated. Each method will give > 0 error for some functions, not for others.>Yes, this was clear to me (as stated in the above>paragraph you quoted).>I was hoping to ?d out if in general, or in a>certain class of functions, or say, in a certain>approximate percentage of the cases it might be>true that Simpson's rule would beat higher order>polynomials in terms of accuracy per number of>calculations.No.Consider, e.g., polynomials f(x) = sum_{j=0}^5 a_j x^j of degree 5, integrated on [0,1]. If you allow 5 function evaluations,Simpson's rule gives you 1/12*f(0)+1/3*f(1/4)+1/6*f(1/2)+1/3*f(3/4)+1/12*f(1)which has an error of a_4/1920 + a_5/768. With the same 5 functionevaluations, the Newton-Cotes rule of order 4 (Bode's rule) gives you7/90*f(0)+16/45*f(1/4)+2/15*f(1/2)+16/45*f(3/4)+7/90*f(1) which has error 0. This Bode's formula will have a smaller error thanSimpson's on int_0^1 x^n dx for any integer n > 3. So there'sno way Simpson's rule can be considered to be optimal.On the other hand, if you allow 5 function evaluations anywhere in the interval, 5-point Gaussian quadrature gives you a formula thatis exact for polynomials of degree up to 9:sum_{j=1}^5 w_j f(x_j) wherew_1 = 161/900-13/1800*70^(1/2), x_1 = -1/42*(245+14*70^(1/2))^(1/2)+1/2w_2 = 161/900+13/1800*70^(1/2), x_2 = -1/42*(245-14*70^(1/2))^(1/2)+1/2w_3 = 64/225, x_3 = 1/2w_4 = 161/900+13/1800*70^(1/2), x_4 = 1/42*(245-14*70^(1/2))^(1/2)+1/2w_5 = 161/900-13/1800*70^(1/2), x_5 = 1/42*(245+14*70^(1/2))^(1/2)+1/2In general, n-point Gaussian quadrature will be exact on polynomialsof degree up to 2n-1, and it's easy to see there can't be an n-point formula that is exact on all polynomials of degree 2n. In this sense,Gaussian quadrature is the optimal method with a ?ed number of function evaluations.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === In practice, it is a trade off between the number of divisions and andnumber of bits of accuracy of your computer.If you keep repeatig, but each time doubling the number of divisions, youwill see the answers home in on the most accurate and then start to diverge.-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft >Is the Simpson's method optimal with respect to the>number of calculations? (i.e., does it yield the>lowest error for a ?ed number of calculations?) No. >I know that there may not exact and absolute answer>to this -- but in general, for non-pathological,>non-super-obscure situations, is it true? No. There is no such thing as an optimal method for a ?ed> number of calculations, unless you severely restrict the set> of possible functions to be integrated. Each method will give> 0 error for some functions, not for others. Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> === |The lawyers for the|Mathematical Association of America also did a lot of work, ?ding|all sorts of precedents where a person was called a scab, a traitor,|and other nasty things in print and the courts let the authors get|away with it.MAA lawyers-- what a concept! :-)Was the issue of whether it was factually correct that he wasa crank ever raised in court?Keith Ramsay === >|The lawyers for the>|Mathematical Association of America also did a lot of work, ?ding>|all sorts of precedents where a person was called a scab, a traitor,>|and other nasty things in print and the courts let the authors get>|away with it.MAA lawyers-- what a concept! :-)Was the issue of whether it was factually correct that he was>a crank ever raised in court?Prof. Dudley will obviously be in a better position to answer than Ican, but judging from the opinion in the 7th Circuit, I would guessthat the answer is no. The case was dismissed on a question of law,that the label crank, as de?ed and used by Prof. Dudley in hisbook, cannot be held to be difamatory (beginning of the 3rd paragraphand 4th paragraph of the opinion:http://www.law.emory.edu/7circuit/jan96/95-2282.html) Matters of fact are for the jury, matters of law are for the judge. Ifthe term is held incapable of being defamatory (as crank was here),then there is no question of fact of whether it was correct or not. Itis immaterial whether the person ?s or not the description, sincethe term cannot cause defamation. The question of whether the personwas factually deserving of the word would be a question of fact, andthus left to a jury... So it was never addressed at all. In fact, theopinion states that ->if<- the label was incorrectly applied, thenthere was a ready remedy available for the other person: This is especially clear where, as in this case, the word is used in a work of scholarship. As we emphasized in the Underwager case, judges are not well equipped to resolve academic controversies, of which a controversy over Cantor's diagonal process is a daunting illustration, and scholars have their own remedies for unfair criticisms of their work--the publication of a rebuttal. Unlike the ordinary citizen, a scholar generally has ready access to the same media by which he is allegedly defamed. If Dudley's criticisms of Dilworth are unsound, Dilworth should be able to publish a rebuttal in the same journal in which he published the publish a rebuttal since its own reputation was impugned by Dudley's charges.(paragraph 7 of the opinion). === === ===It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === = === ==Arturo Magidinmagidin@math.berkeley.edu === >If his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hearthat. Do you know of any examples where such a thing has happened?-- Richard-- FreeBSD rules! === >If his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hear> that. Do you know of any examples where such a thing has happened?-- RichardAll his heirs would have to do is show harm, that is, a tort.If his results were legitimate, then proper recognition would havebeen a bene?, both to him, and I'd think to his family.Like would you rather have it known that a relative was a crank or adiscoverer?James Harris === >If his work ever turns out to have legitimacy then his heirs, assuming>he has any, can sue you as well Mr. Dudley.Really? I'm not very familiar with US law, but I'm surprised to hear>that. Do you know of any examples where such a thing has happened?-- RichardAll his heirs would have to do is show harm, that is, a tort.If his results were legitimate, then proper recognition would have> been a bene?, both to him, and I'd think to his family.Like would you rather have it known that a relative was a crank or a> discoverer?You are getting yourself in a very dangerous situation here. If your argument were right, then all your relatives could sue you because your posts on the internet make it clear to the whole world that you are a complete and utter idiot. Personally, I believe that is just tough and none of your relatives should have the right to sue you for that. === Could you please help me on the following as I simply do not understandhow to attack the problem?Use generating functions to ?d the number of ways to select 14 ballsfrom a Jar holding 100 red balls, 100 blue balls and 100 green balls sothat no fewer than 3 and no more than 10 blue balls are selected. Assumethat the order in which the balls are drawn does not matterPLEASE HELP explain how to do this as I have 3 more to do also....muchappreciated...TIA === > Could you please help me on the following as I simply do not understand> how to attack the problem?Use generating functions to ?d the number of ways to select 14 balls> from a Jar holding 100 red balls, 100 blue balls and 100 green balls so> that no fewer than 3 and no more than 10 blue balls are selected. Assume> that the order in which the balls are drawn does not matterPLEASE HELP explain how to do this as I have 3 more to do also....much> appreciated...TIA You want the coef?ient of x^14 in (x^3 + x^4 + ... + x^10)*(1 + x + x^2 + ...)^2 = x^3*(1 + x + x^2 + ... + x^7)*(1 + x + x^2 + ... )^2.So, you want the coef?ient of x^11 in (1 + x + x^2 + ... + x^7)*(1 + x + x^2 + ... )^2 =[(1 - x^8)/(1 - x)]*[1/((1-x)^2)] =(1 - x^8)/((1 - x)^3) = (1 - x^8)*(1 - x)^(-3).Now, use the generalized binomial theorem and note there aren't toomany ways to get x^11.-- Post in haste; repent at leisure. Paul SperryColumbia, SC (USA) === > So before I can make any money, I have to get past mathematicians, the> dark gatekeepers testing their ability to deny knowledge from the> general public.> James HarrisIf JSH really wants to make money, he should try some ?ld where there is more of it than in mathematics. === >expand the Cunningham tables (a list of factors of 2^n, etc.). Around > Bill, I think you meant /factors of large prime numbers/ ? >Large prime numbers don't have proper factors. > And Dik's name isn't Bill.O, I do not know. I do not have a birth certi?ate here at home. Butat least my passport says it is indeed not. Sheesh, a lucky escape.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [snip]> My experience with patents is that the provisional application which> but I advise him to contact a patent attorney to write it. This costs> me about $400 for this provisional application. The provisional> application protects him for 1 year till he ?es the complete> application. This also gives him time to shop the idea around to> companies. After a year he ?es the complete application. My> experience is that this could cost from $6000-10000 when using a> patent attorney.> I wish him luck as the world needs new ideas.Your understanding of the patent requirements is incorrect. There are no provisional applicationcapabilities. There are disclosure documents which can be ?ed to establish temporal priority, but theydo not permit the future applicant to publish or otherwise publicly disclose his work.Having said that, I think your suggestion that James apply for intellectual property protection is sound --especially if he retains a patent attorney.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I think this is morally wrong. Don't tell James to waste his money.Mocking the town idiot is bad enough, but actually convincing him to jumpdown a well is much worse.And James, If you want to make money in math you should try one of the Clayproblems. Their worth a million dollars each if you prove one. It might bethe only way to make alot of money in math.http://www.claymath.org/Millennium_Prize_Problems/Justin Van Winkle [snip]My experience with patents is that the provisional application which>but I advise him to contact a patent attorney to write it. This costs>me about $400 for this provisional application. The provisional>application protects him for 1 year till he ?es the complete>application. This also gives him time to shop the idea around to>companies. After a year he ?es the complete application. My>experience is that this could cost from $6000-10000 when using a>patent attorney.>I wish him luck as the world needs new ideas. Your understanding of the patent requirements is incorrect. There are noprovisional application> capabilities. There are disclosure documents which can be ?ed toestablish temporal priority, but they> do not permit the future applicant to publish or otherwise publiclydisclose his work. Having said that, I think your suggestion that James apply forintellectual property protection is sound --> especially if he retains a patent attorney. --> There are two things you must never attempt to prove: the unprovable -- and the obvious.> --> Democracy: The triumph of popularity over principle.> --> http://www.crbond.com === > I think this is morally wrong.Since you top posted, I have no idea exactly *what* was said that you regard asmorally wrong.> Don't tell James to waste his money.I only recommended that he get the advice of a patent attorney if he wanted tosecure intellectual property rights. I assume you regard your *free* adviceabout the matter as superior to the advice of a professional. If so, you may beguilty of practicing law without a license.[snip]--There are two things you must never attempt to prove: the unprovable -- and theobvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === I have to mention here that I was giving you advice. So it was reallymeta-advice. And how being *free* applies I'm not sure, since I wouldn'tcharge you for an admonition, as it would be unlikely that anyone would pay.The tone of your reply threw me, I must admit.According to my newsreader I didn't top post. But I'm using a microsoftproduct and as such it might not be working as intended. If I am topJustin Van WinkleI think this is morally wrong. Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong. Don't tell James to waste his money. I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license. [snip] --> There are two things you must never attempt to prove: the unprovable -- and the> obvious.> --> Democracy: The triumph of popularity over principle.> --> http://www.crbond.com === I think this is morally wrong. Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong. Don't tell James to waste his money. I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license.[snip!]Justin Van Winkle === > I think this is morally wrong.> Since you top posted, I have no idea exactly *what* was said that you> regard as>morally wrong.> Don't tell James to waste his money.> I only recommended that he get the advice of a patent attorney if he> wanted to>secure intellectual property rights. I assume you regard your *free*> advice>about the matter as superior to the advice of a professional. If so, you> may be>guilty of practicing law without a license.> [snip!]Justin Van WinkleRegarding patents, a previous poster is correct, a provisionalapplication establishes who was ?st only. However, I think that itis possible in the US to publish up to a year prior to ?ing a patentapplication. After ?ing the provisional application, the personshould obtain a simple non-disclosure agreement from anyone hediscusses it with. The original poster should contact an attorney forgood advice.Now, the market for a method to reliably count primes may besubstantial, perhaps many millions of $/yr. Such a method wouldenable someone to decide the probability of whether he he has triedall possible prime factors of a number for code breaking. Such amethod combined with others could be worth a great deal of money. === ...You do not know the ?ld, yes? > Now, the market for a method to reliably count primes may be > substantial, perhaps many millions of $/yr.You are off by a factor of many millions. > Such a method would > enable someone to decide the probability of whether he he has tried > all possible prime factors of a number for code breaking. Such a > method combined with others could be worth a great deal of money.Such a method would be the most stupid to be done. Trying all possiblefactors of a number for code breaking is about the slowest possiblemethod you can ?d, even if each possible factor is found and triedeach femtosecond.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === [snip]> Now, the market for a method to reliably count primes may be> substantial, perhaps many millions of $/yr. Such a method would> enable someone to decide the probability of whether he he has tried> all possible prime factors of a number for code breaking. Such a> method combined with others could be worth a great deal of money.This doesn't really make sense to me. If you can generate all the possibleprime factors, then you know how many there are. If you can't generatethem, then certainly you can't use the fact that there are more factors tosomehow generate more factors. Seriously, if you know that there must be atleast 10^20 more primes to try, that still leaves you to ?d all theseprimes. If you know that there aren't any more primes to try, then you knowhow many you've tried. (I'm not an expert on this topic, so I may be wayoff.)Now, if you had a function that returned how many primes are less than agiven number that would be something. However, if you must iterate throughevery number less than that given number, your method isn't really thatuseful (unless it is much faster than any current method). If you had analgorithm to factor a number, that would be useful. If that algorithminvolves testing every prime less than sqrt(n) to test the primality of n,then the algorithm is useless (and trivial).In the end, however, you may still have a valid point. But that point restson the assumption that this is the best method to do such work. (or it iscomperable to good methods) There are already pretty good estimates of howmany primes are less than a given number. There are very fast methods tocount how many primes are less than a given number. There are VERY fastmethods to determine the primality of even extremely large numbers, withsome small chance of error. === > [snip] >Now, the market for a method to reliably count primes may be >substantial, perhaps many millions of $/yr. Such a method would >enable someone to decide the probability of whether he he has tried >all possible prime factors of a number for code breaking. Such a >method combined with others could be worth a great deal of money. > This doesn't really make sense to me. If you can generate all the possible > prime factors, then you know how many there are. If you can't generate > them, then certainly you can't use the fact that there are more factors to > somehow generate more factors. Seriously, if you know that there must be at > least 10^20 more primes to try, that still leaves you to ?d all these > primes. If you know that there aren't any more primes to try, then you know > how many you've tried. (I'm not an expert on this topic, so I may be way > off.)If you know that there are 10^20 primes to try it will take you 10^11seconds to try them all when you could try each prime in a nanosecond.That is something like 3000 years. That would even get you moderate40 digit numbers out of reach. > There are very fast methods to > count how many primes are less than a given number. There are VERY fast > methods to determine the primality of even extremely large numbers, with > some small chance of error.There are even VERY fast methods to determine the primality of evenextremely large numbers without a chance of error.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === > [snip]>Now, the market for a method to reliably count primes may be>substantial, perhaps many millions of $/yr. Such a method would>enable someone to decide the probability of whether he he has tried>all possible prime factors of a number for code breaking. Such a>method combined with others could be worth a great deal of money.> This doesn't really make sense to me. If you can generate all the possible> prime factors, then you know how many there are. If you can't generate> them, then certainly you can't use the fact that there are more factors to> somehow generate more factors. Seriously, if you know that there must be at> least 10^20 more primes to try, that still leaves you to ?d all these> primes. If you know that there aren't any more primes to try, then you know> how many you've tried. (I'm not an expert on this topic, so I may be way> off.) If you know that there are 10^20 primes to try it will take you 10^11> seconds to try them all when you could try each prime in a nanosecond.> That is something like 3000 years. That would even get you moderate> 40 digit numbers out of reach. > There are very fast methods to> count how many primes are less than a given number. There are VERY fast> methods to determine the primality of even extremely large numbers, with> some small chance of error. There are even VERY fast methods to determine the primality of even> extremely large numbers without a chance of error.Just to clarify, as I understand it... knowing a number is composite does nottell you the factors, to be commercially useful you have to ?d the factors, otherwisethe ?discovery' is mostly academic.The main techinique is ?ding witnesses to compositeness. Over half of all numbersless than any composite are ?witnesses' to that composite, so to determine a number hasfactors just requires more and more random attempts until a witness is found, unfortanetlythis only establishes a con?ence of primality, you seem to be suggesting a certaintechnique.Herc === This doesn't really make sense to me. If you can generate all thepossible> prime factors, then you know how many there are. If you can't generate> them, then certainly you can't use the fact that there are more factors to> somehow generate more factors. Seriously, if you know that there must beat> least 10^20 more primes to try, that still leaves you to ?d all these> primes. If you know that there aren't any more primes to try, then youknow> how many you've tried. (I'm not an expert on this topic, so I may be way> off.)There is a simple proof that there can be no largest prime. Even my kidsunderstand it (10, 11 and 13). === This is where you are totally off. possible prime factors =/= primenumbers . If you were silly enough to attempt to test primality of a largenumber by trial division, then you would only have to check that the numberwasn't divisable by all the primes less than the square root of the number.So the possible prime factors might be every every prime less than half of anumber, the ones that you have to check to establish primality are thosethat are less than the square root. Obviosly, a if n < m, m cannot be afactor of n.The fact that your kids understand this proof at such a young age seems tobe an attempted slight on my knowledge or intelligence, since I am mucholder and you imply that I do not understand it. It must be ratherembarrassing then to have insulted me based on a complete lack ofunderstanding of the thread. I'll refrain from insulting you, since itwould be pointless and immature to advise you to have your kids double checkyour usenet posts.Justin Van Winkle This doesn't really make sense to me. If you can generate all the> possible>prime factors, then you know how many there are. If you can't generate>them, then certainly you can't use the fact that there are more factorsto>somehow generate more factors. Seriously, if you know that there mustbe> at>least 10^20 more primes to try, that still leaves you to ?d all these>primes. If you know that there aren't any more primes to try, then you> know>how many you've tried. (I'm not an expert on this topic, so I may beway>off.) There is a simple proof that there can be no largest prime. Even my kids> understand it (10, 11 and 13).> === I think this is morally wrong. Since you top posted, I have no idea exactly *what* was said that youregard as> morally wrong. Don't tell James to waste his money. I only recommended that he get the advice of a patent attorney if hewanted to> secure intellectual property rights. I assume you regard your *free*advice> about the matter as superior to the advice of a professional. If so, youmay be> guilty of practicing law without a license.Not if it's free. === [snip]> There is no question of the legitimacy of my method for counting prime> numbers by integrating a partial difference equation.Yes, there is. Partial difference equations are *not* solved by integration. They are solved using the sumcalculus. Integration is is used to ?d anti-derivatives.> There is also no question that I'm the ?st person in recorded> history to ever present such a method.Right now, no one has presented such a method. Your presentation did not involve integration of any kind, muchless of a partial difference equation.> There's also no question that partial difference equations are analogs> to partial differential equations.There are parallels, but not the ones you think.[snip]> So before I can make any money, I have to get past mathematicians, the> dark gatekeepers testing their ability to deny knowledge from the> general public.Your paranoia is showing. No one has *ever* stopped you from ?g your pseudo-mathematics all over theinternet. You are well-known to at least a segment of the readership as an uneducable crank.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === ... > My experience with patents is that the provisional application which > but I advise him to contact a patent attorney to write it. This costs > me about $400 for this provisional application. The provisional > application protects him for 1 year till he ?es the complete > application. This also gives him time to shop the idea around to > companies. After a year he ?es the complete application. My > experience is that this could cost from $6000-10000 when using a > patent attorney. > I wish him luck as the world needs new ideas.That is a large amount of money to invest in something with which you*never* would recuperate your cost, *even if it was new and original*!If he had an astoundingly fast method to count primes, my estimateis that from the sales of the algorithm he would get at most $50-200.If he is lucky. The market for such algorithms is small to non-existing.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === >If he ?es for IP protection, the USPTO will determine >originality. > I don't believe mathematical proofs are patentable. Sadly, software > algorithms *are* patentable in the United States, and so it is unclear > where to draw the distinction.Software is indeed patentable in the US (and in Japan I think). It is*not* patentable in the EU. Laws to allow such patenting have justsome months ago been rejected in the European parliament. So even *if*he had a patent on his software it would have no validity in Europe.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === If A is a retract of B^2 (the disk in R^2), then every continuous map f :A ---> A has a ?ed point.How can I prove this, knowing Brouwer's ?ed point theorem for B^2?Mike === > If A is a retract of B^2 (the disk in R^2), then every continuous map f :> A ---> A has a ?ed point.How can I prove this, knowing Brouwer's ?ed point theorem for B^2?Let r be a retraction of B^2 onto A. Then f r (the composition) mapsB^2 into A (which is a subset of B^2), therefore has a ?ed-point x. But x is in the range of f, hence in A, hence r(x) = x, hence x =f(r(x)) = f(x), QED.This is one of the simplest and most fundamental tricks in ?ed-pointtheory...--Ron Bruck === >For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?>This one has really had me stumped. My ?st guess was that the answer>was no, but I didn't get anywhere trying to prove it. I then tried to>construct such a set: since the set of rationals is countable, we can>write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of>length 1/2^n centred at r_n, let I be the union of the I_n, and let X>be the complement of I in R. Since the total length of the intervals>is at most 1, X is nonempty. As the complement of an open set, X is>closed. But I can't show that X is perfect.Your example might not be perfect. However, it can be ?ed.Instead of 1/2^n, make the length be irrational (so all the intervals haveirrational endpoints), and small enough so that either I_n is contained ineither the union of the previous I_k's or its closure is in the interiorof the complement of that union.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?> Look up the Cantor-Bendixson Theorem. === > For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?> This one has really had me stumped. My ?st guess was that the answer> was no, but I didn't get anywhere trying to prove it. I then tried to> construct such a set: since the set of rationals is countable, we can> write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of> length 1/2^n centred at r_n, let I be the union of the I_n, and let X> be the complement of I in R. Since the total length of the intervals> is at most 1, X is nonempty. As the complement of an open set, X is> closed. But I can't show that X is perfect. So: Is the answer yes or> no? If yes, does my set work as an example? Any hints would be most> welcome.I think it depends on how you enumerate the rationals. For example,if r_1 = 0, r_2 = 3/4, ....., then the point 1/2 lies exactly on the boundaries of I_1 and I_2, and _may_ end up an isolated point in your X.I also believe that multiplying your 1/2^n by some irrational numberwould help, although I've not worked out the details. === > For those who have the book, it is W. Rudin, Principles of> Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:> [A set E is perfect if E is closed and every point of E is a limit> point of E] Is there a nonempty perfect set in R which contains no> rational number?This one has really had me stumped. My ?st guess was that the answer> was no, but I didn't get anywhere trying to prove it. I then tried to> construct such a set: since the set of rationals is countable, we can> write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of> length 1/2^n centred at r_n, let I be the union of the I_n, and let X> be the complement of I in R. Since the total length of the intervals> is at most 1, X is nonempty. As the complement of an open set, X is> closed. But I can't show that X is perfect. So: Is the answer yes or> no? If yes, does my set work as an example? Any hints would be most> welcome.Try showing (1) X is not countable (2) the isolated points of X constitute at most a countable set (3) X {isolated points of X} is closed === >For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?>This one has really had me stumped. My ?st guess was that the answer>was no, but I didn't get anywhere trying to prove it. I then tried to>construct such a set: since the set of rationals is countable, we can>write it as {r_1, r_2, r_3,...}. Let I_n be the open interval of>length 1/2^n centred at r_n, let I be the union of the I_n, and let X>be the complement of I in R. Since the total length of the intervals>is at most 1, X is nonempty. As the complement of an open set, X is>closed. But I can't show that X is perfect. So: Is the answer yes or>no? If yes, does my set work as an example? Any hints would be most>welcome.> Try showing (1) X is not countable> (2) the isolated points of X constitute at most a> countable set> (3) X {isolated points of X} is closedI don't think this will work. Try considering points of X such thatevery neighborhood contains an uncountable in?ity of points of Xinstead. === For those who have the book, it is W. Rudin, Principles of>Mathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?> ... [ contructs open set containing rationals and with measure < 1, takex X as its complement ] ... > Try showing (1) X is not countable> (2) the isolated points of X constitute at most a> countable set> (3) X {isolated points of X} is closedI don't think this will work. Try considering points of X such that> every neighborhood contains an uncountable in?ity of points of X> instead.By Cantor-Bendixson, (2) must be true. (3) seems clear since X, being closed, contains all its accumulation points, and these are also exactly the accumulation points of X{isolated points}.Is there a problem with (1) or C-B too much to take as known? === >For those who have the book, it is W. Rudin, Principles ofMathematical Analysis, chapter 2 ( Basic Topology), problem 18:>[A set E is perfect if E is closed and every point of E is a limit>point of E] Is there a nonempty perfect set in R which contains no>rational number?> ... [ contructs open set containing rationals and with measure < 1,> takex X as its complement ] ...> Try showing (1) X is not countable> (2) the isolated points of X constitute at most a> countable set> (3) X {isolated points of X} is closedI don't think this will work. Try considering points of X such that>every neighborhood contains an uncountable in?ity of points of X>instead.By Cantor-Bendixson, (2) must be true. (3) seems clear since > X, being closed, contains all its accumulation points, and these > are also exactly the accumulation points of X{isolated points}.> Is there a problem with (1) or C-B too much to take as known?The theorem is not listed in the index to the book, so I don't thinkit can be assumed in the problem. <845b431.0311101837.20305be0@posting.google.com> <845b431.0311110702.6d8c40fb@posting.google.com>X-Cise: tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate: admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Punge: Micro$oft === at 07:02 AM, snizpilbor@yahoo.com (Sniz Pilbor) said:>Let P(x) denote the logic statement x < x^2>Let Q(x) denote the logic statement x < f(x)>P and Q are unequal iff there is some real x such that P(x) != Q(x)>Else, they are equalNo. They can be different propositional functions even if their valuesare provably the same.>It seems to me upon inspection (though I keep an open mind to>correctionNo. You haven't kept an open mind.>And since they are equal, if one has a given property, the other>must as well. sniz and SNIZ refer to the same person, so they must be the same.Since the ?st is lower case, the second must be lower case as well.Do you see the fallacy in that?>This is all quite confusing,To you. Others see the distinctions quite clearly.>know that I am just a naive person who humbly desires>to understand these things.Were that the ask you would ask more and assert less.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to spamtrap@library.lspace.org === >The arithmetic mean of two consecutive primes minus the geometric mean of>the same two consecutive primes equals less than 1/2.>What's the name of the original conjecture?For any reals x and y, AM(x,y) - GM(x,y) = (x+y)/2 - sqrt(xy) = (sqrt(x)-sqrt(y))^2/2so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's Conjecture.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's >Conjecture.I didn't know this conjecture. I suppose that a related problem hasbeen studied as well: given the set of numbers 0 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 === >so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's>Conjecture. I didn't know this conjecture. I suppose that a related problem has> been studied as well: given the set of numbers 0 p_{n+1}^a-p_n^a<1 for all n, which is currently the largest number> (provably) known to belong to it?See: http://mathworld.wolfram.com/AndricasConjecture.html> 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 === Right! Maybe too easy. :-)Tapio>The arithmetic mean of two consecutive primes minus the geometric mean of>the same two consecutive primes equals less than 1/2.>What's the name of the original conjecture? For any reals x and y,> AM(x,y) - GM(x,y) = (x+y)/2 - sqrt(xy) = (sqrt(x)-sqrt(y))^2/2> so you're saying sqrt(p_(n+1)) - sqrt(p_n) < 1, which is Andrica's> Conjecture. Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2> === Fish, or cut bait.Solve problems.Borel says almost all sequences are normal and have equal zero and onedensities, combinatorics say almost no sequences do, I think half do.How is reconciled Borel and asymptotic combinatorics?Sets of numbers contain only sets.Ross === > I made the (possibly controversial) step of teaching myself homotopy theory> from GB's Geometry and Topology book, and noticed a shed load of typos. Does> anyone know if they've all been catalogued and whether I could access a list> of them somewhere?>What you think the typos? What's the 4 or 5 simplest? === >I made the (possibly controversial) step of teaching myself homotopytheory>from GB's Geometry and Topology book, and noticed a shed load of typos.Does>anyone know if they've all been catalogued and whether I could access alist>of them somewhere?> What you think the typos? What's the 4 or 5 simplest?What difference does that make? Is there a minimum dif?ulty level beforeprinted typo lists are considered?I don't remember off-hand so I'll have to wait until I get to my of?etomorrow. === > If you allow ?ite numbers constructed from a series of mathematical> formulae, I think Graham's number takes the cake, by far.>FYI,Harvey Friedman's lower bound for n(4) in his block subsequence theorem is>much larger than Graham's number- it involves the Ackermann numbers A(n,n).> let A(n) equal A(n,n)> A(A(A........(A(1)..) where there are A(187,196) A'sAren't Ackermann's functions something like A(1,1)=1+1, A(2,2)=2*2,> A(3,3)=3^3, A(4,4)=(((4^4)^4)^4)^4, and so on? In that case A(187, 196)> must be pretty huge.> I read that webpage about writing big numbers on a piece of paper. If> I am allowed to refer to Ackermann's function only by name, I think I> can write a ?ite natural number that is much larger than Harvey> Friedman's number.Actually,Friedman uses a streamlined Ackermann hierarchy,to base 2,whereA(1,n) is doubling, A(2,n) is 2^n, A(3,n) is 2^^n,etc.In the expression above,he uses the unary form--A(n) = A(n,n).So the startingpoint, A(1) = A(1,1) = 2.And, there are A(187,196) = A(187196,187196) A's in total, which dwarfs theGraham number.How you would Ackermann down that many times,I haven't a clue. === > Can anyone point me toward a good method for evaluating the prolate> spheroidal wave functions?There's a book Computation of Special Functions, by Shan-jie Zhangand Jianming Jin. The programs themselves are available on theweb, in Fortran-77. Here:http://iris-lee3.ece.uiuc.edu/~jjin/routines/ routines.html-- Andrew === Can anyone point me toward a good method for evaluating the prolate>spheroidal wave functions? There's a book Computation of Special Functions, by Shan-jie Zhang> and Jianming Jin. The programs themselves are available on the> web, in Fortran-77. Here:> http://iris-lee3.ece.uiuc.edu/~jjin/routines/routines.html -- > Andrew> === so I'll check out the various things you suggested and see how they go.Gene === I'm looking for some resources to learn a bit more about the gauge integral (Alternately called the Generalised Riemann Integral and the Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd probably prefer lebesgue integration anyway, as it's so much more general) and worth learning more about. Also I'm in the process of (slowly) writing an analysis textbook, and it seems a fairly natural extension of the way I'm introducing Riemann integration, so I thought it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de?ition, and a couple basic facts about it - not much more than I learned from http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't have to pay money for it. :) But if you can reccomend some good books on the subject that would be great as well.Now lets see if this gets through... my college newsserver has been eating a lot of posts recently.David === also:Pfeffer: The Riemann Approach to Integration (CUP)Gordon: The Integrals of Lebesgue, Denjoy, Perron, and Henstock (AMS)-- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === > I'm looking for some resources to learn a bit more about the gauge > integral (Alternately called the Generalised Riemann Integral and the > Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd > probably prefer lebesgue integration anyway, as it's so much more > general) and worth learning more about. Also I'm in the process of > (slowly) writing an analysis textbook, and it seems a fairly natural > extension of the way I'm introducing Riemann integration, so I thought > it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de?ition, and a couple basic facts about it - > not much more than I learned from > http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't > have to pay money for it. :) But if you can reccomend some good books on > the subject that would be great as well.Now lets see if this gets through... my college newsserver has been > eating a lot of posts recently.DavidI think Bartle's and Sherbert's Introduction to Real Analysis is agood choice. I guess that book is not that expensive.Artur === > I'm looking for some resources to learn a bit more about the gauge > integral (Alternately called the Generalised Riemann Integral and the > Henstock-Kurzweil integral or probably a couple others).It seems like an extremely neat idea (although for most purposes I'd > probably prefer lebesgue integration anyway, as it's so much more > general) and worth learning more about. Also I'm in the process of > (slowly) writing an analysis textbook, and it seems a fairly natural > extension of the way I'm introducing Riemann integration, so I thought > it might be worth pursuing to see if it's a good idea to include it.At the moment I know the de?ition, and a couple basic facts about it - > not much more than I learned from > http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .An online treatment would of course be preferable as, well, I wouldn't > have to pay money for it. :) But if you can reccomend some good books on > the subject that would be great as well.I suppose there must be online lecture notes abot this, but I don't knowthem. Try this book: Lee Peng Yee & Rudolf Vyborny Integral: An Easy Approach After Kurzweil and Henstock Cambridge University PressJose Carlos Santos === > I'm looking for some resources to learn a bit more about the gauge > integral (Alternately called the Generalised Riemann Integral and the > Henstock-Kurzweil integral or probably a couple others).> It seems like an extremely neat idea (although for most purposes I'd > probably prefer lebesgue integration anyway, as it's so much more > general) and worth learning more about. Also I'm in the process of > (slowly) writing an analysis textbook, and it seems a fairly natural > extension of the way I'm introducing Riemann integration, so I thought > it might be worth pursuing to see if it's a good idea to include it.> At the moment I know the de?ition, and a couple basic facts about it > - not much more than I learned from > http://www.math.vanderbilt.edu/~schectex/ccc/gauge/ .> An online treatment would of course be preferable as, well, I wouldn't > have to pay money for it. :) But if you can reccomend some good books > on the subject that would be great as well.> I suppose there must be online lecture notes abot this, but I don't know> them. Try this book: Lee Peng Yee & Rudolf Vyborny Integral: An Easy Approach After Kurzweil and Henstock Cambridge University Press> Jose Carlos Santos> Cambridge University Press? Woo hoo! Student discount. :)I thought I'd seen a book by CUP about it, and I was there the other day looking for it without succes, but I was looking under Gauge integration rather than Kurzweil and Henstock. Glad to know I'm not going completely insane. I shall go take a look for it shortly.David === Re: what spin 1/2 is in physical reality Re: Robertsonversus Philipsscrewdriver Re: Archimedes Plutonium NOdtgEMAIL whole entire Universe is just one big atom where dots ofthe electron-dot-cloud are galaxies sci.physics, sci.math, sci.physics.electromag> http://rapfast.petcom.com/~john/Be.GIF> Start a point orbitting a globe at a ?ed radius> from the north to the south pole.> At the same time take this circular orbit and> start it precessing at twice the rate at which> the point is travelling within the orbit.> The point will make one complete trip around the globe> in the east-west or west-east direction (whichever> way you precessed it) by the time it gets> to the south pole. Then it will make another> trip around, cutting all the lines of longitude,> on its way back up to the north pole.> Place another point exactly opposite the> ?st. It follows exactly the same path BUT> it is THE ONLY OTHER POINT which follows that> pathway. Divide the ring into four, then eight, then> sixteen points. You will ?d that sixteen points> completes the galaxy pattern:Not sure why a nice pattern should have any meaning for physicsitself. http://rapfast.petcom.com/~john/galaxypattern.gif> Then consider this; growth of rings of 16> can be directly related to the Periodic Table:> http://rapfast.petcom.com/~john/periodicpattern.GIF THIS is what spin * (Alt0189) is in physical reality.> JohnI have seen the Periodic Table placed into a circular pattern andthereis reallyno connection with a geometric ?ure that has 16 points. One can justas easily seek out a nice pattern with triangles or trapezoids where anumber of 16 comes into play and then claim that it relates to thePeriodic Table.No, John, I do not buy your above. But your above does con?m mysuspicions in the way that you need 2 prime movers. You needorbitingand you needprecession and that ties and ?s with my need for 2 forces asrepresented by2 rectangles inside a circumscribing ellipse (or the 3rd dimensionalequivalents).Both of us need 2 entities, better to call them forces.Yours John is purely a fanciful geometry pattern.Mine however, is more than geometry but tells of the Prime Mover ofspininside or interior where at least 2 forces are diametrical to oneanother. Inside an electron which has spin 1/2 there are the twoforcesof Gravity and Antigravity and they can be visualized as a Philipsheadof a cross where one force is one line of the cross and the otherforceis the other line of the Philips cross.And since forces are in motion the Philips head embedded in eachelectronis spinning. Not that there exists the screwdriver to cause theelectronto turn and spin. But that the interior forces of Gravity andAntigravity themselves generate the spin motion.would havespin 3/2 and then you would have a different Philips head patternbecause you have more than 2 diametrically opposing internal forces.Sorry to say John, yours seems to be no more than a cute mathematicalpattern and nothing of any Physics. At least nothing I can salvage atthe moment.whole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Electron spin of 1/2 is just a myth invented to fudge over bad physics!!!!!It is magnetic ?at is quantised. The hydrogen atom in ground statecontains one quantum ? most of which tightly wraps the electronsorbit. The electrons behaviour is coverned by the laws of induction. In thenext orbital, two ?s ?l the orbit. Either they both tightlly wrapthe orbit, or one tightly wraps it wit the other free to link with other?s. So in an experiment to measure the magentic moment, you loose atleast one unit of magnetic moment.Seehttp://users.powernet.co.uk/bearsoft/ ncqt.pdfAndhttp://users.powernet.co.uk/bearsoft/ ClasAtom.htmwhich will download its original Mathcad ?e.-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> Re: what spin 1/2 is in physical reality Re: Robertson> versus Philipsscrewdriver Re:> Archimedes Plutonium whole entire Universe is just one big atom where dots of> the electron-dot-cloud are galaxies> sci.physics, sci.math, sci.physics.electromaghttp://rapfast.petcom.com/~john/Be.GIF> Start a point orbitting a globe at a ?ed radius>from the north to the south pole.>At the same time take this circular orbit and>start it precessing at twice the rate at which>the point is travelling within the orbit.>The point will make one complete trip around the globe>in the east-west or west-east direction (whichever>way you precessed it) by the time it gets>to the south pole. Then it will make another>trip around, cutting all the lines of longitude,>on its way back up to the north pole.>Place another point exactly opposite the>?st. It follows exactly the same path BUT>it is THE ONLY OTHER POINT which follows that>pathway. Divide the ring into four, then eight, then>sixteen points. You will ?d that sixteen points>completes the galaxy pattern: Not sure why a nice pattern should have any meaning for physics> itself. http://rapfast.petcom.com/~john/galaxypattern.gif>Then consider this; growth of rings of 16>can be directly related to the Periodic Table:>http://rapfast.petcom.com/~john/periodicpattern.GIF> THIS is what spin * (Alt0189) is in physical reality.>John I have seen the Periodic Table placed into a circular pattern and> there> is really> no connection with a geometric ?ure that has 16 points. One can just> as easily seek out a nice pattern with triangles or trapezoids where a> number of 16 comes into play and then claim that it relates to the> Periodic Table. No, John, I do not buy your above. But your above does con?m my> suspicions in the way that you need 2 prime movers. You need> orbiting> and you need> precession and that ties and ?s with my need for 2 forces as> represented by> 2 rectangles inside a circumscribing ellipse (or the 3rd dimensional> equivalents). Both of us need 2 entities, better to call them forces. Yours John is purely a fanciful geometry pattern. Mine however, is more than geometry but tells of the Prime Mover of> spin> inside or interior where at least 2 forces are diametrical to one> another. Inside an electron which has spin 1/2 there are the two> forces> of Gravity and Antigravity and they can be visualized as a Philips> head> of a cross where one force is one line of the cross and the other> force> is the other line of the Philips cross. And since forces are in motion the Philips head embedded in each> electron> is spinning. Not that there exists the screwdriver to cause the> electron> to turn and spin. But that the interior forces of Gravity and> Antigravity themselves generate the spin motion. would have> spin 3/2 and then you would have a different Philips head pattern> because you have more than 2 diametrically opposing internal forces. Sorry to say John, yours seems to be no more than a cute mathematical> pattern and nothing of any Physics. At least nothing I can salvage at> the moment. whole entire Universe is just one big atom where dots> of the electron-dot-cloud are galaxies === > Electron spin of 1/2 is just a myth invented to fudge over bad physics!!!!!> It is magnetic ?at is quantised. The hydrogen atom in ground state> contains one quantum ? most of which tightly wraps the electrons> orbit. The electrons behaviour is coverned by the laws of induction. In the> next orbital, two ?s ?l the orbit. Either they both tightlly wrap> the orbit, or one tightly wraps it wit the other free to link with other> ?s. So in an experiment to measure the magentic moment, you loose at> least one unit of magnetic moment.See> http://users.powernet.co.uk/bearsoft/ncqt.pdf> And> http://users.powernet.co.uk/bearsoft/ClasAtom.htm> which will download its original Mathcad ?e.> -- Looking at your website, seems as though to me that you are claimingmetaphorically to remove an apple from a basket of apples and thenreplacing the removed apple with another apple. So why really bother?M_s as quantum spin and give it a different name of ?. I mayhave missed your point, but it seems to me that all you are doing isshuf?r replacing and not bringing anything new into the picture.If someone did not like Forces and replaced them with kinetic-energy,well, that is do-able but it sheds no new physics, and is just anotherreplacement scheme. Or someone may like momentum instead of forces.I do not remember the physics history exactly, only recall that N, L,M_L werecrucially required and time passed and only later realized that a 4thquantum number M_s was needed.I have 2 ideas to tackle what M_s is in physical reality. In the 1990sdecade I had only the AtomTotality where the nightsky is anelectron-cloud and so I asked myself what would M_s be if the nightskyof galaxies is an electron? Would it be like some spinning topanalogy?But in the 2000s I came up with the Coulomb Uni?ation of Forces. Soinstead of relying only on one theory--- AtomTotality to penetrate themeaning of M_s spin, I now have 2 theories to try to anchor M_s.If M_s is the result of internal forces paired off against one anothersuch as Gravity to Antigravity. If that is the physical meaning ofspin M_s. Then that begs the question of whether N, L, M_L are alsointernal forces paired off against each other. So is N theStrongNuclear Force paired off diametrically to the WeakNuclearForce??Perhaps N, L, M_L as the physics-history noted that they arefundamental but that M_s is not fundamental and a secondary outcome ofN, L, M_LI am going to have to do more thinking on this and perhaps much morerevisiting before satis?d.Archimedes Plutoniumwhole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === Let a and b be two real numbers with a < b and consider a function ffrom [a,b] into the reals which has the intermediate value propertyand is Riemann-integrable. Let F(x) be the integral of f between aand x.Question: Must F be a primitive of f?Jose Carlos Santos === Let a and b be two real numbers with a < b and consider a function f>from [a,b] into the reals which has the intermediate value property>and is Riemann-integrable. Let F(x) be the integral of f between a>and x.Question: Must F be a primitive of f?No. Say a < 0 < b. Let f(x) = sin(1/x) for x <> 0, f(0) = 0.Let g(x) = sin(1/x) for x <> 0, g(0) = 1. Then both f and gare Riemann integrable and have the intermediate value property.But they have the same inde?ite integral, and a function hasat most one derivative.Jose Carlos SantosDavid C. Ullrich === >Let a and b be two real numbers with a < b and consider a function f>from [a,b] into the reals which has the intermediate value property>and is Riemann-integrable. Let F(x) be the integral of f between a>and x.>Question: Must F be a primitive of f?> No. Say a < 0 < b. Let f(x) = sin(1/x) for x <> 0, f(0) = 0.> Let g(x) = sin(1/x) for x <> 0, g(0) = 1. Then both f and g> are Riemann integrable and have the intermediate value property.> But they have the same inde?ite integral, and a function has> at most one derivative.Jose Carlos Santos === While the intuition you have suggested is true, I do not seehow that is related to the maximization problem I am thinking about...The intuition can explain how can one decompose H as V D V^T with V orthonormal, but I cannot see how it can be related to the maximizationproblem I am thinking about.Mind elaborating? :-)> It is well-known that, for a symmetric matrix H of size n by n,> the following optimization problem> Maximize 1/2 x' H x + 1/2 y' H y> subject to ||x|| = ||y|| = 1, x' y = 0> can be solved by the eigenvectors corresponding to the two largest> eigenvalues of H. Here, both x and y are n by 1 vectors.> I am just wondering if there is any geometrical proof of this?> Or, does this property contain any geometrical intuition?> Well, for any two eigenvectors x,y with distinct eigenvalues, we have x' y => 0. So, if all the eigenvalues are distinct, we can separate x and y> (improvised jargon) and the geometric interpretation is pretty clear. If the> largest eigenvalue is duplicated, then the two summands in> 1/2 x' H x + 1/2 y' H y> are the same, so the geometric interpretation is similar to that in the> ?st case.> I'm not a good arm-waver :)> LH === |I'm guessing a complete answer is too much to put in|one post, so what I'd really like to see are|subject, or any good books which discuss some of the|fruitful implications the hypothesis has.It tells us facts about the distribution of primes. One usualway to prove the number of primes Can you give a page reference?>It is page 41 in the last version of those course notes (page 35 for those> using the old version). If you can help me I would be very grateful.Gamma(D(h_i),O_V) is the ring of regular functions on D(h_i).This is a subring of the ?ld of rational functions on V(Milne denotes this ?ld as k(V)). Milne is asserting that,as subrings of k(V), the intersection Gamma(D(h_i),O_V)is Gamma(U,O_V), the ring of regular functions on U.Naturally each regular function on U is regular on D(h_i)so that Gamma(U,O_V) is a subring of, and so a subring of their intersection. On the other hand if f in k(V) lies inall Gamma(D(h_i),O_V), then it is a regular function on U dueto the sheaf property of O_V. (A function regular on an opencover of U is regular on U).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > at least a county, rather than an individual dwelling. According to my dictionary, area can certainly mean that, but it has > numerous other meanings as well, including part of a building. If he does > not wish to be misinterpreted, perhaps he should make his meaning clearer.If you're going to willfully misinterpret commonplace uses of area,he can hardly hope to include your usage of words generally, includingbillion, can he?-- Jesse HughesLOL. How arrogant you are. Now when you realize that I DID proveGoldbach's conjecture and that I proved Fermat's Last Theorem as well,how are you going to feel then? -- James Harris === > According to my dictionary, area can certainly mean that, but it has > numerous other meanings as well, including part of a building. If he does > not wish to be misinterpreted, perhaps he should make his meaning clearer.He means: if you don't know the usage of an area, the best guess isto assume that.... And he's right...You will be right more often (by vastly more) if you make hisassumption than the other assumption. === > You will be right more often (by vastly more) if you make his> assumption than the other assumption.[Slartibartfast] gave a hollow laugh. ?What does it matter? Science has achieved some wonderful things of course, but I'd far rather be happy than right any day.'I am content with my great big billions, and I will continue to use them when appropriate (which is hardly ever, of course!). If others are content with milliscopic billions, then that's entirely up to them. :-)-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > You will be right more often (by vastly more) if you make his> assumption than the other assumption.[Slartibartfast] gave a hollow laugh. ?What does it matter? Science has >achieved some wonderful things of course, but I'd far rather be happy than >right any day.'Ha! You didn't transcribe the (important) next two lines! :-o--dgates@spamfreelinkline.com === > You will be right more often (by vastly more) if you make his> assumption than the other assumption.>[Slartibartfast] gave a hollow laugh. ?What does it matter? Science has>achieved some wonderful things of course, but I'd far rather be happy than>right any day.'Ha! You didn't transcribe the (important) next two lines! :-oThat's right. I didn't. :-)-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > Pretty much everything in Europe and Asia is being replaced by US things> nowadays. Reverse occurrences are rare to the point of non-existence.Ah well, the US is a replacement of American things (remember theamerinds?) by European things.The UK thing is a replacement of the English thing with a Frenchthing. The English were a replacement of a British thing with anGerman thing.The French are the replacement of a Gallic thing with a German thing.And so it goes.Thomas === > Douglas Hofstadter once proposed a game called Hruska*, in which the> object was to obtain the middle score. There was to be a series of> games, in which the winner was not the one achieving the middle score> most often, or least often, but (of course) the middlemost number of> times.IIRC, he, or someone, analysed it in game theory terms, and it turned out>that the optimum strategy even when you allow in?itely high numbers was>only to play numbers from 1 to 5, in a certain proportion.>That wasn't Hruska, but a two player game called IIRC underwhelm, wheretwo players choose a number, and the one with the smallest number gets thatnumber added to her score, unless the larger number is exactly one biggerthan the smaller, in wich case the one with the larger number gets the sumof the two numbers.A three person game can't have such a simple strategy where you just playa sequence of random numbers. The other player will just settle on a pairof consecutive numbers wich come up only rarely in your strategy and thenplay (n,n+1) and then (n+1, 1) both winning nearly half of the games.Every non-losing strategy depends on what the others do.-- Wim Benthem === > times.*named after former Senator Roman Hruska, who once gave a speech> extolling the virtues of mediocre people.> I think Nixon once nominated some judge for the supreme court and the> general (legal) opinion was that he was a mediocre judge. Nixon (or> was it Hruska?--this could have been around the same time) responded> something to the effect that mediocre people need representation too.> I think the judge's name was Carswell.Yes, Harold Carswell, who was Nixon's second choice for the court afterhis ?st nominee, Clement Haynsworth, was rejected. Even if he is mediocre, there are a lot of mediocre judges and lawyers. They are entitled to a little representation, aren't they, and a little chance? We can't have all Brandeises, Cardozos, and Frankfurters, and stuff like that there. - Senator Roman Hruska-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === I have looking at a few web pages dealing with the largest knowncalculated primes and a great deal of computational time is takinginto searching for these numbers and verifying they are primes. I haveseen the Euclid's proof of in?itude primes and it occurs that methat super-large prime numbers can be calculated using the following:p1=2 < p2=3A large prime number, p, can be generated usingp = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbersexceeding currently known largest primes can be generated ratherquickly. For example, we calculate the ?st million primes, multiplythem and just add one.If this is true, then calculating super-large primes should be acomputational trivial task, shouldn't it? Then why the big fuss overcalculating large primes?I suspect that I am missing fundamental here. Can super large primesreally be calculated using the trivial method outlined above? HopeBlueBear === > I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly.Excellent! Why not look at q = (p_1 p_2 p_3 .... p_n) - 1also. That must be prime, by the same argument, and you have a proofthat there are in?itely many twin primes :-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === > I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31Look at the page on Primorial Primes at http://primepages.org/> p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one.In the same way that 2*3*5*7*11*13+1 = 30031 = 59*509 is prime?> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? HopeIt's all at:http://primepages.org/Phil-- Unpatched IE vulnerability: Notepad popupsDescription: Opening popup windows without scriptingReference: http://computerbytesman.com/security/ notepadpopups.htmFollowup: http://msgs.securepoint.com/cgi-bin/get/bugtraq0308/55.html== ==> I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:Others have already given you an example of why your idea won't ?t I is no largest prime (and therefore an in?ite number of primes) is as follows:If there is a largest prime, P, then the number of primes is ?ite. We can therefore form a new number, N, which is one greater than the product of all primes. Therefore, N > P (and, indeed, it would be very much greater than P).When divided by any prime in our list, it leaves remainder 1. Therefore, /either/ it is prime /or/ it is the product of two or more primes, at least one of which is greater than P.In /either/ case, P has been shown not to be the largest prime number.Don't forget the second case! :-)-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === >I have looking at a few web pages dealing with the largest known>calculated primes and a great deal of computational time is taking>into searching for these numbers and verifying they are primes. I have>seen the Euclid's proof of in?itude primes and it occurs that me>that super-large prime numbers can be calculated using the following:Others have already given you an example of why your idea won't ?t I > is no largest prime (and therefore an in?ite number of primes) is as > follows:If there is a largest prime, P, then the number of primes is ?ite. We can > therefore form a new number, N, which is one greater than the product of > all primes. Therefore, N > P (and, indeed, it would be very much greater > than P).When divided by any prime in our list, it leaves remainder 1. Therefore, > /either/ it is prime /or/ it is the product of two or more primes, at least > one of which is greater than P. In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is complete up to P. You've simply proved that the ?ite list ofknown primes is not the list of all primes.If you use Euclid's proof as constructive, then the sets you generate (assuming you have a factorisation oracle) are: {2}{2,3}{2,3,7}{2,3,7,43}{2,3,7,13,43,139}{ 2,3,7,13,43,139,3263443}and thence you ?d the new primes {547,607,1033,31051}, all of which are less than the currently largest known prime 3263443.> Don't forget the second case! :-)Don't forget that if you're pretending that you don't know what the primes are, you can't suddenly pull otherwise well-known features of the prime numbers out of a hat.Phil-- Unpatched IE vulnerability: ADODB.Stream local ?e writingDescription: Planting arbitrary ?es on the local ?e systemExploit: http://ip3e83566f.speed.planet.nl/eeye.html (but unrelated to the EEye exploit) === > Nope. Nowhere have you assumed that> the list of primes known is > complete up to P.Well, he said, ``form a new number, N, which is one greater than the product of_all_ primes [emphasis added].'' It seems to me that this would necessarily bea complete list.I am not sure what you're getting at. If your point is that RichardHeath?ld's proof is not the same as Euclid's, that's one thing. But hisargument seems sound to me. Filling in a little, but I think really rephrasingRichard's argument: If there's a largest prime P, then there are ?itely many primes. If there are ?itely many primes, we can take the product of all primes, and add 1 (call this N). We know 2 is a prime and 3 is a prime, and there are others, so N > P. No prime less than or equal to P (we took the product of all primes, where P was assumed to be the largest prime) divides N (they all leave a remainder of 1). So, any prime that divides N must be greater than P. There must be a prime that divides N. This contradicts P being the largest prime.Euclid's proof, if I recall correctly, is just given a ?ite list of primes,we can show there must be another prime. So that is different from Richard'sproof. But that doesn't mean there's anything wrong with Richard's (apart fromthe claim that it's Euclid's).Your {2, 3, 7} is not ``all primes,'' as Richard speci?d. If we think 7 isthe largest prime, then the list is {2, 3, 5, 7} and any prime that divides2.3.5.7+1 is bigger than 7. He's not required to produce the list. We knowanything less than P is a prime or not a prime; to get a list of all primes, wetake everything less than P that is a prime.John Robertson === >Nope. Nowhere have you assumed that>the list of primes known is >complete up to P.Well, he said, ``form a new number, N, which is one greater than the product of> _all_ primes [emphasis added].'' It seems to me that this would necessarily be> a complete list.Let all known primes be {2,3,7,13,43,139,3263443}, the largest prime being 3263443. > I am not sure what you're getting at. If your point is that Richard> Heath?ld's proof is not the same as Euclid's, that's one thing. But his> argument seems sound to me. Filling in a little, but I think really rephrasing> Richard's argument: If there's a largest prime P, then there are ?itely> many primes. If there are ?itely many primes, we can take the product> of all primes, and add 1 (call this N). We know 2 is a prime and 3 is a prime, and there are> others, so N > P.Yup to here. > No prime less than or equal to P (we took the product> of all primes, where P was assumed to be the largest> prime) divides N (they all leave a remainder of 1).i.e. you want to add the assumption that you know all primes up to P. > So, any prime that divides N must be greater than P. There must be a prime that divides N. This contradicts P being the largest prime.Or it contradicts the assumption that you know all the primes up to P.If you use proof by contradiction with two assumptions, then all you know is that the conjunction of your assumptions is false.> Euclid's proof, if I recall correctly, is just given a ?ite list of primes,> we can show there must be another prime. So that is different from Richard's> proof. Call me a pedant, if you like, as that's what I am. I see it, I call it.> But that doesn't mean there's anything wrong with Richard's (apart from> the claim that it's Euclid's).Your {2, 3, 7} is not ``all primes,'' as Richard speci?d. This is why I asked Richard to de?e all primes. He gave no de?ition.Euclid's gave one, which permits {2, 3, 7}. Which is why I don't view myinsistance that we all know what all primes means to be blathering.Richard's insistance that all primes up to P be known and there be no primesgreater than P is cannot work directly, as is, to prove that there's a prime greater than P, as the contradiction doesn't tell you whether there be no primes greater than P is false or all primes up to P be known is known.Adding extra assumptions is almost always a bad thing to do when performinga proof by contradiction.Phil-- Unpatched IE vulnerability: WMP local ?e bounceDescription: Switching security zone, arbitrary command execution, Reference: http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307 &L=ntbugtraq&F=P&S=&P=6783Exploit: http://www.malware.com/once.again!.html === [...]|> The proof could be expressed|> in terms of a ?ite number of known primes, as Phil seems to have assumed|> it was, but that's not the way Richard expressed it -- he spoke explicitly|> of the product of *all* primes.||If you view it in terms of sets, subsets of Z, my view is perfectly standard |one, it's what Ribenboim calls Euclid's proof, and as such is applicable to |the premise that Richard was assuming.The proof you like is standard, but I don't believe your stance toward hisway of formulating it is standard.|Richard was the one who was|introducing|more assumptions. All I need is a ?ite set of primes. A maximum prime P,|and |the fact that we're in Z, gets me that. Euclid's proof just follows |immediately. Let all primes be all primes in that set, and the connection|is made.Not being as elegant as possible doesn't invalidate a proof. If he wants toprove that there are in?itely many primes, he is free to assume that thereare only ?itely many of them.Starting with that assumption is unnecessary, but it's apparently a commonway to present the proof. For instance, Robert Wisner's _A Panorama ofNumbers_: But Euclid proved that such a thing [all numbers past a certain point being sieved out] does not happen. Here's how he went about it. He said (to himself, and later to the rest of mankind) that if the list of primes has an end, then we could stare at the complete list as 2 3 5 7 11 13 ... P where P is the largest prime, whatever its name. [...] This is essentially the proof of Euclid....The word essentially is his escape. :-) I also have a Ivan Niven bookwhich starts by assuming there is a largest prime, but he doesn't attributethe proof to Euclid.|This is why I was blathering about what all meant. The meaning is plain; all primes means all of them: {n : n is prime}.|> But Richard's conclusion that P has been shown not to be the largest|> prime number is also wrong; what has been shown is that the assumption|> of a ?ite number of primes is wrong.||Yup, which is why I said Nope..But his ?st step was to say that if P is the largest prime, then thereare ?itely many primes. That's a correct deduction. If the conclusion ofthat step then leads to a contradiction, it is valid to conclude that thepremise of it was false too.|> The only reason to assume that|> at least one of the additional primes discovered must be larger than|> P is that we *thought* P was the largest prime, before we discovered them.||Euclid permits me to assume {2,3,7,13,43,139,3263443} is the ?ite set |of primes, with maximum prime P=3263443. In what way does |{2,3,7,13,43,139,3263443}, P=3263443 violate the premise Let P be the |largest prime?3263443 is not the largest prime. Obviously, there's no satisfying theassumption that P is the largest prime. Nor is it possible to satisfythe requirement that these are all the primes while leaving gaps!|In which case, surely Richard, as he originally stated his|argument should permit it too. However, _none_ of the primes that Euclid's |construction discovers is larger than P. If the premise is satis?d, but |the conclusion isn't, then there's a syllogistic error somewhere.But the premise is NOT satis?d. It is typically trickier to evaluate thesoundness of a proof by contradiction, since the assumption *isn't* evercorrect.Compare the proof with what is in Hardy and Wright, which they also callEuclid's proof: Let 2,3,5,...,p be the aggregate of primes up to p, and let q = 2.3.5. ... .p + 1. Then q is not divisible by any of the numbers 2,3,5,...,p. It is therefore either prime, or divisible by a prime between p and q. In either case, there is a prime greater than p, which proves the theorem.It is valid to reason in this way if the set of primes in question is allof the primes up to a given prime p. This is mainly what Richard does.It appears to me, then, that a lot of your objection hinges on the readernot being ready to regard as equivalent the fact that a set of naturalnumbers is ?ite, that it has a largest element P, and that it consists of(all) those elements which are <= P. I think anybody who is unable torecognize these as equivalent is not really ready to be reading Euclid'sproof. And these equivalences continue to hold even when we're consideringa counterfactual condition such as P being the largest prime.his version of it, but since his P is introduced as a hypothetical largestprime, references to it only make sense so long as we are still under theassumption that there is such a P. So in particular, the statement thatthere is a prime greater than P is still among the consequences of P beingthe largest prime. That's valid. True, the same reasoning can be used toshow that there is a prime > P without assuming that P is the largest primenumber.too. It refers to P again, but simply denies the original assumption.|If Richard has simply added let all primes <=P be known to his premise|I wouldn't have jumped on it that way.I don't think this would help. Talking about what primes are known issubjective.Without the subjective reference, what's the alleged extra assumption?That all the primes <= P are actually among the set of all primes? Orvice-versa? It's true that he refers to a list, but I don't see anybetter antecedent for that than his reference to the primes.|However, if he had, then it would |_not_ have been Euclid's proof (and he claimed what followed was Euclid's |proof so I would have jumped on that instead),I think you would have been better off saying that instead. Of course, anerror of a similar nature is in Hardy and Wright. They don't start with acounterfactual assumption, but they do ask us to consider all the primesup to p, and as I understand it Euclid doesn't.I would guess that in fact the typical mathematician would consider all ofthese distinctions to be minor stylistic variations.|and wouldn't (on its own) |have been a proof of the in?itude of primes (as you've added another |assumption, so the proof by contradiction only disproves the _conjunction_|of the assumptions, not either one individually).Are you saying the hypothetical reader is supposed not to recognize thatthe one implies the other?|Ockham has some wise words for moments like this.I'm sure Ockham would have appreciated an elegant argument, but I doubt hewould say that an argument becomes invalid if it contains redundancies.[...]|This is why I asked Richard to de?e all primes. He gave no de?ition.|Euclid's gave one, which permits {2, 3, 7}. Which is why I don't view my|insistance that we all know what all primes means to be blathering.Euclid did not de?e all primes, let alone de?e it to permit gapswhere there happen to be additional primes.|Richard's insistance that all primes up to P be known and there be no primes|greater than P is cannot work directly, as is, to prove that there's a prime |greater than P, as the contradiction doesn't tell you whether there be no |primes greater than P is false or all primes up to P be known is known.||Adding extra assumptions is almost always a bad thing to do when performing|a proof by contradiction.I admit that for someone lacking the mathematical competence to recognizethat a set of natural numbers being ?ite, its having a largest element,and its consisting of precisely those elements <= its maximum element areequivalent, this presentation would not be enough of a proof.I'm just having a hard time seeing any of these critiques as a meaningfulobjections to the proof. We should not be trying to create the impressionthat in mathematics, it's normal to engage in hair-splitting like that.You already have Richard chalking up a new, phony reason to think he's notquite cut out to be a mathematician (not that that necessarily makes anydifference). That strikes me as silly.Keith Ramsay === > Starting with that assumption is unnecessary, but it's apparently a common> way to present the proof. For instance, Robert Wisner's _A Panorama of> Numbers_: But Euclid proved that such a thing [all numbers past a certain point> being sieved out] does not happen. Here's how he went about it. He said> (to himself, and later to the rest of mankind) that if the list of> primes has an end, then we could stare at the complete list as 2> 3> 5> 7> 11> 13> ...> P where P is the largest prime, whatever its name.> [...]> This is essentially the proof of Euclid....The word essentially is his escape. :-) I also have a Ivan Niven book> which starts by assuming there is a largest prime, but he doesn't attribute> the proof to Euclid.Forget the largest prime thing. That's what's confusing everyone.Magnitude of the primes is _utterly_ irrelevant to Euclid's proof.In black and white - and you may quote me on this -EUCLID DID NOT MAKE REFERENCE TO THERE BEING A LARGEST PRIME IN HIS PROOFOF THE INFINITUDE OF PRIMES. (Of course, Euclid had no concept of in?ite, as such, so he didn't word it that way.)The only time he refers to magnitude is that of the set size, not the numeric size of the elements. i.e the set with 4 elements is larger in magnitude than the set with 3 elements.> |This is why I was blathering about what all meant. The meaning is plain; all primes means all of them: {n : n is prime}.I think the meaning is plain. However, I, like Euclid, think that {2,3,7}is a perfectly valid example for what could be posited as a ?ite set of all primes. i.e. in taht case {2,3,7} _is_ all primes. (And that leads to a contradiction, and therefore it isn't.) There seems to be as muc a fundamental misunderstanding of how proof by contradiction works as well as a misunderstanding of Euclid's proof.If you have a problem with {2,3,7} being posited as all primes, then youhave a problem with the mechanics of proof by contradiction.> |> But Richard's conclusion that P has been shown not to be the largest> |> prime number is also wrong; what has been shown is that the assumption> |> of a ?ite number of primes is wrong.> |> |Yup, which is why I said Nope..But his ?st step was to say that if P is the largest prime, then there> are ?itely many primes. That's a correct deduction. If the conclusion of> that step then leads to a contradiction, it is valid to conclude that the> premise of it was false too.Or one of the other premises. Richard introduced other assumptions, not realising that he'd done so.> |> The only reason to assume that> |> at least one of the additional primes discovered must be larger than> |> P is that we *thought* P was the largest prime, before we discovered them.> |> |Euclid permits me to assume {2,3,7,13,43,139,3263443} is the ?ite set > |of primes, with maximum prime P=3263443. In what way does > |{2,3,7,13,43,139,3263443}, P=3263443 violate the premise Let P be the > |largest prime?3263443 is not the largest prime. I'm sorry, I thought we were playing a game of proof by contradiction.As far as I cen tell immediate gainsaying is not a valid move in that game.If the set of all primes is {2,3,7,13,43,139,3263443} then 3263443 _is_ the largest prime. That's a simple unassailable mathematical fact.Euclid's proof does not say: Let PP be the set of all primes, but PP musn't be {2,3,7,13,43,139,3263443}.does it? It says (when reworded in more modern language): Let PP be the set of all primes.Why do you have a problem with {2,3,7,13,43,139,3263443}?Euclid didn't. Kummer didn't. I don't.> Obviously, there's no satisfying the> assumption that P is the largest prime. Nor is it possible to satisfy> the requirement that these are all the primes while leaving gaps!|In which case, surely Richard, as he originally stated his> |argument should permit it too. However, _none_ of the primes that Euclid's > |construction discovers is larger than P. If the premise is satis?d, but > |the conclusion isn't, then there's a syllogistic error somewhere.But the premise is NOT satis?d. It is typically trickier to evaluate the> soundness of a proof by contradiction, since the assumption *isn't* ever> correct.Compare the proof with what is in Hardy and Wright, which they also call> Euclid's proof: Let 2,3,5,...,p be the aggregate of primes up to p, and let q = 2.3.5. ... .p + 1. Then q is not divisible by any of the numbers 2,3,5,...,p. It is> therefore either prime, or divisible by a prime between p and q.> In either case, there is a prime greater than p, which proves> the theorem.That's quite a way from Euclid's formulation.It presupposed that you can generate all primes up to P.Euclid's proof doesn't.I don't care that one can generate all primes up to P trivially, and can prove it can be done pretty trivially, it's just _unnecessary_ as part of a proof of the in?iteness of the set of primes.Euclid's proof was that the set of primes is larger than any ?ite set. At no point did it make reference to the magnitude of any of theelements in the set. (except that primes aren't units, of course.)> It is valid to reason in this way if the set of primes in question is all> of the primes up to a given prime p. This is mainly what Richard does.It appears to me, then, that a lot of your objection hinges on the reader> not being ready to regard as equivalent the fact that a set of natural> numbers is ?ite, that it has a largest element P, and that it consists of> (all) those elements which are <= P. I think anybody who is unable to> recognize these as equivalent is not really ready to be reading Euclid's> proof. And these equivalences continue to hold even when we're considering> a counterfactual condition such as P being the largest prime.No. My objection is to people presupposing they know what all means, when they've not considered what assumptions they've made in order to come up with that meaning.> his version of it, but since his P is introduced as a hypothetical largest> prime, references to it only make sense so long as we are still under the> assumption that there is such a P. So in particular, the statement that> there is a prime greater than P is still among the consequences of P being> the largest prime. That's valid. True, the same reasoning can be used to> show that there is a prime > P without assuming that P is the largest prime> number. too. It refers to P again, but simply denies the original assumption.But he had extra unstated assumptions. That's what I've been jumping on.Repeatedly. Proof by contradiction denies one of the assumptions, but doesn't tell you which one is denied. As I have said repeatedly.> |If Richard has simply added let all primes <=P be known to his premise> |I wouldn't have jumped on it that way.I don't think this would help. Talking about what primes are known is> subjective.Not really, would assigned make you happy? The set of all primes is the set of all known primes in this proof. I've said that repeatedly. The set-theoretic notation for what my sentences expressed would beno different if I included or excluded the word known. I was simply trying to avoid the naked word all as people immediately misinterpret that based on their knowledge about the primes. > Without the subjective reference, what's the alleged extra assumption?> That all the primes <= P are actually among the set of all primes? This is why I jump on people's wording - the above is vacuously true as worded (and uses the naked term all twice, which immedately biases the inexpert reader as to what it might refer to). But yes, that is the assumption. It is possible to deny that clause and still prove the in?iteness of the primes using Euclid's proof.> Or> vice-versa? The proof relied on both directions, but you only assume one, you derive the other quite easily.> It's true that he refers to a list, but I don't see any> better antecedent for that than his reference to the primes.|However, if he had, then it would > |_not_ have been Euclid's proof (and he claimed what followed was Euclid's > |proof so I would have jumped on that instead),I think you would have been better off saying that instead. if he had... . He hadn't, so I didn't.> Of course, an> error of a similar nature is in Hardy and Wright. They don't start with a> counterfactual assumption, but they do ask us to consider all the primes> up to p, and as I understand it Euclid doesn't.Yup, Euclid asks us to consider an arbitrary ?ite list.Euclid, like Kummer (whose proof hasn't been distorted over time)_doesn't_ even require _2_ to be in the list of primes.This comes a shock to many people, but it's God's honest truth. Euclid's proof doesn't presume _any_ particular number is prime, not even 2.When people see Let p1 I would guess that in fact the typical mathematician would consider all of> these distinctions to be minor stylistic variations.|and wouldn't (on its own) > |have been a proof of the in?itude of primes (as you've added another > |assumption, so the proof by contradiction only disproves the _conjunction_> |of the assumptions, not either one individually).Are you saying the hypothetical reader is supposed not to recognize that> the one implies the other?|Ockham has some wise words for moments like this.I'm sure Ockham would have appreciated an elegant argument, but I doubt he> would say that an argument becomes invalid if it contains redundancies.But, as I said before, if they do introduce new assumptions, which this did, then it throws a spanner into the works when it comes to proof by contradiction.With the assumption, you prove A |= B, without the assumption you prove |= B.In order to get |= B from the ?st you need |= A. I.e. it's _not_ a proof of B until you add a proof of A.If you bring Euclid up to date with terminology, then you can't get much more elegant than Euclid's proof, IMHO. Kummer's is the closest to that, but is worded in terms of his ideal numbers (and therefore applies to more general rings, not just the integers).> [...]> |This is why I asked Richard to de?e all primes. He gave no de?ition.> |Euclid's gave one, which permits {2, 3, 7}. Which is why I don't view my> |insistance that we all know what all primes means to be blathering.Euclid did not de?e all primes, let alone de?e it to permit gaps> where there happen to be additional primes.I have no idea what he said in Greek, but the usual translation is assigned primes. He made no reference to gaps, at all.Euclid's construction permits _any_ list of primes. It makes _no_ statement about the magnitude either of any of the primes in the ?ite set or of thenewly proved to exist prime(s). e.g. {3} yeilds {2}. {5} yeilds {2,3}. {3,5} yeilds {2}. (Yup, Euclid's proof doesn't even require a plurality of primes; anyone who told you it did was deceiving you.)> |Richard's insistance that all primes up to P be known and there be no primes> |greater than P is cannot work directly, as is, to prove that there's a prime > |greater than P, as the contradiction doesn't tell you whether there be no > |primes greater than P is false or all primes up to P be known is known.> |> |Adding extra assumptions is almost always a bad thing to do when performing> |a proof by contradiction.I admit that for someone lacking the mathematical competence to recognize> that a set of natural numbers being ?ite, its having a largest element,> and its consisting of precisely those elements <= its maximum element are> equivalent, this presentation would not be enough of a proof.I'm glad I don't have the mathematical competence to think that a set of natural numbers a) being ?ite b) having a largest element c) consisting of precisely those elements <= its maximum elementare equivalent.{2,3,7} satis?s (a), and (b), but does not satisfy (c).So - are you mathematically competent enough to think (a), (b), and (c)are equivalent? > I'm just having a hard time seeing any of these critiques as a meaningful> objections to the proof.What is the proof? The original leaves us with A |= B. That's _not_ a proof of B. Sure, we know that A is provable, but as it stands without aproof of A, we don't have a proof of B. (e.g. all things that are dependent on RH aren't proven yet.) > We should not be trying to create the impression> that in mathematics, it's normal to engage in hair-splitting like that.> You already have Richard chalking up a new, phony reason to think he's not> quite cut out to be a mathematician (not that that necessarily makes any> difference). That strikes me as silly.It's unnecessary to introduce things into a proof that are not necesary for the proof. That strikes me as silly. I saw it, I said it. If it takes hair-splitting to separate an unconditional proof from a conditional proof reliant on an unproved assumption, then split hairs I will.Phil-- Unpatched IE vulnerability: Click hijackingDescription: Pointing IE mouse events at non-IE/system windows === > Let all known primes be > {2,3,7,13,43,139,3263443},> the largest prime being> 3263443.If you change Richard's proof, you can make it wrong. Richard did not say,``take all _known_ primes'', or ``take all primes found by process X.'' Hesaid, ``take _all_ primes.''Your list above is missing a few primes less than 3263443. If you multiply_all_ primes up to 3263443 and add 1, the result will not be divisible by anyprime less than or equal to 3263443.that there are a ?ite number of primes. Not a ?ite number of _known_in principle, take the product of _all_ primes and add 1, and label that N. Then there has to be a prime dividing N. That prime is necessarily larger thanP.I agree that if you take something less than _all_ primes and apply the aboveprocess, that N could be divisible by something less than P. So what?Also, we all understand that this is not Euclid's proof. Again, so what?> No prime less than or equal to P> divides N.> i.e. you want to add the assumption> that you know all primes up to P.I don't see this as adding an assumption. If there's a largest prime, thereare ?itley many primes, and in principle they can be listed. There is noaddtional assumption after that of a largest prime.I did not use the word ``blathering.''John Robertson === >Let all known primes be > {2,3,7,13,43,139,3263443},>the largest prime being>3263443.If you change Richard's proof, you can make it wrong. Richard did not say,> ``take all _known_ primes'', or ``take all primes found by process X.'' He> said, ``take _all_ primes.''Euler permits all primes to be {2,3,7,13,43,139,3263443}.Richard's wording does not explicitly forbid that.My de?ition of all and of is known is such that if there's a ?ite number of primes, the premise to be disproved, then all primes = all known primes, i.e. knowledge simply consists of listing them in the set of primes.I deliberately try to avoid saying all primes, as when I say things likelet {2,3,7} be all the primes, some doofus pipes up but you're missing 5.> Your list above is missing a few primes less than 3263443. If you multiply> _all_ primes up to 3263443 and add 1, the result will not be divisible by any> prime less than or equal to 3263443.that there are a ?ite number of primes. Not a ?ite number of _known_> in principle, take the product of _all_ primes and add 1, and label that N. > Then there has to be a prime dividing N. Yes.> That prime is necessarily larger than> P.No.> I agree that if you take something less than _all_ primes and apply the above> process, that N could be divisible by something less than P. So what?Also, we all understand that this is not Euclid's proof. Again, so what?At the moment it doesn't prove what it was claimed to prove, and thereforeisn't a proof at all. It's a proof of C->F > No prime less than or equal to P> divides N.>i.e. you want to add the assumption>that you know all primes up to P.I don't see this as adding an assumption.My form is take the set of all primes, your/Richard's version is take all primes and exclude the possibility that there can be any prime less than the largest element. If you _don't_ see that as adding an assumption, then there's nothing more I can say. I've tried to explain it _repeatedly_. It's an extra assumption. > If there's a largest prime, there> are ?itley many primes, and in principle they can be listed. There is no> addtional assumption after that of a largest prime.Nope. > I did not use the word ``blathering.''Nor do I say you do. You're not the only responedent in this thread.Phil-- Unpatched IE vulnerability: window.open search injectionDescription: cross-domain scripting, cookie/data/identity theft, command executionExploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm === > If there is a largest prime, P, then the number of primes is ?ite. We can > therefore form a new number, N, which is one greater than the product of > all primes. Therefore, N > P (and, indeed, it would be very much greater > than P).When divided by any prime in our list, it leaves remainder 1. Therefore, > /either/ it is prime /or/ it is the product of two or more primes, at least > one of which is greater than P.> In /either/ case, P has been shown not to be the largest prime number. > Nope. Nowhere have you assumed that the list of primes known is > complete up to P. You've simply proved that the ?ite list of> known primes is not the list of all primes.Well, Richard and Phil are both wrong here. The proof could be expressedin terms of a ?ite number of known primes, as Phil seems to have assumedit was, but that's not the way Richard expressed it -- he spoke explicitlyof the product of *all* primes.But Richard's conclusion that P has been shown not to be the largestprime number is also wrong; what has been shown is that the assumptionof a ?ite number of primes is wrong. The only reason to assume thatat least one of the additional primes discovered must be larger thanP is that we *thought* P was the largest prime, before we discovered them.-- Mark Brader | Sir, your composure baf?. A single counterexampleToronto | refutes a conjecture as effectively as ten... Hands up!msb@vex.net | You have to surrender. -- Imre Lakatos === If there is a largest prime, P, then the number of primes is ?ite. We can > therefore form a new number, N, which is one greater than the product of > all primes. Therefore, N > P (and, indeed, it would be very much greater > than P).> When divided by any prime in our list, it leaves remainder 1. Therefore, > /either/ it is prime /or/ it is the product of two or more primes, at least > one of which is greater than P.> In /either/ case, P has been shown not to be the largest prime number.>Nope. Nowhere have you assumed that the list of primes known is >complete up to P. You've simply proved that the ?ite list of>known primes is not the list of all primes.Well, Richard and Phil are both wrong here. Richard: Phil: Nope.Both wrong? !a AND !!a?> The proof could be expressed> in terms of a ?ite number of known primes, as Phil seems to have assumed> it was, but that's not the way Richard expressed it -- he spoke explicitly> of the product of *all* primes.If you view it in terms of sets, subsets of Z, my view is perfectly standard one, it's what Ribenboim calls Euclid's proof, and as such is applicable to the premise that Richard was assuming. Richard was the one who was introducingmore assumptions. All I need is a ?ite set of primes. A maximum prime P, and the fact that we're in Z, gets me that. Euclid's proof just follows immediately. Let all primes be all primes in that set, and the connectionis made. This is why I was blathering about what all meant. > But Richard's conclusion that P has been shown not to be the largest> prime number is also wrong; what has been shown is that the assumption> of a ?ite number of primes is wrong.Yup, which is why I said Nope..> The only reason to assume that> at least one of the additional primes discovered must be larger than> P is that we *thought* P was the largest prime, before we discovered them.Euclid permits me to assume {2,3,7,13,43,139,3263443} is the ?ite set of primes, with maximum prime P=3263443. In what way does {2,3,7,13,43,139,3263443}, P=3263443 violate the premise Let P be the largest prime? In which case, surely Richard, as he originally stated hisargument should permit it too. However, _none_ of the primes that Euclid's construction discovers is larger than P. If the premise is satis?d, but the conclusion isn't, then there's a syllogistic error somewhere.If Richard has simply added let all primes <=P be known to his premiseI wouldn't have jumped on it that way. However, if he had, then it would _not_ have been Euclid's proof (and he claimed what followed was Euclid's proof so I would have jumped on that instead), and wouldn't (on its own) have been a proof of the in?itude of primes (as you've added another assumption, so the proof by contradiction only disproves the _conjunction_of the assumptions, not either one individually).Ockham has some wise words for moments like this.Phil-- Unpatched IE vulnerability: Web Archive buffer over?cription: Possible automated code execution.Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0303/107.html= === > If Richard has simply added let all primes <=P be known to his premise> I wouldn't have jumped on it that way.Let all primes <=P be known. :-)> Ockham has some wise words for moments like this.I was about to say don't needlessly multiply spellings, but my dictionary agrees with you that Ockham is an acceptable variant of Occam. I ?d this deliciously ironic.-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === >If Richard has simply added let all primes <=P be known to his premise>I wouldn't have jumped on it that way.Let all primes <=P be known. :-)If you assume that then proof by contradiction leads to the possibilty that that assumption is false.I.e. the ?al conclusion is that either there's a new prime >P or there's actually a new prime

Ockham has some wise words for moments like this.I was about to say don't needlessly multiply spellings, but my dictionary > agrees with you that Ockham is an acceptable variant of Occam. I ?d > this deliciously ironic.It's a town name after all, and the town's name was and is Ockham. http://uk2.multimap.com/map/browse.cgi?X=505000&Y=155000& width=500&height=300&client=public&gride=&gridn=&srec=0& coordsys=gb&addr1=&addr2=&addr3=&pc=&scale=100000&advanced=& multimap.x=338&multimap.y=92Some francophiles prefered a more italic rendering, but others, such as the IEP (which I think contains the single most detailed biography of him that I've seen anywhere), list only Ockham.Phil-- Unpatched IE vulnerability: Basic Authentication URL spoo?gDescription: Spoo?g the URL displayed in the Address barReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html== ==> Well, Richard and Phil are both wrong here. The proof could be expressed> in terms of a ?ite number of known primes, as Phil seems to have assumed> it was, but that's not the way Richard expressed it -- he spoke explicitly> of the product of *all* primes.Yes.> But Richard's conclusion that P has been shown not to be the largest> prime number is also wrong; what has been shown is that the assumption> of a ?ite number of primes is wrong. The only reason to assume that> at least one of the additional primes discovered must be larger than> P is that we *thought* P was the largest prime, before we discovered them.Hmmm. I seem to have taken a step for granted. Okay, let me make it more explicit. We have shown that the number we thought was the largest prime cannot be the largest prime, so there must be a bigger one. So now we ?d out what that new largest prime is, and then iterate. We can do this in?itely often (if we have the time to spare).Is that a bit closer to the mark?-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > Hmmm. I seem to have taken a step for granted. Okay, let me make it more > explicit. We have shown that the number we thought was the largest prime > cannot be the largest prime,Yes, but not directly. You have shown that the assumption that there werebe the largest prime.> so there must be a bigger one.No. If you had proved *only* that the number P is not in fact the largestprime, this would *not* prove that a larger prime exists. It would alsobe possible that P is not prime after all, or that *no* largest primeexists.Now in fact we know that primes exist, and we have proved that there aremany, and since we're dealing with integers, from *this* it follows thatthere exists a prime larger than any given number P.In other words, you've taken a valid proof and made it invalid by falselyinserting, as an intermediate step, a fact that you actually only knowto be true because of the ?al conclusion.-- Mark Brader, Toronto Logic is logic. That's all I say.msb@vex.net -- Oliver Wendell Holmes === > Hmmm. I seem to have taken a step for granted. Okay, let me make it more> explicit. We have shown that the number we thought was the largest prime> cannot be the largest prime,Yes, but not directly. You have shown that the assumption that there were> be the largest prime.> so there must be a bigger one.No. If you had proved *only* that the number P is not in fact the largest> prime, this would *not* prove that a larger prime exists. It would also> be possible that P is not prime after all, or that *no* largest prime> exists.But I de?ed P to be the largest prime. Therefore, it /is/ prime, and therefore /either/ no largest prime exists /or/ there is a prime larger than P, which is what I was obviously failing to say.> Now in fact we know that primes exist, and we have proved that there are> many, and since we're dealing with integers, from *this* it follows that> there exists a prime larger than any given number P.In other words, you've taken a valid proof and made it invalid by falsely> inserting, as an intermediate step, a fact that you actually only know> to be true because of the ?al conclusion. A mathematician cut out to be I was not.-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > Hmmm. I seem to have taken a step for granted. Okay, let me make it more> explicit. We have shown that the number we thought was the largest prime> cannot be the largest prime,> Yes, but not directly. You have shown that the assumption that there were> be the largest prime.> so there must be a bigger one.> No. If you had proved *only* that the number P is not in fact the largest> prime, this would *not* prove that a larger prime exists. It would also> be possible that P is not prime after all, or that *no* largest prime> exists.> But I de?ed P to be the largest prime. Therefore, it /is/ prime, and > therefore /either/ no largest prime exists /or/ there is a prime larger > than P, which is what I was obviously failing to say.Summary of Euclid's proof:1) Suppose that there is a largest prime; call it P2) Calculate N = (product of all primes from 2 to P, inclusive) + 13) N has one or more prime factors, all of which must be > P4) #3 contradicts #1, so #1 is wrong, so there is no largest primeYou were trying to conclude that, for any prime P, N is a largerprime. Not so. There is *some* prime larger than P, but it's notnecessarily N; it could just be one of N's factors.Consider: P = 13; N = (2*3*5*7*11*13) + 1 = 3003130031 is not a prime > 13, but its factors (59 and 509) are. === prime.No, I wasn't. I was saying that it might be a larger prime, but that it might be a composite of primes at least one of which is larger than P.-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > Summary of Euclid's proof:1) Suppose that there is a largest prime; call it P> 2) Calculate N = (product of all primes from 2 to P, inclusive) + 1Are you trying to say that the set of all primes has no primes missing? That is vacuously true, and holds for {2,3,7}. If {2,3,7} is the setof all primes, which we are permitted to assume according to Euclid, then {2,3,7} is the set of all primes, bar none.Or are you trying to say and we assume that we know all primes <=P.> 3) N has one or more prime factors, all of which must be > P> 4) #3 contradicts #1, so #1 is wrong, so there is no largest primeOr the assumption in #2 was wrong.I never thought that Euclid's proof would be hard for people to grasp. I think I can only recommend that people look at Kummer's proof instead, as that's even simpler.Phil-- Unpatched IE vulnerability: DNSError folder disclosureDescription: Gaining access to local security zonesReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/52.html== ==Correct me if I'm wrong, but isn't the proof much simpler if you saysomething like:1) Assume P is the largest prime.2) Calculate P!+1 (I.E. the product of all numbers from 1 to P, plus one)3) Dividing that number by any number <= P will give a remainder of 14) Therefore, P!+1 is either prime or a multiple of a prime > PSaSW, Willem (at stack dot nl)-- made in the above text. For all I know I might be drugged or something.. No I'm not paranoid. You all think I'm paranoid, don't you !#EOT === > Correct me if I'm wrong, but isn't the proof much simpler if you say> something like:1) Assume P is the largest prime.> 2) Calculate P!+1 (I.E. the product of all numbers from 1 to P, plus one)> 3) Dividing that number by any number <= P will give a remainder of 1> 4) Therefore, P!+1 is either prime or a multiple of a prime > PThat's ?e. (Pedantically you could change the conclusion to either prime or a multiple of primes >P, but what you've said is correct, and suf?ient for the proof.)Phil-- Unpatched IE vulnerability: Basic Authentication URL spoo?gDescription: Spoo?g the URL displayed in the Address barReference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html== ==But I think I got a case. :-) ]>I have looking at a few web pages dealing with the largest known>calculated primes and a great deal of computational time is taking>into searching for these numbers and verifying they are primes. I have>seen the Euclid's proof of in?itude primes and it occurs that me>that super-large prime numbers can be calculated using the following: Others have already given you an example of why your idea won't ?t> there is no largest prime (and therefore an in?ite number of primes) is> as follows:If there is a largest prime, P, then the number of primes is ?ite. We> can therefore form a new number, N, which is one greater than the product> of all primes. Therefore, N > P (and, indeed, it would be very much> greater than P).When divided by any prime in our list, it leaves remainder 1. Therefore,> /either/ it is prime /or/ it is the product of two or more primes, at> least one of which is greater than P.> In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes and then adding 1. To multiply all primes together, I must ?st have a complete list of primes. If the list is not complete, then I cannot perform the multiplication. the primes are, you can't suddenly pull otherwise well-known> features of the prime numbers out of a hat.It's not necessary to know all the primes in order to construct the argument. It is suf?ient to know that it is /in theory/ possible to know them, e.g. by sieving. It is only necessary to know all the primes if you wish to perform the multiplication in real life.-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === [...]|> If there is a largest prime, P, then the number of primes is ?ite. We|> can therefore form a new number, N, which is one greater than the product|> of all primes. Therefore, N > P (and, indeed, it would be very much|> greater than P).|> |> When divided by any prime in our list, it leaves remainder 1. Therefore,|> /either/ it is prime /or/ it is the product of two or more primes, at|> least one of which is greater than P.|>|> In /either/ case, P has been shown not to be the largest prime number.|> |> Nope.Yes it has.|> Nowhere have you assumed that the list of primes known is|> complete up to P.He doesn't need to assume that all the primes up to P are known for theargument to work. He just assumes (for the sake of proof by contradiction)that P is the largest prime.|On the contrary, I formed N by multiplying all primes and then adding 1. To |multiply all primes together, I must ?st have a complete list of primes. |If the list is not complete, then I cannot perform the multiplication.There are two versions of the proof. One proceeds like you just did,starting with a (false) assumption that there is a largest prime. Theother just starts by considering an arbitrary ?ite set of primes. If Iremember correctly, the latter is more like Euclid's own proof. The crux isthat the N you get by adding one to the product is always divisible by aprime not in the set. It's perhaps worth noting that you're not reallyusing the assumption that P is the largest prime, then, until at the endwhen you contradict it.proofs as if it was a proof by contradiction, by starting it with anassumption that the conclusion of the proposition was false, and deducinga contradiction in each case where the result was shown true. This is astylistic weakness to avoid.In this case starting with the assumption that P is the largest prime doesnot seem so bad; at least it seems more popular. It doesn't really play afunctional role, though, and it's perhaps worth doing it the other way tomake it more obvious that we're seeing as well a method for ?ding a primeoutside of any ?ite set of primes.Keith Ramsay === > But I think I got a case. :-) ]I hope you've got a hot-shot lawyer...Your premise is:> If there is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.> then the number of primes is ?ite. We> can therefore form a new number, N, which is one greater than the product> of all primes. Therefore, N > P (and, indeed, it would be very much> greater than P). When divided by any prime in our list, it leaves remainder 1. Therefore,> /either/ it is prime /or/ it is the product of two or more primes, at> least one of which is greater than P.> In /either/ case, P has been shown not to be the largest prime number.Nope. Nowhere have you assumed that the list of primes known is>complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de?ed all in this context. Nowhere do you say anything about having a contiguous set of primes from 2 to P. The only thing you've said is that your set of primes has a maximumelement P. Full stop. There were no further premises -- look above if you don't believe me.Anyway, your maximum prime P is a red herring and completely unnecessary - why did you even introduce it in the ?st place?> and then adding 1. To > multiply all primes together, I must ?st have a complete list of primes.You must have a list of primes. What does ?complete' mean to you in this context? If you wish to multiply together all primes between 2 and P then you must know /a priori/ all the primes between 2 and P. How do you know that you know all those primes? If you wish to use that as an assumption your argument you must state so. You didn't.But why would you want to?Let the complete list of primes be {2}.The multiplication/addition yields 3. The factorisation oracle tells you that that's prime. Try again - let the complete list of primes be {2,3}The multiplication/addition yeilds 7. The factorisation oracle tells you that that's prime.Try again - let the complete list of primes be {2,3,7}Do you permit me to do the above?Are you going to claim that I must insert 5 into that set?If so, then why? Would Euclid insist that I inserted 5? What would bethe justi?ation?If not, then your argument seems to be too transient and ?kle to be able to be countered rationally.> If the list is not complete, then I cannot perform the multiplication.Erm??? If the assumed ?ite set of primes is {2,3,7}, then the multiplication is trivial, and yields the new prime 43. Can you not multiply 2, 3, and 7 together? Completeness is not a clearly de?ed in this context. At this point we _don't know_ that 5 is prime. Unless your book of axioms says oh, you may assume 5 is prime whenever you like, but _my_ book of axioms differs from yours in that regard.The same for _any_ ?ite set of primes. Euclid's proof simply requires that one assume there's a ?ite set of primes, it does _not_ ask one to assume that this set has any other properties. Why do you wish to give it extra unnecesary properties? In particular, properties that rely on an external or prior prime generation?As I said before, and gave an example which you for some reasonsnipped, if you use the Euclid method as a constructive prime generator you very quickly end up generating primes smaller than the previously known maximum. (On the assumption that this generator is the only generator you have.)>Don't forget that if you're pretending that you don't know what>the primes are, you can't suddenly pull otherwise well-known>features of the prime numbers out of a hat.It's not necessary to know all the primes in order to construct the > argument. It is suf?ient to know that it is /in theory/ possible to know > them, e.g. by sieving. It is only necessary to know all the primes if you > wish to perform the multiplication in real life.That makes it sound like you've missed the entire point of Euclid's proof.Euclid's proof shows that you never need to to _any multiplications at all_ in real life.As an _existance_ proof, Euclid's algorithm is non-constructive.If you're trying to use it as a _constructive_ algorithm, why the heck do you want/need to invoke a second constructive algorithm (the one that you used to generate all primes up to P)?Phil-- Unpatched IE vulnerability: ?e-protocol proxyDescription: cross-domain scripting, cookie/data/identity theft, command executionExploit: http://safecenter.net/liudieyu/WsOpenFileJPU/ WsOpenFileJPU-MyPage.HTM === There appears to be some incomprehensible logic in this thread.No-one is claiming that Euclid's method can generate all primes. Butit appears that someone IS saying that Richard is claiming this. He'snot.Let us all agree that if we are ever to multiply all primes up to P,we need a way of generating all of them, and knowing that they reallyare all prime, and that there are none omitted. I'm sure even Euclidhimself would agree with this - so would any rules-lawyer.So let's start. We'll keep it simple with digits we can all manage,and work our way up.A. Euclid says there is no largest prime, because for any givenprime P, a larger one Q can be found. (Well, not exactly that, but Idon't speak Greek, and certainly not the Greek of Euclid's time.)B. It's not hard to see that once Q has been found it can be treatedas though it were P, and its own Q found, and the new Q can again betreated as though it were P, and a further Q found, and so on. So ifthe ?st Q is found, we know there will be an in?ite chain ofgreater and greater prime numbers.C. Euclid then de?es a number N, which is one more than theproduct of all primes up to and including P. Note that Euclid neverrequres the value of N to be known.D. He states that either N (which is larger than P) is prime, orthat it has at least one prime factor M which is larger than P. Notethat Euclid never requires the value of M to be known.up to and including P, as they all give a remainder of 1.F. If N is composite, its prime factors are greater than P. Therecould be just one prime factor M, in which case N is a square number. More likely, there are two or more prime factors. Any one of thesefactors will do for M, and thus for Q. Because M is larger than P, Qis larger than P.G. If N is not composite, it's prime (as it's greater than 1), so Nwill do for Q. Because N is larger than P, Q is larger than P.H. Because neither the value of N nor M need be known, the value ofQ need not be known, but we do know that is is prime and that it islarger than P.If Euclid's method is true, is should work for ANY prime number - evensmall ones. So let us start at P = 5. Note, we could have startedanywhere. 5 is small enough that calculations are easy, without beingtrivial.Using a sieve of Eratosthenes, we know the primes up to and including5 are {2,3,5}. The number N Euclid's method calculates is 31. Usinga seive, we know 31 is prime, so Euclid is correct.Let's work with the next highest prime after 5, i.e. P = 7. N = 211, which is prime.Next, P = 11. N = 2311, which is prime.Next, P = 13. N = 30031, which is NOT prime as it is 59 * 509. ButEuclid claims that either N is prime OR it has at least one primefactor M greater than P. Either 59 or 509 will do for M. They areboth prime, and both greater than P (13).Next, P = 17. N = 223092871, which is NOT prime as it is 19 * 97 *277. M could be any of the three factors.So lets's check (adding in the smaller primes): P N N is prime? M exits? Q? Q > P? 2 3 true false Q = N = 3 true 3 7 true false Q = N = 7 true 5 31 true false Q = N = 31 true 7 211 true false Q = N = 211 true 11 2311 true false Q = N = 2311 true 13 30031 false true (59) Q = M = 59 true 17 510511 false true (19) Q = M = 19 trueSo what happens when we reach the last KNOWN value of P where we alsoKNOW all previous values. For any higher values of P (such as a knownmersenne prime) we will not be able to calculate N.Well, it doesn't matter, as Euclid never required us to know allprimes smaller than P. It is suf?ient to know than N is not evenlydivisible by any prime smaller than P or by P itself, simply becauseof the way N is calculated (see paragraphs E and C).We can then follow through paragraphs F, G and H to know that Euclidwas correct.The red herring that Phil threw in about how do we get 5 into thelist of prime numbers is nonsense. No-one ever claimed the only wayto get the list of prime numbers was to generate N from each P bysubstituting the previous N as the current P.Phil has his P - (N => P) - (N => P) - (N => P) chain {2,3,7,31}. Sure that chain doesn't include 5. Nor does it include 11, 13, 17,19, 23 or 29. Nor 37, 41, 43 and a whole lot more. And Phil seems tothink that 5 therefore needs special dispensation to be allowed as aprime number.But Phil: What justi?ation do you have for starting at 2? Why is2 a prime number, and 5 not?ObPuzzle: The de?ition of a prime number is a counting number thathas exactly two factors - 1 and itself (which is why 1 is consideredas not being prime [which is odd, given the derivation of the wordprime] - neither is it composite, as those numbers have a factorother than 1 or itself). If we glibly change the de?ition fromcounting number to integer, how many prime numbers are there? What are they (if any)? === > [am.> But I think I got a case. :-) ]I hope you've got a hot-shot lawyer...He might end up as just a shot lawyer at this rate...Your premise is:> If there is a largest prime, P,and that's all, as can be deduced from the then and therefores> hereafter.Right. And the point of the premise (and the argument) is to take us to reductio ad absurdum - i.e. to show that there is no largest prime.>Nope. Nowhere have you assumed that the list of primes known is>complete up to P.On the contrary, I formed N by multiplying all primesErm, says who? You've not de?ed all in this context.Haven't I? Okay, guilty, your honour. all means every prime between 2 and P, including 2 and P.> Anyway, your maximum prime P is a red herring and completely> unnecessary - why did you even introduce it in the ?st place?I was merely explaining to the OP that Euclid's proof of the in?itude of primes is no use in attempting to construct a prime.> and then adding 1. To> multiply all primes together, I must ?st have a complete list of> primes.You must have a list of primes. What does ?complete' mean to you in> this context? If you wish to multiply together all primes between 2> and P then you must know /a priori/ all the primes between 2 and P.Well, I can get that list by starting at 2, using your rather handy factorisation oracle, and iterating up to and including P. As I said before, and gave an example which you for some reason> snipped,Brevity, my dear chap. Mere brevity.> if you use the Euclid method as a constructive prime> generator you very quickly end up generating primes smaller than> the previously known maximum. (On the assumption that this> generator is the only generator you have.)I wasn't trying to construct primes. I was trying to demonstrate why you can't construct them in that way.-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === Richard Heath?ld and Phil Carmody write:> If there is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.> then the number of primes is ?ite. We> can therefore form a new number, N, which is one greater than the product> of all primes. ...> Nope. Nowhere have you assumed that the list of primes known is> complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de?ed all in this context. Nowhere do you > say anything about having a contiguous set of primes from 2 to P. Phil, stop blathering about de?ing ?all'. You misread or mis-interpreted Richard's wording, and you've been caught.> The only thing you've said is that your set of primes has a maximum> element P. Full stop. There were no further premises -- look above if > you don't believe me.We're dealing with positive integers. If there is a largest one in aset of them, then that set is ?ite, as Richard said, and if it's?ite, then a product of all its members exists. It is not necessaryto specify an algorithm to construct the set (although in this case itwould be simple to do so), nor to de?e ?all' in this context.-- Mark Brader, Toronto | Professor, I think I have a counterexample.msb@vex.net | That's all right; I have two proofs. === > Richard Heath?ld and Phil Carmody write:> If there is a largest prime, P, and that's all, as can be deduced from the then and therefores hereafter.> then the number of primes is ?ite. We> can therefore form a new number, N, which is one greater than the product> of all primes. ...>Nope. Nowhere have you assumed that the list of primes known is>complete up to P.On the contrary, I formed N by multiplying all primes Erm, says who? You've not de?ed all in this context. Nowhere do you >say anything about having a contiguous set of primes from 2 to P. Phil, stop blathering about de?ing ?all'. You misread or mis-> interpreted Richard's wording, and you've been caught.I did not. He introuced concepts which were unnecessary, and I claim harmful. >The only thing you've said is that your set of primes has a maximum>element P. Full stop. There were no further premises -- look above if >you don't believe me.We're dealing with positive integers. If there is a largest one in a> set of them, then that set is ?ite, as Richard said, and if it's> ?ite, then a product of all its members exists. Since when has that been at issue? I've even said as much myself. You must have misread.> It is not necessary> to specify an algorithm to construct the set (although in this case it> would be simple to do so), nor to de?e ?all' in this context.Nor do I say it is. You must have misread.Phil-- Unpatched IE vulnerability: Click hijackingDescription: Pointing IE mouse events at non-IE/system windows === > It's not necessary to know all the primes in order to construct the > argument. It is suf?ient to know that it is /in theory/ possible to know > them, e.g. by sieving. It is only necessary to know all the primes if you > wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest prime! === >It's not necessary to know all the primes in order to construct the >argument. It is suf?ient to know that it is /in theory/ possible to know >them, e.g. by sieving. It is only necessary to know all the primes if you >wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest prime!But the (practical) largest prime is time dependant -- as time passes and better computers and programs come into existence, the largest know prime keeps getting larger. With, as yet, no end in sight. === > It's not necessary to know all the primes in order to construct the> argument. It is suf?ient to know that it is /in theory/ possible to> know them, e.g. by sieving. It is only necessary to know all the primes> if you wish to perform the multiplication in real life.So, since you can't sieve all of them there is a (practical) largest> prime!Great! Please share...-- Richard Heath?ld : binary@eton.powernet.co.ukUsenet is a strange place. - Dennis M Ritchie, 29 July 1999.C FAQ: http://www.eskimo.com/~scs/C-faq/top.htmlK&R answers, C books, etc: http://users.powernet.co.uk/eton === > Great! Please share...10^80 - 3? === > I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? HopeIs 30031 a prime? === In sci.math, Christian Bau I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? HopeIs 30031 a prime?30031 = 2*3*5*7*11*13 + 1 = 59 * 509Regardless of whether the number is prime or not, one always getsa prime not in the original set, though.As it is, RSA's algorithm requires generation of largeprimes and, in order to crack it, the factorizationof a product of two large primes (the product, not theprimes, is given as part of the public key). Large primegeneration isn't all that dif?ult unless one gets stuckin a desert (pick a random odd number N of the requisitesize; if it's prime, stop; otherwise add 2 to N and loop).A known desert is just after a factorial; it's obviousthat none of N! + 2, N! + 3, ..., N! + N can be prime.Or one can take the lcm of the numbers 1 through N, forthe same effect.Another interesting generator is x^2 + x + 41. The ?sttime this fails is x = 40, as one can verify using asimple program.-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === [...] p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one. If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?There are at least three problems in relying on this constructionto produce primes:Firstly in the main real-life application of large prime numbers,namely cryptography, the primes must be hard (in practice impossible)to guess, and for this they must be taken from a large/random ?pool'of possibilities. Relying on a simple recursive formula would meanyour decrypter would ?d it trivial to run through all practicalpossibilities.That brings us to the next problem - Values produced by the abovemethod very soon become impractically large, although you couldgeneralize it by spreading the factors between two factors, i.e.p_{n+1} = p_1.p_3.p_4...p_m + p_2.p_5...p_n. This also increasesthe pool of values, which somewhat ameliorates the ?st problem.But I've saved the worst snag until last - Values obtained by yourconstruction, and the two-product extension of it, are not alwaysprime!- --John R Ramsden (jr@adslate.com)--- Eternity is a long time, especially towards the end. Woody Allen === > it occurs that me> that super-large prime numbers can be calculated using the following: p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = .... Alas, it usually doesn't work. 2*3*5*7*11*13+1 = 30031 = 59*509: not prime; See http://www.asahi-net.or.jp/~KC2H-MSM/mathland/matha1/matha102 .htm http://www.research.att.com/projects/OEIS?Anum=A014545 Subtracting one at the end is also a good try, and usually fails to produce a prime.--Don Reble djr@nk.ca === I have looking at a few web pages dealing with the largest knowncalculated primes and a great deal of computational time is takinginto searching for these numbers and verifying they are primes. I haveseen the Euclid's proof of in?itude primes and it occurs that methat super-large prime numbers can be calculated using the following:p1=2 < p2=3A large prime number, p, can be generated usingp = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbersexceeding currently known largest primes can be generated ratherquickly. For example, we calculate the ?st million primes, multiplythem and just add one.If this is true, then calculating super-large primes should be acomputational trivial task, shouldn't it? Then why the big fuss overcalculating large primes?I suspect that I am missing fundamental here. Can super large primesreally be calculated using the trivial method outlined above? HopeBlueBear === > I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one.It seems that you haven't really understood Euclid's proof then. Thenew number constructed does not need to be prime; the point is that itis itself a product of primes and all of its factors must be newprimes. 2*3*5*7*11*13+1=59*509, as was mentioned earlier on thisthread. Also, 2*3*5*7-1=11*19. In either case, 59 and 509 are not in{2,3,5,7,11,13} and 11 and 19 are not in {2,3,5,7}. So long as yourset of primes is ?ite, you can extend it. Therefore the full set,which is not extensible by de?ition, must be in?ite.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?But it's not trivial. See my examples, which are very standard andwell-known.I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? HopeYou're welcome.BlueBear---- David === > I have looking at a few web pages dealing with the largest known> calculated primes and a great deal of computational time is taking> into searching for these numbers and verifying they are primes. I have> seen the Euclid's proof of in?itude primes and it occurs that me> that super-large prime numbers can be calculated using the following:p1=2 < p2=3> A large prime number, p, can be generated using> p = (p1 * p2) + 1 = 7p1=2 < p2=3 < p3=5> p = (p1*p2*p3) + 1 = 31p1=2 < p2=3 < p3=5 < p4=7> p = (p1*p2*p3*p4) + 1 = 211p1=2 < p2=3 < p3=5 ..... < pn=...> p = (p1*p2*p3*p4*....*pn) + 1 = ....It seems to me that using the above method, super large prime numbers> exceeding currently known largest primes can be generated rather> quickly. For example, we calculate the ?st million primes, multiply> them and just add one.> If this is true, then calculating super-large primes should be a> computational trivial task, shouldn't it? Then why the big fuss over> calculating large primes?I suspect that I am missing fundamental here. Can super large primes> really be calculated using the trivial method outlined above? HopeBlueBear2*3*5*7*11*13 + 1 = 30031 = 59*509 === This is just to thank you for your help!Paul>Dear All,> I would like to know whether it is possible to de?e the limit of a>function f as x tends to +oo, where f is de?ed on R^+, except on an>in?ite countable number of points of R^+ (for instance, except on>the set of integers).or for instance, something like gamma(-x) ...I think you can safely use the standard limit de?itions:> For all P>0, there is a Q such that for all x of dom(f):> x > Q ==> f(x) < -P> For all e>0, there is a Q such that for all x of dom(f):> x > Q ==> |f(x)-b| < e> For all P>0, there is a Q such that for all x of dom(f):> x > Q ==> f(x) > PI think these de?itions work ?e for the functions you have in mind.> Of course the gamma(-x) would have no limit for x --> +ooDirk Vdm === >The set of non-normal numbers has Lebesgue measure zero. >If you pick a real uniformly at random from the interval [0, 1] >then with probability 1 you have picked a normal number. >This leads many to take the position that if some number pops up >somewhere & there's no good reason to think it's not normal - some >number like pi, or like the reciprocal of the number you get when >you replace every other digit in the decimal expansion of pi with >a zero - then it's safe to assume that the number is normal.I started thinking about this and the same result could possibly beextended to tell something about the reciprocals of normal numbers.Please inform if I've made a logical error with the de?ition ofalmost all.---De?ition:A normal number whose reciprocal is non-normal is called a seminormalnumber. A number whose reciprocal is normal is called an inverse-normal number.Theorem: Almost all real numbers are inverse-normal.Proof:Assume two subsets of R called R_1 and R_2 so that for each r inR{0}, r is either in R_1 or R_2 and its reciprocal 1/r is in theother one. This divides the reals to two subsets and creates thebijection:T: R_1->R_2, T(r) = 1/rbetween them. Now, assume the set of all seminormal numbers in R_1called S_1. We call the image of S_1:U_1 = {T(s) | s in S_1}By de?ition the set U_1 contains only non-normal numbers. Sincealmost all reals are normal, we know that almost all numbers in R_2don't belong in U_1.Since there is a bijection between S_1 and U_1, we know that almostall numbers in R_1 don't belong in S_1 either, so we have proved thatalmost all numbers in R_1 aren't semi-normal. We now use the sameargument to obtain a similar result for R_2.Since almost all numbers in R_1 union R_2 = R{0} aren't non-normaleither, the only option is that almost all numbers in R are in factinverse-normal. === > [...]De?ition:A normal number whose reciprocal is non-normal is called a seminormal>number. A number whose reciprocal is normal is called an inverse->normal number.Theorem: Almost all real numbers are inverse-normal.Proof: [omitted]>The reciprocal is a measure automorphism on the nonzero reals. Given that almost all reals are normal, doesn't the conclusion follow immediately? One might need to invoke absolute continuity of measures.-- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === >Peer review limitations:--reviewer subjectivity and agenda>--constraints of scienti? dogma and orthodoxy>--by de?ition, *peer* reviewers are usually limited to the same>paradigm and philosophical traditionsFalse. Counter-example: me.I'm sure you're in a class quite by yourself. === >To draw truth tables and calculate prime implicants,>feel free to download from> http://users.pandora.be/vdmoortel/dirk/Boole/boole.html> (carful: you'll have to replace C0 with C and C1 with D) This is really a very useful program.>I have just uploaded a new version (1.1) of the program.It had a little bug. Please replace.Dirk Vdm === > in the following boolean equation simpli?ation: R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ])>The best and my 2nd ??al' version, has only four disjunctions.I doubt there's anything shorter.R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ]) = (C1'+C0'+A'+B')(C1'+A +B)(C1+C0'+A')(C1+A+B')(C1'+A+B)( C1+A+B' ) = A + (C1'+B)(C1+B') = A + C1'B' + BC1(C1'+C0'+A'+B')(C1+C0[Paragr aph]+A') = C0' + A' + (C1' + B)C1 = C0' + A' + B'C1R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ]) = (C1'+C0'+A'+B')(C1'+A +B)(C1+C0'+A')(C1+A+B') = (A' + C0' + B'C1)(A + C1'B' + BC1) = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A = A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1 + AC0'C1'B' + AC0'BC1 + C0'A + B'C1A = A'C1'B' + A'BC1 + C0'A + B'C1A = A'(B'C1' + BC1) + A(C0' + B'C1)---- === >in the following boolean equation simpli?ation:> R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ])> The best and my 2nd ??al' version, has only four disjunctions.> I doubt there's anything shorter. R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ])> = (C1'+C0'+A'+B')(C1'+A +B)(C1+C0'+A')(C1+A+B') (C1'+A+B)(C1+A+B') = A + (C1'+B)(C1+B')> = A + C1'B' + BC1> (C1'+C0'+A'+B')(C1+C0[Paragraph ]+A') = C0' + A' + (C1' + B)C1> = C0' + A' + B'C1 R = (C1'+C0'+A'+B')(C1'+A +B)(C1'+C0+A+B)(C1+C0'+A')(C1+A+B[Paragraph ])> = (C1'+C0'+A'+B')(C1'+A +B)(C1+C0'+A')(C1+A+B')> = (A' + C0' + B'C1)(A + C1'B' + BC1)> = A'C1'B' + A'BC1 + C0'A + C0'C1'B' + C0'BC1 + B'C1A = A'C1'B' + A'BC1 + A'C0'C1'B' + A'C0'BC1> + AC0'C1'B' + AC0'BC1 + C0'A + B'C1A> = A'C1'B' + A'BC1 + C0'A + B'C1A> = A'(B'C1' + BC1) + A(C0' + B'C1)Neat.Verifying this, it helped me spot a little bug in my program.Dirk Vdm === I've been disassembling an FPU emulator and as expected, itcontained a signi?ant number of constants. A fair amount of these are very obvious, but a quite a few are complete mysteriesto me. For all but three I ?ured out the (likely) relationship between the successive terms, but as for what they are al used? I hope someone in this group can tell me or point me to a sitethat has information about mathematical series. The constants are:-0.005208333333333333423683514 -1/192+0.000048828124999993646575269 1/20480 *320/3-0.000000544956752147722229412 -1/1835008 *448/5+0.000000006622737896066696605 1/150994944 *576/7-0.000000000084664710789036923 -1/11811160064 *704/9+0.000000000001118158950846881 1/big *832/11-0.000000000000014377930269982 -1/bigger *960/13+0.12451171875+0.498751787292275571965858085+ 0.000262287674813145674657804+0.249050582915605093375140816+ 0.498101165831210186750281632+0.002034561246815280125422448+ 0.249977256935346610686307297+0.482158790777797965568110885+ 0.798637441128639956279457873+0.250000000000000000162630325 1/4+0.041666666666666666576316485 1/24 /6+0.005208333333333335212617098 1/192 /8+0.000520833333333347420733561 1/1920 /10+0.000043402777777744446993657 1/23040 /12+0.000003100198412569815306638 1/322560 /14+0.000000193762401041715562333 1/5160960 /16+0.000000010764578285932073498 1/92897280 /18+0.000000000538227910807861920 1/1857945600 /20+0.000000000024464188541715309 1/408748032000 /22+0.000000000001021270795795864 1/9809952768000 /24+0.000000000000039701152521748 1/255058771968000 /26+0.020833333333333334345255360 1/48 *48/1+0.000781250000000021185310450 1/1280 *80/3+0.000034877232141512120363463 1/28672 *112/5+0.000001695421027976551104493 1/589824 *144/7+0.000000086697509034958444851 1/11534336 *176/9+0.000000004585603636012795109 1/218103808 *208/11+0.000000000246930795093098370 1/4026531840 *240/13+0.000000000015363636849275580 1/73014444032 *272/15Robert-- Robert AH Prinsprino@bigfoot.com === Robert AH Prins contained a signi?ant number of constants. A fair amount of> these are very obvious, but a quite a few are complete mysteries> to me. For all but three I ?ured out the (likely) relationship> between the successive terms, but as for what they are al used?> I hope someone in this group can tell me or point me to a site> that has information about mathematical series.I don't have anything speci? but maybe search on inverse symboliccalculator?> The constants are:... [snipped, see OP]There is this pattern:48 = 2^4 *31280 = 2^8 *528672 = 2^12 *7589824 = 2^16 *911534336 = 2^20 *11218103808 = 2^24 *134026531840 = 2^28 *1573014444032 = 2^32 *17No doubt the software is calculating a logarithm, because these numbers arethe inverses of the coef?ients of x^n in the power series for ln(1+4x).Similarly (with alternating signs)192 = 2^6 *320480 = 2^12 *51835008 = 2^18 *7150994944 = 2^24 *911811160064 = 2^36 *11big = 2^42 *13 (I presume)bigger = 2^48 *15and we are looking at a power series for ln(1+6x).4 = 2 *2!24 = 2^2 *3!192 = 2^3 *4!1920 = 2^4 * 5!23040 = 2^5 *6!322560 = 2^6 *7!etc, and it's calculating an exponential.The isolated constants in the list aren't so easy :)Larry === >Robert AH Prins I've been disassembling an FPU emulator and as expected, it> contained a signi?ant number of constants. A fair amount of> these are very obvious, but a quite a few are complete mysteries> to me. For all but three I ?ured out the (likely) relationship> between the successive terms, but as for what they are al used?> I hope someone in this group can tell me or point me to a site> that has information about mathematical series.I don't have anything speci? but maybe search on inverse symbolic>calculator?> The constants are:>... [snipped, see OP]There is this pattern:>48 = 2^4 *3>1280 = 2^8 *5>28672 = 2^12 *7>589824 = 2^16 *9>11534336 = 2^20 *11>218103808 = 2^24 *13>4026531840 = 2^28 *15>73014444032 = 2^32 *17>No doubt the software is calculating a logarithm, because these numbers are>the inverses of the coef?ients of x^n in the power series for ln(1+4x).Similarly (with alternating signs)>192 = 2^6 *3>20480 = 2^12 *5>1835008 = 2^18 *7>150994944 = 2^24 *9>11811160064 = 2^36 *11>big = 2^42 *13 (I presume)>bigger = 2^48 *15>and we are looking at a power series for ln(1+6x).4 = 2 *2!>24 = 2^2 *3!>192 = 2^3 *4!>1920 = 2^4 * 5!>23040 = 2^5 *6!>322560 = 2^6 *7!>etc, and it's calculating an exponential.The isolated constants in the list aren't so easy :)a debugger is very dif?ult, as the emulated FPU stack anddebugger stack occupy the same storage...)I guess it's time to trace my Carmichael & Smith and see if thatgives me any clues as to how the above series are used to calculate logarithms or powers.Robert-- Robert AH Prinsprino@bigfoot.com === let~ rate 0.002 of decayed tooth patients are a cancer of mouth patient.Until ?st cancer of mouth patient have discovered in any dentist'sconsultation room, we diagnose.?d the expected value whether we treat and you must diagnose the couple ofpatients.---um......i think.....answer is reciprocal of 1/500it is right?? === we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,?d that probability of possibility.----very dif?ult......um....help....me please....... === > we cuts a two place into a wire at random.let length of wire is 1We got the wire of a three piece.when we made a triangle from three wire piece,?d that probability of possibility.----very dif?ult......um....help....me please.......Put the wire on a number line with left end-point at 0 and right end-point at 1. Let c and d be the points at which the wire iscut such that 0 < c < d < 1. Now use the fact that in a trianglethe sum of the lengths of any two sides must be greater than thethird. For example, we must have c + (d - c) > 1 - d. Whenyou get the appropriate inequalities between c and d, plot themin the Cartesian plane with horizontal axes labelled c and d.Then ?d the area of the solution region. This will be yourprobability. === > we cuts a two place into a wire at random. let length of wire is 1 We got the wire of a three piece. when we made a triangle from three wire piece, ?d that probability of possibility. ---- very dif?ult......um....help....me please.......Let say that lenghts of pieces are A,B,C. To form triangle pieces mustsatisfy:(A+B)>C(A+C)>B(B+C)>AThese conditions have another geometric interpretation: A B C0----X-Y----1Let points X,Y be random numbers (let say that they are places where we cutoriginal wire to make three pieces)Then we may restate conditions as:(x+y-x)>1-y <=> 2y>1 <=> y>0.5(x+1-y)>y-x <=>2x-2y+1>0 <=> x-y+0.5>0 <=> y>x+0.5(y-x+1-y)>x <=> 1>2x <=>x<0.5also note that 0<=x<=1, 0<=y<=1.Draw these in x-y plane - the area of part of square ( 0<=x<=1, 0<=y<=1) inwhich all conditions are satis?d represents your probability.Goran === thank...you....very much....i think......if we only use 0-x-y-1in this case, probability is 1/8but, if we use 0-x-y-1, 0-y-x-1in this case, probability is 1/4which of case is right??advice....please....sir~ === > thank...you....very much.... i think......if we only use 0-x-y-1 in this case, probability is 1/8 but, if we use 0-x-y-1, 0-y-x-1 in this case, probability is 1/4 which of case is right?? advice....please....sir~1/4 === thank...you....very much....> i think......if we only use 0-x-y-1> in this case, probability is 1/8> but, if we use 0-x-y-1, 0-y-x-1> in this case, probability is 1/4> which of case is right??> advice....please....sir~> 1/4Problem may be restated - if we have two independent uniformly distributedvariables on [0,1] variables x,y representing distances from one end ofwire, where we cut the wire - what is probabilities that pieces may formtriangle? If we consider only 0-x-y-1 then we have additional constraintthat y>x and space of all possible events are not represented by square butupper left triangle. If we allow ywe cuts a two place into a wire at random.>let length of wire is 1>We got the wire of a three piece.>when we made a triangle from three wire piece,>?d that probability of possibility.I believe what you're saying is that the three pieces must be able toform a genuine triangle. WLOG, assume that the pieces are cut in increasing order of size, so thelength of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], whichalso must be larger than Sqrt[a^2+b^2].Proceed in this fashion.Doug === >WLOG, assume that the pieces are cut in increasing order of size, so the>length of the pieces are a < b < c. Then c = Sqrt[1-a^2-b^2], which>also must be larger than Sqrt[a^2+b^2].Yeah, I messed the details on this one, as I've found by playing aroundwith it some more. It's always a treat to post upon waking. :-PDoug === This night my dream was crucially based on thefact that sec(2x)=(sec(x)+sec(4x))/2.Uhm, I get the nagging feeling that in that stateI'm just as lousy as a mathematician as awake :-)Do me a favor and compute the x for which thatequation is accidentally right before someonegets hurt...-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === > This night my dream was crucially based on the> fact that sec(2x)=(sec(x)+sec(4x))/2.> Uhm, I get the nagging feeling that in that state> I'm just as lousy as a mathematician as awake :-)> Do me a favor and compute the x for which that> equation is accidentally right before someone> gets hurt...>Perhaps you can plot the left function and the right function and see foryourself before you go off and hurt someone? === In sci.math, ?p_alpha@safebunch.com><1068995840.212611@news-1. nethere.net>:> This night my dream was crucially based on the> fact that sec(2x)=(sec(x)+sec(4x))/2.> Uhm, I get the nagging feeling that in that state> I'm just as lousy as a mathematician as awake :-)> Do me a favor and compute the x for which that> equation is accidentally right before someone> gets hurt...>Perhaps you can plot the left function and the right function and see for> yourself before you go off and hurt someone?> There are also multiple solutions -- although one of themshould jump right out and bite the OP (were the value tohave any teeth, that is -- and this particular numeralobviously hasn't hatched any yet :-) ).Consider that 1 = (1 + 1)/2. Now when does sec(2x) = 1?This hint should make solving the problem a dream... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > There are also multiple solutions -- although one of them> should jump right out and bite the OP (were the value to> have any teeth, that is -- and this particular numeral> obviously hasn't hatched any yet :-) ).Just saw that my handy plotting software (MathGV) hassec(x) built-in. Looks like an EPR spectrum on crack :-)-- Hauke Reddmann <:-EX8 For our chemistry workgroup,remove math from the addressFor spamming, remove anything else === In sci.math, Hauke Reddmann:> There are also multiple solutions -- although one of them> should jump right out and bite the OP (were the value to> have any teeth, that is -- and this particular numeral> obviously hasn't hatched any yet :-) ).Just saw that my handy plotting software (MathGV) has> sec(x) built-in. Looks like an EPR spectrum on crack :-)Or an EPR spectrum of someone *on* crack... :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. === > Why can't a conjugate of a subgroup be a proper subset of it?See http://mathworld.wolfram.com/ConjugateSubgroup.html-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft === > Why can't a conjugate of a subgroup be a proper subset of it?> See http://mathworld.wolfram.com/ConjugateSubgroup.htmlSurely the question doesn't deserve an URL?-- Timothy Murphy tel: +353-86-2336090, +353-1-2842366 === > Why can't a conjugate of a subgroup be a proper subset of it?>See http://mathworld.wolfram.com/ConjugateSubgroup.htmlSurely the question doesn't deserve an URL?It doesn't have one - at any rate, if it does, that's not it. That URL just de?es conjugacy & links to some other stuff, it doesn't address the original question.-- === >For any given natural number m, the Goodstein sequence, G(m), only>terminates if m < 4.G(4) terminates, so the argument is invalid. I don't have a formal> proof, but just look at the following and it should be obvious (unless> I made a mistake somewhere):Bravo, indeed - I could not spot any mistake in the reasoning. If Ihad bothered to apply myself to the tedious drudgery of speci?allyworking out the steps you did - instead of attempting to visualisethem generally - I would have, of course, held back archiving myargument.So, my thanks on two counts. Firstly for helping me spot the ?tis not as obvious as I thought it might be); secondly, since I didn't- fortunately - hold back my argument, for helping me bring intobetter focus the issue that triggered my investigation:As I see it, your reasoning that we can see that G(4) terminatesshould, surely, be capable of expression as a formal proof sequence inPA. Now, assuming the standard interpretation of Goodstein's argumentis valid, then, for any given natural number m, we should also,reasonably, be able to see, similarly, that G(m) must terminate -necessarily in 0. Thus, for any given m, we should also be able tomirror this vision, without appeal to trans?ite induction, as aformal proof sequence in PA.In other words, assuming that Goodstein's theorem cannot be formallyproved in PA, there should, reasonably, be some constructivemeta-proof - in Goedel's sense, i.e., one that does not appeal totrans?ite concepts - to the effect that, for any given naturalnumber m, there is some proof sequence in PA that implies that [G(m)]terminates for the numeral [m].Bhup === > For any given natural number m, the Goodstein sequence, G(m), > only terminates if m < 4.Others in the thread have shown your claim is false; however, for more detail about the structure of these sequences, you might want to see the text ?e athttp://r.s.home.mindspring.com/GoodsteinSequenceswhich shows, incidentally, that the Goodstein sequence that starts at 4 has (two) maximum terms equal to N = (j+1)*2^j-1, where j=3*2^27-1, and terminates at 0 on the (2N)th term.--r.e.s. === [...] for more detail about the structure of these> sequences, you might want to see the text ?e at > http://r.s.home.mindspring.com/GoodsteinSequences > [...]r.e.s. === sequences is, indeed, illuminating, and the graphical analysis isfascinating. I shall certainly study them carefully.Bhup === [...] for more detail about the structure of these>sequences, you might want to see the text ?e at >http://r.s.home.mindspring.com/GoodsteinSequences >[...]r.e.s.> === > sequences is, indeed, illuminating, and the graphical analysis is> fascinating. I shall certainly study them carefully.One correction to my post: The graph of the Goodstein sequence is piecewise linear, having a plateau consisting of (N+1)/2 terms-- not 2, of course. Rather, it's the 2nd to last, and 2nd longest linear segment (relative to the number of terms). All the terms in this plateau are equal to N = (j+1)*2^j-1, j=3*2^27-1.--r.e.s. === In response to self...> (ii) Express m by its hereditary representation in base 2;...> For instance, the Goodstein sequence G(4) is given by:G(1, 4) = 1*(2^2) + 0*(2^1) + 0*(2^0) (base 2) > = 4> G(2, 4) = 1*(3^3) + 0*(3^1) + 0*(3^0) - 1 Because in hereditory representation 2 is 2^1, and thereforebumping the base turns it into 3^1.Sorry for any confusion.Phil-- Unpatched IE vulnerability: history.back method cachingDescription: cross-domain scripting, cookie/data/identity theft, command executionExploit: http://www.safecenter.net/liudieyu/RefBack/RefBack-MyPage.HTM === > Hallo,> Ich have given 5 vectors and have to ?d a > base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th> vector through this base.Look up Gram-Schmidt orthogonalization.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === >Hallo,Ich have given 5 vectors and have to ?d a>base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th>vector through this base. Look up Gram-Schmidt orthogonalization.>complicatedfor me. Can you give me at least one really short example for theapplication of this method? Please!Karl.[P.S. To tell the truth, I don't even understand how this method could behelpfulfor me. (:-{ ] === > Hallo,> Ich have given 5 vectors and have to ?d a> base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th> vector through this base.> The vectors are:> ^^^^^^^^^^^^^^^^^^> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);> v3 = (0,1,3,0); v4 = (2,-1,4,-7)> v5 = (5,-8,1,2)> Look up Gram-Schmidt orthogonalization.>complicated> for me. Can you give me at least one really short example for the> application of this method? Please!> Karl.[P.S. To tell the truth, I don't even understand how this method could be> helpful> for me. (:-{ ]> Using Gram-Schmidt here is rather like using a sledge hammer to crack eggs. It is way too much work for the result desired.A much better approach is to try to determine whether each vector, in (v1,v2,v3,v4,v5) is a linear combination of the previous ones.This can be done pretty much by eyeball analysis. One ?ds that v1 is trivially independent,v2 is not dependent on (v1),v3 is depentent on (v1,v2), v3 = 2*v1 - 1*v2v4 is not dependent on (v1,v2) so not on (v1,v2,v3) either, andv5 is dependent on (v1,v2,v3,v4), v5 = v1 + v2 + v4Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed as linear conbinations of basis vectors and v3 and v5 can be expressed less trivially as linear combinations of these basis vectors. === > Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed > as linear conbinations of basis vectors and v3 and v5 can be > expressed less trivially as linear combinations of these basis > vectors.But I'm working in a R^4 room. So how can a 4D-room have a3D-basis? Or did I misunderstand something? === >Thus (v1,v2,v4) is a basis, v1, v2 and v4 can trivially be expressed >as linear conbinations of basis vectors and v3 and v5 can be >expressed less trivially as linear combinations of these basis >vectors.> But I'm working in a R^4 room. So how can a 4D-room have a> 3D-basis? Or did I misunderstand something?Did you mean that you want4ed a basis of the entire space, rather than a basis of the space spanned by the 5 vectors given? If so, that meaning was not clear to me. And such a basis is impossible from just the 5 vectors given.If you did want a basis of the entire space, what is wrong withb1 = (1,0,0,0), b2 = (0,1,0,0), b3 = (0,0,1,0), b4 = (0,0,0,1)as such a basis? === >Hallo,>Ich have given 5 vectors and have to ?d a>base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th>vector through this base.> Look up Gram-Schmidt orthogonalization.> complicated> for me. Can you give me at least one really short example for the> application of this method? Please!> Karl.> [P.S. To tell the truth, I don't even understand how this method could be> helpful> for me. (:-{ ]You are given ?e vectors in R^4. The Gram-Schmidt algorithm allows youto ?d a minimal set of vectors that spans the same subspace as thegiven vectors. If your vectors happen to span the entire space, thenGram-Schmidt gives you a basis for that space.As a bonus, the vectors you get from Gram-Schmidt are orthonormal, whichmakes it very easy to decompose any given vector as a linear combinationof the basis vectors.The important thing is not to get bogged down in computation until youunderstand what the method is doing.The ?st vector is very easy. If the vector v1 is nonzero, then {v1} isa linearly independent set, and therefore all you need to do isnormalize. Take u1 = v1 / ||v1||. That is, divide the ?st vector byits Euclidean norm, and call that u1, which is a unit vector pointing inthe same direction as v1.Next consider your second vector, v2, and its projection along u1, whichis (v2 . u1) u1. If v2 happens to be a scalar multiple of u1, this dotproduct will just give you v2. If it's not a scalar multiple, then thedot product gives you the component of v2 in the direction of u1. Youneed to subtract this from v2 in order to get the orthogonal complement,which is w2 = v2 - (v2 . u1) u1and then normalize, obtaining u2 = w2 / ||w2||.Next consider v3 and its components in the directions of u1 and u2.Since we now have two vectors in our orthonormal set, the orthogonalcomplement is w3 = v3 - (v3 . u1) u1 - (v3 . u2) u2.If w3 is zero, it means v3 is already in the subspace spanned by u1 andu2, and therefore you can ignore it and move on to the next vector. Ifw3 is nonzero, then you normalize it and add it to the basis you arebuilding: u3 = w3 / ||w3||and so on.If it turns out that v5 is a linear combination of the other vectors thatyou started with, then it will reduce to a linear combination in terms ofyour orthonormal basis, and you can easily compute its components. Thenjust substitute what each of the u_i's is in terms of the v_j's, andyou're done.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > Hallo,> Ich have given 5 vectors and have to ?d a> base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th> vector through this base.> Look up Gram-Schmidt orthogonalization.>complicated> for me. Can you give me at least one really short example for the> application of this method? Please!> Karl.> Take a look at-- Paul SperryColumbia, SC (USA) === You are confusing vectors with the algebra of real numbers!!!!!Vectors are mathematical representations of real physical entities. Aproblem is meaningless unless you de?e the nature of the vectors.-- Bruce Harveybruce@bearsoft.co.ukThe Alternative Physics Sitehttp://users.powernet.co.uk/bearsoft> Hallo, Ich have given 5 vectors and have to ?d a> base for V = L(v1, v2, v3, v4, v5). Then I must express the ?th> vector through this base. I think the general approach is to try every possible linear> combination: 1.) a*v1+b*v2+c*v3+d*v4 = v5> or> 2.) a*v1+b*v2+c*v3+d*v5 = v4> or> 3.) a*v1+b*v2+c*v5+d*v4 = v3> or> 4.) a*v1+b*v5+c*v3+d*v4 = v2> or> 5.) a*v5+b*v2+c*v3+d*v4 = v1 The vectors are:> ^^^^^^^^^^^^^^^^^^> v1 = (1,-2,0,3); v2 = (2,-5,-3,6);> v3 = (0,1,3,0); v4 = (2,-1,4,-7)> v5 = (5,-8,1,2) I got the following results: 1.) a + 2c = 1 and b - c = 1 and d = 1> 2.) a + 2b = -3 and c - b = 1 and d = 1> 3.) a - d = 2 and b - d = -1 and c + d = 0> 4.) a - 3d = 2 and b + d = 0 and c + d = -1> 5.) a + d = 0 and 2b - 3d - 1 = 0 and 2c - d - 1 = 0 I have expected to get simple numerical values for a,b,c> and d (like: a = 2,...) but not this. Can you tell me what my> mistake was?> Karl.> === If A is a real nXn (0,1) matrix such that A^m = I (the identitymatrix) for some m, must A be a permutation matrix ? === > If A is a real nXn (0,1) matrix such that A^m = I (the identity> matrix) for some m, must A be a permutation matrix ?Yes. AB = I where B = A^{m-1} has nonnegative integer entries. === Can someone explain to me what's the difference between basis transformation and a similarity transformation.I'm really confused ... === Can someone explain to me what's the difference between > basis transformation and a similarity transformation.A similarity transform of A is any transformation of theform inv(S)*A*S with nonsingular S.According to my Horn and Johnson, every invertiblematrix is a change-of-basis matrix, and every change-of-basis matrix is invertible. Thus, every similaritytransform is a change of basis transform for thesame matrix. - Randy === Arnold> Can someone explain to me what's the difference between> basis transformation and a similarity transformation.No doubt basis transformation refers to a square matrix which describes achange of basis. For any two given basesx={x1,...,xn}y={y1,...,yn}the change of basis from x to y is described by a unique such matrix M.Perforce M has a nonzero determinant and is invertible.The only meaning of similarity transformation that I know of is this onefrom geometry: a mapping f of a Euclidean space into itself such that[f(x),f(y)]=k[x,y] for any x,y and some real constant k>0, where [ , ]refers to the distance between two points. If this doesn't seem to makesense, maybe you could tell us more about the context in which you cameacross the term.LH === Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de?edin such a way that, at every irrational x in [a,b], f is continuousand f(x) = g(x). Then, is it true that the Riemann integral of f over[a,b] equals the integral of g? Since the set of discontinuities of fon [a,b] is at most countable and, therefore, has measure 0, itfollows f is certainly integrable over [a,b], but I couldn't come to aconclusion if its integral needs to equal the integral of g or if itdepends on how we de?e f for the rationals of [a,b].I know that if f is Thomae's function, then the answer is yes and theinegralas vanish, but in this case we have f(x) =0 for everyirrational, which seems to imply a loss of generality when compared tothe function f de?ed above.In case the answer is yes, does it remain yes if we replace theassumption that g is continuous by the weaker one that g is onlyRiemann integrable over [a,b]?Now, suppose in the questions above we replace the set Q cap [a,b], ofthe rationals in [a,b], by an uncountable set E with measure zero?Then, is there anything interesting we can af?m about the integralof f?Amanda === > Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de?ed> in such a way that, at every irrational x in [a,b], f is continuous> and f(x) = g(x). Then, is it true that the Riemann integral of f over> [a,b] equals the integral of g? Since the set of discontinuities of f> on [a,b] is at most countable and, therefore, has measure 0, it> follows f is certainly integrable over [a,b], but I couldn't come to a> conclusion if its integral needs to equal the integral of g or if it> depends on how we de?e f for the rationals of [a,b].In fact, your argument to prove that f is integrable over [a,b] is notcorrect. Indeed, in order that a function de?ed on [a,b] is integrable(in the sense of Riemann), the set of discontinuities must have measure0 *and* f must be bounded. However, if f is supposed to be bounded, yourargument is correct.On the other hand, the integrals of f and g must be equal (stillassuming that f is bounded, of course), since the function f - g isintegrable and takes the value 0 at every irrational form [a,b].> In case the answer is yes, does it remain yes if we replace the> assumption that g is continuous by the weaker one that g is only> Riemann integrable over [a,b]?Sure. I did not use the assumption g is continuous. > Now, suppose in the questions above we replace the set Q cap [a,b], of > the rationals in [a,b], by an uncountable set E with measure zero? > Then, is there anything interesting we can af?m about the integral > of f?I don't think so.Jose Carlos Santos === >Suppose g:[a,b]->R is continuous on [a,b] and let f[a,b]->R be de?ed>in such a way that, at every irrational x in [a,b], f is continuous>and f(x) = g(x). Then, is it true that the Riemann integral of f over>[a,b] equals the integral of g? Since the set of discontinuities of f>on [a,b] is at most countable and, therefore, has measure 0, it>follows f is certainly integrable over [a,b], but I couldn't come to a>conclusion if its integral needs to equal the integral of g or if it>depends on how we de?e f for the rationals of [a,b].In fact, your argument to prove that f is integrable over [a,b] is not> correct. Indeed, in order that a function de?ed on [a,b] is integrable> (in the sense of Riemann), the set of discontinuities must have measure> 0 *and* f must be bounded. However, if f is supposed to be bounded, your> argument is correct.Yes, sure. I forgot to say f was bounded. On the other hand, the integrals of f and g must be equal (still> assuming that f is bounded, of course), since the function f - g is> integrable and takes the value 0 at every irrational form [a,b].How can we prove this?Amanda === >On the other hand, the integrals of f and g must be equal (still>assuming that f is bounded, of course), since the function f - g is>integrable and takes the value 0 at every irrational form [a,b].> How can we prove this?ffffThm: If f is Riemann integrable on [a,b] and = 0 a.e., then int_[a,b] f(x) dx = 0.Proof sketch: Let eps > 0. Then there is an open set U in [a,b] such that m([a,b] U) = 0 and such that |f| < eps everywhere on U. Now [a,b] U is compact, and so can be covered with ?itely many disjoint open intervals, the sum of whose lengths is < eps. The endpoints of these intervals form a partition P of [a,b] such that the upper Riemann sum of |f| < eps*(b-a) + eps*M (or something like that). === Let's have a permutation:(3 5 1 2 4 6)After detaching into cycles we have:(1 3)(2 5 4)(6)The same with transpositions:(1 2 3 4) = (1 4)(1 3)(1 2)My question is how to translate these algorithms into anycomputer-understandable code. It's easy to transform it on a sheet of paper,but implementation seems a little bit harder for me.-- Pawe StobiskiRepublic of Poland === def cycles(permutation): n=len(permutation) accountedFor=[False]*n returnValue=[] while True: i=0 while i 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 === [something not in english!]>You know, whenever one posts source code in a NG that is not>comp.lang.thatlanguage or a subhierarchy of it, it would be a Very>Nice Thing(TM) to specify in which language it is written!It's Python. You can tell because it looks like pseudocodebut it's not. You know, nobody ever complains about not recognizing C.Things are gonna be different after the revolution...>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.miscDavid C. Ullrich === U[YAcute]ytkownik Elaine Jackson napisa w> HTHtranspositions disjoining (c++):Transposition: a class, with some methods and ?lds ( int* for permuation)void Perm::detachTranspositions(){ if (isIdent()) { cout<0) { if (tmp.p[pos-1] > tmp.p[pos]) { swap(tmp.p[pos-1],tmp.p[pos]); X1[--index] = pos-1; X2[index] = pos; } } } while(x != tmp.getPos(x)); } for(x=inv-1; x>=0; x--) cout<<[< Let's have a permutation: (3 5 1 2 4 6) After detaching into cycles we have: (1 3)(2 5 4)(6)I don't understand... Your cycle is already decomposed into a product ofdisjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).-- Maxi === > I don't understand... Your cycle is already decomposed into a product of> disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).>By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case thedecomposition holds.--@+ sur le forum fran.8dais !! === U[YAcute]ytkownik Julien Santini napisa w wiadomoci>I don't understand... Your cycle is already decomposed into a product of>disjoint cycles! (1 3)(2 5 4)(6) is not equal to (3 5 1 2 4 6).> By (3 5 1 2 4 6) he meant p(1)=3, p(2)=5, etc... in which case the> decomposition holds.Right, sorry for lack of any description to my question.-- ===