prof. Charles N. Moore ?
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From: Mark Murray on 27 Jun 2010 05:27 On 27/06/2010 05:33, JSH wrote: >> "But I am the top mathematician in the world." -- James S. Harris > > Might have been drunk. I don't remember saying that at all. A paraphrase of one of your Twitter remarks is "I must admit that a lot of my math is beer-fueled". > I am NOT a mathematician. You claim to be a discoverer of things mathematical. http://en.wikipedia.org/wiki/Mathematician Your primary area of "research" is mathematics. Now read the above link. I quoted "research", because I'd prefer the term "speculation", as you do very little actual research, and a great deal of wild conclusion- jumping. Applying a bit of logic, I'd conclude that you are not a mathematician, but not simply because you say so. >> Which is pretty close to what Tommy wrote. > > I'll give you that but I won't stand by any such idiot utterance, even > if it came from me at some time in the past. It's just not true. I'm > not even a mathematician. > > I have done drunk postings. At the time they seem hilarious. Given your own self-inconsistence, and your obvious inability to recall what you have previously said, I'd have to agree. Now do you understand why so many people don't take you seriously and read your posts for the humour value? M -- Mark "No Nickname" Murray Notable nebbish, extreme generalist.
From: Jesse F. Hughes on 27 Jun 2010 08:56 JSH <jstevh(a)gmail.com> writes: > On Jun 26, 5:10 pm, "Jesse F. Hughes" <je...(a)phiwumbda.org> wrote: >> JSH <jst...(a)gmail.com> writes: >> > On May 3, 12:45 pm, master1729 <tommy1...(a)gmail.com> wrote: >> >> tommy1729 >> >> >> " but it has to be true , because i am the world's top mathematician " JSH >> >> > I never said that. >> >> But you *did* say this: >> >> "Is that possible? Could it be that easy? No way. [...] There must be >> a mistake. Right? >> >> "But I am the top mathematician in the world." -- James S. Harris > > Might have been drunk. I don't remember saying that at all. You wrote it. According to my official archives, it occurred in you blog in December, 2005. > > I am NOT a mathematician. > >> Which is pretty close to what Tommy wrote. > > I'll give you that but I won't stand by any such idiot utterance, even > if it came from me at some time in the past. It's just not true. I'm > not even a mathematician. > > I have done drunk postings. At the time they seem hilarious. I am pretty sure that we all know you've posted drunk. But the blog posting I mention was evidently not done for humor's sake, even though I expect you were drinking at the time. Here's a trip down memory lane for you. (Notice: you edited this blog posting a couple of times, and so the .sig quote appears only as one of a few endings.) At the time, I was an avid reader of your fine blog. ---------------------------------------------------------------------- Counterpoint: Trying to simplify And I do like hyperbole, clearly, from my last post, but I also like to attack my own ideas, as the way I see it, if they're correct, then there are no worries, they will stand. But if they don't, then it's time to edit! I'll gleefully delete posts. The other thing is this result is starting to bug me in terms of the questions that remain about it. Even if I crush it by finding an error that at least is a resolution. But if it's right... Now then, I did try to simplify an earlier, more complex derivation of a general algebraic factorization, but never did with the latest version I've talked about recently, so now I'll do so, and see if it survives, or if it'll be time to delete back: T = (x-n+vz)(vz-x) when k2z2 + nx - x2 = T and z = nv/(v2 - k2) where v does not equal sqrt(k2) to prevent a problem with division by zero. First step is to let f1 = x-n+vz and f2 = vz-x where, of course, T = f1 f2, and now I can just subtract one from the other to eliminate vz, in my first go at it and get f1 - f2 = 2x-n and using f2 = T/f1, I can substitute to get f1 - T/f1 = 2x-n and multiplying both sides by f1 gives f1 2 -(2x-n)f1 - T = 0 so I can solve to get f1 = ((2x-n) +/- sqrt((2x-n)2 + 4T))/2 and obviously 2x-n is some difference of factors of T, but, it turns out that you have degrees of freedom so it's not clear what x and n are at this point. That is, they can still be anything. In thinking about it that result is not at all surprising but is just repeating what follows before, but with the twist that it's clear that choosing x and n determines all the remaining variables in the system. OOPS! That's wrong. Turns out that despite having three equations with 6 variables you have 4 degrees of freedom in one of those fascinating aspects to these equations. I do need to re-read my own research as I think I've noted that before. It also makes it clear that any factorization over complex numbers is available. You can, pick n or you can pick x, and you can pick n and x at this point, as you have three degrees of freedom. You can also add f1 to f2 and get f1 + f2 = 2vz-n and again making the substitution for f2 you get f1 + T/f1 = 2vz-n and multiply both sides by f1 and simplify a bit to get f1 2 -(2vz-n)f1 + T = 0 and solve to find f1 = ((2vz-n) +/- sqrt((2vz-n)2 - 4T))/2 and, again, because of those degrees of freedom, you can just pick, but then again, what if I assume you don't have T factored? I need to go to the variables I used before and that means k2 and v as the ones to be picked, which will leave one free variable, when T is chosen. And that makes substituting out z, with z = nv/(v2 - k2) the logical choice and still wanting to just try things I'll substitute it out with f1 2 -(2vz-n)f1 + T = 0 to get f1 2 -(2nv2/(v2 - k2)-n)f1 + T = 0 and multiplying both sides by (v2 - k2) gives (v2 - k2)f1 2 -(2nv2 - n(v2 - k2))f1 + T(v2 - k2) = 0 and solving now for f1 gives f1 = ((2nv2 - n(v2 - k2)) +/- sqrt((2nv2 - n(v2 - k2))2 -4T(v2 - k2)2))/2(v2 - k2) which I hope is right, and now there's that complexity that I remember from before jumping in, so that it's kind of painful just to look at it all. Note that if v and k2 are chosen then there is still one more variable to pick, before all the degrees of freedom are used up. Goodness. Now I kind of remember why I stepped away from analysis on this system. Irreducible complexity indeed. Head hurts. Time to do something else for a while. Or what if you just use a trivial factorization of T and then pick vz and n with it? Then you get a trivial factorization, as doing so is directly picking f1 and f2 to be trivial. What's more interesting is getting solutions to the system through the other variables, which must be where I got quintics before. I think I would set k2 and v, which I can check by going to earlier posts. Just sort of muddling around here. But that's kind of what you do, just sort of try things until something breaks. Well that's enough for that go at it, as I'm getting a bit bored, and maybe somewhat frustrated. How did I get quintics before? I guess I could look back. Don't worry, I won't give up completely at this point, but will need to come back, and keep fiddling with this thing, which I can do by just coming back to edit this post, and see if I can find a way to simplify the system. As, the nice thing about making a mathematical discovery is that once you have it, you can spend as much time as you like trying to tear it down. Math discoveries are wonderfully tolerant to attack. If you believe you have a real result, attack it! It's great fun attacking your own work, and can be almost as satisfying as finding it, as there is the suspense... What if you ARE wrong? What if you just screwed up and missed something? Then eventaully as the assault moves forward, eventually you may find the key which tears down the entire house of cards, and then, what then? You get to start over, looking for an answer. The process never ends, unless you prove a result absolutely by relying on a complete proof, but I tend to save that for last, and drag out the fun, if I bother to do it at all, which I've done with other research. After all, it's my math. And my math is my tool for my needs. I can do with it what I want. [[Alternate ending: I need to check to be sure. But if that's correct, it's so bizarre that it can be that simple, as then you need just pick v and n and then solve for k2 and you would have 50% chance of factoring any composite T, which means, if this is right, that right there, in a freaking blog, is the secret to breaking RSA encryption. No way. Is that possible? Could it be that easy? No way. If it were, no way government agencies in the United States would just let me stick it on some freaking blog!!! No way. There must be a mistake. Right? But I am the top mathematician in the world. Other mathematicians have for some stupid reason decided to try and fight me, but it seems to me that if they wish to end the world, there's no reason for me to stop them. Just remember, they lied. Not me. If you want to kill someone, look at the people who really betrayed you. ]] [[ Alternate endings: But, of course, how can that be true? I'll re-check the equations later. And, if I really screwed up, I will delete back. And how could I be right? After all, ending the Internet as it currently operates, isn't the sort of thing that some crackpot can do, and mathematicians have clearly left me with that label, so that you know not to listen to me. ]] James Harris -- Jesse F. Hughes "You will never manage to completely convince me that Jews weren't at least a LITTLE bit guilty of irrationnal anti-nazi bias." -- Attributed to humorist Pierre Desproges
From: master1729 on 29 Jun 2010 03:25
JSH : > On May 3, 12:45 pm, master1729 <tommy1...(a)gmail.com> > wrote: > > Gerry Myerson wrote : > > > > > > > > > In article > > > > <a578e10d-30b9-4074-94cd-72fc8e8c1...(a)r11g2000yqa.goog > > > legroups.com>, > > > Tonico <Tonic...(a)yahoo.com> wrote: > > > > > > About the book by Underwood Dudley: I don't > have > > > it. > > > > > The story is on pages 257-258 of that book. > Dudley > > > has an undated > > > newspaper clipping reporting that Moore presented > a > > > proof at an > > > Amer Math Soc meeting in Wellesley, > Massachusetts. > > > Other evidence > > > indicates the clipping is from a midwestern > newspaper > > > during the > > > Second World War. > > > > > Maybe someone has tracked things down and told > Dudley > > > more > > > details. I suppose anyone who really wanted to > know > > > could ask Dudley. > > > > > -- > > > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for > > > email) > > > > so , is it a blunder of dudley , > > There was someone. I remember reading about him > online years ago when > I first decided that twin primes were governed by a > random process. I > remember seeing several sources online talking about > a mathematician > who made that claim, who was pushed out of the > mainstream for it. name = ?? i believe you , but i would still like a name if you remember. > > But recently I could find nothing. NOTHING. > > It's like he got wiped from the face of the earth, > except maybe for > that Dudley reference. he should make a website , that forces him to be found on google. i did the same ... lots of work to do ... maybe a wiki ... the paradox is , you need to give info for free to sell it. like the backcover of a book , but much worse ! > > The main reason for math people to block a random > explanation for twin > primes is money. > > There's a lot more funding with it being an "open" > problem. i guess i got the andrica equivalent of that. i could sell you andrica ... intrested ? > > > > > or was someone of the newspaper taking drugs ? > > > > i still say its a conspiracy ! :) > > > > those good mathematicians are just made-up persons > , just like JSH told us. :p > > If an established mathematician presents positions > well against the > mainstream he can simply be derided. For people who > do not need > funding--I'm entirely self-funded and an NOT a > mathematician--that is > not a way to produce pressure. yes , but this time its just a lie/blunder of the book writer and/or media rather i guess. > > It happens in other fields. In astronomy, Dr. Halton > Arp ended up > fleeing the country to Germany, when he could no > longer get telescope > time here. > true. same with ' green ' science and meds. together with deadline principles its a disaster ! > It's a weakness of established researchers--their > need for funding > from the establishment. > > Private researchers do not have that weakness. indeed. i dont have a weakness. > > > > > tommy1729 > > > > " but it has to be true , because i am the world's > top mathematician " JSH > > I never said that. i got it from jesse f hughes ... and i believe you did ... oh well nevermind. > > > James Harris tommy1729 |