mm-319 Subject: Re: JSH: Basic result, test of current mathematics > Of course, it is possible that is referring to the technical > meaning of symposium (hence the use of that rather than > conference): symposium comes from the Greek symposion/sympinein, to > drink together. Hence a 'convival party', like the parties after > banquets in Greece... I can assure you that, in any talks involving Harris, there would be much drinking. === Subject: Re: JSH: Basic result, test of current mathematics > Although this is merely more tiresome Harris-bashing, it does raise > an interesting possibility. I have poin out to James the curious > coincidence that the 2005 Joint Mathematics Meeting is in his home > state of Georgia, suggesting that he might enjoy presenting his work > there. What hadn't occurred to me is that he might instead try to > organize a mini-symposium on the fundamental defect in the algebraic > integers, or on some other topic of interest to him. Count me in - I'm hoping to present two papers myself, but other than that my schedule is clear. === Subject: Re: JSH: Basic result, test REVISION 2 what theorem is that? (at this stage, I confuse every thing with Fermat's Little Theorem .-) > I think that this follows from the fact that if j is relatively prime to > p-1 then all numbers are jth residues mod p, so x^j == (-y)^j (mod p) =x == -y (mod p). I know that htis is handwaving, but I've a train to > catch. as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler (Don Fowler was the DNC Chair in '96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in '07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! thus saith: but, at the national level, will you be able to maintain the DNC's Any One But George -- unless it's Lyndon! media glut?... are you on the board of GOPpers For Howie Troisieme? http://www.wlym.com/pdf/iclc/communism.pdf --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: JSH: Basic result, test REVISION 2 ... > There may actually be an interesting theorem in here somewhere. > I found that, for j = 3, I could find lots of counterexamples > when f_1 and f_2 were primes congruent to 1 mod 6, but I couldn't > find any when either f_1 or f_2 or both were congruent to 5 mod 6. > Similarly, when j = 5, I could find counterexamples when > f_1 and f_2 were congruent to 1 mod 10, but not otherwise. Not > that I did really extensive searches. I don't see an immediate proof. I think that this follows from the fact that if j is relatively prime to > p-1 then all numbers are jth residues mod p, so x^j == (-y)^j (mod p) = x == -y (mod p). I know that htis is handwaving, but I've a train to > catch. [ I switch from y to -y, because it makes the exposition easier, and assume the primes, f1 and f2, are different.] It has more to do with the multiplicative group mod M (= f1*f2), of the elements coprime to M. It has order phi(M) = (f1 - 1)*(f2 - 1). So when j is a divisor of that number there are examples of x^j = y^j mod M (the order of the elements are divisors of the group order %). So a priory, when j = phi(M) all powers are equal mod M. An easy disproof of James' assertion that has already been given by somebody else. David is right in his assertion. When j is coprime to f2-1, and x != y mod f2, x^j != y^j mod f2, and so x^j != y^j mod M. So with j = 3 and f2 = 5 mod 6, all examples have x = y mod f2, and there are no counter-examples to James' assertion. So the actual theorem reads: Let's have M = f1.f2, with f1 and f2 distinctive primes. If x^j = y^j mod M, and j is coprime to (f1-1), x = y mod f1. If x^j = y^j mod M, and j is coprime to (f2-1), x = y mod f2. I *think* that when j is a divisor of both (f1-1) and of (f2-1), != 1, there are counterexamples, but I do not have an off-hand easy proof. It would be interesting when there are f1 and f2 such that there are no counterexamples in this case. There is an easy extension when M is the product of many different primes. When M is not square-free it becomes a bit more complica. ---- % This is a simple result from group theory. I think it was Euler who first has shown that x^phi(M) = 1 mod M when x coprime to M. This is an extension of Fermat's little theorem. -- === Subject: Re: JSH: Basic result, test REVISION 2 Adjunct Assistant Professor at the University of Montana. > There may actually be an interesting theorem in here somewhere. > I found that, for j = 3, I could find lots of counterexamples > when f_1 and f_2 were primes congruent to 1 mod 6, but I couldn't > find any when either f_1 or f_2 or both were congruent to 5 mod 6. > Similarly, when j = 5, I could find counterexamples when > f_1 and f_2 were congruent to 1 mod 10, but not otherwise. Not > that I did really extensive searches. I don't see an immediate proof. I think that this follows from the fact that if j is relatively prime to > p-1 then all numbers are jth residues mod p, so x^j == (-y)^j (mod p) = x == -y (mod p). I know that htis is handwaving, but I've a train to > catch. >[ I switch from y to -y, because it makes the exposition easier, and >assume the primes, f1 and f2, are different.] >It has more to do with the multiplicative group mod M (= f1*f2), of the >elements coprime to M. It has order phi(M) = (f1 - 1)*(f2 - 1). So when >j is a divisor of that number there are examples of x^j = y^j mod M (the >order of the elements are divisors of the group order %). So a priory, >when j = phi(M) all powers are equal mod M. An easy disproof of James' >assertion that has already been given by somebody else. >David is right in his assertion. When j is coprime to f2-1, and >x != y mod f2, x^j != y^j mod f2, and so x^j != y^j mod M. So with >j = 3 and f2 = 5 mod 6, all examples have x = y mod f2, and there are >no counter-examples to James' assertion. >So the actual theorem reads: > Let's have M = f1.f2, with f1 and f2 distinctive primes. Primes that wear really loud ties? (-: Oh, distinct. (-; > If x^j = y^j mod M, and j is coprime to (f1-1), x = y mod f1. > If x^j = y^j mod M, and j is coprime to (f2-1), x = y mod f2. >I *think* that when j is a divisor of both (f1-1) and of (f2-1), != 1, >there are counterexamples, but I do not have an off-hand easy proof. If j>1 is a divisor of f1-1, then there is an element a of order j in the multiplicative group modulo f1; likewise, there is an element b of order j in the multiplicative group modulo f2. That is, a^j = 1 (mod f1) and b^j = 1 (mod f2), neither a nor b congruent to 1 modulo the corresponding prime. Using the Chinese remainder theorem, we can find an element c modulo M such that c=a (mod f1) and c=b (mod f2). Therefore, c^j = a^j = 1 (mod f1) and c^j = b^j = 1 (mod f2); in particular, c^j = 1^j (mod M). However, c is not congruent to 1 modulo f1 (since it is congruent to a), and is not congruent to 1 modulo f2 (since it is congruent to b), giving a counterexample. If you want to avoid 1, it is not hard to do. Instead of picking to preimages of 1 in the raise to the j-th power map in the multiplicative group of integers modulo f1, pick some other j-th power and pick two noncongruent preimages; do the same modulo f2, and then use the Chinese Remainder Theorem to obtain elements x and y which have the right congruences modulo f1 and modulo f2. >It would be interesting when there are f1 and f2 such that there >are no counterexamples in this case. If both primes are odd, then of course j=2 will always divide both f1-1 and f2-1, so there will be counterexamples with j even. If you restrict to j odd, then you want gcd(f1-1,f2-1) = 2^r for some nonnegative integer r. If one of them is equal to 2, say f1, then every j is coprime to f1-1, so you will always fall within the clauses of the theorem. -- === Subject: Re: JSH: Basic result, test REVISION 2 >So the actual theorem reads: > Let's have M = f1.f2, with f1 and f2 distinctive primes. Primes that wear really loud ties? (-: Oh, distinct. (-; Reminds me to make English my mother tongue . >I *think* that when j is a divisor of both (f1-1) and of (f2-1), != 1, >there are counterexamples, but I do not have an off-hand easy proof. Thanks for the proof (and yes, I wan to avoid 1). James Dolan also mentioned the Chinese remainder theorem. Interesting though, it appears to be uncertain what time it dates from, but apparently the origin is indeed China. (I have seen 1247 and 4th century.) -- === Subject: generalization of a formula Good mornin/afternoon/evening/night there's a problem that i've never completely understood, it's about covariant differenziation, in particular: Let M be a differntiable manifold and f a differentiable real valued function over M. nabla will denote a linear connession over M, nabla : iTalicM times iTalicM --> iTalicM , where times denotes the cartesian product and iTalicM is the space of tangent vector fields over M, for example, given X in iTalicM and p in M, we have X(p) in T_p M , where T_pM is the tangent space of M at p. Let X, Y,...,X^ i ,.., Z, denote vector fields over M then it's obvious that ( nabla f ) X = df (X) (nabla^2 f ) (X, Y) = (nabla nabla f) (X, Y) = = ( nabla (nabla f) ) (X, Y) = = Y( Xf ) -(nabla_Y X) f Is there a closed formula which express (nabla^k f) (X^1 ,... , X^k) in term of X^1, ...X^k and nabla_(X^i) X^j etc? In other words i'd like a generalization of the two relations written above, May you help me please? Regards Tern ternnret@yahoo.it === Subject: Math Problem: Worker Efficiency Two employees are putting orders together for a distribution warehouse. Warren picks 47 orders in 7 hours. The total number of pieces he picks is 538. The total number of locations on his orders is 415. Preet picks 24 orders in 7.5 hours. The total number of pieces he picks is 1169. The total number of locations on his orders is 616. Which employee is more efficient and why? Please show the work. Thank-you, I am stuck with this one and need help. Coupon === Subject: Re: Math Problem: Worker Efficiency > Two employees are putting orders together for a distribution > warehouse. > Warren picks 47 orders in 7 hours. The total number of pieces he > picks is 538. The total number of locations on his orders is 415. > Preet picks 24 orders in 7.5 hours. The total number of pieces he > picks is 1169. The total number of locations on his orders is 616. > Which employee is more efficient and why? Please show the work. > Thank-you, I am stuck with this one and need help. Coupon I already answered this in alt.algebra.help. It is not useful to ask the same question multiple times. Bill === Subject: Re: Math Problem: Worker Efficiency Sorry, I didn't know. > Two employees are putting orders together for a distribution > warehouse. Warren picks 47 orders in 7 hours. The total number of pieces he > picks is 538. The total number of locations on his orders is 415. Preet picks 24 orders in 7.5 hours. The total number of pieces he > picks is 1169. The total number of locations on his orders is 616. Which employee is more efficient and why? Please show the work. Thank-you, I am stuck with this one and need help. Coupon I already answered this in alt.algebra.help. It is not useful to ask the same > question multiple times. > Bill === Subject: Re: Math Problem: Worker Efficiency > Two employees are putting orders together for a distribution > warehouse. > Warren picks 47 orders in 7 hours. The total number of pieces he > picks is 538. The total number of locations on his orders is 415. > Preet picks 24 orders in 7.5 hours. The total number of pieces he > picks is 1169. The total number of locations on his orders is 616. > Which employee is more efficient and why? Please show the work. Define 'efficient'; one such definition should be PiecesPicked * LocationsPicked / (OrdersPicked * timeTaken) [1] Or you could define efficiency as ordersPicked / timeTaken, which would produce a different answer. [1]: (avgPiecesPerOrder * avgLocationsPerOrder)*OrdersPicked / timeTaken -- === Subject: Re: Factoring Conjecture: Triplets >> [...] >> Wow! Another blow to the very foundations of mathematics by James! >Nope. I'm just wrong. No big deal. >> He has discovered a theorem (it must be a theorem, otherwise he would not >> have been able to construct a proof for it) to which you have, so it seems, >> been able to construct a counterexample! >Irrational. Time to tune up the irony detector... this is exactly as rational as the things you've said about flaws in the algebraic integers. >> There must be something SERIOUSLY wrong with CORE MATHEMATICS!! >You really need to get a life. >I had an idea, tossed it out, and it didn't fly. No big deal. Uh, not really. Starting a post with One hallmark of new research is the ability to answer questions that others might not have even considered. I've been playing with a result that I've mentioned before, but became curious enough to post it again, and see if I'm wrong in assuming that what's easy for me to prove, is not so easy for contemporary mathematicians using contemporary mathematics! is not tossing out an idea... >I wonder about some of you. [Too easy...] > ************************ === Subject: Re: Factoring Conjecture: Triplets > One hallmark of new research is the ability to answer questions that > others might not have even considered. I've been playing with a > result that I've mentioned before, but became curious enough to post > it again, and see if I'm wrong in assuming that what's easy for me to > prove, > If it is easy for you to prove, your methods are sorely > deficient. I had an idea. It turned out to be wrong. I had an approach that I thought lead to a proof, so I tossed the idea out there. I've done that many times before and described the process *many* times before. What I want reader to see now is the reality of posters like Nora Baron. > is not so easy for contemporary mathematicians using > contemporary mathematics! ... mainly because it's false ... I've pos that several times only to have to backtrack, and it may > be VERY annoying to some for me to say that, but the central point > that new research brings up interesting new results remains. > Hopefully I got it right finally. > Not yet. > Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integers, f_1, f_2, and integer M, where > M = f_1 f_2 > if you find integers x, y and z, where > x^j = y^j = z^j mod M > where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or > x = y mod f_2 or x = z mod f_2. > At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it. > You're partly right: no one can prove this conjecture with > contemporary mathematics. A conjecture means a *guess*, and part of what I did was save myself time. I'd thought about this conjecture for a while, and thought it was true. If my idea for a path to proof were correct, it'd be outside of the range of contemporary mathematics, but it turns out that I was just going down a wrong path. Now to me, that's not a bad use of the newsgroup!!! I saved myself time, and posters who replied made the decision to reply!!! I didn't make it for them. > Then again, maybe not, but I'm not going to get emotional about it. > Why not? Most people would be very upset to find that > what they very confidently believed to be proofs were in > fact repealy wrong. They would start to wonder if they > were losing it, or if they don't really understand contemporary > mathematics or what a proof actually is. Now I want readers to see what I've been talking about, and why I find posters like Nora Baron so, unsettling. This poster is now clearly engaged in more than just critiquing the idea. Clearly Nora Baron is about more than just trying to set me straight mathematically or question this or that mathematical position. This poster is out to GET ME, and it's something that I find, odd. Why? I put Nora Baron in quotes because the name is apparently a pseudonym, and it's not even clear that Nora Baron is even female. Who is this poster? Anybody know? I'm curious to know if anyone will even post to say that they know who this poster is, without revealing the name. I just wonder if *any* of you know who is the person behind the name. === Subject: Re: Factoring Conjecture: Triplets > One hallmark of new research is the ability to answer questions that > others might not have even considered. I've been playing with a > result that I've mentioned before, but became curious enough to post > it again, and see if I'm wrong in assuming that what's easy for me to > prove, > If it is easy for you to prove, your methods are sorely > deficient. > I had an idea. It turned out to be wrong. > I had an approach that I thought lead to a proof, so I tossed the idea > out there. You are trying to rewrite extremely recent history. You had an idea. You said you had a proof. You sneeringly issued a challenge to the rest of us, as though only you could prove it. It was not presen as a conjecture. It was presen as though you had absolute certainty about it. You said it was easy for you to prove: see above. Another point on this. It is clearly rela to your other failed conjecture/proofs in the last several days. You said they were rela to your work in polynomial factorization. I think not. This has almost nothing to do with polynomial factorization. This is rela to your fumbling around with the RSA challenge. Perhaps you hoped to gain some free help. > I've done that many times before and described the process *many* > times before. > What I want reader to see now is the reality of posters like Nora > Baron. What you see is what you get and ALL you get. > is not so easy for contemporary mathematicians using > contemporary mathematics! ... mainly because it's false ... I've pos that several times only to have to backtrack, and it may > be VERY annoying to some for me to say that, but the central point > that new research brings up interesting new results remains. Hopefully I got it right finally. Not yet. > Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integers, f_1, f_2, and integer M, where M = f_1 f_2 if you find integers x, y and z, where x^j = y^j = z^j mod M where j is a positive integer greater than 1, then it must be true that x = y mod f_1 or x = z mod f_1, or x = y mod f_2 or x = z mod f_2. At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it. You're partly right: no one can prove this conjecture with > contemporary mathematics. > A conjecture means a *guess*, and part of what I did was save myself > time. It wasn't posed as a conjecture. You said it was easy for you to prove. You implied that contemporary methods could not touch it, but *your* methods would work. Again you are trying to rewrite the record. > I'd thought about this conjecture for a while, and thought it was > true. And in fact you said you had an easy proof. That was untrue. You were premature in claiming it. Note that in this case a lot of people, not just me, found counterexamples. If your intent was to demonstrate the superiority of your methods and intellect, just the opposite happened. Any math undergrad could prove you wrong here. What should happen now is, you should say, Gee, maybe I'm not smarter than most of these math guys after all. I should stop being so contempuous. Right. > If my idea for a path to proof were correct, it'd be outside of the > range of contemporary mathematics, but it turns out that I was just > going down a wrong path. > Now to me, that's not a bad use of the newsgroup!!! So why present it as a sneering challenge, as if no one else could do it? Why not just present it as a question? As you sow, so shall you reap! > I saved myself time, and posters who replied made the decision to > reply!!! You could have saved yourself even more time by writing a little program to check it. > I didn't make it for them. You didn't want replies??? Or did you only want replies that (erroneously) agreed with you? Or did you want someone else to do the heavy lifting? There is no way to spin this that puts you in a good light. > Then again, maybe not, but I'm not going to get emotional about it. Why not? Most people would be very upset to find that > what they very confidently believed to be proofs were in > fact repealy wrong. They would start to wonder if they > were losing it, or if they don't really understand contemporary > mathematics or what a proof actually is. > Now I want readers to see what I've been talking about, and why I find > posters like Nora Baron so, unsettling. > This poster is now clearly engaged in more than just critiquing the > idea. > Clearly Nora Baron is about more than just trying to set me straight > mathematically or question this or that mathematical position. > This poster is out to GET ME, and it's something that I find, odd. > Why? Because you're a jerk. > I put Nora Baron in quotes because the name is apparently a > pseudonym, and it's not even clear that Nora Baron is even female. > Who is this poster? Anybody know? I'm curious to know if anyone will > even post to say that they know who this poster is, without revealing > the name. > I just wonder if *any* of you know who is the person behind the name. Credentials clearly mean a lot to you. You judge people by their credentials. I am taking that out of the equation. I make no appeals to authority or credentials. You have to judge me on what I say and can prove. Another point on this. In another thread (Basic etc. REVISION 2, I think) yesterday, I made a conjecture rela to what you star two days ago. There is actually a valid theorem growing out of what you star, but some key qualifications have to be added. 'David Einstein' replied with an incomplete idea for a proof, and Dik Winter replied with an actual proof. You totally ignored it. Over your head, possibly. If it ever gets published (not likely, it is too minor a theorem) you will get no credit for it: you didn't state the theorem and you didn't provide a proof. The point is, you are not actually interes in the math at all. All you care about is puffing up your own ego. Hence the sneering challenges, hence the re-writing of history as above to make yourself look less incompetent. But you never learn. Mistake after mistake after mistake, and you STILL have the delusion that we are all incompetent evil conspirators, and you are the misunderstood mutant boy genius. There is a word for people who cannot learn from mistakes but just continue to make the same ones over and over again. The word is ... stupid. > === Subject: Re: Factoring Conjecture: Triplets if we were to rewrite the History of the Ten-year Programme, we'd just replace all occurence of look at my fab, simple proof, with the C-word. of course, I'd also extirpate my own frigging name! > Another point on this. It is clearly rela to your other > failed conjecture/proofs in the last several days. You said > they were rela to your work in polynomial factorization. > I think not. This has almost nothing to do with polynomial > factorization. This is rela to your fumbling around with > the RSA challenge. Perhaps you hoped to gain some free help. > A conjecture means a *guess*, and part of what I did was save myself > time. > It wasn't posed as a conjecture. You said it was easy for > you to prove. You implied that contemporary methods could not > touch it, but *your* methods would work. Again you are trying > to rewrite the record. --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Factoring Conjecture: Triplets [snip] > I had an idea. It turned out to be wrong. > I had an approach that I thought lead to a proof, so I tossed the idea > out there. That characterization is not just dishonest, it is *blatantly* dishonest. You said you *had* a proof! You claimed the proof was simple for you, but was beyond others. This was false. > I've done that many times before and described the process *many* > times before. And you have been wrong every time. Furthermore, every time you have distor the facts in the hope that you could cast yourself in a better light. All it did was confirm your lack of integrity. You are a provable liar. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: Factoring Conjecture: Triplets >> One hallmark of new research is the ability to answer questions that >> others might not have even considered. I've been playing with a >> result that I've mentioned before, but became curious enough to post >> it again, and see if I'm wrong in assuming that what's easy for me to >> prove, >> If it is easy for you to prove, your methods are sorely >> deficient. >I had an idea. It turned out to be wrong. >I had an approach that I thought lead to a proof, so I tossed the idea >out there. You didn't say you had an idea you thought might lead to a proof, you said you had a proof. Then after people poin out that what you said was wrong you said you had a proof of a correc version. Which turned out to be wrong as well. >I've done that many times before and described the process *many* >times before. Yes. The process is you make a wild guess, probably you write down something that looks to you like a proof, and you announce that you have a proof that nobody else is smart enough to find. >[...] >If my idea for a path to proof were correct, it'd be outside of the >range of contemporary mathematics, Jesus. Even your _errors_ turn out to be indications that you're better than the rest of us at this stuff. Wow - one for the books. >but it turns out that I was just >going down a wrong path. >Now to me, that's not a bad use of the newsgroup!!! >I saved myself time, and posters who replied made the decision to >reply!!! And you also made a fool of yourself. Again. >I didn't make it for them. >> Then again, maybe not, but I'm not going to get emotional about it. >> Why not? Most people would be very upset to find that >> what they very confidently believed to be proofs were in >> fact repealy wrong. They would start to wonder if they >> were losing it, or if they don't really understand contemporary >> mathematics or what a proof actually is. >Now I want readers to see what I've been talking about, and why I find >posters like Nora Baron so, unsettling. >This poster is now clearly engaged in more than just critiquing the >idea. >Clearly Nora Baron is about more than just trying to set me straight >mathematically or question this or that mathematical position. >This poster is out to GET ME, and it's something that I find, odd. >Why? >I put Nora Baron in quotes because the name is apparently a >pseudonym, and it's not even clear that Nora Baron is even female. >Who is this poster? Anybody know? I'm curious to know if anyone will >even post to say that they know who this poster is, without revealing >the name. >I just wonder if *any* of you know who is the person behind the name. Why do you care? What does the identity of Nora Baron have to do with the validity of her comments? > ************************ === Subject: Re: Factoring Conjecture: Triplets > One hallmark of new research is the ability to answer questions that > others might not have even considered. I've been playing with a > result that I've mentioned before, but became curious enough to post > it again, and see if I'm wrong in assuming that what's easy for me to > prove, If it is easy for you to prove, your methods are sorely > deficient. I had an idea. It turned out to be wrong. > I had an approach that I thought lead to a proof, so I tossed the idea > out there. > I've done that many times before and described the process *many* > times before. > What I want reader to see now is the reality of posters like Nora > Baron. is not so easy for contemporary mathematicians using > contemporary mathematics! > ... mainly because it's false ... > I've pos that several times only to have to backtrack, and it may > be VERY annoying to some for me to say that, but the central point > that new research brings up interesting new results remains. Hopefully I got it right finally. > Not yet. > Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integers, f_1, f_2, and integer M, where M = f_1 f_2 if you find integers x, y and z, where x^j = y^j = z^j mod M where j is a positive integer greater than 1, then it must be true that x = y mod f_1 or x = z mod f_1, or x = y mod f_2 or x = z mod f_2. At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it. > You're partly right: no one can prove this conjecture with > contemporary mathematics. A conjecture means a *guess*, and part of what I did was save myself > time. > I'd thought about this conjecture for a while, and thought it was > true. > If my idea for a path to proof were correct, it'd be outside of the > range of contemporary mathematics, but it turns out that I was just > going down a wrong path. > Now to me, that's not a bad use of the newsgroup!!! > I saved myself time, and posters who replied made the decision to > reply!!! > I didn't make it for them. Then again, maybe not, but I'm not going to get emotional about it. > Why not? Most people would be very upset to find that > what they very confidently believed to be proofs were in > fact repealy wrong. They would start to wonder if they > were losing it, or if they don't really understand contemporary > mathematics or what a proof actually is. Now I want readers to see what I've been talking about, and why I find > posters like Nora Baron so, unsettling. > This poster is now clearly engaged in more than just critiquing the > idea. > Clearly Nora Baron is about more than just trying to set me straight > mathematically or question this or that mathematical position. > This poster is out to GET ME, and it's something that I find, odd. > Why? > I put Nora Baron in quotes because the name is apparently a > pseudonym, and it's not even clear that Nora Baron is even female. > Who is this poster? Anybody know? I'm curious to know if anyone will > even post to say that they know who this poster is, without revealing > the name. > I just wonder if *any* of you know who is the person behind the name. > But how do we know you or anyone else on sci.math is who they say they are? For all you know, I could be a hamburger flipper at McDonald's instead of a Meteorologist, like I've said I was in the past. === Subject: Re: Factoring Conjecture: Triplets > One hallmark of new research is the ability to answer questions that > others might not have even considered. I've been playing with a > result that I've mentioned before, but became curious enough to post > it again, and see if I'm wrong in assuming that what's easy for me to > prove, If it is easy for you to prove, your methods are sorely > deficient. > I had an idea. It turned out to be wrong. I had an approach that I thought lead to a proof, so I tossed the idea > out there. I've done that many times before and described the process *many* > times before. What I want reader to see now is the reality of posters like Nora > Baron. > is not so easy for contemporary mathematicians using > contemporary mathematics! > ... mainly because it's false ... > I've pos that several times only to have to backtrack, and it may > be VERY annoying to some for me to say that, but the central point > that new research brings up interesting new results remains. Hopefully I got it right finally. > Not yet. > Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integers, f_1, f_2, and integer M, where M = f_1 f_2 if you find integers x, y and z, where x^j = y^j = z^j mod M where j is a positive integer greater than 1, then it must be true that x = y mod f_1 or x = z mod f_1, or x = y mod f_2 or x = z mod f_2. At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it. > You're partly right: no one can prove this conjecture with > contemporary mathematics. > A conjecture means a *guess*, and part of what I did was save myself > time. I'd thought about this conjecture for a while, and thought it was > true. If my idea for a path to proof were correct, it'd be outside of the > range of contemporary mathematics, but it turns out that I was just > going down a wrong path. Now to me, that's not a bad use of the newsgroup!!! I saved myself time, and posters who replied made the decision to > reply!!! I didn't make it for them. > Then again, maybe not, but I'm not going to get emotional about it. > Why not? Most people would be very upset to find that > what they very confidently believed to be proofs were in > fact repealy wrong. They would start to wonder if they > were losing it, or if they don't really understand contemporary > mathematics or what a proof actually is. > Now I want readers to see what I've been talking about, and why I find > posters like Nora Baron so, unsettling. This poster is now clearly engaged in more than just critiquing the > idea. Clearly Nora Baron is about more than just trying to set me straight > mathematically or question this or that mathematical position. This poster is out to GET ME, and it's something that I find, odd. Why? I put Nora Baron in quotes because the name is apparently a > pseudonym, and it's not even clear that Nora Baron is even female. Who is this poster? Anybody know? I'm curious to know if anyone will > even post to say that they know who this poster is, without revealing > the name. I just wonder if *any* of you know who is the person behind the name. But how do we know you or anyone else on sci.math is who they say they are? > For all you know, I could be a hamburger flipper at McDonald's instead of a > Meteorologist, like I've said I was in the past. > People belong to groups . Mathematicians tend to hang out with other mathematicians. If Nora Baron is a mathematician, then it's quite likely that someone on the newsgroup knows who that poster is. I'm not asking anyone to reveal the poster, just to verify that the poster isn't totally anonymous. Yup, I'm trying to figure out what type of group Nora Baron is part of, because I'm somewhat concerned here. Subject: Re: Factoring Conjecture: Triplets === > >I just wonder if *any* of you know who is the person behind the name. >But how do we know you or anyone else on sci.math is who they say they are? >>For all you know, I could be a hamburger flipper at McDonald's instead of a >>Meteorologist, like I've said I was in the past. > People belong to groups . Mathematicians tend to hang out > with other mathematicians. > If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. > I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. > For reference, I only spend time around more than one other mathematician for about 3 days out of the year. I don't know if any of the mathematicians who know me would read this. I'm more likely to have a student read this. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Factoring Conjecture: Triplets > People belong to groups . Mathematicians tend to hang out > with other mathematicians. > If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. > I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. > Since JSH is unable to attack Nora's logic, he now wants information which will allow him to attack her person. Those who have been following JSH's disgraceful performances for a while may remember his spate of personal attacks on David Ullrich. JSH even star emailing and even phoning David's superiors at his place of work trying to cause trouble. Fortunately, the obvious flaws in JSH's personality made his efforts ineffectual. Now JSH want to start on Nora Baron because she has been so effective in showing him up for what he really is. === Subject: Re: Factoring Conjecture: Triplets > People belong to groups . Mathematicians tend to hang out > with other mathematicians. If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. Since JSH is unable to attack Nora's logic, he now wants information > which will allow him to attack her person. > Those who have been following JSH's disgraceful performances for a while > may remember his spate of personal attacks on David Ullrich. > JSH even star emailing and even phoning David's superiors at his > place of work trying to cause trouble. Fortunately, the obvious flaws in > JSH's personality made his efforts ineffectual. > Now JSH want to start on Nora Baron because she has been so effective in > showing him up for what he really is. I guess I better be careful. The place where I work is easy to find online. === Subject: Re: Factoring Conjecture: Triplets ... > Now JSH want to start on Nora Baron because she has been so effective in > showing him up for what he really is. I guess I better be careful. The place where I work is easy to find online. For me it does not matter. Both the place where I work and the place where I live are easy to find. They are in my signature. But I wonder what the FBI will do when James puts them onto me. -- === Subject: Re: Factoring Conjecture: Triplets > People belong to groups . Mathematicians tend to hang out > with other mathematicians. > If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. > I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. I take what Nora says, about her identity or anything else, at face value because she's demonstra herself to be trustworthy in the past. I distrust virtually everything you say, because you've demonstra yourself *repealy* to be *un*trustworthy. Nora clearly has shown that she's a mathematician with the quality of the work she's pos here. You have shown that you are *not* a mathematician in exactly the same way. -- rs, === Subject: Re: Factoring Conjecture: Triplets > People belong to groups . Mathematicians tend to hang out > with other mathematicians. If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. I think nobody in this newsgroup knows who I am. At least, except for Bob Silverman (who appears not to post anymore), I have met nobody from this group. And Bob Silverman only shortly when he visi my institute (I think not more than 5 minutes). > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. I take what Nora says, about her identity or anything else, at face > value because she's demonstra herself to be trustworthy in the past. You should not. Nora also makes errors. I also make errors. I think everyone in this newsgroup makes the occasional error. In most cases such an error is spot by another reader (Arturo is pretty good at spotting errors), and life goes on. But you'd better not spot an error irrelevant. > I distrust virtually everything you say, because you've demonstra > yourself *repealy* to be *un*trustworthy. That is too much. In general the algebraic manipulations (except for occasionally quite a few false starts) are correct. It is just the conclusions where James fails. > Nora clearly has shown that she's a mathematician with the quality > of the work she's pos here. She has shown that she is educa with quite a bit of mathematics, I do not know whether she is a mathematician or not (what makes one a mathematician?), but again, that is not so very interesting. Fermat also was not a mathematician by profession, it was just a hobby. -- === Subject: Re: Factoring Conjecture: Triplets > People belong to groups . Mathematicians tend to hang out > with other mathematicians. > If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. > I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. > I take what Nora says, about her identity or anything else, at face > value because she's demonstra herself to be trustworthy in the past. > I distrust virtually everything you say, because you've demonstra > yourself *repealy* to be *un*trustworthy. > Nora clearly has shown that she's a mathematician with the quality > of the work she's pos here. You have shown that you are *not* > a mathematician in exactly the same way. Yeah right. What else has Nora Baron pos? Where else has that poster made comments besides my threads? Do you know Wayne Brown? Or are you just babbling here? === Subject: Re: Factoring Conjecture: Triplets >> I take what Nora says, about her identity or anything else, at face >> value because she's demonstra herself to be trustworthy in the past. >> I distrust virtually everything you say, because you've demonstra >> yourself *repealy* to be *un*trustworthy. >> Nora clearly has shown that she's a mathematician with the quality >> of the work she's pos here. You have shown that you are *not* >> a mathematician in exactly the same way. > Yeah right. What else has Nora Baron pos? Where else has that > poster made comments besides my threads? > Do you know Wayne Brown? > Or are you just babbling here? Do I know Wayne Brown? Of course I do; he's me. Oh, I see, you're just indulging your semi-literate habit of addressing people by their full names (and omitting the requisite comma). Do you have any idea how ignorant that makes you appear? To answer your first question, about 30 seconds' searching revealed a couple: <36024859.0311141218.40039bef@posting.google.com> <36024859.0312121757.613e264f@posting.google.com> If you want more you can look them up yourself; far too often you depend on other people to do your work for you. But no references outside your threads are necessary. She's demonstra her competence *and* (BTW, just because you star some of those threads doesn't mean they're yours. Your USENET ignorance is showing once again.) In short, I trust Nora, and I don't trust you. If you took one of your idiotic newsgroup polls on this issue I'm certain you'd find an overwhelming majority who feel the same way. (Not that it means her math is infallible, of course; it just means I trust her not to lie about it, whereas I *expect* you to lie at every opportunity.) -- rs, === Subject: Re: Factoring Conjecture: Triplets > In short, I trust Nora, and I don't trust you. If you took one of > your idiotic newsgroup polls on this issue I'm certain you'd find an > overwhelming majority who feel the same way. (Not that it means her math > is infallible, of course; it just means I trust her not to lie about it, > whereas I *expect* you to lie at every opportunity.) If someone doesn't know the truth, is it possible for him to lie? Gib === Subject: Re: Factoring Conjecture: Triplets > One hallmark of new research is the ability to answer questions that > others might not have even considered. I've been playing with a > result that I've mentioned before, but became curious enough to post > it again, and see if I'm wrong in assuming that what's easy for me to > prove, If it is easy for you to prove, your methods are sorely > deficient. > I had an idea. It turned out to be wrong. I had an approach that I thought lead to a proof, so I tossed the idea > out there. I've done that many times before and described the process *many* > times before. What I want reader to see now is the reality of posters like Nora > Baron. is not so easy for contemporary mathematicians using > contemporary mathematics! > ... mainly because it's false ... I've pos that several times only to have to backtrack, and it may > be VERY annoying to some for me to say that, but the central point > that new research brings up interesting new results remains. > Hopefully I got it right finally. > Not yet. Following from my research on factoring polynomials into > non-polynomial factors I have a rather simple result that given prime > integers, f_1, f_2, and integer M, where > M = f_1 f_2 > if you find integers x, y and z, where > x^j = y^j = z^j mod M > where j is a positive integer greater than 1, > then it must be true that > x = y mod f_1 or x = z mod f_1, or > x = y mod f_2 or x = z mod f_2. > At this point I seriously doubt anyone can prove that conjecture with > contemporary mathematics, which is why I give it, as I think I can > prove it. > You're partly right: no one can prove this conjecture with > contemporary mathematics. > A conjecture means a *guess*, and part of what I did was save myself > time. I'd thought about this conjecture for a while, and thought it was > true. If my idea for a path to proof were correct, it'd be outside of the > range of contemporary mathematics, but it turns out that I was just > going down a wrong path. Now to me, that's not a bad use of the newsgroup!!! I saved myself time, and posters who replied made the decision to > reply!!! I didn't make it for them. Then again, maybe not, but I'm not going to get emotional about it. > Why not? Most people would be very upset to find that > what they very confidently believed to be proofs were in > fact repealy wrong. They would start to wonder if they > were losing it, or if they don't really understand contemporary > mathematics or what a proof actually is. > Now I want readers to see what I've been talking about, and why I find > posters like Nora Baron so, unsettling. This poster is now clearly engaged in more than just critiquing the > idea. Clearly Nora Baron is about more than just trying to set me straight > mathematically or question this or that mathematical position. This poster is out to GET ME, and it's something that I find, odd. Why? I put Nora Baron in quotes because the name is apparently a > pseudonym, and it's not even clear that Nora Baron is even female. Who is this poster? Anybody know? I'm curious to know if anyone will > even post to say that they know who this poster is, without revealing > the name. I just wonder if *any* of you know who is the person behind the name. > But how do we know you or anyone else on sci.math is who they say they are? > For all you know, I could be a hamburger flipper at McDonald's instead of a > Meteorologist, like I've said I was in the past. > People belong to groups . Mathematicians tend to hang out > with other mathematicians. > If Nora Baron is a mathematician, then it's quite likely that > someone on the newsgroup knows who that poster is. > I'm not asking anyone to reveal the poster, just to verify that the > poster isn't totally anonymous. > Yup, I'm trying to figure out what type of group Nora Baron is part > of, because I'm somewhat concerned here. > Ok, I'll agree with that to a certain extent. I think we tend to belong to groups as you suggest, however, not entirely. I have friends in fields TOTALLY different than mine. For instance, I have friends who are english majors and economists. Both of those are totally different from my field, meteorology. === Subject: Re: Factoring Conjecture: Triplets I didn't follow the conjecture. at least, though, he's actually using modular arithmetic, which I didn't think he'd bothered with. > Again there is no shortage of examples: > f_1 = 19, f_2 = 13, M = 247, j = 3: > x = 89 y = 86 z = 72 > x = 92 y = 81 z = 74 > x = 99 y = 85 z = 63 > x = 100 y = 92 z = 55 > x = 106 y = 83 z = 58 > x = 106 y = 96 z = 45 > x = 109 y = 97 z = 41 > etc., etc. > (There are 259 examples like > this with x, y & z <= 300.) as you know, if you've read anyhting that I typed, there was a conspiracy of states that DID vote Gore -- and the Supremes sealed that conspiracy on March 27, 2000, by refusing to hear the appeal in LaRouche v. Fowler. (Don Fowler was the DNC Chair in '96; Sentelle's 3-judge panel made the Voting Rights Act unconsitutitonal. but, hey; it's up for re-auhtorization in '07 !-) I am frankly scared of the touchscreen mentality. just like the Supremes' abrogation of the USA and Florida constitutions foments a pop-culture hatred of the electoral college on the part of some rabid Democrats. ah, so; imagine if North Dakota ... rather, imagine if Wyoming had less than one electoral college vote for president. anyway, every one of us in this debate knows about the Texas cirterium for chad -- it ain't just missing confetti! thus saith: but, at the national level, will you be able to maintain the DNC's Any One But George -- unless it's Lyndon! media glut?... are you on the board of GOPpers For Howie Troisieme? http://www.wlym.com/pdf/iclc/communism.pdf --Give the Gift of Dick Cheeny -- out of office, at last! http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac === Subject: Re: Factoring Conjecture: Triplets > Wow! Another blow to the very foundations of mathematics by James! Gawd they're still at it! === Subject: Re: Factoring Conjecture: Triplets > Why you're doing here the last 2 days just looks > like desperation. It's very easy to write a little > program to check on your conjectures for this kind > of thing. Why not do that before you post? You > just need the attention? You have to admit that one thing JSH is _very_ good at is getting attention. Something about his style makes it very difficult for mathematicians to ignore him. It reminds me of the classic male-female rolling disagreement (the marital kind that lasts years) where she is expressing her feelings and he tries to respond with logic; never the twain shall meet. (Yes, I'm married.) Gib === Subject: Re: Factoring Conjecture: Triplets [snip] > I had an idea, tossed it out, and it didn't fly. No big deal. No big deal, indeed! But this is classic REVISIONIST HISTORY, style!!! Note that you certainly did *not* just have an idea and tossed it out. You claimed you had a 'proof' and wondered if anyone else was as smart as you. Ha! > I wonder about some of you. Do you? Well, James, we no longer wonder about you. We *know*! > James Good-for-a-laugh Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: JSH: Basic result, test REVISED > |> f1=7,f2=13,j=3,x=53,y=75 > || > |Well I'm impressed you found that counterexample in 76 minutes! Could > |you explain how you did it to those of us who don't know where to start? > the chinese remainder theorem makes it easy to answer lots of > questions about arithmetic modulo a product of pairwise relatively > prime factors. i did use a calculator to check my calculations but > mostly i did them in my head and it only took a few minutes to do them > and anyone who used the same trick would probably find a solution > equally fast. > the other trick i used was that after deciding to try j=3 it seemed to > make sense to focus on primes of the form 6n+1 because cubing modulo a > prime of the form 6n+1 can be non-invertible. or something like that. Thanks. Also thanks to Christian - I had considered brute force search, but I guessed that it would take too long. Mark Atherton === Subject: A good basic calculus text? Anybody recommend a good basic calculus book for someone who left High School towards the end of the Eisenhower Administration. Back then I had algebra, geometry , and trig. Not much else and no math since..... === Subject: Re: A good basic calculus text? > Anybody recommend a good basic calculus book for someone who > left High School towards the end of the Eisenhower Administration. > Back then I had algebra, geometry , and trig. Not much else and no > math since..... I like anything by James Stewart. Howard Anton is good as well, but I prefer Stewart. === Subject: JSH: Questioning my conclusions I ended up in a discussion with a poster who seemed to believe that I don't consider the possibility that I'm wrong. I've seen that position put up quite a few times, and I think it's worth addressing, so I'll go back yet again to an example pos by a Rick Decker, a professor at Hamilton College, to go over what I say happens and consider how I might be wrong. Here's some identifying to find Decker's post, which also shows that he is indeed at Hamilton College: Subject: Re: Mathematical consistency, courage In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where I'll now make a slight change to have (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x) + 7(2) and his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). My emphasis has been on the actual constant terms as while it appears fomr that factorization that both are 7 on the left that doesn't add up with what's on the right, so I set x=0 to find that one of the a's equals 0, then, as a^2 + a = 0 and picking a_1(0) = 0, then a_2(1) = -1, so it makes sense I think now to introduce b_2(x), where a_2(x) = b_2(x) - 1 so that at x=0, b_2(0) = 0, and making that substitution gives (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) and now the constant terms match up. Now I've seen posters make a big deal out of even the phrase constant terms so I've now moved to getting rid of them!!! That's easy enough as all I have to do is multiply out to get 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) + 14 and not surprisingly the constant terms match up, so let's delete them off!!! That gives 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x). Now nothing spectacular so far, so let's move to where there's been EXTREME disagreement as my position is that from the constant terms it only makes sense that if you divide both sides of (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) by 7, you end up with (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 if you're to remain in the ring of algebraic integers. That's an important point. Multiplying out as before gives me 25 a_1(x) b_2(x)/7 + 10 a_2(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2 and again getting rid of the constant terms, I have 25 a_1(x) b_2(x)/7 + 10 a_2(x)/7 + 5 b_2(x) = 25x^2 + 30x and from before I had 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x) and dividing both sides of it gives 25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) = 25x^2 + 30x, so the answers the same. Now that might not impress some of you, so I'm going to let x=3, and consider what would happen if somehow there was a *different* way to divide both sides of (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) by 7, so let x=3, then (5a_1(3) + 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 2) and even if you believe that 7 splits up as a *function* of x to divide the left side, you should also believe that at x=3 its factors actually have some definite value. So let's use w_1(3), and w_2(3) for those factors where w_1(3) w_2(3) = 7, and dividing both sides gives (5a_1(3)/w_1(3) + w_2(3))(5b_2(3)/w_2(3) + 2/w_2(3)) = 25(3)^2 + 30(3) + 2 but now there's a problem as 2/w_2(3) isn't in the ring of algebraic integers if w_2(3) isn't a unit, as 2 and 7 are coprime. Now the way I see it w_2(3) and 2/w_2(3) are factors of 2 on the right side, so there's no way to argue with 2/w_2(3) being forbidden in the ring of algebraic integers, but some of you might wish to claim that you can escape having to have such a factor by concluding that (5b_2(3) + 2)/w_2(3) is in the ring. you can simply mush all your numbers together here and have it all work out in the end. That's actually kind of funny as a math position. I'll call it the mushing position. So then from my analysis, for me to be wrong, it must be true for 2/w_2(3) to be in the ring of algebraic integer when w_2(3) is not a unit, which is a contradiction. I just don't accept that mushing position as it's not mathematics. As a sidenote, when x=1, it IS the case that you're pushed out of the ring of algebraic integers, as w_1(1) = w_2(1) = sqrt(7). That looks like (5a_1(1)/sqrt(7) + sqrt(7))(5b_2(1)/sqrt(7) + 2/sqrt(7)) = 25(1)^2 + 30(1) + 2 and it turns out that Rick Decker's example is a special case where that happens with an integer x. Notice that everything follows from figuring out the factors of the constant term with (5a_1(3) + 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 7(2) as once they're identified, you can track what happens to them when you divide both sides by 7. The argument is simple enough that I find myself looking for *emotional* reasons to explain why people keep arguing with me about it. What's especially frustrating to me is that the argument is rather basic, but in response I see people who seem intent on bulldozing through with VERY LONG replies, where they usually try to claim there's some alternate argument supporting their position. Somehow I doubt it'll be different this time, but I like the exercise of going over the argument yet again. === Subject: Re: JSH: Questioning my conclusions > I ended up in a discussion with a poster who seemed to believe that I > don't consider the possibility that I'm wrong. I've seen that > position put up quite a few times, and I think it's worth addressing, > so I'll go back yet again to an example pos by a Rick Decker, a > professor at Hamilton College, to go over what I say happens and > consider how I might be wrong. > Here's some identifying to find Decker's post, which also shows that > he is indeed at Hamilton College: > Subject: Re: Mathematical consistency, courage > In his post Decker claimed to mirror my argument using a quadratic > instead of a cubic, where he has > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where I'll now make a slight change to have > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x) + 7(2) > and his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > My emphasis has been on the actual constant terms as while it appears > fomr that factorization that both are 7 on the left that doesn't add > up with what's on the right, so I set x=0 to find that one of the a's > equals 0, then, as > a^2 + a = 0 > and picking a_1(0) = 0, then a_2(1) = -1, > so it makes sense I think now to introduce b_2(x), where > a_2(x) = b_2(x) - 1 > so that at x=0, b_2(0) = 0, and making that substitution gives > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x) + 7(2) > and now the constant terms match up. Now I've seen posters make a big > deal out of even the phrase constant terms so I've now moved to > getting rid of them!!! > That's easy enough as all I have to do is multiply out to get > 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) + 14 = 7(25x^2 + 30x) + 14 > and not surprisingly the constant terms match up, so let's delete them > off!!! > That gives > 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x). > Now nothing spectacular so far, so let's move to where there's been > EXTREME disagreement as my position is that from the constant terms it > only makes sense that if you divide both sides of > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > by 7, you end up with > (5a_1(x)/7 + 1)(5b_2(x) + 2) = 25x^2 + 30x + 2 > if you're to remain in the ring of algebraic integers. No matter how many times you say obvious or stands to reason or only makes sense it is still true that, in the general case, this is simply false. > That's an important point. And an incorrect one. > Multiplying out as before gives me > 25 a_1(x) b_2(x)/7 + 10 a_2(x)/7 + 5 b_2(x) + 2 = 25x^2 + 30x + 2 > and again getting rid of the constant terms, I have > 25 a_1(x) b_2(x)/7 + 10 a_2(x)/7 + 5 b_2(x) = 25x^2 + 30x > and from before I had > 25 a_1(x) b_2(x) + 10 a_1(x) + 35 b_2(x) = 7(25x^2 + 30x) > and dividing both sides of it gives > 25 a_1(x) b_2(x)/7 + 10 a_1(x)/7 + 5 b_2(x) = 25x^2 + 30x, > so the answers the same. > Now that might not impress some of you, No, we can do simple algebra as well. > so I'm going to let x=3, and > consider what would happen if somehow there was a *different* way to > divide both sides of > (5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2) > by 7, so let x=3, then > (5a_1(3) + 7)(5b_2(3) + 2) = 7(25(3)^2 + 30(3) + 2) > and even if you believe that 7 splits up as a *function* of x to > divide the left side, you should also believe that at x=3 its factors > actually have some definite value. > So let's use w_1(3), and w_2(3) for those factors where w_1(3) w_2(3) > = 7, and dividing both sides gives > (5a_1(3)/w_1(3) + w_2(3))(5b_2(3)/w_2(3) + 2/w_2(3)) = 25(3)^2 + 30(3) > + 2 > but now there's a problem as 2/w_2(3) isn't in the ring of algebraic > integers if w_2(3) isn't a unit, as 2 and 7 are coprime. And we have had this many times before. Yes 2/w_2(3) is not an algebraic integer. No there is no reason why it should be. If a divides (b+c) there is no reason to conclude that a divides c. Example, 2 divides (3+5) but 2 does not divide 5. So from the fact that w_2(3) divides (5b_2(3) +2) we cannot conclude that w_2(3) divides 2. > Now the way I see it w_2(3) and 2/w_2(3) are factors of 2 on the right > side, so there's no way to argue with 2/w_2(3) being forbidden in the > ring of algebraic integers, but some of you might wish to claim that > you can escape having to have such a factor by concluding that > (5b_2(3) + 2)/w_2(3) is in the ring. > you can simply mush all your numbers together here and have it all > work out in the end. > That's actually kind of funny as a math position. I'll call it the > mushing position. That's right. There is no way I can find two algebraic integers b and c such that neither b nor c is divisible by 2 but when I mush b and c together the sum is divisible by 2. > So then from my analysis, for me to be wrong, it must be true for > 2/w_2(3) to be in the ring of algebraic integer when w_2(3) is not a > unit, which is a contradiction. Unless we accept the mushing position. > I just don't accept that mushing position as it's not mathematics. That's right. 2 does not divide (3+5). - William Hughes === Subject: Re: JSH: Questioning my conclusions so, when will your Marketing book be out? if these topics are like crack, then the Experience is a ring; eh? how am i going to break the habit !?! of going over the argument yet again. --ils duces d'Enron! http://tarpley.net/bush8.htm === Subject: Re: JSH: Questioning my conclusions [snip] Hi James, How does it feel to be a stupid idiot? I'd really appreciate your answer, because you are clearly the best example of one to ever grace this newsgroup. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- -- http://www.crbond.com === Subject: Re: JSH: Questioning my conclusions > [snip] > Hi James, > How does it feel to be a stupid idiot? ROTFLOL! :) What is this grade 4? > I'd really appreciate your answer, > because you are clearly the best example of one to ever grace this > newsgroup. Some friendly advice, he's not worth the stress... :) l8r, Mike N. Christoff === Subject: Re: Questioning my conclusions skrev i melding I snip all, I am tired of geting all the treads repea. Mr. Harris. How can you set x = 0 in the beginning of your proof, and still use x as a variable later? Is b(0) =0 or something else? Can b(0) be two things at the same time? And if they are, explain that! === Subject: Re: Questioning my conclusions > I ended up in a discussion with a poster who seemed to believe that I > don't consider the possibility that I'm wrong. I've seen that > position put up quite a few times I wonder why that is?