mm-179 I think I've figured out a way to show basically all of you, including> people who think they don't know any math that mathematicians have> been lying about my work. It's so trivial you *should* wonder why> they thought they could get away with it....> That is, MOST of you will doubt algebra itself rather than consider> that mathematicians, whom you probably don't even personally know,> would lie....> Mathematicians *are* a part of society after all. Why should they> tell the truth now? It'd be like Bush owning up. They can just sit> tight, and be quiet, like so many American citizens or they can out> and out lie, like so many other patriots.James,What are you? A mathematician? That is the name reserved for thosewith an understanding of mathematics. If you are not a mathematician,you in fact do not understand what exactly you are talking about. Instead, you are a magician who pulls numbers out of his hat.If you are in fact a mathematician, wouldn't people be following yourown advice by not listening to you? In such case, you are complainingabout yourself.Deny that you are a mathematician, and admit that you do notunderstand what you are talking about. Otherwise, stop attacking ayourself.-Chris Pitman =I'm saving this to print it out, later,in order to lend some credence to my generally non-analytical commentsto your alleged & continuously-improved proof(of what I refer to as the plausibly 'rst major resultof la noodling de Fermat .-) otherwise,I've come to the conclusion that you are just a big joker,considering the soap-opera of your collected works,as recently posted on another thread --where they threatened you with some hardcore (so-called) elliptical forms & their semistability ... andyou ran-off with your tail between your legs(not that I understood too much of it, either .-)... either that,or you're some sort of rogue agent or other,'shing-about for some ridiculous clue of your fancy (orthat of yo'boss); eh?on the other hand, if you're really serious about this,the only dysaster taht could conceivably result is, a)you'd have a temporary loss of self-esteem,in being denied this woderful attention from a tiny groupof sycophants (why else would we waste our time, herein?), or b)you'd miraculously 'nd one of the 'rst billion-or-so proofs! > My paper Advanced Polynomial Factorization depends on considering a> factor of a polynomials that I call g.> > (Paper linked to at http://groups.msn.com/AmateurMath as usual.)> > And in my paper I start by showing that I can write that as> > g = r + c> > where either r=0, or r changes as the polynomial's value changes,> while c does not.> > Now you can consider all factors of a given polynomial using g's, with> something like> > g_1...g_k = P(x)> > where you have k factors. For instance, for P(x)=x^2 + 2x + 1,> > g_1 = x+1, g_2 = x+1, gives you 2 factors.> > Those are polynomial factors, but I'm generalizing in a simple way to> say that for the factors g, in general, you have an element I call r,> which changes as the independent variable changes, and you have> another element I call c, which does not.> > For my example up above it's easy, as with g_1 = x + 1, x varies as x> varies, while 1 does not.> > Now that's enough that the proof in the paper is straightforward, but> posters have argued with me anyway, with some trying to argue over the> de'nition of polynomial, amazingly enough. However, consider that the g's have an important feature, which is> that when x=0, I have> > g_1...g_k = P(0). For instance, with my simple example, with P(x) = x^2 + 2x + 1, with> x=0,> > P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.> > You see, P(0) gives the constant term, so at x equal 0, the g's must> multiply to give the constant term.> > So then, maybe you still want to believe the mathematicians and> question that I can write g = r + c. Well consider that substituting gives me> > g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives> > r_1...r_k +...+c_1...c_k = P(x), which is> > r_1...r_k +...+P(0) = P(x), > > which means that if you believe the mathematicians then they've> convinced you to doubt algebra itself, as then you must believe that> everything to the left of P(0) above can *maybe* be constant, but also> *maybe* vary as x varies.> > So why would mathematicians argue against such a simple result?> > Two reasons I suggest. First because they wish to disagree with me. > Second because they probably believe that they can get away with it.> > That is, MOST of you will doubt algebra itself rather than consider> that mathematicians, whom you probably don't even personally know,> would lie.> > So where does this lead?> > Well the polynomial I show in the paper is P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf> > which seems to be just complicated enough to give mathematicians room> to lie.> > For instance, you may be saying, HEY, what's with the .95m' when you> had .95x' before??!!!> > Well, there's no rule that says that you have to use the letter x as> the variable label for a polynomial. Also, there are historical> reasons for my usage as it goes back to my work with FLT where x, y> and z are used with x^p + y^p = z^p.> > Finally, the weirder thing is that one poster in particular got a lot> of mileage out of questioning my 'nding the constant term with an> expression like the above by using m=0, as that gives me> > P(0) = 3xy^2 + y^3> > and he got a lot of mileage for YEARS (before I had discovered proof I> want to add and after) by emphasizing that two of the ROOTS of such an> expression considered as a polynomial with respect to x are not> de'ned at that point.> > Well that's easy enough to see as the original expression is> > (v^3+1)x^3 - 3vxy^2 + y^3> > which if you *wish* to see it as a polynomial with respect to x, is of> degree 3, but when v=-1, it's of degree 1, so if you solve for the> roots, you'll get funky stuff. > For mathematicians the situation could be considered one of the worst> possible disasters they can imagine.--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:(FOSSILISATION [McCainanites?] (TM/sic))/BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/ bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 23 -- Le FIN d'HISTOIRE 24 -- L'ORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! =Go home, little man...> I think I've 'gured out a way to show basically all of you, including> people who think they don't know any math that mathematicians have> been lying about my work. It's so trivial you *should* wonder why> they thought they could get away with it.>> Here goes.>> My paper Advanced Polynomial Factorization depends on considering a> factor of a polynomials that I call g.>> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)>> And in my paper I start by showing that I can write that as>> g = r + c>> where either r=0, or r changes as the polynomial's value changes,> while c does not.>> Now you can consider all factors of a given polynomial using g's, with> something like>> g_1...g_k = P(x)>> where you have k factors. For instance, for P(x)=x^2 + 2x + 1,>> g_1 = x+1, g_2 = x+1, gives you 2 factors.>> Those are polynomial factors, but I'm generalizing in a simple way to> say that for the factors g, in general, you have an element I call r,> which changes as the independent variable changes, and you have> another element I call c, which does not.>> For my example up above it's easy, as with g_1 = x + 1, x varies as x> varies, while 1 does not.>> Now that's enough that the proof in the paper is straightforward, but> posters have argued with me anyway, with some trying to argue over the> de'nition of polynomial, amazingly enough.>> However, consider that the g's have an important feature, which is> that when x=0, I have>> g_1...g_k = P(0).>> For instance, with my simple example, with P(x) = x^2 + 2x + 1, with> x=0,>> P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.>> You see, P(0) gives the constant term, so at x equal 0, the g's must> multiply to give the constant term.>> So then, maybe you still want to believe the mathematicians and> question that I can write g = r + c.>> Well consider that substituting gives me>> g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives>> r_1...r_k +...+c_1...c_k = P(x), which is>> r_1...r_k +...+P(0) = P(x),>> which means that if you believe the mathematicians then they've> convinced you to doubt algebra itself, as then you must believe that> everything to the left of P(0) above can *maybe* be constant, but also> *maybe* vary as x varies.>> So why would mathematicians argue against such a simple result?>> Two reasons I suggest. First because they wish to disagree with me.> Second because they probably believe that they can get away with it.>> That is, MOST of you will doubt algebra itself rather than consider> that mathematicians, whom you probably don't even personally know,> would lie.>> So where does this lead?>> Well the polynomial I show in the paper is>> P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf>> which seems to be just complicated enough to give mathematicians room> to lie.>> For instance, you may be saying, HEY, what's with the .95m' when you> had .95x' before??!!!>> Well, there's no rule that says that you have to use the letter x as> the variable label for a polynomial. Also, there are historical> reasons for my usage as it goes back to my work with FLT where x, y> and z are used with x^p + y^p = z^p.>> Finally, the weirder thing is that one poster in particular got a lot> of mileage out of questioning my 'nding the constant term with an> expression like the above by using m=0, as that gives me>> P(0) = 3xy^2 + y^3>> and he got a lot of mileage for YEARS (before I had discovered proof I> want to add and after) by emphasizing that two of the ROOTS of such an> expression considered as a polynomial with respect to x are not> de'ned at that point.>> Well that's easy enough to see as the original expression is>> (v^3+1)x^3 - 3vxy^2 + y^3>> which if you *wish* to see it as a polynomial with respect to x, is of> degree 3, but when v=-1, it's of degree 1, so if you solve for the> roots, you'll get funky stuff.>> Now when I was 'nding the proof of FLT...remember the process took> some years...at times I'd talk of polynomials with respect to other> than m, but I re'ned my discourse as my understanding improved.>> However, people arguing with me did not.>> You may *choose* to believe that they did not because they don't know> enough mathematics to follow, but we're talking about actual> mathematicians here.>> What's more rational?>> I say it's more rational to suppose that they *did* 'gure out that it> worked as described, but also noticed that as long as they disagreed,> no one seemed to call them on making false statements, except me, and> they knew my credibility wasn't so great.>> For most of you, there's probably the belief that there's some funky> higher math involved that your pitiful brain can't follow or you> don't know about, as you may suppose that mathematicians either> wouldn't lie, or they wouldn't lie in such a dumb way where I could> catch them so easily.>> But consider what's in the balance:>> 1. I discovered a proof of Fermat's Last Theorem that's more> available to people in general than most of what mathematicians have> been producing lately.>> 2. Worse I did so having said I'd 'nd it years ago, and having spent> years looking for it posting a lot of my ideas, and getting in insult> battles with posters, quite a few who happened to be mathematicians.>> 3. Then to add insult to injury, I keep questioning mathematicians in> terms of their ethics, and maybe *extremely importantly* I doubt that> Wiles found a proof of Fermat's Last Theorem.>> And those are just highlights as there's more but I think that kind of> gives my point.>> For mathematicians the situation could be considered one of the worst> possible disasters they can imagine.>> EXCEPT, it looks like all they have to do is either stay quiet, or> *claim* I'm wrong.>> Many of you simply believe them, and question algebra itself, which is> rather sad. I'd think at least some of you valued your educations.>> Others of you may 'gure it doesn't matter, maybe because Western> civilization seems to be based on lying anyway, and maybe you 'gure I> should just grow up, accept that everybody lies and move on. And I> don't have to talk about Enron or pedophile Catholic priests or things> in that vein.>> I mean, look at George W. Bush and Iraq. If people can be *killed*> over lies, without consequences to the liars, then what's with some> freaking stupid math?>> Good point.>> Mathematicians *are* a part of society after all. Why should they> tell the truth now? It'd be like Bush owning up. They can just sit> tight, and be quiet, like so many American citizens or they can out> and out lie, like so many other patriots.>> After all, that's so easy, now isn't it?>> Which is why you need to understand why I talk about mathematicians> potentially being prosecuted. Liars don't just stop because the gig> is up, as then, they wouldn't necessarily be liars, then eh?>> It'd be against their *true* natures.>>> James Harris =[snip preamble]> Well consider that substituting gives me> > g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives> > r_1...r_k +...+c_1...c_k = P(x), which is> > r_1...r_k +...+P(0) = P(x), > > which means that if you believe the mathematicians then they've> convinced you to doubt algebra itself, as then you must believe that> everything to the left of P(0) above can *maybe* be constant, but also> *maybe* vary as x varies.How do you mean?The stuff to the left of P(0) appears to ber_1...r_k + c_1.r_2...r_k + r_1.c_2.r_3...r_k + (lots of other terms involving r_i and c_i)In general, this is going to vary with x. That seems reasonable enough,because the r_i can vary with x. For all x, the value of this expressionwill be the same as P(x)-P(0).It's only going to be constant if P(x) is constant. In which case,either all the g_i are constant, or you've chosen some non-polynomialfunction for at least one of the g_i.e.g. for P(x)==1, you might have g_1 = x^2+1, g_2 = 1/(x^2+1)There doesn't seem to be any problem with this.> So why would mathematicians argue against such a simple result?> > Two reasons I suggest. First because they wish to disagree with me. > Second because they probably believe that they can get away with it.> > That is, MOST of you will doubt algebra itself rather than consider> that mathematicians, whom you probably don't even personally know,> would lie.> > So where does this lead?> > Well the polynomial I show in the paper is> > P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf> > which seems to be just complicated enough to give mathematicians room> to lie.I suppose this must have been discussed before, but I don't like yournotation here: you seem to want to talk about a polynomial in m whosecoef'cients are functions of x, y and f. To make this clearer maybeyou could writeP(x,y,f)(m) = (v^3+1)x^3 - 3vxy^2 + y^3or start by de'ningQ(m,x,y,f) = (v^3+1)x^3 - 3vxy^2 + y^3and then say considering Q as a polynomial in m...> For instance, you may be saying, HEY, what's with the .95m' when you> had .95x' before??!!!No, that's 'ne. Although your post would be easier to read if youdidn't change variable half way through.> Well, there's no rule that says that you have to use the letter x as> the variable label for a polynomial. Also, there are historical> reasons for my usage as it goes back to my work with FLT where x, y> and z are used with x^p + y^p = z^p.> > Finally, the weirder thing is that one poster in particular got a lot> of mileage out of questioning my 'nding the constant term with an> expression like the above by using m=0, as that gives me> > P(0) = 3xy^2 + y^3> > and he got a lot of mileage for YEARS (before I had discovered proof I> want to add and after) by emphasizing that two of the ROOTS of such an> expression considered as a polynomial with respect to x are not> de'ned at that point.> > Well that's easy enough to see as the original expression is> > (v^3+1)x^3 - 3vxy^2 + y^3> > which if you *wish* to see it as a polynomial with respect to x, is of> degree 3, but when v=-1, it's of degree 1, so if you solve for the> roots, you'll get funky stuff.Well, goodness knows why you were arguing.But you shouldn't say as a polynomial with respect to x, [it] is ofdegree 3, since (as you go on to explain) there are exceptions.A true statement is, as a polynomial in three variables (v, x and y),the highest power of x appearing is 3. It follows that, as a polynomialwith respect to x, it has degree *at most* 3.More speci'cally,When v =/= -1, it has degree 3.When v = 1 and y =/= 0, it has degree 1.When v = 1 and y = 0 it is the zero polynomial.If you're going to be pedantic, which is a virtue, you shouldremember the last case as well.-- David CollierHampshire, UK => Go home, little man...This is the only home he has...V.-- => I've looked at your paper... now looking at this I'll bite. I may be > getting in over my head but I'll point out the things that don't make > sense to me. I'm open to being enlightened. I think I've 'gured out a way to show basically all of you, including> people who think they don't know any math that mathematicians have> been lying about my work. It's so trivial you *should* wonder why> they thought they could get away with it.> > Either you aren't being clear, or they missed something. It is a strong > charge to suggest that ALL mathematicians are in this great conspiracy > to teach incorrect math. Sort of goes against what they stand for.I say mathematicians but don't speci'cally say ALL mathematicians.I'm talking about a pattern of denial that I'm seeing inmathematicians that I've contacted or observed. > Here goes.> > My paper Advanced Polynomial Factorization depends on considering a> factor of a polynomials that I call g.Should be factor of a polynomial, and not polynomials.> > (Paper linked to at http://groups.msn.com/AmateurMath as usual.)> > And in my paper I start by showing that I can write that as > g = r + c> > where either r=0, or r changes as the polynomial's value changes,> while c does not.> > Questions: is g the polynomial to be factored or a factor? Is r a > polynomial, variable, or constant? Is c a polynomial or constant? What > do you mean about a polynomial's value changing? A polynomial is in the > form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_, it > doesn't change.Hmmm...doesn't sound like you actually looked at the paper.In the paper the ring is algebraic integers; therefore, r and c arealgebraic integers.In the paper it's noted that I'm generalizing beyond polynomialfactors, that is, factors of a polynomial that are themselvespolynomials.In the paper I give an example of g = sqrt(x+1), so it's hard tounderstand how you could miss the truth.I state as included in your post, so it's hard to see how you missedit, but I'll give it again here for emphasis:>> where either r=0, or r changes as the polynomial's value changes,> while c does not.But you state above:What do you mean about a polynomial's value changing? A polynomialis in theform a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,it doesn't change.That's called bait-and-switch as you switched from the polynomial'svalue to the polynomial.I didn't say the polynomial changes. I said its value changes and younoted that in one sentence and switched in the other.Now given, say, the polynomial P(x) = x+1, the polynomial's value atx=0, is 1.It's not rocket science.And for the rest of you, I've seen a troubling tendency from postersand people I know are actual mathematicians to cheat in this way.Over and over and over again, they'll use deceitful tactics, and fromwhat I've seen it works.It's like most of you WANT TO BELIEVE THEM, even when they're trashinglogic, mathematics, and what I'd think are typical ideas aboutfairplay.Mathematicians cheat. Ok, so I'm not saying ALL mathematicians, butI'm noting a trend that I've noticed time and time again.Over and over and over again, mathematicians have cheated. I pointout that they cheat, and then they just keep doing it!!!> Note: at this point I'm fairly confused, but I'll read on.I'll keep reading on as well.> Now you can consider all factors of a given polynomial using g's, with> something like> > g_1...g_k = P(x)> > where you have k factors. For instance, for P(x)=x^2 + 2x + 1,> > g_1 = x+1, g_2 = x+1, gives you 2 factors.> > > This part made sense, except you'd usually note them as g_1(x), g_2(x).I gave a simple example with factors of the polynomial that arethemselves polynomials.But you can't even assume that if P(x) is of degree n that k=n.In fact k can vary out to in'nity.Just like 2 has an in'nite number of factors in the ring of algebraicintegers.Similarly a polynomial like P(x) = x+1 can have an in'nity offactors.It's that simple.Now if that bothers you and you start trying to throw polynomialfactors back at me, I'll look at you the way someone might look at aperson who refuses to believe that numbers like 2 and 3 aren't primein higher rings.Then that person in denial might keep saying things like, but 2(3)=6,and what do you mean .95bout factors of 3?They might ask me if I'm not making up my own de'nition of factor.Basically, yes, I'm saying that an inability to comprehend that likeintegers can have non-integer factors in a ring, and not a 'eld,polynomials can have non-polynomial factors is primitive.Are mathematicians math experts or not?> Those are polynomial factors, but I'm generalizing in a simple way to> say that for the factors g, in general, you have an element I call r,> which changes as the independent variable changes, and you have> another element I call c, which does not.> > So, r is a function, and c is a constant? Why are you talking about > independent variables changing?I'm not worried about whether or not r 'ts the formal de'nition forfunction, and when I've talked about functions in this area there'sjust been more confusion.What matters for the argument is that r changes as x changes, and notspeci'cs about how, while c remains constant.Polynomials represent single independent variable systems.For more generality I said independent variable rather than x.> For my example up above it's easy, as with g_1 = x + 1, x varies as x> varies, while 1 does not.> > so in this case, r=x, c=1?Yup. But it's a trivial case. The method is general enough to handlenot having an explicit representation for r. It merely notes that itchanges, not exactly what it is.> Now that's enough that the proof in the paper is straightforward, but> posters have argued with me anyway, with some trying to argue over the> de'nition of polynomial, amazingly enough.> > Question: does each factor g_i(x) get its own r_i(x)? I'm not looking > at the proof right now, but simply trying to understand what you are > claiming to have prooved. The terminology you are using is not clear to > me, and I can't easily comment on the value or validity of your work > until that is made clear. If you read math papers, you will note that > people specify what each object is, whether it's an integer, polynomial > over the rationals, etc. I don't see these details.Yet in the paper it's speci'cally stated that the ring is algebraicintegers.Therefore, variables like r and c are in that ring.When more information is needed beyond that, I provide it.Again, I see a person saying the damndest things.Name *one* thing in the paper that isn't de'ned that's part of theargument.The *insinuation* that there's something just pisses me off.If there's something GIVE THE EVIDENCE.I'm sick of all these statements without evidence.MATHEMATICIANS NEED TO QUIT CHEATING!!!!!!!!!!!!!!!> However, consider that the g's have an important feature, which is> that when x=0, I have> > g_1...g_k = P(0).> > Don't you mean (g_1...g_k)(0)=P(0)?I'm not interested in a form debate. There's style and there'ssubstance.If you want to argue style over substance why don't you switch tofashion?Those runway models look nice as they sashay down the catwalk,swinging those hips.But if you want to talk mathematics, don't give me style issues.> For instance, with my simple example, with P(x) = x^2 + 2x + 1, with> x=0,> > P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.> > x+1 is not 1. x+1 at x=0 is 1.Yeah, which is why I say that with x=0 above.Hmmm...you're just trying to piss me off, aren't you?I mean, you've brought up so much bull stuff.> You see, P(0) gives the constant term, so at x equal 0, the g's must> multiply to give the constant term.> > So then, maybe you still want to believe the mathematicians and> question that I can write g = r + c.> > Just questioning what it means.Yeah right. I think you're politicking.> Well consider that substituting gives me> > g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives> > This seems to suggest that each g_i gets it's own r_i, c_i.Hell yeah!!!> r_1...r_k +...+c_1...c_k = P(x), which is> > This step is NOT clear. What all goes between the product of r's and > product of c's?Now you don't know how to multiply?Sorry, I'm not giving a class on multiplying out expresssions like (r_1 + c_1)...(r_k + c_k)as I'll let you go to some extent, but not that far. r_1...r_k +...+P(0) = P(x), > > which means that if you believe the mathematicians then they've> convinced you to doubt algebra itself, as then you must believe that> everything to the left of P(0) above can *maybe* be constant, but also> *maybe* vary as x varies.> > Well, based on how you're de'ning things it appears that all of your > r's should be written as r_i(x), not simply r_i. What am I missing? If > r depends on x, please clearly indicate it. If r does not depend on x, > then your initial de'nition of r doesn't make sense. There is > ambiguity in here that appears to be the source of your *maybe*s, but I > think it came from you, not mathematicians.Style questions don't interest me. > So why would mathematicians argue against such a simple result?> > Because it's not clearly stated.Bull.> Two reasons I suggest. First because they wish to disagree with me. > Second because they probably believe that they can get away with it.> > Right now, I don't understand what you're trying to say because there is > too much that you have not de'ned. I cannot disagree with you because > I do not understand you. I cannot accept your claims until I understand > them, however. I've worked through enough math to have headaches from > the strain of maintaining precision in language. Your paper lacks that > precision.Stated without supporting evidence.Yes more of what I've found to be typical behavior frommathematicians.> That is, MOST of you will doubt algebra itself rather than consider> that mathematicians, whom you probably don't even personally know,> would lie.> > No, I just want to know what you're trying to say.Like how you asked about things stated in the paper despite saying atthe top that you read the paper?I've demonstrated in several places that you can't be taken at yourword.> So where does this lead?> > Well the polynomial I show in the paper is> > P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf> > which seems to be just complicated enough to give mathematicians room> to lie. I'd like to know why you introduced v, then de'ned it in terms of m and > f, rather than do the substitution yourself. Also, why are you > apparently leaving x,y,f as unaccounted for variables? What type of > polynomial is P supposed to be?IT'S ALL IN THE PAPER YOU CLAIM TO HAVE READ!!!!!!!!!!!!!!That's it. I'm tired of your game.Yes undergrads, if you thought that mathematicians played by rules,notice how they're trying to change rules of fairplay in dealing withme, a non-mathematician. => I've looked at your paper... now looking at this I'll bite. I may be> getting in over my head but I'll point out the things that don't make> sense to me. I'm open to being only recently started reading this> board, so I am not familiar with your previous dif'culties with> mathematicians.It looks like you and James have been tricked into wastinga few hours of each others time.I really enjoyed James' opening lineand how it gradually evolved into IT'S ALL IN THE PAPER YOU CLAIM TO HAVE READ!!!!!!!!!!!!!! That's it. I'm tired of your game.If this was your intention, very well done! If not, don't take itpersonal - it's a question of medication dosage. There mustbe a balance somewhere. We know it will be unstable, butit will be a balance - for a few weeks.Welcome to the playground of James Harrass.Dirk Vdm => >>I've looked at your paper... now looking at this I'll bite. I may be>>getting in over my head but I'll point out the things that don't make>>sense to me. I'm open to being mathematician. I have only recently started reading this>>board, so I am not familiar with your previous dif'culties with>>mathematicians.> > > It looks like you and James have been tricked into wasting> a few hours of each others time.> > I really enjoyed James' opening line> and how it gradually evolved into> IT'S ALL IN THE PAPER YOU CLAIM TO> HAVE READ!!!!!!!!!!!!!!> That's it. I'm tired of your game.> > If this was your intention, very well done! If not, don't take it> personal - it's a question of medication dosage. There must> be a balance somewhere. We know it will be unstable, but> it will be a balance - for a few weeks.> > Welcome to the playground of James Harrass.> > Dirk Vdm> > I noticed it. I've lurked long enough that his response isn't *entirely* surprising. I suppose I should go try to make sense of his paper now, to see where I'm missing things. It would have been helpful if he had started by posting the paper here instead of posting a link to a pdf. It makes quoting his paper... more challenging. And makes it less accessible.I looked at his paper long enough to get confused by it. I didn't try to make sense of it. I suppose I must defend my honor as a mathematician and start looking for the de'nitions in his paper.-- Will Twentyman =>I've looked at your paper... now looking at this I'll bite. I may be >>getting in over my head but I'll point out the things that don't make >>sense to me. I'm open to being enlightened.> Note: looked at is not the same as reading for understanding.>I think I've 'gured out a way to show basically all of you, including>people who think they don't know any math that mathematicians have>been lying about my work. It's so trivial you *should* wonder why>they thought they could get away with it.>>>>Either you aren't being clear, or they missed something. It is a strong >>charge to suggest that ALL mathematicians are in this great conspiracy >>to teach incorrect math. Sort of goes against what they stand for. I say mathematicians but don't speci'cally say ALL mathematicians.> Distinction noted. My mistake.> I'm talking about a pattern of denial that I'm seeing in> mathematicians that I've contacted or observed.> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)>>And in my paper I start by showing that I can write that as>> g = r + c>>where either r=0, or r changes as the polynomial's value changes,>while c does not.>>>>Questions: is g the polynomial to be factored or a factor? Is r a >>polynomial, variable, or constant? Is c a polynomial or constant? What >>do you mean about a polynomial's value changing? A polynomial is in the >>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_, it >>doesn't change.> > Hmmm...doesn't sound like you actually looked at the paper.I'll be going over it in detail this weekend. Do you mind if I quote it directly in my responses?> In the paper the ring is algebraic integers; therefore, r and c are> algebraic integers.An algebraic integer is a constant. It does not vary. Period. Note: I'm taking my de'nitions of terms from Wolfram's Mathworld. If you are using a different de'nition, please provide it.> In the paper it's noted that I'm generalizing beyond polynomial> factors, that is, factors of a polynomial that are themselves> polynomials.> > In the paper I give an example of g = sqrt(x+1), so it's hard to> understand how you could miss the truth.To be honest, I didn't get past the statement of your lemma. I will attempt to correct that.> I state as included in your post, so it's hard to see how you missed> it, but I'll give it again here for emphasis:> where either r=0, or r changes as the polynomial's value changes,>while c does not.> > > But you state above:> > What do you mean about a polynomial's value changing? A polynomial> is in the> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,> it doesn't change.> > That's called bait-and-switch as you switched from the polynomial's> value to the polynomial.That's called using standard terminology. Precise use of terms is a hallmark of mathematics. A polynomial is an expression. For different values of its variable(s) it evaluates to different values, but that is not the same as saying the polynomial's value changes. I didn't say the polynomial changes. I said its value changes and you> noted that in one sentence and switched in the other.> > Now given, say, the polynomial P(x) = x+1, the polynomial's value at> x=0, is 1.> > It's not rocket science.Correct. But you did NOT change the value of the polynomial, you changed x and you changed P at x. P(x)=x+1, P(0)=1. P's value didn't change, you simply evaluated it for a given value of x.> And for the rest of you, I've seen a troubling tendency from posters> and people I know are actual mathematicians to cheat in this way.If insisting on precise use of terminology is cheating, you might as well give up. We've been trained to be very precise in how words are using, because many (if not all) of us got proofs shredded when we forgot a technicality.[rant deleted]>Note: at this point I'm fairly confused, but I'll read on.> > > I'll keep reading on as well.> > >Now you can consider all factors of a given polynomial using g's, with>something like>> g_1...g_k = P(x)>>where you have k factors. For instance, for P(x)=x^2 + 2x + 1,>> g_1 = x+1, g_2 = x+1, gives you 2 factors.>>>>>This part made sense, except you'd usually note them as g_1(x), g_2(x).> > > I gave a simple example with factors of the polynomial that are> themselves polynomials.> > But you can't even assume that if P(x) is of degree n that k=n.> > In fact k can vary out to in'nity.> > Just like 2 has an in'nite number of factors in the ring of algebraic> integers.> > Similarly a polynomial like P(x) = x+1 can have an in'nity of> factors.> > It's that simple.That's not a problem, but your notation was inconsistent.> Now if that bothers you and you start trying to throw polynomial> factors back at me, I'll look at you the way someone might look at a> person who refuses to believe that numbers like 2 and 3 aren't prime> in higher rings.> > Then that person in denial might keep saying things like, but 2(3)=6,> and what do you mean .95bout factors of 3?> > They might ask me if I'm not making up my own de'nition of factor.> > Basically, yes, I'm saying that an inability to comprehend that like> integers can have non-integer factors in a ring, and not a 'eld,> polynomials can have non-polynomial factors is primitive.> > Are mathematicians math experts or not?Theoretically, yes. On the other hand, my dad is a mathematician by degree only. He has forgotten much of what he knew through lack of use.>Those are polynomial factors, but I'm generalizing in a simple way to>say that for the factors g, in general, you have an element I call r,>which changes as the independent variable changes, and you have>another element I call c, which does not.>>>>So, r is a function, and c is a constant? Why are you talking about >>independent variables changing?> > > I'm not worried about whether or not r 'ts the formal de'nition for> function, and when I've talked about functions in this area there's> just been more confusion.> > What matters for the argument is that r changes as x changes, and not> speci'cs about how, while c remains constant.If r changes as x changes, regardless of how, then either r is a function of x, a relation on x, or not well-de'ned. I'll accept function of x. The reason I ask is that it would be nice if you noted it as r(x), rather than simply as r.> > Polynomials represent single independent variable systems.> > For more generality I said independent variable rather than x.> For generality, I'll assume whatever you stick in () is the independent variable.> >For my example up above it's easy, as with g_1 = x + 1, x varies as x>varies, while 1 does not.>>>>so in this case, r=x, c=1?> > > Yup. But it's a trivial case. The method is general enough to handle> not having an explicit representation for r. It merely notes that it> changes, not exactly what it is.I can deal with this being a special case, I just want the details nailed down.>Now that's enough that the proof in the paper is straightforward, but>posters have argued with me anyway, with some trying to argue over the>de'nition of polynomial, amazingly enough.>>>>Question: does each factor g_i(x) get its own r_i(x)? I'm not looking >>at the proof right now, but simply trying to understand what you are >>claiming to have prooved. The terminology you are using is not clear to >>me, and I can't easily comment on the value or validity of your work >>until that is made clear. If you read math papers, you will note that >>people specify what each object is, whether it's an integer, polynomial >>over the rationals, etc. I don't see these details.> > > Yet in the paper it's speci'cally stated that the ring is algebraic> integers. Therefore, variables like r and c are in that ring.If r varies with x, then it's not an algebraic integer. It might be something you could call an algebraic function, but it's NOT a constant.> > When more information is needed beyond that, I provide it.> > Again, I see a person saying the damndest things.> > Name *one* thing in the paper that isn't de'ned that's part of the> argument.I'll go over your work this weekend and see what I can conclude. I was responding the the content of the post, not what it refers to.> The *insinuation* that there's something just pisses me off.> > If there's something GIVE THE EVIDENCE> I'm sick of all these statements without evidence.> > MATHEMATICIANS NEED TO QUIT CHEATING!!!!!!!!!!!!!!!> Just be sure we ARE cheating before accusing. It does nothing for your credibility. If we are confused, explain. If we have made an error, point it out *clearly*. If we respond to the content of a post that refers to an outside document, don't be surprised if there is a limited amount of cross-referencing.>However, consider that the g's have an important feature, which is>that when x=0, I have>> g_1...g_k = P(0).>>>>Don't you mean (g_1...g_k)(0)=P(0)?> > > I'm not interested in a form debate. There's style and there's> substance.In mathematics, the line between style and substance can be *VERY* 'ne. Don't be so quick to dismiss it. Also, don't be surprised if failure to use the accepted style generates additional confusion. I've seen it too often.> > If you want to argue style over substance why don't you switch to> fashion?> I need too much help shopping for clothes to switch to fashion.> But if you want to talk mathematics, don't give me style issues.What looks like style may be substance. Also, consistent style makes it easier to see the substance.>For instance, with my simple example, with P(x) = x^2 + 2x + 1, with>x=0,>> P(0) = 1 = g_1(g_2), where g_1 = g_2 = x+1 = 1.>>>>x+1 is not 1. x+1 at x=0 is 1.> > > Yeah, which is why I say that with x=0 above.> > Hmmm...you're just trying to piss me off, aren't you?No, I'm trying to get you to be precise.Try writing it as P(x)=g_1(x)*g_2(x) with x=0,P(0)=1=g_1(0)*g_2(0), g_1(0) = g_2(0) = 0+1 = 1The added level of precision will prevent confusion on the part of your reader and make it easier to see the content, and comment intelligently on it. We might even all decide that you're a genius.> > I mean, you've brought up so much bull stuff.You haven't seen how bad it can get. Trust me, the level of detail can get a LOT worse.>You see, P(0) gives the constant term, so at x equal 0, the g's must>multiply to give the constant term.>>So then, maybe you still want to believe the mathematicians and>question that I can write g = r + c.>>>>Just questioning what it means.> > Yeah right. I think you're politicking.If you want us to accept your proof as valid, then we need to understand it. If you don't want to conform your writing to the conventions, don't be surprised if you are scorned. Can you name a 'eld of study that does NOT have rules for presentation of material?>Well consider that substituting gives me>> g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which gives>>>>This seems to suggest that each g_i gets it's own r_i, c_i.> > Hell yeah!!!Good! I understood something :)> r_1...r_k +...+c_1...c_k = P(x), which is>>>>This step is NOT clear. What all goes between the product of r's and >>product of c's?> > Now you don't know how to multiply?No, I'm saying writing the 'rst and last terms can easily lead to erroneous conclusions.> Sorry, I'm not giving a class on multiplying out expresssions like> > (r_1 + c_1)...(r_k + c_k)> > as I'll let you go to some extent, but not that far.Summation and product notations can make it compact without missing anything.> r_1...r_k +...+P(0) = P(x), >>which means that if you believe the mathematicians then they've>convinced you to doubt algebra itself, as then you must believe that>everything to the left of P(0) above can *maybe* be constant, but also>*maybe* vary as x varies.>>>>Well, based on how you're de'ning things it appears that all of your >>r's should be written as r_i(x), not simply r_i. What am I missing? If >>r depends on x, please clearly indicate it. If r does not depend on x, >>then your initial de'nition of r doesn't make sense. There is >>ambiguity in here that appears to be the source of your *maybe*s, but I >>think it came from you, not mathematicians.> > Style questions don't interest me.That is probably why you are having so many con¤icts. If your style is not within the accepted norm, you will be viewed as babbling. Have you ever had an instructor that speaks broken English or has a thick accent? Some students 'ght through that to understand, others hear only babbling. The same thing can happen with style. Some will attempt to decipher, others will dismiss until you adapt to the accepted standards.>So why would mathematicians argue against such a simple result?>>>>Because it's not clearly stated.> > Bull.Actually, that was one of the 'rst responses I saw to your paper. Something to the effect of It's not right, it's not wrong, it's unintelligible. If you are not communicating clearly, your arguments will hold little weight.>Two reasons I suggest. First because they wish to disagree with me. >Second because they probably believe that they can get away with it.>>>>Right now, I don't understand what you're trying to say because there is >>too much that you have not de'ned. I cannot disagree with you because >>I do not understand you. I cannot accept your claims until I understand >>them, however. I've worked through enough math to have headaches from >>the strain of maintaining precision in language. Your paper lacks that >>precision.> > Stated without supporting evidence.That's because what I call precision, you denegrate as style.> Yes more of what I've found to be typical behavior from> mathematicians.So why not play our game by our rules? If you can't win within our rules for communication, it may mean you need to convince us of the problem in HOW we communicate, not WHAT we communicate.>That is, MOST of you will doubt algebra itself rather than consider>that mathematicians, whom you probably don't even personally know,>would lie.>>>>No, I just want to know what you're trying to say.> > Like how you asked about things stated in the paper despite saying at> the top that you read the paper?Looking at a paper is NOT understanding it. I've had to look up 3 terms in your paper and I'm hoping that you are using the same de'nitions I found. If you don't appear to be, I'll note it in my response to your paper.> I've demonstrated in several places that you can't be taken at your> word.I'll leave that statement for others to judge.>So where does this lead?>>Well the polynomial I show in the paper is>> P(m) = (v^3+1)x^3 - 3vxy^2 + y^3, v=-1+mf>>which seems to be just complicated enough to give mathematicians room>to lie.>>>>I'd like to know why you introduced v, then de'ned it in terms of m and >>f, rather than do the substitution yourself. Also, why are you >>apparently leaving x,y,f as unaccounted for variables? What type of >>polynomial is P supposed to be?> > > IT'S ALL IN THE PAPER YOU CLAIM TO HAVE READ!!!!!!!!!!!!!!See notes above.> > That's it. I'm tired of your game.> > Yes undergrads, if you thought that mathematicians played by rules,> notice how they're trying to change rules of fairplay in dealing with> me, a non-mathematician.> > who claim to be logicians or scientists.> > They're not; they're bull artists. > James HarrisAnd you'll notice that he never addressed my last question. James: if you are unwilling to answer questions asking for clari'cation, you are not likely to persuade anyone. It would be more helpful to cite the 'nd what addresses what. May I suggest that in the future you answer questions, rather than insult the person asking them? If I was in some way rude, I appologize. I will do my best to not insult you. However, I *will* ask questions when I don't understand what is being said. It is safer to do that than assume you mean something other than what you intended.I believe I asked it above, but I will ask again: do I have your -- Will Twentyman => >>> I've looked at your paper... now looking at this I'll bite. I may be> getting in over my head but I'll point out the things that don't make> sense to me. I'm open to being mathematician. I have only recently started reading this> board, so I am not familiar with your previous dif'culties with> mathematicians.>>>>>>>> It looks like you and James have been tricked into wasting>> a few hours of each others time.>>>> I really enjoyed James' opening line>> and how it gradually evolved into>> IT'S ALL IN THE PAPER YOU CLAIM TO>> HAVE READ!!!!!!!!!!!!!!>> That's it. I'm tired of your game.>>>> If this was your intention, very well done! If not, don't take it>> personal - it's a question of medication dosage. There must>> be a balance somewhere. We know it will be unstable, but>> it will be a balance - for a few weeks.>>>> Welcome to the playground of James Harrass.>>>> Dirk Vdm>>>>> > I noticed it. I've lurked long enough that his response isn't > *entirely* surprising. I suppose I should go try to make sense of his > paper now, to see where I'm missing things. It would have been helpful > if he had started by posting the paper here instead of posting a link to > a pdf. It makes quoting his paper... more challenging. And makes it > less accessible.> > I looked at his paper long enough to get confused by it. I didn't try > to make sense of it. I suppose I must defend my honor as a > mathematician and start looking for the de'nitions in his paper.I've been lurking long enough to see several well-meaning mathematicians take a lot of trouble to try to understand what JSH is doing, and to help him to clarify it. The outcome has become quite predictable. Rather than follow the advice that he de'ne his terms and follow accepted standards of proof, JSH turns on his would-be helpers and abuses them as liars and cheats. You are asking for the same treatment.Gib => Deny that you are a mathematician, and admit that you do not> understand what you are talking about. Otherwise, stop attacking a> yourself.He has frequently proclaimed that he is not a mathematician. For a rational person this would disqualify him from attempting to do mathematics, but JSH is immune to such logic.Gib => >>I've looked at your paper... now looking at this I'll bite. I may be>>getting in over my head but I'll point out the things that don't make>>sense to me. I'm open to being enlightened.> > have only recently started reading this>>board, so I am not familiar with your previous dif'culties with>mathematicians.That's just sad. It means that maybe you picked up bad habits whichas I've feared are typical for mathematicians.In any event, if you get serious I'll work through it carefully aslong as you don't try any tricks, no bait-and-switch, nor anythingelse not aboveboard. > It looks like you and James have been tricked into wasting> a few hours of each others time.Nope. I don't necessarily post for one person. Undergrads who *can*follow and see what's going on can also get out of mathematics whilethey can.Biology is a good major.> > I really enjoyed James' opening line> and how it gradually evolved into> IT'S ALL IN THE PAPER YOU CLAIM TO> HAVE READ!!!!!!!!!!!!!!> That's it. I'm tired of your game.Yeah, my elation turned to frustration as 'rst I was repeating whatwas in the paper, then dealing with underhanded things likebait-and-switch and 'nally I just got tired of it.> > If this was your intention, very well done! If not, don't take it> personal - it's a question of medication dosage. There must> be a balance somewhere. We know it will be unstable, but> it will be a balance - for a few weeks.> > Welcome to the playground of James Harrass.> > Dirk Vdm> > > > I noticed it. I've lurked long enough that his response isn't > *entirely* surprising. I suppose I should go try to make sense of his > paper now, to see where I'm missing things. It would have been helpful > if he had started by posting the paper here instead of posting a link to > a pdf. It makes quoting his paper... more challenging. And makes it > less accessible.The paper started on the newsgroup.Undergrads the warning is clear. Mathematicians don't play fair, theyseem to run from logic if it doesn't suit them, and when you catchthem, they just come up with more *stuff*.> I looked at his paper long enough to get confused by it. I didn't try > to make sense of it. I suppose I must defend my honor as a > mathematician and start looking for the de'nitions in his paper.What I'm doing is showing that apparently there is little honor as amathematician, which is a modern thing.Mathematicians like Gauss, Euler, Dedekind, and so many otherswouldn't play these bull games, I'm sure.But I think a lot of you got in under a system where lies are not onlytolerated, but expected, where mathematics is just a club, so you canpay your bills, get married, and teach poor unsuspecting students whoget caught up in the grandeur of mathematics past.Those students need to remember that the future keeps happeningdespite the past.Mathematicians today *can* be as contemptibly bad as I've been showingdespite the grand history of the discipline.It apparently is going through a bad spell.James Harris => g_1...g_k = P(0). Don't you mean (g_1...g_k)(0)=P(0)?> > I'm not interested in a form debate. There's style and there's> substance.It's not style, it's precision. Your g_1...g_k = P(0) admits of theinterpretation g_1...g_k is a function that is identially P(0), i.e.,for all x: (g_1....g_k)(x) = P(0).V.-- =No kidding!!>> Go home, little man...>> This is the only home he has...>> V.>> -- =You're an admirable person to try and make sense of his drither. Good luck,Will Twentyman!!!>>>I've looked at your paper... now looking at this I'll bite. I may be>>getting in over my head but I'll point out the things that don't make>>sense to me. I'm open to being a mathematician. I have only recently started reading this>board, so I am not familiar with your previous dif'culties with>>mathematicians.>>> It looks like you and James have been tricked into wasting> a few hours of each others time.>> I really enjoyed James' opening line and how it gradually evolved into> IT'S ALL IN THE PAPER YOU CLAIM TO> HAVE READ!!!!!!!!!!!!!!> That's it. I'm tired of your game.>> If this was your intention, very well done! If not, don't take it> personal - it's a question of medication dosage. There must> be a balance somewhere. We know it will be unstable, but> it will be a balance - for a few weeks.>> Welcome to the playground of James Harrass.>> Dirk Vdm>>> I noticed it. I've lurked long enough that his response isn't> *entirely* surprising. I suppose I should go try to make sense of his> paper now, to see where I'm missing things. It would have been helpful> if he had started by posting the paper here instead of posting a link to> a pdf. It makes quoting his paper... more challenging. And makes it> less accessible.>> I looked at his paper long enough to get confused by it. I didn't try> to make sense of it. I suppose I must defend my honor as a> mathematician and start looking for the de'nitions in his paper.>> --> Will Twentyman> => But I think a lot of you got in under a system where lies are not only> tolerated, but expected, where mathematics is just a club, so you can> pay your bills, get married, and teach poor unsuspecting students who> get caught up in the grandeur of mathematics past.My lecturers actively encourage us to 'nd ¤aws in the stuff presented tous. Frankly, they would be rather disappointed if we believed something justbecause they told us it is true.Jon =>We've been trained to be very precise in how words are >using, ...And words using goodly we is... ;-)-- Thomas Wasell | Does the name Pavlov ring a bell? wasell@bahnhof.se | => > >>We've been trained to be very precise in how words are >>using, ...> > > And words using goodly we is... ;-)> *sigh* I knew I shouldn't have studied under Yoda and Jar-Jar Binks.-- Will Twentyman => > I've been lurking long enough to see several well-meaning mathematicians > take a lot of trouble to try to understand what JSH is doing, and to > help him to clarify it. The outcome has become quite predictable. > Rather than follow the advice that he de'ne his terms and follow > accepted standards of proof, JSH turns on his would-be helpers and > abuses them as liars and cheats. You are asking for the same treatment.> > Gib> > I seem to have received it already. I can hope he will listen, and if not, I can be know that I attempted to help someone. It is up to him whether he accepts help.-- Will Twentyman =>> > > Yeah, my elation turned to frustration as 'rst I was repeating what> was in the paper, then dealing with underhanded things like> bait-and-switch and 'nally I just got tired of it.Notice that you assumed1) I made a bait-and-switch and2) If I did, it was intentional rather than a result of not understanding.>If this was your intention, very well done! If not, don't take it>personal - it's a question of medication dosage. There must>be a balance somewhere. We know it will be unstable, but>it will be a balance - for a few weeks.>>Welcome to the playground of James Harrass.>>Dirk Vdm>>>>>>I noticed it. I've lurked long enough that his response isn't >>*entirely* surprising. I suppose I should go try to make sense of his >>paper now, to see where I'm missing things. It would have been helpful >>if he had started by posting the paper here instead of posting a link to >>a pdf. It makes quoting his paper... more challenging. And makes it >>less accessible.> > > The paper started on the newsgroup.In sci.math, where I picked this up, you posted a link to the paper that do not recall ever seeing it in sci.math, but that could easily be an oversite.-- Will Twentyman => > I can be know that I attempted to help someone. > Note to self: I have GOT to proofread before hitting -- Will Twentyman => >>I've looked at your paper... now looking at this I'll bite. I may be >>getting in over my head but I'll point out the things that don't make >>sense to me. I'm open to being enlightened.> > > Note: looked at is not the same as reading for understanding.Ok. >I think I've 'gured out a way to show basically all of you, including>people who think they don't know any math that mathematicians havebeen lying about my work. It's so trivial you *should* wonder why>they thought they could get away with it.>>>Either you aren't being clear, or they missed something. It is a strong >>charge to suggest that ALL mathematicians are in this great conspiracy >>to teach incorrect math. Sort of goes against what they stand for. I say mathematicians but don't speci'cally say ALL mathematicians.> > > Distinction noted. My mistake.Note that I've been careful to actually have working mathematicians inmind when I say that mathematicians have been lying about my work.Two that readily come to mind are David Ullrich, a professor atOklahoma State University, and Arturo Magidin, whose university I'mnot sure of at this time as I'd have to check a header in a post ofhis, but he obtained his Ph.D from Berkeley University.I've veri'ed that each is associated as given by checking at thewebsites of the respective universities.They are mathematicians as noted. I'm talking about a pattern of denial that I'm seeing in mathematicians that I've contacted or observed.> (Paper linked to at http://groups.msn.com/AmateurMath as usual.)>>And in my paper I start by showing that I can write that as>> g = r + c>where either r=0, or r changes as the polynomial's value changes,>while c does not.>>>>Questions: is g the polynomial to be factored or a factor? Is r a >polynomial, variable, or constant? Is c a polynomial or constant? What >>do you mean about a polynomial's value changing? A polynomial is in the >>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_, it >>doesn't change.> > Hmmm...doesn't sound like you actually looked at the paper.> > I'll be going over it in detail this weekend. Do you mind if I quote it > directly in my responses?No.> In the paper the ring is algebraic integers; therefore, r and c are> algebraic integers. An algebraic integer is a constant. It does not vary. Period. Note: > I'm taking my de'nitions of terms from Wolfram's Mathworld. If you are > using a different de'nition, please provide it.You deleted context. My guess is that you were asking what r and cwere and I noted that because the ring is algebraic integers, they arealgebraic integers.For instance, given P(x)=x+1 in the ring of integers, if you come backto me and ask me what is x, I'll tell you that it is an integer.If you reply that integers do not vary I'll doubt your veracity andassume that you are deliberately trying to deceive.I suggest that you be careful about what you delete out when you makesuch statements.My assessment at this time is that you have been *repeatedly* caught.My own view is that your behavior is typicaly of what I've seen fromother mathematicians.Clearly I see an advantage in pointing out to others--especiallyundergrads--that behavior like yours is, in my experience, typical ofthe mathematical community.Mathematicians cheat.> In the paper it's noted that I'm generalizing beyond polynomial> factors, that is, factors of a polynomial that are themselves> polynomials.> > In the paper I give an example of g = sqrt(x+1), so it's hard to> understand how you could miss the truth.> > To be honest, I didn't get past the statement of your lemma. I will > attempt to correct that.Well you can say To be honest but I think you need to work harder atshowing that you're being honest.> I state as included in your post, so it's hard to see how you missed> it, but I'll give it again here for emphasis: >where either r=0, or r changes as the polynomial's value changes,>while c does not.> > > But you state above:> > What do you mean about a polynomial's value changing? A polynomial> is in the> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,> it doesn't change.> > That's called bait-and-switch as you switched from the polynomial's> value to the polynomial.> > That's called using standard terminology. Precise use of terms is a > hallmark of mathematics. A polynomial is an expression. For different > values of its variable(s) it evaluates to different values, but that is > not the same as saying the polynomial's value changes.Posts on newsgroups are not formal, and there's always a possibilityof misunderstandings.I'm not a mathematician. There are likely to be any number of timeswhen I end up using something that's not exactly the way you may beused to seeing it, but if you're a mathematician, by de'nition you'rea math expert.If I say the value of P(x) = x + 1 varies as x varies, then are yousuggesting you can't 'gure out what I mean, even if it's not exactlywhat you're used to seeing?> > I didn't say the polynomial changes. I said its value changes and you> noted that in one sentence and switched in the other.> > Now given, say, the polynomial P(x) = x+1, the polynomial's value at> x=0, is 1.> > It's not rocket science.> > Correct. But you did NOT change the value of the polynomial, you > changed x and you changed P at x. P(x)=x+1, P(0)=1. P's value didn't > change, you simply evaluated it for a given value of x.Then what changed? You say, ...you changed P at x. Yet above youhave:> What do you mean about a polynomial's value changing? A polynomial> is in the> form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,> it doesn't change.Now I take your claim to be that the value of a polynomial is itsexplicit expression.While I said things like the polynomial's value at x=0.Your claim is that my usage is nonstandard.Ok, I don't care to hold on to a nonstandard usage as my aim is tocommunicate.So then, however it's stated so that I can talk of a polynomial, likex+1, and also talk of the value for some particular x.> And for the rest of you, I've seen a troubling tendency from posters> and people I know are actual mathematicians to cheat in this way.> > If insisting on precise use of terminology is cheating, you might as > well give up. We've been trained to be very precise in how words are > using, because many (if not all) of us got proofs shredded when we > forgot a technicality.> > [rant deleted]Again this is a *post* on a newsgroup and not a formal paper. I'mtrying to explain in a medium that allows questions to be asked incase there is a misunderstanding.Here there seems to be a misunderstanding as to how one talks of thevalues for the explicit expression of a polynomial P(x) for a given x.Now I'm suspicious of you still, but again, I see your post as typicalof what I've come to expect from mathematicians.Which is also why I say you're like lying English professors.I also say you're like bad lawyers who argue over technicalities, asif the point or the truth doesn't matter.I'm not interested in reading further as you've repeatedlydemonstrated behavior I 'nd deceitful and annoying.James Harris =[cut]> Ok, I don't care to hold on to a nonstandard usage as my aim is to> communicate.> > So then, however it's stated so that I can talk of a polynomial, like> x+1, and also talk of the value for some particular x.There are probably many ways. Below is one way. Let A be the ring of algebraic integers. Let P(X) = X+1 be a polynomial over A. Let x be an algebraic integer. Consider the algebraic integer P(x). etc.Another example is your lemma being discussed: Let A be the ring of algebraic integers. Let P(X) be a polynomial in A[X]. Let x be an algebraic integer. Let g be an algebraic integer that is a factor of P(x). Then, I can 'nd algebraic integers r and c such that g = r + c where c is a factor of P(0) and either r = 0 or else x and r are not coprime.Note that the above implies that c depends upon x,whereas you say that c is independent of x. Thus,the above phrasing is not quite what you want.-- Bill Hale => >>>I think I've 'gured out a way to show basically all of you, including>people who think they don't know any math that mathematicians have>been lying about my work. It's so trivial you *should* wonder why>they thought they could get away with it.>>>>Either you aren't being clear, or they missed something. It is a strong >>charge to suggest that ALL mathematicians are in this great conspiracy >>to teach incorrect math. Sort of goes against what they stand for.>>I say mathematicians but don't speci'cally say ALL mathematicians.>>>>>Distinction noted. My mistake.> > > Note that I've been careful to actually have working mathematicians in> mind when I say that mathematicians have been lying about my work.> > Two that readily come to mind are David Ullrich, a professor at> Oklahoma State University, and Arturo Magidin, whose university I'm> not sure of at this time as I'd have to check a header in a post of> his, but he obtained his Ph.D from Berkeley University.> > I've veri'ed that each is associated as given by checking at the> websites of the respective universities.> > They are mathematicians as noted.For the record, I have a Master's Degree in math from the University of Illinois, Urbana-Champaign. My emphasis was in logic, not algebra, so I'll be doing a fair bit of refreshing as I work through this.>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I quote it >>directly in my responses?> > No.>In the paper the ring is algebraic integers; therefore, r and c are>algebraic integers.>>>>An algebraic integer is a constant. It does not vary. Period. Note: >>I'm taking my de'nitions of terms from Wolfram's Mathworld. If you are >>using a different de'nition, please provide it.> > > You deleted context. My guess is that you were asking what r and c> were and I noted that because the ring is algebraic integers, they are> algebraic integers.I will attempt to deal with this carefully in my comments about your paper.> For instance, given P(x)=x+1 in the ring of integers, if you come back> to me and ask me what is x, I'll tell you that it is an integer.> > If you reply that integers do not vary I'll doubt your veracity and> assume that you are deliberately trying to deceive.> > I suggest that you be careful about what you delete out when you make> such statements.I will keep that in mind.>I state as included in your post, so it's hard to see how you missed>it, but I'll give it again here for emphasis:>>where either r=0, or r changes as the polynomial's value changes,>while c does not.>>>But you state above:>>What do you mean about a polynomial's value changing? A polynomial>is in the>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,>it doesn't change.>>That's called bait-and-switch as you switched from the polynomial's>value to the polynomial.>>>>That's called using standard terminology. Precise use of terms is a >>hallmark of mathematics. A polynomial is an expression. For different >>values of its variable(s) it evaluates to different values, but that is >>not the same as saying the polynomial's value changes.> > Posts on newsgroups are not formal, and there's always a possibility> of misunderstandings.> > I'm not a mathematician. There are likely to be any number of times> when I end up using something that's not exactly the way you may be> used to seeing it, but if you're a mathematician, by de'nition you're> a math expert> > If I say the value of P(x) = x + 1 varies as x varies, then are you> suggesting you can't 'gure out what I mean, even if it's not exactly> what you're used to seeing?As stated here, I can understand it. I just wish to be careful to not place assumptions on what you mean that may not be accurate. If it makes me sound like I'm being nitpicky, I'll accept that. I prefer that to blasting you only to 'nd there has been a misunderstanding. The exchange below is a good example of trying to clearly understand what you're saying, even if it sounds like I'm asking annoying and/or stupid questions.>I didn't say the polynomial changes. I said its value changes and you>noted that in one sentence and switched in the other.>>Now given, say, the polynomial P(x) = x+1, the polynomial's value at>x=0, is 1.>>It's not rocket science.>>>>Correct. But you did NOT change the value of the polynomial, you >>changed x and you changed P at x. P(x)=x+1, P(0)=1. P's value didn't >>change, you simply evaluated it for a given value of x.> > > Then what changed? You say, ...you changed P at x. Yet above you> have:> > >What do you mean about a polynomial's value changing? A polynomial>is in the>form a_n*x^n+a_(n-1)*x^(n-1)+...+a_2*x^2+a_1*x+a_0. It simply _is_,>it doesn't change.> > > Now I take your claim to be that the value of a polynomial is its> explicit expression.> > While I said things like the polynomial's value at x=0.> > Your claim is that my usage is nonstandard.> > Ok, I don't care to hold on to a nonstandard usage as my aim is to> communicate.> > So then, however it's stated so that I can talk of a polynomial, like> x+1, and also talk of the value for some particular x.> > [deletia]> > Again this is a *post* on a newsgroup and not a formal paper. I'm> trying to explain in a medium that allows questions to be asked in> case there is a misunderstanding.Understood.> Here there seems to be a misunderstanding as to how one talks of the> values for the explicit expression of a polynomial P(x) for a given x.> > Now I'm suspicious of you still, but again, I see your post as typical> of what I've come to expect from mathematicians. Which is also why I say you're like lying English professors.> > I also say you're like bad lawyers who argue over technicalities, as> if the point or the truth doesn't matter.That's because I've found that technicalities can make the difference between a solid proof and a very pretty piece of garbage. When doing proofs, technicalities frequently matter.> I'm not interested in reading further as you've repeatedly> demonstrated behavior I 'nd deceitful and annoying.I can't promise that I'll say anything in the future that will not be similarly annoying. It is the nature of asking questions that they can become annoying. As for deceit, it suggests I have an agenda of some sort. I'll clearly state my agenda: to understand what you are saying, and determine if what you have said is valid. This could result in demonstrating that you are correct and misunderstod, or that you have made some mistakes that may or may not be 'xable. Either way is 'ne with me. If you believe I have some other agenda that is not appropriate, please state it and I will do my best to avoid it.-- Will Twentyman => I think I've 'gured out a way to show basically all of you, including> people who think they don't know any math that mathematicians have> been lying about my work. It's so trivial you *should* wonder why> they thought they could get away with it.> > Here goes.> > My paper Advanced Polynomial Factorization depends on considering a> factor of a polynomials that I call g.> > (Paper linked to at http://groups.msn.com/AmateurMath as usual.)> soon to be published as Mein Kampf. Twenty years of struggle againstthe world's mathematicians and their lies. => Two that readily come to mind are David Ullrich, a professor at> Oklahoma State University, and Arturo Magidin, whose university I'm> not sure of at this time as I'd have to check a header in a post of> his, but he obtained his Ph.D from Berkeley University.Sheesh, you can't even get the name of the University right! It's TheUniversity of California at Berkeley, or just Berkeley, but *not*Berkeley University!> Posts on newsgroups are not formal, and there's always a possibility> of misunderstandings.If you're going to talk about mathematics you need to use the properconversation in a coffee shop.(BTW, speaking of terminology, I really wish you'd refer to newsgroupTerminology matters in more places than mathematics, you know.)> I'm not a mathematician. There are likely to be any number of times> when I end up using something that's not exactly the way you may be> used to seeing it, but if you're a mathematician, by de'nition you're> a math expert.Idiot. If two mathematicians (or math experts as you prefer tocall them) try to discuss mathematics using non-standard terminology,and without clearly de'ning their non-standard terms *in relation to*standard terminology, then they'll probably confuse *each other*. It'seven *more* important for someone like you to be very careful and precisein de'ning and using your terms. That's why standard terminology existsin the 'rst place!> I'm not interested in reading further as you've repeatedly> demonstrated behavior I 'nd deceitful and annoying.This is James Harris code-speak for I'm completely out of my depth andcan't understand your argument well enough to dispute it intelligently,so I'll just ignore you and hope you go away. Whenever James startsaccusing people of lying or cheating or being deceitful, it's a suresign he knows he's outclassed.-- Wayne Brown | When your tail's in a crack, you improvisefwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock =You know, Harris, you seem to think that anyone who has a degree in ANYTHINGis a liar. What's your pathetic excuse?? Why do you, who have NO FORMALDEGREE IN MATHEMATICS, feel you can critique REAL mathematicians???You are such a loser, Harris!! The time you waste on these forums isamazing. If you could put forth 1/8th the effort into having a REAL lifeinstead of this pathetic research you're doing, you may actually make afriend or two. Until then, you continue to be a waste of human ¤esh...Don't forget your doomsday predictions from about a month or two ago tothose who doubted you... Weren't you planning on killing us all and you weregoing to be the only one left around???Can you really expect anyone to take you seriously??? You're such a joke...>>>I've looked at your paper... now looking at this I'll bite. I may be>>getting in over my head but I'll point out the things that don't make>>sense to me. >>>>Notes: I am a mathematician. I have only recently started readingthis>>board, so I am not familiar with your previous dif'culties with>mathematicians.>> That's just sad. It means that maybe you picked up bad habits which> as I've feared are typical for mathematicians.>> In any event, if you get serious I'll work through it carefully as> long as you don't try any tricks, no bait-and-switch, nor anything> else not aboveboard.>>>> It looks like you and James have been tricked into wasting> a few hours of each others time.>> Nope. I don't necessarily post for one person. Undergrads who *can*> follow and see what's going on can also get out of mathematics while> they can.>> Biology is a good major.>>> I really enjoyed James' opening line> and how it gradually evolved into> IT'S ALL IN THE PAPER YOU CLAIM TO> HAVE READ!!!!!!!!!!!!!!> That's it. I'm tired of your game.>> Yeah, my elation turned to frustration as 'rst I was repeating what> was in the paper, then dealing with underhanded things like> bait-and-switch and 'nally I just got tired of it.>>> If this was your intention, very well done! If not, don't take it> personal - it's a question of medication dosage. There must> be a balance somewhere. We know it will be unstable, but> it will be a balance - for a few weeks.>> Welcome to the playground of James Harrass.>> Dirk Vdm>>> I noticed it. I've lurked long enough that his response isn't> *entirely* surprising. I suppose I should go try to make sense of his> paper now, to see where I'm missing things. It would have been helpful> if he had started by posting the paper here instead of posting a link to> a pdf. It makes quoting his paper... more challenging. And makes it> less accessible.>> The paper started on the newsgroup.>> Undergrads the warning is clear. Mathematicians don't play fair, they> seem to run from logic if it doesn't suit them, and when you catch> them, they just come up with more *stuff*.>> I looked at his paper long enough to get confused by it. I didn't try> to make sense of it. I suppose I must defend my honor as a mathematician and start looking for the de'nitions in his paper.>> What I'm doing is showing that apparently there is little honor as a> mathematician, which is a modern thing.>> Mathematicians like Gauss, Euler, Dedekind, and so many others> wouldn't play these bull games, I'm sure.>> But I think a lot of you got in under a system where lies are not only> tolerated, but expected, where mathematics is just a club, so you can> pay your bills, get married, and teach poor unsuspecting students who> get caught up in the grandeur of mathematics past.>> Those students need to remember that the future keeps happening> despite the past.>> Mathematicians today *can* be as contemptibly bad as I've been showing> despite the grand history of the discipline.>> It apparently is going through a bad spell.>>> James Harris => Don't forget your doomsday predictions from about a month or two ago to> those who doubted you... Weren't you planning on killing us all and you were> going to be the only one left around???I don't remember that. There was the thing about how hehas Army generals just waiting for word from him to invadethe halls of academia. And the thing about how he's goingto press charges and everyone in the math establishment isgoing to jail. And occasionally the thing about how thehuman race is doomed to extinction if we don't accept hisproof as correct (haven't seen that one in a while). But Idon't remember that particular threat or prediction. - Randy => >>>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I quote it >directly in my responses?> > No.> Having spent the weekend going through musty books on algebra... hereis my analysis:BTW, I'm using Google to post this since my usual newsfeed is actingup.Original paper will be quoted, the rest is my comments. There may beformatting issues with the quoted paper since it was a .pdf ratherthan plain HARRIS>Abstract. Algebraic method for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization Approached>Determining the distribution of factors within irrational algebraic>integers has long been considered impossible as it is not possible todo>using Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight thestandard>belief consider the algebraic integers (1 − sqrt(21))/2 and(1+sqrt(21))/2 >which are roots of x^2 + x − 5.This is a minor note, but it should be -1 for both roots. While thishas little impact on the validity of the rest of the work, it should,at the very least, be corrected in your 'nal draft.>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not havenon>unit factors of 5? How do you know?These questions don't make a lot of sense to me, but I'll skip overthat to get to what I believe is the true problem.>In looking to consider distribution of algebraic integer factorswithin>a factorization I'll be using a more complicated example than x 2 +x− 5.>The paper will show that given the factorization, in the ring ofalge->braic integers,>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.And here we run into two problems. First, while it is true that thea's are algebraic integers, it is something that at least deserves acomment. Second, it is far from obvious that coprime is meaningfulon the algebraic integers. Here's the problem: coprime is based onthe de'nition of the least common divisor. The LCD is onlyguaranteed to exist for Principal Ideal Domains and Uniquealgebraic integers are certainly NOT a UFD, and I've seen nothing toindicate anything stronger than that they are a commutative ring. Until someone can point to work that indicates that LCD is de'ned onthe algebraic integers, this paper cannot move forward. The remainderof the paper (which I have NOT checked for accuracy but am includingfor completeness) depends on the concept of coprime being meaningfuland de'ned on the algebraic integers.>First I'll need a simple lemma to generalize beyond factors of apoly->nomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factor>for all x, which means that the value of g for a value .95a' of x is afactor>of P(a), within the ring of algebraic integers, there exists r and csuch>that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).The notation here could be improved. g should be g(x), r should ber(x).>Proof. Let x=0, then g must be a factor of P(0), so at that point c =g.>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r whichvaries>with x, and as r equals 0 when x equals 0 it is not coprime to>x.>I'm not sure step 2 makes sense. I don't see why it makes r coprimeto x.>As an example consider sqrt(x + 1) which is a non polynomial factorof>x+1, and while there are an in'nity of irrational solutions considerthe>rational solution at x=35.Are you sure you want to talk about non-polynomial roots? Whilelegitimate, it changes the topic of discussion from polynomials overthe algebraic integers to algebraic functions over the algebraicintegers.>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,>and c=1. But for different values of x, g and r will vary, while rwill>not be coprime to x.Again, coprime needs to be de'ned for the particular ring underconsideration.>2. Primary Argument>Given>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. Let>P(m) = f^2 ((m^3 f^4 − 3m^2 f^2 + 3m)x^3 − 3( − 1 +mf^2 )xu^2 + u^3 f)>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)>has a factor that is f^2 .Again, coprime must be de'ned on the algebraic integers. Also, thechange of variable and the introduction of additional variables seemsodd. Since it appears that you mean for f and u to be constants, itappears that you actually have P(m,x).>That expression comes from expanding (v^3 +1)x^3 − 3vxy^2 + y^3, using>the substitutions v = − 1+mf^2 , and y = uf, where additionalvariables>provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime inintegers,>as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the ring ofalgebraic>integers.I don't follow what you mean by this at all. Some additionalexplanation would be helpful. What are the factors and how did youcompute them?>Now consider the factorization>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows that>a1a2a3 = m^3 f^6 − 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 − 3m^2f^2 + 3m)>so a1a2a3 = mf^2 (m^2 f^4 − 3mf^2 + 3).>Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.>(Note: The a's are roots of a monic polynomial with algebraic integer>coef'cients so they are algebraic integers.)Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logicof your note could stand some clari'cation. (note: coprime shows upagain)>Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get thefactor>f^2 in the constant term P(0), while one must not equal 0, or f^3would>be the factor.This statement doesn't make a lot of sense either... but I'm nottrying as hard at this point. If the other problems can be 'xed,I'll be happy to examine it more carefully.>Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;therefore,>given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that samefactor>in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 − 3m^2 f^4 + 3m)x^3 − 3( − 1 + mf^2 )xu^2+ u^3 = 65x^3 − 12x + 1>which may be more easily seen from using v = − 1 + mf^2 = 4,y=1>with (v^3 + 1)x^3 − 3vxy^2 + y^3 .>Therefore, with the factorization>65x^3 − 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors beingirrational.Overall analysis: if you can clarify the notation and the logicalconnections, along with your result about factors not being factors,what you probably have is a proof that coprime is not always de'nedon the algebraic integers. = [.sci.skeptic removed from newsgroups.] [.snip.]>>In looking to consider distribution of algebraic integer factors>within>>a factorization I'll be using a more complicated example than x 2 +x>− 5.>>The paper will show that given the factorization, in the ring of>alge->>braic integers,>>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>>one of the a's is coprime to 5, using basic algebraic methods.>>And here we run into two problems. First, while it is true that the>a's are algebraic integers, it is something that at least deserves a>comment. Second, it is far from obvious that coprime is meaningful>on the algebraic integers. Here's the problem: coprime is based on>the de'nition of the least common divisor. The usual notion is not of a least common divisor, but of a->greatest<- common divisor, GCD.Coprime also makes sense for ideals: two ideals are coprime if andonly if there is no prime ideal containing both. If we assume theAxiom of Choice and that our rings have a 1, then this is equivalentto comaximal, no maximal ideal contains both ideals. As such, it ispossible to de'ne coprime for arbitrary commutative rings with 1:two elements, x and y in R, are coprime if and only if (x) and (y) arecoprime; this occurs if and only if (x,y)=(1); this occurs if and onlyif there exist a and b in R such that ax+by=1.>The LCD is only>guaranteed to exist for Principal Ideal Domains and Unique>algebraic integers are certainly NOT a UFD, and I've seen nothing to>indicate anything stronger than that they are a commutative ring.They are a Bezout Domain: every 'nitely generated ideal is principal:so given any two elements x and y, they DO have a GCD, given by anygenerator of the ideal (x,y); however, it is not unique, and there isno natural choice among all of them. On the plus side, any two differby a constant. [.rest deleted.] Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan =Arturo Magidinmagidin@math.berkeley.edu =... >The paper will show that given the factorization, in the ring of > alge- >braic integers, >65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1) >one of the a's is coprime to 5, using basic algebraic methods. > > And here we run into two problems. First, while it is true that the > a's are algebraic integers, it is something that at least deserves a > comment. Second, it is far from obvious that coprime is meaningful > on the algebraic integers. Here's the problem: coprime is based on > the de'nition of the least common divisor.The common de'nition for coprime is that within a ring, two numbersa and b are called co-prime if there are numbers x and y such thata*x + b*y = 1. This is a change from the former de'nition wherea and b are coprime if LCD(a, b) = 1 but the theorem that there arenumbers x and y such that a*x + b*y = LCD(a, b) assures that thede'nitions are the same when both could be applied. So in thealgebraic integers 2 and 3 are coprime because 2 * 2 + 3 * (-1) = 1. Lemma 1.1. Factorization Lemma: >Given a factor g of a polynomial P(x), further de'ned as a factor >for all x, which means that the value of g for a value .95a' of x is a > factor >of P(a), within the ring of algebraic integers, there exists r and c > such >that >g = r + c where r=0, or is not coprime to x, and c is a factor of the constant >term P(0). > > The notation here could be improved. g should be g(x), r should be > r(x).More could be improved, but as written it is wrong. Let me attempt abetter wording (as I think JSH intends it): Lemma 1.1. Factorisation lemma; Let's have a ring R (JSH intends the algebraic integers, but I generalise) and a polynomial P(X) in R[X]. A function g: R => R is de'ned to be a factor of P(X) if for all X in R, g(X) is a divisor of P(X). In that case there is a function r: R => R and a constant c in R such that either r(X) == 0, or r(X) is not coprime to X for all X in R and c divides P(0) in R.However with this formulation the lemma is wrong (this could be done in*all* rings that contain the integers): Let R be the algebraic integers. Let P(X) be the polynomial X + 5. Let g(X) be the function: g(X) = if X != 7: X + 5 = if X = 7: 3this satis'es the conditions: g(7) = 3 and P(7) = 12, for all otherX, g(X) = P(X), so g(X) is a factor of P(X).Because of the de'nition of g(X) for X != 7 we 'nd: c = 5, r(X) = if X != 7: X = if X = 7: -2.But r(7) = -2 is coprime to X = 7. > Proof. Let x=0, then g must be a factor of P(0), so at that point c = g.Agreed. > (1) If when x does not equal 0, g=c, r=0.And so P(X) is a multiple of c. The formulation is awkward... > (2) If when x does not equal 0, g =/= c there must exist r which varies > with x, and as r equals 0 when x equals 0 it is not coprime to x. > > I'm not sure step 2 makes sense. I don't see why it makes r coprime > to x.(You mean *not* coprime.) Indeed, it makes no sense. > Are you sure you want to talk about non-polynomial roots? While > legitimate, it changes the topic of discussion from polynomials over > the algebraic integers to algebraic functions over the algebraic > integers.But indeed, that is what he wants to do, opening a can of worms. > Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5, > and c=1. But for different values of x, g and r will vary, while r will > not be coprime to x. > > Again, coprime needs to be de'ned for the particular ring under > consideration.As it is de'ned, that is no problem. And indeed, the Lemma holdsfor sqrt(x + 1) as factor of x + 1. But I am not sure for whatkind of functions g(x) the lemma holds. Step 2 in the proof is noclari'cation for that, and there are (as shown above) function forwhich it does not hold.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => [...]>>[The algebraic integers] are a Bezout Domain: every 'nitely generated ideal is principal:>so given any two elements x and y, they DO have a GCD, given by any>generator of the ideal (x,y); however, it is not unique, and there is>no natural choice among all of them. On the plus side, any two differ>by a constant.That was a slight Harrisism for differ by a unit, right?************************David C. Ullrich =>>> [...]>>>>[The algebraic integers] are a Bezout Domain: every 'nitely generated ideal is principal:>>so given any two elements x and y, they DO have a GCD, given by any>>generator of the ideal (x,y); however, it is not unique, and there is>>no natural choice among all of them. On the plus side, any two differ>>by a constant.>>That was a slight Harrisism for differ by a unit, right?Well, it was certainly a lapsus and I meant differ by a unit.But just what is a Harrisism? It is not a simple error like theabove, certainly. It should encompass some complete missing the pointof a de'nition, theorem, or something, no? == ==It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) = Magidinmagidin@math.berkeley.edu = [.snip.]>More could be improved, but as written it is wrong. Let me attempt a>better wording (as I think JSH intends it):> Lemma 1.1. Factorisation lemma;> Let's have a ring R (JSH intends the algebraic integers, but I generalise)> and a polynomial P(X) in R[X]. A function g: R => R is de'ned to be> a factor of P(X) if for all X in R, g(X) is a divisor of P(X).> In that case there is a function r: R => R and a constant c in R such> that either r(X) == 0, or r(X) is not coprime to X for all X in R> and c divides P(0) in R.You forgot and g(X)=r(X)+c as functions...>However with this formulation the lemma is wrong (this could be done in>*all* rings that contain the integers):> Let R be the algebraic integers.> Let P(X) be the polynomial X + 5.> Let g(X) be the function:> g(X) = if X != 7: X + 5> = if X = 7: 3>this satis'es the conditions: g(7) = 3 and P(7) = 12, for all other>X, g(X) = P(X), so g(X) is a factor of P(X).>Because of the de'nition of g(X) for X != 7 we 'nd:> c = 5,> r(X) = if X != 7: X> = if X = 7: -2.>But r(7) = -2 is coprime to X = 7.A much easier way to see that the lemma is false in a ring with 1would be to note that if X is a unit, then r(X) cannot be not coprimeto X, and so that if X is a unit that would necessarily imply thatr(X) is identically zero. That means that r(X) would always have to bethe constant zero function, which means that g(X) would have to beconstant. This is clearly false for pretty much most rings. =Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan =Arturo Magidinmagidin@math.berkeley.edu =>>> [...]>>[The algebraic integers] are a Bezout Domain: every 'nitely generated ideal is principal:>so given any two elements x and y, they DO have a GCD, given by any>generator of the ideal (x,y); however, it is not unique, and there is>no natural choice among all of them. On the plus side, any two differ>by a constant.>>>>That was a slight Harrisism for differ by a unit, right?>>Well, it was certainly a lapsus and I meant differ by a unit.>>But just what is a Harrisism? It is not a simple error like the>above, certainly. It should encompass some complete missing the point>of a de'nition, theorem, or something, no?Calling it a Harrisism wasn't really appropriate, because it doesn'tinvolve any sort of awesome cluelessness, just writing one wordwhere you meant another, in an informal context. The only reasonit reminded me of JSH was because of perennial confusion overthe difference between constants and non-constants (no, notthat you were actually confused about anything like that.) Sorry.> == ==>It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> =>>Arturo Magidin>magidin@math.berkeley.edu************************ David C. Ullrich => [.snip.]>> On the plus side, any two differ by a constant.>>That was a slight Harrisism for differ by a unit, right?>>>>Well, it was certainly a lapsus and I meant differ by a unit.>>>>But just what is a Harrisism? It is not a simple error like the>>above, certainly. It should encompass some complete missing the point>>of a de'nition, theorem, or something, no?>>Calling it a Harrisism wasn't really appropriate, because it doesn't>involve any sort of awesome cluelessness, just writing one word>where you meant another, in an informal context. The only reason>it reminded me of JSH was because of perennial confusion over>the difference between constants and non-constants (no, not>that you were actually confused about anything like that.) Sorry.Actually, I was more interested in coming up with a good notion ofHarrisism we can use on the newsgroup... (-: You know, likespoonerism. It seems a bit hard to encompass all the behaviors that could be thusnamed, but we should we able to hash out something. I mean, had Ireplied attacking you for taking me to task on what was clearly just asimple error instead of just noting the error and leaving it at thator something, that would be harris-like behavior...Perhaps:Harrisism: the repeated and dogged blatantly incorrect use of atechnical term, caused by either ignorance or confusion.(So the recent discussion on .95well-ordering the reals' would be a Harrisism...)Any other suggestions? = very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidinmagidin@math.berkeley.edu =thank you. that's how bad I am, that I was always puzzledby that ax + by = 1 de'nition of coprimeness,as recondite as it is ... I think! > Lemma 1.1. Factorisation lemma;> Let's have a ring R (JSH intends the algebraic integers, but I generalise)> and a polynomial P(X) in R[X]. A function g: R => R is de'ned to be> a factor of P(X) if for all X in R, g(X) is a divisor of P(X).> In that case there is a function r: R => R and a constant c in R such> that either r(X) == 0, or r(X) is not coprime to X for all X in R> and c divides P(0) in R.> However with this formulation the lemma is wrong (this could be done in> *all* rings that contain the integers):> Let R be the algebraic integers.> Let P(X) be the polynomial X + 5.> Let g(X) be the function:> g(X) = if X != 7: X + 5> = if X = 7: 3> this satis'es the conditions: g(7) = 3 and P(7) = 12, for all other> X, g(X) = P(X), so g(X) is a factor of P(X).> Because of the de'nition of g(X) for X != 7 we 'nd:> c = 5,> r(X) = if X != 7: X> = if X = 7: -2.> But r(7) = -2 is coprime to X = 7.--UN HYDROGEN (sic; Methanex (TM) reformanteurs) ECONOMIE?...La Troi Phases d'Exploitation de la Protocols des Grises de Kyoto:(FOSSILISATION [McCainanites?] (TM/sic))/BORE/GUSH/NADIR @ http://www.tarpley.net/aobook.htm.Http://www.tarpley.net/ bushb.htm (content partiale, below): 17 -- L'ATTEMPTER de COUP D'ETAT, 3/30/81 23 -- Le FIN d'HISTOIRE 24 -- L'ORDEUR du MONDE NOUVEAU 25 -- THYROID STORK !?! = ... stuff deleted ...>>>>That was a slight Harrisism for differ by a unit, right?>>Well, it was certainly a lapsus and I meant differ by a unit.>>But just what is a Harrisism? It is not a simple error like the>above, certainly. It should encompass some complete missing the point>of a de'nition, theorem, or something, no?>>>>Calling it a Harrisism wasn't really appropriate, because it doesn't>>involve any sort of awesome cluelessness, just writing one word>>where you meant another, in an informal context. The only reason>>it reminded me of JSH was because of perennial confusion over>>the difference between constants and non-constants (no, not>>that you were actually confused about anything like that.) Sorry.> > > Actually, I was more interested in coming up with a good notion of> Harrisism we can use on the newsgroup... (-: You know, like> spoonerism. > > It seems a bit hard to encompass all the behaviors that could be thus> named, but we should we able to hash out something. I mean, had I> replied attacking you for taking me to task on what was clearly just a> simple error instead of just noting the error and leaving it at that> or something, that would be harris-like behavior...> > Perhaps:> > Harrisism: the repeated and dogged blatantly incorrect use of a> technical term, caused by either ignorance or confusion.> Harrisment? I think it 'ts [as well as having a 'tting homophone].> (So the recent discussion on .95well-ordering the reals' would be a Harrisism...)> Oh, you meant Harrisy!> Any other suggestions?> Harrisism would be included in the next edition of the Diagnostic andStatistical Manual of Mental Disorders, DSM-V. A subtype of BorderlinePersonality Disorder. Occasionally compounded with NarcissisticPersonality Disorder.> => It's not denial. I'm just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> => > Arturo Magidin> magidin@math.berkeley.edu > There are also Harristotelian logic (n), a contradiction in terms if ever there was one. It would involve a major effort to catalogue all the variants of this. Virtually any JSH thread will contain prime examples of this, that will, ideally, factor into any serious de'nition. Harrisite (n), a moral relative of the leech, one who depends critically on the assistance of others, but who actively subjects his helpers to abuse. Also, a member of the legion of imagined JSH supporters: JSH is addressing the Harrisites when he makes his appeals to the masses. Harriside (n), the so-called majority of one, although some may imagine a throng on the Harriside. The opposite of JSH's usage of some in the classic some may disagree with this argument. Not to be confused with: Harricide (n), an occasional fantasy of some of the less well self-controlled participants in these threads (several readers have suggested Harricide as a solution for sci.math). Harristic (n), an informal argument (that can only be expressed using utterly trivial (and non-informative) examples. Also (adj) an epithet painting someone's argument as belonging to this category. (The proof of the Advanced Factorization Lemma by JSH was via a Harristic argument). By extension, this usage may be applied to any object that fails to meet even minimal standards for its type, as in That sieve sure makes a Harristic bucket! Harrispective (adj): paying absolutely no regard to the arguments of others (usage: Harrispective of Magidin's argument, we 'nd q to be an algebraic integer unit with its inverse not integral over the rationals). Also (n): a peculiar form of insight leading to the propensity to pay no attention to others' arguments. Harision (n): 'erce, often comically so, verbal attack on the provider of technical assistance. All attempts to correct the argument were met with insult and Harrision.OK, that's all I could come up with. Sorry I couldn't go with thesuggested meaning of Harrisism, however.Dale. =>>>> [.snip.]>> On the plus side, any two differ by a constant.>>>>That was a slight Harrisism for differ by a unit, right?>>Well, it was certainly a lapsus and I meant differ by a unit.>>But just what is a Harrisism? It is not a simple error like the>above, certainly. It should encompass some complete missing the point>of a de'nition, theorem, or something, no?>>>>Calling it a Harrisism wasn't really appropriate, because it doesn't>>involve any sort of awesome cluelessness, just writing one word>>where you meant another, in an informal context. The only reason>>it reminded me of JSH was because of perennial confusion over>>the difference between constants and non-constants (no, not>>that you were actually confused about anything like that.) Sorry.>>Actually, I was more interested in coming up with a good notion of>Harrisism we can use on the newsgroup... (-: That's a lie. (hth)>You know, like spoonerism. >It seems a bit hard to encompass all the behaviors that could be thus>named, but we should we able to hash out something. I mean, had I>replied attacking you for taking me to task on what was clearly just a>simple error instead of just noting the error and leaving it at that>or something, that would be harris-like behavior...>>Perhaps:>>Harrisism: the repeated and dogged blatantly incorrect use of a>technical term, caused by either ignorance or confusion.>>(So the recent discussion on .95well-ordering the reals' would be a Harrisism...)>>Any other suggestions?>> = just very selective about> what I accept as reality.> --- Calvin (Calvin and Hobbes)> =>>Arturo Magidin>magidin@math.berkeley.edu>>******************* *****David C. Ullrich => Will Twentyman>>>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if Iquote it>>directly in my responses?>>> No.>>Having spent the weekend going through musty books on algebra...here is my analysis:Original paper will be quoted, the rest is my comments. There may beformatting issues with the quoted paper since it was a .pdf ratherthan plain text.>ADVANCED method for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization Approached>Determining the distribution of factors within irrational algebraic>integers has long been considered impossible as it is not possible to >dousing Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight the standard>belief consider the algebraic integers (1 - sqrt(21))/2 and>(1+sqrt(21))/2 >which are roots of x^2 + x - 5.This is a minor note, but it should be -1 for both roots. While thishas little impact on the validity of the rest of the work, it should,at the very least, be corrected in your 'nal draft.>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not>have non unit factors of 5? How do you know?These questions don't make a lot of sense to me, but I'll skip overthat to get to what I believe is the true problem.>In looking to consider distribution of algebraic integer factors>within a factorization I'll be using a more complicated example than>x^2 +x - 5.>The paper will show that given the factorization, in the ring of>algebraic integers,>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.And here we run into two problems. First, while it is true that thea's are algebraic integers, it is something that at least deserves acomment. Second, it is far from obvious that coprime is meaningfulon the algebraic integers. Here's the problem: coprime is based onthe de'nition of the least common divisor. The LCD is onlyguaranteed to exist for Principal Ideal Domains and Uniquealgebraic integers are certainly NOT a UFD, and I've seen nothing toindicate anything stronger than that they are a commutative ring.Until someone can point to work that indicates that LCD is de'ned onthe algebraic integers, this paper cannot move forward. The remainderof the paper (which I have NOT checked for accuracy but am includingfor completeness) depends on the concept of coprime being meaningfuland de'ned on the algebraic integers.>First I'll need a simple lemma to generalize beyond factors of a>polynomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factor>for all x, which means that the value of g for a value .95a' of x is>a factor of P(a), within the ring of algebraic integers, there exists>r and c such that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).The notation here could be improved. g should be g(x), r should ber(x).>Proof. Let x=0, then g must be a factor of P(0), so at that point>c = g.>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r which>varies with x, and as r equals 0 when x equals 0 it is not coprime to>x.>I'm not sure step 2 makes sense. I don't see why it makes r coprimeto x.>As an example consider sqrt(x + 1) which is a non polynomial factor>of x+1, and while there are an in'nity of irrational solutions>consider the rational solution at x=35.Are you sure you want to talk about non-polynomial roots? Whilelegitimate, it changes the topic of discussion from polynomials overthe algebraic integers to algebraic functions over the algebraicintegers.>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,>and c=1. But for different values of x, g and r will vary, while r>will not be coprime to x.Again, coprime needs to be de'ned for the particular ring underconsideration.>2. Primary Argument>Given>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. Let>P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 +u^3 f)>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)>has a factor that is f^2 .Again, coprime must be de'ned on the algebraic integers. Also, thechange of variable and the introduction of additional variables seemsodd. Since it appears that you mean for f and u to be constants, itappears that you actually have P(m,x).>That expression comes from expanding (v^3 +1)x^3 - 3vxy^2 + y^3 ,>using the substitutions v = - 1+mf^2 , and y = uf, where additional>variables provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime in>integers, as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the>ring of algebraic integers.I don't follow what you mean by this at all. Some additionalexplanation would be helpful. What are the factors and how did youcompute them?>Now consider the factorization>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows that>a1a2a3 = m^3 f^6 - 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 - 3m^2 f^2 + 3m)>so a1a2a3 = mf^2 (m^2 f^4 - 3mf^2 + 3).>Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.>(Note: The a's are roots of a monic polynomial with algebraic>integer coef'cients so they are algebraic integers.)Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logicof your note could stand some clari'cation. (note: coprime shows upagain)>Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get the>factor f^2 in the constant term P(0), while one must not equal 0, or >f^3would be the factor.This statement doesn't make a lot of sense either... but I'm nottrying as hard at this point. If the other problems can be 'xed,I'll be happy to examine it more carefully.>Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;>therefore, given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that same>factor in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 = 65x^3 -12x + 1>which may be more easily seen from using v = - 1 + mf^2 = 4, y=1>with (v^3 + 1)x^3 - 3vxy^2 + y^3 .>Therefore, with the factorization>65x^3 - 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors being irrational.Overall analysis: if you can clarify the notation and the logicalconnections, along with your result about factors not being factors,what you probably have is a proof that coprime is not always de'nedon the algebraic integers.-- Will Twentyman =Admire you, Will.... wait to see what his reply is.....> Will Twentyman>>>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I> quote it>directly in my responses?>>> No.>>> Having spent the weekend going through musty books on algebra...> here is my analysis:>>> Original paper will be quoted, the rest is my comments. There may be> formatting issues with the quoted paper since it was a .pdf rather> than plain text.>>ADVANCED POLYNOMIAL for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization ApproachedDetermining the distribution of factors within irrational algebraic>integers has long been considered impossible as it is not possible to >do> using Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight the standardbelief consider the algebraic integers (1 - sqrt(21))/2 and>(1+sqrt(21))/2 >which are roots of x^2 + x - 5.>> This is a minor note, but it should be -1 for both roots. While this> has little impact on the validity of the rest of the work, it should,> at the very least, be corrected in your 'nal draft.>>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not>have non unit factors of 5? How do you know?>> These questions don't make a lot of sense to me, but I'll skip over> that to get to what I believe is the true problem.>>In looking to consider distribution of algebraic integer factors>within a factorization I'll be using a more complicated example than>x^2 +x - 5.>The paper will show that given the factorization, in the ring of>algebraic integers,>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.>> And here we run into two problems. First, while it is true that the> a's are algebraic integers, it is something that at least deserves a> comment. Second, it is far from obvious that coprime is meaningful> on the algebraic integers. Here's the problem: coprime is based on> the de'nition of the least common divisor. The LCD is only> guaranteed to exist for Principal Ideal Domains and Unique> algebraic integers are certainly NOT a UFD, and I've seen nothing to> indicate anything stronger than that they are a commutative ring.> Until someone can point to work that indicates that LCD is de'ned on> the algebraic integers, this paper cannot move forward. The remainder> of the paper (which I have NOT checked for accuracy but am including> for completeness) depends on the concept of coprime being meaningful> and de'ned on the algebraic integers.>>First I'll need a simple lemma to generalize beyond factors of a>polynomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factor>for all x, which means that the value of g for a value .95a' of x is>a factor of P(a), within the ring of algebraic integers, there exists>r and c such that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).>> The notation here could be improved. g should be g(x), r should be> r(x).>>Proof. Let x=0, then g must be a factor of P(0), so at that point>c = g.>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r which>varies with x, and as r equals 0 when x equals 0 it is not coprime to>x.>>> I'm not sure step 2 makes sense. I don't see why it makes r coprime> to x.>>As an example consider sqrt(x + 1) which is a non polynomial factor>of x+1, and while there are an in'nity of irrational solutions>consider the rational solution at x=35.>> Are you sure you want to talk about non-polynomial roots? While> legitimate, it changes the topic of discussion from polynomials over> the algebraic integers to algebraic functions over the algebraic> integers.>>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,and c=1. But for different values of x, g and r will vary, while r>will not be coprime to x.>> Again, coprime needs to be de'ned for the particular ring under> consideration.>>2. Primary Argument>Given>65x^3 - 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. Let>P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 +> u^3 f)>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)>has a factor that is f^2 .>> Again, coprime must be de'ned on the algebraic integers. Also, the> change of variable and the introduction of additional variables seems> odd. Since it appears that you mean for f and u to be constants, it> appears that you actually have P(m,x).>That expression comes from expanding (v^3 +1)x^3 - 3vxy^2 + y^3 ,>using the substitutions v = - 1+mf^2 , and y = uf, where additional>variables provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime in>integers, as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the>ring of algebraic integers.>> I don't follow what you mean by this at all. Some additional> explanation would be helpful. What are the factors and how did you> compute them?>>Now consider the factorizationP(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows that>a1a2a3 = m^3 f^6 - 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 - 3m^2 f^2 + 3m)>so a1a2a3 = mf^2 (m^2 f^4 - 3mf^2 + 3).>Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.>(Note: The a's are roots of a monic polynomial with algebraic>integer coef'cients so they are algebraic integers.)>> Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logic> of your note could stand some clari'cation. (note: coprime shows up> again)>>Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>g1 = c = uf>meaning f is a factor of the constant term.Therefore, exactly two of the a's equal 0, when m=0, to get the>factor f^2 in the constant term P(0), while one must not equal 0, or >f^3> would be the factor.>> This statement doesn't make a lot of sense either... but I'm not> trying as hard at this point. If the other problems can be 'xed,> I'll be happy to examine it more carefully.>Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;>therefore, given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that samefactor in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 - 3m^2 f^4 + 3m)x^3 - 3( - 1 + mf^2 )xu^2 + u^3 = 65x^3 -> 12x + 1which may be more easily seen from using v = - 1 + mf^2 = 4, y=1>with (v^3 + 1)x^3 - 3vxy^2 + y^3 .>Therefore, with the factorization>65x^3 - 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors being irrational.>> Overall analysis: if you can clarify the notation and the logical> connections, along with your result about factors not being factors,> what you probably have is a proof that coprime is not always de'ned> on the algebraic integers.>> --> Will Twentyman>>> => ...>The paper will show that given the factorization, in the ring of> alge->braic integers,>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.> > And here we run into two problems. First, while it is true that the> a's are algebraic integers, it is something that at least deserves a> comment. Second, it is far from obvious that coprime is meaningful> on the algebraic integers. Here's the problem: coprime is based on> the de'nition of the least common divisor.> > The common de'nition for coprime is that within a ring, two numbers> a and b are called co-prime if there are numbers x and y such that> a*x + b*y = 1. This is a change from the former de'nition where> a and b are coprime if LCD(a, b) = 1 but the theorem that there are> numbers x and y such that a*x + b*y = LCD(a, b) assures that the> de'nitions are the same when both could be applied. So in the> algebraic integers 2 and 3 are coprime because 2 * 2 + 3 * (-1) = 1.> I saw that as one de'nition, but I'm not sure that's what JSH had in mind. Either way, we now have a de'nition, can he make it work?> Are you sure you want to talk about non-polynomial roots? While> legitimate, it changes the topic of discussion from polynomials over> the algebraic integers to algebraic functions over the algebraic> integers.> > But indeed, that is what he wants to do, opening a can of worms.Not to mention completely changing the topic.> Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5, and c=1. But for different values of x, g and r will vary, while r will> not be coprime to x.> > Again, coprime needs to be de'ned for the particular ring under> consideration.> > As it is de'ned, that is no problem. And indeed, the Lemma holds> for sqrt(x + 1) as factor of x + 1. But I am not sure for what> kind of functions g(x) the lemma holds. Step 2 in the proof is no> clari'cation for that, and there are (as shown above) function for> which it does not hold.I've noticed that he stopped replying to us. Have we perhaps convinced him? It would be nice to know.-- Will Twentyman => I've noticed that he stopped replying to us. Have we perhaps convinced > him? It would be nice to know.Oh, don't worry, he comes and goes in cycles... perhaps he circulatesaround to other newsgroups in a great migration, I'm not sure.In any case, probably in about another week he'll start posting a newseries of rants and will treat these threads as if they never happened. =... > I've noticed that he stopped replying to us. Have we perhaps convinced > him? It would be nice to know.No chance. He just stopped reading this thread and will start anewin another thread.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ = Will Twentyman>>>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I> quote it>>directly in my responses?>>> No.>> Having spent the weekend going through musty books on algebra...> here is my analysis:> > Harris's paper has already been proven to be incorrectby two posters independently. See the Harris thread in sci.mathentitled My Work -- Objective Review, and the poston June 18 by Overall analysis: if you can clarify the notation and the logical> connections, along with your result about factors not being factors,> what you probably have is a proof that coprime is not always de'ned> on the algebraic integers. This is not accurate - the algebraic integers form a Bezoutdomain - suggest you consult some good texts on algebraic number theory. Andrzej => >>>Will Twentyman>>>>>>>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I>>>> quote it>>>>directly in my responses?>>>No.>>>>>>Having spent the weekend going through musty books on algebra...>>here is my analysis:>>>>> > > Harris's paper has already been proven to be incorrect> by two posters independently. See the Harris thread in sci.math> entitled My Work -- Objective Review, and the post> on June 18 by Nora Baron and that on June 22 by W. Dale Hall.> > > notation and the logical>>connections, along with your result about factors not being factors,>>what you probably have is a proof that coprime is not always de'ned>>on the algebraic integers.> > > > This is not accurate - the algebraic integers form a Bezout> domain - suggest you consult some good texts on algebraic > number theory.> AndrzejThe key word was if. I missed a better de'nition of coprime, but had a strong suspicion that the holes were huge.-- Will Twentyman => >>> > > Yeah, my elation turned to frustration as 'rst I was repeating what> was in the paper, then dealing with underhanded things like> bait-and-switch and 'nally I just got tired of it.> > Notice that you assumed> 1) I made a bait-and-switch and> 2) If I did, it was intentional rather than a result of not understanding.What I've noticed is that mathematicians tend to follow a pattern whenthey're trying to deny the correctness of my work. >If this was your intention, very well done! If not, don't take itpersonal - it's a question of medication dosage. There must>be a balance somewhere. We know it will be unstable, but>it will be a balance - for a few weeks.>Welcome to the playground of James Harrass.>Dirk Vdm>>>>>>I noticed it. I've lurked long enough that his response isn't >>*entirely* surprising. I suppose I should go try to make sense of his >paper now, to see where I'm missing things. It would have been helpful >>if he had started by posting the paper here instead of posting a link to >>a pdf. It makes quoting his paper... more challenging. And makes it >>less accessible.> > > The paper started on the newsgroup.> > In sci.math, where I picked this up, you posted a link to the paper that > do not recall ever seeing it in sci.math, but that could easily be an > oversite.It started on the newsgroup.Now let's see about your other posts.James Harris =>> >>>>> >> >> Yeah, my elation turned to frustration as 'rst I was repeating what>> was in the paper, then dealing with underhanded things like>> bait-and-switch and 'nally I just got tired of it.>> >> Notice that you assumed>> 1) I made a bait-and-switch and>> 2) If I did, it was intentional rather than a result of not understanding.>>What I've noticed is that mathematicians tend to follow a pattern when>they're trying to deny the correctness of my work.This is the pattern they tend to follow:1. This doesn't make any sense2. You didn't de'ne this term2a. You keep using the term factor. I do not think it means what you think it means.3. Are you going to stop ranting at mathematicians and actually write some math?4. I guess not.5. Are these algebraic integers, integers, polynomials, or what?6. If you numbered your equations it would make life a lot easier.7. No one is objecting to that speci'c example, it's the general case that we have a problem with. 8. To hell with this. I'm going to get a beer.Alan-- Defendit numerus => >>Hmmm...doesn't sound like you actually looked at the paper.>>>>I'll be going over it in detail this weekend. Do you mind if I quote it >>directly in my responses?> > No.> > > Having spent the weekend going through musty books on algebra... here> is my analysis:Hmmm...so my short paper sent you to books for an entire weekend?This post should be interesting.> BTW, I'm using Google to post this since my usual newsfeed is acting> up.> > Original paper will be quoted, the rest is my comments. There may be> formatting issues with the quoted paper since it was a .pdf rather> than plain text.> Abstract. Algebraic method for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization Approached>Determining the distribution of factors within irrational algebraicintegers has long been considered impossible as it is not possible to> do>using Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight the> standard>belief consider the algebraic integers (1 − sqrt(21))/2 and> (1+sqrt(21))/2 >which are roots of x^2 + x − 5.> > This is a minor note, but it should be -1 for both roots. While this> has little impact on the validity of the rest of the work, it should,> at the very least, be corrected in your 'nal draft.That was pointed out by someone else, I've acknowledged it as amistake, and I've changed it in the paper.>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not have> non>unit factors of 5? How do you know?> > These questions don't make a lot of sense to me, but I'll skip over> that to get to what I believe is the true problem.Well, it looks to me like your comment might be taken as somewhatnegative, but the preamble to the paper can be skipped.>In looking to consider distribution of algebraic integer factors> within>a factorization I'll be using a more complicated example than x 2 +x> − 5.>The paper will show that given the factorization, in the ring of> alge->braic integers,>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.>> > And here we run into two problems. First, while it is true that the> a's are algebraic integers, it is something that at least deserves a> comment. Why should that be commented on here?>Second, it is far from obvious that coprime is meaningful> on the algebraic integers. That was discussed by others in reply to your post.However I'll add that I switched to coprime rather than use factorin this situation because algebraic integers is an incomplete ring sothere is a problem with the term factor if I want to stay in the ring. That is, rather than say that one of the a's is coprime, I might havesaid that only two of the a's have a factor of f, but the ring isincomplete, so that's problematic.That is, the problem with algebraic integers being incomplete meansthat I'd have to knowingly use factor in a way not exactly correct,but coprime still works, so I used it.>Here's the problem: coprime is based on> the de'nition of the least common divisor. The LCD is only> guaranteed to exist for Principal Ideal Domains and Unique> algebraic integers are certainly NOT a UFD, and I've seen nothing to> indicate anything stronger than that they are a commutative ring. > Until someone can point to work that indicates that LCD is de'ned on> the algebraic integers, this paper cannot move forward. The remainder> of the paper (which I have NOT checked for accuracy but am including> for completeness) depends on the concept of coprime being meaningful> and de'ned on the algebraic integers.The problem is that algebraic integers are incomplete as a ring. Whatthe paper shows is that two of the a's should have f as a factor, asthe paper shows them to have f as a factor, but technically, it's nota factor within the ring of algebraic integers.>First I'll need a simple lemma to generalize beyond factors of a> poly->nomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factorfor all x, which means that the value of g for a value .95a' of x is a> factor>of P(a), within the ring of algebraic integers, there exists r and c> such>that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).> > The notation here could be improved. g should be g(x), r should be> r(x).I'm not interested in style issues.You have claimed to be a mathematician, and mathematicians are de'nedto be math experts. The fundamental question about the paper is notstyle but correctness. Minor issues aren't relevant, but so farthat's all you have.For instance, you talked about the sign problem in the preamble, andthen you questioned the use of coprime, and now you talk aboutnotation.These type of issues are the kinds of things I see brought up bypeople I deem aren't interested in the truth.The discussion is not a game. It's not about winning or losing. It'sabout the truth.If I'm correct then there's a problem in mathematics which needs to be'xed.If I'm wrong then you should be able to 'nd more than a style issue.>Proof. Let x=0, then g must be a factor of P(0), so at that point c => g.>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r which> varies>with x, and as r equals 0 when x equals 0 it is not coprime to>x.> > > I'm not sure step 2 makes sense. I don't see why it makes r coprime> to x.The argument isn't complicated but I simpli'ed that section ratherthan deal with it, after I realized it wasn't correct.That is, r is not coprime to x or contains a unit factor of x.With that unit factor things get messy as I'm sure some would jump onthat and claim that since r would always have a unit factor of x, it'smeaningless.Rather than get into an involved explanation with more room forconfusion, I realized that I really was just using the fact that anyfactor of a polynomial can be split up between what's constant andwhat's varying.That is, r changes, but c does not, as x varies.The proof of that is actually trivial as all you do is take g at 0,which gives you c, then r = g-c.Given that c is constant, while g is not, obviously r is not either.>As an example consider sqrt(x + 1) which is a non polynomial factor> of>x+1, and while there are an in'nity of irrational solutions consider> the>rational solution at x=35.> > Are you sure you want to talk about non-polynomial roots? While> legitimate, it changes the topic of discussion from polynomials over> the algebraic integers to algebraic functions over the algebraic> integers.Which would seem to indicate that you haven't read further down in thepaper, or haven't bothered to consider what I said before thissection:First I'll need a simple lemma to generalize beyond factors of apolynomial that are themselves polynomials.Now I'd think that says what you need to know, and besides, what'swith the talk of roots?What are non-polynomial roots?Are you sure you're a mathematician?>Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,>and c=1. But for different values of x, g and r will vary, while r> willnot be coprime to x.> > Again, coprime needs to be de'ned for the particular ring under> consideration.And it sounds to me like maybe you didn't spend a weekend going overmath texts as you repeatedly brought up as an issue something otherposters dismissed immediately.>2. Primary Argument>Given>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. LetP(m) = f^2 ((m^3 f^4 − 3m^2 f^2 + 3m)x^3 − 3( − 1 +> mf^2 )xu^2 + u^3 f)>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)has a factor that is f^2 .> > Again, coprime must be de'ned on the algebraic integers. Also, the> change of variable and the introduction of additional variables seems> odd. Since it appears that you mean for f and u to be constants, it> appears that you actually have P(m,x).Nope. And if you're a mathematician changing variables as I didshouldn't be a problem for you.Again, the question is to the correctness of the paper, and not styleissues.So far, just like before when I got upset, you've spent a lot of timeon nonessential issues, or ones that others quickly refuted.It seems to me that your behavior is consistent with what I've seenfrom posters trying to hide the truth, rather than get to the bottomof things.But the issue is taught mathematics which is wrong.>That expression comes from expanding (v^3 +1)x^3 − 3vxy^2 + y^3> , using>the substitutions v = − 1+mf^2 , and y = uf, where additional> variables>provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime in> integers,>as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the ring of> algebraicintegers.> > I don't follow what you mean by this at all. Some additional> explanation would be helpful. What are the factors and how did you> compute them?I explain to some extent where the expression I use comes from, andthen give a quick analogy. If the analogy confuses you, you can skipit, as you did with the other section that confused you.It's not part of the argument, but merely there to help.>Now consider the factorization>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows that>a1a2a3 = m^3 f^6 − 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 − 3m^2> f^2 + 3m)>so a1a2a3 = mf^2 (m^2 f^4 − 3mf^2 + 3).Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.(Note: The a's are roots of a monic polynomial with algebraic integer>coef'cients so they are algebraic integers.)> > Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logic> of your note could stand some clari'cation. (note: coprime shows up> again)Strange question, given what follows:>Notice that the constant term P(0) is given byP(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get the> factor>f^2 in the constant term P(0), while one must not equal 0, or f^3> would>be the factor. This statement doesn't make a lot of sense either... but I'm not> trying as hard at this point. If the other problems can be 'xed,> I'll be happy to examine it more carefully.In the paper I take a rather complicated expression, consider it as apolynomial with respect to m, meaning that I let only m vary, and then'nd that I can factor it in a not surprising way, given theexpression.That factorization gives me non-polynomial factors, with which I usemy lemma to show that only one of the a's is coprime to f.The factorization I use, isn't terribly surprising, and the argumentfollows rather simply from a rather basic lemma, and the distributiveproperty.>Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;> therefore,given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that same> factor>in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 − 3m^2 f^4 + 3m)x^3 − 3( − 1 + mf^2 )xu^2> + u^3 = 65x^3 − 12x + 1>which may be more easily seen from using v = − 1 + mf^2 = 4,> y=1>with (v^3 + 1)x^3 − 3vxy^2 + y^3 .Therefore, with the factorization>65x^3 − 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors being> irrational.> > Overall analysis: if you can clarify the notation and the logical> connections, along with your result about factors not being factors,> what you probably have is a proof that coprime is not always de'ned> on the algebraic integers.Well I refer you to the statements by others about coprime in thering of algebraic integers.What I can do is take out the use of the term in the paper, and usefactor in the ring of algebraic integers, and note that will lead toa contradiction as the point of the paper is that the ring isincomplete.What I want to impress upon readers is that I'm quite willing to workwith mathematicians to explain the mistake that they're teaching.The ring of algebraic integers is incomplete, it's easy to show, andI've shown it with my paper.If mathematicians are having trouble understanding any part of it, I'mable to explain further.James Harris =[...]| What I can do is take out the use of the term in the paper, and use| factor in the ring of algebraic integers, and note that will lead to| a contradiction as the point of the paper is that the ring is| incomplete.| | What I want to impress upon readers is that I'm quite willing to work| with mathematicians to explain the mistake that they're teaching.If you want to show that a mistake is being taught, you should quote itspeci'cally.| The ring of algebraic integers is incomplete, it's easy to show, and| I've shown it with my paper.| | If mathematicians are having trouble understanding any part of it, I'm| able to explain further.You have a big problem with terminology. You use terms that aren'tused by others in the context in which you use them, such as countingfactors of 5 in algebraic integers. I haven't seen anything I wouldconsider an adequate de'nition of those terms. You don't explainwhy your terms should be considered relevant to the ones mathematiciansare using.I don't know what you think is a fair way for things to work, but theway things actually work, this kind of problem with terminology willabsolutely prevent you from making any headway.In an experiment in communication, I once tried elaborating on such aconcept for you. You usually write as though the number of factors of5 in an algebraic number satis'ed certain axioms which are familiarto me: being a rational number v(x)>=0 associated with each algebraicinteger x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraicintegers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriacintegers x<>0, y<>0, and x+y<>0. Assuming that we have such a functionv, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the rootsr1, r2, and r3 of your cubic, if we take them in the right order.I think this is the best way to try to make sense of your argumentin your advanced polynomial factorization thing.Unfortunately, this known concept is not so directly related todivisibility in the algebraic integers, so even if you did manage toproduce a de'nition on these lines, it wouldn't show mathematiciansare doing anything wrong when they make claims about divisibility inthe algebraic integers as they teach about it in classrooms. They'rejust apples and oranges.Say we de'ne the number of factors of 5 in an algebraic integer x to bethe highest rational number r such that 5^r divides x in the algebraicintegers, i.e. such that there exists an algebraic integer y such that5^r*y = x. I think this de'nition works (i.e., I think there is such ahighest rational power for each algebraic number, although I haven'ttried to write out a proof). If we de'ne it that way, though, then itsimply doesn't have one of the properties I listed above.Keith Ramsay => The ring of algebraic integers is incomplete, it's easy to show, and> I've shown it with my paper.> > If mathematicians are having trouble understanding any part of it, I'm> able to explain further.I don't understand what it means to say, The ring of algebraicintegers is incomplete.I don't claim to be a mathematician (I'm a jigsaw puzzle retailer,actually), but there's considerable evidence that most mathematicianswho've looked can't understand it either. Could you please (either orboth):(a) Post a clear de'nition, of the sort one sees in maths books of acomplete ring(b) Say of some simple rings whether they are complete or not -The ring of integers (Z)The ring of integers modulo 12 (clock arithmetic)The ring of integers modulo 2 (even/odd arithmetic)The ring of integers modulo 7(b') If all of these are complete, can you name another ring otherthan the ring of algebraic integers which is also incomplete?Brian Chandler----------------geo://Sano.Japan.Planet_3Jigsaw puzzles from Japan at:http://imaginatorium.org/shop/ => [...]> | What I can do is take out the use of the term in the paper, and use> | factor in the ring of algebraic integers, and note that will lead to> | a contradiction as the point of the paper is that the ring is> | incomplete.> | > | What I want to impress upon readers is that I'm quite willing to work> | with mathematicians to explain the mistake that they're teaching.> > If you want to show that a mistake is being taught, you should quote it> speci'cally.That's not very sensible if you expect that for a mistake to be taughtfor any length of time it'd have some subtlety, unless you're alsoquestioning the competence of mathematicians.It seems to me that rather than deal with the important question,which is whether or not I'm correct, people often try to introduceweird ad hoc conditions.If you disagree with that assessment, can you please explain why youbelieve there should be something simple and short enough for me toquote? > | The ring of algebraic integers is incomplete, it's easy to show, and> | I've shown it with my paper.> | > | If mathematicians are having trouble understanding any part of it, I'm> | able to explain further.> > You have a big problem with terminology. You use terms that aren't> used by others in the context in which you use them, such as counting> factors of 5 in algebraic integers. I haven't seen anything I would> consider an adequate de'nition of those terms. You don't explain> why your terms should be considered relevant to the ones mathematicians> are using.That statement possibly has something to do with what I now call thepreamble in the paper as it is there to help explain context, butisn't part of the actual argument.Here's what I say, copied from the paper (some editing for format):To highlight the standard belief consider the algebraic integers(-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.While you know that the algebraic integer factors are themselvesfactors of 5, in what way is each a factor of 5? Can either not havenon unit factors of 5? How do you know?There I'm talking about algebraic integers, so one can assume I'mtalking about algebraic integer factors. Given that 5 has algebraicinteger factors, how is what I say nonstandard Keith Ramsay?I'd like you to carefully explain your assertion. The question I'mraising is the possibility that you lied.> I don't know what you think is a fair way for things to work, but the> way things actually work, this kind of problem with terminology will> absolutely prevent you from making any headway.What will prevent me from making headway is if mathematicianscontinually lie.Now then in what way is it improper terminology to talk aboutalgebraic integer factors of 5? > In an experiment in communication, I once tried elaborating on such a> concept for you. You usually write as though the number of factors of> 5 in an algebraic number satis'ed certain axioms which are familiar> to me: being a rational number v(x)>=0 associated with each algebraic> integer x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraic> integers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac> integers x<>0, y<>0, and x+y<>0. Assuming that we have such a function> v, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the roots> r1, r2, and r3 of your cubic, if we take them in the right order.> I think this is the best way to try to make sense of your argument> in your advanced polynomial factorization thing.> > Unfortunately, this known concept is not so directly related to> divisibility in the algebraic integers, so even if you did manage to> produce a de'nition on these lines, it wouldn't show mathematicians> are doing anything wrong when they make claims about divisibility in> the algebraic integers as they teach about it in classrooms. They're> just apples and oranges.There is no divisibility argument within the paper, and I only evenmention roots when the roots are algebraic integers. I've talkedabout x/y in discussing the error in taught mathematics, but did makea post explaining that was when an argument is considered in the 'eldof rationals.Your statement falls ¤at Keith Ramsay, and I think you're just tryingto sound good enough to fool people, rather than trying to get to thetruth.What is clear is that the paper does NOT operate over any 'elds, anddepends only on ring operations.> Say we de'ne the number of factors of 5 in an algebraic integer x to be> the highest rational number r such that 5^r divides x in the algebraic> integers, i.e. such that there exists an algebraic integer y such that> 5^r*y = x. I think this de'nition works (i.e., I think there is such a> highest rational power for each algebraic number, although I haven't> tried to write out a proof). If we de'ne it that way, though, then it> simply doesn't have one of the properties I listed above.> > Keith RamsayYour post isn't coherent mathematically given what I say in the paper. Consider that in the ring of algebraic integers, 5 has algebraicinteger factors, and given algebraic integers r_1 and r_2, where theirproduct is 5, why are you acting as if it's so dif'cult to comprehendthat there must be some distribution of factors of 5?For instance, both could have a factor that is sqrt(5), or one couldhave a factor of 1+2i, while the other had a factor of 1-2i.The question I'm trying to get the reader to explore is, given thatthey are roots of this particular polynomial, could one of them onlyhave unit factors of 5?How do you know?James Harris =... > To highlight the standard belief consider the algebraic integers > (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5. > > While you know that the algebraic integer factors are themselves > factors of 5, in what way is each a factor of 5? Can either not have > non unit factors of 5? How do you know?Note that the polynomial is primitive.... > For instance, both could have a factor that is sqrt(5), or one could > have a factor of 1+2i, while the other had a factor of 1-2i. > > The question I'm trying to get the reader to explore is, given that > they are roots of this particular polynomial, could one of them only > have unit factors of 5? > > How do you know?(Note: with divisor I mean divisor in the ring of algebraic integers.)1. Each divisor of r1 is also a divisor of 5 (because r1 is itself a divisor of 5).2. If all divisors of r1 are units, it is a unit itself (the product of units is a unit).3. If r1 is a unit, 1/r1 is an algebraic integer (de'nition of unit).4. 1/r1 is a root of the primitive polynomial 5x^2 - x - 1, so is not an algebraic integer (we have gone ove this before).So r1 is not a unit and it has non-unit divisors that are also divisorsof 5.Satis'ed?-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ =|> If you want to show that a mistake is being taught, you should quote it|> speci'cally.||That's not very sensible if you expect that for a mistake to be taught|for any length of time it'd have some subtlety, unless you're also|questioning the competence of mathematicians.||It seems to me that rather than deal with the important question,|which is whether or not I'm correct, people often try to introduce|weird ad hoc conditions.||If you disagree with that assessment, can you please explain why you|believe there should be something simple and short enough for me to|quote?so it's possible for someone to be wrong, and yet for you to be unableto point to some one localized place in what they're saying as theplace where the mistake is being made?-- =[...]> For instance, both could have a factor that is sqrt(5), or one could> have a factor of 1+2i, while the other had a factor of 1-2i.> > The question I'm trying to get the reader to explore is, given that> they are roots of this particular polynomial, could one of them only> have unit factors of 5?> > How do you know?may be relevant:http://www.ams.org/new-in-math/cover/ factorization.html``On the other hand, if D = -5, then R is not a UFD.David Bernier => > The ring of algebraic integers is incomplete, it's easy to show, and> I've shown it with my paper.> > If mathematicians are having trouble understanding any part of it, I'm> able to explain further.> > I don't understand what it means to say, The ring of algebraic> integers is incomplete. I don't claim to be a mathematician (I'm a jigsaw puzzle retailer,> actually), but there's considerable evidence that most mathematicians> who've looked can't understand it either. Could you please (either or> both):You'll have better luck getting Cool Giraffe to explain the apes andbananas thread than prying any information out of JSH.> > (a) Post a clear de'nition, of the sort one sees in maths books of a> complete ring> > (b) Say of some simple rings whether they are complete or not -> > The ring of integers (Z)> > The ring of integers modulo 12 (clock arithmetic)> > The ring of integers modulo 2 (even/odd arithmetic)> > The ring of integers modulo 7> > (b') If all of these are complete, can you name another ring other> than the ring of algebraic integers which is also incomplete?> > > Brian Chandler> ----------------> geo://Sano.Japan.Planet_3> Jigsaw puzzles from Japan at:> http://imaginatorium.org/shop/ =>>>>Having spent the weekend going through musty books on algebra... here>>is my analysis:> > Hmmm...so my short paper sent you to books for an entire weekend?> Sorry if I disappointed you. I tend to do more work in logic/discrete math than I do in HARRIS>Abstract. Algebraic method for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization Approached>Determining the distribution of factors within irrational algebraic>integers has long been considered impossible as it is not possible to>>>> do>>>using Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight the>>>> standard>>>belief consider the algebraic integers (1 − sqrt(21))/2 and>>>>(1+sqrt(21))/2 >which are roots of x^2 + x − 5.>>>>This is a minor note, but it should be -1 for both roots. While this>>has little impact on the validity of the rest of the work, it should,>>at the very least, be corrected in your 'nal draft.> > > That was pointed out by someone else, I've acknowledged it as a> mistake, and I've changed it in the paper.Agreed. I was being complete.>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not have>>>> non>>>unit factors of 5? How do you know?>>>>These questions don't make a lot of sense to me, but I'll skip over>>that to get to what I believe is the true problem.> > > Well, it looks to me like your comment might be taken as somewhat> negative, but the preamble to the paper can be skipped.I just prefer clarity in a paper. Some things I will take as my own ignorance, but I should have a reasonable way to go educate myself. These questions appear to be a driving concept behind what you intend to do, but it isn't obvious to me what that concept is. It could be my ignorance, but it didn't appear to be clear what you meant.>In looking to consider distribution of algebraic integer factors>>>> within>>>a factorization I'll be using a more complicated example than x 2 +x>>>> − 5.>>>The paper will show that given the factorization, in the ring of>>>> alge->>>braic integers,>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.>>>>>>And here we run into two problems. First, while it is true that the>>a's are algebraic integers, it is something that at least deserves a>>comment. > > Why should that be commented on here?> Because with a brief comment you can avoid someone ignoring you out of hand. It can easily _appear_ that you have just switched the discussion from algebraic integers to algebraic numbers. _That_ would put everything you are doing into question, or possibly cause someone to ignore the rest of your work as unrelated to the claimed topic of discussion.Put brie¤y, you are putting the burden of responsibility for understanding you on the reader, rather than taking it on yourself to be clear.>Second, it is far from obvious that coprime is meaningful>>on the algebraic integers. > > > That was discussed by others in reply to your post.It has been pointed out that the de'nition of coprime I was using is not optimal. A better version states x and y are coprime in a ring if there is an a and b in the same ring satisfying ax+by=1. This simple statement negates all my comments about the meaningfulness of coprime.> However I'll add that I switched to coprime rather than use factor> in this situation because algebraic integers is an incomplete ring so> there is a problem with the term factor if I want to stay in the ring.> That is, rather than say that one of the a's is coprime, I might have> said that only two of the a's have a factor of f, but the ring is> incomplete, so that's problematic.What do you mean by the term incomplete?> That is, the problem with algebraic integers being incomplete means> that I'd have to knowingly use factor in a way not exactly correct,> but coprime still works, so I used it.If you are using the term factor in a way that is not exactly correct, you probably shouldn't use it. Find another term, or invent one. If you aren't using the words right, then you are not communicating effectively.If I decide to use the word cat to refer to any pet, since the only pets I have are cats, then when I say my neighbor's three cats in the back yard were barking up a storm last night I should _expect_ to have people complain that cats don't bark. >>Here's the problem: coprime is based on>>the de'nition of the least common divisor. The LCD is only>>guaranteed to exist for Principal Ideal Domains and Unique>>algebraic integers are certainly NOT a UFD, and I've seen nothing to>>indicate anything stronger than that they are a commutative ring. >>Until someone can point to work that indicates that LCD is de'ned on>>the algebraic integers, this paper cannot move forward. The remainder>>of the paper (which I have NOT checked for accuracy but am including>>for completeness) depends on the concept of coprime being meaningful>>and de'ned on the algebraic integers.> > > The problem is that algebraic integers are incomplete as a ring. What> the paper shows is that two of the a's should have f as a factor, as> the paper shows them to have f as a factor, but technically, it's not> a factor within the ring of algebraic integers.And this is where the terminology becomes important. If it's a factor, 'ne. If it's not a factor, then it's something else. If you don't use the term at the right time, you can go in circles trying to make things be what they aren't and do what they don't to make the term apply better.> > >First I'll need a simple lemma to generalize beyond factors of a>>>> poly->>>nomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factor>for all x, which means that the value of g for a value .95a' of x is a>>>> factor>>>of P(a), within the ring of algebraic integers, there exists r and c>>>> such>>>that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).>>>>The notation here could be improved. g should be g(x), r should be>>r(x).> > > I'm not interested in style issues.> You have yet to learn the difference between style and precision. Precision preserves the accuracy of what you say. Style perserves the readability of what you say. Both are important parts of clear communication.> You have claimed to be a mathematician, and mathematicians are de'ned> to be math experts. The fundamental question about the paper is not> style but correctness. Minor issues aren't relevant, but so far> that's all you have.I am a mathematician. My expertise is NOT in algebra or number theory. That is why I was looking up terms in books. The problem is the following: when I look things up and read and can't determine correctness because I can't get past the style, the style has become an issue and ceased to be minor.I don't understand what you're trying to say is not a minor issue.> For instance, you talked about the sign problem in the preamble, and> then you questioned the use of coprime, and now you talk about> notation.The sign problem was minor. Notation often is not.> > These type of issues are the kinds of things I see brought up by> people I deem aren't interested in the truth.Perhaps you are missing part of the truth: namely that these issues are signi'cant, not minor.> > The discussion is not a game. It's not about winning or losing. It's> about the truth.> > If I'm correct then there's a problem in mathematics which needs to be> 'xed.> > If I'm wrong then you should be able to 'nd more than a style issue.I always thought math _was_ a game. It has rules you must play by. The problem is, the rules don't seem to be what you want them to be. It would be easier to 'x the style issues and _then_ deal with the substance rather than keep 'ghting to preserve your style.>Proof. Let x=0, then g must be a factor of P(0), so at that point c =>>>> g.>>>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r which>>>> varies>>>with x, and as r equals 0 when x equals 0 it is not coprime to>x.>>>>>I'm not sure step 2 makes sense. I don't see why it makes r coprime>>to x.> > > The argument isn't complicated but I simpli'ed that section rather> than deal with it, after I realized it wasn't correct> > That is, r is not coprime to x or contains a unit factor of x.> > With that unit factor things get messy as I'm sure some would jump on> that and claim that since r would always have a unit factor of x, it's> meaningless.If it gets messy, then you should present it rather than force the reader to try to slog through all of that on their own. Skip simple things that are clear cut. Messy means not obvious a lot of the time.> > Rather than get into an involved explanation with more room for> confusion, I realized that I really was just using the fact that any> factor of a polynomial can be split up between what's constant and> what's varying.> > That is, r changes, but c does not, as x varies.> > The proof of that is actually trivial as all you do is take g at 0,> which gives you c, then r = g-c.> > Given that c is constant, while g is not, obviously r is not either.This I agree with.> > >As an example consider sqrt(x + 1) which is a non polynomial factor>>>> of>>>x+1, and while there are an in'nity of irrational solutions consider>>>> the>>>rational solution at x=35.>>>>Are you sure you want to talk about non-polynomial roots? While>>legitimate, it changes the topic of discussion from polynomials over>>the algebraic integers to algebraic functions over the algebraic>>integers.> > > Which would seem to indicate that you haven't read further down in the> paper, or haven't bothered to consider what I said before this> section: First I'll need a simple lemma to generalize beyond factors of a> polynomial that are themselves polynomials.> > Now I'd think that says what you need to know, and besides, what's> with the talk of roots?> > What are non-polynomial roots?Ah. That should be factor. This is a prime example of saying what you _mean_, not just using a word that's related.> > Are you sure you're a mathematician?Only if a master's degree counts.> > >Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,>and c=1. But for different values of x, g and r will vary, while r>>>> will>>>not be coprime to x.>>>>Again, coprime needs to be de'ned for the particular ring under>>consideration.> > > And it sounds to me like maybe you didn't spend a weekend going over> math texts as you repeatedly brought up as an issue something other> posters dismissed immediately.> > >2. Primary Argument>Given>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. Let>P(m) = f^2 ((m^3 f^4 − 3m^2 f^2 + 3m)x^3 − 3( − 1 +>>>> mf^2 )xu^2 + u^3 f)>>>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)>has a factor that is f^2 .>>>>Again, coprime must be de'ned on the algebraic integers. Also, the>>change of variable and the introduction of additional variables seems>>odd. Since it appears that you mean for f and u to be constants, it>>appears that you actually have P(m,x).> > > Nope. And if you're a mathematician changing variables as I did> shouldn't be a problem for you.> > Again, the question is to the correctness of the paper, and not style> issues.Which is easier: 'xing the style so there are no style issues, or 'ghting to preserve your style when they obscure the ability to determine correctness?> > So far, just like before when I got upset, you've spent a lot of time> on nonessential issues, or ones that others quickly refuted.> > It seems to me that your behavior is consistent with what I've seen> from posters trying to hide the truth, rather than get to the bottom> of things.> > But the issue is taught mathematics which is wrong.> > >That expression comes from expanding (v^3 +1)x^3 − 3vxy^2 + y^3>>>> , using>>>the substitutions v = − 1+mf^2 , and y = uf, where additional>>>> variables>>>provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime in>>>> integers,>>>as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the ring of>>>> algebraic>>>integers.>>>>I don't follow what you mean by this at all. Some additional>>explanation would be helpful. What are the factors and how did you>>compute them?> > > I explain to some extent where the expression I use comes from, and> then give a quick analogy. If the analogy confuses you, you can skip> it, as you did with the other section that confused you.> > It's not part of the argument, but merely there to help.> Now consider the factorization>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows that>a1a2a3 = m^3 f^6 − 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 − 3m^2>>>> f^2 + 3m)>>>so a1a2a3 = mf^2 (m^2 f^4 − 3mf^2 + 3).>Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.>(Note: The a's are roots of a monic polynomial with algebraic integer>coef'cients so they are algebraic integers.)>>>>Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logic>>of your note could stand some clari'cation. (note: coprime shows up>>again)> > > Strange question, given what follows:> > >Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>>>>>g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get the>>>> factor>>>f^2 in the constant term P(0), while one must not equal 0, or f^3>>>> would>>>be the factor.>>>>This statement doesn't make a lot of sense either... but I'm not>>trying as hard at this point. If the other problems can be 'xed,>>I'll be happy to examine it more carefully.> > > In the paper I take a rather complicated expression, consider it as a> polynomial with respect to m, meaning that I let only m vary, and then> 'nd that I can factor it in a not surprising way, given the> expression.> > That factorization gives me non-polynomial factors, with which I use> my lemma to show that only one of the a's is coprime to f.Ok, I just looked at your paper again... where are the non-polynomial factors? I see a ton of polynomial factors, but no non-polynomial factors.> > The factorization I use, isn't terribly surprising, and the argument> follows rather simply from a rather basic lemma, and the distributive> property.> > >Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;>>>> therefore,>>>given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that same>>>> factor>>>in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 − 3m^2 f^4 + 3m)x^3 − 3( − 1 + mf^2 )xu^2>>>> + u^3 = 65x^3 − 12x + 1>>>which may be more easily seen from using v = − 1 + mf^2 = 4,>>>> y=1>>>with (v^3 + 1)x^3 − 3vxy^2 + y^3 .>Therefore, with the factorization>65x^3 − 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors being>>>> irrational.>>>>>>Overall analysis: if you can clarify the notation and the logical>>connections, along with your result about factors not being factors,>>what you probably have is a proof that coprime is not always de'ned>>on the algebraic integers.> > > Well I refer you to the statements by others about coprime in the> ring of algebraic integers.> > What I can do is take out the use of the term in the paper, and use> factor in the ring of algebraic integers, and note that will lead to> a contradiction as the point of the paper is that the ring is> incomplete.> > What I want to impress upon readers is that I'm quite willing to work> with mathematicians to explain the mistake that they're teaching.> > The ring of algebraic integers is incomplete, it's easy to show, and> I've shown it with my paper.> > If mathematicians are having trouble understanding any part of it, I'm> able to explain further.> > > James HarrisSure: what does incomplete mean?-- Will Twentyman =>>>Having spent the weekend going through musty books on algebra... here>>is my analysis:> > Hmmm...so my short paper sent you to books for an entire weekend?> Sorry if I disappointed you. I tend to do more work in logic/discrete > math than I do in Algebra.I didn't say I was disappointed. Actually I *was* intrigued. Then Iread HARRIS>Abstract. Algebraic method for determining distribution of fac->tors within a polynomial factorization, which breaks through what>was seen as a barrier from overinterpretations of Galois Theory.>1. Advanced Polynomial Factorization Approached>Determining the distribution of factors within irrational algebraicintegers has long been considered impossible as it is not possible to>>>> do>>>using Galois Theory. However a simple technique through the intro->duction of more variables makes it possible. To highlight the>>>> standard>>>belief consider the algebraic integers (1 − sqrt(21))/2 and>>>>(1+sqrt(21))/2 >which are roots of x^2 + x − 5.>>>>This is a minor note, but it should be -1 for both roots. While this>>has little impact on the validity of the rest of the work, it should,>at the very least, be corrected in your 'nal draft.> > That was pointed out by someone else, I've acknowledged it as a> mistake, and I've changed it in the paper.> > Agreed. I was being complete.Ok.>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not have>>> non>>>unit factors of 5? How do you know?>>>>These questions don't make a lot of sense to me, but I'll skip over>>that to get to what I believe is the true problem.> > > Well, it looks to me like your comment might be taken as somewhat> negative, but the preamble to the paper can be skipped.> > I just prefer clarity in a paper. Some things I will take as my own > ignorance, but I should have a reasonable way to go educate myself. > These questions appear to be a driving concept behind what you intend to > do, but it isn't obvious to me what that concept is. It could be my > ignorance, but it didn't appear to be clear what you meant.Well I'm going to explain brie¤y but I'm not interested in spending alot of time explaining things to posters, primarily because I have todeal with so many, including a lot whom I consider cheats.Consider r_1 and r_2 algebraic integers where their product is 5. Nowthen, can you imagine different ways in which algebraic integerfactors of 5 might distribute?As given, one easy way for one to be coprime to 5 is for one of themto equal 1.Now both could equal sqrt(5), so neither would be coprime to 5, andboth would have a factor of 5, that is sqrt(5).Got it? >In looking to consider distribution of algebraic integer factors>>>> within>>>a factorization I'll be using a more complicated example than x 2 +x>>>> − 5.>>The paper will show that given the factorization, in the ring of>>>> alge->>>braic integers,65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)one of the a's is coprime to 5, using basic algebraic methods.>>>>>>And here we run into two problems. First, while it is true that the>>a's are algebraic integers, it is something that at least deserves a>comment. > > Why should that be commented on here?> > > Because with a brief comment you can avoid someone ignoring you out of > hand. It can easily _appear_ that you have just switched the discussion > from algebraic integers to algebraic numbers. _That_ would put > everything you are doing into question, or possibly cause someone to > ignore the rest of your work as unrelated to the claimed topic of > discussion.> > Put brie¤y, you are putting the burden of responsibility for > understanding you on the reader, rather than taking it on yourself to be > clear.That's false.The statement begins with,The paper will show that given the factorization, in the ring ofalgebraic integers...So the ring is declared to algebraic integers at the start.So once again, you're shown up, and as I've said repeatedly, yourcomments are typical of what I've seen from mathematicians trying to*hide* the truth.Undergrads need only note the subject line of this thread to considerwhy I keep pointing that out.I want you to see how clear it is that mathematicians *are* lying. >>Second, it is far from obvious that coprime is meaningful>>on the algebraic integers. > > > That was discussed by others in reply to your post.> > It has been pointed out that the de'nition of coprime I was using is > not optimal. A better version states x and y are coprime in a ring if > there is an a and b in the same ring satisfying ax+by=1. This simple > statement negates all my comments about the meaningfulness of coprime.> > However I'll add that I switched to coprime rather than use factor> in this situation because algebraic integers is an incomplete ring so> there is a problem with the term factor if I want to stay in the ring.> That is, rather than say that one of the a's is coprime, I might have> said that only two of the a's have a factor of f, but the ring is> incomplete, so that's problematic.> > What do you mean by the term incomplete?Because I can take an algebraic integer x, prove that within the ringof algebraic integers it has the algebraic integer y (I'm beingredundant with the algebraic integers declarations but you claimed aproblem before), and then moving to the 'eld of rationals show thatx/y is NOT an algebraic integer, which is a contradiction.That is, given that in the ring I can prove x has y as a factor, ithardly makes sense that in checking, by going to the 'eld ofrationals, I 'nd that x/y is not an algebraic integer.So x/y *should* be in the ring, as it is not, the ring is incomplete.> That is, the problem with algebraic integers being incomplete means> that I'd have to knowingly use factor in a way not exactly correct,> but coprime still works, so I used it.> > If you are using the term factor in a way that is not exactly correct, > you probably shouldn't use it. Find another term, or invent one. If > you aren't using the words right, then you are not communicating > effectively.Well I used coprime so I wouldn't have to get into discussion on thesubject!!! > If I decide to use the word cat to refer to any pet, since the only > pets I have are cats, then when I say my neighbor's three cats in the > back yard were barking up a storm last night I should _expect_ to have > people complain that cats don't bark.Your statement is irrelevant, as I've explained in detail why I usedcoprime instead of factor and in fact you brought up an issue withcoprime which was shot down by other posters.My fear is that if it hadn't been, I could have been arguing with youor any number of other posters for some time.But thankfully you bent over when some other posters corrected you.> > >>Here's the problem: coprime is based on>>the de'nition of the least common divisor. The LCD is only>>guaranteed to exist for Principal Ideal Domains and Unique>>algebraic integers are certainly NOT a UFD, and I've seen nothing to>>indicate anything stronger than that they are a commutative ring. >>Until someone can point to work that indicates that LCD is de'ned on>>the algebraic integers, this paper cannot move forward. The remainder>of the paper (which I have NOT checked for accuracy but am including>>for completeness) depends on the concept of coprime being meaningful>>and de'ned on the algebraic integers.> > > The problem is that algebraic integers are incomplete as a ring. What> the paper shows is that two of the a's should have f as a factor, as> the paper shows them to have f as a factor, but technically, it's not> a factor within the ring of algebraic integers. And this is where the terminology becomes important. If it's a factor, > 'ne. If it's not a factor, then it's something else. If you don't use > the term at the right time, you can go in circles trying to make things > be what they aren't and do what they don't to make the term apply better.And I've explained repeatedly, and again I want undergrads to considerthe subject line of this thread. >First I'll need a simple lemma to generalize beyond factors of a>>>> poly->>>nomial that are themselves polynomials.>Lemma 1.1. Factorization Lemma:>Given a factor g of a polynomial P(x), further de'ned as a factor>for all x, which means that the value of g for a value .95a' of x is a>>>> factor>>of P(a), within the ring of algebraic integers, there exists r and c>>>> such>>>that>g = r + c>where r=0, or is not coprime to x, and c is a factor of the constant>term P(0).>>>>The notation here could be improved. g should be g(x), r should be>>r(x). > > I'm not interested in style issues.> > > You have yet to learn the difference between style and precision. > Precision preserves the accuracy of what you say. Style perserves the > readability of what you say. Both are important parts of clear > communication.Yet if you are a mathematician, then by de'nition you are a mathexpert, so the question will arise, are you bringing up an issue thatremoves your ability to comprehend?Yet you say the notation here could be improved, so I think itreasonable to suppose that you *do* understand.I'm not interested in style issues.> You have claimed to be a mathematician, and mathematicians are de'ned> to be math experts. The fundamental question about the paper is not> style but correctness. Minor issues aren't relevant, but so far> that's all you have.> > I am a mathematician. My expertise is NOT in algebra or number theory. > That is why I was looking up terms in books. The problem is the > following: when I look things up and read and can't determine > correctness because I can't get past the style, the style has become an > issue and ceased to be minor.Then point out such an instance. So far you've brought up minorissues which have all turned out to be related to your ignorance, oryour *opinion* that something like notation could be improved.Remember, my point is that mathematicians work *against* 'nding thetruth, so it hardly makes sense for you not to be on your bestbehavior!!!And in fact, part of my point is that if you were to behave, you'dhave to admit the truth, which is that there is an error in taughtmathematics.> I don't understand what you're trying to say is not a minor issue.Then you need to say that in a relevant portion. You did say thatabout the preamble, but that's just an introduction section.So far, you've talked a lot, without saying much of anything, whileI've explained quite a bit.> For instance, you talked about the sign problem in the preamble, and> then you questioned the use of coprime, and now you talk about> notation.> > The sign problem was minor. Notation often is not.So are you going to just whine and whine about notation?> > These type of issues are the kinds of things I see brought up by> people I deem aren't interested in the truth.> > Perhaps you are missing part of the truth: namely that these issues are > signi'cant, not minor.Then *show* why the issues are relevant to the paper.The conclusion of the paper would be a start. Try mightily to thinkof something that might have relevance to the conclusion.Technicalities will just get ripped by me.If you're an expert in mathematics, prove it.> > The discussion is not a game. It's not about winning or losing. It's> about the truth.> > If I'm correct then there's a problem in mathematics which needs to be> 'xed.> > If I'm wrong then you should be able to 'nd more than a style issue.> > I always thought math _was_ a game. It has rules you must play by. The > problem is, the rules don't seem to be what you want them to be. It > would be easier to 'x the style issues and _then_ deal with the > substance rather than keep 'ghting to preserve your style.Yeah right, like I'll move to your turf. Mathematicians cheat. If Istart trying to play in areas where they cheat well, I could betalking about this notation versus that notation for months!!!The issue is an error in taught mathematics.I'd think mathematicians would care more about that than notation!!!But then again, maybe they wouldn't, which is something undergradswould do well to learn now, better sooner than later.>Proof. Let x=0, then g must be a factor of P(0), so at that point c =>>> g.>>>(1) If when x does not equal 0, g=c, r=0.(2) If when x does not equal 0, g =/= c there must exist r which>>>> varies>>>with x, and as r equals 0 when x equals 0 it is not coprime to>x.>>>>I'm not sure step 2 makes sense. I don't see why it makes r coprime>>to x.> > > The argument isn't complicated but I simpli'ed that section rather> than deal with it, after I realized it wasn't correct> > That is, r is not coprime to x or contains a unit factor of x.> With that unit factor things get messy as I'm sure some would jump on> that and claim that since r would always have a unit factor of x, it's> meaningless.> > If it gets messy, then you should present it rather than force the > reader to try to slog through all of that on their own. Skip simple > things that are clear cut. Messy means not obvious a lot of the time.Huh? What I found was that I could use something simpler than I hadthat kept it from getting messy. I decided to go with simpler.> > Rather than get into an involved explanation with more room for> confusion, I realized that I really was just using the fact that any> factor of a polynomial can be split up between what's constant and> what's varying.> > That is, r changes, but c does not, as x varies.> > The proof of that is actually trivial as all you do is take g at 0,> which gives you c, then r = g-c.> > Given that c is constant, while g is not, obviously r is not either.> > This I agree with.Excellent! Then you just got past the lemma!!!> > >As an example consider sqrt(x + 1) which is a non polynomial factor>>>> of>>x+1, and while there are an in'nity of irrational solutions consider>>>> the>>>rational solution at x=35.>>>>Are you sure you want to talk about non-polynomial roots? While>>legitimate, it changes the topic of discussion from polynomials over>>the algebraic integers to algebraic functions over the algebraic>>integers.> > > Which would seem to indicate that you haven't read further down in the> paper, or haven't bothered to consider what I said before this> section:> > First I'll need a simple lemma to generalize beyond factors of a> polynomial that are themselves polynomials.> > Now I'd think that says what you need to know, and besides, what's> with the talk of roots?> > What are non-polynomial roots?> > Ah. That should be factor. This is a prime example of saying what > you _mean_, not just using a word that's related.Ok.> > Are you sure you're a mathematician?> > Only if a master's degree counts.Forget that question as I'll admit I was quick on the gun.The point is that the lemma is there so that I can talk aboutnon-polynomial factors of a polynomial later. That's the point of it.> > >Then I have sqrt(35 + 1) = 6 = 5 + 1; therefore when x=35, g=6, r=5,>and c=1. But for different values of x, g and r will vary, while r>>> will>>>not be coprime to x.>>>>Again, coprime needs to be de'ned for the particular ring under>consideration.> > > And it sounds to me like maybe you didn't spend a weekend going over> math texts as you repeatedly brought up as an issue something other> posters dismissed immediately.> > >2. Primary Argument>Given>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>in the ring of algebraic integers. Let>P(m) = f^2 ((m^3 f^4 − 3m^2 f^2 + 3m)x^3 − 3( − 1 +>>>> mf^2 )xu^2 + u^3 f)>>>Here f is a non unit, non zero algebraic integer coprime to 3 and x,>and u a non unit, non zero algebraic integer coprime to f. Note P(m)>has a factor that is f^2 .>>>>Again, coprime must be de'ned on the algebraic integers. Also, the>>change of variable and the introduction of additional variables seems>>odd. Since it appears that you mean for f and u to be constants, it>appears that you actually have P(m,x).> > > Nope. And if you're a mathematician changing variables as I did> shouldn't be a problem for you.> > Again, the question is to the correctness of the paper, and not style> issues.> > Which is easier: 'xing the style so there are no style issues, or > 'ghting to preserve your style when they obscure the ability to > determine correctness?I changed variables deliberately. It doesn't change the mathematicalargument at all.The point is that the lemma is in general, and the use of x with theI'm not interested in style issues. The point is that there's aproblem in taught mathematics, and it's not about style.> > So far, just like before when I got upset, you've spent a lot of time> on nonessential issues, or ones that others quickly refuted.> > It seems to me that your behavior is consistent with what I've seen> from posters trying to hide the truth, rather than get to the bottom> of things.> > But the issue is taught mathematics which is wrong.> > >That expression comes from expanding (v^3 +1)x^3 − 3vxy^2 + y^3>>>> , using>>>the substitutions v = − 1+mf^2 , and y = uf, where additional>>>> variables>>provide an additional degree of freedom.>Consider that a similar idea can be used to factor 3, prime in>>> integers,>>>as x^2 +7x+10 = 3, allows you to 'nd factors of 3 in the ring of>>>> algebraic>>integers.>>>>I don't follow what you mean by this at all. Some additional>>explanation would be helpful. What are the factors and how did you>>compute them? > I explain to some extent where the expression I use comes from, and> then give a quick analogy. If the analogy confuses you, you can skip> it, as you did with the other section that confused you.> > It's not part of the argument, but merely there to help.> > Now consider the factorization>P(m) = (a1x + uf)(a2x + uf)(a3x + uf)>where multiplying out shows thata1a2a3 = m^3 f^6 − 3m^2 f^4 + 3mf^2 = f^2 (m^3 f^4 − 3m^2>>>> f^2 + 3m)>>>so a1a2a3 = mf^2 (m^2 f^4 − 3mf^2 + 3).>Therefore, at least one of the a's cannot be coprime to m, and at>least one of the a's must equal 0 when m=0.>(Note: The a's are roots of a monic polynomial with algebraic integercoef'cients so they are algebraic integers.)>>>Maybe. Did P(m) have a constant term of 1? if so, uf=1. The logic>>of your note could stand some clari'cation. (note: coprime shows up>>again)> > > Strange question, given what follows:> > >Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.> > g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get the>>>> factor>>>f^2 in the constant term P(0), while one must not equal 0, or f^3>>> would>>>be the factor.>>>>This statement doesn't make a lot of sense either... but I'm not>trying as hard at this point. If the other problems can be 'xed,>>I'll be happy to examine it more carefully. > In the paper I take a rather complicated expression, consider it as a> polynomial with respect to m, meaning that I let only m vary, and then> 'nd that I can factor it in a not surprising way, given the> expression.> > That factorization gives me non-polynomial factors, with which I use> my lemma to show that only one of the a's is coprime to f.> > Ok, I just looked at your paper again... where are the non-polynomial > factors? I see a ton of polynomial factors, but no non-polynomial factors.The polynomial is P(m), which is given. Since m is the onlyindependent variable, those other factors of P(m) are NOT polynomials. They may *look* like polynomials to you, but are in factnon-polynomial factors of P(m), and cannot be considered polynomialsunless you shift variables. Which is why I say they may *look* likepolynomials, when they aren't, given that m is the key variable.> > The factorization I use, isn't terribly surprising, and the argument> follows rather simply from a rather basic lemma, and the distributive> property. >Now as noted before in general P(m) has a factor that is f^2 , and>separating that factor off, gives a constant term coprime to f;>>>> therefore,>>given g1 = a1x + uf>where with m = 0, g1 gives a factor of f it must have that same>>>> factor>>in general, proving that two of the a's have a factor that is f.>Therefore, one factor is coprime to f.>Now letting m=1, f=sqrt(5), where I can let u=1 as its value doesn't>change the a's, I have>(m^3 f^6 − 3m^2 f^4 + 3m)x^3 − 3( − 1 + mf^2 )xu^2>>> + u^3 = 65x^3 − 12x + 1>>>which may be more easily seen from using v = − 1 + mf^2 = 4,>>>> y=1>>>with (v^3 + 1)x^3 − 3vxy^2 + y^3 .Therefore, with the factorization>65x^3 − 12x + 1 = (a1x + 1)(a2x + 1)(a3x + 1)>one of the a's is coprime to 5, which shows where some of the alge->braic integer factors distribute despite the factors being>>>> irrational.>>>>>>Overall analysis: if you can clarify the notation and the logical>>connections, along with your result about factors not being factors,>>what you probably have is a proof that coprime is not always de'ned>>on the algebraic integers.> > > Well I refer you to the statements by others about coprime in the> ring of algebraic integers.> > What I can do is take out the use of the term in the paper, and use> factor in the ring of algebraic integers, and note that will lead to> a contradiction as the point of the paper is that the ring is> incomplete.> > What I want to impress upon readers is that I'm quite willing to work> with mathematicians to explain the mistake that they're teaching.> > The ring of algebraic integers is incomplete, it's easy to show, and> I've shown it with my paper.> > If mathematicians are having trouble understanding any part of it, I'm> able to explain further.> > > James Harris> > Sure: what does incomplete mean?I've explained above, but I'll explain again. I want undergrads tonote that I'm going the distance, and that I'm the person who takes agood deal of trouble to go into detail.The ring of algebraic integers is incomplete in that I can prove thatan algebraic integer x has an algebraic integer factor y, but then inthe 'eld of rationals, considering x/y I 'nd that it is not analgebraic integer, which is a contradiction.The proof is in the paper. The other with x/y has been discussed atlength in various posts.Of course the numbers are NOT called x and y in the paper. They arecalled a_1 and a_2, which must have factors of f, since a_3 is provento be coprime to f, but a_1 a_2 a_3 has a factor of f, which is f^2.James Harris => >>While you know that the algebraic integer factors are themselves>factors of 5, in what way is each a factor of 5? Can either not have>>>>non>>>>>unit factors of 5? How do you know?>>>>These questions don't make a lot of sense to me, but I'll skip over>>that to get to what I believe is the true problem.>>>Well, it looks to me like your comment might be taken as somewhat>negative, but the preamble to the paper can be skipped.>>>>I just prefer clarity in a paper. Some things I will take as my own >>ignorance, but I should have a reasonable way to go educate myself. >>These questions appear to be a driving concept behind what you intend to >>do, but it isn't obvious to me what that concept is. It could be my >>ignorance, but it didn't appear to be clear what you meant.> > > Well I'm going to explain brie¤y but I'm not interested in spending a> lot of time explaining things to posters, primarily because I have to> deal with so many, including a lot whom I consider cheats. Consider r_1 and r_2 algebraic integers where their product is 5. Now> then, can you imagine different ways in which algebraic integer> factors of 5 might distribute? As given, one easy way for one to be coprime to 5 is for one of them> to equal 1.> > Now both could equal sqrt(5), so neither would be coprime to 5, and> both would have a factor of 5, that is sqrt(5).> > Got it?Got it.>In looking to consider distribution of algebraic integer factors>>>>within>>>>>a factorization I'll be using a more complicated example than x 2 +x>>>>− 5.>>>>>The paper will show that given the factorization, in the ring of>>>>alge->>>>>braic integers,>65x^3 − 12x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)>one of the a's is coprime to 5, using basic algebraic methods.>>>>>>And here we run into two problems. First, while it is true that the>>a's are algebraic integers, it is something that at least deserves a>>comment. >>Why should that be commented on here?>>>>>Because with a brief comment you can avoid someone ignoring you out of >>hand. It can easily _appear_ that you have just switched the discussion >>from algebraic integers to algebraic numbers. _That_ would put >>everything you are doing into question, or possibly cause someone to >>ignore the rest of your work as unrelated to the claimed topic of >>discussion.>>>>Put brie¤y, you are putting the burden of responsibility for >>understanding you on the reader, rather than taking it on yourself to be >>clear. > That's false.> > The statement begins with,> > The paper will show that given the factorization, in the ring of> algebraic integers...> > So the ring is declared to algebraic integers at the start.> The de'nition of an algebraic integer is: the root of a monic polynomial with integer coef'cients. You have just introduced a non-monic polynomial. This is likely to make someone pause for a bit trying to see the connection. Why does this seem unlikely?> So once again, you're shown up, and as I've said repeatedly, your> comments are typical of what I've seen from mathematicians trying to> *hide* the truth.> > Undergrads need only note the subject line of this thread to consider> why I keep pointing that out.> > I want you to see how clear it is that mathematicians *are* lying.> >Second, it is far from obvious that coprime is meaningful>>on the algebraic integers. >>>That was discussed by others in reply to your post.>>>>It has been pointed out that the de'nition of coprime I was using is >>not optimal. A better version states x and y are coprime in a ring if >>there is an a and b in the same ring satisfying ax+by=1. This simple >>statement negates all my comments about the meaningfulness of coprime.>>>>>However I'll add that I switched to coprime rather than use factor>in this situation because algebraic integers is an incomplete ring so>there is a problem with the term factor if I want to stay in the ring.> That is, rather than say that one of the a's is coprime, I might have>said that only two of the a's have a factor of f, but the ring is>incomplete, so that's problematic.>>>>What do you mean by the term incomplete?> > > Because I can take an algebraic integer x, prove that within the ring> of algebraic integers it has the algebraic integer y (I'm being> redundant with the algebraic integers declarations but you claimed a> problem before), and then moving to the 'eld of rationals show that> x/y is NOT an algebraic integer, which is a contradiction. That is, given that in the ring I can prove x has y as a factor, it> hardly makes sense that in checking, by going to the 'eld of> rationals, I 'nd that x/y is not an algebraic integer.> > So x/y *should* be in the ring, as it is not, the ring is incomplete.> If y is a factor of x, both in the ring, then there is an r in the ring which satis'es ry=x. r is what you are referring to as x/y. If r is NOT in the ring, y is NOT a factor of x.If you have shown that there is a problem in math, it would be nice if you show which theorem that you relied on in your proof has the error. Otherwise, your proof has the error. You cannot rede'ne what a factor is, which is what you appear to be trying to do. Until you can 'nd which theorem from someone else's work that you used has the ¤aw, all you've done is noted that SOMETHING is 'shy. It's up to you to 'nd where that something 'shy is. Until then, you haven't discovered anything useful.Note: it is obviously the belief of the people who have responded that the something 'shy is located in your work. Until you can locate it elsewhere or address all objections (including counterexamples provided by others), you will have a hard time convincing anyone of anything.[ much of rest deleted ]>>I'm not interested in style issues.>>>>>You have yet to learn the difference between style and precision. >>Precision preserves the accuracy of what you say. Style perserves the >>readability of what you say. Both are important parts of clear >>communication.> > > Yet if you are a mathematician, then by de'nition you are a math> expert, so the question will arise, are you bringing up an issue that> removes your ability to comprehend?> > Yet you say the notation here could be improved, so I think it> reasonable to suppose that you *do* understand.> > I'm not interested in style issues.Nor am I interested in rewriting your paper with all the style issues corrected to see if it still says what you claim it says. If you 'nd that an objectionable trait, I'm sorry.> > >You have claimed to be a mathematician, and mathematicians are de'ned>to be math experts. The fundamental question about the paper is not>style but correctness. Minor issues aren't relevant, but so far>that's all you have.>>>>I am a mathematician. My expertise is NOT in algebra or number theory. >> That is why I was looking up terms in books. The problem is the >>following: when I look things up and read and can't determine >>correctness because I can't get past the style, the style has become an >>issue and ceased to be minor.> > > Then point out such an instance. So far you've brought up minor> issues which have all turned out to be related to your ignorance, or> your *opinion* that something like notation could be improved.When I'm not sure if g is a polynomial or an algebraic integer. That was clari'ed in another post, but was still an obstacle at the time.> > Remember, my point is that mathematicians work *against* 'nding the> truth, so it hardly makes sense for you not to be on your best> behavior!!!Have I been insulting to you? I haven't said what you want to hear, but that does not constitute poor behavior.> > And in fact, part of my point is that if you were to behave, you'd> have to admit the truth, which is that there is an error in taught> mathematics.> When you fail to address points raised (and mine are minor compared to those raised by others), the burden of proof falls back on you.>Proof. Let x=0, then g must be a factor of P(0), so at that point c =>>>>g.>>>>>(1) If when x does not equal 0, g=c, r=0.>(2) If when x does not equal 0, g =/= c there must exist r which>>>>varies>>>>>with x, and as r equals 0 when x equals 0 it is not coprime to>x.>>>>>I'm not sure step 2 makes sense. I don't see why it makes r coprime>>to x.>>>The argument isn't complicated but I simpli'ed that section rather>than deal with it, after I realized it wasn't correct>>That is, r is not coprime to x or contains a unit factor of x.>>With that unit factor things get messy as I'm sure some would jump on>that and claim that since r would always have a unit factor of x, it's>meaningless.>>>>If it gets messy, then you should present it rather than force the >>reader to try to slog through all of that on their own. Skip simple >>things that are clear cut. Messy means not obvious a lot of the time.> > > Huh? What I found was that I could use something simpler than I had> that kept it from getting messy. I decided to go with simpler.You left a gap in your proof that you didn't address.>Rather than get into an involved explanation with more room for>confusion, I realized that I really was just using the fact that any>factor of a polynomial can be split up between what's constant and>what's varying.>>That is, r changes, but c does not, as x varies.>>The proof of that is actually trivial as all you do is take g at 0,>which gives you c, then r = g-c.>>Given that c is constant, while g is not, obviously r is not either.>>>>This I agree with. > Excellent! Then you just got past the lemma!!!> Wrong, I just got past the statement of the lemma. Not the proof.>Are you sure you're a mathematician?>>>>Only if a master's degree counts.> > > Forget that question as I'll admit I was quick on the gun.> > The point is that the lemma is there so that I can talk about> non-polynomial factors of a polynomial later. That's the point of it.> Then 'nish prooving the lemma.>Notice that the constant term P(0) is given by>P(0) = f^2 (3xu^2 + u^3 f)>and also that P(0)/f^2 = 3xu^2 + u^3 f, which is coprime to f.>Then I have the factor of P(m), g1, where g1 = a1x + uf, where a1>is not coprime to m.>>>>>g1 = c = uf>meaning f is a factor of the constant term.>Therefore, exactly two of the a's equal 0, when m=0, to get the>>>>factor>>>>>f^2 in the constant term P(0), while one must not equal 0, or f^3>>>>would>>>>>be the factor.>>>>This statement doesn't make a lot of sense either... but I'm not>>trying as hard at this point. If the other problems can be 'xed,>>I'll be happy to examine it more carefully.>>>In the paper I take a rather complicated expression, consider it as a>polynomial with respect to m, meaning that I let only m vary, and then>'nd that I can factor it in a not surprising way, given the>expression.>>That factorization gives me non-polynomial factors, with which I use>my lemma to show that only one of the a's is coprime to f.>>>>Ok, I just looked at your paper again... where are the non-polynomial >>factors? I see a ton of polynomial factors, but no non-polynomial factors.> > > The polynomial is P(m), which is given. Since m is the only> independent variable, those other factors of P(m) are NOT polynomials.> They may *look* like polynomials to you, but are in fact> non-polynomial factors of P(m), and cannot be considered polynomials> unless you shift variables. Which is why I say they may *look* like> polynomials, when they aren't, given that m is the key variable.> Since all your factors had integers powers on m, your factors were polynomials. Please show one that isn't.-- Will Twentyman = [.snip.]>> If you are using the term factor in a way that is not exactly correct, >> you probably shouldn't use it. Find another term, or invent one. If >> you aren't using the words right, then you are not communicating >> effectively.>>Well I used coprime so I wouldn't have to get into discussion on the>subject!!!Then you are an idiot. Apparently, you are using a term whose meaningyou do not know in order to avoid arguments and discussions. That'snot how mathematics is done, and that is a very far cry from thelogic you claim to love.I commented some months ago that you seem to think that terminologyand de'nitions are like magic spells and incantantions: you do notneed to understand them, you just need to say the words in the rightorder with the right in¤ection, and all your problems will disappearlike magic.If you do NOT know what coprime means, then either ask, or do notuse it.If you are NOT willing to give a FORMAL, PRECISE, CLEAR de'nition ofthe terms you use, then you are not doing math.At this point, you need to provide a FORMAL, PRECISE, CLEAR de'nitionof complete ring, incomplete ring, and factor. The latter,because you are clearly NOT using it in the standard meaning. = expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of 'gures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde'nite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan =Arturo Magidinmagidin@math.berkeley.edu => Because I can take an algebraic integer x, prove that within the ring> of algebraic integers it has the algebraic integer y (I'm being> redundant with the algebraic integers declarations but you claimed a> problem before), and then moving to the 'eld of rationals show that> x/y is NOT an algebraic integer, which is a contradiction.> > That is, given that in the ring I can prove x has y as a factor, it> hardly makes sense that in checking, by going to the 'eld of> rationals, I 'nd that x/y is not an algebraic integer.> > So x/y *should* be in the ring, as it is not, the ring is incomplete.This strikes me as a bit nonsensical.It's kind of like saying that I've proved that in the integers, 2 isa factor of 3, even though 3/2 isn't an integer. If you run across astatement like that, it's not time to declare the integers incomplete,but rather time to take a close look at why you thought 2 was a factorto start with, because it simply isn't.You can't factor a number part-way... unless *both* y and y/x are inthe ring, then y isn't a factor of x, by de'nition of factor. Youmay be able to multiply together y and y/x to get x, but that's notthe same thing. => You can't factor a number part-way... unless *both* y and y/x are in> the ring, then y isn't a factor of x, by de'nition of factor. You> may be able to multiply together y and y/x to get x, but that's not> the same thing.Sorry, that should have been x/y above in both places. Typo. =James Harris spewed all over himself in message:[most babble deleted for brevity]>> Here I'm right, and you're wrong, as the *de'nition* of algebraic> integer leads to an incomplete ring, which I prove mathematically.>> You cannot depend on a *de'nition* in mathematics to prove something> that's been disproven by mathematics.>>> James HarrisTROLL!!!Harris, do the world a favor and give it up!! You are attempting to'educate' math undergrads, when I can almost guarantee they are laughingtheir asses off when you post!! You even sound like some spoiled 5-year oldwhen someone points out that they're wrong. If you want to act like amathematician, talk like a mathematician, INCLUDING using their de'nitions,etc. Otherwise, you're a kook!!If it's entertainment you're trying to provide, you're doing a great job!!If it's education you're trying to provide, you suck!!~Bhuvan =I'm putting this bit 'rst to highlight it:[...]| Consider that in the ring of algebraic integers, 5 has algebraic| integer factors, and given algebraic integers r_1 and r_2, where their| product is 5, why are you acting as if it's so dif'cult to comprehend| that there must be some distribution of factors of 5?I'm not acting as if I'm having any kind of dif'culty. I'm sayingthat *you're* having dif'culty in realizing that there must be somedistribution of factors of 5 is a meaninglessly vague phrase. You'resaying this as if to say, Shouldn't this be true?, and not as ifyou actually had proven it was true.By the standard meanings of the terms, saying an algebraic integer r_1has a factor of 5 in the algebraic integers is just the same assaying it's divisible by 5 in the algebraic integers. That's the sameas saying that there's an algebraic integer t for which r_1=5t.There isn't, and I think you know there isn't. Since neither r_1 norr_2 is divisible by 5, there's no distribution of factors of 5between them.I think you use ill-de'ned terms because on some level you're awarethat using only well-de'ned terms spoils your fun. It's disillusioning.It clear away the fog of obfuscation. Suddenly you're back to reality,and it's the rather ordinary reality that people keep telling you thatyou're in, not the amazing story you wish it was.Back to the top.| > [...]| > | What I can do is take out the use of the term in the paper, and use| > | factor in the ring of algebraic integers, and note that will lead to| > | a contradiction as the point of the paper is that the ring is| > | incomplete.| > || > | What I want to impress upon readers is that I'm quite willing to work| > | with mathematicians to explain the mistake that they're teaching.| >| > If you want to show that a mistake is being taught, you should quote it| > speci'cally.|| That's not very sensible if you expect that for a mistake to be taught| for any length of time it'd have some subtlety, unless you're also| questioning the competence of mathematicians.Quite to the contrary-- it's because mathematicians are generallycompetent that one can ordinarily say very speci'cally what is wrongwhen they make mistakes.What makes a mathematical mistake subtle is not what it takes to pointit out once it's been recognized; it's the dif'culty in recognizingit in the 'rst place.| It seems to me that rather than deal with the important question,| which is whether or not I'm correct, people often try to introduce| weird ad hoc conditions.You created an ad hoc rule that in order to refute a claim of yours,it's not good enough for someone else to prove that the conclusion iswrong. You think a proof of yours can show that a proof of amathematician is wrong, but you appear unwilling to accept thepossibility that a proof from a mathematician can just as well showthat one of yours is wrong (i.e., not really a proof).| If you disagree with that assessment, can you please explain why you| believe there should be something simple and short enough for me to| quote?In mathematics, we try to make each *mathematical* statement have itsown well-de'ned meaning. (There are of course nonmathematical claimsthat we sometimes make, which aren't always precise.) If a sequence ofwell-de'ned mathematical statements is wrong, it's wrong because (atleast) one of the statements is wrong.The key is making sure all the de'nitions are sound.Mathematics that is vaguely enough written that it can be wrongwithout any individual statements in it being de'nitely wrong, isconsidered very bad. It's bad because an author hasn't de'ned theirterms well enough.| > | The ring of algebraic integers is incomplete, it's easy to show, and| > | I've shown it with my paper.| > || > | If mathematicians are having trouble understanding any part of it, I'm| > | able to explain further.| >| > You have a big problem with terminology. You use terms that aren't| > used by others in the context in which you use them, such as counting| > factors of 5 in algebraic integers. I haven't seen anything I would| > consider an adequate de'nition of those terms. You don't explain| > why your terms should be considered relevant to the ones mathematicians| > are using.|| That statement possibly has something to do with what I now call the| preamble in the paper as it is there to help explain context, but| isn't part of the actual argument.|| Here's what I say, copied from the paper (some editing for format):|| To highlight the standard belief consider the algebraic integers| (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.|| While you know that the algebraic integer factors are themselves| factors of 5, in what way is each a factor of 5? Can either not have| non unit factors of 5? How do you know?|| There I'm talking about algebraic integers, so one can assume I'm| talking about algebraic integer factors. Given that 5 has algebraic| integer factors, how is what I say nonstandard Keith Ramsay?|| I'd like you to carefully explain your assertion. The question I'm| raising is the possibility that you lied.No, I haven't lied. You are not using have non unit factors of 5 ina standard way.If at this point in the paper, you meant it in a completely standardway, here's what the meaning would be. Have a factor of 5 issynonymous with have 5 as a factor, or be divisible by 5. Also 5is a non-unit (so that part is redundant).Now obviously we all know you don't mean that. It's possible you meansomething like, shares non-unit factors with 5, meaning that thereare non-unit algebraic integers that are factors of both numbers. Butthis is just guesswork. Any serious writer of mathematics takesresponsibility for taking out the guesswork, NOT forcing the reader tokeep inferring which of various possible meanings is the intended one.The standard usage is also inconsistent with the rest of the paper.You write, for example, ...proving that two of the a's have a factorthat is f at a point where you plainly have NOT shown that f is afactor of any of the a's, in any of the usual ways factor isde'ned.Elsewhere, you refer to an algebraic integer f coprime to x. Thishas no standard meaning in the context you use it, where x is avariable. If you meant coprime in some ring of polynomials, which isclosest to a standard meaning, it would mean that there existspolynomials P and Q such that P*f+Q*x=1. But if f is an algebraicinteger, that's possible (if and) only if f is a unit, with Q=0, andyou also assume in the same sentence that f is a non unit.| > I don't know what you think is a fair way for things to work, but the| > way things actually work, this kind of problem with terminology will| > absolutely prevent you from making any headway.|| What will prevent me from making headway is if mathematicians| continually lie.Mathematicians don't continually lie. What you keep deciding arelies are just disagreements with you.Perhaps the biggest problem of all is that you've reached a pointwhere you don't have anybody you are willing to trust, who could helpyou sort out which of the claims people are making are valid and whichare baloney. So you tend automatically to assume that the thingspeople say that seem wrong to you are baloney of some kind or another.This also leaves you without any very good way to learn how to improveyour mathematical tinkering. We keep telling you different thingswhich _we_ claim would help you avoid pitfalls like you keep fallinginto. But (a) you don't trust us enough to believe we've identi'ed aproblem with what you're doing, and (b) you don't trust us to give youstraight advice on how to avoid it; you suspect us of just trying towaste your time. So you're stuck with only your own inklings of what'sa good way to work with proofs.| Now then in what way is it improper terminology to talk about| algebraic integer factors of 5?I didn't say it was improper to talk about algebraic integer factorsof 5. That has a perfectly well-de'ned meaning. An algebraic integerr is a factor of 5 if it divides 5 in the algebraic integers, meaningthat 5/r is also an algebraic integer.| > In an experiment in communication, I once tried elaborating on such a| > concept for you. You usually write as though the number of factors of| > 5 in an algebraic number satis'ed certain axioms which are familiar| > to me: being a rational number v(x)>=0 associated with each algebraic| > integer x <> 0, where v(5)=1, v(xy)=v(x)+v(y) for any algebraic| > integers x<>0 and y<>0, and v(x+y)>=min{v(x),v(y)} for algebriac| > integers x<>0, y<>0, and x+y<>0. Assuming that we have such a function| > v, then we can show that v(r1)=0 and v(r2)=v(r3)=1/2 for the roots| > r1, r2, and r3 of your cubic, if we take them in the right order.| > I think this is the best way to try to make sense of your argument| > in your advanced polynomial factorization thing.| >| > Unfortunately, this known concept is not so directly related to| > divisibility in the algebraic integers, so even if you did manage to| > produce a de'nition on these lines, it wouldn't show mathematicians| > are doing anything wrong when they make claims about divisibility in| > the algebraic integers as they teach about it in classrooms. They're| > just apples and oranges.|| There is no divisibility argument within the paper, and I only even| mention roots when the roots are algebraic integers.I guess you think of factorization as thoroughly unrelated todivisibility. Note P(m) has a factor that is f^2 has nothing to dowith divisibility of P(m) by f^2, then. This just highlights how faryour terminology departs from standard. Yes, I automatically take []is a factor of [] as a synonym for [] divides [], because that'sthe way it is with any normal usage of the terms. It's only a mistakeif one is dealing with someone like you who departs so far fromstandard usage.| I've talked| about x/y in discussing the error in taught mathematics, but did make| a post explaining that was when an argument is considered in the 'eld| of rationals.|| Your statement falls ¤at Keith Ramsay, and I think you're just trying| to sound good enough to fool people, rather than trying to get to the| truth.People have seldom had a hard time getting at the mathematical truthin these discussions, whenever the actual mathematical question hasbeen well-de'ned. (There was *once* a question about polynomialfactorization over the algebraic integers which required some effort,but that was a rare exception.)The problem is with getting well-de'ne claims. Take your termincomplete ring for example. Your de'nition of incomplete ringfor example amounts essentially to a ring in which one element doesn'tdivide another one, but it *should* divide it. You only imagine thatthe dif'culties created by using this meaninglessly vague term aresomeone else's fault.| What is clear is that the paper does NOT operate over any 'elds, and| depends only on ring operations.You seem to have an odd idea of what one is allowed to say afterhaving declared an argument to be in a ring. The whole point is tomake all of your mathematical statements be well-de'ned, not to obeyarbitrary conventions.When you say that you are working in the algebraic integers, thatprovides us with a certain kind of context, which allows us tounderstand the meaning of certain terms. For example, after you'vesaid that you're working in the algebraic integers, if you say ydivides x, it means y divides x in the algebraic integers.Otherwise it might be ambiguous what it meant, or mean somethingdifferent from what you intended.Saying you're doing this does NOT mean that it magically becomesforbidden to refer to x/y. Certainly it would be incorrect to assumethat x/y is de'ned in the algebraic integers. But there's an obviousmeaning to the expression x/y, so it's legitimate to refer to it (justso long as one realizes that it isn't necessarily an algebraic integeritself). If we're trying to 'gure out whether there exists analgebraic integer z with the property that x=yz, it makes perfectsense for us to consider 'rst the fact that there exists one and onlyone z that satis'es x=yz, and THEN ask whether that z (which existsin the algebraic numbers) is a member of the algebraic integers (whichis the ring we're considering primarily). Saying that x/y is analgebraic integer means just the same thing as saying that thereexists an algebraic integer z such that yz=x.| > Say we de'ne the number of factors of 5 in an algebraic integer x to be| > the highest rational number r such that 5^r divides x in the algebraic| > integers, i.e. such that there exists an algebraic integer y such that| > 5^r*y = x. I think this de'nition works (i.e., I think there is such a| > highest rational power for each algebraic number, although I haven't| > tried to write out a proof). If we de'ne it that way, though, then it| > simply doesn't have one of the properties I listed above.|| Your post isn't coherent mathematically given what I say in the paper.| Consider that in the ring of algebraic integers, 5 has algebraic| integer factors, and given algebraic integers r_1 and r_2, where their| product is 5, why are you acting as if it's so dif'cult to comprehend| that there must be some distribution of factors of 5?Just to repeat, it's not a dif'culty I have; it's your dif'culty insaying what you actually mean by that. And I don't think you have aclear idea of what you mean by that yourself.| For instance, both could have a factor that is sqrt(5), or one could| have a factor of 1+2i, while the other had a factor of 1-2i.|| The question I'm trying to get the reader to explore is, given that| they are roots of this particular polynomial, could one of them only| have unit factors of 5?|| How do you know?If you were using the terms in the standard way, this question wouldbe, How can you tell when one algebraic integer divides anotheralgebraic integer?If you actually de'ned your question, it would become clear eitherthat it's the same as the standard one (and that the answermathematicians give is correct), or that it's a different questionfrom the standard one (so that even if you get a different answer toyour question, it's just irrelevant to whether mathematicians arebeing truthful). I think this has a lot to do with why you simply keepthe question vague; this kind of clarity would take the drama out of it.Keith Ramsay =[snip for brevity]>> You can't factor a number part-way... unless *both* y and y/x are in>> the ring, then y isn't a factor of x, by de'nition of factor. You>> may be able to multiply together y and y/x to get x, but that's not>> the same thing.>>I note your follow-up correction post. My answer still is that you're>*assuming* that algebraic integers are complete, but then I say it's>not, and you try to fault me on your assumption, which I've proven>false.>>How about this?>>Let's say that someone claims that they have a complete ring, which is>a subset of the ring of integers, but they give you 6 as a member, ...Ok. There are exactly four such rings: Z, 2Z, 3Z, and 6Z (with usual addition and multiplication). Of those, only Z is a ring with unity. > ... and>*say* that every member of the ring has a factor of 2.Why would anyone say that? >Now you say, no, that's false. And they argue with you.>>You say, but 6/2 is not in the ring then, as it doesn't have a factor>of 2, and then they give an argument similar to what you gave above,>saying that of course as 2 and 6 are in the ring, and by the>de'nition every member has 2 as a factor!!!In Z, 2 is a factor of 6, because 2, 6, and 6/2 are _all_ in Z. In 2Z, 2 is a _not_ factor of 6, because 6/2 is not in 2Z.In 3Z, 2 is a _not_ factor of 6, because 2 is not in 3Z (even though 6/2 is).In 6Z, 2 is a _not_ factor of 6, because 2 is not in 6Z.>The arguing goes back and forth, back and forth, back and forth.>>But of course 6/2 = 3, so 3 is not in the ring. So you're right, and>they're wrong.You appear to be arguing that 3 _should_ be in the ring, otherwise the ring is incomplete. If so, which of the rings 2Z, 3Z and 6Z do you think is incomplete? They all contain 0; they're all closed under (usual) addition and multiplication. What's the problem?>Here I'm right, and you're wrong, as the *de'nition* of algebraic>integer leads to an incomplete ring, which I prove mathematically.>>You cannot depend on a *de'nition* in mathematics to prove something>that's been disproven by mathematics.Try this: (*) De'nition: The set of contradictory numbers is the set of all odd positive integers, that are divisible by 2.There is nothing wrong with this de'nition. The set of contradictory numbers is perfectly well de'ned. As any de'nition it can't be true or false, right or wrong. The de'nition itself doesn't say anything about the existence of such numbers, so we need a theorem: (*) Theorem: The set of contradictory numbers is empty._This_ can be right or wrong.It the same with algebraic numbers: (*) De'nition: The _set_ of algebraic integers is the set of all complex roots of monic polynomials with integer coef'cients.Note that the algebraic integers aren't _de'ned_ to be a ring. The de'nition doesn't even say anything about the existence of algebraic integers, so we need: (*) Theorem: The set of algebraic integers is not empty.Now, _this_ can be true or false, and so can this: (*) Theorem: The algebraic numbers form a ring under usual addition and multiplication.You (apparently) claim that this theorem is false. The only thing that can possibly mean, is that the algebraic integers are either not closed under addition or not closed under multiplication. When you try to prove that it is false, you appeal to factors. The normal de'nition of factor is: (*) De'nition: y is a factor of x in the ring R, if and only if, *all* of x, y, and x/y are in R.This is what is meant by the word factor. The _only_ way to prove that y is factor of x in the ring of algebraic integers, is to prove that x, y AND x/y are algebraic integers, because that is what the word factor means. _That_ is why 2 is NOT a factor of 6 in 2Z, 3Z and 6Z.-- Thomas Wasell | The heart has its reasons which reason knows nothingwasell@bahnhof.se | of. | -- Blaise Pascal =... > First consider that the de'nition of algebraic integers as the > *roots* of monic polynomials with integer coef'cients does not > speci'cally give you any protection from contradiction.That is true. > The de'nition is over the 'eld of rationals, and it's not clear in > just looking at it, if it will give a complete ring.Indeed, nor is it claimed that it does. > Mathematicians apparently *assumed* it did, and have gone on that > assumption for some time.You have that wrong. It has been *proven* that with this de'nitionthe set of algebraic integers is closed under addition and multiplication.That is, it has been proven that given two algebraic integers, theirsum and their product is also an algebraic integer. (See for instanceVan der Waerden, Algebra.) As the set contains also 0, 1 and the additiveinverse of each element, the set is (from this) actually a ring. > People seem 'xed on the *de'nition* with the assumption, when > there's no proof based on the de'nition that the ring is complete.What do you *mean* with an incomplete ring? The algebraic integersare closed under addition and multiplication and satisfy the followingaxioms:1: a + (b + c) = (a + b) + c2: a + b = b + a3: there is a 0 such that 0 + a = a4: for each a there is a (-a) such that a + (-a) = 05: a * (b + c) = a * b + a * c6: (a * b) * c = a * (b * c)7: a * b = b * a8: there is a 1 such that 1 * a = aand so they form a ring (a commutative ring to be precise). > And there can't be such a proof because I've *proven* it's not > complete, as that de'nition leads to a contradiction.In what way not complete? What of the above axioms does it notsatisfy? Or is it not closed under addition or multiplication? > Go to the following link, 'nd and read my paper Advanced Polynomial > Factorization: > http://groups.msn.com/AmateurMath > and you will see how you can prove that x has y as a factor.I might do that when you do not put it behind some doors. > Consider c=ab, where .95c' is an algebraic integer, and .95a' is an > algebraic integer, but .95b' is not.Yup, entirely possible. So what? So .95a' is not a factor of .95c'.Your proof that for some reason .95a' must be a factor of .95c'contained ¤aws.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ => > I'm putting this bit 'rst to highlight it:Note: Keith Ramsay is pointing out that he has edited my post interms of placement of text, as he has moved a section. > [...]> | Consider that in the ring of algebraic integers, 5 has algebraic> | integer factors, and given algebraic integers r_1 and r_2, where their> | product is 5, why are you acting as if it's so dif'cult to comprehend> | that there must be some distribution of factors of 5?> > I'm not acting as if I'm having any kind of dif'culty. I'm saying> that *you're* having dif'culty in realizing that there must be some> distribution of factors of 5 is a meaninglessly vague phrase. You're> saying this as if to say, Shouldn't this be true?, and not as if> you actually had proven it was true.Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute insome way between r_1 and r_2.For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 inboth r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unitfactors of 5 with r_1.It hardly seems like rocket science.Can you explain why you as a mathematician are having dif'culty inthe area Keith Ramsay?> By the standard meanings of the terms, saying an algebraic integer r_1> has a factor of 5 in the algebraic integers is just the same as> saying it's divisible by 5 in the algebraic integers. That's the same> as saying that there's an algebraic integer t for which r_1=5t.Well I thought by standard usage a factor of 12 is 3.Or I could say that a factor of 5 is 1+2i.Are you saying that is nonstandard usage?> There isn't, and I think you know there isn't. Since neither r_1 nor> r_2 is divisible by 5, there's no distribution of factors of 5> between them.Well, if r_1 has a factor of 5 that is 1+2i, and r_2 has a factor of 5that is 1-2i, then I'd say there it is clear there is a distributionof factors of 5.I've noted before how I talk of factors of and when I mean aspeci'c factor I say that is, for instance, a factor of 5 that is1+2i.> I think you use ill-de'ned terms because on some level you're aware> that using only well-de'ned terms spoils your fun. It's disillusioning.> It clear away the fog of obfuscation. Suddenly you're back to reality,> and it's the rather ordinary reality that people keep telling you that> you're in, not the amazing story you wish it was.You're having problems Keith Ramsay with standard usage, as it lookslike you may be sticking in the word multiple where I use the wordfactor.So it's your fault, not mine.> Back to the top.> > | > [...]> | > | What I can do is take out the use of the term in the paper, and use> | > | factor in the ring of algebraic integers, and note that will lead to> | > | a contradiction as the point of the paper is that the ring is> | > | incomplete.> | > |> | > | What I want to impress upon readers is that I'm quite willing to work> | > | with mathematicians to explain the mistake that they're teaching.> | >> | > If you want to show that a mistake is being taught, you should quote it> | > speci'cally.> |> | That's not very sensible if you expect that for a mistake to be taught> | for any length of time it'd have some subtlety, unless you're also> | questioning the competence of mathematicians.> > Quite to the contrary-- it's because mathematicians are generally> competent that one can ordinarily say very speci'cally what is wrong> when they make mistakes.The mistake is in assuming that de'nition of the ring of algebraicintegers does not lead to contradictions.> What makes a mathematical mistake subtle is not what it takes to point> it out once it's been recognized; it's the dif'culty in recognizing> it in the 'rst place.You don't give a proof. I contend that you can have a subtle mathmistake that is dif'cult to point out when it's been recognized, andindeed it can also be dif'cult to recognize in the 'rst place.> | It seems to me that rather than deal with the important question,> | which is whether or not I'm correct, people often try to introduce> | weird ad hoc conditions.> > You created an ad hoc rule that in order to refute a claim of yours,> it's not good enough for someone else to prove that the conclusion is> wrong. You think a proof of yours can show that a proof of a> mathematician is wrong, but you appear unwilling to accept the> possibility that a proof from a mathematician can just as well show> that one of yours is wrong (i.e., not really a proof).That's not true. I've noted that proofs don't duel, which is thatmath proofs don't contradict each other.Therefore, you cannot claim that a proof of mine is wrong based onanother proof, if neither has an error, as they will NOT contradict.Therefore, if a proof of mine is not a proof, then it must have anerror.Several people have claimed that they have proofs that refute mine,but I've also pointed out that there are other interpretations, evenassuming they have proofs.I've also noted that it's odd that mathematicians would rather try toduel with proofs than simply 'nd an error in my proof. But, ofcourse, as it is a proof, there is no error, which is why I think theytry to 'nd other means.That's not mathematics. That's cheating.> | If you disagree with that assessment, can you please explain why you> | believe there should be something simple and short enough for me to> | quote?> > In mathematics, we try to make each *mathematical* statement have its> own well-de'ned meaning. (There are of course nonmathematical claims> that we sometimes make, which aren't always precise.) If a sequence of> well-de'ned mathematical statements is wrong, it's wrong because (at> least) one of the statements is wrong.> > The key is making sure all the de'nitions are sound.As I've stated before a proof begins with a truth and proceeds bylogical statements to a conclusion that then must be true.Therefore it follows that de'nitions must be sound, or the steps willnot be logical.> Mathematics that is vaguely enough written that it can be wrong> without any individual statements in it being de'nitely wrong, is> considered very bad. It's bad because an author hasn't de'ned their> terms well enough.The term bad is a human term inapplicable in context as mathematicsis about truth.That is, a proof is neither bad nor good, it is true or false. > | > | The ring of algebraic integers is incomplete, it's easy to show, and> | > | I've shown it with my paper.> | > |> | > | If mathematicians are having trouble understanding any part of it, I'm> | > | able to explain further.> | >> | > You have a big problem with terminology. You use terms that aren't> | > used by others in the context in which you use them, such as counting> | > factors of 5 in algebraic integers. I haven't seen anything I would> | > consider an adequate de'nition of those terms. You don't explain> | > why your terms should be considered relevant to the ones mathematicians> | > are using.> |> | That statement possibly has something to do with what I now call the> | preamble in the paper as it is there to help explain context, but> | isn't part of the actual argument.> |> | Here's what I say, copied from the paper (some editing for format):> |> | To highlight the standard belief consider the algebraic integers> | (-1-sqrt(21))/2 and (-1+sqrt(21))/2 which are roots of x^2 + x - 5.> |> | While you know that the algebraic integer factors are themselves> | factors of 5, in what way is each a factor of 5? Can either not have> | non unit factors of 5? How do you know?> |> | There I'm talking about algebraic integers, so one can assume I'm> | talking about algebraic integer factors. Given that 5 has algebraic> | integer factors, how is what I say nonstandard Keith Ramsay?> |> | I'd like you to carefully explain your assertion. The question I'm> | raising is the possibility that you lied.> > No, I haven't lied. You are not using have non unit factors of 5 in> a standard way.Can a number have non unit factors of 5 Keith Ramsay?I simply applied a not to the front.If you claim that a number cannot have non unit factors of 5, then I'dlike to know why.Here your confusion may come from thinking that non unit factors of 5must be 5. However another non unit factor of 5 is 1+21. Yes, 5 is afactor of itself, but there is no reason to assume that the statementnon unit factors of 5 forces one of the the factors to be 5, thoughit may be 5.Now let's say I have something like 25, which has factors of 5. Notice that it *also* has 1+2i as a factor, while it also has 5*5 as afactor.You seem to be 'xed on a more limiting case where factors of 5means *multiples* of 5.But because you read multiples when I say factors of you assumethat I'm wrong.I use the de'nitions precisely, while you interpret.Are you a math expert Keith Ramsay?If so, why am I demonstrably more precise than you?> If at this point in the paper, you meant it in a completely standard> way, here's what the meaning would be. Have a factor of 5 is> synonymous with have 5 as a factor, or be divisible by 5. Also 5> is a non-unit (so that part is redundant).You are incorrect. > Now obviously we all know you don't mean that. It's possible you mean> something like, shares non-unit factors with 5, meaning that there> are non-unit algebraic integers that are factors of both numbers. But> this is just guesswork. Any serious writer of mathematics takes> responsibility for taking out the guesswork, NOT forcing the reader to> keep inferring which of various possible meanings is the intended one.Yet your problem apparently is in adding the term multiple which isnot my fault.It's your failure Keith Ramsay.> The standard usage is also inconsistent with the rest of the paper.> You write, for example, ...proving that two of the a's have a factor> that is f at a point where you plainly have NOT shown that f is a> factor of any of the a's, in any of the usual ways factor is> de'ned.Hmmm...that's an interesting point. What has happened at that pointin the paper is that I've considered the constant term P(0), and theconstant term with f^2 separated off, which is P(0)/f^2 = 3x u^2 + u^3f, and noted that it is coprime to f.Now I then consider g_1 at m=0, where c=g_1, and notice it has afactor of f, and then based on P(0)/f^2 being coprime to f, I havethat r + c, must have a factor of f, which proves that r must have afactor of f that is f.> Elsewhere, you refer to an algebraic integer f coprime to x. This> has no standard meaning in the context you use it, where x is a> variable. If you meant coprime in some ring of polynomials, which is> closest to a standard meaning, it would mean that there exists> polynomials P and Q such that P*f+Q*x=1. But if f is an algebraic> integer, that's possible (if and) only if f is a unit, with Q=0, and> you also assume in the same sentence that f is a non unit.But f while a variable is constant, and so is x. So it is not in anyring of polynomials.Unfortunately, you seem to be 'xated on the *letters* as in seeing anx you may assume that it's varying. Nope. It's constant.That's troubling Keith Ramsay as that's basic algebra. The symbolshave to be understood as de'ned, not based on your experience of howyou may be used to seeing them. > | > I don't know what you think is a fair way for things to work, but the> | > way things actually work, this kind of problem with terminology will> | > absolutely prevent you from making any headway.> |> | What will prevent me from making headway is if mathematicians> | continually lie.> > Mathematicians don't continually lie. What you keep deciding are> lies are just disagreements with you.That's not true. What I've done is explain clearly and in detail.In reply I 'nd people making specious issues, like your claims aboutfactor when you apparently are sticking in multiple.> Perhaps the biggest problem of all is that you've reached a point> where you don't have anybody you are willing to trust, who could help> you sort out which of the claims people are making are valid and which> are baloney. So you tend automatically to assume that the things> people say that seem wrong to you are baloney of some kind or another.How can I take people like you seriously when you have trouble withsuch simple things as factor of?> This also leaves you without any very good way to learn how to improve> your mathematical tinkering. We keep telling you different things> which _we_ claim would help you avoid pitfalls like you keep falling> into. But (a) you don't trust us enough to believe we've identi'ed a> problem with what you're doing, and (b) you don't trust us to give you> straight advice on how to avoid it; you suspect us of just trying to> waste your time. So you're stuck with only your own inklings of what's> a good way to work with proofs.Yet I've caught you in a contradiction with yourself, where youapparently have been inserting the word multiple when I saysomething like factor of 5, so that you read multiples of 5. Ifthat's not what you're doing then, who knows what's going through thatbrain of yours.Readers should consider what follows if they think that's too harsh.> | Now then in what way is it improper terminology to talk about> | algebraic integer factors of 5?> > I didn't say it was improper to talk about algebraic integer factors> of 5. That has a perfectly well-de'ned meaning. An algebraic integer> r is a factor of 5 if it divides 5 in the algebraic integers, meaning> that 5/r is also an algebraic integer.You're contradicting yourself. Go back and read over what you said atthe top of this post. I'm tiring of this exercise in pointing outyour errors.James Harris => Nope. There is no claim that y is a factor as it's *provably* a> factor.*sigh* Could you give me your de'nition of a factor, then? I suspectit's different from what mathematicians usually mean, because the way toprove y is a factor of x is to 'nd some z such that y*z=x, with suitableproperties for all three of them (i.e. all in the algebraic integers,say). If you haven't done this, how can you prove it's a factor?> Rather than deal with the term factor as people get confused, in my> paper Advanced Polynomial Factorization, I have three numbers a_1,> a_2, and a_3, and I prove that a_3 is coprime to a factor I call f.Argh, you profess to want to get away from the term .95factor', and thenyou go and use it right again. Please de'ne your terms rigorously.> Consider c=ab, where .95c' is an algebraic integer, and .95a' is an> algebraic integer, but .95b' is not.> > Now if you include fractions or move to a 'eld like algebraic numbers> that's ok. But I'm talking about .95b' that's not in any way a fraction> or fractional.>> It's like with the ring of evens, and given 6 = 2(3), you have that 3> is outside the ring, as in the ring of evens, 2 is NOT a factor of 6.> > The situation is analogous.> > Do you understand?No, I don't. You seem to be operating under the assumption that allare you claiming that the set of evens is incomplete, because 2should be a factor of 6, but isn't?How is your situation with the algebraic integers distinct from this,if it is distinct? => > > |>> I'm putting this bit 'rst to highlight it:> > Note: Keith Ramsay is pointing out that he has edited my post > in terms of placement of text, as he has moved a section.Note that James Harris has drawn attention to Keith RamsayJim Burns> [...]> | Consider that in the ring of algebraic integers, 5 has > | algebraic integer factors, and given algebraic integers > | r_1 and r_2, where their product is 5, why are you acting > | as if it's so dif'cult to comprehend that there must be > | some distribution of factors of 5?>> I'm not acting as if I'm having any kind of dif'culty. I'm > saying that *you're* having dif'culty in realizing that > there must be some distribution of factors of 5 is a > meaninglessly vague phrase. You're saying this as if to say, > Shouldn't this be true?, and not as if you actually had > proven it was true.[ etc. etc. etc.] =>> >> I'm putting this bit 'rst to highlight it:>>Note: Keith Ramsay is pointing out that he has edited my post in>terms of placement of text, as he has moved a section.> >> [...]>> | Consider that in the ring of algebraic integers, 5 has algebraic>> | integer factors, and given algebraic integers r_1 and r_2, where their>> | product is 5, why are you acting as if it's so dif'cult to comprehend>> | that there must be some distribution of factors of 5?>> >> I'm not acting as if I'm having any kind of dif'culty. I'm saying>> that *you're* having dif'culty in realizing that there must be some>> distribution of factors of 5 is a meaninglessly vague phrase. You're>> saying this as if to say, Shouldn't this be true?, and not as if>> you actually had proven it was true.>>Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in>some way between r_1 and r_2.It makes sense, but (assuming I know what you mean - yes you_are_ using the language in nonstandard ways) it's simply_not_ _true_.Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factorof 5. Does it go with r_1 or r_2?A: Neither (again assuming go with means what I conjectureit does). In the algebraic integers sqrt(5) is _not_ a factor of r_1and it is also _not_ a factor of r_2.Proof for r_1: The minimal polynomial of (1+2i)/sqrt(5) is5x^4 + 6x^2 + 5. So (1+2i)/sqrt(5) is not an algebraic integer.Proof for r_2: Exactly the same.>For instance, if r_1 = r_2 = sqrt(5), you have the same factor of 5 in>both r_1 and r_2, while with r_1=5, r_2 = 1, you have all the non unit>factors of 5 with r_1.>>It hardly seems like rocket science.>>Can you explain why you as a mathematician are having dif'culty in>the area Keith Ramsay?>>> By the standard meanings of the terms, saying an algebraic integer r_1>> has a factor of 5 in the algebraic integers is just the same as>> saying it's divisible by 5 in the algebraic integers. That's the same>> as saying that there's an algebraic integer t for which r_1=5t.>>Well I thought by standard usage a factor of 12 is 3.>>Or I could say that a factor of 5 is 1+2i.>>Are you saying that is nonstandard usage?That's not the usage he's talking about. The problem isthis has a factor of thing. When you say a has a factorof b that _means_ that b _is_ a factor of a. But whenyou say it that's not what you mean, what you meanis that there _is_ a factor of b which is also a factor of a.>> There isn't, and I think you know there isn't. Since neither r_1 nor>> r_2 is divisible by 5, there's no distribution of factors of 5>> between them.>>Well, if r_1 has a factor of 5 that is 1+2i, and r_2 has a factor of 5>that is 1-2i, then I'd say there it is clear there is a distribution>of factors of 5.>>I've noted before how I talk of factors of and when I mean a>speci'c factor I say that is, for instance, a factor of 5 that is>1+2i.And many other people have noted before that the way youuse these terms is simply wrong. Nobody knows why youinsist on speaking your own language - if you were actuallytrying to communicate you'd use words in their standardways instead of requiring the rest of the world to rememberthat when _you_ say a has a factor of b it means somethingdifferent from what it means when everyone else says it.>> I think you use ill-de'ned terms because on some level you're aware>> that using only well-de'ned terms spoils your fun. It's disillusioning.>> It clear away the fog of obfuscation. Suddenly you're back to reality,>> and it's the rather ordinary reality that people keep telling you that>> you're in, not the amazing story you wish it was.>>You're having problems Keith Ramsay with standard usage, as it looks>like you may be sticking in the word multiple where I use the word>factor.>>So it's your fault, not mine.No, his interpretation is _exactly_ what what you said _actually_means. It's _your_ fault that you mean something else, not his.>> Back to the top.>>[...]>>But because you read multiples when I say factors of you assume>that I'm wrong.>>I use the de'nitions precisely, while you interpret.Just the opposite. Saying a has a factor of b _does_ meanthat a is a multiple of b. To say what you mean by the phraseyou should say instead a shares a factor with b or a has afactor in common with b.>Are you a math expert Keith Ramsay?>>If so, why am I demonstrably more precise than you?It's exactly like when you were being demonstrably moreprecise than all the fools who didn't realize that integerswere irrational.>> If at this point in the paper, you meant it in a completely standard>> way, here's what the meaning would be. Have a factor of 5 is>> synonymous with have 5 as a factor, or be divisible by 5. Also 5>> is a non-unit (so that part is redundant).>>You are incorrect.> >> Now obviously we all know you don't mean that. It's possible you mean>> something like, shares non-unit factors with 5, meaning that there>> are non-unit algebraic integers that are factors of both numbers. But>> this is just guesswork. Any serious writer of mathematics takes>> responsibility for taking out the guesswork, NOT forcing the reader to>> keep inferring which of various possible meanings is the intended one.>>Yet your problem apparently is in adding the term multiple which is>not my fault.>>It's your failure Keith Ramsay.>>>[...]>>That's not true. What I've done is explain clearly and in detail.>>In reply I 'nd people making specious issues, like your claims about>factor when you apparently are sticking in multiple.>>> [...]>>Yet I've caught you in a contradiction with yourself, where you>apparently have been inserting the word multiple when I say>something like factor of 5, so that you read multiples of 5. If>that's not what you're doing then, who knows what's going through that>brain of yours.>>Readers should consider what follows if they think that's too harsh.>>> | Now then in what way is it improper terminology to talk about>> | algebraic integer factors of 5?>> >> I didn't say it was improper to talk about algebraic integer factors>> of 5. That has a perfectly well-de'ned meaning. An algebraic integer>> r is a factor of 5 if it divides 5 in the algebraic integers, meaning>> that 5/r is also an algebraic integer.>>>>You're contradicting yourself. No he's not. It's not improper to say that an algebraic integer hasa factor of 5, and he never said it was improper. What _is_improper is to say it has a factor of 5 _when_ that's not whatyou actually _mean_.> Go back and read over what you said at>the top of this post. I'm tiring of this exercise in pointing out>your errors.>>>James Harris************************David C. Ullrich Burns >>> >> >> |>>>> I'm putting this bit 'rst to highlight it:>> >> Note: Keith Ramsay is pointing out that he has edited my post >> in terms of placement of text, as he has moved a section.>>Note that James Harris has drawn attention to Keith RamsaySay, I'm glad you pointed that out.>Jim Burns>>> [...]>> | Consider that in the ring of algebraic integers, 5 has >> | algebraic integer factors, and given algebraic integers >> | r_1 and r_2, where their product is 5, why are you acting >> | as if it's so dif'cult to comprehend that there must be >> | some distribution of factors of 5?>>>> I'm not acting as if I'm having any kind of dif'culty. I'm >> saying that *you're* having dif'culty in realizing that >> there must be some distribution of factors of 5 is a >> meaninglessly vague phrase. You're saying this as if to say, >> Shouldn't this be true?, and not as if you actually had >> proven it was true.>[ etc. etc. etc.]************************David C. Ullrich => > >> I'm putting this bit 'rst to highlight it:>Note: Keith Ramsay is pointing out that he has edited my post in>terms of placement of text, as he has moved a section.> >> [...]>> | Consider that in the ring of algebraic integers, 5 has algebraic>> | integer factors, and given algebraic integers r_1 and r_2, where their>> | product is 5, why are you acting as if it's so dif'cult to comprehend>> | that there must be some distribution of factors of 5?>> >> I'm not acting as if I'm having any kind of dif'culty. I'm saying>> that *you're* having dif'culty in realizing that there must be some>> distribution of factors of 5 is a meaninglessly vague phrase. You're>> saying this as if to say, Shouldn't this be true?, and not as if>> you actually had proven it was true.>>Given that r_1 r_2 = 5, it makes sense that factors of 5 distribute in>some way between r_1 and r_2.> > It makes sense, but (assuming I know what you mean - yes you> _are_ using the language in nonstandard ways) it's simply> _not_ _true_.> > Suppose that r_1 = 1 + 2i and r_2 = 1 - 2i. Now sqrt(5) is a factor> of 5. Does it go with r_1 or r_2?See what I mean sci.skeptic? The issue here is that I'm actuallyusing standard mathematical usage, while several posters apparentlykeep reading factors of 5 as multiples of 5.The problem has to do with their misuse of mathematical terminology.For instance 2 is a factor of 6, but so is 3, and in fact, so is 6.When you say, factor of, it means something that is a factor of thegiven number. And similarly factors of would mean factors of thegiven number.But you have these posters, like David Ullrich who is a math professorat Oklahoma State University, who are lost with rather basicmathematical language to the point that they *keep* arguing *after*I've corrected them. You see their *belief* system apparently is thatas mathematicians they can't be the ones with the error, so amazingly,they simply keep almost mindlessly repeating it.You may guess that they say factors of when they mean multiplesof, but I'm using the proper terminology, in the correct way.So when I say factors of 5, I'm NOT saying multiples of 5. Forinstance, 2 and 3 are factors of 6. So when I say factors of 6 itdoesn't mean multiples of 6. And I emphasize that there is amathematical term multiples of which applies.And in fact, using factors of when you mean multiples of whilecommon, is technically incorrect.But people like Ramsay and Ullrich are unlikely to reply to *correct*their mistakes because the society of sci.math lets people itconsiders part of the society get away with the dumbest mistakes.That's how mathematicians operate as demonstrated before your eyes.And that's how they can have errors in their discipline for years, andyears because they seek to by de'nition have a society that isperfect, when in fact, mathematicians are just people, and people makemistakes.James Harris