mm-422 === Subject: Re: Correction: Error in NOTICE Visiting Assistant Professor at the University of Montana. [.snip.]>You seem to be back to your periodic denial that Magidin & McKinnon were>incorrect in their rediscovery of the result of Cohn, McAdam, and Rush.Too many negations!You mean, You seem to be going back to your periodic CLAIM that [...] were incorrect in [...]or do you mean You seem to be going back to your periodic denial that [...] were CORRECT in [...] === Subject: Re: Correction: Error in NOTICE> [.snip.]>You seem to be back to your periodic denial that Magidin & McKinnon were>incorrect in their rediscovery of the result of Cohn, McAdam, and Rush.> Too many negations!> You mean,> You seem to be going back to your periodic CLAIM that [...] were> incorrect in [...]> or do you mean> You seem to be going back to your periodic denial that [...] were> CORRECT in [...] I had meant his periodic denial of your re-discovery, et cetera.Still working on that proofreading business, === Subject: Re: Correction: Error in NOTICE> My message NOTICE: Error in Currently Taught Mathematics posted June> Here's the relevant portion from the post:> coefficients being monic so that you have> P(x) = (x + a_1)...(x + a_n)> but that is unbced as a complete ring must handle all non monic> polynomials of degree n with integer coefficients to get> P(x) = (a_1 x + b_1)...(a_n x + b_n)> but in fact the a's and b's cannot here always be algebraic integers> as I've shown in my paper Advanced Polynomial Factorization...> algebraic integers for that factorization as it shows a coprimeness> result for a particular family of polynomials, which then can be used> to show a problem in the ring as explained in multiple postings.> My apologies for the error and the lateness in issuing a correction. This is not good enough, for three reasons: 1. As an apology it is lame. You have called me and others here liars for saying the conclusions of Advanced Polynomial Factorization are wrong. You owe us a direct personal apology. I am not a liar. 2. Your APF web page has not changed. You are thus flouting your own apology and admission of error. People who continue to access that site are going to think either that it is correct, and thus learn some erroneous math (admittedly the chances of this are slim), or that you are too proud or stupid to retract it even after you have admitted it is wrong. 3. You persist in claiming the following: The definition of algebraic integers allows one to find algebraic integers a_1, a_2, and a_3, such that they are roots of a monic cubic irreducible over Q with integer coefficients, where one is provably coprime to a prime by you today in this same thread] Let G(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. G(x) is clearly monic. Assume G(x) is irreducible over the rationals. Let u1, u2, and u3 be roots of G(x). Note that by definition, u1, u2, and u3 are algebraic integers. You are claiming that at least one of u1, u2, or u3 is coprime to p. Assume u1 is coprime to p. By standard theory***, there exists an automorphism F12 of the field of algebraic numbers such that: 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(u1) = u2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since u1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*u1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(u1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(u1) = u2. Thus one obtains: r'*u2 + s'*p = 1, which says: u2 and p are coprime in the algebraic integers. Similarly one shows that u3 and p are coprime. Therefore if one of u1, u2, or u3 is coprime to p, then they all are. But u1 * u2 * u3 = - p * v. That is, p divides the product of u1, u2, and u3. Therefore p cannot be coprime to each of u1, u2, and u3. Putting all this together, one concludes that NONE of u1, u2, or u3 can be coprime to p. This directly contradicts your statement which was quoted above. Please feel free to point out any errors in the proof I just gave. You need to do more than just retract your June 10 NOTICE. You need to retract all of Advanced Polynomial Factorization. There is a lot more wrong with it than just a few misprints and algebraic slips. The whole underlying idea is a crock. I have pointed out previously where your logic goes astray but you have given no hint of having understood it. Worse yet: Consider the following snippet from your web page purporting to prove Fermat's Last Theorem: ... exactly two of the a's must have a factor of f^j. That is, you need exactly the same kind of conclusion that you had for APF in your FLT argument. And you absolutely do not have it. You no more have a proof for your ill-defined objects than you do for algebraic integers. But look on the bright side. You get a lot more free space on your website. Nora B.***: For discussion of such automorphisms, see: http://www.math.niu.edu/~beachy/aaol/galois.html, especially Prop. 8.6.2. Or see the excellent textbook Abstract Algebra by John Beachy and William D. Blair. === Subject: Re: Correction: Error in NOTICE> I'll make one reply in this thread, which is this one.> The definition of algebraic integers allows one to find algebraic> integers a_1, a_2, and a_3, such that they are roots of a monic cubic> irreducible over Q with integer coefficients, where one is provably> coprime to a prime factor of the last coefficient. That coprimeness> results leads inevitably to the conclusion that all must be coprime to> that prime factor in the ring of algebraic integers, which is the> contradictory result which proves the incompleteness of the ring.> By incomplete I mean that elements that should be in the ring are> not such that the contradiction arises.> By handle all non-monic polynomials above I meant that the> factorization for all non-monic polynomials should be in the ring.> After that I asserted that my paper Advanced Polynomial Factorization> proved they did not, which is the error corrected by this post, as> actually I prove a coprimeness result.Give up, . You are wrong! You have been proven wrong repeatedly, and spawning new threads won't coveryour tracks. Go back to your sandbox.--There are two things you must never attempt to prove: the unprovable -- and the obvious.----http://www.crbond.com === Subject: Re: Correction: Error in NOTICE>My message NOTICE: Error in Currently Taught Mathematics posted June> No, really? As soon as I saw the subject line, I knew this was a JSH thread :) === Subject: Re: Correction: Error in NOTICE>You seem to be back to your periodic denial that Magidin & McKinnon were>incorrect in their rediscovery of the result of Cohn, McAdam, and Rush.> Too many negations!> You mean,> You seem to be going back to your periodic CLAIM that [...] were> incorrect in [...]> or do you mean> You seem to be going back to your periodic denial that [...] were> CORRECT in [...]> ?>Darn it all, anyhow. I outworded myself. I couldn't fail to disagree with you less! === Subject: Re: Correction: Error in NOTICE>You seem to be back to your periodic denial that Magidin & McKinnon were>incorrect in their rediscovery of the result of Cohn, McAdam, and Rush.> Too many negations!> You mean,> You seem to be going back to your periodic CLAIM that [...] were> incorrect in [...]> or do you mean> You seem to be going back to your periodic denial that [...] were> CORRECT in [...]> ?>Darn it all, anyhow. I outworded myself. > I couldn't fail to disagree with you less! Is the following statement about you true or false: You are not the kind of person who wouldn't oppose the idea of not taking a negative stance against those who do not fail to protest the opposition to legislation that would legalize child pornography? from Scott Campisi, Wake Village, Tex. Contest: A bogus question you would like to sneak into the interview portion of the Miss Universe Pageant) === Subject: cross productI don't understand the proof in my textbook that A, B and AxB form aright-handed triad. Little help? === Subject: Re: cross product > I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help? Assuming that you have a list of three linearly independent vectors in a real 3-space, do you know how to tell whether they form a right handed triad or a left handed triad? === Subject: Re: cross product>I don't understand the proof in my textbook that A, B and AxB form a>right-handed triad. Little help?You realize, I hope, that we don't know what textbook you have, sowe don't know exactly what proof you're referring to. Can you giveus more details of what the proof is, and what you don't understandabout it? === Subject: Re: cross product> I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help?My understanding of a right-handed triad is that if you use your rightthumb, index and middle fingers to represent the directions of A, B and AxBrespectively, it tells you in which direction AxB is.So A = thumb, B = index finger (make an L with these fingers), and AxB isyou middle finger (bent so it sticks out from your palm).So AxB is perpendicular to the plane that the vectors A, B are in. (ie: AxBis perpendicular to both A and B) === Subject: Re: cross product>I don't understand the proof in my textbook that A, B and AxB form a>right-handed triad. Little help?There is another version, IIRC, which says point thumb in thedirection of A, second finger in the direction of B, and then themiddle finger, positioned perpendicular to both these, will show youthe orientation of A x B. If you are not sure of what sense that makes, try it with both handsand visualise the difference. === Subject: Re: cross product> I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help?Is it a proof ? I mean, isn't the right-hand rule just theorientation-definition of the cross product. Sounds like one ofthose stupid prove axiom by using the axiom tasks, e.g Prove that1+2=2+1./TV === Subject: Re: cross productI don't know the proof.But I do know that two vectors are perpendicular if their dot product iszero. You presumably know the formula for the crossproduct a = b x c and theformula for the dot product x=a dot byou could just substitute and confirm (b x c) dot b = a dot b = 0 bysubstitution.> I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help? === Subject: Re: cross product> My understanding of a right-handed triad is that if you use your right> thumb, index and middle fingers to represent the directions of A, B and AxB> respectively, it tells you in which direction AxB is.> So A = thumb, B = index finger (make an L with these fingers), and AxB is> you middle finger (bent so it sticks out from your palm).> So AxB is perpendicular to the plane that the vectors A, B are in. (ie: AxB> is perpendicular to both A and B)It amounts to the same thing, but my text has it like this:Place your right hand so that your fingers point in the direction of A,and when you bend them they rotate toward the direction of B. Thenyour thumb will point in the direction of AxB. That's the right-handrule. === Subject: Re: cross productI know what it means (geometrically), but I don't know how to tell(algebraically) whether it's true or not.| ,|| > I don't understand the proof in my textbook that A, B and AxB form a| > right-handed triad. Little help?| > --| >| > Peace,| > EJ| >| >|| Assuming that you have a list of three linearly independent| vectors in a real 3-space, do you know how to tell whether| they form a right handed triad or a left handed triad? === Subject: Re: cross productThe textbook I'm talking about is Calculus of Several Variables by Robert A.Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) crossproduct UxV is the vector determined by the following (geometric) conditions:1) perpendicular to U and to V2) length is |U|*|V|*|sin(x)|3) U,V,(UxV) form a right-handed triadThe first two parts are amenable to simple algebraic verification. Those partsare clear. But when he gets to part (3) he lapses into hand-waving and loses mecompletely.| >I don't understand the proof in my textbook that A, B and AxB form a| >right-handed triad. Little help?|| You realize, I hope, that we don't know what textbook you have, so| we don't know exactly what proof you're referring to. Can you give| us more details of what the proof is, and what you don't understand| about it?|| | Department of Mathematics http://www.math.ubc.ca/~israel| University of British Columbia| Vancouver, BC, Canada V6T 1Z2| === Subject: Re: cross product> I know what it means (geometrically), but I don't know how to tell> (algebraically) whether it's true or not.Take the three vectors in order, make them the rows of a determinant.If the determinant is positive, it is a right-handed system. Ifnegative, left-handed. If zero, they three vectors are co-planar, soright or left is not defined. === Subject: Re: cross productElaine says...>The textbook I'm talking about is Calculus of Several Variables by Robert A.>Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross>product UxV is the vector determined by the following (geometric) conditions:>1) perpendicular to U and to V>2) length is |U|*|V|*|sin(x)|>3) U,V,(UxV) form a right-handed triad>The first two parts are amenable to simple algebraic verification. Those parts>are clear. But when he gets to part (3) he lapses into hand-waving and loses me>completely.Here's one definition of a right-handed triad: vectors A, B, and Cform a right-handed triad if the triple product A . (B x C) is positive,where . is the scalar product.So compute U . (V x (U x V)). It's pretty easy to see that it is positiveif you use the permutation rule A . (B x C) = C . (A x B)--Daryl McCullough === Subject: Re: cross product> The textbook I'm talking about is Calculus of Several Variables by Robert A.> Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross> product UxV is the vector determined by the following (geometric) conditions:> 1) perpendicular to U and to V> 2) length is |U|*|V|*|sin(x)|> 3) U,V,(UxV) form a right-handed triad> The first two parts are amenable to simple algebraic verification. Those parts> are clear. But when he gets to part (3) he lapses into hand-waving and loses me> completely.I have two comments:(1) For the geometric picture, you know (1) and (2), so you'veidentified (from 1) a line in 3-space, together with (from 2) amagnitude. To get a unique vector, since there are at this stageTWO possibilities, all you need is to establish which of the twovalues is the correct one; that is, all you care about at thispoint is a direction along that line. The right-handed triadjazz is all about taking a [standard, right-handed] screw,aligning it along that line, and rotating it in the directionfrom U towards V The direction that the screw would move ifyou were actually screwing [pardon the unintentional salaciousallusion], gives the direction along that line, to the correctvalue of UxV.You could also imagine a (again, right-hand) threaded rod, alignedwith that line, and a nut that you place on the rod at some point.As you rotate the nut from U towards V, it will move in thedirection you want to choose.(2) The real proof is most likely a hand-wave because it relies onthe right-handedness of your coordinate system; that is (using thei,j,k convention for unit vectors in 3-space) one has this: i x j = k.If the above screw analogy applies to this triple, then yourcoordinate system is right-handed, and your cross-product yieldsa right-handed triad. What the cross-product *does* do is toyield a triple with the same handedness as the basic triple i,j,k.> | >I don't understand the proof in my textbook that A, B and AxB form a> | >right-handed triad. Little help?> |> | You realize, I hope, that we don't know what textbook you have, so> | we don't know exactly what proof you're referring to. Can you give> | us more details of what the proof is, and what you don't understand> | about it?> | === Subject: Re: cross product>The textbook I'm talking about is Calculus of Several Variables by Robert A.>Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross>product UxV is the vector determined by the following (geometric) conditions:>1) perpendicular to U and to V>2) length is |U|*|V|*|sin(x)|>3) U,V,(UxV) form a right-handed triad>The first two parts are amenable to simple algebraic verification. Those parts>are clear. But when he gets to part (3) he lapses into hand-waving In this case that should be OK as long as it's the right hand he's waving... Sorry about that.> and loses me>completely.For the case U=i, V=j, UxV=k you can see it's right-handed. Now imaginestarting there and moving U and V into some other positions by a continuous motion, being careful to avoid U and V ever being parallel.As you do so, UxV also moves continuously (because its componentsare continuous functions of the components of U and V). If yourright hand's thumb points in the direction of U and index finger in the direction of V, UxV will either be in the direction of yourpalm or the back of your hand. But just as UxV moves continuously, so does your hand (the front and back won't suddenly switch). So since it starts out with the palm in the direction of UxV, it will end thatway too.Hope this helps. === Subject: Re: cross product> I know what it means (geometrically), but I don't know how to tell> (algebraically) whether it's true or not.> Take the three vectors in order, make them the rows of a determinant.> If the determinant is positive, it is a right-handed system. If> negative, left-handed. If zero, they three vectors are co-planar, soright or left is not defined.The nice thing about this approach is that it generalizes immediatelyto n dimensions (unless I am forgetting something obvious).fourierr at fastermail dot com-- Posted via http://web2news.com the faster web2news on the web === Subject: Re: cross product> I know what it means (geometrically), but I don't know how to tell> (algebraically) whether it's true or not.> Take the three vectors in order, make them the rows of a determinant.> If the determinant is positive, it is a right-handed system. If> negative, left-handed. If zero, they three vectors are co-planar, soright or left is not defined.> The nice thing about this approach is that it generalizes immediately> to n dimensions (unless I am forgetting something obvious).> fourierr at fastermail dot com> Posted via http://web2news.com the faster web2news on the webIirc, only in dimensions 1, 3 and 7 do cross products exist.GC === Subject: Re: cross product> , Elaine> I know what it means (geometrically), but I don't know how to tell> (algebraically) whether it's true or not.> Take the three vectors in order, make them the rows of a> determinant.> If the determinant is positive, it is a right-handed system. If> negative, left-handed. If zero, they three vectors are> co-planar, soright or left is not defined.> The nice thing about this approach is that it generalizes> immediately> to n dimensions (unless I am forgetting something obvious).> fourierr at fastermail dot com>Iirc, only in dimensions 1, 3 and 7 do cross products exist.>GCNiel doesn't explicitly mention cross-product. HoweverSpivak does and says that the definition of cross-product is sometimes given only on R^3. I suspect that we are talking about two different things. The following is fromSpivak's _Calculus on Manifolds_ : Given v_1, v_2,..., v_(n-1)vectors in R^n we may define a function from R^n to R byf(w) = det(v_1,...,v_(n-1),w). This is multi-linear, alternatingfunction on R^n and therefore is given by for some z inR^n. In this case define the n-dimensional cross-product of n-1 vectors in R^n by v_1 x v_2 x ... x v_(n-1) = z. Notice thatthe cross-product of n-1 vectors is defined rather than the cross-product of two.Spivak uses Niel's method rather than the cross-product(even after the trouble to define it). The method is used to define an orientation for any basis of R^n. In R^3 bases which share the orientation of {(1,0,0),(0,1,0),(0,0,1)} are saidto satisfy the right-hand rule.fourierr at fastermail dot com === Subject: Re: cross product>The textbook I'm talking about is Calculus of Several Variables by Robert A.>Adams. In Section 10.3, Theorem 2 states that the (algebraically defined) cross>product UxV is the vector determined by the following (geometric) conditions:>1) perpendicular to U and to V>2) length is |U|*|V|*|sin(x)|>3) U,V,(UxV) form a right-handed triadJust out of interest, which edition of Calculus of Several Variables isthat? I have both the first and fourth editions, and neither has a Section 10.3; the fourth edition of Calculus: A Complete Course byAdams does talk about cross products in Section 10.3, but it has conditions 1), 2) and 3) as the definition of cross product and Theorem 2 as giving the algebraic formula. === Subject: Re: cross product> , Elaine>I know what it means (geometrically), but I don't know>how to tell>(algebraically) whether it's true or not.> Take the three vectors in order, make them the rows of a> determinant.> If the determinant is positive, it is a right-handed system. If> negative, left-handed. If zero, they three vectors are> co-planar, so>right or left is not defined.> The nice thing about this approach is that it generalizes> immediately> to n dimensions (unless I am forgetting something obvious).> fourierr at fastermail dot com>Iirc, only in dimensions 1, 3 and 7 do cross products exist.>GC> Niel doesn't explicitly mention cross-product. However> Spivak does and says that the definition of cross-product is> sometimes given only on R^3. I suspect that we are> talking about two different things. The following is from> Spivak's _Calculus on Manifolds_ : Given v_1, v_2,..., v_(n-1)> vectors in R^n we may define a function from R^n to R by> f(w) = det(v_1,...,v_(n-1),w). This is multi-linear, alternatingTypo: That should read linear rather than multi-linear.> function on R^n and therefore is given by for some z in> R^n. In this case define the n-dimensional cross-product of n-1> vectors in R^n by v_1 x v_2 x ... x v_(n-1) = z. Notice that> the cross-product of n-1 vectors is defined rather than the> cross-product of two.> Spivak uses Niel's method rather than the cross-product(even> after the trouble to define it). The method is used to define> an orientation for any basis of R^n. In R^3 bases whichAnother typo: That should be ordered basis rather thanjust basis.> share the orientation of {(1,0,0),(0,1,0),(0,0,1)} are said> to satisfy the right-hand rule.To clarify and match up with what Niel said: Given 3 vectorsput them in order into a matrix and take the determinant. Theones with positive determinant all share one orientation and theones with negative determinant share the opposite orientation. My only point was that this approach immediately generalizes ton dimensions-simply replace 3 by n. The ordered basis consistingof {(1,0,...,0),...,(0,...,0,1)} is said to have the usual orientation. The other orientation is the negative or opposite of the usual orientation.> fourierr at fastermail dot comfourierr at fastermail dot com-- Posted via http://web2news.com the faster web2news on the web === Subject: Re: cross productIt helps a little. In fact I thought about that proof, but it's too informal formy taste. I also thought about showing that a matrix M that rotates thingsaround in 3-space preserves cross-products (ie: M(VxW)=MVxMW...I'm surprisednobody suggested this). That's much more formal, but it seems to have moremoving parts than what's warranted. What I'd really like is some way oftranslating the condition right-handed triad into an algebraic condition onthe coordinates of the vectors in question.| >The textbook I'm talking about is Calculus of Several Variables by RobertA.| >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined)cross| >product UxV is the vector determined by the following (geometric) conditions:|| >1) perpendicular to U and to V| >2) length is |U|*|V|*|sin(x)|| >3) U,V,(UxV) form a right-handed triad|| >The first two parts are amenable to simple algebraic verification. Thoseparts| >are clear. But when he gets to part (3) he lapses into hand-waving|| In this case that should be OK as long as it's the right hand he's| waving... Sorry about that.|| > and loses me| >completely.|| For the case U=i, V=j, UxV=k you can see it's right-handed. Now imagine| starting there and moving U and V into some other positions by a| continuous motion, being careful to avoid U and V ever being parallel.| As you do so, UxV also moves continuously (because its components| are continuous functions of the components of U and V). If your| right hand's thumb points in the direction of U and index finger in| the direction of V, UxV will either be in the direction of your| palm or the back of your hand. But just as UxV moves continuously,| so does your hand (the front and back won't suddenly switch). So since| it starts out with the palm in the direction of UxV, it will end that| way too.|| Hope this helps.|| | Department of Mathematics http://www.math.ubc.ca/~israel| University of British Columbia| Vancouver, BC, Canada V6T 1Z2| === Subject: Re: cross product What I'd really like is some way of> translating the condition right-handed triad into an algebraic conditionon> the coordinates of the vectors in question.Maybe what you're looking for is this: vectors {a,b,c}, {d,e,f},and {g,h,i} form a right-handed triad if the determinant of[a b c][d e f][g h i]is positive. Well, this still has 9 parts which could unexpectedly move. Ihope that's not too many.I once had an interesting discussion with the science fiction writerVernor Vinge, who is (I opine) the most brilliant of all sciencefiction writers. I asserted that if we could exchange streams ofzeros and ones with an intelligent culture on a distant star, then wecould eventually communicate everything about ourselves excepthandedness. Vernor disagreed, pointing out that the electromaticwaves that carry the zeros and ones have a built-in handedness. Itis with some trepidation that I mention anything philosophical inthis den of crackpots. === Subject: Re: cross productWhat I'm looking for is some kind of proof that the cross product as you defineit has, in every case, the same geometric relationship to the vectors whosecross product it is.| Elaine says...| >| >The textbook I'm talking about is Calculus of Several Variables by RobertA.| >Adams. In Section 10.3, Theorem 2 states that the (algebraically defined)cross| >product UxV is the vector determined by the following (geometric) conditions:| >| >1) perpendicular to U and to V| >2) length is |U|*|V|*|sin(x)|| >3) U,V,(UxV) form a right-handed triad| >| >The first two parts are amenable to simple algebraic verification. Thoseparts| >are clear. But when he gets to part (3) he lapses into hand-waving and losesme| >completely.|| Here's one definition of a right-handed triad: vectors A, B, and C| form a right-handed triad if the triple product A . (B x C) is positive,| where . is the scalar product.|| So compute U . (V x (U x V)). It's pretty easy to see that it is positive| if you use the permutation rule|| A . (B x C) = C . (A x B)|| --| Daryl McCullough| === Subject: Re: cross product> What I'm looking for is some kind of proof that the cross product as youdefine> it has, in every case, the same geometric relationship to the vectorswhose> cross product it is.I don't quite follow same geometric relationship. Maybe we coulddo it this way: you state a theorem, and if it's true, somebody inthe ng will prove it for you. And for that matter, if it's false,someone will undoubtedly provide a counterexample.The only way I can make sense of same geometric relationship is thetheorem that the cross product is invariant under orthogonaltransformations of R^3, but I seem to recall that you weren't happywith that. === Subject: Re: cross productWill says...>I once had an interesting discussion with the science fiction writer>Vernor Vinge, who is (I opine) the most brilliant of all science>fiction writers. I asserted that if we could exchange streams of>zeros and ones with an intelligent culture on a distant star, then we>could eventually communicate everything about ourselves except>handedness. Vernor disagreed, pointing out that the electromatic>waves that carry the zeros and ones have a built-in handedness.No, they really don't. He might have been thinking about the right-handedrule for electromagnetic waves, which says that for an electromagneticwave travelling in direction K, the following form a right-handed triple(or maybe it's left-handed, I don't remember which) E, B, Kwhere E is the electric field, and B is the magnetic field.However, the direction of the B field is not actually measurable;the most you can measure about the B field is: (1) the plane thatis perpendicular to B and (2) a sense of circulation about thatplane. That sense of circulation is the direction (clockwise ordeflected if it is travelling in the plane perpendicular to B.by the magnetic field at all.)Using the right-hand rule, you can compute a direction forB from the plane and the sense of circulation, but withoutsuch a rule, the direction of B is not specified. So you can'tuse E,B,K to compute handedness unless you already have abuilt-in handedness to compute the right-hand rule.-- === Subject: Re: cross product> Will says...>I once had an interesting discussion with the science fiction writer>Vernor Vinge, who is (I opine) the most brilliant of all science>fiction writers. I asserted that if we could exchange streams of>zeros and ones with an intelligent culture on a distant star, then we>could eventually communicate everything about ourselves except>handedness. Vernor disagreed, pointing out that the electromatic>waves that carry the zeros and ones have a built-in handedness.> No, they really don't. He might have been thinking about the right-handed> rule for electromagnetic waves, which says that for an electromagnetic> wave travelling in direction K, the following form a right-handed triple> (or maybe it's left-handed, I don't remember which)> E, B, K> where E is the electric field, and B is the magnetic field.> However, the direction of the B field is not actually measurable;But you _can_ communicate handedness with electromagnetic waves bymanipulating their polarization: slowly rotate the direction of the E field. === Subject: Re: cross product>Is it a proof ? I mean, isn't the right-hand rule just theorientation-definition of the cross product. Sounds like one of>those stupid prove axiom by using the axiom tasks, e.g Prove that>1+2=2+1.But that has a proof (assuming associativity of addition):1+2=1+(1+1) ... definition of 2 =(1+1)+1 ... associativity =2+1 ... definition of 2 === Subject: Re: cross product orientation-definition of the cross product. Sounds like one of>those stupid prove axiom by using the axiom tasks, e.g Prove that>1+2=2+1.> But that has a proof (assuming associativity of addition):> 1+2=1+(1+1) ... definition of 2> =(1+1)+1 ... associativity> =2+1 ... definition of 2Are these types of proofs really considered as _proofs_ ?I mean, e.g.Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom foraddition)By axiom (a+b)+c=a+(b+c) , QED ?I would like to state the task as Show that (a+b)+c=a+(b+c). === Subject: Re: cross product>Is it a proof ? I mean, isn't the right-hand rule just theorientation-definition of the cross product. Sounds like one of>those stupid prove axiom by using the axiom tasks, e.g Prove that>1+2=2+1.> But that has a proof (assuming associativity of addition):> 1+2=1+(1+1) ... definition of 2> =(1+1)+1 ... associativity> =2+1 ... definition of 2>Are these types of proofs really considered as _proofs_ ?>I mean, e.g.>Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for>addition)>By axiom (a+b)+c=a+(b+c) , QED ?>I would like to state the task as Show that (a+b)+c=a+(b+c).Where in the above proof do you see commutativity used to provecommutativity? Where do you see the result asserted as an axiom?Two axioms are invoked. One is that 2 = 1+1, the other is thataddition is associative. Neither one is the thing being proven. - Randy === Subject: Re: cross product <5o9gfv8hn95ma2h4v7dc9n19elfmnon13k@4ax.comIs it a proof ? I mean, isn't the right-hand rule just theorientation-definition of the cross product. Sounds like one of>those stupid prove axiom by using the axiom tasks, e.g Prove that>1+2=2+1.> But that has a proof (assuming associativity of addition):> 1+2=1+(1+1) ... definition of 2> =(1+1)+1 ... associativity> =2+1 ... definition of 2>Are these types of proofs really considered as _proofs_ ?>I mean, e.g.>Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for>addition)>By axiom (a+b)+c=a+(b+c) , QED ?>I would like to state the task as Show that (a+b)+c=a+(b+c).> Where in the above proof do you see commutativity used to prove> commutativity? Where do you see the result asserted as an axiom?> Two axioms are invoked. One is that 2 = 1+1, the other is that> addition is associative. Neither one is the thing being proven.My point was (originally) that to prove 1+2=2+1 you simply need to stateby the commutative axiom of addition x+y=y+x => 1+2=2+1. In some casesif seems so stupid to prove a property/relation for a formula/theoremthat is build up, constructed and based on that very property/relationthat should be proven.Anyway, thanks for the more rigorous proof of why 1+2=2+1 :), but now Istart to wonder why commutativity of addition is an axiom ? (Eventhoughit's a very obvious relation). I thought the point of having axioms isthat in order to prove them they prove themself, not that you setup anew set of axioms based on other, older (more fundamental) axioms. What Iremember, from class, was that the fundamental axioms could differquite alot from book to book.TV === Subject: Re: cross product> ,> I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help?> --> Peace,> EJ> Assuming that you have a list of three linearly independent > vectors in a real 3-space, do you know how to tell whether > they form a right handed triad or a left handed triad? === =========================My opinion is the following: it's better to teach, in this order,following notions : dot (scalar) product , cross product and thenvector-product.Consider three (free) vectors A=(a_1,a_2,a_3), B=(b_1,b_2,b_3) , C=(c_1,c2,c_3)=c_1*i+c_2*j+c_3*k where i=(1,0,0) ,j=(0,1,0) ,k=(0,0,1).1) The dot (scalar) product A*B=(A,B):=a_1*b_1+a_2*b_2+a_3*b_3:= ||A||.||B||cos(phi) where phi is the angle ,from [0,pi], between vectors A and B.2) The cross-product (A,B,C) is by definition |a_1 a_2 a_3| (A,B,C) := |b_1 b_2 b_3| |c_1 c_2 c_3| DEFINITION : the vectors A,B,C form a right handed triad, if andonly if (A,B,C) > 0 .3) The (vector) -product A x B , A=/=0,B=/=0, is defined to be theuniquevector D with -the norm ||D||=||A||.||B||sin(phi) -the direction of D is such that A*D=0 , B*D=0, that isD is perpendicular on the plane determined by suports of Aand B . -the sense of D is such that (A,B,D) >= 0 .Moreover, when A=0 or B=0 the one define A x B = 0 . |a_2 a_3| |a_3 a_1| |a_1 a_2| A x B= | |*i + | |*j + | |*k = formaly= |b_2 b_3| |b_3 b_1| |b_1 b_2| |i j k | = |a_1 a_2 a_3| |b_1 b_2 b_3|THEOREM 1 . (A,B,C)=A*(B x C) . Proof. According to above formulas it's very easy. THEOREM 2 . (A,B,C)=(B,C,A)=(C,A,B)=-(A,C,B)=-(B,A,C)=-(C,B,A) .Proof. Nothing to prove taking into account properties ofdeterminants.THEOREM 3. Suppose that the non-zero vectors A,B does not have thesame direction ( B=/=k*A , k in R).Then A, B, AxB (in this order) form a right handed triad.Proof. Use above formulas,or theorem 2, and you find that(A,B,A xB)=(AxB,A,B)=(AxB)*(AxB)=||AxB||^2 >0 . THEOREM 4. A,B,C with (A,B,C)=/=0 is a vectorial basis, namely anyvector X may be written as X=a*A+b*B+c*C , a,b,c in R .More precisely a=(X,B,C)/(A,B,C) , b=(A,X,C)/(A,B,C) , c=(A,B,X)/(A,B,C) .It's possible to use also the terminology:Definition. A triplet A,B,C is a positive basis iff (A,B,C) > 0. A,B,C form a negative basis when (A,B,C) < 0 .Alex === ==================Subject: Re: cross product>Is it a proof ? I mean, isn't the right-hand rule just theorientation-definition of the cross product. Sounds like one of>those stupid prove axiom by using the axiom tasks, e.g Prove that>1+2=2+1.> But that has a proof (assuming associativity of addition):> 1+2=1+(1+1) ... definition of 2> =(1+1)+1 ... associativity> =2+1 ... definition of 2> Are these types of proofs really considered as _proofs_ ?> I mean, e.g.> Task : Prove that (a+b)+c=a+(b+c). (which is the associativity axiom for> addition)> By axiom (a+b)+c=a+(b+c) , QED ?> I would like to state the task as Show that (a+b)+c=a+(b+c).If you go back to foundations, it's possible to formally construct a systemwhich the 'axioms' of a complete ordered field._Foundations of Mathematics_ by Ian Stewart and David Tall (OUP) goes intothis in a fairly accessible manner.--MHW === Subject: Re: cross product> If you go back to foundations, it's possible to formally construct asystem> which the 'axioms' of a complete ordered field.That should read which *satisfies* the axioms of a complete ordered field.> _Foundations of Mathematics_ by Ian Stewart and David Tall (OUP) goes into> this in a fairly accessible manner.> MHW === Subject: Re: cross productIf you take the cross product of two linearly independent vectors, you get a third vector which is perpendicular to the other two vectors.The right-hand rule, if you take the thumb (pointing up) and pointerfinger (pointing forward) as the first two vectors, your middle finger (pointing to the left) will be the direction of the vector formed from the cross product.- Stephenhttp://pages.prodigy.net/stephenmorais> I don't understand the proof in my textbook that A, B and AxB form a> right-handed triad. Little help? === Subject: Re: cross product> If you take the cross product of two linearly independent vectors, you > get a third vector which is perpendicular to the other two vectors.> The right-hand rule, if you take the thumb (pointing up) and > pointerfinger (pointing forward) as the first two vectors, your middle > finger (pointing to the left) will be the direction of the vector formed > from the cross product.Sounds obscene. === Subject: Re: cross product> If you take the cross product of two linearly independent vectors, you> get a third vector which is perpendicular to the other two vectors.> The right-hand rule, if you take the thumb (pointing up) and> pointerfinger (pointing forward) as the first two vectors, your middle> finger (pointing to the left) will be the direction of the vector formed> from the cross product.> Sounds obscene.No worse than self-injective modules. === Subject: Differential equation solution?How do I solve the differential equation:a (2 + 3y + y^2) = x y ( y' + y^2 + 3y + 2)where y of course means y(x), and a is a constant, if it has any solutions?Jeremy === Subject: Re: Differential equation solution?Jeremy scribe:> How do I solve the differential equation:> a (2 + 3y + y^2) = x y ( y' + y^2 + 3y + 2)> where y of course means y(x), and a is a constant, if it has any solutions?(...)Some experiments show, that there exist solutions. Aclosed solution I have not got, but it is also aninteresting task to look for the asymptotic behaviour ofsolutions. First some special solutions should be mentioned:y =(ident.)=-1 and -2. Theses are the zero points of2+3y+y^2=0. Solve Your equation explicit toy=(a-x*y)*(2+3y+y^2)/(x*y) and show, thatlim((a-x*y)/(x*y)=-1 for x-->00. With this substitution youwill get an equation with separated variables, which easyshould be solved. Hope, there is no mistake, because I am ina hurry.Another chance seems to be the development of[(1/2)*y^2(x^2/2)] and to compare it with x*y*y.Alfred === Subject: Easy proof of mathematician liesI think I've figured out a way to show basically all of you, includingpeople who think they don't know any math that mathematicians havebeen lying about my work. It's so trivial you *should* wonder whythey thought they could get away with it.Here goes.My paper Advanced Polynomial Factorization depends on considering afactor 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 + cwhere 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, withsomething 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 tosay that for the factors g, in general, you have an element I call r,which changes as the independent variable changes, and you haveanother element I call c, which does not.For my example up above it's easy, as with g_1 = x + 1, x varies as xvaries, while 1 does not.Now that's enough that the proof in the paper is straightforward, butposters have argued with me anyway, with some trying to argue over thedefinition of polynomial, amazingly enough.However, consider that the g's have an important feature, which isthat when x=0, I have g_1...g_k = P(0).For instance, with my simple example, with P(x) = x^2 + 2x + 1, withx=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 mustmultiply to give the constant term.So then, maybe you still want to believe the mathematicians andquestion 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'veconvinced you to doubt algebra itself, as then you must believe thateverything 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 considerthat 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+mfwhich seems to be just complicated enough to give mathematicians roomto lie.For instance, you may be saying, HEY, what's with the 'm' when youhad 'x' before??!!!Well, there's no rule that says that you have to use the letter x asthe variable label for a polynomial. Also, there are historicalreasons for my usage as it goes back to my work with FLT where x, yand z are used with x^p + y^p = z^p.Finally, the weirder thing is that one poster in particular got a lotof mileage out of questioning my finding the constant term with anexpression like the above by using m=0, as that gives me P(0) = 3xy^2 + y^3and he got a lot of mileage for YEARS (before I had discovered proof Iwant to add and after) by emphasizing that two of the ROOTS of such anexpression considered as a polynomial with respect to x are notdefined at that point.Well that's easy enough to see as the original expression is (v^3+1)x^3 - 3vxy^2 + y^3which if you *wish* to see it as a polynomial with respect to x, is ofdegree 3, but when v=-1, it's of degree 1, so if you solve for theroots, you'll get funky stuff.Now when I was finding the proof of FLT...remember the process tooksome years...at times I'd talk of polynomials with respect to otherthan m, but I refined 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 knowenough mathematics to follow, but we're talking about actualmathematicians here.What's more rational?I say it's more rational to suppose that they *did* figure out that itworked as described, but also noticed that as long as they disagreed,no one seemed to call them on making false statements, except me, andthey knew my credibility wasn't so great.For most of you, there's probably the belief that there's some funkyhigher math involved that your pitiful brain can't follow or youdon't know about, as you may suppose that mathematicians eitherwouldn't lie, or they wouldn't lie in such a dumb way where I couldcatch them so easily.But consider what's in the bce:1. I discovered a proof of Fermat's Last Theorem that's moreavailable to people in general than most of what mathematicians havebeen producing lately.2. Worse I did so having said I'd find it years ago, and having spentyears looking for it posting a lot of my ideas, and getting in insultbattles with posters, quite a few who happened to be mathematicians.3. Then to add insult to injury, I keep questioning mathematicians interms of their ethics, and maybe *extremely importantly* I doubt thatWiles found a proof of Fermat's Last Theorem.And those are just highlights as there's more but I think that kind ofgives my point.For mathematicians the situation could be considered one of the worstpossible 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 israther sad. I'd think at least some of you valued your educations.Others of you may figure it doesn't matter, maybe because Westerncivilization seems to be based on lying anyway, and maybe you figure Ishould just grow up, accept that everybody lies and move on. And Idon't have to talk about Enron or pedophile Catholic priests or thingsin 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 somefreaking stupid math?Good point.Mathematicians *are* a part of society after all. Why should theytell the truth now? It'd be like Bush owning up. They can just sittight, and be quiet, like so many American citizens or they can outand 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 mathematicianspotentially being prosecuted. Liars don't just stop because the gigis up, as then, they wouldn't necessarily be liars, then eh?It'd be against their *true* natures. === Subject: Re: Easy proof of mathematician liesX-Program-URL: http://www.beosdevelopers.org/brunsona/pineapplenews.html> 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. Mr. Harris, Thank you for not posting to alt.writing. -X === Subject: Re: Easy proof of mathematician liesI'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 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.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.> 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.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.Note: at this point I'm fairly confused, but I'll read on.> 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).> 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?> 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?> 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> definition 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.> 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)?> 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.> 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.> Well consider that substituting gives me> g_1...g_k = (r_1 + c_1)...(r_k + c_k)= P(x), which givesThis seems to suggest that each g_i gets it's own r_i, c_i.> r_1...r_k +...+c_1...c_k = P(x), which isThis step is NOT clear. What all goes between the product of r's and product of c's?> 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 defining 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 definition 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.> So why would mathematicians argue against such a simple result?Because it's not clearly stated.> 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 defined. 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.> 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.> 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 defined 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?> For instance, you may be saying, HEY, what's with the 'm' when you> had 'x' before??!!!Couldn't care less what your independent variable is. I would like to know how x,y,f depend on m.> 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.FLT specifically defines x,y,z, and p.> Finally, the weirder thing is that one poster in particular got a lot> of mileage out of questioning my finding 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> defined 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.Based on this statement, what do you mean by a polynomial? I suspect you are using a non-traditional meaning.[anti-mathematician discussion deleted]> But consider what's in the bce:> 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.I'd like to read that proof. Where is it located?> 2. Worse I did so having said I'd find 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.Then read his proof and find a flaw in it. I believe it's publicly available.[anti-mathematician discussion deleted]Notes: I am a mathematician. I have only recently started reading this board, so I am not familiar with your previous difficulties with mathematicians. What I can see here is that your mathematics is unclear. It looks like what you're doing should be easy, except I'm not sure what it is. Also, what is your background? Having studied logic, abstract algebra, etc... I find some of your accusations difficult to believe because I am aware of how far mathematicians have gone to make sure what is being taught actually works. Most of that involves precisely defined terms and precise use of notation. This level of precision takes a long time to learn and is part of why peer reviews are common. It's easy to overlook a detail. If we ALL overlooked one, all we ask is that you point it out so that we can fix the mistake.-- Will Twentymanemail: wtwentyman at copper dot net === Subject: Re: Easy proof of mathematician liesjstevh@msn.com ():> 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 === Subject: Re: Easy proof of mathematician liesI'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 first 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,fishing-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 find one of the first ion-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> definition 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 'm' when you> had 'x' 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 finding 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> defined 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 !?! === Subject: Re: Easy proof of mathematician liesGo home, little man...> 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.> 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> definition 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 'm' when you> had 'x' 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 finding 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> defined 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 finding 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 refined 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* figure 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 funkyhigher 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 bce:> 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 find 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 figure it doesn't matter, maybe because Western> civilization seems to be based on lying anyway, and maybe you figure 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. === Subject: Re: Easy proof of mathematician lies[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 whosecoefficients 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 definingQ(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 'm' when you> had 'x' before??!!!No, that's fine. 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 finding 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> defined 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 specifically,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 petic, which is a virtue, you shouldremember the last case as well.-- David Collier