mm-8 === No one is assaulting algebra. They are denying that you have constructed a valid proof. And they are correct. Youare wrong. What is telling here is your monumental incompetence.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Correct. She was DEFINING the values of the a_i(1). The discussion had dealt with what the a_i(0) were, but she simply showed that when m changes from 0 to 1, it is possible for the a_i(m) to change from having the properties you wanted to ones that you did NOT want. Nothing obtuse. Now, do you understand what a function is? Do you perhaps see the importance of de?ing the characteristics of these functions at all values of m if you want them to have certain properties?When you set x=2,f=5,u=1, then you change the nature of the problem. If you would prefer to think of the a_i as varying over m,x,f,u, we can do that but then Nora is still able to choose the values she did. If you don't want people de?ing the a_i's in inconvenient ways, you will have to specify what is legal. Don't be surprised if the speci?ations you provide limit you to cases that don't give the broad result you want to achieve.I have no problem with this.I do. That's why you keeping seeing my posts with a_1(m).It's not for the others, it's to clearly de?e what are the independent variables for the functions a_i. It's so that you know what you're talking about and can clearly communicate it to us.It helps a great deal. How does it hurt?Unlikely. It may be busier but it is now precise. You have clearly indicated that the a_i depend only on m, and that when you change x or y, the a_i will not change. Of course, y and v are both de?ed in terms of f, so there's probably a little more going on. Your introductions of v and y is hiding a relationship.Ones that you can either clearly defend against or concede to as everyone will be discussing the same things. Right now there are no de?itions so people are interpreting the same symbols differently.You are the one that is being dragged into de?ing your symbols.If the a's vary with m, then there must be a way to de?e them for each choice of m. You have NOT done that. You have given what looks like a de?ition but would lead to the a's being constants. Perhaps if you go back and POST your paper HERE we can see if you've changed things enough to address the issues that you feel are valid. I'm working with a copy that is approximately a month old.Ok, I just noticed the below... That connection would have been helpful. Why do you berate me for not connecting something below with that above?Here's the funny thing: Nora did it. It worked. You just don't like it. If she failed to do it correctly, point out the *speci?* ?at family of values? It might help if you informed people about that in advance.It's easier than trying to keep track of 20 short threads. I can mark a thread for future reference. It's harder to mark a lot of threads, and harder to follow the logic between the threads. A concession in one thread is not apparent in another thread.-- Will Twentyman === [deleted]Correction 1: 2 a_1(x) + a_2(x) = 4x.Correction 2: only if a_1(x) = x, a_2(x) = 2x.If you want to allow anything else, your assertion is incorrect. And you consistently want to allow the a_i(x) to be non-polynomials.-- Will Twentyman === Just a statement of fact, and anyway backed up immediatelyby a speci? justi?ation. As for an appeal to the gallery:that seems like an odd complaint coming from you, with yourfrequent indictments of mathematicians as corrupt liars whowant to deceive the innocent naive silent readers of sci.math,alt.math, and alt.math.undergrad. You oversimpli?d things and paid the price. What you have here is that when m = 0, a_1 = a_2 = 0,and therefore a_1 and a_2 are divisible by 5, and 2*a_1 / 5 and 2*a_2 / 5 are both algebraic integers (because both are 0, and conclude that a_1 and a_2 must be divisible by 5 when m takes on nonzero values also. You do notprove this. You do not cite any mathematical principlethat justi?s it. It is clearly just your hunch. Itlooks like a nice pattern, so it must be true. Of course a_1 and a_2 are actually dependent on m. Whenm = 0, they are both 0 and both divisible by 5, buthow does that tell you anything about either of them You have not provided a justi?ation. You simply make thestatement. And you deleted out the rest of my post which dealt withthis. Why? Further, you have not found a valid objection to my proof that your central claim related to this is incorrect. Nor have youresponded at all to Dale Hall's separate proof. To show that Dale is wrong, all you would have to do wouldbe carry out a simple computation. You don't have to understandanything about Galois theory or Gauss' lemma or anythingelse from the algebraic theory of numbers. But you haven'tdone anything at all. Why? Since you deleted my proof that your central claim iswrong, I am appending it again below. Feel free to point outany errors in it, or post a question if there are partsof it that you don't understand. === === === = JSH CLAIM: It is possible to ?d a 3rd degree polynomial with integercoef?ients, monic and irreducible over the rationals, such thatif a1, a2, and a3 are the roots, then at least one of a1, a2, ora3 is coprime (in the algebraic integers) to a prime factor ofthe constant term of the polynomial.DISPROOF OF CLAIM: Let Q(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. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by de?ition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory, there exists an automorphism F12 of the ?ld of algebraic numbers with the following properties *** : 1. F12 leaves the sub?ld of rational numbers ?ed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + 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(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts your claim which was quoted above. Again, please feel free to point out any errors in the proof I just gave. *** Reference on automorphisms: http://www.math.niu.edu/~beachy/aaol/galois.html See especially Proposition 8.6.2 on that page. Or see the excellent textbook, Abstract Algebra, by John Beachy and == Oversimpli?d examples are NOT a substitute for proofs. I would think you would have learned that by now, if nothingelse. Nora B. === sorry for violating my promise, today;tomorrow, it will hoepfully remain in full force. anyway, I had to say that your statement is peculiar, becausean appeal to the gallery can be countered by anyonewho happens to be *in* the gallery (*I* don't know that e.g.);the important part, apparently, was that*you* should perhaps know it, given your extensive 10-year missionto prove la premiere theoreme de Fermat.That everyone else knows is an appeal to the gallery which is alogical fallacy.The a's are de?ed by (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y),where v=-1+mf^2, and I guessed that with so many symbols people wereconfused, so I stuck in some values to lessen the symbol load.--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 === Previously I answered Nora Baron's claim that I was acting on a hunchas I refuted that claim mathematically, and then I deleted out therest as I wanted to shorten the length of posts.This time I'm deleting down to just past her hunch assertion. Nora Baron's comments follow.I'm leaving that in as part of the setup without saying much morehere, as I handle everything below. What's fascinating is how far shegoes in trying to push against algebra, as I'll show below.Well either you're lying or you think you've refuted algebra, as Ishow below.I'd think that you'd prefer that people think you're lying.Nora Baron's comments follow.The factorization exists for all x so why would it go away for aparticular x? Well you're ?hting algebra. Guess what? The algebra wins.See below... It can only be false if you wish to refute algebra Nora Baron. Seebelow... Let me ask you a question, are you claiming that given P(x) = (x+1)(x+2)that if I stick in actual values for x, the factorization is gone?Yes, if I have P(2)=12, it is true that you just see a number, butnotice that P(2) = 3(4).The expression I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)where v=-1+mf^2, and y=uf, and now you are arguing that thefactorization goes away simply because I set x=2?Um, that's not algebra Nora Baron; that's more voodoo math.What I've done is put in actual numbers for x, f, and u, as I set x=2,f=5, and u=1, which gave me a polynomial. But the factorization*still* exists, and your wish that it doesn't--no matter howdesperate--won't change that reality.Why?So you think that factorizations go away if you stick in values?It doesn't seem that I can oversimplify for you Nora Baron.And Nora Baron rattles on a bit more. Where from before her primaryposition is that given a factorization I lose the factorization byactually putting in a value for x, as I used x=2. The key expressionis (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)where v=-1+mf^2, and y=uf, and I put in actual numbers for severalvalues as a simpli?ation to remove room for bogus objections likeNora Baron's here. Yet Nora Baron in one reply simply put in somepicked values for the a's for m=1 in what she thought of as arefutation, and now seems to think that my saying x=2 takes away thefactorization above. That's hardly even voodoo mathematics.I think it's more simply: nonsensical.James Harris === Or you don't understand.Please note: you are factoring 25(5000m^3 - 600m^2 - 126m + 11)Where is x? Answer: nowhere. Now stop talking about it, or stop talking about this polynomial.[deleted]Because you got rid of x. If you see an x up there, get glasses.Notice: P(2)=2(6)=1(12)The uniqueness of the factorization into 2 factors is gone.Does this look anything like the polynomial above? Does it have the same number of variables?You substituted for x but want to keep x. That's voodoo alright.The factorization still exists, but not uniquely. Perhaps I've been wrong all along and you meant a_i(m,x,f,u). If this is the case, what she did *still* works. She de?ed a_i(1,2,5,1). She just didn't do it the way you want.Because otherwise your paper is wrong.See above with P(2)=12.At this point, how are you dealing with her proof? I would think that of all people, you would know better than to delete material and then respond to it. You were very upset when I did that by accident, yet you make sure everyone knows you deleted it.This does not refute her argument. It does not even deal with it. There is something you are welcome to own, however. voodoo mathematics. I'm willing to let you have it. It's all yours. Whatever it is.-- Will Twentyman === Hmmm...which might lead a *rational* reader to suppose you're tryingto claim that it doesn't work with non-polynomial factors.How about this? p(x) = sqrt(2(x^2+2x+1)) = sqrt((a_1+1)(a_2+2))with a_1 a_2 = 2x^2, 2a_1 + a_2 = 4x?Mathematicians, what a crew. But you know what? They take themselvesSO seriously, even when they're ?hting basic algebra.So why would *mathematicians* ?ht algebra? Because they're actuallylike English professors...really stuck-up English professors...whobelieve they're owed a certain amount of respect.But you see, English professors can pooh-pooh a literary work, somathematicians acting like really stuck-up English professors thinkthey can pooh-pooh valid mathematics, which inevitably leads to themattacking the foundations of mathematics because math is funny thatway.James Harris === ... stuff deleted ... ... stuff deleted ... ... stuff deleted ...You obtained the polynomial 65 x^3 - 12 x + 1by substituting v=4, y=1 into that expression, correct?You claim that this factorization: 65 x^3 - 12 x + 1 = (a1 x + 1)(a2 x + 1)(a3 x + 1)with the a's being algebraic integers, has the a's beingcoprime to 5, correct?What you say is wrong, and repeating it doesn't change thefacts. I've lost count of how many times I've posted thisparticular argument, but I don't mind, since I can easilyI have proposed that these polynomials q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104have the property that, for any root z of the polynomial p(x)= x^3 - 12 x^2 + 65,we have q(z)*r(z) = 5 r(z)*s(z) = z.I've shown how this can be veri?d, by doing the followingmultiplications (courtesy of DOE Macsyma):First, here are the products that I'm making claims about: q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325 r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680Next, a couple of products of p(x) = x^3 - 12 x^2 + 65 withpolynomials of degree 1: (64 x + 128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320 (32 x + 72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680Finally, we compare the results and see this: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x,Note that, for any value xo that makes p(xo) = 0,that same value xo will make q(xo)r(xo) = 5, sor(xo) is a factor of 5.That value of xo also makes r(xo)*s(xo) = xo, sor(xo) is a factor of xo.In short, r(xo) becomes a factor of *both* xo and 5.Since r(x) is a polynomial with integral coef?ients,r(xo) is an algebraic integer whenever xo is.It is similarly simple to demonstrate that, wheneverxo is a root of p(x), then r(xo) is a root of thepolynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5:First, expand the polynomial mpr(r(x)): mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5 = (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2 + 315 (8 x^2 - 4 x - 45) + 5 = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520Next, multiply these two polynomials: p(x) = x^3 - 12 x^2 + 65,and w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808to get this: p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520Notice the equality mpr(r(x)) = p(x)*w(x).That means for every value of x, the polynomial you get by computingr(x), then evaluating mpr(x) at that value, is equal to the product ofp(x) and w(x).If xo is a root of p(x), you have p(xo) = 0, so p(xo)*w(xo) = 0, andtherefore r(xo) is a root of mpr(x).Note that there are three such roots of mpr(x), and (taking the threeroots x1,x2,x3 of p(x)), three values r1 = r(x1) ~ 968.67481 r2 = r(x2) ~ -0.01517 r3 = r(x3) ~ 0.34036correspond to the three real roots of mpr(x). As such, their productmust be -5.Your earlier claim that the a's must be coprime to 5 implies that ther's must be units (since they're common factors of ai with 5, in thering of algebraic integers).If you multiply units, even you must realize that the product is againa unit.Therefore, -5 (and equivalently, 5 itself) must be a unit in the ringof algebraic integers.Do you agree or not?If so, then say so. I'll make it easy; here's a form for youto ?l out and post to your full array of newsgroups: I, James S. Harris, af?m my belief that, in the ring of algebraic integers, the (rational) integer 5 [?e] is a unit. James S. Harris.If not, then show me (hey, don't worry about me, show your public!)where my error is. I've done all the multiplication; you can verifyor refute all this very easily, given the ability to multiplyor expand polynomial expressions.Show how highly you value algebra: Do some.It's easy: ordinary polynomials, ordinary polynomial multiplication,nothing up my sleeves, no salesman will call. You don't even needto solve any equations. Just multiply, combine terms, show me wrong!Prove me wrong. Your mama says you can't. ... stuff deleted ...I haven't bothered with your evaluation at x=2, and don't care aboutwhether you think anyone is doing anything.What I do notice is the level of cowardice it must take to sit thereand claim that people are, what was it, ?hting algebra?You don't even ?ht. The algebra I've presented is simple enoughthat you shouldn't evade it. Well, I know that if I had made anactual mistake (as contrasted to the typos that Wayne Brown pickedup), you would be all over it like ?n a pile of poo, but ifsomeone shows up with the actual goods, you're Brave Sir Robin: Brave Sir Robin ran away. Bravely ran away, away! When danger reared its ugly head, He bravely turned his tail and ?es, brave Sir Robin turned about And gallantly he chickened out. Bravely taking to his feet He beat a very brave retreat, Bravest of the brave, Sir Robin! He is packing it in and packing it up And sneaking away and buggering up And chickening out and pissing off home, Yes, bravely he is throwing in the sponge... You might think I'll give up on this.Dale. === Let's stick with the original problem.P(x)=2(x^2 + 2x + 1)P(x)= (a_1(x) + 1)(a_2(x) + 2)a_1(x)=sqrt(x)a_2(x)=2x sqrt(x) - 2x + 6 sqrt(x) - 8 + 8/(sqrt(x)+1)These are valid factorizations in the algebraic functions.a_1(x) a_2(x) = 2x^2 - 2x sqrt(x) + 6x - 8 sqrt(x) + [8 sqrt(x)]/[sqrt(x)+1]Therefor, a_1(x) a_2(x) =/= 2x^2.Put bluntly: I *am* claiming that non-polynomial factors don't have the properties you want them to have. Simply put, non-polynomial factorizations do not behave like polynomial factorizations.-- Will Twentyman === Some of! People should certainly judge for themselves, unless youconsider that playing to the gallery. Again? My point in showing what happened when you substituted 2 for x was in part that it completely changes the problem.You started with statements about factoring a polynomial.You evaluated the polynomial at x = 2. The evaluation issimply an integer. Your proposed factorization was thefactorization of an ordinary integer. If you wanted us tocontinue regarding it as a polynomial factorization, youshould have left it as a polynomial. This point is admittedly peripheral to the main argument here, that is, statements you have made regarding factorizations of polynomials. However I thought it was worth making anyway,to demonstrate how polynomials and their evaluations are different things. Sorry you didn't get it. Nope. You made a statement about a factorization of an ordinaryinteger. You said it could not be of a certain form. I showeddirectly and unambiguously that your statement was wrong. Youdid NOT say that, after you had evaluated the polynomial and obtained an ordinary integer, that you still wanted the factorizationto be that associated with the polynomial. You were, literally,wrong. You tried to get an oversimpli?d example to prove yourpoint. I gave a perfectly valid counterexample. Now you want togo back and say, No, I meant only polynomial-type factorizations!To which I say, Well, then, why didn't you leave it in the formof a polynomial in the ?st place? It doesn't make it any simplerto replace ?x' with ?2' if you are just going to consider the ?2'as a polynomial variable, rather than a constant. Why did you evaluate to an ordinary integer anyway if you are not going to takeadvantage of it? What did you expect to gain? Again, this is peripheral to the main argument. If you don't get my point here and want to think of this as a minor distraction,it's ?e with me, as long as you address the *real* problem asdiscussed below. No, it's just the opposite. If you stick in actual valuesfor x, you may get factorizations which are not consistentwith the polynomial factorization. For example, youlet x = 2. P(x) = 12. The factorization that is consistentwith the polynomial factorization that you just gave is 12 = 3 * 4 = (2 + 1)*(2 + 2).There are, however, other factorizations: for example 2 * 6or 1 * 12. These are NOT consistent with the polynomial factorization. See above. Not at all. You don't lose any of the polynomial factorizationswhen you evaluate. You gain some *new* ones. That is exactly whathappened in what I posted. Nonsense. It's just boring arithmetic. You got it backwards. You don't lose any of the polynomialfactorizations. You just gain some other ones. That is whyyour oversimpli?d example didn't make any sense. It mostcertainly did not prove your point. It went exactly the otherdirection: it showed that factorizations existed which wereexactly of the form you said could not occur. You would have been better off leaving it as a polynomial in x. A truly strange question. I have pointed out an explicit place in your argument where youhave made a mistake. I have pointed out exactly what went wrong with your reasoning. You still apparently do not understand it. That is unfortunate. But in addition to identifying your error [which basically boils down to assumingwithout a shred of proof that the form of a factorization is m = 0], I have a proof that your main conclusion *** cannot be correct ***. It is not a long or dif?ult proof. Most people would react by trying to ?d an error in myproof. You have failed to do so. It stands. I have found your error. You have not found mine, if Ihave one. It's that simple. Worse yet: W. Dale Hall also has a proof, totally inde-pendent from mine, that one of your main claims is false.In his case, all you have to do to check whether he is right or not is carry out a little computation. You havenot done it, or if you have, you are being mighty quietabout it. Dale's proof stands also. Yes, most people would not be happy to have two independent unrefuted proofs that their beloved work waswrong, hanging out there unrefuted. They would try?d where the proofs went wrong. That's why. No, no, no, again: you don't lose factorizations which work for the underlying polynomial when you stick in values. You gain new ones. No, on the contrary. As has often happened with your simplistic little toy examples, you went too far. Itback?ed on you. Whoa here, pardner! Looks like I am going to have to repeat that proof again! I don't want you to lose sight of the fact that this is really your main #1 problem here, and I am sure you wouldn't want other people to overlook it either. So I am putting it back in. Here it is, and again *please* feel free to point out any errors! DISPROOF OF CLAIM Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers, and assume Q(x) is irreducible over the rationals. Assume that c = p * v, where p is a prime and v is an integer. Let a1, a2, and a3 be the roots of Q(x). Note that by de?ition, since Q(x) is monic, a1, a2, and a3 are algebraic integers. Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime to p. This means that there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now, let F12 be an automorphism of the ?ld of algebraic numbers with the following properties: 1. F12(a1) = a2. 2. F12 leaves the sub?ld of rationals invariant, that is, if d is rational, F12(d) = d. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. The existence of such an F12 is an elementary result in Galois theory. See, e.g., Abstract Algebra, by Beachy & Blair. Now apply F12 to both sides of equation [1]: [2] F12(r) * F12(a1) + F12(s) * F12(p) = F12(1). Note that F12(a1) = a2, F12(p) = p, and F12(1) = 1. Let r' = F12(r) and s' = F12(s). Note that r' and s' are algebraic integers. With these substitutions, [2] becomes r' * a2 + s' * p = 1, which immediately implies that a2 and p are coprime. Similarly, use another automorphism F13 to show that a3 and p are also coprime. Therefore: if a1 is coprime to p, so are a2 and a3. That is, if one of the three is coprime to p, they all are. But: you know that a1 * a2 * a3 = -c = -p * v. Therefore p is a divisor of a1 * a2 * a3. Therefore at least one of a1, a2 and a3 is NOT coprime to p. This contradiction shows that the original assumption was false. The original assumption was based on your claim that one of a1, a2, or a3 must be coprime to p. Therefore you were wrong. Again, feel free to poke any holes in this that you can. Nora B. === [...]| DISPROOF OF CLAIM| | Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are| integers, and assume Q(x) is irreducible over the rationals.| Assume that c = p * v, where p is a prime and v is an integer.| Let a1, a2, and a3 be the roots of Q(x). Note that by| de?ition, since Q(x) is monic, a1, a2, and a3 are algebraic| integers.| | Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime| to p.It seems to me that it should be possible to prove this without relyingupon the existence of the automorphism of the algebraic numbers you used.Keith Ramsay === It seems to me that you are both wasting your time if you hope toeducate JSH - I have been following the many JSH initiated threads formaybe all of 2 months or so now and I have not yet seen a instance of him refuting an apparently sensible and logical argument that is contrary to his position.I can't follow the mathematics involved but the standard of discussion involved is simultaneously entertaining and educational: I de?itely learn more as I attempt to follow the sensible discussion and I am most certainly amused by the mental acrobactics that JHS is displaying for public entertainment ... all this and I don't have to worry about paying for my degree (*laughs loudly*) and it comes complete with a free pass to watch the monkey in the zoo!As I have said previously, I seriously hope that you (Keith) and Nora (and Will Twentyman and Arturo Magidin and David Ulrich - sorry if I don't have all the names perfectly correct) will continue to respond seriously and sensibly to JSH as this is fantastic learning stuff for me ... as you have to make your argument and discussion more and more simple in an attempt to show JSH the many ways in which you believe him to be incorrect I learn more and more.My latest instruction, thanks very much to the efforts of Keith and Nora, has been in the area of mathematical proof - I always believed that proofs belonged solely to the realm of geometry (showing one triangle to be the same as the other) and now I have seen how proof pervades even (especially ?) the most esoteric math (not that I believe, looking at other threads in this NG, that this stuff is particularly esoteric).I wonder, in the spirit of JSH, whether that last sentence is syntactically and gramatically correct ?I'm not sure if I have set the NG stuff correctly - if I've done this right then only sci.math will see me ... I don't understand why JSH feels a need to post to sci.*, alt.*, *.fr and so on ... if it's mathematics then it belongs here doesn't it ? (unless it's speci?ally related to undergraduate studies, the de?ition of which varies by country)Ivan McDonagh. === That is true. Hopefully there's some progress being made NoraBaron!!! Well let's consider what you actually *did* which was to put in valuesfor a_1, a_2, and a_3, as if you could just pick them at will.However, above you admit that the factorization still remains evenwhen I put in a value for x. And in fact, the x's and the value ofthe a's are independent anyway, as can be seen from (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y)so your position that my picking a value for x, as I picked x=2,affected the a's is NOT algebra. That is, there is no rational wayyou could have supposed that my picking an actual value for x wouldaffect the a's in such a way that you thought you could just pickvalues for the a's as you did in your post.So I have (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and to simplify in order to lessen theability of posters to confuse because of the symbol load, I set f=5,u=1, and x=2.Of those choices *only* the selection of f=5, affected the value ofthe a's, not x, Nora Baron, so your focus on x is bogus.James Harris === Ah good. You understand that simplifying can introduce extraneous solutions that do not help you analyze your original problem.Or perhaps you don't.It remains, but not uniquely. You appear to have missed the entire point. Put in a value for x and you get *extra* *valid* factorizations. If you don't want them, don't plug a value in for x.Nora pretty clearly discredited that statement in her post.What part of her algebra is incorrect? At what point did she do something where the right side doesn't equal the left side? If she did something that is not algebra, there is a mistake. Where is it?Assigning x a value changed it to a completely different problem with MORE factorizations. One of them was inconvenient. I guess you didn't understand after all. If you don't like the results, point out the speci? error in her work. Otherwise, you are wasting bandwidth.-- Will Twentyman === You are right about that. W. Dale Hall has produced an independentproof for a special case that is simpler to verify. However on balance I think the automorphism argument is the shortest and simplest to understand. Nora B. === I agree that the Galois theory argument is more informative, shortest,and [given even a glimmer of understanding of what Galois theory isall about], simplest to understand. Further, it has the advantage ofadmitting some amount of generalization, which my approach foregoesentirely.That said, I think there is a major hurdle in getting JSH to acceptthe fact that Galois theory is correct, and applicable in the contextof the factorization of polynomials. He seemingly has a huge bug upin an unmentionable ori?e, about the fact that Galois theorycanonically deals with ?lds and ?ld extensions, not realizingthat the ring of integers of a number ?ld enjoys some propertiesthat arbitrary rings do not.I had vainly hoped that JSH would [irrespective of any acknowledgementof the hated source] take a look at the veri?ation of those commonfactors that I gave, via direct polynomial multiplications, verifythose multiplications for himself to see that I wasn't lying, andcome to his senses about his argument.Well, I *did* strongly suspect that my errand was in vain, but Iwanted to give him the bene? of the doubt, and failing that, givehim enough opportunity to demonstrate his unwillingness to face reality.I have apparently been granted that second wish rather than the ?st.Dale === This seems a little ironic. You were claiming previously that by evaluating the polynomial, one would lose the factor-ization associated with it. I pointed out that you had it exactly backwards - that by evaluating the polynomial, in general you introduce some factorizations that are differentfrom the polynomial factorization. Of course you still havethe polynomial factorization as well. Also of course I knewall this. So when you say there's some progress being made,it is evidently on your end, not on mine. Also as I pointed out previously, in showing the factorizationof P(2), I was making the point that it does you no good whatsoeverto try to simplify by substituting in an actual value for x,when you have no intention of considering any other factorizationthan the polynomial factorization. It seemed to me that you didnot understand that little subtlety, but now perhaps you aregetting it. It is not the central point anyway, so I don't really care on this one. No, I didn't pick them at will. I did retain exactly the*form* of the factorization, (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5),and of course I factored in such a way that a1, a2, and a3 wereall divisible by 5. Since when you let x = 2, it was a perfectlyvalid factorization of an ordinary integer, not of a polynomial.You said it could not be factored in this form and I proved thatit could. Again, this is a minor side-issue. If you still don't get, don'tworry about it. The main things you should worry about are what you deleted. I will say them again in a different way that you may possibly be able to understand: 1. You are considering a degree 3 polynomial P(x) which is also a function of m. When m = 0, that polynomial becomes of ?st degree in x. You note that if the factorization is of the form (a1*x + 5)*(a2*x + 5)*(a3*x + 5), then when m = 0, to retain this form for the factorization, two of the a's, say, a1 and a2, must be zero. Of course zero is divisible by 5, so you can say that when m = 0, a1 and a2 are multiples of 5. Up to this point everything is OK. Then you make the great leap. You conclude that not only are a1 and a2 multiples of 5 when m = 0, they must be multiples of 5 for other values of m also. It goes without saying here that for different values of m, the values of a1 and a2 are different. You may consider them as functions of m, and it would make sense to denote them as a1(m) and a2(m). So you are saying: a1(m) and a2(m) are divisible by 5 when m = 0 therefore a1(m) and a2(m) are divisible by 5 for ALL OTHER values of m. Right? So where's the problem? Why is everyone being so obtuse about this? Because you do not have the slightest hint, not the slightest shred, not the faintest wisp of just?ation for the word therefore. It is pure hunch, pure intuition, pure guess. And pure error. It is a false conclusion. You do not cite any general theorem or mathematical principle that justi?s this. You just say it. You think it is obvious and everyone else should just endorse it on the dotted line. Saying it is enough, right? You say it, and as with so many other things you have said, everyone else should just shut up and believe it. No further proof needed, really. It's not enough. It's false. There is another slightly interesting issue here, related to the degeneracy and singularity that occurs when m = 0. I will post something on this later. 2. How do I know it's false? Because W. Dale Hall and I have given separate, independent proofs that your main claim is false. Since you have YET AGAIN deleted out the section of my post that contained my proof, and since you have not previously found any valid objection to it, I am going to give you yet another chance and reproduce it here. I am sure your many fans out there are beginning to be embarrassed by your failure to cope with this, and they will appreciate the fact that you are being given another shot at it. Right, fans ? Not to mention the many future math historians, when they starting writing your biography: === === ==JSH CLAIM: It is possible to ?d a 3rd degree polynomial with integer coef?ients, monic and irreducible over the rationals, such that, if a1, a2, and a3 are the three roots, then at least one of a1, a2 or a3 is coprime in the algebraic integers to a prime integer which divides the constant term of the polynomial.DISPROOF OF CLAIM: Let Q(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. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by de?ition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory, there exists an automorphism F12 of the ?ld of algebraic numbers such that: 1. F12 leaves the sub?ld of rational numbers ?ed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + 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(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts your claim noted above. Please feel free to point out any errors or gaps in the proof I just gave. === === === == Yes, of course the original polynomial factorization is still there.I never said it wasn't. I just pointed out that when you evaluate,you get other factorizations also - factorizations which clearly,obviously violate your claims. Wrong. The original a's were derived from the roots of apolynomial in x. When you chose x = 2, you were abandoningthe polynomial. What you had was no longer a polynomial inx. When you talked about factoring it, you were factoringan ordinary number. You didn't say, OK, now I have an ordinary number, but I still want to factor it as if itwere a polynomial in x. That would have been a silly statement to make, and it would have implied that yoursimpli?ation was really quite pointless. The only sensibleinterpretation is that after you evaluated the polynomial, youwere thinking about factoring an ordinary number. Now youare trying weasel out of that because you see it was obviously false. AGAIN: this is a minor side issue. If you STILL reallydon't get it, stop worrying about it. Your real big-timeproblems are the two I listed above. To summarize: A. I and Dale Hall have found separate proofs that your central claims are incorrect. Unless you can ?d an error in both our proofs, it really doesn't matter much what you say. There is an error in your proofs and an error in your thinking. B. I have found an explicit place in your argument where your thinking is incorrect. I have described it in excruciating detail: see above. So far you have either been unable to understand it or unwilling to understand it. It is conceivable that I am wrong and that you could convince me that your argument actually makes sense. Of course you would also need to show that my proofs and Dale's are incorrect, because they say your CONCLUSION is wrong, regardless of how you got there. However so far you have not made even a feeble attempt on either front. Basically you just say, If it's true when m = 0, it must be true for all m. Period, end of argument. I got news for you. That ain't a proof. Nora B. === Why not just take a simpler approach with James? Just ignore him. Nobody really believes him anyway (with the *possible* exception ofhimself). Why bother entertaining him? (If nothing else, it giveshim the misleading appearance of credibility.) === However the a's in general are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2,so x has *nothing* to do with the value of the a's. Yup, one of them is clearly coprime to 5. The coprimeness resultleads to the conclusion that they all must be coprime to 5 in the ringof algebraic integers, which is what's wacky, and shows a problem withthe ring.That is not correct as you can't just pick values for the a's in thatway. Ok.Wrong as I pointed out above as Nora Baron apparently never realizedthat the a's are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2,so she can't just pick, though you can see she tried. And in fact form=1, where she actually picked values for the a's the cubic is a^3 + 72a - 13825 = 0and none of her picks work. James Harris === And again.a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m)For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5.Perhaps you mean that none of the a's are coprime to 5?Which is trivially obvious and uninteresting after all.Sorry to have bothered you.Phil Nicholson. === You keep saying this as if it's obvious... Let's look at it.P(m)= f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f)distribute in f^2P(m) = (m^3 f^6 - 3m^2 f^4 + 3mf^2) x^3 - 3(-1+mf^2)xu^2 f^2 + u^3 f^3using v=-1+mf^2, v^3+1 is the coef?ient on x^3P(m) = (v^3+1) x^3 - 3vxu^2 f^2 + u^3 f^3This can be viewed as a polynomial in terms of x, or v, or x and v. Viewed as a polynomial in terms of x and reversing it's coef?ients gives the expression you have listed. Here's the catch, keeping x as an unknown RESTRICTS the possible values of the a's.Watch what happens when you plug in f=5, u=1P(m) = (v^3+1) x^3 - 75 vx + 125 where now v=-1+25m.Now plug in x=2P(m) = 8(v^3+1) - 600 v + 125P(m) = 8v^3 - 600 v + 133 where v=-1+25mThis can *not* be viewed as a polynomial in terms of x. It can only be the a's as you have suggested. When you set x=2, you lose information about the a's, and introduce additional values. In this example you can clearly see the difference between having x as a variable, and x=2. You fundamentally change the nature of the problem.Worse, when x=2, P(m) is no longer a monic polynomial under any available interpretation. The roots of this polynomial need not be algebraic integers.You lost information when you chose x, and the available assignments of the a's grew. Inconvenient, but true.Your assertion about the a's only holds true when x is kept as a variable. See my argument above. Perhaps if you clearly de?ed which variables the a's are dependent on we could clear up a lot of this mess. The only thing you've made clear is that they should depend on m. Do they also depend on f? u? x? If so, you must be careful which letters you substitute values in for. It appears that you do NOT wish m to be a function of x. This means that you cannot include a simpli?ation that includes choosing values for x. If you do, you have *over*simpli?d and changed the problem.-- Will Twentyman === [snip]I'm afraid that your hope was truly in vain. In the few instances where James tried to follow up a simplemultiplication of a few binomials, he constistently got the exponents and the signs wrong. He is hopelessly sloppyin his algebra. I think if you want to gain some ground (no promises) you will have to post the exact values of thenumbers a1, a2 and a3, and then derive the expressions which prove they are not coprime to 5 in the ring ofalgebraic integers so that they are public record. I don't think James is disposed to do this or has the ability.It is simpler to for him just to assert that you must be wrong, since his proof is irrefutable. In a previous posthe even said not to bother citing errors in his proof because if any existed *he* would let everyone know aboutit.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Your (and Will's) patience is admirable, but Quixotic. JSH has shown time and time again that he either fails to understand the nature of mathematical proof, or simply does not feel constrained by it. Observing your attempts to enlighten him reminds me of a Simpsons episode where Homer is talking to the dog, but from the dog's end it is just meaningless blah blah blah interspersed with occurrences of the only word the dog understands, his own name.Gib === Being an educator, I like to maintain the belief that all people are capable of learning. James has shown a capacity for thinking about math, so with my (perhaps false but still comforting) belief and his professed willingness to learn, I am willing to explain as long as he listens.I also ?d it fascinating to observe the various ways in which he avoids listening and the various mental gymnastics he displays in defending his version of truth.Finally, it's good for me. I've knocked a lot of rust off the mental gears by following the arguments and presenting my own.-- Will Twentyman === | Why not just take a simpler approach with James? Just ignore him. | Nobody really believes him anyway (with the *possible* exception of| himself). Why bother entertaining him? (If nothing else, it gives| him the misleading appearance of credibility.)By now this suggestion has been made many times, and there hasn'tbeen anything like the kind of general cooperation that it wouldtake to end the discussion. If the only point were to preventpeople from being taken in, I agree just letting him attempt topersuade people would be as effective as arguing with him. Butthere are other reasons why people keep at it.For one thing, sometimes it's interesting to try to convince someonewho's exceptionally resistant to being convinced, and see how it itthat they manage to hang on to their opinions anyway. I think a numberof us just ?d it irritating to see someone making claims we knoware wrong and not being corrected. I won't claim that's a rationalirritation. Some people ?d it entertaining in other ways.The main thing that would convince me to stop would be if I thoughtit would be better for his own well-being not to have me as adistraction from his other issues. It seems possible to me that he'dprefer I stopped discussing his proofs with him too, but I don't know.Keith Ramsay === Perhaps I am a little behind on this discussion, but why can't we justgive James a polynomial and see if he can make his technique work? Working from a polynomial you made yourself around your method (whichif I checked correctly cannot be factored with integers anyway) isquite different than trying to apply it in practice.It also seems that - though a bit of work - the traditional p/qapproach and sign examination are pretty simple. They are also fairlystraightforward to implement programmatically. (I wonder if James hasseen this?) === JSH is intelligent enough to do basic algebra pretty much through the quadratic equation and has learned random facts past that - e.g.,he now knows what algebraic integers are - but he has not absorbed what it means to construct a complete rigorous proof. In the presentcontroversy the problem is that he has an overpowering intuition thatthere must be a formula connecting the roots of a polynomial equationwith the constant term - in fact he is right, in the sense that forpolynomials through 4th degree, there are formulas involving radicalsthat specify the roots in terms of the coef?ients. He believes that such formulas extend to degenerate cases. He then ?ds a formula-typerelationship for a degenerate case, and then assumes that this formula must hold in nondegenerate cases as well. He cannot conceive that innondegenerate cases, the formula generalizes in any but the obvious way. Ideally what we would do to prove this is incorrect is actuallywrite down the roots of his polynomial and show how each root sharesalgebraic integer factors with prime divisors of the constant term. Incredibly enough, this is not completely simple even in the case of aquadratic. In the case he is interested in, the cubic, the formulas for the roots are quite complicated and showing directly that they includealgebraic integer factors of the constant term is a horrible mess. It is much easier to prove it indirectly. That is what I have done using automorphisms and what W. Dale Hall has done in a quite different The automorphism argument is, I think, a nice example of how sometimeshaving some theoretical superstructure can create a shorter, more easilyunderstood path to the answer than a direct frontal assault. Sometimesthe long way round the mountain is actually the shorter than tryingto drill your way through it. JSH essentially refuses to look at either of our arguments. That is the other problem. Although not stupid, he has truly enormous ego investment in his argument. He does not see that it has a ? gap. He thinks that bit about generalizing from the degenerate case is thenatural and obvious thing to do, because he has that simple formula (a_1/5). He would rather conclude that there is something wrong with the de?ition of algebraic integers or there is a ? Galois theory than look really critically and rigorously at his ownproof. However I actually don't think this can continue forever. I think we will eventually ?d a way through his armor. It has happened before. Simpson's dog: my recollection of that is that it came from aFar Side cartoon. In one panel the dog's owner is saying something like No, Ginger! You must not bark in the house, Ginger!Do you hear me, Ginger?, and the dog hears Blah, Ginger! Blahblah blah, Ginger! Blah blah blah, Ginger! I wonder whichcame ?st - Far Side, or the Simpson's version? In any case, yes,JSH has reinvented this also. Nora B. === certainly, but can he Complete the Square? it's not really a property of tetragona per se, butthe diagram will help, either way.(I prefer the lunes proof of the pythagoreean th.; and,I ?ally realized what the spatial analog is,a couple o'weeks, ago .-) --A church-school McCrusade (Blair's ideals?):Harry-the-Mad-Potter want's US to kill Iraqis?...http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/?es/curriculum/Cosmo.PCX== ==(A lot of the usual stuff snipped.)Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3= 0 and conclude that at least one of the a_i's, say a_1, is zero. Putting this into your equation following the line Now setting m=0gives me will then give you5(2 a_2 +5)(2 a_3 + 5) = 25(11),and I can divide both sides by 5. Now, why couldn't it be that a_2 anda_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how youarrive at your statement, And to get that factor that is 25, you must-- Mark Thornquist === Sure. I'm considering non-polynomial factors of P(m), and I have frommy lemma that any such factor can be written as r+c, where r=0, orvaries with m, while c remains constant and is a factor of theconstant term.So with g_1 = (2 a_1 + 5), and since one of the a's MUST equal 0,when m=0, selecting a_1 as the one gives me g_1 = 5,so c=5, and of course r = g_1 - c, and as m varies, r varies, while,of course, at m=0, it also equals 0.Now then I notice that P(m)/25 has a constant term that is coprime to5, as P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11.So when that 25 goes, then a factor has to come out of g_1 as well.That is, looking at P(0)/25 = 11, I have that the constant term iscoprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out.It's like P(m) = g_1 g_2 g_3, and I checked at P(0), to ?d that at that pointg_1=5.But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5goes.But when m doesn't equal 0, that only handles one of the 5's as 25 hastwo factors where each is 5. Therefore, ONE other of the a's must goto 0, when m=0, and it also has a factor that is 5 which separatesout.That's why the lemma is KEY, and is the linchpin of the argument.I've also used P(m) = 25 Q(m) to explain, which can help you byletting you consider factors of Q(m). For instance, if you have h_1 = 2a_1/5 + 1as a factor of Q(m), you can check at m=0, to ?d that h_1=1.But let's say that h_1 = 2a_1/s + 5/swhere s is some non unit factor of 5, but 5/s is not coprime to 5.Then at m=0, you'd have h_1 = 5/s, which contradicts with Q(0)=11.Now you MAY wish to forget that a_1 = 0 from before with P(m) butdoing so is not logical.If someone wishes any portion of what I just showed you expanded on,then please point out which section. I'm quite willing to explain indetail, but I can't read your mind. You need to tell me where you'renot sure, so that I can give you more details at that point.James Harris === How do you ?ure? a_1, a_2, a_3 are non-polynomial functions of m. They don't have to be anything so simple as this.See objection above regarding this. Note: you have never addressed the objections to your assumption that non-polynomial factors behave like polynomial factors.Why? I'll suggest an alternative: a_2 = -2, a_3 = 25. Along with a_1=0 you end up with:g_1=5, g_2=1, g_3=55Not if h_2 = s/5 and h_3 = 11Ok, I'd love to see why non-polynomial a_i must have the product you claimed. I've provided counter-examples elsewhere and you have yet to address them.-- Will Twentyman === Good morning,Is the symmetric algebra generated by diagonal tensors?Tern_ === Let E is a extension ?ld of a ?ld F.Prove that: If u in E is algebraic over F then, u^2 is algebraic over F {1,u^2,(u^2)^2, ... ,(u^2)^(n-1)} is a basis over F. so, {1,u^2,(u^2)^2, ... ,(u^2)^n} is linearly dependent over F. therefore, c_0,c_1,...,c_n are not all zero, where c_0 + c_1*u^2 + c_2*(u^2)^2 + ... + c_n*(u^2)^n=0. so, if we let c_0 + c_1*x + c_2*x^2 + ... + c_n*x^n=f(x), then hence, u^2 is algebraic over F.----------------------------------------------------------I wonder whether this proof is really correct or not.If not, please show me right way.In addition, How can I prove that F(u)=F(u^2) ?If anyone know this, please post reply. === [...]You can't, because it's not true in general.Consider F = Q, u = sqrt(2).-- Jim Heckman === If E is ?ite extension ?ld of a ?ld F and [F(u):F] is odd,then Is it true, in general ?If then, Can you give me some hint for proving ... === Right here, you've assumed what you want to prove. You have to prove that [F(u^2):F] is ?ite.What do you want to do to prove that u^2 is algebraic? I get the idea algebraic and [E:F]=[E:K][K:F]). Not knowing what book you're using (if you are using a book), I don't know if the point of the exercise is to reinforce the theorem (and make you check the assumptions), or if you're just supposed to use the de?ition of algebraic to see if u^2 is algebraic (which is what I do assume in the absence of any other information). That is, this exercise is setting up that nice theorem on the degrees of extensions.Why? (disclaimer below signature)The basic idea here would be to to ?d [F(u):F(u^2)]. You've really only got two choices here. But once you've done the ?st part (u^2 is algebraic), you know that [F(u):F] = [F(u):F(u^2)][F(u^2):F] (if you haven't proven this yet, now is a good time, because it's a very convenient tool), which tells you an awful lot.Jon Miller === do you know such an n exists? this is pretty much what you're trying toprove...this doesn't work as your starting point is wrong.But it looks as though you've already shown that u is algebraic over F iff[F(u):F] is ?ite. (And in fact that [F(u):F] is the degree of the minimalpolynomial of u over F, but we don't need this for what follows). If so,you can make your approach work in a slightly modi?d form...You need to start from what is given... u is algebraic over F, so F(u):F isof ?ite dimension?Also u^2 is in F(u), and so powers of u^2 are all in F(u). But there's alimit to how many independent powers of u^2 there can be in F(u):F (why?),which is where your approach above can naturally come in...HTHMike. === [one top-posting ?ed for free, next one will not be answered]Yes.Another poster already has. Consider that [F(u):F] =[F(u):F(u^2)] [F(u^2):F].-- Jim Heckman === Saturday June 21st 2003 172/193 16925T Y S S E L A N D20 25 19 19 5 12 1 14 4 = 119 Angie was playing piano on the NW corner of 21st Street and 4thAve., I stopped to listen and to collect her stats. She is fromSaskatoon, has produced a CD a few years ago and was working towardanother.189 Angie 27 2 61 58/307 1471Angie 36 Grace 34 Tysseland 119 Angie was born in 61 (Exodus 11), her last name adds to 119. Herlast 11 letters add to 128 (Numbers 11). Her ?st 7 letters add to 61(Exodus 11 and is her year of birth). She has 11 consonants, they addto 137 (the 3x11th prime) and are in positions adding to 118 (the8x11th non-prime).Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 -- -- 58 50 Angie was born on day 58 (the 7 primes up to 17). Her ?st nameadds to 36, or 7 plus the 7th prime (17) plus the 7th non-prime (12),or simply 7+7p+7np. Her middle name adds to 17+17, her given names addtogether for 70. Her last name adds to 7x17, or 7 times the 7th prime.Her common name adds to the 155 verses of Ephesians, Bible Book 49(7x7 and is the 17+17th non-prime). Her last two names add togetherfor 153 (1 through 17 and is the 117th non-prime, it's the terminatingchapter of Numbers and is the number of ?h in the net in John 21).Her full name adds to 189 (the ?st 17 primes minus the ?st 17non-primes). The 7 different letters in her given names add to 57. Theconsonants in her given names add to 49 (7x7 and is the 17+17thnon-prime). In her given names, her consonants exceed her vowels by 28(1 through 7). She has 12 (7th non-prime) letters in all, her 7+7missing letters exceed her 12 (7th non-prime) represented letters by77. She has 7 unrepeated letters, they add to 91 (Leviticus 1, the?st of the chapters to contain the length of 17 verses). Herconsonants exceed her vowels by 85 (5x17 and is 17p+17np, Ruth with 85verses is the 17th shortest Book in the Bible). Her odd valued lettersadd to 107 (Leviticus 17), it's the 28th (1 through 7) prime. Herprimes and squares add together for the 108 verses of Bible Book 59(the 17th prime), it is the 17th prime short of the 167 verses of Book17). Her initials add to 28 (1 through 7). She was born on the 27thday of the month, corresponding to Daniel with 357 (7x17+7x17+7x17)verses.Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 --- --- 440 251 Angie was born on day 58 (the 19th Book of the New Testament). Shehas 19 letters and a last name adding to 119. In her given names, herconsonants exceed her vowels by 28 (19th non-prime). The odd valuedletters in her given names add to 38 (19+19). The 7 different lettersin her given names add to 57 (19+19+19). In all her unrepresentedletters exceed her represented letters by 77 (the primes up to 19).Her consonants add with their positions for 255 (First Samuel 19). Hersquares add to 38 (19+19). Her odd valued letters add to 107 (the 19thprime in non-prime position). Her initials add to 28 (the 19thnon-prime). We meet on the 172nd day of the year (Deuteronomy 19), wemeet with 193 days remaining in the year. Her birthday was 114 (6x19and is 19+19p+q9np) days ago. Tysseland (119) and I had our birthdayswere an average of 119 days ago. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 --- 167 Esther Angie's consonants add to 137, vowels to 52 (37th non-prime). Herrepresented letters add to 137. Her names add to the 25th, 23rd and89th non-primes, together for 137. In her given names, herunrepresented letters exceed her represented letters by 237. She wasborn with 307 days remaining in the year. Angie's consonants add to the 33rd prime, vowels to the 37thnon-prime, together for 70, pretty as her given names add together for70. Angie's last name adds to 119, her common name adds to 155 (the119th non-prime). Angie Tysseland was born on the 27th. The consonants in her givennames are together in positions adding to 27. In her given names, herconsonants plus their positions together exceed her vowels plus theirpositions by 27. The primes in her given names add to 27. In her fullname, her unrepresented letters exceed her represented letters by 77(Exodus 27). Her odd valued letters are in positions adding to 117(Leviticus 27). Her primes and squares together add with theirpositions for 108 (4x27). Her full name adds to 189 (7x27). OldTestament Book 27 and New Testament Book 27 are the major Books ofend-times prophecy.Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Saturday June 21st 2003 172/193 16925T Y S S L A N D20 25 19 19 12 1 14 4 = 119 Angie was playing piano on the NW corner of 21st Street and 4thAve., I stopped to listen and to collect her stats. She is fromSaskatoon, has produced a CD a few years ago and was working towardanother.189 Angie 27 2 61 58/307 1471Angie 36 Grace 34 Tyssland 119 Angie was born in 61 (Exodus 11), her last name adds to 119. Herlast 11 letters add to 128 (Numbers 11). Her ?st 7 letters add to 61(Exodus 11 and is her year of birth).Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 -- -- 58 50 Angie was born on day 58 (the 7 primes up to 17). Her ?st nameadds to 36, or 7 plus the 7th prime (17) plus the 7th non-prime (12),or simply 7+7p+7np. Her middle name adds to 17+17, her given names addtogether for 70. Her last name adds to 7x17, or 7 times the 7th prime.Her common name adds to the 155 verses of Ephesians, Bible Book 49(7x7 and is the 17+17th non-prime). Her last two names add togetherfor 153 (1 through 17 and is the 117th non-prime, it's the terminatingchapter of Numbers and is the number of ?h in the net in John 21).Her full name adds to 189 (the ?st 17 primes minus the ?st 17non-primes). The 7 different letters in her given names add to 57. Theconsonants in her given names add to 49 (7x7 and is the 17+17thnon-prime). In her given names, her consonants exceed her vowels by 28(1 through 7). She has 12 (7th non-prime) letters in all, her 7+7missing letters exceed her 12 (7th non-prime) represented letters by77. She has 7 unrepeated letters, they add to 91 (Leviticus 1, the?st of the chapters to contain the length of 17 verses). Herconsonants exceed her vowels by 85 (5x17 and is 17p+17np, Ruth with 85verses is the 17th shortest Book in the Bible). Her odd valued lettersadd to 107 (Leviticus 17), it's the 28th (1 through 7) prime. Herprimes and squares add together for the 108 verses of Bible Book 59(the 17th prime), it is the 17th prime short of the 167 verses of Book17). Her initials add to 28 (1 through 7). She was born on the 27thday of the month, corresponding to Daniel with 357 (7x17+7x17+7x17)verses.Primes Non-Primes 2 1 3 4 5 6 7 8 11 9 13 10 17 12 19 14 23 15 29 16 31 18 37 20 41 21 43 22 47 24 53 25 --- --- 440 251 Angie was born on day 58 (the 19th Book of the New Testament). Shehas 19 letters and a last name adding to 119. In her given names, herconsonants exceed her vowels by 28 (19th non-prime). The odd valuedletters in her given names add to 38 (19+19). The 7 different lettersin her given names add to 57 (19+19+19). In all her unrepresentedletters exceed her represented letters by 77 (the primes up to 19).Her squares add to 38 (19+19). Her odd valued letters add to 107 (the19th prime in non-prime position). Her initials add to 28 (the 19thnon-prime). We meet on the 172nd day of the year (Deuteronomy 19), wemeet with 193 days remaining in the year. Her birthday was 114 (6x19and is 19+19p+q9np) days ago. Tyssland (119) and I had our birthdayswere an average of 119 days ago. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 --- 167 Esther Angie's consonants add to 137, vowels to 52 (37th non-prime). Herrepresented letters add to 137. Her names add to the 25th, 23rd and89th non-primes, together for 137. In her given names, herunrepresented letters exceed her represented letters by 237. She wasborn with 307 days remaining in the year. Angie's consonants add to the 33rd prime, vowels to the 37thnon-prime, together for 70, pretty as her given names add together for70. Angie's last name adds to 119, her common name adds to 155 (the119th non-prime).Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === Actually it is...T Y S S E L A N D20 25 19 19 5 12 1 14 4 = 119 === Friday June 20th 2003 171/194 16924S L U S A R C Z Y K19 12 21 19 1 18 3 26 25 11 = 155260 Ania 23 12 82 357/8 9440Ania 25 Elzbieta 80 Slusarczyk 155 Ania is from Krakow Poland, she provided stats in the afternoon atthe Extra Food Store on Broadway Ave., it's the street with the penisposter poles. Ania was born on day 357, it's the number of verses in Daniel, theBook is in part about 666. Her last two names add together for 235,the 184th prime (1097) and the 184th non-prime (235) averages 666. Theconsonants in her given names add to 74 (a factor of 666). Her oddvalued letters add to 130 and also her even valued letters add to 130,corresponding to Numbers 13 (the 6th prime). Her names add to 25, 80and 155, these are the 16th, 58th and the 119th non-primes, togetherfor 193 (Bible Book 6 chapter 6). Her ?st and last names average 90(66th non-prime), Exodus contains 1213 verses (66+66+66th prime) andbrings the Bible up to 90 chapters (66th non-prime). Her name adds tothe 260 chapters of the New Testament, New Testament chapter 260 isRevelation 22, it's the terminating chapter of Bible Book 66. Her ?st name adds to 25 and her initials add to 25. Her 357th dayof birth exceeds her 260 valued name by 97 (25th prime). She was bornin the 12th month (Second Kings with 25 chapters). I am 25.84 yearsolder than Ania.389 <-77th prime104 <-77th non-prime 77 <-77---570 The Four 57's ---- 1373 <-220th primeChapter 57 is Exodus 7 with 25 verses Book 57 is Philemon with 25 verses -- -- 41st non-prime 16th non-prime Major Books of End-Times Prophecy (Daniel and Revelation are in partabout 666 while Isaiah contains 66 chapters):Daniel - 357 versesRevelation - 404 verses <-57 plus the 57th prime plus the 57th non-primeIsaiah - 1292 verses <-an average of 19.575757... verses per chapter Ania was born on day 357, her ?st name adds to 25 and herinitials add to 25 (Bible chapter 57 contains 25 verses and also BibleBook 57 contains 25 verses, pretty as 57 and 25 are the 41st and 16thnon-primes, together for 57). Her given names add together for 105,it's the 78th non-prime while 78 in turn is the 57th non-prime (105 isthe 57th non-prime in non-prime position). Her vowels add to 78 (57thnon-prime). Her consonants exceed her vowels by 104 (Leviticus 14 with57 verses), prettier as 77 plus the 77th prime (389) plus the 77thnon-prime (104) adds to 570. She was born on the 23rd, correspondingto Isaiah (the Book contains an average of 19.575757... verses perchapter). I meet Ania and she provides the stats on the 57+57+57th dayof the year. Ania's day, month and year of birth adds to the 117 verses of BibleBook 22 and she has 22 letters. Ania was born on day 357, her day and month of birth adds togetherfor 35. Her last two names add together for 235. Ania was born 349 days closer to the end of the year than to thebeginning of the year, she was born 49 weeks (and 6 days) closer tothe end of the year than to the beginning of the year. Her last nameadds to the 155 verses of Ephesians, Bible Book 49. Non-Primes Non In Prime Primes Positions 1 1 2 4 <- 4 3 6 <- 6 4 8 5 9 <- 9 6 10 7 12 <- 12 8 14 9 1510 1611 18 <- 1812 2013 21 <- 2114 2215 2416 2517 26 <- 26 <-7th non-prime in prime position The nubile sweety was born on day 357 (7x17+7x17+7x17), she wasborn exactly 17+17+17 weeks into the year. Her name adds to the 260chapters of the New Testament, or 10 times the 17th non-prime (26).She was born in the 12th month and has 12 letters in her given names(7th non-prime). Her last name adds to the 155 verses of Bible Book 49(7x7 and is the 17+17th non-prime). She might take my 62 valued lastname and end up with a name adding to the 167 verses of Bible Book 17,it is the 7 primes in prime positions up to the 17th prime. Estherbecomes Queen in Book 17 and Q is the 17th letter of the alphabet. Primes In Prime Primes Positions 1 2 2 3 <- 3 3 5 <- 5 4 7 5 11 <- 11 6 13 7 17 <- 17 8 19 9 2310 2911 31 <- 3112 3713 41 <- 4114 4315 4716 5317 59 <- 59 --- 167 Esther Book 17 After showing Ania evidence that her name was a gift from God, sherewarded my work by requesting my copy of the record of our meeting.But it was my paper, my pen, my calculator and my work, so I told herthat she was not welcomed to it. Starting in 1988 I was repeatedlyarrested and tortured in psychiatric facilities in an effort to makeme shut up about the penis poster poles on Broadway Ave. and to makeme shut up about the Egyptian penises at The Vatican, Whitehouse, infront of Saskatoon City Hall and on the roofs of the churches, and toshut up about the mathematical harmony between people's names,birthdays and numbers in the Bible. Year after year they tortured me,year after year I begged people in the community for assistance to?e country, but they are so cheap and so ignorant that theycan't even offer me a cookie for my work. Protestants and Catholicslobbied my abusive parents to shut me up, then Protestants andCatholics sat on psychiatric appeal panel hearings and repeatedly gave When I complained to people about the way these Christians andthat I was there because I was supposed to be there. And so it is withAnia, she is here on the usenet because she is supposed to be here onthe usenet. And likely even if she did give me $2 for my work, she andher friends and relatives annually and collectively spend thousandsupon thousands of dollars turning pagan fertility symbols (evergreentrees, they stay green all year) into decorated idols. In December,their churches would have a decorated tree at the entrance, anothernext to the pulpit, and yet another downstairs in the lunch room,maybe Ania would attend church once a year to please her mom, and itwould be in late December, and she would comment on how beautiful thetrees are, and will give them money. I said that the penis posterpoles on Broadway Avenue were representations of penises like theEgyptian obelisks on church roofs, I get tortured year after year, andyear after year the Broadway merchants add additional representationsof penises along their street. Then Ania comes to Penis Street and Ishow her some gems, and she rewards my work by wanting my work and ourrecord of our meeting. You recent eastern European, Asian and LatinAmerican immigrants have done much to bring organized crime andviolence to Canada and disrupt whatever peace I am able to ?d here.While back in your home land you hear that millions of dollars havebeen spent on torturing me in Canada, and your only concern is to cometo Canada so you can make some of this money too. Your traditions arein con?ith God's Commandments, and I am on my knees begging Godto honor Exodus 20:5 and Hosea 4:6 as promises, and terminate thelives of Ania and her siblinks. If I hear of the deaths of any of thepeople in her family I will cheer, it will be in accordance toScripture (Psalm 137:9), and I will post these stats again. Fifteenyears of this work and you people reward me with torture, you are anignorant and compassionless piece of crap and I am on my knees beggingpeople are so cheap and ignorant that you don't even have the decencyevidence of who your God is, and that He provides you with your veryname. They tortured me here for years, nobody cared then and nobodycares now, all you care about is raising money so you can ? toIndia in the search of truth. Go to hell!!!Daryl Shawn KabatoffBox 7134Saskatoon SaskatchewanCanadaS7K 4J1Isaiah 45:4, Ephesians 3:15 - God gives you your name!!! === [snip standard Kabatoff irrational rant about his favorite subject -penises]Aha! There's the pitch. Just as I predicted.--Few things are more satisfying to a zealot than condemning everyone elseto hell.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === How do I integratee^ ((1/3200)x^2) between 0 and inf?Havn't done integration by parts in many a year so might need to be writtenin Help for Dummys format :-) === You don't... that function gets bigger and bigger asx does, so the integral from 0 to in?ity doesn'tconverge.If it should have been e^ (-(1/3200)x^2) then considersubstituting some multiple of x and then lookingup a table for the integral of e^(-x^2), which isa standard result.-- Rob. http://www.mis.coventry.ac.uk/~mtx014/ === Sorry should have beene^ ((-1/3200)x^2) between 0 and inf?written === Look up the Gamma function Gamma(n) = integral(0,oo) x^(n-1) e^-x dxProve Gamma(n) Gamma(1-n) = pi/sin pi.nShow Gamma(1/2) = 2 integral(0,oo) e^(-t^2) dtConclude integral(0,oo) e^(-t^2) dt = (sqr pi)/2Transform variable t = x(sqr 2)/80 === There's an even easier function to look up - without wanting to give it away, it's relevant that this was posted to a stats NG.Bob-- Bob O'HaraRolf Nevanlinna InstituteP.O. Box 4 (Yliopistonkatu 5)FIN-00014 University of HelsinkiFinlandMobile: +358 50 599 0540WWW: http://www.RNI.Helsinki.FI/~boh/ === Now I am really confused.... === On Wed, 9 Jul 2003 13:53:31 UTC, Debbie Sewell : Sorry should have been: e^ ((-1/3200)x^2) between 0 and inf?There's a short way and a long way.The short way is to think about what this integral is - it's very nearly the same as somethig which is easy to ?d tabulated.The long way is to ?d the integral of exp(-x^2) over the same range.Which is a wee bit of a pain to do in the general case, but in this case is traditionally done by noting that ifI = int(exp(-x^2), x = 0 .. inf) dxthen I^2 = int(exp(-x^2), x = 0 .. inf) dx int(exp(-y^2), y = 0 .. inf) dyAnd by not being too fussy about orders and limitsI^2 = int(exp(-(x^2 + y^2) dx dywhich is a surface integral. Put it into polar coordinates, note that dx dy = r dr dtheta, ?dle around a bit, get another I to appear and orff yer goes.Ian-- === Indeed, the easy way. I^2 =integral(0,oo) integral(0,oo) e^-(x^2 + y^2) dx dy = integral(0,pi/2) integral(0,oo) e^(-r^2) r dr dt = 1/2 * integral(0,pi/2) integral(0,oo) e^-u du dt = 1/2 * integral(0,pi/2) (-e^-u)|0,oo dt = 1/2 * integral(0,pi/2) dt = pi/4I = (sqr pi)/2integral(-oo,oo) e^(-t^2) dt = sqr pi = Gamma(1/2) = (-1/2)! === OK, we've gone off track here. If you want an answer, it does helpto ask the right question. It may no longer matter, but let me seeif I can clear up the confusion.To recap, we want the expected value of the density f(x) = (x/1600) exp(-(1/3200) x^2). See:http://groups.google.com/groups?selm=6714766d .0307072121.2142515%40posting.google.comThe expected value is integral(0,inf) x f(x) dx, i.e., integral(0,inf) (x^2)/1600 exp(-(1/3200) x^2) dxFor convenience, let a = 1/3200, so the integrand is 2 a x^2 exp(-a x^2). Let t = x sqrt(a). Then dt = dx sqrt(a), so the integral (call it E) is E = integral(0 sqrt(a), inf sqrt(a)) 2 t^2 exp(-t^2) dt/sqrt(a) = (1/sqrt(a)) integral(0,inf) 2 t^2 exp(-t^2) dtLet u = t, and let dv = 2 t exp(-t^2) dt. Then du = dt, and v = -exp(-t^2). So putting the factor 1/sqrt(a) aside for the moment, the integral is u(inf) v(inf) - u(0) v(0) - integral(0,inf) v du = 0 - 0 + integral(0,inf) exp(-t^2) dt = sqrt(pi)/2as pointed out elsewhere in this thread. So the overall result is E = (1/sqrt(a)) sqrt(pi)/2 = sqrt(3200) sqrt(pi)/2,which is approximately 50.Well, I'm pretty sure I'm answering the right question, and I hopeI didn't make any elementary mistakes in working out the solution.For what it's worth,Robert an integral a day Dodier--Science may be described as the art of systematic over-simpli?ation. -- Karl Popper === I need to show that limsup(s(n)+t(n))<=limsup s(n) + lim sup t(n)can I do this by proving that they can be equal or the lhs can be smallerthan the rhs or do I need to prove these two ideas together? === Do both together - there is really not much choice. You might want togive an example where equality holds and one where it does not.-- Paul SperryColumbia, SC (USA) === I have a solution manual that states this problem is associative:Determine whether the binary operation * defned is commutative and whether * is associative...* de?ed on Z by a*b=a-bI can clearly see how to check to see if this problem is commutative because I have 2 numbers.1-2 != 2-1 therefore not commutativeMy question is, how do you check if its associative without 3 numbers?The solution manual said: 2=1-(2-3) != (1-2)-3=-4...where did the 3rd number (or c) come from?Marcos === Never heard of such, would you explain what an associative problem is?I suppose it comes with 1 & 2 but until you tell us what set the binaryoperator operates upon, how are we to know?Who's c? You've yet to introduce her to us.As for a two number counterexample, presuming - is substraction o?tegers. 1-(1-1) = 1-0 = 1 (1-1)-1 = 0-1 = -1 === An operation * on set R is de?ed to be associative if it satis?s the following property:For all numbers a,b,c in R, (a*b)*c = a*(b*c)-- Will Twentyman === Sorry I wasn't speci? enough. I didn't know how to ask the question. I understand what the associative property says. I was looking for an operational procedure to ?ure out if the problem was associative. Lucky for me, I've ?ured out how to work the problem.Binary operation * de?ed on set Z by a*b=a-b where a, b are members of Z.Commutative if and only if a*b=b*a:a-b != b-a // therefore not commutativeAssociative if (a*b)*c=a*(b*c):(a-b)*c ?= a*(b-c) // where a-b and b-c represents a*b(a-b)-c ?= a-(b-c) // where ()-c and a-() represents a*ba-b-c != a-b+c // therefore not associativeMarcos === if x