mm-190 === Subject: Re: JSH: Easy answers, algebraic integer issueBill Dubuque wrote: Simpler: let a,b,c,d = 7,x-1,x+1,x inLEMMA (a,b-c ) = 1 = (b,d) in a gcd domain => (a,b,cd) = 1Even simpler: it's just a special case of Euclid's Lemma, namely (a,b) is coprime to c & d hence to their product i.e. (a,b, c) = 1 via (a,b-c) = 1 by hypothesis (a,b, d) = 1 via (b, d) = 1 by hypothesis-Bill Dubuque> PROOF (a,b,cd) = (a,b,c) via (b,d) = 1, see [1] = (a,b-c,c) via (b,c) = (b-c,c) = 1 via (a,b-c) = 1 [1] http://google.com/groups?selm=y8z65evhler.fsf% 40nestle.ai.mit.eduSubject: Re: linear functional === I'd be surprised if you could prove this: (**) still holds, if you replace M by (say) 2M, so if you can prove |f| >=M, you can as well prove |f| >= 2M etc. -- Or am I missing something? === Subject: Re: linear functionalHow about this. Consider the sequence a_1=M ||x_1||, a_2=a_3 = ... = 0.Then observe |F(x_1)| = |a_1| = M ||x_1||.So there is an x where equality holds, so ||F||=M.> I'd be surprised if you could prove this: (**) still holds, if you > replace M by (say) 2M, so if you can prove |f| >=M, you can as well > prove |f| >= 2M etc. -- Or am I missing something?Subject: Re: linear functional === How about this: a_1 = |x1|, a_k=0 (k>1), x_k=0, k>1then for any scalars b_ : | b_1*a_1 + ... + b_n*a_n | <= 1000 || b_1*x_1 + ... + b_n*x_n |||F(x_1)| = a_1 = |x_1|but I really don't see, how it would follow, that ||F|| = 1000? === Subject: Re: Question about abelian groups Adjunct Assistant Professor at the University of Montana.Z wrote:> Hi everyone, I`ve got a problem I can`t solve.Suppose G is a group and every element a in G has the> property that a^3=e (the identity)> Can G be a non-abelian group??For any odd prime p, you can 'nd a group with the property that everyelement x satis'es x^p = e, and p is nonabelian.The easiest example has order p^3. In your case, a group of order 27will do. Take upper triangular matrices with 1's on the diagonal, andthe remaining three entries taken in the integers modulo 3, undermatrix multiplication. === Subject: Re: Question about abelian groupsZ wrote:> Hi everyone, I`ve got a problem I can`t solve.Suppose G is a group and every element a in G has the> property that a^3=e (the identity)> Can G be a non-abelian group??Yes.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: One simple question> Hi, Let G=(-00,a). Regardless of whether in'nity is real or otherwise, this is an open> interval - and by de'nition, neither a, nor -00 are in it. Similarly, (-3,3], then -3 is not in this interval, but 3 is. Also, you can never have [-00,00] - it is has to be (-00,00)This, of course, depends on what you de'ne 00 to mean. If you are justusing it as shorthand for goes on forever, then of course, you are right.But there are other ways to de'ne 00, as David Cantrell is so fond of(correctly) pointing out. In some of these other ways of de'ning it, itbecomes just another number.Jon Miller === Subject: Re: One simple questionPosted-And-Mailed: yes === wrote:> .... > Let G=(-00,a).> Then, I wonder wheather we can say -00 is in G or not.If we can't say -00 is in G,> Can we say -00 not in G ?.... No. Logically, -00 is in G is not a well-formed formula, because G stands for a set of real numbers and -00 doesn't stand for any real number. It's only when you have a well-formed formula that you can expect it to be either true or false. Ken Pledger. === Subject: Please Help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1BM7VB26839; === I have to sovle a simple vector problem geometricaly and componently.Geometrialy is easy but i am having trouble solving it componently.I end up with:1. x(sin50) = 250(sin(theta))2. x(cos50) = 250(cos(theta))solve for (theta)i do it at all i do is spin around in circles.plz solve and show me how you did it.thanks === Subject: Re: Please Help>1. x(sin50) = 250(sin(theta))>2. x(cos50) = 250(cos(theta))>solve for (theta)i do it at all i do is spin around in circles.What happens if you divide the 'rst equation by the second?-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comAn expense does not have to be required to be considered necessary. -- IRS Form 1040 line 23 instructions === Subject: Re: Please Help> 1. x(sin50) = 250(sin(theta))> 2. x(cos50) = 250(cos(theta))Just divide #1 by #2 and you get:Tan(theta) = Tan(50)Thus theta = 50 and x=250Does this help?Kev === Subject: Groups of even order questionHello everyone,Question: If G is group of even order, then G has anodd number of elements of order 2.Let G have order 2k (k being an interger > 0). Sincethe identity element e is of order 1, then the remainingelements, i.e. G - {e}, must have order 2 or higher. Whathappens when G - {e} only has elements of order greaterthan 2? Then the number of elements of order 2 in G is notodd but even (since it is 0). What gives?Bernd === Subject: Re: Groups of even order question Adjunct Assistant Professor at the University of Montana.Bernd wrote:>Hello everyone,Question: If G is group of even order, then G has an>odd number of elements of order 2.Let G have order 2k (k being an interger > 0). Since>the identity element e is of order 1, then the remaining>elements, i.e. G - {e}, must have order 2 or higher. What>happens when G - {e} only has elements of order greater>than 2? The universe implodes... because it cannot happen.>Then the number of elements of order 2 in G is not>odd but even (since it is 0). What gives?Exactly: a group of even order MUST have elements of order 2. Theanswer to your question is that G-{e} cannot have only elements oforder greater than 2, it must have at least 1 elements of orderexactly 2.-- === === === =It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === === Arturo Magidinmagidin@math.berkeley.edu === Subject: Resolution to Decker Quadratic IssueTurns out there's another approach to prove a problem with the oldconcepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraicintegers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.Importantly, for any integer b, such that it is irreducible over Q,*none* of the solutions for z can be an algebraic integer!Now then, imagine that there exists some algebraic integer b for whichit is reducible over Q, then the root will be a fraction with a 7 or49 in the denominator.Let c/49 be such a root, where c is then an integer, then I'd have(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/49,but for that to be an algebraic integer, at that point, (b + sqrt(b^2 + 28))would have to have a factor that is 49, which is not possible, as itis itself a factor of 7.So consider the root c/7, which gives(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7,which would force (1 + sqrt(-167)) to be coprime to 7.Therefore, (1 + sqrt(-167)) has no factors in common with 7 in thering of algebraic integers!!!QEDJames HarrisDecker Quadratic Source Information------Recently Rick Decker, a professor at Hamilton College, apparentlytrying to refute my research came up with a quadratic example, which Ilike because it's a quadratic, and easier to manipulate than thecubics I've used before.If you wish to see his original post here are some headers which alsoshow that he posts from Hamilton College:Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: Resolution to Decker Quadratic Issue> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraic> integers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.Importantly, for any integer b, such that it is irreducible over Q,> *none* of the solutions for z can be an algebraic integer!> I believe your arithmetic is wrong here, but that's not therelevant point. The relevant point is, you have at last recognized the*possibility* that the roots might share algebraic integerfactors with 7, other than 7 itself or units. Of course that was obvious when x = 1, as Rick Decker showed. The case x = 2 is much harder. I asked over a month ago what factor (1 + sqrt(-167))/2 might have in common with 7. The answer was soon providedby Keith Ramsay: he found a monic 11th degree polynomial whichhad roots that were factors of 7 and also factors of (1 + sqrt(-167))/2. I believe you ignored or misunderstoodRamsay's post. But the key thing here is: *you are asking the right question*.You have opened a door which inevitably leads to the explanation of our side of the argument. The case x = 1 is not an exception. It is the *typical* case.In general the key polynomial is irreducible. The case x = 0is the exception, and everything you have done is an attempt togeneralize from the exceptional case. For *most* cases, including x = 2, it is quite dif'cult to writedown what polynomial the factor of 7 may satisfy, but it can be done. This all goes back to work of Dedekind on class numbers, with the key theorem being the assertion that class numbers are 'nite. But ï'nite' does not necessarily mean ïsmall'! There is no reason that the algebraic integer factor of7 which works should be a root of a quartic polynomial. A rationale for focussing on z as you do above is lacking. But again, the key thing here is: *you are at last thinkingabout the problem in the right way*.------------ The following is another example like Decker's, where theroots and factors are easily calculated when x = 2: De'ne R(x) = 7 * (25*(x^2 - 3x)/2 + 5*(2*x - 1) + 7).[Note that for integer values of x, the term (x^2 - 3x)/2 is always an integer.] If R(x) is factored in the form R(x) = (5*a_1(x) + 7)*(5*a_2(x) + 7),then a_1(x) and a_2(x) are roots of: a^2 - (2*x - 1)*a + 7*(x^2 - 3x)/2. When x = 0, this reduces to a^2 + aso that a_1(0) = 0 and a_2(0) = -1. When x = 2, the polynomial in ïa' becomes a^2 - 3*a - 7. In this case, a_1(2) = (3 + sqrt(37))/2 and a_2(2) = (3 - sqrt(37))/2.These are both algebraic integers, and a_1(2)*a_2(2) = -7, which implies -a_2(2) = 7/a_1(2) and -a_1(2) = 7/a_2(1).Therefore R(2)/7 = -3 = -(5*1 - a_2(2))*(5*1 - a_1(2)). = (-5*1 + [3 + sqrt(37))/2])*(5*1 - [(3 - sqrt(37)/2]).Again, this is a factorization of R(2)/7 as a polynomialin the number 5 in which *all* the coef'cients are algebraic integers. Nora B.> Now then, imagine that there exists some algebraic integer b for which> it is reducible over Q, then the root will be a fraction with a 7 or> 49 in the denominator.Let c/49 be such a root, where c is then an integer, then I'd have(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/49,but for that to be an algebraic integer, at that point, (b + sqrt(b^2 + 28))would have to have a factor that is 49, which is not possible, as it> is itself a factor of 7.So consider the root c/7, which gives(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7,which would force (1 + sqrt(-167)) to be coprime to 7.Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the> ring of algebraic integers!!!QED> James HarrisDecker Quadratic Source Information> ------> Recently Rick Decker, a professor at Hamilton College, apparently> trying to refute my research came up with a quadratic example, which I> like because it's a quadratic, and easier to manipulate than the> cubics I've used before.If you wish to see his original post here are some headers which also> show that he posts from Hamilton College:Subject: Re: Mathematical consistency, courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x). === Subject: Re: Resolution to Decker Quadratic IssueJames Harris wrote:> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.Another approach? You have never proven that there is any problem with thering of algebraic integers using *any* approach. In fact, there is no suchproblem. The problem is in your choice of mathematical operations, yoursloppy logic and inattention to detail -- in short, your ignorance. Allyou have done, and continue to do, is post faulty mathematics. This threadprovides no exception.[snip §awed math with blatant sign errors, unsupported conclusions andcontradictions]James, you are clearly way over your head with the mathematics you areattempting to develop. Was the crash dummy vacany already 'lled?--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 === Subject: Re: Resolution to Decker Quadratic Issue>Turns out there's another approach to prove a problem with the old>concepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraic>integers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2>So z is an algebraic number. Is there any particular reason for itto be an algebraic integer?Writing A = (1+ sqrt(-167))/2 and B = (-b + sqrt(b^2 + 28))/2,A is an algebraic integer, and B is an algebraic integer, and theirproduct AB is an algebraic integer, but there seems to be no reasonto suppose that z = AB/7 is an algebraic integer.>which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0>Yes.>and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0>No. You have introduced two sign errors. This should be196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0.>and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.>and this should be 49 z^4 + 7b z^3 + (42b^2 + 581) z^2 - 42bz + 1764 = 0,which can be divided by 7 to give7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0.I thought you said you'd bought a computer algebra package;why don't you use it?>Importantly, for any integer b, such that it is irreducible over Q,>*none* of the solutions for z can be an algebraic integer!>Yes, even with the corrected polynomial.But why should they be?>Now then, imagine that there exists some algebraic integer b for which>it is reducible over Q, then the root will be a fraction with a 7 or>49 in the denominator.>It is only meaningful to speak of a polynomial as being reducibleor irreducible over Q if each of its coef'cients is in Q, so Iassume you mean for b to be a rational integer, and you mean thatit is reducible over Z.Since the leading coef'cient after correction is 7, I'll take itas if you had said that the root will be a fraction with a 7 inthe denominator. Unfortunately, this is wrong.Firstly, the polynomial is quartic, so this equation has fourroots, one being the value of z you started with and the otherthree being introduced by repeated squaring. Since the product ofthe four roots is 252/7 = 36, they can't all be such fractions.Secondly, if the quartic is reducible over Z then it is the productof two polynomials of lesser degree, with leading coef'cients 1 and 7. The zeros of the polynomial with leading coef'cient 1would be algebraic integers.For example, if b = -6 then the quartic is reducible, as7z^4 - 6z^3 + 299z^2 + 36z + 252 = (z^2 - z + 42)(7z^2 + z + 6),so its four zeros are (1 + sqrt(-167))/2, (1 - sqrt(-167))/2, (-1 + sqrt(-167))/14 and (-1 - sqrt(-167))/14.The 'rst of these is the value of z you started with; it is analgebraic integer. The other three zeros are spurious.>Let c/49 be such a root, where c is then an integer, then I'd have(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/49,but for that to be an algebraic integer, at that point, (b + sqrt(b^2 + 28))would have to have a factor that is 49, which is not possible, as it>is itself a factor of 7.>Since the leading coef'cient is 7, not 49, the above paragraph isirrelevant.>So consider the root c/7, which gives(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7,which would force (1 + sqrt(-167)) to be coprime to 7.Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the>ring of algebraic integers!!!Your argument, that the root must have this form, has been shown tobe faulty, so the above two paragraphs are irrelevant.QED>So what have you shown? Merely that if you take an algebraicinteger AB and divide it by 7 you get an algebraic number that ingeneral is not an algebraic integer. So what?John Roberts-Jones === Subject: Re: Resolution to Decker Quadratic Issue>Turns out there's another approach to prove a problem with the old>concepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraic>integers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 So z is an algebraic number. Is there any particular reason for it> to be an algebraic integer?Writing A = (1+ sqrt(-167))/2 and B = (-b + sqrt(b^2 + 28))/2,> A is an algebraic integer, and B is an algebraic integer, and their> product AB is an algebraic integer, but there seems to be no reason> to suppose that z = AB/7 is an algebraic integer.>which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 Yes.>and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0 No. You have introduced two sign errors. This should be196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0.Oh, ok, thanks!!!>and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0. and this should be 49 z^4 + 7b z^3 + (42b^2 + 581) z^2 - 42bz + 1764 = 0,which can be divided by 7 to give7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0.Ok. Thanks again. > I thought you said you'd bought a computer algebra package;> why don't you use it?>Importantly, for any integer b, such that it is irreducible over Q,>*none* of the solutions for z can be an algebraic integer! Yes, even with the corrected polynomial.> But why should they be?> y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2, which I use with a solution for a_1(x), which is(1 + sqrt(-167))/2, so I get(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z,so z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))and by picking b's in the ring of algebraic integers, which *should*be an algebraic integer, for any algebraic integer b, where I've founda factor in common (1 + sqrt(-167))/2 and 7.That's the key point, as in fact, none exists.>Now then, imagine that there exists some algebraic integer b for which>it is reducible over Q, then the root will be a fraction with a 7 or>49 in the denominator. It is only meaningful to speak of a polynomial as being reducible> or irreducible over Q if each of its coef'cients is in Q, so I> assume you mean for b to be a rational integer, and you mean that> it is reducible over Z.Since the leading coef'cient after correction is 7, I'll take it> as if you had said that the root will be a fraction with a 7 in> the denominator. Unfortunately, this is wrong.Firstly, the polynomial is quartic, so this equation has four> roots, one being the value of z you started with and the other> three being introduced by repeated squaring. Since the product of> the four roots is 252/7 = 36, they can't all be such fractions.That's merely making a stronger position for my case saying no rootover Q exists; therefore, z is NEVER an algebraic integer for thatreason. > Secondly, if the quartic is reducible over Z then it is the product> of two polynomials of lesser degree, with leading coef'cients > 1 and 7. The zeros of the polynomial with leading coef'cient 1> would be algebraic integers.Still at least *one* of the solutions for z would have to be anirreducible fraction with a 7 in the denominator. > For example, if b = -6 then the quartic is reducible, as7z^4 - 6z^3 + 299z^2 + 36z + 252 = (z^2 - z + 42)(7z^2 + z + 6),so its four zeros are (1 + sqrt(-167))/2, (1 - sqrt(-167))/2, > (-1 + sqrt(-167))/14 and (-1 - sqrt(-167))/14.The 'rst of these is the value of z you started with; it is an> algebraic integer. The other three zeros are spurious.Nope. On what basis do you claim they're spurious?The four roots z's are as follows:z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))so in fact the other roots are NOT spurious!!!You can check with your software package to verify.James Harris === Subject: Re: Resolution to Decker Quadratic Issue jstevh@msn.com (James Harris) wrote:>Turns out there's another approach to prove a problem with the old>concepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraic>integers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 So z is an algebraic number. Is there any particular reason for it> to be an algebraic integer?> Writing A = (1+ sqrt(-167))/2 and B = (-b + sqrt(b^2 + 28))/2,> A is an algebraic integer, and B is an algebraic integer, and their> product AB is an algebraic integer, but there seems to be no reason> to suppose that z = AB/7 is an algebraic integer.>which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 Yes.>and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0 No. You have introduced two sign errors. This should be> 196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0.Oh, ok, thanks!!!>and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0. and this should be > 49 z^4 + 7b z^3 + (42b^2 + 581) z^2 - 42bz + 1764 = 0,> which can be divided by 7 to give> 7 z^4 + b z^3 + (6b^2 + 83) z^2 - 6bz + 252 = 0.> I thought you said you'd bought a computer algebra package;> why don't you use it?>Importantly, for any integer b, such that it is irreducible over Q,>*none* of the solutions for z can be an algebraic integer! Yes, even with the corrected polynomial.> But why should they be?> y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2, which I use with a solution for a_1(x), which is(1 + sqrt(-167))/2, so I get(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z,so z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))and by picking b's in the ring of algebraic integers, which *should*> be an algebraic integer, for any algebraic integer b, where I've found> a factor in common (1 + sqrt(-167))/2 and 7.That's the key point, as in fact, none exists.>Now then, imagine that there exists some algebraic integer b for which>it is reducible over Q, then the root will be a fraction with a 7 or>49 in the denominator. It is only meaningful to speak of a polynomial as being reducible> or irreducible over Q if each of its coef'cients is in Q, so I> assume you mean for b to be a rational integer, and you mean that> it is reducible over Z.> Since the leading coef'cient after correction is 7, I'll take it> as if you had said that the root will be a fraction with a 7 in> the denominator. Unfortunately, this is wrong.> Firstly, the polynomial is quartic, so this equation has four> roots, one being the value of z you started with and the other> three being introduced by repeated squaring. Since the product of> the four roots is 252/7 = 36, they can't all be such fractions.That's merely making a stronger position for my case saying no root> over Q exists; therefore, z is NEVER an algebraic integer for that> reason.> Secondly, if the quartic is reducible over Z then it is the product> of two polynomials of lesser degree, with leading coef'cients > 1 and 7. The zeros of the polynomial with leading coef'cient 1> would be algebraic integers.Still at least *one* of the solutions for z would have to be an> irreducible fraction with a 7 in the denominator.> For example, if b = -6 then the quartic is reducible, as> 7z^4 - 6z^3 + 299z^2 + 36z + 252 = (z^2 - z + 42)(7z^2 + z + 6),> so its four zeros are (1 + sqrt(-167))/2, (1 - sqrt(-167))/2, > (-1 + sqrt(-167))/14 and (-1 - sqrt(-167))/14.> The 'rst of these is the value of z you started with; it is an> algebraic integer. The other three zeros are spurious.Nope. On what basis do you claim they're spurious?Because you started by assuming that (1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zso, unless you are claiming equal roots, only the 'rst is relevant. The others are artifacts of the process by which the square roots were eliminated. This is a situation well known to those familiar with algebra but often overlooked by putzers.The four roots z's are as follows:z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))so in fact the other roots are NOT spurious!!!You can check with your software package to verify.> James Harris === Subject: Re: Resolution to Decker Quadratic Issue jstevh@msn.com (James Harris) wrote:> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.Decker put forward the quadratic(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 + x).Letting x=2, you havea_1(2)^2 - a_1(2) + 42 = 0, which gives a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.Now consider the quadratic y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2.Note that root is an algebraic integer factor of 7 for all algebraic> integers y, and b.Now consider(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2which is(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))zand working to eliminate square root terms gives28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0and working still further I get196 z^4 + 28b z^3 + (186b^2 - 2324) z^2 - 500bz + 7056 = 0and I can divide both sides by 4 to 'nally get49 z^4 + 7b z^3 + (42b^2 - 581) z^2 - 125bz + 1764 = 0.As this equation is wrong, as one may check by substitutingz = (1+sqrt(-167))/(b+sqrt(b^2+28)) for z, the rest of JSH's claim, being based on this equation, is irrelevant. === Subject: Re: Resolution to Decker Quadratic IssueThank God everything's 'nally resolved. Now that Fermat's Last Theoremhas 'nally been proven, an ultra-ef'cient prime number generatorcreated, and world hunger eradicated, we can all go home now.Turn the lights off when you leave, Harris.Doug === Subject: Re: Resolution to Decker Quadratic Issue> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers. Decker put forward the quadratic (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)>James,I will agree that there is a bunch of people just teeing off on you, andserving up insults. On the other hand, there is a pretty good bunchof nice people who are trying their best to get you on to the righttrack. I just wish you were able to tell the difference between these twogroups, take the advice of the ones who wish you well, studytheir arguments instead of giving a knee jerk response. Take theiradvice and actually study the area of mathematics that you areinterested in. I think you are capable of doing this, having somefun, and perhaps actually discovering something. Relax, andenjoy life. My impression is that you're not having very muchfun now. I send you my best wishes for a successful future.I also apologize for any snide remarks I might have made.Skip === Subject: Re: Resolution to Decker Quadratic Issue > Turns out there's another approach to prove a problem with the old > concepts about the ring of algebraic integers. > Decker put forward the quadratic > (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x). > Letting x=2, you have > a_1(2)^2 - a_1(2) + 42 = 0, which gives > a_1(2) = (1 + sqrt(-167))/2 as one of two solutions. > Now consider the quadratic Why a quadratic?... > Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the > ring of algebraic integers!!!Keith Ramsay has shown the factor in common with 7 in the ringof algebraic integers. Alas, it is indeed *not* the root ofa quadratic polynomial. It is the root of a polynomial withdegree 11. *And there are good reasons for that.*In general, when you have a number of the form p + sqrt(q) and youtry to 'nd the common factor with some integer n, you may getsomething quite different. For instance, let's look at the factorin common between y = (1 - sqrt(-5)) and 3. It is z = (sqrt(2) - sqrt(-10))/2.You may verify that z.sqrt(2) = y, and z * z' = 3 (z' denotes complexconjugation). Also: z^4 + 4z + 9 = 0, so z is an algebraic integer.There is *no* quadratic polynomial with integer coef'cients of whichz is a root.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Resolution to Decker Quadratic Issue> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> Letting x=2, you have> a_1(2)^2 - a_1(2) + 42 = 0, which gives > a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.> Now consider the quadratic Why a quadratic?Simpler. > ...> Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the> ring of algebraic integers!!!Keith Ramsay has shown the factor in common with 7 in the ring> of algebraic integers. Alas, it is indeed *not* the root of> a quadratic polynomial. It is the root of a polynomial with> degree 11. *And there are good reasons for that.*It doesn't matter as I can use *any* algebraic integer b. > In general, when you have a number of the form p + sqrt(q) and you> try to 'nd the common factor with some integer n, you may get> something quite different. For instance, let's look at the factor> in common between y = (1 - sqrt(-5)) and 3. It is> z = (sqrt(2) - sqrt(-10))/2.> You may verify that z.sqrt(2) = y, and z * z' = 3 (z' denotes complex> conjugation). Also: z^4 + 4z + 9 = 0, so z is an algebraic integer.> There is *no* quadratic polynomial with integer coef'cients of which> z is a root.You can't get a monic polynomial with integer coef'cients with myexample Dik Winter.What I did was show that there is no algebraic integer factor of 7that will work with the Decker quadratic at x=2.James Harris === Subject: Re: Resolution to Decker Quadratic Issue jstevh@msn.com (James Harris) wrote:> (James Harris) writes:> Turns out there's another approach to prove a problem with the old> concepts about the ring of algebraic integers.> Decker put forward the quadratic> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2) > where his a's are roots of > a^2 - (x - 1)a + 7(x^2 + x).> Letting x=2, you have> a_1(2)^2 - a_1(2) + 42 = 0, which gives > a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.> Now consider the quadratic > Why a quadratic?Simpler. > ...> Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the> ring of algebraic integers!!!> Keith Ramsay has shown the factor in common with 7 in the ring> of algebraic integers. Alas, it is indeed *not* the root of> a quadratic polynomial. It is the root of a polynomial with> degree 11. *And there are good reasons for that.*It doesn't matter as I can use *any* algebraic integer b.Again JSH's cavalier attitude towards truth, he says it doesn't matter. > In general, when you have a number of the form p + sqrt(q) and you> try to 'nd the common factor with some integer n, you may get> something quite different. For instance, let's look at the factor> in common between y = (1 - sqrt(-5)) and 3. It is> z = (sqrt(2) - sqrt(-10))/2.> You may verify that z.sqrt(2) = y, and z * z' = 3 (z' denotes complex> conjugation). Also: z^4 + 4z + 9 = 0, so z is an algebraic integer.> There is *no* quadratic polynomial with integer coef'cients of which> z is a root.You can't get a monic polynomial with integer coef'cients with my> example Dik Winter.What I did was show that there is no algebraic integer factor of 7> that will work with the Decker quadratic at x=2.Oh?> James Harris === Subject: Re: Resolution to Decker Quadratic Issue > Now consider the quadratic > Why a quadratic? > Simpler.But losing information. > Therefore, (1 + sqrt(-167)) has no factors in common with 7 in the > ring of algebraic integers!!! > Keith Ramsay has shown the factor in common with 7 in the ring > of algebraic integers. Alas, it is indeed *not* the root of > a quadratic polynomial. It is the root of a polynomial with > degree 11. *And there are good reasons for that.* > It doesn't matter as I can use *any* algebraic integer b.It doesn't matter, because Keith Ramsay has shown the common factorexplicitly. To wit: start with r = (1 + sqrt(-167))/2,now: r^11 = (-1272087893 + 86559857.sqrt(-167))/2 = = (44444 - 111.sqrt(-167))(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2] = = s.t.uwith s = (44444 - 111.sqrt(-167)) t = (298 - 23.sqrt(-167)) u = [(-3 - 7.sqrt(-167)]/2.Now set s', t' and u' the complex conjugates of s, t and u above, andsee: s.s' = 7^11, t.t' = 3^11 and u.u' = 2^11.Now set a = s^(1/11) (it does not matter which of the elevenroots you take), and a' the complex conjugate of that. Setb = t^(1/11) (also here it does not matter which you take) andc = u^(1/11) (here it does matter; there is exactly one of theroots that 'ts the bill).We now have: a.a' = 7 a.b.c = rwith a, a', b and c algebraic integers. (If you had chosen the wrongroot c above, the second equation would yield the product of r and aunit, however it is than easy to divide both sides by that unit.)So a is a factor in common between 7 and (1 + sqrt(-167))/2. Moreover,a is a root of: x^22 - 88888.x^11 + 7^11 = 0 > You can't get a monic polynomial with integer coef'cients with my > example Dik Winter.Not with your example because you start with a quadratic. > What I did was show that there is no algebraic integer factor of 7 > that will work with the Decker quadratic at x=2.What is shown *just above* is an algebraic integer factor of 7 thatwill work with the Decker quadratic at x = 2.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: 2D DTFT symmetry propertyDoes someone knows how to derive the symmetry properties for the 2-D DTFT ofa real signal (i.e. compute X(?1,?2) as function of X(?1, ?2)). It isrequired to prove that X(0, 0), X(?, 0), X(0, ?) and X(?, ?) must be real.Help on that would be very appreciated.thanks,Gabriel === Subject: Re: 2D DTFT symmetry propertyoupsThe ?'s go as follow:Does someone knows how to derive the symmetry properties for the 2-D DTFT ofa real signal (i.e. compute X(w1,w2) as function of X(w1, w2)). It isrequired to prove that X(0, 0), X(pi, 0), X(0, pi) and X(pi, pi) must bereal.> Does someone knows how to derive the symmetry properties for the 2-D DTFTof> a real signal (i.e. compute X(?1,?2) as function of X(?1, ?2)). It is> required to prove that X(0, 0), X(?, 0), X(0, ?) and X(?, ?) must be real. === Subject: Need examplehello,I,m looking a set A which is a subset of realand int(A)=empty and cl(A)=R(whole real #).but, I couldn't think such set A .....If someone know these set A, please post reply.Thanks, === Subject: Re: Need examplewrote:> hello,I,m looking a set A which is a subset of real> and int(A)=empty and cl(A)=R(whole real #).but, I couldn't think such set A .....> If someone know these set A, please post reply.Thanks,The set of rationals or the set of irrationals, assuming the usual topology on the reals.Subject: Re: Need example === Subject: Can you help me solve this riddle? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1CEZlW05963; === I build up castles, I tear down mountains, I make some men blind, Ihelp others to see.what am I? === Subject: Re: vector calculus clue. by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1CKB1T07296; === The cone z^2= x^2+ y^2 has vertex at (0,0,0). If you draw the x,zplane (with y= 0) you get z^2= x^2 or |z|= |x| two lines showing theedges of the cone. The cone will cross the cylinder when z^2= x^2+y^2= a^2: that is in the circles z= a and z= -a. It's projection downto the xy-plane is the circle x^2+ y^2= a^2. If you think of the cone as the surface on which F(x,y,z)= x^2+ y^2-z^2= 0, then grad F= 2xi+ 2yj- 2zk is orthogonal to the cone. Normalizing to the xy-plane by dividing the entire vector by -2z,F/(2z)= -(x/z)i- (y/z)j+ k and its length gives the differential ofsurface area projected down to the xy-plane: That isdS= sqrt(x^2/z^2+y^2/z^2+ 1)dA= sqrt(x^2+ y^2+ z^2)/z dA. Since z^2= x^2+ y^2, this is dS= sqrt(2x^2+ 2y^2)/z dA = sqrt(2)dA.(Yes, a cone is a developable surface- it can be swept out by astraight line pivoting through the vertex and so its area is just aconstant (here sqrt(2)) times the area of its projection down onto thexy-plane.) Since we have to choose z to be positive or negative,sqrt(2) times the area of the circle would be the area of the uppernappe of the cone. In order to get the area of both nappes, multiplyby 2. === Subject: Re: vector calculus clue Kevin Stone wrote:> I've got a cone (z >= 0)> z^2 = x^2 + y^2and a cylinder:> (x-a)^2 + y^2 = a^2I've to 'nd the area of the cone which is inside the cylinder and the clue> given says to use the area of a circle.What is the integral you need to set up for the area of the surface z = f(x,y) over a region R in the plane? Tell us what you've done and what is giving you trouble.